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# Statistics and Data Science: A Modeling Approach

## 7.1 Specifying the Model

### Reviewing the Empty Model

In the previous chapters we introduced the idea of a statistical model as a function that generates a predicted score for each observation. We developed what we called the empty model, which we wrote like this (in GLM notation):

$Y_{i}=b_{0}+e_{i}$

In the case of thumb length, this model states that the DATA (each data point, represented as $$Y_{i}$$, which is each person’s measured thumb length), can be thought of as having two components: MODEL, represented as $$b_{0}$$, which is the mean thumb length for everyone (usually called the Grand Mean); plus ERROR, which is each person’s residual from the model, represented by $$e_{i}$$.

(Note: We use the term Grand Mean to refer to the mean of everyone in the sample in order to distinguish it clearly from other means, such as the mean for males or the mean for females.)

When we use the notation of the General Linear Model, we must define the meaning of each symbol in context. $$Y_{i}$$, for example, could mean lots of different things, depending on what our outcome variable is. But we will always use it to represent the outcome variable. Because we are talking about the outcome of Thumb length here, we can also write the empty model like this:

$Thumb_{i}=b_{0}+e_{i}$

It’s useful to illustrate the null model (or empty model) with our TinyFingers data set. TinyFingers, you will recall, contains six people’s thumb lengths, three males and three females, randomly selected from our complete Fingers data set.

TinyFingers
     Sex Thumb
1 female    56
2 female    60
3 female    61
4   male    63
5   male    64
6   male    68

We could represent the distribution of the six thumb lengths, broken down by Sex, using a faceted histogram (left panel, below). Or, we could use a scatterplot (gf_point(), as shown in the right panel), which might be clearer with such a small data set.

In the scatterplot below, the Grand Mean of the distribution, ignoring sex, is represented by the blue line. The Grand Mean is the model prediction for all observations under the empty model. Whether someone is male or female, their predicted thumb length ($$b_0$$) is 62 millimeters. Each person’s actual measured thumb length ($$Y_{i}$$) would be equal to 62 ($$b_0$$) plus their error or residual from the model prediction ($$e_i$$).

### Adding an Explanatory Variable to the Model

Now let’s add an explanatory variable, Sex, into the model. In the Sex model, which includes sex as an explanatory variable, we model the variation in thumb length with two numbers: the mean for males (65), and the mean for females (59). The model still generates a predicted thumb length for each person, but now the model generates a different prediction for a male than it does for a female.

Error is still measured the same way, as the deviation of each person’s measured thumb length from their predicted thumb length. But this time, the error is calculated from each person’s group mean (male or female) instead of from the Grand Mean (see figure above).

Whereas the empty model was a one-parameter model (we only had to estimate one parameter, the Grand Mean), the Sex model is a two-parameter model. One of the parameters is the mean for males, the other is the mean for females.

There are actually a few ways you could write this model; we will usually write it like this:

$Y_{i}=b_{0}+b_{1}X_{i}+e_{i}$

We can also write it like this:

$Thumb_{i}=b_{0}+b_{1}Sex_{i}+e_{i}$

In this equation, $$Y_{i}$$ is still the thumb length for person i (DATA), and $$e_{i}$$ is still the error for person i (ERROR), i.e., the deviation of the thumb length of person i from the thumb length predicted by the model. The part of the equation in between DATA and ERROR ($$b_{0}+b_{1}X_{i}$$) is the MODEL part of DATA = MODEL + ERROR, and it requires a bit of unpacking.

### Interpreting the GLM Notation for a Two-Group Model

In the two-group model, the model statement, $$b_{0}+b_{1}X_{i}$$, still represents each person’s predicted thumb length. But in this case, the model makes a different prediction for each person depending on their sex, $$X_i$$. This prediction ($$b_{0}+b_{1}X_{i}$$) plus error ($$e_i$$) will equal each person’s actual thumb length ($$Y_{i}$$).

Let’s unpack the GLM notation to see how it generates the two possible model predictions. In the two-group model, $$b_{0}$$ no longer represents the Grand Mean (62), but instead represents the mean for one of the groups. For example, $$b_{0}$$ might represent the mean thumb length for females, which, in the TinyFingers data frame, is 59.

If $$b_{0}$$ represents the mean for females, then $$b_1$$ represents what you would add on to the mean for females in order to get the mean (i.e., the model prediction) for males. As such, $$b_{1}$$ would represent the difference, or increment, in means between females and males. In the TinyFingers data, that increment is 6 mm (meaning that if you add 6 mm to the mean for females you will get the mean for males).

Here is how we would rewrite the model to include the estimated parameters:

$Y_{i}=59+6X_{i}+e_{i}$

You can also write Thumb and Sex instead of writing Y and X:

$Thumb_{i}=59+6Sex_{i}+e_{i}$

Now let’s see how these two parameter estimates (59 and 6) work inside the overall model ($$Y_{i}=b_{0}+b_{1}X_{i}+e_{i}$$).

$$X_{i}$$ is going to represent our explanatory variable, which is Sex. The subscript i reminds us that each person will have their own value on this variable (i.e., each person is either female or male). In order to make the model add up, we code $$X_{i}$$ as 0 for females and 1 for males. (This method of coding the explanatory variable is called “dummy coding”. There are other methods, but this is the one we use in this book.)

Let’s see what would happen if we used this model to predict a female’s thumb length. If someone is female, she would be coded 0 for $$X_{i}$$. $$b_{1}*X_{i}$$ would then be 6 times 0, which means the model would assign her a predicted thumb length of $$b_{0}$$ (which is 59), plus 0. This adds up to 59, the mean for females.

Let’s consider what would happen if we asked this model to predict a male’s thumb length. If someone is male, he would be coded 1 for $$X_{i}$$. A male’s thumb length would be 59 + (6*1), which means the model would assign him a predicted thumb length of 65, the mean for males.

Under this model, as expected, males would all end up with one predicted thumb length (65, the mean for males, or $$b_0+b_1*1$$), and females would all end up with a different predicted thumb length (59, the mean for females, or $$b_0+b_1*0$$).

Note that this does not mean that you have to code the variable in the data frame in any particular way; R takes care of that for you. As long as R knows that the explanatory variable is categorical (e.g., a Factor), you can code Sex with any numbers you choose (e.g., 1 and 2 or 0 and 100), or even with the words male and female. Internally, R will recode the variable as 0 and 1.

#### How Does R Know to Make Female = 0?

The answer to this question is: R doesn’t know; it’s just taking whatever group comes first and making it 0, then taking the next group and making it 1. The “female” group comes first in the variable Sex because it is alphabetically before “male”. It doesn’t really matter, though it will affect the interpretation of the parameter estimate.

Let’s say, just for fun, that you coded male as 0 and female as 1 in your data set. In this case, the best fitting value for $$b_0$$ would be the mean for males, which is 62 mm. Because males are coded as 0, $$X_i$$ would be equal to 0, so you wouldn’t add the second parameter estimate ($$b_1 * 0=0$$).

For females, $$X_i$$ would be equal to 1, which means $$b_1$$ would be added on to 62 to get the model prediction for females. Notice, however, that now the parameter estimate for $$b_1$$ would be negative. 62 plus (-6) would yield the mean for females, which is 56.

### A Final Note on Error

Note that we are broadening our definition of error from the way we thought of it for the empty model. For the empty model, error was the residual from the mean (i.e., the Grand Mean). Now we need to expand our thinking a bit, seeing error as the residual from the predicted score, which may not necessarily be the Grand Mean.

Of course, under both models (empty and two-group) the error is the residual from the predicted score under the model. It so happens that in the empty model, that predicted score is the Grand Mean. In the two-group model, the error is the residual from the male mean if you are male, or the female mean if you are female.

