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https://pressbooks.uiowa.edu/clonedbook/chapter/forced-oscillations-and-resonance/ | 1,726,600,454,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00228.warc.gz | 418,298,081 | 34,330 | # 123 16.8 Forced Oscillations and Resonance
### Summary
• Observe resonance of a paddle ball on a string.
• Observe amplitude of a damped harmonic oscillator.
Sit in front of a piano sometime and sing a loud brief note at it with the dampers off its strings. It will sing the same note back at you—the strings, having the same frequencies as your voice, are resonating in response to the forces from the sound waves that you sent to them. Your voice and a piano’s strings is a good example of the fact that objects—in this case, piano strings—can be forced to oscillate but oscillate best at their natural frequency. In this section, we shall briefly explore applying a periodic driving force acting on a simple harmonic oscillator. The driving force puts energy into the system at a certain frequency, not necessarily the same as the natural frequency of the system. The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force.
Most of us have played with toys involving an object supported on an elastic band, something like the paddle ball suspended from a finger in Figure 2. Imagine the finger in the figure is your finger. At first you hold your finger steady, and the ball bounces up and down with a small amount of damping. If you move your finger up and down slowly, the ball will follow along without bouncing much on its own. As you increase the frequency at which you move your finger up and down, the ball will respond by oscillating with increasing amplitude. When you drive the ball at its natural frequency, the ball’s oscillations increase in amplitude with each oscillation for as long as you drive it. The phenomenon of driving a system with a frequency equal to its natural frequency is called resonance. A system being driven at its natural frequency is said to resonate. As the driving frequency gets progressively higher than the resonant or natural frequency, the amplitude of the oscillations becomes smaller, until the oscillations nearly disappear and your finger simply moves up and down with little effect on the ball.
Figure 3 shows a graph of the amplitude of a damped harmonic oscillator as a function of the frequency of the periodic force driving it. There are three curves on the graph, each representing a different amount of damping. All three curves peak at the point where the frequency of the driving force equals the natural frequency of the harmonic oscillator. The highest peak, or greatest response, is for the least amount of damping, because less energy is removed by the damping force.
It is interesting that the widths of the resonance curves shown in Figure 3 depend on damping: the less the damping, the narrower the resonance. The message is that if you want a driven oscillator to resonate at a very specific frequency, you need as little damping as possible. Little damping is the case for piano strings and many other musical instruments. Conversely, if you want small-amplitude oscillations, such as in a car’s suspension system, then you want heavy damping. Heavy damping reduces the amplitude, but the tradeoff is that the system responds at more frequencies.
These features of driven harmonic oscillators apply to a huge variety of systems. When you tune a radio, for example, you are adjusting its resonant frequency so that it only oscillates to the desired station’s broadcast (driving) frequency. The more selective the radio is in discriminating between stations, the smaller its damping. Magnetic resonance imaging (MRI) is a widely used medical diagnostic tool in which atomic nuclei (mostly hydrogen nuclei) are made to resonate by incoming radio waves (on the order of 100 MHz). A child on a swing is driven by a parent at the swing’s natural frequency to achieve maximum amplitude. In all of these cases, the efficiency of energy transfer from the driving force into the oscillator is best at resonance. Speed bumps and gravel roads prove that even a car’s suspension system is not immune to resonance. In spite of finely engineered shock absorbers, which ordinarily convert mechanical energy to thermal energy almost as fast as it comes in, speed bumps still cause a large-amplitude oscillation. On gravel roads that are corrugated, you may have noticed that if you travel at the “wrong” speed, the bumps are very noticeable whereas at other speeds you may hardly feel the bumps at all. Figure 4 shows a photograph of a famous example (the Tacoma Narrows Bridge) of the destructive effects of a driven harmonic oscillation. The Millennium Bridge in London was closed for a short period of time for the same reason while inspections were carried out.
In our bodies, the chest cavity is a clear example of a system at resonance. The diaphragm and chest wall drive the oscillations of the chest cavity which result in the lungs inflating and deflating. The system is critically damped and the muscular diaphragm oscillates at the resonant value for the system, making it highly efficient.
1: A famous magic trick involves a performer singing a note toward a crystal glass until the glass shatters. Explain why the trick works in terms of resonance and natural frequency.
# Section Summary
• A system’s natural frequency is the frequency at which the system will oscillate if not affected by driving or damping forces.
• A periodic force driving a harmonic oscillator at its natural frequency produces resonance. The system is said to resonate.
• The less damping a system has, the higher the amplitude of the forced oscillations near resonance. The more damping a system has, the broader response it has to varying driving frequencies.
### Conceptual Questions
1: Why are soldiers in general ordered to “route step” (walk out of step) across a bridge?
### Problems & Exercises
1: How much energy must the shock absorbers of a 1200-kg car dissipate in order to damp a bounce that initially has a velocity of 0.800 m/s at the equilibrium position? Assume the car returns to its original vertical position.
2: If a car has a suspension system with a force constant of how much energy must the car’s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m?
3: (a) How much will a spring that has a force constant of 40.0 N/m be stretched by an object with a mass of 0.500 kg when hung motionless from the spring? (b) Calculate the decrease in gravitational potential energy of the 0.500-kg object when it descends this distance. (c) Part of this gravitational energy goes into the spring. Calculate the energy stored in the spring by this stretch, and compare it with the gravitational potential energy. Explain where the rest of the energy might go.
4: Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction (a) How far can the spring be stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is what total distance does it travel before stopping? Assume it starts at the maximum amplitude.
5: Engineering Application: A suspension bridge oscillates with an effective force constant of (a) How much energy is needed to make it oscillate with an amplitude of 0.100 m? (b) If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart of energy each second, how long does it take for the bridge’s oscillations to go from 0.100 m to 0.500 m amplitude?
## Glossary
natural frequency
the frequency at which a system would oscillate if there were no driving and no damping forces
resonance
the phenomenon of driving a system with a frequency equal to the system’s natural frequency
resonate
a system being driven at its natural frequency
### Solutions
1: The performer must be singing a note that corresponds to the natural frequency of the glass. As the sound wave is directed at the glass, the glass responds by resonating at the same frequency as the sound wave. With enough energy introduced into the system, the glass begins to vibrate and eventually shatters.
Problems & Exercises
1:
384 J
3:
(a). 0.123 m
(b). −0.600 J
(c). 0.300 J. The rest of the energy may go into heat caused by friction and other damping forces.
5:
(a)
(b) s | 1,755 | 8,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.957483 |
https://blog.csdn.net/TaiJi1985/article/details/75087742 | 1,558,677,643,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257514.68/warc/CC-MAIN-20190524044320-20190524070320-00291.warc.gz | 400,469,886 | 44,490 | # 2 SVM原理
## 2.1 线性回归
J(w,b)=i(wTxi+byi)2
minw,bJ(w,b)
yi(wTxi+b)>0
# 3 公式推导
## 3.1 点到直线的距离
wTx+b=0
wTx′′+b=0
wT(xx′′)=0
dist(x,h)=|wT||w||(xx)|=|wTx+b|||w||
yi(wTxi+b)>0
dist(xi,h)=yi(wTxi+b)||w||
maxb,w=margin(b,w)使, yi(wTxi+b)>0margin(b,w)=mini=1,...Ndist(xi,h)
1=yi(wTx+b)$1 = y_i(w^Tx+b)$ 。 注意，因为我们要找到最小的 dist(xi,h)$dist(x_i,h)$，对w和b 的放缩不会影响谁是最近的点这个结果。也不会影响超平面的位置因为
wTx+b=0$w^Tx+b=0$(wTx/t+b/t)=0$(w^Tx/t+b/t) = 0$并无分别。
yi(wTxi+b)>=1
margin(b,w)=1||w||
maxb,w1||w||使yi(wTxi+b)>=1
minb,w||w||使yi(wTxi+b)>=1
minb,w12wTw使yi(wTxi+b)>=1
# 转化为二次规划的标准型
uQP(H,f,A,b)minu12uTHu+fTu,aTmu<=bm
u=[bw]
H=[00K0KIK]
, K为w的维度，H是(K+1)*(K+1)的矩阵，除了（1，1）点，其他对角线均为1
H = [
0 0 0 ;
0 1 0 ;
0 0 1
]
f=0KRK×1
yi(wTxi+b)>=1$y_i(w^Tx_i+b) >= 1$修改为 yi(wTxi+b)<=1$-y_i(w^Tx_i+b) <= -1$
Ai=yi[1,xi]$A_i = - y_i [ 1 , x_i]$ , 使得bi=1$b_i = -1$
problem:
min 0.5*x'*H*x + f'*x subject to: A*x <= b
x
X = quadprog(H,f,A,b,Aeq,beq) solves the problem above while
additionally satisfying the equality constraints Aeq*x = beq.
X = quadprog(H,f,A,b,Aeq,beq,LB,UB) defines a set of lower and upper
bounds on the design variables, X, so that the solution is in the
range LB <= X <= UB. Use empty matrices for LB and UB if no bounds
exist. Set LB(i) = -Inf if X(i) is unbounded below; set UB(i) = Inf if
X(i) is unbounded above.
X = quadprog(H,f,A,b,Aeq,beq,LB,UB,X0) sets the starting point to X0.
X = quadprog(H,f,A,b,Aeq,beq,LB,UB,X0,OPTIONS) minimizes with the
default optimization parameters replaced by values in the structure
OPTIONS, an argument created with the OPTIMSET function. See OPTIMSET
for details.
X = quadprog(PROBLEM) finds the minimum for PROBLEM. PROBLEM is a
structure with matrix 'H' in PROBLEM.H, the vector 'f' in PROBLEM.f,
the linear inequality constraints in PROBLEM.Aineq and PROBLEM.bineq,
the linear equality constraints in PROBLEM.Aeq and PROBLEM.beq, the
lower bounds in PROBLEM.lb, the upper bounds in PROBLEM.ub, the start
point in PROBLEM.x0, the options structure in PROBLEM.options, and
solver name 'quadprog' in PROBLEM.solver. Use this syntax to solve at
the command line a problem exported from OPTIMTOOL. The structure
PROBLEM must have all the fields.
[X,FVAL] = quadprog(H,f,A,b) returns the value of the objective
function at X: FVAL = 0.5*X'*H*X + f'*X.
[X,FVAL,EXITFLAG] = quadprog(H,f,A,b) returns an EXITFLAG that
describes the exit condition of quadprog. Possible values of EXITFLAG
and the corresponding exit conditions are
All algorithms:
1 First order optimality conditions satisfied.
0 Maximum number of iterations exceeded.
-2 No feasible point found.
-3 Problem is unbounded.
Interior-point-convex only:
-6 Non-convex problem detected.
Trust-region-reflective only:
3 Change in objective function too small.
-4 Current search direction is not a descent direction; no further
Active-set only:
4 Local minimizer found.
-7 Magnitude of search direction became too small; no further
conditioned.
[X,FVAL,EXITFLAG,OUTPUT] = quadprog(H,f,A,b) returns a structure
OUTPUT with the number of iterations taken in OUTPUT.iterations,
maximum of constraint violations in OUTPUT.constrviolation, the
type of algorithm used in OUTPUT.algorithm, the number of conjugate
gradient iterations (if used) in OUTPUT.cgiterations, a measure of
first order optimality (large-scale algorithm only) in
OUTPUT.firstorderopt, and the exit message in OUTPUT.message.
[X,FVAL,EXITFLAG,OUTPUT,LAMBDA] = quadprog(H,f,A,b) returns the set of
Lagrangian multipliers LAMBDA, at the solution: LAMBDA.ineqlin for the
linear inequalities A, LAMBDA.eqlin for the linear equalities Aeq,
LAMBDA.lower for LB, and LAMBDA.upper for UB.
Reference page in Help browser
## matlab求解SVM
function test_svm()
t1 = 5+4*randn(2,10);
t2 = 20+4*randn(2,10);
X = [t1 t2]
X = [t1 t2];
Y = [ones(10,1) ; -ones(10,1)]
plot(t1(1,:),t1(2,:),'ro');
hold on;
plot(t2(1,:),t2(2,:),'bx');
u = svm(X,Y)
x = [-u(1)/u(2) , 0];
y = [0 , -u(1)/u(3)];
plot(x,y);
end
function u = svm( X,Y )
%SVM Summary of this function goes here
% Detailed explanation goes here
[K,N] = size(X);
u0= rand(K+1,1); % u= [b ; w];
A = - repmat(Y,1,K+1).*[ones(N,1) X'];
b = -ones(N,1);
H = eye(K);
H = [zeros(1,K);H];
H = [zeros(K+1,1) H];
p = zeros(K+1,1);
lb = -10*ones(K+1,1);
rb = 10*ones(K+1,1);
options = optimset; % Options是用来控制算法的选项参数的向量
options.LargeScale = 'off';
options.Display = 'off';
options.Algorithm = 'active-set';
end
ans =
-0.7964 6.3340
6.5654 6.8067
4.4789 5.7348
3.0954 8.4481
-0.4468 6.8201
1.6052 3.6605
7.2111 9.1564
0.5294 10.0426
7.6406 4.7285
4.2191 4.1296
18.7876 20.0922
20.2052 23.3043
26.1079 21.8677
19.1611 22.5008
20.7329 15.8809
23.7969 21.2282
20.5407 22.0610
21.0456 16.2341
19.3506 19.4158
17.8720 26.7284
Y =
1
1
1
1
1
1
1
1
1
1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
Optimization terminated.
u =
2.3951
-0.1186
-0.0590
val =
0.0088 | 1,924 | 4,932 | {"found_math": true, "script_math_tex": 37, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-22 | latest | en | 0.535151 |
http://completewoodworkingplans.com/easy-woodworking-project-plans-sideboard-furniture-plans.html | 1,558,276,047,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254889.43/warc/CC-MAIN-20190519141556-20190519163556-00167.warc.gz | 47,526,826 | 6,856 | Finally, properly preparing your wood surface is super important. It will make a huge difference when it comes times to stain, paint, or finish your wood. There are a lot of tips for wood surface preparation, but sanding the wood is one of the most important steps. And I find it really helpful to do the bulk of my sanding before I start ripping (cutting) and building with my wood since it’s still in whole pieces. You can check out my simple tips for how to sand wood in my how to stain wood tutorial, which is also good to reference if you need to learn how to stain too!
How do you divide 11-3/8-in. (or any other mathematically difficult number) into equal parts without dividing fractions? Simple. Angle your tape across the workpiece until it reads an easily-divisible dimension and make your marks with the tape angled. For example, say you want to divide an 11-3/8-in. board into three equal parts. Angle the tape until it reads 12-in., and then make marks at “4” and “8”. Plus: More measuring tips and tricks.
I saw in the March 2019 Vol36,No.1 Iss 259. I have 2 Christian people very close to me that could likely be getting married soon . When I saw the unity cross on page one this would be the perfect gift from a parent. I would like to make it but need the plans. I do not need 3,000 or 60,000 plans I just wish to purchase the plans for the unity cross. I am a beginner so I need detail plans. Please send me information on ordering just the unity cross plans and where to purchase giant Sequoia and white oak woods. Thank you in advance for your help. I need the plans and information before April.
Woodworkers set up, operate, and tend all types of woodworking machines, such as saws, milling machines, drill presses, lathes, shapers, routers, sanders, planers, and wood-fastening machines. Operators set up the equipment, cut and shape wooden parts, and verify dimensions, using a template, caliper, and rule. After the parts are machined, woodworkers add fasteners and adhesives and connect the parts to form an assembled unit. They also install hardware, such as pulls and drawer slides, and fit specialty products for glass, metal trims, electrical components, and stone. Finally, workers sand, stain, and, if necessary, coat the wood product with a sealer or topcoats, such as a lacquer or varnish.
Clamping mitered edges can be a real hassle because they never seems to line up correctly. The easiest way that I’ve found to get around this process is to use painter’s tape as clamps. First set the pieces so that the outer edges are facing up and tape them edge-to-edge. The flip the pieces over so the beveled edges are facing up and glue them together. Complete the process by taping the last two edges together and let sit until completed. The tape removes easily and the glue won’t attach to the tape, making sanding and finishing very simple. Try this tip with this clever project!
When it comes to woodworking for beginners, there are 6 things that I think are essential to know for how to start woodworking. I’m going to discuss each of these tips in hopefully a really simple way to make it a breeze to understand, so you can get to the fun part of actually starting to woodwork! I wish when I started and was learning how to woodwork that I had a simple beginner tips guide like this one!
Another important factor to be considered is the durability of the wood, especially in regards to moisture. If the finished project will be exposed to moisture (e.g. outdoor projects) or high humidity or condensation (e.g. in kitchens or bathrooms), then the wood needs to be especially durable in order to prevent rot. Because of their oily qualities, many tropical hardwoods such as teak and mahogany are popular for such applications.[9]
# While trying to trace an exact copy of the throat plate for my table saw, I came up with this nifty technique using an ordinary pencil. I just shaved my pencil into a half-pencil by carefully grinding it on my belt sander. The flat edge enables my modified pencil to ride straight up along the edge of the template. It also works great for marking and then shaping inlays for my woodworking projects. — Tim Reese. How to cut circles with a band saw.
Cutting sandpaper is a quick way to dull your scissors or utility knife blade. Instead, I fastened a hacksaw blade to the edge of my workbench. I slipped a washer behind the blade at each of the mounting holes so a sheet of sandpaper to easily slides in behind the blade. I fold the paper where I want to cut, just as a reference. — Kim Boley. Try some of these storage solutions!
There is no cost to use the database. Registering is not required. You should be able to browse the database and click through the links. Having said that, considering the fact there are so many browsers out there being used, this site's software might not allow some visitors to browse, it all depends on the Internet traffic, and your browser's compatibility.
Begin by cutting off a 10-in. length of the board and setting it aside. Rip the remaining 38-in. board to 6 in. wide and cut five evenly spaced saw kerfs 5/8 in. deep along one face. Crosscut the slotted board into four 9-in. pieces and glue them into a block, being careful not to slop glue into the saw kerfs (you can clean them out with a knife before the glue dries). Saw a 15-degree angle on one end and screw the plywood piece under the angled end of the block.
Build your own platform bed frame at your home by following the source linked tutorial given above. The source link also includes more pictures that can help you to build a better bed frame. You can see a step by step set of instructions and guidelines to follow with real life pictures, as well as you can download a PDF file detailing the list of materials and tools you’ll need, know about the length of every board, and most importantly color-coded illustrations of the building process.
The rubber cushion on my old palm sander was wearing thin around the edges. Because of its age, I couldn’t find a replacement pad. As I was drinking my beverage with a foam can cover around it, I realized I could cut the foam to fit the sander and glue it on. I peeled off the old pad, cleaned the metal base and attached the foam with contact cement. Works for clamp-on as well as stick-on sanding squares! You can find can covers at discount and convenience stores. — Allen J. Muldoon
Cut the 6-1/2-in. x 3-in. lid from the leftover board, and slice the remaining piece into 1/4-in.-thick pieces for the sides and end of the box. Glue them around the plywood floor. Cut a rabbet on three sides of the lid so it fits snugly on the box and drill a 5/8-in. hole for a finger pull. Then just add a finish and you’ve got a beautiful, useful gift. If you don’t have time to make a gift this year, consider offering to do something for the person. You could offer to sharpen their knives! Here’s how.
My planer blasts shavings all over the shop floor. I decided to make my own dust chute from 4-in. PVC sewer pipe (which has thinner walls than regular Schedule 40 pipe) and a couple caps. I cut a slit in the pipe and used a heat gun to soften the plastic. That allowed me to open the slit. (Heating PVC releases fumes; ventilation is critical.) I then drilled holes in the flap and screwed it to the planer housing. Finally, I cut a 2-1/2- in. hole in one of the end caps to accept my shop vacuum hose. Works great! — Luis Arce. Here’s what else you can do with PVC pipe.
Hardwoods are separated into two categories, temperate and tropical hardwoods, depending on their origin. Temperate hardwoods are found in the regions between the tropics and poles, and are of particular interest to wood workers for their cost-effective aesthetic appeal and sustainable sources.[9] Tropical hardwoods are found within the equatorial belt, including Africa, Asia, and South America. Hardwoods flaunt a higher density, around 65lb/cu ft as a result of slower growing rates and is more stable when drying.[9] As a result of its high density, hardwoods are typically heavier than softwoods but can also be more brittle.[9] While there are an abundant number of hardwood species, only 200 are common enough and pliable enough to be used for woodworking.[11] Hardwoods have a wide variety of properties, making it easy to find a hardwood to suit nearly any purpose, but they are especially suitable for outdoor use due to their strength and resilience to rot and decay.[9] The coloring of hardwoods ranges from light to very dark, making it especially versatile for aesthetic purposes. However, because hardwoods are more closely grained, they are typically harder to work than softwoods. They are also harder to acquire in the United States and, as a result, are more expensive.[9]
Whether you're new to woodworking or you've been doing it for years, Woodcraft's selection of woodworking projects is one the best places to find your next big project. Whether you're looking to make wooden furniture, pens, toys, jewelry boxes, or any other project in between, the avid woodworker is sure to find his or her next masterpiece here. Find hundreds of detailed woodworking plans with highly accurate illustrations, instructions, and dimensions. Be sure to check out our Make Something blog to learn expert insights and inspiration for your next woodworking project. | 2,084 | 9,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-22 | latest | en | 0.93629 |
http://archive.ambermd.org/201710/0028.html | 1,656,199,103,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00579.warc.gz | 2,986,550 | 3,503 | # Re: [AMBER] Amber16 electric field parameters
From: Pengfei Li <ambermailpengfei.gmail.com>
Date: Tue, 3 Oct 2017 21:10:00 -0500
Hi Alessandro,
> On Oct 2, 2017, at 8:52 AM, Ross Walker <ross.rosswalker.co.uk> wrote:
>
>>
>> On Oct 2, 2017, at 4:18 AM, Alessandro Mariani <alessandro.mariani.uniroma1.it> wrote:
>>
>> Hello,
>> as far as I can get, the conversion is something like:
>> 1 kcal/(mol*A*e) = 4.184 kJ/(mol*A*e) =
>> (4184/1.6022*10**-19)*V/(mol*A) = (4184/1.6022*10**-29)*V/(mol*m)
>> = 2.61*10**32 V/(mol*m)
>>
>> Now i really do not understand why that "mol" is there. What is it supposed
>> to be? Avogadro's number? The number of moles in the box? The number of
>> molecules in the box? Anyways, a conversion factor in 10**32 is
>> astonishing, and even if mol is substituted with Avogadro's number, I
>> get 1 kcal/(mol*A*e)
>> = 4.34*10**8 V/m that is a nonsense to me. Can please someone give me an
>> hint? And, maybe, explain to me why to not use SI units in Amber?
>
I think your calculation has no problem.
1 (kcal/mol*A*e) = 4184.0 / 1.6022 / 6.022 * 10^6 V/m, which is about 4.34 * 10^8 V/m.
In fact, people like to use MV/cm unit under this kind of situation.
1 MV/cm = 1*10^8 V/m
So 1 kcal/(mol*A*e) is about 4.34 MV/cm.
Kind regards,
Pengfei
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Custom Search | 569 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-27 | latest | en | 0.789025 |
http://eprints.utm.my/48121/ | 1,725,878,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00898.warc.gz | 9,810,508 | 5,636 | # Free and mixed convective boundary layer flow of a viscoelastic fluid past a horizontal circular cylinser
Mohd. Kasim, Abdul Rahman (2011) Free and mixed convective boundary layer flow of a viscoelastic fluid past a horizontal circular cylinser. Masters thesis, Universiti Teknologi Malaysia, Faculty of Science.
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## Abstract
The study o f viscoelastic fluid has become increasingly important in the last few years. This is mainly due to its many applications in petroleum drilling, manufacturing o f food and paper, and many other similar activities. In this thesis, the steady free and mixed convective boundary layer flow of a viscoelastic fluid past a horizontal circular cylinder has been studied separately subject to their own constant surface temperature boundary conditions. For the problem of m ixed convection, the study also considered the problem that subjected to constant heat flux boundary conditions. The constitutive equations of viscoelastic fluids usually generate a higher-order derivative term in the momentum equation than equations of Newtonian fluid. Thus, there are insufficient boundary conditions to solve the problems of viscoelastic fluid completely. Therefore, the augmentation o f an extra boundary condition is needed at infinity (far from the wall). In each case, the governing boundary layer equations are first transformed into a non-dimensional form, and then into a set of non similar boundary layer equations which are solved numerically using an efficient im plicit finite-difference method known as K eller-box scheme. Numerical result presented include velocity profiles, temperature profiles, heat transfer characteristics, namely the local heat transfer, local skin friction coefficient and local wall temperature distribution for a w ide range of material paramater K (viscoelastic parameter), prandtl number Pr, and m ixed convection parameter X. In each problem, it is found that velocity distributions decrease when the value of viscoelastic parameter, K increases, whereas the opposite behaviour is observed for the temperature distribution. It is worth mentioning that the results obtained in viscoelastics fluids when the parameter K = 0 (Newtonian fluids) are in excellent agreement with those obtained in viscous fluids (Newtonian fluids).
Item Type: Thesis (Masters) Thesis (Sarjana Sains (Matematik)) - Universiti Teknologi Malaysia, 2011 viscoelastic fluid, petroleum drilling, manufacturing of food Q Science > QA Mathematics Science 48121 Narimah Nawil 15 Oct 2015 01:09 30 May 2018 03:57
Repository Staff Only: item control page | 524 | 2,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.88394 |
https://ebubekirsezer.com/en/english-selection-sort-secerek-siralama/ | 1,621,242,794,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00125.warc.gz | 243,906,335 | 10,654 | Hello, Selection Sort is a sorting algorithm. With Selection Sort, we can sort the array minimum to the maximum or maximum to the minimum. Algorithm, choose the first element of the array as minimum and then in the for loops, we controlled the conditions if there is some minimum value than the or minimum then we change the their places. Algorithm is really simple to implement. Algorithms has complexity and complexity of the Selection Sort can find by the using this formula n2. For the make example about the algorithm, I implement the Selection Sort Algorithm using by the Python and C. In the examples, I sort the array minimum to maximum and maximum to minimum.
C Codes;
```#include<stdio.h>
void selectionSort(int array[],int size){
int min;
for(int i =0;i<size-1;i++){
min = i;
for(int j = i;j<size;j++){
if(array[j]<array[min]){
min = j;
}
}
int temp = array[i];
array[i] = array[min];
array[min] = temp;
}
}
void maxToMin(int array[],int size){
int max;
for(int i=0;i<size;i++){
max = i;
for(int j=i;j<size;j++){
if(array[j]>array[max]){
max= j;
}
}
int temp = array[i];
array[i] = array[max];
array[max] = temp;
}
}
int main(){
int array[] = {1,3,6,5,7,8,2,4};
int n = sizeof(array)/sizeof(array[0]);
selectionSort(array,n);
for(int i =0;i<n;i++){
printf("%d ",array[i]);
}
printf("\n");
maxToMin(array,n);
for(int i =0;i<n;i++){
printf("%d ",array[i]);
}
}```
Python Codes;
```array = [1,3,6,5,7,8,2,4]
for i in range(len(array)):
min = i
for j in range(i+1,len(array)):
if array[min]>array[j]:
min = j
array[i],array[min] = array[min],array[i]
print("Selection Sort")
for i in range(len(array)):
print("%d" %array[i])
for i in range(len(array)):
max = i
for j in range(i+1,len(array)):
if array[j]>array[max]:
max = j
array[i],array[max] = array[max],array[i]
print("Max to Min")
for i in range(len(array)):
print("%d " %array[i])
``` | 529 | 1,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-21 | latest | en | 0.418584 |
http://practicealgebra.net/about/poly_fact01.html | 1,508,575,032,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824675.67/warc/CC-MAIN-20171021081004-20171021101004-00831.warc.gz | 272,052,766 | 1,124 | Suppose you have to factor a polynomial like x2 − 5x + 6. That is, there's nothing in front of the x2, but there can be a linear term and a constant term. You need to find two numbers whose sum is 6 (the constant term) and whose product is −5 (the number in front of the x). The numbers that work are −2 and −3. (It doesn't matter what order we put them in.) Then the two factors are x − 2 and x − 3, so the answer is (x − 2)(x − 3).
• If there's no number in front of the x, then it's just as if there was a 1 there (or a −1, if it was − x).
• If there's no constant term, then it's as if the constant term is 0.
• Finally, if there's no x (just an x2), then it's as if there a 0 in front of the x. | 214 | 700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-43 | latest | en | 0.97486 |
https://books.google.com.eg/books?id=ibg2AAAAMAAJ&qtid=1e7e7815&lr=&hl=ar&source=gbs_quotes_r&cad=5 | 1,575,718,910,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540499389.15/warc/CC-MAIN-20191207105754-20191207133754-00169.warc.gz | 298,591,609 | 5,964 | بحث صور خرائط YouTube الأخبار Gmail Drive تقويم المزيد »
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الكتب الكتب 1 - 10 من 53 فيAC the same multiple of AD, that AB is of the part which is to be cut off from it....
AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off.
The Elements of Euclid: Viz. the First Six Books, with the Eleventh and ... - الصفحة 145
بواسطة Alexander Ingram - 1799 - عدد الصفحات: 351
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## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Robert Simson - 1762 - عدد الصفحات: 466
...ftraight line AC making any angle with AB ; and in AC take any point D, and take AC which is the fam« multiple of AD that AB is of the part which is to...join BC, and draw DE parallel to it. then AE is the fame part of AB that AD is of AC ; that is, AE is the part required to be cut off. Becaufe ED is parallel...
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## The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have ...
Robert Simson - 1775 - عدد الصفحات: 520
...requi- gee N. fed. Let AB be the given ftraight line ; it is required to cut off any part from it. Fiom the point A draw a ftraight line AC making any angle...parallel to one of the fides of the triangle ABC, viz.toBC,asCD is to DA, fois'BE to EA ; and, by compofition b, CA is to AD, as BA to AE : But CA is...
عرض كامل - حول هذا الكتاب
## Elements of Geometry;: Containing the First Six Books of Euclid, with Two ...
John Playfair - 1795 - عدد الصفحات: 400
...and in AC take any point D, and take AC M 4 the Book VI. the fame multiple of AD, that AB i? - f - of the part which is to be cut off from it ; join...one of the fides of the triangle ABC, viz. to BC, a i. 6. as CD is to DA, fo is» BE to EA; and, 6.18. 5. by compofition b, CA is to AD, as BA to, AE...
عرض كامل - حول هذا الكتاب
## Elements of Geometry;: Containing the First Six Books of Euclid, with a ...
Euclid, John Playfair - 1804 - عدد الصفحات: 440
...take AC M 4 fnch Book VI «t. '. b i«. 5. fuch that it fhall contain AD, as oft as AB js to contain the part, which is to be cut off from it ; join BC,...one of the fides of the triangle ABC, viz. to BC, CD : DA : : BE : EA » ; and by compofition b, CA : AD : : BA : AE : But CA is a multiple of AD ; therefore...
عرض كامل - حول هذا الكتاب
## The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The ...
Robert Simson - 1804
...ftraight line AC making any angle with AB; and in AC take any point D, and take AC which is the ftme multiple of AD that AB is of the part which is to...join BC, and draw DE parallel to it. then AE is the fame part of AB that AD is of AC ; that is, AE is the part required to be cut off. Becaufe ED is parallel...
عرض كامل - حول هذا الكتاب
## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1806 - عدد الصفحات: 311
...; in AC take any point D, and take AC sucli that it shall contain AD as" often as AB is to contain the part which is to be cut off from it ; join BC,...to it ; then AE is the part required to be cut off. Because ED is parallel to BC, one of the sides of the triangle ABC, CD : DA : : BE : EAa ; therefore,...
عرض كامل - حول هذا الكتاب
## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid, Robert Simson - 1806 - عدد الصفحات: 518
...draw a straight line AC making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it ; join DC, and draw DE parallel to it : then AE is the part required to be cut off. Because ED is parallel...
عرض كامل - حول هذا الكتاب
## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclid - 1810 - عدد الصفحات: 518
...draw a straight line AC making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to...to it : then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to / \ DA, so...
عرض كامل - حول هذا الكتاب
## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...
Euclides - 1816 - عدد الصفحات: 528
...draw a straight line AC, making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to...to it: Then AE is the part required to be cut off. . Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, so isa...
عرض كامل - حول هذا الكتاب
## Elements of Geometry: Containing the First Six Books of Euclid, with a ...
John Playfair - 1819 - عدد الصفحات: 333
...; and in AC take any point D, and take AC such that it shall contain AD, as oft as AB is to contain the part, which is to be cut off from it ; join BC,...to it : then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, CD : DA : : DK : 1^(2^6.)...
عرض كامل - حول هذا الكتاب | 1,463 | 5,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-51 | latest | en | 0.68941 |
https://solvedlib.com/n/the-natural-length-of-a-spring-is-2m-to-stretch-the-spring,15456521 | 1,653,485,227,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662587158.57/warc/CC-MAIN-20220525120449-20220525150449-00435.warc.gz | 588,512,230 | 18,968 | # The natural length of a spring is 2m. To stretch the spring fromits natural length to 4m requires 20N of
###### Question:
The natural length of a spring is 2m. To stretch the spring from its natural length to 4m requires 20N of work done. How much work is required to stretch the spring from its natural length to 6m
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WILL RATE <3! A 0.717 H inductor is connected in series with a fluorescent lamp to limit the current drawn by the lamp. If the combination is connected to a 61.3 Hz, 145 V line, and if the voltage across the lamp is to be 28.2 V, what is the current in the circuit? (The lamp is a pure resistive ... | 2,816 | 9,753 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.85552 |
https://mtp.tools/converters/exbibit-to-byte-calculator | 1,571,616,373,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00370.warc.gz | 617,359,830 | 5,759 | # Exbibits (Eib) to Bytes (B) calculator
Input the amount of exbibits you want to convert to bytes in the below input field, and then click in the "Convert" button. But if you want to convert from bytes to exbibits, please checkout this tool.
## Formula
Formula used to convert Eib to B:
F(x) = x * 144115188075855872
For example, if you want to convert 15 Eib to B, just replace x by 15 [Eib]:
15 Eib = 15*144115188075855872 = 2161727821137838080 B
## Steps
1. Multiply the amount of exbibits by 144115188075855872.
2. The result will be expressed in bytes.
## Exbibit to Byte Conversion Table
The following table will show the most common conversions for Exbibits (Eib) to Bytes (B):
Exbibits (Eib) Bytes (B)
0.001 Eib 144115188075855.875 B
0.01 Eib 1441151880758558.75 B
0.1 Eib 14411518807585588 B
1 Eib 144115188075855872 B
2 Eib 288230376151711744 B
3 Eib 432345564227567616 B
4 Eib 576460752303423488 B
5 Eib 720575940379279360 B
6 Eib 864691128455135232 B
7 Eib 1008806316530991104 B
8 Eib 1152921504606846976 B
9 Eib 1297036692682702848 B
10 Eib 1441151880758558720 B
20 Eib 2882303761517117440 B
30 Eib 4323455642275676160 B
40 Eib 5764607523034234880 B
50 Eib 7205759403792793600 B
60 Eib 8646911284551352320 B
70 Eib 10088063165309911040 B
80 Eib 11529215046068469760 B
90 Eib 12970366926827028480 B
100 Eib 14411518807585587200 B
## About Exbibits (Eib)
A exbibit is a unit of measurement for digital information and computer storage. The binary prefix exbi (which is expressed with the letters Ei) is defined in the International System of Quantities (ISQ) as a multiplier of 2^60. Therefore, 1 exbibit is equal to 1,024 pebibits and equal to 1,152,921,504,606,846,976 bits (around 1.152 exabits). The symbol commonly used to represent a exbibit is Eib (sometimes as Eibit).
## About Bytes (B)
A byte is a unit of digital information that represents eight (8) bits. It is widely used in computing and in digital communications. The symbol used to represent a byte is B (upper-case letter B). It was designated by the International Electrotechnical Commission (IEC) and Institute of Electrical and Electronics Engineers (IEEE).
The amount of bits that a byte represent changed over the years, but nowadays a byte represents 8 bits. With 1 byte you can represent numbers from 0 to 255.
## FAQs for Exbibit to Byte calculator
### What is Exbibit to Byte calculator?
Exbibit to Byte is a free and online calculator that converts Exbibits to Bytes.
### How do I use Exbibit to Byte?
You just have to insert the amount of Exbibits you want to convert and press the "Convert" button. The amount of Bytes will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Exbibit to Byte work on?
Exbibit to Byte calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 897 | 3,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-43 | latest | en | 0.653379 |
https://www.w3resource.com/machine-learning/tensorflow/python-tensorflow-basic-exercise-10.php | 1,709,205,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00778.warc.gz | 1,029,388,576 | 27,764 | TensorFlow constant and variable operations in Python
TensorFlow constant and variable operations in Python
Python TensorFlow Basic: Exercise-10 with Solution
Write a Python program that creates a TensorFlow constant and a TensorFlow variable. Add, subtract, multiply and divide them and print the result.
Sample Solution:
Python Code:
``````import tensorflow as tf
# Create a TensorFlow constant and a TensorFlow variable
# Tensors are multi-dimensional arrays with a uniform type (called a dtype ).
constant_ts = tf.constant(4.0)
initial_value = 5.0
variable_ts = tf.Variable(initial_value)
# Perform mathematical operations
subtract_result = constant_ts - variable_ts
multiply_result = constant_ts * variable_ts
divide_result = constant_ts / variable_ts
# Print the results
print("Constant:", constant_ts.numpy())
print("Variable (initial value):", initial_value) # Variables are not initialized explicitly in TensorFlow 2.x
print("Subtraction Result:", subtract_result.numpy())
print("Multiplication Result:", multiply_result.numpy())
print("Division Result:", divide_result.numpy())
```
```
Output:
```Constant: 4.0
Variable (initial value): 5.0
Subtraction Result: -1.0
Multiplication Result: 20.0
Division Result: 0.8
```
Explanation:
In the exercise above -
• Create a TensorFlow constant named constant_tensor with the value 5.0.
• Create a TensorFlow variable named variable_tensor with an initial value of 3.0.
• Perform addition, subtraction, multiplication, and division operations between the constant and the variable.
• Initialize the variable using tf.compat.v1.global_variables_initializer().run() (for TensorFlow 2.x, use tf.compat.v1.global_variables_initializer().run()).
• Finally, we print the values of the constant, the initialized variable, and the results of mathematical operations.
Python Code Editor:
What is the difficulty level of this exercise?
| 401 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-10 | longest | en | 0.692686 |
https://mail.scipy.org/pipermail/scipy-user/2010-June/025650.html | 1,495,566,261,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607649.26/warc/CC-MAIN-20170523183134-20170523203134-00280.warc.gz | 778,555,836 | 8,726 | [SciPy-User] re[SciPy-user] moving for loops...
Benjamin Root ben.root@ou....
Tue Jun 8 21:39:11 CDT 2010
```Correction for me as well. To mask out the negative values, use masked
arrays. So we will turn jules_2d into a masked array (second line), then
all subsequent commands will still work as expected. It is very similar to
replacing negative values with nans and using nanmin().
> jules_2d = jules.reshape((-1, 12))
> jules_monthly = np.mean(jules_2d, axis=0)
> print jules_monthly.shape
(12,)
Ben Root
On Tue, Jun 8, 2010 at 7:49 PM, Benjamin Root <ben.root@ou.edu> wrote:
> The np.mod in my example caused the data points to stay within [0, 11] in
> order to illustrate that these are months. In my example, months are
> column, years are rows. In your desired output, months are rows and years
> are columns. It makes very little difference which way you have it.
>
> Anyway, let's imagine that we have a time series of data "jules". We can
> easily reshape this like so:
>
> > jules_2d = jules.reshape((-1, 12))
> > jules_monthly = np.mean(jules_2d, axis=0)
> > print jules_monthly.shape
> (12,)
>
> voila! You have 12 values in jules_monthly which are means for that month
> across all years.
>
> protip - if you want yearly averages just change the ax parameter in
> np.mean():
> > jules_yearly = np.mean(jules_2d, axis=1)
>
> I hope that makes my previous explanation clearer.
>
> Ben Root
>
>
> On Tue, Jun 8, 2010 at 5:41 PM, mdekauwe <mdekauwe@gmail.com> wrote:
>
>>
>> OK...
>>
>> but if I do...
>>
>> In [28]: np.mod(np.arange(nummonths*numyears), nummonths).reshape((-1,
>> nummonths))
>> Out[28]:
>> array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]])
>>
>> When really I would be after something like this I think?
>>
>> array([ 0, 12, 24, 36, 48, 60, 72, 84, 96, 108, 120],
>> [ 1, 13, 25, 37, 49, 61, 73, 85, 97, 109, 121],
>> [ 2, 14, 26, 38, 50, 62, 74, 86, 98, 110, 122]
>> etc, etc
>>
>> i.e. so for each month jump across the years.
>>
>> Not quite sure of this example...this is what I currently have which does
>> seem to work, though I guess not completely efficiently.
>>
>> for month in xrange(nummonths):
>> tmp = jules[xrange(0, numyears * nummonths, nummonths),VAR,:,0]
>> tmp[tmp < 0.0] = np.nan
>> data[month,:] = np.mean(tmp, axis=0)
>>
>>
>>
>>
>> Benjamin Root-2 wrote:
>> >
>> > If you want an average for each month from your timeseries, then the
>> > sneaky
>> > way would be to reshape your array so that the time dimension is split
>> > into
>> > two (month, year) dimensions.
>> >
>> > For a 1-D array, this would be:
>> >
>> >> dataarray = numpy.mod(numpy.arange(36), 12)
>> >> print dataarray
>> > array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3,
>> 4,
>> > 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8,
>> 9,
>> > 10, 11])
>> >> datamatrix = dataarray.reshape((-1, 12))
>> >> print datamatrix
>> > array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> > [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
>> > [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]])
>> >
>> > Hope that helps.
>> >
>> > Ben Root
>> >
>> >
>> > On Fri, May 28, 2010 at 3:28 PM, mdekauwe <mdekauwe@gmail.com> wrote:
>> >
>> >>
>> >> OK so I just need to have a quick loop across the 12 months then, that
>> is
>> >> fine, just thought there might have been a sneaky way!
>> >>
>> >> Really appreciated, getting there slowly!
>> >>
>> >>
>> >>
>> >> josef.pktd wrote:
>> >> >
>> >> > On Fri, May 28, 2010 at 4:14 PM, mdekauwe <mdekauwe@gmail.com>
>> wrote:
>> >> >>
>> >> >> ok - something like this then...but how would i get the index for
>> the
>> >> >> month
>> >> >> for the data array (where month is 0, 1, 2, 4 ... 11)?
>> >> >>
>> >> >> data[month,:] = array[xrange(0, numyears * nummonths,
>> >> nummonths),VAR,:,0]
>> >> >
>> >> > you would still need to start at the right month
>> >> > data[month,:] = array[xrange(month, numyears * nummonths,
>> >> > nummonths),VAR,:,0]
>> >> > or
>> >> > data[month,:] = array[month: numyears * nummonths :
>> nummonths),VAR,:,0]
>> >> >
>> >> > an alternative would be a reshape with an extra month dimension and
>> >> > then sum only once over the year axis. this might be faster but
>> >> > trickier to get the correct reshape .
>> >> >
>> >> > Josef
>> >> >
>> >> >>
>> >> >> and would that be quicker than making an array months...
>> >> >>
>> >> >> months = np.arange(numyears * nummonths)
>> >> >>
>> >> >> and you that instead like you suggested x[start:end:12,:]?
>> >> >>
>> >> >> Many thanks again...
>> >> >>
>> >> >>
>> >> >> josef.pktd wrote:
>> >> >>>
>> >> >>> On Fri, May 28, 2010 at 3:53 PM, mdekauwe <mdekauwe@gmail.com>
>> wrote:
>> >> >>>>
>> >> >>>> Ok thanks...I'll take a look.
>> >> >>>>
>> >> >>>> Back to my loops issue. What if instead this time I wanted to take
>> >> an
>> >> >>>> average so every march in 11 years, is there a quicker way to go
>> >> >>>> doing
>> >> >>>> that than my current method?
>> >> >>>>
>> >> >>>> nummonths = 12
>> >> >>>> numyears = 11
>> >> >>>>
>> >> >>>> for month in xrange(nummonths):
>> >> >>>> for i in xrange(numpts):
>> >> >>>> for ym in xrange(month, numyears * nummonths, nummonths):
>> >> >>>> data[month, i] += array[ym, VAR, land_pts_index[i], 0]
>> >> >>>
>> >> >>>
>> >> >>> x[start:end:12,:] gives you every 12th row of an array x
>> >> >>>
>> >> >>> something like this should work to get rid of the inner loop, or
>> you
>> >> >>> could directly put
>> >> >>> range(month, numyears * nummonths, nummonths) into the array
>> >> >>> of ym and sum()
>> >> >>>
>> >> >>> Josef
>> >> >>>
>> >> >>>
>> >> >>>>
>> >> >>>> so for each point in the array for a given month i am jumping
>> >> through
>> >> >>>> and
>> >> >>>> getting the next years month and so on, summing it.
>> >> >>>>
>> >> >>>> Thanks...
>> >> >>>>
>> >> >>>>
>> >> >>>> josef.pktd wrote:
>> >> >>>>>
>> >> >>>>> On Wed, May 26, 2010 at 5:03 PM, mdekauwe <mdekauwe@gmail.com>
>> >> wrote:
>> >> >>>>>>
>> >> >>>>>> Could you possibly if you have time explain further your comment
>> >> re
>> >> >>>>>> the
>> >> >>>>>> p-values, your suggesting I am misusing them?
>> >> >>>>>
>> >> >>>>> Depends on your use and interpretation
>> >> >>>>>
>> >> >>>>> test statistics, p-values are random variables, if you look at
>> >> several
>> >> >>>>> tests at the same time, some p-values will be large just by
>> chance.
>> >> >>>>> If, for example you just look at the largest test statistic, then
>> >> the
>> >> >>>>> distribution for the max of several test statistics is not the
>> same
>> >> as
>> >> >>>>> the distribution for a single test statistic
>> >> >>>>>
>> >> >>>>> http://en.wikipedia.org/wiki/Multiple_comparisons
>> >> >>>>> http://www.itl.nist.gov/div898/handbook/prc/section4/prc47.htm
>> >> >>>>>
>> >> >>>>> we also just had a related discussion for ANOVA post-hoc tests on
>> >> the
>> >> >>>>> pystatsmodels group.
>> >> >>>>>
>> >> >>>>> Josef
>> >> >>>>>>
>> >> >>>>>> Thanks.
>> >> >>>>>>
>> >> >>>>>>
>> >> >>>>>> josef.pktd wrote:
>> >> >>>>>>>
>> >> >>>>>>> On Sat, May 22, 2010 at 6:21 AM, mdekauwe <mdekauwe@gmail.com>
>> >> >>>>>>> wrote:
>> >> >>>>>>>>
>> >> >>>>>>>> Sounds like I am stuck with the loop as I need to do the
>> >> comparison
>> >> >>>>>>>> for
>> >> >>>>>>>> each
>> >> >>>>>>>> pixel of the world and then I have a basemap function call
>> which
>> >> I
>> >> >>>>>>>> guess
>> >> >>>>>>>> slows it down further...hmm
>> >> >>>>>>>
>> >> >>>>>>> I don't see much that could be done differently, after a brief
>> >> look.
>> >> >>>>>>>
>> >> >>>>>>> stats.pearsonr could be replaced by an array version using
>> >> directly
>> >> >>>>>>> the formula for correlation even with nans. wilcoxon looks
>> slow,
>> >> and
>> >> >>>>>>> I
>> >> >>>>>>> never tried or seen a faster version.
>> >> >>>>>>>
>> >> >>>>>>> just a reminder, the p-values are for a single test, when you
>> >> have
>> >> >>>>>>> many of them, then they don't have the right size/confidence
>> >> level
>> >> >>>>>>> for
>> >> >>>>>>> an overall or joint test. (some packages report a Bonferroni
>> >> >>>>>>> correction in this case)
>> >> >>>>>>>
>> >> >>>>>>> Josef
>> >> >>>>>>>
>> >> >>>>>>>
>> >> >>>>>>>>
>> >> >>>>>>>> i.e.
>> >> >>>>>>>>
>> >> >>>>>>>> def compareSnowData(jules_var):
>> >> >>>>>>>> # Extract the 11 years of snow data and return
>> >> >>>>>>>> outrows = 180
>> >> >>>>>>>> outcols = 360
>> >> >>>>>>>> numyears = 11
>> >> >>>>>>>> nummonths = 12
>> >> >>>>>>>>
>> >> >>>>>>>> # Read various files
>> >> >>>>>>>> fname="world_valid_jules_pts.ascii"
>> >> >>>>>>>> (numpts, land_pts_index, latitude, longitude, rows, cols) =
>> >> >>>>>>>>
>> >> >>>>>>>> fname = "globalSnowRun_1985_96.GSWP2.nsmax0.mon.gra"
>> >> >>>>>>>> jules_data1 = jo.readJulesOutBinary(fname, numrows=15238,
>> >> >>>>>>>> numcols=1,
>> >> >>>>>>>> \
>> >> >>>>>>>> timesteps=132, numvars=26)
>> >> >>>>>>>> fname = "globalSnowRun_1985_96.GSWP2.nsmax3.mon.gra"
>> >> >>>>>>>> jules_data2 = jo.readJulesOutBinary(fname, numrows=15238,
>> >> >>>>>>>> numcols=1,
>> >> >>>>>>>> \
>> >> >>>>>>>> timesteps=132, numvars=26)
>> >> >>>>>>>>
>> >> >>>>>>>> # grab some space
>> >> >>>>>>>> data1_snow = np.zeros((nummonths * numyears, numpts),
>> >> >>>>>>>> dtype=np.float32)
>> >> >>>>>>>> data2_snow = np.zeros((nummonths * numyears, numpts),
>> >> >>>>>>>> dtype=np.float32)
>> >> >>>>>>>> pearsonsr_snow = np.ones((outrows, outcols),
>> >> dtype=np.float32)
>> >> *
>> >> >>>>>>>> np.nan
>> >> >>>>>>>> wilcoxStats_snow = np.ones((outrows, outcols),
>> >> dtype=np.float32)
>> >> >>>>>>>> *
>> >> >>>>>>>> np.nan
>> >> >>>>>>>>
>> >> >>>>>>>> # extract the data
>> >> >>>>>>>> data1_snow = jules_data1[:,jules_var,:,0]
>> >> >>>>>>>> data2_snow = jules_data2[:,jules_var,:,0]
>> >> >>>>>>>> data1_snow = np.where(data1_snow < 0.0, np.nan, data1_snow)
>> >> >>>>>>>> data2_snow = np.where(data2_snow < 0.0, np.nan, data2_snow)
>> >> >>>>>>>> #for month in xrange(numyears * nummonths):
>> >> >>>>>>>> # for i in xrange(numpts):
>> >> >>>>>>>> # data1 =
>> >> >>>>>>>> jules_data1[month,jules_var,land_pts_index[i],0]
>> >> >>>>>>>> # data2 =
>> >> >>>>>>>> jules_data2[month,jules_var,land_pts_index[i],0]
>> >> >>>>>>>> # if data1 >= 0.0:
>> >> >>>>>>>> # data1_snow[month,i] = data1
>> >> >>>>>>>> # else:
>> >> >>>>>>>> # data1_snow[month,i] = np.nan
>> >> >>>>>>>> # if data2 > 0.0:
>> >> >>>>>>>> # data2_snow[month,i] = data2
>> >> >>>>>>>> # else:
>> >> >>>>>>>> # data2_snow[month,i] = np.nan
>> >> >>>>>>>>
>> >> >>>>>>>> # exclude any months from *both* arrays where we have dodgy
>> >> >>>>>>>> data,
>> >> >>>>>>>> else
>> >> >>>>>>>> we
>> >> >>>>>>>> # can't do the correlations correctly!!
>> >> >>>>>>>> data1_snow = np.where(np.isnan(data2_snow), np.nan,
>> >> data1_snow)
>> >> >>>>>>>> data2_snow = np.where(np.isnan(data1_snow), np.nan,
>> >> data1_snow)
>> >> >>>>>>>>
>> >> >>>>>>>> # put data on a regular grid...
>> >> >>>>>>>> print 'regridding landpts...'
