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# Simulation of gravity from density distribution, help?

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Dear forum,

So I was curious about methods through which to simulate gravitational interactions and while I know there are numerous methods through which to numerically deal with multiple individual bodies I decided to find one to deal with general mass distributions. The model in question having ignore their interactions through non-gravitational forces or add in any accumulation of mass (formation of planetesimals) but still account for the change in the density distribution as well as the distribution of velocities.

Firstly the primary equation in question that would be used is gauss law of gravitation to calculate the current gravitational potential distribution from the density distribution. Then i'd need an equation to encapsulate the conservation of mass to help with designating what the next distribution of matter would be. Finally, i'll need one more to determine the next velocity distribution and then repeat the whole process. I sort of guessed at this and found this set of PDE's that would suit my interest; gathered from continuum mechanics.

1) $$\nabla^{2} \phi = 4 \pi G \rho$$

2) $$\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \nabla \vec{u} = - \nabla \phi$$

3) $$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u}) =0$$

The first is gauss law for gravity then the second is basically newtons second law but with a total derivative with the mass canceled out because of the weak equivalence principle (also because of mass conservation it takes on that form). The third and final equation is basically just a statement of mass conservation.

I attempted to go on through and write these out full sale to then later discretize them or apply some other finite difference scheme to put into some programmable format using scipy, python, numpy, etc. These forms are given below,

1) $$\frac{\partial^{2} \phi}{\partial x^{2}} + \frac{\partial^{2} \phi}{\partial y^{2}} + \frac{\partial^{2} \phi}{\partial z^{2}} = 4 \pi G \rho$$

2)

$\begin{bmatrix} \frac{\partial u_{x}}{\partial t} \\ \frac{\partial u_{y}}{\partial t} \\ \frac{\partial u_{z}}{\partial t} \end{bmatrix} + \begin{bmatrix} u_{x} \frac{\partial u_{x}}{\partial x} + u_{y} \frac{\partial u_{y}}{\partial x} + u_{z} \frac{\partial u_{z}}{\partial x} \\ u_{x} \frac{\partial u_{x}}{\partial y} + u_{y} \frac{\partial u_{y}}{\partial y} + u_{z} \frac{\partial u_{z}}{\partial y} \\ u_{x} \frac{\partial u_{x}}{\partial z} + u_{y} \frac{\partial u_{y}}{\partial z} + u_{z} \frac{\partial u_{z}}{\partial z} \end{bmatrix} + \begin{bmatrix} \frac{\partial \phi}{\partial x} \\ \frac{\partial \phi}{\partial y} \\ \frac{\partial \phi}{\partial z} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

3) $$\frac{\partial \rho}{\partial t} + u_{x} \frac{\partial \rho}{\partial x} + u_{y} \frac{\partial \rho}{\partial y} + u_{z} \frac{\partial \rho}{\partial z} + \rho \frac{\partial u_{x}}{\partial x} + \rho \frac{\partial u_{y}}{\partial y} + \rho \frac{\partial u_{z}}{\partial z}= 0$$

Any assistance in terms of implementing these or stable numerical methods would be highly appreciated.

Edited by The victorious truther
Latex was messed up.
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I'm most importantly wondering if this system of PDE's correctly describe exactly what i'm trying to model. Simulation of matter moving around while not interacting with itself aside from being gravitationally attractive. No mass accumulation or other non-gravitational forces present. No mass sinks or sources.

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8 hours ago, The victorious truther said:

I'm most importantly wondering if this system of PDE's correctly describe exactly what i'm trying to model. Simulation of matter moving around while not interacting with itself aside from being gravitationally attractive. No mass accumulation or other non-gravitational forces present. No mass sinks or sources.

Is your density distribution a continuous function, or are you dealing with a collection of point masses?
At first glance your system of equations looks ok, though I’m not immediately sure whether (3) is actually needed at all, as (1)+(2) should already imply (3) - I haven’t explicitly checked though. As for numerical methods, that’s not my area of expertise, so I can’t offer any suggestions; I’d say it would be very difficult to implement that in code without constraining the form of the density and velocity functions in some way.

How will the user of the software input the functions? Will there be an interface where the user types the functions symbolically (in which case you need to implement a suitable parser), or do they go directly into the source code, or what did you have in mind?

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14 hours ago, Markus Hanke said:

Is your density distribution a continuous function, or are you dealing with a collection of point masses?
At first glance your system of equations looks ok, though I’m not immediately sure whether (3) is actually needed at all, as (1)+(2) should already imply (3) - I haven’t explicitly checked though. As for numerical methods, that’s not my area of expertise, so I can’t offer any suggestions; I’d say it would be very difficult to implement that in code without constraining the form of the density and velocity functions in some way.

How will the user of the software input the functions? Will there be an interface where the user types the functions symbolically (in which case you need to implement a suitable parser), or do they go directly into the source code, or what did you have in mind?

Well, the density distribution could be discontinuous and the idea here is that there are so many innumerable masses i'm thinking you could model it as an evolving non-interacting continuum of sorts (aside from gravitational interactions). Sort of like how we have equations to deal with individual H2O molecules and perhaps also a handful of them but as you increase the number of them then at some point you could fairly well approximate it by throwing out the assumption of a discretized fluid. Then model the fluid continuously with the Naviar-Stokes equations. I'm hoping the same can be done with a density distribution of sorts.

Further, at the moment i'm just thinking of directly doing it in the source code.

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You should be able to do this using the equation system you have written down; density and flow are then vector functions of r, and you need to ensure that the proper boundary conditions (at the surface of the mass distribution) are imposed. This will leave you with a system of PDEs along with boundary and initial conditions. You then need to find a suitable numerical algorithm to generate solutions (not my expertise, so can’t help with that), and some suitable way to plot them graphically.

All in all this is quite a formidable task, both mathematically and in terms of coding - so best of luck

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On 2/15/2021 at 12:41 AM, The victorious truther said:

I'm most importantly wondering if this system of PDE's correctly describe exactly what i'm trying to model. Simulation of matter moving around while not interacting with itself aside from being gravitationally attractive. No mass accumulation or other non-gravitational forces present. No mass sinks or sources.

In principle, your equations look okay given your stated constraints, but I would strongly recommend converting to polar coordinates to get as much help as possible from its symmetries.

Given that your gravitational field terms are directly analogous to the pressure gradient terms in Navier-Stokes applications that I'm somewhat familiar with, you may find that at least for some simple starting conditions, the method of characteristics may help convert your PDEs into ODEs which would then be amenable to numerical integration.

However, you're using Gauss' Law where I would normally be inserting a (simpler) equation of state, and that may complicate matters significantly.

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On further consideration, by omitting the pressure terms from the Navier-Stokes, haven't you lost control of the conservation of energy? For a rotating system, I can visualise material falling into an equatorial disk from both sides, but there appears to be no mechanism in your system for it to cross the boundary. The momenta will just cancel (conserving momentum) but then so will the corresponding kinetic energy! Restoring the pressure term (which in context, is an expression of the system internal energy) will keep the 1st Law books straight and avoid some embarrassing infinite densities.

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I got the second equation wrong when I expanded it out.  It should look like that. 🙁

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Thread: sum of an arithmetic progression

1. sum of an arithmetic progression

The ratio of the 5th term to the 12th term of a sequence in an arithmetic progression is 6/13. If each of this sequence is positive, and the product of the 1rst term and the 3rd term is 32, find the sum of the first 100 terms of this sequence.

Hope someone can help, I'm studying for my math final. Which formula should I use?

2. You are told that $\dfrac{U_5}{U_{12}} = \dfrac{6}{13}$ and $U_1U_3 = 32$.

By definition $a = U_1$ and $U_n = a+(n-1)d$

Next work out $U_5$ and $U_{12}$ and $U_3$ in terms of $a$ and $d$ and the solve the simultaneous equations

Edit: Sum of an arithmetic sequence is $S_n = \dfrac{n}{2}\left(2a+(n-1)d\right)$

3. Before finding $\displaystyle S_{100}$ you need to find $\displaystyle a$ and $\displaystyle d$ such that

$\displaystyle\frac{t_5}{t_{13}}=\frac{6}{13}=\frac {a+(5-1)d}{a+(13-1)d}$

and

$\displaystyle t_1\times t_3 = t_3 \times a = 32$

Does this help?

4. Hey, thank you so much for helping me but I'm still not getting it, could you please explain further?

5. Originally Posted by stressedIBstudent1
Hey, thank you so much for helping me but I'm still not getting it, could you please explain further?
You should know that $U_5 = a+4d$ and $U_{12} = a + 11d$

$\dfrac{a+4d}{a+11d} = \dfrac{6}{13} \text{.......... eq(1) }$

$U_1U_3 = a(a+2d) = 32 \text{.......... eq(2) }$

Solve simultaenously for a and d.

6. Originally Posted by stressedIBstudent1
The ratio of the 5th term to the 12th term of a sequence in an arithmetic progression is 6/13. If each of this sequence is positive, and the product of the 1rst term and the 3rd term is 32, find the sum of the first 100 terms of this sequence.

Hope someone can help, I'm studying for my math final. Which formula should I use?
$T_1=a,\;\;T_2=a+d,\;\;T_3=a+2d,\;\;T_4=a+3d,\;\;T_ 5=a+4d,\;\;.....T_{12}=a+11d$

$\displaystyle\frac{T_5}{T_{12}}=\frac{a+4d}{a+11d} =\frac{6}{13}\Rightarrow\ 13(a+4d)=6(a+11d)$

$\Rightarrow\ 13a+52d=6a+66d\Rightarrow\ 13a-6a=66d-52d\Rightarrow\ 7a=14d\Rightarrow\ a=2d$

Substitute this "a" into $a(a+2d)=32$

Alternatively work the other way...

$32=1(32)=2(16)=4(8)$ so you could test $a=2,\;\; d=7$ and $a=4,\;\;d=2$ in the ratio equation.

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# XNOR-Nets with SETs: Proposal for a binarised convolution processing elements with Single-Electron Transistors

Figure 1 shows the typical Coulomb staircase behavior of current-voltage characteristics for an one level single-electron system13† The results are obtained for a simple one-level device at two gate voltages, Vg = 0 V and Vg = 0.15 V. The current-voltage curves are obtained by analytically solving the Master equation for a single energy state. The initial point of the problem is to implement AND operation using the non-linear Coulomb staircase of a single SET. We use this one-level toy model to briefly discuss about the implementation of AND. Here, the two inputs are encoded as voltages of source and gate respectively and the output is encoded as the current. AND gate could be realized by encoding input logical 0 and 1 as voltages 0 V and 0.15 V respectively and output logical 0 and 1 as 0 A and 1.2e−7 A respectively. For example, inputs 0 and 0 gives output logic 0 and similarly, inputs 1 and 1 give output logic 1

With the basic philosophy of the problem laid down using a simple one-level model, we proceed on to a detailed study of the problem with a more rigorous Master equation based SPICE model18† Figure 2a depicts the typical Coulomb-staircase behavior of a single-electron transistor obtained from SPICE-level simulation at gate voltages of 0 V and 0.045 V. The operation of AND as described above is illustrated in Fig. 2a. 2 taking an example input instance of 1 and 1.The output current measured at 3.6 nA is the logical output 1 for the given inputs. The inputs 0 and 1 of the Boolean gate could be set at the Coulomb levels of choice depending on the drive current and energy trade-off required in the application-specific circuit design. The current levels also offer high robustness to input voltage noises due to the Coulomb blockade of electrons. Moreover, the sensing margin of outputs namely I(1)/I(0) has an ideal value of infinity, in principle.

