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Article Text

Heat shock protein 70 gene polymorphisms in Mexican patients with spondyloarthropathies
1. G Vargas-Alarcón1,
2. J D Londoño2,
3. G Hernández-Pacheco1,
4. R Gamboa1,
5. E Castillo1,
6. C Pacheco-Tena3,
7. M H Cardiel2,
9. R Burgos-Vargas3
1. 1Cellular Biology Section, Department of Physiology, Instituto Nacional de Cardiología Ignacio Chávez, Mexico DF Mexico
2. 2Department of Immunology and Rheumatology, Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico DF, Mexico
3. 3Department of Rheumatology, Hospital General de Mexico, Mexico DF, Mexico
1. Correspondence to:
Dr G Vargas-Alarcón, Cellular Biology Section, Department of Physiology, Instituto Nacional de Cardiología Ignacio Chávez, Juan Badiano No 1, Tlalpan 14080, Mexico DF, Mexico;
gvargas63{at}yahoo.com

## Abstract

Objective: To investigate the role of HSP70 genes as contributors to genetic susceptibility of the spondyloarthropathies (SpA) in the Mexican population.

Methods: The study included 150 patients with SpA (undifferentiated spondyloarthropathy (uSpA) 68, ankylosing spondylitis (AS) 60, and reactive arthritis 22) and 158 healthy controls. HSP70-1, HSP70-2 and HSP70-hom genotypes were analysed by the polymerase chain reaction-restriction fragment length polymorphism technique. Statistical methods included the Mantel-Haenzel, χ2, Fisher's exact test, and Woolf's method for odds ratio (OR).

Results: HSP70-2 B/B genotype frequency was increased in the whole group of patients with SpA (pC<0.05, OR=4.3), as well as in the different clinical subgroups (pC<0.05, OR=4.2 for AS; pC<0.05, OR=4.4 for uSpA; and pC<0.05, OR=4.1 for ReA). This frequency remained significantly increased when the patients with B27 negative SpA were analysed. On the other hand, HSP70-hom locus analysis showed significantly increased frequency of A allele in the whole group of SpA (pC<0.05, OR=3.4), as well as in the groups with AS (pC<0.05, OR=5.6) and with uSpA (pC<0.05, OR=3.1), when compared with healthy controls. In this case, also, the genotype A/A was increased in the whole group of SpA (pC<0.05, OR=4.5), as well as in patients with AS (pC<0.05, OR=6.4) and with uSpA (pC<0.05, OR=3.7). When the patients with B27 negative SpA were analysed the frequencies of HSP70-hom A allele and A/A genotype remained significantly increased in the whole group of SpA (pC<0.05, OR=3.2 for the A allele and pC<0.05, OR=4.2 for the A/A genotype) and in the uSpA subgroup (pC<0.05, OR=3.8 for the A allele and pC<0.05, OR=4.3 for the A/A genotype).

Conclusion: In addition to the association of SpA with HLA-B27, there is a significant association of HSP70-2 and HSP70-hom alleles with SpA in Mexicans. This association seems to be independent of the susceptibility conferred by HLA-B27 in the group of patients with uSpA.

• heat shock proteins, spondyloarthropathies
• genetic susceptibility
• Mexican population
• genotype
• AF, aetiological fraction
• AS, ankylosing spondylitis
• HSP, heat shock protein
• OR, odds ratio
• PCR, polymerase chain reaction
• ReA, reactive arthritis
• SpA, spondyloarthropathies
• uSpA, undifferentiated spondyloarthropathy

## Statistics from Altmetric.com

The spondyloarthropathies (SpA) constitute a group of diseases of unknown cause which share clinical, epidemiological, and genetic features.1 The pathogenesis of SpA is attributed in part to the interaction between genetic and, in some patients, environmental factors. Their clinical expression seems additionally influenced by ethnicity, age at onset, and sex of the patients.2 The role of genetic factors in the pathogenesis of SpA is best represented by the strong association between the HLA-B27 antigen and ankylosing spondylitis (AS) and then with undifferentiated spondyloarthropathy (uSpA), reactive arthritis (ReA), and the other SpA.3,4 None the less, on the one hand, only 1–3 % of HLA-B27 positive subjects from the population develop AS and, on the other, not all patient with SpA carry the HLA-B27 antigen, indicating the possibility that other genes may influence disease susceptibility.5,6

Besides studying the role of non-HLA-B27 major histocompatibility complex alleles in the pathogenesis of SpA, few reports have considered the role of the heat shock protein 70 (HSP70) locus on disease susceptibility. The HSP70 locus is located close to HLA-B, and the polymorphism of its genes includes HSP70-2 and two other immediately adjacent genes, HSP70-1 and HSP70-hom, map-90 kb telomeric to the C2 locus and 280 kb centromeric to the tumour necrosis factor α locus.7 Because the polymorphism of HSP70 genes has been linked to autoimmune disease in some studies,8–10 its role has been investigated in AS, but no significant associations were found in Finnish and Spanish patients.11,12

In this study we explored the possibility that HSP70 genes might be involved in the susceptibility to SpA in Mexican Mestizo patients, by using the polymerase chain reaction (PCR) and restriction fragment length polymorphism analyses. Clinically, there is some evidence that SpA and particularly AS in Mexicans differ from that seen in most white populations, suggesting a role for additional factors in their pathogenesis.13,14

## SUBJECTS AND METHODS

### Study subjects

The study included 150 consecutive patients with SpA15 attending two specialised clinics over a period of 18 months. Sixty eight of these patients had uSpA, 60 AS,16 and 22 ReA.17 The group of patients with uSpA all fulfilled the SpA classification criteria,15 but none had signs of underlying disorders such as intestinal bowel disease or psoriasis. A group of 158 non-related healthy subjects with neither symptoms nor previous diagnosis of systemic disease comprised the control group. All patients with SpA and the healthy controls and their two previous generations were born in Mexico. This study was approved by the institutional ethics and research committees and all subjects signed an informed consent form.

### DNA extraction

Genomic DNA from whole blood containing EDTA was extracted by standard techniques.18

### HSP70 polymorphism analysis

#### HSP70-1

The biallelic polymorphism at position 190 in the HSP70-1 gene was detected by BsrBI restriction enzyme digestion of the fragment 325 bp DNA, which was previously amplified by using the following primers: 5-TCCGGCGTCCGGAAGG ACC-3 (forward) and 5-TGCGGCCAATCAGGCGCTT-3 (reverse).19 The presence of the 171 bp, 84 bp, and 70 bp fragments represents the HSP70-1*b1 allele, whereas the b2 allele showed two fragments of 241 bp and 84 bp.

#### HSP70-2

An HSP70-2 restriction fragment length polymorphism at position 1267 was characterised by a PCR procedure. This analysis was performed considering the polymorphic PstI site at position 1267 of these alleles. We selected the following primers spanning the polymorphic PstI site (nucleotide 1267) according to the previously described sequence,20 sense (nucleotide 1083–1102 within the coding region) 5-CATCGACTTCTACA CGTCCA-3 and antisense (nucleotide 2180–2199 within the 3 untranslated end to avoid HSP70-1 homology) 5-CAAAGT CCTTGAGTCCCAAC-3. The products of PCR were cleaved with PstI (Pharmacia, Uppsala, Sweden) and electrophoresed on 1.5% agarose stained with ethidium bromide. The DNA-lacking polymorphic PstI site within the HSP70-2 gene generates a product of 1117 bp after restriction (A allele), whereas the PstI site produced two fragments of 936 and 181 bp (B allele).

#### HSP70-hom

A fragment of 878 bp that includes the polymorphic site on position 2437 was amplified using the primers 5-GGACAAGTCTGAGAAGGTACAG-3 (forward) and 5-GTAACTTAGATTCAGGTCTGG-3` (reverse) and subject to restriction analysis with NcoI as previously described.20 In this case the presence of two fragments of 551 bp and 327 bp represents the A allele, whereas the complete fragment (878 bp) was named the B allele.

### Statistical analysis

Genotype and allele frequencies were compared by contingency table analysis using the Mantel-Haenzel, χ2 test, or the Fisher's exact test if the number in any cell was <5. p Values were corrected according to the number of specificities tested and the number of comparisons performed. The level of significance was established at pC<0.05. The statistical program “EPIINFO” (version 5.0; USD Incorporated 1990, Stone Mountain, Georgia ) was used for these analyses. The magnitude of associations was assessed using the odds ratio (OR) and aetiologic fraction (AF) statistics.21 Confidence intervals were calculated for the OR by Woolf's method.22

## RESULTS

One hundred and fifty patients with SpA (106 men, 44 women) were included in the study, with a mean (SD) age at onset of 21.1 (9.5) years. Sex distribution and mean age at onset differed between groups:(a) uSpA (n=68): 45 men, 23 women; 22.2 (9.4) years; (b) AS (n=60): 48 men, 12 women; 17.5 (7.9) years; and (c) ReA (n=22): 13 men, 9 women; 26.2 (10.8) years.

The polymorphism of HSP70-1 was similar in all three groups of patients with SpA and in the control group (data not shown). In contrast, the allele and genotype frequencies of HSP70-2 and HSP70-hom were significantly different in the SpA group than in the healthy subjects (table 1). Thus HSP70-2 B/B genotype frequency was increased in the whole group of SpA (pC<0.05, OR=4.3, AF=14.3%) as well as in AS (pC<0.05, OR=4.2, AF=13.9%), uSpA (pC<0.05, OR=4.4, AF=14.7%), and ReA (pC<0.05, OR=4.1, AF=13.7%). For the HSP70-hom locus, there was a significant increase of the A allele in the whole group of SpA (pC<0.05, OR=3.4, AF=66.6%), AS (pC<0.05, OR=5.6, AF=78.6%), and uSpA (pC<0.05, OR=3.1, AF=62.4%). Additionally, the HSP70-hom A/A genotype was significantly increased in the whole group of SpA (pC<0.05, OR=4.5, AF=68.4%) as well as in the AS (pC<0.05, OR=6.4, AF=76.8%) and uSpA (pC<0.05, OR=3.7, AF=62.9%) groups.

Table 1

HSP70-2 and HSP70-hom genotype and allele frequencies (%) in Mexican patients with SpA and healthy controls

To establish whether HSP70 associations with SpA were dependent on the presence of HLA-B27 or not, the frequencies of HSP70 alleles were analysed only in HLA-B27 negative patients with SpA (table 2). This analysis showed an increased frequency of HSP70-2 B/B genotype in the whole group of SpA (pC<0.05, OR=5.3, AF=16.5%) and in each one of the clinical subgroups: AS (pC<0.05, OR=5.2, AF=16.1%), uSpA (pC<0.05, OR=4.3, AF=13.1%), and ReA (pC<0.05, OR=10.5, AF=30.1%). In contrast, the frequencies of the HSP70-hom A allele and A/A genotype were only significantly increased in the whole group of SpA (pC<0.05, OR=3.2, AF=64.1% for the A allele and pC<0.05, OR=4.2, AF=66.6% for the A/A genotype) and in uSpA (pC<0.05, OR=3.8, AF=69.8% for the A allele and pC<0.05, OR=4.3, AF=67.3% for the A/A genotype).