No matter how complex our models become, error is always defined as the residual, for each data point, from its predicted score under the model.


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# Quiescent Galaxies 1.5 Billion Years after the Big Bang and Their Progenitors

Abstract : We report two secure ( $z=3.775,4.012$ ) and one tentative (z ≈ 3.767) spectroscopic confirmations of massive and quiescent galaxies through K-band observations with Keck/MOSFIRE and Very Large Telescope/X-Shooter. The stellar continuum emission, absence of strong nebular emission lines, and lack of significant far-infrared detections confirm the passive nature of these objects, disfavoring the alternative solution of low-redshift dusty star-forming interlopers. We derive stellar masses of log(M/M) ∼ 11 and ongoing star formation rates placing these galaxies ≳1-2 dex below the main sequence at their redshifts. The adopted parameterization of the star formation history suggests that these sources experienced a strong ( $\langle \mathrm{SFR}\rangle \sim 1200\mbox{--}3500$ M yr-1) and short (∼50 Myr) burst of star formation, peaking ∼150-500 Myr before the time of observation, all properties reminiscent of the characteristics of submillimeter galaxies (SMGs) at z > 4. We investigate this connection by comparing the comoving number densities and the properties of these two populations. We find a fair agreement only with the deepest submillimeter surveys detecting not only the most extreme starbursts but also more normal galaxies. We support these findings by further exploring the Illustris TNG cosmological simulation, retrieving populations of both fully quenched massive galaxies at z ∼ 3-4 and SMGs at z ∼ 4-5, with number densities and properties in agreement with the observations at z ∼ 3 but in increasing tension at higher redshift. Nevertheless, as suggested by the observations, not all of the progenitors of quiescent galaxies at these redshifts shine as bright SMGs in their past, and, similarly, not all bright SMGs quench by z ∼ 3, both fractions depending on the threshold assumed to define the SMGs themselves.
Keywords :
Document type :
Journal articles

https://hal-insu.archives-ouvertes.fr/insu-03667136
Contributor : Nathalie Pothier Connect in order to contact the contributor
Submitted on : Friday, May 13, 2022 - 10:10:30 AM
Last modification on : Monday, May 16, 2022 - 8:20:25 AM

### Citation

Francesco Valentino, Masayuki Tanaka, Iary Davidzon, Sune Toft, Carlos Gómez-Guijarro, et al.. Quiescent Galaxies 1.5 Billion Years after the Big Bang and Their Progenitors. The Astrophysical Journal, 2020, 889, ⟨10.3847/1538-4357/ab64dc⟩. ⟨insu-03667136⟩

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# Is my solution to finding the Hicksian demand correct? Maximize $x_1^{1/2} + x_2^{1/2}$ subejct to the budget constraint

Maximize $$x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}$$ subejct to the budget constraint $$p_1x_1+p_2x_2=m$$

Setting up the Lagrange and finding the first-order conditions:

$$L(x_1, x_2, \lambda)=x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}+\lambda(p_1x_1+p_2x_2-m)$$

$$\frac{\partial L}{\partial x_1}=\frac{1}{2}x_1^\frac{-1}{2}+\lambda p_1=0$$

$$\frac{\partial L}{\partial x_2}=\frac{1}{2}x_2^\frac{-1}{2}+\lambda p_2=0$$

Equating these,

$$2p_2x_1^\frac{-1}{2}=2p_1x_2^\frac{-1}{2}$$

Solving for $$x_1$$ and $$x_2$$, we get

$$x_1=\frac{p_2^2x_2}{p_1^2}$$

$$x_2=\frac{p_1^2x_1}{p_2^2}$$

Substituting these into the budget constraint and solving to find the Hicksian demand ($$x^*(p,m)$$),

$$p_1x_1+p_2x_2=m$$

$$\frac{p_2^2x_2}{p_1}+p_2x_2=m$$

Solving for $$x_2$$,

$$x_2(\frac{p_2^2}{p_1}+p_2)=m$$

$$x_2^*=\frac{m}{p_2}.\frac{p_2}{p_1+p_2}=\frac{m}{p_1+p_2}$$

We do the same thing for $$x_1$$ and we get

$$x_1^*=\frac{m}{p_1+p_2}$$

We now have the Hickdian demands, substitute these into the objective function to get the indirect utility function and by duality, we know, $$V(p,E(p,u))=u$$,

$$V(p,m)=(\frac{m}{p_1+p_2})^\frac{1}{2}+(\frac{m}{p_1+p_2})^\frac{1}{2}=u$$

Solving this to find m,

$$\frac{2m^\frac{1}{2}}{p_1^\frac{1}{2}+p_2^\frac12}=u$$

$$m^\frac{1}{2}=\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})$$

$$m=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$$

Finally we also know by duality that $$E(p,V(p,m))=m$$, therefore

$$E=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$$

To find the Hicksian demand we use Shephards Lemma (take the partial derivative of the expenditure function)

$$\frac{\partial E}{\partial p_i}=2.\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2}).\frac{1}{2}.\frac{u}{2}.p_1^\frac{-1}{2}$$

Simplifying this

$$h_1^*=\frac{u^2}{4}.p_1^\frac{-1}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})=\frac{u^2}{4}(p_1^\frac{-1}{4}+p_2^\frac{1}{2})$$

Is this correct? (I've tried to make each of my steps really clear so this is easy to follow)

• I dont believe your last equation where $p_1$ disappears. – Jesper Hybel Dec 20 '20 at 20:54
• @JesperHybel You're right! I've changed it now, thanks – CorporateNationalism Dec 20 '20 at 21:11

Here is what I am thinking (and i could have made a mistake since CES algebra is a bit tedious)

First you have the MRS condition from Lagrange which boils down to

$$p_1 x_2^{\rho-1} = p_2 x_1^{\rho-1}$$

from here I always aim to reconstruct the utility function (because it is a constant equal to the max utility or in this instance you know it will satisfy constraint and be u). So to get the utility function to appear I do this

$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} = p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$

I then add $$p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}$$ to both sides

$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} + p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}= p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}+p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$

and isolate

$$p_1^{\frac{\rho}{\rho-1}}( x_2^{\rho} + x_1^{\rho})= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$

utility function appears on LHS and I impose constraint $$p_1^{\frac{\rho}{\rho-1}}u= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$

finding

$$\frac{p_1^{\frac{\rho}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )}u=x_1^{\rho}$$

which reduces to

$$\frac{p_1^{\frac{1}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )^\frac{1}{\rho}}u^\frac{1}{\rho}=x_1$$

Inserting value of $$\rho = 1/2$$ I get

$$\frac{p_1^{-2}}{(p_1^{-1}+p_2^{-1} )^2}u^2=x_1$$


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# Question 14 (2 points) Refer to the triangle below:Gl,5)E6-1,31 1a3,W MntoFind the length of Uhe median (ramivertexleOa 15 0b 5Oe

###### Question:

Question 14 (2 points) Refer to the triangle below: Gl,5) E6-1,31 1a3,W Mnto Find the length of Uhe median (ramivertexle Oa 15 0b 5 Oe 3 d) N Question 15 (2 points)" What is the length ol theline segment thatihas endpoints at (4 -?) and (-7)? Oa m 0b) 7 : 13

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# How do you integrate x^3/(x^2+2x+1) using partial fractions?

