text:
1. ## Nondecreasing Partial Sums

Suppose that...

$
\sum\limits_{n = 1}^\infty {a_n }
$
is an infinite series with $
a_n \geqslant 0
$
for all n. Then each partial sum is greater than or equal to its predecessor because $
s_{n + 1} = s_n + a_n
$

$
s_1 \leqslant s_2 \leqslant s_3 \leqslant \cdot \cdot \cdot \leqslant s_n \leqslant s_{n + 1} \leqslant \cdot \cdot \cdot
$

Would anyone mind explaining to me how $
s_{n + 1} = s_n + a_n
$
is true?

Let's say we have: $
\sum\limits_{n = 1}^\infty {n^2 }
$

Let's write out the first couple terms:

$
\sum\limits_{n = 1}^\infty {n^2 } = 1 + 4 + 9 + \cdot \cdot \cdot + n
$

So....

$
\begin{gathered}
s_{n + 1} = s_n + a_n \hfill \\
s_3 = s_2 + a_2 \hfill \\
14 \ne 5 + 4 \hfill \\
\end{gathered}
$

What am I not getting?

Thank you.

2. Originally Posted by RedBarchetta
Suppose that...

$
\sum\limits_{n = 1}^\infty {a_n }
$
is an infinite series with $
a_n \geqslant 0
$
for all n. Then each partial sum is greater than or equal to its predecessor because $
s_{n + 1} = s_n + a_n
$
That's not correct. It should be $s_{n + 1} = s_n + a_{n+1}$ ( $s_n$ is the sum of the first n terms; you need to add the (n+1)th term to that in order to get the sum of the first (n+1) terms).

3. Thank you. Just a misprint in the book.


text:
In the optimization algorithms we introduced previously, each element of the objective function’s independent variables uses the same learning rate at the same time step for self-iteration. For example, if we assume that the objective function is $$f$$ and the independent variable is a two-dimensional vector $$[x_1, x_2]^\top$$, each element in the vector uses the same learning rate when iterating. For example, in gradient descent with the learning rate $$\eta$$, element $$x_1$$ and $$x_2$$ both use the same learning rate $$\eta$$ for iteration:

(10.7.1)$x_1 \leftarrow x_1 - \eta \frac{\partial{f}}{\partial{x_1}}, \quad x_2 \leftarrow x_2 - \eta \frac{\partial{f}}{\partial{x_2}}.$

In Section 10.6, we can see that, when there is a big difference between the gradient values $$x_1$$ and $$x_2$$, a sufficiently small learning rate needs to be selected so that the independent variable will not diverge in the dimension of larger gradient values. However, this will cause the independent variables to iterate too slowly in the dimension with smaller gradient values. The momentum method relies on the exponentially weighted moving average (EWMA) to make the direction of the independent variable more consistent, thus reducing the possibility of divergence. In this section, we are going to introduce Adagrad, an algorithm that adjusts the learning rate according to the gradient value of the independent variable in each dimension to eliminate problems caused when a unified learning rate has to adapt to all dimensions.

## 10.7.1. The Algorithm¶

The Adadelta algorithm uses the cumulative variable $$\boldsymbol{s}_t$$ obtained from a square by element operation on the mini-batch stochastic gradient $$\boldsymbol{g}_t$$. At time step 0, Adagrad initializes each element in $$\boldsymbol{s}_0$$ to 0. At time step $$t$$, we first sum the results of the square by element operation for the mini-batch gradient $$\boldsymbol{g}_t$$ to get the variable $$\boldsymbol{s}_t$$:

(10.7.2)$\boldsymbol{s}_t \leftarrow \boldsymbol{s}_{t-1} + \boldsymbol{g}_t \odot \boldsymbol{g}_t,$

Here, $$\odot$$ is the symbol for multiplication by element. Next, we re-adjust the learning rate of each element in the independent variable of the objective function using element operations:

(10.7.3)$\boldsymbol{x}_t \leftarrow \boldsymbol{x}_{t-1} - \frac{\eta}{\sqrt{\boldsymbol{s}_t + \epsilon}} \odot \boldsymbol{g}_t,$

Here, $$\eta$$ is the learning rate while $$\epsilon$$ is a constant added to maintain numerical stability, such as $$10^{-6}$$. Here, the square root, division, and multiplication operations are all element operations. Each element in the independent variable of the objective function will have its own learning rate after the operations by elements.

## 10.7.2. Features¶

We should emphasize that the cumulative variable $$\boldsymbol{s}_t$$ produced by a square by element operation on the mini-batch stochastic gradient is part of the learning rate denominator. Therefore, if an element in the independent variable of the objective function has a constant and large partial derivative, the learning rate of this element will drop faster. On the contrary, if the partial derivative of such an element remains small, then its learning rate will decline more slowly. However, since $$\boldsymbol{s}_t$$ accumulates the square by element gradient, the learning rate of each element in the independent variable declines (or remains unchanged) during iteration. Therefore, when the learning rate declines very fast during early iteration, yet the current solution is still not desirable, Adagrad might have difficulty finding a useful solution because the learning rate will be too small at later stages of iteration.

Below we will continue to use the objective function $$f(\boldsymbol{x})=0.1x_1^2+2x_2^2$$ as an example to observe the iterative trajectory of the independent variable in Adagrad. We are going to implement Adagrad using the same learning rate as the experiment in last section, 0.4. As we can see, the iterative trajectory of the independent variable is smoother. However, due to the cumulative effect of $$\boldsymbol{s}_t$$, the learning rate continuously decays, so the independent variable does not move as much during later stages of iteration.

%matplotlib inline
import d2l
import math
from mxnet import nd

# The first two terms are the independent variable gradients
g1, g2, eps = 0.2 * x1, 4 * x2, 1e-6
s1 += g1 ** 2
s2 += g2 ** 2
x1 -= eta / math.sqrt(s1 + eps) * g1
x2 -= eta / math.sqrt(s2 + eps) * g2
return x1, x2, s1, s2

def f_2d(x1, x2):
return 0.1 * x1 ** 2 + 2 * x2 ** 2

eta = 0.4

epoch 20, x1 -2.382563, x2 -0.158591

Now, we are going to increase the learning rate to $$2$$. As we can see, the independent variable approaches the optimal solution more quickly.

eta = 2

epoch 20, x1 -0.002295, x2 -0.000000

## 10.7.3. Implementation from Scratch¶

Like the momentum method, Adagrad needs to maintain a state variable of the same shape for each independent variable. We use the formula from the algorithm to implement Adagrad.

def init_adagrad_states(feature_dim):
s_w = nd.zeros((feature_dim, 1))
s_b = nd.zeros(1)
return (s_w, s_b)

eps = 1e-6
for p, s in zip(params, states):
p[:] -= hyperparams['lr'] * p.grad / (s + eps).sqrt()

Compared with the experiment in Section 10.5, here, we use a larger learning rate to train the model.

data_iter, feature_dim = d2l.get_data_ch10(batch_size=10)
{'lr': 0.1}, data_iter, feature_dim);

loss: 0.243, 0.298 sec per epoch

## 10.7.4. Concise Implementation¶

Using the Trainer instance of the algorithm named “adagrad”, we can implement the Adagrad algorithm with Gluon to train models.

d2l.train_gluon_ch10('adagrad', {'learning_rate': 0.1}, data_iter)

loss: 0.242, 0.302 sec per epoch

## 10.7.5. Summary¶

• Adagrad constantly adjusts the learning rate during iteration to give each element in the independent variable of the objective function its own learning rate.
• When using Adagrad, the learning rate of each element in the independent variable decreases (or remains unchanged) during iteration.

## 10.7.6. Exercises¶

• When introducing the features of Adagrad, we mentioned a potential problem. What solutions can you think of to fix this problem?
• Try to use other initial learning rates in the experiment. How does this change the results?

## 10.7.7. Reference¶

[1] Duchi, J., Hazan, E., & Singer, Y. (2011). Adaptive subgradient methods for online learning and stochastic optimization. Journal of Machine Learning Research, 12(Jul), 2121-2159.

format: plain


text:
# Simplify? sqrt8 xx sqrt(48^3)

$\sqrt{8} \times \sqrt{{48}^{3}} = 384 \sqrt{6}$

#### Explanation:

$\sqrt{8} \times \sqrt{{48}^{3}}$

Because both terms are under a square root sign, we can combine them:

$\sqrt{8 \times {48}^{3}}$

Rather than doing the cube of 48 first, let's see that $8 = {2}^{3}$, and so we can combine again:

$\sqrt{{2}^{3} \times {48}^{3}} = \sqrt{{96}^{3}} = {96}^{\frac{3}{2}}$

I'll go ahead and cube now, then work on the square root after:

$\sqrt{884736} = \sqrt{16384 \times 54}$

(I kept dividing by 4 to find the 16384).

Note that ${2}^{14} = 16384$, and so $\sqrt{{2}^{14}} = {2}^{7} = 128$. Also, $54 = 9 \times 6$ and so $\sqrt{54} = 3 \sqrt{6}$

Putting it together, we get:

$\sqrt{884736} = \sqrt{16384 \times 54} = 128 \times 3 \times \sqrt{6} = 384 \sqrt{6}$


text:
# Document Tree Practice Problem 4

Here is the solution. You can view the page itself here.

<!doctype html>

<meta charset="utf-8"/>
<title>Quiz</title>

Foo

Directions:

One

Two

Three

This is the original diagram


text:
View Single Post

## Can someone verify a computational result for me?

This is for a computational physics class, but isn't a homework question, per se. My friend and I have two different sets of code trying to achieve the same thing and we're getting vastly different answers, orders of magnitude apart. We haven't covered this sort of problem in our intro E&M class yet so we aren't able to confirm a result by hand.

Using a λ0 value of 1e-6 and L = 0.5m I'm getting V = ~0.0995 and he's getting 10,596.

Write a Python code that computes the electrostatic potential of a finite segment of charged wire of length 2L, at a height H=L above the center of the wire. The charge density along the wire varies as λ=λ0cos$\frac{Pi*x}{2L}$
 PhysOrg.com physics news on PhysOrg.com >> Is there an invisible tug-of-war behind bad hearts and power outages?>> Penetrating the quantum nature of magnetism>> Rethinking the universe: Groundbreaking theory proposed in 1997 suggests a 'multiverse'


text:
How many integers satisfy the inequality x(3x-4) is less than or equal to 6x^2-3x+5/10

Why write 5/10 instead of 1/2 or 0.5?

Perhaps you mean (6x^2- 3x+ 5)/10? Saying that $x(3x- 4)\le (6x^2- 3x+ 5)/10$ is the same as saying $10x(3x- 4)\le 6x^2- 3x+ 5$ which in turn is the same as $30x^2- 40x- (6x^2- 3x+ 5)= 24x^2- 37x- 5\le 0$. You can use the quadratic formula to determine where that polynoial is equal to 0. It will be less than 0 between those two values so it is only necessary to determine how many integers there are between the two roots of $24x^2- 37x- 5$.