>> >> >>>>>>>> for i in xrange(numpts):
>> >> >>>>>>>> # exclude the NaN, note masking them doesn't work in
>> the
>> >> >>>>>>>> stats
>> >> >>>>>>>> func
>> >> >>>>>>>> x = data1_snow[:,i]
>> >> >>>>>>>> x = x[np.isfinite(x)]
>> >> >>>>>>>> y = data2_snow[:,i]
>> >> >>>>>>>> y = y[np.isfinite(y)]
>> >> >>>>>>>>
>> >> >>>>>>>> # r^2
>> >> >>>>>>>> # exclude v.small arrays, i.e. we need just less over 4
>> >> >>>>>>>> years
>> >> >>>>>>>> of
>> >> >>>>>>>> data
>> >> >>>>>>>> if len(x) and len(y) > 50:
>> >> >>>>>>>> pearsonsr_snow[((180-1)-(rows[i]-1)),cols[i]-1] =
>> >> >>>>>>>> (stats.pearsonr(x, y)[0])**2
>> >> >>>>>>>>
>> >> >>>>>>>> # wilcox signed rank test
>> >> >>>>>>>> # make sure we have enough samples to do the test
>> >> >>>>>>>> d = x - y
>> >> >>>>>>>> d = np.compress(np.not_equal(d,0), d ,axis=-1) # Keep
>> all
>> >> >>>>>>>> non-zero
>> >> >>>>>>>> differences
>> >> >>>>>>>> count = len(d)
>> >> >>>>>>>> if count > 10:
>> >> >>>>>>>> z, pval = stats.wilcoxon(x, y)
>> >> >>>>>>>> # only map out sign different data
>> >> >>>>>>>> if pval < 0.05:
>> >> >>>>>>>>
>> wilcoxStats_snow[((180-1)-(rows[i]-1)),cols[i]-1]
>> >> =
>> >> >>>>>>>> np.mean(x - y)
>> >> >>>>>>>>
>> >> >>>>>>>> return (pearsonsr_snow, wilcoxStats_snow)
>> >> >>>>>>>>
>> >> >>>>>>>>
>> >> >>>>>>>> josef.pktd wrote:
>> >> >>>>>>>>>
>> >> >>>>>>>>> On Fri, May 21, 2010 at 10:14 PM, mdekauwe <
>> mdekauwe@gmail.com>
>> >> >>>>>>>>> wrote:
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> Also I then need to remap the 2D array I make onto another
>> >> grid
>> >> >>>>>>>>>> (the
>> >> >>>>>>>>>> world in
>> >> >>>>>>>>>> this case). Which again I had am doing with a loop (note
>> >> numpts
>> >> >>>>>>>>>> is
>> >> >>>>>>>>>> a
>> >> >>>>>>>>>> lot
>> >> >>>>>>>>>> bigger than my example above).
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> wilcoxStats_snow = np.ones((outrows, outcols),
>> >> dtype=np.float32)
>> >> >>>>>>>>>> *
>> >> >>>>>>>>>> np.nan
>> >> >>>>>>>>>> for i in xrange(numpts):
>> >> >>>>>>>>>> # exclude the NaN, note masking them doesn't work in
>> >> the
>> >> >>>>>>>>>> stats
>> >> >>>>>>>>>> func
>> >> >>>>>>>>>> x = data1_snow[:,i]
>> >> >>>>>>>>>> x = x[np.isfinite(x)]
>> >> >>>>>>>>>> y = data2_snow[:,i]
>> >> >>>>>>>>>> y = y[np.isfinite(y)]
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> # wilcox signed rank test
>> >> >>>>>>>>>> # make sure we have enough samples to do the test
>> >> >>>>>>>>>> d = x - y
>> >> >>>>>>>>>> d = np.compress(np.not_equal(d,0), d ,axis=-1) # Keep
>> >> all
>> >> >>>>>>>>>> non-zero
>> >> >>>>>>>>>> differences
>> >> >>>>>>>>>> count = len(d)
>> >> >>>>>>>>>> if count > 10:
>> >> >>>>>>>>>> z, pval = stats.wilcoxon(x, y)
>> >> >>>>>>>>>> # only map out sign different data
>> >> >>>>>>>>>> if pval < 0.05:
>> >> >>>>>>>>>>
>> >> wilcoxStats_snow[((180-1)-(rows[i]-1)),cols[i]-1]
>> >> >>>>>>>>>> =
>> >> >>>>>>>>>> np.mean(x - y)
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> Now I think I can push the data in one move into the
>> >> >>>>>>>>>> wilcoxStats_snow
>> >> >>>>>>>>>> array
>> >> >>>>>>>>>> by removing the index,
>> >> >>>>>>>>>> but I can't see how I will get the individual x and y pts
>> for
>> >> >>>>>>>>>> each
>> >> >>>>>>>>>> array
>> >> >>>>>>>>>> member correctly without the loop, this was my attempt which
>> >> of
>> >> >>>>>>>>>> course
>> >> >>>>>>>>>> doesn't work!
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> x = data1_snow[:,:]
>> >> >>>>>>>>>> x = x[np.isfinite(x)]
>> >> >>>>>>>>>> y = data2_snow[:,:]
>> >> >>>>>>>>>> y = y[np.isfinite(y)]
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> # r^2
>> >> >>>>>>>>>> # exclude v.small arrays, i.e. we need just less over 4
>> years
>> >> of
>> >> >>>>>>>>>> data
>> >> >>>>>>>>>> if len(x) and len(y) > 50:
>> >> >>>>>>>>>> pearsonsr_snow[((180-1)-(rows-1)),cols-1] =
>> >> (stats.pearsonr(x,
>> >> >>>>>>>>>> y)[0])**2
>> >> >>>>>>>>>
>> >> >>>>>>>>>
>> >> >>>>>>>>> If you want to do pairwise comparisons with stats.wilcoxon,
>> >> then
>> >> >>>>>>>>> you
>> >> >>>>>>>>> might be stuck with the loop, since wilcoxon takes only two
>> 1d
>> >> >>>>>>>>> arrays
>> >> >>>>>>>>> at a time (if I read the help correctly).
>> >> >>>>>>>>>
>> >> >>>>>>>>> Also the presence of nans might force the use a loop.
>> >> stats.mstats
>> >> >>>>>>>>> has
>> >> >>>>>>>>> masked array versions, but I didn't see wilcoxon in the list.
>> >> >>>>>>>>> (Even
>> >> >>>>>>>>> when vectorized operations would work with regular arrays,
>> nan
>> >> or
>> >> >>>>>>>>> masked array versions still have to loop in many cases.)
>> >> >>>>>>>>>
>> >> >>>>>>>>> If you have many columns with count <= 10, so that wilcoxon
>> is
>> >> not
>> >> >>>>>>>>> calculated then it might be worth to use only array
>> operations
>> >> up
>> >> >>>>>>>>> to
>> >> >>>>>>>>> that point. If wilcoxon is calculated most of the time, then
>> >> it's
>> >> >>>>>>>>> not
>> >> >>>>>>>>>
>> >> >>>>>>>>> Josef
>> >> >>>>>>>>>
>> >> >>>>>>>>>
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> thanks.
>> >> >>>>>>>>>>
>> >> >>>>>>>>>>
>> >> >>>>>>>>>>
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> mdekauwe wrote:
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>> Yes as Zachary said index is only 0 to 15237, so both
>> methods
>> >> >>>>>>>>>>> work.
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>> I don't quite get what you mean about slicing with axis >
>> 3.
>> >> Is
>> >> >>>>>>>>>>> there
>> >> >>>>>>>>>>> a
>> >> >>>>>>>>>>> link you can recommend I should read? Does that mean given
>> I
>> >> >>>>>>>>>>> have
>> >> >>>>>>>>>>> 4dims
>> >> >>>>>>>>>>> that Josef's suggestion would be more advised in this case?
>> >> >>>>>>>>>
>> >> >>>>>>>>> There were several discussions on the mailing lists (fancy
>> >> slicing
>> >> >>>>>>>>> and
>> >> >>>>>>>>> indexing). Your case is safe, but if you run in future into
>> >> funny
>> >> >>>>>>>>> shapes, you can look up the details.
>> >> >>>>>>>>> when in doubt, I use np.arange(...)
>> >> >>>>>>>>>
>> >> >>>>>>>>> Josef
>> >> >>>>>>>>>
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>> Thanks.
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>> josef.pktd wrote:
>> >> >>>>>>>>>>>>
>> >> >>>>>>>>>>>> On Fri, May 21, 2010 at 10:55 AM, mdekauwe <
>> >> mdekauwe@gmail.com>
>> >> >>>>>>>>>>>> wrote:
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>> Thanks that works...
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>> So the way to do it is with np.arange(tsteps)[:,None],
>> that
>> >> >>>>>>>>>>>>> was
>> >> >>>>>>>>>>>>> the
>> >> >>>>>>>>>>>>> step
>> >> >>>>>>>>>>>>> I
>> >> >>>>>>>>>>>>> was struggling with, so this forms a 2D array which
>> >> replaces
>> >> >>>>>>>>>>>>> the
>> >> >>>>>>>>>>>>> the
>> >> >>>>>>>>>>>>> two
>> >> >>>>>>>>>>>>> for
>> >> >>>>>>>>>>>>> loops? Do I have that right?
>> >> >>>>>>>>>>>>
>> >> >>>>>>>>>>>> Yes, but as Zachary showed, if you need the full index in
>> a
>> >> >>>>>>>>>>>> dimension,
>> >> >>>>>>>>>>>> then you can use slicing. It might be faster.
>> >> >>>>>>>>>>>> And a warning, mixing slices and index arrays with 3 or
>> more
>> >> >>>>>>>>>>>> dimensions can have some surprise switching of axes.
>> >> >>>>>>>>>>>>
>> >> >>>>>>>>>>>> Josef
>> >> >>>>>>>>>>>>
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>> A lot quicker...!
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>> Martin
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>> josef.pktd wrote:
>> >> >>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>> On Fri, May 21, 2010 at 8:59 AM, mdekauwe
>> >> >>>>>>>>>>>>>> <mdekauwe@gmail.com>
>> >> >>>>>>>>>>>>>> wrote:
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> Hi,
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> I am trying to extract data from a 4D array and store
>> it
>> >> in
>> >> >>>>>>>>>>>>>>> a
>> >> >>>>>>>>>>>>>>> 2D
>> >> >>>>>>>>>>>>>>> array,
>> >> >>>>>>>>>>>>>>> but
>> >> >>>>>>>>>>>>>>> avoid my current usage of the for loops for speed, as
>> in
>> >> >>>>>>>>>>>>>>> reality
>> >> >>>>>>>>>>>>>>> the
>> >> >>>>>>>>>>>>>>> arrays
>> >> >>>>>>>>>>>>>>> sizes are quite big. Could someone also try and explain
>> >> the
>> >> >>>>>>>>>>>>>>> solution
>> >> >>>>>>>>>>>>>>> as
>> >> >>>>>>>>>>>>>>> well
>> >> >>>>>>>>>>>>>>> if they have a spare moment as I am still finding it
>> >> quite
>> >> >>>>>>>>>>>>>>> difficult
>> >> >>>>>>>>>>>>>>> to
>> >> >>>>>>>>>>>>>>> get
>> >> >>>>>>>>>>>>>>> over the habit of using loops (C convert for my sins).
>> I
>> >> get
>> >> >>>>>>>>>>>>>>> that
>> >> >>>>>>>>>>>>>>> one
>> >> >>>>>>>>>>>>>>> could
>> >> >>>>>>>>>>>>>>> precompute the indices's i and j i.e.
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> i = np.arange(tsteps)
>> >> >>>>>>>>>>>>>>> j = np.arange(numpts)
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> but just can't get my head round how i then use them...
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> Thanks,
>> >> >>>>>>>>>>>>>>> Martin
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> import numpy as np
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> numpts=10
>> >> >>>>>>>>>>>>>>> tsteps = 12
>> >> >>>>>>>>>>>>>>> vari = 22
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> data = np.random.random((tsteps, vari, numpts, 1))
>> >> >>>>>>>>>>>>>>> new_data = np.zeros((tsteps, numpts), dtype=np.float32)
>> >> >>>>>>>>>>>>>>> index = np.arange(numpts)
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> for i in xrange(tsteps):
>> >> >>>>>>>>>>>>>>> for j in xrange(numpts):
>> >> >>>>>>>>>>>>>>> new_data[i,j] = data[i,5,index[j],0]
>> >> >>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>> The index arrays need to be broadcastable against each
>> >> other.
>> >> >>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>> I think this should do it
>> >> >>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>> new_data = data[np.arange(tsteps)[:,None], 5,
>> >> >>>>>>>>>>>>>> np.arange(numpts),
>> >> >>>>>>>>>>>>>> 0]
>> >> >>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>> Josef
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> --
>> >> >>>>>>>>>>>>>>> View this message in context:
>> >> >>>>>>>>>>>>>>>
>> >> http://old.nabble.com/removing-for-loops...-tp28633477p28633477.html
>> >> >>>>>>>>>>>>>>> Sent from the Scipy-User mailing list archive at
>> >> Nabble.com.
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>> _______________________________________________
>> >> >>>>>>>>>>>>>>> SciPy-User mailing list
>> >> >>>>>>>>>>>>>>> SciPy-User@scipy.org
>> >> >>>>>>>>>>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>> _______________________________________________
>> >> >>>>>>>>>>>>>> SciPy-User mailing list
>> >> >>>>>>>>>>>>>> SciPy-User@scipy.org
>> >> >>>>>>>>>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>>
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>> --
>> >> >>>>>>>>>>>>> View this message in context:
>> >> >>>>>>>>>>>>>
>> >> http://old.nabble.com/removing-for-loops...-tp28633477p28634924.html
>> >> >>>>>>>>>>>>> Sent from the Scipy-User mailing list archive at
>> >> Nabble.com.
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>>> _______________________________________________
>> >> >>>>>>>>>>>>> SciPy-User mailing list
>> >> >>>>>>>>>>>>> SciPy-User@scipy.org
>> >> >>>>>>>>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>>>>>>>
>> >> >>>>>>>>>>>> _______________________________________________
>> >> >>>>>>>>>>>> SciPy-User mailing list
>> >> >>>>>>>>>>>> SciPy-User@scipy.org
>> >> >>>>>>>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>>>>>>
>> >> >>>>>>>>>>>>
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>>
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> --
>> >> >>>>>>>>>> View this message in context:
>> >> >>>>>>>>>>
>> >> http://old.nabble.com/removing-for-loops...-tp28633477p28640656.html
>> >> >>>>>>>>>> Sent from the Scipy-User mailing list archive at Nabble.com.
>> >> >>>>>>>>>>
>> >> >>>>>>>>>> _______________________________________________
>> >> >>>>>>>>>> SciPy-User mailing list
>> >> >>>>>>>>>> SciPy-User@scipy.org
>> >> >>>>>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>>>>
>> >> >>>>>>>>> _______________________________________________
>> >> >>>>>>>>> SciPy-User mailing list
>> >> >>>>>>>>> SciPy-User@scipy.org
>> >> >>>>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>>>
>> >> >>>>>>>>>
>> >> >>>>>>>>
>> >> >>>>>>>> --
>> >> >>>>>>>> View this message in context:
>> >> >>>>>>>>
>> >> http://old.nabble.com/removing-for-loops...-tp28633477p28642434.html
>> >> >>>>>>>> Sent from the Scipy-User mailing list archive at Nabble.com.
>> >> >>>>>>>>
>> >> >>>>>>>> _______________________________________________
>> >> >>>>>>>> SciPy-User mailing list
>> >> >>>>>>>> SciPy-User@scipy.org
>> >> >>>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>>
>> >> >>>>>>> _______________________________________________
>> >> >>>>>>> SciPy-User mailing list
>> >> >>>>>>> SciPy-User@scipy.org
>> >> >>>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>>
>> >> >>>>>>>
>> >> >>>>>>
>> >> >>>>>> --
>> >> >>>>>> View this message in context:
>> >> >>>>>>
>> >> http://old.nabble.com/removing-for-loops...-tp28633477p28686356.html
>> >> >>>>>> Sent from the Scipy-User mailing list archive at Nabble.com.
>> >> >>>>>>
>> >> >>>>>> _______________________________________________
>> >> >>>>>> SciPy-User mailing list
>> >> >>>>>> SciPy-User@scipy.org
>> >> >>>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>>
>> >> >>>>> _______________________________________________
>> >> >>>>> SciPy-User mailing list
>> >> >>>>> SciPy-User@scipy.org
>> >> >>>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>>
>> >> >>>>>
>> >> >>>>
>> >> >>>> --
>> >> >>>> View this message in context:
>> >> >>>>
>> http://old.nabble.com/removing-for-loops...-tp28633477p28711249.html
>> >> >>>> Sent from the Scipy-User mailing list archive at Nabble.com.
>> >> >>>>
>> >> >>>> _______________________________________________
>> >> >>>> SciPy-User mailing list
>> >> >>>> SciPy-User@scipy.org
>> >> >>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>>
>> >> >>> _______________________________________________
>> >> >>> SciPy-User mailing list
>> >> >>> SciPy-User@scipy.org
>> >> >>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>>
>> >> >>>
>> >> >>
>> >> >> --
>> >> >> View this message in context:
>> >> >>
>> http://old.nabble.com/removing-for-loops...-tp28633477p28711444.html
>> >> >> Sent from the Scipy-User mailing list archive at Nabble.com.
>> >> >>
>> >> >> _______________________________________________
>> >> >> SciPy-User mailing list
>> >> >> SciPy-User@scipy.org
>> >> >> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >>
>> >> > _______________________________________________
>> >> > SciPy-User mailing list
>> >> > SciPy-User@scipy.org
>> >> > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >> >
>> >> >
>> >>
>> >> --
>> >> View this message in context:
>> >> http://old.nabble.com/removing-for-loops...-tp28633477p28711581.html
>> >> Sent from the Scipy-User mailing list archive at Nabble.com.
>> >>
>> >> _______________________________________________
>> >> SciPy-User mailing list
>> >> SciPy-User@scipy.org
>> >> http://mail.scipy.org/mailman/listinfo/scipy-user
>> >>
>> >
>> > _______________________________________________
>> > SciPy-User mailing list
>> > SciPy-User@scipy.org
>> > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >
>> >
>>
>> --
>> View this message in context:
>> http://old.nabble.com/removing-for-loops...-tp28633477p28824023.html
>> Sent from the Scipy-User mailing list archive at Nabble.com.
>>
>> _______________________________________________
>> SciPy-User mailing list
>> SciPy-User@scipy.org
>> http://mail.scipy.org/mailman/listinfo/scipy-user
>>
>
>
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``` | 8,757 | 27,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-22 | longest | en | 0.776852 |
http://www.opengl.org/discussion_boards/archive/index.php/t-163294.html | 1,411,330,143,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657135930.79/warc/CC-MAIN-20140914011215-00056-ip-10-234-18-248.ec2.internal.warc.gz | 714,931,052 | 4,584 | PDA
View Full Version : Only 42 triangles can be emitted by Geometry Shader
Ren
10-12-2007, 11:23 PM
I'm making a simple geometry program, which keep emitting triangles. I can get at most 42 triangles from a single Geometry Shader execution.
I understand that maximum vertices can be emitted is 256 and maximum components is 1024. I think the limitation in my case it probably the component. If 8 components are used for each vertex (4 for position and 4 for color), then I would only be able to get 1024/8 = 128 vertices, which is roughly 42 triangles.
However, is it possible to just use four components for a vertex? i.e a vertex will only have a position attribute but no color? I tried that in GS, I only set gl_Position varying variable. However, I still get 42 triangles.
So, does anyone know why, or does anyone know how to tell GS to only emit 4-component vertices not 8 component vertices
Here's a screen shot
http://www.lhviolinist.com/images/42_tri.png
Vertex
void main(){
gl_Position = gl_ModelViewProjectionMatrix * gl_Vertex;
gl_FrontColor = vec4(1, 1, 0, 1); // change vertex color to yellow
}Geometry
#version 120
void main(){
float centerX, centerY;
for(int x = 0; x < 10; x++){
for(int y = 0; y < 10; y++){
centerX = -0.9 + 0.2 * x;
centerY = 0.9 - 0.2 * y;
gl_Position = vec4(centerX - 0.1, centerY - 0.1, 0.0, 1);
EmitVertex();
gl_Position = vec4(centerX, centerY + 0.1, 0.0, 1);
EmitVertex();
gl_Position = vec4(centerX + 0.1, centerY - 0.1, 0.0, 1);
EmitVertex();
EndPrimitive();
}
}
}Fragment
void main(){
gl_FragColor = vec4(1, 1, 0, 1);
}OpenGL setup
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <GL/glew.h>
#include <GL/glut.h>
// Vertex and fragment shader handle
GLuint vs, gs, fs;
// GLSL program handle
GLuint program;
// read contents of a text file
char * readFile(const char * filePath){
char * text;
long chars;
FILE * fp = fopen(filePath, "r");
fseek(fp, 0L, SEEK_END); // Position to end of file
chars = ftell(fp);
rewind(fp);
text = (char*)calloc(chars, sizeof(char));
fclose(fp);
printf("%s\n", text);
return text;
}
// This function calculates FPS
void fps(){
static int fps = 0;
static float previousTime = 0.0f;
static char strFPS[20] = {0};
float currentTime = (glutGet(GLUT_ELAPSED_TIME) * 0.001f);
++fps; // Increment the FPS counter
if( currentTime - previousTime > 1.0f ){
previousTime = currentTime;
sprintf(strFPS, "FPS: %d", fps);
glutSetWindowTitle(strFPS);
fps = 0.0f;
}
}
// Initialise GLSL
void initGLSL(){
const char * src; // shader source code
char * glVersionString;
GLuint compileResult;
char infoLog[200];
glVersionString = (char *)glGetString(GL_VERSION);
printf("OpenGL Version: %s\n", glVersionString);
printf("GLSL Version: %s\n", glVersionString);
free(src);
free(src);
free(src);
// Check if there are any compile errors
if(compileResult == GL_TRUE){
}else{
printf("%s", infoLog);
}
// Check if there are any compile errors
if(compileResult == GL_TRUE){
}else{
printf("%s", infoLog);
}
if(compileResult == GL_TRUE){
}else{
printf("%s", infoLog);
}
program = glCreateProgram(); // Create GLSL program object
// Geometry Shader input and output type
glProgramParameteriEXT(program, GL_GEOMETRY_INPUT_TYPE_EXT, GL_TRIANGLES);
glProgramParameteriEXT(program, GL_GEOMETRY_OUTPUT_TYPE_EXT, GL_TRIANGLE_STRIP);
glProgramParameteriEXT(program, GL_GEOMETRY_VERTICES_OUT_EXT, 1024);
if(compileResult == GL_TRUE){
}else{
}
glUseProgram(program); // finally, tell OpenGL to use this GLSL program
}
void display(){
float centerX, centerY;
int x, y;
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glNormal3f(0, 0, 1);
// The color has been set to red, see vertex program, which changes the color to yellow
glColor3f(1, 0, 0);
glBegin(GL_TRIANGLES);
glVertex3f(1, 1, 1);
glVertex3f(1, 1, 1);
glVertex3f(1, 1, 1);
glEnd();
glutSwapBuffers();
fps();
}
void reshape(int w, int h) {
//Set up an orthographic view of the objects
glViewport(0, 0, w, h);
glMatrixMode(GL_PROJECTION);
glOrtho(-15, 15, -15, 15, 1, 1);
glMatrixMode(GL_MODELVIEW);
}
// render as fast as possible
void idle(){
glutPostRedisplay();
}
int main(int argc, char **argv){
//glut initialisation
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGB | GLUT_DEPTH);
glEnable(GL_DEPTH_TEST); // Enable hidden surface removal
// create window
glutInitWindowSize(500, 500);
glutInitWindowPosition(0, 0);
glutCreateWindow("Hello GLSL");
// Initialise GLUT
glutDisplayFunc(display);
glutReshapeFunc(reshape);
glutIdleFunc(idle);
glewInit(); // must initialise GLEW context before creating shader
initGLSL(); //
//Background colour set to black
glClearColor(0.0, 0.0, 0.0, 1.0);
glutMainLoop();
// clean up, all shaders and GLSL program object must be manually deleted
glDeleteProgram(program);
}
RGHP
10-13-2007, 07:15 AM
Hello
The model of your target? maybe a bug in the Linu driver(you are using ubuntu right?).
RGHP
10-13-2007, 07:18 AM
By 'target' I mean your card. Sorry, but in spanish(my native language) card is 'tarjeta' :) .
Ren
10-13-2007, 08:42 PM
I've tried this on Windows Vista, still get only 42 triangles | 1,501 | 5,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2014-41 | latest | en | 0.666003 |
https://it.mathworks.com/matlabcentral/cody/problems/37-pascal-s-triangle/solutions/801266 | 1,591,135,688,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426801.75/warc/CC-MAIN-20200602193431-20200602223431-00098.warc.gz | 383,819,485 | 15,639 | Cody
# Problem 37. Pascal's Triangle
Solution 801266
Submitted on 5 Jan 2016 by George-Felix Leu
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 = 0; correct = [1]; assert(isequal(pascalTri(n),correct))
y = 1
2 Pass
%% n = 1; correct = [1 1]; assert(isequal(pascalTri(n),correct))
y = 1 1
3 Pass
%% n = 2; correct = [1 2 1]; assert(isequal(pascalTri(n),correct))
y = 1 2 1
4 Pass
%% n = 3; correct = [1 3 3 1]; assert(isequal(pascalTri(n),correct))
y = 1 3 3 1
5 Pass
%% n = 10; correct = [1 10 45 120 210 252 210 120 45 10 1]; assert(isequal(pascalTri(n),correct))
y = 1 10 45 120 210 252 210 120 45 10 1 | 285 | 750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-24 | latest | en | 0.50948 |
http://mathcentral.uregina.ca/QQ/database/QQ.09.97/fill1.html | 1,555,791,681,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530040.33/warc/CC-MAIN-20190420200802-20190420222802-00274.warc.gz | 112,504,678 | 1,735 | Date: Tue, 14 Oct 1997 18:19:29 -0400
Sender: David
Subject: Middle School Level Math Question (HELP!!)
I am a teacher in Massachusettes. We have been diagramming numbers such as two to the third (a 3-D cube). One of my students asked me how you would diagram two to the fourth. I have searched through all of my teachers books and cannot seem to find the answer to this question. Is there a way to diagram this? If there is, how would you do this?
Your help would be greatly appreciated.
Thank you,
David
Hi David,
I would use a tree diagram to illustrate powers. For example for powers of 2 start with a tree with two branches.
To diagram 2^2=2*2 double the number of paths through the tree by adding a second level of branches to the tree.
For 2^3=2(2^2) again double the number of paths through the tree by adding a thrid level.
And for 2^4=2(2^3).
Conceptually you can continue this process but the tree soon becomes unmanagable.
Powers with a base other than 2 can also be described with a tree diagram, for example
and
Trees can be useful in many areas of mathematics. For example the tree for 2^3 can be used to describe all possibilities when a coin is tossed three times.
Cheers,
Penny
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https://help.scilab.org/docs/6.0.0/pt_BR/stdevf.html | 1,716,558,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00634.warc.gz | 254,129,182 | 3,714 | Change language to:
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See the recommended documentation of this function
stdevf
standard deviation
Syntax
```s=stdevf(x,fre)
s=stdevf(x,fre,'r') or s=stdevf(x,fre,1)
s=stdevf(x,fre,'c') or s=stdevf(x,fre,2)```
Arguments
x
real or complex vector or matrix
fre
real or complex vector or matrix
Description
This function computes the standard deviation of the values of a vector or matrix `x`, each of them counted with a frequency given by the corresponding values of the integer vector or matrix `fre` who has the same type of `x`.
For a vector or matrix `x`, s=stdevf(x,fre) (or `s=stdevf(x,fre,'*')` returns in scalar `s` the standard deviation of all the entries of `x`, each value counted with the multiplicity indicated by the corresponding value of `fre`.
`s=stdevf(x,fre,'r')` (or, equivalently, `s=stdevf(x,fre,1)`) returns in each entry of the row vector `s` of type 1xsize(x,'c') the standard deviation of each column of `x`, each value counted with the multiplicity indicated by the corresponding value of `fre`.
`s=stdevf(x,fre,'c')` (or, equivalently, `s=stdevf(x,fre,2)`) returns in each entry of the column vector `s` of type size(x,'c')x1 the standard deviation of each row of `x`, each value counted with the multiplicity indicated by the corresponding value of `fre`.
Examples
```x=[0.2113249 0.0002211 0.6653811;0.7560439 0.9546254 0.6283918]
fre=[1 2 3;3 4 3]
m=stdevf(x,fre)
m=stdevf(x,fre,'r')
m=stdevf(x,fre,'c')```
Bibliography
Wonacott, T.H. & Wonacott, R.J.; Introductory Statistics, fifth edition, J.Wiley & Sons, 1990.
Report an issue << stdev Descriptive Statistics variance >> | 515 | 1,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.628424 |
https://holooly.com/solutions/the-ball-bearing-of-weight-w-travels-over-the-edge-a-with-velocity-v-_-a-determine-the-speed-at-which-it-rebounds-from-the-smooth-inclined-plane-at-b-take-e-0-8-given-w-0-2-lb-%CE%B8/ | 1,603,232,194,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874340.10/warc/CC-MAIN-20201020221156-20201021011156-00052.warc.gz | 356,530,455 | 16,313 | ## Question:
The ball bearing of weight W travels over the edge A with velocity ${ v }_{ A }$. Determine the speed at which it rebounds from the smooth inclined plane at B. Take e = 0.8. Given: W = 0.2 lb , θ = 45 deg , e = 0.8 , ${ v }_{ A }$ = 3 ft/s , g = ${ 32.2ft/s }^{ 2 }$ | 98 | 280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 3, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-45 | latest | en | 0.789765 |
https://rpchurchill.com/wordpress/posts/2016/04/19/fixing-the-thermodynamic-functions-for-saturated-water-as-a-function-of-temperature/ | 1,716,755,625,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00656.warc.gz | 440,289,954 | 11,209 | Fixing the Thermodynamic Functions for Saturated Water As A Function of Temperature
I’ve updated five of the thermodynamic property functions vs. temperature. Note that most of these define properties only up to a bit above 600 °F and, more importantly, to 1800 psi. The values for saturated water are defined up to 705.44 °F and 3204 psi. I only fit the curves to the values I did because I was modeling systems in a boiling water reactor (BWR), which runs at not more than 1200 psi. Pressurized water reactors (PWR), by contrast, run up to around 2000 psi. It would be a good idea to extend all of the curves vs. both temperature and pressure to their maximum values.
The specific volume functions were both updated.
The enthalpy of vapor function was updated.
The entropy of vaporization and vapor functions were also updated.
Of further note was that I found a more advanced version of the curve fit program I used back when. It not only had more features and was a bit easier to use but also supported extra terms for polynomial powers of four through seven. It’s generally better to use polynomial functions than it is to use more expensive functions like square root and natural logarithm, especially if the multiplications are written out efficiently.
The automatically generated function code did not make use of the technique, but it’s possible to write out a polynomial function of the form:
Result = A + Bx + Cx2 + Dx3 + Ex4
in code as:
Result = A + x * (B + x * (C + x * (D + E * x)))
This has the advantage of requiring far fewer multiplications and additions than the method I have used.
I’ll have to do some reverse engineering to figure out what I was doing with the different versions of the curve fit program before putting together a new one. I had four or five versions altogether, the last of which seemed to be incomplete.
This entry was posted in Tools and methods and tagged , . Bookmark the permalink. | 424 | 1,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.949368 |
https://math.stackexchange.com/questions/1027746/how-to-prove-this-gamma-function-identity | 1,561,001,188,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.98/warc/CC-MAIN-20190620024754-20190620050754-00193.warc.gz | 516,209,311 | 35,600 | # How to prove this gamma function identity?
Reading Landau and Lifshitz "Quantum Mechanics. Non-relativistic theory", I've come across an identity, which after being a bit simplified, reads
$$\left|\frac{((a+b)\Gamma(2a)\Gamma(-(a+b))^2}{(a-b)\Gamma(-2a)\Gamma(a-b)^2}\right|=\left|\frac{\sin(\pi(a-b))}{\sin(\pi(a+b))}\right|,\tag1$$
and apparently holds for pure imaginary $a$ and $b$. The book says the left hand side can be calculated to equal right hand side using the reflection formula:
$$\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}.\tag2$$
But I fail to see how I could actually do this calculation. Plotting the difference I see that this actually is false for real $a$ and $b$, so the reflection formula, which AFAIK works for all $x\in\mathbb C$ seems to be not enough to get this result.
So, how do I prove $(1)$, or even better, derive RHS from LHS?
The identity holds for pure imaginary $a$ and $b$. So, let $a=i u$ and $b=i v$ where $u$ and $v$ are real. The identity to be proved is : $$\left|\frac{(i(u+v)\Gamma(2iu)\Gamma(-i(u+v))^2}{i(u-v)\Gamma(-2iu)\Gamma(i(u-v))^2}\right|=\left|\frac{\sin(\pi i(u-v))}{\sin(\pi i(u+v))}\right|$$
At first look, I don't see any interest to use the reflexion relationship. It is much easier to use the formula giving the modulus : $$|\Gamma{(iy)}|=\sqrt{\frac{\pi}{y \sinh(\pi y)}}$$ With this formula, remplace $y$ by $(u+v)$ or $(u-v)$ and put it back into the left term above. The simplification leads to $\frac{\sinh(\pi(u-v))}{\sinh(\pi(u+v))}$
With $\sinh(y)=-i\sin(i y)$ we have: $\frac{\sinh(\pi(u-v))}{\sinh(\pi(u+v))}=\frac{\sin(\pi i(u-v))}{\sin(\pi i(u+v))}$ which is the right term of the equation and so, it is proved.
• Wow! In fact the original variables were indeed sum and difference multiplied by $i$, it was me who changed them into $a$ and $b$ trying to work out the result. – Ruslan Nov 18 '14 at 19:19
• You seem to have lost a minus in $\sinh(y)=-i\sin(iy)$. – Ruslan Nov 18 '14 at 19:47 | 633 | 1,972 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-26 | latest | en | 0.856212 |
https://www.jiskha.com/display.cgi?id=1352948053 | 1,498,648,224,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323604.1/warc/CC-MAIN-20170628101910-20170628121910-00087.warc.gz | 858,483,421 | 3,766 | # math
posted by .
given the digits 0,1,2,3,4,5,6, and 7 find the number of possibilities in each category four digit numbers, odd three digit numbers and three digit numbers without repeating digits.
• math -
In each case we can assume that no whole number starts with a zero, and no repeating digits for all 3 cases.
number of 4 digits numbers = 6x6x5x4 = 720
number of 3 digit odd numbers, (must end in an odd) = 5x5x3 = 75
number of 3 digit numbers = 6x6x5 = 180 | 141 | 472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-26 | longest | en | 0.747839 |
https://au.mathworks.com/matlabcentral/answers/714588-changing-length-of-pendulum-while-it-is-in-motion | 1,716,080,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00228.warc.gz | 94,782,636 | 27,298 | # Changing Length of Pendulum while it is in motion
7 views (last 30 days)
Yan Koon Ang on 12 Jan 2021
Answered: Humberto Ramos on 8 Mar 2021
Hi
I have seen many simulations of pendulums oscillating found on matlab. One such example can be found in this link. The link provides the code to simulate and visualise the oscillation of the pendulum by setting up different initial fixed parameters. I was wondering if it is possible to simulate the length of the rod, attached to the pendulum, changing while the pendulum is in motion( i.e the length of the rod should be able to increase/decrease such that it reaches the target length specified over a fixed period of time). This would also mean that the rate of change of the length of the rod can also be controlled.
Currently, using a ode45 solver, I am able to plot the path of the pendulum's oscillation. This is done through specifying the initial conditions, which are fixed. Hence, is there a way that the length of the rod attached to the pendulum be modified during the oscillation?
KSSV on 12 Jan 2021
Are you talking about spring pendulum?
Humberto Ramos on 8 Mar 2021
Mischa Kim on 12 Jan 2021
Dear Yan Koon Ang,
this is possible. In the example you are referring to in your question you would have to make the time variable t available in the ode file and then you could do something like this:
function xdot = Equation(t,x)
% Set the input values in order to pass onto ode45
% [...]
L = x(5) * (1 + 0.5*t); % replace by whatever behavior/equation you need
Yan Koon Ang on 12 Jan 2021
Hi Mischa Kim,
I am new to this, hence could you please elaborate on how I could modify the code to allow the time variable " t " available in the ode file (Animation.m file).
function Animation(ivp, duration, fps, movie, arrow)
% [...]
nframes=duration*fps; % Number of Frames
% Solving the Equation using ode45
sol=ode45(@Equation,[0 duration], ivp);
t = linspace(0,duration,nframes);
y = deval(sol,t);
Mischa Kim on 12 Jan 2021
The first changes you need to make are in the Equation() function, as pointed out above. Just browsing through the example files in the link you provided you need to adapt the Anmiation() function accordingly:
% Position and Angular velocity
phi = y(1,:)';
dtphi = y(2,:)';
L = ivp(5)*(1 + 0.5*t); % replace by whatever behavior/equation you need
% To set the Range os Phase plane, time vs. depl plots
L is used further below to scale the axis. range is a scalar, L a vector, so, e.g.
range = 1.1*L(1);
There may be more changes to get you exactly what you need, but this should get you started.
Humberto Ramos on 8 Mar 2021
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Start Hunting! | 716 | 2,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-22 | latest | en | 0.891337 |
https://www.sitepoint.com/premium/courses/interview-prep-insertion-sort-3023/ | 1,716,299,768,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00287.warc.gz | 880,694,104 | 15,518 | # Interview Prep: Insertion Sort
4.0 Average Customer Rating
## Brush up on this algorithm that is sometimes asked about in interviews
• 8 Videos
• 0 hours 23 minutes
• 63 Happy Students
## About the course
Description
Insertion sort is a commonly taught algorithm, and some interviewers like to ask about it. As you’ll see in this class, it’s an easy algorithm to understand and implement. You’ll learn the algorithmic complexity and understand why the algorithm is rarely used.
I’ll give a walkthrough of the source code, which I also provide in my repository: github.com
Also watch my course on Binary Search, insertion sort makes use of that algorithm.
Project
I want you to write an insertion sort function. Choose your favorite language, or the one for your interview, and implement a function that sorts a list of values. Use the algorithm I present in this class.
In the example Python implementation, the signature looks like this:
`````````py
def insertion_sort( items : List[ComparableT] ) -> List[ComparableT]:
`````````
Verify Sort Algorithm
As a bonus project, write the verify sort algorithm. This is an essential bit of code for unit tests. It also explores concepts that will help you during your interviews.
Verify sort takes an unsorted list and a sorted list. It verifies that the sorted list is indeed the sorted version of the unsorted one.
• 8 Videos
• 0 hours 23 minutes
• 63 Happy Students
## Courses Outline
Lesson 1: Insertion Sort: Design and Implementation
Free
Introduction
0:33
Algorithm Design
4:46
Algorithm Complexity
3:43
Verify Sort Algorithm
4:49
Insert Sort Code
3:31
Stable Sort
2:34
Verify Sort Code
2:54
Conclusion
0:17
• 8 Videos
• 0 hours 23 minutes
• 63 Happy Students | 411 | 1,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-22 | latest | en | 0.838589 |
https://in.mathworks.com/matlabcentral/profile/authors/1887879?detail=cody | 1,660,131,380,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00115.warc.gz | 319,092,665 | 23,658 | Community Profile
# ChrisR
Last seen: Today Active since 2015
I teach courses in water resources and environmental fluid mechanics.
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https://www.physicsforums.com/threads/fine-and-hyperfine-splitting.107989/ | 1,544,963,657,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827727.65/warc/CC-MAIN-20181216121406-20181216143406-00615.warc.gz | 1,002,252,160 | 12,679 | # Homework Help: Fine and Hyperfine splitting
1. Jan 25, 2006
### Henk
I'm having some problems with the Fine and Hyperfine splitting (FS and HFS). For example the 3s3p Configuration, I understand the FS:
*****************************_______Mj=1
************1P________J=1*****_______ Mj=0
*****************************_______ Mj=-1
*
*
************3P______ Is split into J=0,1,2 and they are split into the Mj levels
*
*
3s3p_____
But where then is the HFS. I understand that I and J couple to F and that gives MF-levels but I don't know where to put in the picture above.
edit: the picture isn't very clear, sorry had to add some stars.
Last edited: Jan 25, 2006
2. Jan 25, 2006
### Gokul43201
Staff Emeritus
What is the degeneracy of the Mj levels in the absense of the HFS ?
Perhaps, I should ask this first : Do you know where to put the hyperfine levels for the 1s orbital ?
3. Jan 26, 2006
### Henk
For the 1s Orbital: L=0, S=1/2 thus J=1/2. That means the level has FS for J = 1/2 and J = -1/2. I don't know I so I can't give the F values however I suppose that the J-levels are splitted into the F-levels. And that if there is an external magnetic field (weak) the F-levels are splitted into 2F+1 Mf levels. Is this correct? | 375 | 1,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-51 | latest | en | 0.928897 |
https://optimization.mccormick.northwestern.edu/index.php?title=Adaptive_robust_optimization&diff=prev&oldid=4582 | 1,670,638,279,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00670.warc.gz | 477,737,281 | 15,454 | # Difference between revisions of "Adaptive robust optimization"
Author: Woo Soo Choe (ChE 345 Spring 2015)
Steward: Dajun Yue, Fengqi You
## Methodology
In order to investigate how Adaptive Robust Optimization problem, numerous techniques may be used. However, given the scope of this page, only three of the techniques will be introduced. The three algorithms are Bender's Decomposition, Trilevel Optimization, and column-and-Constraint Generation Algorithm and for the Benders Decomposition and Trilevel . When using Benders Decomposition approach, the algorithm essentially breaks down the original problem into the outer and inner problems. Once the problem is divided into two parts, the outer problem is solved using the Benders Decomposition and the inner problem is solved using the Outer Approximation. The detailed steps are as follows.
Benders Decomposition
The Outer Problem: Benders Decomposition
Step 1: Initialize, by denoting the lower bound as $LB = - \infty$ and the upper bound as $UB=\infty$ and set the iteration count as $C=0$. Then choose the termination tolerance $\epsilon$.
Step 2: Solve the master problem
$\begin{array}{llr} max_{x,\zeta} &c^T x + \zeta \\ s.t. &Fx \le f &\\ &{\zeta} \ge -h^T \alpha_l + (Ax-g)^T \beta_l + d_l^T \lambda_l , \forall \le C \end{array}$
In this case, $(x_c, \zeta_c)$ denote the optimum solution.
Step 3: Update the lower bound $LB=c^T x_c + \zeta_c$
Step 4: Increase $C$, the iteration count by 1
Step 5: Solve $I(x_c)$, the inner problem and denote the optimal solution as $(d_c, \alpha_c, \beta_c, \lambda_c)$. Update $UB=min(UB, c^T x_c + I(x_c))$, where $UB$ stands for the upper bound.
Detailed procedure of Step 5 is as follows.
if $UB-LB \ge \epsilon$ then
Go to step 2
else
Calculate $y_c$, the dispatch variable given $x_c$ and $d_c$
end
The Inner Problem : Outer Approximation
Step 1: Initialize by using the commitment decision from the outer problem $x_c$ from the outer problem. Then, find an initial uncertainty realization $d_1 \in \mathbb{D}$, set the lower bound $LOA = - \infty$ and the upper bound $UOA = \infty$, set iteration count j=1 and then termination tolerance which is denoted as $\theta$
Step 2: Solve the sub-problem below.
$\begin{array}{llr} S(x_c,d_j) = max -h^T \alpha + (Ax_c - g ) ^T \Beta + d_j^T \lambda \\ s.t. & -H^T \alpha - B^T \beta + J^T \lambda = b \\ &alpha \ge 0, \beta \ge 0 \\ \end{array}$
In the inner problem, the optimal solution is denoted as $(\alpha_j, \beta_j, \lambda_j)$. Furthermore, define $d_j^T \lambda_j$ as $L_j(d_j,\lambda_j) + (d-d_j)^T \lambda_j + (\lambda - \lambda_j)^T d_j$. Then, update $LOA = max (S(x_c,d_j)), LOA)$
Step 3: Solve the master problem
$\begin{array}{llr} U(d_j, \lambda_j) = max -h^T \alpha + (Ax_c-g)^T \beta + \zeta \\ s.t &\zeta \le L_i (d_i, \lambda_i), \forall \le j \\ &-H^T \alpha -B^T \beta +J^T \lambda =b \\ &Dd \le k &\\ &\alpha \ge 0 , \beta \ge 0 \\ \end{array}$
Increase the iteration of j by 1. While the optimal solution is denoted as $d_j, \alpha_j, \beta_j, \lambda_j, \zeta)$, update the upper bound as $UOA = min(UOA, U(d_j, \lambda_j))$
if $UB - LB \ge 0$ then
Go to Step 2
else
Return optimal solution as the output
end
As seen from the algorithms, Benders Decomposition divides an Adaptive Robust Optimization problem into outer and inner problems and solves them using two algorithms. While this may cause some confusion for ones who have no previous exposures to Benders Decomposition, the approach to solving the outer problem is also called the Benders Decomposition and the approach to solving the inner problem is called the Outer Approximation method. Fundamentally, this algorithm works by first solving the outer problem until $UB -LB \ge \epsilon$ condition is met and then use the $x. \zeta$ from the outer problem to plug into the inner problem and solve for the optimum solution until $UB -LB \ge \ 0$ condition is met. This method has an advantage over traditional Robust Optimization in a sense that it does not sacrifice as much optimality in the solution at the cost of obtaining a conservative answer. Unfortunately, Benders Decomposition method has three problems. First problem is the fact that the master problem relies on the dual variables of the inner and outer problems, which means that the sub-problems cannot have integer variables. Second problem is that the solution does not guarantee a global optimal solution, and it means that the algorithm may not return the absolute worst case scenario before returning the solution. Third problem is that it takes a long time to compute the answer and this might pose a problem when solving a large scale problem.
In order to resolve this issue, another algorithm called Trilevel Optimization was proposed by Bokan Chen of Iowa University. Before iterative Trilevel Optimization algorithm applied, the problem needs to be reformulated in an appropriate form as shown below.
Failed to parse(PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \begin{array}{llr} min\limits_x c^T x + b^T y \\ s.t &Fx \ le f \\ &max_d & b^T y \\ s.t &Dd \le k \\ &min\limits_y b^T y \\ s.t &Ax + By \le g \\ &Hy \le h \\ &Jy = d \\ \end{array}
Equation $Fx \ le f$ represents the first constraints on the first stage commitment variables and equation $Dd \le k$ represents the uncertainty set of demand. Equation $Ax + By \le g$ represents the constraints that couples the commitment and dispatch variables. $Hy \le h$ constrains the dispatch variable and $Jy =d$ is the constraint that couples the demand variables and the dispatch variables. Then, the reformulated model can be further refined into the following model.
$\begin{array}{llr} \min_{x,\phi} c^T x + \phi &\\ \text{s.t.} Fx \le f &\\ & \phi \ge b^T y, \forall y \in \mathbb{Y_D} \end{array}$
When $\mathbb{Y_D} = \big\{ y| Ax + By \le g Hy \le h \big\} \cap \big\{y | Jy = d, d \in \mathbb{D} \big\}$. Assuming $\Omega \subset \mathbb{D}$, we have $\mathbb{Y_\Omega} = \big\{ y| Ax + By \le g Hy \le h \big\} \cap \big\{y | Jy = d, d \in \Omega \big\}$. This implies $\mathbb{Y}_\Omega \subset \mathbb{Y_D}$. This relaxes the problem into the following form.
$\begin{array}{llr} \min_{x,\phi} c^T x + \phi &\\ \text{s.t.} &Fx \le f &\\ &\phi \ge b^T y, \forall y \in \mathbb{Y_\Omega} \end{array}$
This allows the trilevel problem to be split into the master problem and a sub-problem. Following is the relaxation of the master problem M of the trilevel problem as follows.
$\begin{array}{llr} \min_{x,\phi} c^T x + \phi &\\ \text{s.t.}&Fx \le f &\\ &\phi \ge b^T y, \forall y \in \mathbb{Y_D} \end{array}$
The master problem M is a relaxation of the trilevel problem as follows:
$\begin{array}{llr} \min_{x,\phi} c^T x + \phi &\\ \text{s.t.}&Fx \le f &\\ &\phi \ge b^T y^i, \forall i = 1,...,| \Omega | &\\ &H y^i \le h, \forall i = 1, ..., \ \Omega | &\\ &Ax + By^i \le g, \forall i = 1,..., | \Omega | &\\ &J y^i = d^i, \forall i = 1,..., | \Omega | \end{array}$
Following is the bilevel sub-problem which yields the dispatch cost under the worst-case scenario
$\begin{array}{llr} \max\limits_d b^T y &\\ \text{s.t.} &Dd \le k &\\ &min_y b^T y &\\ &s.t. Hy \le h &\\ &By \le g - Ax &\\ &Jy = d &\\ \end{array}$
An Iterative Algorithm For The Trilevel Optimization Problem Optimization Problem
Step 1: Initialize by denoting lower bound as $LB= -\infty$ and the upper bound as $UB= \infty$. Then create and empty set $\Omega$.
Step 2: Solve the master problem M. Where the solution of the problem is $(x^M, \zeta^M, y^M)$. Then update the lower bound of the algorithm $LB=max(LB, c^Tx^M+\zeta^M)$.
Step 3: Solve the sub-problem $S(x^M)$. The solution to the problem is $(y^S, d^S)$. Update the upper bound which is $UB=min(UB, c^Tx^M+b^Ty^S)$ and the set $\Omega = d^8 \cap \Omega$.
if Failed to parse(unknown function '\g'): UB-LB \g 0
then
Go to Step 2
else Find $argmax_i b^T y^i$. Calculate the total cost \zeta = c^T x^M + b^T y^i, return the optimal solution as $x^M, d^i, y^i, \zeta$
end
## Example
In order to illustrate how Adaptive Robust Optimization works, a numerical example is given in this section. This example involves 3 factories and 5 customers and a detailed information if provided through the table below.
Before solving the problem, the basic set up is as follows.
$\begin{array}{llr} & v_j = \min\limits_{i \in O(y)} \big\{c_{ij}\big\} \\ & w_{ij} \begin{cases} 0 &i \in O(y) \\ max_{i \in C(y)} \big\{(v_j - c_{ij}),0 \big\} &i \in C(y) \\ \end{cases} \end{array}$
In this case, $v_{ij}$ are the dual variables associated with the demand constraints and $w_{ij}$ represent the dual variables associated with the setup constraints. Furthermore the dual variable can be represented as $u$ and it means the combination of $(v,w)$. From the proposition, the following Benders cut is derived.
$\beta_y(y)=u(b-By)+f^ty$
For this specific problem, the Benders cut can be rewritten as follows.
$\beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i$
Returning to the problem, we denote factory 1, factory 2, and factory 3 as $y_1, y_2, y_3$ and to start the problem, we only assume factory 1 is open and in this case, $v_j = min_{i \in O(y)} \big\{ c_{ij} \big\} , j=1,...,m$ would become $v_j=(2,3,4,5,7)$. Based on the proposition, $w_{ij}$ may be found as follows. $\begin{array}{llr} w_{1j}=0 \\ w_{2j}=(0,0,3,3,1) \\ w_{3j}=(0,0,2,4,4) \end{array}$
Then, solving Benders Cut, we get the following result.
$\begin{array}{llr} \beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i \\ \beta_y(y)=2+3+4+5+7+(2-0)y_1+(3-(3+3+1))y_2+(3-(2+4+4+4))y_3 \\ \beta_y(y)=21+2y_1-4y_2-7y_3 \end{array}$
From this, the upper bound on the solution, 23, obtained. Then, the Benders cut is used to solve the master problem and by inserting the Benders cut into the master problem, we get the problem in the following form.
$\begin{array}{llr} min z \\ s.t. &z \ge 21+2y_1-4y_2-7y_3 \\ &y \in \mathbb{B}^3 \end{array}$
In the above problem, the optimal solution is y=(0,1,1), meaning it is best to keep factory 1 closed and open factories 2 and 3. This yield a solution of 10, which becomes new y and one more iteration of the algorithm may be done with this. When the $v_j$ and $w_{ij}$ are found again with the solution, following values are obtained
$\begin{array}{llr} v_j = (4,3,1,1,3) \\ w_{1j} = (2,0,0,0,0)\\ w_{2j}=w{3j}=0 \end{array}$
Then the Benders cut was calculated again as follows.
