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Conical Pendulum, find tension, radial force, speed, period, and angular speed

1. Oct 15, 2012

LibGirl17

1. The problem statement, all variables and given/known data
A metal sphere is attached to the end of a string and then set in motion such that it rotates in a horizontal circle as shown in the sketch. The metal sphere has a mass of 0.5kg and the radius of the circle is .2m.

1)Find the tension in the string
3)Determine the speed of the rotating mass
4)Determine how long it takes for the sphere to complete one rotation, i.e. determine its period
5)What is the sphere's angular speed in rad/seconds?

2. Relevant equations
Assuming the metal sphere is on the left side of the circle;
F_netx=Tcos(θ)=ma_x
F_nety=Tsin(θ)-mg=ma_y=0
θ(angle between x-axis and string)=arccos(.2)

3. The attempt at a solution
I believe I found everything except angular velocity, but the period I found also seems too short. Any input would be appreciated!
1)T=mg/sin(θ)=5.001N
2)Tsin(θ)=4.899N
3)Tsin(θ)=mv^2/r, 4.899=.5v^2/.2, v=1.3999
4)Δt=Δs/v=2∏(.2)/1.3999=.8977s
5)I know ω=dθ/dt, but I'm not sure what to do with this...

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Last edited: Oct 15, 2012
2. Oct 15, 2012

cepheid

Staff Emeritus
Welcome to PF LibGirl17!

This looks good. It doesn't really matter where in the circle the sphere is. You can always resolve the tension into two components, the "vertical" component, and the "radially-inward" component (the component of the tension that points towards the centre of the circle). The latter lies in the plane of the circle and is thus "horizontal" (although its direction changes, this doesn't matter, because it is always radially-inward).

With a cone, you typically measure the angle from the vertical, in which case the horizontal component would have the sine factor, and the vertical component would have the cosine factor. I'm not sure where you're measuring the angle from here. EDIT: I see that you're using the angle from the horizontal, which means that the vertical component is Tsin(theta), and the horizontal component is Tcos(theta). That means you should use Tcos(theta) for the centripetal force below.

This seems like the right general method. The radial component of the tension is equal to the centripetal force, so use that to solve for v.

As for the angular velocity: in this case, the rotation rate is constant, so it's also true that ω = Δθ/Δt. You know how long it takes the object to go through an angular displacement of Δθ = 2π radians (one full rotation), because you computed the period.

Last edited: Oct 15, 2012


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## Net Present Value - Net Profit Value

In finance, the net present value (NPV) or net present worth (NPW) is a measurement of profit calculated by subtracting the present values (PV) of cash outflows (including initial cost) from the present values of cash inflows over a period of time. Incoming and outgoing cash flows can also be described as benefit and cost cash flows, respectively.

## Common pitfalls

Solved: Gaston Company Is Considering A Capital Budgeting .... Source : www.chegg.com

• If, for example, the Rt are generally negative late in the project (e.g., an industrial or mining project might have clean-up and restoration costs), then at that stage the company owes money, so a high discount rate is not cautious but too optimistic. Some people see this as a problem with NPV. A way to avoid this problem is to include explicit provision for financing any losses after the initial investment, that is, explicitly calculate the cost of financing such losses.
• Another common pitfall is to adjust for risk by adding a premium to the discount rate. Whilst a bank might charge a higher rate of interest for a risky project, that does not mean that this is a valid approach to adjusting a net present value for risk, although it can be a reasonable approximation in some specific cases. One reason such an approach may not work well can be seen from the following: if some risk is incurred resulting in some losses, then a discount rate in the NPV will reduce the effect of such losses below their true financial cost. A rigorous approach to risk requires identifying and valuing risks explicitly, e.g., by actuarial or Monte Carlo techniques, and explicitly calculating the cost of financing any losses incurred.
• Yet another issue can result from the compounding of the risk premium. R is a composite of the risk free rate and the risk premium. As a result, future cash flows are discounted by both the risk-free rate as well as the risk premium and this effect is compounded by each subsequent cash flow. This compounding results in a much lower NPV than might be otherwise calculated. The certainty equivalent model can be used to account for the risk premium without compounding its effect on present value.
• Another issue with relying on NPV is that it does not provide an overall picture of the gain or loss of executing a certain project. To see a percentage gain relative to the investments for the project, usually, Internal rate of return or other efficiency measures are used as a complement to NPV.
• Non-specialist users frequently make the error of computing NPV based on cash flows after interest. This is wrong because it double counts the time value of money. Free cash flow should be used as the basis for NPV computations.

## History

Net present value as a valuation methodology dates at least to the 19th century. Karl Marx refers to NPV as fictitious capital, and the calculation as "capitalising," writing:

The forming of a fictitious capital is called capitalising. Every periodically repeated income is capitalised by calculating it on the average rate of interest, as an income which would be realised by a capital at this rate of interest.

In mainstream neo-classical economics, NPV was formalized and popularized by Irving Fisher, in his 1907 The Rate of Interest and became included in textbooks from the 1950s onwards, starting in finance texts.

## Alternative capital budgeting methods

• Adjusted present value (APV): adjusted present value, is the net present value of a project if financed solely by ownership equity plus the present value of all the benefits of financing.
• Accounting rate of return (ARR): a ratio similar to IRR and MIRR
• Cost-benefit analysis: which includes issues other than cash, such as time savings.
• Internal rate of return (IRR): which calculates the rate of return of a project while disregarding the absolute amount of money to be gained.
• Modified internal rate of return (MIRR): similar to IRR, but it makes explicit assumptions about the reinvestment of the cash flows. Sometimes it is called Growth Rate of Return.
• Payback period: which measures the time required for the cash inflows to equal the original outlay. It measures risk, not return.
• Real option: which attempts to value managerial flexibility that is assumed away in NPV.
• Equivalent annual cost (EAC): a capital budgeting technique that is useful in comparing two or more projects with different lifespans.

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# How do you graph  y=5-cosx?

Oct 9, 2016

graph{5-cosx [-10, 10, 3, 7.5]}

#### Explanation:

First think about the graph of $y = \cos x$, which is as follows:
graph{cosx [-10, 10, -1.5, 1.5]}

So then the graph of $y = - \cos x$ will be the above graph inverted:
graph{-cosx [-10, 10, -1.5, 1.5]}

And so $y = 5 - \cos x$ will be the above graph moved 5 units up
graph{5-cosx [-10, 10, 3, 7.5]}


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## Algebra: A Combined Approach (4th Edition)

Take the inequality: $5 \lt 20$ as an eample. When multiplying the inequality by $-1$, $-5 \lt -20$. This is false as -5 should be greater than -20. So this is only true when the inequality sign is reversed. $-5 \gt -20$ A larger magnitude of a positive value is larger than a smaller magnitude of a positive value. However, a smaller magnitude of a negative value is larger than a larger magnitude of a negative value. Hence, the inequality sign is reversed when multiplying or dividing by a negative value. See "question 84" as another example.


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# In how many ways can some or all of the $5$ distinct coins be put into $8$ pockets?

In how many ways can some or all of the $5$ distinct coins be put into $8$ pockets?

Could this be modeled as the problem of "In how many ways N distinct items be put into r distint groups (where some groups may be empty)?"

My instructor is of the opinion that this problem should be modeled as like for every coin we have $9$ options to put it into one of the 8 pockets or don't and so the answer is $9^5$,however in this case I think $-1$ is necessary to avoid the case when none is selected but he thinks we don't need that.

There is also a different model that I think could be the solution which is as a summation of $8^1 + 8^2 + \cdots + 8^5$ pertaining to how many ways $1$ coin could be distributed in $8$ pockets then $2$ coins in $8$ pockets ... till $5$ coins in $8$ pockets.(assuming pockets are distinct in every case)

But as the answer is not given I can't be sure which is correct,(instructor haven't disclosed it either),could any body tell me which one is correct,if none is correct,then how exactly I am supposed to count this?

-
Are the pockets also distinct? (Your first rephrasing is incorrect because the problem stipulates that you don't need to use all of the coins.) –  Qiaochu Yuan Aug 31 '11 at 20:38
@Qiaochu Yuan:I am not sure of that either,yes I realized it that's why I think of the third summation approach. –  Quixotic Aug 31 '11 at 20:39
The "different model" is incorrect, because it assumes the coins are indistinguishable but the pockets are distinguishable. Whether or not the pockets are distinguishable is unclear, but the problem explicitly says the coins are different, so there are not 8 ways to distribute one coin into the pockets. There's either 5 (if the pockets are all the same; one way for each of the different coins), or there are 40 (if the pockets are different; eight ways for each possible choice of coin). Either way, it's not $8^1$. –  Arturo Magidin Aug 31 '11 at 20:41
@Arturo Magidin:May be I am missing something,but we can distribute $1$ distinct items into $8$ distinct groups in $8^1$ (where some groups may be empty) isn't? –  Quixotic Aug 31 '11 at 20:48
@FoolForMath: If you have a penny, a nickel, a dime, a quarter, and a half-dollar (5 distinct coins), and you have 8 pockets, in how many ways can you distribute one coin into the pockets? You could have the penny in pocket 1, or pocket 2, or ... or pocket 5; or you could have the nickel in pocket 1, or in pocket 2;... or you could have the dime in pocket 1, or in pocket 2, or... , ..., or you could have the half-dollar in pocket 8. That's 40 different ways. If all the pockets are the same, then you first pick which coin you will be putting into a pocket (5 ways), then put it in. –  Arturo Magidin Aug 31 '11 at 20:59

If I understand correctly, for each of $\binom{5}{k}$ choices of $k$ coins, you will have $8^k$ maps from those $k$ coins to the $8$ pockets. If that is the case, then the answer would be $$\sum_{k=0}^5\binom{5}{k}8^k = (1+8)^5$$ by the binomial theorem.
Another way of looking at this, is to add a ninth hidden pocket for the coins that don't appear in the visible $8$ pockets. That is, you are asking how many ways to put $5$ coins into $9$ pockets, or $9^5$.
If you don't want to allow $8$ empty (visible) pockets, then the answer would be $9^5-1$.


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Why do current-feedback amplifiers have a higher bandwidth?

The advice when dealing with amplifying high-frequency signals always seems to be "use a current-feedback amplifier, not an op amp". (see 1, comments on 2, 3 and comments on its question) But why is this? What about current-feedback amplifiers makes them inherently faster than conventional voltage-feedback amplifiers?

I've also seen it said that their gain is largely independent of frequency, as compared to op amps which have a gain-bandwidth product that is relatively constant, meaning their gain is limited at high frequencies. Is this related to the high bandwidth, or is it a separate advantage of the CFA?

