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# z-value

(noun)

## Definition of z-value

the standardized value of an observation found by subtracting the mean from the observed value, and then dividing that value by the standard deviation; also called z-score

Source: Boundless Learning - CC BY-SA 3.0

## Examples of z-value in the following topics:

• ### Using the Normal Curve

• The z-value tells us how many standard deviations, or "how many sigmas," the variable is from its respective mean.Areas Under the CurveTo calculate the probability that a variable is within a range, we have to find the area under the curve.
• The intersection of the 6th row and 2nd column is 0.6950, which tells us that there is a 69.50% percent chance that a variable is less than 0.51 sigmas (or standard deviations) above the mean.A common mistake is to look up a z-value in the table and simply report the corresponding entry, regardless of whether the problem asks for the area to the left or to the right of the z-value.
• The table only gives the probabilities to the left of the z-value.
• Since the total area under the curve is 1, all we need to do is subtract the value found in the table from 1.
• What if we wished instead to find out the probability that a value falls below a z-value of -0.51, or 0.51 standard deviations below the mean?
• The normal curve is used to find the probability that a value falls within a certain standard deviation away from the mean.
• ### Estimating a Population Proportion

• This is computed slightly differently, using the formula $\hat{p}\pm z^{*}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where z* is the upper critical value of the standard normal distribution.
• In the above example, if we wished to calculate p with a confidence of 95%, we would use a Z-value of 1.960 (found using a critical value table: ), and we would find p to be estimated as 0.52$\pm$0.04896.
• ### Change of Scale

• This conversion process is called standardizing or normalizing.Standard scores are also called z-values, z-scores, normal scores, and standardized variables.
• The use of "Z" is because the normal distribution is also known as the "Z distribution".
• They are most frequently used to compare a sample to a standard normal deviate (standard normal distribution, with μ = 0 and σ = 1).The z-score is only defined if one knows the population parameters.
•  The absolute value of z represents the distance between the raw score and the population mean in units of the standard deviation. z is negative when the raw score is below the mean, positive when above.A key point is that calculating z requires the population mean and the population standard deviation, not the sample mean or sample deviation.
• In cases where it is impossible to measure every member of a population, a random sample may be used.The Z value measures the sigma distance of actual data from the average and provides an assessment of how off-target a process is operating.


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## Rank Nonsense

Posted in Bad Statistics, Education, Politics with tags , , , , , on September 8, 2016 by telescoper

It’s that time of year when international league tables (also known as “World Rankings”)  appear. We’ve already had the QS World University Rankings and the Shanghai (ARWU) World University Rankings. These will soon be joined by the Times Higher World Rankings, due out on 21st September.

A lot of people who should know a lot better give these league tables far too much attention. As far as I’m concerned they are all constructed using extremely suspect methodologies whose main function is to amplify small statistical variations into something that looks significant enough to justify constructing  a narrative about it. The resulting press coverage usually better reflects a preconceived idea in a journalist’s head than any sensible reading of the tables themselves.

A particularly egregious example of this kind of nonsense can be found in this week’s Guardian. The offending article is entitled “UK universities tumble in world rankings amid Brexit concerns”. Now I make no secret of the fact that I voted “Remain” and that I do think BrExit (if it actually happens) will damage UK universities (as well as everything else in the UK). However, linking the changes in the QS rankings to BrExit is evidently ridiculous: all the data were collected before the referendum on 23rd June anyway! In my opinion there are enough good arguments against BrExit without trying to concoct daft ones.

In any case these tables do not come with any estimate of the likely statistical variation from year to year in the metrics used to construct them, which makes changes impossible to interpret. If only the compilers of these tables would put error bars on the results! Interestingly, my former employer, the University of Sussex, has held its place exactly in the QS rankings between 2015 and 2016: it was ranked 187th in the world in both years. However, the actual score corresponding to these two years was 55.6 in 2015 and 48.4 in 2016. Moreover, Cambridge University fell from 3rd to 4th place this year but its score only changed from 98.6 to 97.2. I very much doubt that is significant at all, but it’s mentioned prominently in the subheading of the Guardian piece:

Uncertainty over research funding and immigration rules blamed for decline, as Cambridge slips out of top three for first time.

Actually, looking closer, I find that Cambridge was joint 3rd in 2015 and is 4th this year. Over-interpretation, or what?

To end with, I can’t resist mentioning that the University of Sussex is in the top 150 in the Shanghai Rankings for Natural and Mathematical Sciences this year, having not been in the top 200 last year. This stunning improvement happened while I was Head of School for Mathematical and Physical Sciences so it clearly can not be any kind of statistical fluke but is entirely attributable to excellent leadership. Thank you for your applause.

## The Rising Stars of Sussex Physics

Posted in Bad Statistics, Biographical, Education with tags , , , , on July 28, 2016 by telescoper

This is my penultimate day in the office in the School of Mathematical and Physical Sciences at the University of Sussex, and a bit of news has arrived that seems a nice way to round off my stint as Head of School.

It seems that Physics & Astronomy research at the University of Sussex has been ranked as 13th in western Europe and 7th in the UK by leading academic publishers, Nature Research, and has been profiled as one of its top-25 “rising stars” worldwide.

I was tempted to describe this rise as ‘meteoric’ but in my experience meteors generally fall down rather than rise up.

Anyway, as regular readers of this blog will know, I’m generally very sceptical of the value of league tables and there’s no reason to treat this one as qualitatively any different. Here is an explanation of the (rather curious) methodology from the University of Sussex news item:

The Nature Index 2016 Rising Stars supplement identifies the countries and institutions showing the most significant growth in high-quality research publications, using the Nature Index, which tracks the research of more than 8,000 global institutions – described as “players to watch”.

The top 100 most improved institutions in the index between 2012 and 2015 are ranked by the increase in their contribution to 68 high-quality journals. From this top 100, the supplement profiles 25 rising stars – one of which is Sussex – that are already making their mark, and have the potential to shine in coming decades.

The institutions and countries examined have increased their contribution to a selection of top natural science journals — a metric known as weighted fractional count (WFC) — from 2012 to 2015.

Mainly thanks to a quadrupling of its physical sciences score, Sussex reached 351 in the Global 500 in 2015. That represents an 83.9% rise in its contribution to index papers since 2012 — the biggest jump of any UK research organisation in the top 100 most improved institutions.

It’s certainly a strange choice of metric, as it only involves publications in “high quality” journals, presumably selected by Journal Impact Factor or some other arbitrary statistical abominatio,  then taking the difference in this measure between 2012 and 2015  and expressing the change as a percentage. I noticed one institution in the list has improved by over 4600%, which makes Sussex’s change of 83.9% seem rather insignificant…

But at least this table provides some sort of evidence that the investment made in Physics & Astronomy over the last few years has made a significant (and positive) difference. The number of research faculty in Physics & Astronomy has increased by more than 60%  since 2012 so one would have been surprised not to have seen an increase in publication output over the same period. On the other hand, it seems likely that many of the high-impact papers published since 2012 were written by researchers who arrived well before then because Physics research is often a slow burner. The full impact of the most recent investments has probably not yet been felt. I’m therefore confident that Physics at Sussex has a very exciting future in store as its rising stars look set to rise still further! It’s nice to be going out on a high note!

## The 3.5 keV “Line” that (probably) wasn’t…

Posted in Bad Statistics, The Universe and Stuff with tags , , , , , , , on July 26, 2016 by telescoper

About a year ago I wrote a blog post about a mysterious “line” in the X-ray spectra of galaxy clusters corresponding to an energy of around 3.5 keV. The primary reference for the claim is a paper by Bulbul et al which is, of course, freely available on the arXiv.

The key graph from that paper is this:

The claimed feature – it stretches the imagination considerably to call it a “line” – is shown in red. No, I’m not particularly impressed either, but this is what passes for high-quality data in X-ray astronomy!

Anyway, there has just appeared on the arXiv a paper by the Hitomi Collaboration describing what are basically the only set of science results that the Hitomi satellite managed to obtain before it fell to bits earlier this year. These were observations of the Perseus Cluster.

Here is the abstract:

High-resolution X-ray spectroscopy with Hitomi was expected to resolve the origin of the faint unidentified E=3.5 keV emission line reported in several low-resolution studies of various massive systems, such as galaxies and clusters, including the Perseus cluster. We have analyzed the Hitomi first-light observation of the Perseus cluster. The emission line expected for Perseus based on the XMM-Newton signal from the large cluster sample under the dark matter decay scenario is too faint to be detectable in the Hitomi data. However, the previously reported 3.5 keV flux from Perseus was anomalously high compared to the sample-based prediction. We find no unidentified line at the reported flux level. The high flux derived with XMM MOS for the Perseus region covered by Hitomi is excluded at >3-sigma within the energy confidence interval of the most constraining previous study. If XMM measurement uncertainties for this region are included, the inconsistency with Hitomi is at a 99% significance for a broad dark-matter line and at 99.7% for a narrow line from the gas. We do find a hint of a broad excess near the energies of high-n transitions of Sxvi (E=3.44 keV rest-frame) – a possible signature of charge exchange in the molecular nebula and one of the proposed explanations for the 3.5 keV line. While its energy is consistent with XMM pn detections, it is unlikely to explain the MOS signal. A confirmation of this interesting feature has to wait for a more sensitive observation with a future calorimeter experiment.

And here is the killer plot:

The spectrum looks amazingly detailed, which makes the demise of Hitomi all the more tragic, but the 3.5 keV is conspicuous by its absence. So there you are, yet another supposedly significant feature that excited a huge amount of interest turns out to be nothing of the sort. To be fair, as the abstract states, the anomalous line was only seen by stacking spectra of different clusters and might still be there but too faint to be seen in an individual cluster spectrum. Nevertheless I’d say the probability of there being any feature at 3.5 keV has decreased significantly after this observation.

P.S. rumours suggest that the 750 GeV diphoton “excess” found at the Large Hadron Collider may be about to meet a similar fate.