Apr 27, 2017

$\int {x}^{3} / \left({x}^{2} + 2 x + 1\right) \mathrm{dx} = {x}^{2} - 2 x + 3 \ln | x + 1 | + \frac{1}{x + 1}$

#### Explanation:

${x}^{3} / \left({x}^{2} + 2 x + 1\right) = \frac{x \left({x}^{2} + 2 x + 1\right) - 2 \left({x}^{2} + 2 x + 1\right) + 3 x + 2}{{x}^{2} + 2 x + 1}$

= $x - 2 + \frac{3 x + 2}{{x}^{2} + 2 x + 1}$. Let

$\frac{3 x + 2}{{x}^{2} + 2 x + 1} = \frac{3 x + 2}{x + 1} ^ 2 \equiv \frac{A}{x + 1} + \frac{B}{x + 1} ^ 2$

i.e. $3 x + 2 = A \left(x + 1\right) + B$

Now if $x = 0$, we have $A + B = 2$ and if $x = - 1$, we have $B = - 1$ and hence $A = 3$

Therefore $\frac{3 x + 2}{{x}^{2} + 2 x + 1} = \frac{3 x + 2}{x + 1} ^ 2 \equiv \frac{3}{x + 1} - \frac{1}{x + 1} ^ 2$

and ${x}^{3} / \left({x}^{2} + 2 x + 1\right) = x - 2 + \frac{3}{x + 1} - \frac{1}{x + 1} ^ 2$

hence $\int {x}^{3} / \left({x}^{2} + 2 x + 1\right) \mathrm{dx} = \int \left(x - 2 + \frac{3}{x + 1} - \frac{1}{x + 1} ^ 2\right) \mathrm{dx}$

= ${x}^{2} - 2 x + 3 \ln | x + 1 | + \frac{1}{x + 1}$

Apr 27, 2017

${\left(x + 1\right)}^{2} / 2 - 3 \left(x + 1\right) + 3 \ln | x + 1 | + \frac{1}{x + 1} + C .$

#### Explanation:

Though the Problem is requred to be solved using Partial Fractions,

as a Second Method, we show that the same can be solved

easily without its use.

Let, $I = \int {x}^{3} / \left({x}^{2} + 2 x + 1\right) \mathrm{dx} = \int {x}^{3} / {\left(x + 1\right)}^{2} \mathrm{dx} .$

Sbst.ing, $x + 1 = t , \text{ so that, } \mathrm{dx} = \mathrm{dt} , \mathmr{and} , x = t - 1.$

$\therefore I = \int {\left(t - 1\right)}^{3} / {t}^{2} \mathrm{dt} ,$

$= \int \frac{{t}^{3} - 1 - 3 t \left(t - 1\right)}{t} ^ 2 \mathrm{dt} ,$

$= \int \frac{{t}^{3} - 3 {t}^{2} + 3 t - 1}{t} ^ 2 \mathrm{dt} ,$

$= \int \left\{{t}^{3} / {t}^{2} - 3 {t}^{2} / {t}^{2} + 3 \frac{t}{t} ^ 2 - \frac{1}{t} ^ 2\right\} \mathrm{dt} ,$

$= \int \left\{t - 3 + \frac{3}{t} - \frac{1}{t} ^ 2\right\} \mathrm{dt} ,$

$= {t}^{2} / 2 - 3 t + 3 \ln | t | + \frac{1}{t} ,$

$\Rightarrow I = {\left(x + 1\right)}^{2} / 2 - 3 \left(x + 1\right) + 3 \ln | x + 1 | + \frac{1}{x + 1} + C .$

Enjoy Maths.!

N.B.: $I = {x}^{2} / 2 + x + \frac{1}{2} - 3 x - 3 + 3 \ln | x + 1 | + \frac{1}{x + 1} + C ,$

$= {x}^{2} / 2 - 2 x - \frac{5}{2} + 3 \ln | x + 1 | + \frac{1}{x + 1} + C ,$

$= {x}^{2} / 2 - 2 x + 3 \ln | x + 1 | + \frac{1}{x + 1} + c , w h e r e , c = C - \frac{5}{2} ,$ as

Respected Shwetank Mauria has derived!


text:
# probability (selection from a collection of 100)

Suppose a microprocessor is chosen at random from a collection of 100. Assume 20 are Intels and 80 are AMDs. Also assume that 10 of the Intels are 2.0 GHz. What is the probability that the selected microprocessor is 2.0 GHz given that it is an Intel?

-

Alternatively, you can use the formula $$P(A|B) = \frac{P(A \cap B)}{P(B)}.$$ Here, the event $B$ is that the microprocessor is an Intel, and the event $A$ is that it is a $2.0$ Ghz processor. To find the probability of $A$ given $B$, you must determine the probability that a randomly chosen processor is both Intel and $2.0$ Ghz (what proportion of the processors posess both these properties?) and divide that by the probability that the processor is Intel.


text:
Statistics

# Formula Review

StatisticsFormula Review

### 5.1Continuous Probability Functions

Probability density function (pdf) f(x):

• f(x) ≥ 0
• The total area under the curve f(x) is one.

Cumulative distribution function (cdf): P(Xx)

### 5.2The Uniform Distribution

X = a real number between a and b (in some instances, X can take on the values a and b). a = smallest X, b = largest X

X ~ U(a, b)

The mean is $μ= a+b 2 . μ= a+b 2 .$

The standard deviation is

Probability density function: $f(x)= 1 b−a f(x)= 1 b−a$ for $a≤X≤b a≤X≤b$

Area to the left of x: P(X < x) = (xa)$( 1 b−a ) ( 1 b−a )$

Area to the right of x: P(X > x) = (bx)$( 1 b−a ) ( 1 b−a )$

Area between c and d: P(c < x < d) = (base)(height) = (dc)$( 1 b−a ) ( 1 b−a )$

Uniform: X ~ U(a, b) where a < x < b

• pdf: $f( x )= 1 b−a f( x )= 1 b−a$ for a ≤ x ≤ b
• cdf: P(Xx) = $x−a b−a x−a b−a$
• mean µ = $a+b 2 a+b 2$
• standard deviation σ $= (b−a) 2 12 = (b−a) 2 12$
• P(c < X < d) = (dc)$( 1 b–a ) ( 1 b–a )$

### 5.3The Exponential Distribution (Optional)

Exponential: X ~ Exp(m) where m = the decay parameter

• pdf: f(x) = me(–mx) where x ≥ 0 and m > 0
• cdf: P(Xx) = 1 – e(–mx)
• mean µ = $1 m 1 m$
• standard deviation σ = µ
• percentile k: k = $ln(1−AreaToTheLeftOfk) (−m) ln(1−AreaToTheLeftOfk) (−m)$
• P(X > x) = e(–mx)
• P(a < X < b) = e(–ma)e(–mb)
• Memoryless property: P(X > x + k|X > x) = P (X > k)
• Poisson probability: with mean λ
• k! = k*(k−1)*(k−2)*(k−3)*…3*2*1


text:
14.2.1 Ubuntu

Four kinds of sound recording network samples are available as shown in Table 14.2. The left column indicates network files and the right column indicates processing contents.

Table 14.2: Sound recording network list.
 Network file name Processing content demoALSA.n ALSA sound recording demoWS_8ch.n Sound recording for eight channels with RASP demoRASP24_8ch.n Sound recording for eight channels with RASP24

For ALSA devices, you can use a shell script demoALSA.sh for recording various number of frames, number of channels, and device names. For example, if you want to record 200 frames from an 2-ch audio device whose name is plughw:1,0, you can use this script as shown below:

> ./demoALSA.sh 200 2 plughw:1,0
UINodeRepository::Scan()
Scanning def /usr/lib/flowdesigner/toolbox
done!
Building network  :MAIN

<Bool 1 >

Figure 14.2: Execution example of demoALSA.sh.

Note that the difference between these networks is only the parameters of AudioStreamFromMic module. The DEVICE parameter (plughw:1,0, 127.0.0.1, etc.) changes according to an environment. Therefore, an user has to set correct value.