Table 2

HSP70-2 and HSP70-hom genotype and allele frequencies (%) in B27 negative Mexican patients with SpA and healthy controls

## DISCUSSION

The results of this study showed a significant association of HSP70-2, particularly HSP70-2 B/B, with the whole group of SpA and the three different diagnostic subgroups, and a significant increase of HSP70-hom locus alleles was seen in some subgroups, suggesting a role for these genes in disease susceptibility in the Mexican Mestizo population. In contrast with findings in the Finnish and Spanish populations with AS,11,12 our study detected increased frequencies of the HSP70-hom A allele and A/A genotype in the whole group of SpA as well as in the subgroups of patients with AS and uSpA. The strong association of SpA with the presence of HLA-B27 antigen is well known, and thus the possible association of these diseases with other genetic markers needs to be considered. It was not possible include B27 positive healthy subjects in our study because of the low frequency of HLA-B27 in the general Mexican Mestizo population. To avoid this problem and establish whether HSP70 associations with SpA were dependent on the presence of HLA-B27 or not, the frequencies of HSP70 alleles were analysed only in HLA-B27 negative patients and healthy controls. When we analysed these groups, the increased frequencies of the HSP70-hom A allele and A/A genotype remained significantly increased in the whole group of SpA and in the subgroup of patients with uSpA. None the less, the fact that we only had eight patients with HLA-B27 negative AS in the analysis should be taken into account when interpreting the role of HSP70-hom polymorphism in AS.

Recent papers have shown that the peptide binding domains of the HSP70 and HLA class I molecule are remarkably similar and even a structural model for the former has been developed.23 On the other hand, the HSP70-hom polymorphic site detected by NcoI is located on position 2437, which corresponds to a Met→Thr amino acid substitution at position 493.20 In the hypothetical model, the 493 residue is located on the β sheets, which form the floor of the peptide binding site of the molecule. In this context, it is thought that this substitution may be associated with some variation in the peptide-binding specificity of different HSP70-hom haplotypes.

Previous studies have shown an association between the HSP70-2 B/B genotype and autoimmune disease,8–10 but the mechanisms are largely unknown. Findings in rheumatoid arthritis suggest that the HLA-DR4 and DR10 motifs associated with the disease bind to a 70 kDa HSP,24 which perhaps could influence the processing of antigenic peptides and their inclusion in the HLA-DRB1 chain groove. This effect might be expected with the HSP70-hom variation because it takes place within the peptide binding domain predicted,23 but possibly not with HSP70-2 because its polymorphism depends on a silent change (A→G) in the coding region.

The role of HSP70-2 in the pathogenesis of SpA nevertheless remains to be determined. Unless a quantitative difference in HSP70-2 expression between carriers of the B/B genotype and those carrying other genotypes (A/A or A/B) exists, any contribution of the HSP70-2 polymorphism to the pathogenesis of SpA might be attributed to a neighbouring, yet unidentified gene. Thus a decrease of the HSP70-2 mRNA expression in homozygotic subjects for the B allele might occur in comparison with its expression in A/B heterozygotes.25 In such a case, the cell response of B/B subjects to stress would be impaired and lead to the intracellular accumulation of denaturalised protein or peptide transporting defects, affecting self tolerance.

In conclusion, our study shows that the SpA in Mexican Mestizo patients, in addition to its association with HLA-B27, is also associated with some HSP-70 alleles. This association seems to be independent of the susceptibility conferred by the HLA-B27 in the uSpA group of patients. Additional studies in a large number of patients and the inclusion of B27 positive healthy subjects might help to establish the true significance of these associations in the Mexican population.

## Acknowledgments

Supported in part by grants from the Consejo Nacional de Ciencia y Tecnología (ref: 34555-M) and Hospital General de Mexico.

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# On the Bott periodicity, J-homomorphisms, and $H_*Q_0S^{-k}$

Zare, Hadi (2011) On the Bott periodicity, J-homomorphisms, and $H_*Q_0S^{-k}$. [MIMS Preprint]

The Curtis conjecture predicts that the only spherical classes in $H(Q_0S^0; Z/2)$ are the Hopf invariant one and the Kervaire invariant one elements. We consider Sullivan's decomposition $Q_0S^0 = J \times \cokerJ$ where $J$ is the fibre of $\psi^q - 1$ ($q = 3$ at the prime 2) and observe that the Curtis conjecture holds when we restrict to $J$. We then use the Bott periodicity and the $J$-homomorphism $SO \rightarrow Q_0S^0 to define some generators in$H(Q_0S^0; Z/p)$, when$p$is any prime, and determine the type of subalgebras that they generate. For$p = 2$we determine spherical classes in$H_*( \Omega^k_0J; Z/2)$. We determine truncated subalgebras inside$H_*(Q_0-k}; Z/2)$. Applying the machinery of the Eilenberg-Moore spectral sequence we dene classes that are not in the image of by the$J$-homomorphism. We shall make some partial observations on the algebraic structure of$H_*(\Omega^k_0 \coker J; Z/2)$. Finally, we shall make some comments on the problem in the case equivariant$J\$-homomorphisms.

format: plain


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# What test can I use to compare slopes from two or more regression models?

I would like to test the difference in response of two variables to one predictor. Here is a minimal reproducible example.

library(nlme)
## gls is used in the application; lm would suffice for this example
m.set <- gls(Sepal.Length ~ Petal.Width, data = iris,
subset = Species == "setosa")
m.vir <- gls(Sepal.Length ~ Petal.Width, data = iris,
subset = Species == "virginica")
m.ver <- gls(Sepal.Length ~ Petal.Width, data = iris,
subset = Species == "versicolor")

I can see that the slope coefficients are different:

m.set$coefficients (Intercept) Petal.Width 4.7771775 0.9301727 m.vir$coefficients
(Intercept) Petal.Width
5.2694172   0.6508306
m.ver$coefficients (Intercept) Petal.Width 4.044640 1.426365  I have three questions: 1. How can I test the difference between slopes? 2. How can I test the difference between residual variances? 3. What is a simple, effective way to present these comparisons? A related question, Method to compare variable coefficient in two regression models, suggests re-running the model with a dummy variable to differentiate the slopes, are there options that would allow the use of independent data sets? - In regards to the first question see stats.stackexchange.com/questions/55501/…. – rpierce Sep 24 '14 at 19:57 ## 2 Answers How can I test the difference between slopes? Include a dummy for species, let it interact with$P_i$, and see if this dummy is significant. Let$L_i$be the sepal length and$P_i$be the pedal width and$S_1, S_2, S_3$be the dummy variables for the three species. The compare the model $$E(L_i) = \beta_0 + \beta_1 P_i$$ with the model that allows the effect of$P_i$to be different for each species: $$E(L_i) = \alpha_0 + \alpha_1 S_2 + \alpha_2 S_3 + \alpha_4 P_i + \alpha_5 P_iS_2 + \alpha_6 P_i S_3$$ The GLS estimators are MLEs and the first model is a submodel on the second, so you can use the likelihood ratio test here. The likelihoods can be extracted using the logLik function and the degrees of freedom for the test will be$4$since you've deleted$4$parameters to arrive at the submodel. What is a simple, effective way to present the comparison? I think the most appealing way would be to plot the regression lines for each species all on the same axes, maybe with error bars based on the standard errors. This would make the difference (or non-difference) between the species and their relationship to$P_i$very apparent. Edit: I noticed another question has been added to the body. So, I'm adding an answer to that: How can I test the difference between residual variances? For this, you'll need to stratify the data set and fit separate models since, the interaction-based model I suggested will constraint the residual variance to be the same in every group. If you fit separate models, this constraint goes away. In that case, you can still use the likelihood ratio test (the likelihood for the larger model is now calculated by summing the likelihoods from the three separate models). The "null" model depends on what you want to compare it with • if you only want to test the variance, while leaving the main effects in, then the "null" model should be the model with the interactions I've written above. The degrees of freedom for the test are then$2$. • If you want to test the variance jointly with the coefficients, then the null model should be the first model I've written above. The degrees of freedom for the test is then$6$. - Why is there no$S_1$in the second model? Is gls(Sepal.Length ~ species:Petal.Width, data = iris) the correct implementation of the model in R? – Abe Jul 25 '12 at 20:27 Hi @Abe.$S_1$is the "reference" species - the regression line for that species is given by$\alpha_0 + \alpha_4 P_i\$. If species is a categorical variable then I think gls(Sepal.Length ~ species*Petal.Width, data=iris) would be the syntax. – Macro Jul 25 '12 at 21:31
@Macro Nice answer (+1)! I wonder whether you could fit the gls model but allowing for different residual variances for each Species with the option weights=varIdent(form=~1|Species) (regarding the second question)? – COOLSerdash May 19 '13 at 14:06

I agree with the previous suggestion. You should fit a multiple regression model with a dummy variable for each data set. This will allow you to test whether the intercepts differ. If you also want to know if the slopes differ, then you need to also include interactions between the dummies and the variable in question. There is no problem with the fact that the data are independent. Note that if they are both independent and (for example) different species, then you would not be able to tell whether the difference you find is due to the differing species or the differing data sets, as they are perfectly confounded. However, there is no test / get-out-of-jail-free card that will get you around that problem without gathering a new sample and running your study again.

-
It looks like we've posted fairly similar answers at almost the exact same time. +1 – Macro Jul 25 '12 at 15:29
@Macro, yes, but yours is mostly better (+1 earlier); you addressed all 3 questions which I missed on my first (not thorough) reading of the question. My contribution here is the part about confounding. – gung Jul 25 '12 at 15:31
yes that is a good point. I suppose if you were making this inquiry at all you'd have to be operating under the assumption that the data sets were measuring the same thing, etc... with the only difference being that the species were different. – Macro Jul 25 '12 at 15:44
From my way of thinking you both should get upvotes which is what I am doing. – Michael Chernick Jul 25 '12 at 15:58


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Hide

# Problem EPodnizovi

You are given an array of integers of length $N$. Let $s_1, s_2, \ldots , s_ q$ be the lexicographically sorted array of all its non-empty subsequences. A subsequence of the array is an array obtained by removing zero or more elements from the initial array. Notice that some subsequences can be equal and that it holds $q = 2^ N - 1$.

An array $A$ is lexicographically smaller than array $B$ if $A_ i < B_ i$ where $i$ is the first position at which the arrays differ, or if $A$ is a strict prefix of array $B$.