## The Distribution of Cauchy

Posted in Bad Statistics, The Universe and Stuff with tags , , , , , on April 6, 2016 by telescoper

Back into the swing of teaching after a short break, I have been doing some lectures this week about complex analysis to theoretical physics students. The name of a brilliant French mathematician called Augustin Louis Cauchy (1789-1857) crops up very regularly in this branch of mathematics, e.g. in the Cauchy integral formula and the Cauchy-Riemann conditions, which reminded me of some old jottings aI made about the Cauchy distribution, which I never used in the publication to which they related, so I thought I’d just quickly pop the main idea on here in the hope that some amongst you might find it interesting and/or amusing.

What sparked this off is that the simplest cosmological models (including the particular one we now call the standard model) assume that the primordial density fluctuations we see imprinted in the pattern of temperature fluctuations in the cosmic microwave background and which we think gave rise to the large-scale structure of the Universe through the action of gravitational instability, were distributed according to Gaussian statistics (as predicted by the simplest versions of the inflationary universe theory).  Departures from Gaussianity would therefore, if found, yield important clues about physics beyond the standard model.

Cosmology isn’t the only place where Gaussian (normal) statistics apply. In fact they arise  fairly generically,  in circumstances where variation results from the linear superposition of independent influences, by virtue of the Central Limit Theorem. Thermal noise in experimental detectors is often treated as following Gaussian statistics, for example.

The Gaussian distribution has some nice properties that make it possible to place meaningful bounds on the statistical accuracy of measurements made in the presence of Gaussian fluctuations. For example, we all know that the margin of error of the determination of the mean value of a quantity from a sample of size $n$ independent Gaussian-dsitributed varies as $1/\sqrt{n}$; the larger the sample, the more accurately the global mean can be known. In the cosmological context this is basically why mapping a larger volume of space can lead, for instance, to a more accurate determination of the overall mean density of matter in the Universe.

However, although the Gaussian assumption often applies it doesn’t always apply, so if we want to think about non-Gaussian effects we have to think also about how well we can do statistical inference if we don’t have Gaussianity to rely on.

That’s why I was playing around with the peculiarities of the Cauchy distribution. This distribution comes up in a variety of real physics problems so it isn’t an artificially pathological case. Imagine you have two independent variables $X$ and $Y$ each of which has a Gaussian distribution with zero mean and unit variance. The ratio $Z=X/Y$ has a probability density function of the form

$p(z)=\frac{1}{\pi(1+z^2)}$,

which is a Cauchy distribution. There’s nothing at all wrong with this as a distribution – it’s not singular anywhere and integrates to unity as a pdf should. However, it does have a peculiar property that none of its moments is finite, not even the mean value!

Following on from this property is the fact that Cauchy-distributed quantities violate the Central Limit Theorem. If we take $n$ independent Gaussian variables then the distribution of sum $X_1+X_2 + \ldots X_n$ has the normal form, but this is also true (for large enough $n$) for the sum of $n$ independent variables having any distribution as long as it has finite variance.

The Cauchy distribution has infinite variance so the distribution of the sum of independent Cauchy-distributed quantities $Z_1+Z_2 + \ldots Z_n$ doesn’t tend to a Gaussian. In fact the distribution of the sum of any number of  independent Cauchy variates is itself a Cauchy distribution. Moreover the distribution of the mean of a sample of size $n$ does not depend on $n$ for Cauchy variates. This means that making a larger sample doesn’t reduce the margin of error on the mean value!

This was essentially the point I made in a previous post about the dangers of using standard statistical techniques – which usually involve the Gaussian assumption – to distributions of quantities formed as ratios.

We cosmologists should be grateful that we don’t seem to live in a Universe whose fluctuations are governed by Cauchy, rather than (nearly) Gaussian, statistics. Measuring more of the Universe wouldn’t be any use in determining its global properties as we’d always be dominated by cosmic variance

## The Insignificance of ORB

Posted in Bad Statistics with tags , , , on April 5, 2016 by telescoper

A piece about opinion polls ahead of the EU Referendum which appeared in today’s Daily Torygraph has spurred me on to make a quick contribution to my bad statistics folder.

The piece concerned includes the following statement:

David Cameron’s campaign to warn voters about the dangers of leaving the European Union is beginning to win the argument ahead of the referendum, a new Telegraph poll has found.

The exclusive poll found that the “Remain” campaign now has a narrow lead after trailing last month, in a sign that Downing Street’s tactic – which has been described as “Project Fear” by its critics – is working.

The piece goes on to explain

The poll finds that 51 per cent of voters now support Remain – an increase of 4 per cent from last month. Leave’s support has decreased five points to 44 per cent.

This conclusion is based on the results of a survey by ORB in which the number of participants was 800. Yes, eight hundred.

How much can we trust this result on statistical grounds?

Suppose the fraction of the population having the intention to vote in a particular way in the EU referendum is $p$. For a sample of size $n$ with $x$ respondents indicating that they hen one can straightforwardly estimate $p \simeq x/n$. So far so good, as long as there is no bias induced by the form of the question asked nor in the selection of the sample which, given the fact that such polls have been all over the place seems rather unlikely.

A little bit of mathematics involving the binomial distribution yields an answer for the uncertainty in this estimate of $p$ in terms of the sampling error:

$\sigma = \sqrt{\frac{p(1-p)}{n}}$

For the sample size of 800 given, and an actual value $p \simeq 0.5$ this amounts to a standard error of about 2%. About 95% of samples drawn from a population in which the true fraction is $p$ will yield an estimate within $p \pm 2\sigma$, i.e. within about 4% of the true figure. In other words the typical variation between two samples drawn from the same underlying population is about 4%. In other other words, the change reported between the two ORB polls mentioned above can be entirely explained by sampling variation and does not at all imply any systematic change of public opinion between the two surveys.

I need hardly point out that in a two-horse race (between “Remain” and “Leave”) an increase of 4% in the Remain vote corresponds to a decrease in the Leave vote by the same 4% so a 50-50 population vote can easily generate a margin as large as  54-46 in such a small sample.

Why do pollsters bother with such tiny samples? With such a large margin error they are basically meaningless.

I object to the characterization of the Remain campaign as “Project Fear” in any case. I think it’s entirely sensible to point out the serious risks that an exit from the European Union would generate for the UK in loss of trade, science funding, financial instability, and indeed the near-inevitable secession of Scotland. But in any case this poll doesn’t indicate that anything is succeeding in changing anything other than statistical noise.

Statistical illiteracy is as widespread amongst politicians as it is amongst journalists, but the fact that silly reports like this are commonplace doesn’t make them any less annoying. After all, the idea of sampling uncertainty isn’t all that difficult to understand. Is it?

And with so many more important things going on in the world that deserve better press coverage than they are getting, why does a “quality” newspaper waste its valuable column inches on this sort of twaddle?

## More fine structure silliness …

Posted in Bad Statistics, The Universe and Stuff with tags , on March 17, 2016 by telescoper

Wondering what had happened to claims of a spatial variation of the fine-structure constant?

Well, they’re still around but there’s still very little convincing evidence to support them, as this post explains…

I noticed this paper by Pinho & Martins on astro ph today (accepted to Phys Lett B) concerning the alleged spatial variation of the fine structure constant; I say alleged but from reading this paper you’d think it was a settled debate with only the precise functional form of the final spatial model left to be decided.  In this latest instalment the authors propose to consider what updates can be made to the parameters of the spatial dipole model given 10 new quasar absorption datapoints (along 7 unique sight lines) drawn from post-Webb et al. studies published in the recent literature, with “the aim of ascertaining whether the evidence for the dipolar variation is preserved”.  Which, since they don’t consider the possibility of systematic errors* in the Webb et al. dataset, it is … since 10 data points with slightly lower standard measurement errors—and supposedly lower systematic errors—cannot trump the couple of hundred original measurements…

View original post 303 more words

## A Bump at the Large Hadron Collider

Posted in Bad Statistics, The Universe and Stuff with tags , , , on December 16, 2015 by telescoper

Very busy, so just a quickie today. Yesterday the good folk at the Large Hadron Collider announced their latest batch of results. You can find the complete set from the CMS experiment here and from ATLAS here.

The result that everyone is talking about is shown in the following graph, which shows the number of diphoton events as a function of energy:

Attention is focussing on the apparent “bump” at around 750 GeV; you can find an expert summary by a proper particle physicist here and another one here.

It is claimed that the “significance level” of this “detection” is 3.6σ. I won’t comment on that precise statement partly because it depends on the background signal being well understood but mainly because I don’t think this is the right language in which to express such a result in the first place. Experimental particle physicists do seem to be averse to doing proper Bayesian analyses of their data.

However if you take the claim in the way such things are usually presented it is roughly equivalent to a statement that the odds against this being a real detection are greater that 6000:1. If any particle physicists out there are willing to wager £6000 for £1 of mine that this result will be confirmed by future measurements then I’d happily take them up on that bet!

P.S. Entirely predictably there are 10 theory papers on today’s ArXiv offering explanations of the alleged bump, none of which says that it’s a noise feature..

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# 5.2. Procedures and Functions¶

We have seen string functions like lower() which returns a new string with all lowercase characters. Functions return a value. The Screen() and Turtle() in the code below both create objects and return them, so they are functions. You can also have procedures which do some action, but don’t return anything. The forward(75) and left(90) below are both procedures since they do an action, but don’t return anything.

Note

Some Python books don’t make a distinction between procedures and functions and will call both of these functions. In this book we are using function only for named code that returns a value and procedure for named code that doesn’t return a value.

Note

In these next problems we will draw a letter similar to how you might do it without lifting your pen or pencil. See the numbers and arrows on the letters for the order that the lines are drawn.

Mixed up programs

The following program uses a turtle to draw a capital T as shown below but the lines are mixed up. The program should do all necessary set-up: import the turtle module, get the space to draw on, and create the turtle. After that the turtle should draw the lines in the numbered order as shown in the picture on the left. Drag the needed blocks of statements from the left column to the right column and put them in the right order. Then click on Check Me to see if you are right. You will be told if any of the lines are in the wrong order or are the wrong blocks.

Mixed up programs

The following program uses a turtle to draw a capital A as shown below, but the lines are mixed up. The program should do all necessary set-up: import the turtle module, get the space to draw on, and create the turtle. After that the turtle should draw the lines in the numbered order as shown in the picture on the left. Drag the needed blocks of statements from the left column to the right column and put them in the right order. Then click on Check Me to see if you are right. You will be told if any of the lines are in the wrong order or are the wrong blocks.