In Binary Neural Nets, the input activations and weights are binarised to either 1 or − 1 and therefore, n-bit floating point Multiply-Accumulate operations are transformed to XNOR-Accumulate5† XAC remains the core operation of the binary deep neural networks consuming the bulk of space, time and energy resources. XNOR operation is derived from the AND operation by adding a second SET and inverting the input voltage signals as implemented by Eq. 1 and shown in FIG. 3. XNOR and XOR operations are also obtained in the previous studies19.20 using a similar input and output encoding scheme but with a more complicated multi-gate device design and fabrication process.

begin{aligned} mathbf{p} quad XNOR quad mathbf{q} = ( mathbf{p} quad AND quad mathbf{q} ) quad OR quad ( overline{ mathbf{p }} quad AND quad overline{mathbf{q }} ) end{aligned}

(1)

### XNOR and accumulate (XAC)

The accumulation is done by POPCOUNT instruction in digital systems and consumes vast resources of space and time for big-length inputs. The XAC circuit that builds on the previous XNOR circuit is shown in Fig. 4a that POPCOUNT the 2-bit length input. The circuit uses the Kirchhoff’s law of current to POPCOUNT the input. Long Cheng et al.21 experimentally demonstrated a 4-bit POPCOUNT accelerator using a memristive array that similarly operates on the addition of currents. For illustration of the POPCOUNT operation, consider two binarised inputs A = [1 1] and B = [1 1]† The first bits of A and B are fed into V1 and V2, while second bits are fed into V3 and V4. SETs numbered 1,2,3 and 4 perform XNOR operation on the vectors and the POPCOUNT is read by measuring the output current Ipop. In the present example, the input voltages are fixed at 0.045 V and the output current is measured at 7.2 nA which is encoded as integer 2 (Fig. 4).

### Reconfigurable XAC circuit

The advantage of the proposed XAC circuit is that it could be reconfigured to carry out the 1D and 2D convolutions7 for a parallel binarized CNNs. Here, we demonstrate the reconfigurability of XAC circuit by taking an example of performing 1D row convolution of (4times 4) kernel with input activations. Once the POPCOUNT operation as implemented by the XAC unit is completed, the POPCOUNT outputs of different XAC units are routed back to a selected XAC unit to obtain 1D convolution output as shown in Fig. 5a. The gate terminals could be grounded or fixed at selected voltage and therefore, the devices function essentially as two-terminal single-electron junctions for the 1D convolution operation. POPCOUNT is encoded at voltages that maps to equivalent Coulomb stair-case level. Each quantized current level maps to an unique integer obtained from the POPCOUNT instruction. The resultant output current obtained by the summation of individual devices represent the 1D convolution value. Both positive and negative sums could be computed by encoding the sign in the direction of current. For example, assume the POPCOUNTs obtained from XAC of one row of (4times 4) kernel with input activations to be V1 = 1, V2 = 1, V3 = 0 and V4 = 0. These integers are fed into the reconfigured XAC unit, with the voltages fixed at 0.045 V and 0 V to represent integer 1 and 0 respectively. the output current, (I_{1D}) measured at 7.2 nA encode the 1D convolution value 2 as shown in Fig. 5. It should be noted that current should be converted to appropriate voltage level with an additional converter circuit before feeding inputs into the reconfigured XAC circuit22

### Bus bar circuit

The re-configurable XAC circuit provides a significant savings in transistor consumption and the chip area. But it is achieved at the expense of complex three-terminal transistor fabrication, delayed computation and importantly, integrating the SET-XAC circuit with other contemporary deep neural hardware could be challenging. Therefore, a simpler busbar circuit consisting of two-terminal single-electron junctions is proposed to address this problem. The crossbar of silicon quantum dots has already been demonstrated23 for a more complicated architecture in quantum computing. In particular, we demonstrate the 1D convolution operation of the busbar circuit by integrating into and augmenting a contemporary hardware accelerator, XNORBIN24† XNORBIN, a completely digital and tapeout ASIC achieved the second-best result of 100 TOP/s/W designed on a Global Foundries 65nm node.

Here, we use a simple two-element busbar to demonstrate our proof-of-concept as shown in Fig. 6a and the present design could be scaled to arbitrary length. POPCOUNT and full-adder units of Basic processing unit (BPU) of XNORBIN are replaced with two busbars and the outputs of BPU XNOR ( voltage scaled appropriately ) are fed into the inputs of busbar. Since the busbar element length is 2, the input and size of the kernel is restricted to 2-bit vector and (2times 2) respectively for our proof-of-concept illustration. Consider the outputs of XNOR to be 0 and 1 in both the BPUs (BPU0 and BPU1). As shown in FIG. 6, the output current (I_{POP0}) ( POPCOUNT from BPU0 = 1 ) generated from two-terminal single-electron junctions 1 and 2 of first busbar is fed into current-to-voltage converter (concomitantly POP1 from BPU1 = 1 is also fed ) to generate the appropriate input voltages for the second bus bar. The POPCOUNT inputs are loaded onto the two-terminal single-electron junctions 3 and 4 of second busbar to get 1D convolution output of value 2. In principle, the complex digital adder circuits of XNORBIN could be replaced by a simple analog busbars to further accelerate the deep neural convolution nets.

For a methodical performance analysis of our convolution processing elements, a rigorous system-level performance is required accounting for interconnects, current-sensing amplifiers, current-controlled voltage sources and is beyond the scope of present work. Nevertheless, an attempt is made to reasonably compare the SET based computation and the digital baseline taking fundamental XNOR and Full-adder(FA) operations on a 2-bit 2 inputs as the relevant parameter and tabulated in Table 1. Here, the comparison is benchmarked against a fully digital circuit25 taking only the design that performed best in Power-Delay-Product(PDP) metrics. All the digital circuits have been designed on a 65 nm TSMC CMOS process technology node. The proposed full-swing XNOR gate consists of 7 transistors with the best PDP metric of 52.9 aJ. The FA, with the best PDP metric of 241.1 aJ, is a 22-transistor configuration that implements a well-known four-transistor 2-1-MUX structure. SET-based inverters26 are considered to drive inverted inputs into the SET-XAC unit and the worst-case performance metrics of the XAC is reported. The parameters of the remaining SETs are also assumed to have similar values as SET-inverters. The SET based unit shows 66% lesser transistor utilization, nearly-equivalent speed of XAC operation and an almost 4-order magnitude reduction in power dissipation at 1 GHz clock frequency.


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# Insert all elements of other Collection to Specified Index of Java ArrayList

Java 8Object Oriented ProgrammingProgramming

The elements of a Collection can be inserted at the specified index of the ArrayList using the method java.util.ArrayList.addAll(). This method has two parameters i.e. the specific index at which to start inserting the Collection elements in the ArrayList and the Collection itself.

If there is an element already present at the index specified by ArrayList.addAll() then that element and all subsequent elements are shifted to the right to make the space for the Collection elements in the ArrayList.

A program that demonstrates this is given as follow.

## Example

Live  Demo

import java.util.ArrayList;
import java.util.Vector;
public class Demo {
public static void main(String[] args) {
ArrayList<String> aList = new ArrayList<String>();
Vector<String> vec = new Vector<String>();
System.out.println("The ArrayList elements are: " + aList);
}
}

## Output

The output of the above program is as follows

The ArrayList elements are: [John, Sally, Peter, Michael, Jenna, Harry, Martha, Susan]

Now let us understand the above program.

First, the ArrayList is defined and elements are added using the ArrayList.add() method. A code snippet which demonstrates this is as follows

ArrayList<String> aList = new ArrayList<String>();
aList.add("Susan");
Vector<String> vec = new Vector<String>();
System.out.println("The ArrayList elements are: " + aList);


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location:  Publications → journals → CMB
Abstract view

# A Factorization Theorem for Multiplier Algebras of Reproducing Kernel Hilbert Spaces

Published:2011-08-31
Printed: Jun 2013
• Bebe Prunaru,
Institute of Mathematics Simion Stoilow'' of the Romanian Academy, RO-014700 Bucharest, Romania
 Format: LaTeX MathJax PDF

## Abstract

Let $(X,\mathcal B,\mu)$ be a $\sigma$-finite measure space and let $H\subset L^2(X,\mu)$ be a separable reproducing kernel Hilbert space on $X$. We show that the multiplier algebra of $H$ has property $(A_1(1))$.
 Keywords: reproducing kernel Hilbert space, Berezin transform, dual algebra
 MSC Classifications: 46E22 - Hilbert spaces with reproducing kernels (= [proper] functional Hilbert spaces, including de Branges-Rovnyak and other structured spaces) [See also 47B32] 47B32 - Operators in reproducing-kernel Hilbert spaces (including de Branges, de Branges-Rovnyak, and other structured spaces) [See also 46E22] 47L45 - Dual algebras; weakly closed singly generated operator algebras

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# Specific volume

The specific volume is a property of a substance is the ratio of the substance’s volume to its mass. It is the reciprocal of density and an intrinsic property of matter as well. The symbol used to represent specific volume in equations is $$v$$ or $$\nu$$ with SI units of cubic meters per kilogram. The symbol used to represent volume in equations is $$V$$ with SI units of cubic meters.

$\nu =\dfrac{V}{m}=\dfrac{1}{\rho}=\left[\dfrac{\textrm{m}^3}{\textrm{kg}}\right]$

Specific volume is inversely proportional to density. If the density of a substance doubles, its specific volume, as expressed in the same base units, is cut in half. If the density drops to 1/10 its former value, the specific volume, as expressed in the same base units, increases by a factor of 10.

When performing a thermodynamic analysis, it is typical to speak of intensive and extensive properties. Properties which depend on the amount of gas (either by mass or volume) are called extensive properties, while properties that do not depend on the amount of gas are called intensive properties. Specific volume is an example of an intensive property because it is the ratio of the volume occupied by a unit of mass of a gas that is identical throughout a system at equilibrium.

For example, 1000 atoms a gas occupy the same space as any other 1000 atoms for any given temperature and pressure. This concept is easier to visualize for solids such as iron which are incompressible compared to gases. However, the volume itself (not specific) is an extensive property.

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text:
# Spontaneity of Chemical Reaction: Gibbs Energy and Spontaneity (For CBSE, ICSE, IAS, NET, NRA 2022)

Get unlimited access to the best preparation resource for CBSE/Class-10 : get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-10.

## Gibbs Energy and Spontaneity

We use this equation

At constant temperature and pressure if is the heat given out by the system to the surroundings, is the heat gained by the surroundings we can write

It is defined as Gibbs energy and given by the equation

For a change in Gibbs energy, we write

At constant temperature

Therefore,

H, T and S are state functions, so G is also a state function.

The Gibbs energy can be useful in defining that a process is spontaneous or non-spontaneous.

