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# Similarity of two discrete fourier tranforms?

In climate modelling, you’re looking for a models that can adequately portray the Earth’s climate. This includes showing patterns that are semi-cyclical: things like the El Nino Southern Oscillation. But model verification occurs generally over relatively short time periods, where there is decent observational data (last ~150 years). This means that your model could be displaying the right patterns, but be out of phase, such that linear comparisons, like correlation, will not pick up that the model is performing well..

Discrete Fourier transforms are commonly used to analyse climate data (here’s an example), in order to pick up such cyclic patterns. Is there any standard measure of the similarity of two DFTs, that could be used as a verification tool (ie. a comparison between the DFT for the model, and the one for the observations)?

Would it make sense to take the integral of the minimum of the two area-normalised DFTs (using absolute real values)? I think this would result in a score $x\in[0,1]$, where $x=1\implies$ exactly the same patterns, and $x=0\implies$totally different patterns. What might the drawbacks of such a method be?

Spectral coherence, if used correctly would do it. Coherence is computed at each frequency-and hence is a vector. Hence, a sum of a weighted coherence would be a good measure. You would typically want to weight the coherences at frequencies that have a high energy in the power spectral density. That way, you would be measuring the similarities at the frequencies that dominate the time series instead of weighting the coherence with a large weight, when the content of that frequency in the time series is negligible.

So, in simple words- the basic idea is to find the frequencies at which the amplitude(energy) in the signals are high(interpret as the frequencies that dominantly constitute each signal) and then to compare the similarities at these frequencies with a higher weight and compare the signals at the rest of the frequencies with a lower weight.

The area which deals with questions of this kind is called cross-spectral analysis.
http://www.atmos.washington.edu/~dennis/552_Notes_6c.pdf is an excellent introduction to cross-spectral analysis.

Optimal Lag:
Also look at my answer over here: How to correlate two time series, with possible time differences

This deals finding the optimal lag, using the spectral coherence. R has functions to compute the power spectral densities, auto and cross correlations, Fourier transforms and coherence. You have to right code to find the optimal lag to obtain the max. weighted coherence. That said, a code for weighting the coherence vector using the spectral density must also be written. Following which you can sum up the weighted elements and average it to get the similarity observed at the optimal lag.

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# Definition:Riemann Integral/Historical Note

## Historical Note on Riemann Integral

Consider the Riemann sum:

$\displaystyle \sum_{i \mathop = 1}^n \ f \left({c_i}\right) \ \Delta x_i$

Historically, the definite integral was an extension of this type of summation such that:

The finite difference $\Delta x$ is instead the infinitely small difference $\rd x$
The finite sum $\Sigma$ is instead the sum of an infinite number of infinitely small quantities: $\int$

Hence the similarity in notation:

 $\displaystyle \sum_a^b \ f \left({x}\right) \, \Delta x$ $\to$ $\displaystyle \int_a^b f \left({x}\right) \rd x$ as $\Delta x \to \rd x$

The notion of "infinitely small" does not exist in the modern formulation of real numbers. Nevertheless, this idea is sometimes used as an informal interpretation of the definite integral.

Riemann established this definition in his paper Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe of $1854$, on the subject of Fourier series.


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# [Mathematica-vs-R] Handwritten digits recognition by matrix factorization

GROUPS:

## Introduction

This MathematicaVsR project is for comparing Mathematica and R for the tasks of classifier creation, execution, and evaluation using the MNIST database of images of handwritten digits.

Here are the bases built with two different classifiers:

• Singular Value Decomposition (SVD)

• Non-Negative Matrix Factorization (NNMF)

Here are the confusion matrices of the two classifiers:

• SVD

• NNMF

The blog post "Classification of handwritten digits" (published 2013) has a related more elaborated discussion over a much smaller database of handwritten digits.

## Concrete steps

The concrete steps taken in scripts and documents of this project follow.

1. Ingest the binary data files into arrays that can be visualized as digit images.
• We have two sets: 60,000 training images and 10,000 testing images.
1. Make a linear vector space representation of the images by simple unfolding.

2. For each digit find the corresponding representation matrix and factorize it.

3. Store the matrix factorization results in a suitable data structure. (These results comprise the classifier training.)

• One of the matrix factors is seen as a new basis.
1. For a given test image (and its linear vector space representation) find the basis that approximates it best. The corresponding digit is the classifier prediction for the given test image.

2. Evaluate the classifier(s) over all test images and compute accuracy, F-Scores, and other measures.

## Scripts

There are scripts going through the steps listed above:

## Documents

The following documents give expositions that are suitable for reading and following of steps and corresponding results.

## Observations

### Ingestion

I figured out first in R how to ingest the data in the binary files of the MNIST database. There were at least several online resources (blog posts, GitHub repositories) that discuss the MNIST binary files ingestion.

After that making the corresponding code in Mathematica was easy.

### Classification results

Same in Mathematica and R for for SVD and NNMF. (As expected.)

### NNMF

NNMF classifiers use the MathematicaForPrediction at GitHub implementations: NonNegativeMatrixFactorization.m and NonNegativeMatrixFactorization.R.

### Parallel computations

Both Mathematica and R have relatively simple set-up of parallel computations.

### Graphics

It was not very straightforward to come up in R with visualizations for MNIST images. The Mathematica visualization is much more flexible when it comes to plot labeling.

## Going further

### Comparison with other classifiers

Using Mathematica's built-in classifiers it was easy to compare the SVD and NNMF classifiers with neural network ones and others. (The SVD and NNMF are much faster to built and they bring comparable precision.)

It would be nice to repeat that in R using one or several of the neural network classifiers provided by Google, Microsoft, H2O, Baidu, etc.

### Classifier ensembles

Another possible extension is to use classifier ensembles and Receiver Operation Characteristic (ROC) to create better classifiers. (Both in Mathematica and R.)

### Importance of variables

Using classifier agnostic importance of variables procedure we can figure out :

• which NNMF basis vectors (images) are most important for the classification precision,

• which image rows or columns are most important for each digit, or similarly

• which image squares of a, say, 4x4 image grid are most important.

1 year ago
4 Replies
 Daniel Lichtblau 1 Vote Anton,This is very nice. One remark is that I've had some success using the Fourier Sine or Cosine transform (I'll use FCT in referring to this below) as a preprocessor. The idea is to use it on the image matrix, then extract a submatrix of say the lowest 10x10 components and use only that in the SVD phase. This offers three advantages.(1) It improves speed. The FCT is fast enough that it is not a bottleneck, and by reducing size of the input to SVD that becomes much faster.(2) It cuts down the memory footprint. FCT does not really take much memory since it can be done one at a time. SVD is done on the entire training ensemble and on a large matrix it will thus require more memory relative to a smaller one.(3) The extra level of dimension reduction via FCT improves the outcome. Depending on fine tuning of parameter settings (how many FCT components, how many singular values, how many lookup values to use in a majority rules classifier) it can hit 97.8% correct recognition. Also if we look for the 5 nearest neighbors, my recollection is that nearly always at least one will be the correct digit.Details of all this will be in the SYNASC 2016 proceedings (final version was submitted yesterday, and I think it comes out in January or so). The code for the MNIST example is not very long so I'll include it below. Since the images are coarse (28x28) I also sharpen them. That turns out to be the biggest speed bottleneck actually (takes several minutes whereas the rest totals less than a minute). Also I use the fourth Fourier Sine transform. I can get similar results with the second Cosine transform, if memory serves correctly. keep = 31; dn = 10; dst = 4; Clear[nearestImages, processInput]; nearestImages[ilist_, vals_, dn_, dnum_, keep_] := Module[ {idata, images = ilist, dcts, top, topvecs, uu, ww, vv, udotw, norms}, idata = Map[ImageData, images]; dcts = Map[FourierDST[# - Mean[Flatten[#]], dnum] &, idata]; top = dcts[[All, 1 ;; dn, 1 ;; dn]]; topvecs = Map[Flatten, top]; topvecs = Map[# - Mean[#] &, topvecs]; {uu, ww, vv} = SingularValueDecomposition[topvecs, keep]; udotw = uu.ww; norms = Map[Sqrt[#.#] &, udotw]; udotw = udotw/norms; udotw = Join[udotw, Transpose[{Log[norms]}], 2]; {Nearest[udotw -> vals, Method -> "KDTree"], vv}] processInput[ilist_, vv_, dn_, dnum_] := Module[ {idata, images = ilist, dcts, top, topvecs, tdotv, norms}, idata = Map[ImageData, images]; dcts = Map[FourierDST[# - Mean[Flatten[#]], dnum] &, idata]; top = dcts[[All, 1 ;; dn, 1 ;; dn]]; topvecs = Map[Flatten, top]; topvecs = Map[# - Mean[#] &, topvecs]; tdotv = topvecs.vv; norms = Map[Sqrt[#.#] &, tdotv]; tdotv = tdotv/norms; tdotv = Join[tdotv, Transpose[{Log[norms]}], 2]; tdotv] trainingBytes = Import[ "http://yann.lecun.com/exdb/mnist/train-images-idx3-ubyte.gz" , "Byte"]; trainingImages = Map[Image[Partition[#, 28]] &, Partition[Drop[trainingBytes, 16], 28^2]]; testBytes = Import[ "http://yann.lecun.com/exdb/mnist/t10k-images-idx3-ubyte.gz" , "Byte"]; testImages = Map[Image[Partition[#, 28]] &, Partition[Drop[testBytes, 16], 28^2]]; trainingLabels = Drop[Import[ "http://yann.lecun.com/exdb/mnist/train-labels-idx1-ubyte.gz", "Byte"], 8]; testLabels = Drop[Import[ "http://yann.lecun.com/exdb/mnist/t10k-labels-idx1-ubyte.gz", "Byte"], 8]; func = Sharpen[#, 12] &; trainingImages = Map[func, trainingImages]; testImages = Map[func, testImages]; guesses[nf_, tvecs_, n_] := Module[ {nbrs, counts}, nbrs = Map[nf[#, n] &, tvecs]; counts = Map[Tally, nbrs]; counts = Map[Reverse, Map[SortBy[#[[2]] &], counts]]; counts[[All, 1, 1]]] correct[guess_, actual_] /; Length[guess] == Length[actual] := Count[guess - actual, 0] correct[__] := \$Failed AbsoluteTiming[{nf, vv} = nearestImages[trainingImages, trainingLabels, dn, dst, keep]; testvecs = processInput[testImages, vv, dn, dst];] (* Out[121]= {4.248808, Null} *) AbsoluteTiming[ guessed = guesses[nf, testvecs, 5];] correct[guessed, testLabels] (* Out[122]= {13.071579, Null} Out[123]= 9781 *)
 Thank you for your code, Dan! And sorry for my delayed reply -- I was too busy analyzing Trump tweets. Some comments follow.1. I used NNs based image classifiers for couple of my consultancy projects. In my opinion, the use of NNs methods is underestimated in image classification.2. Using the function CrossTabulate I visualized the confusion matrices for several experiments using different values for keep. (The variable keep is not defined in your code.) Pretty good overall accuracy is achieved with keep = 40 and reasonably fast.3. Instead of using Sharpen I was thinking to do the opposite -- use Gaussian blur.4. It might be a good idea to look in ROC application to your approach. (I might do that later this week.)Here is experimental code for over a set of keep values: Table[( trainingTime = AbsoluteTiming[ {nf, vv} = nearestImages[trainingImages, trainingLabels, dn, dst, keep]; testvecs = processInput[testImages, vv, dn, dst]; ][[1]]; classificationTime = AbsoluteTiming[guessed = guesses[nf, testvecs, 5];][[1]]; ColumnForm[{ Row[{"keep = ", keep}], Row[{"Training time, (s): ", trainingTime}], Row[{"Classification time, (s): ", classificationTime}], Row[{"Overall accuracy: ", correct[guessed, testLabels]/Length[testLabels] // N}], MatrixForm@CrossTabulate[Transpose[{guessed, testLabels}]] }] ), {keep, {5, 10, 15, 20, 30, 40, 60, 100}}] And here is the output:Note that increasing keep after 30 does not produce significantly better results.
 Daniel Lichtblau 1 Vote Anton,I had keep=31;. Once upon a time I had also tried different preprocessing, including blurring. My best result came with sharpening but I cannot say I tried all variations at the same settings and maybe blurring is better for the MNIST suite. It might also be the case that one should use regular, sharpened, and blurred versions in the training stage.Your nice charts are giving me a case of "output envy".


