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Pełnotekstowe zasoby PLDML oraz innych baz dziedzinowych są już dostępne w nowej Bibliotece Nauki.
Zapraszamy na https://bibliotekanauki.pl

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• Artykuł - szczegóły

Studia Mathematica

2004 | 160 | 2 | 185-200

On the compact approximation property

EN

Abstrakty

EN
We show that a Banach space X has the compact approximation property if and only if for every Banach space Y and every weakly compact operator T: Y → X, the space
𝔈 = {S ∘ T: S compact operator on X}
is an ideal in 𝔉 = span(𝔈,T) if and only if for every Banach space Y and every weakly compact operator T: Y → X, there is a net $(S_γ)$ of compact operators on X such that $sup_{γ}||S_{γ}T|| ≤ ||T||$ and $S_{γ} → I_{X}$ in the strong operator topology. Similar results for dual spaces are also proved.

185-200

wydano
2004

Twórcy

autor
• Department of Mathematics, Agder University College, Serviceboks 422, 4604 Kristiansand, Norway
autor
• Department of Mathematics, Agder University College, Serviceboks 422, 4604 Kristiansand, Norway
autor
• Department of Mathematics, Agder University College, Serviceboks 422, 4604 Kristiansand, Norway


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# How to test which combination of product attributes generates better sales performance?

I am looking for help in identifying what is the best statistical approach to address the following market research question:

(1) I have a dataset from an online website that helps users rent their vacation homes.

For each vacation home I have information on:

• Their attributes (predictor variables): location (city and distance from city center), # of rooms, guest rating (10 stars),price, etc.
• Their sales performance (dependent variables): conversion rate, # of nights sold, sales volume in $(2) The website rents vacation homes mainly for US and Canada customers Question: I would like to be able to identify which combinations of predictor variables are associated with better performance, in order to be more targeted in gathering new homes for the site. Because most of the website users (renters) originate from the US and Canada (and I have a sense that they have different preferences) I would be especially interested in understanding which combinations of attributes work best for each country? I imagine that for Canadian customers you could have 3 combinations of attributes that would work best. For instance, one combination could be: price$50-\$75/day, 2 rooms, far from the city center, etc...

Any help would be really appreciated.

-

There are a number of techniques you can use, but I always have a tendency to lean towards using Classification and Regression trees (rpart in R). They are simple yet robust as it can handle non-linear data and yields trees which are intuitive.

It's worth noting that the caret package mentioned above is a very powerful package to compare a number of models once you have the general structure of your data identified and think that is a great R library as well.

-

I would start with regressing (using OLS) the conversion rate and/or sales volume on the attributes separately for US and Canada (non-numeric attributes would have to be converted to binary flags first). Those variables that have highest absolute value of t-value (ratio of parameter value to its standard error) are the most important ones. One thing to watch out for is multi-collinearity (remove variables that have largest Variance Inflation Factor one by one until multi-collinearity is under control)

Ordinary Least Squares (OLS) regression comes with many tools including Excel where it is known as "Multiple Regression".

One could use more complex techniques as well such as decision trees etc. R package caret provides a convenient framework for that (but can have a steep learning curve if you are not familiar with R)

-

format: plain


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A summary of the automatic evaluation metric for natural language generation (NLG) applications.

The human evaluation considers the aspects of adequacy, fidelity, and fluency, but it is quite expensive.

• Adequacy: Does the output convey the same meaning as the input sentence? Is part of the message lost, added, or distorted?
• Fluency: Is the output good fluent English? This involves both grammatical correctness and idiomatic word choices.

Thus, a useful metric for automatic evaluation in NLG applications holds the promise, such as machine translation, text summarization, image captioning, dialogue generation, poetry/story generation, etc.

Goals for evaluation metrics:

• Low cost: reduce time and money cost.
• Tunable: automatically optimize system performance towards metric.
• Meaningful: give the intuitive interpretation of quality.
• Consistent: repeated use of metric should provide the same results.
• Correct: The metric must rank better systems higher.

Evaluation task: for NLG tasks, given candidate hypothesis and human references, compute the similarity between them and provide the corresponding score.

# BLEU (ACL 2002)

Designing for machine translation evaluation, BiLingual Evaluation Understudy (BLEU) [1] measures the $n$-gram overlap between translation output and reference, by computing the precision for n-grams of size 1 to N (N=4).

BLEU take the geometric mean of the test corpus’s modified $n$-gram precision scores $p_n$, using $n$-grams up to length $N$ and positive weights $w_n$ summing to one. To compute the modified $n$-gram precision, all candidate $n$-gram counts and their corresponding maximum reference counts are collected. The candidate counts are clipped by their corresponding reference maximum value, summed and divided by the total number of candidate $n$-grams. In other words, truncating the $n$-gram counts in candidate with the corresponding maximum in the reference.

Then multiply by an exponential brevity penalty (BP) factor, given $c$ as the length of the candidate translation and $r$ as the effective reference corpus length (summing the best match lengths, i.e., the closest reference sentence length, forr each candidate sentence in the corpus).

Thus,

where $N=4$, and $w_n = 1/N$ is usually uniform weights. That is,

1. The BLEU score ranges from 0 to 1.
2. 1-gram tends to satisfy adequacy, whereas large $n$-gram accounts for fluency.
3. BLEU treats all systems similarly and uses multiple human translators with different styles to circumvent the variety differences such as “East Asian economy” with the reference “economy of East Asia”[1].

# METEOR (WMT@ACL 2004)

Meteor[2] evaluates translation candidates in a more flexible way and calculates the sentence-level similarity scores. For a hypothesis-reference pair, the alignment is constructed by exhaustively identifying all possible matches between the sentences as follows:

1. Exact: identical surface forms.
2. Stem: same stem words (using Snoball Stemmer toolkit).
3. Synonym: share membership in any synonym set in WordNet database.
4. Paraphrase: match if they are paraphrases.

The final alignment solution using a beam search follows:

1. Each word in each sentence is covered by zero or one match.
2. Maximize the number of covered words across both sentences.
3. Minimize the number of chunks, where a chunk is a series of contiguous and identically ordered words in both sentences.
4. Minimize the sum of absolute distances between match start indices in the two sentences.

Content and function words are identical in the hypothesis $(h_c, h_f)$ and reference $(r_c, r_f)$ accoring to the function word list (any word whose relative frequency above 10^-3). For matches $m_i$, count the number of content and function words in hypothesis $m_i(h_c), m_i(h_f)$ and reference $m_i(r_c), m_i(r_f)$. Calculate weighted precision and recall using matcher weights $(w_i,\cdots,w_n)$ and content-function word weight $\delta$:

To account for gaps and differences in word order, a fracmentation penalty uses the total number of match words $m$ and number of chunks ($ch$):

The Meteor score is calculated:

where parameters ${ \alpha, \beta, \gamma, \delta, (w_i, \cdots, w_n)}$ are tuned to maximize correlatin with human judgements.

# ROUGE (ACL 2004)

Recall-Oriented Understudy for Gisting Evaluation (ROUGE)[3] is an automatic evaluation metric for text summarization, consisting of ROUGE-N, ROUGE-L, ROUGE-W, and ROUGE-S.

## ROUGE-N: N-gram Co-Occurrence Statistics

ROUGE-N is an n-gram recall between a candidate summary and a set of reference summaries.

where $\textrm{count}_\textrm{match} (\textrm{n-gram})$ represent the maximum number of n-grams co-occuring in candidate summry and a set of reference summaries. It is consistent with the intuition that a candidate summary that is more similar to many reference summaries is preferred.

vs. BLEU: BLEU is a precision-based measure whereas ROUGE-N is recall-based.

## ROUGE-L: Longest Common Subsequence

### Sentence-level LCS

Given a reference summary $X$ of length $m$, and a candidate summary sentence $Y$ of length $n$, the sentence level Longest Common Subsequence (LCS) is:

### ROUGE-L

ROUGE-L takes the sentence-level normalized pairwise LCS.

## ROUGE-W: Weighted LCS

It measures weighted LCS-based statistics that favor consecutive LCS.

## ROUGE-S: Skip-bigram Co-Occurrence Statistics

ROUGE-S considers the skip-bigram-based F-measure:

where skip-bigram means any pair of words in the sentence regardless of the order and distance.

# CIDEr (CVPR 2015)

Consensus-based Image Description Evaluation (CIDEr)[4] is proposed for evaluating the candidate sentence $c_i$ compared with reference $S_i = \{ s_{i1}, \cdots, s_{im} \}$ for image $I_i$ image captioning. All words in both candidate and reference sentences are first mapped to their stem or root forms.

To penalize the commonly occurred words across all images, TF-IDF weighting for each $n$-gram is applied. Let the count that an $n$-gram $\omega_k$ occurs in a reference sentence $s_{ij}$ be $h_k(s_{ij})$, and that of candidate sentence $c_i$ be $h_k(s_{ij})$. The TF-IDF weighting $g_k (s_{ij})$ for each $n$-gram $\omega_k$ using:

where $\Omega$ is the vocabulary for all $n$-grams and $I$ is the image set. The second term (IDF) measures the word saliency by discounting popular words that are likely to be less visually informative.

The CIDErn score for $n$-grams uses the average cosine similarity between the candidate sentence and reference sentences:

where $\mathbf{g^n}(c_i)$ is a vector formed by $g_k(c_i)$ corresponding to all $n$-grams of length $n$. So is $\mathbf{g^n}(s_{ij})$. $w_n$ takes the uniform weights $1/N$, $N=4$.

# SPICE (ECCV 2016)

Previous evaluation methods are all $n$-gram based, sensitive to $n$-gram overlap. However, $n$-gram overlap is neither necessary nor sufficient for two sentences to convey the same meaning.

Semantic Propositional Image Caption Evaluation (SPICE)[5] measures the semantic meanings rather than $n$-gram based methods.

Given a candidate caption $c$ and a set of reference captions $S = \{ s_1, \cdots, is_m \}$ associated with an image $I$, the goal is to compute the score between $c$ and $S$. SPICE builds scene graph using Stanford Scene Graph Parser. A Probabilistic Context-Free Grammar (PCFG) dependency parser and additional linguistic rules are applied.

Given a set of object class $C$, a set of relation typles $R$, aset of attribute types $A$, a caption $c$, we parse $c$ to a scene graph:

where $O(c) \subseteq C$ is the set of object in $c$, $E(c) \subseteq O(c) \times R \times O(c)$ is the set of hyper-edges of relations, $K(c) \subseteq O(c) \times A$ ia the set of attributes of objects.

Define the function $T$ that returns logical tuples from a scene graph as:

The presicion $P$, recall $R$, SPICE are computed as:

where $\otimes$ denotes the binary matching that returns tha matching tuples in two scene graphs, using the WordNet synonym matching approach of Meteor.

SPICE is also bounded between 0 and 1.

# References

1. 1.Papineni, K., Roukos, S., Ward, T., & Zhu, W. (2001). Bleu: a Method for Automatic Evaluation of Machine Translation. ACL.
2. 2.Denkowski, M.J., & Lavie, A. (2014). Meteor Universal: Language Specific Translation Evaluation for Any Target Language. WMT@ACL.
3. 3.Lin, C. (2004). ROUGE: A Package For Automatic Evaluation Of Summaries. ACL 2004.
4. 4.Vedantam, R., Zitnick, C.L., & Parikh, D. (2015). CIDEr: Consensus-based image description evaluation. 2015 IEEE Conference on Computer Vision and Pattern Recognition (CVPR), 4566-4575.
5. 5.Anderson, P., Fernando, B., Johnson, M., & Gould, S. (2016). SPICE: Semantic Propositional Image Caption Evaluation. ECCV.