You might note that $24x^2- 37x- 5=(8x+1)(3x-5)$
You might note that $24x^2- 37x- 5=(8x+1)(3x-5)$


text:
# The relationship between the number of support vectors and the number of features

I ran an SVM against a given data set, and made the following observation: If I change the number of features for building the classifier, the number of resulting support vectors will also be changed.

I would like to know how to explain this kind of scenario.

• What was the type and style of those extra features? Where they look-a-like variants of the extisting features, or some more fresh features that you thought may have extra resolving power? – Philip Oakley Aug 8 '12 at 9:13
• This is a document classification problem, and the extra features are just words. I used unigram to build feature space. – user3269 Aug 8 '12 at 13:21
• Given @marc's answer, which did the change go, did the number of vectors rise with the number of features, or the reverse. – Philip Oakley Aug 9 '12 at 8:40
• @Phillip, my original response was wrong. I think the edited answer is accurate now. – Marc Shivers Aug 10 '12 at 1:13

If you look at the optimization problem that SVM solves:

$\min_{\mathbf{w},\mathbf{\xi}, b } \left\{\frac{1}{2} \|\mathbf{w}\|^2 + C \sum_{i=1}^n \xi_i \right\}$

s.t. $y_i(\mathbf{w}\cdot\mathbf{x_i} - b) \ge 1 - \xi_i, ~~~~\xi_i \ge 0,$ for all $i=1,\dots n$

the support vectors are those $x_i$ where the corresponding $\xi_i \gt 0$. In other words, they are the data points that are either misclassified, or close to the boundary.

Now let's compare the solution to this problem when you have a full set of features, to the case where you throw some features away. Throwing a feature away is functionally equivalent to keeping the feature, but adding a contraint $w_j=0$ for the feature $j$ that we want to discard.

When you compare these two optimization problems, and work through the math, it turns out there is no hard relationship between the number of features and the number of support vectors. It could go either way.

It's useful to think about a simple case. Imagine a 2-dim case where your negative and positive features are clustered around (-1,-1) and (1,1), respectively, and are separable with a diagonal separating hyperplane with 3 support vectors. Now imagine dropping the y-axis feature, so your data in now projected on the x-axis. If the data are still separable, say at x=0, you'd probably be left with only 2 support vectors, one on each side, so adding the y-feature would increases the number of support vectors. However, if the data are no longer separable, you'd get at least one support vector for each point that's on the wrong side of x=0, in which case adding the y-feature would reduce the number of support vectors.

So, if this intuition is correct, if you're working in very high-dimensional feature spaces, or using a kernel that maps to a high dimensional feature space, then your data is more likely to be separable, so adding a feature will tend to just add another support vector. Whereas if your data is not currently separable, and you add a feature that significantly improves separability, then you're more likely to see a decrease in the number of support vectors.


text: The normality in products with a countably compact factor It is known that the product $\omega_1 \times X$ of $\omega_1$ with an $M_1$-space may be nonnormal. In this paper we prove that the product $\kappa \times X$ of an uncountable regular cardinal $\kappa$ with a paracompact semi-stratifiable space is normal if{f} it is countably paracompact. We also give a sufficient condition under which the product of a normal space with a paracompact space is normal, from which many theorems involving such a product with a countably compact factor can be derived. Categories:54B19, 54D15, 54D20


text:
Judging from this the map_update_interval is set quite high. This may lead to slower map updates.
Does your odometry give you reasonable movement data? In general GMapping updates the map after some minimal translational or angular movement of the robot. This is controlled by the linearUpdate and angularUpdate paramters. (See: GMapping parameters) If no movement is detected (by the odom -> base_link transform), no new scans are integrated into the map.


text:
Search
Research Article

# High Prevalence of Schistosoma japonicum Infection in Carabao from Samar Province, the Philippines: Implications for Transmission and Control

• Catherine.Gordon@qimr.edu.au

Affiliations: Molecular Parasitology Laboratory, Infectious Diseases Division, Queensland Institute of Medical Research, Brisbane, Australia, Infectious Disease Epidemiology Unit, School of Population Health, University of Queensland, Brisbane, Australia

X
• Affiliation: Department of Immunology, Research Institute of Tropical Medicine, Manila, Philippines

X
• Affiliations: Molecular Parasitology Laboratory, Infectious Diseases Division, Queensland Institute of Medical Research, Brisbane, Australia, Infectious Disease Epidemiology Unit, School of Population Health, University of Queensland, Brisbane, Australia, Griffith Health Institute, Griffith University, Queensland, Australia

X
• Affiliation: Department of Immunology, Research Institute of Tropical Medicine, Manila, Philippines

X
• Affiliation: Department of Immunology, Research Institute of Tropical Medicine, Manila, Philippines

X
• Affiliation: Molecular Parasitology Laboratory, Infectious Diseases Division, Queensland Institute of Medical Research, Brisbane, Australia

X
• Affiliation: Griffith Health Institute, Griffith University, Queensland, Australia

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• Affiliation: Molecular Parasitology Laboratory, Infectious Diseases Division, Queensland Institute of Medical Research, Brisbane, Australia

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• Published: September 20, 2012
• DOI: 10.1371/journal.pntd.0001778

## Abstract

Schistosoma japonicum is endemic in the Philippines, China and Indonesia, and infects more than 40 mammalian host species, all of which can act as reservoirs of infection. In China, water buffaloes have been shown to be major reservoirs of human infection. However, in the Philippines, carabao have not been considered important reservoir hosts for S. japonicum due to the low prevalence and infection intensities reported, the only exception being a qPCR-based study indicating 51% of carabao were S. japonicum-positive. However, the low prevalence found for the same animals when using conventional copro-parasitological techniques means that there is still confusion about the role of carabao in the transmission of schistosomiasis japonicum. To address this inconsistency, and to shed light on the potential role of carabao in the transmission of S. japonicum in the Philippines, we undertook a pilot survey, collecting fecal samples from animals in Western Samar Province and we used a combination of molecular and copro-parasitological techniques to determine the prevalence and intensity of S. japonicum. We found a high prevalence of S. japonicum in the carabao using a validated real-time PCR (qPCR) and a copro-parasitological tool, the formalin-ethyl acetate sedimentation (FEA-SD) technique. A much lower prevalence of S. japonicum was recorded for the same fecal samples using conventional PCR, the Kato-Katz technique and miracidial hatching. These results suggest that, due to their low diagnostic sensitivity, traditional copro-parasitological techniques underestimate infection in carabao. The use of FEA-SD and qPCR provides a more accurate diagnosis. Based on these findings, the role of bovines in the transmission of S. japonicum appears to be more important in the Philippines than previously recognized, and this may have significant implications for the future control of schistosomiasis there, particularly as, in contrast with previous surveys, we found an unprecedented high prevalence of S. japonicum in humans.

## Author Summary

Schistosomiasis japonica is endemic in China, the Philippines and parts of Indonesia. Extensive research efforts have shown that water buffaloes are major reservoir hosts for transmission of Schistosoma japonicum in China. In contrast, comparatively fewer studies have been undertaken on the reservoir hosts for schistosomiasis japonica in the Philippines, although several reports have indicated that water buffaloes (carabao) are unimportant in transmission. Here, we undertook a pilot survey, using a new diagnostic tool (formalin-ethyl acetate sedimentation; FEASD), combined with real-time PCR (qPCR) for the detection of S. japonicum eggs in fecal samples, and report high prevalence of S. japonicum in carabao from Samar province. This study suggests that water buffaloes in the Philippines, as in China, are major reservoir hosts for the transmission of schistosomiasis. Future integrated control, including interventions targeting bovines, would likely reduce the prevalence of infection in humans.

### Introduction

Schistosoma japonicum is the causative agent of schistosomiasis in China, the Philippines and Indonesia. In the Philippines, 10 of 16 regions (administrative divisions) have reported cases of clinical schistosomiasis, with 6.7 million people living in the endemic areas [1][3]. Furthermore, within these endemic areas, 1.8 million people are considered to be directly exposed to potential infection through their daily lifestyle and contact with water including farming, fishing, domestic activities (bathing and washing) and recreation [1], [2]. Currently, there are 560,000 known cases of schistosomiasis in the Philippines [4].

Unlike African schistosomiasis, schistosomiasis japonica is a zoonosis, naturally infecting more than 40 mammalian host species [5][7]. Large mammals, such as bovines, excrete substantial amounts of feces and, therefore, have a high contamination index. Daily fecal excretion rates have been estimated to be as high as 25 kg for cattle and ≥50–60 kg for water buffaloes (Bubalis bubalis) [8]. In China, a number of drug intervention trials have demonstrated the importance of bovines in the transmission of S. japonicum to humans [8][12]. As a result, water buffaloes in China are well recognized as the major reservoir hosts for schistosomiasis in the lake and marshland areas, contributing up to 75% of subsequent transmission to humans [8][12].

In contrast to China, limited research has been undertaken in the Philippines to determine whether water buffaloes (carabao; Bubalus bubalis carabanesis) act similarly as major reservoir hosts. The majority of information comes from studies undertaken more than 10 years ago, which suggested instead a central role for rodents in S. japonicum transmission [13][16]. Additionally, dogs and cats have also been implicated as potentially important hosts, while, based on low prevalence data, carabao have been considered as less important [5], [7], [17], [18]. For example, on Samar, the easternmost island in the Visayas Group, previous studies using copro-parasitological techniques have demonstrated very low prevalence (3.8%) of S. japonicum infection in carabao using the Kato-Katz method (KK) [18]. In direct contrast, a more recent study has shown that 51.5% of carabao on the nearby island of Leyte were test-positive for S. japonicum using real-time PCR (qPCR), although few or no eggs/hatched miracidia were found using the KK, the Miracidial Hatching Technique (MHT) and the Danish Laboratory Bilharziasis (DBL) technique [19]. Despite these inconsistencies, the investigators of this study suggested that the actual prevalence of S. japonicum infection in carabao may be much higher than previously considered, and that the true role and impact that these carabao have on schistosome transmission in the Philippines should be re-evaluated [19].

Accordingly, in this pilot study, we have used a combination of qPCR, conventional PCR (PCR), modified KK, MHT and a novel copro-parasitological tool, the formalin-ethyl acetate sedimentation (FEA-SD) technique, recently developed in China for improving the detection of S. japonicum eggs in the feces from bovines and other ruminant hosts (Gordon et al., 2012, accompanying paper). We report a high prevalence of S. japonicum in carabao and humans on the island of Samar, based on both the presence of fecal eggs and the quantification of S. japonicum DNA in fecal samples. We discuss the implications of these results in terms of the transmission of schistosomiasis japonica and its future control in the Philippines.

