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Q-space imaging

source: Sosnovik, D. E., Wang, R., Dai, G., Reese, T. G., & Wedeen, V. J. (2009). Diffusion MR tractography of the heart. Journal of Cardiovascular Magnetic Resonance, 11(1), 1-15. [link]

Q-space imaging provides the empirical distribution of diffusion.

Callaghan proposed the idea of "q-space imaging" [1], which can be compared to the k-space relation. The k-space encoding (coordinates) is defined by the readout gradient, whereas q-space encoding is defined by the diffusion gradient. The k-space signals have a Fourier relation with the spatial distribution of spins, whereas the q-space signals also have a Fourier relation with the "ensemble average propagator" (EAP).

$S(q) = \int P(R) e^{2\pi iqR} dR$

Q-space imaging estimates the "ensemble average propagator" (EAP). EAP, in short, is a 3-dimensional probability density function of diffusion displacement. The terms "ensemble" and "average" indicate that EAP provides an averaged estimation of the diffusion environment (regardless of its heterogeneity). The term propagator indicates the function considers only the displacement of the diffusing spins. It does not discriminate on the absolute location of the spins.

Q-space imaging is a nonparametric approach to study the diffusion distribution. It is thus categorized under model-free methods because it does not assume a particular diffusion model (e.g. tensor model, multiple-tensor model...).

The q value is defined as follows:

$b = (2 \pi q) ^{2} \Delta$

q = gGd/2p and thus

The bandwidth resolution relation in Fourier transform indicates that higher q band width results in higher resolution in R. For example, a DSI 101 sequence has b-max = 4,000 and min b-value=307 s/mm2, diffusion time = 80 ms (diffusion distance=38 micron). The q interval = 12 mm-1. Thus Rmax = 1/2(q interval) = 41 microns.

Significance

QSI provides a rich information for characterizing the diffusion environment. It has been used to study the microscopic geometry of the tissues such as the axonal diameters, pore diameters.

Pros
1. No model violation problem in DTI
2. Provide rich diffusion information for resolving crossing fibers and accessing their integrity.

Cons

1. The number of sampling in q-space imaging is limited by the scanning time. Simultaneous acquisition of k-space and q-space imaging results in 6D data, making q-space imaging more intractable.
2. Due to low SNR of diffusion signal, QSI data are often acquired with limited diffusion gradient strength. This suffers substantial truncation artifact.
3. The q-space imaging assumes that the diffusion gradient encoding duration is sufficiently smaller than diffusion time. This often results in long TE. A feasible approach is using stimulated echo.
4. EAP often has to be transformed to another form to make it more feasible for analysis.

Recital

The Fourier transform of diffusion signals provides the ensemble average propagator, which is the density distribution of the diffusion displacement.

Diffusion orientation distribution function (dODF)

fiber geometry     Signals            EAP (PDF)          dODF

One way to analyze EAP (3D PDF) is by projecting the values on a unit sphere to calculate the diffusion orientation distribution function (dODF), which is defined as the probability distribution of the diffusion spins on a sphere.

One should note that there are several ways to project the EAP to a sphere, and thus the calculation methods of dODF can be different. For example,

none-weighted infinite-upper-bound dODF:

$\Psi (\hat{u}) = z\int_{0}^{\infty }P(r\hat{u})dr$

The psi function is the dODF. u is a unit orientation vector. P is the EAP. r is the displacement distance.

r-squared weighted infinite-upper-bound dODF:

$\Psi (\hat{u}) = z\int_{0}^{\infty }P(r\hat{u})r^2dr$

The upper bound of the integral can be finite, e.g., none-weighted finite-upper-bound dODF:

$\Psi (\hat{u}) = z\int_{0}^{L}P(r\hat{u})dr$

QBI [2], DSI [3], and GQI [4] have adopted different definitions to compute the dODF.

QBI used the none-weighted infinite-upper-bound ODF (only an approximation, see [5] for discussion).
DSI used r-squared infinite-upper-bound weighted dODF
GQI uses a finite version of none-weighted or r-squared weighted ODF.

To calculate the dODF, QBI used Funk-Radon transform to approximate the dODF.
DSI first used inverted Fourier transform to calculate EAP, and then the EAP was then integrally radially to estimate the calculates dODF by integrating EAP. However, the EAP is often too noisy, and DSI reconstruction has to add a hanning filter to smooth it (see Reconstruction(DTI, QBI, DSI, GQI, QSDR) for detail).

Significance

dODF provides a model-free approach to study diffusion characteristics and resolve crossing fibers.

Pros
1. dODF is a model free approach to resolve crossing fiber.
2. dODF can provide anisotropy measures.

Cons

1. QBI can only be applied to HARDI (uniform b-value), DSI can only be applied to grid scheme (if re-griding is not used). They are limited by specific diffusion sampling schemes.
2. dODF has limited angular resolution to resolve cross fibers.
3. dODF does not have "radial information" to further explore its biophysical meanings.

Spin Distribution Function (SDF)

The dODF was formulated as a probability distribution. It can be further scaled by spin density to quantify the amount of diffusing spin at a different orientation. This idea gives rise to GQI, which allows for computing the SDF directly from q-space signals by a simple analytical relation between them.

$\Psi (\hat{u})= z_0 \sum_{ \forall S} S(\hat{g},b)H( \sigma \sqrt{6Db} <\hat{g},\hat{u}>)$
z0: a constant that unifies spin density with respect to the amount of water.
S: diffusion signals
b: b-value
sigma: the upper limit of the diffusion distance.
H(.): is a basis function. Using a sinc function as the H(.) makes SDF a none-weighted version of the ODF (also known as GQI0, the non-weighted version of GQI). It is also possible to calculate GQI1 (the r-weighted version) and GQI2 ( the r-squared weighted version, see [4] for definition) using different basis function.

It is noteworthy that SDF scaled with spin density, and thus SDF can be viewed as a resampling of spin density along the radial direction. This turns it into quantity measurement, not a probabilistic measurement. By definition, SDF quantifies the density of the diffusing spins with displacement less than L (finite upper bound) and oriented at u. This is different from the dODF from QBI and DSI because dODF was originally defined as a probability density function and the sum of its support is equal to one.  The quantity of diffusing water at axonal direction is defined as the quantitative anisotropy.

The parameter sigma allows for exploring the radial information and calculating the restricted diffusion given any length variables.

QA is less susceptible to the partial volume of crossing fibers

The diffusion phantom examines three types of the partial volume effect: (a) fibers/free water, (b) fibers crossing, and (c) fibers/non-diffusive materials.

Studies have shown that QA is less susceptible to the partial volume effect:

A: FA map
B: GFA map
C: QA at horizontal
D: QA at vertical
source: Yeh, Fang-Cheng, et al. "Deterministic diffusion fiber tracking improved by quantitative anisotropy." (2013): e80713.

QA has specific pattern in each individuals

source: http://journals.plos.org/ploscompbiol/article?id=10.1371/journal.pcbi.1005203
B: QA from repeat scans shows identical pattern for each individuals C: FA does not have much difference between normal subjects.

Significance

SDF provides a density-based measurement that can be correlated with biological characteristics. It can be calculated from any diffusion sampling schemes (e.g. DTI, DSI, HARDI, multishell) and also allows for quantifying restricted diffusion.

Pros
SDF provides a absolute measurement for diffusion quantity.

Cons

SDF has limited angular resolution to resolve cross fibers.
Spin density has arbitrary unit. How to normalize SDF to allows inter-subject comparability is challenging.

Diffusion sampling scheme

The diffusion-weighted images can be acquired with a variety of combinations of gradient sensitization settings (schemes). The classification is often based on the spatial distribution of gradient encoding vectors (Gx,Gy,Gz).

Single-shell

(Gx,Gy,Gz) distributes equally on a sphere, where |G| is constant. It is noteworthy that there is no perfect tessellation of a unit sphere (except icosahedron), and thus it is impossible to keep sampling vectors perfectly equal distributed. Certain vectors may inevitably more close to each other, causing bias.

Multi-shell

A combination of more than two single shells is termed multi-shell. Multi-shell scheme shares the same distribution problem as the single shell. In addition, the distance between shell points can be highly nonuniform and add more bias into the scheme.

Grid

The most popular non-shell scheme is the grid scheme, also known as the diffusion spectrum imaging (DSI) scheme. (Gx,Gy,Gz) distributes on the Cartesian points within a sphere, where |G| is less than a maximum magnitude value. The major advantage of a grid scheme is its equal distance between sampling points. The limitation is that not every modeling method can work on grid sampling data. Other non-shell scheme includes body-center-cubic and face-center-cubic.

How to choose the diffusion scheme?

Single shell maximizes the sensitive to orientation, whereas the grid scheme is a balance between orientation and radial information. The choice is thus depending on the application. If the study concerns only axonal direction, then the single shell is the most efficient approach. If the study involves pathological condition, then grid scheme gives the best coverage on both orientation and radial information.

Questions

1. Can we reconstruct HARDI and multishell data by DSI? Why or why not?
2. Can we reconstruct multishell and DSI data using QBI? Why or why not?
3. A study proposes to use EAP to study a microscopic tissue with a tiny tube structure. How to setup the q-space imaging parameter? (e.g. maximum b-value, diffusion time, ..etc.)
4. A study proposes to use EAP to study non-restricted diffusion. How to setup the q-space imaging parameter?

Exercise
1. Reconstruct diffusion data with QBI, DSI, GQI (see instructions in Reconstruction(DTI, QBI, DSI, GQI, QSDR)). Evaluate the result by inspecting the dODF and axonal direction map.
2. Change the QBI, DSI, and GQI reconstruction parameters and observe how this affect the smoothness/sharpness of the dODF.
3. Remove one or more DWI and reconstruct DSI, QBI, or GQI? Does it affect the result?
4. Load the sample images provided from Sample Images. Could you tell what diffusion scheme was used by the b-table?

