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# Support Vector Machines (SVM) maximum margin hyperplane use

I have been approaching to Support Vector Machines in this period, and I need a clarification about the use of the maximum margin hyperplane.

From a labeled point training set, we train a model that will correctly classify the future test set points. To do this, we search for a best hyperplane that divide the training set points into two classes (True, False). So we use kernel and etc to get this goal.

But, when we've gotten our best hyperplane, what do we do with it?

Do we just use it to classify new data points, by verifying which ones are "on this side" and which ones "on the other side"?

I've understood how to get the famous best hyperplane, but I need some clarification on what to do next with it...

Thanks!

-
Yes, the optimal hyperplane gives you a classifier that you use to classify new data points: $$f(x) = sgn (<w, x> + b)$$ –  Leo May 4 '12 at 14:59
I don't know much about support vector machines but this looks mor elike plain old linear discriminant analysis. In LDA you find the hyperplane in the feature space that best separates the two classes and then use that hyperplane to predict new cases. So your guess as to what to do with the hyperplane is exactly what is done in LDA. I am assuming that the term "maximum margin hyperplane" is synonymous with "best separating hyperplane." –  Michael Chernick May 4 '12 at 15:50
Yes, keep in mind that if you find the hyperplane that perfectly divides your test points, it may not generalize well, it might be overfitted. –  karenu May 9 '12 at 22:01

There are two common ways of utilizing the maximum-margin hyperplane of a trained SVM.

(1) Prediction for new data points

Based on a given training dataset, an SVM hyperplane is fully specified by its slope $w$ and an intercept $b$. (These variable names derive from a tradition established in the neural-networks literature, where the two respective quantities are referred to as 'weight' and 'bias.') As noted before, a new data point $x \in \mathbb{R}^d$ can then be classified as

\begin{align} f(x) = \textrm{sgn}\left(\langle w,x \rangle + b \right) \end{align}

where $\langle w,x \rangle$ represents the inner product. Thanks to the Karush-Kuhn-Tucker complementarity conditions, the discriminant function can be rewritten as

\begin{align} f(x) = \sum_{i \in SV} \alpha_i \langle x_i, x \rangle + b, \end{align}

where the hyperplane is implicitly encoded by the support vectors $x_i$, and where $\alpha_i$ are the support vector coefficients. The support vectors are those training data points which are closest to the separating hyperplane. Thus, predictions can be made very efficiently, since only inner products (alternatively: kernel functions) between some training points and the test point have to be evaluated.

Some have suggested also considering the distance between a new data point $x$ and the hyperplane, as an indicator of how confident the model was in its prediction. However, it is important to note that hyperplane distance itself does not afford inference; there is no probability associated with a new prediction, which is why an SVM is sometimes referred to as a point classifier. If probabilistic output is desired, other classifiers may be more appropriate, e.g., the SVM's probabilistic cousin, the relevance vector machine (RVM).

(2) Reconstructing feature weights

There is another way of putting an SVM model to use. In many classification analyses it is interesting to examine which features drove the classifier, i.e., which features played the biggest role in shaping the separating hyperplane. Given a trained SVM model with a linear kernel, these feature coefficients $w_1, \ldots, w_d$ can be reconstructed easily using

\begin{align} w = \sum_{i=1}^n y_i \alpha_i x_i \end{align}

where $x_i$ and $y_i$ represent the $i^\textrm{th}$ training example and its corresponding class label.

An important caveat of this approach is that the resulting feature weights are simple numerical coefficients without inferential quality; there is no measure of confidence associated with them. Thus, we cannot readily argue that some features were 'more important' than others, and we cannot infer that a feature with a particularly low coefficient was 'not important' in the classification problem. In order to allow for inference on feature weights, we would need to resort to more general-purpose approaches, such as the bootstrap, a permutation test, or a feature-selection algorithm embedded in a cross-validation scheme.

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text:
Geometry Level 3

Triangle $$ABC$$ has $$\angle BAC = 20 ^\circ$$ and $$\angle ACB = 75 ^\circ$$. $$P$$ is a point in the interior of triangle $$ABC$$. Points $$D, E$$ and $$F$$ lie on $$BC, CA$$ and $$AB$$, respectively, such that $$A, E, P, F$$ are concyclic and $$C, D, P, E$$ are concyclic. What is the measure (in degrees) of $$\angle DPF$$?

Details and assumptions

Points are concyclic if they lie on a circle.

×

format: plain


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# Учебно-методическое пособие по английскому языку для вузов. Кислякова Н.В - 10 стр.

Составители:

Рубрика:

• ## Иностранный язык

10
VI. Insert prepositions where necessary.
1. Elements are represented ... chemical symbols.
2. Each element is made ... atoms.
3. Matter is composed ... one or more elements.
4. A chemical unit cant be broken ....
5. An acid releases hydrogen ions when mixed ... water.
VII. Which words on the right side can combine with the word on the left side?
1. element
water 2. solution
3. substance
4. process
1. unit
chemical 2. storage
3. reaction
4. symbol
1. solution
living 2. acid
3. sentence
4. cell
them working in pairs.
1. Matter is something that has a mass and takes up space.
2. Each element is represented by a chemical symbol.
3. The element has a set of properties that distinguishes it from other elements.
4. The formation of a solution is a physical change.
5. There are two kinds of substances: soluble and insoluble.
VI. Insert prepositions where necessary.
1. Elements are represented ... chemical symbols.
2. Each element is made ... atoms.
3. Matter is composed ... one or more elements.
4. A chemical unit cant be broken ....
5. An acid releases hydrogen ions when mixed ... water.

VII. Which words on the right side can combine with the word on the left side?

1. element
water                          2. solution
3. substance
4. process

1. unit
chemical                       2. storage
3. reaction
4. symbol

1. solution
living                         2. acid
3. sentence
4. cell

them working in pairs.
1. Matter is something that has a mass and takes up space.
2. Each element is represented by a chemical symbol.
3. The element has a set of properties that distinguishes it from other elements.
4. The formation of a solution is a physical change.
5. There are two kinds of substances: soluble and insoluble.

10


text:
# User talk:Darwish

Thanks for contributing! I don't think you can do better than ${\displaystyle O(n\log n)}$ though. People on the UVa leaderboards usually optimize for I/O! Larry 17:46, 24 March 2010 (UTC)
Thanks for the encouragement! It seems yes, no one has done better than ${\displaystyle O(n\log n)}$ on the boards; after small modifications to code (without optimizing I/O), I was able to get a rank!


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# I Maximum work done by a body in an external medium

#### hgandh

I am reading Landau's Vol 5 on Statistical Physics and have trouble grasping some concepts in Section 20.
Let us now consider a different formulation of the maximum-work problem. Let a body be in an external medium whose temperature $T_0$ and pressure $P_0$ differ from the temperature $T$ and pressure $P$ of the body. The body can do work on some object, assumed thermally isolated both from the medium and from the body. The medium, together with the body in it and the object on which work is done, forms a closed system. The volume and energy of the medium are so large that the change in these quantities due to processes involving the body does not lead to any appreciable change in the temperature and pressure of the medium, which may therefore be regarded as constant.
If I understand this correctly, the body and the medium are in direct contact and can exchange work and heat while the object can only exchange work with the body.
A transition in which the body does the maximum work $|R|_{max}$ corresponds to a reverse transition which requires the external source to do the minimum work $R_{min}$.
Thus the total change $\Delta E$ in the energy of the body in some (not necessarily small) change in its state consists of three parts: the work $R$ done by the body the external source, the work done by the medium, and the heat gained from the medium...
$$\Delta E =R + P_0 \Delta V_0 - T_0 \Delta S_0$$
Since the total volume of the medium and the body remains constant, $\Delta V_0 = -\Delta V$, and the law of increase of entropy shows that $\Delta S + \Delta S_0 \geq 0$...
$$R \geq \Delta E - T_0 \Delta S +P_0 \Delta V$$
The equality occurs for a reversible process....
$$R_{min} = \Delta E - T_0 \Delta S +P_0 \Delta V$$
So the minimum work is realized during a reversible process which means that the system is in thermodynamic equilibrium at every instant, right? That means the body and the medium must have the same temperature and pressure, right? However, Landau then goes on to say...
If the body is in an equilibrium state at every instant during the process (but not, of course, in equilibrium with the medium), then for infinitesimal changes in its state formula (20.2) may be written differently. Substituting $dE = T dS - P dV$ in $dR_{min}-T_0 dS + P_0 dV$, we find
$$dR_{min}= (T-T_0) dS - (P-P_0) dV$$
I don't understand what Landau is doing here. Shouldn't $dR_{min}$ be equal to zero here since it is only for reversible processes so $T=T_0, P=P_0$? Yet, this equation seems to be valid when the body is not equilibrium with the medium. What am I not seeing? Another confusion comes when Landau writes:
Two important cases may be noted. If the volume and temperature of the body remain constant, the latter being equal to the temperature of the medium, (20.2) gives $R_{min}=\Delta (E-TS)$, or
$$R_{min}=\Delta F$$
i.e. the minimum work is equal to the change in the free energy of the body. Secondly, if the temperature and pressure of the body are constant and equal to $T_0$ and $P_0$, we have
$$R_{min} = \Delta \Phi$$
i.e. the work done by the external source is equal to the change in the thermodynamic potential of the body.
It should be emphasized that in both these particular cases the body concerned must be one not in equilibrium, so that its state is not defined by $T$ and $V$ (or $P$) alone; otherwise the constancy of these quantities would mean no process could occur at all. We must consider, for example, a chemical reaction in a mixture or reacting substances, a process of dissolution, or the like.
Again I am confused here. As I am understanding correctly, all of the equations involving $R_{min}$ only apply to reversible processes but Landau says the bodies must not be in equilibrium. Can somebody help me understand these concepts?

Related Classical Physics News on Phys.org

#### tnich

Homework Helper
If I understand this correctly, the body and the medium are in direct contact and can exchange work and heat while the object can only exchange work with the body.
I haven't read your whole post, yet, but right away I noticed that, in contrast to your assumption, the text says: "The body can do work on some object, assumed thermally isolated both from the medium and from the body." If the object is thermally isolated from the medium, then it cannot exchange heat with it.

#### hgandh

I haven't read your whole post, yet, but right away I noticed that, in contrast to your assumption, the text says: "The body can do work on some object, assumed thermally isolated both from the medium and from the body." If the object is thermally isolated from the medium, then it cannot exchange heat with it.
The body and the object are two different things. The body is in an external medium and can exchange both work and heat. The body can only exchange work with the object. So you have 3 things here: the medium, body and the object.

#### tnich

Homework Helper
The body and the object are two different things. The body is in an external medium and can exchange both work and heat. The body can only exchange work with the object. So you have 3 things here: the medium, body and the object.
You are right. I misunderstood your comment.