$\begin{array}{llr} \beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i \\ \beta_y(y)= (4+3+1+1+3)+(2-2)y_1+(3-0)y_2+(3-0)y_3 \\ \beta_y(y)=12+3y_2+3y_3 \end{array}$
From this, we get a new upper bound which is 18 and the master problem looks like:
$\begin{array}{llr} min z \\ s.t. &z \ge 21+2y_1-4y_2-7y_3 \\ &z \ge 12+ 3y_2+3y_3 \\ &y \in \mathbb{B}^3 \end{array}$
As the solution to the problem, we get $y=(0,0,1)$. Then, we repeat the iteration process, which yields:
$\begin{array}{llr} v_j = (5,4,2,1,3) \\ w_{1j} = (3,1,0,0,0) \\ w_{2j} = (1,1,1,0,0) \\ w_{3j} = 0 \end{array}$
Then the Benders cut becomes:
$\begin{array}{llr} \beta_y(y)=\sum_{i=1}^m v_j + \sum_{i=1}^n (f_i - \sum_{j=1}^m w_{ij}) y_i \\ \beta_y(y)=(5+4+2+1+3)+(2-4)y_1+(3-3)y_2+(3-0)y_3 \\ \beta_y(y)=15-2y_1+3y_3 \end{array}$
Now, there is no better upper bound and the master problem becomes:
$\begin{array}{llr} min z \\ s.t. &z \ge 21+2y_1-4y_2-7y_3 \\ &z \ge 12+ 3y_2+3y_3 \\ &z \ge 15-2y_1+3y_3 &y \in \mathbb{B}^3 \end{array}$
This yields the optimal solution of $y=(1,0,1)$ and the new lower bound is 16. when the iteration process is repeated until the upper and lower bound are the same, we obtain the optimal solution value of 16 and come to the decision of opening factories 1 and 3 only.
As seen from the iterative procedure, Trilevel Optimization also breaks an optimization problem into smaller parts and use iterative algorithm to close in the difference between the upper and the lower bound. However, the Trilevel Optimization addresses the issues with the Benders Decomposition approach.
## Model Formulation
Adaptive Robust Optimization implements different techniques to improve on the original static robust optimization by incorporating multiple stages of decision into the algorithm. Currently, in order to minimize the complexity of algorithm, most of the studies on adaptive robust optimization have focused on two-stage problems. Generally, Adaptive Robust Optimization may be formulated in various different forms but for simplicity, Adaptive Robust Optimization in convex case was provided.
$\begin{array}{llr} \max\limits_{x\in \mathit{S}} &f(x) + \max\limits_{b\in \mathit{B}} Q(x,b) &\\ \end{array}$
In the equation $x$ is the first stage variable and $y$ is the second stage variable, where S and Y are all the possible decisions, respectively.$b$ represents a vector of data and when $\mathit{B}$ represents uncertainty set.
In order for the provided convex case formulation to work, the case must satisfy five conditions:
1. $\mathit{S}$ is a nonempty convex set
2. $f(x)$ is convex in $x$
3. $\mathit{Y}$ is a nonempty convex set
4. $h(y)$ is convex in $y$
5. For all i=1,...,n, $H_i (x,y,b)$ is convex in $(x,y), \forall b \in \mathit{B}$
Clearly, not every Adaptive Robust Optimization problem may be solved using exactly one model. However, key features that need to be present in a model of Adaptive Robust Optimization are the variables which respectively represent the multiple stages, uncertainty sets whether in ellipsoidal form, polyhedral form, or other novel way, and general layout of the problem which solves for the minimum loss at the worst case scenario. Furthermore, another key feature is that second stage variables are not known. Another form of Adaptive Robust Optimization formulation is provided below.
$\begin{array}{llr} \ min_x c^T x + \max\limits_{d\in \mathbb{D}} \min\limits_{y\in {\Omega}} b^T y &\\ \text{s.t.} Fx \le f &\\ &{\Omega} (x,d)= \big\{y: Hy \le h, Ax+By \le g, Jy=d \big\} &\\ &\mathbb{D} = \big\{ d: Dd \le k \big\} \end{array}$
Similarly as in the first formulation provided, $x$ and $y$ represent the first stage variable and the second stage variable respectively. In this case the, $\mathbb{D}$ is the polyhedron uncertainty set of demand $d$and $\Omega$ represents the uncertainty set for the second stage variable $y$. In this case, H, A, B, g, J, D, and k are numerical parameters which could represent different parameters under different circumstances.
## Introduction
Traditionally, robust optimization has solved problems based on static decisions which are predetermined by the decision makers. Once the decisions were made, the problem was solved and whenever a new uncertainty was realized, the uncertainty was incorporated to the original problem and the entire problem was solved again to account for the uncertainty.[1] Generally, robust optimization problem is formulated as follows.
In the equation $x\epsilon\mathbb{R}^n$ is a vector of decision variables and $f_o,f_i$are functions and are the uncertainty parameters which take random value in the uncertainty sets Failed to parse(unknown function '\subseteqmathbb'): \mathcal{U}_i\subseteqmathbb{R}^k . When robust optimization is utilized to solve a problem, three implicit assumptions are made.
1. All entries need in the decision vector$x$ get specific numerical values prior to the realization of the actual data.
2. When the real data is within the range of the uncertainty set $\mathcal{U}$, the decision maker is responsible for the result obtained through the robust optimization algorithm
3. The constraints are hard and the violation of the constraints may not be tolerated when the real data is within the uncertainty set $mathcal{U}$
The three assumptions grant robust optimization technique immunity from uncertainties. There are other types of optimization techniques such as Stochastic Optimization which may be used to handle problems with uncertainties. However, because Stochastic Optimization has its own drawback because it requires the probability distribution of the events. By having the decision makers make guesses about the probability distribution, Stochastic Optimization method often yield results that are less conservative than the ones by Robust Optimization method.
Robust Optimization certainly may have advantages over other optimization methods, but unfortunately, most robust optimization problems for real life applications require multiple stages to account for uncertainties and traditional static robust has shown limitations. In order to improve the pre-existing technique, Adaptive Robust Optimization was studied and advances in the field was made to address the problems which could not be easily handled with previous methods.[2] | 4,990 | 16,855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 107, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-49 | latest | en | 0.788896 |
https://sprachlogik.blogspot.com/2012/01/structured-modal-operators.html | 1,721,011,414,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514655.27/warc/CC-MAIN-20240715010519-20240715040519-00505.warc.gz | 472,548,722 | 26,277 | ## Monday 16 January 2012
### Structured Modal Operators
Propositions have modal characters and truth-values. For now, we will distinguish two modal characters and two truth-values: necessary character, contingent character, truth and falsity.
Necessary character is what necessarily true and necessarily false propositions have in common. Contingent character is what contingent truths and mere possibilities have in common.
In effect, the modal operator 'Necessarily' (box), ascribes necessary character and truth to a proposition. 'Contingently' ascribes contingent character and truth. 'Necessarily, it is not the case that' and 'It is impossible that' ascribe necessary character and falsity. 'It is merely possible that' ascribes contingent character and falsity.
But not all modal operators ascribe a particular character/truth-value pair. Some merely rule out certain combinations. For example, 'Possibly' merely rules out the combination of necessary character and falsity.
(NB that I am here talking about what are often called alethic modal operators, rather than modal operators in a more general formal setting in which these claims only hold for certain choices of accessibility relation.)
It is common to see the following list of four modal operators presented, sometimes as though it were exhaustive: possibility, necessity, contingency and impossibility.
But reflect again that, of these four modalities, possibility is an odd one out, since it is non-commital on truth-value. Also, note that systems have been developed where other operators, e.g. one for non-contingency, are taken as primitive.
This can give rise to an uneasy, lost feeling. Are the usual four modal operators just a hodge-podge? What modal operators are there (could there be)? Is there a systematic way of producing them all? And is there then a systematic way of determining logical relations between them?
In this post, I try to begin answering these questions.
The notation
The notation I want to introduce here can be said to stand to the box, the diamond and such symbols roughly as truth-tables stand to truth-functional connectives. (Or instead of truth-tables, Wittgenstein's ab-notation, Venn diagrams, or the shuttle diagrams pioneered by Martin Gardner and extended by Gregory Landini.)
We have said that 'Necessarily', 'Contingently', 'It is impossible that' and 'It is merely possible that' all ascribe a particular character/truth-value pair, or: they all rule out all but one character/truth-value pair.
We can represent operators as matrices containing four cells, one for each character/truth-value combination. We can then mark the fields representing pairs which are ruled out by the operator in question. A blank canvas, not representing any modal operator, looks like this:
(The box and diamond here represent modal characters.)
The four aforementioned operators then look like this:
We can also consider the class of operators which rule out two character/truth-value pairs:
And finally the class of operators which rule out just one character/truth-value pair:
A syntactical test for implication
For any two modal operators A and B, Ap implies Bp iff all boxes crossed in B are crossed in A. (This could license a simple rule of cross-elimination.)
Then, rules could be given allowing detachment of the truth-operator, conversion of the falsity operator to negation, attachment of the truth-operator, possibility operator, etc.
Flipping and inversion
An operator can be negated by inverting its markings. Its operand can be negated in effect by flipping the operator's marking vertically. The dual of an operator can be obtained by inversion and flipping.
Relation to model theory
For now, atoms are treated as, in effect, simply being assigned a truth-value and a modal character in the semantics, but this can be brought into connection with the standard Kripke semantics for modal logic. Given an S5 frame, for example, an atom's having necessary character (at a world, if you like) amounts to its truth-value being invariant across all worlds. Contingent character amounts to its not being so invariant.
To do
Among other things: study iteration of operators. Iteration will raise philosophical issues about the application of the formalism. These will turn, at least in part, on how propositions are individuated. Similarly, a case could be made for distinguishing a third character, impossible character, when propositions are individuated in a fine-grained way such that our proposition 'Hesperus isn't Phosphorus' is not the same as the Babylonian. (Our version has impossible character. Theirs, necessary.)
1. Why *arent't* you considering modal operators in the more general formal setting involving accessibility relations?
2. In a word, because that setting gives one a greater multiplicity or generality than is required for alethic modal notions such as metaphysical/subjunctive necessity, possibility etc. and a priori necessity (a priority), possibility, etc.
In a possible worlds semantics for those notions, one would (for instance) always want the accessibility relation to be reflexive. (I need to think more about transitivity and symmetry in this connection - it connects to questions about the individuation of propositions.) So, while I do intend to say more in future about how the notation here can be given a full-blown possible worlds semantics, the whole point in a sense was to start with something more specific and purpose-built, to better enable a clear overview of the space of metaphysical/subjuctive modal operators (or, equally, modal operators based on the notion of a priority) without raising separate issues simultaneously. (These modalities could be brought together under the heading 'alethic', although I think the original motivation for that label might be somewhat confused.)
3. What kind of necessity is alethic necessity, then? I suppose my thinking is that answering this question would imply a specific possible-worlds structure, automatically answering all of the questions mentioned in the first section of your post. More generally, I'm fairly well convinced that accessibility-relation type analysis is all that can be said of the English notion of "necessary" (that is, specific English usage will imply whatever accessibility relation and world structure is convenient for the conversation).
On the other hand, I'm curious about causality and the kind of possible world semantics needed for causal reasoning via counterfactuals. In particular, must we believe in those as "true possibilities" (realism concerning possible worlds)? I guess my tentative answer is "no", but I'm generally worried that this has some troublesome implications. (Rejecting the merely possible, in general, feels like it might remove a bunch of needed structure.)
4. Your approach looks quite interesting. Would it perhaps be possible to combine with some of Moretti's ideas about the geometry of modalities?
http://alessiomoretti.perso.sfr.fr/GeometryForModalities.pdf
5. Thanks very much for the interest, Rael, and for the link. This paper you've pointed me to looks rich and exotic, not to mention difficult! I'll have to take a good look some time before I have an answer for you. Meanwhile, any hints from you regarding possible points of contact are welcome.
6. Exotic indeed. What intrigues me is that your approach is very clear and logical, but that i don't see how to make sense of it in terms of the modal square of oppositions. We can take the extended square on p. 3 in Morretti (the S5 one to the right), and interpret the point to the left as T and the one to the right as F. This is I think quite accepted, not wild and exotic as many other things in Moretti's paper. (Now, to compare with your diagrams, we must remeber that the diamond in the square (hexagon) is what you call contingent character, whereas your diamond is contingency as opposed to necessity.)
What bothers me is that I don't see the easy way of translating your scheme into the hexagon, or the hexagon into your scheme, that surely must exist. Perhaps you can find the right way to do this.
7. - Sorry, this is confused: "Now, to compare with your diagrams, we must remeber that the diamond in the square (hexagon) is what you call contingent character, whereas your diamond is contingency as opposed to necessity."
8. I'll stop spamming you now (for today at least). Just a final thing: what I would like to accommodate your scheme to is the modal octagon as presented on p. 13 in this piece by Beziau:
http://www.jyb-logic.org/papers/risingsquare.pdf
(Take p (extreme left) in this diagram to correspond to T, and ~p (extreme right) to correspond to F. Note also that Beziau doesn't draw lines for implications (e.g. subalternation), as is normally done in these figures.)
Your "contingent character" is the lowest point of the octagon, "necessary character" is the highest point". So the highest point, the lowest, the left and the right are the four parameters that you use for constructing your matrices. However, there is still something I don't grasp...
9. This isn't spam! Thanks kindly and don't hesitate to comment again. I will have a look at the Beziau piece soon.
10. All right! I've been looking around to see if there is something similar to your approach. It seems that in the early days of modal logic, people had some hope for a truth-functional account based on matrices of this sort. But they ran into problems; see pp. 3-4 here:
http://ocw.mit.edu/courses/linguistics-and-philosophy/24-244-modal-logic-fall-2009/lecture-notes/MIT24_244F09_lec03.pdf
But the idea lives on in some systems:
Kearns, Modal Sematics without Possible Worlds, Journal of Symbolic Logic, No 1 1981.
Beziau (again), 2011, esp. p. 4:
http://www.jyb-logic.org/papers/la4.pdf
Fariñas del Cerro & Herzig 2011: http://commonsensereasoning.org/2011/papers/DelCerro.pdf
These systems start from necessary truth, contingent truth, contingent falsity and necessary falsity. They tend to use matrices too. Whether the further moves they make are compatible with your project I don't know. I look forward to see how you will develop the idea, which is nice and intuitive.
11. Thanks for the references. I'd seen the Kearns piece before. The Stalnaker notes seem particularly illuminating.
'I look forward to see how you will develop the idea' - in terms of formal developments, I have no idea what if anything I'll do next, but stay tuned. (If I knew who you were I'd say 'I'll keep you posted'!)
In any case, the basic strategy of clearly separating the issue of modal character from that of truth-value will be a key ingredient in the book I am working on, the object of which is to investigate and clarify the topic of metaphysical or subjunctive modality.
12. I am struck by something in the Stalnaker notes which I would not go along with (at least not without qualification). He writes that 'Necessitation of a necessary truth should indeed yield a necessary truth: BOX(1) = 1.'
I think that on one natural and important way of individuating propositions, which I will in this comment call 'the internal mode of individuation', this does not hold. Consider the necessary a posteriori truth, 'Hesperus = Phosphorus'.
On the internal mode of individuation (where sameness of extension/truth-value is not required for sameness of content), instances or tokens of this same proposition(-type) can come out true in some environments (such as ours) and false in others (such as a Twin Earth on which the morning star there is not the evening star there). This hangs together with this proposition's a posteriori status: its truth-value (on the internal mode of individuation) is not an internal property of it.
So, individuating proposition-types internally in this way, the claim that (our tokens of) 'Hesperus is Phosphorus' is necessary amounts to something like this conjunction:
(Our tokens of) 'Hesperus is Phosphorus' are of necessary character AND true.
And this claim itself will not be necessary, but contingent.
(The way I have expressed all this is slightly confusing, due to type/token issues, but I think and hope that a clearer and more explicit terminology will fix this.)
13. Are your boxes Boolean. If they were then the field that is rules out is a 0 and we can represent your squares as Boolean strings i.e.
1100 Falsity
0011 Truth
0001 Nec
0010 Cont
0100 Imposs
1000 Mere Poss
1010 Chont.Char.
0101 Nec.char
1001 Nec. Or Mere pos
1001 Cont. OR Imposs
1101 Non-cont
1110 Non-nec
Are your boxes Boolean. If they were then the field that is ruled out is a 0 and we can represent your squares as a 16 valued logic with 4-tuple Boolean strings i.e.
1100 Falsity
0011 Truth
0001 Nec
0010 Cont
0100 Imposs
1000 Mere Poss
1010 Chont.Char.
0101 Nec.char
1001 Nec. Or Mere pos
1001 Cont. OR Imposs
1101 Non-cont
1110 Non-nec
Is this compatible with you box notation?
Here is something very similar to your work.
http://www.semantic-cube.com/modalities16v.png | 2,883 | 13,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.93762 |
https://socratic.org/questions/5a78e361b72cff3e6e48fd98 | 1,575,655,123,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00001.warc.gz | 547,209,874 | 6,149 | # Question #8fd98
Feb 5, 2018
Like terms have the same variable, and the same power.
#### Explanation:
$- 4$, $- 5$, and $90$ are all like terms because they don't have a variable. These are called 'constants'.
So, add the $- 4$ and $- 5$ to the $90$ to get $99$. Note these are added as positive values since they're negative on the left-hand side of the equation.
Then you have
$9 x = 99$
Divide both sides by $9$ to get
$x = 11$
You divide since it's the inverse operation for multiplication.
Check your work by plugging $11$ back into your equation for variable $x$.
$- 4 - 9 + 9 \cdot 11 = 90$ | 183 | 610 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-51 | latest | en | 0.935694 |
http://core-cms.prod.aop.cambridge.org/core/journals/journal-of-mechanics/article/isogeometric-analysis-of-the-dual-boundary-element-method-for-the-laplace-problem-with-a-degenerate-boundary/386600CDE1C671AD4BF757C3E1D6AC4B | 1,575,611,960,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484815.34/warc/CC-MAIN-20191206050236-20191206074236-00542.warc.gz | 35,784,285 | 43,638 | Home
# Isogeometric Analysis of the Dual Boundary Element Method for the Laplace Problem With a Degenerate Boundary
## Abstract
In this paper, we develop the isogeometric analysis of the dual boundary element method (IGA-DBEM) to solve the potential problem with a degenerate boundary. The non-uniform rational B-Spline (NURBS) based functions are employed to interpolate the geometry and physical function. To deal with the rank-deficiency problem due to the degenerate boundary, the hypersingular integral equation is introduced to promote the full rank for the influence matrix in the dual BEM. Finally, three numerical examples are given to verify the accuracy of our proposed method. Both circular and square domains subjected to the Dirichlet boundary condition are considered. The engineering problem containing a degenerate boundary is considered, e.g., a seepage flow problem with a sheet pile. Numerical results of the IGA-DBEM agree well with these of the exact solution and the original dual boundary element method.
## Corresponding author
*Corresponding author (jtchen@mail.ntou.edu.tw)
## References
Hide All
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20 Chen, Y.Z., Lin, X.Y.Regularity condition and numerical examination for degenerate scale problem of BIE for exterior problem of plane elasticity,” Engineering Analysis with Boundary Elements, 32, pp. 811823 (2008).
21 Chen, J.T., Wu, C.S., Chen, K.H. and Lee, Y.T.Degenerate scale for analysis of circular plate using the boundary integral equations and boundary element method,” Computational Mechanics, 38, pp. 3349 (2006).
22 Ursell, F.Irregular frequencies and the motion of floating bodies,” Journal of Fluid Mechanics, 105, pp. 143156 (1981).
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26 Chen, J.T., Lin, S.R., Chen, K.H., Chen, I.L. and Chyuam, S.W.Eigenanalysis for membranes with stringers using conventional BEM in conjunction with SVD technique,” Computer Methods in Applied Mechanics and Engineering, 192, pp. 12991322 (2003).
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28 De Lacerda, L.A., Wrobel, L.C., Power, H. and Mansur, W.J.A novel boundary integral formulation for three-dimensional analysis of thin acoustic barriers over an impedance plane,” Journal of the Acoustical Society of America, 104, pp. 671678 (1998).
29 Portela, A., Aliabadi, M.H. and Rooke, D.P.The dual boundary element method: effective implementation for crack problems,” International Journal for Numerical Methods in Engineering, 33, pp. 12691287 (1992).
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35 Chen, K.H., Chen, J.T., Chou, C.R. and Yueh, C.Y.Dual boundary element analysis of oblique incident wave passing a thin submerged breakwater,” Engineering Analysis with Boundary elements, 26, pp.917928 (2002).
36 Chen, K.H., Chen, J.T., Lin, S.Y. and Lee, Y.T.Dual boundary element analysis of normal incident wave passing a thin submerged breakwater with rigid, absorbing, and permeable boundaries,” Journal of Waterway, Port, Costal and Ocean Engineering, ASCE, 130, pp.179190 (2004).
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40 Chen, J.T., Lin, S.R., Chen, K.H., Chen, I.L., and Chyuan, S.W.Eigenanalysis for membranes with stringers using conventional BEM in conjunction with SVD technique,” Computer Methods in Applied Mechanics and Engineering, 192, pp. 12991322 (2003).
41 Sun, F.L., Dong, C.Y., Yang, H.S., “Isogeometric boundary element method for crack propagation based on Bézier extraction of NURBS,” Engineeing Analysis with Boundary Elements, 99, pp. 7688 (2019).
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46 Chen, J.T., Chen, Y.W., “Dual boundary element analysis using complex variables for potential problems with or without a degenerate boundary,” Engineeing Analysis with Boundary Elements, 24, pp. 671684 (2000).
# Isogeometric Analysis of the Dual Boundary Element Method for the Laplace Problem With a Degenerate Boundary
## Metrics
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https://discourse.julialang.org/t/get-tuple-length-from-type/32483 | 1,653,175,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00597.warc.gz | 260,820,097 | 8,799 | # Get tuple length from type?
Is there a built-in function to get the length of a tuple as a `Val{N}`?
Basically I’m looking for something that would be of the form:
``````tuplen(::Type{Tuple{}}) = Val{0}()
tuplen(::Type{Tuple{T}}) where T = Val{1}()
tuplen(::Type{Tuple{T1, T2}}) where {T1, T2} = Val{2}()
tuplen(::Type{Tuple{T1, T2, T3}}) where {T1, T2, T3} = Val{3}()
...
``````
In case this is an XY problem - what I’m actually trying to do is convert an array-of-tuples representation to a tuple-of-arrays representation, and with the above definition I can do:
``````"""
Convert an Array-of-Tuples to a Tuple-of-Arrays.
"""
toa(aot) = ntuple(i->getindex.(aot, i), tuplen(eltype(aot)))
``````
without the above I could use `length(first(aot))` to get the length but that assumes the array is not empty.
edit: as a nested XY problem, I just realized that Plots can take vectors of tuples natively and I don’t need to transform my data to tuple-of-vectors after all. Carry on…
2 Likes
Yup–it’s actually a simple one-liner:
``````julia> tuple_len(::NTuple{N, Any}) where {N} = Val{N}()
tuple_len (generic function with 1 method)
julia> tuple_len((1, 2, 3))
Val{3}()
julia> tuple_len((1, "hello", π))
Val{3}()
julia> tuple_len((1, "hello", π, [1]))
Val{4}()
``````
This works because tuples (unlike all other types) are variadic, so you can dispatch on `NTuple{Any, N}` to match all tuples of length N.
10 Likes
Ah, thanks for that. I’d forgotten that tuple types are special.
2 Likes
So why isn’t it `length()`? Already defined in Base?
Sorry, that isn’t the same thing as I just realized. `length()` already exists in base for tuples,
but returns an integer for the length.
Also it’s not defined on the `Tuple` type, only tuple values:
``````julia> length(Tuple{Float64, Int64})
ERROR: MethodError: no method matching length(::Type{Tuple{Float64,Int64}})
Closest candidates are:
length(::Core.SimpleVector) at essentials.jl:593
``````
Right.
It’s also worth noting that the compiler is pretty good at dealing with tuple lengths as regular old integers using the built-in length function. For example, we can write a function that looks like it might be type-unstable:
``````julia> function maybe_type_unstable(x::Tuple)
if length(x) == 2
return 1.0
else
return "hello world"
end
end
maybe_type_unstable (generic function with 1 method)
``````
but the compiler is smart enough to figure out the value of `length(x)` at compile time and infer the result:
``````julia> @code_warntype maybe_type_unstable((1, 2))
Variables
#self#::Core.Compiler.Const(maybe_type_unstable, false)
x::Tuple{Int64,Int64}
Body::Float64
...
``````
``````julia> @code_warntype maybe_type_unstable((1, 2, 3))
Variables
#self#::Core.Compiler.Const(maybe_type_unstable, false)
x::Tuple{Int64,Int64,Int64}
Body::String
...
``````
6 Likes
Sorry to came so late after the fight, but we finally don’t have a solution to get the length of a tuple type ?
Do you mean besides defining the one-liner function marked as a solution?
Yep this function gets the length of a tuple, not from the tuple type ad was requested at the beginning (and as it seems the OP is referenced) . Actually I found a solution asking another question there How can i index a tuple type? so nvm
Type{Tuple{…}} Is not the same as NTuple{…}
The solution in the other thread may be efficient, but is probably an implementation detail that cannot be relied upon.
As it is said in the accepted answer:
So you can use `NTuple{N, T}` to match any `Tuple{...}`. The only matter is that you want the function to take tuple types instead of tuples. The solution below is correct and do not rely in any implementation details, while it may be ugly and inefficient.
``````julia> tuple_type_length(x) = (n = -1; while !(x <: NTuple{n+=1, Any}); end; return n)
tuple_type_length (generic function with 1 method)
julia> tuple_type_length(Tuple{Int, Char})
2
julia> tuple_type_length(Tuple{Int, Char, String})
3
``````
1 Like
One can avoid internals by using
``````julia> fieldcount(Tuple{Int,Float64})
2
``````
2 Likes
Seems a good solution, I did not know `fieldcount`. My only worry is the caveat in the documentation:
Get the number of fields that an instance of the given type would have. An error is thrown if the type is too abstract to determine this.
Kinda hard to know beforehand what is a “too abstract” type.
Any concrete type should be fine with `fieldtype`. Even many incomplete types with defined-but-type-unspecified fields such as `Complex{T} where T`, `Tuple{T,T} where T`, and `Tuple{Tuple{Vararg}}` will work.
Examples of types that will fail are `Tuple` (with no further elaboration), `Tuple{Vararg}`, and `NTuple{N,Any} where N`. The number of fields that these types have is undefined.
So `fieldcount` should work with almost any type that actually has a number of fields. One might be able to cook up some pathological corner cases, but I haven’t thought of one yet. If you find one that fails that you think shouldn’t, you can consider opening a bug report.
2 Likes
Is there anything wrong with this variant of the accepted solution (sligthly adapted to work on Tuple types instead of Tuple instances)?
``````julia> tuple_len(::Type{<:NTuple{N, Any}}) where {N} = Val{N}()
tuple_len (generic function with 1 method)
julia> x = (1, "hello", 42.0)
(1, "hello", 42.0)
julia> tuple_len(typeof(x))
Val{3}()
``````
2 Likes
I believe there is no problem, but I would adopt the simpler `fieldcount` solution now that I am aware of it, unless you want the code to fail in non-tuple types.
OK, but `fieldcount` gives a plain integer result, as opposed to the initial requirement that the result be a `Val`-wrapped value.
I agree that, since constant propagation is likely to work well in many cases, `fieldcount` would probably a very good and standard way to do this (but in those cases, the question then becomes: why not simply `length`?)
Is there any api guarantee that an NTuple will always have exactly N fields? Using `fieldcount` seems a bit ‘hacky’.
1 Like
`length` works if you have an instance of the object, true, but I thought the idea was to have a function that would work directly over the type?
Also, yes, `fieldtype` does not return a `Val` but neither does `N` in the parameter type annotation, both give `Int` values which can after be wrapped inside `Val`. It may be that one solution is recognized as type-stable by the compiler (i.e., recognizes the static mapping between input and output types) and the other solution isn’t, but this is speculation and should be checked.
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# Maths Syllabus watch
1. Hello, just interested to know if any of this stuff is in P1-6?
Matrices: Canonical Forms, Gaussian Elimination, Cayley-Hamilton Theorem, matrix decompositions.
Differential Equations: General solution methods for 2nd order Ordinary Differential Equations.
Mathematical Structure: Mathematical proof, set theory (i.e. Boolean algebra), group theory (i.e. proving Fermat's Little Theorem).
Numerical Methods: Solving differential equations using Runga-Kutta methods, integration with Simpson's Rule,
Complex Numbers: complex mappings (incl. conformal mapping)
2. Bloody hell, that looks complicated!
I think nearly all of that is Uni Level only; however:
"Differential Equations: General solution methods for 2nd order Ordinary Differential Equations." - This may appear in P6, I think I read something about it, where it appeared in P6. Maybe not though.
3. Second order DEs of the form a(d2y/dx2)+b(dy/dx)+cy = f(x) are in P4. Mathematical proof is covered in P1-6 but there is not a huge emphasis upon proof in the A-level. Matrices are covered in P6. However, most of the stuff you mention is above a-level.
4. (Original post by shiny)
Hello, just interested to know if any of this stuff is in P1-6?
Matrices: Canonical Forms, Gaussian Elimination, Cayley-Hamilton Theorem, matrix decompositions.
Differential Equations: General solution methods for 2nd order Ordinary Differential Equations.
Mathematical Structure: Mathematical proof, set theory (i.e. Boolean algebra), group theory (i.e. proving Fermat's Little Theorem).
Numerical Methods: Solving differential equations using Runga-Kutta methods, integration with Simpson's Rule,
Complex Numbers: complex mappings (incl. conformal mapping)
I think in MEI P6 you will find Gaussian Elimination and the cayley-hamilton theorem.
As mikesgt2 has said you learn 2nd order ODE solutions in p4 edexcel.
Proof, including by induction, is done in p6 edexcel, you should probably read up a bit on set theory any way, and group theory is in mei p6. How do you use group theory to prove fermats little theorem? I've always done it by a direct method.
I don't think you'll find Runga-Kutta in anything, you might find simpsons rule in mei or aqa. Complex mappings are dealt with a bit in edexcel p6. (transformations from one set of complex numbers to another)
5. (Original post by bono)
Bloody hell, that looks complicated!
I think nearly all of that is Uni Level only; however:
"Differential Equations: General solution methods for 2nd order Ordinary Differential Equations." - This may appear in P6, I think I read something about it, where it appeared in P6. Maybe not though.
No, differential equations is P3 mate. Only first order though.
6. (Original post by shiny)
, set theory (i.e. Boolean algebra), group theory (i.e. proving Fermat's Little Theorem).
LOL...the proof is over a hundred pages long.
7. (Original post by Ralfskini)
LOL...the proof is over a hundred pages long.
That's the last theorem
8. The proof of Fermat's Little Theorem comes from the fact that for a finite cyclic group G of order n, there is exactly one subgroup of order d for every d which is a divisor of n.
9. This is interesting since all the stuff I quoted was compulsory in the A-Level Further Maths syllabus I did!
10. (Original post by shiny)
The proof of Fermat's Little Theorem comes from the fact that for a finite cyclic group G of order n, there is exactly one subgroup of order d for every d which is a divisor of n.
Hmm, perhaps I still don't follow. How can conclude then that a^p-1 = 1 mod p where p is a prime that doesn't divide a? I've never really studied group theory properly so can't really see how this works...
11. (Original post by theone)
Hmm, perhaps I still don't follow. How can conclude then that a^p-1 = 1 mod p where p is a prime that doesn't divide a? I've never really studied group theory properly so can't really see how this works...
Let d be the order of a (where a is in the finite cyclic group Z which contains p-1 elements). From the theorem I just quoted, d divides p-1 or p-1 = qd ...
Hence,
a^(p-1) (mod p) = (a^d)^q (mod p) = 1^q (mod p) = 1 mod p
12. (Original post by shiny)
Let d be the order of a (where a is in the finite cyclic group Z which contains p-1 elements). From the theorem I just quoted, d divides p-1 or p-1 = qd ...
Hence,
a^(p-1) (mod p) = (a^d)^q (mod p) = 1^q (mod p) = 1 mod p
Ah, thanks for that (Far easier than my method of proving it )
13. (Original post by theone)
Ah, thanks for that (Far easier than my method of proving it )
What did you do? Induction?
14. (Original post by shiny)
What did you do? Induction?
Imagine the numbers a, 2a, 3a ... (p-1)a. where a is coprime to p
Now if any of these are congruent mod p, then we have the difference is 0 mod p. But the difference is going to be na where n < p. Now both of these are coprime to p, so the difference can not be 0 mod p. So none of the numbers are congruent mod p.
So the numbers are have different congruences mod p. Therefore they must, in some order, equate to 1, 2, 3, 4... p-1 mod p.
Taking the product. (p-1)! a^p-1 = (p-1)! mod p. Dividing through gives a^p-1 = 1 mod p.
15. (Original post by theone)
Imagine the numbers a, 2a, 3a ... (p-1)a. where a is coprime to p
Now if any of these are congruent mod p, then we have the difference is 0 mod p. But the difference is going to be na where n < p. Now both of these are coprime to p, so the difference can not be 0 mod p. So none of the numbers are congruent mod p.
So the numbers are have different congruences mod p. Therefore they must, in some order, equate to 1, 2, 3, 4... p-1 mod p.
Taking the product. (p-1)! a^p-1 = (p-1)! mod p. Dividing through gives a^p-1 = 1 mod p.
I don't think I have ever thought of it quite like that.
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https://blogs.sas.com/content/author/rickwicklin/page/39/ | 1,624,540,203,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00394.warc.gz | 147,785,260 | 13,242 | # Author
Distinguished Researcher in Computational Statistics
Rick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of PROC IML and SAS/IML Studio. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.
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If you are a statistical programmer, sooner or later you have to compute a confidence interval. In the SAS/IML language, some beginning programmers struggle with forming a confidence interval. I don't mean that they struggle with the statistics (they know how to compute the relevant quantities), I mean that they | 1,758 | 8,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-25 | latest | en | 0.91218 |
http://www.appstate.edu/~whiteheadjc/eco2030/games/duopoly/index.htm | 1,540,242,237,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515539.93/warc/CC-MAIN-20181022201445-20181022222945-00403.warc.gz | 407,782,250 | 1,933 | Duopoly Game
We are going to play a card game in which everybody will be randomly matched with someone else in the class. I will give each of you a pair of playing cards, one red card (hearts or diamonds) and one black card (clubs or spades). The numbers or faces on the cards do not matter, just the color. You will be asked to play one of these cards by holding it to your chest (so we can see that you have made your decision but not what that decision is).
You and your partner are competing firms operating in a market with only two firms -- a duopoly. You can choose to engage in price competition or you can cooperate and charge high prices. If you want to charge a low price then you should play the red card. If you want to charge a high price then you should play the black card.
Your profits are determined by the card that you play and by the card played by the person matched with you. There are three possible outcomes:
• Both firms play the black card (i.e., charge a high price), split the market and earn positive profits.
• Both firms play the red card (i.e., charge a low price), split the market and earn positive profits.
• One firm plays the red card (i.e., charges a low price) when the other firm plays a black card (i.e., charges a high price). This increases the market share of the firm with the low price and increases profits. The firm that plays the black card loses market share and loses profits.
All profits are hypothetical, except as noted below.
After you choose which card to play, hold it to your chest. We then tell you who you are matched with, and you can each reveal the card that you played. Record your earnings on the record sheet. Please note that in round 2 you will be matched with a different person, and payoffs will change. In round 3, you will be matched with a different person and payoffs change again, but you get to play with him/her in the remaining periods.
After we finish all periods, I will pick one person randomly and pay that person either (a) X% of his or her total profits, in cash (or an IOU that pays the 10-year t-note daily market interest rate) or (b) a fabulous prize. All earnings for everyone else are hypothetical.
Payoffs:
Source: Charles A. Holt and Monica Capra, "Classroom Games: A Prisoner's Dilemma" Journal of Economic Education, 31, 3, 229-236, Summer 2000. | 535 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-43 | latest | en | 0.967497 |
https://glasp.co/youtube/p/counterfeit-gold-coins-riddle-don-t-memorise | 1,718,945,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00383.warc.gz | 248,778,144 | 82,110 | # Counterfeit gold coins Riddle | Don't Memorise | Summary and Q&A
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December 27, 2020
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Infinity Learn NEET
Counterfeit gold coins Riddle | Don't Memorise
## TL;DR
Zeki uses a strategic approach to solve a puzzle involving 10 bags of gold coins with one bag containing defective coins, ultimately securing his place in a competition.
## Key Insights
• 🥖 The puzzle involves 10 bags of gold coins, with one bag containing counterfeit coins that weigh 9 grams instead of 10 grams.
• 🏋️ Zeki's strategy involves assuming a small sample of bags, removing an unequal number of coins from each bag, and using the weighing machine to determine the total weight.
• 🥖 By applying this strategy to the 10 bags, Zeki can identify the bag with counterfeit coins by using the least number of bags.
## Transcript
zeki is selected from his college to participate in an intercollegiate personality competition this screening round for this competition is logical reasoning zeki and his opponents have to solve a puzzle given to them in this screening round to enter the competition let's first look at the puzzle there are 10 bags of gold coins each bag contains 10... Read More
### Q: How does Zeki approach the puzzle involving the 10 bags of gold coins?
Zeki begins by assuming a sample of three bags and removing a different number of coins from each bag. He then uses the weighing machine to determine the total weight, allowing him to identify the bag with counterfeit coins.
### Q: Why does Zeki remove an unequal number of coins from each bag?
Zeki realizes that if he were to remove an equal number of coins from each bag, the total weight would always be 60 grams, regardless of whether any bags contain counterfeit coins. By selecting an unequal number of coins, he can identify the bag with the defective coins based on the difference in total weight.
### Q: What strategy does Zeki use to solve the puzzle with the 10 bags of gold coins?
Zeki divides the bags into two sets, with nine bags in one set and one bag in the other. He applies the strategy he used for the sample of three bags to the set of nine bags, allowing him to identify the bag with counterfeit coins. If the weighing machine displays the weight as 450 grams, he knows that the tenth bag contains the defective coins.
### Q: Why does Zeki choose the ratio of nine is to one in dividing the bags?
By dividing the bags into the ratio of nine is to one, Zeki ensures that he can determine the bag with counterfeit coins without using the weighing machine a second time. This ratio provides an optimal solution, guaranteeing his place in the competition.
## Summary & Key Takeaways
• Zeki is selected for an intercollegiate personality competition where participants must solve a puzzle involving 10 bags of gold coins, one of which contains counterfeit coins that weigh 9 grams instead of 10 grams.
• Zeki's strategy involves assuming a sample of three bags, removing a different number of coins from each bag, and using the weighing machine to determine the total weight. Through this process, he can identify which bag contains the defective coins.
• By applying this strategy to the 10 bags, Zeki successfully finds the bag with counterfeit coins by using the least number of bags, securing his place in the competition. | 707 | 3,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.949945 |
http://reference.wolfram.com/legacy/v5/Built-inFunctions/NumericalComputation/NumberRepresentation/FurtherExamples/ContinuedFraction.html | 1,511,137,181,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805881.65/warc/CC-MAIN-20171119234824-20171120014824-00395.warc.gz | 249,391,348 | 7,266 | This is documentation for Mathematica 5, which was
based on an earlier version of the Wolfram Language.
Further Examples: ContinuedFraction The result of ContinuedFraction is given as a list. In[1]:= Out[1]= The continued fraction simplifies to the original fraction. In[2]:= Out[2]= This gives the continued fraction representation of a quadratic number. The sublist represents the repeated part. In[3]:= Out[3]= You can go back to the number with FromContinuedFraction. In[4]:= Out[4]= This gives the first few terms of the continued fraction representation of a transcendental number. In[5]:= Out[5]= Here is a special case of Pell's equation. In a Diophantine equation like this the parameter m and the variables x and y are assumed to be integers. In[6]:= PellSolve gives the least positive solution for x and y when m is not a perfect square. In[7]:= Here is the solution when m is . In[8]:= Out[8]= In[9]:= Out[9]= In[10]:= See also FromContinuedFraction. | 234 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-47 | latest | en | 0.856254 |
https://www.jiskha.com/display.cgi?id=1260723314 | 1,503,284,793,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00336.warc.gz | 918,630,351 | 4,300 | posted by .
The monthly salaries of a sample of 100 employees were rounded to the nearest ten dollars. They ranged from a low of \$1,040 to a high of \$1,720. If we want to condense the data into seven classes, what is the most convenient class interval?
A. \$50
B. \$100
C. \$150
D. \$200
E. none of the above
The difference between the low and the high is \$680. If we want to divide that difference into seven classes, \$680/7 is close to \$100. An interval of \$100 would make the most sense here.
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More Similar Questions | 784 | 2,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-34 | latest | en | 0.916102 |
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• maths yr 8 trigonometry questions | 1,039 | 4,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.948282 |
http://www.jiskha.com/display.cgi?id=1225237252 | 1,495,853,255,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608765.79/warc/CC-MAIN-20170527021224-20170527041224-00469.warc.gz | 682,335,463 | 3,672 | # pre algebra
posted by on .
simplify
8(n-1)-10n
• pre algebra - ,
First-Distribute:
8(n-1)= 8xn-8(1)= 8n-8.
Now, subtract 10n so you have
-2n-8
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Post a New Question | 89 | 283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-22 | latest | en | 0.70151 |
https://statmd.wordpress.com/2013/11/09/page-rev-bayes-we-found-statistical-irregularities-in-a-randomized-controlled-trial/ | 1,500,813,175,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424559.25/warc/CC-MAIN-20170723122722-20170723142722-00445.warc.gz | 717,347,805 | 33,881 | ## Page Rev Bayes – we found statistical irregularities in a randomized controlled trial
The Bayesian counterpart to the frequentist analysis of the Randomized Controlled Trial is in many aspects more straightforward than the Bayesian analysis. One starts with a prior probability about the probability of a patient being assigned to each of the three arms and combines it with the (multinomial) likelihood of observing a given assignment pattern in the 240 patients enrolled in the study. Bayes theorem gives the posterior probability quantifying our belief about the magnitudes of the unknown assignment probabilities. Note that testing the strict equality is bound to lead us straight to the arms of the Lindley paradox so that a different approach is likely to be more fruitful. Specifically, we specify a maximum tolerable threshold for the difference between the maximum and the minimum probability of being assigned the trial arms (let’s say 1-5%) and we directly calculate the probability for this difference (“probability of foul play”).
In the absence of prior evidence for (or against) foul play we use a non-informative prior in which all possible values of assignment probabilities are equally plausible. This (Dirichlet) prior corresponds to a prior state of knowledge in which three individuals were randomized and all three ended up in different treatment arms. Under this prior, the posterior distribution is itself a Dirichlet distribution with parameters equal to the number of individuals actually assigned to each arm+1. The following R code may then be used to calculate the probability of foul play, as previously defined i.e.
``` event<-c(105,70,65)
set.seed(1234);
r<-rdirichlet(10000,event+1);
res0<-mean(apply(r,1,function(x,tol)
I(abs(max(x)-min(x))<=tol),0.01))
res0*100
```
This probability comes down to 0.4% which is numerically close to the frequentist answer, yet with a more intuitive interpretation: based on the observed trial sizes and a numerical tolerance for the maximum tolerable difference in assignment probability the odds for “foul play” are 249:1.
Increasing the tolerance will obviously decrease these odds, but in such a case we would be willing to tolerate larger differences in assignment probabilities. Although these results are mathematically trivial (and non-controversial), the plot will become more convoluted when one proceeds to use them to make a declaration of “foul play”. For in that case, a decision needs to be made which has to consider not only the probability of the uncertain events: “foul play” v.s. “not foul play” but also the consequences for the journal, the study investigators and the scientific community at large. At this level one would need to decide whether the odds of 249:1 are high enough or not for subsequent action to be taken. But this consideration will take us to the realm of decision theory (and it already 11pms).
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https://www.convertunits.com/from/micron/to/attometer | 1,620,672,220,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00544.warc.gz | 749,582,444 | 12,883 | ## ››Convert micron to attometre
micron attometer
How many micron in 1 attometer? The answer is 1.0E-12.
We assume you are converting between micron and attometre.
You can view more details on each measurement unit:
micron or attometer
The SI base unit for length is the metre.
1 metre is equal to 1000000 micron, or 1.0E+18 attometer.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between microns and attometers.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of micron to attometer
1 micron to attometer = 1000000000000 attometer
2 micron to attometer = 2000000000000 attometer
3 micron to attometer = 3000000000000 attometer
4 micron to attometer = 4000000000000 attometer
5 micron to attometer = 5000000000000 attometer
6 micron to attometer = 6000000000000 attometer
7 micron to attometer = 7000000000000 attometer
8 micron to attometer = 8000000000000 attometer
9 micron to attometer = 9000000000000 attometer
10 micron to attometer = 10000000000000 attometer
## ››Want other units?
You can do the reverse unit conversion from attometer to micron, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Micron
a metric unit of length equal to one millionth of a meter
## ››Definition: Attometre
The SI prefix "atto" represents a factor of 10-18, or in exponential notation, 1E-18.
So 1 attometre = 10-18 metre.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 521 | 1,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-21 | latest | en | 0.805997 |
http://fmaths.com/test08.php | 1,603,738,112,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891624.95/warc/CC-MAIN-20201026175019-20201026205019-00362.warc.gz | 38,939,443 | 2,654 | Grade 8 Final test of the 8th Grade Good luck!Question 1How much is ? Question 2What do you find when you factorize ? Question 3What are the solutions of the ininequality ? Question 4How is called a function written ? An exponential function An affine function A linear functionQuestion 5What do you find if you write with only one square root the number ? Question 6What are the solutions of the system ? x = - 8 and y = 10 x = 1 and y = 1 x = - 3 and y = 5Question 7What is the median number of 3, 3, 2, 0 ? 2 2,5 3Question 8A mountain is 400m high. If you unfold your arm which measures 1 meter, you succeed in hiding it with your hand which measures 0,06m. How far is the mountain? 6,2 kms 6,7 kms 6,9 kmsQuestion 9In which angle can you see a church 30m high which is 100m away from you? 17° 21° 72°Question 10In a mark, we have A(1;1) and B(2;7). How much is the length AB? 6 | 276 | 881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-45 | latest | en | 0.899181 |
https://buthowto.com/how-to-find-the-exponential-regression-equation-on-desmos | 1,660,567,308,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572174.8/warc/CC-MAIN-20220815115129-20220815145129-00691.warc.gz | 163,838,180 | 11,053 | # How to find the exponential regression equation on Desmos
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Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. (e.g., if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function.)
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Desmos Graphing Calculator. Kuta Software Algebra II. ... They will collect data and use exponential regression (manually and using technology) to model and analyze the function. They will evaluate the strength of their model and compare their models to peers models.
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Students can use graph paper and/or graphing calculators or Desmos. To engage students more effectively in discourse around the lines of best fit, the teacher could use technology (such as ClassPad or Desmos) to show all the different lines of best fit the students have identified layered on the scatterplot.
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This will draw in the Least Squares Regression Line and give you the slope (m) and y-intercept (b).How to Find . a Quadratic Regression: Input your data. In the next line, under your table, type in y 1 ~a x 1 2 +b x 1 +c . This will draw in the quadratic curve.How to Find . an Exponential Regression: Input your data.
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Class 10
Chapter 12 Class 10 - Electricity
## Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Β
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### Transcript
Case 1 Resistance = R Potential Difference = V Case 2 Resistance = R Potential Difference = π/2 By Ohmβs law, V = I R π/π
= I β΄ I = π/π
Current in Case 1 I1 = π/π
Current in Case 2 I2 = π/2 Γ 1/π
= π/2π
Now, πΌ_1/πΌ_2 = (π/π
)/(π/2π
) πΌ_1/πΌ_2 = π/π
Γ 2π
/π πΌ_1/πΌ_2 = 2 1/2 πΌ_1 = πΌ_2 πΌ_2 = 1/2 πΌ_1 β΄ When voltage becomes half, current also becomes half.
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Title: Chapter 5: Mining Frequent Patterns, Association and Correlations
1
Chapter 5 Mining Frequent Patterns, Association
and Correlations
• Basic concepts and a road map
• Efficient and scalable frequent itemset mining
methods
• Mining various kinds of association rules
• From association mining to correlation analysis
• Constraint-based association mining
• Summary
2
What Is Frequent Pattern Analysis?
• Frequent pattern a pattern (a set of items,
subsequences, substructures, etc.) that occurs
frequently in a data set
• First proposed by Agrawal, Imielinski, and Swami
AIS93 in the context of frequent itemsets and
association rule mining
• Motivation Finding inherent regularities in data
• What products were often purchased together?
Beer and diapers?!
• What are the subsequent purchases after buying a
PC?
• What kinds of DNA are sensitive to this new drug?
• Can we automatically classify web documents?
• Applications
• Basket data analysis, cross-marketing, catalog
design, sale campaign analysis, Web log (click
stream) analysis, and DNA sequence analysis.
3
Why Is Freq. Pattern Mining Important?
• Discloses an intrinsic and important property of
data sets
• Forms the foundation for many essential data
• Association, correlation, and causality analysis
• Sequential, structural (e.g., sub-graph) patterns
• Pattern analysis in spatiotemporal, multimedia,
time-series, and stream data
• Classification associative classification
• Cluster analysis frequent pattern-based
clustering
• Data warehousing iceberg cube and cube-gradient
• Semantic data compression fascicles
4
Basic Concepts Frequent Patterns and Association
Rules
• Itemset X x1, , xk
• Find all the rules X ? Y with minimum support and
confidence
• support, s, probability that a transaction
contains X ? Y
• confidence, c, conditional probability that a
transaction having X also contains Y
Let supmin 50, confmin 50 Freq. Pat.
A3, B3, D4, E3, AD3 Association rules A ?
D (60, 100) D ? A (60, 75)
5
Closed Patterns and Max-Patterns
• A long pattern contains a combinatorial number of
sub-patterns, e.g., a1, , a100 contains (1001)
(1002) (110000) 2100 1 1.271030
sub-patterns!
• Solution Mine closed patterns and max-patterns
• An itemset X is closed if X is frequent and there
exists no super-pattern Y ? X, with the same
support as X (proposed by Pasquier, et al. _at_
ICDT99)
• An itemset X is a max-pattern if X is frequent
and there exists no frequent super-pattern Y ? X
(proposed by Bayardo _at_ SIGMOD98)
• Closed pattern is a lossless compression of freq.
patterns
• Reducing the of patterns and rules
6
Closed Patterns and Max-Patterns
• Exercise. DB lta1, , a100gt, lt a1, , a50gt
• Min_sup 1.
• What is the set of closed itemset?
• lta1, , a100gt 1
• lt a1, , a50gt 2
• What is the set of max-pattern?
• lta1, , a100gt 1
• What is the set of all patterns?
• !!
7
Chapter 5 Mining Frequent Patterns, Association
and Correlations
• Basic concepts and a road map
• Efficient and scalable frequent itemset mining
methods
• Mining various kinds of association rules
• From association mining to correlation analysis
• Constraint-based association mining
• Summary
8
Scalable Methods for Mining Frequent Patterns
• The downward closure property of frequent
patterns
• Any subset of a frequent itemset must be frequent
• If beer, diaper, nuts is frequent, so is beer,
diaper
• i.e., every transaction having beer, diaper,
nuts also contains beer, diaper
• Scalable mining methods Three major approaches
• Apriori (Agrawal Srikant_at_VLDB94)
• Freq. pattern growth (FPgrowthHan, Pei Yin
_at_SIGMOD00)
• Vertical data format approach (CharmZaki Hsiao
_at_SDM02)
9
Apriori A Candidate Generation-and-Test Approach
• Apriori pruning principle If there is any
itemset which is infrequent, its superset should
not be generated/tested! (Agrawal Srikant
_at_VLDB94, Mannila, et al. _at_ KDD 94)
• Method
• Initially, scan DB once to get frequent 1-itemset
• Generate length (k1) candidate itemsets from
length k frequent itemsets
• Test the candidates against DB
• Terminate when no frequent or candidate set can
be generated
10
The Apriori AlgorithmAn Example
Supmin 2
Database TDB
L1
C1
1st scan
C2
C2
L2
2nd scan
C3
L3
3rd scan
11
Association rules
L3
L1
L2
• Min_confidence 80, the association rules are
shown as follows.
• A?C, B?E, E?B,
• B,C?E, C,E?B
12
The Apriori Algorithm
• Pseudo-code
• Ck Candidate itemset of size k
• Lk frequent itemset of size k
• L1 frequent items
• for (k 1 Lk !? k) do begin
• Ck1 candidates generated from Lk
• for each transaction t in database do
• increment the count of all candidates in
Ck1 that are
contained in t
• Lk1 candidates in Ck1 with min_support
• end
• return ?k Lk
13
Important Details of Apriori
• How to generate candidates?
• Step 1 self-joining Lk
• Step 2 pruning
• How to count supports of candidates?
• Example of Candidate-generation
• L3abc, abd, acd, ace, bcd
• Self-joining L3L3
• abcd from abc and abd
• acde from acd and ace
• Pruning
• acde is removed because ade is not in L3
• C4abcd
14
How to Generate Candidates?
