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The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
(Elsevier, 2018-01)
This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
(Elsevier, 2018-06)
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV
(Elsevier, 2018-05)
We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2018-01)
We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ...
Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV
(Springer Berlin Heidelberg, 2018-07-16)
Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range |η| < 0.8 ...
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Learning Objectives Find the macro equilibrium using algebra
In the income-expenditure model, the equilibrium occurs at the level of GDP where aggregate expenditures equal national income (or GDP). We can identify this equilibrium using algebra as well as graphically. Given algebraic equations for the aggregate expenditure line and the income=expenditure line, the point where they cross can be readily calculated.
USING AN ALGEBRAIC APPROACH TO THE EXPENDITURE-OUTPUT MODEL
Imagine an economy defined by the following:
C = 140 + 0.9 (Yd).
This is the consumption function where 140 is autonomous consumption, 0.9 is the marginal propensity to consume, and Yd is disposable (i.e. after tax income).
Yd = Y- T, where Y is national income (or GDP) and T = Tax Revenues = 0.3Y; note that 0.3 is the average income tax rate.
I = Investment = 400
G = Government spending = 800
X = Exports = 600
M = Imports = 0.15Y
Step 1. Determine the aggregate expenditure function. Using the numbers from above, it is:
AE = C + I + G + X – M AE = 140 + 0.9(Y – T) + 400 + 800 + 600 – 0.15Y Step 2. The equation for the 45-degree line is the set of points where GDP or national income on the horizontal axis is equal to aggregate expenditure on the vertical axis. Thus, the equation for the 45-degree line is: AE = Y. Step 3. The next step is to solve these two equations for Y (or AE, since they will be equal to each other). Substitute Y for AE:
Y = AE = 140 + 0.9(Y – T) + 400 + 800 + 600 – 0.15Y
Step 4. Insert the term 0.3Y for the tax rate T. This produces an equation with only one variable, Y. Step 5. Work through the algebra and solve for Y.
Y = 140 + 0.9(Y – 0.3Y) + 400 + 800 + 600 – 0.15Y Y = 140 + 0.9Y –0.27Y + 1800 – 0.15Y Y = 1940 + 0.48Y Y – 0.48Y = 1940 0.52Y = 1940 [latex]\displaystyle\frac{0.52\text{Y}}{0.52}[/latex] = [latex]\displaystyle\frac{1940}{0.52}[/latex] Y = 3730
This algebraic framework is flexible and useful in predicting how economic events and policy actions will affect real GDP.
Say, for example, that because of changes in the relative prices of domestic and foreign goods, the marginal propensity to import falls to 0.1. Calculate the equilibrium output when the marginal propensity to import is changed to 0.10.
Y = 140 + 0.9(Y – 0.3Y) + 400 + 800 + 600 – 0.1Y Y = 1940 – 0.53Y 0.47Y = 1940 Y = 4127
Alternatively, suppose because of a surge of business confidence, investment rises to 500. Calculate the equilibrium output.
Y = 140 + 0.9(Y – 0.3Y) + 500 + 800 + 600 – 0.15Y Y = 2040 + 0.48Y Y – 0.48Y = 2040 0.52Y = 2040 Y = 3923 Exercise: Consumption in the Income-Expenditure Model
National Income Taxes After-tax income Consumption I + G + X Imports Aggregate Expenditures $300 $236 $400 $500 $600 $700 Step 1. Calculate the amount of taxes for each level of national income (reminder: GDP = national income) for each level of national income using the following as an example:
[latex]\begin{array}{lr}\text{National Income (Y)}&\$300\\\text{Taxes = 0.2 or 20%}&\times0.2\\\text{Tax amount (T)}&\$60\end{array}[/latex]
Step 2. Calculate after-tax income by subtracting the tax amount from national income for each level of national income using the following as an example:
[latex]\begin{array}{lr}\text{National income minus taxes}&\$300\\&-\$60\\\text{After-tax income}&\$240\end{array}[/latex]
Step 3. Calculate consumption. The marginal propensity to save is given as 0.1. This means that the marginal propensity to consume is 0.9, since MPS + MPC = 1. Therefore, multiply 0.9 by the after-tax income amount using the following as an example:
[latex]\begin{array}{lr}\text{After-tax Income}&\$240\\\text{MPC}&\times0.9\\\text{Consumption}&\$216\end{array}[/latex]
Step 4. Consider why the table shows consumption of $236 in the first row. As mentioned earlier, the Keynesian model assumes that there is some level of consumption even without income. That amount is $236 – $216 = $20. Step 5. There is now enough information to write the consumption function. The consumption function is found by figuring out the level of consumption that will happen when income is zero. Remember that:
[latex]\text{C}=\text{Consumption when national income is zero}+\text{MPC (after-tax income)}[/latex]
Let C represent the consumption function, Y represent national income, and T represent taxes.
[latex]\begin{array}{lcl}\text{C}&=&\$20+0.9\left(\text{Y}-\text{T}\right)\\&=&\$20+0.9\left(\$300-\$60\right)\\&=&\$236\end{array}[/latex]
Step 6. Use the consumption function to find consumption at each level of national income. Step 7. Add investment (I), government spending (G), and exports (X). Remember that these do not change as national income changes: Step 8. Find imports, which are 0.2 of after-tax income at each level of national income. For example:
[latex]\begin{array}{lr}\text{After-tax income}&\$240\\\text{Imports of 0.2 or 20% of Y}-\text{T}&\times0.2\\\text{Imports}&\$48\end{array}[/latex]
Step 9. Find aggregate expenditure by adding C + I + G + X – I for each level of national income. Your completed table should look like this:
National Income (Y) Tax = 0.2 × Y (T) After-tax income (Y – T) Consumption C = $20 + 0.9(Y – T) I + G + X Minus Imports (M) Aggregate Expenditures AE = C + I + G + X – M $300 $60 $240 $236 $200 $48 $388 $400 $80 $320 $308 $200 $64 $444 $500 $100 $400 $380 $200 $80 $500 $600 $120 $480 $452 $200 $96 $556 $700 $140 $560 $524 $200 $112 $612 Step 10. Answer the question: What is equilibrium? Equilibrium occurs where AE = Y. This table shows that equilibrium occurs where national income equals aggregate expenditure at $500. Step 11. Find equilibrium mathematically, knowing that national income is equal to aggregate expenditure.Step 10. Answer the question: What is equilibrium? Equilibrium occurs where AE = Y. The table shows that equilibrium occurs where national income equals aggregate expenditure at $500.
[latex]\begin{array}{rcl}\text{Y}&=&\text{AE}\\&=&\text{C}+\text{I}+\text{G}+\text{X}-\text{M}\\&=&\$20+0.9\left(\text{Y}-\text{T}\right)+\$70+\$80+\$50-0.2\left(\text{Y}-\text{T}\right)\\&=&\$220+0.0\left(\text{Y}-\text{T}\right)-0.2\left(\text{Y}-\text{T}\right)\end{array}[/latex]
Since T is 0.2 of national income, substitute T with 0.2 Y so that:
[latex]\begin{array}{rcl}\text{Y}&=&\$220+0.9\left(\text{Y}-0.2\text{Y}\right)-0.2\left(\text{Y}-0.2\text{Y}\right)\\&=&\$220+0.9\text{Y}-0.18\text{Y}-0.2\text{Y}+0.04\text{Y}\\&=&\$220+0.56\text{Y}\end{array}[/latex]
Solve for Y.
[latex]\begin{array}{rcl}\text{Y}&=&\$220+0.56\text{Y}\\\text{Y}-0.56\text{Y}&=&\$220\\0.44\text{Y}&=&\$220\\\frac{0.44\text{Y}}{0.44}&=&\frac{\$200}{0.44}\\\text{Y}&=&\$500\end{array}[/latex]
Step 12. Answer this question: Why is a national income of $300 not an equilibrium? At national income of $300, aggregate expenditures are $388. Step 13. Answer this question: How do expenditures and output compare at this point? Aggregate expenditures cannot exceed output (GDP) in the long run, since there would not be enough goods to be bought. Try It
Try It
This question allow you to get as much practice as you need, as you can click the link at the top of the question (“Try another version of this question”) to get a new question. Practice until you feel comfortable doing the question.
Try It
These questions allow you to get as much practice as you need, as you can click the link at the top of the first question (“Try another version of these questions”) to get a new set of questions. Practice until you feel comfortable doing the questions.
Try It
These questions allow you to get as much practice as you need, as you can click the link at the top of the first question (“Try another version of these questions”) to get a new set of questions. Practice until you feel comfortable doing the questions.
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Update: The MathJax Plugin for TiddlyWiki has a new home: https://github.com/guyru/tiddlywiki-mathjax Some time ago I came across MathJax, a nifty, Javascript based engine for displaying TeX and LaTeX equations. It works by “translating” the equation to MathML or HTML+CSS, so it works on all modern browsers. The result isn’t a raster image, like in most LaTeX solutions (e.g. MediaWiki), so it’s scales with the text around it. Furthermore, it’s quite easy to integrate as it doesn’t require any real installation, and you could always use MathJax’s own CDN, which makes things even simpler.
I quickly realized MathJax will be a perfect fit for TiddlyWiki which is also based on pure Javascript. It will allow me to enter complex formulas in tiddlers and still be able to carry my wiki anywhere with me, independent of a real TeX installation. I searched the web for an existing MathJaX plugin for TiddlyWiki but I came up empty handed (I did find some links, but they referenced pages that no longer exist). So I regarded it as a nice opportunity to begin writing some plugins for TiddlyWiki and created the MathJaxPlugin which integrates MathJax with TiddlyWiki.
As I don’t have an online TiddlyWiki, you’ll won’t be able to import the plugin, instead you’ll have to install it manually (which is pretty simple).
Start by creating a new tiddler named
MathJaxPlugin, and tag with
systemConfig (this tag will tell TiddlyWiki to execute the JS code in the tiddler, thus making it a plugin. Now copy the following code to the tiddler content:
/***
|''Name:''|MathJaxPlugin|
|''Description:''|Enable LaTeX formulas for TiddlyWiki|
|''Version:''|1.0.1|
|''Date:''|Feb 11, 2012|
|''Source:''|http://www.guyrutenberg.com/2011/06/25/latex-for-tiddlywiki-a-mathjax-plugin|
|''Author:''|Guy Rutenberg|
|''License:''|[[BSD open source license]]|
|''~CoreVersion:''|2.5.0|
!! Changelog
!!! 1.0.1 Feb 11, 2012
* Fixed interoperability with TiddlerBarPlugin
!! How to Use
Currently the plugin supports the following delemiters:
* """\(""".."""\)""" - Inline equations
* """$$""".."""$$""" - Displayed equations
* """\[""".."""\]""" - Displayed equations
!! Demo
This is an inline equation \(P(E) = {n \choose k} p^k (1-p)^{ n-k}\) and this is a displayed equation:
\[J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m + \alpha + 1)}{\left({\frac{x}{2}}\right)}^{2 m + \alpha}\]
This is another displayed equation $$e=mc^2$$
!! Code
***/
//{{{
config.extensions.MathJax = {
mathJaxScript : "http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML",
// uncomment the following line if you want to access MathJax using SSL
// mathJaxScript : "https://d3eoax9i5htok0.cloudfront.net/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML",
displayTiddler: function(TiddlerName) {
config.extensions.MathJax.displayTiddler_old.apply(this, arguments);
MathJax.Hub.Queue(["Typeset", MathJax.Hub]);
}
};
jQuery.getScript(config.extensions.MathJax.mathJaxScript, function(){
MathJax.Hub.Config({
extensions: ["tex2jax.js"],
"HTML-CSS": { scale: 100 }
});
MathJax.Hub.Startup.onload();
config.extensions.MathJax.displayTiddler_old = story.displayTiddler;
story.displayTiddler = config.extensions.MathJax.displayTiddler;
});
config.formatters.push({
name: "mathJaxFormula",
match: "\\\\\\[|\\$\\$|\\\\\\(",
//lookaheadRegExp: /(?:\\\[|\$\$)((?:.|\n)*?)(?:\\\]|$$)/mg,
handler: function(w)
{
switch(w.matchText) {
case "\\[": // displayed equations
this.lookaheadRegExp = /\\\[((?:.|\n)*?)(\\\])/mg;
break;
case "$$": // inline equations
this.lookaheadRegExp = /\$\$((?:.|\n)*?)(\$\$)/mg;
break;
case "\\(": // inline equations
this.lookaheadRegExp = /\\\(((?:.|\n)*?)(\\\))/mg;
break;
default:
break;
}
this.lookaheadRegExp.lastIndex = w.matchStart;
var lookaheadMatch = this.lookaheadRegExp.exec(w.source);
if(lookaheadMatch && lookaheadMatch.index == w.matchStart) {
createTiddlyElement(w.output,"span",null,null,lookaheadMatch[0]);
w.nextMatch = this.lookaheadRegExp.lastIndex;
}
}
});
//}}}
After saving the tiddler, reload the wiki and the MathJaxPlugin should be active. You can test it by creating a new tiddler with some equations in it:
This is an inline equation $$P(E) = {n \choose k} p^k (1-p)^{ n-k}$$ and this is a displayed equation:
\[J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m + \alpha + 1)}{\left({\frac{x}{2}}\right)}^{2 m + \alpha}\]
Which should result in the tiddler that appears in the above image.
Update 2011-08-19: Removed debugging code from the plugin.
Changelog 1.0.1 (Feb 11, 2012 Applied Winter Young’s fix for interoperability with other plugins (mainly TiddlerBarPlugin
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I have the following problem : I'm calculating the sample covariance matrix in the frequency domain ( $y_{k}$ is the FFT of a time domain $k_{th}$ symbol vector signal , basically a simulated received signal) as follows:
$$\mathbf{R}=\frac{1}{N_{f}}\sum_{k=0}^{N_{f}-1}y_{k}y_{k}^{H}$$
Well , the next step on my algorithm is to solve a optimization process, as an essential part of it I need to compute the eigenvalues by doing SVD ,method of powers, etc... in MATLAB of the following equation:
$$(\mathbf{R}^{-1}\mathbf{A})$$
To not get into too many details because I believe my issue comes from a numerical problem and many insights of the actual algortihm / context aren't needed .Let's assume, $A$ is simply a predefined matrix that I compute.
The REAL ISSUE appears now because $\mathbf R$ is ill-conditioned as MATLAB tells me so. So the inverse procedure seems to be failing and the eigenvalues I'm obtaining are incredibly small due to this issue (In fact I only get 1 eigenvalue different than zero). The dimensions of $\mathbf R$ are typically large ( since they are compressed depends on the actual compression ratio I'm using but let's say $32\times 32$ for example).
One approach to solve this problem I found is to use diagonal loading as:
$$(\left(\mathbf R+\sigma\mathbf I\right)^{-1}\mathbf A)\quad\text{with}\quad\sigma > 0$$
This seems to be solving the problem, the eigenvalues are now scaled due to this background "noise". My question is how can I obtain the truly well scaled eigenvalues because , later on, in my algorithm these eigenvalues will serve as weights since I'm considering them as an actual power estimate.
Note : I've been playing with the
cond() function in MATLAB for $\sigma= 0.05$ ,
cond(R+sigma*I)= 2 , which is not bad I believe.
Feel free to ask more questions about the problem. But I think my question relates to a purely numerical issue involving eigenvalues and covariances matrices.
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(Sorry was asleep at that time but forgot to log out, hence the apparent lack of response)
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
Yes you can (since $k=\frac{2\pi}{\lambda}$). To convert from path difference to phase difference, divide by k, see this PSE for details
http://physics.stackexchange.com/questions/75882/what-is-the-difference-between-phase-difference-and-path-difference
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I am stuck on finding the 'right' mediator in weak interactions. Consider the following reactions.
1:
$$\mu^+\rightarrow \bar{\nu}_\mu + e^+ +\nu_e$$
This is mediated by the vector boson $W^+$.
2:
$$\pi^-\rightarrow \bar{\nu}_\mu+\mu^-$$
This is mediated by the vector boson $W^-$.
3:
$$\nu_\mu +e^- \rightarrow \nu_\mu + e^-$$
This is mediated by the neutral current $Z^0$.
My question is: is there a systematic way of finding the vector boson or mediator in weak interactions? It might be very trivial but I do not see it. I want to understand this since I want to draw Feynman diagrams well.
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Happy Pi day to all non-American applied mathematicians and scientists (like me) who make approximations!
This is in fact TeX related as I have a question, or perhaps a puzzle. How would I make the equation
$\pi \approx \today$
give the output that makes sense, given the appropriate formatting of today's date. Furthermore, suppose the equation were typeset tomorrow, one might like it to output `\pi \neq \today', if you see what I mean. And if it were typeset on March 14, it might give yet a different answer. I'm afraid I don't have any MWEs for these.
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9.10. Predicting Pizza Prices - Linear Regression¶
Linear regression is probably one of the most widely used algorithms in data science, and many other sciences. One of the best things about linear regression is that it allows us to learn from things that we know or use observations and measurements of things we know to make predictions about new things. These predictions might be about the likelihood of a person buying a product online or the chance that someone will default on their loan payments. To start we are going to use an even simpler example predicting the price of pizza based on its diameter.
I have made an extensive study of the pizza places in my neighborhood and here is a table of observations of pizza diameters and their price.
Diameter
Price
6
7
8
9
10
13
14
17.5
18
18
Your first task is to put the data into a spreadsheet and make a scatter plot of the diameter versus the price. What you can see pretty easily from this graph is that as the diameter of the pizza goes up, so does the price.
If you were to draw a straight line through the points that came as close as possible to all of them, it would look like this:
The orange line called the trend-line or the regression line is our best guess at a line that describes the data. This is important because we can come up with an equation for the line that will allow us to predict the y value (price) for any given x value (diameter). Linear regression is all about finding the best equation for the line.
How do we do that? There are actually several different ways we can come up with the equation for the line. We will look at two different solutions, one is a closed form equation that will work for any problem like this in just two dimensions. The second is a solution that will allow us to generalize the idea of a best fit line to many dimensions!
Recall the equation for a line that you learned in algebra: \(y = mx + b\) What we need to do is to determine values for m and b. One way we can do that is to simply guess! And keep refining our guesses until we get to a point where we are not really getting any better. You may think this sounds kind of stupid, but it is actually a pretty fundamental part of many machine learning algorithms.
You may also be wondering how we decide what it means to “get better”? In the case of our pizza problem we have some data to work with, and so for a given guess for m and b we can compare the calculated y (price) against the known value of y and measure our error. For example: Suppose we guess that b = 5 and that m = .8 for a diameter of 10 we get y = .7 x 10 + 5 or 12. Checking against our table the value should be 13 so our error is our known value minus our predicted value 13-12 or 1. If we try the same thing for a diameter of 8 we get y = .7 x 8 + 5 or 10.6 The error here is 9 - 10.6 or -1.6.
Add a column to the spreadsheet that contains the predicted price for the pizza using the diameter as the x value and using a slope of .7 and intercept of 5.
Now plot the original set of data along with the this new table of data. Make the original one color and your calculated table another color. Experiment with some different guesses for the slope and intercept to see how it works.
Next lets add another column to the table where we include the error. Now we have our ‘predicted values’ and a bunch of error measurements. One common way we combine these error measurements together is to compute the Mean Squared Error (MSE) This is easy to compute because all we have to do is square each of our errors, add them up and then divide by the number of error terms we have. Why do we square them first? Well, did you notice that in our example one of the errors was positive and one was negative, but when we add together both positive and negative numbers they tend to cancel each other out making our final mean value smaller. So we square them to be sure they are all positive. We call this calculation of the MSE an
objective function. In many machine learning algorithms our goal is to minimize the objective function. That is what we want to do here, we want to find the value for m and b that minimizes the error.
Add two cells to your spreadsheet where you can try different values for the slope and intercept. Also update the column where you compute a value for the price to use the values from these cells rather than the hardcoded values os .7 and 5.
Q-1: Using a slope of 2.5 and an intercept of 0.5 what is the MSE?
Now lets make use of the Solver functionality to search for and find the best values for the slope and intercept. Make sure that you have the Frontline Systems Solver add-on installed for Google Sheets. If you haven’t used solver before you may want to take a look at Optimization with Solver. Setting up solver for this problem doesn’t even have any constraints. What we want to do is minimize the MSE value, by changing the values for slope and intercept. Note that because we are squaring the errors this is a non-linear problem and will require the Standard LSGRG Nonlinear solver. Now set up the solver and run it for the pizza problem.
Q-2: Fill in the values Solver found for the slope: and intercept
If you are having any trouble, your setup should look like this.
9.10.1. Closed form Solution¶
The closed form solution to this problem is known to many science students.
slope = \(\frac{\sum{(x_i-\bar{x})(y_i - \bar{y})}}{\sum{(x_i - \bar{x})^2}}\)
intercept = \(\bar{y} - slope \cdot \bar{x}\)
Lets use the closed form solution to calculate values for the slope and intercept. To do this you will need to calculate a value for \(\bar{x}\) and \(\bar{y}\) that is the average value for both x and y. You can add two columns to do the calculation of \(y_i - \bar{y}\) and \(x_i - \bar{x}\)
Q-3: What values do you get for the slope and intercept?
9.10.2. The Payoff - Supervised Learning¶
The payoff from this exercise with Solver is that we have “learned” values for the slope and intercept that will allow us to predict the price of any pizza! If your friend calls you up and says “I just ate a 7 inch pizza, guess how much it cost?” You can quickly do the math of 1.97 + 0.98 x 7 and guess $8.83! Won’t they be amazed!?
In the world of machine learning, using the sample data for pizza along with a solver like algorithm for finding the values for the slope and intercept are called
supervised learning. That is because we are using the known values for the prices of different pizzas along with their diameters to help correct our algorithm and come up with a value for the slope and intercept. The values that the learns, are called our model. This model is pretty simple because it just uses two numbers and the formula for a line. But don’t let the simplicity fool you, Regression is one of the most commonly used algorithms in a data scientists arsenal.
In the next section we’ll make a fancier model that uses more data to do a better job of making predictions. If you want to try your hand at writing your own learning algorithm you can do that in the optional section below.
9.10.3. A simple Machine Learning Approach (Optional)¶
To do this we will follow these steps:
Pick a random value for m and b
Compute the MSE for all our known points
Repeat the following steps 1000 times 1. Make m slightly bigger and recompute the MSE does that make it smaller? If so then use this new value for m. If it doesn’t make MSE smaller than make m slightly smaller and see if that helps. 1. Make b slightly bigger and recompute the MSE does that make it smaller? If so then use this new value for b and go back to step 3a. If not then try a slightly smaller b and see if that makes the MSE smaller if so keep this value for b and go back to step 3a.
After repeating the above enough times we will be very close to the best possible values for m and b. We can now use these values to make predictions for other pizzas where we know the diameter but don’t know the price.
Let’s develop some intuition for this whole thing by writing a function and trying to minimize the error.
You will write three functions
compute_y(x, m, b),
compute_all_y(list_of_x) This shoudl use
compute_y and
compute_mse(list_of_known, list_of_predictions)
Next write a function that systematically tries different values for m and b in order to minimize the MSE. Put this function in a loop for 1000 times and see what your value is for m and b at the end.
Congratulations! You have just written your first “machine learning” algorithm. One fun thing you can do is to save the MSE at the end of each time through the loop then plot it. You should see the error go down pretty quickly and then level off or go down very gradually. Note that the error will ever go to 0 because the data isn’t perfectly linear. Nothing in the real world is!
At this point your algorithms ability to ‘learn’ is limited by how much you change the slope and intercept values each time through the loop. At the beginning its good to change them by a lot but as you get closer to the best answer its better to tweak them by smaller and smaller amounts. Can you adjust your code above to do this?
For two dimensional data there is even a closed form solution to this problem that one could derive using a bit of calculus. It is worthwhile to have the students do this to see that their solution is very very close to the solution you get from a simple formula that slope = covariance / variance and intercept = avg(y) - slope * avg(x). Write a function that will calculate the slope and intercept using this method and compare the slope and intercept with your previous error.
Lesson Feedback During this lesson I was primarily in my... Comfort Zone Learning Zone Panic Zone Completing this lesson took... Very little time A reasonable amount of time More time than is reasonable Based on my own interests and needs, the things taught in this lesson... Don't seem worth learning May be worth learning Are definitely worth learning For me to master the things taught in this lesson feels... Definitely within reach Within reach if I try my hardest Out of reach no matter how hard I try
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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Proceedings of the Thirty-Second Conference on Learning Theory, PMLR 99:2173-2174, 2019.
Abstract
We study the linear contextual bandit problem with finite action sets. When the problem dimension is $d$, the time horizon is $T$, and there are $n \leq 2^{d/2}$ candidate actions per time period, we (1) show that the minimax expected regret is $\Omega(\sqrt{dT \log T \log n})$ for every algorithm, and (2) introduce a Variable-Confidence-Level (VCL) SupLinUCB algorithm whose regret matches the lower bound up to iterated logarithmic factors. Our algorithmic result saves two $\sqrt{\log T}$ factors from previous analysis, and our information-theoretical lower bound also improves previous results by one $\sqrt{\log T}$ factor, revealing a regret scaling quite different from classical multi-armed bandits in which no logarithmic $T$ term is present in minimax regret. Our proof techniques include variable confidence levels and a careful analysis of layer sizes of SupLinUCB on the upper bound side, and delicately constructed adversarial sequences showing the tightness of elliptical potential lemmas on the lower bound side.
@InProceedings{pmlr-v99-li19b,title = {Nearly Minimax-Optimal Regret for Linearly Parameterized Bandits},author = {Li, Yingkai and Wang, Yining and Zhou, Yuan},booktitle = {Proceedings of the Thirty-Second Conference on Learning Theory},pages = {2173--2174},year = {2019},editor = {Beygelzimer, Alina and Hsu, Daniel},volume = {99},series = {Proceedings of Machine Learning Research},address = {Phoenix, USA},month = {25--28 Jun},publisher = {PMLR},pdf = {http://proceedings.mlr.press/v99/li19b/li19b.pdf},url = {http://proceedings.mlr.press/v99/li19b.html},abstract = {We study the linear contextual bandit problem with finite action sets. When the problem dimension is $d$, the time horizon is $T$, and there are $n \leq 2^{d/2}$ candidate actions per time period, we (1) show that the minimax expected regret is $\Omega(\sqrt{dT \log T \log n})$ for every algorithm, and (2) introduce a Variable-Confidence-Level (VCL) SupLinUCB algorithm whose regret matches the lower bound up to iterated logarithmic factors. Our algorithmic result saves two $\sqrt{\log T}$ factors from previous analysis, and our information-theoretical lower bound also improves previous results by one $\sqrt{\log T}$ factor, revealing a regret scaling quite different from classical multi-armed bandits in which no logarithmic $T$ term is present in minimax regret. Our proof techniques include variable confidence levels and a careful analysis of layer sizes of SupLinUCB on the upper bound side, and delicately constructed adversarial sequences showing the tightness of elliptical potential lemmas on the lower bound side. }}
%0 Conference Paper%T Nearly Minimax-Optimal Regret for Linearly Parameterized Bandits%A Yingkai Li%A Yining Wang%A Yuan Zhou%B Proceedings of the Thirty-Second Conference on Learning Theory%C Proceedings of Machine Learning Research%D 2019%E Alina Beygelzimer%E Daniel Hsu%F pmlr-v99-li19b%I PMLR%J Proceedings of Machine Learning Research%P 2173--2174%U http://proceedings.mlr.press%V 99%W PMLR%X We study the linear contextual bandit problem with finite action sets. When the problem dimension is $d$, the time horizon is $T$, and there are $n \leq 2^{d/2}$ candidate actions per time period, we (1) show that the minimax expected regret is $\Omega(\sqrt{dT \log T \log n})$ for every algorithm, and (2) introduce a Variable-Confidence-Level (VCL) SupLinUCB algorithm whose regret matches the lower bound up to iterated logarithmic factors. Our algorithmic result saves two $\sqrt{\log T}$ factors from previous analysis, and our information-theoretical lower bound also improves previous results by one $\sqrt{\log T}$ factor, revealing a regret scaling quite different from classical multi-armed bandits in which no logarithmic $T$ term is present in minimax regret. Our proof techniques include variable confidence levels and a careful analysis of layer sizes of SupLinUCB on the upper bound side, and delicately constructed adversarial sequences showing the tightness of elliptical potential lemmas on the lower bound side.
Li, Y., Wang, Y. & Zhou, Y.. (2019). Nearly Minimax-Optimal Regret for Linearly Parameterized Bandits. Proceedings of the Thirty-Second Conference on Learning Theory, in PMLR 99:2173-2174
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The definition of a limit continues on in our learning limits and how they are applied. Specifically we are going to look at some proofs in this section. With a precise definition we can now prove many limit properties.
Definition of a Limit
The \(\delta\) value will always depend on the \(\epsilon\) value. That is important to remember in order to understand what is happening in these limits. Once you find a value that works in your problem then smaller values of that \(\delta\) also work.
Problem 1.
For the given function f(x) and values L,c, and \(\epsilon\) > 0 find the largest open interval about c on which the inequality |f(x)-L| < \(\epsilon\) holds. Then determine the largest value for \(\delta\) >0 such that 0<|x-c|<\(\delta\)\(\rightarrow\) |f(x)-L|<\(\epsilon\).
f(x) = \(\frac{1}{x}\), L= 0.125, c = 8, \(\epsilon\) = 0.015
Solve |f(x) – L| < \(\epsilon\) to find the largest interval containing c on which the inequality holds.
Write the inequality without absolute value. Substitute \(\frac{1}{x}\) for f(x), 0.1215 for L, and 0.015 for \(\epsilon\).
|f(x)-L| < \(\epsilon\)
|\(\frac{1}{x}\) – 0.125| < 0.015
-0.015 < \(\frac{1}{x}\) – 0.125 < 0.015
Simplify by adding 0.125 to all expressions.
-0.015 < \(\frac{1}{x}\) – 0.125 < 0.015
0.11 < \(\frac{1}{x}\) < 0.14
Simplify further by taking the reciprocals of all terms. Be sure to reverse the inequalities. Make sure you check that the rounded endpoints still allow the inequality |f(x)-L| < \(\epsilon\) to hold true.
9.0909 > x > 7.1429
Since the interval 7.1429 < x < 9.0909 is not centered on c=8, \(\delta\) is the distance from 8 to closer endpoint of the interval. The value of \(\delta\) is then 0.8571.
Problem 2.
For the given function \(f(x)\), the point \(c\), and a positive number \(\epsilon\), find \(L=lim f(x)\). Then find a number \(\delta\) > 0 such that for all x, 0<|x-c|<\(\delta\)\(\rightarrow\)|\(f(x)-L\)| < \(\epsilon\).
\(f(x)=8-3x, c=4, \epsilon=0.03\)
Notice that \(f(x)\) is a linear function. Since a linear function is a simple polynomial function and it is defined for all \(x\), the limit \(L\) = lim is the value of \(f(x)\) at \(c\).
Evaluate the function at \(c=4\).
\(L=8-3(4)=-4\)
To find \(\delta\), begin by solving the inequality \(|f(x)-L|<\epsilon\) to find an open interval \((a,b)\) containing \(c\) on which the inequality holds for all \(x\neq c\).
Remove the absolute value sign and rewrite the inequality as a compound inequality.
\(|(8-3x)-(-4)| < 0.03\)
\(-0.03 < (8-3x) – (-4) < 0.03\)
Simplify the center expression and isolate the x-term.
\(-0.03 < -3x+12 < 0.03\)
\(-12.03 < -3x < -11.97\)
Then isolate \(x\) in the center. Notice that the direction of the inequalities has been changed.
\(4.01 > x > 3.99\)
Therefore, for \(x\) in the interval \((3.99,4.01)\), the inequality \((8-3x)-(-4)|<0.03\) holds. Now choose a positive value for \(\delta\) that places the open interval \((c-\delta,c+\delta)\) centered on \(c\) inside the interval \((3.99,4.01)\).
The distance to the first endpoint is \(4-3.99=0.01\).
The distance to the second endpoint is \(4.01 -4=0.01\).
The largest possible value for \(\delta\) is \( 0.01\).
Therefore, for all \(x\) satisfying \(0<|x-4|<0.01\), the inequality \(|(8-3x)-(-4)|<0.03\) holds.
Problem 3.
Give an \(\epsilon – \delta\) proof of the limit fact.
\(lim_{x \to 0}(5x+1)=1\)
We begin by giving the precise meaning of a limit.
To say that \(lim f(x)=L\) means that for each \(\epsilon>0\) there is a corresponding \(\delta>0\) such that \(|f(x)-L|<\epsilon\), provided that \(0<|x-c|<\delta\).
To establish the proof, we first perform a preliminary analysis to find the appropriate choice of \(\delta\).
Let \(\epsilon\) be any positive number.
We must produce a \(\delta > 0\) such that \( 0<|x-0|<\delta \rightarrow |(5x+1) -1 < \epsilon\).
By simplifying the inequality on the right, we will determine the value of \(\delta\) needed for the inequality on the left.
\(|(5x+1)-1| < \epsilon \Leftrightarrow |5x| < \epsilon \)
\(\Leftrightarrow |5| |x| < \epsilon\)
\(\Leftrightarrow 5|x| < \epsilon\)
\(\Leftrightarrow |x| < \frac{\epsilon}{5}\).
Now we see that we should choose \(\delta=\frac{\epsilon}{5}\). We can now construct the formal proof of the statement \(\lim_{x \to 0}(5x+1)=1\).
Let \(\epsilon > 0\) be given. Choose \(\delta=\frac{\epsilon}{5}\). Then \(0<|x-0|<\delta\) implies the following chain of equalities and an inequality.
\(|(5x+1)-1| = |5x|\) Simplify inside the absolute value bars.
\( = |5| |x|\) Rewrite as product of absolute values.
\( = 5 |x| \) Simplify.
\( =5 |x-0|\) Rewrite expression inside absolute value bars.
\( < 5\delta\) Apply the condition \(0<|x-0|<\delta\).
\( = \epsilon \) Substitute \(\delta=\frac{\epsilon}{5}\).
The result of the above chain of equalities and an inequality is that \(|(5x+1)-1|< \epsilon\).
Therefore, we have proven that \(\lim_{x \to 0}(5x+1)=1\).
Conclusion
The definition of a limit has some nice proofs that can establish. If you go over these problems slowly the ideas behind limits will begin to emerge. While this section is short it is essential to understanding limits and beginning calculus in general.
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Proceedings of the Thirty-Second Conference on Learning Theory, PMLR 99:696-726, 2019.
Abstract
We propose the first contextual bandit algorithm that is parameter-free, efficient, and optimal in terms of dynamic regret. Specifically, our algorithm achieves $\mathcal{O}(\min\{\sqrt{KST}, K^{\frac{1}{3}}\Delta ^{\frac{1}{3}}T^{\frac{2}{3}}\})$ dynamic regret for a contextual bandit problem with $T$ rounds, $K$ actions, $S$ switches and $\Delta$ total variation in data distributions. Importantly, our algorithm is adaptive and does not need to know $S$ or $\Delta$ ahead of time, and can be implemented efficiently assuming access to an ERM oracle. Our results strictly improve the $\mathcal{O} (\min \{S^{\frac{1}{4}}T^{\frac{3}{4}}, \Delta^{\frac{1}{5}}T^{\frac{4}{5}}\})$ bound of (Luo et al., 2018), and greatly generalize and improve the $\mathcal{O}(\sqrt{ST})$ result of (Auer et al., 2018) that holds only for the two-armed bandit problem without contextual information. The key novelty of our algorithm is to introduce {\it replay phases}, in which the algorithm acts according to its previous decisions for a certain amount of time in order to detect non-stationarity while maintaining a good balance between exploration and exploitation.
@InProceedings{pmlr-v99-chen19b,title = {A New Algorithm for Non-stationary Contextual Bandits: Efficient, Optimal and Parameter-free},author = {Chen, Yifang and Lee, Chung-Wei and Luo, Haipeng and Wei, Chen-Yu},booktitle = {Proceedings of the Thirty-Second Conference on Learning Theory},pages = {696--726},year = {2019},editor = {Beygelzimer, Alina and Hsu, Daniel},volume = {99},series = {Proceedings of Machine Learning Research},address = {Phoenix, USA},month = {25--28 Jun},publisher = {PMLR},pdf = {http://proceedings.mlr.press/v99/chen19b/chen19b.pdf},url = {http://proceedings.mlr.press/v99/chen19b.html},abstract = {We propose the first contextual bandit algorithm that is parameter-free, efficient, and optimal in terms of dynamic regret. Specifically, our algorithm achieves $\mathcal{O}(\min\{\sqrt{KST}, K^{\frac{1}{3}}\Delta ^{\frac{1}{3}}T^{\frac{2}{3}}\})$ dynamic regret for a contextual bandit problem with $T$ rounds, $K$ actions, $S$ switches and $\Delta$ total variation in data distributions. Importantly, our algorithm is adaptive and does not need to know $S$ or $\Delta$ ahead of time, and can be implemented efficiently assuming access to an ERM oracle. Our results strictly improve the $\mathcal{O} (\min \{S^{\frac{1}{4}}T^{\frac{3}{4}}, \Delta^{\frac{1}{5}}T^{\frac{4}{5}}\})$ bound of (Luo et al., 2018), and greatly generalize and improve the $\mathcal{O}(\sqrt{ST})$ result of (Auer et al., 2018) that holds only for the two-armed bandit problem without contextual information. The key novelty of our algorithm is to introduce {\it replay phases}, in which the algorithm acts according to its previous decisions for a certain amount of time in order to detect non-stationarity while maintaining a good balance between exploration and exploitation.}}
%0 Conference Paper%T A New Algorithm for Non-stationary Contextual Bandits: Efficient, Optimal and Parameter-free%A Yifang Chen%A Chung-Wei Lee%A Haipeng Luo%A Chen-Yu Wei%B Proceedings of the Thirty-Second Conference on Learning Theory%C Proceedings of Machine Learning Research%D 2019%E Alina Beygelzimer%E Daniel Hsu%F pmlr-v99-chen19b%I PMLR%J Proceedings of Machine Learning Research%P 696--726%U http://proceedings.mlr.press%V 99%W PMLR%X We propose the first contextual bandit algorithm that is parameter-free, efficient, and optimal in terms of dynamic regret. Specifically, our algorithm achieves $\mathcal{O}(\min\{\sqrt{KST}, K^{\frac{1}{3}}\Delta ^{\frac{1}{3}}T^{\frac{2}{3}}\})$ dynamic regret for a contextual bandit problem with $T$ rounds, $K$ actions, $S$ switches and $\Delta$ total variation in data distributions. Importantly, our algorithm is adaptive and does not need to know $S$ or $\Delta$ ahead of time, and can be implemented efficiently assuming access to an ERM oracle. Our results strictly improve the $\mathcal{O} (\min \{S^{\frac{1}{4}}T^{\frac{3}{4}}, \Delta^{\frac{1}{5}}T^{\frac{4}{5}}\})$ bound of (Luo et al., 2018), and greatly generalize and improve the $\mathcal{O}(\sqrt{ST})$ result of (Auer et al., 2018) that holds only for the two-armed bandit problem without contextual information. The key novelty of our algorithm is to introduce {\it replay phases}, in which the algorithm acts according to its previous decisions for a certain amount of time in order to detect non-stationarity while maintaining a good balance between exploration and exploitation.
Chen, Y., Lee, C., Luo, H. & Wei, C.. (2019). A New Algorithm for Non-stationary Contextual Bandits: Efficient, Optimal and Parameter-free. Proceedings of the Thirty-Second Conference on Learning Theory, in PMLR 99:696-726
This site last compiled Sat, 17 Aug 2019 00:05:37 +0000
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1 Fractional control actions 1.1 Time domain
In the following, we illustrate the effects of the basic fractional-order control actions in the time domain [1]. For this we consider the response of the system
where $K$ is the gain of the differentiator/integrator of order $\alpha$, such that $-1\leqslant \alpha \leqslant 1$.
In the following figure the response of a fractional-order integrator to a square input signal is shown.
In Figure 2 it is possible to observe the response of a fractional-order differentiator to a trapezoidal input signal.
1.2 Frequency domain
For the fractional-order components the following is achieved by varying the value of $\alpha$ in the range $\alpha\in[-1, 0]$ for the fractional-order integrator and in the range $\alpha\in[0, 1]$ for the fractional-order differentiator it is possible to achieve the following:
A constant increment in the slope of the magnitude curve that is $20\alpha$ dB/dec. A constant delay in the phase plot that is $({\pi}/{2})\alpha$.
It can be seen, that in both the time domain and in the frequency domain additional flexibility of the control actions is introduced. Based on this it is possible to design robust fractional-order controllers.
2 Fractional controllers
In this section, we introduce two fractional-order controllers which are derived from classical controllers by means of generalization of the integral and differential components.
2.1 Fractional PID controller
The fractional PID controller was introduced by Podlubny [2]. The controller is called the PI$^{\lambda}$D$^{\mu}$ controller
1 since it has an integrator of order $\lambda$ and a differentiator of order $\mu$. It has been confirmed, that this type of controller offers superior performance compared to the classical PID controller [1, 3, 4].
The control action of the parallel form PI$^{\lambda}$D$^{\mu}$ controller can be expressed in the time domain as follows:
where $e(t)$ is the error signal. In the Laplace domain, assuming zero initial conditions, the fractional-order PID controller has the following form:
The case $\lambda=\mu=1$ corresponds to the classical, integer-order PID controller. In Figure 3 a comparison of the frequency domain characteristics of such a controller to those of the fractional PID controller with integrator/differentiator orders of $\lambda=\mu=0.5$ is shown.
2.2 Fractional lead-lag compensator
Lead-lag compensators are a well-known type of feedback controller that is widely used in practice. A fractional-order lead-lag compensator has the following transfer function [1]:
where $\alpha$ is the fractional power, $1/\lambda=\omega_{z}$ and $1/(x\lambda)=\omega_{p}$ are the zero and pole frequencies, respectively, when $\alpha \gt 0$. Two cases are possible:
$\alpha \gt 0$: the controller in (4) corresponds to a fractional-order lead compensator, whose behavior is similar to that of a PD$^{\mu}$ controller; $\alpha \lt 0$: the controller in (4) corresponds to a fractional-order lag compensator, whose behavior is similar to that of a PI$^{\lambda}$ controller.
A survey of tuning methods for both PI$^{\lambda}$D$^{\mu}$ controllers and fractional lead-lag compensators can be found in [1, 5]. Additionally, methods suitable for practical implementations of fractional-order controllers are provided in e.g. [6, 7].
Other fractional-order control techniques are elaborated upon. Once they become relevant to the FOMCON project, a summary will be included on this page.
References
[1] C. A. Monje, Y. Chen, B. Vinagre, D. Xue, and V. Feliu, Fractional-order Systems and Controls: Fundamentals and Applications, ser. Advances in Industrial Control. Springer Verlag, 2010. [2] I. Podlubny, “Fractional-order systems and PI$^{\lambda}$D$^{\mu}$-controllers,” IEEE Transactions on Automatic Control, vol. 44, no. 1, pp. 208–214, 1999. [3] M. Čech and M. Schlegel, “The fractional-order PID controller outperforms the classical one,” in Process control 2006. Pardubice Technical University, 2006, pp. 1–6. [4] Y. Luo and Y. Chen, “Fractional-order [proportional derivative] controller for robust
motion control: Tuning procedure and validation,” in Proc. ACC ’09. American Control Conference, 2009, pp. 1412–1417. [5] C. Monje, B. Vinagre, V. Feliu, and Y. Chen, “Tuning and auto-tuning of fractional order controllers for industry
applications,” Control Engineering Practice, vol. 16, no. 7, pp. 798–812, 2008. [6] Y. Q. Chen, I. Petráš, and D. Xue, “Fractional order control – A tutorial,” in Proc. ACC ’09. American Control Conference, 2009, pp. 1397–1411. [7] I. Petráš, “Practical Aspects of Tuning and Implementation of Fractional-Order
Controllers,” ASME Conference Proceedings, vol. 2011, no. 54808, pp. 75–84, 2011. 1Notation PI$^{\lambda}$D$^{\delta}$ is also used
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Category:Convergence Let $T = \left({S, \tau}\right)$ be a topological space.
Let $\left \langle {x_n} \right \rangle_{n \in \N}$ be an infinite sequence in $S$.
$\forall U \in \tau: \alpha \in U \implies \left({\exists N \in \R_{>0}: \forall n \in \N: n > N \implies x_n \in U}\right)$ Subcategories
This category has the following 14 subcategories, out of 14 total.
A ► Absolute Convergence (2 C, 5 P) C I ► Infinite Products (5 C, 47 P) R U ► Uniform Convergence (1 C, 19 P) Pages in category "Convergence"
The following 47 pages are in this category, out of 47 total.
C Cauchy Product of Absolutely Convergent Series Cauchy Sequence Converges on Real Number Line Convergence in Indiscrete Space Convergence of Limsup and Liminf Convergence of Sequence in Discrete Space Convergence of Sequence in Discrete Space/Corollary Convergent Sequence in Normed Division Ring is Bounded Convergent Sequence in Set of Integers Convergent Sequence in Set of Integers/Corollary Convergent Subsequence in Closed Interval Convergent Subsequence of Cauchy Sequence in Normed Division Ring D E F L Limit of Positive Real Sequence is Positive Limit of Sequence is Limit of Real Function Limit of Subsequence equals Limit of Real Sequence Limit of Subsequence equals Limit of Sequence Limit of Subsequence equals Limit of Sequence/Metric Space Limit of Subsequence equals Limit of Sequence/Real Numbers Logarithm of Convergent Product of Real Numbers Logarithm of Divergent Product of Real Numbers Logarithm of Infinite Product of Real Numbers
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Electrical resistivity and conductivity is an important property for materials. Different materials have different conductivity and resistivity. Electrical conductivity is based on electrical transport properties. These can be measured with multiple techniques by using a variety of instruments. If electricity easily flows through a material, that material has high conductivity. Some materials that have high conductivity include copper and aluminum. Electrical conductivity is the measure of how easily electricity flows through a material.
Conductivity vs Resistivity
Conductivity and resistivity are inversely proportional to each other. When conductivity is low, resistivity is high. When resistivity is low, conductivity is high. The equation is as follows:
\[ \rho = \dfrac{1}{\sigma}\]
where
Resistivity is represented by \(\rho\) and is measured in Ohm-meters(\(Ωm\)), Conductivity is represented by \( \sigma \) and is measured in Siemens(\(1/Ωm\)).
Since conductivity is the measure of how easily electricity flows, electrical resistivity measures how much a material resists the flow of electricity.
Electrical Transport Properties
Simply put, electricity is the movement of electrons through a material. As electrons move through a material, it comes into contact with atoms in the material. Collisions slow the electrons down. Each collision increases the resistivity of the material. The easier the electrons continue through a material, the fewer collisions that take place and the higher the conductivity.
When temperature increases, the conductivity of metals usually decreases, while the conductivity of semiconductors increases. This of course assumes that the material is homogenous, which is not always the case. You can calculate resistivity using the following equation
\[\dfrac{E}{J} = ρ\]
As you already read, ρ is the symbol for resistivity. \(E\) is the electric field and has units of Volts per meter (V/m). J is the current density and has units of amps per meter squared (A/m2). The electric field is calculated by dividing the Voltage by the length, l, that voltage is applied.
\[E=\dfrac{V}{l}\]
The current density is calculated by the equation below
\[J=\dfrac{I}{A}\]
I is the current and is divided by the cross sectional area, A, over which the current flows.