### Materials and Methods

#### Ethics Statement

Informed written consent was received from all human participants in the study and ethical approval for human work given from Ethics Committee of the Research Institute of Tropical Medicine (RITM), Manila, and the Queensland Institute of Medical Research (QIMR) Human and Animal Research Ethics Committees (Project Number P288). The conducts and procedures involving animal experiments were approved by the Animals Ethics Committee of the Queensland Institute of Medical Research (Project Number P288) and the Ethics Committee of the Research Institute of Tropical Medicine (RITM), Manila. Informed written consent was received from all animal owners to collect samples. This study was performed in accordance with the recommendations of the Australian code of practice for the care and use of animals for scientific purposes, 2004.

#### Study design

We carried out a cross-sectional survey in Samar Province, the Philippines, to determine the S. japonicum infection status in carabao and humans using a range of diagnostic techniques. This study took place in September 2010. Primary endpoints were carabao and human prevalence and intensity of infection; secondary end points were sensitivity and specificity of the diagnostic techniques employed.

#### Study area

The study was undertaken in three barangays (villages); Cantaguic, Diaz and Hinugacuan and one dairy farm, near Hinugacuan, all in Samar Province. The barangays had previously been part of an earlier survey in 2003 [5], [17], [18], [20], after which praziquantel-treatment of infected humans was carried out. In the intervening years, clinical mass treatment was also performed in these barangays, although details of the treatment coverage were not recorded. As far as we are aware, there has been no treatment of bovines or other animals in these locations. Bovines from the four locations were sampled for fecal examination. Stool samples were collected from 44 bovines. We also collected 52 human fecal samples from Cantaguic barangay.

#### Study procedures

Carabao fecal samples were collected either rectally by trained veterinary personnel or from freshly deposited stools collected from individual animals by the carabao owners. Two separate stool samples were sought from all carabao on different days. For human samples, individuals were given stool cups which were collected over a period of four days. Two stool samples on separate days were sought from all participants.

#### Copro-parasitology techniques

All carabao fecal samples were subjected to analysis using a modified KK, the formalin-ethyl acetate sedimentation method (FEA-SD), and PCR and qPCR techniques. Additionally, the MHT was performed on 21 stool samples from 19 individual carabao. All human samples were subjected to testing using KK, PCR and qPCR arrays.

##### Modified Kato-Katz (KK).

A modified KK was performed on all collected bovine stool samples, according to a published protocol [21]. Briefly, the collected stool sample was pressed through gauze and the non-retained material used to fill a standard volume template representing approximately 50 mg of fecal material. This material was then transferred to a glass slide where cellophane soaked in glycerin was pressed against a firm surface to spread the stool evenly which was then viewed under a microscope. Two slides per stool sample were read blind and independently by four microscopists.

##### Miracidial hatching test (MHT).

The MHT was performed as previously described [22] with minor modifications. Briefly, ~50 g of bovine stool were homogenized and sieved. The sediment was collected, placed into a triangular flask filled with fresh, unchlorinated water (pH 7.8) and left in a well-lit room at 24–30°C. The neck of the flask was illuminated with a lamp and examined after 4, 8, 12 and 24 hours to detect the presence of hatched miracidia.

##### Formalin-ethyl acetate sedimentation technique (FEA-SD).

The recently developed FEA-SD method (Gordon et al., accompanying paper) was used with some minor modifications. Briefly, 50 g of bovine stool were sieved sequentially through a 40–60 mesh (Tyler scale) (234–380 µm (pore opening size)) sieve and then a 260 mesh (61 µm) sieve. The sediment was washed into a 50 ml falcon tube and mixed by shaking and sedimented in the tube by leaving for 30 min. The supernatant was removed and the sediment re-suspended in 10% (v/v) formalin and natural sedimentation repeated twice more. After the final sedimentation step, the 50 ml tube was topped up to 50 ml and vortexed. Ten ml of this final suspension was transferred to two 15 ml tube (5 ml suspension per tube), centrifuged for 10 min at 1800× g, the supernatant removed, and the pellet re-suspended in 7 ml of 10% (v/v) formalin and 2 ml of 100% ethyl acetate (v/v), and then centrifuged at 500× g for 10 min. This resulted in a four layer sedimentation, of which the top three layers were removed. Water was added to each tube to adjust the volume up to a minimum of 5 ml and an equal volume of 10% (w/v) potassium hydroxide solution was added. After vortexing, the tubes were left overnight at 37°C. The next day, the suspension was centrifuged at 1800× g for 10 min, the resultant pellet was washed once with water and then re-suspended in 6 ml of 10% (v/v) formalin solution. The solution was examined under a microscope following the pipetting of 200 µl of the suspension on to a glass slide. Two ×200 µl aliquots were examined, and eggs counted by each of four independent microscopists (1.6 ml suspension in total).

#### Molecular analysis

##### DNA extraction from fecal samples.

Genomic DNA was isolated from bovine and human fecal samples using QIAamp mini stool mini kits (QIAGEN, Hilden, Germany). DNA concentration was determined using a NanoDrop 2000 (Thermo Scientific) and all samples were diluted to 50 ng/µl for analysis.

##### Conventional polymerase chain reaction (PCR).

The primers used for the PCR were employed in a previously reported assay [23], which amplifies fragments of the cytochrome oxidase 2 (cox2) and NADH dehydrogenase 6 (nad6) mitochondrial genes, and are specific for S. japonicum. A reaction mix of 18 µl was prepared containing 11.25 µl H2O, 2 µl kit provided buffer, 1.0 mM MgCl2, 800 µM deoxyribonucleotide triphosphate (dNTPs), 1.25 units/reaction Gold AmpliTaq (Applied Biosystems, Foster City, CA) and 5 µM of each primer. To this mix, 2 µl isolated template DNA (50 ng/µl) were added in a total reaction volume of 20 µl. The PCR was performed on a thermocycler (Mastercycler, Eppendorf) under previously described conditions [23], and PCR products were resolved by electrophoresis on 2% (w/v) agarose gels and stained with ethidium bromide.

##### Real-time PCR (qPCR).

The primers employed were of a previously reported [19], [24] PCR assay which amplifies a fragment of the NADH dehydrogenase I (nad1) mitochondrial gene. Reaction mixtures of 22 µl were prepared containing 12.2 µl SYBR Green (Invitrogen), 5 µM of each primer and 7.8 µl of H2O. To this, 2 µl of DNA (50 ng/µl) template was added for a total reaction volume of 24 µl. The PCR conditions were as follows: 50°C initialization for 2 min, 95°C denaturation for 10 min, followed by 45 cycles of denaturation at 95°C for 15 sec, annealing at 60°C for 60 sec, extension at 72°C for 90 sec and a final dissociation phase at 60–95°C. The PCR was performed using a conventional thermocycler standard (Corbett RotorGene 6000). The results were quantified for in relation to eggs per gram [EPG] (1–120 eggs per gram) using Ct (cycle threshold) scores. Melt curve analysis was performed after each qPCR. To determine the equivalent egg number relating to Ct scores, a series of seeding and dilution experiments were performed. For seeding experiments, stool samples known to be negative for S. japonicum were seeded with a known number of eggs and DNA extracted from these seeded samples. A standard curve using the dilutions and the results of the seeding experiments was then created to determine a range of Ct scores which corresponded to a known number of eggs. This curve was then used to compare the unknown samples from the study and to determine egg numbers.

Amplicons for both cPCR and qPCR were sequenced, thereby confirming each DNA target sequence.

#### Data management and data analyses

Microsoft Excel and SAS software (SAS Institute, Cary, NC) were used for data analyses. Diagnostic sensitivity and specificity were calculated using the following formulae:

The bovine contamination index (BCI) was derived using the published formula [25]:
A conservative approach was taken when deriving the BCI by using 25 kg of feces excreted per day in our calculations; thus the value 25, 000 was used. It has been shown in China that bovines can excrete 25–50 kg of bovine feces per day.

### Results

#### Molecular analysis (qPCR)

The seeding experiments showed that the qPCR was sufficiently sensitive to detect a single S. japonicum egg. Dilution experiments showed that the qPCR could detect as little as 14 pg (approximately 30% of the DNA in one egg). This value was obtained by extracting DNA from a known number of eggs and calculating for the content of one egg.

#### Prevalence

The prevalence of S. japonicum infection in carabao was high (95.45%; 95% CI 89.05–100%) when the results achieved using all diagnostic techniques used were combined (Table 1). The prevalence of S. japonicum infection obtained from the FEA-SD and qPCR techniques were 93.18% (95% CI 85.43–100%) and 90.91% (95% CI 82.07–99.75%), respectively, and proved to be the most sensitive for detecting S. japonicum in carabao, whereas the prevalence estimated based on the use of KK, MHT and conventional PCR were 25% (95% CI 11.68–38.32%), 19.05% (95% CI 0.86–41.24%) and 31.82% (95% CI 17.49–46.14), respectively. MHT was performed on 21 samples from 19 carabao (Table 1). Prevalence was also calculated by barangay using FEA-SD and qPCR data (Table 2).

The prevalence of S. japonicum in humans varied depending on the technique used, with the highest prevalence, 92.31% (95% CI 84.82–99.8%), estimated using qPCR and the lowest (30.77%, (95% CI 17.79–43.74%)) using KK (Table 3). A prevalence of 75% (95% CI 62.83–87.17%) was detected by conventional PCR (Table 3). Human samples test-positive by qPCR, but negative by KK, were examined again by microscopy, after sieving, sedimenting and KOH digestion of each sample. S. japonicum eggs were found in all of the samples examined by one of the authors (CAG).

#### Intensity

The intensity of S. japonicum infection in carabao (geometric mean eggs per gram (GMEPG) feces) was determined using the modified KK, FEA-SD and qPCR methods (Table 1). The GMEPG obtained by qPCR was calculated from the average and standard deviation of Ct scores compared with a standard curve prepared using a DNA dilution series. Analysis of the FEA-SD results gave lower overall GMEPG than the modified KK and qPCR for carabao (Table 1).

The intensity of S. japonicum infection by age, sex and individual barangay and the dairy farm for each carabao was also calculated using the FEA-SD and qPCR data (Table 2). The intensity of infection by FEA-SD was highest in carabao from the dairy farm (1.82), with the lowest values in Diaz (0.63) and Hinugacuan (0.55) (Table 2). Infection intensity did not vary with the age or sex of the carabao (data not shown).

For the samples from humans, a GMEPG of 16.54 (95% CI 10.06–27.18) was found with the qPCR and 4.90 (95% CI 2.48–9.67) with the KK technique; no differences in intensity of infection according to age or sex were recorded.

#### Sensitivity and specificity of the copro-parasitological and molecular techniques

Sensitivity and specificity were calculated for each of the techniques used, with a combination of any positive results from all techniques used as the reference value. For carabao, qPCR and FEA-SD provided the highest sensitivities of 95.25% (95% CI 88.82–100) and 97.62% (95% CI 93.01–100), respectively, whereas the conventional PCR (33.33%; 95% CI 19.07–47.59), KK (26.19%; 95% CI 12.89–39.49) and MHT (21.05%; 95% CI 2.72–39.38; based on the test results for 19 fecal samples) had lower sensitivity. All techniques had a specificity of 100%.