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# Representing the same game with different extensive form representations

I have a question related to how to represent a game in extensive form.

Consider a game that starts with nature choosing one of two states, with equal probability, with the outcome only known to player 1 (but player 2 has a prior on the state of nature given by $$q$$). Both players 1 and 2 then act simultaneously.

We know that a simultaneous move game can be represented in extensive form by including an information set for one of the players that indicates they do not know the action of the other player. This is represented by the following extensive form.

Since the moves of the players are simultaneous, we can exchange the order in the tree. So, an equivalent representation is as follows.

How do we denote player 2's prior information in the above? It seems unfair to treat player 2's uncertainty as a single information set since it's due to two sources of uncertainty: one because of nature's move, and the other just because the players are moving simultaneously. It seems that we should have some notion of nested information sets for player 2 to make this distinction, as follows.

Where player 2's belief in its information set is structured in the sense that the first two nodes must have probability $$q$$ and the last two assigned probability $$1-q$$; player 2 cannot distinguish between the nodes within the two pairs of nodes, represented by the dotted lines.

Question: Is this correct? How does one usually represent this structure in extensive form games?

• Does player 1 know player 2's prior is parameterized by $q$ instead of the actual $\frac12$? – Herr K. Aug 23 '19 at 22:41
• Hi @HerrK.! I've updated the post such that the player 2's belief lines up with nature's probability. – jonem Aug 24 '19 at 2:23

There is no need to separately mark the left-two and right-two nodes in P2's information set. Bayes' rule guarantees that the first two nodes will be reached with probability $$q$$ and the last two nodes with $$1-q$$.
Based on your second game tree, suppose P1 plays $$A$$ with probability $$p$$ at his left info set. Then the probability that the play reaches either of the first two nodes at P2's info set is $$$$qp+q(1-p)=q.$$$$


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## Physics: Principles with Applications (7th Edition)

The pressure difference is due to the depth of water. Use equation 10–3b, and round the answer to 2 significant figures. $$\Delta P=\rho g \Delta h$$ $$\Delta h= \frac{\Delta P}{\rho g}$$ $$=\frac{(85 mm-Hg)((133 N/m^2)/(1mm-Hg))}{(1.00 \times 10^3 kg/m^3)(9.80 m/s^2)} \approx 1.2m$$ His maximum depth would be about 1.2 m.


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# Differential branching fraction and angular analysis of $\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-$ decays

[to restricted-access page]

## Abstract

The differential branching fraction of the rare decay $\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-$ is measured as a function of $q^{2}$, the square of the dimuon invariant mass. The analysis is performed using proton-proton collision data, corresponding to an integrated luminosity of $3.0 fb^{-1}$, collected by the LHCb experiment. Evidence of signal is observed in the $q^2$ region below the square of the $J/\psi$ mass. Integrating over $15 < q^{2} < 20 GeV^2/c^4$ the branching fraction is measured as $d\mathcal{B}(\Lambda^{0}_{b} \rightarrow \Lambda \mu^+\mu^-)/dq^2 = (1.18 ^{+ 0.09} _{-0.08} \pm 0.03 \pm 0.27) \times 10^{-7} ( GeV^{2}/c^{4})^{-1}$, where the uncertainties are statistical, systematic and due to the normalisation mode, $\Lambda^{0}_{b} \rightarrow J/\psi \Lambda$, respectively. In the $q^2$ intervals where the signal is observed, angular distributions are studied and the forward-backward asymmetries in the dimuon ($A^{l}_{\rm FB}$) and hadron ($A^{h}_{\rm FB}$) systems are measured for the first time. In the range $15 < q^2 < 20 GeV^2/c^4$ they are found to be $A^{l}_{\rm FB} = -0.05 \pm 0.09 (stat) \pm 0.03 (syst)$ and $A^{h}_{\rm FB} = -0.29 \pm 0.07 (stat) \pm 0.03 (syst)$.

## Figures and captions

 Invariant mass distribution of the $\Lambda ^0_ b \rightarrow { J \mskip -3mu/\mskip -2mu\psi \mskip 2mu} \Lambda$ candidates selected with the neural network requirement used for the high- $q^2$ region. The (black) points show data, combining downstream and long candidates, and the solid (blue) line represents the overall fit function. The dotted (red) line represents the combinatorial and the dash-dotted (brown) line the peaking background from $B ^0 \rightarrow { J \mskip -3mu/\mskip -2mu\psi \mskip 2mu} K ^0_{\rm\scriptscriptstyle S}$ decays. figure1.pdf [22 KiB] HiDef png [207 KiB] Thumbnail [157 KiB] *.C file Invariant mass distribution of the $\Lambda ^0_ b \rightarrow \Lambda \mu ^+\mu ^-$ candidates, integrated over the region $15.0 < q^2 < 20.0$ ${\mathrm{ Ge V^2 /}c^4}$ together with the fit function described in the text. The points show data, the solid (blue) line is the overall fit function and the dotted (red) line represents the combinatorial background. The background component from $B ^0 \rightarrow K ^0_{\rm\scriptscriptstyle S} \mu ^+\mu ^-$ decays, (brown) dashed line, is barely visibile due to the very low yield. figure2.pdf [16 KiB] HiDef png [211 KiB] Thumbnail [175 KiB] *.C file Invariant mass distributions of $\Lambda ^0_ b \rightarrow \Lambda \mu ^+\mu ^-$ candidates, in eight $q^2$ intervals, together with the fit function described in the text. The points show data, the solid (blue) line is the overall fit function and the dotted (red) line represents the combinatorial background component. figure3.pdf [43 KiB] HiDef png [847 KiB] Thumbnail [711 KiB] *.C file Total relative efficiency, $\varepsilon_{\mathrm{rel}}$, between $\Lambda ^0_ b \rightarrow \Lambda \mu ^+\mu ^-$ and $\Lambda ^0_ b \rightarrow { J \mskip -3mu/\mskip -2mu\psi \mskip 2mu} \Lambda$ decays. The uncertainties are the combination of both statistical and systematic components, and are dominated by the latter. figure4.pdf [15 KiB] HiDef png [127 KiB] Thumbnail [147 KiB] *.C file Measured $\Lambda ^0_ b \rightarrow \Lambda \mu ^+\mu ^-$ branching fraction as a function of $q^2$ with the predictions of the SM [15] superimposed. The inner error bars on data points represent the total uncertainty on the relative branching fraction (statistical and systematic); the outer error bar also includes the uncertainties from the branching fraction of the normalisation mode. figure5.pdf [20 KiB] HiDef png [126 KiB] Thumbnail [123 KiB] *.C file Angular efficiencies as a function of (left) $\cos\theta_\ell$ and (right) $\cos\theta_h$ for (upper) long and (lower) downstream candidates, in the interval $15 < q^2 < 20$ ${\mathrm{ Ge V^2 /}c^4}$ , obtained using simulated events. The (blue) line shows the fit that is used to model the angular acceptance in the fit to data. figure6a.pdf [16 KiB] HiDef png [130 KiB] Thumbnail [123 KiB] *.C file figure6b.pdf [16 KiB] HiDef png [139 KiB] Thumbnail [137 KiB] *.C file figure6c.pdf [16 KiB] HiDef png [139 KiB] Thumbnail [140 KiB] *.C file figure6d.pdf [17 KiB] HiDef png [156 KiB] Thumbnail [152 KiB] *.C file Angular distributions as a function of (left) $\cos \theta_\ell$ and (right) $\cos \theta_h$, for candidates in the integrated $15 < q^2 <20$ ${\mathrm{ Ge V^2 /}c^4}$ interval with the overall fit function overlaid (solid blue). The (red) dotted line represents the combinatorial background. figure7a.pdf [14 KiB] HiDef png [143 KiB] Thumbnail [119 KiB] *.C file figure7b.pdf [14 KiB] HiDef png [146 KiB] Thumbnail [133 KiB] *.C file Measured values of (left) the leptonic and (right) the hadronic forward-backward asymmetries in bins of $q^2$ . Data points are only shown for $q^2$ intervals where a statistically significant signal yield is found, see text for details. The (red) triangle represents the values for the $15 < q^2 < 20$ ${\mathrm{ Ge V^2 /}c^4}$ interval. Standard Model predictions are obtained from Ref. [17]. figure8a.pdf [15 KiB] HiDef png [114 KiB] Thumbnail [127 KiB] *.C file figure8b.pdf [15 KiB] HiDef png [118 KiB] Thumbnail [134 KiB] *.C file Two-dimensional 68 % CL region (black) as a function of $A_{\rm FB}^\ell$ and $f_{\rm L}$. The shaded area represents the region where the PDF is positive over the complete $\cos \theta_{\ell}$ range. The best fit point is given by the (blue) star. figure9.pdf [13 KiB] HiDef png [99 KiB] Thumbnail [93 KiB] *.C file Two-dimensional 68 % CL regions (black) as a function of $A_{\rm FB}^\ell$ and $f_{\rm L}$. The shaded areas represent the regions in which the PDF is positive over the complete $\cos \theta_{\ell}$ range. The best fit points are indicated by the (blue) stars. figure10a.pdf [14 KiB] HiDef png [114 KiB] Thumbnail [106 KiB] *.C file figure10b.pdf [14 KiB] HiDef png [117 KiB] Thumbnail [111 KiB] *.C file figure10c.pdf [14 KiB] HiDef png [113 KiB] Thumbnail [106 KiB] *.C file figure10d.pdf [13 KiB] HiDef png [102 KiB] Thumbnail [96 KiB] *.C file figure10e.pdf [13 KiB] HiDef png [107 KiB] Thumbnail [102 KiB] *.C file Animated gif made out of all figures. PAPER-2015-009.gif Thumbnail