Resistivity vs Resistance
Resistivity and resistance are two different things. Resistivity does not depend on size or shape. Resistance, however, does. You can calculate resistance with the equation below.
\[ R=\dfrac{V}{I} \]
R refers to resistance and is measured in Ω. \(V\) is the voltage and is measured in volts. I measures the current and its unit is amps (A).
References Levy, Peter M., and Shufeng Zhang. "Electrical Conductivity of Magnetic Multilayered Structures." Physical Review Letters65.13 (1990): 1643-646. Print. Problems What is the current density of a material with a resistivity of 12.Ωm and an electric field of 64.V/m? If the voltage of 6V is passed through a substance with a radius of 2m and a length of 3m, what is the electric field? What is the electric field of a material when the current is equal to 25A, the resistance is measured to be 78Ω, the current density equals 24A/m2, and the length the current flows is 100m? A material has a voltage of 150V and width of 24m. The material also has a current of 62A and travels a distance of 5m. What is the conductivity? A metal originally has an electron colliding with every fifth atom and increases from a temperature of 6K to 100K. A semiconductor originally has an electron colliding with every fifth atom and increases from a temperature of 6K to 100K. What material will have a greater resistivity? Why? Answers to Problems:
1. E/J = ρ ---> J=E/ρ = 64V/m /12Ωm = 5.33A/m
2
2. E=V/l = 6V/3m = 2V/m
3. E=V/l
V=IR ---> E=IR/l = 25A x 78Ω/100m = 19.5V/m
4. E/J = ρ
E=V/l
J=I/A ---> ρ=(V/l)/(I/A) = (150V/5m)/(62A/(24m x 5m) = 58Ωm
ρ = 1/σ ---> 1/ρ = σ = 1/58Ωm
5. The material that has the greatest resistivity is the metal because as temperature increases metals are more likely to increase in resistivity and semiconductors usually decrease in resistivity as temperature increase.
Contributors Michael Ford (UCD) and Alexandra Christman (UCD)
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The Lagrangian formulation of mechanics will be useful later when we study the Feynman path integral. For our purposes now, the Lagrangian formulation is an important springboard from which to develop another useful formulation of classical mechanics known as the
Hamiltonian formulation. The Hamiltonian of a system is defined to be the sum of the kinetic and potential energies expressed as a function of positions and their conjugate momenta. What are conjugate momenta?
Recall from elementary physics that momentum of a particle, \(P_i\), is defined in terms of its velocity \(\dot {r}_i\) by
\[ p_i = m_i \dot {r} _i\]
In fact, the more general definition of conjugate momentum, valid for any set of coordinates, is given in terms of the
Lagrangian:
\[ p_i = \frac {\partial L}{\partial \dot {r} _i } \]
Note that these two definitions are equivalent for Cartesian variables. In terms of Cartesian momenta, the kinetic energy is given by
\[ K = \sum _{i=1}^N \frac {P^2_i}{2 m_i} \]
Then, the Hamiltonian, which is defined to be the sum, \(K + U \), expressed as a function of positions and momenta, will be given by
\[ H (p, r) = \sum _{i=1}^N \frac {P^2_i}{2m_i} + U ( r_1, \cdots , r_N) = H (p, r)\]
where \(p \equiv p_1, \cdots , p_N \) . In terms of the Hamiltonian, the equations of motion of a system are given by
Hamilton's equations:
\[ \dot {r} _i = \frac {\partial H}{\partial p_i} \dot {p}_i = - \frac {\partial H}{\partial r_i} \]
The solution of Hamilton's equations of motion will yield a trajectory in terms of positions and momenta as functions of time. Again, Hamilton's equations can be easily shown to be equivalent to Newton's equations, and, like the Lagrangian formulation, Hamilton's equations can be used to determine the equations of motion of a system in any set of coordinates.
The Hamiltonian and Lagrangian formulations possess an interesting connection. The Hamiltonian can be directly obtained from the Lagrangian by a transformation known as a
Legendre transform. We will say more about Legendre transforms in a later lecture. For now, note that the connection is given by
\[ H (p, r) = \sum _{i=1}^N p_i \cdot \dot {r}_i - L (r , \dot {r} ) \]
which, when the fact that \(\dot {r}_i = \frac {p_i}{m_i} \) is used, becomes
\[H (p, r)= \sum _{i=1}^N p_i \cdot \frac {p_i}{m_i} - \sum _{i=1}^N \frac {1}{2} m_i \left (\frac {p_i}{m_i} \right )^2 + U (r_1, \cdots , r_N )\]
\[= \sum _{i=1}^N \frac {P_i^2}{2m_i} + U(r_1, \cdots , r_N ) \]
Because a system described by conservative forces conserves the total energy, it follows that Hamilton's equations of motion conserve the total Hamiltonian. Hamilton's equations of motion conserve the Hamiltonian
\[ H (p (t), r (t) ) = H (p(0), r (0) ) = E \]
Proof \(H = \text {const} \Rightarrow \frac {dH}{dt} = 0 \)
\[\frac {dH}{dt}= \sum _{i=1}^N \left ( \frac {\partial H}{\partial r_i} \cdot \dot {r} _i + \frac {\partial H}{\partial p_i} \cdot \dot {p} _i \right ) \]
\[=\sum _{i=1}^N \left ( \frac {\partial H}{\partial r_i} \cdot \frac {\partial H}{\partial p_i} - \frac {\partial H}{\partial p_i} \cdot \frac {\partial H}{\partial r_i} \right ) = 0 \]
QED. This, then, provides another expression of the law of conservation of energy.
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Mumford’s drawing has a clear emphasis on the vertical direction. The set of all vertical lines corresponds to taking the fibers of the natural ‘structural morphism’ : $\pi~:~\mathbf{spec}(\mathbb{Z}[t]) \rightarrow \mathbf{spec}(\mathbb{Z}) $ coming from the inclusion $\mathbb{Z} \subset \mathbb{Z}[t] $. That is, we consider the intersection $P \cap \mathbb{Z} $ of a prime ideal $P \subset \mathbb{Z}[t] $ with the subring of constants.
Two options arise : either $P \cap \mathbb{Z} \not= 0 $, in which case the intersection is a principal prime ideal $~(p) $ for some prime number $p $ (and hence $P $ itself is bigger or equal to $p\mathbb{Z}[t] $ whence its geometric object is contained in the vertical line $\mathbb{V}((p)) $, the fiber $\pi^{-1}((p)) $ of the structural morphism over $~(p) $), or, the intersection $P \cap \mathbb{Z}[t] = 0 $ reduces to the zero ideal (in which case the extended prime ideal $P \mathbb{Q}[x] = (q(x)) $ is a principal ideal of the rational polynomial algebra $\mathbb{Q}[x] $, and hence the geometric object corresponding to $P $ is a horizontal curve in Mumford’s drawing, or is the whole arithmetic plane itself if $P=0 $).
Because we know already that any ‘point’ in Mumford’s drawing corresponds to a maximal ideal of the form $\mathfrak{m}=(p,f(x)) $ (see last time), we see that every point lies on precisely one of the set of all vertical coordinate axes corresponding to the prime numbers ${~\mathbb{V}((p)) = \mathbf{spec}(\mathbb{F}_p[x]) = \pi^{-1}((p))~} $. In particular, two different vertical lines do not intersect (or, in ringtheoretic lingo, the ‘vertical’ prime ideals $p\mathbb{Z}[x] $ and $q\mathbb{Z}[x] $ are comaximal for different prime numbers $p \not= q $).
That is, the structural morphism is a projection onto the “arithmetic axis” (which is $\mathbf{spec}(\mathbb{Z}) $) and we get the above picture. The extra vertical line to the right of the picture is there because in arithmetic geometry it is customary to include also the archimedean valuations and hence to consider the ‘compactification’ of the arithmetic axis $\mathbf{spec}(\mathbb{Z}) $ which is $\overline{\mathbf{spec}(\mathbb{Z})} = \mathbf{spec}(\mathbb{Z}) \cup { v_{\mathbb{R}} } $.
Yuri I. Manin is advocating for years the point that we should take the terminology ‘arithmetic surface’ for $\mathbf{spec}(\mathbb{Z}[x]) $ a lot more seriously. That is, there ought to be, apart from the projection onto the ‘z-axis’ (that is, the arithmetic axis $\mathbf{spec}(\mathbb{Z}) $) also a projection onto the ‘x-axis’ which he calls the ‘geometric axis’.
But then, what are the ‘points’ of this geometric axis and what are their fibers under this second projection?
We have seen above that the vertical coordinate line over the prime number $~(p) $ coincides with $\mathbf{spec}(\mathbb{F}_p[x]) $, the affine line over the finite field $\mathbb{F}_p $. But all of these different lines, for varying primes $p $, should project down onto the same geometric axis. Manin’s idea was to take therefore as the geometric axis the affine line $\mathbf{spec}(\mathbb{F}_1[x]) $, over the virtual field with one element, which should be thought of as being the limit of the finite fields $\mathbb{F}_p $ when $p $ goes to one!
How many points does $\mathbf{spec}(\mathbb{F}_1[x]) $ have? Over a virtual object one can postulate whatever one wants and hope for an a posteriori explanation. $\mathbb{F}_1 $-gurus tell us that there should be exactly one point of size n on the affine line over $\mathbb{F}_1 $, corresponding to the unique degree n field extension $\mathbb{F}_{1^n} $. However, it is difficult to explain this from the limiting perspective…
Over a genuine finite field $\mathbb{F}_p $, the number of points of thickness $n $ (that is, those for which the residue field is isomorphic to the degree n extension $\mathbb{F}_{p^n} $) is equal to the number of monic irreducible polynomials of degree n over $\mathbb{F}_p $. This number is known to be $\frac{1}{n} \sum_{d | n} \mu(\frac{n}{d}) p^d $ where $\mu(k) $ is the Moebius function. But then, the limiting number should be $\frac{1}{n} \sum_{d | n} \mu(\frac{n}{d}) = \delta_{n1} $, that is, there can only be one point of size one…
Alternatively, one might consider the zeta function counting the number $N_n $ of ideals having a quotient consisting of precisely $p^n $ elements. Then, we have for genuine finite fields $\mathbb{F}_p $ that $\zeta(\mathbb{F}_p[x]) = \sum_{n=0}^{\infty} N_n t^n = 1 + p t + p^2 t^2 + p^3 t^3 + \ldots $, whence in the limit it should become
$1+t+t^2 +t^3 + \ldots $ and there is exactly one ideal in $\mathbb{F}_1[x] $ having a quotient of cardinality n and one argues that this unique quotient should be the unique point with residue field $\mathbb{F}_{1^n} $ (though it might make more sense to view this as the unique n-fold extension of the unique size-one point $\mathbb{F}_1 $ corresponding to the quotient $\mathbb{F}_1[x]/(x^n) $…)
A perhaps more convincing reasoning goes as follows. If $\overline{\mathbb{F}_p} $ is an algebraic closure of the finite field $\mathbb{F}_p $, then the points of the affine line over $\overline{\mathbb{F}_p} $ are in one-to-one correspondence with the maximal ideals of $\overline{\mathbb{F}_p}[x] $ which are all of the form $~(x-\lambda) $ for $\lambda \in \overline{\mathbb{F}_p} $. Hence, we get the points of the affine line over the basefield $\mathbb{F}_p $ as the orbits of points over the algebraic closure under the action of the Galois group $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p) $.
‘Common wisdom’ has it that one should identify the algebraic closure of the field with one element $\overline{\mathbb{F}_{1}} $ with the group of all roots of unity $\mathbb{\mu}_{\infty} $ and the corresponding Galois group $Gal(\overline{\mathbb{F}_{1}}/\mathbb{F}_1) $ as being generated by the power-maps $\lambda \rightarrow \lambda^n $ on the roots of unity. But then there is exactly one orbit of length n given by the n-th roots of unity $\mathbb{\mu}_n $, so there should be exactly one point of thickness n in $\mathbf{spec}(\mathbb{F}_1[x]) $ and we should then identity the corresponding residue field as $\mathbb{F}_{1^n} = \mathbb{\mu}_n $.
Whatever convinces you, let us assume that we can identify the non-generic points of $\mathbf{spec}(\mathbb{F}_1[x]) $ with the set of positive natural numbers ${ 1,2,3,\ldots } $ with $n $ denoting the unique size n point with residue field $\mathbb{F}_{1^n} $. Then, what are the fibers of the projection onto the geometric axis $\phi~:~\mathbf{spec}(\mathbb{Z}[x]) \rightarrow \mathbf{spec}(\mathbb{F}_1[x]) = { 1,2,3,\ldots } $?
These fibers should correspond to ‘horizontal’ principal prime ideals of $\mathbb{Z}[x] $. Manin proposes to consider $\phi^{-1}(n) = \mathbb{V}((\Phi_n(x))) $ where $\Phi_n(x) $ is the n-th cyclotomic polynomial. The nice thing about this proposal is that all closed points of $\mathbf{spec}(\mathbb{Z}[x]) $ lie on one of these fibers!
Indeed, the residue field at such a point (corresponding to a maximal ideal $\mathfrak{m}=(p,f(x)) $) is the finite field $\mathbb{F}_{p^n} $ and as all its elements are either zero or an $p^n-1 $-th root of unity, it does lie on the curve determined by $\Phi_{p^n-1}(x) $.
As a consequence, the localization $\mathbb{Z}[x]_{cycl} $ of the integral polynomial ring $\mathbb{Z}[x] $ at the multiplicative system generated by all cyclotomic polynomials is a principal ideal domain (as all height two primes evaporate in the localization), and, the fiber over the generic point of $\mathbf{spec}(\mathbb{F}_1[x]) $ is $\mathbf{spec}(\mathbb{Z}[x]_{cycl}) $, which should be compared to the fact that the fiber of the generic point in the projection onto the arithmetic axis is $\mathbf{spec}(\mathbb{Q}[x]) $ and $\mathbb{Q}[x] $ is the localization of $\mathbb{Z}[x] $ at the multiplicative system generated by all prime numbers).
Hence, both the vertical coordinate lines and the horizontal ‘lines’ contain all closed points of the arithmetic plane. Further, any such closed point $\mathfrak{m}=(p,f(x)) $ lies on the intersection of a vertical line $\mathbb{V}((p)) $ and a horizontal one $\mathbb{V}((\Phi_{p^n-1}(x))) $ (if $deg(f(x))=n $).
That is, these horizontal and vertical lines form a coordinate system, at least for the closed points of $\mathbf{spec}(\mathbb{Z}[x]) $.
Still, there is a noticeable difference between the two sets of coordinate lines. The vertical lines do not intersect meaning that $p\mathbb{Z}[x]+q\mathbb{Z}[x]=\mathbb{Z}[x] $ for different prime numbers p and q. However, in general the principal prime ideals corresponding to the horizontal lines $~(\Phi_n(x)) $ and $~(\Phi_m(x)) $ are not comaximal when $n \not= m $, that is, these ‘lines’ may have points in common! This will lead to an exotic new topology on the roots of unity… (to be continued).
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Distributions Introduction
This section will begin to formalize the connection between random variables, probability density functions, and population parameters. We generally use language like the random variable $X$ follows a named distribution, which has a probability density function defined by, possibly many, parameters. Hence, the word distribution in some sense is just the name the binds random variables, probability density functions, and parameters together.
Warm Up
Before we look at two common named population parameters, let’s introduce a few new words that we’ll use throughout this section.
support: The set of values a random variable might assume, and equally the set of values a random variable’s probability density function is defined over. For example, the support of $X \sim \text{Uniform}(1, 6)$ is the integers from $1$ to $6$ inclusive. expected value: A population-level measure of center for a random variable. For example, $3.5$ for a fair die. variance: A population-level measure of variability for a random variable. standard deviation: The square root of the variance. Random Variables
A random variable is a function from a set of all possible outcomes, named the sample space, to exactly one real number. We often assume that random variables follow named distributions, e.g. $Y \sim \text{Uniform}(a, b)$ where $a < b$, or $X \sim \text{Bernoulli}(p)$ where $p \in [0, 1]$. Named distributions are common because they often abstractly represent processes in the world worth measuring. Based on the outcome of the process of interest, we calculate probabilities for a random variable that follows a specific distribution.
The Uniform distribution represents well rolling die. Much of the probabilities surrounding gambling are found by assuming random variables follow various Uniform distributions. Ignoring payouts, roulette is essentially a random variable $X \sim \text{Uniform}(1, 36)$.
The Bernoulli distirbution represents well any process that has two mutually exclusive outcomes with a fixed probability of “success.” Anything from unfair coins to the outcomes of elections are modeled with Bernoulli random variables.
These are not the only random variables, nor are random variables restricted to countable outcomes. Discrete random variables are restrictued to countable outcomes and continous random variables are the extension to uncountable outcomes. Discrete random variables take on non-negative mass or probability at single points in the support of the random variable, and thus have probability mass functions. On the other hand, continuous random variables have probability density functions, since zero mass occurs at distinct points in the support of the random variable. These lecture notes will only use the name probability density functions, even when referring to discrete random variables.
Before providing a long list of some other common named distributions, we will discuss the mean and variance of a random variable. These quantities describe a measure of center and a measure of spread of random variables. Recall, statistics uses data to estimate population parameters. The mean and variance of a random variable are two of the more commonly estimated quantities that describe a population. With a data set in hand, the sample mean (add up all the data divide by the number of data points) is an approximation of the mean of a random variable. With a data set in hand, the measure of spread called the variance is an approximation of the variance of a random variable.
import numpy as npimport pandas as pdimport bplot as bpbp.LaTeX()bp.dpi(300)
Mean of a Random Variable
Think back to our discrete random variable that represented rolling a single fair die, $X \sim \text{Uniform}(1, 6)$. We formalized the mathematical notation $P(X \in {2,4,6}) = 1/2$ by imagining rolling the same fair die an infinite number of times and dividing the number of times either $2, 4$, or $6$ turns up by the total number of rolls. Next, we will formalize, in a similar notion, the idea of the mean of a random variable.
The expected value describes a measure of center of a random variable. This is related to but not exactly, the same thing as, the sample mean where you add up all the numbers and divide by how every many numbers there are. The expected value does not describe data. The expected value instead describes a measure of center of the probaility density function for a random variable.
For the discrete random variable $X \sim \text{Uniform}(1, 6)$ the probability density function is displayed below. More generally, as uniform implies sameness, mathematically the probability density function is the same for all arguments
for $x \in \{a, a+1, \ldots, b-1, b\}$. Notice that the random variable is only defined for integer values between $a$ and $b$ inclusive. These values make up the
support of the random variable. Think of the support as the values for which the probability density function is positive.
x = np.arange(1, 7)fx = 1 / (6 - 1 + 1)df = pd.DataFrame({'x': x, 'f': fx})bp.point(df['x'], df['f'])bp.labels(x='$x$', y='uniform$(x|1,6)$', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x11f3db080>
Example
Since population mean describes a measure of center, and the probability density function takes on the same value $1/6$ at each value in the support ${1, 2, 3, 4, 5, 6}$, the expected value must be the value in the middle of the support, namely $3.5$. Formally, we read $\mathbb{E}(X) = 3.5$ as the
expected value of the random variable $X$ is $3.5$. As the sample mean is to data, the expected value is to a random variable.
More formally, the expected value of $X \sim \text{Uniform}(a, b)$ is
In R, we can apply this formula to $X \sim \text{Uniform}(1,6)$,
a = 1; b = 6x = np.arange(1, 6 + 1)fx = 1 / (b - a + 1)sum(x * fx) # E(X)
3.5
Notice that the we are simply weighting each value in the support of the random variable by the probability density function evaluated at each value in the support. The expected value is to be thought of as the value you’d get by taking the sample mean of the outcomes produced by infinitely rolling a fair die. Let’s approximate this process in R,
N = int(1e3)die = np.random.choice(x, size=N)flips = np.arange(1, N + 1)df = pd.DataFrame({'flip': flips, 'm': np.cumsum(die)/flips})bp.line(df['flip'], df['m'])bp.line_h(y=3.5, color='black')bp.labels(x='flip', y='$\hat{E}(X)$', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x12152bba8>
DEFINITION. Let $X \sim F$, where $F$ is the name of a distribution. The expected value of a random variable is
The fancy integral here is just to remind you that for discrete random variables, the integral becomes a sum, as above, and for continuous random variables the integral stays. In both cases, the sum/integral ranges over the support of the random variable and the summand/integrand is the product of $x$ and the probability density function.
Variance and Standard Deviation of a Random Variable
Where the mean is a measure of center of a random variable, the variance is a measure of spread. Specifically, the variance measures squared distance from the mean, again weighted by the probability density function.
DEFINITION. Let $X \sim F$, where $F$ is the name of a distribution function with mean $\mu = \mathbb{E}(X)$, the variance of $X$ is DEFINITION. Let $X \sim F$, where $F$ is the name of a distribution function with variance $\mathbb{V}(X)$, the standard deviation of $X$ is
The standard deviation is another measure of spread, like the variance, but the standard deviation is in the same units as the mean.
Example
In Python, we can apply this formula to $X \sim \text{Uniform}(1,6)$ by first calculating the expected value $\mathbb{E}(X)$,
a = 1; b = 6x = np.arange(1, 6 + 1)fx = 1 / (b - a + 1)m = sum(x * fx)v = sum(np.power(x - m, 2) * fx)np.sqrt(v)
1.707825127659933
Example
Assume $X \sim \text{Bernoulli}(p)$. If you work through the math for the variance, then you end up at an equation that we can almost make sense of.
Let’s plot this.
p = np.linspace(0, 1, 101)fp = p * (1 - p)bp.curve(p, fp)bp.labels(x='$p$', y='$f(p)$', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x1216598d0>
We can infer that the variance is maximized at p = 0.5; the math isn’t too difficult, try. This says that for a coin has the most variability (e.g. bounces back and forth between heads and tails the most) when it is a fair coin. This makes sense. An extremely biased coin, say $p = 0.001$ will very often come up tails, and rarely fall on heads.
It takes some time to understand the fancy integral notation above. There are a few ideas that you should keep in mind when working with the definitions of $\mathbb{E}(X)$ and $\mathbb{V}(X)$:
if the random variable $X$ is discrete then the integral is really a sum, the summand/integrand is defined as the product of the function that shows up in front of $dF(x)$ times the probability density function of $X$, and the summand/integrand is evaluated at each value in the support of the random variable $X$. Common Distributions
There are a few moving pieces to keep in mind while scanning or referencing the list of distributions below.
Each distribution presented is a generally accepted statistical abstraction of a common process in the world. Different sets of the real numbers $\mathbb{R}$ will support different real world processes. For instance, when measuring time until an event (Exponential), only positively valued real numbers are relevant. When measuring counting events (Poisson), only positive integers are relevant. This idea is known as the
supportof the probability density function. The support describes the set of possible values a random variable might take on.
The variables following the pipe in the function signature are generally referred to as parameters. For instance, $a$ and $b$ of the uniform distribution define the support of the random variable. For a discrete uniform random variable meant to describe the process of rolling a die, $a = 1$ and $b = 6$. These parameters constitue some, but not all, of the population parameters that are estimated from a sample. The parameter $\beta$ in the Exponential distribution describes the rate at which events occur in time or space.
These lecture notes focus on the likelihood function as a means to estimate population parameters from a sample. This method will work simply for any probability density function who’s support does not depend on the parameters. However, the maximum likelihood estimator for the uniform distribution will take more careful thought than it will direct application of calculus.
Discrete Distributions Uniform
For $X \sim \text{Uniform}(a, b)$ where $a < b$, $X$ has probability density function
for $x \in \{a, a+1, \ldots, b - 1, b \}$.
Bernoulli
For $X \sim \text{Bernoulli}(p)$ where $p \in [0, 1]$, $X$ has probability density function
for $x \in \{0, 1\}$.
Geometric
For $X \sim \text{Geometric}(p)$ where $p \in [0, 1]$, $X$ has probability density function
for $x \in \{1, 2, \ldots \}$.
Binomial
For $X \sim \text{Binomial}(K, p)$ where $K \in \mathbb{N}
\{0\}$ and $p \in [0, 1]$, $X$ has probability density funciton
for $x \in \{0, 1, \ldots, K \}$.
Poisson
For $X \sim \text{Poisson}(\lambda)$ where $\lambda > 0$, $X$ has probability density function
for $x \in \mathbb{N}$.
Continuous Distributions Uniform
For $X \sim \text{Uniform}(a, b)$ where $a < b$, $X$ has probability density function
for $x \in [a, b]$.
Beta
For $X \sim \text{Beta}(\alpha, \beta)$ where $\alpha > 0$ and $\beta > 0$, $X$ has probability density function
for $x \in [0, 1]$.
Exponential
For $X \sim \text{Exponential}(\beta)$ where $\beta > 0$, $X$ has probability density function
for $x \geq 0$.
Gamma
For $X \sim \text{Gamma}(\alpha, \beta)$ where $\alpha > 0$ and $\beta > 0$, $X$ has probability density function
for $x \geq 0$.
Normal
For $X \sim \text{Normal}(\mu, \sigma)$ where $\mu \in \mathbb{R}$ and $\sigma > 0$, $X$ has probability density function
for $x \in \mathbb{R}$.
Student-t
For $X \sim \text{StudentT}(\nu, \mu, \sigma)$ where $\nu > 0$, $\mu \in \mathbb{R}$, and $\sigma > 0$, $X$ has probability density function
for $x \in \mathbb{R}$.
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Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.Let's denote the width of a sample by $h$ where $$h\rightarrow0$$Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can a...
@Ultradark You can try doing a finite difference to get rid of the sum and then compare term by term. Otherwise I am terrible at anything to do with primes that I don't know the identities of $\pi (n)$ well
@Silent No, take for example the prime 3. 2 is not a residue mod 3, so there is no $x\in\mathbb{Z}$ such that $x^2-2\equiv 0$ mod $3$.
However, you have two cases to consider. The first where $\binom{2}{p}=-1$ and $\binom{3}{p}=-1$ (In which case what does $\binom{6}{p}$ equal?) and the case where one or the other of $\binom{2}{p}$ and $\binom{3}{p}$ equals 1.
Also, probably something useful for congruence, if you didn't already know: If $a_1\equiv b_1\text{mod}(p)$ and $a_2\equiv b_2\text{mod}(p)$, then $a_1a_2\equiv b_1b_2\text{mod}(p)$
Is there any book or article that explains the motivations of the definitions of group, ring , field, ideal etc. of abstract algebra and/or gives a geometric or visual representation to Galois theory ?
Jacques Charles François Sturm ForMemRS (29 September 1803 – 15 December 1855) was a French mathematician.== Life and work ==Sturm was born in Geneva (then part of France) in 1803. The family of his father, Jean-Henri Sturm, had emigrated from Strasbourg around 1760 - about 50 years before Charles-François's birth. His mother's name was Jeanne-Louise-Henriette Gremay. In 1818, he started to follow the lectures of the academy of Geneva. In 1819, the death of his father forced Sturm to give lessons to children of the rich in order to support his own family. In 1823, he became tutor to the son...
I spent my career working with tensors. You have to be careful about defining multilinearity, domain, range, etc. Typically, tensors of type $(k,\ell)$ involve a fixed vector space, not so many letters varying.
UGA definitely grants a number of masters to people wanting only that (and sometimes admitted only for that). You people at fancy places think that every university is like Chicago, MIT, and Princeton.
hi there, I need to linearize nonlinear system about a fixed point. I've computed the jacobain matrix but one of the elements of this matrix is undefined at the fixed point. What is a better approach to solve this issue? The element is (24*x_2 + 5cos(x_1)*x_2)/abs(x_2). The fixed point is x_1=0, x_2=0
Consider the following integral: $\int 1/4*(1/(1+(u/2)^2)))dx$ Why does it matter if we put the constant 1/4 behind the integral versus keeping it inside? The solution is $1/2*\arctan{(u/2)}$. Or am I overseeing something?
*it should be du instead of dx in the integral
**and the solution is missing a constant C of course
Is there a standard way to divide radicals by polynomials? Stuff like $\frac{\sqrt a}{1 + b^2}$?
My expression happens to be in a form I can normalize to that, just the radicand happens to be a lot more complicated. In my case, I'm trying to figure out how to best simplify $\frac{x}{\sqrt{1 + x^2}}$, and so far, I've gotten to $\frac{x \sqrt{1+x^2}}{1+x^2}$, and it's pretty obvious you can move the $x$ inside the radical.
My hope is that I can somehow remove the polynomial from the bottom entirely, so I can then multiply the whole thing by a square root of another algebraic fraction.
Complicated, I know, but this is me trying to see if I can skip calculating Euclidean distance twice going from atan2 to something in terms of asin for a thing I'm working on.
"... and it's pretty obvious you can move the $x$ inside the radical" To clarify this in advance, I didn't mean literally move it verbatim, but via $x \sqrt{y} = \text{sgn}(x) \sqrt{x^2 y}$. (Hopefully, this was obvious, but I don't want to confuse people on what I meant.)
Ignore my question. I'm coming of the realization it's just not working how I would've hoped, so I'll just go with what I had before.
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The question is long because of the demonstrations I give, but the problem is simple, so bear with me for a minute.
I am trying to derive the dispersion relation of a semi-infinite system using Euler-Lagrange equations, and I started with the simplest case of a semi-infinite string. However, the result, as I show below, is $\omega^2 = -c_0^2 \gamma^2$ instead of the correct $\omega^2 = c_0^2 \gamma^2$.
My question is: can I use Euler-Lagrange equations to derive the dispersion relation of infinite and semi-infinite vibrating systems? I've seen it being used only for finite systems to calculate their natural frequencies. If the method is correct for semi-infinite and infinite systems, what did I do wrong in the string case?
I can't think of a reason why I shouldn't use it, but since it doesn't work for a string, how could it possibly be correct for more complicated media and geometries? And yet, it does work for a beam in pure bending, and this makes me think that either I did something wrong in the string case, or that it never works and the correct result for the beam is just a coincidence.
Considering transverse displacement $y = y(x,t) = A(t)X(x)$ of a point in a taut string, the kinetic energy $K$ of the string is $K = \frac{1}{2}\int{\rho\dot y^2}\text dx$, where $\rho$ is the string's linear mass density. As for the potential energy $U$, Graff
[1, p.20] considers it as the work done by the tension $T$ as the string changes its length from $l$ to $l'$ when it vibrates. Therefore,
$$ l' = \int \text d s = \int \sqrt{1 + y_x^2} \, \text dx \approx \int1 + \frac 1 2 y_x^2 \, \text dx = l + \frac 1 2 \int y_x^2 \,\text dx $$ $$ U = T\Delta l = T(l' - l) = \frac T 2 \int y_x^2 \, \text dx $$
Making the Lagrangian $L = K - U$, and taking $A(t)$ as the generalized coordinate, the Euler-Lagrange equation becomes
$$ \frac{\text d}{\text dt}\left(\frac {\partial L}{\partial \dot q}\right) = \frac {\partial L}{\partial q} $$ $$ \frac{\text d}{\text dt}\left(\frac {\partial K}{\partial \dot A}\right) = -\frac {\partial U}{\partial A} $$ $$ \rho \ddot A(t) \int X(x) \, \text dx = - A(t) T \int (X'(x))^2 \, \text dx \tag{1} $$
Now, differentiating w.r.t $x$, and making $y$ an harmonic vibration, i.e. $\ddot A(t) = -\omega^2 A(t)$, $X(x) = \exp(-\text i \gamma x)$, and $c_0^2 = T/\rho$, the relationship between the frequency $\omega$ and the wavenumber $\gamma$ then comes out of $(1)$ as
$$ \omega^2 = -c_0^2 \gamma^2, $$ which is incorrect and can be verified by substituting $y(x,t)$ into the wave equation and making the same assumptions as above.
Now, if I try a Euler-Bernoulli beam in pure bending, taking the potential energy as $$ U = \frac 1 2 \int \frac{M^2}{EI} \, \text dx = \frac 1 2 \int \frac{(-EI y_{xx})^2}{EI} \, \text dx = \frac 1 2 \int EI y_{xx}^2 \, \text dx $$ and supposing again that $y = A(t) \exp(-\text i \gamma x)$, the dispersion relation is $$ \omega^2 = a^2 \gamma^4, \ \ \ a^2 = EI/\rho $$ which is correct, with $E$ and $I$ being the beam's Young modulus and cross section's second moment of area, respectively.
[1] Karl F. Graff. "Wave motion in elastic solids". 1st ed. Oxford Press, 1975.
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Hi, Can someone provide me some self reading material for Condensed matter theory? I've done QFT previously for which I could happily read Peskin supplemented with David Tong. Can you please suggest some references along those lines? Thanks
@skullpatrol The second one was in my MSc and covered considerably less than my first and (I felt) didn't do it in any particularly great way, so distinctly average. The third was pretty decent - I liked the way he did things and was essentially a more mathematically detailed version of the first :)
2. A weird particle or state that is made of a superposition of a torus region with clockwise momentum and anticlockwise momentum, resulting in one that has no momentum along the major circumference of the torus but still nonzero momentum in directions that are not pointing along the torus
Same thought as you, however I think the major challenge of such simulator is the computational cost. GR calculations with its highly nonlinear nature, might be more costy than a computation of a protein.
However I can see some ways approaching it. Recall how Slereah was building some kind of spaceitme database, that could be the first step. Next, one might be looking for machine learning techniques to help on the simulation by using the classifications of spacetimes as machines are known to perform very well on sign problems as a recent paper has shown
Since GR equations are ultimately a system of 10 nonlinear PDEs, it might be possible the solution strategy has some relation with the class of spacetime that is under consideration, thus that might help heavily reduce the parameters need to consider to simulate them
I just mean this: The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
@ooolb Even if that is really possible (I always can talk about things in a non joking perspective), the issue is that 1) Unlike other people, I cannot incubate my dreams for a certain topic due to Mechanism 1 (consicous desires have reduced probability of appearing in dreams), and 2) For 6 years, my dream still yet to show any sign of revisiting the exact same idea, and there are no known instance of either sequel dreams nor recurrence dreams
@0celo7 I felt this aspect can be helped by machine learning. You can train a neural network with some PDEs of a known class with some known constraints, and let it figure out the best solution for some new PDE after say training it on 1000 different PDEs
Actually that makes me wonder, are the space of all coordinate choices more than all possible moves of Go?
enumaris: From what I understood from the dream, the warp drive showed here may be some variation of the alcuberrie metric with a global topology that has 4 holes in it whereas the original alcuberrie drive, if I recall, don't have holes
orbit stabilizer: h bar is my home chat, because this is the first SE chat I joined. Maths chat is the 2nd one I joined, followed by periodic table, biosphere, factory floor and many others
Btw, since gravity is nonlinear, do we expect if we have a region where spacetime is frame dragged in the clockwise direction being superimposed on a spacetime that is frame dragged in the anticlockwise direction will result in a spacetime with no frame drag? (one possible physical scenario that I can envision such can occur may be when two massive rotating objects with opposite angular velocity are on the course of merging)
Well. I'm a begginer in the study of General Relativity ok? My knowledge about the subject is based on books like Schutz, Hartle,Carroll and introductory papers. About quantum mechanics I have a poor knowledge yet.
So, what I meant about "Gravitational Double slit experiment" is: There's and gravitational analogue of the Double slit experiment, for gravitational waves?
@JackClerk the double slits experiment is just interference of two coherent sources, where we get the two sources from a single light beam using the two slits. But gravitational waves interact so weakly with matter that it's hard to see how we could screen a gravitational wave to get two coherent GW sources.
But if we could figure out a way to do it then yes GWs would interfere just like light wave.
Thank you @Secret and @JohnRennie . But for conclude the discussion, I want to put a "silly picture" here: Imagine a huge double slit plate in space close to a strong source of gravitational waves. Then like water waves, and light, we will see the pattern?
So, if the source (like a Black Hole binary) are sufficent away, then in the regions of destructive interference, space-time would have a flat geometry and then with we put a spherical object in this region the metric will become schwarzschild-like.
if**
Pardon, I just spend some naive-phylosophy time here with these discussions**
The situation was even more dire for Calculus and I managed!
This is a neat strategy I have found-revision becomes more bearable when I have The h Bar open on the side.
In all honesty, I actually prefer exam season! At all other times-as I have observed in this semester, at least-there is nothing exciting to do. This system of tortuous panic, followed by a reward is obviously very satisfying.
My opinion is that I need you kaumudi to decrease the probabilty of h bar having software system infrastructure conversations, which confuse me like hell and is why I take refugee in the maths chat a few weeks ago
(Not that I have questions to ask or anything; like I said, it is a little relieving to be with friends while I am panicked. I think it is possible to gauge how much of a social recluse I am from this, because I spend some of my free time hanging out with you lot, even though I am literally inside a hostel teeming with hundreds of my peers)
that's true. though back in high school ,regardless of code, our teacher taught us to always indent your code to allow easy reading and troubleshooting. We are also taught the 4 spacebar indentation convention
@JohnRennie I wish I can just tab because I am also lazy, but sometimes tab insert 4 spaces while other times it inserts 5-6 spaces, thus screwing up a block of if then conditions in my code, which is why I had no choice
I currently automate almost everything from job submission to data extraction, and later on, with the help of the machine learning group in my uni, we might be able to automate a GUI library search thingy
I can do all tasks related to my work without leaving the text editor (of course, such text editor is emacs). The only inconvenience is that some websites don't render in a optimal way (but most of the work-related ones do)
Hi to all. Does anyone know where I could write matlab code online(for free)? Apparently another one of my institutions great inspirations is to have a matlab-oriented computational physics course without having matlab on the universities pcs. Thanks.
@Kaumudi.H Hacky way: 1st thing is that $\psi\left(x, y, z, t\right) = \psi\left(x, y, t\right)$, so no propagation in $z$-direction. Now, in '$1$ unit' of time, it travels $\frac{\sqrt{3}}{2}$ units in the $y$-direction and $\frac{1}{2}$ units in the $x$-direction. Use this to form a triangle and you'll get the answer with simple trig :)
@Kaumudi.H Ah, it was okayish. It was mostly memory based. Each small question was of 10-15 marks. No idea what they expect me to write for questions like "Describe acoustic and optic phonons" for 15 marks!! I only wrote two small paragraphs...meh. I don't like this subject much :P (physical electronics). Hope to do better in the upcoming tests so that there isn't a huge effect on the gpa.
@Blue Ok, thanks. I found a way by connecting to the servers of the university( the program isn't installed on the pcs on the computer room, but if I connect to the server of the university- which means running remotely another environment, i found an older version of matlab). But thanks again.
@user685252 No; I am saying that it has no bearing on how good you actually are at the subject - it has no bearing on how good you are at applying knowledge; it doesn't test problem solving skills; it doesn't take into account that, if I'm sitting in the office having forgotten the difference between different types of matrix decomposition or something, I can just search the internet (or a textbook), so it doesn't say how good someone is at research in that subject;
it doesn't test how good you are at deriving anything - someone can write down a definition without any understanding, while someone who can derive it, but has forgotten it probably won't have time in an exam situation. In short, testing memory is not the same as testing understanding
If you really want to test someone's understanding, give them a few problems in that area that they've never seen before and give them a reasonable amount of time to do it, with access to textbooks etc.
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@Mathphile I found no prime of the form $$n^{n+1}+(n+1)^{n+2}$$ for $n>392$ yet and neither a reason why the expression cannot be prime for odd n, although there are far more even cases without a known factor than odd cases.
@TheSimpliFire That´s what I´m thinking about, I had some "vague feeling" that there must be some elementary proof, so I decided to find it, and then I found it, it is really "too elementary", but I like surprises, if they´re good.
It is in fact difficult, I did not understand all the details either. But the ECM-method is analogue to the p-1-method which works well, then there is a factor p such that p-1 is smooth (has only small prime factors)
Brocard's problem is a problem in mathematics that asks to find integer values of n and m for whichn!+1=m2,{\displaystyle n!+1=m^{2},}where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.== Brown numbers ==Pairs of the numbers (n, m) that solve Brocard's problem are called Brown numbers. There are only three known pairs of Brown numbers:(4,5), (5,11...
$\textbf{Corollary.}$ No solutions to Brocard's problem (with $n>10$) occur when $n$ that satisfies either \begin{equation}n!=[2\cdot 5^{2^k}-1\pmod{10^k}]^2-1\end{equation} or \begin{equation}n!=[2\cdot 16^{5^k}-1\pmod{10^k}]^2-1\end{equation} for a positive integer $k$. These are the OEIS sequences A224473 and A224474.
Proof: First, note that since $(10^k\pm1)^2-1\equiv((-1)^k\pm1)^2-1\equiv1\pm2(-1)^k\not\equiv0\pmod{11}$, $m\ne 10^k\pm1$ for $n>10$. If $k$ denotes the number of trailing zeros of $n!$, Legendre's formula implies that \begin{equation}k=\min\left\{\sum_{i=1}^\infty\left\lfloor\frac n{2^i}\right\rfloor,\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\right\}=\sum_{i=1}^\infty\left\lfloor\frac n{5^i}\right\rfloor\end{equation} where $\lfloor\cdot\rfloor$ denotes the floor function.
The upper limit can be replaced by $\lfloor\log_5n\rfloor$ since for $i>\lfloor\log_5n\rfloor$, $\left\lfloor\frac n{5^i}\right\rfloor=0$. An upper bound can be found using geometric series and the fact that $\lfloor x\rfloor\le x$: \begin{equation}k=\sum_{i=1}^{\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor\le\sum_{i=1}^{\lfloor\log_5n\rfloor}\frac n{5^i}=\frac n4\left(1-\frac1{5^{\lfloor\log_5n\rfloor}}\right)<\frac n4.\end{equation}
Thus $n!$ has $k$ zeroes for some $n\in(4k,\infty)$. Since $m=2\cdot5^{2^k}-1\pmod{10^k}$ and $2\cdot16^{5^k}-1\pmod{10^k}$ has at most $k$ digits, $m^2-1$ has only at most $2k$ digits by the conditions in the Corollary. The Corollary if $n!$ has more than $2k$ digits for $n>10$. From equation $(4)$, $n!$ has at least the same number of digits as $(4k)!$. Stirling's formula implies that \begin{equation}(4k)!>\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\end{equation}
Since the number of digits of an integer $t$ is $1+\lfloor\log t\rfloor$ where $\log$ denotes the logarithm in base $10$, the number of digits of $n!$ is at least \begin{equation}1+\left\lfloor\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)\right\rfloor\ge\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right).\end{equation}
Therefore it suffices to show that for $k\ge2$ (since $n>10$ and $k<n/4$), \begin{equation}\log\left(\frac{\sqrt{2\pi}\left(4k\right)^{4k+\frac{1}{2}}}{e^{4k}}\right)>2k\iff8\pi k\left(\frac{4k}e\right)^{8k}>10^{4k}\end{equation} which holds if and only if \begin{equation}\left(\frac{10}{\left(\frac{4k}e\right)}\right)^{4k}<8\pi k\iff k^2(8\pi k)^{\frac1{4k}}>\frac58e^2.\end{equation}
Now consider the function $f(x)=x^2(8\pi x)^{\frac1{4x}}$ over the domain $\Bbb R^+$, which is clearly positive there. Then after considerable algebra it is found that \begin{align*}f'(x)&=2x(8\pi x)^{\frac1{4x}}+\frac14(8\pi x)^{\frac1{4x}}(1-\ln(8\pi x))\\\implies f'(x)&=\frac{2f(x)}{x^2}\left(x-\frac18\ln(8\pi x)\right)>0\end{align*} for $x>0$ as $\min\{x-\frac18\ln(8\pi x)\}>0$ in the domain.
Thus $f$ is monotonically increasing in $(0,\infty)$, and since $2^2(8\pi\cdot2)^{\frac18}>\frac58e^2$, the inequality in equation $(8)$ holds. This means that the number of digits of $n!$ exceeds $2k$, proving the Corollary. $\square$
We get $n^n+3\equiv 0\pmod 4$ for odd $n$, so we can see from here that it is even (or, we could have used @TheSimpliFire's one-or-two-step method to derive this without any contradiction - which is better)
@TheSimpliFire Hey! with $4\pmod {10}$ and $0\pmod 4$ then this is the same as $10m_1+4$ and $4m_2$. If we set them equal to each other, we have that $5m_1=2(m_2-m_1)$ which means $m_1$ is even. We get $4\pmod {20}$ now :P
Yet again a conjecture!Motivated by Catalan's conjecture and a recent question of mine, I conjecture thatFor distinct, positive integers $a,b$, the only solution to this equation $$a^b-b^a=a+b\tag1$$ is $(a,b)=(2,5).$It is of anticipation that there will be much fewer solutions for incr...
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Definition:Relation Induced by Partition Definition
Let $S$ be a set.
Let $\mathcal R \subseteq S \times S$ be the relation defined as:
$\forall \tuple {x, y} \in S \times S: \tuple {x, y} \in \mathcal R \iff \exists T \in \Bbb S: \set {x, y} \subseteq T$ Then $\mathcal R$ is the (equivalence) relation induced by (the partition) $\Bbb S$. Also known as
Some sources refer to this as the
(equivalence) relation defined by (the partition) $\Bbb S$. Also see
It is proved in Relation Induced by Partition is Equivalence that:
$\mathcal R$ is unique $\mathcal R$ is an equivalence relation on $S$.
Hence $\Bbb S$ is the quotient set of $S$ by $\mathcal R$, that is:
$\Bbb S = S / \mathcal R$ Sources 1960: Paul R. Halmos: Naive Set Theory... (previous) ... (next): $\S 7$: Relations 1965: Seth Warner: Modern Algebra... (previous) ... (next): $\S 10$ 1977: Gary Chartrand: Introductory Graph Theory... (previous) ... (next): Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $16$
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Here is a way with aligned minipages:\documentclass[a4paper, 10pt, oneside]{memoir}% Page layout\setlrmarginsandblock{1.7cm}{8.5cm}{*}\setulmarginsandblock{1.7cm}{2.5cm}{*}\setmarginnotes{0.5cm}{\dimexpr(\stockwidth-\textwidth-4.4cm)}{1em}\checkandfixthelayout\chapterstyle{bianchi}\usepackage{titlesec}\usepackage{titletoc}\usepackage{hyperref}\...
Here are two ways, depending on whether you have fixed width cells or not:for a standard cell (column specifier r, l or c), you can use the makecell command,for a fixed width cell, use the m{some length} column typeA demo of both:\documentclass[12pt]{article}\usepackage[utf8]{inputenc}\usepackage{fourier}\usepackage{array}\usepackage{makecell}\...
You can use TABstacks. Shown here in 3 ways, depending on the desired equation-number vertical alignment.\documentclass{article}\usepackage{tabstackengine}\stackMath\setstackgap{L}{14pt}\begin{document}\begin{equation}\alignCenterstack{a+b+c+d+e+&f+g+h \\=& j + k + l + m +n\\=& j' + k' + l'+ m' +n'}\end{equation}\begin{...
Here are three possibilities. In the first, you alignthe = signs with another symbol of the first line, The second uses the optional argument of the \MoveEqLeft command from mathtools, and the third nests the aligned environment in a gathered environment (to fine-tune the placement of w= w.r.t. the first line, you can add to the latter some \hspace)....
You should rotate with the option origin=c, rather than adding spaces by hand. Adding \mathstrut will ensure correct alignment.Use an array for the alignments:\documentclass{article}\usepackage{amsmath,array,graphicx}\newcommand{\rotaterelation}[1]{\rotatebox[origin=c]{90}{$\mathstrut#1$}}\begin{document}\begin{equation}\begin{array}[t]{ @{} c *{...