• At the most fundamental because in a bipolar circuit, we can switch currents faster than voltages (all other things being equal). – Peter Smith Apr 8 '20 at 14:14
• analog.com/media/en/training-seminars/tutorials/MT-034.pdf – Andy aka Apr 8 '20 at 14:22
• Voltage feedback opamps have high impedance inputs, CFAs have low impedance inputs, at least the one taking feedback, equivalent to the top stage of a cascode, with similar speed advantages – Neil_UK Apr 8 '20 at 14:44
• Where did you read this? Currently there are voltage feedback OP-amps available with GBW in the multiple gigahertz. 20 years ago you might have needed a CFA for the applications where you'd now use those op-amps, though. – The Photon Apr 8 '20 at 15:03
• @ThePhoton Multiple different answers on this website, in fact. – Hearth Apr 8 '20 at 15:09

I think, the explanation is as follows:

• Voltage opamps (if they are unity-gain stable) are internally compensated, which means: Their open-loop gain has a pretty small 3-dB cutoff frequency (20...200 Hz). As a consequence (with 20dB drop per decade) and - let`s say Aoo=100 dB - the transit frequency is app. (2...20) Mhz.

This is necessary because the resistor ratio in the feedback loop determines (a) the desired closed-loop gain as well as (b) the loop gain which is responsible for stability properties. Hence, both gains (closed-loop gain and loop gain) are directly coupled and cannot be set independent to each other.

• Current-Feedback-Amplifiers (CFA): In contrast to the voltage opamps, the loop gain is determined not by the gain-setting ratio of the feedback resistors, but by the value of the feedback resistor (between output and inv. input) only.

Therefore, the loop gain can be set independent on the closed-loop gain at a value that allows good and stable operation. Hence, the amplifier is not required to be fully compensated. The 3dB cutoff of the open-loop gain and, with it, the transit frequency can be designed much larger than for an opamp.

Here are the closed-loop gain expressions:

• Opamp: Acl=[Ai]*[1/(1+Ai/Ao(jw))] with A(ideal)=Ai=(1+R2/R1)

• CFA : Acl=[Ai]*[1/(1+R2/Ztr(jw))] with A(ideal)=Ai=(1+R2/R1) and Ztr(jw)=transfer impedanz

Comment to CFA: Because the stability can be ensured by proper selection of the feedback resistor R2, in the data sheets an "optimum" value for R2 is specified (recommended). The closed-loop gain can be set with R1.

Answer to the last question: The bandwidth of any amplifier with feedback is always set by the loop gain (which, for opamps, is closely related to the closed-loop gain). For CFAs however, the loop gain is constant (set by the fedback resistor R2). Therefore, the closed-loop bandwidth is also constant and there is no dependence on the closed-loop gain which is set by the resistor ratio.

If a circuit can avoid Miller Effect, so the precious signal charge is not wasted in fighting internal junction (collector-base, drain-gate) charge demands, the bandwidth soars.

For example, examine the schematic of the UA715 operational amplifier, designed in the late 1960s.

The input differential pair is CASCODED, minimizing the Miller Effect, and producing a fast-settling circuit.

Notice the input capacitance is seldom specified on opamps.

And when spec'd, the operating conditions are not made clear.

And when slewing, the Cin can be much smaller than when linearly responding and settling.

Consider a non-inverting amplifier configuration using a voltage feedback amplifier. The gain can be shown to be

$$\frac{1 + R_1 / R_2}{1 + (1 + R_1 / R_2)/A_\mathrm{OL}} = A_\mathrm{CL} \frac{1}{1 + \frac{A_\mathrm{CL}}{A_\mathrm{OL}}}.$$

Hence the attenuation factor is dependent on the closed loop gain.

For a non-inverting amplifier using a current feedback amplifier with an open loop transconductance $$\T(f)\$$, the closed loop gain can be shown to be approximately

$$\frac{1 + R_1/R_2}{1 + R_1/T} = A_\mathrm{CL}\frac{1}{1 + R_1/T}.$$

Hence the attenuation factor is dependent on $$\R_1\$$. However, the closed loop gain can be controlled only with $$\R_2\$$ while keeping $$\R_1\$$ constant. Thus the bandwidth can be independent of the closed loop gain.

• user110971...could it be that you have mixed 6dB with 3dB? – LvW Apr 10 '20 at 9:17
• @LvW Remember it is voltage. An attenuation of 0.5 in voltage is 6 dB in power. – user110971 Apr 10 '20 at 9:21
• user110971...so you consider the ratio Acl/Aol as real (and not as imaginary)? To me Aol should be considered with a -20dB/dec slope (and 90deg phase shift) – LvW Apr 10 '20 at 9:34
• @LvW Good point. Yes, I agree. I edited out the 6 dB part. You shouldn’t look at a single point of the frequency response in isolation anyway. You should consider the entire thing. I don’t know why I put it there. – user110971 Apr 10 '20 at 9:47


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MathSciNet bibliographic data MR507912 14F05 (32G13) Trautmann, Günther Moduli for vectorbundles on ${\bf P}\sb{n}({\bf C})$${\bf P}\sb{n}({\bf C})$. Math. Ann. 237 (1978), no. 2, 167–186. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


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## Antipodal points after Vîlcu

Posted in Math by pgadey on 2013/03/26

I’ve been thinking a lot about convex bodies in ${{\mathbb R}^3}$ lately. This post is going to be a write up of a useful lemma in the paper: Vîlcu, Constin, On Two Conjectures of Steinhaus, Geom. Dedicata 79 (2000), 267-275.

Let ${S}$ be a centrally symmetric convex body in ${{\mathbb R}^3}$. Let ${d(x,y)}$ denote thes intrinsic metric of ${S}$ and ${D = \sup_{x,y} d(x,y)}$ its intrinsic diameter. For a point ${x \in S}$ we write ${\bar{x}}$ for its image under the central symmetry.

Lemma 1 (Vîlcu) If ${d(x,y) = D}$ then ${y = \bar{x}}$.

This lemma says that if a pair realizes the inner diameter of a centrally symmetric convex body, the pair has to be centrally symmetric. This aligns well with our intuition about the sphere and cube, for example.

Proof: Let ${d(x,y) = D}$ and suppose, for contradiction, that ${y \neq \bar{x}}$. Pick some length minimizing geodesic ${\gamma_{x\bar{x}}}$ connecting ${x}$ to ${\bar{x}}$. Let ${\Gamma = \gamma_{x\bar{x}} \cup \bar{\gamma}_{x\bar{x}}}$ denote the concatenation of the paths ${\gamma_{x\bar{x}}}$ and ${\bar{\gamma}_{x\bar{x}}}$.

We check that ${\Gamma}$ is self-intersection free. Certainly ${\gamma_{x\bar{x}}}$ is self-intersection free, because it is a minimizing geodesic. Suppose that ${p \in \gamma_{x\bar{x}} \cap \bar{\gamma}_{x\bar{x}}}$. Then we have two minimizing geodesics from ${x}$ to ${\bar{x}}$ intesecting at a point on their interior. Hence they must coincide, a contradiction. Thus ${\Gamma}$ is self-intersection free and hence ${\Gamma}$ seperates ${S}$ into two open regions. Let ${S \setminus \Gamma = S_1 \cup S_2}$. We know ${S_2 = \bar{S}_1}$ since the central symmetry has to swap the components.

Suppose, for contradiction, that ${y \in \Gamma}$. Then ${d(x,y) \leq d(x,\bar{x})}$. If ${d(x,y) = d(x,\bar{x})}$ then ${y = \bar{x}}$, contradicting our hypothesis on ${y}$. If ${d(x,y) < d(x, \bar{x})}$ then we contradict the maximality of the diameter. Without loss of generality, take ${y \in S_1}$. We have ${\bar{y} \in \bar{S}_1}$. Take a minimizing geodesic ${\gamma_{y\bar{y}}}$ joining ${y}$ to ${\bar{y}}$. We have that ${\gamma_{y\bar{y}} \cap \Gamma \neq \emptyset}$ by the Jordan curve theorem. Take ${z \in \gamma_{y\bar{y}} \cap \Gamma}$. Suppose ${z \in \gamma_{x\bar{x}}}$.

The triangle inequality gives us:

$\displaystyle \begin{array}{rcl} d(x,y) & \leq & d(x,z) + d(z,y)\\ d(\bar{x},\bar{y}) & \leq & d(\bar{x},z) + d(z,\bar{y}) \end{array}$

Thus:

$\displaystyle d(x,y) + d(\bar{x},\bar{y}) \leq d(x,z) + d(z,\bar{x}) + d(y, z) + d(z, \bar{y}) = d(x,\bar{x}) + d(y,\bar{y})$

The maximality of the diameter and the equality ${d(x,y) = d(\bar{x},\bar{y})}$ then give:

$\displaystyle d(x,y) + d(\bar{x}, \bar{y}) = d(x,\bar{x}) + d(y,\bar{y})$

Suppose that ${d(x,y) < d(x,z) + d(z,y)}$ and ${d(\bar{x}, \bar{y}) < d(\bar{x},z) + d(z,\bar{y})}$. We then contradict the equality above. Thus, one of the two strict inequalities is an equality. Suppose without loss of generality that ${d(x,y) = d(x,z) + d(z,y)}$. Let ${\gamma_{xy}}$ denote a minimizing geodesic segment connecting ${x}$ to ${y}$.

We then have that ${\gamma_{x\bar{x}} \subset \gamma_{xy}}$, otherwise we’ll have two geodesics diverging from one another at a point. We then form ${\gamma'_{xy}}$ by removing ${\gamma_{x\bar{x}}}$ from ${\gamma_{xy}}$ and replacing it with ${\bar{\gamma}_{x\bar{x}}}$. We then have that ${\gamma_{xy}}$ and ${\gamma'_{xy}}$ diverge at a point, a contradiction. $\Box$