With the samples from humans, qPCR was the most sensitive (94.0%; 95% CI 87.42–100), followed by PCR (78.0%; 95% CI 66.52–89.48) and KK (32.0%; 95% CI 19.07–44.93).

#### Bovine contamination index (BCI)

We used 25 kg as a conservative estimate for the total daily fecal output for carabao, compared with 25–50 kg reported in the literature for Chinese water buffaloes [8]. Our calculations indicated each carabao excreted, on average, 41,750 S. japonicum eggs per day into the environment (Table 4).

### Discussion

In this pilot study, we show that the prevalence of S. japonicum infection in carabao in Samar is substantially higher than reported previously in this and other regions of the Philippines. A recent study on the island of Leyte reported a prevalence of 52% in carabao by qPCR [19], whereas the results obtained by the same group using the MHT, KK and DBL techniques on carabao from Samar detected less than 4% prevalence [19]. A previous study from Leyte using the merthiolate iodine formalin concentration technique (MIFC) and circumoval precipitin test (COPT) found a combined prevalence of 9.8% in carabao [26], whereas another survey of 50 Samar barangays [5] revealed very low prevalence (2.1%) in carabao using the DBL (Danish Bilharzias Laboratory) technique. Surprisingly, previous studies of the same three barangays that we surveyed in Samar indicated that all carabao present were test-negative for S. japonicum infection [5], [17], [18], [20]. This contrasts to our findings showing that the majority of carabao in Cantaguic, Diaz, Hinugacuan, and the dairy farm (near Hinugacuan) were infected with S. japonicum, including 100% of carabao at Diaz and the dairy farm, resulting in an overall prevalence of 95.45%. This high prevalence was obtained due to the diagnostic sensitivity of the FEA-SD and qPCR techniques employed, with the traditional KK technique being far less sensitive.

The sensitivity of each of the diagnostic methods used in this study varied substantially, with the KK technique and MHT proving to be the least sensitive. It is recognized that the sensitivity of the KK increases with the number of stool samples taken and the number of slides examined per sample [27], [28]. In the present study, two stool samples and two slides per sample were assessed, such that it is likely that increasing the number of slides examined would have improved the sensitivity of the method. Additionally, we were unable to effectively control the temperature or water pH for the MHT under the prevailing field conditions in Samar, both of which are critical for optimal performance of the test.

The use of qPCR is likely to result in a less accurate egg count than physical counting of eggs, as only approximations can be made with the technique after using a standard curve of known DNA concentrations (which correspond to a known number of eggs). Additionally for the qPCR method, only ~200 µg of stool were used for the analysis, which is much lower than that for the FEA-SD which uses a final volume equivalent to 10 g of stool. The modified KK showed a higher intensity of infection than the FEA-SD due to the lack of sensitivity of the KK. The KK usually detects only high intensity infections which may represent only a small proportion of the total animals infected [28][32]. The GMEPG is calculated using the positive samples only, thus giving a skewed GMEPG result for the KK. Because the FEA-SD is more sensitive than the KK, it detects the high and low intensity infections, giving a more accurate assessment of estimation of current infection intensity.

The bovine contamination index (BCI) indicated that, on average, each carabao released more than 40,000 S. japonicum eggs into the environment daily, noting that the BCI was calculated conservatively using the lower end of the published daily fecal output rate for water buffaloes [8]. In comparison, Chinese water buffaloes have been shown to excrete ~120,588 eggs per day (28.7 million S. japonicum eggs across 238 infected bovines) [9], [10]. In the Philippines, the habitats for the Oncomelania snails comprise streams, springs, canals, swampy areas, grass lands and rice fields, where carabao, other bovines and farmers work [17]. In the barangays, we observed that carabao were tethered with long ropes and were allowed to graze in communal areas, in or near rice paddy fields, and often grazed in or near water sources, particularly rivers. During the wet season (June–October for Manila and Northern Philippines; December–March for eastern Visayas and Mindanao) in the Philippines, bovines are either allowed to continue grazing in the rice fields or they are fed cut grass [33].

Based on the results obtained in this survey, we would recommend using either qPCR or FEA-SD in future surveys of large animals, such as bovines, for S. japonicum, as these techniques have a much higher sensitivity than the MHT or the KK and should thus provide much better estimates of prevalence. An additional benefit of the FEA-SD is that it can be performed in the field. While the procedure requires the use of a centrifuge, this equipment is available in many regional laboratories in the Philippines, whereas thermocyclers for PCR are unlikely to be readily available. The FEA-SD technique is also considerably less expensive per sample at 68 cents per test, compared with $8.8 ($US) for qPCR when calculated based on the costs of the reagents and consumable materials required.

The overall prevalence of S. japonicum infection in carabao was high, suggesting that these animals play a greater role in transmission and human infection in the Philippines than previously appreciated. As shown in China, the treatment of bovines with praziquantel [9], [10], or their removal and replacement with motorized tractors [34], as part of a comprehensive control strategy based on interventions to reduce the rate of transmission of S. japonicum infection to snails, can be highly effective, resulting in a reduction in transmission and human infection.

The detection of eggs in the stool by microscopy is still the standard for diagnosis. Despite the sensitivity of molecular and immunological techniques, eggs still need to be visualized for validation of these techniques. In the Leyte study [19], which estimated 51% prevalence of S. japonicum in bovines, the qPCR was not validated against egg counts. Indeed, eggs were found in <4% of samples using microscopic-based methods. In contrast, we employed the qPCR and the newly developed FEA-SD tool for more precise quantification of S. japonicum eggs in carabao feces, and by extensive microscopic examination of human fecal samples by two independent microscopists for the presence of eggs.

Recent studies in the Philippines have reported a low overall prevalence of schistosomiasis in previously classified endemic provinces [35], [36]. However, these surveys included non-endemic barangays, bringing down the overall prevalence for a province. Schistosomiasis is a focal disease and geographically close barangays can have varying infection status [37]. In this study, we found an unprecedentedly high prevalence of S. japonicum infection in humans, as well as in bovines This finding may be due to the insensitivity of the diagnostic tests used in previous studies and the low infection intensities, particularly in humans [5], [26], [35], [36], [38]. Despite the high prevalence recorded in humans, their precise involvement in the transmission of S. japonicum in the Philippines remains to be fully established. Additional surveys in Samar and other endemic areas of the Philippines should now be undertaken to determine the schistosomiasis-endemic status, in terms of human and animal infection prevalence and intensity. Nevertheless, the high S. japonicum BCI and prevalence data that we report for carabao suggest that, in the Philippines, they act, as in China [11][15], as important reservoir hosts. As such, future integrated control, including interventions targeting bovines, would likely reduce human incidence there.

### Acknowledgments

The authors sincerely acknowledge the assistance and work of members of the Immunology Laboratory at the Research Institute of Medical Research (RITM), Manila, and regional staff from the municipalities of Gandara and San Jose.

### Author Contributions

Conceived and designed the experiments: CAG DJG DPM GNG AGR. Performed the experiments: CAG LPA BJ. Analyzed the data: CAG. Contributed reagents/materials/analysis tools: CAG DJG DPM RMO LPA. Wrote the paper: CAG DPM.

### References

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29. 29. Habtamu K, Degarege A, Ye-Ebiyo Y, Erko B (2011) Comparison of the Kato-Katz and FLOTAC techniques for the diagnosis of soil-transmitted helminth infections. Parasitol Int 60: 398–402. doi: 10.1016/j.parint.2011.06.020
30. 30. Kongs A, Marks G, Verle P, van der Stuyft P (2001) The unreliability of the Kato-Katz technique limits its usefullness for evaluating S. mansoni infections. Trop Med Int Health 6: 163–169. doi: 10.1046/j.1365-3156.2001.00687.x
31. 31. Pontes LA, Oliveira MC, Katz N, Dias-Neto E, Rabello A (2003) Comparison of a polymerase chain reaction and the kato-katz technique for diagnosing infection with Schistosoma mansoni. Am J Trop Med Hyg 68: 652–656.
32. 32. Zhang Y-Y, Luo J-P, Liu Y-M, Wang Q-Z, Chen J-H, et al. (2009) Evaluation of Kato-Katz examination method in three areas with low-level endemicity of schistosomiasis japonica in China: A Bayesian modeling approach. Acta Trop 112: 16–22. doi: 10.1016/j.actatropica.2009.05.020
33. 33. Gray GD, Copland RS, Copeman DB, editors (2008) Overcoming liver fluke as a constraint to ruminant production in South-East Asia. Canberra: Australian Centre for International Agricultural Research.
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35. 35. Leonardo LR, Rivera P, Saniel O, Villacorte E, Crisostomo B, et al. (2008) Prevalence survey of schistosomiasis in Mindanao and the Visayas, the Philippines. Parasitol Int 57: 246–251. doi: 10.1016/j.parint.2008.04.006
36. 36. Leonardo LR, Rivera P, Saniel O, Villacorte E, Labanan MA, et al. (2012) A national baseline prevalence survey of Schistosomiasis in the Philippines using stratified two-step systematic cluster sampling design. Journal of Tropical Medicine In press. doi: 10.1155/2012/936128
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38. 38. Lewert RM, Yogore MG Jr, Blas BL (1979) Schistosomiasis japonicum in Barrio San Antonio, Basey, Samar, the Philippines. Am J Trop Med Hyg 28: 1312–1321. doi: 10.1016/s0140-6736(06)69440-3

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text:
# Merge PDFs, fix page numbering and compress

Say there is a book that has been split into several PDFs. I want to do two things:

1. Merge the PDFs and reduce the size of the resulting PDF.
2. Fix the page numbering.

(1) can be done with pdftk or Ghostscript. For me, Ghostscript creates considerably smaller PDFs and is the only program (of the two) available on my PC.

Example:

gs -dBATCH -dNOPAUSE -sDEVICE=pdfwrite -dCompatibilityLevel=1.4 -dPDFSETTINGS=/screen -q -sOutputFile=output.pdf input_1.pdf input_2.pdf [...] input_n.pdf

Why is (2) a problem at all? Because different books start counting page numbers on different pages whereas your PDF reader starts counting at page 1 if you don't do anything. This can be fixed with an empty document, the package pdfpages, \setcounter, and \pagenumbering.

Example:

\documentclass{article}
\usepackage{pdfpages}
\usepackage{hyperref}
\begin{document}
\pagenumbering{Roman}
\setcounter{page}{4}
\includepdf[pages=-]{input_1_with_TOC.pdf}
\pagenumbering{arabic}
\includepdf[pages=-]{input_2.pdf}
\includepdf[pages=-]{input_3.pdf}
% ...
\end{document}

My problem is now: I want to fix the page numbering and create a small PDF but calling pdflatex and then Ghostscript creates a PDF that is considerably bigger than a PDF that has been merged with Ghostscript.