Here I use TABstacks. I also kept the same vertical alignment with respect to the equation number as the OP's original query (but that can be easily changed).\documentclass{article}\usepackage{amsmath,amssymb,rotating,tabstackengine}\TABstackMath\TABbinary\begin{document}\newcommand{\verteq}{\rotatebox[origin=c]{90}{$\mkern1mu=$}}\newcommand{\...
Probably the following is close to what you want to achieve:\documentclass{article}\usepackage{caption}\usepackage{graphicx}\usepackage{pdflscape}\usepackage{adjustbox}\usepackage{xcolor}\begin{document}%\begin{landscape}\begin{figure}\begin{adjustbox}{minipage=0.45\linewidth,frame=1pt 5pt}\includegraphics[width=\linewidth]{example-image-a}\...
I didn't figure out why you define two different \formule commands. I estimate, than only one would suffice:\documentclass{article}\usepackage{mathtools,amssymb}\mathtoolsset{showonlyrefs}\usepackage{array}\newcommand{\formule}[2]{\par\medskip\noindent%\begin{tabular*}{\linewidth}{@{}>{\centering$\displaystyle}p{\dimexpr0.34\linewidth-2\...
I think the central problem with your equation is not that the column vector looks puny next to the matrices, but that you're using \displaystyle (and, implicitly) \limits inside the matrices. Speacking for myself, creating super-tall column vectors -- as you do in your screenshot and as is done in the first solution shown below -- does not good at all. In ...
Modify below of=entry to below=of entry, and change node distance=10mm to node distance=1mm. below of=.. is deprecated syntax, where the distance is calculated between node centers, while with below=of .. (requires the positioning library) the distance is calculated between node borders by default.I also removed the use of \tikzstyle, as that is considered ...
Like this?You only need to add (in your simple example) anchor=north to styles of nodes:\documentclass[10pt]{article}\usepackage[edges]{forest}% Document\begin{document}\begin{figure}[hbt!]\centering\begin{forest}for tree={ % style of tree nodesdraw, semithick, rounded corners,align = center,...
There's no need for tabularx here.The following uses \hfills to spread the content across the page in the same way tabularx does. In order to vertically align different sized images, you can capture the differing elements inside boxes (\bigbox and \smallbox below) from which you can extract their respective heights. This allows you to raise them into ...
See if the following solution is acceptable to you:\documentclass{report}\usepackage{tabularx}\usepackage[export]{adjustbox}\usepackage{subcaption}\usepackage{tikz}\usetikzlibrary{arrows.meta}\newcommand\bigrightArrow{\tikz[baseline=-3pt]{\draw[line width=12pt,-{Triangle[length=16pt, width=24pt]}, gray](0,0) -- ++ (32pt,0);}...
You need to capture the widths of the widest elements in each of your equations. Then you can use those widths to impose alignment between elements that aren't as wide.Below I use a slight modification to eqparbox via \eqmathbox[<tag>][<align>]{<math>} which stores the maximum width of each <tag>ged box with varying <math> ...
I can't reproduce the huge space. However, it's possible to move up the equation number by reducing the display's width.Also, you're misusing cases: the alignment point is for stating conditions, not for aligning the right-hand side of equations.This is solved by nesting aligned in cases.Next, split the overlong expression with the help of mathtools ...
Your equation is too wide, causing the number to be placed below, rather than adjacent to, the equation. You must strive to break the long (full-width) line into multiple, shorter lines.Also, as David noted, there is no need for the outer align, as you only have one row to align. I replaced with the equation environment.I propose doing this by way of ...
In your code you are using c as column definition which does not support line breaks. That means the rows height will not exceed one line and so vertical alignment is neglectable.In case of using fixed column widths with p{...} which supports line breaks this issue can be solved by using the array packe and m{...} for column definition.\documentclass[...
Using, appropriately the package nicematrix using the due spacings (\mkern), you can obtain a similar result.\documentclass{article}\usepackage{amssymb}\usepackage{nicematrix}\NiceMatrixOptions{code-for-last-row=\scriptstyle}\begin{document}Let $m\in\mathbb{N}^{\times}$. For $j=1,\ldots,m,$ define$e_j\mathrel{\mathop:}=\mkern-10mu \begin{pNiceArray}{...
As already mentioned in my previous comments, I'd split up the single table into five different tables. This will make the code much longer but (at least in my opinion) a lot more readable and easier to adjust.In the following example, I have also used the automated row numbering approach. I have also introduced a new, centered fixed width column type. In ...
One way is to use [anchor=base] and add a \vphantom{X} node on the left to get the correct vertical spacing and a \hphantom{\widetilde{X} to get the correct horizontal spacing:\node [anchor=base] at (0,0) {$\widetilde{X}$};\node [anchor=base] at (0,0) (input) {$\vphantom{X}\hphantom{\widetilde{X}}$};which yields:Note:I replaced the \tikzstyle ...
My proposal, as alternative to the best answers, is the use the package blkarray.\documentclass[12pt]{article}\usepackage{mathtools,blkarray}\begin{document}\[\begin{blockarray}{ccccccccc}\begin{block}{c(ccccccc)c}e_{j} \coloneqq \mkern-5mu& 0, & \ldots, & 0, & 1, & 0, & \ldots, & 0 & \mkern-5mu \in K^{m}\\\end{block}...
Here's a solution which employs the amssymb and mathtools packages.If you would like to push the (j) term a bit lower than in the picture shown above, simply change {1} to {1\mathstrut}.\documentclass{article}\usepackage{mathtools,amssymb}\begin{document}Let $m\in\mathbb{N}^{\times}$. For $j=1,\dots,m$ define\[e_j := (0,\dots,0,\underset{\mathclap{(...
Here is an example:\documentclass{article}\usepackage{amsmath}\begin{document}$E_j := \underset{(j)}{(0,\dots,0,1,0,\dots,0)}$\end{document}EDIT:For asymmetric case:e_j := (0,\dots,\underset{(j-1)}{0},\underset{(j)}{1},0,\dots,0)
For example:\documentclass{article}\usepackage[T1]{fontenc}\usepackage[utf8]{inputenc}\usepackage{stackengine}% these are from https://www.tug.org/TUGboat/tb22-4/tb72perlS.pdf\def\clap#1{\hbox to 0pt{\hss#1\hss}}\begin{document}$(0,\dots,0,\ensurestackMath{\stackunder{1}{\clap{$(i)$}}},0,\dots,0)$\end{document}(The \clap is a bit harsh; see ...
The solution lies in keeping subfigures as rows (for easy alignment), and using the addtocounter command to adjust numbering of subfigures manually.\documentclass[sigplan,anonymous,review,10pt]{acmart}\usepackage{listings}\usepackage{graphicx}\usepackage{subcaption}\begin{document}\begin{figure*}[htbp]\centering\begin{subfigure}[b]{\textwidth}...
The symbols for \leq and < have different height. You can force the latter to be the same height as the former by using \vphantom, but \mathrel has to surround the construction. Also {<}\vphantom{\leq} should be used in order to avoid spurious spacing.\documentclass[12pt,a4paper]{article}\usepackage{amsmath}\newcommand\leqhyp{%\overset{\mathrm{...
Based on the answer of Stefan Kottwitz I wrote this little snippet to make it a little easier to implement:\newcommand{\tabfigure}[2]{\raisebox{-.5\height}{\includegraphics[#1]{#2}}}To be used as:\documentclass{article}\usepackage{mwe}\newcommand{\tabfigure}[2]{\raisebox{-.5\height}{\includegraphics[#1]{#2}}}\begin{document}\begin{tabular}{cc}...
In good typography, the distance between two consecutive lines of text on a page should not depend on how tall the letters they contain are (well, this doesn’t hold true only in good typography!). For this reason, (La)TeX places their baselines at a fixed distance (for example, 12 points, or about 4.2mm), independently of the actual height of the glyphs ...
As explained in the comments, booktabs was designed to avoid using vertical rules in tables: as it adds some padding around horizontal lines, they normally cannot intersect vertical lines. I propose to replace booktabs with package \boldline which defines variable thickness horizontal and vertical lines. Padding of horizontal lines can be emulated with the ...
I'd set the verbatim content in a box (via lrbox) before using it (via \usebox):\documentclass{beamer}\begin{document}\begin{frame}[t,fragile]\begin{columns}[T]\begin{column}{0.45\textwidth}\begin{verbatim}Lorem ipsum dolor sit\end{verbatim}\end{column}\begin{column}{0.45\textwidth}\includegraphics[width=\...
To have the \forall and the \in vertically aligned, I used a 3 columns alignedat (due to the difference in width between t and x). Further, as newtx produced error messages on my system, I replaed them with fourier:\documentclass[a4paper,12pt]{article}\usepackage{mathtools,amssymb}\usepackage{fourier}%\usepackage{newtxtext}%\usepackage[libertine]{...
Here's a solution that employs an array environment to align the elements of the two rows of conditioning information.Note that I use a vertical bar to denote "given that" or "conditional on". If you prefer using a colon, you should input it as :, not as \colon.\documentclass[a4paper,12pt]{article}\usepackage{mathtools,array,newtxtext,newtxmath}\...
Writing \eta:=\eta(x,t) has no mathematical meaning whatsoever. Since apparently D is a subset of the plane, functions over D are two-variable by definition; how you call the variables is completely irrelevant.I wouldn't align the two final intervals. Around the colon I would add some additional space because of the split line on the right.\documentclass[...
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NTSGrad Spring 2018/Abstracts
This page contains the titles and abstracts for talks scheduled in the Spring 2018 semester. To go back to the main NTSGrad page, click here.
Contents Jan 23
Solly Parenti Rankin-Selberg L-functions
What do you get when you cross an Eisenstein series with a cuspform? An L-function! Since there's no modular forms course this semester, I will try to squeeze in an entire semester's course on modular forms during the first part of this talk, and then I'll explain the Rankin-Selberg method of establishing analytic continuation of certain L-functions.
Jan 30
Wanlin Li Intersection Theory on Modular Curves
My talk is based on the paper by François Charles with title "FROBENIUS DISTRIBUTION FOR PAIRS OF ELLIPTIC CURVES AND EXCEPTIONAL ISOGENIES". I will talk about the main theorem and give some intuition and heuristic behind it. I will also give a sketch of the proof.
Feb 6
Dongxi Ye Modular Forms, Borcherds Lifting and Gross-Zagier Type CM Value Formulas
During the course of past decades, modular forms and Borcherds lifting have been playing an increasingly central role in number theory. In this talk, I will partially justify these by discussing some recent progress on some topics in number theory, such as representations by quadratic forms and Gross-Zagier type CM value formulas.
Feb 20
Ewan Dalby The Cuspidal Rational Torsion Subgroup of J_0(p)
I will define the cuspidal rational torsion subgroup for the Jacobian of the modular curve J_0(N) and try to convince you that in the case of J_0(p) it is cyclic of order (p-1)/gcd(p-1,12).
Feb 27
Brandon Alberts A Brief Introduction to Iwasawa Theory
A bare bones introduction to the subject of Iwasawa theory, its main results, and some of the tools used to prove them. This talk will serve as both a small taste of the subject and a prep talk for the upcoming Arizona Winter School.
Mar 13
Solly Parenti Do You Even Lift?
Theta series are generating functions of the number of ways integers can be represented by quadratic forms. Using theta series, we will construct the theta lift as a way to transfer modular(ish) forms between groups.
Mar 20
Soumya Sankar Finite Hypergeometric Functions: An Introduction Finite Hypergeometric functions are finite field analogues of classical hypergeometric functions that come up in analysis. I will define these and talk about some ways in which they are useful in studying important number theoretic questions. Apr 3
Brandon Alberts Certain Unramified Metabelian Extensions Using Lemmermeyer Factorizations
We use conditions on the discriminant of an abelian extension [math]K/\mathbb{Q}[/math] to classify unramified extensions [math]L/K[/math] normal over [math]\mathbb{Q}[/math] where the (nontrivial) commutator subgroup of [math]\text{Gal}(L/\mathbb{Q})[/math] is contained in its center. This generalizes a result due to Lemmermeyer stating that the quadratic field of discriminant [math]d[/math], [math]\mathbb{Q}( \sqrt{d})[/math], has an unramified extension [math]M/\mathbb{Q}( \sqrt{d})[/math] normal over [math]\mathbb{Q}[/math] with [math]\text{Gal}(M/\mathbb{Q}( \sqrt{d})) = H_8[/math] (the quaternion group) if and only if the discriminant factors [math]d = d_1 d_2 d_3[/math] into a product of three coprime discriminants, at most one of which is negative, satisfying [math]\left(\frac{d_i d_j}{p_k}\right) = 1[/math] for each choice of [math]\{i, j, k\} = \{1, 2, 3\}[/math] and prime [math]p_k | d_k[/math].
Apr 10
Niudun Wang Nodal Domains of Maass Forms
Hecke-Maass cusp forms on modular surfaces produce nodal lines that divide the surface into disjoint nodal domains. I will briefly talk about this process and estimate the number of nodal domains as the eigenvalues vary.
Apr 17
Qiao He An Introduction to Automorphic Representations
Automorphic representation is a powerful tool to study L-functions. For me, Tate's marvelous thesis is the real beginning of the whole theory. So I will start with Tate's thesis, which is really the automorphic representation of [math]GL_1[/math]. Then I will talk about how to generalize Tate's idea to higher dimensions and explain some ideas behind Langlands program. If there is still time left, I will also mention the trace formula and use it to prove the classical Poisson summation formula.
Apr 23
Iván Ongay Valverde Definability of Frobenius Orbits and a Result on Rational Distance Sets
In this talk I will present a paper by Héctor Pastén. We will talk about the meaning of definability in a ring and how having a formula that identifies Frobenius orbits can help you show an analogous case of Hilbert's tenth problem (the one asking for an algorithm that tells you if a diophantine equation is solvable or not). Finally, if time permits, we will do an application that solves the existence of a dense set in the plane with rational distances, assuming some form of the ABC conjecture. This last question was proposed by Erdös and Ulam.
Apr 24
Brandon Boggess Moving from Local to Global
What do problems over local fields tell us about global problems?
May 1
Qiao He An Introduction to Automorphic Representations - Part II
Last time I talked about Tate's thesis, which is actually the theory of automorphic representation of GL_1. This time I will continue. First, I will give the definition of automorphic representation, and use Hecke characters and modular forms to motivate the definition. Then I will explain some classical results about automorphic representation, and discuss how automorphic representations are related to L-functions.
May 8
Sun Woo Park Parametrization of elliptic curves by Shimura curves
Let f be a weight-2 newform on [math]\Gamma_0(N)[/math]. Given a fixed isogeny class of semistable elliptic curves over [math]\mathbb{Q}[/math], for some [math]N[/math] there exists a distinguished element [math]A[/math] of the isogeny class such that [math]A[/math] is the strong modular curve attached to f. In fact, [math]A[/math] is a quotient of [math]J_0(N)[/math] by an abelian variety, from which we can obtain a covering map [math]\pi: X_0(N) \rightarrow A [/math]. Based on Ribet and Takahashi’s paper, I will discuss the properties of the covering map as well as its generalization to Shimura curves.
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Now showing items 1-10 of 27
Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
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Under the CEV model the stock price has the following dynamics:
$dS_t=\mu S_tdt+\sigma S_t^\gamma dW_t$, where $\sigma\geq0, $ $\gamma\geq0$.
According to Wikipedia, if $\gamma <1$ the volatility of the stock increases as the price falls.
But why is this true? Shouldn't be the exponent negative in order to have an inverse relationship between stock price and the volatility term?
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We have the following definition about convergence in a normed space:
"Let $(x_n)_{n=1}^\infty$ be a sequence in a normed space $(X,\|\cdot\|)$. We say that $x_n\to x$ in $X$ if, $$d(x_n,x)\equiv \|x-x_n\|\to 0$$ as $n\to \infty.$"
My questions:
How is it that we read this statement, and consequently understand it? In particular, what does it mean to talk about the norm (magnitude or length) of the difference of two sequences? Take the case that $x$ is some non-constant sequence $x_m$.
I understand that a metric is induced by a norm, but how much of a distinction should be made between a norm and a metric
withinthe context of a normed space? A normed space is automatically a metric space, and we talk about a metric $d$ being induced by the norm $d(x,y)=\|x-y\|$. Should I think of them, withinthe context of a normed space, as simply being the exact same, or should one take more care and try to distinguish them still?
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In O'searcoid,
Metric Spaces, he provides the following example of a metric space:
Suppose C is a circle and, for each $a,b ∈ C$, define $d(a,b)$ to be the distance along the line segment from $a$ to $b$. Then $d$ is a metric on $C$.
I have decided to confirm that his example is, indeed, a metric (here might be a good place to admit that I am weak in trigonometry).
According to wikipedia, the length of a line segment (or chord) can be written as:
$$d(a,b) = 2 r \sin{ \dfrac{ \theta }{2} }$$
where $r$ denotes the radius of $C$ and $\theta$ the angle formed between points $a,b$ with centre.
From here, it can clearly be seen that if $a = b$ then $d = 0$ since $\theta = 0$. Further, since $\theta \ge 0$, it follows that $d \ge 0$. Finally, $d(a,b) = d(b,a)$ since the angle is determined by the relation of these points with the centre of the circle. Thus, all that remains is the triangle inequality.
Take three points in $C$, $a,b,c$. The triangle inequality states that:
$$ d ( a,b ) \le d( a,c ) + d( c,b ) $$
Thus the following must satisfy:
$$ 2 r \sin{ \dfrac{ \theta_{ab} }{2} } \le 2 r \sin{ \dfrac{ \theta_{ac} }{2} } + 2 r \sin{ \dfrac{ \theta_{cb} }{2} }$$
From here, it may be deduced that $\theta_{ab} = \theta_{ac} + \theta_{cb} $ since they essentially just "split" the angle into two parts.
Removing like terms from the triangle inequality yields:
$$ \sin{ \dfrac{ \theta_{ab} }{2} } \le \sin{ \dfrac{ \theta_{ac} }{2} } + \sin{ \dfrac{ \theta_{cb} }{2} }$$
A wolfram calculation demonstrates that an alternative form for an equation in the form $ \sin{ \frac{x}{2} } + \sin{ \frac{y}{2} }$ is $2 \sin{(\frac{x}{4}+\frac{y}{4})} \cos{(\frac{x}{4}-\frac{y}{4})}$. A further wolfram calculation shows that an equation in the form $ \sin{ \frac{x}{2} }$ may be written as $2 \sin{ \frac{x}{4} } \cos{ \frac{x}{4} }$ I can rewrite the triangle inequality as:
$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{( \frac{ \theta_{bc} }{4} + \frac{ \theta_{cb}}{4} )} \cos{( \frac{ \theta_{bc} }{4} - \frac{ \theta_{cb}}{4} )} $$
And using the fact that $\theta_{ab} = \theta_{ac} + \theta_{cb} $, this can be written:
$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{( \frac{ \theta_{ab} - \theta_{cb} }{4} + \frac{ \theta_{ab} - \theta_{bc}}{4} )} \cos{( \frac{ \theta_{ab} - \theta_{cb} }{4} - \frac{ \theta_{ac} - \theta_{bc}}{4} )} $$
And this simplifies to:
$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ - \theta_{ab} } {4} } $$
And this $ \cos{-x} = \cos{x} $, we end up with:
$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } $$
From what I can see, this also demonstrates that:
$$ \sin{ \frac{x}{2} } = \sin{ \frac{y}{2} } + \sin{ \frac{z}{2} } $$
where $x = y + z$ (just the original triangle inequality).
So based upon this I decided to use some test values. However, I am finding that the triangle inequality (equality) is not holding for the calculations I put in. Thus, there must be an error in the proof somewhere.
Thus my question is simply: What am I doing wrong?
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One Mean
Consider a dataset about $N = 54$ cars sampled from the year $1993$. Of interest is the (unknown population) mean miles per gallon, $\mu$. We assume
The parameter $\sigma$ might be of interest to some, but not us now. By choosing this model we are implicitly assuming that $\mathbb{E}(Y) = \mu$, where $\mu$ is a constant function that does not depend on any other characteristics about the population of cars from $1993$.
Assuming the mean $\mu$ is constant is an unrealistic assumption if you think too long about this problem. Nevertheless, this is a common assumption because of its simplicity. In situations where simplicity is what you want, assuming the mean $\mu$ is constant across all cars from $1993$ is fine. On the other hand, such simplicity is not always desired. More complex models that relate mean miles per gallon to, say, weight or the drivetrain type are explored in the following sections of this chapter.
The following code reads in the dataset, plots the $\texttt{mpgCity}$ data, and calculates an estimate of the population mean, $\hat{\mu}$ based on the observed MPG for each car.
import numpy as npimport pandas as pdimport bplot as bpfrom scipy.optimize import minimizebp.LaTeX()bp.dpi(300)
cars = pd.read_csv("https://raw.githubusercontent.com/roualdes/data/master/cars.csv")
def ll(mu, y): d = y - mu return np.sum(d * d)mu_hat = minimize(ll, 10, args=(cars['mpgCity']), method="BFGS")['x']
bp.density(cars['mpgCity'])bp.rug(mu_hat)
[<matplotlib.lines.Line2D at 0x12260bd68>]
The estimate $\hat{\mu}$ is our best guess of the population mean city MPG for cars from $1993$. What we don’t yet have is any measure of uncertainty surrounding this best guess. Though this number seems reasonable, it’s possible this number could have shown up by pure chance. We’ll next produce a confidence interval for the true population mean city MPG for cars from $1993$.
The following code uses the library $\texttt{boot}$ to perform the random sampling (uniformly and with replacement) from the original data. The same strategy that’s discussed in Chapter Bootstrap is used here. Sampling will happen over the indices of our original data. The function $\texttt{boot::boot}$ requires that we write a function that takes two arguments, the data to calculate our statistic of interest on and a vector of indices.
R = 999N = cars['mpgCity'].sizemus = np.full((R,), np.nan)for r in range(R): idx = np.random.choice(N, N) mus[r] = minimize(ll, 10, args=(cars['mpgCity'][idx]), method="BFGS")['x'] mu_p = np.percentile(mus, [10, 90])
The standard conclusion from this goes as follows. We are $90\%$ confident that the true population mean city MPG for cars from $1993$ is between $21.9$ and $24.8$.
Notice though this confidence interval carries with it a number of assumptions. The same assumption from before is carried forward, that the true population mean does not vary by any other population characteristics. Further, the bounds of the confidence interval are randomly produced. If you re-run the code above, you’ll likely get slightly different numbers. So don’t get too carried away with carrying decimal places, $2$ is a good general policy.
bp.density(mus)bp.rug(mu_p, color='black')bp.rug(np.percentile(mus, [50]))bp.labels(x='Mean MPG city', y='Density', size=18)
<matplotlib.axes._subplots.AxesSubplot at 0x1258c62e8>
When interpreting the estimated mean, and the lower and upper bound of the confidence interval, it’s important to remember that these statistics are referring to the population mean, not specific data points. Specifically, it is
not the case that $90\%$ of all observed data (within this sample nor any future sample) will fall within this interval.
A strict interpretation of this interval says that $90\%$ of an infinite collection of confidence intervals, calculated by and your infinite number of friends, will contain the true population mean. That is $90\%$ of these (hypothetical) confidence intervals will capture the true population mean. This indeed suggests that $10\%$ of the hypothetical confidence intervals will not capture the true population mean.
This subtle difference highlights the difference between estimating a population mean, for which it is reasonable to assume a Normal distribution, and fitting a distribution (via its parameters) to a dataset.
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Now showing items 1-10 of 26
Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ...
Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-06)
The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ...
Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC
(Springer, 2017-01)
The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ...
Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC
(Springer, 2017-06)
We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ...
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Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.Let's denote the width of a sample by $h$ where $$h\rightarrow0$$Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can a...
@Ultradark You can try doing a finite difference to get rid of the sum and then compare term by term. Otherwise I am terrible at anything to do with primes that I don't know the identities of $\pi (n)$ well
@Silent No, take for example the prime 3. 2 is not a residue mod 3, so there is no $x\in\mathbb{Z}$ such that $x^2-2\equiv 0$ mod $3$.
However, you have two cases to consider. The first where $\binom{2}{p}=-1$ and $\binom{3}{p}=-1$ (In which case what does $\binom{6}{p}$ equal?) and the case where one or the other of $\binom{2}{p}$ and $\binom{3}{p}$ equals 1.
Also, probably something useful for congruence, if you didn't already know: If $a_1\equiv b_1\text{mod}(p)$ and $a_2\equiv b_2\text{mod}(p)$, then $a_1a_2\equiv b_1b_2\text{mod}(p)$
Is there any book or article that explains the motivations of the definitions of group, ring , field, ideal etc. of abstract algebra and/or gives a geometric or visual representation to Galois theory ?
Jacques Charles François Sturm ForMemRS (29 September 1803 – 15 December 1855) was a French mathematician.== Life and work ==Sturm was born in Geneva (then part of France) in 1803. The family of his father, Jean-Henri Sturm, had emigrated from Strasbourg around 1760 - about 50 years before Charles-François's birth. His mother's name was Jeanne-Louise-Henriette Gremay. In 1818, he started to follow the lectures of the academy of Geneva. In 1819, the death of his father forced Sturm to give lessons to children of the rich in order to support his own family. In 1823, he became tutor to the son...
I spent my career working with tensors. You have to be careful about defining multilinearity, domain, range, etc. Typically, tensors of type $(k,\ell)$ involve a fixed vector space, not so many letters varying.
UGA definitely grants a number of masters to people wanting only that (and sometimes admitted only for that). You people at fancy places think that every university is like Chicago, MIT, and Princeton.
hi there, I need to linearize nonlinear system about a fixed point. I've computed the jacobain matrix but one of the elements of this matrix is undefined at the fixed point. What is a better approach to solve this issue? The element is (24*x_2 + 5cos(x_1)*x_2)/abs(x_2). The fixed point is x_1=0, x_2=0
Consider the following integral: $\int 1/4*(1/(1+(u/2)^2)))dx$ Why does it matter if we put the constant 1/4 behind the integral versus keeping it inside? The solution is $1/2*\arctan{(u/2)}$. Or am I overseeing something?
*it should be du instead of dx in the integral
**and the solution is missing a constant C of course
Is there a standard way to divide radicals by polynomials? Stuff like $\frac{\sqrt a}{1 + b^2}$?
My expression happens to be in a form I can normalize to that, just the radicand happens to be a lot more complicated. In my case, I'm trying to figure out how to best simplify $\frac{x}{\sqrt{1 + x^2}}$, and so far, I've gotten to $\frac{x \sqrt{1+x^2}}{1+x^2}$, and it's pretty obvious you can move the $x$ inside the radical.
My hope is that I can somehow remove the polynomial from the bottom entirely, so I can then multiply the whole thing by a square root of another algebraic fraction.
Complicated, I know, but this is me trying to see if I can skip calculating Euclidean distance twice going from atan2 to something in terms of asin for a thing I'm working on.
"... and it's pretty obvious you can move the $x$ inside the radical" To clarify this in advance, I didn't mean literally move it verbatim, but via $x \sqrt{y} = \text{sgn}(x) \sqrt{x^2 y}$. (Hopefully, this was obvious, but I don't want to confuse people on what I meant.)
Ignore my question. I'm coming of the realization it's just not working how I would've hoped, so I'll just go with what I had before.
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I'm writing an exercise book, and in some exercises I'll have several numbered equations in an align environment, e.g.,
(1) x^2+x+1=0, (2) ax^3+2x=0, (3) x^5=2i,
(4) x^4=-i, (5) x-1/x=2, (6) cosh(x)=3.
(I'll want them to be aligned with respect to the number).
I'm using a macro
\newcommand*\ExoEq{\refstepcounter{ExoEq}(\theExoEq)~}
so that the previous example is typed as:
\begin{align*}\ExoEq&x^2+x+1=0,&\ExoEq&ax^3+2x=0,&\ExoEq&x^5=2i,\\\ExoEq&x^4=-i,&\ExoEq&x-\frac1x=2,&\ExoEq&\cosh(x)=3.\end{align*}
All is good so far, except that I sometimes need to refer to an equation, e.g.,
In Equation~(2), we assume that $a\neq0$.
The thing is when I want to use
\label and
\eqref, I get weird results in the align environment: here is a minimal example that shows what goes wrong:
\documentclass{amsart}\usepackage{amsthm}\usepackage[pdftex]{hyperref}\parindent0pt\newtheoremstyle{myexercise}{\baselineskip}{\baselineskip}{}{}{\bfseries}{.}{ }{\thmname{#1}\ \thmnumber{#2}}%\theoremstyle{myexercise}\newtheorem{Exo}{Exercise}[section]\newcounter{MyCounter}\newcommand*\Number{\refstepcounter{MyCounter}(\theMyCounter)~}\let\oldlabel\label\begin{document}\section{Exercises}\begin{Exo}\begin{align*}\Number\oldlabel{L1}&x^2+x=2,&\Number\oldlabel{L2}&x^3+x^2=0.\end{align*}Equation~\eqref{L1} is very nice, and Equation~\eqref{L2}is lovely too.\end{Exo}\end{document}
Without the hyperref package, I get the exercise reference, instead of the equation reference. With the hyperref package, I get an error on the second latex pass.
I'm using a
\let\oldlabel\label trick, as the align* environment redefines
\label, and it just doesn't work with
\label instead.
How can I work around this annoying thing to have it work?
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I need to give a lot of quite basic background to this question because I think (at least from conversing with fellow graduate students) that most mathematicians have not
really thought about fractions for a long time. I think that there is an interesting germ of an idea in here somewhere, but I cannot exactly pinpoint it. Essentially there seems to be two canonical ways to solve division problems and there does not seem to be a "natural isomorphism" relating the two ways. I am interested in framing this duality formally: is there a "categorification" of the rational numbers where this duality can be precisely framed?
I TA a class for future elementary school teachers. The idea is to go back and really understand elementary school mathematics at a deep level. Hopefully this understanding gets passed on to the next generation.
We were discussing division of fractions. Rather than say "Yours is not to wonder why, just invert and multiply", we try to make sense of this question physically, and then use reasoning to solve the problem. Take (3/4) / (2/3). When doing this there seems to be two reasonable interpretations:
1) 3/4 of a cup of milk fills 2/3 of a container. How much milk (in cups) does it take to fill the entire container?
This is a "How many in each group" division problem, analogous to converting 6/3 into the question "If I have six objects divided into three equal groups, how many objects will be in each group?"
The solution that stares you in the face if you draw a picture of this situation is the following: 3/4 of a cup fills 2 thirds of a container. That means there must be 3/8 cups of milk in each third of the container. One container must have 9/8 cups of milk then, because this is 3 of these thirds. Note that the solution involved first dividing and then multiplying.
2) I have 3/4 cups of milk, and I have bottles which each hold 2/3 of a cup. How many bottles can I fill?
This is a "How many groups" division problem, analogous to converting 6/3 into the question "If I have six objects divided into groups of two, how many groups do I have?"
The solution suggested by this situation is the following: 3/4 of a cup of milk is actually 9 twelfths of a cup . Each twelfth is an eighth of a bottle. So I have 9/8 of a bottle. This solution involved first multiplying and then dividing.
Now I come to my question. This pattern persists! Every real world example of a "how many in each group" division problem suggests a solution by first dividing and then multiplying, whereas each "How many groups" division problem involves first multiplying and then dividing. It seems that solving the problem in the other order does not admit a conceptual realization in terms of the original problem. This is interesting to me! It suggests that the two solution methods are fundamentally different somehow. The standard approach to rational numbers (natural numbers get grothendieck grouped into integers, which get ring of fractioned into rational numbers) ignores this kind of distinction. Is there a "categorification" of the rational numbers which preserves the duality between these two types of question?
UPDATE 1: In the category of sets, if you wanted to express $(\frac{6}{2})(3) = \frac{(6)(3)}{2}$ you would have to do something like this:
Let $A$ be a set with 6 elements, $B$ a set with 3 elements, $\sim$ an equivalence relation on A where each equivalence class has 2 elements, and $\cong$ an equivalence relation on $A \times B$ where each equivalence class has 2 elements. Then there is no canonical morphism from $(A/ \sim) \times B \to (A \times B)/ \cong$. This seems to explain things somewhat on the level of integers, but we are talking fractions here.
UPDATE 2: Qiaochu points out in a comment to his answer that the order of operations is not the most essential thing here. You can solve the first problem by observing that 9/4 cups of milk fill 2 containers, so 9/8 must fill one. Torsors give a formal distinction between the two situations, but it still feels like UPDATE 1 should go through in a suitable category of "fractional sets".
UPDATE 3: For a very nice discussion of the ideas related to Theo's answer see http://golem.ph.utexas.edu/category/2008/12/groupoidification_made_easy.html
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The question
I consider the Laplacian $\Delta = \partial_1^2 + \partial_2^2 + \partial_3^2$ in $\mathbb{R}^3$. By the "standard" fundamental solution of the Laplacian, I mean the function
$$ \displaystyle E(x) = C|x|^{-1} $$
where $C$ is some normalization constant.
I would like to know if one can construct a fundamental solution of the Laplacian that decays faster than any rational function at infinity. I have an idea how one could maybe construct this, but I'm not sure if my idea is correct as the idea of a suspicious fundamental solution seems highly suspicious to me. In the following, I sketch my idea: An attempt at an answer
I already know that one can construct a rapidly decaying parametrix $F$ by multiplying $\hat E \sim |x|^{-2}$ with a cutoff function $\chi$ supported in the complement of the unit ball and then taking the inverse Fourier transform. Then, one has $$x^\alpha F(x) = \int D^\alpha(\chi\hat E)(\xi)e^{ix\xi}\text{d}\xi $$ and the right hand side is absolutely convergent for large $\alpha$. This proves the decay at infinity.
I want to modify this construction in order to turn the parametrix $F$ into a proper fundamental solution. To do this, I want to use a simplified version of the construction by Hörmander (see "Distribution" by Duistermaat and Kolk, Theorem 17.11). Letting $\eta$ be a unit vector, I set
$$ G(x) = (2\pi)^{-n} \int \frac{1- \chi(\xi)}{2\pi i}\int_{|z| = 100} \frac{e^{ix\cdot(\xi + z \eta)}}{|\xi + z\eta|^2}\frac{\text{d}z}{z}\text{d}\xi\,.$$
With Cauchy's integral theorem, one can show that $$\tilde E = F + G$$ is indeed a fundamental solution. The idea here is that we avoid the singularity at $0$ by taking a contour integral around it. It seems as if one can use the same trick that one used for $F$ to show the rapid decay of $G$. This suggest that one can indeed find a rapidly deacying fundamental solution.
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For fixed $m = 0, 1, 2, ...$ $$f_m(k) = \prod_{j=1}^{m}(k+j).$$ Some examples of $f_m(k)$ are as following: $$f_0(k) = 1, \quad f_1(k) = (k+1), \quad f_2(k) = (k+1)(k+2).$$
The $s_m(n)$ is defined as following: $$s_m(n) = \sin\left(\frac{t}{2}\right)\sum_{k=0}^nf_m(k)\sin(k+0.5)t,\qquad t\in[0,\pi].$$
The $s_m(n)$ can also be defined as following: $$s_m(n) = \sum_{j=0}^n\frac{(-4)^j}{(2j+1)!}\left(\sum_{k=j}^n\frac{f_m(k)(2k+1)(k+j)!}{(k-j)!}\right)x^{2j+2},\qquad x\in[0,1].$$
I want to prove $$|s_m(n)| \le f_m(n), \forall x ~or ~t$$
I am sure the inequality holds but I am unable to prove it. I used MATLAB and verified the inequality for some values of $m$ and $n$ as presented below:
\begin{array}{ccccccccc} n & \max(s_0(n)) & f_0(n) & \max(s_1(n)) & f_1(n) & \max(s_2(n))& f_2(n) & \max(s_3(n)) & f_3(n)\\ 0 & 1.00 & 1 & 1.00 & 1 & 2.00 & 2 & 6.00 & 6 \\ 1 & 1.00 & 1 & 1.53 & 2 & 4.17 & 6 & 18.00 & 24 \\ 2 & 1.00 & 1 & 2.07 & 3 & 8.00 & 12 & 42.00 & 60 \\ 3 & 1.00 & 1 & 2.60 & 4 & 12.46 & 20 & 78.30 & 120 \\ 4 & 1.00 & 1 & 3.13 & 5 & 18.03 & 30 & 132.00 & 210 \end{array}
Any help will be greatly appreciated.
PS: Please refer to this question. I asked for inductive proof so that I could use induction steps in the above inequality. But I did not get one.
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There are many question about life and physics in higher dimensions. but is there any physical thing (i mean things like force, momentum, speed,...) that cannot exist in 2D world? could 2D world have life forms(not necessarily the life form we have) or there is something used by life that needs at least 3 spatial dimensions?
Let us start with Newtonian mechanics. For any point particle, position, speed, acceleration, momentum and force would still make sense, with two components instead of three. But the question is then how those forces would be modified.
Let us look at gravity for example. If we were to assume the same form for the equation the gravitational field $g$ is solution of,
$$\DeclareMathOperator{\div}{div} \iint_S g.dS = -4\pi M$$
where $S$ is a closed surface and $M$ is the mass inside $S$, then instead of getting a force in $1/r^2$, we would get a force in $1/r$, or equivalently a gravitational potential in $\log r$, in the classic two-body problem. This leads to very different a phenomenology, as such a force field does not result in close trajectories (Bertrand's theorem). So the Earth, for example, would keep "precessing" around the Sun, never coming back to the same point. The Ephemerides of the Solar system would be a lot of fun!
The equivalent Quantum mechanical problem of the Hydrogen atom, would also have a potential in $\log r$ if we assume the same form of Maxwell equations. This result in unbounded energy level: they go to $+\infty$ as the quantum number $n$ goes to $+\infty$, which completely changes the phenomenology of ionisation and conduction. I will find some interesting references when I get back to my desk!
Yes, life can exist in a 2D world. This was done by a long simulation experiment called Conway's Game of Life where signs of life were shown by tiles on a computer screen(search it up for details).
It is true that life would be vastly different and in a 2D world the laws of physics would have to adapt to allow formation of molecules etc.( e.g. In a 2D space, if laws did not modify, elliptical orbits would be impossible) Remember this, just because a mathematical framework for this exists does not mean it is truly something physical. The models we create should fit what we observe to create a satisfactory model in physics.
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How to Implement the Fourier Transformation from Computed Solutions
We previously learned how to calculate the Fourier transform of a rectangular aperture in a Fraunhofer diffraction model in the COMSOL Multiphysics® software. In that example, the aperture was given as an analytical function. The procedure is a bit different if the source data for the Fourier transformation is a computed solution. In this blog post, we will learn how to implement the Fourier transformation for computed solutions with an electromagnetic simulation of a Fresnel lens.
Fourier Transformation with Fourier Optics
Implementing the Fourier transformation in a simulation can be useful in Fourier optics, signal processing (for use in frequency pattern extraction), and noise reduction and filtering via image processing. In Fourier optics, the Fresnel approximation is one of the approximation methods used for calculating the field near the diffracting aperture. Suppose a diffracting aperture is located in the (x,y) plane at z=0. The diffracted electric field in the (u,v) plane at the distance z=f from the diffracting aperture is calculated as
where, \lambda is the wavelength and E(x,y,0), \ E(u,v,f) account for the electric field at the (x,y) plane and the (u,v) plane, respectively. (See Ref. 1 for more details.)
In this approximation formula, the diffracted field is calculated by Fourier transforming the incident field multiplied by the quadratic phase function {\rm exp}\{-i\pi (x^2+y^2)/(\lambda f)\}.
The sign convention of the phase function must follow the sign convention of the time dependence of the fields. In COMSOL Multiphysics, the time dependence of the electromagnetic fields is of the form {\rm exp}(+i\omega t). So, the sign of the quadratic phase function is negative.
Fresnel Lenses
Now, let’s take a look at an example of a Fresnel lens. A Fresnel lens is a regular plano-convex lens except for its curved surface, which is folded toward the flat side at every multiple of m \lambda/(n-1) along the lens height, where
m is an integer and n is the refractive index of the lens material. This is called an m th-order Fresnel lens.
The shift of the surface by this particular height along the light propagation direction only changes the phase of the light by 2m \pi (roughly speaking and under the paraxial approximation). Because of this, the folded lens fundamentally reproduces the same wavefront in the far field and behaves like the original unfolded lens. The main difference is the diffraction effect. Regular lenses basically don’t show any diffraction (if there is no vignetting by a hard aperture), while Fresnel lenses always show small diffraction patterns around the main spot due to the surface discontinuities and internal reflections.
When a Fresnel lens is designed digitally, the lens surface is made up of discrete layers, giving it a staircase-like appearance. This is called a multilevel Fresnel lens. Due to the flat part of the steps, the diffraction pattern of a multilevel Fresnel lens typically includes a zeroth-order background in addition to the higher-order diffraction.
Why are we using a Fresnel lens as our example? The reason is similar to why lighthouses use Fresnel lenses in their operations. A Fresnel lens is folded into m \lambda/(n-1) in height. It can be extremely thin and therefore of less weight and volume, which is beneficial for the optics of lighthouses compared to a large, heavy, and thick lens of the conventional refractive type. Likewise, for our purposes, Fresnel lenses can be easier to simulate in COMSOL Multiphysics and the add-on Wave Optics Module because the number of elements are manageable.
Modeling a Focusing Fresnel Lens in COMSOL Multiphysics®
The figure below depicts the optics layout that we are trying to simulate to demonstrate how we can implement the Fourier transformation, applied to a computed solution solved for by the
Wave Optics, Frequency Domain interface. Focusing 16-level Fresnel lens model.
This is a first-order Fresnel lens with surfaces that are digitized in 16 levels. A plane wave E_{\rm inc} is incident on the incidence plane. At the exit plane at z=0, the field is diffracted by the Fresnel lens to be E(x,y,0). This process can be easily modeled and simulated by the
Wave Optics, Frequency Domain interface. Then, we calculate the field E(u,v,f) at the focal plane at z=f by applying the Fourier transformation in the Fresnel approximation, as described above.
The figures below are the result of our computation, with the electric field component in the domains (top) and on the boundary corresponding to the exit plane (bottom). Note that the geometry is not drawn to scale in the vertical axis. We can clearly see the positively curved wavefront from the center and from every air gap between the saw teeth. Note that the reflection from the lens surfaces leads to some small interferences in the domain field result and ripples in the boundary field result. This is because there is no antireflective coating modeled here.
The computed electric field component in the Fresnel lens and surrounding air domains (vertical axis is not to scale). The computed electric field component at the exit plane. Implementing the Fourier Transformation from a Computed Solution
Let’s move on to the Fourier transformation. In the previous example of an analytical function, we prepared two data sets: one for the source space and one for the Fourier space. The parameter names that were defined in the Settings window of the data set were the spatial coordinates (x,y) in the source plane and the spatial coordinates (u,v) in the image plane.
In today’s example, the source space is already created in the computed data set, Study 1/Solution 1
(sol1){dset1}, with the computed solutions. All we need to do is create a one-dimensional data set, Grid1D {grid1}, with parameters for the Fourier space; i.e., the spatial coordinate u in the focal plane. We then relate it to the source data set, as seen in the figure below. Then, we define an integration operator
intop1 on the exit plane.
Settings for the data set for the transformation.
The intop1 operator defined on the exit plane (vertical axis is not to scale).
Finally, we define the Fourier transformation in a 1D plot, shown below. It’s important to specify the data set we previously created for the transformation and to let COMSOL Multiphysics know that u is the destination independent variable by using the
dest operator.
Settings for the Fourier transformation in a 1D plot.
The end result is shown in the following plot. This is a typical image of the focused beam through a multilevel Fresnel lens in the focal plane (see Ref. 2). There is the main spot by the first-order diffraction in the center and a weaker background caused by the zeroth-order (nondiffracted) and higher-order diffractions.
Electric field norm plot of the focused beam through a 16-level Fresnel lens. Concluding Remarks
In this blog post, we learned how to implement the Fourier transformation for computed solutions. This functionality is useful for long-distance propagation calculation in COMSOL Multiphysics and extends electromagnetic simulation to Fourier optics.
Next Steps
Download the model files for the Fresnel lens example by clicking the button below.
Read More About Simulating Wave Optics Simulating Holographic Data Storage in COMSOL Multiphysics How to Simulate a Holographic Page Data Storage System How to Implement the Fourier Transformation in COMSOL Multiphysics References J.W. Goodman, Introduction to Fourier Optics, The McGraw-Hill Company, Inc. D. C. O’Shea, Diffractive Optics, SPIE Press.
Comments (10) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science TAGS CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
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I actually got this part and I got $16$ choose $2$, which would be $120$. The part I didn't get which wouldn't fit into the title was in how many of these solutions is $x_1 \geq 1$, $x_2 \geq 2$,and $x_3 \geq 3$? I'm not very good at combinatorics so I don't really know. I think the way to approach it would be saying how many solutions where $x < 1 +$ solutions where $x < 2 +$ solutions where $x < 3$ and subtract that from $120$, but I don't know how to do that.
Your solution to determining how many solutions the equation $$x_1 + x_2 + x_3 = 14 \tag{1}$$ has in the nonnegative integers is correct. To determine how many solutions equation 1 if $x_1 \geq 1$, $x_2 \geq 2$, and $x_3 \geq 3$, let \begin{align*} y_1 & = x_1 - 1\\ y_2 & = x_2 - 2\\ y_3 & = x_3 - 3 \end{align*} Then $y_1, y_2, y_3$ are nonnegative integers. Substituting $y_1 + 1$ for $x_1$, $y_2 + 2$ for $x_2$, and $y_3 + 3$ for $x_3$ in equation 1 yields \begin{align*} y_1 + 1 + y_2 + 2 + y_3 + 3 & = 14\\ y_1 + y_2 + y_3 & = 8 \end{align*} which is an equation in the nonnegative integers with $$\binom{8 + 2}{2} = \binom{10}{2} = 45$$ solutions.