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Lab 3
Get started now
## Objectives * Translate UML class diagrams into source code * Construct software using a interface * Employ aggregation and composition in the development of software ## Assignment In this lab, you will write a program for managing the parts needed to assemble a machine. Your program will provide a complete bill of materials including both the weight and cost for each part. To facilitate a wide variety of parts, your program should make use of a generic Part interface. This interface will allow generic information about a part to be accessed: * The part's name * The total cost of the part * The total weight of the part * The bill of materials for a part There are a few kinds of parts that the system supports: 1. Nuts 2. Bolts 3. Sheet metal 4. Duplicate parts 5. Assemblies of parts Nuts, bolts, and sheet metal are basic parts. Duplicate parts and assemblies are built out of other pre-existing parts. ## Assignment Details A class diagram for the program is shown in the figure below:
[![UML Class Diagram](parts.gif)](parts.gif)
 Implement all of the classes shown in the diagram above based on the discussion that follows. The arrows with black diamonds simply indicate that one class has a reference to another, as is already shown in the member variables of the class diagrams for Assembly and Duplicate part. * A nut costs $\$0.50 \times size $, where size is the inner diameter of the nut in inches. A nut weights$ 0.01 lbs \times (size)^2 $. * A bolt costs$ \$1 \times diameter \times length$, where $diameter$ is the diameter of the threads in inches and $length$ is the overall length of the bolt in inches. A bolt weighs $0.05 lbs \times (diameter)^2 \times length$. * Sheet metal costs $\$0.50 \times thickness \times width \times length $, where each of the dimensions is measured in inches. Sheet metal weighs$ 0.1 lbs \times thickness \times width \times length $. Bolts, nuts, and sheet-metal print a simple bill of materials including the name of the part in the title and the dimensions, cost, and weight of the part. For example, this is a Bill of Materials for a Sheet:  ========================== 12.0x12.0x0.25 sheet ========================== Length: 12.0 inches Width: 12.0 inches Thickness: 0.25 inches Cost:$18.00 Weight: 3.6 lbs  A duplicate part represents count identical parts. Its cost and weight are both count times the cost and weight of the single part. However, if there are at least 5 identical parts, there is a 5% discount off the cost of the duplicate parts, and if there are at least 10 identical parts, the discount increases to 10% discount off the cost of the duplicate parts. The duplicate part stores a single reference to another object describing just one of the identical parts. A duplicate part's Bill of Materials includes both a summary section and the bill of materials for the part that is duplicated. The summary section includes the duplicate part's name on a title line. Below the title line are the name, cost, and weight of the individual part, along with the number of copies of that part. It prints the total cost and weight for the duplicate part. After the summary comes the complete bill of materials for the individual part. For example, the fan option in the program [MachineDriver.java](MachineDriver.java) is a duplicate part consisting of five identical blades. When the printBillOfMaterials() is called on the dupicate part, it prints:
 Building a fan ========================== 5 8.0x6.0x0.0625 sheets ========================== Duplicate part: 8.0x6.0x0.0625 sheet Copies: 5 Individual cost: $1.50 Individual weight: 0.3 lbs Total cost:$7.12 Total weight: 1.5 lbs ========================== 8.0x6.0x0.0625 sheet ========================== Length: 8.0 inches Width: 6.0 inches Thickness: 0.0625 inches Cost: $1.50 Weight: 0.3 lbs
 An assembly consists of multiple different parts. Both the cost and weight of the assembly are determined by adding the cost and weight of each part in the assembly. However, the assembly has an additional construction cost which is$0.25 per sub-part in the assembly. The assembly maintains a List of all the parts that are stored in it. A duplicate part or an assembly may be a sub-part of another part. For example, in the cube option in the program [MachineDriver.java](MachineDriver.java), the cube consists of two parts: a set of metal sheets and several sets of nut-bolt pairs. The set of sheets is actually a duplicate part, consisting of six identical pieces of sheet metal. The sets of nut-bolt pairs are also a duplicate part, consisting of 36 identical nut-bolt pairs. Each nut-bolt pair is an assembly, consisting of both a nut and a bolt. This hierarchy of parts is illustrated in the diagram below.
[![Parts Hierarchy](partsHierarchy.png)](partsHierarchy.png)
 When a duplicate part or assembly prints the bill of materials, it first prints a summary of each part (including name, cost, and weight for each part), then prints the full bill of materials for each part. For example, the bill of materials for the cube in [MachineDriver.java](MachineDriver.java) is:  ========================== Cube ========================== Part: 36 Nut-Bolt Pairs Cost: $36.45 Weight: 0.247 lbs Part: 6 12.0x12.0x0.25 sheets Cost:$102.60 Weight: 21.6 lbs Total cost: $139.55 Total weight: 21.848 lbs ========================== 36 Nut-Bolt Pairs ========================== Duplicate part: Nut-Bolt Pair Copies: 36 Individual cost:$1.12 Individual weight: 0.007 lbs Total cost: $36.45 Total weight: 0.247 lbs ========================== Nut-Bolt Pair ========================== Part: 0.25x2.0 bolt Cost:$0.50 Weight: 0.006 lbs Part: 0.25 inch nut Cost: $0.12 Weight: 0.001 lbs Total cost:$1.12 Total weight: 0.007 lbs ========================== 0.25x2.0 bolt ========================== Diameter: 0.25 inches Length: 2.0 inches Cost: $0.50 Weight: 0.006 lbs ========================== 0.25 inch nut ========================== Diameter: 0.25 inches Cost:$0.12 Weight: 0.001 lbs ========================== 6 12.0x12.0x0.25 sheets ========================== Duplicate part: 12.0x12.0x0.25 sheet Copies: 6 Individual cost: $18.00 Individual weight: 3.6 lbs Total cost:$102.60 Total weight: 21.6 lbs ========================== 12.0x12.0x0.25 sheet ========================== Length: 12.0 inches Width: 12.0 inches Thickness: 0.25 inches Cost: $18.00 Weight: 3.6 lbs  As shown in the example above, format numeric values cleanly. In particular, dollar amounts should always include two places after the decimal. You may add new options to the example program, but do not edit the options provided in the example. Let the single printBillOfMaterials() call print everything. ## Appendix: Shovel option As a final example, the shovel option in the example program produces an assembly consisting of three different parts.  ========================== Shovel ========================== Part: 30.0x3.0x0.125 sheet Cost:$5.62 Weight: 1.125 lbs Part: 10.0x8.0x0.25 sheet Cost: $10.00 Weight: 2 lbs Part: 0.125x0.5 bolt Cost:$0.06 Weight: 0 lbs Total cost: $16.44 Total weight: 3.125 lbs ========================== 30.0x3.0x0.125 sheet ========================== Length: 30.0 inches Width: 3.0 inches Thickness: 0.125 inches Cost:$5.62 Weight: 1.125 lbs [... more sub-parts ...]

Thursday, 07-Dec-2017 13:42:48 CST


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# What is the complexity of the fastest method of k-coloring any graph? [closed]

I heard brute-force is the only method. Is there any other way? Is there a way to prove that the complexity cannot be exponential?

There are non-obvious improvements over simple brute-force search for $k$-coloring (and for many other NP-hard problems). The obvious approach would take roughly $k^n$ time, but one can do it in time $O(2^n \cdot poly(n))$, and for $k = 3$, one can get $O(1.33^n)$. Algorithms in this area involve simple but interesting applications of dynamic programming and inclusion-exclusion, and for state-of-the-art results can involve intricate case analysis. One good starting point: these lecture notes (with references) by Uri Feige
which also include open problems; e.g., can one get running time $O(2^n \cdot poly(n))$ as above with polynomial space usage?


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# Looking for a function

#### katkat

Does anyone know a function t(x) satisfying the following conditions.

as x -> 0 we have

(1) t(x)-> oo

(2) x t(x) -> 0

(3) x t(x) exp(t(x)) -> 0

(4) x t(x)^2 exp(t(x)) -> oo

I have "two extremes" as

t(x) = sqrt(-log x) satisfies (1-3) but x t(x)^2 exp(t(x)) -> 0

t(x) = -log( -x log(x))

satisfies (1-2) and (4), but x t(x) exp(t(x)) -> c >0

I have been searching desperately.

#### whozum

What level course is this?

#### steven187

hello

if your really desperate you should try ploting it through mathematica or some similar program and watch what happens as x goes to infinity it shouldnt be too hard, a few guesses and you should be able to get your solution

#### rachmaninoff

This is in the reals, right? And am I misreading this:
as x -> 0 we have

(1) t(x)-> oo

(2) x t(x) -> 0
or are (1) and (2) mutually exclusive?

#### whozum

rachmaninoff said:
This is in the reals, right? And am I misreading this:

or are (1) and (2) mutually exclusive?
I want to agree with you but first I want to verify that the only functions that tend to infinity as x-> 0 are ones where the denominator has a quicker tendency to infinity than the numerator.

#### joeboo

rachmaninoff said:
This is in the reals, right? And am I misreading this:

or are (1) and (2) mutually exclusive?

$$t(x) = \frac{1}{\sqrt{\vert x \vert}}$$

Am I missing something?

#### whozum

$$\lim_{x\rightarrow 0} \frac{x}{\sqrt{|x|}} \neq 0$$

#### rachmaninoff

edit: I misread the OP, ignore this post.

Last edited by a moderator:

#### shmoe

Homework Helper
whozum said:
$$\lim_{x\rightarrow 0} \frac{x}{\sqrt{|x|}} \neq 0$$
It's zero. When x>0 it's sqrt(x). It appears he's only looking at right handed limts (hence his log's), but for this function the two-sided limit is also zero, it's -sqrt(x) for x<0.

These limits are as x->0, none of them exclude any others.

t(x) = -log( -x log(x)) is hopeful. In condition 3, you're looking at [-log(x)-log(-log(x))]/(-log(x)). You'd like to speed up the growth of the denominator without affecting the numerator signifigantly. Try fiddling with the inside log of t(x)

#### whozum

I thought of this last night, does it have to be continuous?
And also, for a limit to exist, the function must approach the value from both sides, however since the log function can't take negative values, there is a whole in his domain for t(x) = -log( -x log(x)), correct?

#### shmoe

Homework Helper
The log(x) is what made me think he was considering right hand limits. Finding a function satisfying those limits from the right is enough in any case, just put |x| where you see x and you've got your two sided limits as well. There's nothing about continuity as he's written it, just a limit. In fact the first condition rules on continuity at 0. The t(x) = -log( -x log(x)) will work, with some minor alterations.

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text:
Why Do Planets Revolve Faster When they Are Closer to Their Parent Star?

According to Kepler's second law: A planet sweeps out equal areas in equal times (A simple definition). This means that if the orbit is somewhat eccentric, the planet will move faster when on the closest approach to it's parent star, and slowest when at aphelion. So why does this happen, is the curvature of space-time greater closer to the star?

-

Velocity is a form of kinetic energy, while height within a gravity well is a form of potential energy. For an orbiting body, conservation of energy will keep the total energy constant.

So as a planet moves away from the parent star, it loses velocity and gains potential energy. As it moves closer, it trades the potential energy back for velocity. The point of lowest potential has the highest velocity and the point of highest potential has the lowest velocity.

-
This isn't wrong, but it doesn't address Kepler's second law that was part of the question. –  Stan Liou Jan 1 at 23:11
The reason is that gravity is a radial force, and so conserves angular momentum $\mathbf{L} = \mathbf{r}\times m\mathbf{v}$, where $\mathbf{r}$ is the vector to the planet from the star and $\mathbf{v}$ is its velocity.
First, remember that the cross product of two vectors $\mathbf{a}$ and $\mathbf{b}$ has a magnitude that's equal to the area of a parallelogram with those sides, and hence is twice the area of the triangle with those sides:
If the planet is at position $\mathbf{r}$ and you wait a small amount of time $\mathrm{d}t$, in that time it will be displaced by $\mathrm{d}\mathbf{r} = \mathbf{v}\,\mathrm{d}t$. Think of the area $\mathrm{d}A$ it sweeps out in that time as the area of the triangle defined by its old position $\mathbf{r}$ and its displacement $\mathrm{d}\mathbf{r}$, and therefore its rate of change of the swept area is: $$\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2}\left|\mathbf{r}\times\mathbf{v}\right| = \frac{1}{2m}|\mathbf{L}|\text{,}$$ which is constant by conservation of angular momentum. Since the rate of increase swept area does not change, a line joining the star and planet sweeps out equal areas in equal times.


text:
Orthogonal Sets in an Inner Product Space

# Orthogonal Sets in an Inner Product Space

 Definition: Let $H$ be an inner product space. Two elements $x, y \in H$ are said to be Orthogonal denoted $x \perp y$ if $\langle x, y \rangle = 0$.
 Definition: Let $H$ be an inner product space, $x \in H$, and $S \subseteq H$. Then $x$ is said to be Orthogonal to $S$ denoted $x \perp S$ if $\langle x, y \rangle = 0$ for every $y \in S$.