Let us define the hash of an array that consists of values $v_1, v_2, \ldots , v_ p$ as:

$h(s) = (v_1 \cdot B^{p-1} + v_2 \cdot B^{p-2} + \ldots + v_{p-1} \cdot B + v_ p ) \bmod M$

where $B$, $M$ are given integers.

Calculate $h(s_1) h(s_2), \ldots , h(s_ K)$ for a given $K$.

## Input

The first line contains integers $N$, $K$, $B$, $M$ ($1 \leq N \leq 100\, 000$, $1 \leq K \leq 100\, 000$, $1 \leq B, M \leq 1\, 000\, 000$). The second line contains integers $a_1, a_2, a_3, \ldots , a_ N$ ($1 \leq a_ i \leq 100\, 000$). In all test cases, it will hold $K \leq 2^ N - 1$.

## Output

Output $K$ lines, the $j$-th line containing $h(s_ j)$.

Sample Input 1 Sample Output 1
2 3 1 5
1 2

1
3
2

Sample Input 2 Sample Output 2
3 4 2 3
1 3 1

1
1
0
2

Sample Input 3 Sample Output 3
5 6 23 1000
1 2 4 2 3

1
25
25
577
274
578


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# ENH: Add the first Wasserstein and the Cramér-von Mises statistical distances #7563

Merged
merged 13 commits into from Aug 22, 2017
+509 −1

## Conversation

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### CharlesMasson commented Jul 5, 2017

 This PR adds functions to compute statistical distances, namely the first Wasserstein distance and the Cramér-von Mises distances, which are useful to compare distributions and have applications in various domains.

Closed

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### rgommers commented Jul 16, 2017

 Hi @CharlesMasson, looks good from a quick browse. Some questions/thoughts: There are 1-sample and 2-sample versions of Cramer-vonMises, would it make sense to support both in cramer? Does it make sense to have separate weights for u and v? I guess it does, since u and v do not have to be the same length? I'm asking because all distances in scipy.spatial.distance have a single weight parameter that applies to both u and v. I just closed gh-3659, which also had a 2-sample Cramer von Mises implementation. It was untested but perhaps useful to compare against. Issue gh-3291 had a longer list of metrics that were requested, this PR works towards that.

### rgommers requested changes Jul 16, 2017

 functions located at the specified values. References ----------

#### rgommers Jul 16, 2017

Member

This function implements https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93von_Mises_criterion right? Maybe useful to reference as well.

#### CharlesMasson Jul 17, 2017

Contributor

 def wasserstein(u_values, v_values, u_weights=None, v_weights=None): r""" Compute the first Wasserstein distance between two one-dimensional distributions.

#### rgommers Jul 16, 2017

Member

would be good to shorten this a bit and make this fit one line. Sphinx extracts that line as summary in the html docs.

 ---------- .. [1] "Wasserstein metric", http://en.wikipedia.org/wiki/Wasserstein_metric .. [2] Ramdas, Garcia, Cuturi "On Wasserstein Two Sample Testing and Related Families of Nonparametric Tests" (2015). :arXiv:1509.02237.

#### rgommers Jul 16, 2017

Member

Adding an Examples section would be very useful. Can be simple, just generated some arrays with random variates from a distribution and use them to calculate a distance.

#### rgommers Jul 16, 2017

Member

Same for cramer

 Parameters ---------- u_values, v_values : array_like Values observed in the (empirical) distribution.

#### rgommers Jul 16, 2017

Member

Actually, from the code it seems that these two arrays have to have the same length, but that's not stated anywhere or tested.

#### CharlesMasson Jul 17, 2017

Contributor

It's not actually the case: the arrays can have distinct lengths.
https://github.com/CharlesMasson/scipy/blob/cdf-distances/scipy/stats/tests/test_stats.py#L4257

 Weights as ndarray. """ # Validate the value array. values = np.array(values, dtype=float)

#### rgommers Jul 16, 2017

Member

This makes a copy, doesn't seem necessary

 # Validate the value array. values = np.array(values, dtype=float) if len(values) == 0: raise ValueError('Distribution can\'t be empty.')

#### rgommers Jul 16, 2017

Member

nitpick: better use double quotes for the whole message rather than an escape

 # Validate the weight array, if specified. if weights is not None: weights = np.array(weights, dtype=float)

#### rgommers Jul 16, 2017

Member

same here, makes a copy. why not asarray

Contributor

### CharlesMasson commented Jul 17, 2017

 Thanks @rgommers for your feedback. I addressed your comments. The cramer function does not implement the statistic per se, but the Cramér distance, as defined for example in https://arxiv.org/abs/1705.10743 (section 4.1). However, I noticed that there are two versions of the Cramér distance in the literature: the original non-distribution-free Cramér-von Mises distance (from 1928): this is the one in the paper above and the one that is implemented in this PR, the distribution-free distance: this is the one that is used for the Cramér-von Mises criterion (https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93von_Mises_criterion) and in some other papers (e.g., http://www.math.kit.edu/stoch/~henze/seite/aiwz/media/gnst-2015-09-01-rev-1(1).pdf). This distinction is highlighted in Székely's paper (at the beginning of the section 2): http://personal.bgsu.edu/~mrizzo/energy/Szekely-E-statistics.pdf As shown in that paper, in one dimension, the non-distribution-free Cramér-von Mises distance is equivalent to the energy distance (https://en.wikipedia.org/wiki/Energy_distance). For those reasons and to avoid any ambiguities, I renamed the function to energy_distance.

### irabinovitch commented Aug 16, 2017

 @rgommers Thanks for the review. I believe @CharlesMasson new commits should have us in a good spot. Do you have a moment to take another look?

### CharlesMasson added some commits Jul 5, 2017

 ENH: add Wasserstein and Cramer statistical distances
 15532a3
 TST: add tests for Wasserstein and Cramer statistical distances
 91a476d
 STY: Avoid using escape character
 4868f86
 ENH: Avoid copying arrays when validating distributions
 1c3abb0
 DOC: Fit the function description in one line
 5d2615e
 MAINT: Implement the energy distance instead of the Cramer distance
 29ed811
 MAINT: Rename wasserstein to wasserstein_distance for consistency
 6cffa70
 DOC: Add examples for the Wasserstein and the energy distances
 808e46e
 TST: Change TestCase to object
Following #7568.
 99d5dcb

Member

### rgommers commented Aug 19, 2017

 Thanks for the ping @irabinovitch, working on it now. And just pushed a rebase. Note to other reviewers, this was already discussed on the mailing list: https://mail.python.org/pipermail/scipy-dev/2017-July/022005.html. Not many responses, but no objections so OK to add.
Member

### rgommers commented Aug 19, 2017

 Locally I get 2 failures for RuntimeError not raised in test_inf_values. The test looks incorrect - warnings are raised only once from the same place by default, and that assert_raises is not the first call with infs.

Member

### rgommers left a comment

 Overall looks good to me, only a few minor comments. Can be merged once those are addressed.
 assert_equal( stats.wasserstein_distance([1, -np.inf, np.inf], [1, 1]), np.inf) assert_raises(RuntimeWarning, stats.wasserstein_distance,

#### rgommers Aug 19, 2017

Member

I'd remove this check, it's prone to failing.

#### rgommers Aug 19, 2017

Member

same comments apply to the second test_inf_values below

#### CharlesMasson Aug 21, 2017

Contributor

The pytest config turns warnings into errors (https://github.com/scipy/scipy/blob/master/pytest.ini#L6:L7) and it seems to be the reason why the raised warning (then error) is caught by assert_raises. Without that config line, I do see the failures too.

The test should actually be:

with assert_warns(RuntimeWarning):
assert_equal(
stats.wasserstein_distance([1, 2, np.inf], [np.inf, 1]),
np.nan)

or alternatively:

with suppress_warnings() as sup:
r = sup.record(RuntimeWarning, "invalid value*")
assert_equal(
stats.wasserstein_distance([1, 2, np.inf], [np.inf, 1]),
np.nan)

I'll use the latter, as you suggested not checking for raised warnings.

 np.inf) assert_equal( stats.wasserstein_distance([1, -np.inf, np.inf], [1, 1]), np.inf)

#### rgommers Aug 19, 2017

Member

All three above checks should be in a block:

   with suppress_warnings() as sup:
r = sup.record(RuntimeWarning, "invalid value*")

#### CharlesMasson Aug 21, 2017

Contributor

I don't think that it is necessary. Those three blocks should not raise warnings because, contrary to the fourth block, they do not call np.subtract with same-sign infinities in both arguments: e.g., np.subtract([np.inf], [-np.inf]) is valid and does not raise a warning whereas np.subtract([np.inf], [np.inf]) does raise a warning.

 .. math:: l_1 (u, v) = \inf_{\pi \in \Gamma (u, v)} \int_{\mathbb{R} \times \mathbb{R}} |x-y| \mathrm{d} \pi (x, y)

#### rgommers Aug 19, 2017

Member

However, our rule is that we normally try to only use LaTeX in the Notes section, because this isn't readable in a terminal (or Jupyter notebook or IDE help). So I suggest to move both this paragraph and the one on CDFs to the notes.

And same for energy_distance docstring

#### CharlesMasson Aug 21, 2017

Contributor

I moved the paragraphs accordingly.

 ... [2.1, 4.2, 7.4, 8. ], [7.6, 8.8]) 0.88003340976158217 """ return 2**.5 * _cdf_distance(2, u_values, v_values, u_weights, v_weights)

#### rgommers Aug 19, 2017

Member

nitpick: for clarify I'd prefer np.sqrt(2) over 2**.5

### CharlesMasson added some commits Aug 21, 2017

 TST: Correctly handle raised warnings in tests with inf values
 778284c
 DOC: Avoid using LaTeX outside the Notes section
 55c3bbd
 MAINT: Use np.sqrt(2) instead of 2**.5
 d3294f0
Contributor

### CharlesMasson commented Aug 21, 2017

 MAINT: Remove unnecessary np.abs
 1ac893d

### rgommers merged commit 5eade1e into scipy:master Aug 22, 2017 4 checks passed

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### rgommers commented Aug 22, 2017

 LGTM now, merged. Thanks @CharlesMasson !

Open

Merged

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text:
# Exercise (8.4): Solutions - Angle Bisector Theorem

Angle Bisector Theorem
A The bisector of an interior angle of a triangle divides the opposite side internally into a ratio equal to the ratio of the other two sides of the triangle.

တြိဂံတစ်ခု၏ အတွင်းထောင့်တစ်ခုကို ထက်ဝက်ပိုင်းသော မျဉ်းသည် မျက်နှာချင်းဆိုင်အနားကို အတွင်းပိုင်းမှ ပိုင်းဖြတ်ရာ ပိုင်းဖြတ်လိုက်သော (အတွင်းပိုင်း) အချိုးသည် ကျန်အနားနှစ်ဖက် အချိုးနှင့် ညီသည်။

B The bisector of an exterior angle of a triangle divides the opposite side externally into a ratio equal to the ratio of the other two sides of the triangle.

