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FM11 39 - h 2 What is project Ps NPV What is its IRR Its...
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h. 2. What is project P’s NPV? What is its IRR? Its MIRR? Answer: Here is the time line for the cash flows, and the NPV: 0 1 2 | | | 10% -800,000 5,000,000 -5,000,000 NPV P = -\$386,776.86. We can find the NPV by entering the cash flows into the cash flow register, entering i = 10, and then pressing the NPV button. However, calculating the IRR presents a problem. With the cash flows in the register, press the IRR button. An hp-10b financial calculator will give the message "error-soln." This means that project P has multiple IRRs . An HP-17B will ask for a guess. If you guess 10%, the calculator will produce IRR = 25%. If you guess a high number, such as 200%, it will produce the second IRR, 400% 1 . The MIRR of project P = 5.6%, and is found by computing the discount rate that equates the terminal value (\$5.5 million) to the present value of cost (\$4.93 million). h. 3. Draw project P's NPV profile. Does project P have normal or non-normal cash flows? Should this project be accepted? Answer:
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# VWVortex
1. As long as the version on the site is the same as the one you have, you are up to date.
i'm getting into space no problem.But i cant get my angle right to keep me in orbit...
I made a thread on this over in Aviation and Space. I was having trouble too orbiting but I was trying to build my rocket too big. Start small and work your way up. Obviously I hope you are using the Map/Orbital view by hitting the "M" key. The orbital basics once you get up are that you effect the opposite side of the orbit:
To increase periapsis : You have to be at apoapsis... speed up pointing in orbital direction (indicated by the empty yellow circle on gyro instrument)
To increase apoapsis : You have to be at periapsis... speed up pointing in orbital direction (indicated by the empty yellow circle on gyro instrument)
To decrease periapsis : You have to be at apoapsis... slow down by pointing in opposite orbital direction (indicated by the yellow circle with X in it on gyro instrument)
To decrease apoapsis : You have to be at periapsis... slow down by pointing in opposite orbital direction (indicated by the yellow circle with X in it on gyro instrument)
Adjusting an orbit is a series of, point toward or away from orbital direction from either apoapsis or periapsis, and doing slow burns (sometimes literally bumping the throttle). Use the time acceleration to move between the apoapsis and periapsis.
***Also, remember that the faster/further your orbit the faster/further changes are made. While at 300km orbit, a 1-2 second burn at lowest throttle can be 10+km. When you get to 2,000km orbit, that 1-2 second burn can be 100+ km. It takes a whole lot of energy to break gravity, but to maintain or change orbit, it takes very little. I think it takes less than 1 fuel tank to get from a ~200km orbit to the Mun (11,400 km).
Don't forget to use the Wiki: http://kerbalspaceprogram.com/wiki/i...itle=Main_Page
3. Success on orbiting,Started out kinda of lop sided.Took me a 3 stage liquid to get up there.the last stage is on the command center so i have some thrust to keep orbit in check.Now i'm going to re enter and safely land!
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
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https://www.civilengineeringx.com/earthquake-engineering/probabilistic-seismic-demand-analysis-through-a-reliability-based-design-approach/
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Earthquake Engineering
# Probabilistic seismic demand analysis through a reliability-based design approach
The reliability of a structural system or lifeline may be referred to as the ability of the system or its components to perform their required functions under stated conditions for a specified period of time. Because of uncertainties in loading and capacity, the subject usually includes probabilistic methods and is often made through indices such as a safety index or the probability of the failure of the structure or lifeline.
2.1.1. Reliability index and failure
To evaluate the seismic performance of the structures, performance functions are defined.
Let us assume that z=g(x1, x2, …,xn) is taken as a performance function. As such, failure or damage occurs when z
Pf=P[z
Simply assume that z=EDP-C where EDP stands for Engineering Demand Parameter and C is the seismic capacity of the structure.
Damage or failure in a structural system or lifeline occurs when the Engineering Demand Parameter exceeds the capacity provided. For example, in a bridge structural damage may refer to the unseating of the deck, the development of a plastic hinge at the bottom of piers or damage due to the pounding of the decks to the abutments, etc.
Given that EDP and C are random parameters having the expected or mean values of μEDP and μC and standard deviation of σEDP and σC, the “safety index” or “reliability index”, β, is defined as:
It has been observed that the random variables such as “EDP” or “C” follow normal or lognormal
distribution. Accordingly, the performance function, z, also will follow the same
distribution. Accordingly, probability of failure (or damage occurrence) may be expressed as
a function of safety index, as follows:
## Engineering demand parameters
The Engineering Demand Parameters describe the response of the structural framing and the non-structural components and contents resulting from earthquake shaking. The parameters are calculated by structural response simulations using the IMs and corresponding earthquake motions. The ground motions should capture the important characteristics of earthquake ground motion which affect the response of the structural framing and nonstructural components and building contents. During the loss and risk estimation studies, the EDP with a greater correlation with damage and loss variables must be employed.
The EDPs were categorized in the ATC 58 task report as either direct or processed [9]. Direct EDPs are those calculated directly by analysis or simulation and contribute to the risk assessment through the calculation of P[EDP | IM]; examples of direct EDPs include interstory drift and beam plastic rotation. Processed EDPs – for example, a damage index – are derived from the values of direct EDPs and data on component or system capacities.
Processed EDPs could be considered as either EDPs or as Damage Measures (DMs) and, as such, could contribute to risk assessment through P[DM | EDP]. Direct EDPs are usually introduced in codes and design regulations. For example, the 2000 NEHRP Recommended Provisions for Seismic Regulations for Buildings and Other Structures introduces the EDPs presented in Table 2 for the seismic design of reinforced concrete moment frames [12-13]:
Processed EDPs are efficient parameters which could serve as a damage index during loss and risk estimation for structural systems and facilities. A Damage Index (DI), as a singlevalued damage characteristic, can be considered to be a processed EDP [10]. Traditionally, DIs have been used to express performance in terms of a value between 0 (no damage) and 1 (collapse or an ultimate state). An extension of this approach is the damage spectrum, which takes on values between 0 (no damage) and 1 (collapse) as a function of a period. A detailed summary of the available DIs is available in [14].
Park and Angin [15] developed one of the most widely-known damage indices. The index is a linear combination of structural displacement and hysteretic energy, as shown in the equation:
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# Question 2D Get contact time and collider position when multiple collision in 1 physics fram
Discussion in 'Physics' started by Luwyliscious, Feb 3, 2023.
1. ### Luwyliscious
Joined:
Mar 21, 2021
Posts:
3
When multiple contacts happen during 1 physics frame, the continuous detection collision solver seems to be able to detect all contacts, a "OnCollisionEnter2D" is triggered for every colliders and GetContacts retrieve the list of every contact that happened during the physics frame.
The ContactPoint2D struct doesn't containt any time and the collider.transform / otherCollider.transform only contains the position of the colliders at the end of the physics frame. It seems impossible to get the positions and time that the Box2D collision solver had to calculate in order to properly calculate the result of the physics frame.
I'm trying to achieve the following :
• Smoothly interpolate position of a fast moving object even when multiple collision happens in a frame
• Be able to override the collision response of a collider while still using unity's Phyics2D system
If you override the bounce (velocity/position) of a fast moving rigidbody2D in the OnCollisionEnter2D method, the colliding object will actually already have bounce at the velocity calculated by the physics engine. As a result, if you then modify the velocity, the resulting motion will not be as precise as if the desired velocity was applied exactly at the point of contact and at the moment of contact.
The provided interpolation for rigidbody2D only interpolate between the last known and newly calculated position, ignoring any collision that would have happened during the physics frame.
In order to achieve my goals, I tried to :
1. Manually simulate using Physics2d.Simulate()
2. Gather all collisions that happened during this physics frame using OnCollisionEnter2D()
3. Order the collisions as best as I can and calculate the object's position at each contact based on previous velocity (I only considered circle for this approach)
4. Detect the first collision for which I want to override the collision response (if any)
5. Redo simulation up to that point
6. Set the desired velocity to my rigidbody
7. Conclude the simulation for the remaining time (Recursivly loop if many custom collision in 1 physics frame)
Then I use the calculated contacts and object positions to feed an interpolation script.
My current implementation sometimes fails to order the contacts and properly position the colliders at each collision since I only have the normal impulse and contact point to help me figure the rigidbody's trajectory.
In my current implementation, the added precision in collision response does help with the feel and "fairness" of a prototype i'm working on,and the accurate interpolation makes the motions feel more believable, especially when timescale is reduced (slowmo), but i'm having a hard time making it 100% reliable.
Before I continue down that rabbit hole, is there any known way to access the actual contact time / position of colliders at the point of contact that were calculated by the continuous collision solver when multiple collision happens in 1 physics frame?
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Cm To M
# 4620 cm to m4620 Centimeters to Meters
cm
=
m
## How to convert 4620 centimeters to meters?
4620 cm * 0.01 m = 46.2 m 1 cm
A common question is How many centimeter in 4620 meter? And the answer is 462000.0 cm in 4620 m. Likewise the question how many meter in 4620 centimeter has the answer of 46.2 m in 4620 cm.
## How much are 4620 centimeters in meters?
4620 centimeters equal 46.2 meters (4620cm = 46.2m). Converting 4620 cm to m is easy. Simply use our calculator above, or apply the formula to change the length 4620 cm to m.
## Convert 4620 cm to common lengths
UnitLengths
Nanometer46200000000.0 nm
Micrometer46200000.0 µm
Millimeter46200.0 mm
Centimeter4620.0 cm
Inch1818.8976378 in
Foot151.57480315 ft
Yard50.5249343832 yd
Meter46.2 m
Kilometer0.0462 km
Mile0.0287073491 mi
Nautical mile0.0249460043 nmi
## What is 4620 centimeters in m?
To convert 4620 cm to m multiply the length in centimeters by 0.01. The 4620 cm in m formula is [m] = 4620 * 0.01. Thus, for 4620 centimeters in meter we get 46.2 m.
## Alternative spelling
4620 cm to m, 4620 cm in m, 4620 Centimeter to Meter, 4620 Centimeter in Meter, 4620 Centimeter to m, 4620 Centimeter in m, 4620 Centimeters to Meters, 4620 Centimeters in Meters, 4620 cm to Meter, 4620 cm in Meter, 4620 Centimeters to Meter, 4620 Centimeters in Meter, 4620 Centimeters to m, 4620 Centimeters in m
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A137225 Triangle T(k,q) of minimal q-Niven numbers: smallest number such that the sum of its digits in base q equals k, 2<=q<=k+1. 0
1, 3, 2, 7, 5, 3, 15, 8, 7, 4, 31, 17, 11, 9, 5, 63, 26, 15, 14, 11, 6, 127, 53, 31, 19, 17, 13, 7, 255, 80, 47, 24, 23, 20, 15, 8, 511, 161, 63, 49, 29, 27, 23, 17, 9, 1023, 242, 127, 74, 35, 34, 31, 26, 19, 10, 2047, 485, 191, 99, 71, 41, 39, 35, 29, 21, 11, 4095, 728, 255 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS H. Fredricksen, E. J. Ionascu, F. Luca, P. Stanica, Minimal Niven numbers, arXiv:0803.0477 [math.NT] FORMULA T(k,2)=A000225(k). T(k,k+1)=2k-1. Conjecture: T(k,3)=A062318(k), verified up to k=23. EXAMPLE T(8,4) =47 because 47, written 233 in base q=4, is the smallest number with digit sum 2+3+3=8=k in base q=4. The triangle reads T(k,q), k=1,2,..., 2<=q up to the diagonal, after which the values stay constant: 1 1 1 1 1 1 1 1 1 3 2 2 2 2 2 2 2 2 7 5 3 3 3 3 3 3 3 15 8 7 4 4 4 4 4 4 31 17 11 9 5 5 5 5 5 63 26 15 14 11 6 6 6 6 127 53 31 19 17 13 7 7 7 255 80 47 24 23 20 15 8 8 511 161 63 49 29 27 23 17 9 1023 242 127 74 35 34 31 26 19 ... MAPLE sd := proc(n, b) local i ; add(i, i=convert(n, base, b)) ; end: T := proc(k, q) local a; for a from 1 do if sd(a, q) = k then RETURN(a) ; fi ; od: end: for k from 1 to 20 do for q from 2 to k+1 do printf("%d, ", T(k, q)) ; od: od: CROSSREFS Cf. A005349, A052491. Sequence in context: A011384 A128140 A213579 * A213777 A118834 A255547 Adjacent sequences: A137222 A137223 A137224 * A137226 A137227 A137228 KEYWORD base,easy,nonn,tabl AUTHOR R. J. Mathar, Mar 07 2008 STATUS approved
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Last modified November 29 04:38 EST 2021. Contains 349416 sequences. (Running on oeis4.)
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# Rotational Dynamics: Calculating a Moment of Inertia
• Haths
In summary: I = 2 \pi \, h \, \frac{m}{\pi R^2 h} \int_0^R r^3 dr = \frac{2}{4} m R^2 = \frac{1}{2} m R^2In summary, the moment of inertia of a solid disk is equal to half its mass times the radius squared. This is found by summing up all the small pieces of the disk using integration, where the small piece's mass is calculated using its volume and density.
Haths
I feel like a div. Quite simply I've forgotten how to use integration to calculate a moment of inertia (MOI). Ok I want to calculate the MOI of say a solid disk, about an axis perpendicular to the plane of the disk, running through its center of mass.
I remember that for any point particle in the disk its MOI is going to be:
mr2
I can show this because if we consider a point mass at some distance r from a point of rotation, it will still subscribe to F=ma in the tangential direction. Knowing that:
T= I [alpha]
Where T is the torque, I the MOI, and alpha the angular acceleration.
T=Fr
Therefore:
T=mar
As
[alpha] = a / r
mar=I a/r
I=mr2
Hence I'm certain of that. Thus I can appreciate that for a solid disk I am looking at a summation of the MOI's between the axis and the edge of the disk. Hence I need an integration factor such that;
IT= Integr0{ dm/dr r2 } dr
I think. Yes? Hence I need an expression for the rate of change of mass with respect to radial displacement. Hence I need an area integral, and this is where I'm left blank, because I have no idea how to express the circle's mass with respect to increasing radius. Unless...
I make my integral:
IT= Integ2PI0{ dm/dr r2 } d[theta]
Where dm/dr = r2PI
Therefore dm = 1/3 r3PI |r0 hence:
IT= Integ2PI0{ 1/3 r3 PI r2 } d[theta]
= 1/3 PI r5 [theta] |2PI0
= 2/3 PI2 r5
Which I'm certain is wrong. I'm not asking why it's wrong. But how should I construct my dm/dr and integrate that through with reasoning why that's a good method. Because I have a feeling I'm forgetting something fundamental in this part of the calculation.
Cheers,
Haths
Moment of inertia is found from summing each mass element times the square of the distance from the axis of rotation to the mass element. Typically, the mass element is expressed as the product of density and a volume (or area) element. If the density is uniform it can be moved outside the integrand and the volume element is expressed in terms of the radius and can be readily integrated. For a thin disk using polar coordinates is easiest and gives an area element of r*dr*d{theta}. The limits of double integration are from zero to R (the radius of the disk) and from zero to 2pi for theta. The result is the density times pi times radius to the fourth power all divided by two. But the density is the total mass, M, divided by the area (pi times radius squared). So I = (MR^2)/2.
Yeah I already knew what the MOI was. Ok
So your saying I can express my Integral as:
I = [rho] Integ2PI0 [ Integr0{ r } dr ] d[theta]
+++Quick working+++
1/2 r^2
eval?
1/2 r^2 [theta]
eval?
r^2 PI
*[rho]
=m only?...
+++/ends+++
I don't get it. I can see that intergrating r*dr*d{theta} is esentailly summing up all the possition vectors of the point masses* in the circular disk (which was my initail question answered). But my quick working came out wrong. *Tries again long hand, non-mentally*
I still come out with I=m only.
Perhaps my intergration skills are bad. Ok First I intergrate r with respect to dr and get;
1/2 r2 to be evaluated between r and 0. Yes?
This is still just 1/2 r2
Then I intergrate this with respect to d[theta] and evaluate between 2PI and 0 and get;
1/2 2PI r2
Now I multiply this by [rho] which is m / PI r2
Which leaves me with m. Ok so where is my mistake here? Is it that I didn't start with r2 in my integral *penny drops*. It's Integ Integ{ r2*r*dr*d[theta]} dr d[theta]
Because r*dr*d[theta] is dm/dr. Yes ok. I'm I right with this? Because *mental calculation* r^3 => 1/4r^4 when multiplied 2PI: 2/4=1/2, PI drops out with Rho.
Cheers man, that was a bad moment when I forgot how to this.
I = [rho] Integ2PI0 [ Integr0{ r2 dm/dr } dr ] d[theta] where dm/dr = r dr d[theta]
*That's the learning point I take away.
Haths
You're post(s) are really hard to read for me (maybe it's because it's late lol...) but I'm going to give it a try.
The moment of inertia can be calculated as:
$$I = \int r^2 dm$$
Here, dm is the mass of a small piece of your disk at radius r from the center of mass (origin).
The mass of the disk is unknown, but you do know that it will be equal to
$$m = \rho V$$
(density times volume)
So the mass of a small piece of disk (volume dV) is
$$dm = \rho dV$$
(Note that rho could still depend on the radius, but I assume it is a uniform disk which means rho is constant)
So
$$I = \int \rho \, r^2 \, dV = \rho \int r^2 dV$$
Now, what do you know about the volume of a small piece of your disk (dV)?
If you take a circular segment of your disk (a loop around your disk), and take its volume as dV, then it is not very hard to see that its volume is
$$dV = 2 \pi \, h \, r \, dr$$
This is just the circumference of your loop (2 pi r) times the width of the loop (dr) times the height (h) which gives you the volume.
Entering this into the equation for I we get:
$$I = \rho \int r^2 2 \pi \, h \, r \, dr = 2 \pi \, h \rho \int r^3 dr$$
Now if you look at the density \rho again, that is also equal to the mass divided by the total volume, which is just the area of the disk times its height:
$$\rho = \frac{m}{V} = \frac{m}{\pi R^2 h}$$
To sum over the entire disk, you need your loops to run with a radius from r = 0 to r = R (radius of the disk), so we integrate from 0 to R:
$$I = 2 \pi \, h \rho \int r^3 dr = \frac{2 \pi h m}{\pi R^2 h} \int_0^R r^3 dr = \frac{2m}{R^2} \frac{R^4}{4} = \frac{mR^2}{2}$$
Sorry if I knew how to display the equations like you'd done then I'd of been able to make it easier to read.
It's alright I understand. I wasn't sure on how to express dm/dr, at first though, I was thinking of strips of [delta]r decreasing in height on the y-axis. Which left me rather confused. So then I thought that you simply intergrate the formula for the area of the shape with respect for r. But of course it's not that.
Once chrisk reminded me that it's not the area formula, you are actually using intergration to express the area as part of this calculation (the penny dropped). Then your dm/dr is an expression which when intergrated expresses the area. Then [rho] is used to link the mass of the object into this, and yes if [rho] was a function of r then you can't take that out of the integral.
Anyhow thanks people,
Haths
## 1. What is rotational dynamics?
Rotational dynamics is a branch of physics that deals with the behavior of objects that are rotating or moving in a circular motion. It involves understanding the relationship between force, torque, and motion in rotational systems.
## 2. What is a moment of inertia?
Moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is similar to mass in linear motion and depends on the distribution of mass around the axis of rotation.
## 3. How is a moment of inertia calculated?
The moment of inertia of an object can be calculated by summing the products of each particle's mass and its distance from the axis of rotation squared. Alternatively, it can be calculated using the parallel axis theorem, which takes into account the object's shape and the distance between the object's center of mass and the axis of rotation.
## 4. Why is calculating a moment of inertia important?
Calculating a moment of inertia is important because it helps determine how easy or difficult it is to change an object's rotational motion. It also plays a crucial role in understanding the behavior of objects in rotational systems, such as spinning tops, wheels, and gyroscopes.
## 5. What are some real-life applications of rotational dynamics?
Rotational dynamics has many real-life applications, including designing and analyzing the performance of machines and vehicles that involve rotational motion, such as engines, turbines, and propellers. It is also essential in sports, such as figure skating, gymnastics, and diving, where rotational motion is a key component of the movements.
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# How does density relate to volume?
## How does density relate to volume?
Density and Volume are inversely proportional to each other. Mass and volume are not the same. When density increases, volume decreases. When volume increases, density decreases.
### What is the relationship between mass density and volume?
Mass, volume and density are three of an object’s most basic properties. Mass is how heavy something is, volume tells you how big it is, and density is mass divided by volume.
#### What is the relationship between density and particle arrangement?
All matter contains particles. The difference between the different states of matter is how the particles are arranged: in a liquid – particles are tightly packed but free to move past each other….Solids, liquids and gases.
Material Density in grams per cubic centimetre (g/cm 3)
Water 1
Air 0.00129
Why does density decrease when volume increases?
As you heat something up, the volume usually increases because the faster moving molecules are further apart. Since volume is in the denominator, increasing the volume decreases the density.
What is the difference between density mass and volume?
Volume – How much space an object or substance takes up. Mass – Measurement of the amount of matter in an object or substance. Density – How much space an object or substance takes up (its volume) in relation to the amount of matter in that object or substance (its mass).
## Does more density mean less volume?
Density is a measure of mass per unit of volume. Density is a measure of mass per volume. An object made from a comparatively dense material (such as iron) will have less volume than an object of equal mass made from some less dense substance (such as water).
### Does density decrease when volume increases?
If volume increases without an increase in mass, then the density decreases (Fig. 2.2 A to 2.2 C). Adding additional matter to the same volume also increases density, even if the matter added is a different type of matter (Fig.
#### How does density relate to the volume of a material?
If the particles of a material are stored very neatly, and closer together, then more particles will be able to fit into the volume. Since particles have mass, the more particles you can fit into the volume, the more the material will weigh. Density is just a measurement of how much mass is in a given volume.
How does mass and volume relate to matter?
All particles occupy space. or volume. As a gas a few particles of matter can occupy a large degree of space. liquid particles occupy more space generally as the particles increase in kinetic energy or temperature.
Why are more particles able to fit into a volume?
If the particles of a material are stored very neatly, and closer together, then more particles will be able to fit into the volume. Since particles have mass, the more particles you can fit into the volume, the more the material will weigh.
## How to calculate the density of a product?
Designers will estimate the weight of a product by multiplying the volume by the density of a material, this is very useful if a product had to be designed within a certain weight limit. You can also identify a material by measuring its volume and mass. Density is calculated simply by dividing the mass (m) of an object by its volume (V).
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https://studysoup.com/tsg/1897/physics-principles-with-applications-6-edition-chapter-8-problem-54p
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×
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 8 - Problem 54p
Get Full Access to Physics: Principles With Applications - 6 Edition - Chapter 8 - Problem 54p
×
# (II) A diver (such as the one shown in Fig. 8-28) can
ISBN: 9780130606204 3
## Solution for problem 54P Chapter 8
Physics: Principles with Applications | 6th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
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Physics: Principles with Applications | 6th Edition
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3
Problem 54P
(II) A diver (such as the one shown in Fig. ) can reduce her moment of inertia by a factor of about when changing from the straight position to the tuck position. If she makes rotations in when in the tuck position, what is her angular speed (rev/s) when in the straight position?
Step-by-Step Solution:
Step-by-step solution
Step 1 of 3
CONCEPT
Consider the diver and platform as a system. There is no external torque acting on the system to change the angular velocity. So the angular momentum is conserved.
Therefore, by applying the conservation of angular momentum to the diver and platform system, we can calculate the final angular speed of the person.
Step 2 of 3
Step 3 of 3
#### Related chapters
Unlock Textbook Solution
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https://logic-masters.de/Raetselportal/Raetsel/zeigen.php?id=000J36
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## Insight Out (Chaos Construction)
(Eingestellt am 24. Juli 2024, 21:57 Uhr von sfushidahardy)
This time the arrows see the number of regions, rather than the length of regions. This puzzle should be more difficult than my previous two chaos constructions, but still smooth if you ask the right questions! I hope you enjoy it.
• Chaos construction: every row, column, and orthogonally connected region contains the digits 1~9 exactly once each. The region boundaries must be determined.
• Arrows: the digit in a cell with arrows is the number of distinct regions seen in the direction of one of the arrows, including the cell itself. If there are multiple arrows in a cell, each one must see the same number of regions. To clarify, if the line-of-sight leaves and re-enters the same region, this does not count as a new region.
Solve Insight Out on Sudokupad.
The following image will hopefully clarify any questions about the ruleset:
Lösungscode: The digits in row 7, with a forward-slash at region boundaries. For example, 123/45/6789.
Gelöst von Fool on Hill, someguy209, SKORP17, dogfarts, BirdyChirp, Steven R, Grothenlace, bansalsaab, appletrapezoid, henrypijames, Voidslime, han233ing, marcmees, dmm90, yttrio, flyingdragon, ZornsLemon, ... happyteo, japanoise_breakfast, Pink Flag, harwiltz, GertVonnegut, Tompzini, Jorrr2, misko, koba1917, Ziboo, Jesper, Paletron, pillowss, Counterfeitly, steeto, SudokuHero, zakkai, lmdemasi
Komplette Liste
### Kommentare
am 29. Juli 2024, 07:20 Uhr von Tank
Just started doing 4* puzzles, loved this one. Challenging but satisfying
am 29. Juli 2024, 06:55 Uhr von Tank
Just started doing 4* puzzles, loved this one. Challenging but satisfying
am 27. Juli 2024, 13:24 Uhr von Franjo
The first few digits got in quickly, but you have to be very careful afterwards. I made a silly mistake while running on, broke the puzzle and wasn’t able to detect my fault. So I had to start again. But it was a pleasure to solve this gem. Thank you very much for creating and sharing this masterpiece.
am 26. Juli 2024, 03:46 Uhr von yttrio
Super fun puzzle with some great region-building deductions!
am 25. Juli 2024, 18:42 Uhr von Voidslime
Love it! So many barriers and breakthroughs. I love when a puzzle stays interesting throughout
Zuletzt geändert am 25. Juli 2024, 18:04 Uhr
am 25. Juli 2024, 18:03 Uhr von henrypijames
This puzzle isn't harder, but rather significantly easier than the previous ones - slightly below 3 stars for me.
am 24. Juli 2024, 22:16 Uhr von someguy209
Great use of sudoku in a puzzle where you'd least expect it!
Schwierigkeit: Bewertung: 96 % Gelöst: 43 mal Beobachtet: 1 mal ID: 000J36
Lösungscode:
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# Government Consumption Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Malawi
2010: 18.43041 Percent (+ see more)
Annual, Not Seasonally Adjusted, KGPPPGMWA156NUPN, Updated: 2012-09-17 12:02 PM CDT
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: kg
Source: University of Pennsylvania
Release: Penn World Table 7.1
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Government Consumption Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Malawi, Percent, Not Seasonally Adjusted (KGPPPGMWA156NUPN)
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
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Suggested Citation
``` University of Pennsylvania, Government Consumption Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Malawi [KGPPPGMWA156NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/KGPPPGMWA156NUPN/, November 26, 2014. ```
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# sum
sum
/sum/, n., v., summed, summing.
n.
1. the aggregate of two or more numbers, magnitudes, quantities, or particulars as determined by or as if by the mathematical process of addition: The sum of 6 and 8 is 14.
2. a particular aggregate or total, esp. with reference to money: The expenses came to an enormous sum.
3. an indefinite amount or quantity, esp. of money: to lend small sums.
4. a series of numbers or quantities to be added up.
5. an arithmetical problem to be solved, or such a problem worked out and having the various steps shown.
6. the full amount, or the whole.
7. the substance or gist of a matter, comprehensively or broadly viewed or expressed: the sum of his opinions.
8. concise or brief form: in sum.
9. Math.
a. the limit of the sequence of partial sums of a given infinite series.
b. union (def. 10a).
10. a summary.
v.t.
11. to combine into an aggregate or total (often fol. by up).
12. to ascertain the sum of, as by addition.
13. to bring into or contain in a small compass (often fol. by up).
v.i.
14. to amount (usually fol. by to or into): Their expenses summed into the thousands.
15. sum up,
a. to reckon: We summed up our assets and liabilities.
b. to bring into or contain in a brief and comprehensive statement; summarize: to sum up the case for the prosecution.
c. to form a quick estimate of: I summed him up in a minute.
[1250-1300; (n.) ME summe < L summa sum, n. use of fem. of summus highest, superl. of superus (see SUPERIOR); (v.) ME summen ( < OF summer) < ML summare, deriv. of summa]
Syn. 1. See number.
* * *
Universalium. 2010.
Synonyms:
### Look at other dictionaries:
• Sum 41 — Sum 41 … Википедия
• Sum 41 — Datos generales Origen Ajax, Ontario, Canadá … Wikipedia Español
• Sum 41 — lors du West Palm Beach Warped Tour, en 2010 Pays d’origine … Wikipédia en Français
• Sum — Sum, n. [OE. summe, somme, OF. sume, some, F. somme, L. summa, fr. summus highest, a superlative from sub under. See {Sub }, and cf. {Supreme}.] 1. The aggregate of two or more numbers, magnitudes, quantities, or particulars; the amount or whole… … The Collaborative International Dictionary of English
• sum — [sum; ] for n. 6 [ so͞om] n. [ME somme < MFr < L summa, fem. of summus, highest, superl. < base of super: see SUPER ] 1. an amount of money [a sum paid in reparation] 2. the whole amount; totality; aggregate [the sum of our experience] 3 … English World dictionary
• SUM — Saltar a navegación, búsqueda SUM Información personal Origen Santiago, Chile … Wikipedia Español
• sum — I (tally) noun compendium, essence, figure, gist, idea conveyed, meaning, score, substance, summary II (total) noun aggregate amount, all, entirety, everything, gross amount, sum total, the whole, totality, wholeness associated concepts: sum paid … Law dictionary
• Sum'ay — (auch Sama ay u.ä.; altsüdarabisch s1mʿy) war ein Königreich und späterer Stamm im Westen des altsüdarabischen Reiches Saba, im jemenitischen Hochland. Der Name rührt vom Stamm Sama bzw. dessen Stammesgott Sama, der etwa im 4. Jahrhundert v. Chr … Deutsch Wikipedia
• Šum — Шум Shumi … Deutsch Wikipedia
• šum — šȗm m <N mn šúmovi> DEFINICIJA neodređeni zvuk umjerene jačine koji nastaje nepravilnim titrajem, treperenjem zvučnih valova različite duljine i postojanosti [šum potoka; šum u glavi; šum u primanju emisije radija; šum srca, pat.]… … Hrvatski jezični portal
• Sum — steht für: eine Universität in Moskau, siehe Staatliche Universität für Management Eine Verwaltungseinheit in mongolischsprachigen Gebieten, siehe Sum (Verwaltungsgliederung) Für diese Verwaltungseinheit in der Mongolei siehe Sum (Mongolei) Für… … Deutsch Wikipedia
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https://rdrr.io/cran/circular/src/R/aov.circular.R
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R/aov.circular.R In circular: Circular Statistics
Documented in aov.circularprint.aov.circular
```#############################################################
# #
# Original Splus: Ulric Lund #
# E-mail: ulund@calpoly.edu #
# #
#############################################################
#############################################################
# #
# aov.circular function #
# Author: Claudio Agostinelli #
# E-mail: claudio@unive.it #
# Date: August, 10, 2006 #
# Version: 0.2-1 #
# #
# Copyright (C) 2006 Claudio Agostinelli #
# #
#############################################################
aov.circular <- function(x, group, kappa=NULL, method=c("F.test", "LRT"), F.mod=TRUE, control.circular=list()) {
method <- match.arg(method)
# Handling missing values
ok <- complete.cases(x, group)
x <- x[ok]
group <- group[ok]
if (length(x)==0 | length(table(group)) < 2) {
warning("No observations or no groups (at least after removing missing values)")
return(NULL)
}
if (is.circular(x)) {
datacircularp <- circularp(x)
} else {
datacircularp <- list(type="angles", units="radians", template="none", modulo="asis", zero=0, rotation="counter")
}
dc <- control.circular
if (is.null(dc\$type))
dc\$type <- datacircularp\$type
if (is.null(dc\$units))
dc\$units <- datacircularp\$units
if (is.null(dc\$template))
dc\$template <- datacircularp\$template
if (is.null(dc\$modulo))
dc\$modulo <- datacircularp\$modulo
if (is.null(dc\$zero))
dc\$zero <- datacircularp\$zero
if (is.null(dc\$rotation))
dc\$rotation <- datacircularp\$rotation
x <- conversion.circular(x, units="radians", zero=0, rotation="counter", modulo="2pi")
attr(x, "class") <- attr(x, "circularp") <- NULL
result <- AovCircularRad(x, group, kappa=NULL, method, F.mod)
result\$mu <- conversion.circular(circular(result\$mu), dc\$units, dc\$type, dc\$template, dc\$modulo, dc\$zero, dc\$rotation)
result\$mu.all <- conversion.circular(circular(result\$mu.all), dc\$units, dc\$type, dc\$template, dc\$modulo, dc\$zero, dc\$rotation)
result\$call <- match.call()
class(result) <- "aov.circular"
return(result)
}
AovCircularRad <- function(x, group, kappa=NULL, method, F.mod) {
### x must be in radians, modulo 2pi
ns <- tapply(x, group, FUN=length)
resultant <- tapply(x, group, FUN=function(x) RhoCircularRad(x)*length(x))
mean.dirs <- tapply(x, group, FUN=MeanCircularRad)
kappas <- tapply(x, group, FUN=function(x) MlevonmisesRad(x)[4])
grps <- length(resultant)
n <- length(group)
if (method=="F.test"){
if (!is.null(kappa))
warning("Specified value of kappas is not used in the F-test")
sum.res <- sum(resultant)
df <- c(grps-1, n-grps, n-1)
SS <- c(sum.res - res.all, n-sum.res, n-res.all)
MS <- SS/df
if (F.mod==TRUE) {
stat <- (1+3/(8*kappa.all))*MS[1]/MS[2]
} else {
stat <- MS[1]/MS[2]
}
p.value <- 1-pf(stat, grps-1,n-grps)
} else {
SS <- NA
MS <- NA
if (is.null(kappa))
kappa <- kappa.all
stat1 <- 1-1/(4*kappa)*A1(kappa)*(sum(1/ns)-1/n)
stat2 <- 2*kappa*sum(resultant*(1-cos(mean.dirs-mean.dir.all)))
stat <- stat1*stat2
df <- grps-1
p.value <- 1-pchisq(stat, df)
}
result <- list()
result\$mu <- mean.dirs
result\$mu.all <- mean.dir.all
result\$kappa <- kappas
result\$kappa.all <- kappa.all
result\$rho <- resultant
result\$rho.all <- res.all
result\$method <- method
result\$df <- df
result\$SS <- SS
result\$MS <- MS
result\$statistic <- stat
result\$p.value <- p.value
return(result)
}
#############################################################
# #
# print.aov.circular function #
# Author: Claudio Agostinelli #
# E-mail: claudio@unive.it #
# Date: April, 30, 2005 #
# Version: 0.2-1 #
# #
# Copyright (C) 2005 Claudio Agostinelli #
# #
#############################################################
print.aov.circular <- function(x, digits = max(3, getOption("digits") - 3), ...) {
cat("\nCall:\n",deparse(x\$call),"\n\n",sep="")
if (x\$method=="F.test") {
result.matrix <- cbind(x\$df, x\$SS, x\$MS, c(x\$statistic,NA,NA), c(x\$p.value,NA,NA))
dimnames(result.matrix) <- list(c("Between","Within","Total"),c("df", "SS", "MS", "F", "p"))
cat("\n", "Circular Analysis of Variance: High Concentration F-Test", "\n", "\n")
print(result.matrix, digits=digits)
cat("\n \n")
} else {
cat("\n", "Circular Analysis of Variance: Likelihood Ratio Test", "\n", "\n")
cat(" df: ", format(x\$df, digits=digits), "\n ChiSq: ", format(x\$statistic, digits=digits), "\n p.value:", format(x\$p.value, digits=digits), "\n \n")
}
invisible(x)
}
```
Try the circular package in your browser
Any scripts or data that you put into this service are public.
circular documentation built on April 27, 2022, 1:06 a.m.
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http://math.stackexchange.com/questions/48288/something-basic-in-l-adic-properties-of-the-partition-function-paper
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# Something basic in “l-adic properties of the partition function” paper
I am trying to understand the basic result in this paper:
http://www.aimath.org/news/partition/folsom-kent-ono.pdf
My problem is with the example at the end of page 2. I understand it's supposed to be a particular instance of theorem 1.1 with $l=13, b_1=1,b_2=3,m=1$, but I can't see how they pass from $p(\frac{l^{b_2}n+1}{24})$ (which might be a fraction for all I know, since it seems like it's equal to $\frac{13^3n+1}{24}$) to $p(13^3n+1007)$.
Where did the 1007 come from? Where did the division by 24 disappear?
-
## 1 Answer
They define $p(\alpha)=0$ if $\alpha$ is not an integer. So you only have to care about those $n$ for which $13^3n+1$ is divisble by 24.
Reducing mod 24 this means that we need $13n +1 \equiv 0 {\pmod {24}}$ or equivalently $13n \equiv 23 \pmod {24}$ or equivalently that $n \equiv 11 \pmod {24}$. Now if you parametrize these numbers i.e $n= 24N+11$ and plug that in you get $$(13^3(24N+11)+1)/24=13^3N+1007$$.
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https://www.hackmath.net/en/math-problems/basic-functions?page_num=97
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# Basic functions - problems - page 97
1. A perineum
A perineum string is 10% shorter than its original string. The first string is 24, what is the 9th string or term?
2. 15 teachers
15 teachers teach for a combined amount of 128 days over a period of 64 days. What is this expressed as a percentage?
3. Price inflation
This year the average wage grew by 4.5%. Prices rose by 2%. How much did the real wage increase in the given year?
How many 1/4 cup servings are in 2 and 1/3 cups of lemonade?
5. Mba studium
At MBA school, fourth-year students can choose from three optional subjects: a) mathematical methods, b) social interaction, c) management Each student studies one of these subjects. The mathematical methods studied 28 students, the social interaction 27.
6. Thomas
Thomas lives 400 meters away from Samko, Robo from Thomas also 400 m and Samko from Robo 500. Anton lives 300 meters away from Robo further as Samko. How far away lives Anton from Rob?
7. Employees
The company employs 1 440 employees (men and women). For over-average results, the premiums were 18.75% of all men and 22.5% of all women. 20% of employees were rewarded with premiums. How many men and how many women are employed in the company?
8. Divide
Divide area of rectangles with dimensions 32m and 10m by the ratio 7: 9. What area corresponds to a smaller section?
9. A triangle
A triangle has an angle that is 63.1 other 2 are in ratio of 2:5 What are the measurements of the two angles?
10. Diofant equation
250x + 120y = 5640
11. Adding mixed numerals
3 3/4 + 2 3/5 + 5 1/2 Show your solution.
12. Report 2
A School reports students to teacher ratio of 6:1. If there are 45 teachers in the School, how many students are there?
13. Domains of functions
F(x)=x2-7x and g(x)=5-x2 Domain of (fg)(x) is. .. . . Domain of (f/g)(x). ..
14. Bitoo and Reena
Bitoo ate 3/5 part of an apple and the remaining part was eaten by his sister Reena. How much part of an apple did Renna eat? Who had the larger share? By how much?
15. Coal mine
The monthly plan of 17,000 tons of coal exceeded the mine by 1/25. How many tonnes of coal have been harvested from the mine above plan?
16. Calculate 2
Calculate the largest angle of the triangle whose side are 5.2cm, 3.6cm, and 2.1cm
17. Tallest and shortest
Jenn is the tallest on the team. He is 1 1/2 times as tall as the shortest girl whom is 4 1/4 feet tall. How tall is Jebb?
18. Journey 5
A man has to do a journey of 84km in 3 hours. He travels the first 30km at 20km/hr. At what rate must he travel the remaining distance to complete his journey on time?
19. Erica
Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric, how much will she have left?
20. Length subtracting
Express in mm: 5 3/10 cm - 2/5 mm
Do you have an interesting mathematical problem that you can't solve it? Enter it, and we can try to solve it.
To this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox.
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# Physics - centripetal acceleration
posted by .
What is the centripetal acceleration of a satellite orbiting Saturn at the location exactly on Saturn radius above the surface of Saturn
• Physics - centripetal acceleration -
R = 2r
m v^2/R = G m M /R^2
so
v^2/R = Ac = G M/(4r^2)
where M is Saturn mass
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > caucvgprlemnkj GIF version
Theorem caucvgprlemnkj 6764
Description: Lemma for caucvgpr 6780. Part of disjointness. (Contributed by Jim Kingdon, 23-Oct-2020.)