#### sponteous

I agree, I found this section in Landau and Lifshitz very confusing for the same reasons as the OP when I read it a month or two ago. I just reread it to make sure, and I still find it confusing. It would certainly seem that for a reversible process the temperature of the body and the medium would have to be the same, or only infinitesimally different. It also isn't obvious to me that the work the medium does on the body is $P_0 \Delta V_0$ if the pressures are different. Although, now that I think of it, perhaps the thermally isolated object can also exchange compression/expansion work with the body, and that would account for the discrepancy.

I think what the authors are trying to do is make the analysis as general as possible. For example, to show that change in free energy can give a lower bound to the work required to drive a process even if the system itself is not at constant temperature during the process. As long as the surrounding medium is at constant temperature $T_0$, and the initial and final temperatures of the system (body) are both $T=T_0$, then the total work done on the system is at least $$W \geq \Delta A = \Delta (U-TS).$$ The body is allowed to assume other temperatures in between. This follows from the Clausius Theorem: $$\Delta S \geq \int_A^B dq/T_0 .$$ Since the temperature of the surroundings, $T_0$ is constant, it comes outside the integral and you get $$T_0 \Delta S \geq Q = \Delta U - W$$, where $W$ is the total work done on the system. This can be rearranged to get $$W \geq \Delta U - T_0 \Delta S = \Delta(U-TS),$$ The last equality comes from assuming that the initial and final temperatures of the body are the same as the surroundings.

This is more general than the usual statement that the change in free energy gives a lower bound for _isothermal_ processes. However, it looks like L&L are trying to show something even more general than this, and I don't follow their analysis.

#### DrDu

It says explicitly that the system doesnt have to be in equilibrium with the surrounding medium. Reversible means here only that the system and source of work can be returned to their initial state, not that all the heat will return from the surrounding. Maybe a trivial example is that of the body driving a paddle in the system, with so much heat transferred to the medium that T and V remain constant.

#### sponteous

It says explicitly that the system doesnt have to be in equilibrium with the surrounding medium. Reversible means here only that the system and source of work can be returned to their initial state, not that all the heat will return from the surrounding. Maybe a trivial example is that of the body driving a paddle in the system, with so much heat transferred to the medium that T and V remain constant.
I am trying to picture this setup but I don't quite follow. Are you saying that some of the work performed by the body is used to stir the surrounding medium with a paddle?

At any rate, I haven't encountered that use of "reversible" before. I thought reversible meant everything, including the surroundings and any lifted weights, can be returned to their initial state. But I'm still new to thermodynamics so that doesn't mean much.

"Maximum work done by a body in an external medium"

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text:
ansancle
317 9

I am trying to set the contents of an image using org.zkoss.zul.Image::setSrc in my Java code. Every possible permutation of the path to the image file has failed to load the image.

Just to be sure, I added the following into my .zul file and the image showed right up.

<image src="/stylesheets/images/zones/speakers_32x32.png"/>

Can someone please help me figure out how to load an image from the filesystem in the java code using a path. I even wrote a small app that allowed me to try paths by entering them into a page and it would update the image using setSrc and the path, everything I tried failed including the full path with http:// etc....

Im sure this is simple but my searches failed to produce an answer,

Thanks
Andrew

delete retag edit

## 2 Replies

dennis
3669 1 6
http://www.javaworld.com....

ZK add "Application Context" to src if the attribute start with "/" , it is very common usage in J2EE.
form the view of HTML , it will become <IMAGE src="/APP_CONCTXT/stylesheets/images/zones/speakers_32x32.png" />

The image file in file system should be pleased at file:YOU_APPLICATION_FOLDER/stylesheets/images/zones/speakers_32x32.png

ansancle
317 9

Figured it out, so I might as well post it for somebody else to see if need be.

I already new that the path to the image needs to be from the root, but my mistake was in not just setting the path into the setSrc method on button but instead creating an image, setting the src on that, then using the Button.setImageContent(Image.getImageContent());

For example :
In my application WEB-ROOT/zul is the root, my images are under WEB-ROOT/zul/stylesheets/images, so the path that you set into the setSrc method on Button is /stylesheets/images/imageName.png, etc....
The second part of this is I was initially creating an Image, setting the src on the Image, then using that Image to call Button.setImageContent(Image.getImageContent()) - This didn't work.
Just using the correct path to the image and calling setSrc on the Button works.

[hide preview]


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# The Limited Effect of Reforms on Growth

NOTE: The Growth Economics Blog has moved sites. Click here to find this post at the new site.

I said in my last post that transitional growth is slow, and therefore changing potential GDP – as many of the recent Cato Growth proposals would do – could not add much to the growth rate of GDP in the near term.

There were several questions that came up in the comments, so let me try to be more clear about distinguishing between influences of trend growth and short-run shocks. Output in period ${t+1}$ is

$\displaystyle y_{t+1} = (1+g)y_t + (1+g)\lambda (y^{\ast}_t - y_t) \ \ \ \ \ (1)$

where the first term on the right is the normal trend growth rate, and the second term is the additional transitional growth that occurs because the economy is not at potential GDP, ${y^{\ast}_t}$.

We need to distinguish between changes in potential GDP and changes in current GDP. Let’s take the above equation, plug in ${\lambda=0.02}$, and then use it to iterate forward from period 0 (today) until some arbitrary period ${t}$. You get

$\displaystyle y_t = (1+g)^t \left[(1-0.98^t)y^{\ast}_0 + 0.98^t y_0 \right]. \ \ \ \ \ (2)$

In period ${t}$, GDP will have grown by a factor of ${(1+g)^t}$ due to trend growth in GDP. The term in the brackets shows the cumulative effect of having ${y_0 \neq y^{\ast}_0}$ in the initial period. The 0.98 terms are just ${1-.02}$, and capture the changing role of this transitional growth over time. Note that as ${t}$ goes up, ${0.98^t}$ goes to zero and the effect of initial GDP ${y_0}$ falls to nothing. As ${t}$ gets big, the economy reaches potential GDP.

Now let’s assume that period 0 is 2014. Potential GDP is 17 trillion and actual GDP is 16 trillion, and the trend growth rate is 2%. Let’s consider two alternative policies to enact today that take effect in 2015.

• Policy A: a short run spending surge sufficient to make GDP in 2015 equal to potential GDP. Policy A immediately eliminates the gap between actual and potential GDP, but has no other long term effect.
• Policy B: raises potential GDP by 1 trillion dollars, but adds no immediate spending to GDP. The effect on potential GDP is permanent.

For Policy A, GDP in 2015 (period 1) is

$\displaystyle y_1 = (1.02)^1\left[(1-0.98)\times 17 + 0.98 \times 17 \right] = 17.34. \ \ \ \ \ (3)$

The growth rate of GDP from 2014 to 2015 is ${(17.34 - 16)/16 = 0.084}$ or about 8.4%. That’s a massive GDP growth rate for a developed economy like the US. But it is a one-time shock to the growth rate. From 2015 to 2016, and from 2016-2017, and every year thereafter, the growth rate will be exactly 2% because the economy is precisely back on trend. Policy A gives a one-year gigantic boost to the growth rate.

What about Policy B? GDP in 2015 here is

$\displaystyle y_1 = (1.02)^1\left[(1-0.98)\times 18 + 0.98 \times 16 \right] = 16.36. \ \ \ \ \ (4)$

This is nearly 1 trillion less than Policy A. The growth rate of GDP from 2014 to 2015 is ${(16.36 - 16)/16 = 0.023}$. As the prior post noted, reforms that raise potential GDP don’t have big effects on growth rates. But while the effect on growth is small, it is persistent. From 2015-2016, the growth rate of GDP will be roughly…0.023. It’s actually minutely smaller than from 2014-2015, but rounding makes them look the same. It will take a few years before the growth rate declines appreciably. Fifty years from now the growth rate will still be almost 0.021. Changing potential GDP, like with Policy B, is like turning an oil tanker with a tug boat. It doesn’t go fast, but it goes on for a long time.

So is Policy B worse than Policy A? It depends entirely on your time preferences. In 2015 GDP under Policy A is nearly 1 trillion dollars higher than with policy B. But 100 years from now, GDP will be nearly 1 trillion dollars higher with Policy B. We can actually figure out how soon it will be before Policy B passes Policy A. Set

$\displaystyle (1.02)^t \left[(1-0.98^t)17 + 0.98^t 17 \right] = (1.02)^t \left[(1-0.98^t)18 + 0.98^t 16 \right] \ \ \ \ \ (5)$

and solve for ${t}$. This turns out to be roughly 34 years from now, in 2048. It takes a long, long, time for changes in potential GDP to really pay off. If you want to increase the level of GDP in the near term, and hence raise near-term growth rates by implication, then you have to, you know, boost GDP. GDP is a measure of current spending, so raising GDP means raising current spending. There isn’t a trick to get around this.

Now, could I be underselling Policy B as a near-term boost to growth rates and GDP? Let’s consider a few possibilities:

• I’m underestimating the size of ${\lambda}$. As I mentioned last time, there is lots of empirical evidence that this is pretty small. But okay, let’s make ${\lambda = 0.05}$, more than double my 0.02 value. Now in 2015 policy B yields GDP of 16.4 trillion and a growth rate of 2.6%. Yes, it helps policy B, but doesn’t get it anywhere close to Policy A. It is still 14 years before GDP under Policy B is larger than under Policy A.
• I’m underestimating the boost to potential GDP that Policy B can deliver. So let’s ask, given ${\lambda = 0.05}$, how much would ${y^{\ast}_0}$ have to go up to match the 8.4% growth rate of Policy A? Potential GDP would have to jump to roughly 36 trillion, meaning it has to roughly **double** in size thanks to the policy. I think it is totally fair to say that this is implausible in a country like the US.
• But China was able to do it. Right, when China opened up, made reforms, etc.., it was able to raise its potential GDP by a large amount. You could probably plausibly argue that it raised potential GDP by a factor of something like 8-10. But the rapid growth in China over the last 30 years is not some victory lap for good state-led policy reforms, it’s a testament to just how screwed up Maoism was as an economic system. [Egad! An institutions explanation!]
• What if Policy B raised the trend growth rate, ${g}$? If it changed ${g}$ appreciably, then Policy B would be something really special. Let’s review for a moment a few of the changes that did not change the long-run growth rate in the US: the introduction of electricity, the income tax, the Great Depression, the New Deal, Medicare, higher tax rates, the Cold War, the oil crisis, lower tax rates, de-regulation, the IT revolution, and New Coke. There have been shifts in the level of potential GDP, such as the IT revolution shifting up potential GDP and inducing a period of relatively rapid transitional growth in the 1990’s. But it’s hard – if not impossible if you take Chad Jones‘ semi-endogenous growth idea seriously – to fundamentally alter ${g}$. It is dictated by changes in the scale of the global economy, not by policy effects within the US.

I’m all for policy reforms that raise potential GDP, and several of those proposed in the Cato forum would probably do that. We might want to undertake several of them at once to counteract the drags on potential GDP that Robert Gordon has outlined.

