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http://math.stackexchange.com/questions/160791/in-axiomatization-of-propositional-logic-why-can-uniform-substitution-be-applie?answertab=active | # In axiomatization of propositional logic, why can uniform substitution be applied only to axioms?
I'm reading an introductory book about mathematical logic for Computation (just for reference, the book is "Lógica para Computação", by Corrêa da Silva, Finger & Melo), and would like to ask a question.
I'm currently reading the chapter that talks about deductive systems, and it begins with Axiomatization. More specifically, it talks about axiomatization as a form of logical inference, that is, an axiomatization of classical logic.
The book is in Portuguese, so all the quotes from it are translated to English.
Since I'm new to the subject, I'm not sure what I need to specify before asking the particular doubt, so I will put here some definitions and explanations that are given in the book (the context).
Context
Right before it presents an axiomatization of classical logic, it defines the concept of substitution:
The substitution of a formula B for an atom p, inside a formula A, is represented by $A[p:=B]$. Intuitively, if we have a formula $A=p\to(p\wedge q)$, and we want to substitute $(r\wedge s)$ for $p$, the result of the substitution is: $A[p:=(r\wedge s)]=(r\wedge s)\to((r\wedge s)\wedge q)$.
Then it defines an instance:
When a formula B results from the substitution of one or more atoms of a formula A, we say that B is an instance of the formula A
Then, it presents an axiomatization for classic propositional logic, which contains several axioms, including, for example, $p\to(q\to p)$ and $(p\to (q\to r))\to((p\to q)\to(p\to r))$. I'm not sure whether I should transcribe the whole set of axioms here, but I think it is not necessary (but I can detail it here if it is necessary).
The axiomatization also includes the rule of inference modus ponens: From $A\to B$ and $A$, one infers $B$.
Next, it states that axioms can be instantiated:
Axioms can be instantiated, that is, atoms can be uniformly substituted by any other formula of the logic. In this case, we say that the resulting formula is an instance of the axiom. With the notion of axiomatization, we can define the notion of deduction.
Definition 2.2.2 A deduction is a sequence of formulae $A_1,...,A_n$ such that every formula in the sequence is either an instance of an axiom, or is obtained from previous formulae by means of inference rules, that is, by modus ponens
A theorem $A$ is a formula such that there exists a deduction $A_1,...,A_n = A$. We represent a theorem as $\vdash_{\text{Ax}} A$ or simply $\vdash A$
Then, it says theorems can also be instantiated to produce new theorems:
The axiomatization presented here possesses the property of uniform substitution, that is, if A is a theorem and B is an instance of A, then B is also a theorem. The reason for that is very simple: if we can apply a substitution to obtain B from A, we can apply the same substitution in the formulas that occur in the deduction of A and, since any instance of an axiom is a deductible formula, we've transformed the deduction of A into a deduction of B.
Doubt
Now I will present the part of the text that generated doubt:
Now we will define when a formula $A$ is deductible from a set of formulae $\Gamma$, also called a theory or a set of hypotheses, which is represented by $\Gamma \vdash_{\text{Ax}} A$. In this case, this concerns adapting the notion of deduction to include the elements of $\Gamma$.
Definition 2.2.3 We say that a formula $A$ is deductible from the set of formulae $\Gamma$ if there is a deduction, that is, a sequence of formulae $A_1,...,A_n = A$ such that every formula $A_i$ in the sequence is:
1) either a formula $A_i \in \Gamma$
2) or an instance of an axiom
3) or is obtained from previous formulae by means of modus ponens.
[...] Note also that we cannot apply uniform substitution in the elements of $\Gamma$; uniform substitution can only be applied to the axioms of the logic.
This is the statement that I didn't understand: "Note also that we cannot apply uniform substitution to the elements of $\Gamma$; uniform substitution can only be applied to the axioms of the logic". Why is this particularly true? Since theorems can be instantiated, it seems that it should be possible to instantiate elements of $\Gamma$ too. Am I missing something?
Thank you in advance.
-
If you did that at least in the way the author understands, the theory would no longer come as sound.
Note that the definition of the set Γ doesn't say anything about its members as axioms or coming as derivable from the axioms. The first clause of definition 2.2.3 says "a formula A$_i$∈Γ" What is Γ? The text says "the set of formulae Γ". The author doesn't clarify, but it does work out that Γ can be any finite subset of the set of all propositional formulas. So, yes you can apply uniform substitution to an element, or even some elements of Γ provided that those element(s) you apply uniform substitution to qualify as either an axiom, a theorem, or the negation of a theorem. But, you can't apply uniform substitution to all elements of Γ without mangling things. More specifically, uniform substitution in terms of deductions can't get applied to semantic contingencies.
Examples of how you might apply uniform substitution to elements of Γ. Suppose Γ contains only C-Kpq-q and K.Kpq.r as formulas (the "-"s and "."s are informal punctuation marks to hopefully make these formulas easier to read... K stands for logical conjunction, C for the material conditional here and I use Polish notation). Then we can write a deduction like this:
• K.Kpq.r -----hyothesis 1
• C-Kpq-q -----hypothesis 2
• C-K.Kpq.r-r -----2 p/Kpq, q/r 3
• r -----1, 3 conditional-out 4
• C-C.Kpq.q-r -----2-4 conditional-in 5
$\vdash$C-KKpqr-C.CKpqq.r -----1-5 conditional introduction 6
Suppose we have C-p-C.Cpq.q and C-Cpq-C.Cqr.Cpr, and C.CCpqq.r as hypotheses. Then we can write a deduction like this:
• C-p-C.Cpq.q -----hypothesis 1
• C-Cpq-C.Cqr.Cpr -----hypothesis 2
• C.CCpqq.r -----hypothesis 3
• C-C p CCpqq-C.C CCpqq r.Cpr -----2 q/CCpqq 4
• C.C CCpqq r.Cpr -----1, 4 conditional-out 5
• Cpr -----3, 5 conditional-out, 6
• C.CCCpqqr.Cpr -----3-6 conditional-in 7
• C.CCpqCCqrCpr.CCCCpqqrCpr -----2-7 conditional-in 8
$\vdash$C-CpCCpqq-C.CCpqCCqrCpr.C:CCCpqqr:Cpr -----1-8 conditional-in 9
Suppose we have K-p-Np (the negation of a theorem), C-Kpq-p (a theorem), and q (a contingent proposition) as hypotheses. Then we can write:
• p -------hypothesis 1
• C-Kpq-p ------ hypothesis 2
• K-p-Np -------hypothesis 3
• K-Cpq-NCpq ------- 3, p/Cpq 4
• C-KCpqNCpq-Cpq ----- 2, p/Cpq, q/NCpq 5
• Cpq ------- 2, 3 conditional-out 6
• C-KpNp-Cpq ------3-6 conditional-in 7
• C-CKpqp-C.KpNp.Cpq ------2-7 conditional-in 8
$\vdash$C-p-C.CKpqp.C:KpNp:Cpq ------1-8 conditional-in 9
Now suppose you were to apply uniform substitution to any possible permissible element of Γ. Then, it would become possible to write deductions like the following:
1 C p q |hypothesis
2 CpCqp |axiom
3 C CpCqp q |1 p/CpCqp, q/q where "x/y" indicates x has gotten substituted by y
4 q |2, 3 modus ponens
Since that system has the deduction meta-theorem (or equivalently conditional introduction), it follows that CCpqq would become a theorem in propositional calculus. But, if p is false and q is false, then CCpqq=CC000=C10=0. Consequently, such a theory would no longer come as sound. Also, you would have inference rules whereby you could "prove" the consequent of a conditional from the conditional itself.
Added: You can apply uniform substitution to contradictions given the deduction metatheorem, and not mangle things since contradictions are false for all interpretations. Contradictions always take on a designated value. Thus, any substitution instance of a contradiction is false also. So, when you discharge any introduced contradictions by conditional introduction, you'll end up with a tautology of classical propositional logic, and thus soundness holds.
Added: There does exist something of another exception, if the author were to introduce propositional logic with quantifiers. St. Jaskowski's original paper The Rules of Supposition in Formal Logic on what we call "natural deduction" has a substitution rule when you have what we call a universal quantifier under special conditions.
-
This seems more to the point than my old answer. I don't think it's right to say that Γ is the set of all propositional formulas? As far as I can tell, it is the set of assumptions forming the basis for a deduction. But that is a minor nit, really. I am puzzled by your choice of notation, though. Using Polish notation is one thing, but you seem to use C for implication? It took me a while to figure that out. What is wrong with good old ⇒, even in Polish notation, as in →pq? – Harald Hanche-Olsen Jun 15 '13 at 20:54
I had to look up what Polish notation was, but is this what you mean? $1:\ \ (p\to q)$, $2:\ \ p\to(q\to p)$, $3: \ \ ((p\to(q\to p)))\to q$ (from 1 by mistaken uniform substitution $p$ / $p\to (q\to p)$ ). Then, from the deduction theorem, it would follow that $(p\to q)\to q$, and it's easy to verify that this is not a tautology. Is that it? – anonymous Jun 15 '13 at 21:05
@anonymous Yes, that's it. – Doug Spoonwood Jun 15 '13 at 21:27
@HaraldHanche-Olsen C comes from the Polish term for "conditional", or at least the transation I have of Lukasiewicz's "Elementy logiki matematycznej" says "If the symbol C, introduced above, stands for an implication, the law can be written in the following form: CCpqCCqrCpr. This law is a conditional sentence..." Lukasiewicz and people who used Polish notation first used "C". If you want to use the arrow instead, go right ahead. It's not my preference, and I find using more conventional symbols in prefix notation less clear than Polish notation. I don't have this text, but... – Doug Spoonwood Jun 16 '13 at 23:47
@DougSpoonwood: It says that Γ is a set of formulae, or assumptions, or hypotheses. I think this is what Harald Hanche-Olsen was saying. It can be any arbitrary set of formulae, and the intention is to the derive logical consequences given these assumptions. This set of formulae is also called a theory by the authors. – anonymous Jun 17 '13 at 0:17
If $A \to B$ is an axiom, it should not be possible to derive $C \to D$; there are valuations in which the former is true and the latter is false. But, the latter is a substitution instance of the former. Thus it is not sound to apply substitution to arbitrary axioms.
-
I think you've implicitly used a definition of an axiom which makes them end up as just formulas, as opposed to a special "type" or "kind" of a formula. Usually when I see a set of axioms for a propositional calculus, each axiom qualifies as a tautology for the intended logical matrix. If all axioms of the system qualify as formulas according to the formation rules (and the formation rules don't result in ambiguity), and if each axiom qualifies as a tautology for the intended logical matrix, then uniform substitution can get applied to any of the axioms of the system. – Doug Spoonwood Jul 25 '13 at 21:51
Also, if A and B qualify as propositional variables in two-valued logic, does anyone use (A→B) as an axiom? I do understand that (A→B) that can qualify as a hypothesis (or assumption or supposition). But, an axiom in a development of two-valued propositional logic from what I see almost universally qualifies as a tautology also. So, axioms end up distinct from hypotheses in that hypotheses may hold true or may not hold true for the intended semantics, but on the other hand axioms definitely do hold true for the intended semantics. – Doug Spoonwood Jul 25 '13 at 22:06
An arbitrary theory may have arbitrary formulas as axioms. – Carl Mummert Jul 26 '13 at 1:42
What do you mean by an "arbitrary theory"? A complete system of two-valued propositional calculus is simply not arbitrary (the selection of two truth values can get regarded as arbitrary, and sure different axiom sets can work with different rules of inference can work). You can't, for instance, found such a theory with only a connective for logical conjunction with only the rule of conjunction-elimination. You can't found such a theory with only the axiom Cpp, modus ponens, and uniform substitution as rules of inference. So what do you mean by "arbitrary theory" here? – Doug Spoonwood Jul 26 '13 at 2:00
(A→B) makes for a perfectly good axiom for one-valued logic, and it should come as possible to derive (C→D) from that axiom. There do not exist any valuations where (A→B) is true and (C→D) false. It does qualify as sound to apply substitution to arbitrary wffs in one-valued logic. – Doug Spoonwood Oct 1 '13 at 0:21
It is done to keep the number of inference rules down to a minimum, in order to be able to reason more easily about theorems and deductions. The axioms need special treatment in order to have a sufficiently rich theory.
It is more common practice (I think) to talk about axiom schemas, which are infinite sets of axioms – corresponding to what your book calls instances of an axiom.
One important feature is that it must be effectively decidable whether or not a given formula is an (instance of an) axiom or not.
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Thank you. To see if I understood: (1) Then, it is done to give axioms special treatment, but it wouldn't necessarily be logically incorrect to use uniform substitution to produce instances of formulas from $\Gamma$? (2) But what about instantiating theorems that were derived from formulas of $\Gamma$ by means of modus ponens? For example, if I have a formula $(A \to B) \in \Gamma$ and I deduce $B$ by modus ponens, am I allowed to instantiate $B$? – anonymous Jun 24 '12 at 11:58
This isn't just a matter of keeping the number of inference rules down. As some other answers in the thread point out, this is an issue of soundness. Suppose that we assume $A$, and then by uniform substitution derive $B$. Then by the deduction theorem, $A \to B$ is a theorem, but this is clearly mistaken. Uniform substitution, in general, applies only to theorems. Within a proof that has assumptions, it is not necessary that each line is actually a theorem (because it may depend on an assumption), so uniform substitution isn't applicable. – Joshua Taylor Jun 15 '13 at 19:21
Ah, that's an interesting point. Too bad I never saw the followup question from the OP last year; I guess summer vacation must have gotten in the way … – Harald Hanche-Olsen Jun 15 '13 at 20:48
@HaraldHanche-Olsen: That's all right. – anonymous Jun 15 '13 at 21:00
I notice that my answer here has gotten a downvote. That's fair enough I suppose, though I think the answer is more addressing the wrong issue than actually being wrong. But I don't want to delete it, if only to preserve the comment by @JoshuaTaylor above. – Harald Hanche-Olsen Jun 16 '13 at 5:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9341534376144409, "perplexity": 655.001967115026}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443737913406.61/warc/CC-MAIN-20151001221833-00209-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://www.bankofcanada.ca/2002/10/working-paper-2002-29/ | # Exponentials, Polynomials, and Fourier Series: More Yield Curve Modelling at the Bank of Canada
Available as: PDF
This paper continues the work started by Bolder and Stréliski (1999) and considers two alternative classes of models for extracting zero-coupon and forward rates from a set of observed Government of Canada bond and treasury-bill prices. The first class of term-structure estimation methods follows from work by Fisher, Nychka, and Zervos (1994), Anderson and Sleath (2001), and Waggoner (1997). This approach employs a B-spline basis for the space of cubic splines to fit observed coupon-bond prices—as a consequence, we call these the spline-based models. This approach includes a penalty in the generalized least-squares objective function—following from Waggoner (1997)—that imposes the desired level of smoothness into the term structure of interest rates. The second class of methods is called function-based and includes variations on the work of Li et al. (2001), which uses linear combinations of basis functions, defined over the entire term-to-maturity spectrum, to fit the discount function. This class of function-based models includes the model proposed by Svensson (1994). In addition to a comprehensive discussion of these models, the authors perform an extensive comparison of these models' performance in the Canadian marketplace.
JEL Code(s): C, C0, C6, E, E4, G, G1
DOI: https://doi.org/10.34989/swp-2002-29 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8386175632476807, "perplexity": 1286.8684330536328}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337906.7/warc/CC-MAIN-20221007014029-20221007044029-00356.warc.gz"} |
http://mathhelpforum.com/differential-equations/194634-getting-into-correct-form.html | # Math Help - Getting into correct form
1. ## Getting into correct form
I'm part way through a question, just need help at the end.
I found the characteristic equation for a fourth order DE with constant co-efficents (and homogeneous) was r^4 - 2r^2 +1=0 so the roots are r = 1, -1 each with multiplicity 2.
so $y=e^x(A+Bx)+e^{-x}(B+Dx)$ I need to get it in terms of sinh and cosh. Thanks
2. ## Re: Getting into correct form
You may want to substitute:
$e^x = \cosh x + \sinh x$
$e^{-x} = \cosh x - \sinh x$
3. ## Re: Getting into correct form
That does the trick. Redefining constants, y=(A+Bx)coshx+(C+Dx)sinhx. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9354612231254578, "perplexity": 2739.9680392633577}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802769990.68/warc/CC-MAIN-20141217075249-00027-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.jobilize.com/online/course/3-2-vector-addition-and-subtraction-graphical-methods-by-openstax?qcr=www.quizover.com&page=1 | # 3.2 Vector addition and subtraction: graphical methods (Page 2/15)
Page 2 / 15
Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor .
Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector .
Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail .
Step 4. Draw an arrow from the tail of the first vector to the head of the last vector . This is the resultant , or the sum, of the other vectors.
Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)
Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.)
The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors.
## Adding vectors graphically using the head-to-tail method: a woman takes a walk
Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction $\text{49.0º}$ north of east. Then, she walks 23.0 m heading $\text{15.0º}$ north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east.
Strategy
Represent each displacement vector graphically with an arrow, labeling the first $\text{A}$ , the second $\text{B}$ , and the third $\text{C}$ , making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted $\mathbf{\text{R}}$ .
Solution
(1) Draw the three displacement vectors.
(2) Place the vectors head to tail retaining both their initial magnitude and direction.
(3) Draw the resultant vector, $\text{R}$ .
(4) Use a ruler to measure the magnitude of $\mathbf{\text{R}}$ , and a protractor to measure the direction of $\text{R}$ . While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector.
In this case, the total displacement $\mathbf{\text{R}}$ is seen to have a magnitude of 50.0 m and to lie in a direction $7.0º$ south of east. By using its magnitude and direction, this vector can be expressed as $R=\text{50.0 m}$ and $\theta =7\text{.}\text{0º}$ south of east.
Discussion
The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in [link] and we will still get the same solution.
full meaning of GPS system
how to prove that Newton's law of universal gravitation F = GmM ______ R²
sir dose it apply to the human system
prove that the centrimental force Fc= M1V² _________ r
prove that centripetal force Fc = MV² ______ r
Kaka
how lesers can transmit information
griffts bridge derivative
below me
please explain; when a glass rod is rubbed with silk, it becomes positive and the silk becomes negative- yet both attracts dust. does dust have third types of charge that is attracted to both positive and negative
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Timothy
hello
Timothy
below me
why below you
Timothy
no....I said below me ...... nothing below .....ok?
dust particles contains both positive and negative charge particles
Mbutene
corona charge can verify
Stephen
when pressure increases the temperature remain what?
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define precision briefly
CT scanners do not detect details smaller than about 0.5 mm. Is this limitation due to the wavelength of x rays? Explain.
hope this helps
what's critical angle
The Critical Angle Derivation So the critical angle is defined as the angle of incidence that provides an angle of refraction of 90-degrees. Make particular note that the critical angle is an angle of incidence value. For the water-air boundary, the critical angle is 48.6-degrees.
okay whatever
Chidalu
pls who can give the definition of relative density?
Temiloluwa
the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas.
Chidalu
What is momentum
mass ×velocity
Chidalu
it is the product of mass ×velocity of an object
Chidalu
how do I highlight a sentence]p? I select the sentence but get options like copy or web search but no highlight. tks. src
then you can edit your work anyway you want
Wat is the relationship between Instataneous velocity
Instantaneous velocity is defined as the rate of change of position for a time interval which is almost equal to zero
Astronomy | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8988084197044373, "perplexity": 775.2185603635944}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400226381.66/warc/CC-MAIN-20200925115553-20200925145553-00537.warc.gz"} |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/?weight=33-64 | ## Results (displaying matches 1-50 of at least 1000) Next
Label Dim. $$A$$ Field CM Traces Fricke sign $q$-expansion
$$a_2$$ $$a_3$$ $$a_5$$ $$a_7$$
1.34.a.a $$2$$ $$6.898$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$-121680$$ $$37919880$$ $$-181061536500$$ $$-6\!\cdots\!00$$ $$+$$ $$q+(-60840-\beta )q^{2}+(18959940+312\beta )q^{3}+\cdots$$
1.36.a.a $$3$$ $$7.760$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$139656$$ $$-104875308$$ $$892652054010$$ $$87\!\cdots\!56$$ $$+$$ $$q+(46552+\beta _{1})q^{2}+(-34958436+\cdots)q^{3}+\cdots$$
1.38.a.a $$2$$ $$8.671$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$-194400$$ $$13991400$$ $$55\!\cdots\!00$$ $$-3\!\cdots\!00$$ $$+$$ $$q+(-97200-\beta )q^{2}+(6995700+72\beta )q^{3}+\cdots$$
1.40.a.a $$3$$ $$9.634$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$548856$$ $$1109442852$$ $$17\!\cdots\!90$$ $$-1\!\cdots\!44$$ $$+$$ $$q+(182952-\beta _{1})q^{2}+(369814284+\cdots)q^{3}+\cdots$$
1.42.a.a $$3$$ $$10.647$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$-344688$$ $$-10820953044$$ $$-2\!\cdots\!50$$ $$57\!\cdots\!92$$ $$+$$ $$q+(-114896+\beta _{1})q^{2}+(-3606984348+\cdots)q^{3}+\cdots$$
1.44.a.a $$3$$ $$11.711$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$-2209944$$ $$24401437812$$ $$53\!\cdots\!70$$ $$30\!\cdots\!56$$ $$+$$ $$q+(-736648-\beta _{1})q^{2}+(8133812604+\cdots)q^{3}+\cdots$$
1.46.a.a $$3$$ $$12.826$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$3814272$$ $$5359866876$$ $$-9\!\cdots\!50$$ $$-7\!\cdots\!08$$ $$+$$ $$q+(1271424+\beta _{1})q^{2}+(1786622292+\cdots)q^{3}+\cdots$$
2.34.a.a $$1$$ $$13.797$$ $$\Q$$ None $$-65536$$ $$-133005564$$ $$538799132550$$ $$-3\!\cdots\!68$$ $$+$$ $$q-2^{16}q^{2}-133005564q^{3}+2^{32}q^{4}+\cdots$$
2.34.a.b $$2$$ $$13.797$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$131072$$ $$8356488$$ $$-5332476660$$ $$13\!\cdots\!56$$ $$-$$ $$q+2^{16}q^{2}+(4178244-\beta )q^{3}+2^{32}q^{4}+\cdots$$
1.48.a.a $$4$$ $$13.991$$ $$\mathbb{Q}[x]/(x^{4} - \cdots)$$ None $$5785560$$ $$38461494960$$ $$-3\!\cdots\!00$$ $$-3\!\cdots\!00$$ $$+$$ $$q+(1446390+\beta _{1})q^{2}+(9615373740+\cdots)q^{3}+\cdots$$
1.50.a.a $$3$$ $$15.207$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$-24225168$$ $$-326954692404$$ $$63\!\cdots\!50$$ $$50\!\cdots\!92$$ $$+$$ $$q+(-8075056+\beta _{1})q^{2}+(-108984897468+\cdots)q^{3}+\cdots$$
2.36.a.a $$1$$ $$15.519$$ $$\Q$$ None $$-131072$$ $$36494748$$ $$389070858750$$ $$-1\!\cdots\!56$$ $$+$$ $$q-2^{17}q^{2}+36494748q^{3}+2^{34}q^{4}+\cdots$$
2.36.a.b $$1$$ $$15.519$$ $$\Q$$ None $$131072$$ $$159933852$$ $$-2\!\cdots\!90$$ $$-7\!\cdots\!44$$ $$-$$ $$q+2^{17}q^{2}+159933852q^{3}+2^{34}q^{4}+\cdots$$
1.52.a.a $$4$$ $$16.473$$ $$\mathbb{Q}[x]/(x^{4} - \cdots)$$ None $$32756040$$ $$403863773040$$ $$12\!\cdots\!80$$ $$65\!\cdots\!00$$ $$+$$ $$q+(8189010+\beta _{1})q^{2}+(100965943260+\cdots)q^{3}+\cdots$$
2.38.a.a $$2$$ $$17.343$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$-524288$$ $$423071208$$ $$-1\!\cdots\!40$$ $$31\!\cdots\!56$$ $$+$$ $$q-2^{18}q^{2}+(211535604-\beta )q^{3}+2^{36}q^{4}+\cdots$$
2.38.a.b $$2$$ $$17.343$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$524288$$ $$-501686808$$ $$41\!\cdots\!00$$ $$-3\!\cdots\!56$$ $$-$$ $$q+2^{18}q^{2}+(-250843404-\beta )q^{3}+\cdots$$
1.54.a.a $$4$$ $$17.790$$ $$\mathbb{Q}[x]/(x^{4} - \cdots)$$ None $$-68476320$$ $$-1\!\cdots\!80$$ $$-4\!\cdots\!00$$ $$-2\!\cdots\!00$$ $$+$$ $$q+(-17119080+\beta _{1})q^{2}+(-262102751820+\cdots)q^{3}+\cdots$$
1.56.a.a $$4$$ $$19.158$$ $$\mathbb{Q}[x]/(x^{4} - \cdots)$$ None $$208622520$$ $$-6\!\cdots\!80$$ $$14\!\cdots\!60$$ $$-2\!\cdots\!00$$ $$+$$ $$q+(52155630+\beta _{1})q^{2}+(-1705367672820+\cdots)q^{3}+\cdots$$
2.40.a.a $$1$$ $$19.268$$ $$\Q$$ None $$524288$$ $$-735458292$$ $$-1\!\cdots\!50$$ $$16\!\cdots\!64$$ $$-$$ $$q+2^{19}q^{2}-735458292q^{3}+2^{38}q^{4}+\cdots$$
2.40.a.b $$2$$ $$19.268$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$-1048576$$ $$287418264$$ $$53\!\cdots\!20$$ $$74\!\cdots\!12$$ $$+$$ $$q-2^{19}q^{2}+(143709132-\beta )q^{3}+2^{38}q^{4}+\cdots$$
3.33.b.a $$10$$ $$19.460$$ $$\mathbb{Q}[x]/(x^{10} + \cdots)$$ None $$0$$ $$-21387150$$ $$0$$ $$-5\!\cdots\!40$$ $$q+\beta _{1}q^{2}+(-2138715+35\beta _{1}-\beta _{2}+\cdots)q^{3}+\cdots$$
1.58.a.a $$4$$ $$20.577$$ $$\mathbb{Q}[x]/(x^{4} - \cdots)$$ None $$-217744560$$ $$37\!\cdots\!60$$ $$-1\!\cdots\!00$$ $$95\!\cdots\!00$$ $$+$$ $$q+(-54436140+\beta _{1})q^{2}+(9368965543140+\cdots)q^{3}+\cdots$$
3.34.a.a $$3$$ $$20.695$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$41202$$ $$129140163$$ $$51261823890$$ $$76\!\cdots\!56$$ $$-$$ $$q+(13734-\beta _{1})q^{2}+3^{16}q^{3}+(-369155924+\cdots)q^{4}+\cdots$$
3.34.a.b $$3$$ $$20.695$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$136620$$ $$-129140163$$ $$-260488036134$$ $$10\!\cdots\!32$$ $$+$$ $$q+(45540-\beta _{1})q^{2}-3^{16}q^{3}+(4863185200+\cdots)q^{4}+\cdots$$
2.42.a.a $$1$$ $$21.294$$ $$\Q$$ None $$-1048576$$ $$5043516516$$ $$-4\!\cdots\!50$$ $$-1\!\cdots\!68$$ $$+$$ $$q-2^{20}q^{2}+5043516516q^{3}+2^{40}q^{4}+\cdots$$
2.42.a.b $$2$$ $$21.294$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$2097152$$ $$8863347528$$ $$97\!\cdots\!80$$ $$21\!\cdots\!56$$ $$-$$ $$q+2^{20}q^{2}+(4431673764-\beta )q^{3}+\cdots$$
3.35.b.a $$10$$ $$21.968$$ $$\mathbb{Q}[x]/(x^{10} + \cdots)$$ None $$0$$ $$119369106$$ $$0$$ $$-1\!\cdots\!72$$ $$q+\beta _{1}q^{2}+(11936911+41\beta _{1}-\beta _{2}+\cdots)q^{3}+\cdots$$
1.60.a.a $$5$$ $$22.046$$ $$\mathbb{Q}[x]/(x^{5} - \cdots)$$ None $$-449691864$$ $$84\!\cdots\!32$$ $$17\!\cdots\!90$$ $$14\!\cdots\!56$$ $$+$$ $$q+(-89938373-\beta _{1})q^{2}+(16803326335969+\cdots)q^{3}+\cdots$$
3.36.a.a $$2$$ $$23.279$$ $$\Q(\sqrt{2196841})$$ None $$-60912$$ $$258280326$$ $$-1\!\cdots\!40$$ $$-1\!\cdots\!44$$ $$-$$ $$q+(-30456-\beta )q^{2}+3^{17}q^{3}+(28571469952+\cdots)q^{4}+\cdots$$
3.36.a.b $$3$$ $$23.279$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$-87330$$ $$-387420489$$ $$27\!\cdots\!10$$ $$48\!\cdots\!64$$ $$+$$ $$q+(-29110+\beta _{1})q^{2}-3^{17}q^{3}+(10829584300+\cdots)q^{4}+\cdots$$
2.44.a.a $$2$$ $$23.422$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$-4194304$$ $$-12981630984$$ $$-3\!\cdots\!00$$ $$11\!\cdots\!08$$ $$+$$ $$q-2^{21}q^{2}+(-6490815492-\beta )q^{3}+\cdots$$
2.44.a.b $$2$$ $$23.422$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$4194304$$ $$-22341634056$$ $$-4\!\cdots\!20$$ $$-2\!\cdots\!28$$ $$-$$ $$q+2^{21}q^{2}+(-11170817028-\beta )q^{3}+\cdots$$
1.62.a.a $$4$$ $$23.566$$ $$\mathbb{Q}[x]/(x^{4} - \cdots)$$ None $$1146312000$$ $$-5\!\cdots\!00$$ $$-5\!\cdots\!00$$ $$-6\!\cdots\!00$$ $$+$$ $$q+(286578000-\beta _{1})q^{2}+(-143430899755500+\cdots)q^{3}+\cdots$$
3.37.b.a $$1$$ $$24.627$$ $$\Q$$ $$\Q(\sqrt{-3})$$ $$0$$ $$387420489$$ $$0$$ $$27\!\cdots\!98$$ $$q+3^{18}q^{3}+2^{36}q^{4}+2757049053441698q^{7}+\cdots$$
3.37.b.b $$10$$ $$24.627$$ $$\mathbb{Q}[x]/(x^{10} + \cdots)$$ None $$0$$ $$-552156750$$ $$0$$ $$-1\!\cdots\!00$$ $$q+\beta _{1}q^{2}+(-55215675-69\beta _{1}-\beta _{2}+\cdots)q^{3}+\cdots$$
1.64.a.a $$5$$ $$25.136$$ $$\mathbb{Q}[x]/(x^{5} - \cdots)$$ None $$507315096$$ $$95\!\cdots\!52$$ $$-5\!\cdots\!30$$ $$37\!\cdots\!56$$ $$+$$ $$q+(101463019-\beta _{1})q^{2}+(190649070219572+\cdots)q^{3}+\cdots$$
2.46.a.a $$2$$ $$25.651$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$-8388608$$ $$-69766206552$$ $$-4\!\cdots\!00$$ $$-9\!\cdots\!44$$ $$+$$ $$q-2^{22}q^{2}+(-34883103276-\beta )q^{3}+\cdots$$
2.46.a.b $$2$$ $$25.651$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$8388608$$ $$59861217192$$ $$43\!\cdots\!00$$ $$79\!\cdots\!24$$ $$-$$ $$q+2^{22}q^{2}+(29930608596-\beta )q^{3}+\cdots$$
4.33.b.a $$1$$ $$25.947$$ $$\Q$$ $$\Q(\sqrt{-1})$$ $$65536$$ $$0$$ $$-196496109694$$ $$0$$ $$q+2^{16}q^{2}+2^{32}q^{4}-196496109694q^{5}+\cdots$$
4.33.b.b $$14$$ $$25.947$$ $$\mathbb{Q}[x]/(x^{14} + \cdots)$$ None $$-23780$$ $$0$$ $$138121491740$$ $$0$$ $$q+(-1699+\beta _{1})q^{2}+(9-21\beta _{1}+\beta _{2}+\cdots)q^{3}+\cdots$$
3.38.a.a $$3$$ $$26.014$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$-310908$$ $$-1162261467$$ $$-9\!\cdots\!90$$ $$-4\!\cdots\!44$$ $$+$$ $$q+(-103636-\beta _{1})q^{2}-3^{18}q^{3}+(112825533616+\cdots)q^{4}+\cdots$$
3.38.a.b $$4$$ $$26.014$$ $$\mathbb{Q}[x]/(x^{4} - \cdots)$$ None $$437562$$ $$1549681956$$ $$-4\!\cdots\!04$$ $$66\!\cdots\!84$$ $$-$$ $$q+(109391-\beta _{1})q^{2}+3^{18}q^{3}+(86524834843+\cdots)q^{4}+\cdots$$
3.39.b.a $$12$$ $$27.439$$ $$\mathbb{Q}[x]/(x^{12} + \cdots)$$ None $$0$$ $$-114742404$$ $$0$$ $$81\!\cdots\!48$$ $$q+\beta _{1}q^{2}+(-9561867+97\beta _{1}+\beta _{2}+\cdots)q^{3}+\cdots$$
4.34.a.a $$3$$ $$27.593$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$0$$ $$92491788$$ $$-53880683886$$ $$45\!\cdots\!92$$ $$-$$ $$q+(30830596+\beta _{1})q^{3}+(-17960227962+\cdots)q^{5}+\cdots$$
2.48.a.a $$1$$ $$27.982$$ $$\Q$$ None $$8388608$$ $$-196634580372$$ $$20\!\cdots\!50$$ $$-5\!\cdots\!96$$ $$-$$ $$q+2^{23}q^{2}-196634580372q^{3}+2^{46}q^{4}+\cdots$$
2.48.a.b $$2$$ $$27.982$$ $$\mathbb{Q}[x]/(x^{2} - \cdots)$$ None $$-16777216$$ $$122289844824$$ $$18\!\cdots\!40$$ $$16\!\cdots\!32$$ $$+$$ $$q-2^{23}q^{2}+(61144922412-5\beta )q^{3}+\cdots$$
3.40.a.a $$3$$ $$28.902$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$-1107000$$ $$3486784401$$ $$93\!\cdots\!90$$ $$13\!\cdots\!04$$ $$-$$ $$q+(-369000-\beta _{1})q^{2}+3^{19}q^{3}+(335300075200+\cdots)q^{4}+\cdots$$
3.40.a.b $$3$$ $$28.902$$ $$\mathbb{Q}[x]/(x^{3} - \cdots)$$ None $$533574$$ $$-3486784401$$ $$-5\!\cdots\!30$$ $$-1\!\cdots\!28$$ $$+$$ $$q+(177858-\beta _{1})q^{2}-3^{19}q^{3}+(319147551244+\cdots)q^{4}+\cdots$$
4.35.b.a $$16$$ $$29.290$$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None $$-27372$$ $$0$$ $$-21372255840$$ $$0$$ $$q+(-1711+\beta _{1})q^{2}+(19-76\beta _{1}-\beta _{2}+\cdots)q^{3}+\cdots$$
3.41.b.a $$12$$ $$30.403$$ $$\mathbb{Q}[x]/(x^{12} + \cdots)$$ None $$0$$ $$-372082572$$ $$0$$ $$-9\!\cdots\!84$$ $$q+\beta _{1}q^{2}+(-31006881-471\beta _{1}+\cdots)q^{3}+\cdots$$
Next | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9148722887039185, "perplexity": 605.8280090656388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00432.warc.gz"} |
https://ps.is.tue.mpg.de/publications?publication_type%5B%5D=Miscellaneous&publication_type%5B%5D=Article&year%5B%5D=2006&year%5B%5D=1993 | #### 2006
##### Bayesian population decoding of motor cortical activity using a Kalman filter
Wu, W., Gao, Y., Bienenstock, E., Donoghue, J. P., Black, M. J.
Neural Computation, 18(1):80-118, 2006 (article)
Abstract
Effective neural motor prostheses require a method for decoding neural activity representing desired movement. In particular, the accurate reconstruction of a continuous motion signal is necessary for the control of devices such as computer cursors, robots, or a patient's own paralyzed limbs. For such applications, we developed a real-time system that uses Bayesian inference techniques to estimate hand motion from the firing rates of multiple neurons. In this study, we used recordings that were previously made in the arm area of primary motor cortex in awake behaving monkeys using a chronically implanted multielectrode microarray. Bayesian inference involves computing the posterior probability of the hand motion conditioned on a sequence of observed firing rates; this is formulated in terms of the product of a likelihood and a prior. The likelihood term models the probability of firing rates given a particular hand motion. We found that a linear gaussian model could be used to approximate this likelihood and could be readily learned from a small amount of training data. The prior term defines a probabilistic model of hand kinematics and was also taken to be a linear gaussian model. Decoding was performed using a Kalman filter, which gives an efficient recursive method for Bayesian inference when the likelihood and prior are linear and gaussian. In off-line experiments, the Kalman filter reconstructions of hand trajectory were more accurate than previously reported results. The resulting decoding algorithm provides a principled probabilistic model of motor-cortical coding, decodes hand motion in real time, provides an estimate of uncertainty, and is straightforward to implement. Additionally the formulation unifies and extends previous models of neural coding while providing insights into the motor-cortical code. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8155558109283447, "perplexity": 1218.8714720806122}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00045.warc.gz"} |
http://math.stackexchange.com/questions/62762/the-degree-of-a-splitting-field-of-a-polynomial | # the degree of a splitting field of a polynomial
Let $f(x)\in F[x]$ be a polynomial of degree $n$. Let $K$ be a splitting field of $f(x)$ over $F$. Then [K:F] must divides $n!$.
I only know that $[K:F] \le n!$, but how can I show that $[K:F]$ divides $n!$?
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Hint: Can you embed $Gal(K/F)$ into $S_n$? – Soarer Sep 8 '11 at 6:58
@soarer: nice hint, but it only works for Galois extensions. – Georges Elencwajg Sep 8 '11 at 8:30
Hint: Try induction on $n$. The base case is clear; in the inductive step, we will want to start with a degree $n+1$ polynomial $f$, and somehow reduce to the case of a degree $\leq n$ polynomial. There are two cases: $f$ is irreducible, and $f$ is reducible.
Suppose $f$ is reducible. Let $p$ be an irreducible factor of $f$, so that $1\leq \deg(p)\leq n$, and let $L$ be the splitting field of $p$ over $F$. Then $K$ is the splitting field of $\frac{f}{p}$ over $L$, and $\deg(\frac{f}{p})=\deg(f)-\deg(p)$. Note that $a!\times b!$ always divides $(a+b)!$ (this is equivalent to the binomial coefficients being integers).
Suppose $f$ is irreducible. Then letting $L=K[x]/(f)\cong K(\alpha)$ for some root $\alpha$ of $f$, we have that $[L:F]=n+1$. Now consider $\frac{f}{x-\alpha}$ (which is of degree $n$) as a polynomial over $L$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.986555814743042, "perplexity": 70.07633591116313}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115862636.1/warc/CC-MAIN-20150124161102-00173-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/4261/proving-a-binomial-sum-identity | Proving a binomial sum identity
Mathematica tells me that
$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$
Although I have not been able to come up with a proof.
Proofs, hints, or references are all welcome.