A sample network consists of two sub-networks (MAIN, MAIN_LOOP). MAIN (subnet) and MAIN_LOOP (iterator) have one and six modules, separately. Figure 14.3 and 14.4 shows MAIN (subnet) and MAIN_LOOP (iterator). It is simple network configuration in which the audio waveforms collected in the AudioStreamFromMic module are selected in ChannelSelector , and written in files in SaveWavePCM . The SaveWavePCM module connected to OUTPUT1 saves a file with eight channel signal. The other SaveWavePCM module saves eight files with each channel signal.

Iterate is a module that indicates that the execution is repeated for the number of times set by the user. Designate the number of frames to be collected and adjust sound recording time.

There are five properties for the MAIN_LOOP node in MAIN (subnet). Table 14.3 shows a list of them. SAMPLING_RATE and GET_FRAMES are important. Each parameter value is set according to the values in the table. Set value of GET_FRAMES is int :ARG1 here. This means the monobasic argument of demoALSA_2ch.n casted to the integral type and substituted. The number of sound recording frames is given to this argument in demoALSA_2ch.n in demoALSA_2ch.sh. Designate sound recording time length as the number of frames acquired and the actual sound recording time length is expressed in sec as follows.

 $\displaystyle$ $\displaystyle$ $\displaystyle ( LENGTH + (GET\_ FRAMES - 1) * ADVANCE ) / SAMPLING\_ RATE$ (1)
Table 14.3: Parameter list of MAIN_LOOP
 Parameter name Type Set value Unit Description ADVANCE 160 [pt] Shift Length LENGTH 512 [pt] FFT length SAMPLING_RATE 16000 [Hz] Sampling frequency GET_FRAMES int :ARG1 Number of frames for sound recording DOWHILE Blank

14.2.1.1 Stereo recording with ALSA-based devices

Execute demoALSA_2ch.sh in the Record directory. After the execution, one stereo file rec_all_0.wav and two monaural files rec_each_0.wav, rec_each_1.wav are generated. When replaying these files, the recorded content can be confirmed.

Some computer rarely do not accept stereo recording so an error may occur. In the case that the sound recording cannot be performed well even though the computer is able to accept stereo recording, confirm the check items for monaural sound recording. Five nodes are included in this sample. There is one node in MAIN (subnet) and are four nodes in MAIN_LOOP (iterator). It is simple network configuration in which the audio waveforms collected in the AudioStreamFromMic module are written in a file in SaveWavePCM . Data formats between nodes are uniformed through MatrixToMap . Iterate is a node that indicates that the execution is repeated for the number of times set by the user. Designate the number of frames to be collected and adjust sound recording time. What is different from the monaural sound recording network is the only CHANNEL_COUNT property of the AudioStreamFromMic node. It is 1 for monaural recording and 2 for stereo recording.

Recording cannot be performed well, perform the following checks.

1. Confirm if the sound is replayed or recorded with software other than HARK. If the sound cannot be replayed or recorded, the OS and driver, and audio interface may not be properly set or connected so check them.

2. The microphones are properly connected. Confirm if the plugs are not unplugged or loosened and connect them properly.

3. Check if the microphone terminals of the computer accept plug in power. If they do not accept plug in power, it is necessary to supply a power to the microphones. For battery type microphones, set batteries and switch on. For battery box type microphones, connect battery boxes and switch on.

4. When more than two audio interfaces are connected, remove audio interfaces after the second ones before recording the sound. If recording cannot be performed with one audio interface, change the property of demo.n. For the DEVICE property of the AudioStreamFromMicmodule, try setting of 0,0 plughw: 0,1 plughw: 0,2.

Parameters of AudioStreamFromMic are shown in Table 14.4. In this node, designate a sound recording device. Designate a sound recording device in this node. In this demo, the number of sound recording channels is 2ch so that a lot of users can check the operation, and ALSA is designated for DEVICETYPE and plughw: 0,0 is designated for DEVICE.

Table 14.4: Parameter list of AudioStreamFromMic
 Parameter name type Set value Unit Description LENGTH LENGTH [pt] FFT length ADVANCE ADVANCE [pt] Shift length CHANNEL_COUNT 2 [ch] Number of sound recording channel SAMPLING_RATE 16000 [Hz] Sampling frequency DEVICETYPE ALSA Device type DEVICE plughw:0,0 Device name

14.2.1.2 8ch sound recording with radio RASP

Execute demoWS_8ch.sh in a Record directory, after setting several items. Although 127.0.0.1 is designated as an IP address of the radio RASP in demoWS_8ch.n, it is necessary to change this to an appropriate IP address. If configuration of FPAA for the radio RASP is not completed yet, complete it here. See the document of radio RASP for the method to execute ws_fpaa_config. After the execution, one file rec_all_0.wav and eight monaural files rec_each_0.wav, rec_each_1.wav, ..., rec_each_7.wav are generated. When replaying these files, the recorded content can be confirmed.

When recording cannot be performed well, perform the following checks.

1. Check if the microphones are properly connected. Check if the plugs are not unplugged or loosened and connect them properly.

2. The network connection is established with the IP address of RASP. Check the network connection with ping.

3. A correct IP address of RASP is set to AudioStreamFromMic .

4. Initialization of FPAA of RASP is completed.

5. Confirm if the sound is replayed or recorded with software other than HARK. If the sound cannot be replayed or recorded, the OS and driver, and audio interface may not be properly set or connected so check them.

Parameters of AudioStreamFromMic are shown in Table 14.5. Designate a sound recording device in this node. Designate a multi-channel sound recording device. Set 16ch to the number of sound recording channel, designate WS, which indicates radio RASP, in DEVICETYPE and an IP address in DEVICE. In order to execute the sample, it is necessary to change the address to that of the actual radio RASP.

Note that when WS is used, CHANNEL_COUNT must be 16. Therefore, when a user want to record eight channel sound with WS, use ChannelSelector to select eight channels from 16 channels. In this sample, from 0 to 7 channels are selected.

Table 14.5: Parameter list of AudioStreamFromMic
 Parameter Name Type set value Unit Description LENGTH LENGTH [pt] FFT length ADVANCE ADVANCE [pt] Shift length CHANNEL_COUNT 16 [ch] Number of channels for sound recording SAMPLING_RATE 16000 [Hz] Sampling frequency DEVICETYPE WS Device type DEVICE 127.0.0.1 Device name

14.2.1.3 8ch sound recording with RASP24

Execute demoRASP24_8ch.n in the Record directory. After the execution, one file rec_all_0.wav and eight monaural files rec_each_0.wav, rec_each_1.wav, ..., rec_each_7.wav are generated. When replaying these files, the recorded content can be confirmed.

When recording cannot be performed well, perform the following checks.

1. Check if the microphones are properly connected. Check if the plugs are not unplugged or loosened and connect them properly.

2. The network connection is established with the IP address of RASP24. Check the network connection with ping.

3. A correct IP address of RASP24 is set to AudioStreamFromMic .

4. Confirm if the sound is replayed or recorded with software other than HARK. If the sound cannot be replayed or recorded, the OS and driver, and audio interface may not be properly set or connected so check them.

Parameters of AudioStreamFromMic are shown in Table 14.6. RASP24 can record sound with 16 bit or 32 bit. When perform 16 bit and 32 bit recording, use RASP24-16 and RASP24-32, separately. Designate a sound recording device in this node. Designate a multi-channel sound recording device. Set 16ch to the number of sound recording channel and IP address in DEVICE. In order to execute the sample, it is necessary to change the address to that of the actual radio RASP24.

Note that a parameter GAIN is added when RASP24 is used as shown in Figure 14.5.