## Tables and captions

 Number of $\Lambda ^0_ b \rightarrow { J \mskip -3mu/\mskip -2mu\psi \mskip 2mu} \Lambda$ decays in the long and downstream categories found using the selection for low- and high- $q^2$ regions. Uncertainties shown are statistical only. Table_1.pdf [37 KiB] HiDef png [59 KiB] Thumbnail [27 KiB] tex code Signal decay yields ($N_\mathrm{S}$) obtained from the mass fit to $\Lambda ^0_ b \rightarrow \Lambda \mu ^+\mu ^-$ candidates in each $q^2$ interval together with their statistical significances. The yields are the sum of long and downstream categories with downstream decays comprising $\sim 80 \%$ of the total yield. The $8-11$ and $12.5-15$ ${\mathrm{ Ge V^2 /}c^4}$ $q^2$ intervals are excluded from the study as they are dominated by decays via charmonium resonances. Table_2.pdf [38 KiB] HiDef png [88 KiB] Thumbnail [47 KiB] tex code Systematic uncertainties as a function of $q^2$ , assigned for yields and efficiencies. Values reported are the sums in quadrature of all contributions evaluated within each category. Table_3.pdf [46 KiB] HiDef png [76 KiB] Thumbnail [40 KiB] tex code Measured differential branching fraction of $\Lambda ^0_ b \rightarrow \Lambda \mu ^+\mu ^-$ , where the uncertainties are statistical, systematic and due to the uncertainty on the normalisation mode, $\Lambda ^0_ b \rightarrow { J \mskip -3mu/\mskip -2mu\psi \mskip 2mu} \Lambda$ , respectively. Table_4.pdf [55 KiB] HiDef png [102 KiB] Thumbnail [53 KiB] tex code Measured values of leptonic and hadronic angular observables, where the first uncertainties are statistical and the second systematic. Table_5.pdf [53 KiB] HiDef png [76 KiB] Thumbnail [40 KiB] tex code Differential branching fraction of the $\Lambda ^0_ b \rightarrow \Lambda \mu ^+\mu ^-$ decay relative to $\Lambda ^0_ b \rightarrow { J \mskip -3mu/\mskip -2mu\psi \mskip 2mu} \Lambda$ decays, where the uncertainties are statistical and systematic, respectively. Table_6.pdf [68 KiB] HiDef png [101 KiB] Thumbnail [51 KiB] tex code

## Supplementary Material [file]

 figure11.pdf [22 KiB] HiDef png [227 KiB] Thumbnail [132 KiB] *C file figure12.pdf [19 KiB] HiDef png [93 KiB] Thumbnail [103 KiB] *C file figure13.pdf [15 KiB] HiDef png [180 KiB] Thumbnail [163 KiB] *C file figure14.pdf [14 KiB] HiDef png [91 KiB] Thumbnail [102 KiB] *C file

Created on 24 August 2019.Citation count from INSPIRE on 24 August 2019.


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Documentation

## Troubleshooting Multivariate Normal Regression

This section provides a few pointers to handle various technical and operational difficulties that might occur.

### Biased Estimates

If samples are ignored, the number of samples used in the estimation is less than `NumSamples`. Clearly the actual number of samples used must be sufficient to obtain estimates. In addition, although the model parameters `Parameters` (or mean estimates `Mean`) are unbiased maximum likelihood estimates, the residual covariance estimate `Covariance` is biased. To convert to an unbiased covariance estimate, multiply `Covariance` by

`$\text{Count}/\left(\text{Count}-1\right),$`

where `Count` is the actual number of samples used in the estimation with `Count` ≤ `NumSamples`. None of the regression functions perform this adjustment.

### Requirements

The regression functions, particularly the estimation functions, have several requirements. First, they must have consistent values for `NumSamples`, `NumSeries`, and `NumParams`. As a rule, the multivariate normal regression functions require

and the least-squares regression functions require

`$\text{Count}×\text{NumSeries}\le \text{NumParams},$`

where `Count` is the actual number of samples used in the estimation with

`$\text{Count}\le \text{NumSamples}\text{.}$`

Second, they must have enough nonmissing values to converge. Third, they must have a nondegenerate covariance matrix.

Although some necessary and sufficient conditions can be found in the references, general conditions for existence and uniqueness of solutions in the missing-data case, do not exist. Nonconvergence is usually due to an ill-conditioned covariance matrix estimate, which is discussed in greater detail in Nonconvergence.

### Slow Convergence

Since worst-case convergence of the ECM algorithm is linear, it is possible to execute hundreds and even thousands of iterations before termination of the algorithm. If you are estimating with the ECM algorithm regularly with regular updates, you can use prior estimates as initial guesses for the next period's estimation. This approach often speeds up things since the default initialization in the regression functions sets the initial parameters b to zero and the initial covariance C to be the identity matrix.

Other improvised approaches are possible although most approaches are problem-dependent. In particular, for mean and covariance estimation, the estimation function `ecmnmle` uses a function `ecmninit` to obtain an initial estimate.

### Nonrandom Residuals

Simultaneous estimates for parameters b and covariances C require C to be positive-definite. So, the general multivariate normal regression routines require nondegenerate residual errors. If you are faced with a model that has exact results, the least-squares routine `ecmlsrmle` still works, although it provides a least-squares estimate with a singular residual covariance matrix. The other regression functions fail.

### Nonconvergence

Although the regression functions are robust and work for most “typical” cases, they can fail to converge. The main failure mode is an ill-conditioned covariance matrix, where failures are either soft or hard. A soft failure wanders endlessly toward a nearly singular covariance matrix and can be spotted if the algorithm fails to converge after about 100 iterations. If `MaxIterations` is increased to 500 and display mode is initiated (with no output arguments), a typical soft failure looks like this.

This case, which is based on 20 observations of five assets with 30% of data missing, shows that the log-likelihood goes linearly to infinity as the likelihood function goes to 0. In this case, the function converges but the covariance matrix is effectively singular with a smallest eigenvalue on the order of machine precision (`eps`).