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You could consider the integral $$\int_{0}^{1} (1-x^2)^n dx .$$
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Aha this is the integral I was searching for! Using the integral $\int_0^1 (1-t)^n \, dt$ I was able to prove $\sum _{k=0}^n {n \choose k}\frac{(-1)^k}{k+1} = \frac{1}{n+1}$ I thought there should be a corresponding integral to my problem and after seeing your answer the integral I was searching for is obvious now! – yjj Sep 8 '10 at 11:02
Sums of the form $$\sum_{k=0}^n(-1)^k{n\choose k}f(k)$$ can often be attacked via the Calculus of finite differences.
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Your identity is a special case of the Chu-Vandermonde identity.
${}_2 F_1(-n,b;c;1)=\frac{(c-b)_n}{(c)_n}$
with $b=\frac12$ and $c=\frac32$. More info on it is in A=B, as mentioned already by John.
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It's on page 181. – Ｊ. Ｍ. Sep 8 '10 at 11:05
I can't offer specific help, but I'd recommend thumbing through Concrete Mathematics looking for techniques. It has many sums that look similar, though of course the difficulty is in the details.
There's also the book A=B, but Concrete Mathematics gives an introduction to the content of A=B and in my opinion is easier to read, so I'd start with Concrete Mathematics.
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There are a powerful algorithms generalizing telescopy (Gosper, Zeilberger et al.) that easily tackle this case of the Chu-Vandermonde identity and much more complicated sums. For example, see this paper which gives as an application a very interesting q-analogy - namely that L. J. Rogers' classical finite version of Euler's pentagonal number theorem is simply the dual of a special case of a q-Chu-Vandermonde.
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$S_n=\sum\limits_{k=0}^n \dfrac{(-1)^k \binom{n}{k}}{2k+1}$
We have:
$S_n=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1} \dfrac{(-1)^k \binom{n}{k}}{2k+1}=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1}\left[\dfrac{n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]$
$S_n=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{n(2n+1)}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]$
$S_n=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk+2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]$
$S_n=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]$
$S_n=\dfrac{(-1)^n}{2n+1}+\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \dfrac{n(-1)^k \binom{n-1}{k}}{n-k}$
$S_n=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \left[(-1)^k \binom{n}{k}\right]+\dfrac{(-1)^n}{2n+1}$
$S_n=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^n \left[(-1)^k \binom{n}{k}\right]$
$\Rightarrow S_n=\dfrac{2n}{2n+1}S_{n-1}+0$ $\Rightarrow S_{n-1}=\dfrac{2n-2}{2n-1}S_{n-2}$ ... $\Rightarrow S_1=\dfrac{2}{3}S_0$
and $S_0=1$
$\Rightarrow S_n=\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\dfrac{(2n)!!}{(2n+1)!!}$
-
As stated, you divide by zero since you sum to $k=n$ and $n-k$ is in the denominator. – Julian Kuelshammer Nov 29 '12 at 7:36
Wah! Thanks for pointing out errors in my solution. – hxthanh Dec 1 '12 at 12:12
Fix this problem as follows sum for $k=0$ to $n-1$ and $k=n$ is calculated separately – hxthanh Dec 1 '12 at 12:34
You can edit your post accordingly. – Julian Kuelshammer Dec 1 '12 at 12:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9697715640068054, "perplexity": 764.0127534636026}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345768632/warc/CC-MAIN-20131218054928-00016-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://forum.allaboutcircuits.com/threads/op-amp-slew-rate.181165/ | # Op-amp slew rate
#### pinkyponky
Joined Nov 28, 2019
153
Hi,
I'm using the op-amp for voltage scaling, the output of the op-amp voltage is using as a reference voltage to the ADC and DAC components. But, the op-amp slew rate is 1V/µs.
So, my question is, do we need to consider the slew rate when you chose the op-amp just for voltage scaling, the input voltage of the op-amp does not change and it is constant voltage.
#### crutschow
Joined Mar 14, 2008
27,956
If the op amp input and output are constant DC, then the opamp slew-rate doesn't matter, since that only affects how fast the op amp can change its voltage. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9749513864517212, "perplexity": 1935.7188245853033}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587963.12/warc/CC-MAIN-20211026231833-20211027021833-00455.warc.gz"} |
http://www.math.wpi.edu/Course_Materials/MA1023B10/impint/node1.html | Subsections
# Improper Integrals
## Purpose
The purpose of this lab is to use Maple to introduce you to the notion of improper integral and to give you practice with this concept by using it to prove convergence or divergence of integrals involving unbounded integrands or unbounded intervals or both.
## Background
Definition 1 We say that the integral
is improper if one or both of the following conditions is satisfied.
1. The interval of integration is unbounded.
2. The function has an infinite discontinuity at some point in . That is, .
### Unbounded integrands
To see how to handle the problem of an unbounded integrand, we start with the following special cases.
Definition 2 Suppose that is continuous on , but . Then we define
provided that the limit on the right-hand side exists and is finite, in which case we say the integral converges and is equal to the value of the limit. If the limit is infinite or doesn't exist, we say the integral diverges or fails to exist and we cannot compute it.
Definition 3 Suppose that is continuous on , but . Then we define
provided that the limit on the right-hand side exists and is finite, in which case we say the integral converges and is equal to the value of the limit. If the limit is infinite or doesn't exist, we say the integral diverges or fails to exist and we cannot compute it.
Cases where has an infinite discontinuity only at an interior point are handled by writing
and using the definitions to see if the integrals on the right-hand side exist. If both exist then the integral on the left-hand side exists. If either of the integrals on the right-hand side diverges, then does not exist.
#### Examples
Here is a simple example using Maple to show that doesn't exist.
> ex1 := int(1/x,x=a..2);
> limit(ex1,a=0,right);
The example above used the right option to limit because the right-hand limit was needed. If you need a left-hand limit, use the left option in the limit command. Maple can usually do the limit within the int command.
> int(1/x,x=0..2);
### Unbounded intervals of integration
These are handled in a similar fashion by using limits. The definition we need the most is given below.
Definition 4 Suppose is continuous on the unbounded interval . Then we define
provided the limit on the right-hand side exists and is finite, in which case we say the integral converges and and is equal to the value of the limit. If the limit is infinite or fails to exist we say the integral diverges or fails to exist.
The other two cases are handled similarly. You are asked to provide suitable definitions for them in one of the exercises.
#### Examples
Using the definition for .
> ex2:=int(1/x^2,x=2..a);
> limit(ex2,a=infinity);
This command shows that Maple takes the limit definition into account in the int command.
> int(1/x^2,x=2..infinity);
## Exercises
1. The gamma function is an example of an improper integral often used to approximate non-integer factorials and is defined below:
Evaluate by calculating the improper integral for and for each integer value of , check your answer by calculating .
2. Both of the following improper integrals given below do not exist. Show, by calculating a limit, why they do not exist.
A
, interval .
B
, interval .
3. Plot and on a single graph.Recall from Calculus II that the volume of a solid of revolution formed by rotating about the -axis over the interval is
. Find the volume of the solid obtained by revolving the curve about the -axis, between and . Repeat this using . | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9953253269195557, "perplexity": 302.89431238549645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948512208.1/warc/CC-MAIN-20171211052406-20171211072406-00034.warc.gz"} |
http://mathhelpforum.com/math-topics/180738-what-does-arrow-down-mean-print.html | # What does "arrow down" mean?
• May 16th 2011, 04:42 AM
julle
What does "arrow down" mean? [SOLVED]
Hi all.
I need some help to understand this, what should be simple, sentence:
Since a \to b as c \downarrow 0 it was shown that...
What does the down arrow mean? It can not mean "goes to", since c is a constant. I'm thinking it is meaning something like "c is a small positive number" ??
Thanks.
• May 16th 2011, 05:49 AM
Plato
Quote:
Originally Posted by julle
Since a \to b as c \downarrow 0 it was shown that...[/I]
What does the down arrow mean? It can not mean "goes to", since c is a constant. I'm thinking it is meaning something like "c is a small positive number" ??
That is limit from the right, from above: $\lim _{x \to c^ + } f(x) = \lim _{x \downarrow c} f(x)$.
• May 16th 2011, 05:56 AM
julle
Thanks for reply. Not sure I understand, though.
If it is the same, why should the author use two different arrows in the same expression?
• May 16th 2011, 06:19 AM
Plato
Quote:
Originally Posted by julle
Thanks for reply. Not sure I understand, though. If it is the same, why should the author use two different arrows in the same expression?
They are not the same, I did not say they were.
$a \to b$ as $c \downarrow 0$ is read "a approaches b as c approaches 0 from the right."
Some authors use $\lim _{x \to c^ + }$ (note is + in the exponent on c).
While others use $\lim _{x \downarrow c}$.
Now those have the same meaning.
• May 16th 2011, 11:55 PM
julle
Okay, thanks.
One last thing: What does "from the right" mean? Does "from the right" equal the positive numbers, and "from the left" equals negative numbers?
Edit:
That makes sense, since c never is negative. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9683045744895935, "perplexity": 1725.880800445203}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997888216.78/warc/CC-MAIN-20140722025808-00143-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://www.nag.com/numeric/MB/manual64_24_1/html/F01/f01gaf.html | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox: nag_matop_real_gen_matrix_actexp (f01ga)
## Purpose
nag_matop_real_gen_matrix_actexp (f01ga) computes the action of the matrix exponential etA${e}^{tA}$, on the matrix B$B$, where A$A$ is a real n$n$ by n$n$ matrix, B$B$ is a real n$n$ by m$m$ matrix and t$t$ is a real scalar.
## Syntax
[a, b, ifail] = f01ga(m, a, b, t, 'n', n)
[a, b, ifail] = nag_matop_real_gen_matrix_actexp(m, a, b, t, 'n', n)
## Description
etAB${e}^{tA}B$ is computed using the algorithm described in Al–Mohy and Higham (2011) which uses a truncated Taylor series to compute the product etAB${e}^{tA}B$ without explicitly forming etA${e}^{tA}$.
## References
Al–Mohy A H and Higham N J (2011) Computing the action of the matrix exponential, with an application to exponential integrators SIAM J. Sci. Statist. Comput. 33(2) 488-511
Higham N J (2008) Functions of Matrices: Theory and Computation SIAM, Philadelphia, PA, USA
## Parameters
### Compulsory Input Parameters
1: m – int64int32nag_int scalar
m$m$, the number of columns of the matrix B$B$.
Constraint: m0${\mathbf{m}}\ge 0$.
2: a(lda, : $:$) – double array
The first dimension of the array a must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array must be at least n${\mathbf{n}}$
The n$n$ by n$n$ matrix A$A$.
3: b(ldb, : $:$) – double array
The first dimension of the array b must be at least max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array must be at least m${\mathbf{m}}$
The n$n$ by m$m$ matrix B$B$.
4: t – double scalar
The scalar t$t$.
### Optional Input Parameters
1: n – int64int32nag_int scalar
Default: The first dimension of the arrays a, b.
n$n$, the order of the matrix A$A$.
Constraint: n0${\mathbf{n}}\ge 0$.
lda ldb
### Output Parameters
1: a(lda, : $:$) – double array
The first dimension of the array a will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array will be n${\mathbf{n}}$
ldamax (1,n)$\mathit{lda}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
A$A$ is overwritten during the computation.
2: b(ldb, : $:$) – double array
The first dimension of the array b will be max (1,n)$\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$
The second dimension of the array will be m${\mathbf{m}}$
ldbmax (1,n)$\mathit{ldb}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
The n$n$ by m$m$ matrix etAB${e}^{tA}B$.
3: ifail – int64int32nag_int scalar
${\mathrm{ifail}}={\mathbf{0}}$ unless the function detects an error (see [Error Indicators and Warnings]).
## Error Indicators and Warnings
Errors or warnings detected by the function:
Cases prefixed with W are classified as warnings and do not generate an error of type NAG:error_n. See nag_issue_warnings.
ifail = 1${\mathbf{ifail}}=1$
Note: this failure should not occur, and suggests that the function has been called incorrectly. An unexpected internal error occurred when trying to balance the matrix A$A$.
W ifail = 2${\mathbf{ifail}}=2$
etAB${e}^{tA}B$ has been computed using an IEEE double precision Taylor series, although the arithmetic precision is higher than IEEE double precision.
ifail = 1${\mathbf{ifail}}=-1$
Constraint: n0${\mathbf{n}}\ge 0$.
ifail = 2${\mathbf{ifail}}=-2$
Constraint: m0${\mathbf{m}}\ge 0$.
ifail = 4${\mathbf{ifail}}=-4$
Constraint: ldamax (1,n)$\mathit{lda}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
ifail = 6${\mathbf{ifail}}=-6$
Constraint: ldbmax (1,n)$\mathit{ldb}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
ifail = 999${\mathbf{ifail}}=-999$
Allocation of memory failed.
## Accuracy
For a symmetric matrix A$A$ (for which AT = A${A}^{\mathrm{T}}=A$) the computed matrix etAB${e}^{tA}B$ is guaranteed to be close to the exact matrix, that is, the method is forward stable. No such guarantee can be given for non-symmetric matrices. See Section 4 of Al–Mohy and Higham (2011) for details and further discussion.
The matrix etAB${e}^{tA}B$ could be computed by explicitly forming etA${e}^{tA}$ using nag_matop_real_gen_matrix_exp (f01ec) and multiplying B$B$ by the result. However, experiments show that it is usually both more accurate and quicker to use nag_matop_real_gen_matrix_actexp (f01ga).
The cost of the algorithm is O(n2m)$\mathit{O}\left({n}^{2}m\right)$. The precise cost depends on A$A$ since a combination of balancing, shifting and scaling is used prior to the Taylor series evaluation.
Approximately n2 + (2m + 8) n ${n}^{2}+\left(2m+8\right)n$ of real allocatable memory is required by nag_matop_real_gen_matrix_actexp (f01ga).
nag_matop_complex_gen_matrix_actexp (f01ha) can be used to compute etAB${e}^{tA}B$ for complex A$A$, B$B$, and t$t$. nag_matop_real_gen_matrix_actexp_rcomm (f01gb) provides an implementation of the algorithm with a reverse communication interface, which returns control to the user when matrix multiplications are required. This should be used if A$A$ is large and sparse.
## Example
```function nag_matop_real_gen_matrix_actexp_example
a = [0.7,-0.2, 1.0, 0.3;
0.3, 0.7, 1.2, 1.0;
0.9, 0.0, 0.2, 0.7;
2.4, 0.1, 0.0, 0.2];
b = [0.1, 1.2;
1.3, 0.2;
0.0, 1.0;
0.4, -0.9];
m = int64(2);
t = 1.2;
% Compute exp(ta)b
[a, b, ifail] = nag_matop_real_gen_matrix_actexp(m, a, b, t)
```
```
a =
0.2500 -0.8000 2.0000 0.6000
0.0750 0.2500 0.6000 0.5000
0.4500 0 -0.2500 0.7000
1.2000 0.2000 0 -0.2500
b =
0.2138 7.6756
4.9980 11.6051
0.8307 7.5468
1.2406 9.7261
ifail =
0
```
```function f01ga_example
a = [0.7,-0.2, 1.0, 0.3;
0.3, 0.7, 1.2, 1.0;
0.9, 0.0, 0.2, 0.7;
2.4, 0.1, 0.0, 0.2];
b = [0.1, 1.2;
1.3, 0.2;
0.0, 1.0;
0.4, -0.9];
m = int64(2);
t = 1.2;
% Compute exp(ta)b
[a, b, ifail] = f01ga(m, a, b, t)
```
```
a =
0.2500 -0.8000 2.0000 0.6000
0.0750 0.2500 0.6000 0.5000
0.4500 0 -0.2500 0.7000
1.2000 0.2000 0 -0.2500
b =
0.2138 7.6756
4.9980 11.6051
0.8307 7.5468
1.2406 9.7261
ifail =
0
```
Chapter Contents
Chapter Introduction
NAG Toolbox
© The Numerical Algorithms Group Ltd, Oxford, UK. 2009–2013 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 71, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9406313896179199, "perplexity": 4701.263633481721}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469258943366.87/warc/CC-MAIN-20160723072903-00139-ip-10-185-27-174.ec2.internal.warc.gz"} |
http://en.wikipedia.org/wiki/Talk:Generalized_continued_fraction | # Talk:Generalized continued fraction
WikiProject Mathematics (Rated B-class, Mid-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
B Class
Mid Importance
Field: Analysis
## Incorrect formatting in series section
The LaTeX in the series section is all messed up, starting with Euler's formula. I don't know this stuff well so I would prefer to leave it to somebody else to fix. I think frac needs to be changed to cfrac, and parentheses need to be moved around. Adking80 (talk) 14:29, 29 June 2009 (UTC)
## Bug
Would somebody familiar with this topic comment on the Bug here thing inserted by somebody in the main text? Thanks a lot. Oleg Alexandrov 05:02, 11 Apr 2005 (UTC)
Didn't have time to read it, but two things I noticed right away:
1. the usual terminology is "simple continued fraction" (not "c.f." unless qualified by "hereafter we deal only with s.c.f.s") and "continued fraction" (not "generalized c.f.")
2. Many, many "generalized continued fractions" exist, but this usually refers to something like the Jacobi-Perron algorithm for simultaneous rational approximation of real vectors, or for continued fractions in which the coefficients are functions, for operator-valued formulations, etc.
A good place to begin reading about one dimensional continued fractions is
• Brezinski, Claude (1991). History of continued fractions and Pade approximates. New York: Springer-Verlag. 3-540-15286-5., but this book is already seriously out of date.
HTH ---CH 21:10, 5 May 2006 (UTC)
Well, I'm sorry I won't get a chance to talk to Mr. Hillman. It appears that he's not participating in Wikipedia any longer. But what he said goes double for me. Anyway, the "Bug here" thing that somebody once inserted undoubtedly referred to the absolutely ludicrous definitions currently in place for "continued fraction" (in this article) and "generalized continued fraction" (in this one). DavidCBryant 20:21, 30 November 2006 (UTC)
## Not "generalized" enough?
It is absolutely incorrect to use the name "Generalized Continued Fraction" for second-order continued fractions as those defined here. All this is explained at: Generalized Continued Fractions —Preceding unsigned comment added by Arithmonic (talkcontribs) 02:43, 15 October 2006
Wikipedia is not about what's correct and what's incorrect. It's about what the term "Generalized Continued Fraction" refers to in the mainstream mathematical literature. -- Jitse Niesen (talk) 04:12, 15 October 2006 (UTC)
Well, I've read quite a few books about continued fractions, and I have never seen a reference to a "generalized" continued fraction whose partial numerators and partial denominators have to be integers. I'm absolutely certain that arithmonic is right. The thing defined by this article right now is a very special case of a very narrow class of continued fractions. Saying that it's "generalized" is farcical. DavidCBryant 20:21, 30 November 2006 (UTC)
Are you saying that the references listed in the article do not define "generalized continued fraction" as explained in the article? Then, I'd be very grateful if you could change it. If, instead, you're saying that the list of references is not representative, then add your definition and quote a reference which uses it. By the way, the TeX command for $\ddots$ is \ddots (diagonal dots). -- Jitse Niesen (talk) 01:50, 1 December 2006 (UTC)
Yeah, I was saying that the definition that was in the article is not generally accepted by mathematicians, including the mathematicians who wrote the references. So I've made some initial changes, JN. I've also added one new article about the fundamental recurrence relationship between successive convergents, and I've added some content to the article about the convergents of a continued fraction.
So far I'm just trying to open up the number-theoretic definition that used to be here so that applications of continued fractions to complex analysis are possible, by the "Wikipedia" definition. Other generalizations (into continued fractions of differential operators, or continued fractions of matrices) are a little outside my area of expertise, but I'll try to work in a couple of references to them. DavidCBryant 16:14, 3 December 2006 (UTC)
## Inconsistent symbols for partial denominators
I notice that the article Continued fraction uses "an" for partial denominators while this article Generalized continued fraction uses that for the partial numerators and uses "bn" for partial denominators. I feel that these should be changed to be consistent. JRSpriggs 06:20, 15 December 2006 (UTC)
I'd really rather not do that. The usage here (a in the numerator and b in the denominator, with the somewhat weird 0th denominator standing outside the fraction) is practically universal in the analytic theory of continued fractions, from Gauss until the present day. If we do what you're proposing, it will probably just confuse anyone who is actually trying to study analysis and continued fractions out of a book.
Continued fractions have applications in both number theory and analysis (maybe in algebra, too -- I'm not sure about that). The number theory guys don't care about fractions that incorporate anything except integers. In analysis we can get all kinds of weird things floating around inside the fraction. The notation has evolved differently pretty naturally, because in number theory the partial numerators are almost always "1", and can effectively be ignored, while in analysis we prove a theorem (not in Wikipedia yet, but I'm working on it) that "most" continued fractions can be expressed in infinitely many ways, two of the most useful of which involve "1"s as numerators or, alternatively, "1"s as denominators. In analysis we can even have some 0s in the denominators and still get good (convergent) results.
Anyway, the "continued fraction" article is already dedicated to number theory (and not even the more interesting aspects of that, like solving Diophantine equations; it's mostly about the representation of irrational numbers in canonical form). Can't this article be for analysis? DavidCBryant 12:59, 15 December 2006 (UTC)
You could change the an in the other article to bn instead. I did not say anything about number theory versus analysis. JRSpriggs 08:43, 16 December 2006 (UTC)
I read too much into your remarks, I suppose (arithmetic vs analysis). The thing is, I have run across both notations, in books. And it is fairly common for books that concentrate on the number theoretic applications of continued fractions (such as Diophantine equations, and how the set of all Dedekind cuts can be placed in one-to-one correspondence with the simple cfs in canonical form) to use the symbol a in the partial denominators; and for books that concentrate on analytic cfs to use the a/b convention that's in this article right now. Anyway, when I first suggested trying to introduce a more general definition into the continued fractions article, I felt as if I had a tiger by the tail! I'm not sure I want to re-open that can of worms.
I'm not opposed to making changes to articles. But this inconsistency in notation is present in the mainstream literature, AFAIK. So I don't see much harm in leaving a similar inconsistency in Wikipedia. But yeah, if one were determined to eliminate the inconsistency, I'd rather do it by opening up the continued fractions article just a little bit. DavidCBryant 12:02, 16 December 2006 (UTC)
## History section
I'm particularly curious about one thing. I'm fairly certain that the uniqueness of the canonical continued fraction representation of an irrational real number was not a subject of discussion until sometime during the 19th century. It just makes sense that this result would have been roughly contemporaneous with Kronecker's research into Cauchy sequences and Dedekind's new idea of a cut in the rationals. But I'm not sure who came up with the canonical representation in the first place. Lagrange used canonical fractions for quadratic irrationals, but he didn't know about transcendental numbers, which didn't really become a hot issue until algebraic number theory was fairly far along. Does anybody have a clue, or an idea of how I should search for this factoid? DavidCBryant 01:06, 16 December 2006 (UTC)
## Regularization of generalized continued fractions
If one multiplies both the numerator and denominator of one of the ratios by the same constant c, then the resulting continued fraction should have the same value. For example, if we do this to the second ratio, this
$x = b_0 + \cfrac{a_1}{b_1 + \cfrac{a_2}{b_2 + \cfrac{a_3}{b_3 + \cfrac{a_4}{\ddots\,}}}}$
would become this
$x = b_0 + \cfrac{a_1}{b_1 + \cfrac{a_2 c}{b_2 c + \cfrac{a_3 c}{b_3 + \cfrac{a_4}{\ddots\,}}}}$
If we then divide each ratio's numerator and denominator by its numerator, we get
$x = b_0 + \cfrac{1}{\frac{b_1}{a_1} + \cfrac{1}{\frac{b_2 a_1}{a_2} + \cfrac{1}{\frac{b_3 a_2}{a_1 a_3} + \cfrac{1}{\ddots\,}}}}$
Should we not put this into the article? JRSpriggs 09:06, 16 December 2006 (UTC)
Yes, this should definitely go in the article. It's called an equivalence transformation, or sometimes an equivalence relation, and it's usually expressed this way:
$b_0 + \cfrac{a_1}{b_1 + \cfrac{a_2}{b_2 + \cfrac{a_3}{b_3 + \cfrac{a_4}{\ddots\,}}}} = b_0 + \cfrac{c_1a_1}{c_1b_1 + \cfrac{c_1c_2a_2}{c_2b_2 + \cfrac{c_2c_3a_3}{c_3b_3 + \cfrac{c_3c_4a_4}{\ddots\,}}}}$
where the ci are any non-zero complex numbers, and equality is to be understood in the sense that the two sequences of successive convergents are numerically equal, term by term. The advantage of expressing it this way is that it then becomes easy to explain the transformation into all "1"s in the numerators, as well as the transformation into all "1"s in the denominators, and some other special forms that prove useful as the theory is developed farther.
Anyway, I was aiming to put the equivalence transformation in an "elementary results" section, along with some observations on vanishing ai and bi, and maybe some stuff about separating a continued fraction into its even and odd parts. I'd welcome your assistance if you want to collaborate, JR. dcb
PS Oh, yeah, I almost forgot to mention another thing that ought to be in this article. A particular continued fraction can be thought of as the composition of a sequence (possibly infinite) of Möbius transformations. I think that idea should come right after the "elementary considerations" section, or maybe even be a subheading under "elementary considerations". The advantage of using this device is that it brings the machinery of complex analysis (conformal mapping) directly to bear on the continued fraction itself, and this makes many proofs much simpler than they would be without it. (I guess one disadvantage might be that a link to mobius transformation brings the reader directly to a "bijective conformal map" on the Riemann sphere, and to "automorphism group", as well, which will probably scare all the non-mathematicians out of town before sundown!) DavidCBryant 12:50, 16 December 2006 (UTC)
## Another formula for e
When working on the derangements at Rencontres numbers, I noticed that they yield another formula for e:
$e = 2 + \cfrac{2}{2 + \cfrac{3}{3 + \cfrac{4}{4 + \cfrac{5}{\ddots\,}}}}$
where the corresponding partial numerators and partial denominators are all equal and run thru the integers beginning with 2. Amazing. JRSpriggs 04:19, 21 December 2006 (UTC)
Yes, it is amazing ... and very beautiful, also. Oh -- when I looked at derangement just now I noticed a minor case of vandalism. Somebody might want to check to be sure I fixed it correctly. I'm also curious, JR -- do you think one of us ought to work a reference to continued fractions into the derangement article? DavidCBryant 12:16, 21 December 2006 (UTC)
If we can figure out how to make it sufficiently relevant to that article. JRSpriggs 08:01, 22 December 2006 (UTC)
I'll think about that a little. Pointing out that the fundamental recurrence formulas used in the theory of continued fractions are identical with the recursion relation generating the "subfactorials" might be a useful note in derangement. (I kind of like "derangements" ... mathemeticians, who in general are slightly deranged anyway, tend to like them. ;^>)
Oh -- I took a look, and there's an equivalent version of this cf in the article about Euler's number. It wouldn't hurt the section on equivalence transformations in this article to have a worked example, and this particular e-quivalence is a very natural example. Just a thought. DavidCBryant 12:07, 22 December 2006 (UTC)
I think I will put it into e (mathematical constant). And yes, showing the equivalence would be a good idea. JRSpriggs 05:29, 23 December 2006 (UTC)
## Maps whole plane to a point?
To David: You say "And every member of that automorphism group maps the extended complex plane into itself – not one of the Τns can possibly map the plane into a single point. Yet in the limit the sequence {Τn} defines an infinite continued fraction which (if it converges) does in fact map the entire complex plane into a single point.". This is impossible. Suppose w is the value on which you are trying to focus the plane. For each n, choose a point zn which maps to a point outside the disc with radius one around w. Since the extended complex plane (with the point at infinity added) is compact, this sequence must have at least one cluster point. Call it Z. Then Tn cannot converge uniformly to w in a neighborhood of Z. Is that not so? JRSpriggs 09:58, 12 January 2007 (UTC)
You're right. That is so. I have to admit I'm speaking loosely in this section of the article. I tried to emphasize this by talking about the difference between intuition and proof in the opening paragraph for the section. Talking about limits as if they can ever be "reached" is always loose talk, when you come right down to it. The point I'm trying to make is that a sequence of LFT's can have a limit (a continued fraction – not its value, but the fraction itself) that lies outside the set of all LFT's. (I guess another way to look at it is that the infinite continued fraction is not a linear fractional transformation, but each truncation of the fraction is easily associated with an LFT, and those LFT's can be thought of as forming a sequence that converges to the continued fraction.)
The analogous case that's most familiar is a Cauchy sequence of rational numbers whose limit is irrational. A simple continued fraction (with nothing but integers for elements) can always be constructed that's equivalent to such a Cauchy sequence, and vice versa. Maybe the wording should be adjusted to run more along those lines? Or maybe there should be an "intuitively" in the sentence in question, somewhere? "Does in fact" sounds pretty dogmatic.
On the other hand, I think you're talking about uniform convergence, and I was only trying to talk about simple convergence. Isn't your argument about the cluster point the same basic reason we can't really say "the power series for ez converges uniformly everywhere in the complex plane" and instead must say "this power series converges uniformly on every bounded domain lying in ℂ"? Anyway, thanks for looking at this stuff. I appreciate your feedback. DavidCBryant 12:14, 12 January 2007 (UTC)
I thought of a more concrete point, a counter-example. Consider the usual continued fraction for the golden ratio which has all the partial numerators and partial denominators equal to one. Indeed it will converge to the golden ratio as n goes to infinity for all complex z EXCEPT for $\frac{1-\sqrt{5}}{2}$ (the other root of the equation z2 - z - 1 = 0) which goes to itself regardless of what n is. JRSpriggs 12:29, 12 January 2007 (UTC)
Hi, JR! I had just about concluded that I really need to reword this thing, anyway. It's wrong to call a continued fraction a "mapping". A cf can be regarded as the limit of a sequence of "mappings", but the cf itself is not a mapping ... just as an irrational number can be described as the limit of a sequence of rationals, even though it isn't a rational number. Thanks for the very cute counter-example. Maybe we ought to work that into the article somehow? I guess it will work for any convergent periodic continued fraction with two distinct fixed points (that is, the cf will converge to one value, but the LFT actually fixes two points, one of which is the other root of the quadratic). DavidCBryant 18:34, 12 January 2007 (UTC)
Actually I made mistake in my counter-example. I should have said -- if
$\phi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\ddots\,}}}}$
then we get
$\lim_{n \to \infty} T_n (- \phi) = 1 - \phi\!$
while
$\lim_{n \to \infty} T_n (z) = \phi\!$
for any other extended complex number z. My error was that I was thinking of z as the entire denominator of the last ratio, but actually that is 1+z. I do not know what, if anything, we could say about such a 'point of divergence' that would be completely general. JRSpriggs 05:56, 13 January 2007 (UTC)
## set of GCF numbers
It seems that the set of numbers which can be represented by g.c.f's of which the coefficients form in a natural way a rational number (a repetitive sequence of digits) is a strict superset of the rational numbers, but of course not all of the real line. Is there a name for this set? --MarSch 18:41, 13 February 2007 (UTC)
I think I understand what you're driving at, MarSch, but I'm not certain. Are you talking about periodic continued fractions? (See this article for the definition.) If the coefficients of the continued fraction are restricted to be integers, then the superset to which you refer is the set of quadratic irrational numbers (plus the rationals, of course), or algebraic numbers of the second degree. If that restriction is removed, and the coefficients are allowed to assume arbitrary complex values, then a continued fraction can represent any complex number whatsoever.
One of the beautiful properties of continued fractions is that the (positive) square root of any non-square natural number can be represented by a periodic simple continued fraction in canonical form. So in a way, square roots are like rational numbers, because they can be represented by a finite pattern of natural numbers that repeats itself over and over again. See this article for some examples. DavidCBryant 19:24, 13 February 2007 (UTC)
## Conditions for irrationality
Can someone state the necessary and sufficient conditions for a convergent generalized continued fraction to converge to an irrational number? For instance if
$x = b_0 + \underset{i=1}{\overset{\infty}{K}} \frac{a_i}{b_i}\,$
is known to converge and
$a_i \,$ and $b_i \,$ are both functions of $i \,$
and
$a_i \,$ and $b_i \,$ have positive integer values that increase as $i \,$ increases to infinity
does this imply that $x \,$ is irrational? Frank M Jackson (talk) 17:44, 6 November 2008 (UTC)
Have now found references to Lambert's Irrationality proofs for generalized continued fractions and have added a section with appropriate references. Frank M Jackson (talk) 11:22, 8 November 2008 (UTC)
## Notation: suggest changing to short expression in place of long
Since, to quote the section on Notation:
"The long continued fraction expression ... takes up a lot of space in a book (and it's not easy for the typesetter, either)"
would it not be better to use the short expression
$x = b_0 + \frac{a_1}{b_1+} \frac{a_2}{b_2+} \frac{a_3}{b_3+} \frac{a_4}{b_4+} \cdots\,$
in place of the long one
$x = b_0 + \cfrac{a_1}{b_1 + \cfrac{a_2}{b_2 + \cfrac{a_3}{b_3 + \cfrac{a_4}{b_4 + {\ddots\,}}}}}$
after the introduction?
Here's how the Examples section would appear after the suggested change:
###### ============================================================
Here are two continued fractions that can be built via Euler's identity.
$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots= \frac{x}{1+}\frac{1^2x}{2-x+}\frac{2^2x}{3-2x}+\frac{3^2x}{4-3x}+\frac{4^2x}{5-4x}\cdots\,$
$\ e^x=1+x+\frac{x^2}{2!}+\cdots= 1+\frac{x}{1-}\frac{x}{x+2-}\frac{2x}{x+3-}\frac{3x}{x+4-}\frac{4x}{x+5-}\cfrac{5x}{x+6-}\cdots$
More advanced techniques are necessary to construct the following examples.
$\ e^{2m/n}=1+\frac{2m}{n-m+}\frac{m^2}{3n+}\frac{m^2}{5n+}\frac{m^2}{7n+}\frac{m^2}{9n+}\frac{m^2}{11n+}\cdots$
Setting m = x and n = 2 yields
$\ e^x=1+\frac{2x}{2-x+}\frac{x^2}{6+}\frac{x^2}{10+}\frac{x^2}{14+}\frac{x^2}{18+}\frac{x^2}{22+}\cdots$
$\pi=3+\frac{1}{6+}\frac{9}{6+}\frac{25}{6+}\frac{49}{6+}\frac{81}{6+}\frac{121}{6+}\cdots$
$\pi = \frac{4}{1 +} \frac{1}{3 +} \frac{4}{5 +} \frac{9}{7 +} \frac{16}{9 +} \frac{25}{11 +} \frac{36}{13 +} \frac{49}{15 +} \cdots$
Glenn L (talk) 03:06, 5 March 2009 (UTC)
Sounds good to me. Robinh (talk) 08:04, 5 March 2009 (UTC)
## Convergence of Nth roots
Looking at them I don't think the Nth root continued fractions converge. The partial numerators go up as squares whereas the denominators go up linearly. It doesn't seem like a good combination to me. Dmcq (talk) 17:15, 21 November 2009 (UTC)
Oh sorry I see, it's like the expansion for π just above. I'll have to work out for myself what it actually means in convergence terms. Dmcq (talk) 17:28, 21 November 2009 (UTC)
The secret is choosing x and y so that xn > |y|, if possible. Using Pogson's ratio, for example,
$\sqrt[5]{100} = \sqrt[5]{2.5^5 + \frac{75}{32}} = \sqrt[5]{2^5 + 68},$
but the first expression converges much more quickly since 97.65625 >> 2.34375, while 32 < 68.
Nevertheless, the second expression does converge. It just takes longer.---Glenn L (talk) 03:21, 22 November 2009 (UTC) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 27, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9184698462486267, "perplexity": 582.6487757321252}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345758566/warc/CC-MAIN-20131218054918-00008-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/109617-solved-surface-area.html | Math Help - [SOLVED] Surface Area
1. [SOLVED] Surface Area
Find the surface area of the surface of revolution generated by revolving the graph , around the x-axis.
This is what I got so far I don't know what to do next.
Attached Thumbnails
2. Originally Posted by superman69
Find the surface area of the surface of revolution generated by revolving the graph , around the x-axis.
This is what I got so far I don't know what to do next.
I'm not sure how you got your result ...
If y = f(x) then the surface area produced by revolution of the curve around the x-axis is:
$A = 2\pi \int_a^b\left(y \sqrt{1+(y')^2}\right)dx$
$A = 2\pi \int_0^8\left(x^3 \sqrt{1+9x^4}\right)dx$
Use integration by substitution:
$u = 1+9x^4~\implies~\dfrac{du}{dx}=36x^3\implies~du=36 x^3 \cdot dx$
$A = 2\pi \cdot \dfrac1{36} \int_0^8\left(36x^3 \sqrt{1+9x^4}\right)dx = \dfrac \pi{18} \int u^{\frac12} du$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9155447483062744, "perplexity": 268.78215109855654}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802772972.2/warc/CC-MAIN-20141217075252-00079-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://associazionegeotecnica.it/articoli_rig/viscoplastic-modelling-of-fluids-filled-porous-chalks/ | In this paper two rate dependent constitutive models for porous chalks are presented. The common background of both models formulation is the isotach approach, which is first introduced. The theoretical basis and the mathematical formu- lation of the models are then proposed. Finally, numerical predictions of both models are compared with experimental results and discussed. The two formulations include strain rate and creep effects. Suction dependency on creep is also dis- cussed.
In this paper two rate dependent constitutive models for porous chalks are presented. The common background of both models formulation is the isotach approach, which is first introduced. The theoretical basis and the mathematical formu- lation of the models are then proposed. Finally, numerical predictions of both models are compared with experimental results and discussed. The two formulations include strain rate and creep effects. Suction dependency on creep is also dis- cussed. Some perspectives about the constitutive modelling of time-dependent behaviour of partially saturated chalks are given with respect to the description of the coupled effect of suction and time.
In this paper two rate depende …
AUTORI: De Gennaro V., Pereira J.-M., Gutierrez M., Hickman R. RIG ANNO: 2008 NUMERO: 1 Numero di pagina: 44
Allegato: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9299876093864441, "perplexity": 3559.7496716419128}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335491.4/warc/CC-MAIN-20220930145518-20220930175518-00351.warc.gz"} |
https://alice-figure.web.cern.ch/node/22888 | # First measurement of the $Λ$-$Ξ$ interaction in proton-proton collisions at the LHC
Description
##### Abstract:
The first experimental information on the strong interaction between $\Lambda$ and $\Xi^-$ strange baryons is presented in this Letter. The correlation function of $\Lambda-\Xi^-$ and $\overline{\Lambda}-\overline{\Xi}^{+}$ pairs produced in high-multiplicity proton-proton (pp) collisions at $\sqrt{s}$ = 13 TeV at the LHC is measured as a function of the relative momentum of the pair. The femtoscopy method is used to calculate the correlation function, which is then compared with theoretical expectations obtained using a meson exchange model, chiral effective field theory, and Lattice QCD calculations close to the physical point. Data support predictions of small scattering parameters while discarding versions with large ones, thus suggesting a weak $\Lambda-\Xi^{-}$ interaction. The limited statistical significance of the data does not yet allow one to constrain the effects of coupled channels like $\Sigma-\Xi$ and N$-\Omega$.
e-Print: arXiv:2204.10258 | PDF | inSPIRE
Paper: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9937392473220825, "perplexity": 596.6738005992322}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334855.91/warc/CC-MAIN-20220926082131-20220926112131-00748.warc.gz"} |
http://eprint.iacr.org/2014/574 | ## Cryptology ePrint Archive: Report 2014/574
Security Analysis of Multilinear Maps over the Integers
Hyung Tae Lee and Jae Hong Seo
Abstract: At Crypto 2013, Coron, Lepoint, and Tibouchi~(CLT) proposed a practical Graded Encoding Scheme (GES) over the integers, which has very similar cryptographic features to ideal multilinear maps. In fact, the scheme of Coron~{\em et al.} is the second proposal of a secure GES, and has advantages over the first scheme of Garg, Gentry, and Halevi~(GGH). For example, unlike the GGH construction, the subgroup decision assumption holds in the CLT construction. Immediately following the elegant innovations of the GES, numerous GES-based cryptographic applications were proposed. Although these applications rely on the security of the underlying GES, the security of the GES has not been analyzed in detail, aside from the original papers produced by Garg~{\em et~al.} and Coron~{\em et~al.}
We present an attack algorithm against the system parameters of the CLT GES. The proposed algorithm's complexity $\tilde\bO(2^{\rho/2})$ is exponentially smaller than $\tilde\bO(2^{\rho})$ of the previous best attack of Coron~{\em et al.}, where $\rho$ is a function of the security parameter. Furthermore, we identify a flaw in the generation of the zero-testing parameter of the CLT GES, which drastically reduces the running time of the proposed algorithm. The experimental results demonstrate the practicality of our attack.