Hypotheses
Ref Expression
caucvgpr.f (𝜑𝐹:NQ)
caucvgpr.cau (𝜑 → ∀𝑛N𝑘N (𝑛 <N 𝑘 → ((𝐹𝑛) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑛) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )))))
caucvgprlemnkj.k (𝜑𝐾N)
caucvgprlemnkj.j (𝜑𝐽N)
caucvgprlemnkj.s (𝜑𝑆Q)
Assertion
Ref Expression
caucvgprlemnkj (𝜑 → ¬ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆))
Distinct variable group: 𝑘,𝐹,𝑛
Allowed substitution hints: 𝜑(𝑘,𝑛) 𝑆(𝑘,𝑛) 𝐽(𝑘,𝑛) 𝐾(𝑘,𝑛)
Proof of Theorem caucvgprlemnkj
Dummy variables 𝑎 𝑏 𝑓 𝑔 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 ltsonq 6496 . . . 4 <Q Or Q
2 ltrelnq 6463 . . . 4 <Q ⊆ (Q × Q)
31, 2son2lpi 4721 . . 3 ¬ (𝑆 <Q (𝐹𝐽) ∧ (𝐹𝐽) <Q 𝑆)
4 simprl 483 . . . . . . 7 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾))
5 caucvgpr.cau . . . . . . . . . . . 12 (𝜑 → ∀𝑛N𝑘N (𝑛 <N 𝑘 → ((𝐹𝑛) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑛) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )))))
6 breq1 3767 . . . . . . . . . . . . . 14 (𝑛 = 𝑎 → (𝑛 <N 𝑘𝑎 <N 𝑘))
7 fveq2 5178 . . . . . . . . . . . . . . . 16 (𝑛 = 𝑎 → (𝐹𝑛) = (𝐹𝑎))
8 opeq1 3549 . . . . . . . . . . . . . . . . . . 19 (𝑛 = 𝑎 → ⟨𝑛, 1𝑜⟩ = ⟨𝑎, 1𝑜⟩)
98eceq1d 6142 . . . . . . . . . . . . . . . . . 18 (𝑛 = 𝑎 → [⟨𝑛, 1𝑜⟩] ~Q = [⟨𝑎, 1𝑜⟩] ~Q )
109fveq2d 5182 . . . . . . . . . . . . . . . . 17 (𝑛 = 𝑎 → (*Q‘[⟨𝑛, 1𝑜⟩] ~Q ) = (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))
1110oveq2d 5528 . . . . . . . . . . . . . . . 16 (𝑛 = 𝑎 → ((𝐹𝑘) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) = ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))
127, 11breq12d 3777 . . . . . . . . . . . . . . 15 (𝑛 = 𝑎 → ((𝐹𝑛) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) ↔ (𝐹𝑎) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))))
137, 10oveq12d 5530 . . . . . . . . . . . . . . . 16 (𝑛 = 𝑎 → ((𝐹𝑛) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) = ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))
1413breq2d 3776 . . . . . . . . . . . . . . 15 (𝑛 = 𝑎 → ((𝐹𝑘) <Q ((𝐹𝑛) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) ↔ (𝐹𝑘) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))))
1512, 14anbi12d 442 . . . . . . . . . . . . . 14 (𝑛 = 𝑎 → (((𝐹𝑛) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑛) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q ))) ↔ ((𝐹𝑎) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))))
166, 15imbi12d 223 . . . . . . . . . . . . 13 (𝑛 = 𝑎 → ((𝑛 <N 𝑘 → ((𝐹𝑛) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑛) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )))) ↔ (𝑎 <N 𝑘 → ((𝐹𝑎) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))))))
17 breq2 3768 . . . . . . . . . . . . . 14 (𝑘 = 𝑏 → (𝑎 <N 𝑘𝑎 <N 𝑏))
18 fveq2 5178 . . . . . . . . . . . . . . . . 17 (𝑘 = 𝑏 → (𝐹𝑘) = (𝐹𝑏))
1918oveq1d 5527 . . . . . . . . . . . . . . . 16 (𝑘 = 𝑏 → ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) = ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))
2019breq2d 3776 . . . . . . . . . . . . . . 15 (𝑘 = 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ↔ (𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))))
2118breq1d 3774 . . . . . . . . . . . . . . 15 (𝑘 = 𝑏 → ((𝐹𝑘) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ↔ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))))
2220, 21anbi12d 442 . . . . . . . . . . . . . 14 (𝑘 = 𝑏 → (((𝐹𝑎) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))) ↔ ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))))
2317, 22imbi12d 223 . . . . . . . . . . . . 13 (𝑘 = 𝑏 → ((𝑎 <N 𝑘 → ((𝐹𝑎) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))) ↔ (𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))))))
2416, 23cbvral2v 2541 . . . . . . . . . . . 12 (∀𝑛N𝑘N (𝑛 <N 𝑘 → ((𝐹𝑛) <Q ((𝐹𝑘) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )) ∧ (𝐹𝑘) <Q ((𝐹𝑛) +Q (*Q‘[⟨𝑛, 1𝑜⟩] ~Q )))) ↔ ∀𝑎N𝑏N (𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))))
255, 24sylib 127 . . . . . . . . . . 11 (𝜑 → ∀𝑎N𝑏N (𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))))
26 caucvgprlemnkj.k . . . . . . . . . . . 12 (𝜑𝐾N)
27 caucvgprlemnkj.j . . . . . . . . . . . 12 (𝜑𝐽N)
28 breq1 3767 . . . . . . . . . . . . . 14 (𝑎 = 𝐾 → (𝑎 <N 𝑏𝐾 <N 𝑏))
29 fveq2 5178 . . . . . . . . . . . . . . . 16 (𝑎 = 𝐾 → (𝐹𝑎) = (𝐹𝐾))
30 opeq1 3549 . . . . . . . . . . . . . . . . . . 19 (𝑎 = 𝐾 → ⟨𝑎, 1𝑜⟩ = ⟨𝐾, 1𝑜⟩)
3130eceq1d 6142 . . . . . . . . . . . . . . . . . 18 (𝑎 = 𝐾 → [⟨𝑎, 1𝑜⟩] ~Q = [⟨𝐾, 1𝑜⟩] ~Q )
3231fveq2d 5182 . . . . . . . . . . . . . . . . 17 (𝑎 = 𝐾 → (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ) = (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))
3332oveq2d 5528 . . . . . . . . . . . . . . . 16 (𝑎 = 𝐾 → ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) = ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
3429, 33breq12d 3777 . . . . . . . . . . . . . . 15 (𝑎 = 𝐾 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ↔ (𝐹𝐾) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))
3529, 32oveq12d 5530 . . . . . . . . . . . . . . . 16 (𝑎 = 𝐾 → ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) = ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
3635breq2d 3776 . . . . . . . . . . . . . . 15 (𝑎 = 𝐾 → ((𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ↔ (𝐹𝑏) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))
3734, 36anbi12d 442 . . . . . . . . . . . . . 14 (𝑎 = 𝐾 → (((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))) ↔ ((𝐹𝐾) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))))
3828, 37imbi12d 223 . . . . . . . . . . . . 13 (𝑎 = 𝐾 → ((𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))) ↔ (𝐾 <N 𝑏 → ((𝐹𝐾) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))))
39 breq2 3768 . . . . . . . . . . . . . 14 (𝑏 = 𝐽 → (𝐾 <N 𝑏𝐾 <N 𝐽))
40 fveq2 5178 . . . . . . . . . . . . . . . . 17 (𝑏 = 𝐽 → (𝐹𝑏) = (𝐹𝐽))
4140oveq1d 5527 . . . . . . . . . . . . . . . 16 (𝑏 = 𝐽 → ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) = ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
4241breq2d 3776 . . . . . . . . . . . . . . 15 (𝑏 = 𝐽 → ((𝐹𝐾) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ↔ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))
4340breq1d 3774 . . . . . . . . . . . . . . 15 (𝑏 = 𝐽 → ((𝐹𝑏) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ↔ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))
4442, 43anbi12d 442 . . . . . . . . . . . . . 14 (𝑏 = 𝐽 → (((𝐹𝐾) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))) ↔ ((𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))))
4539, 44imbi12d 223 . . . . . . . . . . . . 13 (𝑏 = 𝐽 → ((𝐾 <N 𝑏 → ((𝐹𝐾) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))) ↔ (𝐾 <N 𝐽 → ((𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))))
4638, 45rspc2v 2662 . . . . . . . . . . . 12 ((𝐾N𝐽N) → (∀𝑎N𝑏N (𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))) → (𝐾 <N 𝐽 → ((𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))))
4726, 27, 46syl2anc 391 . . . . . . . . . . 11 (𝜑 → (∀𝑎N𝑏N (𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))) → (𝐾 <N 𝐽 → ((𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))))
4825, 47mpd 13 . . . . . . . . . 10 (𝜑 → (𝐾 <N 𝐽 → ((𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))))
4948imp 115 . . . . . . . . 9 ((𝜑𝐾 <N 𝐽) → ((𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))
5049simpld 105 . . . . . . . 8 ((𝜑𝐾 <N 𝐽) → (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
5150adantr 261 . . . . . . 7 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
521, 2sotri 4720 . . . . . . 7 (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))) → (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
534, 51, 52syl2anc 391 . . . . . 6 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
54 ltanqg 6498 . . . . . . . 8 ((𝑓Q𝑔QQ) → (𝑓 <Q 𝑔 ↔ ( +Q 𝑓) <Q ( +Q 𝑔)))
5554adantl 262 . . . . . . 7 ((((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) ∧ (𝑓Q𝑔QQ)) → (𝑓 <Q 𝑔 ↔ ( +Q 𝑓) <Q ( +Q 𝑔)))
56 caucvgprlemnkj.s . . . . . . . 8 (𝜑𝑆Q)
5756ad2antrr 457 . . . . . . 7 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → 𝑆Q)
58 caucvgpr.f . . . . . . . . 9 (𝜑𝐹:NQ)
5958, 27ffvelrnd 5303 . . . . . . . 8 (𝜑 → (𝐹𝐽) ∈ Q)
6059ad2antrr 457 . . . . . . 7 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝐹𝐽) ∈ Q)
61 nnnq 6520 . . . . . . . . 9 (𝐾N → [⟨𝐾, 1𝑜⟩] ~QQ)
62 recclnq 6490 . . . . . . . . 9 ([⟨𝐾, 1𝑜⟩] ~QQ → (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ) ∈ Q)
6326, 61, 623syl 17 . . . . . . . 8 (𝜑 → (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ) ∈ Q)
6463ad2antrr 457 . . . . . . 7 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ) ∈ Q)
65 addcomnqg 6479 . . . . . . . 8 ((𝑓Q𝑔Q) → (𝑓 +Q 𝑔) = (𝑔 +Q 𝑓))
6665adantl 262 . . . . . . 7 ((((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) ∧ (𝑓Q𝑔Q)) → (𝑓 +Q 𝑔) = (𝑔 +Q 𝑓))
6755, 57, 60, 64, 66caovord2d 5670 . . . . . 6 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝑆 <Q (𝐹𝐽) ↔ (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ))))
6853, 67mpbird 156 . . . . 5 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → 𝑆 <Q (𝐹𝐽))
69 nnnq 6520 . . . . . . . . 9 (𝐽N → [⟨𝐽, 1𝑜⟩] ~QQ)
70 recclnq 6490 . . . . . . . . 9 ([⟨𝐽, 1𝑜⟩] ~QQ → (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ) ∈ Q)
7127, 69, 703syl 17 . . . . . . . 8 (𝜑 → (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ) ∈ Q)
7271ad2antrr 457 . . . . . . 7 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ) ∈ Q)
73 ltaddnq 6505 . . . . . . 7 (((𝐹𝐽) ∈ Q ∧ (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ) ∈ Q) → (𝐹𝐽) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
7460, 72, 73syl2anc 391 . . . . . 6 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝐹𝐽) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
75 simprr 484 . . . . . 6 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)
761, 2sotri 4720 . . . . . 6 (((𝐹𝐽) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) → (𝐹𝐽) <Q 𝑆)
7774, 75, 76syl2anc 391 . . . . 5 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝐹𝐽) <Q 𝑆)
7868, 77jca 290 . . . 4 (((𝜑𝐾 <N 𝐽) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝑆 <Q (𝐹𝐽) ∧ (𝐹𝐽) <Q 𝑆))
7978ex 108 . . 3 ((𝜑𝐾 <N 𝐽) → (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) → (𝑆 <Q (𝐹𝐽) ∧ (𝐹𝐽) <Q 𝑆)))
803, 79mtoi 590 . 2 ((𝜑𝐾 <N 𝐽) → ¬ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆))
811, 2son2lpi 4721 . . 3 ¬ (((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆𝑆 <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
82 opeq1 3549 . . . . . . . . . . . 12 (𝐾 = 𝐽 → ⟨𝐾, 1𝑜⟩ = ⟨𝐽, 1𝑜⟩)
8382eceq1d 6142 . . . . . . . . . . 11 (𝐾 = 𝐽 → [⟨𝐾, 1𝑜⟩] ~Q = [⟨𝐽, 1𝑜⟩] ~Q )
8483fveq2d 5182 . . . . . . . . . 10 (𝐾 = 𝐽 → (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ) = (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))
8584oveq2d 5528 . . . . . . . . 9 (𝐾 = 𝐽 → (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) = (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
86 fveq2 5178 . . . . . . . . 9 (𝐾 = 𝐽 → (𝐹𝐾) = (𝐹𝐽))
8785, 86breq12d 3777 . . . . . . . 8 (𝐾 = 𝐽 → ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ↔ (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q (𝐹𝐽)))
8887anbi1d 438 . . . . . . 7 (𝐾 = 𝐽 → (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) ↔ ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q (𝐹𝐽) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)))
8988adantl 262 . . . . . 6 ((𝜑𝐾 = 𝐽) → (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) ↔ ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q (𝐹𝐽) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)))
9054adantl 262 . . . . . . . . 9 ((𝜑 ∧ (𝑓Q𝑔QQ)) → (𝑓 <Q 𝑔 ↔ ( +Q 𝑓) <Q ( +Q 𝑔)))
91 addclnq 6473 . . . . . . . . . 10 ((𝑆Q ∧ (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ) ∈ Q) → (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∈ Q)
9256, 71, 91syl2anc 391 . . . . . . . . 9 (𝜑 → (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∈ Q)
9365adantl 262 . . . . . . . . 9 ((𝜑 ∧ (𝑓Q𝑔Q)) → (𝑓 +Q 𝑔) = (𝑔 +Q 𝑓))
9490, 92, 59, 71, 93caovord2d 5670 . . . . . . . 8 (𝜑 → ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q (𝐹𝐽) ↔ ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))
9594adantr 261 . . . . . . 7 ((𝜑𝐾 = 𝐽) → ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q (𝐹𝐽) ↔ ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))
9695anbi1d 438 . . . . . 6 ((𝜑𝐾 = 𝐽) → (((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q (𝐹𝐽) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) ↔ (((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)))
9789, 96bitrd 177 . . . . 5 ((𝜑𝐾 = 𝐽) → (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) ↔ (((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)))
981, 2sotri 4720 . . . . 5 ((((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) → ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)
9997, 98syl6bi 152 . . . 4 ((𝜑𝐾 = 𝐽) → (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) → ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆))
100 ltaddnq 6505 . . . . . . 7 ((𝑆Q ∧ (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ) ∈ Q) → 𝑆 <Q (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
10156, 71, 100syl2anc 391 . . . . . 6 (𝜑𝑆 <Q (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
102 ltaddnq 6505 . . . . . . 7 (((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∈ Q ∧ (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ) ∈ Q) → (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
10392, 71, 102syl2anc 391 . . . . . 6 (𝜑 → (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
1041, 2sotri 4720 . . . . . 6 ((𝑆 <Q (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))) → 𝑆 <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
105101, 103, 104syl2anc 391 . . . . 5 (𝜑𝑆 <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
106105adantr 261 . . . 4 ((𝜑𝐾 = 𝐽) → 𝑆 <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
10799, 106jctird 300 . . 3 ((𝜑𝐾 = 𝐽) → (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) → (((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆𝑆 <Q ((𝑆 +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))))
10881, 107mtoi 590 . 2 ((𝜑𝐾 = 𝐽) → ¬ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆))
1091, 2son2lpi 4721 . . 3 ¬ (𝑆 <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)
11056ad2antrr 457 . . . . . . 7 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → 𝑆Q)
11163ad2antrr 457 . . . . . . 7 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ) ∈ Q)
112 ltaddnq 6505 . . . . . . 7 ((𝑆Q ∧ (*Q‘[⟨𝐾, 1𝑜⟩] ~Q ) ∈ Q) → 𝑆 <Q (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
113110, 111, 112syl2anc 391 . . . . . 6 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → 𝑆 <Q (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )))
114 simprl 483 . . . . . . 7 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾))
115 breq1 3767 . . . . . . . . . . . . . 14 (𝑎 = 𝐽 → (𝑎 <N 𝑏𝐽 <N 𝑏))
116 fveq2 5178 . . . . . . . . . . . . . . . 16 (𝑎 = 𝐽 → (𝐹𝑎) = (𝐹𝐽))
117 opeq1 3549 . . . . . . . . . . . . . . . . . . 19 (𝑎 = 𝐽 → ⟨𝑎, 1𝑜⟩ = ⟨𝐽, 1𝑜⟩)
118117eceq1d 6142 . . . . . . . . . . . . . . . . . 18 (𝑎 = 𝐽 → [⟨𝑎, 1𝑜⟩] ~Q = [⟨𝐽, 1𝑜⟩] ~Q )
119118fveq2d 5182 . . . . . . . . . . . . . . . . 17 (𝑎 = 𝐽 → (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ) = (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))
120119oveq2d 5528 . . . . . . . . . . . . . . . 16 (𝑎 = 𝐽 → ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) = ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
121116, 120breq12d 3777 . . . . . . . . . . . . . . 15 (𝑎 = 𝐽 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ↔ (𝐹𝐽) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))
122116, 119oveq12d 5530 . . . . . . . . . . . . . . . 16 (𝑎 = 𝐽 → ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) = ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
123122breq2d 3776 . . . . . . . . . . . . . . 15 (𝑎 = 𝐽 → ((𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ↔ (𝐹𝑏) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))
124121, 123anbi12d 442 . . . . . . . . . . . . . 14 (𝑎 = 𝐽 → (((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q ))) ↔ ((𝐹𝐽) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))))
125115, 124imbi12d 223 . . . . . . . . . . . . 13 (𝑎 = 𝐽 → ((𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))) ↔ (𝐽 <N 𝑏 → ((𝐹𝐽) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))))
126 breq2 3768 . . . . . . . . . . . . . 14 (𝑏 = 𝐾 → (𝐽 <N 𝑏𝐽 <N 𝐾))
127 fveq2 5178 . . . . . . . . . . . . . . . . 17 (𝑏 = 𝐾 → (𝐹𝑏) = (𝐹𝐾))
128127oveq1d 5527 . . . . . . . . . . . . . . . 16 (𝑏 = 𝐾 → ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) = ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
129128breq2d 3776 . . . . . . . . . . . . . . 15 (𝑏 = 𝐾 → ((𝐹𝐽) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ↔ (𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))
130127breq1d 3774 . . . . . . . . . . . . . . 15 (𝑏 = 𝐾 → ((𝐹𝑏) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ↔ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))
131129, 130anbi12d 442 . . . . . . . . . . . . . 14 (𝑏 = 𝐾 → (((𝐹𝐽) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))) ↔ ((𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))))
132126, 131imbi12d 223 . . . . . . . . . . . . 13 (𝑏 = 𝐾 → ((𝐽 <N 𝑏 → ((𝐹𝐽) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))) ↔ (𝐽 <N 𝐾 → ((𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))))
133125, 132rspc2v 2662 . . . . . . . . . . . 12 ((𝐽N𝐾N) → (∀𝑎N𝑏N (𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))) → (𝐽 <N 𝐾 → ((𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))))
13427, 26, 133syl2anc 391 . . . . . . . . . . 11 (𝜑 → (∀𝑎N𝑏N (𝑎 <N 𝑏 → ((𝐹𝑎) <Q ((𝐹𝑏) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )) ∧ (𝐹𝑏) <Q ((𝐹𝑎) +Q (*Q‘[⟨𝑎, 1𝑜⟩] ~Q )))) → (𝐽 <N 𝐾 → ((𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))))
13525, 134mpd 13 . . . . . . . . . 10 (𝜑 → (𝐽 <N 𝐾 → ((𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))))
136135imp 115 . . . . . . . . 9 ((𝜑𝐽 <N 𝐾) → ((𝐹𝐽) <Q ((𝐹𝐾) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))))
137136simprd 107 . . . . . . . 8 ((𝜑𝐽 <N 𝐾) → (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
138137adantr 261 . . . . . . 7 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
1391, 2sotri 4720 . . . . . . 7 (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ (𝐹𝐾) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))) → (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
140114, 138, 139syl2anc 391 . . . . . 6 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
1411, 2sotri 4720 . . . . . 6 ((𝑆 <Q (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) ∧ (𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q ))) → 𝑆 <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
142113, 140, 141syl2anc 391 . . . . 5 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → 𝑆 <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )))
143 simprr 484 . . . . 5 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)
144142, 143jca 290 . . . 4 (((𝜑𝐽 <N 𝐾) ∧ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)) → (𝑆 <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆))
145144ex 108 . . 3 ((𝜑𝐽 <N 𝐾) → (((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆) → (𝑆 <Q ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆)))
146109, 145mtoi 590 . 2 ((𝜑𝐽 <N 𝐾) → ¬ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆))
147 pitri3or 6420 . . 3 ((𝐾N𝐽N) → (𝐾 <N 𝐽𝐾 = 𝐽𝐽 <N 𝐾))
14826, 27, 147syl2anc 391 . 2 (𝜑 → (𝐾 <N 𝐽𝐾 = 𝐽𝐽 <N 𝐾))
14980, 108, 146, 148mpjao3dan 1202 1 (𝜑 → ¬ ((𝑆 +Q (*Q‘[⟨𝐾, 1𝑜⟩] ~Q )) <Q (𝐹𝐾) ∧ ((𝐹𝐽) +Q (*Q‘[⟨𝐽, 1𝑜⟩] ~Q )) <Q 𝑆))
Colors of variables: wff set class Syntax hints: ¬ wn 3 → wi 4 ∧ wa 97 ↔ wb 98 ∨ w3o 884 ∧ w3a 885 = wceq 1243 ∈ wcel 1393 ∀wral 2306 ⟨cop 3378 class class class wbr 3764 ⟶wf 4898 ‘cfv 4902 (class class class)co 5512 1𝑜c1o 5994 [cec 6104 Ncnpi 6370
Copyright terms: Public domain W3C validator
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Home > Standard Error > R Standard Error Of Mean
# R Standard Error Of Mean
## Contents
Of course deriving confidence intervals around your data (using standard deviation) or the mean (using standard error) requires your data to be normally distributed. Datacamp offers a free interactive introduction to R coding tutorial as an additional resource. You'd have to sample with replacement for that, i.e. share|improve this answer answered Mar 4 '15 at 17:22 gung 74.4k19161310 add a comment| up vote 3 down vote If I wanted to calculate the standard error of the mean weights, my review here
Already over 100,000 people took this free tutorial to sharpen their R coding skills. So... > sqrt(var(nums,na.rm=T)/samp.size(nums)[1]) # your result may differ 2.954936 ...you can use it like this. Just as in the case of finding the p values in previous chapter we have to use the pmin command to get the number of degrees of freedom. Spacing is optional, but I think it makes it a bit easier to understand if you use some indenting here.
## Plotrix Standard Error
Not the answer you're looking for? So from the command line... > setwd("Rspace") # if you've created this directory > rm(list=ls()) # clean out the workspace > ls() character(0) > nums = rnorm(25, mean=100, sd=15) # create After you hit the Enter key, R will see that you are defining a function, and it will give you the + prompt, meaning "tell me more." Type an open curly Number (pop.) Group I 10 3 300 Group II 10.5 2.5 230 Comparison 2 Mean Std.
1. I usually do.
2. If provided it's interpreted as sd of the population.
4. Mangiafico Search Contents Introduction Purpose of this book The Handbook for Biological Statistics About the author About R Obtaining R A Few Notes to Get Started with R
5. Object Oriented Programming 17.
6. print(with(PlantGrowth, tapply(weight, group, mean))) with(PlantGrowth, aov(weight ~ group)) -> aov.out print(summary.aov(aov.out)) print(summary.lm(aov.out)) Pull down File and choose Save.
7. The sem is going to be calculated on a data object--a vector in this case--so we have to pass the data to the function, and that is the point of "(x)".
what does one mean by numerical integration is too expensive? Do this. > x [1] 22 39 50 25 18 > mean(x) [1] 30.8 See? of connection –forecaster Jan 21 '15 at 0:01 3 @forecaster Tom didn't say stderr calculates the standard error, he was warning that this name is used in base, and John Ggplot Standard Error I typically use se.
Our sem function is good enough, but if there are missing values in the data vector, sem() will choke. > nums[20] = NA # create a missing value > sem(nums) [1] When math and english collide! Here's what they should have said when they were first thinking about this. "Hey! The estimator $\widehat{\sigma_{p}}$ is (almost) unbiased, so you can improve this estimator using all your samples by taking the mean of $\widehat{\sigma_{p}}$ over the 20 samples.
Can anyone identify the city in this photo? Standard Error In Excel Is there a way to eat rice with your fingers, in front of Westerners? more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed The script has created the variables "x" and "y" in your workspace (and has erased any old objects you had by that name--sorry).
## R Standard Error Of Regression
Details MeanSE calculates the standard error of the mean defined as: \frac{σ}{√{n}} σ being standard deviation of x and n the length of x. How to explain leaving a job for a huge ethical/moral issue to a potential employer - without REALLY explaining it When math and english collide! Plotrix Standard Error You have to be careful here. R Standard Error Lm Then choose to open "script2.txt" (or "script2.R", whatever!).
You split 100 into 20 partitions, and got the standard deviation of the mean of the parts of this partition. http://caribtechsxm.com/standard-error/r2-standard-error.php The standard deviation of a sample, divided by $\sqrt N$, is also an estimate of the standard error of the mean. Case Study II: A JAMA Paper on Cholesterol R Tutorial Docs » 9. We will refer to group one as the group whose results are in the first row of each comparison above. R Aggregate Standard Error
It depends. The commands to find the confidence interval in R are the following: > a <- 5 > s <- 2 > n <- 20 > error <- qnorm(0.975)*s/sqrt(n) > left <- And who came up with that convoluted syntax? http://caribtechsxm.com/standard-error/r-glm-standard-error.php When to use standard deviation?
sd the standard deviation of x. Plotrix R In R, something like that should work : sdEstimator <- function(sample) { return(sqrt((1-n/N)*1/n) * N / N-1 * sd(sample)) } And assuming you put all sdEstimator values in sdEstVector, your new However, nothing was echoed to your Console because you didn't tell it to print().
## To get to it, pull down the File menu and choose New Script (New Document on a Mac).
Value the standard error as numeric value. You got the standard deviation of the partition. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Standard Error R Studio What's the point of Pauli's Exclusion Principle if time and space are continuous?
Join them; it only takes a minute: Sign up In R, how to find the standard error of the mean? It remains that standard deviation can still be used as a measure of dispersion even for non-normally distributed data. If you just hit the Enter key at this point, your function is done. http://caribtechsxm.com/standard-error/r-help-standard-error.php Take the tapply() function for example.
Again we assume that the sample mean is 5, the sample standard deviation is 2, and the sample size is 20. Example Standard error example ### -------------------------------------------------------------- ### Standard error example, p. 115 ### -------------------------------------------------------------- Input =(" Stream Fish Mill_Creek_1 76 Mill_Creek_2 102 North_Branch_Rock_Creek_1 12 North_Branch_Rock_Creek_2 39 Rock_Creek_1 55 Rock_Creek_2 93 R news and tutorials contributed by (580) R bloggers Home About RSS add your blog! Don't like it?
It's good programming practice if you think you might need a reminder later of what the heck it is you've done here! > ?describe No documentation for 'describe' in specified packages I have changed this answer to reflect that. –John Jan 13 '14 at 14:02 2 Tom, NO stderr does NOT calculate standard error it displays display aspects. Can unconnected inputs make an IC get warm? Don't understand what a file extension is?
Standard error of the mean It is a measure of how precise is our estimate of the mean. #computation of the standard error of the mean sem<-sd(x)/sqrt(length(x)) #95% confidence intervals of For each comparison there are two groups. We use a 95% confidence level and wish to find the confidence interval.
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# IC Engine Terminology | Nomenclature | Basic Definitions used In Ic Engine
## IC Engine Terminology | Nomenclature | Basic Definitions used In IC Engine
IC ENGINE TERMINOLOGY:
The following terms/Nomenclature associated with an engine are explained for the better understanding of the working principle of the IC engines.
1. Bore (B) :
The nominal inside diameter of the engine cylinder is called bore.
2. Top dead center (T.D.C.)
In a reciprocating engine the piston moves to and fro motion in the cylinder. When the piston moves upper direction in the cylinder, a point at which the piston comes to rest or change its direction known as top dead center. It is situated at top end of cylinder.
3. Bottom dead center (B.D.C.)
When the piston moves in downward direction, a point at which the piston come to rest or change its direction known as bottom dead center. It is situated in bottom side of cylinder.
4. Stroke (L)
The distance travelled by the piston from TDC to BDC is called stroke. In other words, the maximum distance travelled by the piston in the cylinder in one direction is known as stroke.
It is equal to twice the radius of the crank.
5. Total volume of cylinder (Vtotal)
It is the volume of cylinder when the piston is at bottom dead center. Generally, it is measure in centimeter cube (c.c.).
6. Clearance volume of cylinder (Vclearance)
It is the volume of cylinder when the piston is at top dead center.
Clearance volume: the space above the piston head at the TDC, and is denoted by Vc:
Volume of the cylinder: V = Vc + Vs
7. Swept or displace volume (Vswept)
It is the volume which swept by the piston. The difference between total volume and clearance volume is known as swept volume.
Swept volume = Total volume – Clearance volume
8. Compression ratio
The ratio of maximum volume to minimum volume of cylinder is known as the compression ratio. It is 8 to 12 for spark ignition engine and 12 to 24 for compression ignition engine.
Compression ratio = Total volume / Clearance volume
R= (Vc + Vs) / Vc
9. Stroke bore ratio
Stroke bore ratio is the ratio of bore (diameter of cylinder) to length of stroke. It is generally equal to one for small engine and less than one for large engine.
Stroke bore ratio = inner diameter of cylinder / length of stroke
10. Mean effective pressure
The average pressure acting upon the piston is known as mean effective pressure. It is given by the ratio of the work done by the engine to the total volume of engine.
Mean effective pressure = Work done by engine / Total volume of cylinder
11. Piston
Area (A) – The area of circle of diameter equal to the cylinder bore is called as piston area. It is
denoted by the letter ‘A’ and is expressed in cm2.
12. Mean piston speed: the distance traveled by the piston per unit of time:
Mean piston speed (m/s ) = 2lN/60
where, l= length of strokes (mm)
N= Number of crankshaft revolution per minute (rpm).
Sachin Thorat
Sachin is a B-TECH graduate in Mechanical Engineering from a reputed Engineering college. Currently, he is working in the sheet metal industry as a designer. Additionally, he has interested in Product Design, Animation, and Project design. He also likes to write articles related to the mechanical engineering field and tries to motivate other mechanical engineering students by his innovative project ideas, design, models and videos.
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# How to compute data imbalance?
I have to compare amplitudes of two signal ("ideal " signal and "estimated" signal) and want to compute an amplitude imbalance
For my calculation I used the following equation: $$A = \frac{\left | A_{max} - A_{aver}\right |}{A_{aver}} \cdot 100%$$
as I do if I need compute a voltage or a current imbalance.
I need the value in dB-scale, so I didnt multiply by 100%... and $$20 \cdot \log_{10}A$$.
Ma result is -25 dB ( the result should be less than 0,5dB, p 20-21 (65)). I think I should use for amplitude imbalance another math equation.
Could someone explain how imbalance should be computed?
Compute the dB of each and then subtract the two to get the difference in dB which should be very small. You can get the same result from converting the ratio of the two to dB directly.
For dB of voltage quantities of AC sources that are arbitrary waveforms you would use the rms value, so the difference between two different sources in dB using both approaches would be:
$$20Log_{10}(A_1) - 20Log_{10}(A_2)$$
$$20Log_{10}(A_1/A_2)$$
Where $$A_1$$ is the rms value for source 1 and $$A_2$$ is the rms value for source 2. The rms value can never be negative or complex. Since we are using a ratio, if the two waveforms are similar (both sinusoids for example), then any other consistent metric of amplitude could be used, such as peak amplitude instead of rms).
From the link the OP gave, the intention here is to derive an amplitude and phase imbalance specification and for that the relationship between SNR and imbalance is that the SNR will be the normalized error vector. For example, if the error vector normalized to ideal was 0.1 radians, the SNR would be $$20Log_{10}(0.1) = 20$$ dB. For small angle the amplitude error and phase error would have the same effect, so if the error vector was 0.1 due to amplitude imbalance, we would also get a 20 dB SNR. (Universally, it is the magnitude of the error vector). An error vector of 0.1 dB is an amplitude imbalance of $$20Log_{10}(1\pm 0.1) = +0.8$$ or $$-0.9$$ dB.
• I have an "ideal " amplitude of the signal and "estimated" amplitude of the signal. I can compute only one rms value Apr 20, 2022 at 5:20
• i calculated the subtract of the two of them: amplitude of the "ideal"signal and amplitude of the "estimated" signal (from -0.0506 to 0.0482) , and convert it in dB $20 \cdot \log_{10} (\Delta A)$. In dB i got the complex numbers. Apr 20, 2022 at 7:24
• The rms value cannot be negative or complex. If the "ideal" is a sinusoid, divide the peak value by $\sqrt{2}$ to get rms. You don't do $\Delta A$, but the ratio of the rms values as I described, or take the dB of each one separately and then subtract the dB results. Apr 20, 2022 at 11:37
• Assuming you have actual samples of each one, then compute the standard deviation of the samples. If there is no DC offset, the standard deviation is the rms value. Or compute the variance and then use $10Log_{10}()$ given the variance is a power quantity. All equivalent. Apr 20, 2022 at 11:40
• $\Delta A = A_{estimated} - A_{ideal}$. I want to determine a deviation. The magnitude of estimated signal is found to be slighly varing around the magnitude of the ideal signal. I need to comput the deviation. I thiught I need to work with substract of the magnitudes Apr 20, 2022 at 12:18
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View the step-by-step solution to:
# PatientInfo Program *The user(employee at the doctor's office) inputs: the patient's first, middle and last name. the patient's smoking status: (Y/N)...
PatientInfo Program
*The user(employee at the doctor's office) inputs:
• the patient's first, middle and last name. the patient's smoking status: (Y/N)
• the patient's DOB family history of heart disease: (Y/N)
• the patient's sex the patient's 3 cholesterol levels: Total
• the patient's height in inches cholesterol, HDL, and LDL
• the patient's weight in pounds
*if the month was input as an integer, your program will use a switch statement to get the name of the month. If the month was input as a String, your program will use a switch statement to get the month number.
*calculate the BMI for the patient
*calculate the patient's weight in kg (1kg = 0.45359237lbs)
*calculate the patient's age (you can use integer literals to represent the current date) *evaluate the total cholesterol based on age and gender. See charts. (low/normal/high) evaluate the LDL based on age and gender. See charts. (normal/high) evaluate the HDL based on age and gender. See charts. (low/normal)
*Show the patient's first and middle initial followed by their last name in capital letters
*Show the patient's DOB in 2 ways (ex: 1/1/1990 and January 1, 1990)
*Show the patient's age
*Show the patient's height in the form of feet and inches (ex: 6'1")
*Show the patient's weight in pounds (*on the same line show the patient's weight in kilograms) - limit the precision to 2 places after the decimal
*Show the patient's BMI - limit the precision to 1 place after the decimal
*Show the patient's Total Cholesterol (show the evaluation of the Total Cholesterol: high/normal)
*Show the patient's HDL
(extra: show the eval of the HDL: normal/high)
*Show the patient's LDL
(extra: show the eval of the LDL: low/normal)
*Ask the user if the program should process an interpretation of the information:
*If they type Y (not case sensitive), show interpretation of:
*the BMI (underweight/normal/overweight) based on the chart
*If the patient is underweight or overweight, print a message "Referral to registered dietitian recommended."
* if the patient has any of the following risk factors for heart disease: over 60, is a smoker, has a family history of heart disease, BMI suggests they are overweight, total cholesterol is high, LDL is high, or HDL is low. Print "Patient may be at risk of heart disease". Print each reason the patient may be at risk of heart disease.
BMI <18.5: underweight
BMI between 18.5 and 25: normal
BMI > 25: overweight
if the patient is under 19:
type of Cholesterol Healthy Level
Total Cholesterol Less than 170mg/DL
LDL Less than 100mg/DL
HDL More than 45mg/dl
Men age 20 or older:
type of Cholesterol Healthy Level
Total Cholesterol 125 to 200mg/DL
LDL Less than 100mg/DL
HDL 40mg/dl or higher
Women age 20 or older:
type of Cholesterol Healthy Level
Total Cholesterol 125 to 200mg/DL
LDL Less than 100mg/DL
HDL 50mg/dl or higher
import java.util.Scanner;
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https://questions.examside.com/past-years/jee/question/an-ideal-gas-occupies-a-volume-of-2m3-at-a-pressure-of-3-jee-main-physics-units-and-measurements-thkzp3yjeferiu7g
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1
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
An ideal gas occupies a volume of 2m3 at a pressure of 3 $$\times$$ 106 Pa. The energy of the gas is :
A
6 $$\times$$ 104 J
B
9$$\times$$ 106 J
C
3 $$\times$$ 102 J
D
108 J
2
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and the of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is :
A
K1 + K2
B
$${{{K_1} + 3{K_2}} \over 4}$$
C
$${{{K_1} + {K_2}} \over 2}$$
D
$${{2{K_1} + 3{K_2}} \over 5}$$
3
JEE Main 2019 (Online) 12th January Morning Slot
+4
-1
For the given cyclic process CAB as shown for a gas, the work done is :
A
1 J
B
10 J
C
5 J
D
30 J
4
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180oC and B upto ToC, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is
A
200oC
B
270oC
C
230oC
D
250oC
EXAM MAP
Medical
NEET
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https://www.solutioninn.com/the-engine-of-the-lightweight-plane-is-supported-by-struts
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# The engine of the lightweight plane is supported by struts
The engine of the lightweight plane is supported by struts that are connected to the space truss that makes up the structure of the plane. The anticipated loading in two of the struts is shown. Express each of these forces as a Cartesian vector.
Given:
F1 = 400 lb
F2 = 600 lb
a = 0.5 ft
b = 0.5 ft
c = 3.0 ft
d = 2.0 ft
e = 0.5 ft
f = 3.0 ft
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# Viscosity grooves benchmark#
This benchmark was designed by Dave May and this section was contributed by Cedric Thieulot.
The domain is a two-dimensional Cartesian box of size $$L\times L$$. The velocity and pressure fields are given by
\begin{split}\begin{aligned} u(x,y) &=& x^3 y + x^2 + xy + x, \\ v(x,y) &=& -\frac{3}{2}x^2y^2 - 2xy - \frac{1}{2}y^2 - y, \\ p(x,y) &=& x^2y^2 + xy + 5 + p_0, \end{aligned}\end{split}
where $$p_0$$ is a constant to be determined based on the type of pressure normalization. The viscosity is chosen to be
$\eta(x,y)=-\sin(p)+1+\epsilon = -\sin (x^2y^2 + xy + 5) + 1 + \epsilon,$
where $$\epsilon$$ controls the viscosity contrast. It is easy to verify that the flow is incompressible as the velocity field satisfies $$\nabla\cdot \mathbf u = 0$$. The right hand side term of the Stokes equation is obtained by inserting the expressions for velocity, pressure and viscosity in the momentum conservation equation, see Thieulot [2019] for details. The velocity, pressure and right hand side magnitude are shown in Fig. 196 for $$L=3$$ and $$\epsilon=0.1$$.
The $$p_0$$ constant can be determined by requiring that the pressure is normalized over the volume of the domain:
$\int_\Omega p dV= \int_0^L\int_0^L p(x,y) \, dx dy = \int_0^L\int_0^L (x^2y^2+xy+5)\, dx \, dy + \int_0^L \int_0^L p_0 \, dx \, dy =0.$
It then follows that:
$p_0 =- \frac{1}{L^2} \int_0^L\int_0^L (x^2y^2+xy+5) dx dy = -\frac{L^4}{9}-\frac{L^2}{4} - 5.$
As seen in Fig. 197, the value of $$\epsilon$$ controls the viscosity field amplitude: when the $$\sin$$ term of the viscosity takes value 1, the viscosity is then equal to $$\epsilon$$; when the $$\sin$$ is equal to $$-1$$, the viscosity is then $$2+\epsilon$$. In other words, the ratio between maximal and minimal viscosity in the domain is of the order $$\frac{2}{\epsilon}$$.
Another interesting aspect of this benchmark is the fact that increasing the domain size adds complexity to it as it increases the number of low viscosity zones and the spacing between them decreases.
The velocity and pressure errors (in the $$L_2$$ norm) are measured for $$L=1,2,3$$, global refinement levels 3 to 9 (resolutions $$8\times 8$$ to $$512\times 512$$) and $$\epsilon=10^{-1},10^{-2},10^{-3}$$. Fig. 198 shows the velocity and pressure error convergence as a function of the mesh size for $$\epsilon=0.1$$ (results are identical for the other two $$\epsilon$$ values). The expected convergence rates (cubic convergence for velocity and quadratic for pressure) are recovered for the $$1\times 1$$ domain at all resolutions. These rates are recovered for the $$2\times 2$$ domain for resolutions above level 6. We find that the multitude of low viscosity bands in the upper right corner of the $$3\times 3$$ domain will require a refinement level larger than 9 to recover the optimal convergence rates.
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# Subjective probabilities
Also found in: Medical, Encyclopedia.
## Subjective probabilities
Probabilities that are determined subjectively (for example, on the basis of judgment rather than statistical sampling).
## Subjective Probabilities
One's personal judgment about the outcome of a particular investment. At its most extreme, relying on subjective probabilities represents the exact opposite of a formula plan, which seeks to eliminate all personal judgments and biases from investing. By definition, one cannot calculate subjective probabilities because they are based ultimately on one's feeling. Most investors rely on a combination of subjective probabilities and more objective calculations when making investment decisions.
References in periodicals archive ?
Objective information about individuals known and used or known and not used by insurers and information privately held by individuals affect objective and subjective probabilities about having an accident next year.
Uncertain and subjective probabilities can be granulated in different terms like: improbable or doubtful, but we will draw our attention to special form of probability granulation, focusing on point value with inherent uncertainty: around 30%, around 75%, etc.
An entrepreneur thinking of launching a new product does not calculate subjective probabilities and then maximize expected utility.
The approach is a way of emphasizing the relationship between useful subjective probabilities and experimental probabilities.
(41) The most popular candidate for rigorously evaluating and combining related evidence is the Bayesian approach, utilizing subjective probabilities.
Note that Good is saying merely that all probabilities can be interpreted as subjective probabilities, not that all probabilities must be interpreted as subjective probabilities.
As discussed in the previous section, from a subjetivist interpretation, this result justifies the reasons that lead to associate -inteasubjectively- the probability of certain events to the frequency observed in similar events, replacing the unknown objective probabilities and the concept of independence by subjective probabilities and exchangeability (23).
If investors set a club's stock price prior to a game, [V.sub.0], as the weighted average of postgame values, with weights equal to their subjective probabilities of game outcomes, [prob.sup.subj.sub.win], [prob.sup.subj.sub.draw], and [prob.sup.subj.sub.loss], then [V.sub.0] can be expressed as
in Teacher Preparation (Frederic Gourdeau) [Presented in both French and English]; (11) On the Origins of Dynamic Number in the Breakdown of Structural, Metaphoric, and Historic Conceptions of Human Mathematics (Nicholas Jackiw and Nathalie Sinclair); (12) Analysis of Resources Mobilized by a Teacher and Researcher in Designing/Experimenting/Reflecting Upon Modelling Sequences Based on Elementary Combinatorics and Aimed at Introducing 7th Graders to Modelling (Souleymane Barry); (13) Being (Almost) a Mathematician: Teacher Identity Formation in Post-Secondary Mathematics (Mary Beisiegel); (14) Subjective Probabilities Derived from the Perceived Randomness of Sequences of Outcomes (Egan J.
Referring to mistakes such as the assessment about Himalayan glaciers, the review noted that such recommendations that involved subjective probabilities should be based on the principle of " high agreement, much evidence." " Scientific uncertainty is best communicated by indicating the nature, amount, and quality of studies on a particular topic, as well as the level of agreement among studies," it said.
More specifically, the subjective probabilities inherent in odds and associated with particular information features (for example, a horse's PP) can be compared with the objective probability of success (as revealed, ex post, by race outcomes).
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# Hamiltonian path
This article is about the overall graph theory concept of a Hamiltonian path. For the specific problem of determining whether a Hamiltonian path or cycle exists in a given graph, see Hamiltonian path problem.
A Hamiltonian cycle in a dodecahedron. Like all platonic solids, the dodecahedron is Hamiltonian.
The Herschel graph is the smallest possible polyhedral graph that does not have a Hamiltonian cycle.
In the mathematical field of graph theory, a Hamiltonian path (or traceable path) is a path in an undirected or directed graph that visits each vertex exactly once. A Hamiltonian cycle (or Hamiltonian circuit) is a Hamiltonian path that is a cycle. Determining whether such paths and cycles exist in graphs is the Hamiltonian path problem, which is NP-complete.
Hamiltonian paths and cycles are named after William Rowan Hamilton who invented the icosian game, now also known as Hamilton's puzzle, which involves finding a Hamiltonian cycle in the edge graph of the dodecahedron. Hamilton solved this problem using the icosian calculus, an algebraic structure based on roots of unity with many similarities to the quaternions (also invented by Hamilton). This solution does not generalize to arbitrary graphs.
However, despite being named after Hamilton, Hamiltonian cycles in polyhedra had also been studied a year earlier by Thomas Kirkman, who, in particular, gave an example of a polyhedron without Hamiltonian cycles.[1]
## Definitions
A Hamiltonian path or traceable path is a path that visits each vertex exactly once. A graph that contains a Hamiltonian path is called a traceable graph. A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices.
A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a cycle that visits each vertex exactly once (except for the vertex that is both the start and end, which is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph.
Similar notions may be defined for directed graphs, where each edge (arc) of a path or cycle can only be traced in a single direction (i.e., the vertices are connected with arrows and the edges traced "tail-to-head").
A Hamiltonian decomposition is an edge decomposition of a graph into Hamiltonian circuits.
## Properties
Any Hamiltonian cycle can be converted to a Hamiltonian path by removing one of its edges, but a Hamiltonian path can be extended to Hamiltonian cycle only if its endpoints are adjacent.
All Hamiltonian graphs are biconnected, but a biconnected graph need not be Hamiltonian (see, for example, the Petersen graph).[3]
An Eulerian graph G (a connected graph in which every vertex has even degree) necessarily has an Euler tour, a closed walk passing through each edge of G exactly once. This tour corresponds to a Hamiltonian cycle in the line graph L(G), so the line graph of every Eulerian graph is Hamiltonian. Line graphs may have other Hamiltonian cycles that do not correspond to Euler tours, and in particular the line graph L(G) of every Hamiltonian graph G is itself Hamiltonian, regardless of whether the graph G is Eulerian.[4]
A tournament (with more than two vertices) is Hamiltonian if and only if it is strongly connected.
The number of different Hamiltonian cycles in a complete undirected graph on n vertices is (n 1)! / 2 and in a complete directed graph on n vertices is (n 1)!. These counts assume that cycles that are the same apart from their starting point are not counted separately.
## Bondy–Chvátal theorem
The best vertex degree characterization of Hamiltonian graphs was provided in 1972 by the BondyChvátal theorem, which generalizes earlier results by G. A. Dirac (1952) and Øystein Ore. Both Dirac's and Ore's theorems can also be derived from Pósa's theorem (1962). Hamiltonicity has been widely studied with relation to various parameters such as graph density, toughness, forbidden subgraphs and distance among other parameters.[5] Dirac and Ore's theorems basically state that a graph is Hamiltonian if it has enough edges.
Bondy–Chvátal theorem operates on the closure cl(G) of a graph G with n vertices, obtained by repeatedly adding a new edge uv connecting a nonadjacent pair of vertices u and v with degree(v) + degree(u) ≥ n until no more pairs with this property can be found.
Bondy–Chvátal theorem
A graph is Hamiltonian if and only if its closure is Hamiltonian.
As complete graphs are Hamiltonian, all graphs whose closure is complete are Hamiltonian, which is the content of the following earlier theorems by Dirac and Ore.
Dirac (1952)
A simple graph with n vertices (n ≥ 3) is Hamiltonian if every vertex has degree n / 2 or greater.
Ore (1960)
A graph with n vertices (n ≥ 3) is Hamiltonian if, for every pair of non-adjacent vertices, the sum of their degrees is n or greater (see Ore's theorem).
The following theorems can be regarded as directed versions:
Ghouila-Houiri (1960)
A strongly connected simple directed graph with n vertices is Hamiltonian if every vertex has a full degree greater than or equal to n.
Meyniel (1973)
A strongly connected simple directed graph with n vertices is Hamiltonian if the sum of full degrees of every pair of distinct non-adjacent vertices is greater than or equal to 2n − 1.
The number of vertices must be doubled because each undirected edge corresponds to two directed arcs and thus the degree of a vertex in the directed graph is twice the degree in the undirected graph.
A simple graph with n vertices has a Hamiltonian path if, for every non-adjacent vertex pairs the sum of their degrees and their shortest path length is greater than n.[6]
The above theorem can only recognize the existence of a Hamiltonian path in a graph and not a Hamiltonian Cycle. A similar sufficiency condition for Hamiltonian cycles is introduced by Kaykobad.[7] It follows-
A simple biconnected graph with n vertices is Hamiltonian if, for every non-adjacent vertex pair the sum of their degrees and their shortest path length is greater than or equal to n+1 with strict inequality holding for at least one pair of vertices.
## Existence of Hamiltonian cycles in planar graphs
Theorem (Whitney, 1931)
A 4-connected planar triangulation has a Hamiltonian cycle.
Theorem (Tutte, 1956)
A 4-connected planar graph has a Hamiltonian cycle.
## Notes
1. Biggs, N. L. (1981), "T. P. Kirkman, mathematician", The Bulletin of the London Mathematical Society, 13 (2): 97–120, doi:10.1112/blms/13.2.97, MR 608093.
2. Eric Weinstein. "Biconnected Graph". Wolfram MathWorld.
3. Balakrishnan, R.; Ranganathan, K. (2012), "Corollary 6.5.5", A Textbook of Graph Theory, Springer, p. 134, ISBN 9781461445296.
4. Gould, Ronald J. (July 8, 2002). "Advances on the Hamiltonian Problem - A Survey" (PDF). Emory University. Retrieved 2012-12-10.
5. Rahman, M. S.; Kaykobad, M. (April 2005). "On Hamiltonian cycles and Hamiltonian paths". Information Processing Letters. 94: 37–41. doi:10.1016/j.ipl.2004.12.002.
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14-Alexandrite
How to use valid range constraint for string??
Hi All,
I would like to know, how to use "valid range" constraint for String attribute.
I've defined: From aa To bb.
As a result, I'm able to create business object that has 'ac' as a value of attribute, but 'ca' isn't acceptable.
Could somebody explain it in simple words?
Krzysztof
1 ACCEPTED SOLUTION
Accepted Solutions
21-Topaz I
(To:Gucio)
Hi @Gucio
it is similar to numbers.
range from 11 to 22 is not just 11, 12, 21, 22 but it is 11-19 and 20 - 22
an alphabet range is a-z for one position of a character. same as a number 0-9.
Is it more clear?
With range "aaa - ccc" you can still write "cbz" because it is in specific range.
PetrH
4 REPLIES 4
21-Topaz I
(To:Gucio)
Hi @Gucio
What do you expect?
Your result definition is from aa to az and from ba to bb
if you expect a range aa, ab, ba, bc then you should use different constraint for example the Legal value list
Hope this help
PetrH
14-Alexandrite
(To:HelesicPetr)
Hi Petr,
I'm still not able to catch it.
I was expected to achieve result similar to your example with Legal Value.
I was thought, that giving boundaries starting from 'aa' and finishing at 'bb' will allow user to fill one of four options: 'aa' or 'ab' or 'ba' or bb'. Why 'az' and 'ba' did you mention as result?
Going deeper, making valid range form 'aaa' to 'ccc' in my mind should allow typing 'aaa' or 'aab' ..... up to 'ccc'.
Would you be so kind as to clarify the answer?
Krzysztof
21-Topaz I
(To:Gucio)
Hi @Gucio
it is similar to numbers.
range from 11 to 22 is not just 11, 12, 21, 22 but it is 11-19 and 20 - 22
an alphabet range is a-z for one position of a character. same as a number 0-9.
Is it more clear?
With range "aaa - ccc" you can still write "cbz" because it is in specific range.
PetrH
14-Alexandrite
(To:HelesicPetr)
@Petr
now it is clear 🙂
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# cycloid
cycloid, the curve generated by a point on the circumference of a circle that rolls along a straight line. If r is the radius of the circle and θ (theta) is the angular displacement of the circle, then the polar equations of the curve are x = r(θ - sin θ) and y = r(1 - cos θ).
The points of the curve that touch the straight line are separated along the line by a distance equal to 2πr, which is the circumference of the circle, indicating one complete revolution of the circle. The curve is periodic, which means that it repeats in an identical pattern for each cycle, or length of the line, that is equal to 2πr.
One variant of the simple cycloid is the curtate cycloid, for which the curve falls below the line at the cusps, making retrograde loops in which the curve moves in the direction opposite to that of the rolling circle.
The prolate cycloid is similar to the simple cycloid except that the curve has no cusps and does not intersect the line. The prolate is formed by a point on a radius less than that of the rolling circle, such as a point on the spoke of a wheel.
For the case of a circle rolled along outside the circumference of another circle, an epicycloid is formed. For a circle rolled along inside the circumference of another circle, a hypocycloid is formed. See also brachistochrone.
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# Using Multinomial with an underlying model for the probabilities
Hi all,
I’m trying to formulate a simple example in PyMC3 of using a Multinomial distribution to infer the probabilities of a die throw, except where the side probabilities are linked by a specific model with some lesser number of parameters. In the example below it is just a linearly decreasing probability with some characteristic slope m, which is the parameter to infer. I’ve got it working using the code below by setting up a Uniform random variable for the prior and then making a Deterministic variable for each side probability. But I don’t really like having all the terms written out individually, especially if one would extend the number of possible states it would get very ugly. I was thinking of using a custom DensityDist to specify the probabilities, but I can’t quite figure out if that is feasible or not. Or possibly I could write a custom distribution, but also it is not readily apparent how to do that. Does anyone have any insight on how to improve the code below? Thanks!
``````import numpy as np
import pymc3 as mc
def randomDataLinearDiceModel(p, n):
# Calculate the cumulative probabilities.
c = np.cumsum(p)
# Generate a random sequence of dice rolls.
r = np.random.rand(n)
x = np.zeros((n,), dtype=np.uint8)
for i in range(n):
x[i] = np.argmax(r[i]<c)
# Get the counts for each die side.
(counts,edges) = np.histogram(x,np.arange(0-0.5,len(p)+0.5,1.0))
centers = (edges[:-1]+edges[1:])/2
return centers,counts
def bayesianInferenceLinearDiceModel(p, n, counts):
with mc.Model() as model:
# Prior distributions.
m = mc.Uniform('m', lower=0, upper=0.05)
p0 = mc.Deterministic('p0', (1/6) + 3*m)
p1 = mc.Deterministic('p1', (1/6) + 2*m)
p2 = mc.Deterministic('p2', (1/6) + 1*m)
p3 = mc.Deterministic('p3', (1/6) - 1*m)
p4 = mc.Deterministic('p4', (1/6) - 2*m)
p5 = mc.Deterministic('p5', (1/6) - 3*m)
# Likelihood.
counts_likelihood = mc.Multinomial('counts', n=n, p=[p0,p1,p2,p3,p4,p5], observed=counts)
# Perform the sampling.
trace = mc.sample(2000, tune=1000)
return trace
m=0.045
p = np.array([1/6+3*m,1/6+2*m,1/6+1*m,1/6-1*m,1/6-2*m,1/6-3*m])
n=10000
sides,rolls = randomDataLinearDiceModel(p, n)
trace = bayesianInferenceLinearDiceModel(p, n, rolls)
print("Posterior distribution summary:")
print(mc.summary(trace))
print("Finished.")
``````
``````def bayesianInferenceLinearDiceModel(p, n, counts):
lst = np.concatenate((np.arange(3,0,-1), np.arange(-1,-4,-1)))
with mc.Model() as model:
# Prior distributions.
m = mc.Uniform('m', lower=0, upper=0.05)
p = mc.Deterministic('p', 1/6 + lst*m)
# Likelihood.
counts_likelihood = mc.Multinomial('counts', n=n, p=p, observed=counts)
# Perform the sampling.
trace = mc.sample(2000, tune=1000)
return trace
m=0.045
p = np.array([1/6+3*m,1/6+2*m,1/6+1*m,1/6-1*m,1/6-2*m,1/6-3*m])
n=10000
sides,rolls = randomDataLinearDiceModel(p, n)
trace = bayesianInferenceLinearDiceModel(p, n, rolls)
print("Posterior distribution summary:")
print(mc.summary(trace))
print("Finished.")
``````
1 Like
@nze22 solution looks good to me!