But we can’t be fooled into thinking that any of them would make a really appreciable difference to economic growth today. You can revolutionize education, or corporate taxation, or urban planning, or immigration all you want, but the gains those changes induce will take decades to manifest themselves.

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## 5 thoughts on “The Limited Effect of Reforms on Growth”

1. Very nice exposition; thank you for this!

Two broad questions:

1) In this model, does potential GDP automatically change over time? (This isn’t *immediately* clear to me from the math.) Or, without any policies to change y*, is y*(t+1)=y*(t)?

If the latter, than how, in equation (3) can y(1=2015)=17.34, which is above y*(0=2014)=17? (Unless I’m misunderstanding something, Policy A seems to have overshot its goal of increasing GDP to potential!)

Could you say more about the empirics of growth in potential GDP, perhaps which is not captured in this model?

I’m thinking here of the charts in the secular stagnation literature (e.g. Larry Summers’s Fig. 1a on p28 of the CEPR book: http://www.voxeu.org/content/secular-stagnation-facts-causes-and-cures), where y* was growing steadily prior to 2007, but estimates of its level since have fallen below its pre-2007 trend. (I note that the figure roughly for US y(2014)=$17T, and y*(2014) estimated *in 2014* at$18T; but that the y* prediction *from 2007* would have y*(2014)=\$19T; these correspond roughly with the values you are using in your calculations, answering another question I had about the relative effect of the drop in y* growth since the 2007.)

2) Re historical changes that did not affect “g”: was the (first) Industrial Revolution possibly one change that *did* increase trend growth?

What about the ‘second’ Industrial Revolution?

It seems clear to me that you would hold that the (supposed) ‘Third’ IR (the ‘digital’ revolution) *didn’t* affect g; is that correct?

Thanks again,
=Peter

• Peter
1) Yes, potential GDP is growing at the rate g every period. That’s why potential is 17.34 in period 1 rather than 17.

In terms of the empirics, the thing to remember is that the data just show us the observed change in GDP, or (y_{t+1} – y_t)/y_t. We don’t know how much of that is due to changes in potential, changes in the growth rate g (I’d argue for none, but it could happen), and just random shocks to GDP. And if someone like Summers shows that GDP is growing more slowly since 2007, that doesn’t mean that g has fallen. It almost certainly just means that potential GDP has fallen, and transitional growth is particularly slow.

2) Right, something like that is roughly true. We don’t see g get above 0.0001 until around 1750-1850 depending on exactly where you measure it, and g doesn’t get up to the 1.8-2.0 percent range until the late 1800’s for Europe and the US. But since then it’s been particularly constant. The big difference between those earlier periods and now is really in the population response to income per capita. In the past, it was positive, so that acted as a natural check on GDP growth per capita. Now, population growth tends to decline with income per capita, and that helps keep the rate constant.

2. This is great – thanks.

Does anyone know of any papers which attempt to estimate the effect of reforms on potential GDP by looking at the (small) effects on the growth rates? Something along the lines of a formal version of the bullet point above which begins with “But China was able to do it…”?

• Sean – no, I don’t know of any. And the problem is that even if someone tried, statistically you will never be able to see the effects through all the “noise” in growth rates. We don’t have enough statistical power. The effect sizes we are looking for are small, which means we’d need a gigantic sample to detect the effects. We’ve got 140 years of GDP growth data, and maybe 280 observations at the quarterly frequency. Just not enough to see the effects (positive or negative) of a policy reform on growth rates.


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How to subset ExpressionSet based on vector of sample names
0
0
Entering edit mode
3.2 years ago
nkabo ▴ 60

I have an ExpressionSet object composed of 37 samples for HCC and cirrhosis situation and I would like to subset it according to names that I specified in two vectors. As a result, I want to have 2 expression arrays (one set is for HCC and other is for cirrhosis) after I subset this ExpressionSet. In order to subset the ExpressionSet, I have tried several methods but I could not get the samples and features at the same time.

expset_forall is ExpressionSet and names_HCC and names_cirr are the character vectors containing the names of samples.

This is expset_forall:

ExpressionSet (storageMode: lockedEnvironment)

assayData: 20962 features, 37 samples

protocolData
sampleNames: GSM437457.CEL.gz GSM437458.CEL.gz ... GSM437493.CEL.gz (37 total)...

I have tried:

eset_forHCC = expset_forall[, sampleNames(expset_forall) %in% names_HCC]

it gives error of "incorrect number of dimensions"

then I tried:

eset_forHCC= exprs(expset_forall[expset_forall@protocolData$sampleNames==names_HCC,] dim(eset_forHCC) [1] 0 37  At last, I tried to subset it by reaching via pData: levels(pData(expset_forall)$sampleNames)

it gives "NULL"

As eset_forHCC, I expect the output:

ExpressionSet (storageMode: lockedEnvironment)

assayData: 20962 features, 17 samples

element names: exprs, se.exprs

protocolData

sampleNames: GSM437458.CEL.gz GSM437459.CEL.gz ... GSM437493.CEL.gz (17 total)

As eset_forcirr, I expect the output:

ExpressionSet (storageMode: lockedEnvironment)

assayData: 20962 features, 17 samples

element names: exprs, se.exprs

protocolData

sampleNames: GSM437460.CEL.gz GSM437459.CEL.gz ... GSM437491.CEL.gz (17 total)

R ExpressionSet subset Bioconductor • 1.5k views
1
Entering edit mode

Once you get the ExpressionSet in a data frame object you can try to perform subset() a base function or filter() from dplyr package.

0
Entering edit mode

Does:

eset_forHCC= exprs(expset_forall)[,sampleNames(expset_forall) %in% names_HCC]

not work?

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Entering edit mode

Thank you for your reply, it works but it gives a matrix I should have an ExpressionSet.

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Entering edit mode
eset_forHCC = expset_forall[, sampleNames(expset_forall) %in% names_HCC]

This is correct and recommended way to get the subset. Could you recheck it?

Also, see that the output of sampleNames(expset_forall) %in% names_HCC is as intended.

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Entering edit mode

Thank you for your reply, I checked it again and it works fine but it gives an expression matrix, I would like to have it as ExpressionSet.


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# Math Help - hhhellllpp????????

1. ## hhhellllpp????????

Show that AuB=Bua for any sets A and B.......

2. Originally Posted by kodirl
Show that AuB=Bua for any sets A and B.......
I have told you what you can use to prove that in similar your question.

Use Logic rules.

3. Originally Posted by kodirl
Show that AuB=Bua for any sets A and B.......
I recommend to show that,
$(A\cup B)\subseteq (B\cup A)$
$(B\cup A)\subseteq (A\cup B)$
Then,
$A=B$


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# Hacking Tube

## 0CTF 2015 Quals -- geo-newbie

Talentyange gives lots of tedious apks and you know how bad he is now. Let's try some interesting geography knowledge.

nc 202.112.26.111 29995 / nc 202.112.28.118 29995

So basically we just connect to the server, and it will ask us a bunch of questions about geography. We'll have to pass 3 levels (75 questions) to get the flag.

For level0, it asked us for the alpha2 code of a country. For example, if they gave us "JAPAN", we'll have to send "JP" back to the server. To pass this level, I download a json file from the internet and use it as the alpha2 code database. Just extract the country name and send the corresponding alpha2 code. After answering 20 questions, we pass level0.

For level1, it gave us a location (ex. Mount Fuji), and we'll have to answer the country (in alpha2 code format, ex. Mount Fuji --> JP) that contains the location. To pass the level, I use the google API to solve the problems. Extract the location and pass it as the "address" parameter, the API will respond with a json format response. Parse the response and get the correct country
name. Notice that sometimes google API will respond a wrong country(mostly it respond US instead of the correct country), so we will need to handle some special cases (by hard-code the correct answer in the script). This level has 50 questions, solve them all and we'll be able to reach level2.

And for the last level, the server gave us a river or a moutain (ex. Congo River, Andes Mountains....etc), and we'll have to answer all the countries that the given river/mountain run across. To pass this level, I just hard-code all the answer in the script. First we'll have to reach level2 several times, collect as much rivers/mountains as we can, then google the answer, hard-code them in the script. I add some conditions in my script. If the program encounters a river/mountain that it already has the answer in the script, it will send the answer to the server. Or else, it'll switch to the interactive mode , and I'll take control of the situation -- by google & send the answer with my own hand (you know, speed mode activated!)

Here's the script I used to solve the challenge:

After answering all the questions, we get the flag: 0CTF{eNj0y_geography_l0v3_7hE_w0lRd}


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# Parameter very close to 1 gives me NAN

Dear users,

I have a CES parameter alpha_d which is very close to 1 (0.99999999999919). The simulations give me NAN. This causes a problem for Dynare because in some of my FOCs I have the (1-alpha_d) denominator, which causes me NaN. What should I do ?

Thank you.

Best regards

There is nothing you can do about this. For all numerical purposes 0.99999999999919 is equal to 1. Usually, there exists a limit if the coefficient goes to 1, but you cannot rely on a computer to compute that limit. That needs to be done analytically.

1 Like

Hi, If \alpha_d is the elasticity of substitution (you don’t give any details) you can probably switch to a Cobb Douglas. This can be done with a condition in your code if you use different calibrations for this parameter in the same .mod file.

Best,
Stéphane.

Thank you @stepan-a

Alpha_d is a CES distribution parameter, hence the issue…

But then cannot you just rewrite the model by removing one production factor (if it’s a production function)?

Best,
Stéphane.

Dear @stepan-a,

Here is the issue. I have a CES function that combines durables and energy goods:

Initially alpha_d is equal to 0.97 with sigma_es being equal to 0.75. So there is no issue here.

I want to perform senstivity analysis on that sigma_es (the elasticity of sub between the two goods) and fix its value value close to perfect complementarity, let’s say 0.1. But that value of sigma_es, gives me a value for alpha_d of 0.999999999919. My whole model is built to see the interaction between consumers durables and energy. So taking out energy from the CES function would discredit the model.

I think I’m just going to lower sigma_es to 0.45 for the sensitivity analysis so alpha-d does not get close too 1.

Thank you.

best regards

As @stepan-a and I said: you are moving extremely close to a limit case here: perfect complementarity. In that case, you end up with a Leontief production function. This leads to numerical problems. There are two ways to proceed:

1. Consider the exact limit case by computing it analytically instead of relying on numerics (what @stepan-a suggested in the form of a macro-processor case distinction)
2. Stay further away from the limit in order to avoid numerical issues (what you suggested)
1 Like

Note that in the perfect complementarity case (with a Leontieff technology) you will have to use a min function, which is available in dynare language, but only makes sense with a perfect foresight model.With a local approximation you would always stay along a branch of the Leontieff function, this is the same issue as with the ZLB (non differentiable function). I do not know which steady state restrictions you use for calibrating \alpha_d, but did you try alternative formulation of the CES production (for instance as the one used in Barro Sala-i-Martin’s textbook in section 1.3.4)?

