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# Srednicki equation 7.7

vaibhavtewari
Hello everyone,

I will be glad if someone can explain how equation 7.7

$$\tilde{x}(E)$$ = $$\tilde{q}(E)$$ + $$\frac{\tilde{f}(E)}{E^2-\omega^2+i\epsilon}$$

is a shift by constant, heres the link for the book

http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

thanks

LAHLH
Hi, I'm working through this book at the moment too, and have received lots of excellent help here, so would be glad to try and help (although being a newbie at this stuff too I may not be completley correct).

I believe in chapter 7, the path integral has a measure Dq, i.e. it is with respect to variations of the configuration space (integrating over all possible configurations the particle could traverse, including the the classically realised one, but also every single other possible. Later these will be field configurations)

If we make the change in 7.7 $$\tilde{x}(E)= \tilde{q}(E)+ \frac{\tilde{f}(E)}{E^2-\omega^2+i\epsilon}$$, then the last term is a constant as far as $$\int Dq$$ is concerned as it doesn't depend on the dynamical variable q. It's like shifting everyone of the infinite possible trajectories you're integrating over by a constant.

This is more or less the same as changing variables in a normal integral $$\int dx$$, by having $$y \rightarrow x-\chi$$. Clearly dy=dx, the same thing is going on in the path integral case, it's just complicated by integrating over all the paths.

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vaibhavtewari
Thank you for replying, I am glad I am not the only one struggling trough qft,

vaibhavtewari
I wil be glad if you can also explain why eq 7.17 is not just

$$\frac{1}{i^2}[G(t_2-t_1)G(t_4-t_3)]$$

LAHLH
Well it ultimately comes down to the Leibniz product rule (and making sure you don't set f=0 until after the differentiation, so you have to differentiate the exponential as part of the product when you go onto the next derivative etc). Acting with the three $$\delta$$ derivatives on $$\langle 0 \mid 0 \rangle_f$$ (as given explicitly in 7.11), you will get a range of terms, but then you set f to 0 only the ones in 7.17 survive.

I hope that helps, it's hard to explain without writing it out explicitly which would probably take forever in latex. But it really is just the product rule, and making sure you only set f=0 at the end, so you don't lose any terms.

LAHLH
I will do part of the process for the 3 point function (which will be zero but will show the details of how this works anyway)

$$\langle 0 \mid TQ(t_1)Q(t_2)Q(t_3) \mid 0 \rangle= \frac{1}{i}\frac{\delta}{\delta f(t_1)}\frac{1}{i}\frac{\delta}{\delta f(t_2)}\frac{1}{i}\frac{\delta}{\delta f(t_3)} \langle0\mid0\rangle_f$$

$$\langle 0 \mid TQ(t_1)Q(t_2)Q(t_3) \mid 0 \rangle= \frac{1}{i}\frac{\delta}{\delta f(t_1)}\frac{1}{i}\frac{\delta}{\delta f(t_2)}\frac{1}{i}\frac{\delta}{\delta f(t_3)} \langle0\mid0\rangle_f$$

$$\langle0\mid0\rangle_f=exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'})]$$

So acting with first derivative:

$$\frac{1}{i}\frac{\delta}{\delta f(t_3)}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'})]= \{\frac{1}{i} \frac{i}{2} \int dt dt^{'} \delta (t-t_3) G(t-t^{'})f(t^{'})+\frac{1}{i}\frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})\delta (t^{'}-t_3) ) \}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ]$$

This is equal to

$$\{\frac{1}{2} \int dt^{'} G(t_3-t^{'})f(t^{'})+\frac{1}{2} \int dt f(t)G(t-t_3) \}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ]$$

Now since $$t, t^{'}$$ are dummy variables and since G is symmetric in it's arguments (because of the modulo) we can write this simply as

$$\int dt^{'} G(t_3-t^{'})f(t^{'})exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ]$$

This is how Srednicki gets line 2 of 7.16.

Now you want to act on this in the same way with the $$\frac{1}{i}\frac{\delta}{\delta f(t_2)}$$

So now you just use the product rule again not forgetting exp term yet which hasn't gone to one as we haven't yet set f=0.

This leads you to the next line:

$$\frac{1}{i}\frac{\delta}{\delta f(t_2)} \int dt^{'} G(t_3-t^{'})f(t^{'})exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] = \{ G(t_3-t_2)+\int dt^{'} G(t_3-t^{'})f(t^{'})\int dt^{''} G(t_2-t^{''})f(t^{''})\}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ]$$

Note this is equation 7.16 (only with t2, t3 instead of t1,t2) explcitly, my second term being what he calls "term with fs", which dies when we set f=0 at the end.

Finally we can hit this with the last derivative as follows:

$$\frac{1}{i}\frac{\delta}{\delta f(t_1)}\{ G(t_3-t_2)+\int dt^{'} G(t_3-t^{'})f(t^{'})\int dt^{'} G(t_2-t^{'})f(t^{'})\}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ]$$

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LAHLH
Which gives:
$$\{ G(t_3-t_2)\int dt^{'} G(t_1-t^{'})f(t^{'})+G(t_3-t_1)\int dt^{'} G(t_2-t^{'})f(t^{'})$$

$$+G(t_2-t_1)\int dt^{'} G(t_3-t^{'})f(t^{'})+\int dt^{'} G(t_3-t^{'})f(t^{'})\int dt^{''} G(t_2-t^{''})\int dt^{'} G(t_1-t^{'}) \} exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ]$$

Notice that setting f to zero now, will kill everything, this is why all odd number correlation functions are vanishing. Hopefully I haven't made too many mistakes and you can carry out this process to get the 4 point correlation function. You should be left with Srednicki's three terms as the only ones without f's then you can set f=0 to kill all others.

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vaibhavtewari
wonderful, thankyou very much...

I am grateful you helped me in both the problems


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January 30, 2016

# Rory's Prime

One day I decided I want to discover a prime number of my very own.

After a bit of research, and 203.53 quadrillion floating-point operations later, I became the proud father of:

## 5687 · 21376451 + 1

a 414,357 digit-long titanic Proth Prime! As of 1/29/2016 06:20:31 CDT, this is the 2,261st largest prime ever discovered!

Raw decimal representation: rorys-prime-decimal.txt

This Proth Prime Search was done in collaboration with the Proth Search project. This search looks for primes in the form of k · 2n + 1, with the condition 2n > k. These are often called Proth primes.

Thanks to the PrimeGrid project and BOINC for making this pretty easy to get started!

format: plain


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# How can I remove the year label of the misc type in biblatex?

I would like to remove the year label, if it is undefined. What I get:

EC No 197/38 2012 (n.d.)

But it should be:

EC No 197/38 2012

In the bibliography I get this, but it is correct for me:

EC No 197/38 2012: Waste electrical and electronic equipment (WEEE), 4 July 2012, Commission Regulation.

\documentclass{book}
\usepackage[style=authoryear-icomp]{biblatex}
\usepackage{filecontents}
\begin{filecontents}{biblio.bib}
@Misc{EC4July2012,
Title                    = {Waste electrical and electronic equipment ({WEEE}), 4 July 2012},
Author                   = {{EC No 197/38 2012}},
Type                     = {Commission Regulation},
}
\end{filecontents}

\nocite{*}

\begin{document}
Something \autocite{EC4July2012}.
\printbibliography

\end{document}

• The proper way would be to have the date (in your 4 July 2012) case in the year field in the bibtex entry. – Guido May 21 '14 at 11:35
• @Guido The problem is that it is a European Comission Regulation. And in the short form citation should be "EC No 197/38 2012". The form and style of laws are difficult to make correct in this case. – Xeo May 21 '14 at 11:41
• If you have more of these rules/regulations/laws that need special handling, it might be a good idea to create a new entry type for those (a starting point might be List of People analog to Bibliography and How can I create entirely new data types with BibLaTeX/Biber?), if you can provide a prototype of what your citations contain and how they should look like in citations and bibliographies, we might be able to help you here. – moewe May 22 '14 at 5:17
• Of course you can also just redefine a cite command to always omit dates for @miscs: \renewbibmacro*{cite:labelyear+extrayear}{% \ifentrytype{misc} {} {\iffieldundef{labelyear} {} {\printtext[bibhyperref]{% \printfield{labelyear}% \printfield{extrayear}}}}} (modulo line breaks and comment characters, if you want that, I can write up an answer). – moewe May 22 '14 at 5:21
• @moewe It is working with your redefinition, but now i have space before parenthesis in all cites. Author x and Author y ( 2014 ) or (Author y et. al. 2002 )? – Xeo May 23 '14 at 19:44

Handling of laws, regulations etc. that are supposed to be cited in a very particular way isn't all that easy. If you have several of such documents you might want to consider creating dedicated entrytypes for easy customisation (see How can I create entirely new data types with BibLaTeX/Biber? on how to do that, and How to implement a jurisdiction biblatex entry type? & How to cite a section of a Constitution for a more thought through and an ad-hoc solution; as well as What is the proper way to cite a legal document? for a general discussion).

In your case, there is the obvious ad-hoc solution of suppressing years for @miscs:

\renewbibmacro*{cite:labelyear+extrayear}{%
\ifentrytype{misc}
{}
{\iffieldundef{labelyear}
{}
{\printtext[bibhyperref]{%
\printfield{labelyear}
\printfield{extrayear}}}}}


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# Proof of equal cardinality $|\Bbb N \times\Bbb N \times\Bbb N| = |\Bbb N|$

How do I prove that the following sets have equal cardinality?