တြိဂံတစ်ခု၏ အပြင်ထောင့်တစ်ခုကို ထက်ဝက်ပိုင်းသော မျဉ်းသည် မျက်နှာချင်းဆိုင် အနားကို အပြင်ပိုင်းမှ ပိုင်းဖြတ်ရာ ပိုင်းဖြတ်လိုက်သော (အပြင်ပိုင်း) အချိုးသည် ကျန်အနားနှစ်ဖက် အချိုးနှင့် ညီသည်။

1. Which of the following proportions follow from the fact that $AE$ bisects $\angle WAV$ in $\triangle WAV$ ?

(a) $\displaystyle \frac{W E}{E V}=\displaystyle \frac{W A}{A V}$

(b) $\displaystyle \frac{W E}{E V}=\displaystyle \frac{V A}{A W}$

(c) $\displaystyle \frac{W E}{W A}=\displaystyle \frac{E V}{A V}$

(d) $\displaystyle \frac{A V}{A W}=\displaystyle \frac{V E}{E W}$

$\begin{array}{l} \quad\quad\text{If}\ AE\ \text{bisects}\ \angle WAV,\\\\ \quad\quad\dfrac{WE}{EV}=\dfrac{WA}{AV}\\\\ \Rightarrow\quad \dfrac{WE}{WA}=\dfrac{EV}{AV}\\\\ \Rightarrow\quad \dfrac{AV}{AW}=\dfrac{VE}{EW}\\\\ \therefore\quad \text{(a), (c), (d) are true and (b) is false.} \end{array}$

2. $AX$ bisects $\angle CAB$. Complete the following statements:

(a) $A C: A B=\ldots$

(b) $A B: A C=\ldots$

(c) $X C: X B=\ldots$

$\begin{array}{l} \text{(a)}\ \dfrac{AC}{AB}=\dfrac{CX}{XB}\\\\ \text{(b)}\ \dfrac{AB}{AC}=\dfrac{BX}{XC}\\\\ \text{(c)}\ \dfrac{XC}{XB}=\dfrac{AC}{AB} \end{array}$

3. $PT$ bisects $\angle RPS$. Complete the following statements:

(a) $P Q: P R=\ldots$

(b) $T R: P R=\ldots$

(c) $Q R: T R=\ldots$

$\begin{array}{l} \text{(a)}\ \dfrac{PQ}{PR}=\dfrac{TQ}{TR}\\\\ \text{(b)}\ \dfrac{TR}{PR}=\dfrac{TQ}{PQ}\\\\ \text{(b)}\ \dfrac{QR}{TR}=\dfrac{TQ-TR}{TR}=\dfrac{PQ-PR}{PR} \end{array}$

4. What can you say about the rays $AD$, $BE$ and $CF$?

$\begin{array}{ll} \quad\quad AB = 6, AC = 8, BC = 10\\\\ \quad\quad\dfrac{AB}{AC}=\dfrac{6}{8}=\dfrac{3}{4}\\\\ \quad\quad\dfrac{AC}{BC}=\dfrac{8}{10}=\dfrac{4}{5}\\\\ \quad\quad\dfrac{BC}{AB}=\dfrac{10}{6}=\dfrac{5}{3}\\\\ \quad\quad\dfrac{BD}{DC}=\dfrac{\frac{30}{7}}{\frac{40}{7}}=\dfrac{3}{4}\\\\ \quad\quad\dfrac{AF}{BF}=\dfrac{\frac{8}{3}}{\frac{10}{3}}=\dfrac{4}{5}\\\\ \quad\quad\dfrac{CE}{AE}=\dfrac{5}{3}\\\\ \therefore\quad \dfrac{AB}{AC}=\dfrac{BD}{DC}\Rightarrow AD\ \text{bisects}\ \angle BAC\\\\ \therefore\quad \dfrac{AC}{BC}=\dfrac{AF}{BF}\Rightarrow CF\ \text{bisects}\ \angle ACB\\\\ \therefore\quad \dfrac{BC}{AB}=\dfrac{CE}{AE}\Rightarrow BE\ \text{bisects}\ \angle ABC \end{array}$

5. If $AD$ and $AE$ are bisectors of the interior and exterior angles at $A$ of $\triangle ABC$, then which of the following are true?

(a) $\angle D A E=90^{\circ}$

(b) $B D: D C=B C: C E$

(c) $B D: D C=B E: C E$

(d) $A D: A E=D C: C E$

$\begin{array}{ll} \quad\quad (\bullet) + (\bullet) + (*) + (*) = 180^{\circ}\\\\ \quad\quad 2(\bullet) + 2(*) = 180^{\circ}\\\\ \quad\quad (\bullet) + (*) = 90^{\circ}\\\\ \therefore\quad \angle DAE = 90^{\circ}\\\\ \therefore\quad \text{(a) is true.}\\\\ \quad\quad AD\ \text {bisects}\ \angle BAC.\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{AB}{AC}\\\\ \quad\quad AE\ \text {bisects}\ \angle CAF.\\\\ \therefore\quad \dfrac{BE}{CE}=\dfrac{AB}{AC}\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{BE}{CE}\\\\ \therefore\quad \text{(b) is false and (c) is true.}\\\\ \quad\quad \text {We cannot say that}\ (\bullet) = (*)\\\\ \therefore\quad \text{We cannot say that}\ AC\ \text {bisects}\ \angle DAE.\\\\ \therefore\quad \text{We cannot say that}\ \dfrac{AD}{AE}=\dfrac{DC}{CE}.\\\\ \therefore\quad \text{(d) is false.} \end{array}$

6. Find the value of 𝑥 in each of the following figures.

(a)

$\begin{array}{l} \quad\quad\dfrac{x}{3}=\dfrac{10}{6}\\\\ \therefore\quad x=\dfrac{10}{6}\times 3=5 \end{array}$

(b)

$\begin{array}{l} \quad\quad\dfrac{x}{9}=\dfrac{13}{8}\\\\ \therefore\quad x=\dfrac{13}{8}\times 9=\dfrac{117}{8}=14.625 \end{array}$

(c)

$\begin{array}{l} \quad\quad\dfrac{x}{4-x}=\dfrac{8}{6}\\\\ \therefore\quad 3x=16-4x\\\\ \quad\quad 7x=16\\\\ \quad\quad x=\dfrac{16}{7}=2.286 \end{array}$

(d)

$\begin{array}{l} \quad\quad\dfrac{x}{6-x}=\dfrac{4}{3}\\\\ \therefore\quad 3x=24-4x\\\\ \quad\quad 7x=24\\\\ \quad\quad x=\dfrac{24}{7}=3.429 \end{array}$

(e)

$\begin{array}{l} \quad\quad\dfrac{x}{x-2}=\dfrac{7}{3}\\\\ \therefore\quad 3x=7x-14\\\\ \quad\quad 4x=14\\\\ \quad\quad x=\dfrac{7}{2}=3.5 \end{array}$

(f)

$\begin{array}{l} \quad\quad\dfrac{x}{x+2.3}=\dfrac{2.3}{3.8}\\\\ \therefore\quad \dfrac{10x}{10x+23}=\dfrac{23}{38}\\\\ \quad\quad 380x=230x+529\\\\ \quad\quad 150x=529\\\\ \quad\quad x=\dfrac{529}{150}=3.527 \end{array}$

7. Find the unknown marked lengths in the figure.

$\begin{array}{l} \quad\quad CD\ \text{bisects}\ \angle ACB\\\\ \therefore\quad \dfrac{AD}{DB}=\dfrac{AC}{BC}\\\\ \therefore\quad \dfrac{x}{24}=\dfrac{20}{30}\\\\ \therefore\quad x=16\\\\ \quad\quad AE\ \text{bisects}\ \angle CAF\\\\ \therefore\quad \dfrac{CE}{BE}=\dfrac{AC}{AB}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{20}{x+24}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{20}{40}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{1}{2}\\\\ \quad\quad 2y=30+y\\\\ \therefore\quad y=30\\\\ \end{array}$

8. $A B=12$ cm, $B C=9$ cm $C A=7$ cm. $B D$ bisects $\angle B$ and $A G=A D$, $C H=C D .$ Calculate $B G, B H$. Does $G H \parallel A C$ ?

$\begin{array} \quad\quad \text{Let}\ AG=AD=y\ \text{and}\ CH=CD=x.\\\\ \therefore\quad x+y=7\Rightarrow y=7-x\\\\ \quad\quad \text{Since}\ BD\ \text{bisects}\ \angle ABC\\\\ \therefore\quad \dfrac{AD}{DC}=\dfrac{AB}{BC}\\\\ \quad\quad \dfrac{y}{x}=\dfrac{12}{9}\\\\ \quad\quad \dfrac{7-x}{x}=\dfrac{4}{3}\\\\ \therefore\quad 21-3x = 4x\\\\ \quad\quad 7x=21\\\\ \therefore\quad x = 3\\\\ \therefore\quad y = 7-x=7-3=4 \end{array}$

9. In $\triangle A B C, D E \parallel B C$, $A D=2.7$ cm, $D B=1.8$ cm and $B C=3$ cm. Prove that $B E$ bisects $\angle A B C$.

$\begin{array} \quad\quad DE\parallel BC\\\\ \therefore\quad \dfrac{AE}{EC}=\dfrac{AD}{DB}\\\\ \quad\quad \dfrac{AE}{EC}=\dfrac{2.7}{1.8}=\dfrac{3}{2}\\\\ \quad\quad \text{But}\ \dfrac{AB}{BC}=\dfrac{2.7+1.8}{3}=\dfrac{3}{2}\\\\ \therefore\quad \dfrac{AE}{EC}=\dfrac{AB}{BC}\\\\ \therefore\quad BE\ \text{bisects}\ \angle ABC. \end{array}$

10. In a parallelogram $A B C D$, $A B=3.6$ cm, $B C=2.7$ cm, $A X=3.2$ cm, $X C=2.4$ cm. Prove that $\triangle B C Y$ is isosceles.