Best,
Stéphane.

1 Like

Yes thank you. It works now.

Best regards


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# Features for time series classification

I consider the problem of (multiclass) classification based on time series of variable length $T$, that is, to find a function $$f(X_T) = y \in [1..K]\\ \text{for } X_T = (x_1, \dots, x_T)\\ \text{with } x_t \in \mathbb{R}^d ~,$$ via a global representation of the time serie by a set of selected features $v_i$ of fixed size $D$ independent of $T$, $$\phi(X_T) = v_1, \dots, v_D \in \mathbb{R}~,$$ and then use standard classification methods on this feature set. I'm not interested in forecasting, i.e. predicting $x_{T+1}$. For example, we may analyse the way a person walks to predict the gender of the person.

What are the standard features that I may take into account ? In example, we can obviously use the mean and variance of the serie (or higher order moments) and also look into the frequency domain, like the energy contained in some interval of the Discrete Fourier Transform of the serie (or Discrete Wavelet Transform).

## Simple statistical features

• Means in each of the $d$ dimensions
• Standard deviations of the $d$ dimensions
• Skewness, Kurtosis and Higher order moments of the $d$ dimensions
• Maximum and Minimum values

## Time serie analysis related features

• The $d \times d-1$ Cross-Correlations between each dimension and the $d$ Auto-Correlations
• Orders of the autoregressive (AR), integrated (I) and moving average (MA) part of an estimated ARIMA model
• Parameters of the AR part
• Parameters of the MA part

## Frequency domain related features

See Morchen03 for a study of energy preserving features on DFT and DWT

• frequencies of the $k$ peaks in amplitude in the DFTs for the detrended $d$ dimensions
• $k$-quantiles of these DFTs
• Emile, This question is similar to one I just posed (stats.stackexchange.com/questions/51475/…). Would you be able to post any R code for the DFT features? – B_Miner Mar 6 '13 at 15:43
• Is there any method based on shapelets for variable length time series? – Simone Jun 9 '15 at 4:18

Emile, I think the features listed in your answer are pretty good starting points, though as always, I think some domain expertise (or at least a good long think) about your problem is equally important.

You may want to consider including features calculated from the derivatives (or integrals) of your signal. For example, I would wager that rapid acceleration/deceleration is a reasonably good predictors of accident-prone driving. That information is obviously still present in the position signal, but it's not nearly as explicit.

You may also want to consider replacing the Fourier coefficients with a wavelet or wavelet packet representation. The major advantage of wavelets is that they allow you to localize a feature in both frequency and time, while the traditional Fourier coefficients are restricted to only time. This might be particularly useful if your data contains components that switch on/off irregularly or has square wave-like pulses that can be problematic for Fourier methods.

As the other answers suggested, there is a huge number of time series characteristics that can be used as potential features. There are simple features such as the mean, time series related features such as the coefficients of an AR model or highly sophisticated features such as the test statistic of the augmented dickey fuller hypothesis test.

# Comprehensive Overview over possible time series features

The python package tsfresh automates the extraction of those features. Its documentation describes the different calculated features. You can find the page with the calculated features here.

Disclaimer: I am one of the authors of tsfresh.

I suggest you, instead of using classic approaches for extracting hand-engineered features, utilise autoencoders. Autoencoders plays an important role in the feature extraction of deep learning architecture.

The autoencoder tries to learn a function $f(X_T)≈X_T$. In other words, it is trying to learn an approximation to the identity function, so as to output $\hat X_T$ that is similar to $X_T$.

The identity function seems a particularly trivial function to be trying to learn; but by placing constraints on the network, such as by limiting the number of hidden units, we can discover interesting structure about the data.

In this way, your desired $ϕ(X_T)=v_1,…,v_D \in R$ will be equivalent to the output values of the middlemost layer in a deep autoencoder, If you limit the number of hidden units in the middlemost to $D$.

Additionally, you can use many flavors of autoencoder for finding the best solution to your problem.

The linked paper will be somewhat enlightening, since it is interested in the more or less the same issue in another context.

Paper abstract (in the Internet Archive)

Paper PDF

Depending on the length of your time series, the usual approach is to epoch the data into segments, e.g. 10 secs.

However, often prior to breaking the time-series into segments it is necessary to perform some preprocessing such as filtering and artifact rejection. You can then compute a variety of features such as those based on frequency (i.e. take an FFT for each epoch), time (e.g. mean, variance etc of the time-series in that epoch) or morphology, (i.e. the shape of the signal/time-series in each epoch).

Usually the features used to classify segments (epochs) of a time-series/signal are domain-specific but Wavelet/Fourier analysis are simply tools to allow you examine your signal in the frequency/time-frequency domains rather than being features in themselves.

In a classification problem each epoch will have a class label e.g. 'happy' or 'sad', you would then train a classifier to distinguish between 'happy' and 'sad' epochs using the 6 features calculated for each epoch.

In the event that each time series represents a single case for classification, you need to calculate each feature across all samples of the time series. The FFT is only relevant here if the signal is linear time invariant (LTI), i.e. if the signal can be considered to be stationary over the whole time series, if the signal is not stationary over the period of interest, a wavelet analysis may be more appropriate. This approach will mean that each time series will produce one feature vector and will constitute one case for classification.

• I've always felt that breaking time series, a naturally continuous process, into discrete periods (or epochs), results in a loss of information. Unless there is natural epochs in the series, how does one choose epochs? Likely epochs can be found that fits any desired outcome. – Cam.Davidson.Pilon Feb 25 '13 at 16:02
• Not sure that an epoch can be found to suit every outcome but for any practical problem With a non-stationary signal you need to find someway to take into account temporal variations (if a signal/time-series adequately described analytically or is linear time invariant this not necessary). The epoch length is again domain specific but usually chosen to be short enough that the signal is stationary under the time period (epoch) of interest. – BGreene Feb 25 '13 at 16:10
• BGreene, could you explain what is the benefit of partitioning the serie into epochs for selecting features used in classification ? I see this partitioning as a helper for later computations (i.e. FFT), but not as something related to the selection of features themselves. Maybe it is related to the "morphology" you mentioned. – Emile Feb 25 '13 at 16:20
• The partitioning into epochs has nothing to do with feature selection. If you have a long signal (e.g. 10 hours of recording sampled @100Hz) then in order to examine how the signal changes over time you need to break it into epochs. In a classification problem each epoch will have a class label e.g. 'happy' or 'sad', you would then train a classifier to distinguish between 'happy' and 'sad' epochs using the 6 features calculated for each epoch. – BGreene Feb 25 '13 at 16:39
• Ok:) It is definitely unrelated to my original question. I consider the case where the class labels the whole serie. I'm going to edit my question to add a clarifying example. – Emile Feb 25 '13 at 16:52


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Glaucoma is a group of eye diseases causing optic nerve damage. The optic nerve carries images from the retina, which is the specialized light sensing tissue, to the brain so we can see. In glaucoma, eye pressure plays a role in damaging the delicate nerve fibers of the optic nerve. When a significant number of nerve fibers are damaged, blind spots develop in the field of vision. Once nerve damage and visual loss occur, it is permanent. Most people don’t notice these blind areas until much of the optic nerve damage has already occurred. If the entire nerve is destroyed, blindness results. Glaucoma is a leading cause of blindness in the world, especially in older people. Early detection and treatment by your ophthalmologist are the keys to preventing optic nerve damage and vision loss from glaucoma..- Kellogg Eye Center
By Melissa Wilson, Anchor & Health Reporter
HOUSTON (FOX 26) –
Glaucoma is one of the leading causes of blindness. The problem is, it can damage your vision so gradually, you may not notice any loss of vision.
Early detection and new advances help to diagnose it. It’s a good idea to get an exam every year to check for it. If your doctor believes you have vision loss, you should consider being seen twice a year. “We test the pressure of the eye. Lower interocular pressure protects the nerve fibers, or the wire, that connects the eyes to the brain.
There is not one single cause for glaucoma. We could say high interocular pressure or high pressure in the eye causes glaucoma. The question is: what causes the pressure to be high in the first place? What I find personally is that lymphatic drainage may be blocked in the upper torso when you have glaucoma. There are other theories from chronic inflammation at the drainage system, that drains the water out of the eye, to other types of glaucoma that are related to anatomy. There are many types of glaucoma, but 90% of all glaucomas are related to pressure.
Dr. Arroyo says this is what raises your risks of glaucoma:

• African-American, Irish, Russian, Japanese, Hispanic, Inuit, or Scandinavian descent
• Over age 40
• Family history of glaucoma                                                                                Read & Watch the video: http://www.myfoxhouston.com/story/29116194/detecting-glaucoma-early                                                                                Source: My Fox Houston
On


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# What would be the impact on tides if the earth had no tilt?

I would like to know what would be the impacts/effects on the tides if earth were to have no tilt (perpendicular to plane of ecliptic)?

• What's the actual problem that you are trying to solve? – 410 gone Aug 28 '16 at 14:57
• What other assumptions are you considering (i.e. does the Earth still have a moon whose orbit is tilted about 5° from the ecliptic, that that moon is about the same size; or do we assume that there is no moon, or that, since the Earth was never knocked on its side, that the moon orbits in the ecliptic)? – BillDOe Aug 28 '16 at 23:12
• The question is too general to be answerable. In the mid-oceans the tidal bulge is very small. It is greatly amplified close to land, and is highly sensitive to the continental slope and land configuration. To answer in full would require running a global program on a super-computer. There's nothing wrong with running 'what-if' models, but there has to be a point. 'Zero tilt' will never happen, so is there a point to this? – Gordon Stanger Aug 29 '16 at 2:13
• in my science class, we have to write a report and one of the sections is focussed on if the earth had a differnt tilt, such as no tilt, 90 degree tilt etc; we then have to do research on this. I chose no tilt, and could not find any information on this, so that is why I asked. In this case, i am assuming that the earth still as a moon, and that moon does orbit in the ecliptic. I do agree though, it is kind of pointless, as it will never happen. -Kait- – Kait Aug 29 '16 at 6:51
• I think the question has some clear answers as can be seen below. "What if" questions are extremely useful as they can help simplify very complex problems. In this case, it helps understand the way the different tidal constituents are originated and their relative effects. – arkaia Aug 29 '16 at 16:00

A simple way to estimate the implications is to simplify the problem. If the orbits of Earth, Sun and Moon were circular and in the same plane, and the Earth had no tilt, the only remaining tidal constituents would be $M_2$ and $S_2$ (also the overtides and combination tides: $M_4$, $MS_4$...). The rest of the tidal constituents can be expressed as linear combinations of their rates of change based on the remaining forcings: orbital eccentricity and inclination and axial tilt.