You might also wanna look into softmax regression if you need something more generalizable - e.g. you have to estimate the intercepts also, or the intercept and/or slopes vary by category, etc.
Do we need to pass in the previously calculated values for `m` and `p` to the model, since they’re RVs in the model?
So, for clarity maybe:
``````def bayesianInferenceLinearDiceModel(n, counts):
lst = np.concatenate((np.arange(3,0,-1), np.arange(-1,-4,-1)))
with mc.Model() as model:
# Prior distributions.
m = mc.Uniform('m', lower=0, upper=0.05)
p = mc.Deterministic('p', 1/6 + lst*m)
# Likelihood.
counts_likelihood = mc.Multinomial('counts', n=n, p=p, observed=counts)
# Perform the sampling.
trace = mc.sample(2000, tune=1000)
return trace
m=0.045
p = np.array([1/6+3*m,1/6+2*m,1/6+1*m,1/6-1*m,1/6-2*m,1/6-3*m])
n=10000
_, counts = randomDataLinearDiceModel(p, n)
trace = bayesianInferenceLinearDiceModel(n, counts)
print("Posterior distribution summary:")
print(mc.summary(trace))
print("Finished.")
``````
?
First, thanks for the reply. I didn’t realize that the deterministic RV would automatically be expanded to the correct dimensions. That said, I now have the two following versions, one of which uses Deterministic variables and one which just does Theano math directly, which both work correctly. But I don’t have a good feel for if one approach is better for performance in PyMC3. As in if I were using a 1000 side dice would making 1000 deterministic RVs be bad compared to manipulating the probabilities using Theano. And in that case would it be better to make a custom distribution class, similar to how the Dirichlet can be used to specify the p values for Multinomial, or does it not matter? Thanks for any insight anyone might have!
``````def linearDiceProbabilities_1(m):
return 1/6 + np.array([3,2,1,-1,-2,-3])*m
def linearDiceProbabilities_2(m):
return mc.Deterministic('p', 1/6 + np.array([3,2,1,-1,-2,-3])*m)
def bayesianInferenceLinearDiceModel(p, n, counts):
with mc.Model() as model:
# Prior distributions.
m = mc.Uniform('m', lower=0, upper=0.05)
p = linearDiceProbabilities_1(m)
#p = linearDiceProbabilities_2(m)
# Likelihood.
counts_likelihood = mc.Multinomial('counts', n=n, p=p, observed=counts)
# Perform the sampling.
trace = mc.sample(2000, tune=1000)
return trace
``````
@clausherther Yes, that’s a good point. That was just leftover code.
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# m02 - 11
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Joined: 18 Aug 2009
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### Show Tags
10 Nov 2009, 22:43
Is $$p^2 \gt q^2$$ ?
1. $$p \gt 0$$
2. $$q \gt 0$$
If test creator reaches the stage of modifying/replacing the questions, I think this will be a good candidate.
IMO, too easy to be in the elite list (Q asks for comparison between P & Q and the options give individual information... doesn't even look like a trap). Comments?
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### Show Tags
11 Nov 2009, 06:11
It certainly is not a difficult question. It is part of the math tests here, but I don't think it is on an "elite list" is it? If so, where is this list?
The question tests basic understanding of properties of squares. Makes sure the person remembers that the values of each could be negative until each statement rules that possibility out.
gmattokyo wrote:
Is $$p^2 \gt q^2$$ ?
1. $$p \gt 0$$
2. $$q \gt 0$$
If test creator reaches the stage of modifying/replacing the questions, I think this will be a good candidate.
IMO, too easy to be in the elite list (Q asks for comparison between P & Q and the options give individual information... doesn't even look like a trap). Comments?
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### Show Tags
11 Nov 2009, 06:41
This question has already been discussed here:
m02-q11-ds-69415.html
Please don't create duplicate topics. You're welcome to look through this thread before posting another question in this forum.
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### Show Tags
11 Nov 2009, 06:43
jallenmorris wrote:
It certainly is not a difficult question. It is part of the math tests here, but I don't think it is on an "elite list" is it? If so, where is this list?
The question tests basic understanding of properties of squares. Makes sure the person remembers that the values of each could be negative until each statement rules that possibility out.
gmattokyo wrote:
Is $$p^2 \gt q^2$$ ?
1. $$p \gt 0$$
2. $$q \gt 0$$
If test creator reaches the stage of modifying/replacing the questions, I think this will be a good candidate.
IMO, too easy to be in the elite list (Q asks for comparison between P & Q and the options give individual information... doesn't even look like a trap). Comments?
By elite list I meant, in general, the other questions are tricky and high quality...
Ran into a similar level Q in m03 25. I suppose these questions act as a filler and give some respite and a brief break to breathe
Are positive integers $$P$$ , $$Q$$ , and $$R$$ equal?
1. $$P = Q$$
2. $$Q^2 = R^2$$
oops! sorry to create duplicate thread... will take care to search carefully in future
Re: m02 - 11 [#permalink] 11 Nov 2009, 06:43
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# Electronic Circuits I - Tutorial 03 Diode Applications I
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1 Electronic Circuits I - Tutoril 03 Diode Applictions I -1 / 9 -
2 T & F # Question 1 A diode cn conduct current in two directions with equl ese. F 2 When reverse-bised, diode idelly ppers s short. F 3 A bsic hlf-wve rectifier consists of one diode. T 4 The diode in hlf-wve rectifier conducts for hlf the input cycle. T 5 Ech diode in full-wve rectifier conducts for the entire input cycle. F MCQ1 # Question 1 When diode is forwrd-bised nd the bis voltge is incresed, the forwrd current will 2 When diode is reverse-bised nd the bis voltge is incresed, the reverse current (ssuming the prcticl model) will 3 When diode is forwrd-bised nd the bis voltge is incresed, the voltge cross the diode (ssuming the complete model) will 4 If the forwrd current in diode is decresed, the diode voltge (ssuming the complete model) will 5 If the input voltge in Figure 2 28 is incresed, the pek inverse voltge cross the diode will c b A -2 / 9 -
3 6 If the frequency of the input voltge in Figure 2 36 is incresed, the output voltge will c MCQ 2 # Question Answer 1 1. The term bis mens c () the rtio of mjority crriers to minority crriers (b) the mount of current cross diode (c) dc voltge is pplied to control the opertion of device (d) neither (), (b), nor (c) 2 When diode is forwrd-bised, d () the only current is hole current (b) the only current is electron current (c) the only current is produced by mjority crriers (d) the current is produced by both holes nd electrons 3 For silicon diode, the vlue of the forwrd-bis voltge typiclly b () must be greter thn 0.3 V (b) must be greter thn 0.7 V (c) depends on the width of the depletion region (d) depends on the concentrtion of mjority crriers 4 A diode is normlly operted in d () reverse brekdown (b) the forwrd-bis region (c) the reverse-bis region (d) either (b) or (c) 5 The V-I curve for diode shows () the voltge cross the diode for given current (b) the mount of current for given bis voltge (c) the power dissiption (d) none of these 6 In the prcticl diode model, () the brrier potentil is tken into ccount -3 / 9 -
4 (b) the forwrd dynmic resistnce is tken into ccount (c) none of these (d) both () nd (b) 7 The verge vlue of hlf-wve rectified voltge with pek vlue of 200 V is () 63.7 V (b) V (c) 141 V (d) 0 V 8 The pek vlue of the input to hlf-wve rectifier is 10 V. The pproximte pek vlue of the output is () 10 V (b) 3.18 V (c) 10.7 V (d) 9.3 V 7 The verge vlue of full-wve rectified voltge with pek vlue of 75 V is () 53 V (b) 47.8 V (c) 37.5 V (d) 23.9 V 8 The totl secondry voltge in center-tpped full-wve rectifier is 125 V rms. Neglecting the diode drop, the rms output voltge is () 125 V (b) 177 V (c) 100 V (d) 62.5 V D b d Problems Problem 1 Assume tht the diode in Figure 2 18() fils open. Wht is the voltge cross the diode nd the voltge cross the limiting resistor? Answer 1 Vd=10V, VRlimit-0; -4 / 9 -
5 Problem 2 Determine the verge vlue of the hlf-wve voltge if its pek mplitude is 12 V. Answer V Proble m 3 Answer 3 Q4 To forwrd-bis diode, to which region must the positive terminl of voltge source be connected? 4 p region -5 / 9 -
6 Q5 Explin how to generte the forwrd-bis portion of the chrcteristic curve. 5 To generte the forwrd bis portion of the chrcteristic curve, connect voltge source cross the diode for forwrd bis nd plce n mmeter in series with the diode nd voltmeter cross the diode. Slowly increse the voltge from zero nd plot the forwrd voltge versus the current. Q6 Drw the output voltge wveform for ech circuit in Figure nd include the voltge vlues 6-6 / 9 -
7 Q7 Clculte the verge vlue of hlf-wve rectified voltge with pek vlue of 200 V V Problem 8 Wht diode PIV rting is required to hndle pek input of 160 V in Figure 2 36? Answer V including diode drop Q9 9 A power-supply trnsformer hs turns rtio of 5:1. Wht is the secondry voltge if the primry is connected to 120 V rms source? 24 V rms -7 / 9 -
8 Q01 Find the verge vlue of ech voltge in Figure 01 () 1.59 V (b) 63.7 V (c) 16.4 V (d) 10.5 V -8 / 9 -
9 Q00 Clculte the pek voltge cross ech hlf of center-tpped trnsformer used in full-wve rectifier tht hs n verge output voltge of 120 V V -9 / 9 -
### Electronic Circuits I - Tutorial 03 Diode Applications I
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## Scoot Cards.pdf - Section 2: Practice
Scoot Cards.pdf
Scoot Cards.pdf
Unit 4: Adding and Subtracting Fractions
Lesson 13 of 17
## Big Idea: Scoot on over!
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60 minutes
### Erin Doughty
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# How to re-parametrize posterior function?
How can I write the unnormalized posterior
$f(p_1, p_2 | Y) = (z_1-1)*log(p_1) + (n_1-z_1-1)*log(1-p_1) + (z_2-1)*log(p_2) + (n_2-z_2-1)*log(1-p_2)$
in terms of the log odds-ratio $\alpha$ and the log odds-product $\eta$?
$\alpha = log \left(\frac{p_2/(1-p_2)}{p_1/(1-p_1}\right)$ and $\eta = log \left(\frac{p_2}{1-p_2} \times \frac{p_1}{1-p_1}\right)$?
where z_1, n_1, z_2, and n_2 are constants.
-
Note that we have $$\alpha = \log p_2 - \log(1-p_2) - \log p_1 + \log(1-p_1),$$ $$\eta=\log p_2 -\log (1-p_2)+\log p_1 -\log(1-p_1).$$ Then $$\eta-\alpha = 2(\log p_1 - \log (1-p_1),$$ $$\eta +\alpha=2(\log p_2 -\log(1-p_2).$$
EDIT: previously I wrote that your unnormalized posterior couldn't be written in terms of $\alpha$ and $\eta$. However I just noticed that $$\frac{\eta - \alpha}{2}=\log \frac{p_1}{1-p_1}$$ so that $$p_1 = \frac{e^{(\eta-\alpha)/2}}{e^{(\eta-\alpha)/2}+1},$$ and similarly $$\frac{\eta + \alpha}{2}=\log \frac{p_2}{1-p_2}$$ so that $$p_2 = \frac{e^{(\eta+\alpha)/2}}{e^{(\eta+\alpha)/2}+1}.$$ Now since your unnormalized posterior is a function of $p_1$ and $p_2$, and of the constants $z_1,n_1,z_2,n_2$, you can using the above expressions write the posterior in terms of $\alpha$ and $\eta$.
Note to the OP: please reread this, after the EDIT, since I figured out you can rewrite as you wanted in terms of $\alpha,\eta$. – coffeemath Dec 9 '12 at 2:55
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Solved
Posted on 2011-02-24
962 Views
I've downloaded the Excel Morefunc Addin which contains a formula to create a list of unique values based on a list (assume column A).
In Cell B2, I have a formula that reads {=uniquevalues(\$A\$2:\$A\$500,1)}, however this only shows the first value when you drag the formula down; thus not allowing you to create a list of unique values. Any help would be amazing!
Thanks
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Question by:Golfer219
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Accepted Solution
dlmille earned 500 total points
ID: 34977233
First, select a range to hold your unique values, then type your formula, and hit ctrl-shift-enter when done. The results will go into that originally selected range. That's how you create an array output.
See attached where I've done that. The yellow area was the original area I selected, then I entered the =UniqueValues() function, then I hit CTRL-SHIFT-ENTER and presto - the unique values turned up in the yellow area.
Enjoy!
Dave
Unique-r1.xlsx
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##### how do i solve and expression with exponents
Algebra Tutor: None Selected Time limit: 1 Day
x to the 10 over x to 2? would it be x to the 5 or x to the 8
Oct 1st, 2015
X/10 /X/2 = This would be x to the 5 after you cross-multiply
Oct 1st, 2015
well in the book it is one of those correct the error snd it says that that is the answer that is wrong.
Oct 1st, 2015
and it says the other alternative is right?
Oct 1st, 2015
...
Oct 1st, 2015
...
Oct 1st, 2015
Dec 9th, 2016
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Prove that the concatenation function is continuous
• docnet
In summary, the problem asks to prove the continuity of the function (f*g) with f and g being continuous functions in [0,1] and [1,2] respectively, and f(1)=g(1). To prove this, it is necessary to use the universal property of quotient spaces. However, the solution provided is not correct as it does not define the equivalence relation on the correct set. The correct approach would be to use the standard epsilon-delta definition of continuity for each case of x in [0,2].
docnet
Gold Member
Homework Statement
prove the continuity of the function
Relevant Equations
f,g
Let f be continuous in [0,1] and g be continuous in [1,2] and f(1)=g(1). prove that
$$(f*g)= \begin{cases} f(t), 0\leq t\leq 1\\ g(t), 1\leq t \leq2 \end{cases}$$
is continuous using the universal property of quotient spaces.
Let ##f:[0,1]→X## and ##g:[1,2]→Y##
f and y are continuous, thus for open sets in [0,1] and open sets in [1,2], their images are also open in X and Y, hence X and Y are topological spaces.
f(1)∈X is an element in X and g(1)∈Y is an element in Y.
let W be a discrete union of X and Y, which itself is a topological space
##f(1)=g(1)## is an equivalence relation on W.
Then (f*g) is a projection map from W onto W/(f(1) ~g(1)) and hence continuous by the universal property of quotient spaces
Last edited:
docnet said:
Homework Statement:: prove the continuity of the function
Relevant Equations:: f,g
Let f be continuous in [0,1] and g be continuous in [1,2] and f(1)=g(1). prove that
$$(f*g)= \begin{cases} f(t), 0\leq t\leq 1\ g(t), 1\leq t \leq2 \end{cases}$$
is continuous using the universal property of quotient spaces.
Let f:[0,1]→X and g:[1,2]→Y
f and y are continuous, thus ...
Thus is wrong here. A function is continuous if the pre-image of an open set is open. That is: ##f## is continuous if ##f^{-1}(U) \subseteq [0,1]## is always open when ##U\subseteq X## is open.
docnet said:
... for open sets in [0,1] and open sets in [1,2], their images are also open in X and Y, hence X and Y are topological spaces.
This would be the definition of an open function, i.e. a function that maps open sets to open sets. Open functions do not need to be continuous and continuous functions do not need to be open.
docnet said:
f(1)∈X is an element in X and g(1)∈Y is an element in Y.
let W be a discrete union of X and Y.
What does discrete here mean? You can either consider ##X\cup Y## as subspaces of ##\mathbb{R}## or ##X\times Y## as direct product of two topological spaces. I assume you meant the latter since the rest of the vocabulary fits better on a direct product than a simple union.
docnet said:
f(1)∈X=g(1)∈Y is an equivalence relation on W.
A relation ##R## is a subset of a direct (Cartesian) product of two sets. In this case, I assume you meant a subset of ##W=X\times Y##. An equivalence relation is a special relation. You defined ##R=\{(f(1),g(1))\}## which is only a single point. This is trivially an equivalence relation, but useless, too.
Maybe you meant ##R=\{(x,y)\in W\,|\,\exists \,t\in [0,1] \,\exists \,s\in [1,2]\, : \,x=f(t)\,\wedge\,y=f(s)\,\wedge\,f(1)=g(1)\}## but this is a wild guess.
Another possibility is to consider all such functions ##f,g##. Then ##\mathcal{W}:=\{f\,|\,f:[0,1]\longrightarrow X\} \times \{g\,|\,g:[1,2]\longrightarrow Y\}## would be the direct product and all pairs ##\mathcal{R}:=\{(f,g)\in \mathcal{W}\,|\,f(1)=g(1)\}\subseteq \mathcal{W}## the relation. But this is neither reflexive (##(f,f)\not\in \mathcal{W}##) nor symmetric (##(g,f)\not\in \mathcal{W}##), hence no equivalence relation.
I think I know what you are trying to do, but I have difficulties to impose an equivalence relation on concatenated fuctions, will say my ideas are too complicated. Do you have a precise definition of your relation, as a subset of some direct product? ##W## should be a set of functions, not just ##X\times Y## since what would ##f,g## be?
docnet said:
Then (f*g) is a projection map from W onto W/(f(1) ~g(1)) and hence continuous by the universal property of quotient spaces
docnet
There is another problem: If ##X\ni f(1)=g(1) \in Y## how do ##X## and ##Y## share a point? We could pick a point in both (##x_1\in X,y_1 \in Y##) and clue them together (##x_1=y_1##) in a space ##X\cup Y,## but what if the functions change?
FactChecker said:
I think that your attempt has gone wrong. You can use the standard ##\epsilon##, ##\delta## definition of continuity. For any ##x \in [0,2]## and ##\epsilon \gt 0 ##, divide the proof into 3 cases: ##x \in [0,1)##, ##x=0##, ## x \in (1,2]## to prove that there is a ##\delta## for each case.
The problem explicitly says not to do this.
I think the biggest problem with the op is that f and g almost certainly are intended to have the same codomain. There is no X and Y. The thing you need to define an equivalence relation on is the domain, not the codomain.
WWGD, docnet and FactChecker
Office_Shredder said:
The problem explicitly says not to do this.
Oh! I stand corrected. I didn't read the full sentence. I will try to delete or correct my posts.
Office_Shredder said:
The problem explicitly says not to do this.
I think the biggest problem with the op is that f and g almost certainly are intended to have the same codomain. There is no X and Y. The thing you need to define an equivalence relation on is the domain, not the codomain.
yes. the original problem says to do exactly as you say. I am sorry, i did not know that writing the problem differently would cause this issue!
PeroK
So in this case, the domain you probably want to focus on is something that is a union of [0,1] and [1,2] with an equivalence relation where the two 1s are equal. You have to be pretty careful with notation here because the two 1s aren't equal in the original domain!
fresh_42 said:
Thus is wrong here. A function is continuous if the pre-image of an open set is open. That is: f is continuous if f−1(U)⊆[0,1] is always open when U⊆X is open.
so i was trying to start from the fact that f and g are given to be continuous
fresh_42 said:
This would be the definition of an open function, i.e. a function that maps open sets to open sets. Open functions do not need to be continuous and continuous functions do not need to be open.
Is it true that a map f:X->Y is continuous if for all open subsets U in Y (w.r.t. the topology on Y), the inverse image is an open subset of X (w.r.t. to the topology on X)?
fresh_42 said:
What does discrete here mean? You can either consider X∪Y as subspaces of R or X×Y as direct product of two topological spaces. I assume you meant the latter since the rest of the vocabulary fits better on a direct product than a simple union.
my professor taught us that disjoint unions can be thought of as ##U\times \{0\} \cup V\times \{1\}##. if U and V are both subsets of Y, and there is an equivalence relation on Y s. t. a u in U is similar to a v in V, then U and V belong in the same equivalence class?
I am trying to figure out how to use this to solve
it seems that the concatenation glues the two images of [0,1] under ##γ_1## and ##γ_2## together. but i am not sure how to establish that it is equivalent to a continuous map ##gp## so that ##g:X/\sim\rightarrow Y## and ##p:X\rightarrow X\sim##.
You shouldn't be thinking of gluing the images together, you should be thinking of gluing the domains together. The equivalence relationship is on the domain, not the image of the function.
docnet said:
Is it true that a map f:X->Y is continuous if for all open subsets U in Y (w.r.t. the topology on Y), the inverse image is an open subset of X (w.r.t. to the topology on X)?
Yes. The crucial point is inverse images. You defined it with images, in which case we speak of an open map. You can even use it as a mnemonic! Open maps map open sets to open sets. Continuity is the other way around.
The book has ##X=Y## which makes a difference, because ##f(1)=g(1)## makes sense in this case. Also both functions are already defined on the entire interval ##[0,1]##. This means both functions live in the same space, namely the space of all continuous functions from ##[0,1]## to ##X##. With the split into two separate intervals in post #1 came the difficulties to define a relation. Now we have:
$$W:=\{f:[0,1]\longrightarrow X\,|\,f\text{ is continuous}\} \text{ and } f\sim g :\Longleftrightarrow f(1/2)=g(1/2)$$
which is an equivalence relation on ##W## and we can speak of ##W/\sim.##
Last edited:
docnet
fresh_42 said:
Yes. The crucial point is inverse images. You defined it with images, in which case we speak of an open map. You can even use it as a mnemonic! Open maps map open sets to open sets. Continuity is the other way around.
The book has ##X=Y## which makes a difference, because ##f(1)=g(1)## makes sense in this case. Also both functions are already defined on the entire interval ##[0,1]##. This means both functions live in the same space, namely the space of all continuous functions from ##[0,1]## to ##X##. With the split into two separate intervals in post #1 came the difficulties to define a relation. Now we have:
$$W:=\{f:[0,1]\longrightarrow X\,|\,f\text{ is continuous}\} \text{ and } f\sim g :\Longleftrightarrow f(1/2)=g(1/2)$$
which is an equivalence relation on ##W## and we can speak of ##W/\sim.##
I think this is likely too advanced for the level of the question; to do a quotient on a function space, with points being functions. I think points in the quotient are more likely to be "standard" points, aka Real Numbers. EDIT: Besides, we would also need to know the topology on the infinite-dimensional topological space of functions. What are the open sets, in order to decide if the inverse image is continuous.
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You are starting with two functions defined on $[0,1]$ and you want to end up with another function defined on $[0,1]$. That suggests taking two disjoint copies of $[0,1]$ (for example $[0,1] \times \{0,1\}$), and acting on each copy with a different function, giving you a function $g : [0,1] \times \{0,1\} \to X$. You then use an euivalence relation to glue the endpoints of the copies of $[0,1]$ together in such a way as to get a function $f$ from the quotient space (which is homeomorphic to $[0,1]$) to $X$. Then $g = f \circ p$ and the universal property is used to show (by contradiction) that if $g$ is continuous then so is $f$.
docnet
pasmith said:
You are starting with two functions defined on $[0,1]$ and you want to end up with another function defined on $[0,1]$. That suggests taking two disjoint copies of $[0,1]$ (for example $[0,1] \times \{0,1\}$), and acting on each copy with a different function, giving you a function $g : [0,1] \times \{0,1\} \to X$. You then use an euivalence relation to glue the endpoints of the copies of $[0,1]$ together in such a way as to get a function $f$ from the quotient space (which is homeomorphic to $[0,1]$) to $X$. Then $g = f \circ p$ and the universal property is used to show (by contradiction) that if $g$ is continuous then so is $f$.
Is a map from ##[0,1]## to ##[0,1] \times \{0,1\}## a quotient map?
Is this map a projection map?
Is the quotient space ##[0,1] \times \{0,1\}##?
Is a map from the quotient space ##[0,1] \times \{0,1\}## to ##X## a quotient map?
Is this map also a projection map?
Is there a name for a function from ##[0,1]## to ##[0,1] \times \{0,1\}## to ##X## ?
1. What is the concatenation function?
The concatenation function is a mathematical operation that combines two or more strings or sequences of characters into a single string or sequence. It is denoted by the symbol "||" or by using a dot between the strings (e.g. "hello" || "world" = "helloworld").
2. How is the continuity of the concatenation function defined?
The continuity of the concatenation function is defined as the ability to manipulate the input variables by a small amount, resulting in a small change in the output. In other words, if the inputs are close together, the outputs will also be close together.
3. Why is it important to prove that the concatenation function is continuous?
Proving the continuity of the concatenation function is important because it ensures that the function can be used in various mathematical operations and that the results will be accurate. It also allows for easier analysis and understanding of the function's behavior.
4. What are the steps to prove the continuity of the concatenation function?
To prove the continuity of the concatenation function, we must show that for any two inputs, x and y, and a small change in the inputs, ε, the change in the output is also small. This can be done by using the definition of continuity and the properties of the concatenation function.
5. Can the continuity of the concatenation function be proven in all cases?
Yes, the continuity of the concatenation function can be proven in all cases as long as the inputs are valid strings or sequences. However, the specific method of proving continuity may vary depending on the specific properties and constraints of the function.
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https://schoollearningcommons.info/question/14-mrs-tuddu-sold-her-golden-for-rs-3125-at-the-gain-of-12-whole-1-by-2-find-the-cost-of-the-rin-19827054-61/
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## 14. Mrs Tuddu sold her golden for Rs 3125 at the gain of 12 whole 1 by 2% find the cost of the ring. iska answer hai Rs.2776
Question
14. Mrs Tuddu sold her golden for Rs 3125 at the gain of 12 whole 1 by 2% find the cost of the ring.
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# Uniform boundedness Schrödinger operator eigenfunctions with Dirichlet conditions
I would like to ask a question with possibly a reference. If we have a Schrödinger operator $$-\Delta+V$$ on an interval $$[0,L]$$ with $$V$$ continous and Dirichlet conditions, can we state that the eigenfunctions of such operator are uniformly bounded, i.e. there exists $$M>0$$ such that the eigenfunctions $$\{\phi_n\}_n$$ satisfy
$$\begin{equation*} \sup_n \lvert\lvert \phi_n \rvert\rvert_\infty\leq M \end{equation*}$$ ?
Yes, this follows because asymptotically, as $$|z|\to\infty$$, the solutions of $$-y''+Vy=zy$$ look like those of the free equation $$V\equiv 0$$, and the eigenfunctions of $$-y''=zy$$, $$\phi_n=(2/(L\pi ) )^{1/2}\sin n\pi x/L$$, are uniformly bounded.
In fact, they are uniformly bounded not just in $$n$$, but also in the potential $$V$$ as long as we impose a uniform bound on $$\|V\|_1$$.
(I'm assuming here that you normalize your eigenfunctions as usual, $$\|\phi_n\|_2=1$$, and then you are asking about $$\|\phi_n\|_{\infty}$$.)
• The book by Poschel and Trubowitz discusses asymptotics in some detail (though with an $L^2$ assumption on $V$, which isn't necessary; you can adapt the proofs if required) and it's what I usually quote when I use these things, but there are many other references, so keep searching if you don't like this (Marchenko's book is another option). Commented May 15, 2022 at 20:25
• @Christian Remling Christian, do you know examples where $\sup_n \|\phi_n\|_\infty$ really depends on $\|V\|_1$ ($V \geq 0$)? Commented May 16, 2022 at 17:10
• @GiorgioMetafune: I think one way of doing this is to take $V=0$ on $(0,a)$ and $V=N\to\infty$ on $(a,L)$. In the limit $N\to\infty$, this will simulate a Dirichlet boundary condition at $x=a$, and if $a$ is small, then $\|\phi_n\|_{\infty}/\|\phi_n\|_2$ will become large. Commented May 16, 2022 at 17:22
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# Train distance physics problem
Salman
I'm stuck... help someone!
here is the problem:
Two trains are traveling on the same track. One is going West at 25 km/h and the other is going East at 30 km/h. When they are 100 km apart an eagle takes off from the first train and flies at 50 km/h towards the other train. On reaching the other train it reverses direction and goes off to the first train, and keeps on flying between the two.
Assuming that the eagle takes no time in reversing direction and that its speed remains constant calculate :
i) the distance traveled by the eagle until the two trains collide.
ii) the number of time the eagle touches the two trains
p.s:I tried my hand on the problem but didn't succeed.All I can conclude from my calculations is that the process is never ending and a an infinite series is formed by the distance traveled by the eagle everytime it flies from one train to another.I know the answer to the first part will be finite but I can not really find a method to sum all the terms of the sequence bcoz it's not a normal AP or GP.Someone help me please!
bomba923
For question (i), by collision I assume the trains are traveling towards each other. Anyway,
i) When the trains are 100 km apart, they will collide in
$$t = \frac{{100km}}{{25\frac{{km}}{{hr}} + 30\frac{{km}}{{hr}}}} = \frac{{20}}{{11}}hr$$
In $\left( 20/11 \right)$ hr, the distance the eagle travels will be
$$50\frac{{km}}{{hr}} \cdot \frac{{20}}{{11}}hr = \frac{{1000}} {{11}}km \approx \boxed{91 km}$$
Last edited:
Homework Helper
Don't forget the finite size of the eagle! :)
Homework Helper
As pointed out- the simple way to do this problem is calculate the time until the two trains collide and divide by the speed of the eagle! Of course, if one were really determined to do it the hard way, one might calculate lengths of the individual flights back and forth and add- getting an infinite geometric series.
There is a story that someone posed this question to VonNeumann who thought for about 15 seconds and then gave the correct answer. The person laughed and said "You know, some people try to do that as an infinite series". VonNeumann looked puzzled ans said "But I did it as an infinite series!"
Homework Helper
I've always thought von Neumann was pulling the guy's leg!
neutrino
bomba923 said:
For question (i), by collision I assume the trains are traveling towards each other. Anyway,
i) When the trains are 100 km apart, they will collide in
$$t = \frac{{100km}}{{25\frac{{km}}{{hr}} + 30\frac{{km}}{{hr}}}} = \frac{{20}}{{11}}hr$$
In $\left( 20/11 \right)$ hr, the distance the eagle travels will be
$$50\frac{{km}}{{hr}} \cdot \frac{{20}}{{11}}hr = \frac{{100}} {{11}}km \approx \boxed{9km}$$
Shouldn't it be a little over 10 times that distance?
Homework Helper
Tide said:
I've always thought von Neumann was pulling the guy's leg!
He probably was. He was well known for that!
(Though not as quite as well known for treating everyone around him as a git as Feynman!)
bomba923
neutrino said:
Shouldn't it be a little over 10 times that distance?
Yes, and so I updated the post
I have a long procedure for the second problem, but it concludes that:
*If $t_n$ represents the time spent on $n$ quantity of flights, then
$$t_n = \frac{{100km + t_{n - 1} \left( {50\tfrac{{km}}{{hr}} - v_n } \right)}}{{50\tfrac{{km}}{{hr}} + v_{n - 1} }}$$
where
$$t_0 = 0 \, {hr} , \;v_n = \frac{5}{2}\left[ {11\tfrac{{km}}{{hr}} + \left( { - 1} \right)^n } \right]$$
*(Feel free to omit the units for mathematical clarity )
I haven't bothered to simplify this (sorry), but the number of flights will be the value of $n$ for which
$$t _ n = \frac{20}{11} \, {hr}$$
Salman
bomba923 said:
Yes, and so I updated the post
I have a long procedure for the second problem, but it concludes that:
*If $t_n$ represents the time spent on $n$ quantity of flights, then
$$t_n = \frac{{100km + t_{n - 1} \left( {50\tfrac{{km}}{{hr}} - v_n } \right)}}{{50\tfrac{{km}}{{hr}} + v_{n - 1} }}$$
where
$$t_0 = 0 \, {hr} , \;v_n = \frac{5}{2}\left[ {11\tfrac{{km}}{{hr}} + \left( { - 1} \right)^n } \right]$$
*(Feel free to omit the units for mathematical clarity )
I haven't bothered to simplify this (sorry), but the number of flights will be the value of $n$ for which
$$t _ n = \frac{20}{11} \, {hr}$$
thanx people for all ur help but @ bomba923 the sequence u have given here is a recurrence relation(if I'm not wrong) but I don't know how to solve these sort of relations!(bcoz I haven't learned it at school )If u could please solve for $$n$$ I would be really thankful to u!
Best regards,
Salman.
Homework Helper
It's infinity times. Why? The eagle flies faster than both trains. Say it starts from train 1, flies to train 2. Train 1 and train 2 is $$\Delta s_1$$(m) apart. Because the eagle flies faster than train 1, it will meet train 2 before train 1 does. That means, when it reaches train 2, train 1 has not reached train 2 yet. And they are $$\Delta s_2$$(m) apart. And the flies flies back to train 1, it reaches train 1 before train 2. At the time it reaches train 1, the 2 trains are $$\Delta s_3$$(m) apart, and so on... The distance between the two trains decreases, but the eagle will never touch one of the two train at the exact time the other train touches it (the eagle is faster than 2 trains).
So you notice:$$\Delta s_1 > \Delta s_2 > \Delta s_3 > \Delta s_4 > ... > \Delta s_n$$(m)
The distance between two trains gets smaller and smaller, the eagle completes its loop faster.
We have:
$$\lim_{n \to \infty} \Delta s_n = 0$$
So the eagle touches the two train 'infinity' times.
Viet Dao,
Last edited:
bomba923
VietDao29 said:
It's infinity times. Why? The eagle flies faster than both trains. Say it starts from train 1, flies to train 2. Train 1 and train 2 is $$\Delta s_1$$(m) apart. Because the eagle flies faster than train 1, it will meet train 2 before train 1 does. That means, when it reaches train 2, train 1 has not reached train 2 yet. And they are $$\Delta s_2$$(m) apart. And the flies flies back to train 1, it reaches train 1 before train 2. At the time it reaches train 1, the 2 trains are $$\Delta s_3$$(m) apart, and so on... The distance between the two trains decreases, but the eagle will never touch one of the two train at the exact time the other train touches it (the eagle is faster than 2 trains).
So you notice:$$\Delta s_1 < \Delta s_2 < \Delta s_3 < \Delta s_4 < ... < \Delta s_n$$(m)
The distance between two trains gets smaller and smaller, the eagle completes its loop faster.
We have:
$$\lim_{n \to \infty} \Delta s_n = 0$$
So the eagle touches the two train 'infinity' times.
Viet Dao,
Only if you assume the eagle has no size
As Tide said
Tide said:
Don't forget the finite size of the eagle! :)
Please don't squish the bird between the trains!
Anyway, if you provide some specifications as to the eagle's size, you CAN set a finite number of flights in between...that is, until the trains are so close as to squish the eagle, and he flies away...
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Party Edition Bidita
In-process control applications, the Z esteem gives an evaluation of how off-focus on a procedure is working.
Examination of scores estimated on various scales: ACT and SAT
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The z-score for understudy B is {\displaystyle z={x-\mu \over \sigma }={24-21 \over 5}=0.6}{\displaystyle z={x-\mu \over \sigma }={24-21 \over 5}=0.6}
Since understudy A has a higher z-score than understudy B, understudy A performed better contrasted with other test-takers than did understudy B.
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Proceeding with the case of ACT and SAT scores, on the off chance that it very well may be additionally expected that both ACT and SAT scores are ordinarily conveyed (which is roughly right), at that point, the z-scores might be utilized to ascertain the level of test-takers who got lower scores than understudies An and B.
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In the head segments examination, “Factors estimated on various scales or a typical scale with broadly varying reaches are frequently institutionalized.”
Relative significance of factors in numerous relapse: Standardized relapse coefficients
Institutionalization of factors before various relapse investigations is, at times, utilized as a guide to interpretation.[10] (page 95) express the accompanying.
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# A Physics Puzzle - Uphill vs. Downhill Putts
## Recommended Posts
So I was thinking about the difference between uphill, level, and downhill putts. I found this paper , and decided to make a quick simulation. You give it stimp, grade, and putt speed, and it finds the rollout of your putt.
Here's the puzzle:
I'm on a green where my level putt rolls 12ft, and my 4% grade uphill putt rolls 7.5 ft. How far does the same putt go if it's hit down the 4% grade???
Hint: the stimp is 12 (Augusta National?).
Initial Speed % Grade Rollout of Putt 6 ft/s Level 12 ft 6 ft/s 4% Uphill 7.5 ft 6 ft/s 4% Downhill ???
Any guesses?
19.5 feet
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It's all about gravity, size of ball, friction of the grass, temperature and humidity, angle of the dangle, and so forth. And about 11 feet past the hole, there you will find the ball.
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Originally Posted by SCfanatic35
19.5 feet
Too lazy to look it up.....can you show your work? Thks!
College physics was a loooooong time ago.
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12ft + 7.5ft = 19.5ft
No idea what the answer is, just trying to throw out what seems to me a logical possible answer to the question. The uphill putt stopped 4.5 feet short of the hole. So I assumed the downhill putt wouldn't stop 4.5 feet past the hole, it must be more because of the downhill. I didn't think it would go double the distance, so I just added the two lengths of the other two putts. IDK
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Is there a conveyor belt involved?
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How much grain? Which way is the grain? Most of the time the grain will lean downhill, thus increasing the speed of the downhill putt even as it kills the pace on the uphill putt.
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If a putt goes 7.5 feet uphill, and goes 12 feet on level ground. Then that is a 4.5 feet difference. Since potential energy is the same, downhill versus uphill, if you hit the same initial speed, your looking at the putt going 16.5 feet.
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Hmmm...4% grade. 12 on the Stimp. Hit the putt with the same force as a 12 foot level putt.
Answer: Until it hits the fringe or the rough on the other side of the green (however far away that might be). Ha ha!
Calculated from hole number 4 at the course in Trenton Ga. Probably about a 4% grade that if you hit the ball hard enough to go a foot on level ground you might as well go back to the bag and get your wedge for the next shot, because you're not going to be putting. If you just touch the ball enough to move it an inch you might get lucky and only have a 5 footer back up the hill for the next putt.
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According to the USGA: "Physical Qualities of the Green The USGA suggests that at least a 2-foot radius surrounding the hole “should be as nearly level as possible and of uniform grade.” The hole shouldn’t be placed on a steep slope on which a missed putt from above the hole will roll a long distance past the cup. “ A player above the hole should be able to stop the ball at the hole ,” according to Rule 15-3(iii). Additionally, the hole shouldn’t be located on a former hole’s spot until the old location has healed completely." Of course there are diabolical greenskeepers who thrive on torturing the players. So to the OP, it should stop AT the hole.... IF you hit it on line- it will stop at the bottom of the cup. If not, work on your putting. :-D
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Uphill you have two things stopping the ball. Gravity and Friction. Without friction it would eventually stop and come back the other way.
Downhill only friction is acting on it. Obviously without friction it would roll forever much as it does at the Masters at times.
The point is it will not roll past the hole downhill the same distance it stopped short uphill.
Don't have the exact numbers right now. I'd have to dust off my slide rule. Who am I kidding, I mean my TI-86.
29 feet
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Stimpmeter downhill, St(dn)
St(dn) = 0.82 ft / (CF - sin Φ cos Φ)
12ft St CF = 0.82 / 12
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Originally Posted by MS256
Hmmm...4% grade. 12 on the Stimp. Hit the putt with the same force as a 12 foot level putt.
Answer: Until it hits the fringe or the rough on the other side of the green (however far away that might be). Ha ha!
Calculated from hole number 4 at the course in Trenton Ga. Probably about a 4% grade that if you hit the ball hard enough to go a foot on level ground you might as well go back to the bag and get your wedge for the next shot, because you're not going to be putting. If you just touch the ball enough to move it an inch you might get lucky and only have a 5 footer back up the hill for the next putt.
You say that jokingly, however, I don't think you are that far off. When I took my Aimpoint class I learned that 4% is basically the steepest a green can be where a ball will sit. Anything steeper than that and you have "rolloff." I have no idea what the actual answer is, but I would suspect it is way higher than any of the answers given so far.
I also think that there is not enough info to get the answer. Like Hendog said, gravity and friction are the only outside forces acting on it after the putter hits it. I think we need to know the friction factor of the green to figure it out.
My wild guess ... 45 feet.
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How close are we to Lake Merced?
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Quote:
Originally Posted by MS256
Hmmm...4% grade. 12 on the Stimp. Hit the putt with the same force as a 12 foot level putt.
Answer: Until it hits the fringe or the rough on the other side of the green (however far away that might be). Ha ha!
Calculated from hole number 4 at the course in Trenton Ga. Probably about a 4% grade that if you hit the ball hard enough to go a foot on level ground you might as well go back to the bag and get your wedge for the next shot, because you're not going to be putting. If you just touch the ball enough to move it an inch you might get lucky and only have a 5 footer back up the hill for the next putt.
You say that jokingly, however, I don't think you are that far off. When I took my Aimpoint class I learned that 4% is basically the steepest a green can be where a ball will sit. Anything steeper than that and you have "rolloff." I have no idea what the actual answer is, but I would suspect it is way higher than any of the answers given so far.
I also think that there is not enough info to get the answer. Like Hendog said, gravity and friction are the only outside forces acting on it after the putter hits it. I think we need to know the friction factor of the green to figure it out.
Seems like we should be able to calculate friction since it's the only unknown in the uphill putt scenario. I.e., gravity is a constant, % grade is a constant (at least we're assuming it is), and we know initial speed and total distance - so we have an algebra equation with friction as the only unknown. Also, friction is basically stimp, which has been given.
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A flat 12' putt on stimp 12 goes 12 feet. According to the AimPoint app a 12 foot putt needs to be hit with 20.7 feet of speed (equivalent to flat ground) to go 12 feet. Ditto for the downhill putt, 4.5 feet. Stimp 12 is pretty fast, and 4% is pretty steep.
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Interesting... we have answers ranging from 16.5 ft to ball-rolls-forever.
I had this idea because I recently realized how I always go long on downhill putts, but I'm rarely short on uphill putts. It not obvious to me how much I need to change my stroke for an uphill putt vs a downhill putt, as compared to a straight putt of the same length.
I posted this as a fun experiment so people could give a general guess as to what as to what an steep-ish Augusta green might do. I linked the paper, so anyone could work it out if they really wanted to, but that wasn't really the intent. Maybe I'll wait til tonight to post the simulation result.
Quote:
Is there a conveyor belt involved?
Why, yes. Actually I forgot to mention this "green" is inside the cabin of a lifting off 747!
Originally Posted by Fourputt
How much grain? Which way is the grain? Most of the time the grain will lean downhill, thus increasing the speed of the downhill putt even as it kills the pace on the uphill putt.
The equations assume grain has no effect...I'm just going by what the simulation spits out.
I also think that there is not enough info to get the answer. Like Hendog said, gravity and friction are the only outside forces acting on it after the putter hits it. I think we need to know the friction factor of the green to figure it out.
Like sacm3bill said, everything you need to know about friction should be wrapped up in stimp. When you roll the ball down the stimpmeter on flat ground, it goes 12 feet. That tells you how much friction there is.
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# How to Perform a Breusch-Pagan Test in Excel
A Breusch-Pagan Test is used to determine if heteroscedasticity is present in a regression analysis.
This tutorial explains how to perform a Breusch-Pagan Test in Excel.
### Example: Breusch-Pagan Test in Excel
For this example we will use the following dataset that describes the attributes of 10 basketball players.
We will fit a multiple linear regression model using rating as the response variable and points, assists, and rebounds as the explanatory variables. Then we will perform a Breusch-Pagan Test to determine if heteroscedasticity is present in the regression.
Step 1: Perform multiple linear regression.
Along the top ribbon in Excel, go to the Data tab and click on Data Analysis. If you don’t see this option, then you need to first install the free Analysis ToolPak.
Once you click on Data Analysis, a new window will pop up. Select Regression and click OK. Fill in the necessary arrays for the response variables and the explanatory variables, then click OK.
This produces the following output:
Step 2: Calculate the squared residuals.
Next, we will calculate the predicted values and the squared residuals for each response value. To calculate the predicted values, we will use the coefficients from the regression output:
We will use the same formula to obtain each predicted value:
Next, we will calculate the squared residuals for each prediction:
We will use the same formula to obtain each squared residual:
Step 3: Perform a new multiple linear regression using the squared residuals as the response values.
Next, we will perform the same steps as before to conduct multiple linear regression using points, assists, and rebounds as the explanatory variables, except we will use the squared residuals as the response values this time. Here is the output of that regression:
Step 4: Perform the Breusch-Pagan Test.
Lastly, we will perform the Breusch-Pagan Test to see if heteroscedasticity was present in the original regression.
First we will calculate the Chi-Square test statistic using the formula:
X2 = n*R2new
where:
n = number of observations
R2new = R Square of the “new” regression in which the squared residuals were used as the response variable.
In our example, X2 = 10 * 0.600395 = 6.00395.
Next, we will find the p-value associated with this test statistic. We can use the following formula in Excel to do so:
=CHISQ.DIST.RT(test statistic, degrees of freedom)
In our case, the degrees of freedom is the number shown for df of regression in the output. In this case, it’s 3. Thus, our formula becomes:
=CHISQ.DIST.RT(6.00395, 3) = 0.111418.
Because this p-value is not less than 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that heteroscedasticity is present in the original regression model.
May 13, 2024
April 25, 2024
April 19, 2024
## One Reply to “How to Perform a Breusch-Pagan Test in Excel”
1. Chris Wright says:
Nice explanation.
One small suggestion to make the process a little simpler: check the ‘Residuals’ box in the ‘Regression’ window before running the original regression in Step #1. This will provide a list of predicted y values and, more importantly, residuals. Then, you can just square those in Step #2. Your version is good practice for generating predicted values, though.
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# Nonlinear Contact Stress Analysis of Bolted Poles
Document Sample
``` Nonlinear Contact Stress Analysis of Bolted Poles
Hengfeng Chen,Senlin Huang
TECO FA Global Wuxi R&D Center, Wuxi TECO Electric&Machinery Co.Ltd,
Jiangsu, China
Abstract
ANSYS has powerful functions of mechanical analysis, such as contact stress analysis, which can include
structure large deformation and material nonlinear property. In electric machine, such as salient
synchronous motor with bolted pole configurations, structural forces (centrifugal force and bolt pull force)
and magnetic forces (torque) are applied on the poles, and contact stress of interfaces between poles and
rims becomes very complicated, which is the key consideration of the bolted pole design. In this paper, we
carry out a nonlinear contact stress analysis of the bolted pole using ANSYS. The results give us a good
guide of bolted pole design.
Introduction
Comparing to dovetail pole structure, bolted pole configuration is much easier to be installed, and is widely
used in large salient pole motor. For large motor with very low speed, centrifugal force is not large, while
magnetic pull torque may be very large and becomes a key consideration. Even the magnetic torque may be
large enough to pull the pole off the rim. Therefore, for bolted pole configuration, bolt pull force is very
important, which withstands not only the centrifugal force, but also the magnetic pull torque. The bolt pull
force must be applied properly, and will not be small to cause pole pulled off and large to make pole or rim
yielding. In the following, some nonlinear contact stress analyses of the bolted pole are carried out using
ANSYS. The results give us a good guide for bolted poles design.
Geometry and FEA model
A 3D model of a large synchronous motor with 18 salient poles is shown in Figure 1. Each laminated pole
is bolted by 8 bolts on the spider rim. The stud and spider rim help to fix pole. Since the poles are
symmetric in circumference and long distance in axial direction, a 2D 1/18 FEA model (one pole) is
created. As the contact area between the poles and the spider rim is high stress and strain region we care,
element size of it should be small enough to make contact stress converge. Table 1 shows material
properties of pole, stud and spider rim.
Figure 1. FEA model
Table 1 Material properties
Material Properties
Item Elastic Tangent
Density Possion’s Yield Strength
Modulus Modulus
(Kg/m3) Ratio (MPa)
(MPa) (MPa)
Pole
7870 0.29 75269 241.38 784.08
(lamination)
Bolt 7870 0.29 97244 370 478.81
Spider Rim 7870 0.26 76597 248.28 771.10
Nonlinear Contact Stress Analysis
As we know, contact problem is highly nonlinear and requires significant computer resources to solve, and
it presents two significant difficulties. First, we generally do not know the regions of contact until we've
run the problem. Contact surfaces can come into and go out of contact with each other in a largely
unpredictable and abrupt manner. Second, most contact problems need to account for friction. Frictional
response can be chaotic, making solution convergence difficult. Therefore, solution convergence is the
main consideration of contact problem
In the following analyses, two parameters, i.e. contact stiffness and element size, are modified to make
stress solution convergence. For each FEA model with certain element size, contact stiffness should be
modified from 0.01 to 1 until the solution is convergent.
The following contact options are set for all analyses.
• Contact face: inner face of pole, Conta172 element
• Target face: outer face of spider rim, Targe169 element
• Unsymmetrical contact
• Friction coefficient =0.2
• Default options otherwise specified
Contact stress for minimal and maximal bolt pretension
As stated above, the pole is fixed by bolts with pretension that cannot be small to cause pole pulled off and
large to make pole or rim yielding. With minimal and maximal pretension, the motor should run normally
in rated condition and 20% over speed condition and in the case we want to know how large the contact
stress is. The minimal and maximal pretensions are 320272 N per bolt and 651998 N per bolt, respectively.
As table 2 shows, for each element size, after many contact stiffness modifications, the contact solution
arrives at convergence and a maximal contact stress can be obtained. Also, when the element size becomes
smaller, the maximal contact stress is convergent and almost unvaried. Figure 2 shows the contact stress for
minimal pretension, and figure 3 shows the contact stress for maximal pretension.