Reference

[1] Callaghan, P.T., Principles of Nuclear Magnetic Resonance Microscopy. 1994: Oxford University Press.
[2] Tuch, David S. "Q‐ball imaging." Magnetic Resonance in Medicine 52.6 (2004): 1358-1372.
[3] Wedeen, Van J., et al. "Mapping complex tissue architecture with diffusion spectrum magnetic resonance imaging." Magnetic Resonance in Medicine 54.6 (2005): 1377-1386.
[4] Yeh, Fang-Cheng, Van Jay Wedeen, and Wen-Yih Isaac Tseng. "Generalized-sampling imaging." Medical Imaging, IEEE Transactions on 29.9 (2010): 1626-1635. [pdf]
[5] Barnett, Alan. "Theory of Q‐ball imaging redux: Implications for fiber tracking."Magnetic Resonance in Medicine 62.4 (2009): 910-923.


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##### Activity2.5.3

Suppose that we work for a company that produces baked goods, including cakes, donuts, and eclairs. Our company operates two plants, called Plant 1 and Plant 2. In one hour of operation,

• Plant 1 produces 10 cakes, 50 donuts, and 30 eclairs.

• Plant 2 produces 20 cakes, 30 donuts, and 30 eclairs.

1. If plant 1 operates for $$x_1$$ hours and Plant 2 for $$x_2$$ hours, how many cakes $$C$$ does the company produce? How many donuts $$D\text{?}$$ How many eclairs $$E\text{?}$$

2. We define a matrix transformation $$T(\xvec) = \threevec{C}{D}{E}$$ where $$\threevec{C}{D}{E}$$ represents the number of baked goods produced when the plants are operated for times $$\xvec=\twovec{x_1}{x_2}\text{.}$$ If $$T(\xvec) = A\xvec\text{,}$$ what are the dimensions of the matrix $$A\text{?}$$

3. Find the vector $$T\left(\twovec{1}{0}\right)$$ and the vector $$T\left(\twovec{0}{1}\right)$$ and use your results to write the matrix $$A\text{.}$$

4. If we operate Plant 1 for 40 hours and Plant 2 for 50 hours, how many baked goods have we produced?

5. Suppose the marketing department says we need to produce 1500 cakes, 4700 donuts, and 3300 eclairs. Is it possible to meet this order? If so, how long should the two plants operate?

6. Let's now consider the needed ingredients:

• Each cake requires 4 units of flour and and 2 units of sugar.

• Each donut requires 1 unit of flour and 1 unit of sugar.

• Each eclair requires 1 units of flour and 2 units of sugar.

Suppose we make $$C$$ cakes, $$D$$ donuts, and $$E$$ eclairs. How many units of flour $$F$$ are required? How many units of sugar $$S\text{?}$$

7. Write a matrix $$B$$ that defines the matrix transformation $$R\left(\threevec{C}{D}{E}\right) = \twovec{F}{S}\text{.}$$

8. If Plant 1 operates for 30 hours and Plant 2 operates for 20 hours, how many units of flour and sugar are required?

9. We can consider the matrix transformation $$P(\xvec) = \twovec{F}{S}$$ that tells us how many units of flour and sugar are required when we operate the plants for $$x_1$$ and $$x_2$$ hours. Find the matrix that defines the transformation $$P\text{.}$$

in-context

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# Decimal Sum Dilly-Dallying

Level pending

Let f(n) denote the digit sum of integer n when written in decimal notation. What is fff(4444^4444)?

×


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Torsiversion is amedical condition in which a tooth is rotated away from its normalposition about the long axis of the root. For example, the firstpremolar in the figure is not only displaced slightly in thenegative direction, it is also rotated clockwise from its properorientation. To correct this condition, an orthodontist might usean archwire and a bracket to apply both a force and a torque to thetooth. In the case shown, the forces and produce a net force in the positive direction of 1.7 and a torque about the center of the tooth (theorigin of the coordinate system) of 0.0097 .

The question asks to find the forces F1 and F2. I am not surehow to set up the equations. Thanks.


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# What is 51/284 as a decimal?

## Solution and how to convert 51 / 284 into a decimal

51 / 284 = 0.18

To convert 51/284 into 0.18, a student must understand why and how. Both are used to handle numbers less than one or between whole numbers, known as integers. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. Now, let's solve for how we convert 51/284 into a decimal.

## 51/284 is 51 divided by 284

Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators and Denominators. This creates an equation. Now we divide 51 (the numerator) into 284 (the denominator) to discover how many whole parts we have. This is how we look at our fraction as an equation:

### Numerator: 51

• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Overall, 51 is a big number which means you'll have a significant number of parts to your equation. 51 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. Let's take a look below the vinculum at 284.

### Denominator: 284

• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 284 is a large number which means you should probably use a calculator. And it is nice having an even denominator like 284. It simplifies some equations for us. Ultimately, don't be afraid of double-digit denominators. Next, let's go over how to convert a 51/284 to 0.18.

## Converting 51/284 to 0.18

### Step 1: Set your long division bracket: denominator / numerator

$$\require{enclose} 284 \enclose{longdiv}{ 51 }$$

Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.

### Step 2: Extend your division problem

$$\require{enclose} 00. \\ 284 \enclose{longdiv}{ 51.0 }$$

Uh oh. 284 cannot be divided into 51. So we will have to extend our division problem. Add a decimal point to 51, your numerator, and add an additional zero. This doesn't add any issues to our denominator but now we can divide 284 into 510.

### Step 3: Solve for how many whole groups you can divide 284 into 510

$$\require{enclose} 00.1 \\ 284 \enclose{longdiv}{ 51.0 }$$

We can now pull 284 whole groups from the equation. Multiply this number by 284, the denominator to get the first part of your answer!

### Step 4: Subtract the remainder

$$\require{enclose} 00.1 \\ 284 \enclose{longdiv}{ 51.0 } \\ \underline{ 284 \phantom{00} } \\ 226 \phantom{0}$$

If your remainder is zero, that's it! If there is a remainder, extend 284 again and pull down the zero

### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.

In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.

### Why should you convert between fractions, decimals, and percentages?

Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. And the same is true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But they all represent how numbers show us value in the real world. Here are examples of when we should use each.

### When you should convert 51/284 into a decimal

Dining - We don't give a tip of 51/284 of the bill (technically we do, but that sounds weird doesn't it?). We give a 17% tip or 0.18 of the entire bill.

### When to convert 0.18 to 51/284 as a fraction

Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct.

### Practice Decimal Conversion with your Classroom

• If 51/284 = 0.18 what would it be as a percentage?
• What is 1 + 51/284 in decimal form?
• What is 1 - 51/284 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.18 + 1/2?


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This week I was training for a interview for Toptal by doing some Codility Lessons. One I've found particularly difficult and also intriguing was the MinAbsSum from Lesson 17 - Dynamic Programming.

The problem consisted in taking an array $A$ containing $N$ numbers where $A_n \in [-100,100]$ and returning the minimum absolute value we could get by summing or subtracting all numbers of $A$.

## My Approach

I've tried this three times, the first I've run out of time because I was doing other things at the same time and also was getting ready for bed. Then I woke up in the other day and went to the university thinking about the way I was trying to solve and then started to code before my Software Engineering class. However, When my teacher arrived at the classroom I've stopped. That afternoon I went over my code again, discovered what I've done wrong and corrected it. And then, finally, my Dynamic Programming approach worked.

However, it was not the most efficient. The requested time complexity was $O(N*max(|A|)^2)$, and space-complexity $O(N * \Sigma_{i=0}^{n}|A_i|)$, and my solution had both space-complexity and time-complexity equals to $O(N * \Sigma_{i=0}^{n}{(|A_i|)})$. However since my solution was completely different from all the others I've seen in the internet I wanted to share here.

### Defining the recursive function

Well, to do a Dynamic Programming approach we first have to define our recursive function. What I've did was redefining the problem, so instead of finding the most close to zero we can get, I've generalized the call to be the most close to any $K \in [0, \Sigma_{i=0}^{n}{|A_i|}]$ we could get after doing $i$ operations.

The main idea is, to get to a absolute value of $K$, at step $i$, we can either do that by summing the last value with $A_i$, when that step tried to get close to $K-A_i$ or we can do that by subtracting $A_i$ from our best attempt to get close to $K+A_i$.

So our best solution function, let's call it $bs(A,i,K)$, where $A$ is the array, $i$ is the step and $K$ is where we are trying to get can be defined by:

\begin{align} \text{bs}(A,i,K) = \text{mostCloseTo}( & \\ & K, \\ & |\text{bs}(A,i+1,|K+A_i|)-A_i|, \\ & |\text{bs}(A,i+1,|K-A_i|)+A_i| \\ ) \end{align}

And the $mostCloseTo(K,a,b)$ function returns the value between $a$ and $b$ that are the closest to $K$.

### The code

To do the code is just the same thing for all DP algorithms, to avoid stack overflow, we have to unwrap the recursion. After that this is my final code:

### The results

I did not passed the performance tests, however some of those tests were off by just a few milisseconds. Maybe by little optimization we can achieve this with the same algorithm. The only one that went really wrong was the medium2 with 6+ seconds.

### Other solutions

Well, the other solutions I've found on the internet is the official Codility Solution and blog post. The way they did was by reducing the problem in a sub-multiset problem, really similar to the Knapsack Problem, which is a well known problem that is solved in polinomial time with Dynamic Programmin. It's better than mine but in general, I liked my idea because is more closely related to the actual problem.


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# diss.ACF

0th

Percentile

##### Autocorrelation-based Dissimilarity

Computes the dissimilarity between two time series as the distance between their estimated simple (ACF) or partial (PACF) autocorrelation coefficients.

Keywords
~kwd1 , ~kwd2
##### Usage
diss.ACF(x, y, p = NULL, omega=NULL, lag.max=50)
diss.PACF(x, y, p = NULL, omega=NULL, lag.max=50)
##### Arguments
x

Numeric vector containing the first of the two time series.

y

Numeric vector containing the second of the two time series.

p

If not NULL, sets the weight for the geometric decaying of the autocorrelation coefficients. Ranging from 0 to 1.

lag.max

Maximum number of simple or partial autocorrelation coefficients to be considered.

omega

If not NULL, completely specifies the weighting matrix for the autocorrelation coefficients. p is ignored if omega is used.