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# Sensory Neuroscience: Hearing and speech/Speech

Speech perception is a hot topic because it is critically important to human communication. As well, there are some controversial claims being made. For example, it may be the case that we have two auditory systems: one for speech, one for non-speech.

## The nature of speech soundsEdit

The stream of speech sounds is relatively unbroken during normal, fluent speech. There is no space between the words - detecting word boundaries is a major problem for psycholinguistics.

The International Phonetic Alphabet is an alphabet which does uniquely identify every speech sound by a single character. IPA symbols will be used extensively in this section; you would do well to familiarize yourself with the symbols, and their use.

### DefinitionsEdit

phoneme
The smallest unit of linguistically distinctive sound. Phonemes are not the same thing as letters, morphemes, nor syllables.
letter
An element of written language. It is critically important that there is no ${\displaystyle 1:1}$  relationship between letters of the alphabet and the phonemes of the language.
morpheme
The smallest linguistic unit that has semantic meaning. Work is a morpheme; -ed is another morpheme. Worked is two morphemes.
utterance
A complete unit of spoken language.

### Structure of languageEdit

There are four rough levels of the analysis of language. Beginning from the highest level:

• Syntactic - the utterance has a grammatical structure
• Semantic - a string of phonemes has a particular meaning
• Phonetic - a sound is a string of phonemes
• Acoustic - a sound has certain properties

While the bottom-up processes are obvious, there is significant top-down influence as well. For example, the phonemic restoration effect is very robust, and similar processes can repair input at the level of individual phonemes, words, and even sentences.

If you replace a single phoneme in a sentence with white noise, or a cough, or silence, listeners can still identify the word.

In the sentence "I scream, you scream, we all scream for ice cream." there is no difference between the /aɪ/ in "I scream" and in "ice cream" - the sound is exactly the same in both cases, yet the word boundary is perceived to be in a different location based solely on context effects.

If you insert a semantically unrelated sentence into a paragraph-length discourse, listeners will not get it (but will continue with normal perception after the sentence).

## Measuring speechEdit

### Speech spectrogramEdit

A speech spectrogram of "I owe you."

A speech spectrogram (careful with terminology here!) graphs frequency vs time & the intensity (amplitude, loudness) is coded by darkness. This shows qualitatively how speech sounds unfold over time. Note that consonants are noisy (not loud! but broadband in frequency). Vowels, in contrast have prototyped horizontal bands of energy. These bands are called formants. Near consonants, one or more of the formant curves - this is a formant transition. Most vowels are steady-state; they do not change over time (monophthongs). Some vowel sounds are dipthongs - they transition from one spectrum to another over the course of the vowel.

However, the speech spectrogram isn't very helpful since it is not quantitative. For that, we can use a speech spectrum (careful with terminology!).

### Speech spectrumEdit

A speech spectrum plots energy in dB SPL vs frequency for a short period of time: on the order of 40ms. This is essentially a cross-sectional slice of a speech spectrogram. But now you can't see how the sound unfolds over time!

### Waterfall plotEdit

A waterfall plot is essentially a series of speech spectrums over time, laid out in 3D. This overcomes the limitations of speech spectrograms (shows speech unfolding over time, but isn't quantitative) and also overcomes the limitations of speech spectrums (highly qualitative, but shows only a short snippet of the speech stream).


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All issues

Experimental study of the dynamics of single and connected in a lattice complex-valued mappings: the architecture and interface of author’s software for modeling

pdf (4611K)

The paper describes a free software for research in the field of holomorphic dynamics based on the computational capabilities of the MATLAB environment. The software allows constructing not only single complex-valued mappings, but also their collectives as linearly connected, on a square or hexagonal lattice. In the first case, analogs of the Julia set (in the form of escaping points with color indication of the escape velocity), Fatou (with chaotic dynamics highlighting), and the Mandelbrot set generated by one of two free parameters are constructed. In the second case, only the dynamics of a cellular automaton with a complex-valued state of the cells and of all the coefficients in the local transition function is considered. The abstract nature of object-oriented programming makes it possible to combine both types of calculations within a single program that describes the iterated dynamics of one object.

The presented software provides a set of options for the field shape, initial conditions, neighborhood template, and boundary cells neighborhood features. The mapping display type can be specified by a regular expression for the MATLAB interpreter. This paper provides some UML diagrams, a short introduction to the user interface, and some examples.

The following cases are considered as example illustrations containing new scientific knowledge:

1) a linear fractional mapping in the form $Az^{n} +B/z^{n}$, for which the cases $n=2$, $4$, $n>1$, are known. In the portrait of the Fatou set, attention is drawn to the characteristic (for the classical quadratic mapping) figures of <>, showing short-period regimes, components of conventionally chaotic dynamics in the sea;

2) for the Mandelbrot set with a non-standard position of the parameter in the exponent $z(t+1)\Leftarrow z(t)^{\mu }$ sketch calculations reveal some jagged structures and point clouds resembling Cantor's dust, which are not Cantor's bouquets that are characteristic for exponential mapping. Further detailing of these objects with complex topology is required.

Keywords: holomorphic dynamics, nonlinear dynamics, cellular automata, experimental mathematics, fractal, MATLAB, iterated mappings
Citation in English: Matyushkin I.V., Rubis P.D., Zapletina M.A. Experimental study of the dynamics of single and connected in a lattice complex-valued mappings: the architecture and interface of author’s software for modeling // Computer Research and Modeling, 2021, vol. 13, no. 6, pp. 1101-1124
Citation in English: Matyushkin I.V., Rubis P.D., Zapletina M.A. Experimental study of the dynamics of single and connected in a lattice complex-valued mappings: the architecture and interface of author’s software for modeling // Computer Research and Modeling, 2021, vol. 13, no. 6, pp. 1101-1124
DOI: 10.20537/2076-7633-2021-13-6-1101-1124

Full-text version of the journal is also available on the web site of the scientific electronic library eLIBRARY.RU

The journal is included in the Russian Science Citation Index

The journal is included in the List of Russian peer-reviewed journals publishing the main research results of PhD and doctoral dissertations.

International Interdisciplinary Conference "Mathematics. Computing. Education"

The journal is included in the RSCI

Indexed in Scopus


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# Why is acylation giving the major product compared to alkylation in the reaction between benzene and chloroacetyl chloride?

Why are we getting this products as major and the other as minor ?

According to my knowledge, the carbocation formed on the $$\alpha$$-carbon of the carbonyl chloride is more stable than the carbocation formed on the doubly bonded carbon.

Shouldn't the major and minor product be reversed ?

• Then your knowledge is incorrect - acyl cations are way more stabilised then alkyl. Jun 3 at 16:00
• @Mithoron how ?? the carbocation is formed on doubly bonded carbon. so how is it stable ? Jun 4 at 8:39
• Did you bother you read Safdar's answer? Not the "not an answer" you marked as accepted, which missed the point entirely... Jun 4 at 12:45
• @Mithoron i never saw the edit. now it is better and does explain the concept Jun 5 at 5:28

## 2 Answers

This can be explained by comparing the stability of the carbocations formed in Friedel-Crafts acylation and Friedel-Crafts alkylation using chloroacetylchloride. This is because the rate determining step in both reactions is the formation of the carbocation.

As you can see, the acyl carbocation is resonance stabilized, whereas the carbocation formed in the alkylation process is very weakly resonance stabilized.[1]

According to supporting information to J. Chem. Educ. 2012, 89 (1), 147–149[2]

This o-toluidine selective attack at the acyl carbon atom of the bifunctional reagent is illustrative of the significantly greater reactivity of nucleophiles toward acylation, compared to alkylation

Therefore the acylation gives you the major product compared to the Friedel-Crafts alkylation product.

I was able to find a reference to your a similar reaction in J Flow Chem 2019, 9 (1), 35–42 with difluorobenzene instead of benzene.

A series of experiments were performed reacting a solution of 1,3-difluorobenzene (6) and aluminum trichloride ($$\ce{AlCl3}$$) in nitromethane [...] however increasing the temperature and time increased conversion to $$22–74\%$$ (entries 2–6). At $$\pu{90 °C}$$ (entry 5) the product was accompanied by the 2-acylated regioisomer. To further improve the reaction, the amounts of chloroacetyl chloride (7) and $$\ce{AlCl3}$$ were increased.These modifications gave the highest conversion, $$79\%$$ conversion [...]

$$\begin{array}{c|ccc} \hline \bf{Entry} & \bf{Residence\,time\,(min)}& \bf{Temp\,(^\circ C)} & \bf{Conversion\,(\%)} \\ \hline 1 & 5 & 50 & 0 \\ 2 & 5 & 60 & 22 \\ 3 & 10 & 60 & 45 \\ 4 & 15 & 60 & 46 \\ 5 & 15 & 90 & 77 \\ 6 & 25 & 70 & 74 \\ 7 & 25 & 80 & 77 \\ 8 & 25 & 80 & 79 \\ \hline \end{array}$$

Another paper that talks about the acetylation of benzene and toluene is J. Am. Chem. Soc. 1971, 93 (25), 6964–6967 which includes multiple reagents including chloroacetyl chloride.

### Reference:

1. Creary, X. Stabilization of α-keto cations by carbonyl conjugation. J. Am. Chem. Soc. 1981, 103 (9), 2463–2465. DOI: 10.1021/ja00399a078.
2. Demare, P.; Regla, I. Synthesis of Two Local Anesthetics from Toluene: An Organic Multistep Synthesis in a Project-Oriented Laboratory Course. J. Chem. Educ. 2012, 89 (1), 147–149. DOI: 10.1021/ed100838a.

No the products shouldn’t be reversed as, to put simply, acylation is preferred over alkylation.

This is a conclusion based on various experimental results and there are many reasons that can be given for the same.

As said in the concluding lines of this article from Chemistry Libretexts,

Friedel-Crafts Acylations offer several synthetic advantages over Friedel-Crafts Alkylation. These advantages provide greater control over the production of reaction products

I looked up online but couldn’t find any relevant reason particular to the acylation of chloroacetyl chloride, as asked in your question. But I did find some explanations to why acylation is preferred over alkylation:

1. In friedel crafts aklylation there is a chance of polyalkyation. Because the alkyl group increases benzene ring electron density and ring becomes activated to another elelctrophilic substitution.... thus polyalkyation occurs….But in Fridel crafts acylation there is no chance of polyacylation because the acyl group (RC=O) decreases benzene ring electron density and henceforth the second acylation does not take place.

2. During acylation reaction the carbon chain of the acyl halide does not rearrange and hence there's no possibility of formation of Isomeric ketone compound and thus 100% product yield is obtained in acylation reaction.

This is an answer to a question asked on Quora.

Again, these are all just explanations given to explain the experimental results and might not be conclusive or convincing enough.

• Sorry for a misunderstanding. "This answer doesn't explain why acylation takes place over alkylation but rather that the product of acylation is preferred compared to the products generated by alkylation when each reaction takes place in isolation."No the products shouldn’t be reversed as, to put simply, acylation is preferred over alkylation", what is the proof for this statement? Jun 3 at 8:15


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# Tag Info

## New answers tagged vision

0

This is an old question, but I'm very surprised no one has mentioned this: You CAN'T see red-green. According to multiple websites, some studies showed that the human eye cannot see both red and green simultaneously, as the red cone and the green cone send out signals that cancel each other out. Similarly, blue-yellow is equally impossible to perceive. ...