This is pretty easy to calculate through Wolfram Alpha: Here is an example of an input that will solve this for you. The answer provided is not entirely true though, because of your constraints. We must exclude the solution $12+1+1$, which does not have $x_2 \geq 2$. We thus subtract $1$ from the provided result of 16 to get a final answer of $\color{red}{15}$. For clarification I have provided a list of the solutions here. If you want further info check out this page, although the topic of integer partitions is fairly complicated. If you have further questions I can try to help point you in the right direction, although I am not the best source of info on this topic; if you desire to learn more I recommend creating a new question specifically geared towards this.
$$ 11 + 2 + 1 = 14\\ 10 + 3 + 1 = 14\\ 10 + 2 + 2 = 14\\ 9 + 4 + 1 = 14\\ 9 + 3 + 2 = 14\\ 8 + 5 + 1 = 14\\ 8 + 4 + 2 = 14\\ 8 + 3 + 3 = 14\\ 7 + 6 + 1 = 14\\ 7 + 5 + 2 = 14\\ 7 + 4 + 3 = 14\\ 6 + 6 + 2 = 14\\ 6 + 5 + 3 = 14\\ 6 + 4 + 4 = 14\\ 5 + 5 + 4 = 14$$
Note: In the event you also desire for the condition $x_1 \neq x_2 \neq x_3$ to hold it appears we get $10$ solutions
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Answer
\begin{equation} N_{m}=\frac{n-1}{2}+1=\frac{n-1+2}{2}=\frac{n+1}{2} \end{equation}
Work Step by Step
We suppose we have an odd number of items, like $15 .$ Our task is to find the position for the middle item. Here's how we can find the position for a 15 -element list: $15=14+1=2 \cdot 7+1=7+1+7$ Here we can see that we have one single element in the middle because it is an odd number and odd numbers can be written in the form $2 \cdot m+1 .$ It is obvious that the middle element is the $8^{t h}$ element which has 7 elements before and 7 after it. We can do the same to a couple of other odd numbers: $\quad \quad \bullet 23=2 \cdot 11+1,$ with $12^{t h}$ as the middle element $\quad \quad \bullet 75=2 \cdot 37+1,$ with $38^{t h}$ as the middle element The number of comparisons required to find the middle item for each of the above lists is equal to the middle's position: $8,12,38 .$ We can write an expression to find this middle item $N_{m}$ in a $n$ -element list, where $n$ is odd: $N_{m}=\frac{n-1}{2}+1=\frac{n-1+2}{2}=\frac{n+1}{2}$
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Hi, Can someone provide me some self reading material for Condensed matter theory? I've done QFT previously for which I could happily read Peskin supplemented with David Tong. Can you please suggest some references along those lines? Thanks
@skullpatrol The second one was in my MSc and covered considerably less than my first and (I felt) didn't do it in any particularly great way, so distinctly average. The third was pretty decent - I liked the way he did things and was essentially a more mathematically detailed version of the first :)
2. A weird particle or state that is made of a superposition of a torus region with clockwise momentum and anticlockwise momentum, resulting in one that has no momentum along the major circumference of the torus but still nonzero momentum in directions that are not pointing along the torus
Same thought as you, however I think the major challenge of such simulator is the computational cost. GR calculations with its highly nonlinear nature, might be more costy than a computation of a protein.
However I can see some ways approaching it. Recall how Slereah was building some kind of spaceitme database, that could be the first step. Next, one might be looking for machine learning techniques to help on the simulation by using the classifications of spacetimes as machines are known to perform very well on sign problems as a recent paper has shown
Since GR equations are ultimately a system of 10 nonlinear PDEs, it might be possible the solution strategy has some relation with the class of spacetime that is under consideration, thus that might help heavily reduce the parameters need to consider to simulate them
I just mean this: The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
@ooolb Even if that is really possible (I always can talk about things in a non joking perspective), the issue is that 1) Unlike other people, I cannot incubate my dreams for a certain topic due to Mechanism 1 (consicous desires have reduced probability of appearing in dreams), and 2) For 6 years, my dream still yet to show any sign of revisiting the exact same idea, and there are no known instance of either sequel dreams nor recurrence dreams
@0celo7 I felt this aspect can be helped by machine learning. You can train a neural network with some PDEs of a known class with some known constraints, and let it figure out the best solution for some new PDE after say training it on 1000 different PDEs
Actually that makes me wonder, are the space of all coordinate choices more than all possible moves of Go?
enumaris: From what I understood from the dream, the warp drive showed here may be some variation of the alcuberrie metric with a global topology that has 4 holes in it whereas the original alcuberrie drive, if I recall, don't have holes
orbit stabilizer: h bar is my home chat, because this is the first SE chat I joined. Maths chat is the 2nd one I joined, followed by periodic table, biosphere, factory floor and many others
Btw, since gravity is nonlinear, do we expect if we have a region where spacetime is frame dragged in the clockwise direction being superimposed on a spacetime that is frame dragged in the anticlockwise direction will result in a spacetime with no frame drag? (one possible physical scenario that I can envision such can occur may be when two massive rotating objects with opposite angular velocity are on the course of merging)
Well. I'm a begginer in the study of General Relativity ok? My knowledge about the subject is based on books like Schutz, Hartle,Carroll and introductory papers. About quantum mechanics I have a poor knowledge yet.
So, what I meant about "Gravitational Double slit experiment" is: There's and gravitational analogue of the Double slit experiment, for gravitational waves?
@JackClerk the double slits experiment is just interference of two coherent sources, where we get the two sources from a single light beam using the two slits. But gravitational waves interact so weakly with matter that it's hard to see how we could screen a gravitational wave to get two coherent GW sources.
But if we could figure out a way to do it then yes GWs would interfere just like light wave.
Thank you @Secret and @JohnRennie . But for conclude the discussion, I want to put a "silly picture" here: Imagine a huge double slit plate in space close to a strong source of gravitational waves. Then like water waves, and light, we will see the pattern?
So, if the source (like a Black Hole binary) are sufficent away, then in the regions of destructive interference, space-time would have a flat geometry and then with we put a spherical object in this region the metric will become schwarzschild-like.
if**
Pardon, I just spend some naive-phylosophy time here with these discussions**
The situation was even more dire for Calculus and I managed!
This is a neat strategy I have found-revision becomes more bearable when I have The h Bar open on the side.
In all honesty, I actually prefer exam season! At all other times-as I have observed in this semester, at least-there is nothing exciting to do. This system of tortuous panic, followed by a reward is obviously very satisfying.
My opinion is that I need you kaumudi to decrease the probabilty of h bar having software system infrastructure conversations, which confuse me like hell and is why I take refugee in the maths chat a few weeks ago
(Not that I have questions to ask or anything; like I said, it is a little relieving to be with friends while I am panicked. I think it is possible to gauge how much of a social recluse I am from this, because I spend some of my free time hanging out with you lot, even though I am literally inside a hostel teeming with hundreds of my peers)
that's true. though back in high school ,regardless of code, our teacher taught us to always indent your code to allow easy reading and troubleshooting. We are also taught the 4 spacebar indentation convention
@JohnRennie I wish I can just tab because I am also lazy, but sometimes tab insert 4 spaces while other times it inserts 5-6 spaces, thus screwing up a block of if then conditions in my code, which is why I had no choice
I currently automate almost everything from job submission to data extraction, and later on, with the help of the machine learning group in my uni, we might be able to automate a GUI library search thingy
I can do all tasks related to my work without leaving the text editor (of course, such text editor is emacs). The only inconvenience is that some websites don't render in a optimal way (but most of the work-related ones do)
Hi to all. Does anyone know where I could write matlab code online(for free)? Apparently another one of my institutions great inspirations is to have a matlab-oriented computational physics course without having matlab on the universities pcs. Thanks.
@Kaumudi.H Hacky way: 1st thing is that $\psi\left(x, y, z, t\right) = \psi\left(x, y, t\right)$, so no propagation in $z$-direction. Now, in '$1$ unit' of time, it travels $\frac{\sqrt{3}}{2}$ units in the $y$-direction and $\frac{1}{2}$ units in the $x$-direction. Use this to form a triangle and you'll get the answer with simple trig :)
@Kaumudi.H Ah, it was okayish. It was mostly memory based. Each small question was of 10-15 marks. No idea what they expect me to write for questions like "Describe acoustic and optic phonons" for 15 marks!! I only wrote two small paragraphs...meh. I don't like this subject much :P (physical electronics). Hope to do better in the upcoming tests so that there isn't a huge effect on the gpa.
@Blue Ok, thanks. I found a way by connecting to the servers of the university( the program isn't installed on the pcs on the computer room, but if I connect to the server of the university- which means running remotely another environment, i found an older version of matlab). But thanks again.
@user685252 No; I am saying that it has no bearing on how good you actually are at the subject - it has no bearing on how good you are at applying knowledge; it doesn't test problem solving skills; it doesn't take into account that, if I'm sitting in the office having forgotten the difference between different types of matrix decomposition or something, I can just search the internet (or a textbook), so it doesn't say how good someone is at research in that subject;
it doesn't test how good you are at deriving anything - someone can write down a definition without any understanding, while someone who can derive it, but has forgotten it probably won't have time in an exam situation. In short, testing memory is not the same as testing understanding
If you really want to test someone's understanding, give them a few problems in that area that they've never seen before and give them a reasonable amount of time to do it, with access to textbooks etc.
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Using the Product Rule to Simplify Square Roots
To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the
product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite [latex]\sqrt{15}[/latex] as [latex]\sqrt{3}\cdot \sqrt{5}[/latex]. We can also use the product rule to express the product of multiple radical expressions as a single radical expression. A General Note: The Product Rule for Simplifying Square Roots
If [latex]a[/latex] and [latex]b[/latex] are nonnegative, the square root of the product [latex]ab[/latex] is equal to the product of the square roots of [latex]a[/latex] and [latex]b[/latex].
How To: Given a square root radical expression, use the product rule to simplify it. Factor any perfect squares from the radicand. Write the radical expression as a product of radical expressions. Simplify. Example 2: Using the Product Rule to Simplify Square Roots
Simplify the radical expression.
[latex]\sqrt{300}[/latex] [latex]\sqrt{162{a}^{5}{b}^{4}}[/latex] Solution [latex]\begin{array}{cc}\sqrt{100\cdot 3}\hfill & \text{Factor perfect square from radicand}.\hfill \\ \sqrt{100}\cdot \sqrt{3}\hfill & \text{Write radical expression as product of radical expressions}.\hfill \\ 10\sqrt{3}\hfill & \text{Simplify}.\hfill \\ \text{ }\end{array}[/latex] [latex]\begin{array}{cc}\sqrt{81{a}^{4}{b}^{4}\cdot 2a}\hfill & \text{Factor perfect square from radicand}.\hfill \\ \sqrt{81{a}^{4}{b}^{4}}\cdot \sqrt{2a}\hfill & \text{Write radical expression as product of radical expressions}.\hfill \\ 9{a}^{2}{b}^{2}\sqrt{2a}\hfill & \text{Simplify}.\hfill \end{array}[/latex] Try It 2
Simplify [latex]\sqrt{50{x}^{2}{y}^{3}z}[/latex].
How To: Given the product of multiple radical expressions, use the product rule to combine them into one radical expression. Express the product of multiple radical expressions as a single radical expression. Simplify. Example 3: Using the Product Rule to Simplify the Product of Multiple Square Roots
Simplify the radical expression.
[latex]\sqrt{12}\cdot \sqrt{3}[/latex]
Solution
[latex]\begin{array}{cc}\sqrt{12\cdot 3}\hfill & \text{Express the product as a single radical expression}.\hfill \\ \sqrt{36}\hfill & \text{Simplify}.\hfill \\ 6\hfill & \end{array}[/latex]
Try It 3
Simplify [latex]\sqrt{50x}\cdot \sqrt{2x}[/latex] assuming [latex]x>0[/latex].
Using the Quotient Rule to Simplify Square Roots
Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the
quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. We can rewrite [latex]\sqrt{\frac{5}{2}}[/latex] as [latex]\frac{\sqrt{5}}{\sqrt{2}}[/latex]. A General Note: The Quotient Rule for Simplifying Square Roots
The square root of the quotient [latex]\frac{a}{b}[/latex] is equal to the quotient of the square roots of [latex]a[/latex] and [latex]b[/latex], where [latex]b\ne 0[/latex].
How To: Given a radical expression, use the quotient rule to simplify it. Write the radical expression as the quotient of two radical expressions. Simplify the numerator and denominator. Example 4: Using the Quotient Rule to Simplify Square Roots
Simplify the radical expression.
[latex]\sqrt{\frac{5}{36}}[/latex]
Solution
[latex]\begin{array}{cc}\frac{\sqrt{5}}{\sqrt{36}}\hfill & \text{Write as quotient of two radical expressions}.\hfill \\ \frac{\sqrt{5}}{6}\hfill & \text{Simplify denominator}.\hfill \end{array}[/latex]
Try It 4
Simplify [latex]\sqrt{\frac{2{x}^{2}}{9{y}^{4}}}[/latex].
Example 5: Using the Quotient Rule to Simplify an Expression with Two Square Roots
Simplify the radical expression.
[latex]\frac{\sqrt{234{x}^{11}y}}{\sqrt{26{x}^{7}y}}[/latex]
Solution
[latex]\begin{array}{cc}\sqrt{\frac{234{x}^{11}y}{26{x}^{7}y}}\hfill & \text{Combine numerator and denominator into one radical expression}.\hfill \\ \sqrt{9{x}^{4}}\hfill & \text{Simplify fraction}.\hfill \\ 3{x}^{2}\text{ }\hfill & \text{Simplify square root}.\hfill \end{array}[/latex]
Try It 5
Simplify [latex]\frac{\sqrt{9{a}^{5}{b}^{14}}}{\sqrt{3{a}^{4}{b}^{5}}}[/latex].
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This task is more complex than the task to solve a quadratic equation, for example, and one must master a significant portion of a textbook – such as Georgi's textbook – and perhaps something beyond it to have everything he needs.
For the 8-dimensional representation of $SU(3)$, things simplify because it's the "adjoint rep" of $SU(3)$ – the vector space that formally coincides with the Lie algebra itself. And the action of the generator $G_i$ on the basis vector $V_j=G_j$ of the adjoint representation is given by $$ G_i (V_j) = [G_i,V_j]= \sum_k f_{ij}{}^k G_k $$This implies that the structure constants $f$ directly encode the matrix elements of the generator $G_i$ with respect to the adjoint representation – $j$ and $k$ label the row and the column, respectively.
The structure constants $f$ determining the commutators may be extracted from all the roots. The whole mathematical structure is beautiful but the decomposition of the generators under the Cartan subalgebra has several pieces, and therefore an even greater number of different types of "pairs of pieces" that appear as the commutators.
Some ($r$, rank) of the generators $G_i$ are identified with the Cartan generators $u_a$. The rest of the generators $G_j$ are uniquely associated with all the roots.
If you only have the Cartan matrix, you effectively have the inner products of the simple roots only. You first need to get all the roots, and those are connected with the $d-r$ (dimension minus rank) root vectors $r_j$.
The commutators of two Cartan generators vanish, $$[h_i,h_j]=0$$The commutator of a Cartan generator with a non-Cartan generator is given by$$[h_i,G_{r(j)}] = r_i G_{r(j)}$$because we organized the non-Cartan generators as simultaneous eigenstates under all the Cartan generators. Finally, the commutator$$[G_{r(i)},G_{r(j)}]$$is zero if $r_i=r_j$. It is a natural linear combination of the $h_i$ generators if the root vectors obey $r_i=-r_j$. If $r_i+r_j$ is a vector that isn't a root vector, the commutator has to vanish. And if $r_i+r_j$ is a root vector but $r_i\neq \pm r_j$, then the commutator is proportional to $G_{r(i)+r(j)}$ corresponding to this "sum" root vector. The coefficient (mostly sign) in front of this commutator is subtle.
Once you have all these commutators, you have effectively restored all the structure constants $f$, and therefore all the matrix entries with respect to the adjoint representation.
To find matrix elements for a general representation is much more complex. You must first figure out what the representations are. Typically, you want to start with the fundamental (and/or antifundamental) rep, and all others may be obtained as terms composing a direct sum decomposition of tensor products of many copies of the fundamental (and/or antifundamental, if it is different) representation.
All the representations may be obtained from the weight lattice which is a subset of the root lattice and is similar. In fact, the weight lattice is the dual (in the "form" vector space sense) of the root lattice under the natural inner product.
In practice, physicists don't ever do the procedures in this order because that's not how Nature asks us to solve problems. We learn how to deal with the groups we need – which, at some moment, includes all the compact simple Lie groups as the "core" (special unitary, orthogonal, symplectic, and five exceptional), and we learn the reps of these Lie groups – the obvious fundamental ones, the adjoint, and the pattern how to get the more complicated ones.
I am afraid that it doesn't make any sense to fill in any "gaps" if you would need to elaborate upon something because in this way, one would gradually be forced to write another textbook on Lie groups and representation theory as this answer, and I don't think that such a work would be appropriately rewarded – even this work wasn't. ;-)This post imported from StackExchange Physics at 2015-11-01 21:04 (UTC), posted by SE-user Luboš Motl
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Let $T$ be a torus. We have a parameterization by $((c+a \cdot cos(v))cos(u),(c+a\cdot cos(v),a\cdot sin(v))$ for $u,v \in [0,2\pi)$. The first fundamental form is given by $E=(c+a\cdot cos(v))^{2}, F=0, G=a^2$
and the second fundamental form is given by $e=-(c+a\cdot cos(v))cos(v),f=0,g=-a$. Since gaussian curvature is equal to determinante of the 2. fundamental form divided by 1. first fundamental, the gaussian curvature is equal to $\frac{cos(v)\cdot a }{a^2(c+a\cdot cos(v))}$
Now I want to integrate the gaussian curvature over T.
Do I integrate the gaussian curvature over T correctly: $\int_{[0,2\pi)^2}\frac{cos(v)\cdot a }{a^2(c+a\cdot cos(v))} du dv$?
This integral seems to me quite uncomputable.
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A
tetrahedral snake, sometimes called a Steinhaus snake, is a collection of tetrahedra, linked face to face.
Steinhaus showed in 1956 that the last tetrahedron in the snake can never be a translation of the first one. This is a consequence of the fact that the group generated by the four reflexions in the faces of a tetrahedron form the free product $C_2 \ast C_2 \ast C_2 \ast C_2$.
For a proof of this, see Stan Wagon’s book The Banach-Tarski paradox, starting at page 68.
The thread $(3|3)$ is the
spine of the $(9|1)$-snake which involves the following lattices \[ \xymatrix{& & 1 \frac{1}{3} \ar@[red]@{-}[dd] & & \\ & & & & \\ 1 \ar@[red]@{-}[rr] & & 3 \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 1 \frac{2}{3} \\ & & & & \\ & & 9 & &} \] It is best to look at the four extremal lattices as the vertices of a tetrahedron with the lattice $3$ corresponding to its point of gravity.
The congruence subgroup $\Gamma_0(9)$ fixes each of these lattices, and the arithmetic group $\Gamma_0(3|3)$ is the conjugate of $\Gamma_0(1)$
\[ \Gamma_0(3|3) = \{ \begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}.\begin{bmatrix} a & b \\ c & d \end{bmatrix}.\begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & \frac{b}{3} \\ 3c & 1 \end{bmatrix}~|~ad-bc=1 \} \] We know that $\Gamma_0(3|3)$ normalizes the subgroup $\Gamma_0(9)$ and we need to find the moonshine group $(3|3)$ which should have index $3$ in $\Gamma_0(3|3)$ and contain $\Gamma_0(9)$.
So, it is natural to consider the finite group $A=\Gamma_0(3|3)/\Gamma_9(0)$ which is generated by the co-sets of
\[ x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix} \qquad \text{and} \qquad y = \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix} \] To determine this group we look at the action of it on the lattices in the $(9|1)$-snake. It will fix the central lattice $3$ but will move the other lattices.
Recall that it is best to associate to the lattice $M.\frac{g}{h}$ the matrix
\[ \alpha_{M,\frac{g}{h}} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \] and then the action is given by right-multiplication.
\[
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}.x=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] That is, $x$ corresponds to a $3$-cycle $1 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 1$ and fixes the lattice $9$ (so is rotation around the axis through the vertex $9$).
To compute the action of $y$ it is best to use an alternative description of the lattice, replacing the roles of the base-vectors $\vec{e}_1$ and $\vec{e}_2$. These latices are projectively equivalent
\[ \mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \quad \text{and} \quad \mathbb{Z} \vec{e}_1 \oplus \mathbb{Z} (\frac{g’}{h} \vec{e}_1 + \frac{1}{h^2M} \vec{e}_2) \] where $g.g’ \equiv~1~(mod~h)$. So, we have equivalent descriptions of the lattices \[ M,\frac{g}{h} = (\frac{g’}{h},\frac{1}{h^2M}) \quad \text{and} \quad M,0 = (0,\frac{1}{M}) \] and we associate to the lattice in the second normal form the matrix \[ \beta_{M,\frac{g}{h}} = \begin{bmatrix} 1 & 0 \\ \frac{g’}{h} & \frac{1}{h^2M} \end{bmatrix} \] and then the action is again given by right-multiplication.
In the tetrahedral example we have
\[ 1 = (0,\frac{1}{3}), \quad 1\frac{1}{3}=(\frac{1}{3},\frac{1}{9}), \quad 1\frac {2}{3}=(\frac{2}{3},\frac{1}{9}), \quad 9 = (0,\frac{1}{9}) \] and \[ \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix}.y = \begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix},\quad \begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix} \] That is, $y$ corresponds to the $3$-cycle $9 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 9$ and fixes the lattice $1$ so is a rotation around the axis through $1$.
Clearly, these two rotations generate the full rotation-symmetry group of the tetrahedron
\[ \Gamma_0(3|3)/\Gamma_0(9) \simeq A_4 \] which has a unique subgroup of index $3$ generated by the reflexions (rotations with angle $180^o$ around axis through midpoints of edges), generated by $x.y$ and $y.x$.
The moonshine group $(3|3)$ is therefore the subgroup generated by
\[ (3|3) = \langle \Gamma_0(9),\begin{bmatrix} 2 & \frac{1}{3} \\ 3 & 1 \end{bmatrix},\begin{bmatrix} 1 & \frac{1}{3} \\ 3 & 2 \end{bmatrix} \rangle \]
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Dini Lipschitz functions for the Dunkl transform in the Space \(\mathrm {L}^{2}(\mathbb {R}^{d},w_{k}(x)dx)\) 799 Downloads Citations Abstract
Using a generalized spherical mean operator, we obtain an analog of Theorem 5.2 in Younis (J Math Sci 9(2),301–312 1986) for the Dunkl transform for functions satisfying the \(d\)-Dunkl Dini Lipschitz condition in the space \(\mathrm {L}^{2}(\mathbb {R}^{d},w_{k}(x)dx)\), where \(w_{k}\) is a weight function invariant under the action of an associated reflection group.
KeywordsDunkl transform Dunkl kernel Generalized spherical mean operator Mathematics Subject Classification42B37 1 Introduction and preliminaries
Younis Theorem 5.2 [13] characterized the set of functions in \(\mathrm {L}^{2}(\mathbb {R})\) satisfying the Cauchy Lipschitz condition by means of an asymptotic estimate growth of the norm of their Fourier transforms, namely, we have the following
Theorem 1.1 1.
\(\Vert f(x+h)-f(x)\Vert _{2}=O\left( \frac{h^{\alpha }}{(\log \frac{1}{h})^{\beta }}\right) \quad as\, h\longrightarrow 0, ~0<\alpha <1, \beta \ge 0\),
2.
\(\int _{|x|\ge r}|\mathcal {F}(f)(x)|^{2}dx=O\left( \frac{r^{-2\alpha }}{(\log r)^{2\beta }}\right) \quad as \,r \longrightarrow +\infty \),
In this paper, we obtain an analog of Theorem 1.1 for the Dunkl transform on \(\mathbb {R}^{d}\). For this purpose, we use a generalized spherical mean operator. We point out that similar results have been established in the Bessel transform [4].
We consider the Dunkl operators \(\mathrm {D}_{i}\); \(1\le i\le d\), on \(\mathbb {R}^{d}\), which are the differential-difference operators introduced by Dunkl in [6]. These operators are very important in pure mathematics and in physics. The theory of Dunkl operators provides generalizations of various multivariable analytic structures, among others we cite the exponential function, the Fourier transform and the translation operator. For more details about these operators see [5, 6, 7]. The Dunkl Kernel \(E_{k}\) has been introduced by Dunkl in [8]. This Kernel is used to define the Dunkl transform.
Let \(\mathrm {R}\) be a root system in \(\mathbb {R}^{d}\), \(W\) the corresponding reflection group, \(\mathrm {R}_{+}\) a positive subsystem of \(\mathrm {R}\) (see [5, 7, 9, 10, 11]) and \(k\) a non-negative and \(W\)-invariant function defined on \(\mathrm {R}\).
Proposition 1.2 1.
\(E_{k}(z,0)=1\).
2.
\(E_{k}(z,w)=E_{k}(w,z)\).
3.
\(E_{k}(\lambda z,w)= E_{k}(z,\lambda w)\).
4.For all \(\nu =(\nu _{1},\ldots ,\nu _{d})\in {\mathbb N}^{d},~ x\in {\mathbb R}^{d},~ z\in {\mathbb C}^{d}\), we havewhere$$\begin{aligned} |\partial _{z}^{\nu }E_{k}(x,z)|\le |x|^{|\nu |}exp(|x||Re(z)|), \end{aligned}$$In particular$$\begin{aligned} \partial _{z}^{\nu }=\frac{\partial ^{|\nu |}}{\partial z_{1}^{\nu _{1}}\ldots \partial z_{d}^{\nu _{d}}},\quad |\nu |=\nu _{1}+\cdots +\nu _{d}. \end{aligned}$$for all \(x,z\in {\mathbb R}^{d}.\)$$\begin{aligned} |\partial _{z}^{\nu }E_{k}(ix,z)|\le |x|^{\nu }, \end{aligned}$$ Lemma 1.3 1.
\(|j_{p}(x)|\le 1\),
2.
\(1-j_{p}(x)=O(1),~x\ge 1\).
3.
\(1-j_{p}(x)=O(x^{2});~0\le x \le 1\).
Lemma 1.4 Proof
(Analog of lemma 2.9 in [3]) \(\square \)
Proposition 1.5 2 Dini Lipschitz condition Definition 2.1 Definition 2.2 Remark Theorem 2.3
Let \(\alpha >1\). If \(f\in Lip(\alpha ,\gamma )\), then \(f\in lip(1,\gamma )\).
Proof Theorem 2.4
If \(\alpha <\beta \), then \(Lip(\alpha ,0)\supset Lip(\beta , 0)\) and \(lip(\alpha , 0)\supset lip(\beta , 0)\).
Proof
We have \(0\le h \le 1\) and \(\alpha <\beta \), then \(h^{\beta }\le h^{\alpha }\).
Then the proof of this theorem. \(\square \)
Theorem 2.5
Let \(f, g\in \mathrm {L}_{k}^{2}(\mathbb {R}^{d})\) such that \(\mathrm {M}_{h}(fg)(x)=\mathrm {M}_{h}f(x)\mathrm {M}_{h}g(x)\). If \(f,g\in Lip(\alpha , \gamma )\), then \(fg\in Lip(\alpha ,\gamma )\).
Proof 3 New results on Dini Lipschitz class Theorem 3.1 Proof
Analog of the theorem 3.1, we obtain this theorem
Theorem 3.2
Now, we give another the main result of this paper analog of theorem 1.1.
Theorem 3.3 1.
\(f\in Lip(\alpha ,\gamma )\)
2.
\(\int _{|\xi |\ge s}|\widehat{f}(\xi )|^{2}w_{k}(\xi )d\xi =O\left( \frac{s^{-2\alpha }}{(\log s)^{2\gamma }}\right) ~as~ s \rightarrow +\infty \)
Proof References 1. 2. 3. 4.Daher, R., El Hamma, M.: Bessel transform of \((k, \gamma )\)-Bessel Lipschitz functions, Hindawi Publishing Corporation. J. Math., vol. 2013, Article ID 418546Google Scholar 5. 6. 7.Dunkl, C.F.: Hankel transforms associated to finite reflection groups. In: Proceedings of Special Session on Hypergeometric Functions in Domains of Positivity. Jack Polynomials and Applications (Tampa, 1991), Contemp. Math. 138(1992), 123–138 (1991)Google Scholar 8. 9. 10. 11. 12. 13. Copyright information Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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Lorenz equations part II: "randomly" rotated homoclinic orbits and chaotic trajectories
1.
Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260
$x' =s(y-x), \quad y'= Rx -y-xz, \quad z'= xy -qz,$
where $s$, $R$, and $q$ are positive parameters. We show by a purely analytic proof that for each non-negative integer $N$, there are positive parameters $s, q, $ and $R$ such that the Lorenz system has homoclinic orbits associated with the origin (i.e., orbits that tend to the origin as $t\to \pm \infty$) which can rotate around the $z$-axis $N/2$ times; namely, the $x$-component changes sign exactly $N$ times, the $y$-component changes sign exactly $N+1$ times, and the zeros of $x$ and $y$ are simple and interlace.
Mathematics Subject Classification:34C3. Citation:Xinfu Chen. Lorenz equations part II: "randomly" rotated homoclinic orbits and chaotic trajectories. Discrete & Continuous Dynamical Systems - A, 1996, 2 (1) : 121-140. doi: 10.3934/dcds.1996.2.121
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Sergey Gonchenko, Ivan Ovsyannikov.
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Homoclinic trajectories and chaotic behaviour in a piecewise linear oscillator.
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Cristina Lizana, Leonardo Mora.
Lower bounds for the Hausdorff dimension of the geometric Lorenz attractor: The homoclinic case.
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Homoclinic orbits for a class of asymptotically quadratic Hamiltonian systems.
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Francesca Alessio, Vittorio Coti Zelati, Piero Montecchiari.
Chaotic behavior of rapidly oscillating Lagrangian systems.
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A.V. Borisov, A.A. Kilin, I.S. Mamaev.
Reduction and chaotic behavior of point vortices on a plane and a sphere.
[10]
C. Foias, M. S Jolly, I. Kukavica, E. S. Titi.
The Lorenz equation as a metaphor for the Navier-Stokes equations.
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G. A. Leonov.
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Yingxiang Xu, Yongkui Zou.
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delay differential equations.
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Night Side Maneuvers
We can minimize
night light pollution (NLP) by turning the thinsat as we approach eclipse. The goal will be to perform one complete rotation of the thinsat per orbit, with it perpendicular to the sun on the day-side of the earth, but turning it by varying amounts on the night side.
Another advantage of the turn is that if thinsat maneuverability is destroyed by radiation or a collision on the night side, it will come out of night side with a slow tumble that won't be corrected. The passive radar signature of the tumble will help identify the destroyed thinsat to other thinsats in the array, allowing another sacrificial thinsat to perform a "rendezvous and de-orbit". If the destroyed thinsat is in shards, the shards will tumble. The tumbling shards ( or a continuously tumbling thinsat ) will eventually fall out of the normal orbit, no longer get J_2 correction, and the thinsat orbit will "eccentrify", decay, and reenter. This is the fail-safe way the arrays will reenter, if all active control ceases.
Maneuvering Thrust and Satellite Power
Neglecting tides, the synodic angular velocity of the m288 orbit is \Large\omega = 4.3633e-4 rad/sec = 0.025°/s. The angular acceleration of a thinsat is 13.056e-6 rad/sec
2 = 7.481e-4°/s 2 with a sun angle of 0°, and 3.740e-4°/s 2 at a sun angle of 60°. Because of tidal forces, a thinsat entering eclipse will start to turn towards sideways alignment with the center of the earth; it will come out of eclipse at a different velocity and angle than it went in with.
If the thinsat is rotating at \omega and either tangential or perpendicular to the gravity vector, it will not turn while it passes into eclipse. Otherwise, the tidal acceleration is \ddot\theta = (3/2) \omega^2 \sin 2 \delta where \delta is the angle to the tangent of the orbit. If we enter eclipse with the thinsat not turning, and oriented directly to the sun, then \delta = 30° .
Three Strategies and a Worst Case Failure Mode
There are many ways to orient thinsats in the night sky, with tradeoffs between light power harvest, light pollution, and orbit eccentricity. If we reduce power harvest, we will need to launch more thinsats to compensate, which makes more problems if the system fails. I will present three strategies for light harvest and nightlight pollution. The actual strategies chosen will be blend of those.
Tumbling
If things go very wrong, thinsats will be out of control and tumbling. In the long term, the uncontrolled thinsats will probably orient flat to the orbital plane, and reflect very little light into the night sky, but in the short term (less than decades), they will be oriented in all directions. This is equivalent to mapping the reflective area of front and back (2 π R
2 ) onto a sphere ( 4 π R 2 ). Light with intensity I shining onto a sphere of radius R is evenly reflected in all directions uniformly. So if the sphere intercepts π R 2 I units of light, it scatters e I R 2/4 units of light (e is albedo) per steradian in all directions. While we will try to design our thinsats with low albedo ( high light absorption on the front, high emissivity on the back), we can assume they will get sanded down and more reflective because of space debris, and they will get broken into fragments of aluminum with shiny edges, adding to the albedo. Assume the average albedo is 0.5, and assume the light scattering is spherical for tumbling.
Source for the above animation: g400.c
Three design orientations
All three orientations shown are oriented perpendicularly in the daytime sky.
Max remains perpendicular in the night sky, Min is oriented vertically in the night sky, and Zero is edge on to the terminator in the night sky. All lose orientation control and are tilted by tidal forces in eclipse - the compensatory thrusts are not shown. Min and Zero are accelerated into a slow turn before eclipse, so they come out of the eclipse in the correct orientation. In all cases, there will probably be some disorientation and sun-seeking coming out of eclipse, until each thinsat calibrates to the optimum inertial turn rate during eclipse. So, there may be a small bit of sky glow at the 210° position, directly overhead at 2am and visible in the sky between 10pm and 6am.
Max NLP: Full power night sky coverage, maximum night light pollution
The most power is harvested if the thinsats are always oriented perpendicular to the sun. During the half of their orbit into the night sky, there will be some diffuse reflection to the side, and some of that will land in the earth's night sky. The illumination is maximum along the equator. For the M288 orbit, about 1/6th of the orbit is eclipsed, and 1/2 of the orbit is in daylight with the diffuse (Lambertian) reflection scattering towards the sun and onto the day side of the earth. Only the two "horns" of the orbit, the first between 90° and 150° (6pm to 10pm) and the second between 210° and 270° (2am to 6am) will reflect night into the light sky. The light harvest averages to 83% around the orbit.
This is the worst case for night sky illumination. Though it is tempting to run thinsats in this regime, extracting the maximum power per thinsat, it is also the worst case for eccentricity caused by light pressure, and the thinsats must be heavier to reduce that eccentricity.
Min NLP: Partial night sky coverage, some night light pollution
This maneuver will put some scattered light into the night sky, but not much compared to perpendicular solar illumination all the way into shadow. In the worst case, assume that the surface has an albedo of 0.5 (typical solar cells with an antireflective coating are less than 0.3) and that the reflected light is entirely Lambertian (isotropic) without specular reflections (which will all be away from the earth). At a 60° angle, just before shadow, the light emitted by the front surface will be 1366W/m
2 × 0.5 (albedo) × 0.5 ( cos 60° ), and it will be scattered over 2π steradians, so the illumination per steradian will be 54W/m 2-steradian just before entering eclipse.
Estimate that the light pollution varies from 0W to 54W between 90° and 150° and that the average light pollution is half of 54W, for 1/3 of the orbit. Assuming an even distribution of thinsat arrays in the constellation, that is works out to an average of 9W/m
2-steradian for all thinsats in M288 orbit.
The full moon illuminates the night side of the equatorial earth with 27mW/m
2 near the equator. A square meter of thinsat at 6400km distance produces 9W/6400000 2 or 0.22 picowatts per m 4 times the area of all thinsats. If thinsat light pollution is restricted to 5% of full moon brightness (1.3mW/m 2), then we can have 6000 km 2 of thinsats up there, at an average of 130 W/m 2, or about 780GW of thinsats at m288. That is about a million tons of thinsats.
The orientation of the thinsat over a 240 minute synodic m288 orbit at the equinox is as follows, relative to the sun:
time min
orbit degrees
rotation rate
sun angle
Illumination
Night Light
0 to 60
0° to 90°
0 ~ \Large\omega
0°
100%
0W
60 to 100
90° to 150°
1 ~ \Large\omega
0° to 60°
100% to 50%
0W to 54W
100 to 140
150° to 210°
4 ~ \Large\omega
60° to 300°
Eclipse
0W
140 to 180
210° to 270°
1 ~ \Large\omega
300° to 0°
50% to 100%
54W to 0W
180 to 240
270° to 0°
0 ~ \Large\omega
0°
100%
0W
The angular velocity change at 0° takes 250/7.481 = 33.4 seconds, and during that time the thinsat turns 0.42° with negligible effect on thrust or power. The angular velocity change at 60° takes 750/3.74 = 200.5 seconds, and during that time the thinsat turns 12.5°, perhaps from 53.7° to 66.3°, reducing power and thrust from 59% to 40%, a significant change. The actual thrust change versus time will be more complicated (especially with tidal forces), but however it is done, the acceleration must be accomplished before the thinsat enters eclipse.
The light harvest averages 78% around the orbit.
Zero NLP: Partial night sky coverage, no night light pollution
In this case, in the night half of the sky the edge of the thinsat is always turned towards the terminator. As long as the thinsats stay in control, they will never produce any nighttime light pollution, because the illuminated side of the thinsat is always pointed away from the night side of the earth. The average illumination fraction is around 68%.
The orientation of the thinsat over a 240 minute synodic m288 orbit at the equinox is as follows, relative to the sun:
time min
orbit degrees
average rotation rate
sun angle
Illumination
Night Light
0 to 60
0° to 90°
0 \Large\omega
0°
100%
0W
60 to 100
90° to 150°
1.5 \Large\omega
0° to 90°
100% to 0%
0W
100 to 140
150° to 210°
3 \Large\omega Start at 3.333\Large\omega
90° to 270°
Eclipse
0W
140 to 180
210° to 270°
1.5 \Large\omega
270° to 0°
0% to 100%
0W
180 to 240
270° to 0°
0 \Large\omega
0°
100%
0W
Pedants and thinsat programmers take note: The actual synodic orbit period is 240 minutes and 6.57 seconds long; that results in 2190.44 rather than 2191.44 sidereal orbits per year, accounting for the annual apparent motion of the sun around the sky.
The light harvest averages 67% around the orbit.
Why would a profit maximizing operator settle for 67% when 83% was possible? Infrared-filtering thinsats reduce launch weight and can use the Zero NLP flip to increase the minimum temperature of a thinsat during eclipses. An IR filtering thinsat in maximum night light pollution mode will have the emissive backside pointed at 2.7K space when it enters eclipse; the thinsat temperature will drop towards 20K if it cannot absorb the 64W/m
2 of 260K black body radiation reaching it from the earth through the 3.5μm front side infrared filter. The thinsat will become very brittle at those temperatures, and the thermal shock could destroy it. If the high thermal emissivity back side is pointed towards the 260K earth, the temperature will drop to 180K - still challenging, but the much higher thermal mobility may heal atomic-scale damage.
Details of the Zero NLP maneuver
In the night sky, assuming balanced thrusters, only tidal forces act on the thinsat, \ddot\theta = -(3/2) \omega^2 \sin( 2 \theta . Integrating numerically from the proper choice of initial rotation rate (10/9ths the average, due to tidal decceleration) we go from 30 degrees "earth relative tilt" at the beginning of eclipse, to 150 degrees tilt at the end of eclipse, with the night sky sending 260K infrared at the back side throughout the manuever. The entire disk of the earth is always in view, and always emits the same amount of infrared to a given radius, but the Lambertian angle of absorption changes.
Power versus angle
Night light pollution versus hour
The night light pollution for 1 Terawatt of thinsats at M288. Mirror the graph for midnight to 6am. Some light is also put into the daytime sky (early morning and late afternoon), but it will be difficult to see in the glare of sunlight. The Zero NLP option puts no light in the night sky, so that curve is far below the bottom of this graph.
Source for the above two graphs: nl02.c
Note to Astronomers
Yes, we will slightly occult some of your measurements, though we won't flash your images like Iridium. We will also broadcast our array ephemerides, and deliver occultation schedules accurate to the microsecond, so you can include that in your luminosity calculations. Someday, server sky is where you will perform those calculations, rather than in your own coal powered (and haze producing) computer.
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For some strange reason, this:
\hat{\dot{\bm{\phi}}}
prints the dot and the hat slightly off the left of the phi letter. Any ideas if this can be fixed? thanks. Using bm package for bold symbols.
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$\mkern4mu\hat{\mkern-4mu\dot{\bm{\phi}}}$$\skew{3}{\hat}{\skew{3}{\dot}{\bm{\phi}}}$
As to
why this happens: TeX handles accents over single characters differently, using the
\skewchar of the font. Thus you get nicely aligned math accents only over single characters, and
\dot{\bm{\phi}} is already a composed symbol.
If you want double accents, you can use the
amsmath or the
accents package. However, both packages don't work together well with
bm, so the simple
\hat{\dot{\bm{\phi}}} won't work with
\usepackage{accents}. It took me forever to find out what happens, but here's a fix:
\documentclass{minimal}\usepackage{bm,accents}\makeatletter\newcommand\accbm[1]{\use@mathgroup{\M@OMS}{5}{\bm{#1}}}\makeatother\begin{document}$\hat{\dot{\accbm{\phi}}}$\end{document}
This gives you . Thus, whenever you want to put accents over
\bmed characters, you just have to use
\accbm. I should add a few words of caution: 1. It works with
\M@OMS, but I don't know if it's the Right Way. 2. In this example, it works with font family
5 in the
\use@mathgroup, but I don't know if it'll always work with
5.
UPDATE: Philippe Goutet's great answer to my question How do I identify the encoding and family of a new math font? provides a solution that determines the font family automatically.
It's easier to get accent positioning right if the accents are bold as well as the base.Also if
\bm does get confused by its argument adding an extra set of
{} stops it trying to disassemble the math construct and instead it sets the entire argumant as a sing;e expression (using
\boldmath, essentially) this is a lot simpler (but slower).Note the extra set of braces, and the better accent positioning in row 9.
\documentclass{article}\usepackage{amsmath}\usepackage{bm}\begin{document}1 $\mathcal{A} \bm{\mathcal{A}}$2 $\hat{\mathcal{A}} \bm{\hat{\mathcal{A}}}$3 $\mathcal{\hat{A}} \bm{\mathcal{\hat{A}}}$4 $\hat{\hat{\mathcal{A}}} \bm{\hat{\hat{\mathcal{A}}}}$5 $\mathcal{\hat{\hat{A}}} \bm{\mathcal{\hat{\hat{A}}}}$6 $\phi \bm{\phi}$7 $\hat{\phi} \bm{\hat{\phi}}$8 $\hat{\hat{\phi}} \bm{\hat{\hat{\phi}}}$9 $\hat{\hat{\phi}} \bm{{\hat{\hat{\phi}}}}$\end{document}
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Current browse context:
hep-th
Change to browse by: Bookmark(what is this?) Mathematics > Differential Geometry Title: Spectral sections, twisted rho invariants and positive scalar curvature
(Submitted on 23 Sep 2013 (v1), last revised 25 Apr 2014 (this version, v3))
Abstract: We had previously defined the rho invariant $\rho_{spin}(Y,E,H, g)$ for the twisted Dirac operator $\not\partial^E_H$ on a closed odd dimensional Riemannian spin manifold $(Y, g)$, acting on sections of a flat hermitian vector bundle $E$ over $Y$, where $H = \sum i^{j+1} H_{2j+1} $ is an odd-degree differential form on $Y$ and $H_{2j+1}$ is a real-valued differential form of degree ${2j+1}$. Here we show that it is a conformal invariant of the pair $(H, g)$. In this paper we express the defect integer $\rho_{spin}(Y,E,H, g) - \rho_{spin}(Y,E, g)$ in terms of spectral flows and prove that $\rho_{spin}(Y,E,H, g)\in \mathbb Q$, whenever $g$ is a Riemannian metric of positive scalar curvature. In addition, if the maximal Baum-Connes conjecture holds for $\pi_1(Y)$ (which is assumed to be torsion-free), then we show that $\rho_{spin}(Y,E,H, rg) =0$ for all $r\gg 0$, significantly generalizing our earlier results. These results are proved using the Bismut-Weitzenb\"ock formula, a scaling trick, the technique of noncommutative spectral sections, and the Higson-Roe approach. Submission historyFrom: Varghese Mathai [view email] [v1]Mon, 23 Sep 2013 09:48:25 GMT (23kb) [v2]Sat, 2 Nov 2013 20:05:59 GMT (326kb,D) [v3]Fri, 25 Apr 2014 20:53:53 GMT (313kb,D)
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does this series converge/diverge conditionally or absolutly
$\sum_{n=2}^{\infty} (-1)^n \cdot \frac{\sqrt{n}}{(-1)^n + \sqrt{n}} \cdot \sin(\frac{1}{\sqrt{n}})$
i can use the facts that:
$\lim_{x\to0}\frac{\sin x}{x} =1 $;
$\sin(\frac{1}{\sqrt{n}})$ is monotone decreasing
my problem is the alternating $-1$'s in the denominator, i dont know how to deal with that, though the fraction looks similar to an $e$ based limit
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Why does the heat capcity not jump at an energy threshold
To understand why heat capacity rises continuously and not step-wise consider a simple harmonic oscillator with eigenenergies $\epsilon_n = \hbar \omega \left(n + \frac 1 2\right)$. In the following $k_B = 1$ and $\beta = 1/T$.
The partition function is given by:$$ Z = \sum_{n=0}^\infty e^{-\beta \epsilon_n} = e^{-\frac 1 2 \beta \hbar \omega} \frac{e^{-\beta \hbar \omega}}{1 - e^{-\beta\hbar\omega}} = e^{-\frac 1 2 \beta \hbar \omega} \frac{1}{e^{\beta \hbar \omega} - 1}. $$
The energy in the system is then given by:$$E = \frac 1 Z \sum_n \epsilon_n e^{\beta \epsilon_n} = -\partial_\beta \ln(Z) = \frac 1 2 \hbar \omega + \frac{\hbar \omega e^{\beta\hbar\omega}}{e^{\beta\hbar\omega}-1} = \frac 1 2 \hbar \omega + \frac{\hbar\omega}{1 - e^{-\beta\hbar\omega}}.$$
The heat capacity will be given by $C = \frac{dE}{dT}$:$$C = \partial_T E = (\partial_T \beta) \partial_\beta E = \frac 1 {T^2} \frac{\hbar^2\omega^2e^{-\beta\hbar\omega}}{(1 - e^{-\beta\hbar\omega})^2}.$$While this term looks weird, so you could (a) plot it or (b) consider limit cases (and argue with continuity and monotonicity in between). But one can easily see that this term is continuous (so there no discrete jump in the heat capacity).
If $T \ll \hbar\omega$, the argument of the exponential gets large, so we can neglect it compared to the 1 in the denominator, this gives an expression$$C = \frac 1 {T^2} \hbar^2 \omega^2 e^{-\hbar\omega/T}.$$Which gets exponentially small.
If $T \gg \hbar\omega$, the argument of the exponential is close to zero, so we can Taylor expand the exponential, giving:$$C = \frac 1 {T^2} \hbar^2 \omega^2 \frac{1 - \beta\hbar\omega}{\big(1-(1 - \beta\hbar\omega) \big)^2} \approx 1.$$Which coincides with the well known result: In the high temperature limit there is one $T$ of energy per oscillatory degree of freedom.
Why does this hold in general? Below an excitation threshold the energy will have a contribution $\frac 1 Z e^{-\beta \Delta E} \Delta E$, this gives no zero contribution when $T < \Delta E$, but is rather exponentially suppressed when $T \ll \Delta E$ and then the coefficient goes to one as $T$ reaches and surpasses $T \approx \Delta E$.
As a final remark, there are special situations where the heat capacity can jump: Phase transitions.
Why does the heat capacity go to zero?
In the following we will always normalize the ground state energy of the system to 0 for simplicity.