For example, consider the space $\mathbb{R}^2$ and the vector $(1, 1) \in \mathbb{R}^2$. Let $(x, y) \in \mathbb{R}^2$ and consider the inner product:

(1)
\begin{align} \quad \langle (1, 1), (x, y) \rangle = 1 \cdot x + 1 \cdot y = x + y \end{align}

Then $(1, -1)$ orthogonal to $(1, 1)$, that is:

(2)
\begin{align} \quad (1, 1) \perp (1, -1) \end{align}

Furthermore, if $S = \{ (x, y) \in \mathbb{R}^2 : x + y = 0 \}$, then:

(3)
\begin{align} \quad (1, 1) \perp S \end{align}
 Definition: Let $H$ be an inner product space and $S \subseteq H$. Then the Orthogonal Set of $S$ denoted $S^{\perp}$ is the set of all vectors $x \in H$ that are orthogonal to $S$, that is, $S^{\perp} = \{ x \in H : x \perp S \}$.


text:
Browse Questions

# How many electrons make up one coulomb of negative charge?

Can you answer this question?

$(C) 6.25 \times 10^{18}$


text: The study of top-quark production and decay is central in the LHC physics programme, allowing precise tests of the Standard Model and offering a window on possible new physics. Accurate theoretical predictions for this process are therefore crucial. We report on a new calculation of the next-to-next-to-leading order QCD radiative corrections to the production of top-quark pairs at hadron colliders. The calculation is performed by using the $q_T$-subtraction method to handle and cancel infrared singular contributions at intermediate stages of the computation, and it represents its first complete application to the hadroproduction of a colourful high-mass system at next-to-next-to-leading order. We discuss the calculation of the additional soft contributions needed to implement $q_T$ subtraction for this process, and we show first numerical results.


text:
# Observational Evidence for High Neutronization in Supernova Remnants: Implications for Type Ia Supernova Progenitors

 dc.contributor.author Martínez-Rodríguez, Héctor dc.contributor.author Badenes, Carles dc.contributor.author Yamaguchi, Hiroya dc.contributor.author Bravo, Eduardo dc.contributor.author Timmes, F. X. dc.contributor.author Miles, Broxton J. dc.contributor.author Townsley, Dean M. dc.contributor.author Piro, Anthony L. dc.contributor.author Mori, Hideyuki dc.contributor.author Andrews, Brett dc.contributor.author Park, Sangwook dc.date.accessioned 2018-08-24T13:17:22Z dc.date.available 2018-08-24T13:17:22Z dc.date.issued 2017-07-01 dc.identifier.citation Martínez-Rodríguez, H., et al. (2017): Observational Evidence for High Neutronization in en_US Supernova Remnants: Implications for Type Ia Supernova Progenitors. The Astrophysical Journal, 843(1). dc.identifier.uri http://ir.ua.edu/handle/123456789/3768 dc.description.abstract The physical process whereby a carbon–oxygen white dwarf explodes as a Type Ia supernova (SN Ia) remains en_US highly uncertain. The degree of neutronization in SN Ia ejecta holds clues to this process because it depends on the mass and the metallicity of the stellar progenitor, and on the thermodynamic history prior to the explosion. We report on a new method to determine ejecta neutronization using Ca and S lines in the X-ray spectra of Type Ia supernova remnants (SNRs). Applying this method to Suzaku data of Tycho, Kepler, 3C 397, and G337.2−0.7 in the Milky Way, and N103B in the Large Magellanic Cloud, we find that the neutronization of the ejecta in N103B is comparable to that of Tycho and Kepler, which suggests that progenitor metallicity is not the only source of neutronization in SNe Ia. We then use a grid of SN Ia explosion models to infer the metallicities of the stellar progenitors of our SNRs. The implied metallicities of 3C 397, G337.2−0.7, and N103B are major outliers compared to the local stellar metallicity distribution functions, indicating that progenitor metallicity can be ruled out as the origin of neutronization for these SNRs. Although the relationship between ejecta neutronization and equivalent progenitor metallicity is subject to uncertainties stemming from the 12C + 16O reaction rate, which affects the Ca/S mass ratio, our main results are not sensitive to these details. dc.format.mimetype application/pdf en_US dc.subject atomic data en_US dc.subject nuclear reactions, nucleosynthesis, abundances en_US dc.subject ISM: supernova remnants en_US dc.subject X-rays: ISM en_US dc.title Observational Evidence for High Neutronization in Supernova Remnants: Implications for Type Ia Supernova Progenitors en_US dc.type text en_US


text:
# Resting state fMRI and Dynamic Time Wrapping

Resting state fMRI provides neural measurements of the functional relationship between areas of the brain. Rs-fMRI data is particularly useful for investigation of clinical populations. It allows for investigation of the disruption brain networks without the added complexity of variation associated with task-related brain activation Plitt et al., 2015, Smith, 2009. It may be applied in the investigation of mental states, memory and the recall of events, clinical populations, among others Fox, 2010, Shirer et al., 2012. Rs-fMRI has been shown to be highly reproducible and provides data sets that can be easily compared across studies Franco et al., 2013, Shehzad, 2009.

The synchronized spontaneous low frequency fluctuations of the BOLD signal within a task-free environment (resting state) are known to represent the functional connections of different brain areas.

#### Approaches used to compute functional connectivity:

1. Seed based correlation analysis (SCA), a whole-brain connectivity map is obtained by iteratively calculating the linear correlation between the time series of a seed region or voxel and every other region or voxel specified using correlation coefficient as the measure of similarity.
2. Independent component analysis (ICA),
• data driven approach. identifies a predefined number of spatial templates and corresponding time courses to model the data
• activities of the individual voxel are linear combination of these component time series.
• determine the main source or sources that affect the BOLD time course of the corresponding voxel, thus resulting in a fuzzy clustering of voxels.
• Each independent source defines a component, therefore clusters arise not only from actual neural synchronization, but from other signal sources, like measurement noise and artifacts.
3. other

#### How do we look at Resting state fMRI now? What could go wrong?

1. The correlation of low frequency fluctuations on resting-state fMRI arises from fluctuations in blood oxygenation or flow.
2. Traditional resting-state network concept is based on calculating linear dependence of spontaneous low frequency fluctuations of the BOLD signals of different brain areas, which assumes temporally stable zero-lag synchrony across regions. It is a manifestation of functional connectivity of the brain (Biswal et al., 1995).
3. To investigate brain connectivity, a correlation is calculated for the average of the time series of the regions of interest. The correlation is used to build a connectivity matrix.
4. However the brain functional connectivity exhibits dynamic changes and a complex time-lag structure, which cannot be captured by the static zero-lag correlation analysis.
5. Bold signal are not same for all brain regions, example default mode network ( what is default mode network) shows negatively correlated bold signals in normal human begin. Get a better example please something is missing

#### Resting state fMRI data preprocessing technique involved:

• slice time correction
• motion correction
• normalization of voxel intensity
• Nuisance signal removal
• low frequency drift (linear trend and quadratic trend)

#### Why DTW, what property of DTW make it more desirable measure for analysis of fMRI signal?

1. Originally developed for time series analysis and classification of speech signals and have proven results in the field, this would mean that DTW should be able to handle autocorrelation induced by noise component with the fMRI signal just like the speech. (inference considering the similarity of data).
2. DTW distance measure are less sensitive to linearly combined noise in the data.
3. DTW, performs non linear wrapping on the compared time series, this is can be exploited to correct the non stationary time lags introduced by the dynamic transformation of brain states.
4. Non linear wrapping, would also help in negating the shape distortion between brain regions due to variability in neurovascular coupling /hemodynamic response function (I believe both are same but still check this out)

### What is Dynamic Time Wrapping?

DTW (Sakoe and Chiba, 1978) is a family of algorithms which compute the local stretch or compression to apply to the time axes of two time series in order to optimally map one onto the other. DTW outputs the remaining cumulative distance between the two and, if desired, the mapping itself (warping function). In another word they can compute the similarity between time series which may vary in time (i.e. wrap in time).

DTW is widely used e.g. for classification and clustering tasks in Econometrics, Chemometrics and general time series mining. Originally designed for speech recognition, Basically DTW finds the optical global alignment between two time series by exploiting temporal distortion between them.

Multiple studies as confirmed that for the problem involving time series data for example a classification task, algorithm exploiting the DTW to wrap the invariance among the signal is hard to beat. [1][2]

#### Things to be considered while implementing or using DTW:

1. Z Normalizing
2. just-in-time normalization
3. early abandoning techniques
4. Applying endpoint constrain
5. setting up wrapping window

#### Computing DTW,

Compute the distance matrix(cost matrix) between every possible pair of points between the two time series. Any possible wrapping between two time series will be a path through the computed cost matrix.

$DTW(q,c) = min \Bigg( \frac{\sqrt{\sum_{k=1}^{K}w_k)}}{K} \Bigg)$, here w is the wrapping constant.

The optimal (minimum) path or wrapping between two time series will provide us with the DTW, which can be obtained using the below recursive function.

$\gamma(i,j) = distance(q_i,c_i) + min( \gamma(i-1,j-1), \gamma(i-1,j), \gamma(1,j-1))$

Its important to note that DTW is symmetric and has constancy of self similarity but does not follow positivity (separation) and triangular inequality. This would be the best way to say DTW is a distance measure and not a metric. Now interesting if we increase the sample space to an enormous level, we would be able to find lots of A,B, and C which would follow triangular inequality(just read this over the internet try to find the source). Mathematically in the limit, as w tends to zero DTW is a metric, interesting isn’t it?

#### Reference

[1] Bagnall, A., Lines, J., Bostrom, A., Large, J., & Keogh, E. (2017). The great time series classification bake off: a review and experimental evaluation of recent algorithmic advances. Data Mining and Knowledge Discovery, 31(3), 606-660.

[2] Bagnall, A., Lines, J., Bostrom, A., Large, J., & Keogh, E. (2017). The great time series classification bake off: a review and experimental evaluation of recent algorithmic advances. Data Mining and Knowledge Discovery, 31(3), 606-660.


text:
• # question_answer Direction: These questions are based on following graphs. Classification of appeared candidates in a competitive test from different States and qualified candidates from those States. Total appeared candidates = 45,000. Total qualified candidates = 9000. What is the ratio between number of candidates qualified from States B and D together and the number of candidates appeared from State C respectively ? A) 8 : 37 B) 11 : 12       C) 37 : 40 D) 7 : 37

(c) No. of qualified candidates from States B and D =1440 +1890 = 3330 No. of appeared candidates from State C = 3600 3330 $\therefore$Required ratio = $\frac{3330}{3600}$ = 37:40


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# Extended real number line

Jump to: navigation, search

In mathematics, the affinely extended real number system is obtained from the real number system by adding two elements: + ∞ and – ∞ (read as positive infinity and negative infinity respectively). These new elements are not real numbers. It is useful in describing various limiting behaviors in calculus and mathematical analysis, especially in the theory of measure and integration. The affinely extended real number system is denoted $\overline{\mathbb{R}}$ or [–∞, +∞] or ℝ ∪ {–∞, +∞}.

When the meaning is clear from context, the symbol +∞ is often written simply as .