##### Details

Performs the weighted Euclidean distance between the simple autocorrelation ( dist.ACF) or partial autocorrelation ( dist.PACF ) coefficients. If neither p nor omega are specified, uniform weighting is used. If p is specified, geometric wights decaying with the lag in the form $p(1-p)^i$ are applied. If omega ($\Omega$) is specified, $$d(x,y) = {\{ ( \hat{\rho}_{x} - \hat{\rho}_{y} )^t \bm{\Omega} (\hat{\rho}_{x} - \hat{\rho}_{y} ) \}}^\frac{1}{2}$$ with $\hat{\rho}_{x}$ and $\hat{\rho}_{y}$ the respective (partial) autocorrelation coefficient vectors.

##### Value

The computed distance.

##### References

Galeano, P. and Pe<U+00F1>a, D. (2000). Multivariate analysis in vector time series. Resenhas, 4 (4), 383--403.

Montero, P and Vilar, J.A. (2014) TSclust: An R Package for Time Series Clustering. Journal of Statistical Software, 62(1), 1-43. http://www.jstatsoft.org/v62/i01/.

diss.COR

• diss.ACF
• diss.PACF
##### Examples
# NOT RUN {
## Create three sample time series
x <- cumsum(rnorm(100))
y <- cumsum(rnorm(100))
z <- sin(seq(0, pi, length.out=100))
## Compute the distance and check for coherent results
diss.PACF(x, y)
diss.ACF(x, z)
diss.PACF(y, z)
#create a dist object for its use with clustering functions like pam or hclust
diss( rbind(x,y,z), "ACF", p=0.05)

# }

Documentation reproduced from package TSclust, version 1.2.4, License: GPL-2

### Community examples

Looks like there are no examples yet.


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# Week 7

Example 1. A shopkeeper bought two TV sets at Rs. 10,000 each. He sold one at a profit 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.

Example 2. A shopkeeper sold two TV sets at Rs. 10,000 each. He sold one at a profit 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

The questions given above are not same! First, make sure of it.

The first question is different from the second only with respect to the terms bought and sold. The rest all are same. Due to these terms, how the behaviour of a question of profit and loss will change, that is the main point of discussion here. Let’s see the solution.

Solution 1. In this question, the C.P of each of the two TV sets is given (so that we know the total C.P) and we need to find the total S.P.

Cost Price (C.P) of the first and second TV sets are Rs.10,000 each. Therefore, total cost price = Rs. (10,000 + 10,000) = Rs.20,000. To find the total profit/ loss in this transaction, we need to find the total Selling Price (S.P). We know the following formulas for finding S.P:

In the first case, C.P = Rs.10,000 and profit = 10%. Hence

In the second case, C.P = Rs.10,000 and loss = 10%. Hence

Therefore, total S.P = Rs. (11,000 + 9,000) = Rs. 20,000.

Since total C.P = total S.P, there is no profit or loss in this transaction.

Alternatively, we can find the total profit in the first transaction and the total loss in the second transaction. Then we will compare the total profit and loss.

Profit occurred in the first transaction

Loss occurred in the second transaction

Since the amount of profit and loss are same, there is no profit or loss in this transaction.

As another alternative, we can use the unitary method to solve the given question. Just agree to the facts: in case of profit, if C.P is Rs.100 then S.P is Rs.110 and in case of loss, if C.P is Rs.100 then S.P is Rs.90. We need to find the respective S.P using this.

In the first case,

If C.P is Rs.100 then S.P is Rs.110

If C.P is Rs.1 then S.P is Rs.(110/100) = Rs.(11/10)

If C.P is Rs.10,000 then S.P is Rs.(11/10) × 10,000 = Rs.11,000.

In the second case,

If C.P is Rs.100 then S.P is Rs.90

If C.P is Rs. 1 then S.P is Rs. (90/100) = Rs. (9/10)

If C.P is Rs. 10,000 then S.P is Rs. (9/10) × 10,000 = Rs. 9,000.

As before, total C.P is Rs. 10,000 + Rs. 10,000 = Rs. 20,000 and total S.P is Rs. (11,000 + 9,000) = Rs. 20,000.

Since total C.P = total S.P, there is no profit or loss in this transaction.

Solution 2. The second example is totally different from the first one. In this question, the S.P of each of the two TV sets is given (so that we know the total S.P) and we need to find the total C.P.

Selling Price (S.P) of the first and second TV sets are ₹10,000 each. Therefore, total selling price = Rs. (10,000 + 10,000) = Rs. 20,000. To find the total profit/ loss in this transaction, we need to find the total Cost Price (C.P). We know the following formulas for finding C.P:

In the first case, S.P = Rs. 10,000 and profit = 10%. Hence

In the second case, S.P = Rs. 10,000 and loss = 10%. Hence

Since total C.P > total S.P, there is a loss in this transaction.

As another alternative, we can use the unitary method to solve the given question. Here, we need to find the respective C.P using this.

In the first case,

If S.P is Rs. 110 then C.P is Rs. 100

If S.P is Rs. 1 then C.P is Rs. (100/110) = Rs. (10/11)

In the second case,

If S.P is Rs. 90 then C.P is Rs. 100

If S.P is Rs. 1 then C.P is Rs. (100/90) = Rs. (10/9)

Since total C.P > total S.P, there is a loss in this transaction.


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Illumination-adaptive skin color detection method. (Chinese)
J. Comput. Appl. 31, No. 5, 1233-1236 (2011).
1
The cellular competitive decision algorithm for the degree-constrained minimum spanning tree problem. (Chinese)
J. Shanghai Second Polytech. Univ. 28, No. 3, 207-213 (2011).
2
Research on the establishment of distributed storage virtual geographic environment based on lustre file system. (English)
Zhou, Qihai (ed.), Theoretical and mathematical foundations of computer science. Second international conference, ICTMF 2011, Singapore, May 5‒6, 2011. Selected papers. Berlin: Springer (ISBN 978-3-642-24998-3/pbk; 978-3-642-24999-0/ebook). Communications in Computer and Information Science 164, 329-336 (2011).
3
Microgroup mining on TSina via network structure and user attribute. (English)
Tang, Jie (ed.) et al., Advanced data mining and applications. 7th international conference, ADMA 2011, Beijing, China, December 17‒19, 2011. Proceedings, Part II. Berlin: Springer (ISBN 978-3-642-25855-8/pbk). Lecture Notes in Computer Science 7121. Lecture Notes in Artificial Intelligence, 138-151 (2011).
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Relative self-localization estimation for indoor mobile robot. (English)
Park, James J. (ed.) et al., Information anf technoloy convergence and services (ITCS 2011) and intelligent robotics, automations, telecommunications facilities, and applications (IRoA 2011). Selected papers based on the presentations at the conferences, Gwanju, Korea, October 20‒22, 2011. Berlin: Springer (ISBN 978-94-007-2597-3/hbk; 978-94-007-2598-0/ebook). Lecture Notes in Electrical Engineering 107, 429-431 (2012).
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Development of a deep ocean master-slave electric manipulator control system. (English)
Jeschke, Sabina (ed.) et al., Intelligent robotics and applications. 4th international conference, ICIRA 2011, Aachen, Germany, December 6‒8, 2011. Proceedings, Part II. Berlin: Springer (ISBN 978-3-642-25488-8/pbk). Lecture Notes in Computer Science 7102. Lecture Notes in Artificial Intelligence, 412-419 (2011).
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Development of a networked multi-agent system based on real-time ethernet. (English)
Jeschke, Sabina (ed.) et al., Intelligent robotics and applications. 4th international conference, ICIRA 2011, Aachen, Germany, December 6‒8, 2011. Proceedings, Part I. Berlin: Springer (ISBN 978-3-642-25485-7/pbk). Lecture Notes in Computer Science 7101. Lecture Notes in Artificial Intelligence, 356-365 (2011).
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Diagnosis method of combing feature extraction based on time-frequency analysis and intelligent classifier. (English)
Gong, Zhiguo (ed.) et al., Web information systems and mining. International conference, WISM 2011, Taiyuan, China, September 24‒25, 2011. Proceedings, Part I. Berlin: Springer (ISBN 978-3-642-23970-0/pbk). Lecture Notes in Computer Science 6987, 419-425 (2011).
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Two identities involving the cubic partition function $\text{\cyr Ya}$. (English)
Int. J. Number Theory 7, No. 6, 1615-1626 (2011).
9
Architecture design of variable block size motion estimation for full and fast search algorithms in H.264/AVC. (English)
Comput. Electr. Eng. 37, No. 3, 285-299 (2011).
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Policy-centric protection of OS kernel from vulnerable loadable kernel modules. (English)
Bao, Feng (ed.) et al., Information security practice and experience. 7th international conference, ISPEC 2011, Guangzhou, China, May 30 ‒ June 1, 2011. Proceedings. Berlin: Springer (ISBN 978-3-642-21030-3/pbk). Lecture Notes in Computer Science 6672, 317-332 (2011).
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Dancer: disease-annotated chromatin epigenetics resource. (English)
Nucleic Acids Res. 39, No. Database-Issue, 889-894 (2011).
12
Phylopro: a web-based tool for the generation and visualization of phylogenetic profiles across eukarya. (English)
Bioinformatics 27, No. 6, 877-878 (2011).
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Integrating offline analysis and online protection to defeat buffer overflow attacks. (English)
Burmester, Mike (ed.) et al., Information security. 13th international conference, ISC 2010, Boca Raton, FL, USA, October 25‒28, 2010. Revised selected papers. Berlin: Springer (ISBN 978-3-642-18177-1/pbk). Lecture Notes in Computer Science 6531, 409-415 (2011).
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Assessment of MODIS thermal emissive band on-orbit calibration (English)
IEEE T. Geoscience and Remote Sensing 49, No. 6-2, 2415-2425 (2011).
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GF(2^{n}) subquadratic polynomial basis multipliers for some irreducible trinomials (English)
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Cross-market financial risk analysis: an agent-based computational finance (English)
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Architecture design of variable block size motion estimation for full and fast search algorithms in H.264/AVC (English)
Computers & Electrical Engineering 37, No. 3, 285-299 (2011).
18
Policy-centric protection of OS kernel from vulnerable loadable kernel modules (English)
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Diagnosis method of combing feature extraction based on time-frequency analysis and intelligent classifier (English)
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CBSE (Science) Class 11CBSE
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# Emission Transitions in the Paschen Series End at Orbit N = 3 and Start from Orbit N and Can Be Represented as V = 3.29 × 1015 (Hz) Calculate the Value of N If the Transition is Observed at 1285 Nm. Find the Region of the Spectrum. - CBSE (Science) Class 11 - Chemistry