1. $|\Bbb N \times\Bbb N \times\Bbb N| = |\Bbb N|$ ($|\Bbb N \times\Bbb N| = |\Bbb N|$ also for that matter)
2. $|\Bbb Z \times\Bbb Z| = |\Bbb Z|$
3. $|\Bbb R \times\Bbb R| = |\Bbb R|$

Thank you!

-
What did you try? Did you try searching the site? – Asaf Karagila Jan 5 '13 at 18:19
yes I tried searching the site – Yechiel Labunskiy Jan 5 '13 at 18:26
Look here, here, here, here, and here; among them they answer all of your questions. – Brian M. Scott Jan 5 '13 at 18:41
Thank you Brian! Guess I didn't know what to search for. – Yechiel Labunskiy Jan 5 '13 at 18:54
Wie ought to have something like an abstract duplicate "When is $|X\times X|=|X|$?" – Hagen von Eitzen Jan 6 '13 at 10:34

It is easy to produce a sequence that includes all the elements in $\Bbb N\times\Bbb N\times\Bbb N$.

E.g. $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(2,0,0)$, $(1,1,0)$, $(1,0,1)$, $(0,2,0)$, $(0,1,1)$, $(0,0,2)$, $(3,0,0)$, $(2,1,0)$, $(2,0,1)$,...

This can be adjusted to the case of $\Bbb Z\times\Bbb Z\times\Bbb Z$ and/or any finite number of copies of $\Bbb N$ or $\Bbb Z$, or even $\Bbb Q$.

To show that $|\Bbb R|=|\Bbb R\times\Bbb R|$ requires, of course, a different argument.

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How do I map that sequence to |N| with a one-to-one and an unto function? – Yechiel Labunskiy Jan 5 '13 at 18:55
Well, you map $1$ to the first element, $2$ to the second element, $3$ to the third element and so on. Recall that a sequence in a set $X$ is just a function $\Bbb N\rightarrow X$. If a sequence contains exactly once all the elements of $X$, then the sequence, thought as a function, is bijective. – Andrea Mori Jan 5 '13 at 19:12

Note that having size $|N|$ means something is infinite but enumerable; meaning, roughly, that you can come up with a way of listing all the elements in a single row.

Note: I am abiding by $N = \{0, 1, 2, 3, \ldots\}$.

To show $|N \times N| = |N|$, you could start by listing all the elements $(m, n)$ with $m,n \in N$ such that $m + n \leq 0$. Next, list all the elements with $m+n \leq 1$, and so forth. In this way, you produce a list of the elements; a list of the form:

$a_0, a_1, a_2, a_3 \ldots$ which can be matched up with elements of $N$ using $a_n \rightarrow n$.

The case for $N \times N \times N$ is similar, as is the case for anything $Z$ related.

Your third problem is a bit more involved, but one place to start is by thinking about decimal expansions. I won't say anything more unless you post what you have tried thus far.

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Pairs like (2,2) and (3,1) will both map to 4 which means it's not one-to-one, am I missing something? Don't really know where to start on R x R. Thanks – Yechiel Labunskiy Jan 5 '13 at 18:29
The key is to come up with an organized way of listing all of them. One way to avoid the sort of problem you have described is to always list $(a,b)$ before $(c,d)$ when $a + b = c + d$ and $a < c$. So the list would begin: $(0, 0), (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (0, 3), (1, 2), (2, 1), (3, 0), (0, 4), (1, 3), (2, 2), (3, 1)$ Of course, a similar approach can be used to demonstrate that $\mathbb{Q}$ is countable. – Benjamin Dickman Jan 6 '13 at 10:22

The function:$$f:\mathbb{Z}^+\times\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow \mathbb{Z}^+$$ that sends $(i,j,k)$ to $(2(2j-1)2^{i-1}-1)2^{k-1}$ is a bijection.

-


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# Probability - Conditional Probability

There are 15 tennis balls in a box, of which 9 have NOT previously been used. Three of the balls are randomly chose, played with, and then RETURNED TO THE BOX. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.

Seems pretty simple, right?

Attempted Solution:

Event A - 1st round balls have never been used before
Event B - 2nd round balls have never been used before

We want the probability of events A and B both occuring.

Prob(A and B) = Prob(B given A) * Prob(A)

Prob(A) = (9C3)/(15C3) = 84/455

Prob(B given A) = (6C3)/(15C3) = 20/455

Prob(A and B) = .008114962

Book says answer is .0893

EDIT: I got it now. I misunderstood the question.

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# Visualizing the Metropolis Algorithm

April 2, 2016

Let’s say you’re doing some sort of Bayesian analysis. You’ll have a prior P(θ) over your model parameters θ. You get some data D, and you want to update on this data to get the posterior P(θ|D), the updated distribution of the model parameters given D. Let’s wheel out Bayes’ theorem and work out how to calculate P(θ|D), which for convenience we’ll call π(θ):

π(θ)=P(θ|D)=P(D|θ)P(θ)P(D)

If we can calculate P(D|θ) by using some sort of loss function, computing the numerator P(D|θ)P(θ) is relatively straightforward. And once we have the numerator, we’ve specified our posterior π(θ) up to the normalization constant P(D).

Computing the normalization constant is trickier. P(D) is the probability of seeing this data in the model, which means we have to integrate over all possible values of θ:

P(D)=ΘP(D|θ)P(θ)dθ

In most cases1, this won’t have a closed-form solution, and deterministic numerical integration can scale poorly with increasing dimensionality.

## Monte Carlo integration

Let’s assume we can’t easily compute this integral; we can turn to Monte Carlo methods to estimate it instead. If we can directly draw from the posterior distribution π(θ), we can simply compute the density of π(θ) at a set of uniformly distributed values θ1,,θN that cover a broad range of the parameter space for θ:

P(D)=ΘP(D|θ)P(θ)dθ1NNi=1P(D|θ(i))

By the law of large numbers, our estimate will converge to the true distribution as N goes to infinity. But this only works if we can directly draw from π(θ). Many Bayesian models use arbitrarily complex distributions over θ that we can’t easily sample.

## Importance sampling

Luckily, we can use importance sampling to help us approximate π(θ), by sampling from a somewhat arbitrary distribution q(θ), and then correcting for the fact that we’re sampling from q(θ) and not π(θ):

P(D)=P(D|θ)P(θ)dθ=P(D|θ)P(θ)q(θ)q(θ)dθ

In theory, all we’ve done is divide and multiply by q(θ); it shouldn’t really matter what q(θ) is2. But now, instead of trying to draw samples from π(θ), we can draw them from q(θ), which can be some distribution that’s easy to sample from. We can think of this as sampling from q(θ), and correcting the sample by multiplying by its importance weight, given by:

w(θ)=P(D|θ)P(θ)q(θ)

Now, as before, we can use q(θ) to create a random set of samples for Monte Carlo integration:

P(D)=w(θ)q(θ)dθ1NNi=1w(θ(i))

Importance sampling is a very useful tool, but starts to become ineffective for θ of 6 dimensions. In an high-dimensional parameter space, if you try to ensure that q(θ) has support everywhere that π(θ) is large, q(θ) won’t have enough probability mass to properly explore the space, and your realisations will be scattered too sparsely to be of any significance.

## Markov chain Monte Carlo

Markov chain Monte Carlo (MCMC) involves creating a Markov process with a stationary distribution π(θ), and then running the simulation long enough to produce a chain that converges close enough to π(θ) for our purposes.

One family of MCMC methods uses the Metropolis algorithm, which is incredibly simple to implement. It can be thought of as a random walk through the parameter space that’s weighted to converge to the target distribution. Here’s a simple visualization3 of the Markov chains the Metropolis algorithm produces when it’s run on the Rosenbrock function:

Try reducing the variance to 0.001 and choosing a relatively low number of iterations. The acceptance rate will approach 100%, and as it does, the chain will reduce to a random walk.

Now try setting the variance around 0.5, and increasing the number of iterations around 10,000. You’ll see that regions of probability space with greater density4 are more likely to be visited by the chain.

## The Metropolis algorithm

The Metropolis algorithm is really simple to implement; it can be written in a few lines of code. Given a starting point θt1, a single iteration of the algorithm looks like:

1. Perturb θt1 with a proposal distribution qt(θ|θt1) to produce a proposal θ. For the Metropolis algorithm, this proposal distribution must be symmetric, so qt(θa|θb)=qt(θb|θa), for all θa, θb, and t.
2. Calculate the ratio of posterior densities: rπ=π(θ)π(θt1) We can see that even if we can only calculate the unnormalized posterior, computing the ratio here will cause P(D) to cancel out, and we’ll get a properly normalized term.
3. Set θt: θt={θwith probabilitymin(1,r)θt1otherwise

function metropolis(dist, startPoint, variance, iterations) {
let current = startPoint;
let chain = [current];
var oldLik = dist(...current);

// Perturbs a number using a normal distribution
const perturb = x => x + normalRandom(0, variance);

// Construct chain
for (var i=0; i < iterations; i++) {
const candidate = map(perturb, current);
const newLik = dist(...candidate);
const acceptProbability = newLik / oldLik;

if (Math.random() < acceptProbability) {
// Accept candidate
oldLik = newLik;
current = candidate;
}
chain.push(current);
}
return chain;
};

In this case, the proposal distribution is the normal distribution centered at 0 with a user-selected variance, and our “posterior distribution” is actually just the Rosenbrock function. We also don’t need to calculate min(1,r), because Math.random() always produces a value that’s less than 1.