1

Sadly, you're not likely to get any better chart than the one in the link you provided unless you can track down the OEM of the cameras supplied to the various phone manufacturers. However, the reason your camera can respond to ("see") IR remote controllers and the like is that the IR filters are not 100% blocking -- and the IR emitters in the remotes are ...

0

If we could 'see' a lot more of the EM spectrum, then yes, the rainbow would be wider, and we would have a larger vocabulary to name the 'colors'. All the wavelengths of the EM would be present, but our eyes would've a limit to its discerning power. For example, $a_0$ nm to $a_1$ nm would be one 'color'. So, we won't see all the wavelengths, but the rainbow ...

Top 50 recent answers are included


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Javascript is required
1
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
Let C be the set of all complex numbers. Let

S1 = {z$$\in$$C : |z $$-$$ 2| $$\le$$ 1} and

S2 = {z$$\in$$C : z(1 + i) + $$\overline z$$(1 $$-$$ i) $$\ge$$ 4}.

Then, the maximum value of $${\left| {z - {5 \over 2}} \right|^2}$$ for z$$\in$$S1 $$\cap$$ S2 is equal to :
A
$${{3 + 2\sqrt 2 } \over 4}$$
B
$${{5 + 2\sqrt 2 } \over 2}$$
C
$${{3 + 2\sqrt 2 } \over 2}$$
D
$${{5 + 2\sqrt 2 } \over 4}$$
2
JEE Main 2021 (Online) 27th July Morning Shift
+4
-1
Let C be the set of all complex numbers. Let

$${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\}$$

$${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\}$$ and

$${S_3} = \{ z \in C||z - \overline z | \ge 8\}$$.

Then the number of elements in $${S_1} \cap {S_2} \cap {S_3}$$ is equal to :
A
1
B
0
C
2
D
Infinite
3
JEE Main 2021 (Online) 22th July Evening Shift
+4
-1
Let n denote the number of solutions of the equation z2 + 3$$\overline z$$ = 0, where z is a complex number. Then the value of $$\sum\limits_{k = 0}^\infty {{1 \over {{n^k}}}}$$ is equal to
A
1
B
$${4 \over 3}$$
C
$${3 \over 2}$$
D
2
4
JEE Main 2021 (Online) 20th July Morning Shift
+4
-1
If z and $$\omega$$ are two complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$\arg (z) - \arg (\omega ) = {{3\pi } \over 2}$$, then $$\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$ is :

(Here arg(z) denotes the principal argument of complex number z)
A
$${\pi \over 4}$$
B
$$- {{3\pi } \over 4}$$
C
$$- {\pi \over 4}$$
D
$${{3\pi } \over 4}$$
Policy


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# Series $\sum\limits_{n=1}^\infty \frac{1}{\cosh(\pi n)}= \frac{1}{2} \left(\frac{\sqrt{\pi}}{\Gamma^2 \left( \frac{3}{4}\right)}-1\right)$

I was playing around with Mathematica and found that

$$\sum_{n=1}^\infty\frac1{\cosh(\pi n)} = \frac12\left(\frac{\sqrt{\pi}}{\Gamma \left(\tfrac34\right)^2}-1\right)$$

Does anybody know how to prove this manually?

-

Here's a short sketch which I might further expand on later.

Consider the equivalent Lambert series

$$2\sum_{k=1}^\infty \frac{q^k}{1+q^{2k}}$$

with $q=\exp(-\pi)$.

This sum, through the use an appropriate geometric series, can also be expressed as

$$2\sum_{j=0}^\infty (-1)^j\frac{q^{2j+1}}{1-q^{2j+1}}$$

This sum can be expressed in terms of the Jacobi theta function $\vartheta_3(0,q)$. Using formula 57 in the previously given link, we have the expression

$$\frac{\vartheta_3(0,q)^2-1}{2}\tag{1}$$

Formula 45 in that same link gives the relation

$$\vartheta_3(0,\exp(-\pi))=\frac{\sqrt[4]\pi}{\Gamma(3/4)}\tag{2}$$

Substituting $(2)$ into $(1)$ gives the identity in the OP.

If you're interested in hyperbolic sums like these, you will want to see these three papers.

-


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# Homework Help: Absolute Value Inequalities

1. Sep 18, 2010

### priscilla98

1. The problem statement, all variables and given/known data

Solve (x+2)/(x+1) -2 < 0

2. Relevant equations

LCD: x + 1

3. The attempt at a solution

(x+2-2x-2)/(x+1) < 0

(-x )/(x+1) < 0

But somehow, I got the wrong answer. The answer is (4, 7]. But I don't know where I went wrong.

2. Sep 18, 2010

### priscilla98

I tried everything I could to answer this problem. Am i right that the lcd is x + 1? Because when i multiply - 2 by x + 1, i end up having to cross out + 2 and -2.

3. Sep 18, 2010

### eumyang

I'm confused because you title this thread "Absolute Value Inequalities," but your problem does not have any absolute value symbols. Can you double check the problem, please? Is the problem this?
$$\frac{x+2}{x+1} -2 < 0$$
Because that's what I'm interpreting.

Furthermore, according to WolframAlpha, the solution to what I typed above is
(-∞, -1) U (0, ∞), not (4, 7].

69

4. Sep 18, 2010

### priscilla98

Re: Rational Inequalities

Your right, its not absolute inequalities. It's rational equalities.

Okay, thanks a lot for answering this question. I really do appreciate it.


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Differences between reflexives and projectives modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:41:00Z http://mathoverflow.net/feeds/question/7490 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules Differences between reflexives and projectives modules Francisco Perdomo 2009-12-01T19:44:16Z 2010-01-07T19:14:15Z
Let R be a normal noetherian domain.

What is the difference between a finitely generated reflexive module and a finitely generated projective module?

Can anybody recommend any references or make any suggestions about this?

Finitely generated projective modules can be identified with idempotents matrix...

Finitely generated projective modules correspond with vector bundles over topological space...

Are there similar results about reflexives modules?
 http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7496#7496 Answer by Gonçalo Marques for Differences between reflexives and projectives modules Gonçalo Marques 2009-12-01T20:09:44Z 2009-12-01T20:09:44Z
According to Lam's "Lectures on Modules and Rings" every f.g. projective module is reflexive. See page 55, exercise 7.
 http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7503#7503 Answer by Mariano Suárez-Alvarez for Differences between reflexives and projectives modules Mariano Suárez-Alvarez 2009-12-01T20:48:41Z 2009-12-01T21:18:14Z
This is not an answer to your question, but I found it a few days ago looking for something else and did strike me as quite curious... :

According to E. E. Enochs [A note on reflexive modules. Pacific J. Math. Volume 14, Number 3 (1964), 879-881.] a free module of infinite countable rank over a discrete valuation ring is reflexive iff the ring is not complete.

A nice related example to keep in mind in this context is that a free abelian group of infinite countable rank is reflexive.
 http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7510#7510 Answer by Mariano Suárez-Alvarez for Differences between reflexives and projectives modules Mariano Suárez-Alvarez 2009-12-01T21:50:01Z 2009-12-01T21:50:01Z
More to the point: if $A$ is a left noetherian ring of global dimension at most $2$, then every finitely generated reflexive left $A$-module is projective. Indeed, if $M$ is a finitely generated module and $$P_1\to P_0\to M\to 0$$ is a projective presentation by finitely generated projectives, applying the functor $(\mathord-)^*=\hom_A(\mathord-,A)$ we get an exact sequence $$0\to M^*\to P_0^*\to P_1^*\to E\to 0,$$ when $E$ is just the cokernel of the map $P_0^*\to P_1^*$. Since $\mathrm{pdim}\,E\leq 2$ and since $P_0^*$ and $P_1^*$ are projective, the kernel of $P_0^*\to P_1^*$, namely $M^*$, must be projective. This shows that the dual of finitely generated module is projective, so a finitely generated reflexive, being isomorphic to the dual of its dual module, is projective.
 http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7519#7519 Answer by Graham Leuschke for Differences between reflexives and projectives modules Graham Leuschke 2009-12-01T22:48:32Z 2009-12-01T22:48:32Z
In general, there are many many more reflexive modules than projective modules. As pointed out in another answer, every second syzygy is reflexive, and not necessarily projective (unless your ring happens to be regular of dimension at most $2$). Thus the converse of that answer is, I think, more interesting.

In particular, assume that $R$ is (semi)local. Then there are only finitely many indecomposable projectives. If there are finitely many indecomposable reflexives, then $R$ must have dimension at most $2$, and thus finite Cohen-Macaulay type (which is a very rare condition).

The moral is that reflexivity is a (very) much weaker condition than projectivity.
 http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/7588#7588 Answer by VA for Differences between reflexives and projectives modules VA 2009-12-02T16:08:54Z 2009-12-02T16:08:54Z
Well, the answer is well known of course. For a finitely generated module over a commutative normal Noetherian domain TFAE

M is reflexive

M is torsion-free and equals the intersection of its localizations at the codimension 1 primes

M satisfies Serre's condition S2 and its support is Spec R.

M is the dual of a f.g. module N

As you say, a finite projective module is the same as a locally free sheaf on Spec R. Similarly, a finite reflexive module is the same as the push forward of a locally free sheaf from an open subset U of Spec R whose complement has codimension $\ge2$.

So for an easy example take a Weil divisor D which is not Cartier, the associated divisorial sheaf (corresponding to an R-module of rank 1) is reflexive, not projective. Your example with a line on a quadratic cone is of this form.

This stuff is standard and used all the time in higher-dimensional algebraic geometry around the Minimal Model Program. For an old reference covering some of this, see e.g. Bourbaki, Chap.7 Algebre commutative, VII.
 http://mathoverflow.net/questions/7490/differences-between-reflexives-and-projectives-modules/11060#11060 Answer by David Speyer for Differences between reflexives and projectives modules David Speyer 2010-01-07T19:14:15Z 2010-01-07T19:14:15Z
I asked a very similar question a few months ago and got some very good answers.


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# A.A. Agrachev, S. Scholtes, D. Pallaschke, On Morse theory for piecewise smooth functions

Title: On Morse theory for piecewise smooth functions
Authors: A.A. Agrachev, S. Scholtes, D. Pallaschke
Journal title: Dynamical and Control Systems
Year: 1997
Volume: 3
Pages: 449-469
Abstract:

Lower level sets of continuous selections of $C^2$ functions defined on a smooth manifold in the vicinity of a nondegenerate critical point in the sense of [11] are studied. It is shown that the lower level set is homotropy equivalent to the join of the lower level sets of the smooth and the nonsmooth part, respectively, of the corresponding normal form. Some generalized Morse inequalities are deduced from this result.


text:
# Definition:Zermelo-Fraenkel Axioms

The Zermelo-Fraenkel axioms are the most well-known basis for axiomatic set theory.

There is no standard numbering for them, and their exact formulation varies. Certain axioms can in fact be derived from other axioms, so their status as "axioms" can be questioned.

The axioms are as follows:

## The Axiom of Extension

Two sets are equal iff they contain the same elements:

$\forall x: \left({x \in A \iff x \in B}\right) \iff A = B$

The order of the elements in the sets is immaterial.