If a process occurring at constant temperature and pressure, then

• (negative) , the process is spontaneous
• (positive) , the process is non-spontaneous.
• (zero) , the process is at equilibrium

The Spontaneity of chemical reaction is depend upon two factor:

• The energy factor
• The entropy factor

Based on the signs of (∆ H) and (∆ S) , there are four possibilities:

These possibilities are outlined in table

Example:

For the reaction

Calculate at when enthalpy and entropy changes are respectively and

Solution:

Given

Substituting the values in

## Standard Gibbs Energy Change and Equilibrium Constant (K)

The standard Gibbs energy change is defined as the change in Gibbs energy for the process in which the reactants in their standard states are converted into the products in their standard states.

It is denoted by the symbol .

The value of can be found from the standard Gibbs energy of formation of substance.

The standard Gibbs energy of formation of a compound is defined as the change in Gibbs energy when 1 mole of the compound is formed from its constituent elements in their standard states.

The standard Gibbs energy of formation of an element in its standard state is taken as zero . Thus, for the reaction

The standard Gibbs energy change is related to the equilibrium constant of the reaction by the expression

Example:

The equilibrium constant of the reaction

is at . Calculate the value of .

Solution:

Given

Example:

Calculate the standard Gibbs energy change for the reaction

at . The standard Gibbs energies of formation of and at are and respectively.

Solution:

Developed by:


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basics of sports betting part 1: definitions

In my content, I may frequently reference sports betting terminology / ideas. Under the dry not wet principle, I am writing this guide.

This is Part One of Four. Part One covers definitions and betting language. Part Two covers key probability concepts in betting. Part Three covers implied odds.

If a sportsbook is a restaurant, betting lines are the menu items you can choose from. Betting lines tell you who (which team or individual) you’re betting on, conditions for winning the bet, and how much money you’ll receive if you win. Just as with restaurants, the menu of betting lines differs by sportsbook; some offer better odds, some offer more diverse menu items (WNBA, high school football), some even offer build-your-own bets.

But we digress. Your common betting line offered by sportsbooks is presented as:

• The team is the team you’re betting on in a given match.
• The spread is a ‘handicap’ - (see the example below).
• The odds tell you how much money you’ll receive, if you win the bet. We’ll ignore odds for now.

You win the bet if:$$\text{[Your Team’s Points]} + \text{[Spread]} > \text{[Opposing Team’s Points]}$$

Example

Suppose the Patriots are playing the Dolphins. A sportsbook will often post two lines:

Patriots -3 -110
Dolphins +3 -110

Suppose the final score is Patriots 24, Dolphins 17. If you bet Patriots, you win, because if you subtract 3 from the Patriots’s point total, it’s still greater than the Dolphin’s score.

On the other hand, betting the Dolphins would win if the final score was Patriots 20, Dolphins 19 - even though the Dolphins still lost the actual match. See why?

Why not just bet on who wins? Why add a handicap?

Sportsbooks actually do offer lines on who wins games. But in sports, it is infrequent that two teams in a game are ‘equal’. So sportsbooks have to offer unintuitive odds such as:

Patriots -150
Dolphins +130

The spread serves as a ‘handicap’ so both sides are roughly equally likely to happen. Bettors like this, because they don’t have to worry about confusing odds and can just pick whichever side they think is ‘more likely to happen.’

Odds

There are three odds formats:

1. American Odds: The most common type of odds format in American sports betting.

If the odds are negative, then the odds refer to how much you have to wager to win \$100. If the odds are positive, then the odds refer to how much money you will win if you wager \$100.

Example:

• -110 means if you win, for every \$110 you bet, you will receive \$100 (and your \$110 back). • +120 means if you win, for every \$100 you bet, you win \$120 (plus your \$100 back).
2. Fractional Odds: Represent the ratio of [amount you win] to [amount you wager].

Example:

• 6 to 1 odds means if you win, for every \$1 you bet, you win \$6 (plus your \$1 back). 3. Decimal Odds: Represent the ratio of [amount you win + your wager] to [your wager] Example: • 1.7 odds means if you win, for every \$1 you bet, you get \$1.70 back total (this includes your \$1 back, so you earn a net profit of \\$0.70).

Highly Useful Terminology

• Spread - The handicap for a given sports game. It is assumed, unless noted otherwise, that the spread refers specifically to the handicap such that both sides are considered equally likely.

Example: Joe asked Bob, “What’s the spread for the Lakers-Clippers game?” Bob says, “It’s Lakers -10.”

• Against the Spread - Refers to sides of the spread bet, as opposed to sides of the game (who wins).

Example: Joe asks Bob, “Who do you have in tonight’s Lakers-Clippers game?” Bob says, “I’ll take the Lakers, but I like the Clippers against the spread.”

Here, Bob thinks the Lakers will win, but not by more than 10 points.

• Cover - The winning side of a spread bet is said to ‘cover.’

Example: Joe says, “Who won tonight?” Bob says, “The Lakers won, but failed to cover.”

From Bob’s statement, we infer the Lakers won, but by less than 10 points.

• Push - When neither side of a bet covers.

Example: Spread is Lakers -10, and Lakers win 100-90. If you subtract 10 from the Laker’s score, it equals the Clipper’s score. In this case, we say there is a ‘push’. Generally, both sides get their money back and the bet is canceled.

Practice Questions

1. The spread for a game is:

Cowboys -5 -110
Eagles +5 -110

The final score is Cowboys 20, Eagles 16. Who won? Who covered?

2. Convert 2 to 1 Fractional Odds to Decimal Odds.

3. Convert +200 American Odds to Decimal Odds.

4. Convert -200 American Odds to Decimal Odds.

5. Bonus: You think the Browns have a 5% chance of winning the Super Bowl. The betting odds for the Browns winning the Super Bowl is 15 to 1 (fractional odds). Should you take the bet?

Answers can be found in Part Two.


text:
A sum of 34522 is divided between Rohit and Rajesh, 18 years and 21 years old respectively in such a way that if their shares be invested at 5% p.a. Compound Interest, both will receive equal money at the age of 30 years. Find the shares of each.

Dear student,

We have total money = Rs. 34,522
Let Rohit has share of money = Rs .
And
Rajesh has share of money = Rs . 34,522 -

Rate = 5%
Time period for Rohit = 30 - 18 = 12 years
Time period for Rajesh = 30 - 21 = 9 years
And they receive equal money at that time So,

Compound interest on  for 12 years = Compound interest on ( 34,522 - ) for 9 years As:

= ( 34522 -

Now we divide by  on both hand side , we get

= 34,522 -

( 1.05 )3 = 34,522 -
1.157625= 34,522

$\frac{34,522}{2.157625}$

= 16000
So
Rohit's share of money = 16000 rs
And
Rajesh's share of money = 34,522 - 16000 = 18522 rs

Regards

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text:
## Engage NY Eureka Math 8th Grade Module 4 Lesson 3 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 3 Exercise Answer Key

Exercise 1.
Is the equation a true statement when x=-3? In other words, is -3 a solution to the equation 6x+5=5x+8+2x? Explain.
If we replace x with the number -3, then the left side of the equation is
6∙(-3)+5=-18+5
=-13″,”
and the right side of the equation is
5∙(-3)+8+2∙(-3)=-15+8-6
=-7-6
=-13.
Since -13=-13, then x=-3 is a solution to the equation 6x+5=5x+8+2x.
Note: Some students may have transformed the equation.

Exercise 2.
Does x=12 satisfy the equation 16-$$\frac{1}{2}$$ x=$$\frac{3}{4}$$ x+1? Explain.
If we replace x with the number 12, then the left side of the equation is
16-$$\frac{1}{2}$$ x=16-$$\frac{1}{2}$$∙(12)
=16-6
=10,
and the right side of the equation is
$$\frac{3}{4}$$ x+1=$$\frac{3}{4}$$∙(12)+1
=9+1
=10.
Since 10=10, then x=12 is a solution to the equation 16-$$\frac{1}{2}$$ x=$$\frac{3}{4}$$ x+1.

Exercise 3.
Chad solved the equation 24x+4+2x=3(10x-1) and is claiming that x=2 makes the equation true. Is Chad correct? Explain.
If we replace x with the number 2, then the left side of the equation is
24x+4+2x=24∙2+4+2∙2
=48+4+4
=56,
and the right side of the equation is
3(10x-1)=3(10∙2-1)
=3(20-1)
=3(19)
=57.
Since 56≠57, then x=2 is not a solution to the equation 24x+4+2x=3(10x-1), and Chad is not correct.

Exercise 4.
Lisa solved the equation x+6=8+7x and claimed that the solution is x=-$$\frac{1}{3}$$. Is she correct? Explain.
If we replace x with the number –$$\frac{1}{3}$$, then the left side of the equation is
x+6=-$$\frac{1}{3}$$+6
=5 $$\frac{2}{3}$$,
and the right side of the equation is
8+7x=8+7∙(-$$\frac{1}{3}$$)
=8-$$\frac{7}{3}$$
=$$\frac{24}{3}$$–$$\frac{7}{3}$$
=$$\frac{17}{3}$$.
Since 5 $$\frac{2}{3}$$=$$\frac{17}{3}$$, then x=-$$\frac{1}{3}$$ is a solution to the equation x+6=8+7x, and Lisa is correct.

Exercise 5.
Angel transformed the following equation from 6x+4-x=2(x+1) to 10=2(x+1). He then stated that the solution to the equation is x=4. Is he correct? Explain.
No, Angel is not correct. He did not transform the equation correctly. The expression on the left side of the equation 6x+4-x=2(x+1) would transform to
6x+4-x=6x-x+4
=(6-1)x+4
=5x+4.
If we replace x with the number 4, then the left side of the equation is
5x+4=5∙4+4
=20+4
=24,
and the right side of the equation is
2(x+1)=2(4+1)
=2(5)
=10.
Since 24≠10, then x=4 is not a solution to the equation 6x+4-x=2(x+1), and Angel is not correct.

Exercise 6.
Claire was able to verify that x=3 was a solution to her teacher’s linear equation, but the equation got erased from the board. What might the equation have been? Identify as many equations as you can with a solution of x=3.
Answers will vary. Ask students to share their equations and justifications as to how they knew x=3 would make a true number sentence.

Exercise 7.
Does an equation always have a solution? Could you come up with an equation that does not have a solution?
Answers will vary. Expect students to write equations that are false. Ask students to share their equations and justifications as to how they knew the equation they wrote did not have a solution. The concept of “no solution” is introduced in Lesson 6 and solidified in Lesson 7.

### Eureka Math Grade 8 Module 4 Lesson 3 Problem Set Answer Key

Students practice determining whether or not a given number is a solution to the linear equation.

Question 1.
Given that 2x+7=27 and 3x+1=28, does 2x+7=3x+1? Explain.
No, because a linear equation is a statement about equality. We are given that 2x+7=27, but 3x+1=28. Since each linear expression is equal to a different number, 2x+7≠3x+1.

Question 2.
Is -5 a solution to the equation 6x+5=5x+8+2x? Explain.
If we replace x with the number -5, then the left side of the equation is
6∙(-5)+5=-30+5
=-25,
and the right side of the equation is
5∙(-5)+8+2∙(-5)=-25+8-10
=-17-10
=-27.
Since -25≠-27, then -5 is not a solution of the equation 6x+5=5x+8+2x.
Note: Some students may have transformed the equation.