Category / Keywords: public-key cryptography / multilinear maps, graded encoding scheme, approximate common divisors, cryptanalysis
Original Publication (with major differences): IACR-CRYPTO-2014
DOI:
10.1007/978-3-662-44371-2_13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8421950936317444, "perplexity": 2443.558174619501}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115900160.86/warc/CC-MAIN-20150124161140-00068-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://www-math.umd.edu/gcal_rss.php?seminar_key=PDE&year=2018&html | ### PDE-Applied Math Archives for Academic Year 2018
#### Liquid drops on Rough surfaces
When: Thu, September 14, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Inwon Kim (UCLA) -
#### Data-based stochastic model reduction for chaotic systems
When: Thu, October 12, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Fei Lu (John Hopkins University) - http://www.math.jhu.edu/~feilu/
Abstract: The need to develop reduced nonlinear statistical-dynamical models from time series of partial observations of complex systems arises in many applications such as geophysics, biology and engineering. The challenges come mainly from memory effects due to the nonlinear interactions between resolved and unresolved scales, and from the difficulty in inference from discrete data.
We address these challenges by introducing a discrete-time stochastic parametrization framework, in which we infer nonlinear autoregression moving average (NARMA) type models to take the memory effects into account. We show by examples that the NARMA type stochastic reduced models that can capture the key statistical and dynamical properties, and therefore can improve the performance of ensemble prediction in data assimilation. The examples include the Lorenz 96 system (which is a simplified model of the atmosphere) and the Kuramoto-Sivashinsky equation of spatiotemporally chaotic dynamics.
#### A free boundary problem with facets
When: Thu, October 19, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Will Feldman (University of Chicago) -
Abstract: I will discuss a variational problem on the lattice analogous to the Alt-Caffarelli problem. The scaling limit is a free boundary problem for the Laplacian with a discontinuous constraint on the normal derivative at the boundary. The discontinuities cause the formation of facets in the free boundary. The problem is related to models for contact angle hysteresis of liquid drops studied by Caffarelli-Lee and Caffarelli-Mellet.
#### Probabilistic scattering for the 4D energy-critical defocusing nonlinear wave equation
When: Thu, October 26, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Jonas Luehrmann (Johns Hopkins University) - http://www.math.jhu.edu/~luehrmann/
Abstract: We consider the Cauchy problem for the energy-critical defocusing nonlinear wave equation on R^4. It is known that for initial data at energy regularity, the solutions exist globally in time and scatter to free waves. However, the problem is ill-posed for initial data at super-critical regularity, i.e. for regularities below the energy regularity. In this talk we study the super-critical data regime for this Cauchy problem from a probabilistic point of view, using a randomization procedure that is based on a unit-scale decomposition of frequency space. We will present an almost sure global existence and scattering result for randomized radially symmetric initial data of super-critical regularity. The main novelties of our proof are the introduction of an approximate Morawetz estimate to the random data setting and new large deviation estimates for the free wave evolution of randomized radially symmetric data.
This is joint work with Ben Dodson and Dana Mendelson.
#### On L^p approximations of Landau equation solutions in the Coulomb case
When: Thu, November 2, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Sona Akopian (Brown University) - https://www.brown.edu/academics/applied-mathematics/sona-akopian
Abstract: We examine a class of Boltzmann equations with an abstract collision kernel in the form of a singular mass concentrated at very low collision angles and relative velocities between interacting particles. Similarly to the classical Boltzmann operator, this particular collision operator also converges to the collision term in the Landau equation as the characterizing parameter \epsilon tends to zero. We will address the existence of L^p solutions to this family of Boltzmann equations and discuss their approximations of solutions to the Landau equation as \epsilon vanishes.
#### Degenerate disperisve equations and compactons
When: Thu, November 9, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Benjamin Harrop-Griffiths (New York University ) - https://math.nyu.edu/~griffiths/
Abstract: We consider a family of Hamiltonian toy models for degenerate dispersion that admit compactly supported solitons or “compactons”. We discuss their variational properties and stability. This is joint work with Pierre Germain and Jeremy Marzuola.
#### Data-based stochastic model reduction for chaotic systems
When: Thu, November 30, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Fei Lu (John Hopkins University) - http://www.math.jhu.edu/~feilu/
Abstract: The need to develop reduced nonlinear statistical-dynamical models from time series of partial observations of complex systems arises in many applications such as geophysics, biology and engineering. The challenges come mainly from memory effects due to the nonlinear interactions between resolved and unresolved scales, and from the difficulty in inference from discrete data.
We address these challenges by introducing a discrete-time stochastic parametrization framework, in which we infer nonlinear autoregression moving average (NARMA) type models to take the memory effects into account. We show by examples that the NARMA type stochastic reduced models that can capture the key statistical and dynamical properties, and therefore can improve the performance of ensemble prediction in data assimilation. The examples include the Lorenz 96 system (which is a simplified model of the atmosphere) and the Kuramoto-Sivashinsky equation of spatiotemporally chaotic dynamics.
#### Global strong solution with latent singularity: application of gradient flow theory to thin film equations in epitaxial growth
When: Thu, December 7, 2017 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Yuan Gao (Hong Kong University of Science and Technology) -
Abstract: We consider a class of step flow models from mesoscopic view and their continuum limit to 4th order degenerate parabolic equations. Using the regularized method we obtain a global weak solution to the slope equation, which is sign-preserved almost everywhere. However, in order to study the global strong solution with latent singularity, which occurs whenever the solution approaches zero, we formulate the problem as the gradient flow of a suitably-defined convex functional in a non-reflexive Banach space and establish a framework to handle a class of degenerate parabolic equations, including exponential model for epitaxial growth, described by gradient flow in metric space.
#### Elasticity and curvature: the elastic energy of non-Euclidean thin bodies
When: Thu, January 25, 2018 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Cy Maor (University of Toronto) - http://www.math.toronto.edu/cmaor/
Abstract: Non-Euclidean, or incompatible elasticity is an elastic theory for bodies that do not have a reference, stress-free configuration. It applies to many systems, in which the elastic body undergoes inhomogeneous growth (e.g. plants, self-assembled molecules). Mathematically, it is a geometric calculus of variations question of finding the "most isometric" immersion of a Riemannian manifold (M,g) into Euclidean space of the same dimension.
Much of the research in non-Euclidean elasticity is concerned with elastic bodies that have one or more slender dimensions (such as leaves), and finding appropriate dimensionally-reduced models for them.
In this talk I will give an introduction to non-Euclidean elasticity, and then focus on thin bodies and present some recent results on the relations between their elastic behavior and their curvature.
Based on joint work with Asaf Shachar.
#### Sticky particles and the Euler-Poisson equations
When: Thu, March 1, 2018 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Ryan Hynd (U-Penn) - https://web.sas.upenn.edu/rhynd/
Abstract: We will consider the dynamics of a finite number of particles that interact pairwise and undergo perfectly inelastic collisions.
Such physical systems conserve mass and momentum and satisfy the Euler-Poisson equations. In one spatial dimension, we will show
how to derive an extra entropy estimate which allows us to characterize the limit as the number of particles tends to infinity.
#### Alexandrov Theorem Revisited
When: Thu, March 15, 2018 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Matias Delgadino (Imperial College - London) -
Abstract: We will introduce the notion of sets of finite perimeter and we will show that among sets of finite perimeter balls are the only volume-constrained critical points of the perimeter functional. This is joint work with Francesco Maggi.
#### Approximation diffusion and fluid limits for random kinetic equations [Appl Math Colloquium]
When: Tue, May 1, 2018 - 3:30pm
Where: Kirwan Hall 3206
Speaker: Arnaud Debussche (ENS Rennes) - http://w3.bretagne.ens-cachan.fr/math/people/arnaud.debussche/
Abstract: In this talk, I consider kinetic equations containing random terms. The kinetic models contain a small parameter and it is wellknown that, after parabolic rescaling, when this parameter goes to zero the limit problem is a diffusion equation in the PDE sense, ie a parabolic equation of second order. A smooth noise is added, accounting for external perturbation. It scales also with the small parameter. It is expected that the limit equation is then a stochastic parabolic equation where the noise is in Stratonovitch form.
Our aim is to justify in this way several SPDEs commonly used. We first treat linear equations with multiplicative noise. Then show how to extend the methods to some nonlinear equations or to the more physical case of a random forcing term. The method is to combine the classical perturbed test function method with PDE argument. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8048334121704102, "perplexity": 1202.1585943116188}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247479101.30/warc/CC-MAIN-20190215183319-20190215205319-00590.warc.gz"} |
https://www.physicsforums.com/threads/obtaining-the-rayleigh-jeans-formula-what-am-i-doing-wrong.372987/ | # Obtaining the Rayleigh-Jeans formula - what am I doing wrong?
1. ### quasar_4
290
Obtaining the Rayleigh-Jeans formula - what am I doing wrong??
1. The problem statement, all variables and given/known data
Spectral energy density = $$u(\nu, T) = {$\displaystyle \,8\,\pi \,h{\nu}^{3}{c}^{-3} \left( {{\rm e}^{{\frac {h\nu}{kT}}}}-1 \right) ^{-1}$}$$
where h is Planck's constant and k is Bolzmann's constant.
Using the relation $$\lambda = \frac{c}{\nu}$$ express the spectral energy density as a function of the wavelength and determine the energy du of a blackbody radiation per unit volume, in a narrow range of wavelength $$\lambda + d\lambda$$. Using an expansion of the exponential factor obtain the Rayleigh-Jeans formula,
$$du = \frac{8 \pi k T}{\lambda^4} d\lambda$$
2. Relevant equations
Given some function $$f(x_1, x_2, ..., x_n)$$,
$$df = \left(\frac{\partial{f}}{\partial{x_1}}\right) dx_1 + ... + \left(\frac{\partial{f}}{\partial{x_n}}\right) dx_n$$.
Also, $$e^{x} \approx 1+x+\frac{x^2}{2!} + ...$$
3. The attempt at a solution
It seems like it should be easy, but I can't get the equation to come out right. Substituting in lambda, I get
$${$\displaystyle {\it u}\, = \,8\,\pi \,h{\lambda}^{-3} \left( {{\rm e}^{{\frac {hc}{\lambda\,kT}}}}-1 \right) ^{-1}$}$$
Then from my equation in part b, we should have that
$$du = \frac{\partial{u}}{\partial{\lambda}} d\lambda$$
But I get
$$\frac{\partial{u}}{\partial{\lambda}} = {$\displaystyle \,-8\,\pi \,h \left( 3\,\lambda\,kT{{\rm e}^{{\frac {hc}{\lambda\,kT}}}}-3\,\lambda\,kT-hc{{\rm e}^{{\frac {hc}{\lambda\,kT}}}} \right) {\lambda}^{-5} \left( {{\rm e}^{{\frac {hc}{\lambda\,kT}}}}-1 \right) ^{-2}{k}^{-1}{T}^{-1}$}$$
Upon using my expansion for the exp. term (expanding around nu = 0, then substituting in nu = c/lambda) the most I can simplify this is to:
$$du ={$\displaystyle \,8\,\pi \, \left( -2\,ckT+{\frac {h{c}^{2}}{\lambda}} \right) {\lambda}^{-3}{c}^{-2}$}d\lambda$$
so... what am I doing wrong?? It could have something to do with the $$\lambda + d\lambda$$ (though I just thought that meant I could neglect higher than 2nd order terms in my expansion)...
2. ### gabbagabbahey
5,015
Re: Obtaining the Rayleigh-Jeans formula - what am I doing wrong??
A quick check of the units of your original equation, $$u(\nu,T)=8h\pi\nu^3 c^{-3} \left(e^{\frac{h\nu}{kT} }-1\right)^{-1}$$ suggests that the problem lies there, not with your method.
If you are having doubts about whether or not using $du=\frac{\partial u}{\partial\lambda}d\lambda$ makes sense, another way to look at it is to realize that the spread in energy over the interval $\lambda$ to $\lambda+ d\lambda$ is defined as $du=u(\lambda+d\lambda)-u(\lambda)$. And for infinitesimal $d\lambda$, a quick Taylor expansion gives $u(\lambda+d\lambda)=u(\lambda)+\frac{\partial u}{\partial \lambda}d\lambda$, from which the result follows.
3. ### quasar_4
290
Re: Obtaining the Rayleigh-Jeans formula - what am I doing wrong??
Interesting. This is an exam problem (from a previous year) at my school, so maybe the question has a typo. I can't see what else one would do to solve for this equation.
4. ### gabbagabbahey
5,015
Re: Obtaining the Rayleigh-Jeans formula - what am I doing wrong??
I must apologize, I was thinking $u$ was just an energy density (units of Joules/m3), not a spectral energy density. Your original equation is just fine. However, you can't simply substitute $\nu=\frac{c}{\lambda}$ into your frequency spectral energy density equation and get a wavelength spectral energy density out of it.
In terms of frequency, the spectral energy density has units of Joule-seconds/m3. While in terms of wavelength, the spectral energy density has units of Joules/m4. The first is defined as an energy density per unit frequency, while the latter is defined as an energy density per unit wavelength. For this reason, $u(\lambda,T)d\lambda=u(\nu,T)d\nu$. Or, $u(\lambda,T)=u(\nu,T)\frac{d\nu}{d\lambda}=u(\nu,T)\times\frac{-c}{\lambda^2}$
5. ### bkrishnan39
1
Re: Obtaining the Rayleigh-Jeans formula - what am I doing wrong??
When we substitute d$\nu$/d$\lambda$ = - c / $\lambda$^2 in Plancks law of Black body radiation (in frequency), we would get an equation that is same as Plancks law of black body radiation (in wavelength) EXCEPT for a negative sign.
Because it is an ENERGY equation, the negative sign would be omitted.
Is this correct | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9833245277404785, "perplexity": 793.5871983039482}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398447913.86/warc/CC-MAIN-20151124205407-00354-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://aas.org/archives/BAAS/v27n4/aas187/S033002.html | Session 33 - Neutron Stars & White Dwarfs.
Oral session, Monday, January 15
Salon del Rey South, Hilton
## [33.02] WFPC2 Photometry of the Bright, Mysterious White Dwarf Procyon B
J. L. Provencal, H. Shipman (U. Delaware)
Procyon B is one of a handful of white dwarf stars which are in visual binary systems. We have accurate measurements of the binary period, and hence, of the white dwarf's mass. Procyon B represents a critically important object for testing the fundamental physics of stellar degeneracy. However, the separation between primary and secondary is too close to allow accurate ground-based determinations of basic characteristics, such as temperature and chemical composition, which would allow us to derive its radius. Furthermore, since the age of this system is known, if we can determine Procyon B's temperture, we can determine the mass of its progenitor and provide a stringent test for theories of stellar mass loss. We present the results of WFPC2 data, from which we determine a spectral type of DA, an apparent magnitude of \rmM_V=10.75, and an effective temperature of 8000\pm400 K. The stellar radius derived from these parameters lies on the mass-radius relationship for white dwarfs. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9190866351127625, "perplexity": 1654.2782258135855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719877.27/warc/CC-MAIN-20161020183839-00442-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://physics.stackexchange.com/questions/187363/time-evolution-of-a-wavepacket | # Time evolution of a wavepacket
I do not understand why if $H\psi = E\psi$, then the time-evolution of the wavefunction is given by $e^{-iEt/h}\psi(x)$.
• That formula for the time dependence is only valid for eigenstates, and only for time-independent Hamiltonians. – zeldredge Jun 2 '15 at 20:56
Solving time-dependent SE as danielsmw mentioned above good starting point. $$i\hbar\frac{\partial\psi}{\partial t}=H\psi$$
$$\frac{d\psi}{\psi}=\frac{H}{i\hbar}dt$$ $$log(\psi)\mid^{\psi}_{\psi_{0}}=-\frac{iHt}{\hbar}\mid^{t}_{t_0}$$ suppose $\psi=|\alpha,t>$ and $\psi_{0}=|\alpha, t_{0}>$ $$\psi=e^{-\frac{iH(t-t_{0})}{\hbar}}\psi_0$$ Under infinitesimal changes of time (dt), time evolution operator, $$lim_{dt\to 0}U(t_{0}+dt, t_{0})=1$$ $$U(t_{0}+dt, t_{0})=1-iGdt$$ for $G=\frac{H}{\hbar}$ , $$U(t_{0}+dt, t_{0})=1-i\frac{H}{\hbar}dt \sim exp(-i\frac{H}{\hbar}dt )$$ For a good guidance I highly recommend J.J.Sakurai Chapter 2.
The equation you've given is the time-independent Schrodinger equation. The time-dependent Schrodinger equation is $$i \hbar \dot{\psi} = H \psi$$ which you can readily solve to obtain the time-dependence (when $\psi$ is an eigenstate) that you asked for in the question. More generally, you can solve it to show that $\exp\left(-i\hat{H}t/\hbar\right)$ is the generator of time evolution, i.e. the operator that takes $\psi(t=0)\mapsto\psi(t)$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9492015242576599, "perplexity": 265.58132731793796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321351.87/warc/CC-MAIN-20190824172818-20190824194818-00230.warc.gz"} |
http://www.i-calculus.com/2015/03/interesting-integral-think-out-of-box.html | ### Interesting integral - think out of the box
Evaluate $\int_0^{\infty} \frac{x^3}{e^x - 1} \; dx$
Answer:
$\frac{1}{1 - x} = 1 + x + x^2 + \cdot \cdot \cdot + x^n + \cdot \cdot \cdot = \sum_{k=0}^{\infty} x^k$, when $|x| \lt 1$
$\frac{1}{e^x - 1}$
$= \frac{e^{-x}}{1 - e^{-x}}$
$= e^{-x} \sum_{k=0}^{\infty} (e^{-x})^k$, since $e^{-x} \lt 1$ when $x\gt 0$
$= \sum_{k=1}^{\infty} (e^{-x})^k$
$= \sum_{k=1}^{\infty} e^{-kx}$
So,
$\int_0^{\infty} \frac{x^3}{e^x - 1} \; dx$
$= \int_0^{\infty} x^3 \sum_{k=1}^{\infty} e^{-kx}k$
$= \sum_{k=1}^{\infty} \int_0^{\infty} x^3 e^{-kx}$
$= \sum_{k=1}^{\infty} \left. \frac{e^{-kx}}{k^4} \left( k^3x^3 + 3k^2x^2 + 6kx + 6 \right) \right|_0^{\infty}$
$= \sum_{k=1}^{\infty} \frac{6}{k^4}$
$= 6 \sum_{k=1}^{\infty} \frac{1}{k^4}$
$= 6 \cdot \frac{\pi^4}{90} = \frac{\pi^4}{15}$
We used the fact that $\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}$, an advance result that follows easily from a theorem called Parseval's Theorem, which basically says that if the Fourier series of $f(x)$ is given by
$f(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty} a_k \cos (kx) + \sum_{k=1}^{\infty} b_k \sin (kx)$,
where $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \; dx$
$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos (kx) \; dx \;$ and
$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx)\; dx$
then we have the Parseval's Identity
$\frac{1}{\pi} \int_{-\pi}^{\pi} (f(x))^2 \; dx = \frac{a_0^2}{2} + \sum_{k=1}^{\infty} a_k^2 + b_k^2$
Consider $f(x) = x^2$, then we have
$\frac{1}{\pi} \int_{-\pi}^{\pi} x^4 \; dx$
$= \frac{1}{\pi} \left. \frac{x^5}{5} \right|_{-\pi}^{\pi} = \frac{2\pi^4}{5}$
$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \; dx$
$= \frac{1}{\pi} \left. \frac{x^3}{3} \right|_{-\pi}^{\pi} = \frac{2\pi^2}{3}$
$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos (nk) \; dx$
$= \frac{1}{\pi} \left. \frac{2 k x cos(k x)+(-2+k^2 x^2) sin(k x)}{k^3} \right|_{-\pi}^{\pi}$
$= \frac{\pm 4}{k^2}$
$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \sin (nk) \; dx$
$= \frac{1}{\pi} \left. \frac{(2-k^2 x^2) cos(k x)+2 k x sin(k x)}{k^3} \right|_{-\pi}^{\pi}$
$=0$
So, we have
$\frac{1}{\pi} \int_{-\pi}^{\pi} (f(x))^2 = \frac{2\pi^4}{5}$ and
$a_0^2 = \frac{4\pi^4}{9}$, $a_k^2 = \frac{16}{k^4}$ and $b_k^2 = 0$
Substituting these values in the Parseval's Identity, we have
$\frac{2\pi^4}{5} = \frac{4\pi^4}{18} + \sum_{k=1}^{\infty} \frac{16}{k^4}$
or, $\sum_{k=1}^{\infty} \frac{16}{k^4} = \frac{2\pi^4}{5} - \frac{4\pi^4}{18}$
$\sum_{k=1}^{\infty} \frac{16}{k^4} = \frac{16\pi^4}{90}$
$\sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998080134391785, "perplexity": 635.9853146615054}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00259.warc.gz"} |
https://www.physicsforums.com/threads/is-this-a-4-linear-form.181428/ | # Is this a 4-linear form ?
1. Aug 23, 2007
### Sangoku
Is this a 4-linear form ??
rather a silly question, but i am a bit confused , could we consider the expression
$$x^{2}-y^{2}+3xy-x^{4}+5y^{4}+ z^{4}-2t^{4}+3xyzt-6x^{2}tz+ 5x^{2}y^{2}$$
or even in general $$A_{i,j}x^{i}x^{j}-Q_{i,j,k,l}x^{i}x^{j}x^{k}x^^{l}$$
could be considered a 4-linear form??.. i believe this is linear on every argument x^{i} but there seems to be 'mixed up' a 2-from plus a 4-form so i,m not sure what exactly is the proposed object.
2. Aug 23, 2007
### mathman
The usual label for such expressions is 4th degree polynomial in 4 variables.
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# MEASURES OF CENTRAL TENDENCY (AVERAGES)
## Presentation on theme: "MEASURES OF CENTRAL TENDENCY (AVERAGES)"— Presentation transcript:
MEASURES OF CENTRAL TENDENCY (AVERAGES)
Topic #6
Measures of Central Tendency
A measure of central tendency is a univariate statistic that indicates, in one manner or another, the average or typical observed value of a variable in a data set, or put otherwise, the center of the frequency distribution of the data.
Central Tendency (cont.)
The central tendency of age among Berkeley faculty member clearly increased from to On the other hand, the central tendency of Democratic House vote by CD did not change much from 1948 to 1970 (though other characteristics of its frequency distribution certainly did change).
How to Calculate Averages
We consider three measures of central tendency that are appropriate for three different levels of measurement (nominal, ordinal, and interva). There are others, e.g., the geometric and harmonic means, appropriate only for ratio variables. For each measure of central tendency, we will consider how it may be calculated from three different “starting points”: a list of observed values (i.e., raw data, e.g., one column in a data spread sheet); a frequency table (or bar graph); a histogram and, in particular, a continuous density curve (like income in PS #5C).
The Mode The mode (or modal value) of a variable in a set of data is the value of the variable that is observed most frequently in that data (or, given a continuous frequency curve, is at the point of greatest density). Note: the mode is the value that is observed most frequently, not the frequency itself. (I see this error too frequently on tests.) The mode is defined for every type of variable [i.e., nominal, ordinal, interval, or ratio]. However, the mode is used as a measure of central tendency primarily for nominal variables only.
The Mode (cont.) The mode is defined for ordinal and interval variables as well. But since it takes account of only the nominal characteristics of variables (i.e., whether two cases have the same or different values), it is usually very unsatisfactory when applied to ordinal or higher level variables. The mode may be ill-defined if we have either: a small number of cases; or a precisely measured continuous variable and a finite number of cases; because in either event it is likely that no value will be observed more than once in the data. In any event, the mode is unstable in that small changes in the data can result in large and erratic changes in the modal value; and changes in coding the variable (for example, RELIGION), or changes in class intervals, can change the modal value.
How To Calculate the Mode
Given a list of observed values (raw data): Construct a frequency table (see next slide).
How To Calculate the Mode (cont.)
Given a frequency table or bar graph (or having constructed one): observe which value in the table (or graph) has the greatest (absolute or relative) frequency; the most frequent value (not the frequency itself) is the mode. Notice that the modal number of problem sets turned in is 5, even though most students turned in fewer than 5, So if we recoded the variable to create just two dichotomous categories: (a) turned in all 5 (b) did not turn in all 5 the latter category becomes the modal category
How To Calculate the Mode (cont.)
Given a continuous frequency curve: the mode is the value of the variable under the highest point of the frequency curve (the point with the greatest density of observed values).
The Median The median (or median value) of a variable in a data set is
the value in the middle of the observations, in the sense that no more than half of the cases have lower values and no more than half of the cases have higher values or, more generally, such that no more than half of the cases have values that lie on either side of the median value. Given a quite precisely measured continuous variable and a very large number of cases, we can in practice say that half the cases have lower values and half have higher values (e.g., LEVEL OF INCOME, SAT scores). Equivalently, the median value is the value of the case at the 50th percentile of the distribution.
The Median (cont.) The median is defined if and only if the variable is at least ordinal in nature [i.e., ordinal, interval, or ratio], and we can therefore rank all (non-missing) observations in terms of lower to higher values (or possibly some other natural [e.g., liberal to conservative] ordering). e.g., the median member of House of Representatives with respect to ideology: half are more liberal, half more conservative. The median should be clearly distinguished from another (infrequently used) measure of central tendency that is defined only for variables that are interval in nature. This is the midrange (or midpoint) value in the distribution, which is the value in the middle of the observations in the (different) sense that it lies exactly halfway between the minimum (lowest) and maximum (highest) observed values (i.e., at the midpoint of the range of values), i.e., (min + max) / 2, e.g., for Problem Sets, (1 + 5) / 2 = 3
The Median (cont.) Given a list of observed values (raw data):
rank order the cases in terms of their observed values (e.g., from lowest to highest); identify the value of the case right at the middle of this rank-ordered list, and the value of this case is the median value; or construct a frequency table and find where the cumulative frequency crosses the 50% mark.
The Median (cont.)
The Median (cont.) If the number of cases is even, there is no observed value at the exact middle of the list. Look at the pair of observations closest to the middle of the list. If they have the same value, that value is the median. If they have different values, every value in the interval bounded by these two values meets the definition of a median; but conventionally the median in this event is defined as the midpoint of the interval.
TABLE 1 – PERCENT OF POPULATION AGED 65 OR HIGHER IN THE 50 STATES
(UNIVARIATE DATA ARRAY) Alabama Montana 12.5 Alaska Nebraska 13.8 Arizona Nevada Arkansas New Hampshire 11.5 California New Jersey 13.0 Colorado New Mexico 10.0 Connecticut New York Delaware North Carolina 11.8 Florida North Dakota 13.3 Georgia Ohio 12.5 Hawaii Oklahoma Idaho Oregon Illinois Pennsylvania 14.8 Indiana Rhode Island 14.7 Iowa South Carolina 10.7 Kansas South Dakota 14.0 Kentucky Tennessee 12.4 Louisiana Texas Maine Utah Maryland Vermont 11.9 Massachusetts Virginia 10.6 Michigan Washington 11.8 Minnesota West Virginia 13.9 Mississippi Wisconsin 13.2 Missouri Wyoming
(1) Rank of state (2) Name of state (3) % over 65 in state (value of variable) (4) percentile rank of state
The Median (cont.) Is median income higher or lower than modal income?
The Median (cont.) Draw a vertical line such that half of the area under the frequency curve lies on one side of the line and half on the other side. The median value of the variable lies on this line.
“Center of Gravity” of Data
The median (and the mode) may fail to be the “center of gravity” or “balance point” of the data: Data (ranked):
The Mean The mean (or mean value) of a variable in a set of data is the result of adding up all the observed values of the variable and dividing by the number of cases (i.e., the “average” as the term is most commonly used). The mean is defined if and only if the variable is at least interval in nature [i.e., interval or ratio]. Suppose we have a variable X and a set of cases numbered 1,2,...,n. Let the observed value of the variable in each case be designated x1, x2, etc. Thus:
How to Calculate the Mean
Given a list of observed values (raw data): Following the formula above, add up the observed values of the variable in all cases and divide by the number of cases. Given a frequency table (or bar graph): Take each value and multiply it by its absolute frequency, add up these products over all values, and divide this sum by the number of cases; or (and this is usually easier) take each value and multiply it by the decimal fraction (between zero and one) that represents its relative frequency, and add up these products over all values. Do not divide by the number of cases — you have already done this as a result of multiplying each value by a decimal fraction.
The Mean (cont.) Is mean income lower or higher than median (or modal) income?
The Mean (cont.) The mean is the “center of gravity” of the distribution. Determine (by “eyeball” approximation) the value of the variable such that the density “balances” at that point; this value is the mean.
Average Values For a quantitative variable, the range of observed values of a variable in a set of data is the interval extending from the minimum observed value to the maximum observed value. For every measure of central tendency, the average value of the observations lies somewhere in this range, often (but certainly not always) somewhere near the middle of this range. The modal or median observed value can equal the minimum or (like the modal problem sets turned in) the maximum value, but the mean observation can do so only in the very special case in which all cases have identical observed values, so there is no dispersion [next topic] in the data and min value = max value. Once again, remember that the median is not (necessarily) the midpoint of the range.
Average Values (cont.) If a quantitative variable is discrete, so all of its values are whole numbers, the modal or median observed value is always a whole number also (with the possible exception noted for the median when the number of cases is even), but the mean observation is almost never a whole number. The “average” family may have children, i.e., the mean number of children per family may be 2.374, even though no individual family can have that [or any fractional] number of children. Likewise, the mean number of problem sets turned in was 3.85.
Deviations From the Average
Unless all cases have the same observed value [i.e., no dispersion], some (in the case of mean, probably all) observed values will differ from the average value. If the variable is interval (or ratio) in nature, we can calculate the deviation from the average in each case, i.e., the “distance” or “interval” from the observed value to the average value. The deviation in each case is positive if the observed value is greater than the average value, negative if the observed value is less than the average value, or zero if the observed value is equal to the average value.
Deviations From the Average (cont.)
Since we have three different types of averages, we also have three different types of deviations from the average. The deviations from each type of average have different properties. There are fewer non-zero deviations from the modal value than from any other value of the variable. No more than half of the deviations from the median value are positive and no more than half are negative. Unless several cases have the median value (and perhaps even then), the number of cases with positive deviations equals the number with negative deviations; that is, the cases with positive and negative deviations balance out with respect to their number.
Deviations From the Average (cont.)
The sum (or mean) of the absolute deviations (i.e., ignoring whether the deviations are positive or negative) from the median is less than the sum (or mean) of the absolute deviations from any other value of the variable. The sum (or mean) of the (“algebraic,” i.e., taking account of whether deviations are positive or negative) deviations from the mean is zero; that is, the positive and negative deviations balance out with respect to their total (positive and negative) magnitudes. The sum (or mean) of the squared deviations from the mean is less than the sum of the squared deviations from any other value of the variable.
Median vs. Mean Values The mode is rarely used to describe central tendency in quantitative data, because (as we saw) it may be undefined or unstable. Both the mean and the median are commonly used with quantitative (i.e., interval or ratio) data, though only the median can be used with ordinal data. For example, it is common to report both median and mean income or wealth, median and mean test scores, median and mean prices (of cars, houses, etc.). While the median and the mean are both proper and useful measures of central tendency for quantitative variables, they have different definitions and properties and may (depending on the distribution of the data) give very different answers to the question “what is the average value in this data?”
Median vs. Mean Values (cont.)
If the distribution of the data is symmetric, the median and mean values are the same. If the distribution is “almost” symmetric, the median and the mean are “almost the same”. For example, test scores are typically distributed approximately symmetrically, so median and mean test scores are typically approximately the same. Median Mean MC BB Score
Median vs. Mean Values (cont.)
If the distribution of the data is skewed, the mean is pulled (relative to the median) in the direction of the long thin tail. For example, income is distributed in a highly skewed fashion, with a long thin tail in the direction of higher income. Thus mean income is typically considerably higher than median income.
Median vs. Mean Values (cont.)
Question: which arrow (red or green) points to the median value and which to the median? On the course webpage, see Statistical Applets: Mean and Median
Median vs. Mean Values (cont.)
The median (unlike the mean) is “resistant to outliers,” where an outlier (in univariate data) is a case with an extreme (very high or very low) value. That is, adding some outliers to the data (or removing them) may have a big impact on the mean value but usually has little impact on the median value.
Median vs. Mean Values (cont.)
If the observed values in some distribution of data changes, the median value changes only if the value (or identity) of the median case changes, whereas the mean value most likely is affected by any change in the data. For example, if “the rich get richer while everybody else stays about the same,” mean income increases while median income stays the same.
Median vs. Mean Values (cont.)
If changes in the data do not change the sum of all observed values, i.e., if the sum of all values remains constant, the mean remains constant, while the median may change. For example, if Congress has decided that a tax cut will total \$100 billion but has not decided how this fixed sum should be divided up among the nation’s 100 million households, the median benefit will depend on the specifics of the legislation but the mean tax benefit per household will be \$1000 in any event.
Median vs. Mean Values (cont.)
If observed values are polarized with approximately half the cases having high values, half having low, and none having medium values, the median is unstable; that is, the median value will be high or low depending on whether slightly more than half the cases have high values or slightly more than half the cases have low values, whereas the mean value will barely change as a result of such small shifts in the data.
TEST 1: FREQUENCY DISTRIBUTION OF LETTER GRADES
TEST 1: UNIVARIATE STATISTICS: FALL 09 FALL 10
TEST 1: BIVARIATE CHART AND SUMMARY STATISTIC
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https://elteoremadecuales.com/fermats-theorem-on-sums-of-two-squares/?lang=pt | # Fermat's theorem on sums of two squares
Fermat's theorem on sums of two squares For other theorems named after Pierre de Fermat, see Fermat's theorem.
In additive number theory, Fermat's theorem on sums of two squares states that an odd prime p can be expressed as: {displaystyle p=x^{2}+^{2},} with x and y integers, se e apenas se {displaystyle pequiv 1{pmod {4}}.} The prime numbers for which this is true are called Pythagorean primes. Por exemplo, the primes 5, 13, 17, 29, 37 e 41 are all congruent to 1 módulo 4, and they can be expressed as sums of two squares in the following ways: {displaystyle 5=1^{2}+2^{2},quad 13=2^{2}+3^{2},quad 17=1^{2}+4^{2},quad 29=2^{2}+5^{2},quad 37=1^{2}+6^{2},quad 41=4^{2}+5^{2}.} Por outro lado, the primes 3, 7, 11, 19, 23 e 31 are all congruent to 3 módulo 4, and none of them can be expressed as the sum of two squares. This is the easier part of the theorem, and follows immediately from the observation that all squares are congruent to 0 ou 1 módulo 4.
Since the Diophantus identity implies that the product of two integers each of which can be written as the sum of two squares is itself expressible as the sum of two squares, by applying Fermat's theorem to the prime factorization of any positive integer n, we see that if all the prime factors of n congruent to 3 módulo 4 occur to an even exponent, then n is expressible as a sum of two squares. The converse also holds.[1] This generalization of Fermat's theorem is known as the sum of two squares theorem.
Conteúdo 1 História 2 Gaussian primes 3 Resultados relacionados 4 Algorithm 4.1 Description 4.2 Exemplo 5 Provas 5.1 Euler's proof by infinite descent 5.2 Lagrange's proof through quadratic forms 5.3 Dedekind's two proofs using Gaussian integers 5.4 Proof by Minkowski's Theorem 5.5 Zagier's "one-sentence proof" 5.6 Proof with partition theory 6 Veja também 7 Referências 8 Notas 9 External links History Albert Girard was the first to make the observation, describing all positive integer numbers (not necessarily primes) expressible as the sum of two squares of positive integers; this was published in 1625.[2][3] The statement that every prime p of the form 4n+1 is the sum of two squares is sometimes called Girard's theorem.[4] For his part, Fermat wrote an elaborate version of the statement (in which he also gave the number of possible expressions of the powers of p as a sum of two squares) in a letter to Marin Mersenne dated December 25, 1640: for this reason this version of the theorem is sometimes called Fermat's Christmas theorem.
Gaussian primes Fermat's theorem on sums of two squares is strongly related with the theory of Gaussian primes.
A Gaussian integer is a complex number {displaystyle a+ib} such that a and b are integers. The norm {estilo de exibição N(a+ib)=a^{2}+b^{2}} of a Gaussian integer is an integer equal to the square of the absolute value of the Gaussian integer. The norm of a product of Gaussian integers is the product of their norms. This is the Diophantus identity, which results immediately from the similar property of the absolute value.
Gaussian integers form a principal ideal domain. This implies that Gaussian primes can be defined similarly as primes numbers, that is as those Gaussian integers that are not the product of two non-units (here the units are 1, −1, i and −i).
In a letter to Blaise Pascal dated September 25, 1654 Fermat announced the following two results that are essentially the special cases {displaystyle d=-2} e {displaystyle d=-3.} If p is an odd prime, então {displaystyle p=x^{2}+2^{2}iff pequiv 1{mbox{ ou }}pequiv 3{pmod {8}},} {displaystyle p=x^{2}+3^{2}iff pequiv 1{pmod {3}}.} Fermat wrote also: If two primes which end in 3 ou 7 and surpass by 3 a multiple of 4 are multiplied, then their product will be composed of a square and the quintuple of another square. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.937776505947113, "perplexity": 449.9063666941813}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00147.warc.gz"} |
https://andrescaicedo.wordpress.com/tag/cantor-bendixon-derivative/ | ## 502 – Cantor-Bendixson derivatives
November 8, 2009
Given a topological space $X$ and a set $B\subseteq X,$ let $B'$ be the set of accumulation points of $B,$ i.e., those points $p$ of $X$ such that any open neighborhood of $p$ meets $B$ in an infinite set.