## The Axiom of Existence

There exists a set that has no elements:

$\exists x: \forall y: \left({\neg \left({y \in x}\right)}\right)$

## The Axiom of Pairing

For any two sets, there exists a set to which only those two sets are elements:

$\forall A: \forall B: \exists x: \forall y: \left({y \in x \iff y = A \lor y = B}\right)$

Thus it is possible to create a set containing any two sets that you have already created.

Otherwise known as the Axiom of the Unordered Pair.

The Axiom of Pairing can be deduced as a consequence of:

$(1): \quad$ The Axiom of Infinity and the Axiom of Replacement: see Axiom of Pairing from Infinity and Replacement
$(2): \quad$ The Axiom of Powers and the Axiom of Replacement: see Axiom of Pairing from Powers and Replacement.

## The Axiom of Subsets

For every set and every condition, there corresponds a set whose elements are exactly the same as those elements of the original set for which the condition is true.

Because we cannot quantify over functions, we need an axiom for every condition we can express. Therefore, this axiom is sometimes called an axiom scheme, as we introduce a lot of similar axioms.

This axiom scheme is formally stated as follows:

For any function of propositional logic $P \left({y}\right)$, we introduce the axiom:

$\forall z: \exists x: \forall y: \left({y \in x \iff \left({y \in z \land P \left({y}\right)}\right)}\right)$

This means that if you have a set, you can create a set that contains some of the elements of that set, where those elements are specified by stipulating that they satisfy some (arbitrary) condition.

## The Axiom of Unions

For every collection of sets, there exists a set (the sum or union set) that contains all the elements that belong to at least one of the sets in the collection:

$\forall A: \exists x: \forall y: \left({y \in x \iff \exists z: \left({z \in A \land y \in z}\right)}\right)$

## The Axiom of Powers

For every set, there exists a collection of sets that contains amongst its elements all the subsets of the given set.

$\forall x: \exists y: \left({\forall z: \left({z \in y \iff \left({w \in z \implies w \in x}\right)}\right)}\right)$

## The Axiom of Infinity

There exists a set containing a set with no elements and the successor of each of its elements.

$\exists x: \left({\left({\exists y: y \in x \land \forall z: \neg \left({z \in y}\right)}\right) \land \forall u: u \in x \implies u^+ \in x}\right)$

## The Axiom of Replacement

For any function $f$ and subset $S$ of the domain of $f$, there is a set containing the image $f \left({S}\right)$.

More formally, let us express this as follows:

Let $P \left({y, z}\right)$ be a propositional function, which determines a function.

That is, we have $\forall y: \exists x: \forall z: \left({ P \left({y, z}\right) \iff x = z }\right)$.

Then we state as an axiom:

$\forall w: \exists x: \forall y: \left({y \in w \implies \left({ \forall z: \left({ P \left({y, z}\right) \implies z \in x }\right) }\right) }\right)$

Alternatively, the two above statements may be combined into a single (somewhat unwieldy) expression:

$\left({ \forall y: \exists x: \forall z: \left({ P \left({y, z}\right) \implies x = z }\right) }\right) \implies \forall w: \exists x: \forall y: \left({y \in w \implies \forall z: \left({ P \left({y, z}\right) \implies z \in x }\right) }\right)$

## The Axiom of Foundation

For all non-empty sets, there is an element of the set that shares no element with the set.

That is:

$\forall S: \left({ \left({\exists x: x \in S}\right) \implies \exists y \in S: \forall z \in S: \neg \left({z \in y}\right) }\right)$

The antecedent states that $S$ is not empty.

Otherwise known as the Axiom of Regularity.

The above axioms taken together as a system, but without the Axiom of Choice below, is called ZF set theory. The validity of the AC is still debated.

## The Axiom of Choice

For every set of non-empty sets, we can provide a mechanism for choosing one element of each element of the set.

$\displaystyle \forall s: \left({\varnothing \notin s \implies \exists \left({f: s \to \bigcup s}\right): \forall t \in s: f \left({t}\right) \in t}\right)$

That is, one can always create a choice function for selecting one element from each member of the set.

The system of the axioms of ZF set theory in combination with the Axiom of Choice is known as ZFC set theory; ZF plus Choice.

## Source of Name

This entry was named for Ernst Zermelo and Abraham Fraenkel.


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# Matrix of linear transformation

Matrix of linear transformation (urgent)

Identify the matrix of the transformation for the following:

a) (x,y,z) = (2x-y+4z,x+y-z,x-z)

b) (x,y) = (x,2x)

c) (x,y,z) = (x-2y,3x-6y)

Here are my attempts
a)
2,-1, 4
1, 1,-1
1, 0,01

b)
1,0
2,0

c)

1,-2, 0
3,-6, 0

Are theses correct? Thanks in advance:)

## Answers and Replies

Mark44
Mentor

Identify the matrix of the transformation for the following:

a) (x,y,z) = (2x-y+4z,x+y-z,x-z)

b) (x,y) = (x,2x)

c) (x,y,z) = (x-2y,3x-6y)

Here are my attempts
a)
2,-1, 4
1, 1,-1
1, 0,01

b)
1,0
2,0

c)

1,-2, 0
3,-6, 0

Are theses correct? Thanks in advance:)

In the first one you have what looks like a typo. The last term in the third row should be -1. You probably hit the wrong key. The other two are fine.

Ye sorry that should be -1 :) Next the question asks to find the basis for the Kernal for each. How do I go about doing this?

Mark44
Mentor

That's kernel. The kernel is the subspace consisting of all vectors that are solutions to the equation Ax = 0. Start with that.

HallsofIvy
Science Advisor
Homework Helper

By the way, it is a very bad idea to write something like:
"(x,y,z) = (2x-y+4z,x+y-z,x-z)"

That says the two are equal which is NOT what you mean. What you mean is that the linear transformation maps (changes) (x, y, z) into (2x- y+ 4z, x+ y- z, x- z) which might be written
T(x,y,z) = (2x-y+4z,x+y-z,x-z)

or simply (x,y,z) => (2x-y+4z,x+y-z,x-z) where the "arrow", =>, indicates the change.

The kernel of the matrix $$\displaystyle \begin{bmatrix}2 & -1 & 4 \\ 1 & 1 & -1 \\ 1 & 0 & -1\end{bmatrix}$$ is the set of vectors, $$\displaystyle \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$ such that
$$\displaystyle \begin{bmatrix}2 & -1 & 4 \\ 1 & 1 & -1 \\ 1 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$$

which is the same as the equations 2x- y+ 4z= 0, x+ y- z= 0, x- z= 0, which you could have got from the original definition of the linear transformation. Solve those equations for x, y, and z. Notice that one "obvious" solution is x= y= z= 0, the "trivial" solution. If there are other, non-trivial solutions, then there will be an infinite number of them. It is easy to show that the kernel is a subspace of the vector space- if u and v are non-zero members of the kernel of a linear transformation, so is any linear combination of them.

Sorry halls of ivy but ur post goes a little strange in the middle:)

Ok so I have worked out the follwowing:

a) 2x-y+4z =0
b) x+y-z=0
c) x-z=0

therefore

x = z

equation b) becomes:

y = 0

and a) becomes:

6x=0

so us the basis of kernal x(000)??


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# Early Two-Author Math Papers

The middle of the twentieth-century featured several famous papers with two authors. For example, Eilenberg and Mac Lane's papers introducing categories and Eilenberg-MacLane spaces appeared in 1945. The Feit-Thompson Odd Order Theorem appeared in 1962. Atiyah and Singer published their index theorem in 1963.

I can't think of any important papers with two or more authors before the Eilenberg-Mac Lane collaboration, which could just be a lacuna in my historical knowledge. My question is: what are the first math papers with two or more authors? (A subsidiary question is: why were collaborations so rare before that?)

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Hardy-Littlewood? –  Olivier Feb 20 '11 at 20:33
Whitehead-Russell –  Balazs Feb 20 '11 at 20:37
The question is subjective - among all the multi-author papers, how do we determine which was the first important one? If we're just compiling a list of early multi-author papers, the question should be community wiki. –  Gerry Myerson Feb 20 '11 at 22:47
-1, because I don't like this kind of question. –  Theo Johnson-Freyd Feb 21 '11 at 2:35
This thread is long enough. I've started a meta discussion: tea.mathoverflow.net/discussion/965/oldest-joint-paper Please vote for this comment so that it appears above the fold. –  Loop Space Feb 21 '11 at 9:59

The famous paper of Dedekind and Weber:

R. Dedekind, H. Weber: Theorie der algebraischen Funktionen einer Veraendlichen, J. Reine Angew. Math 92 (1882) 181-290.

is the first place where the points of a Riemann surface are described in terms of ideals of the ring of functions. To put this into context, Dedekind had only invented the notion of ideal a few years earlier. They also give an algebraic proof of the Riemann-Roch theorem.

I think the analogy between function fields and number fields started here.

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Already Gauss, in his unpublished part of the Disquisitiones [published later by Dedekind in Gauss's Werke), observed that the proof of unique factorization in $F_p[X]$ is very similar to that in ${\mathbb Z}$. This analogy was extended to the quadratic reciprocity law and forms over $F_p[x]$ by various writers such as Schoenemann, Serret, Heine, and Dedekind way before Dedekind-Weber. –  Franz Lemmermeyer Feb 21 '11 at 16:48
Dedekind and Weber's paper developed the analogy in more depth than the earlier authors. –  KConrad Feb 22 '11 at 2:48

The Brill--Noether paper appeared in 1874: "Ueber die algebraischen Functionen und ihre Anwendung in der Geometrie". Math. Annalen 7, 269–316.

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I never realized that the Noether of "Brill-Noether" referred to Max Noether, not Emmy Noether. –  user332 Feb 20 '11 at 21:51
@Rex: 1874 is a bit early for Emmy Noether, gifted as she was. –  Jim Humphreys Feb 20 '11 at 22:57
Yes, of course. I had not been aware that Brill-Noether theory was that old. –  user332 Feb 20 '11 at 23:09
the famous "brill noether" number occurs already in riemann for r=1. hence "brill noether" theory is even older than they are. –  roy smith Feb 21 '11 at 3:15

Abraham Ecchellensis (Ibrahim Al-Haqilani) and Giovanni Alfonso Borelli published in 1661 a Latin translation of the 5th, 6th and 7th books of the Conics by Apollonius of Perga. Both Ecchellensis's and Borelli's names are on the title page, a highly unusual feature at the time.

For more details, see the paper Authorship and Teamwork Around the Cimento Academy by Domenico Bertoloni Meli, available here.

-
link is not working? –  Mark Sapir Feb 21 '11 at 14:43
Weird. It works for me. –  arsmath Feb 21 '11 at 14:49
indiana.edu/~hpscdept/Fac-MeliPublications.shtml (go to Research articles and look for a paper published in 2001 in Early Science and Medicine). –  Did Feb 21 '11 at 15:24
That's a really interesting paper. It directly addresses the question of how collaboration worked differently in the 1600s. –  arsmath Feb 21 '11 at 22:24
I don't think a jointly prepared translation of someone else's earlier work is in the spirit of the OP's question. –  KConrad Feb 22 '11 at 2:52

Aufgaben und Lehrsätze, erstere aufzulösen, letztere zu beweisen by N.H. Abel, Th. Clausen, J. Steiner, published in 1827, Journal für die reine und angewandte Mathematik. Volume 1827, Issue 2, Pages 286–292. I think every paper by Abel must be important.