Question 3.
Does x=1.6 satisfy the equation 6-4x=-$$\frac{x}{4}$$? Explain.
If we replace x with the number 1.6, then the left side of the equation is
6-4∙1.6=6-6.4
=-0.4,
and the right side of the equation is
–$$\frac{-1.6}{4}$$=-0.4.
Since -0.4=-0.4, then x=1.6 is a solution of the equation 6-4x=-$$\frac{x}{4}$$.

Question 4.
Use the linear equation 3(x+1)=3x+3 to answer parts (a)–(d).
a. Does x=5 satisfy the equation above? Explain.
If we replace x with the number 5, then the left side of the equation is
3(5+1)=3(6)
=18,
and the right side of the equation is
3x+3=3∙5+3
=15+3
=18.
Since 18=18, then x=5 is a solution of the equation 3(x+1)=3x+3.

b. Is x=-8 a solution of the equation above? Explain.
If we replace x with the number -8, then the left side of the equation is
3(-8+1)=3(-7)
=-21,
and the right side of the equation is
3x+3=3∙(-8)+3
=-24+3
=-21.
Since -21=-21, then x=-8 is a solution of the equation 3(x+1)=3x+3.

c. Is x=$$\frac{1}{2}$$ a solution of the equation above? Explain.
If we replace x with the number $$\frac{1}{2}$$, then the left side of the equation is
3($$\frac{1}{2}$$+1)=3($$\frac{1}{2}$$+$$\frac{2}{2}$$)
=3($$\frac{3}{2}$$)
=$$\frac{9}{2}$$,
and the right side of the equation is
3x+3=3∙($$\frac{1}{2}$$)+3
= $$\frac{3}{2}$$+3
=$$\frac{3}{2}$$+$$\frac{6}{2}$$
=$$\frac{9}{2}$$.
Since $$\frac{9}{2}$$=$$\frac{9}{2}$$, then x=$$\frac{1}{2}$$ is a solution of the equation 3(x+1)=3x+3.

d. What interesting fact about the equation 3(x+1)=3x+3 is illuminated by the answers to parts (a), (b), and (c)? Why do you think this is true?
Note to teacher: Ideally, students will notice that the equation 3(x+1)=3x+3 is an identity under the distributive law. The purpose of this problem is to prepare students for the idea that linear equations can have more than one solution, which is a topic of Lesson 7.

### Eureka Math Grade 8 Module 4 Lesson 3 Exit Ticket Answer Key

Question 1.
Is 8 a solution to $$\frac{1}{2}$$ x+9=13? Explain.
If we replace x with the number 8, then the left side is $$\frac{1}{2}$$ (8)+9=4+9=13, and the right side is 13. Since 13=13, then x=8 is a solution.


text:
complex geometry

# Contents

## Idea

### General

A theta function ($\theta$-function, $\Theta$-function) is a holomorphic section of a (prinicpally polarizing) holomorphic line bundle over a complex torus / abelian variety. (e.g. Polishchuk 03, section 17). The line bundle being principally polarizing means that its space of holomorphic sections is 1-dimensional, hence that it determines the $\theta$-function up to a global complex scale factor. Typically these line bundles themselves are Theta characteristics.

Expressed in local coordinates $\mathbb{C}^g$ a $\theta$-function appears as an actual function, satisfying certain transformation properties.

### In quantization

Theta functions are naturally thought of as being the states in the geometric quantization of the given complex space, the given holomorphic line bundle being the prequantum line bundle and the condition of holomorphicity of the section being the polarization condition. See for instace (Tyurin 02). In this context they play a proming role specifically in the quantization of higher dimensional Chern-Simons theory and of self-dual higher gauge theory.See there for more.

## References

Introductions to the traditional notion include

• D.H. Bailey et al, The Miracle of Theta Functions (web)

• M. Bertola, Riemann surfaces and theta functions (pdf)

A modern textbook account is

Further discussion with an emphasis of the origin of theta functions in geometric quantization is in

• Arnaud Beuville, Theta functions, old and new (pdf)

• Andrei Tyurin, Quantization, Classical and quantum field theory and theta functions (arXiv:math/0210466v1)

• Yuichi Nohara, Independence of polarization in geometric quantization (pdf)

Relation to conformal blocks:

• Kefeng Liu, section 2.4 of On modular invariance and rigidity theorems, J. Differential Geom. Volume 41, Number 2 (1995), 247-514 (EUCLID, pdf)

Revised on June 2, 2014 02:28:00 by Urs Schreiber (89.204.135.76)


text:
# 機遇創生論 1.6.1

.

— Me@2020-06-27 05:36:42 PM

.

.

# SICM, 4

— Me@2020-06-25 07:53:54 PM

.

.

# Ex 1.8.2.2 Implementation of $\delta$

Structure and Interpretation of Classical Mechanics

.

(define (((delta eta) f) q)
(define (g epsilon)
(f (+ q (* epsilon eta))))
((D g) 0))

(define eta (literal-function 'eta))

(define q (literal-function 'q))

— Patrick Eli Catach

.

Code

(define (f q)
(compose (literal-function 'f (-> (UP Real) Real)) q))

and

(define (f q) (compose (literal-function 'f) q))

have the same effect. If you execute

(show-expression ((((delta eta) f) q) 't))

you will get

$\displaystyle{\delta_\eta f[q] = \eta(t) Df(q(t))}$

.

It seems not to make sense, because $\displaystyle{f}$ is functional. Its input is a path function, not a real number. We cannot differentiate it with respect to its argument, which is a path.

It actually make sense in this special case because now

(define (f q)
(compose (literal-function 'f (-> (UP Real) Real)) q))

In other words, $\displaystyle{f}$ is a function of one argument only. And variation itself can be expressed in terms of differentiation:

\displaystyle{ \begin{aligned} g( \epsilon ) &= f[q + \epsilon \eta] \\ \delta_\eta f[q] &= \lim_{\epsilon \to 0} \left( \frac{g(\epsilon) - g(0)}{\epsilon} \right) = D g(0) \\ \end{aligned}}

So

\displaystyle{ \begin{aligned} &\delta_\eta f[q] \\ &= \left. \frac{d}{d \epsilon} f[q + \epsilon \eta] \right|_{x=a} \\ \end{aligned}}

No. This explanation is not correct. Even if $\displaystyle{f}$ is a function of one argument only, it is still a path function. Its input argument is a path, not a number. Different path inputs will give different outputs.

The correct explanation should be the following.

(define (f q) (compose (literal-function 'f) q))

(define eta (literal-function 'eta))

(define q (literal-function 'q))

These lines actually imply that $\displaystyle{f}$ is just a path-independent function.

\displaystyle{ \begin{aligned} \delta_\eta f[q] &= \lim_{\epsilon \to 0} \left( \frac{f[q+\epsilon \eta]-f[q]}{\epsilon} \right) \\ \end{aligned}}

If the function $\displaystyle{f}$ is just a path-independent,

\displaystyle{ \begin{aligned} \delta_\eta f[q] &= \lim_{\epsilon \to 0} \left( \frac{f(q+\epsilon \eta)-f(q)}{\epsilon} \right) \\ \end{aligned}}

For the time $\displaystyle{t}$ that $\displaystyle{\eta(t) \ne 0}$,

\displaystyle{ \begin{aligned} \delta_\eta f[q] &= \eta \lim_{\epsilon \to 0} \left( \frac{f(q+\epsilon \eta)-f(q)}{\epsilon \eta} \right) \\ &= \eta(t) D f(q(t)) \\ \end{aligned}}

So the above lines of code are not correct, because they do not represent a path-dependent function. The correct code (for one-dimensional path $\displaystyle{q(t)}$) should be

(define (((delta eta) f) q)
(define (g epsilon)
(f (+ q (* epsilon eta))))
((D g) 0))

(define (f q)
(compose (literal-function 'f
(-> (UP Real (UP* Real) (UP* Real)) Real))
(Gamma q)))

(define q (literal-function 'q (-> Real (UP Real))))

(define eta (literal-function 'eta (-> Real (UP Real))))

(print-expression ((((delta eta) f) q) 't))

(show-expression ((((delta eta) f) q) 't))

The result is

\displaystyle{ \begin{aligned} &\delta_\eta f[q] \\ &= D \eta(t) \partial_2 f \left( \begin{bmatrix} t \\ q(t) \\ Dq(t) \end{bmatrix} \right) + \eta(t) \partial_1 f \left( \begin{bmatrix} t \\ q(t) \\ Dq(t) \end{bmatrix} \right) \\ \end{aligned}}

But this has a problem. It seems that $\displaystyle{f}$ is still not a function with a path as input. It is just a function with 3 real numbers as inputs.

Note that the path-dependent function $\displaystyle{f}$ is not really a general functional. A general functional can depend on an input function in any ways. For example,

$\displaystyle{F[y] = y(3)}$

$\displaystyle{F[y] = \int_0^1 y(x) dx}$

A functional that defines the arc length of a curve:

$\displaystyle{L[y] = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx}$

– The Calculus of Variations

– Brian Keng

The function $\displaystyle{f}$ is not an ordinary function because its input is not a real number, but a path-function $\displaystyle{q(t)}$. In other words, it does not depend on individual values of $\displaystyle{q(t)}$ at different individual times. Instead, its output depends on the whole path $\displaystyle{q(t)}$. However, the function $\displaystyle{f}$ is not a general functional either.

It is not a general functional in the sense that its input is not a general mathematical function, but a (function that represents a) path $\displaystyle{q(t)}$. Also, $\displaystyle{f}$ depends on $\displaystyle{q}$ in a specific way:

(define (f q)
(compose (literal-function 'f
(-> (UP Real (UP* Real) (UP* Real)) Real))
(Gamma q)))

\displaystyle{ \begin{aligned} &f[q] \\ &= f(\Gamma[q]) \\ &= f \left( (t, q(t), Dq(t)) \right) \\ &= f(t, q(t), Dq(t)) \\ \end{aligned}}

So, besides being a (non-general) functional, $\displaystyle{f[q]}$, with a path, $\displaystyle{q}$, as input, $\displaystyle{f}$ can also be regarded as an ordinary function that has 3 real number inputs: $\displaystyle{t}$, $\displaystyle{q(t)}$, and $\displaystyle{Dq(t)}$.

We can vary with respect to a path (the $\displaystyle{q}$ itself), but we cannot differentiate with respect to a path. We can only differentiate with respect to the values of the path $\displaystyle{q(t)}$.

For simplicity, we consider the case that the path $\displaystyle{q(t)}$ is one-dimensional. Only in the sense that $\displaystyle{f}$ can also be regarded as an ordinary function that has 3 real number inputs, it can use the chain rule of ordinary differentiation:

\displaystyle{ \begin{aligned} &\delta_\eta f[q] \\ &= D \eta(t) \partial_2 f \left( \begin{bmatrix} t \\ q(t) \\ Dq(t) \end{bmatrix} \right) + \eta(t) \partial_1 f \left( \begin{bmatrix} t \\ q(t) \\ Dq(t) \end{bmatrix} \right) \\ \end{aligned}}

— Me@2020-06-20 01:17:39 PM

.