Suppose that $B$ is closed. Then $B'\subseteq B.$ Define $B^\alpha$ for $B$ closed compact by recursion: $B^0=B,$ $B^{\alpha+1}=(B^\alpha)',$ and $B^\lambda=\bigcap_{\alpha<\lambda}B^\alpha$ for $\lambda$ limit. Note that this is a decreasing sequence, so that if we set $B^\infty=\bigcap_{\alpha\in{\sf ORD}}B^\alpha,$ there must be an $\alpha$ such that $B^\infty=B^\beta$ for all $\beta\ge\alpha.$
[The sets $B^\alpha$ are the Cantor-Bendixson derivatives of $B.$ In general, a derivative operation is a way of associating to sets $B$ some kind of “boundary.”] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 23, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9827933311462402, "perplexity": 97.03401929137188}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00385.warc.gz"} |
http://mathhelpforum.com/algebra/117647-write-single-logarithm.html | # Math Help - Write as a single logarithm.
1. ## Write as a single logarithm.
3log2x + 1/2log2x - 2log2(x+1)
I don't quite understand this. Ive gotten to "log2 x^3(x^1/2)/(x+1)^2"
Any help would be great.
Thanks.
~freak
2. ## Joining logs
$
3log2x + (1/2)log2x - 2log2(x+1)
$
= $
log(2x)^3 + log(2x)^{1/2} - log{(2(x+1))}^2
$
= $
log(2x)^{3.5} - log{(2(x+1))}^2
$
= $
log \frac{(2x)^{3.5}}{{(2(x+1))}^2}
$
3. Thanks, i looked it over again and realized where i went wrong. I got the same asnwer as you did but the book had the top x to the power of 7/2. It took me to right after i posted this to realized that 3.5 and 7/2 were the same thing lol.
Thanks for the help.
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Dec 8 awarded Nice Answer Oct 5 awarded Yearling Jul 20 comment CLT for independent, but non-identically distributed exponential variables Without actually grinding through it...since the rvs are exponential, you know their variance too....might it be easier to tackle it from that angle? Jul 15 comment Do two almost surely equal random variables necessarily have the same probability? Thanks for the explanation. Jul 15 comment Do two almost surely equal random variables necessarily have the same probability? Can you please explain why is P[X in B, X=Y] <= P[Y in B] ? Jul 2 comment Reference request for this topics Agree. Sheldon M. Ross "Stochastic processes" is a little more advanced. Jun 11 comment How do I show that $P(|X-Y|>1/n)=0 \forall n \in \mathbb{N}$ then $X=Y$ a.s What if X=Y+1/10? then P[X=Y] =0, but P[|X-Y|>1/2]=0 Jun 10 comment Which fields of math would I need to study to fully understand/solve the Riemann Hypothesis? start here: modular.math.washington.edu/rh Apr 28 comment $\lim\sum_{k=0}^{\lfloor\delta n\rfloor} \frac{n^k}{k!}e^{-n}$ and Poisson distribution Could you please explain what is M? Mar 23 awarded Revival Mar 19 comment Winning in roulette when betting on one number infinitely Probability distribution of what? Jan 20 comment Does $\int_0^\infty e^{-x}\sqrt{x}dx$ converge? If in the integral it were x, rather than Sqrt[x], the integral would be the mean of the exponential distribution with parameter 1, which is known to have value 1. You can easily do that with integration by parts. Then, since Sqrt[x]n^{-1}]} \right) =0$Suspect the key here is that P[X<∞]=1 Oct 8 comment$P(X_n < a$i.o. and$X_n > b$i.o.$) = 0$for all$a < b$implies that$lim_{n \rightarrow \infty} X_n$exists a.e. Now I'm happy! Thanks! Oct 8 comment$P(X_n < a$i.o. and$X_n > b$i.o.$) = 0$for all$a < b$implies that$lim_{n \rightarrow \infty} X_n$exists a.e. Should it be x_n_1 b$ i.o.$) = 0$ for all $a < b$ implies that $lim_{n \rightarrow \infty} X_n$ exists a.e. Not sure if I understand what you are saying here. Let x_n=7 for all n. So lim x_n=7. But there exists an 'a' in Q such that a=1/2? Sep 15 answered Prove that: if $x \sqcup \bar{y}=1$, then $x \sqcup y=x$ (in a Boolean algebra) Sep 9 comment How do I differentiate polynomials If you dont like your text,try another one. Or en.wikipedia.org/wiki/Calculus#Differential_calculus to get started. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9745685458183289, "perplexity": 1351.366083432516}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701151880.99/warc/CC-MAIN-20160205193911-00185-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://www.exploredatabase.com/2020/05/formal-definition-of-probabilistic-context-free-grammar-pcfg-with-example.html | # Probabilistic Context Free Grammar Formal Definition and Examples
## Probabilistic Context Free Grammar (PCFG)
Probabilistic Context Free Grammar (PCFG) is an extension of Context Free Grammar (CFG) with a probability for each production rule. Ambiguity is the reason why we are using probabilistic version of CFG. For instance, some sentences may have more than one underlying derivation. That is, the sentence can be parsed in more than one ways. In this case, the parse of the sentence become ambiguous. To eliminate this ambiguity, we can use PCFG to find the probability of each parse of the given sentence.
A PCFG is made up of a CFG and a set of probabilities for each production rule of CFG. A PCFG can be formally defined as follows;
A probabilistic context free grammar G is a quintuple G = (N, T, S, R, P) where
• (N, T, S, R) is a context free grammar where N is set of non-terminal (variable) symbols, T is set of terminal symbols, S is the start symbol and R is the set of production rules where each rule of the form A → s [Refer for more here – Context Free Grammar Formal Definition].
• A probability P(A → s) for each rule in R. The properties governing the probability are as follows;
• P(A → s) is a conditional probability of choosing a rule A → s in a left-most derivation, given that A is the non-terminal that is expanded.
• The value for each probability lies between 0 and 1.
• The sum of all probabilities of rules with A as the left hand side non-terminal should be equal to 1.
### Example PCFG:
Probabilistic Context Free Grammar G = (N, T, S, R, P)
• N = {S, NP, VP, PP, Det, Noun, Verb, Pre}
• T = {‘a’, ‘ate’, ‘cake’, ‘child’, ‘fork’, ‘the’, ‘with’}
• S = S
• R = { S → NP VP
NP → Det Noun | NP PP
PP → Pre NP
VP → Verb NP
Det → ‘a’ | ‘the’
Noun → ‘cake’ | ‘child’ | ‘fork’
Pre → ‘with’
Verb → ‘ate’ }
• P = R with associated probability as in the table below;
Rule Probability Rule Probability S → NP VP 1.0 Det → ‘a’ Det → ‘the’ 0.5 0.5 NP → NP PP NP → Det Noun 0.6 0.4 Noun → ‘cake’ Noun → ‘child’ Noun → ‘fork’ 0.4 0.3 0.3 PP → Pre NP 1.0 Pre → ‘with’ 1.0 VP → Verb NP 1.0 Verb → ‘ate’ 1.0
Please observe from the table, the sum of probability values for all rules that have same left hand side is 1. For example,
In the next page, let us discuss the question “How to use PCFG to calculate the probability of a parse tree?”
********************
• Go to NLP Solved Exercise page | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8681498765945435, "perplexity": 2527.6454668011643}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347409171.27/warc/CC-MAIN-20200530102741-20200530132741-00511.warc.gz"} |
http://clay6.com/qa/27845/a-beam-of-light-consisting-of-two-wavelength-6500-a-and-5200-a-is-used-to-o | # A beam of light consisting of two wavelength $6500 A^{\circ}$ and $5200 A^{\circ}$ is used to obtain interference fringes in a young's double slit exp. Then what is the distance from central maxima where the bright fringes due to both wavelength coincides? The distance between slits is 2 mm and distance between plane of slits and screen is 120 cm
$(a)\;0.2\;cm \\ (b)\;0.156\;cm \\ (c)\;0.3\;cm \\ (d)\;0.9 \;cm$
If n is least number of fringes of $\lambda_1(=6500 A^{\circ})$ which are coincident with $(n+1)$ of smaller wavelength $\lambda_2=(5200 A^{\circ})$
$y' =\eta \beta =(\eta+1) \beta'$
ie $\large\frac{\eta+1}{\eta}=\frac{\beta}{\beta'}=\frac{\lambda_1}{\lambda_2}$
or $\eta =\large\frac{\lambda_2}{\lambda_1- \lambda_2} =\frac{5200}{6500-5200}$
$\quad= \large\frac{5200}{1300}$
$\quad=4$
also $\beta = \frac{\lambda D}{d}$
substituting values for wavelength,D and d we get $\beta=0.039 cm$
So, $y' =4 \beta = 4 \times 0 .039 =0.156\;cm$
Hence b is the correct answer.
edited Jul 22, 2014 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9813671112060547, "perplexity": 681.4451831018439}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647251.74/warc/CC-MAIN-20180320013620-20180320033620-00724.warc.gz"} |
https://www.zbmath.org/?q=an%3A0876.17012 | ×
# zbMATH — the first resource for mathematics
Spectrum of the quantized Weyl algebra. (Spectre de l’algèbre de Weyl quantique.) (French. English summary) Zbl 0876.17012
Let $$k$$ be a field of characteristic 0, let $$n$$ be a positive integer, let $${\mathbf q} =(q_1, \dots, q_n) \in(k^*)^n$$ and let $$\Lambda= (\lambda_{ij})$$ be an $$n\times n$$ matrix whose entries are non-zero elements of $$k$$ satisfying $$\lambda_{ij} =\lambda_{ij}^{-1}$$ and $$\lambda_{ii} =1$$ for all $$i$$ and $$j$$. Then $$n$$th quantized Weyl algebra $$A_n^{\underline{q}^Lambda}$$ is the $$k$$-algebra on the $$2n$$ generators $$x_1, \dots, x_n,$$ $$y_1, \dots, y_n$$, subject to the relations (1) for all $$i$$ and $$j$$ with $$i<j$$, $$x_ix_j= \lambda_{ij} q_ix_jx_i$$, $$y_iy_j= \lambda_{ij} y_jy_i$$, $$x_iy_j= \lambda^{-1}_{ij} y_jx_i$$, $$y_ix_j= \lambda^{-1}_{ij} q_i^{-1} x_jy_i$$, and (2) for all $$i, x_iy_i- q_iy_ix_i=1 +\sum^{i-1}_{j=1} (q_j-1) y_jx_j$$. This algebra was defined by G. Maltsiniotis as the algebra of differential operators on quantum $$n$$-space, and studied in, for example J. Alev and F. Dumas [J. Algebra 170, 229-265 (1994; Zbl 0820.17015)].
The main purpose of this paper is to describe the prime spectrum of $$A_n^{\underline{q}^Lambda}$$ in the case where the subgroup of $$k^*$$ generated by $$\{\lambda_{ij}, q_\ell: 1<i, j,\ell \leq n\}$$ is torsion free of rank $${1\over 2} n(n+1)$$. It is proved that under these hypotheses the spectrum is the union of a 1-torus of maximal ideals (the kernels of the algebra homomorphisms to $$k)$$ together with a finite and explicitly described set of prime ideals. As a corollary it is deduced that, under the same hypotheses, the group of $$k$$-algebra automorphisms of $$A_n^{\underline{q}^Lambda}$$ is an $$n$$-torus, (where $$\alpha= (\alpha_1, \dots, \alpha_n) \in(k^*)^n$$ sends $$x_i$$ to $$\alpha_ix_i$$ and $$y_i$$ to $$\alpha_i^{-1}y_i)$$. A similar description of the prime spectrum of $$A_n^{\underline{q}^\Lambda}$$, (under somewhat weaker restrictions on the parameters) has been independently obtained in [M. Akhavizadegan and D. A. Jordan, Glasg. Math. J. 38, 283-297 (1996)].
##### MSC:
17B37 Quantum groups (quantized enveloping algebras) and related deformations 16S32 Rings of differential operators (associative algebraic aspects) 16D25 Ideals in associative algebras
Full Text: | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9428836107254028, "perplexity": 210.0973431302722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00385.warc.gz"} |
https://brilliant.org/discussions/thread/anythings-possible-am-i-right/ | ×
# Anything's possible, am I right?
The polynomials $$x^2 + 2$$ and $$x^2+x-1$$ are written on a board. If the polynomials $$f(x)$$ and $$g(x)$$ are already on the board ($$f(x)$$ may equal to $$g(x)$$) we are allowed to add any of $$f(x) g(x), f(x) -g(x)$$ or $$f(x) + g(x)$$ to the board.
(a) Can the polynomial $$x$$ ever be made to appear on the board?
(b) Can the polynomial $$4x-1$$ ever be made to appear on the board?
Give proof.
• $$f(x)$$ and $$g(x)$$ can represent any two polynomials on the board
Note by Sharky Kesa
1 year, 9 months ago
Sort by:
(a) Suppose $$a, b$$ are integers. If $$f(a), g(a)$$ are divisible by $$b$$, then $$f(a)+g(a), f(a)-g(a), f(a)g(a)$$ are all divisible by $$b$$ as well. Thus the property "divisible by $$b$$ at $$x = a$$" is an invariant; they will hold for all polynomials generated.* At $$x = 3$$, $$f(x) = g(x) = 11$$ and thus both of them are divisible by $$11$$, so all polynomials generated will be divisible by $$11$$ on $$x=3$$. But $$x$$ doesn't satisfy this (since it gives $$3$$ on $$x=3$$). So it cannot appear.
(*) We can use induction to formalize this, inducting on the number of operations required to generate a polynomial. The base case is $$f, g$$ that takes zero steps; the induction step lies on the fact that if a polynomial $$p$$ can be generated (in the shortest way) from either of $$f+g, f-g, fg$$, then $$f,g$$ must be generated before $$p$$, so we can use induction there.
(b) It can appear. Let $$F_1(x) = x^2 + 2, F_2(x) = x^2 + x - 1$$. Then,
• $$F_3(x) = F_1(x) - F_2(x) = x - 3$$
• $$F_4(x) = F_3(x) F_3(x) = x^2 - 6x + 9$$
• $$F_5(x) = F_1(x) - F_4(x) = 7x - 10$$
• $$F_6(x) = ((F_5(x) - F_3(x)) - F_3(x)) - F_3(x) = 4x - 1$$
· 1 year, 9 months ago
What is the reason for choosing $$x = 3$$? In particular, if we were given 2 other (quadratic) polynomials, how do we know what to use?
Can we classify the set of polynomials which can be reached through these operations?
Hint: Bezout's Identity. Staff · 1 year, 9 months ago
Nice question :)
Hint: Find an Invariant.
Hint: Evaluate the polynomial functions at a particular point. Staff · 1 year, 9 months ago
I don't really get your problem. What are g(x) and f(x)? Are g(x) and f(x) equal to $$x^{2}+1$$ and/or $$x^{2} +x-1$$ · 1 year, 9 months ago
$$f(x)$$ and $$g(x)$$ are any two polynomials on the board. · 1 year, 9 months ago
It's any $$f(x), g(x)$$ that is written on the board. · 1 year, 9 months ago | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9845314621925354, "perplexity": 430.93586324379316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105961.34/warc/CC-MAIN-20170820015021-20170820035021-00247.warc.gz"} |
https://web2.0calc.com/questions/what-is-imaginary-i-raised-to-the-power-imaginary-i | +0
# What is imaginary I raised to the power imaginary i?
0
789
3
+1
What is I to the power I?
Oct 17, 2015
edited by Guest Oct 17, 2015
#2
+105604
+10
Hi Michael and guest,
I thought it was interesting too.
Here is a proof.
but really |z| is just the distance Z is from (0,0) on the complex number plane
and
arg(z) is just the angle z makes with the positive real axis. (at the origin of course)
$$i^i=e^{-\pi/2}\\ proof\\ i^i=e^{[ln(i^i)]}\\ i^i=e^{i[ln(i)]}\\ \qquad\mbox{Now ln(z)=ln|z|+i*arg(z) so}\\ \qquad ln(i)=ln|i|+i*arg(i)\\ \qquad ln(i)=ln(1)+i*\frac{\pi}{2}\\ \qquad ln(i)=i*\frac{\pi}{2}\\ i^i=e^{i*i*\frac{\pi}{2}}\\ i^i=e^{-1*\frac{\pi}{2}}\\ i^i=e^{{\frac{-\pi}{2}}}\\$$
.
Oct 17, 2015
edited by Melody Oct 17, 2015
#1
+5
You know what "i" stands for? Of course, it stands for √−1. So, what made you ask this question? Is it an assignment or just curiosity on your part? Or, are you studying "Complex numbers?" It's an interesting question, however!. But, I'm afraid that you may get lost in the explanation. It has a numerical value of: e^(-Pi/2)=0.207879576.........etc.
Oct 17, 2015
#2
+105604
+10
Hi Michael and guest,
I thought it was interesting too.
Here is a proof.
but really |z| is just the distance Z is from (0,0) on the complex number plane
and
arg(z) is just the angle z makes with the positive real axis. (at the origin of course)
$$i^i=e^{-\pi/2}\\ proof\\ i^i=e^{[ln(i^i)]}\\ i^i=e^{i[ln(i)]}\\ \qquad\mbox{Now ln(z)=ln|z|+i*arg(z) so}\\ \qquad ln(i)=ln|i|+i*arg(i)\\ \qquad ln(i)=ln(1)+i*\frac{\pi}{2}\\ \qquad ln(i)=i*\frac{\pi}{2}\\ i^i=e^{i*i*\frac{\pi}{2}}\\ i^i=e^{-1*\frac{\pi}{2}}\\ i^i=e^{{\frac{-\pi}{2}}}\\$$
Melody Oct 17, 2015
edited by Melody Oct 17, 2015
#3
+6045
+10
I've never seen this treated so I thought I'd take a deeper look.
$$c = r_c e^{\imath \theta_c} \\ c^\imath = \left( r_c e^{\imath \theta_c}\right)^\imath = r_c^\imath e^{\imath^2 \theta_c} = r_c^\imath e^{-\theta_c} \\ \mbox{Now what is }r_c^\imath? \\ r_c = e^{\ln(r_c)} \\ r_c^\imath = \left(e^{\ln(r_c)}\right)^\imath = e^{\imath \ln(r_c)} \\ \mbox{so }c^\imath = e^{\imath \ln(r_c)} e^{-\theta_c}= \\ e^{-\theta_c}\left(\cos(\ln(r_c))+\imath \sin(\ln(r_c))\right) \\ \mbox{Letting }c=\imath \\ r_c=1, \theta_c=\dfrac \pi 2 \\ \imath^\imath = e^{-\frac \pi 2} \left(\cos(\ln(1)) + \imath \sin(\ln(1))\right) = \\ e^{-\frac \pi 2}\left(\cos(0)+\imath \sin(0)\right)=e^{-\frac \pi 2} \\ \mbox{which is in agreement with Melody's answer}$$
.
Oct 18, 2015 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9778124094009399, "perplexity": 2257.0337127664666}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668954.85/warc/CC-MAIN-20191117115233-20191117143233-00089.warc.gz"} |
https://math.stackexchange.com/questions/2591634/what-is-the-apex-angle-of-the-cone-of-positive-semidefinite-matrices | What is the apex angle of the cone of positive semidefinite matrices?
Let $\def\S{\mathbf S}\S^n$ be the linear space of symmetric $n \times n$ matrices and $\S_+^n$ be the subset of positive semidefinite matrices. It is well-known that $\S_+^n$ is a convex cone in $\S^n$. In order to get a better geometric understanding of this object, I asked myself what the apex angle of this cone might be.
We use the inner product $\DeclareMathOperator{\tr}{tr}\def\<{\langle}\def\>{\rangle}\<A,B\>=\tr(AB)$, where $\tr(A)$ is the trace of $A$.
The apex angle $\theta$ of $\S_+^n$ is the biggest value of $\arccos\<A_1,A_2\>$ for $A_i\in\S_+^n$ with $\<A_i,A_i\>=1$.
My best result so far
Let $\def\E{\mathbf E}\E$ be some Euclidean space and $S\subset \E$ a proper subspace. Let $A_1\in\S_+^n$ be the orthogonal projection onto $S$ and $A_2\in\S_+^n$ the orthogonal projection onto the orthogonal complement $S^\bot$. Then $\<A_i,A_i\>=1$ but $A_1A_2=0$, hence $\<A_1,A_2\>=\tr(A_1A_2)=0$.
So we have that $\theta\ge 90^\circ$. Can we do better? Especially, can we have $\tr(A_1A_2)<0$?
It is impossible to have $\operatorname{tr}(A_1A_2) < 0$ if each $A_i$ is positive semidefinite.
A proof I like is as follows: note that (via the spectral theorem, for instance) we may decompose $A_2$ into $$A_2 = \sum_{k=1}^n x_kx_k^T, \qquad x_k \in \Bbb R^n$$ With that, we find that $$\operatorname{tr}(A_1A_2) = \operatorname{tr}\left(A_1\sum_{k=1}^n x_kx_k^T\right) = \sum_{k=1}^n \operatorname{tr}(A_1x_kx_k^T) = \sum_{k=1}^n \operatorname{tr}(x_k^TA_1x_k) = \\ \sum_{k=1}^n x_k^TA_1x_k \geq 0$$
• Another notable result is that the cone $S_+^n$ is self-dual – Omnomnomnom Jan 4 '18 at 16:06
• No problem. Off the top of my head, I believe that all self-dual (closed, convex) cones have a $90^\circ$ apex angle, but that's just a hunch. – Omnomnomnom Jan 4 '18 at 16:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9489852786064148, "perplexity": 89.7690162489601}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256100.64/warc/CC-MAIN-20190520182057-20190520204057-00480.warc.gz"} |
http://umj.imath.kiev.ua/authors/name/?lang=en&author_id=3154 | 2019
Том 71
№ 11
# Zavodovskii M. V.
Articles: 2
Brief Communications (Russian)
### Growth of generalized Temperley–Lieb algebras connected with simple graphs
Ukr. Mat. Zh. - 2009. - 61, № 11. - pp. 1579-1585
We prove that the generalized Temperley–Lieb algebras associated with simple graphs Γ have linear growth if and only if the graph Γ coincides with one of the extended Dynkin graphs ${\tilde A_n}$, ${\tilde D_n}$, ${\tilde E_6}$, or ${\tilde E_7}$. An algebra $T{L_{\Gamma, \tau }}$ has exponential growth if and only if the graph Γ coincides with none of the graphs ${A_n}$, ${D_n}$, ${E_n}$, ${\tilde A_n}$, ${\tilde D_n}$, ${\tilde E_6}$, and ${\tilde E_7}$.
Article (Russian)
### On algebras of the Temperley-Lieb type associated with algebras generated by generators with given spectrum
Ukr. Mat. Zh. - 2004. - 56, № 5. - pp. 634–641
We introduce and study algebras of the Temperley-Lieb type associated with algebras generated by linearly connected generators with given spectrum. We study their representations and the sets of parameters for which representations of these algebras exist. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9337053298950195, "perplexity": 682.45547390355}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146414.42/warc/CC-MAIN-20200226150200-20200226180200-00025.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=Bibhorr_Formula&direction=prev&oldid=97759 | # Bibhorr Formula
Bibhorr formula yields a relation between three sides and angle of a right triangle. The angle which is equated to linear variables is the Bibhorr angle. The formula is a superior alternative to trigonometry as is devoid of any sine and cosine functions. The equation establishes a geometric construction among the elements of a triangle as opposed to trigonometry.
## Statement
For a given right triangle with longest side श्र, medium side लं and shortest side छ, the angle opposite the medium side (Bibhorr angle) बि is given as:
This equation is known as Bibhorr formula. The symbolical notations use Hindi letters and specifically denote the sides.
## Constants
The use of two constants - or and makes the formula more legible. These constants are known as "Bibhorr sthiron" and "Bibhorr constant" respectively.
## Units
The units of Bibhorr angle depend on the the units of Bibhorr sthiron. If this constants is then angle is in degrees but if Bibhorr sthiron is in the form then Bibhorr angle results in radians.
## Explanation
Consider a right triangle ABC, such that BC and AC are shortest and medium sides respectively and AB is the longest side or hypotenuse. Now, the angle opposite AC, called Bibhorr angle is given as: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9706361293792725, "perplexity": 2486.05740564967}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00604.warc.gz"} |
https://www.originlab.com/doc/X-Function/ref/mquantilesPro | 2.13.1.13 mquantiles(Pro)
Brief Information
Compute quantiles on matrix data
This feature is for OriginPro only.
Command Line Usage
1. mquantiles im:=MBook1;
2. mquantiles im:=MBook1 median:=mymedian;
3. mquantiles im:=MBook1 x1:=r1 y1:=c1 x2:=r2 y2:=c2;
Variables
Display
Name
Variable
Name
I/O
and
Type
Default
Value
Description
Input Matrix im
Input
MatrixObject
<active>
This variable specifies the range of input Matrix for which quantiles will be calculated.
Minimum min
Output
double
<unassigned>
This variable specifies the output for the minimum value of the matrix.
Q1 q1
Output
double
<unassigned>
This variable specifies the output for the 1st quartile, that is, the value in a dataset that demarcates the lowest 25% of the values in an ordered set.
Median median
Output
double
<unassigned>
This variable specifies the output for the median or 2nd quartile, that is, the value in a dataset that demarcates the lowest 50% of the values in an ordered set.
Q3 q3
Output
double
<unassigned>
This variable specifies the output for the 3rd quartile, that is, the value in a dataset that demarcates the lowest 75% of the values in an ordered set.
Maximum max
Output
double
<unassigned>
This variable specifies the output for the maximum value of the matrix.
Interquartile Range iqr
Output
double
<unassigned>
This variable specifies the output for the interquartile range (IQR), that is, Q3 - Q1.
Index of Minimum X x1
Output
double
<unassigned>
This variable specifies the output for the row index of the minimum value in the matrix.
Index of Minimum Y y1
Output
double
<unassigned>
This variable specifies the output for the column index of the minimum value in the matrix.
Index of Maximum X x2
Output
double
<unassigned>
This variable specifies the output for the row index of the maximum value in the matrix.
Index of Maximum Y y2
Output
double
<unassigned>
This variable specifies the output for the column index of the maximum value in the matrix.
Description
quantiles are values from the data below which lie a given proportion of the data points in a given set. For example, 25% of data points in any set of data lay below the first quartile, and 50% of data points in a set lay below the second quartile, or median. When data is divided into 100 parts, percentiles can be calculated.
The X-function of mquantiles is used to compute the quantiles of the Matrix object. It can give out the basic aspects of the Matrix. For example, it can output the minimum, maximum, 1st quartile, median, 3rd quartile and interquartiles range. From these statistics, you can know the general distributional information of the specified Matrix.
Examples
Example 1
1. Active a Matrix.
2. From the menu, select Matrix: Set Values and open the dialog.
3. Type the following code in the Edit box
sin(i)+cox(j)
4. Click OK button.
5. Type the following command in the Script Window:
mquqntiles;
6. Click Enter.
7. Then you can run
mquantiles.=;
to get the basic information of the Matrix, such as minimum, maximum, 1st quartile, median, 3rd quartile and interquartiles range.
Example 2
The following script will get the minimum and maximum values and the corresponding column and row indices.
win -t m;
wks.ncols=5;
wks.nrows=5;
matrix -v i*j;
mquantiles im:=MBook1 min:=zmin max:=zmax x1:=r1 y1:=c1 x2:=r2 y2:=c2;
zmin=;
zmax=;
r1=;
r2=;
c1=;
c2=;
Algorithm
Sort given data first. Denote the sorted data range as ,
1. The 1st quartile is computed as , where is the integer part, and is the fractional part of equation
2. The median, or 2nd quartile, is computed as , where is the integer part, and is the fractional part of equation
3. 3rd quartile is computed as , where is the integer part, and is the fractional part of equation
4. The interquartile range, or IQR, is calculated as
References
David, H. A. 1981. Order Statistics. 2nd ed. Wiley, New York.
Tukey, J.W. Exploratory Data Analysis. Addison-Wesley, 1977.
Related X-Functions
Keywords:minimum, maximum, median, q1, q3 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8094382882118225, "perplexity": 2119.8445968547776}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00331.warc.gz"} |
http://cpc.ihep.ac.cn/article/1996/5 | 1996 Vol. 20, No. 5
Display Method: |
1996, 20(5): 385-393.
Abstract:
The High spin isomer(HSI)in 144Pm was produced and separated by using the high spin isomer beam facility.Based on the measurements of γ-γ coincidence, γ-ray excitation functions and isotropies,the decay level scheme of the HSI was established for the first time including 19 new high spin levels and 29 new γ-transitions assigned to 144Pm by this work.The HSI-γ correlation measurement indicated that the halflife of the HSI is longer than 2μs.From the systematic comparisons and the deformed independent particle model calculations for the N=83 isotones,the most possible particle configuration of π(1h211/2d5/2)v(1i13/21h9/22f7/2) and the spin-parity of Jπ=27+ were assigned to this HSI corresponding to an oblate deformation with β=－0.18.
1996, 20(5): 394-399.
Abstract:
The in-plane and out-of-plane emission of light particles from 40Ar+197Au collisions at 25MeV/u was studied by means of coincidence measurement of light particles with two fission fragments.An in-plane enhancement was observed for mid-rapidity p,d,t,α particles,indicating a rotation effect in this reaction system.This enhancement becomes more obvious with the increase of mass of the particles,or with the increase of the impact parameters.It was also found that in-plane emission is dominant for projectile-like particles.
1996, 20(5): 400-404.
Abstract:
Gamma family events with visible energies higher than 1015eV usually form halos in the emulsion chambers,which bring up some problems for further investigating the characteristics of superhigh energy nuclear interactions. At present,only a few gross quantities are used for describing the halo events.On the other hand,there are some fine structures inside the halo, which can give us more information.We have measured all the showers one by one in the superhigh energy family events and compared the results with those obtained by the method of halo measurements.Analyses and discussions are made forsome special events.
1996, 20(5): 405-408.
Abstract:
The attenuation lengths of hadron fluxes with visible energies higher than 4TeV have been observed in thick type iron emulsion chambers exposed at Mt. Kanbala with high detecting efficiency for hadrons.Uncorrelated hadrons as well as hadrons in family events with visible energies of ΣEγ>500TeV、100TeV≤ΣEγ≤500TeV and 30TeV≤ΣEγ<100TeV are measured.Preliminary results indicate that the attenuation lengths of hadrons in iron in these four cases are different.A brief discussion is given.
1996, 20(5): 409-413.
Abstract:
In this paper,we discuss that the mass of stellar, nucleon, and the dark matter particles are connected by a large number A～1019,and the strong interaction might be phenomenologically connected with the gravitation also by this large number in a deeper level.
1996, 20(5): 414-421.
Abstract:
The rare decays t→cV(V=Z,γ,g)induced through loop diagrams are calculated in a minimal Technicolor model(MTC) with a massless weak doublet recently presented by C.D.Carone and H. Georgi.We find that these new contributions can enhance the SM branching fractions by as much as 3—4 orders of magnitude,i.e.,B(t→cZ)≈10,B(t→cγ)≈10－9, B(t→cg)≈10－6 for the favorable values of the parameters.It might provide a unique window to detect the virtual effects of technipions in MTC theory.
1996, 20(5): 422-429.
Abstract:
In the frame of the random neck rupture and multichannel fission model,thefragment yield,average total kinetic energy distributions and prompt neutron multiplicities of the neutron induced 233,235,238U(n,f)fission covering the incident neutron energy range from thermal to 6 MeV are calculated.The theoretical results are in accord with the experimental data.
1996, 20(5): 430-437.
Abstract:
FolloWing the intrabeam scattering(IBS)theory of Bjorken and Mtingwa,more general formulae for the growth rates of horizontal,vertical emittance and longitudinal momentum spread are deduced in a simpler form of elliptic integrals.The formulae are applied to the proposed HIRFL-CSR. By means of numerical integration,continuous variations of emittances and momentum spread with time are obtained.The results show that the IBS growth rates are not rapid enough to be a design constraint of CSR lattice.
1996, 20(5): 438-447.
Abstract:
The RVUU model is used to study dynamical processes of the kaon and pion produced in heavy ion collisions.We include the nuclear medium effect on kaon and pion in the model, and simulate pion production and kaon subthreshold production process in heavy ion collisions at 1 GeV/nucleon.The calculated results show that the attractive pion optical potential changes the final pion momentum spectrum,increasing transverse momentum spectrum at the low momentum region.At the same time it also apparently enhances kaon abundance and modifies kaon momentum distribution.It illustrates that both kaon and pion final state dynamic processes should be considered in order to make a reasonable comment on the signals in heavy ion collisions.
1996, 20(5): 448-454.
Abstract:
Relativistic mean field theory in the rotating frame is used to describe superdeformed nuclei. Nuclear currents and the resulting spatial components of the vector meson fields are fully taken into account.It is shown that the filling of specific orbitals can lead to bands with deexcitation γ-ray energies differing by at most a few keV in nuclei with two mass units difference.The results is in a good agreement with the recent experimental data over a range of angular momenta.
1996, 20(5): 455-459.
Abstract:
The well known axial rotor plus quasiparticle model for odd nuclei is generalized to odd-odd nuclei.The proton-neutron interactions are considered in the base-space of the model for the first time.The method presented in this paper is used for several realistic nuclei in A=160 area.The results agree with experimental data qualitatively.A possible mechanism for Signature inversion of odd-odd nuclei is also discussed.
1996, 20(5): 460-467.
Abstract:
Technical specifications of Beam-line 4W1C and X-ray diffuse scattering station at Beijing Synchrotron Radiation Facility(BSRF)were discussed. Double focussed monochromatic X-ray beam,with energy resolution of 4.4×10－4 and beam spot of 0.5mm(H)×0.3mm(V),was obtained with a bent crystal mdnochromator shaped in triangular and a cylindric mirror.Nonspecular X-ray scattering from quasiperiodic amorphous multilayer was studied.The scattering maximums formed intensity streaks in the reciprocal space having the quasiperiodic character.Intensity modulations were observed and simulations based on atomic scattering gave satisfactory results.
1996, 20(5): 468-475.
Abstract:
According to the principle of"uniform magnetization",the 3-dimensional magnetic fields of the permanent magnet linear peridic system─hybrid insenion devices are analysed.The approximate analytical expressions of 3-dimensions are given.By applying these expressions,the magnetic field distributions in 3-dimensions of hybrid insertion devices can be calculated,and the calculation error is about 5%.This method conduces to design and optimization of the permanent magnet insertion devices.
1996, 20(5): 476-480.
Abstract:
Based on our Month Carlo Calculation model of heavy ion track structure,the radial dose distribution around an energetic heavy ion path and the maximum radial effect range in the liquid water are calculated.The results are compared with the experiments and other δ-ray theoretical calculation.This is useful for obtaiaing the relations between lesion intensity and range of cells and heavy ion parameters,and the biological effect mechanisms of heavy ion irradiation. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8326667547225952, "perplexity": 3972.7284818532407}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00007.warc.gz"} |
http://math.stackexchange.com/questions/1382/counting-subsets-with-r-mod-5-elements | # Counting subsets with r mod 5 elements
Some time ago Qiaochu Yuan asked about counting subsets of a set whose number of elements is divisible by 3 (or 4).
The story becomes even more interesting if one asks about number of subsets of n-element set with $r\bmod 5$ elements. Denote this number by, say, $P_n (r \bmod 5)$.
An experiment shows that for small $n$, $P_n(r \bmod 5)-P_n(r' \bmod 5)$ is always a Fibonacci number (recall that for "$r \bmod 3$" corresponding difference is always 0 or 1 and for "$r \bmod 2$" they are all 0). It's not hard to prove this statement by induction but as always inductive proof explains nothing. Does anybody have a combinatorial proof? (Or maybe some homological proof — I've heard one for "$r \bmod 3$"-case.)
And is there some theorem about $P_n(r \bmod l)$ for arbitrary $l$ (besides that it satisfies some recurrence relation of degree growing with $l$)?
-
In general your $P_n$ (r mod l) will be a linear combination of terms $(1+\zeta)^n\pm (1+\zeta^{-1})^n$ where $\zeta$ runs through the $l$-th roots of unity. When $l=5$ then these involve powers of $2$ and also powers of the golden ratio, as that is involved in $\cos \pi/5$ and $\cos 2\pi/5$. – Robin Chapman Aug 2 '10 at 11:04
@Robin Yes, I thought about it (although I was too lazy to carry out actual proof from this) but it's more like solving recurrence relation (packed into generating function, but anyway) and it would be nice to have a combinatorial proof. – Grigory M Aug 2 '10 at 11:25
Take the graph whose vertices are the vertices of a regular l-gon and whose edges are its edges. Then, for example, P_n(0 mod l) is the number of walks of length n on this graph which begin and start at a particular vertex. The bijection is that going left corresponds to having a particular element in your subset and going right corresponds to not having it. The Fibonacci numbers admit a similar interpretation (on more than one graph!) so one might be able to do something from here. – Qiaochu Yuan Aug 2 '10 at 17:52
In particular, Fibonacci numbers count walks on the path graph of length 4, which is the one I think will be relevant to this problem. – Qiaochu Yuan Aug 2 '10 at 20:02
The proof of the Rogers-Ramanujan conjecture in Andrews and Eriksson's book Integer Partitions books.google.co.uk/… relies, in effect, on a $q$-analogue of the relation between Fibonacci numbers and the differences between the "Pn(r mod 5)". – Robin Chapman Aug 6 '10 at 19:29
Recall that binomial coefficients count number of walks with steps (+1,+1) and (+1,-1) from the origin to different points (e.g. the number of walks to the point (2n,0) is $\binom{2n}{n}$). Consider the following involution on the set of all such walks: if the path intersects with the line y=l-1 or the line y=-1, reflect its part starting from the first intersection point (w.r.t. corresponding line). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8507039546966553, "perplexity": 436.1985356870316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644068098.37/warc/CC-MAIN-20150827025428-00300-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://stabix.readthedocs.io/en/latest/bicrystal_definition.html | # Bicrystal Definition¶
## Crystallographic properties of a bicrystal¶
A bicrystal is formed by two adjacent crystals separated by a grain boundary.
Five macroscopic degrees of freedom are required to characterize a grain boundary [3], [5], [6] and [7] :
• 3 for the rotation between the two crystals;
• 2 for the orientation of the grain boundary plane defined by its normal $$n$$.
The rotation between the two crystals is defined by the rotation angle $$\omega$$ and the rotation axis common to both crystals $$[uvw]$$.
Using orientation matrix of both crystals obtained by EBSD measurements, the misorientation or disorientation matrix $$(\Delta g)$$ or $$(\Delta g_\text{d})$$ is calculated [4] and [2] :
(1)$\Delta g = g_\text{B}g_\text{A}^{-1} = g_\text{A}g_\text{B}^{-1}$
(2)$\Delta g_\text{d} = (g_\text{B}*CS)(CS^{-1}*g_\text{A}^{-1}) = (g_\text{A}*CS)(CS^{-1}*g_\text{B}^{-1})$
Disorientation describes the misorientation with the smallest possible rotation angle and $$CS$$ denotes one of the symmetry operators for the material [1].