Also:

Nouvelles formules analogues aux séries de Taylor et de Maclaurin. by B. Clapeyron G. Lamé, in Journal für die reine und angewandte Mathematik. Volume 1830, Issue 6, Pages 40–44.

I do not know who was Clapeyron, but Lamé is well known. These two papers may be the first collaborative journal papers because Crelle was the first math journal (or one of the first journals). But this site: http://www.daviddarling.info/encyclopedia/B/Bernoulli.html talks about a joint paper in astronomy produced by Bernoulli family. At that time (~1730) astronomy and mathematics were not that far apart.

-
Clapeyron is probably the Physicist: en.wikipedia.org/wiki/Beno%C3%AEt_Paul_%C3%89mile_Clapeyron In France, he seems mostly remembered for work in Thermodynamics. –  Thierry Zell Feb 20 '11 at 22:43
@Thierry: Of course, I only just realized who he was. Before now, I only saw Russian transcription of his name. Thanks! –  Mark Sapir Feb 20 '11 at 22:54
The "Aufgaben und Lehrsätze" is not a joint paper, it is a collection of problems to be solved by the readers, of which the first four were submitted by Abel, the next few by Clausen etc. –  Franz Lemmermeyer Feb 21 '11 at 17:32
As for the Bernoulli paper, wikipedia (mathematik.ch/mathematiker/daniel_bernoulli.php) has this to say: "Daniel Bernoulli submitted an entry for the Grand Prize of the Paris Academy for 1734 giving an application of his ideas to astronomy. This had unfortunate consequences since Daniel's father, Johann Bernoulli, also entered for the prize and their entries were declared joint winners of the Grand Prize." –  Franz Lemmermeyer Feb 21 '11 at 17:37
@Franz: you are right about Abel's paper, but was it a joint venture or the journal combined three papers together? That is why I included the second paper in my answer. That one seems really collaborative. –  Mark Sapir Feb 21 '11 at 17:42

Sturm and Liouville published joint papers in 1836-1837.

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The paper springerlink.com/content/k28765tu63333677/fulltext.pdf on the development of Sturm--Liouville theory says they published only one joint paper on S--L theory, in 1837. The paper they published together in 1836, also mentioned in that article, was about counting complex roots of a polynomial inside a contour. The bibliography lists a paper by Colladon and Sturm from 1834 about compression of liquids. –  KConrad Feb 22 '11 at 4:03

Here is one from the eighteenth century. There must be others.

Theoria motuum lunae, nova methodo pertractata una cum tabulis astronomicis, unde ad quodvis tempus loca lunae expedite computari possunt incredibili studio atque indefesso labore trium academicorum: Johannis Alberti Euler, Wolffgangi Ludovici Krafft, Johannis Andreae Lexell. Opus dirigente Leonhardo Eulero acad. scient. Borussicae directore vicennali et socio acad. Petrop. Parisin. et Lond. Petropoli, typis academiae imperialis scientiarum. 1772.

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Translation, anyone? "Theory of the motion of the moon...something" –  David Roberts Feb 21 '11 at 8:09
Here's my clumsy translation: 'A theory of the movements of the moon, treated by means of a single new method with astronomical tables, by which the position of the moon at any given time may be easily calculated thanks to the marvellous studies and indefatigable efforts of three academicians.' –  Finn Lawler Feb 21 '11 at 15:11
Euler's son, Krafft and Lexell were among the "scribes" responsible for writing up Euler's work after he had become blind. –  Franz Lemmermeyer Feb 21 '11 at 17:40
I don't think a jointly prepared write-up of another (blind) person's work is in the spirit of the OP's question. –  KConrad Feb 22 '11 at 2:54
Lexell published 66 papers, only 4 of them were coauthored with Euler. He was eventually appointed as Euler's successor. I don't think he should be labelled a "scribe." Wikipedia has an article about Lexell: en.wikipedia.org/wiki/Anders_Johan_Lexell. –  Michael Renardy Feb 22 '11 at 13:15

In differential geometry a fundamental collaborative paper is: Ricci, Gregorio, Levi-Civita, Tullio (March 1900). "Méthodes de calcul différentiel absolu et leurs applications". Mathematische Annalen (Springer) 54 (1–2): 125–201. doi:10.1007/BF01454201.

Its digitalization is freely available here.

Edit: I replaced the link to a pay-walled copy with another to the GDZ repository.

-

Very important joint work one of whose authors is a somewhat well-known mathematician, but not a mathematics paper was the construction of the first electromagnetic telegraph by C. F. Gauss and W. E. Weber (1833), and then:

Gauss, Carl Friedrich; Weber, Wilhelm Eduard (1840). Atlas Des Erdmagnetismus: Nach Den Elementen Der Theorie Entworfen. Leipzig: Weidmann'sche Buchhandlung.

Don't know if that counts, but I don't know of any joint work of Gauss in pure mathematics...

-

## protected by Scott Morrison♦Jul 7 '13 at 0:00

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text:
# Make Pythagoras Proud

Geometry Level 3

Using the image above, find $N$.

(The image is not given to scale.)

×


text:
# What time complexity is more significant? [closed]

A certain algorithm executes $$n$$ operations of three types: insert, delete, and find. We know that $$n/10$$ of the operations are inserts, and the rest are deletes and finds.

You are given two implementations of the algorithm. The first one implements insert in worst case $$\Theta(n)$$, and the amortized cost of every operation (including insert) is $$\Theta(1)$$. The second implementation implements insert in worst case $$\Theta(\log n)$$, and the amortized cost of every operation (including insert) is $$\Theta(\log n)$$.

Which implementation would you recommend, and why?

• In order to recommend an implementation, you have to tell us when is one implementation preferred over another for you. It's hard to compare different choices without any criteria for comparison. – Yuval Filmus Sep 12 at 6:57

## 1 Answer

You are asking for a practical answer, so you need a practical problem.

State your practical problem. Then examine what would be the effect of using one implementation vs. the other. Based on this examination, you can choose which implementation is more acceptable to you.


text:
Can someone give me a clear explanation on this topic? [duplicate]

What is Divergence and Curl of Magnetic field?

Divergence is a scalar quantity that tells how much a vector field is pointing into or out of a point. For magnetic fields, because there are no magnetic monopoles, the divergence of a magnetic field is zero everywhere (Gauss' Law for Magnetism).

Curl is a vector quantity that tells how a vector field is curving around an axis through a point, with the direction of that axis given by the direction of the curl vector. For magnetic fields, curl is related to the current density and rate of change of the electric field at that point (Maxwell-Ampère Law).


text:
Oct 15 2015
2 min

testthat 0.11.0

testthat 0.11.0 is now available on CRAN. Testthat makes it easy to turn your existing informal tests into formal automated tests that you can rerun quickly and easily. Learn more at http://r-pkgs.had.co.nz/tests.html. Install the latest version with:

install.packages("testthat")

In this version:

• New expect_silent() ensures that code produces no output, messages, or warnings. expect_output(), expect_message(), expect_warning(), and expect_error() now accept NA as the second argument to indicate that there shouldn’t be any output, messages, warnings, or errors (i.e. they should be missing)
f <- function() {
print(1)
message(2)
warning(3)
}
expect_silent(f())
#> Error: f() produced output, warnings, messages

expect_warning(log(-1), NA)
#> Error: log(-1) expected no warnings:
#> *  NaNs produced

• Praise gets more diverse thanks to Gabor Csardi’s praise package, and you now also get random encouragment if your tests don’t pass.

• testthat no longer muffles warning messages. This was a bug in the previous version, as warning messages are usually important and should be dealt with explicitly, either by resolving the problem or explicitly capturing them with expect_warning().

• Two new skip functions make it easier to skip tests that don’t work in certain environments: skip_on_os() skips tests on the specified operating system, and skip_on_appveyor() skips tests on Appveyor.

There were a number of other minor improvements and bug fixes. See the release notes for a complete list.

A big thanks goes out to all the contributors who made this release happen. There’s no way I could be as productive without the fantastic commmunity of R developers who come up with thoughtful new features, and who discover and fix my bugs!

Tags:


text:
# What are the different types of rational numbers?

Mar 27, 2018

Integers, whole numbers, counting/natural numbers

#### Explanation:

Integers can be negative or positive. They cannot be decimals/fractions/percentages.

Examples of integers:

$- 3 , 4 , 56 , - 79 , 82 , 0$

Whole numbers include 0, but they cannot be negative. They cannot be decimals/fractions/percentages.

Examples of whole numbers:

$3 , 4 , 56 , 79 , 82 , 0$

Counting/natural numbers are the order in which we count. They are positive whole numbers, but do not include zero (we do not count by saying 0, 1, 2, 3, etc.).

Examples of counting/natural numbers:

$1 , 2 , 3 , 4 , 5 , 6$


text:
- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnline Submission ㆍMy Manuscript - For Reviewers - For Editors
 On conformal and quasi-conformal curvature tensors of an $N(k)$-quasi Einstein manifold Commun. Korean Math. Soc. 2012 Vol. 27, No. 2, 317-326 https://doi.org/10.4134/CKMS.2012.27.2.317Published online June 1, 2012 Aliakbar Hosseinzadeh and Abolfazl Taleshian Mazandaran University, Mazandaran University Abstract : We consider $N(k)$-quasi Einstein manifolds satisfying the conditions $C(\xi,X).S=0$, $\tilde{C}(\xi,X).S=0$, $\bar{P}(\xi,X).C=0$, $P(\xi,X).\tilde{C}=0$ and $\bar{P}(\xi,X).\tilde{C}=0$ where $C$, $\tilde{C}$, $P$ and $\bar{P}$ denote the conformal curvature tensor, the quasi-conformal curvature tensor, the projective curvature tensor and the pseudo projective curvature tensor, respectively. Keywords : $k$-nullity distribution, quasi Einstein manifold, $N(k)$-quasi Einstein manifold, conformal curvature tensor, quasi-conformal curvature tensor, projective curvature tensor, pseudo projective curvature tensor MSC numbers : 53C25 Downloads: Full-text PDF


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## Boundary Layers and Domain Decomposition for Radiative Heat Transfer and Diffusion Equations: Applications to Glass Manufacturing Processes

• In this paper domain decomposition methods for radiative transfer problems including conductive heat transfer are treated. The paper focuses on semi-transparent materials, like glass, and the associated conditions at the interface between the materials. Using asymptotic analysis we derive conditions for the coupling of the radiative transfer equations and a diffusion approximation. Several test cases are treated and a problem appearing in glass manufacturing processes is computed. The results clearly show the advantages of a domain decomposition approach. Accuracy equivalent to the solution of the global radiative transfer solution is achieved, whereas computation time is strongly reduced.