Table 14.6: Parameter list of AudioStreamFromMic
 Parameter name type Set value Unit Description LENGTH LENGTH [pt] FFT length ADVANCE ADVANCE [pt] Shift length CHANNEL_COUNT 16 [ch] Number of sound recording channel SAMPLING_RATE 16000 [Hz] Sampling frequency DEVICETYPE RASP24-16 Device type GAIN 0dB GAIN DEVICE 127.0.0.1 Device name

When a user set GAIN parameter, be careful about clipping.


text:
## Efficient approximation of functions of some large matrices by partial fraction expansions. (arXiv:1709.06351v2 [math.NA] UPDATED)

Some important applicative problems require the evaluation of functions $\Psi$ of large and sparse and/or \emph{localized} matrices $A$. Popular and interesting techniques for computing $\Psi(A)$ and $\Psi(A)\mathbf{v}$, where $\mathbf{v}$ is a vector, are based on partial fraction expansions. However, some of these techniques require solving several linear systems whose matrices differ from $A$ by a complex multiple of the identity matrix $I$ for computing $\Psi(A)\mathbf{v}$ or require inverting sequences of matrices with the same characteristics for computing $\Psi(A)$. Here we study the use and the convergence of a recent technique for generating sequences of incomplete factorizations of matrices in order to face with both these issues. The solution of the sequences of linear systems and approximate matrix inversions above can be computed efficiently provided that $A^{-1}$ shows certain decay properties. These strategies have good parallel potentialities. Our claims are confirmed b查看全文

## Solidot 文章翻译

 你的名字 留空匿名提交 你的Email或网站 用户可以联系你 标题 简单描述 内容 Some important applicative problems require the evaluation of functions $\Psi$ of large and sparse and/or \emph{localized} matrices $A$. Popular and interesting techniques for computing $\Psi(A)$ and $\Psi(A)\mathbf{v}$, where $\mathbf{v}$ is a vector, are based on partial fraction expansions. However, some of these techniques require solving several linear systems whose matrices differ from $A$ by a complex multiple of the identity matrix $I$ for computing $\Psi(A)\mathbf{v}$ or require inverting sequences of matrices with the same characteristics for computing $\Psi(A)$. Here we study the use and the convergence of a recent technique for generating sequences of incomplete factorizations of matrices in order to face with both these issues. The solution of the sequences of linear systems and approximate matrix inversions above can be computed efficiently provided that $A^{-1}$ shows certain decay properties. These strategies have good parallel potentialities. Our claims are confirmed b


text:
# Plotting an ellipse given axial ratio and orientation

I have already looked at this answer but couldn't figure out how to use it in my case, particularly due to needing to include the orientation. I currently have a value for my axial ratio and a value for my orientation, and that is all. The size is irrelevant due to prior normalization. I feel like I am just missing something in the other question, but don't know what it is, this kind of math isn't my forte. Thanks in advance.

• "a value for my orientation" - by this, you mean the angle of the major axis from the horizontal? – J. M. will be back soon Mar 31 '16 at 6:24
• Yes, that is what I have. – Adrian Smith Mar 31 '16 at 20:13

a = 5; b = 3;


text:
# Publisher Object¶

Publisher Object Properties
Value Type Description
id string

An exchange specific identifier comprised of the the Supplier exchange name, and the publisher ID appended following an underscore. For example, if a Supplier such as Rubicon sends the publisher ID "25", then the value of the field becomes rubicon_25.

name* string Publisher name, for example "AAP"
cat* array of string Array of IAB content categories for the publisher. ["IAB1", "IAB2-3"]
domain* string The highest level domain of the publisher, e.g. example.com

Note

Fields marked with an asterisk (*) are optional and may not be sent in each request.

## Publisher Object Example¶

{
"publisher":{
"id":"SSPid_12345",
"name":"Publisher A"
}
}


text:
# String tension, is this the correct formula?

1. Feb 26, 2004

### ant0

I read in a book that tension = 1/(pa'), where a' = alpha prime.

Is this the correct formula for string tension and if so what is p?

2. Feb 26, 2004

### lethe

in my book, it is

$$T=\frac{1}{2\pi\alpha'}$$

$\pi$ is the ratio of the circumference of a circle to its diameter, approximately 3.14. and if your source differs by a factor of 2 from mine, i wouldn't worry about it.

3. Feb 26, 2004

### ant0

ah, p must be pi then, would that make sense for a closed string then?


text:
Problem with dimensionality

1. Jun 20, 2004

alexepascual

In the following equation:
$$\left\langle {\phi } \mathrel{\left | {\vphantom {\phi \Psi }} \right. \kern-\nulldelimiterspace} {\Psi } \right\rangle = \int {\phi ^ * } \left( x \right)\psi \left( x \right)dx$$
my understanding would have been that the bracket represents a probability amplitude. But the dx on the integral gives it a dimension of length. OK, the square of the absolute value of the braket is not a probability but a probability density. Wouldn't then the units be probability over unit of length?
I am a little confused. Any help will be appreciated.
By the way, this equation is from Feynman Lectures, Vol III, 16-6

Last edited: Jun 20, 2004
2. Jun 21, 2004

turin

Don't forget about normalization. To interpret it as a probability, you are already assuming that it is normalized. In other words, implicit in the interpretation is that an integral over all of space has been done and divided by.

3. Jun 21, 2004

alexepascual

If both state vectors are normalized, so would the bracket, as well as the integral on the right. Wouldn't it?. Well, I mean they would reflect the correct value which would have an absolute value typically less than one. On the other hand, doesn't normalization only affect the absolute value of the vector and not any units?. As far as I know, the state vector as well as a bracket are dimensionless. Now, it appears that the integral would give you a dimensionless quantity only if you drop the units of length from dx. If you do that, everything would be fine and the right side of the equation would be dimensionally consistent with the left.
Now, in a different situation of integration, such as W= integral F*dx , or W= integral P*dv, you do consider the dimension of the diferential. As a matter of fact, in these cases you have to consider it in order for the equation to be dimensionally consistent. So, I would assume that in physics in general, when you do an integration, if you are integrating over some variable that has units, you keep the units and use them in your integration.
In the quantum mechanics case I posted, the only apparent way to keep it dimensionally correct would be to add a factor of (1/unit of length) in front of the integral. I wonder if there is some convention where this factor is ommited even it would be needed to make the equation dimensionally consistent.

4. Jun 22, 2004

Your mistake is supposing that the wave functions dont have dimensions. They do. The dimensionality is correct. Take a square well problem. The wave functions (sin(nx) and cos(nx)) must have normalization inversely proportional to the square root of a lenght constant. This way the product of two wave functions has dimensionality of L to the minus one, keeping the dimensionality of the scalar product correct. If you think about it you will see that normalization constants have to have dimensionality. In momentum representation it's 1 divided by the square root of momentum units and in space representation one divided by the square root of space units.keep it up.

ciao!

Last edited: Jun 22, 2004
5. Jun 22, 2004

somy

can you explain more.I didn't understand your explanation.

6. Jun 22, 2004

hey somy

WHen constructing the solutions to the infinite square well, we get from the time independent Schrödinger equation + boundary value considerations that the wave functions are sin(n*pie*x/L), where L is the size of the well. THe probability of finding the particle inside the well is 100%, the integral of the wave function squared must be equal to one. THus the earlier given solution is not complete, we must modify it with a factor wich is the inverse square root of the value of the above integral. THus, the scalar product of the two wave funcions has the dimensions of nothing, as expected, because both wave functions are multiplied by constants with dimensionality of length to the minus one half.

7. Jun 22, 2004

alexepascual

Thanks Tavi,
Your explanation is very clear and it solves my problem.
Now I can continue reading chapters 16 and 20 from Feynman's Lectures.