Explicitely, in the case of an solid state system there is no continuous translation or rotation that can be excited, the only thing you can store energy in are phonons and electrons, that phonons lead to the $\propto T^3$ behaviour is simple to prove (in metals, electrons actually dominate for $T \to 0$ with a heat-capacity $\propto T$): The lowest lying phonon mode is acoustic, that is has linear dispersion and no gap (this can be seen as a direct consequence of the Goldstone theorem: Phonons are the Goldstone modes of the broken translation symmetry), this means that in 3d the density of states has the form $z \propto \epsilon^2$ for small $\epsilon$, the heat capacity is then given by:\begin{align*}C &= \partial_T \int_0^\infty d\epsilon\, z(\epsilon) \epsilon n_B(\beta \epsilon) \propto \partial_T \int_0^\infty d\epsilon\, \epsilon^2 \epsilon e^{-\beta\epsilon} \\ &\propto \partial_T T^4 \int_0^\infty dx\, x^3 n_B(x) \propto T^3.\end{align*}Where $n_B$ is the Bose distribution and the significant step was the substitution $x = \beta \epsilon$ (that is $\epsilon = Tx$). This also shows that for a system without an excitation gap the heat capacity can go to zero as $T \to 0$, but it will not be exponentially small as $T \to 0$ but rather follow a power law. A similar argument will work for translations of gas molecules as well (they have a spectrum $\epsilon \propto k^2$, so $z(\epsilon) \propto \epsilon^{1/2}$, which will give $C \propto T^{3/2}$). With Bosons there are some complications due to Bose-Einstein-Condensation (which can be fixed by considering the ground state separately). With Fermions you get in trouble when reaching the degenerate Fermi gas regime (there the heat capacity will behave as $\propto T$ as only electrons close to the Fermi level can be excited).
As a final remark, the requirement that the heat capacity goes to zero for $T \to 0$ is an alternative formulation of the third law of thermodynamics. So it should not be too surprising it holds.
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Notice that $A$ is upper triangular, so you can just set $L = {\rm I}_2$, $U = A$.
Without that, we can do some computation. I denote elements of $L$ as $l_{ij}$ and elements of $U$ as $u_{ij}$ (both for $i=1,2$ and $j=1,2$).
If the first row of $L$ is zero, then the first row of $A$ is zero. So, $l_{11} \ne 0$, i.e., $l_{11} = 1$. By definition, $l_{12} = 0$ and $u_{21} = 0$. We now have:\begin{align*}0 &= l_{11}u_{11} + l_{12}u_{21} = u_{11} &\Rightarrow \quad u_{11} = 0, \\a &= l_{11}u_{12} + l_{12}u_{22} = u_{12} &\Rightarrow \quad u_{12} = a, \\0 &= l_{21}u_{11} + l_{22}u_{21} = 0, \\b &= l_{21}u_{12} + l_{22}u_{22} = l_{21}a + l_{22}u_{22}.\end{align*}Since there are no more constraints, we can choose $l_{21} = 0$, $l_{22} = 1$, $u_{22} = b$ (because such choice is convenient). Finally, we get the solution from the begining of this answer:$$\begin{bmatrix} 0 & a \\ 0 & b \end{bmatrix} = A = LU = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & a \\ 0 & b \end{bmatrix}.$$However, a different choice of $l_{21}, l_{22}, u_{22}$ would have produced a different LU factorization of $A$.
To address your concern:
Doesn't this contradict the theorem for $LU$ factorization which says for a matrix $A$: If the leading principal minors of $A$ are all nonsingular, then $A$ has an $LU$ decomposition. Clearly, the leading principal minors of $A$ are singular.
It says "if $A$ has something, then it has LU". It doesn't say that $A$ doesn't have LU if some of the leading minors are singular.
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Reynolds Number (Re) - Blayne Sarazin Contents Reynolds Number
The Reynolds number is a dimensionless quantity in fluid mechanics that is used to help predict flow patterns in different fluid flow situations. The Reynolds Number serves as a guide to the laminar-turbulent transition in a particular flow situation,
1 and for the scaling of similar but different-sized flow situations. The Reynolds number is often used to predict the velocity at which a certain fluid flow turns turbulent, while it can also be used to determine what state of flow the fluid in question is currently under. Calculation of the Reynolds number depends heavily on what type of fluid is being utilized, as well as through what type of channel (i.e. pipe flow, duct, open channel, etc.) this fluid is travelling. Figure 1 is a good example of a fluid experiencing all three types of flows: laminar at the bottom, transitional near the middle of the stream (if only very briefly), and turbulent flow towards the top.
The concept was first introduced in 1851 by George Gabriel Stokes,
2 however it was named by Arnold Sommerfield in 1908 3 after Osborne Reynolds, who popularized its use in 1883. 4 Definition
The Reynolds Number is defined as the ratio of inertial forces to viscous forces in a flowing fluid. It is used in many fluid flow correlations and is used to describe the boundaries of fluid flow regimes (laminar, transitional and turbulent).
1 Viscous force is what tends to keep the layers moving smoothly. When these forces are sufficiently high, this removes any disturbances from the flow and we see what we call laminar flow. However, as velocity increases, inertia forces increase and particles are pushed out of the smoother path. This causes disturbances within the flow, and will eventually lead to what we call turbulent flow. 11 Determining whether a flow is experiencing laminar or turbulent flow is quite simple. Reynold's number tells us everything we need to know about the behavior of a given flow scenario.
Reynolds number is determined via the following equation:
[math] \mathrm{Re} = \frac{\rho v L}{\mu} [/math]
where,
[math]\rho[/math] is the density of the fluid (SI Units: kg/m^3) v is the velocity of the fluid (SI Units: m/s) L is the characteristic length of the fluid. This varies depending on through what the fluid is flowing (SI Units: m) [math] \mu [/math] is the dynamicviscosity of the fluid (SI Units: Ns/m 2)
You will sometimes see Reynolds number in a simplified version such as this:
[math] \mathrm{Re} = \frac{v L}{\nu} [/math]
where [math]\nu[/math] is simply the
kinematic viscosity of the acting fluid (SI Units: m 2/s). ([math] \nu = \mu/\rho) [/math]
The Reynolds Number can be used to determine if flow is laminar, transient or turbulent.
5 The flow is laminar - when Re < 2300 transient - when 2300 < Re < 4000 turbulent - when Re > 4000 Flow Types
Reynolds number can tell us the behavior of the flow we are analyzing. Each flow phase corresponds with a specific range of Reynolds numbers. The flow types can be broken down into the following three branches:
Laminar
Laminar flow is the flow that corresponds with low velocities and Reynolds numbers less than 2300.
5 In this type of flow, the fluid flows in parallel layers, with no disruption between the layers. 7 At low enough velocities, the fluid will tend to flow without lateral mixing, while adjacent layers simply slide past one another. This can be especially important in microfluidics when you do not want lateral mixing. These phenomenon are displayed well by the streamlines depicted in the laminar flow case of Figure 2. Note that the flow is very clean and without disturbance. Furthermore, laminar flow is entirely reversible. This means that we can always return to almost exactly where we began when the flow started. The following video expertly displays this phenomenon (skip ahead to 20 seconds): Transitional (transient)
Transitional or transient flow is the phase of flow that occurs between laminar and turbulent flow, and corresponds to Reynolds numbers that land between 2300 and 4000
5. In this type of flow, there is a mixture of laminar and turbulent flows present. As Reynolds number increase from 2300 to ~4000, there are an increasing amount of disturbances appearing within the flow. Turbulent
Turbulent flow is the most common form of flow in nature, and corresponds to the Reynolds numbers higher than a value of 4000
5. Turbulent flow is ultimately described as chaotic and unpredictable, and is often seen with fluids at high velocities. The flow undergoes irregular fluctuations, or mixing, and continuously changes magnitude and direction 6. As can be seen from figure 2, the sphere in the upper portion of the figure has steady streamlines in front of it, but severe eddy (vortex) formation behind it. Since turbulent flow is extremely hard to measure compared to laminar flow, experimental tools such as a hot wire probe 10 must be used in order to obtain good results. A hot wire probe is a device that has a very fast response time, with a probe that can respond to temperature changes within 1 millisecond, which makes this tool a good candidate for measuring flows (such as air) that experience constant, rapid change as we see in turbulent flow. Flow Scenarios
Calculation of the Reynolds number for a given fluid depends on several things such as speed, density, and viscosity. Two of these are properties of the fluid itself and are usually pretty easy to look up. Speed is often a given, or is one of the things you are trying to solve for if initially given a Reynolds number. However, the characteristic length is something that changes based upon what the fluid is flowing through. The following sections will describe how calculation of characteristic length changes, and will show the equations to use in each of these situations.
Pipe Flow
Let's say you have water flowing through a pipe with a diameter of 25 millimeters at a speed of 5 meters per second. If we take a look at the equation for Reynolds number defined above, we simply need to replace to the characteristic length term with a D, which denotes the diameter of the pipe. The equation then looks as such:
[math] \mathrm{Re} = \frac{\rho v D}{\mu} [/math]
where D is the diameter of the pipe (SI units: m)
For this specific scenario, the equation with the aforementioned values plugged in becomes:
[math] \mathrm{Re} = \frac{(1000kg/m^3)(5 m/s)(0.025 m)}{1.002 Ns/m^2} = 124.75 [/math]
This is laminar flow based on the previously defined boundaries. To create turbulent flow, the velocity or diameter of the pipe would have to be increased. Increasing either would easily force the flow to approach turbulence.
Duct
You can imagine a duct as either a square or rectangular structure through which the fluid is flowing. For this, the calculation of Reynolds number changes slightly due to the fact that we use something called the
hydraulic diameter is the characteristic length. 5 The hydraulic diameter, [math]D_h[/math], is defined by the following equation 5:
[math] \mathrm{D_h} = \frac{4A}{U} [/math]
where,
[math] D_h [/math] is the hydraulic diameter of the duct. A is the area of the duct. U is the wettedperimeter of the duct.
While the U term may be somewhat confusing because it is not simply just perimeter, wetted perimeter just means the cross-sectional perimeter that is in contact with the acting fluid. For a square, this U term becomes [math] 2*(a+b) [/math]
So the [math] D_h [/math] term turns into [math] \mathrm{D_h} = \frac{4*(a*b)}{2*(a+b)} [/math] which simplifies into the following equation we use to calculate the hydraulic diameter:
[math] \mathrm{D_h} = \frac{2*(a*b)}{a+b} [/math]
Therefore, our original equation for Reynold's number takes the following form when were are dealing with a rectangular duct:
[math] \mathrm{Re} = \frac{\rho v D_h}{\mu} [/math]
Open Channel
An open channel or duct behaves similarly to the previously mentioned closed duct. The only thing that changes in this case is the wetted perimeter. The following problem explores the small difference between open and closed ducts:
The perfect example of an open duct that transports a fluid is an aqueduct, which was used to transport water for various reasons. Imagine we have an aqueduct with the dimensions shown in Figure 3 and we want to calculate the Reynold's number for water flowing at 25 meters per second.
The hydraulic diameter for this case is simply:
[math] \mathrm{D_h} = \frac{4*(2*1)}{2+1+1} = 2 [/math]
Note that the wetted perimeter for an open channel differs slightly because there are only three walls used when calculating the hydraulic diameter. This is the only difference between open and closed ducts.
[math] \mathrm{Re} = \frac{(1000kg/m^3)(25 m/s)(2 m)}{1.002 Ns/m^2} = ~ 49900 [/math]
Therefore, the flow within this duct is fully-developed turbulent flow.
Reynolds Number in Microscopic Flow
As far as Reynold's number and microfluidics is concerned, flow is almost always laminar. Meaning that almost without exception, microfluidics experience extremely small Reynolds numbers. The reasons the Reynolds numbers are kept so low is a result of low velocities and small channel sizes. Typical velocities range from 1 micrometer per second up to 1 centimeter per second, while channel radii ranges from 1-100 micrometers.
12 Constrained by these dimensions, the Reynolds number lands safely within the laminar regime, the viscous forces almost dominate over the inertial forces, resulting in smooth, laminar flow. In Physiology (Hemodynamics)
Hemodynamics is a word specifically used to address blood flow within the human body.
14 When considering how Reynolds number works in the body, we must first address a few important things about flow within the body. When dealing with the circulatory system, the following key differences must be addressed: 14 Blood is a non-Newtonian fluid, meaning that blood viscosity is not constant. Flow in the body is pulsatile. In this case, the flow moves forward in pulses. Furthermore, between the pulses the flow actually reverses direction for a very short period of time. Blood vessels are elastic "pipes" whose shapes and diameters constantly change.
Calculating Reynolds number can only be done locally, and is not representative of the flow everywhere in the body at that time. For this application, one would treat arteries like pipes. Further measurements may be necessary to attain flow velocity. With that being said, blood flow in the body is generally laminar, with the exception usually occurring in the ascending aorta where this flow can be disrupted and turn turbulent.
15 The location of the ascending aorta can be seen in Figure 5. This turbulent flow can also occur in large arteries at branch points as well as across stenotic heart valves. Ideally, the critical Reynolds number is high enough such that turbulent flow is not common in the circulatory system. Turbulent blood flow within the circulatory system is never a good thing, with turbulent flow being linked to heart murmurs 15 and aneurysm formation at arterial branch points. 16 References
1. neutrium.net/fluid_flow/reynolds-number/
2. Stokes, George (1851). "On the Effect of the Internal Friction of Fluids on the Motion of Pendulums". Transactions of the Cambridge Philosophical Society. 9: 8–106. 3. Sommerfeld, Arnold (1908). "Ein Beitrag zur hydrodynamischen Erkläerung der turbulenten Flüssigkeitsbewegüngen (A Contribution to Hydrodynamic Explanation of Turbulent Fluid Motions)". International Congress of Mathematicians . 3: 116–124 4. Reynolds, Osborne (1883). "An experimental investigation of the circumstances which determine whether the motion of water shall be direct or sinuous, and of the law of resistance in parallel channels". Philosophical Transactions of the Royal Society. 174 (0): 935–982 5. www.engineeringtoolbox.com/reynolds-number-d_237.html 6. abyss.uoregon.edu/~js/glossary/turbulent_flow.html 7. Batchelor, G. (2000). Introduction to Fluid Mechanics 8. www.nuclear-power.net/nuclear-engineering/fluid-dynamics/laminar-flow-viscous/ 9. en.wikipedia.org/wiki/Reynolds_number 10. web.mst.edu/~cottrell/ME240/Resources/Fluid_Flow/Fluid_flow.pdf 11. http://www.uobabylon.edu.iq/eprints/paper_2_2117_1369.pdf 12. Squires, Todd (2004). "Microfluidics: Fluid Physics at the Nanoliter Scale". Review of Modern Physics. 77; 977. 13. Cheng, D. (2007). "Laminar Flow in Microfluidic Channels". Expo 2007. Department of Electrical and Computer Engineering, University of Wisconsin-Madison. 14. http://www.sci.utah.edu/~macleod/bioen/be6000/notes/L09-hemo.pdf 15. http://www.cvphysiology.com/Hemodynamics/H007 16. Foutrakis GN1, Yonas H, Sclabassi RJ. "Saccular aneurysm formation in curved and bifurcating arteries". AJNR Am J Neuroradiol. 1999 Aug;20(7):1309-17.
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Maryam Aliakbarpour,Themis Gouleakis,John Peebles,Ronitt Rubinfeld,Anak Yodpinyanee;
Proceedings of the Thirty-Second Conference on Learning Theory, PMLR 99:34-82, 2019.
Abstract
In this work, we consider the sample complexity required for testing the monotonicity of distributions over partial orders. A distribution $p$ over a poset is {\em monotone} if, for any pair of domain elements $x$ and $y$ such that $x \preceq y$, $p(x) \leq p(y)$. To understand the sample complexity of this problem, we introduce a new property called \emph{bigness} over a finite domain, where the distribution is $T$-big if the minimum probability for any domain element is at least $T$. We establish a lower bound of $\Omega(n/\log n)$ for testing bigness of distributions on domains of size $n$. We then build on these lower bounds to give $\Omega(n/\log{n})$ lower bounds for testing monotonicity over a matching poset of size $n$ and significantly improved lower bounds over the hypercube poset. We give sublinear sample complexity bounds for testing bigness and for testing monotonicity over the matching poset. We then give a number of tools for analyzing upper bounds on the sample complexity of the monotonicity testing problem.
@InProceedings{pmlr-v99-aliakbarpour19a,title = {Towards Testing Monotonicity of Distributions Over General Posets},author = {Aliakbarpour, Maryam and Gouleakis, Themis and Peebles, John and Rubinfeld, Ronitt and Yodpinyanee, Anak},booktitle = {Proceedings of the Thirty-Second Conference on Learning Theory},pages = {34--82},year = {2019},editor = {Beygelzimer, Alina and Hsu, Daniel},volume = {99},series = {Proceedings of Machine Learning Research},address = {Phoenix, USA},month = {25--28 Jun},publisher = {PMLR},pdf = {http://proceedings.mlr.press/v99/aliakbarpour19a/aliakbarpour19a.pdf},url = {http://proceedings.mlr.press/v99/aliakbarpour19a.html},abstract = {In this work, we consider the sample complexity required for testing the monotonicity of distributions over partial orders. A distribution $p$ over a poset is {\em monotone} if, for any pair of domain elements $x$ and $y$ such that $x \preceq y$, $p(x) \leq p(y)$. To understand the sample complexity of this problem, we introduce a new property called \emph{bigness} over a finite domain, where the distribution is $T$-big if the minimum probability for any domain element is at least $T$. We establish a lower bound of $\Omega(n/\log n)$ for testing bigness of distributions on domains of size $n$. We then build on these lower bounds to give $\Omega(n/\log{n})$ lower bounds for testing monotonicity over a matching poset of size $n$ and significantly improved lower bounds over the hypercube poset. We give sublinear sample complexity bounds for testing bigness and for testing monotonicity over the matching poset. We then give a number of tools for analyzing upper bounds on the sample complexity of the monotonicity testing problem. }}
%0 Conference Paper%T Towards Testing Monotonicity of Distributions Over General Posets%A Maryam Aliakbarpour%A Themis Gouleakis%A John Peebles%A Ronitt Rubinfeld%A Anak Yodpinyanee%B Proceedings of the Thirty-Second Conference on Learning Theory%C Proceedings of Machine Learning Research%D 2019%E Alina Beygelzimer%E Daniel Hsu%F pmlr-v99-aliakbarpour19a%I PMLR%J Proceedings of Machine Learning Research%P 34--82%U http://proceedings.mlr.press%V 99%W PMLR%X In this work, we consider the sample complexity required for testing the monotonicity of distributions over partial orders. A distribution $p$ over a poset is {\em monotone} if, for any pair of domain elements $x$ and $y$ such that $x \preceq y$, $p(x) \leq p(y)$. To understand the sample complexity of this problem, we introduce a new property called \emph{bigness} over a finite domain, where the distribution is $T$-big if the minimum probability for any domain element is at least $T$. We establish a lower bound of $\Omega(n/\log n)$ for testing bigness of distributions on domains of size $n$. We then build on these lower bounds to give $\Omega(n/\log{n})$ lower bounds for testing monotonicity over a matching poset of size $n$ and significantly improved lower bounds over the hypercube poset. We give sublinear sample complexity bounds for testing bigness and for testing monotonicity over the matching poset. We then give a number of tools for analyzing upper bounds on the sample complexity of the monotonicity testing problem.
Aliakbarpour, M., Gouleakis, T., Peebles, J., Rubinfeld, R. & Yodpinyanee, A.. (2019). Towards Testing Monotonicity of Distributions Over General Posets. Proceedings of the Thirty-Second Conference on Learning Theory, in PMLR 99:34-82
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Difference between revisions of "Probability Seminar"
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+ + + Revision as of 09:45, 1 May 2019 Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 14, Timo Seppäläinen, UW-Madison
Title:
Geometry of the corner growth model
Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah).
February 21, Diane Holcomb, KTH
Title:
On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Title: Quantitative homogenization in a balanced random environment
Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison).
Wednesday, February 27 at 1:10pm Jon Peterson, Purdue
Title:
Functional Limit Laws for Recurrent Excited Random Walks
Abstract:
Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina.
March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the
character ratio:
$$ \text{trace}(\rho(g))/\text{dim}(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant
rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM). April 4, Philip Matchett Wood, UW-Madison
Title:
Outliers in the spectrum for products of independent random matrices
Abstract: For fixed positive integers m, we consider the product of m independent n by n random matrices with iid entries as in the limit as n tends to infinity. Under suitable assumptions on the entries of each matrix, it is known that the limiting empirical distribution of the eigenvalues is described by the m-th power of the circular law. Moreover, this same limiting distribution continues to hold if each iid random matrix is additively perturbed by a bounded rank deterministic error. However, the bounded rank perturbations may create one or more outlier eigenvalues. We describe the asymptotic location of the outlier eigenvalues, which extends a result of Terence Tao for the case of a single iid matrix. Our methods also allow us to consider several other types of perturbations, including multiplicative perturbations. Joint work with Natalie Coston and Sean O'Rourke.
April 11, Eviatar Procaccia, Texas A&M Title: Stabilization of Diffusion Limited Aggregation in a Wedge.
Abstract: We prove a discrete Beurling estimate for the harmonic measure in a wedge in $\mathbf{Z}^2$, and use it to show that Diffusion Limited Aggregation (DLA) in a wedge of angle smaller than $\pi/4$ stabilizes. This allows to consider the infinite DLA and questions about the number of arms, growth and dimension. I will present some conjectures and open problems.
April 18, Andrea Agazzi, Duke
Title:
Large Deviations Theory for Chemical Reaction Networks
Abstract: The microscopic dynamics of well-stirred networks of chemical reactions are modeled as jump Markov processes. At large volume, one may expect in this framework to have a straightforward application of large deviation theory. This is not at all true, for the jump rates of this class of models are typically neither globally Lipschitz, nor bounded away from zero, with both blowup and absorption as quite possible scenarios. In joint work with Amir Dembo and Jean-Pierre Eckmann, we utilize Lyapunov stability theory to bypass this challenges and to characterize a large class of network topologies that satisfy the full Wentzell-Freidlin theory of asymptotic rates of exit from domains of attraction. Under the assumption of positive recurrence these results also allow for the estimation of transitions times between metastable states of this class of processes.
April 25, Kavita Ramanan, Brown
Title:
Beyond Mean-Field Limits: Local Dynamics on Sparse Graphs
Abstract: Many applications can be modeled as a large system of homogeneous interacting particle systems on a graph in which the infinitesimal evolution of each particle depends on its own state and the empirical distribution of the states of neighboring particles. When the graph is a clique, it is well known that the dynamics of a typical particle converges in the limit, as the number of vertices goes to infinity, to a nonlinear Markov process, often referred to as the McKean-Vlasov or mean-field limit. In this talk, we focus on the complementary case of scaling limits of dynamics on certain sequences of sparse graphs, including regular trees and sparse Erdos-Renyi graphs, and obtain a novel characterization of the dynamics of the neighborhood of a typical particle. This is based on various joint works with Ankan Ganguly, Dan Lacker and Ruoyu Wu.
Friday, April 26, Colloquium, Van Vleck 911 from 4pm to 5pm, Kavita Ramanan, Brown
Title:
Tales of Random Projections
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
Tuesday , May 7, Van Vleck 901, 2:25pm, Duncan Dauvergne (Toronto)
Title: The directed landscape
Abstract: I will describe the construction of the full scaling limit of (Brownian) last passage percolation: the directed landscape. The directed landscape can be thought of as a random scale-invariant `directed' metric on the plane, and last passage paths converge to directed geodesics in this metric. The directed landscape is expected to be a universal scaling limit for general last passage and random growth models (i.e. TASEP, the KPZ equation, the longest increasing subsequence in a random permutation). Joint work with Janosch Ormann and Balint Virag.
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I am reading this resource to learn statistical mechanics: http://blancopeck.net/Statistics.pdf
I am trying to learn about the partition function, which as I understand it, is equal to the number of legal configurations, in some of these randomized systems that we study to simulate molecular dynamics (hard disks, clothes pins on a line, etc.).
In general, it makes sense to me that the partition function can be related to the probability of acceptance of a configuration via the following:
$$Z(\eta) = Z(0) \,P_\text{accept}(\eta)$$
However, on page 99, in equation 2.8, I see the following form of that partition function:
$$Z(\eta) \approx V^N \exp[-2(N-1)\eta] \, .$$
It seems to me like this equation is only for use with hard disks, since at the bottom of page 98, it says at the beginning of the paragraph, "We shall now determine $P_\text{accept}(\eta)$ for the hard-disk system". Is this only for hard disks?
Then, I see another expression for the partition function on page 270, equation 6.5:
$$Z(\eta) = (L-2 N \sigma)^N $$
From the explanation given in the book, this equation 6.5 seems to be true only for the case when we are using Monte Carlo sampling.
Is the partition function algorithm-dependent, or configuration-dependent? Equation 2.8 is dependent on hard disks which is a configuration, which makes me think it is configuration dependent, but then equation 6.5 makes me think it is algorithm-dependent, since it is derived from using Monte Carlo algorithm.
Which one is it?
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Kaushar Ali
Articles written in Journal of Earth System Science
Volume 119 Issue 6 December 2010 pp 753-762
Surface snow and lake water samples were collected at different locations around Indian station at Antarctica, Maitri, during December 2004-March 2005 and December 2006-March 2007.Samples were analyzed for major chemical ions. It is found that average pH value of snow is 6.1. Average pH value of lake water with low chemical content is 6.2 and of lake water with high chemical content is 6.5.The Na
+ and Cl − are the most abundantly occurring ions at Antarctica. Considerable amount of SO$^{2-}_{4}$ is also found in the surface snow and the lake water which is attributed to the oxidation of DMS produced by marine phytoplankton.Neutralization of acidic components of snow is mainly done by NH$^{+}_{4}$ and Mg 2+. The Mg 2+, Ca 2+ and K + are nearly equally effective in neutralizing the acidic components in lake water.The NH$^{+}_{4}$ and SO$^{2-}_{4}$ occur over the Antarctica region mostly in the form of (NH 4)2SO 4.
Volume 121 Issue 2 April 2012 pp 373-383
Data on surface ozone concentration compiled for a 10-year period from 1990 to 1999 for Pune and Delhi are analyzed in terms of its frequency distribution, annual trend, diurnal variation and its relation with various meteorological and chemical parameters. It is found that the surface ozone concentration range showing highest frequency of occurrence at Pune is 0–5 ppb during winter and post-monsoon seasons and 15–20 ppb and 5–10 ppb during summer and monsoon seasons, respectively. It is 0–5 ppb at Delhi during all the seasons. The surface ozone concentration has shown a decreasing trend at Pune during the observational period with an average rate of decrease of 1.54 ppb/year. On the other hand, there is no trend whatsoever in the variation of surface ozone concentration at Delhi. Minimum value of surface ozone occurs before sunrise and maximum in the afternoon hours. Regression analyses of surface ozone with maximum temperature (𝑟 = 0.46 for Pune and 0.51 for Delhi, significant at more than 0.1%) and NO
2 at respective locations indicate that surface ozone at these locations is mainly produced by photochemistry. Transport mechanism is also understood to have contributed significantly to the total concentration of ozone. Inverse relationship obtained between surface ozone concentration and relative humidity indicates that major photochemical paths for removal of ozone become effective when humidity increases at these locations.
Volume 128 Issue 4 June 2019 Article ID 0088 Research Article
Physical characterisation of PM$_{2.5}$, PM$_{10}$ and surface ozone measured during the period from 17 July to 21 August 2012 at four strategic locations in and around the Lumbini Protected Zone, Nepal, is done to assess air quality of the region and understand qualitatively source mechanisms of these pollutants. The measurement locations are Panditarama Lumbini International Vipassana Meditation Centre, Parsahawa, Bhairahawa and Tilaurakot, representing monastic, industrial, urban and control areas, respectively. The overall average concentration of PM$_{2.5}$ at these locations is $\sim$19 $\pm$ 12, 35 $\pm$ 13, 35 $\pm$ 11 and 25 $\pm$ 6 $\mu$g/m$^{3}$ and of PM$_{10}$ is $\sim$25 $\pm$ 11, 103 $\pm$ 41, 58 $\pm$ 15 and 32 $\pm$ 7 $\mu$g/m$^{3}$, respectively. PM$_{2.5}$ never crosses the safe limit of the National Ambient Air Quality Standards of Nepal (NNAAQS) in the monastic and control areas but either crosses the NNAAQS occasionally or remains in its vicinity at the other two locations. The PM$_{10}$ concentration frequently exceeds the safe limit in the industrial area but not in the other remaining areas. The analysis indicates the dominance of the impact of local sources and boundary layer thickness on the atmospheric loadings of the particulate matter. The daily average mixing ratio of surface ozone remains normally low at all the four observational sites although the mixing ratio of ozone at Panditarama Lumbini International Vipassana Meditation Centre is much lower than the NNAAQS but higher than that observed at Tilaurakot.
Current Issue
Volume 128 | Issue 8 December 2019
Click here for Editorial Note on CAP Mode
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X-rays
In order to make metal radioactive one have to turn it into another element or isotope. This can be performed only with high-energy particles (including photons).
X-rays can be produces if an electron enters metal with very high speed in two ways:
deceleration radiation (Bremsstrahlung) an atom absorbs part of the electron's kinetic energy, moves to one of the excited states and moves back to ground state emitting a high-energy photon
In any case the energy of the incident radiation (or particles) must be comparable to the energy of X-ray or gamma photons. This is far from cellular phone frequency range for sure.
Shields and cages
Solid metallic shield reflects electromagnetic waves back. The material of the shield is important since it should have good conductivity. Most of metals works well. As far as I know, copper and gold are the best especially for high frequencies (microwaves).
If the frequency is quite low there is no need for solid shield.
The effective area that reflects the wave is proportional to $\frac{\lambda^2}{4\pi}$, where $\lambda$ is the wavelength. It can be much larger than the antenna size. So metallic lattice works well if the distance between the wires is lower than $\lambda$. GSM phones uses frequencies about 1 GHz which corresponds to $\lambda\approx$ 30 cm.
For low frequencies the diffraction effects are important. If the size of the shield is comparable to $\lambda$ the radiation can just bypass the obstacle as it happens with sound. In this case metallic box is the best solution. It can be a cage with appropriate cell size. There should be no big holes like doors and windows.
Grounding
Grounding removes charges from the outside surface of Faraday cage, but if you are inside there is no way to determine whether it is grounded or not. This is more concerned with safety.
Shape of the antenna
This is important if you need something more interesting than just screening.
Using the effect of Bragg diffraction, it is possible to build a shield that reflects one frequency and does not affect others (this will work only for some directions). The shape of the shield also allows to control polarization of the radiation. Edit 1. Answering the questions in the comments
Is the 'energy' of incident radiation proportional to the frequency of transmitted EMF waves, or what people commonly also call 'radiated power' measured in Watts/meter-sq (as well) ?
There are two energy characteristics for EMWs:
Intensity - the amount of energy incident on unit area per unit time (measured in W/m$^2$). It describes total power of the radiation. Energy of quantum - the energy of single photon (measured in Joules). It is equal to product of Planck's constant and frequency: $h\nu$ or $\hbar\omega$. This value is very important in quantum mechanics since quantum system can absorb only integer number of photons. If one photon is not enough to change system state then even 1000 photons will just pass through with no effect.
if EMF radiation that is several hundred/thousand times in excess of international norms, could cause the X-ray generation
This is possible if you have a system that can collect energy of low-frequency radiation and turns it to something else. For example, if EMW induces plasma discharge between some metallic details and electrons collect enough energy before collision there can be X-rays (I'm not sure such situation is possible).
Edit 2. Answering the questions in the comments
would the lattice structure/size computation be good enough if I base it on the highest frequency (thus get the smallest lattice size needed) ?
Also, does it matter if the material (s.a. common steel mesh) has the cross-over joints fused or insulated from each other ?
If lattice period is smaller than $\lambda$ then it should work as a good screen.
Since you need computations and optimization it's better to ask someone who specializes in electrodynamics and antenna theory. May be it's better to ask this as a separate question. Edit 3. Answering the questions in the comments
is it possible to make practical application of Bragg diffraction to cause destructive interference of the EMF wave, when the waves are for large no. of different carrier waves, and clustered around 4-5 group of central frequencies ?
This can be done with multiple Bragg mirrors one for each frequency. AFAIK it is done for infrared radiation.
Apart from Bragg diffraction are there other ways in which the EMF can be reduced / nullified in a small region (say within a radius of 5-6 meters), where the EMF energy is captured using an antenna, converted to electrical energy, and converted to heat/light
Single antenna affects EMF only within $\lambda$ distance. 5 meters is too much for 1 GHz which corresponds to $\lambda\approx$ 30 cm.
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Search
Now showing items 1-10 of 18
J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
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D-Wave Systems Inc. builds annealing devices that exhibit quantum effects. These devices as well as other physical systems implement a programmable Ising model of the form.
\begin{equation}
H = \sum_i c_i Z_i+\sum_{ ij} c_{ij}Z_iZ_j \end{equation
Where zed is the familiar Pauli matrix.
The approach I took program D-Wave’s computer was to encode universal computation into its ground state by relying on specific Hamiltonians that embed logical Boolean operations. This is what I’m going to briefly tell you about here.
Different methods also exist to emulate many-body interactions with two-body Hamiltonians, see the section on many-body to two-body decomposition gadget Hamiltonians. We have sometimes called the Ising formulation of logical functions:
penalty functions (penalty Hamiltonians) classical/energy cost functions Ising model many-body decompositions ground state spin logic gates for adiabatic quantum computing classical or exact gadgets
We’ve done various projects that have relied on our development of different embeddings and we’ve focused on developing methods to treat the efficiency of the embedding directly.
Embedding Boolean circuits into Ising spins
The following paper developed the methods to synthesize classical circuits into the ground states of such a system. In slightly different words, it encodes universal computation into the ground states of Ising lattices.
Non-perturbative k-body to two-body commuting conversion Hamiltonians and embedding problem instances into Ising spinsJacob Biamonte Physical Review A 77, 052331 (2008)
This work realized all of the basic gates and logical operations and included methods to emulate higher-order couplings with the use of only two-body interactions together with ancillary, slack spins. We later revisited the ideas again, this time focusing on the new results which emerged and their relationship, as well as Hamiltonian symmetry.
Ground State Spin Logic James Whitfield, Mauro Faccin and Jacob Biamonte European Physics Letters 99, 57004 (2012)
The idea goes like this. We use diagonal Hamiltonians of $$N$$ spins like this
\begin{equation} H = \sum_i c_i \sigma_i+\sum_{ ij} c_{ij} \sigma_i\sigma_j+\sum_{ijk} c_{ijk}\sigma_i\sigma_j\sigma_k+… \label{eq:H1} \end{equation} with $$\sigma \equiv \sigma^z$$ defined by $$\sigma=\ket{0}\bra{0}-\ket{1}\bra{1}$$. Since the eigenvalues of $$\sigma$$ are $$\pm 1$$, we identify Boolean variable, $$x\in\{0,1\}$$, with $$(\id-\sigma)/2$$ instead of $$\sigma$$ itself. The subscript of each $$\sigma$$ indicates which spin the operator acts on. Terms such as $$\sigma_i\sigma_j$$ are understood as the tensor product $$\sigma_i\otimes\sigma_j$$.
Limiting the Hamiltonian above to
two-spin interactions yields an experimentally relevant tunable Ising Hamiltonian which continues to be the primary focus.
The idea of ground state spin logic is to embed Boolean functions, $$f:\{0,1\}^n\rightarrow\{0,1\}^m$$, into the ground state
subspace, $$\LL(H_{f(\mathbf{x})})$$, of spin Hamiltonian $H_{f(\mathbf{x})}(\sigma_i,\sigma_j,\cdots,\sigma_k)$ acting on the spins $\sigma_i$, $\sigma_j$, \ldots, $\sigma_k$. As an example, consider the universal \NAND~gate defined by $\NAND(x,y)=\bar x\lor\bar y$. The corresponding Hamiltonian, $H_{\bar x\lor\bar y}(\sigma_1,\sigma_2,\sigma_3)$, should have the following ground state subspace \begin{eqnarray} \LL(H_{\bar x\lor\bar y}) &=& \Span\{\ket{x}\ket{y}\ket{{\bar{x}\lor \bar{y}}}\}\\ &=& \Span\{\ket{001},\ket{011},\ket{101},\ket{110}\}\nonumber \end{eqnarray} Using the $\sigma$ matrices, such a Hamiltonian is given in \cite{Crosson10} as, \begin{equation} H_{\bar{x}\lor\bar{y}}(\sigma_1,\sigma_2,\sigma_3)= 2\id+\left( \id + \sigma_1+\sigma_2 -\sigma_1\sigma_2\right)\sigma_3 \label{eq:nand1} \end{equation} This construction uses a three-spin interaction which can be replaced using the same number of spins and only two-spin interactions. This was done in~\cite{JDB08,Gu12} by penalizing and rewarding certain interactions such that the ground state subspace is not altered while the higher energy eigenstates are. Physical Review A 77, 052331 (2008) Non-perturbative k-body to two-body commuting conversion Hamiltonians and embedding problem instances into Ising spins Jacob Biamonte Physical Review A 77, 052331 (2008)
Abstract. An algebraic method has been developed which allows one to engineer several energy levels including the low-energy subspace of interacting spin systems. By introducing ancillary qubits, this approach allows k-body interactions to be captured exactly using 2-body Hamiltonians. Our method works when all terms in the Hamiltonian share the same basis and has no dependence on perturbation theory or the associated large spectral gap. Our methods allow problem instance solutions to be embedded into the ground energy state of Ising spin systems. Adiabatic evolution might then be used to place a computational system into it’s ground state.
European Physics Letters 99, 57004 (2012) Ground State Spin Logic James Whitfield, Mauro Faccin and Jacob Biamonte European Physics Letters 99, 57004 (2012)
Abstract. Designing and optimizing cost functions and energy landscapes is a problem encountered in many fields of science and engineering. These landscapes and cost functions can be embedded and annealed in experimentally controllable spin Hamiltonians. Using an approach based on group theory and symmetries, we examine the embedding of Boolean logic gates into the ground-state subspace of such spin systems. We describe parameterized families of diagonal Hamiltonians and symmetry operations which preserve the ground-state subspace encoding the truth tables of Boolean formulas. The ground-state embeddings of adder circuits are used to illustrate how gates are combined and simplified using symmetry. Our work is relevant for experimental demonstrations of ground-state embeddings found in both classical optimization as well as adiabatic quantum optimization.
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Abstract
We prove that almost any pair of real numbers $\alpha,\beta$, satisfies the following inhomogeneous uniform version of Littlewood’s conjecture: $$\begin{align}\label{C1abst}\tag{C1} \forall \gamma,\delta\in\mathbb{R},\quad \liminf_{|n|\to\infty} \left|n\right|\langle n\alpha-\gamma \rangle\langle n\beta-\delta\rangle=0, \end{align}$$ where $\langle\cdot\rangle$ denotes the distance from the nearest integer. The existence of even a single pair that satisfies statement (C1), solves a problem of Cassels from the 50’s. We then prove that if $1,\alpha,\beta$ span a totally real cubic number field, then $\alpha,\beta$, satisfy (C1). This generalizes a result of Cassels and Swinnerton-Dyer, which says that such pairs satisfy Littlewood’s conjecture. It is further shown that if $\alpha,\beta$ are any two real numbers, such that $1,\alpha,\beta$, are linearly dependent over $\mathbb{Q}$, they cannot satisfy (C1). The results are then applied to give examples of irregular orbit closures of the diagonal group of a new type. The results are derived from rigidity results concerning hyperbolic actions of higher rank commutative groups on homogeneous spaces.
[B] D. Berend, "Multi-invariant sets on tori," Trans. Amer. Math. Soc., vol. 280, iss. 2, pp. 509-532, 1983.
@article {B, MRKEY = {716835},
AUTHOR = {Berend, Daniel}, TITLE = {Multi-invariant sets on tori}, JOURNAL = {Trans. Amer. Math. Soc.}, FJOURNAL = {Transactions of the American Mathematical Society}, VOLUME = {280}, YEAR = {1983}, NUMBER = {2}, PAGES = {509--532}, ISSN = {0002-9947}, CODEN = {TAMTAM}, MRCLASS = {11K06 (11K55 28D10 54A15)}, MRNUMBER = {0716835}, MRREVIEWER = {G{é}rard Rauzy}, DOI = {10.2307/1999631}, ZBLNUMBER = {0532.10028}, } [Ba] E. S. Barnes, "The inhomogeneous minima of indefinite quadratic forms," J. Austral. Math. Soc., vol. 2, pp. 9-10, 1961/1962.
@article {Ba, MRKEY = {0124296},
AUTHOR = {Barnes, E. S.}, TITLE = {The inhomogeneous minima of indefinite quadratic forms}, JOURNAL = {J. Austral. Math. Soc.}, FJOURNAL = {Australian Mathematical Society. Journal. Series A. Pure Mathematics and Statistics}, VOLUME = {2}, YEAR = {1961/1962}, PAGES = {9--10}, ISSN = {0263-6115}, MRCLASS = {10.25}, MRNUMBER = {0124296}, MRREVIEWER = {R. P. Bambah}, ZBLCOMMENT = {BIBPROC: YEAR doesn't match found ZBLNUMBER}, ZBLNUMBER = {0097.26202}, } [Bac] P. Bachmann, Die Arithmetik der Quadratischen FormenLeipzig and Berlin: Teubner, 1923.
@misc{Bac,
author={Bachmann, P.}, TITLE={Die Arithmetik der Quadratischen Formen}, PUBLISHER={Teubner}, ADDRESS={Leipzig and Berlin}, YEAR={1923}, NOTE={especially Kap. 12 (Die zerlegbaren formen)}, } [Bu] Y. Bugeaud, "Multiplicative Diophantine approximation," in Dynamical Systems and Diophantine Approximation, Paris: Soc. Math. France.
@incollection{Bu,
author={Bugeaud, Y.}, TITLE={Multiplicative Diophantine approximation}, BOOKTITLE={Dynamical {{S}}ystems and {{D}}iophantine {{A}}pproximation}, NOTE={proceedings of conference held at the Institute Henri Poincaré, Sem. et Congress, to appear}, PUBLISHER={Soc. Math. France}, ADDRESS={Paris}, } [BugeaudEtAlSrinking] Y. Bugeaud, S. Harrap, S. Kristensen, and S. Velani, "On shrinking targets for $\Bbb Z^m$ actions on tori," Mathematika, vol. 56, pp. 193-202, 2010.
@article{BugeaudEtAlSrinking,
author={Bugeaud, Y. and Harrap, S. and Kristensen, S. and Velani, S.}, TITLE={On shrinking targets for {$\Bbb Z^m$} actions on tori}, JOURNAL={Mathematika}, VOLUME={56}, YEAR={2010}, PAGES={193--202}, MRNUMBER={2678024}, } [Ca] J. W. S. Cassels, An Introduction to the Geometry of Numbers, New York: Springer-Verlag, 1997.
@book {Ca, MRKEY = {1434478},
AUTHOR = {Cassels, J. W. S.}, TITLE = {An {{I}}ntroduction to the {{G}}eometry of {{N}}umbers}, SERIES = {Classics Math.}, NOTE = {corrected reprint of the 1971 edition}, PUBLISHER = {Springer-Verlag}, ADDRESS = {New York}, YEAR = {1997}, PAGES = {viii+344}, ISBN = {3-540-61788-4}, MRCLASS = {11Hxx}, MRNUMBER = {1434478}, ZBLNUMBER = {0866.11041}, } [Ca2] J. W. S. Cassels, "The inhomogeneous minimum of binary quadratic, ternary cubic and quaternary quartic forms," Proc. Cambridge Philos. Soc., vol. 48, pp. 72-86, 1952.
@article {Ca2, MRKEY = {0047709},
AUTHOR = {Cassels, J. W. S.}, TITLE = {The inhomogeneous minimum of binary quadratic, ternary cubic and quaternary quartic forms}, JOURNAL = {Proc. Cambridge Philos. Soc.}, VOLUME = {48}, YEAR = {1952}, PAGES = {72--86}, MRCLASS = {10.0X}, MRNUMBER = {0047709}, ZBLNUMBER={0046.04601}, MRREVIEWER = {R. Hull}, } [Cerri] J. -P. Cerri, "Inhomogeneous and Euclidean spectra of number fields with unit rank strictly greater than $1$," J. Reine Angew. Math, vol. 592, pp. 49-62, 2006.
@article{Cerri,
author={Cerri, J.-P.}, TITLE={Inhomogeneous and {E}uclidean spectra of number fields with unit rank strictly greater than $1$}, JOURNAL={J. Reine Angew. Math}, VOLUME={592}, YEAR={2006}, PAGES={49--62}, MRNUMBER={2222729}, } [CaSD] J. W. S. Cassels and H. P. F. Swinnerton-Dyer, "On the product of three homogeneous linear forms and the indefinite ternary quadratic forms," Philos. Trans. Roy. Soc. London. Ser. A., vol. 248, pp. 73-96, 1955.
@article {CaSD, MRKEY = {0070653},
AUTHOR = {Cassels, J. W. S. and Swinnerton-Dyer, H. P. F.}, TITLE = {On the product of three homogeneous linear forms and the indefinite ternary quadratic forms}, JOURNAL = {Philos. Trans. Roy. Soc. London. Ser. A.}, FJOURNAL = {Philosophical Transactions of the Royal Society of London. Series A. Mathematical and Physical Sciences}, VOLUME = {248}, YEAR = {1955}, PAGES = {73--96}, ISSN = {0080-4614}, MRCLASS = {10.0X}, MRNUMBER = {0070653}, ZBLNUMBER={0065.27905}, MRREVIEWER = {J. F. Koksma}, DOI = {10.1098/rsta.1955.0010}, } [D] H. Davenport, "Indefinite binary quadratic forms, and Euclid’s algorithm in real quadratic fields," Proc. London Math. Soc., vol. 53, pp. 65-82, 1951.
@article {D, MRKEY = {0041883},
AUTHOR = {Davenport, H.}, TITLE = {Indefinite binary quadratic forms, and {E}uclid's algorithm in real quadratic fields}, JOURNAL = {Proc. London Math. Soc.}, FJOURNAL = {Proceedings of the London Mathematical Society. Second Series}, VOLUME = {53}, YEAR = {1951}, PAGES = {65--82}, ISSN = {0024-6115}, MRCLASS = {10.0X}, MRNUMBER = {0041883}, ZBLNUMBER={0045.01402 }, MRREVIEWER = {R. Hull}, DOI = {10.1112/plms/s2-53.1.65}, } [F] H. Furstenberg, "Disjointness in ergodic theory, minimal sets, and a problem in Diophantine approximation," Math. Systems Theory, vol. 1, pp. 1-49, 1967.
@article {F, MRKEY = {0213508},
AUTHOR = {Furstenberg, Harry}, TITLE = {Disjointness in ergodic theory, minimal sets, and a problem in {D}iophantine approximation}, JOURNAL = {Math. Systems Theory}, FJOURNAL = {Mathematical Systems Theory. An International Journal on Mathematical Computing Theory}, VOLUME = {1}, YEAR = {1967}, PAGES = {1--49}, ISSN = {0025-5661}, MRCLASS = {28.70 (10.00)}, MRNUMBER = {021350}, ZBLNUMBER={0146.28502}, MRREVIEWER = {W. Parry}, DOI = {10.1007/BF01692494}, } [KleinbockBadlyAppSysJNT] D. Kleinbock, "Badly approximable systems of affine forms," J. Number Theory, vol. 79, pp. 83-102, 1999.
@article{KleinbockBadlyAppSysJNT,
author={Kleinbock, D.}, TITLE={Badly approximable systems of affine forms}, JOURNAL={J. Number Theory}, VOLUME={79}, YEAR={1999}, PAGES={83--102}, MRNUMBER={1724255}, } [LW] E. Lindenstrauss and B. Weiss, "On sets invariant under the action of the diagonal group," Ergodic Theory Dynam. Systems, vol. 21, iss. 5, pp. 1481-1500, 2001.