## Helping our intuition

Interestingly, just observing the algorithm operate in low dimensions gave me some intuition about its behaviour in higher dimensions.

### Proposal distribution

The proposal distribution can greatly affect the convergence of the chain. If the proposal distribution tends to produce very small steps5, the chain will traverse the probability space very slowly.

In our visualization, we can sort of see that the start point is in a rather featureless region of probability density, which means if we take tiny steps, π(θ) is very likely to be close to π(θt1). The acceptance ratio will hover around 1, which means almost every proposal will be accepted, and the behaviour of the chain will be decided almost entirely by the proposal distribution.

If our proposal distribution makes very large steps, however, most of the proposals will be rejected. You can observe this behaviour by making the variance very large and noticing that the acceptance rate tends to drop. This means that you’re basically doing importance sampling, and you’ll need a prohibitive number of iterations to properly explore the probability space.

In practice, we generally target acceptance rates of around 30%.

### Burn-in

A chain is likely to spend its initial states trying to find some region of high density, and this groping search is unlikely to approximate the target distribution. It’s standard practice to “burn-in” a chain by discarding the first, say, 50% of the chain, and only checking the remainder for convergence. This behaviour can be observed in our visualization if you set the variance to about 0.01; the first portion of the chain is generally spent wandering around the uninteresting portions of the probability space.

## Results

Once we’re satisfied with the convergence of the chain, we can use it as an approximation for our target distribution. Here’s how the Metropolis algorithm approximates the Rosenbrock function:

At a low number of iterations, the chain doesn’t really explore much of the probability space, so we get a rather noisy look at a portion of the target distribution. If we increase the number of iterations to the hundred thousand range, we can see the familiar Rosenbrock arch shape emerge.

Increasing our iterations to the millions6 (hold down the Shift key while dragging!) will give a clearer picture of the function, with a lot of detail around the global maximum at the top right of the arch.

## Why does it converge?

We’ve got an intuitive understanding of why the chain is more likely to explore regions of higher density, but we don’t have a proof- even an informal one- that the Markov chain will converge to the target distribution. Let’s take a look at one now.

The Metropolis algorithm aims to create a Markov chain with π(θ) as its stationary distribution. Because we’ve selected our proposal distribution to be symmetric, our chain is irreducible and aperiodic by construction. For reasons that I won’t get into here, this means that it must have a unique stationary distribution. But why is this stationary distribution π(θ)?

For an unspecified point θa, the distribution of it having successor θb is given by integrating over all possible values of θa:

π(θa)qt(θb|θa)dθa

We’ve chosen qt(θ) to be symmetric:

π(θa)qt(θb|θb)=π(θb)qt(θa|θb)

so we can substitute this in to get

π(θa)qt(θb|θa)dθa=π(θb)qt(θa|θb)dθa=π(θb)

because of course qt(θa|θb)dθa=1.

This means that the probability density of the chain moving to θb, given that it’s at θa, is given by the density of the distribution π(θb)!

1. Usually this is only feasible in simple Bayesian models, particularly those with conjugate priors.

2. Provided we didn’t introduce any singularities.

3. You can drag up and down on the text boxes to change the numbers. Holding Shift while dragging will increase the step size by 10x, while holding Control while dragging will reduce it by 10x.

4. In this case, the greatest density is around the top right of the bend.

5. Which happens in our visualization if the variance is very low

6. This can take a while. The algorithm itself runs on a web worker so that the page doesn’t lock up while it’s running, but the final graphing needs to happen on the UI thread and can freeze up the page for a short time.


text:
| English | 简体中文 |

# 669. Trim a Binary Search Tree

## Description

Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.

Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.

Example 1:

Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]

Example 2:

Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]

Example 3:

Input: root = [1], low = 1, high = 2
Output: [1]

Example 4:

Input: root = [1,null,2], low = 1, high = 3
Output: [1,null,2]

Example 5:

Input: root = [1,null,2], low = 2, high = 4
Output: [2]

Constraints:

• The number of nodes in the tree in the range [1, 104].
• 0 <= Node.val <= 104
• The value of each node in the tree is unique.
• root is guaranteed to be a valid binary search tree.
• 0 <= low <= high <= 104


text:
## k-Interval-filament graphs

### Author: Fanica Gavril

ABSTRACT
For a fixed $k$, an oriented graph is $k-transitive$ if it is acyclic and for every directed path $p=u1-> u2-> ...->uk+2$ with $k+2$ vertices, $p$ induces a clique if each of the two subpaths $u1->u2->...->uk+1$ and $u2->...->uk+2$ induces a clique. We describe an algorithm to find a maximum weight clique in a $k$-transitive graph.

Consider a hereditary family $G$ of graphs. A graph $H(V,E)$ is called $G-k-mixed$ if its edge set can be partitioned into two disjoint subsets $E1$, $E2-E3$ such that $H(V,E1) \in$G$,$H(V,E2)$is transitive,$H(V,E2-E3)$is$k$-transitive and for every three distinct vertices$u,v,w$if$u->v \in E2$and$(v,w) \in E1$then$(u,w) \in E1$. The letter$G$is generic and can be replaced by names of specific families. We show that if the family$G$has a polynomial time algorithm to find a maximum clique, then, under certain restrictions there exists a polynomial time algorithm to find a maximum clique in a family of$G-k$-mixed graphs. Let$I$be a family of intervals on a line$L$in a plane$PL$such that every two intersecting intervals have a common segment. In$PL$, above$L$, construct to each interval$i(v) \in I$a filament$v $connecting its two endpoints, such that for every two filaments$u,v$having$u \bigcap v \not = empty_set $and disjoint intervals$i(u)<i(v)$, there are no$k$filaments$w$with$i(u)<i(w)<i(v)$which intersect neither$u$nor$v$, are mutually disjoint and have mutually disjoint intervals. This is a$family of k-interval-filaments$and its intersection graph is a$k-interval-filament graph$; their complements are ($k$-transitive) mixed graphs and have a polynomial time algorithm for maximum cliques. Now, when two filaments$u,v$do not intersect because$u \subset v$, and between$u$and$v$there are at most$k-1$non-intersecting filaments$w1,...,wk-1$such that$wi \subset wi+1$and intersect neither$u$nor$v$, we attach to each one of$u,v$a branch in the space above$PL$such that the two branches intersect. This is a$ family of general-k-interval-filaments$and its intersection graph is a$general-k-interval-filament graph$; their complements are ($k$-transitive)-$k\$-mixed graphs.

Paper Available at: ftp://dimacs.rutgers.edu/pub/dimacs/TechnicalReports/TechReports/2004/2004-30.pdf

DIMACS Home Page


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# Is it not recommendable to use a spectrum analyzer in its limit frequencies?

Suppose we have a spectrum analyzer whose range of frequencies goes from $x \ \mathrm{Hz}$ to $y \ \mathrm{Hz}$. Is there any problem that could arise from measuring signals that are near those frequencies? Any kind of "border effect", or something of that nature.

For instance, Keysight's N9320B states in the datasheet that the frequency range starts at 9 kHz. If I want to measure a sinusoid at 9 kHz, would there by any problem or uncertainty due to the fact that this frequency is at the limit of the SA operating range? (I'm not interested in just that SA, but I just put it here to show what I mean by 'frequency range'.)

• Depends on the nature of the limit. If it's a filter roll-off (especially in the DC-block), might be okay for non-critical use. If it's an actual limit of what the instrument will display, then not being able to see both sides of the signal will leave you with questions about what you are looking at - in fact, even a slight frequency error might put it off the display entirely. Conversely if it's a limit of the center frequency you may well get half a normal display span below. Remember also that 9 KHz is audio frequency - there are other options there, including a quality sound card. – Chris Stratton Jul 28 '18 at 22:18
• @ChrisStratton Thanks for your comments. Do you think that there is a way of knowing the nature of the limits? Also, what do you mean by "a filter roll-off"? I mean, I know what the roll-off of a filter is, but I didn't understand how that would put a limit on frequency range here. Doesn't it depend just on the oscillator and the range of frequencies it covers? (Sorry if this is too basic, I'm not too experienced with spectrum analyzers!) – Tendero Jul 28 '18 at 22:24
• You might judge the quality of a SA's VCO stability by its lower-frequency spec (9 kHz for example). You can easily see drift and phase noise down there. You cannot tell from the display if your input signal is drifting, phase-noisy, or if the analyzer's VCO is drifting, phase-noisy. It is very easy to build a stable 9 kHz oscillator as a test signal. Any drift or phase noise you see displayed is the result of SA's VCO. A decent crystal oscillator test signal is OK too. – glen_geek Jul 28 '18 at 22:39
• @glen_geek So would the border frequency present, in general, more phase noise than frequencies in the middle of the oscillating range? – Tendero Jul 28 '18 at 22:46
• What's your actual goal? The instrument you link to is hugely expensive to buy or even rent, so much so that normally it would not be considered without project resources correlated to a far higher level of familiarity with these topics. If you are looking at low frequency signals, this is not the right choice, unless most of your need is elsewhere. – Chris Stratton Jul 28 '18 at 22:46

There are several limits to the low frequency behaviour of spectrum analysers. Some are easy to intuit, some less so. Newer digital instruments are slightly different to the older analogue instruments in what they display.