## Motivation

### Limits

We often wish to describe the behavior of a function $f(x)$, as either the argument $x$ or the function value $f(x)$ gets "very big" in some sense. For example, consider the function

$f(x) = x^{-2}.$

The graph of this function has a horizontal asymptote at y = 0. Geometrically, as we move farther and farther to the right along the $x$-axis, the value of $\frac{1}{x^2}$ approaches 0. This limiting behavior is similar to the limit of a function at a real number, except that there is no real number to which $x$ approaches.

By adjoining the elements $+\infty$ and $-\infty$ to $\mathbb{R}$, we allow a formulation of a "limit at infinity" with topological properties similar to those for $\mathbb{R}$.

To make things completely formal, the Cauchy sequences definition of $\mathbb{R}$ allows us to define $+\infty$ as the set of all sequences of rationals which, for any $K>0$, from some point on exceed $K$. We can define $-\infty$ similarly.

### Measure and integration

In measure theory, it is often useful to allow sets which have infinite measure and integrals whose value may be infinite.

Such measures arise naturally out of calculus. For example, in assigning a measure to $\mathbb{R}$ that agrees with the usual length of intervals, this measure must be larger than any finite real number. Also, when considering infinite integrals, such as

$\int_1^{\infty}\frac{dx}{x}$

the value "infinity" arises. Finally, it is often useful to consider the limit of a sequence of functions, such as

$f_n(x) = \begin{cases} 2n(1-nx), & \mbox{if } 0 \le x \le \frac{1}{n} \\ 0, & \mbox{if } \frac{1}{n} < x \le 1\end{cases}$

Without allowing functions to take on infinite values, such essential results as the monotone convergence theorem and the dominated convergence theorem would not make sense.

## Order and topological properties

The affinely extended real number system turns into a totally ordered set by defining $-\infty\leq a \leq +\infty$ for all $a$. This order has the desirable property that every subset has a supremum and an infimum: it is a complete lattice.

This induces the order topology on $\overline{\mathbb{R}}$. In this topology, a set $U$ is a neighborhood of $+\infty$ if and only if it contains a set $\{x : x > a\}$ for some real number $a$, and analogously for the neighborhoods of $-\infty$. $\overline\mathbb R$ is a compact Hausdorff space homeomorphic to the unit interval $[0, 1]$. Thus the topology is metrizable, corresponding (for a given homeomorphism) to the ordinary metric on this interval. There is no metric which is an extension of the ordinary metric on $\mathbb{R}$.

With this topology the specially defined limits for $x$ tending to $+\infty$ and $-\infty$, and the specially defined concepts of limits equal to $+\infty$ and $-\infty$, reduce to the general topological definitions of limits.

## Arithmetic operations

The arithmetic operations of $\mathbb{R}$ can be partially extended to $\overline\mathbb{R}$ as follows:

\begin{align} a + \infty = +\infty + a & = +\infty, & a & \neq -\infty \\ a - \infty = -\infty + a & = -\infty, & a & \neq +\infty \\ a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\ a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\ \frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\ \frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\ \frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0) \end{align}

For exponentiation, see Exponentiation#Limits of powers. Here, "$a + \infty$" means both "$a + (+\infty)$" and "$a - (-\infty)$", while "$a - \infty$" means both "$a - (+\infty)$" and "$a + (-\infty)$".

The expressions $\infty - \infty, 0 \times (\pm\infty)$ and $\pm\infty/\pm\infty$ (called indeterminate forms) are usually left undefined. These rules are modeled on the laws for infinite limits. However, in the context of probability or measure theory, $0 \times \pm\infty$ is often defined as $0$.

The expression $1/0$ is not defined either as $+\infty$ or $-\infty$, because although it is true that whenever $f(x) \to 0$ for a continuous function $f(x)$ it must be the case that $1/f(x)$ is eventually contained in every neighborhood of the set $\{-\infty, +\infty\}$, it is not true that $1/f(x)$ must tend to one of these points. An example is $f(x) = (\sin x)/x$ (as $x$ goes to infinity). (The modulus $|1/f(x)|$, nevertheless, does approach $+\infty$.)

## Algebraic properties

With these definitions $\overline\mathbb{R}$ is not a field, nor a ring, and not even a group or semigroup. However, it still has several convenient properties:

• $a + (b + c)$ and $(a + b) + c$ are either equal or both undefined.
• $a + b$ and $b + a$ are either equal or both undefined.
• $a \times (b \times c)$ and $(a \times b) \times c$ are either equal or both undefined.
• $a \times b$ and $b \times a$ are either equal or both undefined
• $a \times (b + c)$ and $(a \times b) + (a \times c)$ are equal if both are defined.
• If $a \leq b$ and if both $a + c$ and $b + c$ are defined, then $a + c \leq b + c$.
• If $a \leq b$ and $c > 0$ and if both $a \times c$ and $b \times c$ are defined, then $a \times c \leq b \times c$.

In general, all laws of arithmetic are valid in $\overline\mathbb{R}$ as long as all occurring expressions are defined.

## Miscellaneous

Several functions can be continuously extended to $\overline\mathbb{R}$ by taking limits. For instance, one defines $\exp(-\infty) = 0, \exp(+\infty) = +\infty, \ln(0) = -\infty, \ln(+\infty) = +\infty$ etc.

Some discontinuities may additionally be removed. For example, the function $1/x^2$ can be made continuous (under some definitions of continuity) by setting the value to $+\infty$ for $x = 0$, and $0$ for $x = +\infty$ and $x = -\infty$. The function $1/x$ can not be made continuous because the function approaches $-\infty$ as $x$ approaches 0 from below, and $+\infty$ as $x$ approaches $0$ from above.

Compare the real projective line, which does not distinguish between $+\infty$ and $-\infty$. As a result, on one hand a function may have limit $\infty$ on the real projective line, while in the affinely extended real number system only the absolute value of the function has a limit, e.g. in the case of the function $1/x$ at $x = 0$. On the other hand

$\lim_{x \to -\infty}{f(x)}$ and $\lim_{x \to +\infty}{f(x)}$

correspond on the real projective line to only a limit from the right and one from the left, respectively, with the full limit only existing when the two are equal. Thus $e^x$ and $\arctan(x)$ cannot be made continuous at $x = \infty$ on the real projective line.


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# Is a vector space containing values in $\mathbb{R}^n$ itself a subspace of $\mathbb{R}^n$?

Such that if $$W$$ is a subspace of $$ℝ^n$$ and $$V=$${$$v∈ ℝ^n | v ∉ W$$}, is $$V$$ also a subspace of $$ℝ^n$$?

I think that it is, considering that if all elements of $$V$$ are in the set $$ℝ^n$$ that would mean that $$V$$ has to be as well, but another student disagrees. Why is it/is it not?

• You state your reason for thinking that $V$ is a subspace as “if all elements of $V$ are in $\mathbb{R}^n$ that would mean that $V$ has to be as well”. That sounds to me like you are confusing subspace with subset. You are correct that $V$ is a subset of $\mathbb{R}^n$. A subspace is a subset that satisfies certain properties. To check if a subset is a subspace, you have to check if it satisfies those properties. – Joe Oct 1 '19 at 2:49

$$W$$ is a subspace, so it contains the zero vector $$\mathbf{0}$$. Hence $$V$$ cannot contain $$\mathbf{0}$$, so it cannot be a vector space.
Here's a slightly different take, if this helps. Take $$W$$ to be the $$x$$-axis in $$\mathbb{R}^2$$. Then your $$V$$ consists of all vectors off the $$x$$-axis. Now, $$(1,1)$$ and $$(-1,-1)$$ are both vectors in $$V$$. But their sum is $$\mathbf{0}$$, which isn't in $$V$$ (because it is in $$W$$). So, $$V$$ isn't closed under addition.
• Your very definition of $V$ tells me that $\mathbf{0} \notin V$. If $\mathbf{0} \in V$ then this means $\mathbf{0} \notin W$, which is silly. Have you looked at any concrete examples? Consider $W$ as the $x$-axis inside $\mathbb{R}^2$. – Randall Oct 1 '19 at 0:42


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# Expression to disable internet to speed up Quantity

There's a way to disable internet or WolframAlpha connectivity. I know Quantity caches unit lookups, but I have reasons that I need to disable internet connectivity. Can someone help me with that expression?

• \$AllowInternet = False will disable internet in its entirety. There may be some internal function to apply this specifically to Quantity, or some other way to tackle this problem, but I don't know what that would be personally. – user6014 Jan 22 '18 at 16:20
• Thank you! That's intuitive enough. – Eric William Smith Jan 22 '18 at 16:28


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# How to write this Norm constraint effieiciently?

(Dipak Narayanan) #1

I have a known Binary Matrix \bf A of size M\times N.
Let \bf B (also a binary matrix of the same size) be an optimization variable.

I want \bf B to be as close as possible. The difference between \bf A and \bf B should be minimal.

The following constraint is working and giving me the desired output. First I vectorize the matrices and use the following constraint.

       norm (A(:)-B(:),1)<=4;

But the problem is that CVX is taking much longer time to give me the output.

How can I resolve this issue? Can I impose the constraint in different ways so that it is efficiently solved?

(Mark L. Stone) #2

Well it is a binary optimization, so that can potentially be very time consuming to solve - in the worst case, for n not very large, it may not complete before the sun dies. I believe CVX presents this problem to the solver as a Binary ILinear programming problem, which is a type of MILP, Are there any constraints (other than binary restriction)? If so, does the solver have difficulty finding a feasible solution or difficulty closing the MIP gap to optimality?

.Options I can think of are

1. Use this norm and consider changing solver options using cvx_solver_settings http://cvxr.com/cvx/doc/solver.html#advanced-solver-settings. Thus could be adjusting parameters which control solution strategy, or specifying a larger than default MIP gap

2. Try another solver. I think your choices under CVX are Gurobi and Mosek.

3. Try a different norm, such as ‘fro’ or ‘inf’

4. Choose a means of accessing the solver which allows you to specify an initial (starting) value.for the optimization variable. CVX does not let you do this. Perhaps this will help if you know a good, but not optimal solution. Or if there are constraints, and the solver called by CVX is having difficulty finding any feasible solution.

(Dipak Narayanan) #3

Different norms does not provide the expected results. Gurobi is solving much faster! However, is there any other way (formulation) I can define the same constraint, minimizing the mismatch between two matrices?

(Mark L. Stone) #4

Changing the norm changes the measure of closeness being optimized. it is up to you to decide what is a good measure of closeness for your purposes.


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# SAGECAL [1]¶

This chapter is a step by step description of the SAGECAL method for self-calibration of LOFAR data. The mathematical framework of the algorithm can be found in Yatawatta et al. 2009 and Kazemi et al. 2011. The contents are applicable to the latest version (0.3.8) of SAGECAL and older versions are obsolete. The latest source code can be obtained from http://sagecal.sf.net.

## Introduction¶

The acronym SAGECAL stands for Space Alternating Generalized Expectation Maximization Calibration. The Expectation Maximization is used as a solution to maximum likelihood estimation to reduce the computational cost and speed of convergence. The commonly used Least Squared calibration method involves the inversion of a matrix corresponding to the full set of unknown parameters; where the convergence to a local minimum impels a slow speed of convergence and significant computational cost. The SAGE algorithm, on the other hand, allows to compute a direct estimation of subsets unknown parameters, providing faster convergence and/or reduced computational costs. If K is the number of sources and N are the stations, the computational cost scales as $${\mathcal O} ((KN)^2)$$ for the Least Squared, while is $${\mathcal O} (K N^2)$$ for the Expectation Maximization.