@article {LW, MRKEY = {1855843},
AUTHOR = {Lindenstrauss, Elon and Weiss, Barak}, TITLE = {On sets invariant under the action of the diagonal group}, JOURNAL = {Ergodic Theory Dynam. Systems}, FJOURNAL = {Ergodic Theory and Dynamical Systems}, VOLUME = {21}, YEAR = {2001}, NUMBER = {5}, PAGES = {1481--1500}, ISSN = {0143-3857}, MRCLASS = {22E40 (22D40 37A15)}, MRNUMBER = {1855843}, ZBLNUMBER={1073.37006}, MRREVIEWER = {S. G. Dani}, DOI = {10.1017/S0143385701001717}, } [Ma] G. Margulis, "Problems and conjectures in rigidity theory," in Mathematics: Frontiers and Perspectives, Providence, RI: Amer. Math. Soc., 2000, pp. 161-174.
@incollection {Ma, MRKEY = {1754775},
AUTHOR = {Margulis, Gregory}, TITLE = {Problems and conjectures in rigidity theory}, BOOKTITLE = {Mathematics: {{F}}rontiers and {{P}}erspectives}, PAGES = {161--174}, PUBLISHER = {Amer. Math. Soc.}, ADDRESS = {Providence, RI}, YEAR = {2000}, MRCLASS = {22E40 (37C85 37D20 53C24)}, MRNUMBER = {1754775}, ZBLNUMBER={0952.22005}, MRREVIEWER = {A. I. Danilenko}, } [Mau] F. Maucourant, "A nonhomogeneous orbit closure of a diagonal subgroup," Ann. of Math., vol. 171, iss. 1, pp. 557-570, 2010.
@article {Mau, MRKEY = {2630049},
AUTHOR = {Maucourant, Fran{ç}ois}, TITLE = {A nonhomogeneous orbit closure of a diagonal subgroup}, JOURNAL = {Ann. of Math.}, FJOURNAL = {Annals of Mathematics. Second Series}, VOLUME = {171}, YEAR = {2010}, NUMBER = {1}, PAGES = {557--570}, ISSN = {0003-486X}, CODEN = {ANMAAH}, MRCLASS = {22Exx}, MRNUMBER = {2630049}, ZBLNUMBER={1192.22006}, DOI = {10.4007/annals.2010.171.557}, } [R] M. Ratner, "Raghunathan’s topological conjecture and distributions of unipotent flows," Duke Math. J., vol. 63, iss. 1, pp. 235-280, 1991.
@article {R, MRKEY = {1106945},
AUTHOR = {Ratner, Marina}, TITLE = {Raghunathan's topological conjecture and distributions of unipotent flows}, JOURNAL = {Duke Math. J.}, FJOURNAL = {Duke Mathematical Journal}, VOLUME = {63}, YEAR = {1991}, NUMBER = {1}, PAGES = {235--280}, ISSN = {0012-7094}, CODEN = {DUMJAO}, MRCLASS = {22E40 (22D40 28D10)}, MRNUMBER = {1106945 }, MRREVIEWER = {Gopal Prasad}, DOI = {10.1215/S0012-7094-91-06311-8}, ZBLNUMBER = {0733.22007}, } [Mc] C. T. McMullen, "Minkowski’s conjecture, well-rounded lattices and topological dimension," J. Amer. Math. Soc., vol. 18, iss. 3, pp. 711-734, 2005.
@article {Mc, MRKEY = {2138142},
AUTHOR = {McMullen, Curtis T.}, TITLE = {Minkowski's conjecture, well-rounded lattices and topological dimension}, JOURNAL = {J. Amer. Math. Soc.}, FJOURNAL = {Journal of the American Mathematical Society}, VOLUME = {18}, YEAR = {2005}, NUMBER = {3}, PAGES = {711--734}, ISSN = {0894-0347}, MRCLASS = {11H31 (11E57 11J83 55M10)}, MRNUMBER = {2138142}, MRREVIEWER = {Stefan K{ü}hnlein}, DOI = {10.1090/S0894-0347-05-00483-2}, ZBLNUMBER = {1132.11034}, } [Shah91] N. A. Shah, "Uniformly distributed orbits of certain flows on homogeneous spaces," Math. Ann., vol. 289, pp. 315-334, 1991.
@article{Shah91,
author={Shah, N. A.}, TITLE={Uniformly distributed orbits of certain flows on homogeneous spaces}, JOURNAL={Math. Ann.}, VOLUME={289}, YEAR={1991}, PAGES={315--334}, MRNUMBER={1092178}, } [Sh] U. Shapira, "On a generalization of Littlewood’s conjecture," J. Mod. Dyn., vol. 3, iss. 3, pp. 457-477, 2009.
@article {Sh, MRKEY = {2538476},
AUTHOR = {Shapira, Uri}, TITLE = {On a generalization of {L}ittlewood's conjecture}, JOURNAL = {J. Mod. Dyn.}, FJOURNAL = {Journal of Modern Dynamics}, VOLUME = {3}, YEAR = {2009}, NUMBER = {3}, PAGES = {457--477}, ISSN = {1930-5311}, MRCLASS = {37A17 (11J20 37A45)}, MRNUMBER = {2538476}, MRREVIEWER = {Thomas Ward}, DOI = {10.3934/jmd.2009.3.457}, ZBLNUMBER = {1185.37008}, } [Ts] M. Einsiedler and J. Tseng, Badly approximable systems of affine forms.
@misc{Ts,
author={Einsiedler, M. and Tseng, J.}, TITLE={Badly approximable systems of affine forms}, NOTE={preprint}, } [TW] G. Tomanov and B. Weiss, "Closed orbits for actions of maximal tori on homogeneous spaces," Duke Math. J., vol. 119, pp. 367-392, 2003.
@article{TW,
author={Tomanov, G. and Weiss, B.}, TITLE={Closed orbits for actions of maximal tori on homogeneous spaces}, JOURNAL={Duke Math. J.}, VOLUME={119}, YEAR={2003}, PAGES={367--392}, MRNUMBER={1997950}, } [TsengBadlyAppSysJNT] J. Tseng, "Badly approximable affine forms and Schmidt games," J. Number Theory, vol. 129, pp. 3020-3025, 2009.
@article{TsengBadlyAppSysJNT,
author={Tseng, J.}, TITLE={Badly approximable affine forms and {S}chmidt games}, JOURNAL={J. Number Theory}, VOLUME={129}, YEAR={2009}, PAGES={3020--3025}, MRNUMBER={2560849}, }
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First of all it's good to graph the functions $(a)$ (red) and $(b)$ (blue):
The second step is to fund points of intersection, so we find all solutions of
$$\begin{cases}r=|4\cos(2\theta)|\\r=|4\sin(2\theta)|\\\end{cases}$$
We are taking absolute values because $r$ may be negative and this might complicate things. The solutions are:
$$r=2\sqrt2,\quad\theta\in\left\{\pm\tfrac18\pi,\pm\tfrac38\pi,\pm\tfrac58\pi,\pm\tfrac78\pi\right\}$$
It's enough to find area between $\theta=0$ and $\theta=\tfrac{\pi}{8}$ and multiply it by $8$:
$$S=8\cdot\left(\frac12\int_{0}^{\pi/8}r_a^2\,d\theta-\frac12\int_{0}^{\pi/8}r_b^2\,d\theta\right)$$
where $r_a=4\cos(2\theta),r_b=4\sin(2\theta)$. Simplifying:
$$S=8\cdot16\cdot\left(\frac12\int_{0}^{\pi/8}(\cos(2\theta))^2\,d\theta-\frac12\int_{0}^{\pi/8}(\sin(2\theta))^2\,d\theta\right)$$$$=64\int_{0}^{\pi/8}(\cos(2\theta)^2-\sin(2\theta)^2)\,d\theta$$$$=64\int_{0}^{\pi/8}\cos(4\theta)\,d\theta$$$$=64/4\big[\sin(4\theta)\big]_0^{\pi/8}=16$$
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I think there are two issues here. The first is that
LogicalExpand is for expanding
logical expressions, like so:
In[1]:= LogicalExpand[(a || b) && (b || c)]
Out[1]= (a || b) && (b || c)
Second, you can actually work directly with power series and expand them about $ \infty $, as follows:
In[2]:= series = Series[f[(a L /x)], {x, Infinity, 4}];
In Mathematica, this is a
SeriesData object that represents the series in a compact way, but you can do many normal operations on it. Using
TeXForm to get nice output for SE yields:
$$ f(0)+\frac{a L f'(0)}{x}+\frac{a^2 L^2 f''(0)}{2 x^2}+\frac{a^3 f^{(3)}(0) L^3}{6 x^3}+\frac{a^4 f^{(4)}(0) L^4}{24 x^4}+O\left(\left(\frac{1}{x}\right)^5\right) $$
However, for the rest of the problem, I'll use your definition in terms of $ w = \frac{1}{x} $. Let's use your
DE definition and substitute in the series for
f[a L w] before taking the derivatives using
Block:
ode = Block[{F = Series[f[a*L*w], {w, 0, 4}]},
Simplify[(1/16)*((16*a^2*L^2*(1 + a*L*w)^4*W^2)/((1 + 2*a*L*w)^2*
(1 + 2*a*L*w*(1 + a*L*w))^2) - (a^2*L^2*W*(4*I + W))/
(1 + a*L*w)^2 + (8*(-1 - 2*a*L*w + 2*a^3*L^3*w^3)*(I*a*L*W))/
(w*(1 + a*L*w)*(1 + 2*a*L*w)*(1 + 2*a*L*w*(1 + a*L*w))))*
D[F, {w, 0}] + (1/16)*((8*I*a*L*W)/(1 + a*L*w) +
(8*(-1 - 2*a*L*w + 2*a^3*L^3*w^3)*(4*(1 + a*L*w)))/
(w*(1 + a*L*w)*(1 + 2*a*L*w)*(1 + 2*a*L*w*(1 + a*L*w))))*
D[F, {w, 1}] + D[F, {w, 2}] == 0]];
The Mathematica output is kinda gross, so I'll
TeXForm it again:
$$ \frac{a L \left(-2 f'(0)-\frac{1}{2} i f(0) W\right)}{w}+\frac{1}{16} a^2 L^2 \left(-16 f''(0)+64 f'(0)+15 f(0) W^2+20 i f(0) W\right)+\frac{1}{16} a^3 L^3 w \left(W^2 \left(15 f'(0)-62 f(0)\right)-4 i W \left(-f''(0)-3 f'(0)+8 f(0)\right)+64 \left(f''(0)-f'(0)\right)\right)+\frac{1}{96} a^4 L^4 w^2 \left(W^2 \left(45 f''(0)-372 f'(0)+942 f(0)\right)+4 i W \left(4 f^{(3)}(0)+3 f''(0)-36 f'(0)+66 f(0)\right)+16 \left(f^{(4)}(0)+12 f^{(3)}(0)-24 f''(0)+24 f'(0)\right)\right)+O\left(w^3\right)=0 $$
As you can see, this equation is given in terms of $ \left\{f(0),f'(0),f''(0),f^{(3)}(0),f^{(4)}(0)\right\} $. Assuming $ f(0) $ and $ f'(0) $ are fixed (since it's a second-order ODE), we can reduce
ode for non-zero $ a $ and $ L $:
reduced = Assuming[{a != 0, L != 0},
Refine[Reduce[ode, Array[Derivative[#1][f][0] & , 3, 2]]]];
Which comes out to
$$ \left(f'(0)=0\land f(0)=0\land f''(0)=0\land f^{(4)}(0)=-i f^{(3)}(0) (W-12 i)\right)\lor \left(f(0)\neq 0\land W=\frac{4 i f'(0)}{f(0)}\land f''(0)=\frac{1}{4} i (15 W+4 i) f'(0)\land f^{(4)}(0)=-\frac{1}{64} i \left(64 f^{(3)}(0) W-768 i f^{(3)}(0)+675 W^3 f'(0)+1848 i W^2 f'(0)+8688 W f'(0)+1152 i f'(0)\right)\right) $$
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Strictly Monotone Mapping with Totally Ordered Domain is Injective Theorem
Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a strictly monotone mapping.
Then $\phi$ is injective. Proof
\(\displaystyle x, y\) \(\in\) \(\displaystyle S\) \(\, \displaystyle \land \, \) \(\displaystyle x\) \(\ne\) \(\displaystyle y\) \(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(\prec_1\) \(\displaystyle y\) \(\, \displaystyle \lor \, \) \(\displaystyle y\) \(\prec_1\) \(\displaystyle x\) Trichotomy Law \(\displaystyle \leadsto \ \ \) \(\displaystyle \map \phi x\) \(\prec_2\) \(\displaystyle \map \phi y\) \(\, \displaystyle \lor \, \) \(\displaystyle \map \phi y\) \(\prec_2\) \(\displaystyle \map \phi x\) $\phi$ is strictly monotone \(\displaystyle \leadsto \ \ \) \(\displaystyle \map \phi x\) \(\ne\) \(\displaystyle \map \phi y\) Definition of $\prec_2$
Hence the result.
$\blacksquare$
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1Department of Mathematics, Yazd University, 89195-741, Yazd, Iran.
2Department of Mathematics, Yazd University, Yazd, Iran.
Receive Date: 04 October 2016,Revise Date: 18 September 2017,Accept Date: 19 September 2017
Abstract
The distinguishing number (resp. index) $D(G)$ ($D'(G)$) of a graph $G$ is the least integer $d$such that $G$ has an vertex labeling (resp. edge labeling) with $d$ labels that is preserved only by a trivialautomorphism. For any $n \in \mathbb{N}$, the $n$-subdivision of $G$ is a simple graph $G^{\frac{1}{n}}$ which is constructed by replacing each edge of $G$ with a path of length $n$.The $m^{th}$ power of $G$, is a graph with same set of vertices of $G$ and an edge between two vertices if and only if there is a path of length at most $m$ between them in $G$. The fractional power of $G$, is the $m^{th}$ power of the $n$-subdivision of $G$, i.e., $(G^{\frac{1}{n}})^m$ or $n$-subdivision of $m$-th power of $G$, i.e., $(G^m)^{\frac{1}{n}}$. In this paper we study the distinguishing number and the distinguishing index of the natural and the fractional powers of $G$. We show that the natural powers more than one of a graph are distinguished by at most three edge labels. We also show that for a connected graph $G$ of order $n \geqslant 3$ with maximum degree $\Delta (G)$, and for $k\geqslant 2$, $D(G^{\frac{1}{k}})\leqslant \lceil \sqrt[k]{\Delta (G)} \rceil$. Finally we prove that for $m\geqslant 2$, the fractional power of $G$, i.e., $(G^{\frac{1}{k}})^m$ and $(G^m)^{\frac{1}{k}}$ are distinguished by at most three edge labels.
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In 1877, Richard Dedekind discovered one of the most famous pictures in mathematics : the black&white tessellation of the upper half-plane in hyperbolic triangles. Recall that the group $SL_2(\mathbb{Z}) $ of all invertible 2×2 integer matrices with determinant $1$ acts on the upper halfplane via
$\begin{bmatrix} a & b \\\ c & d \end{bmatrix}. z = \frac{az+b}{cz+d} $
and as minus the identity matrix acts trivially, it is really an action of the
modular group $\Gamma = PSL_2(\mathbb{Z}) $. Any black or white triangle in the Dedekind-tessellation is a fundamental domain for the action of the extended modular group $\Gamma^{\ast}$, generated by $\Gamma$ and the morphism $z \mapsto – \overline{z}$. . Dedekind showed that the union of any back and white region is a fundamental domain for the action of the modular group. For example, the ‘usual’ fundamental domain is the union of the top middle black and white regions in Dedekinds picture. Having this tessellation before you is essential if you have to wade through the heavy notation of the important paper by Ravi Kulkarni “An arithmetic-geometric method in the study of the subgroups of the modular group”, which is what we aim to do now. Applications will be given in future posts in this series.
At some points (such as $i $) two black and two white regions are coming together, we call such points
even vertices and they form the $\Gamma^{\ast} $-orbit of $i $. At other points (such as $\rho = e^{\frac{\pi i}{3}} $) three black and white regions are meeting and we call such points odd vertices (they form the $\Gamma^{\ast} $-orbit of $\rho $). The $\Gamma^* $-orbit of $\infty $ consists of the rational numbers and they are called the cusps.
Now, for the edges. There are three types of edges :
even edges connecting a cusp and an even vertex (they form the $\Gamma^{\ast} $-orbit of the line $~(\infty,i) $), odd edges connecting a cusp to an odd vertex (the translates of $~(\infty,\rho) $) and finally f-edges (f for finite) connecting an odd and even vertex (the $\Gamma^{\ast} $-orbit of the arc $~(\rho,i) $).
The geodesics (the semi-circles and the vertical lines) are made of edges and they come in two types :
even lines are complete geodesics which are unions of two even edges (such as the semi-circle $~(0,1) $) and odd lines (such as the semi-circle $~(-1,1) $) are complete geodesics which are unions of two f-edges and two odd edges. Remark that the vertical lines are even if they pass through an integer and odd when they go through a half-integer. The modular group $\Gamma $ acts transitively on the even (resp. odd) lines.
If we write rational numbers in reduced form $\frac{a}{b} $ (and if we agree to write integers as $\frac{n}{1} $ and $\infty=\frac{1}{0} $) then, if a geodesic has endpoints $\frac{a}{b} $ and $\frac{c}{d} $ it is an even line iff $| ad-bc |=1 $ and an odd line iff $| ad-bc | = 2 $.
This notation was set up to define the notion of a
special polygon which is a connected polygonal region $P $ in the upper-halfplane such that its boundary $\partial P $ consists entirely of even and odd edges (so no f-edges) together with a side-pairing satisfying the following requirements Even edges in $\partial P $ come in pairs and each such pair forms an even line. Odd edges in $\partial P $ come in pairs and each pair meets at an odd vertex where they make an internal angle of $\frac{2\pi}{3} $. Any odd edge e is side-paired to a different odd edge f which makes on internal angle $\frac{2\pi}{3} $ with e. If e and f are even edges in $\partial P $ forming an even line, then either e is side-paired to f or else e,f form a free sideand is side-paired to a different such free side. $0,\infty $ are among the vertices of $P $.
The
sides of P are : the odd edges on the boundary, the free sides and the even edges on non-free sides. The vertices of P are the intersections of adjacent sides.
For example, the region inside the thick edges is a special polygon. Its boundary consists of 8 even edges (two on the 4 complete geodesics : the vertical lines at 0 and 1 and the semi-circles $~(\frac{1}{2},\frac{2}{3}) $ and $~(\frac{2}{3},1) $) and 2 odd edges the arc-fragments in the lower left corner, the leftmost being part of the semi-circle $~(0,\frac{1}{3}) $, the other part of the semi-circle $~(\frac{1}{4},\frac{1}{2}) $.
We have several option for the side-pairing, the only forded pairing being the two odd edges which have to be paired. For the even edges we can either consider 0,2 or 4 of the geodesics as free sides and pair these, or we can have 0,2 or 4 non-free sides and then we have to pair up the two even edges making such a non-free side.
The number of sides of the special polygon depends on the number of free sides chosen. For 0 free sides, there are 10 sides and vertices. For 2 free sides, there are 8 sides and vertices and for 4 free sides we have 6 sides and vertices.
Special polygons are a combinatorial gadget to describe the subgroups of finite index in the modular group $PSL_2(\mathbb{Z}) $. Later, we will connect this notion to _quilts_ which are special ‘dessins d’enfants’ and to generalized _Farey sequences_. This will then allow us to find explicit generators of the subgroups.
Some technical issues : if some of the latex-pictures don’t show up nicely it often helps to resize the browser-window and resize it back. The drawing of the special polygon was made using the LaTeX-package MFPIC which is an easy to use interface to MetaPost.
Reference
Ravi S. Kulkarni “An arithmetic-geometric method in the study of the subgroups of the modular group” Amer. J. Math. 113 (1991) 1053-1133
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Switched Three Phase Slot Antenna Drive Note, 2015 April 21: This will not work without modification. Different frequencies will scatter at different angles. With a signal bandwidth of 1 GHz, and a thinsat 20 cm wide, and a beam angle of 45°, there is a ±234 psec difference at the edges of the array, ±0.117 radians at 500 MHz, ±6.7° of aiming error for the "edge" frequency components. If we can add a frequency-dependent phase component for each antenna, as well as feed from multiple locations with skewed phase for each data channel, perhaps this can be fixed.
A server sky array produces a few watts of transmit power, divided by more than 10 million antenna slots on all the thinsats. That means less than a microwatt per antenna slot - even with 3KΩ slot impedance, that is less than 100mV peak to peak, which is difficult to make efficiently from a 1 volt power rail. Digitally synthesizing a thinsat's broadcast signal at every die is also power-hungry.
Instead, imagine that many chips on a thinsat have synthesis and amplification capability, but only a few are turned on, driving a net of three phase 70 GHz signal rails feeding all 350 chips. Every chip is hooked up to those three rails, and has a three way selector switch between 0°, 120°, and 240° phased versions of the broadcast signal. The switched signal is fed (perhaps through a capacitor) to the antenna slot.
Ideally, each slot would be phased exactly for the signal it is supposed to transmit, given the angle and relative distance to the distant target, computing the "reciprocal wave vector". If the actual phase deviates from the ideal by angle θ, there will be a cos
2(θ) loss of contribution to the power reaching the target, the rest of the power will be scattered off-target. But our targets are individual square-meter receivers, so all of the power not reaching the target is wasted anyway. Feeding an antenna slot with out-of-phase power is a win if it saves system power, and averages out to a useful broadcast beam.
In detail, the center-of-beam power is the square of the sum of the voltage (or current) amplitudes of all the slots in the array. The in-phase amplitude is cos(θ) and the average in phase amplitude is:
{average} = { \Large { 1 \over {\theta} } {\int_0}^{\theta} } \cos({\theta}) d{\theta}
{average} = { \Large { { \sin({\theta}) } \over {\theta} } }
The array power is proportional to the square of that average, so
{power}~{efficiency} = { \Large { \left( { \sin({\theta}) } \over {\theta} \right)^2} }
For a 3 phase system, θ is π/3 (half of the total angular "sweep" of 120°). Here's the average amplitudes and efficiencies for various numbers of phases:
phases
Sweep
theta (radians)
mean amplitude
power efficiency
sidelobes
3
120°
π/3
0.827
0.683
-3.3 dB
4
90°
π/4
0.900
0.810
-6.3 dB
5
72°
π/5
0.935
0.875
-8.5 dB
6
60°
π/6
0.955
0.912
-10.2 dB
The sidelobes in the table above are not the total sidelobes, just the additional sidelobe energy due to the slot selection scheme. Far more sidelobe energy will result because the array is sparse, though this per-thinsat sidelobe energy will spread some sidelobe energy beyond the 20 milliradian (200 km) main Airy disk central lobe associated with thinsat size.
Assuming 380 μW per thinsat transmitted, the difference in power loss between three phases and six phases is 141 microwatts; it will cost far more than that to distribute three more phases around the thinsat.
At a gigabit per second, a 9000 byte jumbo frame requires 72 microseconds to transmit; during that interval, the array (orbiting at 4650 m/s) moves 33.4 centimeters. With a 300 meter diameter ground spot (to -3dB), we could transmit 900 frames with the same phasing before recalculation, every 65 milliseconds (about a ping time). Practically, we can recompute phase every 10 milliseconds per slot and stay within 0.1 dB of optimum.
The 100 meter diameter array rotates every orbit, and thinsats move as fast as 2 cm/second north/south while it does so. Over the same 10 milliseconds, the thinsats change their position (and hence phasing) by 0.36 mm - this is a significant fraction of the 4.3 mm wavelength, but it is "common mode" to the whole thinsat. We can take care of this doppler-like movement by adjusting the phase of the common signal rails by 30° over the course of the movement, rather than re-selecting the slot feeds. That common mode phase adjustment will depend on the dot product of the azimuth vector of the target and relative velocity vector of the thinsat to the array. If a multiplexed array is simultaneously sending to multiple ground targets using multiple bus groups, each three phase bus must be adjusted separately for each dot product.
Computing a two bit value (three phases or off) for 1400 slots per thinsat 100 times per second is a really cheap calculation. Even if radiation hits upset the calculation or the bit settings for one slot selector, the "sparkle noise" will not reduce main beam amplitude noticably for a whole array. Of course, the calculation must include the position of each antenna and the current shape of the (possibly vibrating) thinsat, the precise direction of the target (within a microradian), temperature, switch resistance and voltage, and many other complications, but this is just another application of the transistor hose.
Summing two vectors
Look at the phase circle. The sum of two vectors produces another on the black line between them, and results in a six phase system. If the summing is efficient (say through a pair of capacitors, while maintaining the impedance to the phase buses) then our power efficiency goes up, and sidelobes drop by 7 dB.
As part of the 100 Hz rate configuration cycle, we will configure and balance the RF buses. These will be driven from multiple points and have stubs, some of them terminating on buses interrupted by micrometeoroid punchthroughs or dead chips. The parameters used to select and calibrate the drivers feeding the buses will be precomputed and stored in EPROM, and recomputed every time some unpredictable accident damages the network, and also during slack operation times.
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Dr Eligio Lisi (INFN, Bari, Italy)
15/03/2015, 08:30
Theory
Ordinary
The status of known and unknown three-neutrino parameters will be briefly reviewed, providing an introduction to subsequent talks in the neutrino session.
Dr Barbara Caccianiga (INFN, Sezione di Milano)
15/03/2015, 08:55
Experiment
Ordinary
Borexino is a large volume liquid scintillator detector mainly devoted to the study of solar neutrinos. It is located under the Gran Sasso mountain in Italy and has been taking data since 2007. Thanks to its low background and its large mass, Borexino has been able of performing an almost complete spectroscopy of solar neutrinos. In this talk I will focus on the latest result published by...
Rachel Carr (Columbia University)
15/03/2015, 09:15
Experiment
Ordinary
In 2011, Double Chooz became the first reactor-based experiment to indicate a nonzero value of the neutrino mixing parameter $\theta_{13}$. This observation was made with a single detector located approximately 1 km from the two cores of the Chooz Nuclear Power Plant in northeastern France. Since then, the Double Chooz collaboration has increased the precision of its single-detector...
BEI-ZHEN HU (Department of Physics, National Taiwan University)
15/03/2015, 09:40
Experiment
Ordinary
The Daya Bay reactor neutrino experiment announced the discovery of a non-zero value of $\sin^22\theta_{13}$ with significance better than 5 $\sigma$, in 2012. The experiment is continuing to improve the precision of $\sin^22\theta_{13}$ and explore other physics topics. In this talk, I will show the current oscillation and mass-squared difference results which is base on the combination of...
Dr Shunichi Mine (University of California, Irvine)
15/03/2015, 10:25
Experiment
Ordinary
I will summarize the most recent nucleon decay and sterile neutrino results from Super-Kamiokande (SuperK) experiment.
Dr Anna Hayes (Los Alamos National Lab)
15/03/2015, 10:45
Theory
Ordinary
In this talk I will discuss the origin of the reactor neutrino anomaly, the physics issues involved, and the uncertainties. Since the suggestion of an anomaly, a second puzzle has arisen with the reactor antineutrino spectra, namely the appearance of a shoulder in the spectra measured at Daya Bay, RENO, and Double Chooz. The shoulder is predicted within some nuclear database analyses, but...
Mr Jordan Myslik (University of Victoria)
15/03/2015, 11:10
Experiment
Ordinary
The T2K ("Tokai to Kamioka") experiment is a long-baseline neutrino oscillation experiment in Japan. A beam of muon neutrinos or muon anti-neutrinos is produced at the Japan Proton Accelerator Research Complex (J-PARC) in Tokai. The unoscillated neutrino flux is measured by the near detector complex 280 m from the proton target, and the oscillated neutrino flux is measured by the far detector,...
Dr Amine Ahriche (U. Jijel)
15/03/2015, 17:00
Theory
Ordinary
We consider classes of standard model (SM) extensions with a scalar representation (charged singlets, triplet and/or 5-plet) and three generation fermionic representations (right handed neutrinos, triplets or 5-plets). In these models, the neutrino masses are generated at three loops, which provide an explanation for their smallness, and the lightest neutral fermion, is a dark matter...
Dr Azusa GANDO (Research Center for Neutrino Science, Tohoku University)
15/03/2015, 17:20
Experiment
Ordinary
KamLAND-Zen searches for neutrinoless double beta decay with Xe-136 loaded liquid scintillator. After purification of Xe and liquid scintillator, we started 2nd phase of the experiment on December, 2013. We report latest result and future prospects of KamLAND-Zen.
Dr Jacobo Lopez Pavon (SISSA)
15/03/2015, 17:40
Theory
Ordinary
I will study the simplest extension of the SM that can account for neutrino masses: the Type-I seesaw with 2 and 3 right-handed neutrinos. The model introduces a New Physics scale, M, which is often assumed to be much larger than the electroweak scale. However, it is presently unconstrained and the light neutrino masses and mixing can be generated for any value of M above eV. Paying special...
Dr Sandhya Choubey (Harish-Chandra Research Institute)
15/03/2015, 18:00
Theory
Ordinary
Discovery of the third mixing angle theta13 has opened up the possibility of discovering the neutrino mass hierarchy. There are a number of near future experiments proposed (some already funded) which aim at determining the mass hierarchy. In this talk I will review the prospects of discovering the neutrino mass hierarchy in the next generation neutrino experiments.
Ms Tessa Johnson (Indiana University)
15/03/2015, 18:20
Experiment
Ordinary
EXO-200 is an experiment searching for neutrinoless double beta decay using a time projection chamber with 175 kg of liquid xenon enriched in $^{136}$Xe. The observation of this process would indicate that the neutrino is a Majorana fermion and lepton number is not a conserved quantity, and would allow for the calculation of the absolute mass of the neutrino. The low radioactive background...
Prof. Prabhu K Raina (IIT Ropar, india)
15/03/2015, 19:05
Theory
Ordinary
Confirmation of neutrino oscillations in different experiments has established the massive character of neutrino and served as a strong evidence to look beyond the well-accepted standard model of particles. Most of the unified theories thus evolved are based on Majorana character of neutrinos. At present most promising and only experiment to settle the Majorana/Dirac nature of neutrino that is...
Dr Julian Heeck (Université Libre de Bruxelles)
15/03/2015, 19:25
Theory
Ordinary
We discuss the various incarnations of a gauged B-L symmetry: 1) as an unbroken symmetry, it features Dirac neutrinos, neutrinogenesis to create the baryon asymmetry, and a potentially light Z' boson; 2) broken by two units, we obtain the standard case of Majorana neutrinos, seesaw and thermal leptogenesis; 3) broken by four units, we find Dirac neutrinos with lepton-number-violating...
Alexis Haesler (Unige)
15/03/2015, 19:45
Experiment
Ordinary
The NA61/SHINE experiment at the CERN SPS has a rich physics pro- gramme. Beside the ion and the cosmic ray programmes, hadron production measurements are conducted for precise prediction of conventional accelera- tor neutrino beams. The neutrino physics program started with a dedicated hadron production campaign for the T2K experiment. The success of this data taking has proven the...
Ms Cheryl Patrick (Northwestern University)
15/03/2015, 20:05
Experiment
Ordinary
Fermilab's MINERvA experiment is designed to make precision measurements of neutrino scattering cross sections on a variety of materials. After introducing the MINERvA detector, I will explain why these measurements are so important to the current neutrino program. I will then describe several recently published results that are already being applied by the neutrino community to improve their...
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Forcing and construction schemes 19 Downloads Abstract
We investigate forcing and independence questions relating to construction schemes. We show that adding \(\kappa\geq\omega_{1} \) Cohen reals adds a capturing construction scheme. We study the weaker structure of
n-capturing construction schemes and show that it is consistent to have n-capturing construction schemes but no ( n + 1)-capturing construction schemes. We also study the relation of n-capturing with the m-Knaster hierarchy and show that \({\rm MA}_{\omega1}({\rm K}_{m}) \) and n-capturing are independent if \(n \leq m\) and incompatible if \(n>m\). Key words and phrasesconstruction scheme Knaster hierarchy Cohen real Mathematics Subject Classification03E05 03E35 03E65 Preview
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Notes Acknowledgement
The authors thank Osvaldo Guzmán for pointing out a nontrivial error on a previous version of Definition 3.1.
References 1. 2. 3.Kenneth Kunen, Set Theory – An Introduction To Independence Proofs, Studies in logic and the foundations of mathematics, North Holland (1980)Google Scholar 4. 5. 6. 7. 8.
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Find all the functions $f:\mathbb{R}→\mathbb{R}$ such that $f(mx+c)=mf(x)+c$, $m≠1$.
I know that $f(x)=x$ and $f(x)=c/(1-m)$ are two solutions. But to completely solve it I have no idea. Can we completely solve it using elementary mathematics (without assuming additional conditions) ?
Find all the functions $f:\mathbb{R}→\mathbb{R}$ such that $f(mx+c)=mf(x)+c$, $m≠1$.
Let $g(x)=f(x+\frac c{1-m})-\frac c{1-m}$. Then we have the functional equation $g(mx)=mg(x)$. If $|m|>1$, we can pick any two function $g_0\colon(-|m|,-1]\cup [1,|m|)\to \mathbb R$ and let $$ g(x)=\begin{cases}0&\text{if $x=0$}\\ m^kg_0(m^{-k}x)&\text{if $x\ne0$, $k=\lfloor\log_{|m|}|x|\rfloor$} \end{cases}$$ The case $0<|m|<1$ is equivalent to $g(m')=m'g(x)$ with $|m'|=\frac1{|m|}>1$.
The case $m=-1$ is solved by picking $g_1\colon(0,\infty)\to\mathbb R$ arbitrary and letting $$ g(x)=\begin{cases}0&\text{if $x=0$}\\ g_1(x)&\text{if $x>0$}\\ -g_1(x)&\text{if $x<0$}\end{cases}$$
Remark: We can use arbitrary functions $g_0,g_1$ above. If we wanbt $g$ (and hence $f$) to be continuous for example, we must make sure that $g_0$ resp. $g_1$ is continuous and "fits" at the ends, i.e. $\lim_{x\to \pm m}g_0(x)=mg_0(\pm1)$ $\lim_{x\to 0}g_1(x)=0$
Note that the two special solutions you found come from $g(x)=0$ and $g(x)=x$
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Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example:
We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:
We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.
With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm.
Example 6: Expanding Logarithms Using Product, Quotient, and Power Rules
Rewrite [latex]\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)[/latex] as a sum or difference of logs.
Solution
First, because we have a quotient of two expressions, we can use the quotient rule:
Then seeing the product in the first term, we use the product rule:
Finally, we use the power rule on the first term:
Try It 6
Expand [latex]\mathrm{log}\left(\frac{{x}^{2}{y}^{3}}{{z}^{4}}\right)[/latex].
Example 7: Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression
Expand [latex]\mathrm{log}\left(\sqrt{x}\right)[/latex].
Solution
[latex]\begin{cases}\mathrm{log}\left(\sqrt{x}\right)\hfill & =\mathrm{log}{x}^{\left(\frac{1}{2}\right)}\hfill \\ \hfill & =\frac{1}{2}\mathrm{log}x\hfill \end{cases}[/latex]
Try It 7
Expand [latex]\mathrm{ln}\left(\sqrt[3]{{x}^{2}}\right)[/latex].
Q & A Can we expand [latex]\mathrm{ln}\left({x}^{2}+{y}^{2}\right)[/latex]? No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Example 8: Expanding Complex Logarithmic Expressions
Expand [latex]{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)[/latex].
Solution
We can expand by applying the Product and Quotient Rules.
Try It 8
Expand [latex]\mathrm{ln}\left(\frac{\sqrt{\left(x - 1\right){\left(2x+1\right)}^{2}}}{\left({x}^{2}-9\right)}\right)[/latex].
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Given an ensemble with many members, each member having a different phase space vector \(x\) corresponding to a different microstate, we need a way of describing how the phase space vectors of the members in the ensemble will be distributed in the phase space. That is, if we choose to observe one particular member in the ensemble, what is the probability that its phase space vector will be in a small volume \(dx\) around a point \(x\) in the phase space at time \(t\). This probability will be denoted
\[ f (x,t) dx \]
where \(f (x, t)\) is known as the
phase space probability density or phase space distribution function. It's properties are as follows:
\[f (x, t ) \ge (0\]
\(\int dx f (x, t)\) = Number of members in the ensemble
Theorem: Liouville's Theorem The total number of systems in the ensemble is a constant. What restrictions does this place on \(f (x, t ) \)? For a given volume \(\Omega \) in phase space, this condition requires that the rate of decrease of the number of systems from this region is equal to the flux of systems into the volume.
Let \(\hat {n} \) be the unit normal vector to the surface of this region.
Figure 1: Divergence through a surface
The flux through the small surface area element, \(dS\) is just \(\hat {n}\cdot \dot {x}f(x,t)dS \). Then the total flux out of volume is obtained by integrating this over the entire surface that encloses \(\Omega \):
\[\int dS \hat {n} \cdot (\dot {x} f(x,t)) = \int_{\Omega}\nabla _{x}\cdot(\dot{x} f(x,t))\]
which follows from the divergence theorem. \(\nabla _x\) is the \(6N\) dimensional gradient on the phase space
\[\nabla _x = \left(\frac {\partial}{\partial p_1},\cdots,\frac {\partial}{\partial p_N}, \frac {\partial}{\partial r_1},\cdots, \frac {\partial}{\partial r_N}\right)\]
\[=\left(\nabla_{p_1},\cdots,\nabla_{p_N},\nabla_{r_1},\cdots,\nabla_{r_N}\right)\]
On the other hand, the rate of decrease in the number of systems out of the volume is
\[-\frac {d}{dt}\int_{\Omega} d{x}f(x,t) = -\int_{\Omega}d{x} \frac {\partial}{\partial t} f(x,t)\]
Equating these two quantities gives
\[ \int_{\Omega} dx \nabla_{x} \cdot (\dot{x}f( x,t)) = -\int_{\Omega} d{x} \frac {\partial}{\partial t}f(x,t)\]
But this result must hold for any arbitrary choice of the volume \(\Omega \), which we may also allow to shrink to 0 so that the result holds locally, and we obtain the local result:
\[ \frac {\partial}{\partial t}f(x,t) + \nabla_{x} \cdot (\dot {x} f(x,t)) = 0 \]
But
\[ \nabla _{x} \cdot (\dot {x} f (x,t)) = \dot {x} \cdot \nabla _x f(x,t) + f(x,t) \nabla _{x} \cdot \dot {x}\]
This equation resembles an equation for a ``hydrodynamic'' flow in the phase space, with \(f (x, t )\) playing the role of a density. The quantity \(\nabla _x \cdot \dot {x} \), being the divergence of a velocity field, is known as the phase space compressibility, and it does not, for a general dynamical system, vanish. Let us see what the phase space compressibility for a Hamiltonian system is:
\[\nabla_{x}\cdot\dot{x} = \sum_{i=1}^{N}\left[\nabla _{p_i} \cdot \dot {p} _i + \nabla_{r_i}\cdot \dot{r}_i \right] \]
However, by Hamilton's equations:
\[ \dot {p}_i = -\nabla_{r_i}H\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\dot{r}_i = \nabla_{p_i}H\]
Thus, the compressibility is given by
\[ \nabla_{x}\cdot\dot{x} =\sum_{i=1}^N \left[-\nabla _{p_i} \cdot \nabla _{r_i}H +\nabla_{r_i}\cdot\nabla_{p_i}H\right] = 0 \]
Thus, Hamiltonian systems are incompressible in the phase space, and the equation for \(f (x, t )\) becomes
\[ \frac {\partial}{\partial t}f(x,t) + \dot{x}\cdot \nabla_{x}f(x,t) = \frac {df}{dt} = 0\]
which is Liouville's equation, and it implies that \(f (x, t )\) is a conserved quantity when \(x\) is identified as the phase space vector of a particular Hamiltonian system. That is, \(f (x_t, t)\) will be conserved along a particular trajectory of a Hamiltonian system. However, if we view \(x\) is a
fixed spatial label in the phase space, then the Liouville equation specifies how a phase space distribution function \(f (x, t )\) evolves in time from an initial distribution \(f (x, 0 )\).
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Part 6 in the quest for the hydrogen molecule
In this part I will present, as promised, a derivation of the famous Schrödinger equation.
As I have shown in the last part, with our approach so far we are able to calculate important properties of physical systems already. Usually these properties are calculated from the Schrödinger equation, which is assumed at the start to do quantum mechanics. To show you that this is not a necessary starting point, I will now do the opposite: I will derive the Schrödinger equation from our assumption that \(\langle x|p\rangle=e^{ixp/\hbar}\).
Consider a particle in any kind of position-dependent potential \(V(x)\). The energy of the particle is
$$E=\frac{p^2}{2m} + V(x).$$
We write down the energy equation to determine the energy states \(|\psi\rangle\):
$$\hat E |\psi\rangle = E |\psi\rangle.$$
This is all just the same as we did in the previous part. This time we will project the energy equation on position states \(\langle x|\):
$$\langle x|\hat E |\psi\rangle = E\langle x|\psi\rangle=E\psi(x),$$
where we define a function \(\psi(x) \equiv \langle x|\psi\rangle\). Again, we split the equation into the kinetic and potential energy terms:
$$\langle x|\hat E |\psi\rangle = \langle x|\hat E_k |\psi\rangle+\langle x|\hat E_p|\psi\rangle.$$
Kinetic energy
The kinetic energy term of the energy equation needs a little bit of mathematics. The trick is to insert a projection on momentum states \(|p\rangle\)
and a projection on position states \(|x’\rangle\):
$$\langle x|\hat E_k |\psi\rangle=\langle x| \frac{\hat{p}^2}{2m} |\psi\rangle=\\
\int_{-\infty}^{+\infty}\! dp \int_{-\infty}^{+\infty}\! dx’ \langle x| \frac{\hat{p}^2}{2m}| p \rangle \langle p| x’ \rangle \langle x’|\psi\rangle.$$
We can now remove the hats from the operators \(\hat p\) and start filling in the known expressions for \(\langle x|p\rangle\) and \(\langle x|\psi\rangle\):
$$\langle x| \frac{\hat{p}^2}{2m}| p \rangle \langle p| x’ \rangle \langle x’|\psi\rangle=\\
\frac{p^2}{2m} \langle x| p \rangle \langle p| x’ \rangle \langle x’|\psi\rangle=\\ \frac{p^2}{2m} e^{ip(x-x’)/\hbar}\psi(x’).$$
Now comes the main trick: I integrate over \(x’\) using integration by parts twice. First time:
$$\frac{p^2}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi(x’)=\\
\frac{i\hbar p}{2m} \bigg[e^{ip(x-x’)/\hbar}\psi(x’)\bigg]_{-\infty}^{+\infty}- \frac{i\hbar p}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi'(x’),$$
where \(\psi'(x)\) denotes the derivative of the function \(\psi(x)\). We reason that the value of \(\psi(x)\) at \(x = \pm \infty\) should be 0 since it should not be physically relevant for the problem. We can always put the system we are studying in a very large box from which the particles cannot escape, and it should give the same results. Therefore I drop the first term. I continue with the second integration by parts, again dropping the \(\psi'(x)|_{-\infty}^{+\infty}\) term:
$$-\frac{i\hbar p}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi'(x’)=\\
-\frac{\hbar^2}{2m} \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi^″ (x’).$$
To get rid of the two integrals, I can take the reverse steps of what I did before, replacing the integrals by the projection on states and dropping that projection since it doesn’t really do anything.
$$-\frac{\hbar^2}{2m} \int_{-\infty}^{+\infty}\! dp \int_{-\infty}^{+\infty}\! dx’ e^{ip(x-x’)/\hbar}\psi^″ (x’)=\\
-\frac{\hbar^2}{2m}\int_{-\infty}^{+\infty}\! dp \int_{-\infty}^{+\infty}\! dx’ \langle x|p\rangle \langle p| x’\rangle\psi^″ (x’)=\\ -\frac{\hbar^2}{2m}\int_{-\infty}^{+\infty}\! dx’ \langle x|x’\rangle\psi^″ (x’)=\\ -\frac{\hbar^2}{2m} \psi^″ (x),$$
where the last step is explained in part 5.
Potential energy
This time, the potential energy term is the easiest:
$$\langle x|\hat E_p|\psi\rangle=\langle x|V(\hat{x})|\psi\rangle.$$
As we established in part 5, an operator like \(\hat{x}\) may also operate on the projected states on the left, so we can continue with
$$\langle x|V(\hat{x})|\psi\rangle=
V(x)\langle x|\psi\rangle= V(x)\psi(x).$$ Results
Putting the kinetic and potential energy terms together again, we find
$$-\frac{\hbar^2}{2m} \psi^″ (x)+V(x)\psi(x)=E \psi(x),$$
which is called the time-independent Schrödinger equation.
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"A good notation has a subtlety and suggestiveness which at times make it almost seem like a live teacher."— Bertrand Russell "We could, of course, use any notation we want; do not laugh at notations; invent them, they are powerful. In fact, mathematics is, to a large extent, invention of better notations."— Richard Feynman
"
By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and in effect increases the mental power of the race." — Alfred North Whitehead
Notation is perhaps one of the most important aspects of mathematics. The right choice of notation can make a concept clear as day; the wrong choice can make extracting its meaning hopeless. Of course, one great thing about notation is that even if there's a poor choice of notation out there (such as \(\left[x\right]\) or \(\pi\)), often someone comes along and creates a better one (such as \(\lfloor x\rfloor\) for the floor function or multiples of tau, \(\tau\approx 6.28318\), for radian measure of angles).
Which brings me to one such poor choice of notation, one that I believe needs fixing: the rather asymmetrical notation of powers, roots, and logarithms.
Here we have three very closely related concepts — both roots and logarithms are ways to invert exponentiation, the former returning the base and the latter returning the exponent. And yet their notation couldn't be more different:
\[2^3=8\\
\sqrt[3]{8}=2\\
\log_2{8}=3\]This always struck me as annoyingly inelegant. Wouldn't it be nice if these notations bore at least some resemblance to each other?
After giving it some thought, I believe I have found a possible solution. As an alternative to writing \(\log_2 {8}\), I propose the following notation:
This notation makes use of a reflected radical symbol, such that the base of the logarithm is written in a similar manner to the index of a radical but below the "point" (the pointy part of the radical symbol), and the argument of the logarithm is written "inside". The use of this notation has a number of advantages:
The symmetry between the normal radical for roots and the reflected radical for logarithms highlights both their similarities and their differences — each one "undoes" an exponential expression, but each one gives a different part of the expression (the base and the exponent, respectively.) The radical symbol can be looked at as a modified lowercase letter "r". (This may actually be the origin of the symbol, where the "r" stands for radix, the Latin word for "root".) In a similar way, the new symbol for logarithms resembles a capital "L". The placement of the "small number" and the "point" can take on a secondary spatial meaning: The "small number" represents a piece of information we knowabout an exponential expression, and its placement indicates which partwe know. For a root, the "small number" is on top, so we know the. exponent For a logarithm, the "small number" is on bottom, so we know the. base The symbol seems to "point" to the piece of information that we are looking for. For a root, the "point" is pointing downward, so we are looking for the. base For a logarithm, the "point" is pointing upward, so we are looking for the. exponent Looking at the image above, the new notation seems to say "We know the baseis 2, so what's the exponentthat will get us to 8?" Similarly, the expression \(\sqrt[3]{8}\) now can be interpreted as saying "We know the exponentis 3, so what's the basethat will get us to 8?"