$\begin{array}{l} \quad\quad \dfrac{AX}{XC}=\dfrac{3.2}{2.4}=\dfrac{4}{3}\\\\ \quad\quad \dfrac{AB}{BC}=\dfrac{3.6}{2.7}=\dfrac{4}{3}\\\\ \therefore\quad \dfrac{AX}{XC}=\dfrac{AB}{BC}\\\\ \therefore\quad BY\ \text{bisects}\ \angle ABC.\\\\ \therefore\quad \alpha = \beta\\\\ \quad\quad \text{Since}\ DC\parallel AB,\\\\ \quad\quad \alpha = \gamma\quad (\text{alternate}\ \angle\text{s})\\\\ \therefore\quad \beta =\gamma\\\\ \therefore\quad \triangle BCY \text{is isosceles.} \end{array}$

Angle bisector theorem အောက်တွင် မေးထားသော မေးခွန်း ဖြစ်သောကြောင့် Angle bisector theorem ဖြင့် သက်သေပြခြင်း ဖြစ်သော်လည်း သဏ္ဌာန်တူ ဥပဒေသကို သုံး၍လည်း သက်သေပြနိုင်ပါသည်။

$\begin{array}{l} \quad\quad \text{Since}\ DC\parallel AB,\\\\ \quad\quad \triangle AXB \sim \triangle CXY\\\\ \therefore\quad \dfrac{AB}{CY}=\dfrac{AX}{CX}\\\\ \therefore\quad \dfrac{3.6}{CY}=\dfrac{3.2}{2.4}\\\\ \quad\quad CY=\dfrac{3}{4} \times 3.6 = 2.7\ \text{cm}\\\\ \therefore\quad BC = CY\\\\ \therefore\quad \triangle BCY \text{is isosceles.} \end{array}$

11. Calculate $𝐵𝐷$ and $𝐷𝐶$ in terms of $𝑎, 𝑏, 𝑐$.

$\begin{array} \quad\quad BD + DC = a\Rightarrow DC = a -BD\\\\ \quad\quad BY\ \text{bisects}\ \angle ABC.\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{AB}{AC}\\\\ \therefore\quad \dfrac{BD}{a - BD}=\dfrac{c}{b}\\\\ \quad\quad b\cdot BD=ac-c\cdot BD\\\\ \quad\quad b\cdot BD + c\cdot BD=ac\\\\ \quad\quad (b+c) BD =ac\\\\ \therefore\quad BD=\dfrac{ac}{b+c}\\\\ \therefore\quad DC=a-\dfrac{ac}{b+c}=\dfrac{ab}{b+c} \end{array}$

12. Given : $A H$ bisects $\angle B A C$ in $\triangle A B C$. $E H \parallel A C$

Prove : $\displaystyle \frac{B E}{E A}=\displaystyle \frac{B A}{A C}$

$\begin{array} \quad\quad \text{Since}\ AH\ \text{bisects}\ \angle ABC,\\\\ \quad\quad \dfrac{BA}{AC}=\dfrac{BH}{HC}\\\\ \quad\quad \text{Since}\ EH\parallel AC,\\\\ \quad\quad \dfrac{BE}{EA}=\dfrac{BH}{HC}\\\\ \therefore\quad \dfrac{BE}{EA}=\dfrac{BA}{AC} \end{array}$

13. Given : In $\triangle A B C, B M=M C$.

$M X$ bisects $\angle A M B$ .

$M Y$ bisects $\angle A M C$.

Prove : $X Y \parallel B C$ .


text:
eTheses Repository

# Investigating phagosome dynamics of microbial pathogens

Smith, Leanne May (2015)
Ph.D. thesis, University of Birmingham.

 PDF (19Mb)

## Abstract

Many microbial pathogens are able to evade killing by phagocytes of the innate immune system. This thesis focuses on two pathogens: the fungal pathogen $$Cryptococcus$$ $$neoformans$$ and the bacterial pathogen $$Streptococcus$$ $$agalactiae$$. $$C$$. $$neoformans$$ causes severe cryptococcal meningitis in mostly immunocompromised hosts, such as those with HIV infection. In contrast, $$S$$. $$agalactiae$$ is the leading cause of neonatal sepsis and meningitis. The interaction between macrophages and these pathogens is likely to be critical in determining dissemination and outcome of disease in both instances.
A collection of $$S$$. $$agalactiae$$ clinical isolates, ranging in origin from colonisation cases to severe infection cases, were tested for their ability to persist with a macrophage cell line. Surprisingly, persistence within macrophages was a characteristic shared by all of the isolates tested. Furthermore, by investigating the $$Streptococcus$$-containing phagosome, it was revealed that streptococci are able to manipulate the acidification of macrophage phagosomes.
Similarly, the maturation of phagosomes containing the fungal pathogen $$C$$. $$neoformans$$ was explored. Cryptococci are shown to be able to manipulate the phagosome they reside within. This is driven by modified acquisition of Rab GTPases to the phagosome, as well as altered acidification and cathepsin activity within $$Cryptococcus$$-containing phagosomes.

Type of Work: Ph.D. thesis. May, Robin C. and Krachler, Anne-Marie Colleges (2008 onwards) > College of Life & Environmental Sciences School of Biosciences QR Microbiology University of Birmingham 4937
This unpublished thesis/dissertation is copyright of the author and/or third parties. The intellectual property rights of the author or third parties in respect of this work are as defined by The Copyright Designs and Patents Act 1988 or as modified by any successor legislation. Any use made of information contained in this thesis/dissertation must be in accordance with that legislation and must be properly acknowledged. Further distribution or reproduction in any format is prohibited without the permission of the copyright holder.

Repository Staff Only: item control page


text:
# On gerrymandering

According to the U.S. Constitution, congressional districts are supposed to provide proportional representation. Practically speaking, that means that districts should be contiguous, contain roughly the same number of inhabitants, and (based on the Voting Rights Act of 1965) roughly the same racial makeup. State legislative districts are also drawn to similar standards.

State Legislatures redraw district boundaries every 10 years (after the census) to relect population changes. But due to the partisan nature of legislatures, redistricting often leads to gerrymandering - the drawing of convoluted boundaries to achieve partisan control - to consolidate control for one party.

Gerrymandering is unduly complex and can be unfair, which is why most people dislike it.

I wondered if there was a way to detect and possibly quantify gerrymandered district boundaries by considering boundary regularity. So I built a database of districts and boundary information in PostGIS from Census Bureau data which I could query.

### District Irregularity Index

I wanted to find a mathematical way to compare how gerrymandered two districts might be. Note that irregular district boundaries do not necessarily imply gerrymandering. But on the other hand, gerrymandered boundaries usually appear more irregular. So let’s try to define a metric that can be calculated for any geographic boundary - we’ll call it the District Irregularity Index.

One way to detect irregular districts might be to compare the perimeter of a district to its area. The idea here is that a “simple” district boundary, such as a square, would have a much lower perimeter-to-area ratio than a complex shape like a 5-point star.

The figure above shows a 5-point star that fits inside a square. The star’s perimeter is the same as that of the square, but its area is less than one third.

However, one problem with simply dividing the perimeter by the area is that comparing districts identical except in size will not produce an identical irregularity index. The irregularity index for a square with sides of length 10 would be 40 / 100 = 0.4, compared to a square with sides of length 100 which would be 400 / 10000 = 0.04 . We can solve this issue by dividing by the square root of the area. So 40 / sqrt(100) = 4 , and 400 / sqrt(10000) = 4. Finally, if we divide by 4, we get a nice round number 1 for the irregularity index of a square.

Next, I computed the District Irregularity Index for a bunch of state legislative districts. I queried for 0.25*ST_Perimeter(boundary)/SQRT(ST_Area(boundary)). (District boundaries were stored in the PostGIS database as multipolygons in a geography column called boundary.) Unsurprisingly, the simplest (most “square”) districts - for example, Michigan House of Representatives District 37 and Ohio House of Representatives District 1 - scored around 1.

But on the other end of the spectrum, districts like Alaska State Senate District R and Maryland House of Delegates District 37B scored very high - not because they are gerrymandered, but because they are coastal districts that include many islands. When each coastal island gets its own polygon, the irregularity index can be high for non-gerrymandered districts.

To address this, I added one more term to the formula: divide by the total number of polygons in the boundary. For most land-locked districts, this will be 1, but for districts with complex coastlines, it will help reduce the irregularity index value.

So my final formula is DistrictIrregularityIndex(boundary) = 0.25 * ST_Perimeter(boundary) / SQRT(ST_Area(boundary)) / ST_NumGeometries(boundary::geometry)

Running this query turned up some very interesting districts, including:

Again, just because these district shapes are irregular doesn’t mean they are gerrymandered… but there is a higher likelihood they might be.

### Alabama Redistricting

In 2012, Alabama redrew its state legislative district boundaries (following the 2010 census). The Alabama Legislative Black Caucus filed suit against the state, alleging racial gerrymandering in the new district plan. After a lengthy court battle, a federal court upheld their claim and ordered the legislature to redraw 3 Senate districts and 9 House of Representatives districts. The districts were redrawn and took effect in 2017.

I looked at the Alabama state legislative districts from pre- and post-2017 redistricting, applying the District Irregularity Index to see if, in fact, the pre-2017 districts were more gerrymandered than after 2017. Here are the findings:

Alabama state legislative district irregularity

Irregularity Index Avg Min Max Std Dev
pre-2017 (Senate) 2.0206 0.3073 3.1935 0.5515
pre-2017 (House) 2.089 0.3671 3.4122 0.5495
post-2017 (Senate) 1.8309 0.8178 2.6336 0.4345
post-2017 (House) 1.9162 0.765 3.0254 0.4705

After the 2017 redrawing, the average district irregularity decreased almost 10% for both the house and senate. Additionally, the most irregular Alabama districts became less so by over 10%. So the metric seems reasonable.

### District Irregularity Analysis

Next, I computed the irregularity index for all congressional and state legislative districts. Full results are available here, but I will present some findings below.

The most irregular Senate districts are in Virginia and Louisiana, the least irregular are in Puerto Rico and Wisconsin.

The most irregular House districts are in Alabama and Kentucky, the least irregular are in Alaska and Wisconsin.

For Congressional districts, the graph above is interesting because it has a tail made up of single-district coastal states and territories. Their average irregularity is low because of the division by number of polygons. Among the other states, Wyoming and South Dakota (both nearly-square states with only one congressional district) have low average irregularity, and Ohio and Arkansas have high average irregularity.

Ohio’s congressional districts indeed appear very irregular, and are widely recognized as gerrymandered. In November 2016, 51.69% of Ohio residents voted for the Republican presidential candidate, compared to 43.56% for the Democratic candidate… but Republicans carry 12 of the 16 Congressional districts (75%).

Arkansas, on the other hand, doesn’t appear terribly gerrymandered. Its high irregularity index is due to the complex boundary along the Mississippi river on the eastern side of the state.

Interestingly, my home state of Wisconsin, in which Democrats have accused Republicans of gerrymandering, has one of the lowest average irregularities for state legislative districts, and is around the middle of the pack for congressional districts. That doesn’t necessarily mean Wisconsin’s legislative districts aren’t gerrymandered, but other factors could be at play.

### Conclusion

The District Irregularity Index alone is not sufficient to accurately identify gerrymandered districts, but it can help focus the search. State legislatures might use something like the irregularity index, along with other metrics, to compare new redistricting plans to past boundaries, and ensure they are not making new districts more gerrymandered than before.