The modulation of the amplitude $A_M$ of the main constituent wave, $\eta_M = A_M cos (\omega_M t)$, by another constituent of amplitude $A_c$ is: $\eta_t= (A_M+A_ccos(\omega_c t))cos(\omega_Mt)$. Expressing it as a linear combinations of the rates of change gives $\eta_t= A_Mcos(\omega_Mt)+{A_c\over2}cos(\omega_M-\omega_c)t+{A_c\over2}cos(\omega_M+\omega_c)t$, such that the original amplitude remains unchanged, while two extra components with slightly different frequencies appear.

In the case of the effect of the axial tilt, it modulates the main frequencies based on the solar parameters (the inclination of the Moon orbit is not affected by the axial tilt). The period of the solar declination is a sidereal year (365.2564 days). In the case of the modulation of the $M_2$, the resulting periodicities are 1 cycle/solar day ± 1 cycle/sidereal year. The two resulting constituents are $P_1$ and $K_1$. $P_1$ is known as the solar declinational diurnal constituent. The case of $K_1$ is slightly more complex as the inclination of the Moon also results in a contribution at this frequency. The rest of the tidal constituents are also modified accordingly, but their amplitudes are much smaller.

In short, the removal of the axial tilt will result in the removal of the $P_1$ tidal constituent (and some other minor ones) and an amplitude decrease in the $K_1$ constituent.

$K_1$ amplitude and phase from AVISO. The maximum $K_1$ amplitudes of around 0.4m are found south of Alaska and west of the Kamchatka Peninsula in the Sea of Okhotsk. The amplitude of the $P_1$ constituent is about half of $K_1$ with a maximum of around 0.2m in the Sea of Okhotsk. Overall, the effect of the tilt on tides is pretty minor relative to the rest of the effects.

I cannot answer your hypothetical question, but the tilt does probably affect the tides and there have been some studies on the topic. Hopefully they can inspire to further ideas and speculations. It appears that the main outcome from the report should be for you to understand how the moon, the Sun and the Earth control the tide and how the tidal force works.

Change in axial tilt over time from Laskar (1986)

We cannot remove the tilt completely, but we can look at the constant axial tilt change in the Milankovitch cycles. Some believe that the impact of this 2.4° change is rather large. Clive Best blog gives the main arguments and suggests some good references. See also this discussion and this paper on tidal change and glaciation.

However, it's difficult to specify how much of the impact that is directly related to the tilt change, as sea level change and glaciation also affects and interacts with the tides pattern.

This model from Wilmes and Green (2014) suggests much higher tides at high latitudes during LGM, 21.000 years ago. The model is based on changes in sea-level and glaciation.

From the astronomical data, it however does appear that the impact (tractional acceleration) of the Moon was much higher at large latitudes when the obliquity of the tilt is large (see figure) and this resulted in higher tides at latitudes where glaciation forms.

With less or no tilt, tides at large latitudes should be even smaller and maybe contribute to less breakup of ice sheets.


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# Integration Short Cuts

Integration has all sorts of useful applications. For example, you might have been given a function definition that tells you how quickly students come up with excuses to avoid doing extra homework and want to work out the total number of excuses that have been made in the past 5 days. Integration will help you out here as it gives you a way to find the area under the 'rate of making excuses' curve between day 0 and day 5:
$\displaystyle{\text{Total Excuses} = \int_0^5 \frac{d\text{Excuse}}{dt} \; dt}$.
This area represents the total number of excuses made over these 5 days.

Other applications might include finding volumes of solids, areas under more usual graphs like the one in the picture (no you don't really have to cut the area up into little squares and count them), lengths of arcs, velocities of objects whose acceleration functions have been given, distances travelled, and centres of mass. Physicists really know how to exploit a good integral!

So that you don't have to continually calculate integrals using limits of sums of shapes that you can fit under, or over (or around) the curve, there's a list of integrals of common functions that you can look up. There are also rules that can help you to calculate the antiderivatives of wonkier functions, and a handy little theorem called the Fundamental Theorem of Calculus that actually lets you evaluate definite integrals, and retrieve useful information like the total number of excuses from the rate of excuses curve. Why not have a look at the article about definite integrals for more information on this useful topic?

In this article, we'll give you a brief overview of the rules of integration, as well as the antiderivatives of some common functions. We'll also work through a few examples so that you can get a better idea of how the common integrals and rules of integration are applied. If you're interested in finding out more, have a look at the articles on integration by substitution and integration by parts. Oh and, by the way, integration is sometimes viewed as the opposite of differentiation, so you might see an indefinite integral (one without limits) referred to as an antiderivative in many books.

### Some Common Antiderivatives (Integrals)

 Function description Integral expression Antiderivative Constant Function ($k$ is for 'konstant') $\displaystyle{\int k \; dx}$ $kx + C$ Linear Variable $\displaystyle{\int x \; dx}$ $\dfrac{x^2}{2} + C$ Square $\displaystyle{\int x^2 \; dx}$ $\dfrac{x^3}{3} + C$ Reciprocal $\displaystyle{\int \dfrac{1}{x}\;dx}$ $\ln |x| + C$ Exponentials $\displaystyle{\int e^x \;dx}$ $e^x + C$ Exponentials $\displaystyle{\int a^x \;dx}$ $\dfrac{a^x}{\ln (a)} + C$ Logarithms $\displaystyle{\int \ln (x) \;dx}$ $x\ln (x) - x + C$ Trig functions (don't forget $x$ is measured in radians!) $\displaystyle{\int \sin x \; dx}$ $-\cos x + C$ $\displaystyle{\int \cos x \; dx}$ $\sin x + C$ $\displaystyle{\int \sec^2 x \; dx}$ $\tan x + C$

### Rules of Integration

And now for the rules that help you stitch the above antiderivatives together to come up with the integrals of some really impressive functions. This is great fun at parties! We'll start with the basics, and deal with the more complicated ones like integration by substitution and integration by parts in later articles.

RuleIntegral expressionTransformed Integral
Multiplication by a constant $\displaystyle{\int kf(x) \; dx}$ $\displaystyle{k\int f(x) \;dx}$
Power Rule ($n \neq -1$) $\displaystyle{\int x^n \; dx}$ $\dfrac{x^{n + 1}}{n + 1} + C$
Sum of Integrals $\displaystyle{\int (f + g)(x) \; dx}$ $\displaystyle{ \int f(x) \; dx + \int g(x) \; dx}$
Difference of Integrals $\displaystyle{\int (f - g)(x) \; dx}$ $\displaystyle{\int f(x) \; dx - \int g(x) \; dx}$
Mathematicians have a special name for the first rule and the last two rules. They refer to these properties as the linearity of the integral. Basically, all this means is that integrals behave the way you want them to when it comes to multiplication by products, sums and differences. You can pull the constant out the front of the integral, and you can split the integral up into bite sized pieces across the plus and minus signs. The property is called linearity because it's the same property that straight line functions have. If $f(x)$ is a straight line function, then $f(kx + y) = kf(x) + f(y)$ just like $\displaystyle{\int kf(x) + g(x) \; dx = k \int f(x) \; dx + \int g(x) \; dx}$.

Unfortunately, products don't play quite so nicely, but then they weren't so friendly with derivatives either. Techniques such as integration by substitution and integration by parts will help us to deal with the integrals of products of functions.

## Examples of applying the rules

Let's finish by using the rules to find the integrals of some functions. Don't forget that the table of common antiderivatives will be a big help here.

### Example 1

Find the integral of $\displaystyle{\sec^2 (x)}$.

In the table of common antiderivatives, the antiderivative of $\sec^2 (x)$ is listed as $\tan (x) + C$. So

$\displaystyle{\int \sec^2(x)\; dx= \tan (x) + C}$.

### Example 2: applying the power rule

Evaluate $\displaystyle{\int x^4 \; dx}$.

Use the power rule with $n = 4$:

$\displaystyle{ \int x^4\; dx = \dfrac{x^5}{4 + 1} + C = \dfrac{x^5}{5} + C}$.
Watch out! Lots of students get confused here and apply the power rule for derivatives. Try to remember: when you're integrating, the power of $x$ goes up. When you're differentiating, the power of $x$ goes down ("d" for "differentiation").

### Example 3: more on the power rule - fractional indices

Evaluate $\displaystyle{\int \sqrt{x}\; dx}$.

Remember your index laws: $\sqrt{x} = x^{\frac{1}{2}}$, so we can use the power rule here.

$\displaystyle{\int \sqrt{x} \; dx = \int x^{\frac{1}{2}}\; dx = \dfrac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C = \dfrac{x^{1.5}}{1.5} + C}.$

### Example 4:constant multiplication

Find $\displaystyle{\int 7x^3\; dx}$.

Bring the $7$ out to the front of the integral:

$\displaystyle{\int 7x^3 \; dx = 7\int x^3\; dx}.$
Now use the power rule on the $x^3$:
$\displaystyle{\int 7x^3 \; dx = 7\int x^3\; dx = 7 \left( \dfrac{x^4}{4} + C\right) = \dfrac{7x^4}{4} + C'},$
where $C' = 7C$.

### Example 5: the integral of a sum is the sum of the integrals

Find $\displaystyle{\int \sec^2 x + x^2 \; dx}.$

$\displaystyle{\int \sec^2 x + x^2 \; dx = \int \sec^2 x\; dx + \int x^2 \; dx}.$
Now use the table of common integrals to work out the integral of each part:
$\displaystyle{\int \sec^2 x + x^2 \; dx = \int \sec^2 x\; dx + \int x^2 \; dx = \tan x + \dfrac{x^3}{3} + C.}$

### Example 6: the integral of a difference is the difference of the integrals

Find $\displaystyle{\int \cos x - e^x\; dx}$.

$\displaystyle{\int \cos x - e^x \; dx = \int \cos x\; dx - \int e^x \; dx}.$.
Now use the table of common integrals to work out the integral of each bit:
$\displaystyle{\int \cos x - e^x \; dx = \int \cos x\; dx - \int e^x \; dx = \sin x - e^x + C}$.

### Example 7: using all the rules together (exploiting the linearity of the integral)

Find the integral of $\displaystyle{\int 5\sin x + 3x^3 - e^x \; dx}$.

Use the sum and difference rules to split the integral into three parts:

$\displaystyle{\int 5\sin x + 3x^3 - e^x \; dx = \int 5\sin x\; dx + \int 3x^3 \; dx - \int e^x\; dx}$.
Next use the rule about constant multiplication to pull the constants out to the front of the integrals:
$\displaystyle{\int 5\sin x + 3x^3 - e^x \; dx = \int 5\sin x\; dx + \int 3x^3 \; dx - \int e^x\; dx = 5 \int \sin x \; dx + 3 \int x^3 \; dx - \int e^x \; dx}$.
Finally, use the power rule and common integrals to evaluate each part:
$\displaystyle{\int 5\sin x + 3x^3 - e^x \; dx = 5 \int \sin x \; dx + 3 \int x^3 \; dx - \int e^x \; dx = - 5 \cos x + \frac{3x^4}{4} - e^x + C }$.
So, we've exploited the linearity of the integral to calculate:
$\displaystyle{\int 5\sin x + 3x^3 - e^x \; dx = - 5 \cos x + \frac{3x^4}{4} - e^x + C }$.