For the function `ecmnmle`, a hard error looks like this:

```> In ecmninit at 60 In ecmnmle at 140 ??? Error using ==> ecmnmle Full covariance not positive-definite in iteration 218. ```

From a practical standpoint, if in doubt, test your residual covariance matrix from the regression routines to ensure that it is positive-definite. This is important because a soft error has a matrix that appears to be positive-definite but actually has a near-zero-valued eigenvalue to within machine precision. To do this with a covariance estimate `Covariance`, use `cond(Covariance)`, where any value greater than `1/eps` should be considered suspect.

If either type of failure occurs, however, note that the regression routine is indicating that something is probably wrong with the data. (Even with no missing data, two time series that are proportional to one another produce a singular covariance matrix.)


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## A contribution towards a Distributed Java Virtual Machine

Authors

João F. Ferreira

Abstract

The work described in this report is part of PPC-VM project (Portable Parallel Computing based on Virtual Machines). The PPC-VM project aims to build an environment to support the development and execution of parallel applications that efficiently execute on a wide range of computing platforms, based on virtual machines. This work contributes to PPC-VM project with a parallel computing environment, which aims to simplify the implementation of parallel applications and to test the paradigms/methodologies developed within the PPC-VM project. The parallel computing environment uses the Java programming language and it provides two components: a skeleton catalog implemented as an abstract class library and an au- tomatic object distribution platform, based on source code generation. The former helps programmers creating parallel applications, while the latter transparently distributes objects in a parallel distributed environment.

Keywords

parallel computing, parallel programming, automatic object distribution, skeletons, Java

Bibtex entry
@techreport{jff*05:distjvm,
author = {Ferreira, Jo\~{a}o  F.}
title = {A contribution towards a Distributed Java Virtual Machine},


text:
# How do you express  (x^3 +x^2+2x+1) / [(x^2+1) (x^2+2)] in partial fractions?

May 26, 2017

The answer is $= \frac{x}{{x}^{2} + 1} + \frac{1}{{x}^{2} + 2}$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{{x}^{3} + {x}^{2} + 2 x + 1}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)} = \frac{A x + B}{{x}^{2} + 1} + \frac{C x + D}{{x}^{2} + 2}$

$= \frac{\left(A x + B\right) \left({x}^{2} + 1\right) + \left(C x + D\right) \left({x}^{2} + 1\right)}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)}$

The denominators are the same, we compare the numerators

${x}^{3} + {x}^{2} + 2 x + 1 = \left(A x + B\right) \left({x}^{2} + 2\right) + \left(C x + D\right) \left({x}^{2} + 1\right)$

Coeficients of ${x}^{3}$

$1 = A + C$

Let $x = 0$, $\implies$, $1 = 2 B + D$

Coefficients of ${x}^{2}$

$1 = B + D$

Coefficients of $x$

$2 = 2 A + C$

Solving for $A$, $B$, $C$ and $D$ from the 4 equations

$2 = 2 A + 1 - A$, $\implies$, $A = 1$

$C = 1 - A = 1 - 1 = 0$

$1 = 2 B + 1 - B$, $\implies$, $B = 0$

$1 = B + D$, $\implies$, $D = 1$

Therefore,

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


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# Quantum Zeno effect and unstable particles

Is it possible to increase indefinitely the lifetime of unstable particles by applying the quantum Zeno effect? Is there a bound from theoretical principles about the maximum extension one can get in a unstable energy level in general from continuous measurement alone?

-

Yes, it is possible, but this effect is hard to calculate precisely, and has no chance of experimental realization. You can see this happen in the Unruh effect, when you accelerate a particle fast enough, the lifetime increases, due to the constant measurement by the thermal bath (or in the rest frame, due to the constant particle emission).

Exponential decay which is interrupted by a measurement is still exponential. If you have a particle which is decaying according to Fermi's golden rule, the probability of finding decay products at time t is exponentially growing in t, with a rate proportional to the matrix element squared integrated over the final states. This means that if you are working on time-scales where Fermi's rule is applicable, measuring whether that particle has decayed or not will do nothing to the decay rate.

The way Fermi's golden rule is derived, it is from different states cancelling in amplitude, over a time-scale comparable to the inverse amount of energy violation in the state. So when $\Delta t \gg {1\over \Delta E}$, Fermi's rule applies, and you get exponential decay.

Quantum zeno effect happens when you muck up the states in intermediate times, before Fermi's golden rule has a chance to become valid. So the time between interactions must be comparable to the inverse energy difference, not to the decay time. These are very different when you have a small coupling, so it is easy to mislead yourself. For example, if you want to stabilize the neutron, you need to measure it on inverse MeV timescales, which is the time it takes to cross 1000 nuclear diameters, or once every $10^{-20}s$, not once every seven minutes. This rate of interaction is experimentally hopeless (let alone keeping it up for seven minutes).

Similarly, for a muon decay, you need to measure once every inverse 100 MeV's. Likewise for pion decay. For proton decay, same thing, so we aren't wrecking proton decay by zeno effect by putting it in a vat of water.

There aren't any pairs of unstable particles which are separated in energy by less than the eletron mass, so for realistic zeno effect experiments, you are limited to atomic or perhaps nuclear energy levels.

-
what if the interaction/measurement is continuous, why have the abitrary timeslicing? or you need to deliver a certain amount of measurement power (energy over time) in order for the effect to work? –  lurscher Aug 1 '12 at 7:10
@lurscher: You need one quantum to interact to carry information out. –  Ron Maimon Aug 1 '12 at 7:30
duh, of course! –  lurscher Aug 1 '12 at 7:32
"when you accelerate a particle fast enough, the lifetime increases, due to the constant measurement by the thermal bath" - do you presume that a particle on a planet's surface also experience a thermal bathe because it gets prolonged lifetime? What is the origin of such thermal bath? Hawking radiation? –  Anixx Dec 18 '12 at 1:52
"You need one quantum to interact to carry information out." - what is the box is a passive detector of the decay products? –  Anixx Dec 18 '12 at 2:01


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## De Brogile wavelengths

$\lambda=\frac{h}{p}$

Semi Yoon
Posts: 60
Joined: Fri Sep 28, 2018 12:27 am

### De Brogile wavelengths

How can we classify if a particle has any measurable wavelike properties? Why aren't we able to measure wavelike properties of a moving particle if it is too small?

MinaChoi_3H
Posts: 31
Joined: Fri Sep 28, 2018 12:25 am

### Re: De Brogile wavelengths

We can measure the wavelike properties of particles but not for everyday objects because the mass is too big. That is why the equation works best for small particles like electrons, because we will have to divide by the object's mass. In the example with the car we calculated it's wavelength to be 1.64 x 10^-38m. But that is an EXTREMELY small wavelength. This would require very advanced and sensitive technology to detect such a small wavelength, but to measure something like that is currently impossible. We can measure small wavelengths but anything 10^-38 is so minute and indistinguishable we are unable to detect it.


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• Substantial experimental and theoretical efforts worldwide are devoted to explore the phase diagram of strongly interacting matter. At LHC and top RHIC energies, QCD matter is studied at very high temperatures and nearly vanishing net-baryon densities. There is evidence that a Quark-Gluon-Plasma (QGP) was created at experiments at RHIC and LHC. The transition from the QGP back to the hadron gas is found to be a smooth cross over. For larger net-baryon densities and lower temperatures, it is expected that the QCD phase diagram exhibits a rich structure, such as a first-order phase transition between hadronic and partonic matter which terminates in a critical point, or exotic phases like quarkyonic matter. The discovery of these landmarks would be a breakthrough in our understanding of the strong interaction and is therefore in the focus of various high-energy heavy-ion research programs. The Compressed Baryonic Matter (CBM) experiment at FAIR will play a unique role in the exploration of the QCD phase diagram in the region of high net-baryon densities, because it is designed to run at unprecedented interaction rates. High-rate operation is the key prerequisite for high-precision measurements of multi-differential observables and of rare diagnostic probes which are sensitive to the dense phase of the nuclear fireball. The goal of the CBM experiment at SIS100 (sqrt(s_NN) = 2.7 - 4.9 GeV) is to discover fundamental properties of QCD matter: the phase structure at large baryon-chemical potentials (mu_B > 500 MeV), effects of chiral symmetry, and the equation-of-state at high density as it is expected to occur in the core of neutron stars. In this article, we review the motivation for and the physics programme of CBM, including activities before the start of data taking in 2022, in the context of the worldwide efforts to explore high-density QCD matter.
• ### Low energy magnetic excitations in the spin-orbital Mott insulator Sr$_2$IrO$_4$(1312.4847)

March 20, 2014 cond-mat.str-el
We report a high-field electron spin resonance study in the sub-THz frequency domain of a single crystal of Sr$_2$IrO$_4$ that has been recently proposed as a prototypical spin-orbital Mott insulator. In the antiferromagnetically (AFM) ordered state with noncollinear spin structure that occurs in this material at $T_{\rm N} \approx 240$ K we observe both the "low" frequency mode due to the precession of weak ferromagnetic moments arising from a spin canting, and the "high" frequency modes due to the precession of the AFM sublattices. Surprisingly, the energy gap for the AFM excitations appears to be very small, amounting to 0.83 meV only. This suggests a rather isotropic Heisenberg dynamics of interacting Ir$^{4+}$ effective spins despite the spin-orbital entanglement in the ground state.
• ### Influence of antisymmetric exchange interaction on quantum tunneling of magnetization in a dimeric molecular magnet Mn6(0810.0616)

We present magnetization measurements on the single molecule magnet Mn6, revealing various tunnel transitions inconsistent with a giant-spin description. We propose a dimeric model of the molecule with two coupled spins S=6, which involves crystal-field anisotropy, symmetric Heisenberg exchange interaction, and antisymmetric Dzyaloshinskii-Moriya exchange interaction. We show that this simplified model of the molecule explains the experimentally observed tunnel transitions and that the antisymmetric exchange interaction between the spins gives rise to tunneling processes between spin states belonging to different spin multiplets.
• ### Energy Level Lifetimes in the Single-Molecule Magnet Fe_8 : Experiments and Simulations(0712.0931)

Dec. 6, 2007 cond-mat.mes-hall
We present pump-probe measurements on the single-molecule magnet Fe_8 with microwave pulses having a length of several nanoseconds. The microwave radiation in the experiments is located in the frequency range between 104 GHz and 118 GHz. The dynamics of the magnetization of the single Fe_8 crystal is measured using micrometer-sized Hall sensors. This technique allows us to determine the level lifetimes of excited spin states, that are found to be in good agreement with theoretical calculations. The theory, to which we compare our experimental results, is based on a general spin-phonon coupling formalism, which involves spin transitions between nearest and next-nearest energy levels. We show that good agreement between theory and experiments is only obtained when using both the Delta m_S = +-1 transition as well as Delta m_S = +-2, where Delta m_S designates a change in the spin quantum number m_S. Temperature dependent studies of the level lifetimes of several spin states allow us finally to determine experimentally the spin-phonon coupling constants.
• ### Pump-Probe Experiments on the Single-Molecule Magnet Fe8 : Measurement of Excited Level Lifetimes(0708.3024)

Aug. 22, 2007 cond-mat.mes-hall
We present magnetization measurements on the single molecule magnet Fe8 in the presence of pulsed microwave radiation. A pump-probe technique is used with two microwave pulses with frequencies of 107 GHz and 118 GHz and pulse lengths of several nanoseconds to study the spin dynamics via time-resolved magnetization measurements using a Hall probe magnetometer. We find evidence for short spin-phonon relaxation times of the order of one microsecond. The temperature dependence of the spin-phonon relaxation time in our experiments is in good agreement with previously published theoretical results. We also established the presence of very short energy diffusion times, that act on a timescale of about 70 ns.