Note

Discuss topics in this section with classmates.


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• question_answer Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero) A) Adiabatic process : $\Delta U=-\,W$ B) Cyclic process : $q=-\,W$ C) Isothermal process : $q=-W$ D) Isochoric process : $\Delta U=q$

 1st law of thermodynamics $\Delta E=q+w$ 1. Adiabatic process $q=0,(\Delta E=w)or\,(\Delta U=w)$ 2. Cyclic process $\Delta E=0,(q=-w)$ 3. Isothermal process $\Delta E=n{{C}_{V}}\Delta T.(\Delta T=0)$ $\Delta E=0,(q=-w)$ 4. Isochoric process $\Delta V=0$ $W=P\Delta V=0$ $\Delta E=q.$


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Last edited by Mikat
Monday, November 9, 2020 | History

6 edition of Signal processing with alpha-stable distributions and applications found in the catalog.

# Signal processing with alpha-stable distributions and applications

Written in English

Subjects:
• Signal processing -- Mathematics.,
• Distribution (Probability theory)

• Edition Notes

Classifications The Physical Object Statement Chrysostomos L. Nikias, Min Shao. Series Adaptive and learning systems for signal processing, communications, and control Contributions Shao, Min. LC Classifications TK5102.9 .N543 1995 Pagination xiii, 168 p. : Number of Pages 168 Open Library OL1270310M ISBN 10 047110647X LC Control Number 95001482

In this paper a novel method for informative frequency band selection for local damage detection is presented. Local damage in bearings/gearbox provides specific vibration signature, i.e. train of impulses with cycle related to fault frequency. The proposed approach is based on the α-stable distribution, which is an extension of the Gaussian by: Advanced Signal Processing and Coordination Applied to Electronic Support Measures Navy SBIR - Topic N NAVAIR Nikias, C. and Shao, M. “Signal processing with alpha-stable distributions and applications”. New York: Wiley-Interscience, ISBN: 0 .

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### Signal processing with alpha-stable distributions and applications by Chrysostomos L. Nikias Download PDF EPUB FB2

This book offers, for the first time, a full and lucid introduction to a very useful type of these non-Gaussian models, namely those specified by alpha-stable distributions. Signal Processing with Alpha-Stable Distributions and Applications presents a long-awaited survey of the statistical properties, methods, and applications of symmetrical Cited by: The book is intended as an introduction to the use of alpha-stable distribution models by researchers and engineers with a background in signal processing.

Each chapter is followed by problems that require an understanding of the concepts presented and the ability to map the techniques to real-world situations. (source: Nielsen Book Data) Summary Presenting a survey of the statistical properties, methods and applications of symmetrical alpha-stable distributions, this study emphasizes the practical rather than the theoretical aspects of signal processing.

Alpha-Stable Distributions in Signal Processing of Audio Signals Preben Kidmose, Department of Mathematical Modelling, Section for Digital Signal Processing, Technical University of Denmark, BuildingDK Lyngby, Denmark Abstract First, we propose two versions of a sliding window, block based parameter estimator for estimating the.

Signal processing with alpha-stable distributions and applications. [Chrysostomos L Nikias; Min Shao] Stable Models for Impulsive Noise -- Signal Detection in Stable Noise -- Current and Future Trends in Signal Processing with Alpha-Stable Distributions\/span>\"@ en\/a> ; \u00A0 # Adaptive and learning systems for signal.

This book offers, for the first time, a full and lucid introduction to a very useful type of these non–Gaussian models, namely those specified by alpha–stable distributions. Signal Processing with Alpha–Stable Distributions and Applications presents a long–awaited survey of the statistical properties, methods, and applications of Book Edition: 1.

Signal Processing with Alpha-stable Distributions and Applications by Min Shao,available at Book Depository with free delivery worldwide.4/5(1).

In many practical applications, there is a natural upper bound that truncates the prob-ability tail. This paper derives estimators for the truncated Pareto distribution, investigates their properties, and illustrates a way to check for fit.

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III Regression, Time Series, Signal Processing and Stable Processes 11 Signal Processing with Stable Distributions This book describes the basic facts about univariate and multivariate stable distributions, with an emphasis on practical applications. Part I focuses on univariate stable Size: KB. Wideband array signal processing with alpha-stable distributions. To help engineers better understand the stable models and develop methodologies for their applications in signal processing. Signal Processing With Alpha Stable Distributions And Applications Adaptive And Cognitive Dynamic Systems Signal Processing Learning Communications And Control. The Signal Processing Laboratory has had long involvement in audio and music processing. Early work in sound restoration here in the 's led to the founding of the successful company CEDAR Audio Ltd. which produces DSP equipment for remastering and enhancement of sound in the recording, broadcast and forensic industries. Nikias, “Signal Processing with Alpha-Stable Distributions and Applications,” Proceedings of the IEEE Workshop on Nonlinear Signal and Image Processing, Neos Marmaras, Halkidiki, Greece, June Alpha‐Stable distributions have received wide interest in the signal processing community and became state of the art models for impulsive noise and internet traffic in the last 20 years since the influential paper of Nikias and Shao in Book review Full text access Signal processing with alpha-stable distributions and applications: C.L. Nikias and Min Shoa (): Wiley, ISBN x, £pp. Walter Krämer. Advances in nonlinear signal processing A paper on using stable distributions in signal processing. It shows that significant improvement can be made in the presence of heavy tailed noise. ( mb) Modeling financial data with stable distributions A paper from on using stable distributions in finance (with some later corrections). Alpha-Stable Matrix Factorization Article (PDF Available) in IEEE Signal Processing Letters 22(12) September with 67 Reads How we measure 'reads'. Book Description. Nonlinear signal and image processing methods are fast emerging as an alternative to established linear methods for meeting the challenges of increasingly sophisticated applications. Advances in computing performance and nonlinear theory are making nonlinear techniques not only viable, but practical. Signal processing with alpha-stable distributions and applications: C.L. Nikias and Min Shoa (): Wiley, ISBN x, [pound sign]pp. Author & abstract Download. Signal Processing with Alpha-Stable Distributions and Applications presents a long-awaited survey of the statistical properties, methods, and applications of symmetrical alpha-stable distributions. Its emphasis on practical rather than theoretical aspects will appeal to both researchers and practicing engineers in signal processing. xiii, p.: ill. ; 25 cm. Signal. Signal Processing with Alpha-Stable Distributions and Applications (Adaptive and Cognitive Dynamic Systems: Signal Processing, Learning, Communications and Control) by Chrysostomos L. Nikias, Min Shao. 作者: Chrysostomos L. Nikias Min Shao isbn: 书名: Signal Processing with Alpha-Stable Distributions and Applications 定价: USD 出版社: Wiley-Interscience 装帧: Hardcover 出版年: >. the Gaussian is the class of -stable distributions. InNikias and Shao [1] gave an introductory review of -stable distributions from a statistical signal processing viewpoint that was followed by a book from the same authors in [2]. Alpha-stable distributions have been used to model diverseFile Size: 1MB. This paper describes an improved Constant False Alarm Rate (CFAR) ship detection algorithm in spaceborne synthetic aperture radar (SAR) image based on Alphastable distribution model. Typically, the CFAR algorithm uses the Gaussian distribution model to describe statistical characteristics of a SAR image background clutter. However, the Gaussian distribution is only Cited by: Adaptive Signal Processing presents the next generation of algorithms that will produce these desired results, with an emphasis on important applications and theoretical advancements. This highly unique resource brings together leading authorities in the field writing on the key topics of significance, each at the cutting edge of its own area. This book describes the basic facts about univariate and multivariate stable distributions, with an emphasis on practical applications. Part I focuses on univariate stable Size: KB. Nolan, J. (), Fitting data and assessing goodness of fit with stable distributions, in Proceedings of the Conference on Applications of Heavy Tailed Distributions in Eco- nomics, Engineering and Statistics, American University, Washington, DC, June 3–5, Cited by: C. Nikias and M. Shao, Signal processing with alpha-stable distributions and applications. Adaptive and learning systems for signal processing, communications, and control, Chicester: Wiley, Google ScholarAuthor: Ludvík Tesař, Anthony Quinn. Can one use a non-Gaussian$\alpha\$-stable distribution (e.g. the Cauchy distribution) instead, and still construct a Stack Exchange Network Stack Exchange network consists of Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Digital Signal Process () Coefficient of relationship for two symmetric alpha-stable variables when alpha is in the interval (1,2]. Communications in Statistics - Theory and MethodsCited by: IEEE TRANSACTIONS ON SIGNAL PROCESSING, VOL.

46, NO. 6, JUNE Analytic Alpha-Stable Noise Modeling in a Poisson Field of Interferers or Scatterers Jacek Ilow, Member, IEEE, and Dimitrios Hatzinakos, Member, IEEE Abstract— This paper addresses non-Gaussian statistical mod-eling of interference as a superposition of a large number of small.

Alpha stable noise, also called Lévy noise, was put forward by Lévy when he studied the Generalized Central Limit Theorem. Compared with Gaussian distribution, alpha stable distribution is better when applied to describe non-Gaussian signals with impulsive properties and a heavy tail, because alpha stable distribution fits these properties better than Gaussian Cited by: 8.

Breiman's probability book solve the general version of this problem (consider stable only, not symmetric). However, it uses infinite divisible distributions and lot of calculations.

So I was wondering if there is a short argument for the symmetric case. My try: Note that using the equal in distribution relation. 'Foundations of Signal Processing is a pleasure to read. Drawing on the authors’ rich experience of research and teaching of signal processing and signal representations, it provides an intellectually cohesive and modern view of the subject from the.

Alpha-Stable distributions in MATLAB The following gives a brief introduction to the Levy alpha-stable distribtuion and some MATLAB functions I've written pertaining to this distribtuion. There are several issues which make dealing with these distribtuions difficult, including infinite means and variances, and the fact that the pdf or cdf.

This book details recent advances in nonlinear theory and methods and explores an array of modern signal and image processing applications. The first several chapters focus on nonlinear signal processing theory, targeting three critical areas: filter analysis, nonlinear filter class design, and signal analysis.