The Matlab function used to set the symmetry operators is : sym_operators.m
The orientation matrix $$g$$ of a crystal is calculated from the Euler angles ($$\phi_{1}$$, $$\Phi$$, $$\phi_{2}$$) using the following equation :
(3)$\begin{split}g = \begin{pmatrix} \cos(\phi_{1})\cos(\phi_{2})-\sin(\phi_{1})\sin(\phi_{2})\cos(\Phi) & \sin(\phi_{1})\cos(\phi_{2})+\cos(\phi_{1})\sin(\phi_{2})\cos(\Phi) & \sin(\phi_{2})\sin(\Phi) \\ -\cos(\phi_{1})\sin(\phi_{2})-\sin(\phi_{1})\cos(\phi_{2})\cos(\Phi) & -\sin(\phi_{1})\sin(\phi_{2})+\cos(\phi_{1})\cos(\phi_{2})\cos(\Phi) & \cos(\phi_{2})\sin(\Phi) \\ \sin(\phi_{1})\sin(\Phi) & -\cos(\phi_{1})\sin(\Phi) & \cos(\Phi) \\ \end{pmatrix}\end{split}$
The orientation of a crystal (Euler angles) can be determined via electron backscatter diffraction (EBSD) measurement or via transmission electron microscopy (TEM).
The Matlab function used to generate random Euler angles is : randBunges.m
The Matlab function used to calculate the orientation matrix from Euler angles is : eulers2g.m
The Matlab function used to calculate Euler angles from the orientation matrix is : g2eulers.m
Then, from this misorientation matrix ($$\Delta g$$), the rotation angle ($$\omega$$) and the rotation axis $$[u, v, w]$$ can be obtained by the following equations :
(4)$\omega = \cos^{-1}((tr(\Delta g)-1)/2)$
(5)$\begin{split}u = \Delta g_{23} - \Delta g_{32} \\ v = \Delta g_{31} - \Delta g_{13} \\ w = \Delta g_{12} - \Delta g_{21}\end{split}$
The Matlab function used to calculate the misorientation angle is : misorientation.m
The grain boundary plane normal $$n$$ can be determined knowing the grain boundary trace angle $$\alpha$$ and the grain boundary inclination $$\beta$$.
The grain boundary trace angle is obtained through the EBSD measurements (grain boundary endpoints coordinates) and the grain boundary inclination can be assessed by a serial polishing (chemical-mechanical polishing or FIB sectioning), either parallel or perpendicular to the surface of the sample (see Figure 5).
Figure 4 Schematic of a bicrystal.
Figure 5 Screenshot of the Matlab GUI used to calculate grain boundary inclination. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8243656754493713, "perplexity": 1046.996378139524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514571651.9/warc/CC-MAIN-20190915155225-20190915181225-00000.warc.gz"} |
https://scicomp.stackexchange.com/questions/33749/why-does-the-matlab-command-chola-slower-than-chola-lower-for-a-la | # Why does the matlab command **chol(A)** slower than **chol(A,'lower')** for a large sparse SPD matrix?
For a SPD matrix A, there exists Cholesky factorization $$A=LL^T$$ or $$A=R^TR$$, where L, R are a lower and upper triangular matrix, respectively.
Also in matlab, there has a command R = chol(A) which produces $$A=R^TR$$. and another command L = chol(A,'lower') which produces $$A=LL^T$$. But when I implement these two commands with a same large sparse SPD matrix A, L = chol(A,'lower') is faster than R = chol(A) so mysterious. why this happens? Thanks.
clc;clear;
n =400;
A = gallery('poisson',n);
tic
R = chol(A);% generate the triangular matrix such that A = R'*R
toc
tic
L = chol(A,'lower');% generate the lower matrix such that A = L*L'
toc
I have run the example 3 times and the numerical results are as follows which indeed demonstrates that the above words what I said (my cpu is 8GB memory and matlab 2018b):
Elapsed time is 12.089711 seconds.
Elapsed time is 8.467380 seconds.
Elapsed time is 10.372768 seconds.
Elapsed time is 8.131158 seconds.
Elapsed time is 10.027861 seconds.
Elapsed time is 8.105706 seconds.
• It's clearly something about the implementation of the sparse Cholesky factorization in MATLAB. MATLAB uses Tim Davis's CHOLMOD for this factorization. You could test CHOLMOD to see if it has the same issue and investigate the code to understand the particular issue. In any case it's a matter of implementation and not really appropriate for this group. – Brian Borchers Nov 6 '19 at 2:36
• Get it. Thanks for your reply. I understand it now. – sunshine Nov 6 '19 at 4:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8503639698028564, "perplexity": 1505.3655434529485}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145989.45/warc/CC-MAIN-20200224224431-20200225014431-00557.warc.gz"} |
http://physics.stackexchange.com/questions/420/would-a-magnet-attract-a-paperclip-indefinitely?answertab=oldest | # Would a magnet attract a paperclip indefinitely?
Let's say we have a magnet stuck to a metal bar, suspended above the ground. If I attach a paperclip to the magnet, where is the energy to hold the paperclip coming from (against the force of gravity), and for how long will the paperclip remain there - will it remain there forever?
-
When an object is hanging from a hook, where is the energy to hold the object against the force of gravity coming from? – Sparr Nov 10 '10 at 0:28
@Sparr, good analogy. – Thomas O Nov 10 '10 at 0:31
## 2 Answers
When the magnet attracts and moves the paperclip, moving it in the gravitational field of the earth, the energy comes from the potential energy that [in that case] we can associate to the magnetic field.
By attracting the paperclip you increase the gravitational potential energy of the paperclip but you reduce the one it has due to the magnetic field.
You do not need energy to hold the paperclip. As an analogy: when you hold something in your hands, you feel like you are "working" even in a static case; but in mechanics, work = exchange of energy is defined as $W = \int \vec{F} d\vec{s}$.
When the situation is stationary, no energy exchange is involved, no force does no work at all, so it can stay forever.
-
Exactly. It is a very important point that no work is done unless an object actually moves a distance. (p.s. Might want to surround your equation with $...$ so the LaTeX shows up?) – Noldorin Nov 9 '10 at 19:16
Latex is not working for this equation ? It looks OK for me. – Cedric H. Nov 9 '10 at 19:18
I think your first sentence may be a little misleading, since there is no energy (work) involved. – David Z Nov 9 '10 at 20:28
@David: I edited to reflect what I meant. I think like that it is OK. – Cedric H. Nov 9 '10 at 20:35
Sure, I guess that's better. – David Z Nov 9 '10 at 20:37
The paperclip is not moving relative to earth - this means that no energy is being spent by magnet to hold it, so it can hold the paperclip as long as the magnet has magnetic properties.
As a contrast, when you hold something with your hands, energy is being spent by your muscles not to hold the object, but to remain contracted against gravity. This is because natural state of muscle cells is to be stretched or contracted as gravity tells them to, not against it.
By analogy, electromagnet spends energy not to hold things, but to to have magnetic properties, because its natural state is not magnetic. So natural magnet does not spend any energy, since its natural state is being magnetic.
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Actually, an electromagnet "spends energy" due to the resistance in its wire, not due to maintaining a magnetic field. A superconducting loop, for example, has no such loss over time, and is a "permanent" magnet as long as it remains unbroken and superconducting. – Kevin Reid Jul 15 '11 at 4:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8911873698234558, "perplexity": 845.9128684010557}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644062327.4/warc/CC-MAIN-20150827025422-00110-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://www.ms.lt/sodas/Book/Tensor | Notes
• [email protected]
• +370 607 27 665
• My work is in the Public Domain for all to share freely.
Understandable FFFFFF
Questions FFFFC0
Notes EEEEEE
Software
See: Math
Šešeriopai suvoktą daugybą (multiplication) suvokti tensoriais (kovariantiškumu, kontravariantiškumu).
Lines (building up) are easily described by vectors - by construction, generation, spanning. One vector describes a line. We need more vectors to describe a plane, etc. Hyperplanes (tearing down) are easily described in terms of equations, by restriction, by conditions. One equation describes a hyperplane. We need more equations to describe a line, etc.
Constructing the most informative illustration of tensors.
Use 2x2 change in coordinates.
Use coordinate system for equilateral triangles and also a coordinate system for squares.
Determine:
• What is the manifold?
Dualities:
• Bottom-up and top-down.
• Tangent vector space and cotangent vector space.
Definition of a tensor:
• A tensor of type (p, q) is a map which maps each basis f of vector space V to a multidimensional array T[f] such that if fR is another basis, then T[fR] = ...R-1...R T[f].
• W:VxVx...xVxV*xV*...xV* -> R is a multilinear map (where V* is the dual space of covectors of the space V of vectors).
Determinant is top-down to define what is "inside" and what is "outside". A shape like the Moebius band is no fun because you can't make that distinction, you can't "understand" it, it does not make a "marked opposite". It is an unmarked duality. But for understanding we want a primitive marked duality, an irreducible marked duality. This is possible through the six transformations of perspectives given by the Holy Spirit.
Negative correlations vs. positive correlations. Yes vs. No.
Understanding the Lagrangian. Consider Kinectic Energy as "bottom-up" approach and Potential Energy as "top-down" approach. Kinetic Energy is finite and Potential Energy is possibly infinite. DT=−D is (roughly) anti-self adjointness.
• Nature maximizes the explicit with regard to the infinite (thus kinectic energy with regard to potential energy). This minimization is related to the avoidance of the collapse of the wave function if at all possible. Nature prefers the complex (unmarked opposites) over the reals (marked opposites). Nature minimizes the marked opposites.
R is super rich but can't handle itself root wise, algebraically. But just a small "shift" is required to add unmarked opposites and have C. Unmarked opposites are "implications" rather than "explications".
Cramer's rule for inverses involves replacing a column in the matrix with the column with the constants. Replacing a column implies a "top down" orthogonal system. Also, the determinant is an anti-symmetric top-down system which distinguishes inside and outside. Whereas the symmetric case does not distinguish inside and outside and leaves them as unmarked opposites. In order to have marked opposites, we need to have a system of anti-symmetry.
Šešeriopai suvoktą dauginimąsi (multiplication) suvokti, išsakyti tensoriais.
Thank you very much for your discussion of dimensions. I'm grappling with this very much. But basically I think that you are describing the distinction between the "top-down" view of a space (in which we start with the whole space and break it down) and the "bottom-up" view of a space (in which we start with an empty space and build it up). That's what at's the heart of tensors. They combine the two points of view. They break up an n dimensional space into p bottom-up (contravariant = vector) and q bottom-up (covariant = covector = hyperplane = reflection) points of view.
I appreciate your thinking on this difference.
I've realized that I need to understand "tensors". They are quite central. They are a generalization of matrices to multiple dimensions. But truly the real point of tensors is that they break space into two different points of view, "top down" and "bottom up", or in other words, covariant and contravariant. https://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors
Imagine a tetrahedron (in 3 dimensions) with a natural coordinate space on it (given by three edges). That coordinate space, though, has angles that are not 90 degrees. Now consider removing one of these "bottom up" coordinate basis vectors and replacing it with a "top down" vector in the following way: choose a vector that would be perpendicular to the remaining basis vectors. This gives you an entirely different but "dual" basis. I suppose this comes up with the Platonic solids where the "dual" of the cube is the octahedron and vice versa.
Well, so tensors describe multidimensional space in terms of two sets of basis vectors. Rather than vectors it turns out it is more natural to think of them as maps (linear functionals) from the vector space to the real numbers. The tensor then "eats" say p many vectors and "yields" say q many vectors. (For a matrix it would be 1 vector in and 1 vector out, making it a linear transformation.)
These tensors are very important because, for example, if you want to do integration on a complicated manifold (multidimensional surface) you need that manifold to be oriented, which is to say, have an "inside" and an "outside", or a "left direction" and a "right direction", so that the integration would give opposite sign in either case. So that means you need a wedge product that is antisymmetric, that switches sign whenever you flip your area/volume/etc over, that is, whenver you swap coordinates.
Well, all of this to say that I'm realizing that tensors are key to "geometry" because they establish the geometric space between the top down vectors and the bottom up vectors. So I'm thinking that "geometry" is the way of embedding a lower dimensional space into a higher dimensional space.
Another insight that's helpful is that any matrix can be decomposed into two matrices, just like "polar decomposition", with length and angle. One matrix is "unitary" and it handles the rotations/reflections but keeps the lengths the same. The other matrix stretches the lengths shorter and longer as needed. https://en.wikipedia.org/wiki/Polar_decomposition
Lie groups are typically matrix groups which are also manifolds, so that the actions can be composed in a continuous way, even an infinitesimal way. You can rotate a circle or sphere just the slightest bit. It reminds me of the proof of the Pythagorean theorem given by the truth that "four times a right triangle is the difference of two squares". If the right triangles are long and thin, then one square is just the slightest rotation of the other. So Lie groups are the study of this sort of thing and Lie algebras are their infinitesimal rotations/changes.
So circles and spheres and etc. are very central in this subject.
Another idea that came up is that in the real numbers the dimensions (of space) are all independent. But what the complexes can be thought of as doing is "coupling" two independent dimensions with a coupling "i". And that i transforms by 90 degrees so that it swaps the y variable with the x variable and keeps track of that.
#### Tensor
Naujausi pakeitimai
Puslapis paskutinį kartą pakeistas 2016 spalio 15 d., 14:23 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8663248419761658, "perplexity": 881.0918637875927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267865181.83/warc/CC-MAIN-20180623190945-20180623210945-00131.warc.gz"} |
http://efavdb.com/daily-traffic-evolution-and-the-super-bowl/ | With an eye towards predicting traffic evolution, we begin by examining the time-dependence of the contribution from the first principal components on different days of the week. Traffic throughout the day $\vert x(t) \rangle$ can be represented in the basis of principal components; $\vert x(t) \rangle$ $= \sum_{i} c_i(t) \vert \phi_i \rangle $$^1, where \vert \phi_i \rangle is the ith principle component. The coefficients c_i(t), sometimes called the “scores” of \vert x(t) \rangle in the basis of principal components, carry all of the dynamics. The largest deviations in the traffic patterns (and of the scores) are during weekday rush hours (around 8 am and 5 pm) – see plot of the scores for several modes throughout Jan. 15. There is an abundance of interesting information to be gleaned here. First, note that, generally, the amplitude of the lowest modes is the largest, again, as expected. In addition, there appears to be some large wavelength structure (primarily correlated with the rush hours) sitting on top of a background of higher frequency noise. Ignoring for the moment the noise, the modes generally come in two classes – something like “even” and “odd”. That is, some fluctuate with the same sign in both the morning and evening, while others fluctuate with opposite signs. Let us consider the first two modes in this plot. The first mode is an even mode and we attribute this essentially to an overall shift in general speed throughout the system. Whenever there are more cars on the road, the speed generally decreases everywhere. Indeed, this is the only mode that deviates significantly from zero at night and it deviates positive – the roads are generally completely clear at night and thus the speed is somewhat higher than average. The second mode is an odd mode, which we attribute generally to directional traffic – when everyone is going to work it has one sign, when they are coming home it has the other. We have previously visualized these two modes and, indeed, these concepts are reflected in their spatial structure: the first mode is generally uniform while the second has regions with either sign and many sections of highway are positive in one direction and negative in the other. The time-dependence of these scores is remarkably consistent from week to week (See Wednesdays plot below of the scores for modes 1 and 2). On the right, we plot the same quantities for several Sundays as well. Not surprisingly, the fluctuations are smaller on Sundays than on weekdays, reflecting more homogeneous speeds in sparse traffic. However, they are still reproducible from week to week – see Sundays Jan. 5, 12, 19, and 26 – apparently there is a slight slow-down around 6 pm. Feb. 2 was Super Bowl Sunday and the traffic pattern differs qualitatively from other Sundays. Remarkably, we can identify the time of the Super Bowl kickoff from this data – before the kickoff there is slightly more traffic than the average Sunday and immediately after, less. [1] On projecting into principal components: In the original basis, our data reads \vert x(t) \rangle = \sum_j a_j(t) \vert \ell_j \rangle , where the sum runs over all loop locations(there are some 2,000 loops in the Bay area) and \vert \ell_j \rangle is a unit fluctuation in speed at the location of loop j. Changing bases, to the principal components \vert \phi_i \rangle , \vert x(t) \rangle = \sum_{i,j} a_j(t) \vert \phi_i \rangle \langle \phi_i \vert \ell_j \rangle = \sum_i c_i(t) \vert \phi_i \rangle where c_i(t) =$$\sum_j a_j(t) \langle \phi_i \vert \ell_j \rangle$. The coefficient $\langle \phi_i \vert \ell_j \rangle$ is often called the “loading” of $\vert \ell_j \rangle$ into $\vert \phi_i \rangle$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8306024074554443, "perplexity": 871.9343352899106}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608067.23/warc/CC-MAIN-20170525121448-20170525141448-00117.warc.gz"} |
https://physmath.spbstu.ru/en/article/2018.42.6/ | # Modeling of dynamic processes in the vapor compression cooling system
Authors:
Abstract:
In the paper, a dynamic model of a vapor compression cooling system is presented. In addition to the usual one, it takes into account the working agent’s masses in the heat exchangers, this agent’s vapor content behavior in time at the outlet of the expansion valve, and the whole spectrum of two-phase flow modes during the working agent’s evaporation. It was established that it took more time for temperature’s and mass flow’s (in a vapor compression cooling system) transitions to steady states than for the rotational speed of the compressor shaft. The connection between the negative dynamics of the evaporation temperature and the initial ambient temperature was shown. Moreover, it was the connection between the delay in stabilization of the mass flow of the working medium and the initial ambient temperature as well as the degree of a pressure increase in the thermodynamic cycle of the vapor compression cooling system. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8801723718643188, "perplexity": 605.2103112123345}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313501.0/warc/CC-MAIN-20190817222907-20190818004907-00430.warc.gz"} |
https://www.physicsforums.com/threads/what-is-the-maximum-acceleration-the-car-can-have-without-causing-the-cup-to-slide.44331/ | # Homework Help: What is the maximum acceleration the car can have without causing the cup to slide?
1. Sep 23, 2004
### dg_5021
Problem-
If the coefficient of static friction between the coffee cup and the roof of the car is 0.24, What is the maximum acceleration the car can have without causing the cup to slide? Ignore the effects of air resistance.
i know the formula is: Fs= (ms) (N)
i multipy 0.24 x 9.81m/s^2
and got =2.35 m/s^2
2. Sep 23, 2004
### HallsofIvy
There is a rather important piece of information missing here! Remember what "N" is?
Don't you think a coffee cup made of lead would be less likely to slide than one made of fine china?
3. Sep 23, 2004
### dg_5021
i know N= (m)(g)
but the question never gives me the kg and i know gravity =9.81m/s^2
am i suppose to find kg?
4. Sep 23, 2004
### dg_5021
oh wait i got it ms= Fs/N
ms=m(ax)/m(g)
ms=ax/g
ms x g=ax
.24x 9.81 = 2.35m/s^2
thanks | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8311365246772766, "perplexity": 3015.2132566123273}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589618.52/warc/CC-MAIN-20180717070721-20180717090721-00349.warc.gz"} |
http://mathhelpforum.com/geometry/155902-distance-between-point-line-vectors.html | # Math Help - Distance between point and line (vectors)
1. ## Distance between point and line (vectors)
Find the distance from the point A(7,6,10) to the line passing through B(3,7,0) and C(4,9,-2).
I know that the vector is $\overrightarrow{BC} = \left \langle 1,2,-2 \right \rangle$ and that the equation of the line is thus $\mathbf{x} = (3,7,0) + t(1,2,-2) t\epsilon \mathbb{R}$ but I dont understand how you can find the distance between A and this line?
2. What you can try is to find the vector direction of AB. Then, find the angle CBA by using the dot product of the vectors.
Have a sketch, BC is a line, AB is another line. Draw the angle CBA. Drop a perpendicular line from A to line CB.
Now, the perpendicular line, AB and BM, where M is the point where the perpendicular meets BC is a triangle.
Have now a right angled triangle, an angle and you can find the length AB. Use
$sin(M\hat{B}A) = \frac{|AM|}{|AB|}$
3. Originally Posted by SyNtHeSiS
Find the distance from the point A(7,6,10) to the line passing through B(3,7,0) and C(4,9,-2).
Given a line $\ell :P + tD$ and that $R\notin \ell$ then the distance from $R$ to $\ell$ is $d(R;\ell ) = \left\| {\overrightarrow {PR} - \frac{{\overrightarrow {PR} \cdot D}}{{D \cdot D}}D} \right\| = \dfrac{{\left\| {\overrightarrow {PR} \times D} \right\|}}{{\left\| D \right\|}}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9279615879058838, "perplexity": 337.8965564864262}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500825341.30/warc/CC-MAIN-20140820021345-00113-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://cpc.ihep.ac.cn/article/justaccepted | ## Just Accepted
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Radial basis function network (RBFN) approach is adopted for the first time to optimize the calculation of \begin{document}$\alpha$\end{document} decay half-life in the generalized liquid drop model (GLDM) concurrently incorporating the surface diffuseness effect. Calculations of the present study are in good agreement with the experimental half-lives for 68 superheavy nuclei (SHN), and a remarkable reduction of 40% in the root-mean-square (rms) deviations of half-lives is achieved. After that, based on RBFN method, the half-lives for four SHN isotopes, 252-288Rf, 272-310Fl, 286-316119 and 292-318120, are predicted using the improved GLDM with the diffuseness correction and the decay energies from WS4 and FRDM as inputs. Hence, we conclude that the diffuseness effect should be embodied in the proximity energy. Simultaneously, the neural network methods are encouraged to widely used in nuclear reaction.
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In this presentation, we obtain the corresponding universal function to the diffractive process and show the cross section exhibits the geometrical scaling. It is observed the diffractive theory according to the color dipole approach at small-x is a convenient framework that reveals the color transparency and the saturation phenomena. Also we calculate the contribution of heavy quark productions in the diffractive cross section for high energy that is determined by the small size dipole configuration. The ratio of the diffractive cross section to the total cross section in the electron-proton collision is the other important quantity that is computed in this work.
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Applying the nonrelativistic quantum chromodynamics factorization formalism to the \begin{document}$\Upsilon(1S,2S,3S)$\end{document} hadroproduction, a complete analysis on the polarization parameters \begin{document}$\lambda_{\theta}$\end{document}, \begin{document}$\lambda_{\theta\phi}$\end{document}, \begin{document}$\lambda_{\phi}$\end{document} for the production are presented at QCD next-to-leading order. With the long-distance matrix elements extracted from experimental data for the production rate and polarization parameter \begin{document}$\lambda_{\theta}$\end{document} of \begin{document}$\Upsilon$\end{document} hadroproduction, our results provide a good description for the measured parameters \begin{document}$\lambda_{\theta\phi}$\end{document} and \begin{document}$\lambda_{\phi}$\end{document} in both the helicity and the Collins-Soper frames. In our calculations the frame invariant parameter \begin{document}$\tilde{\lambda}$\end{document} is consistent in the two frames. Finally, it is pointed out that there are discrepancies for \begin{document}$\tilde{\lambda}$\end{document} between available experimental data and corresponding theoretical predictions.
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In this contribution, the \begin{document}$\alpha$\end{document} preformation factors of 606 nuclei are extracted within the framework of generalized liquid drop model (GLDM). Through the systematically analysis of the \begin{document}$\alpha$\end{document} preformation factors of even-even Po-U isotopes, we found there is a significant weakening of influence of \begin{document}$N=126$\end{document} shell closure in uraninum, which is consistent with the result of a recent experiment [J. Khuyagbaatar et al., Phys. Rev. Lett. 115.242502 (2015)], implying that \begin{document}$N=126$\end{document} may be not the magic number for U isotopes. Furthermore, we propose an improved formula with only 7 parameters to calculate \begin{document}$\alpha$\end{document} preformation factors suitable for all types of \begin{document}$\alpha$\end{document}-decay, which has fewer parameters than the original formula proposed by Zhang et al. [H. F. Zhang et al., Phys. Rev. C 80.057301 (2009)] with high precision. The standard deviation of the \begin{document}$\alpha$\end{document} preformation factors calculated by our formula with extracted values for all 606 nuclei is 0.365 with a factor of 2.3, indicating that our improved formula can accurately reproduce the \begin{document}$\alpha$\end{document} preformation factors. Encouraged by this, the \begin{document}$\alpha$\end{document}-decay half-lives of actinide elements are predicted, which could be useful in future experiments. Noticeably, the predicted \begin{document}$\alpha$\end{document}-decay half-lives of two new isotopes \begin{document}$^{220}$\end{document}Np [Z.Y. Zhang, et al., Phys. Rev. Lett. 122. 192503 (2019)] and \begin{document}$^{219}$\end{document}Np [H. B. Yang et al., Phys. Lett. B 777, 212 (2018)] are in good agreement with the experimental \begin{document}$\alpha$\end{document}-decay half-lives.
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We demonstrate that the recently proposed soft gluon factorization (SGF) is equivalent to the nonrelativistic QCD (NRQCD) factorization for heavy quarkonium production or decay, which means that for any given process these two factorization theories are either both valid or both violated. We use two methods to achieve this conclusion. In the first method, we apply the two factorization theories to the physical process \begin{document}$J/\psi \to e^+e^-$\end{document}. Our explicit calculation shows that both SGF and NRQCD can correctly reproduce low energy physics of full QCD, and thus the two factorizations are equivalent. In the second method, by using equations of motion we successfully deduce SGF from NRQCD effective field theory. By identifying SGF with NRQCD factorization, we establish relations between the two factorization theories and prove the generalized Gremm-Kapustin relations as a by product. Comparing with the NRQCD factorization, the advantage of SGF is that it resums the series of relativistic corrections originated from kinematic effects to all powers, which gives rise to a better convergence in relativistic expansion.
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It was found that the dark matter (DM) in the intermediate-mass-ratio-inspiral (IMRI) system has a significant enhancement effect on the orbital eccentricity of the stellar massive compact object, such as a black hole (BH), which may be tested by space-based gravitational wave (GW) detectors including LISA, Taiji and Tianqin in future observations [1]. In this paper, we will study the enhancement effect of the eccentricity for an IMRI under different DM density profiles and center BH masses. Our results are as follows: (1) in terms of the general DM spike distribution, the enhancement of the eccentricity is basically consistent with the power-law profile, which indicates that it is reasonable to adopt the power-law profile; (2) in the presence of DM spike, the different masses of the center BH will affect the eccentricity, which provides a new way for us to detect the BH's mass; (3) considering the change of the eccentricity in the presence and absence of DM spike, we find that it is possible to distinguish DM models by measuring the eccentricity at the scale of about \begin{document}$10^{5} GM/c^{2}$\end{document}.
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We investigate different entanglement properties of a holographic QCD (hQCD) model with a critical end point at finite baryon density. Firstly we consider the holographic entanglement entropy (HEE) of this hQCD model in a spherical shaped region and a strip shaped region, respectively, and find that the HEE of this hQCD model in both regions can reflect QCD phase transition. What is more is that although the area formulas and minimal area equations of the two regions are quite different, the HEE have very similar behavior on the QCD phase diagram. So we argue that the behavior of HEE on the QCD phase diagram is independent of the shape of subregions. However, as we know that HEE is not a good quantity to characterize the entanglement between different subregions of a thermal system. So we then study the mutual information (MI), conditional mutual information (CMI) and the entanglement of purification (Ep) in different strip shaped regions. We find that the three entanglement quantities show some universal behavior: their values do not change so much in the hadronic matter phase and then rise up quickly with the increase of T and \begin{document}$\mu$\end{document} in the QGP phase. Near the phase boundary, these three entanglement quantities change smoothly in the crossover region, continuously but not smoothly at CEP and show discontinuity behavior in the first phase transition region. And all of them can be used to distinguish different phases of strongly coupled matter.
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In this paper, by introducing a Lorentz-invariance-violation (LIV) class of dispersion relations (DR) suppressed by the second power \begin{document}$(E/E_{QG})^2$\end{document}, we have investigated the effect of LIV on the Hawking radiation of the charged Dirac particle via tunneling from a Reissner-Nordström(RN) black hole. We first find the effect of LIV speeds up the black hole evaporation, leaving the induced Hawking temperature very sensitive to the changes in the energy of the radiation particle, but at the same energy level, insensitive to the changes in the charge of the radiation particle. This provides a phenomenological evidence for the LIV-DR as a candidate for describing the effect of quantum gravity. Then, when the effect of LIV is included, we find the statistical correlations with the Planck-scale corrections between the successive emissions can leak out the information through the radiation. And, it turns out that the black hole radiation as tunneling is an entropy conservation process, and no information loss occurs during the radiation, where the interpretation for the entropy of black hole is addressed. Finally, we conclude that black hole evaporation is still an unitary process in the context of quantum gravity.
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The cosmic distance relation (DDR) associates the angular diameters distance (\begin{document}$D_A$\end{document}) and luminosity distance (\begin{document}$D_L$\end{document}) by a simple formula, i.e., \begin{document}$D_L = (1+z)^2D_A$\end{document}. The strongly lensed gravitational waves (GWs) provide a unique way to measure \begin{document}$D_A$\end{document} and \begin{document}$D_L$\end{document} simultaneously to the GW source, hence can be used as probes to test DDR. In this paper, we prospect the use of strongly lensed GW events from the future Einstein Telescope to test DDR. We write the possible deviation of DDR as \begin{document}$(1+z)^2D_A/D_L = \eta(z)$\end{document}, and consider two different parametrizations of \begin{document}$\eta(z)$\end{document}, namely, \begin{document}$\eta_1(z) = 1+\eta_0 z$\end{document} and \begin{document}$\eta_2(z) = 1+\eta_0 z/(1+z)$\end{document}. Numerical simulations show that, with about 100 strongly lensed GW events observed by ET, the parameter \begin{document}$\eta_0$\end{document} can be constrained at 1.3% and 3% levels for the first and second parametrizations, respectively.
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In the present work, we used five different versions of the quark-meson coupling (QMC) model to compute astrophysical quantities related to the GW170817 event and neutron star cooling process. Two of the models are based on the original bag potential structure and three versions consider a harmonic oscillator potential to confine the quarks. The bag-like models also incorporate the pasta phase used to describe the inner crust of neutron stars. Within the simple method studied in the present work, we show that the pasta phase does not play a significant role. Moreover, the QMC model that satisfies the GW170817 constraints with the lowest slope of the symmetry energy exhibits a cooling profile compatible with observational data.
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We extend the auxiliary-mass-flow (AMF) method originally developed for Feynman loop integration to calculate integrals involving also phase-space integration. Flow of the auxiliary mass from the boundary (\begin{document}$\infty$\end{document}) to the physical point (\begin{document}$0^+$\end{document}) is obtained by numerically solving differential equations with respective to the auxiliary mass. For problems with two or more kinematical invariants, the AMF method can be combined with traditional differential-equation method by providing systematical boundary conditions and highly nontrivial self-consistent check. The method is described in detail with a pedagogical example of \begin{document}$e^+e^-\rightarrow \gamma^* \rightarrow t\bar{t}+X$\end{document} at NNLO. We show that the AMF method can systematically and efficiently calculate integrals to high precision.
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Recent low-redshift observations give value of the present-time Hubble parameter \begin{document}$H_{0}\simeq 74\;\rm{km s}^{-1} \rm{Mpc}^{-1}$\end{document}, roughly 10% higher than the predicted value \begin{document}$H_{0}=67.4\;\rm{km s}^{-1}\rm{Mpc}^{-1}$\end{document} from Planck's observations of the Cosmic Microwave Background radiation (CMB) and the \begin{document}$\Lambda$\end{document}CDM model. Phenomenologically, we show that by adding an extra component X with negative density in the Friedmann equation, it can relieve the Hubble tension without changing the Planck's constraint on the matter and dark energy densities. For the extra negative density to be sufficiently small, its equation-of-state parameter must satisfy \begin{document}$1/3\leq w_{X}\leq1$\end{document}. We propose a quintom model of two scalar fields that realizes this condition and potentially alleviate the Hubble tension. One scalar field acts as a quintessence while another “phantom” scalar conformally couples to matter in such a way that viable cosmological scenario can be achieved. The model depends only on two parameters, \begin{document}$\lambda_{\phi}$\end{document} and \begin{document}$\delta$\end{document} which represent rolling tendency of the self-interacting potential of the quintessence and the strength of conformal phantom-matter coupling respectively. The toy quintom model with \begin{document}$H_{0}=73.4\;\rm{km s}^{-1}\rm{Mpc}^{-1}$\end{document} (Quintom I) gives good Supernovae-Ia luminosity fits, decent \begin{document}$r_{\rm BAO}$\end{document} fit, but slightly small acoustic multipole \begin{document}$\ell_{A}=285.54$\end{document}. Full parameter scan reveals that quintom model provide better model than the \begin{document}$\Lambda$\end{document}CDM model in certain region of the parameter space, \begin{document}$0.02<\delta<0.10, \Omega_{m}^{(0)}<0.31$\end{document}, while significantly relieving Hubble tension even though not completely resolving it. A benchmark quintom model, Quintom II, is presented as an example.
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Experimental data on \begin{document}$R(D^{(*)})$\end{document}, \begin{document}$R(K^{(*)})$\end{document} and \begin{document}$R(J/\psi)$\end{document}, provided by different collaborations, show sizable deviations from the standard model (SM) predictions. To describe these anomalies many new physics scenarios have been proposed. One of them is leptoquark model with introducing the vector and scalar leptoquarks coupling simultaneously to the quarks and leptons. To look for similar possible anomalies in baryonic sector, we investigate the effects of a vector leptoquark \begin{document}$U_3 (3,3, \frac{2}{3})$\end{document} on various physical quantities related to the tree-level \begin{document}$\Lambda_b \rightarrow \Lambda_c \ell ~ \overline{\nu}_\ell$\end{document} decays (\begin{document}$\ell=\mu, ~\tau$\end{document}), which proceed via \begin{document}$b \rightarrow c~\ell ~ \overline{\nu}_\ell$\end{document} transitions at quark level. We calculate the differential branching ratio, forward-backward asymmetry and longitudinal polarizations of lepton and \begin{document}$\Lambda_{c}$\end{document} baryon at \begin{document}$\mu$\end{document} and \begin{document}$\tau$\end{document} lepton channels in leptoquark model and compare their behavior with respect to \begin{document}$q^2$\end{document} with the predictions of the SM. In the calculations we use the form factors calculated in full QCD as the main inputs and take into account all the errors coming from the form factors and model parameters. It is observed that, at \begin{document}$\tau$\end{document} channel, the \begin{document}$R_A$\end{document} fit solution to data related to the leptoquark model sweeps some regions out of the SM band but it has a considerable intersection with the SM predictions. The \begin{document}$R_B$\end{document} type solution gives roughly the same results with the those of the SM on \begin{document}$DBR(q^2)-q^2$\end{document}. At \begin{document}$\mu$\end{document} channel, the leptoquark model gives consistent results with the SM predictions and existing experimental data on the behavior of \begin{document}$DBR(q^2)$\end{document} with respect to \begin{document}$q^2$\end{document}. As far as the \begin{document}$q^2$\end{document} behavior of the \begin{document}$A_{FB}(q^2)$\end{document} is concerned, the two types of fits in leptoquark model for \begin{document}$\tau$\end{document} and the predictions of this model at \begin{document}$\mu$\end{document} channel give exactly the same results as the SM. We also investigate the behavior of the parameter \begin{document}$R(q^2)$\end{document} with respect to \begin{document}$q^2$\end{document} and the value of \begin{document}$R(\Lambda_c)$\end{document} both in vector leptoquark and SM models. Both types fit solutions lead to results that deviate considerably from the SM predictions on \begin{document}$R(q^2)- q^2$\end{document} as well as \begin{document}$R(\Lambda_c)$\end{document}. Future experimental data on \begin{document}$R(q^2)- q^2$\end{document} as well as \begin{document}$R(\Lambda_c)$\end{document}, which would be available after measurements on \begin{document}$\Lambda_b \rightarrow \Lambda_c \tau ~ \overline{\nu}_\tau$\end{document} channel, will be very helpful. Any experimental deviations from the SM predictions in this channel will strengthen the importance of the tree-level hadronic weak transitions as good probes of the new physics effects beyond the SM (BSM).
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We present a dark matter model to explain the excess events in the electron recoil data recently reported by the Xenon1T experiment. In our model, dark matter \begin{document}$\chi$\end{document} annihilates into a pair of on-shell particles \begin{document}$\phi$\end{document} which subsequently decay into \begin{document}$\psi \psi$\end{document} final state; \begin{document}$\psi$\end{document} interacts with electron to generate the observed excess events. Due to the mass hierarchy, the velocity of \begin{document}$\psi$\end{document} can be rather large and can have an extended distribution, which provides a good fit to the electron recoil energy spectrum. We estimated the flux of \begin{document}$\psi$\end{document} from dark matter annihilations in the galaxy and further determined the interaction cross section which is sizable but small enough to allow \begin{document}$\psi$\end{document} to penetrate the rocks to reach the underground labs.
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It was claimed by the author that black holes can be considered as topological insulators. They both have boundary modes and those boundary modes can be described by an effective BF theory. In this paper, we analyze the boundary modes on the horizon of black holes with the methods developed for topological insulators. Firstly the BTZ black hole is analysed, and the results are compatible with the previous works. Then we generalize those results to Kerr black holes. Some new results are obtained: dimensionless right- and left-temperature can be defined and have well behaviors both in Schwarzschild limit \begin{document}$a\rightarrow 0$\end{document} and in extremal limit \begin{document}$a\rightarrow M$\end{document}. Upon the Kerr/CFT correspondence, we can associate a central charge \begin{document}$c=12 M r_+$\end{document} with an arbitrary Kerr black hole. We can identify the microstates of the Kerr black hole with the quantum states of this scalar field. From this identification we can count the number of microstates of the Kerr black hole and give the Bekenstein-Hawking area law for the entropy.
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The problem of the deuteron interaction with lithium nuclei which treated as the systems of two coupled pointlike clusters is formulated to calculate d+Li reaction's cross sections. The d+Li reaction mechanism is described using the Faddeev theory for the three-body problem of deuteron-nucleus interaction. This theory is slightly extended for calculation of as stripping processes 6Li(d,p)7Li, 7Li(d,p)8Li, 6Li(d,n)7Be and 7Li(d,n)8Be well as fragmentation reactions yielding tritium, \begin{document}$\alpha$\end{document} -particles, and continuous neutrons and protons in the initial deuteron kinetic-energy region \begin{document}$E_d=0.5-20$\end{document} MeV. The phase shifts found for \begin{document}$d+^6$\end{document} Li and \begin{document}$d+^7$\end{document} Li elastic scattering, as part of the simple optic model with a complex central potential, were used to find the cross sections for the 6Li \begin{document}$(d,\gamma_{M1})^8{\rm{Be}}$\end{document} and 7Li \begin{document}$(d,\gamma_{E1})^9{\rm{Be}}$\end{document} radiation captures. The three-body dynamics role is also summarized to demonstrate its significant influence within \begin{document}$d+^7$\end{document} Li system.
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Qualities of nucleons, such as the fundamental parameter mass, in extreme conditions might be modified relative to the isolate ones. We show the ratio of the EMC-effect tagged nucleon mass to that of the free one (\begin{document}$m^{\ast}/m$\end{document}), which are derived from nuclear structure function ratio between heavy nuclei and deuterium measured in electron Deep Inelastic Scattering (DIS) reaction in 0.3\begin{document}$\leqslant x\leqslant$\end{document}0.7. The increase of \begin{document}$m^{\ast}/m$\end{document} with \begin{document}$A^{-1/3}$\end{document} is phenomenological interpreted via the release of color-singlet cluster formed by sea quarks and gluons in bound nucleons holding high momentum in the nucleus, from which the mass and fraction of non-nucleonic components in nuclei are deduced. The mass of color-singlet cluster released from per short range correlated (SRC) proton in high momentum region (\begin{document}$k>$\end{document} 2 fm\begin{document}$^{-1}$\end{document}) is extracted to be 16.890\begin{document}$\pm$\end{document}0.016 MeV/c\begin{document}$^{2}$\end{document}, which is an evidence of the possible indication of a light neutral boson and quantized mass of matter.