Table 2 Maximal Contact Stress (MPa)
Element size (mm)
Boundary Conditions
2 1 0.6 0.4 0.3 0.2
Minimal pretension
112 158 185 215 224 227
Running speed
Minimal pretension
98.2 144 168 192 203 207
20% over speed
Maximal pretension
166 216 248 283 290 288
Running speed
Maximal pretension
158 209 241 276 284 281
20% over speed
227Mpa (Running speed) 207Mpa (20% over speed)
Figure 2. Contact Stress (for minimal pretension)
288Mpa (Running speed) 281Mpa (20% over speed)
Figure 3. Contact Stress (for maximal pretension)
Maximal torque calculation
For minimal and maximal pretension, we want to know how large the pull torque is that the motor can
withstand. However, the maximal torque calculation is an inverse problem. Given a torque, a contact
analysis is carried out, and the contact status of the system is determined. When the contact status is open
but near contact, the maximal torque is obtained. Therefore, the calculation is trial running and time-
consuming.
For minimal pretension, after many torque trials, when a torque (23872601 N.m) is applied on the pole, the
contact status is shown as figure 4. The pole and the spider rim are closely abrupt from contact. Also the
contact stress is very little (6.74MPa) as figure 5 shows.
Figure 4. Contact Status and Slide (2-closed and sliding, 1-open but near contact)
Figure 5. Von Mises Stress (Max. 88Mpa) and Contact Stress (Max. 6.74 MPa)
Also, for maximal pretension, after many torque trials, when a torque (73453808 N.m) is applied on the
pole, the contact status is shown as figure 6. The pole and the spider rim are closely abrupt from contact.
Also the contact stress is very little (61.2MPa) as figure 7 shows.
Figure 6. Contact Status and Slide (2-closed and sliding, 1-open but near contact)
Figure 7. Von Mises Stress (Max. 187Mpa) and Contact Stress (Max. 61.2 MPa)
Conclusion
Through nonlinear contact stress analysis, the minimal and maximal bolt pretension can be obtained, and
also the maximal torque can be determined. These can help to apply proper pretension on the bolt, and
determine the proper speed and mechanical power of a large motor.
Acknowledgement
We would like to thank colleagues, especially Dr. Peter Zhong of TECO FA Global R&D Center. Their
strong supports give us motivation to accomplish this paper.
References
1. ANSYS Help Document
2. Bingnan Sun, Tao Hong, etc. 1998. Elastoplastic Mechanical in Engineering, Zhejiang University
Press: Zhejiang (Chinese), pp.98-101.
```
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### Download Free Practical Algebra A Self Teaching Guide Second Edition Book in PDF and EPUB Free Download. You can read online Practical Algebra A Self Teaching Guide Second Edition and write the review.
Practical Algebra If you studied algebra years ago and now need a refresher course in order to use algebraic principles on the job, or if you're a student who needs an introduction to the subject, here's the perfect book for you. Practical Algebra is an easy and fun-to-use workout program that quickly puts you in command of all the basic concepts and tools of algebra. With the aid of practical, real-life examples and applications, you'll learn: * The basic approach and application of algebra to problem solving * The number system (in a much broader way than you have known it from arithmetic) * Monomials and polynomials; factoring algebraic expressions; how to handle algebraic fractions; exponents, roots, and radicals; linear and fractional equations * Functions and graphs; quadratic equations; inequalities; ratio, proportion, and variation; how to solve word problems, and more Authors Peter Selby and Steve Slavin emphasize practical algebra throughout by providing you with techniques for solving problems in a wide range of disciplines--from engineering, biology, chemistry, and the physical sciences, to psychology and even sociology and business administration. Step by step, Practical Algebra shows you how to solve algebraic problems in each of these areas, then allows you to tackle similar problems on your own, at your own pace. Self-tests are provided at the end of each chapter so you can measure your mastery.
A sharp mind, like a healthy body, is subject to the same ruleof nature: Use it or lose it Need a calculator just to work out a 15 percent service charge?Not exactly sure how to get the calculator to give you the figureyou need? Turn to this revised and updated edition of All the MathYou'll Ever Need, the friendliest, funniest, and easiest workoutprogram around. In no time, you'll have total command of all the powerfulmathematical tools needed to make numbers work for you. In adollars-and-cents, bottom-line world, where numbers influenceeverything, none of us can afford to let our math skills atrophy.This step-by-step personal math trainer: Refreshes practical math skills for your personal andprofessional needs, with examples based on everydaysituations. Offers straightforward techniques for working with decimals andfractions. Demonstrates simple ways to figure discounts, calculatemortgage interest rates, and work out time, rate, and distanceproblems. Contains no complex formulas and no unnecessary technicalterms.
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John Broskie's Guide to Tube Circuit Analysis & Design
30 Nov 2010
# Cathode-Coupled Amplifier (+ Aikido CF)
Cathode-Coupled Amplifier (CCA)
I have covered this circuit many, many times before. Background Blogs:
The last entry is actually a good place to start, as it shows several ways to overcome the problem of dissimilar cathode-to-plate voltages in a CCA. This problem is the CCA's one big hassle to overcome. In fact, if the using CCA circuit didn't entail dealing with dissimilar cathode-to-plate voltages, I am sure that this topology would be much more popular, as it has much to offer, such as low input capacitance and no phase inversion and a high-impedance feedback port.
In the schematic above, on the left, we see two triodes that share a common cathode resistor and separate B+ voltages, but the same DC grid voltage (0V). If we wish to use a single B+ voltage, then we must alter the cathode-to-grid voltages to compensate for the dissimilar cathode-to-plate voltages, as shown in the schematic on the right.
Actually, the unspoken assumption here is that the two triodes share the same current draw. In fact, this is a precondition worth retaining, as the identical idle current helps linearize the two tube's gain. If we are, however, willing to run different idle current draws, then we can use the same grid-bias voltage for both triodes. An inspection of a triode's plate curves shows how this can be done.
Note the -2V grid line and where it intercepts the 100V and 150V plate voltages; with 100V on the plate of this triode, the conduction equals 3mA; with 150V, 7.6mA. Yet, both plate voltages rely on the same -2V of grid voltage. Applying these two operational points in one circuit is straightforward enough: just add the 3mA and 7.6mA together to find the correct common-cathode resistor value and use just the 3mA idle current amount for determining the plate resistor value.
The downside to this technique is an increased distortion figure. Furthermore, the right triode will be expected to drive the external load resistance and capacitance, but its idle current is less than half the left triode's current conduction, unfortunately. (On the other hand, if the right triode is only expected to drive a cathode follower's grid, then the weaker current is much less of an issue.)
If we wish to keep the equal idle currents and the single B+ voltage, then we will have to give the triodes differing grid-bias voltages. Once again, an inspection of the plate curves shows how this is done.
This time we hold the plate current steady (for both triodes), which the single green horizontal line indicates. At 5.2mA, the green intersects the two plate voltages used in following cathode-coupled amplifier design example; namely, 150V and 300V. The grid voltage intercepts occur at -3V and -9.5V, as marked by the red grid-voltage lines. In other words, the left triode will see a 300V differential and will need to see a grid voltage of -9.5V to draw an idle current of 5.2mA, while the right triode will experience only a 150V differential and will need to see a grid voltage of only -3V to draw an idle current of 5.2mA.
Okay, I know many are scratching their heads right now, as the right triode is seeing a +6.5V grid bias voltage, not -3V, as the graph's inspection implied. Remember, the cathode-coupled amplifier's triode share a common cathode connection, so the common cathode voltage becomes our reference. Relative to the +9.5V cathode voltage, the left triode's grid is at -9.5V and the right triode's grid is at -3V, as +6.5V minus +9.5V equals -3V. Now that everyone is on the same page, the next question is: What happened to the negative power supply rail?
A negative power supply could be used, but many tube fanciers just don't do negative power supplies and the relatively high common cathode voltage (+9.5V) means that, in a line-stage amplifier, more than enough input voltage headroom exists, as line-level input signals seldom exceed 2Vpk. Okay, but where do we get a 10.2mA constant-current source? We could build one out of discrete solid-state devices, such as a FET and a source resistor; or we could use a constant-current source IC, such as the LM334 or the new Linear Technology LT3092 constant-current source.
The LT3092 is an extremely interesting constant-current source that only requires two resistors to set its idle current. Because the two resistors tie together at one end, the LT3092 is considered a two-terminal constant-current source, which can be floated or ground or B+ referenced. It can withstand up 40V and source up to 200mA. I haven't actually had any hands on experience with the LT3092 yet, but I am eager to try some tests on it.
On the other hand, because the common cathode voltage is so high, we can get away with using a 913-ohm common cathode resistor, as 9.5V / 10.4mA = 913 ohms, which will make the solid-state-fearing types happy. The last unanswered question is: How do we establish the +6.5V bias voltage for the right triode's grid? Ah, finally we arrive at the meaty bits. We could use a two-resistor voltage divider that spans from ground to the B+ connection. While this would work well enough, this arrangement is not optimal. A better setup is to terminate the voltage divider's top resistor into the right triode's plate, rather than the B+ connection.
Why is this arrangement better? If nothing else, we will not have to deal with power-supply noise leaking into the cathode-coupled amplifier's output, as the Aikido cathode follower's output should be dead quiet. Moreover, in this setup, the voltage-divider resistors constitute a DC feedback mechanism that works to keep the correct bias voltage needed to establish a plate voltage equal to half the B+ voltage on the right triode's plate. Why is this important?
The cathode-coupled amplifier, however, needs a bit more help to make a first-rate line-stage amplifier, as its output impedance is too high and PSRR is weak (almost nonexistent). Adding an Aikido cathode follower to the CCA input stage makes a complete, high-quality line-stage amplifier that offers low distortion, low output impedance, wide bandwidth, no phase inversion, and a stellar PSRR figure.
The Aikido cathode follower DC couples with the CCA stage, and the two-resistor voltage divider defined by resistors R5 & R6 provides the DC feedback loop to keep the Aikido cathode follower output centered at half the B+ voltage. The second voltage divider resistor string, R11 & R22, injects the required sampling of power-supply noise into the Aikido cathode follower's bottom triode's grid to null the power-supply noise at the Aikido cathode follower's output. The diode D11 is a safety device that protects the Aikido cathode follower top triode at start up, when the tubes are cold and not conducting; once the tubes are hot, the diode falls out of the circuit, as it will no longer be forward biased.
The LM334 constant-current source is optional, as resistor R3 can be used by itself (the LT3092 cannot be used on the CCA PCB). On the other hand, the LM334 constant-current source does reduce the distortion substantially (-20dB improvement), but it does make the CCA a more sensitive to changes in the B+ voltage and will require more circuitry tweaking to obtain as low a noise figure as the common-cathode resistor achieves, such as lowering reisitor R9's value a tad relative to R8, say 270 ohms to 300 ohms. On the other hand, the current source does produce more voltage gain from the CCA, about +3dB more with a 12AU7.
Finding the right values for resistors R5 & R6 is easy enough. As we all know, a triode's amplification factor is simply a measure of the relative effectiveness of the grid over the plate in controlling the current flow through the triode. Thus, a triode with a mu of 10 holds a grid that is 10 times more efficient than its plate. So an increase in plate voltage of 10 volts can be countered by a decrease of 1V on the grid. In other words, we can maintain a fixed idle current in spite of a 10V increase in plate voltage by moving the grid 1V more negative.
If we inspect the plate curves from the previous examples, we will find the triode's mu equal to about 23.1. If we then divide the 150V difference in plate voltages by 23.1, we get 6.49V, which is close to the 6.5V difference in grid voltages (9.5V and 3V) that we gleaned from plate curves. Thus, we need a voltage that yield a 1/mu voltage division; so R6 should be (mu – 1) times bigger than R5. This wonderful trick works because the voltage from ground to the Aikido cathode follower's out is equal to the voltage across the plate resistor, R4, which is also the difference in plate voltages.
Finding R11's value takes a bit more work, as the old Aikido formula does not apply here. If a constant-current source is used to load the common cathodes, then the cathode-coupled amplifier offers almost zero power supply rejection, so all of the power-supply noise must be fed to the Aikido cathode follower's bottom triode's grid. Thus, if the LM334 CCS is used in place of the simple common-cathode resistor, R3, then the CCA's plate resistor's value must equal B+/Iccs and R11 must be replaced by a jumper wire and R8's value must be slightly greater than R9's value to null the power-supply noise from the CCA's output. How much greater? It depends on the triode's mu and rp. The following formula establishes the correct ratio between resistors R8 & R9, when a constant-current source is used.
With a common-cathode resistor and 6CG7s, in contrast, there is a 17% reduction in power-supply noise at the cathode-coupled amplifier's output. I used the following values in my own CCA setup (all tubes are 6CG7s):
If a 12AU7 is used as the input tube, use a 931-ohm common-cathode resistor and set R11 to 15k and R6 to 175k. Wait a minute! Don't you first need to know what the B+ voltage is? No. That's the amazing thing about this topology—these values work perfectly with a B+ voltage that spans 100V to 400V. On the other hand, if a CCS is used in place of the common-cathode resistor, then the CCA's plate resistor's value must equal B+/Iccs. While testing my CCA prototype, I was thrilled to see the output stage always center to within 1V of the half-the-B+ target, as spun the variac's AC output voltage up and down.
The PSRR figure, even without the LM334 CCS, was stellar and distortion quite low, about 0.1% with 1Vpk at 1kHz into a 47k load. I wondered if the DC feedback loop was not also contributing AC feedback, so I tried the line-stage amplifier with resistor R5 bypassed, so the small AC signal present would be shorted to ground. The results were interesting; with the bypass capacitor in place, the distortion almost imperceptibly budged upwards. in other words, very little negative feedback obtains. This makes sense, as the line-stage amplifier's open loop gain would have to be greater than the input triode's mu for negative feedback to take hold. With the 6CG7s, the AC gain is only 5.2 (or +14dB).
Relatively Low-Voltage CCA Line Amplifier
The input triode in the CCA sees the biggest cathode-to-plate voltage differential, so care should be taken to not to exceed the triode's maximum cathode-to-plate voltage. This brings up the controversial issue of the 6DJ8's actual cathode-to-plate voltage limit. Most tube manuals state that the limit is 130Vdc, an amazingly low voltage for a triode. I believe this voltage only refers to the tube under cascode use, not in a grounded-cathode amplifier or cathode follower. For example, the JJ E88CC specifications sheet states that its maximum plate voltage is 220Vdc:
E88CC R. F. DOUBLE TRIODE Base: NOVAL Uf = 6,3 V If = 365 mA Typical characteristic: Ua = 90 V Ug = -1,3 V Ia = 15 mA S = 12,5 mA/V Ri = 2,6 kO µ = 33 Limiting values: Ua0 = 550 V Ua(la=0) = 400 V Ua = 220 V Ua(War<0,8 W) = 250 V PaR = 1,5 W Wg1R = 0,03 W Ik = 20 mA Ug = -100 V Rg = 1 MO U+k/f- = 120 V U-k/f+ = 60 V Rk/f = 20 kO
See Blog 62 for more information. So before proceeding, WARNING DO NOT ATTEMPT THIS AT HOME, UNLESS YOU REALLY KNOW WHAT YOU ARE DOING. I tried the 6DJ8 in the CCA, with a B+ voltage of only 150V. (I started with a 120Vac power transformer, which rectified up to about 165Vdc, which was then dropped to 150Vdc through resistors R12 and R17.) Nothing exploded or arced. Furthermore, the sound was surprisingly fine and the gain was close to +20dB. The lower B+ voltage allowed me to use much larger-valued capacitors in the high-voltage power supply (220µF/200V rather than 47µF/450V for C7 & C8 and 270µF/200V rather than 150µF/400V for C5), which greatly reduced the ripple the Aikido cathode follower had to wisk away. In other words, it was dead quiet. I used the following resistor values:
LM334 constant-current source
The National Semiconductor LM334 is a well established device. It can be bought in four different packages, but the CCA PCB is configured for the TO-92, three-lead package.
Specifications V+ to V- Forward Voltage LM134/LM234/LM334 40V LM234-3/LM234-6 30V V+ to V- Reverse Voltage 20V R Pin to V- Voltage 5V Set Current (max) 10mA Power Dissipation 400mW ESD Susceptibility 2000V Operating Temperature Range LM134 -55°C to +125°C LM234/LM234-3/LM234-6 -25°C to +100°C LM334 0°C to +70°C Soldering Information TO-92 Package (10 sec.) 260°C
The LM334 requires only one resistor, Rset, to establish its idle current.
The following graph shows the turn-on voltages and bias currents set by various Rset resistor values.
Note that 10mA, the optimal value for most CCA setups, does not receive a resistor value, but we can readily see that 6.8 ohms is the correct value. (The LM334 datasheet goes into much more detail, but for tube work, 6.8 ohms is close enough.) Although 10mA is the LM334's maximum current flow, the device sees less than 10V with most tubes, such as the 6CG7 and 12AU7, so the device's dissipation is usually less than 100mW, well below its 400mW limit. Nonetheless, it is a good idea to attach a small heatsink to the IC, as it better ensures an accurate idle current.
The CCA Power Supplies
The CCA PCB is an All-in-One affair, wherein the audio and power-supply circuitry reside on the one board. The B+ power supply is a simple design that uses cascading pi filters to reduce ripple and rectification nastiness.
THe CCA PCB can accept either a center-tapped or non-center-tapped high-voltage power transformer. (In my own CCA setup, I used a 240Vac non-center-tapped Hammond power transformer.)
The CCA PCB holds the heater power supply as well and it is regulated.
As can be seen, the power supply can accept either full-wave bridge rectifier circuit or a full-wave voltage doubler rectifier configuration. When used as a full-wave bridge rectifier circuit, the two power supply filtering capacitors are placed in parallel by orienting their positive leads to where the heatsink sits; and the secondary attaches to the two encircled AC pads.
Configured as a voltage doubler, these capacitors placed in series by being rotated 90 degrees clockwise, so the positive leads point to the center-tap (CT) pad at the bottom of the PCB; the transformer secondary attaches to both the single AC pad in between capacitors C13 and C14 and AC pad that feeds rectifier D10 and D8; and D7, D9, C9, C11 are left off the PCB. If used as a full-wave center-tap circuit, the two heater capacitors, C13 & C14, are placed in parallel by orienting their positive leads to where the heatsink sits; and the secondary attaches to the two encircled AC pads while the secondary center-tap attaches to the CT pad.
I used a 12Vac/3A transformer, as I like being able to switch the heater power supply on independently from the B+ power supply. By the way, the CCA can easily be built upside down, wherein the tube sockets are soldered to the top of the PCB, but everything else attaches to the PCB's bottom. Why? This allows the tubes to protrude through holes in the top of the chassis. The usual roadblock is the heater voltage regulator, but on the CCA PCB, I placed redundant solder pads on the bottom for the LD1085.
The heater regulator uses the LD1085 low-dropout, adjustable, three-pin, 3A voltage regulator. The regulator can be set to an output voltage between 6V to 25V, but the assumption is that a 12Vdc output voltage will be used for the heaters, so that 6.3V heater tubes (like the 6FQ7 and 6DJ8) or 12.6V tubes (like the 12AU7 or 12BH7) can be used.
Both voltage types can be used exclusively, or simultaneously; for example a 6GC7 for the input tube and an ECC99 for the output tube. Thus, if the input tubes (V1 and V2) are 6CG7s and the output tubes (V3 and V4) are 12BH7s and the heater regulator output voltage is 12Vdc, then use jumpers J2, J4, and J6.
More RMAF Details
Last time, I mentioned several of the loudspeakers that I liked at the Rocky Mountain Audio Festival, which prompted a friend to ask why I didn't mention any new tube gear. Well, I did see and hear some interesting tube amplifiers. But first let me describe what happened when I first arrived at the show. I walked into a room filled with \$200,000 worth of audio gear and I listened to an excellent solid-state power amp through some excellent speakers that cost twice as much as my minivan. How did it sound? Marvelous. So much so that I began to worry about tubes having a harder time competing with so strong a contender.
Imagine a wife or a girlfriend being dragged into a car part store by her significant other. Her expectations of having a good time are low, so she is startled see an impossibly handsome man standing behind the counter. "Maybe I should shop here more often," she says to herself. Upon approaching the counter, she is disappointed to see that the gorgeous face was only printed on a life-sized poster cutout of a famous race car driver and that the actual living fellow behind the counter was only okay looking, with a bad haircut and in need of a shave. Well, this is how I felt, for as pretty as the picture the solid-state amplifier painted was it never felt real. And even when the tube power amplifier rendered a tarnished sonic picture, zits and pockmarks included, the sound still maintained a believability about it. In a nutshell, this is why tubes still matter.
With a Name Like Schiit, You have to Sound Good
I had great time talking to Jason and Rina at the Schiit table. Jason Stoddard is a TCJer and he even used my software in designing his tube headphone amplifier, the Valhalla. Check out the specifications on this cool tube-based headphone amplifier (actually, it is hot to the touch and draws 40W from the wall socket).
I just do not understand how Schiit can make a such an amazing headphone amplifier, an amplifier that is made in America (not China), that uses such high-quality parts (Alps, Nichicon, Wima, Dale, and Neutrik), that uses internal power transformers and not a switcher wallwart, that comes with a 5-year warranty, that sports four tubes and does not use a solid-state output buffer, and that sells for only \$349. Amazing stuff.
Schiit also makes a solid-state headphone amplifier, the Asgard (with a name like Asgard...), that sells for only \$249. From the Schiit website:
"The Asgard is a fully discrete, Class A, single-ended FET headphone amplifier with no overall feedback and a noninverting circuit topology. Its high-current design makes it uniquely suitable for low-impedance headphones."
If I didn't already own about ten headphone amplifiers, I would certainly buy a Valhalla from Schiit. My only wish is that they paint their boxes flat-black enamel paint. Here's why:
Typical Emissivity of Various Surfaces Material Finish Emissivity Silver – Polished 0.02 Copper - commercial polished 0.03 Aluminum – polished 0.04 Tin – bright 0.04 Aluminum – rough 0.06 Copper – machined 0.07 Nickel Plate - dull finish 0.11 Stainless Steel - alloy 316 0.28 Steel - rolled sheet 0.55 Copper - thick oxide coating 0.78 Steel - oxidized 0.78 Aluminum - anodized (any color) 0.80 Paints & Lacquers - gloss finish 0.89 Paints & Lacquers - flat finish 0.94
Taller feet would also prove effective in keeping the little amplifier cooler.
At the Other \$\$\$ Extreme
I saw a lot of interesting audio products at the RMAF and when I asked the selling price, the answer was usually \$20k, \$30k, \$40k, \$50K, \$60k, \$70k... Maybe it was my eyebrows doing somersaults that gave away my astonishment, for the the next thing I heard, over and over, was "Sure it's a lot of money, but don't think that we are making any money off of it." I never knew that such munificence existed in high-end audio. When I think about zero-profit enterprises, \$10k record clamps and \$70k speakers never come to mind. Or, perhaps, I got it wrong, maybe what I was being told was that they haven't sold any of \$70k speakers and they were still searching for their first customer. Nonetheless, it seems to me that if a speaker does not sound better than a Quad 2805 (less than \$10k), it should not cost more than a Quad 2805. Either way, there was a lot of super expensive gear on display, even if no one was making any money selling it. Strange, very strange.
So, the only thing I can conclude is that the nice people at Schiit must be making a killing. Of course, ours is a bad economy and I heard from many American firms that if it weren't for the export market, they would be out of business. I also encountered near universal scorn and derision for high-end cable manufacturers. I can understand the logic behind such animus, as it must be painful to see \$20 worth of wire selling for more than your power amplifier or loudspeaker or turntable... These cable guys run a markup that would make a jewelry store owner blush. Sure, cables can prove important, but they should never cost more than the rest of your system. Or am I getting this wrong? Would \$20 speakers hooked up with \$10,000 cables sound much better than \$10,000 speakers hooked up with \$20 cables?
New Books
The RMAF was not just about audio gear; other items were on display and for sale, such as records, CDs, and books. Finding any new audio-related books is amazing enough, but finding two exceptional new audio books a treasure find. The first is by Robert Cordell and it is titled, Designing Audio Power Amplifiers.
This handsome 600-page book covers aspects of solid-state power amplifier design, from the basics to the secrets only the masters know. The following table of contents gives you a good idea of much material is held between the covers:
Part 1: Audio Power Amplifier Basics
1. Introduction
2. Power Amplifier Basics
3. Power Amplifier Design Evolution
4. Negative Feedback Compensation and Slew Rate
5. Amplifier Classes, Output Stages and Efficiency
6. Summary of Amplifier Design Considerations
Part 2: Advanced Power Amplifier Design
7. Input and VAS Circuits
8. DC Servos
9. Advanced Forms of Feedback Compensation
10. Output Stage Design and Crossover Distortion
11. MOSFET Power Amplifiers
12. Error Correction
13. Other Sources of Distortion
Part 3: Real World Design Considerations
14. Output Stage Thermal Design and Stability
15. Safe Area and Short Circuit Protection
16. Power Supplies and Grounding
17. Clipping Control and Civilized Amplifier Behavior
18. Interfacing the Real World
Part 4: Simulation and Measurement
19. SPICE Simulation
20. SPICE Models and Libraries
21. Audio Instrumentation
22. Distortion and its Measurement
23. Other Amplifier Tests
Part 5: Topics in Amplifier Design
24. The Negative Feedback Controversy
25. Amplifiers without Negative Feedback
26. Balanced and Bridged Amplifiers
27. Integrated Circuit Power Amplifiers and Drivers
Part 6: Class D Audio Amplifiers
28. Class D Audio Amplifiers
29. Class D Design Issues
30. Alternative Class D Modulators
31. Class D Measurement, Performance and Efficiency
As soon as I have finished reading Bob's book, I will give it a proper review; but if you have any interest in solid-state or hybrid power design—or audio design in general, as much of what Mr. Cordell writes about also applies to phono stages and electronic crossovers—then be sure to drop many hints to your loved ones that you would enjoy receiving this book for Christmas. Well written, thoroughly researched, beautifully illustrated, highly recommended.
The second book is not a book, at least not in the typical sense, as it a bookzine. A magazine as thick as a book, but ad free—this is the answer to the question What is Linear Audio? Jan Didden, famous for his many articles in Audio Amateur and audioXpress, is the editor and publisher. He has brought out Volume 0 (yes, zero) and filled it with a who's who of audio design:
From the Linear Audio website:
HowLinear Audio is a printed publication, published in a schedule that will be determined by available content of the required quality level. A publication of minimum two issues per year (April and September) of at least 120 pages of technical content is anticipated. I call it a 'bookzine'. Linear Audio will be sold on-line through www.linearaudio.net
By whom - Linear Audio will be published by Linear Audio Publishing from The Netherlands. Content will be selected from submissions from professional and serious amateurs and DIY-ers alike, the sole criterion being the quality and timeliness of the content.
For whom - Linear Audio is meant for anyone who is interested in technical audio developments or who wants to contribute to them. Being employed in audio engineering is not a requirement at all. Happy reading, happy writing, happy building and happy listening!
Loudspeaker, solid-state, and tube design articles abound in this 169 page bookzine. And those 169 pages are ad free. Another way of looking at it that most magazines run about 50% content and 50% ads; thus, Linear Audio would have been 338 pages long had it been filled with ads. In addition, regular magazines engage in funny math to convince their advertisers that they are getting their money's worth, such as printing twice as many magazines as they have subscribers. In other words, the ads pay for magazine's printing. With Linear Audio, the reader pays for it all. Sure that is all very nice, but can't I just read about audio off the net and pay nothing? Indeed, but high-quality articles like these deserve to be printed, which would certainly cost more in paper and toner or ink than the cost of this beautifully put together bookzine.
In short, also highly recommended.
Next Time
I have to come clean, none of what I just wrote was what I had intended to write. I planned on writing about OTL output stages and split-load phase splitters. I will try to catch up soon. In the mean time, be sure to check out the new CCA PCB and kit (only \$45 for the PCB and 20-page user guide and only \$99 for the PCB and all the parts resistors, capacitors, rectifiers, LDO regulator, LM334 CCS, tube sockets, standoffs and O-rings, except the tubes and coupling capacitors) at the GlassWare Yahoo store. Thanks.
//JRB
Kit User Guide PDFs
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E-mail from GlassWare Customers
Hi John,
I received the Aikido PCB today - thank you for the first rate shipping speed.
Wanted to let you know that this is simply the best PCB I have had in my hands, bar none. The quality is fabulous, and your documentation is superb. I know you do this because you love audio, but I think your price of \$39 is a bit of a giveaway! I'm sure you could charge double and still have happy customers.
Looking forward to building the Aikido, will send some comments when I'm done!
Thank you, regards
Gary
Mr Broskie,
I bought an Aikido stereo linestage kit from you some days ago, and I received it just this Monday. I have a few things to say about it. Firstly, I'm extremely impressed at the quality of what I've been sent. In fact, this is the highest quality kit I've seen anywhere, of anything. I have no idea how you managed to fit all this stuff in under what I paid for it. Second, your shipping was lightning-quick. Just more satisfaction in the bag, there. I wish everyone did business like you.
Sean H.
9-Pin & Octal PCBs
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## matplotlib-users
[Matplotlib-users] Customizing Quiver From: Ryan May - 2008-06-27 22:30:20 ```Hi, Has anyone ever tried to modify quiver to change away from plotting arrows? In my quest to make matplotlib useful the meteorological community at large, I'm trying to be able to plot what we refer to as wind barbs. Basically, instead of an arrow whose size indicates the magnitude of the vector field, a wind barb consists of a line whose alignment indicates the direction of the vector field (like the arrow) and has a number of lines or triangles along this shaft to indicate wind speed. It's a little odd and domain specific, but it's a must for the meteorologist. We like it because, unlike the arrows, you can pick off the actual wind speed at a point. See here for an example: http://www.rap.ucar.edu/weather/info/about_windbarb.html http://www.rap.ucar.edu/weather/surface/displaySfc.php?region=abi&endDate=20080627&endTime=-1&duration=0 Has anyone ever tried this (I'm pretty sure I know the answer)? Failing that, can anyone give me an idea of how difficult it might be for me to tweak quiver to do this? Thanks, Ryan -- Ryan May Graduate Research Assistant School of Meteorology University of Oklahoma ```
Re: [Matplotlib-users] Customizing Quiver From: Eric Firing - 2008-06-28 01:48:17 ```Ryan, Quiver is not quite the right starting point, although parts of it might work--and actually it could be bent into shape. The problem with attempting a direct transformation of arrows into wind barbs is that the arrows are formed with a fixed number of vertices, so everything is done with nice, regular ndarrays. The natural way to make wind barbs would be with the number of vertices depending on the wind speed. One could probably get around this by using the maximum number of vertices, corresponding to the max wind speed, and then letting some of the barbs be zero-length. Another difference might be that wind barbs would best be drawn with lines, not with patches. It might be worthwhile to start with the Quiver class and modify it as needed. I certainly would not try to include Windbarb functionality as an option or extension of Quiver. At most, one might be able to pull some common functionality out into a base class for both. Something I keep wanting to get to is an ellipse variant of quiver; that would be much closer to the present quiver than a windbarb version would. In any case, if you can come up with a good windbarb class, that would be great. I expect Jeff Whitaker would like to add that to basemap, too. Eric Ryan May wrote: > Hi, > > Has anyone ever tried to modify quiver to change away from plotting > arrows? In my quest to make matplotlib useful the meteorological > community at large, I'm trying to be able to plot what we refer to as > wind barbs. Basically, instead of an arrow whose size indicates the > magnitude of the vector field, a wind barb consists of a line whose > alignment indicates the direction of the vector field (like the arrow) > and has a number of lines or triangles along this shaft to indicate wind > speed. It's a little odd and domain specific, but it's a must for the > meteorologist. We like it because, unlike the arrows, you can pick off > the actual wind speed at a point. See here for an example: > > http://www.rap.ucar.edu/weather/info/about_windbarb.html > http://www.rap.ucar.edu/weather/surface/displaySfc.php?region=abi&endDate=20080627&endTime=-1&duration=0 > > Has anyone ever tried this (I'm pretty sure I know the answer)? Failing > that, can anyone give me an idea of how difficult it might be for me to > tweak quiver to do this? > > Thanks, > > Ryan > ```
Re: [Matplotlib-users] Customizing Quiver From: Ryan May - 2008-06-28 01:57:12 ```I kind of figured that was the case, upon looking at the code and realizing that the only commonality was in the broad purpose of both representations. Otherwise, there's really no commonality in the details or implementation. I just wanted to see what the "experts" thought before I embarked. Between this and getting a working SkewT and the meteorological community will be all set! Ryan Eric Firing wrote: > Ryan, > > Quiver is not quite the right starting point, although parts of it might > work--and actually it could be bent into shape. The problem with > attempting a direct transformation of arrows into wind barbs is that the > arrows are formed with a fixed number of vertices, so everything is done > with nice, regular ndarrays. The natural way to make wind barbs would > be with the number of vertices depending on the wind speed. One could > probably get around this by using the maximum number of vertices, > corresponding to the max wind speed, and then letting some of the barbs > be zero-length. > > Another difference might be that wind barbs would best be drawn with > lines, not with patches. > > It might be worthwhile to start with the Quiver class and modify it as > needed. I certainly would not try to include Windbarb functionality as > an option or extension of Quiver. At most, one might be able to pull > some common functionality out into a base class for both. > > Something I keep wanting to get to is an ellipse variant of quiver; that > would be much closer to the present quiver than a windbarb version would. > > In any case, if you can come up with a good windbarb class, that would > be great. I expect Jeff Whitaker would like to add that to basemap, too. > > Eric > > > Ryan May wrote: >> Hi, >> >> Has anyone ever tried to modify quiver to change away from plotting >> arrows? In my quest to make matplotlib useful the meteorological >> community at large, I'm trying to be able to plot what we refer to as >> wind barbs. Basically, instead of an arrow whose size indicates the >> magnitude of the vector field, a wind barb consists of a line whose >> alignment indicates the direction of the vector field (like the arrow) >> and has a number of lines or triangles along this shaft to indicate >> wind speed. It's a little odd and domain specific, but it's a must >> for the meteorologist. We like it because, unlike the arrows, you can >> pick off the actual wind speed at a point. See here for an example: >> >> http://www.rap.ucar.edu/weather/info/about_windbarb.html >> http://www.rap.ucar.edu/weather/surface/displaySfc.php?region=abi&endDate=20080627&endTime=-1&duration=0 >> >> >> Has anyone ever tried this (I'm pretty sure I know the answer)? >> Failing that, can anyone give me an idea of how difficult it might be >> for me to tweak quiver to do this? >> >> Thanks, >> >> Ryan >> > -- Ryan May Graduate Research Assistant School of Meteorology University of Oklahoma ```
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%% %% Ein Beispiel der DANTE-Edition %% %% %% Copyright (C) 2010 Herbert Voss %% %% It may be distributed and/or modified under the conditions %% of the LaTeX Project Public License, either version 1.3 %% of this license or (at your option) any later version. %% %% See http://www.latex-project.org/lppl.txt for details. %% %% %% ==== % Show page(s) 1 %% \documentclass[]{exaarticle} \pagestyle{empty} \setlength\textwidth{352.81416pt} \setlength\parindent{0pt} \StartShownPreambleCommands \usepackage{pstricks-add} \def\Lissa{t dup 2 RadtoDeg mul cos 3.5 mul exch 6 mul RadtoDeg sin 3.5 mul} \StopShownPreambleCommands \begin{document} \psset{unit=0.75cm} \begin{pspicture}(-4,-5)(4,5)% SEKANTENsteigung \parametricplot[plotpoints=500,linewidth=1.5pt]{0}{3.141592}{\Lissa} \multido{\r=0.000+0.314}{11}{% \psplotTangent[arrows=<->,linecolor=black!40]{\r}{1.5}{\Lissa}} \multido{\r=0.157+0.314}{11}{% \psplotTangent[arrows=<->,linecolor=black!40]{\r}{1.5}{\Lissa}} \rput(0,4.5){Sekantensteigung} \end{pspicture}\hfill% % \def\LissaAlg{3.5*cos(2*t)|3.5*sin(6*t)} \def\LissaAlgDer{-7*sin(2*t)|21*cos(6*t)} \begin{pspicture}(-4,-5)(4,5)% TANGENTENsteigung \parametricplot[algebraic,plotpoints=500, linewidth=1.5pt]{0}{3.141592}{\LissaAlg} \multido{\r=0.000+0.314}{11}{% \psplotTangent[algebraic,arrows=<->,linecolor=black!40]{\r}{1.5}{\LissaAlg} } \multido{\r=0.157+0.314}{11}{% \psplotTangent[algebraic,arrows=<->,linecolor=black!40,% Derive=\LissaAlgDer]{\r}{1.5}{\LissaAlg} } \rput(0,4.5){Tangentensteigung} \end{pspicture} \end{document}
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Surds Simplest Form
Simplify each of the following surds:√112 =√ Simplify each of the following surds:√50 =√ Simplify each of the following surds:√150 =√ Simplify each of the following surds:√45 =√ Simplify each of the following surds:√48 =√ Simplify each of the following surds:√275 =√ Simplify each of the following surds:√500 =√ Simplify each of the following surds:√486 =√ Simplify each of the following surds:√891 =√ Simplify each of the following surds:√648 =√ Simplify each of the following surdsinto their simplest form:3√2 + 5√2 =√ Simplify each of the following surdsinto their simplest form:9√3 - 11√3 =√ Simplify each of the following surdsinto their simplest form:4√2 - 7√2 +√2 =√ Simplify each of the following surdsinto their simplest form:⅔√7 - ¾√7 = -√ Simplify each of the following surdsinto their simplest form:√18 + √32 =√ Simplify each of the following surdsinto their simplest form:√96 - √54 =√ Simplify each of the following surdsinto their simplest form:√245 + √45 =√ Simplify each of the following surdsinto their simplest form:√432 - √243 =√ √567 + √320 - √112 - √20 =Simplify each of the following surdsinto their simplest form:√+√ 2√450 + 9√12 - 7√48 - 3√98 =Simplify each of the following surdsinto their simplest form:√-√
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# 5.10 Peak Analysis with Labtalk
## Summary
This tutorial demonstrates using labtalk for peak analysis.
## What You Will Learn
This tutorial will show you how to use Labtalk script to:
• Subtract a line from source data
• Find the highest peaks of each dataset
• Calculate the area, height and FWHM of the highest peak
• Create a graph of the quantities of the peak
• Create a theme of Peak Analyzer and run batch processing with the pre-saved theme
## Steps
### Example 1: Peak Analysis with Scripts
1. Import the data file Waterfall.dat under <Origin EXE folder>\Samples\Graphing folder with the associated import filter into the workbook Waterfall.
2. // Create a newbook and name it Waterfall (both long and short name)
newbook name:="Waterfall" option:=lsname;
// Define the file name and filter name
string fname$= system.path.program$ + "Samples\Graphing\Waterfall.dat";
string filtername$= "waterfall.oif"; // Use the impfile x-function to import the specified file with the import filter impfile fname:=fname$ filtername:=filtername$location:=data orng:=[Waterfall]1!; 3. Use a loop and the subtract_line X-Function to subtract a simple baseline of end points connection. 4. //Create new sheet named as SubtractedData newsheet name:=SubtractedData; //Define the newly created worksheet(current active) as a range variable range sdwk = !; //Active the "Waterfall" worksheet by its name page.active$ = "Waterfall";
//Define an integer to hold the number of columns in Waterfall sheet
int nc = wks.nCols;
//Find out the columns for excitation wavelength = 600, 610, 620, 630 ect
//Use the loop and subtract_line x-function to subtract baseline
//The first and last data points are found and connected linearly
//This line will be subtracted as baseline from source data
//Subtracted data will be named by excitation wavelength of source data
int nexc;
loop(ii, 2, nc)
{
range ry = [Waterfall]Waterfall!col($(ii)); int nWave = %(ry[D1]$);
if (mod(nWave, 10)==0)
{
nexc = nexc + 1;
range rx = [Waterfall]Waterfall!col(1);
range rr = [Waterfall]SubtractedData!(1,$(nexc)); subtract_line iy:=ry x1:=rx[1] /* Specifies the X value of the first point*/ y1:=ry[1] /* Specifies the Y value of the first point*/ x2:=rx[$(wks.maxRows)] /* Specifies the X value of the last point*/
y2:=ry[$(wks.maxRows)] /* Specifies the Y value of the last point*/ oy:=rr; //Rename the column long names in SubtractedData sheet //Use the User Defined Parameter Wavelength in Waterfall sheet rr[L]$ = ry[D1]$; //add a new column for next loop wks.addcol(); } else { continue; } } 5. Create new sheet for summary report 6. // Create new sheet for summary report and define ranges newsheet cols:=7 xy:="XYYYYYY" name:="Summary"; range rWaveLength = 1, rCntrInd = 2, rLtInd =3, rRtInd = 4; range rArea = 5, rCntr = 6, rHt = 7, rFWHM = 8; rWaveLength[L]$ = "WaveLength";
rCntrInd[L]$= "Peak Center Index"; rLtInd[L]$ = "Peak Left Index";
rRtInd[L]$= "Peak Right Index"; rArea[L]$ = "Peak Area";
rCntr[L]$= "Peak Center"; rHt[L]$ = "Peak Height";
rFWHM[L]$= "FWHM"; 7. To calculate the quantities of the highest peak of the data, we should find the highest peak with the pkfind X-Function and then calculate the quantities of the peak with the integ1 X-Function 8. //active the SubtractedData sheet page.active$="SubtractedData";
//add columns for temp results of the finded peaks
loop(cc, 1, 2)
{
wks.addcol();
};
loop(ii, 2, nexc)
{
//find the highest peak
range rData = col($(ii)); range pcenter=$(nexc+1), pleft= $(nexc+2), pright=$(nexc+3);
pkfind rData filter:=num value:= 1 ocenter:=pcenter oleft:=pleft oright:=pright; //find a highest peak
//copy the result from the temp columns to summary sheet
wrcopy iw:=SubtractedData ow:=Summary c1:=nexc+1 c2:=nexc+3 r1:=1 r2:=1 dc1:=2 dr1:= ii-1;
//compute the quantities of the highest peak with integ1
int n1 = pleft[1]; //specify the peak left index
int n2 = pright[1]; //specify the peak right index
range rint = rData[$(n1):$(n2)]; // range of data to be integrated
double aa, cc, hh, fw;
integ1 iy:=rint area:=aa x0:=cc y0:=hh dx:=fw oy:=<optional>;
//put the quantities to summary sheet
rArea[$(ii-1)] = aa; rCntr[$(ii-1)] = cc;
rHt[$(ii-1)] = hh; rFWHM[$(ii-1)] =fw;
//copy the longname of calculated data to summary sheet
rWaveLength[$(ii-1)] = %(rData[L]$);
}
9. Create a graph of the quantities as a function of excitation wavelength with the plotxy X-Function
10. //create a graph of the quantities as a function of excitation wavelength
plotxy iy:=(rWaveLength, rArea) plot:=202 ogl:=[<new>];
plotxy iy:=(rWaveLength, rHt) plot:=202 ogl:=[<new>];
plotxy iy:=(rWaveLength, rFWHM) plot:=202 ogl:=[<new>];
### Example 2: Running Batch Processing with pre-saved Theme
1. Drag and drop the prepared theme,0-PA-BatchPA.ois, under the <Origin EXE folder>\Samples\Batch Processing folder into Origin.
2. Open the project file with data for peak analysis
3. // Open an Origin Project file
string fname$= system.path.program$ + "Samples\Batch Processing\Batch Peak Analysis.opj";
doc -o %(fname\$); // Abbreviation of ''document -open''
4. Perform the peak analysis batch processing with the pre-saved theme with the pamultiY X-Function
5. paMultiY iy:=2:end theme:="BatchPA" append:=4 dataid:="My Parameter";
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Tutorial 5 Solutions
# Tutorial 5 Solutions - TUTORIAL 5 SOLUTIONS Accounting for...
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1 TUTORIAL 5 SOLUTIONS Accounting for Income Tax Picker Problem 8.5 – Oakley Ltd Change in tax rate Adjusting journal entry: Dr. Deferred Tax Liability 18 200 Cr. Deferred Tax Asset 7 400 Cr. Income Tax Expense 10 800 (being recognition of change of tax rate) Working DTA DTL Opening balance at previous tax rate \$29 600 \$72 800 Divide by previous tax rate ÷ 40% ÷ 40% Equals opening temporary differences balances \$74 000 \$182 000 Multiply by new tax rate x 30% x 30% Opening balance at new tax rate \$22 200 \$54 600 Adjustment for change in tax rate (7,400) (18 200) (Difference between opening balances at previous decrease decrease and new tax rates) Alternative Calculation Opening balance at previous tax rate \$29 600 \$72 800 Adjustment for change in tax rate: (40-30)/40 (7 400) (18 200) Adjusted opening balance \$22 200 \$54 600
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2 Current tax Oakley Ltd Current Tax Worksheet (for year ended 30 June 2010) Accounting profit \$920 000 Add: Impairment - goodwill \$20 000 Amortisation – development expenditure 64 000 Depreciation - buildings 29 000 Depreciation - plant 70 000 Warranty expense
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## Tutorial 5 Solutions - TUTORIAL 5 SOLUTIONS Accounting for...
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## Re: Cannot enter a parament for JMP stepwise regression
New Contributor
Joined:
Dec 5, 2017
I'm doing a stepwise regression to study the effect of some parameters on salaries. So:
Dependent variable is: salary
Independent variable are: gender, yrs_in_rank, URM_NonURM, Professor, AssociateProfessor, Assistant Professor, Instructor, and Lecturer (Professor, associate Professor, Assistant Professor, Instructor and Lecture are the five ranks of the study pool, they are recoded as dummy variable based one source column)
When I try to enter the above parameters, most parameters are entered except Lecturer. In fact, I can only enter four out of five ranks. I dont know why. I suspect it might be due to collinearity, but how could JMP know they are collinear variable? Is there a way I could enter all of five ranks into the analysis?
Thank you!
2 ACCEPTED SOLUTIONS
Accepted Solutions
Highlighted
Community Trekker
Joined:
Jan 29, 2015
Solution
This is a common issue that comes up. JMP always leaves out one category from the model - it has to, or the model will be over-identified (in lay terms, if we know an inidividual is not in the first 4 classes, they must be in the 5th, so that information would be superfluoous - in more mathematical terms, the matrix would be singular and not invertible). If you want to see the full set of categories, click on the red arrow at the regression output and ask for the Expanded Estimates. The missing category will have a coefficient that, when added to all the other category coefficients, sums to zero.
Community Trekker
Joined:
Jan 29, 2015
Solution
Try putting in one variable for rank instead of dummy variables for each rank. With one column for rank (with values Prof, Assoc, Asst, Inst, Lect) and expanded estimates you will get what you want. Putting in separate dummy variables will only permit you to put 4 of the 5 ranks and the effect of the Lecturer rank is buried within the intercept in the regression model (with each of the other rank coefficients interpreted as the effect of that rank either present or absent: Lecturer is the baseline that these are compared with the way you have it formulated.
5 REPLIES
Highlighted
Community Trekker
Joined:
Jan 29, 2015
Solution
This is a common issue that comes up. JMP always leaves out one category from the model - it has to, or the model will be over-identified (in lay terms, if we know an inidividual is not in the first 4 classes, they must be in the 5th, so that information would be superfluoous - in more mathematical terms, the matrix would be singular and not invertible). If you want to see the full set of categories, click on the red arrow at the regression output and ask for the Expanded Estimates. The missing category will have a coefficient that, when added to all the other category coefficients, sums to zero.
New Contributor
Joined:
Dec 5, 2017
Thank you dale_lehman for your very clear explanation. I understand and agree what you said about why the 5th variable cannot be entered. I understand statistically it doesn’t make sense to include the 5th variable if it can be linearly explained by the other 4. I'm still having trouble to explain the effect of 5th variable, in this case, it's the effect of lecturer rank. Can you give me some suggestions how to interpret this regression results, especially the effect of rank lecturer? Any advice/comment/suggestion would be much appreciated.
Also I tried to expand the estimates, it shows expanded results for selected parameter, and the 5th variables’ estimate still missing.
Community Trekker
Joined:
Jan 29, 2015
Solution
Try putting in one variable for rank instead of dummy variables for each rank. With one column for rank (with values Prof, Assoc, Asst, Inst, Lect) and expanded estimates you will get what you want. Putting in separate dummy variables will only permit you to put 4 of the 5 ranks and the effect of the Lecturer rank is buried within the intercept in the regression model (with each of the other rank coefficients interpreted as the effect of that rank either present or absent: Lecturer is the baseline that these are compared with the way you have it formulated.
New Contributor
Joined:
Dec 5, 2017
Thanks a lot dale_lehman. Now I understand and know how to do it.
Very appreciate it! Thanks again.
Community Member
Joined:
Dec 7, 2017
Thank you so much I fixed my problem.
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27/10/2022
Are electro magnets expensive?
Compared to permanent magnets, electromagnets can be, size for size, less expensive. The electromagnet is usually cheaper because fewer materials are used to make it. Depending on your needs, an Electromagnet’s magnetic strength can be altered for the same design size.
What is the strongest electro magnet?