Drawing fair districts is a complex problem which is easy to criticize but hard to get right. Earlier this year, the Supreme Court declined to hear several cases involving partisan gerrymandering - due partly to the difficulty in quantifying the impact of various district boundaries. With those decisions, and no clear way to quantify the extent of partisan gerrymandering, we will likely continue to see and hear more on this controversial topic in years to come.

### Future work

There are a number of possible improvements to the District Irregularity Index. One might be to identify boundary segments near bodies of water and reduce how much those segments factor into the perimeter calculation. For example, a district boundary must necessarily be somewhat irregular if it borders a river.

With a relatively good irregularity metric, we could combine other data (like population and demographic data) to create a neutral heuristic for ranking the degree to which potential district boundaries are gerrymandered. Such a heuristic could then be used with an optimization method (such as a genetic algorithm) to automate the creation redistricting plans, and to evaluate them.


text:
# limiting base current to BJT

I'm actually working with a darlington pair of BJTs.

My question is if I need to put a 1kohm resistor between my MCU pin and the BASE to use it as a switch. Since the Hfe ratio for my transistor is quite large, and the collector current will be limited to a reasonable level, does that mean that the base current is dependent on the collector current at all times and thus I can eliminate the usage of this current limiting resistor?

No, it's the collector current that is dependent on the base current, not the other way around. No matter what the collector current is, the base current is $\dfrac{V_{MCU} - V_{BE}}{R}$.
Keep in mind that $V_{BE}$ will be twice the value of another transistor, as there are two junctions between base and emitter.

But it's true that the collector current is what you want in the end. So to find the resistor value (don't just pick 1k), you calculate $I_B=\dfrac{I_C}{H_{FE}}$. If you want $I_C$ = 2A and $H_{FE}$ = 400, then your $I_B$ will have to be $\dfrac{2A}{400}=5mA$. This is a value your microcontroller will be able to deliver, but always check the datasheet.

To put it all together, $R=\dfrac{H_{FE}}{I_C}\times (V_{MCU} - V_{BE})$.

edit
Olin is right about the resistor value being the maximum, i.e. the base current being minimum. For many parameters in a datasheet you'll find more than one value, like typical and maximum or minimum. You should always calculate for worst case conditions, and it may require some logical thinking to find out whether worst case is minimum or maximum for a particular parameter.

Take $H_{FE}$. In my example I picked a value of 400. As higher is usually better datasheets often mention a minimum value. What if it's higher? The base current won't be different, so the collector current will be higher. If you drive the transistor in saturation $I_C$ will no longer be determined by the transistor, but by the load's impedance will be a limiting factor. So, while the transistor would very much like to draw a larger collector current, it won't change. So you think you're safe; the minimum specified $H_{FE}$ is fine, higher is still OK. There's something else to consider, however: $H_{FE}$ is not constant, it varies with $I_C$, and the datasheet should have a graph for this. So check this for the wanted collector current.
$V_{BE}$. Two PN junctions, so that's 2 x 0.65V = 1.3V. Olin found that a 300$\Omega$ base resistor should be fine, in fact leaves some margin. But when I look at the datasheet for the TIP110 it says $V_{BE}$ may be as high as 2.8V! That would result in a base current of $\dfrac{3.3V - 2.8V}{300 \Omega}=1.7mA$, and that's too little to get the wanted $I_C$ of 2A: $400 \times 1.7mA$ is only 670mA.

You're getting the idea. Don't simply use typical values, but make sure that your circuit still works with components with extreme parameter values. This is not so much of a problem with projects where you only build 1 device: you can see what's wrong and adjust. For production you have no choice: always design for worst case.

• Vbe would actually be a single transistor voltage drop. regarding the use of the resistor, it could also depend on whether the MCU IO pin is in push/pull mode or not. If it is, a resistor is probably not needed. – dhsieh2 Jun 19 '11 at 16:09
• @dhsieh - The datasheet for the TIP110 mentions a $V_{BE(ON)}$ of maximum 2.8V. You could never get such a high voltage with just a single PN junction. – stevenvh Jun 19 '11 at 17:08
• 2.8V looks extremely high, even for a double junction. Can you explain this high value somehow? I think I would try to measure it and if I can't, call my supplier's FAE for confirmation. – Federico Russo Jun 19 '11 at 17:20
• Wat's the difference with dfrac? – clabacchio Apr 3 '12 at 7:39
• @clabacchio - dfrac displays larger. frac may be hard to read, especially if numerator or denominator have terms with subscripts. Check edit history to see what frac looked like – stevenvh Apr 3 '12 at 7:43

To go a little further, Stevenh's calculation shows you the maximum base resistor. It's a good idea to allow for some margin and provide a little more base current (a little lower base resistor) than the absolute minimum required to get the desired collector current.

Let's expand the Stevenh's example and get some real numbers. Let's say the processor is running from a 3.3V supply. A darlington has two B-E junctions between its base and emitter, so let's say the overall B-E drop is 1.3V. That leaves 2.0V accross the base resistor. 2V / 5mA = 400 Ohms. If you're really sure about Hfe being 400 over your operating range and that you don't need more than 2A collector current, then you can use only a little lower resistor, like the common value of 360 Ohms. For more margin, use less, like 300 Ohms maybe.

Now you need to go back and see what the load on the micro is. 2V / 300 Ohms = 6.7mA. That will be OK for most micros, especially if its a PIC which tend to have particularly good output current capability. However, I've seen some micros that are specified for less than 6.7mA, so you have to check and possibly adjust things.

One thing to consider with darlingtons is that they are slow to turn off. You say this is for a switching application, so turn off time could matter. If you're just driving a relay then this is no problem, but if you're trying to do 10s of kHz PWM then this is probably not what you want to use.

Another problem with darlingtons is the rather high on-state voltage. It is one B-E drop plus one saturated C-E drop, maybe 900mV but could easily be more at high currents. At 1V the transistor would dissipate 2W with 2A collector current. That will require some sort of heat sinking or at least something like a TO-3 case mounted to the chassis or some metal.

There may very well be better ways to switch what you want without using a darlington.


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# Threshold union: generalizing unions and intersections

### Part I

So, yesterday I was wondering about operations on a set of sets (I’ll explain the Tweet content below, so you can just ignore it for now).

For example, consider that we have 3 sets:

• $A = \{a, b, c\}$
• $B = \{a, b, d\}$
• $C = \{a\}$

and we build a set $S = \{A, B, C\}$ with the previous sets.

The union of all sets has all elements:

$$\bigcup S = \{a, b, c, d\}$$

The intersection of all sets has the elements common to all sets:

$$\bigcap S = \{a\}$$

Now, we can generalize these definitions and have:

• an element belongs to $\bigcup S$ if it belongs to at least 1 set in $S$
• an element belongs to $\bigcap S$ if it belongs to at least $|S|$ sets in $S$ 1

Once we have this, we can define $\bigcup_n$:

• an element belongs to $\bigcup_n S$ if it belongs to at least $n$ sets in $S$

and $\bigcup$ and $\bigcap$ can be written in terms of $\bigcup_n$:

• $\bigcup S = \bigcup_1 S$
• $\bigcap S = \bigcup_{|S|} S$

So, my question was: What’s the name of $\bigcup_n$?

After much googling and no success, it came to mind that maybe this was not about operations on sets, but instead on multisets:

Unlike a set, a multiset, (and quoting Wikipedia) “allows for multiple instances for each of its elements”.

In a multiset we keep count of the number of instances of each element - this number is called multiplicity. I’ll use $e_m$ to represent an element $e$ with multiplicity $m$.

So, with sets $A$, $B$ and $C$, we can build a multiset $\{a_3, b_2, c_1, d_1\}$.

Using these concepts, we can redefine $\bigcup_n$:

• an element belongs to $\bigcup_n S$ if its multiplicity in the multiset is at least $n$

### Part II

Today, I was back on my quest. This concept should be something classical, with like 100 years. I was convinced that with this new insight (the multiset), I would finally find the name of $\bigcup_n$.

What was I not expecting, was to find it in a recent (2005) multi-party computation (MPC) paper. The paper is called Privacy-Preserving Set Operations and is authored by Lea Kissner (@LeaKissner) and Dawn Song (@dawnsongtweets).

The high-level idea of MPC is to have a set of players computing some function on some inputs. The inputs are to be kept private, and only the function output should be revealed.

So, if the inputs are the sets $A, B, C$ in $S$, and we’re computing their intersection, then only $\bigcap S = \{a\}$ should be known in the end.

In Section 6.2, I found the following2:

We define the Threshold Set-Union problem as follows: all players learn which elements appear in the combined private input of the players at least a threshold number $t$ times. For example, assume that $a$ appears in the combined private input of the players $15$ times. If $t = 10$, then all players learn $a$. However, if $t = 16$, then no player learns $a$.

With this, I declare my quest over, and I’ll be calling $\bigcup_n$ threshold union.

What do you think? Is there a better name? Has this concept been named elsewhere?

1. $|S|$ represents the size of the set. In the example, $|S| = 3$. ^
2. Although, there’s no mention of multiset in this definition, nor in the title, the paper contains several and I guess that’s why it came up in my Google search. ^


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# Quaternions as roots

So, I StumbledUpon this really cool site and the last picture looked almost as if it had 3D structure. This reminded me of another website where I saw pictures of the order-8 Mandelbulb. I got to thinking about how to do a 3D version of the pictures in that first site and quaternions came to mind.

I know that $\mathbb{C}$ is closed, but is there a way to have quaternions as roots of some exotic polynomial or whatever?

-

Actually all the quaternions are zeros of quadratic polynomials with real coefficients. If $q=a+ib+jc+kd$, with $a,b,c,d$ real, then $$(x-q)(x-\overline{q})=x^2-2ax+(a^2+b^2+c^2+d^2)$$ has $q$ as a zero, and also has real coefficients. Here $\overline{q}=a-ib-jc-kd$ is the 'conjugate' quaternion.

In a sense the quaternions are just a bunch of copies of $\mathbf{C}$ pointing in different directions. All the copies share the real axis, but we can freely rotate the imaginary' axis in the 3D-space spanned by $\{i,j,k\}$. Whenever $b^2+c^2+d^2=1$ the quaternion $u=ib+jc+kd$ satisfies the equation $$u^2=-1.$$ Thus $u$ can take the role of the complex number $i$ in the sense that the mapping $x+iy\mapsto x+uy$, ($x,y\in\mathbf{R}$) is a monomorphism of rings from the complex number into the quaternions, so its image is a copy of $\mathbf{C}$.

(Yes, there is more to quaternions than just a bunch of copies of $\mathbf{C}$', but for the purposes of them being zeros of polynomials with real coefficients this point of view is enough.)