• ### Magnetization dynamics in the single-molecule magnet Fe8 under pulsed microwave irradiation(cond-mat/0701324)

Jan. 15, 2007 cond-mat.mes-hall
We present measurements on the single molecule magnet Fe8 in the presence of pulsed microwave radiation at 118 GHz. The spin dynamics is studied via time resolved magnetization experiments using a Hall probe magnetometer. We investigate the relaxation behavior of magnetization after the microwave pulse. The analysis of the experimental data is performed in terms of different contributions to the magnetization after-pulse relaxation. We find that the phonon bottleneck with a characteristic relaxation time of 10 to 100 ms strongly affects the magnetization dynamics. In addition, the spatial effect of spin diffusion is evidenced by using samples of different sizes and different ways of the sample's irradiation with microwaves.


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# I cannot seem to translate any of my models on the z axis using a model translation

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### #1cineclick  Members   -  Reputation: 133

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Posted 21 February 2013 - 08:23 AM

I had a question up on stackoverflow where a user and myself were able to dig through an issue, and determined it must be an issue with either how I am transposing my matrices, or multiplying them.

To summarize the problem I'm having, I can set a model at a fixed z position without any problem by setting the actual model origin z coordinate value to any non-zero value, and this works, however, if I try to create a model with with an origin of [0,0,0] and use a model matrix to transform it, I can transform the x and y coordinates with no problems, but if my z value is not between 0.0f and -0.9999f it just disappears.  pivotnig (the user that was helping me on stackoverflow) had suggested it may be a matrix multiplication order issue, so I tried reversing the order, and just experimenting with different orderings, which mostly just made my scene disappear.

To jump back an idea for a second, another suspicion was that I maybe shouldn't be transposing the matrices before sending them to the vertex shader.  The reason I am is because every example I have came across on msdn and some other places have always sent the matrices to the vertex shader after a XMMatrixTranspose().

Another thing that I haven't tried because I do not have a great understanding of it is using the camera's inverse matrix.  I noticed that some samples I looked at used it, but some did not.  I don't know if the issue resides in the camera matrix, however, because while running the application, I can move the camera without any problems, and the scene seems to response appropriately in terms of distance and projection.

I have no found any other person with a similar issue in DirectX 11, and any similar issues otherwise are with versions of DirectX where the matrices were managed internally.

For the sake of convenience, here are my code snippets that I'm possibly having trouble with - I can upload other snippets if anyone thinks the problem is elsewhere.

I'll admit that I'm not particularly skilled in 3d math, so if anyone has a very mathy answer, can you try and simplify it as much as possible for me please.

### #2GuardianX  Crossbones+   -  Reputation: 1486

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Posted 21 February 2013 - 09:21 AM

Are you sure about what do you want to do in this code:

void TransformMatrix::Rotate( float x, float y, float z, float ang )
{
if ( x * ang != 0.0f ) rotate *= XMMatrixRotationX( x * ang );
if ( y * ang != 0.0f ) rotate *= XMMatrixRotationY( y * ang );
if ( z * ang != 0.0f ) rotate *= XMMatrixRotationZ( z * ang );
}

Don't you want to rotate around vector with x, y, z coordinates actually? Because if so, rotating around each orthogonal axis for angle ang is not the same as rotating around specified vector for angle ang and you should use XMMatrixRotationAxis instead.

Also, I couldn't find any code, responsible for view and projection matrix creation.

To jump back an idea for a second, another suspicion was that I maybe shouldn't be transposing the matrices before sending them to the vertex shader.  The reason I am is because every example I have came across on msdn and some other places have always sent the matrices to the vertex shader after a XMMatrixTranspose().

You must use XMMatrixTranspose(), because matrices in shaders are transposed version of XNA SSE matrices by default.

Also make sure that you constant buffer structure has the same member's order as your shader constant buffer:

cbuffer ProjectionBuffer : register(b0)
{
matrix p;
matrix m;
matrix v;
};

Edited by GuardianX, 21 February 2013 - 09:21 AM.

### #3cineclick  Members   -  Reputation: 133

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Posted 21 February 2013 - 10:18 AM

Ah, your last point may have fixed it.  I could have sworn that they were the same order, but I hadn't really checked it or thought about it.  Thanks for that catch, I'll test it out soon.

That rotation function was mostly put there out of frustration; I was originally using XMMatrixRotationAxis, but was having some issues with rotation that was a result from some other odd behaviour I was having a while ago.  I haven't been doing a lot of rotation with my models yet so I didn't get around to fixing it.  Thanks for the catch on that one, too, I can mark it as a todo task for me.

### #4Burnt_Fyr  Members   -  Reputation: 1201

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Posted 21 February 2013 - 12:15 PM

What vertex shader are you using? objects disappearing as you describe lead me to believe that perhaps your projection is not correct, as alluded to by GuardianX. Projective space spans from 0->1 ranging from near plane to far plane, anything outside of that range will not be visible.

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# mutual positions of vectors

In this entry, we work within a Euclidean space $E$.

1. 1.

Two non-zero Euclidean vectors $\vec{a}$ and $\vec{b}$ are said to be , denoted by  $\vec{a}\parallel\vec{b}$,  iff there exists a real number $k$ such that

 $\vec{a}=k\vec{b}.$

Since both $\vec{a}$ and $\vec{b}$ are non-zero, $k\neq 0$.  So $\parallel$ is a binary relation on on  $E\!\smallsetminus\!\{\vec{0}\}$  and called the parallelism.  If  $k>0$,  then $a$ and $b$ are said to be in the same direction, and we denote this by  $\vec{a}\upuparrows\vec{b}$;  if  $k<0$,  then $a$ and $b$ are said to be in the opposite or contrary directions, and we denote this by  $\vec{a}\downarrow\uparrow\vec{b}$.

Remarks

• Actually, the parallelism is an equivalence relation on  $E\!\smallsetminus\!\{\vec{0}\}$.  If the zero vector $\vec{0}$ were allowed along, then the relation were not symmetric ($\vec{0}=0\vec{b}$  but not necessarily  $\vec{b}=k\vec{0}$).

• When two vectors $\vec{a}$ and $\vec{b}$ are not parallel to one another, written  $\vec{a}\nparallel\vec{b}$,  they are said to be diverging.

2. 2.

Two Euclidean vectors $\vec{a}$ and $\vec{b}$ are , denoted by  $\vec{a}\perp\vec{b}$,  iff

 $\vec{a}\cdot\vec{b}=0,$

i.e. iff their scalar product vanishes.  Then $\vec{a}$ and $\vec{b}$ are normal vectors of each other.

Remarks

• We may say that $\vec{0}$ is perpendicular to all vectors, because its direction is and because  $\vec{0}\cdot\vec{b}=0$.

• Perpendicularity is not an equivalence relation in the set of all vectors of the space in question, since it is neither reflexive nor transitive.

3. 3.

The angle $\theta$ between two non-zero vectors $\vec{a}$ and $\vec{b}$ is obtained from

 $\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}.$

The angle is chosen so that  $0\leqq\theta\leqq\pi$.

 Title mutual positions of vectors Canonical name MutualPositionsOfVectors Date of creation 2013-03-22 14:36:24 Last modified on 2013-03-22 14:36:24 Owner pahio (2872) Last modified by pahio (2872) Numerical id 25 Author pahio (2872) Entry type Definition Classification msc 15A72 Related topic AngleBetweenTwoLines Related topic DirectionCosines Related topic OrthogonalVectors Related topic PerpendicularityInEuclideanPlane Related topic MedianOfTrapezoid Related topic TriangleMidSegmentTheorem Related topic CommonPointOfTriangleMedians Related topic FluxOfVectorField Related topic NormalOfPlane Defines parallel Defines parallelism Defines perpendicular Defines perpendicularity Defines diverging Defines normal vector


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## Suppose that the lengths of human pregnancies are normally distributed with a mean of 265 days and a standard deviation of 16 days. Complete

Question

Suppose that the lengths of human pregnancies are normally distributed with a mean of 265 days and a standard deviation of 16 days. Complete the following statements.
A) Approximately _____ pregnancies have lengths between 217 days and 313 days.
A) 68%
B) 75%
C) 95%
D) 99.7%

B) Approximately 68% of pregnancies have lengths between _____ days and _____ days.

in progress 0
2 weeks 2021-07-20T10:40:29+00:00 1 Answers 2 views 0


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Two identical planets, each with a mass of 10 24 kg, orbit around the midpoint between the two...

Question:

Two identical planets, each with a mass of {eq}10^{24} kg {/eq}, orbit around the midpoint between the two planets. If the distance between the two planets is {eq}2 \times 10^8 m {/eq} (as measured from their centers), how fast is each of the planets moving in their orbit? (EXPLAIN STEP BY STEP)

Orbital Velocity

Orbital velocity is defined as the velocity required by a heavenly body to orbit other massive body. If the mass of the massive body is M ans orbital radius is r then orbital velocity is given by

{eq}\begin{align} v = \sqrt{\frac{GM}{r}} \end{align} {/eq}

Let us recap important information from the question

• Mass of each planet {eq}m = 10^{24} \ kg {/eq}
• Distance between two planet {eq}d = 2 \times 10^8 \ m {/eq}

Here we will assume that both the planet are orbiting the center of mass of the two particle system and whole mass is concentrated at the center of mass i.e. mid point as both the particle are identical. Now total mass on the center of mass

• Mass {eq}M = m+ m = 2m = 2\times 10^{24} \ kg {/eq}
• Distance of each planet from midpoint i.e. orbital radius {eq}r = d/2 = 10^8 \ m {/eq}

Now orbital velocity is given by

{eq}\begin{align} v = \sqrt{\frac{GM}{r}} \end{align} {/eq}

{eq}\begin{align} v & = \sqrt{\frac{6.67 \times 10^{-11} \ N.m^2.kg^{-2} \times 2 \times 10^{24} \ kg }{10^{8} \ m}} \\ & = 1154.99 \ m/s \end{align} {/eq}

{eq}\begin{align} v & \approx 1155 \ m/s \end{align} {/eq}


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# saturation

Create a saturation nonlinearity estimator object

## Syntax

NL = saturation
NL = saturation('LinearInterval',[a,b])

## Description

NL = saturation creates a default saturation nonlinearity estimator object for estimating Hammerstein-Wiener models. The linear interval is set to [NaN NaN]. The initial value of the linear interval is determined from the estimation data range during estimation using nlhw. Use dot notation to customize the object properties, if needed.

NL = saturation('LinearInterval',[a,b]) creates a saturation nonlinearity estimator object initialized with linear interval, [a,b].

Alternatively, use NL = saturation([a,b]).

## Object Description

saturation is an object that stores the saturation nonlinearity estimator for estimating Hammerstein-Wiener models.

Use saturation to define a nonlinear function $y=F\left(x,\theta \right)$, where y and x are scalars, and θ represents the parameters a and b that define the linear interval, [a,b].

The saturation nonlinearity function has the following characteristics:

For example, in the following plot, the linear interval is [-4,3].

The value F(x) is computed by evaluate(NL,x), where NL is the saturation object.

For saturation object properties, see Properties.

## Examples

collapse all

NL = saturation;

Specify the linear interval.

NL.LinearInterval = [-4,5];

z = iddata(y,u,0.2,'Name','Two tank system');
z1 = z(1:1000);

Create a saturation object with lower limit of 0 and upper limit of 5.

InputNL = saturation('LinearInterval',[0 5]);

Estimate model with no output nonlinearity.

m = nlhw(z1,[2 3 0],InputNL,[]);

Create an iddata object.

z = iddata(y,u,0.02,'Name','Motorized Camera','TimeUnit','s');

z is an iddata object with 6 inputs and 2 outputs.

Specify the model orders and delays.

Orders = [ones(2,6),ones(2,6),ones(2,6)];

Specify the same nonlinearity estimator for each input channel.

InputNL = saturation;

Specify different nonlinearity estimators for each output channel.

Estimate the Hammerstein-Wiener model.

sys = nlhw(z,Orders,InputNL,OutputNL);

To see the shape of the estimated input and output nonlinearities, plot the nonlinearities.

plot(sys)

Click on the input and output nonlinearity blocks on the top of the plot to see the nonlinearities.

## Input Arguments

collapse all

Linear interval of the saturation, specified as a 2–element row vector of doubles.

The saturation nonlinearity is initialized at the interval [a,b]. The interval values are adjusted to the estimation data by nlhw. To remove the lower limit, set a to -Inf. The lower limit is not adjusted during estimation. To remove the upper limit, set b to Inf. The upper limit is not adjusted during estimation.