BBS uses the Least Squared algorithm to solve the Measurement Equation. The maximum number of directions for which solve using directional gains with BBS is currently limited to 5 or 6 directions; this is a consequence of the limited computing power available in CEP2 and CEP3. The typical LOFAR Field customarily requires to solve for a number of directions generally higher than 5 or 6, this is due to the wide field of view (FOV), variable beam pattern and ionospheric errors. SAGECAL has been tested to solve to a maximum of 300 directions (in the GPU EoR cluster not in CEP2 or CEP3) and 512 stations. The version installed in the CEP clusters is not optimized so that other users can also use the same node while SAGECAL is running; e.g. a total of about 50 directions have been successfully tested in CEP1 cluster.

## Using SAGECAL¶

### Data preparation¶

Before running SAGECAL, you need to take a few precautions. First of all, the data format is the Measurement Set (MS), so no extra conversions are necessary. Averaging in time is not essential, but it is definitely recommended in order to work with smaller datasets (you can average down to e.g. 10 seconds). If data have been already demixed, this step is not needed. The MS can have more than one channel. Also it is possible to calibrate more than one MS together in SAGECAL. This might be useful to handle situations where the data is very noisy. An interesting note about SAGECAL: If the conventional demixing could not be performed on your data (e.g. if the A-team was too close to the target source) you will be able to successfully remove the A-team sources using this new algorithm by working on averaged data(!). Another procedure you will need apply to your data before SAGECAL is to calibrate them in the standard way with BBS, solving for the four G Jones elements as well as for the element beam. After BBS, you will need to flag the outliers using DPPP. You are now ready to start the algorithm.

All the programs related to the SAGECAL can be found in /opt/cep/sagecal/bin/ or see /home/sarod/bin for the latest build.

### Model¶

Make an image of your MS (using casapy or awimager). You can use Duchamp to create a mask for the image. To create a sky model, you can adopt buildsky for point sources and shapelet_gui in case of extended emission (see Sky model construction using Shapelets). Note that you are free to use any other source finder (like PyBDSF) if you feel more confident with it. A new functionality has been implemented to transform the the BBS model into the model format used by SAGECAL. The important is that in the sky model any source name starting with ‘S’ indicates shapelet, ‘D’ a disk, ‘R’ a ring, ‘G’ a Gaussian, while others keys are point sources. In the following, we report a few useful sky models examples:

## name h m s d m s I Q U V spectral_index RM extent_X(rad) extent_Y(rad)
P1C1 0 12 42.996 85 43 21.514 0.030498 0 0 0 -5.713060 0 0 0 0 115039062.0
P5C1 1 18 5.864 85 58 39.755 0.041839 0 0 0 -6.672879 0 0 0 0 115039062.0

# A Gaussian mjor,minor 0.1375,0.0917 deg diameter, pa 43.4772 deg
G0  5 34 31.75 22 00 52.86 100 0 0 0 0.00 0 0.0012  0.0008 -2.329615801 130.0e6

D01 23 23 25.67 58 48 58 80 0 0 0 0 0 0.000715 0.000715 0 130e6

R01 23 23 25.416 58 48 57 70 0 0 0 0 0 0.00052 0.00052 0 130e6

# A shapelet ('S3C61MD.fits.modes' file must be in the current directory)
S3C61MD 2 22 49.796414 86 18 55.913266 0.135 0 0 0 -6.6 0 1 1 0.0 115000000.0

Note that it is also possible to have sources with 3rd order spectra (with -F 1 option). Here is such an example:

##  name h m s d m s I Q U V spectral_index0 spectral_index1 spectral_index2
PJ1C1 18 53 33.616 86 10 19.559 0.008594 0 0 0 -5.649676 -2.0 -60.0 0 0 0 0 152391463.2

But all the sources should have either 1st order or 3rd order spectra, mixing is not allowed.

The number of directions you want to solve for are described in the cluster file. In here, one or more sources corresponding to the patch of the skymodel you need to correct for are combined together as the following example:

## cluster_id chunk_size source1 source2 ...
0 1 P0C1 P0C2
1 3 P11C2 P11C1 P13C1
2 1 P2C1 P2C2 P2C3

Note: comments starting with a ‘#’ are allowed for both sky model and cluster files.

### SAGECAL¶

SAGECAL will solve for all directions described in the cluster file and will subtract the sources listed in the sky model. Note that if you are not interested in the residual data, putting negative values for cluster_id will not subtract the corresponding sources from data. Run SAGECAL as follows:

sagecal -d my.MS -s my_skymodel -c my_clustering -t 120 -p my_solutions

This will read the data from the DATA column of the MS and write the calibrated data to the CORRECTED_DATA column. If these columns are not present, you have to create them first.

If you need to calibrate more than one MS together, first create a text file with all the MS names, line by line. Then run

sagecal -f MS_names.txt -s my_skymodel -c my_clustering -t 120 -p my_solutions

Running sagecal -h will provide the additional options, some of them are:

-F sky model format: 0: LSM, 1: LSM with 3 order spectra : default 0
-I input column (DATA/CORRECTED_DATA) : default DATA
-O ouput column (DATA/CORRECTED_DATA) : default CORRECTED_DATA
-e max EM iterations : default 3
-g max iterations  (within single EM) : default 2
-l max LBFGS iterations : default 10
-m LBFGS memory size : default 7
-n no of worker threads : default 6
-t tile size : default 120
-x exclude baselines length (lambda) lower than this in calibration : default 0

-k cluster_id : correct residuals with solution of this cluster : default -99999
-j 0,1,2... 0 : OSaccel, 1 no OSaccel, 2: OSRLM, 3: RLM:  4: RTR, 5: RRTR: default 0

Use a solution interval (e.g. -t 120) that is big enough to get a decent solution and not too big to make the parameters vary too much (about 20 minutes per solution is a reasonable value).

In case of both bright and faint sources in the model, you might need to use different solution intervals for different clusters. In order to do that you need to define the values of the second column of the cluster file in such a way that the cluster with the longest solution interval is 1. While the cluster with shorter solution interval will be equal to n, where n is the number of times the longer solution interval is divided. This means that if -t 120 is used to select 120 timeslots, cluster 0 will find a solution using the full 120 timeslots, while cluster 1 will solve for every 120/3=40 timeslots. The option -k will allow to correct the residuals using the solutions calculated for a specific direction which is defined by the cluster_id. The -k option is analogue to the correct step in BBS. See the simulations section later in this chapter for more detailed use of this.

You are now ready to image the residual data. Successively, run “restore” (see Sky model construction using Shapelets) on the residual image before updating the sky model and starting another loop of SAGECAL. SAGECAL cycles can be done till you are satisfied by the end product.

### Robustness¶

Many have experienced flux loss of weaker background sources after running any form of directional calibration. There is a new algorithm in SAGECAL that minimizes this. To enable this, use -j 2 option while running SAGECAL. The theory can be found in Kazemi and Yatawatta, 2013. Also for best speed and robustness, it is recommended to use -j 5. Also for number of stations greater than 64, -j 5 option is faster and less memory consuming.

### Simulations¶

After running SAGECAL, you get solutions for all the directions used in the calibration. Using these solutions, it is possible to correct the data for each direction and make images of that part of the sky. This will in theory enable to image the full field of view with correct solutions applied to each direction. This is a brief description on how to do it. We will use an example to illustrate this. Assume you have run SAGECAL as below:

sagecal -d my.MS -s my_skymodel -c my_clustering -t 120 -p my_solutions

Now you have the my_solutions file that contains solutions for all the directions given in my_clustering file. The -a 1 option or -a 2 option enables simulation mode in SAGECAL. If -a 1 is given, the sky model given by my_skymodel and my_clustering is simulated (with gains taken from solutions if -p my_solutions is given) and written to the output data. If -a 2 is given, the model given by my_skymodel and my_clustering is simulated (with the gains if -p my_solutions is given), and then added to the input data (-I option) and written to the output data (-O option).

There is one additional thing you can do while doing a simulation: correction for any particular direction. By using -k cluster_id , you can select any cluster number from the my_clustering file to use as the solutions that are used to correct the data. If the cluster_id specified is not present in the my_clustering file, no corrections will be applied.

Simulating the full sky model back to the data might be too much in some cases. It is also possible to simulate only a subset of clusters back to the data. This is done by specifying a list of clusters to ignore during the simulation, using -z ignore_list option. Here, ignore_list is a text file with the cluster numbers (one per line) to ignore. Note also that even while clusters are ignored, their solutions can still be used to correct the data (so these options are independent from each other).

Thus, by cleverly using the -a, -k and -z options, you can get an image that is fully corrected (using SAGECAL solutions) for all directional errors and moreover, simulations take only a fraction of the time taken for calibration.

## Distributed calibration¶

It is possible to use SAGECAL to calibrate a large number of subbands, distributed across many computers. This is done by using OpenMPI and SAGECAL together; see sagecal-mpi -h for further information. Here is a simple example:

### MPI¶

Using MPI, you can run a program across a network of computers. The basic way to run any program with MPI is

mpirun -np N -hostfile machines.txt (your program) (program options)

where -np gives the number of processes (can be different from the number of computers) and -hostfile gives the computer names to use (machines.txt includes the computer names to use). Always recommended to use the full path for the program to run. If you are only using one computer, you can skip the hostfile option.

### Initial setup¶

Let’s say there are 10 subbands, SB1.MS, SB2.MS, to SB10.MS in one computer. First you need to create a text file (say mslist.txt)

SB1.MS
SB2.MS
...
SM10.MS

that includes all the MS names, one per each line. Since we are running on one computer, no need to create a hostfile.

### SAGECAL options¶

Apart from the usual command line options used in running stand alone SAGECal, there are a few special options for running it in a disributed way.

• The number of ADMM iterations is given by -A option.
• The type of polynomial in freqyency to use as a regularizer is given by -Q option.
• The number of freqyency terms in the polynomial is given by -P option.
• The regularization (penalty) factor is given by the -r option.

### Running¶

Since there are 10 subbands, we need to invoke 10+1 SAGECal processes, this is done by selecting -np 11. The simplest way to run is

mpirun -np 11 /home/sarod/bin/sagecal-mpi -A 10 -P 2 -r 10  -s my_skymodel -c my_clustering -t 120 -p my_solutions

this will calibrate each subband the usual way, but the solutions will be much better because information across the frequency range covered by the subbands is used. Note that almost all options used for standalone SAGECal can still be used here.

## References¶

1. Yatawatta et al., “Radio interferometric calibration using the SAGE algorithm}”, IEEE DSP, Marco Island, FL, Jan. 2009.
1. Kazemi and S. Yatawatta et al., “Radio interferometric calibration using the SAGE algorithm”, MNRAS, 414-2, 2011.
1. Kazemi and S. Yatawatta, “Robust radio interferometric calibration using the T-distribution”, MNRAS, 2013.