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We perform the potential analysis for the holographic Schwinger effect in a deformed \begin{document}$AdS_5$\end{document} model with conformal invariance broken by a background dilaton. We evaluate the static potential by analyzing the classical action of a string attaching the rectangular Wilson loop on a probe D3 brane sitting at an intermediate position in the bulk AdS space. We observe that the inclusion of chemical potential tends to enhance the production rate, reverse to the effect of confining scale. Also, we calculate the critical electric field by Dirac-Born-Infeld (DBI) action.
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Recently, an action principle for the \begin{document}$D\rightarrow4$\end{document} limit of the Einstein-Gauss-Bonnet gravity has been proposed. It is a special scalar-tensor theory that belongs to the family of Horndeski gravity. It also has a well defined \begin{document}$D\rightarrow3$\end{document} and \begin{document}$D\rightarrow2$\end{document} limit. In this work, we examine this theory in three and four dimensions in Bondi-Sachs framework. In both three and four dimensions, we find that there is no news function associated to the scalar field, which means that there is no scalar propagating degree of freedom in the theory. In four dimensions, the mass-loss formula is not affected by the Gauss-Bonnet term. This is consistent with the fact that there is no scalar radiation. However, the effects of the Gauss-Bonnet term are quite significant in the sense that they arise just one order after the integration constants and also arise in the quadrupole of the gravitational source.
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The strangeonium-like \begin{document}$s\bar{s}g$\end{document} hybrids are investigated from lattice QCD in the quenched approximation. In the Coulomb gauge, spatially extended operators are constructed for \begin{document}$1^{--}$\end{document} and \begin{document}$(0,1,2)^{-+}$\end{document} states with the color octet \begin{document}$s\bar{s}$\end{document} component being separated from the chromomagnetic field strength by spatial distances \begin{document}$r$\end{document}, whose matrix elements between the vacuum and the corresponding states are interpreted as Bethe-Salpeter (BS) wave functions. In each of the \begin{document}$(1,2)^{-+}$\end{document} channels, the masses and the BS wave functions are reliably derived. The \begin{document}$1^{-+}$\end{document} ground state mass is around 2.1-2.2 GeV, and that of \begin{document}$2^{-+}$\end{document} is around 2.3-2.4 GeV, while the masses of the first excited states are roughly 1.4 GeV higher. This mass splitting is much larger than the expectation of the phenomenological flux-tube model or constituent gluon model for hybrids, which is usually a few hundred MeV. The BS wave functions with respect to \begin{document}$r$\end{document} show clear radial nodal structures of non-relativistic two-body system, which imply that \begin{document}$r$\end{document} is a meaningful dynamical variable for these hybrids and motivate a color halo picture of hybrids that the color octet \begin{document}$s\bar{s}$\end{document} is surrounded by gluonic degrees of freedom. In the \begin{document}$1^{--}$\end{document} channel, the properties of the lowest two states comply with those of \begin{document}$\phi(1020)$\end{document} and \begin{document}$\phi(1680)$\end{document}. We have not obtained convincing information relevant to \begin{document}$\phi(2170)$\end{document} yet, however, we argue that whether \begin{document}$\phi(2170)$\end{document} is a conventional \begin{document}$s\bar{s}$\end{document} meson or a \begin{document}$s\bar{s}g$\end{document} hybrid within the color halo scenario, the ratio of partial decay widths \begin{document}$\Gamma(\phi \eta)$\end{document} and \begin{document}$\Gamma (\phi \eta')$\end{document} observed by BESIII can be understood by the mechanism of hadronic transition of a strangeonium-like meson along with the \begin{document}$\eta-\eta'$\end{document} mixing.
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Abstract: Providing a possible connection between neutrino emission and gravitational-wave (GW) bursts is im- portant to our understanding of the physical processes that occur when black holes or neutron stars merge. In the Daya Bay experiment, using the data collected from December 2011 to August 2017, a search has been performed for electron-antineutrino signals coinciding with detected GW events, including GW150914, GW151012, GW151226, GW170104, GW170608, GW170814, and GW170817. We used three time windows of ±10 s, ±500 s, and ±1000 s relative to the occurrence of the GW events, and a neutrino energy range of 1.8 to 100 MeV to search for correlated neutrino candidates. The detected electron-antineutrino candidates are consistent with the expected background rates for all the three time windows. Assuming monochromatic spectra, we found upper limits (90% confidence level) on electron-antineutrino fluence of (1.13 − 2.44)×1011 cm−2 at 5 MeV to 8.0×107 cm−2 at 100 MeV for the three time windows. Under the assumption of a Fermi-Dirac spectrum, the upper limits were found to be (5.4 − 7.0)×109 cm−2 for the three time windows.
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The decay \begin{document}$t \to c V$\end{document} (\begin{document}$V=\gamma,~Z,~g$\end{document}) process in the mirror twin Higgs models with the colorless top partners are studied in this paper. We found that the branching ratios of these decays can in some parameter spaces alter the standard model expectations greatly and may be detectable according to the currently precision electroweak measurements. Thus, the constraints on the model parameters may be obtained from the branching fraction of the decay processes, which may serve as a robust detection to this new physics model.
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Inspired by the hypothesis of the black hole molecule, with the help of the Hawking temperature, entropy and the thermodynamics curvature of the black hole, we propose a new measure of the relation between the interaction and the thermal motion of molecules of the AdS black hole as a preliminary and coarse-grained description. The measure enables us to introduce a dimensionless ratio to characterize this relation and show that there is indeed competition between the interaction among black hole molecules and their thermal motion. For the charged AdS black hole, below the critical dimensionless pressure, there are three transitions between the interaction state and the thermal motion state. While above the critical dimensionless pressure, there is only one transition. For the Schwarzschild-AdS black hole and five-dimensional Gauss-Bonnet AdS black hole, there is always a transition between the interaction state and the thermal motion state.
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The Jiangmen Underground Neutrino Observatory (JUNO) features a 20 kt multi-purpose underground liquid scintillator sphere as its main detector. Some of JUNO's features make it an excellent experiment for \begin{document}$^8$\end{document}B solar neutrino measurements, such as its low-energy threshold, its high energy resolution compared to water Cherenkov detectors, and its much larger target mass compared to previous liquid scintillator detectors. In this paper we present a comprehensive assessment of JUNO's potential for detecting \begin{document}$^8$\end{document}B solar neutrinos via the neutrino-electron elastic scattering process. A reduced 2 MeV threshold on the recoil electron energy is found to be achievable assuming the intrinsic radioactive background \begin{document}$^{238}$\end{document}U and \begin{document}$^{232}$\end{document}Th in the liquid scintillator can be controlled to 10\begin{document}$^{-17}$\end{document} g/g. With ten years of data taking, about 60,000 signal and 30,000 background events are expected. This large sample will enable an examination of the distortion of the recoil electron spectrum that is dominated by the neutrino flavor transformation in the dense solar matter, which will shed new light on the inconsistency between the measured electron spectra and the predictions of the standard three-flavor neutrino oscillation framework. If \begin{document}$\Delta m^{2}_{21} = 4.8\times10^{-5}\; (7.5\times10^{-5})$\end{document} eV\begin{document}$^{2}$\end{document}, JUNO can provide evidence of neutrino oscillation in the Earth at the about 3\begin{document}$\sigma$\end{document} (2\begin{document}$\sigma$\end{document}) level by measuring the non-zero signal rate variation with respect to the solar zenith angle. Moreover, JUNO can simultaneously measure \begin{document}$\Delta m^2_{21}$\end{document} using \begin{document}$^8$\end{document}B solar neutrinos to a precision of 20% or better depending on the central value and to sub-percent precision using reactor antineutrinos. A comparison of these two measurements from the same detector will help understand the current mild inconsistency between the value of \begin{document}$\Delta m^2_{21}$\end{document} reported by solar neutrino experiments and the KamLAND experiment.
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The minimal \begin{document}${\rm{U}}(1)_{\rm{{B-L}}}$\end{document} extension of the Standard Model (B-L-SM) offers an explanation for neutrino mass generation via a seesaw mechanism as well as contains two new physics states such as an extra Higgs boson and a new \begin{document}$Z'$\end{document} gauge boson. The emergence of a second Higgs particle as well as a new \begin{document}$Z^\prime$\end{document} gauge boson, both linked to the breaking of a local \begin{document}${\rm{U}}(1)_{\rm{{B-L}}}$\end{document} symmetry, makes the B-L-SM rather constrained by direct searches at the Large Hadron Collider (LHC) experiments. We investigate the phenomenological status of the B-L-SM by confronting the new physics predictions with the LHC and electroweak precision data. Taking into account the current bounds from direct LHC searches, we demonstrate that the prediction for the muon \begin{document}$\left(g-2\right)_\mu$\end{document} anomaly in the B-L-SM yields at most a contribution of approximately \begin{document}$8.9 \times 10^{-12}$\end{document} which represents a tension of \begin{document}$3.28$\end{document} standard deviations, with the current \begin{document}$1\sigma$\end{document} uncertainty, by means of a \begin{document}$Z^\prime$\end{document} boson if its mass lies in a range of \begin{document}$6.3$\end{document} to \begin{document}$6.5\; {\rm{TeV}}$\end{document} , within the reach of future LHC runs. This means that the B-L-SM, with heavy yet allowed \begin{document}$Z^\prime$\end{document} boson mass range, in practice does not resolve the tension between the observed anomaly in the muon \begin{document}$\left(g-2\right)_\mu$\end{document} and the theoretical prediction in the Standard Model. Such a heavy \begin{document}$Z^\prime$\end{document} boson also implies that the minimal value for a new Higgs mass is of the order of 400 GeV.
Published:
Abstract:
In this paper we study the symmetry energy and the Wigner energy in the binding energy formula for atomic nuclei. We extract simultaneously the \begin{document}$I^2$\end{document} symmetry energy and Wigner energy coefficients by using the double difference of "experimental" symmetry-Wigner energies, based on the binding energy data of nuclei with \begin{document}$A \geq 16$\end{document}. Our study of the triple difference formula and the "experimental" symmetry-Wigner energy suggests that the macroscopic isospin dependence of binding energies is well explained by the \begin{document}$I^{2}$\end{document} symmetry energy and the Wigner energy, and further considering the \begin{document}$I^{4}$\end{document} term in the binding energy formula does not substantially improve the calculation result.
Published:
Abstract:
Recently a novel four-dimensional Einstein-Gauss-Bonnet (4EGB) theory of gravity was proposed by Glavan and Lin [D. Glavan and C. Lin, Phys. Rev. Lett. 124, 081301 (2020)] which includes a regularized Gauss-Bonnet term by using the re-scalaring of the Gauss-Bonnet coupling constant \begin{document}$\alpha \to \alpha/(D-4)$\end{document} in the limit \begin{document}$D\to 4$\end{document}. This theory also has been reformulated to a specific class of the Horndeski theory with an additional scalar degree of freedom and to a spatial covariant version with a Lagrangian multiplier which can eliminate the scalar mode. Here we study the physical properties of the electromagnetic radiation emitted from a thin accretion disk around the static spherically symmetric black hole in the 4EGB gravity. For this purpose, we assume the disk is in a steady-state and in hydrodynamic and thermodynamic equilibrium so that the emitted electromagnetic radiation is a black body spectrum. We study in detail the effects of the Gauss-Bonnet coupling constant \begin{document}$\alpha$\end{document} in 4EGB gravity on the energy flux, temperature distribution, and electromagnetic spectrum of the disk. It is shown that with the increases of the parameter \begin{document}$\alpha$\end{document}, the energy flux, temperature distribution, and electromagnetic spectrum of the accretion disk all increases. Besides, we also show that the accretion efficiency increases as the growth of the parameter \begin{document}$\alpha$\end{document}. Our results indicate that the thin accretion disk around the static spherically symmetric black hole in the 4EGB gravity is hotter, more luminosity, and more efficient than that around a Schwarzschild black hole with the same mass for a positive \begin{document}$\alpha$\end{document}, while it is cooler, less luminosity, and less efficient for a negative \begin{document}$\alpha$\end{document}.
Published:
Abstract:
The heavy quark effective theory vastly reduces the weak-decay form factors of hadrons containing one heavy quark. Many works attempt to apply this theory to the multiple heavy quarks hadrons directly. In this paper, we examine this confusing application by the instantaneous Bethe-Salpeter method from phenomenological respect, and give the numerical results for the \begin{document}$B_c$\end{document} decays to charmonium where the final states including \begin{document}$1S$\end{document}, \begin{document}$1P$\end{document}, \begin{document}$2S$\end{document} and \begin{document}$2P$\end{document}. Our results indicate that the form factors parameterized by a single Isgur-Wise function deviate seriously from the full ones, especially involving the excited states. The relativistic corrections (\begin{document}$1/m_Q$\end{document} corrections) require the introduction of more non-perturbative universal functions, similarly to the Isgur-Wise function, which are the overlapping integrals of the wave functions with the relative momentum between the quark and antiquark.
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Abstract:
In this article, we study the ground states and the first radial excited states of the flavor antitriplet heavy baryon states \begin{document}$\Lambda_Q$\end{document} and \begin{document}$\Xi_Q$\end{document} with the spin-parity \begin{document}$J^P={1\over 2}^{+}$\end{document} by carrying out the operator product expansion up to the vacuum condensates of dimension \begin{document}$10$\end{document} in a consistent way. We observe that the higher dimensional vacuum condensates play an important role, and obtain very stable QCD sum rules with variations of the Borel parameters for the heavy baryon states for the first time. The predicted masses \begin{document}$6.08\pm0.09\,{\rm{GeV}}$\end{document}, \begin{document}$2.78\pm0.08\,{\rm{GeV}}$\end{document} and \begin{document}$2.96\pm0.09\,{\rm{GeV}}$\end{document} for the first radial excited states \begin{document}$\Lambda_b(2{\rm{S}})$\end{document}, \begin{document}$\Lambda_c(2{\rm{S}})$\end{document} and \begin{document}$\Xi_c(2{\rm{S}})$\end{document} respectively are in excellent agreement with the experimental data and support assigning the \begin{document}$\Lambda_b(6072)$\end{document}, \begin{document}$\Lambda_c(2765)$\end{document} and \begin{document}$\Xi_c(2980/2970)$\end{document} to be the first radial excited states of the \begin{document}$\Lambda_b$\end{document}, \begin{document}$\Lambda_c$\end{document} and \begin{document}$\Xi_c$\end{document}, respectively, the predicted mass \begin{document}$6.24\pm0.07\,{\rm{GeV}}$\end{document} for the \begin{document}$\Xi_b(2{\rm{S}})$\end{document} can be confronted to the experimental data in the future.
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Abstract:
The problem of the flat limits of the scalar and spinor fields on the de Sitter expanding universe is considered in the traditional adiabatic vacuum and in the new rest frame vacuum we proposed recently where the frequencies are separated in the rest frames as in special relativity. It is shown that only in the rest frame vacuum the Minkowskian flat limit can be reached naturally for any momenta while in the adiabatic vacuum this limit remains undefined in the rest frames where the momentum vanishes. An important role is played by the phases of the fundamental solutions in the rest frame vacuum which must be regularized in order to obtain the desired Minkowskian flat limits. This procedure fixes the phases of the scalar mode functions and Dirac spinors leading to their definitive expressions derived here. The physical consequence is that in the rest frame vacuum the flat limits of the one-particle operators are just the corresponding operators of special relativity.
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Abstract:
We present a dispersive representation of the \begin{document}$\gamma N\rightarrow \pi N$\end{document} partial-wave amplitude based on unitarity and analyticity. In this representation, the right-hand-cut contribution responsible for \begin{document}$\pi N$\end{document} final-state-interaction effects are taken into account via an Omnés formalism with elastic \begin{document}$\pi N$\end{document} phase shifts as inputs, while the left-hand-cut contribution is estimated by invoking chiral perturbation theory. Numerical fits are performed in order to pin down the involved subtraction constants. It is found that good fit quality can be achieved with only one free parameter and the experimental data of the multipole amplitude \begin{document}$E_{0}^+$\end{document} in the energy region below the \begin{document}$\Delta(1232)$\end{document} are well described. Furthermore, we extend the \begin{document}$\gamma N\rightarrow \pi N$\end{document} partial-wave amplitude to the second Riemann sheet so as to extract the couplings of the \begin{document}$N^\ast(890)$\end{document}. The modulus of the residue of the multipole amplitude \begin{document}$E_{0}^+$\end{document} (\begin{document}${\rm S_{11}pE}$\end{document}) is \begin{document}$2.41\rm{mfm\cdot GeV^2}$\end{document} and the partial width of \begin{document}$N^*(890)\to\gamma N$\end{document} at the pole is about \begin{document}$0.369\ {\rm MeV}$\end{document}, which is almost the same as the one of the \begin{document}$N^*(1535)$\end{document} resonance, indicating that \begin{document}$N^\ast(890)$\end{document} strongly couples to \begin{document}$\pi N$\end{document} system.
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Abstract:
We investigate the evolution of abundance of the asymmetric thermal Dark Matter when its annihilation rate at chemical decoupling is boosted by the Sommerfeld enhancement. Then we discuss the effect of kinetic decoupling on relic abundance of asymmetric Dark Matter when the interaction rate depends on the velocity. Usually the relic density of asymmetric Dark Matter is analyzed in the frame of chemical decoupling. Indeed after decoupling from the chemical equilibrium, asymmetric Dark Matter particles and anti--particles were still in kinetic equilibrium for a while. It has no effect on the case of s−wave annihilation since there is no temperature dependence in this case. However, the kinetic decoupling has impacts for the case of p−wave annihilation and Sommerfeld enhanced s− and p−wave annihilations. We investigate in which extent the kinetic decoupling affects the relic abundances of asymmetric Dark Matter particle and anti--particle in detail. We found the constraints on the cross section and asymmetry factor by using the observational data of relic density of Dark Matter.
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Abstract:
We investigate the bulk viscosity of strange quark matter in the framework of equivparticle model, where analytical formulae are obtained for certain temperature ranges and can be readily applied to those with various quark mass scalings. In the case of adopting a quark mass scaling with both linear confinement and perturbative interactions, the obtained bulk viscosity increases by \begin{document}$1 \sim 2$\end{document} orders of magnitude comparing with bag model scenarios. Such an enhancement is mainly due to the large quark equivalent masses adopted in the equivparticle model, which essentially attribute to the strong interquark interactions and are related to the dynamical chiral symmetry breaking. Due to the large bulk viscosity, the predicted damping time of oscillations for canonical 1.4 \begin{document}${\rm{M}}_\odot$\end{document} strange star is less than one millisecond, which is faster than previous findings. Consequently, the obtained \begin{document}$r$\end{document}-mode instability window for the canonical strange stars well accommodates the observational frequencies and temperatures for pulsars in the low-mass X-ray binaries (LMXBs).
Published:
Abstract:
Motivated by the problem of expanding single-trace tree-level amplitude of Einstein-YangMills theory to the BCJ basis of Yang-Mills amplitudes, we present an alternative expansion formula in the gauge invariant vector space. Starting from a generic vector space consisting of polynomials of momenta and polarization vectors, we define a new sub-space as gauge invariant vector space by imposing constraints of gauge invariant conditions. To characterize this sub-space, we compute its dimension and construct an explicit gauge invariant basis from it. We propose an expansion formula in the gauge invariant basis with expansion coefficients being linear combinations of Yang-Mills amplitude, manifesting the gauge invariance of both expansion basis and coefficients. With help of quivers, we compute the expansion coefficients via differential operators and demonstrate the general expansion algorithm by several examples.
Published: , doi: 10.1088/1674-1137/44/5/055101
Abstract:
We construct an alternative uniformly accelerated reference frame based on 3+1 formalism in adapted coordinate. It is distinguished with Rindler coordinate that there is time-dependent redshift drift between co-moving observers. The experimentally falsifiable distinguishment might promote our understanding of non-inertial frame in laboratory. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9295012354850769, "perplexity": 1105.4980873845457}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107879362.3/warc/CC-MAIN-20201022082653-20201022112653-00495.warc.gz"} |
https://www.physicsforums.com/threads/black-holes-and-degenerate-pressure.859844/ | # B Black Holes and Degenerate Pressure
Tags:
1. Feb 29, 2016
### lavinia
Naive reading on the web says that stellar collapse is halted by quantum mechanical processes called "degenerate pressures" that arise when gravity tries to force fermions such as electrons or neutrons into the same quantum state. White dwarfs are propped up by electrons, neutron stars by neutrons, and it is speculated that other quantum mechanical plasmas may stabilize larger collapsing stars e.g quark plasmas. See for instance the Wikipedia articles on White Dwarfs and Neutron Stars. Also the Hubble telescope seems to have detected evidence of an anomalous neutron star that might really be a quark star.
From this it seems that matter tries to do its best to resist gravitational crunch but at some point gives up the battle. Why is this? Is it that beyond a certain limit there are no longer any more types of elementary particles that can counter stellar collapse? Or is is just that no matter what goes on in the quantum world General Relativity says a singularity is inevitable if the collapsing star is big enough? It seems a bit peculiar that Relativity predicts a singularity without any reference whatsoever to these quantum pressures.
Finally, as the singularity forms what happens to matter on its way down? Does it go through a series of stages of attempted resistance first say forming an electron plasma , then a neutron, then perhaps a quark and then others or is the process completely different?
2. Feb 29, 2016
### phyzguy
One way to look at it is that as the material becomes more compressed, it gets denser and stiffer. This means the speed of sound increases. The speed of sound in neutron stars is estimated to be about 2/3 the speed of light. So matter cannot get much denser and stiffer than neutron star matter or the speed of sound would exceed the speed of light, which relativity says is impossible. So even if we do not know the details of matter interactions at these extremely high pressures and densities, we know that a body much larger than a neutron star cannot resist gravitational collapse. I'm not sure if we know exactly where this limit is (where no material could resist collapse, no matter how strong, or the speed of sound in the material would exceed the speed of light), but it is somewhere around 3-5 times the mass of the sun.
3. Feb 29, 2016
### lavinia
OK. That makes sense from the point of view of Relativity. But what about the Pauli Exclusion Principle? Does it just stop working?
4. Feb 29, 2016
### phyzguy
I don't think anyone knows. If we could answer questions like that, we would have a working theory of quantum gravity, which we don't have. Others may have a better answer.
5. Feb 29, 2016
### dendros
Another possibility is that beyond a certain threshold, the collapsed matter's particles becomes bosonic.
6. Mar 9, 2016
### nikkkom
Pauli exclusion principle does not make it absolutely impossible to cram more particles into a fixed volume. It only requires that every new added particle must have a different state from all other already present particles. Which usually means it needs to have higher energy. Thus, a newly added particle needs to be moving faster. (Which, in turn, makes it exert pressure, causing "degenerate pressure").
Since it is always possible to have higher energy (possible energy levels are not bounded from above), neutron star can become denser and denser as matter is added to it. (in fact, calculations show that it even _shrinks_, it does not stay the same size). At some point it becomes a black hole. As matter collapses while BH is forming, even below event horizon, Pauli exclusion principle still works: particles go to higher and higher energy levels.
The point of singularity is, of course, problematic, because there ehergy of particles will go to infinity. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8767763376235962, "perplexity": 472.99415429389694}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645824.5/warc/CC-MAIN-20180318145821-20180318165821-00595.warc.gz"} |
https://www.physicsforums.com/threads/isotropy-in-torus-topology.425222/ | # Isotropy in torus-topology
1. Aug 30, 2010
### tom.stoer
My question is if a torus admits coordinates which guarantuee isotropy.
Background:
In cosmology one usually assumes homogenity and isotropy of the universe. These principles are respected by most prominent cosmological models. Now what about the torus universe? Is there a choice of coordinates for which isotropy is guarantueed?
Using standard torus coordinates via embedding obviously violates isotropy. Using a square with opposite edges identified and cartesian coordinates on this square violates isotropy as well.
My feeling is that isotropy is always violated, i.e. that the torus topology does not allow for a geometry which respects isotropy. My reasoning goes as follows: using the square (cube, ...) one immediately sees that for a straight curve parallel to the edges of the square the curve always closes with winding number 1. A curve not parallel to the edges will close with winding number >1 (in rational cases) or it will never close (in irrational cases). But of course this is only one counter example for one specific geometry, not a general proof.
So my question is if there is a geometry on a torus which respects isotropy.
Last edited: Aug 30, 2010
2. Aug 30, 2010
### Office_Shredder
Staff Emeritus
In general, how is isotropy defined? Do you have to follow geodesics?
3. Aug 30, 2010
### tom.stoer
Good question.
The general statement is that "all directions are equivalent". I would suggest to define "directions" via geodesics (= generalized straight lines which are equivalent to shortest connection of twopoints in the absence of torsion; in the presence of torsion the equivalence breaks down).
4. Aug 30, 2010
### tom.stoer
This what Wikipedia says:
"In mathematics, an isotropic manifold is a manifold in which the geometry doesn't depend on directions. A simple example is the surface of a sphere.
...
A homogeneous space can be non-isotropic (for example, a flat torus !!!), in the sense that an invariant metric tensor on a homogeneous space may not be isotropic."
So according to Wikipedia (w/o proof) flat geometry on a torus is not compatible with isotropy.
5. Aug 30, 2010
### lavinia
the Wikipedia link is a stub and explains nothing. Can you define isotropic in rigorous mathematical terms?
Last edited: Aug 30, 2010
6. Aug 30, 2010
### tom.stoer
A space is called homogenous if its properties do not depend on the location. A space is called isotropic if (at each point) its properties do not depend on the direction (one looks).
I have no rigorous mathematical definition, but the example of the torus should make it clear: a torus with flat geometry is definitly homogeneous, i.e. its geometric properties do not depend on the location (it looks flat everywhere). In addition it looks isotropic locally. But globally its properties do depend on the direction. Its properties do not change along a straight curve, but there are curves with different winding properties depending on the direction of the curve.
Perhaps a thesaurus may help: isotropy - the property of being isotropic; having the same value when measured in different directions.
7. Aug 30, 2010
### lavinia
the Wikipepedia article is a stub. Can you define isotropic rigorously?
8. Aug 30, 2010
### tom.stoer
I saw that! As I said I have no rigorous definition. You can check the following link, perhaps it becomes clearer.
"Within mathematics, isotropy has a few different meanings:
(1) Isotropic manifolds: Some manifolds are isotropic, meaning that the geometry on the manifold is the same regardless of direction. A similar concept is homogeneity. A manifold can be homogeneous without being isotropic. But if it is inhomogeneous, it is necessarily anisotropic."
9. Aug 31, 2010
### tom.stoer
I guess I have both the definition of isotropy and the explanation why the torus fails to be isotropic.
A friend of mine found the following definition (Lee: "Riemann Manifolds")
"..given a point p \in M, M is isotropic at p if there exists a Lie group G acting smoothly on M by isometries such that the isometry subgroup G_p \subset G acts transitively on the set of unit vectors in T_p M."
I think that it's is clear what it means: given any p and two unit vectors x and y in T_M(p) one can find a rotation g which sends x to y, g(x) = y. Therefore the action is transitive locally i.e. for each p. But a rotation on the torus cannot be extended globally in a smooth manner. This becomes intuitively clear using a quadratic chart with cartesian coordinates covering nearly all of the (flat) 2-torus = leaving out only a small strip along the edges; within the square everything is fine, but in the small strip one has to "unwind" the rotation (which breaks local flatness of the torus and causes anisotropy within the strip) or one has to abandon smoothness at all.
So the crucial fact is GLOBAL transitivity and smoothness.
10. Aug 31, 2010
### lavinia
So projective space is isotropic.
the set of lattice preserving isometries of Euclidean space is what?
I guess it is just the group of invertible integer matrices of determinant +-1
Last edited: Aug 31, 2010
11. Aug 31, 2010
### lavinia
such spaces must have constant curvature - so which hyperbolic manifolds if any are isotropic? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8872554898262024, "perplexity": 919.8624323791728}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00461.warc.gz"} |
https://www.math.princeton.edu/events/positive-mass-conjecture-2017-10-26t163010 | # Positive Mass Conjecture
-
Yangyang Li, Princeton University
Fine Hall 110
In this talk, I will talk about the positive mass conjecture, which, roughly speaking, asserts that the total mass of an isolated physical object with positive local energy density must be nonnegative. I will begin with the ADM formalism in general relativity and the history of positive mass (energy) conjecture. Then, I will mainly discuss two different proofs of the conjecture by Yau and Schoen (1979, for n=3) and by Witten (1981, for spin manifolds). Time permitting, I will talk about its application and its recent progress. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.916882336139679, "perplexity": 643.9807796462876}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00312.warc.gz"} |
https://yuji.wordpress.com/2009/08/31/django-on-windows-setting-path-environment-variable/ | # Django/Python on Windows — Setting PATH Environment Variable
## What is the PATH?
The PATH environment variable is in general a list of directories that the operating system will search through when programs attempt to execute a command.
For example, say we go into the windows command prompt or Ubuntu’s bash shell and type in “foobar” – the operating system will attempt to locate an executable (program) named “foobar” in the folders listed in the PATH. Replace “foobar” with “python” or “django-admin.py” and you start to understand why this is useful.
## How do we set the PATH?
To set the PATH on Vista or Windows 7:
Push the windows button and type “environment variable” on the search bar and select “Edit environment variables for your account”.
Non Vista/Windows 7 users just right click on My Computer, go to Properties, click Advanced Properties, then the Environment Variables button to get to the same place.
In the Environment Variables screen, look for the “Path” variable under the “System Variables” section on the bottom. If it doesn’t exist, just make a new one.
Edit or enter in folder paths separated by ; — Append whatever you like to the string ending with a “;”
Do not delete the default path – append to it.
Path2;Path2;Path3;
Done!
## 19 thoughts on “Django/Python on Windows — Setting PATH Environment Variable”
1. This post seems often searched for.. is the question not answered? :3
2. Shaon Bhuiyan says:
Thank you Yuji, I was looking for a way to do this.
3. Pingback: gcc issue
4. rumman says:
I tried this but it doesn’t work.Can any one help me ?
C:\Program Files\Common Files\Microsoft Shared\Windows Live;C:\Program Files\Common Files\ArcSoft\Bin;%SystemRoot%\system32;%SystemRoot%;%SystemRoot%\System32\Wbem;%SYSTEMROOT%\System32\WindowsPowerShell\v1.0\;C:\Program Files\Dell\DW WLAN Card;C:\Program Files\Windows Live\Shared;
E:\install soft\Python\Python install\python.exe
1. rumman says:
added that and tried to test it in cmd , it shows this
C:\Users\BDR-User>python
‘python’ is not recognized as an internal or external command,
operable program or batch file.
2. Yeah, if you add the parent directory to the PATH, and restart cmd, typing python should launch python.exe. Did you restart?
1. rumman says:
Hey , yuji thanks for your help I was able to add the path, unfortunately I was trying to add the path in another directory but i installed in some other directory, once i tried in the correct directory the problem was solved. Thanks again 🙂
5. Curtis says:
Anyone interested in modifying the PATH or any other environment variable in Windows should check out “Rapid Environment Editor”.
6. 0cool says:
I did that but It is showing an error!
but when I set my command prompt to python dir .. its working!
7. Thank you very much in explaining how to append Path. I didnt know I can set up multiple paths. That helps my Python and Django installations now. Cheers!!!
8. I don’t understand what the path needs to be set to. Also I don’t have an option of editing the path because the edit button is greyed out. Thanks for the help but I’m still stuck : (
1. The path needs to be set to the directory of any executable scripts you want access to from the command line. Say you have a script foo.py – if you put the directory it’s in into the path, then you can execute foo from any command line.
If you can’t edit it – then that’s a different problem. Do you need admin rights or something?
2. CCC says:
I had the same problem, but there’s another way to get there. On your desktop, right-click My Computer. (If you don’t have that icon on your desktop, go to the Desktop via Windows Explorer.) Select Properties. Then click Advanced System Settings on the left, and then click the Environment Variables button at the bottom. Hopefully you’ve figured all that out by now, and this comment will just be useful to the next person to come along.
9. BB says:
Thank you! Works like charm! I did need to close the cmd and re-open it!
10. This is exactly what I was looking for, however I can’t access my system variables, only the user variables (I have an admin account on my windows 7). Anybody got an idea why? And how I can get access? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8021365404129028, "perplexity": 2543.215029806883}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549436330.31/warc/CC-MAIN-20170728042439-20170728062439-00528.warc.gz"} |
https://www.physicsforums.com/threads/prove-finite-dimensional-normed-vector-space-is-differentiable.383194/ | # Prove Finite Dimensional Normed Vector Space is Differentiable
• Thread starter cassiew
• Start date
• #1
6
0
## Homework Statement
Let V be a finite dimensional normed vector space and let U= L(V)*, the set of invertible elements in L(V). Show, f:U-->U defined by f(T)= T-1 is differentiable at each T in U and moreover,
Df(T)H = -T-1HT-1
where Df(T)= f'(T).
## Homework Equations
Apparently these propositions are supposed to help (which I've already proved and can use):
1.) Let V be a normed vector space and suppose T is an element of L(V). If ||T||<1, then the sequence Sn=sum(from j=0 to n) of Tj converges. Moreover, I-T is invertible, (Sn) converges to (I-T)-1, and ||(I-T)-1||<= 1/(1-||T||).
2.) Suppose V is a normed vector space and T, an element of L(V), is invertible. If S is an element of L(V) and ||T-S||<||T-1||-1, then S is invertible.
## The Attempt at a Solution
I've already proven that the function q:V-->V where q(T)=T2 is differentiable at each T and Dq(T)H= TH+HT, and I tried to prove this one using the same technique, but it's getting me nowhere.
## Answers and Replies
• #2
352
0
First try differentiating $$f$$ at $$I$$; your proposition 1 should help with this. Then see if you can find a way to transfer that computation to any $$T \in U$$.
• #3
6
0
Nevermind, I think I figured it out.
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13K | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9849517345428467, "perplexity": 3170.6403728169953}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178364764.57/warc/CC-MAIN-20210302190916-20210302220916-00376.warc.gz"} |
https://projecteuclid.org/euclid.ade/1356651826 | ### Semilinear elliptic equations and systems with measure data: existence and a priori estimates
#### Abstract
We give existence results and a priori estimates for a semilinear elliptic problem of the form \begin{equation*} \left\{ \begin{array}{l} -\Delta w=w^{Q}+\mu ,\qquad \text{in \thinspace }\Omega , \\ w=\lambda ,\qquad \qquad \qquad \quad \text{on }\partial \Omega , \end{array} \right. \end{equation*} where $Q>0,$ and $\mu$ and $\lambda$ are nonnegative Radon measures in $\Omega$ and $\partial \Omega ,$ with $\int_{\Omega }\rho \,d\mu <+\infty ,$ where $\rho$ is the distance to $\partial \Omega .$ We extend the results to the case of systems \begin{equation*} \left\{ \begin{array}{l} -\Delta u=v^{p}+\mu ,\qquad -\Delta v=u^{q}+\eta ,\qquad \text{in }\Omega , \\ u=\lambda ,\qquad v=\kappa ,\qquad \text{on }\partial \Omega , \end{array} \right. \end{equation*} with $p,q>0,$ and the same assumptions on $\eta$ and $\kappa .$
#### Article information
Source
Adv. Differential Equations, Volume 7, Number 3 (2002), 257-296.
Dates
First available in Project Euclid: 27 December 2012 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9993471503257751, "perplexity": 575.3437676979282}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578548241.22/warc/CC-MAIN-20190422075601-20190422101601-00440.warc.gz"} |
https://www.physicsforums.com/threads/good-contour-integral-reference.258215/ | # Good contour integral reference?
1. Sep 21, 2008
### quasar_4
Can anyone recommend a good introduction to contour integrals for someone not taking complex analysis? We are doing these integrals in a physics class and I'm terribly confused. I know that I have to choose contours that "go around" my poles, but I don't understand how to do this (I can't seem to visualize it at all). I am also a bit confused on finding the residues.
Any tips/resources/infos would be great!
Thanks!
2. Sep 22, 2008
### cellotim
Have you looked at Visual Complex Analysis by Tristan Needham? If you are a visual person, then you'll like it. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8771302103996277, "perplexity": 502.0690821871248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267159570.46/warc/CC-MAIN-20180923173457-20180923193857-00189.warc.gz"} |
https://sprayers101.com/pressure-spray/ | # How does Pressure Affect Spray?
This very short article is all about the info-graphic.
Pressure is integral to how nozzles perform: Lower pressures reduce nozzle rate, increase median droplet size, and typically increase spray fan angle. Higher pressures increase nozzle rate, reduce median droplet size and typically reduce spray fan angle. In extreme cases, too low a pressure can collase the fan angle enough to reduce overlap and compromise coverage (Watch this, and this and if you’re still interested, this).
Pressure affects all aspects of spray quality. Using a flat fan nozzle as an example, a lower pressure increases the median droplet diameter, reduces the droplet count, reduces the nozzle rate and typically reduces the spray angle. Alternately, a higher pressure decreases the median droplet diameter, increases the droplet count, increases the nozzle rate and typically increases the spray angle. Always plan to operate a nozzle in the middle of its recommended range so it can handle small changes in pressure during spraying (such as from a rate controller, or changing PTO speeds on hilly terrain). Don’t operate an air induction nozzle below 30 psi, even if it’s rated for lower in the manufacturer’s table.
Pressure can be used to make minor changes to rate while spraying. This is how rate-controllers work to compensate for changes in ground speed and maintain a constant overall rate per hectare or per acre. However, pressure should not be used to make significant changes to nozzle rates. It takes a 4x change in pressure for a 2x change in rate, so it’s inefficient. Increased pressure increases nozzle wear and pressure changes may negatively impact the spray quality if the nozzle is pushed outside its recommended range.
It is far better to simply switch nozzles when a significant change in rate is required.
Sending | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8286347985267639, "perplexity": 2786.274706090248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496671260.30/warc/CC-MAIN-20191122115908-20191122143908-00279.warc.gz"} |
http://mathhelpforum.com/trigonometry/156770-solve-equation-2-093-x-sin-x.html | # Math Help - solve the equation 2.093 = x- sin(x)
1. ## solve the equation 2.093 = x- sin(x)
Hello every body
can you please tell me how to solve the following equation
$2.093 = x-sin(x)$
$2.093 = x-sin(x)$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9361065030097961, "perplexity": 1450.391484595998}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736678818.13/warc/CC-MAIN-20151001215758-00012-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://www.chegg.com/homework-help/questions-and-answers/stepladder-negligible-weight-constructed-shown-apainter-mass-70-kg-stands-ladder-3-meters--q414053 | A stepladder of negligible weight is constructed as shown. Apainter of mass 70 kg stands on the ladder 3 meters from thebottom. Assuming that the floor is frictionless, find (a) thetension in the horizontal bar connecting the two halves of theladder, (b) the normal forces at A and B, and (c) the components ofthe reaction force at the single hinge C that the left half of theladder exerts on the right half. (Suggestion: Treat the ladder as asingle object, but also treat each half of the ladderseparately) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9731968641281128, "perplexity": 1007.3787143053556}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783397428.37/warc/CC-MAIN-20160624154957-00071-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/217672/harmonic-function | # harmonic function
Let $f$ be a real function with $\Delta f=0$ on an open ball $B_{2n}(y)\subset\mathbb{R}^N$.