One common limit is the high pass nature of the input DC blocking filter, which is generally no more than a series capacitor. That's quite simple. The filter might be -3dB at 9kHz, or it might be -1dB or even -0.1dB. It's fairly easy to see the level, and sort it out with calibration.

For the other limits, you need a block diagram for a typical spectrum analyser. This shows the simplified front end of a typical superheterodyne spectrum analyser.

This one goes up to 1GHz, and you can see from the diagram down to nominally DC.

The bandpass filter at 2GHz provides the selectivity. This is the filter that is switched to the various bandwidths that the analyser will provide, the 10MHz, 1MHz, 100kHz, 10kHz settings. For the larger bandwidths, it's a real switchable LC bandpass filter. For the smaller bandwidths, it's a system that uses further mix-down to lower IFs. In the modern digital analysers, the last IF is left fairly high frequency and wideband, and is digitised with FFT used for the final filters.

For a 0.5GHz input, the VTLO is set to 2.5GHz, producing a 2GHz IF.
For a 1GHz input, it's set to 3GHz, producing the same 2GHz IF.
For a nominally DC input, it's set to 2GHz.

On old analogue instruments, the sweep controlling the LO frequency could be set to sweep 'below' DC, for an input frequency below 0Hz. What you'd see on the display is the negative frequency for the first part of the sweep, a 'DC pip' at DC, and then the normal positive frequency range above that. Digital synthesiser based spectrum analysers tend to limit the display so you don't see either.

The input mixer, while it provides isolation between all its ports in theory, in practice 'leaks' some of the LO signal out to the 2GHz IF. This is the origin of the 'DC pip' mentioned above, aka 'LO leak'. The leakage can be quite severe. A not uncommon specification for RF spectrum analysers was for -40dBm signal at the input mixer, which tended to mean -50dBm to the IF. With a typical LO drive of +13dBm to +17dBm, you can see that even 60dB LO to IF isolation would still mean the DC pip was more than the expected signal. This sort of isolation needs active balance in the first mixer, and is still a struggle to achieve stably.

The LO leak causes two problems

1) Phase noise on the LO signal lifts the noise floor for input frequencies close to it, it reduces the dynamic range from the bottom. That's why 9kHz is an often seen lower limit. Just low enough to sneak into the audio range, high enough to get away from the worst of the LO phase noise.

2) The LO leak gets into the 2GHz IF, as a signal of similar size to the wanted signal. Generally the first element in the 2GHz IF is an LC filter, wide enough for the widest expected IF, perhaps 40MHz width. This filters out the LO leak for input frequencies above 20MHz. The distortion of the analyser is now controlled by the single tuned signal getting into the IF's amplifiers. This is generally the distortion that's specified for the instrument.

At input frequencies less than 20MHz, the LO leak now appears full strength at the IF amplifiers, reducing the signal handling of the amplifiers and increasing their distortion. This reduces the dynamic range from above. As mentioned above, the 2GHz IF is actually a system, using progressively lower frequency IFs and smaller bandwidth filters. There will therefore be a progressive degradation in dynamic range as the smaller bandwidth means that filters can no longer protect the detector from seeing the LO leak signal. Finally at DC, the LO gets through to the detectors, and the spectrum analyser can no longer distinguish the input signal.


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# Veritas Prep 10 Year Anniversary Promo Question #10

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Veritas Prep 10 Year Anniversary Promo Question #10 Critical Reasoning

Sept 21, Friday is the last day for the Veritas Prep 10 Year Anniversary Promo. Today, we will be giving away 2 Courses in the Verbal Challenge and 2 winners will be announced

One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course ($1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it. Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time. Good Luck! May the best and fastest win! College football recruiting services rank incoming players on a scale of 1-star (not a highly sought-after prospect) to 5-star (considered to be the best players). Recently a service attempted to validate its rankings by assigning star ratings to players upon completion of their careers to determine the accuracy of the initial rankings. The survey averaged the post-career ratings of each player and found that 5-star players’ final average was 4.46, compared with 3.98 for 4-stars and 3.11 for 3-stars. This suggests that the rankings services do not effectively judge high-end talent as well as they judge players in the middle of the range. Which of the following identifies a problem with the service’s attempt to validate its rankings? (A) Players at certain positions might be harder to judge at a younger age than players at other positions (B) A five-star scale does not allow the most elite players to overperform their initial ranking (C) Players may change positions over their careers and be judged at multiple different positions (D) Some players transfer to different schools and therefore need to change their playing styles (E) Because of differences in strength training programs at different schools, players may develop at different rates _________________ Founder of GMAT Club US News Rankings progression - last 10 years in a snapshot - New! Just starting out with GMAT? Start here... Need GMAT Book Recommendations? Best GMAT Books Co-author of the GMAT Club tests GMAT Club Premium Membership - big benefits and savings If you have any questions you can ask an expert New!  Veritas Prep GMAT Discount Codes Magoosh Discount Codes Kaplan GMAT Prep Discount Codes Founder Affiliations: AS - Gold, HH-Diamond Joined: 04 Dec 2002 Posts: 14546 Location: United States (WA) GMAT 1: 750 Q49 V42 GPA: 3.5 Followers: 3778 Kudos [?]: 23559 [1] , given: 4551 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:00 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED Winners: igloo and rajathpanta Official Explanation: Answer is B This problem demonstrates a type of data flaw that you might not expect – which is exactly what the author of a hard question wants, for you to accept a faulty conclusion because you did not critically examine the data. What’s the flaw here? In a five-point scale, a predicted ‘4’ has multiple outcomes: a 5 (exceeding expectations and bringing the group average up), a 4 (meeting expectations, keeping the average the same), or a 3 or below (underperforming the prediction, bringing the average down). But a predicted ‘5’ can only live up to expectations – it cannot exceed them. Anything other than a 5 brings the expected average down. So while the experts’ misevaluations of a 4-star prospect might be balanced out by some 5s to counter 3s, their evaluations of a 5-star prospect all count in the same downward direction. Choice B exposes this flaw – the experts may have missed just as often with 3 and 4 star players, but because they could miss on either side of the predicted value the average is close to what they predicted.For your study, this is important – especially with the Integrated Reasoning section included, the GMAT will feature plenty of problems in which statistics are used improperly, to draw invalid conclusions. Be skeptical of statistics and train yourself to look for data flaws, both on the GMAT and in business in general. _________________ Founder of GMAT Club US News Rankings progression - last 10 years in a snapshot - New! Just starting out with GMAT? Start here... Need GMAT Book Recommendations? Best GMAT Books Co-author of the GMAT Club tests GMAT Club Premium Membership - big benefits and savings Intern Joined: 20 Sep 2012 Posts: 5 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:02 Ans is C. Rating may be different at diff time.s Director Status: Final Countdown Joined: 17 Mar 2010 Posts: 563 Location: India GPA: 3.82 WE: Account Management (Retail Banking) Followers: 17 Kudos [?]: 282 [0], given: 75 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:03 bb wrote: Veritas Prep 10 Year Anniversary Promo Question #10 Critical Reasoning Sept 21, Friday is the last day for the Veritas Prep 10 Year Anniversary Promo. Today, we will be giving away 2 Courses in the Verbal Challenge and 2 winners will be announced One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course ($1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html

To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it.
Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time.

Good Luck! May the best and fastest win!

College football recruiting services rank incoming players on a scale of 1-star (not a highly sought-after prospect) to 5-star (considered to be the best players).

Recently a service attempted to validate its rankings by assigning star ratings to players upon completion of their careers to determine the accuracy of the initial rankings. The survey averaged the post-career ratings of each player and found that 5-star players’ final average was 4.46, compared with 3.98 for 4-stars and 3.11 for 3-stars. This suggests that the rankings services do not effectively judge high-end talent as well as they judge players in the middle of the range.

Which of the following identifies a problem with the service’s attempt to validate its rankings?

(A) Players at certain positions might be harder to judge at a younger age than players at other positions
(B) A five-star scale does not allow the most elite players to overperform their initial ranking
(C) Players may change positions over their careers and be judged at multiple different positions
(D) Some players transfer to different schools and therefore need to change their playing styles
(E) Because of differences in strength training programs at different schools, players may develop at different rates

c wins- different time, the ranking differs and consolidated ranking hence differs
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21 Sep 2012, 09:04
Answer C.Players may change positions over their careers and will be judged at multiple different positions

All others are out of scope
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21 Sep 2012, 09:04
C
The initial player's position in college may not be the same as the position after the player retire.
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21 Sep 2012, 09:04
(C) Players may change positions over their careers and be judged at multiple different positions.