$Rev: 13581$


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Biome Height Tweaker Mod is a small mod for adjusting the height and variance of Biomes in the game.  Usage Note: You must configure this mod. If you don’t configure it, all aspects of world generation will remain the default, thus this mod...[Read More]

Biome Height Tweaker Mod is a small mod for adjusting the height and variance of Biomes in the game.  Usage Note: You must configure this mod. If you don’t configure it, all aspects of world generation will remain the default, thus this mod...[Read More]

Biome Height Tweaker Mod is a small mod for adjusting the height and variance of Biomes in the game.  Usage Note: You must configure this mod. If you don’t configure it, all aspects of world generation will remain the default, thus this mod...[Read More]

Biome Height Tweaker Mod is a small mod for adjusting the height and variance of Biomes in the game.  Usage Note: You must configure this mod. If you don’t configure it, all aspects of world generation will remain the default, thus this mod...[Read More]

Biome Height Tweaker Mod is a small mod for adjusting the height and variance of Biomes in the game.  Usage Note: You must configure this mod. If you don’t configure it, all aspects of world generation will remain the default, thus this mod...[Read More]

Biome Height Tweaker Mod is a small mod for adjusting the height and variance of Biomes in the game.  Usage Note: You must configure this mod. If you don’t configure it, all aspects of world generation will remain the default, thus this mod...[Read More]

Biome Height Tweaker Mod is a small mod for adjusting the height and variance of Biomes in the game.  Usage Note: You must configure this mod. If you don’t configure it, all aspects of world generation will remain the default, thus this mod...[Read More]


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Algebra (sort Of)

• Jun 9th 2009, 02:07 AM
Awsom Guy
Algebra (sort Of)
Hi Everyone
This is for my physics assignment but it is maths
I have a formula n1sin1Ө=n2sin2Ө
Where:
n1 = 1
n2 =?
sin1Ө = 44
sin2Ө = 48

An my answer should be or close to 1.48. I can't get it.
Help would be good.
Thanks.(Nod)
• Jun 9th 2009, 02:26 AM
Ruun
Remember that the image of the function $sin(x)$ is always between $[-1,1]$ so $sin (1\theta)$ and $sin (2\theta)$ can't be 44 and 48 !.
• Jun 9th 2009, 02:27 AM
Awsom Guy
Quote:

Originally Posted by Ruun
Remember that the image of the function $sin(x)$ is always between $[-1,1]$ so $sin (1\theta)$ and $sin (2\theta)$ can't be 44 and 48 !.

That doesn't make sense
• Jun 9th 2009, 02:40 AM
yeongil
The range of the sine function is [-1, 1]; meaning, when you take the sine of any angle it's going to be a number between -1 and 1, so to say that
$\sin 1\theta = 44$ and $\sin 2\theta = 48$ is impossible. Can you double-check the assignment and see how exactly the problem is worded?

01
• Jun 9th 2009, 02:41 AM
Ruun
• Jun 9th 2009, 02:47 AM
Ruun
Oh, maybe I've just made a mistake with the image word, in spanish range and image (rango e imagen) are equivalent, so maybe I just didn't make my point clearly. In that case, my apologies.
• Jun 9th 2009, 02:51 AM
Awsom Guy
Maybe my results are wrong, but even then I should get close to it because my results were quite good.(Nod)


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# User:Ninetiger

My name is Xiao Liu.

I'm a PhD. student in Math of University of Toronto.

## Orbital stability

Definition: Let $U=U(t,x)$ be a solitary wave solution of the PDE. We say that $U(t)$ is orbitally stable for the PDE if for any $\epsilon>0$ there exists $\delta>0$ such that if $u_0\in H^1(\mathbb{R})$ satisfies $||u_0-U(0)||_{H^1}<\delta$, then the solution $u(t)$ of the PDE with initial data $u(0)=u_0$ exists globally in time and satisfies $$\sup_{t\geq 0} \inf_{(\theta,y)\in\mathbb{R}^2}||u(t)-e^{i\theta} U(t,.-y)||_{H^1}<\epsilon.$$ Otherwise, $U(t)$ is said to be orbitally unstable.

This definition is from the paper Stability of solitary waves for derivative nonlinear Schrodinger equation (2006) by Colin and Ohta.

Here is the definition of orbital stability on wikipedia:orbital stability

## Regularity and Fourier transform

Last week (Jan 27), Jim showed some relation between regularity and Fourier transform. I try to explain his idea by Matlab.

Consider three functions:

$f_1(x)=8-|x|,x\in(-8,8)$ and $f_1(x)=0$ otherwise.

$f_2(x)=8((\frac{x}{8})^4-2(\frac{x}{8})^2+1),x\in[-8,8]$ and $f_2(x)=0$ otherwise.

$f_3(x)=e^{-x^2}.$

Then we have $f_1(x)\in \mathbf{C}(\mathbb{R})$, $f_2(x)\in \mathbf{C}^1(\mathbb{R})$ and $f_2(x)\in \mathbf{C}^{\infty}(\mathbb{R})$.

Here is the loglog graph of $f_1(x),f_2(x)$ and $f_3(x)$ after Fourier Transform:

From the graph, we find that the $f_2(x)$ decays fast than $f_1(x)$, and $f_3(x)$, which is smooth, decays fastest, as we expect.

Thanks Xiao! I am happy to see these examples which help to develop our intuition about how to understand regularity properties of functions when viewing their Fourier transform. The log-frequency versus log-Fourier-transform-modulus plots I described in class on 27 January are called Bode plots in engineering, especially appearing in discussions of filter designs. Colliand 10:07, 1 February 2011 (EST)


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# Installation from source code¶

Compiling GammaLib from source is the recommended method to install the package in case that you do not use Anaconda (if you use Anaconda, please follow the instructions at Installing via Anaconda, and better do not try to compile the source code against your Anaconda Python installation). You also need to compile from source if you want to use the development version of GammaLib.

Note

These installation instructions apply to gammalib-1.1.0 and later. gammalib-1.1.0 was built, installed and tested successfully on Debian, Ubuntu, Mandriva, OpenSUSE, Scientific Linux, CentOS, Mac OS X, FreeBSD, and OpenSolaris (using gcc), so hopefully it also installs on your distribution. Windows is not supported. In case you encounter problems, please post a report on the bug tracker or send an e-mail to the mailing list.

## Prerequisites¶

GammaLib should compile on every modern Unix system and on Mac OS X, Windows is not supported. To enable support for FITS file handling the cfitsio library from HEASARC needs to be installed. cfitsio exists on many Linux distributions as a prebuilt library, so in general you can directly install it using your package manager. Make sure that you install the library and the development package, as the cfitsio header file (which usually comes only in the development package) is needed to compiling GammaLib.

If cfitsio is not available as a prebuilt package, or if you encounter some problems with the prebuilt package, cfitsio can be downloaded from http://heasarc.gsfc.nasa.gov/fitsio and installed from the source files (for details, please refer to the installation instructions on the HEASARC site). We recommend installation of cfitsio in the default GammaLib install directory as a shared library by typing

$./configure --prefix=/usr/local/gamma$ make shared
$make install  GammaLib can also benefit from the presence of the readline library that provides line-editing and history capabilities for text input (GammaLib offers however full functionality without having readline installed). readline (which depends on ncurses) is available on most system as a prebuilt library, so also here we recommend to use your package manager to install the libraries if they are not already there. Also here, the readline (and ncurses) development packages are required, so that the header files become available. If readline and ncurses are not available as a prebuilt package, or if you encounter some problems with the prebuilt packages, both libraries can be downloaded from http://ftp.gnu.org/gnu/ and installed from the source files. We recommend to put also those in the default GammaLib install directory /usr/local/gamma. GammaLib includes a Python module that is compatible with Python 2 (version 2.3 or higher) and Python 3 (all versions). To generate the Python module, the Python development package including the Python.h header file needs to be installed on your system. ## Building GammaLib¶ To build GammaLib from the source files, download the latest release tarball, save it in an appropriate location, and type $ tar xvfz gammalib-x.y.z.tar.gz

where x.y.z is the version number of the library. Step in the created directory using

$cd gammalib-x.y.z  and build GammaLib by typing $ ./configure
$make  at the operating system prompt. The configuration command customizes the Makefiles for your particular system, the make command compiles the source files and builds the C++ library and Python module. Type ./configure and not simply configure to ensure that the configuration script in the current directory is used and not some other system-wide configuration script. Note To accelerate compilation you may type $ make -j8

which will instruct the automake system to perform a parallelized compilation, where 8 specifies the maximum number of parallel threads (set this number to about twice the number of cores you have available).

By default, the install directory is set to /usr/local/gamma. To change the install directory, provide an optional --prefix argument, for example

$./configure --prefix=/usr/local  If cfitsio and/or readline is not installed in a standard location for libraries (such as /usr/lib or /usr/local/lib), you can specify the appropriate location(s) using the LDFLAGS (for the library directory) and CPPFLAGS (for the include directory) options $ ./configure LDFLAGS='-L/opt/local/lib' CPPFLAGS='-I/opt/local/include'

You can find a full list of configuration options using

$./configure --help  ## Testing GammaLib¶ Before you install GammaLib you should test the C++ library and Python module by typing $ make check

This will execute an extensive testing suite that should terminate with

===================
All 21 tests passed
===================

or

============================================================================
Testsuite summary for gammalib 1.1.0
============================================================================
# TOTAL: 21
# PASS:  21

Eventually, loading the shared cfitsio and/or readline libraries may fail during the test if the libraries are not located in standard locations. In this case, add the library directories to the LD_LIBRARY_PATH environment variables (DYLD_LIBRARY_PATH on Mac OS X), e.g.

$export LD_LIBRARY_PATH=/opt/local/lib:$LD_LIBRARY_PATH

## Installing GammaLib¶

Now you are ready to install GammaLib by typing

$make install  at the operating system prompt. You may need to prepend a sudo in case that you need administrator privileges to access the install directory. If you do not have such privileges, chose an install directory that you can access using the --prefix option. ## Setting up your environment¶ Before using GammaLib you have to setup some environment variables. This will be done automatically by an initialisation script that will be installed in the bin directory. Assuming that you have installed GammaLib in the default directory /usr/local/gamma you need to add the following to your $HOME/.bashrc or $HOME/.profile script on a Linux machine: export GAMMALIB=/usr/local/gamma source$GAMMALIB/bin/gammalib-init.sh

If you use C shell or a variant then add the following to your $HOME/.cshrc or $HOME/.tcshrc script:

setenv GAMMALIB /usr/local/gamma
source $GAMMALIB/bin/gammalib-init.csh  ## Installing documentation¶ Note The documentation of the latest GammaLib release can be found at http://cta.irap.omp.eu/gammalib/. The documentation corresponding to the git development branch can be found at http://cta.irap.omp.eu/gammalib-devel/. You can however also install the GammaLib documentation locally on your machine, and this section describes how to do that. The GammaLib documentation is shipped together with the source code and will be installed in the directory $(prefix)/share/doc/gammalib/html, where $(prefix) is the installation base path, by default /usr/local/gamma. This comprises user documentation and code documentation. To build the user documentation you need the Sphinx reStructuredText documentation generator installed (see http://sphinx-doc.org/rest.html for more information). Code documentation is based on Doxygen, which also needs to be installed on your system (see http://dogygen.org to download and install the Doxygen package). Doxygen will scan the source files for code annotations, and compiles a complete documentation of the implemented C++ classes in a set of html pages. In addition, it will create a set of man files that can be accessed using the man command. To build and install all documentation, type the following $ ./configure
$make doc$ make install

To build only user documentation, type

$./configure$ make sphinx
$make install  and to build only Doxygen documentation, type $ ./configure
$make doxygen$ make install

The Doxygen documentation will be installed into $(prefix)/share/doc/gammalib/html/doxygen and can be browsed using a regular web browser. To check man support, type for example $ man GObservations

and you should see the documentation for the GObservations C++ class.


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# Summary of Bullet Lists discussion

I strongly agree. In fact, anything enclosed between some sort of brackets should be kept in one unit, i.e. without blank line in the middle. If for some reason such content must have blank lines, and so be divided in two units, please balance brackets in each unit by adding matching brackets in a comment, as fx

`[[ <!--]]--> ...`

with similar comments for other kinds of brackets. Unbalanced brackets (including ordinary parentheses) creates lots of problems for translators.