ConceptEvidence for the Quantized Electronic Energy Levels - Atomic Spectra

#### Question

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as = 3.29 × 1015 (Hz) [1/32 – 1/n2]

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

#### Solution 1

Wavelength of transition = 1285 nm

= 1285 × 10–9 m (Given)

v = 3.29 xx 10^(15)(1/3^2 - 1/n^2) (Given)

Since v = c/lambda

= (3.0 xx 10^8)/(1285xx10^(-9)m)

ν = 2.33 × 1014 s–1

Substituting the value of ν in the given expression,

3.29 xx 10^(15)(1/9 - 1/n^2) = 2.33 xx 10^14

1/9 - 1/n^2 = (2.33xx10^14)/(3.29xx10^15)

1/9 - 0.7082 xx 10^(-1) = 1/n^2

=> 1/n^2 = 1.1 xx 10^(-1) - 0.7082 xx 10^(-1)

1/n^2 = 4.029 xx 10^(-2)

n = sqrt(1/(4.029 xx 10^(-2)))

n = 4.98

n ≈ 5

Hence, for the transition to be observed at 1285 nm, n = 5.

The spectrum lies in the infra-red region.

#### Solution 2

ν = c/λ = 3.0×10ms-1/1285×10-9 m = 3.29×1015 (1/3- 1/n2)

⇒ 1/n= 1/9 - (3.0×10ms-1/1285×10-9 m)×(1/3.29×1015) = 0.111-0.071 = 0.04 = 1/25

⇒ n= 25

⇒ n = 5

The radiation corresponding to 1285 nm lies in the infrared region.

Is there an error in this question or solution?

#### APPEARS IN

Solution Emission Transitions in the Paschen Series End at Orbit N = 3 and Start from Orbit N and Can Be Represented as V = 3.29 × 1015 (Hz) Calculate the Value of N If the Transition is Observed at 1285 Nm. Find the Region of the Spectrum. Concept: Evidence for the Quantized Electronic Energy Levels - Atomic Spectra.
S


text:
Boundary Element Analysis of Three-Dimensional Exponentially Graded Isotropic Elastic Solids
R. Criado; J.E. Ortiz; V. Manti c; L.J. Gray; and F. Par ' \i s

doi:10.3970/cmes.2007.022.151
 Source CMES: Computer Modeling in Engineering & Sciences, Vol. 22, No. 2, pp. 151-, 2007 Download Full length paper in PDF format. Size = 562,879 bytes Keywords functionally graded materials, boundary element method, three-dimensional elasticity, So\discretionary {-}{}{}mi\discretionary {-}{}{}glia\discretionary {-}{}{}na identity, fundamental solution in tractions. Abstract A numerical implementation of the Somigliana identity in displacements for the solution of 3D elastic problems in exponentially graded isotropic solids is presented. An expression for the fundamental solution in displacements,$U_{j\ell }$, was deduced by Martin et al. (\textit {Proc. R. Soc. Lond. A}, \textbf {458}, pp. 1931--1947, 2002). This expression was recently corrected and implemented in a Galerkin indirect 3D BEM code by Criado et al. (\textit {Int. J. Numer. Meth. Engng.}, 2008). Starting from this expression of$U_{j\ell }$, a new expression for the fundamental solution in tractions$T_{j\ell }$ has been deduced in the present work. These quite complex expressions of the integral kernels$U_{j\ell }$ and$T_{j\ell }$ have been implemented in a collocational direct 3D BEM code. The numerical results obtained for 3D problems with known analytic solutions verify that the new expression for$T_{j\ell }$ is correct. Excellent accuracy is obtained with very coarse boundary element meshes, even for a relatively high grading of elastic properties considered.


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# The Unapologetic Mathematician

## Deriving Physics from Maxwell’s Equations

It’s important to note at this point that we didn’t have to start with our experimentally-justified axioms. Maxwell’s equations suffice to derive all the physics we need.

In the case of Faraday’s law, we’re already done, since it’s exactly the third of Maxwell’s equations in integral form. So far, so good.

Coulomb’s law is almost as simple. If we have a point charge $q$ it makes sense that it generate a spherically symmetric, radial electric field. Given this assumption, we just need to calculate its magnitude at the radius $r$. To do this, set up a sphere of that radius around the point; Gauss’ law in integral form tells us that the flow of $E$ out through this sphere is the total charge $q$ inside. But it’s easy to calculate the integral, getting

$\displaystyle4\pi r^2\lvert E(\lvert r\rvert)\rvert=\frac{q}{\epsilon_0}$

or

$\displaystyle\lvert E(\lvert r\rvert)\rvert=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

which is the magnitude given by Coulomb’s law.

To get the Biot-Savart law, we can use Ampère’s law to calculate the magnetic field around an infinitely long straight current $I$. We again argue on geometric grounds that the magnitude of the magnetic field should only depend on the distance from the current and should point directly around the current. If we set up a circle of radius $r$ then, the total circulation around the circle is, by Ampère’s law:

$\displaystyle2\pi r\lvert B(\lvert r\rvert)\rvert=\mu_0I$

or

$\displaystyle\lvert B(\lvert r\rvert)\rvert=\frac{\mu_0}{2\pi}\frac{I}{r}$

Now, we can compare this to the last time we computed the magnetic field of the straight infinite current by integrating the Biot-Savart law directly and got essentially the same answer.

Finally, we can derive conservation of charge from Ampère’s law, with Maxwell’s correction by taking its divergence:

$\displaystyle\nabla\cdot(\nabla\times B)=\mu_0\nabla\cdot J+\epsilon_0\mu_0\frac{\partial}{\partial t}(\nabla\cdot E)$

The quantity on the left is the divergence of a curl, so it automatically vanishes. Meanwhile, Gauss’ law tells us that $\epsilon\nabla\cdot E=\rho$, so we conclude

$\displaystyle0=\mu_0\left(\nabla\cdot J+\frac{\partial\rho}{\partial t}\right)$

or

$\displaystyle\nabla\cdot J+\frac{\partial\rho}{\partial t}=0$

which is the “continuity equation” expressing the conservation of charge.

The importance is that while we originally derived Maxwell’s equations from four experimentally-justified laws, those laws are themselves essentially derivable from Maxwell’s equations. Thus any reformulation of Maxwell’s equations is just as sufficient a basis for all of electromagnetism as our original physical axioms.

February 3, 2012 -


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## University Calculus: Early Transcendentals (3rd Edition)

We take $\delta=1$ here.
$$a=1 \hspace{1cm} b=7\hspace{1cm}c=2$$ Here we need to find $\delta\gt0$ such that for all $x$, $0\lt |x-2|\lt\delta$ then $1\lt x\lt 7$ The sketch is shown below. The distance from $c=2$ to the nearer endpoint of $(1,7)$ is $1$. So if we take $\delta=1$ or any smaller positive value, then $0\lt |x-2|\lt1$, and $x$ will only range in $(1,3)$, as shown by the red lines in the sketch, thereby $x$ is always placed between $1$ and $7$ as requested. In other words, $$0\lt |x-2|\lt1\hspace{2cm}1\lt x\lt 7$$


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# Massively scalable Sinkhorn distances via the Nystr om method Massively scalable Sinkhorn distances

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### Text of Massively scalable Sinkhorn distances via the Nystr om method Massively scalable Sinkhorn distances

• Massively scalable Sinkhorn distances via the Nyström

method

Jason Altschuler∗

MIT jasonalt@mit.edu

Francis Bach†

INRIA - ENS - PSL francis.bach@inria.fr

Alessandro Rudi†

INRIA - ENS - PSL alessandro.rudi@inria.fr

Jonathan Niles-Weed‡

NYU jnw@cims.nyu.edu

Abstract

The Sinkhorn “distance,” a variant of the Wasserstein distance with entropic regu- larization, is an increasingly popular tool in machine learning and statistical inference. However, the time and memory requirements of standard algorithms for computing this distance grow quadratically with the size of the data, making them prohibitively expensive on massive data sets. In this work, we show that this challenge is surpris- ingly easy to circumvent: combining two simple techniques—the Nyström method and Sinkhorn scaling—provably yields an accurate approximation of the Sinkhorn distance with significantly lower time and memory requirements than other approaches. We prove our results via new, explicit analyses of the Nyström method and of the stabil- ity properties of Sinkhorn scaling. We validate our claims experimentally by showing that our approach easily computes Sinkhorn distances on data sets hundreds of times larger than can be handled by other techniques.

∗Supported in part by NSF Graduate Research Fellowship 1122374. †Supported in part by the European Research Council (grant SEQUOIA 724063). ‡Supported in part by the Josephine de Kármán Fellowship.