## What about all those constants?

You'll have noticed that we've added a constant to the end of each integral we've calculated. If you've done integration at school, your teacher will have insisted that you add constants to indefinite integrals. For each indefinite integral, we've found a family of functions that could be differentiated to give the integrand (the thing we're integrating). The effect of different values of the constant is to move the graph of the function (and all its values) up and down the $xy$-plane. Sometimes a problem will give you a condition that the antiderivative must satisfy, such as $F(0) = 0$. You can plug this condition into the formula for the anti-derivative to work out what the constant is.

### Example:

Find the function $\displaystyle{F(x) = \int x^3 \;dx}$ that satisfies $F(2) = 5$.

First find the indefinite integral using the power rule:

$\displaystyle{F(x) = \int x^3 \; dx = \frac{x^4}{4} + C}$.
Next, plug in $F(2) = 5$ and solve to find the value of $C$:
\begin{align*} F(2) = 5 &= \dfrac{2^4}{4} + C \\ &=\dfrac{16}{4} + C\\ &= 4 + C\\ C &= 1. \end{align*}.
So $F(x) = \dfrac{x^4}{4} + 1$.

### Description

Calculus is the branch of mathematics that deals with the finding and properties of derivatives and integrals of functions, by methods originally based on the summation of infinitesimal differences. The two main types are differential calculus and integral calculus.

### Environment

It is considered a good practice to take notes and revise what you learnt and practice it.

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# Solving A Factorial Inequality Without Trial And Error

How does one solve the following inequality for n, without trial and error, and assuming n can only be an integer?

The inequality is $(n+1)!-1>10^9$.

I want to find the minimum value of n such that $(n+1)!-1>10^9$. How does one do this without graphing the inequality, or using a calculator? I figured it out throwing trial and error on a calculator, but I desire a more elegant solution, one that gets to the answer algebraically, and without a calculator or graph. Any suggestions?

• This inequality does not have an algebraic solution. Trial and error is necessary. The number of trials can be pretty small if you're clever about using the information you learn from each failure. You could edit the question to show us your method and ask if there's a more efficient one. (Question: why is $n$ raised to the first power? Is that a typo?) – Ethan Bolker Sep 5 '18 at 13:27
• @EthanBolker I see. Thank you. – user590211 Sep 5 '18 at 13:28
• @EthanBolker I had no method though, I literally just brute forced my way to an answer. What is one more efficient way you suggest? – user590211 Sep 5 '18 at 13:31
• Do we want the least positive integer or the least value that satisfies this? Because they need two very different methods – Rhys Hughes Sep 5 '18 at 13:56
• The Gamma function (coincides with the factorial on positive integers) has no known elementary inverse. See here: math.stackexchange.com/questions/931846/…. – Adrian Keister Sep 7 '18 at 0:07

A possible way would be to use upper and lower bound to the factorial function. There are of course very precise ones, but you said you don't want to use a calculator, so we take simple ones:

$$\left(\frac{n}{2}\right)^{\frac{n}{2}}\leq n!\leq n^n$$

(I leave the simple proofs to you, as they are quite insightful).

Then we first want to solve the easier inequation $m!>10^9$. We immediately see that we must have $m>9$. This already gives you some starting point. An upper bound on the other hand is 20. It is still somewhat bruteforce.

A better approximation is $n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ which will help you more to get a good starting guess.

• If we ignore the square root in Stirling, one can notice that $27 \approx 10e$ so $27! \approx 10^{27}$. That can be handy in estimating factorials. The square root gives another factor $10$ but that is not much in this range. – Ross Millikan Sep 6 '18 at 0:00

If you know that $10!$ is between $3$ and $4$ million, then it seems that you would need to go three terms further in the factorial sequence (multiplying by $11$, $12$, $13$) to first exceed $1$ billion.

• @Deepesh Meena, Fair enough. I would still start with the highest factorial I knew as a starting point and then try to figure out (or estimate) how many more factors I might need. – paw88789 Sep 5 '18 at 13:46

Observe that between $1$ and $10$, multiplying $1$ and $10$, $2$ and $9$, etc., every pairwise product contributes at least a factor of $10$. Thus $10!$ is at least $10^5$ (but of course you can deduce quite easily that it must actually be at least $10^6$). Then multiplying by $11$ and $12$ contributes about $10^2$. Now we know that $12!>10^8$. So the largest possible value of $n+1$ is now $13$, and you only have to try $12!$ to show that it is indeed smaller than $10^9$. Minimal brute force involved!


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High-redshift JWST predictions from IllustrisTNG: dust modelling and galaxy luminosity functions
ABSTRACT The James Webb Space Telescope (JWST) promises to revolutionize our understanding of the early Universe, and contrasting its upcoming observations with predictions of the Λ cold dark matter model requires detailed theoretical forecasts. Here, we exploit the large dynamic range of the IllustrisTNG simulation suite, TNG50, TNG100, and TNG300, to derive multiband galaxy luminosity functions from z = 2 to z = 10. We put particular emphasis on the exploration of different dust attenuation models to determine galaxy luminosity functions for the rest-frame ultraviolet (UV), and apparent wide NIRCam bands. Our most detailed dust model is based on continuum Monte Carlo radiative transfer calculations employing observationally calibrated dust properties. This calibration results in constraints on the redshift evolution of the dust attenuation normalization and dust-to-metal ratios yielding a stronger redshift evolution of the attenuation normalization compared to most previous theoretical studies. Overall we find good agreement between the rest-frame UV luminosity functions and observational data for all redshifts, also beyond the regimes used for the dust model calibrations. Furthermore, we also recover the observed high-redshift (z = 4–6) UV luminosity versus stellar mass relation, the H α versus star formation rate relation, and the H α luminosity function at z = 2. The bright end more »
Authors:
; ; ; ; ; ; ; ; ; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10104829
Journal Name:
Monthly Notices of the Royal Astronomical Society
Volume:
492
Issue:
4
Page Range or eLocation-ID:
5167 to 5201
ISSN:
0035-8711
National Science Foundation
##### More Like this
1. ABSTRACT We present predictions for high redshift (z = 2−10) galaxy populations based on the IllustrisTNG simulation suite and a full Monte Carlo dust radiative transfer post-processing. Specifically, we discuss the H α and H β + $[\rm O \,{\small III}]$ luminosity functions up to z = 8. The predicted H β + $[\rm O \,{\small III}]$ luminosity functions are consistent with present observations at z ≲ 3 with ${\lesssim} 0.1\, {\rm dex}$ differences in luminosities. However, the predicted H α luminosity function is ${\sim }0.3\, {\rm dex}$ dimmer than the observed one at z ≃ 2. Furthermore, we explore continuum spectral indices, the Balmer break at 4000 Å; (D4000) and the UV continuum slope β. The median D4000 versus specific star formation rate relation predicted at z = 2 is in agreement with the local calibration despite a different distribution pattern of galaxies in this plane. In addition, we reproduce the observed AUV versus β relation and explore its dependence on galaxy stellar mass, providing an explanation for the observed complexity of this relation. We also find a deficiency in heavily attenuated, UV red galaxies in the simulations. Finally, we provide predictions for the dust attenuation curves of galaxies at z = 2−6 and investigate their dependence on galaxy colours andmore »
2. ABSTRACT

We post-process galaxies in the IllustrisTNG simulations with skirt radiative transfer calculations to make predictions for the rest-frame near-infrared (NIR) and far-infrared (FIR) properties of galaxies at z ≥ 4. The rest-frame K- and z-band galaxy luminosity functions from TNG are overall consistent with observations, despite ${\sim}0.5\, \mathrm{dex}$ underprediction at z = 4 for MK ≲ −25 and Mz ≲ −24. Predictions for the JWST MIRI observed galaxy luminosity functions and number counts are given. Based on theoretical estimations, we show that the next-generation survey conducted by JWST can detect 500 (30) galaxies in F1000W in a survey area of $500\, {\rm arcmin}^{2}$ at z = 6 (z = 8). As opposed to the consistency in the UV, optical, and NIR, we find that TNG, combined with our dust modelling choices, significantly underpredicts the abundance of most dust-obscured and thus most luminous FIR galaxies. As a result, the obscured cosmic star formation rate density (SFRD) and the SFRD contributed by optical/NIR dark objects are underpredicted. The discrepancies discovered here could provide new constraints on the sub-grid feedback models, or the dust contents, of simulations. Meanwhile, although the TNG predicted dust temperature and its relations with IR luminosity and redshiftmore »

3. ABSTRACT

The ionizing photon escape fraction [Lyman continuum (LyC) fesc] of star-forming galaxies is the single greatest unknown in the reionization budget. Stochastic sightline effects prohibit the direct separation of LyC leakers from non-leakers at significant redshifts. Here we circumvent this uncertainty by inferring fesc using resolved (R > 4000) Lyman α (Lyα) profiles from the X-SHOOTER Lyα survey at z = 2 (XLS-z2). With empirically motivated criteria, we use Lyα profiles to select leakers ($f_{\mathrm{ esc}} > 20{{\ \rm per\ cent}}$) and non-leakers ($f_{\mathrm{ esc}} < 5{{\ \rm per\ cent}}$) from a representative sample of >0.2L* Lyman α emitters (LAEs). We use median stacked spectra of these subsets over λrest ≈ 1000–8000 Å to investigate the conditions for LyC fesc. Our stacks show similar mass, metallicity, MUV, and βUV. We find the following differences between leakers versus non-leakers: (i) strong nebular C iv and He ii emission versus non-detections; (ii) [O iii]/[O ii] ≈ 8.5 versus ≈3; (iii) Hα/Hβ indicating no dust versus E(B − V) ≈ 0.3; (iv) Mg ii emission close to the systemic velocity versus redshifted, optically thick Mg ii; and (v) Lyα fesc of ${\approx} 50{{\ \rm per\ cent}}$ versus ${\approx} 10{{\ \rm per\ cent}}$. The extreme equivalent widths (EWs) in leakers ([O iii]+$\mathrm{ H}\beta \approx 1100$ Å rest frame)more »

4. Abstract

We investigate the relationship between dust attenuation and stellar mass (M*) in star-forming galaxies over cosmic time. For this analysis, we compare measurements from the MOSFIRE Deep Evolution Field survey atz∼ 2.3 and the Sloan Digital Sky Survey (SDSS) atz∼ 0, augmenting the latter optical data set with both UV Galaxy Evolution Explorer (GALEX) and mid-infrared Wide-field Infrared Survey Explorer (WISE) photometry from the GALEX-SDSS-WISE Catalog. We quantify dust attenuation using both spectroscopic measurements of Hαand Hβemission lines, and photometric measurements of the rest-UV stellar continuum. The Hα/Hβratio is used to determine the magnitude of attenuation at the wavelength of Hα,AHα. Rest-UV colors and spectral energy distribution fitting are used to estimateA1600, the magnitude of attenuation at a rest wavelength of 1600 Å. As in previous work, we find a lack of significant evolution in the relation between dust attenuation andM*over the redshift rangez∼ 0 toz∼ 2.3. Folding in the latest estimates of the evolution ofMdust, (Mdust/Mgas), and gas surface density at fixedM*, we find that the expectedMdustand dust mass surface density are both significantly higher atz∼ 2.3 than atz∼ 0. These differences appear at odds with the lack of evolution in dust attenuation. To explain the striking constancymore »