How would I show
$$\int\limits_{B_n(y)}|Df|^2(z)dz\leq Cn\int\limits_{\partial B_n(y)}|Df|^2(z)d\sigma(z)$$
for some constant C, where $\sigma$ is surface measure?
-
Suppose $z\in B_{n}(y)$. Write $z=y+r\sigma$, where $r\in[0,1]$ and $\sigma\in\partial B_{n}(0)$. Hence $dz=nr^{N-1}$.
Now, we have for $D_{i}f=\frac{\partial f}{\partial x_{i}}$
In the inequality, i used the fact that $f$ is harmonic which implies that $D_if$ is harmonic and hence must attains its maximum on $\partial B_{n}(y)$. We conclude that $$\sum_{i=1}^N \int_{B_{n}(y)}(D_if(z))^{2}dz\leq\sum_{i=1}^n\frac{n}{N}\int_{\partial B_{n}(y)}(D_if(z))^2d\sigma(z)$$
Therefore, $$\int_{B_{n}(y)}|\nabla f(z)|^{2}dz\leq\frac{n}{N}\int_{\partial B_{n}(y)}|\nabla f(z)|^2d\sigma(z)$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9908590316772461, "perplexity": 95.28313185885125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500831565.57/warc/CC-MAIN-20140820021351-00246-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/173702/projective-and-injective-modules-direct-sums-and-products | # Projective and injective modules; direct sums and products
I need two counterexamples.
First, a direct sum of $R$-modules is projective iff each one is projective. But I need an example to show that, “an arbitrary direct product of projective modules need not be a projective module.”
If I let $R= \mathbb Z$ then $\mathbb Z$ is a projective $R$-module, but the direct product $\mathbb Z \times \mathbb Z \times \cdots$ is not free, hence it is not a projective module. We have a theorem which says that every free module over a ring $R$ is projective. Am I correct?
Second, a direct product of $R$-modules is injective iff each one is injective but I need an example to show that the direct sum of injective modules need not be injective.
-
"its not free ,hence it is not projective" All free modules are projective, but not all projective modules are free. – Alex Becker Jul 21 '12 at 21:25
I rewrote most of this. Let me know if I made any mistakes. Please try to take more care in your writing and formatting, for my sake! I second Alex's observation. – Dylan Moreland Jul 21 '12 at 21:29
@AlexBecker Correct, but over a PID a module is projective iff it is free, and $\mathbb{Z}$ is a PID, so maybe this was a hidden step in OP's argument. – Derek Allums Jul 21 '12 at 21:34
So the example is right????? i need an example on injective you didn't answer my question – Miss Independent Jul 21 '12 at 21:36
The direct product of infinitely many copies of $\mathbb{Z}$ is indeed not projective, but the reason you give is incorrect. You know that we always have that free implies projective, and that the module here is not free. But from $P\to Q$ and $\neg P$ you cannot conclude $\neg Q$: if it rains, then you get wet; that does not mean that if it doesn't rain, then you don't get wet (maybe you fall into a pool?) (cont) – Arturo Magidin Jul 21 '12 at 22:02
As for the first question: yes, $P = \prod_{i=1}^{\infty} \mathbb{Z}$ is a direct product of free $\mathbb{Z}$-modules which is not free. Since $\mathbb{Z}$ is a PID, $P$ is also not projective. The proof that $P$ is not free is nontrivial, but I believe it has already been given either here or on MathOverflow.
As for the second question: the Bass-Papp Theorem asserts that a commutative ring $R$ is Noetherian iff every direct sum of injective $R$-modules is injective. Thus every non-Noetherian ring carries a counterexample. The proof of the result -- given for instance in $\S 8.9$ of these notes -- is reasonably constructive: if
$I_1 \subsetneq I_2 \subsetneq \ldots \subsetneq I_n \subsetneq \ldots$
is an infinite properly ascending chain of ideals of $R$, then for all $n$ let $E_n = E(R/I_n)$ be the injective envelope (see $\S 3.6.5$ of loc. cit.) of $R/I_n$, and let $E = \bigoplus_{n=1}^{\infty} E_n$. Then $E$ is a direct sum of injective modules and (an argument given in the notes shows) that $E$ is not itself injective.
-
Thank you very much – Miss Independent Jul 21 '12 at 23:16
Is the dual statement also true? Namely $R$ is left/right Noetherian (or maybe Artinian) iff every direct product of projective left/right $R$-modules is projective. – Leon Oct 20 '13 at 19:00
@Leon: No (for Noetherian rings). The countable product of copies of $\mathbb{Z}$ is not a projective (equivalently, free) $\mathbb{Z}$-module. See e.g. Theorem 2.4 in math.uga.edu/~pete/Math8030_Exercises.pdf. – Pete L. Clark Oct 21 '13 at 1:48
@Leon: But, yes, a commutative ring is Artinian iff every product of projective modules is projective. This is a 1960 theorem of Chase. So far as I can see this result is hard to find in standard references, but e.g. it appears as an exercise on p. 161 of T.Y. Lam's Lectures on Modules and Rings. – Pete L. Clark Oct 21 '13 at 2:57
@Leon: The theorem takes place in the non-commutative case, but the result is a bit more complicated there. I recommend you take a look at Lam's text. – Pete L. Clark Oct 21 '13 at 15:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9417844414710999, "perplexity": 184.07334843766276}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768276.101/warc/CC-MAIN-20141217075248-00016-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://mathhelpforum.com/number-theory/118630-fibonacci-property-proof-help.html | # Math Help - Fibonacci Property Proof help
1. ## Fibonacci Property Proof help
Hi,
I was hoping that someone could help me with some proofs for the Fibonacci sequence.
1) If p is prime then $F_p \equiv \ \frac{5}{p} (mod p)$
2) If p is prime then $2F_{p+1} \equiv \ 1+ \frac{5}{p} (mod p)$
Thanks!
2. Hmm?
2 is prime. $F2 = 1 \equiv 1 ~ (mod ~ 2)$
3 is prime. $F3 = 1+1 = 2 \equiv 2 ~ (mod ~ 3)$
Are you sure this is the intended question?
3. I took the equation directly from the sheet because I am having the same difficulties and currently just completely confused. I should have mentioned this in the first post but it is going off of the assumption that $F_1 = 1 , F_2 = 1 , F_3=2, F_4=3,..., F_n = F_{n-1} + F_{n-2}$ for $n \geq \ 3$ So I don't know if that changes anything.
4. Isn't it supposed to be: $F_p\equiv{\left(\frac{5}{p}\right)_L}(\bmod.p)$ where $\left(.\right)_L$ is the Legendre Symbol ?
I give you the ingredients: Binet's Formula,the Binomial Theorem, Fermat's Little Theorem and Euler's Criterion.
5. Yes I think you are right about it being a Legendre Symbol. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9132530689239502, "perplexity": 330.16224124443715}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657130067.43/warc/CC-MAIN-20140914011210-00285-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
http://tlainc.com/best-supplements-uygoeh/ac0bea-upper-triangular-matrix-properties | Matrices transpire naturally in a system of simultaneous equations. Let us discuss the definition, properties and some examples for the upper triangular matrix. The numbers in the matrix are known as the elements, or entries, of the matrix. A matrix can always be transformed into row echelon form by a series of row operations, and a matrix in row echelon form is upper-triangular. Required fields are marked *. Apart from these two, there are some special form matrices, such as; Download BYJU’S app and enjoy learning with us. Similar arguments show that L n (lower triangular n × n matrices) and D n (diagonal n × n matrices) are also subspaces of M n n . The bottom two blocks, however, might pose more of an issue. Therefore, a square matrix which has zero entries below the main diagonal, are the upper triangular matrix and a square matrix which has zero entries above the main diagonal of the matrix is considered as lower triangular one. In the upper triangular matrix we have entries below the main diagonal (row $$i$$ greater than column $$j$$) as zero. the determinant of a triangular matrix is the product of the entries on the diagonal, detA = a 11a 22a 33:::a nn. A square matrix for which all the entries below the main diagonal are 0 is said to be upper triangular. The term matrix was first introduced by an English mathematician named James Sylvester during the19th-century. Special matrices Definition A square matrix is upper-triangular if all entries below main diagonal are zero. Taking transposes leads immediately to: Corollary If the inverse L 1 of an lower triangular matrix L exists, A = 0000 0000 0000 analogous definition for a lower-triangular matrix A square matrix whose oDefinition ff-diagonal entries are all zero is called a diagonal matrix. So it's 7 times minus 6 which is equal to minus 42. Examples of Upper Triangular Matrix Question 1) What is the Application of Matrices? If the upper-block consists of zeros, we call such a matrix lower-triangular. If we multiply two upper triangular, it will result in an upper triangular matrix itself. 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In the next slide, we shall prove: Theorem If the inverse U 1 of an upper triangular matrix U exists, then it is upper triangular. If either two rows or two columns are identical, the determinant equals zero. As we have known, what are matrices earlier and how they are helpful for mathematical calculations. Elleuch, S., Mnif, M.: Essential approximate point spectra for upper triangular matrix of linear relations. & a_{2n} \\ 0 & 0 & a_{33} & …. Pro Lite, Vedantu If all the elements of a row (or column) are zeros, then the value of the determinant is zero. Upper Triangular Matrix Watch more videos at https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Er. n look alright, since the top row of an upper triangular matrix does not have to contain any zeros. When two rows are interchanged, the determinant changes sign. This is done using matrices. If we change the number of rows and columns within a matrix, we can construct such buildings. Its transpose is upper triangular. They are named as Unitriangular matrix, Strictly Triangular Matrix, and Atomic Triangular Matrix. Since the transpose does not change the diagonal elements, then and . Given below are some detailed applications of matrices: Encryption: In encryption, we use matrices to scramble the data for security purposes, basically to encode or to decode the data. An upper triangular matrix with elements f[i,j] above the diagonal could be formed in versions of the Wolfram Language prior to 6 using UpperDiagonalMatrix[f, n], which could be run after first loading LinearAlgebraMatrixManipulation.. A strictly upper triangular matrix is an upper triangular matrix having 0s along the diagonal as well, i.e., for . In applications such as Adobe Photoshop uses matrices to process linear transformations to represent images. Engineers also use matrices for Fourier analysis, Gauss Theorem, to find forces in the bridge, etc. The different types of matrices are row and column matrix, zero or null matrix, singleton matrix, vertical and horizontal matrix, square matrix, diagonal matrix, scalar matrix, identity matrix, equal matrix, triangular matrix, singular, and non-singular matrix, symmetric matrix, skew-symmetric matrix, hermitian matrix, skew-hermitian matrix, orthogonal matrix, idempotent matrix, involuntary matrix, and nilpotent matrix. The UpperTriangularSolver object solves UX = B for X when U is a square, upper-triangular matrix with the same number of rows as B. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Matrices are also helpful in taking seismic surveys. Let us have a look. In hospitals, matrices are used for medical imaging, CAT scans, and MRI’s. If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal. This is a square matrix, which has 3 rows and 3 columns. Your email address will not be published. We know that a matrix is made of rows and columns. Usually the buildings that we see are straight but sometimes architects construct buildings with a little changed the outer structure, for example, the famous Burj Khalifa, etc. They use the 3d matrix to a 2d matrix to switch it into the different objects as per requirement. In economics and business studies, a matrix is used to study the trends of a business, shares, to create business models, etc. Note that these are all the eigenvalues of A since A is a 3×3matrix. 1.6.2 Triangular Matrices Recall that a square matrix for which all the entries above the main diagonal are 0 is said to be lower triangular. ITo determine if every number in a set is nonzero, we can multiply them. A triangular matrix is a square matrix where all its entries above the principal diagonal or below the principal diagonal are zero. Properties of Upper Triangular Matrix If we add two upper triangular matrices, it will result in an upper triangular matrix itself. $$\begin{bmatrix} 1 & -1 \\ 0 & 2 \\ \end{bmatrix}$$, $$\begin{bmatrix} 1 & 2 & 4 \\ 0 & 3 & 5 \\ 0 & 0 & 6 \\ \end{bmatrix}$$, $$\begin{bmatrix} 31 & -5 & 14 \\ 0 & 20 & -15 \\ 0 & 0 & 45 \\ \end{bmatrix}$$. The encoding and decoding of the data can be done with the help of a key that is generated by matrices. And it's that easy. Sorry!, This page is not available for now to bookmark. Note that upper triangular matrices and lower triangular matrices must be square matrices. Animation: Matrices can make animations more precise and perfect. If we add two upper triangular matrices, it will result in an upper triangular matrix itself. If the lower-block consists of zeros, we call such a matrix upper-triangular. Pro Lite, Vedantu If we multiply any scalar quantity to an upper triangular matrix, then the matrix still remains as upper triangular. Get Interactive and fun related educational videos and have happy learning. Mathematically, we say that A = [a Also, the matrix which has elements above the main diagonal as zero is called a lower triangular matrix. They are named after Karl Hessenberg. In physics, we use matrices in the study of electrical circuits, optics, and quantum mechanics. Let s take an (n-1) x (n-1) upper triangular matrix for which the eigenvalues are the diagonal elements. A matrix that is both upper and lower triangular is called a diagonal matrix. The upper triangular matrix will remain an upper triangular matrix if inversed. Matrices are used to modify or reconstruct the object, in 3d space. In a similar vein, a matrix which is both normal(meaning A*A= AA*, where A*is the conjugate transpose) and triangular is also diagonal. •Inverse exists only if none of the diagonal element is zero. Theorem 3.2.1 If A is an n×n upper or lower triangular matrix, then det(A) = a11a22a33 ... the rows of a matrix also hold for the columns of a matrix. Definition A matrix is upper triangular if and only if whenever. If we multiply two upper triangular, it will result in an upper triangular matrix itself. Therefore, we can say that matrices play a dominant role in calculations especially when it comes to solving the problems using Kirchoff’s laws of voltage and current. See the picture below. The transpose of an upper triangular matrix is a lower triangular matrix and vice versa. Now stick a (n x 1) column vector on the right and fill in the bottom 1..n-1 elements with 0 s. We now have an n x n upper triangular matrix. Also, if we multiply two upper triangular matrices, the result will be an upper triangular matrix. are upper-triangular, while the matrices Geology: Matrices are also helpful in taking seismic surveys. The inverse of the upper triangular matrix remains upper triangular. & a_{nn} \end{bmatrix}\). Determinant after row operations. A lower triangular matrix is a square matrix in which all entries above the main diagonal are zero (only nonzero entries are found below the main diagonal - in the lower triangle). Theorem 1.7.1 (a) The transpose of a lower triangular matrix is upper triangular, and the trans- pose of an upper triangular matrix is lower triangular. Acta Math. Symmetric Matrix and Skew Symmetric Matrix, Difference Between Upper and Lower Motor Neuron, Introduction and Characteristics of Management, Vedantu A matrix can be defined as a set of numbers that are arranged in rows and columns to create a rectangular array. Let B=P−1AP. Thus, in an upper triangular matrix all the elements below the main diagonal (i.e., those whose column index is less than the row index) are zero. A triangular matrix is a matrix that is an upper triangular matrix or lower triangular matrix. The determinant is equal to 7 times minus 2 times 1 times 3. With matrices, a resistor conversion of electrical energy into another useful energy is also possible. To be exact, an upper Hessenberg matrix has zero entries below the first subdiagonal, and a lower Hessenberg matrix has zero entries above the first superdiagonal. The transpose of the upper triangular matrix is a lower triangular matrix, U. The closure property in U n for scalar multiplication also holds, since any scalar multiple of an upper triangular matrix is again upper triangular. Question 2) How Many Types of Matrices are There? & …. A matrix that has all its entries above the principal diagonal as zero is called the lower triangular matrix. The transpose of an upper triangular matrix will be a lower triangular matrix, UT = L. The matrix will remain an upper triangular matrix if it is multiplied to a scalar quantity. No need to compute determinant. Hence, U n is a subspace of M n n . & a_{3n} \\ . The transposeof an upper triangular matrix is a lower triangular matrix and vice versa. Determinants of block matrices: Block matrices are matrices of the form M = A B 0 D or M = A 0 C D with A and D square, say A is k k and D is l l and 0 - a (necessarily) l k matrix with only 0s. Two matrices say A and B will be equal to one another if both of them possess an equal number of rows and columns. Since B is an upper triangular matrix, its eigenvalues are diagonal entries 1,4,6. Chemical engineering requires perfectly calibrated computations that are obtained from matrix transformations. In particular, the properties P1–P3 regarding the effects that elementary row operations have on the determinant Set of all strictly upper triangular matrices is a subgroup of general linear group; 2×2 invertible upper triangular matrices form a subgroup of general linear group; Exhibit symmetric group as a subgroup of a general linear group; Compute the number of invertible 2×2 matrices over Z/(2) Special linear group is a subgroup of general linear group IFor upper triangular matrices, the rank is the number of nonzero entries on the diagonal. Entries on the main diagonal and above can be any number (including zero). $\begin{bmatrix}5 &5 &8 \\0 &3 &10 \\0 &0 &8 \end{bmatrix}$ $\begin{bmatrix}-1 &7 &3 \\0 &6 &1 \\0 &0 &5 \end{bmatrix}$ $\begin{bmatrix}3 &0 &3 \\0 &7 &-1 \\0 &0 &2 \end{bmatrix}$. Games Especially 3Ds: Matrices are used to modify or reconstruct the object, in 3d space. Construction: Usually the buildings that we see are straight but sometimes architects construct buildings with a little changed the outer structure, for example, the famous Burj Khalifa, etc. Under certain conditions, we can also add and multiply matrices as individual entities, to give rise to important mathematical systems known as matrix algebras. Answer 2) There are many different types of matrices. Example of an upper triangular matrix: 1 0 2 5 0 3 1 3 0 0 4 2 0 0 0 3 By the way, the determinant of a triangular matrix is calculated by simply multiplying all its diagonal elements. We know that a matrix is made of rows and columns. \\ 0 & 0 & 0 & …. The upper triangular matrix can also be called a right triangular matrix and the lower triangular matrix can also be called a left triangular matrix. Indeed, the diagonal subdivides the matrix into two blocks: one above the diagonal and the other one below it. •Can be computed from first principles: Using the definition of an Inverse. The matrix "L" is lower triangular. & . Specifically, comparing the two matrices above, for U 1 n AU n to be upper triangular, l 1RT 1 =~0, and RT AR n 1 must itself be an n 1-dimensional square upper triangular matrix. Therefore, we have completely justified the … A triangular matrix (upper or lower) is invertible if and only if no element on its principal diagonal is 0. Sci. The transpose of a lower triangular matrix is an upper triangular matrix and the transpose of an upper triangular matrix is a lower triangular matrix. Because matrix equations with triangular matrices are easier to solve, they are very important in numerical analysis . Since A and B=P−1AP have the same eigenvalues, the eigenvalues of A are 1,4,6. This is done using matrices. Ser. B Engl. It goes like this: the triangular matrix is a square matrix where all elements below the main diagonal are zero. Matrices are also used in electronics networks, airplanes, and spacecraft. It helps us in the calculation of battery power outputs. The determinants of upper and lower non-singular matrices are the products of their diagonal elements. 33(4), 1187–1201 (2013) zbMATH MathSciNet CrossRef Google Scholar Other Uses: Matrices are also used in electronics networks, airplanes, and spacecraft. The inverse of upper/lower triangular matrices, such as Adobe Photoshop uses matrices to process linear transformations to images!, it will result in an upper triangular matrix the difference between triangular!, Arthur Cayley, a resistor conversion of electrical circuits, optics, and quantum mechanics, triangular! Adding two upper triangular matrices, there are 3 more special types of matrices follows that all the elements then. Will result in an upper triangular matrix remains upper triangular matrix the of! Data can be done with the help of a key that is generated by matrices & &! That some matrices, the diagonal elements the upper-block consists of zeros, we can see the between... Us in the calculation of battery power outputs the principal diagonal are 0 is said to be upper triangular remains... Helps us in the bridge, etc rows and columns called as right matrix! Result will be an upper triangular matrix itself blocks: one above the diagonal element is zero or reconstruct object! A matrix, Strictly triangular matrix 6 which is both symmetric and triangular diagonal! Which the eigenvalues of a since a and B=P−1AP have the same eigenvalues the... It will result in an upper triangular matrix itself determinant changes sign and have happy learning and how are! \\ 0 & 0 & a_ { nn } \end { bmatrix } \ ) or entries, the! These triangular matrices are the diagonal and above can be any number ( including zero ), upper triangular matrix properties triangular itself... The determinants of upper triangular matrix square matrices get Interactive and fun educational... Lot of concepts related to matrices be calling you shortly for Your Online Counselling session &... Below the principal diagonal as zero forces in the upper triangular matrix properties of electrical circuits, optics, quantum... Not be published a lower triangular matrix the transpose of an upper triangular matrix itself answer 1 what... Diagonal entries 1,4,6 per requirement if the lower-block consists of zeros, we such! Another if both of them possess an equal number of rows and columns a triangular is. And have happy learning a 2d matrix to switch it into the different objects per. As we have known, what are matrices earlier and how they are very important in numerical.. And spacecraft & Subtraction of two matrices, Your email address will not be published triangular if! To modify or reconstruct the object, in 3d space seismic surveys there... Are helpful for mathematical calculations a rectangular array or entries, of the diagonal elements called the triangular! See the difference between upper triangular matrix the inverse of upper/lower triangular matrix are as... Known, what are matrices earlier and how they are named as Unitriangular matrix, and.... Minus 2 times 1 times 3 1,42,62, that is, 1,16,36 transpose does not change the elements! Your email address will not be published 3d space triangular, it will result in an triangular... Two matrices, the determinant equals the product of entries down the diagonal... Matrix Addition & Subtraction of two matrices say a and B=P−1AP have the same eigenvalues, the determinant sign. Of rows and columns Many types of matrices hospitals, matrices are used for medical,... Diagonal elements, or entries, of the upper triangular matrix be defined as set... Introduced by an English mathematician named James Sylvester during the19th-century a key that is an triangular... Hospitals, matrices are also helpful in taking seismic surveys then and upper lower... Matrix where all elements below the main diagonal are zero analysis, Gauss,. Ever imagine a square matrix where all elements below the main diagonal are zero, might pose of. Us discuss the definition of an upper triangular matrix is a matrix its! Has all its entries above the principal diagonal is 0, however, might pose more of an upper matrix! Group dances diagonal are zero scramble the upper triangular matrix properties can be done with the of... Numbers in the calculation of battery power outputs of battery power outputs are diagonal entries.... And B will be a lower triangular matrix for which all the eigenvalues of are. Us discuss the definition, properties and some examples for the upper triangular will... Naturally in a system of simultaneous equations get Interactive and fun related videos! Invertible if and only if whenever the object, in 3d space triangular form, the changes. The bottom two blocks, however, might pose more of an issue,... Product of entries down the main diagonal as zero is called a triangular... Are the diagonal and above can be any number ( including zero ) to complicated... Used for medical imaging, CAT scans, and quantum mechanics, triangular. Representation, we can multiply them are obtained from matrix transformations be square matrices or below the main diagonal how... The inverse of the upper triangular matrices •Inverse of an upper triangular matrix for which the eigenvalues of a that. Bridge, etc educational videos and have happy learning is a square where! Object, in 3d space are interchanged, the determinant changes sign happy., what are matrices earlier and how they are helpful for mathematical calculations Strictly triangular matrix, we can such..., Your email address will not be published a and B=P−1AP have the eigenvalues., this page is not available for now to bookmark 3 rows and 3 columns elements below main... A subspace of M n n an English mathematician named James Sylvester the19th-century! None of the diagonal and the other one below it has all its entries above the main diagonal are.... Counsellor will be calling you shortly for Your Online Counselling session numbers are. Either two rows are interchanged, the determinant equals zero scramble the data we known! The transpose of an issue from first principles: Using the definition of upper., Arthur Cayley, a resistor conversion of electrical circuits, optics, Atomic. Some matrices, it will result in an upper triangular matrix whereas the lower triangular matrix has... Will result in an upper triangular matrices, a resistor conversion of electrical circuits,,. Represent images we change the number of rows and columns to create a rectangular array by.... James Sylvester during the19th-century are used to modify or reconstruct the object, in 3d space between triangular... Transposeof an upper triangular matrix, we call such a matrix can be done with the help a... Diagonal or below the principal diagonal or below the principal diagonal are 0 is to... To one another if both of them possess an equal number of rows 3... Are very important in numerical analysis Definition a square matrix, and MRI ’ s linear to! In upper triangular matrix is in upper triangular, it will result in an upper triangular matrix diagonal! And vice versa like this: the triangular matrix the encoding and of! If the upper-block consists of zeros, we can multiply them determinant changes.. Times minus 2 times 1 times 3 blocks: one above the principal diagonal as zero called... Also use matrices in the study of electrical circuits, optics, and quantum.... Encryption, we call such a matrix is upper-triangular if all entries below the main diagonal the... Switch it into the different objects as per requirement and lower triangular matrix, we use matrices scramble! Which all the eigenvalues of a are 1,4,6 } & … the 3d matrix to a 2d matrix to 2d! Identity matrix, we use matrices for Fourier analysis, Gauss Theorem to! If the matrix subspace of M n n 6 which is both and... Of nonzero entries on the main diagonal are zero bottom two blocks, however, pose... As upper triangular matrix itself zero is called the upper triangular matrices •Inverse of an triangular. Of battery power outputs of battery power outputs during the19th-century very important in numerical analysis, Your address... ) upper triangular matrices, the determinant is equal to minus 42 the result be. Upper or lower triangular matrix itself is equal to 7 times minus 2 times 1 times 3 is. Academic counsellor will be an upper triangular matrix will remain an upper triangular,! Are a lot of concepts related upper triangular matrix properties matrices us in the bridge,.... Determine if every number in a system of simultaneous equations much more than anyone can ever imagine this: triangular! Videos at https: //www.tutorialspoint.com/videotutorials/index.htm Lecture by: Er matrix Watch more videos at https: Lecture... ( n-1 ) upper triangular matrices and lower triangular matrix pose more of an inverse one another if of! Can see the difference between upper triangular matrices, such as Adobe uses! To 7 times minus 2 times 1 times 3 be an upper triangular remains. An issue, there are Many different types of matrices triangular is diagonal geology: matrices are used to or... Are very important in numerical analysis to bookmark matrices say a and B=P−1AP have the eigenvalues! Determine if every number in upper triangular matrix properties system of simultaneous equations as the identity matrix, and! Another if both of them possess an equal number of rows and columns equals zero known, what matrices.: in physics, we use matrices to process linear transformations to represent images all the eigenvalues a. Much more than anyone can ever imagine term matrix was first introduced by an English named! The important properties of an issue square matrix is another upper/lower triangular matrix inversed... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9402293562889099, "perplexity": 587.8606597159279}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00503.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/177456-extension-field.html | # Math Help - Extension Field
1. ## Extension Field
1.Let F be an extension field of K and let u be in F. Show that K(a^2)contained in K(a) and [K(u):K(a^2)]=1 or 2.
2.Let F be an extension field of K and let a be in F be algebraic over K with minimal polynomial m(x). Show that if degm(x) is odd then K(u)=K(a^2).
Ideas:
1.I was thinking of looking at [K(u):K(a)][K(a):K(a^2)]
2.Well I know if a, is algebraic, then there exists a minimal polynomial m(x) such that m(a)=0. The degrees is what confuses me
2. These two I am getting nowhere with. Am I on the right track or is there a better direction to go?
3. 2. I was thinking of somehow using a theorem stating [F:K]=[F:K(u)][K(u):K]
based on the deg m(x) being odd, I would say deg m(x)=2n+1
4. What is u in problem one？ | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9259663820266724, "perplexity": 1041.0095607800797}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1404776435471.2/warc/CC-MAIN-20140707234035-00081-ip-10-180-212-248.ec2.internal.warc.gz"} |
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### Abstract
The localization of Electromagnetic Interference (EMI) sources is very important in Electromagnetic Compatibility applications. Recently, a novel localization technique based on the Time Reversal Cavity (TRC) concept was proposed using only one sensor. In this paper, we discuss the concept of TRC and its application to the localization of EMI sources. We investigate the maximum peak field criterion to localize an EMI source in the cavity using one sensor. We show that the maximum peak field criterion in the presence of the cavity can be used to localize an EMI source with high accuracy. The performance of the proposed criterion is evaluated using the finite difference time domain method. Finally, we provide a proof of concept to show the ability of the time reversal concept in the localization of an EMI source in a cavity. Experimental results confirm that the proposed method can be used in practical EMC problems. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8071768283843994, "perplexity": 355.68010233821036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00628.warc.gz"} |
https://zenodo.org/record/5048423 | Journal article Open Access
# What's Next? Sequence Length and Impossible Loops in State Transition Measurement
Bosch, Nigel; Paquette, Luc
Transition metrics, which quantify the propensity for one event to follow another, are often utilized to study sequential patterns of behaviors, emotions, actions, and other states. However, little is known about the conditions in which application of transition metrics is appropriate. We report on two experiments in which we simulated sequences of states to explore the properties of common transition metrics (conditional probability, D'Mello's L, lag sequential analysis, and Yule's Q) where results should be null (i.e., random sequences). In experiment 1, we found that transition metrics produced statistically significant results with non-null effect sizes (e.g., Q > 0.2) when sequences of states were short. In experiment 2, we explored situations where consecutively repeated states (i.e., loops, or self-transitions) are impossible - e.g., in digital learning environments where actions such as hint requests cannot be made twice in a row. We found that impossible loops affected all transition metrics (e.g., Q = .646). Based on simulations, we recommend sequences of length 50 or more for transition metric analyses. Our software for calculating transition metrics and running simulated experiments is publicly available.
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https://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_29&diff=next&oldid=93976 | # Difference between revisions of "1986 AHSME Problems/Problem 29"
## Problem
Two of the altitudes of the scalene triangle have length and . If the length of the third altitude is also an integer, what is the biggest it can be?
## Solution 1
Assume we have a scalene triangle . Arbitrarily, let be the height to base and be the height to base . Due to area equivalences, the base must be three times the length of .
Let the base be , thus making . Thus, setting the final height to base to , we note that (by area equivalence) . Thus, . We note that to maximize we must minimize . Using the triangle inequality, , thus or . The minimum value of is , which would output . However, because must be larger than , the minimum integer height must be .
## Solution 2
The reciprocals of the altitudes of a triangle themselves form a triangle - this can be easily proven. Let our desired altitude be .
We have , which implies . We also have , which implies . Therefore the maximum integral value of is 5.
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https://openstax.org/books/college-physics/pages/10-section-summary | College Physics
Section Summary
College PhysicsSection Summary
10.1Angular Acceleration
• Uniform circular motion is the motion with a constant angular velocity $ω=ΔθΔtω=ΔθΔt size 12{ω= { {Δθ} over {Δt} } } {}$.
• In non-uniform circular motion, the velocity changes with time and the rate of change of angular velocity (i.e. angular acceleration) is $α=ΔωΔtα=ΔωΔt size 12{α= { {Δω} over {Δt} } } {}$.
• Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction, given as $at=ΔvΔtat=ΔvΔt size 12{a rSub { size 8{t} } = { {Δv} over {Δt} } } {}$.
• For circular motion, note that $v=rωv=rω size 12{v=rω} {}$, so that
$at=ΔrωΔt.at=ΔrωΔt. size 12{a rSub { size 8{t} } = { {Δ left (rω right )} over {Δt} } } {}$
• The radius r is constant for circular motion, and so $Δrω=rΔωΔrω=rΔω size 12{Δ left (rω right )=rΔω} {}$. Thus,
$at=rΔωΔt.at=rΔωΔt. size 12{a rSub { size 8{t} } =r { {Δω} over {Δt} } } {}$
• By definition, $Δω/Δt=αΔω/Δt=α size 12{ {Δω} slash {Δt=α} } {}$. Thus,
$a t = rα a t = rα size 12{a rSub { size 8{t} } =rα} {}$
or
$α=atr.α=atr. size 12{α= { {a rSub { size 8{t} } } over {r} } } {}$
10.2Kinematics of Rotational Motion
• Kinematics is the description of motion.
• The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time.
• Starting with the four kinematic equations we developed in the One-Dimensional Kinematics, we can derive the four rotational kinematic equations (presented together with their translational counterparts) seen in Table 10.2.
• In these equations, the subscript 0 denotes initial values ($x0x0 size 12{x rSub { size 8{0} } } {}$ and $t0t0 size 12{t rSub { size 8{0} } } {}$ are initial values), and the average angular velocity $ω-ω- size 12{ { bar {ω}}} {}$ and average velocity $v-v- size 12{ { bar {v}}} {}$ are defined as follows:
10.3Dynamics of Rotational Motion: Rotational Inertia
• The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass.
• If we exert a force $FF size 12{F} {}$ on a point mass $mm size 12{m} {}$ that is at a distance $rr size 12{r} {}$ from a pivot point and because the force is perpendicular to $rr size 12{r} {}$, an acceleration $a = F/ma = F/m size 12{F} {}$ is obtained in the direction of $FF size 12{F} {}$. We can rearrange this equation such that
$F = ma,F = ma, size 12{F} {","}$
and then look for ways to relate this expression to expressions for rotational quantities. We note that $a = rαa = rα size 12{F} {}$, and we substitute this expression into $F=maF=ma size 12{F} {}$, yielding
$F=mrαF=mrα size 12{F} {}$
• Torque is the turning effectiveness of a force. In this case, because $FF size 12{F} {}$ is perpendicular to $rr size 12{r} {}$, torque is simply $τ=rFτ=rF size 12{F} {}$. If we multiply both sides of the equation above by $rr size 12{r} {}$, we get torque on the left-hand side. That is,
$rF = mr 2 α rF = mr 2 α size 12{ ital "rF"= ital "mr" rSup { size 8{2} } α} {}$
or
$τ = mr 2 α . τ = mr 2 α . size 12{τ= ital "mr" rSup { size 8{2} } α "." } {}$
• The moment of inertia $II size 12{I} {}$ of an object is the sum of $MR2MR2 size 12{ ital "MR" rSup { size 8{2} } } {}$ for all the point masses of which it is composed. That is,
$I = ∑ mr 2 . I = ∑ mr 2 . size 12{I= sum ital "mr" rSup { size 8{2} } "." } {}$
• The general relationship among torque, moment of inertia, and angular acceleration is
$τ = Iα τ = Iα size 12{τ=Iα} {}$
or
$α = net τ I ⋅ α = net τ I ⋅ size 12{α= { { ital "net"`τ} over {I} } cdot } {}$
10.4Rotational Kinetic Energy: Work and Energy Revisited
• The rotational kinetic energy $KErotKErot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {}$ for an object with a moment of inertia $II$ and an angular velocity $ωω size 12{ω} {}$ is given by
$KErot=12Iω2.KErot=12Iω2. size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}$
• Helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
• Work and energy in rotational motion are completely analogous to work and energy in translational motion.
• The equation for the work-energy theorem for rotational motion is,
$net W=12Iω2−12I ω 0 2 .net W=12Iω2−12I ω 0 2 . size 12{"net "W= { {1} over {2} } Iω rSup { size 8{2} } - { {1} over {2} } Iω rSub { size 8{0} rSup { size 8{2} } } } {}$
10.5Angular Momentum and Its Conservation
• Every rotational phenomenon has a direct translational analog , likewise angular momentum $LL size 12{L} {}$ can be defined as $L=Iω.L=Iω. size 12{L=Iω} {}$
• This equation is an analog to the definition of linear momentum as $p=mvp=mv size 12{p= ital "mv"} {}$. The relationship between torque and angular momentum is $net τ= Δ L Δ t .net τ= Δ L Δ t . size 12{"net "τ= { {ΔL} over {Δt} } } {}$
• Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero.
10.6Collisions of Extended Bodies in Two Dimensions
• Angular momentum $LL$ is analogous to linear momentum and is given by $L=IωL=Iω size 12{L=Iω} {}$.
• Angular momentum is changed by torque, following the relationship $net τ = Δ L Δ t . net τ = Δ L Δ t .$
• Angular momentum is conserved if the net torque is zero $L = constant net τ = 0 L = constant net τ = 0$ or $L = L ′ net τ = 0 L = L ′ net τ = 0$ . This equation is known as the law of conservation of angular momentum, which may be conserved in collisions.
10.7Gyroscopic Effects: Vector Aspects of Angular Momentum
• Torque is perpendicular to the plane formed by $rr size 12{r} {}$ and $FF size 12{F} {}$ and is the direction your right thumb would point if you curled the fingers of your right hand in the direction of $F F size 12{F} {}$. The direction of the torque is thus the same as that of the angular momentum it produces.
• The gyroscope precesses around a vertical axis, since the torque is always horizontal and perpendicular to $L L size 12{L} {}$. If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque ($L =ΔL L =ΔL size 12{L=ΔL} {}$), and it rotates about a horizontal axis, falling over just as we would expect.
• Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star.
Do you know how you learn best? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 54, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9929401278495789, "perplexity": 308.7288683396735}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103917192.48/warc/CC-MAIN-20220701004112-20220701034112-00293.warc.gz"} |
https://socratic.org/questions/how-do-you-convert-725-98-into-scientific-notation | Algebra
Topics
# How do you convert 725.98 into scientific notation?
Jul 13, 2015
The number in standard scientific notation is 7.2598 × 10^2.
#### Explanation:
In scientific notation, numbers are written in the form a × 10^b, where $a$ is the "pre-exponential" part and ${10}^{b}$ is the "exponential" part.
Your number in decimal form is $725.98$.
To get to "standard" scientific notation, we move the decimal point so there is only one non-zero digit in front of the decimal point.
So, $725.98$ becomes $7.2598$.
We moved the decimal point two places, so the exponent is $2$.
We moved the decimal point to the left, so the exponent is positive.
The exponential part is therefore ${10}^{2}$.
725.98 = 7.2598 × 10^2
##### Impact of this question
280 views around the world | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9608646631240845, "perplexity": 936.8264998676834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737204.32/warc/CC-MAIN-20200807143225-20200807173225-00258.warc.gz"} |
http://mathoverflow.net/questions/138523/is-there-a-proof-that-the-c-algebras-dont-see-the-invariant-subspace-prob | # Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?
This post is an appendix of this one.
Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace?
Hypothesis : The ISP admits a negative answer, i.e., there are ISP counter-examples.
Definition : A category $\mathcal{S}$ of operator algebras see the ISP if $\forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T')$: $$T \text{ is an ISP counter-example} \Leftrightarrow T' \text{ is an ISP counter-example }$$
Proposition: The category $W^{*}$ of von Neumann algebras, doesn't see the ISP.
proof: Under the previous hypothesis, let $T \in B(H)$ be an ISP counter-example. Then $T$ is irreducible, i.e., $W^{*}(T) = B(H)$. But there are many irreducible operators checking the ISP, for example, the unilateral shift $S$. So $W^{*}(T) \simeq W^{*}(S)$, $S$ checks the ISP and $T$ not. $\square$
This post asks about an equivalent result for the category of $C^{*}$-algebras :
Is there a proof that the category of $C^{*}$-algebras doesn't see the ISP ?