Bitter electromagnets have been used to achieve the strongest continuous manmade magnetic fields on earth―up to 45 teslas, as of 2011.
Where can you find electromagnets?
Electromagnets are found in doorbells, hard drives, speakers, MagLev trains, anti-shoplifting systems, MRI machines, microphones, home security systems, VCRs, tape decks, motors, and many other everyday objects.
How much weight can an electromagnet lift?
The electromagnet can lift a mass of 3.1 kg.
Do electromagnets need a battery?
An electromagnet is a magnet that can be turned on and off. In this experiment, the battery is a source of electrons. When you connect the wire to the battery, the electrons flow through the wire. If there is not a complete circuit, the electrons will not flow.
How do you make a super strong electromagnet at home?
To make a super-strong electromagnet, you will need the following materials:
1. A static stack.
2. Thin-coated copper wire.
3. A large iron nail (approximately 3 inches in length)
4. Dry cell batteries.
5. Electrical tape (or insulating tape).
6. Iron filings, paper clips, or other small magnetic objects.
How many Teslas are in a fridge magnet?
According to the National High Magnetic Field Laboratory of Florida State University, a fridge magnet is about 0.001 tesla. To put that in perspective, the Earth’s magnetic field is about 0.00005 tesla and an average MRI magnet measures 1.5 tesla.
How do you make a super strong permanent magnet?
Take two magnets put one North pole and one South pole on the middle of the iron. Draw them towards its ends, repeating the process several times. Take a steel bar, hold it vertically, and strike the end several times with a hammer, and it will become a permanent magnet.
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area : PS Archive
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04 Jun 2009, 10:35
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In triangle ABC, D is the middle point of AC, E is the middle point of BC, and F is the middle point of CD (Not AB). What is the ratio of the area of ABC to that of FBC?
Senior Manager
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04 Jun 2009, 12:50
Thats a toug one!!
Could you please post source / OA / explanation?
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### Show Tags
10 Jun 2009, 11:11
I sketched it like an isosceles triangle.
Area of ABC:
Let base = AC = X
Area ABC = 1/2 (X) h
Area of FBC:
Let base = FC = [AC / 2] / 2 = [X/2]/2 = X/4
Area of FBC = 1/2 (X/4) h
Compare:
1/2 (X) h : 1/2 (X/4) h
X : X/4
4X : 1X
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### Show Tags
10 Jun 2009, 11:32
IMO Area od ABC : Area of FBC = 4:1
My approach was....
Let Area of Abc be X.
=> Area of BDC = x/2 (since BD divides ABC into 2 equal halves)
=> Area of BFC = x/4 (since BF divides BDC into 2 equal halves)
Therefore Area of ABC : Area of BFC = x:x/4 = 4:1
Wats the OA??
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### Show Tags
10 Jun 2009, 14:12
Pls. post the OA.This is a good and a tricky question.
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### Show Tags
11 Jun 2009, 01:50
The answer is 4:1 only if AB = BC. If you divide the triangle in half ABD and DBC share one side, DB, while one side of each triangle, AD & DC, has the same lenght. The ratio of the remaining side of each triangle, AB and BC, can have an infinite number of values. Since you can compute the area of a triangle using the lenght of all three sides the ratio of the area the two triangles must have an infinite number of values, depending of what kind of triangle that is originally used in the question (which is not mentioned here).
Please take note that English is not my native language
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### Show Tags
11 Jun 2009, 06:34
Hi,
I will try to explain my solution:
Since DE connects the centers of AC and BC then it is parralel to AB and is 1/2 AB. Then Triangle DEC has area 1/4 of the area of ABC.
Now Triangle DFC has the same area as triangle FCE since DF=FE and the height is same.
Triangle FCE has the same area as triangle FBE since BE=EC and height is the same.
So triangles DFC, FCE and FBE have same areas.
Then the area of FBC= area DCE and the ratio is 4/1
Regards
Re: area [#permalink] 11 Jun 2009, 06:34
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Search 75,602 tutors
1 0
## show how to solve absolute value equations
Show how to write the equation in Standard Form
|4x+5|-7-=0
you have an extra (-) there, so i assume you mean |4x+5| - 7 = 0
first step is to solve for the absolute expression...
|4x+5| = 7
now that we have this, imagine what this would mean: |y|=7
this means either y=7 or y=-7 (same as -y=7)
we use this same idea...
4x+5 = 7 or 4x+5 = -7
solving the first one, we get 4x=2 which leads to x = 1/2
and the second one gets us 4x=-12 which leads to x = -3
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A171064 G.f.: -x*(x-1)*(1+x)/(1-x-7*x^2-x^3+x^4). 1
0, 1, 1, 7, 15, 64, 175, 631, 1905, 6433, 20224, 66529, 212625, 692119, 2226799, 7217728, 23284815, 75343591, 243328225, 786800449, 2542156800, 8217744577, 26556314401, 85835882791, 277405671375, 896595420736, 2897714688751 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS The member k=7 of a family of sequences starting 0,1,1,k with recurrence a(n) = a(n-1)+k*a(n-2)+a(n-3)-a(n-4). LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Hugh Williams, R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277 Index entries for linear recurrences with constant coefficients, signature (1,7,1,-1). FORMULA a(n) = +a(n-1) +7*a(n-2) +a(n-3) -a(n-4). The roots (r1..r4) of the characteristic polynomials for this "family" of sequences have the following form (not simplified) for k= 1,2,3,4,5,6.... r1=(sqrt(4*k+10+2*sqrt(4*k+9))+sqrt(4*k-6+2*sqrt(4*k+9)))/4. r2=(sqrt(4*k+10+2*sqrt(4*k+9))-sqrt(4*k-6+2*sqrt(4*k+9)))/4. r3=(-sqrt(4*k+10-2*sqrt(4*k+9))-sqrt(4*k-6-2*sqrt(4*k+9)))/4. r4=(-sqrt(4*k+10-2*sqrt(4*k+9))+sqrt(4*k-6-2*sqrt(4*k+9)))/4. For k=1,2,3, r3 and r4 are complex . Closed-form (not simplified) is as follows for all k (note:for k1-k3 set r3 and r4 =0 and round a(n) to nearest integer): a(n)=sqrt(4*k+9)/(4*k+9)*(((r1)^n+(r2)^n)-((r3)^n+(r4)^n)). [Tim Monahan, Sep 17 2011] MATHEMATICA CoefficientList[Series[-x*(x - 1)*(1 + x)/(1 - x - 7*x^2 - x^3 + x^4), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 19 2012 *) PROG (MAGMA) I:=[0, 1, 1, 7]; [n le 4 select I[n] else Self(n-1) + 7*Self(n-2) + Self(n-3) - Self(n-4): n in [1..30]]; // Vincenzo Librandi, Dec 19 2012 CROSSREFS Cf. A116201 (k=1), A105309 (k=2), A152090 (k=3), A007598 (k=4), A005178 (k=5), A003757 (k=6). Sequence in context: A187986 A039789 A279882 * A042313 A058206 A219523 Adjacent sequences: A171061 A171062 A171063 * A171065 A171066 A171067 KEYWORD nonn,easy AUTHOR R. J. Mathar, at the request of R. K. Guy, Sep 03 2010 STATUS approved
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Last modified August 14 04:13 EDT 2020. Contains 336477 sequences. (Running on oeis4.)
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《Z的悲剧》
# A honoka和格点三角形
```#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
ll n,m;
int main()
{
cin>>n>>m;
ll a = (m-1)%mod*m%mod*(n-2)%mod*2%mod;
ll b = (m-2)%mod*m%mod*(n-1)%mod*2%mod;
ll c = (n-1)%mod*n%mod*(m-2)%mod*2%mod;
ll d = (n-2)%mod*n%mod*(m-1)%mod*2%mod;
//ll re1 = (m-2)%mod*(n-1)%mod*4%mod;
//ll re2 = (n-2)%mod*(m-1)%mod*4%mod;
//ll re = (re1+re2)%mod;
ll sum = (a+b+c+d)%mod;
//sum = (sum-re)%mod;
cout<<sum<<endl;
//cout<<(a%mod+b%mod+c%mod+d%mod-re1%mod-re2%mod)%mod<<endl;
system("pause");
return 0;
}```
# B kotori和bangdream
```#include<iostream>
#include<cstdlib>
using namespace std;
typedef long long ll;
int main()
{
//ios::sync_with_stdio(false);
//cin.tie(NULL);
//cout.tie(NULL);
double n,x,a,b;
cin>>n>>x>>a>>b;
double ans = a*x/100.0+b-b*x/100.0;
ans = ans * n;
printf("%.2f\n",ans);
system("pause");
return 0;
}```
# D hanayo和米饭
```#include<iostream>
#include<cstdlib>
#include<cstring>
using namespace std;
typedef long long ll;
ll n;
ll a[100001];
int main()
{
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin>>n;
//memset(a,0,sizeof(a));
for(ll i = 1;i<n;++i)
{
int x;
cin>>x;
a[x]++;
}
ll flag ;
for(ll i=1;i<=n;++i)
{
if(a[i]==0)
{
flag = i;
break;
}
}
cout<<flag<<endl;
//system("pause");
return 0;
}```
# E rin和快速迭代
```#include<iostream>
using namespace std;
typedef long long ll;
ll n;
ll count(ll n)
{
ll s = 1;
for(ll i =2;i*i <= n;++i)
{
if(n%i==0)
{
ll a = 0;
while(n%i==0)
{
n = n/i;
a++;
}
s = s*(a+1);
}
}
if(n>1) s = s * 2;
return s;
}
int main()
{
cin>>n;
ll num = 1;
while(count(n)!=2)
{
n = count(n);
num++;
}
cout<<num<<endl;
system("pause");
return 0;
}```
# G eli和字符串
```#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
ll n,k;
int main()
{
cin>>n>>k;
string s;
cin>>s;
int l,r,ans;
ans = INF;
int a[100]={0};
l = r = 0;
while(r<s.size())
{
a[s[r]-'a']++;
while(a[s[r]-'a']>=k)
{
ans = min(ans,r-l+1);
a[s[l]-'a']-=1;
l++;
}
r++;
}
if(ans==INF)
cout<<-1<<endl;
else
cout<<ans<<endl;
system("pause");
return 0;
}```
# H nozomi和字符串
```#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
int a[200009];
int n,k;
bool check(int x)
{
for(int i=x;i<=n;i++)
{
int ans = a[i]-a[i-x];//求零的数量也就是操作次数
if(ans<=k||x-ans<=k)//如果0的数量比K小，那就在区间的右侧；
return 1;
}
return 0;//否则在区间左侧
}
int main()
{
cin >> n >> k;
string s;
cin>>s;
a[1] = s[0]-'0';
/*前缀和*/
for(int i=1;i<n;++i)
{
a[i+1]=a[i]+s[i]-'0';
//cout<<a[i]<<endl;
}
/*二分*/
int l = 1,r = n,mid;
while(l<=r)
{
mid = (l+r)>>1;
if(check(mid))
{
if(!check(mid+1))
break;
l = mid+1;
}
else
r = mid-1;
}
cout << mid <<endl;
system("pause");
return 0;
}```
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• ### 51Nod 1004 n^n的末位数字(日常复习快速幂，莫名的有毒，卡mod值)
1004 n^n的末位数字 题目来源： Author Ignatius.L (Hdu 1061) 基准时间限制：1 秒 空间限制：131072 KB 分值: 5...
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# Thread: Normal distributions in Mathematica
1. Hello, I'm fairly new to programming and I have an upcoming due date for an assignment. I'm implementing a recursion based on normal (or Gaussian) variables. R is very useful here because it has a built-in function for generating normal distributions. The problem is, R seems to be way more difficult to work with than Mathematica (and there are fewer resources available), so I'd really rather do this in Mathematica. The trade-off is that in Mathematica I'd have to construct my own normal variables from scratch.
Is there any simple way for me to just generate a list of 500 normally distributed numbers and then just "copy and paste" this into Mathematica? Ideally I'd like to end up in Mathematica with 5 sets of 500 normally distributed numbers with each of the sets indexed by u_1, u_2, ..., u_500.
2.
3. Check the RandomReal function.
4. Originally Posted by oldfold
Hello, I'm fairly new to programming and I have an upcoming due date for an assignment. I'm implementing a recursion based on normal (or Gaussian) variables. R is very useful here because it has a built-in function for generating normal distributions. The problem is, R seems to be way more difficult to work with than Mathematica (and there are fewer resources available), so I'd really rather do this in Mathematica. The trade-off is that in Mathematica I'd have to construct my own normal variables from scratch.
Is there any simple way for me to just generate a list of 500 normally distributed numbers and then just "copy and paste" this into Mathematica? Ideally I'd like to end up in Mathematica with 5 sets of 500 normally distributed numbers with each of the sets indexed by u_1, u_2, ..., u_500.
Box
The above describes a method to generate pairs of independent normal random variables, given pairs of random numbers.
5. Originally Posted by mathman
Box
The above describes a method to generate pairs of independent normal random variables, given pairs of random numbers.
Okay that doesn't look too bad. Just two questions:
1. Can I just substitute every mention of U_1 and U_2 with "RandomReal[]"?
2.Do I use Z_0 or Z_1 for my recursion based on normal variables? Some combination of both? Should I just make sure that each of Z_0 and Z_1 is used an equal number of times as the input for my recursion?
6. Mathematica also has a NormalDistribution... distribution, so there's no need to use a transformation, and actually using the Box-Muller method would be a pain since it works on pairs at a time.
7. Originally Posted by MagiMaster
Mathematica also has a NormalDistribution... distribution, so there's no need to use a transformation, and actually using the Box-Muller method would be a pain since it works on pairs at a time.
Wow, thank you so much. I thought I tried to look this up earlier but I think my using "Gaussian Distribution" as my key words rather than "Normal Distribution" had me doomed from the start of my search.
8. I'm surprised nothing came up as the two are synonymous.
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# Many lawmakers are discouraged from seeking offices because
Author Message
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Many lawmakers are discouraged from seeking offices because [#permalink]
### Show Tags
14 Jul 2004, 12:03
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Many lawmakers are discouraged from seeking offices because "past indiscretations, if they were exposed to the public, would prove to be embarrassing"
A. s/a
B. their past indiscretations would prove embarrassing if exposed to the public
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### Show Tags
14 Jul 2004, 12:37
boksana wrote:
Many lawmakers are discouraged from seeking offices because past indiscretations, if they were exposed to the public, would prove to be embarrassing
B. their past indiscretations would prove embarrassing if exposed to the public
A is like: ...lawmakers...because past indiscretions would prove to be embrassing. "THEY" doesn't have clear referent.
B is straight
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### Show Tags
14 Jul 2004, 12:42
B it is.
it is "indiscretion", not "indiscretation"
In A, "they" is not required whereas in B, "their" makes it clear that it the "indiscretions of those lawmakers"
_________________
Best Regards,
Paul
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### Show Tags
14 Jul 2004, 16:03
Thanks, i was not sure for my choice
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### Show Tags
15 Jul 2004, 00:54
yeah... I would opt for B too because of clearer pronouns.
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15 Jul 2004, 00:54
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# Tispe
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Offline Last Active Yesterday, 04:40 AM
### Rectilinear Texture Warping - Building Warp Maps
13 December 2013 - 06:12 AM
Hi
I have read Paul Rosens white paper on RTW shadow mapping many times but I have a hard time understanding how exactly the warp maps are converted from 1-D blurred importance maps to the 1-D warp maps. The source code is confusing to me aswell.
Key terms: Shadow map, warp maps, super cells, importance map, output frame.
http://www.cspaul.com/wiki/doku.php?id=publications:Rosen.2012.I3D
The paper does not suggest the importance map resolution the demo uses, only that "resolution only needs to be large enough to detect the existence of the smallest features of interest" to shadow properly. It also states that "The warping maps are composed of a set of super-cells at equal or lower resolution than the base texture."
Questions:
Is the importance map in the demo the same or lower resolution than the shadow map?
When collapsing the importance maps to 1-D, does it keep the same width (i.e 512x512 becomes 512)?
What resolution does the warp map have (looks like 28 super cells), does this mean that it is a 32 pixel long 1D texture?
How do you then collapse a 512px importance map down to a 32px warp map?
The warp map is colored red, blue and black. Is the Red channel displacing in the positive direction and blue displacing in the negative direction?
The "GetWarp(k)" formula to produce the 32 super cell values is kinda tricky to get the picture of. It sums upto the k'th super cell and divides by the total importance. Do we do this sum over the 512px importance map? Do we sum every 16 pixels in the 1-D importance map to create a super cell and divide by the total importance? Kinda hard to follow where and what we sum up.
### Multiple shaders in the same HLSL file (no effect framework)
06 November 2013 - 03:54 PM
Hi again
Currently I have one .hlsl file for the vertex shader and one .hlsl file for my pixel shader as resources. I do not use the effect framework. I am going to add more shaders to my game and instead of creating more files I thought I could just place new shaders into the same files I already use. But then if that is possible, the globals/constants would then be the same for both shaders. And if then, do I use more then one constant table, or just one per .hlsl file?
```D3DXCompileShaderFromResource(
NULL,
NULL, //macro's
NULL, //includes
"vs_main", //main function
&code, //compiled operations
&debugcode, //errors
&VSConstantTable);
```
Is this possible, and how many Constant Tables should I need?
Later on when I am happy with the shaders, I will compile them offline and pack the shader binary as a resource. Any things I need to consider at this stage aswell?
### Render-To-Texture and Z-buffer issues
05 November 2013 - 01:43 PM
Hello
I have followed the this guide http://www.two-kings.de/tutorials/dxgraphics/dxgraphics16.html to render to a texture.
I have no trouble rendering to the backbuffer. But when I swap out the backbuffer with a surface from a texture the z buffer does not work. I keep everything else the same except the render target. I can draw the colors, view/perspective just fine, but there are no depth test going on even though all states are the same.
The clear function returns with no errors and every draw is within a begin() and end(), just like with the backbuffer as target.
```device->Clear(0, NULL, D3DCLEAR_TARGET | D3DCLEAR_ZBUFFER, D3DCOLOR_XRGB(255, 255, 255), 1.0f, 0);
```
```device->CreateTexture(512, 512,
1, D3DUSAGE_RENDERTARGET, D3DFMT_X8R8G8B8, //D3DFMT_R32F
```
What could cause this?
### Shadow mapping: D3DUSAGE_DEPTHSTENCIL or D3DUSAGE_RENDERTARGET?
01 November 2013 - 07:03 AM
Hi
```pD3DDev->CreateTexture(texWidth, texHeight, 1,
D3DUSAGE_DEPTHSTENCIL, D3DFMT_D24S8, D3DPOOL_DEFAULT,
&pTex);```
Note that you must create a corresponding color surface to go along with your
depth surface since Direct3D requires you to set a color surface / z surface pair
when doing a SetRenderTarget(). If you’re not using the color buffer for
anything, it’s best to turn off color writes when rendering to it using the
D3DRS_COLORWRITEENABLE renderstate to save bandwidth.
```g_pd3dDevice->CreateTexture( SHADOW_MAP_SIZE, SHADOW_MAP_SIZE, 1,
D3DUSAGE_RENDERTARGET, D3DFMT_R32F,
NULL )```
What is appropiate to use?
Seems like the first alternative you need a color surface but you turn off color renderstate to save bandwidth.
But the second alternative you don't???
### Windowed mode beyond desktop resolution issue
29 October 2013 - 05:05 PM
Hello
If I set my game to a windowed resolution (1920x1080) which makes the window(client area + border area) larger than the monitor resolution(1920x1080) then Windows will automatically "crop" down the window size to fit inside the desktop. This causes the client area to be compressed causing image distortions and positioning errors.
Is there a way to allow "larger then desktop" resolution windows without this client area compression?
PARTNERS
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# Calculating the day of the week for 2011
4 posts / 0 new
#1 15 May, 2011 - 14:04
Offline
Joined: 5 years 10 months ago
#### Calculating the day of the week for 2011
One of the memory skills I use is Dominic O’Brien’s method for calculating the day of the week for a given date. It’s part mnemonics, combined with basic addition and subtraction. I’m not going to present the full method here, but I will present a partial method you can use for calculating the day of the week for 2011. It’s really easy.
To simplify things, though, the first concept you need to be aware of is called Casting Out 7s. Our calculation is based on the seven day week, so 1 though 7 are the only numbers we need. If you can count by 7s, 0, 7, 14, 21, 28, 35, then you’re halfway there. Casting out 7s means that if you have a number like 23, you cast out 7, 14, 21, leaving you with a two remainder. I like to think of it as reducing…23 is reduced to 2. 19 is reduced to 5. 33 is also reduced to 5. In fact, you can easily see this on a calendar
Second, our reduction answers, 1 though 7, represent the days of the week. 1 is Sunday, 4 Wednesday, 7 is Saturday. 13 is reduced to 6 which is Friday. 24 is reduced to 3 which is Tuesday. If any of that doesn’t make sense, then give the above calendar a good look-see. All the information you need is there.
Now, here’s the memory part. You have to remember a number for each month.
January 6
February 2
March 2
April 5
May 0
June 3
July 5
August 1
September 4
October 6
November 2
December 4
The math is really simple. You add the month number (above) to the day of the month, cast out 7s, and your answer is the day of the week for this year.
January 1st. 6+1=7 That was a Saturday.
October 25th. 6+25=31. Reduce that to 3. That will be a Tuesday.
August 22nd. 1+22=23 Reduce that to 2. That will be a Monday.
December 25th. 4+25=29. Reduce that to 1. Christmas will be on a Sunday.
Tips: 7 also equals 0. If the day of the month is 7, 14, 21, or 28, go head and call it 0. When you cast 7s out of 21, you reduce all the way to 0.
Cast out numbers at any time. In the above example August 22nd, I’d reduce the 22 to 1 just out of habit, so the math becomes 1+1=2=Monday.
The full system requires you to remember a number, 0-6, for all years from 00-99, and a number for the centuries. By a happy coincidence, the numbers for 2011 are both zero, which is why for this year we only need to add the month number and the day of the month. (I’m going to post a followup about calculating for 2012, which is a slightly bit trickier since it’s a leap year.)
If you have any questions, let me know.
--Larry
15 May, 2011 - 15:38
Offline
Joined: 5 years 11 months ago
I've always used the Doomsday Algorithm.
17 May, 2011 - 01:50
Offline
Joined: 1 year 11 months ago
I haven't tried Dominic O'Brien's method yet. I learned a method from the book Mind Performance Hacks, though haven't practiced it much. It is described here:
http://www.ludism.org/mentat/CalendarFeat
The Doomsday Algorithm looks interesting:
http://en.wikipedia.org/wiki/Doomsday_rule
17 May, 2011 - 10:14
Offline
Joined: 5 years 10 months ago
I can recommend Arthur Benjamin's "Secrets of Mental Math." Lots of good tricks and techniques, including calculating the day of the week. Here's a link to his TED Talk: http://www.ted.com/talks/arthur_benjamin_does_mathemagic.html.
-cvstuart
Learn memory techniques for free! Just click the "Sign up" button below to create an account and we'll send you an email with some tips on how to get started.
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# Cave 75 Sketch and the the 3 1 intervals maximum the wherc points . the function pue hoolima local incninsing
###### Question:
cave 75 Sketch and the the 3 1 intervals maximum the wherc points . the function pue hoolima local incninsing , points_ decreasing; Hint: COncestigape du 1 the sign col-
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##### Ioeetl InrendioCAarettt MabulitdcanuiiAnal <Fa Daal ]Lont41?UneReFaltJTlekino ol Lura_tn1ulieek (to ) drutiEdelWcietn MeleOf ln Ia9l Unaenaa ol /1 Coudo LneWmllte Mapa0+7j]mln &rui LEATlnnaMiomeemet|Nontlu [alttdn Jdcule{Dnatio-RCeeIad
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##### PHYS 1402 (SUMMER 2)2020)(5003-5004) Course Home <CH 20 HW Problem 20.67 A50 mg bad with a...
PHYS 1402 (SUMMER 2)2020)(5003-5004) Course Home <CH 20 HW Problem 20.67 A50 mg bad with a charge of 2.1 Crests on a table. A second bead with a charge of 53 nC is directly above the first bead and is slowly lowered toward it Part A What is the closes the centers of the two beads can be brought t...
##### 15 A certain bacteria doubles in size every 15 minutes. If we started with an initial population of 100 cells, afier how long will the population reach 5000 cells?
15 A certain bacteria doubles in size every 15 minutes. If we started with an initial population of 100 cells, afier how long will the population reach 5000 cells?...
##### Unit vectors Define the points $P(-4,1), Q(3,-4),$ and $R(2,6) .$ Carry out the following calculations. Find two vectors parallel to $\overrightarrow{Q P}$ with length 4.
Unit vectors Define the points $P(-4,1), Q(3,-4),$ and $R(2,6) .$ Carry out the following calculations. Find two vectors parallel to $\overrightarrow{Q P}$ with length 4....
##### Determine whether or not the statement (3r € 2)(z? +I = 6) is true: Write the pegation_ of this statement using quantifiers
Determine whether or not the statement (3r € 2)(z? +I = 6) is true: Write the pegation_ of this statement using quantifiers...
##### A triangle has two corners of angles (pi )/4 and (7 pi)/12 . What are the complement and supplement of the third corner?
A triangle has two corners of angles (pi )/4 and (7 pi)/12 . What are the complement and supplement of the third corner?...
##### Example 10.2- The lifetime (in Years) of a product is Y 7xo3 where X Exp(4)_ Find the pdlf of Example 10.3- The pelf of X is fx(r) = 01" where 0 < I < [ and 0 > 0. Find the pclf 20 In( 1)
Example 10.2- The lifetime (in Years) of a product is Y 7xo3 where X Exp(4)_ Find the pdlf of Example 10.3- The pelf of X is fx(r) = 01" where 0 < I < [ and 0 > 0. Find the pclf 20 In( 1)...
##### How do you find (f^-1)'(a) if f(x)= x/sqrt{x-3} and a = 4?
How do you find (f^-1)'(a) if f(x)= x/sqrt{x-3} and a = 4?...
##### IMI 1 93 9 1 1 7 2 0 8 1 1 HW 1 L 1 1 I H I W L 1 1 3 L 3 1 0 1
IMI 1 93 9 1 1 7 2 0 8 1 1 HW 1 L 1 1 I H I W L 1 1 3 L 3 1 0 1...
##### An 8.0-kg oak block, X, on a horizontal, rough oak surface isattached by a light string that passes over a light, frictionlesspulley to a hanging 4.0-kg mass, Y. The magnitude of the force offriction on block X is 24 N. What is the magnitude of theacceleration of the system? ( mc010-1.jpg = 9.81 m/s2[down])
An 8.0-kg oak block, X, on a horizontal, rough oak surface is attached by a light string that passes over a light, frictionless pulley to a hanging 4.0-kg mass, Y. The magnitude of the force of friction on block X is 24 N. What is the magnitude of the acceleration of the system? ( mc010-1.jpg = 9.81...
##### Computer Architecture (Urgent Reply Plz)
Q1.Write the cost, speed,capacity and policy features or comparisons of all types ofRAM.Q2."Cache Memory and RAID" bestresearch papers (10) links and also presentations in the form ofslides ( in MS Power Point.)Q#3.In ALU,(Multiply, Add,Subtrat) is necessary just like Adder?...
##### Table B.11 presents data on the quality of Pinot Noir wine.a. Build a regression model relating quality $y$ to flavor $x_{4}$ that incorporates the region information given in the last column. Does the region have.an impact on wine quality?b. Perform a residual analysis for this model and comment on model adequacy.c. Are there any outliers or influential observations in this data set?d. Modify the model in part a to include interaction terms between flavor and the region variables. Is this model
Table B.11 presents data on the quality of Pinot Noir wine. a. Build a regression model relating quality $y$ to flavor $x_{4}$ that incorporates the region information given in the last column. Does the region have.an impact on wine quality? b. Perform a residual analysis for this model and comment ...
##### Some chemical compounds are listed in the first column of the table below. Each compound is...
Some chemical compounds are listed in the first column of the table below. Each compound is soluble in water. Imagine that a few tenths of a mole of each compound is dissolved in a liter of water. The important chemical species that would be present in this solution are written in the second column ...
##### Gypsum on heating gives:(a) $mathrm{CaS}+mathrm{O}_{2}$(b) $mathrm{CaO}+mathrm{SO}_{3}$(c) $mathrm{CaSO}_{4}-1 / 2 mathrm{H}_{2} mathrm{O}$(d) $mathrm{CaSO}_{4}$
Gypsum on heating gives: (a) $mathrm{CaS}+mathrm{O}_{2}$ (b) $mathrm{CaO}+mathrm{SO}_{3}$ (c) $mathrm{CaSO}_{4}-1 / 2 mathrm{H}_{2} mathrm{O}$ (d) $mathrm{CaSO}_{4}$...
##### N.A. is brought to the ED for management of intentional drug overdose. He ingested 56 tables...
N.A. is brought to the ED for management of intentional drug overdose. He ingested 56 tables of Amoxicillin (prescribed 875 mg po BID for 28 days for a sinus infection) and 30 tablets of Seconal (prescribed 100 mg capsules po QHS prn for sleep). His respirations are slow and shallow, and he is non-r...
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Problem Set
# Practice Problems
Lesson 6. Unit 5. Grade 6 Open Up Resources
Open Up ResourcesVaries
This Problem Set is a part of the Lesson 6, Unit 5, Grade 6. Practice Problems 1. One gallon of gasoline in Buffalo, New York costs \$2.29. In Toronto, Canada, one liter of gasoline costs \$0.91. There are 3.8 liters in one gallon. How much does one gallon of gas cost in Toronto? Round your answer to the nearest cent. Is the cost of gas greater in Buffaalo or in Toronto? How much greater?
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Make up to \$500 this semester by taking notes for StudySoup as an Elite Notetaker
# Prove the second absorption law from Table 1 by showing
## Problem 13E Chapter 2.2
Discrete Mathematics and Its Applications | 7th Edition
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Discrete Mathematics and Its Applications | 7th Edition
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19
5
Problem 13E
Prove the second absorption law from Table 1 by showing that if A and B are sets, then, A ? (A U B) =, A.
Step-by-Step Solution:
SOLUTION Step 1A and B are two sets Step 2 We have to prove
Step 3 of 4
Step 4 of 4
##### ISBN: 9780073383095
Since the solution to 13E from 2.2 chapter was answered, more than 228 students have viewed the full step-by-step answer. The answer to “Prove the second absorption law from Table 1 by showing that if A and B are sets, then, A ? (A U B) =, A.” is broken down into a number of easy to follow steps, and 25 words. Discrete Mathematics and Its Applications was written by and is associated to the ISBN: 9780073383095. The full step-by-step solution to problem: 13E from chapter: 2.2 was answered by , our top Math solution expert on 06/21/17, 07:45AM. This textbook survival guide was created for the textbook: Discrete Mathematics and Its Applications, edition: 7th. This full solution covers the following key subjects: absorption, Law, prove, Sets, showing. This expansive textbook survival guide covers 101 chapters, and 4221 solutions.
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# Determine if a line intersects a plane where 2 points for line, 3 points for plane
• 03-25-2012
Aaron Xia
Determine if a line intersects a plane where 2 points for line, 3 points for plane
Hi, how can I determine if a line(given (x1,y1,z1),(x2,y2,z2) on the line) intersects a plane(given (x3,y3,z3),(x4,y4,z4), (x5,y5,z5) on the plane):confused:
Thanks helping.
• 03-25-2012
oogabooga
line - Google Search intersect plane
• 03-25-2012
Cynic
While it's not hard, if you don't know cross products, dot products and parametric equations of lines, it's a long lecture.
http://softsurfer.com/Archive/algori...ithm_0104B.htm
I use this site along with the book "3D Math Primer for Graphics and Game Development" as my primary sources.
• 03-26-2012
Aaron Xia
Quote:
Originally Posted by Cynic
While it's not hard, if you don't know cross products, dot products and parametric equations of lines, it's a long lecture.
Intersection of Lines and Planes
I use this site along with the book "3D Math Primer for Graphics and Game Development" as my primary sources.
It helps a lot! Thanks for sharing!
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# In the given figure, ABCD is a rectangle in which diag.
Question:
In the given figure, $A B C D$ is a rectangle in which diag. $A C=17 \mathrm{~cm}, \angle B C A=\theta$ and $\sin \theta=\frac{8}{17}$.
Find
(i) the area of rect. ABCD,
(ii) the perimeter of rect. ABCD.
Solution:
Given: In $\Delta A B C$,
$A C=17 \mathrm{~cm}$
$\sin \theta=\frac{8}{17}$
Since, $\sin \theta=\frac{P}{H}$
$\Rightarrow P=8$ and $H=17$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow 8^{2}+B^{2}=17^{2}$
$\Rightarrow B^{2}=289-64$
$\Rightarrow B^{2}=225$
$\Rightarrow B=15$
Therefore,
$A B=8 \mathrm{~cm}$ and $B C=15 \mathrm{~cm}$
Therefore,
(i) Area of rectangle $A B C D=A B \times B C$
$=8 \times 15$
$=120 \mathrm{~cm}^{2}$
(ii) Perimeter of rectangle $A B C D=2(A B+B C)$
$=2(8+15)$
$=2(23)$
$=46 \mathrm{~cm}$
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How to splitAt a scala list using shapeless
Facebook and Stack Exchange are now working together to support the Facebook developer community. Facebook engineers participate here along with the best Facebook developers in the world. If you have a technical question about Facebook, this is the best place to ask.
I'm trying to split a list of size S at N, where it is known that N, M sum up to S. This does not compile:
``````def splitIt[N <: Nat,
M <: Nat,
S <: Nat](u: Sized[List[Int], N] {type A = N},
v: Sized[List[Int], M] {type A = M},
t: Sized[List[Int], S] {type A = S})(implicit sum: SumAux[N, M, S]): Unit = {
val z = t.splitAt[N]
}
``````
Errors
``````No implicit view available from List[Int] => scala.collection.GenTraversableLike[S,List[Int]].
not enough arguments for method sizedOps: (implicit evidence\$2: List[Int] => scala.collection.GenTraversableLike[S,List[Int]])shapeless.SizedOps[S,List[Int],S]. Unspecified value parameter evidence\$2.
``````
Final correct version
``````def splitIt[N <: Nat,
M <: Nat, S <: Nat](u: Sized[List[Int], N] {type A = Int},
v: Sized[List[Int], M] {type A = Int},
t: Sized[List[Int], S] {type A = Int})(implicit sum: DiffAux[S, N, M], toInt: ToInt[N]): Unit = {
val z = t.splitAt[N]
}
``````
-
1 Answer
The `type` `A` needs to be the type of the elements of the list, not the size argument again. That's why it's trying to convert to a `GenTraversableLike[A, List[Int]]`. You need to set `A` to `Int` in each case.
-
+1, but this is only the first step—knowing `SumAux[N, M, S]` doesn't get you the `Diff[S, N]` you need to split the list (unfortunately), and you'll also have to add a `ToInt` for `N`. – Travis Brown Dec 24 '12 at 14:25
Changed the `A`s and added a `toInt` implicit. How to add the `Diff`? – Peteris Dec 24 '12 at 14:28
@Peteris: You can remove the `SumAux` (unless you need it for some other reason in your real code) and add either `Diff[S, N]` or `DiffAux[S, N, M]`. – Travis Brown Dec 24 '12 at 14:33
All good now, thanks. – Peteris Dec 24 '12 at 14:38
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# description of waves
0 pts ended
You see lightning strike a mountaintop, and 2.80 s later you hear the accompanying thunder. How far away is the mountaintop? The speed of sound is 340 m/s, and the speed of light is 3.00*10^8 m/s.
THE ANSWER TO THIS SAMPLE PROBLEM IS 0.952 km... How do i get this answer... YES THIS IS THE ANSWER... IT WAS GIVEN IN THE BOOK...
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rectification notice of excessive oxygen content of coal-fired boilers
• Home|
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• rectification notice of excessive oxygen content of coal-fired boilers
## rectification notice of excessive oxygen content of coal-fired boilers
### Why Oxygen Levels Matter in High Efficiency Boilers
Jun 15, 1994 · Unit One is a pulverized coal-fired, dry bottom boiler with an opposed firing configuration. Low NOx burners are used to control NOx emissions. Normal operating temperature in the combustion zone is approximately 2,000 - 3,000 degF. The boiler produces 4,500,000 lb/hr of steam for a General Electric steam turbine generator rated at 688 MWe.
### Combustion Efficiency and Excess Air
Nitrogen content = 1.40 Oxygen content = 6.05 GCV of Coal = 3401 kCal/kg Step 1. Find theoretical air requirement Theoretical air required for complete combustion = [(11.6 X C) + {34.8 X ( X /8)} + (4.35 X S)] /100 kg/kg of coal = 4.74 kg/kg of coal Step 2. Find theoretical % at theoretical conditions = moles= +
### Boiler-Tuning Basics, Part I - POWER Magazine
Oct 10, 2017 · There is a diminishing returns effect in oxygen content and performance for the unit. Too much available oxygen will lead to inefficient combustion. Over time, the unit will require unscheduled maintenance. Too little available oxygen causes unstable combustion and increased emissions. This will lead to flame loss faults and, once again, unscheduled maintenance.
### Guidelines for NOx Control by Combustion Modification for
Boiler-Tuning Basics, Part I - POWER Magazine
### Oxygen Control | CleanBoiler.org
Results of Boiler Tuning and Optimization • Improved excess oxygen, CO, NO X and FEGT balance 5 10 15 20 25 30 35 40 45 50 55 5 10 15 20 25 30 35 0 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 5 10 15 20 25 30 35 40 45 50 55 5 10 15 20 25 30 35 1500 17505 10 200015 202250 25 250030 35 40 45
### Energy Audit of Coal Fired Thermal Power Plant
Typically, excess air levels on coal-fired utility boilers vary from 18 to 25 percent (3.5 to 5% Q£ in the flue gas). Only recently has it been found possible to reduce oxygen levels on pulverized coal-fired boilers down to 1 to 2 percent to minimize NOX emissions (and possibly increase efficiency) in short term tests.
### Guidelines for Industrial Boiler Performance Improvement
Oxygen Control | CleanBoiler.org
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In Super Auto Pets, players build teams of cute emoji animals then send them into battle (think Pokémon, not dog-fighting).
Gameplay starts in a shop, where you're given a certain amount of gold and can buy either friends (animals) or food (bonuses).
Screenshots courtesy of Team Wood Games
After building a team, you send it into battle, which proceeds automatically:
The emoji-laden art style hides surprising strategic depth and a constantly-evolving metagame.
If you've never played it, I encourage you to try a round or two: you can play for free on Itch, without even creating an account.
(this writeup will make more sense if you're familiar with the basic mechanics)
After playing many rounds over the course of a few months, I decided to build a game simulator and do some science. The state space explodes quickly, so I limited myself to asking questions about the very first turn.
This writeup explains how to do deterministic exploration of the state space, then presents my results (including the best possible team for the first turn).
Generating every possible team
Available actions in the shop
The first turn (like every turn) starts in the shop, where you're given 10 gold and a set of things you could buy. Here's what that looks like:
You have a few choices:
• Buy an animal for 3 gold and place it on an empty space
• Combine two animals on your team of the same species
• Buy a food for 3 gold and give it to an animal on your team
• Reroll the shop, generating a new set of animals and foods
• Change the order of your team
• Sell an animal on your team, receiving 1 gold
You can take actions until you run out of gold, or choose to go battle at any point (even with an empty team, which is not recommended!).
Every animal has two statistics: health (❤️) and attack (⚔️)., which determine how it performs in battle. You can boost these statistics by buying food: for example, an apple 🍎 gives +1 to each stat.
In addition, many animals have actions that occur when they are bought or sold.
As one example, an otter will give a random friend on your team [+1, +1] when it is bought. This is where things get tricky, because it's non-deterministic: you can end up with a different team depending on which friend gets the bonus!
Exploring a non-deterministic space
If you want to fully explore the state space, you need to either
• Run many random trials, or
• Search the space deterministically
The former is much easier to implement. You can imagine a trait that lets you roll a random dice in a particular range:
``````pub trait Dice {
fn roll(&mut self, range: std::ops::Range<usize>) -> usize;
}
``````
Then, you can implement this trait for anything that implements `rand::Rng`:
``````impl<R: rand::Rng> Dice for R {
fn roll(&mut self, range: std::ops::Range<usize>) -> usize {
rand::Rng::gen_range(self, range)
}
}
``````
Here's a very simple game which uses this `Dice` trait:
``````fn play<R: Dice>(r: &mut R) {
let a = r.roll(0..6);
println!("Roll: {}", a);
if a == 5 {
let b = r.roll(0..6);
println!(" Bonus roll: {}", b);
}
}
``````
Finally, we can play this game a few times and see what we get:
``````fn main() {
for _ in 0..10 {
play(&mut rng)
}
}
``````
``````Roll: 3
Roll: 4
Roll: 1
Roll: 2
Roll: 0
Roll: 0
Roll: 0
Roll: 4
Roll: 5
Bonus roll: 5
Roll: 2
``````
This `play` function is very easy to write, but may need many iterations to fully explore the state space. Here's a modification that records how many tries it takes; I see something like 73 iterations to find all 11 possible combinations of dice rolls.
Searching the state space deterministically is much harder to write than our straightforward `play` function: you need to track your current state, generate all possible new states, explore them one by one...
(I'd encourage you to try this and see how it looks for this simple function)
To deterministically explore the state space while preserving easy-to-read code, I came up with an interesting idea: what if we store the state of the "random number generator", then proceed deterministically through the possible random numbers?
For example, if the game asks us for a random number in the range `0..6`, we'd generate `0` on the first run, `1` on the second, and so on.
Here's what that looks like:
``````#[derive(Debug)]
pub struct DeterministicDice {
initialized: bool,
index: usize,
data: Vec<(usize, std::ops::Range<usize>)>,
}
impl DeterministicDice {
pub fn new() -> Self {
Self {
initialized: false,
index: 0,
data: vec![],
}
}
pub fn next(&mut self) -> bool {
if !self.initialized {
self.initialized = true;
true
} else {
while let Some((mut v, r)) = self.data.pop() {
v += 1;
if v >= r.end {
continue;
} else {
self.data.push((v, r));
break;
}
}
self.index = 0;
!self.data.is_empty()
}
}
}
impl Dice for DeterministicDice {
fn roll(&mut self, range: std::ops::Range<usize>) -> usize {
let out = if let Some((v, r)) = self.data.get_mut(self.index) {
assert!(*r == range);
assert!(range.contains(v));
*v
} else {
self.data.push((range.start, range.clone()));
range.start
};
self.index += 1;
out
}
}
``````
The `DeterministicDice::next()` function advances to the next state, or returns `false` if we've fully explored the space.
Using the `DeterministicDice` instead of `thread_rng()`, here's our new `main`:
``````fn main() {
let mut rng = DeterministicDice::new();
while rng.next() {
play(&mut rng);
}
}
``````
Now, the output generates all 11 possibilities in exactly 11 iterations:
``````Roll: 0
Roll: 1
Roll: 2
Roll: 3
Roll: 4
Roll: 5
Bonus roll: 0
Roll: 5
Bonus roll: 1
Roll: 5
Bonus roll: 2
Roll: 5
Bonus roll: 3
Roll: 5
Bonus roll: 4
Roll: 5
Bonus roll: 5
``````
This is neat, because it means you can write simple code that makes "random" selections, but still explore the entire state space deterministically!
(I'm sure this idea exists elsewhere, but I haven't seen it before; send me prior art and I'll add it to this post!)
Exploring Shop Space
With this tool, we can explore the entirety of shop space and build every possible team. Like another recent writeup, this becomes an exercise in pruning and limiting possibilities.
For example, shops are always kept sorted. This has no impact on gameplay, but reduces the starting shop space from 1458 to a mere 330.
Our main loop randomly selects and applies one of the shop actions, or halts if the selected action is invalid. We wrap this in the `DeterministicDice` RNG to fully explore the available actions.
In between actions, the set of active shop states (i.e. the combination of shop animals, shop food, gold, and current team) are deduplicated.
Here's the number of active shops over time, along with the number of unique teams that we've created:
As you can see, we start with 330 shop states, i.e. every possible result of a sorted reroll. The number of active states balloons to 429K by action 4, then shrinks back down; by action 10, we have exhausted all possibilities.
This is a lot of state, but it's not ridiculous. However, this is far from naive: there are a bunch of tricks to keep it low (consider that simply rerolling for 1 gold, 10 times, would explode the space to 33010, or 15315789852644490000000000).
First, deduplication can be pretty aggressive: for example, if we've already seen a particular shop state and had more gold when we last saw it, then we can discard the current branch, since the previous exploration has strictly more possibilities.
The second big optimization is to strictly limit shop rerolling, which always creates 330 new branches. Since we're exploring every possible branch, there's no use in rerolling twice in a row: even if this particular branch doesn't contain an particular species in the shop, a different branch will have it.
Therefore, we should only reroll if the shop has empty slots.
(I also allow for rerolling if the shop has animals with non-standard stats, which can happen: sellng a duck adds +1 health to every shop animal. This makes a tiny different to the number of active shops, but still generates the same total number of teams, so I'm not sure if there's a situation where it matters.)
The last optimization is to also sort the team when deduplicating shop states. This is less obvious, because team order matters in battle. The trick is to store two separate deduplicated sets:
• Shop states, which have sorted teams to reduce the state space
• Teams, which include every possible order (but don't include shop animals / food / gold)
After each shop action, we insert every possible team permutation into the set of teams, but only store the sorted shop state in the set of next shops.
The result is 3514 teams. We can discard 90 of them as obviously bad, because they contain fewer than 3 animals with no compensating bonus stats or equipment; this includes the empty team, which wins no battles (although it can tie if facing another empty team!).
This leaves us with 3424 teams to evaluate.
Battle
Since we have 3424 teams, doing a battle per pair requires 11723776 battles.
However, there are also non-deterministic situations that arise in battle: for example, a mosquito randomly chooses an enemy to attack at the beginning of the round.
We can use the same `DeterministicDice` to fully evaluate every battle outcome between a pair of teams. To explore every battle outcome, we end up simulating 24946805 battles, recording the win/lose/draw record for every pair of teams.
Summing the scores gives a total win rate across the entire tournament:
The mean win rate is 35.7%, and the median is 33%.
Results
Our best team, with a 94.6% win rate:
🦗 🦟 🦟 ❤️ 1 ❤️ 2 ❤️ 2 ⚔️ 2 ⚔️ 2 ⚔️ 2
(shown in the same orientation as the game, i.e. the cricket will fight last)
The second-best team, with a 93.4% win rate, is
🐴 🦗 🦗 ❤️ 2 ❤️ 1 ❤️ 1 ⚔️ 1 ⚔️ 3 ⚔️ 3
There's a ton of random facts that I can pull from this data. I've presented a few below, but feel free to run the code and make new discoveries for yourself!
The best team with two friends
🦗 🦦 ❤️ 2 ❤️ 3 ⚔️ 3 ⚔️ 4
(71% win rate)
The best team with one friend
🐷 🍯 ❤️ 5 ⚔️ 3
(50% win rate; all hail the Big Pig)
The worst team with three friends
🦆 🦦 🐴 ❤️ 1 ❤️ 1 ❤️ 2 ⚔️ 4 ⚔️ 2 ⚔️ 1
(3.9% win rate)
Notes on implementation and future work
I've published the code on Github, and encourage folks to mess with it and send me any interesting discoveries.
The code is about 1.2KLOC of Rust, so it's fairly approachable.
It takes about 37 seconds to run, broken down as follows:
• Generating all teams: 31.5 seconds
• Simulating all battles: 5.25 seconds
• Aggregating stats: 0.25 seconds
Intermediate results are cached in local files on disk (`teams.binz` and `scores.binz`), to make it easier to test small changes to the final analysis.
Cases where I've simplified the logic are marked with `XXX`; these are typically assumptions which work for the first turn but could break down later in the game.
I only implemented animals from the Free-To-Play pack; testing with other packs would be a fun extension of this work. The code is up to date as of the February 10th patch; later patches could invalidate these results!
There's also one hilarious corner case which I didn't implement. It is possible, on the very first turn, to do the following:
• Buy three of the same tier 1 unit (leaving 1 gold)
• Combine into a level 2 unit (which spawns a Tier 2 animal in the shop)
• Sell the level 2 unit (receiving 2 gold), then buy the new animal
In other words, given a sufficiently lucky shop, you can go into battle on turn 1 with with a Tier 2 unit, which normally wouldn't be available until a later round.
I decided not to implement all of the new logic for Tier 2 units, but PRs are welcome if you want to take it on!
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| 2.578125
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.939226
|
https://responsivedesigntool.com/qa/question-how-far-can-you-go-on-18-gallons-of-gas.html
| 1,606,456,412,000,000,000
|
text/html
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crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00542.warc.gz
| 477,013,762
| 8,824
|
# Question: How Far Can You Go On 18 Gallons Of Gas?
## How many miles can you drive with 12 gallons of gas?
For example, if your car averages 25 miles per gallon on the highway and has a 12-gallon fuel tank, its range is 25 x 12 = 300 miles..
## How much does it cost to drive a mile?
Based on Driving 15,000 miles annuallySmall SedanSedan AverageCost Per Mile44.9 cents59.6 centsCost Per Year\$6,735\$8,946
## How do you calculate cost per mile?
To calculate the “cost per mile,” divide the cost by the number of miles you drove that month. To determine the total monthly cost per mile, simply add the fixed and variable costs.