8. Jun 24, 2004

alexepascual

Tavi,
I haven't had time to do much reading, but I was thinking about the following. The expression <x|psi> is equivalent to psi(x). Now, psi(x) has units, according to your post. This would make me think that <x|psi> also has units. But in that case where do the units come from?. I was not aware of bras or kets having units. I guess the ket psi by itself could not come with any units as it is a state vector that can be represented in different bases. Could the units be in the <x| bra? I tought this bra just had a label, which is the value of x, and a bare complex number without units.
Maybe you can clarify this for me. I'll appreciate it.

9. Jun 25, 2004

Staff Emeritus
Why do you say $$\psi(x)$$ has units?
Here's what QM says about $$\psi$$:

- It's a square summable function forming a point in a complex Hilbert space
- It satisfies the Schroedinger equation
- $$\psi\bar{\psi}$$ maps into the configuration space of an observable and gives the probability for the observable to be in the state indicated by the point on the configuration space.

None of this allows the wave function to assume the attributes of a real measurement in spacetime. It requires the action of a self-adjoint operator to create a number that can do that.

10. Jun 25, 2004

alexepascual

Thanks for your response. If you look to the previous posts you'll see what my concern is. But in short: when integrating the wave function over all space, if properly normalized, we are supposed to get just the number one. But the integral has the differential element dx which has units of volume, ( length in one-dimensional case). So Tavi suggested that the wave function has units of lenght to the minus 1/2 (one dimensional case again) . I can accept that. But I had some doubts as to how this concept translates when using Dirac notation. I never suggested you could get a measurable out of a wave function without using an operator.

11. Jun 25, 2004

Suppose [psi> represents the state an electron in a given potential. We know that <x[psi> times it's complex conjugate represents the probability density of the electron being at x, not the probability of finding it there. The probability of finding it there is
P(x)dx=<x[psi>·<psi[x>dx=<x[psi>^2dx.
Therefore, psi(x) must have units because a probability, by definition can't have any.
You know that a state can be expressed in terms of the eigenvectors of a hermitian operator. The position operator,x, has eigenvectors [x>, so psi(x) is the coefficient of the expansion of [psi> in terms of the proper base of x. THat's why <x[psi> squared is the probability density of measuring the position of the electron and finding it at x.

12. Jun 25, 2004

alexepascual

Tavi,
I am not questioning that $$\psi(x)$$ has units. I accepted that.
My concern is, in the bracket <x|psi> , where are the units hiding?
So, see, my problem comes when expressing things in dirac notation, not when expressed as a wave function.
Perhaps I am looking at this the wrong way. I usually think of the bracket as an inner product. Maybe in this case I should look at it as just the components in x-space of |psi>, together with its units.

Last edited: Jun 25, 2004
13. Jun 25, 2004

Eye_in_the_Sky

Solution to Problem with dimensionality

Any ket |f> belonging to the Hilbert space (regardless of whether or not |f> is normalized) is dimensionless. A "generalized" ket, such as |x>, however, is not a member the Hilbert space ... and it has dimension. If x has dimension L, then |x> has dimension L^(-1/2), and likewise for the corresponding "generalized" bra.

--------------------------

You can convince yourself of this by writing

<x|x'> = delta(x-x')

and observing that

Integral { delta(x-x') dx }

is dimensionless, implying that delta(x-x') has dimension 1/L.

--------------------------

In this way, a function f(x) = <x|f> gets the dimension it needs.

Similarly, <x|p>, where |p> is a "generalized" eigenket of the momentum operator, gets the dimension of reciprocal square-root of action.

14. Jun 25, 2004

alexepascual

Thanks a lot Eye_in_the_Sky. Your explanation is easy to understand and straight-to-the-point. I think it completely answers my question.
Thanks again,
Alex

15. Jun 27, 2004

turin

I guess I'm not appreciated at all around here.

16. Jun 28, 2004

alexepascual

Turin,
Why do you say that?
You posted the first response to this post and I answered the same day. Your post was apparently correct but it did not totally answer my question. When Tavi replied, that solved part of my puzzle, but some questions remained, which were addressed by Eye_in_the_Sky. The response by SelfAdjoint was also correct but was not answering my question.
Some times it is hard to understand what a person asking for help on a topic is really looking for. Part of this may be because the original poster has not communicated the problem in enough detail. Part might be that the person responding is looking at the subject from a different angle, which is satisfactory for him but not for the original poster.
You should not take it personally. I appreciate all responses to my posts and I remember you very well from an interesting discussion we had in another thread.
I view this forum as a collaborative effort where we are all learning from each other. In my physics classes, where instructors graded on a curve, I found that there was more competition than collaboration. Here I have enjoyed a different atmosphere, where collaboration prevails over competition.
I also think that due to the complexity of the subject and its abstract character, it is not always easy to communicate effectively, partly also due to the fact that different people have different angles of approach which others may not find satisfactory.
I, personally like to think in terms of pictures, and I like to always connect the abstract features of the theory to concrete examples. There are people who are perfectly comfortable exploring all the mathematical apparatus without making contact with "reality". That is their preference and I respect it.
As an added benefit of this forum, we get exposed to different views which we may not find useful today but may gradually start having more appeal in the future. All the disagreement and difficulties in communication may not be that counterproductive after all.
Well, Turin, you are always welcome and certainly appreciated.
-Alex Pascual-

17. Jun 28, 2004

Alex, this has nothing to do with the thread, but I find it remarcable you can think pictorialy about QM. I've tried that aproach because it's always been easier for me that way in other subjects, but failed completely. I cannot attack QM prloblems without use of mathematical weapons, I envy you. In fact, come to think of it, I guess I wont be able to adress any problem in modern physics pictorialy. I guess I'm a bit depressed 'cause I can't for the life of me grasp the spirit of Misner/Thorne/Wheelers gravitation. How can they say 2-forms are like honycombs!!???Any GR expert fell up to explain?

18. Jun 28, 2004

robphy

I posted these links on this other thread regarding visualizing field lines
http://www.ee.byu.edu/ee/forms/
http://www.lgep.supelec.fr/mse/perso/ab/IEEEJapan2.pdf [Broken] [see p. 10]

In addition, you may find this useful
http://clifford-algebras.org/Beijing2000/page127-154.pdf [Broken]
http://physics.syr.edu/courses/vrml/electromagnetism/references.html [Broken]
I believe the field of 2-forms visualized as a honeycomb is due to Jan A Schouten.

(This is now off-topic for this thread.)

Last edited by a moderator: May 1, 2017
19. Jun 28, 2004

alexepascual

Tavi,
The fact that I have as a preference visualizing things, does not mean that I am being successful in doing so in quantum mechanics. But I try. Memorizing rules to manipulate symbols on a piece of paper does not give me the feeling that I am really learning about nature. On the other hand, when I have a good explanation of the math in terms of "things" in the real world, that helps. With respect to pictures, even seing the state vector as an arrow, and a basis as a set of 3-D coordinate axes gives me a more intuitive description. When you do the math, I like to know what is actually happening to the state. Is anything happening to it or are we simply changin our way to describe it?. I am not really asking here, I am saying that every time that's the kind of question I ask.
In this thread, though, my question was purelly mathematical, which means that it did not concern what is happening to the state but how we use the symbols to describe it.
In this respect I have my suspicion that the explanation given by you and Eye-in-the-Sky, (which I am sure represents common knowledge in traditional quantum mechanics) may not be the only one, or the only possible way to do things. I think this may be a matter of convention, an I am not knowlegeable enough about the subject to challenge the conventional approach, but I could venture an opinion to see what you guys think. I'll do that my next post.
With respect to the Gravitation book, I haven't read it. Probably I will within the next two years, but I think it is kind of heavy reading.