When the interval is [NaN NaN], the initial value of the linear interval is determined from the estimation data range during estimation using nlhw.

Example: [-2 1]

## Properties

 LinearInterval Linear interval of the saturation, specified as a 2–element row vector of doubles. Default: [NaN NaN]

## Output Arguments

collapse all

Saturation nonlinearity estimator object, returned as a saturation object.


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시간 제한메모리 제한제출정답맞힌 사람정답 비율
5 초 128 MB78301833.962%

## 문제

Byteasar is going to watch a meteor shower tonight. He has read an exact prognosis of its route. He knows that the shower will be composed of n meteors and the i-th meteor will appear at ti-th second after midnight, i.e., it will be visible between moments ti - 1 and ti. For each meteor, Byteasar is only interested in watching it the whole time it is visible.

Byteasar will watch the sky from a nearby hill. He is not going to do it with a naked eye. On the hill there is a telescope which perfectly suits an amateur's needs. However, it is not for free. The telescope accepts coins. For each c inserted bythalers it allows to watch the sky for exactly c seconds. Unfortunately the telescope is a bit shabby, so after an insertion of a coin one must wait for r seconds before anything is visible. Every time only a single coin is accepted. Thus, if you insert a c coin, the next coin will be accepted only after at least r + c seconds.

Byteasar has m coins in his pocket with the values of c1, ..., cm bythalers. He is going to use them to pay for the telescope. He wonders what is the maximum number of meteors he can observe with the telescope.

## 입력

The first line of the input contains three integers n, m and r (1 ≤ n ≤ 100, 1 ≤ m ≤ 10, 1 ≤ r ≤ 108). The second line is composed of n integers t1, ..., tn given in the increasing order (1 ≤ ti ≤ 108). The third line contains m integers c1, ..., cm (1 ≤ ci ≤ 108).

## 출력

The only line of the output should contain one integer, i.e., the maximum number of meteors that Byteasar can observe with the telescope.

## 예제 입력 1

7 3 2
1 3 6 7 8 12 13
2 5 1

## 예제 출력 1

6

## 힌트

In the figure above black rectangles show seconds during which a meteor is visible. If Byteasar wants to see 6 meteors, he should insert his coins in moments -2, 2 and 9 in the following order of values: 1, 5 and 2 bythalers.


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Meredith Lee

# R 4 Beginners

An exploration of data science as taught in R for Data Science by Hadley Wickham and Garrett Grolemund. This blog is meant to be a helpful addition to your own journey through the book. No prior programming experience necessary!

## Chapter 5: Data Transformation

In a perfect world, any data we need would come to us in exactly the way we need it: tidy, complete with all of the variables we want to analyze, and everything descriptively named. Alas, in this world such a thing rarely happens, and so we usually have to do some work to our data before we can analyze it. That’s where data transformation comes in! The package that handles most data transformation in the tidyverse is dplyr. There are a variety of “verbs” in dpylr that allow you to manipulate data in various ways.

### filter()

The filter() function does basically what you’d expect it to do given the name– it filters the data based on criteria that you specify, returning a new data frame with only the rows that meet those criteria. The simplest form of a filtering operation is this:

new_data <- filter(DATA, COMPARISON OPERATION)

For a “real” example, we’ll use the flights data frame from the nycflights13 package (so this time you’ll need to load both the tidyverse and nycflights13 packages before getting started). Take a look at the flights data frame first to see what data it contains. Once you’ve done that, you can decide what subset of data you’d like to look at. For example, let’s say we wanted to just look at flights that left New York City in July:

july_flights <- filter(flights, month == 7)

If you run this code, you’ll see that the new data frame july_flights contains many fewer rows than flights, and if you look at the month column, it is always “7” (which corresponds to July). You aren’t limited to only one criterion either! Try running this code and see what happens:

nye_flights <- filter(flights, month == 12, day == 31)

This is useful because if you’re only interested in flights that left NYC on New Year’s Eve, this is an easy way to isolate those data. This is simple in theory, but it can get tricky in practice because of the strangeness that is involved in using comparison operators and logical operators.

#### Comparison and logical operators

In the code above, the argument where we set our filtering criteria (month == 7) is a comparison operator, or relational operator. Comparison operators are basically asking questions like “are these two things the same?”, or “is x greater than y?”. There are several comparison operators that are important to know:

• == : equal to (be sure to use the == and not =, which does a different thing in R and isn’t used to compare values– you’ll get an error if you try to use =.)
• != : not equal to
• < : less than
• > : greater than
• <= : less than or equal to
• >= : greater than or equal to

There is another slightly unusual comparison operator near(), which works like == except that it’s a “safer” option when comparing floating point numbers (decimals) in which the computer’s non-infinite precision makes a comparison that should work not work. In R4DS they use the example of sqrt(2)^2 == 2, which should obviously evaluate to TRUE but instead evaluates to FALSE, and so near() is required to get TRUE. Read the near() documentation for more information.

More complex operations can be constructed using logical operators, or boolean operators:

• & : AND (as in x&y, x AND y)
• | : OR (as in x|y, x OR y)
• ! : NOT (as in !x, NOT x)

In the nye_flights example above, the & is already built into the function, but you could also write it like this:

nye_flights <- filter(flights, month == 12 & day == 31)

Try another operation, like month == 4 | month == 5 and see what happens! Just a note, if you’re using anything that isn’t &, you’ll need to be explicit. And read up more on logical operators, they can be tricky (it’s also a good idea to look at your data when you do operations like this to make sure you’re getting what you think you’re getting).

One more thing to note is that almost any operation you try to do in which missing values (NA in R parlance) are involved will result in an unknown value. You can use is.na() to look for missing values (for example, if x <- NA, the operation x == NA will evaluate to NA, while is.na(x) evaluates to TRUE).

### arrange()

The arrange() function reorders the rows of a data frame using columns that you choose (the first column you specify will get priority, then the second, and so on). For example, the flights data frame is already ordered in ascending order of month, followed by day, but say you wanted to look at the most delayed flights (in terms of arrival time) for each month:

arrange(flights, month, desc(arr_delay))

Since each column that you specify after the first will break ties in the first column (in this case month), you end up with the flights in each month reordered in descending order from the most delayed arrival to the least. If instead you wanted the flights that had the smallest arrival delay first, you’d omit the desc() function for arr_delay (the default is to sort in ascending order, as month is in this example).

### select()

The select() function is a way to narrow down your data frame to just the columns that you want, which is especially useful if your data set has dozens or even hundreds of columns but you’re only interested in a few. The first argument must be the data frame you’re taking the columns from, followed by the columns you want from that data frame (they will appear in the new data frame in the order you list them). You can list all of the columns explicitly (e.g. select(flights, year, month, day, dep_time)), use subsetting notation (select(flights, year:dep_time)), or use the minus sign to give you all of the columns except the ones you specify (so select(flights, -(carrier:dest) would give you every column except carrier, tailnum, origin, and dest; you can think of it as a way to drop columns).

And just as a reminder, it’s good practice to be explicit whenever possible. Subsetting notation can save a lot of time (especially if you want a lot of columns) but it’s possible that “future” you will come back to your code in 6 months and not know exactly what “past” you was trying to accomplish! Worse, if you’re using data from a source that isn’t yours, the structure of the data set could change without you realizing it, and your subsetting will no longer be doing what you think it’s doing. If you need to use subsetting, do yourself a favor and 1) make good notes about exactly what you’re trying to do (I’ll talk more about how to do that in the next chapter), and 2) take a peek at the data after your manipulations, so you know what you have.

select() is often used with a variety of matching functions:

• starts_with(): Lets you match a string to the beginning of the column names. For example, select(flights, starts_with("dep")) would return a data frame with only the columns dep_time and dep_delay. Not super useful, but you get the idea.
• ends_with(): Same as starts_with(), except it matches the end instead of the beginning.
• contains(): Looks for the string anywhere in the columns names. For example, select(flights, contains("time")) returns a data frame with the columns dep_time, sched_dep_time, arr_time, sched_arr_time, air_time, and time_hour.
• num_range(): Looks for column names with a range of numbers. The flights data set doesn’t have any columns like this, but if for example you had columns names like layover_1, layover_2, and layover_3 you could get these columns with select(flights, num_range("layover_", 1:3))
• matches(): Looks for column names that match a regular expression. Don’t worry about this for now, we’ll come back to regular expressions later (they’re powerful, but tricky).
• everything(): The select() function can also be used to reorder columns, and everything() is useful in this case (since select() only keeps the columns you specify). If you wanted all of the columns in flights but instead wanted the flight number, the origin, and the destination at the beginning, you would run select(flights, flight, origin, dest, everything()).

### mutate()

dplyr’s mutate() is the function used to add new columns to your data frame. This is most useful when you want to calculate new values based on the data in the data frame, like if your data set gives you columns for distance and time for a cohort of marathon runners, but you need speed for your analysis. You can use mutate() to calculate a speed column from the data in the distance and time columns. Going back to our flights data set, one thing that might make the data difficult to work with are the time columns (dep_time, sched_dep_time, arr_time, and sched_arr_time), where the time is in a 24 hour system. You can’t perform mathematical operations on time values in this form generally, so if we want to do calculations with them we’ll need to change them to a continuous format, such as “minutes since midnight”. Since the times are in the format “hhmm” (meaning, 10:00 am would be shown as 1000, and 2:30 pm would be 1430), you can divide by 100 to get the number of hours since midnight, then multiply by 60 to get those hours in minutes, then add the remainder (the minutes) from the first operation to the result from the second. For example:

• 4:45 pm would be represented as 1645 in the flights data set.
• 1645/100 is 16 hours with a remainder of 45 minutes.
• There are 960 minutes in 16 hours (16*60).
• Therefore, the total number of minutes since midnight would be 960+45=1005.

The great thing about mutate() is that once we know the operations we need, we can compute an entirely new column using the values of an existing column (or another data source, if there is a way to link the two data frames, but that’s a problem for another day). So let’s go back and create four new columns, changing the times I listed above to minutes since midnight. Since mutate() (like all of the transformation functions we’ve talked about) returns a new data frame, we can assign it to a new variable. A couple of arithmetic operators that are particularly useful here are %/%, which returns the integer result of a division operation, and %% which returns the remainder (this is called the modulo operator).

The last thing we need to take into account is midnight. In our new columns, since we want the time to be measured as minutes from midnight, midnight itself should clearly be 0. In fact, in a 24 hour clock midnight is represented as 2400, which according to our method above would calculate midnight to be 1440 minutes from midnight. We can correct for this by using the operation x %% 1440, where x represents our previous calculation to get the hours and minutes. This is because when both x and y are positive integers and x < y, then x %% y = x. If we say that x is time in minutes from midnight, then by definition all values for x will be less than 1440. Therefore, x %% 1440 = x in every case except at midnight. At midnight x = 1440, and 1440 %% 1440 = 0.