• Suppose the items in Lk-1 are listed in an order
• Step 1 self-joining Lk-1
• insert into Ck
• select p.item1, p.item2, , p.itemk-1, q.itemk-1
• from Lk-1 p, Lk-1 q
• where p.item1q.item1, , p.itemk-2q.itemk-2,
p.itemk-1 lt q.itemk-1
• Step 2 pruning
• forall itemsets c in Ck do
• forall (k-1)-subsets s of c do
• if (s is not in Lk-1) then delete c from Ck
15
Challenges of Frequent Pattern Mining
• Challenges
• Multiple scans of transaction database
• Huge number of candidates
• Tedious workload of support counting for
candidates
• Improving Apriori general ideas
• Reduce passes of transaction database scans
• Shrink number of candidates
• Facilitate support counting of candidates
16
Partition Scan Database Only Twice
• Any itemset that is potentially frequent in DB
must be frequent in at least one of the
partitions of DB
• Scan 1 partition database and find local
frequent patterns
• Scan 2 consolidate global frequent patterns
• A. Savasere, E. Omiecinski, and S. Navathe. An
efficient algorithm for mining association in
large databases. In VLDB95
17
Partition approach
• Key idea If X is a large itemset in database D,
which is divided into n partitions p1, p2, , pn,
then X must be a large itemset in at least one of
the n partitions. (Prove by contrapositive.)
• The partition algorithm first scans partitions
pi, for i1 to n, to find the set of all local
large itemsets in pi, denoted as Lpi.
• Let CG be the union of Lpi, for i1 to n. Then CG
is a superset of the set of all large itemsets in
D.
• Finally, the algorithm scans each partition for
the second time to calculate the support of each
itemset in CG and to find out which candidate
itemsets are really large itemsets in D.
• Thus, only two scans are needed to find all the
large itemsets in D.
18
Example-Partition
. . .
. . .
P2
Pn
P1
. . .
LP2
LP1
LPn
• CGLp1?Lp2 ? ? Lpn
19
DHP Reduce the Number of Candidates
• A k-itemset whose corresponding hashing bucket
count is below the threshold cannot be frequent
• Candidates a, b, c, d, e
• Hash entries ab, ad, ae bd, be, de
• Frequent 1-itemset a, b, d, e
• ab is not a candidate 2-itemset if the sum of
count of ab, ad, ae is below support threshold
• J. Park, M. Chen, and P. Yu. An effective
hash-based algorithm for mining association
rules. In SIGMOD95
20
Sampling for Frequent Patterns
• Select a sample of original database, mine
frequent patterns within sample using Apriori
• Scan database once to verify frequent itemsets
found in sample, only borders of closure of
frequent patterns are checked
• Example check abcd instead of ab, ac, , etc.
• Scan database again to find missed frequent
patterns
• H. Toivonen. Sampling large databases for
association rules. In VLDB96
21
Sampling approach
• The sampling algorithm first takes a random
sample of the database D, and finds the set of
large itemsets (S) in the sample using a smaller
min_support.
• Then, the algorithm calculates the negative
border set Bd-(S) which is the set of minimal
itemsets X that is not in S.
• The algorithm scans D to check if c is a large
itemset in D, for each itemset c?S?Bd-(S).
• (If there is no large itemset in Bd-(S), the
algorithm has found all the large itemsets.
Otherwise, the algorithm constructs a set of
candidate itemsets by expanding S?Bd-(S)
recursively until Bd-(S) is empty.)
• The algorithm needs only one scan over D.
22
D
• Scan S to find all possible candidates.
• Scan D to find all the large itemsets.
• The algorithm needs only one scan over D.
S
23
Example-Sampling
• Let RA,B, ,F and assume the large itemsets S
is
• A,B,C,F,A,B,A,C
• A,F,C,F,A,C,F.
• The negative border set Bd-(S) is
D,E,B,C,B,F.
• Theorem Given an attribute set X and a random
sample s of size
• the probability that error e(X,s) gt? is at most
?, where e(X,s) is the error that X is a large
itemset in D but not in sample s.
24
Example Sampling
• Let RA, B, C, D, E, F and assume the large
itemsets S is
• A,B,C,F,A,B,A,C
• A,F,C,F,A,C,F.
• The negative border set Bd-(S) is
D,E,B,C,B,F.
25
Bottleneck of Frequent-pattern Mining
• Multiple database scans are costly
• Mining long patterns needs many passes of
scanning and generates lots of candidates
• To find frequent itemset i1i2i100
• of scans 100
• of Candidates (1001) (1002) (110000)
2100-1 1.271030 !
• Bottleneck candidate-generation-and-test
• Can we avoid candidate generation?
26
Mining Frequent Patterns Without Candidate
Generation
• Grow long patterns from short ones using local
frequent items
• abc is a frequent pattern
• Get all transactions having abc DBabc
• d is a local frequent item in DBabc ? abcd is
a frequent pattern
27
Construct FP-tree from a Transaction Database
TID Items bought (ordered) frequent
items 100 f, a, c, d, g, i, m, p f, c, a, m,
p 200 a, b, c, f, l, m, o f, c, a, b,
m 300 b, f, h, j, o, w f, b 400 b, c,
k, s, p c, b, p 500 a, f, c, e, l, p, m,
n f, c, a, m, p
min_support 3
• Scan DB once, find frequent 1-itemset (single
item pattern)
• Sort frequent items in frequency descending
order, f-list
• Scan DB again, construct FP-tree
F-listf-c-a-b-m-p
28
Benefits of the FP-tree Structure
• Completeness
• Preserve complete information for frequent
pattern mining
• Never break a long pattern of any transaction
• Compactness
• Reduce irrelevant infoinfrequent items are gone
• Items in frequency descending order the more
frequently occurring, the more likely to be
shared
• Never be larger than the original database (not
count node-links and the count field)
• For Connect-4 DB, compression ratio could be over
100
29
Partition Patterns and Databases
• Frequent patterns can be partitioned into subsets
according to f-list
• F-listf-c-a-b-m-p
• Patterns containing p
• Patterns having m but no p
• Patterns having c but no a nor b, m, p
• Pattern f
• Completeness and non-redundency
30
Find Patterns Having P From P-conditional Database
• Starting at the frequent item header table in the
FP-tree
• Traverse the FP-tree by following the link of
each frequent item p
• Accumulate all of transformed prefix paths of
item p to form ps conditional pattern base
Conditional pattern bases item cond. pattern
base c f3 a fc3 b fca1, f1, c1 m fca2,
fcab1 p fcam2, cb1
31
From Conditional Pattern-bases to Conditional
FP-trees
• For each pattern-base
• Accumulate the count for each item in the base
• Construct the FP-tree for the frequent items of
the pattern base
m-conditional pattern base fca2, fcab1
f 4 c 4 a 3 b 3 m 3 p 3
All frequent patterns relate to m m, fm, cm, am,
fcm, fam, cam, fcam
f4
c1
b1
b1
c3
?
?
p1
a3
b1
m2
p2
m1
32
Recursion Mining Each Conditional FP-tree
Cond. pattern base of am (fc3)
Cond. pattern base of cm (f3)
f3
cm-conditional FP-tree
Cond. pattern base of cam (f3)
f3
cam-conditional FP-tree
33
A Special Case Single Prefix Path in FP-tree
• Suppose a (conditional) FP-tree T has a shared
single prefix-path P
• Mining can be decomposed into two parts
• Reduction of the single prefix path into one node
• Concatenation of the mining results of the two
parts
?
34
Mining Frequent Patterns With FP-trees
• Idea Frequent pattern growth
• Recursively grow frequent patterns by pattern and
database partition
• Method
• For each frequent item, construct its conditional
pattern-base, and then its conditional FP-tree
• Repeat the process on each newly created
conditional FP-tree
• Until the resulting FP-tree is empty, or it
contains only one pathsingle path will generate
all the combinations of its sub-paths, each of
which is a frequent pattern
35
Scaling FP-growth by DB Projection
• FP-tree cannot fit in memory?DB projection
• First partition a database into a set of
projected DBs
• Then construct and mine FP-tree for each
projected DB
• Parallel projection vs. Partition projection
techniques
• Parallel projection is space costly
36
Partition-based Projection
• Parallel projection needs a lot of disk space
• Partition projection saves it
37
FP-Growth vs. Apriori Scalability With the
Support Threshold
Data set T25I20D10K
38
FP-Growth vs. Tree-Projection Scalability with
the Support Threshold
Data set T25I20D100K
39
Why Is FP-Growth the Winner?
• Divide-and-conquer
• decompose both the mining task and DB according
to the frequent patterns obtained so far
• leads to focused search of smaller databases
• Other factors
• no candidate generation, no candidate test
• compressed database FP-tree structure
• no repeated scan of entire database
• basic opscounting local freq items and building
sub FP-tree, no pattern search and matching
40
Implications of the Methodology
• Mining closed frequent itemsets and max-patterns
• CLOSET (DMKD00)
• Mining sequential patterns
• FreeSpan (KDD00), PrefixSpan (ICDE01)
• Constraint-based mining of frequent patterns
• Convertible constraints (KDD00, ICDE01)
• Computing iceberg data cubes with complex
measures
• H-tree and H-cubing algorithm (SIGMOD01)
41
MaxMiner Mining Max-patterns
• 1st scan find frequent items
• A, B, C, D, E
• 2nd scan find support for
• AB, AC, AD, AE, ABCDE
• BC, BD, BE, BCDE
• CD, CE, CDE, DE,
• Since BCDE is a max-pattern, no need to check
BCD, BDE, CDE in later scan
• R. Bayardo. Efficiently mining long patterns from
databases. In SIGMOD98
Potential max-patterns
42
CLOSET Mining Closed Itemsets by Pattern-Growth
• Itemset merging if Y appears in every occurrence
of X, then Y is merged with X
• Sub-itemset pruning if Y ? X, and sup(X)
sup(Y), X and all of Xs descendants in the set
enumeration tree can be pruned
• Hybrid tree projection
• Bottom-up physical tree-projection
• Top-down pseudo tree-projection
• Item skipping if a local frequent item has the
same support in several header tables at
different levels, one can prune it from the
• Efficient subset checking
43
CHARM Mining by Exploring Vertical Data Format
• Vertical format t(AB) T11, T25,
• tid-list list of trans.-ids containing an
itemset
• Deriving closed patterns based on vertical
intersections
• t(X) t(Y) X and Y always happen together
• t(X) ? t(Y) transaction having X always has Y
• Using diffset to accelerate mining
• Only keep track of differences of tids
• t(X) T1, T2, T3, t(XY) T1, T3
• Diffset (XY, X) T2
• Eclat/MaxEclat (Zaki et al. _at_KDD97), VIPER(P.
Shenoy et al._at_SIGMOD00), CHARM (Zaki
Hsiao_at_SDM02)
44
Visualization of Association Rules Plane Graph
45
Visualization of Association Rules Rule Graph
46
Visualization of Association Rules (SGI/MineSet
3.0)
47
Chapter 5 Mining Frequent Patterns, Association
and Correlations
• Basic concepts and a road map
• Efficient and scalable frequent itemset mining
methods
• Mining various kinds of association rules
• From association mining to correlation analysis
• Constraint-based association mining
• Summary
48
Mining Various Kinds of Association Rules
• Mining multilevel association
• Miming multidimensional association
• Mining quantitative association
• Mining interesting correlation patterns
49
Mining Multiple-Level Association Rules
• Items often form hierarchies
• Flexible support settings
• Items at the lower level are expected to have
lower support
• Exploration of shared multi-level mining (Agrawal
Srikant_at_VLB95, Han Fu_at_VLDB95)
50
Multi-level Association Redundancy Filtering
• Some rules may be redundant due to ancestor
relationships between items.
• Example
• milk ? wheat bread support 8, confidence
70
• 2 milk ? wheat bread support 2, confidence
72
• We say the first rule is an ancestor of the
second rule.
• A rule is redundant if its support is close to
the expected value, based on the rules
ancestor.
51
Mining Multi-Dimensional Association
• Single-dimensional rules
• Multi-dimensional rules ? 2 dimensions or
predicates
• Inter-dimension assoc. rules (no repeated
predicates)
• age(X,19-25) ? occupation(X,student) ?
• hybrid-dimension assoc. rules (repeated
predicates)
coke)
• Categorical Attributes finite number of possible
values, no ordering among valuesdata cube
approach
• Quantitative Attributes numeric, implicit
ordering among valuesdiscretization, clustering,
52
Mining Quantitative Associations
• Techniques can be categorized by how numerical
attributes, such as age or salary are treated
• Static discretization based on predefined concept
hierarchies (data cube methods)
• Dynamic discretization based on data distribution
(quantitative rules, e.g., Agrawal
Srikant_at_SIGMOD96)
• Clustering Distance-based association (e.g.,
Yang Miller_at_SIGMOD97)
• one dimensional clustering then association
• Deviation (such as Aumann and Lindell_at_KDD99)
• Sex female gt Wage mean7/hr (overall mean
9)
53
Static Discretization of Quantitative Attributes
• Discretized prior to mining using concept
hierarchy.
• Numeric values are replaced by ranges.
• In relational database, finding all frequent
k-predicate sets will require k or k1 table
scans.
• Data cube is well suited for mining.
• The cells of an n-dimensional
• cuboid correspond to the
• predicate sets.
• Mining from data cubescan be much faster.
54
Quantitative Association Rules
• Proposed by Lent, Swami and Widom ICDE97
• Numeric attributes are dynamically discretized
• Such that the confidence or compactness of the
rules mined is maximized
• 2-D quantitative association rules Aquan1 ?
Aquan2 ? Acat
association rules
to form
general
rules using a 2-D grid
• Example
age(X,34-35) ? income(X,30-50K) ?
55
Mining Other Interesting Patterns
• Flexible support constraints (Wang et al. _at_
VLDB02)
• Some items (e.g., diamond) may occur rarely but
are valuable
• Customized supmin specification and application
• Top-K closed frequent patterns (Han, et al. _at_
ICDM02)
• Hard to specify supmin, but top-k with lengthmin
is more desirable
• Dynamically raise supmin in FP-tree construction
and mining, and select most promising path to mine
56
Chapter 5 Mining Frequent Patterns, Association
and Correlations
• Basic concepts and a road map
• Efficient and scalable frequent itemset mining
methods
• Mining various kinds of association rules
• From association mining to correlation analysis
• Constraint-based association mining
• Summary
57
Interestingness Measure Correlations (Lift)
• play basketball ? eat cereal 40, 66.7 is
• The overall of students eating cereal is 75 gt
66.7.
• play basketball ? not eat cereal 20, 33.3 is
more accurate, although with lower support and
confidence
• Measure of dependent/correlated events lift
58
Are lift and ?2 Good Measures of Correlation?
• if 85 of customers buy milk
• Support and confidence are not good to represent
correlations
• So many interestingness measures? (Tan, Kumar,
Sritastava _at_KDD02)
59
Which Measures Should Be Used?
• lift and ?2 are not good measures for
correlations in large transactional DBs
• all-conf or coherence could be good measures
(Omiecinski_at_TKDE03)
• Both all-conf and coherence have the downward
closure property
• Efficient algorithms can be derived for mining
(Lee et al. _at_ICDM03sub)
60
Chapter 5 Mining Frequent Patterns, Association
and Correlations
• Basic concepts and a road map
• Efficient and scalable frequent itemset mining
methods
• Mining various kinds of association rules
• From association mining to correlation analysis
• Constraint-based association mining
• Summary
61
Constraint-based (Query-Directed) Mining
• Finding all the patterns in a database
autonomously? unrealistic!
• The patterns could be too many but not focused!
• Data mining should be an interactive process
• User directs what to be mined using a data mining
query language (or a graphical user interface)
• Constraint-based mining
• User flexibility provides constraints on what to
be mined
• System optimization explores such constraints
for efficient miningconstraint-based mining
62
Constraints in Data Mining
• Knowledge type constraint
• classification, association, etc.
• Data constraint using SQL-like queries
• find product pairs sold together in stores in
Chicago in Dec.02
• Dimension/level constraint
• in relevance to region, price, brand, customer
category
• Rule (or pattern) constraint
• small sales (price lt 10) triggers big sales
(sum gt 200)
• Interestingness constraint
• strong rules min_support ? 3, min_confidence
? 60
63
Constrained Mining vs. Constraint-Based Search
• Constrained mining vs. constraint-based
search/reasoning
• Both are aimed at reducing search space
• Finding all patterns satisfying constraints vs.
finding some (or one) answer in constraint-based
search in AI
• Constraint-pushing vs. heuristic search
• It is an interesting research problem on how to
integrate them
• Constrained mining vs. query processing in DBMS
• Database query processing requires to find all
• Constrained pattern mining shares a similar
philosophy as pushing selections deeply in query
processing
64
Anti-Monotonicity in Constraint Pushing
TDB (min_sup2)
• Anti-monotonicity
• When an intemset S violates the constraint, so
does any of its superset
• sum(S.Price) ? v is anti-monotone
• sum(S.Price) ? v is not anti-monotone
• Example. C range(S.profit) ? 15 is anti-monotone
• Itemset ab violates C
• So does every superset of ab
65
Monotonicity for Constraint Pushing
TDB (min_sup2)
• Monotonicity
• When an intemset S satisfies the constraint, so
does any of its superset
• sum(S.Price) ? v is monotone
• min(S.Price) ? v is monotone
• Example. C range(S.profit) ? 15
• Itemset ab satisfies C
• So does every superset of ab
66
Succinctness
• Succinctness
• Given A1, the set of items satisfying a
succinctness constraint C, then any set S
satisfying C is based on A1 , i.e., S contains a
subset belonging to A1
• Idea Without looking at the transaction
database, whether an itemset S satisfies
constraint C can be determined based on the
selection of items
• min(S.Price) ? v is succinct
• sum(S.Price) ? v is not succinct
• Optimization If C is succinct, C is pre-counting
pushable
67
The Apriori Algorithm Example
Database D
L1
C1
Scan D
C2
C2
L2
Scan D
C3
L3
Scan D
68
Naïve Algorithm Apriori Constraint
Database D
L1
C1
Scan D
C2
C2
L2
Scan D
C3
L3
Constraint SumS.price lt 5
Scan D
69
The Constrained Apriori Algorithm Push an
Anti-monotone Constraint Deep
Database D
L1
C1
Scan D
C2
C2
L2
Scan D
C3
L3
Constraint SumS.price lt 5
Scan D
70
The Constrained Apriori Algorithm Push a
Succinct Constraint Deep
Database D
L1
C1
Scan D
C2
C2
L2
Scan D
not immediately to be used
C3
L3
Constraint minS.price lt 1
Scan D
71
Converting Tough Constraints
TDB (min_sup2)
• Convert tough constraints into anti-monotone or
monotone by properly ordering items
• Examine C avg(S.profit) ? 25
• Order items in value-descending order
• lta, f, g, d, b, h, c, egt
• If an itemset afb violates C
• So does afbh, afb
• It becomes anti-monotone!
72
Strongly Convertible Constraints
• avg(X) ? 25 is convertible anti-monotone w.r.t.
item value descending order R lta, f, g, d, b, h,
c, egt
• If an itemset af violates a constraint C, so does
every itemset with af as prefix, such as afd
• avg(X) ? 25 is convertible monotone w.r.t. item
value ascending order R-1 lte, c, h, b, d, g, f,
agt
• If an itemset d satisfies a constraint C, so does
itemsets df and dfa, which having d as a prefix
• Thus, avg(X) ? 25 is strongly convertible
73
Can Apriori Handle Convertible Constraint?
• A convertible, not monotone nor anti-monotone nor
succinct constraint cannot be pushed deep into
the an Apriori mining algorithm
• Within the level wise framework, no direct
pruning based on the constraint can be made
• Itemset df violates constraint C avg(X)gt25
• Since adf satisfies C, Apriori needs df to
assemble adf, df cannot be pruned
• But it can be pushed into frequent-pattern growth
framework!
74
Mining With Convertible Constraints
• C avg(X) gt 25, min_sup2
• List items in every transaction in value
descending order R lta, f, g, d, b, h, c, egt
• C is convertible anti-monotone w.r.t. R
• Scan TDB once
• remove infrequent items
• Item h is dropped
• Itemsets a and f are good,
• Projection-based mining
• Imposing an appropriate order on item projection
• Many tough constraints can be converted into
(anti)-monotone
TDB (min_sup2)
75
Handling Multiple Constraints
• Different constraints may require different or
even conflicting item-ordering
• If there exists an order R s.t. both C1 and C2
are convertible w.r.t. R, then there is no
conflict between the two convertible constraints
• If there exists conflict on order of items
• Try to satisfy one constraint first
• Then using the order for the other constraint to
mine frequent itemsets in the corresponding
projected database
76
What Constraints Are Convertible?
77
Constraint-Based MiningA General Picture
78
A Classification of Constraints
79
Chapter 5 Mining Frequent Patterns, Association
and Correlations
• Basic concepts and a road map
• Efficient and scalable frequent itemset mining
methods
• Mining various kinds of association rules
• From association mining to correlation analysis
• Constraint-based association mining
• Summary
80
Frequent-Pattern Mining Summary
• Frequent pattern miningan important task in data
mining
• Scalable frequent pattern mining methods
• Apriori (Candidate generation test)
• Projection-based (FPgrowth, CLOSET, ...)
• Vertical format approach (CHARM, ...)
• Mining a variety of rules and interesting
patterns
• Constraint-based mining
• Mining sequential and structured patterns
• Extensions and applications
81
Frequent-Pattern Mining Research Problems
• Mining fault-tolerant frequent, sequential and
structured patterns
• Patterns allows limited faults (insertion,
deletion, mutation)
• Mining truly interesting patterns
• Surprising, novel, concise,
• Application exploration
• E.g., DNA sequence analysis and bio-pattern
classification
• Invisible data mining | 7,382 | 27,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-22 | latest | en | 0.824625 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese771.htm | 1,695,763,874,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00301.warc.gz | 683,352,549 | 3,934 | ### 3.771 $$\int \frac{(a+b x)^4}{(a^2-b^2 x^2)^3} \, dx$$
Optimal. Leaf size=10 $\frac{x}{(a-b x)^2}$
[Out]
x/(a - b*x)^2
________________________________________________________________________________________
Rubi [A] time = 0.0070928, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {627, 34} $\frac{x}{(a-b x)^2}$
Antiderivative was successfully verified.
[In]
Int[(a + b*x)^4/(a^2 - b^2*x^2)^3,x]
[Out]
x/(a - b*x)^2
Rule 627
Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))
Rule 34
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(d*x*(a + b*x)^(m + 1))/(b*(m + 2)), x] /
; FreeQ[{a, b, c, d, m}, x] && EqQ[a*d - b*c*(m + 2), 0]
Rubi steps
\begin{align*} \int \frac{(a+b x)^4}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac{a+b x}{(a-b x)^3} \, dx\\ &=\frac{x}{(a-b x)^2}\\ \end{align*}
Mathematica [A] time = 0.0035456, size = 10, normalized size = 1. $\frac{x}{(a-b x)^2}$
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x)^4/(a^2 - b^2*x^2)^3,x]
[Out]
x/(a - b*x)^2
________________________________________________________________________________________
Maple [B] time = 0.044, size = 29, normalized size = 2.9 \begin{align*}{\frac{a}{b \left ( bx-a \right ) ^{2}}}+{\frac{1}{b \left ( bx-a \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^4/(-b^2*x^2+a^2)^3,x)
[Out]
1/b*a/(b*x-a)^2+1/b/(b*x-a)
________________________________________________________________________________________
Maxima [A] time = 1.00771, size = 27, normalized size = 2.7 \begin{align*} \frac{x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^4/(-b^2*x^2+a^2)^3,x, algorithm="maxima")
[Out]
x/(b^2*x^2 - 2*a*b*x + a^2)
________________________________________________________________________________________
Fricas [A] time = 1.65899, size = 39, normalized size = 3.9 \begin{align*} \frac{x}{b^{2} x^{2} - 2 \, a b x + a^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^4/(-b^2*x^2+a^2)^3,x, algorithm="fricas")
[Out]
x/(b^2*x^2 - 2*a*b*x + a^2)
________________________________________________________________________________________
Sympy [B] time = 0.357719, size = 17, normalized size = 1.7 \begin{align*} \frac{x}{a^{2} - 2 a b x + b^{2} x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**4/(-b**2*x**2+a**2)**3,x)
[Out]
x/(a**2 - 2*a*b*x + b**2*x**2)
________________________________________________________________________________________
Giac [A] time = 1.19041, size = 15, normalized size = 1.5 \begin{align*} \frac{x}{{\left (b x - a\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^4/(-b^2*x^2+a^2)^3,x, algorithm="giac")
[Out]
x/(b*x - a)^2 | 1,345 | 3,364 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-40 | latest | en | 0.285636 |
https://www.mathworks.com/matlabcentral/answers/1563276-how-plot-4d-data?s_tid=srchtitle | 1,653,565,146,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00474.warc.gz | 996,258,186 | 24,902 | How plot 4D data?
100 views (last 30 days)
Ali Almakhmari on 14 Oct 2021
Commented: Kevin Holly on 14 Oct 2021
Lets say that I have the following data and I would like to plot the function f that is dependent on x, y, and z, In other words, f(x, y, z). How can I do that? x, y, z, and f(x, y, z) are variables that are eventually made of 10 by 10 by 10, for example:
x = 0.1:0.01:0.19;
y = 0.1:0.01:0.19;
z = 0.1:0.01:0.19;
f = x.*exp(x.^2 + y.^2 + z.^2);
I tried doing this and it failed:
[X, Y, Z] = meshgrid(x, y, z);
surf(X, Y, Z, f);
colorbar;
KSSV on 14 Oct 2021
x = 0.1:0.01:0.19;
y = 0.1:0.01:0.19;
z = 0.1:0.01:0.19;
[x,y,z] = meshgrid(x,y,z) ;
f = x.*exp(x.^2 + y.^2 + z.^2);
figure
hold on
for i = 1:size(x,3)
surf(x(:,:,i),y(:,:,i),z(:,:,i),f(:,:,i))
end
view(3)
Kevin Holly on 14 Oct 2021
ah, you beat me and had a better result
x = 0.1:0.01:0.19;
y = 0.1:0.01:0.19;
z = 0.1:0.01:0.19;
[X, Y, Z] = meshgrid(x, y, z);
f = X.*exp(X.^2 + Y.^2 + Z.^2);
slice(f,5,5,5)
colorbar
R2020b
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Start Hunting! | 478 | 1,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-21 | latest | en | 0.831654 |
http://stackoverflow.com/questions/19485414/changing-to-tail-recursive-sml | 1,449,040,208,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448399455473.15/warc/CC-MAIN-20151124211055-00331-ip-10-71-132-137.ec2.internal.warc.gz | 221,417,917 | 19,141 | Changing to Tail Recursive SML
Ok, so I'm trying to change this function into Tail Recursive. The Definition I have of Tail Recursive is to use a "Local Helper Function" to accumulate my answer and return it without calling the primary function recursively.
these functions work properly.
fun same_string(s1 : string, s2 : string) =
s1 = s2
fun all_except_option (name, []) = NONE
| all_except_option (name, x::xs)=
case same_string (x , name) of
true => SOME xs
| false => case all_except_option(name,xs) of
NONE => NONE
| SOME z => SOME(x::z)
fun get_substitutions1 ([],name2) = [] (*get_substitutions2 is same but tail recursive *)
| get_substitutions1 (x::xs,name2) =
case all_except_option (name2,x) of
NONE => get_substitutions1 (xs,name2)
| SOME z => z @ get_substitutions1(xs,name2)
So here are my attempts at tail recursion which do not work and I think I am missing something fairly basic that I am overlooking due to my lack of experience in SML.
fun get_substitutions2 (lst,name3) =
let fun aux (xs,acc) =
case all_except_option(name3,x::xs) of
NONE => aux(xs, acc)
| SOME z => aux(xs, z::acc)
in
aux(lst,[])
end
and
fun get_substitutions2 (lst,name3) =
let fun aux (xs,acc) =
case all_except_option(name3,x::xs) of
NONE => aux(xs, acc)
| SOME z => aux(xs, z@acc)
in
aux(lst,[""])
end
Both "get_substitutions" functions are supposed to do the same thing. compare String1 to string list list, return single list made up of all lists containing String1 minus String1.
My attempts at using Tail Recursion have resulted in the following error.
Error: unbound variable or constructor: x
uncaught exception Error
raised at: ../compiler/TopLevel/interact/evalloop.sml:66.19-66.27
../compiler/TopLevel/interact/evalloop.sml:44.55
../compiler/TopLevel/interact/evalloop.sml:296.17-
Here are a few examples of calling get_substitutions2:
get_substitutions2 ([["foo"],["there"]], "foo"); (* = [] *)
get_substitutions2 ([["fred","fredrick","freddie","F","freddy"],["Will","William","Willy","Bill"]],"Bill"); (* = ["Will","William","Willy"] *)
get_substitutions2 ([["a","b"],["a","c"],["x","y"]], "a"); (* = ["c","b"] *)
-
The error is coming up because (BIG SURPRISE!) x is unbound in your function. You don't define x anywhere, so the compiler doesn't know what it is. I think you just want xs, not x::xs. However, after spending about 10 minutes looking at your code, I can't figure out exactly what you're trying to do, and I don't think you have any idea what you're doing. For starters, your recursive helper function acc is missing a base case. You should also know if you should use @ (concatenate) or :: (prepend) based on the type of z, and what you want for your result. – DaoWen Oct 21 '13 at 3:07
Thank you for comment. You are right, I have little idea of what I am doing because I am being given extreme basics and being asked to make intermediate code in a language I am unfamiliar with. This would be why I asked for help. I am having difficulty resolving how x::xs seems to be synonymous with hd xs, and also for combining lists as seen in function get_substitutions1. – user2865449 Oct 21 '13 at 3:26
I don't think the language is so much the problem as the paradigm. You don't seem very familiar with a lot of functional programming concepts (recursion, pattern matching, cons-lists, etc). You should try to ask more specific questions. You might have better luck chatting with the people in #sml@irc.freenode.net, since you can actually have a discussion there at a decent pace, unlike leaving comments on stackoverflow. – DaoWen Oct 21 '13 at 3:33
Pattern matching is exactly what I am attempting to figure out. you mentioned the Base Case. While this is simple with numbers as I was shown. With strings the only way I know to start is with an empty string, and build onto it for the answer. Which is what I was attempting to do. Not everyone can take "hey 1+1=2" and from that spell Obfuscate. Thank you for your time. – user2865449 Oct 21 '13 at 3:42
From the code you posted, I can't even tell that you're working with a string. But, again, posting comments here is really too slow to have a real discussion. You should go to the sml channel on freenode.net (#sml@irc.freenode.net). I'm in there right now, and there are lots of other people that probably actually use SML regularly there too (I prefer other functional languages like Clojure LISP and Scala to ML). If you don't have an IRC client, you can use their online interface: webchat.freenode.net – DaoWen Oct 21 '13 at 3:46
1 Answer
You need to use the same patterns you had for get_substitutions1 in your aux function definition:
fun get_substitutions2 (lst,name3) =
let fun aux ([],acc) = acc (* BASE CASE *)
| aux (x::xs,acc) = (* BINDING x IN PATTERN *)
case all_except_option(name3,x) of
NONE => aux(xs, acc)
| SOME z => aux(xs, z@acc)
in
aux(lst,[])
end
- | 1,305 | 4,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-48 | latest | en | 0.72015 |
http://www.cs.duke.edu/csed/alice/aliceInSchools/workshop11/twoday.php | 1,506,060,957,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688671.43/warc/CC-MAIN-20170922055805-20170922075805-00022.warc.gz | 450,940,307 | 5,076 | Alice Programming
Duke University, Durham, NC
## Workshop Schedule June 22-23, 2011
This is a two day followup workshop on Alice for previous Alice participants to come together to learn about new Alice materials and to talk about how they are using Alice at their schools.
Wednesday, June 22, Day 1
• Workshop from 10:30am - 4pm with half hour lunch break. (5 hrs)
• Welcome/Introductions (pdf) (pptx)
Gave an overview of the work we have been doing at Duke the past three years.
• Subscribe to the Alice listserv. For teachers only, a great way to ask questions and get answers about what you and your students are doing with Alice. One teacher recommended not getting in digest format, as you may not get attached Alice worlds.
• Link to Alice page that describes how to build 3D models (not easy) and also to the newest Alice models
• Presentations/Discussions by students and participants
• Nate Schuler - He teaches Alice in AP Gov and European History. In both courses, students work with Alice for 9 weeks and have certain requirements that have to be in their Alice world. His students start to learn Alice as he goes through about a 45 minute tutorial with them. He has developed a rubric. He wants to be able to do better "vocal narration".
• There was some discussion about whether you could make an object that you cannot drive through (for collision detection.
• Collision Detection. Liz mentioned that there is a Alice function "threshold of object" that checks if you are anywhere near the boundry box of the object.
• Melissa demoed her worlds: Multiplication table, Matrix multiplication band Permuations.
• Someone asked if there was an addition game where you would show 3 objects and 4 objects and then show the sum is 7 objects.
• There was a discussion on randling sampling. Create a world with people who have additional new properties (such as gender). Customize each person and add them all to an array. Randomly call a number and have a person appear (something like 45% of the time female and 55% of the time male).
• Heidi talked about her course which is an elective course that uses just Alice. She mostly works with at risk kids.
• Peggy showed her Distributive Property world.
• Frances mentioned she had trouble installing Alice so she has not done much of it. Next year her school will be a one-to-one school, each kid will get a Mac laptop. She wants ideas on how to use Alice. She wants to do something with Photograph.
• Susan Demoed the Camera Angles worlds a teacher did from the 2008 workshop. It is on this page.
• Liz showed her three science worlds including the barchart object she created. She also demoed her Treasure hunt adventure game and the Pinata game. These are all on the tutorials web page.
• Linda teaches Spanish at high school level. Looking for ways to use Alice.
• Chitra demoed her ice cream shop world and her toystory world.
• Ideas
• On tutorial website, for a particular tutorial, link to tutorials that you need to do as prerequisite.
• Need to create an Alice general guide or link to one someone else has created. Might include basics with arrays, describing what are the new methods/functions that come with them.
• Need a tutorial on working with numbers. Mention the "int as a string" that will display a decimal number as an int. Possibly create an invisible object with lots of math functionality that you could drop into any world.
• Include in the general tutorial how to do accents in foreign languages.
Tuesday, June 23, 2011, Day 2
• Workshop from 9am - 4pm with one hour lunch break. (6 hrs)
• 9:00am VCL Discussion - Richard Lucic
• What software is available for K-12 students to use? How many students can use the software at the same time.
• How does one get software installed for use? What is the cost? How does licensing work? For example it would be helpful to have access to the most recent version of Microsoft Office.
• One person mentioned that it would be helpful if students could take online exams on it. For example a national spanish exam.
• Kathleen talked about how they use Alice, mostly in clubs during the day. Students can take a club once and a club meets once a week for 4 weeks. She said this coming year, Alice in included in a list of possible presentations that students can pick from to use for projects. She said it would help if she could make a video/screencast of some of the tutorials. We may be able to use Camtasia to make a video of some of the basic getting started tutorials. The Video could be made into chapters (break into shorter parts if it is too long).
• Tom and Don talked about how they have been using Alice with both High School students and College students and they showed several demos. Some of their Alice worlds are very short about a particular topic. One student project was a "cat"apult and one was a penguin memory game. They use Alice at the high school level in a full course.
• Joe Mack talked about some of the work he has done teaching programming and science to kids.
• Fill out Survey form. Hand out Certificates
Misc Items
• Ideas for worlds to build
• Angry birds like game that will focus on the sling shot and shooting. Or create a world where you try to shoot a cannon.
• New objects to make
• Buildings with rooms inside, doors and windows that can move.
• Can you make an object (rectangular box) that uses collision resolution so that you can't move through walls?
• Dice with 6 sides
• Card object and or deck of 52 cards. | 1,214 | 5,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-39 | latest | en | 0.959581 |
http://www.caclubindia.com/experts/cenvat-credit-and-pla-1002789.asp | 1,369,253,652,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702444272/warc/CC-MAIN-20130516110724-00026-ip-10-60-113-184.ec2.internal.warc.gz | 376,716,871 | 13,565 | Home > Experts > Excise > Cenvat credit and pla
# Cenvat credit and pla(Excise)
This query is : Resolved
( Author )
16 June 2012
Dear Sir,
Kindly see my following example:
There is start of plant & there is 'No' opening Cenvat Credit.
Excise duty of inputs purchased- 3,00,000
Excise duty of Sale during month-5,00,000
Is I am saying right that , now we have Cenvat credit of Rs. 2,00,000.
Now, Suppose every month , the equation is such that , excise duty of sales is never greater than (excise duty of purchase + opening credit).
Will one have to pay any excise duty during whole year. as we have cenvat credit for whole year.
If, Yes . why. as we have credit
If,No. why. as I have seen that some pay duty in PLA although they have credit.they do like following, let's Suppose :
opening 50,000
ed of purchases 3,00,000
------------
3,50,000
ed of sales 3,00,000
-----------
50,000
The ed of sales entered above is by substracting ed on sales of whole month say rs. 500000-excise duty paid for month through PLA RS. 2,00,000.
I want to know why they paid rs. 200000 through PLA . how the figure is arrived at?
I think it should be like this.
opening 50,000 + ed purchase 3,00,000 - ed on sales rs. 5,00,000= Rs. 1,50,000 (Net payable)
How this Pla came into and there is still rs. 50,000 credit in my former example.
Kindly give the reply as soon as possible as these concepts are very urgent for me.
Ajay Raghatate
( Expert )
18 June 2012
Dear Rahul,
Plz explain clearly, what do u want to say.
You need to be the querist or approved CAclub expert to take part in this query .
Similar Resolved Queries : | 456 | 1,621 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2013-20 | latest | en | 0.970374 |
https://www.physicsforums.com/threads/11th-grade-oscillation-problem.642274/ | 1,537,740,547,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159820.67/warc/CC-MAIN-20180923212605-20180923233005-00516.warc.gz | 833,219,265 | 18,083 | # Homework Help: 11th grade Oscillation problem
1. Oct 8, 2012
### squareroot
1. The problem statement, all variables and given/known data
A pendulum with the lenght L is suspended by a inclined wall at an β angle.The pendulum is deviated with an angle of 2*β and then set free.Determine the oscillation period(T) of the pendulum considering that all the clashes are perfectly elastic.
2. Relevant equations
Here are some relevant equations
T=2$\pi$*$\sqrt{\frac{L}{g}}$
T=2$\pi$*$\sqrt{\frac{m}{k}}$
and all the quations from mechanical oscillations
3. The attempt at a solution
Nothing relevant yet.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Oct 8, 2012
anyone?
3. Oct 9, 2012
### squareroot
Amost forgot , a picture for you.
File size:
15.5 KB
Views:
143
4. Oct 9, 2012
### bayan
since it is 11th grade problem I assume the angle is small such that sin θ=θ. With this being the case T mainly depends on L.
You are told that the collision with the wall is elastic which means it rebounds at same speed, this is same as if the wall was not there.
The only effect the wall has is it changes the oscillation period to 3/4 ( just pic one oscillation as 8 parts, you are missing 2 from wall to left side max height)
Find the oscillation period if the wall was not there and take the effect of wall into account.
Hope this helps
5. Oct 9, 2012
### squareroot
Thank you.I've initially thought of what you said, it's observable , but my problem stands in proving that mathematically.
6. Oct 9, 2012
### TSny
Hi, squareroot. Have you studied the mathematical expression for the position of the pendulum as a function of time? That would be the most accurate way to determine the time to go from release to hitting the wall. Taking 3/8 of the period of the unblocked pendulum will not be very accurate. That's because it does not take the pendulum the same amount of time to swing through each eighth of a cycle.
If you've studied the relationship between simple harmonic motion and uniform circular motion, then you might see if you can use that relationship to get the answer.
7. Oct 9, 2012
### squareroot
I've been fussing with this one for 1 hour , still coudn't figure it out and i need it for tomorrow.Ahhh!
8. Oct 9, 2012
### TSny
If you go to this link, you will see how simple harmonic motion is a "projection" of uniform circular motion.
http://www.physics.uoguelph.ca/tutorials/shm/phase0.html
Think of the blue ball as moving back and forth in simple harmonic motion similar to a pendulum swinging back and forth. The red ball travels at a constant speed around the circle. The interesting thing here is that the blue ball, which does not move with constant speed, is always directly under the red ball.
Imagine that when the blue ball is at the far right, that corresponds to the release of the pendulum. Where would you locate a wall on the line of motion of the blue ball so that it would correspond to the wall in your problem?
How far around the circle has the red ball traveled when it is over the wall?
9. Oct 9, 2012
### squareroot
Yes , i understood that , as I said it's intuitive, but i can't prove it mathematically.It is there where my problems are.
10. Oct 9, 2012
### TSny
When you say you are trying to prove it mathematically, you will need to know more than just the formula for the period of a pendulum. What other mathematical equations have you studied that relate to pendulum motion?
11. Oct 9, 2012
### squareroot
well , we can find out how much the red ball has traveled around the portion of the circle "behind" the wall with the formula i attached below , where r is the length of the string(L) and n I THINK(but i m not sure) is alpha.but from here on , i'm lost.I also have the answer sheet and it says the answer is $\frac{4}{3}$*$\pi$*$\sqrt[]{\frac{l}{g}}$
Here are some more formulas that i know related to oscillation
v=ωA*cos(ωt+θ)
y=Asin(ωt+θ)
a=-ω$^{2}$Asin(ωt+θ)
where v= velocity , y=elongation , A-amplitude(maximum elongation) and ω-angular speed , θ-initial phase
T=2$\pi$$\sqrt{\frac{m}{k}}$
ω=2$\pi$*$\nu$ , $\nu$=$\frac{1}{T}$
k=m*ω*ω
T=2$\pi$$\sqrt{\frac{l}{g}}$
Kinetic energy=m*v*v/2 , potential energy=k*y*y/2
#### Attached Files:
• ###### Screen Shot 2012-10-09 at 8.15.22 PM.png
File size:
570 bytes
Views:
125
12. Oct 9, 2012
### TSny
OK. So, here's an equation that will tell you the position, y, at any time, t. For a pendulum you may think of y as replaced by the angle, $\phi$, that the string makes with the vertical. So,
$\phi$ = Asin(ωt+θ).
Think about what you would use for the amplitude A in this equation.
Think about the value of $\phi$ at t = 0. This will help you determine the phase angle θ.
Finally, think about the value of $\phi$ at the time the pendulum hits the wall. That will help you find the time at which the pendulum hits the wall.
13. Oct 9, 2012
### TSny
If you want to use the graphical approach, study the picture attached. What fraction of a circle has the red ball traveled by the time the blue ball hits the green wall?
#### Attached Files:
• ###### Pendulum with Wall.jpg
File size:
26.6 KB
Views:
121
14. Oct 9, 2012
### squareroot
So, from the 2 relation that you've given me, for i finding A i thought of this , sin(2alpha)=A/l , but it is specified that 2alpha<6 degrees which means that 2alpha=A/l and A is 2alpha*l so by replaing this in the equation 2alpha=Asin(ωt+θ) at t=0 that means that
2alpha=2alpha*l*sinθ so θ=1/l (thinking that sinθ≈θ)
moving to the second relation where θ=alpha(when tha ball hits the wall)
alpha=Asin(ωt+θ) so that alpha=2alpha*l* sin(ωt+1/l) so that t=$\frac{-1}{2lω}$
But how do i get from this to the period?
15. Oct 9, 2012
### TSny
In the equation $\phi$ = A sin(ωt+θ), A is the amplitude or the maximum amount that $\phi$ would have during the motion. So, how does A relate to the angle $\alpha$ shown in the figure of the pendulum that you posted?
16. Oct 9, 2012
### squareroot
Well the maximum angle is 2α in the picture of the problem , so at t=0 , when you release the ball the angle is 2α, i figured that ϕ is your notation for α.And when the ball hits the wall the angle is α.
A is the maximum distance from the point of equilibrum so the max distance is reached when the object is at the angle of 2α and we (in class)take the distance from the point of equillibrum to the ball as a straight line(for simplicity) so sin2α=A/l, where l is the hypotenuse and A is the adjacent.
17. Oct 9, 2012
### TSny
Right, the amplitude A is 2α. Also, $\phi$ = 2α at t = 0. Try to use this information to figure out $\theta$.
[Got a doctor's appointment, so I need to skip out for now.]
18. Oct 9, 2012
### squareroot
Oh , man , i need it until tomorrow and its 9.30 PM in my country.Boy am I dead....
19. Oct 10, 2012
### squareroot
I managed to get another day , i have until tomorrow.I l start working as soon as i get home.
20. Oct 10, 2012
### TSny
As the pendulum swings, think of y = A sin (ωt+θ) as describing the horizontal motion of the pendulum. So, y represents the horizontal position relative to the equilibrium position.
At the moment the pendulum is released, what is the value of y? You won’t be able to give a numerical value, but how would you express y in terms of A at the moment of release?
What would the value of t be at the moment of release?
Substitute these values of y and t into the equation and see what you discover. | 2,097 | 7,531 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-39 | latest | en | 0.934817 |
https://www.doubtnut.com/question-answer/in-any-triangle-abc-find-the-least-value-of-r1-r2-r3-r-642546375 | 1,642,937,818,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00433.warc.gz | 768,992,401 | 86,066 | # In any triangle ABC, find the least value of (r_(1) + r_(2) + r_(3))/(r)
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We have already proved that (1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3)) = (1)/(r) <br> Now using A.M. ge H.M. we get <br> (r_(1) + r_(2) + r_(3))/(3) ge (3)/((1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3))) = 3r <br> rArr (r_(1) + r_(2) + r_(3))/(r) ge 9 | 233 | 658 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-05 | longest | en | 0.844862 |
http://mcthings.createaforum.com/support/how-to-construct-a-signed-short-from-bytes/ | 1,603,276,751,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876307.21/warc/CC-MAIN-20201021093214-20201021123214-00350.warc.gz | 64,752,499 | 6,346 | ### Author Topic: How to construct a Signed Short from Bytes (Read 38 times)
#### Nick_W
• Full Member
• Posts: 215
##### How to construct a Signed Short from Bytes
« on: September 02, 2016, 04:59:14 pm »
I had some problems making a signed Short from two bytes (in a ListOfBytes). As a Short is signed, I thought the result would be signed by default - well it isn't!
This was my code:
Code: [Select]
` //**************************************************************************/ // // @brief Reads a signed 16 bit value over I2C // //**************************************************************************/ Public Function readS16(reg As Byte) As Integer //should return Short Dim result As Short = 0 Dim value As ListOfByte = read2X8(reg) If value <> Nothing Then result = (value(0) << 8) | value(1) //result = value.ExtractShort(0, Endianness.Big) //not working End If Return result End Function`
So first ExtractShort() doesn't work (although this does exactly what I want - I think). The above always returns an unsigned Short (as Integer). Even if you change the return type to Short, it still returns an unsigned Short (or Integer, not sure which).
To make it a Signed Short (or Integer) I had to manually convert it to two's compliment:
Code: [Select]
` //**************************************************************************/ // // @brief Reads a signed 16 bit value over I2C // //**************************************************************************/ Public Function readS16(reg As Byte) As Integer //should return Short Dim result As Short = 0 Dim value As ListOfByte = read2X8(reg) If value <> Nothing Then result = (value(0) << 8) | value(1) //result = value.ExtractShort(0, Endianness.Big) //not working End If //convert value to 2's compliment If (value(0) > 0x7f) Then result = (~result + 1) * (-1) End If Return result End Function`
Just thought I would point this out for anyone manipulating bytes out there. Anyone know how to use ExtractShort() or why it doesn't work in the above?
Thanks, | 535 | 2,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-45 | latest | en | 0.555462 |
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# Assignment problem:Bayville elementary school
Bayville has built a new elementary school, increasing the town's total to four schools- Addision, Beeks, Canfield, and Daley. Each has a capacity of 400 students. The school board wants to assign children to schools so that their travel time by bus is as short as possible. The school board has partitioned the town into five districts conforming to population density- north, south, east, west, and central. The average bus travel time from each district to each school is shown as follows:
Travel Time (mins)
District Addison Beeks Canfield Daley Student Population
North 12 23 35 17 250
South 26 15 21 27 340
East 18 20 22 31 310
West 29 24 35 10 210
Central 15 10 23 16 290
Determine the number of children that should be assigned from each district to each school to minimize total student travel time.
#### Solution Summary
This posting contains solution to following problem on Assignment method.
\$2.19 | 240 | 992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-50 | longest | en | 0.943311 |
https://ubt.opus.hbz-nrw.de/frontdoor/index/index/searchtype/authorsearch/author/%22Welker%2C+Kathrin%22/rows/100/start/0/subjectfq/PDE+Constraints/docId/740 | 1,726,118,134,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00248.warc.gz | 558,318,909 | 6,875 | • search hit 1 of 1
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## Efficient PDE Constrained Shape Optimization in Shape Spaces
### Effiziente PDE beschränke Formoptimierung in Formenräumen
• Shape optimization is of interest in many fields of application. In particular, shape optimization problems arise frequently in technological processes which are modelled by partial differential equations (PDEs). In a lot of practical circumstances, the shape under investigation is parametrized by a finite number of parameters, which, on the one hand, allows the application of standard optimization approaches, but, on the other hand, unnecessarily limits the space of reachable shapes. Shape calculus presents a way to circumvent this dilemma. However, so far shape optimization based on shape calculus is mainly performed using gradient descent methods. One reason for this is the lack of symmetry of second order shape derivatives or shape Hessians. A major difference between shape optimization and the standard PDE constrained optimization framework is the lack of a linear space structure on shape spaces. If one cannot use a linear space structure, then the next best structure is a Riemannian manifold structure, in which one works with Riemannian shape Hessians. They possess the often sought property of symmetry, characterize well-posedness of optimization problems and define sufficient optimality conditions. In general, shape Hessians are used to accelerate gradient-based shape optimization methods. This thesis deals with shape optimization problems constrained by PDEs and embeds these problems in the framework of optimization on Riemannian manifolds to provide efficient techniques for PDE constrained shape optimization problems on shape spaces. A Lagrange-Newton and a quasi-Newton technique in shape spaces for PDE constrained shape optimization problems are formulated. These techniques are based on the Hadamard-form of shape derivatives, i.e., on the form of integrals over the surface of the shape under investigation. It is often a very tedious, not to say painful, process to derive such surface expressions. Along the way, volume formulations in the form of integrals over the entire domain appear as an intermediate step. This thesis couples volume integral formulations of shape derivatives with optimization strategies on shape spaces in order to establish efficient shape algorithms reducing analytical effort and programming work. In this context, a novel shape space is proposed.
• Formoptimierung ist in vielen Anwendungsbereichen von großem Interesse. Insbesondere entstehen Formoptimierungsprobleme in technologischen Prozessen, die mit Hilfe von partiellen Differentialgleichungen (PDEs) modelliert werden. In vielen praktischen Anwendungen ist die zu untersuchende Form durch endlich viele Parameter charakterisiert. Einerseits ermöglicht dies die Anwendung von Standardoptimierungsansätzen, andererseits wird der Raum der erreichbaren Formen unnötig begrenzt. Einen Weg dieses Dilemma zu umgehen liefert das Formenkalkül. Bisher wurden die auf dem Formenkalkül basierenden Formoptimierungprobleme jedoch hauptsächlich mit Gradientenabstiegsverfahren gelöst. Die im Allgemeinen vorhandene Unsymmetrie der zweiten Formableitung beziehungsweise der Form-Hesse-Matrix ist ein Grund hierfür. Generell sind Formenräume keine linearen Räume, wodurch eine große Lücke zwischen Formoptimierung und der herkömmlichen PDE-beschränkten Optimierung entsteht. Diese Lücke gilt es zu schließen. Falls kein linearer Raum in Frage kommt beziehungsweise vorhanden ist, dann fällt die nächst beste Wahl auf eine Riemannsche Mannigfaltigkeit. Möchte man auf Riemannschen Mannigfaltigkeiten optimieren, so muss man mit dem Riemannschen Form-Gradienten beziehungsweise der Form-Hesse-Matrix arbeiten. Die Riemannsche Form-Hesse-Matrix besitzt im Gegensatz zu der herkömmlichen Form-Hesse-Matrix die häufig gewünschte Eigenschaft der Symmetrie. Solch eine Hesse-Matrix gibt Auskunft über die Wohlgestelltheit eines Optimierungsproblems und definiert die hinreichenden Optimalitätsbedingungen. Im Allgemeinen wird sie verwendet, um gradientenbasierte Formoptimierungsverfahren zu beschleunigen. Diese Arbeit befasst sich mit PDE-beschränkten Formoptimierungsproblemen, welche als Optimierungsprobleme auf Riemannschen Mannigfaltigkeiten betrachtet werden, um effiziente Methoden auf Formenräumen für diese bereitzustellen. Es werden Lagrange-Newton, sowie quasi-Newton Ansätze für PDE-beschränkte Probleme in dieser Arbeit formuliert. In diesen Methoden wird die Hadamard-Form der Formableitung verwendet. Diese Hadamard-Form ist ein Oberflächenintegral und oftmals sehr aufwendig und zeitintensiv herzuleiten. Als Zwischenergebnis dieser Herleitung erhält man ein Volumenintegral, welches aus mehreren Gesichtspunkten viel attraktiver ist als ein Oberflächenintegral. Daher beschäftigt sich diese Arbeit zusätzlich mit der Frage, wie man dieses Volumenintegral in den in dieser Arbeit entwickelten Formoptimierungsmethoden verwenden kann, um Effizienz durch Reduzierung analytischer Arbeit und des Programmieraufwandes zu erreichen. In diesem Zusammenhang wird ein neuer Formenraum definiert.