20. Jun 28, 2004

alexepascual

Robphy,
I read a little about Geometric Algebra a few weeks ago and it sounded interesting. It wonder how much it is used in quantum mechanics and what are it's advantages. It appears it may be a better way to conceptualize some operations such as products of vectors, but I am not sure about it yet. I have that in my to-do list of things to read. There is so much to learn and so little time!

21. Jun 28, 2004

alexepascual

Tavi and Eye_in_the_Sky.
Your posts have completely answered my original question. But, as I say earlier, I have my suspicion that this mathematical scheme may represent a convention and not a unique way of doing the math.
To be more explicit: In the scheme described by you, the kets corresponding to eigenvectors of a continous variable's operator carry with them units that are the units of the variable to the -1/2. I understand that these kets are special as they represent "limits" or things not properly within the Hilbert space. Still, handling the units like this seems a bit artificial.
It may seem too audacious of me to venture opinions about these things while I am still learning and have not read everything about it. But anyway I'll say it, as you, having more knowledge about it may even mention some thinking along the same lines by other people.
I am somewhat aware that Von Newman had objections about the Dirac delta function, but I don't know the details behind his argument.
On the other hand, it appears kind of obvious that many of the infinities we get in quantum mechanics are the result of considering certain variables as continuous when they are actually discrete.
I don't know much about quantum loop gravity and superstring theory, but I think I remember reading that they address this problem.
Going back to QM. if we were to consider all variables as discrete, then the scheme you mentioned would have to be changed as it implies treating differently continuous varables from discrete ones. For instance, the spin up and spin down unit vectors, as far as I know don't carry any units, while the position eigenvectors do.
It seems to me, and here is where I may be venturing too far with my little knowledge, that a more natural way to do the math would be (taking the position eigenvectors as example) to keep them dimensionless and to add a denominator with the square root of the space units to the dx in the integration. This way, we would be doing the integration over a dimensionless domain, which would be consistent with the way we do summation for discrete variables. This scheme would also make it easier to transition to a treatment where all variables are discrete.
It seems to me that in quantum mechanics we are dealing with two big domains or languages: one being the Hilbert space, in which all the classical variables dissapear and are transformed into mere lables for unit vectors, and the other domain is that of classical variables which we multiply, divide etc. and keep track of the units by dimensional analysis. The operators corresponding to measurables would allow us to "extract" the classical variable with it's units from the state vector living in Hilbert space.
If we look at it this way, when we are in Hilbert space, the numbers we manipulate should not have any units, they should only be related to the square root of a probability and a phase factor. And perhaps it should not matter wether the Hilbert space is rigged or not. Then when we "translate" to our usual "domain" or "language" of classical variables, then the units are reconstructed from the lables.

22. Jun 28, 2004

Eye_in_the_Sky

From Another Angle - Seeing it Deeper

[Note: This post is intended to provide an alternative argument to that given in the post entitled Solution to Problem with dimensionality in which the Dirac delta function was used.]

Consider an observable for which the associated operator is given by

A = Sigma_n { a(n) P(n) } ,

where a(n) are the eigenvalues and P(n) are the corresponding eigenprojections. If the observable has dimensions, then the eigenvalues
a(n) will have those very same dimensions, while, on the other hand, the eigenprojections P(n) are (always) dimensionless.

For a measurement of position, the operator

X = Integral { x |x><x| dx }

is an idealization. The operator which corresponds more precisely to the actual physical measurement has a form just like A above, where

a(n) = x(n) , the "location" of each detector

and

P(n) = Integral_[region(n)] { |x><x| dx } ,

where "region(n)" is the interval over which the detector "detects".

Clearly, "|x><x| dx" must be dimensionless.

[Alex, later on today, I hope to reply to your most recent post.]

Last edited: Jun 29, 2004
23. Jun 28, 2004

turin

Regarding the honeycomb: can you picture the difference between ∂μ and dxμ ?

alexepascual,
Regarding units in QM: I think it is more important for the eigenvalues to have the appropriate units. Then, when your observable operates on the spin state, for instance, you get an hbar. Come to think of it, eigenvalues themselves have units, so how can that be explained with the rotation picture?
Regarding normalization: you simply divide the state vector by the square root of the self inner product. This inner product is an integration over all of the domain of interest, so it will pick up whatever units are used to describe that domain. For the spin states the units are kind of like angular values, so they should be unitless I reckon.

24. Jun 29, 2004

alexepascual

Eye_in_the_Sky:
Thanks for your post. I see you use the projector for discretized states, and an integral to obtain each of the discrete states.
I guess this approach makes sense. In the end you get to the same conclusion as in your previous post, which is nice. But I was already convinced after your post using the Dirac delta function.
I have looked for references to "generalized" kets in the books but I haven't found anything. Would they use a different terminology?. Or maybe none of the books I have from the library deal with this topic in detail. (Sakurai, Shankar, Merzbacher). I also have Feynman's Lectures and Griffiths.
You mention in your first post that a definite momentum ket would carry units of reciprocal square-root of action. I don't understand that. Being a momentum ket, shouldn't it have to have units of reciprocal of square root of momentum?. I am not really disputing what you are saying. I am sure you are right. It is just that I don't understand it.

25. Jun 29, 2004

alexepascual

Turin:
I agree with you that in general state vectors don't have units as they represent the square root of a probability + a phase (for each component).
But, as Tavi_Boada and Eye_in_the_sky have shown us, the wave function, as well as the eigenvectors of the operators of continuous variable do need some units (not the units of the variable in question) in order for the equations to be dimensionally consistent. I am not very happy with this way of doing the math, but I accept it. The books that I have seen appear to ingnore this issue.
If we limit ourselves to the case of discrete variables, where we use summations instead of integrals, then the kets are all dimensionless, like the kets for spin.
It is only when we apply a measurable's operator that we get eigenvalues together with their units (as you mention). But the units we were talking about are not these units, but the reciprocal square root of the measurable's units which are needed only for eigenkets of continuous variables.
The subject of getting units when applying the operator would fit more in my previous thread on the meaning of operators than here. But as you mention using a rotation picture used to interpret this, I'll give you my opinion. You could interpret the measurement process as a rotation of the state vector to one of the eigenvectors of the measurement operator. But I think this would have to be done "after the fact", once you know to which eigenvector it went. This picture would not give you the probability of obtaining the eigenvector, but it would give you the new eigenvector and you would not have to normalize it as it would keep it's unit modulus. In using this "rotration picture" you would stay within the realm of Hilbert space, and no measurable operators would be used (just a unitary operator to perform the rotation).
I think using measurement operators presents a different picture. You get the eigenvalues which are not in Hilbert space but are embedded in the operator. So, as I say in a previous post in this thread, I think operators provide a translation from the language of Hilbert space, where the quantities being manipulated are just complex dimensionless numbers, to the language of classical mechanics, where we use the typical dimensioned variables.
Form what I understand, using operators goes hand-in-hand with the projection picture, where besides getting the probabilities, you also get the eigenvalues which are the numbers used in classical mechanics.
I suppose there is a lot of thinking that can be done about all these things, but most people just accept them like they are presented in the books, without questioning or trying to visualize them a different way.
The way I see it, QM is usually presented as a dogma and it requires a lot of faith on your part to accept it.

Last edited: Jun 29, 2004


text:
Polytropic Piston-cylinder & entropy

1. Dec 30, 2012

George32

1. The problem statement, all variables and given/known data
We have a piston in a cylinder containing Helium. The initial states are:
P1=150kPa
T1= 20°C
V1=0.5m3

Following a polytropic process, the final states are:
P2=400kPa
T2=140°C
V2 is unknown.

We're also given R=2.0769 kPa.m3.(kg.K)-1
And Cp=5.1926kJ.(kg.K)-1
As well as the temperature of the surrounding being 20°C

We are then asked to find:
a) The entropy change inside the system
b) The entropy change in the surroundings
c) Is the process Irreversible, Reversible or Impossible?