## How long will 5 gallons of gas last?
About 3- 5 months in a sealed container, or 6 – 8 months with fuel stabilizer added. Recommended maximum storage is 1 year.
## How far will 2 gallons of gas get you?
As a general rule, most cars have about 2.5 gallons left in the tank when the gas light comes on. So depending on how many miles you get per gallon, you can probably go anywhere between 30-60 miles.
## How do you calculate how many miles per gallon?
CalculateGet the miles traveled from the trip odometer, or subtract the original odometer reading from the new one.Divide the miles traveled by the amount of gallons it took to refill the tank. The result will be your car’s average miles per gallon yield for that driving period.
## How far can you drive after gas light comes on?
Generally speaking, you will be able to drive between 30 and 50 (0r more) miles even after the light comes on. Which means that even if the gas light comes on, it may not be time to panic yet.
## Does cruise control save gas?
Use the Cruise Control On long stretches of highway driving, cruise control can save fuel by helping your car maintain a steady speed. However, this efficiency is lost on steep hills where the cruise control tries to maintain even speeds. In hilly terrain, it is best to turn off the cruise control.
## How far can you drive on 0 miles to empty?
According to the chart, you can expect to get anywhere from 30 miles to more than 100 miles on a nearly empty tank, depending on the car. Of course, the real-life numbers will vary based on how you drive, your car’s condition, and other variables, so Your Mechanic emphasizes they’re all just rough approximations.
## How many miles does 1 gallon of gas get you?
You would need to divide 1000 miles by 50 gallons of gas. That would equal 20; therefore, you traveled 20 miles for every 1 gallon of gas. Your gas mileage would be 20 mpg (miles per gallon). You may wish to use an online auto mpg fuel cost calculator.
## How far can I go on a full tank of gas?
For example, assume your vehicle’s gas tank can hold 20 gallons of gas. 23 x 20 = 460 miles. This figure represents the distance you can expect to go in your vehicle on a tank of gas.
## Is it bad to run a car on low fuel?
It’s still bad for your car to let the gas light come on, however. … “If you run the car low on fuel consistently, you can wear out the fuel pump prematurely, over-stressing it and making it hotter,” the fuel experts at Bell Performance explain.
## What is the average mpg?
Projected to hit 25.2 mpg for 2017 Reuters reports that the U.S. car and truck fleet averaged 24.7 mpg in total, up 0.1 miles per gallon over 2015. Based on current projections, 2017 is expected to set an even more significant record, hitting an average of 25.2 mpg.
## How many gallons of gas are in my car?
The amount of gas a car holds depends on the size of the car. Smaller cars generally have gas tanks that hold 12 gallons worth of gas, while larger cars can hold 15 or 16 gallons. For the purpose of this story, let’s say gas costs \$3.85 a gallon.
## Is a 400 mile tank good?
Most gas cars don’t make it 400 miles Basically, you need a gas tank size of 13 gallons and an average mpg of 31 to hit – just barely – the 400-mile range mark. At a tank size of 14 gallons, you need to hit 29 mpg, at 15 gallons, 27 mpg, and at 16 gallons, 25 mpg.
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| 3.359375
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https://www.freestatistics.org/blog/index.php?v=date/2010/Nov/14/t12897733069mzz658rdm2guhg.htm/
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| 12,243
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## Free Statistics
of Irreproducible Research!
Author's title
Author*The author of this computation has been verified*
R Software Modulerwasp_histogram.wasp
Title produced by softwareHistogram
Date of computationSun, 14 Nov 2010 22:22:59 +0000
Cite this page as followsStatistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?v=date/2010/Nov/14/t12897733069mzz658rdm2guhg.htm/, Retrieved Mon, 22 Jul 2024 21:02:45 +0000
Statistical Computations at FreeStatistics.org, Office for Research Development and Education, URL https://freestatistics.org/blog/index.php?pk=94661, Retrieved Mon, 22 Jul 2024 21:02:45 +0000
QR Codes:
Original text written by user:
IsPrivate?No (this computation is public)
User-defined keywords
Estimated Impact208
Family? (F = Feedback message, R = changed R code, M = changed R Module, P = changed Parameters, D = changed Data)
- [Histogram] [Bad example of Hi...] [2010-09-25 09:28:23] [b98453cac15ba1066b407e146608df68]
F P [Histogram] [Task 3 - Histogram] [2010-10-02 20:26:59] [19f9551d4d95750ef21e9f3cf8fe2131]
- PD [Histogram] [WS 6 Yt skewness ...] [2010-11-14 22:22:59] [fca744d17b21beb005bf086e7071b2bb] [Current]
- R D [Histogram] [WS 6 Linear Regre...] [2011-11-13 00:40:52] [f5fdea4413921432bb019d1f20c4f2ec]
- RM [Histogram] [] [2011-11-14 19:02:38] [74be16979710d4c4e7c6647856088456]
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Post a new message
Dataseries X:
255
280.2
299.9
339.2
374.2
393.5
389.2
381.7
375.2
369
357.4
352.1
346.5
342.9
340.3
328.3
322.9
314.3
308.9
294
285.6
281.2
280.3
278.8
274.5
270.4
263.4
259.9
258
262.7
284.7
311.3
322.1
327
331.3
333.3
321.4
327
320
314.7
316.7
314.4
321.3
318.2
307.2
301.3
287.5
277.7
274.4
258.8
253.3
251
248.4
249.5
246.1
244.5
243.6
244
240.8
249.8
248
259.4
260.5
260.8
261.3
259.5
256.6
257.9
256.5
254.2
253.3
253.8
255.5
257.1
257.3
253.2
252.8
252
250.7
252.2
250
251
253.4
251.2
255.6
261.1
258.9
259.9
261.2
264.7
267.1
266.4
267.7
268.6
267.5
268.5
268.5
270.5
270.9
270.1
269.3
269.8
270.1
264.9
263.7
264.8
263.7
255.9
276.2
360.1
380.5
373.7
369.8
366.6
359.3
345.8
326.2
324.5
328.1
327.5
324.4
316.5
310.9
301.5
291.7
290.4
287.4
277.7
281.6
288
276
272.9
283
283.3
276.8
284.5
282.7
281.2
287.4
283.1
284
285.5
289.2
292.5
296.4
305.2
303.9
311.5
316.3
316.7
322.5
317.1
309.8
303.8
290.3
293.7
291.7
296.5
289.1
288.5
293.8
297.7
305.4
302.7
302.5
303
294.5
294.1
294.5
297.1
289.4
292.4
287.9
286.6
280.5
272.4
269.2
270.6
267.3
262.5
266.8
268.8
263.1
261.2
266
262.5
265.2
261.3
253.7
249.2
239.1
236.4
235.2
245.2
246.2
247.7
251.4
253.3
254.8
250
249.3
241.5
243.3
248
253
252.9
251.5
251.6
253.5
259.8
334.1
448
445.8
445
448.2
438.2
439.8
423.4
410.8
408.4
406.7
405.9
402.7
405.1
399.6
386.5
381.4
375.2
357.7
359
355
352.7
344.4
343.8
338
339
333.3
334.4
328.3
330.7
330
331.6
351.2
389.4
410.9
442.8
462.8
466.9
461.7
439.2
430.3
416.1
402.5
397.3
403.3
395.9
387.8
378.6
377.1
370.4
362
350.3
348.2
344.6
343.5
342.8
347.6
346.6
349.5
342.1
342
342.8
339.3
348.2
333.7
334.7
354
367.7
363.3
358.4
353.1
343.1
344.6
344.4
333.9
331.7
324.3
321.2
322.4
321.7
320.5
312.8
309.7
315.6
309.7
304.6
302.5
301.5
298.8
291.3
293.6
294.6
285.9
297.6
301.1
293.8
297.7
292.9
292.1
287.2
288.2
283.8
299.9
292.4
293.3
300.8
293.7
293.1
294.4
292.1
291.9
282.5
277.9
287.5
289.2
285.6
293.2
290.8
283.1
275
287.8
287.8
287.4
284
277.8
277.6
304.9
294
300.9
324
332.9
341.6
333.4
348.2
344.7
344.7
329.3
323.5
323.2
317.4
330.1
329.2
334.9
315.8
315.4
319.6
317.3
313.8
315.8
311.3
Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 1 seconds R Server 'George Udny Yule' @ 72.249.76.132
\begin{tabular}{lllllllll}
\hline
Summary of computational transaction \tabularnewline
Raw Input & view raw input (R code) \tabularnewline
Raw Output & view raw output of R engine \tabularnewline
Computing time & 1 seconds \tabularnewline
R Server & 'George Udny Yule' @ 72.249.76.132 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=94661&T=0
[TABLE]
[ROW][C]Summary of computational transaction[/C][/ROW]
[ROW][C]Raw Input[/C][C]view raw input (R code) [/C][/ROW]
[ROW][C]Raw Output[/C][C]view raw output of R engine [/C][/ROW]
[ROW][C]Computing time[/C][C]1 seconds[/C][/ROW]
[ROW][C]R Server[/C][C]'George Udny Yule' @ 72.249.76.132[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=94661&T=0
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=94661&T=0
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Summary of computational transaction Raw Input view raw input (R code) Raw Output view raw output of R engine Computing time 1 seconds R Server 'George Udny Yule' @ 72.249.76.132
Frequency Table (Histogram) Bins Midpoint Abs. Frequency Rel. Frequency Cumul. Rel. Freq. Density [230,240[ 235 3 0.008333 0.008333 0.000833 [240,250[ 245 17 0.047222 0.055556 0.004722 [250,260[ 255 41 0.113889 0.169444 0.011389 [260,270[ 265 32 0.088889 0.258333 0.008889 [270,280[ 275 20 0.055556 0.313889 0.005556 [280,290[ 285 38 0.105556 0.419444 0.010556 [290,300[ 295 37 0.102778 0.522222 0.010278 [300,310[ 305 21 0.058333 0.580556 0.005833 [310,320[ 315 22 0.061111 0.641667 0.006111 [320,330[ 325 25 0.069444 0.711111 0.006944 [330,340[ 335 20 0.055556 0.766667 0.005556 [340,350[ 345 24 0.066667 0.833333 0.006667 [350,360[ 355 12 0.033333 0.866667 0.003333 [360,370[ 365 7 0.019444 0.886111 0.001944 [370,380[ 375 7 0.019444 0.905556 0.001944 [380,390[ 385 7 0.019444 0.925 0.001944 [390,400[ 395 4 0.011111 0.936111 0.001111 [400,410[ 405 7 0.019444 0.955556 0.001944 [410,420[ 415 3 0.008333 0.963889 0.000833 [420,430[ 425 1 0.002778 0.966667 0.000278 [430,440[ 435 4 0.011111 0.977778 0.001111 [440,450[ 445 5 0.013889 0.991667 0.001389 [450,460[ 455 0 0 0.991667 0 [460,470] 465 3 0.008333 1 0.000833
\begin{tabular}{lllllllll}
\hline
Frequency Table (Histogram) \tabularnewline
Bins & Midpoint & Abs. Frequency & Rel. Frequency & Cumul. Rel. Freq. & Density \tabularnewline
[230,240[ & 235 & 3 & 0.008333 & 0.008333 & 0.000833 \tabularnewline
[240,250[ & 245 & 17 & 0.047222 & 0.055556 & 0.004722 \tabularnewline
[250,260[ & 255 & 41 & 0.113889 & 0.169444 & 0.011389 \tabularnewline
[260,270[ & 265 & 32 & 0.088889 & 0.258333 & 0.008889 \tabularnewline
[270,280[ & 275 & 20 & 0.055556 & 0.313889 & 0.005556 \tabularnewline
[280,290[ & 285 & 38 & 0.105556 & 0.419444 & 0.010556 \tabularnewline
[290,300[ & 295 & 37 & 0.102778 & 0.522222 & 0.010278 \tabularnewline
[300,310[ & 305 & 21 & 0.058333 & 0.580556 & 0.005833 \tabularnewline
[310,320[ & 315 & 22 & 0.061111 & 0.641667 & 0.006111 \tabularnewline
[320,330[ & 325 & 25 & 0.069444 & 0.711111 & 0.006944 \tabularnewline
[330,340[ & 335 & 20 & 0.055556 & 0.766667 & 0.005556 \tabularnewline
[340,350[ & 345 & 24 & 0.066667 & 0.833333 & 0.006667 \tabularnewline
[350,360[ & 355 & 12 & 0.033333 & 0.866667 & 0.003333 \tabularnewline
[360,370[ & 365 & 7 & 0.019444 & 0.886111 & 0.001944 \tabularnewline
[370,380[ & 375 & 7 & 0.019444 & 0.905556 & 0.001944 \tabularnewline
[380,390[ & 385 & 7 & 0.019444 & 0.925 & 0.001944 \tabularnewline
[390,400[ & 395 & 4 & 0.011111 & 0.936111 & 0.001111 \tabularnewline
[400,410[ & 405 & 7 & 0.019444 & 0.955556 & 0.001944 \tabularnewline
[410,420[ & 415 & 3 & 0.008333 & 0.963889 & 0.000833 \tabularnewline
[420,430[ & 425 & 1 & 0.002778 & 0.966667 & 0.000278 \tabularnewline
[430,440[ & 435 & 4 & 0.011111 & 0.977778 & 0.001111 \tabularnewline
[440,450[ & 445 & 5 & 0.013889 & 0.991667 & 0.001389 \tabularnewline
[450,460[ & 455 & 0 & 0 & 0.991667 & 0 \tabularnewline
[460,470] & 465 & 3 & 0.008333 & 1 & 0.000833 \tabularnewline
\hline
\end{tabular}
%Source: https://freestatistics.org/blog/index.php?pk=94661&T=1
[TABLE]
[ROW][C]Frequency Table (Histogram)[/C][/ROW]
[ROW][C]Bins[/C][C]Midpoint[/C][C]Abs. Frequency[/C][C]Rel. Frequency[/C][C]Cumul. Rel. Freq.[/C][C]Density[/C][/ROW]
[ROW][C][230,240[[/C][C]235[/C][C]3[/C][C]0.008333[/C][C]0.008333[/C][C]0.000833[/C][/ROW]
[ROW][C][240,250[[/C][C]245[/C][C]17[/C][C]0.047222[/C][C]0.055556[/C][C]0.004722[/C][/ROW]
[ROW][C][250,260[[/C][C]255[/C][C]41[/C][C]0.113889[/C][C]0.169444[/C][C]0.011389[/C][/ROW]
[ROW][C][260,270[[/C][C]265[/C][C]32[/C][C]0.088889[/C][C]0.258333[/C][C]0.008889[/C][/ROW]
[ROW][C][270,280[[/C][C]275[/C][C]20[/C][C]0.055556[/C][C]0.313889[/C][C]0.005556[/C][/ROW]
[ROW][C][280,290[[/C][C]285[/C][C]38[/C][C]0.105556[/C][C]0.419444[/C][C]0.010556[/C][/ROW]
[ROW][C][290,300[[/C][C]295[/C][C]37[/C][C]0.102778[/C][C]0.522222[/C][C]0.010278[/C][/ROW]
[ROW][C][300,310[[/C][C]305[/C][C]21[/C][C]0.058333[/C][C]0.580556[/C][C]0.005833[/C][/ROW]
[ROW][C][310,320[[/C][C]315[/C][C]22[/C][C]0.061111[/C][C]0.641667[/C][C]0.006111[/C][/ROW]
[ROW][C][320,330[[/C][C]325[/C][C]25[/C][C]0.069444[/C][C]0.711111[/C][C]0.006944[/C][/ROW]
[ROW][C][330,340[[/C][C]335[/C][C]20[/C][C]0.055556[/C][C]0.766667[/C][C]0.005556[/C][/ROW]
[ROW][C][340,350[[/C][C]345[/C][C]24[/C][C]0.066667[/C][C]0.833333[/C][C]0.006667[/C][/ROW]
[ROW][C][350,360[[/C][C]355[/C][C]12[/C][C]0.033333[/C][C]0.866667[/C][C]0.003333[/C][/ROW]
[ROW][C][360,370[[/C][C]365[/C][C]7[/C][C]0.019444[/C][C]0.886111[/C][C]0.001944[/C][/ROW]
[ROW][C][370,380[[/C][C]375[/C][C]7[/C][C]0.019444[/C][C]0.905556[/C][C]0.001944[/C][/ROW]
[ROW][C][380,390[[/C][C]385[/C][C]7[/C][C]0.019444[/C][C]0.925[/C][C]0.001944[/C][/ROW]
[ROW][C][390,400[[/C][C]395[/C][C]4[/C][C]0.011111[/C][C]0.936111[/C][C]0.001111[/C][/ROW]
[ROW][C][400,410[[/C][C]405[/C][C]7[/C][C]0.019444[/C][C]0.955556[/C][C]0.001944[/C][/ROW]
[ROW][C][410,420[[/C][C]415[/C][C]3[/C][C]0.008333[/C][C]0.963889[/C][C]0.000833[/C][/ROW]
[ROW][C][420,430[[/C][C]425[/C][C]1[/C][C]0.002778[/C][C]0.966667[/C][C]0.000278[/C][/ROW]
[ROW][C][430,440[[/C][C]435[/C][C]4[/C][C]0.011111[/C][C]0.977778[/C][C]0.001111[/C][/ROW]
[ROW][C][440,450[[/C][C]445[/C][C]5[/C][C]0.013889[/C][C]0.991667[/C][C]0.001389[/C][/ROW]
[ROW][C][450,460[[/C][C]455[/C][C]0[/C][C]0[/C][C]0.991667[/C][C]0[/C][/ROW]
[ROW][C][460,470][/C][C]465[/C][C]3[/C][C]0.008333[/C][C]1[/C][C]0.000833[/C][/ROW]
[/TABLE]
Source: https://freestatistics.org/blog/index.php?pk=94661&T=1
Globally Unique Identifier (entire table): ba.freestatistics.org/blog/index.php?pk=94661&T=1
As an alternative you can also use a QR Code:
The GUIDs for individual cells are displayed in the table below:
Frequency Table (Histogram) Bins Midpoint Abs. Frequency Rel. Frequency Cumul. Rel. Freq. Density [230,240[ 235 3 0.008333 0.008333 0.000833 [240,250[ 245 17 0.047222 0.055556 0.004722 [250,260[ 255 41 0.113889 0.169444 0.011389 [260,270[ 265 32 0.088889 0.258333 0.008889 [270,280[ 275 20 0.055556 0.313889 0.005556 [280,290[ 285 38 0.105556 0.419444 0.010556 [290,300[ 295 37 0.102778 0.522222 0.010278 [300,310[ 305 21 0.058333 0.580556 0.005833 [310,320[ 315 22 0.061111 0.641667 0.006111 [320,330[ 325 25 0.069444 0.711111 0.006944 [330,340[ 335 20 0.055556 0.766667 0.005556 [340,350[ 345 24 0.066667 0.833333 0.006667 [350,360[ 355 12 0.033333 0.866667 0.003333 [360,370[ 365 7 0.019444 0.886111 0.001944 [370,380[ 375 7 0.019444 0.905556 0.001944 [380,390[ 385 7 0.019444 0.925 0.001944 [390,400[ 395 4 0.011111 0.936111 0.001111 [400,410[ 405 7 0.019444 0.955556 0.001944 [410,420[ 415 3 0.008333 0.963889 0.000833 [420,430[ 425 1 0.002778 0.966667 0.000278 [430,440[ 435 4 0.011111 0.977778 0.001111 [440,450[ 445 5 0.013889 0.991667 0.001389 [450,460[ 455 0 0 0.991667 0 [460,470] 465 3 0.008333 1 0.000833
par1 <- as.numeric(par1)if (par3 == 'TRUE') par3 <- TRUEif (par3 == 'FALSE') par3 <- FALSEif (par4 == 'Unknown') par1 <- as.numeric(par1)if (par4 == 'Interval/Ratio') par1 <- as.numeric(par1)if (par4 == '3-point Likert') par1 <- c(1:3 - 0.5, 3.5)if (par4 == '4-point Likert') par1 <- c(1:4 - 0.5, 4.5)if (par4 == '5-point Likert') par1 <- c(1:5 - 0.5, 5.5)if (par4 == '6-point Likert') par1 <- c(1:6 - 0.5, 6.5)if (par4 == '7-point Likert') par1 <- c(1:7 - 0.5, 7.5)if (par4 == '8-point Likert') par1 <- c(1:8 - 0.5, 8.5)if (par4 == '9-point Likert') par1 <- c(1:9 - 0.5, 9.5)if (par4 == '10-point Likert') par1 <- c(1:10 - 0.5, 10.5)bitmap(file='test1.png')if(is.numeric(x[1])) {if (is.na(par1)) {myhist<-hist(x,col=par2,main=main,xlab=xlab,right=par3)} else {if (par1 < 0) par1 <- 3if (par1 > 50) par1 <- 50myhist<-hist(x,breaks=par1,col=par2,main=main,xlab=xlab,right=par3)}} else {plot(mytab <- table(x),col=par2,main='Frequency Plot',xlab=xlab,ylab='Absolute Frequency')}dev.off()if(is.numeric(x[1])) {myhistn <- length(x)load(file='createtable')a<-table.start()a<-table.row.start(a)a<-table.element(a,hyperlink('histogram.htm','Frequency Table (Histogram)',''),6,TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'Bins',header=TRUE)a<-table.element(a,'Midpoint',header=TRUE)a<-table.element(a,'Abs. Frequency',header=TRUE)a<-table.element(a,'Rel. Frequency',header=TRUE)a<-table.element(a,'Cumul. Rel. Freq.',header=TRUE)a<-table.element(a,'Density',header=TRUE)a<-table.row.end(a)crf <- 0if (par3 == FALSE) mybracket <- '[' else mybracket <- ']'mynumrows <- (length(myhist$breaks)-1)for (i in 1:mynumrows) {a<-table.row.start(a)if (i == 1)dum <- paste('[',myhist$breaks[i],sep='')elsedum <- paste(mybracket,myhist$breaks[i],sep='')dum <- paste(dum,myhist$breaks[i+1],sep=',')if (i==mynumrows)dum <- paste(dum,']',sep='')elsedum <- paste(dum,mybracket,sep='')a<-table.element(a,dum,header=TRUE)a<-table.element(a,myhist$mids[i])a<-table.element(a,myhist$counts[i])rf <- myhist$counts[i]/ncrf <- crf + rfa<-table.element(a,round(rf,6))a<-table.element(a,round(crf,6))a<-table.element(a,round(myhist$density[i],6))a<-table.row.end(a)}a<-table.end(a)table.save(a,file='mytable.tab')} else {mytabreltab <- mytab / sum(mytab)n <- length(mytab)load(file='createtable')a<-table.start()a<-table.row.start(a)a<-table.element(a,'Frequency Table (Categorical Data)',3,TRUE)a<-table.row.end(a)a<-table.row.start(a)a<-table.element(a,'Category',header=TRUE)a<-table.element(a,'Abs. Frequency',header=TRUE)a<-table.element(a,'Rel. Frequency',header=TRUE)a<-table.row.end(a)for (i in 1:n) {a<-table.row.start(a)a<-table.element(a,labels(mytab)\$x[i],header=TRUE)a<-table.element(a,mytab[i])a<-table.element(a,round(reltab[i],4))a<-table.row.end(a)}a<-table.end(a)table.save(a,file='mytable1.tab')}
| 6,796
| 14,485
|
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| 2.53125
| 3
|
CC-MAIN-2024-30
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latest
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en
| 0.505473
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http://woodworkingplans2013.com/woodworking-projects-and-plans-woodcraft-woodworking.html
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text/html
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crawl-data/CC-MAIN-2019-13/segments/1552912203378.92/warc/CC-MAIN-20190324063449-20190324085449-00285.warc.gz
| 236,527,825
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|
Another awesome thing about this coffee table is that it is also has a storage unit. So you can store drinks, and other stuff in the half barrel of your table and then close or open it whenever you need. Pete has also constructed a video for this tutorial for which you can find the link below. It illustrates the same process in a video guide that shows you the exact process to be followed while building this whiskey barrel coffee table.
Have you got an old whiskey barrel at home that you haven’t used for ages? If yes, this project is for you. You can make a really beautiful coffee table from that old whiskey barrel in a few easy steps. Apart from a coffee table, whiskey barrels can also be used to build several other furniture items. But that is a talk for later. Here, we will discuss how to make a coffee table from a whiskey barrel.
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.
This plan is probably the easiest plan ever added in the list. The one who is working on this project, don’t need any professional skills but just knowing some basics of woodworking will be enough for this DIY. You will get step by step detailed process of this tutorial in the source linked tutorial. This tutorial will surely help you to build this plan quickly.
Finding affordable lumber has always been a mainstay for woodworkers, and when you tie our dwindling natural resources into the conversation the time is right to look at milling your own lumber. This seven-part weekly video series takes you through how to find lumber, how to operate a sawmill, details on types of sawing methods, stickering and drying and ultimately advice on using a mill as part of a business. Learn what you need to know to understand Milling Your Own Lumber.
##### Figure out a simple setup for your woodworking space. You don’t need a fancy and expensive workshop or garage to start woodworking. In fact, we’ve never had a workshop or garage (though I do dream of having one haha). In our current town home, we always setup a temporary workshop table in our backyard with a pair of sawhorses and a plywood board from the home improvement store.
Why would you buy a costly platform bed from Ikea or somewhere else when you can make one yourself at home? Oh yes, you can. A bed is the most common furniture piece used in the house and probably the costliest one. Wouldn’t it be great if you could just make a bed of your own, without having to spend many bucks for buying one? So I am here sharing a great tutorial to help you to build a nice comfy platform bed that you can use anywhere in the house.
| 633
| 2,980
|
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| 3.515625
| 4
|
CC-MAIN-2019-13
|
latest
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en
| 0.937792
|
https://www.bartleby.com/questions-and-answers/3x2-2hu-3uaxpercent3d-4hu-dy-o-y-xv3x-cx-o-v-3xy-x-cx-o-y-3xly-x-cx-o-y-3xy-x-cx/4665cc47-98c7-4a83-a507-2195c245df2f
| 1,627,770,519,000,000,000
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text/html
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crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00484.warc.gz
| 657,265,015
| 34,199
|
# (3x2 - 2ху+ 3у?)аx%3D 4ху dy O (y- x)³(v+3x) = Cx³ O (v +3x)°(y- x)= Cx³ O (y-3x)°ly+ x) = Cx² O (y-3x)(y+ x)³ = Cx²
Question
Solve and determine what type of DE.
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Tagged in
Math
Calculus
| 168
| 476
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| 2.53125
| 3
|
CC-MAIN-2021-31
|
latest
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en
| 0.909088
|
http://thegolfnewsnet.com/golfnewsnetteam/2016/07/08/what-do-dots-golf-scorecard-mean-handicap-strokes-51402/
| 1,475,231,077,000,000,000
|
text/html
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crawl-data/CC-MAIN-2016-40/segments/1474738662159.54/warc/CC-MAIN-20160924173742-00161-ip-10-143-35-109.ec2.internal.warc.gz
| 264,606,367
| 21,276
|
Albatrossities
If you've ever showed up to compete in a golf tournament of any kind, there's a good chance you've received your scorecard for the day and seen dots on it. You might have been perplexed to see some dots on your card, and you were wondering what they meant.
Well, the dots on your scorecard are used to indicate the number of handicap strokes you get on that hole. They're on the card to let you know those holes so you can play accordingly and make doing the math of your net score -- your gross score minus your handicap -- easier.
Here's how the dots are placed on the card:
• If your handicap index translates to a 12 on the golf course you're playing, then you'll have one dot in your score box on the card for the 12 hardest holes on the course. You won't get any dots on the card for the other six holes.
• If you're getting 20 strokes that day, then you'll have one dot in the score box for 16 holes and two dots in the score box for the two hardest holes on the golf course. After you get 18 strokes on a golf course, the 19th handicap stroke is applied to the No. 1 handicap hole on a course, and it follows from there on to 36 strokes, the maximum handicap.
At the end of the round, you're going to have two scores, a gross score that is your actual score for the round and your net score, which is your gross score minus your handicap strokes.
| 318
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| 3.09375
| 3
|
CC-MAIN-2016-40
|
longest
|
en
| 0.956109
|
https://sheetnatives.com/how-to-use-power-function-in-google-sheets/
| 1,680,253,400,000,000,000
|
text/html
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crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00046.warc.gz
| 575,054,435
| 16,526
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# How to Use POWER function in Google Sheets in 2020?
POWER functionReturns a number raised to a power.
Sample Usage
POWER(4,0.5)
POWER(A2,B2)
POWER(2,5)
Syntax
POWER(base, exponent)
base – The number to raise to the exponent power.
If base is negative, exponent must be an integer.
exponent – The exponent to raise base to.
SQRTPI: Returns the positive square root of the product of Pi and the given positive number.
SQRT: Returns the positive square root of a positive number.
LOG10: Returns the the logarithm of a number, base 10.
LOG: Returns the the logarithm of a number given a base.
LN: Returns the the logarithm of a number, base e (Euler’s number).
GAMMALN: Returns the the logarithm of a specified Gamma function, base e (Euler’s number).
EXP: Returns Euler’s number, e (~2.718) raised to a power.
Examples
| 212
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| 2.59375
| 3
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CC-MAIN-2023-14
|
latest
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en
| 0.65358
|
https://www.theburningofrome.com/blog/what-is-the-logo-with-3-lines/
| 1,716,998,474,000,000,000
|
text/html
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crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00775.warc.gz
| 878,060,910
| 40,190
|
What is the logo with 3 lines?
What is the logo with 3 lines?
The triple bar, ≡, is a symbol with multiple, context-dependent meanings. It has the appearance of an equals sign ⟨=⟩ sign with a third line.
What brand has three lines?
Three stripes is a trademark of Adidas consisting of three parallel lines, which typically feature along the side of Adidas apparel. Adidas was known for this branding early in its history, with its owner, Adolf Dassler, describing it as “The three stripe company”.
Which of the following words means a symbol small design adopted by an Organisation to identify its product ?
logos
nounplural noun logos. A symbol or other small design adopted by an organization to identify its products, uniform, vehicles, etc.
What do 3 lines mean?
The triple bar, ≡, is a symbol with multiple, context-dependent meanings. It has the appearance of a “=” sign with a third line. In mathematics it sometimes used a symbol for congruence. Particularly, in number theory, it has the meaning of modular congruence: if N divides a − b.
What is an R logo?
The R symbol indicates that this word, phrase or logo is a registered trademark for the product or service. It must only be used in the case of registered trademarks and by the owner or licensee. It also must only be used in the regions in which you possess a valid trademark registration.
What is the Lululemon logo?
“The lululemon name was chosen in a survey of 100 people from a list of 20 brand names and 20 logos. The logo is actually a stylized ‘A’ that was made for the name ‘athletically hip’, a name which failed to make the grade.”
What are the 7 different types of logos?
The different kinds of logos can be placed into seven categories: emblems, pictorial marks, logotypes, lettermarks, abstract logos, mascot logos, and combination logos. Let’s analyze each category.
What does the 3 horizontal lines mean?
A symbol with three horizontal line segments ( ) resembling the equals sign is used to denote both equality by definition (e.g., means is defined to be equal to ) and congruence (e.g., means 13 divided by 12 leaves a remainder of 1–a fact known to all readers of analog clocks).
What are the reviews on the 3 line phone?
User rating, 4.5 out of 5 stars with 17 reviews. User rating, 4.4 out of 5 stars with 479 reviews. User rating, 4.4 out of 5 stars with 46 reviews.
Which is the best brand of 3M products?
3M™ Novec™ products are smart, safe and sustainable. These high-performing, fluid-based products are designed to protect, clean, cool and enable new electronics products and processes. Protect hearing and ensure clear communications in high noise environments with 3M™ PELTOR™ Two-Way Headsets and Earmuffs.
How big is the Anker powerline 3 Lightning cable?
New! Anker – PowerLine III Flow USB-C to Lightning Cable 6-ft – Green New! Anker – PowerLine III Flow USB-C to Lightning Cable 6-ft – Purple New! Anker – PowerLine III Flow USB-C to Lightning Cable 6-ft – White New! Anker – PowerLine III Flow USB-C to Lightning Cable 6-ft – Black New! Anker – PowerLine III Flow USB-C to Lightning Cable 6-ft – Pink
Which is the best brand for licensing options?
LICENSING OPTIONS Rank Brand Brand Value 1-Yr Value Change Brand Revenue 1 Apple \$241.2 B 17% \$260.2 B 2 Google \$207.5 B 24% \$145.6 B 3 Microsoft \$162.9 B 30% \$125.8 B 4 Amazon \$135.4 B 40% \$260.5 B
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# Trouble converting definite integrals to Riemann's and back
• B
can i request anyone to please show me the step by step with specific explanations? thank you! i saw this on stackexchange, and the steps shown are really fuzzy to me :(
#### Attachments
• convert to definite integral.png
2.9 KB · Views: 384
## Answers and Replies
mathman
In general $\int_0^1 f(x)dx$ can be approximated by a Riemann sum $\frac{1}{n}\sum_{k=1}^n f(\frac{k-1/2}{n})$. For large n, -1/2 can be ignored.
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# Riddle #335
## As small as your thumb, I am light in the air
As small as your thumb, I am light in the air. You may hear me before you see me, but trust that I'm there. What am I?
A hummingbird.
73.37 %
## An ancient invention
What is the ancient invention that allows people to see through walls?
A window.
76.73 %
The more you take, the more you leave behind. What am I?
Footsteps.
76.04 %
## You go on red and stops on green
When it comes to me, you go on red and stops on green. What am I?
Watermelon.
75.88 %
## The more you take away
The more you take away, the bigger I become. What am I?
Hole.
75.86 %
## Wordplay
What word in the English language does the following: the first two letters signify a male, the first three letters signify a female, the first four letters signify a great man, the first six letters signify a drug, while the entire world signifies a great woman. What is the word?
Heroine
75.84 %
## You bury me when I am alive
You bury me when I am alive, and dig me up when I die. What am I?
A plant.
75.84 %
## Gets around a lot
What has a tongue, cannot walk, but gets around a lot?
A shoe.
75.63 %
## Up and down
What goes up and down the stairs without moving?
A rug.
75.62 %
## How many apples?
If there're 3 apples and you take away 2, how many do you have?
If you take 2 apples, than you have of course 2.
75.53 %
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# Power Formula: Derivation of Power formula, Examples
We all are very familiar with power. Also, it helps us to measure the energy that we use to do the work. In this topic, we will discuss what is power and what is Power formula, its derivation and solved examples.
## Power
Power refers to the quantity of work done by a body. Also, quantity work has to do with a force causing a displacement. Moreover, work has nothing to do with the amount of time, which this force acts to cause the displacement. In addition, there are times when we do work quickly and there are times when we do work rather slowly.
For example, a trail hiker (One who selects an easier path up the mountain) can elevate her/his body a few meters in a short amount of time. On the other hand, a rock climber might take relatively more time to elevate her/his body up a few meters next to the side of a cliff.
Although, both of them do the same work but the rock climber might take more time than the trail hiker. Also, the quantity that has something to do with the rate at which a certain amount of work is done is called power. Besides, in the above case, the hiker has greater power rating than the rock climber.
### Power Formula
We can refer to power as the rate at which we do work. Also, it is the ratio of work and time. Besides, we calculate it mathematically using this formula or equation.
### Derivation of Power formula
Power = unit of measure (Watt)
W = work done by the body
t = time taken to do the work
Moreover, the standard unit of measuring power is Watt. According to the power formula, a unit of power is equal to a unit of work divided by a unit of time. Also, a watt is equal to Joule/ second.
Besides, due to historical reasons, power is occasionally used to describe the power delivered by a machine. In addition, one horsepower equals 750 Watt.
Suppose that there is a car engine of 40 horsepower that could accelerate from 0 mi/hr to 60 mi/hr in 16 seconds. Moreover, if there is a car that has four times the horsepower that could do the same work in one-fourth the time. That means a 160 horsepower engine could accelerate the car from 0 mi/hr to 60 mi/hr in just 4 seconds.
Here the point is that for some amount of work, power and time are inversely proportional. Moreover, it means that a more powerful engine can do the same work in less time.
### Solved Examples on Power Formula
Example 1
Suppose that Mr. X elevates his body of 80 kg up by a 2.0-meter stairwell in 1.8 seconds. Also, in this case, then we could calculate Mr. X power rating. Besides, we must assume that Mr. X must apply an 800 Newton downward force upon the stairs to elevate his body. In addition, by doing this the stairs would push upward on Mr. X body with just enough force to lift his body up the stairs.
Besides, we can assume that the angle between the force of the stairs on Mr. X and Mr. X’s displacement is 0 degrees. Then power can be resolute as shown below:
Solution: Power = Work/Time
W = 871 Watts.
So, Mr.X power rating is 871 Watts.
Example 2
Calculate the power that a person requires to lift an object to a height of 8 m in 10 seconds. Also, the mass of the object is 10 kg. Besides, take g = 10 m/s^2
Solution: For calculating work first find the work done by the person which is equal to the potential energy at that height
Work (W) = mgh = 10 × 8 × 10 = 400J
Power = Work/Time = 800/10 = 80 J/s.
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Hong Kong
Stage 1 - Stage 3
# Ratio of Sides in Right-Angled Triangles
Lesson
Right-angled triangle
In the above right-angled triangle $c$c is the the hypotenuse and $\angle RPQ$RPQ is $90^\circ$90°
Now if we call $\angle PQR=\theta$PQR=θ and we can identify the opposite and adjacent sides with respect to that angle, (we did this here, if you need to refresh). So $b$b is the opposite side, and $a$a is the adjacent side.
A ratio is a statement of a mathematical relationship between two objects, often represented as a fraction. Various ratios of the following can be constructed from the right-angled triangle with respect to angle $\theta$θ.
$\frac{Opposite}{Adjacent}=\frac{b}{a}$OppositeAdjacent=ba
$\frac{Adjacent}{Hypotenuse}=\frac{a}{c}$AdjacentHypotenuse=ac
$\frac{Opposite}{Hypotenuse}=\frac{b}{c}$OppositeHypotenuse=bc
#### Examples
##### Question 1
Considering the angle $\theta$θ, what is the value of the ratio $\frac{Adjacent}{Hypotenuse}$AdjacentHypotenuse ?
Think: First we need to identify which sides are the adjacent and hypotenuse with respect to angle theta. I can see that $BA$BA is the hypotenuse, $AC$AC is the opposite side and $BC$BC is the adjacent.
Do: $\frac{Adjacent}{Hypotenuse}$AdjacentHypotenuse = $\frac{BC}{AB}=\frac{5}{13}$BCAB=513
##### Question 3
Consider the angle $\theta$θ.
What is the value of the ratio $\frac{Opposite}{Adjacent}$OppositeAdjacent?
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# Chapter 4: Managing Your Money Lecture notes Math 1030 Section D
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## Transcription
1 Section D.1: Loan Basics Definition of loan principal For any loan, the principal is the amount of money owed at any particular time. Interest is charged on the loan principal. To pay off a loan, you must gradually pay down the principal. Thus, in general, every payment should include all the interest you owe plus some amount that goes toward paying off the principal. Ex.1 Suppose you borrow \$1200 at an annual interest rate APR= 12% (or 1% per month). At the end of the first month, you owe interest in the amount of If you paid only this \$12 in interest, you would still owe \$1200. That is, the loan principal would still be \$1200. In that case you would owe the same \$12 in interest the next month and this can go on forever. If you hope to make progress in paying off the loan, you need to pay part of the principal as well as interest. For example, suppose that you paid \$200 toward your loan principal each month, plus the current interest. At the end of the first month, you would pay \$200 toward principal, plus \$12 for the 1% interest you owe: Because you have paid \$200 toward principal, your new loan principal would be At the end of the second month, you would again pay \$200 toward principal and 1% interest The table shows how the calculations continue until the loan is paid after 6 months. AFTER N MONTHS PRIOR PRINCIPAL INTEREST TOTAL PAYMENT NEW PRINCIPAL 1 \$1200 1% 1200 = = 212 \$ \$1000 1% 1000 = = 210 \$800 3 \$800 1% 800 = = 208 \$600 4 \$600 1% 600 = = 206 \$400 5 \$400 1% 400 = = 204 \$200 6 \$200 1% 200 = = 202 \$0 1
2 Installment loan and loan payment formula There is nothing wrong with this method of paying off a loan, but most people prefer to pay the same total amount each month because it makes planning a budget easier. A loan that you pay off with equal regular payments is called installment loan (or amortized loan). The regualr payment amount can be computed using the loan payment formula: ( ) P APR PMT = where PMT = regular payment amount P = starting loan principal (amount borrowed) APR = annual percentage rate (as a decimal) n = number of payment periods per year Y = loan term in years [ 1 Ex.2 What is the regular payment amount in Example 1? ( n 1 + APR n ) ( ny ) ] 2
3 Principal and interest payments. Because the loan principal is gradually paid down with the installment payments, the interest due each month must also decline gradually. Thus, because the payments remain the same, the amount paid toward principal each month gradually rises. Therefore, the portions of installment loan payments going toward principal and toward interest vary as the loan is paid down. Early in the loan term, the portion going toward interest is relatively high and the portion going toward principal is relatively low. As the term proceeds, the portion going toward interest gradually decreases and the portion going toward principal gradually increases. Ex.3 Student loan. Suppose you have student loans totaling \$7500 when you graduate from college. The interest rate is APR = 9% and the term is 10 years. What are your monthly payments? How much will you pay over the lifetime of the loan? What is the total interest you will pay on the loan? 3
4 Ex.4 Principal and interest payments. For the loan in Example 3, calculate the portions of your payments that go to principal and to interest during the first 3 months. 4
5 Choices of rate and term Choices of rate and term. You will usually have several choices of interest rate and loan term when seeking a loan. Thus, you will have to evaluate your choices and make the decision that is the best for your personal situation. Ex.5 You need a \$6000 loan to buy a used car. Your bank offers a 3-year loan at 8%, a 4-year loan at 9%, and a 5-year loan at 10%. Calculate your monthly payments and total interest over the loan term with each option. 5
6 Section D.2: Credit Cards Credit card loans Credit card loans differ from installment loans in that you are not required to pay off your balance in any set period of time. Instead, you are required to make only a minimum monthly payment that generally covers all the interest but very little principal. As a result, it takes a very long time to pay off your credit card loan in a particular amount of time. You should use the loan payment formula to calculate the necessary payments. Ex.6 You have a credit card balance of \$2300 with an annual interest rate of 21%. You decide to pay off your balance in 1 year. How much will you need to pay each month? Assume you make no further credit card purchases. 6
7 Ex.7 Paul has gotten into credit card trouble. He has a balance of \$9500 and just lost his job. His credit card company allows him to suspend his payments until he finds a new job, but continues to charge interest. If it takes him a year to find a new job and his credit card company charges interest of APR = 21% compounded daily, how much will he owe when he starts his new job? 7
8 Section D.3: Mortgages Mortgages One of the most popular types of installment loans is designed specifically to help you buy a home. It is called mortgage. Mortgage interest rates generally are lower than interest rates on other types of loans because your home itself serves as a payment guarantee. If you fail to make your payments, the lender (usually a bank or a mortgage company) can take possession of your home and sell it to cover the amount loaned to you. There are several considerations in getting a home mortgage. First, the lender will probably require a down payment, typically 10% or 20% of the purchase price. Then the lender will loan you the rest of the money needed to purchase the home. Most lenders also charge fees, or closing costs, at the time you take out the loan. Closing costs can be substantial and may vary significantly between lenders, so you should be sure that you understand them. There are two types of closing costs: Direct fees, such as fees for getting the home appraised and checking your credit history, for which the lender charges a fixed dollar amount. These fees typically range from a few hundred dollars to a couple thousand dollars. Fees charged as points, where each point is 1% of the loan amount. If you are seeking a home mortgage, be sure to keep the following considerations in mind as you compare lenders: What interest rate and down payment are required for the loan? What closing costs will be charged? Watch out for fine print, such as prepayment penalties, that may make the loan more expansive than it seems on the surface. 8
9 Ex.8 You need a loan of \$100,000 to buy your new home. The bank offers a choice of a 30-year loan at an APR of 8% or a 15-year loan at 7.5%. Compare your monthly payments and total loan cost under the two options. Assume that the closing costs are the same in both cases and therefore do not affect the choice. 9
10 Ex.9 Great Bank offers a 100,000, 30-year, 8% loan with closing costs of \$500 plus 2 points. Big Bank offers a lower rate of 7.9%, but with closing costs of \$1000 plus 2 points. Evaluate the two options. 10
11 Ex.10 Continuing Example 9, suppose you have decided to go with Great Bank s lower closing costs. You learn that Great Bank actually offers two options for a 30-year loan: an 8% interest rate with 2 points or a 7.5% rate with 4 points. Evaluate your options. 11
12 Adjustable Rate Mortgages Fixed and adjustable rate mortgages Till now all examples had fixed rate mortgage in which you are guaranteed that the interest rate will not change over the life of the loan. A fixed rate mortgage is advantageous for you because your monthly payments never change. However, it poses a risk to the lender. Lenders can lessen the risk of rising interest rate by charging higher rates for longer-term loans. That is why rates generally are higher for 30-year loans than for 15-year loans. But an even lower-risk strategy for the lender is an adjustable rate mortgage (ARM) in which the interest rate you pay changes whenever prevailing rates change. Because of the reduced long-term risk to lenders, ARM generally have much lower initial interest rates than fixed interest rate loans. Most ARMs guarantee their starting interest rate for the first 6 months or 1 year, but interest rates in subsequent years move up or down to reflect prevailing rates. Most ARMs also include a rate cap that cannot be exceeded. Making a decision between a fixed rate loan and an ARM can be one of the most important financial decision of your life!!! Ex.11 You have a choice between a 30-year fixed rate loan at 8% and an ARM with a first-year rate of 5%. Neglecting compounding and changes in principal, estimate your monthly savings with the ARM during the first year on a \$100,000 loan. Suppose that the ARM rate rises to 11% by the fourth year. How will your payments be affected? 12
### Chapter 22: Borrowings Models
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### Finding the Payment \$20,000 = C[1 1 / 1.0066667 48 ] /.0066667 C = \$488.26
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## Averaging stock purchases
If prices are low, then your set investment amount will buy more shares. This is different from buying a stock several times and averaging the prices you paid. You
5 Mar 2020 4 investment strategies that can help you take advantage of low stock averaging is that it helps an investor buy low—when investments are Stock average calculator calculates the average cost of your stocks when you purchase Enter the detail price for the stock you buy and sell to get the return on 1 Mar 2020 Dollar-cost averaging takes place when a fixed dollar amount of a certain investment is purchased regularly, no matter what its share price is. 16 Dec 2019 However, rather than trying the futile task of timing the markets, you buy the asset at predetermined price points. There are both pros and cons to
## Averaging down is an investment strategy that involves buying more of a stock after its price declines, which lowers its average cost. A simple example: Let's say
If prices are low, then your set investment amount will buy more shares. This is different from buying a stock several times and averaging the prices you paid. You 1 Oct 2019 You can choose to buy your investments all at once — lump sum investing — or begin an investment schedule — dollar cost averaging. Now imagine that the price of the stock falls to \$1 per share and the investor purchases 1 additional share at the new price. The average price of the stocks in the Average down refers to a situation in which an investor purchases additional shares as a stock's price falls in an effort to add value to a portfolio. more Short Selling
### By using Dollar Cost Averaging investment method, you can buy more shares/ units Amount available for security purchase = Customer's regular investment
Dollar cost averaging—spreading an investor's stock purchases evenly over dollar amount in stocks, adjusting the number of shares purchased as stock 23 Mar 2016 To do that you would take \$2,500 out of your stock holdings each month and move it into bonds. (To keep things simple, I'm leaving investment 22 May 2018 But many of those sitting on cash missed out on huge gains in stocks. lump- sum investment over a year against 12 monthly purchases
### Average Cost Calculator You can use an average cost calculator to determine the average share price you paid for a security with multiple buys. This can be handy when averaging in on a stock purchase or determining your cost basis .
Dollar cost averaging—spreading an investor's stock purchases evenly over dollar amount in stocks, adjusting the number of shares purchased as stock 23 Mar 2016 To do that you would take \$2,500 out of your stock holdings each month and move it into bonds. (To keep things simple, I'm leaving investment 22 May 2018 But many of those sitting on cash missed out on huge gains in stocks. lump- sum investment over a year against 12 monthly purchases 4 Oct 2017 If the stock you purchased drops, don’t try to buy more shares to bring down your average buying price. Investors often try to cover their 15 Oct 2018 Dollar cost averaging is an investment strategy where an investor buys a fixed dollar amount of a security at regular intervals regardless of the 1 May 2017 Dollar-Cost Averaging is buying stocks in smaller, constant chunks of money over time, as opposed to buying all the stocks at once. By spreading
## 1 Mar 2020 Dollar-cost averaging takes place when a fixed dollar amount of a certain investment is purchased regularly, no matter what its share price is.