The use of families of generalised Gaussian and alpha-stable distributions for modelling image wavelet Parallel Computation Booth Multiplier Design for Digital Signal Processing Applications free download Abstract Multiplication is an important fundamental function in .Statistical characterization of acoustic signals is a pre-processing technique aiming at the definition of signal observables, to be used as input data of appropriately defined inverse problems aiming at the estimation of critical parameters of the marine : Michael I.

Taroudakis.IEEE TRANSACTIONS ON SIGNAL PROCESSING, VOL. 53, NO. 3, MARCH The Stability Test for Symmetric Alpha-Stable Distributions Ramon F. Brcich, Member, IEEE, D. Robert Iskander, Senior Member, IEEE, and Abdelhak M. Zoubir, Senior Member, IEEE Abstract—Symmetric alpha-stable distributions are a popular statistical model for heavy-tailed .


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Outlook: HIGHWAY CAPITAL PLC is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Buy
Time series to forecast n: 24 Feb 2023 for (n+1 year)
Methodology : Statistical Inference (ML)

Abstract

HIGHWAY CAPITAL PLC prediction model is evaluated with Statistical Inference (ML) and Stepwise Regression1,2,3,4 and it is concluded that the LON:HWC stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Buy

Key Points

1. Can neural networks predict stock market?
2. Fundemental Analysis with Algorithmic Trading
3. What are buy sell or hold recommendations?

LON:HWC Target Price Prediction Modeling Methodology

We consider HIGHWAY CAPITAL PLC Decision Process with Statistical Inference (ML) where A is the set of discrete actions of LON:HWC stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Stepwise Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Statistical Inference (ML)) X S(n):→ (n+1 year) $\begin{array}{l}\int {e}^{x}\mathrm{rx}\end{array}$

n:Time series to forecast

p:Price signals of LON:HWC stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

LON:HWC Stock Forecast (Buy or Sell) for (n+1 year)

Sample Set: Neural Network
Stock/Index: LON:HWC HIGHWAY CAPITAL PLC
Time series to forecast n: 24 Feb 2023 for (n+1 year)

According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Buy

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

IFRS Reconciliation Adjustments for HIGHWAY CAPITAL PLC

1. When an entity designates a financial liability as at fair value through profit or loss, it must determine whether presenting in other comprehensive income the effects of changes in the liability's credit risk would create or enlarge an accounting mismatch in profit or loss. An accounting mismatch would be created or enlarged if presenting the effects of changes in the liability's credit risk in other comprehensive income would result in a greater mismatch in profit or loss than if those amounts were presented in profit or loss
2. If a financial instrument is designated in accordance with paragraph 6.7.1 as measured at fair value through profit or loss after its initial recognition, or was previously not recognised, the difference at the time of designation between the carrying amount, if any, and the fair value shall immediately be recognised in profit or loss. For financial assets measured at fair value through other comprehensive income in accordance with paragraph 4.1.2A, the cumulative gain or loss previously recognised in other comprehensive income shall immediately be reclassified from equity to profit or loss as a reclassification adjustment.
3. At the date of initial application, an entity shall use reasonable and supportable information that is available without undue cost or effort to determine the credit risk at the date that a financial instrument was initially recognised (or for loan commitments and financial guarantee contracts at the date that the entity became a party to the irrevocable commitment in accordance with paragraph 5.5.6) and compare that to the credit risk at the date of initial application of this Standard.
4. Paragraph 5.7.5 permits an entity to make an irrevocable election to present in other comprehensive income subsequent changes in the fair value of particular investments in equity instruments. Such an investment is not a monetary item. Accordingly, the gain or loss that is presented in other comprehensive income in accordance with paragraph 5.7.5 includes any related foreign exchange component.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

Conclusions

HIGHWAY CAPITAL PLC is assigned short-term Ba1 & long-term Ba1 estimated rating. HIGHWAY CAPITAL PLC prediction model is evaluated with Statistical Inference (ML) and Stepwise Regression1,2,3,4 and it is concluded that the LON:HWC stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Buy

LON:HWC HIGHWAY CAPITAL PLC Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementCBaa2
Balance SheetCaa2B1
Leverage RatiosCaa2Baa2
Cash FlowBaa2Baa2
Rates of Return and ProfitabilityB1Baa2

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

Prediction Confidence Score

Trust metric by Neural Network: 81 out of 100 with 852 signals.

References

1. Mnih A, Hinton GE. 2007. Three new graphical models for statistical language modelling. In International Conference on Machine Learning, pp. 641–48. La Jolla, CA: Int. Mach. Learn. Soc.
2. Zeileis A, Hothorn T, Hornik K. 2008. Model-based recursive partitioning. J. Comput. Graph. Stat. 17:492–514 Zhou Z, Athey S, Wager S. 2018. Offline multi-action policy learning: generalization and optimization. arXiv:1810.04778 [stat.ML]
3. R. Sutton, D. McAllester, S. Singh, and Y. Mansour. Policy gradient methods for reinforcement learning with function approximation. In Proceedings of Advances in Neural Information Processing Systems 12, pages 1057–1063, 2000
4. Bierens HJ. 1987. Kernel estimators of regression functions. In Advances in Econometrics: Fifth World Congress, Vol. 1, ed. TF Bewley, pp. 99–144. Cambridge, UK: Cambridge Univ. Press
5. Chernozhukov V, Chetverikov D, Demirer M, Duflo E, Hansen C, et al. 2018a. Double/debiased machine learning for treatment and structural parameters. Econom. J. 21:C1–68
6. O. Bardou, N. Frikha, and G. Pag`es. Computing VaR and CVaR using stochastic approximation and adaptive unconstrained importance sampling. Monte Carlo Methods and Applications, 15(3):173–210, 2009.
7. Burkov A. 2019. The Hundred-Page Machine Learning Book. Quebec City, Can.: Andriy Burkov
Frequently Asked QuestionsQ: What is the prediction methodology for LON:HWC stock?
A: LON:HWC stock prediction methodology: We evaluate the prediction models Statistical Inference (ML) and Stepwise Regression
Q: Is LON:HWC stock a buy or sell?
A: The dominant strategy among neural network is to Buy LON:HWC Stock.
Q: Is HIGHWAY CAPITAL PLC stock a good investment?
A: The consensus rating for HIGHWAY CAPITAL PLC is Buy and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of LON:HWC stock?
A: The consensus rating for LON:HWC is Buy.
Q: What is the prediction period for LON:HWC stock?
A: The prediction period for LON:HWC is (n+1 year)


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Performing p-value Hypothesis Tests

This blog is aimed at students and focuses on how to perform a hypothesis test using the p-value approach.

What is a p-value?

The p-value is the probability of getting a sample statistic as extreme or more extreme than the one obtained if the null hypothesis is true.

For example, if you were testing the claim that a population mean was less that 25 and your sample of 144 observations had a mean of 21 with a standard deviation of 2, the p-value is the probability that a sample mean is less than 21 under those conditions.

The lower this probability is, the less likely it is that the null hypothesis is true.

Step 1 – State the null and alternate hypotheses.

• List the null hypothesis $H_0$ and the alternate hypothesis $H_1$.
• The null hypothesis is assumed to be true, and we look for evidence (a low p-value) that the alternate hypothesis is true.
• The null hypothesis always contains equality, and the alternate hypothesis will always contain one of these three symbols: $<, >, \ne$.

Step 2 – Establish a level of significance.

• The level of significance, $\alpha$, is the equivalent of reasonable doubt in a jury trial.
• If the p-value is less than $\alpha$, then we consider this to be strong evidence that the null hypothesis is false (Reject $H_0$) and the alternate hypothesis is true (Support $H_1$).
• In homework problems the level of significance is provided, but when performing your own hypothesis test choose a level of significance that corresponds to the consequences of making a wrong decision. Choosing $\alpha = 0.05$ is pretty standard, and you have a 5% chance of rejecting a null hypothesis that is actually true. If the consequences of making such an error are high (someone could get sick, you could be sued, …), consider using $\alpha = 0.01$.

Step 3 – Choose a test statistic.

• The test statistic is a formula we use to calculate the p-value for a hypothesis test.
• The formula we use depends on the test we are performing. For instance, if you were performing a 1-mean t test, the test statistic would be:
$t = \frac {\bar{x} - \mu}{s/\sqrt{n}}$
• In a later blog I will share advice on how to select the correct test statistic.

Step 4 – Establish the decision rule.

• The decision rule is a rule that determines when the null hypothesis will be rejected.
• When using the p-value approach, the decision rule is always “Reject $H_0$ if p-value < $\alpha$.”

Step 5 – Calculations, Decision about $H_0$, Conclusion

• By Hand
To calculate the p-value, begin by calculating the test statistic.
Use that test statistic and the appropriate probability table, along with the direction of the test suggested by $H_1$, to calculate the p-value.
• By Technology
Whether you are using StatCrunch, Excel, Minitab, the TI-83 or TI-84 calculator, or some other statistical package, the p-value will be calculated for you. (In today’s world, you should be using technology to find p-values!)
• If the p-value is less than the level of significance, we reject $H_0$ and support $H_1$. Otherwise we fail to reject $H_0$ (we cannot conclude $H_0$ is false) and we fail to support $H_1$ (we cannot conclude that $H_1$ is true).
• Finish by writing a conclusion about the claim you were testing.

Summary

I hope this is helpful to you. In a later blog I will go over how to choose the appropriate test (and test statistic). I will also go over how to perform hypothesis tests using StatCrunch. I’ll conclude by saying that StatCrunch is finally allowing me to teach this course the way it should be taught. If you want to see what StatCrunch is all about for yourself, check out their web site at http://StatCrunch.com .

-George

I am a math instructor at College of the Sequoias in Visalia, CA. Each Wednesday I post an article related to General Teaching on my blog. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website – http://georgewoodbury.com


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# Getting started with ikiwiki

## Installation

This is pretty simple, but I want the version from jessie/sid. After installation, there are bunch of helper scripts to setup some easy use cases for ikiwiki.

But if you move things about you might manually touch your ikiwiki setup file, and regenerate the wiki. The webserver just needs to render static pages, unless you want to edit things via a browser, in which case you need to set up your webserver to use cgi.