How do I create PDFs as small as with (1) while having (2)?

EDIT: Forgot to add hyperref to the LaTeX code.

-
Is TeX really the best tool for this job? There are PDF libraries for many scripted languages that could renumber pages –  Matthew Leingang May 1 '13 at 21:04
If you are dealing with the pdf outputs only, there is no feasible way to modify the page numbers of the pagination. From the pdf's point of view they are just digit (or character!) sequences that happen to be positioned at some peculiar place of the page. –  Hagen von Eitzen May 1 '13 at 21:04
@HagenvonEitzen: There are the page numbers in the text ("on the paper") and there are the page numbers that are shown by your PDF reader. I want to change the latter. –  Throwaway 1352 May 1 '13 at 22:32
GhostScript likely reduces the size of the PDFs because it reduces the resolution of included bitmap images. Do you really want that? –  Martin Schröder May 3 '13 at 11:20
@Martin Schröder: Yes, I want to reduce the resolution. Otherwise I would have used Ghostscript with -dPDFSETTINGS=/preprint or only pdfpages. –  Throwaway 1352 May 3 '13 at 14:25

1. Merge with Ghostscript.
2. Use pdfpages on the Ghostscript output like this:
\documentclass{article}
\usepackage{pdfpages}
\usepackage{hyperref}

\begin{document}
\pagenumbering{Roman}
\setcounter{page}{4}


text:
# Math Help - Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

1. ## Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

test y=ax^n and y=Ae^kx
1). i need to investigate a power model in the form y=ax^n by plotting my raw data and log x vs log y
2). i need to investigate a exponential modle in the form y=Ae^kx by plotting the raw data and x vs log y

i now need to use graphs and correlation coefficients to determine the equation of the model that fits your Data.?????

how do i find the equation... my text book only gose briefly and i missed the class due sports, i asked my teachers and friends but they only confused me more.
thank you for the fast help

2. ## Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

What is your "raw data"? Presummably, each data point is (x, y). Find the two logarithms, and plot point (log(x), log(y)). Do those points seem to be close to a staright line?
The point is that if y= ax^n then lpg(y)= log(ax^n)= nlog(x)+ log(a) where log(y) is a linear function of log(x).

Similarly, if y= ae^(kx) the log(y)= kx+ log(a) where log(y) is a linear function of x.

3. ## Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

Hi thanks for being fast,
yes the raw data is in x and y x being time and y the difference in temperature as apart of a table we were given.
what i don't under stand is how do i find A and n in the fist and A and k in the next??

4. ## Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

and every time i graph with my graphics cal it doesn't pass the x axis

5. ## Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

Is there any reason why "not passing the x-axis" is a problem? If you are referring to finding the x-intercept, you can extend the line until it does cross.

As I said before, the whole point is to try to get, in the first case, log(y)= A log(x)+ B, and, in the second case, log(y)= Ax+ B.

"A", in each case, is the slope of the line: choose two points, (log(x1), log(y1) and (log(x2), log(y2)), to find the slope $\frac{log(y2)- log(y1)}{log(x2)- log(x1)}$ and (x1, log(y1)), (x2, log(y2)) to find the slope $\frac{log(y2)- log(y1)}{x2- x1}$. You can find "B" by using that A with one point: B= log(y)- Alog(x) or B= log(y)- Ax.

6. ## Re: Cooling the silly coffee !!!!! emmidiate help Due in a few Hours

Thank you for that


text:
## Tuesday, March 11, 2014

### Blogging My Genome, episode 3: data wrangling

After learning about Illumina's Understand Your Genome (UYG) program at ASHG 2013, I decided to sign up to get my genome sequenced. This is the third in a series of blog posts I'm writing about my own adventure in very personal genomics.

Along with my clinical report, Illumina delivered a portable hard drive (pictured) containing the underlying next-generation sequencing data. Given my computational genomics background, this was the product I'd most been looking forward to receiving. In fact, after picking it up, I found myself hastily biking five miles through the rain to get it to a computer!

The drive is protected using TrueCrypt, with the encryption key e-mailed separately. I installed the TrueCrypt on my Linux workstation, mounted the drive, and started to look around:

PG0001312-BLD is the identifier Illumina's lab assigned to my case. In the root folder we have copies of my PDF clinical report and a file with the 5,390 variants in my genome that were interpreted by Illumina's CLIA lab, in TSV format. (It's an irritating truth that the field of bioinformatics is largely based on tab-delimited file formats - but that's another blog post.) As mentioned last time, this only includes single-nucleotide variants in the exons of 1,600 genes with currently-known medical significance.

In the Variations folder we have files containing all of the variants detected across my genome in VCF format (also tab-delimited). This is the main product of Illumina's bioinformatics pipeline, Isaac, which takes the jumble of reads from the sequencing instrument through to variant calls, preceding clinical interpretation.

The Assembly folder contains the mother lode: a 66 GiB BAM file containing all of the reads, associated quality information, and their estimated positions of origin, or mappings, with respect to the human reference genome assembly. While this file is really big, it's actually just half the size I'd been expecting, and I'll get to why in a future post.

The BAM file is an intermediate product of the Isaac bioinformatics pipeline, from which the input data is still recoverable. It'll be my goal to get that raw data out and redo much of the bioinformatics from scratch. Not because there's anything wrong with Isaac, but rather because that'll be the best way for me - and, I hope, readers of this blog - to learn about the process. Analyzing this raw data would typically be a daunting challenge, because of the technical expertise required, the sheer quantity of the data, and the computational horsepower needed to work with it in an exploratory fashion. Fortunately however, I happen to be a bridge officer on the world's most powerful genome informatics platform.

### To the cloud!

I created a DNAnexus project called "My Genome", and then used our command-line client to begin uploading the entire contents of the portable hard drive.

I set this running just before going to dinner, and it'd completed by the time I woke up the next morning. As a result, all my genome data now lives safely in the cloud.

### FASTQ regeneration

To begin the bioinformatics, I'd like to get all my sequencing reads into FASTQ format, which contains the sequence and quality information, but not the mappings also encoded in the BAM file. FASTQ isn't truly the raw data that comes straight from the sequencing instrument, but the steps leading to it are relatively uncontroversial. Picard has a utility to extract FASTQ data from the BAM file, which I wrapped to run on DNAnexus. I can upload this to my project and execute it like so:

~/src/blogging-my-genome$dx build -f bam_to_fastq ~/src/blogging-my-genome$ dx run bam_to_fastq -i bam=:/Assembly/PG0001312-BLD.bam

Producing:

These eight FASTQ files break down into four pairs. Each group is a pair owing to the use of paired-end sequencing, a neat trick to squeeze more information from a given quantity of reads. There are four such pairs corresponding to four lanes on two HiSeq flowcells  used to sequence my sample. Here are the first eight lines of C2KC2ACXX_1_6_none_1.fastq:

~\$ dx cat C2KC2ACXX_1_6_none_1.fastq.gz | gunzip -c | head -n 8 | nl
1  @C2KC2ACXX_1:6:1101:4:0/1
2  AGTTNCCATATTAATTTTAGGATGAATTTTATTATGTCTGCAACAAAAACAACAACAAAATGTGGCCAGGCATGGTGGCTCATGCCTGTAATCCCAGCAC
3  +
4  BBBF!0BFFFFFFIIFIIIIIIIIIIIIIIFIIFIFFFIIIFIIIIFIFFFFFI<FFBF7FFBBFIIIIIFFFFFFBFFBF<BBBFBBB<<<BBB<<B'<
5  @C2KC2ACXX_1:6:1101:6:0/1
6  TTTCNAGCTTCCAGTATGTTGTCGCCNCCNNTCCTTTCTNGCNNNNNNNAGGATTTGTATTTTNNNNNNNNNNNNNNNNNNNNNNNNNNNNANNNNNNNN
7  +
8  BBBF!0<BB0FBBFIIIFFIFFFBBF!0<!!00<B'<BB!00!!!!!!!0000'70<<'7<BB!!!!!!!!!!!!!!!!!!!!!!!!!!!!'!!!!!!!!

Line 1 is a read identifier, line 2 is the 100 base pair (bp) DNA sequence of the read, line 3 marks the end of the sequence, and line 4 presents the Phred quality scores for each position in the read. Lines 5-8 repeat this for a second read, and so on for the remainder of this 35 GiB (uncompressed) FASTQ file. You can see that the second read wasn't very good, with a large proportion of no-call positions (N) and the lowest quality scores (!).

Each read in C2KC2ACXX_1_6_none_1.fastq has a corresponding mate pair in C2KC2ACXX_1_6_none_2.fastq, and so on for the three remaining groups. The eight FASTQ files contain a grand total of 1,306,309,136 reads, or 131 Gbp - around 40 times the (haploid) human reference genome size of 3.2 Gbp. Such high coverage is typical for clinical-grade genome sequencing.

Next time: I'll generate my own mappings of these reads back onto the reference genome.

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## Category Archives: Uncategorized

1

### Unit normal to a radial graph over sphere

Consider $\Omega\subset \mathbb{S}^n$ is a domain in the sphere. $S$ is a radial graph over $\Omega$.

$\boldsymbol{F}(x)=\{v(x)x:x\in \Omega\}$

What is the unit normal to this radial graph?

Suppose $\{e_1,\cdots,e_n\}$ is a smooth local frame on $\Omega$. Let $\nabla$ be the covariant derivative on $\mathbb{S}^n$. Tangent space of $S$ consists of $\{\nabla_{e_i}\boldsymbol{F}\}_{i=1}^n$ which are

$\nabla_{e_i}\boldsymbol{F}=v(x)e_i+e_i(v)\cdot x$

In order to get the unit normal, we need some simplification. Let us assume $\{e_i\}$ are orthonormal basis of the tangent space of $\Omega$ and $\nabla v=e_1(v)e_1$. Then

$\nabla_{e_1}\boldsymbol{F}=v(x)e_1+e_1(v)x, \quad \nabla_{e_i}\boldsymbol{F}=v(x)e_i, \quad i\geq 2$

Then we obtain an orthonormal basis of the tangent of $S$

$\{\frac{1}{\sqrt{v^2+|\nabla v|^2}}\nabla_{e_1}\boldsymbol{F},\nabla_{e_2}\boldsymbol{F},\cdots,\nabla_{e_n}\boldsymbol{F}\}$

We are able to get the normal by projecting $x$ to this subspace

$\nu=x-\frac{1}{v^2+|\nabla v|^2}\langle x,\nabla_{e_1}\boldsymbol{F}\rangle\nabla_{e_1}\boldsymbol{F}=\frac{v^2x-v\nabla v}{v^2+|\nabla v|^2}.$

After normalization, the (outer)unit normal can be written

$\frac{vx-\nabla v}{\sqrt{v^2+|\nabla v|^2}}$

Remark: Guan, Bo and  Spruck, Joel. Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.