27 Nov 2019 The concept of rupee cost averaging lies in averaging out the cost at which you buy units of a mutual fund. The equity markets have always
Dollar cost averaging is a method of accumulating assets by purchasing a fixed dollar An automatic investment plan ensures that the predetermined amount is This method determines your transaction's cost basis by taking the average cost that you buy mutual fund shares each month through an automatic investment Instead of buying your stocks right away, in a single purchase you split it into some smaller transactions. For instance, instead of spending \$12,000 at once, you
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# act05 - Do NOT write on these sheets or take them with you...
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Do NOT write on these sheets or take them with you! The next class needs them too! PHYS-1200 PHYSICS II SPRING 2005 Class 5 Activity: Simple Circuits In this experiment, you will investigate the equivalent resistance of several combinations of resistors. You have been supplied with three resistors. Set the multimeter as an ohmmeter, and measure the resistance of each of the resistors. On your paper, record your results as shown below. To help keep track of the resistors, call the two brown resistors R 1 and R 2 , and call the one with colored bands R 3 . Record the resistance of each on your paper as follows: R 1 = _____________ R 2 = _____________ R 3 = _____________ units units units Now, connect the resistors as shown in each diagram and determine the equivalent resistance by calculation and by measurement. Then record the results on your paper, arranged as shown below. DO NOT WRITE ON THIS PAPER! Calculated
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## This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
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# Opposition of kings in endgame
Dec 28, 2013, 3:22 PM 3
Jeddah, John, Daisy & Marcus
Welcome, Daisy, and come again! and thankyou, Marcus, for introducing us.
Jeddah took on a grand teaching position for the afternoon, talking for over 4 hours and playing over the options for both white & black. Hearing you has reminded me that that's the best way of teaching, or the one I can handle, anyway: relaxed discussion, one to one.
Gathered about my laptop, we tried also a couple of videos from chess.com, where I have recently bought a membership which permits that. I thought that wasn't so successful though: too much background noise in cafe; not enuf oomf in computer; no PAUSE button; not responsive to individual needs. Got any thoughts? More videos, done a bit differently?
While Daisy moved towards delivering Jeddah a coup de grace, Marcus & I sat down to look over several things. We played out one tiny part of that most fundamental of chess-positions: king vs. king and one pawn.
(I've made a mistake, here, on white's move 3. See comment below.) This position introduces a term you will find useful in late endgame positions: opposition of kings. By this opposition, kings are both limited in their options of legal moves. Neither can move toward the other. When one king has his back to a wall, he has only two moves. So "opposition" helps you chivvy an opponent's king backwards in the following position, and then to give checkmate:
On this blue board, black's aim is to keep out of "opposition" with white to move. On moves 1 & 8, being in opposition first reduces mobility for the black king, and then reduces his height by a head.
The same is partly true of the position on the brown board. White uses opposition to reduce black's options. White wants the black king one side of the pawn's advance, so that black cannot stop that advance. Opposition of kings enables this - right until the very end. This means that the difference in which side is to move on the brown board above is crucial. But in such positions, black can also use opposition to get a draw. Let's see what happens if it's still white to move in that position on the brown board, not black. Because at the very last move below, black wants to be in opposition with the black king required to move, but with nowhere to go, at all:
This is a draw by stalemate.
Look at how carefully black must play on the 3rd move! Only one move will do; the others lose. By accurate play, black has saved half a point. Thus, in endgames the options become fewer and fewer, until we reach positions such as the ones here. Technique and knowledge become important to the chess-player in ways that are different from earlier in a game.
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Definition of
# Value
Mathematics: a number, or the result of a calculation.
Example: 3 × 4 gives the value of 12.
Money: how much something is worth.
Example: the value of this coin is one dollar.
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# Search by Topic
#### Resources tagged with Working systematically similar to Which Numbers? (2):
Filter by: Content type:
Age range:
Challenge level:
### There are 340 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### A First Product Sudoku
##### Age 11 to 14 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### How Old Are the Children?
##### Age 11 to 14 Challenge Level:
A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?"
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Seven Pots of Plants
##### Age 7 to 11 Challenge Level:
There are seven pots of plants in a greenhouse. They have lost their labels. Perhaps you can help re-label them.
### The Moons of Vuvv
##### Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Different Deductions
##### Age 7 to 11 Challenge Level:
There are lots of different methods to find out what the shapes are worth - how many can you find?
### Route Product
##### Age 7 to 11 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### Ordered Ways of Working Upper Primary
##### Age 7 to 11 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps if you have an ordered approach.
### Button-up Some More
##### Age 7 to 11 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Curious Number
##### Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
##### Age 7 to 14 Challenge Level:
I added together some of my neighbours house numbers. Can you explain the patterns I noticed?
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### More and More Buckets
##### Age 7 to 11 Challenge Level:
In this challenge, buckets come in five different sizes. If you choose some buckets, can you investigate the different ways in which they can be filled?
### Octa Space
##### Age 7 to 11 Challenge Level:
In the planet system of Octa the planets are arranged in the shape of an octahedron. How many different routes could be taken to get from Planet A to Planet Zargon?
### Pouring the Punch Drink
##### Age 7 to 11 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
##### Age 11 to 14 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Eight Queens
##### Age 7 to 11 Challenge Level:
Place eight queens on an chessboard (an 8 by 8 grid) so that none can capture any of the others.
### 1 to 8
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line.
##### Age 5 to 11 Challenge Level:
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
### Plates of Biscuits
##### Age 7 to 11 Challenge Level:
Can you rearrange the biscuits on the plates so that the three biscuits on each plate are all different and there is no plate with two biscuits the same as two biscuits on another plate?
### More Magic Potting Sheds
##### Age 11 to 14 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Wag Worms
##### Age 7 to 11 Challenge Level:
When intergalactic Wag Worms are born they look just like a cube. Each year they grow another cube in any direction. Find all the shapes that five-year-old Wag Worms can be.
### Half Time
##### Age 5 to 11 Challenge Level:
What could the half time scores have been in these Olympic hockey matches?
### Seating Arrangements
##### Age 7 to 11 Challenge Level:
Sitting around a table are three girls and three boys. Use the clues to work out were each person is sitting.
### Two Dots
##### Age 7 to 11 Challenge Level:
Place eight dots on this diagram, so that there are only two dots on each straight line and only two dots on each circle.
### Team Scream
##### Age 7 to 11 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
### Make Pairs
##### Age 7 to 11 Challenge Level:
Put 10 counters in a row. Find a way to arrange the counters into five pairs, evenly spaced in a row, in just 5 moves, using the rules.
### Crack the Code
##### Age 7 to 11 Challenge Level:
The Zargoes use almost the same alphabet as English. What does this birthday message say?
### Count the Trapeziums
##### Age 7 to 11 Challenge Level:
How many trapeziums, of various sizes, are hidden in this picture?
### Bunny Hop
##### Age 7 to 11 Challenge Level:
What is the smallest number of jumps needed before the white rabbits and the grey rabbits can continue along their path?
### Broken Toaster
##### Age 7 to 11 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Twinkle Twinkle
##### Age 7 to 14 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### Possible Pieces
##### Age 5 to 11 Challenge Level:
Can you create jigsaw pieces which are based on a square shape, with at least one peg and one hole?
### Room Doubling
##### Age 7 to 11 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Pasta Timing
##### Age 7 to 11 Challenge Level:
Nina must cook some pasta for 15 minutes but she only has a 7-minute sand-timer and an 11-minute sand-timer. How can she use these timers to measure exactly 15 minutes?
### Ice Cream
##### Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Snails' Trails
##### Age 7 to 11 Challenge Level:
Alice and Brian are snails who live on a wall and can only travel along the cracks. Alice wants to go to see Brian. How far is the shortest route along the cracks? Is there more than one way to go?
### Finding Fifteen
##### Age 7 to 11 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### Arrangements
##### Age 7 to 11 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### Polo Square
##### Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Zargon Glasses
##### Age 7 to 11 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Dart Target
##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
### Family Tree
##### Age 7 to 11 Challenge Level:
Use the clues to find out who's who in the family, to fill in the family tree and to find out which of the family members are mathematicians and which are not.
### Centred Squares
##### Age 7 to 11 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### Plate Spotting
##### Age 7 to 11 Challenge Level:
I was in my car when I noticed a line of four cars on the lane next to me with number plates starting and ending with J, K, L and M. What order were they in?
##### Age 7 to 11 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Symmetry Challenge
##### Age 7 to 11 Challenge Level:
Systematically explore the range of symmetric designs that can be created by shading parts of the motif below. Use normal square lattice paper to record your results.
### Ancient Runes
##### Age 7 to 11 Challenge Level:
The Vikings communicated in writing by making simple scratches on wood or stones called runes. Can you work out how their code works using the table of the alphabet?
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
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# The CENTRE for EDUCATION
## in MATHEMATICS and COMPUTING
cemc.uwaterloo.ca
Cayley Contest
Thursday, February 20, 2014
(in North America and South America)
Friday, February 21, 2014
(outside of North America and South America)
Time: 60 minutes 2013 University of Waterloo
Calculators are permitted
Instructions
1. Do not open the Contest booklet until you are told to do so.
2. You may use rulers, compasses and paper for rough work.
3. Be sure that you understand the coding system for your response form. If you are not sure,
ask your teacher to clarify it. All coding must be done with a pencil, preferably HB. Fill in
circles completely.
4. On your response form, print your school name and city/town in the box in the upper right
corner.
5. Be certain that you code your name, age, sex, grade, and the Contest you are
writing in the response form. Only those who do so can be counted as eligible
students.
6. This is a multiple-choice test. Each question is followed by ve possible answers marked
A, B, C, D, and E. Only one of these is correct. After making your choice, ll in the
appropriate circle on the response form.
7. Scoring: Each correct answer is worth 5 in Part A, 6 in Part B, and 8 in Part C.
There is no penalty for an incorrect answer.
Each unanswered question is worth 2, to a maximum of 10 unanswered questions.
8. Diagrams are not drawn to scale. They are intended as aids only.
9. When your supervisor tells you to begin, you will have sixty minutes of working time.
Do not discuss the problems or solutions from this contest online for the next 48 hours.
The name, grade, school and location, and score range of some top-scoring students will be
and location, and score of some top-scoring students may be shared with other mathematical
organizations for other recognition opportunities.
Scoring: There is no penalty for an incorrect answer.
Each unanswered question is worth 2, to a maximum of 10 unanswered questions.
Part A: Each correct answer is worth 5.
1. The value of 2000 80 + 200 120 is
(A) 2000 (B) 1600 (C) 2100 (D) 1860 (E) 1760
2. If (2)(3)(4) = 6x, what is the value of x?
(A) 2 (B) 6 (C) 12 (D) 3 (E) 4
3. In the diagram, three line segments intersect as shown.
The value of x is
(A) 40 (B) 60 (C) 80
(D) 100 (E) 120
40
60
x
4. At 2 p.m., Sanjay measures the temperature
to be 3
## C. He measures the temperature
every hour after this until 10 p.m. He plots
the temperatures that he measures on the
graph shown. At what time after 2 p.m. does
he again measure a temperature of 3
C?
(A) 9 p.m. (B) 5 p.m. (C) 8 p.m.
(D) 10 p.m. (E) 7 p.m.
2
4
6
8
10
2 4 6 8
Time of day (p.m.)
T
e
m
p
e
r
a
t
u
r
e
(
C
)
Temperature in Waterloo
10
0
5. If 2n + 5 = 16, the expression 2n 3 equals
(A) 8 (B) 10 (C) 18 (D) 14 (E) 7
6. When the numbers 3,
5
2
and
## 10 are listed in order from smallest to largest, the
list is
(A) 3,
5
2
,
10 (B)
5
2
, 3,
10 (C)
10,
5
2
, 3 (D)
5
2
,
10, 3 (E) 3,
10,
5
2
7. Meg started with the number 100. She increased this number by 20% and then
increased the resulting number by 50%. Her nal result was
(A) 120 (B) 240 (C) 187.5 (D) 200 (E) 180
8. In the diagram, PQR has RPQ = 90
, PQ = 10, and
QR = 26. The area of PQR is
(A) 100 (B) 120 (C) 130
(D) 60 (E) 312
P
Q
R
26
10
9. In a group of ve friends:
Amy is taller than Carla.
Dan is shorter than Eric but taller than Bob.
Eric is shorter than Carla.
Who is the shortest?
(A) Amy (B) Bob (C) Carla (D) Dan (E) Eric
10. Consider the following owchart:
INPUT Subtract 8 Divide by 2 Add 16 OUTPUT
If the OUTPUT is 32, the INPUT must have been
(A) 16 (B) 28 (C) 36 (D) 40 (E) 32
Part B: Each correct answer is worth 6.
11. A line intersects the positive x-axis and positive y-axis,
as shown. A possible equation of this line is
(A) y = 2x + 7 (B) y = 4 (C) y = 3x 5
(D) y = 5x 2 (E) y = 2x + 3
x
y
12. If x = 2y and y = 0, then (x y)(2x +y) equals
(A) 5y
2
(B) y
2
(C) 3y
2
(D) 6y
2
(E) 4y
2
13. In a factory, Erika assembles 3 calculators in the same amount of time that Nick
assembles 2 calculators. Also, Nick assembles 1 calculator in the same amount of time
that Sam assembles 3 calculators. How many calculators in total can be assembled by
Nick, Erika and Sam in the same amount of time as Erika assembles 9 calculators?
(A) 30 (B) 24 (C) 27 (D) 81 (E) 33
14. Storage space on a computer is measured in gigabytes (GB) and megabytes (MB),
where 1 GB = 1024 MB. Julia has an empty 300 GB hard drive and puts 300 000 MB
of data onto it. How much storage space on the hard drive remains empty?
(A) 72 MB (B) 720 MB (C) 7200 MB (D) 7.2 GB (E) 72 GB
15. In the 4 4 grid shown, each of the four symbols has a
dierent value. The sum of the values of the symbols in
each row is given to the right of that row. What is the
value of ?
(A) 5 (B) 6 (C) 7
(D) 8 (E) 9
26
24
27
33
16. The table shows the results of a poll which asked each student how many hamburgers
he or she ate at the end of year class party.
Number of hamburgers 0 1 2 3 4
Number of students 12 14 8 4 2
What is the average (mean) number of hamburgers eaten per student?
(A) 1.8 (B) 2 (C) 1.25 (D) 2.5 (E) 8
17. A circle with area 36 is cut into quarters and three
of the pieces are arranged as shown. What is the
perimeter of the resulting gure?
(A) 6 + 12 (B) 9 + 12 (C) 9 + 18
(D) 27 + 12 (E) 27 + 24
18. At the post oce, Sonita bought some 2 stamps and she bought ten times as many
1 stamps as 2 stamps. She also bought some 5 stamps. She did not buy any
other stamps. The total value of the stamps that she bought was 100 . How many
stamps did Sonita buy in total?
(A) 66 (B) 30 (C) 44 (D) 63 (E) 62
19. Two dierent numbers are randomly selected from the set {3, 1, 0, 2, 4} and then
multiplied together. What is the probability that the product of the two numbers
chosen is 0?
(A)
1
10
(B)
1
5
(C)
3
10
(D)
2
5
(E)
1
2
20. If wxyz is a four-digit positive integer with w = 0, the layer sum of this integer equals
wxyz +xyz +yz +z. For example, the layer sum of 4089 is 4089+089+89+9 = 4276.
If the layer sum of wxyz equals 2014, what is the value of w +x +y +z?
(A) 12 (B) 15 (C) 11 (D) 13 (E) 10
Part C: Each correct answer is worth 8.
21. In the diagram, the shape consists of seven identical cubes
with edge length 1. Entire faces of the cubes are attached
to one another, as shown. What is the distance between
P and Q?
(A)
20 (B)
26 (C)
14
(D)
18 (E)
30
P
Q
22. A ve-digit positive integer is created using each of the odd digits 1, 3, 5, 7, 9 once
so that
the thousands digit is larger than the hundreds digit,
the thousands digit is larger than the ten thousands digit,
the tens digit is larger than the hundreds digit, and
the tens digit is larger than the units digit.
How many such ve-digit positive integers are there?
(A) 12 (B) 8 (C) 16 (D) 14 (E) 10
23. Three friends are in the park. Bob and Clarise are standing at the same spot and Abe
is standing 10 m away. Bob chooses a random direction and walks in this direction
until he is 10 m from Clarise. What is the probability that Bob is closer to Abe than
Clarise is to Abe?
(A)
1
2
(B)
1
3
(C)
1
(D)
1
4
(E)
1
6
24. For each positive integer n, dene S(n) to be the smallest positive integer divisible
by each of the positive integers 1, 2, 3, . . . , n. For example, S(5) = 60. How many
positive integers n with 1 n 100 have S(n) = S(n + 4) ?
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13
25. Point P is on the y-axis with y-coordinate greater than 0 and less than 100. A circle
is drawn through P, Q(4, 4) and O(0, 0). How many possible positions for P are there
so that the radius of this circle is an integer?
(A) 2 (B) 68 (C) 66 (D) 65 (E) 67
2
0
1
4
C
a
y
l
e
y
C
o
n
t
e
s
t
(
E
n
g
l
i
s
h
)
The CENTRE for EDUCATION
in MATHEMATICS and COMPUTING
cemc.uwaterloo.ca
For students...
Thank you for writing the 2014 Cayley Contest!
In 2013, more than 65 000 students around the world registered to
write the Pascal, Cayley and Fermat Contests.
Encourage your teacher to register you for the Galois Contest
which will be written in April.
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https://coq.gitlab.io/zulip-archive/stream/237977-Coq-users/topic/match.20goal.20as.20lambda.html
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## Stream: Coq users
### Topic: match goal as lambda
#### choukh (Aug 20 2021 at 08:42):
Is there a way to match goal and turn the whole to a lambda expression.
#### choukh (Aug 20 2021 at 08:43):
Suppose I have a goal: n = n. I want to turn it to lambda n, n = n
(deleted)
#### choukh (Aug 20 2021 at 08:47):
I want it because I need the lambda expression to construct other term for proof
#### choukh (Aug 20 2021 at 08:50):
If the goal is not only about n, I should be able to specify which to bind to the lambda
#### choukh (Aug 20 2021 at 08:52):
I want to make it a Ltac so I dont need to copy and paste the goal every time
#### Kenji Maillard (Aug 20 2021 at 08:52):
ok, my first try didn't work, here is something that may correspond to what you want using 2nd order patterns in match goal
``````Goal forall n m p : nat, n = n.
Proof.
intros n m p.
(* choose the variable to capture*)
revert n.
match goal with
| [ |- forall n, @?G n ] => pose (z := G)
end.
(* revert the goal to the original state *)
intro n.
``````
#### Kenji Maillard (Aug 20 2021 at 08:54):
My first tentative didn't introduce the variable in the goal, but 2nd order pattern matching does not seem to accept variables bound in the context (e.g a pattern `[n : _ |- @?G n]`). Does anyone know why it is the case ?
#### choukh (Aug 20 2021 at 09:03):
Thank you. But what if n is already used in hypothesis?
#### choukh (Aug 20 2021 at 09:04):
Oh I got some idea to try
#### Kenji Maillard (Aug 20 2021 at 09:08):
Something like this ?
``````Goal forall n m p : nat, n = n -> n = n.
Proof.
intros n m p H.
refine (let k := n in _).
revert k.
match goal with
| [ |- let k := _ in @?G k ] => pose (z := G)
end.
hnf. (* remove the let ; or intro k; unfold k; clear k. to be more precise *)
``````
#### Olivier Laurent (Aug 20 2021 at 11:52):
@choukh is `pattern` what you are looking for?
with for example:
``````pattern n.
match goal with | |- ?t _ => pose (f := t) end.
``````
#### choukh (Aug 20 2021 at 12:34):
Yes! It's exactly what I need. Thank you very much!
Last updated: Jun 18 2024 at 21:01 UTC
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iCub-main
A short introduction to iDyn
iDyn is a library for computing kinematics and dynamics of serial-links chains of revolute joints and iCub limbs.
It started as an extension of the iKin library, with the purpose of including the dynamic description of the links, and gradually evolved until considering the whole body dynamics. Thus, starting from links (from iKinLink to iDynLink) the library provides classes for links, chain of links, and generic limbs. Of course, iCub limbs are included, with the specific kinematics and dynamics data already set (by default, the CAD parameters).
The library also provides a Newton-Euler based method to compute forces, moments, and joint torques, for each link in single or multiple interconnected chains, in presence or not of FT sensors, placed anywhere in the chain (e.g. in the middle of the chain).
The main purposes of the library are:
• computation of all iCub joint torques
• computation of external wrenches acting at the end-effector of the arms/legs of iCub
The modules using iDyn and performing these operations are WholeBodyTorqueObserver and WrenchObserver, which are the bases for the Torque Interface, Torque Control, Impedance Control, Zero-force control, external contact detection and so on. The complete and detailed documentation of iDyn code is available here.
# Library modules
The library modules are:
• RecursiveNewtonEuler: force/torque computation in a chain using Newton-Euler recursive formulation, both in the forward/backward sense
• iDynInv: estimation of force/moment measured by a virtual FT sensor placed everywhere in a chain
• iDynFwd: retrieve joint torques of limbs with single FT sensor measurements
• iDynBody: whole body dynamics, i.e. interconnection of multiple chains
• iTransform: projection of forces along the chain
# Core classes
The base classes describing serial links chains are:
• iDynChain: a chain of serial links, with kinematic and dynamic properties
• iDynLimb: a generic limb, described by a chain
Note that iDynLink and iDynChain are derived from the corresponding classes (iKinLink and iKinChain) in the iKin library, so they provide both kinematic and dynamic data and methods. iDynLimb is, as in iKin, basically a redefinition of iDynChain methods, used to protect some chain data. iDynLink is an extension of iKinLink, since it adds the dynamic parameters of the link:
• (HC,COM) : the center of mass, i.e. a roto-translational matrix defining the COM with respect to the link frame
• (m): the link mass, concentrated in the COM
• (I): the inertia matrix of the link
and the other characteristic variables used for dynamics:
• (dq/dAng, ddq/d2q/d2Ang) : joint velocity & acceleration
• (w,dw): link angular velocity & acceleration
• (ddp,ddpC): link and its COM linear acceleration
• (F, Mu): link force, moment
• (Tau): joint torque
An iDynChain is basically a list of links; the main difference with respect to iKin is that blocked links contribute to the dynamics, so there's no need to have a second list to fasten dynamic computations. However, since iDynChain inherits from iKinChain, blocked links are considered for the Jacobian computation, which is unchanged. One of the advantages of iDyn is the possibility to interconnect multiple limbs and computing also a "shared" Jacobian. To this purpose, the concept of "node" must be introduced.
iDynNode represents a virtual node where multiple limbs are connected, and may exchange kinematic and wrench information among them. Multiple limbs can be attached to the Node, but at least one must be attached.
The mutual exchange between node and limbs (full duplex) is managed used a RigidBodyTransformation class, containing the roto-translational matrix which describes the connection, and the type of flow'' of kinematic and wrench variables: from limb to node or from node to limb. One limb can be attached to a node at the base or at the end-effector: this, combined with the information flows, affects the computations in particular the application of the Newton-Euler algorithm while solving the limbs dynamic.
If a FT sensor is present inside a kinematic chain, its measurements can be exploited to initialize the wrench phase of the Newton-Euler recursive algorithm, starting from the sensor instead of the end-effector of the chain (or the final link of the chain). A iDynSensor class is then used: a generic class, which attaches a FT sensor into a iDynChain. The attach'' is specified by the link hosting the sensor, the roto-translational matrix defining the sensor frame with respect to the link frame, and the dynamical parameters of the "sub-link" between the link frame and the sensor frame (the portion of link between the sensor and the end of the link).
# iCub limbs
iDyn is a generic library that could be used for any robot, but of course iCub parts/limbs are already available to the user, pre-configured with the iCub CAD parameters. A generic iCub limb name is: iCub + part + Dyn where part is: Arm, Torso, Leg, Eye, EyeNeckRef, InertialSensor, NeckInertial. For legs and arms the part can be also followed by the tag "NoTorso": the reason is that in iKin the arm limb includes the torso links in the chain, whereas in iDyn one may choose to work on the arm only; in a similar way, legs in iKin are referred to the torso matrix, whereas in iDyn one can choose to have them without reference to the torso.
# Using iDyn
In order to use the iDyn library, make sure that the following steps are done:
1. Update YARP and iCub repositories.
2. Compile YARP and iCub (always a good practice).
3. Try first the tutorial examples, located in
icub-tutorials/src/iDyn
iDyn depends only on YARP and iKin. In order to include iDyn and use it in your software, make sure you include the following line in the CMakeLists.txt:
Of course if you are developing outside the iCub project (e.g. in contrib), you may also add these lines:
FIND_PACKAGE(YARP)
FIND_PACKAGE(ICUB)
In your code the iDyn namespace must be declared, and the header files needed must be included:
#include <iCub/iDyn/iDyn.h>
...
using namespace iCub::iDyn;
# Basic Newton-Euler algorithm
The Newton-Euler recursive algorithm is basically a set of computations, consisting in two steps: the so called "kinematic" and "wrench phase". In the first, the kinematic variables of the link (angular velocity/acceleration, linear acceleration, and so on) are computed, whereas in the latter the force, moment and joint torque are computed. Both phases must be initialized properly, with kinematic information (usually at the base of the chain) and external wrench information (usually at the end-effector of the chain).
Classic Newton-Euler computations
Kinematic phase (Forward):
initialized by .
Wrench phase (Backward):
initialized by .
The detailed description of the algorithm can be found in: L. Sciavicco, B. Siciliano, Modelling and Control of Robot Manipulators, 2nd Edition, Springer-Verlag, 2000.
Note
iDyn provides Newton-Euler recursive algorithm computations for the classical case, but also for all possible cases, i.e. when the external kinematic and wrench information are set at the beginning or at the end of the chain, or whenever external force measurements (coming from a FT sensor inside the chain) are available.
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0
# What is the percent for 1.04?
Updated: 9/19/2023
Wiki User
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104%
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Q: What is the percent for 1.04?
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### What is 71 percent of 104 percent?
71 percent of 104 percent = 73.84% 0.71 * 104% = 73.84%
### What is 25 percent of 104?
25 percent of 104 is 26.
104 = 10,400%
### What is 75 percent of 104?
75% of 104= 75% * 104= 0.75 * 104= 78
### What is 65 percent of 104?
65% of 104= 65% * 104= 0.65 * 104= 67.6
### What is 27 percent of 104?
27% of 104= 27% * 104= 0.27 * 104= 28.08
### What 52 is what percent of 104?
52 is what percent of 104:= 52 / 104= 0.5Converting decimal to a percentage:0.5 * 100 = 50%
### What percent of 104 is 52?
52 is what percent of 104:= 52 / 104= 0.5Converting decimal to a percentage:0.5 * 100 = 50%
104% is 1.04
### What is 4 percent of 104?
To find 4 percent of a number, multiply that number by 0.04. In this instance, 0.04 x 104 = 4.16. Therefore, 4 percent of 104 is equal to 4.16.
### What is 104 percent of 35?
104% of 35 = 104% * 35 = 1.04 * 35 = 36.4
### How do you get 104 from 65 percent?
Take 104/0.65 = 160.
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# Calling all Columbia Applicants for 2009!
new topic post reply Update application status
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Manager
Joined: 26 Jan 2009
Posts: 141
Schools: Yale SOM 2011
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
05 Feb 2009, 16:42
bostonsparky wrote:
smokedpotatoes2 wrote:
Has Columbia shown some love to any R2 applicants?
Define 'R2'. Do you mean the Regular Decision round as opposed to Early Decision?
Yes RD, I know its rolling but I meant those who submited on 1/07 (sorry for calling in R2)
SP
Senior Manager
Joined: 20 Apr 2008
Posts: 316
Schools: HBS Class of 2011
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
05 Feb 2009, 16:52
smokedpotatoes2 wrote:
bostonsparky wrote:
smokedpotatoes2 wrote:
Has Columbia shown some love to any R2 applicants?
Define 'R2'. Do you mean the Regular Decision round as opposed to Early Decision?
Yes RD, I know its rolling but I meant those who submited on 1/07 (sorry for calling in R2)
SP
Columbia definitely showed some love to GClubbers!
Current Student
Joined: 21 Aug 2008
Posts: 348
Schools: Fuqua '11
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
05 Feb 2009, 16:52
You can check on page 1 to see the admit/dings results. So far there are 2 admits from ED and 2 from RD(stamp / jj) from what I can discern. Lets hope there isnt a quota.
I got bored near the end of work today so I ended up running a simulation for myself based on 1000 trials if Id get into any of Kellogg, Wharton or Columbia (used the 2007-8 admit rates for each, aka I gave myself a 16% shot for Wharton, 19% for Kellogg, 15% for CBS). I got into at least one around 42% of the time...
Manager
Joined: 23 Jun 2008
Posts: 138
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
06 Feb 2009, 16:08
Has anyone gotten interview invite from Round 2 submission - 1/7?
Manager
Joined: 22 Oct 2008
Posts: 204
Schools: Chicago Booth (Re-app), LBS, UCLA Anderson, Oxford Said, INSEAD (withdrawn)
WE 1: Investment Management
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
Updated on: 09 Feb 2009, 07:37
balboa wrote:
Has anyone gotten interview invite from Round 2 submission - 1/7?
I am curious about that. Any interview invites?
Originally posted by levfin2003 on 06 Feb 2009, 17:18.
Last edited by levfin2003 on 09 Feb 2009, 07:37, edited 1 time in total.
Current Student
Joined: 09 Apr 2008
Posts: 197
Schools: Chicago Booth '11
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
06 Feb 2009, 18:12
Not that I've seen. I wouldn't expect them to come out until later in Feb. I'm guessing results will be done in March and their turnaround time from interview to decision is pretty quick.
Current Student
Joined: 27 Mar 2008
Posts: 415
Schools: Kellogg Class of 2011
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
06 Feb 2009, 22:41
Had been travelling for the last 2 weeks, so hadn't checked my status. Just found out that I got dinged. I think I was pretty ready to take a 2 year break from New York anyway.
Good luck to you guys still waiting.
Manager
Joined: 23 Jun 2008
Posts: 138
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
07 Feb 2009, 07:48
did you apply 1st or 2nd round?
Current Student
Joined: 09 Apr 2008
Posts: 197
Schools: Chicago Booth '11
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
07 Feb 2009, 08:25
Balboa--you can track all that info on the first page. Yellow was RD
Senior Manager
Joined: 20 Apr 2008
Posts: 316
Schools: HBS Class of 2011
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
08 Feb 2009, 16:15
balboa wrote:
Has anyone gotten interview invite from Round 2 submission - 1/7?
A few of the people who submitted on 1/6 received interviews. I don't recall seeing anybody from 1/7 get an interview yet. The admissions committee is definitely getting close to the many 1/7ers though!
Intern
Joined: 04 Feb 2009
Posts: 17
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 08:13
I had my interview this past week. I submitted for R2. I sent in my application on 12/30 and received initial notification 2 weeks ago. Had a bit of difficulty scheduling my interview with an Ambassador so there was a bit of a delay.
Now just waiting / sweating bullets
Manager
Joined: 23 Jun 2008
Posts: 138
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 08:55
Awesome....good luck buddy!....as for the rest of us this waiting game is a test of my patience.
Intern
Joined: 04 Feb 2009
Posts: 17
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 09:47
balboa wrote:
Awesome....good luck buddy!....as for the rest of us this waiting game is a test of my patience.
Thanks for the support! Best of luck to you as well
SVP
Joined: 04 Dec 2007
Posts: 1688
Schools: Kellogg '11
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 14:38
Well just heard back from CBS - it's a ding for me.
Good luck to everyone still waiting!
Current Student
Joined: 21 Aug 2008
Posts: 348
Schools: Fuqua '11
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 14:51
isa wrote:
Well just heard back from CBS - it's a ding for me.
Good luck to everyone still waiting!
Same here I am afraid Brutal day. Although considering I've had the same result as you on 4 out of 4 schools I am that much more optimistic for Kellogg
SVP
Joined: 04 Dec 2007
Posts: 1688
Schools: Kellogg '11
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 14:56
bostonsparky wrote:
isa wrote:
Well just heard back from CBS - it's a ding for me.
Good luck to everyone still waiting!
Same here I am afraid Brutal day. Although considering I've had the same result as you on 4 out of 4 schools I am that much more optimistic for Kellogg
wow we do have similar results don't we? I'm keeping fingers crossed for you for Kellogg!
VP
Joined: 22 Oct 2006
Posts: 1422
Schools: Chicago Booth '11
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 15:07
isa wrote:
Well just heard back from CBS - it's a ding for me.
Good luck to everyone still waiting!
sorry to hear that...you can't get into every school I guess
Senior Manager
Joined: 20 Apr 2008
Posts: 316
Schools: HBS Class of 2011
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
09 Feb 2009, 15:09
I am sorry guys. But look on the bright side, both of you have other fantastic options!
Intern
Joined: 30 Dec 2008
Posts: 14
Schools: R2: HBS, Columbia, Wharton, BU (\$\$)
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
10 Feb 2009, 15:10
This is weird. I got an email saying my status had changed, but when I logged in it looked just like it has since it went complete a few weeks ago. Has anyone else had this happen?
Application Status
Status Code PEND - Pending Decision
Message Your application is complete and has been placed under review .
You can expect to hear from us with either a final decision or an invitation to interview within 12 weeks from this date.
Please keep in mind that an invitation to interview can come at any point in the evaluation process.
Submitted Date 05 Jan 2009
Fee Received Date 05 Jan 2009
Current Student
Joined: 10 Nov 2008
Posts: 177
Re: Calling all Columbia Applicants for 2009! [#permalink]
### Show Tags
10 Feb 2009, 15:32
I would check back in an hour. Sometimes the system takes longer to update I guess. Good luck though.
Re: Calling all Columbia Applicants for 2009! &nbs [#permalink] 10 Feb 2009, 15:32
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# Calling all Columbia Applicants for 2009!
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# Events & Promotions
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# What is Coordinate chart: Definition and 17 Discussions
In topology, a branch of mathematics, a topological manifold is a topological space which locally resembles real n-dimensional Euclidean space. Topological manifolds are an important class of topological spaces, with applications throughout mathematics. All manifolds are topological manifolds by definition. Other types of manifolds are formed by adding structure to a topological manifold (e.g. differentiable manifolds are topological manifolds equipped with a differential structure). Every manifold has an "underlying" topological manifold, obtained by simply "forgetting" the added structure.
View More On Wikipedia.org
1. ### I Time dilation vs Differential aging vs Redshift
Hi, I would like to ask for a clarification about the terms time dilation vs differential aging vs gravitational redshit. As far as I can tell, time dilation is nothing but the rate of change of an object's proper time ##\tau## w.r.t. the coordinate time ##t## of a given coordinate chart (aka...
2. ### I ##SL(2,\mathbb R)## Lie group as manifold
Hi, consider the set of the following parametrized matrices $$\begin{bmatrix} 1+a & b \\ c & \frac {1 + bc} {1 + a} \\ \end{bmatrix}$$ They are member of the group ##SL(2,\mathbb R)## (indeed their determinant is 1). The group itself is homemorphic to a quadric in ##\mathbb R^4##. I believe...
3. ### I Schwarzschild spacetime in Kruskal coordinates
As explained here in Kruskal coordinates the line element for Schwarzschild spacetime is: $$ds^2 = \frac{32 M^3}{r} \left( – dT^2 + dX^2 \right) + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$$ My simple question is: why in the above line element are involved 5 coordinates and not just...
4. ### I Wald synchronous reference frame proof
Hi, on Wald's book on GR there is a claim at pag. 43 about the construction of synchronous reference frame (i.e. Gaussian coordinate chart) in a finite region of any spacetime. In particular he says: $$n^b\nabla_b (n_aX^a)=n_aX^b\nabla_b \, n^a$$Then he claims from Leibnitz rule the above equals...
5. ### B Further Understanding Simultaneity Conventions
Summary Almost a year ago, I created a post titled “Understanding the phrase 'simultaneity convention'”. The answers included requirements for defining a simultaneity convention. But some simultaneity conventions, while meeting all the requirements, still appear problematic. What am I missing...
6. ### I Einstein Definition of Simultaneity for Langevin Observers
Hi, reading this old thread Second postulate of SR quiz question I'd like to ask for a clarification on the following: Here the Einstein definition of simultaneity to a given event on the Langevin observer's worldline locally means take the events on the 3D spacelike orthogonal complement to...
7. ### I Clarification about submanifold definition in ##\mathbb R^2##
Hi, a clarification about the following: consider a smooth curve ##γ:\mathbb R→\mathbb R^2##. It is a injective smooth map from ##\mathbb R## to ##\mathbb R^2##. The image of ##\gamma## (call it ##\Gamma##) is itself a smooth manifold with dimension 1 and a regular/embedded submanifold of...
8. ### I Coord. Time Vector Field: Schwarzschild vs Gullstrand-Painleve
Hi, I was reading this insight schwarzschild-geometry-part-1 about the transformation employed to rescale the Schwarzschild coordinate time ##t## to reflect the proper time ##T## of radially infalling objects (Gullstrand-Painleve coordinate time ##T##). As far as I understand it, the vector...
15. ### Vectors in Tangent Space to a Manifold Independent of Coordinate Chart
In Nakahara's book, "Geometry, Topology and Physics" he states that it is, by construction, clear from the definition of a vector as a differential operator [itex] X[\itex] acting on some function [itex]f:M\rightarrow\mathbb{R}[\itex] at a point [itex]p\in M[\itex] (where [itex]M[\itex] is an...
16. ### Do Coordinate Charts Alone Identify a Unit Circle?
Hi. I have been looking at the coordinate charts for the unit circle x^2 + y^2 = 1. In the notes I have the circle is split into 4 coordinate charts the first being - ##U_1## : x>0 , ##A_1## = y (PS without the symbols tab I have used A for the letter phi ) There are 3...
17. ### Coordinate Chart on Manifold: What is $\mathbb{R}^{n}$?
In defining a coordinate chart, \left ( U,\phi \right ), U \in M, \phi : U \to \mathbb{R}^{n}, on a manifold M, what exactly is \mathbb{R}^{n}: the set of all n-tuples, a topological space, a metric space, a vector space, Euclidean space conceived of as an inner product space, Euclidean...