### Additional Services
Author: Kathrin Welker urn:nbn:de:hbz:385-10247 https://doi.org/10.25353/ubtr-xxxx-6575-788c/ Roland Herzog Volker H. Schulz Doctoral Thesis English 2017/01/16 Universität Trier Universität Trier, Fachbereich 4 2016/09/15 2017/01/16 Formenräume; Formoptimierung; PDE BeschränkungenPDE Constraints; Shape Optimization; Shape Spaces Optimierung; Partielle Differentialgleichung Fachbereich 4 / Mathematik 5 Naturwissenschaften und Mathematik / 51 Mathematik / 510 Mathematik 35-XX PARTIAL DIFFERENTIAL EQUATIONS / 35Qxx Equations of mathematical physics and other areas of application [See also 35J05, 35J10, 35K05, 35L05] / 35Q93 PDEs in connection with control and optimization 49-XX CALCULUS OF VARIATIONS AND OPTIMAL CONTROL; OPTIMIZATION [See also 34H05, 34K35, 65Kxx, 90Cxx, 93-XX] / 49Mxx Numerical methods [See also 90Cxx, 65Kxx] / 49M15 Newton-type methods 49-XX CALCULUS OF VARIATIONS AND OPTIMAL CONTROL; OPTIMIZATION [See also 34H05, 34K35, 65Kxx, 90Cxx, 93-XX] / 49Qxx Manifolds [See also 58Exx] / 49Q10 Optimization of shapes other than minimal surfaces [See also 90C90] 57-XX MANIFOLDS AND CELL COMPLEXES (For complex manifolds, see 32Qxx) / 57Nxx Topological manifolds / 57N25 Shapes [See also 54C56, 55P55, 55Q07] 65-XX NUMERICAL ANALYSIS / 65Kxx Mathematical programming, optimization and variational techniques / 65K10 Optimization and variational techniques [See also 49Mxx, 93B40]
\$Rev: 13581 \$ | 1,679 | 6,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.907852 |
https://kclibrary.bibliocommons.com/item/show/2133048120 | 1,542,219,010,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742253.21/warc/CC-MAIN-20181114170648-20181114191808-00006.warc.gz | 656,409,894 | 29,002 | # The Mathematics of Derivatives Securities With Applications in MATLAB
eBook - 2012
Rate this:
Quantitative Finance is expanding rapidly. One of the aspects ofthe recent financial crisis is that, given the complexity offinancial products, the demand for people with high numeracy skillsis likely to grow and this means more recognition will be given toQuantitative Finance in existing and new course structuresworldwide. Evidence has suggested that many holders of complexfinancial securities before the financial crisis did not havein-house experts or rely on a third-party in order to assess therisk exposure of their investments. Therefore, this experienceshows the need for better understanding of risk associate withcomplex financial securities in the future.
The Mathematics of Derivative Securities with Applications inMATLAB provides readers with an introduction to probability theory,stochastic calculus and stochastic processes, followed bydiscussion on the application of that knowledge to solve complexfinancial problems such as pricing and hedging exotic options,pricing American derivatives, pricing and hedging under stochasticvolatility and an introduction to interest rates modelling.
The book begins with an overview of MATLAB and the variouscomponents that will be used alongside it throughout the textbook.Following this, the first part of the book is an in depthintroduction to Probability theory, Stochastic Processes and ItoCalculus and Ito Integral. This is essential to fully understandsome of the mathematical concepts used in the following part of thebook. The second part focuses on financial engineering and guidesthe reader through the fundamental theorem of asset pricing usingthe Black and Scholes Economy and Formula, Options Pricing throughEuropean and American style options, summaries of Exotic Options,Stochastic Volatility Models and Interest rate Modelling. Topicscovered in this part are explained using MATLAB codes showing howthe theoretical models are used practically.
Authored from an academic's perspective, the book discussescomplex analytical issues and intricate financial instruments in away that it is accessible to postgraduate students with or withouta previous background in probability theory and finance. It iswritten to be the ideal primary reference book or a perfectcompanion to other related works. The book uses clear and detailedmathematical explanation accompanied by examples involving realcase scenarios throughout and provides MATLAB codes for a varietyof topics.
Publisher: Chichester, West Sussex, UK ; Hoboken : John Wiley & Sons Inc., 2012.
ISBN: 9780470683699
9781119973409
Characteristics: 1 online resource (xii, 236 pages).
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There are no quotes for this title yet. | 571 | 3,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-47 | latest | en | 0.902678 |
https://fastresearchers.com/calculate-the-correlation-coefficient-between-each-of-the-independent-variables-and-the-variable-diabetes/ | 1,679,602,792,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00676.warc.gz | 288,321,881 | 19,146 | # Calculate the correlation coefficient between each of the independent variables and the variable—diabetes.
Please use the data, StatCrunch, and your knowledge of statistics to answer the questions below.. Download the provided CDC data into StatCrunch (located at the bottom of these directions). You will utilize this data in weeks 3 and 8 for your Healthcare Applications Assignment.
Part I
Calculate the correlation coefficient between each of the independent variables and the variable—diabetes. What does this value tell us about the relationship between each of the independent variables and diabetes?
Run a regression using diabetes as the dependent variable and smoking as the independent variable.
Is there a statistically significant relationship between poverty rates and diabetes? (Use the T-stat or P-value.) Explain.
Write out the regression line calculated using the data.
Interpret b1 (the slope coefficient).
Suppose a state raises the tobacco tax and the rate of smoking in the state falls from 21% to 18%. What would the decrease be in the rate of diabetes based on the regression results?
Part II
Suppose you work for a local hospital. Your manager requests that you use the CDC data on diabetes and smoking, obesity rate, physical activity, and poverty rate to present a compelling argument to the board regarding the importance of creating state initiatives to reduce diabetes rates. You must present statistical data in a written report to the board. Write an APA formatted paper, double spaced, 2-4 pages in length, with a cover page and references page addressing the following:
Include a minimum of four types of statistical tests to make your case. Examples include hypothesis tests, confidence intervals, correlations, regressions, etc. You may use statistical tests that you have completed in weeks 3 and 8 or develop your own.
Include a minimum of three graphs with your report. Examples include scatter plots, histograms, regression lines, box-plots, etc. You may use graphs that you have completed in weeks 3 and 8 or develop your own.
Write an essay using these statistics and graphs to make a convincing argument for the need to create an initiative to reduce the rate of diabetes. In your essay, explain the relationship between diabetes and obesity, smoking, physical activity, and poverty rates utilizing statistical data, tests, and graphs. Be sure to explain your results and how these results show there is a need to create an initiative to reduce the diabetes rates.
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-Timely delivery | 633 | 3,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-14 | longest | en | 0.914369 |
https://coq.inria.fr/doc/v8.13/stdlib/Ltac2.Notations.html | 1,716,694,885,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00658.warc.gz | 162,782,779 | 7,028 | # Library Ltac2.Notations
Constr matching
Ltac2 Notation "lazy_match!" t(tactic(6)) "with" m(constr_matching) "end" :=
Pattern.lazy_match0 t m.
Ltac2 Notation "multi_match!" t(tactic(6)) "with" m(constr_matching) "end" :=
Pattern.multi_match0 t m.
Ltac2 Notation "match!" t(tactic(6)) "with" m(constr_matching) "end" :=
Pattern.one_match0 t m.
Goal matching
Ltac2 Notation "lazy_match!" "goal" "with" m(goal_matching) "end" :=
Pattern.lazy_goal_match0 false m.
Ltac2 Notation "multi_match!" "goal" "with" m(goal_matching) "end" :=
Pattern.multi_goal_match0 false m.
Ltac2 Notation "match!" "goal" "with" m(goal_matching) "end" :=
Pattern.one_goal_match0 false m.
Ltac2 Notation "lazy_match!" "reverse" "goal" "with" m(goal_matching) "end" :=
Pattern.lazy_goal_match0 true m.
Ltac2 Notation "multi_match!" "reverse" "goal" "with" m(goal_matching) "end" :=
Pattern.multi_goal_match0 true m.
Ltac2 Notation "match!" "reverse" "goal" "with" m(goal_matching) "end" :=
Pattern.one_goal_match0 true m.
Tacticals
Ltac2 orelse t f :=
match Control.case t with
| Err e => f e
| Val ans =>
let (x, k) := ans in
Control.plus (fun _ => x) k
end.
Ltac2 ifcatch t s f :=
match Control.case t with
| Err e => f e
| Val ans =>
let (x, k) := ans in
Control.plus (fun _ => s x) (fun e => s (k e))
end.
Ltac2 fail0 (_ : unit) := Control.enter (fun _ => Control.zero (Tactic_failure None)).
Ltac2 Notation fail := fail0 ().
Ltac2 try0 t := Control.enter (fun _ => orelse t (fun _ => ())).
Ltac2 Notation try := try0.
Ltac2 rec repeat0 (t : unit -> unit) :=
Control.enter (fun () =>
ifcatch (fun _ => Control.progress t)
(fun _ => Control.check_interrupt (); repeat0 t) (fun _ => ())).
Ltac2 Notation repeat := repeat0.
Ltac2 dispatch0 t (head, tail) :=
match tail with
| None => Control.enter (fun _ => t (); Control.dispatch head)
| Some tacs =>
let (def, rem) := tacs in
Control.enter (fun _ => t (); Control.extend head def rem)
end.
Ltac2 Notation t(thunk(self)) ">" "[" l(dispatch) "]" : 4 := dispatch0 t l.
Ltac2 do0 n t :=
let rec aux n t := match Int.equal n 0 with
| true => ()
| false => t (); aux (Int.sub n 1) t
end in
aux (n ()) t.
Ltac2 Notation do := do0.
Ltac2 Notation once := Control.once.
Ltac2 progress0 tac := Control.enter (fun _ => Control.progress tac).
Ltac2 Notation progress := progress0.
Ltac2 rec first0 tacs :=
match tacs with
| [] => Control.zero (Tactic_failure None)
| tac :: tacs => Control.enter (fun _ => orelse tac (fun _ => first0 tacs))
end.
Ltac2 Notation "first" "[" tacs(list0(thunk(tactic(6)), "|")) "]" := first0 tacs.
Ltac2 complete tac :=
let ans := tac () in
Control.enter (fun () => Control.zero (Tactic_failure None));
ans.
Ltac2 rec solve0 tacs :=
match tacs with
| [] => Control.zero (Tactic_failure None)
| tac :: tacs =>
Control.enter (fun _ => orelse (fun _ => complete tac) (fun _ => solve0 tacs))
end.
Ltac2 Notation "solve" "[" tacs(list0(thunk(tactic(6)), "|")) "]" := solve0 tacs.
Ltac2 time0 tac := Control.time None tac.
Ltac2 Notation time := time0.
Ltac2 abstract0 tac := Control.abstract None tac.
Ltac2 Notation abstract := abstract0.
Base tactics
Note that we redeclare notations that can be parsed as mere identifiers as abbreviations, so that it allows to parse them as function arguments without having to write them within parentheses.
Enter and check evar resolution
Ltac2 enter_h ev f arg :=
match ev with
| true => Control.enter (fun () => f ev (arg ()))
| false =>
Control.enter (fun () =>
Control.with_holes arg (fun x => f ev x))
end.
Ltac2 intros0 ev p :=
Control.enter (fun () => Std.intros ev p).
Ltac2 Notation "intros" p(intropatterns) := intros0 false p.
Ltac2 Notation intros := intros.
Ltac2 Notation "eintros" p(intropatterns) := intros0 true p.
Ltac2 Notation eintros := eintros.
Ltac2 split0 ev bnd :=
enter_h ev Std.split bnd.
Ltac2 Notation "split" bnd(thunk(with_bindings)) := split0 false bnd.
Ltac2 Notation split := split.
Ltac2 Notation "esplit" bnd(thunk(with_bindings)) := split0 true bnd.
Ltac2 Notation esplit := esplit.
Ltac2 exists0 ev bnds := match bnds with
| [] => split0 ev (fun () => Std.NoBindings)
| _ =>
let rec aux bnds := match bnds with
| [] => ()
| bnd :: bnds => split0 ev bnd; aux bnds
end in
aux bnds
end.
Ltac2 Notation "exists" bnd(list0(thunk(bindings), ",")) := exists0 false bnd.
Ltac2 Notation "eexists" bnd(list0(thunk(bindings), ",")) := exists0 true bnd.
Ltac2 Notation eexists := eexists.
Ltac2 left0 ev bnd := enter_h ev Std.left bnd.
Ltac2 Notation "left" bnd(thunk(with_bindings)) := left0 false bnd.
Ltac2 Notation left := left.
Ltac2 Notation "eleft" bnd(thunk(with_bindings)) := left0 true bnd.
Ltac2 Notation eleft := eleft.
Ltac2 right0 ev bnd := enter_h ev Std.right bnd.
Ltac2 Notation "right" bnd(thunk(with_bindings)) := right0 false bnd.
Ltac2 Notation right := right.
Ltac2 Notation "eright" bnd(thunk(with_bindings)) := right0 true bnd.
Ltac2 Notation eright := eright.
Ltac2 constructor0 ev n bnd :=
enter_h ev (fun ev bnd => Std.constructor_n ev n bnd) bnd.
Ltac2 Notation "constructor" := Control.enter (fun () => Std.constructor false).
Ltac2 Notation constructor := constructor.
Ltac2 Notation "constructor" n(tactic) bnd(thunk(with_bindings)) := constructor0 false n bnd.
Ltac2 Notation "econstructor" := Control.enter (fun () => Std.constructor true).
Ltac2 Notation econstructor := econstructor.
Ltac2 Notation "econstructor" n(tactic) bnd(thunk(with_bindings)) := constructor0 true n bnd.
Ltac2 specialize0 c pat :=
enter_h false (fun _ c => Std.specialize c pat) c.
Ltac2 Notation "specialize" c(thunk(seq(constr, with_bindings))) ipat(opt(seq("as", intropattern))) :=
specialize0 c ipat.
Ltac2 elim0 ev c bnd use :=
let f ev (c, bnd, use) := Std.elim ev (c, bnd) use in
enter_h ev f (fun () => c (), bnd (), use ()).
Ltac2 Notation "elim" c(thunk(constr)) bnd(thunk(with_bindings))
use(thunk(opt(seq("using", constr, with_bindings)))) :=
elim0 false c bnd use.
Ltac2 Notation "eelim" c(thunk(constr)) bnd(thunk(with_bindings))
use(thunk(opt(seq("using", constr, with_bindings)))) :=
elim0 true c bnd use.
Ltac2 apply0 adv ev cb cl :=
Ltac2 Notation "eapply"
cb(list1(thunk(seq(constr, with_bindings)), ","))
cl(opt(seq("in", ident, opt(seq("as", intropattern))))) :=
apply0 true true cb cl.
Ltac2 Notation "apply"
cb(list1(thunk(seq(constr, with_bindings)), ","))
cl(opt(seq("in", ident, opt(seq("as", intropattern))))) :=
apply0 true false cb cl.
Ltac2 default_on_concl cl :=
match cl with
| None => { Std.on_hyps := Some []; Std.on_concl := Std.AllOccurrences }
| Some cl => cl
end.
Ltac2 pose0 ev p :=
enter_h ev (fun ev (na, p) => Std.pose na p) p.
Ltac2 Notation "pose" p(thunk(pose)) :=
pose0 false p.
Ltac2 Notation "epose" p(thunk(pose)) :=
pose0 true p.
Ltac2 Notation "set" p(thunk(pose)) cl(opt(clause)) :=
Std.set false p (default_on_concl cl).
Ltac2 Notation "eset" p(thunk(pose)) cl(opt(clause)) :=
Std.set true p (default_on_concl cl).
Ltac2 assert0 ev ast :=
enter_h ev (fun _ ast => Std.assert ast) ast.
Ltac2 Notation "assert" ast(thunk(assert)) := assert0 false ast.
Ltac2 Notation "eassert" ast(thunk(assert)) := assert0 true ast.
Ltac2 enough_from_assertion(a : Std.assertion) :=
match a with
| Std.AssertType ip_opt term tac_opt => Std.enough term (Some tac_opt) ip_opt
| Std.AssertValue ident constr => Std.pose (Some ident) constr
end.
Ltac2 enough0 ev ast :=
enter_h ev (fun _ ast => enough_from_assertion ast) ast.
Ltac2 Notation "enough" ast(thunk(assert)) := enough0 false ast.
Ltac2 Notation "eenough" ast(thunk(assert)) := enough0 true ast.
Ltac2 default_everywhere cl :=
match cl with
| None => { Std.on_hyps := None; Std.on_concl := Std.AllOccurrences }
| Some cl => cl
end.
Ltac2 Notation "remember"
c(thunk(open_constr))
na(opt(seq("as", ident)))
pat(opt(seq("eqn", ":", intropattern)))
cl(opt(clause)) :=
Std.remember false na c pat (default_everywhere cl).
Ltac2 Notation "eremember"
c(thunk(open_constr))
na(opt(seq("as", ident)))
pat(opt(seq("eqn", ":", intropattern)))
cl(opt(clause)) :=
Std.remember true na c pat (default_everywhere cl).
Ltac2 induction0 ev ic use :=
let f ev use := Std.induction ev ic use in
enter_h ev f use.
Ltac2 Notation "induction"
ic(list1(induction_clause, ","))
use(thunk(opt(seq("using", constr, with_bindings)))) :=
induction0 false ic use.
Ltac2 Notation "einduction"
ic(list1(induction_clause, ","))
use(thunk(opt(seq("using", constr, with_bindings)))) :=
induction0 true ic use.
Ltac2 generalize0 gen :=
enter_h false (fun _ gen => Std.generalize gen) gen.
Ltac2 Notation "generalize"
gen(thunk(list1(seq (open_constr, occurrences, opt(seq("as", ident))), ","))) :=
generalize0 gen.
Ltac2 destruct0 ev ic use :=
let f ev use := Std.destruct ev ic use in
enter_h ev f use.
Ltac2 Notation "destruct"
ic(list1(induction_clause, ","))
use(thunk(opt(seq("using", constr, with_bindings)))) :=
destruct0 false ic use.
Ltac2 Notation "edestruct"
ic(list1(induction_clause, ","))
use(thunk(opt(seq("using", constr, with_bindings)))) :=
destruct0 true ic use.
Ltac2 Notation "simple" "inversion"
arg(destruction_arg)
pat(opt(seq("as", intropattern)))
ids(opt(seq("in", list1(ident)))) :=
Std.inversion Std.SimpleInversion arg pat ids.
Ltac2 Notation "inversion"
arg(destruction_arg)
pat(opt(seq("as", intropattern)))
ids(opt(seq("in", list1(ident)))) :=
Std.inversion Std.FullInversion arg pat ids.
Ltac2 Notation "inversion_clear"
arg(destruction_arg)
pat(opt(seq("as", intropattern)))
ids(opt(seq("in", list1(ident)))) :=
Std.inversion Std.FullInversionClear arg pat ids.
Ltac2 Notation "red" cl(opt(clause)) :=
Std.red (default_on_concl cl).
Ltac2 Notation red := red.
Ltac2 Notation "hnf" cl(opt(clause)) :=
Std.hnf (default_on_concl cl).
Ltac2 Notation hnf := hnf.
Ltac2 Notation "simpl" s(strategy) pl(opt(seq(pattern, occurrences))) cl(opt(clause)) :=
Std.simpl s pl (default_on_concl cl).
Ltac2 Notation simpl := simpl.
Ltac2 Notation "cbv" s(strategy) cl(opt(clause)) :=
Std.cbv s (default_on_concl cl).
Ltac2 Notation cbv := cbv.
Ltac2 Notation "cbn" s(strategy) cl(opt(clause)) :=
Std.cbn s (default_on_concl cl).
Ltac2 Notation cbn := cbn.
Ltac2 Notation "lazy" s(strategy) cl(opt(clause)) :=
Std.lazy s (default_on_concl cl).
Ltac2 Notation lazy := lazy.
Ltac2 Notation "unfold" pl(list1(seq(reference, occurrences), ",")) cl(opt(clause)) :=
Std.unfold pl (default_on_concl cl).
Ltac2 fold0 pl cl :=
let cl := default_on_concl cl in
Control.enter (fun () => Control.with_holes pl (fun pl => Std.fold pl cl)).
Ltac2 Notation "fold" pl(thunk(list1(open_constr))) cl(opt(clause)) :=
fold0 pl cl.
Ltac2 Notation "pattern" pl(list1(seq(constr, occurrences), ",")) cl(opt(clause)) :=
Std.pattern pl (default_on_concl cl).
Ltac2 Notation "vm_compute" pl(opt(seq(pattern, occurrences))) cl(opt(clause)) :=
Std.vm pl (default_on_concl cl).
Ltac2 Notation vm_compute := vm_compute.
Ltac2 Notation "native_compute" pl(opt(seq(pattern, occurrences))) cl(opt(clause)) :=
Std.native pl (default_on_concl cl).
Ltac2 Notation native_compute := native_compute.
Ltac2 Notation "eval" "red" "in" c(constr) :=
Std.eval_red c.
Ltac2 Notation "eval" "hnf" "in" c(constr) :=
Std.eval_hnf c.
Ltac2 Notation "eval" "simpl" s(strategy) pl(opt(seq(pattern, occurrences))) "in" c(constr) :=
Std.eval_simpl s pl c.
Ltac2 Notation "eval" "cbv" s(strategy) "in" c(constr) :=
Std.eval_cbv s c.
Ltac2 Notation "eval" "cbn" s(strategy) "in" c(constr) :=
Std.eval_cbn s c.
Ltac2 Notation "eval" "lazy" s(strategy) "in" c(constr) :=
Std.eval_lazy s c.
Ltac2 Notation "eval" "unfold" pl(list1(seq(reference, occurrences), ",")) "in" c(constr) :=
Std.eval_unfold pl c.
Ltac2 Notation "eval" "fold" pl(thunk(list1(open_constr))) "in" c(constr) :=
Std.eval_fold (pl ()) c.
Ltac2 Notation "eval" "pattern" pl(list1(seq(constr, occurrences), ",")) "in" c(constr) :=
Std.eval_pattern pl c.
Ltac2 Notation "eval" "vm_compute" pl(opt(seq(pattern, occurrences))) "in" c(constr) :=
Std.eval_vm pl c.
Ltac2 Notation "eval" "native_compute" pl(opt(seq(pattern, occurrences))) "in" c(constr) :=
Std.eval_native pl c.
Ltac2 change0 p cl :=
let (pat, c) := p in
Std.change pat c (default_on_concl cl).
Ltac2 Notation "change" c(conversion) cl(opt(clause)) := change0 c cl.
Ltac2 rewrite0 ev rw cl tac :=
let cl := default_on_concl cl in
Std.rewrite ev rw cl tac.
Ltac2 Notation "rewrite"
rw(list1(rewriting, ","))
cl(opt(clause))
tac(opt(seq("by", thunk(tactic)))) :=
rewrite0 false rw cl tac.
Ltac2 Notation "erewrite"
rw(list1(rewriting, ","))
cl(opt(clause))
tac(opt(seq("by", thunk(tactic)))) :=
rewrite0 true rw cl tac.
coretactics
Ltac2 exact0 ev c :=
Control.enter (fun _ =>
match ev with
| true =>
let c := c () in
Control.refine (fun _ => c)
| false =>
Control.with_holes c (fun c => Control.refine (fun _ => c))
end
).
Ltac2 Notation "exact" c(thunk(open_constr)) := exact0 false c.
Ltac2 Notation "eexact" c(thunk(open_constr)) := exact0 true c.
Ltac2 Notation "intro" id(opt(ident)) mv(opt(move_location)) := Std.intro id mv.
Ltac2 Notation intro := intro.
Ltac2 Notation "move" id(ident) mv(move_location) := Std.move id mv.
Ltac2 Notation reflexivity := Std.reflexivity ().
Ltac2 symmetry0 cl :=
Std.symmetry (default_on_concl cl).
Ltac2 Notation "symmetry" cl(opt(clause)) := symmetry0 cl.
Ltac2 Notation symmetry := symmetry.
Ltac2 Notation "revert" ids(list1(ident)) := Std.revert ids.
Ltac2 Notation assumption := Std.assumption ().
Ltac2 Notation etransitivity := Std.etransitivity ().
Ltac2 clear0 ids := match ids with
| [] => Std.keep []
| _ => Std.clear ids
end.
Ltac2 Notation "clear" ids(list0(ident)) := clear0 ids.
Ltac2 Notation "clear" "-" ids(list1(ident)) := Std.keep ids.
Ltac2 Notation clear := clear.
Ltac2 Notation refine := Control.refine.
extratactics
Ltac2 absurd0 c := Control.enter (fun _ => Std.absurd (c ())).
Ltac2 Notation "absurd" c(thunk(open_constr)) := absurd0 c.
Ltac2 subst0 ids := match ids with
| [] => Std.subst_all ()
| _ => Std.subst ids
end.
Ltac2 Notation "subst" ids(list0(ident)) := subst0 ids.
Ltac2 Notation subst := subst.
Ltac2 Notation "discriminate" arg(opt(destruction_arg)) :=
Std.discriminate false arg.
Ltac2 Notation discriminate := discriminate.
Ltac2 Notation "ediscriminate" arg(opt(destruction_arg)) :=
Std.discriminate true arg.
Ltac2 Notation ediscriminate := ediscriminate.
Ltac2 Notation "injection" arg(opt(destruction_arg)) ipat(opt(seq("as", intropatterns))):=
Std.injection false ipat arg.
Ltac2 Notation "einjection" arg(opt(destruction_arg)) ipat(opt(seq("as", intropatterns))):=
Std.injection true ipat arg.
Auto
Ltac2 default_db dbs := match dbs with
| None => Some []
| Some dbs =>
match dbs with
| None => None
| Some l => Some l
end
end.
Ltac2 default_list use := match use with
| None => []
| Some use => use
end.
Ltac2 trivial0 use dbs :=
let dbs := default_db dbs in
let use := default_list use in
Std.trivial Std.Off use dbs.
Ltac2 Notation "trivial"
use(opt(seq("using", list1(thunk(constr), ","))))
dbs(opt(seq("with", hintdb))) := trivial0 use dbs.
Ltac2 Notation trivial := trivial.
Ltac2 auto0 n use dbs :=
let dbs := default_db dbs in
let use := default_list use in
Std.auto Std.Off n use dbs.
Ltac2 Notation "auto" n(opt(tactic(0)))
use(opt(seq("using", list1(thunk(constr), ","))))
dbs(opt(seq("with", hintdb))) := auto0 n use dbs.
Ltac2 Notation auto := auto.
Ltac2 new_eauto0 n use dbs :=
let dbs := default_db dbs in
let use := default_list use in
Std.new_auto Std.Off n use dbs.
Ltac2 Notation "new" "auto" n(opt(tactic(0)))
use(opt(seq("using", list1(thunk(constr), ","))))
dbs(opt(seq("with", hintdb))) := new_eauto0 n use dbs.
Ltac2 eauto0 n p use dbs :=
let dbs := default_db dbs in
let use := default_list use in
Std.eauto Std.Off n p use dbs.
Ltac2 Notation "eauto" n(opt(tactic(0))) p(opt(tactic(0)))
use(opt(seq("using", list1(thunk(constr), ","))))
dbs(opt(seq("with", hintdb))) := eauto0 n p use dbs.
Ltac2 Notation eauto := eauto.
Ltac2 Notation "typeclasses_eauto" n(opt(tactic(0)))
dbs(opt(seq("with", list1(ident)))) := Std.typeclasses_eauto None n dbs.
Ltac2 Notation "typeclasses_eauto" "bfs" n(opt(tactic(0)))
dbs(opt(seq("with", list1(ident)))) := Std.typeclasses_eauto (Some Std.BFS) n dbs.
Ltac2 Notation typeclasses_eauto := typeclasses_eauto.
Congruence
Ltac2 f_equal0 () := ltac1:(f_equal).
Ltac2 Notation f_equal := f_equal0 ().
now
Ltac2 now0 t := t (); ltac1:(easy).
Ltac2 Notation "now" t(thunk(self)) := now0 t. | 5,063 | 16,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | latest | en | 0.52344 |
https://www.digglicious.com/tips-for-writing/how-do-you-solve-logarithmic-functions-step-by-step/ | 1,675,607,945,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00283.warc.gz | 753,260,085 | 9,669 | ## How do you solve logarithmic functions step by step?
Solving Logarithmic Equations
1. Step 1: Use the rules of exponents to isolate a logarithmic expression (with the same base) on both sides of the equation.
2. Step 2: Set the arguments equal to each other.
3. Step 3: Solve the resulting equation.
5. Solve.
Type the number you’re working with into your graphing or scientific calculator. For example, type “1000.” Press the “Log” button on your calculator. The number you immediately see is the exponent for the original number you entered.
How do you enter logarithms into a calculator?
To do this using most simple scientific calculators, enter the number, press the inverse (inv) or shift button, then. press the log (or ln) button.
### How do you do log2 on a calculator?
Filed under Difficulty: Easy, TI-83 Plus, TI-84 Plus, TI-89, TI-92 Plus. For example, to evaluate the logarithm base 2 of 8, enter ln(8)/ln(2) into your calculator and press ENTER.
How do you do log base E on a calculator?
The power to which the base e (e = 2.718281828…….)…To do this using most simple scientific calculators,
1. enter the number,
2. press the inverse (inv) or shift button, then.
3. press the log (or ln) button. It might also be labeled the 10x (or ex) button.
## How to solve logarithmic function?
y = log b x. Then the logarithmic function is given by; f (x) = log b x = y, where b is the base, y is the exponent, and x is the argument. The function f (x) = log b x is read as “log base b of x.”. Logarithms are useful in mathematics because they enable us to perform calculations with very large numbers.
## How to type logarithms into a calculator?
Your calculator may have simply a ln (or log (button, but for this formula you only need one of these: For example, to evaluate the logarithm base 2 of 8, enter ln (8)/ln (2) into your calculator and press ENTER. You should get 3 as your answer. Try it for yourself!
How do calculators calculate logarithms?
,therefore
• ,therefore
• (between and )
• ### How to calculate logarithms manually?
ln xy= y ln x. log = log x1/y= (1/y )log x. ln = ln x1/y=(1/y)ln x. Example 9:log 5.0 x 106= log 5.0 + log 106= 0.70 + 6 = 6.70. Hint: This is an easy way to estimate the log of a number in scientific notation! Example 10: log (154/25) = log 154 – log 25 = 2.188 – 1.40 = 0.788 = 0.79 (2 sig. fig.) | 660 | 2,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-06 | latest | en | 0.840032 |
https://www.theunitconverter.com/nautical-miles-to-kilometers-conversion/60-nautical-miles-to-kilometers.html | 1,623,921,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629632.54/warc/CC-MAIN-20210617072023-20210617102023-00188.warc.gz | 893,021,847 | 16,226 | # Nautical Miles to Kilometers Conversion
## 60 Nautical Miles to Kilometers Conversion - Convert 60 Nautical Miles to Kilometers (Nm to km)
### You are currently converting Distance and Length units from Nautical Miles to Kilometers
60 Nautical Miles (Nm)
=
111.12 Kilometers (km)
Nautical Miles : The nautical mile (symbol M, NM or nmi) is a unit of length, defined as 1,852 meters (approximately 6,076 feet). It is a non-SI unit used especially by navigators in the shipping and aviation industries, and also in polar exploration.
Kilometers : The kilometer (SI symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
### Length conversion table
millimeter (mm) centimeter (cm) meter (m) kilometer (km) inch (in) foot / feet (ft) yard (yd) mile (mi)
1 millimeter (mm) 1 0.1 0.001 0.000001 0.03937 0.003281 0.0010936 0.0000006214
1 centimeter (cm) 10 1 0.01 0.00001 0.3937 0.03281 0.010936 0.000006214
1 meter (m) 1000 100 1 0.001 39.37 3.281 1.0936 0.0006214
1 kilometer (km) 1000000 100000 1000 1 39370 3281 1093.6 0.6214
1 inch (in) 25.4 2.54 0.0254 0.0000254 1 0.08333 0.02778 0.000015783
1 foot / feet (ft) 304.8 30.48 0.3048 0.0003048 12 1 0.33333 0.0001894
1 yard (yd) 914.4 91.44 0.9144 0.0009144 36 3 1 0.0005682
1 mile (mi) 1609344 160934 1609.3 1.6093 63360 5280 1760 1
1 nautical mile (nmi) 1852000 185200 1852 1.852 72913 6076 2025.4 1.1508
We decided to round some conversion factors to fit this table. Therefore, some of these values are not accurate, but they still have reasonable accuracy. | 613 | 1,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-25 | latest | en | 0.700447 |
https://stats.stackexchange.com/questions/tagged/cointegration | 1,657,159,873,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00119.warc.gz | 570,547,677 | 73,288 | # Questions tagged [cointegration]
Two or more non-stationary, integrated variables are cointegrated if there exists a linear combination of those variables which is integrated of a lower order, e.g. stationary.
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### R: Johansen test for two variables
I am trying to replicate work from this paper, specifically examination of the rationality of inflation and inflation expectations when both series are non-stationary I(1). I need to apply Johansen ...
5 views
### Panel Cointegration when some individual series in the panels are stationary
I'm studying a few panels which are comprised of series, of which some seem to be stationary. This conclusion was reached after using the CIPS test (Pesaran, 2007) as well as ADF and PP Fisher-type ...
• 21
18 views
### Significance of parameter on cointegrating vector
I have been reading Section 6.2 (page 96) of this manual, where a procedure called "fully modified least squares" is discussed, and was developed on a paper of Phillips and Hansen (1990) . ...
• 101
13 views
### How to prove that cointegration tests between two time series are invalid in the presence of structural breaks in both time series?
How to prove that cointegration tests on $two$ time series are invalid in the presence of structural breaks in both time series? Wouldn't the presence of a structural break in all the time series ...
• 285
1 vote
8 views
### Testing cointegration in a VECM with autocorrelation?
I have four I(1) (monthly) variables. AIC is used to determine the number of lags. A VECM is estimated by ML-method and the Eigevalues imply that there are two cointegrating vectors. Now the problem ...
31 views
### Non-stationary time series data and OLS regression (with controls)
I am using 2000-2020 quarterly data and want to test if an appreciation of the real exchange rate leads to a decrease in manufacturing output of a country. However, my data seems to be non-stationary ...
15 views
### Constant in Cointegrating Relationship
If I had a cointegrating relationship $c_t = \beta_0 + y_t$, is there a difference between the interpretation of $\beta_0$ and the interpretation of the constant estimated in a simple static ...
1 vote
27 views
### How powerful is cointegration test?
I am performing Engle-Granger cointegration test on my data as below ...
• 65
1 vote
26 views
### Johansen procedure shows cointegration r=1, but ect is not significant?
I have 6 variables, all of them I(1). I tested for cointegration and got a significant result for r=1, so I decided to estimate a VECM. The problem is now that the ECTs of the VECM are not significant....
1 vote
16 views
### Eigenvalues of Johansen Trace Test
I'm currently taking a course in time series and have been struggling with understanding the Johansen trace test. Specifically, the calculation of the eigenvalues for the Likelihood ratio statistic. ...
17 views
### Update cointegration vector
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### Johansen Cointegration Test at Levels or First Differences in a VAR Model
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### Simplified Version of the Error Correction Model
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### Interpretation of Cointegration Test Results
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### What is the coefficient that shows whether there is a positive or negative relationship between variables in vector error correction models?
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### Analyzing relationships in 5 time series (cointegration)
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### Why does the prediction of a VAR dgp diverge from the test set?
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### Estimating cointegrating equations in VEC models
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### Cointegration with Multiple Structural Break in R
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### Panel Data Short-Run & Long-Run estimators
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### Is a random walk cointegrated with its own lag?
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### Can a linear combination of an I(2) and I(1) variable be I(1)?
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[–] 0 points1 point (6 children)
Why are the PLL moves named the way they are? Why is it “F” perm and “Ab”? (I figured perm = permutation, but I was unsure of the lettering)
[–]It should not hurt if you relax and use lube 0 points1 point (0 children)
By the way, you should check out Parity PLLs - those only possible on 4x4, Square-1, and cuboids - or if you messed up solving cross/F2L.
https://speedcubedb.com/a/SQ1/SQ1LinParityPLL
They have O (all 4 edges move by 1 step in a circle), W (2 diag corners swapped), X, Q, S and other cool letters.
[–] 1 point2 points (0 children)
its what the arrows in the corner-edge cycle look like
[–]It should not hurt if you relax and use lube 1 point2 points (0 children)
I think it's just the general shape of the swaps.
It is obvoius and easy logic for most of them.
A = triangle of corners, T = t-shape, J = j-shape, N = diagonal, V = 2 blocks pointing at each other. Y = spear shape, Z = zigzag of edges. G = weird freaking shape.
And the rest are using the most similar letter from the leftovers.
Can't call (+ swap) a T-perm, let's call it H instead because the sides are solved.
Two lines near each other? F. Two lines separated? E.
A line and a dot? R - line on the bottom left, dot on the bottom right.
And sure, we have a lot of letters and could've chosen some different ones, but that's a good thing - it uses only 13 letters and each of them is very memorable. And mirrors are separated by adding a/b letters.
On the other side, OLLs are really messed up because at best, cases that look similar, have numbers in a row like 3 and 4.
But for some of them having the same shape on top doesn't mean they're even solved the same way. But not always! T-shapes are 33 and 45, edge OLL are 28 and 57, line shapes are 51-52 55-56 with two easy corner-shapes tucked in the middle between them.
But it's legacy, so people keep just checking the number of the PLL they're looking for, saying that number, and then people look in the table what shape that number corresponds to.
That's why they're sometimes described as H-shape (literally H shape), H-swap (2 headlights), Pi-OLL (headlights and sidelights) and so on. Really confusing.
[–]Sub-41 4x4 Yau(CN) 28.58 PB single, 35.12 PB Ao5 0 points1 point (2 children)
I’m guessing it’s something to do with how the pieces move? (Which is probably false cuz H perm)
[–]Sub-16 (Roux) 0 points1 point (1 child)
Well there not going to name the opposite edge swap egg sperm (x perm)
[–]Sub-41 4x4 Yau(CN) 28.58 PB single, 35.12 PB Ao5 0 points1 point (0 children)
Well, yeah, but it doesn’t explain why they chose H instead
Maybe it’s because M and Z uses the M layers, so the R and L layers are like the parallel lines in H and Z. H is more commonly used than Z, so they labelled H perm as H perm and Z perm as Z perm cuz H perm is better than Z perm
[–] 4 points5 points (1 child)
did a super long 4x4 session today and broke all my PBs, from single(36.04) to ao100(49.98)!
[–] 1 point2 points (0 children)
me crying with much slower average and single...
[–]Sub-40 (Roux), PB: 25 1 point2 points (0 children)
Phoenix Patterson now #2 in the world for 2x2 average.
[–] 1 point2 points (10 children)
For those in Canada that ordered from Cubezz, did you have to pay taxes and duties when it came at your door or there was no charge?
[–]2012ONGR01 Sub-8 2 points3 points (0 children)
Never got taxes and duties. My orders have been less than 100 USD over the years.
[–] 0 points1 point (3 children)
whats cubezz
[–] 0 points1 point (0 children)
One of the China cube shops in the wiki list of known-good cube shops, Where to buy cubes.
Very good prices on cubes and free worldwide shipping.
The other 2 popular China shops, Picubeshop and Ziicube, have better prices but no free shipping.
[–] 0 points1 point (1 child)
A chinese site for cubes. I normally prefer sites like speedcubeshop or others like that, cause the service is impeccable, but for the clock, its really cheap on cube zz so i wanted to try
[–] 0 points1 point (0 children)
k thanks
[–]ao12: 1:53 (Roux) 1 point2 points (1 child)
I didn’t have any taxes or anything when I ordered from Cubezz last.
[–] 0 points1 point (0 children)
Thanks!
[–]pineapple on pizza is good -1 points0 points (2 children)
i use cancube and cubing out loud so no idea
[–] 0 points1 point (1 child)
I normally use these sites too as well as speedcube shop since their lubricants fit me perfect, but clock is half the price so I was tempted to try it out!
[–]pineapple on pizza is good 0 points1 point (0 children)
i see i see
[–] 0 points1 point (2 children)
If I take my cube apart and put it back together in a solved position is there any way it can be unsolvable?
[–]pineapple on pizza is good 8 points9 points (0 children)
no, because it's solved lmao
[–]Sub-16 (Roux) 2 points3 points (0 children)
No if it is solved and then scrambled it will still be solvable
[–]Sub-15 (CFOP) 2 points3 points (2 children)
Just made skewb & 3x3 finals for my first time! Also got 5 PRs! Very happy!
<image>
[–]Sub-16 (Roux) 0 points1 point (1 child)
I surprised you actually made it to skewb finals. At my last comp a 6 average didn’t even guarantee a final spot
[–]Sub-15 (CFOP) 0 points1 point (0 children)
It was only top 12 so I barely made it😬 it wasn't the most competitive comp
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 2 points3 points (0 children)
I've been out of town for over a week and solving on my old main, the WRM 21 with springs. After a few days I was surprised both by how regularly I was able to get sub 20 solves (at least one every day) and by quickly it got gunked up, how much slower it is than my 3nado pioneer, and how much more force every turn took. It was much more tiring to solve than I'm used to, and I had to do less solves and take more breaks. At the start of the trip I was averaging 2s slower than my regular average, and by the end of the trip my Ao100 was back to normal, 25s.
Got home last night and immediately got hit with a stomach bug, so I sat on a couch and did a lot of solves on the 3nado today. Had a bit of adjustment to do then I beat my PB Ao50 6 times, and my PB Ao12 once!
New PB's:
Ao12 - 23.68 --> 23.20 Ao50 - 24.93 --> 24.08
[–] 10 points11 points (13 children)
YALL I FINALLY GOT 19S PB!!!!! THAT F2L WAS SO EASY AND SEXY
[–]Sub-16 (Roux) 3 points4 points (12 children)
Nice! I see a sub 19 solve in your future.
[–] 0 points1 point (11 children)
I hope so! After learning full PLL and OLL, I'll definitely be able to do that
[–]It should not hurt if you relax and use lube 1 point2 points (10 children)
Disclaimer. You don't need full OLL.
You can learn it if you like, after you're done with full PLL, but it's recommended that you get better at planning full cross, F2L, 2-look OLL, full PLL, and just get really good at recognition and high-quality execution (good finger tricks, so miminal regrips/pauses, etc).
if you want to learn them, that's pretty easy - all of the algs are mostly just sexy move, sledge, sune, and such. But because they're so short and easy, learning full OLL also makes less of a difference.
I am slow, but still I only got from 30 second averages to 20 second average, by just doing a lot of solves. OLL doesn't make you suddenly fast. It's more of a high-tier improvement.
[–] 0 points1 point (7 children)
I should've went with the ROUX route since my turning isn't the fastest but it's too late.
[–]Sub-16 (Roux) 0 points1 point (3 children)
It’s never too late to change methods. I’ve changed so many times now. I should probably stop though
[–] 0 points1 point (2 children)
Na I prefer CFOP in general.
[–]Sub-16 (Roux) 0 points1 point (1 child)
Remember can always switch to nautilus
[–] -1 points0 points (0 children)
u should search up how to do petrus lol
[–]one day, i’ll need to buy ketchup 0 points1 point (2 children)
It’s absolutely not too late! I was ~55 seconds with CFOP when I made the switch, and now I’m ~35. Sean Villanueva was sub 15 with CFOP, and is now sub-8 with Roux. Try it out. Kian’s tutorials are the best for beginners.
[–] 0 points1 point (0 children)
Na I like CFOP more too so yeah. My hands hurt by too much M moves. I prefer Algorithms even tho I'm not the fastest at some algs.
[–] 1 point2 points (0 children)
My turning is not the fastest so I think its better for me to actually learn full OLL too. I'm only doing it because it's fun not just for the sake of getting less time. Learning new algorithm is actually quite fun.
[–]17 average (CFOP 3.5LLL) PB 11.05 0 points1 point (0 children)
lol you barely even need full PLL I'm averaging mid 16s without it (although I admit that this is a terrible idea for actually improving and I'm just stubborn)
[–]Sub 25(CFOP) 2LOLL 3/4 PLL learned 1 point2 points (4 children)
i currently have a QiYi MS 2x2. I have a comp in a few months and im looking to see if i need to upgrade anything. would it be a good idea to upgrade to an RS2M? i avg 6.3s
[–] 0 points1 point (2 children)
Make sure it's the newer RS2 M Evolution, and not the older RS2 M (discontinued, but still in stock in some shops).
[–]Sub 25(CFOP) 2LOLL 3/4 PLL learned 0 points1 point (1 child)
yes the evolution didnt know they discontinued the previous iteration
[–] 0 points1 point (0 children)
Some sites, like Ziicube and Cubezz, call it the RS2 M V2 and say nothing about Evolution.
Others, like SpeedCubeShop, Cubicle, Picubeshop, call it the RS2 M Evolution and say nothing about V2.
Same cube, tho.
[–]Sub-16 (Roux) 0 points1 point (0 children)
As I said before cube doesn't matter too much. If you want to upgrade to a cube that you think suits you better than go for it. I can get my usual times on some random 2x2 from amazon.
[–]Sub 25(CFOP) 2LOLL 3/4 PLL learned 1 point2 points (13 children)
registering for a comp soon. since its a few months out(also im waiting for registration to open), i have plenty of time to practice/learn new events. this particular one is 2x2, 3x3, 4x4, OH, pyraminx, clock, and skewb. so i dont have a pyraminx or sqewb and im looking for some solid, budget friendly puzzles. what are some recommendations?
side note: at my first comp last month i noticed everyone had super expensive cubes. a few people i talked to told me not to waste money on expensive 3x3s. i currently have a RS3M 2020 and i love it. i dont think its necessary to upgrade but should i?
[–]Sub-16 (Roux) 1 point2 points (7 children)
About upgrading your cube, do what suits you. I was at a comp last week and I had the cheapest cube or one of the cheapest. Everyone around me had Gans, Tornado v3's Weilong wr Maglevs. I had a Meilong 3M. I can attest that cube doesn't matter and just use what you like.
And some pyraminx's and skewbs
for pyraminx, the most recomended one nowadays is the weilong pyraminx, used to be the yuxin little magic before its quality control went down. A qiyi ms, will still be good for a beginner
And skewb. The moyu rs skewb is a good skewb for a good price.
[–]Sub 25(CFOP) 2LOLL 3/4 PLL learned 0 points1 point (6 children)
is there a cheaper pyraminx? i was looking for something around 10\$
[–]2H: sub 16 OH: sub 20 (Roux) 0 points1 point (1 child)
The RS pyraminx is also very nice for its price.
[–]Sub 25(CFOP) 2LOLL 3/4 PLL learned 0 points1 point (0 children)
thats the price range i was looking at. thanks
[–]Sub-16 (Roux) 0 points1 point (3 children)
a yuxin one is still good but quality control went down and a qiyi ms is also good as I said previously
[–]Sub 25(CFOP) 2LOLL 3/4 PLL learned 0 points1 point (2 children)
yeah qiyi makes good stuff for their budget series. ive heard people say that there is better budget options now so they arent as good as some others.
[–]Sub-16 (Roux) 0 points1 point (1 child)
That’s true for all the nxn’s. But the 2x2 and the pyraminx are better compared to their counterparts
[–]Sub 25(CFOP) 2LOLL 3/4 PLL learned 0 points1 point (0 children)
yeah their 2x2 is pretty good. i might go for the rs pyraminx but ill consider the qiyi
[–]Sub-37s - PB: 25.05 - (Roux x2y CN) - PB Ao5: 31.78 1 point2 points (4 children)
Moyu RS for Skewb is current top budget option. Same for Pyraminx (not maglev).
Tornado V3 M is current top cube independent of price. But it is very different from RS3M. If you abuse corner cutting, then be aware nothing corner cuts like RS3M.
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 0 points1 point (0 children)
My tornado V3 corner cuts more than any RS3M I have, but doesn't feel like it's built to be abused in the same way the RS3M does. It's sturdy, but the RS3M feels like you can abuse it, but feels worse in every other way IMO
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (2 children)
Eh, I’d argue. Any WRM or Tornado v3 cut just as much. Tengyun v1 and v2 too.
[–]Sub-37s - PB: 25.05 - (Roux x2y CN) - PB Ao5: 31.78 0 points1 point (1 child)
To me they are all great in their different ways. But tornado v3 was used for the WR as well. Tie with GAN. So meant as a way to say GAN is not necessarily better.
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (0 children)
I'm answering to "nothing corner cuts like RS3M". This is not true.
[–]Sub-18 (CFOP) PB:11.71 0 points1 point (0 children)
After having the ys3m for awhile now, I went back to the super bc. I think I prefer the super bc over the ys3m, something about it just does it for me!
[–] 0 points1 point (5 children)
What can you do if you ever strip a tension adjustment screw? Is there anything you can do other than buying a new cube? Hasn't happened to me yet but I'm always stressing out over it happening whenever I adjust the tension.
[–] 0 points1 point (3 children)
Strip the phillips head of the screw, or strip the threads in the core?
It's a metal screw going into plastic, so the core getting stripped can happen.
SpeedCubeReview has a simple fix for stripped cores - https://www.youtube.com/watch?v=SvaAFopZV4Q
[–] 0 points1 point (2 children)
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 0 points1 point (0 children)
You can use a thick rubber band placed between the head of the screw and the tip of the screwdriver if it strips slightly. I don't use the screwdrivers included with cubes to adjust my tensions, I use a well-fit full size screwdriver where it's easier to apply some downward pressure and torque, this helps prevent stripping the screw head too
[–] 0 points1 point (0 children)
For that, you just have to make sure you use a screwdriver that fits the screw.
[–]Sub-16 (Roux) 0 points1 point (0 children)
So you stripped the screw head I'm assuming. I used a Dremel and carved out a space for a flat head screwdriver and used that on my mgc square-1.
[–]Sub-12 PB-7.52 (CFOP) 2 points3 points (1 child)
New fullstep PB:
``````L U L' F L' F2 B' R2 U' F2 B' U2 B D2 B' L2 F R2
x2 y // inspection
F R' F' D U' R' F R // cross
U' R U' R' // 1st pair
U2 L' U' L // 2nd pair
y U' R U R' U2 R U' R' U R U' R' // 3rd pair
y U2 R' F R F' R U' R' // 4th pair
U R' U' F U R U' R' F' R // OLL
U' R U R' U' R' F R2 U' R' U' R U R' F' // PLL
61STM / 7.52sec =8.11TPS
``````
[–]all rounder | 2019rego03 0 points1 point (0 children)
Damn 8tps is pretty fast, nice
[–] 0 points1 point (2 children)
I've solved 3x3 (working towards learning CFOP, I have the C and F down lmao) and 2x2 I was able to figure out on my own. I decided to get a Megaminx as my next puzzle, maybe it was too much of a jump in difficulty. I don't know whether solving it intuitively is even remotely realistic but if so, can anyone give me some small hints to point me in the right direction?
[–]It should not hurt if you relax and use lube 2 points3 points (0 children)
Don't worry. Megaminx is basically a long 3x3. Unless you get seriously into it and start learning more and more algs - which is also what happens with 3x3, so you should be fine.
Except Megaminx 4-look is already pretty long, so you might want to start with even smaller alg amount.
First 2 Layers and Second 2 layers are similar to 3x3 or, after some practice, even easier because of the extra freedom you have.
Last layer edge orientation is literally the same as 3x3 - same alg to flip same 2 edges.
If you do beginner methods, then corners are basically identical too.
If you're learning orient and permute algs, there's more of those than on 3x3, but a lot of them are similar (just do different things because of the 5-corner shape instead of 4)
I personally for now just do the last layer exactly how beginner method called 8355 does it for 3x3 - where you just do Sune (RUR'U RU2R') to swap 3 edges, then a bunch of (RUR'U') sexy moves to move corners around and orient them.