### Principle curvature of translator

Suppose ${\Sigma\subset\mathbb{R}^{3}}$ is a translator in ${e_{3}}$ direction. Denote ${V=e_{3}^T}$ and ${A}$ is the second fundamental form of ${\Sigma}$. Then it is well know that

$\displaystyle \Delta A+\nabla_VA+|A|^2A=0 \ \ \ \ \ (1)$

Choose a local orthonormal frame ${\{\tau_1,\tau_2\}}$ such that ${A(\tau_1,\tau_1)=\lambda}$, ${A(\tau_2,\tau_2)=\mu}$ and ${A(\tau_1,\tau_2)=0}$ in the neighborhood of some point ${p}$. We want to change (1) to some expression on ${\lambda}$ or ${\mu}$. To do that, we need to apply both sides of (1) to ${\tau_1,\tau_1}$, getting

$\displaystyle (\nabla_VA)(\tau_1,\tau_1)=\nabla_V \lambda-2A(\nabla_V \tau_1,\tau_1)=\nabla_V\lambda \ \ \ \ \ (2)$

where we have used ${\nabla_V\tau_1\perp \tau_1}$. Similarly

$\displaystyle (\nabla_{\tau_k} A)(\tau_1,\tau_1)=\nabla_{\tau_k}\lambda$

$\displaystyle (\nabla_{\tau_k}A)(\tau_1,\tau_2)=(\lambda-\mu)\langle \nabla_{\tau_k}\tau_1,\tau_2\rangle$

Now let us calculate the Laplacian of second fundamental form

$\displaystyle (\nabla_{\tau_k}^2A)(\tau_k,\tau_1)=\tau_k(\nabla_{\tau_k}\lambda)-2(\nabla_{\tau_k}A)(\nabla_{\tau_k}\tau_1,\tau_1)-(\nabla_{\tau_k}\tau_k)\lambda$

$\displaystyle =\tau_1(\nabla_{\tau_1}\lambda)-2\frac{(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}{\lambda-\mu}.$

Then

$\displaystyle (\Delta A)(\tau_1,\tau_1)=(\nabla_{\tau_k}^2A)(\tau_k,\tau_1)= \Delta \lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}.$

Combining all the above estimates to (1), we get

$\displaystyle \Delta \lambda+\nabla_V\lambda+|A|^2\lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}=0.$

where we write ${A_{12,k}=(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}$. Using this equation and the other one on ${\mu}$, one can derive that if ${\Sigma}$ is mean convex then it is actually convex.

### Laplacian on graph

Suppose ${\Omega\subset \mathbb{R}^n}$ is a domain and ${\Sigma=graph(F)\subset \mathbb{R}^{n+1}}$ is a hypersurface, where ${F=F(u_1,\cdots,u_n)}$ is a function on ${\Omega}$. Define ${f=f(u_1,\cdots,u_n)}$ on ${\Omega}$. Then ${f}$ also can be considered as a function on ${\Sigma}$. How do we understand ${\Delta_\Sigma f}$?

Denote ${\partial_i=\frac{\partial}{\partial{u_i}}}$ for short. If we pull the metric of ${\mathbb{R}^{n+1}}$ back to ${\Omega}$, denote as ${g}$, then

$\displaystyle g_{ij}=g(\partial_i,\partial_j)=\delta_{ij}+F_{u_i}F_{u_j},\quad g^{ij}=\delta_{ij}-\frac{F_{u_i}F_{u_j}}{W^2},\quad \det g=W^2$

where ${W=\sqrt{1+|\nabla F|^2}}$ and ${F_{u_i}=\frac{\partial F}{\partial u_i}}$. Then one can use the local coordinate to calculate ${\Delta_\Sigma f}$

$\displaystyle \Delta_\Sigma f=\frac{1}{\sqrt{\det g}}\partial_{u_i}\left(\sqrt{\det g}\,g^{ij}f_{u_j}\right)$

$\displaystyle =g^{ij}f_{u_iu_j}+f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}$

also one can see from another definition of Laplacian

$\displaystyle \Delta_\Sigma f=g^{ij}Hess(f)(\partial_i,\partial_j)=g^{ij}[\partial_j(\partial_i(f))-(\nabla_{\partial_j}\partial_i)f]$

$\displaystyle =g^{ij}f_{u_iu_j}-g^{ij}(\nabla_{\partial_j}\partial_i)f$

By using the expression of ${g^{ij}}$ stated above, we can calculate

$\displaystyle \frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=\frac{-F_{u_j}F_{u_{i}u_{k}}g^{ik}}{W^2}$

It follows from the definition of tangential derivative on ${\Sigma}$, see, that

$\displaystyle \langle\nabla^\Sigma f, e_{n+1}\rangle=g^{ij}f_{u_{i}}F_{u_{j}}=f_{u_{i}}F_{u_{i}}-\frac{f_{u_i}F_{u_i}|\nabla F|^2}{W^2}=\frac{f_{u_{i}}F_{u_{i}}}{W^2}$

then

$\displaystyle f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=-\langle \nabla^\Sigma f,e_{n+1}\rangle F_{u_iu_k}g^{ik}=-\langle \nabla^\Sigma f,e_{n+1}\rangle HW$

where ${H}$ is the mean curvature of the ${\Sigma}$. Combining all the above calculations,

$\displaystyle \Delta_\Sigma f=g^{ij}f_{u_iu_j}-\langle \nabla^\Sigma f,e_{n+1}\rangle HW$

### One example of blowing up corner

Consider ${u(x,y)=\sqrt{x^2+y^4}}$ on ${\mathbb{R}^2_+=\{(x,y):x\geq 0, y\geq 0\}}$. Notify ${\mathbb{R}^2_+}$ has a corner at the origin and ${u}$ is not smooth at the origin. ${u\approx x}$ when ${x\geq y^2}$ and ${u\approx y^2}$ when ${x\leq y^2}$. We want to resolve ${u}$ by blowing up the origin through a map ${\beta}$. After blowing up, ${W}$ looks like the following picture.

Denote ${W=[\mathbb{R}^2_+,(0,0)]}$ and ${\beta:W\rightarrow \mathbb{R}^2_+}$ is the blow down map. On ${W\backslash lb}$(near A), ${\beta}$ takes the form

$\displaystyle \beta_1(\xi_1,\eta_1)=(\xi_1^2,{\xi_1\eta_1})$

where ${\xi_1}$ is the boundary defining function for ff and ${\eta_1}$ is boundary defining function for rb. Similarly on ${W\backslash rb}$(near B), ${\beta}$ takes the form

$\displaystyle \beta_2(\xi_2,\eta_2)=(\xi_2\eta^2_2,\eta_2)$

where ${\xi_2}$ is a bdf for lb and ${\eta_2}$ is a bdf for ff. One can verify that ${\beta}$ is a diffeomorphism ${\mathring{W}\rightarrow \mathring{\mathbb{R}}^2_+}$. Let ${w=\beta^* u}$. Then ${w}$ is a polyhomogeneous conormal function on ${W}$. Its index can be denoted ${(E,F,H)}$ correspond to lb, ff and rb.

$\displaystyle E=\{(n,0)\}, F=\{(2n,0)\}, H=\{(2n,0)\}$

Suppose ${\pi_1}$ is the projection to ${x}$ coordinate. Consider ${f=\pi_1\circ \beta:W\rightarrow \mathbb{R}_+}$, ${f}$ is actually a ${b-}$fibration. Then the push forward map ${f_*}$ maps ${w}$ to a polyhomogeneous function on ${\mathbb{R}^+}$.

$\displaystyle f_*w=\pi_* u=\int_0^\infty u(x,y)dy$

In order to make ${u}$ is integrable, let us assume ${u}$ support ${x\leq 1}$ and ${y\leq 1}$. What is the index for ${f_* w}$ on ${\mathbb{R}^+}$?

$\displaystyle \int_0^1\sqrt{x^2+y^4}dy=\int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy+\int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy$

For the first integral, letting ${y^2/x=t}$

$\displaystyle \int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy=x\sqrt{x}\int_0^1\sqrt{1+t^4}dt=c_0x\sqrt{x}$

For the second integral, letting ${x/y^2=t}$

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=\frac{1}{2}x\sqrt{x}\int_x^1t^{-\frac{5}{2}}\sqrt{t^2+1}dt$

Since the Taylor series

$\displaystyle \sqrt{1+t^2}=1+\frac{1}{2}t^2-\frac{1}{8}t^4+\cdots,\quad \text{for }|t|<1$

Consequently

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=a_0+a_1x+a_2x\sqrt{x}+\cdots$

Combining all the above analysis, the index for ${f_*w}$ is ${\{(n,0)\}\cup \{\frac{n}{2},0\}}$

From another point of view, the vanishing order of ${f}$ on each boundary hypersurface of ${W}$ are ${e_f(lb)=1}$, ${e_f(\text{ff})=2}$ and ${e_f(rb)=0}$. ${f}$ maps lb and ff to the boundary of ${\mathbb{R}^+}$. Therefore the index of ${f_*w}$ is contained in

$\displaystyle \frac{1}{e_f(lb)}E\overline{\cup}\frac{1}{e_f(\text{ff})}F=E\overline{\cup}\frac{1}{2}F$

Remark: Daniel Grieser, Basics of ${b-}$Calculus.

### Stereographic projection from center

Suppose we are doing stereographic projection at the center. Namely consider the following map

$\displaystyle \phi:\mathbb{S}^2\rightarrow \mathbb{R}^2$

$\displaystyle (x,y,z)\mapsto\frac{(u_1,u_2,-1)}{\lambda}$

where ${\lambda=\sqrt{u_1^2+u_2^2+1}}$. Then one can see ${u_1=\frac{x}{z}}$, ${u_2=\frac{y}{z}}$. Under this stereographic projection, a strip will be equivalent to some lens domain on the sphere.