### Nested Directives

This took me a little while to understand, but to use the !more directive, you need to wrap most of your text inside it. The trick is to do

[[!more   text=<<FOO

... your long post and commands here

FOO]]

BUT, I find that tags won't autocreate themselves unless they are outside the block


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# Circle Question

Printable View

• Aug 3rd 2007, 11:57 AM
charrie berri
Circle Question
Graph the circle (x+2)^2 + (y-1)^2= 25
• Aug 3rd 2007, 12:18 PM
earboth
Quote:

Originally Posted by charrie berri
Graph the circle (x+2)^2 + (y-1)^2= 25

Determine the equations of the tangents with slope of 3/4 =S

Hello,

1. It is a circle with the centre C(-2, 1) and the radius 5.

2. The radius is perpendicular to the tangent in the tangent point. Thus the radius has the slope -4/3 and has to pass the centre:

$r: y=-\frac{4}{3}x-\frac{5}{3}$

Calculate the intersection of this line and the circle and you'll get the tangent points:

$(x+2)^2+\left(-\frac{4}{3}x-\frac{5}{3}-1\right)^2 = 25$. After a few transformation you get:
$\frac{25}{9}x^2 + \frac{100}{9}x - \frac{125}{9}=0~\Longrightarrow~ x^2+4x-5=0$. Solve for x. Plug in the values into the equation of the straight line r. The tangent points are:
$T_1(1, -3),~~T_2(-5, 5)$

3. You know the slope of the tangent and one point of the tangent. Use the point-slope-formula.
$t_1:y = \frac{3}{4}x-\frac{15}{4}$

$t_2:y = \frac{3}{4}x+\frac{35}{4}$
• Aug 3rd 2007, 10:25 PM
earboth
Quote:

Originally Posted by charrie berri
Graph the circle (x+2)^2 + (y-1)^2= 25

Hello,

here is another approach:

You allways find 2 parallel tangents to a circle. The equation of these tangents differ only in the constant summand. With your values the tangents have the form:

$t: y=\frac{3}{4}x + C$

Now calculate the intersection points of these tangents and the circle. Plug in the tangent term into the equation of the circle:

$(x+2)^2+(\frac{3}{4}x + C-1)^2=25$ . Expand the brackets and after a few steps you'll get the equation:

$25 \cdot x^2 + 8 \cdot x \cdot (3 \cdot C + 5) + 16C - 32C = 320$. Solve for x:

$x=\frac{4\left(\sqrt{-16C^2+80C+525}-3C-5\right)}{25}~~\vee~~$ $~~ x=-\frac{4\left(\sqrt{-16C^2+80C+525}-3C-5\right)}{25}$

You get only one intersetion point (that is the tangent point) if the radicand equals zero:

$\sqrt{-16C^2+80C+525}=0$. Solve for C. You'll get:

$C=\frac{35}{4}~~\vee~~C=-\frac{15}{4}$ Plug in this value into the equation of the parallel lines and you have the result you know from my previous post.
• Aug 4th 2007, 10:56 PM
earboth
Quote:

Originally Posted by charrie berri
Graph the circle (x+2)^2 + (y-1)^2= 25

Hello,

in my previous post I mentioned that the centre of the circle is C(-2, 1) and the radius is 5. I have some difficulties to imagine what difficulties you could have to draw a circle if all data are present :confused:

But nevertheles here it is:

By the way: Don't delete your problem when you have received an answer. Other members of the forum could benefit from your problem and the way it is solved.


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Exponential integral

1. Apr 29, 2009

zetafunction

if we suppose that f(x) is Real then what condition should we impose to the integral

$$g(u) = \int_{-\infty}^{\infty}dx e^{iuf(x)}$$

in order to exists for every 'u' no matter if it gives a function or a distribution (i.e in case f(x)=x then we have a dirac delta)

also le us suppose we must calculate f(x) so $$g(u) = \int_{-\infty}^{\infty}dx e^{iuf(x)}$$ is correct for a certain given function or idstribution g(u)

my idea is that if we could expand for big 'u' the integral into a divergent series

$$g(u) = \sum _{n=0}^{\infty}a_{n}u^{-n}h(u)$$

for a certain h(u)

also if we knew the expression for f(x) and its inverse f(-1)(x) could we then make the change of variable x=f(-1)(r) in our integral equation to transform it into a Fourier integral ?


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When V is an affine algebraic set, we say that the quotient K$x₁, ..., xₙ$/I(V) is the affine algebra of V. I've always had trouble grasping the notion of quotients in group/ring theory and I really don't understand what these are. According to Wikipedia, the elements of a quotient R / I are equivalence classes with respect to the equivalence relation a\~b iff a-b ∈ I. By this logic, the elements of K$x₁, ..., xₙ$/I(V) should be equivalence classes of n-variable polynomials in which the equivalence class of a polynomial f is the set of polynomials g such that f(v) = g(v) for all v in I(V). My problem with this is that the lecture notes seem to treat K$x₁, ..., xₙ$/I(V) as a set of polynomials, not as a set of equivalence classes of polynomials. If we are to see the quotient as a set of polynomials, then **which** polynomials belong to K$x₁, ..., xₙ$/I(V)?
You can represent each equivalent class of polynomial with a polynomial, and this suffices in practice. Because the intuition behind the quotient is that 2 polynomials in the same equivalent class would evaluate to the same value on the affine variety (since we are quotienting out by all the polynomials that evaluate to identically 0 on the variety). So in the appropriate context (e.g. you're thinking about the variety itself, not its embedding in an affine space), then it's fine to conflate the polynomial with its equivalence class.
It's probably easiest to see this by example. Let's take a quadric curve in C^2 defined by

x^2 + y^2 - 1 = 0

which we cill call V.
Polynomials on C^2 are given by the ring C[x,y]. What about polynomials on V? Clearly, any polynomial on C^2 can be restricted to a polynomial on V, but this restriction isn't injective. For example, the polynomial x^2 + y^2 and the polynomial 1 both take the same value on V (hence are the same polynomial on V) but are clearly different when considering all of C^2 .

More generally, two polynomials p(x,y) and q(x,y) are the same on V if their difference is zero on V. This means (by some basic AG theorem) that

p(x,y)-q(x,y) =  f(x,y)(x^2 + y^2 - 1)

i.e. p(x,y)-q(x,y) lies in the ideal I(V).

Conversely, take a polynomial p(x,y) defined only on V. We want to write down a formula for this polynomial in terms of x and y, so we want to express it as a polynomial on all of C^2. As we have seen, we have many inequivalent choices of expressions for p(x,y), but all of them reduce to the same polynomial when restricted to V.

Hence, if we're talking about p(x,y) as a function on V, it is precise to instead refer to the set of all polynomials in two variables that restrict to the same function on V. If we want a formula, we may pick one from this set (the "representative"), but all choices are equivalent on V.

(This is why some people write p(x,y) for the polynomial on C^2 , and p(x,y) + I(V) for the polynomial on V, since any expression of the form p(x,y)+ i(x,y) for i in I(V) restricts to the same polynomial on V)


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## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 4 years ago A positive integer is picked randomly from 61 to 70, inclusive. What is the probability that it is prime?

• This Question is Closed
1. anonymous

pls explain the concept of prime no

2. anonymous

The numbers from 61 to 70 are: 61, 62, 63, 64, 65, 66, 67, 68, 69, 70 Prime numbers: 61, 63, 67, 69

3. anonymous

do you know what "prime number " means?

4. anonymous

i m confused

5. hoblos

the prime numbers are 61 & 67 only

6. anonymous

lol prime is not odd!

7. anonymous

prime means only divisors are 1 and itself, a whole number that you cannot factor

8. hoblos

so the probability is 2/10 = 1/5

9. anonymous

we can try the numbers one by one and check

10. anonymous

okay,pratu u were mistaken

11. anonymous

61, 62, 63, 64, 65, 66, 67, 68, 69, 70 61 is prime because it is not the product of any two numbers. think about the multiplication table. 61 does not appear as an entry there

12. anonymous

62 is even, so it is divisible by 2, not prime. and neither are 64,66,68,70 for the same reason

13. anonymous

63 is divisible by 3 because $63=3\times 21$

14. anonymous

why did u take such a big no?

15. anonymous

65 is obviously divisible by 5 67 is prime and finally 69 is divisible by 3

16. anonymous

I think a better question is how do you know if a number is a prime. We know that all even numbers are not prime. But which of the odd ones are prime. I guess the question is how would you calculate if a number is or is not prime? For example, 3335679 ?

17. anonymous

true

18. anonymous

Can this only be done by brute force. i.e. take each integer and start dividing by the smaller integers and then keep going until you reach an answer without a remainder? We need a mathematician.

19. anonymous

well sorry i m a lost in maths

20. anonymous

Yeah, I made a mistake and had some internet problem in the middle so couldn't reply. The others are right though.

21. hoblos

some notes that may help: if the number is even then it is divisible by 2 if the sum of digits of the number is divisible by 3 then the whole number is divisible by 3 if the number ends with 0 or 5 then it is divisible by 5 out of these cases you have to try dividing by PRIME numbers until you reach an answer without a remainder

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Browse Questions

# To obtain a p-type semiconductor germanium is doped with

$(a)\;Gallium \\ (b)\;Boron \\ (c)\;both\;gallium\;and\; Boron \\ (d)\;Neither\;gallium\;and\; Boron$

To obtain p-type semiconductor the doping element must be trivalent .
Both Gallium and Boron are trivalent and so can be used to dope Germanium to get p-type semiconductor.
Hence c is the correct answer.


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Feeds:
Posts

## On the Zakharov’s weak turbulence theory for capillary waves

In this paper with M.-B. Tran, we construct solutions to the following weak turbulence kinetic equation for capillary waves (cf. Hasselmann ’62, Zakharov ’67):

\displaystyle \begin{aligned} \partial_tf + 2 \nu |k|^2 f \ = \ Q[f] \end{aligned}

## Prandtl’s layer expansions for steady Navier-Stokes

In 1904, Prandtl conjectured that slightly viscous flows can be decomposed into the inviscid flows away from the boundary and a so-called Prandtl’s layer near the boundary. While various instabilities indicate the failure of the conjecture for unsteady flows (for instance, see Grenier 2000), recently with Y. Guo, we are able to prove that the conjecture holds for certain steady Navier-Stokes flows; see our paper which is to appear on Annals of PDEs.