5. Abstract We present Ly α and ultraviolet (UV)-continuum luminosity functions (LFs) of galaxies and active galactic nuclei (AGNs) at z = 2.0–3.5 determined by the untargeted optical spectroscopic survey of the Hobby–Eberly Telescope Dark Energy Experiment (HETDEX). We combine deep Subaru imaging with HETDEX spectra resulting in 11.4 deg 2 of fiber spectra sky coverage, obtaining 18,320 galaxies spectroscopically identified with Ly α emission, 2126 of which host type 1 AGNs showing broad (FWHM > 1000 km s −1 ) Ly α emission lines. We derive the Ly α (UV) LF over 2 orders of magnitude covering bright galaxies and AGNs in log L Ly α / [ erg s − 1 ] = 43.3 – 45.5 (−27 < M UV < −20) by the 1/ V max estimator. Our results reveal that the bright-end hump of the Ly α LF is composed of type 1 AGNs. In conjunction with previous spectroscopic results at the faint end, we measure a slope of the best-fit Schechter function to be α Sch = − 1.70 − 0.14 + 0.13 , which indicates that α Sch steepens from z = 2–3 toward high redshift. Our UV LF agrees well with previous AGN UVmore »


text:
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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01rn3011488
 Title: Molecular mechanisms of Us9-mediated anterograde transport of alphaherpes viruses Authors: Kratchmarov, Radomir Advisors: Enquist, Lynn Department: Molecular Biology Class Year: 2013 Abstract: Alphaherpes viruses infect the nervous system of their hosts and undergo long-distance transport in neuronal axons during different steps of the life cycle. Active viral replication at the neuronal soma produces progeny virions, which must undergo anterograde transport down the axon to facilitate anterograde spread within the host. Throughout this work, the molecular mechanisms underlying anterograde transport are explored using pseudorabies virus infection as a model system. Several functional domains in the viral protein Us9, which is essential for anterograde transport, were characterized through a convergence of methodologies including live-cell imaging of fluorescent viruses as well as in vitro spread and biochemical assays. In Chapter 1, GFP-Us9 fusion proteins are employed to characterize the role of a dityrosine motif in the protein in anterograde transport of virion structural components. The dityrosine motif was required for anterograde neuron-to-cell spread in vitro as well as for axonal targeting of virion structural components. GFP-Us9 fusion proteins are then employed further in Chapter 2 to characterize the Us9 diserine motif. Interestingly, unlike the dityrosine motif, mutagenesis of the diserine motif resulted in only a modest defect in spread and was found to modulate the efficiency of anterograde transport. In Chapter 3, the requirement of Us9 for transport of viral membrane proteins is assessed for particles that do not constitute mature virions. Interestingly, we established that the viral glycoprotein M is capable of undergoing anterograde transport independently of Us9. A summation of our current understanding of the anterograde transport mechanism, known as the Married Model, is then presented in Chapter 4 through a critical analysis of experimental techniques in published works. Finally, Appendix A contains a preliminary investigation of glycoprotein E functionality with respect to Us9-Kif1A interactions and anterograde transport of virions. Together, these results expand our understanding of Us9 functionality and biochemical properties. Extent: 137 pages URI: http://arks.princeton.edu/ark:/88435/dsp01rn3011488 Access Restrictions: Walk-in Access. This thesis can only be viewed on computer terminals at the Mudd Manuscript Library. Type of Material: Princeton University Senior Theses Language: en_US Appears in Collections: Molecular Biology, 1954-2020

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# Hectogon

A hectogon

A hectogon or hectagon is a shape with 100 sides and 100 corners.

## Regular hectogon

All sides of a regular hectogon are the same length. Each corner is 176.4°. All corners added together equal 17640°.

### Area

The amount of space a regular hectogon takes up is

The area of a regular hectogon of side length a is given by

${\displaystyle A=25a^{2}\cot {\frac {\pi }{100}}\simeq 795.513a^{2}}$


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# What does the Fundamental Theorem of Algebra say about the number of complex zeros of a polynomial function?

I was watching the Khan Academy video on the Fundamental Theorem of Algebra when I got confused by something that Sal Khan states. From what I understand, the Theorem says that the complex zeros of a polynomial function always come in pairs because they are conjugates of each other. But exactly what would happen with a 3rd degree polynomial with no real zeros? Why can you not have 3 complex zeros with a 3rd degree polynomial but it is possible with a 4th degree polynomial?

Finally, what then would the zeros of a 3rd degree polynomial with no real zeros be?

• That statement is only true for polynomials with real coefficients, so you need to be careful. – Thomas Andrews Sep 18 '14 at 4:10
• You can have a polynomial with three complex zeroes. Try $(x^2+1)(x-i)$. The point is that if the coefficients are real, then one needs a complex conjugate to get rid of the imaginary parts in the coefficients. – Isomorphism Sep 18 '14 at 4:12
• If you have a moment, please check out the video from 4:38 on and you will see what I am confused about. – Cherry_Developer Sep 18 '14 at 4:15
• The video is very sloppy, because shortly after declaring that polynomials can have complex coefficients they repeat the (unqualified) claim that non-real complex roots always come in conjugate pairs. That's just wrong. – Marc van Leeuwen Sep 18 '14 at 9:41

Why can you not have $3$ complex zeros with a $3$rd degree polynomial, but it is possible with a $4$th degree polynomial?

Observation: You can, if the coefficients are themselves complex.

Hint: Odd exponents maintain the sign, and even ones suppress it. Notice the obvious difference between the graphics of cubic and quartic $($or even quadratic$)$ functions as $x\to\pm\infty$. It is clear from their plots that for every even-order polynomial there is a big-enough free term $($positive or negative$)$ such that the graphic will no longer intersect the horizontal axis, meaning that our poly-nomial will have no roots. But this does not apply to odd-order polynomials, which always have at least one real root, since they always span from $-\infty$ to $+\infty$, or vice-versa.

If you have a polynomial with real coefficients, then complex roots always come in conjugate pairs. It is however altogether possible that you could a construct a cubic polynomial with three complex roots -- just take $(x-z_1)(x-z_2)(x-z_3)$ for any complex $z_1,z_2,z_3$. However you will find that when you expand this polynomial out, the coefficients will not be real, unless you picked 3 real roots, or two complex conjugate root and one real root.

The reason why any polynomial with real coefficients of odd degree (including cubics) must have at least one real root is because the highest power term dominates when the variable $x$ gets large. Assuming the coefficient of the highest term $x^n$ is 1, since $(-x)^n=-x^n$ when $n$ is odd, then when $x\to\infty$ and $-\infty$, then your polynomial will tend to $\infty$ and $-\infty$ respectively. Somewhere in between it must cross the $x$ axis by the intermediate value theorem, giving us a real root.

The reason why complex roots come in conjugate pairs, is because complex conjugation respects addition and multiplication. More precisely, if $z=x+iy$ and $w=a+bi$, and $\overline{z}$ represents the complex conjugate of $z$, then:

\begin{align*}\overline{z+w}&=\overline{x+iy+a+bi} \\ &=\overline{(x+a)+(y+b)i} \\ &=(x+a)-(y+b)i \\ &=(x-yi)+(a-bi) \\ &=\overline{z}+\overline{w} \end{align*}

and \begin{align*} \overline{zw}&=\overline{(x+iy)(a+bi)}\\ &=\overline{(xa-yb)+(xb+ya)i} \\ &=(xa-yb)-(xb+ya)i \\ &=(x-yi)(a-bi) \\ &=\overline{z}\cdot\overline{w} \end{align*}

Repeated application of the second identity gives $\overline{z^n}=(\overline{z})^n$. So suppose we know that $x=w$ is a root of a polynomial

$$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$

where $a_1,\ldots,a_n$ are real numbers. Then

\begin{align*} p(\overline{w})&=a_n\overline{w}^n+a_{n-1}\overline{w}^{n-1}+\cdots+a_1\overline{w}+a_0 \\ &= a_n\overline{w^n}+a_{n-1}\overline{w^{n-1}}+\cdots+a_1\overline{w}+a_0 \\ &= \overline{a_nw^n}+\overline{a_{n-1}w^{n-1}}+\cdots+\overline{a_1w}+\overline{a_0} \\ &= \overline{a_nw^n+a_{n-1}w^{n-1}+\cdots+a_1w+a_0} \\ &= \overline{p(w)} \\ &= 0 \end{align*}

meaning that $\overline{w}$, the complex conjugate was also a root, and therefore complex roots come in conjugate pairs. Note that it was crucial that the $a_i$ were all real, because then they are equal to their complex conjugate.

Any degree $3$ polynomial with real coefficients has at least one real zero. In fact any polynomial of odd degree with real coefficients has at least one real zero.

For simplicity consider the polynomial $p(x)=x^3+bx^2+cx+d$. When $x$ is very large negative, $p(x)$ is negative. When $x$ is very large positive, $p(x)$ is positive. So, by the Intermediate Value Theorem, somewhere in between we have $p(x)=0$.

• But what happens to a 3rd degree polynomial with no real zeros? – Cherry_Developer Sep 18 '14 at 4:06
• @Cherry_Developer There aren't any. – Thomas Andrews Sep 18 '14 at 4:09
• A degree $3$ polynomial with real coefficients always has at least one real zero. Of course if the polynomial has some non-real coefficients, then there may be no real zero. Exanple: $x^3+ix^2+x+i$. – André Nicolas Sep 18 '14 at 4:10
• But then, the statement from Khan is not true - the complex roots don't come in pairs. So Khan is clearly talking about polynomials with real coefficients. @AndréNicolas – Thomas Andrews Sep 18 '14 at 4:11
• So a polynomial of the 4th degree with no real zeros has 4 complex zeros. What roots does the polynomial of the 3rd degree have? – Cherry_Developer Sep 18 '14 at 4:14

It seems to me that either Mr. Khan or you are overlooking (for want of a better word) this well-known result:

Every odd-degree poynomial equation with real coefficients has a real zero.

This implies in particular that there is no such thing as a 3rd degree polynomial equation (with real coefficients) with no real zeros.