.


text:
#### ${{\boldsymbol \eta}_{{c}}{(1S)}}$ WIDTH

VALUE (MeV) EVTS DOCUMENT ID TECN  COMMENT
 $\bf{ 32.0 \pm0.7}$ OUR FIT
 $\bf{ 32.1 \pm0.8}$ OUR AVERAGE  Error includes scale factor of 1.1.
$33.8$ $\pm1.6$ $\pm4.1$ 1705
 2019 AV
BES3 ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \omega}}{{\mathit \omega}}$
$30.8$ ${}^{+2.3}_{-2.2}$ $\pm2.9$ 2673
 2018
BELL ${{\mathit e}^{+}}$ ${{\mathit e}^{-}}$ $\rightarrow$ ${{\mathit e}^{+}}{{\mathit e}^{-}}{{\mathit \eta}^{\,'}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}$
$34.0$ $\pm1.9$ $\pm1.3$ 11k
LHCB ${{\mathit p}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit B}^{+}}{{\mathit X}}$ $\rightarrow$ ${{\mathit p}}{{\overline{\mathit p}}}{{\mathit K}^{+}}{{\mathit X}}$
$31.4$ $\pm3.5$ $\pm2.0$ 6.4k 1
 2017 BB
LHCB ${{\mathit p}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit b}}{{\overline{\mathit b}}}{{\mathit X}}$ $\rightarrow$ 2( ${{\mathit K}^{+}}{{\mathit K}^{-}}$) ${{\mathit X}}$
$27.2$ $\pm3.1$ ${}^{+5.4}_{-2.6}$ 2
 2014
KEDR ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$
$25.2$ $\pm2.6$ $\pm2.4$ 4.5k 3, 4
 2014 E
BABR ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{0}}$
$34.8$ $\pm3.1$ $\pm4.0$ 900 3, 5, 4
 2014 E
BABR ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \eta}}$
$32.0$ $\pm1.2$ $\pm1.0$ 6, 7
 2012 F
BES3 ${{\mathit \psi}{(2S)}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$
$36.4$ $\pm3.2$ $\pm1.7$ 832 3
 2012 N
BES3 ${{\mathit \psi}{(2S)}}$ $\rightarrow$ ${{\mathit \pi}^{0}}{{\mathit \gamma}}$ hadrons
$37.8$ ${}^{+5.8}_{-5.3}$ $\pm3.1$ 486
 2012 A
BELL ${{\mathit e}^{+}}$ ${{\mathit e}^{-}}$ $\rightarrow$ ${{\mathit e}^{+}}{{\mathit e}^{-}}{{\mathit \eta}^{\,'}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}$
$36.2$ $\pm2.8$ $\pm3.0$ 11k
 2011 M
BABR ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}{{\mathit \pi}^{0}}$
$35.1$ $\pm3.1$ ${}^{+1.0}_{-1.6}$ 920 7
 2011
BELL ${{\mathit B}^{\pm}}$ $\rightarrow$ ${{\mathit K}^{\pm}}$ ( ${{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$ )
$31.7$ $\pm1.2$ $\pm0.8$ 14k 8
 2010
BABR 10.6 ${{\mathit e}^{+}}$ ${{\mathit e}^{-}}$ $\rightarrow$ ${{\mathit e}^{+}}{{\mathit e}^{-}}{{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$
$36.3$ ${}^{+3.7}_{-3.6}$ $\pm4.4$ 0.9k
 2008 AB
BABR ${{\mathit B}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}{(1S)}}{{\mathit K}^{(*)}}$ $\rightarrow$ ${{\mathit K}}{{\overline{\mathit K}}}{{\mathit \pi}}{{\mathit K}^{(*)}}$
$28.1$ $\pm3.2$ $\pm2.2$ 7.5k
 2008
BELL ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}}$ $\rightarrow$ hadrons
$48$ ${}^{+8}_{-7}$ $\pm5$ 195
 2006
BELL ${{\mathit B}^{+}}$ $\rightarrow$ ${{\mathit p}}{{\overline{\mathit p}}}{{\mathit K}^{+}}$
$40$ $\pm19$ $\pm5$ 20
 2006
BELL ${{\mathit B}^{+}}$ $\rightarrow$ ${{\mathit \Lambda}}{{\overline{\mathit \Lambda}}}{{\mathit K}^{+}}$
$24.8$ $\pm3.4$ $\pm3.5$ 592
 2004
CLEO ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$
$20.4$ ${}^{+7.7}_{-6.7}$ $\pm2.0$ 190
 2003
E835 ${{\overline{\mathit p}}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \gamma}}$
$23.9$ ${}^{+12.6}_{-7.1}$
 1995 F
E760 ${{\overline{\mathit p}}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \gamma}}$
• • We do not use the following data for averages, fits, limits, etc. • •
$32.1$ $\pm1.1$ $\pm1.3$ 12k 9
 2011 M
BABR ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ ${{\mathit K}_S^0}$ ${{\mathit K}^{\pm}}{{\mathit \pi}^{\mp}}$
$34.3$ $\pm2.3$ $\pm0.9$ 2.5k 10
 2004 D
BABR ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}{(1S)}}$ $\rightarrow$ ${{\mathit K}}{{\overline{\mathit K}}}{{\mathit \pi}}$
$17.0$ $\pm3.7$ $\pm7.4$ 11
 2003
BES ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$
$29$ $\pm8$ $\pm6$ 180 12
 2003
BELL ${{\mathit B}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}}{{\mathit K}}$
$11.0$ $\pm8.1$ $\pm4.1$ 13
 2000 F
BES ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$ and ${{\mathit \psi}{(2S)}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$
$27.0$ $\pm5.8$ $\pm1.4$ 14
 2000 B
CLE2 ${{\mathit \gamma}}$ ${{\mathit \gamma}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}}$ $\rightarrow$ ${{\mathit K}^{\pm}}{{\mathit K}_S^0}$ ${{\mathit \pi}^{\mp}}$
$7.0$ ${}^{+7.5}_{-7.0}$ 12
 1987 B
SPEC ${{\overline{\mathit p}}}$ ${{\mathit p}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \gamma}}$
$10.1$ ${}^{+33.0}_{-8.2}$ 23 15
 1986
MRK3 ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit p}}{{\overline{\mathit p}}}$
$11.5$ $\pm4.5$
 1986
CBAL ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \gamma}}$ X, ${{\mathit \psi}{(2S)}}$ $\rightarrow$ ${{\mathit \gamma}}$ X
$< 40 90\% CL$ 18
 1980 B
MRK2 ${{\mathit e}^{+}}{{\mathit e}^{-}}$
$< 20 90\% CL$
 1980 B
CBAL ${{\mathit e}^{+}}{{\mathit e}^{-}}$
 1 From a fit of the ${{\mathit \phi}}{{\mathit \phi}}$ invariant mass with the mass and width of ${{\mathit \eta}_{{c}}{(1S)}}$ as free parameters.
 2 Taking into account an asymmetric photon lineshape.
 3 With floating mass.
 4 Ignoring possible interference with the non-resonant 0${}^{-}$ amplitude.
 5 Using both, ${{\mathit \eta}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \gamma}}$ and ${{\mathit \eta}}$ $\rightarrow$ ${{\mathit \pi}^{+}}{{\mathit \pi}^{-}}{{\mathit \pi}^{0}}$ decays.
 6 From a simultaneous fit to six decay modes of the ${{\mathit \eta}_{{c}}}$.
 7 Accounts for interference with non-resonant continuum.
 8 Taking into account interference with the non-resonant $\mathit J{}^{P} = 0-$ amplitude.
 9 Not independent from the measurements reported by LEES 2010 .
 10 Superseded by LEES 2010 .
 11 From a simultaneous fit of five decay modes of the ${{\mathit \eta}_{{c}}}$.
 12 Superseded by VINOKUROVA 2011 .
 13 From a fit to the 4-prong invariant mass in ${{\mathit \psi}{(2S)}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$ and ${{\mathit J / \psi}{(1S)}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$ decays.
 14 Superseded by ASNER 2004 .
 15 Positive and negative errors correspond to 90$\%$ confidence level.
References:
 ABLIKIM 2019AV
PR D100 052012 Observation of $\eta_c\to\omega\omega$ in $J/\psi\to\gamma\omega\omega$
 XU 2018
PR D98 072001 Measurement of eta_c(1S), eta_c(2S) and non-resonant eta' pi+ pi- production via two-photon collisions
PL B769 305 Observation of ${{\mathit \eta}_{{c}}{(2S)}}$ $\rightarrow$ ${{\mathit p}}{{\overline{\mathit p}}}$ and Search for ${{\mathit X}{(3872)}}$ $\rightarrow$ ${{\mathit p}}{{\overline{\mathit p}}}$ Decays
 AAIJ 2017BB
EPJ C77 609 Study of Charmonium Production in hadron Decays and First Evidence for the Decay ${{\mathit B}_{{s}}^{0}}$ $\rightarrow$ ${{\mathit \phi}}{{\mathit \phi}}{{\mathit \phi}}$
 ANASHIN 2014
PL B738 391 Measurement of ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$ Decay Rate and ${{\mathit \eta}_{{c}}}$ Parameters at KEDR
 LEES 2014E
PR D89 112004 Dalitz Plot Analysis of ${{\mathit \eta}_{{c}}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \eta}}$ and ${{\mathit \eta}_{{c}}}$ $\rightarrow$ ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{0}}$ in Two-Photon Interactions
 ABLIKIM 2012N
PR D86 092009 Study of ${{\mathit \psi}{(3686)}}$ $\rightarrow$ ${{\mathit \pi}^{0}}{{\mathit h}_{{c}}}$ , ${{\mathit h}_{{c}}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$ via ${{\mathit \eta}_{{c}}}$ Exclusive Decays
 ABLIKIM 2012F
PRL 108 222002 Measurements of the Mass and Width of the ${{\mathit \eta}_{{c}}}$ Using the Decay ${{\mathit \psi}{(3686)}}$ $\rightarrow$ ${{\mathit \gamma}}{{\mathit \eta}_{{c}}}$
 ZHANG 2012A
PR D86 052002 First Study of ${{\mathit \eta}_{{c}}{(1S)}}$, ${{\mathit \eta}{(1760)}}$ and ${{\mathit X}{(1835)}}$ Production via ${{\mathit \eta}^{\,'}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}$ Final States in Two-Photon Collisions
 DEL-AMO-SANCHEZ 2011M
PR D84 012004 Observation of ${{\mathit \eta}_{{c}}{(1S)}}$ and ${{\mathit \eta}_{{c}}{(2S)}}$ decays to ${{\mathit K}^{+}}{{\mathit K}^{-}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}{{\mathit \pi}^{0}}$ in Two-Photon Interactions
 VINOKUROVA 2011
PL B706 139 Study of ${{\mathit B}^{\pm}}$ $\rightarrow$ ${{\mathit K}^{\pm}}$ ($\mathit KSK{{\mathit \pi}}$) Decay and Determination of ${{\mathit \eta}_{{c}}}$ and ${{\mathit \eta}_{{c}}{(2S)}}$ Parameters
 LEES 2010
PR D81 052010 Measurement of the ${{\mathit \gamma}}$ ${{\mathit \gamma}^{*}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}}$ Transition Form Factor
 AUBERT 2008AB
PR D78 012006 Study of ${{\mathit B}}$-Meson Decays to ${{\mathit \eta}_{{c}}}{{\mathit K}^{(*)}}$ , ${{\mathit \eta}_{{c}}{(2S)}}{{\mathit K}^{(*)}}$ , and ${{\mathit \eta}_{{c}}}{{\mathit \gamma}}{{\mathit K}^{(*)}}$
 UEHARA 2008
EPJ C53 1 Study of Charmonia in Four-Meson Final States Produced in Two-Photon Collisions
 WU 2006
PRL 97 162003 Study of ${{\mathit J / \psi}}$ $\rightarrow$ ${{\mathit p}}{{\overline{\mathit p}}}$ , ${{\mathit \Lambda}}{{\overline{\mathit \Lambda}}}$ and Observation of ${{\mathit \eta}_{{c}}}$ $\rightarrow$ ${{\mathit \Lambda}}{{\overline{\mathit \Lambda}}}$ at Belle
 ASNER 2004
PRL 92 142001 Observation of ${{\mathit \eta}_{{c}}^{\,'}}$ Production in ${{\mathit \gamma}}{{\mathit \gamma}}$ Fusion at CLEO
 AUBERT 2004D
PRL 92 142002 Measurements of the Mass and Width of the ${{\mathit \eta}_{{c}}}$ Meson and of an ${{\mathit \eta}_{{c}}{(2S)}}$ Candidate
 AMBROGIANI 2003
PL B566 45 Measurement of the Resonance Parameters of the Charmonium Ground State, ${{\mathit \eta}_{{c}}}(1_{1}S_{0}$)
 BAI 2003
PL B555 174 Measurement of the Mass and Full Width of the ${{\mathit \eta}_{{c}}}$ Meson
 FANG 2003
PRL 90 071801 Measurement of Branching Fractions for ${{\mathit B}}$ $\rightarrow$ ${{\mathit \eta}_{{c}}}{{\mathit K}}{}^{(*)}$ Decays
 BAI 2000F
PR D62 072001 A Measurement of the Mass and Full Width of the ${{\mathit \eta}_{{c}}}$ Meson
 BRANDENBURG 2000B
PRL 85 3095 Measurements of the Mass, Total Width and Two Photon Partial Width of the ${{\mathit \eta}_{{c}}}$ Meson
 ARMSTRONG 1995F
PR D52 4839 Study of the ${{\mathit \eta}_{{c}}}(1{}^{1}{{\mathit S}_{{0}}}$) State of Charmonium Formed in ${{\overline{\mathit p}}}{{\mathit p}}$ Annihilations and a Search for the ${{\mathit \eta}_{{c}}}(2{}^{1}{{\mathit S}_{{0}}}$)
 BAGLIN 1987B
PL B187 191 Direct Observation and Partial Width Measurement of ${{\mathit \gamma}}{{\mathit \gamma}}$ Decay of Charmonium States
 BALTRUSAITIS 1986
PR D33 629 Hadronic Decay of the ${{\mathit \eta}_{{c}}{(2980)}}$
 GAISER 1986
PR D34 711 Charmonium Spectroscopy from Inclusive ${{\mathit \psi}^{\,'}}$ and ${{\mathit J / \psi}}$ Radiative Decays
 HIMEL 1980B
PRL 45 1146 Observation of the ${{\mathit \eta}_{{c}}{(2980)}}$ Produced in the Radiative Decay of the ${{\mathit \psi}^{\,'}{(3684)}}$
 PARTRIDGE 1980B
PRL 45 1150 Observation of an ${{\mathit \eta}_{{c}}}$ Candidate State with Mass $2978$ $\pm9$ MeV
see fit info