[–] 5 points6 points (0 children)
Starting to do some solves again on 3x3 after 2 years yesterday, and I'm averaging around 22 seconds. I used to average around 17 secs before I stopped, but 22 is still good for barely any derusting
[–]Sub-13 (CFOP, dot OLL’s suck) 2 points3 points (2 children)
I know lookahead is something you can’t really learn from tips, but anyone got tips for lookahead on big cubes or megaminx? I could reduce a lot of time on mega solves by not pausing to look for pieces
[–]A cuber is secretly a screwdriver collector 1 point2 points (0 children)
Don't turn slow on big cubes. Turn as fast as you can; it doesn't really effect ease of lookahead. Try put what your solving in a place so that you can see the front faces, know where partially solved parts are and their orientation, so you don't need to look at it again to add pieces to it, and don't be afraid to rotate if you need to
[–]Sub-16 (Roux) 1 point2 points (0 children)
go ahead and rotate while you are turning
[–]blindfolded solving is where the fun begins 8 points9 points (2 children)
[–]Sub-11 (CFOP) 0 points1 point (0 children)
did I just see virtual Rubik's Magic or am I interpreting something incorrectly
[–]pineapple on pizza is good 1 point2 points (0 children)
damn nice
[–] 3 points4 points (3 children)
It started as a joke, I customize the colour scheme of my cube. Red across green, blue and yellow and white with black. I've been solving for maybe halve an hour and I don't know why but it feels so intuïtief. I might stick with this for a while.
[–]It should not hurt if you relax and use lube 2 points3 points (1 child)
We need more cubes with different color schemes to stop posts about incorrect color schemes.
[–] 0 points1 point (0 children)
Definitely. With this option, I don't even mind to pay 90 euros for a Gan Cube, haha.
[–] 2 points3 points (0 children)
I wish I could see scramblers reaction to this cube
[–] 1 point2 points (2 children)
I currently own the rs3m and I decided to try some stuff I had at home to lube the cube. Turns out it didn't turn well. So I tried to clean the lube of using disinfected wipes and just regalur wet paper towels yet the gummy and slow fell remaind. Is there a better way to clean the cube and if I just by a regular cube lube and put it in, will it solve the problem?
[–] 0 points1 point (0 children)
I had an old cyclone boys which i lubed with machine oil, and to clean it i took apart, leaving the core and screws aside. Then put every plastic piece in soapy water overnight then left to dry next day and it was good again after some real silicone lube. But i dont think this is safe for the magnets on any modern cube so maybe use only warm water as by the other reply.
[–]Sub-16 (Roux) 1 point2 points (0 children)
What did you lube the cube with? Some substances may be harmful to the cube like petroleum. The disenfecting wipes may have also caused some damage depending on the brand. use some warm water and towels. If it doesn't work then your cube has likely been damaged in a way. An actual cube lube may work to lessen the issue.
[–]PB: 6.77sec basically sub-10 CFOP 6 points7 points (1 child)
I got new pb! Finally that lucky solve I’ve been waiting for. And I rolled it and got pb Ao5: 8.11 1. 8,19 2. (6,77) 3. (9,87) 4. 7,60 5. 8,53 Here is pb reconstruction:
6.77 B D2 B2 R' D2 F2 R B2 L2 R D2 U2 F R2 D' L' U' L F' U2
Cross// x2 R2 F2 D R’ (4)
1st pair// R U’ R’ U’ L’ U’ L (7)
2nd pair// U R U’ R’ (4)
3rd pair// y U R’ U’ R (4)
4th pair// U’ R’ U2 R2 U R2 U R (8)
OLL// S’ U2 R U R’ U R U2 R’ S (10) (small lightning bolt)
PLL// skip
37:6,77= 5,5 tps
[–] 0 points1 point (0 children)
interesting OLL alg
[–]all rounder | 2019rego03 12 points13 points (2 children)
I dont think I've ever had a 5x5 avarage this consistent lol, pb 1:01.08 ao5
1. 1:00.61
2. (1:01.67)
3. (1:00.24)
4. 1:01.09
5. 1:01.53
[–]Sub-41 4x4 Yau(CN) 28.58 PB single, 35.12 PB Ao5 3 points4 points (1 child)
The next ao5 shall be a sub-1min
[–]all rounder | 2019rego03 2 points3 points (0 children)
I hope so :')
[–]17 average (CFOP 3.5LLL) PB 11.05 2 points3 points (2 children)
Just hit 2k solves on my 2x2 session, without beating my 1.61 in comp at all haha, do have a 1.23 from a previous session that I lost though
[–] 2 points3 points (1 child)
Backup/export those sessions so you don't lose them.
[–]17 average (CFOP 3.5LLL) PB 11.05 0 points1 point (0 children)
Yeh I’ve been doing that since
[–]Sub-16 (CFOP 2LLL CN) 9 points10 points (6 children)
For anyone who thinks they are plateauing because they haven't improved much over like 5000 solves. I just checked some crazy big averages in my csTimer session (where I do virtually all my timed solves since December 2019). I am sub-16 ao20000. 20 freaking thousand solves. And it's been like 35k solves since I first got sub-16 ao1000.
https://imgur.com/a/9lvPWXq
u/DiDiHD
[–]Sub-20 (CFOP) 4 points5 points (4 children)
Thanks for sharing! Tbh, seeing the number of solves you do, made me realize I have nothing to complain about. You're definitely one a huge motivation for me and I admire the amount of work you put in
[–]Sub-16 (CFOP 2LLL CN) 3 points4 points (3 children)
Well, I enjoy cubing. It's not like I'll reach sub-15 and will suddenly be free or fulfilled or anything. I hate when I suck mostly because the solves are less enjoyable, like my mind feels clouded.
[–]Sub-20 (CFOP) 0 points1 point (2 children)
Oh totally understand what you mean. It's just a milestone and I can finally try going after other hobbies without a bad feeling (or at least other events like learning blind lol)
[–]Sub-16 (CFOP 2LLL CN) 5 points6 points (1 child)
But that's the opposite of what I said. If you cube just to reach this milestone and feel "entitled" to other hobbies, then it can feel like a chore. Want a different hobby, go for it, you don't have to cube unless you want to.
And if you're anything like me, it can take years to get from 17.15 PB ao1000 to sub-15. Don't put other hobbies you like for years.
[–]Sub-20 (CFOP) 1 point2 points (0 children)
Well, guess you're right. It's complicated haha I mean I still doing for enjoyment but still set myself targets. It's the same for all my hobbies, can't "just do it" , always want to get better
[–]Sub-41 4x4 Yau(CN) 28.58 PB single, 35.12 PB Ao5 2 points3 points (0 children)
Looks at my school friend who was sub-15 in mid-last year and is now faster than me (I've lost track of how many sub-9 ao5s he has)
[–] 3 points4 points (14 children)
What are some interesting things I can learn on 3x3. I’m looking for something after my exams end.
[–] 1 point2 points (1 child)
You could improve 3x3 -
Steps for improving with CFOP, and when to work on what.
How to get faster at 3x3
or you could learn OH (one handed) -
https://jperm.net/oh
or 3BLD (blindfolded) -
https://jperm.net/bld
Improving BLD
or FMC (fewest moves challenge) -
Improving FMC
Or start looking into getting different puzzles -
2x2, 4x4, 5x5, pyraminx, skewb, megaminx, square-1, clock, etc.
Most Recommended Puzzles
[–] 1 point2 points (0 children)
I actually have a 4x4 which I will be learning. I think FMC is the most appealing.
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (11 children)
[–] 1 point2 points (10 children)
I average about 50s (not to mention it was a cheap cube which gets frequent lock ups)
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (9 children)
What can you do? Do you know F2L? Keyhole? Solving pairs in back slots? Edge orientation? (I assume that you use beginner's or CFOP, but if you use Roux, I'm sure there are some interesting things to learn too.)
[–] 1 point2 points (8 children)
Yeah I use CFOP. I was thinking to learn full pll and then oll.
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (7 children)
Well, if it's your idea of "interesting", then go on of course. Full PLL sure, full OLL, I don't recommend learning before you're really comfortable with PLL.
[–] 1 point2 points (6 children)
It’s not really, that’s why I was looking for something else. I did see some article about intuitive 3-style.
[–] 2 points3 points (3 children)
3 style is an advanced 3BLD (blindfolded) method.
If you want to start with BLD, you have to start at the beginning and learn OP/OP and M2/OP, first - https://jperm.net/bld
[–] 0 points1 point (2 children)
Yeah I know OP. I just don’t understand if we are going to replace it later, then why don’t we start learning from itself.
[–]It should not hurt if you relax and use lube 0 points1 point (1 child)
Same way as why you teach 3x3 to beginner starting with "this is called Right side, this is called Front, and this is how you draw a daisy" and not with "here's 50 F2L algs, 50 OLL algs and 25 PLL algs"
If you know OP, the next logical step would still be to understand M2 and Orozco. Both are useful for the general understanding.
M2 is edges-only, and has some useful tricks.
Orozco is edges, corners, or both.
Ororzo is reusing one of the sub-sets of full 3-style to solve 1 piece at a time instead of 2. So you can learn it to give you an idea about 3-style. (There are variants depending on which subset it uses)
Then you can start inventing your own commutators to solve 2 pieces at once and fall back to M2/orozco when you can't.
[–]Sub-16 (CFOP 2LLL CN) 0 points1 point (1 child)
Well, cross and F2L are more interesting imo and bring more benefit.
Commutators are cool if you can understand them. I tried a few times but I suck, it never stuck with me.
FMC is interesting. 3BLD maybe.
[–] 1 point2 points (0 children)
Commutators are interesting. I asked about FMC a few days back and that’s actually decent.
[–]Sub-16 (CFOP 2LLL CN) 18 points19 points (7 children)
I beat my PB ao100!! 14.63 → 14.61.
[–]Sub-14(ish) CFOP | PB 9.38 | Sub-21 Roux 3 points4 points (2 children)
Awesome!!
You'll be leaving me in the dust pretty soon!
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (1 child)
Not that it's my goal to put anyone in the dust though lol.
Interesting that I beat you at single (mine is 9.21) but you seem to be a couple seconds faster than me at the average - at least if we use similar metrics (I'm sub-16 by my best rolling ao1000/ao2000/ao5000/ao10000, the last one feeling totally crazy to me and sub-15 by my best rolling ao100/ao200/ao500).
[–]Sub-14(ish) CFOP | PB 9.38 | Sub-21 Roux 1 point2 points (0 children)
I time my solves on too many devices so my ao2000s tend to be spread across 6 months or more, but 4 out of my last 5 ao100 have been between 13.7 and 13.9. However I still have moments where I'm solving badly and end up well in the mid 14. (Hence the "ish"). I tend to only time myself in specific situations though, when im fresh-ish and dedicate some time to it (e.g. on the 2.5h train commute to work twice a week). So I might not be capturing my "ouch" times.
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 2 points3 points (0 children)
Yeah!!!
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 2 points3 points (0 children)
GG!
[–]Sub-25 (CFOP, 3LLL DCN) | Single: 18.70 2 points3 points (1 child)
Wooah congrats!
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (0 children)
Thanks! Can't wait to get sub-15 ao1000.
[–]ao12: 1:53 (Roux) 7 points8 points (2 children)
Spent last evening learning to solve the pyraminx fast. Now I can do it in under a minute, instead of the 20+ minutes it took me to do intuitively (which was a fun challenge, but no good for competition). I need to practice the algs for last three edges more, as those haven’t quite stuck yet. Pyra’s a fun little puzzle; it solves so quickly compared to a 3x3 which makes it a nice change of pace.
[–]Sub-13 (CFOP, dot OLL’s suck) 1 point2 points (0 children)
I disliked pyraminx for a while, but it’s a very fun puzzle to grind
[–] 2 points3 points (0 children)
Nice keep practicing and you'll sub 15 in no time
[–]Sub-20 (CFOP) 6 points7 points (6 children)
OMG after 3 years I'm finally back to my PR ao1000 of 17,15s
From now on it counts. Now I'll see how fast I really am
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (5 children)
Oh wow. I don't think I remember your story, did you have a huge break before?
[–]Sub-20 (CFOP) 7 points8 points (4 children)
Kinda. Been cubing for like 12 years now I think, but always just casually. (started when the Zhanchi/Gushong V2 were the best cubes). In 2020 I went to my first comp where I barely averaged 20s while I already averaged about 17s at home.
I then started practicing more seriously, where I sometimes was able to dip below 15s but never constantly reach it.
Over the span of multiple months I wasn't able to get faster. It broke me. I swear it made me cry sometimes.
Practiced until summer 2022 where I still wasn't noticable faster with still around 17s.
Now, after a comp two weeks ago, I said, that I finally want to reach sub 15. So I've been practicing a lot since then. Finally I'm seeing an increased number of sub 15 times.
I swear, if I ever reach sub 15s globally, it will be my greatest achievement in life. I had times I thought, maybe I'm just physically not able to do it or something lol.
Edit: sorry for the stupid long story
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 1 point2 points (2 children)
You'll get there if you keep pushing for it! It'll be hard but so worth it
[–]Sub-20 (CFOP) 3 points4 points (1 child)
Thank you! I'll keep pushing. Always had the mindset that you could achieve everything if you want to. But this is something I've been struggling so much with, especially when I see all the kids getting past this barrier without much thought.
Most people being sub 15 said, well they just practiced. Only people being around the 12s - sub 10 mark can tell me what exactly they did to practice
[–]Low 9 (CFOP) Sub 40 (4x4) 0 points1 point (0 children)
Have you ever thought about posting some solve videos and asking for a critique? If you don't want to post yourself, you can also just film yourself and compare your solves with other people's critiques. Most of the stuff people point out in critiques is pretty similar anyway.
My advice would probably just be to find some aspect of your solves that you aren't good at and just work on that.
I started cubing like 10 years ago and came back to the hobby 2 years ago doing this and have had a pretty decent progression ever since.
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (0 children)
I love long stories, thanks for telling :) Keep pushing that ao1000!
[–]Sub-20 (CFOP) 2 points3 points (0 children)
Been using my GTS2M for years now, switched back from my WRM Maglev, because that is too fast, but mainly because I like the feeling more. did a thousand solves the last few weeks and my GTS2M gets faster and faster lol. I sped up so much, it gets hard to control lol
[–] 9 points10 points (11 children)
I got my first 12 sec avg of 12. I still use 3LLL so it's quite a milestone
[–] 1 point2 points (0 children)
Congrats!
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (9 children)
Nice! Why not learn OLL tho?
[–]A cuber is secretly a screwdriver collector 1 point2 points (1 child)
Please, never ask me that question... At least I know most of it, right? 😂
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (0 children)
No promises 😈
[–] 2 points3 points (6 children)
I did learn a few of the easier ones but never ended up learning most of them. It's just that so many algorithms together scared me a bit lol. I'm gonna start learning them seriously soon, and the next goal is to become colour neutral and get a get 10 single
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (5 children)
For color neutrality, it may be a bit late, really hard to master it at your level. But dual color neutrality should be much easier.
I'm pretty sure you'll learn that OLL easily if you decide to. Just make sure to use an OLL trainer, and when learning, watch the pairs and look for triggers. OLL is really easy, the hardest part is to stop confusing the algs - trainers help with that.
[–] 2 points3 points (4 children)
Yeah I think you're right about the neutrality, but I'll give it a shot, if not ig dual may not be that bad, white and yellow.
I'll train the olls with the trainer a few each day and practice enough to not confuse them
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 2 points3 points (2 children)
I mean, Yiheng just beat the WR average as a dual color neutral solver on white and yellow, and he barely ever solves on yellow cross. I doubt you need to go full CN at this point, but going dual CN might be useful
[–]Sub-20 (CFOP) 0 points1 point (1 child)
Yiheng is dual? From his comp solves it looked like he was white cross only lol
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 0 points1 point (0 children)
He solved yellow cross for one of the solves in the WR, I think solve 1 or 2
[–]Sub-14 1 point2 points (0 children)
How fast are you rn
I switched from white to CN at sub-14 and that’s around when people say you can’t anymore
You can also learn OLL while you do CN stuff
[–] 1 point2 points (1 child)
What should i learn? Hey guys im sub 24, know full pll, quarter of oll, color neutral from white, yellow and green My cross is from 1 (good, 2 - 3 is average) to 4 secs (very bad) F2L with cross takes me from 11 (very good, averagely about 13-15 secs) to 18 seconds (very very bad) And everything including oll pll averegely 23-24 secs
[–] 1 point2 points (0 children)
[–] 2 points3 points (2 children)
Attempt of solving the curvy copter with out any assistance: state everything solved except corners of the last layer, I have to figure out ang alg to permute and another to orient the final step
[–]Sub-20 (CFOP) 0 points1 point (1 child)
Nice 😎 jumbling or no?
[–] 1 point2 points (0 children)
Not this time, I want to learn and understand how the pieces move and interact, probably I would do 2 more solves with out jumbling,
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (5 children)
For anyone who hates cross planning :)
R' F' B2 D F2 R2 F2 U R2 D' L2 F2 D B' R F' L2 B R2 F
[–] 0 points1 point (1 child)
not bad but im more used to white cross
im not gonna deny i average 15 and i got a 12.08 on this scramble tho
[–]Sub-16 (CFOP 2LLL CN) 0 points1 point (0 children)
I got a 12 too.
[–] 1 point2 points (0 children)
I got a pb of 18.61 lol
[–]Sub-20 (CFOP) 1 point2 points (0 children)
completely f'ed it up, ended up with a 14,47 lol
[–]Sub-41 4x4 Yau(CN) 28.58 PB single, 35.12 PB Ao5 1 point2 points (0 children)
Watch me fail to get a time faster than my PB (anyways, it wouldn’t have counted if it was PB lol)
Edit: yep, got a 7.42, over a second away from my PB
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 8 points9 points (13 children)
50.59 ao5!
new pb a day later
[–]Sub-16 (CFOP 2LLL CN) 3 points4 points (1 child)
Good job :) You are at a stage when you can improve even in one hundred solves. Enjoy it :)
spoken as someone who's been sub-16 for more than a year
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 5 points6 points (0 children)
you actually predicted me beating my record by 11 sec
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 4 points5 points (10 children)
49.01 now!
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 3 points4 points (1 child)
47.24 holy cow what am i doing
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 4 points5 points (0 children)
46.07
what.
[–]Sub-28 (Beginner CFOP) PB: 11 (at school) 2 points3 points (7 children)
nice! keep getting these times and PRACTICE. A. LOT.
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 3 points4 points (6 children)
dude i got 46.07 now
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 1 point2 points (5 children)
45*
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 1 point2 points (4 children)
44.17
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 1 point2 points (3 children)
41.80
what the f**k
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 2 points3 points (2 children)
39.99
i am in shambles
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 0 points1 point (0 children)
This PB streak is wild! Way to go!
[–]Sub-41 4x4 Yau(CN) 28.58 PB single, 35.12 PB Ao5 1 point2 points (0 children)
I have never seen someone break this many PBs in a row, huge congrats
[–]Sub-28 (Beginner CFOP) PB: 11 (at school) 4 points5 points (1 child)
thanks you guys so much for helping me out on this journey to get sub-30, i thank you guys so much because my dream was getting a solved rubik's cube. then sub-60 and then sub-30!! also f2l is pretty cool btw i would tell my past me that f2l is so good. also i'm learning pll and learnt Ja perm in 5 minutes!!!!!!!! for measures my normal time to memorize any algorithm is 30 to 1hour STRAIGHT, but still you guys are awesome in another level :)
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (0 children)
Good job!
Don't forget to do kind of a spaced practice of your algs. You learned the Ja perm, take a break, repeat it a few times, take a break again. Return to it multiple times during the day, and you'll remember it well by the end of the day.
[–]Sub-16 (CFOP 2LLL CN) 3 points4 points (14 children)
Having two of the same cubes is actually pretty fun. I have my "old" Tornado Pioneer (which now has all of the clicky edges and also loose magnets in the corner feet) set on 4 for distance, 5 for maglev, and lubed with Gan No.2. My "quiet" Tornado Pioneer is set to 4 for distance, 1 for maglev, and lubed with Angstrom lubes. And I also have a Standard which is also quiet, set to 1 for distance, 1 for springs and lubed with Gan No.2. It's fun switching between them.
Also, I updated my conclusive Tornado v3 post because I don't think I like the WRM more than Tornado Standard anymore. For \$25, you can go for the WRM 2021 or Tornado Standard, it's personal preference. Tornado Standard is worth getting as a "budget flagship".
[–] 2 points3 points (13 children)
Can you post a picture of your whole collection?
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (12 children)
I'm so lazy to do that actually. I don't have nice-looking display shelves with cubes. I keep them in a drawer plus some cheap cubes in separate box. To make a picture, I'd have to take everything out and arrange the cubes so that they could be discernible lol.
[–] 2 points3 points (11 children)
Aight fair. How many cubes you approximately have?
[–]Sub-16 (CFOP 2LLL CN) 3 points4 points (10 children)
all right, all right... https://imgur.com/a/hPYWKOw
[–] 1 point2 points (0 children)
This is such a great collection you have. Hopefully in a few years I'll be able to build a collection like that when I start working. But right now I'm resisting the urge to dip into my savings and get more cubes. 😂
[–]Sub-45 CFOP, PB 26.58, Ao5 34.70 2 points3 points (1 child)
oh wow we have the same drawer
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (0 children)
Ikea Alex drawers are really popular :)
[–] 1 point2 points (6 children)
Holy sh*t
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (5 children)
I have no idea how many it is. I guess, more than 50 but less than 100?
[–] 2 points3 points (4 children)
I have only like 6 lol
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (3 children)
I'm an adult, so I don't have to beg my parents for cubes, and I love being versed in cubes, reviewing them, giving tips on which cube to get, switching between cubes. For someone with a steady income, cubing is a really cheap hobby.
[–] 2 points3 points (2 children)
How long have you been cubing tho?
[–]Sub-18 PB 10.99 (CFOP) 9 points10 points (2 children)
Im finally sub 20!! I haven't been able to cube much lately due to school tasks but for the psat two days my schedule wasn't busy so I was able to cube for hundreds of solves. Managed to bring down my inconsistent sub 21 avg to a very consistent sub 20 avg. I keep only getting 19s solves lately and 18s and below are much rarer.
[–]Sub-25 (CFOP 3LLL), PB 17.15, DCN 0 points1 point (0 children)
Nice!! Sub 20 is my goal for end of the year, I'm at a 25s average now. Getting lower averages is slower and harder the faster you are, good work!!
[–]Sub-23(CFOP-3LLL) DCN 0 points1 point (0 children)
damn im at sub 21 and it feels so close yet so far away. this has been the longest I've been stuck at a barrier, I feel so impatient.
[–]Sub-16 (CFOP 2LLL CN) 5 points6 points (5 children)
What a weird solve, free pairs everywhere.
B2 L B2 F2 D2 F2 R U2 L' D2 U2 R F' U' R' D L2 D B' U R2
``````y2 // inspection
D' R' F2 L2 D2 // Yellow cross (5)
R U' R' U y' R' U R // Blue Orange Pair (8)
L F' U F L' // Orange Green Pair (5)
d' L' U L2 U' L2' U L // Red Blue & Red Green Pairs (8)
U2 R' U' R' F R F' U R // OLL 46
U R U' R U R' D R D' R U' D R2 U R2 D' R2 U' // PLL - V
``````
52STM / 11.55sec =4.5TPS
view at CubeDB.net
[–] 2 points3 points (1 child)
Wait i just saw that 2 pair simultaneously part….Crazy shit🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (0 children)
Yeah. I'm not great at lookahead, but I noticed those two pairs and managed to cancel between them. I know that if two pairs are located like this, you can insert one and then the other right away, the first one won't break the second one.
[–]It should not hurt if you relax and use lube 4 points5 points (1 child)
PLL skip if you do those last 2 pairs in the opposite order
[–]Sub-16 (CFOP 2LLL CN) 1 point2 points (0 children)
Nice!
I also think I didn't notice the free pair in inspection. Otherwise, I'd plan solving it first and would miss everything else.
[–] 1 point2 points (0 children)
Ohhh coool😁😂
[–]Sub-16 (CFOP 2LLL CN) 2 points3 points (23 children)
I'm having a weird urge to buy a Gan 11 M (not Pro). I don't know what kind of dark magic Gan uses, but I keep wanting their cubes even if I end up not using them very much 🙈 This cube I want for its ridiculous low weight of just 56g (and no, I don't want the non-magnetic 50g Gan 11 Air lol).
Doctor, is there any hope for me?
Does anyone have this cube?
[–]Sub-13 (CFOP, dot OLL’s suck) 1 point2 points (0 children)
I kind of have the impulse to get the gan x and the yuxin little magic 3x3s. Idk why about the ylm, I just want it, but the gan x just felt good at my comp. I prob would t use it much, but it felt better than my 11 m pro
[–]Sub-13 (CFOP, dot OLL’s suck) 1 point2 points (1 child)
Take out the corner magnets on your 11 m pro and you got an 11 m | 13,339 | 43,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-14 | latest | en | 0.943614 |
http://www.transfer-kaufmann.com/e-equivalent-physics-piece-2/ | 1,580,018,178,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687725.76/warc/CC-MAIN-20200126043644-20200126073644-00374.warc.gz | 286,523,414 | 41,683 | # What is E equal to in Physics? Aspect 2
What is E equal to in Physics? It is usually a substantial respond to for the question of “what is a drive of gravity?” It is a closing reply to to the predicament that may be however a thriller in Physics.
The correct notion of E is incredibly hard to determine. It’s because its not one thing, but an abstraction of many concepts, every single of which can be explained in several different ways. www.gurudissertation.net We have to know the primary difference involving a definition and an abstraction so as to have an understanding of E.
The actual theory of E is often a mathematical expression, and that is defined given that the same kind of dilemma that is definitely requested so as http://www.bu.edu/fitrec/fitness/pilates/ to determine a perform in arithmetic. So, precisely what is E equal to in Physics?
The clarification of this idea is a tiny laborious to find out. It is actually detailed as follows. Presume about a free-falling human overall body of mass m which is in a condition of equilibrium.
Let us assume that the Newtonian law of Gravity states that mass is attracted to gravity, that is certainly proportional towards products on the masses m along with the acceleration g. That is potential should the gravitational possible vigor offers a specific amount of electrical power with the scenario of absolutely free slipping. Then again, this isn’t achievable when the mass just isn’t falling including a resistance is encountered. In such a situation, the legislation of gravity needs to be altered.
Well, this is less complicated to learn if we expect of the true object falling. We all know from elementary physics that when an item that is not slipping is subjected to an outward drive, like gravity, that object shall be pushed inside path within the drive.
However, if you want to get to the item, the gravitational would-be vitality have got to be changed into kinetic stamina, which the object would then must go inside of the path on the drive. Put simply, when the item which is falling is subjected to outward pressure, it should go inside of the route belonging to the drive.
It is a breeze to view which the legislation of Newton is strictly the very same concept, only that it is not expressed as being the gravitational likely electricity. That’s, the power of gravity that we contact E is in truth an outward power.
As a consequence, it happens to be a fascinating approach to check the laws and regulations of Physics as well as their romance for the equations which explain the dynamics from the Earth and its technique of geologic processes. The useful element is that, much like the regulations of mechanics and calculus, the laws and regulations of physics also implement into the Earth as being a total. Being a consequence, we will make use of the equations and laws of physics to provide a more accurate description of what comes about inside Earth, quite simply, what happens inside of the Universe.
One fundamental principle to remember is as Einstein stated, “in each and every step there’s an assumption of some variety. This is not a foolproof methodology of contemplating.”
If we have a look at E as E=mc2, then the equation could very well be minimized to your linear type. We can easily get started with by plotting this way to view the way it pertains to the equations from the equations belonging to the equations of your equations of E=mc2. Then, we could repeat the process to the other equations of E=mc2.
0 replies | 715 | 3,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-05 | latest | en | 0.964706 |
http://us.metamath.org/mpeuni/mzpincl.html | 1,638,927,623,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00604.warc.gz | 73,807,254 | 8,336 | Mathbox for Stefan O'Rear < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > mzpincl Structured version Visualization version GIF version
Theorem mzpincl 36812
Description: Polynomial closedness is a universal first-order property and passes to intersections. This is where the closure properties of the polynomial ring itself are proved. (Contributed by Stefan O'Rear, 4-Oct-2014.)
Assertion
Ref Expression
mzpincl (𝑉 ∈ V → (mzPoly‘𝑉) ∈ (mzPolyCld‘𝑉))
Proof of Theorem mzpincl
Dummy variables 𝑓 𝑔 𝑎 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 mzpval 36810 . 2 (𝑉 ∈ V → (mzPoly‘𝑉) = (mzPolyCld‘𝑉))
2 mzpclall 36805 . . . . 5 (𝑉 ∈ V → (ℤ ↑𝑚 (ℤ ↑𝑚 𝑉)) ∈ (mzPolyCld‘𝑉))
3 intss1 4462 . . . . 5 ((ℤ ↑𝑚 (ℤ ↑𝑚 𝑉)) ∈ (mzPolyCld‘𝑉) → (mzPolyCld‘𝑉) ⊆ (ℤ ↑𝑚 (ℤ ↑𝑚 𝑉)))
42, 3syl 17 . . . 4 (𝑉 ∈ V → (mzPolyCld‘𝑉) ⊆ (ℤ ↑𝑚 (ℤ ↑𝑚 𝑉)))
5 simpr 477 . . . . . . . . 9 (((𝑉 ∈ V ∧ 𝑓 ∈ ℤ) ∧ 𝑎 ∈ (mzPolyCld‘𝑉)) → 𝑎 ∈ (mzPolyCld‘𝑉))
6 simplr 791 . . . . . . . . 9 (((𝑉 ∈ V ∧ 𝑓 ∈ ℤ) ∧ 𝑎 ∈ (mzPolyCld‘𝑉)) → 𝑓 ∈ ℤ)
7 mzpcl1 36807 . . . . . . . . 9 ((𝑎 ∈ (mzPolyCld‘𝑉) ∧ 𝑓 ∈ ℤ) → ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ 𝑎)
85, 6, 7syl2anc 692 . . . . . . . 8 (((𝑉 ∈ V ∧ 𝑓 ∈ ℤ) ∧ 𝑎 ∈ (mzPolyCld‘𝑉)) → ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ 𝑎)
98ralrimiva 2961 . . . . . . 7 ((𝑉 ∈ V ∧ 𝑓 ∈ ℤ) → ∀𝑎 ∈ (mzPolyCld‘𝑉)((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ 𝑎)
10 ovex 6638 . . . . . . . . 9 (ℤ ↑𝑚 𝑉) ∈ V
11 snex 4874 . . . . . . . . 9 {𝑓} ∈ V
1210, 11xpex 6922 . . . . . . . 8 ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ V
1312elint2 4452 . . . . . . 7 (((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ (mzPolyCld‘𝑉) ↔ ∀𝑎 ∈ (mzPolyCld‘𝑉)((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ 𝑎)
149, 13sylibr 224 . . . . . 6 ((𝑉 ∈ V ∧ 𝑓 ∈ ℤ) → ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ (mzPolyCld‘𝑉))
1514ralrimiva 2961 . . . . 5 (𝑉 ∈ V → ∀𝑓 ∈ ℤ ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ (mzPolyCld‘𝑉))
16 simpr 477 . . . . . . . . 9 (((𝑉 ∈ V ∧ 𝑓𝑉) ∧ 𝑎 ∈ (mzPolyCld‘𝑉)) → 𝑎 ∈ (mzPolyCld‘𝑉))
17 simplr 791 . . . . . . . . 9 (((𝑉 ∈ V ∧ 𝑓𝑉) ∧ 𝑎 ∈ (mzPolyCld‘𝑉)) → 𝑓𝑉)
18 mzpcl2 36808 . . . . . . . . 9 ((𝑎 ∈ (mzPolyCld‘𝑉) ∧ 𝑓𝑉) → (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ 𝑎)
1916, 17, 18syl2anc 692 . . . . . . . 8 (((𝑉 ∈ V ∧ 𝑓𝑉) ∧ 𝑎 ∈ (mzPolyCld‘𝑉)) → (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ 𝑎)
2019ralrimiva 2961 . . . . . . 7 ((𝑉 ∈ V ∧ 𝑓𝑉) → ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ 𝑎)
2110mptex 6446 . . . . . . . 8 (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ V
2221elint2 4452 . . . . . . 7 ((𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ (mzPolyCld‘𝑉) ↔ ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ 𝑎)
2320, 22sylibr 224 . . . . . 6 ((𝑉 ∈ V ∧ 𝑓𝑉) → (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ (mzPolyCld‘𝑉))
2423ralrimiva 2961 . . . . 5 (𝑉 ∈ V → ∀𝑓𝑉 (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ (mzPolyCld‘𝑉))
2515, 24jca 554 . . . 4 (𝑉 ∈ V → (∀𝑓 ∈ ℤ ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ (mzPolyCld‘𝑉) ∧ ∀𝑓𝑉 (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ (mzPolyCld‘𝑉)))
26 vex 3192 . . . . . . . . 9 𝑓 ∈ V
2726elint2 4452 . . . . . . . 8 (𝑓 (mzPolyCld‘𝑉) ↔ ∀𝑎 ∈ (mzPolyCld‘𝑉)𝑓𝑎)
28 vex 3192 . . . . . . . . 9 𝑔 ∈ V
2928elint2 4452 . . . . . . . 8 (𝑔 (mzPolyCld‘𝑉) ↔ ∀𝑎 ∈ (mzPolyCld‘𝑉)𝑔𝑎)
30 mzpcl34 36809 . . . . . . . . . . 11 ((𝑎 ∈ (mzPolyCld‘𝑉) ∧ 𝑓𝑎𝑔𝑎) → ((𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ (𝑓𝑓 · 𝑔) ∈ 𝑎))
31303expib 1265 . . . . . . . . . 10 (𝑎 ∈ (mzPolyCld‘𝑉) → ((𝑓𝑎𝑔𝑎) → ((𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ (𝑓𝑓 · 𝑔) ∈ 𝑎)))
3231ralimia 2945 . . . . . . . . 9 (∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑎𝑔𝑎) → ∀𝑎 ∈ (mzPolyCld‘𝑉)((𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ (𝑓𝑓 · 𝑔) ∈ 𝑎))
33 r19.26 3058 . . . . . . . . 9 (∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑎𝑔𝑎) ↔ (∀𝑎 ∈ (mzPolyCld‘𝑉)𝑓𝑎 ∧ ∀𝑎 ∈ (mzPolyCld‘𝑉)𝑔𝑎))
34 r19.26 3058 . . . . . . . . 9 (∀𝑎 ∈ (mzPolyCld‘𝑉)((𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ (𝑓𝑓 · 𝑔) ∈ 𝑎) ↔ (∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 · 𝑔) ∈ 𝑎))
3532, 33, 343imtr3i 280 . . . . . . . 8 ((∀𝑎 ∈ (mzPolyCld‘𝑉)𝑓𝑎 ∧ ∀𝑎 ∈ (mzPolyCld‘𝑉)𝑔𝑎) → (∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 · 𝑔) ∈ 𝑎))
3627, 29, 35syl2anb 496 . . . . . . 7 ((𝑓 (mzPolyCld‘𝑉) ∧ 𝑔 (mzPolyCld‘𝑉)) → (∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 · 𝑔) ∈ 𝑎))
37 ovex 6638 . . . . . . . . 9 (𝑓𝑓 + 𝑔) ∈ V
3837elint2 4452 . . . . . . . 8 ((𝑓𝑓 + 𝑔) ∈ (mzPolyCld‘𝑉) ↔ ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 + 𝑔) ∈ 𝑎)
39 ovex 6638 . . . . . . . . 9 (𝑓𝑓 · 𝑔) ∈ V
4039elint2 4452 . . . . . . . 8 ((𝑓𝑓 · 𝑔) ∈ (mzPolyCld‘𝑉) ↔ ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 · 𝑔) ∈ 𝑎)
4138, 40anbi12i 732 . . . . . . 7 (((𝑓𝑓 + 𝑔) ∈ (mzPolyCld‘𝑉) ∧ (𝑓𝑓 · 𝑔) ∈ (mzPolyCld‘𝑉)) ↔ (∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 + 𝑔) ∈ 𝑎 ∧ ∀𝑎 ∈ (mzPolyCld‘𝑉)(𝑓𝑓 · 𝑔) ∈ 𝑎))
4236, 41sylibr 224 . . . . . 6 ((𝑓 (mzPolyCld‘𝑉) ∧ 𝑔 (mzPolyCld‘𝑉)) → ((𝑓𝑓 + 𝑔) ∈ (mzPolyCld‘𝑉) ∧ (𝑓𝑓 · 𝑔) ∈ (mzPolyCld‘𝑉)))
4342a1i 11 . . . . 5 (𝑉 ∈ V → ((𝑓 (mzPolyCld‘𝑉) ∧ 𝑔 (mzPolyCld‘𝑉)) → ((𝑓𝑓 + 𝑔) ∈ (mzPolyCld‘𝑉) ∧ (𝑓𝑓 · 𝑔) ∈ (mzPolyCld‘𝑉))))
4443ralrimivv 2965 . . . 4 (𝑉 ∈ V → ∀𝑓 (mzPolyCld‘𝑉)∀𝑔 (mzPolyCld‘𝑉)((𝑓𝑓 + 𝑔) ∈ (mzPolyCld‘𝑉) ∧ (𝑓𝑓 · 𝑔) ∈ (mzPolyCld‘𝑉)))
454, 25, 44jca32 557 . . 3 (𝑉 ∈ V → ( (mzPolyCld‘𝑉) ⊆ (ℤ ↑𝑚 (ℤ ↑𝑚 𝑉)) ∧ ((∀𝑓 ∈ ℤ ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ (mzPolyCld‘𝑉) ∧ ∀𝑓𝑉 (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ (mzPolyCld‘𝑉)) ∧ ∀𝑓 (mzPolyCld‘𝑉)∀𝑔 (mzPolyCld‘𝑉)((𝑓𝑓 + 𝑔) ∈ (mzPolyCld‘𝑉) ∧ (𝑓𝑓 · 𝑔) ∈ (mzPolyCld‘𝑉)))))
46 elmzpcl 36804 . . 3 (𝑉 ∈ V → ( (mzPolyCld‘𝑉) ∈ (mzPolyCld‘𝑉) ↔ ( (mzPolyCld‘𝑉) ⊆ (ℤ ↑𝑚 (ℤ ↑𝑚 𝑉)) ∧ ((∀𝑓 ∈ ℤ ((ℤ ↑𝑚 𝑉) × {𝑓}) ∈ (mzPolyCld‘𝑉) ∧ ∀𝑓𝑉 (𝑔 ∈ (ℤ ↑𝑚 𝑉) ↦ (𝑔𝑓)) ∈ (mzPolyCld‘𝑉)) ∧ ∀𝑓 (mzPolyCld‘𝑉)∀𝑔 (mzPolyCld‘𝑉)((𝑓𝑓 + 𝑔) ∈ (mzPolyCld‘𝑉) ∧ (𝑓𝑓 · 𝑔) ∈ (mzPolyCld‘𝑉))))))
4745, 46mpbird 247 . 2 (𝑉 ∈ V → (mzPolyCld‘𝑉) ∈ (mzPolyCld‘𝑉))
481, 47eqeltrd 2698 1 (𝑉 ∈ V → (mzPoly‘𝑉) ∈ (mzPolyCld‘𝑉))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 384 ∈ wcel 1987 ∀wral 2907 Vcvv 3189 ⊆ wss 3559 {csn 4153 ∩ cint 4445 ↦ cmpt 4678 × cxp 5077 ‘cfv 5852 (class class class)co 6610 ∘𝑓 cof 6855 ↑𝑚 cmap 7809 + caddc 9891 · cmul 9893 ℤcz 11329 mzPolyCldcmzpcl 36799 mzPolycmzp 36800 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-8 1989 ax-9 1996 ax-10 2016 ax-11 2031 ax-12 2044 ax-13 2245 ax-ext 2601 ax-rep 4736 ax-sep 4746 ax-nul 4754 ax-pow 4808 ax-pr 4872 ax-un 6909 ax-cnex 9944 ax-resscn 9945 ax-1cn 9946 ax-icn 9947 ax-addcl 9948 ax-addrcl 9949 ax-mulcl 9950 ax-mulrcl 9951 ax-mulcom 9952 ax-addass 9953 ax-mulass 9954 ax-distr 9955 ax-i2m1 9956 ax-1ne0 9957 ax-1rid 9958 ax-rnegex 9959 ax-rrecex 9960 ax-cnre 9961 ax-pre-lttri 9962 ax-pre-lttrn 9963 ax-pre-ltadd 9964 ax-pre-mulgt0 9965 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3or 1037 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1878 df-eu 2473 df-mo 2474 df-clab 2608 df-cleq 2614 df-clel 2617 df-nfc 2750 df-ne 2791 df-nel 2894 df-ral 2912 df-rex 2913 df-reu 2914 df-rab 2916 df-v 3191 df-sbc 3422 df-csb 3519 df-dif 3562 df-un 3564 df-in 3566 df-ss 3573 df-pss 3575 df-nul 3897 df-if 4064 df-pw 4137 df-sn 4154 df-pr 4156 df-tp 4158 df-op 4160 df-uni 4408 df-int 4446 df-iun 4492 df-br 4619 df-opab 4679 df-mpt 4680 df-tr 4718 df-eprel 4990 df-id 4994 df-po 5000 df-so 5001 df-fr 5038 df-we 5040 df-xp 5085 df-rel 5086 df-cnv 5087 df-co 5088 df-dm 5089 df-rn 5090 df-res 5091 df-ima 5092 df-pred 5644 df-ord 5690 df-on 5691 df-lim 5692 df-suc 5693 df-iota 5815 df-fun 5854 df-fn 5855 df-f 5856 df-f1 5857 df-fo 5858 df-f1o 5859 df-fv 5860 df-riota 6571 df-ov 6613 df-oprab 6614 df-mpt2 6615 df-of 6857 df-om 7020 df-wrecs 7359 df-recs 7420 df-rdg 7458 df-er 7694 df-map 7811 df-en 7908 df-dom 7909 df-sdom 7910 df-pnf 10028 df-mnf 10029 df-xr 10030 df-ltxr 10031 df-le 10032 df-sub 10220 df-neg 10221 df-nn 10973 df-n0 11245 df-z 11330 df-mzpcl 36801 df-mzp 36802 This theorem is referenced by: mzpconst 36813 mzpproj 36815 mzpadd 36816 mzpmul 36817
Copyright terms: Public domain W3C validator | 5,204 | 7,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-49 | latest | en | 0.172925 |
https://mrswillskindergarten.com/product/100s-chart-mystery-puzzles-kindergarten-bundle/ | 1,670,602,926,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711417.46/warc/CC-MAIN-20221209144722-20221209174722-00355.warc.gz | 452,578,737 | 92,288 | Sale!
# 100’s Chart Mystery Puzzles Kindergarten-Bundle
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## Product Description
100’s Chart Mystery Puzzles with Differentiation!
DOWNLOAD the FREE FALL SAMPLE to see if this product is right for you!
What Are 100’s Chart Mystery Puzzles?
These differentiated 100’s chart mystery puzzles are such a fun way for students to practice a variety of counting skills and number recognition! Digital and printable versions of each puzzle as well as differentiated puzzles and response sheets provide the teacher with several activities!
What is included in the 100’s Chart Mystery Puzzle Seasons Bundle?
This Season’s Bundle of differentiated 100’s chart mystery puzzles include 12 puzzles for each season (Fall, Winter, Spring, and Summer) for a total of 48 different puzzles! Mystery puzzles are such a fun way for students to practice a variety of counting skills and number recognition. Digital and printable versions of each puzzle as well as differentiated puzzles and response sheets provide the teacher with several activities!
Digital Version
*The digital puzzles can be used in PowerPoint installed on devices or they can be uploaded to Google Drive and used as Google slides.* As students scroll through each slide in the presentation they will see a number from the 100’s number chart to color on their response sheet.
Here’s what’s included:
● 12 digital ordered number puzzles for each season (48 total)
● 12 digital randomized number puzzles for differentiated instruction for each season (48 total)
● 8 different response sheets for each mystery puzzle with missing numbers for students to practice skip-counting by two’s, fives, and tens. In addition, the response pages include randomized missing numbers for advanced students. That’s a total of 384 different response sheets!
● Answer keys for each of the 48 different pictures
Printable Version
The printable puzzles can be used in numerous ways. Simply print, cut, and laminate the 100’s number chart squares provided. The squares fit nicely in pocket charts for whole class or small group activities. Printable 100’s number mats are also provided in this resource so students can complete the mystery puzzles on a table or on the floor during table or center times!
Here’s what’s included:
● Printable mystery puzzle pieces for each of the puzzles (48 puzzles)
● Complete mystery puzzle mat for students to place the pieces
● 7 additional mystery puzzle mats with missing numbers for advanced students or end of the year instruction
(The response sheets and answer keys are the same for the digital and printable puzzles.)
*An index with direct links to each puzzle is also included in the product for ease of access!
How can I use the 100’s Chart Mystery Puzzles?
Here are some of the ways the puzzles can be used:
Warm-Up & Early FinishersResponse sheets can be given to the students to complete as a warm-up activity by having students fill in the missing numbers on the differentiated response sheets. Then students can color in and solve the mystery puzzles (digital or print version) after completing other work.
Center TimeThe puzzles are great for center time. Students can work independently or with small groups to complete a puzzle on the floor using the printable 100’s chart mats by placing the colored squares on the corresponding numbers. Response sheets can be colored after the puzzle has been completed.
Table TimeThe differentiated response sheets and puzzles can provide great feedback for the teacher during table time. Students can work together in small groups with the teacher to solve the puzzle and practice specific math strategies such as counting, skip-counting, number recognition, and number patterns.
Game IdeasBy using the 100’s chart floor mats, students can complete the mystery puzzles in teams or small groups. Give each student a few of the printable squares and break the class into teams. Teams can add squares to their puzzles as a relay race or after answering comprehension questions!
Incentive IdeasUse a pocket chart or tape the floor mat in a visible location. Students can earn numbers for their mystery puzzle with good behavior! Once a puzzle is completed students can receive a class party or reward.
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Interested in having Deedee speak at your event? Submit the form below. | 894 | 4,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | longest | en | 0.896652 |
https://www.physicsforums.com/threads/wordy-physics-on-forces-pulleys-and-friction.642991/ | 1,542,886,464,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746205.96/warc/CC-MAIN-20181122101520-20181122123520-00162.warc.gz | 959,258,760 | 15,775 | # Homework Help: Wordy Physics on Forces, Pulleys, and Friction!
1. Oct 10, 2012
### capn awesome
Hi. I'm new here, and decided to sign up now that physics is actually difficult for me in first year university... Hope to find the help I need, and help those that I can. Anyway..
I'm having trouble with these wordy physics problems...
1) A light rope passing over a light frictionless pulley is fastened to a platform and a man on the platform holds the other end of the rope. The man pulls on the rope with sufficient force to give himself and the platform an upward acceleration of 0.91m/s^2. Find the tension in the rope and the reaction between the man's feet and platform on which he stands. The mass of the man and the platform are 72kg and 36 kg respectively.
So I tried drawing my FBD, where acceleration points upwards from the platform, and tension points upwards on each side of the pulley. Fg points downwards from the platform. Finally there is an applied force, Fp, opposing tension on the side with the man. I stated that F = ma, added the two masses together to find the force, and divided that by two to get the tension. Am I on the right track...? And then what is it asking me to do for the "reaction between the man's feet and the platform"?
Fp = a(m1+m2)
Ftension = Fp/2
2) A long plank weighing 142N slides on a level frictionless surface with initial speed = 7.3m/s moving the direction of its length. A block weighing 35.5N with initial speed = 0m/s is set down on the plank. If coefficient of kinetic friction between the plank and block is 0.5, what is the acceration of the plank and block?
Also, what is the length of the skid mark made by the block on the plank?
So far, I have found the masses of each using the given weights, found the force of kinetic friction by multiplying the coefficient with Fn. Where would I go with that to find the deceleration of the plank? I assume the masses will have to combine, but what does net force equal in order to solve for acceleration?
mass of plank = m1 = 14.5kg
mass of block = m2 = 3.6kg
kinetic friction = μFn = 17.8N
2. Oct 11, 2012
### Simon Bridge
Welcome to PF;
The purpose of these problems is to get you used to extracting the important information from a written or spoken description. It is to train up your judgement.
For (1)
The applied force of the man pulling on the rope is what creates the tension - as far as the man-mass is concerned, there is just the tension pulling up.
When the man pulls on the rope, the platform rises, and presses against his feet.
This is the reaction force they are talking about - you need to include it on your free body diagrams. You should end up with two equations with T (tension) and R (reaction) as your unknowns.
For (2)
... never assume - you will just make an *** out of u and me ;)
Did you try drawing the free-body diagrams for the different masses?
For both of them: you have been combining the masses too soon.
Write the expressions for ƩF=ma for each mass separately.
Let the math tell you how the masses combine.
3. Oct 11, 2012
### capn awesome
"The applied force of the man pulling on the rope is what creates the tension - as far as the man-mass is concerned, there is just the tension pulling up."
What exactly do you mean by this? If the man and platform were stationary instead of accelerating, would there be two different forces of tension? In other words, is there only one tension force because he is applying a force instead of just countering the force of gravity?
Even still, wouldn't the tension simply oppose the applied pull on the rope, working out to the same equation just not halved? The man and the platform move as one unit with the same upward acceleration... I must really be missing something crucial here, because I am not seeing how the forces being calculated separately do me any good.
4. Oct 11, 2012
### Simon Bridge
I mean that when you do a free body diagram you only consider the forces on the body.
The applied force is not applied to the man (it actually contributes to the reaction force at the feet via the rope.)
Tension can be tricky to think about: if you and a mate are having a tug of war, and you both pull on your end of the rope with force 50N - what is the tension in the rope?
(a) 0 (b) 50N (c) 100N (d) none of these
Did you try it? You get two equations with two unknowns.
5. Oct 11, 2012
### capn awesome
Is it 100N..? Conceptually, if I was pulling on the rope myself with the other end attached to a fixed point, there would be 50N of tension pulling the other way. If someone on the other end is pulling with the same force, there is still tension in the rope.
For the pulley question, I calculated F = ma for the man and the platform separately to get 32.8N and 65.5N.. Is this what you meant? I don't see how those values help me when you can add them together to get an almost equal force by adding the masses and solving.
For question 2)
I calculated the force of friction between the plank and the block, as it should be the only thing affecting both of their accelerations.
Ff = μFn
Ff = μ(m-block)(g)
Ff = (0.5)(3.62)(9.8)
Ff = 17.7N
Then dividing that value by each mass to get the acceleration of each separate item.
Friction is causing the plank to slow down in the opposite direction of its motion, so it is negative.
a-plank = -17.7/14.5 = -1.2m/s^2
Friction is causing the block to accelerate on the board.
a-block = 17.7/3.62 = 4.9m/s^2
Did I do this correctly? And if so, how would I find the length of the skid mark left by the block? I know it would be the elapsed time between the block hitting the plank to the time where static friction takes over for kinetic friction.. Not sure what formulas I could use here.
6. Oct 11, 2012
### Simon Bridge
And here is your problem ... it is 50N.
Proof - you can fix the rope to a post stuck in the ground anywhere along it's length without changing the physics. Each person is still pulling at 50N and each feels a 50N tension pulling back.
Dunno - I didn't crunch the numbers. That's your job. But there should be a summation sign before the F.
If the man has mass M, and the platform mass m, the tension is T, the reaction force is R, show me the equations from the free body diagrams.
For (2)
I still cannot tell if you got it right because you did not show enough of your working.
I'll give you an example of what I'm expecting:
The plank (weight W) is moving under the influence of only one force, friction (f) which acts against the motion of the plank so:
$\sum F=m_pa_p \Rightarrow -f=Wa_p/g$ ... eq.(1)
Friction is given by the weight (w) of the block so $f=\mu w$ ... eq.(2)
... and you need a third equation which you get from the other fbd.
You seem to have got the essential lesson that friction does not act opposite to the speed.
The "length of the skid mark" part is about your understanding of how friction works. | 1,717 | 6,925 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-47 | latest | en | 0.947687 |
http://mathhelpforum.com/advanced-math-topics/5053-riemannian-geometry.html | 1,529,887,741,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00166.warc.gz | 191,559,638 | 10,801 | # Thread: Riemannian geometry
1. ## Riemannian geometry
Hello.
Let M,N be a connected smooth riemannian manifolds.
I define the metric as usuall, the infimum of lengths of curves between the two points.
(the length is defined by the integral of the norm of the velocity vector of the curve).
Suppose phi is a homeomorphism which is a metric isometry.
I wish to prove phi is a diffeomorphism.
Please, anyone who can help.
Thanks in advance,
Roey
2. Pick a point $\displaystyle q\in N$. There is an $\displaystyle \epsilon>0$ such that $\displaystyle \exp_q$ is a diffeomorphism of the ball $\displaystyle B(0,\epsilon)=B_{\epsilon}^N\subset T_qN$ (the tangent space) into $\displaystyle N$. Since $\displaystyle \phi$ is a homeomorphism, $\displaystyle \phi^{-1}(\exp_q(B_{\epsilon}^N))$ is an open set in $\displaystyle N$ and $\displaystyle \exists \ p\in\phi^{-1}({q})\bigcap\phi^{-1}(\exp_q(B_{\epsilon}^N))$.