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データ解析基礎論 回帰分析３
```dat <- read.csv("http://www.matsuka.info/data_folder/hwsk8-17-6.csv")
dat.lm<-lm(ani~otouto, dat)
plot(hatvalues(dat.lm))
dat.lm\$residuals
x = dat\$otouto
mean.x = mean(x)
h = 1/length(x)+(x-mean.x)^2/sum((x - mean.x)^2)
hatvalues(dat.lm)
e.std = e/(slm\$sigma*sqrt(1-h))
rstandard(dat.lm)
(e^2*h)/(2*slm\$sigma^2*(1-h)^2)
cooks.distance(dat.lm)
plot(dat)
dat.lm<-lm(sales~., data=dat)
install.packages("DAAG")
library(DAAG)
vif(dat.lm)
lm.price<-lm(price~material+design+dump, data=dat)
p.rsq = summary(lm.price)\$r.squared
vif.p = 1/(1-p.rsq)
summary(dat.lm)
plot(dat,pch=20,cex=2)
dat.lm1<-lm(absence~interest,dat)
dat.lm2<-lm(study~interest+knowledge,dat)
```
院 認識情報解析
```library(rjags)
source("http://peach.l.chiba-u.ac.jp/course_folder/HDI_revised.txt")
x = dat\$height
y = dat\$weight
modelS.txt = "
data{
Ntotal <- length(y)
xm <- mean(x)
ym <- mean(y)
xsd <- sd(x)
ysd <- sd(y)
for (i in 1:length(y)){
zx[i] <- (x[i] - xm)/xsd
zy[i] <- (y[i] - ym)/ysd
}
}
model{
for (i in 1:Ntotal){
zy[i] ~ dt(zbeta0 + zbeta1 * zx[i], 1/zsigma^2, nu)
}
zbeta0 ~ dnorm(0, 1/10^2)
zbeta1 ~ dnorm(0, 1/10^2)
zsigma ~ dunif(1.03E-3, 1.03E+3)
nu <- nuMinusOne + 1
nuMinusOne ~ dexp(1/29.0)
#Transfrom to original scale:
beta1 <- zbeta1 * ysd/xsd
beta0 <- zbeta0 * ysd + ym -zbeta1*xm*ysd/xsd
sigma <- zsigma * ysd
}
"
writeLines(modelS.txt, "modelS.txt")
dataList = list(x = x, y = y)
jagsModel =jags.model("modelS.txt", data = dataList, n.chains = 3, n.adapt = 500)
update(jagsModel, 500)
codaSamples = coda.samples(jagsModel, variable.names = c("beta1", "beta0", "sigma","zbeta0","zbeta1"), n.iter = ((10000*1)/1), n.adapt = 500)
mcmcMat<-as.matrix(codaSamples)
plot(dat\$height,dat\$weight,xlab="height",ylab ='weight',pch=20,col="orange",cex=5)
par(mfrow=c(2,2))
HDI.plot(mcmcMat[,1])
plot(mcmcMat[,2],mcmcMat[,1], xlab='B1',ylab="B0",pch=20, col='orange')
plot(mcmcMat[,1],mcmcMat[,2], xlab='B0',ylab="B1",pch=20, col='orange')
HDI.plot(mcmcMat[,2])
par(mfrow=c(2,2))
HDI.plot(mcmcMat[,4])
plot(mcmcMat[,5],mcmcMat[,4], xlab='B1',ylab="B0",pch=20, col='orange')
plot(mcmcMat[,4],mcmcMat[,5], xlab='zB0',ylab="zB1",pch=20, col='orange')
HDI.plot(mcmcMat[,5])
n.mcmc = nrow(mcmcMat)
par(mfrow=c(1,1))
temp.x = c(0,80)
temp.y = mcmcMat[1,1]+mcmcMat[1,2]*temp.x
plot(temp.x,temp.y,type='l',lwd=2,col="orange",xlab="height",ylab='weight')
idx = sample(1:n.mcmc,100)
for (i.plot in 1:length(idx)){
temp.y = mcmcMat[idx[i.plot],1]+mcmcMat[idx[i.plot],2]*temp.x
lines(temp.x,temp.y,lwd=2,col="orange")
}
points(dat\$height,dat\$weight,pch=20,cex=4)
par(mfrow=c(1,1))
n2plot = 100
temp.x = c(0,80)
mean.set = seq(50,80,5)
idx=sample(1:nrow(mcmcMat),n2plot)
plot(x,y,xlim=c(50,80),ylim=c(0,400))
for (i_sample in 1:n2plot) {
temp.y = mcmcMat[idx[i_sample],1]+mcmcMat[idx[i_sample],2]*temp.x
lines(temp.x,temp.y,lwd=2,col="orange")
for (i.means in 1:length(mean.set)){
means = mcmcMat[idx[i_sample],1]+mcmcMat[idx[i_sample],2]*mean.set[i.means]
y.seq = seq(0,400,length.out=1000)
dens.y=dt((y.seq-means)/mcmcMat[idx[i_sample],3],29)
dens.y=dens.y/max(dens.y)
lines(dens.y,y.seq,type='l',col="orange")
lines(-dens.y+mean.set[i.means],y.seq,type='l',col="orange")
}
}
points(x,y,pch=20,cex=3)
model.txt = "
data{
Ntotal <- length(y)
xm <- mean(x)
ym <- mean(y)
xsd <- sd(x)
ysd <- sd(y)
for (i in 1:length(y)){
zx[i] <- (x[i] - xm) / xsd
zy[i] <- (y[i] - ym) / ysd
}
}
model{
for (i in 1:Ntotal){
zy[i] ~ dt( zbeta0[s[i]] + zbeta1[s[i]] * zx[i], 1 / zsigma^2, nu)
}
for (j in 1:Nsubj){
zbeta0[j] ~ dnorm( zbeta0mu, 1/(zbeta0sigma)^2)
zbeta1[j] ~ dnorm( zbeta1mu, 1/(zbeta1sigma)^2)
}
zbeta0mu ~ dnorm(0, 1/(10)^2)
zbeta1mu ~ dnorm(0, 1/(10)^2)
zsigma ~ dunif(1.0E-3, 1.0E+3)
zbeta0sigma ~ dunif(1.0E-3, 1.0E+3)
zbeta1sigma ~ dunif(1.0E-3, 1.0E+3)
nu <- nuMinusOne + 1
nuMinusOne ~ dexp(1/29.0)
for (j in 1:Nsubj){
beta1[j] <- zbeta1[j] * ysd / xsd
beta0[j] <- zbeta0[j] * ysd + ym - zbeta1[j] * xm * ysd / xsd
}
beta1mu <- zbeta1mu * ysd / xsd
beta0mu <- zbeta0mu * ysd + ym -zbeta1mu * xm * ysd / xsd
sigma <- zsigma * ysd
}
"
writeLines(model.txt, "model.txt")
x = dat\$X
y = dat\$Y
s = dat\$Subj
dataList = list(x = x , y = y ,s = s , Nsubj = max(s))
jagsModel =jags.model("model.txt", data = dataList, n.chains = 3, n.adapt = 500)
update(jagsModel, 500)
codaSamples = coda.samples(jagsModel, variable.names = c("beta1mu", "beta0mu", "sigma", "beta0","beta1"),
n.iter = ((10000*1)/1), n.adapt = 500)
mcmcMat<-as.matrix(codaSamples)
par(mfrow=c(1,1))
temp.x = c(40,90)
temp.y = mcmcMat[1,"beta0mu"]+mcmcMat[1,"beta1mu"]*temp.x
plot(temp.x,temp.y,type='l',lwd=2,col="orange",xlab="height",ylab='weight',xlim=c(40,90),ylim=c(40,260))
idx = sample(1:nrow(mcmcMat),100)
for (i.plot in 1:length(idx)){
temp.y = mcmcMat[idx[i.plot],"beta0mu"]+mcmcMat[idx[i.plot],"beta1mu"]*temp.x
lines(temp.x,temp.y,lwd=2,col="orange")
}
points(dat\$X,dat\$Y,pch=20,cex=4)
par(mfrow=c(5,5))
temp.x = c(40,90)
for (i.s in 1:25){
temp.y = mcmcMat[1,i.s]+mcmcMat[1,i.s+26]*temp.x
plot(temp.x,temp.y,type='l',lwd=2,col="orange",xlab="height",ylab='weight',xlim=c(40,90),ylim=c(40,260))
idx = sample(1:nrow(mcmcMat),100)
for (i.plot in 1:length(idx)){
temp.y = mcmcMat[idx[i.plot],i.s]+mcmcMat[idx[i.plot],i.s+26]*temp.x
lines(temp.x,temp.y,lwd=2,col="orange")
}
points(dat\$X[which(dat\$Subj==i.s)],dat\$Y[which(dat\$Subj==i.s)],pch=20,cex=2)
}
# ch 18
par(mfrow=c(1,1))
plot(dat[,c(2,5,8)],pch=20,cex=3)
y = dat\$SATT
x = as.matrix(cbind(dat\$Spend,dat\$PrcntTake))
dataList = list(x = x ,y = y, Nx = dim(x)[2], Ntotal = dim(x)[1])
model.txt = "
data{
ym <- mean(y)
ysd <- sd(y)
for (i in 1:Ntotal){
zy[i] <- (y[i] - ym) / ysd
}
for (j in 1:Nx){
xm[j] <- mean(x[,j])
xsd[j] <- sd(x[,j])
for (i in 1:Ntotal){
zx[i,j] <- (x[i,j] - xm[j])/xsd[j]
}
}
}
model{
for( i in 1:Ntotal){
zy[i] ~ dt( zbeta0 + sum( zbeta[1:Nx] * zx[i, 1:Nx]), 1/zsigma^2, nu)
}
zbeta0 ~ dnorm(0, 1/2^2)
for ( j in 1:Nx){
zbeta[j] ~ dnorm(0, 1/2^2)
}
zsigma ~ dunif(1.0E-5, 1.0E+1)
nu <- nuMinusOne + 1
nuMinusOne ~ dexp(1/29.0)
beta[1:Nx] <- ( zbeta[1:Nx] / xsd[1:Nx] ) * ysd
beta0 <- zbeta0 * ysd + ym - sum( zbeta[1:Nx] * xm[1:Nx] / xsd[1:Nx]) * ysd
sigma <- zsigma * ysd
}
"
writeLines(model.txt, "model.txt")
jagsModel =jags.model("model.txt", data = dataList, n.chains = 3, n.adapt = 500)
update(jagsModel, 500)
codaSamples = coda.samples(jagsModel, variable.names = c("beta[1]", "beta[2]", "beta0", "sigma"), n.iter = ((10000*1)/1), n.adapt = 500)
mcmcMat<-as.matrix(codaSamples)
plot(as.data.frame(mcmcMat))
par(mfrow=c(1,2))
HDI.plot(mcmcMat[,2])
HDI.plot(mcmcMat[,3])
x = as.matrix(cbind(dat\$Spend,dat\$PrcntTake,dat\$Spend*dat\$PrcntTake))
dataList = list(x = x ,y = y, Nx = dim(x)[2], Ntotal = dim(x)[1])
jagsModel =jags.model("model.txt", data = dataList, n.chains = 3, n.adapt = 500)
update(jagsModel, 500)
codaSamples = coda.samples(jagsModel, variable.names = c("beta[1]", "beta[2]", "beta[3]", "beta0", "sigma"), n.iter = ((10000*1)/1), n.adapt = 500)
mcmcMat<-as.matrix(codaSamples)
par(mfrow=c(1,3))
HDI.plot(mcmcMat[,2])
HDI.plot(mcmcMat[,3])
HDI.plot(mcmcMat[,4])
empHDI(mcmcMat[,2])
par(mfrow=c(1,1))
perT = seq(0,80,5)
plot(x=c(perT[1],perT[1]),empHDI(mcmcMat[,2]+mcmcMat[,4]*perT[1]),
xlim=c(-5,85),ylim=c(-15,35),type='l',lwd=3,col="orange",
xlab = c("value on Percent Take"),ylab="Slope on Spend")
for (i.plot in 2:length(perT)){
lines(c(perT[i.plot],perT[i.plot]),empHDI(mcmcMat[,2]+mcmcMat[,4]*perT[i.plot]),
lwd=3,col="orange")
}
abline(h=0,lwd=2,lty=2)
par(mfrow=c(1,1))
perT = seq(0,10,0.5)
plot(x=c(perT[1],perT[1]),empHDI(mcmcMat[,3]+mcmcMat[,4]*perT[1]),
xlim=c(-0.5,11),ylim=c(-5,1),type='l',lwd=3,col="orange",
xlab = c("value on Spent"),ylab="Slope on %Take")
for (i.plot in 2:length(perT)){
lines(c(perT[i.plot],perT[i.plot]),empHDI(mcmcMat[,3]+mcmcMat[,4]*perT[i.plot]),
lwd=3,col="orange")
}
abline(h=0,lwd=2,lty=2)
```
認知情報解析 実習問題
```# 修正済み
# 時間がかかる場合は、繰り返し回数を減らしてください。
# 効率が悪いと思うので、適宜変更してください。
temp.state = expand.grid(loc1 = 0:2,loc2=0:2,loc3=0:2,
loc4 = 0:2,loc5=0:2,loc6=0:2,
loc7 = 0:2,loc8=0:2,loc9=0:2)
temp.state = = expand.grid(rep(list(0:2),9))
n.ones = rowSums(temp.state == 1 )
n.twos = rowSums(temp.state == 2 )
omitTwo = which(n.ones < n.twos)
omitOne = which((n.ones-1 ) > n.twos)
omitUniq = unique(c(omitOne, omitTwo))
state = temp.state[-omitUniq,]
poss.act = apply(state, 1, function(x) which(x==0))
temp.win = matrix(1:9,3)
win.idx = matrix(c(temp.win[1,],temp.win[2,],temp.win[3,],
temp.win[,1],temp.win[,2],temp.win[,3],
diag(temp.win),
diag(temp.win[3:1,])),ncol=3,byrow=T)
idx1 = c()
idx2 = c()
idxCM = c()
for (i.win in 1:nrow(win.idx)){
idx.temp = apply(state, 1, function(x) sum(x[win.idx[i.win,]]==1)==3)
idx1 = c(idx1, which(idx.temp))
idxCM.temp = apply(state, 1, function(x) sum(x[win.idx[i.win,]]==1)==2)
idxCM = c(idxCM, which(idxCM.temp))
idx.temp = apply(state, 1, function(x) sum(x[win.idx[i.win,]]==2)==3)
idx2 = c(idx2, which(idx.temp))
}
n0=apply(state,1,function(x) length((which(x==0))))
tie = which(n0==0)
Q = list()
V = list()
rew.sum = list()
rew.count = list()
policy = list()
for (i.state in 1:nrow(state)){
Q[[i.state]] = rep(0,length(poss.act[[i.state]]))
V[[i.state]] = rep(0,length(poss.act[[i.state]]))
rew.sum[[i.state]] = rep(0,length(poss.act[[i.state]]))
rew.count[[i.state]] = rep(0,length(poss.act[[i.state]]))
policy[[i.state]] = rep(1/length(poss.act[[i.state]]),length(poss.act[[i.state]]))
}
R.W = 10
R.T = 5
R.L = -10
gamma = 1
epsilon = 0.05
eta = 1
ck.result <- function(st.idx, idx1, idx2, tie){
term = F
rew = 0
result = "not terminal"
if (match(st.idx ,idx1, nomatch = 0)!=0){
rew = R.W
term = T
result = "win"
} else if (match(st.idx ,idx2, nomatch = 0)!=0){
rew = R.L
term = T
result = "lose"
} else if (match(st.idx ,tie, nomatch = 0)!=0){
rew = R.T
term = T
result = "tie"
}
return(list(rew = rew, term = term, result = result))
}
n.rep = 10000
game.hist = rep(0,n.rep)
# main loop
for (i.rep in 1:n.rep){
st.idx = 1
term = F
state.temp = rep(0,9)
state.hist1 = c()
state.hist2 = c()
repeat {
# playing game
if (length(poss.act[[st.idx]])==1){
act1 = poss.act[[st.idx]]
} else{
p.act = exp(eta*Q[[st.idx]])/sum(exp(eta*Q[[st.idx]]))
act1 = sample(poss.act[[st.idx]],1, prob = p.act)
}
state.hist1 = rbind(state.hist1,c(st.idx, act1))
state.temp[act1] = 1
st.idx = which(apply(state, 1, function(x) sum(x==state.temp) )==9)
res = ck.result(st.idx, idx1, idx2, tie)
if (res\$term == T){
rew = res\$rew
break
}
p.act = exp(eta*Q[[st.idx]])/sum(exp(eta*Q[[st.idx]]))
act2 = sample(poss.act[[st.idx]],1, prob = policy[[st.idx]])
state.hist2 = rbind(state.hist2,c(st.idx, act2))
state.temp[act2] = 2
st.idx = which(apply(state, 1, function(x) sum(x==state.temp) )==9)
res = ck.result(st.idx, idx1, idx2, tie)
if (res\$term == T){
rew = res\$rew
break
}
}
# update Q & policy
game.hist[i.rep] = res\$result!="lose"
n.st = nrow(state.hist1)
#print(res\$result)
if (i.rep%%100==0){print(i.rep)}
for (i.st in 1:n.st){
act.idx = which(poss.act[[state.hist1[i.st,1]]]==state.hist1[i.st,2])
rew.sum[[state.hist1[i.st,1]]][act.idx] = rew.sum[[state.hist1[i.st,1]]][act.idx] + rew
rew.count[[state.hist1[i.st,1]]][act.idx] = rew.count[[state.hist1[i.st,1]]][act.idx] + 1
Q[[state.hist1[i.st,1]]][act.idx] = rew.sum[[state.hist1[i.st,1]]][act.idx] / rew.count[[state.hist1[i.st,1]]][act.idx]
}
}
# plotting results
library(pracma)
game.hist.smooth = movavg(game.hist, 400, type="s")
plot(game.hist.smooth,type='l')
```
2019 データ解析基礎論A 回帰分析2
```# two sample t-test
boxplot(dat\$ani, dat\$otouto, col = c("skyblue", "orange"),ylab = "Score",
axnt ="n")
axis(1, at = c(1,2),labels = c("Ani", "Otouto"))
t.test(dat\$ani, dat\$otouto, var.equal=T)
dat2<-data.frame(score=c(dat\$ani,dat\$otouto),
order=c(rep("ani",10),rep("otouto",10)))
plot(dat2\$score~as.numeric(dat2\$order), pch=20, xlab="order",
ylab="score", xlim=c(0.5,2.5), cex=2, xaxt="n")
axis(1, at = c(1,2),labels = c("Ani", "Otouto"))
dat2.lm<-lm(score~order,data=dat2)
abline(dat2.lm,col='red',lwd=3)
# one sample t-test
dat.D = dat\$ani - dat\$otouto
boxplot(dat.D,col="skyblue",ylab="Difference")
t.test(dat.D)
plot(dat.D~rep(1,10), pch=20, xlab="", ylab="Difference", cex=3, xaxt="n")
dat.D.lm<-lm(dat.D~1)
abline(dat.D.lm,col='red',lwd=3)
# LM with 3 or more categories
dat2<-data.frame(result=dat\$result, method=dat\$method,
c1=c(rep(0,8),rep(1,8),rep(0,16)),
c2=c(rep(0,16),rep(1,8),rep(0,8)),
c3=c(rep(0,24),rep(1,8)))
dat2.lm<-lm(result~c1+c2+c3,data=dat2)
dat.lm <- lm(result~method, dat)
dat3<-data.frame(result=dat\$result,
c1=c(rep(-3,8), rep(1,24)),
c2=c(rep(0,8),rep(-2,8),rep(1,16)),
c3=c(rep(0,16),rep(-1,8),rep(1,8)))
# trend analysis
plot(dat\$result~dat\$duration,data=dat[dat\$method=="method.X",])
result<-dat\$result[dat\$method=="method.X"]
CL=c(rep(-3,5),rep(-1,5),rep(1,5),rep(3,5))
CQ=c(rep(-1,5),rep(1,5),rep(1,5),rep(-1,5))
CC=c(rep(-3,5),rep(1,5),rep(-1,5),rep(3,5))
trend.lm<-lm(result~CL+CQ+CC)
trend.lm2 <- lm(result~duration, contrasts=list(duration = "contr.poly"),data=dat.x)
summary(trend.lm2)
# trend analysis - numeric predictor
set.seed(15)
x = runif(50,0,30)
y = x^0.5 + rnorm(50,9,0.2)
dat.xy = data.frame(y=y,x=x, xsq = x^2)
plot(x,y, xlab = "Hours Studies", ylab="score",
pch=20, cex=2, col="orange")
summary(lm(y~x+xsq, data=dat.xy))
# regression diagnostic
dat.lm01<-lm(sales~price, data=as.data.frame(scale(dat)))
plot(dat.lm01,which=1)
plot(dat.lm01,which=2)
norm.vars=rnorm(300)
qqnorm(norm.vars)
qqline(norm.vars,col='red',lwd=2)
unif.vars=runif(300)
qqnorm(unif.vars)
qqline(unif.vars,col='green',lwd=2)
plot(sort(unif.vars),sort(unif.vars))
plot(sort(norm.vars),sort(unif.vars))
par(mfrow=c(2,2))
plot(dat.lm01)
plot(dat)
dat.lm<-lm(sales~., data=dat)
install.packages("DAAG")
library(DAAG)
vif(dat.lm)
lm.price<-lm(price~material+design+dump, data=dat)
p.rsq = summary(lm.price)\$r.squared
vif.p = 1/(1-p.rsq)
```
院：認識情報解析 T-test & ANOVA
```source("http://www.matsuka.info/univ/course_folder/HDI_revised.txt")
dat2sd <- dat[dat\$Group=="Smart Drug",]
dataList <- list(y = dat2sd\$Score, Ntotal = nrow(dat2sd),
meanY = mean(dat2sd\$Score), sdY = sd(dat2sd\$Score))
model.txt = "
model {
for (i in 1:Ntotal){
y[i] ~ dnorm(mu, 1/sigma^2)
}
mu ~ dnorm(meanY, 1/(100*sdY)^2)
sigma ~ dunif(sdY/1000, sdY*1000)
}"
writeLines(model.txt, "model.txt")
parameters = c( "mu" , "sigma")
library("rjags")
jagsModel = jags.model( "model.txt", data=dataList, n.chains=3, n.adapt=500 )
update( jagsModel , n.iter=1000)
codaSamples = coda.samples( jagsModel , variable.names=parameters, n.iter=10000, thin=5)
traceplot(codaSamples)
gelman.plot(codaSamples)
mcmcMat<-as.matrix(codaSamples)
HDI.plot(mcmcMat[,1])
HDI.plot(mcmcMat[,2])
hist(dat2sd\$Score, probability = T)
x.temp = seq(40,220,0.1)
n.plot = 100
rand.order = sample(1:nrow(mcmcMat),n.plot)
for (i.plot in 1:n.plot){
y.temp = dnorm(x.temp, mean = mcmcMat[rand.order[i.plot],1],
sd = mcmcMat[rand.order[i.plot],2])
lines(x.temp, y.temp, type='l',col="orange")
}
#y = c(-2:2,15);Ntotal = length(y);meanY = mean(y); sdY = sd(y)
#dataList = list(y= y, Ntotal=Ntotal, meanY = meanY, sdY = sdY)
dataList <- list(y = dat2sd\$Score, Ntotal = nrow(dat2sd),
meanY = mean(dat2sd\$Score), sdY = sd(dat2sd\$Score))
model.txt ="
model {
for ( i in 1:Ntotal ) {
y[i] ~ dt( mu , 1/sigma^2 , nu )
}
mu ~ dnorm( meanY , 1/(100*sdY)^2 )
sigma ~ dunif( sdY/1000 , sdY*1000 )
nu <- nuMinusOne+1
nuMinusOne ~ dexp(1/29)
}"
writeLines(model.txt, "model.txt")
parameters = c( "mu" , "sigma", "nu")
jagsModel = jags.model( "model.txt", data=dataList, n.chains=3, n.adapt=1000 )
update( jagsModel , n.iter=1000)
codaSamples = coda.samples( jagsModel , variable.names=parameters, n.iter=10000, thin=5)
#traceplot(codaSamples)
#gelman.plot(codaSamples)
mcmcMat<-as.matrix(codaSamples)
HDI.plot(mcmcMat[,1])
HDI.plot(mcmcMat[,2])
dataList <- list(y = dat\$Score, Ntotal = nrow(dat),
meanY = mean(dat\$Score), sdY = sd(dat\$Score),
Ngroup = length(unique(dat\$Group)),
group = dat\$Group)
model.txt ="
model {
for ( i in 1:Ntotal ) {
y[i] ~ dt( mu[group[i]] , 1/sigma[group[i]]^2 , nu )
}
for (j in 1:Ngroup){
mu[j] ~ dnorm( meanY , 1/(100*sdY)^2 )
sigma[j] ~ dunif( sdY/1000 , sdY*1000 )
}
nu <- nuMinusOne+1
nuMinusOne ~ dexp(1/29)
}"
writeLines(model.txt, "model.txt")
parameters = c( "mu" , "sigma", "nu")
jagsModel = jags.model( "model.txt", data=dataList, n.chains=3, n.adapt=1000 )
update( jagsModel , n.iter=1000)
codaSamples = coda.samples( jagsModel , variable.names=parameters, n.iter=10000, thin=5)
#traceplot(codaSamples)
#gelman.plot(codaSamples)
mcmcMat<-as.matrix(codaSamples)
HDI.plot(mcmcMat[,1]-mcmcMat[,2])
HDI.plot(mcmcMat[,2])
y=dat\$Longevity
yMean = mean(y)
ySD = sd(y)
Ntotal = length(y)
MeanSD2gamma <- function( mean, sd ) {
shape = mean^2 / sd^2
rate = mean / sd^2
return(data.frame(shape,rate))
}
ModeSD2gamma <- function( mode, sd ) {
rate = ( mode + sqrt( mode^2 + 4 * sd^2 ) )/( 2 * sd^2 )
shape = 1 + mode * rate
return(data.frame(shape,rate))
}
temp.gamma = ModeSD2gamma( mode=sd(y)/2 , sd=2*sd(y) )
gShape = temp.gamma\$shape
gRate = temp.gamma\$rate
x=as.numeric(dat\$CompanionNumber)
Ngroup = length(unique(x))
xc=dat\$Thorax
xcMean=mean(xc)
dataList = list(
y = y ,
x = x ,
Ntotal = Ntotal ,
Ngroup = Ngroup ,
yMean = yMean ,
ySD = ySD ,
gShape = gShape,
gRate = gRate
)
model.txt="
model {
for ( i_data in 1:Ntotal ) {
y[ i_data ] ~ dnorm( a0 + a[ x[ i_data ] ], 1/ySigma^2)
}
ySigma ~ dunif( ySD/100, ySD*10 )
a0 ~ dnorm( yMean, 1/(ySD*5)^2)
for ( i_group in 1:Ngroup ) {
a[ i_group ] ~ dnorm(0.0, 1/aSigma^2)
}
aSigma ~ dgamma( gShape, gRate )
for ( i_group in 1:Ngroup ) { m[ i_group ] <- a0 + a[ i_group ] }
b0 <- mean( m[ 1:Ngroup ] )
for (i_data in 1:Ngroup ) { b[ i_data ] <- m[ i_data ] - b0 }
}
"
writeLines(model.txt, "model.txt")
parameters = c( "b0" , "b" , "ySigma", "aSigma" )
jagsModel = jags.model( "model.txt", data=dataList, n.chains=3, n.adapt=500 )
update( jagsModel , n.iter=1000)
codaSamples = coda.samples( jagsModel , variable.names=parameters, n.iter=10000, thin=5)
mcmcMat<-as.matrix(codaSamples)
# fig 19.3
plot(x,dat\$Longevity,xlim=c(0,5.1),ylim=c(0,120))
for (i_c in 1:100) {
sd=mcmcMat[i_c,"ySigma"]
lim=qnorm(c(0.025,0.975))
means=mcmcMat[i_c,2:6]+mcmcMat[i_c,"b0"]
for (i_group in 1:5){
lower=means[i_group]+lim[1]*sd
upper=means[i_group]+lim[2]*sd
yseq=seq(lower,upper,length.out = 100)
dens.y=dnorm((yseq-means[i_group])/sd)
dens.y=dens.y/max(dens.y)*0.75
lines(-dens.y+i_group,yseq,type='l',col="orange")
}
}
dataList = list(
y = y ,
x = x ,
xc = xc ,
Ntotal = Ntotal ,
Ngroup = Ngroup ,
yMean = yMean ,
ySD = ySD ,
xcMean = xcMean ,
acSD = sd(xc) ,
gShape = gShape,
gRate = gRate
)
model.txt="
model {
for ( i_data in 1:Ntotal ) {
y[ i_data ] ~ dnorm( mu[i_data], 1/ySigma^2)
mu [ i_data] <- a0 + a[ x[ i_data ] ] + ac*( xc[ i_data ] - xcMean )
}
ySigma ~ dunif( ySD/100, ySD*10 )
a0 ~ dnorm( yMean, 1/(ySD*5)^2)
for ( i_group in 1:Ngroup ) {
a[ i_group ] ~ dnorm(0.0, 1/aSigma^2)
}
aSigma ~ dgamma( gShape, gRate )
ac ~ dnorm(0, 1/(2*ySD/acSD)^2 )
b0 <- a0 + mean( a[ 1:Ngroup ] ) - ac*xcMean
for (i_data in 1:Ngroup ) { b[ i_data ] <- a[ i_data ] - mean( a[1:Ngroup]) }
}"
writeLines(model.txt, "model.txt")
parameters = c( "b0" , "b" , "ySigma", "aSigma", "ac" )
jagsModel = jags.model( "model.txt", data=dataList, n.chains=3, n.adapt=500 )
update( jagsModel , n.iter=5000)
codaSamples = coda.samples( jagsModel , variable.names=parameters, n.iter=10000, thin=5)
mcmcMat<-as.matrix(codaSamples)
# fig 19.5 only "none0" will be plotted
plot(Longevity~Thorax,xlim=c(0.6,1),ylim=c(0,120),data=dat[dat\$CompanionNumber=="None0",],type='n')
Thorax.value=c(0.65,0.8,0.95)
for (i_c in 1:100) {
# regression lines
xs=c(0.5,1.1)
ys=mcmcMat[i_c,"b0"]+mcmcMat[i_c,3]+mcmcMat[i_c,"ac"]*xs
lines(xs,ys,col="orange")
# density
sd=mcmcMat[i_c,"ySigma"]
lim=qnorm(c(0.025,0.975))
means=mcmcMat[i_c,"b0"]+mcmcMat[i_c,3]+mcmcMat[i_c,"ac"]*Thorax.value
for (i_thorax in 1:3){
lower=means[i_thorax]+lim[1]*sd
upper=means[i_thorax]+lim[2]*sd
yseq=seq(lower,upper,length.out = 100)
dens.y=dnorm((yseq-means[i_thorax])/sd)
dens.y=dens.y/max(dens.y)*0.05
lines(-dens.y+Thorax.value[i_thorax],yseq,type='l',col="orange")
}
}
abline(v=Thorax.value,col='green')
points(Longevity~Thorax,xlim=c(0.6,1),ylim=c(0,120),data=dat[dat\$CompanionNumber=="None0",])
y=dat\$Y
yMean = mean(y)
ySD = sd(y)
Ntotal = length(y)
x=as.numeric(dat\$Group)
Ngroup = length(unique(x))
model.txt = "
model {
for ( i_data in 1:Ntotal ) {
y[ i_data ] ~ dt( a0 + a[ x[ i_data ] ], 1/ySigma[x[i_data]]^2, nu)
}
nu <- nuMinusOne + 1
nuMinusOne ~ dexp(1/29)
for (i_group in 1:Ngroup) {
ySigma[i_group]~dgamma(ySigSh,ySigR)
}
ySigSh <- 1+ySigMode * ySigR
ySigR <- ((ySigMode + sqrt(ySigMode^2+4*ySigSD^2))/(2*ySigSD^2))
ySigMode ~ dgamma(gShape,gRate)
ySigSD ~ dgamma(gShape,gRate)
a0 ~ dnorm( yMean, 1/(ySD*10)^2)
for ( i_group in 1:Ngroup ) {
a[ i_group ] ~ dnorm(0.0, 1/aSigma^2)
}
aSigma ~ dgamma( gShape, gRate )
for ( i_group in 1:Ngroup ) {
m[ i_group ] <- a0 + a[ i_group ]
}
b0 <- mean( m[ 1:Ngroup ] )
for (i_data in 1:Ngroup ) {
b[ i_data ] <- m[ i_data ] - b0
}
}"
writeLines(model.txt, "model.txt")
dataList = list(
y = y ,
x = x ,
Ntotal = Ntotal ,
Ngroup = Ngroup ,
yMean = yMean ,
ySD = ySD ,
gShape = gShape,
gRate = gRate
)
parameters = c( "b0" , "b" , "ySigma", "aSigma" ,"nu","ySigMode","ySigSD")
jagsModel = jags.model( "model.txt", data=dataList, n.chains=3, n.adapt=500 )
update( jagsModel , n.iter=1000)
codaSamples = coda.samples( jagsModel , variable.names=parameters, n.iter=10000, thin=5)
mcmcMat<-as.matrix(codaSamples)
# plotting
plot(x,y,xlim=c(0,4.1),ylim=c(70,130))
n2plot=100
idx=sample(1:nrow(mcmcMat),n2plot)
for (i_sample in 1:n2plot) {
i_c = idx[i_sample]
lim=qnorm(c(0.025,0.975))
means=mcmcMat[i_c,2:6]+mcmcMat[i_c,"b0"]
for (i_group in 1:4){
sd=mcmcMat[i_c,9+i_group]
lower=means[i_group]+lim[1]*sd
upper=means[i_group]+lim[2]*sd
yseq=seq(lower,upper,length.out = 100)
dens.y=dnorm((yseq-means[i_group])/sd)
dens.y=dens.y/max(dens.y)*0.75
lines(-dens.y+i_group,yseq,type='l',col="orange")
}
}
```
認知情報解析演習a Monte Carlo 01
```north=c(1:3,15,1:10)
east=2:15;east[ c(3,7,11)]=c(3,7,11)
south=c(5:15,12:14)
west=c(15,1:13);west[ c(4,8,12)]=c(4,8,12)
trM=cbind(north,east,south,west)
r=-1;P=rep(0.25,4);V = rep(0,14)max.iter = 10000;
state.count=rep(0,15)
for (i.iter in 1:max.iter){
state = sample(1:14,1)
state.seq = state
while(state!=15){
action = sample(1:4,1,prob = P)
state.seq = c(state.seq,trM[state,action])
state = trM[state,action]
}
uniq.seq = unique(state.seq)
for (i.uniq in 1:(length(uniq.seq)-1)){
first.visit = which(state.seq == uniq.seq[i.uniq])[1]
V[uniq.seq[i.uniq]] = V[uniq.seq[i.uniq]] + r*(length(state.seq)-first.visit-1)
}
state.count[uniq.seq] = state.count[uniq.seq] + 1
}
V = matrix(c(0,V/state.count[1:14],0),nrow=4)
```
2019データ解析基礎論a 回帰分析
```dat <- read.csv("http://www.matsuka.info/data_folder/hwsk8-17-6.csv")
plot(dat\$otouto,dat\$ani,pch=20, cex =3,
xlab = "score of younger brother",
ylab = "score of elder brother")
dat.lm <- lm(ani~otouto, data=dat)
summary(dat.lm)
abline(dat.lm, col = 'red',lwd = 2.5)
dat.reg1<-lm(sales~material,data=dat)
dat.reg2<-lm(sales~price,data=dat)
dat.reg3<-lm(sales~design,data=dat)
dat.regALL<-lm(sales~material+price+design,data=dat)
dat.regALL<-lm(sales~.,data=dat)
t.test(dat\$ani, dat\$otouto, var.equal=T)
dat2<-data.frame(score=c(dat\$ani,dat\$otouto),order=c(rep("ani",10),rep("otouto",10)))
plot(dat2\$score~as.numeric(dat2\$order), pch=20, xlab="order",
ylab="score", xlim=c(0.5,2.5), cex=2,xaxt="n")
axis(1,c(1,2),c("ani","otouto"))
dat2.lm<-lm(score~order,data=dat2)
abline(dat2.lm,col='red',lwd=3)
dat.D = dat\$ani - dat\$otouto
boxplot(dat.D,col="skyblue",ylab="Difference")
t.test(dat.D)
plot(dat.D~rep(1,10),pch=20,xlab="",ylab="Difference",cex=3)
dat.D.lm<-lm(dat.D~1)
abline(dat.D.lm,col='red',lwd=3)
```
院：認知情報解析
```y = sample(c(rep(1,15), rep(0,35)))
Ntotal=length(y)
datalist = list(y=y,Ntotal=Ntotal)
source("http://www.matsuka.info/univ/course_folder/HDI_revised.txt")
library(rjags)
txt = "
model {
for ( i_data in 1:Ntotal ) {
y[ i_data ] ~ dbern( theta )
}
theta ~ dbeta( 1, 1 )
}"
writeLines(txt, "~/model.txt")
jagsModel = jags.model(file="~/model.txt",
update(jagsModel,n.iter=1000)
codaSamples=coda.samples(jagsModel,variable.names=c("theta"),n.iter=5000)
mcmcMat<-as.matrix(codaSamples)
HDI.plot(mcmcMat)
traceplot(codaSamples)
autocorr.plot(codaSamples,type='l')
gelman.plot(codaSamples)
y1 = sample(c(rep(1,6), rep(0,2)))
y2 = sample(c(rep(1,2), rep(0,5)))
y = c(y1,y2)
s = c(rep(1,8),rep(2,7))
Ntotal=length(y)
datalist = list(y = y, Ntotal = Ntotal, s = s, Nsubj = 2)
txt = "
model {
for ( i_data in 1:Ntotal ) {
y[ i_data ] ~ dbern( theta[s[i_data]] )
}
for ( i_s in 1:Nsubj) {
theta[i_s] ~ dbeta( 2, 2 )
}
}"
writeLines(txt, "~/model.txt")
jagsModel = jags.model(file="~/model.txt",
update(jagsModel,n.iter=1000)
codaSamples=coda.samples(jagsModel,variable.names=c("theta"),n.iter=5000)
mcmcMat<-as.matrix(codaSamples)
# checking MCMC
HDI.plot(mcmcMat[,2])
traceplot(codaSamples)
autocorr.plot(codaSamples)
gelman.plot(codaSamples)
x.temp = seq(0,150,0.1)
g1 = dgamma(x.temp, shape =0.01, rate =0.01)
plot(x.temp,g1,type='l')
g2 = dgamma(x.temp, shape =1.56, rate = 0.0312)
plot(x.temp,g2,type='l')
g3 = dgamma(x.temp, shape =6.25, rate = 0.125)
plot(x.temp,g2,type='l')
txt = "
model {
for ( i_data in 1:Ntotal ) {
y[ i_data ] ~ dbern( theta[s[i_data]] )
}
for ( i_s in 1:Nsubj) {
theta[i_s] ~ dbeta( omega*(kappa-2)+1,(1-omega)*(kappa-2)+1)
}
omega ~ dbeta(1,1)
kappa <- kappaMinusTwo + 2
kappaMinusTwo ~ dgamma(0.01, 0.01)
}"
writeLines(txt, "~/model.txt")
y=dat\$y
s=as.numeric(dat\$s)
Ntotal=length(dat\$y)
Nsubj=length(unique(s))
datalist = list(y=y,s=s,Ntotal=Ntotal,Nsubj=Nsubj)
update(jagsModel,n.iter=1000)
codaSamples=coda.samples(jagsModel,variable.names=c("theta","omega","kappa"),n.iter=5000)
mcmcMat<-as.matrix(codaSamples)
z = dat\$Hits
N = dat\$AtBats
s = dat\$PlayerNumber
c = dat\$PriPosNumber
Ncat = 9
Nsubj = 948
datalist = list(z = z, N=N, s=s, c=c, Ncat=Ncat, Nsubj =Nsubj )
txt = "
model {
for (i_s in 1:Nsubj) {
z[i_s] ~ dbin( theta[i_s], N[i_s])
theta[i_s] ~ dbeta(omega[c[i_s]]*(kappa[c[i_s]]-2)+1,
(1-omega[c[i_s]])*(kappa[c[i_s]]-2)+1)
}
for (i_c in 1:Ncat) {
omega[i_c] ~ dbeta(omegaO*(kappaO-2)+1,(1-omegaO)*(kappaO-2)+1)
kappa[i_c] <- kappaMinusTwo[i_c]+2
kappaMinusTwo[i_c] ~ dgamma(0.01,0.01)
}
omegaO ~ dbeta(1,1)
kappaO <- kappaMinusTwoO+2
kappaMinusTwoO ~ dgamma(0.01,0.01)
}"
writeLines(txt, "~/model.txt")
update(jagsModel,n.iter=1000)
codaSamples=coda.samples(jagsModel,variable.names=c("theta","omega","kappa"),n.iter=5000)
mcmcMat<-as.matrix(codaSamples)
HDI.plot(mcmcMat[," omega[1]"]) #pitcher
HDI.plot(mcmcMat[," omega[2]"]) #catcher
HDI.plot(mcmcMat[," omega[1]"]) #1st base
```
2019 基礎実習MA01
```# random number generators
x=rnorm(n=1,mean=100,sd=15)
y=runif(n=3,min=1,max=10)
N = 10000
# N = 1000
random.data = rnorm(N, mean=0, sd=1)
hist(random.data, nclass = 50, col = "navy", xlab = "Data",
probability = T, main = "Histogram of Random Data")
# density of generated data
dens = density(random.data)
lines(dens, col = "orange", lwd = 4)
# theoretical density
x = seq(-4,4,0.1)
true.norm = dnorm(x, mean = 0, sd = 1)
lines(x,true.norm, col = "green", lty = 3, lwd = 4)
legend("topleft",c("empirical", "theoretical"), lty = c(1,3),
col = c('orange','green'),lwd=4)
# random sampling
sample(1:10,3)
sample(c(“gu”,“choki”,“pa”),1)
sample(1:10)
sample(0:1, 10, replace=T)
# flow control
for (i_loop in 1:5){print(i_loop)}
counter <- 1
while(counter<=10){
print(counter)
counter<-counter+1
}
counter <- 1
while(counter^2 <= 10){
print(c(counter, counter^2))
counter<-counter+1
}
affil<-"cogsci"
if (affil=="cogsci") {
print("you are wonderful")
}
affil<-"phil"
if (affil=="cogsci") {
print("you are wonderful")
} else {
print("still, you are wonderful")
}
v1=1633;v2=355;
repeat {
r=v1%%v2
print(paste('v1 =',v1,v2 = ',v2, remainder = ',r))
v1=v2;v2=r
if (r==0){ break}
}
counter=6
repeat{
print(counter)
counter = counter + 1
if(counter>5){break}
}
counter=6
repeat{
if(counter>5){break}
print(counter)
counter+counter+1
}
```
認知情報解析06/05の演習問題
```# initalization
p.win =0.4; p.lose = 0.6; P = c(p.lose, p.win);
R = c(rep(0,100), 1); V = rep(0,101);
gamma = 1; tol = 1e-10; counter=0
cols = c("red","skyblue",'orange','black')
par(mfrow=c(1,2))
# value iteration
repeat{
counter = counter+1
delta = 0
for (i.state in 2:100) {
v <- V[i.state]
temp.v = rep(0, (i.state - 1))
for (i.bet in 1:(i.state - 1)) {
lower.B = i.state - i.bet
upper.B = min(i.state + i.bet, 101)
temp.v[i.bet] = sum(P * (R[c(lower.B, upper.B)] + gamma * V[c(lower.B, upper.B)]))
V[i.state] = max(temp.v)
}
delta <- max(abs(v-V[i.state]), delta)
}
# plotting results
if (counter==1){
plot(V[2:100], type='l', col=cols[1], lwd=2, xlab="capital", ylab="value")
} else {
if (counter<4){
lines(V[2:100], type='l', col=cols[counter], lwd=2)
}
}
if (delta < tol){break}
}
# end of value iteration
lines(V[2:100],type='l', col=cols[4], lwd=2)
legend("topleft", c("1st sweep","2nd sweep", "3rd sweep","32nd sweep") ,col=cols, lwd=2)
# identifying optimal action
policy = rep(0,101)
for (i.state in 2:100) {
temp.v = rep(0, (i.state - 1))
for (i.bet in 1:(i.state - 1)) {
lower.B = i.state - i.bet
upper.B = min(i.state + i.bet, 101)
temp.v[i.bet] = sum(P * (R[c(lower.B, upper.B)] + gamma * V[c(lower.B, upper.B)]))
policy[i.state] = which.max(round(temp.v,4))
}
}
barplot(policy,xlab="capital",ylab="Optimal action")
```
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CC-MAIN-2021-31
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https://www.educationtask.com/strategy-to-teach-children-to-subtract.html
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# Strategy to teach children to subtract
Subtraction is one of the mathematical operations that children first learn, after addition. It is a very useful operation since it allows them to determine how many elements remains in a set after subtracting a part, without having to recount the whole set. So you can quickly know, for example, how many candies are left after eating a few or how much travel time is missing if 2 hours have passed from the 4 hours planned.
In addition, learning to subtract not only will be useful in their daily lives but also help them to solve more complex arithmetic operations later, such as division or equations. A very effective strategy to teach children to subtract is to use educational resources, such as subtraction cards.
Five recommendations to teach children to subtract with subtraction cards:
### 1. Wait for the right moment
To teach children content for those who are not yet prepared can set a negative precedent and make them lose motivation to learn. Therefore, one of the most important aspects to teach them to subtract is to be able to detect the most appropriate moment. In a general sense, children will be prepared to assimilate this content when they already know how to add with relative ease and have matured enough to assimilate new concepts.
### 2. Explain what the subtraction tabs are
Most children like to understand the why of things; in this way they are motivated to discover new concepts. Only if the children understand the usefulness of the subtraction, they will find a meaning in the subtraction cards. Therefore, before showing the children the subtraction cards, it is important that you explain what the subtraction and the subtraction cards consist of, which are nothing more than cards with different subtraction operations.
### 3. Start with simple subtraction cards
At the beginning, learning to subtract can be difficult for children, at least until they understand what this operation consists of. To facilitate the process, you can start working with simple subtraction cards, preferably those in which the numbers are represented by objects, animals or fruits since these are more attractive from the visual point of view. In this way, the children will find it easier to represent the subtraction operations and find the solution.
### 4. Use the subtraction cards as a game
Through the game, children not only have fun but also learn. Therefore, it is recommended to use the game as a tool to teach children to learn to subtract. For example, you can organize a game session in which the protagonists are the subtraction cards and the children, along with the rest of the family members; you pass different cards to complete the results. You can also create a kind of contest in which each has a subtraction card and must complete it in the shortest possible time and committing the least amount of errors.
### 5. Encourage them to create their own subtraction cards
Encouraging children to create their own subtraction cards can be an excellent idea to motivate them to master this arithmetic operation. To start you can ask them to reproduce other cards, adding or adding elements to each number, and later, you can allow them to create their own cards. To make it more fun, instead of writing the numbers, they can represent them with the figures they like most, balls, dolls or pears.
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https://www.math-forums.com/threads/a-combinatorial-question.440721/
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| 895,530,147
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|
# a combinatorial question
Discussion in 'Scientific Statistics Math' started by analyst41, Nov 19, 2011.
1. ### analyst41Guest
You have n objects and m slots, n >m .
A "presentation" is a set of m distinct objects out of the n placed in
the m slots.
It is required that object j be presented count(j) times for j =
1,2,....n.
Is there a simple method to determine of a set of count values is
feasible and if so lay out a method to obtain a feasible set of
presentations.
Thanks.
analyst41, Nov 19, 2011
2. ### Butch MalahideGuest
If I understand the problem, you're asking about a matrix of 0's and
1's with prescribed row and column sums. Namely, if k = [count(1) +...
+ count(n)]/m, you want a k-by-n matrix of 0's and 1's with all row
sums equal to m, and the j-th column sum equal to count(j). If that's
your problem, you should look up the Gale-Ryser theorem.
Butch Malahide, Nov 19, 2011
Ask a Question
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# 3.5 Two dimensional problems Cylindrical symmetry Conformal mapping.
## Presentation on theme: "3.5 Two dimensional problems Cylindrical symmetry Conformal mapping."— Presentation transcript:
3.5 Two dimensional problems Cylindrical symmetry Conformal mapping
Laplace operator in polar coordinates
Example: Two half pipes
Conformal Mapping Is there a simple solution?
For two-dimensional problems complex analytical function are a powerful tool of much elegance. x iy Maps (x,y) plane onto (u,v) plane. For analytical functions the derivative exists. Examples:
Analytical functions obey the Cauchy-Riemann equations which imply that g and h obey the Laplace equation, If g(x,y) fulfills the boundary condition it is the potential. If h(x,y) fulfills the boundary condition it is the potential.
g and h are conjugate. If g=V then g=const gives the equipotentials and h=const gives the field lines, or vice versa. If F(z) is analytical it defines a conformal mapping. A conformal transformation maps a rectangular grid onto a curved grid, where the coordinate lines remain perpendicular. Cartesian onto polar coordinates: Example Polar onto Cartesian coordinates: Full plane z w
A corner of conductors
Edge of a conducting plane equipotentials field lines
Parallel Plate Capacitor
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https://www.ps.uni-saarland.de/extras/fol-trakh-ext/website/Undecidability.TRAKHTENBROT.utils.html
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(**************************************************************)
(* Copyright Dominique Larchey-Wendling * *)
(* *)
(* * Affiliation LORIA -- CNRS *)
(**************************************************************)
(* This file is distributed under the terms of the *)
(* CeCILL v2 FREE SOFTWARE LICENSE AGREEMENT *)
(**************************************************************)
Require Import List Arith Lia Bool Eqdep_dec.
From Undecidability.Shared.Libs.DLW.Utils
Require Import utils_list finite.
Set Implicit Arguments.
(* * Utilities for the Full Trakhtenbrot project *)
(* ** Proof irrelevance *)
Fact eq_bool_pirr (b1 b2 : bool) (H1 H2 : b1 = b2) : H1 = H2.
Proof. apply UIP_dec, bool_dec. Qed.
Fact eq_nat_uniq (n : nat) (H : n = n ) : H = eq_refl.
Proof. apply UIP_dec, eq_nat_dec. Qed.
Fact eq_nat_pirr (x y : nat) (H1 H2 : x = y) : H1 = H2.
Proof. apply UIP_dec, eq_nat_dec. Qed.
(* ** Type casting *)
Definition cast (P : nat -> Type) n k (v : P n) (H : n = k) : P k := eq_rect _ P v _ H.
Arguments cast {P n k} v H.
Lemma cast_refl P n (v : P n) : cast v eq_refl = v.
Proof. reflexivity. Qed.
Lemma cast_fun P Q (f : forall n, P n -> Q n) n k (v : P n) (H : n = k) :
f _ (cast v H) = cast (f _ v) H.
Proof. subst; auto. Qed.
Tactic Notation "solve" "ite" :=
match goal with _ : ?x < ?y |- context[if le_lt_dec ?y ?x then _ else _]
=> let G := fresh in destruct (le_lt_dec y x) as [ G | _ ]; [ exfalso; lia | ]
end.
(* ** functions as graphs/relations *)
Section graphs.
Variable (X Y : Type) (R : X -> Y -> Prop).
Definition graph_tot := forall x, ex (R x).
Definition graph_fun := forall x y1 y2, R x y1 -> R x y2 -> y1 = y2.
Definition is_graph_function := graph_fun /\ graph_tot.
Hypothesis (H1 : finite_t Y)
(H3 : forall x y, { R x y } + { ~ R x y }).
Definition graph_tot_reif : is_graph_function -> { f | forall x y, R x y <-> y = f x }.
Proof.
intros (H2 & H4).
destruct finite_t_dec_choice with (3 := H4) as (f & Hf); auto.
exists f; intros x y; split.
+ intros H; generalize H (Hf x); apply H2.
+ intros ->; auto.
Qed.
End graphs.
Section graphs_equiv.
Variable (X Y : Type) (R : X -> Y -> Prop) (equiv : Y -> Y -> Prop).
Infix "≈" := equiv (at level 70, no associativity).
Definition graph_equiv_tot := forall x, ex (R x).
Definition graph_equiv_fun := forall x y1 y2, R x y1 -> R x y2 -> y1 y2.
Definition is_graph_equiv_function := graph_equiv_fun /\ graph_equiv_tot.
Hypothesis (H1 : finite_t Y)
(H3 : forall x y, { R x y } + { ~ R x y })
(H6 : forall x y1 y2, y1 y2 -> R x y2 -> R x y1).
Definition graph_equiv_function_reif : is_graph_equiv_function -> { f | forall x y, R x y <-> y f x }.
Proof.
intros (H2 & H4).
destruct finite_t_dec_choice with (3 := H4) as (f & Hf); auto.
exists f; intros x y; split.
+ intros H; generalize H (Hf x); apply H2.
+ intros H; generalize (Hf x); apply H6; auto.
Qed.
End graphs_equiv.
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| 2.546875
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CC-MAIN-2024-22
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latest
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en
| 0.303724
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https://www.airmilescalculator.com/distance/dax-to-wua/
| 1,606,486,895,000,000,000
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# Distance between Dazhou (DAX) and Wuhai (WUA)
Flight distance from Dazhou to Wuhai (Dazhou Heshi Airport – Wuhai Airport) is 587 miles / 944 kilometers / 510 nautical miles. Estimated flight time is 1 hour 36 minutes.
Driving distance from Dazhou (DAX) to Wuhai (WUA) is 803 miles / 1292 kilometers and travel time by car is about 14 hours 3 minutes.
## Map of flight path and driving directions from Dazhou to Wuhai.
Shortest flight path between Dazhou Heshi Airport (DAX) and Wuhai Airport (WUA).
## How far is Wuhai from Dazhou?
There are several ways to calculate distances between Dazhou and Wuhai. Here are two common methods:
Vincenty's formula (applied above)
• 586.875 miles
• 944.483 kilometers
• 509.980 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
• 588.153 miles
• 946.540 kilometers
• 511.091 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).
## Airport information
A Dazhou Heshi Airport
City: Dazhou
Country: China
IATA Code: DAX
ICAO Code: ZUDX
Coordinates: 31°17′59″N, 107°30′0″E
B Wuhai Airport
City: Wuhai
Country: China
IATA Code: WUA
ICAO Code: ZBUH
Coordinates: 39°47′36″N, 106°47′57″E
## Time difference and current local times
There is no time difference between Dazhou and Wuhai.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 111 kg (245 pounds).
## Frequent Flyer Miles Calculator
Dazhou (DAX) → Wuhai (WUA).
Distance:
587
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
587
Round trip?
| 510
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| 2.59375
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CC-MAIN-2020-50
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latest
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en
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https://en.m.wikipedia.org/wiki/Strike_(bowling)
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# Strike (bowling)
In bowling, a strike means that all of the pins have been knocked down on the first ball roll of a frame. On a bowling scoresheet, a strike is marked by an "X".[3]
With a proper hook, the ball only contacts the 1, 3, 5 and 9 pins (sequentially tinted red) to achieve a strike. All the other pins are knocked down in a chain reaction called pin scatter and this is commonly known as a perfect strike.
Front view:[1] ball impacts center pocket at "board 17.5"—found by a USBC pin-carry study[2] to maximize strike probability. The ideal impact point is closer to the center of the head pin than most people think.[1]
Top view: A bowling ball impacting the head pin at a point found to be optimum to achieve a strike (for a right-handed release).[2] Many believe—wrongly—that the ideal "pocket" is more "between" the 1 pin and 3 pin.[4] Entry angles of 0°, 2°, 4° and 6° are illustrated.
In American nine-pin bowling, a ringer is an equivalent term for knocking down all pins on the first ball of the frame (known as a full house).
## Scoring
A ten-pin bowling score sheet showing how a strike is scored
Bowling scores are generally linearly proportional to strike frequency, with substantial variance based on whether the strikes are consecutive, and based on the number of open frames versus spares.
When all ten pins are knocked down with the first ball roll (called a strike and typically rendered as an "X" on a score sheet), a player is awarded ten points, plus a bonus of whatever is scored with the next two rolls (not necessarily the next two frames). In this way, the points scored for the two rolls after the strike are counted twice.
Frame 1, ball 1: 10 pins (strike)
Frame 2, ball 1: 3 pins
Frame 2, ball 2: 6 pins
The total score from these throws is:
• Frame one: 10 + (3 + 6) = 19
• Frame two: 3 + 6 = 9
TOTAL = 28
An easier non-standard method of scoring a strike is to score the strike with a flat 10 points and then add 1 to the multiplier of the next two rolls. Thus, the scoring of the above example would play out as below:
• Frame one: 10
• Frame two: (3 x 2) + (6 x 2) = 18
TOTAL = 28
Strike scoring works similarly for five-pin bowling, except strikes are worth 15 points rather than 10 (as the pins are scored with the values of 2, 3, 5, 3, and 2).
## Consecutive strikes
A series of two strikes is known as a "double" (or a "Barney Rubble" to rhyme), and a series of three is known as a "turkey" (sometimes a "sizzling turkey" on the first three frames). Any longer string of strikes is referred to by a number affixed to the word "bagger," as in "four-bagger" for four straight strikes, also known as a "hambone", likely derived from bowling's early days when foodstuffs were awarded to winners of competitions.[5]
When a player is "on the strikes", a string is often referenced by affixing "in a row" to the number of consecutive strikes. A string of six strikes is sometimes called a "six pack" or a "sixer".[6] A string of six and nine strikes are also known as a "wild turkey" and a "golden turkey" respectively. Any string of strikes starting in the first frame or ending "off the sheet" (where all of a bowler's shots from a certain frame to the end of the game strike) are often called the "front" or "back" strikes, respectively (e.g. the 'front nine' for strikes in frames 1-9, or the 'back six' for strikes in frames 7, 8, and 9 with a turkey in the tenth). Twelve strikes in a row is a perfect game; 36 straight strikes constitutes a 900 series. Due to the difficulty of achieving a game of 300 or a series of 900, many bowling alleys maintain 300 and 900 club plaques.
Multiple strikes would be scored like so:
Frame 1, ball 1: 10 pins (strike)
Frame 2, ball 1: 10 pins (strike)
Frame 3, ball 1: 4 pins
Frame 3, ball 2: 2 pins
The score from these throws is:
• Frame one: 10 + (10 + 4)= 24
• Frame two: 10 + (4 + 2) = 16
• Frame three: 4 + 2 = 6
TOTAL = 46
With the simpler non-standard system of scoring, the above example would be scored as below:
• Frame one: 10
• Frame two: 10 x 2 = 20
• Frame three: (4 x 3) + (2 x 2) = 16
TOTAL = 46
The most points that can be scored in one frame is 30 points (10 for the original strike, plus strikes in the two following frames)
The most points that can be scored in one game is 300 points which is a perfect game.
A player who bowls a strike in the tenth (final) frame is awarded two extra balls so as to allow the awarding of bonus points. If both these balls also result in ten pins knocked down each, a total of 30 points (10 + 10 + 10) is awarded for the frame. These bonus points only count as the bonus for the strike and not on their own.
## Publications
• Benner, Donald; Mours, Nicole; Ridenour, Paul (2009). "Pin Carry Study: Bowl Expo 2009" (Slide show presentation). bowl.com. USBC, Equipment Specifications and Certifications Division. Archived (PDF) from the original on December 7, 2010.
• Freeman, James; Hatfield, Ron (July 15, 2018). Bowling Beyond the Basics: What's Really Happening on the Lanes, and What You Can Do about It. BowlSmart. ISBN 978-1 73 241000 8.
## References
1. ^ a b Freeman & Hatfield 2018, Chapter 10 ("The Pocket Isn't the Pocket... and It's Nowhere Near Where You Think It Is").
2. ^ a b
3. ^ Bowling-Tips.org (2013). "How to Bowl a Strike". Bowling-Tips.org. Retrieved 2014-04-01.
4. ^ Freeman & Hatfield 2018, Chapter 8 ("Why Does My Ball Hook?").
5. ^ Jeff Goodger (2013). "Strings of Strikes". About.com. Retrieved 2014-04-01.
6. ^ PBA (2012). "Professional Bowlers Association: Bowling Lingo". Professional Bowlers Association. Retrieved 2012-12-01.
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