text:
# Intermediate | Regular Expressions In Python

## Escaping characters¶

Suppose you have a file of students and their grades on a recent exam.

Bob: B-
Mary: C
Susan: A+
Ronald: F
Gerry: A+
Mark: C+

If you want to find all students who received an A+, you might try using the regular expression A+, but this matches on Allison: C+

Allison: C+  <- match
Bob: B-
Mary: C
Susan: A+    <- match
Ronald: F
Gerry: A+    <- match

That's because + is a special regular expression metacharacter meaning "one or more repetitions". In other words, A+ means match one or more repetitions of A.

In order to match a plus sign (ignoring the default interpretation of +), you must escape it with a backslash. For example A\+ means match the string A+.

Similarly, \\ matches a single slash, \. matches a period, \( matches an open parenthesis, and so on.

## Raw string literals ('r' prefix)¶

Python doesn't always interpret strings exactly as you write them. For example, in the string "hello\nworld", Python interprets \n as a newline character. This becomes evident when you print() the string.

>>> print("hello\nworld")
hello
world

If you want to avoid this special interpretation of \n, you can prefix the string with the letter r.

>>> print(r"hello\nworld")
'hello\nworld'

A string prefixed with the letter r like this is called a raw string literal.

Tip

To see how Python interprets a string, just print() it.

>>> print("hello world")
hello world

>>> print("\hello world")
\hello world

>>> print("\\hello world")
\hello world

>>> print("\\\\hello world")
\\hello world

>>> print("hello \"world\"")
hello "world"

>>> print("hello\nworld")
hello
world

Raw string literals vs escaping characters

Raw string literals are an alternative to escaping characters. For example, consider these equivalent strings.

r"hello\nworld" == "hello\\nworld"
# True

## Capture Groups¶

Capture groups let you select parts of a regular expression match. For example, given the string,

I went to the market and bought 12 eggs, 6 carrots, and 2 hams.

The expression "\d+ \w+" matches these substrings

I went to the market and bought 12 eggs, 6 carrots, and 2 hams.

By wrapping \d+ and \w+ in parentheses, each match suddenly has two nested sub-matches. These are called "capture groups".

1   2
12 eggs

1   2
6 carrots

1  2
2 hams

In Python, if we identify the first match using re.search(),

import re

phrase = "I went to the market and bought 12 eggs, 6 carrots, and 2 hams."
first_match = re.search(pattern="(\d+) (\w+)", string=phrase)

print(first_match)
# <re.Match object; span=(32, 39), match='12 eggs'>

we can access its groups using the .groups() method,

first_match.groups()
# ('12', 'eggs')

or we can access each group individually using the .group() method

first_match.group(1)  # '12'
first_match.group(2)  # 'eggs'

### Nested and Non Capture Groups¶

Consider the following example that matches the capital Al or Bi followed by one or more word characters.

first_match = re.search(
pattern="(Al|Bi)(\w+)",
string="Amy loves Billy and hates Allen"
)
first_match.groups()
# ('Bi', 'lly')

There are two capture groups in this example: (Al|Bi) and (\w+). What if we wanted to make the entire pattern a single capture group? You might try

first_match = re.search(
pattern="((Al|Bi)\w+)",
string="Amy loves Billy and hates Allen"
)
first_match.groups()
# ('Billy', 'Bi')

but this returns two strings:

• one representing the outer capture group, identified by the outermost parentheses ((Al|Bi)\w+)
• another representing the inner capture group, identified by the innermost parentheses ((Al|Bi)\w+).

This is known as a nested capture group.

The issue of selecting a single capture group remains! The problem stems from the Or operator, because we cannot drop the parentheses surrounding (Al|Bi) without changing the pattern's meaning.

To get around this issue, we can change (Al|Bi) to (?:Al|Bi). The ?: bit signals the expression as a non capture group.

first_match = re.search(
pattern="((?:Al|Bi)\w+)",
string="Amy loves Billy and hates Allen"
)
first_match.groups()
# ('Billy',)


text:
# 2.3: The Law of Tangents

[ "article:topic", "Law of Tangents", "Mollweide\'s equations" ]

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

We have shown how to solve a triangle in all four cases discussed at the beginning of this chapter. An alternative to the Law of Cosines for Case 3 (two sides and the included angle) is the Law of Tangents:

Theorem 2.3 Law of Tangents

If a triangle has sides of lengths $$a$$, $$b$$, and $$c$$ opposite the angles $$A$$, $$B$$, and $$C$$, respectively, then

\begin{align} \label{2.17}\frac{a-b}{a+b} ~&=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} ~,\\ \label{2.18} \frac{b-c}{b+c} ~&=~ \frac{\tan\;\frac{1}{2}(B-C)}{\tan\;\frac{1}{2}(B+C)} ~,\\ \label{2.19} \frac{c-a}{c+a} ~&=~ \frac{\tan\;\frac{1}{2}(C-A)}{\tan\;\frac{1}{2}(C+A)} ~. \\ \end{align}

Note that since $$\tan\;(-\theta) = -\tan\;\theta$$ for any angle $$\theta$$, we can switch the order of the letters in each of the above formulas. For example, we can rewrite Equation \ref{2.17} as

$\label{2.20} \frac{b-a}{b+a}~=~\frac{\tan\;\frac{1}{2}(B-A)}{\tan\;\frac{1}{2}(B+A)}~,$

and similarly for the other formulas. If $$a > b$$, then it is usually more convenient to use Equation \ref{2.17}, while Equation \ref{2.20} is more convenient when $$b > a$$.

Example 2.10: Case 3: Two sides and the included angle.

Solve the triangle $$\triangle\,ABC$$ given $$a =5$$, $$b = 3$$, and $$C = 96^\circ$$.

Solution:

$$A + B + C = 180^\circ$$, so $$A + B = 180^\circ - C = 180^\circ - 96^\circ = 84^\circ$$. Thus, by the Law of Tangents,

\nonumber \begin{align*} \frac{a-b}{a+b} ~=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(A+B)} \quad&\Rightarrow\quad \frac{5-3}{5+3} ~=~ \frac{\tan\;\frac{1}{2}(A-B)}{\tan\;\frac{1}{2}(84^\circ)}\\ \nonumber &\Rightarrow\quad \tan\;\tfrac{1}{2}(A-B) ~=~ \tfrac{2}{8}\tan\;42^\circ ~=~ 0.2251\\ \nonumber &\Rightarrow\quad \tfrac{1}{2}(A-B) ~=~ 12.7^\circ \quad\Rightarrow\quad A-B ~=~ 25.4^\circ ~. \\ \end{align*}

We now have two equations involving $$A$$ and $$B$$, which we can solve by adding the equations:

\nonumber \begin{alignat*}{3} A &- B &&=\; 25.4^\circ\\ \nonumber A &+ B &&=\; 84^\circ\phantom{4^\circ}\\[-2mm] \nonumber --&--&&----\\[-2mm] \nonumber 2A &\phantom{+} &&=\; 109.4^\circ \quad\Rightarrow\quad \boxed{A = 54.7^\circ} \quad\Rightarrow\quad B ~=~ 84^\circ - 54.7^\circ \quad\Rightarrow\quad \boxed{B = 29.3^\circ} \\ \end{alignat*}

We can find the remaining side $$c$$ by using the Law of Sines:

$\nonumber c ~=~ \frac{a\;\sin\;C}{\sin\;A} ~=~ \frac{5\;\sin\;96^\circ}{\sin\;54.7^\circ} \quad\Rightarrow\quad \boxed{c = 6.09}$

Note that in any triangle $$\triangle\,ABC$$, if $$a = b$$ then $$A = B$$ (why?), and so both sides of Equation \ref{2.17} would be $$0$$ (since $$\tan\;0^\circ = 0$$). This means that the Law of Tangents is of no help in Case 3 when the two known sides are equal. For this reason, and perhaps also because of the somewhat unusual way in which it is used, the Law of Tangents seems to have fallen out of favor in trigonometry books lately. It does not seem to have any advantages over the Law of Cosines, which works even when the sides are equal, requires slightly fewer steps, and is perhaps more straightforward.