I was wondering, by the way, if a

`<br />`

could replace a completely blank line in case it really is needed?

16:36, 30 September 2010


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set temporary directory - Maple Help

Home : Support : Online Help : Programming : eBookTools Package : eBookTools/SetTempDir

eBookTools[SetTempDir] - set temporary directory

 Calling Sequence SetOutputDir(directory)

Parameters

 directory - path to a directory

Description

 • The SetTempDir command sets the location in which to store temporary files.

Examples

 > $\mathrm{with}\left(\mathrm{eBookTools}\right):$
 > $\mathrm{book}:=\mathrm{NewBook}\left("ProgrammingGuide","Maple Programming Guide","Maplesoft, a division of Waterloo Maple Inc.","2012"\right):$
 > $\mathrm{AddChapter}\left(\mathrm{book},1,"Chapter01.mw"\right):$

Set output folder to be eBook folder on your desktop:

 > $\mathrm{SetTempDir}\left(\mathrm{cat}\left(\mathrm{kernelopts}\left(\mathrm{homedir}\right),\mathrm{kernelopts}\left(\mathrm{dirsep}\right),"Desktop",\mathrm{kernelopts}\left(\mathrm{dirsep}\right),"eBook"\right)\right)$
 ${"/home/maple/Desktop/eBook"}$ (1)
 > $\mathrm{CreatePDF}\left(\mathrm{book}\right)$


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# Homework Help: Evaluate a limit

1. Jun 6, 2012

### KiwiKid

1. The problem statement, all variables and given/known data
Evaluate $\lim_{x \rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}$

2. Relevant equations
The limit laws and the basic rules of algebra.

3. The attempt at a solution
I'm completely stuck right at the beginning. I know I have to find a way to get the denominator to be not-zero, but, well, there doesn't seem to be a way to do that (of course there, I just don't see it). Does anyone have a tip to get me started?

2. Jun 6, 2012

### Pifagor

Hi, are you familiar with l'Hôpital's rule? In this case, you can derive both the enumerator and the denominator, and then take the limit. If this was new to you, check it out on Wikipedia, for example, it's neat!

Regards,
Pifagor

3. Jun 6, 2012

### KiwiKid

Hm, let me see if I get this.

When x goes to 1, both the numerator and the denominator go to zero, therefore, l'Hopital's rule applies if numerator-prime and denominator-prime exist. Deriving the numerator and demoninator ultimately gives $\frac{2\sqrt{x}}{3\sqrt[3]{x^2}}$. (Right?)

Intuitively, I can see the limit evaluates to $\frac{2}{3}$ as $x \rightarrow 1$, but I don't quite see how I can algebraically justify turning those different roots in the numerator and denominator into a $\frac{1}{1}$.

4. Jun 6, 2012

### Infinitum

You can solve this without the L'Hospital method too. Try factorizing numerator using $a^2-b^2$ and denominator using $a^3-b^3$ expansions.

Last edited: Jun 6, 2012
5. Jun 6, 2012

### KiwiKid

Ah, I see. I tried that method quickly, and ended up with the same numerator over the sixth root of x times the third root of x minus one... So, yeah.

6. Jun 6, 2012

### Infinitum

Edited : It works.

7. Jun 6, 2012

### Pifagor

Intuitively, I can see the limit evaluates to $\frac{2}{3}$ as $x \rightarrow 1$, but I don't quite see how I can algebraically justify turning those different roots in the numerator and denominator into a $\frac{1}{1}$.[/QUOTE]

The thing is, you now have a continuous function in both parts, so the quotient is continuous because the the limit of the denominator is nonzero. Insert x=1, and voilá!

Pifagor

8. Jun 6, 2012

### Curious3141

Substitute x = y6, then factorise both numerator and denominator. Find the limit of the expression as y tends to 1.

9. Jun 6, 2012

### KiwiKid

That's a great method, and - seeing as how my book had just explained how introducing a new variable can sometimes make something easier to solve - probably exactly what I'm looking for. Thanks.

10. Jun 6, 2012

### KiwiKid

*kicks himself* You're right. I totally forgot just substituting x with 1, instead assuming x could be anything. Oops.


text:
GamingForum.to provides links to other sites on the internet and doesn't host any files itself.