So, given all of those considerations, we can get our four new columns like this:

flight_mins <- mutate(flights, dep_mins = (dep_time %/% 100 * 60 + dep_time %% 100) %% 1440,
sched_dep_mins = (sched_dep_time %/% 100 * 60 + sched_dep_time %% 100) %% 1440,
arr_mins = (arr_time %/% 100 * 60 + arr_time %% 100) %% 1440,
sched_arr_mins = (sched_arr_time %/% 100 * 60 + sched_arr_time %% 100) %% 1440)

I know that was a lot! I include it because it was tricky for me to figure out (the above example is actually exercise 1 in section 5.5.2), but it also includes a lot of good info about using mutate(). Play around with some other arithmetic operators, see what you come up with. There are the usual suspects like addition, subtraction, multiplication, etc. but there are also logs (useful if your data span many orders of magnitude, and are therefore hard to work with or visualize), ranking (such as min_rank()), and offsets (such as lead() and lag()). Use the “help” tab in RStudio to see what all of these operators do.

And don’t forget about the order of operations when you’re doing your arithmetic!

### summarise() and group_by()

summarise(), as the name suggests, give you a summary of the data that you pass into it, using a calculation such as mean(), median(), standard deviation (sd()), minimum(min()), maximum(max()), count(n()), etc. For example, try running this in your console:

summarise(flights, ave_arr_delay = mean(arr_delay, na.rm = TRUE))

You should see a table with a single value, the mean arrival delay (the na.rm argument is telling the function to ignore missing values– it is FALSE by default, but trying to calculate the mean with missing values will return an unknown value).

This is not super useful on its own, but combined with group_by() it becomes quite powerful, because it allows you to do group-wise analysis. Say we want to compare the average arrival delay by carrier. Try running this in your console:

by_carrier <- group_by(flights, carrier)
summarise(by_carrier, delay = mean(arr_delay, na.rm = TRUE))

You can add more summary functions as additional arguments if you want; they’ll be added as additional columns in the summary data frame. Some of the most useful summary functions are listed in the help for summarise().

#### Counting

When performing summary statistics on a data set, it is important to be aware that the number of observations can affect how informative your conclusions are. This is because small data sets have more variation, and so with larger sample sizes, the variation in the summary calculation (like mean) will decrease. Actually, there are several factors that go into choosing an appropriate sample size for a study, but for initial data transformations it’s just important to look at the sample size before drawing conclusions based on a summary statistic.

There are a few functions that return the number of observations (or count) that work in slightly different ways:

• n(): The simplest function– it takes no argument, and just returns the group size (this is often used within a summarise() call).
• sum(!is.na()): Gives the group count, excluding missing values.
• n_distinct(): Counts distinct (unique) values.
• count(): Basically a way to combine group_by() and n() in one step, if all you want is a group-wise count.

### Putting It All Together

As you might imagine, if you have a large data set you might need to perform many transformations to get your data into the form that you need. The way we’ve been writing code above, you would need to assign each new data frame to an object at each step, even though you won’t need those objects ever again! Seems like a waste, right? Luckily, there is a solution for this! It’s called the pipe.

The pipe comes from a package called magrittr, which is loaded with the tidyverse. The pipe is great because 1) you don’t have to come up with a bunch of names for stuff you won’t need, and 2) it makes your code more readable. Let’s try an example. Let’s say I want to look at the airline carriers that are “best” at making up time in flight after a delayed departure:

time_gain_by_carrier <- mutate(flights, time_gain = dep_delay - arr_delay) %>%
group_by(carrier) %>%
summarise(mean(time_gain, na.rm = TRUE))

That’s two names you don’t have to think of! The output of each step in the pipe goes straight into the next, so that each intermediate step doesn’t need to be stored in its own object. The pipe is very powerful and if you continue to work with data in R you will use it a lot.


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# Four Identical Rods Ab, Cd, Cf and De Are Joined as Shown in Following Figure . the Length, Cross-sectional Area and Thermal Conductivity of Each Rod Are L, a and K Respectively. - Physics

ConceptHeat Transfer Conduction

#### Question

Four identical rods AB, CD, CF and DE are joined as shown in following figure . The length, cross-sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperature T1, T2 and T3 respectively. Assuming no loss of heat to the atmosphere, find the temperature at B.

#### Solution

Let the temperature at junction B be T.
Let q1q2 and q3 be the heat currents, i.e. rate of flow of heat per unit time in AB, BCE and BDF, respectively.
From the diagram, we can see that
q1 = q2 + q3
The rate of flow of heat is given by
q = (KA DeltaT)/l
Using this tn the above equation, we get
(A (T_1 - T))/l = (KA  ( T + T_3))/(31/2)  + (KA (T-T_2))/(31/2)

⇒ T_1 - T = (2(T  - T_3))/3+(2(T- T_2))/3

⇒ 3 (T1 - T) = 2T - 2T3 + 2T - 2T2

⇒ T = (-3T_1 + 2( T_2 +T_3))/7

Is there an error in this question or solution?

#### APPEARS IN

Solution Four Identical Rods Ab, Cd, Cf and De Are Joined as Shown in Following Figure . the Length, Cross-sectional Area and Thermal Conductivity of Each Rod Are L, a and K Respectively. Concept: Heat Transfer - Conduction.
S


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Ideals-algebraic set

Notice that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3) = \{ (t,t^2,t^3) \mid t \in \mathbb{C}\}$$

$$I(V(Y-X^2,Z-X^3)) = \langle Y-X^2,Z-X^3 \rangle$$

Finally, prove that the ideal $\langle Y-X^2,Z-X^3 \rangle$ is a prime ideal of $\mathbb{C}[X,Y,Z]$. Conclude that the algebraic set $V(Y-X^2,Z-X^3)$ is irreducible.

Could you give me some hints to solve the above exercise?

EDIT 1: Could we show like that, that in $\mathbb{C}[X,Y,Z]$:
$$V(Y-X^2,Z-X^3) = \{ (t,t^2,t^3) \mid t \in \mathbb{C}\}\ ?$$

$$V(Y-X^2, Z-X^3) = \{(a,b,c) \in \mathbb{C}^3 \mid b-a^2=0, c-a^3=0 \Rightarrow b=a^2, c=a^3\} = \{(t, t^2, t^3) \mid t \in \mathbb{C}\}$$

EDIT 2:
Could we show like that, that: $I(V(Y-X^2,Z-X^3)) = \langle Y-X^2,Z-X^3 \rangle$ ?

\begin{align} I(V(Y-X^2, Z-X^3)) &= I(\{(t, t^2, t^3) \mid t \in \mathbb{C}\}) \\ &= \{f(X,Y,Z) \in \mathbb{C}[X,Y,Z] \mid f(t,t^2,t^3)=0\} \\ &\overset{*}{=} \{(Y-X^2) \cdot g(X,Y,Z)+(Z-X^3) \cdot h(X,Y,Z) \mid g,h \in \mathbb{C}[X,Y,Z]\} \\ &= \langle Y-X^2, Z-X^3 \rangle \end{align}

At $(*)$ can we say it like that because we know from $V(Y-X^2, Z-X^3)=\{(t,t^2,t^3) \mid t \in \mathbb{C}\}$ that $(t,t^2,t^3)$ is a solution of $Y-X^2=0$ and $Z-X^3=0$ ?

EDIT 3:Let the homomorphism $\phi: \mathbb{C}[x,y,z] \to \mathbb{C}[x]$, with $\phi(x)=x, \phi(y)=x^2, \phi(z)=x^3, \phi(j)=j, \forall j \in \mathbb{C}$

We consider the polynomial $p(x,y,z)=\sum_{k,l,m=0}^n a_{klm} x^k (y-x^2)^l(z-x^3)^m$.

Then, $\phi(p)=\sum_{k,l,m=0}^n \phi(a_{klm}) (\phi(x))^k (\phi(y)-(\phi(x))^2)^l (\phi(z)-(\phi(x))^3)^m=a_{n00}x^n$

$ker(\phi)$ contains all the elements from $\mathbb{C}[x,y,z]$, die auf 0, durch die \phi, abgebildet werden.

So, it contains the polynomials with $a_{n00}=0$, so the non-zero elements of the polynomial must contain $y-x^2$ or $z-x^3$, so $ker \phi=\langle y-x^2,z-x^3 \rangle$.

Actually,

Let $p(x,y,z) \in \langle y-x^2,z-x^3 \rangle$.

Then, $p(x,y,z)=p_1(x,y,z) \cdot (y-x^2) +p_2(x,y,z) \cdot (z-x^3), p_1(x,y,z), p_2(x,y,z) \in \mathbb{C}[x,y,z]$

We have that $\phi(p(x,y,z))=\phi(p_1(x,y,z) \cdot (y-x^2) +p_2(x,y,z) \cdot (z-x^3))=0[/itex], also [itex] p(x,y,z) \in ker(\phi) \Rightarrow \langle y-x^2,z-x^3 \rangle \subseteq ker(\phi)$

Let $p(x,y,z) \in ker(\phi)$. Then, $\phi(p(x,y,z))=0$. The terms of the polynomial are of the form $a_{ij}(y-x^2)^i (z-x^3)^j$,where at least one of $i,j>0$.

So, $p(x,y,z)$ is a sum of terms of $\langle y-x^2,z-x^3 \rangle$. So, $p(x,y,z) \in \langle y-x^2,z-x^3 \rangle$, and so $ker(\phi) \subseteq \langle y-x^2,z-x^3 \rangle$.

So, we conclude that $ker(\phi)=\langle y-x^2,z-x^3 \rangle$.

Since $\phi(p(x,y,z))=a_{n00}x^n \in \mathbb{C}[x], \forall p(x,y,z) \in \mathbb{C}[x,y,z]$,we have that $im(\phi)= \mathbb{C}[x]$.

So, we have that: $\mathbb{C}[x,y,z]/ \langle y-x^2,z-x^3\rangle \cong \mathbb{C}[x]$.

Is it right so far? Do we have to show that $\mathbb{C}[x]$ is an integral domain?

Solutions Collecting From Web of "Ideals-algebraic set"

Edit 1 is correct.

Edit 2 has a flaw at $(*)$: no, you can’t say that; you have to prove that $f(t,t^2,t^3)=0$ for all $t\in\mathbb C$ implies $f\in (Y-X^2,Z-X^3)$. (This is closely related to prove that $(Y-X^2,Z-X^3)$ is a prime ideal, and in the reasoning I’ve posted below $f(t,t^2,t^3)=0$ reduces to $q(t)=0$ for all $t\in\mathbb C$, so $q=0$.)
You could also use the Hilbert’s Nullstellensatz and this gives you $I(V(Y-X^2,Z-X^3))=\sqrt{(Y-X^2,Z-X^3)}$, but then you have to prove that the ideal $(Y-X^2,Z-X^3)$ is radical in order to get what you want. In fact it is prime, and in some sense all gravitates around this as you can immediately notice and this is what you try to prove in

Edit 3 I can’t see why you write a polynomial like this: $$p(X,Y,Z)=\sum_{k,l,m=0}^n a_{klm} X^k (Y-X^2)^l(Z-X^3)^m.$$ (For instance, how to write $XY$ in that form?)
A better strategy is to use the division algorithm and first write $$p(X,Y,Z)=(Y-X^2)f(X,Y,Z)+r(X,Z)$$ with $\deg_Yr<1$, that is, $Y$ doesn’t occur in $r$, and then $$r(X,Z)=(Z-X^3)g(X,Z)+q(X)$$ with $\deg_Zq<1$ and thus $Z$ doesn’t occur in $q$, so $q$ is a polynomial in $X$ only.
Now, if $p\in\ker\phi$ we get $p(X,X^2,X^3)=0$, so $q(X)=0$ and you get $$p(X,Y,Z)=(Y-X^2)f(X,Y,Z)+(Z-X^3)g(X,Z).$$ This shows that $p\in(Y-X^2,Z-X^3)$.
As a consequence you get $$\mathbb C[X,Y,Z]/(Y-X^2,Z-X^3)\simeq\mathbb C[X]$$ which is an integral domain (no need to prove this in my opinion), and therefore $(Y-X^2,Z-X^3)$ is a prime ideal.

1. $V(y-x^2, z-x^3) = \{(a,b,c) \in \mathbb{C}^3 \mid b-a^2=0, c-a^3=0\} = \{ (a, b, c) \in \mathbb{C}^3 | b=a^2, c=a^3\} = \{(t, t^2, t^3) \mid t \in \mathbb{C}\}$.

2. $I(V(y-x^2,z-x^3)) = \{ f(x, y, z) \in \mathbb{C}[x, y, z] | f(t, t^2, t^3) = 0 \}$ (using 1). Obviously $(y – x^2, z – x^3) \subseteq I(V(y-x^2,z-x^3))$. Now we will show that if $f(x, y, z) \in \mathbb{C}[x, y, z]$ is such that $f(t, t^2, t^3) = 0$ then $f(x, y, z) \in (y – x^2, z – x^3)$. Consider the polynomial $y – x^2$ as a polynomial in $y$ over the ring $\mathbb{C}[x, z]$. Since it is a monic polynomial in $y$, we can use division algorithm. We get $f(x, y, z) = f_1(x, y, z)(y – x^2) + f_2(x, z)$. Since the degree of $y – x^2$ is 1 as a polynomial in $y$, $f_2(x, z)$ is a polynomial in $x$ and $z$ only (over $\mathbb{C}$)[Hence the notation is justified]. Similarly, considering the polynomial $z – x^3$ as a polynomial in $z$ over the ring $\mathbb{C}[x]$, we get $f_2(x, y, z) = f_3(x, y, z)(z – x^3) + f_4(x)$. Using the same reasoning as above, $f_4(x)$ is a polynomial in $x$ alone (over $\mathbb{C}$). So, $f(x, y, z) = f_1(x, y, z)(y – x^2) + f_3(x, y, z)(z – x^3) + f_4(x)$. Now $f(t, t^2, t^3) = 0 \Rightarrow f_4(t) = 0 \Rightarrow f(x, y, z) \in (y – x^2, z- x^3)$.

3. To prove the ideal $(y – x^2, z – x^3)$ is a prime ideal in the ring $\mathbb{C}[x, y, z]$, let’s define a $\mathbb{C}$-algebra homomorphism $\phi : \mathbb{C}[x, y, z] \rightarrow \mathbb{C}[t]$ by $x \mapsto t, y \mapsto t^2, z \mapsto t^3.$ This is a surjective ring homomorphism. Using the same arguments as in (2), we can prove that ker$\phi = (y – x^2, z – x^3)$. Thus by first isomorphism theorem of rings, the ideal $(y – x^2, z – x^3)$ is a prime ideal.

4. The algebraic set $Y := V(y – x^2, z – x^3)$ is irreducible is equivalent of saying that the ideal $(y – x^2, z – x^3)$ is prime. Suppose $Y$ is not irreducible. Then there is two non-empty proper closed subsets $Y_1, Y_2$ of $Y$ with $Y = Y_1 \cup Y_2$. Thus $I(Y) = I(Y_1 \cup Y_2) = I(Y_1) \cap I(Y_2).$ Since $I(Y)$ is prime, this means either $I(Y_1) = I(Y)$ or $I(Y_2) = I(Y) \Rightarrow Y = Y_1$ or $Y = Y_2$.


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## Boltzmann type control of opinion consensus through leaders

Giacomo Albi,  Lorenzo Pareschi, Mattia Zanella.
(06/05/2014  arXiv:1405.0736)

The study of formations and dynamics of opinions leading to the so called opinion consensus is one of the most important areas in mathematical modeling of social sciences. Following the Boltzmann type control recently introduced in [G. Albi, M. Herty, L. Pareschi arXiv:1401.7798], we consider a group of opinion leaders which modify their strategy accordingly to an objective functional with the aim to achieve opinion consensus. The main feature of the Boltzmann type control is that, thanks to an instantaneous binary control formulation, it permits to embed the minimization of the cost functional into the microscopic leaders interactions of the corresponding Boltzmann equation. The related Fokker-Planck asymptotic limits are also derived which allow to give explicit expressions of stationary solutions. The results demonstrate the validity of the Boltzmann type control approach and the capability of the leaders control to strategically lead the followers opinion.