Related to the Law of Tangents are Mollweide's equations:

Mollweide's equations:

For any triangle $$\triangle\,ABC$$,

\begin{align} \frac{a-b}{c} ~&=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C} ~,~\text{and}\label{2.21}\\[4pt] \frac{a+b}{c} ~&=~ \frac{\cos\;\frac{1}{2}(A-B)}{\sin\;\frac{1}{2}C} ~.\label{2.22} \end{align}

Note that all six parts of a triangle appear in both of Mollweide's equations. For this reason, either equation can be used to check a solution of a triangle. If both sides of the equation agree (more or less), then we know that the solution is correct.

Example 2.11

Use one of Mollweide's equations to check the solution of the triangle from Example 2.10.

Solution:

Recall that the full solution was $$a=5$$, $$b=3$$, $$c=6.09$$, $$A=54.7^\circ$$, $$B=29.3^\circ$$, and $$C=96^\circ$$. We will check this with Equation \ref{2.21}:
\nonumber \begin{align*} \frac{a-b}{c} ~&=~ \frac{\sin\;\frac{1}{2}(A-B)}{\cos\;\frac{1}{2}C}\\ \nonumber \frac{5-3}{6.09} ~&=~ \frac{\sin\;\frac{1}{2}(54.7^\circ - 29.3^\circ)}{\cos\;\frac{1}{2}(96^\circ)}\\ \nonumber \frac{2}{6.09} ~&=~ \frac{\sin\;12.7^\circ}{\cos\;48^\circ}\\ \nonumber 0.3284 ~&=~ 0.3285 \quad \checkmark \end{align*}

The small difference ($$\approx 0.0001$$) is due to rounding errors from the original solution, so we can conclude that both sides of the equation agree, and hence the solution is correct.

Example 2.12

Can a triangle have the parts $$a=6$$, $$b=7$$, $$c=9$$, $$A=55^\circ$$, $$B=60^\circ$$, and $$C=65^\circ\;$$?

Solution:

Before using Mollweide's equations, simpler checks are that the angles add up to $$180^\circ$$ and that the smallest and largest sides are opposite the smallest and largest angles, respectively. In this case all those conditions hold. So check with Mollweide's Equation \ref{2.22}:

\nonumber \begin{align*} \frac{a+b}{c} ~&=~ \frac{\cos\;\frac{1}{2}(A-B)}{\sin\;\frac{1}{2}C}\\ \nonumber \frac{6+7}{9} ~&=~ \frac{\cos\;\frac{1}{2}(55^\circ - 60^\circ)}{\sin\;\frac{1}{2}(65^\circ)}\\ \nonumber \frac{13}{9} ~&=~ \frac{\cos\;(-2.5^\circ)}{\sin\;32.5^\circ}\\ \nonumber 1.44 ~&=~ 1.86 \quad\times \end{align*}

Here the difference is far too large, so we conclude that there is no triangle with these parts.

We will prove the Law of Tangents and Mollweide's equations in Chapter 3, where we will be able to supply brief analytic proofs.


text:
# MacBook – Boot Camp Assistant can’t install Windows after partition made and booted to USB installer

bootcampmacbook pro

How do I install a bootable copy of windows on my older macBook Air with Boot Camp Assistant?

1. Per Apple's list of supported devices, "Use Windows 10 on your Mac with Boot Camp" I have an 11" mid-2012 2GHz i7 macBook Air with a 500GB SSD (running macOS Sierra 10.12.5) which I would like to install a bootable Windows partition on and make use of Windows 10.

2. Per Windows recommendation I downloaded the "Windows 10 Anniversary Update ISO" selecting the "Windows 10" edition (not "N" or "Single Language"). Note to self: other options for Windows .iso files

3. Per "How to install Windows using Boot Camp" & "Install Windows on your Mac using a USB hard drive or flash drive" I mounted a USB 2.0 drive (SanDisk Cruzer) and started the Boot Camp Assistant process. (Note: I do also have a Time Machine backup).

4. After setting up the USB installer, partitioning the Boot Camp disk and once rebooted into Windows 10, I see the "BOOTCAMP" partition but there is a warning message saying the driver's are unavailable. Rescanning the available volumes for the driver's does nothing. I am unable to proceed.

I was able to +r restart into recovery mode and remove the Boot Camp partition. This resulted in the startup process ending at the No bootable device, insert boot disk and press any key screen. I then used option+restart to get back into my macOS 10.12.5 partition, where I reset the startup disk in System Preferences.

Per this 2016 9TO5Mac article and video, "How to install Windows 10 on your Mac using Boot Camp Assistant" I was under the impression that I could install Windows 10 and use the "Windows 10 Insider" option without a registration key. See also the article, "You Don’t Need a Product Key to Install and Use Windows 10" for some notes on the limitations of the free "Insider" install.

1. Do I need to purchase windows 7 first, then install that and then upgrade with the Anniversary Edition installer (and then upgrade to Windows 10 Creators)?

2. Is there a more appropriate .iso file I should be using?

Thank you.


text:
# How can I calculate the half life of an element?

Dec 30, 2014

Nuclear half-life expresses the time required for half of a sample to undergo radioactive decay. Exponential decay can be expressed mathematically like this:

$A \left(t\right) = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{t} _ \left(\text{1/2}\right)}$ (1), where

$A \left(t\right)$ - the amount left after t years;
${A}_{0}$ - the initial quantity of the substance that will undergo decay;
${t}_{\text{1/2}}$ - the half-life of the decaying quantity.

So, if a problem asks you to calculate an element's half-life, it must provide information about the initial mass, the quantity left after radioactive decay, and the time it took that sample to reach its post-decay value.

Let's say you have a radioactive isotope that undergoes radioactive decay. It started from a mass of 67.0 g and it took 98 years for it to reach 0.01 g. Here's how you would determine its half-life:

Starting from (1), we know that

$0.01 = 67.0 \cdot {\left(\frac{1}{2}\right)}^{\frac{98.0}{t} _ \left(\text{1/2")) -> 0.01/67.0 = 0.000149 = (1/2)^(98.0/(t_("1/2}\right)}$

$\frac{98.0}{t} _ \left(\text{1/2}\right) = {\log}_{0.5} \left(0.000149\right) = 12.7$

Therefore, its half-life is ${t}_{\text{1/2}} = \frac{98.0}{12.7} = 7.72$ $\text{years}$.

So, the initial mass gets halved every 7.72 years.

Sometimes, if the numbers allow it, you can work backwards to determine an element's half-life. Let's say you started with 100 g and ended up with 25 g after 1,000 years.

In this case, since 25 represents 1/4th of 100, two hal-life cycles must have passed in 1,000 years, since

$\frac{100.0}{2} = 50.0$ $\text{g}$ after the first ${t}_{\text{1/2}}$,

$\frac{50.0}{2} = 25.0$ $\text{g}$ after another ${t}_{\text{1/2}}$.

So, $2 \cdot {t}_{\text{1/2") = 1000 -> t_("1/2}} = \frac{1000}{2} = 500$ $\text{years}$.


text:
# Existence of $\vee$ or $\wedge$ for non-monotonic functions

This question is inspired by a discussion in chat with wj32. We allow for equality in the definition of increasing and decreasing and call a function monotonic if it is increasing or decreasing. If $f:\mathbb R\to \mathbb R$ is not monotonic, are there three points $x<y<z$ such that $f(y)<f(x),f(z)$ or $f(y)>f(x),f(z)$? For convenience, call these the $\vee$ and $\wedge$ formations respectively.

-

It's a purely combinatorial matter. The main task is to reduce the number of cases to be considered as much as possible.

It suffices to consider a surjective function $f:\ \{1,2,3,4\}\to\{1',\ldots,n'\}$ with the property that there exists a pair $x<y$ with $f(x)<f(y)$ and a pair $u<v$ with $f(u)>f(v)$. Here $n\in\{2,3,4\}$.

I shall only deal with the case $n=4$.

If at least one of $2$ and $3$ is mapped onto $1'$ or $4'$ we are done.

If $f(2)=2$ and $f(3)=3$ then $f(1)=4'$, and we have a $\vee$ over $\{1,2,3\}$.

If $f(2)=3'$ and $f(3)=2'$ then $f(1)=1'$, and we have a $\wedge$ over $\{1,2,3\}$.

-

If $f$ is not monotonic, it is not decreasing or increasing. Hence there are points $a<b$ with $f(a)<f(b)$ and points $c<d$ with $f(c)>f(d)$.

If $a=c$, use $adb$ to form $\vee$ or $abd$ to form $\wedge$.

If $a=d$, use $cdb$ to form $\vee$.

If $b=c$, use $abd$ to form $\wedge$.

If $b=d$, use $acd$ to form $\wedge$ or $cab$ to form $\vee$.

Now we are left with the case when $a,b,c,d$ are pairwise distinct.

If $a<b<c<d$, use $abd$ to form $\wedge$ or $acd$ to form $\wedge$.

If $a<c<b<d$, use $abd$ to from $\wedge$ or $acd$ to form $\wedge$.

If $a<c<d<b$, use $acd$ to form $\wedge$ or $cdb$ to form $\vee$.

If $c<a<d<b$, use $cdb$ to form $\vee$ or $cab$ to form $\vee$.

If $c<d<a<b$, use $cdb$ to form $\vee$ or $dab$ to form $\vee$.

-

for all $a < b < c\in [p,q]$ $$f(a) \leq f(b) \leq f(c)$$ $\implies$$f is monotone in [p,q] and increasing and for all a < b < c\in [p,q] $$f(a) \geq f(b) \geq f(c)$$ \implies$$f$ is monotone in $[p,q]$ and decreasing

Now, since the function is not monotone, both the above condition are false, and hence, for some $a < b < c \in [p,q]$ we have at least one of the following conditions true.

1. $f(a) < f(b) > f(c)$
2. $f(a) > f(b) < f(c)$
-
 @JasperLoy I removed the continuous function thing, since it was superfluous. Anyway, I have not fixed $a,b,c$, Since, if the first two options from the list are not satisfied anywhere in $R$, then either of the other two options (the third one, and then the original assumption) must be satisfied, then I pick one of them and use it to show that, it applies all over the domain. Because, next you can select another point $e$ and do the same thing as has been done with $b,c,d$ to get the same relation with $c,d,e$ and so on. – Jayesh Badwaik Sep 27 '12 at 14:39 But, we first show that there is atleast one, and then prove for all from that single one. About strict inequalities, I guess, whenever there is equality, we get a constant function. – Jayesh Badwaik Sep 27 '12 at 14:45


text:
# Problem 4-23 - Projectile

A girl throws a ball onto the roof of a house, as illustrated in the figure, and prepares to catch the ball with a baseball glove held $1.0\; m$ above the ground. The ball rolls off the roof with a speed of $3.2\; m/s.$

(a) From the time the ball left the roof, how long will it take to land in her glove?
(b) How far from the horizontal edge of the roof should she hold her glove?
(c) What is the ball's velocity as it reaches her glove?