Footnotes

 [1] The authors of this chapter are Emanuela Orru and Sarod Yatawatta.


text:
# I Why this triple integral equals zero?

1. Sep 5, 2016

### topito2

Hello everyone, I have the this inquiry:

if I compute de following integral:
http://micurso.orgfree.com/Picture1.jpg
by numerical methods I get cero as a result. I used Maxima and Mathematica and their functions for numerical integration give me an answer equal to cero.
But, if I apply transformation with spherical coordinates the result turns out to be equal to pi^2 / 40.
I can't seem to find an explanation for this. Could any of you guys give me a hand with this? Thank you in advance.

2. Sep 5, 2016

### Staff: Mentor

Well we'd need to see how you transformed the integral and its limits into spherical coordinates before we could even attempt to say what happened.

One question I had was whether you transformed the dxdydz properly or not.

https://en.m.wikipedia.org/wiki/Volume_element

Where you can see that dXdydz transforms into:

$dV = \rho^2 sin \theta d\rho d\theta d\phi$

Last edited: Sep 5, 2016
3. Sep 6, 2016

### Staff: Mentor

The thread title asks why this is zero, which is quite easy to show. The first two integrations, over z and y, are trivial to do, and the result is an odd function of x.

4. Sep 6, 2016

### topito2

Thank you so much for your answers, I rechecked what I did this morning and I found out what happens when calculating the integral in spherical coordinates.

Again, thank you so much for your time and cooperation, you guys rock!!!


text:
### Azimuthal asymmetries of back-to-back $\pi^\pm-(\pi^0,\eta,\pi^\pm)$ pairs in $e^+e^-$ annihilation

H. Li, A. Vossen, H. Aihara, D. M. Asner, V. Aulchenko, T. Aushev, V. Babu, I. Badhrees, A. M. Bakich, J. Bennett, V. Bhardwaj, T. Bilka, J. Biswal, A. Bobrov, M. Bračko, M. Campajola, L. Cao, D. Červenkov, V. Chekelian, A. Chen, B. G. Cheon, H. E. Cho, K. Cho, Y. Choi, S. Choudhury … & V. Zhulanov
This work reports the first observation of azimuthal asymmetries around the thrust axis in $e^+e^-$ annihilation of pairs of back-to-back charged pions in one hemisphere, and $\pi^0$ and $\eta$ mesons in the opposite hemisphere. These results are complemented by a new analysis of pairs of back-to-back charged pions. The $\pi^0$ and $\eta$ asymmetries rise with the relative momentum $z$ of the detected hadrons as well as with the transverse momentum with respect to the thrust...
This data repository is not currently reporting usage information. For information on how your repository can submit usage information, please see our documentation.


text:
Matrix Main Diagonal

(Redirected from Matrix Diagonal)
A Matrix Main Diagonal of an $n \times m$ matrix is a vector $v$ where $v_i = A_{i,i}$.
• In linear algebra, the main diagonal (sometimes leading diagonal or major diagonal or primary diagonal or principal diagonal) of a matrix $A$ is the collection of entries $A_{i,j}$ where $i$ is equal to $j$.


text:
Please subscribe to the official Codeforces channel in Telegram via the link https://t.me/codeforces_official. ×

A. Finding Sasuke
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Naruto has sneaked into the Orochimaru's lair and is now looking for Sasuke. There are $T$ rooms there. Every room has a door into it, each door can be described by the number $n$ of seals on it and their integer energies $a_1$, $a_2$, ..., $a_n$. All energies $a_i$ are nonzero and do not exceed $100$ by absolute value. Also, $n$ is even.

In order to open a door, Naruto must find such $n$ seals with integer energies $b_1$, $b_2$, ..., $b_n$ that the following equality holds: $a_{1} \cdot b_{1} + a_{2} \cdot b_{2} + ... + a_{n} \cdot b_{n} = 0$. All $b_i$ must be nonzero as well as $a_i$ are, and also must not exceed $100$ by absolute value. Please find required seals for every room there.

Input

The first line contains the only integer $T$ ($1 \leq T \leq 1000$) standing for the number of rooms in the Orochimaru's lair. The other lines contain descriptions of the doors.

Each description starts with the line containing the only even integer $n$ ($2 \leq n \leq 100$) denoting the number of seals.

The following line contains the space separated sequence of nonzero integers $a_1$, $a_2$, ..., $a_n$ ($|a_{i}| \leq 100$, $a_{i} \neq 0$) denoting the energies of seals.

Output

For each door print a space separated sequence of nonzero integers $b_1$, $b_2$, ..., $b_n$ ($|b_{i}| \leq 100$, $b_{i} \neq 0$) denoting the seals that can open the door. If there are multiple valid answers, print any. It can be proven that at least one answer always exists.

Example
Input
2
2
1 100
4
1 2 3 6

Output
-100 1
1 1 1 -1

Note

For the first door Naruto can use energies $[-100, 1]$. The required equality does indeed hold: $1 \cdot (-100) + 100 \cdot 1 = 0$.

For the second door Naruto can use, for example, energies $[1, 1, 1, -1]$. The required equality also holds: $1 \cdot 1 + 2 \cdot 1 + 3 \cdot 1 + 6 \cdot (-1) = 0$.


text:
## Metric

Hi All;
I need The solution of the following question"
Let $X$ be a set of measurable functions from a measurable set $I$ to $\Re$ , then we defined a metric $d$ on $X$ By
$d(f,g)=\inf_{\varepsilon\longrightarrow0}(\varepsi lon+m(t:|f(t)-g(t)|\geq \varepsilon))$,
Then prove that d is a metric on $X$.


text:
# Blowing down exceptional curves

Let X be a Noetherian separated scheme. Let E ⊂ X be an effective Cartier divisor such that there is an isomorphism E → P1k where k is a field. Then we say E is an exceptional curve of the first kind if the normal sheaf of E in X has degree -1 on E over k.

You can get an example of the situation above by starting with a Noetherian separated scheme Y and a closed point y such that the local ring of Y at y is a regular local ring of dimension 2 and taking the blowup b : X → Y of y and taking E to be the exceptional divisor.

Conversely, if E ⊂ X is gotten in this manner we say that E can be contracted.

The following questions have been bugging me for a while now.

Question 1: Given an exceptional curve E of the first kind on a separated Noetherian scheme X is there a contraction of E?

Question 2: Given an exceptional curve E of the first kind on a separated Noetherian scheme X is there a contraction of E but where we allow Y to be an algebraic space?

Question 3: Suppose that Y is a separated Noetherian algebraic space and that y is a closed point of Y such that the henselian local ring of Y at y is regular of dimension 2. Is there an open neighbourhood of y which is a scheme?

Question 4: With assumptions as in Question 3 assume moreover that the blow up of Y in y is a scheme. Then is Y a scheme?

In these questions the answer is positive if we assume that X or Y is of finite type over an excellent affine Noetherian scheme (and I think in the literature somewhere; I’d be thankful for references).

But… it might be interesting and fun to try and find counter examples for the general statements. Let me know if you have one!


text:
+0

# Having trouble?

0
226
3

The asymptotes of a hyperbola are $$y = 2x - 3$$ and $$y = 17 - 2x.$$ Also, the hyperbola passes through the point $$(4,7).$$ Find the distance between the foci of the hyperbola.

Aug 5, 2019

#1
+24978
+3

The asymptotes of a hyperbola are $$y = 2x - 3$$ and $$y = 17 - 2x$$.
Also, the hyperbola passes through the point $$(4,7)$$.
Find the distance between the foci of the hyperbola.

Formula hyperbola:

$$\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ asymptote}} & 2x - 3 &=& y \\ & 2x - 3-y &=& 0 \\\\ \hline \mathbf{2. \text{ asymptote}} & 17 - 2x &=& y \\ & 17 - 2x-y &=& 0 \\\\ \hline \text{hyperbola: } & (2x - 3-y)(17 - 2x-y) + c &=& 0 \\\\ P(x=4,y=7) & (8-3-7)(17-8-7)+c &=& 0 \\ & (-2)(2)+c &=& 0 \\ & c &=& 4 \\ \mathbf{\text{hyperbola: }} &\mathbf{ (2x - 3-y)(17 - 2x-y) + 4} &\mathbf{=}& \mathbf{0} \\\\ & (2x - 3-y)(17 - 2x-y) + 4 &=& 0 \\ & 34x-4x^2-2xy-51+6x+3y-17y+2xy+y^2+4 &=& 0 \\ & -4x^2+40x+y^2-14y &=& 47 \\ &\ldots \\ & \mathbf{ \dfrac{(x-5)^2}{1} - \dfrac{(y-7)^2}{4}} &=& \mathbf{ 1 } \\ & \boxed{\mathbf{\text{Formula hyperbola:} } \dfrac{ (x-h)^2 }{a^2}- \dfrac{ (x-k)^2 }{b^2} = 1 } \\\\ & a^2 = 1 \\ & \mathbf{a= 1} \\\\ & b^2 = 4 \\ & \mathbf{b= 2} \\ \hline \end{array}$$

$$\begin{array}{|l|rcll|} \hline \text{Focus}_1: & F_1(h+ae,k) \\ \text{Focus}_2: & F_2(h-ae,k) \\ \hline \text{Foci distance}: & && h+ae-(h-ae) \\ & &=&h+ae-h+ae \\ & &=& 2ae \quad | \quad e=\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2a\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=&\mathbf{ 2 \sqrt{a^2+b^2}} \quad | \quad a^2=1, \quad b^2= 4 \\ & &=& 2 \sqrt{1+4} \\ & &\mathbf{=}& \mathbf{2\sqrt{5} } \\ \hline \end{array}$$

The distance between the foci of the hyperbola is $$\mathbf{2\sqrt{5} }$$

Aug 5, 2019
#2
+934
+1

Wow heureka! You have a greatly presented solution!!!

dgfgrafgdfge111  Aug 5, 2019
#3
+24978
+1

Thank you dgfgrafgdfge111

heureka  Aug 5, 2019


text:
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# Price Discrimination: Airline Tickets

Have you ever tried buying airline tickets with your family or friends? Most of the time the prices for these tickets are changing. Prices might increase when your try to book a weekend flight or if you add check-in luggage. The constant changing of airline tickets is no coincidence. There is a perfectly understandable explanation for this. Read on to find out what it is!

## What is price discrimination?

To understand the price changes for airline tickets, we need to understand what price discrimination means.

As hinted by the term, price discrimination relates to when the same or similar goods/services are sold at different prices.

Price discrimination is a selling strategy that charges customers different prices for the same product or service based on what the seller thinks they can get consumers to agree to.

There are three types of price discrimination:

1. First-degree price discrimination.
2. Second-degree price discrimination.
3. Third-degree price discrimination.

### First-degree price discrimination

First-degree price discrimination is also known as the purest form of price discrimination or perfect price discrimination. This is when a seller charges consumers the maximum possible price they are willing to pay.

Fig. 1 - First-Degree Price Discrimination

As Figure 1 illustrates, the producers get all the total surplus. In this case, there is no consumer surplus.

### Second-degree price discrimination

Second-degree price discrimination occurs when a seller charges a different price for different quantities consumed. Examples of this are loyalty cards rewards or discounts for consumers, so that they can purchase more.