This options identifies a situation which calls into question the current rating processes' validity.
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21 Sep 2012, 09:05
bb wrote:

Veritas Prep 10 Year Anniversary Promo Question #10 Critical Reasoning

Sept 21, Friday is the last day for the Veritas Prep 10 Year Anniversary Promo. Today, we will be giving away 2 Courses in the Verbal Challenge and 2 winners will be announced

One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course ($1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it. Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time. Good Luck! May the best and fastest win! College football recruiting services rank incoming players on a scale of 1-star (not a highly sought-after prospect) to 5-star (considered to be the best players). Recently a service attempted to validate its rankings by assigning star ratings to players upon completion of their careers to determine the accuracy of the initial rankings. The survey averaged the post-career ratings of each player and found that 5-star players’ final average was 4.46, compared with 3.98 for 4-stars and 3.11 for 3-stars. This suggests that the rankings services do not effectively judge high-end talent as well as they judge players in the middle of the range. Which of the following identifies a problem with the service’s attempt to validate its rankings? (A) Players at certain positions might be harder to judge at a younger age than players at other positions (B) A five-star scale does not allow the most elite players to overperform their initial ranking (C) Players may change positions over their careers and be judged at multiple different positions (D) Some players transfer to different schools and therefore need to change their playing styles (E) Because of differences in strength training programs at different schools, players may develop at different rates Definitely B. B talks about highly rated 5 star players and says they cannot over perform their initial ranking. The last statement of stimulus says the same. Last edited by igloo on 21 Sep 2012, 09:25, edited 1 time in total. Intern Joined: 16 Sep 2012 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 22 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:06 Answer is C: Correct comparison of rankings A: Comparison of players, not ranking B, D, E: Irrelevant Director Joined: 24 Aug 2009 Posts: 507 Schools: Harvard, Columbia, Stern, Booth, LSB, Followers: 17 Kudos [?]: 691 [0], given: 276 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:06 Answer is C because of changing situation a player may play outside there area _________________ If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth -Game Theory If you have any question regarding my post, kindly pm me or else I won't be able to reply Senior Manager Status: Prevent and prepare. Not repent and repair!! Joined: 13 Feb 2010 Posts: 274 Location: India Concentration: Technology, General Management GPA: 3.75 WE: Sales (Telecommunications) Followers: 9 Kudos [?]: 90 [0], given: 282 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:06 I think B is the best answer. The rest all are not in scope. We are only worried about why there is a deviation for 5 star rating and B gives the best answer supporting that. _________________ I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+ Last edited by rajathpanta on 21 Sep 2012, 09:18, edited 1 time in total. Manager Joined: 31 Oct 2011 Posts: 50 Concentration: General Management, Entrepreneurship GMAT 1: 710 Q50 V35 GPA: 3.4 WE: Accounting (Commercial Banking) Followers: 0 Kudos [?]: 18 [0], given: 7 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:07 Answer is C (C) Players may change positions over their careers and be judged at multiple different positions it explains why the ranking is not consistent. Intern Joined: 31 Aug 2012 Posts: 11 Location: United States Concentration: General Management, Finance Schools: Sloan '15 GMAT 1: 640 Q44 V34 WE: Architecture (Telecommunications) Followers: 0 Kudos [?]: 11 [0], given: 7 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:08 Acc. to me, OA is B. Because this is the only option which is specific for five-star candidates others apply to all candidates Intern Joined: 24 Aug 2012 Posts: 23 Location: India GMAT 1: 700 Q51 V34 Followers: 0 Kudos [?]: 8 [0], given: 3 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:09 Answer is E A, B C &D are not relevant as the system compares initial and final scores and not when the athletes peak in their careers. Manager Joined: 24 Jul 2011 Posts: 76 Location: India Concentration: Strategy, General Management GMAT 1: 670 Q49 V33 WE: Asset Management (Manufacturing) Followers: 3 Kudos [?]: 131 [0], given: 5 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:10 Ans: C (A) Players at certain positions might be harder to judge at a younger age than players at other positions >> irreelvent (B) A five-star scale does not allow the most elite players to overperform their initial ranking >> wrong (C) Players may change positions over their careers and be judged at multiple different positions >> correct (D) Some players transfer to different schools and therefore need to change their playing styles >> irrelevent (E) Because of differences in strength training programs at different schools, players may develop at different rates >> irrelevent _________________ My mantra for cracking GMAT: Everyone has inborn talent, however those who complement it with hard work we call them 'talented'. +1 Kudos = Thank You Dear Are you saying thank you? Intern Joined: 20 Sep 2012 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:10 Answer is D players might change style and it may affect rating by the end of career. Founder Affiliations: AS - Gold, HH-Diamond Joined: 04 Dec 2002 Posts: 14546 Location: United States (WA) GMAT 1: 750 Q49 V42 GPA: 3.5 Followers: 3778 Kudos [?]: 23559 [0], given: 4551 Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink] ### Show Tags 21 Sep 2012, 09:17 I do not see any answers/comments in your reply. Please edit or repost. FYI: Edited comments will be counted at the time of the last edit. igloo wrote: bb wrote: Veritas Prep 10 Year Anniversary Promo Question #10 Critical Reasoning Sept 21, Friday is the last day for the Veritas Prep 10 Year Anniversary Promo. Today, we will be giving away 2 Courses in the Verbal Challenge and 2 winners will be announced One quant and one verbal question will be posted each day starting on Monday Sept 17th at 10 AM PST/1 PM EST and the first person to correctly answer the question and show how they arrived at the answer will win a free Veritas Prep GMAT course ($1,650 value). Winners will be selected and notified by a GMAT Club moderator. For more questions and details please check here: veritas-prep-10-year-anniversary-giveaway-138806.html

To participate, please make sure you provide the correct answer (A,B,C,D,E) and explanation that clearly shows how you arrived at it.
Winners will be announced the following day at 10 AM Pacific/1 PM Eastern Time.

Good Luck! May the best and fastest win!

College football recruiting services rank incoming players on a scale of 1-star (not a highly sought-after prospect) to 5-star (considered to be the best players). Recently a service attempted to validate its rankings by assigning star ratings to players upon completion of their careers to determine the accuracy of the initial rankings. The survey averaged the post-career ratings of each player and found that 5-star players’ final average was 4.46, compared with 3.98 for 4-stars and 3.11 for 3-stars. This suggests that the rankings services do not effectively judge high-end talent as well as they judge players in the middle of the range.

Which of the following identifies a problem with the service’s attempt to validate its rankings?

(A) Players at certain positions might be harder to judge at a younger age than players at other positions
(B) A five-star scale does not allow the most elite players to overperform their initial ranking
(C) Players may change positions over their careers and be judged at multiple different positions
(D) Some players transfer to different schools and therefore need to change their playing styles
(E) Because of differences in strength training programs at different schools, players may develop at different rates

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Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink]

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21 Sep 2012, 09:18
B
Problem with the service's attempt: High score may not be proportional to their expertise!!

Last edited by vinc on 21 Sep 2012, 09:22, edited 1 time in total.
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Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink]

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21 Sep 2012, 09:21
Ans=C
C serves the problem with the ranking and the completion of career.
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Re: Veritas Prep 10 Year Anniversary Promo Question #10 [#permalink]

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21 Sep 2012, 09:23
Ans is (B).

Explaination:
A 5-star player can be rated at a lower star rating later on but never above a 5-star rating.
A 4-star rated player can be assigned 3 of lower star ratings (viz. 1-star, 2-star & 3-star) but only one star rating above what it was originally judged (viz. only 5-star)
Re: Veritas Prep 10 Year Anniversary Promo Question #10   [#permalink] 21 Sep 2012, 09:23

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I just started exploring the ‘active learning’ topic. It’s a very handy tool when the number of data points to build a model is limited and labelling new points is costly. It allows to determine which points should be labelled next to bring the most gain in model performance. In this post I will cover some of my small experiments in this area.

Caution!

If you’re interested in ready-to-use tools for active learning, this post might not be for you - I don’t cover any framework here. It’s all about fun (for me) and building some intuitions.

I will not describe active learning’s basis ideas - if you’re interested in this checkout Wikipedia page - https://en.wikipedia.org/wiki/Active_learning_(machine_learning).

library(knitr)
library(tidyr)
library(dplyr)
library(ggplot2)
library(FSelectorRcpp)
knitr::opts_chunk$set(cache = FALSE, warning = FALSE, message = FALSE, eval = TRUE) To make some simulations, it’s good to have some data. I grabbed a dataset from https://archive.ics.uci.edu/ml/index.php. Below there’s a code to download and unzip the data into tmp directory. # https://stackoverflow.com/questions/16474696/read-system-tmp-dir-in-r gettmpdir <- function() { tm <- Sys.getenv(c('TMPDIR', 'TMP', 'TEMP')) d <- which(file.info(tm)$isdir & file.access(tm, 2) == 0)
if (length(d) > 0)
tm[[d[1]]]
else if (.Platform$OS.type == 'windows') Sys.getenv('R_USER') else '/tmp' } dataDir <- file.path(gettmpdir(), "data") dir.create(dataDir, showWarnings = FALSE, recursive = TRUE) dataZip <- file.path(dataDir, "bank-data.zip") dataUrl <- "https://archive.ics.uci.edu/ml/machine-learning-databases/00222/bank.zip" download.file(dataUrl, dataZip) unzip(dataZip, exdir = dataDir) dataPath <- file.path(dataDir, "bank-full.csv") message("Data path: ", dataPath) Then the data needs to be prepared. Nothing special here, just simple reading data into R session and splitting it into train and test sets. # read data and split into test/traings sets library(readr) allData <- readr::read_delim(dataPath, delim = ";", col_types = readr::cols( age = col_double(), job = col_character(), marital = col_character(), education = col_character(), default = col_character(), balance = col_double(), housing = col_character(), loan = col_character(), contact = col_character(), day = col_double(), month = col_character(), duration = col_double(), campaign = col_double(), pdays = col_double(), previous = col_double(), poutcome = col_character(), y = col_character() )) allData <- allData %>% mutate(row_id = row_number()) %>% select(-month, -job) %>% # I have a problem with those two columns during # training phase so I removed them mutate(y = y == "yes") # transform to TRUE/FALSE # split the data into train/test sets. set.seed(123) idx <- sample.int(nrow(allData), nrow(allData)*0.2) trainAll <- allData[-idx,] testAll <- allData[idx,] For my ‘AL’ experiment I made a special workhorse function which handles nearly everything. As the last argument it takes a special function get_new_idx which returns the rows_id from trainAll data.frame to be added to train set in the next round. This simulates the active learning scheme. Data points selected by the get_new_idx would go to the oracle to be annotated. As a model performance score I’m using AUC. #' @param get_new_idx function which returns the selected rows #' indexes to be labelled by the oracle. In this function the 'active learning' #' logic resides. make_active_learning_path <- function(trainAll, testAll, nstart = 500, n = 50, k = 50, get_new_idx) { # init training data set by selecting randomly nstart rows from # 'unlabelled' data idxInit <- tibble(row_id = sample(trainAll$row_id, nstart))
train <- inner_join(trainAll, idxInit, by = "row_id")
trainAll <- anti_join(trainAll, idxInit, by = "row_id")

aucRes <- rep(0, n)

for(i in 1:n) {

# build a classification model using simple logistics regression
fit <- glm(y ~ ., data = train %>% select(-row_id), family = "binomial")