[1337] 1770. Maximum Score from Performing Multiplication Operations
 LeetCode Guest   LeetCode N/A Posts Threads Likes Credits 1770. Maximum Score from Performing Multiplication Operations You are given two integer arrays nums and multipliers of size n and m respectively, where n >= m. The arrays are 1-indexed. You begin with a score of 0. You want to perform exactly m operations. On the ith operation (1-indexed), you will: Choose one integer x from either the start or the end of the array nums. Add multipliers[i] * x to your score. Remove x from the array nums. Return the maximum score after performing m operations.   Example 1: Input: nums = [1,2,3], multipliers = [3,2,1] Output: 14 Explanation: An optimal solution is as follows: - Choose from the end, [1,2,3], adding 3 * 3 = 9 to the score. - Choose from the end, [1,2], adding 2 * 2 = 4 to the score. - Choose from the end, [1], adding 1 * 1 = 1 to the score. The total score is 9 + 4 + 1 = 14. Example 2: Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6] Output: 102 Explanation: An optimal solution is as follows: - Choose from the start, [-5,-3,-3,-2,7,1], adding -5 * -10 = 50 to the score. - Choose from the start, [-3,-3,-2,7,1], adding -3 * -5 = 15 to the score. - Choose from the start, [-3,-2,7,1], adding -3 * 3 = -9 to the score. - Choose from the end, [-2,7,1], adding 1 * 4 = 4 to the score. - Choose from the end, [-2,7], adding 7 * 6 = 42 to the score. The total score is 50 + 15 - 9 + 4 + 42 = 102.   Constraints: n == nums.length m == multipliers.length 1 <= m <= 103 m <= n <= 105 -1000 <= nums[i], multipliers[i] <= 1000 [TOC] Solution Overview As per the problem description, we have to do m operations and find the maximum score. At every operation, we have to select $i^{th}$ integer from multipliers and multiply it with an integer x from nums. The integer x can be chosen from either the start or the end of the nums. And then we have to remove that integer from nums. At first glance, a greedy approach looks promising. In step i, out of nums[start] and nums[end], we can pick the integer x that maximizes x * multipliers[i]. This greedy approach works well for one of the given examples. nums = [1,2,3], multipliers = [3,2,1] Taking Decision ‣ From multipliers, we have 3, nums is [1, 2, 3], from 3 * 1 and 3 * 3, pick 3, add 3 * 3 = 9. ‣ From multipliers, we have 2, nums is [1, 2], from 2 * 1 and 2 * 2, pick 2, add 2 * 2 = 4. ‣ From multipliers, we have 1, nums is [1], add 1 * 1 = 1. Total Score is 9+4+1=14, which is correct However, it fails for the second example. nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6] Taking Decision ‣ From multipliers, we have 10, nums is [-5,-3,-3,-2,7,1], from (-10) * (-5) and (-10) * 1, pick -5, add (-10) * (-5) = 50. ‣ From multipliers, we have -5, nums is [-3,-3,-2,7,1], from (-5) * (-3) and (-5) * 1, pick -3, add (-5) * (-3) = 15. ‣ From multipliers, we have 3, nums is [-3,-2,7,1], from 3 * (-3) and 3 * 1, pick 1, add 3 * 1 = 3. ‣ From multipliers, we have 4, nums is [-3,-2,7], from 4 * (-3) and 4 * 7, pick 7, add 4 * 7 = 28. ‣ From multipliers, we have 6, nums is [-3,-2], from 6 * (-3) and 6 * (-2), pick -2, add 6 * (-2) = -12. Total Score is 50+15+3+28+(-12)=84 which isn't optimal. 102 is Optimal Solution as given in Problem Example. The logical intuition of why it is not optimal can be deduced from the following two cases: Greedy is short-sighted. For global optimum, we pick local optimum. But picking this Local Optimum may restrict greater positive product afterward. nums = [-10,1,1000,1,1,100], multipliers = [1,1,1] If we pick 100 over -10, we would never ever be able to collect 1000. There are only three elements in multipliers, and we can collect 1000 by taking the left integers only. But selecting 100 at the first point restricts it. Moreover, what if both ends of nums are identical? We don't know which one to favor. One may yield another score, another may yield a very different score. nums = [2, 1000, -1, 2], multipliers = [1, 1] a. if we select Left 2 first, then at the next step, there would be a contest between Left 1000 and Right 2. As per approach we now would select left 1000, obtaining 1002 as the answer. b. if we select Right 2 first, then at the next step, there would be a contest between Left 2 and Right -1. As per approach we now would select left 2, obtaining 4 as the answer. Thus, these examples suggest that we have to look for all two possible combinations at each step: Select nums[start], now problem reduces to another subproblem with nums being nums[start+1] to nums[end] and multipliers being multipliers[i+1] to multipliers[m-1]. Moreover, the number of operations is lessened by one. Select nums[end], now problem reduces to another subproblem with nums being nums[start] to nums[end-1] and multipliers being multipliers[i+1] to multipliers[m-1]. Moreover, the number of operations is lessened by one. We then have to solve these subproblems successively and at each step return the maximum of two possible answers. When all operations are done, we can return 0. Approach 1: Brute Force Intuition As explained above, at each step we have to reduce the problem to two subproblems, with one less operation. And then we have to repeatedly solve these subproblems. This hints at recursion. Now, for solving each subproblem, we essentially need three things nums: the remaining integers to be considered multipliers: the remaining multipliers to be considered op: number of operations done. Minute Improvement: We need not pass multipliers. We are only interested in its i$^{th}$ element. And this i is closely related to op. If we have done 0 operations, this implies we have to do the next operation from multipliers[0]. If we have done 1 operation, this implies we have to do the next operation from multipliers[1]. And so on. Note: There are two possible definitions of op. However, logically they are equivalent. One is to define op as the number of operations done. If we have done 0 operation, this implies we have to do the next operation from multipliers[0]. If we have done 1 operation, this implies we have to do the next operation from multipliers[1]. And so on. In this case, the terminating condition is op == m, when all operations are done. Another is to define op as the number of operations remaining, then the indexing of multipliers and base condition will change. If we have m-0 operations remaining, this implies we have to do next operation from multipliers[0]. If we have m-1 operation remaining, this implies we have to do next operation from multipliers[1]. And so on. In this case, terminating condition is op == 0, when no operation is remaining. We have defined op as the number of operations done. Algorithm Define a helper function that takes two arguments nums, and op. If we are done with all operations, op, return 0. Otherwise, 3.1 One time multiply multipliers[op] with nums[0]. Now, solve the subproblem with nums[1:] and op+1. Add the result of the subproblem to the product. 3.2 Another time multiply multipliers[op] with nums[-1]. Now, solve the subproblem with nums[:-1] and op+1. Add the result of the subproblem to the product. Return the maximum of two results. Call the helper function with nums and op=0 as parameters indicating we have done zero operations so far! Implementation Note: It is likely to give TLE since the time complexity is too high. Note that in python string slicing creates another copy. This consumes a lot of memory. We can reduce it too likewise we removed the passing of the multipliers array. In themultipliers array, we were interested in the left pointer, we obtained it very easily with the op parameter. In the nums array, we were interested in both left and right pointers. Thus, instead of passing the entire nums, we can pass these pointers. Note : It is likely to give TLE because the algorithm is not efficient. Complexity Analysis Let $M$ be the size of multipliers, the same as the number of operations. Time complexity: $O(2^M)$. This can be calculated using the fact that at every step we are reducing the problem of size $M$ to two subproblems of size $M-1$, for doing so we are doing constant time operations (increasing/decreasing left, right and op, and multiplying with y are constant time operations). Thus, the recurrence relation is $T(M)=2T(M-1)+O(1)$, which can be solved using Master Theorem and the result is $O(2^M)$. Another way of analyzing is that at each step, we branch two sub-tree. The height of the recursion tree will be $M$. Thus, there will be $2^M$ nodes in the recursion tree. Space complexity: $O(M)$, the recursion stack will take $O(M)$ space. Approach 2: Recursive Dynamic Programming Intuition We can notice that we may need to solve for a particular (left, right, op) state multiple times. Example: nums = [a, b, c, d, e] and multipliers = [u, v, w, x, y] The tree indicates that from two different paths, we have reached a common state/sub-problem x with c vs d. Thus, we have repeated states. Hence, we can solve this once, and can use its result. Thus, Dynamic Programming. Select best from all possible states and instead of computing again and again, save what you have already computed. Memoizing the pre-visited states while trying all the possible scenarios will reduce the complexity. To determine a state, we essentially need 3 things left: specify we have used left integers from the left side of nums so far. Next, we may use nums[left] right: specify we have used right integers from the right side of nums so far. Next, we may use nums[right] op: number of operations done. Hence, there are 3 state variables, left, right, and op. Thus, it's a 3D Dynamic Programming problem. And to memoize it, we may need a 3D array. If there are n state variables, then we need an array of at most n dimensions. However, with mathematics, we can reduce these 3 state variables to 2. Can you think of how to do that? Try this multiple choice question! (Note that len(nums) will be constant in this approach, since we are not modifying nums and passing indices instead) !mcq! { "description": " The right is related to op, left and len(nums) as", "answer": 1, "choices": [ { "description": "len(nums)-1-left" }, { "description": "len(nums)-1-(op-left)" }, { "description": "len(nums)-1-op" }, { "description": "op-(left-len(nums))" } ], "explanation": "Now, there are len(nums) elements in nums. If we have accomplished op operations, and the left-pointer is ahead by left, means there are op - left operations from the right side. Thus, right should be behind n-1 by op-left. Thus, this formula." } !mcq! Therefore, we can define a state with only two state variables op and left. We will use dp to denote the state in the following. dp[op][left] stores the maximum possible score after we have done op total operations and used left numbers from the left/start side. From this state, we have two options Select Left: Number of operations will advance by one. And, so does the left pointer. Thus, we will multiply mulitpliers[op] and nums[left] (since we have selected from left), and add this product to (optimal) result of state dp[op+1][left+1]. Select Right: Then also the number of operations will advance by one. Then, we will multiply mulitpliers[op] with nums[n-1-(op-left)] (since we have selected from right), and add this product to (optimal) result of state dp[op+1][left] (Now, left will not increment since number has not been chosen from left). Select maximum of results obtained by selecting from Left, and Right. If op == m, means we have performed m operations, add 0. The base Condition of Recursion. Let $\text{mul}$ represent multipliers, and $\text{nums}$ represent nums. Then we can have the following equation! $$dp[op][left] = \begin{cases} \text{max}\bigg((\text{mul}[op]\cdot \text{nums}[left]) + dp[op+1][left+1],\\quad\quad\quad(\text{mul}[op]\cdot \text{nums}[n-1-(op-left)]) + dp[op+1][left]\bigg), & \text{if op \neq m } \[2ex] 0, & \text{if op=m} \end{cases}$$ Algorithm Initialize variable m as the size of multipliers and n as the size of nums. Now, m will help us in determining the number of operations, and with help of n, we can calculate the right pointer. Create a dictionary memo to memoize states. Define a dp function that takes two arguments op, and left. If we are done with all operations, i.e. op == m, return 0. If we have already computed and memoized the state, return the value from the dictionary. Otherwise, 6.1. One time multiply multipliers[op] with nums[left]. Now, solve the subproblem with left+1 and op+1 as parameters. Add the result of the subproblem to the product. 6.2. Another time multiply multipliers[op] with nums[(n-1)-(op-left)]. The index points to the right pointer. Now, solve the subproblem with left and op+1 as parameters. Add the result of the subproblem to the product. Memoize the maximum of 6.1 and 6.2 as the result of state (op, left). Return the memoized result of state op, left. Call the dp function with op = 0 and left = 0. Implementation Note: The python code of top-down DP may give Time Limit Exceeded/Memory Limit Exceeded since it has a larger constant factor. Complexity Analysis Let $M$ be the size of multipliers, the same as the number of operations. Time complexity: $O(M^2)$. op can vary from 0 to M-1. Now, in two recursive calls that we are making, one time we are incrementing left, alongwith op. Other time, we are not incrementing left, but incrementing op. So, left is atmost op. Thus, left also varies from 0 to M-1. So, there are $O(M^2)$ such pairs for computing. Space complexity: $O(M^2)$, the memo will store atmost $M^2$ such pairs! Approach 3: Iterative Dynamic Programming Intuition Using the same equation, we can solve this problem using iterative dynamic programming. We will start from the base condition, and then we will compute the optimal result for each state. We need to convert function dp(operation, left) to dp[op][left]. Hence, we need to use a 2D array. But, we only need those cells where op >= left. Hence, we only need the bottom-right triangle, where left will always be less than or equal to op. As you can see in the diagram, red cells are invalid. Since we know the base condition, we can start from op = m and fill it up. Moreover, since the number of operations at any stage is greater than or equal to the integer chosen from left (op >= left), the left will drop from op to 0. Algorithm Create a 2D array dp of size m+1 by m+1. Reason being op can vary from 0 to m, and so does left. Initialized all elements to 0. By this initialization, we have filled the base condition, that when we have done m operations, and this operation will add nothing to the result. Iterate over op from m-1 to 0. For each op, iterate over left from op to 0. Using the equation, compute the optimal result for state dp[op][left]. We have formulated the problem in such a way that dp[0][0] will force us to do m operations in a row. In other words, dp[0][0] stores the optimal answer. Hence, we can return dp[0][0] at the end. Implementation Complexity Analysis Let $M$ be the size of multipliers, the same as the number of operations. Time complexity: $O(M^2)$ op varies from M-1 to 0, and left varies from to op to 0. This is equivalent to iterating half matrix of order $M\times M$. So, we are computing $O(\frac{M^2}{2})$ states. Space complexity: $O(M^2)$, as evident from dp array. Approach 4: Space Optimized Dynamic Programming Intuition On carefully eyeing mathematical formula and iterative code, we can say that for computing present row $dp[op]$, we need next row $dp[op+1]$ ONLY. Therefore, we can have 1D Array only. We will fill the dp array from bottom to top, saving the next row in a temporary array, and then computing the present row using the saved temporary array. Interview Tip: Always spend time on forming and analyzing Dynamic Programming Equation. For Dynamic Programming, forming an equation requires time. Writing code isn't tough in most cases. Moreover, by analyzing the equation, quite a few times, we can solve the problem using less space. Algorithm Create a 1D array dp of size m+1. Initialize all elements to 0. Iterate over op from m-1 to 0. Since dp array will be modified using next row, and this next row result is saved in dp only, we need to use a temporary array to store the next row. Thus, save all current entries of dp in next_row. For each op, iterate over left from op to 0. Using the equation, compute the optimal result for state dp[left]. Return dp[0] at the end. Implementation Complexity Analysis Let $M$ be the size of multipliers, the same as the number of operations. Time complexity: $O(M^2)$ op varies from M-1 to 0, and left varies from to op to 0. This is equivalent to iterating half matrix of order $M\times M$. So, we are computing $O(\frac{M^2}{2})$ states. Space complexity: $O(M)$, since we have used dp array of size M. https://leetcode.com/articles/maximum-sc...perations/ SPOTS: SPOTS:   •
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text:
On Optimal Secure Message Transmission by Public Discussion

Authors
Type
Preprint
Publication Date
Nov 07, 2009
Submission Date
Jan 15, 2009
Source
arXiv
In a secure message transmission (SMT) scenario a sender wants to send a message in a private and reliable way to a receiver. Sender and receiver are connected by $n$ vertex disjoint paths, referred to as wires, $t$ of which can be controlled by an adaptive adversary with unlimited computational resources. In Eurocrypt 2008, Garay and Ostrovsky considered an SMT scenario where sender and receiver have access to a public discussion channel and showed that secure and reliable communication is possible when $n \geq t+1$. In this paper we will show that a secure protocol requires at least 3 rounds of communication and 2 rounds invocation of the public channel and hence give a complete answer to the open question raised by Garay and Ostrovsky. We also describe a round optimal protocol that has \emph{constant} transmission rate over the public channel.


text:
## PART A

( Answer All Questions Each carries 15 Marks )

1 a) If f (z) is an analytic function with constant modulus , then prove that f (z) is a constant.
b). If w=Φ+iΨ represents the complex potential of an electric field and determine the function Φ .
2.

3   a. Find the bilinear transformation which maps the point -1, i, 1 of the z-plane  on to the points 1,i,-1  Of  the w-plane respectively
b. Show that under the transformation
the real axis in the z-plane is mapped into the  circle │w│=1.

PART B
(Answer any TWO, Each carries 15 Marks)

4.
5.

6.

PART C

(Answer All Questions Each Carries 20 Marks)

7  a. Express v=(2,7,-4) in R3 as a linear combination of the u1=(1,2,0), u2=(1,3,2)  and u3=(0,1,3)  (10 mark)
b.Diagonalize the matrix (10 mark)
8 a. Find the rank and then find the basis for the row space and column space
(15 mark)
b. Are the following sets of vectors linearly independent?
[4  -1   3], [ 0   8    1],  [ 1   3   -5],[2   6   1]       (5 mark)

9 a. Solve the linear system by Gauss elimination method. (10 mark)
b. Find Eigen values and Eigenvectors of the following Matrix (10 mark)


text:
Lemma 15.84.7. Let $I$ be an ideal of a ring $R$. Let $K$ be an object of $D(R)$ with $H^ i(K) = 0$ for $i \not\in \{ -1, 0\}$. Let $R \to R'$ be a ring map. If $K$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5 with respect to $(R, I)$, then $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$ satisfies the equivalent conditions (1), (2), and (3) of Lemma 15.84.5 with respect to $(R', IR')$

Proof. We may assume $K$ is represented by a two term complex $K^{-1} \to K^0$ with $K^0$ free such that for any $a \in I$ the map $a : K^{-1} \to K^{-1}$ is equal to $h_ a \circ d_ K^{-1}$ for some map $h_ a : K^0 \to K^{-1}$. By Lemma 15.84.6 we see that $\tau _{\geq -1}(K \otimes _ R^\mathbf {L} R')$ is represented by $K^\bullet \otimes _ R R'$. Then of course for every $a \in I$ we see that $a \otimes 1 : K^{-1} \otimes _ R R' \to K^{-1} \otimes _ R R'$ is equal to $(h_ a \otimes 1) \circ (d_ K^{-1} \otimes 1)$. Since the collection of maps $K^{-1} \otimes _ R R' \to K^{-1} \otimes _ R R'$ which factor through $\text{d}_ K^{-1} \otimes 1$ forms an $R'$-module we conclude. $\square$

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