Fig. 2 - Second-Degree Price Discrimination

As figure 2 illustrates, firms initially put the price of a particular good or service at P1. The discounted price is P2. Consumers benefit from the lower price and that increases their consumer surplus and firms also benefit from the extra revenue.

### Third-degree price discrimination

Third-degree price discrimination occurs when a seller charges a different price to different customer groups. A classic example is some cinemas that often have different prices for children, students, adults, and the elderly.

Fig. 3 - Third-Degree Price Discrimination

Figure 3 illustrates third-degree price discrimination. The whole market is split into two categories: elastic and inelastic consumers. Firms charge elastic consumers the lowest price. For example, cinemas charge students a lower price since they have less disposable income and are more sensitive to price changes. Inelastic consumers are charged the highest price. In our cinema example, adults tend to be charged the highest price as they have more disposable income and are less sensitive to price changes compared to other groups.

## Types of pricing strategies adopted for airline tickets

Airlines use a combination of price discrimination and dynamic pricing in the pricing strategy for airline tickets. In the airline industry, airlines practice price discrimination in a number of ways:

1. How far the travel date is: a ticket that is bought in advance is usually cheaper compared to a ticket bought at the last minute.
2. Travel peak times: travelling during the weekends tends to be more expensive than travelling during weekdays. Travelling during the summer or Easter holidays also tends to be more expensive.
3. Unsocial hours: travelling during the early mornings and late nights is cheaper, as fewer people tend to travel at those times.
4. Extras: customers usually have to pay more for seats (in the same class) with more legroom to for check-in luggage.

### Airline tickets and third-degree price discrimination

In the airline industry, customers are usually grouped into two customer groups: leisure and business. This demonstrates that the airline industry tends to practice third-degree price discrimination.

Grouping is done this way due to the distinct behaviours between the two groups. Customers who are travelling for the purpose of leisure are more likely to be sensitive to price and so would tend to book in advance to get cheaper tickets. Leisure travellers take advantage of 'advanced purchase' discounts, as the prices of tickets gradually get more expensive as the day of departure approaches. This is also because leisure travellers would have more elastic demand due to their sensitivity to price.

People who travel for business purposes, however, need more flexibility in comparison to those who travel for leisure and so would book only a few days in advance. Airlines would take advantage of their more inelastic demand and charge at higher prices.

## Airline dynamic pricing

A demand-based pricing strategy that solely relies on grouping passengers as either ‘leisure or business travellers’ is not the most effective discrimination as it does not accurately capture the consumer landscape.

That is why airlines today use what is known as a dynamic pricing strategy. This strategy takes into account real-time data on consumer buying patterns.

Dynamic pricing is a technique of pricing a product according to current market conditions.

With the use of dynamic pricing, the prices of tickets change in real-time based on the market data. This involves things such as customer behaviour, competitor prices, and popular events such as music festivals (like Coachella). Other factors like fuel prices, ground handling, and the cost of carrying an additional traveller are also filtered into the prices of tickets.

## Airline price discrimination example: Southwest Airlines

One real-life example of price discrimination in the airline industry is Southwest Airlines.

A randomly chosen one way trip from Philadelphia, PA (PHL) to Los Angeles, CA (LAX) on Tuesday, January 18 (a weekday) would cost between $366 -$456 US dollars. However, on Saturday, January 22 (a weekend) the same trip would cost between $127 -$217.

Weekdays tickets tend to be more expensive for this specific route, as it is a common assumption that customers who buy a weekday ticket are travelling for business purposes. They have inelastic demand as they must be in Los Angeles for work. They have no other option, so they are more willing to pay a higher price.

On the other hand, weekend tickets are usually cheaper because people who travel during the weekend do so for leisure. If prices are too high they would opt for cheaper alternatives. Therefore, they have elastic demand so airlines charge them lower prices.

It is also relevant to note that similar to other airlines, Southwest charges different prices at different times of the day: early morning and late-night flights are less expensive due to less demand.

## Economic effects of price discrimination in the airline market

The engagement of the airline market in price discrimination has both positive and negative effects on all economic agents: consumers, firms, governments, and third parties.

### Positive effects

Some of the positive effects of price discrimination are:

• Capacity. Charging at different prices enables airlines to make better use of their capacity. This means that they are able to ensure that customers use each flight optimally. It prevents waste in the form of insufficient seats being occupied in a flight. This can also have environmental benefits, as airlines would try to fill up flights, selling seats at the best price possible. In other words, it spreads out demand and prevents flight congestion.
• Revenue. If airlines only charged one high set price, they would lose elastic consumers. Being able to separate consumers based on different conditions allows them to continue to make revenue, even if the price is not at the profit maximising point.
• Lower price. Elastic consumers are charged a lower price compared to inelastic consumers. They benefit from the lower price. This lower price increases their consumer surplus and utility.
• Higher profits. Airlines benefit from inelastic consumers who are willing to pay higher fees. They receive higher profits which they can later use to re-invest back for research that will help them improve their ticket pricing strategy.

### Negative effects

Some of the negative effects of price discrimination are:

• Loss in consumer surplus. Consumer surplus is extracted from inelastic consumers when airlines charge higher prices. Airlines benefit as they make supernormal profits, but inelastic consumers lose out.
• Potentially unfair. Price inelastic consumers, such as business travellers may not all be high-income earners. This may also be unfair and inequitable if they are charged a higher price. But we could argue that most companies wi¡ould pay for the tickets instead of the direct consumer.
• Administration costs. It costs the airlines a lot of money to separate the market and carry out extensive research into dynamic pricing. If these administrative costs are too high, airlines could pass it onto consumers through higher prices.

## Price Discrimination: Airline Tickets - Key takeaways

• Price discrimination is a selling strategy that charges customers different prices for the same product or service, based on what the seller thinks they can get consumers to agree to.
• The three types of price discrimination are first-degree price discrimination, second-degree price discrimination, and third-degree price discrimination.
• The types of pricing strategy in the airline industry are price discrimination and dynamic pricing.
• Dynamic pricing is a technique of pricing a product according to current market conditions.
• Price discrimination in the airline industry includes:
1. Time of buying the ticket - a ticket that is bought in advance is usually cheaper.
2. Travel peak times such as weekends are more expensive.
3. Unsocial hours are cheaper, as fewer people tend to travel at those times.
4. Extras - tickets with check-in luggage and more legroom are more expensive.

## Frequently Asked Questions about Price Discrimination: Airline Tickets

Price discrimination is a selling strategy that charges customers different prices for the same product or service based on what the seller thinks they can get consumers to agree to.

Airlines do this in order to maximise their profits.

Price discrimination is generally not harmful to the economy. However, it can be harmful to the economy when it leads to misdistribution of resources to the point that output and profits are not maximised.

Third-degree price discrimination is when a seller charges a different price to different customer groups. This is done between business and leisure travellers in the airline industry.

## Final Price Discrimination: Airline Tickets Quiz

Question

What is price discrimination?

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Price discrimination is a selling strategy that charges customers different prices for the same product or service, based on what the seller thinks they can get consumer to agree to.

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What are the types of price discrimination?

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The three types of price discrimination are: first-degree price discrimination, second-degree price discrimination and third-degree price discrimination.

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What is first-degree price discrimination?

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First degree discrimination is also referred to as the purest form of price discrimination or perfect price discrimination, as the seller would try to charge the consumer at the maximum possible price for each unit consumed.

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What is second-degree price discrimination?

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Second-degree price discrimination is the one which occurs when a company charges a different price for different quantities consumed, such as discounts given for more purchases.

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What is third-degree price discrimination?

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Third-degree price discrimination on the other hand occurs when a seller charges a different price to different customer groups. a classic example is a cinema which often has different prices for children, students, adults and the elderly.

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What type of pricing strategy do airlines adopt for their tickets?

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Airlines use a combination of price discrimination and dynamic pricing in the pricing strategy for airline tickets.

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In what ways is price discrimination practiced in the airline industry?

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1. Time of buying the ticket - a ticket that is bought in advance is usually cheaper.
2. Travel peak times such as weekends are more expensive.
3. Unsocial hours are cheaper, as less people tend to travel at those times.
4. More is paid for seats (within the same class) with more leg room.
5. People are charged extra for having check-in luggage. For instance, local airline might require people to pay for luggage allowance.

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In the airline industry, customers are usually grouped into two customer groups: ____________.

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leisure and business

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What is dynamic pricing?

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Dynamic pricing is technique of pricing a product according to current market conditions.

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What distinguishes dynamic pricing from other pricing strategies?

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The use of dynamic pricing, causes the prices of tickets change in real time based on the market data. Enabling a more efficient pricing of tickets.

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What type of pricing discrimination is used in Southwest Airlines?

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Third-degree price discrimination.

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How does third-degree price discrimination divide the market?

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It divides the market into inelastic and elastic consumers.

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What are the positive effects of price discrimination in the airline industry?

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Some of the positive effects for price discrimination are:

• Capacity. Charging at different prices enables airlines to make better use of capacity, meaning that they are able to ensure that each flight is optimally used by customers. It prevents waste in the form of insufficient seats being occupied in a flight. This can also have environmental benefits, as airlines would try to fill up flights, selling seats at the best price possible. In other words, it spreads out demand and prevents flight congestion.
• Revenue. If airlines only charged one high set price, they would loose elastic consumers. Being able to separate consumers based on different conditions allows them to continue to make revenue, even if the price is not at the profit maximising point.
• Lower prices. Elastic consumers are charged a lower price compared to inelastic consumers. They benefit from the lower price. This lower price increases their consumer surplus and utility.
• Higher profits. Airlines benefit from inelastic consumers who are willing to pay higher fees. They receive higher profits which they can later use to re-invest back for research that will help them improve their ticket pricing strategy.

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What are the negative effects of price discrimination in the airline industry?

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Some of the negative effects for price discrimination are:

• Loss in consumer surplus. Consumer surplus is extracted from inelastic consumers when airlines charge higher prices. Airlines benefit as they make supernormal profit, but inelastic consumers loose out.
• Potentially unfair. Price inelastic consumers, such as business travellers may not all be high income earners. This may also be seen as unfair and inequitable if they are charged a higher price. But it can be argued that it is their company who pays for their ticket rather than they themselves.
• Administration costs. It costs the airlines a lot of money to separate the market and carry out extensive research into dynamic pricing. If this admin costs are too high, airlines could pass it onto consumers through higher prices.

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text:
# Temperature and Mass Flow effect on Pressure

by samshree
Tags: effect, flow, mass, pressure, temperature
 PF Gold P: 1,491 There are models for this, particularly if you can ensure the tank is adiabatic (no heat transfer in or out of the tank). In that case you can treat the problem as isentropic and use the ideal gas law and the isentropic relations to get what you are after. For example, you know one form of the ideal gas law is ##p = \rho R T## where ##R## is the specific gas constant. For an isentropic process in an ideal gas, you have the following relationships: $$\dfrac{p_2}{p_1} = \left(\dfrac{\rho_2}{\rho_1}\right)^{\gamma} = \left(\dfrac{T_2}{T_1}\right)^{\gamma/(\gamma-1)}$$ where ##\gamma = c_p/c_v##. That should get you where you need to go.