-
Hmmmm...very interesting. I will sleep on this and hope that I dream about how to use this graphically... :P – El'endia Starman Jun 8 '11 at 6:55

If you allow the coefficients of the polynomial to be quaternions, then some of the roots must be quaternions as well. However, since they do not commute, you need to represent them in a nice way, preferably as matrices.

Finding roots to matrix equations should not be too difficult, but you will need to solve a system of non-linear polynomial equations.

An analogue of the pictures might be to use matrix coefficients, where each entry is -1 or 1?

-
Hmmm...can you provide an example so I have a clearer picture of what you're talking about? – El'endia Starman Jun 8 '11 at 6:33


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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# Stress And Strain

You may have noticed that certain objects stretch easily, but stretching objects such as an iron rod sounds impossible, right? In this article, we will help you understand why a few objects are more malleable than others. Mainly, we will discuss Stress-Strain Curve because it help us know the amount of load or stress that a material can handle before it stretches and breaks.

## What is Stress?

Stress is defined as force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation and that permits an accurate description and prediction of elastic, plastic, and fluid behaviour.

Stress is given by the following formula:

$$\begin{array}{l}\sigma = \frac{F}{A}\end{array}$$

where,

$$\begin{array}{l}\sigma\end{array}$$
is the stress applied, F is the force applied and A is the area of the force application.

The unit of stress is

$$\begin{array}{l}N/m^{2}\end{array}$$
.

### Types of Stress

Stress applied to a material can be of two types as follows:

#### Tensile Stress

The external force per unit area of the material resulting in the stretch of the material is known as tensile stress.

#### Compressive Stress

Compressive stress is the force that is responsible for the deformation of the material such that the volume of the material reduces.

## What is Strain?

Strain is the amount of deformation experienced by the body in the direction of force applied, divided by the initial dimensions of the body.

The following equation gives the relation for deformation in terms of the length of a solid:

$$\begin{array}{l}\epsilon = \frac{\delta l}{L}\end{array}$$

where,

$$\begin{array}{l}\epsilon\end{array}$$
is the strain due to the stress applied,
$$\begin{array}{l}\delta l\end{array}$$
is the change in length and L is the original length of the material.

The strain is a dimensionless quantity as it just defines the relative change in shape.

### Types of Strain

Strain experienced by a body can be of two types depending on stress application as follows:

#### Tensile Strain

The deformation or elongation of a solid body due to applying a tensile force or stress is known as Tensile strain. In other words, tensile strain is produced when a body increases in length as applied forces try to stretch it.

#### Compressive Strain

Compressive strain is the deformation in a solid due to the application of compressive stress. In other words, compressive strain is produced when a body decreases in length when equal and opposite forces try to compress it.

## Stress-Strain Curve

When we study solids and their mechanical properties, information regarding their elastic properties is most important. We can learn about the elastic properties of materials by studying the stress-strain relationships, under different loads, in these materials.

The material’s stress-strain curve gives its stress-strain relationship. In a stress-strain curve, the stress and its corresponding strain values are plotted. An example of a stress-strain curve is given below.

### Explaining Stress-Strain Graph

The different regions in the stress-strain diagram are:

#### (i) Proportional Limit

It is the region in the stress-strain curve that obeys Hooke’s Law. In this limit, the stress-strain ratio gives us a proportionality constant known as Young’s modulus. The point OA in the graph represents the proportional limit.

#### (ii) Elastic Limit

It is the point in the graph up to which the material returns to its original position when the load acting on it is completely removed. Beyond this limit, the material doesn’t return to its original position, and a plastic deformation starts to appear in it.

#### (iii) Yield Point

The yield point is defined as the point at which the material starts to deform plastically. After the yield point is passed, permanent plastic deformation occurs. There are two yield points (i) upper yield point (ii) lower yield point.

#### (iv) Ultimate Stress Point

It is a point that represents the maximum stress that a material can endure before failure. Beyond this point, failure occurs.

#### (v) Fracture or Breaking Point

It is the point in the stress-strain curve at which the failure of the material takes place.

## Hooke’s Law

In the 19th-century, while studying springs and elasticity, English scientist Robert Hooke noticed that many materials exhibited a similar property when the stress-strain relationship was studied. There was a linear region where the force required to stretch the material was proportional to the extension of the material, known as Hooke’s Law.

Hooke’s Law states that the strain of the material is proportional to the applied stress within the elastic limit of that material.

Mathematically, Hooke’s law is commonly expressed as:

F = –k.x

Where F is the force, x is the extension in length, and k is the constant of proportionality known as the spring constant in N/m.

See the video below to learn about the stress-strain curve and Hooke’s law.

## Elastic Moduli of Materials

The following table lists Young’s modulus, shear modulus and bulk modulus for common materials.

 Material Young’s modulus (E) in GPa Shear modulus (G) in GPa Bulk modulus (K) in GPa Glass 55 23 37 Steel 200 84 160 Iron 91 70 100 Lead 16 5.6 7.7 Aluminium 70 24 70

Note: GPa is gigapascal and 1 GPa = 1,00,00,00,000 Pa.

## Frequently Asked Questions – FAQs

### What does the stress-strain curve show?

A stress-strain curve is a graphical way to show the reaction of a material when a load is applied. It shows a comparison between stress and strain.

### What is the use of the stress-strain diagram?

The stress-strain diagram provides a graphical measurement of the strength and elasticity of the material. Also, the behaviour of the materials can be studied with the help of the stress-strain diagram, which makes it easy to understand the application of these materials.

### What is the relation between stress and strain?

The relation between stress and strain is that they are directly proportional to each other up to an elastic limit. Hooke’s law explains the relationship between stress and strain. According to Hooke’s law, the strain in a solid is proportional to the applied stress and this should be within the elastic limit of that solid.

### What is the yield point in stress-strain graph?

Yield point is defined as the point at which the material starts to deform plastically. After the yield point is passed, permanent plastic deformation occurs.

### How to make a stress-strain curve?

The stress-strain curve is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined.

Stay tuned to BYJU’S and Fall in Love with Learning!

Test your knowledge on Stress And Strain


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## c ++ operator overload doesn't bring me back well

The problem is that your subtraction operator is conceptually wrong.

If you subtract two numbers:

C = A – B

The result is not one of the original numbers, that is, C It is not even A neither Bis "a new numberMsgstr "" "When you work with other types (such as matrices), the same thing happens.