# calculate AUC
res <- predict(fit, newdata = testAll, type="response")
aucRes[i] <- suppressMessages(pROC::auc(testAll$y, res, )) # select new indexes which will be added to the training set newIdx <- get_new_idx(trainAll, train, fit, k) trainNew <- inner_join(trainAll, newIdx, by = "row_id") train <- bind_rows(train, trainNew) # remove selected indexes from available 'unlabelled' set. trainAll <- anti_join(trainAll, newIdx,by = "row_id") } return(aucRes) } ## First experiment - the most uncertain points vs random sample. In the first attempt I’ll use a function which selects data points for which the model is the most uncertain - in the binary classification task those will be the case where the estimated probability is closest to 0.5: get_new_idx_most_uncertain <- function(trainAll, train, fit, k) { predTrainLeftout <- predict(fit, newdata = trainAll, type="response") tr <- trainAll %>% mutate(predTrainLeftout = predTrainLeftout) %>% arrange(abs(predTrainLeftout - 0.5)) tr %>% select(row_id) %>% head(k) } The second function selects the rows at random. There’s nothing fancy in here: get_new_idx_random <- function(trainAll, train, fit, k) { trainAll %>% sample_n(k, replace = FALSE) %>% select(row_id) } Let’s run the first two experiments: # utility function to transform pbreplicate result into data.frame transform_run <- function(x) { xx <- t(x) colnames(xx) <- 1:ncol(xx) rownames(xx) <- 1:nrow(xx) res <- bind_cols(tibble(iter = 1:nrow(xx)), as.data.frame(xx)) res <- pivot_longer(res, cols = c(-iter), names_to = "round", values_to = "AUC") res } ## performing 20 replications of each simulations set.seed(123) mostUncertain <- pbapply::pbreplicate( 50, make_active_learning_path(trainAll, testAll, get_new_idx = get_new_idx_most_uncertain) ) %>% transform_run %>% mutate(Type = "1. Most uncertain") set.seed(123) allRandom <- pbapply::pbreplicate( 50, make_active_learning_path(trainAll, testAll, get_new_idx = get_new_idx_random) ) %>% transform_run %>% mutate(Type = "0. Random") Some code to visualize the result: make_plot <- function(result, addRibbon = FALSE) { result2 <- result %>% group_by(round, Type) %>% summarise( AUC_Mean = mean(AUC), SD = sd(AUC), q025 = quantile(AUC, probs = 0.025), q975 = quantile(AUC, probs = 0.975)) p <- ggplot(result2 %>% mutate(round = as.integer(round))) + geom_line(aes(round, AUC_Mean, color = Type), size = 1.5) + theme_bw() if(addRibbon) { p <- p + geom_ribbon(aes(round, ymax = q025, ymin = q975, fill = Type), alpha = 0.2) } return(p) } And there it is (each round means additional k samples added to training set, the higher the curve is located the better). Here I was really surprised, because I expected that selecting the points for which the model is the most uncertain would be much better than random sampling, but the opposite is true! What could be the reason? make_plot(bind_rows(mostUncertain, allRandom)) I think for this data set the answer is imbalance in the training data. The most uncertain point is not the 50% but ~11%, because if we would not have any model (and we would just use the percentage of yes answers) we should assume that the probability of yes is around 11%, not 50%. mean(allData$y)
## [1] 0.1169848

So, let’s adjust the rows selecting function to take care of the class imbalance (the only change is to replace 0.5 with mean(train$y)): get_new_idx_most_uncertain2 <- function(trainAll, train, fit, k) { predTrainLeftout <- predict(fit, newdata = trainAll, type="response") tr <- trainAll %>% mutate(predTrainLeftout = predTrainLeftout) %>% arrange(abs(predTrainLeftout - mean(train$y)))
}

set.seed(123)
mostUncertain2 <- pbapply::pbreplicate(
50,
make_active_learning_path(trainAll, testAll, get_new_idx = get_new_idx_most_uncertain2)
) %>% transform_run %>% mutate(Type = "2. Most uncertain - sample based")

As I was expecting, the result looks much better than the first version, but it’s still worse than random selection.

make_plot(bind_rows(mostUncertain, mostUncertain2, allRandom))

It seems unintuitive, but it makes sense. To score which points should be taken into account I took only raw probability estimates, completely ignoring errors from the model. For one point the estimate might be 26% +- 5% and for another 26% +- 20%. In my case both points are treated equally, which seems wrong. See an example here:

set.seed(123)
fit <- glm(y ~ ., data = trainAll %>% select(-row_id) %>% sample_n(500), family = "binomial")

pred <- predict(fit, se.fit = TRUE)
pred <- tibble(link = pred$fit, se.fit = pred$se.fit) %>% arrange(desc(link))

pred[51:54,] %>%
mutate_all(function(x) round(x, digits = 3)) %>%
kable()
-1.161 0.938 0.239 0.445
-1.167 0.679 0.237 0.380
-1.172 0.837 0.236 0.417
-1.178 0.926 0.235 0.437

Some in the next experiment I will select those points where the standard error is the biggest. See the code below:

get_new_idx_most_uncertain3_stdErr_based <- function(trainAll, train, fit, k) {
stdErr <- predict(fit, newdata = trainAll, se.fit = TRUE)$se.fit tr <- trainAll %>% mutate(stdErr = stdErr) %>% arrange(desc(stdErr)) tr %>% select(row_id) %>% head(k) } set.seed(123) mostUncertain3StdErrBased <- pbapply::pbreplicate( 50, make_active_learning_path( trainAll, testAll, get_new_idx = get_new_idx_most_uncertain3_stdErr_based) ) %>% transform_run %>% mutate(Type = "3. Most uncertain - std.err based") The idea looked promising, but the reality is tough. This is the worst strategy from all four. Selecting by random is still the best. After some mediation on this result I conclude that this might not be the best idea, because it selects the most noisy points which probably bring much more noise than a good quality signal. make_plot(bind_rows(mostUncertain, mostUncertain2, allRandom, mostUncertain3StdErrBased)) My last idea for now is to remove points with very high standard error. To do this, I’m filtering out everything points with a standard error greater than its 97.5% quantile. get_new_idx_most_uncertain3_stdErr_trimmed <- function(trainAll, train, fit, k) { stdErr <- predict(fit, newdata = trainAll, se.fit = TRUE)$se.fit
tr <- trainAll %>% mutate(stdErr = stdErr) %>% arrange(desc(stdErr))
tr <- tr %>% filter(stdErr < quantile(stdErr, probs = 0.975))

}

set.seed(123)
mostUncertain3StdErrTrimmed <- pbapply::pbreplicate(
50,
make_active_learning_path(
trainAll, testAll,
get_new_idx = get_new_idx_most_uncertain3_stdErr_trimmed)
) %>% transform_run %>% mutate(Type = "4. Most uncertain - std.err trimmed")

Aaand… This is still bad solution. Not as bad as the previous one, but still.

make_plot(
bind_rows(mostUncertain, mostUncertain2,
allRandom, mostUncertain3StdErrBased,
mostUncertain3StdErrTrimmed))

## Summary

Active learning is an interesting idea. It’s very exciting that simple and crude solutions do not work very well. It’s really a place where good theory should thrive (or at least be better than random sampling:)). I will probably do a little more experiments in this area to build more intuitions before digging into proper, well-founded methods.


text:
# Potential difference

Potential difference

In physics, the potential difference or p.d. between two points is the difference of the points' scalar potential, equivalent to the line integral of the field strength between the two points.

There are various types of potential difference related to the amount of energy required to move an object from one place to another against various types of forces.

* in electrodynamics, the potential difference corresponds to the amount of work that would need to be done on a unit electric charge to move it from one point to the other against an electric field. [ [http://www.sasked.gov.sk.ca/docs/physics/u3b23phy.html Physics 30 - Electricity - Electric Potential Difference ] ] P.d. is synonymous with voltage [ [http://www.powerstream.com/1922/battery_1922_WITTE/batteryfiles/definitions.htm Battery - Definitions ] ] and is measured in volts. It is the line integral of the electric field strength between two points. [Pitt, V.H. (ed.), "The Penguin Dictionary of Physics", Penguin Books Ltd., 1977, ISBN 0140510710] Also, if "v" is the p.d. in volts, "w" is the work in joules (J), and "q" is the charge in coulombs (C), then:
*:$v = \left\{ w over q \right\} \left\{mathrm\left\{J\right\} over mathrm\left\{C mathrm\left\{or\right\} \left\{w over q\right\} mathrm\left\{V\right\}$
*:main|Voltage
* In mechanics, the "gravitational potential difference" between two points on Earth is related to the energy that would be required to move a unit mass from one point to the other against the Earth's gravitational field. Units: joules per kilogram.
* In fluid systems the "potential difference" is the difference in pressure. Units: pascals.
* In thermal systems the "potential difference" is the difference in temperature. Units: kelvins.