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• Contents

1 Introduction 3 1.1 Our contributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Prior work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Outline of paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Main result 5 2.1 Preliminaries and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.2 Main result and proposed algorithm . . . . . . . . . . . . . . . . . . . . . . 7

3 Kernel approximation via the Nyström method 10 3.1 Preliminaries: Nyström and error in terms of effective dimension . . . . . . 10 3.2 Adaptive Nyström with doubling trick . . . . . . . . . . . . . . . . . . . . . 11 3.3 General results: data points lie in a ball . . . . . . . . . . . . . . . . . . . . 12 3.4 Adaptivity: data points lie on a low dimensional manifold . . . . . . . . . . 13

4 Sinkhorn scaling an approximate kernel matrix 16 4.1 Using an approximate kernel matrix . . . . . . . . . . . . . . . . . . . . . . 17 4.2 Using an approximate Sinkhorn projection . . . . . . . . . . . . . . . . . . . 17 4.3 Proof of Theorem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Proof of Theorem 1 18

6 Experimental results 19

A Pseudocode for subroutines 22 A.1 Pseudocode for Sinkhorn algorithm . . . . . . . . . . . . . . . . . . . . . . . 22 A.2 Pseudocode for rounding algorithm . . . . . . . . . . . . . . . . . . . . . . . 23

B Omitted proofs 24 B.1 Stability inequalities for Sinkhorn distances . . . . . . . . . . . . . . . . . . 24 B.2 Bregman divergence of Sinkhorn distances . . . . . . . . . . . . . . . . . . . 25 B.3 Hausdorff distance between transport polytopes . . . . . . . . . . . . . . . . 25 B.4 Miscellaneous helpful lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . 26 B.5 Supplemental results for Section 3 . . . . . . . . . . . . . . . . . . . . . . . 27

B.5.1 Full proof of Theorem 3 . . . . . . . . . . . . . . . . . . . . . . . . . 27 B.5.2 Full proof of Theorem 4 . . . . . . . . . . . . . . . . . . . . . . . . . 29 B.5.3 Additional bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

C Lipschitz properties of the Sinkhorn projection 33

References 34

2

• 1 Introduction

Optimal transport is a fundamental notion in probability theory and geometry (Villani, 2008), which has recently attracted a great deal of interest in the machine learning com- munity as a tool for image recognition (Li et al., 2013; Rubner et al., 2000), domain adaptation (Courty et al., 2014, 2017), and generative modeling (Arjovsky et al., 2017; Bousquet et al., 2017; Genevay et al., 2016), among many other applications (see, e.g., Kolouri et al., 2017; Peyré and Cuturi, 2017).

The growth of this field has been fueled in part by computational advances, many of them stemming from an influential proposal of Cuturi (2013) to modify the definition of optimal transport to include an entropic penalty. The resulting quantity, which Cu- turi (2013) called the Sinkhorn “distance”1 after Sinkhorn (1967), is significantly faster to compute than its unregularized counterpart. Though originally attractive purely for computational reasons, the Sinkhorn distance has since become an object of study in its own right because it appears to possess better statistical properties than the unregular- ized distance both in theory and in practice (Genevay et al., 2018; Montavon et al., 2016; Peyré and Cuturi, 2017; Rigollet and Weed, 2018; Schiebinger et al., 2019). Computing this distance as quickly as possible has therefore become an area of active study.

We briefly recall the setting. Let p and q be probability distributions supported on at most n points in Rd. We denote by M(p,q) the set of all couplings between p and q, and for any P ∈ M(p,q), we denote by H(P ) its Shannon entropy. (See Section 2.1 for full definitions.) The Sinkhorn distance between p and q is defined as

Wη(p,q) := min P∈M(p,q)

∑ ij

Pij‖xi − xj‖22 − η−1H(P ) , (1)

for a parameter η > 0. We stress that we use the squared Euclidean cost in our formulation of the Sinkhorn distance. This choice of cost—which in the unregularized case corresponds to what is called the 2-Wasserstein distance (Villani, 2008)—is essential to our results, and we do not consider other costs here. The squared Euclidean cost is among the most common in applications (Bousquet et al., 2017; Courty et al., 2017; Forrow et al., 2018; Genevay et al., 2018; Schiebinger et al., 2019).

Many algorithms to compute Wη(p,q) are known. Cuturi (2013) showed that a simple iterative procedure known as Sinkhorn’s algorithm had very fast performance in practice, and later experimental work has shown that greedy and stochastic versions of Sinkhorn’s algorithm perform even better in certain settings (Altschuler et al., 2017; Genevay et al., 2016). These algorithms are notable for their versatility: they provably succeed for any bounded, nonnegative cost. On the other hand, these algorithms are based on matrix manipulations involving the n × n cost matrix C, so their running times and memory requirements inevitably scale with n2. In experiments, Cuturi (2013) and Genevay et al. (2016) showed that these algorithms could reliably be run on problems of size n ≈ 104.

Another line of work has focused on obtaining better running times when the cost matrix has special structure. A preeminent example is due to Solomon et al. (2015), who focus on the Wasserstein distance on a compact Riemannian manifold, and show

1We use quotations since it is not technically a distance; see (Cuturi, 2013, Section 3.2) for details. The quotes are dropped henceforth.

3

• that an approximation to the entropic regularized Wasserstein distance can be obtained by repeated convolution with the heat kernel on the domain. Solomon et al. (2015) also establish that for data supported on a grid in Rd, significant speedups are possible by decomposing the cost matrix into “slices” along each dimension (see Peyré and Cuturi, 2017, Remark 4.17). While this approach allowed Sinkhorn distances to be computed on significantly larger problems (n ≈ 108), it does not extend to non-grid settings. Other proposals include using random sampling of auxiliary points to approximate semi-discrete costs (Tenetov et al., 2018) or performing a Taylor expansion of the kernel matrix in the case of the squared Euclidean cost (Altschuler et al., 2018). These approximations both focus on the η →∞ regime, when the regularization term in (1) is very small, and do not apply to the moderately regularized case η = O(1) typically used in practice. Moreover, the running time of these algorithms scales exponentially in the ambient dimension, which can be very large in applications.

1.1 Our contributions

We show that a simple algorithm can be used to approximate Wη(p,q) quickly on massive data sets. Our algorithm uses only known tools, but we give novel theoretical guarantees that allow us to show that the Nyström method combined with Sinkhorn scaling provably yields a valid approximation algorithm for the Sinkhorn distance at a fraction of the running time of other approaches.

We establish two theoretical results of independent interest: (i) New Nyström ap- proximation results showing that instance-adaptive low-rank approximations to Gaussian kernel matrices can be found for data lying on a low-dimensional manifold (Section 3). (ii) New stability results about Sinkhorn projections, establishing that a sufficiently good approximation to the cost matrix can be used (Section 4).

1.2 Prior work

Computing the Sinkhorn distance efficiently is a well studied problem in a number of com- munities. The Sinkhorn distance is so named because, as was pointed out by Cuturi (2013), there is an extremely simple iterative algorithm due to Sinkhorn (1967) which converges quickly to a solution to (1). This algorithm, which we

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# Math Help - Joint distribution

1. ## Joint distribution

I am given X has distribution f(x)= (1/σ)*e^(-x/σ) and Y has the same distribution but with μ and y in place of sigma and x. I am also given Z=min{X,Y} and W={1 if Z=X and 0 if Z=Y}.
I am told to find the joint distribution of Z and W and also prove Z and W are independent.
I am stuck on this one. Does it break down into two cases (Z=X and Z=Y)?

2. Originally Posted by kman320
I am given X has distribution f(x)= (1/σ)*e^(-x/σ) and Y has the same distribution but with μ and y in place of sigma and x. I am also given Z=min{X,Y} and W={1 if Z=X and 0 if Z=Y}.
I am told to find the joint distribution of Z and W and also prove Z and W are independent.
I am stuck on this one. Does it break down into two cases (Z=X and Z=Y)?
kman320,

Are you sure of your problem statement? It looks to me like Z and W are dependent.

3. Its seems they are, but he gives a hint. Show the prob. that Z<z given W=i (for i=1,2) is equal to the prob. Z<z.
Show P(Z<z l W=i)=P(Z<z) .
How will I go about doing the first step of finding the joint dist. of Z & W?

4. Anyone?


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# Condition for Superradiance in Higher-dimensional Rotating Black Holes

Authors
Type
Published Article
Publication Date
Apr 16, 2005
Submission Date
Mar 22, 2005
Identifiers
DOI: 10.1016/j.physletb.2005.04.022
Source
arXiv
It is shown that the superradiance modes always exist in the radiation by the $(4+n)$-dimensional rotating black holes. Using a Bekenstein argument the condition for the superradiance modes is shown to be $0 < \omega < m \Omega$ for the scalar, electromagnetic and gravitational waves when the spacetime background has a single angular momentum parameter about an axis on the brane, where $\Omega$ is a rotational frequency of the black hole and $m$ is an azimuthal quantum number of the radiated wave.


text:
# Maximize

Geometry Level 3

Given two real numbers $$x, y$$ satisfying the relation $$x^{2}+y^{2}=1$$. Then find the maximum value of

$\large\dfrac{1}{x^{2}+3xy+5y^{2}}$

×


text:
# Induced EMF Question

## Homework Statement

A rectangular wire loop, 24mm by 36mm, is placed in a magnetic field. The flux density perpendicular to the loop is decreasing in magnitude at a rate of 0.2Wb/$m^{2}$/s. Calculate the induced emf around the loop.

## Homework Equations

The problem is I don't know which equation applies to this question. I know that 0.2Wb/$m^{2}$/s = $\frac{d\phi}{dt}$ but as far as I was concerned from what I have read so far is that you would need to know the number of loops to get the answer to this question.

## The Attempt at a Solution

I know that the answer is 0.17mV. I managed to get this answer by multiplying the dimensions together 24mm x 36mm = 8.64x$10^{-4}$ then multiply this by 0.2. But I don't understand why this this is the method, if someone could explain this for me I would greatly appreciate it

## Answers and Replies

Let's look at the equation in question: $\int$$\vec{B}$$\cdot$d$\vec{A}$ = - N*d$\phi$/dt

Since you're examining magnitudes, let's forget the minus sign. You have B, and the problem states there is one rectangular loop, so you know N. The induced EMF is defined by d$\phi$/dt, so all that's missing is ∫dA. That's just A, which you calculated.

That's why the given B multiplied by the A you calculated gives your answer, and why N = 1.

1 person
As always, it's best to start with the general form of the applicable physical principle. In this case, what is the relationship between induced emf and magnetic flux?

1 person
Ah I see where I have gone wrong! Was just a matter of reading the question properly Thanks a lot guys!

rude man
Homework Helper
Gold Member
Let's look at the equation in question: $\int$$\vec{B}$$\cdot$d$\vec{A}$ = - N*d$\phi$/dt

?? Don't look at it too long ....

Edit: Oops! I forgot the d/dt on the L.H.S.!!

rude man
Homework Helper
Gold Member
Edit: Oops! I forgot the d/dt on the L.H.S.!!

Also the N ...

B is a function of N, so it isn't necessary to include on the L.H.S. It is implied.

rude man
Homework Helper
Gold Member
B is a function of N, so it isn't necessary to include on the L.H.S. It is implied.