A hint is that we need to find n (the polytropic constant)

2. Relevant equations
PV=RT (1)
$TdS=dh - VdP$ (2)
$dh=CpdT$ (3)
$PV^n = Constant$ (4)

3. The attempt at a solution
I have a solution for part (a) which comes out at -0.2546kJ/K.

I got this after rearranging (1) to give $$\frac{V}{T} = \frac{R}{P}$$ and filling (3) and the new (1) into (2) to give $$ds=\frac{CpdT}{T} - \frac{RdP}{P}$$

Integrating each term between their respective limits to give $$\Delta s = Cp(ln (\frac{T2}{T1})) - R(ln(\frac{P2}{P1}))$$

Now for the process to be polytropic it has to be quasi-static which doesn't necessarily mean reversible but is (as I understand it) usually reversible, so the entropy change of the surroundings would have to be 0.2546 at least. However I don't know how I would calculate this, and the third part makes me question if I should get an answer lower and it will be an impossible process.

Any pointers on how to proceed, as well as perhaps a confirmation of using the correct process so far, would be great, thanks :)

G

2. Dec 30, 2012

Staff: Mentor

Why don't you start out by calculating the number of moles of gas in the cylinder using the data for the initial state? Now that you know the number of moles, why don't you use the data for the final state to calculate the volume in the final state? Once you know the initial and final pressures and volumes, you can calculate the polytropic exponent n.

3. Dec 30, 2012

George32

I did calculate n, I used the ideal gas law to find the volume for the second state and then used logs, but I don't see how that helps find the entropy of the surroundings? I will redo the numbers using your method to check I have the right n.

I'm using google and "Thermodynamics: An Engineering Approach" by Cengel and Boles, by the way.

4. Jan 1, 2013

Staff: Mentor

Suppose you could show that the process was reversible. Could you then determine the change in entropy of the surroundings? What value did you get for n?

5. Jan 2, 2013

George32

My value for n (polytropic exponent) came out as 1.538 (4s.f.)

If you could show the process were externally reversible, then the entropy generation would have to be 0 and so entropy change of the surroundings would be equal to positive of the entropy change of the system.

The issue is with showing the system is externally reversible. The process is quasi-static and so internally reversible, however this means that the system's temperature increase is smooth and as it ends up 120K above the surroundings, for the system to be totally reversible, no heat could be transferred across that temperature difference. As I understand it.

I know that the net entropy transfer + the entropy generation = change in entropy of the system. So I know one of three variables ( change in the system). I'm required to find the net entropy transfer (assuming that none of the entropy generated contributes to entropy transferred to the surroundings). Finding the entropy generated will answer the last part of the question.

As the surroundings are at constant temperature, I *think* I can say that $$\Delta S = \frac{Qsurr.}{Tsurr.}$$ but now my issue is that I don't know Qsurr. Calculating Qsurr (Taken as the heat transferred to the surroundings) is now the focus of my reading.

After going over my work I noticed a slight miscalculation. My working above wasn't clear and so I was confusing myself. The result I gave for entropy should be specific entropy, so kJ/(kgK)

6. Jan 2, 2013

Staff: Mentor

I'm also not sure what to do with part B. Your equation for ΔS for the surroundings doesn't seem to be correct, because we have no idea what is going on in the surroundings (in addition to heat transfer to or from 20C). But you can calculate Q from the first law. The first step is to calculate the work done on the surroundings (i.e., the PdV work). The second step is to calculate the change in internal energy (dU). Suppose you calculate Q and it comes out positive. What does that tell you about the answer to question c?

This is kind of a hokey problem, because it is hard to understand how the final equilibrium temperature of the system can be 140 C if the surroundings are at 20 C. If the system is in thermal contact with the surroundings, the system at 140 C cannot be an equilibrium state.

7. Jan 4, 2013

George32

Okie-Dokie.

So I have a result (I think!), let's see what you think. To me it makes sense.

Using the 1st law$$dU=dQ-dW,$$ and the change in energy dU of the surroundings will be equal to that of the system (just in the opposite direction). Compute dW using ∫pdv and p=CV-n and rearranging to get $$\frac{dW=mR(T2-T1}{1-n}$$ which gives the work done by the system as negative (good so far) 57.09kJ.

Now this is a little more scratchy, but we'll see how it goes. using $$cv=\frac{du}{dT}$$ and separating the variables and integrating to find du as cv(T2-T1) we can find dU by then multiplying by the mass (which we have already worked out when finding n, or at least I had) to get dU=46.08kJ.

So for the system we know dU and dW so dQ should be easy enough, I got it as -11.01kJ, which makes sense as that's 11.01kJ of thermal energy being transferred to the surroundings. Then saying that the change in entropy is equal to the heat energy transferred over the temperature, and assuming the temperature of the surroundings stay constant at 20°C, $$\Delta S = \frac{11.01}{293} = 0.03758$$

So then the net change in entropy is the sum of the entropy changes, which is 0.03758 - 0.0313780, which comes out as 0.0061986 kJ/K which is positive but quite small. I therefore took this as being irreversible, as I don't quite know the scale of entropies, as to whether or not this was close enough to 0 to be regarded as reversible.

Thanks for the help Chestermiller!

N.B. I had to use the polytropic exponent n to calculate the work done by the gas, so I managed to get that in, let's hope that was what they meant!

8. Jan 4, 2013

Staff: Mentor

There is no reason why the change in internal energy for the surroundings should be equal to minus the change in internal energy for the system. But that isn't really relevant to our discussion.

I checked your values for the work done by the surroundings on the system, and for the change in internal energy of the system, and got the same results. This confirms that the heat removed from the system during the process was 11 kJ. A surroundings at 20 C could accomplish this heat removal. It still isn't clear to me if it is possible to calculate the change in entropy of the surroundings without knowing more details about what is happening in the surroundings (but I doubt it). But the surroundings temperature remaining constant at 20 C while the average temperature of the system was getting higher suggests an irreversible process to me, since it suggests the presence of temperature non-uniformities within the system during the process, which is a clearcut symptom of irreversibility.

Chet

9. Jan 4, 2013

George32

Yes, sorry, I got myself confused between Total energy and Internal Energy. However the rest should still stand and I'm glad you came to the same conclusions :)

10. Jan 4, 2013

Staff: Mentor

Yes. I have an addition to what I said in my previous post. If we regard the surroundings as a separate system, then the Clausius Inequality tells us that the entropy change for the surroundings is greater than Q/T. So we at least know that the entropy change for the surroundings is at the very least, .03758 J/K. This definitely confirms that the process is irreversible. Even though we don't know the exact entropy change of the surroundings, we have still been able to confirm that this particular process is irreversible.

You did very well at working out the details of this problem.

Chet

11. Jan 4, 2013

George32

Thank you, and thanks for the help! Let's just hope I don't forget before the exams!


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text:
+0

# Just something I noticed...

0
202
5

Nobody is answering https://web2.0calc.com/questions/help-pls-1-question but questions around it (below and above) have been answered...

Apr 28, 2019

#1
#2
+1

Don't you mean https://web2.0calc.com/questions/help-pls-1-question ?  The link didn't work...

Guest Apr 28, 2019
#3
+6046
+1

$$\text{Think about it a moment and it should be clear that the range of }f(x) = \lfloor x\rfloor - x \text{ is }\\ f(x) \in (-1,0]$$

.
Apr 28, 2019
#4
+81
-1

The question quality is a little low...

This type of thing also happens a lot on this forum, either people do not know how to solve the problem, or they did not see it.

Apr 29, 2019
#5
+107060
+1

Well now it is answered by both Rom (here) and me (on the original post)

Melody  May 1, 2019