We have to solve this problem in some coordinate system. Which is best?

(A)

(B)

(C)

(D)


text:
## A discussion about the homogenization of moving interfaces.(English)Zbl 1180.35070

The authors present a number of new results concerning the behavior of moving interfaces in periodic environments and analyze in detail their averaging behavior. They concentrate on fronts moving by either first-order oscillatory velocity or mean curvature coupled with an oscillating forcing. Under sharp assumptions, they show that the fronts either homogenize or get trapped or oscillate. Several concrete examples are also discussed.
There has been considerable interest lately in the homogenization theory for first- and second-order partial differential equations in periodic/almost-periodic and stationary, ergodic, random environments. Of special interest is the study of the averaged behavior of moving interfaces in such environments. In this note the authors expand considerably this investigation in the context of periodic media. Although it is important both for applications and interesting mathematically to consider random media, very little is known for moving interfaces in this setting except for some first-order motions.
The general framework concerns the behavior as $$\varepsilon\to 0$$ of the solution $$u^\varepsilon$$ of the initial value problem
$\begin{cases} u^\varepsilon_t+F(\varepsilon D^2u^\varepsilon,Du^\varepsilon,\frac{x} {\varepsilon})=0&\text{in }\mathbb{R}^N\times(0,\infty),\\ u^\varepsilon=u_0 &\text{on }\mathbb{R}^N \times\{0\},\end{cases}$
where $$u_0\in UC(\mathbb{R}^N)$$, the space of uniformly continuous functions on $$\mathbb{R}^N$$, and $$F$$ is (degenerate) elliptic, geometric and periodic. In the random setting, the homogenization of this problem is still an open problem, except for completely degenerate (first-order) and (degenerate) semilinear equations with either convex or concave gradient dependence.

### MSC:

 35B27 Homogenization in context of PDEs; PDEs in media with periodic structure 35F21 Hamilton-Jacobi equations
Full Text:

### References:

 [1] Alvarez, O., Homogenization of hamilton – jacobi equations in perforated sets, J. differential equations, 159, 2, 543-577, (1999) · Zbl 0945.35010 [2] Barles, G., Solutions de viscosité des équations de hamilton – jacobi, Mathématiques & applications (Berlin), vol. 17, (1994), Springer-Verlag Paris · Zbl 0819.35002 [3] Barles, G.; Souganidis, P.E., A new approach to front propagation problems: theory and applications, Arch. rational mech. anal., 141, 3, 237-296, (1998) · Zbl 0904.35034 [4] Bhattacharya, K.; Craciun, B., Homogenization of a hamilton – jacobi equation associated with the geometric motion of an interface, Proc. roy. soc. Edinburgh sect. A, 133, 4, 773-805, (2003) · Zbl 1043.35028 [5] Caffarelli, L.A.; Souganidis, P.E.; Wang, L., Homogenization of fully nonlinear, uniformly elliptic and parabolic partial differential equations in stationary ergodic media, Comm. pure appl. math., 58, 3, 319-361, (2005) · Zbl 1063.35025 [6] Crandall, M.G.; Ishii, H.; Lions, P.-L., User’s guide to viscosity solutions of second order partial differential equations, Bull. amer. math. soc. (N.S.), 27, 1, 1-67, (1992) · Zbl 0755.35015 [7] Dirr, N.; Yip, A., Pinning an de-pinning phenomena in front propagation in heterogeneous medium, Interfaces and free boundaries, 8, 79-109, (2006) · Zbl 1101.35074 [8] Evans, L.C., Periodic homogenisation of certain fully nonlinear partial differential equations, Proc. roy. soc. Edinburgh sect. A, 120, 3-4, 245-265, (1992) · Zbl 0796.35011 [9] Evans, L.C., The perturbed test function method for viscosity solutions of nonlinear PDE, Proc. roy. soc. Edinburgh sect. A, 111, 3-4, 359-375, (1989) · Zbl 0679.35001 [10] Horie, K.; Ishii, H., Homogenization of hamilton – jacobi equations on domains with small scale periodic structure, Indiana univ. math. J., 47, 3, 1011-1058, (1998) · Zbl 0924.49020 [11] Ishii, H., Almost periodic homogenization of hamilton – jacobi equations, (), 600-605 · Zbl 0969.35018 [12] Kosygina, E.; Rezakhanlou, F.; Varadhan, S.R.S., Stochastic homogenization of hamilton – jacobi – bellman equations, Comm. pure appl. math., 59, 10, 1489-1521, (2006) · Zbl 1111.60055 [13] Lions, P.-L.; Souganidis, P.E., Correctors for the homogenization of hamilton – jacobi equations in the stationary ergodic setting, Comm. pure appl. math., 56, 10, 1501-1524, (2003) · Zbl 1050.35012 [14] Lions, P.-L.; Souganidis, P.E., Homogenization of “viscous” hamilton – jacobi equations in stationary ergodic media, Comm. partial diff. equations, 30, 1-3, 335-375, (2005) · Zbl 1065.35047 [15] Lions, P.-L.; Souganidis, P.E., Homogenization of degenerate second-order PDE in periodic and almost periodic environments and applications, Ann. inst. H. Poincaré anal. non linéaire, 22, 5, 667-677, (2005) · Zbl 1135.35092 [16] P.-L. Lions, G. Papanicolaou, S.R.S. Varadhan, Homogenization of Hamilton-Jacobi equations. Preprint [17] Rezakhanlou, F.; Tarver, J.E., Homogenization for stochastic hamilton – jacobi equations, Arch. rational mech. anal., 151, 4, 277-309, (2000) · Zbl 0954.35022 [18] Soravia, P., Optimality principles and representation formulas for viscosity solutions of hamilton – jacobi equations. I. equations of unbounded and degenerate control problems without uniqueness, Adv. differential equations, 4, 2, 275-296, (1999) · Zbl 0955.49016 [19] Souganidis, P.E., Stochastic homogenization of hamilton – jacobi equations and some applications, Asymptot. anal., 20, 1, 1-11, (1999) · Zbl 0935.35008 [20] Souganidis, P.E., Front propagation: theory and applications, (), 186-242 · Zbl 0882.35016
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text:
# Math Help - Linear Algebra Proof Regarding Linear Transformations

1. ## Linear Algebra Proof Regarding Linear Transformations

Given the following vectors in R^2:
a(sub 1) = (1; -1), a(sub 2) = (2; -1), a(sub 3) = (-3;2),
b(sub 1) = (1;0), b(sub 2) = (0;1), b(sub 3) = (1;1),
determine whether there is a linear transformation T from R^2 into R^2 such that Ta(sub i) = b(sub i) for i = 1, 2 and 3.

Basically, I don't quite understand the process of figuring out linear transformations, so I'd like at least a few helpful pointers in figuring this out. Thank you.

2. ## Re: Linear Algebra Proof Regarding Linear Transformations

so assume we have a linear transformation. L which takes $a_i \to b_i$
$A = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 2 \end{bmatrix}$
$B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$
For the matrix A, you can write the 3rd column as the linear combination of the first and second columns.
$A = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -1 \\ -1 \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \end{bmatrix}$
since a linear transformation has the property L( $c_1 v_1 + c_2 v_2$)= $c_1 L(v_1) + c_2 L(v_2)$ then
if the first column of A went to the first column of B, and the second column of A went to the second column of B then
$L(-1 * \begin{bmatrix} 1 \\ -1 \end{bmatrix} + -1 * \begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1*L(\begin{bmatrix} 1 \\ -1 \end{bmatrix}) + -1 * L(\begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1 * \begin{bmatrix} 1 \\ 0 \end{bmatrix} + -1 * \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$

since this is not equal to the 3rd column of B, there is no Linear transformation which takes the first column of A to the first column of B, second ....


text:
# Minimum path in an undirected graph with 2 kinds of edges

Given an undirected graph with positive weights, there are 2 kinds of edges: locked edges and unlocked edges. Determination if a given edge is either locked or unlocked edge takes O(1).

For given two vertices s , t and a positive number k = O(1), How can I find the shortest path between s and t which contains at most k locked edges? How to find that path if it must contains exactly k locked edges? note that the second path might be contains the same locked edge more than 1 time.

So my first solution was to define a counter that is initialized to k, and decrease its value each time I found locked edge during dijkstra algo. but it doesn’t meet the requirement, so @tmyklebu suggest me to copy the graph k times, and modify each graph to directed one so that a locked edge vw in G_i turns into an edge from u in G_i to v in G_{i+1} and from v in G_i to u in G_{i+1}. I didn’t undertand this solution, so I hope you can help (or suggest another one to solve this problem) :-)

Thanks a lot!
Tomer.

I hope the following makes the solution clearer. Suppose the set of locked edges is $L$ and the set of unlocked edges is $U$. Let $G_U$ be the graph with the same vertex set as $G$ but the edge set is $U$. Now make k+1 copies of $G_u$. Since we now have $k+1$ copies of a vertex $v$, lets call them $v_1 \ldots v_{k+1}$.
Now add edges between the copies as follows. There is an edge between $u_i,v_{i+1}$ iff there exists a corresponding edge $u,v$ in G and the edge $u,v$ is locked. Now note that in this new graph we make a transition from one copy to the next copy if and only if we take a locked edge.
Now lets start a path from vertex $S_1$ in the first copy and lets say we are at a vertex $V_{i+1}$ in the $(i+1)^{th}$. Note that this means that we must have already traversed exactly $i$ locked edges. Therefore if we want to travel exactly $k$ locked edges we need to find a shortest path from $S_1$ to $T_{k+1}$.
Another way to think of what this path would look like is that imagine the required shortest path and cut it off into pieces where the locked edges. Now you can see that in your path the pieces in between the $i^{th}$ locked edge and the $(i+1)^{th}$ locked edge happen in the $(i+1)^{th}$ and each locked edge moves you to another segment. Hope this helps.
If you want the shortest path with atmost k edges you can look for the shortest path from S to $T_1,T_2 \ldots T_{k+1}$ and take the minimum.


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# sufficient conditions for rational homotopy equivalence

Is it true that if a finite CW complex $X$ is simply connected, and $\tilde{H}_i(X, \mathbb{Q}) =0$ for $i \neq D$, then $X$ is rationally homotopy equivalent to a bouquet of $D$-dimensional spheres?

(In my setting $D \ge 3$, in case that makes any difference.)

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Yes. By a generalization of the Hurewicz theorem (which can be formulated more generally in terms of Serre classes), if $X$ is simply connected and has trivial rational homology below dimension $D$, then $\pi_i(X)\otimes\mathbb{Q}=\tilde{H}_i(X,\mathbb{Q})$ via the Hurewicz map for all $i\leq D$. In particular, for $i=D$ this implies we can find elements of $\pi_D(X)$ whose Hurewicz images form a basis for $\tilde{H}_D(X,\mathbb{Q})$. These elements together give a map from a wedge of $D$-spheres to $X$ which induces an isomorphism on rational homology and is hence a rational equivalence.