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# zbMATH — the first resource for mathematics

On the Wolff potential and quasilinear elliptic equations involving measures. (English) Zbl 0860.35041
Annales Academiæ Scientiarum Fennicæ. Mathematica. Dissertationes. 104. Helsinki: Suomalainen Tiedeakatemia. 71 p. (1996).
The author studies weighted degenerated elliptic equations (1) $$-\text{div} A(x, \nabla u) = \mu$$, where $$\mu$$ is a nonnegative Radon measure, $$A(x,\xi) \cdot \xi \approx w(x) |\xi |^p$$, $$1<p< \infty$$, and the weight function is $$p$$-admissible in the terminology of J. Heinonen, T. Kilpeläinen, and O. Martio [Nonlinear Potential Theory of Degenerate Elliptic Equations, Oxford (1993; Zbl 0780.31001)] (for instance, Muckenhoupt’s $$A_p$$ weights will do).
The main result is a substitute for the lack of Riesz representation theorem: if $$u$$ is a nonnegative solution of (1) in the ball $$B(x,3r)$$, pointwise properly defined, then the following estimate holds $c_1 W(x,r) \leq u(x) \leq c_2 \inf_{B(x,r)} u+ c_3 W(x,2r);$ here $$c_j$$’s are constants not depending on $$u$$, $$\mu$$ or $$r$$, and $W(x,r) = \int_0^r \left(t^p {\mu\bigl(B(x,t)\bigr) \over w \bigl(B(x,t) \bigr)} \right)^{1/(p-1)} {dt \over t}$ is the weighted version of the Wolff potential. This estimate generalizes that of T. Kilpeläinen and J. Malý [Acta Math. 172, No. 1, 137-161 (1994; Zbl 0820.35063)], and it is used to prove the necessity of Wiener’s criterion for the regularity of boundary points for the Dirichlet problem. Other applications of the potential estimate are also given. Moreover, uniqueness and weaker solutions of (1) are also discussed.

##### MSC:
 35J70 Degenerate elliptic equations 31C15 Potentials and capacities on other spaces 31C45 Other generalizations (nonlinear potential theory, etc.) 35J67 Boundary values of solutions to elliptic equations and elliptic systems


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HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS

## Circumference of Ellipse

The circumference is the distance around the ellipse.

## $\Pi \left[3\left(a+b\right)-\sqrt{\left(a+3b\right)\left(b+3a\right)}\right]$

Here,a = semimajor axis b = semiminor axis

ENTER THE VARIABLES TO BE USED IN THE FORMULA

Similar formulas which you may find interesting.


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Is it OK to set up more than one account?

+ 3 like - 0 dislike
221 views

My only answer on the site was down voted twice which now makes my reputation negative (damn! ;) ), which means I can no longer comment on stuff which requires a reputation of 0. I'm OK with this if that's the new rule since on physics stack exchange they require a 50 reputation to comment everywhere.

To get around this, I was thinking of setting up a new account and want to know if this is within the rules here?

asked Aug 4, 2014 in Support
recategorized Aug 4, 2014

+ 1 like - 0 dislike

Yes, we allow you to post anonymously, so we should also allow you to make a pen-name character. This can be useful for allowing a person to discuss controversial things without impacting their reputation in general.

But please don't make a new account in this particular case, as this amount of downvoting is extremely small, and trivial to undo--- either delete your answer, or answer a few other questions to zero out the rep loss.

If you just make a new account, you still need to do these things to get the small amount of rep for priviliges.

answered Aug 5, 2014 by (7,550 points)

Are you suggesting to have two accounts, one for the reviews section and one for the Q&A section? I am not sure if I correctly understand this answer.

Certainly, the lack of a privilege should not be circumvented by sockpuppeting ...

(At present, there seems to be a bug concerning commenting the own posts when not fulfilling the permission, I have already pinged polarkernel about it)

I am saying that if Vladimir Kalitivianski decides to have two accounts, one of which is names "Vladimir Kalitvianski" and is devoted to discussing atomic form-factors, antenna physics, Newtonian physics, and so on, and a second sock-puppet account named "VK Renormalization Warrior" with negative rep, which only comments according to the negative reputation accrued to this manifestation, we should treat them as two different users for purposes of reputation accrual (not for voting).

Making a character and speaking in character is an established way to do controversial things in science. You can even make a "Bourbaki", where 6 people have the same username, and together get a lot of rep through their publications, which does not translate to any one individual.

I don't see the point of restricting such activity, as we can judge the contributions of each user, human, collective, or sock, individually, and give them priviledge according to their activity.

I don't think it can be abused, if these users are non-voting. It does require a "nonvoting user", however, so that people can't sockpuppet votes.

@RonMaimon I agree, besides, it should be obvious that sockpuppeting votes is not allowed, and there is polarkernel's vote reversal plugin to reverse a user's votes in the last few (can be set to anything) hours, which is to be used in case of sockpuppet voting.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.


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# incubator-shiro-dev mailing list archives

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From "Grzegorz Borkowski (JIRA)" <j...@apache.org>
Subject [jira] Created: (JSEC-19) In the quick start guide add information about default realm, based on default-jsecurity-users.properties
Date Thu, 16 Oct 2008 07:54:46 GMT
In the quick start guide add information about default realm, based on default-jsecurity-users.properties
---------------------------------------------------------------------------------------------------------

Key: JSEC-19
URL: https://issues.apache.org/jira/browse/JSEC-19
Project: JSecurity
Reporter: Grzegorz Borkowski

For new user it is nor clear where the users mentioned in quick start guide come from, so
it should be updated to make it clear:

> Regarding default users settings - that's interesting. I found the
> properties file, it is actually called
> org\jsecurity\realm\text\default-jsecurity-users.properties. This can be
> useful for jumpstarting, but it should be also clearly explained that it is
> installed by deafult! Please at least update the QuickStart article on
> JSecurity page to mention this fact - I was really puzzled how it can work
> without defining users and roles somewhere!

--
This message is automatically generated by JIRA.
-
You can reply to this email to add a comment to the issue online.

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text:
Lebowitz - Penrose Limit for Continuum Particle Systems

#### I. Merola

2000, v.6, Issue 3, 329-344

ABSTRACT

We consider a particle system in $d\geq 2$ dimensions with an attractive Kac potential whose scaling parameter is $\gamma$ and a repulsive Kac potential with parameter $\gamma^{{1}/{2}}$. They have the same form as the corresponding potentials in [J. Lebowitz, A.E. Mazel and E. Presutti, Liquid-vapour phase transitions for systems with finite range interactions, J. Stat. Phys., 1999, v.94, N 5/6, pp.955-1025], but there the scaling parameter was $\gamma$ for both of them. We compute the Lebowitz - Penrose limit for the pressure and prove that the minimizers of the limit variational problem are spatially constant. The limit phase diagram has a liquid-vapour, of van der Waals type phase transition, as in the mentioned paper.

Keywords: continuum particles system,Kac potential,superstability,mean field theory,liquid-gas phase transition


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# Fundamental group of $X$, a CW complex is isomorphic to the fundamental group of its 2-skeleton

I'm trying to show that if $$X$$ is a CW-complex, then $$\pi_1(X) = \pi_1(X^2)$$ where $$X^2$$ is the 2-skeleton.

I found the following proposition in Hatcher's book:

Proposition 1.26.

(a) If $$Y$$ is obtained from $$X$$ by attaching $$2$$-cells as described above, then the inclusion $$X \hookrightarrow Y$$ induces a surjection $$\pi_1 (X, x_0) \to \pi_1 (Y,x_0)$$ whose kernel is $$N$$. Thus $$\pi_1 (Y) \approx \pi_1 (X)/N$$.

(b) If $$Y$$ is obtained from $$X$$ by attaching $$n$$-cells for a fixed $$n > 2$$, then the inclusion $$X \hookrightarrow Y$$ induces an isomorphism $$\pi_1 (X, x_0) \approx \pi_1 (Y,x_0)$$.

(c) For a path-connected cell complex $$X$$ the inclusion of the $$2$$-skeleton $$X^2 \hookrightarrow X$$ induces an isomorphism $$\pi_1 (X^2,x_0) \approx \pi_1 (X,x_0)$$.

(In (a), $$N$$ is a normal subgroup of $$\pi_1(X,x_0)$$.)

I was wondering if there's a more "direct" proof, or if I should just follow this proposition. Thanks in advance!

The CW-flavoured argument uses the relative version of CW approximation:

Let $$X$$ and $$Y$$ be CW complexes, $$A\subset X$$ a subcomplex. If $$f\colon X \to Y$$ is a continuous function that is cellular on $$A$$ then there is a homotopy $$H\colon X\times I \to Y$$ such that $$H_0 = f$$, $$H_t(a) = f(a)$$ for all $$a\in A$$ and $$t\in I$$, and $$H_1$$ is cellular.

Let $$X$$ be a CW complex, and $$\iota \colon X^n \to X$$ the inclusion of its $$n$$-skeleton (we omit the subscript on $$\iota$$ for the sake of notation). We want to show that $$\iota_*\colon \pi_1(X^2) \to \pi_1(X)$$ is an isomorphism. Suppose $$S^1$$ is given a CW structure so that the basepoint is a $$0$$-cell.

If $$f\colon S^1 \to X$$ is basepoint preserving, then by relative CW approximation there is a basepoint preserving homotopy between $$f$$ and a pointed cellular function $$\tilde{f}\colon S^1 \to X$$. By cellularity the image of $$\tilde{f}$$ is in $$X^1$$, so $$\iota_*\colon \pi_1(X^2)\to \pi_1(X)$$ is surjective.

Now suppose $$f\colon S^1 \to X^2$$ is a pointed map such that $$[\iota\circ f] = 0 \in \pi_1(X)$$, i.e. $$[f]$$ is in the kernel of $$\iota_*$$. Without loss of generality we can assume that $$f$$ is a cellular map. If we consider any basepoint-preserving null-homotopy $$H\colon S^1 \times I \to X$$ of $$\iota\circ f$$, then by relative CW approximation (note that $$H$$ is cellular on the subcomplex $$(X\times \{0\}) \cup (\{x_0\} \times I) \subset X \times I$$) $$H$$ is homotopic to a basepoint-preserving null-homotopy $$\tilde{H}$$ of $$\iota\circ f$$ which is cellular. In particular the image of this null-homotopy is in $$X^2$$ so in fact $$[f] = 0 \in \pi_1(X^2)$$, therefore $$\iota_*$$ is injective.

Note: an essentially identical argument shows that $$\pi_n(X) \cong \pi_n(X^{n+1})$$ for all $$n\geq 0$$, as an exercise you should write out the details.

Edit: Further note: after looking at Hatcher's proof of this proposition, it seems more elementary than the full force of CW approximation, even though I feel like CW approximation is the "conceptual" way to answer your specific question about CW complexes.

• Thanks for the clear answer! So basically if I want a more direct result I should use CW approximation, but if I just want to use basic constructions on CW complexes I have to make use of that proposition, for example? – SantiMontouliu May 23 at 19:35
• I think the methods used in the proofs of that proposition are a bit more elementary than CW approximation, and they work in a more general context (I had assumed they would just be using CW approximation but it looks like they aren't). In my opinion CW approximation is the "conceptual" way of seeing the result you want, but Hatcher's proposition I think has less proof. – William May 23 at 20:11