Let us pull the standard metric of ${\mathbb{S}^2}$ to the ${\mathbb{R}^2}$. For the following statement, we will always omit ${\phi^*}$ and ${\phi_*}$. Calculation shows,

$\displaystyle dx=\left(\frac{1}{\lambda}-\frac{u_1^2}{\lambda^3}\right)du_1-\frac{u_1u_2}{\lambda^3}du_2=z(-1+x^2)du_1+xyzdu_2$

$\displaystyle dy=-\frac{u_1u_2}{\lambda^3}du_1+\left(\frac{1}{\lambda}-\frac{u_2^2}{\lambda^3}\right)du_2=xyzdu_1+z(-1+y^2)du_2$

$\displaystyle dz=\frac{1}{\lambda^3}(u_1du_1+u_2du_2)=z^2(xdu_1+ydu_2)$

therefore

$\displaystyle dx^2+dy^2+dz^2=z^2(1-x^2)du_1^2-2xyz^2du_1du_2+z^2(1-y^2)du^2_2$

$\displaystyle =\frac{1}{\lambda^2}(\delta_{ij}-\frac{u_iu_j}{\lambda^2})du_idu_j$

here we used the fact that ${(x,y,z)\in \mathbb{S}^2}$. Suppose ${\bar\nabla}$ is the connection on ${\mathbb{S}^2}$ equipped with the standard metric. We want to calculate ${\bar \nabla_{\partial_{u_i}}\partial_{u_j}}$. To that end, it is better to use the ${x,y,z}$ coordinates in ${\mathbb{R}^3}$

$\displaystyle \partial_{u_1}=z(x^2-1)\partial_x+xyz\partial_y+xz^2\partial_z$

$\displaystyle \partial_{u_2}=xyz\partial_x+z(y^2-1)\partial_y+yz^2\partial_z$

Since we know

$\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=\left(\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}\right)^T=\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}-\langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle\partial_r$

where ${T}$ means the tangential part to ${\mathbb{S}^2}$ and ${\partial_r=x\partial_x+y\partial_y+z\partial_z}$ is the unit normal to ${\mathbb{S}^2}$. Using the connection in ${\mathbb{R}^3}$, we get

$\displaystyle \nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}=z^2\left[3x(x^2-1)\partial_x+y(3x^2-1)\partial_y+z(3x^2-1)\partial_z\right]$

$\displaystyle \langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle=z^2(x^2-1)$

One can verify from the above equalities that

$\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=2xz\partial_{u_1}$

Similarly

$\displaystyle \bar \nabla_{\partial_{u_1}}\partial_{u_2}=yz\partial_{u_1}+xz\partial_{u_2}$

$\displaystyle \bar\nabla_{\partial u_2}\partial_{u_2}=2yz\partial_{u_2}$

### Scalar curvature of cylinder

Consider the scalar curvature ${R_g}$ of a cylinder ${\mathbb{R}\times \mathbb{S}^{n-1}}$ with the standard product metric ${g_{prod}}$, where ${n\geq 2}$.

First approach, the cylinder is diffeomorphic to the puncture plane,

$\displaystyle \phi:\mathbb{R}^n\backslash\{0\}\longmapsto \mathbb{R}\times \mathbb{S}^{n-1}$

$\displaystyle x\rightarrow (\log|x|,\frac{x}{|x|})$

It is easy to verify that ${(\phi^{-1})^*g_{prod}=|x|^{-2}|dx|^2}$. One can calculate ${R_g}$ from the conformal change formula

$\displaystyle R_g=-\frac{4(n-1)}{n-2}|x|^{-\frac{n+2}{2}}\Delta |x|^{-\frac{n-2}{2}}=(n-2)(n-1)$

Second approach, remember the Ricci curvature splits on product manifold with product metric. Therefore

$\displaystyle R_g=R_{g_{S^n}}=(n-2)(n-1)$

From the above, we know that if ${n=2}$, then ${R_g=0}$ while ${R_g>0}$ if ${n\geq 3}$. I always get confused with this two cases. There is a nice way to see this result for ${n=2}$. ${\mathbb{S}^1\times \mathbb{R}}$ can be covered locally isometrically by ${\mathbb{R}^2}$ with Euclidean metric. Therefore ${R_g=0}$ in this case. However, ${\mathbb{R}^n}$ should cover ${\mathbb{S}^1\times \mathbb{R}^{n-1}}$ instead of ${\mathbb{R}\times \mathbb{S}^{n-1}}$.

### Unique continuation property on the boundary

I am writing a theorem proved by Jin Zhiren in his thesis.

Suppose $\Omega$ is a smooth domain in $\mathbb{R}^n$, $x_0\in \partial \Omega$ and $u$ is a harmonic function in $\Omega$. If there exists $A, b>0$ such that

$\displaystyle |u(x)|\leq Ae^{-\frac{b}{|x-x_0|}}\quad x\in \Omega$

for $|x-x_0|$ small, then $u=0$. If $n=2$, the same conclusion holds for the solutions of a general second order linear elliptic equation.

A borderline example for this theorem is $u$ be the real part of $e^{-1/z^\alpha}$, $\alpha\in (0,1)$. $u$ is harmonic in the right half plane and $u\leq Ae^{-1/|x|^\alpha}$ and consequently $D^\beta u(0)=0$.

### Newton tensor

Suppose ${A:V\rightarrow V}$ is a symmetric endomorphism of vector space ${V}$, ${\sigma_k}$ is the ${k-}$th elementary symmetric function of the eigenvalue of ${A}$. Then

$\displaystyle \det(A+tI)=\sum_{k=0}^n \sigma_k t^{n-k}$

One can define the ${k-}$th Newton transformation as the following

$\displaystyle \det(A+tI)(A+tI)^{-1}=\sum_{k=0}^{n-1}T_k(A)t^{n-k-1}$

This means

$\displaystyle \det(A+tI)=\sum_{k=0}^{n-1}T_k(A+tI)t^{n-k-1}$

$\displaystyle =T_0 t^n+\sum_{k=0}^{n-2}(A\cdot T_k(A)+T_{k+1}(A))t^{n-k-1}+T_{n-1}(A)$

By comparing coefficients of ${t}$, we get the relations of ${T_k}$

$\displaystyle T_0=1,\quad A\cdot T_k(A)+T_{k+1}(A)=\sigma_{k+1}I,\, 0\leq k\leq n-2\quad T_{n-1}(A)=\sigma_n$

Induction shows

$\displaystyle T_{k}(A)=\sigma_kI-\sigma_{k-1}A+\cdots+(-1)^kA^k$

For example

$\displaystyle T_1(A)=\sigma_1I-A$

$\displaystyle T_2(A)=\sigma_2-\sigma_1A+A^2$

One of the important property of Newton transformation is that: Suppose ${F(A)=\sigma_k(A)}$, then

$\displaystyle F^{ij}=\frac{\partial F}{\partial A_{ij}}=T_{k-1}^{ij}(A)$

The is because

$\displaystyle \frac{\partial }{\partial A_{ij}}\det(A+tI)=\det(A+tI)((A+tI)^{-1})_{ij}.$

If ${A\in \Gamma_k}$, then ${T_{k-1}(A)}$ is positive definite and therefore ${F}$ is elliptic.

Remark: Hu, Z., Li, H. and Simon, U. . Schouten curvature functions on locally conformally flat Riemannian manifolds. Journal of Geometry, 88(1${-}$2), (2008), 75${-}$100.

### Self-shrinker and polynomial volume growth

Proposition: If $M$ is an entire graph of at most polynomial volume growth and $H=\langle X,\nu\rangle$, namely $M$ is a self-shrinker. Then $M$ is a plane.

Proof: Suppose

$\displaystyle v=\frac{1}{\langle \nu, w\rangle}$

Then one can derive the following equation

$\displaystyle \Delta v=\langle\nabla v,X\rangle+|A|^2v+2v^{-1}|\nabla v|^2$

Multiplying both sides by $e^{-|X|^2/2}$ and integration on $M$, which makes sense because of the polynomial volume growth, we get

$\int_M (\Delta v-\langle\nabla v,X\rangle)e^{-\frac{|X|^2}{2}}d\mu=\int_M (|A|^2v+2v^{-1}|\nabla v|^2)e^{-\frac{|X|^2}{2}}d\mu$

However, integration by parts shows the LHS is zero. Thus $A\equiv v\equiv 0$, $M$ must be a plane.

### Some calculations of sigma_2

On four-manifold ${(M^4,g_0)}$, we define Shouten tensor

$\displaystyle A = Ric-\frac 16 Rg$

and Einstein tensor and gravitational tensor

$\displaystyle E=Ric - \frac 14 Rg\quad S=-Ric+\frac{1}{2}Rg$

Suppose ${\sigma_2}$ is the elemantary symmetric function

$\displaystyle \sigma_2(\lambda)=\sum_{i

Thinking of ${A}$ as a tensor of type ${(1,1)}$. ${\sigma_2(A)}$ is defined as ${\sigma_2}$ applied to eigenvalues of ${A}$. Then

$\displaystyle \sigma_2(A)= \frac{1}{2}[(tr_g A)^2-\langle A, A\rangle_g] \ \ \ \ \ (1)$

Notice ${A=E+\frac{1}{12}Rg}$. Easy calculation reveals that

$\displaystyle \sigma_2(A)=-\frac{1}{2}|E|^2+\frac{1}{24}R^2 \ \ \ \ \ (2)$

Under conformal change of metric ${g=e^{2w}g_0}$, we have

$\displaystyle R= e^{-2w}(R_0-6\Delta_0 w-6|\nabla_0 w|^2) \ \ \ \ \ (3)$

$\displaystyle A=A_0-2\nabla^2_0 w+2dw\otimes dw-|\nabla_0w|^2g_0 \ \ \ \ \ (4)$

$\displaystyle S=S_0+2\nabla_0^2w-2\Delta_0wg_0-2dw\otimes dw-|\nabla_0 w|^2g_0 \ \ \ \ \ (5)$

We want to solve the equation ${\sigma_2(A)=f>0}$, which is equivalent to solve

$\displaystyle \sigma_2(A_0-2\nabla^2_0w+2dw\otimes dw-|\nabla_0w|^2g_0)=f$

This is an fully nonlinear equation of Monge-Ampere type. Under local coordinates, the above equation can be treated as

$\displaystyle F(\partial_i\partial_j w,\partial_kw,w,x)=f$

where ${F(p_{ij},v_k,s,x):\mathbb{R}^{n\times n}\times\mathbb{R}^n\times\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}}$. This equation is elliptic if the matrix ${\left(\frac{\partial F}{\partial p_{ij}}\right)}$ is positive definite. In order to find that matrix, we need the linearized operator

$\displaystyle L[\phi]=\frac{\partial F}{\partial p_{ij}}(\nabla_0^2\phi)_{ij}=\frac{d}{dt}|_{t=0}F(\partial_i\partial_j w+t\partial_i\partial_j\phi,\partial_kw,w,x) \ \ \ \ \ (6)$

Using the elementary identity

$\displaystyle \frac{d}{dt}\rvert_{t=0}\sigma_2(H+tG)=tr_gH\cdot tr_gG-\langle H, G\rangle_g. \ \ \ \ \ (7)$

for any fixed matrix ${H}$ and ${G}$. Now plug in ${H=A}$ is Schouten tensor and ${B=-2\nabla_0^2\phi}$. One can calculate them as

$\displaystyle tr_g H\cdot tr_g G=\langle \frac{1}{3}Rg, G\rangle_g \ \ \ \ \ (8)$

Then we get

$\displaystyle L[\phi]=\langle S,G\rangle_g=-2\langle S,\nabla^2_0\phi\rangle_g$


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# Equivalence of Definitions of Lower Set

## Theorem

The following definitions of the concept of Lower Set are equivalent:

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $U \subseteq S$.