This notation would obviously not make much of a difference for seasoned mathematicians who are perfectly comfortable with the \(\log\) and \(\ln\) functions. But from a pedagogical standpoint, the reflected radical, with its multi-layered meaning and auto-mnemonic properties, could help students become more comfortable with a concept that many look at as just meaningless manipulation of symbols.
When I first came up with this reflected-radical notation, I had originally imagined that it should
When I first came up with this reflected-radical notation, I had originally imagined that it should
replacethe current notation. However, after some feedback from various people and some further consideration, I think a better course of action would be to have this notation be used alongsidethe current notation, much in the way that we have multiple notations for other concepts in math (such as the many ways to write derivatives). However, I would suggest that, if it were to become commonplace*, this notation would be best to use when first introducing the concept in schools. The current notation isn't wrongper se — it's just not very evocative of the underlying concept. Anything that can better elucidate that concept can't be a bad thing when it comes to students learning mathematics!
It may seem like a
radicalidea.
But it's a
logicalone. * Of course, for this notation to become commonplace, somebody would need to figure out how to replicate it in LaTeX. Any takers?
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Mumford’s drawing has a clear emphasis on the vertical direction. The set of all vertical lines corresponds to taking the fibers of the natural ‘structural morphism’ : $\pi~:~\mathbf{spec}(\mathbb{Z}[t]) \rightarrow \mathbf{spec}(\mathbb{Z}) $ coming from the inclusion $\mathbb{Z} \subset \mathbb{Z}[t] $. That is, we consider the intersection $P \cap \mathbb{Z} $ of a prime ideal $P \subset \mathbb{Z}[t] $ with the subring of constants.
Two options arise : either $P \cap \mathbb{Z} \not= 0 $, in which case the intersection is a principal prime ideal $~(p) $ for some prime number $p $ (and hence $P $ itself is bigger or equal to $p\mathbb{Z}[t] $ whence its geometric object is contained in the vertical line $\mathbb{V}((p)) $, the fiber $\pi^{-1}((p)) $ of the structural morphism over $~(p) $), or, the intersection $P \cap \mathbb{Z}[t] = 0 $ reduces to the zero ideal (in which case the extended prime ideal $P \mathbb{Q}[x] = (q(x)) $ is a principal ideal of the rational polynomial algebra $\mathbb{Q}[x] $, and hence the geometric object corresponding to $P $ is a horizontal curve in Mumford’s drawing, or is the whole arithmetic plane itself if $P=0 $).
Because we know already that any ‘point’ in Mumford’s drawing corresponds to a maximal ideal of the form $\mathfrak{m}=(p,f(x)) $ (see last time), we see that every point lies on precisely one of the set of all vertical coordinate axes corresponding to the prime numbers ${~\mathbb{V}((p)) = \mathbf{spec}(\mathbb{F}_p[x]) = \pi^{-1}((p))~} $. In particular, two different vertical lines do not intersect (or, in ringtheoretic lingo, the ‘vertical’ prime ideals $p\mathbb{Z}[x] $ and $q\mathbb{Z}[x] $ are comaximal for different prime numbers $p \not= q $).
That is, the structural morphism is a projection onto the “arithmetic axis” (which is $\mathbf{spec}(\mathbb{Z}) $) and we get the above picture. The extra vertical line to the right of the picture is there because in arithmetic geometry it is customary to include also the archimedean valuations and hence to consider the ‘compactification’ of the arithmetic axis $\mathbf{spec}(\mathbb{Z}) $ which is $\overline{\mathbf{spec}(\mathbb{Z})} = \mathbf{spec}(\mathbb{Z}) \cup { v_{\mathbb{R}} } $.
Yuri I. Manin is advocating for years the point that we should take the terminology ‘arithmetic surface’ for $\mathbf{spec}(\mathbb{Z}[x]) $ a lot more seriously. That is, there ought to be, apart from the projection onto the ‘z-axis’ (that is, the arithmetic axis $\mathbf{spec}(\mathbb{Z}) $) also a projection onto the ‘x-axis’ which he calls the ‘geometric axis’.
But then, what are the ‘points’ of this geometric axis and what are their fibers under this second projection?
We have seen above that the vertical coordinate line over the prime number $~(p) $ coincides with $\mathbf{spec}(\mathbb{F}_p[x]) $, the affine line over the finite field $\mathbb{F}_p $. But all of these different lines, for varying primes $p $, should project down onto the same geometric axis. Manin’s idea was to take therefore as the geometric axis the affine line $\mathbf{spec}(\mathbb{F}_1[x]) $, over the virtual field with one element, which should be thought of as being the limit of the finite fields $\mathbb{F}_p $ when $p $ goes to one!
How many points does $\mathbf{spec}(\mathbb{F}_1[x]) $ have? Over a virtual object one can postulate whatever one wants and hope for an a posteriori explanation. $\mathbb{F}_1 $-gurus tell us that there should be exactly one point of size n on the affine line over $\mathbb{F}_1 $, corresponding to the unique degree n field extension $\mathbb{F}_{1^n} $. However, it is difficult to explain this from the limiting perspective…
Over a genuine finite field $\mathbb{F}_p $, the number of points of thickness $n $ (that is, those for which the residue field is isomorphic to the degree n extension $\mathbb{F}_{p^n} $) is equal to the number of monic irreducible polynomials of degree n over $\mathbb{F}_p $. This number is known to be $\frac{1}{n} \sum_{d | n} \mu(\frac{n}{d}) p^d $ where $\mu(k) $ is the Moebius function. But then, the limiting number should be $\frac{1}{n} \sum_{d | n} \mu(\frac{n}{d}) = \delta_{n1} $, that is, there can only be one point of size one…
Alternatively, one might consider the zeta function counting the number $N_n $ of ideals having a quotient consisting of precisely $p^n $ elements. Then, we have for genuine finite fields $\mathbb{F}_p $ that $\zeta(\mathbb{F}_p[x]) = \sum_{n=0}^{\infty} N_n t^n = 1 + p t + p^2 t^2 + p^3 t^3 + \ldots $, whence in the limit it should become
$1+t+t^2 +t^3 + \ldots $ and there is exactly one ideal in $\mathbb{F}_1[x] $ having a quotient of cardinality n and one argues that this unique quotient should be the unique point with residue field $\mathbb{F}_{1^n} $ (though it might make more sense to view this as the unique n-fold extension of the unique size-one point $\mathbb{F}_1 $ corresponding to the quotient $\mathbb{F}_1[x]/(x^n) $…)
A perhaps more convincing reasoning goes as follows. If $\overline{\mathbb{F}_p} $ is an algebraic closure of the finite field $\mathbb{F}_p $, then the points of the affine line over $\overline{\mathbb{F}_p} $ are in one-to-one correspondence with the maximal ideals of $\overline{\mathbb{F}_p}[x] $ which are all of the form $~(x-\lambda) $ for $\lambda \in \overline{\mathbb{F}_p} $. Hence, we get the points of the affine line over the basefield $\mathbb{F}_p $ as the orbits of points over the algebraic closure under the action of the Galois group $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p) $.
‘Common wisdom’ has it that one should identify the algebraic closure of the field with one element $\overline{\mathbb{F}_{1}} $ with the group of all roots of unity $\mathbb{\mu}_{\infty} $ and the corresponding Galois group $Gal(\overline{\mathbb{F}_{1}}/\mathbb{F}_1) $ as being generated by the power-maps $\lambda \rightarrow \lambda^n $ on the roots of unity. But then there is exactly one orbit of length n given by the n-th roots of unity $\mathbb{\mu}_n $, so there should be exactly one point of thickness n in $\mathbf{spec}(\mathbb{F}_1[x]) $ and we should then identity the corresponding residue field as $\mathbb{F}_{1^n} = \mathbb{\mu}_n $.
Whatever convinces you, let us assume that we can identify the non-generic points of $\mathbf{spec}(\mathbb{F}_1[x]) $ with the set of positive natural numbers ${ 1,2,3,\ldots } $ with $n $ denoting the unique size n point with residue field $\mathbb{F}_{1^n} $. Then, what are the fibers of the projection onto the geometric axis $\phi~:~\mathbf{spec}(\mathbb{Z}[x]) \rightarrow \mathbf{spec}(\mathbb{F}_1[x]) = { 1,2,3,\ldots } $?
These fibers should correspond to ‘horizontal’ principal prime ideals of $\mathbb{Z}[x] $. Manin proposes to consider $\phi^{-1}(n) = \mathbb{V}((\Phi_n(x))) $ where $\Phi_n(x) $ is the n-th cyclotomic polynomial. The nice thing about this proposal is that all closed points of $\mathbf{spec}(\mathbb{Z}[x]) $ lie on one of these fibers!
Indeed, the residue field at such a point (corresponding to a maximal ideal $\mathfrak{m}=(p,f(x)) $) is the finite field $\mathbb{F}_{p^n} $ and as all its elements are either zero or an $p^n-1 $-th root of unity, it does lie on the curve determined by $\Phi_{p^n-1}(x) $.
As a consequence, the localization $\mathbb{Z}[x]_{cycl} $ of the integral polynomial ring $\mathbb{Z}[x] $ at the multiplicative system generated by all cyclotomic polynomials is a principal ideal domain (as all height two primes evaporate in the localization), and, the fiber over the generic point of $\mathbf{spec}(\mathbb{F}_1[x]) $ is $\mathbf{spec}(\mathbb{Z}[x]_{cycl}) $, which should be compared to the fact that the fiber of the generic point in the projection onto the arithmetic axis is $\mathbf{spec}(\mathbb{Q}[x]) $ and $\mathbb{Q}[x] $ is the localization of $\mathbb{Z}[x] $ at the multiplicative system generated by all prime numbers).
Hence, both the vertical coordinate lines and the horizontal ‘lines’ contain all closed points of the arithmetic plane. Further, any such closed point $\mathfrak{m}=(p,f(x)) $ lies on the intersection of a vertical line $\mathbb{V}((p)) $ and a horizontal one $\mathbb{V}((\Phi_{p^n-1}(x))) $ (if $deg(f(x))=n $).
That is, these horizontal and vertical lines form a coordinate system, at least for the closed points of $\mathbf{spec}(\mathbb{Z}[x]) $.
Still, there is a noticeable difference between the two sets of coordinate lines. The vertical lines do not intersect meaning that $p\mathbb{Z}[x]+q\mathbb{Z}[x]=\mathbb{Z}[x] $ for different prime numbers p and q. However, in general the principal prime ideals corresponding to the horizontal lines $~(\Phi_n(x)) $ and $~(\Phi_m(x)) $ are not comaximal when $n \not= m $, that is, these ‘lines’ may have points in common! This will lead to an exotic new topology on the roots of unity… (to be continued).
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A long while ago I promised to take you from the action by the modular group $\Gamma=PSL_2(\mathbb{Z})$ on the lattices at hyperdistance $n$ from the standard orthogonal laatice $L_1$ to the corresponding ‘monstrous’ Grothendieck dessin d’enfant.
Speaking of dessins d’enfant, let me point you to the latest intriguing paper by Yuri I. Manin and Matilde Marcolli, ArXived a few days ago Quantum Statistical Mechanics of the Absolute Galois Group, on how to build a quantum system for the absolute Galois group from dessins d’enfant (more on this, I promise, later).
Where were we?
We’ve seen natural one-to-one correspondences between (a) points on the projective line over $\mathbb{Z}/n\mathbb{Z}$, (b) lattices at hyperdistance $n$ from $L_1$, and (c) coset classes of the congruence subgroup $\Gamma_0(n)$ in $\Gamma$.
How to get from there to a dessin d’enfant?
The short answer is: it’s all in Ravi S. Kulkarni’s paper, “An arithmetic-geometric method in the study of the subgroups of the modular group”, Amer. J. Math 113 (1991) 1053-1135.
It is a complete mystery to me why Tatitscheff, He and McKay don’t mention Kulkarni’s paper in “Cusps, congruence groups and monstrous dessins”. Because all they do (and much more) is in Kulkarni.
I’ve blogged about Kulkarni’s paper years ago:
– In the Dedekind tessalation it was all about assigning special polygons to subgroups of finite index of $\Gamma$.
– In Modular quilts and cuboid tree diagram it did go on assigning (multiple) cuboid trees to a (conjugacy class) of such finite index subgroup.
– In Hyperbolic Mathieu polygons the story continued on a finite-to-one connection between special hyperbolic polygons and cuboid trees.
– In Farey codes it was shown how to encode such polygons by a Farey-sequence.
– In Generators of modular subgroups it was shown how to get generators of the finite index subgroups from this Farey sequence.
The modular group is a free product
\[ \Gamma = C_2 \ast C_2 = \langle s,u~|~s^2=1=u^3 \rangle \] with lifts of $s$ and $u$ to $SL_2(\mathbb{Z})$ given by the matrices \[ S=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},~\qquad U= \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \]
As a result, any permutation representation of $\Gamma$ on a set $E$ can be represented by a $2$-coloured graph (with black and white vertices) and edges corresponding to the elements of the set $E$.
Each white vertex has two (or one) edges connected to it and every black vertex has three (or one). These edges are the elements of $E$ permuted by $s$ (for white vertices) and $u$ (for black ones), the order of the 3-cycle determined by going counterclockwise round the vertex.
Clearly, if there’s just one edge connected to a vertex, it gives a fixed point (or 1-cycle) in the corresponding permutation.
The ‘monstrous dessin’ for the congruence subgroup $\Gamma_0(n)$ is the picture one gets from the permutation $\Gamma$-action on the points of $\mathbb{P}^1(\mathbb{Z}/n \mathbb{Z})$, or equivalently, on the coset classes or on the lattices at hyperdistance $n$.
Kulkarni’s paper (or the blogposts above) tell you how to get at this picture starting from a fundamental domain of $\Gamma_0(n)$ acting on teh upper half-plane by Moebius transformations.
Sage gives a nice image of this fundamental domain via the command
FareySymbol(Gamma0(n)).fundamental_domain()
Here’s the image for $n=6$:
The boundary points (on the halflines through $0$ and $1$ and the $4$ half-circles need to be identified which is indicaed by matching colours. So the 2 halflines are identified as are the two blue (and green) half-circles (in opposite direction).
To get the dessin from this, let’s first look at the interior points. A white vertex is a point in the interior where two black and two white tiles meet, a black vertex corresponds to an interior points where three black and three white tiles meet.
Points on the boundary where tiles meet are coloured red, and after identification two of these reds give one white or black vertex. Here’s the intermediate picture
The two top red points are identified giving a white vertex as do the two reds on the blue half-circles and the two reds on the green half-circles, because after identification two black and two white tiles meet there.
This then gives us the ‘monstrous’ modular dessin for $n=6$ of the Tatitscheff, He and McKay paper:
Let’s try a more difficult example: $n=12$. Sage gives us as fundamental domain
giving us the intermediate picture
and spotting the correct identifications, this gives us the ‘monstrous’ dessin for $\Gamma_0(12)$ from the THM-paper:
In general there are several of these 2-coloured graphs giving the same permutation representation, so the obtained ‘monstrous dessin’ depends on the choice of fundamental domain.
You’ll have noticed that the domain for $\Gamma_0(6)$ was symmetric, whereas the one Sage provides for $\Gamma_0(12)$ is not.
This is caused by Sage using the Farey-code
\[ \xymatrix{ 0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_1 & \frac{1}{5} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & 1} \]
One of the nice results from Kulkarni’s paper is that for any $n$ there is a symmetric Farey-code, giving a perfectly symmetric fundamental domain for $\Gamma_0(n)$. For $n=12$ this symmetric code is
\[
\xymatrix{ 0 \ar@{-}[r]_1 & \frac{1}{6} \ar@{-}[r]_2 & \frac{1}{4} \ar@{-}[r]_3 & \frac{1}{3} \ar@{-}[r]_4 & \frac{1}{2} \ar@{-}[r]_4 & \frac{2}{3} \ar@{-}[r]_3 & \frac{3}{4} \ar@{-}[r]_2 & \frac{5}{6} \ar@{-}[r]_1 & 1} \]
It would be nice to see whether using these symmetric Farey-codes gives other ‘monstrous dessins’ than in the THM-paper.
Remains to identify the edges in the dessin with the lattices at hyperdistance $n$ from $L_1$.
Using the tricks from the previous post it is quite easy to check that for any $n$ the monstrous dessin for $\Gamma_0(n)$ starts off with the lattices $L_{M,\frac{g}{h}} = M,\frac{g}{h}$ as below
Let’s do a sample computation showing that the action of $s$ on $L_n$ gives $L_{\frac{1}{n}}$:
\[
L_n.s = \begin{bmatrix} n & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} \]
and then, as last time, to determine the class of the lattice spanned by the rows of this matrix we have to compute
\[
\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -n \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -n \end{bmatrix} \]
which is class $L_{\frac{1}{n}}$. And similarly for the other edges.2 Comments
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One of the most intriguing consequences of Bell’s theorem is the idea that one can do
experimental metaphysics: to take some eminently metaphysical concepts such as determinism, causality, and free will, and extract from them actual experimental predictions, which can be tested in the laboratory. The results of said tests can then be debated forever without ever deciding the original metaphysical question.
It was with such ideas in mind that I learned about the Sleeping Beauty problem, so I immediately thought: why not simply do an experimental test to solve the problem?
The setup is as follows: you are the Sleeping Beauty, and today is Sunday. I’m going to flip a coin, and hide the result from you. If the coin fell on heads, I’m going to give you a sleeping pill that will make you sleep until Monday, and terminate the experiment after you wake up. If it falls on tails instead, I’m going also to give you the pill that makes you sleep until Monday, but after your awakening I’m going to give you a second pill that erases your memory and makes you sleep until Tuesday. At each awakening I’m going to ask you: what is the probability1 that the coin fell on tails?
There are two positions usually defended by philosophers:
$p(T) = 1/2$. This is defended by Lewis and Bostrom, roughly because before going to sleep the probability was assumed to be one half (i.e. that the coin is fair), and by waking up you do not learn anything you didn’t know before, so the probability should not change. $p(T) = 2/3$. This is defended by Elga and Bostrom, roughly because the three possible awakenings (heads on Monday, tails on Monday, and tails on Tuesday) are indistinguishable from your point of you, so you should assign all of them the same probability. Since two of them have the coin fallen on tails, the probability of tails must be two-thirds.
Well, seems like the perfect question to answer experimentally, no? Give drugs to people, and ask them to bet on the coins being heads or tails. See who wins more money, and we’ll know who is right! There are, however, two problems with this experiment. The first is that it is not so easy to erase people’s memories. Hitting them hard on the head or giving them enough alcohol usually does the trick, but it doesn’t work reliably, and I don’t know where I could find volunteers that thought the experiment was worth the side effects (brain clots or a massive hangover). And, frankly, even if I did find volunteers (maybe overenthusiastic philosophy students?), these methods are just too grisly for my taste.
Luckily a colleague of mine (Marie-Christine) found an easy solution: just demand people to place their bets in advance. Since they are not supposed to be able to know in which of the three awakenings they are, it makes no sense for them to bet differently in different awakenings (in fact, they should even be be unable to bet differently on different awakenings without access to a random number generator. If they have one in their brains is another question). So if you decide to bet on heads, and then “awakes” on Tuesday, too bad, you have to do the bad bet anyway.
With that solved, we get to the second problem: it is not rational to ever bet on heads. If you believe that the probability is $1/2$ you should be indifferent between heads and tails, and if you believe that the probability is $2/3$ you should definitely bet on tails. In fact, if you believe that the probability is $1/2$ but have even the slightest doubt that your reasoning is correct, you should bet on tails anyway just to be on the safe side.
This problem can be easily solved, simply by biasing the coin a bit towards heads, such that the probability of heads (if you believed in $1/2$) is now slightly above one half, while keeping the probability of tails (if you believed in $2/3$) still above one half. To calculate the exact numbers we use a neat little formula from Sebens and Carroll, which says that the probability of you being the observer labelled by $i$ within a set of observers with identical subjective experiences is
\[ p(i) = \frac{w_i}{\sum_j w_j}, \] where $w_i$ is the Born-rule weight of your situation, and the $w_j$ are the Born-rule weights of all observers in the subjectively-indistinguishable situation.
Let’s say that the coin has a (objective, quantum, given by the Born rule) probability $p$ of falling on heads. The probability of being one of the tail observers is then simply the sum of the Born-rule weight of the Monday tail observer (which is simply $1-p$) with the Born-rule weight of the Tuesday tail observer (also $1-p$), divided by the sum of the Born-rule weights of all three observers ($1-p$, $1-p$, and $p$), so
\[ p(T) = \frac{2(1-p)}{2(1-p) + p}.\] For elegance, let’s make this probability be equal to the objective probability of the coin falling on heads, so that both sides of the philosophical dispute will bet on their preferred solution with the same odds. Solving $p = (2 – 2p)/(2-p)$ gives us then \[ p = 2-\sqrt{2} \approx 0.58,\] which makes the problem quantum, and thus on topic for this blog, since it features the magical $\sqrt2$.2
With all this in hand, time to do the experiment. I gathered 17 impatient hungry physicists in a room, and after explaining them all of this, I asked them to bet on either heads or tails. The deal was that the bet was a commitment to buy, in each awakening, a ticket that would pay them 1€ in case they were right. Since the betting odds were set to be $0.58$, the price for each ticket was 0.58€.
After each physicist committed to a bet, I ran my biased quantum random number generator (actually just the function rand from Octave with the correct weighting), and cashed the bets (once when the result was heads, twice when the result was tails).
There were four possible situations: if the person betted on tails and the result was tails, they paid me 1.16€ for the tickets and got 2€ back, netting 0.84€ (this happened 4 times). If the person betted on heads and the result was tails, they paid me 1.16€ again, but got nothing back, netting -1.16€ (this happened 2 times). If the person betted on tails and the result was heads, they paid me 0.58€ for the ticket and got nothing back, netting -0.58€ (this happened 4 times). Finally, if the person betted on heads and the result was heads, they paid 0.58€ for the ticket and got 1€ back, netting 0.42€ (this happened once).
So on average the people who betted on tails profited 0.13€, while the people who betted on heads lost 0.61€. The prediction of the $2/3$ theory was that they should profit nothing when betting on tails, and lose 0.16€ when betting on heads. The prediction of the $1/2$ theory was the converse: who bets on tails loses 0.16€, while who beats on heads breaks even. In the end the match was not that good, but still the data clearly favours the $2/3$ theory. Once again, physics comes to the rescue of philosophy, solving experimentally a long-standing metaphysical problem!
Speaking more seriously, of course the philosophers knew, since the first paper on the subject, that the experimental results would be like this, and that is why nobody bothered to do the experiment. They just thought that this was not a decisive argument, as the results are determined by how you operationalise the Sleeping Beauty problem, and the question was always about what is the correct operationalisation (or, on other words, what probability is supposed to be). Me, I think that whatever probability is, it should be something with a clear operational meaning. And since I don’t know any natural operationalisation that will give the $1/2$ answer, I’m happy with the $2/3$ theory.
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A
tetrahedral snake, sometimes called a Steinhaus snake, is a collection of tetrahedra, linked face to face.
Steinhaus showed in 1956 that the last tetrahedron in the snake can never be a translation of the first one. This is a consequence of the fact that the group generated by the four reflexions in the faces of a tetrahedron form the free product $C_2 \ast C_2 \ast C_2 \ast C_2$.
For a proof of this, see Stan Wagon’s book The Banach-Tarski paradox, starting at page 68.
The thread $(3|3)$ is the
spine of the $(9|1)$-snake which involves the following lattices \[ \xymatrix{& & 1 \frac{1}{3} \ar@[red]@{-}[dd] & & \\ & & & & \\ 1 \ar@[red]@{-}[rr] & & 3 \ar@[red]@{-}[rr] \ar@[red]@{-}[dd] & & 1 \frac{2}{3} \\ & & & & \\ & & 9 & &} \] It is best to look at the four extremal lattices as the vertices of a tetrahedron with the lattice $3$ corresponding to its point of gravity.
The congruence subgroup $\Gamma_0(9)$ fixes each of these lattices, and the arithmetic group $\Gamma_0(3|3)$ is the conjugate of $\Gamma_0(1)$
\[ \Gamma_0(3|3) = \{ \begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}.\begin{bmatrix} a & b \\ c & d \end{bmatrix}.\begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & \frac{b}{3} \\ 3c & 1 \end{bmatrix}~|~ad-bc=1 \} \] We know that $\Gamma_0(3|3)$ normalizes the subgroup $\Gamma_0(9)$ and we need to find the moonshine group $(3|3)$ which should have index $3$ in $\Gamma_0(3|3)$ and contain $\Gamma_0(9)$.
So, it is natural to consider the finite group $A=\Gamma_0(3|3)/\Gamma_9(0)$ which is generated by the co-sets of
\[ x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix} \qquad \text{and} \qquad y = \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix} \] To determine this group we look at the action of it on the lattices in the $(9|1)$-snake. It will fix the central lattice $3$ but will move the other lattices.
Recall that it is best to associate to the lattice $M.\frac{g}{h}$ the matrix
\[ \alpha_{M,\frac{g}{h}} = \begin{bmatrix} M & \frac{g}{h} \\ 0 & 1 \end{bmatrix} \] and then the action is given by right-multiplication.
\[
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{1}{3} \\ 0 & 1 \end{bmatrix}.x = \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & 1 \end{bmatrix}.x=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] That is, $x$ corresponds to a $3$-cycle $1 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 1$ and fixes the lattice $9$ (so is rotation around the axis through the vertex $9$).
To compute the action of $y$ it is best to use an alternative description of the lattice, replacing the roles of the base-vectors $\vec{e}_1$ and $\vec{e}_2$. These latices are projectively equivalent
\[ \mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \quad \text{and} \quad \mathbb{Z} \vec{e}_1 \oplus \mathbb{Z} (\frac{g’}{h} \vec{e}_1 + \frac{1}{h^2M} \vec{e}_2) \] where $g.g’ \equiv~1~(mod~h)$. So, we have equivalent descriptions of the lattices \[ M,\frac{g}{h} = (\frac{g’}{h},\frac{1}{h^2M}) \quad \text{and} \quad M,0 = (0,\frac{1}{M}) \] and we associate to the lattice in the second normal form the matrix \[ \beta_{M,\frac{g}{h}} = \begin{bmatrix} 1 & 0 \\ \frac{g’}{h} & \frac{1}{h^2M} \end{bmatrix} \] and then the action is again given by right-multiplication.
In the tetrahedral example we have
\[ 1 = (0,\frac{1}{3}), \quad 1\frac{1}{3}=(\frac{1}{3},\frac{1}{9}), \quad 1\frac {2}{3}=(\frac{2}{3},\frac{1}{9}), \quad 9 = (0,\frac{1}{9}) \] and \[ \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix}.y = \begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix},\quad \begin{bmatrix} 1 & 0 \\ \frac{2}{3} & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{9} \end{bmatrix}. y = \begin{bmatrix} 1 & 0 \\ \frac{1}{3} & \frac{1}{9} \end{bmatrix} \] That is, $y$ corresponds to the $3$-cycle $9 \rightarrow 1 \frac{1}{3} \rightarrow 1 \frac{2}{3} \rightarrow 9$ and fixes the lattice $1$ so is a rotation around the axis through $1$.
Clearly, these two rotations generate the full rotation-symmetry group of the tetrahedron
\[ \Gamma_0(3|3)/\Gamma_0(9) \simeq A_4 \] which has a unique subgroup of index $3$ generated by the reflexions (rotations with angle $180^o$ around axis through midpoints of edges), generated by $x.y$ and $y.x$.
The moonshine group $(3|3)$ is therefore the subgroup generated by
\[ (3|3) = \langle \Gamma_0(9),\begin{bmatrix} 2 & \frac{1}{3} \\ 3 & 1 \end{bmatrix},\begin{bmatrix} 1 & \frac{1}{3} \\ 3 & 2 \end{bmatrix} \rangle \]
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I was trying to evaluate this product$$ \sin (1^\circ) \sin (2^\circ) \sin (3^\circ) ... \sin (88^\circ) \sin (89^\circ) \sin (90^\circ) $$But I think it got weird results with this
Product[Sin[i], {i, pi/180, pi/2}]Can anyone suggest me something??
I was trying to evaluate this product$$ \sin (1^\circ) \sin (2^\circ) \sin (3^\circ) ... \sin (88^\circ) \sin (89^\circ) \sin (90^\circ) $$But I think it got weird results with this
You just have to modify the syntax a little.
piis not a predefined symbol. All predefined constants start with
capitalletters, and for the number $\pi$ that means you should enter
Pi.
The
Productsyntax is probably also not what you want: with the previous correction you would have a range specification of
{i, Pi/180, Pi/2}. But
ihere will still be incremented in
integersteps, so there are only two terms in the product. To get a product where
ivaries in discrete steps that are
notintegers, you have to add a fourth element to the range specification,
Example:
result = Product[Sin[i], {i, Pi/180, Pi/2, Pi/180}]
Finally, you may not be happy with the result of this line either, because it is printed with exact numbers in the arguments of the sine functions. This is because Mathematica tries to be as general as possible, and your input had only exact numbers in it. Exact numbers are symbols like
Pibut also the number
180above. To get a
numericalresult you have to state somewhere that you want "floating-point" output. This can be done by giving the inputas floating-point numbers as in
180.0.
Alternatively, you can take the exact value
resultand convert it to a numerical answer using
N,
example:
N[result]
$1.53268\times 10^{-26}$
In addition to Jens' answer, there is a built-in constant
Degree which is equal to
Degree == Pi/180 so you could simplify your code slightly by doing something like
result = Product[Sin[i Degree], {i,90}]
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Area of Square Contents Theorem Integer Side Length
In the case where $L = 1$, the statement follows from the definition of area.
If $L \in \N, L > 1$, then we can divide the square into smaller squares, each of side length one.
Since there will be $L$ squares of side length one on each side, it follows that there will be $L \cdot L = L^2$ squares of side length one.
Thus, the area of the square of side length $L$ is $L^2 \cdot 1 = L^2$.
$\Box$
Rational Side Length
If $L$ is a rational number, then:
$\exists p, q \in \N: L = \dfrac p q$
From the integer side length case, its area equals $p^2$.
Divide the sides into $q$ equal parts.
It follows arithmetically:
$A = \dfrac {p^2} {q^2} = \paren {\dfrac p q}^2 = L^2$
$\Box$
Irrational Side Length
Let $L$ be an irrational number.
Then from Rationals Dense in Reals we know that within an arbitrarily small distance $\epsilon$ from $L$, we can find a rational number less than $L$ and a rational number greater than $L$.
In formal terms, we have:
$\forall \epsilon > 0: \exists A, B \in \Q_+: A < L < B: \left|{A - L}\right| < \epsilon, \left|{B - L}\right| < \epsilon$
Thus:
$\displaystyle \lim_{\epsilon \to 0^+} A = L$ $\displaystyle \lim_{\epsilon \to 0^+} B = L$
Since a square of side length $B$ can contain a square of side length $L$, which can in turn contain a square of side length $A$, then:
$\operatorname {area} \Box B \ge \operatorname {area} \Box L \ge \operatorname {area}\Box A$
By the result for rational numbers:
$\operatorname {area}\Box B = B^2$ $\operatorname {area}\Box A = A^2$
We also note that:
$\displaystyle \lim_{B \to L} B^2 = L^2 = \lim_{A \to L} A^2$
Thus:
$\displaystyle \lim_{B \to L} \operatorname {area} \Box B = \lim_{B \to L} B^2 = L^2$ $\displaystyle \lim_{A \to L} \operatorname {area} \Box A = \lim_{A \to L} A^2 = L^2$
Finally:
$L^2 \ge \operatorname {area}\Box L \ge L^2$
so:
$\operatorname {area}\Box L = L^2$
$\blacksquare$
By definition, $AB : EF = L : 1$.
From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\Box ABCD$ to $\Box EFGH$ is the duplicate ratio of the ratio of $AB$ to $EF$.
Thus by definition of duplicate ratio:
$\Box ABCD : \Box EFGH = \left({AB : EF}\right)^2$
That is:
$\dfrac {\Box ABCD} {\Box EFGH} = \left({\dfrac L 1}\right)^2 = L^2$
That is, the area of $\Box ABCD$ has $L^2$ as many units as $\Box EFGH$.
Hence the result.
$\blacksquare$
This square is equivalent to the area under the graph of $f \left({x}\right) = a$ from $0$ to $a$.
Thus from the geometric interpretation of the definite integral, the area of the square will be the integral:
$\displaystyle A = \int_0^a a \, \mathrm d l$
Thus:
\(\displaystyle A\) \(=\) \(\displaystyle \int_0^a a \, \mathrm d l\) \(\displaystyle \) \(=\) \(\displaystyle \left[{l \cdot a}\right]_0^a\) Integral of Constant \(\displaystyle \) \(=\) \(\displaystyle a \cdot a - 0 \cdot a\) \(\displaystyle \) \(=\) \(\displaystyle a^2\)
$\blacksquare$
Warning This proof is circular.
That fact is in turn derived from this one.
However, this demonstration neatly parallels the integration based proofs of the areas of other figures, for example Area of Circle.
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I think you made a small typo. The sum in the denominator should be running from $1$ to $k-1$, not $k$. Edit: nevermind you fixed it :)
Fitting $k-1$ independent regressions, you get estimates for $\{\theta_j\}$ with $j=1, \ldots, k-1$. The $k-1$ models are parametrized as:$$P(Y=j|X^{(i)}) = \frac{\exp(\theta_j^TX^{(i)})}{1+ \exp(\theta_j^T X^{(i)})}.$$Looking at each row, the dependent variable will either be a succes ($=j$) or not $(\neq j)$. You're basically ignoring all the other outcomes here.
On the other hand if you fit a multinomial logistic regression, it will enforce, as @MaartenBuis says, the constraint that all the probabilities sum to $1$. You're also assuming, probably more correctly, that each dependent variable can be any value from $1$ to $k$, and it will follow a multinomial distribution with the parameter vector being dependent on the $X^{(i)}$ data. In other words, the model assumed is$$P(Y=j|X^{(i)}) = \frac{\exp(\theta_j^TX^{(i)})}{1+ \sum_{m=1}^{k-1}\exp(\theta_m^T X^{(i)})}$$with $j=1,\ldots,K-1$, and you still get the same amount of coefficient estimates. But you can see that $$\sum_{i=1}^k P(Y=i|X^{(i)}) = \sum_{i=1}^{k-1} \frac{\exp(\theta_j^TX^{(i)})}{1+ \sum_{m=1}^{k-1}\exp(\theta_m^T X^{(i)})} + \frac{1}{1+\sum_{m=1}^{k}\exp(\theta_m^T X^{(i)})} = 1.$$
I can't prove when the estimates between these two procedures will be close or not, but I can tell you that these two procedures are assuming totally different models to be true.
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Learning Objectives
To describe the relationship between solute concentration and the physical properties of a solution. To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.
Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration.
Osmosis can be demonstrated using a U-tube like the one shown in Figure \(\PageIndex{1}\), which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (\(\Pi\)) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution.
The osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation:
where
\(M\) is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), \(R\) is the ideal gas constant, and \(T\) is the absolute temperature.
As shown in Example \(\PageIndex{1}\), osmotic pressures tend to be quite high, even for rather dilute solutions.
Example \(\PageIndex{1}\): Yeast Cells
When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% \(NaCl\) by mass; the solution density is 1.02 g/mL at 25°C.
Calculate the osmotic pressure of a 4.0% aqueous \(NaCl\) solution at 25°C.
If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C?
Given: concentration, density, and temperature of \(NaCl\) solution; internal osmotic pressure of cell Asked for: osmotic pressure of \(NaCl\) solution and concentration of glycerol needed Strategy: Calculate the molarity of the \(NaCl\) solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles. Use Equation \ref{eq1} to calculate the osmotic pressure of the solution. Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation \ref{eq1} to calculate the molarity of glycerol needed to create this osmotic pressure. Solution:
A The solution contains 4.0 g of \(NaCl\) per 100 g of solution. Using the formula mass of \(NaCl\) (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity:
\[\begin{align} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} onumber \\[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) onumber \\[4pt] &= 0.70\; M\; NaCl onumber \end{align} onumber\]
Because 1 mol of \(NaCl\) produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M.
B Now we can use Equation \ref{eq1} to calculate the osmotic pressure of the solution:
\[\begin{align*} \Pi &=MRT \\[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K) \\[4pt] &= 34 \;atm \end{align*}\]
C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{eq1} for the molarity corresponding to this osmotic pressure:
\[\begin{align*} M &=\dfrac{\Pi}{RT} \\[4pt] &= \dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})} \\[4pt] &= 1.1 \;M \;glycerol \end{align*}\]
In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the \(NaCl\) solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol.
Exercise \(\PageIndex{1}\)
Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them.
Answer: 24 atm
Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins.
The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure \(\PageIndex{2}\), which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells.
In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves.
Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure \(\PageIndex{3}\), applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats.
Summary
When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through.
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A Matlab toolbox to analyze grain boundary inclination from SEM images¶
First of all, download the source code of the Matlab toolbox.
This toolbox helps to find the grain boundary inclination from two micrographs from serial polishing. At least three marks such as microindents are needed for registration of the images.
Examples of micrographs from serial polishing.
To get started with this toolbox, clone the repository, then run Matlab, and cd into the folder containing this README file. Then add the package path to the Matlab search path by typing “path_management”. Finally you can start the launcher by typing demo or A_gui_gbinc at the Matlab command prompt.
How to use the toolbox ?¶ Run the function A_gui_gbinc.m. Select your first image before serial polishing. Do the calibration to get the factor scale. Do the edge detection. Repeat the same operation for the second image obtained after serial polishing. Do the overlay : If control points don’t exist (it’s the case for the 1st time), a window appears and it is possible to define control points. Define 3 control points per images. Select a point on the figure on the left, then on the figure on the right, and repeat this operation 2 times. Close the window for the selection of control points (Ctrl+W). Control points are saved in .mat file (in the same folder than the 1st picture loaded). If the control points are not satisfying, delete them and redo the step 6 to set new control points and to get a new overlay. Save the overlay in the same folder than the 1st picture loaded (as a screenshot.png) (optional). Do the measurement of the distance between edges (Vickers faces) or ridges of a unique Vickers indent (see Figure 43). Do the measurement of the distance between edges of a unique grain boundary. The value of the grain boundary inclination is finally given in degrees.
Calculation of the thickness of removed material after polishing(29)¶\[h = \frac{d}{\tan(90 - \alpha)}\]
With \(d\) the distance between edges (Vickers faces) or ridges of a unique Vickers indent (obtained before and after polishing), and \(\alpha\) the angle between the Vickers indent and the surface of the sample (see Figure 43).
Calculation of grain boundary inclination(30)¶\[GB_{inc} = \tan \left(\frac{d_{GB}}{h}\right)\]
With \(d_{GB}\) the distance between grain boundary traces (obtained before and after polishing), and \(h\) the thickness of removed material after polishing calculated
Note
Images should have the same scale factor.
Note
Distances and grain boundary inclination values are obtained with the mean scale factor of the two images.
See also¶ Acknowledgements¶
Parts of this work were supported under the NSF/DFG Materials World Network program (DFG ZA 523/3-1 and NSF-DMR-1108211).
Keywords¶
Matlab ; Graphical User Interface (GUI) ; Grain Boundaries ; Polycrystalline Metals ; Grain Boundary Inclination ; Serial Polishing ; Scanning Electron Microscope (SEM).
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I'm probably missing something elementary here, but I guess the only way to be sure is to ask here.
Now, I have encountered a situation where given an nth-degree polynomial $p_n(z)$ with complex coefficients, and a positive real number $\rho$, I need to find the value(s) of $\theta$, $0\leq\theta<2\pi$, such that the value of $|p_n(\rho\exp(i\theta))|$ is minimized (i.e., find the lowest point of the absolute value of a complex polynomial around a radius $\rho$ circle). I know about the usual methods for univariate minimization (golden section, Brent's method, Newton"s method), but I am wondering if there may be special methods that can be used that are more efficient, given that the function to be minimized can be turned into a "trigonometric polynomial". Or would finding these minima be of the same level of difficulty as finding the roots of the polynomial itself?
Thus far, the only simplification I have been able to come up with is that if all the coefficients of $p_n(z)$ are real, I can restrict the search for the optimal $\theta$ in the interval $[0,\pi]$, since $p_n(\bar{z})=\overline{p_n(z)}$. A "grid search", using FFT to evaluate the polynomial at equispaced points around the circle was one idea I thought of, but it seemed wasteful of effort since I have been unable to find a way to reuse the effort done by FFT when the number of points around the circle is doubled.
In short: might there be an easier, more obvious way I am missing?
Addendum:
The application where I'm considering this procedure as a subroutine operates as follows:
The complex polynomial and an initial estimate of $\rho$ are given. The minimization procedure finds the value of $\theta$ where the objective function is minimized; if there is more than one possible $\theta$, the value nearest to the positive real axis is taken (this is the rather ad hocportion of the application I'm looking at).
The tentative $\theta$ is subjected to an "oracle" that
a. if a success flag is returned, the algorithm exits, else
b. a smaller value of $\rho$ is computed through another black-box procedure, and we return to step 2.
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I try to solve the following recursion for $n \in \mathbb{N}$.
$r_i = r_{i-1} - \frac{1}{2} \cdot \sqrt{1 - \frac{4\pi^2\cdot r_{i-1}^2}{n^2} \cdot \cos^2 \left(\frac{\pi}{n}\right)}$
$r_0 = \frac{n}{2\pi}$
I translated it into the following code for Mathematica:
RSolve[{g[x]==g[x-1]- 1/2 Sqrt[1 - 4 Pi^2 g[x-1]^2/n^2 (Cos[Pi/n])^2], g[0]==n/(2 Pi)}, g[x], x]
However, Mathematica cannot interpret this and I get the input as result. Is there any mistake from my side or is Mathematica not able to solve this?
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Part 5 in the quest for the hydrogen molecule
Last time we investigated the energy states of a free particle. That is of course a bit of a boring example: the particle does not really have any behavior except that it flies in one direction or another. This time we will derive our first results on how quantum mechanics leads to special bound states, which is a very important step towards understanding the hydrogen molecule.
Also we will have to roll up our sleeves, because we will have to start doing some tough mathematics do derive some real results. Do not be get discouraged, because mathematics is the language of the universe, and we need it to get to some tangible results now that we have established our first principles.
We will study the case of a single particle that is under the influence of a position-dependent potential \(V(x)\). The total energy of the particle is given by
$$E=\frac{p^2}{2m} + V(x).$$
We will have to choose a simple potential \(V(x)\) to work with. This is a very important choice, and for this post I would like to choose the “finite square well”,
Using the notation of Iverson brackets, the potential energy of the particle at position \(x\) is given by:
$$V(x) =-V_0[|x|<L].$$
As we did in the previous post, we will solve the energy equation to determine the energy states \(|\psi\rangle\):
$$\hat E |\psi\rangle = E |\psi\rangle.$$
We choose to express everything in terms of momentum states \(|p\rangle\). To make that more clear, we project both the left side and the right side on these momentum states, as was explained in part 3:
$$\langle p|\hat E |\psi\rangle = \langle p|E |\psi\rangle=E\phi(p),$$
where we have written the projection on the momentum states as a function which we call \(\phi(p)\equiv\langle p|\psi\rangle\).
A few mathematical tools
Before we continue, let’s develop a few formulas that will be useful. First, we should also remember from part 3 that we can always project a state on the full set of either the position or the momentum states. Let me repeat those formulas here:
$$| \psi \rangle = \int_{-\infty}^{+\infty} \langle x| \psi \rangle |x \rangle dx,\\
| \psi \rangle = \int_{-\infty}^{+\infty} \langle p| \psi \rangle |p \rangle dp.$$
Note that there is an interesting case when \(| \psi \rangle = | p’ \rangle\):
$$| p’ \rangle = \int_{-\infty}^{+\infty} \langle p| p’ \rangle |p \rangle dp,$$
and since the left hand side and right hand side have to be equal, we conclude that \(\langle p| p’ \rangle =0\) whenever \(p \neq p’\). In fact, \(\langle p| p’ \rangle \) satisfies
$$ \int_{-\infty}^{+\infty} f(p)\langle p| p’ \rangle dp=f(p’),$$
for any function \(f(p)\), which will come in handy.
Solving the energy equation
Great, now we are ready to calculate. Get ready for some serious mathematics! What I would like to calculate is for a state \(|\psi\rangle\) that satisfies the energy equation:
what energy \(E\) it might have, and what is its projection on the momentum states \(\phi(p)\)
First I split the kinetic and potential energy terms,
$$\langle p|\hat E |\psi\rangle = \langle p|\hat E_k |\psi\rangle+\langle p|\hat E_p|\psi\rangle.$$
Kinetic energy
The first term, the kinetic energy, is relatively simple. To calculate it, I will insert a projection on momentum states \(|p’\rangle\).
$$
\langle p|\hat E_k |\psi\rangle=\langle p|\frac{\hat p^2}{2m} |\psi\rangle=\\ \int_{-\infty}^{+\infty}\! dp’\langle p|\frac{\hat p^2}{2m} |p’ \rangle \langle p’| \psi \rangle=\\ \int_{-\infty}^{+\infty}\!dp’ \langle p|\frac{p’^2}{2m} |p’ \rangle \langle p’| \psi \rangle =\\ \int_{-\infty}^{+\infty}\!dp’ \frac{p’^2}{2m} \langle p|p’ \rangle \langle p’| \psi \rangle=\\ \frac{p^2}{2m} \langle p| \psi \rangle=\frac{p^2}{2m} \phi(p).$$
Or, to summarize, the \(\hat p\) operator is also allowed to operate on the projected-to states \(\langle p|\) to the left of it.
Potential energy
Good. Now I will work on the potential term. I will use a projection on position states \(|x\rangle\):
$$\langle p|\hat E_p|\psi\rangle=\langle p|V(\hat x)|\psi\rangle=\\
\int_{-\infty}^{+\infty} \!dx\langle p|V(\hat x)\langle x| \psi \rangle |x \rangle=\\ \int_{-\infty}^{+\infty} \!dx\langle p|V(x)\langle x| \psi \rangle |x \rangle=\\ -V_0\int_{-L}^{+L} dx\langle p|x \rangle\langle x| \psi \rangle .$$
Whoa! Now it becomes clear why the square well potential is such a convenient potential! Unfortunately, this term still contains an expression \(\langle x| \psi \rangle \), while the kinetic term was expressed in terms of \(\phi(p)\). To bring the terms together again, we add another projection here, this time on momentum states \(|p’\rangle\):
$$
-V_0\int_{-L}^{+L} \!dx\langle p|x \rangle\langle x| \psi \rangle =\\ -V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\langle p|x \rangle\langle x|p’ \rangle\langle p’| \psi \rangle .$$
Now we remember from part 3 our postulate:
$$\langle x|p \rangle = e^{i x p / \hbar},$$
which actually has a twin brother:
$$\langle p|x \rangle = e^{-i x p / \hbar}.$$
Using also by definition \(\langle p’| \psi \rangle=\phi(p’)\), we manage to arrive at a quite normal mathematical equation:
$$
\langle p|\hat E_p|\psi\rangle=-V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\,e^{-i x p / \hbar}e^{i x p’ / \hbar}\phi(p’) =\\ -V_0\int_{-\infty}^{+\infty}\!dp’\int_{-L}^{+L} \!dx\,e^{i x (p’-p) / \hbar}\phi(p’) .$$
The integration over \(x\) is something we can actually do!
$$
\int_{-L}^{+L} \!dx\,e^{i x (p’-p) / \hbar} =\\ \frac{1}{i(p’-p)}e^{i x (p’-p) / \hbar} \Biggr|_{-L}^{+L}=\\ \frac{1}{i(p’-p)}\bigg[e^{i L (p’-p) / \hbar}-e^{i L (p’-p) / \hbar}\bigg]=\\ \frac{2\sin[L (p’-p) / \hbar]}{p’-p},$$
arriving at a final expression for the potential energy:
$$\langle p|\hat E_p|\psi\rangle=-V_0\int_{-\infty}^{+\infty}\!dp’\frac{2\sin[L (p’-p) / \hbar]}{p’-p}\phi(p’).$$
Final equation
In the end, the full energy equation thus becomes
$$\frac{p^2}{2m}\phi(p)-2V_0\int_{-\infty}^{+\infty}\!dp’\,\frac{\sin[L (p’-p) / \hbar]}{p’-p}\phi(p’) =E\phi(p),$$
which we have to solve for the energy value \(E\) and the function \(\phi\). How to proceed?