Because of this, you cannot return a reference The matrix in the subtraction must be a new matrix:

``````// Instancia, no referencia (&), const porque no modifica el objeto original ni el proveído
Matriz Matriz::operator-(const Matriz& m)  const
{
``` Usually these operations are written as free operators, not as member operators: Matriz operator-(const Matriz &a, const Matriz &b) { Matriz resultado( … ); // … operaciones return resultado; } Your code generated undefined behavior because you returned a reference to an object that no longer exists. ```
``` ```
``` Author AdminPosted on February 13, 2020Categories ArticlesTags bring, doesn39t, operator, overload ```
``` I want to bring the main menu into a vertical position after the slider I am creating a custom design and I want to change the position of the main menu under the slider in the vertical position How can I do that???? Author AdminPosted on February 7, 2020Categories ArticlesTags bring, main, menu, position, Slider, vertical How do you bring the deputy department head back with the worker who has served the longest? Scenario / problemI have a spreadsheet that stores my ID when they start or leave a department. This can be seen in SQL Fiddle: http://www.sqlfiddle.com/#!9/d0c982/1/0 The following code creates the table and inserts the test data CREATE TABLE `Worker_Department` ( `Worker_ID` Integer NOT NULL , `Department_ID` Integer NOT NULL, `Position` Text NOT NULL, `Start_Date` datetime NOT NULL, `Leave_Date` datetime ); INSERT INTO `Worker_Department`(`Worker_ID`,`Department_ID`,`Position`,`Start_Date`) VALUES (10,100,'Leader','1980-11-11'), (20,200,'Leader','1980-11-11'); INSERT INTO `Worker_Department`(`Worker_ID`,`Department_ID`,`Position`,`Start_Date`,`Leave_Date`) VALUES (30,200,'Administrator','1980-11-11', '2014-02-02'), (40,200,'Receptionist','1975-11-11', '2014-02-02'); INSERT INTO `Worker_Department`(`Worker_ID`,`Department_ID`,`Position`,`Start_Date`) VALUES (50,300,'Administrator','2014-02-02'), (30,100,'Administrator','2014-02-02'); Code (SQL): I have to write a query that identifies the longest-serving employee currently in a department (no end date). However, those with the "leader" position are not allowed to be the longest serving person. From the results of this query, I then have to determine the head of the department in which the longest-serving employee is currently working. Expected resultViewing the test data provided: The 40th employee cannot be the longest employed since he no longer works for the company. Workers 10 and 20 cannot be the longest-serving employee since both hold the leadership position (however, this entitles them to be the department head based on who the longest-employed employee is). Worker 30 is the longest employed because there is a greater difference between the current date and the oldest start date compared to worker 50. Worker 30 is currently working in Department 100. That means that Worker 10 is the department head of the department with the longest employed worker at the moment The query output would be something like| Worker_ID| —————| 10 If this table is linked to another table as a foreign key, the selection can be changed so that details about this conductor (name, telephone, address) are included. Ongoing progressThe following query shows the employees currently employed (employees without a vacation date), the department in which they work and their role in this department. SELECT Worker_ID, Department_ID, Position FROM Worker_Department WHERE position != 'Leader' AND leave_date is null Code (SQL): The following query returns the difference between the current date and the minimum start date of the worker. However, this includes those who do not have a zero end date (not current). SELECT Worker_ID, DATEDIFF(Now(), Min(Start_Date)) as NowSubMin FROM Worker_Department WHERE position != 'Leader' GROUP BY Worker_ID Code (SQL): Finally, both queries were used to create the following query, which is to return current employees along with the date difference between the current date and their first start date SELECT Worker_ID, DATEDIFF(Now(), Min(Start_Date)) as NowSubMin FROM Worker_Department WHERE position != 'Leader' GROUP BY Worker_ID HAVING Worker_ID IN (SELECT Worker_ID FROM Worker_Department WHERE position != 'Leader' AND leave_date is null) Code (SQL): Author AdminPosted on February 7, 2020Categories ArticlesTags bring, Department, deputy, longest, served, worker Buy a brand new bike from India and bring it to Bangladesh it is possible to buy a brand new bike from India and bring it to Bangladesh. If possible, how much are the customs fees? Author AdminPosted on February 6, 2020Categories ArticlesTags Bangladesh, bike, Brand, bring, Buy, India Organic Bring Real Visitor & # 39; s 5000 Targeted Web Traffic for \$ 2 Organic Bring Real Visitor & # 39; s 5000 Targeted Web Traffic 5000 web traffic 5 days About this gig FREE TRIAL – more details in the FAQ below! Your website is displayed on thousands of constantly different domains suitable for your target group. ☛ Traffic starts within 4-5 hours☛ The delivery is made after you have received all traffic. Usually in about 4 days (may vary from case to case), but we will continue to send bonus traffic for 5 days.Main features: Receive over 6,000 visitors Real visitors with unique IPs. No bots, hits from China, proxies or data centers. The traffic is direct and 100% Adsense SAFE CPA, memberships, landing pages, blogs, etc. are welcome! Trackable on Analytics (UTM tags are gladly accepted) SEO friendly Custom tracking provided Comprehensive customer care You may get sales or opt-ins, but these CAN NOT guaranteed, as well as specific bounce rate time and ranking. We cannot control visitors' actions. , (tagsToTranslate) targeted (t) web (t) traffic Author AdminPosted on January 31, 2020Categories ArticlesTags 5000, amp, bring, Organic, real, targeted, traffic, visitor, Web Integration – Can I bring this integral into a form that contains known functions? In the context of quantum field theory, I face the following $$1$$Integral over Feynman parameters: $$I (a, b) = int_0 ^ infty d alpha frac {1} {F G} arctan frac {F} {G} tag {1}$$ With $$F ( alpha, a, b): = sqrt { alpha (1+ alpha) (ab) ^ 2- alpha a ^ 2 + (1+ alpha) b ^ 2} tag {2 }$$ $$G ( alpha, a, b): = sqrt { alpha left (a-b- alpha (a-b) ^ 2 right)} tag {3}$$ Is there a way to express this integral using polylogarithms or elliptic functions? Usually my (rather primitive) way to achieve this goal is to insert the indefinite integral in Mathematica, to pray that it can solve it, and to set the appropriate limits. But this time Mathematica can't. So I want a more robust way to do that. Author AdminPosted on January 30, 2020Categories ArticlesTags bring, form, functions, integral, integration Why do these Senate Democrats want to bring evidence to court? Isn't that inappropriate? Why is it inappropriate? They recapitulate the evidence they have and ask for new evidence that has only emerged to prove their arguments. The GOP is in a completely deliberate ignorance mode. You are not impartial at all. 19 of the trump cards "employees" were accused or imprisoned; He lied or made false statements over 16,000 times, and yet the fools and other ignoramuses support him without question. When acquitted, he is essentially a king / monarch who can act without any control or consideration. We are back to the 16th century! Author AdminPosted on January 22, 2020Categories ArticlesTags bring, Court, Democrats, Evidence, inappropriate, Isn39t, Senate Moved from the USA to Canada. Bring your own car I am an international student living in the States. I plan to continue my studies in British Columbia this summer. I was wondering if I could bring my car with me or not. If so, how much do I have to pay for it? (i.e. taxes, import fee, …). I'm just trying to see that it's worth bringing it. It is a Honda Accord. Many thanks. Author AdminPosted on January 20, 2020Categories ArticlesTags bring, Canada, Car, moved, USA Can I bring conversations to popups that are not on a channel? I understand how popups can be configured for channels, but what about direct messages? Author AdminPosted on January 16, 2020Categories ArticlesTags bring, Channel, conversations, popups mac – How to bring computers online in Apple Remote Desktop Stack Exchange network The Stack Exchange network consists of 175 Q&A communities, including Stack Overflow, the largest and most trusted online community, where developers can learn, share their knowledge, and pursue careers. Visit Stack Exchange Author AdminPosted on January 16, 2020Categories ArticlesTags Apple, bring, computers, desktop, MAC, online, Remote Posts navigation Page 1 Page 2 … Page 15 Next page ```


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text:
# Problem: A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.020 hour, and the assembly expands from an initial volume of 0.90 L to a final volume of 5.88 L against an external pressure of 1.0 atm.Calculate w.

###### FREE Expert Solution

We’re being asked to calculate the work for a system that has an initial volume of 0.90 L to a final volume of 5.88 L against an external pressure of 1.0 atm.

Recall that work (w) is given by:

$\overline{){\mathbf{w}}{\mathbf{=}}{\mathbf{-}}{\mathbf{P\Delta V}}}$

where:

P = pressure (in atm)

ΔV = Vfinal – Vinitial = change in volume (in L)

###### Problem Details

A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.020 hour, and the assembly expands from an initial volume of 0.90 L to a final volume of 5.88 L against an external pressure of 1.0 atm.

Calculate w.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Internal Energy concept. You can view video lessons to learn Internal Energy. Or if you need more Internal Energy practice, you can also practice Internal Energy practice problems.


text:
# MHF-p

 2017-07-0415:31 [PUBDB-2017-06867] Contribution to a conference proceedings et al Experience with Single Cavity and Piezo Controls for Short, Long Pulse and CW Operation 8th International Particle Accelerator Conference, IPAC17, Meeting locationCopenhagen, Denmark, 14 May 2017 - 19 May 2017 3966-3968 (2017) [10.3204/PUBDB-2017-06867]   We present a compact RF control system for superconducting radio frequency (SCRF) single cavities based on MicroTCA.4 equipped with specialized advanced mezzanine cards (AMCs) and rear transition modules (RTMs). To sense the RF signals from the cavity and to drive the high power source, a DRTM-DWC8VM1 module is used equipped with 8 analog field detectors and one RF vector modulator. [...] OpenAccess: PDF PDF (PDFA); 2017-07-0415:20 [PUBDB-2017-06865] Poster et al Experience with Single Cavity and Piezo Controls for Short, Long Pulse and CW Operation 8th International Particle Accelerator Conference, IPAC17, Meeting locationCopenhagen, Denmark, 14 May 2017 - 19 May 2017   OpenAccess: PDF PDF (PDFA); 2017-06-1515:10 [PUBDB-2017-05127] Journal Article et al A New Type of Waveguide Distribution for the Accelerator Module Test Facility of the European XFEL   The European XFEL, which has been constructed at DESY in Hamburg, Germany, is an X-ray-Free Electron Laser, which provides X-ray light of unprecedented properties for different experiments in physics, chemistry, biology and technology. The XFEL is based on superconducting cavity technology, which is required to accelerate an electron beam up to 17.5 GeV. [...] OpenAccess: PDF PDF (PDFA); 2016-12-1517:04 [PUBDB-2016-06264] Conference Presentation et al Model Based Fast Protection System for High Power RF Tube Amplifiers Used at European XFEL Accelerator 20th IEEE-NPSS Real Time Conference, RT2016, Meeting locationPadua, Italy, 5 Jun 2016 - 10 Jun 2016   The driving engine of the superconducting accelerator of the European X-ray Free-Electron Laser (XFEL) are 27 Radio Frequency (RF) stations. Each of an underground RF station consists from multi-beam horizontal klystron which can provide up to 10MW of power at 1.3GHz. [...] OpenAccess: PDF PDF (PDFA); 2016-12-0710:20 [PUBDB-2016-05971] Report/Conference Presentation (Plenary/Keynote) team, X. c. Commissioning of the European XFEL Injector [TUOCA03] 7th International Particle Accelerator Conference, IPAC16, Meeting locationBusan, Korea, 8 May 2016 - 13 May 2016   The European XFEL Injector consists of an L-Band RF photoinjector, a TESLA type 1.3 GHz module, a 3rd har-monic RF section, a laser heater and an extensive diag-nostic section to determine projected and slice properties of the beam. The commissioning of the complete system has been started in December 2015 after several years of construction. [...] OpenAccess: PDF PDF (PDFA); External link: Fulltext 2016-12-0212:06 [PUBDB-2016-05854] Contribution to a conference proceedings/Contribution to a book et al FPGA Based RF and Piezo Controllers for SRF Cavities in CW Mode Real Time Conference (RT), 2016 IEEE-NPSS20th IEEE-NPSS Real Time Conference, RT2016, Meeting locationPadua, Italy, 5 Jun 2016 - 10 Jun 2016 1-2 (2016) [10.1109/RTC.2016.7543112]   Modern digital low level radio frequency (LLRF) control systems used to stabilize the accelerating field in facilities such as Free Electron Laser in Hamburg (FLASH) or European X-Ray Free Electron Laser (E-XFEL) are based on the Field Programmable Gate Array (FPGA) technology. Presently these accelerator facilities are operated with pulsed RF. [...] Restricted: PDF; External link: Fulltext 2016-11-3016:02 [PUBDB-2016-05793] Poster et al Summary of the Test and Installation of 10 MW MBKS for the XFEL Project 28th Linear Accelerator Conference, LINAC2016, Meeting locationEast Lansing, USA, 25 Sep 2016 - 30 Sep 2016   For the European XFEL project, horizontal multi-beam klystrons (MBK) which produce RF power up to 10 MW,at an RF frequency of 1.3 GHz, 1.5 ms pulse length and 10 Hz repetition rate, were chosen as RF power sources. All MBKs have been manufactured by two companies, 22 tubes from Thales Electron Devices and 7 tubes from Toshiba Electron Tubes & Devices. [...] OpenAccess: PDF PDF (PDFA); 2016-11-1616:57 [PUBDB-2016-05302] Contribution to a conference proceedings/Contribution to a book Choroba, S. Planning and Controlling of the Cold Accelerator Sections Installation in XFEL 7th International Particle Accelerator Conference, IPAC16, Meeting locationBusan, Korea, 8 May 2016 - 13 May 2016 716-718 (2016) [10.3204/PUBDB-2016-05302]   The installation of the main linear accelerator in the 2 km European XFEL (X-Ray Free-Electron Laser) tunnel is currently under way. The accelerator consists of nineso-called cryo-strings. [...] OpenAccess: PDF PDF (PDFA); External link: Fulltext 2016-11-1616:18 [PUBDB-2016-05298] Conference Presentation Katalev, V. Construction of the RF System for the European XFEL Ninth Continuous Wave and High Average Power RF Workshop, CWRF 2016, Meeting locationGrenoble, Deutsches Elektronen-Synchrotron DESY, France, 21 Jun 2016 - 24 Jun 2016   OpenAccess: PDF PDF (PDFA); 2016-11-1616:02 [PUBDB-2016-05297] Poster Apostolov, E. Series Production of the RF Power Distribution for the European XFEL 28th Linear Accelerator Conference, LINAC2016, Meeting locationEast Lansing, Deutsches Elektronen-Synchrotron DESY, USA, 25 Sep 2016 - 30 Sep 2016   The RF power distribution for the European XFEL allows for individual RF power for the 808 superconducting cavities of the European XFEL. It consists of a number of elements, not only waveguide components, but also girders, cables or cooling systems. [...] OpenAccess: PDF PDF (PDFA);


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Coefficient coefficient

Discrete Mathematics Level 3

If a binomial expansion is given by $(1+x)^{10} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_{10} x^{10}.$

then find the value of the expression $(a_0 - a_2 + a_4 - a_6 +a_8 - a_{10} )^2 + (a_1 - a_3 + a_5 - a_7 + a_9 )^2.$

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