In some engineering fields, "potential" is sometimes described as the 'across variable', whereas flux is the 'through variable'. The product of the flux and the potential difference is the power, which is the time rate of change of energy.

References

Wikimedia Foundation. 2010.

### Look at other dictionaries:

• potential difference — elektrinių potencialų skirtumas statusas T sritis Standartizacija ir metrologija apibrėžtis Apibrėžtį žr. priede. priedas( ai) Grafinis formatas atitikmenys: angl. electric potential difference; potential difference vok. elektrische… …   Penkiakalbis aiškinamasis metrologijos terminų žodynas

• potential difference — potencialų skirtumas statusas T sritis automatika atitikmenys: angl. potential difference; voltage difference vok. Potentialdifferenz, f; Potentialunterschied, m rus. разница потенциалов, f; разность потенциалов, f pranc. différence des… …   Automatikos terminų žodynas

• potential difference — potencialų skirtumas statusas T sritis fizika atitikmenys: angl. potential difference vok. Potentialdifferenz, f; Potentialunterschied, m rus. разность потенциалов, f pranc. différence de potentiel, f …   Fizikos terminų žodynas

• potential difference — potencialų skirtumas statusas T sritis chemija apibrėžtis Dydis, lygus elektrinio lauko darbui, kurio reikia perkelti vienetinį teigiamą elektros krūvį iš vieno lauko taško į kitą. atitikmenys: angl. potential difference rus. разность потенциалов …   Chemijos terminų aiškinamasis žodynas

• potential difference — potential dif·fer·ence dif (ə )rən(t)s n the difference in electrical potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other compare… …   Medical dictionary

• potential difference — difference in electrical tension between two points (Electricity) …   English contemporary dictionary

• potential difference — noun the difference in electrical charge between two points in a circuit expressed in volts • Syn: ↑electric potential, ↑potential, ↑potential drop, ↑voltage • Hypernyms: ↑electrical phenomenon • Hyponyms: ↑ …   Useful english dictionary

• potential difference — noun the difference in potential energy between two points in an electric field; the difference in charge between two points in an electrical circuit; voltage …   Wiktionary

• potential difference — noun Date: 1892 the difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other …   New Collegiate Dictionary

• potential difference — noun Physics the difference of electrical potential between two points …   English new terms dictionary


text:
# Game Theory forest tree dashed ellipse between various nodes

I currently have a Forest binary tree as follows:

This is produced by the code:

\documentclass{standalone}

\usepackage{standalone}
\usepackage[T1]{fontenc}
\usepackage{graphicx}
\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing}
\usepackage{xcolor}
\usepackage{forest}
\usepackage{amsmath}
\usepackage{MnSymbol}
\usepackage{commath}

\forestset{
ptree/.style={
for tree={
grow'=0,
parent anchor=children,
child anchor=parent,
l=2.4cm,
s sep=0.5cm,
fit=rectangle
},
before typesetting nodes={
for tree={
split option={content}{:}{content, my edge label},
},
},
},
my edge label/.style={
if={
> O_= {n'}{1}
}{
edge label={node [midway, fill=white, font=\color{red}\footnotesize] {#1} }
}{
edge label={node [midway, fill=white, font=\color{red}\footnotesize] {#1} }
},
}
}

\begin{document}
\begin{forest}
ptree
[1
[2: 1
[1: 1
[2: 1
[{(1,-1)}: 1]
]
[{(-1,1)}: 2]
]
[1: 2
[{(-1,1)}: 1]
]
]
[2: 2
[1: 1
[{(-1,1)}: 1]
]
[{(1,-1)}: 2]
]
]
\end{forest}
\end{document}

This is perfect for what I need it do, except I’m hoping to pushed dashed ellipses encompassing arbitrary ellipses, often with multiple in the same tree. I put an example of what I mean below.

As well, to make it easier to code rapidly (since I’m using this for lecture notes), I was hoping to simply modify the forest tree to look like:

\begin{forest}
ptree
[1
[2: 1
[1: 1: A
[2: 1
[{(1,-1)}: 1]
]
[{(-1,1)}: 2]
]
[1: 2: A
[{(-1,1)}: 1]
]
]
[2: 2: B
[1: 1: B
[{(-1,1)}: 1]
]
[{(1,-1)}: 2]
]
]
\end{forest}

where A and B will be used to designate two separate ellipses. As well, occasionally, these ellipses may encompass three or more nodes, so I am hoping they can expand as needed.

Any help changing the \forestset would be very much appreciated!

• Please attribute code and provide a link to the source. It is rude not to and a link is often enormously helpful to potential helpers.
– cfr
May 5 '18 at 1:47
• @cfr I've put a link to the original question where I got the code! Sorry about that! May 5 '18 at 2:32

A rather similar question has been answered here, but I am not sure it works with forest and, more importantly, rotates the ellipses. For the latter, I am using one proposal of this answer. I'd love to use fit but that fails when the nodes are just separated vertically. Here I draw the ellipses around a given node and another node that gets specified in the relative node syntax, see page 8 of the forest manual.

\documentclass[tikz,border=3.14pt]{standalone}
\usepackage[T1]{fontenc}
\usetikzlibrary{decorations.pathreplacing,shapes.geometric,calc}
\usepackage{xcolor}
\usepackage{forest}
\usepackage{amsmath}
\usepackage{MnSymbol}
\usepackage{commath}

\forestset{
ptree/.style={
for tree={
grow'=0,
parent anchor=children,
child anchor=parent,
l=2.4cm,
s sep=0.5cm,
fit=rectangle
},
before typesetting nodes={
for tree={
split option={content}{:}{content, my edge label},
},
},
},
my edge label/.style={
if={
> O_= {n'}{1}
}{
edge label={node [midway, fill=white, font=\color{red}\footnotesize] {#1} }
}{
edge label={node [midway, fill=white, font=\color{red}\footnotesize] {#1} }
},
},
myellipse/.style={ tikz={% from https://tex.stackexchange.com/a/186673/121799
\path let
\p1 = ($()-(#1)$),
\p2 = ($(.north east)-(.south west)$),
\p3 = ($(#1.north east)-(#1.south west)$),
\n2 = {veclen(\p2)},
\n3 = {veclen(\p3)},
\n1 = {veclen(\p1)+0.5*(\n2+\n3)}
in %\pgfextra{\typeout{\n1,\n2,\n3}}
() -- (#1)
node[midway, sloped, draw, ellipse,dashed,
minimum width=\n1, minimum height={max(\n2,\n3)}] {};
}}
}

\begin{document}
\begin{forest}
ptree
[1
[2: 1
[1: 1
[2: 1
[{(1,-1)}: 1]
]
[{(-1,1)}: 2]
]
[1: 2,myellipse=!s
[{(-1,1)}: 1]
]
]
[2: 2
[1: 1,myellipse=!u
[{(-1,1)}: 1]
]
[{(1,-1)}: 2]
]
]
\end{forest}
\end{document}

EDIT: Replaced the hardcoded additional distance by the size of the respective nodes.

• Thank you so much! This is exactly what I was looking for! May 4 '18 at 19:40
• @FelixJen You can get the requested syntax: you just need to split to three and collect the third bits into a register which gets turned into an ellipse in a later stage or cycle.
– cfr
May 5 '18 at 1:50

Another solution to use istgame package (version 2.0). You can simply use the macro \xtInfosetO to draw sloped oval type information sets.

\documentclass{standalone}
\usepackage{istgame}

\begin{document}

\begin{istgame}
\setistgrowdirection'{east}
\tikzset{oval node/.style={ellipse node,draw=none}}
\xtdistance{25mm}{25mm}
\istroot(0)[null node]<180>{1}+25mm..40mm+
\istb{1}[fill=white,red] \istb{2}[fill=white,red] \endist
\istrooto(1)(0-1){2}
\istb{1}[fill=white,red] \istb{2}[fill=white,red] \endist
\istrooto(2)(0-2){2}
\istb{1}[fill=white,red] \istb{2}[fill=white,red]{(-1,1)} \endist
\istrooto(3)(1-1){1}
\istb{1}[fill=white,red] \istb{2}[fill=white,red]{(-1,1)} \endist
\istrooto(4)(1-2){1}
\istb{1}[fill=white,red]{(-1,1)} \endist
\istrooto(5)(2-1){1}
\istb{1}[fill=white,red]{(-1,1)} \endist
\istrooto(6)(3-1){2}
\istb{1}[fill=white,red]{(-1,1)} \endist
\setxtinfosetlayer{main}
\xtInfosetO[dashed,thick](3)(4)(1.5em)
\xtInfosetO[ellipse,dashed,thick](2)(5)(2em)
\end{istgame}
\end{document}


text:
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# Evaluate the indefinite integral. $\displaystyle \int x(2x + 5)^8 \, dx$