Now, since $\displaystyle \phi$ is an isometry, for all geodesics $\displaystyle \gamma-\delta,\delta)\rightarrow N$ with $\displaystyle \gamma(0)=\phi(p)$, for the geodesic $\displaystyle \beta-\eta,\eta)\rightarrow M$ with $\displaystyle \beta(0)=p$ and $\displaystyle {\rm d}\phi_p(\beta'(0))=\gamma'(0)$, we have that $\displaystyle \gamma=\phi\circ\beta.$
This means $\displaystyle ({\rm d}\phi_p)^{-1}$ exists (for all of the tangent space, as $\displaystyle N$ is complete); and since $\displaystyle M$ is complete, we can consider the map $\displaystyle \psi:\exp_q(B_{\epsilon}^N)\rightarrow\exp_p(B_{ \epsilon}^M)$ defined by $\displaystyle \psi=\exp_p\circ({\rm d}\phi_p)^{-1}\circ(\exp_q(B_{\epsilon}^N))^{-1}.$ This is a diffeomorphism, and inverse to the (restricted map) $\displaystyle \phi:\exp_p(B_{\epsilon}^M)\rightarrow\exp_q(B_{ \epsilon}^N)$. So $\displaystyle \phi$ is a diffeomorphism.
Note. I do not see why this would not work if $\displaystyle \phi$ was only a local isometry.
Note 2. Thin Lizzy - Whiskey in the jar
3. Sorry, stupid me - use the argument about $\displaystyle ({\rm d}\phi)^{-1}$ to get just $\displaystyle \phi$ to be a diffeo. No need to make our lives harder by computing inverses. Sorry again, I was... checking my signature out and was carried away
4. M,N are not complete
5. What!!
I can see how the hypotheses on M can be relaxed, by following the same argument. But no completeness for N, well... There is no way I can see it happen, and my guess is there is no simple way to do this | 724 | 2,462 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-26 | latest | en | 0.781511 |
https://www.passeidireto.com/arquivo/60182529/density | 1,550,654,506,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494694.1/warc/CC-MAIN-20190220085318-20190220111318-00531.warc.gz | 939,237,187 | 14,943 | Density - Química
15 pág.
# Density
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```CHEM 121L
General Chemistry Laboratory
Revision 3.2
Density Determinations
To learn about intensive physical properties.
To learn how to measure the density of substances.
To learn how to characterize a substance using its density.
To learn different methods for measuring volume.
In this laboratory exercise, we will determine the density of a Leaded Brass Alloy as well as the
densities of a series of aqueous Table Sugar (C12H22O11) solutions. The density of an aqueous
Table Sugar solution of unknown composition will also be determined. This last measurement
will then be compared with the measured densities of the Table Sugar solutions of known
composition to elucidate the Weight Percentage Table Sugar in the unknown sample.
Physical properties of substances are sub-categorized according to whether they are intensive or
extensive. Extensive physical properties depend on the amount of substance being considered,
whereas intensive physical properties do not. Some examples are:
Extensive Intensive
Mass Density
Volume Color
Surface Area Melting Point
Boiling Point
Vapor Pressure
Surface Tension
Conductivity
Intensive properties are particularly important because every pure substance has its own unique
set of intensive physical and chemical properties which distinguish it from all other substances.
And, even if the substance is not pure, its intensive physical properties can be useful in
characterizing it. As an example, two metals which have a very similar appearance can be
distinguished by considering their intensive physical properties:
color: silver white color: steel gray
melting pt: 320.9
o
C melting pt: 1890
o
C
boiling pt: 765
o
C boiling pt: 2482
o
C
density: 8.642 g/cm
3
density: 7.20 g/cm
3
conductivity: 1.38x10
5
cm
-1 conductivity: 7.74x104 cm-1-1
P a g e | 2
A substance’s density is a particularly important intensive physical property of that substance.
The density of a substance, or object, is defined as the ratio of its mass to volume:
density =
mass
volume
(Eq. 1)
In effect, the density is a measure of the substance’s compactness.
This property is important because it is extremely sensitive to the substance's environmental
conditions and its composition. This is illustrated by looking at the variation in the density of
Water with temperature and the variation in the density of Fuming Sulfuric Acid, Sulfuric Acid
(H2SO4) infused with Sulfur Trioxide (SO3), with composition:
Water
Temperature [
o
C] Density [g/mL]
0 0.99987
3.98
*
1.00000
10 0.99973
20
25
*
0.99823
0.99704
30 0.99567
40 0.99224
50 0.98807
60 0.98324
70 0.97781
80 0.97183
90 0.96534
100 0.95838
Fuming Sulfuric Acid
% H2SO4 Density [g/mL]
100 1.839
95 1.862
90 1.880
85 1.899
80 1.915
75 1.934
70 1.952
As can be noted from the above data, Water has a maximum density of 1.00000 g/mL at 3.98
o
C,
and the density decreases as the temperature rises. This is as is to be expected since as the
temperature increases the added thermal energy causes the molecules comprising the sample to
"bounce around" more, therefore requiring an increased volume, resulting in a lower density. It
should be noted, that unless we are performing high precision work, the density of water, near
Room Temperature, can be taken as about 1.00 g/mL.
P a g e | 3
Because the density is a distinguishing characteristic of a given substance, it can be helpful in
identifying it. Consider solutions of Table Sugar in Water. As the Table Sugar becomes more
concentrated, the density of the solution increases. (This is the result of the tight interaction
between the Sugar and Water molecules.) By knowing how the concentration affects the density,
we can determine the Table Sugar concentration in a given sample by simply measuring its
density. This basic procedure has several important industrial applications. Two common
examples are the determination of electrolyte concentration in automobile batteries and the
determination of antifreeze concentration in automobile cooling systems. In both cases, the
density of each solution is measured with a hydrometer. The hydrometer is calibrated to give the
concentration of the desired substance in terms of the solution density.
In a similar fashion, as the solution composition of Leaded Brass, primarily a mixture of Copper
(Cu) and Zinc (Zn), with a little added Lead (Pb) to improve its machinability, is varied the
density will likewise vary.
In this laboratory, we will first determine the density of Brass 306 by a direct measurement of its
geometric parameters. We will compare this measured density with the accepted value in order
to judge the accuracy of this technique. We will then establish the relationship between Table
Sugar concentration in Water and solution density; i.e. we will establish a calibration curve
relating the measured density and concentration. Finally, we will measure the density of a Table
Sugar solution whose concentration is unknown. Use of the prepared calibration curve will
allow us to establish the concentration of the Sugar solution.
In all cases, density measurements require an independent determination of the mass and volume
for a sample of the substance under consideration. Mass measurements are quite easy. The pan
balance was one of the earliest instruments employed by the medieval alchemists; current
analytical balances easily measure mass to a tenth of a milligram (0.0001 g) precision and
beyond. It is the measurement of volume which is difficult.
For liquids, several methods are available for measuring volume. Beakers and Erlenmeyer Flasks
are frequently graduated with volume markings. However, these markings should be considered
only approximate. Graduated Cylinders offer a significant improvement over Beakers and Flasks
for volume measurements. (Think about why Graduated Cylinders are tall and narrow instead of
short and fat. The design has specific implications for the precision of the device.) But once
again, the precision of this device is still fairly low. For work requiring significant accuracy, a
Volumetric Flask is required.
P a g e | 4
A pipette is a high precision volume measuring device used to transfer a given volume of liquid
from one container to another. Volumetric Pipettes generally have a large bulb in the middle
with a single mark along a narrow neck. By contrast, Measuring Pipettes are typically graduated
much like a Graduated Cylinder.
(Volumetric Pipettes)
This is the device we will use to make our needed volume measurements of liquids.
For very high precision density measurements, a pycnometer (fr. Greek puknos, meaning density)
is used to make the needed volume measurement. The pycnometer (one style is pictured below)
is calibrated by determining the mass of Water required to fill it at a given temperature. The
tabulated density of Water is then used to convert this mass to the volume of the device. The
device can then be used to measure out a known volume of the liquid.
http://www2.volstate.edu/chem/1110/Density.htm
For regular solids, a simple ruled measuring device (ruler, caliper, etc.) can be used to determine
the geometric parameters of the solid, and these parameters can be used to determine``` | 1,716 | 7,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-09 | longest | en | 0.844942 |
https://www.physicsforums.com/threads/ancient-egyptian-2-n-table-solved-4k-yrs-ago.161079/ | 1,702,173,206,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00745.warc.gz | 912,003,637 | 17,981 | # Ancient Egyptian 2/n-Table: Solved 4k Yrs Ago?
• DrKareem
In summary, the conversation discussed the Egyptian 2/n-table problem which involves writing 2/N as "egyptian fractions," and the conjecture that the Egyptians had a method for finding these fractions. The conversation also touched on the debate between using inverted integers versus regular fractions and the Erdos-Strauss conjecture. The participants expressed interest in exploring ancient mathematics and finding out more about the problem and its possible solutions.
#### DrKareem
One of my former teachers claimed that he had solved an ancient problem (4k y/o) called Egyptian 2/n-table. He told me (a while ago) that he submitted his proof to historia mathematica but i didn't communicate with him since and didn't know what happened next. I'm wondering if anyone could show me where i can find resources on the problem, and if anyone who has subscription to that journal if it is published yet or not. Cheers
What is the conjuncture exactly?
? This is not a conjuncture at all.
$$\frac{3}{2} + \frac{1}{2} = 2$$
Dividing both side by n, we get
$$\frac{3}{2n} + \frac{1}{2n} = \frac{2}{n}$$
Since $$3 | n$$, $$\frac{3}{2n} = \frac{1}{x}$$ where $$x$$ is some positive integer and $$x = \frac{2n}{3}$$. Letting $$2n = y$$ We have
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{n}$$, which is what needed to be shown.
Last edited:
The conjecture is not the formulas themselves, but trying to discover the method the Egyptians may have used to create them, and therefore to find out what mathematics they knew (not what we know).
Werg22 said:
? This is not a conjuncture at all.
$$\frac{3}{2} + \frac{1}{2} = 2$$
Dividing both side by n, we get
$$\frac{3}{2n} + \frac{1}{2n} = \frac{2}{n}$$
Since $$3 | n$$, $$\frac{3}{2n} = \frac{1}{x}$$ where $$x$$ is some positive integer and $$x = \frac{2n}{3}$$. Letting $$2n = y$$ We have
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{n}$$, which is what needed to be shown.
Can you explain why you are making the claim that 3|n? That is certainly not true for all n.
For some reason I had understood that it was meant for multiples of 3 only My bad.
robert Ihnot said:
This deals with writing 2/N as "egyptian fractions," i.e., as inverted positive integers. This situation deals with, for example, 2/3=1/2+1/6. See http://www.math.buffalo.edu/mad/Ancient-Africa/mad_ancient_egyptroll2-n.html.
The page that provided states that the conjecture is solved? (however, i still don't fully understand the conjecture yet). Anyways I'm still curious...
Dr.Kareem: The page that provided states that the conjecture is solved? (however, i still don't fully understand the conjecture yet).
My own partial understanding of this is that some people think that the Egyptians had a real method of finding these fractions. In the article itself is stated: There are nearly 30,000 ways of representing the numbers 2/q for 2< q < 102. For those where the formula is used, how was the number a chosen?
However, this matter can be looked at from the other side, some have been amazed that the Egyptians did not seem to understand our ordinary use of fractions such as a/p where 1<a<p, insisting generally upon resolving everything in terms of inverted integers. This view hardly suggests that the Egyptians had any really efficient method of working with fractions.
The other side seems to be that the builders of the pyramids were advanced and must have had some general and efficient method of finding, "Best Fractions," what ever that means, considering the multitude of choices. I assume the conjecture goes along the lines of this view.
There is some argument that in the case, for example, of distributing 5 sacks of grain among 8 people that it would be very difficult to give each person 5/8 of a sack, but it would be easier to give 1/2 sack per person, and then divide 1 sack into 8 parts, giving each person an additional 8th.
The Greeks at first adopted the Egyptian system, just as they did for Plane Geometry, but consider:
2/101 = 1/101 + 1/202 + 1/303 + 1/606. Well if I was to distribute 2 sacks among 101 people, I would just suppose that is about 1/50 a sack per person, and hope to have a little left over for the last person. Or if not, open up another sack to serve the last person.
NOTE: The "Erdos-Strauss conjecture" (ESC) is the statement that for any integer n > 1 there are integers a, b, and c with 4/n = 1/a + 1/b + 1/c; a > 0, b > 0, c > 0.
Last edited:
i love these systems, i'll have fun finding them all out. for fun i found a new twist on casting out in pi, maybe not new but still interesting.
all of the angles of a triangle are cast out
Last edited:
Thanks robert Ihnot for this explanation. I guess this is the first time i get exposed to ancient mathematics. I hope i can get the time to fiddle with them before my finals. I'll also try to contact my prof soon and see if his paper about them was published or not. | 1,320 | 4,958 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-50 | latest | en | 0.929433 |
https://www.reference.com/math/graph-inequalities-34317d79f3d54148 | 1,485,021,308,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281162.88/warc/CC-MAIN-20170116095121-00323-ip-10-171-10-70.ec2.internal.warc.gz | 967,902,976 | 20,792 | Q:
# How do you graph inequalities?
A:
### Quick Answer
The first step in graphing an inequality is to draw the line that would be obtained, if the inequality is an equation with an equals sign. The next step is to shade half of the graph.
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## Keep Learning
### Full Answer
To draw the line determined by the inequality, write the inequality with an equals sign. Then find two points on the line. An easy way to do this is to locate the x- and y-intercepts, if they exist. To find the x-intercept, set y = 0, solve the equation for x and plot the resulting point on the x-axis. If no x-intercept exists because the equation describes a horizontal line, find the y-intercept by setting x = 0. Then locate another point on the line, such as x = 2.
Once two points are located, draw a line through them. If the inequality uses the sign for "less than or equal to" or "greater than or equal to," the line should be solid. If the inequality uses the sign for "less than" or "greater than," the line should be dotted to indicate that points on the line are not included in the range of points specified by the inequality.
Finally, determine which part of the graph to shade. This is done by testing a point that is not on the line. For example, if the origin of the graph is not on the line, set x = 0 and y = 0 and see if the resulting inequality is true. If it is, the part of the graph that contains that point should be shaded. If the resulting inequality is false, the part of the graph on the other side of the line should be shaded.
Learn more about Data Graphs
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https://math.answers.com/Q/Is_132_a_prime_number. | 1,670,202,891,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711001.28/warc/CC-MAIN-20221205000525-20221205030525-00334.warc.gz | 424,233,883 | 47,151 | 0
Is 132 a prime number.
Wiki User
2016-08-15 15:54:24
No. 132 has the factors 2,2,3,11, i.e. 2x2x3x11 = 132
Prime numbers have only 1 and the Prime number as factors.
Wiki User
2016-08-15 15:54:24
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22 Reviews | 106 | 260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | latest | en | 0.850866 |
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What is Quantitative Aptitude - Arithmetic Ability?
Quantitative Aptitude - Arithmetic Ability test helps measure one's numerical ability, problem solving and mathematical skills. Quantitative aptitude - arithmetic ability is found in almost all the entrance exams, competitive exams and placement exams. Quantitative aptitude questions includes questions ranging from pure numeric calculations to critical arithmetic reasoning. Questions on graph and table reading, percentage analysis, categorization, simple interests and compound interests, clocks, calendars, Areas and volumes, permutations and combinations, logarithms, numbers, percentages, partnerships, odd series, problems on ages, profit and loss, ratio & proportions, stocks &shares, time & distance, time & work and more .
Every aspirant giving Quantitative Aptitude Aptitude test tries to solve maximum number of problems with maximum accuracy and speed. In order to solve maximum problems in time one should be thorough with formulas, theorems, squares and cubes, tables and many short cut techniques and most important is to practice as many problems as possible to find yourself some tips and tricks in solving quantitative aptitude - arithmetic ability questions.
Wide range of Quantitative Aptitude - Arithmetic Ability questions given here are useful for all kinds of competitive exams like Common Aptitude Test(CAT), MAT, GMAT, IBPS and all bank competitive exams, CSAT, CLAT, SSC Exams, ICET, UPSC, SNAP Test, KPSC, XAT, GRE, Defence, LIC/G IC, Railway exams,TNPSC, University Grants Commission (UGC), Career Aptitude test (IT companies), Government Exams and etc.
• Volume and Surface Area
Q:
Find the odd term from the given series of numbers ?
125, 127, 130, 135, 142, 153, 165 ?
A) 153 B) 165 C) 142 D) 127
Explanation:
From the beginning, the next term comes by adding prime numbers in a sequence of 2, 3, 5, 7, 9, 11, 13... to its previous term. But 165 will not be in the series as it must be replaced by 166 since 153+13 = 166.
46 240503
Q:
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A) 1/2 B) 3/5 C) 9/20 D) 8/15
Explanation:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E)/n(S) = 9/20.
425 240457
Q:
If selling price is doubled, the profit triples. Find the profit percent ?
A) 100% B) 200% C) 300% D) 400%
Explanation:
Let the C.P be Rs.100 and S.P be Rs.x, Then
The profit is (x-100)
Now the S.P is doubled, then the new S.P is 2x
New profit is (2x-100)
Now as per the given condition;
=> 3(x-100) = 2x-100
By solving, we get
x = 200
Then the Profit percent = (200-100)/100 = 100
Hence the profit percentage is 100%
1066 230859
Q:
Find the next number in te given series ?
5, 25, 61, 113, ...
A) 142 B) 181 C) 156 D) 179
Explanation:
Here the series follow the rule that
SO NEXT IS,
27 230091
Q:
Insert the missing number.
7, 26, 63, 124, 215, 342, (....)
A) 391 B) 421 C) 481 D) 511
Explanation:
Numbers are (23 - 1), (33 - 1), (43 - 1), (53 - 1), (63 - 1), (73 - 1) etc.
So, the next number is (83 - 1) = (512 - 1) = 511.
480 228775
Q:
A student multiplied a number by 3/5 instead of 5/3, What is the percentage error in the calculation ?
A) 54 % B) 64 % C) 74 % D) 84 %
Explanation:
Let the number be x.
Then, ideally he should have multiplied by x by 5/3. Hence Correct result was x * (5/3)= 5x/3.
By mistake he multiplied x by 3/5 . Hence the result with error = 3x/5
Then, error = (5x/3 - 3x/5) = 16x/15
Error % = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 %
1068 224849
Q:
What is the next number in this series ?
4, 6, 12, 18, 30, 42, 60, ?
A) 71 B) 72 C) 73 D) 69
Explanation:
With close observation, you will note that each number in the list is in the middle of two prime numbers. Thus:
4 is in the middle of 3 and 5, 6 is in the middle of 5 and 7, 12 is in the middle of 11 and 13, 18 is in the middle of 17 and 19, 30 is in the middle of 29 and 31. 42 is in the middle of 41 and 43, 60 is in the middle of 59 and 61.
Therefore, the next number would be the one that is in the middle of the next two prime numbers, which is 72 (which is in the middle of 71 and 73).
16 219315
Q:
Find the missing number in the following Series ?
18, 97, 26, 91, ? , 83
A) 32 B) 36 C) 30 D) 34
Explanation:
The given series follows the pattern that,
18, 97, 26, 91, ?, 83
=> 18 + 8 = 26
=> 97 - 8 = 91
=> 26 + 8 = 34
=> 91 - 8 = 83
Hence, the missing number is 34. | 1,458 | 4,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-49 | latest | en | 0.890638 |
https://askdev.io/questions/995010/complex-roots-of-xx-math-0 | 1,660,534,143,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00743.warc.gz | 136,634,096 | 9,873 | # Complex roots of $z^3 + \bar{z} = 0$
I'm searching for the intricate roots of $z^3 + \bar{z} = 0$ making use of De Moivre.
Some recommended increasing both sides by z first, yet that appears incorrect to me as it would certainly add a root (and also I would not recognize which root was the added ).
I saw that $z=a+bi$ and also there exists $\theta$ such that the trigonometric depiction of $z$ is $\left ( \sqrt{a^2+b^2} \right )\left ( \cos \theta + i \sin \theta \right )$.
It appears that $-\bar{z} = -\left ( \sqrt{a^2+b^2} \right )\left ( \cos (-\theta) + i \sin (-\theta) \right )$
However, my trig is rather corroded and also I'm not fairly certain where to go from below.
12
2022-07-25 20:39:40
Source Share
Write $z = re^{i\theta}$. Then you are trying to solve $r^3e^{3i\theta} + re^{-i\theta} = 0$, which is the same as $$r^3e^{3i\theta} = -re^{-i\theta}$$ Note that $-1 = e^{i\pi}$, so the above is equivalent to $$r^3e^{3i\theta} = re^{i(\pi - \theta)}$$ Comparing magnitudes, you have $r^3 = r$, which is solved by $r = 0$ and $1$, and comparing arguments you must have $3\theta = \pi - \theta + 2\pi k$ for some integer $k$ (when $r \neq 0$). Thus for some integer $k$ you have $$\theta = {\pi \over 4} + k{\pi \over 2}$$ There are four values of $\theta$ in $[0,2\pi)$ that satisfy this, namely ${\pi \over 4}, {3\pi \over 4}, {5\pi \over 4}$, and ${7\pi \over 4}$. Thus the complex numbers satisfying your original equation are $0, e^{i {\pi \over 4}}, e^{i {3\pi \over 4}}, e^{i {5\pi \over 4}}$, and $e^{i {7\pi \over 4}}$. In rectangular coordinates these are $0$ and $\pm {1 \over \sqrt{2}} \pm {i \over \sqrt{2}}$.
3
2022-07-25 22:31:45
Source
EDIT in view of the comments bellow by JimConant and PeterTaylor. If there is still any error the fault is mine.
This is an alternative solution to the trigonometric one. We will use the algebraic method. Let $z=x+iy$. We have $$\begin{eqnarray*} 0 &=&z^{3}+\overline{z} \\ 0 &=&\left( x+iy\right) ^{3}+\left( x-iy\right) \\ &=&x^{3}+x-3xy^{2}+i\left( 3x^{2}y-y^{3}-y\right) \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=x(x^{2}+1-3y^{2}) \\ 0=y(y^{2}+1-3x^{2}). \end{array} \right. \end{eqnarray*}\tag{1}$$
One of the roots is $$x_{1}=y_{1}=0.\tag{1a}$$ The remaining real roots satisfy the system
$$\begin{eqnarray*} \left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=y^{2}+1-3x^{2} \end{array} \right. &\Leftrightarrow &\left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=\frac{1}{3}+\frac{1}{3}x^{2}+1-3x^{2} \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=4-8x^{2} \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=1-2y^{2} \\ 0=1-2x^{2}. \end{array} \right. \end{eqnarray*} \tag{2}$$
The last system means that $$y=\pm x\tag{3}$$ and that
$$\begin{eqnarray*} x &=&\pm\frac{1}{2}\sqrt{2}, \\ y &=&\pm\frac{1}{2}\sqrt{2}. \end{eqnarray*}\tag{3a}$$
Combining the above results, we conclude that the following five complex numbers
$$z_{1} =0,\tag{4}$$ $$z_{2} =\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2},\quad z_{3} =-\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2}, \tag{5}$$ $$z_{4} =\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2},\quad z_{5} =-\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2}, \tag{6}$$
are the solutions of the given equation $$z^{3}+\overline{z}=0.\tag{7}$$
3
2022-07-25 22:24:22
Source
OK, I'm going to take a stab at this.
Given: $z^3 + \bar{z} = 0$
Therefore $z^3 = -\bar{z}$ and $|z^3| = |-\bar{z}|$ .
We have that $|-\bar{z}|=|\bar{z}|=|z|$ therefore $|z^3|=|z|$ and $|z||z||z|=|z|$.
Therefore $|z|=0$ or $|z|=1$.
Case 1: Assume $|z|=0$ then if
$z=a+bi\leftrightarrow|z|=\sqrt{a^2+b^2}=0\leftrightarrow a=0 \wedge b=0\leftrightarrow a+bi=0$
thus $z = 0$.
Case 2: Assume $|z|=1$
Let's multiply by $z$ and we get $z^4 = -z\bar{z}$. We see that $z\bar{z} = |z|^2$ so we get $z^4 = -( |z|^2 )$ so from the above either:
therefore $z^4 = -|z| = -1$
The trigonometric representation of $-1$ is $1*( \cos \pi + i \sin \pi )$ so according to De Moivre:
$z^4 = r^4(\cos 4\theta + i \sin 4\theta ) = 1*( \cos \pi + i \sin \pi )$
These are two complex numbers in trigonometric form so:
$r^4 = 1$ and $4\theta = \pi + 2\pi*k$ or
$r=1$ and $\theta = \frac{\pi + 2\pi*k}{4}$ and each solution has the form:
$z_k = \cos( \frac{\pi + 2\pi*k}{4} ) + i \sin (\frac{\pi + 2\pi*k}{4})$
for $0\leq k \leq 3$.
Which together with Case 1 gives the following values for $z$:
$0,\pm\frac{1+i}{\sqrt{2}},\pm\frac{1-i}{\sqrt{2}}$
1
2022-07-25 22:22:33
Source
That is a comment to the comment "should there be only 3 roots?" in Zarrax answer. Actually, the questions is "why isn´t there 9 roots?". This is the right question to ask since the intersection of the two curves in $\mathbb{R}^2$ $$x^3-3xy^2 +x = 0$$ $$3x^2y-y^3 -y = 0$$ is your solution set (just expand out $z^3+\bar{z}$). Now, the intersection of two degree 3 equations should have $3x3= 9$ solutions (by Bezout´s theorem). The 4 roots we are missing at "infinity" or in the $\mathbb{C}^2$ plane. We could apply the same thing to the function $z^2+z$. As complex polynomial, we should expect 2 roots. As the real system $(x,y) \rightarrow (x^2-y^2+x,2xy+y)$ we expect four real roots.
2
2022-07-25 21:39:11
Source | 2,115 | 5,218 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-33 | latest | en | 0.85325 |
https://kr.mathworks.com/matlabcentral/answers/1439069-how-to-patch-upperbound-and-lowerbound?s_tid=prof_contriblnk | 1,660,215,798,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00484.warc.gz | 350,021,747 | 27,162 | # How to patch upperbound and lowerbound
조회 수: 2(최근 30일)
alpedhuez 2021년 8월 23일
댓글: Star Strider 2021년 8월 23일
But I am not yet sure of how it works. (Partly due to the fact that https://www.mathworks.com/help/matlab/ref/patch.html only has graphic examples.)
Suppose I have
plot(T.date,upperbound)
hold on
plot(T.date,lowerbound)
Then how can one fill in upperbound and lowerbound?
##### 댓글 수: 1표시숨기기 없음
alpedhuez 2021년 8월 23일
I have tried https://www.mathworks.com/matlabcentral/answers/180829-shade-area-between-graphs but does not seem to work. Maybe it is because x axis is a datetimie variable?
댓글을 달려면 로그인하십시오.
### 채택된 답변
Star Strider 2021년 8월 23일
Try something like this —
T = table(datetime('now')+days(1:14).', rand(14,1), rand (14,1)+1, 'VariableNames',{'date','lowerbound','upperbound'})
T = 14×3 table
date lowerbound upperbound ____________________ __________ __________ 24-Aug-2021 15:54:18 0.3007 1.7723 25-Aug-2021 15:54:18 0.68026 1.703 26-Aug-2021 15:54:18 0.4398 1.6127 27-Aug-2021 15:54:18 0.62549 1.237 28-Aug-2021 15:54:18 0.44381 1.6517 29-Aug-2021 15:54:18 0.71737 1.3115 30-Aug-2021 15:54:18 0.41553 1.1505 31-Aug-2021 15:54:18 0.56936 1.405 01-Sep-2021 15:54:18 0.064302 1.329 02-Sep-2021 15:54:18 0.85725 1.5955 03-Sep-2021 15:54:18 0.15449 1.7525 04-Sep-2021 15:54:18 0.91195 1.8573 05-Sep-2021 15:54:18 0.10671 1.4731 06-Sep-2021 15:54:18 0.41436 1.6558
figure
patch([T.date; flipud(T.date)], [T.lowerbound; flipud(T.upperbound)], 'r', 'FaceAlpha',0.5)
grid
The patch function until recently did not work with datetime arrays, so it was necessary to use fill instead, with the same syntax as demonstrated here.
.
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Star Strider 2021년 8월 23일
As always, my pleasure!
.
댓글을 달려면 로그인하십시오.
R2021a
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A040656 Continued fraction for sqrt(683). 0
26, 7, 2, 4, 3, 1, 1, 25, 1, 1, 3, 4, 2, 7, 52, 7, 2, 4, 3, 1, 1, 25, 1, 1, 3, 4, 2, 7, 52, 7, 2, 4, 3, 1, 1, 25, 1, 1, 3, 4, 2, 7, 52, 7, 2, 4, 3, 1, 1, 25, 1, 1, 3, 4, 2, 7, 52, 7, 2, 4, 3, 1, 1, 25, 1, 1, 3, 4, 2, 7, 52, 7, 2, 4, 3, 1, 1, 25, 1, 1, 3, 4, 2, 7, 52, 7, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1). FORMULA a(n)=(1/1274)*{-3982*(n mod 14)-342*[(n+1) mod 14]+295*[(n+2) mod 14]+22*[(n+3) mod 14]-69*[(n+4) mod 14]+113*[(n+5) mod 14]+2297*[(n+6) mod 14]-2071*[(n+7) mod 14]+113*[(n+8) mod 14]+295*[(n+9) mod 14]+204*[(n+10) mod 14]-69*[(n+11) mod 14]+568*[(n+12) mod 14]+4208*[(n+13) mod 14]}-26*[C(2*n,n) mod 2], with n>=0. - Paolo P. Lava, May 18 2009 MAPLE with(numtheory): Digits := 300: convert(evalf(sqrt(683)), confrac); MATHEMATICA ContinuedFraction[Sqrt[683], 120] (* Harvey P. Dale, Jun 13 2011 *) CROSSREFS Sequence in context: A040657 A309073 A277215 * A070661 A040655 A163987 Adjacent sequences: A040653 A040654 A040655 * A040657 A040658 A040659 KEYWORD nonn,cofr,easy AUTHOR STATUS approved
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Last modified August 20 20:50 EDT 2019. Contains 326155 sequences. (Running on oeis4.) | 773 | 1,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-35 | latest | en | 0.505349 |
https://www.zigya.com/study/book?class=12&board=nbse&subject=Chemistry&book=Chemistry+I&chapter=Solutions&q_type=&q_topic=Colligative+Properties+and+Determination+of+Molar+Mass&q_category=&question_id=CHEN12043923 | 1,547,748,813,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659056.44/warc/CC-MAIN-20190117163938-20190117185938-00082.warc.gz | 1,010,038,089 | 16,934 | Why is vapour pressure of solution of glucose in water is lower than that of water? from Chemistry Solutions Class 12 Nagaland Board
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Why is vapour pressure of solution of glucose in water is lower than that of water?
The vapour pressure of pure solvent decrease when a non- volatile solute is added to the solvent this is because on adding the solute, a fewer number of water molecules are present at the surface which can evaporate as some of the surface area is occupied by non- volatile solute molecules thereby decreasing the vapour pressure of the solution thus the vapour pressure of the solution of the glucose in water lower than that of water.
136 Views
Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Mass % of benzene
Mass% of carbon tetrachloride = 100 - 15.28
= 84.72%
1703 Views
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg 0.25 of molal aqueous solution.
Solution:
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Mol. mass of urea ${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}$
= 14 + 2 + 12 + 16 + 14 + 2
=
Molality (m) =
or Moles of solute
= 0.25 x 0.25 = 0.625
Mass of urea
= Moles of solute x Molar mass
= 0.625 x 60 = 37.5 g
1475 Views
Calculate the molarity of each of the following solution (a) 30 g of Co(NO3)2.6H2O in 4.3 L solution (b) 30 mL of 0.5 MH2SO4 diluted to 500 mL.
solution;
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Moles of $\mathrm{Co}\left(\mathrm{NO}{\right)}_{3}.6{\mathrm{H}}_{2}\mathrm{O}$
Volume of solution = 4.3 L
Molarity,
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=$\frac{0.5×30}{1000}$mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M
844 Views
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
(a) 20% (mass/mass) means that 20 g of KI is present in 80 g of water.
Therefore, Moles of KI in solution
moles of KI = 20/166 =0.12mol
moles of water =80/18 =4.44mol
therefore, mole fraction of KI
=
1010 Views
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,
897 Views | 893 | 2,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-04 | latest | en | 0.863451 |
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If you have values on an Excel Worksheet that you need to permanently increase, or decrease you can use Paste Special.
Increase/Decrease Excel Values by Percentage
Let's say you have a list of values in A1:A100 and you need to increase these values all by 15%. Here is how;
1. Enter the number 1.15 into any blank cell and then Copy it
2. Now select the range A1:A100 and go to Edit>Paste Special
3. Choose Values from under Paste and then Multiply under Operation and click OK.
All value will now have increased by 15%
To decrease the values by 15% you would simply Enter =1 - 0.15 in any blank cell as apposed to 1.15. This would result in 0.85 which when used in steps 2 and 3 above, would result in values being decreased by 15%
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FREE Excel Help | 431 | 1,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-09 | latest | en | 0.81359 |
https://www.airmilescalculator.com/distance/bgr-to-mht/ | 1,670,457,247,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711221.94/warc/CC-MAIN-20221207221727-20221208011727-00397.warc.gz | 662,112,110 | 17,854 | # How far is Manchester, NH from Bangor, ME?
Distance between Bangor (Bangor International Airport) and Manchester (Manchester–Boston Regional Airport) is 184 miles / 295 kilometers / 160 nautical miles. Estimated flight time is 50 minutes.
Driving distance from Bangor (BGR) to Manchester (MHT) is 227 miles / 365 kilometers and travel time by car is about 4 hours 23 minutes.
184
Miles
295
Kilometers
160
Nautical miles
## Distance from Bangor to Manchester
There are several ways to calculate distance from Bangor to Manchester. Here are two common methods:
Vincenty's formula (applied above)
• 183.601 miles
• 295.477 kilometers
• 159.545 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 183.420 miles
• 295.186 kilometers
• 159.388 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Bangor to Manchester?
Estimated flight time from Bangor International Airport to Manchester–Boston Regional Airport is 50 minutes.
## What is the time difference between Bangor and Manchester?
There is no time difference between Bangor and Manchester.
## Flight carbon footprint between Bangor International Airport (BGR) and Manchester–Boston Regional Airport (MHT)
On average flying from Bangor to Manchester generates about 52 kg of CO2 per passenger, 52 kilograms is equal to 115 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Bangor to Manchester
Shortest flight path between Bangor International Airport (BGR) and Manchester–Boston Regional Airport (MHT).
## Airport information
Origin Bangor International Airport
City: Bangor, ME
Country: United States
IATA Code: BGR
ICAO Code: KBGR
Coordinates: 44°48′26″N, 68°49′41″W
Destination Manchester–Boston Regional Airport
City: Manchester, NH
Country: United States
IATA Code: MHT
ICAO Code: KMHT
Coordinates: 42°55′57″N, 71°26′8″W | 524 | 2,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | latest | en | 0.852435 |
https://www.coursehero.com/file/6207961/SomePracticeProblems/ | 1,516,273,241,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887224.19/warc/CC-MAIN-20180118091548-20180118111548-00507.warc.gz | 887,484,324 | 24,536 | SomePracticeProblems
# SomePracticeProblems - UNIVERSITY OF TEXAS AT AUSTIN EE...
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Unformatted text preview: UNIVERSITY OF TEXAS AT AUSTIN EE 351K - P ROBABILITY & R ANDOM P ROCESSES I FALL 2010 EXAM 1 WEDNESDAY, SEPTEMBER 29, 2010 Name: Email: • You have 75 minutes for this exam. • The exam is closed book and closed notes. You are allowed to have one standard letter-sized sheet, two sides, of handwritten notes. • Calculators, laptop computers, Palm Pilots, two-way e-mail pagers, etc. may not be used. • Write your answers in the spaces provided. • Please show all of your work. Answers without appropriate justification will receive very little credit. If you need extra space, use the back of the previous page. Problem 1 ( 16 pnts ): Problem 2 ( 12 pnts ): Problem 3 ( 12 pnts): Problem 1: (16 pnts) Alice has two boxes, blue and red, of chocolates; each box has n pieces of chocolate. Every day she eats one piece of chocolate. She picks this either from the blue box with probability p , or from the red box with probability 1- p ; the choices are independent from one day to the next. Let K denote the first day at which either one of the boxes first becomes empty – i.e. the day its last chocolate is eaten. Note that this empty box could be either blue or red. K is a random variable. Find the PMF (probability mass function) of K . Hint: What is the probability of the event { K = n + k }∩{ Blue box first to be empty } Problem 2: (12 pnts) There are n balls and n bins. For each ball, a random bin is chosen and the ball is thrown into that bin – all choices are independent. (a) (4 pnts) Let X be the number of bins with exactly one ball. Find E [ X ] . (b) (4 pnts) What is the probability that no bin is empty ? (c) (4 pnts) What is the probability that only one of the bins is occupied ? Problem 3: (12 pnts) A biological experiment starts with a single bacterium. At the end of one hour, it either splits into two with probability p , or dies with probability 1- p . If it splits, each of its children proceeds in the same way – at the end of one more hour, each one either splits into two w.p. p or dies w.p. 1- p . And so on for their children, and their children’s children ... (a) (6 pnts) Let X be the number of bacteria at the end of two hours. Find the PMF of X . (b) (6 pnts) Suppose N is the random number of bacteria at some time, with PMF p N ( n ) . Let M be the random number one hour later. Find an expression for the PMF p M ( m ) in terms of p N ( n ) ....
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Ask a homework question - tutors are online | 798 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-05 | latest | en | 0.883557 |
https://studylib.net/doc/6640694/leveled-indicators | 1,537,699,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159193.49/warc/CC-MAIN-20180923095108-20180923115508-00083.warc.gz | 623,260,524 | 12,797 | # Leveled Indicators
```Outcome SS6.2
Extend and apply understanding of perimeter of polygons, area of rectangles, and volume of right rectangular
prisms (concretely, pictorially, and symbolically) including:
relating area to volume
comparing perimeter and area
comparing area and volume
generalizing strategies and formulae
analyzing the effect of orientation
solving situational questions.
[CN, PS, R, V]
a) Generalize formulae and strategies for determining the perimeter of polygons, including rectangles and
squares.
b) Generalize a formula for determining the area of rectangles.
c) Explain, using models, the relationship between the area of the base of a right rectangular prism and
the volume of the same 3-D object.
d) Generalize a rule (formula) for determining the volume of right rectangular prisms.
e) Analyze the effect of orientation on the perimeter of polygons, area of rectangles, and volume of right
rectangular prisms.
f) Solve a situational question involving the perimeter of polygons, the area of rectangles, and/or the
volume of right rectangular prisms.
g) Critique the following statements using concrete or pictorial models:
“For any two right rectangular prisms, the one with the greater volume will be the prism that has
the greatest base area”.
“For any two rectangles, the rectangle with the greatest perimeter will also have the greatest
area”.
Level
Scale
Pre-Requisite
Knowledge
Descriptor
Students who are
not able to be
independently
successful with
level 1 questions
will be given an E.
Indicators
Student-Friendly
Language
of whole and decimal
numbers.
Linear measurement.
NN6 – Checkpoint 11
The student understands that a
measurement unit is an amount,
rather than an object, a shape,
or a mark on a scale.
NN6 – Checkpoint 12
The student uses multiplication
to find perimeter, area, and
volume.
1
B - Beginning
There is a partial
understanding of
some of the simpler
details and
processes.
Prior knowledge is
understood.
2
A–
Approaching
No major errors or
omissions regarding
the simpler details or
processes, but
assistance may be
required with the
complex processes.
3
M – Meeting
No major errors or
omissions regarding
any of the
information and/or
processes that were
explicitly taught.
This is the target
level for proficiency.
Knowledge and
Comprehension
Students who are
successful with
level 1 questions
or those who are
successful with
level 1 or 2
questions with
assistance will be
given a B.
Applying and
Analysing
Students who are
able to be
successful with
level 1 and level 2
questions, or
those who are
successful with
higher-level
questions with
assistance, will be
given an A.
Evaluating and
Creating
Students who are
independently
successful with
level 3 or level 4
questions are
given an M.
a) Generalize formulae and
strategies for determining
the perimeter of
polygons, including
rectangles and squares
I know how to find the
perimeter of any given
shape.
b) Generalize a formula for
determining the area of
rectangles.
c) Explain, using models, the
relationship between the
area of the base of a right
rectangular prism and the
volume of the same 3-D
object.
I understand the formula
for the area of a rectangle.
d) Generalize a rule
(formula) for determining
the volume of right
rectangular prisms.
e) Analyze the effect of
orientation on the
perimeter of polygons,
area of rectangles, and
volume of right
I understand the formula
for the volume of a
rectangular object.
I understand the
differences and
relationships between
perimeter and area.
I understand the
relationship between the
area of the base of an
object and the volume.
I know that no matter how
an object is sitting, the
perimeter, area and
volume will remain the
same.
rectangular prisms.
f) Solve a situational
question involving the
perimeter of polygons,
the area of rectangles,
and/or the volume of
right rectangular prisms.
g) Critique the following
statements using concrete or
pictorial models:
“For any two right
rectangular prisms,
the one with the
greater volume will be
the prism that has the
greatest base area”.
“For any two
rectangles, the
rectangle with the
greatest perimeter
will also have the
greatest area”.
4
performance, indepth inferences
and applications go
beyond what was
explicitly taught.
Students
successful at level
supplementary
to their
achievement in
I can solve problems
involving perimeter, area
and volume.
I can apply my knowledge
of perimeter, area, and
volume to real-life
problems or occurrences.
Outcome SS6.2
Extend and apply understanding of perimeter of polygons, area of rectangles, and volume of right rectangular
prisms (concretely, pictorially, and symbolically) including:
relating area to volume
comparing perimeter and area
comparing area and volume
generalizing strategies and formulae
analyzing the effect of orientation
solving situational questions.
Meeting
Approaching
I understand the
formula for the
area of a rectangle.
Beginning
I know how to find the
perimeter of any given
shape.
I understand the
differences and
relationships
between perimeter
and area.
I understand the
relationship
between the area
of the base of an
object and the
volume.
I understand the
formula for the
volume of a
rectangular
object.
I know that no
matter how an
object is sitting,
the perimeter,
area and volume
will remain the
same.
I can solve
problems
involving
perimeter, area
and volume.
I can apply my
knowledge of
perimeter, area,
and volume to
real-life problems
or occurrences.
``` | 1,400 | 5,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-39 | latest | en | 0.823744 |
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27 replies to this topic
### #1 bonanova
bonanova
bonanova
• Moderator
• 6141 posts
• Gender:Male
• Location:New York
Posted 23 March 2008 - 08:48 AM
My 1000th post.
Maybe I should give my keyboard a rest.
Anyway Happy Easter, and here's a little gem for you.
Take a number N, reverse its digits and add it to N.
Repeat if necessary. Eventually you will get a palindrome.
For example, starting with 39, we have 39 + 93 = 132.
Then 132 + 231 = 363 = palindrome.
After a moment's thought, this is not surprising.
But sometimes you do have to be patient.
Consider the starting number 89:
```89 ------> 159487405
187 | 664272356
968 | 1317544822
1837 | 3602001953
9218 | 7193004016
17347 | 13297007933
91718 | 47267087164
173437 | 93445163438
907808 | 176881317877
1716517 | 955594506548
8872688 | 1801200002107
17735476 | 8813200023188 = palindrome!
85189247 --->```
• 0
Vidi vici veni.
### #2 grey cells
grey cells
Senior Member
• Members
• 2807 posts
Posted 23 March 2008 - 09:43 AM
My 1000th post.
Maybe I should give my keyboard a rest.
Anyway Happy Easter, and here's a little gem for you.
Take a number N, reverse its digits and add it to N.
Repeat if necessary. Eventually you will get a palindrome.
For example, starting with 39, we have 39 + 93 = 132.
Then 132 + 231 = 363 = palindrome.
After a moment's thought, this is not surprising.
But sometimes you do have to be patient.
Consider the starting number 89:
```89 ------> 159487405
187 | 664272356
968 | 1317544822
1837 | 3602001953
9218 | 7193004016
17347 | 13297007933
91718 | 47267087164
173437 | 93445163438
907808 | 176881317877
1716517 | 955594506548
8872688 | 1801200002107
17735476 | 8813200023188 = palindrome!
85189247 --->```
HAPPY EASTER and HAPPY 1000th POST DAY BONONOVA.
And what a mind-boggling post that is.
• 0
### #3 neida
neida
• Members
• 148 posts
Posted 23 March 2008 - 03:16 PM
HAPPY EASTER and HAPPY 1000th POST DAY BONONOVA.
And what a mind-boggling post that is.
Nice post, but I would be amazed if anyone can prove this!!
There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.
For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.
• 0
### #4 grey cells
grey cells
Senior Member
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• 2807 posts
Posted 23 March 2008 - 03:28 PM
Nice post, but I would be amazed if anyone can prove this!!
There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.
For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.
Hi neida.I just wanted to compliment bononova on his 1000th post.You bet I don't intend to prove it .Atleast not now.But your challenge is definitely an interesting one . Maybe I can prove the contrary. But putting all maybe's aside , please post your answer on The palindromic puzzle of yours . It is frustrating to know the question and not the answer for 2 whole days.So please post your answer soon.Sorry for the initial lack of courtesy.(forgot to include please).
Edited by grey cells, 23 March 2008 - 03:33 PM.
• 0
### #5 sunrise
sunrise
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• Members
• 14 posts
Posted 23 March 2008 - 03:29 PM
Nice post, but I would be amazed if anyone can prove this!!
There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.
For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.
Wow!
I was looking for some proof by induction. I found out that some numbers can be expressed as sum of two numbers, n and n', such that n' is obtained by revesing the digits of n, and others which cannot be expressed as such. Both classes contain palindromes. 111 is a palindrome which cannot be expresses as sum of n and n'.
However, the above post has helped me in deciding to abandon this problem now.
• 0
### #6 chanakyavg
chanakyavg
Junior Member
• Members
• 23 posts
Posted 23 March 2008 - 04:07 PM
Nice post, but I would be amazed if anyone can prove this!!
There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.
For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.
Neida, I am not a Math major, the last time I read Math was almost 17/18 years ago. My knowledge rotates mostly around common sense.
I looked up what "Lychrel numbers" were and it showed up on Wikipedia. With respect to 196, they seem to stick with the defenition of how to rotate numbers and get palindromes, but with the rest they dont seem to. Can you explain why? (As i said, i am not a math major)
`A Lychrel number is a natural number which cannot form a palindrome through the iterative process of repeatedly reversing its base 10 digits and adding the resulting numbers.`
Example:
`# 56 becomes palindromic after one iteration: 56+65 = 121.# 57 becomes palindromic after two iterations: 57+75 = 132, 132+231 = 363.# 59 is not a Lychrel number since it becomes a palindrome after 3 iterations: 59+95 = 154, 154+451 = 605, 605+506 = 1111`
Now their Example for 196:
`196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986, 1495, 1497, 1585, 1587, 1675, 1677, 1765, 1767, 1855, 1857, 1945, 1947, 1997`
Should it not go like this instead?
`196 887 1675 7436....`
I know it amounts to the same, but not having studied Abstract Math, I do not understand this. please explain or point to a good source to read up on this.
Thanks. (PM me if u need my email address so we can take this offline. Thanks)
• 0
### #7 unreality
unreality
Senior Member
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Posted 23 March 2008 - 04:23 PM
Why would you take this offline? The whole point is to discuss it here
my idea:
if the number is made up of digits that in total add up to less than 10 (ie, add up to a single digit), than you get it in the first round:
2+1 = 3, which is a single digit, so:
21 + 12 = 33
4+1+1+2+1 = 9, single digit, so:
41121 + 12114 = 53235
1117 adds up to 10, but it doesnt necessarily mean it WONT be a palindrome on the first round:
1117 + 7111 = 8228
1218 + 8121 = 9339
So that leads to a more general (and obvious) rule, that if the corresponding flip-pairs each add up to a single digit, there is a palindrome in the first round, because there are no carry-overs
for example, take 72133641
its flip pairs are 7&1, 2&4, 1&6, and 3&3. All of those add up to single digits, so there are no carry-overs, hence a perfect palindrome after round 1
note you can take this number in two ways:
716
716 + 617 = 1333
or 0716 + 6170 = 6886
This can obviously be done to any number to manipulate it so that its flip-pairs all add up to single digits... so bonanova, what's your rule on zeroes?
btw congrats on post #1000 ;D
• 0
### #8 neida
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Posted 23 March 2008 - 06:05 PM
Now their Example for 196:
196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986, 1495, 1497, 1585, 1587, 1675, 1677, 1765, 1767, 1855, 1857, 1945, 1947, 1997
Should it not go like this instead?
196 887 1675 7436....
Hi chanakyavg - I've just looked at the wikipedia page and I think you've just misread the page a little. That first list of numbers above is provided as a list of suspected Lychrel numbers (or "candidates") rather than an example of how 196 progresses.
• 0
### #9 chanakyavg
chanakyavg
Junior Member
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Posted 23 March 2008 - 06:15 PM
Hi chanakyavg - I've just looked at the wikipedia page and I think you've just misread the page a little. That first list of numbers above is provided as a list of suspected Lychrel numbers (or "candidates") rather than an example of how 196 progresses.
Thanks, i read that page again after ur post, seems like I did mis-read it.
• 0
### #10 ALFRED
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Posted 23 March 2008 - 10:45 PM
bona - congrats on the 1,000th post.
I'm afraid I don't have much to contribute to this puzzle. I do however, have a theory I'd like to contribute to the thread in the hopes that someone else might be able to run with it. I do this while running the risk of giving away the answer to one of my own posts but seeing as how this one is so much more intriguing I think it's worth it.
I worked as an accountant for a little while after college and before I left that career path running and screaming I did pick up a little trick every good accountant uses. Whenever you're reconciling an account and you end up with a difference, the first thing you do is divide the difference by 9. If the answer is divisible by 9, there's a 99% chance the reason you're off is a transposition error. A lot of people (especially those who are good at number puzzles) outside the accounting world may already know of this math fact but I have also learned that if you ever have a coincidence between two numbers there's a 99% chance it's not a coincidence. And seeing as how this puzzle has a lot to do with number transpositions I can't help but think this little math fact might not just be a coincidence.
For example:
86 - 68 = 18/9 = 2
651 - 165 = 486/9 = 54
7824 - 8274 = -450/9 = -50
Notice it doesn't even matter what sort of transposition takes place, as long as all the digits of the original number are still present, the difference will be a multiple of 9.
Now I know the OP has a strict forward to backward transposition and the numbers are added to together instead of subtracted but like I said, I can't help but feel that this coincidence is more than a coincidence. I'd love to know if anyone has any thoughts on this.
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https://datascience.stackexchange.com/questions/41335/entropy-loss-from-collapsing-merging-two-categories/42255 | 1,638,899,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00410.warc.gz | 257,149,350 | 32,951 | # Entropy loss from collapsing/merging two categories
Suppose I am counting occurrences in a sequence. For a classical example, let's say I'm counting how many of each kind of car comes down a highway.
After keeping tally for a while, I see there are thousands of models. But only a handful show up frequently, whereas there are many that show up once or only a few times (iow the histogram resembles exponential decay). When thinking about the statistics of this situation, it hardly seems to matter that I saw this one obscure car this one time as opposed to another obscure car - it doesn't seem like it conveys much information either way. So what if I collapse all the rare models into a single category like "others", to make the data easier to store. How much information will I lose?
I've gotten as far as reducing the problem to a smaller one and finding an upper bound.
• Collapsing 3 categories A, B and C into one category D is the same as first collapsing categories A and B into category E, then collapsing E and C into F. F will be exactly the same as D. So the final information loss is not path-dependent, and it is sufficient for us to solve information loss from collapsing 2 categories. The result should easily generalize to n categories.
• For 2 categories, we can re-encode the sequence so that each occurrence of categories A and B is instead recorded as C. However, for each occurrence of C, an additional bit is recorded that shows whether that C came about from A or B. This re-encoding incurs zero information loss. Erasing these bits would effectively collapse A and B into C. Therefore the average information loss from collapsing categories A and B is (1 bit) * ((number of occurrences of A) + (number of occurrences of B)).
Is my logic above correct? Is my upper bound accurate? What is the lower bound/exact solution?
I ended up coming up with my own solution. I'm not sure if it's correct so I won't mark it as an answer.
Let's call the original distribution P. Suppose we will be collapsing A and B into a new category X. The entropy of P will be: $$S_P = - \sum P_i\ln P_i$$ where $$i$$ is A, B, C, ... Notably, $$P_A > 0, P_B > 0, P_X = 0$$ .
After collapsing, we get a new distribution $$Q$$. Now $$Q_A=Q_B=0$$ and $$Q_X =Q_A+Q_B$$. This will have entropy $$S_Q$$.