## Inviscid limit for Navier-Stokes in a rough domain

In this paper with Gérard-Varet, Lacave, and Rousset, we prove the inviscid limit of Navier-Stokes flows in domains with a rough or oscillating boundary. Precisely, we study the 2D incompressible Navier-Stokes flows with small viscosity ${\nu}$, posed on the following rough domain:

## Graduate student seminar: Kinetic theory of gases

Lat week, I gave a graduate student seminar, whose purpose is to introduce to first and second year graduate students (at Penn State) an active and beautiful topics of research, and suggest a few possible ideas for students’ presentation later in the semester. Here are slides of my talk, which focuses on Kinetic Theory of Gases, a topics that I will teach as a graduate topics course, next fall (2017).

## The Maxwell-Boltzmann approximation for ion kinetic modeling

In this short paper with C. Bardos, F. Golse, and R. Sentis, we are aimed to justify the Maxwell-Boltzmann approximation from kinetic models, widely used in the literature for electrons density distribution; namely, the relation

$\displaystyle n_e =e^{q_{e} \phi / \theta }$

in which ${n_e, q_e, \theta,\phi}$ denote the electrons density, the elementary charge, the electron temperature, and the electric potential.

## Uniform lower bound for solutions to a quantum Boltzmann

With Minh-Binh Tran, we establish uniform lower bounds, by a Maxwellian, for positive radial solutions to a Quantum Boltzmann kinetic model for bosons; see arXiv preprint.

## Instabilities in the mean field limit

Together with Daniel Han-Kwan, we recently wrote a paper, entitled “Instabilities in the mean field limit”, to be published on the Journal of Statistical Physics (see here for a preprint). I shall explain our main theorem below.


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## mitchelsewbaran 2 years ago Which of the following is an exact value of cos 15° ?

1. mitchelsewbaran

2. Azteck

$\large \cos15=\cos(60-45)$ Double angle results. $\large\cos(60-45)=\cos60\cos45+\sin60\sin45$

3. Azteck

Now find out for cos15 after I gave you a headstart


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# Express the Following Decimal in the Form P/Q: 4.Bar7 - Mathematics

#### Question

Express the following decimal in the form p/q: 4.bar7

#### Solution

Let x=4.bar7

x=4+0.bar7

Let y = 0.bar7=0.777...

⇒ 10y = 7.777...

⇒ 10y = 7 + 0.777...

⇒ 10y = 7 + y

⇒ 10y - y = 7

⇒ 9y = 7

⇒ y = 7/9

therefore0.bar7=7/9

Therefore,

x=4+7/9

x=(36+7)/9

x=43/9

Hence, 4.bar7=43/9

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Solution for Mathematics for Class 9 (2018 (Latest))
Chapter 1: Number Systems
1.3 | Q: 2.6 | Page no. 22
Express the Following Decimal in the Form P/Q: 4.Bar7 Concept: Introduction of Real Number.


text:
# Math Help - word problem equations

1. ## word problem equations

I hope you guys can help me... because it's easy but for some reason I can't figure it out? -.-

1) Two hundred people attend a concert. The total receipts are $4500. If one student ticket costs$5, and one adult ticket costs $30, find the number of students who attend the concert? 2) A farmer raises some pigs and chickens. Among the pigs and chickens there are 24 heads and 64 feet. How many pigs and chickens are there? Substitution only please. Come to think of it... maybe you guys can just do one and I can use your steps from the first one to do second 2. from question 1, u can see that if x is number of student and y is number of adults then 200 = x + y, and so$4500 = $5x +$30y then use the substitution x = 200 - y
and go from there

3. Let $x=$the Number of tickets sold

$5x+30x=4500$
$\Rightarrow 35x=4500$
$\Rightarrow x=128.57$ Round up if you like
$(128.57)(5)=642.85$

Roughly 642 students attended the concert assuming my equation is correct.

4. OZzman u havent taken into acount the adults also attending!
and it says in total only 200 ppl attended

5. Oh yea you're right.

6. $4500=5(200-y)+30y$
$4500=1000-5y+30y$
$4500=1000+25y$
$3500=25y$
$y=140$

$200-140=60$
$60$ Students attended

7. Originally Posted by mathmonster
from question 1, u can see that if x is number of student and y is number of adults then 200 = x + y,
and so $4500 =$5x + \$30y then use the substitution x = 200 - y
and go from there
Where did you get 200 from.

8. Alright, so for question #2, there are 8 pigs and 16 chickens.

Now, can you guys help me with the harder questions? XD

Solve using substitution

1)
x-5y=8
4x+26=10

2)
3x+2y=16
3y-x=13

--

Solve using elimination

1)
2u=3v
u+2v=7

2)
-4x+3y=17
3x-2y=10

--

1) Given two numbers whose difference is 20 such that twice on number equals three times the other. Find the numbers.

Thanks!

9. Substitution #2

$3x+2y=16$

$3y-x=13\quad\Rightarrow -x=(13-3y)\quad\Rightarrow x=-13+3y$

$3(-13+3y)+2y=16\quad\Rightarrow -39+9y+2y=16\quad\Rightarrow 11y=55\quad\Rightarrow y=5$

$3(5)-x=13\quad\Rightarrow 15-x=13\quad\Rightarrow -x=-2\quad\Rightarrow x=2$

Elimination #1

$2u=3v\quad\Rightarrow 2u-3v=0$

$u+2v=7\quad\Rightarrow (u+2v=7)-2\quad\Rightarrow -2u-4v=-14$

Now add the equations $2u-3v=0$ and the $-2u-4v=14$ the u's will cross out leaving v as the only variable

$-7v=-14\quad\Rightarrow v=2$

Now plug v back into the original equation to solve for the other variable $u+2(2)=7\quad\Rightarrow u+4=7\quad\Rightarrow u=3$

10. Can you help me do the other ones too? Because I have looked at them and they are nowhere near similar...

[EDIT] Wait I got the substitution one #2. Now just need elimination and that other question...

11. Originally Posted by Rocher
...

1) Given two numbers whose difference is 20 such that twice on number equals three times the other. Find the numbers.
The first number is x
the second number is y.

Then you know: x - y = 20 (from this equation you know that x > y)
and 2x = 3y

Use substitution: y = x - 20

2x = 3(x-20) Expand the bracket and collect like terms, x at the LHS.

-x = -60 that means: x = 60 and therefore y = 40.

12. Okay thank you now I just need the final question


text:
# College Physics : Laws of Thermodynamics and General Concepts

## Example Questions

### Example Question #1 : College Physics

A  steel sword is cooled from  to  by dipping it in  of water that is at

If the specific heat of  the steel is  and the specific heat of water is , what will be the temperature of the water after the steel is cooled?

Explanation:

This is a simple pairing of two  equations.

We will make the left side for the steel and the right side for water. Thus:

Solving for  of water yields , and certainly the water gained heat not lost it, so the final temperature of the water is

### Example Question #1 : College Physics

A steel beam is  long at a temperature of . On a hot day, the temperature reaches . What is the change in the beam's length due to thermal expansion given that the thermal expansion coefficient for steel is

Explanation:

We need to use the equation for thermal expansion in order to solve this problem:

We are given:

for the thermal expansion constant

for the initial length of the steel.

We need to calculate which is the the difference of the two temperatures.

Now we have enough information to solve for the change in the length of steel.

### Example Question #1 : Laws Of Thermodynamics And General Concepts

What is the change in entropy for  of ice initially at  that melts to water and warms to the temperature of the atmosphere at ?

Latent heat of fusion for water:

Specific heat of water:

Specific heat of ice:

Explanation:
• We must account for two separate processes since the ice is melting and then warming. The basic formula to calculate the change in entropy at a constant temperature when the ice melts is:

Where  is the change in entropy,  is the change in heat energy, and  is the temperature at which the change of state for the ice occurs. Now we substitute in an equation for  for when ice melts: . Our entropy equation now becomes:

Now we plug in our known values into the equation:

Mass

Latent heat of fusion

Temperature

And we get the change in entropy for the ice melting at a constant temperature, which is

• Now we need to calculate the change in entropy for the liquid water warming from  to .

The basic formula to calculate the change in entropy at a changing temperature when the water warms is:

Next we substitute in an equation for when there is a temperature change, but not a change in state. That equation is: so our entropy equation now becomes:

next we integrate this equation to get:

Next we plug in the values we know and solve for the change in entropy:

Final temperature of water

Initial temperature of water

Specific heat of water

Mass of water

Now we get the change in entropy for the water warming to be

We add our two entropy values together to find the total change in entropy to be:


text:
Pointer aliasing is a common pitfall:

warning: dereferencing type-punned pointer will break strict-aliasing rules [-Wstrict-aliasing]

From gcc man page:

-fstrict-aliasing

Allow the compiler to assume the strictest aliasing rules applicable to the language being compiled. For C (and C++), this activates optimizations based on the type of expressions. In particular, an object of one type is assumed never to reside at the same address as an object of a different type, unless the types are almost the same. For example, an “unsigned int” can alias an “int”, but not a “void*” or a “double”. A character type may alias any other type.