Numerical calculation
Unfortunately there is no analytical solution. But we can solve this equation numerically. The first thing to do is to apply some substitutions, to remove irrelevant parameters. We use.
$$
\tilde{p} \equiv \frac{L}{\hbar}p,\\ \tilde{V} \equiv \frac{2mL^2}{\hbar^2}V_0,\\ \tilde{E} \equiv \frac{2mL^2}{\hbar^2} E,\\ \tilde{\phi}(\tilde{p})\equiv \phi(p).$$
Resulting in
$$\tilde{p}^2\tilde{\phi}(\tilde{p})-2\tilde{V}\int_{-\infty}^{+\infty}\!d\tilde{p}’\,\frac{\sin(\tilde{p}’-\tilde{p})}{\tilde{p}’-\tilde{p}}\tilde{\phi}(\tilde{p}’) =\tilde{E}\tilde{\phi}(\tilde{p}) ,$$
As we can see, the solutions to the energy value \(\tilde{E}\) only depend on a single input parameter \(\tilde{V}\). Next step to solve this numerically is to replace the integral by a summation using \(\tilde{p} = n\Delta \tilde{p}\) and \(\tilde{\phi}_n=\tilde{\phi}(n\Delta \tilde{p})\):
$$(\Delta \tilde{p})^2n^2\tilde{\phi}_n-2\tilde{V}\sum_{n’=-\infty}^{+\infty}\frac{\sin[\Delta \tilde{p}(n’-n)]}{n’-n}\tilde{\phi}_{n’} =\tilde{E}\tilde{\phi}_n,$$
which makes it look like a linear algebra problem,
$$\sum_{n’}K(n, n’)\phi_{n’}=E\phi_n,$$
with
$$K(n, n’)=((\Delta \tilde{p})^2n^2-2\tilde{V} \Delta \tilde{p})[n=n’]-2\tilde{V}\frac{\sin[\Delta \tilde{p}(n’-n)]}{n’-n}[n \neq n’],$$
which can be solved by Octave.
Results Free particle
First we look at the case \(\tilde{V}=0\). We expect to retrieve the energy spectrum of a free particle, since the potential well is not really there in that case. Indeed, we find the following spectrum:
All energy states have a positive energy. The reason that the energy spectrum is discrete rather than continuous is that we have replaced the integral by a summation (the distance between the states depends on the parameter \(\Delta \tilde{p}\)), but this does show nicely that the states get farther apart for higher energies.
First bound state
Moving on to a nonzero potential for the well, let’s pick \(\tilde{V}=0.5\). Here is the energy spectrum:
Remarkable! A single state was added with a negative energy. This is the energy of the
bound state, and the most remarkable thing is that there is only one bound state. This does not align with our normal experience of the world. We think there are always many ways to put a marble in a shallow but wide cup: we can put it on the left or on the right, or in the middle, rolling with a little speed or not. What we learn is that this is in fact not true for very shallow cups: there is only one state, and the marble in the cup cannot carry any information.
This bound state has energy -2.1754. How does its projection on momentum states \(\phi(p)=\langle p| \psi \rangle\) and its projection on position states \(\phi(x)=\langle x| \phi \rangle\) look like?
As we can see, the bound state can be described as a mix of momentum states, or as a mix of position states. For the latter there is the unexpected discovery that the particle in the bound state does not actually stay within the limits of the well at \(x = \pm 1\). It is a strange world we live in!
Other bound states
Increasing the potential \(\tilde{V}\) lowers the energy of the first bound state, and at some point a second bound state appears. Here is a nice animation to illustrate this:
At \(\tilde{V}=3\) there are three bound states. For illustration, here are the projections on momentum states and on position states:
Conclusions
I have shown that a particle in a well has a number of discrete bound states, in contrast to the continuum of unbound states. That is an important conclusion, it is essentially where “quantum mechanics” derives its name from!
I can calculate the energy of these bound states, and also how these bound states can be written as a superposition of either momentum or position states. With respect to the momentum states, it can be noted that we have a superposition of many momentum states, and not just two of them, as one might expect. With respect to the position states, it is surprising that the particle position is not limited to within the well, although the particle in the bound state cannot escape the well.
But to me, the most remarkable thing is that we have derived all these experimentally verifiable conclusions from very few propositions. In fact we have skipped large parts of what is usually considered to be essential for understanding quantum mechanics: measurements, uncertainties and probabilities, normalization, time-evolution, the Schrödinger equation. It has been very interesting, to study this, but we are certainly not finished. In the next part, we will derive the Schrödinger equation.
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Just as we studied special types of sequences, we will look at special types of series. Recall that an
arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, [latex]d[/latex]. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first [latex]n[/latex] terms of an arithmetic series as:
We can also reverse the order of the terms and write the sum as
If we add these two expressions for the sum of the first [latex]n[/latex] terms of an arithmetic series, we can derive a formula for the sum of the first [latex]n[/latex] terms of any arithmetic series.
Because there are [latex]n[/latex] terms in the series, we can simplify this sum to
We divide by 2 to find the formula for the sum of the first [latex]n[/latex] terms of an arithmetic series.
A General Note: Formula for the Sum of the First n Terms of an Arithmetic Series
An
arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the sum of the first [latex]n[/latex] terms of an arithmetic sequence is How To: Given terms of an arithmetic series, find the sum of the first [latex]n[/latex] terms. Identify [latex]{a}_{1}[/latex] and [latex]{a}_{n}[/latex]. Determine [latex]n[/latex]. Substitute values for [latex]{a}_{1}\text{, }{a}_{n}[/latex], and [latex]n[/latex] into the formula [latex]{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}[/latex]. Simplify to find [latex]{S}_{n}[/latex]. Example 2: Finding the First n Terms of an Arithmetic Series
Find the sum of each arithmetic series.
[latex]\text{5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32}[/latex] [latex]\text{20 + 15 + 10 +}\ldots{ + -50}[/latex] [latex]\sum _{k=1}^{12}3k - 8[/latex] Solution We are given [latex]{a}_{1}=5[/latex] and [latex]{a}_{n}=32[/latex].Count the number of terms in the sequence to find [latex]n=10[/latex].
Substitute values for [latex]{a}_{1},{a}_{n}\text{\hspace{0.17em},}[/latex] and [latex]n[/latex] into the formula and simplify.[latex]\begin{array}{l}\begin{array}{l}\hfill \\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \end{array}\hfill \\ {S}_{10}=\frac{10\left(5+32\right)}{2}=185\hfill \end{array}[/latex]
We are given [latex]{a}_{1}=20[/latex] and [latex]{a}_{n}=-50[/latex].Use the formula for the general term of an arithmetic sequence to find [latex]n[/latex].[latex]\begin{array}{l}{a}_{n}={a}_{1}+\left(n - 1\right)d\hfill \\ -50=20+\left(n - 1\right)\left(-5\right)\hfill \\ -70=\left(n - 1\right)\left(-5\right)\hfill \\ 14=n - 1\hfill \\ 15=n\hfill \end{array}[/latex]
Substitute values for [latex]{a}_{1},{a}_{n}\text{,}n[/latex] into the formula and simplify.[latex]\begin{array}{l}\begin{array}{l}\\ {S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\end{array}\hfill \\ {S}_{15}=\frac{15\left(20 - 50\right)}{2}=-225\hfill \end{array}[/latex]
To find [latex]{a}_{1}[/latex], substitute [latex]k=1[/latex] into the given explicit formula.[latex]\begin{array}{l}{a}_{k}=3k - 8\hfill \\ \text{ }{a}_{1}=3\left(1\right)-8=-5\hfill \end{array}[/latex]
We are given that [latex]n=12[/latex]. To find [latex]{a}_{12}[/latex], substitute [latex]k=12[/latex] into the given explicit formula.[latex]\begin{array}{l}\text{ }{a}_{k}=3k - 8\hfill \\ {a}_{12}=3\left(12\right)-8=28\hfill \end{array}[/latex]
Substitute values for [latex]{a}_{1},{a}_{n}[/latex], and [latex]n[/latex] into the formula and simplify.[latex]\begin{array}{l}\text{ }{S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ {S}_{12}=\frac{12\left(-5+28\right)}{2}=138\hfill \end{array}[/latex]
Use the formula to find the sum of each arithmetic series.
Try It 2
[latex]\text{1}\text{.4 + 1}\text{.6 + 1}\text{.8 + 2}\text{.0 + 2}\text{.2 + 2}\text{.4 + 2}\text{.6 + 2}\text{.8 + 3}\text{.0 + 3}\text{.2 + 3}\text{.4}[/latex]
Try It 3
[latex]\text{13 + 21 + 29 + }\dots \text{+ 69}[/latex]
Try It 4
[latex]\sum _{k=1}^{10}5 - 6k[/latex]
Example 3: Solving Application Problems with Arithmetic Series
On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?
Solution
This problem can be modeled by an arithmetic series with [latex]{a}_{1}=\frac{1}{2}[/latex] and [latex]d=\frac{1}{4}[/latex]. We are looking for the total number of miles walked after 8 weeks, so we know that [latex]n=8[/latex], and we are looking for [latex]{S}_{8}[/latex]. To find [latex]{a}_{8}[/latex], we can use the explicit formula for an arithmetic sequence.
We can now use the formula for arithmetic series.
She will have walked a total of 11 miles.
Try It 5
A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned?
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Definition:Ordering/Definition 1 Definition
Let $S$ be a set.
An
ordering on $S$ is a relation $\mathcal R$ on $S$ such that:
\((1)\) $:$ $\mathcal R$ is reflexive \(\displaystyle \forall a \in S:\) \(\displaystyle a \mathop {\mathcal R} a \) \((2)\) $:$ $\mathcal R$ is transitive \(\displaystyle \forall a, b, c \in S:\) \(\displaystyle a \mathop {\mathcal R} b \land b \mathop {\mathcal R} c \implies a \mathop {\mathcal R} c \) \((3)\) $:$ $\mathcal R$ is antisymmetric \(\displaystyle \forall a \in S:\) \(\displaystyle a \mathop {\mathcal R} b \land b \mathop {\mathcal R} a \implies a = b \)
Symbols used to denote a general ordering relation are usually variants on $\preceq$, $\le$ and so on.
On $\mathsf{Pr} \infty \mathsf{fWiki}$, to denote a general ordering relation it is recommended to use $\preceq$ and its variants:
$\preccurlyeq$ $\curlyeqprec$ $\leqslant$ $\leqq$ $\eqslantless$ The symbol $\subseteq$ is universally reserved for the subset relation. $a \preceq b$
can be read as:
$a$ precedes, or is the same as, $b$.
Similarly:
$a \preceq b$
can be read as:
$b$ succeeds, or is the same as, $a$. If, for two elements $a, b \in S$, it is not the case that $a \preceq b$, then the symbols $a \npreceq b$ and $b \nsucceq a$ can be used.
It is not demanded of an ordering $\preceq$, defined in its most general form on a set $S$, that
every pair of elements of $S$ is related by $\preceq$. They may be, or they may not be, depending on the specific nature of both $S$ and $\preceq$.
It is wise to be certain of what is meant.
As a consequence, on $\mathsf{Pr} \infty \mathsf{fWiki}$ we resolve any ambiguity by reserving the terms for the objects in question as follows: Also see Results about orderingscan be found here. Sources 1960: Paul R. Halmos: Naive Set Theory... (previous) ... (next): $\S 14$: Order 1964: W.E. Deskins: Abstract Algebra... (previous) ... (next): $\S 1.2$: Definition $1.7 \ \text {(b)}$ 1965: Seth Warner: Modern Algebra... (previous) ... (next): $\S 14$ 1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis... (previous) ... (next): $\S 3.1$: Partially ordered sets 1968: Ian D. Macdonald: The Theory of Groups... (previous) ... (next): Appendix: Elementary set and number theory 1972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis... (previous) ... (next): $\S 3$: Equivalence relations and quotient sets: Binary relations 1975: T.S. Blyth: Set Theory and Abstract Algebra... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations 1975: T.S. Blyth: Set Theory and Abstract Algebra... (previous) ... (next): $\S 7$ 1982: P.M. Cohn: Algebra Volume 1(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory(2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.5$: Relations 2008: Paul Halmos and Steven Givant: Introduction to Boolean Algebras... (previous) ... (next): Appendix $\text{A}$: Set Theory: Order
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I would like to use charter for text and formula symbols, excerpt for greek letters, these I'd like to do in palatino (
mathpazo). I've tried around a lot with
\DeclareSymbolFont,
\DeclareMathSymbol and so on. But to be honest, I'm not close to a solution.
A minimal (not working) code example follows:
\documentclass{article}%\usepackage[charter]{mathdesign}\DeclareSymbolFont{greeksymbols}{OML}{ppl}{m}{it}\DeclareMathSymbol{\pii}{\mathalpha}{greeksymbols}{25}\DeclareMathSymbol{\chii}{\mathalpha}{greeksymbols}{31}\DeclareMathSymbol{\mui}{\mathalpha}{greeksymbols}{22}\begin{document}{\Huge $\pii \chii \mui $}{\Huge $\pi \chi \mu $}\end{document}
I've you could give any hints, this would help me a lot!
Thanks!
EK
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Problem Authoring Best Practices
Below is a summary of the stylistic guidelines for "cleaning" and uniformizing WeBWorK problems that were adopted during the WeBWorK Programming Workshop at MSRI on 20-24 May 2004. These guidelines apply to problems intended for the National Problem Library.
The items listed below cover the most common issues. Those who are writing problems for the first time should use National Problem Library problems as templates to insure a good working model.
Guidelines
Answer boxes should be labeled. If a question asks for the derivative of f(x), then "f'(x) = [Answer Box]" is appropriate. If there isn't an obvious label, then "Answer: [Answer Box]" should be used. Also, if the answer implicitly includes units that aren't to be included in the answer box, then they should appear after the box, e.g. "Answer: [Answer Box] furlongs".
All answer boxes should be large enough to actually hold the answer without scrolling. This is easily verified by pasting the correct answer into the box.
Use complete sentences. Check for grammatical and spelling errors. When using words with different singular and plural forms, check the associated number and use the appropriate form rather than e.g. "$n bicycle(s)".
Use display equations e.g.
\[...\]
only when necessary and appropriate.
In most cases, numerical answers should be coded using
Compute() e.g.
$ans = Compute("$a*cos($b)") instead of
$ans = $a*cos($b). This way, if a correct answer is
7*sin(pi/4), when a student checks the "Show Answer" box after the deadline then she will see
7*sin(pi/4) instead of
4.949747468305832. The exception to this rule is when the answer would be extremely complicated/messy, in which case it may be better to use the numerical form.
Use "pi" instead of the number 3.14159... and "e" instead of the number 2.71828... MathObjects recognizes these constants and they will be rendered using the appropriate symbols.
All characters and expressions that would be in math mode in a text or research article should be in math mode in a WeBWorK problem.
Use
$f->reduce for all displayed polynomials where there is the potential for coefficients of 0, 1, or -1. This will eliminate the unnecessary 1's and -1's, and will drop terms with a coefficient of 0.
No hints or comments referring to specific texts or other works should be used.
Unnecessary calls to packages in the "loadMacros" portion of a problem should not be used. You should always use "PGstandard.pl", "MathObjects.pl", and "PGcourse.pl".
Use the MathObjects
$obj->cmp method to get the appropriate answer evaluators for your answer. Old answer evaluators that have been superseded should not be used.
Style Recommendations Don't load a macro unless you actually use it. Use descriptive variable names like
$xintfor an x-intercept (instead of just
$x)
Separate blocks of code using extra space and dividers like ############################# # Main text Layout the code visually so that it is easily readable by other humans. This is a long line of text.$PAR It's not so easy to see the paragraph break. This is a long line of text. $BR $BR (a) Now it's easy to see the paragraph break! Don't use
$SPACE$SPACE$SPACE$SPACEto indent. If you want a paragraph to be indented, use
$PARbefore it. If you don't want it to be indented, use two line breaks
$BR $BR.
Put the info about the textbook, chapter, section, problem number, etc. in the problem tagging section, notinside of a text block
BEGIN_TEXT ... END_TEXT.
Give graphs captions. Make graphs big enough in HTML and the PDF hardcopy. Check that the graph's resolution, font size, line thickness, etc. are all readable. When using a graph with more than a few lines of text, use the
ColumnTable(column 1, column 2, options)routine provided by the macro
unionTables.plto provide a two column format. Put the text in column 1 and the graph in column 2. If you have multiple graphs, use a table of graphs.
Use left justification most of the time, and center sparingly. Make your math formulas readable using
\displaystyle:
\( \displaystyle \frac{a}{b} \) \( \displaystyle \frac{ \displaystyle\frac{a}{b} }{ \displaystyle\frac{c}{d} } \) Test your problem in images, and MathJAX modes. Don't use
\dfrac{}{}because jsMath will doesn't understand that command.
If you were writing a textbook and would put something in TeX's math mode, then put it in math mode. Make sure the PDF output works and everything is readable and usable. Use multiple contexts and text blocks when appropriate Don't overuse bold, don't use ALL CAPS, and DON'T SHOUT AT YOUR READER! For questions with multiple parts, use extra space between parts and keep each part together as a block. This will make it easier to read and use. Write comments for other instructors. For example,
COMMENT('MathObject version');or
COMMENT('MathObject version. Not randomized.');
2007 document from Sam Hathaway TrainingAuthors
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Bug introduced in 8 or earlier and persists through 11.0.1 or later
I am trying to plot the following single variable function:
$$\small\frac{180 \sqrt{\pi ^2-625 t} \left(\pi ^2 (36 t-25)-1500 t \left(45 t-\sqrt{900 t-\pi ^2}-15\right)\right)}{\pi ^4 \sqrt{2500 t-\pi ^2}}+\tan \left(2 \sqrt{\pi ^2-625 t}\right)$$
with code:
Plot[(180 Sqrt[π^2-625 t] (π^2 (-25+36 t)-1500 t (-15+45 t-Sqrt[-π^2+900 t])))/(π^4 Sqrt[-π^2+2500 t])+Tan[2 Sqrt[π^2-625 t]],{t,0.01,0.016},ImageSize->800,(*AxesStyle->{{Directive[Red, 12],Arrowheads[.035],Thick},{Directive[Red, 12],Arrowheads[.035],Thick}},*)PlotStyle->Blue]
However, I have to comment the
AxesStyle line above in order to obtain normal image size:
or it will look like shrinking version below:
What is the reason? How can I customize the axes style for my case?
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MAP (Maximum a posteriori) Inference is a form of Bayesian inference that deals with MAP queries. The goal of a MAP query is to find the most likely assignment for all of the non-evidence random variables given the evidence variables, this is also called the Most Probable Explanation (MPE). This is similar to Bayesian learning, but the computational complexity of MAP Inference is less than that of Bayesian learning. [Russell, 2010, p.804]
Given a set of random variables, $\chi$, and a set of evidence variables $E$ <blockquote>$W = \chi - E$</blockquote>
The MAP query finds the most likely assignment to the query variables $W$ given the evidence variables $E$. <blockquote>$MAP( W | E) = \underset{w~\in~W}{argmax}~P( w | E)$</blockquote>
If there are several such assignments that are equal, then the MAP query can either return the set of all possible assignments or return an arbitrarily chosen assignment. [Koller, 2009, p.26]
The result to an MAP query is not the same thing as finding the most likely assignment to each individual variable, as is shown in the example.
The MAP query is sometimes referred to as the MAP hypothesis, where the evidence variables, $E$, are referred to as data, $D$, and the assignments to the query variables, $W$, are referred to as possible hypotheses, $h$, from a hypothesis space, $H$. The MAP hypothesis is found by the equation <blockquote>$h_{MAP} = \underset{h~\in~H}{argmax}~P( h | D )$ :$= \underset{h~\in~H}{argmax}~\frac{P( D | h ) P ( h )}{P( D )}$ :$= \underset{h~\in~H}{argmax}~P( D | h ) P ( h )$</blockquote>
If the $P(h)$ term is dropped in the previous equation, denoting a uniform prior probability among all hypotheses, then we get the
Maximum Likelihood (ML) hypothesis.[Mitchell, 1997, p.157]
This example comes from Koller [Koller, 2009, p.26]
Conditional Probability Table for $B$
$A$ $b^0$ $b^1$ $a^0$ 0.1 0.9 $a^1$ 0.5 0.5
Probability Table for $A$
$a^0$ $a^1$ 0.4 0.6
We find that $MAP(A) = a^1$ from the Probability Table for $A$. However both assignments to $B$ given $A = a^1$ give a probability of $0.6 * 0.5 = 0.3$, while the assignment of $(a^0, b^1)$ gives a probability of $0.4 * 0.9 = 0.36$.
This gives us the following $MAP(A,B) = (a^0, b^1) \ne (\underset{a~\in~A}{argmax}~P(a), \underset{b~\in~B}{argmax}~P(b))$
The Marginal MAP query is very similar to the MAP query, except that it loosens the constraint that $W$ must contain all non-evidence variables.
Given a set of random variables, $\chi$, a set of evidence variables $E$, and a set of query variables $W$ <blockquote>$Z = \chi - W - E$</blockquote>
The MAP query finds the most likely assignment to the query variables $W$ given the evidence variables $E$. <blockquote>$MAP( W | E) = \underset{w~\in~W}{argmax}~\sum_Z{P( w, z | E)}$</blockquote>
Exact inference is not a simple task. For full MAP it is NP-complete, and for the case of a marginal MAP it is at least that, if not worse. In the worst case we have to generate the entire joint distribution, integrating out anything irrelevant to our query (in the case of a marginal MAP query), and find the entry with the highest probability. This combination of difficult tasks makes for an almost unavoidable exponential increase in complexity. Luckily this is for the worst case, and often the problems we actually run into are of a much more manageable size.
Another unfortunate fact is that even though we can use an approximation algorithm with good results, it does not remove the NP-hardness of the inference. [Koller, 2009, p.288]
MAP inference is useful in many different situations, but the main idea is that MAP inference gives us the most likely explanation for the evidence rather than the most probable assignment to a single variable given the evidence, in isolation from any other variables.
Some examples include:
This discussion is simplified by using the, equivalent, factor representation of a joint distribution. Our task is to find $\xi^{MAP}$, the most likely assignment to the variables in $\chi$. $P_{\Phi}(\chi)$ is a distribution defined via a set of factors $\Phi$ and an unnormalized density $\tilde{P_{\Phi}}$. $Z$ is a normalizing constant.
<blockquote>$\xi^{MAP} = \underset{\xi}{argmax}~P_{\Phi}(\xi)$ :$= \underset{\xi}{argmax}~\frac{1}{Z}\tilde{P_{\Phi}}(\xi)$ :$= \underset{\xi}{argmax}~\tilde{P_{\Phi}}(\xi)$</blockquote>
This is often called the
max-product form. This maximizes the unnormalized probability calculated by taking the product of the factors in $\Phi$ evaluated at $\xi$. This does not give us an actual probability unless we also calculate the normalizing constant.
The
max-sum form is also very useful to avoid some of the issues associated with multiplying floating point numbers.
<blockquote>$\xi^{MAP} = \underset{\xi}{argmax}~log~\tilde{P_{\Phi}}(\xi)$</blockquote>
Another benefit to the
max-sum form is the easy translation into an integer programming problem.
The marginal MAP query is computationally more intensive since it also involves marginalizing the non-query non-evidence variables in addition to the maximization involved in the full MAP query. $Y$ is the set of query variables, $W$ is the set non-query non-evidence variables.
<blockquote>$y^{margMAP} = \underset{y}{argmax}~P_{\Phi}(y)$ :$= \underset{y}{argmax}~\sum_W{\tilde{P_{\Phi}}(y, W)}$</blockquote>
MAP inference is an optimization problem, while that brings an added complexity there are also many algorithms already designed to handle optimization problems. MAP queries are a useful tool for many applications, but the potential intractability of some of the MAP queries we would like to make gives us the need to pause and consider the difficulty of the query we are making before we try and employ an algorithm to do so.
There is an extra difficulty associated with the marginal MAP queries that makes even approximation techniques difficult. Another issue is that the results to an MAP query need not be the same as the results to a marginal MAP query, similarly the results to one marginal MAP query need not be the same as the results to another marginal MAP query even if the query variables in one query are a subset of the query variables in the other query.
Even though MAP inference is problematic in some situations, analysis can be used to find other methods that will still give the appropriate MAP assignment (or hypothesis) without ever using Bayesian procedures.[Mitchell, 1997, p.164]
MAP inference is also more efficient than Bayesian learning, but it isn't as accurate. This trade-off comes from the fact that MAP does not take into account any uncertainty about the assignment to the query variables but the Bayesian learning method does. [Russell, 2010, p.804]
Does MAP inference always give the same result as finding the most likely assignment to each variable individually? Explain your answer using a small example.
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I am interested in deriving the convergence rate of the smallest eigenvalue of a sequence of random matrices with diverging dimension. More precisely, let $W_n(r)$ represent an $n$-dimensional standard brownian motion at time $r$, and define $\lambda_1(A)$ as the minimum eigenvalue of A. Then, I would like to know how fast does $\lambda_1\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right)$ converge to zero in probability, as $n \to \infty$.
First of all, the reason that I believe that the minimum eigenvalue converges to zero in probability is given by the following argument, which was kindly explained to me by fellow MathOverflow user Nate Eldredge. Let $E = \lbrace e_1,\ldots,e_n\rbrace$ represent the collection of basis vectors spanning $\mathbb{R}^n$. Then, for any $\epsilon > 0$,
$\mathbb{P}\left(\lambda_1\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right) > \epsilon \right) = \mathbb{P}\left(\underset{x \in \mathbb{R}^n\setminus \lbrace 0 \rbrace}{\inf}\frac{x^\prime\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right)x}{x^\prime x} > \epsilon \right) \leq \mathbb{P}\left(\underset{x \in E}{\inf} x^\prime\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right)x > \epsilon \right) = \mathbb{P}\left(\underset{1 \leq i \leq n}{\min} \int_0^1 W_{n,i}^2(r)dr > \epsilon \right) = \mathbb{P}\left(\int_0^1 W_{n,1}^2(r)dr > \epsilon \right)^n \to 0,$
as $n \to \infty$. The last inequality follows from the $i.i.d.$-ness of the elements in $W_n(r) = (W_{n,1}(r),\ldots,W_{n,n}(r))^\prime$. Then, what I would like to know essentially is if there exists an increasing function in $n$, say $f(n)$, such that $\mathbb{P}\left(f(n)\times\lambda_1\left(\int_0^1 W_n(r)W_n^\prime(r)dr\right) > \epsilon \right) \to 1$, and, if so, how does $f(n)$ look like?
Here is my thought process so far:
From Theorem 10 in Tolmatz (2002), it can be shown that $\mathbb{P}\left(n^a \times \int_0^1 W_{n,1}^2(r)dr > \epsilon \right)^n \to 1$, such that $f(n) = n^a$ would be sufficient to show that the upper bound that I used before converges to zero in probability slower than $n^a$. Unfortunately, what I am really after is showing that the minimum eigenvalue converges to zero in probability
sufficiently slow. Hence, if I could find an analytically manageable lower bound for the minimum eigenvalue and show the rate at which it goes to zero (from above), my problem would be solved. The minimum eigenvalue bounds that I could find, such as the one based on Gershgorin's circle theorem, are not strict enough, as they lead to negative lower bounds.
Hope somebody can help me with this issue. Any suggestions, either in the form of a nice minimum eigenvalue bound or perhaps in terms of a different strategy to derive my answer, would be sincerely appreciated.
Best,
Etienne
Cited article: Tomatz (2002) On the distribution of the square integral of the Brownian Bridge.
The Annals of Probability, 30, 253-269
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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Maths loves symbols. Everyone loves emoji. It’s 2017 and time we brought the two together. To get you started, here are our
top ten emoji for use in mathematics! 10.
Don’t leave home without one: it’s the nifty
45° set square. What better reminder is there that $$\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \qquad \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \qquad \tan\left(\frac{\pi}{4}\right) = 1$$ ? 9.
Perfect for popping over a letter to make it a vector, it’s the
bow and arrow:
8.
You can’t argue with this, it’s the
QED proof square:
$\displaystyle\implies \Re[s] = \frac{1}{2}$ ◼️
7.
Favoured emoji of all academics:
Do a PhD! Earn a massive £16k/year while your friends in the City earn £50k! ?
6.
Invaluable for those stupid Facebook ambiguous arithmetic questions your aunt likes to share every year, it’s the
basic arithmetic operations which exist in Emoji for some reason despite the fact that they’re in the normal character set anyway:
6 ➖ 1 ✖️ 0 ➕ 2 ➗ 2 = ???
5.
Mathematicians’ pens are their only tool. Keep yours locked up to stop petty theft:
Caution! Pen thieves operate in this area ?
4.
When you’re pretty sure something is true but can’t be bothered to figure out why, tell the reader to use their
thinking face. Similar to proof by intimidation. Combines well with an implies arrow:
3.
Excellent for making up statistics when you can’t be bothered to load up Excel:
As we can see from this very large data set, on average, every colour is green ?
2.
Horrible page of algebra which Wolfram Alpha tells you simplifies to zero? No need to show why, just state the result with
sassy information desk woman!
$\displaystyle\frac{-x^3 y^5 + x^2 y^3 + x^2 + 7 x y + 9 y^2}{y^2} – \frac{(x+3y)^2}{y^2} – x^2 y + (xy)^3 – \frac{x}{y} = \cdots = 0$ ?♀️
1.
You spin me round, round, baby, right round; left-right and top-bottom. It’s everyone’s favourite embedding of $\mathbb{S}\times\mathbb{S}$: the
torus!
…something something topology, something something pop maths. ?♉️☕️
0.
You’ll know when to use it.
?
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Here we want to give an easy mathematical bootstrap argument why solutions to the time independent 1D Schrödinger equation (TISE) tend to be rather nice. First formally rewrite the differential form$$-\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) + V(x) \psi(x) ~=~ E \psi(x) \tag{1}$$into the int...
[Some time travel comments] Since in the previous paragraph, we have explained how travelling to the future will not necessary result in you to arrive in the future that is resulted as if you have never time travelled (via twin paradox), what is the reason that the past you travelled back, has to be the past you learnt from historical records :?
@0ßelö7 Well, I'd omit the explanation of the notation on the slide itself, and since there seems to be two pairs of formulae, I'd just put one of the two and then say that there's another one with suitable substitutions.
I mean, "Hey, I bet you've always wondered how to prove X - here it is" is interesting. "Hey, you know that statement everyone knows how to prove but doesn't bother to write down? Here is the proof written down" significantly less so
Sorry I have a quick question: For questions like this physics.stackexchange.com/questions/356260/… where the accepted answer clearly does not answer the original question what is the best thing to do; downvote, flag or just leave it?
So this question says express $u^0$ in terms of $u^j$ where $u$ is the four-velocity and I get what $u^0$ and $u^j$ are but I'm a bit confused how to go about this one? I thought maybe using the space-time interval and evaluating for $\frac{dt}{d\tau}$ but it's not workin out for me... :/ Anyone give me a quickie starter please? :p
Although a physics question, this is still important to chemistry. The delocalized electric field is related to the force (and therefore the repulsive potential) between two electrons. This in turn is what we need to solve the Schrödinger Equation to describe molecules. Short answer: You can calculate the expectation value of the corresponding operator, which comes close to the mentioned superposition. — Feodoran13 hours ago
If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?
@0ßelö7 I just looked back at chat and noticed Phase's question, I wasn't purposefully ignoring you - do you want me to look over it? Because I don't think I'll gain much personally from reading the slides.
Maybe it's just me having not really done much with Eigenbases but I don't recognise where I "put it in terms of M's eigenbasis". I just wrote it down for some vector v, rather than a space that contains all of the vectors v
If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?
Honey, I Shrunk the Kids is a 1989 American comic science fiction film. The directorial debut of Joe Johnston and produced by Walt Disney Pictures, it tells the story of an inventor who accidentally shrinks his and his neighbor's kids to a quarter of an inch with his electromagnetic shrinking machine and throws them out into the backyard with the trash, where they must venture into their backyard to return home while fending off insects and other obstacles.Rick Moranis stars as Wayne Szalinski, the inventor who accidentally shrinks his children, Amy (Amy O'Neill) and Nick (Robert Oliveri). Marcia...
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I use the notation introduced in the paper: The parameters of ElGamal are $(G, q, g)$ where $q$ is prime, $G$ is a cyclic group of order $q$ and $g$ is a generator. In the paper, the authors use "exponential ElGamal" such that they have an additive homomorphism, i.e., the message is represented as "exponent" of $g$, i.e., as $g^m$. With $h=g^x$ being the public key a ciphertext of a message $m\in Z_q$ is $(c_1,c_2)=(g^y,h^y\cdot g^m)$ and it is clear that by componentwise multiplication of two ciphertexts for messages $m_1$ and $m_2$, the additive homomorphism gives a ciphertext for $m_1+m_2 \mod q$.
Note that the message space is $Z_q$ and we are working with
integers modulo $q$, i.e., the set $\{0,\ldots,q-1\}$.
Now how to handle negative integers $-x$?
There are two ways how you can see this:
If for $-x$ that the absolute value of $x$ is in the set $\{0,\ldots,q-1\}$, i.e., you have that $0\leq |x|<q$. Then, you can view $-x$ as the additive inverse of $x$ in the group $(Z_q,+)$ which has order $q$. This means $-x$ is an element $a$ for which it holds that $x+a\equiv 0 \pmod q$. Consequently, the value of $a$ is $p-x$.For example: Let $q=5$ and we want to know $-2$, i.e., the additive inverse of $2$ in $Z_5$, then we have $-2=5-2=3$, since $2+3\equiv 0 \pmod 5$, so $-2$ is equal to $3$.
Now what if the absolute value of $x$ is not in the set $\{0,\ldots,q-1\}$? For instance, how to handle $-16$ in case of the integers modulo $5$?.
We also can view integers modulo $q$, i.e., the set $\{0,\ldots,q-1\}$, as a set of representatives of the $q$ residue classes modulo $q$. More precisely, $0$ is a representative for the class $[0]$ of all integers having remainder $0$ when divided by $q$, $1$ a representative for class $[1]$, etc. If you encounter a negative integer as defined above, then you simply add $q$ until the value lies within the set $\{0,\ldots,q-1\}$ since this will obviously not change the class.
Back to our example, this means that given $-16$ you add $5$ and find that $-16+5=-11$, $-11+5=-6$, $-6+5=-1$ and finally, $-1+5=4$. Consequently $-16\equiv 4 \pmod 5$ and you can represent $-16$ as $4$.
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I know, you think that
Multi Layer Perceptron seems to be very similar to the Perceptron. And yeah, it is but wait ! It’s not completely only that but lots of things are to be going under hood. To be frank, it is some what difficult to understand the underlying concepts in the Neural Networks at the beginning. But learning how they work really saves you from lots of bottleneck problems you may face later.
So as I said, Multi Layer Perceptron is a very large topic to cover in a single post. I accept that there are many blogs out there which explains this in a single post but trust me you will see a lot more,full packaged meal here. In order to completely understand the way they work, just follow the way this blog turns up.
What is Multi Layer Perceptron?
Okay! It is similar to Perceptron but instead of having only a single layer we have multi layers. That’s it? No there is more. First have a look at it.
So I guess you have an idea now. We can see that it consists of
3 layers namely
Input layer, Hidden layer, Output layer. It is a very simple neural network with 3 layers. We can increase the number of layers based on the requirement. For now, let’s stick to the native
3 layer approach to better understand them.
What are these layers?
The Input layer takes in the input, the Output layer outputs the result, that’s it. But since we have multi layer approach things gonna be more critical from now on. The reason to it is the Hidden layer. We can have as many hidden layers but only one input and output layers as you might have figured it out why?
How do we predict/classify?
Remember the Perceptron? Yeah!! It is the same way we go with this but some what deep as it consists of one more layer !! We have the weights $W_1$ and $W_2$ associated for the layers. Suppose that our input is of dimension $N \times D$, where $N$ is the
no.of.samples and $D$ is the
no.of.features. And our output can be classified into $K$ classes. And
no.of.hidden units be $M$. Then we have the shapes as follows
$X \in (N \times D)$
$W_1 \in (D \times M)$
$W_2 \in (M \times K)$
$Y \in (N \times K)$
So to predict the output we use
feedforword method, which is simply a bottom-top approach as just as with perceptron. Therefore,
$ Z_1 = X \cdot W_1$
$ a_1 = g(Z_1)$
$Z_2 = a_1 \cdot W_2$
$a_2 = g(Z_2)$
where $g$ is the
activation function. Normally sigmoid in majority of the cases. i.e,
$$ g(z) = \frac{1}{1+e^{-z}}$$
And that’s it the output we required is above. It is just the $ a_2$.
How do we learn the weights?
I’m glad you asked. But this is not as easy as we did with the perceptron, but remember that it is the same way we did with the perceptron. All the problem arises is that we have multiple layers now. So we need to implement a method called Backpropagation which made the neural network training a lot more easier.
Backpropagation?
It is the way of calculating
gradients and to learn the weights. Assume that we have our cost function $J$ which is to be minimized. Also we need that cost function to be minimized with respect to $W_1$ and $W_2$. Since we have a layered approach it becomes difficult to handle things at once. So we apply backpropagation which simply follows the chain rule i.e, . “Never stop until you reach the target”
Therefore, given that $a_2$ as output and $Y$ to be true labels, the cost function is defined as follows.
$$J = \frac{1}{2N}\sum (a_2-Y)^2$$
and we need to find $W_1$ and $W_2$ which minimizes the cost $J$, i.e $ \frac{\partial J}{\partial W_1}$ and $ \frac{\partial J}{\partial W_2}$.
We need to apply
chain rule to find the gradients.
$$ \frac{\partial}{\partial W_2}J = \frac{\partial J}{\partial a_2} \frac{\partial a_2}{\partial Z_2} \frac{\partial Z_2}{\partial W_2}$$
Simplifying,
$$\frac{\partial J}{\partial a_2} = \frac{\partial}{\partial a_2} \frac{1}{2N} \sum (a_2-Y)^2 = (a_2 – Y)$$
$$ \frac{\partial a_2}{\partial Z_2} = \frac{\partial }{\partial Z_2}g(Z_2) = \bar g(Z_2)$$
in case of
sigmoid,
$$ \bar g(Z_2) = a_2(1- a_2)$$
and,
$$ \frac{\partial Z_2}{\partial W_2} = \frac{\partial}{\partial W_2} (a_1 \cdot W_2) = a_1$$
combining all these,
$$ \frac{\partial}{\partial W_2} J ={a_1}^T \cdot (a_2 -Y)a_2(1-a_2)$$
which can be rewritten as,
$$\frac{\partial}{\partial W_2} J = {a_1}^T \cdot \delta_2 $$
where $ \delta_2 = (a_2 -Y)a_2(1-a_2)$
The same can be achieved for $W_1$ gradient using the
chain rule. Therfore,
$$ \frac{\partial}{\partial W_1}J = \frac{\partial J}{\partial a_2} \frac{\partial a_2}{\partial Z_2} \frac{\partial Z_2}{\partial a_1}\frac{\partial a_1}{\partial Z_1}\frac{\partial Z_1}{\partial W_1}$$
We can observe some common terms in both $\frac{\partial J}{W_2}$ and $\frac{\partial J}{W_1}$ i.e, $\frac{\partial J}{\partial a_2} \frac{\partial a_2}{\partial Z_2}$. Excluding them we can derive the others as follows.
$\frac{\partial Z_2}{\partial a_1} = W_2$
$\frac{\partial a_1}{\partial Z_1} = \bar g(Z_1) = a_1(1-a_1)$
$\frac{\partial Z_1}{\partial W_1} = X$
So finally we can combine all them as shown below.
$$ \frac{\partial}{\partial W_1}J = X^T \cdot [[a_1(1-a_1)] [(a_2-Y)a_2(1-a_2)] \cdot W^T]$$
The terms are ordered in such a way to preserve the same dimensions as $W_1$.
And the above equation can be rewritten as,
$$\frac{\partial}{\partial W_1}J = X^T \cdot \delta_1$$
where, $\delta_1 = (\delta_2 \cdot W^T) \bar g(Z_1)$
That’s all the backpropagation, it is as simple as that. All you need to remember is that
. “Never stop until you reach the target” How to update weights?
Oh! It’s the same way we had done before. We simply subtract the gradients from the weights, as simple as that. Therefore,
$W_m = W_m – \alpha \frac{\partial J}{\partial W_m}$
where $\alpha$ is the
learning rate. Is that all?
Unfortunately no. There is even more to discuss and I’m sorry for the very long post. But as I said earlier this post is simply a part of that what we intended to learn. So hold on, take a break, relax and come back.
Yes! We need to learn what are the ways to make our neural network learn (training)? I know the one we have implemented earlier is only a one among them. So what are they?
Gradient Descent (one we have seen earlier) Stochastic Gradient Descent (SGD) Batch Gradient Descent Gradient Descent?
So gradient descent is just what we have seen with the perceptron here the training happens as described below.
for each_iteration in total_iterations{ feedforward the input (all samples) find the gradients update the weights }
So that means we update the weights only after processing all the inputs and we repeat this process up to certain stop criterion is met, here in this case the
no.of.iterations. It is pretty slow as for each update it has to process all the inputs and takes more time to converge.
Stochastic Gradient Descent?
Well! It is nothing but a simple gradient descent but the only change being the we update our weights after processing
each sample rather than all the samples. Therfore, randomly shuffle the samples for each_sample in total_samples{ feedforward the input find the gradients update the weights }
Remember to
shuffle the samples randomly as it may help in converging the cost function faster. Unlike Gradient Descent the cost here is not constantly decreases rather it fluctuates rapidly but note that it is much more faster than the simple gradient descent and it allows you for the online learning.
As you can see from the above image that the gradient descent converges more
smoothly than stochastic gradient descent. Most of the times we use stochastic gradient descent for a
no.of.iterations so that it would converge to global minimum rather that fluctuating at the local minimum. Though the
SGD is fast it may not converge to global minimum sometimes but with gradient descent it is guaranteed after several iterations. Batch Gradient Descent?
Okay! It is a
hybrid model for Gradient Descent and Stochastic Gradient Descent. All we do here is we split the samples into batches and train them with Gradient Descent. So it follows gradient descent while splitting the samples without replacement into batches tends to be using stochastic gradient descent. Therfore, randomly shuffle the samples split the samples into batches for each_batch in total_batches{ feedforward the batchu find the gradients update weights }
While the above shown process can be fair enough to converge to global minimum as fast as possible. It is recommended to repeat the above through certain
no.of.iterations for more accurate results.
The above image is enough to explain the difference between them gradient descent and stochastic gradient descent. Though it shows that SGD has not converged but it may converge after several iterations. Also note that the SGD converged so fast at beginning and begin to fluctuate a lot so to make it more robust batch gradient descent is preferable.
Anything more?
Yeah ! I will introduce one more term called Softmax.
Softmax is just like our sigmoid function but it is applied at the output layer to extract the probability of outcomes in classification problems. Though we have sigmoid function it doesn’t gives us the probabilistic approach and hence Softmax is always preferable when our tasks are associated with classification with more than
2 classes involved. So softmax can be defined as below.
$$Softmax(z)_j = \frac{e^{z_j}}{\sum_{k=1}^{K}e^{z_k}}$$
On observing the above equation, we can deduce that $\sum_{k=1}^{K}Softmax(z_k) = 1$. So which is the
rule of thumb in probability theory.
And also the need for discussing Cross Entropy is a must.
Cross Entropy?
Don’t get scared by the name. It’s just an another way of computing
cost. So far we have been using Least Squares Error method to find the cost. But that would not be good when dealing with the classification problems. So the cross entropy is introduced. It is nothing but the logistic loss. Therefore,
$$J = -\frac{1}{N}\sum_{n=1}^{N}[y_n log (\bar y_n) + (1-y_n)log (1-\bar y_n)]$$
where $y$ and $\bar y$ are the
true and predicted outputs.
Remember that we don’t need to worry about the $\frac{\partial J}{\partial W_k}$ since the
derivative of the cross entropy yields the same result as before. But it is worth noting that using cross entropy in classification problems is a good idea. Okay, drive me to code!!
Not so fast, atleast i’m
tired. We will plug the concepts we have learned in this post to build our own neural network from scratch and then we explore which is a good library to build neural networks more easily. And in the next posts we shall see how to apply the learned knowledge to Keras recognizethe hand written digits and later extend to facial expression recognitionboth in images and live streaming video.
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Real Analysis Exchange Real Anal. Exchange Volume 34, Number 2 (2008), 501-520. Divergence in Measure of Rearranged Multiple Orthononal Fourier Series Abstract
Let $\{\varphi_n(x)$, $n=1,2,\dots\}$ be an arbitrary complete orthonormal system (ONS) on the interval $I:=[0,1)$ that consists of a.e. bounded functions. Then there exists a rearrangement $\{ \varphi_{\sigma_1(n)}$, $n=1,2, \dots\}$ of the system $\{\varphi_n(x)$, $n=1,2,\dots\}$ that has the following property: for arbitrary nonnegative, continuous and nondecreasing on $[0,\infty)$ function $\phi(u)$ such that $u\phi (u)$ is a convex function on $[0,\infty)$ and $\phi (u) = o(\ln u)$, $u \to \infty$, there exists a function $f \in L(I^2)$ such that $\int_{I^2} | f(x,y) |$ $\phi( | f(x,y) | )\;dx\; dy \infty$ and the sequence of the square partial sums of the Fourier series of $f$ with respect to the double system $\{ \varphi_{\sigma_1 (m)}(x)\varphi_{\sigma_1 (n)}(y)$, $m,n \in\N \}$ on $I^2$ is essentially unbounded in measure on $I^2$.
Article information Source Real Anal. Exchange, Volume 34, Number 2 (2008), 501-520. Dates First available in Project Euclid: 29 October 2009 Permanent link to this document https://projecteuclid.org/euclid.rae/1256835201 Mathematical Reviews number (MathSciNet) MR2569201 Citation
Getsadze, Rostom. Divergence in Measure of Rearranged Multiple Orthononal Fourier Series. Real Anal. Exchange 34 (2008), no. 2, 501--520. https://projecteuclid.org/euclid.rae/1256835201
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