Then it's implied in ø on the right-hand side as well, so you'd have to get rid of it there.

N needs to be on the left and right.
The relation is ∫B.dA = ø, and emf = - N dø/dt.


text:
# Use only the last sample as the posterior in MCMC

I am new to statistics.

After an MCMC sampler warmed up, the posterior is better estimated as the mean of several samples. (e.g. related question: http://stats.stackexchange.com//questions/56077)

However, in my MCMC sampler, after it reaches the equilibrium (because $\hat{R}=1.0$ in Stan), samples of the parameters still present a fairly large STD. Do the large STD means that the chain hasn't converged yet, even if $\hat{R}=1.0$?

In this case, I do not want to use the mean of the parameter samples, but only use one sample (say the 1000th) of the parameters as the posterior. Is it the right way to do?

P.S. "The $\hat{R}$ statistic measures the ratio of the average variance of samples within each chain to the variance of the pooled samples across chains, if all chains are at equilibrium, these will be the same and $\hat{R}$ will be one" -- Stan Manual

-

Keep in mind that in Bayesian inference, you are looking at the entire range of values in the posterior, rather than point-estimate summaries. After convergence, MCMC samples are samples drawn from that posterior density. Using just the latest sample is the same as picking a random point from the posterior and using that as your sample estimate.

For an intuitive explanation of why this isn't great, imagine that your posterior is a standard normal distribution. The normal distribution has support over the whole real line, so drawing a random sample from the standard normal distribution could be any real number -- though, it will likely be within 2 units of zero. But if you're using MCMC, you likely don't know the distribution of the posterior, so the only information you have is just some value. You have no way to adjudge whether this value is "large" or "small" or "typical."

You would have more information if you had, say, 1000 draws from that posterior. You would know where the high- and low-density regions are. This is why people keep all of their posterior draws.

The fact that some individual parameter estimates have very large variance does not mean that the chains haven't converged -- it just means that there is a large amount of uncertainty about the posterior values of the parameter in the model given the data and the prior. This isn't necessarily bad -- some quantities are not precisely estimated. But perhaps your priors are too diffuse, or your model is not fitting the data well.

The key part from the definition of $\hat{R}$ is that this metric is a ratio of variance: we have evidence that the chain has converged to its stationary distribution because the ratio of variance in independent chains is roughly 1. The ratio can be 1 even if the variance is large in absolute terms.

-
Thanks very much, this answer really helps! There is just one point I'm still a bit confused : $\hat{R}=1$ means samples of different chains have similar variances (but the absolute variance value could be large), so why we call such a state 'stationary'? I mean, can the absolute variance shrink if we continue sampling? – Shockley Aug 1 '14 at 18:01
Standard errors (due to MC sampling) will shrink, for the same reason that adding more respondents to a survey will shrink margin of error. But continuing to draw values from the posterior won't change the posterior variance. I think you're hung up on the idea that stationary means that there is a single, "correct" value. Bayesian inference is much more interested in parameters' variability, which is to say the posterior distribution. – General Abrial Aug 1 '14 at 18:56


text:
# [Tugindia] Colour Background

E. Krishnan ekmath at md5.vsnl.net.in
Wed Aug 13 20:04:20 CEST 2003

On Tue, 12 Aug 2003, Abdullah Sowkaar wrote:

> I need your suggestions for the following

> 1. A text should come inside a oval box with colored background.

The easiest way is to use the "pstcol" package (with the "usenames" option
so that you can use  all the "named" colors) as in:

\begin{center}
\psovalbox*[fillcolor=Yellow]{%
\parbox{0.5\textwidth}{%
\color{Red}
\begin{itemize}
\item The first item
\item The second item
\item The third and last item
\end{itemize}}}
\end{center}

if you want an actual ellipse or

\begin{center}
\psframebox*[fillcolor=Yellow,framearc=1,framesep=10pt]{%
\parbox{0.5\textwidth}{%
\color{Red}
\begin{itemize}
\item The first item
\item The second item
\item The third and last item
\end{itemize}}}
\end{center}

if you need only a "capsule".

> 2. How to control  thickness of the borders of ovalbox.

If you go the "pstricks", this is set by the parameter "linewidth"

> 3. Whats the use of \thicklines or \thinlines, where it has been defined and
> how to control all those things.

These declarations decide the thickness of lines currently in effect. The
default is often decided by the environment.

--
Krishnan


text:
# Corresponding Energy Levels for Normalised Eigenfunctions

1. Jun 9, 2012

### Ryomega

1. The problem statement, all variables and given/known data

Particle of mass m in a 1D infinite square well is confined between 0 ≤ x ≤ a

Given that the normalised energy eigenfunction of the system is:

Un(x) = ($\frac{2}{a}$)$\frac{1}{2}$ sin ($\frac{nx\pi}{a}$)

where n = 1, 2, 3...

what are the corresponding energy levels?

2. Relevant equations

Normalisation: 1 = $\int$ lψl2 dx (from negative to positive infinity)

Energy Level for infinite well: En = $\frac{h^2}{2m}$ ($\frac{n}{2a}$)2

Where h = reduced planck constant (sorry couldn't find it)

3. The attempt at a solution

At first I thought, I could just plug in 1, 2, 3 in Un(x)
But then I realised the question said that Un(x) is normalised.
Does this mean I have to reverse normalise Un(x)? If so, is this order right?

Make Un(x) = 1
Differentiate ($\frac{2}{a}$)$\frac{1}{2}$ sin ($\frac{nx\pi}{a}$)
Then square root whatever the result.

Am I going in the right direction or am I missing something out?

Thank you!

2. Jun 9, 2012

### Muphrid

You don't have to unnormalize the wavefunction. You already have the answer in the $E_n$. The intent of the problem is for you to compute $E_n$ from just the wavefunction and the Hamiltonian, instead of just taking the formula from your notes or book.

In other words, remember that
$$\hat H |U_n \rangle = E_n |U_n \rangle$$

Because, well, that's the basic content of the time-independent Schrodinger equation.

3. Jun 9, 2012

### Ryomega

oh... lol. That's simple enough!

So basically apply the hamiltonian to Un(x)


text:
# cupy.linalg.eigvalsh¶

cupy.linalg.eigvalsh(a, UPLO='L')[source]

Calculates eigenvalues of a symmetric matrix.

This method calculates eigenvalues a given symmetric matrix. Note that cupy.linalg.eigh() calculates both eigenvalues and eigenvectors.

Note

Currenlty only 2-D matrix is supported.

Parameters: a (cupy.ndarray) – A symmetric 2-D square matrix. UPLO (str) – Select from 'L' or 'U'. It specifies which part of a is used. 'L' uses the lower triangular part of a, and 'U' uses the upper triangular part of a. Returns eigenvalues as a vector. cupy.ndarray

Warning

This function calls one or more cuSOLVER routine(s) which may yield invalid results if input conditions are not met. To detect these invalid results, you can set the linalg configuration to a value that is not ignore in cupyx.errstate() or cupyx.seterr().


text:
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In how many different ways can a group of 8 people be divide

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In how many different ways can a group of 8 people be divide  [#permalink]

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24 Oct 2009, 03:20
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In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A, 90
B. 105
C. 168
D. 420
E. 2520
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24 Oct 2009, 17:50
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The answer to the question is 105, not 2520. I posted a solution on another forum, which I'll paste here:

________

Think of this question:

A group of eight tennis players will be divided into four teams of two. One team will play in the Olympics, one in Wimbledon, one in the Davis Cup and one in the US Open. In how many different ways can the teams be selected?

Here, the order of the teams themselves clearly matters. If we choose {A,B} to go to the Olympics, and {C,D} to go to Wimbledon, that's clearly different from sending {C,D} to the Olympics and {A,B} to Wimbledon. The answer to this question is exactly the answer you give above:

-you have 8C2 choices for the Olympics team;
-you have 6C2 choices for the Wimbledon team;
-you have 4C2 choices for the Davis Cup team;
-you have 2C2 (one) choice for the US Open team.

Multiply these to get the answer: 8C2*6C2*4C2*2C2 = (8*7/2)(6*5/2)(4*3/2)(2*1/2) = 2520.

Note that the question I've just asked above is different from the question in the original post. In this question:

A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

the order of the teams does not matter. If we choose, say, these teams:

{A,B}, {C,D}, {E,F}, {G,H}

that's exactly the same set of teams as these:

{C,D}, {A,B}, {G,H}, {E,F}

Because the order of the teams themselves does not matter, we must divide by 4! = 24, the number of different orders we can put the four teams in, because all 24 different orders are in fact the same set of teams. So the answer is 2520/4! = 105.
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Re: combination (groups and that stuff...)  [#permalink]

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24 Oct 2009, 16:14
4 teams can be chosen in following ways,
= 8C2 * 6C2 * 4C2* 2C2
=(8!/6!*2!)*(6!/4!*2!)*(4!/2!*2!)*(2!/2!*0!)

Solving it further you will get
= (28) * (15) * (6) * (1)
= 2520

E is ans
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24 Oct 2009, 16:20
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1
Another approach:

Number of ways 8 people can be arranged = $$8!$$
Number of different teams possible = $$8!/2^4 = 2520$$
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Re: combination (groups and that stuff...)  [#permalink]

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24 Oct 2009, 17:56
19
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I think there is a little problem with the solutions above:

Does the order matters? Think not.
(1,2)(3,4)(5,6)(7,8) should be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

So 2520 should be divided by 4!=105.

I know there is a formula to determine:

A. the number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important.
B. the number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important.

Basically hgp2k used here the second one, but these formulas are not needed for GMAT and there is an easier way to solve this problem, well at least I solve this way and find it easier:

For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one pair left.

So we have 7*5*3*1=105

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Re: combination (groups and that stuff...)  [#permalink]

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24 Oct 2009, 17:58
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Sorry, didn't notice that Ian posted the right solution just 6 min before I did.
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Re: In how many different ways  [#permalink]

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12 Jan 2012, 07:06
10
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manalq8 wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

90
105
168
420
2520

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups

There is also direct formula for this:

1. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is not important is $$\frac{(mn)!}{(n!)^m*m!}$$.

2. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\frac{(mn)!}{(n!)^m}$$

Hope it helps.
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Re: In how many different ways can a group of 8 people be  [#permalink]

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01 Mar 2013, 06:07
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

Hi Bunuel,

Can u tell y have u considered 7 ways above?

Also,in the previous method im not clear why we need to divide by 4!.
Is it like say for eg:how many ways can we arrange the word AEEB so we need to consider 4!/2! since EE is repeated?
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Re: In how many different ways can a group of 8 people be  [#permalink]

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01 Mar 2013, 20:09

Suppose you were choosing a person to pair with person 1. You could form the following pairs
(1,2),(1,3)(1,4),(1,5),(1,6),(1,7)(1,8)
That's a total of 7 choices possible. Hence, one needs to choose 7.

Hope this helps!

shreerajp99 wrote:
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

Hi Bunuel,

Can u tell y have u considered 7 ways above?

Also,in the previous method im not clear why we need to divide by 4!.
Is it like say for eg:how many ways can we arrange the word AEEB so we need to consider 4!/2! since EE is repeated?
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Re: In how many different ways can a group of 8 people be  [#permalink]

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01 Mar 2013, 21:05
1
shreerajp99 wrote:
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

Hi Bunuel,

Can u tell y have u considered 7 ways above?

Also,in the previous method im not clear why we need to divide by 4!.
Is it like say for eg:how many ways can we arrange the word AEEB so we need to consider 4!/2! since EE is repeated?

Let me try explain what I thought:
Person 1 to pair with someone - have 7 choices (out of 8)
Person 2 to pair with somone - have 5 choices (out of remaining 6 people, note the 2nd person is also included in remaining 6)
.......Likewise

Divide by 4!, you are close to correct, it is to avoid repeats of similar groups. Since order of the the chosen groups does not matter here (Person 1, Person 2) is same as (Person 2, Person 1) - that means as per the formula we have number of groups which includes these repeats, to negate those we divide by 4! to get a realistic number with no such repeats

I tried, hope it is clear.....

Thanks
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Re: In how many different ways can a group of 8 people be  [#permalink]

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03 Mar 2013, 05:31
Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)

Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand?

I am messing this up conceptually and want to correct this mistake
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Re: In how many different ways can a group of 8 people be  [#permalink]

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03 Mar 2013, 23:02
2
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manimgoindowndown wrote:
Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)

Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand?

I am messing this up conceptually and want to correct this mistake

Check out this post: http://www.veritasprep.com/blog/2011/11 ... ke-groups/

It discusses two different questions:
1. Distributing 12 different chocolates equally among 4 boys (similar to splitting 8 people in 4 distinct teams with 2 people each - your question)
2. Distributing 12 different chocolates equally in 4 stacks (similar to splitting 8 people in 4 teams of 2 people each - the original question)

See if grouping makes sense thereafter.
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Re: In how many different ways can a group of 8 people be  [#permalink]

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05 Mar 2013, 18:02
VeritasPrepKarishma wrote:
manimgoindowndown wrote:
Hmm Bunuel something I did quickly, and forgot the context to do this in (this doesn't happen as frequently) was to select by team? How would I approach the problem then. IE instad of 8C2, I started with 4C1 and started to proceed from there, with the hopes of multiplying by 2! to account for the different arrangements we could have within each team (but not by 4! to account for the order of these different teams)

Why would 4C1 not be appropriate in this case? Is it because those 4 teams aren't set beforehand?

I am messing this up conceptually and want to correct this mistake

Check out this post: http://www.veritasprep.com/blog/2011/11 ... ke-groups/

It discusses two different questions:
1. Distributing 12 different chocolates equally among 4 boys (similar to splitting 8 people in 4 distinct teams with 2 people each - your question)
2. Distributing 12 different chocolates equally in 4 stacks (similar to splitting 8 people in 4 teams of 2 people each - the original question)

See if grouping makes sense thereafter.

That example and wording was extremely confusing and frustrating. I am still trying to see what language prompted the difference in permutation vs combination.

Maybe my brain has been fried this week (lots of pracatice, one full length CAT), but the big thing in this problem is you DO NOT account for order TWICE
the 4! is for the team
and the 2! is within every single combination for each pair
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Re: In how many different ways can a group of 8 people be  [#permalink]

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05 Mar 2013, 19:09
manimgoindowndown wrote:
That example and wording was extremely confusing and frustrating. I am still trying to see what language prompted the difference in permutation vs combination.

Maybe my brain has been fried this week (lots of pracatice, one full length CAT), but the big thing in this problem is you DO NOT account for order TWICE
the 4! is for the team
and the 2! is within every single combination for each pair

Yes, because in this question, the groups are not distinct. You have to split 8 people in 4 groups.
You can split them like this: (A, B), (C, D), (E, F), (G, H)
or like this: (G, H), (A, B), (C, D), (E, F)
they are the same split. They are not assigned to group1, group2, group3, group4.
Say the total number of ways we get = N

Now, if we change the question and say that we have 8 people and we need to divide them into 4 teams: Team 1, Team 2, Team 3 and Team 4

Then, one split is this: Team 1 = (A, B); Team 2 = (C, D), Team 3 = (E, F), Team 4 = (G, H)
and another split is: Team 1 = (G, H), Team 2 = (A, B); Team 3 = (C, D), Team 4 = (E, F)

These two cases were the same in our original question but if the teams/groups are distinct, the two cases are distinct. Now, the total number of ways = N*4!
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Re: combination (groups and that stuff...)  [#permalink]

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18 Nov 2013, 17:36
1
First the no of ways in which 2 can e selected fro 8 peoples;
= 8C2 * 6C2 * 4C2* 2C2
=(8!/6!*2!)*(6!/4!*2!)*(4!/2!*2!)*(2!/2!*0!)
=2520

Second the no of ways in which 2 can be arranged among 4 groups ; since the arrangement will vary hence simple factorial of will be the choice which is 4!

So finally you need to divide the former by later to get the answer
=2520/4!
=105
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Re: In how many different ways can a group of 8 people be  [#permalink]

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17 Jul 2017, 03:54
manalq8 wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A. 90
B. 105
C. 168
D. 420
E. 2520

Let's select the 4 groups of 2 individuals each

8C2 * 6C2 * 4C2 * 2C2

But wait... There is some problem...

Let 8 individuals are
A B C D E F G H

Case 1: selected 4 groups of 2 are respectively

A&B C&D E&F G&H

i.e. A&B us first group, C&D is second group etc

Case 2: selected 4 groups of 2 are respectively

C&D A&B G&H E&F respectively

So case 1 and 2 are the same except that order of the groups has been accounted for

The arrangements of 4 groups can be done in 4! Which needs to be excluded from calculation

Hence divide the result by 4!

So final result = 8C2 * 6C2 * 4C2 * 2C2 / 4! = 105

Hope this helps!!!
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Re: In how many different ways can a group of 8 people be  [#permalink]

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30 Aug 2017, 06:22
Bunuel wrote:
manalq8 wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

90
105
168
420
2520

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups

There is also direct formula for this:

1. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is not important is $$\frac{(mn)!}{(n!)^m*m!}$$.

2. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\frac{(mn)!}{(n!)^m}$$

Hope it helps.

How did you use the formula? I did not get the answer using formula.
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Posts: 64216
In how many different ways can a group of 8 people be  [#permalink]

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30 Aug 2017, 09:45
gayatriv21 wrote:
Bunuel wrote:
manalq8 wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

90
105
168
420
2520

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups

There is also direct formula for this:

1. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is not important is $$\frac{(mn)!}{(n!)^m*m!}$$.

2. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\frac{(mn)!}{(n!)^m}$$

Hope it helps.

How did you use the formula? I did not get the answer using formula.

We are dividing mn = 8 people into m = 4 teams of n = 2 people each.

Use the first formula, sine the order does not matter: $$\frac{(mn)!}{(n!)^m*m!}=\frac{8!}{(2!)^4*4!}=105$$.

P.S. You don;t really to memorize this formula for the GMAT.
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Updated on: 22 Aug 2019, 03:22
noboru wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A, 90
B. 105
C. 168
D. 420
E. 2520

Number of ways can a group of 8 people be divided into 4 teams of 2 people each = $$^8C_2*^6C_2*^4C_2*^2C_2/4!$$ = 28*15*6*1 /4! = 105

IMO B
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Originally posted by Kinshook on 22 Aug 2019, 03:15.
Last edited by Kinshook on 22 Aug 2019, 03:22, edited 1 time in total.
CEO
Joined: 03 Jun 2019
Posts: 2899
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Re: In how many different ways can a group of 8 people be divide  [#permalink]

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22 Aug 2019, 03:21
noboru wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

A, 90
B. 105
C. 168
D. 420
E. 2520

Number of ways can a group of 8 people be divided into 4 teams of 2 people each = $$\frac{^8C_2*^6C_2*^4C_2*^2C_2}{4!} = \frac{28*15*6*1}{4!} = 105$$

IMO B
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Re: In how many different ways can a group of 8 people be divide   [#permalink] 22 Aug 2019, 03:21

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text:
Difference between sudden force and impulsive force? [duplicate]

What is the difference between a sudden force which continues to act on the body, and an impulsive force?