Pay special attention to code like this:

  union a_union {
int i;
double d;
};

int f() {
union a_union t;
t.d = 3.0;
return t.i;
}

The practice of reading from a different union member than the one most recently written to (called “type-punning”) is common. Even with -fstrict-aliasing, type-punning is allowed, provided the memory is accessed through the union type. So, the code above works as expected. However, this code might not:

  int f() {
union a_union t;
int* ip;
t.d = 3.0;
ip = &t.i;
return *ip;
}

Similarly, access by taking the address, casting the resulting pointer and dereferencing the result has undefined behavior, even if the cast uses a union type, e.g.:

  int f() {
double d = 3.0;
return ((union a_union *) &d)->i;
}

The -fstrict-aliasing option is enabled at levels -O2, -O3, -Os.


text:
CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
-   OpenFOAM (https://www.cfd-online.com/Forums/openfoam/)
-   -   localEuler in rhoCentralFoam (https://www.cfd-online.com/Forums/openfoam/91199-localeuler-rhocentralfoam.html)

 praveen August 3, 2011 05:49

localEuler in rhoCentralFoam

In order to do steady state calculations with rhoCentralFoam, I am trying to use localEuler ddt scheme. From what I understood, this requires specifying rDeltaT as a volScalarField. I have made following additions to rhoCentralFoam in version 2.0.0

createField.H:
Code:

volScalarField rDeltaT (     IOobject     (         "rDeltaT",         runTime.timeName(),         mesh,         IOobject::NO_READ,         IOobject::AUTO_WRITE     ),     mesh,     1/dimensionedScalar("maxDeltaT", dimTime, GREAT),     zeroGradientFvPatchScalarField::typeName );
setrDeltaT.H:
Code:

{                  // Set the reciprocal time-step from the local Courant number     rDeltaT.dimensionedInternalField() = max     (          1/dimensionedScalar("maxDeltaT", dimTime, maxDeltaT),         fvc::surfaceSum(amaxSf)().dimensionedInternalField()       /(maxCo*mesh.V())     );      // Update tho boundary values of the reciprocal time-step     rDeltaT.correctBoundaryConditions();         }
rhoCentralFoam.C:
Code:

        surfaceScalarField aphiv_pos(phiv_pos - aSf);         surfaceScalarField aphiv_neg(phiv_neg + aSf);         // Reuse amaxSf for the maximum positive and negative fluxes         // estimated by the central scheme         amaxSf = max(mag(aphiv_pos), mag(aphiv_neg));         #include "compressibleCourantNo.H"         #include "readTimeControls.H"         //#include "setDeltaT.H"         #include "setrDeltaT.H"         runTime++;         Info<< "Time = " << runTime.timeName() << nl << endl;
Then in system/fvSchemes, I specify

Code:

ddtSchemes {     default        localEuler  rDeltaT; }
The code compiles and runs but blows up after some iterations. With the usual rhoCentralFoam it runs fine, but it takes forever to converge. I use a maxCo of 0.1 and it is a subsonic flow. Limiters are also used (Gamma scheme). I could not find anything on localEuler in the forums so I want to know if anybody has succeeded in using it, or if you can find any mistake in my modifications.

 ramhari September 10, 2011 14:18

@praveen

http://www.cfd-online.com/Forums/ope...time-step.html

In the above thread, check my last post (which is working fine for me )...U may get some idea : )

Cheers!
Hariram

 eric.m.tridas January 9, 2012 15:06

Praveen,

Have you had any more success implementing the localEuler ddt scheme? I am very interested in this aspect as my simulations using rhoCentralFoam are take ages to compute.

Thanks,

-Eric

 laurensvd February 23, 2012 11:31

Im not sure if the poster is still following this but if you are, where did you get that definition for rDeltaT?

Code:

rDeltaT.dimensionedInternalField() = max (             1/dimensionedScalar("maxDeltaT", dimTime, maxDeltaT), fvc::surfaceSum(amaxSf)().dimensionedInternalField()        /(maxCo*mesh.V())    );
If you look at the compressible courant number definition it is formulated differently:
Code:

surfaceScalarField amaxSfbyDelta     (         mesh.surfaceInterpolation::deltaCoeffs()*amaxSf     );     CoNum = max(amaxSfbyDelta/mesh.magSf()).value()*runTime.deltaTValue();
I am trying to implement this latter equation to get rDeltaT but I am unable to get the syntax right to find the amaxSfbyDelta/mesh.magSf() for each cell. How can I find the maximum number of that value (which is I Presume calculated over the cell faces).
In other words, is there a command that finds the maximum surface value of a cell?

My current (not completely correct) implementation for rDeltaT is the following:
Quote:
 // Set the reciprocal time-step from the local Courant number surfaceScalarField amaxSfbyDelta ( mesh.surfaceInterpolation::deltaCoeffs()*amaxSf ); rDeltaT.dimensionedInternalField() = max ( 1/dimensionedScalar("maxDeltaT", dimTime, maxDeltaT), fvc::surfaceSum(mag(amaxSfbyDelta/mesh.magSf()))/(2.0*maxCo) );
Since I did not know how to find the maximum surface value of a cell I performed a surface sum of the magnitude and then divided it by 2 to get approximately the same effect as the initial definition of the compressible courant number.

When you look at the interFoamLTS source code you can also find how to implement smoothing over the rDeltaT field which you might consider implementing if you have big differences in timestep between neighboring cells.

 praveen October 21, 2012 02:19

Suppose is maximum wave speed in the face normal direction across the face between cell i and cell j. Let be area of this face. Then time step in cell i should be

where the sum is over all face neighbouring cells of cell i.

Introducing a courant number maxCo, we get

The time step condition can be seen by applying first order upwind scheme to a scalar convection equation.

 All times are GMT -4. The time now is 15:09.


text:
# 1999A&A...341..329P

 other querymodes : Identifierquery Coordinatequery Criteriaquery Referencequery Basicquery Scriptsubmission TAP Outputoptions Help

 Query : 1999A&A...341..329P

1999A&A...341..329P - Astronomy and Astrophysics, volume 341, 329-347 (1999/1-2)

HST images and properties of the most distant radio galaxies.

PENTERICCI L., ROETTGERING H.J.A., MILEY G.K., McCARTHY P., SPINRAD H., VAN BREUGEL W.J.M. and MACCHETTO F.

Abstract (from CDS):

With the Hubble Space Telescope we have obtained images of 9 of the most distant radio galaxies. The galaxies, which have redshifts between z=2.3 and z=3.6, were observed with the WFPC2 camera in a broad band filter (F606W or F707W, roughly equivalent to V or R-band), corresponding to the near ultraviolet emission in the rest frame of the radio galaxies. The total observing time was 2 orbits per object. In this paper we present the images overlayed on VLA radio maps of comparable resolution. We also present previously unpublished images, taken from the HST archive, of two other high redshift radio galaxies, observed through similar broad band filters. We find that on the scale of the HST observations there is a wide variety of morphological structures of the hosting galaxies: most objects have a clumpy, irregular appearance, consisting of a bright nucleus and a number of smaller components, suggestive of merging systems. Some observed structures could be due (at least partly) to the presence of dust distributed through the galaxies. The UV continuum emission is generally elongated and aligned with the axis of the radio sources, however the characteristics of the alignment effect'' differ from case to case, suggesting that the phenomenon cannot be explained by a single physical mechanism. We compare the properties of our radio galaxies with those of the UV dropout galaxies and conclude that (i) the most massive radio galaxies may well evolve from an aggregate of UV dropout galaxies and (ii) high redshift radio galaxies probably evolve into present day brightest cluster galaxies.

Journal keyword(s): galaxies: individual: TX 1707+1051 - galaxies: individual: MRC 2104-242 - galaxies: formation - galaxies: clusters: general - galaxies: active - cosmology: early Universe

Nomenclature: Fig.9: [PRM99] 1707+105 A (Nos A-C).

 Number of rows : 24
N Identifier Otype ICRS (J2000)
RA
ICRS (J2000)
DEC
Mag U Mag B Mag V Mag R Mag I Sp type #ref
1850 - 2022
#notes
1 NGC 547 GiP 01 26 00.6208982520 -01 20 42.432452268   13.16 13.34     ~ 331 3
2 PMN J0214-1158 rG 02 14 17.4 -11 58 47           ~ 59 1
3 NVSS J031812-253509 rG 03 18 12.00 -25 35 09.0   23.6 23.8     ~ 75 2
4 4C 41.17 rG 06 50 52.098 +41 30 30.53           ~ 323 1
5 NVSS J083053+191314 rG 08 30 53.44 +19 13 15.7           ~ 70 1
6 NVSS J090530+340756 rG 09 05 30.1 +34 07 57   24.0 23.800 23.500   ~ 154 1
7 PMN J0945-2429 rG 09 45 32.7 -24 28 49           ~ 94 1
8 NAME Spiderweb Galaxy AGN 11 40 48.3 -26 29 09     22.00     ~ 265 1
9 4C 03.24 rG 12 45 38.4 +03 23 18           ~ 72 1
10 2C 1066 SyG 12 56 57.9 +47 20 20   22.0       ~ 273 0
11 3C 286 Sy1 13 31 08.2883506368 +30 30 32.960091564   17.51 17.25     ~ 4075 2
12 4C 24.28 rG 13 48 14.890 +24 15 50.34     17.7     ~ 45 1
13 4C -00.54 rG 14 13 15.2 -00 23 01           ~ 47 0
14 3C 324.0 Sy1 15 49 48.9 +21 25 38   21.5       ~ 304 0
15 [RLM94] 1707+105 2 Rad 17 10 06.42 +10 31 05.3           ~ 2 0
16 4C 10.48 rG 17 10 06.50 +10 31 06.0           ~ 37 1
17 3C 356 QSO 17 24 20.2 +50 57 14   21.5 21.5     ~ 219 1
18 LEDA 2830476 Sy1 20 27 59.5 -21 40 56   22.4       ~ 73 1
19 LEDA 2830749 rG 21 06 58.16 -24 05 11.3   23.0       ~ 72 1
20 4C 19.71 rG 21 44 07.5 +19 29 11           ~ 41 1
21 4C 28.58 rG 23 51 59.23 +29 10 29.0           ~ 41 1
22 [PRM99] 1707+105 A G ~ ~           ~ 2 0
23 [PRM99] 1707+105 C G ~ ~           ~ 2 0
24 [PRM99] 1707+105 B G ~ ~           ~ 2 0

To bookmark this query, right click on this link: simbad:objects in 1999A&A...341..329P and select 'bookmark this link' or equivalent in the popup menu

2022.06.25-04:02:24


text:
Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > DETAIL:

### Paper:

TR10-041 | 11th March 2010 19:52

#### Improved Algorithms for Unique Games via Divide and Conquer

TR10-041
Authors: Sanjeev Arora, Russell Impagliazzo, William Matthews, David Steurer
Publication: 11th March 2010 21:03
We present two new approximation algorithms for Unique Games. The first generalizes the results of Arora, Khot, Kolla, Steurer, Tulsiani, and Vishnoi who give polynomial time approximation algorithms for graphs with high conductance. We give a polynomial time algorithm assuming only good local conductance, i.e. high conductance for small subgraphs. The second algorithm runs in mildly exponential time, $e^{\alpha n}$, but makes no assumptions about the underlying constraint graph. As the completeness approaches $1$ (completeness $1-\epsilon$), the constant $\alpha$ in the running time rapidly approaches $0$ ($\alpha = exp(-\Omega(1/\epsilon))$.) The value of the solutions returned by these algorithms depend only on the completeness of the Unique Game and either the local conductance or the allowed running time respectively. In particular, the performance of these algorithms does not depend on the number of labels in the Unique Game.


text:
Mathematics

Vertically opposite angle of $\angle XVW$ is _______.