The entropy we lost by collapsing is $$S_P-S_Q = P_A \ln P_A + P_B \ln P_B - Q_X \ln Q_X$$. All other terms, such as $$P_C\ln P_C$$ for category $$C$$ which is unaffected, are present in both distributions and cancel out. | 614 | 2,533 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-49 | latest | en | 0.956193 |
http://www.4answered.com/questions/view/5ba9a/What39s-the-efficiency-and-quality-of-this-shuffling-algorithm | 1,534,514,784,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212323.62/warc/CC-MAIN-20180817123838-20180817143838-00095.warc.gz | 465,656,138 | 8,124 | ## What's the efficiency and quality of this shuffling algorithm?
Question!
This recent question about sorting randomly using C# got me thinking about the way I've sometimes shuffled my arrays in Perl.
``````@shuffled = sort { rand() <=> rand() } @array;
``````
The proposed solution in the mentioned question is Fisher-Yates shuffle, which works in a linear time.
The question is: how efficient is my snippet and is such shuffle "really" random?
By : Tuminoid
``````
@shuffled = map {
\$_->[1]
} sort {
\$a->[0] <=> \$b->[0]
} map {
[ rand(), \$_ ]
} @array;
``````
By : PP.
For one thing, you know that no matter the comparator you use sort() can't possibly be faster than O(n log n). So even if the shuffle it performs is fair, its performance will be worse.
So is the shuffle fair? It's obviously not fair for some (easy to analyze) sorting algorithms. Consider a simple bubble sort - in order for an element to move from one end to the other, the comparison function has to evaluate positive for n consecutive calls - a 1 in 2^n probability for what should be a 1 in n event. For quick sort, it's hard to analyze and it's possible that ends up being fair. But if it's important that it be right, do it the right way.
By : bmm6o
The perl documentation on `sort` says this
The comparison function is required to behave. If it returns inconsistent results (sometimes saying \$x[1] is less than \$x[2] and sometimes saying the opposite, for example) the results are not well-defined.
So it's a bad idea to do that.
ETA: I just did a benchmark. On a 100000 element array, using a FY-shuffle is more than 10 times faster too. | 405 | 1,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-34 | latest | en | 0.92595 |
https://www.mapleprimes.com/questions/217869-Why-Wouldnt-Subs-Work | 1,723,689,491,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00786.warc.gz | 684,450,370 | 24,087 | # Question:Why wouldn't subs work?
## Question:Why wouldn't subs work?
Maple 2016
Hello people in maple primes
I have a question, which is about the matrix shown in http://www.mapleprimes.com/questions/217852-HOW-I-Convert-Root-Of-In-To-Another-Common-Form
Why can't C below be shown with beta?
A := Matrix(3, 3, [[-a, a, 0], [0, 0, -sqrt(l*b*c*(j+k))/(j+k)], [2*j*sqrt(l*b*c*(j+k))/((j+k)*l), 2*k*sqrt(l*b*c*(j+k))/((j+k)*l), -c]]);
B:=subs(l*b*c*(j+k)=alpha,A);
C:=subs(j*alpha^(1/2) = beta,B);
e:=subs(alpha^(1/2) = gamma,B);
Best wishes.
taro
| 206 | 557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.745715 |
https://www.unitconverters.net/flow-molar/mol-second-to-gigamol-second.htm | 1,519,268,977,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813883.34/warc/CC-MAIN-20180222022059-20180222042059-00403.warc.gz | 986,038,625 | 3,280 | Home / Flow - Molar Conversion / Convert Mol/second to Gigamol/second
# Convert Mol/second to Gigamol/second
Please provide values below to convert mol/second [mol/s] to gigamol/second [Gmol/s], or vice versa.
From: mol/second To: gigamol/second
### Mol/second to Gigamol/second Conversion Table
Mol/second [mol/s]Gigamol/second [Gmol/s]
0.01 mol/s1.0E-11 Gmol/s
0.1 mol/s1.0E-10 Gmol/s
1 mol/s1.0E-9 Gmol/s
2 mol/s2.0E-9 Gmol/s
3 mol/s3.0E-9 Gmol/s
5 mol/s5.0E-9 Gmol/s
10 mol/s1.0E-8 Gmol/s
20 mol/s2.0E-8 Gmol/s
50 mol/s5.0E-8 Gmol/s
100 mol/s1.0E-7 Gmol/s
1000 mol/s1.0E-6 Gmol/s
### How to Convert Mol/second to Gigamol/second
1 mol/s = 1.0E-9 Gmol/s
1 Gmol/s = 1000000000 mol/s
Example: convert 15 mol/s to Gmol/s:
15 mol/s = 15 × 1.0E-9 Gmol/s = 1.5E-8 Gmol/s | 335 | 775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-09 | longest | en | 0.394014 |
https://www.coursehero.com/file/25350643/Lecture-22-S15-CBE-217pptx/ | 1,544,701,582,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824675.15/warc/CC-MAIN-20181213101934-20181213123434-00252.warc.gz | 838,635,953 | 127,272 | Lecture #22 S15 CBE 217.pptx
# Lecture #22 S15 CBE 217.pptx - CBE 217 Lecture#22 Energy...
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CBE 217 Lecture #22 – April 21, 2015 Energy balances 7.4-7.6 1
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Flow Work/Shaft Work Shaft work: done by the process fluid on a moving part within the system Flow work: work done by the fluid at the system outlet minus system inlet
Example An incompressible fluid flows through a horizontal straight pipe. Friction of the fluid within the pipe causes a small amount of heat to be transferred from the fluid. To compensate, flow work must be done on the fluid to move it through the system (so that ) 1. How are V in and V out related? 2. How are P in and P out related?
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Specific Properties Intensive properties: Extensive properties: Specific properties: an intensive property obtained by dividing an extensive property by a total amount Specific volume Specific enthalpy
Specific Enthalpy Example The specific internal energy of helium at a given 300K and 1 atm is 3800 J/mole. Calculate the specific enthalpy, knowing that the specific molar volume at the T and P is 24.63 L/mole.
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Yo, let me tell you something about how zoologists use stats to make sense of their data 🐵📈. First of all, zoologists collect a ton of data about animals and their behavior. We’re talking about everything from the size of an animal’s habitat to the number of times it blinks in a minute. That’s a … Read more
## WHAT ARE SOME STATISTICAL METHODS THAT STUDENTS CAN USE TO ANALYZE THEIR RESULTS
Yo, students! 🙌🏼 If you want to analyze your results statistically, there are a bunch of methods you can use. Let me break down a few for you. First off, we’ve got the mean. This is just the average of all your data points. You add ’em up and divide by the number of data … Read more
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## ONE CRITICAL THINKING STRATEGY OR TOOL TO ANALYZE THE SITUATION
Critical thinking is a skill that is highly valued in today’s society. It involves the ability to analyze information, evaluate arguments, and make reasoned judgments. It is a process that helps individuals to identify and challenge assumptions, and to consider alternative perspectives. Critical thinking is not just about being skeptical, but about being open-minded and … Read more
## CAN YOU PROVIDE AN EXAMPLE OF HOW THE NEOCLASSICAL MODEL IS USED TO ANALYZE ECONOMIC POLICIES
The neoclassical model is a theoretical framework used in economics that emphasizes the role of markets and individual behavior in determining economic outcomes. It assumes that individuals are rational and seek to maximize their utility or satisfaction, and that markets are efficient in allocating resources to their most productive uses. This model has been used … Read more
## HOW DOES GAME THEORY HELP ECONOMISTS ANALYZE OLIGOPOLIES
Game theory is a powerful tool that economists use to analyze the behavior of firms in oligopolistic markets. Oligopolies are markets characterized by a small number of firms that dominate the market. In these markets, firms have a great deal of market power, which they can use to influence prices and output levels. Game theory … Read more
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Crime data analysis is an essential task for law enforcement agencies, policymakers, and researchers. Statistical techniques are commonly used to analyze crime data to identify patterns, trends, and potential risk factors. In this answer, we will discuss some of the statistical techniques used to analyze crime data. Descriptive Statistics: Descriptive statistics are used to summarize … Read more
## WHAT ARE SOME COMMON MATHEMATICAL TOOLS AND TECHNIQUES USED TO ANALYZE MODELS
Mathematical models are used to represent complex systems and phenomena, whether they are physical, biological, social or economic. To analyze these models, mathematical tools and techniques are applied to extract meaningful insights and predictions from the model. In this article, we will discuss some of the most common mathematical tools and techniques used to analyze … Read more
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## CAN YOU PROVIDE AN EXAMPLE OF HOW THE OFFER CURVE CAN BE USED TO ANALYZE THE TERMS OF TRADE
The offer curve is a graphical representation of the different quantities of a good that a country is willing to export at various prices. It is used in international trade theory to analyze the terms of trade between two countries. The terms of trade refer to the ratio at which a country’s exports exchange for … Read more | 905 | 4,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-23 | latest | en | 0.884882 |
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Learn measures of dispersion test MCQs: if arithmetic mean is considered as average of deviations then resultant measure is considered as, with choices close end deviation, mean absolute deviation, mean deviation, and variance deviation for bachelor of business administration. Practice merit scholarships assessment test, online learning average deviation measures quiz questions for competitive assessment in business majors .
MCQ: If arithmetic mean is considered as average of deviations then resultant measure is considered as
1. close end deviation
2. mean absolute deviation
3. mean deviation
4. variance deviation
B
MCQ: If set of observations is 11, 13, 15, 12, 16, 18, 19, 14, 20, 17 and absolute mean deviation is 12 then percentage of coefficient of mean absolute deviation is
1. 47.41%
2. 57.41%
3. 67.41%
4. 77.41%
D
MCQ: Mean absolute deviation is 5 and arithmetic mean is 110 then coefficient of mean absolute deviation is
1. 1.054
2. 0.045
3. 0.054
4. 0.064
B
MCQ: Number of patients who visited cardiologists are as 63, 57, 51, 65 in four days then absolute mean deviation (approximately) is
1. 8 patents
2. 4 patients
3. 10 patients
4. 15 patients
B
MCQ: If in a formula, mean absolute deviation is numerator and arithmetic mean is denominator then resultant value is classified as
1. coefficient of mean deviation
2. coefficient of absolute quartile deviation
3. coefficient of quartile range deviation
4. coefficient of mean absolute deviation
D | 452 | 1,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-30 | latest | en | 0.858641 |
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# help with physical science questions
Asked May 25, 2012, 04:26 PM — 2 Answers
The amount of power measured for a 120 lb person running up 10ft (w = 1200 lb x ft) of stairs over a 6 second time frame is
2 Answers
Thato maine Posts: 1, Reputation: 1 New Member #2 Jun 26, 2012, 07:45 PM
A box rests at the bottom of a 30degree inclined plane. A constant force of 245N moves the box at constant speed 5m higher up the smooth plane. What is the weight of the box ?
ebaines Posts: 10,136, Reputation: 5589 Expert #3 Jun 27, 2012, 06:11 AM
Quote:
Originally Posted by squirrel_1 The amount of power measured for a 120 lb person running up 10ft (w = 1200 lb x ft) of stairs over a 6 second time frame is
Power is equal to work per unit time. You already know the work done - so simply divide by the time takesn to get power in terms of lb-ft/second.
Quote:
Originally Posted by Thato maine A box rests at the bottom of a 30degree inclined plane. A constant force of 245N moves the box at constant speed 5m higher up the smooth plane. What is the weight of the box ?
Thato - please don't tag a new question onto an old thread, but rather you should post a new question.
The force applied to the box must counteract the weight of the box in order for it to move at constant velocity - meaning no acceleration so the sum of forces acting on the box =ma = 0. Can you tell us the component of the box's weight resisting movement in the direction of the inclined plane? This must equal the force being applied - then you can set up a equation and solve for 'm'.
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What is the kinetic energy of an object that has a mass of 12 kilograms and moves With a velocity of 10 m/s? | 607 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2013-20 | latest | en | 0.888535 |
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# Three convex lens problem
## Recommended Posts
# Three convex lens of focal length 10 cm each are mounted coaxially as shown in figure. An object O is placed at a distance of 20 cm from L1 and the final image formed is at a distance of 20 cm from L3. What is the distance L1L2?
Please help me with this problem. If the distance L2L3 was given this problem could be easily solved. But without that value, how to solve this problem?
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If the distance L2L3 was given this problem could be easily solved. But without that value, how to solve this problem?
Looking at the problem, there is only one position L2 can be: Exactly half way between L1 and L3 by symetry. Thus, L1L2 = L2L3.
If you need any more help, let me know.
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"as shown in the figure"...
perhaps you can post a figure here?
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"as shown in the figure"...
perhaps you can post a figure here?
i am sorry, i typed it by mistake.
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Looking at the problem, there is only one position L2 can be: Exactly half way between L1 and L3 by symetry. Thus, L1L2 = L2L3.
If you need any more help, let me know.
Is there an expression to find the equivalent focal length of 3 lenses placed coaxially?
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if im not mistaken, you can probably calculate the focal length of the first 2 lens, then treat it as one lens, and calculate the focal length combining with the third one.
this is probably the equation you'll need:
http://scienceworld.wolfram.com/physics/ThinLensDoublet.html
as optics is not my area of physics....
well, probably you'll need some algebraic manipulations of this equation.
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# AM3.12345 - Chapter 3 Life Insurance Life insurance is a contract(policy between a life insurance company(insurer and an individual(policyholder or
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Chapter 3 – Life Insurance Life insurance is a contract (policy) between a life insurance company (insurer) and an individual (policyholder or insured) The policyholder agrees to make a single payment or a series of payments to the insurer (these payments are called gross premiums) In return for these premiums, the insurer agrees to pay a fixed sum (called a death benefit) upon the death of the insured, payable to a designated person (this person is called the beneficiary)
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In this chapter, we will be interested in calculating the actuarial present value of future benefits: NSP = apv of future benefits The NSP will be calculated in such a way as to be necessary and sufficient to provide the stated future benefits to the insured, given the following underlying assumptions:
1. The death benefit is paid at the end of the year in which death occurs 2. Premiums are immediately invested by the insurer and earn interest at exactly the assumed rate of interest 3. Deaths among policyholders occur exactly according to the adopted mortality table
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Whole Life Insurance (section 3.2) A whole life insurance is issued to ( x ) and it provides a payment of \$1 at the end of the year in which ( x ) dies, regardless of when death occurs actuarial symbol: A x = apv of future benefit
Values of 1000 A x are given in Table 2 Example 3.2.1 Determine the NSP for \$100,000 worth of whole life insurance issued to a man age (a) 20 (b) 40. Solution to 3.2.1
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## This note was uploaded on 04/14/2010 for the course ACSCI 2053 taught by Professor Kopp during the Spring '09 term at UWO.
### Page1 / 26
AM3.12345 - Chapter 3 Life Insurance Life insurance is a contract(policy between a life insurance company(insurer and an individual(policyholder or
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Ask a homework question - tutors are online | 566 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-13 | longest | en | 0.921721 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_INN/TRS/Zantema/z21.trs.Thm26:POLO_7172_DP:LIM.html.lzma | 1,718,208,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00856.warc.gz | 96,249,380 | 2,513 | Term Rewriting System R:
[x, y, z]
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
F(f(a, b), x) -> F(a, f(a, x))
F(f(a, b), x) -> F(a, x)
F(f(b, a), x) -> F(b, f(b, x))
F(f(b, a), x) -> F(b, x)
F(x, f(y, z)) -> F(f(x, y), z)
F(x, f(y, z)) -> F(x, y)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Forward Instantiation Transformation`
Dependency Pairs:
F(f(b, a), x) -> F(b, x)
F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, x)
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), x) -> F(a, f(a, x))
Rules:
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
F(f(a, b), x) -> F(a, x)
one new Dependency Pair is created:
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳Forward Instantiation Transformation`
Dependency Pairs:
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(b, a), x) -> F(b, x)
Rules:
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
F(f(b, a), x) -> F(b, x)
one new Dependency Pair is created:
F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳FwdInst`
` ...`
` →DP Problem 3`
` ↳Forward Instantiation Transformation`
Dependency Pairs:
F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))
F(f(b, a), x) -> F(b, f(b, x))
F(x, f(y, z)) -> F(x, y)
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
Rules:
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Strategy:
innermost
On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule
F(x, f(y, z)) -> F(x, y)
five new Dependency Pairs are created:
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(b, a), f(y', z)) -> F(f(b, a), y')
F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))
F(f(b, a), f(f(y'''', z''''), z)) -> F(f(b, a), f(y'''', z''''))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳FwdInst`
` ...`
` →DP Problem 4`
` ↳Polynomial Ordering`
Dependency Pairs:
F(f(b, a), f(f(y'''', z''''), z)) -> F(f(b, a), f(y'''', z''''))
F(f(b, a), f(y', z)) -> F(f(b, a), y')
F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
F(f(b, a), x) -> F(b, f(b, x))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))
Rules:
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Strategy:
innermost
The following dependency pairs can be strictly oriented:
F(f(b, a), f(f(y'''', z''''), z)) -> F(f(b, a), f(y'''', z''''))
F(f(b, a), f(y', z)) -> F(f(b, a), y')
F(f(a, b), f(f(y'''', z''''), z)) -> F(f(a, b), f(y'''', z''''))
F(f(a, b), f(y', z)) -> F(f(a, b), y')
F(f(a, b), f(y'', z'')) -> F(a, f(y'', z''))
F(x'', f(f(y'', z''), z)) -> F(x'', f(y'', z''))
F(f(b, a), f(y'', z'')) -> F(b, f(y'', z''))
Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(b) = 0 POL(a) = 0 POL(f(x1, x2)) = 1 + x1 + x2 POL(F(x1, x2)) = 1 + x1 + x2
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳FwdInst`
` ...`
` →DP Problem 5`
` ↳Instantiation Transformation`
Dependency Pairs:
F(f(b, a), x) -> F(b, f(b, x))
F(f(a, b), x) -> F(a, f(a, x))
F(x, f(y, z)) -> F(f(x, y), z)
Rules:
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Strategy:
innermost
On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule
F(x, f(y, z)) -> F(f(x, y), z)
two new Dependency Pairs are created:
F(a, f(y', z')) -> F(f(a, y'), z')
F(b, f(y', z')) -> F(f(b, y'), z')
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳FwdInst`
` →DP Problem 2`
` ↳FwdInst`
` ...`
` →DP Problem 6`
` ↳Remaining Obligation(s)`
The following remains to be proven:
Dependency Pairs:
F(a, f(y', z')) -> F(f(a, y'), z')
F(f(a, b), x) -> F(a, f(a, x))
F(b, f(y', z')) -> F(f(b, y'), z')
F(f(b, a), x) -> F(b, f(b, x))
Rules:
f(f(a, b), x) -> f(a, f(a, x))
f(f(b, a), x) -> f(b, f(b, x))
f(x, f(y, z)) -> f(f(x, y), z)
Strategy:
innermost
Innermost Termination of R could not be shown.
Duration:
0:00 minutes | 2,292 | 5,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.588681 |
https://www.lmfdb.org/EllipticCurve/Q/87120/fn/ | 1,632,071,862,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00139.warc.gz | 885,804,025 | 13,994 | Properties
Label 87120.fn Number of curves $4$ Conductor $87120$ CM no Rank $1$ Graph
Related objects
Show commands: SageMath
sage: E = EllipticCurve("fn1")
sage: E.isogeny_class()
Elliptic curves in class 87120.fn
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
87120.fn1 87120fz3 $$[0, 0, 0, -45012, 3674891]$$ $$488095744/125$$ $$2582935938000$$ $$[2]$$ $$207360$$ $$1.3676$$
87120.fn2 87120fz4 $$[0, 0, 0, -39567, 4597274]$$ $$-20720464/15625$$ $$-5165871876000000$$ $$[2]$$ $$414720$$ $$1.7142$$
87120.fn3 87120fz1 $$[0, 0, 0, -1452, -14641]$$ $$16384/5$$ $$103317437520$$ $$[2]$$ $$69120$$ $$0.81830$$ $$\Gamma_0(N)$$-optimal
87120.fn4 87120fz2 $$[0, 0, 0, 3993, -98494]$$ $$21296/25$$ $$-8265395001600$$ $$[2]$$ $$138240$$ $$1.1649$$
Rank
sage: E.rank()
The elliptic curves in class 87120.fn have rank $$1$$.
Complex multiplication
The elliptic curves in class 87120.fn do not have complex multiplication.
Modular form 87120.2.a.fn
sage: E.q_eigenform(10)
$$q + q^{5} + 2q^{7} - 2q^{13} - 6q^{17} - 4q^{19} + O(q^{20})$$
Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rrrr} 1 & 2 & 3 & 6 \\ 2 & 1 & 6 & 3 \\ 3 & 6 & 1 & 2 \\ 6 & 3 & 2 & 1 \end{array}\right)$$
Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels. | 620 | 1,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-39 | latest | en | 0.524317 |
http://au.metamath.org/mpegif/cdlemk37.html | 1,521,872,078,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649931.17/warc/CC-MAIN-20180324054204-20180324074204-00659.warc.gz | 29,175,536 | 14,611 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cdlemk37 Unicode version
Theorem cdlemk37 31327
Description: Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 18-Jul-2013.)
Hypotheses
Ref Expression
cdlemk4.b
cdlemk4.l
cdlemk4.j
cdlemk4.m
cdlemk4.a
cdlemk4.h
cdlemk4.t
cdlemk4.r
cdlemk4.z
cdlemk4.y
cdlemk4.x
Assertion
Ref Expression
cdlemk37
Distinct variable groups: ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,, ,
Allowed substitution hints: (,) () (,)
Proof of Theorem cdlemk37
StepHypRef Expression
1 cdlemk4.b . . 3
2 cdlemk4.l . . 3
3 cdlemk4.j . . 3
4 cdlemk4.m . . 3
5 cdlemk4.a . . 3
6 cdlemk4.h . . 3
7 cdlemk4.t . . 3
8 cdlemk4.r . . 3
9 cdlemk4.z . . 3
10 cdlemk4.y . . 3
11 cdlemk4.x . . 3
121, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11cdlemk36 31326 . 2
13 simp11l 1068 . . . . 5
14 hllat 29777 . . . . 5
1513, 14syl 16 . . . 4
16 simp22l 1076 . . . . 5
17 simp11 987 . . . . . 6
18 simp13l 1072 . . . . . 6
19 simp13r 1073 . . . . . 6
201, 5, 6, 7, 8trlnidat 30586 . . . . . 6
2117, 18, 19, 20syl3anc 1184 . . . . 5
221, 3, 5hlatjcl 29780 . . . . 5
2313, 16, 21, 22syl3anc 1184 . . . 4
24 simp3l 985 . . . . . . . . 9
25 simp3r1 1065 . . . . . . . . 9
261, 5, 6, 7, 8trlnidat 30586 . . . . . . . . 9
2717, 24, 25, 26syl3anc 1184 . . . . . . . 8
281, 3, 5hlatjcl 29780 . . . . . . . 8
2913, 16, 27, 28syl3anc 1184 . . . . . . 7
30 simp21 990 . . . . . . . . . 10
312, 5, 6, 7ltrnat 30553 . . . . . . . . . 10
3217, 30, 16, 31syl3anc 1184 . . . . . . . . 9
331, 5atbase 29703 . . . . . . . . 9
3432, 33syl 16 . . . . . . . 8
35 simp12l 1070 . . . . . . . . . . 11
366, 7ltrncnv 30559 . . . . . . . . . . 11
3717, 35, 36syl2anc 643 . . . . . . . . . 10
386, 7ltrnco 31132 . . . . . . . . . 10
3917, 24, 37, 38syl3anc 1184 . . . . . . . . 9
401, 6, 7, 8trlcl 30577 . . . . . . . . 9
4117, 39, 40syl2anc 643 . . . . . . . 8
421, 3latjcl 14430 . . . . . . . 8
4315, 34, 41, 42syl3anc 1184 . . . . . . 7
441, 4latmcl 14431 . . . . . . 7
4515, 29, 43, 44syl3anc 1184 . . . . . 6
469, 45syl5eqel 2486 . . . . 5
476, 7ltrncnv 30559 . . . . . . . 8
4817, 24, 47syl2anc 643 . . . . . . 7
496, 7ltrnco 31132 . . . . . . 7
5017, 18, 48, 49syl3anc 1184 . . . . . 6
511, 6, 7, 8trlcl 30577 . . . . . 6
5217, 50, 51syl2anc 643 . . . . 5
531, 3latjcl 14430 . . . . 5
5415, 46, 52, 53syl3anc 1184 . . . 4
551, 2, 4latmle1 14456 . . . 4
5615, 23, 54, 55syl3anc 1184 . . 3
5710, 56syl5eqbr 4203 . 2
5812, 57eqbrtrd 4190 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wa 359 w3a 936 wceq 1649 wcel 1721 wne 2565 wral 2664 class class class wbr 4170 cid 4451 ccnv 4834 cres 4837 ccom 4839 cfv 5411 (class class class)co 6038 crio 6499 cbs 13420 cple 13487 cjn 14352 cmee 14353 clat 14425 catm 29677 chlt 29764 clh 30397 cltrn 30514 ctrl 30571 This theorem is referenced by: cdlemk38 31328 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1552 ax-5 1563 ax-17 1623 ax-9 1662 ax-8 1683 ax-13 1723 ax-14 1725 ax-6 1740 ax-7 1745 ax-11 1757 ax-12 1946 ax-ext 2383 ax-rep 4278 ax-sep 4288 ax-nul 4296 ax-pow 4335 ax-pr 4361 ax-un 4658 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3or 937 df-3an 938 df-tru 1325 df-ex 1548 df-nf 1551 df-sb 1656 df-eu 2256 df-mo 2257 df-clab 2389 df-cleq 2395 df-clel 2398 df-nfc 2527 df-ne 2567 df-nel 2568 df-ral 2669 df-rex 2670 df-reu 2671 df-rmo 2672 df-rab 2673 df-v 2916 df-sbc 3120 df-csb 3210 df-dif 3281 df-un 3283 df-in 3285 df-ss 3292 df-nul 3587 df-if 3698 df-pw 3759 df-sn 3778 df-pr 3779 df-op 3781 df-uni 3974 df-iun 4053 df-iin 4054 df-br 4171 df-opab 4225 df-mpt 4226 df-id 4456 df-xp 4841 df-rel 4842 df-cnv 4843 df-co 4844 df-dm 4845 df-rn 4846 df-res 4847 df-ima 4848 df-iota 5375 df-fun 5413 df-fn 5414 df-f 5415 df-f1 5416 df-fo 5417 df-f1o 5418 df-fv 5419 df-ov 6041 df-oprab 6042 df-mpt2 6043 df-1st 6306 df-2nd 6307 df-undef 6500 df-riota 6506 df-map 6977 df-poset 14354 df-plt 14366 df-lub 14382 df-glb 14383 df-join 14384 df-meet 14385 df-p0 14419 df-p1 14420 df-lat 14426 df-clat 14488 df-oposet 29590 df-ol 29592 df-oml 29593 df-covers 29680 df-ats 29681 df-atl 29712 df-cvlat 29736 df-hlat 29765 df-llines 29911 df-lplanes 29912 df-lvols 29913 df-lines 29914 df-psubsp 29916 df-pmap 29917 df-padd 30209 df-lhyp 30401 df-laut 30402 df-ldil 30517 df-ltrn 30518 df-trl 30572
Copyright terms: Public domain W3C validator | 2,559 | 4,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-13 | latest | en | 0.111776 |
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All articles SUM How to do Sumif by Year
# How to do Sumif by Year
We can use the SUMIFS Function to sum selected cells based on the values of the particular year we wish to add. The steps below will walk through the process.
Figure 1: How to do Sum if by Year
## Syntax
`=SUMIFS(sum_range,date_range,”>=”&DATE(year,1,1),date_range,”<=”&DATE(year,12,31)`
## Formula
`=SUMIFS(\$B\$4:\$B\$11,\$A\$4:\$A\$11,">="&DATE(D4,1,1),\$A\$4:\$A\$11,"<="&DATE(D4,12,31))`
## Setting up the Data
• We will set up the data by inputting the Dates into Column A
• We will input the expenditure for each day in Column B
• Column D contains the years where expenditure was incurred
• Column E is where the formula will return the Total expenditure for 2017, 2018, and 2019
Figure 2: Setting up the Data
## Sumif by Year
• We will click on Cell E4
• We will insert the formula below into the cell
`=SUMIFS(\$B\$4:\$B\$11,\$A\$4:\$A\$11,">="&DATE(D4,1,1),\$A\$4:\$A\$11,"<="&DATE(D4,12,31))`
• We will press the enter key
Figure 3: Sum of Expenditure for 2017
• We will click on Cell E4 again
• We will double-click on the fill handle (the small plus sign at the bottom right of Cell E4) and drag down to copy the formula into the other cells
Figure 4: Total Expenditure for 2017, 2018, and 2019
## Explanation
`=SUMIFS(\$B\$4:\$B\$11,\$A\$4:\$A\$11,">="&DATE(D4,1,1),\$A\$4:\$A\$11,"<="&DATE(D4,12,31))`
`=SUMIFS(sum_range,date_range,”>=”&DATE(year,1,1),date_range,”<=”&DATE(year,12,31)`
In this formula, the First DATE section is use to instruct that only cells greater than or equal to the DATE specified in Cell D4 are summed. This date is 2017/1/1, where 2017 is the year, 1 is the month, and the last 1 is the day.
For the Last Date section in the formula, it instructs that only cells lesser than or equal to the DATE specified in Cell D4 are summed. This date is 2017/12/31.
After meeting the criteria for the date_range (\$A\$4:\$A\$11), the values in the adjacent cells in Column B are searched for and summed.
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https://www.jobilize.com/calculus/course/5-4-integration-formulas-and-the-net-change-theorem-by-openstax?qcr=www.quizover.com&page=6 | 1,600,485,429,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189928.2/warc/CC-MAIN-20200919013135-20200919043135-00369.warc.gz | 943,419,692 | 23,016 | # 5.4 Integration formulas and the net change theorem (Page 7/8)
Page 7 / 8
For a given motor vehicle, the maximum achievable deceleration from braking is approximately 7 m/sec 2 on dry concrete. On wet asphalt, it is approximately 2.5 m/sec 2 . Given that 1 mph corresponds to 0.447 m/sec, find the total distance that a car travels in meters on dry concrete after the brakes are applied until it comes to a complete stop if the initial velocity is 67 mph (30 m/sec) or if the initial braking velocity is 56 mph (25 m/sec). Find the corresponding distances if the surface is slippery wet asphalt.
In dry conditions, with initial velocity ${v}_{0}=30$ m/s, $D=64.3$ and, if ${v}_{0}=25,D=44.64.$ In wet conditions, if ${v}_{0}=30,$ and $D=180$ and if ${v}_{0}=25,D=125.$
John is a 25-year old man who weighs 160 lb. He burns $500-50t$ calories/hr while riding his bike for t hours. If an oatmeal cookie has 55 cal and John eats 4 t cookies during the t th hour, how many net calories has he lost after 3 hours riding his bike?
Sandra is a 25-year old woman who weighs 120 lb. She burns $300-50t$ cal/hr while walking on her treadmill. Her caloric intake from drinking Gatorade is 100 t calories during the t th hour. What is her net decrease in calories after walking for 3 hours?
225 cal
A motor vehicle has a maximum efficiency of 33 mpg at a cruising speed of 40 mph. The efficiency drops at a rate of 0.1 mpg/mph between 40 mph and 50 mph, and at a rate of 0.4 mpg/mph between 50 mph and 80 mph. What is the efficiency in miles per gallon if the car is cruising at 50 mph? What is the efficiency in miles per gallon if the car is cruising at 80 mph? If gasoline costs $3.50/gal, what is the cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at 80 mph? Although some engines are more efficient at given a horsepower than others, on average, fuel efficiency decreases with horsepower at a rate of $1\text{/}25$ mpg/horsepower. If a typical 50-horsepower engine has an average fuel efficiency of 32 mpg, what is the average fuel efficiency of an engine with the following horsepower: 150, 300, 450? $E\left(150\right)=28,E\left(300\right)=22,E\left(450\right)=16$ [T] The following table lists the 2013 schedule of federal income tax versus taxable income. Federal income tax versus taxable income Taxable Income Range The Tax Is … … Of the Amount Over$0–$8925 10%$0
$8925–$36,250 $892.50 + 15%$8925
$36,250–$87,850 $4,991.25 + 25%$36,250
$87,850–$183,250 $17,891.25 + 28%$87,850
$183,250–$398,350 $44,603.25 + 33%$183,250
$398,350–$400,000 $115,586.25 + 35%$398,350
>$400,000$116,163.75 + 39.6% $400,000 Suppose that Steve just received a$10,000 raise. How much of this raise is left after federal taxes if Steve’s salary before receiving the raise was $40,000? If it was$90,000? If it was \$385,000?
[T] The following table provides hypothetical data regarding the level of service for a certain highway.
Highway Speed Range (mph) Vehicles per Hour per Lane Density Range (vehicles/mi)
>60 <600 <10
60–57 600–1000 10–20
57–54 1000–1500 20–30
54–46 1500–1900 30–45
46–30 1900 2100 45–70
<30 Unstable 70–200
1. Plot vehicles per hour per lane on the x -axis and highway speed on the y -axis.
2. Compute the average decrease in speed (in miles per hour) per unit increase in congestion (vehicles per hour per lane) as the latter increases from 600 to 1000, from 1000 to 1500, and from 1500 to 2100. Does the decrease in miles per hour depend linearly on the increase in vehicles per hour per lane?
3. Plot minutes per mile (60 times the reciprocal of miles per hour) as a function of vehicles per hour per lane. Is this function linear?
a.
b. Between 600 and 1000 the average decrease in vehicles per hour per lane is −0.0075. Between 1000 and 1500 it is −0.006 per vehicles per hour per lane, and between 1500 and 2100 it is −0.04 vehicles per hour per lane. c.
The graph is nonlinear, with minutes per mile increasing dramatically as vehicles per hour per lane reach 2000.
For the next two exercises use the data in the following table, which displays bald eagle populations from 1963 to 2000 in the continental United States.
Population of breeding bald eagle pairs
Year Population of Breeding Pairs of Bald Eagles
1963 487
1974 791
1981 1188
1986 1875
1992 3749
1996 5094
2000 6471
[T] The graph below plots the quadratic $p\left(t\right)=6.48{t}^{2}-80.3\phantom{\rule{0.2em}{0ex}}1t+585.69$ against the data in preceding table, normalized so that $t=0$ corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of p over $\left[0,37\right].$
[T] The graph below plots the cubic $p\left(t\right)=0.07{t}^{3}+2.42{t}^{2}-25.63t+521.23$ against the data in the preceding table, normalized so that $t=0$ corresponds to 1963. Estimate the average number of bald eagles per year present for the 37 years by computing the average value of p over $\left[0,37\right].$
$\frac{1}{37}{\int }_{0}^{37}p\left(t\right)dt=\frac{0.07{\left(37\right)}^{3}}{4}+\frac{2.42{\left(37\right)}^{2}}{3}-\frac{25.63\left(37\right)}{2}+521.23\approx 2037$
[T] Suppose you go on a road trip and record your speed at every half hour, as compiled in the following table. The best quadratic fit to the data is $q\left(t\right)=5{x}^{2}-11x+49\text{,}$ shown in the accompanying graph. Integrate q to estimate the total distance driven over the 3 hours.
Time (hr) Speed (mph)
0 (start) 50
1 40
2 50
3 60
As a car accelerates, it does not accelerate at a constant rate; rather, the acceleration is variable. For the following exercises, use the following table, which contains the acceleration measured at every second as a driver merges onto a freeway.
Time (sec) Acceleration (mph/sec)
1 11.2
2 10.6
3 8.1
4 5.4
5 0
[T] The accompanying graph plots the best quadratic fit, $a\left(t\right)=-0.70{t}^{2}+1.44t+10.44,$ to the data from the preceding table. Compute the average value of $a\left(t\right)$ to estimate the average acceleration between $t=0$ and $t=5.$
Average acceleration is $A=\frac{1}{5}{\int }_{0}^{5}a\left(t\right)dt=-\frac{0.7\left({5}^{2}\right)}{3}+\frac{1.44\left(5\right)}{2}+10.44\approx 8.2$ mph/s
[T] Using your acceleration equation from the previous exercise, find the corresponding velocity equation. Assuming the final velocity is 0 mph, find the velocity at time $t=0.$
[T] Using your velocity equation from the previous exercise, find the corresponding distance equation, assuming your initial distance is 0 mi. How far did you travel while you accelerated your car? ( Hint: You will need to convert time units.)
$d\left(t\right)={\int }_{0}^{1}|v\left(t\right)|dt={\int }_{0}^{t}\left(\frac{7}{30}{t}^{3}-0.72{t}^{2}-10.44t+41.033\right)dt=\frac{7}{120}{t}^{4}-0.24{t}^{3}-5.22{t}^{3}+41.033t.$ Then, $d\left(5\right)\approx 81.12$ mph $\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{sec}\approx 119$ feet.
[T] The number of hamburgers sold at a restaurant throughout the day is given in the following table, with the accompanying graph plotting the best cubic fit to the data, $b\left(t\right)=0.12{t}^{3}-2.13{t}^{3}+12.13t+3.91,$ with $t=0$ corresponding to 9 a.m. and $t=12$ corresponding to 9 p.m. Compute the average value of $b\left(t\right)$ to estimate the average number of hamburgers sold per hour.
Hours Past Midnight No. of Burgers Sold
9 3
12 28
15 20
18 30
21 45
[T] An athlete runs by a motion detector, which records her speed, as displayed in the following table. The best linear fit to this data, $\ell \left(t\right)=-0.068t+5.14\text{,}$ is shown in the accompanying graph. Use the average value of $\ell \left(t\right)$ between $t=0$ and $t=40$ to estimate the runner’s average speed.
Minutes Speed (m/sec)
0 5
10 4.8
20 3.6
30 3.0
40 2.5
$\frac{1}{40}{\int }_{0}^{40}\left(-0.068t+5.14\right)dt=-\frac{0.068\left(40\right)}{2}+5.14=3.78$
#### Questions & Answers
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apa itu?
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hello please anyone with calculus PDF should share
Which kind of pdf do you want bro?
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You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
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Im not in college but this will still help
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with the help of different formulas and Rules. we use formulas according to given condition or according to questions
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fg[[(45)]]²+45⅓x²=100 | 3,028 | 9,564 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 36, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-40 | latest | en | 0.930455 |
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Zbl 1244.47046
Djolović, Ivana; Malkowsky, Eberhard
The Hausdorff measure of noncompactness of operators on the matrix domains of triangles in the spaces of strongly $C_{1}$ summable and bounded sequences.
(English)
[J] Appl. Math. Comput. 216, No. 4, 1122-1130 (2010). ISSN 0096-3003
Let $\omega$ be the space of all complex sequences $x=\left( x_{k}\right) _{k=1}^{\infty }$, $A=\left( a_{nk}\right) _{n,k=1}^{\infty }$ be an infinite matrix of complex numbers and $X$ a subset of $\omega$. The set $$X_{A}=\left\{ x\in \omega :Ax=\sum_{k=1}^\infty a_{nk} {x_{k}}\in X\right\}$$ is called the matrix domain of $A$ in $X$. Given any sets $X$ and $Y$ in $\omega$, let $\left( X,Y\right)$ denote the class of all matrices $A$ such that $X\subset Y_{A}$. Let $T=\left( t_{nk}\right) _{n,k=1}^{\infty }$ be a triangle, that is, $t_{nk}=0$ for $k>n$ and $t_{nn}\neq 0$ $\left( n=1,2, \dots\right)$, let $e\in \omega$ be the sequence with $e_{k}=1$ for all $k$ and $1\leq p<\infty$. The sets of strongly $C_{1}$-summable and bounded sequences $$w_{0}^{p}=\left\{ x\in \omega : \lim_{n \to \infty}\left( \frac{1}{n}\sum_{k=1}^n \left\vert x_{k}\right\vert ^{p}\right) =0\right\} ,$$ $$w^{p}=\left\{ x\in \omega :x-\xi \cdot e\in w_{0}^{p}\text { for some complex number }\xi \right\}$$ and $$w_{\infty}^{p}=\left\{ x\in \omega :\underset{n}\to{\sup }\left( \frac{1}{n} \sum_{k=1}^n \left\vert x_{k}\right\vert ^{p}\right) <\infty \right\}$$ were defined and studied by {\it I. J. Maddox} [On Kuttner's theorem'', J. Lond. Math. Soc. 43, 285--290 (1968; Zbl 0155.38802)]. \par In the paper under review, the authors apply the Hausdorff measure of noncompactness to characterize the classes of compact operators given by infinite matrices $A\in \left( X,Y\right)$, where $X$ is one of spaces $\left( w_{0}^{p}\right) _{T},\left( w^{p}\right) _{T}$ or $\left( w_{\infty }^{p}\right) _{T}$ and $Y$ is the space $c_{0}$ of null sequences or the space $c$ of convergent sequences. Moreover, they give sufficient conditions for the compactness of operators $A\in \left( X,Y\right)$ when $X$ is again one of spaces $\left( w_{0}^{p}\right) _{T},\left( w^{p}\right) _{T}$ or $\left( w_{\infty }^{p}\right) _{T}$ and the final $Y$ is the space $l_{\infty }$ of bounded sequences.
[Giulio Trombetta (Arcavaceta di Rende)]
MSC 2000:
*47H08
40C05 Matrix methods in summability
40J05 Summability in abstract structures
Keywords: strongly bounded and summable sequences; matrix transformations; compact operators; Cesàro method
Citations: Zbl 0155.38802
Highlights
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How this would affect concept of factor-price equalization
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In the discussion of empirical results on the Heckscher-Ohlin model, we noted that recent work suggests that the efficiency of factors of production seems to differ internationally. Explain how this would affect the concept of factor-price equalization.
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https://physics.stackexchange.com/questions/573741/why-is-power-caused-by-the-driving-force | 1,718,995,263,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00537.warc.gz | 392,105,716 | 38,692 | # Why is Power caused by the driving force?
Say we have a car on a rough inclined plane, the maximum constant velocity at which the car can travel is $$V$$, the driving force is $$F_d$$ and the frictional force is $$F$$.
Shouldn't the power exerted by the engine be used to:
1. Overcome the $$F$$ and $$mg\sin\theta$$ component of weight.
2. Cause the car to move at velocity $$V$$.
Would it not then make sense that the power of the engine should be the product of $$V$$ and $$Fnet$$ in the direction of motion?
I am at an extreme loss at where my reasoning fails as apparently the power of the engine is due to only $$DF$$, which does not make sense to me as if the power is due to only $$F_d$$ would that not imply that the overall speed would be less due to the resistive forces acting on the car?
• A picture of a free body diagram would definitely help. In addition, somewhat more detail in your question would help. Commented Aug 17, 2020 at 1:36
The power of the engine is defined as the rate of work done by the engine. Similarly, the power supplied by friction is the rate of work done by the frictional force, and the power supplied by gravity is the rate of work done by the gravitational force. The net power acting on the vehicle is '$$F_{net}\cdot v$$' which is the reusultant of the individual powers acting on it. | 325 | 1,337 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-26 | latest | en | 0.975809 |
https://www.dataunitconverter.com/gigabit-per-day-to-gibibit-per-second | 1,716,623,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00229.warc.gz | 613,338,837 | 16,955 | # Gbit/Day to Gibps → CONVERT Gigabits per Day to Gibibits per Second
expand_more
info 1 Gbit/Day is equal to 0.0000107791964654569272641782407407407406 Gibps
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
Gbit/Day
## Gigabits per Day (Gbit/Day) Versus Gibibits per Second (Gibps) - Comparison
Gigabits per Day and Gibibits per Second are units of digital information used to measure storage capacity and data transfer rate.
Gigabits per Day is a "decimal" unit where as Gibibits per Second is a "binary" unit. One Gigabit is equal to 1000^3 bits. One Gibibit is equal to 1024^3 bits. There are 1.073741824 Gigabit in one Gibibit. Find more details on below table.
Gigabits per Day (Gbit/Day) Gibibits per Second (Gibps)
Gigabits per Day (Gbit/Day) is a unit of measurement for data transfer bandwidth. It measures the number of Gigabits that can be transferred in one Day. Gibibits per Second (Gibps) is a unit of measurement for data transfer bandwidth. It measures the number of Gibibits that can be transferred in one Second.
## Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps) Conversion - Formula & Steps
The Gbit/Day to Gibps Calculator Tool provides a convenient solution for effortlessly converting data rates from Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps). 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 (Gigabit) and target (Gibibit) data units.
Source Data Unit Target Data Unit
Equal to 1000^3 bits
(Decimal Unit)
Equal to 1024^3 bits
(Binary Unit)
The conversion from Data per Day to Second can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps) can be expressed as follows:
diamond CONVERSION FORMULA Gibps = Gbit/Day x 10003 ÷ 10243 / ( 60 x 60 x 24 )
Now, let's apply the aforementioned formula and explore the manual conversion process from Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Gibibits per Second = Gigabits per Day x 10003 ÷ 10243 / ( 60 x 60 x 24 )
STEP 1
Gibibits per Second = Gigabits per Day x (1000x1000x1000) ÷ (1024x1024x1024) / ( 60 x 60 x 24 )
STEP 2
Gibibits per Second = Gigabits per Day x 1000000000 ÷ 1073741824 / ( 60 x 60 x 24 )
STEP 3
Gibibits per Second = Gigabits per Day x 0.931322574615478515625 / ( 60 x 60 x 24 )
STEP 4
Gibibits per Second = Gigabits per Day x 0.931322574615478515625 / 86400
STEP 5
Gibibits per Second = Gigabits per Day x 0.0000107791964654569272641782407407407406
Example : By applying the previously mentioned formula and steps, the conversion from 1 Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps) can be processed as outlined below.
1. = 1 x 10003 ÷ 10243 / ( 60 x 60 x 24 )
2. = 1 x (1000x1000x1000) ÷ (1024x1024x1024) / ( 60 x 60 x 24 )
3. = 1 x 1000000000 ÷ 1073741824 / ( 60 x 60 x 24 )
4. = 1 x 0.931322574615478515625 / ( 60 x 60 x 24 )
5. = 1 x 0.931322574615478515625 / 86400
6. = 1 x 0.0000107791964654569272641782407407407406
7. = 0.0000107791964654569272641782407407407406
8. i.e. 1 Gbit/Day is equal to 0.0000107791964654569272641782407407407406 Gibps.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Gigabits per Day to Gibibits per Second using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Gigabit ?
A Gigabit (Gb or Gbit) is a decimal unit of digital information that is equal to 1,000,000,000 bits and it is commonly used to express data transfer speeds, such as the speed of an internet connection and to measure the size of a file. In the context of data storage and memory, the binary-based unit of gibibit (Gibit) is used instead.
arrow_downward
#### What is Gibibit ?
A Gibibit (Gib or Gibit) is a binary unit of digital information that is equal to 1,073,741,824 bits and is defined by the International Electro technical Commission(IEC). The prefix 'gibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'gigabit' (Gb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps)
Apply the formula as shown below to convert from 1 Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps).
A B C
1 Gigabits per Day (Gbit/Day) Gibibits per Second (Gibps)
2 1 =A2 * 0.931322574615478515625 / ( 60 * 60 * 24 )
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Gigabits per Day (Gbit/Day) to Gibibits per Second (Gibps) Conversion
You can use below code to convert any value in Gigabits per Day (Gbit/Day) to Gigabits per Day (Gbit/Day) in Python.
gigabitsperDay = int(input("Enter Gigabits per Day: "))
gibibitsperSecond = gigabitsperDay * (1000*1000*1000) / (1024*1024*1024) / ( 60 * 60 * 24 )
print("{} Gigabits per Day = {} Gibibits per Second".format(gigabitsperDay,gibibitsperSecond))
The first line of code will prompt the user to enter the Gigabits per Day (Gbit/Day) as an input. The value of Gibibits per Second (Gibps) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Gibibits(Gibit) are there in a Gigabit(Gbit)?expand_more
There are 0.931322574615478515625 Gibibits in a Gigabit.
#### What is the formula to convert Gigabit(Gbit) to Gibibit(Gibit)?expand_more
Use the formula Gibit = Gbit x 10003 / 10243 to convert Gigabit to Gibibit.
#### How many Gigabits(Gbit) are there in a Gibibit(Gibit)?expand_more
There are 1.073741824 Gigabits in a Gibibit.
#### What is the formula to convert Gibibit(Gibit) to Gigabit(Gbit)?expand_more
Use the formula Gbit = Gibit x 10243 / 10003 to convert Gibibit to Gigabit.
#### Which is bigger, Gibibit(Gibit) or Gigabit(Gbit)?expand_more
Gibibit is bigger than Gigabit. One Gibibit contains 1.073741824 Gigabits.
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what is the probablity the second mouse will be white?
there are 4 white mice and 6 brown mice in a cage. Molly will take out 2 mice at random. If the first mouse she takes out is brown.
a. 2/5
b. 4/9
c.1/2
d.4/5
Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
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1
After taking out the first mouse (a brown one), there are 9 mice remaining in the cage, 4 white ones and 5 brown ones. The probability, then, that the second mouse will be a white one is computed as follows:
P = (Number of White Mice Remaining in the Cage) / (Total Number of Mice Remaining in the Cage)
Let me know what you get.
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
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Notice that the answer will depend on the manner she takes mice out. If she takes them one after another, then Robert's answer is correct. If she takes both together (with two hands, for example) then the probability is:
6/10*4/9=4/15 | 289 | 1,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-26 | latest | en | 0.903354 |
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My question is in regards to systems of ordinary differential equations. One of my research topics right now involves working with some complicated coupled ODEs used to model ecological stuff. Without getting into the details, the model I am working on now has a bad tendency to diverge for some initial conditions. I have thought and searched for a while now, but have been unable to come up with some good ways of determining when a system of ODEs will have certain solutions that diverge. Does anyone know of some literature or techniques that might be useful for me? I would very much appreciate any help you could provide. Thank you very much.
EDIT:
To avoid further misunderstanding, the type of equations I am considering are of the form:
$$\frac{d X_i}{dt} = F_i(\mathbf{X)}$$
When I say that a solution diverges, I mean that given a set of initial conditions, the solution will diverge in time. As for what I mean by diverge, I mean that solutions tend to infinity as t approaches infinity. So solutions that settle in on a point, or go through some oscillatory behavior, even chaotic oscillatory behavior, would be considered non-divergent.
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Recognitions: Gold Member Science Advisor Staff Emeritus Strictly speaking, a divergent series does not define a function. So I am not at all sure what you mean by a divergent series being a solution. It sounds to me like you are asking for conditions in which the "series method" of solution is unstable.
Divergent series are solutions as formal series, and sometimes they define functions, but in sectors, of course not in discs. See Borel summability and multisummablity.
## Existence of divergent solutions to system of ODEs
My apologizes for not making my question as clear as it should have been. All the systems of equations I am working with are of the form:
$$\frac{d X_i}{dt} = F_i(\mathbf{X)}$$
So when I say that a solution diverges, I mean that given a set of initial conditions, the solution will diverge in time. As for what I mean by diverge, I mean that solutions tend to infinity as t approaches infinity. So solutions that settle in on a point, or go through some oscillatory behavior, even chaotic oscillatory behavior, would be considered non-divergent.
I hope this clarifies what I mean. I'll append this to the opening post.
Quote by vega12 My apologizes for not making my question as clear as it should have been. All the systems of equations I am working with are of the form: $$\frac{d X_i}{dt} = F_i(\mathbf{X)}$$ So when I say that a solution diverges, I mean that given a set of initial conditions, the solution will diverge in time. As for what I mean by diverge, I mean that solutions tend to infinity as t approaches infinity. So solutions that settle in on a point, or go through some oscillatory behavior, even chaotic oscillatory behavior, would be considered non-divergent. I hope this clarifies what I mean. I'll append this to the opening post.
Well that just sounds like non-linear dynamics in general and the solution diverges as you described when it finds itself outside an attractor's basin of attraction.
Recognitions:
Gold Member
Quote by vega12 My apologizes for not making my question as clear as it should have been. All the systems of equations I am working with are of the form: $$\frac{d X_i}{dt} = F_i(\mathbf{X)}$$ So when I say that a solution diverges, I mean that given a set of initial conditions, the solution will diverge in time. As for what I mean by diverge, I mean that solutions tend to infinity as t approaches infinity. So solutions that settle in on a point, or go through some oscillatory behavior, even chaotic oscillatory behavior, would be considered non-divergent. I hope this clarifies what I mean. I'll append this to the opening post.
Are you solving the equations numerically, using some kind of numerical integration package? If so, and you are using an explicit integrator, you may want to switch to an implicit integrator.
You can get an idea of the stability of the solution by looking at the eigenvalues of the Jacobian matrix, evaluated at the initial conditions. If any of the eigenvalues have a positive real part, the solution will be unstable, irrespective of the numerical integration method.
Tags diverge, ode, solution, system
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