What would be respective speeds of the body just after time= 0?

marked as duplicate by AccidentalFourierTransform, John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 22 '16 at 7:03

In both cases, the speed of the body at time $t=0$ would be zero, if it was initially at rest. Then you will ask, what is the difference between an impulsive force and an "ordinary force" which keeps acting continuously?
• @ABC123 We can't give an exact value without knowing the impulsive force as a function of time. But we might compare their value. First draw the acceleration time graphs for impulsive force and ordinary force cases. Take some time $t$ and find the area under the graph for both cases. Whichever has greater area will have greater velocity! – FreezingFire Apr 22 '16 at 6:52
• Calculating a limit here tells you that as you approach $t=0$, you get the velocity approaching zero, in both cases. So a limit is useless here. But to answer your question, take the graphs in my answer. You can find a value of $t>0$ where the area under both the graphs is equal (meaning equal velocity). Before that time, the velocity would actually be more for the non impulsive force. After that, the impulsive force would have greater velocity. – FreezingFire Apr 22 '16 at 10:17


text:
# 009C Sample Final 3, Problem 5

Consider the function

${\displaystyle f(x)=e^{-{\frac {1}{3}}x}}$

(a) Find a formula for the  ${\displaystyle n}$th derivative  ${\displaystyle f^{(n)}(x)}$  of  ${\displaystyle f}$  and then find  ${\displaystyle f'(3).}$

(b) Find the Taylor series for  ${\displaystyle f(x)}$  at  ${\displaystyle x_{0}=3,}$  i.e. write  ${\displaystyle f(x)}$  in the form

${\displaystyle f(x)=\sum _{n=0}^{\infty }a_{n}(x-3)^{n}.}$
Foundations:
The Taylor polynomial of ${\displaystyle f(x)}$  at ${\displaystyle a}$ is

${\displaystyle \sum _{n=0}^{\infty }c_{n}(x-a)^{n}}$ where ${\displaystyle c_{n}={\frac {f^{(n)}(a)}{n!}}.}$

Solution:

(a)

Step 1:
We have
${\displaystyle f'(x)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-{\frac {1}{3}}x},}$
${\displaystyle f''(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{2}e^{-{\frac {1}{3}}x},}$
and
${\displaystyle f^{(3)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{3}e^{-{\frac {1}{3}}x}.}$
If we compare these three equations, we notice a pattern.
Thus,
${\displaystyle f^{(n)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-{\frac {1}{3}}x}.}$
Step 2:
Since
${\displaystyle f'(x)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-{\frac {1}{3}}x},}$
we have
${\displaystyle f'(3)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-1}.}$

(b)

Step 1:
Since
${\displaystyle f^{(n)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{3}e^{-{\frac {1}{3}}x},}$
we have
${\displaystyle f^{(n)}(3)={\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-1}.}$
Therefore, the coefficients of the Taylor series are
${\displaystyle c_{n}={\frac {{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-1}}{n!}}.}$
Step 2:
Therefore, the Taylor series for  ${\displaystyle f(x)}$  at  ${\displaystyle x_{0}=3}$  is
${\displaystyle \sum _{n=0}^{\infty }{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}{\frac {1}{e(n!)}}(x-3)^{n}.}$

(a)    ${\displaystyle f^{(n)}(x)={\bigg (}-{\frac {1}{3}}{\bigg )}^{n}e^{-{\frac {1}{3}}x},~f'(3)={\bigg (}-{\frac {1}{3}}{\bigg )}e^{-1}}$
(b)    ${\displaystyle \sum _{n=0}^{\infty }{\bigg (}-{\frac {1}{3}}{\bigg )}^{n}{\frac {1}{e(n!)}}(x-3)^{n}}$


text:
Home » » $$_{11} ^{11}Na + _0^1n → _{11}^{24}Na$$The reaction above is an example of

# $$_{11} ^{11}Na + _0^1n → _{11}^{24}Na$$The reaction above is an example of

### Question

$$_{11} ^{11}Na + _0^1n → _{11}^{24}Na$$

The reaction above is an example of

### Options

A) nuclear fission

B) nuclear fusion

C) artificial transmutation

D) beta decay

The correct answer is C.

## Dicussion (1)

• The conversion of an element into another element by bombarding it with a fundamental particle is known as Artificial Transmutation.


text:
Lemma 84.13.5. In Situation 84.3.3. If $K, K' \in D(\mathcal{C}_{total})$. Assume

1. $K$ is cartesian,

2. $\mathop{\mathrm{Hom}}\nolimits (K_ i[i], K'_ i) = 0$ for $i > 0$, and

3. $\mathop{\mathrm{Hom}}\nolimits (K_ i[i + 1], K'_ i) = 0$ for $i \geq 0$.

Then any map $K \to K'$ which induces the zero map $K_0 \to K'_0$ is zero.

Proof. Consider the objects $X_ n$ and the Postnikov system $Y_ n$ associated to $K$ in Lemma 84.13.4. As $K = \text{hocolim} Y_ n[n]$ the map $K \to K'$ induces a compatible family of morphisms $Y_ n[n] \to K'$. By (1) and Lemma 84.12.9 we have $X_ n = g_{n!}K_ n$. Since $Y_0 = X_0$ we find that $K_0 \to K'_0$ being zero implies $Y_0 \to K'$ is zero. Suppose we've shown that the map $Y_ n[n] \to K'$ is zero for some $n \geq 0$. From the distinguished triangle

$Y_ n[n] \to Y_{n + 1}[n + 1] \to X_{n + 1}[n + 1] \to Y_ n[n + 1]$

we get an exact sequence

$\mathop{\mathrm{Hom}}\nolimits (X_{n + 1}[n + 1], K') \to \mathop{\mathrm{Hom}}\nolimits (Y_{n + 1}[n + 1], K') \to \mathop{\mathrm{Hom}}\nolimits (Y_ n[n], K')$

As $X_{n + 1}[n + 1] = g_{n + 1!}K_{n + 1}[n + 1]$ the first group is equal to

$\mathop{\mathrm{Hom}}\nolimits (K_{n + 1}[n + 1], K'_{n + 1})$

which is zero by assumption (2). By induction we conclude all the maps $Y_ n[n] \to K'$ are zero. Consider the defining distinguished triangle

$\bigoplus Y_ n[n] \to \bigoplus Y_ n[n] \to K \to (\bigoplus Y_ n[n])[1]$

for the homotopy colimit. Arguing as above, we find that it suffices to show that

$\mathop{\mathrm{Hom}}\nolimits ((\bigoplus Y_ n[n])[1], K') = \prod \mathop{\mathrm{Hom}}\nolimits (Y_ n[n + 1], K')$

is zero for all $n \geq 0$. To see this, arguing as above, it suffices to show that

$\mathop{\mathrm{Hom}}\nolimits (K_ n[n + 1], K'_ n) = 0$

for all $n \geq 0$ which follows from condition (3). $\square$

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text:
# Calculate: (2)/(x^2-9)=(4)/(x^2-5x+6)-(2)/(x^2-4)

## Expression: $\frac{ 2 }{ {x}^{2}-9 }=\frac{ 4 }{ {x}^{2}-5x+6 }-\frac{ 2 }{ {x}^{2}-4 }$

Determine the defined range

$\begin{array} { l }\frac{ 2 }{ {x}^{2}-9 }=\frac{ 4 }{ {x}^{2}-5x+6 }-\frac{ 2 }{ {x}^{2}-4 },& \begin{array} { l }x≠-3,& x≠3,& x≠2,& x≠-2\end{array}\end{array}$

Move the expression to the left-hand side and change its sign

$\frac{ 2 }{ {x}^{2}-9 }-\frac{ 4 }{ {x}^{2}-5x+6 }+\frac{ 2 }{ {x}^{2}-4 }=0$

Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ {x}^{2}-5x+6 }+\frac{ 2 }{ {x}^{2}-4 }=0$

Write $-5x$ as a difference

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ {x}^{2}-2x-3x+6 }+\frac{ 2 }{ {x}^{2}-4 }=0$

Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ {x}^{2}-2x-3x+6 }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Factor out $x$ from the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ x \times \left( x-2 \right)-3x+6 }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Factor out $-3$ from the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ x \times \left( x-2 \right)-3\left( x-2 \right) }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Factor out $x-2$ from the expression

$\frac{ 2 }{ \left( x-3 \right) \times \left( x+3 \right) }-\frac{ 4 }{ \left( x-2 \right) \times \left( x-3 \right) }+\frac{ 2 }{ \left( x-2 \right) \times \left( x+2 \right) }=0$

Write all numerators above the least common denominator $\left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right)$

$\frac{ 2\left( x-2 \right) \times \left( x+2 \right)-4\left( x+3 \right) \times \left( x+2 \right)+2\left( x+3 \right) \times \left( x-3 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Use $\left( a-b \right)\left( a+b \right)={a}^{2}-{b}^{2}$ to simplify the product

$\frac{ 2\left( {x}^{2}-4 \right)-4\left( x+3 \right) \times \left( x+2 \right)+2\left( x+3 \right) \times \left( x-3 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Distribute $-4$ through the parentheses

$\frac{ 2\left( {x}^{2}-4 \right)+\left( -4x-12 \right) \times \left( x+2 \right)+2\left( x+3 \right) \times \left( x-3 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Use $\left( a-b \right)\left( a+b \right)={a}^{2}-{b}^{2}$ to simplify the product

$\frac{ 2\left( {x}^{2}-4 \right)+\left( -4x-12 \right) \times \left( x+2 \right)+2\left( {x}^{2}-9 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Distribute $2$ through the parentheses

$\frac{ 2{x}^{2}-8+\left( -4x-12 \right) \times \left( x+2 \right)+2\left( {x}^{2}-9 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Simplify the expression

$\frac{ 2{x}^{2}-8-4{x}^{2}-8x-12x-24+2\left( {x}^{2}-9 \right) }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Distribute $2$ through the parentheses

$\frac{ 2{x}^{2}-8-4{x}^{2}-8x-12x-24+2{x}^{2}-18 }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Collect like terms

$\frac{ 0-8-8x-12x-24-18 }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Calculate the sum of the negative numbers

$\frac{ 0-50-8x-12x }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Collect like terms

$\frac{ 0-50-20x }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

Removing $0$ doesn't change the value, so remove it from the expression

$\frac{ -50-20x }{ \left( x+3 \right) \times \left( x-3 \right) \times \left( x-2 \right) \times \left( x+2 \right) }=0$

When the quotient of expressions equals $0$, the numerator has to be $0$

$-50-20x=0$

Move the constant to the right-hand side and change its sign

$-20x=50$

Divide both sides of the equation by $-20$

$\begin{array} { l }x=-\frac{ 5 }{ 2 },& \begin{array} { l }x≠-3,& x≠3,& x≠2,& x≠-2\end{array}\end{array}$

Check if the solution is in the defined range

\begin{align*}&x=-\frac{ 5 }{ 2 } \\&\begin{array} { l }x=-2 \frac{ 1 }{ 2 },& x=-2.5\end{array}\end{align*}

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