SOLUTION
$\angle YVZ$.

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Subjects 10
Chapters 88
Enrolled Students 86

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text:
# What do the symbols in Cross Validated mean?

I'm new to the field of statistics. While reading a book, I came across some symbols or code-like text (shown below) which I was not able to interpret. I tried Google, but it couldn't help either. I remember seeing similar text in this site and so I'm posting this question here. What is this text and how to read this? Please give some pointers to understand this kind of text.

\begin{eqnarray}
\mbox{output} & = & \left\{
\begin{array}{ll}
0 & \mbox{if } \sum_j w_j x_j \leq \mbox{ threshold} \\
1 & \mbox{if } \sum_j w_j x_j > \mbox{ threshold}
\end{array} \right. \tag{1}
\end{eqnarray}

Please feel free to edit the question or move it the right place if this is not the right place to ask.

• I just edited the formatting in your question... but now that I re-read it I think I may have answered the question by doing it... Maybe you just wanted to know what $\LaTeX$ is?
– nico
Mar 4 '14 at 7:26
• @Glen_b: note that the original question had some (unformatted) LaTeX in it... as said in my previous comment maybe I should not have edited it... waiting for the OP to explain if it's a problem of math or not...
– nico
Mar 4 '14 at 7:29
• this is the book from where I copy pasted that code neuralnetworksanddeeplearning.com/chap1.html Mar 4 '14 at 7:34
• StrikeR, that all renders properly for me (it looks like mathematics not LaTeX code). My guess is you're blocking javascript in some way (I run noscript myself and have encountered the problem many times, so that's why I'm guessing that's your problem with that page) Mar 4 '14 at 7:36
• @StrikeR: do you see those as code such as \begin{eqnarray} or do you actually see some mathematical notation? In the former case you should probably check you have JavaScript running correctly in your browser. That site (as this one) uses a JS engine called MathJax to render LaTeX code into math.
– nico
Mar 4 '14 at 7:37
• @Glen_b I have checked that link opening in other computer and it is being displayed in the form of mathematical formulae instead of this code. So, it must be the javascript problem, I'll try to look into that. And I rolled back the question as you asked. Mar 4 '14 at 8:16


text:
# How do you solve the inequality |2t-9|<=3?

##### 1 Answer
Apr 4, 2016

$| 2 t - 9 | \le 3 \setminus \iff - 3 \le 2 t - 9 \le 3 \setminus \iff 6 \le 2 t \le 12 \setminus \iff 3 \le t \le 6 \setminus \iff t \setminus \in \left[3 , 6\right] .$


text:
# graph avoiding cycle of length 4

let $G$ be a simple graph on $n$ vertices such that $G$ has no cycle of length $4$. show that $e(G)\le \frac{n}{4}(1+\sqrt{4n-3})$ where $e(G)$ denotes the number of the edges of the graph $G$.

-
Posting questions in the imperative like this is generally considered impolite. Please refer to meta.math.stackexchange.com/questions/1803/… . –  Brad Dec 8 '11 at 16:41
It might be of interest for you to think about graphs which have cycles of all lengths (pancyclic) graphs as well as those which are missing a cycle of exactly one length (almost pancyclic) - see: en.wikipedia.org/wiki/Pancyclic_graph for a discussion of these ideas. –  Joseph Malkevitch Dec 8 '11 at 20:49

I included a proof of what you want in a question of mine where I asked for a tighter bound if the graph avoids also cycles of length 3: (I use the convention $m:=e(G)$)

I try to show if $G$ has no subgraph $C_4$, then $m \leq \frac{n}{4}(1+\sqrt{4n-3})$. We can consider the amount $k$ of subgraphs of $G$ that look like $K_{1,2}$ (a "cherry"). Certainly if we (double) count those, considering a vertice in the center of the cherry, we get:

$$k=\sum_{v\in V}\binom{\text{deg}(v)}{2}$$

Now if we don't allow squares a pair of vertices $\{v,w\}$ can be endpoints of only one cherry, otherwise we would have a cycle, therefore

$$\binom{n}{2} \geq k=\sum_{v\in V}\binom{\text{deg}(v)}{2}$$

Then due to convexity of $\binom{x}{2}$ and the Jensen-Inequality one might conclude now that $m \leq \frac{1}{4} \cdot \left(\sqrt{n^2 (4n-3)}+n\right)$.

This is a link to Jensen's inequality in case you don't know it.

Edit: As the operator asked for more details:

Define $\varphi(x):=\binom{n}{2}$, it is easy to see that $\varphi$ is convex. Therefore we have that

$$\varphi\left(\frac{\sum x_i}{n}\right) \le \frac{\sum \varphi (x_i)}{n} \iff \sum \varphi (x_i) \geq n\cdot \varphi\left(\frac{\sum x_i}{n}\right)$$

And therefore

$$\binom{n}{2} \geq \sum_{v\in V}\binom{\text{deg}(v)}{2} \geq n\cdot \binom{\frac{1}{n}\sum_{v\in V}\text{deg}(v)}{2}=n\cdot\binom{\frac{2m}{n}}{2}$$

Because the sum over all degrees of the nodes corresponds to twice the amount of edges. Expanding the binomial on both sides yields your result.

-
I know Jensen's function, but would you please show me how to apply it here? every time I use it on this question I get $m\le \frac{n}{4}(1+2\sqrt{n})$. –  Goodarz Mehr Dec 8 '11 at 17:18
@goodarzmehr I tried to be a bit more explicit. I hope you understand it now. –  Listing Dec 8 '11 at 17:36
now I undrestand, thanks a lot :) –  Goodarz Mehr Dec 9 '11 at 18:39
no problem :P, good that you understood it –  Listing Dec 9 '11 at 19:58


text:
## HW Problem A15

$c=\lambda v$

Hannah Lee 2F
Posts: 105
Joined: Thu Jul 11, 2019 12:15 am

### HW Problem A15

1A.15) In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line.

Is there any way to solve this without the Rydberg equation, using the way we solved it in class, by using En = final - En = initial? Or would we need to use the Rydberg equation for this unit?

005391550
Posts: 92
Joined: Thu Jul 25, 2019 12:15 am

### Re: HW Problem A15

if you derived En(final) - En(initial) it would do the same thing as the Rydberg equation since En = -hR/n^2. so technically you can do either and still get the same result!

Angus Wu_4G
Posts: 95
Joined: Fri Aug 02, 2019 12:15 am

### Re: HW Problem A15

I answered this question in a different discussion thread, but I'll repost my explanation below again:

You would first use the wavelength you have and find the energy change using E=(hc)/wavelength.

Once you have the energy change, use the equation E=(-hR)/n^2 with n=1 to find the energy of the electron at n=1.

Once you have the energy change and the energy of the electron at n=1, you can set up this equation:

(Energy of electron at n=final) - (Energy of electron at n=1) = Energy change

Then finally, solve for n final. According to the solution manual, n should be 3.

As for the Rydberg equation, it is the same as the equation that was given to us in lecture, you could use either one, both will yield the same answer. If you do use the Rydberg Equation though, you have to be careful with your n initial and n final values and substitute then correctly or else you might get a negative energy.


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# When to use Heat Diffusivity eqn and when to use Fourier's law to find temperature distribution?

Let's say that there is a circular conical section that has diameter $D=.25x$ without any heat generation and I need to find the temperature distribution.

Originially I thought I could use the heat diffusivity equation at steady state to find the temperature distribution. The differential equation would be:

$$\frac{d}{dx}(k\frac{dT}{dx})=0$$

I am looking at the solution to the example in the book and they use Fourier's Law $$q_{x}=-kA\frac{dT}{dx}$$ and their result is $T(x)=T_{1}-\frac{4q}{\pi a^{2}k}(\frac{1}{x_{1}}-\frac{1}{x_{2}})$

Why do they use one as opposed to the other? Will the two methods produce the same result?

The reason I ask is because they also provide a derivation for the temperature distribution of a plane wall with no heat generation and they use the heat diffusivity equation

## 1 Answer

Both methods should thus give you the same result. The diffusion equation for heat can be derived from Fourier's law.

For a one-dimensional problem, you can do that easily yourself, by taking a inifinitesimal part of the rod of size $dx$ and realizing that $q_x$ at either side of this part should be the same (since there is no heat source or sink in this piece). Taking the limit for $dx \rightarrow 0$, will give you the diffusion equation. If your problem becomes more complex, the diffusion equation will be easier to solve.


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# Determine whether $f$ is even, odd, or neither. If you have a graphing calculator, use it to check your answer visually.$f(x) = x | x |$

## Odd function.

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all right. We want to determine if this function is even on or neither. And so remember that for odd functions, opposite X values have opposite. Why values? And so the grafts have origin, symmetry and for even functions opposite X values have the same y value. And so the grafts have y axis symmetry. So what we want to do is determine if f of the opposite of X is going to be the same as or the opposite of, or neither compared to f of X. So we substitute the opposite of X in the function, and we get the opposite of X times the absolute value of the opposite of X. Now realize that the absolute value of the opposite of X is just the same thing as the absolute value of X X, and the opposite will be the same distance from zero, so we can simplify it. Now that looks like the opposite of what we originally had. So we've just found that f of the opposite of X is equal to the opposite of F of X, which means this function is odd. Now we can grab it on a calculator and see the origin symmetry. So we type it into the calculator and we look at the graph and there you see that origin symmetry.

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