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## Kinetic description of optimal control problems and applications to opinion consensus

Giacomo Albi, Michael Herty, Lorenzo Pareschi
(30/01/2014 arXiv:1401.7798)

In this paper an optimal control problem for a large system of interacting agents is considered using a kinetic perspective. As a prototype model we analyze a microscopic model of opinion formation under constraints. For this problem a Boltzmann-type equation based on a model predictive control formulation is introduced and discussed.

In particular, the receding horizon strategy permits to embed the minimization of suitable cost functional into binary particle interactions. The corresponding Fokker-Planck asymptotic limit is also derived and explicit expressions of stationary solutions are given. Several numerical results showing the robustness of the present approach are finally reported.

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## Asymptotic Preserving time-discretization of optimal control problems for the Goldstein-Taylor model

(01/08/2013 arXiv:1307.8303)
We consider the development of implicit-explicit time integration schemes for optimal control problems governed by the Goldstein-Taylor model. In the diffusive scaling this model is a hyperbolic approximation to the heat equation. We investigate the relation of time integration schemes and the formal Chapman-Enskog type limiting procedure.
For the class of stiffly accurate implicit-explicit Runge-Kutta methods (IMEX) the discrete optimality system also provides a stable numerical method for optimal control problems governed by the heat equation. Numerical examples illustrate the expected behavior.
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## Stability Analysis of Flock and Mill rings for 2nd Order Models in Swarming

G. Albi, D. Balagué, J. A. Carrillo,  J. von   Brecht                                                                                (19/04/2013  arxiv.org/abs/1304.5459)

We study the linear stability of flock and mill ring solutions of two individual based models for biological swarming. The individuals interact via a nonlocal interaction potential that is repulsive in the short range and attractive in the long range. We relate the instability of the flock rings with the instability of the ring solution of the first order model. We observe that repulsive-attractive interactions lead to new configurations for the flock rings such as clustering and fattening formation. Finally, we numerically explore mill patterns arising from this kind of interactions together with the asymptotic speed of the system.

Related page with video & simulations.

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## Modeling self-organized systems interacting with few individuals: from microscopic to macroscopic dynamics

Giacomo Albi, Lorenzo Pareschi

(27 /09/2012 arXiv:1210.1172)

In nature self-organized systems as flock of birds, school of fishes or herd of sheeps have to deal with the presence of external agents such as predators or leaders which modify their internal dynamic. Such situations take into account a large number of individuals with their own social behavior which interact with a few number of other individuals acting as external point source forces. Starting from the microscopic description we derive the kinetic model through a mean-field limit and finally the macroscopic system through a suitable hydrodynamic limit.

Related page with videos & simulations.

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## Binary interaction algorithms for the simulation of flocking and swarming dynamics

Giacomo Albi, Lorenzo Pareschi
(04/03/2012 arXiv:1203.0721)

Microscopic models of flocking and swarming takes in account large numbers of interacting individuals. Numerical resolution of large flocks implies huge computational costs. Typically for $N$  interacting individuals we have a cost of $O(N^2)$. We tackle the problem numerically by considering approximated binary interaction dynamics described by kinetic equations and simulating such equations by suitable stochastic methods. This approach permits to compute approximate solutions as functions of a small scaling parameter at a reduced complexity of $O(N)$ operations. Several numerical results show the efficiency of the algorithms proposed.


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Chapter 34

• History of athletic training and performance.
• Enhanced exercise ability (V̇o2max > 40 mL/kg/min).
• Increased left ventricular mass by echocardiography.

The concept of the athlete’s heart is one that has been postulated for almost 100 years, promulgating the idea that myocardial hypertrophy could be a purely physiologic phenomenon. Media attention to the sudden deaths of widely known athletes has helped focus attention on the important distinction between pathologic cardiac hypertrophy and physiologic hypertrophy and the upper limits of the latter.

The adaptations of the human body to physical training involve (but are not confined to) the cardiovascular system. The exercise-related changes in other organ systems influence the cardiovascular response to exercise. It is important for the physician to be familiar with the physiologic responses to physical training in order to distinguish them from similar changes that can occur with cardiovascular disease.

Different forms of exercise produce a number of physiologic responses. Also, cardiovascular responses to short-term training and prolonged training differ. Exercise generally takes two basic physiologic forms—dynamic and static, or isometric, exercise—although most athletic activities are a variable combination of both forms of exercise. Dynamic exercise constitutes an alteration in the length of skeletal muscle with comparatively little change in muscle tension. Static exercise is essentially the reverse—that is, a marked alteration in skeletal muscle tension with little or no change in muscle length. Distance running is a classic example of dynamic exercise; weight lifting is a classic example of static exercise.

The morphologic and physiologic consequences of dynamic and isometric exercise are significant and may simulate changes associated with cardiac disease. The normal limits of changes that are due to athletic conditioning require careful identification. Awareness of these limits improves the physician’s ability to determine the end points at which normal anatomy and physiology become clinical disease.

Maron BJ et al. The heart of trained athletes: cardiac remodeling and the risks of sports, including sudden death. Circulation. 2006 Oct 10;114(15):1633–44.  [PubMed: 17030703]

The acute cardiovascular responses to exercise are specific and vary with different forms of exercise (Figure 34–1). There are also specific adaptive responses to exercise, particularly to dynamic exercise. In particular, the adaptive change in heart rate from an alteration in vagal parasympathetic tone defines the normal physiologic range; as noted earlier, this may be initially misinterpreted as representative of cardiovascular disease.

###### Figure 34–1.

Cardiovascular response to exercise. A: Response to dynamic exercise progressively increasing workload to maximal oxygen consumption. B: Response to static handgrip contraction at 30% maximal voluntary contraction. ABP, systolic, mean and diastolic arterial blood pressures; HR, heart rate; Q, cardiac output; SV, stroke volume; TPR, total peripheral resistance; V̇o2, oxygen consumption.

(Reprinted with permission from Mitchell JH, Raven PB. Cardiovascular adaption to physical activity. In: Bouchard ...

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## Key Concepts

Operation Notation Expression Read as Result
$\text{Division}$ $\div$

$\frac{a}{b}$

$b\overline{)a}$

$a/b$

$12\div 4$

$\frac{12}{4}$

$4\overline{)12}$

$12/4$

$\text{Twelve divided by four}$ $\text{the quotient of 12 and 4}$
• Division Properties of One
• Any number (except $0$) divided by itself is one.

$a\div a=1$

• Any number divided by one is the same number.

$a\div 1=a$

• Division Properties of Zero
• Zero divided by any number is $0$.

$0\div a=0$

• Dividing a number by zero is undefined.

$a\div 0$ undefined

• Divide whole numbers.
1. Divide the first digit of the dividend by the divisor. If the divisor is larger than the first digit of the dividend, divide the first two digits of the dividend by the divisor, and so on.
2. Write the quotient above the dividend.
3. Multiply the quotient by the divisor and write the product under the dividend.
4. Subtract that product from the dividend.
5. Bring down the next digit of the dividend.
6. Repeat from Step $1$ until there are no more digits in the dividend to bring down.
7. Check by multiplying the quotient times the divisor.

## Glossary

dividend
When dividing two numbers, the dividend is the number being divided.
divisor
When dividing two numbers, the divisor is the number dividing the dividend.
quotient
The quotient is the result of dividing two numbers.


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# Why can one ignore the real part of a complex number because of Fourier transform “only using the rotational part”?

Why can one ignore the real part of a complex number because of Fourier transform "only using the rotational part"? Particularly, what property of Fourier transform is this?

• Quoting the accepted answer by Kaz: "What's happening is that $\sigma$ is being ignored because it is assumed to be zero. The reason for it is that we are looking at the response of the system to periodic (and thus non-decaying) sinusoidal signals, whereby Laplace conveniently reduces to Fourier along the imaginary axis. The real axis in the Laplace domain represents exponential decay/growth factors that pure signals do not have, and which Fourier does not model." – Saucy O'Path Aug 17 '18 at 16:16
• Physically $\omega$ encodes frequency of pure oscillation while $\sigma$ encodes dissipation. A common simplifying assumption in electronics is that the dissipation is negligible so you can deal with pure oscillation. – Ian Aug 17 '18 at 16:16
• @mavavilj More or less what you are doing in both cases is writing a function as a "linear combination" of a continuum of exponentials with different coefficients. The Laplace transform with a positive real part of $s$ corresponds to finding the coefficients of exponentials which are decaying (and oscillating) as you go forward in time. The Fourier transform with real $\omega$ corresponds to exponentials with no real part, which are purely oscillating. – Ian Aug 17 '18 at 16:20


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# Simplest complex curves with isomorphic Jacobian

Hi.

What are the simplest examples to have in mind of non-isomorphic smooth (complex) algebraic curves with isomorphic jacobian variety?

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It is not easy to give equations of non-isomorphic Riemann surfaces whose Jacobians are isomorphic as (unpolarized) Abelian varieties.

Some explicit examples can be found in the paper by Everett W. Howe Infinite Families of Pairs of Curves Over $\mathbb{Q}$ with Isomorphic Jacobians, J. London Math. Soc. (2005) 72 (2), 327-350.

A preprint version of the paper is also available in the arXiv website, see http://arxiv.org/abs/math.AG/0304471. In Section 8 you can find the examples you are looking for.

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Thanks Francesco! But this does not exactly answers to the question: what are the SIMPLEST examples... But one can also argue that the question is not rigourously stated, I agree. –  BadaBing Apr 24 '13 at 12:44
You are welcome. Right, it is not clear what "simplest" should mean in this case. At any rate, in Howe's paper there are examples given by pairs of genus $2$ curves, which is of course the lowest possible genus. In this sense, such examples might be considered "simple". –  Francesco Polizzi Apr 24 '13 at 12:50

Slightly tangential -- the following paper of Bjorn Poonen offers some interesting computational perspectives on curves of genus at least $2$ which is as Francesco mentions, the lowest possible genus where such examples exist. This seems like it might be relevant.

http://www-math.mit.edu/~poonen/papers/ants2.pdf

I think the following is an interesting question to ask over a fixed number field $K$: Can one make tables of genus $2$ curves over $K$ ordered in some lexicographic fashion by the invariants? Perhaps the curves would be modelled by singular plane curves $y^2 = f(x)$ where $f$ has degree $5$ or $6$. The first pair $(X,Y)$ with isomorphic Jacobian variety to occur in this table might satisfy the request that the pair $(X,Y)$ be the "simplest" such pair to be defined over $K$.

Poonen notes that not every not every hyperelliptic curve has a model of the form $y^2 = f(x)$ over $k$ a field, because the quotient of the curve by its hyperelliptic involution might be a twist of $\mathbb{P}^1$ i.e. birational over $k$ to a conic in $\mathbb{P}^2$ without a $k$-rational point. However when $g$ is even, it seems to turn out (I'm not sure why) that the $k$-form of $\mathbb{P}^1$ has a $k$-point and the hyperelliptic curve has a model $y^2 = f(x)$, so for $g = 2$ we should be okay.

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