## $\frac{1}{40}(2 x+5)^{10}-\frac{5}{36}(2 x+5)^{9}+C$

Integrals

Integration

### Discussion

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### Video Transcript

Okay. The first thing we know is that we can make our you equivalent to two X plus five, which means our detox is 1/2. Do you remember? The derivative of five is simply zero. It's a constant. It doesn't moderate this context. Okay, Now this means we have the interval of humanness. Five over too times you to the eighth terms. 1/2 do you, which is simplified toe 1/4 times the integral of you to the ninth, minus five. You to the eighth D'You Using the power rule, we increased the exponents by wand. Then we divide by the new exponents. So you do. The ninth becomes you to the 10th divide by the new Explain it. Like I said. Now back. Substitute that you is to expose five. And don't forget your plus c

Integrals

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text:
# 12.1.1: Wave Basics

Waves generally begin as a disturbance of some kind, and the energy of that disturbance gets propagated in the form of waves. We are most familiar with the kind of waves that break on shore, or rock a boat at sea, but there are many other types of waves that are important to oceanography:

• Internal waves form at the boundaries of water masses of different densities (i.e. at a pycnocline), and propagate at depth. These generally move more slowly than surface waves, and can be much larger, with heights exceeding 100 m. However, the height of the deep wave would be unnoticeable at the surface.
• Tidal waves are due to the movement of the tides. What we think of as tides are basically enormously long waves with a wavelength that may span half the globe (see section 11.1). Tidal waves are not related to tsunamis, so don’t confuse the two.
• Tsunamis are large waves created as a result of earthquakes or other seismic disturbances. They are also called seismic sea waves (section 10.4).
• Splash waves are formed when something falls into the ocean and creates a splash. The giant wave in Lituya Bay that was described in the introduction to this chapter was a splash wave.
• Atmospheric waves form in the sky at the boundary between air masses of different densities. These often create ripple effects in the clouds (Figure $$\PageIndex{1}$$).

There are several components to a basic wave (Figure $$\PageIndex{2}$$):

• Still water level: where the water surface would be if there were no waves present and the sea was completely calm.
• Crest: the highest point of the wave.
• Trough: the lowest point of the wave.
• Wave height: the distance between the crest and the trough.
• Wavelength: the distance between two identical points on successive waves, for example crest to crest, or trough to trough.
• Wave steepness: the ratio of wave height to length (H/L). If this ratio exceeds 1/7 (i.e. height exceeds 1/7 of the wavelength) the wave gets too steep, and will break.

There are also a number of terms used to describe wave motion:

• Period: the time it takes for two successive crests to pass a given point.
• Frequency: the number of waves passing a point in a given amount of time, usually expressed as waves per second. This is the inverse of the period.
• Speed: how fast the wave travels, or the distance traveled per unit of time. This is also called celerity (c), where

c = wavelength  x  frequency

Therefore, the longer the wavelength, the faster the wave.

Although waves can travel over great distances, the water itself shows little horizontal movement; it is the energy of the wave that is being transmitted, not the water. Instead, the water particles move in circular orbits, with the size of the orbit equal to the wave height (Figure $$\PageIndex{3}$$). This orbital motion occurs because water waves contain components of both longitudinal (side to side) and transverse (up and down) waves, leading to circular motion. As a wave passes, water moves forwards and up over the wave crests, then down and backwards into the troughs, so there is little horizontal movement. This is evident if you have ever watched an object such as a seabird floating at the surface. The bird bobs up and down as the wave pass underneath it; it does not get carried horizontally by a single wave crest.

The circular orbital motion declines with depth as the wave has less impact on deeper water and the diameter of the circles is reduced. Eventually at some depth there is no more circular movement and the water is unaffected by surface wave action. This depth is the wave base and is equivalent to half of the wavelength (Figure $$\PageIndex{4}$$). Since most ocean waves have wavelengths of less than a few hundred meters, most of the deeper ocean is unaffected by surface waves, so even in the strongest storms marine life or submarines can avoid heavy waves by submerging below the wave base.

When the water below a wave is deeper than the wave base (deeper than half of the wavelength), those waves are called deep water waves. Most open ocean waves are deep water waves. Since the water is deeper than the wave base, deep water waves experience no interference from the bottom, so their speed only depends on the wavelength:

where g is gravity and L is wavelength in meters. Since g and π are constants, this can be simplified to:

Shallow water waves occur when the depth is less than 1/20 of the wavelength. In these cases, the wave is said to “touch bottom” because the depth is shallower than the wave base so the orbital motion is affected by the seafloor. Due to the shallow depth, the orbits are flattened, and eventually the water movement becomes horizontal rather than circular just above the bottom. The speed of shallow water waves depends only on the depth:

where g is gravity and d is depth in meters. This can be simplified to:

Intermediate or transitional waves are found in depths between ½ and 1/20 of the wavelength. Their behavior is a bit more complex, as their speed is influenced by both wavelength and depth. The speed of an intermediate wave is calculated as:

which contains both depth and wavelength variables.


text:
# Implicit & Logarithmic differentiation problems.

• Oct 23rd 2012, 11:21 PM
jamesrb
Implicit & Logarithmic differentiation problems.
#1. Find the derivative of: $e^x^y=sin(y^2)$

$lne^x^y=lnsin(y^2)$
$xylne=lnsin(y^2)$
$xy*1=lnsin(y^2)$
$y+xy'=lnsin(y^2)$
I need to use 2 chain rules on $lnsin(y^2)$
$y+xy'=[{\frac{1}{sin(y^2)}]*[cos(y^2)*(2yy')]$
$y+xy'=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}$
$y=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}-xy'$
$y=y'[\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}}-x]$
$y'=\frac{y}{\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}-x}}$

I'm sure there is something wrong in there :)

#2. Find the derivative using logartihmic differentiation:
$y=\frac{x(x+1)^3}{(3x+1)^2}$
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
$lny=[ln(x(x+1)^3]-[ln(3x+1)^2]$
Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.
• Oct 23rd 2012, 11:45 PM
earboth
Re: Implicit & Logarithmic differentiation problems.
Quote:

Originally Posted by jamesrb
[B]...

#2. Find the derivative using logartihmic differentiation:
$y=\frac{x(x+1)^3}{(3x+1)^2}$
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
$lny=[ln(x(x+1)^3]-[ln(3x+1)^2]$
Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.

You were on the right track:

$lny=ln\frac{x(x+1)^3}{(3x+1)^2} = \underbrace{\ln(x)+3\ln(x+1)}_{numerator}-\overbrace{2\ln(3x+1)}^{denominator}$

Now the differentiation of the RHS is quite simple.
• Oct 24th 2012, 12:01 AM
jamesrb
Re: Implicit & Logarithmic differentiation problems.
Quote:

Originally Posted by earboth
You were on the right track:

$lny=ln\frac{x(x+1)^3}{(3x+1)^2} = \underbrace{\ln(x)+3\ln(x+1)}_{numerator}-\overbrace{2\ln(3x+1)}^{denominator}$

Now the differentiation of the RHS is quite simple.

Thanks, I think I got it:
$y'=y(\frac{1}{x}+\frac{3}{x+1}-\frac{6}{3x+1})$
• Oct 24th 2012, 06:12 AM
Prove It
Re: Implicit & Logarithmic differentiation problems.
Quote:

Originally Posted by jamesrb
#1. Find the derivative of: $e^x^y=sin(y^2)$

$lne^x^y=lnsin(y^2)$
$xylne=lnsin(y^2)$
$xy*1=lnsin(y^2)$
$y+xy'=lnsin(y^2)$
I need to use 2 chain rules on $lnsin(y^2)$
$y+xy'=[{\frac{1}{sin(y^2)}]*[cos(y^2)*(2yy')]$
$y+xy'=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}$
$y=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}-xy'$
$y=y'[\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}}-x]$
$y'=\frac{y}{\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}-x}}$

I'm sure there is something wrong in there :)

#2. Find the derivative using logartihmic differentiation:
$y=\frac{x(x+1)^3}{(3x+1)^2}$
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:
$lny=[ln(x(x+1)^3]-[ln(3x+1)^2]$
Or should I use a combination of the quotient, product and multiple chain rules on what I had before:
$lny=ln\frac{x(x+1)^3}{(3x+1)^2}$
You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.

I think in Question 1, attempting logarithmic differentiation is a waste of time.

\displaystyle \begin{align*} e^{x\,y} &= \sin{\left( y^2 \right)} \\ \frac{d}{dx} \left( e^{x\,y} \right) &= \frac{d}{dx}\left[ \sin{\left( y^2 \right)} \right] \\ e^{x\,y} \, \frac{d}{dx} \left( x\, y\right) &= \frac{d}{dy}\left[ \sin{\left( y^2 \right)} \right] \frac{dy}{dx} \\ e^{x\,y} \left( x\,\frac{dy}{dx} + y \right) &= 2y \cos{\left( y^2 \right)} \, \frac{dy}{dx} \\ x\, e^{x\,y} \, \frac{dy}{dx} + y \, e^{x\,y} &= 2y\cos{\left( y^2 \right)} \, \frac{dy}{dx} \\ y\, e^{x\,y} &= 2y\cos{\left( y^2 \right)} \, \frac{dy}{dx} - x\, e^{x\,y} \, \frac{dy}{dx} \\ y\, e^{x\,y} &= \left[ 2y\cos{\left( y^2 \right)} - x\, e^{x\,y} \right] \frac{dy}{dx} \\ \frac{y\, e^{x\,y}}{2y\cos{\left( y^2 \right)} - x \, e^{x\,y} } &= \frac{dy}{dx} \end{align*}