Then the following are equivalent:

 $(1)$ $:$ $\displaystyle \forall u \in U: \forall s \in S: s \preceq u \implies s \in U$ $(2)$ $:$ $\displaystyle U^\preceq \subseteq U$ $(3)$ $:$ $\displaystyle U^\preceq = U$

where $U^\preceq$ is the lower closure of $U$.

## Proof

By the Duality Principle, it suffices to prove that:

$(1^*)$, $(2^*)$ and $(3^*)$ are equivalent

where these are the dual statements of $(1)$, $(2)$ and $(3)$, respectively.

By Dual Pairs, it can be seen that these dual statements are as follows:

 $(1^*)$ $:$ $\displaystyle \forall u \in U: \forall s \in S: u \preceq s \implies s \in U$ $(2^*)$ $:$ $\displaystyle U^\succeq \subseteq U$ $(3^*)$ $:$ $\displaystyle U^\succeq = U$

Their equivalence is proved on Equivalence of Definitions of Upper Set.

$\blacksquare$


text:
1. Home
2. Grade 9 Math
3. Scale Factors and Similarity

Similar triangles

Similar triangles

Learn how to determine similar triangles by using side-side-side (SSS) similarity, side-angle-side (SAS) similarity and the properties of similar triangles to solve questions. It’s always a good idea to refresh your memory on scale factors and proportions because we will be using them a lot in this lesson.


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# Obtaining a key that's using ice algorithm

I have a text file that has been processed by the SNOW steganography tool that uses ICE encryption. However, I do not know the key that would enable me to decrypt and retrieve the message hidden.

What tools are available, that are proven to have worked that would help me in retrieving the key?

Edited: instead of asking "what ways are there", I've changed to "what tools are there"

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Brute forcing the password, since no useful cryptanalysis of ICE is known.

I'm really more of a theory guy, so I don't know of any tools for this, but there are some resources that I've found.

http://security.stackexchange.com/questions/1376/where-can-i-find-good-dictionaries-for-dictionary-attacks answers its title's question, and cryptospecs.googlecode.com/svn/trunk/symmetrical/specs/ice.pdf gives the details of the ICE algorithms. (It might be easier to pull the relevant code from SNOW rather than implementing ICE directly.) www.darkside.com.au/snow/description.html gives somewhat of a description of how passwords are converted into keys; I suspect that they then pad the right with zeros or with a one followed by zeroes, although it would be good to look at SNOW's code to be sure.

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Thanks Ricky for the swift response! I appreciated it so much! What you said makes sense. And, I got my dictionaries from a link among the list of links in the link you provided and managed to obtain a key for a jpg that has encrypted files in it by using the stegbreak tool. However, for snow, there seems to be no known tools to perform a dictionary attack. It appears that it is impossible to detect when you have successfully retrieved the correct data. I am really at my wits end on how to resolve this issue of obtaining the 'iced' key... –  user4981 Feb 8 '13 at 6:56
Ummm... I don't see how detecting "when you have successfully retrieved the correct data" is any different with SNOW than with any kdf-cipher combination. $\:$ See crypto.stackexchange.com/questions/700/…. –  Ricky Demer Feb 8 '13 at 8:52
Thanks Ricky! Unfortunately, I am really inexperienced with this. I don't see how cryptanalysis would help in obtaining the key. And it seems that the information provided is mostly theoretical and not applied in terms of running programs :\ Therefore, is there any existing tool that allows me to use a rainbow table or perhaps other means, to obtain the key on snow? Thanks! –  user4981 Feb 10 '13 at 17:42
Conceivably, a brilliant cryptanalysis could guarantee almost-instant recovery of the key from a guess of a specific two of the plaintext blocks for a known ciphertext such that the hamming distance between the guess for those blocks and the actual plaintext for those blocks in at most 15. $\:$ (I'm not aware of any actual result anywhere near that for any computer-era cipher that was ever thought useful. $\:$ Also, as I mentioned in my answer, no useful cryptanalysis of ICE is known.) $\;\;$ –  Ricky Demer Feb 11 '13 at 6:56
I'm pretty sure there aren't any exiting tools that will do what you want. $\:$ –  Ricky Demer Feb 11 '13 at 6:56


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# Math Help - [SOLVED] Poles, order of it and residue

1. ## [SOLVED] Poles, order of it and residue

I must prove that the singularities of $f$ are poles. I also must determine their order and the corresponding residue.

$f(z)=\frac{z}{\cos z}$.

My attempt :
I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
So I wrote $\cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$.
So if I can find the order of the zeros of $\frac{1}{f(z)}$ I'm done.
I've found out that $\frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...$
The first term of the series really bothers me. If it wasn't there then $\frac{1}{f(z)}$ would be equal to $z\cdot h(z)$ with $h(0) \neq$ $0$ and so $z=0$ would be a $0$ of order $1$ of $\frac{1}{f}$, thus a pole of order $1$ of $f$. But of course I should take in count the $\frac{1}{z}$ term and look for $z=k \frac{\pi}{2}$, $k \in \mathbb{Z}$, but I'm stuck.

2. Originally Posted by arbolis
I must prove that the singularities of $f$ are poles. I also must determine their order and the corresponding residue.

$f(z)=\frac{z}{\cos z}$.

My attempt :
I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
So I wrote $\cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$.
So if I can find the order of the zeros of $\frac{1}{f(z)}$ I'm done.
I've found out that $\frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...$
The first term of the series really bothers me. If it wasn't there then $\frac{1}{f(z)}$ would be equal to $z\cdot h(z)$ with $h(0) \neq$ $0$ and so $z=0$ would be a $0$ of order $1$ of $\frac{1}{f}$, thus a pole of order $1$ of $f$. But of course I should take in count the $\frac{1}{z}$ term and look for $z=k \frac{\pi}{2}$, $k \in \mathbb{Z}$, but I'm stuck.
f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

The singularities are $z = z_m = \frac{(2m+1)\pi}{2}$ where m is an integer.

So you need to show that $\lim_{z \rightarrow z_m} (z - z_m)^n f(z)$ is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

eg. see - Wolfram|Alpha

3. Originally Posted by mr fantastic
f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

The singularities are $z = z_m = \frac{(2m+1)\pi}{2}$ where m is an integer.

So you need to show that $\lim_{z \rightarrow z_m} (z - z_m)^n f(z)$ is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

eg. see - Wolfram|Alpha
I wasn't able to solve the problem using your tip.
Here's what I've done :
Considering the singularity at $z=\frac{\pi}{2}$, the series expansion of $\cos z$ around this singularity (singularity for $f$) is $1+ \left ( z- \frac{\pi}{2} \right )^2 h(z)$ where $h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-...$ so $h\left ( \frac{\pi}{2} \right ) \neq 0$.
--------------------------------------------------------------------
I've a side-note question : From this can I conclude that the order of the pole of $r(z)=\frac{1}{\cos z}$when $z=\frac{\pi}{2}$ is 2?
--------------------------------------------------------------------
Now I wrote that the pole of $f$ occurs when $\frac{1}{f}$ has a zero. That is, when $\frac{1}{z}\left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]=0$. I go nowhere.

No let me try again you way : $\lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left ( z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]$.
For $n=1$, the limit is worth $0$. Does that mean the order of the pole of $f$ at $z=\frac{\pi}{2}$ is $1$, and that I've successfully solved the exercise?

4. Originally Posted by arbolis
I wasn't able to solve the problem using your tip.
Here's what I've done :
Considering the singularity at $z=\frac{\pi}{2}$, the series expansion of $\cos z$ around this singularity (singularity for $f$) is $1+ \left ( z- \frac{\pi}{2} \right )^2 h(z)$ where $h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-...$ so $h\left ( \frac{\pi}{2} \right ) \neq 0$.
--------------------------------------------------------------------
I've a side-note question : From this can I conclude that the order of the pole of $r(z)=\frac{1}{\cos z}$when $z=\frac{\pi}{2}$ is 2?
--------------------------------------------------------------------
Now I wrote that the pole of $f$ occurs when $\frac{1}{f}$ has a zero. That is, when $\frac{1}{z}\left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]=0$. I go nowhere.

No let me try again you way : $\lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left ( z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]$.
For $n=1$, the limit is worth $0$. Does that mean the order of the pole of $f$ at $z=\frac{\pi}{2}$ is $1$, and that I've successfully solved the exercise?
The link I posted shows that the limit exists and is finite. Therefore there is a simple pole. You should consider the general limit $\lim_{z \rightarrow z_m} (z - z_m) f(z)$ to show that f(z) has a simple pole at $z = z_m$.

5. Ah I see, I should find that the limit is worth $-\frac{\pi}{2}$ instead of $0$.
Thanks once again.


text: Title Measurement of the $t\bar{t}$ production cross section in the all-jet final state in pp collisions at $\sqrt{s}=7TeV$Measurement of the $t\bar{t}$ production cross section in the all-jet final state in pp collisions at $\sqrt{s}=7TeV$ Author Chatrchyan, S. Khachatryan, V. Sirunyan, A. M. Bansal, M. Bansal, S. Cornelis, T. de Wolf, E.A. Janssen, X. Luyckx, S. Mucibello, L. Roland, B. Rougny, R. Selvaggi, M. van Haevermaet, H. Van Mechelen, P. Van Remortel, N. Van Spilbeeck, A. et al. Faculty/Department Faculty of Sciences. Physics Research group Elementary Particle Physics Department of Physics Publication type article Publication 2013Bristol, 2013 Subject Physics Source (journal) Journal of high energy physics. - Bristol Volume/pages (2013):5, p. 1-35 ISSN 1126-6708 1029-8479 ISI 000321374400065 Carrier E Target language English (eng) Full text (Publishers DOI) Affiliation University of Antwerp Abstract A measurement is presented of the t (t) over bar production cross section (sigma(t (t) over bar)) in proton-proton collisions at a centre-of-mass energy of 7 TeV, in the all-jet final state that contains at least six jets, two of which are tagged as originating from b quarks. The data correspond to an integrated luminosity of 3.54 fb(-1), collected with the CMS detector at the LHC. The cross section is determined through an unbinned maximum likelihood fit of background and t (t) over bar signal to the reconstructed mass spectrum of t (t) over bar candidates in the data, in which events are subjected to a kinematic fit assuming a t (t) over bar -> W(+)bW(-)(b) over bar -> 6 jets hypothesis. The measurement yields sigma(t (t) over bar) = 139 +/- 10 (stat.) +/- 26 (syst.) +/- 3 (lum.) pb, a result consistent with those obtained in other t (t) over bar decay channels, as well as with predictions of the standard model. Full text (open access) https://repository.uantwerpen.be/docman/irua/390b68/5576.pdf E-info http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000321374400065&DestLinkType=RelatedRecords&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000321374400065&DestLinkType=FullRecord&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000321374400065&DestLinkType=CitingArticles&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 Handle