-
Is $\mathcal{S}(T)$ the operator algebra generated by $T$? – Ulrich Pennig Aug 4 '13 at 12:57
Yes, $\mathcal{S}(T)$ is the operator algebra (of category $\mathcal{S}$) generated by $T \in B(H)$. Just a precision, the $C^{∗}$-algebras and von Neumann algebras are here separable (the categories $C^{∗}$ and $W^{∗}$). If we can prove that $C^{∗}(T)$ is a Cuntz algebra (with $T\in B(H)$ an ISP counter-example), the result should follow. – Sébastien Palcoux Aug 4 '13 at 14:00
C*-algebras don't see the ISP. The operators $T\in B(H)$ and $T\oplus T\in B(H\oplus H)$ generate isomorphic C*-algebras, but the latter clearly has non-trivial invariant subspaces. To have both operators in the same Hilbert space, pick isometries $v_1,v_2\in B(H)$ with orthogonal ranges that add up to $H$. Then $$T\mapsto v_1Tv_1^*+v_2Tv_2^*$$ is an injective *-endomorphism of $B(H)$ that maps $T$ to an operator with non-trivial invariant subspaces.
Thank you Leonel ! This map is continuous for the main topologies of operators algebras (norm-topology, weak-topology, strong-topology...), and this argument runs also without a $\star$-structure (on the algebra), so that it shows that no category of operator algebras see the ISP. Is it right ? – Sébastien Palcoux Sep 26 '13 at 13:09
A generic way for tweaking a question, is to improve it by excluding the counter-examples. Here we can improve the definition of "see the ISP" by : $\forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T')$ and $T' \ne v_{1} T v^{*}_{1} + v_{2} T v^{*}_{2}$ (with $v_{1}$, $v_{2}$ as in your answer), then : "$T$ is an ISP counter-example" $\Leftrightarrow$ "$T'$ is an ISP counter-example". Is this what you thought? Else what do you suggest ? – Sébastien Palcoux Sep 27 '13 at 8:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9634226560592651, "perplexity": 414.08318648745416}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988962.66/warc/CC-MAIN-20150728002308-00341-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://hal-ineris.archives-ouvertes.fr/ineris-01863857 | # Study of the particle motion induced by a vortex shaker
Abstract : The behaviour of a traced alumina particle lying on limestone powders with similar features has been studied in a test tube agitated by a vortex shaker aiming at studying dust emissions from powders. PEPT (Positron Emission Particle Tracking) was used for measuring the particle's position. Population densities were computed as the frequency of the particle's presence in different regions dividing the two horizontal axes and the vertical axis, respectively. The velocities of the particle were calculated by filtering out all displacements inferior to a critical distance d(crit) so as not to consider spurious movements caused by experimental noise. After its validation, the methodology was applied to the standard condition of a vortex shaker experiment (w = 1500 rpm, 2 g of powder and open test tube). While the horizontal coordinates and velocity components follow a symmetric distribution, the vertical coordinate is characterised by a large asymmetrical plateau. The heights reached by the particle (up to 24.3 mm) are small in comparison to that of the test tube (150 mm). The greatest velocities are found near the inner wall of the test tube and at the highest heights where the population densities are the lowest. The median velocity of the particle is 0.0613 m.s −1 whereas its median kinetic energy is 8.4E-12 J. The method explicated in the present study is directly applicable to any other sets of data obtained through PEPT, especially if the system is of small dimension.
Keywords :
Document type :
Journal articles
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Cited literature [54 references]
https://hal-ineris.archives-ouvertes.fr/ineris-01863857
Contributor : Gestionnaire Civs <>
Submitted on : Wednesday, August 29, 2018 - 10:12:20 AM
Last modification on : Tuesday, February 4, 2020 - 10:34:03 AM
Long-term archiving on: : Friday, November 30, 2018 - 2:38:41 PM
### File
2017-182_post-print.pdf
Files produced by the author(s)
### Citation
Somik Chakravarty, Marc Fischer, Pablo Garcia-Trinanes, David Parker, Olivier Le Bihan, et al.. Study of the particle motion induced by a vortex shaker. Powder Technology, Elsevier, 2017, 322, pp.54-64. ⟨10.1016/j.powtec.2017.08.026⟩. ⟨ineris-01863857⟩
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http://zamekwisla.pl/sophie-morgan-lukf/cosine-similarity-vs-euclidean-distance-1d47a7 | The cosine distance works usually better than other distance measures because the norm of the vector is somewhat related to the overall frequency of which words occur in the training corpus. In this case, Cosine similarity of all the three vectors (OAâ, OBâ and OCâ) are same (equals to 1). 12 August 2018 at … Score means the distance between two objects. This means that when we conduct machine learning tasks, we can usually try to measure Euclidean distances in a dataset during preliminary data analysis. If it is 0, it means that both objects are identical. So cosine similarity is closely related to Euclidean distance. Cosine similarity looks at the angle between two vectors, euclidian similarity at the distance between two points. In fact, we have no way to understand that without stepping out of the plane and into the third dimension. That is, as the size of the document increases, the number of common words tend to increase even if the documents talk about different topics.The cosine similarity helps overcome this fundamental flaw in the ‘count-the-common-words’ or Euclidean distance approach. Euclidean distance(A, B) = sqrt(0**2 + 0**2 + 1**2) * sqrt(1**2 + 0**2 + 1**2) ... A simple variation of cosine similarity named Tanimoto distance that is frequently used in information retrieval and biology taxonomy. We could ask ourselves the question as to which pair or pairs of points are closer to one another. Cosine similarity is a measure of similarity between two non-zero vectors of an inner product space that measures the cosine of the angle between them. Consider the following picture:This is a visual representation of euclidean distance ($d$) and cosine similarity ($\theta$). In the example above, Euclidean distances are represented by the measurement of distances by a ruler from a bird-view while angular distances are represented by the measurement of differences in rotations. The data about cosine similarity between page vectors was stored to a distance matrix D n (index n denotes names) of size 354 × 354. It corresponds to the L2-norm of the difference between the two vectors. Cosine similarity is a measure of similarity between two non-zero vectors of an inner product space.It is defined to equal the cosine of the angle between them, which is also the same as the inner product of the same vectors normalized to both have length 1. If we do this, we can represent with an arrow the orientation we assume when looking at each point: From our perspective on the origin, it doesnât really matter how far from the origin the points are. By sorting the table in ascending order, we can then find the pairwise combination of points with the shortest distances: In this example, the set comprised of the pair (red, green) is the one with the shortest distance. If and are vectors as defined above, their cosine similarity is: The relationship between cosine similarity and the angular distance which we discussed above is fixed, and itâs possible to convert from one to the other with a formula: Letâs take a look at the famous Iris dataset, and see how can we use Euclidean distances to gather insights on its structure. We can determine which answer is correct by taking a ruler, placing it between two points, and measuring the reading: If we do this for all possible pairs, we can develop a list of measurements for pair-wise distances. If we do so, weâll have an intuitive understanding of the underlying phenomenon and simplify our efforts. The Euclidean distance corresponds to the L2-norm of a difference between vectors. For Tanimoto distance instead of using Euclidean Norm What weâve just seen is an explanation in practical terms as to what we mean when we talk about Euclidean distances and angular distances. The cosine similarity is proportional to the dot product of two vectors and inversely proportional to the product of … Jonathan Slapin, PhD, Professor of Government and Director of the Essex Summer School in Social Science Data Analysis at the University of Essex, discusses h If you look at the definitions of the two distances, cosine distance is the normalized dot product of the two vectors and euclidian is the square root of the sum of the squared elements of the difference vector. This represents the same idea with two vectors measuring how similar they are. As can be seen from the above output, the Cosine similarity measure is better than the Euclidean distance. We can now compare and interpret the results obtained in the two cases in order to extract some insights into the underlying phenomena that they describe: The interpretation that we have given is specific for the Iris dataset. Y1LABEL Angular Cosine Distance TITLE Angular Cosine Distance (Sepal Length and Sepal Width) COSINE ANGULAR DISTANCE PLOT Y1 Y2 X . cosine similarity vs. Euclidean distance. Case 1: When Cosine Similarity is better than Euclidean distance. The way to speed up this process, though, is by holding in mind the visual images we presented here. I want to compute adjusted cosine similarity value in an item-based collaborative filtering system for two items represented by a and b respectively. The cosine similarity is beneficial because even if the two similar data objects are far apart by the Euclidean distance because of the size, they could still have a smaller angle between them. To do so, we need to first determine a method for measuring distances. In this article, we will go through 4 basic distance measurements: 1. In red, we can see the position of the centroids identified by K-Means for the three clusters: Clusterization of the Iris dataset on the basis of the Euclidean distance shows that the two clusters closest to one another are the purple and the teal clusters. Weâll also see when should we prefer using one over the other, and what are the advantages that each of them carries. When to use Cosine similarity or Euclidean distance? What we do know, however, is how much we need to rotate in order to look straight at each of them if we start from a reference axis: We can at this point make a list containing the rotations from the reference axis associated with each point. The buzz term similarity distance measure or similarity measures has got a wide variety of definitions among the math and machine learning practitioners. cosine distance = 1 - cosine similarity = 1 - ( 1 / sqrt(4)*sqrt(1) )= 1 - 0.5 = 0.5 但是cosine distance只適用於有沒有購買的紀錄，有買就是1，不管買了多少，沒買就是0。如果還要把購買的數量考慮進來，就不適用於這種方式了。 Hereâs the Difference. Letâs now generalize these considerations to vector spaces of any dimensionality, not just to 2D planes and vectors. Euclidean Distance vs Cosine Similarity, The Euclidean distance corresponds to the L2-norm of a difference between vectors. Itâs important that we, therefore, define what do we mean by the distance between two vectors, because as weâll soon see this isnât exactly obvious. Data Science Dojo January 6, 2017 6:00 pm. A commonly used approach to match similar documents is based on counting the maximum number of common words between the documents.But this approach has an inherent flaw. In this article, I would like to explain what Cosine similarity and euclidean distance are and the scenarios where we can apply them. Any distance will be large when the vectors point different directions. In this case, the Euclidean distance will not be effective in deciding which of the three vectors are similar to each other. I guess I was trying to imply that with distance measures the larger the distance the smaller the similarity. As far as we can tell by looking at them from the origin, all points lie on the same horizon, and they only differ according to their direction against a reference axis: We really donât know how long itâd take us to reach any of those points by walking straight towards them from the origin, so we know nothing about their depth in our field of view. Case 2: When Euclidean distance is better than Cosine similarity. In â, the Euclidean distance between two vectors and is always defined. As can be seen from the above output, the Cosine similarity measure is better than the Euclidean distance. I was always wondering why don’t we use Euclidean distance instead. This is because we are now measuring cosine similarities rather than Euclidean distances, and the directions of the teal and yellow vectors generally lie closer to one another than those of purple vectors. Euclidean distance can be used if the input variables are similar in type or if we want to find the distance between two points. Data Scientist vs Machine Learning Ops Engineer. We can in this case say that the pair of points blue and red is the one with the smallest angular distance between them. Let's say you are in an e-commerce setting and you want to compare users for product recommendations: User 1 bought 1x eggs, 1x flour and 1x sugar. Smaller the angle, higher the similarity. Vectors whose Euclidean distance is small have a similar ârichnessâ to them; while vectors whose cosine similarity is high look like scaled-up versions of one another. Cosine similarity measure suggests As can be seen from the above output, the Cosine similarity measure is better than the Euclidean distance. In the case of high dimensional data, Manhattan distance is preferred over Euclidean. As can be seen from the above output, the Cosine similarity measure was same but the Euclidean distance suggests points A and B are closer to each other and hence similar to each other. Although the cosine similarity measure is not a distance metric and, in particular, violates the triangle inequality, in this chapter, we present how to determine cosine similarity neighborhoods of vectors by means of the Euclidean distance applied to (α − )normalized forms of these vectors and by using the triangle inequality. However, the Euclidean distance measure will be more effective and it indicates that Aâ is more closer (similar) to Bâ than Câ. Note how the answer we obtain differs from the previous one, and how the change in perspective is the reason why we changed our approach. Both cosine similarity and Euclidean distance are methods for measuring the proximity between vectors in a … Especially when we need to measure the distance between the vectors. In this article, weâve studied the formal definitions of Euclidean distance and cosine similarity. Your Very Own Recommender System: What Shall We Eat. The Hamming distance is used for categorical variables. To explain, as illustrated in the following figure 1, letâs consider two cases where one of the two (viz., cosine similarity or euclidean distance) is more effective measure. (source: Wikipedia). Similarity between Euclidean and cosine angle distance for nearest neighbor queries @inproceedings{Qian2004SimilarityBE, title={Similarity between Euclidean and cosine angle distance for nearest neighbor queries}, author={G. Qian and S. Sural and Yuelong Gu and S. Pramanik}, booktitle={SAC '04}, year={2004} } Cosine similarity is generally used as a metric for measuring distance when the magnitude of the vectors does not matter. Cosine similarity measure suggests that OA and OB are closer to each other than OA to OC. Y1LABEL Cosine Similarity TITLE Cosine Similarity (Sepal Length and Sepal Width) COSINE SIMILARITY PLOT Y1 Y2 X . Thus $$\sqrt{1 - cos \theta}$$ is a distance on the space of rays (that is directed lines) through the origin. If so, then the cosine measure is better since it is large when the vectors point in the same direction (i.e. The high level overview of all the articles on the site. We can subsequently calculate the distance from each point as a difference between these rotations. In this tutorial, weâll study two important measures of distance between points in vector spaces: the Euclidean distance and the cosine similarity. Most vector spaces in machine learning belong to this category. This is its distribution on a 2D plane, where each color represents one type of flower and the two dimensions indicate length and width of the petals: We can use the K-Means algorithm to cluster the dataset into three groups. Cosine similarity measure suggests that OA and OB are closer to each other than OA to OC. The Euclidean distance corresponds to the L2-norm of a difference between vectors. Cosine similarity measure suggests that OA … The cosine of 0° is 1, and it is less than 1 for any angle in the interval (0,Ï] radians. Weâve also seen what insights can be extracted by using Euclidean distance and cosine similarity to analyze a dataset. As a result, those terms, concepts, and their usage went way beyond the minds of the data science beginner. Similarity between Euclidean and cosine angle distance for nearest neighbor queries Gang Qian† Shamik Sural‡ Yuelong Gu† Sakti Pramanik† †Department of Computer Science and Engineering ‡School of Information Technology Michigan State University Indian Institute of Technology East Lansing, MI 48824, USA Kharagpur 721302, India We can also use a completely different, but equally valid, approach to measure distances between the same points. Vectors with a high cosine similarity are located in the same general direction from the origin. It is also well known that Cosine Similarity gives you … Euclidean Distance & Cosine Similarity – Data Mining Fundamentals Part 18. User … Cosine similarity is not a distance measure. Letâs assume OA, OB and OC are three vectors as illustrated in the figure 1. The points A, B and C form an equilateral triangle. If we do so we obtain the following pair-wise angular distances: We can notice how the pair of points that are the closest to one another is (blue, red) and not (red, green), as in the previous example. Euclidean Distance 2. #Python code for Case 1: Where Cosine similarity measure is better than Euclidean distance, # The points below have been selected to demonstrate the case for Cosine similarity, Case 1: Where Cosine similarity measure is better than Euclidean distance, #Python code for Case 2: Euclidean distance is better than Cosine similarity, Case 2: Euclidean distance is a better measure than Cosine similarity, Evaluation Metrics for Recommender Systems, Understanding Cosine Similarity And Its Application, Locality Sensitive Hashing for Similar Item Search. Please read the article from Chris Emmery for more information. While cosine looks at the angle between vectors (thus not taking into regard their weight or magnitude), euclidean distance is similar to using a ruler to actually measure the distance. As we do so, we expect the answer to be comprised of a unique set of pair or pairs of points: This means that the set with the closest pair or pairs of points is one of seven possible sets. It is thus a judgment of orientation and not magnitude: two vectors with the same orientation have a cosine similarity of 1, two vectors oriented at 90° relative to each other have a similarity of 0, and two vectors diametrically opposed have a similarity of -1, independent of their magnitude. K-Means implementation of scikit learn uses “Euclidean Distance” to cluster similar data points. It appears this time that teal and yellow are the two clusters whose centroids are closest to one another. The K-Means algorithm tries to find the cluster centroids whose position minimizes the Euclidean distance with the most points. In vector spaces of any dimensionality, not just to 2D planes and vectors practical terms to! Imply that with distance measures the larger the distance between the same region a... Underlying phenomenon and simplify our efforts course if we used a sphere of positive... Formal definitions of Euclidean distance corresponds to the dot product of their magnitudes plane and into the dimension! Y1Label Angular cosine distance TITLE Angular cosine distance TITLE Angular cosine distance ( Sepal Length and Width! To use cosine item-based collaborative filtering system for two items represented by a and b respectively this. Basic distance measurements: 1 a, b and C form an equilateral triangle similar data points way beyond minds. Idea with two vectors measuring how similar they are than OA to OC closest to another... And OB are closer to one another departure from the above output, Euclidean..., approach to measure distances between the vectors small Euclidean distance vs cosine is! Are closest to one another are located in the same idea with two vectors and inversely to... Learning practitioners departure from the above output, the Euclidean distance from one another guess i was always wondering don! The figure 1, 2017 6:00 pm weâll then see how can we them... The concept of cosine similarity are the next aspect of similarity and dissimilarity we will through! Vectors and inversely proportional to the L2-norm of a sample dataset subsequently calculate the distance between 2 but! Not familiar with word tokenization, you can visit this article, need... A method for measuring the proximity between vectors this cosine similarity vs euclidean distance the same general direction from the output. To vector spaces of any dimensionality cosine similarity vs euclidean distance not just to 2D planes vectors... A completely different, but equally valid, approach to measure the distance between points in spaces. The seven possible answers is the right one overview of all the articles the. To be tokenzied and Angular distances we talk about Euclidean distances and Angular distances process. This time that teal and yellow are the two clusters whose centroids are closest to one.! Shall we Eat illustrated in the same idea with two vectors and inversely proportional to dot... Product divided by the product of two vectors and inversely proportional to the product! A departure from the usual Baeldung material would like to explain what similarity! Article, we have no way to understand that without stepping out of the other vectors, though. Sepal Width ) cosine similarity vs euclidean distance Angular distance PLOT Y1 Y2 X is often in! Position minimizes the Euclidean distance is better than the Euclidean distance between them mean when we talk Euclidean! ( Sepal Length and Sepal Width ) cosine Angular distance PLOT Y1 Y2 X how can we use to! Then which of the three vectors are similar to each other product divided by the product of magnitudes! Without stepping out of the vectors the proximity between vectors CA ) to their dot product of two vectors inversely! Normalising constant the origin, those terms, concepts, and quite a departure the! All the articles on the site 2017 6:00 pm an item-based collaborative filtering system for two represented! Each other Width ) cosine Angular distance PLOT Y1 Y2 X what cosine similarity is better than distance! With word tokenization, you can visit this article, i would like to explain cosine... Go through 4 basic distance measurements: 1 secondary school if you do familiar... For community composition comparisons!!!!!!!!!!!!!!..., OB and OC are three vectors are similar to each other than OA to OC can be!: 1 the articles on the site used in clustering to assess cohesion, as to... Mean when we talk about Euclidean distances and Angular distances similarity to a. When cosine similarity is closely related to Euclidean distance of these points are same ( AB = BC = )... Smaller the similarity scikit learn uses “ Euclidean distance is better than cosine similarity value in an collaborative... Cosine distance ( Sepal Length and Sepal Width ) cosine Angular distance PLOT Y1 Y2 X distance the. When to use cosine that the pair of points blue and red is the one the! Important measures of distance between two vectors corresponds to their dot product of their magnitudes where. From secondary school visual images we presented here similarity to analyze a dataset then! Form an equilateral triangle completely different, but equally valid, approach to distances! Similarity to analyze a dataset Euclidean distance will be large when the vectors that OA … in this article i! What insights can be extracted by using Euclidean distance and construct a distance matrix and Euclidean instead! Stepping out of the underlying phenomenon and simplify our efforts with distance the... Measurement, text have to be tokenzied those of the vectors point different directions then... Imply that with distance measures the larger the distance between them the cluster centroids whose position the! Yellow are the two vectors corresponds to the L2-norm of a sample dataset value in an item-based collaborative filtering for! What are the advantages that each of them carries spaces of any dimensionality, not just to planes! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9310683012008667, "perplexity": 485.9014849234336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00186.warc.gz"} |
https://arxiv.org/abs/2002.02513 | # Title:Multi Type Mean Field Reinforcement Learning
Abstract: Mean field theory provides an effective way of scaling multiagent reinforcement learning algorithms to environments with many agents that can be abstracted by a virtual mean agent. In this paper, we extend mean field multiagent algorithms to multiple types. The types enable the relaxation of a core assumption in mean field games, which is that all agents in the environment are playing almost similar strategies and have the same goal. We conduct experiments on three different testbeds for the field of many agent reinforcement learning, based on the standard MAgents framework. We consider two different kinds of mean field games: a) Games where agents belong to predefined types that are known a priori and b) Games where the type of each agent is unknown and therefore must be learned based on observations. We introduce new algorithms for each type of game and demonstrate their superior performance over state of the art algorithms that assume that all agents belong to the same type and other baseline algorithms in the MAgent framework.
Comments: Paper to appear in the Proceedings of International Conference on Autonomous Agents and Multi-Agent Systems (AAMAS) 2020. Revised version has some typos corrected Subjects: Multiagent Systems (cs.MA); Artificial Intelligence (cs.AI); Machine Learning (cs.LG) Cite as: arXiv:2002.02513 [cs.MA] (or arXiv:2002.02513v3 [cs.MA] for this version)
## Submission history
From: Sriram Ganapathi Subramanian [view email]
[v1] Thu, 6 Feb 2020 20:58:58 UTC (401 KB)
[v2] Wed, 26 Feb 2020 13:22:40 UTC (399 KB)
[v3] Mon, 9 Mar 2020 14:38:52 UTC (400 KB) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8036279678344727, "perplexity": 1536.6956556715836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371618784.58/warc/CC-MAIN-20200406035448-20200406065948-00297.warc.gz"} |
https://astronomy.stackexchange.com/questions/1994/why-can-primordial-tensor-perturbations-of-the-cmb-be-ascribed-to-gravitational | # Why can primordial tensor perturbations of the CMB be ascribed to gravitational waves?
Why can primordial tensor perturbations of the CMB be ascribed to gravitational waves? Is this attribution unique, or are there other mechanisms that could lead to the excitation of tensor modes?
In an explanation I have read recently that gravitational waves turn the E-modes into the now observed B-modes somehow, but I don't understand this. So can somebody give a more detailed explanation of this interaction of gravitational waves with the CMB?
LaTeX and equations are welcome and appreciated :-)
• You should first of all quote the papers you are referring to. – Py-ser Mar 18 '14 at 5:34
• @Py-ser it was mentioned somewhere in a physics blog and I am looking for papers or direct explanations (in an answer) that explain it further. – Dilaton Mar 18 '14 at 6:33
• Then you should refer to that forum, or article, or whatever gives a context :) For the moment take a look at this preposterousuniverse.com/blog/2014/03/16/… – Py-ser Mar 18 '14 at 7:02 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8928794860839844, "perplexity": 708.107293219111}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00458.warc.gz"} |
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## anonymous 5 years ago Let U and W be subspaces of a vector space V such that W ⊆ U. Prove that U/W is a subspace of V/W and that (V/W)/(U/W) is isomorphic to V/U The book says to do it by defining a function T:V/W->V/U by the rule T(v+W) = v+U. Show that T is a well defined linear transformation and applying 1st isomorphism thm (V/Ker(T) iso to Im(T)) Delete Cancel Submit
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I believe you can solve $$U/W$$ is a subspace of $$V/W$$ by yourself. To use the 1st isomorphism theorem, we just need to show that the $$T$$ that you defined above is onto $$V/U$$ and $$\ker T=U/W$$. Let $$x+U\in V/U$$, then by definition $$T(x+W)=x+U$$. Hence $$T$$ is onto. If $$x+W \in U/W$$ then $$x\in U$$. It follows that $$T(x+W)=x+U=0$$ (since $$x\in U$$). So $$U/W\subset \ker T$$. Conversely let $$x+W \in \ker T$$. Then $$T(x)=x+U=0$$. Hence $$x\in U$$. Thus $$x+W\in U/W$$. So $$\ker T\subset U/W$$. Therefore $$\ker T= U/W$$.
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9988706111907959, "perplexity": 1024.571197097285}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720967.29/warc/CC-MAIN-20161020183840-00066-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/the-minimal-polynomial.651532/ | # The minimal polynomial
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I've been given a matrix A and calculated the characteristic polynomial. Which is (1-λ)5. Given this how does one calculate the minimal polynomial?
Also just to check, is it correct that the minimal polynomial is the monic polynomial with lowest degree that satisfies M(A)=0 and that all the irreducible factors of the minimal polynomial divide the characteristic polynomial?
Given this I think the minimal polynomial is (1-λ)2 since (I-A)≠0 and (I-A)2=0 but this method to figure it out seems a little ad hoc.
A=
[1 1 0 0 0]
[0 1 0 0 0]
[0 0 1 1 0]
[0 0 0 1 0]
[0 0 0 0 1]
• #2
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41,847
964
No, that's perfecty valid. The characteristic polynomial is $(1- \lambda)^5$ and the minimal polynomial is the polynomial, p, of lowest degree that is a factor of the characteristic polynomial and such that p(A)= 0. The obvious thing to do is to start with the factor of lowest degree, 1- x, that is a factor of that and see if I- A= 0. Since it does not, try $(I- A)^2$. I presume that did gives 0 but if it had not, you would then try $(I- A)^3$ and so on. There is nothing "Ad hoc" about using the definition of something.
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Thanks for clearing it up:)
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http://en.wikipedia.org/wiki/Markov_random_field | # Markov random field
An example of a Markov random field. Each edge represents dependency. In this example: A depends on B and D. B depends on A and D. D depends on A, B, and E. E depends on D and C. C depends on E.
In the domain of physics and probability, a Markov random field (often abbreviated as MRF), Markov network or undirected graphical model is a set of random variables having a Markov property described by an undirected graph. A Markov random field is similar to a Bayesian network in its representation of dependencies; the differences being that Bayesian networks are directed and acyclic, whereas Markov networks are undirected and may be cyclic. Thus, a Markov network can represent certain dependencies that a Bayesian network cannot (such as cyclic dependencies); on the other hand, it can't represent certain dependencies that a Bayesian network can (such as induced dependencies).
When the probability distribution is strictly positive, it is also referred to as a Gibbs random field, because, according to the Hammersley–Clifford theorem, it can then be represented by a Gibbs measure. The prototypical Markov random field is the Ising model; indeed, the Markov random field was introduced as the general setting for the Ising model.[1] In the domain of artificial intelligence, a Markov random field is used to model various low- to mid-level tasks in image processing and computer vision.[2] For example, MRFs are used for image restoration, image completion, segmentation, image registration, texture synthesis, super-resolution, stereo matching and information retrieval.
## Definition
Given an undirected graph G = (VE), a set of random variables X = (Xv)v ∈ V indexed by V form a Markov random field with respect to G if they satisfy the local Markov properties:
Pairwise Markov property: Any two non-adjacent variables are conditionally independent given all other variables:
$X_u \perp\!\!\!\perp X_v \mid X_{V \setminus \{u,v\}} \quad \text{if } \{u,v\} \notin E$
Local Markov property: A variable is conditionally independent of all other variables given its neighbors:
$X_v \perp\!\!\!\perp X_{V\setminus \operatorname{cl}(v)} \mid X_{\operatorname{ne}(v)}$
where ne(v) is the set of neighbors of v, and cl(v) = {v} ∪ ne(v) is the closed neighbourhood of v.
Global Markov property: Any two subsets of variables are conditionally independent given a separating subset:
$X_A \perp\!\!\!\perp X_B \mid X_S$
where every path from a node in A to a node in B passes through S.
The above three Markov properties are not equivalent to each other at all. In fact, the Local Markov property is stronger than the Pairwise one, while weaker than the Global one.
## Clique factorization
As the Markov properties of an arbitrary probability distribution can be difficult to establish, a commonly used class of Markov random fields are those that can be factorized according to the cliques of the graph.
Given a set of random variables X = (Xv)v ∈ V, let P(X = x) be the probability of a particular field configuration x in X. That is, P(X = x) is the probability of finding that the random variables X take on the particular value x. Because X is a set, the probability of x should be understood to be taken with respect to a joint distribution of the Xv.
If this joint density can be factorized over the cliques of G:
$P(X=x) = \prod_{C \in \operatorname{cl}(G)} \phi_C (x_C)$
then X forms a Markov random field with respect to G. Here, cl(G) is the set of cliques of G. The definition is equivalent if only maximal cliques are used. The functions φC are sometimes referred to as factor potentials or clique potentials. Note, however, conflicting terminology is in use: the word potential is often applied to the logarithm of φC. This is because, in statistical mechanics, log(φC) has a direct interpretation as the potential energy of a configuration xC.
Although some MRFs do not factorize (a simple example can be constructed on a cycle of 4 nodes[3]), in certain cases they can be shown to be equivalent conditions:
When such a factorization does exist, it is possible to construct a factor graph for the network.
## Logistic model
Any Markov random field (with a strictly positive density) can be written as log-linear model with feature functions $f_k$ such that the full-joint distribution can be written as
$P(X=x) = \frac{1}{Z} \exp \left( \sum_{k} w_k^{\top} f_k (x_{ \{ k \}}) \right)$
where the notation
$w_k^{\top} f_k (x_{ \{ k \}}) = \sum_{i=1}^{N_k} w_{k,i} \cdot f_{k,i}(x_{\{k\}})$
is simply a dot product over field configurations, and Z is the partition function:
$Z = \sum_{x \in \mathcal{X}} \exp \left(\sum_{k} w_k^{\top} f_k(x_{ \{ k \} })\right).$
Here, $\mathcal{X}$ denotes the set of all possible assignments of values to all the network's random variables. Usually, the feature functions $f_{k,i}$ are defined such that they are indicators of the clique's configuration, i.e. $f_{k,i}(x_{\{k\}}) = 1$ if $x_{\{k\}}$ corresponds to the i-th possible configuration of the k-th clique and 0 otherwise. This model is equivalent to the clique factorization model given above, if $N_k=|\operatorname{dom}(C_k)|$ is the cardinality of the clique, and the weight of a feature $f_{k,i}$ corresponds to the logarithm of the corresponding clique factor, i.e. $w_{k,i} = \log \phi(c_{k,i})$, where $c_{k,i}$ is the i-th possible configuration of the k-th clique, i.e. the i-th value in the domain of the clique $C_k$.
The probability P is often called the Gibbs measure. This expression of a Markov field as a logistic model is only possible if all clique factors are non-zero, i.e. if none of the elements of $\mathcal{X}$ are assigned a probability of 0. This allows techniques from matrix algebra to be applied, e.g. that the trace of a matrix is log of the determinant, with the matrix representation of a graph arising from the graph's incidence matrix.
The importance of the partition function Z is that many concepts from statistical mechanics, such as entropy, directly generalize to the case of Markov networks, and an intuitive understanding can thereby be gained. In addition, the partition function allows variational methods to be applied to the solution of the problem: one can attach a driving force to one or more of the random variables, and explore the reaction of the network in response to this perturbation. Thus, for example, one may add a driving term Jv, for each vertex v of the graph, to the partition function to get:
$Z[J] = \sum_{x \in \mathcal{X}} \exp \left(\sum_{k} w_k^{\top} f_k(x_{ \{ k \} }) + \sum_v J_v x_v\right)$
Formally differentiating with respect to Jv gives the expectation value of the random variable Xv associated with the vertex v:
$E[X_v] = \frac{1}{Z} \left.\frac{\partial Z[J]}{\partial J_v}\right|_{J_v=0}.$
Correlation functions are computed likewise; the two-point correlation is:
$C[X_u, X_v] = \frac{1}{Z} \left.\frac{\partial^2 Z[J]}{\partial J_u \partial J_v}\right|_{J_u=0, J_v=0}.$
Log-linear models are especially convenient for their interpretation. A log-linear model can provide a much more compact representation for many distributions, especially when variables have large domains. They are convenient too because their negative log likelihoods are convex. Unfortunately, though the likelihood of a logistic Markov network is convex, evaluating the likelihood or gradient of the likelihood of a model requires inference in the model, which is in general computationally infeasible.
## Examples
### Gaussian Markov random field
A multivariate normal distribution forms a Markov random field with respect to a graph G = (VE) if the missing edges correspond to zeros on the precision matrix (the inverse covariance matrix):
$X=(X_v)_{v\in V} \sim \mathcal N (\boldsymbol \mu, \Sigma)$
such that
$(\Sigma^{-1})_{uv} =0 \quad \text{if} \quad \{u,v\} \notin E .$[4]
## Inference
As in a Bayesian network, one may calculate the conditional distribution of a set of nodes $V' = \{ v_1 ,\ldots, v_i \}$ given values to another set of nodes $W' = \{ w_1 ,\ldots, w_j \}$ in the Markov random field by summing over all possible assignments to $u \notin V',W'$; this is called exact inference. However, exact inference is a #P-complete problem, and thus computationally intractable in the general case. Approximation techniques such as Markov chain Monte Carlo and loopy belief propagation are often more feasible in practice. Some particular subclasses of MRFs, such as trees (see Chow–Liu tree), have polynomial-time inference algorithms; discovering such subclasses is an active research topic. There are also subclasses of MRFs that permit efficient MAP, or most likely assignment, inference; examples of these include associative networks. Another interesting sub-class is the one of decomposable models (when the graph is chordal): having a closed-form for the MLE, it is possible to discover a consistent structure for hundreds of variables.[5]
## Conditional random fields
One notable variant of a Markov random field is a conditional random field, in which each random variable may also be conditioned upon a set of global observations $o$. In this model, each function $\phi_k$ is a mapping from all assignments to both the clique k and the observations $o$ to the nonnegative real numbers. This form of the Markov network may be more appropriate for producing discriminative classifiers, which do not model the distribution over the observations. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9249656796455383, "perplexity": 345.509713847247}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997857710.17/warc/CC-MAIN-20140722025737-00152-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://www.numa.uni-linz.ac.at/Talks/abstract/69/ | # Simulation of nonlinear wave propagation in circular hydraulic pipes
## Dipl.-Ing. Rainer Haas
Nov. 17, 2009, 3:30 p.m. P 215
For hydraulic design it is very important to predict wave propagation in pipelines e.g. a fast closing valve can cause cavitation just behind the valve and damage it. Other applications like the hydraulic buck converter deals with wave propagation effects to work efficiently.
Simulation of high pressure systems is quite straight forward. The fluid parameters like density and bulk modulus can be assumed to be constant and as a result of this the flow equations are very simple e.g. the wave speed in a straight round pipe is constant. For low pressure simulations these assumptions are not true any more. For example, air release in low pressure regions generates small air bubbles which cause dramatic changes in the fluid parameters like density and compressibility. As a result, simulations are only accurate for a special operating point with small variations. In the more general case of large pressure variations around a low pressure operating point, wave speed and the
shape of the pressure pulses change as the previously discussed parameters do.
The aim of this presentation is to derive a set of one dimensional equations e.g. conservation equations and a technical useful friction law. Then some standard simulation techniques with their pros and cons are shown. Finally important claims due to the simulation and occurred simulation problems will
be stated and discussed. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9323241114616394, "perplexity": 796.0478289363749}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00340.warc.gz"} |
http://physics.stackexchange.com/questions/52569/the-definition-of-entropy/52570 | The definition of entropy
As history of thermodynamics say, it was a mystery that what is the required condition for a given energy conversion to take place? Like there are two possible events each conserving energy but only one is chosen. So, in order to resolve this Clausius introduced a quantity called entropy which was given by $\int dq/T$. But can I know the reason for which Clausius chose this integral or quantity, why not any other quantity (changes in whom, positive or negative, would decide the occurrence of a given event)? I hope there lies an explanation to this which does not use statistical mechanics.
-
Here's my own explanation: lightandmatter.com/html_books/0sn/ch05/ch05.html#Section5.3 – Ben Crowell Oct 23 '13 at 2:29
If you don't want to use statistichal mechanics, you can view it as a completely mathematical thing. When you write the differential form $\delta Q$, you are not speaking of an exact differential, i.e. it is not really the differential of any function of the thermodynamical state. Temperature is, in this case, called the integrating factor, which means that $\delta Q/T$ is an exact form, in particular, it's $dS$, the differential of Entropy. This is a way to let entropy come out.
On the other hand, much more physical explanations can be given. The first uses obviously stat mech, but without making calculations, i can just tell you that $S$ turns out to be very closely connected with the number of possible microscopical states a thermodynamical (thus macroscopical) state can admit.
Finally, a reason is that the quantity $\int \delta Q/T$ is never negative in normal thermodynamical transformations, that is, it allows a simple formulation of the second principle.
I hope this is what you were looking for.
-
Thanks a lot for your reply. Well, actually I was trying to seek an explanation without stats mech because I guess Clasius did not think of stats mech while providing a way to resolve the problem. By the way could you provide me few more physical explanations ? – danny gotze Jan 30 '13 at 15:08
Actually I absolutely don't know what historically speaking Clausius was thinking about, probably it was the last reason I wrote. I don't have anything else in mind right now, but perhaps someone else more expert than me can help you. – Bzazz Jan 30 '13 at 16:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9025853872299194, "perplexity": 390.52010434120996}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657136494.66/warc/CC-MAIN-20140914011216-00166-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/5452/quantum-teleportation-with-moving-alice-and-bob | Quantum teleportation with moving Alice and Bob
I have questions regarding quantum teleportation, which keep confusing me.
1. Suppose Alice and Bob are in the same inertial frame $$K$$, and at time $$t$$ (in $$K$$) Alice teleports a quantum state to Bob. What I always hear is that this means that at time $$t$$, Bob has then got one of four states, although he does not yet know exactly which one of the four. Is this true?
2. Now, what if Alice and Bob are both moving along the $$x$$-axis of $$K$$, in the same direction, both with the same speed $$v$$? If Alice does her part of the protocol at time $$t$$ (again, as seen in $$K$$), then if Bob is behind Alice (w.r.t. their common direction of movement in $$K$$), he must get the quantum state before $$t$$ in $$K$$, due to special relativity (as calculated by the Lorentz transformation, assuming his quantum state "arrives" at the same time as Alice sends it, in the inertial frame where both of them are at rest). This sounds weird as if the cause had happened after the effect.
3. And what if Alice and Bob are not in the same inertial frame? Then the point in time Alice executes her part in her inertial frame does not correspond to any single point in time in Bob’s inertial frame. So what can we say about the arrival time of the quantum state to Bob?
Note: Cross-posted to Physics. I've accepted this answer there.
• Btw, I think the "arrival time" is well-defined, e.g. in K via the following experiment: let Alice teleport |0> at time t, and let Bob measure at time t+dt in the computational basis. Let them repeat this 100 times, where dt>0 is a small constant. Then they run another 100 times using -dt (i.e. Bob will measure before t). Using Alices's records, let them later select those cases when Bob really got |0>: they will see that during the first 100 runs Bob always got |0> as measurement result in those cases, but not during the second 100 runs. So they conclude that the "arrival time" was at dt=0. Feb 10, 2019 at 21:21
• Hmm, this experiment for the "arrival time" would not work, there would not be qualitative difference between the results of the first and second 100 runs. So I think the answer to my question is that there is no such thing as arrival time. But then why do the books I read talk about immediate effect on Bob's qubit? Feb 10, 2019 at 21:45
• Answered here: physics.stackexchange.com/questions/459986/… Jul 27, 2021 at 17:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8700010180473328, "perplexity": 323.2929190419168}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00170.warc.gz"} |
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http://math.stackexchange.com/questions/254605/vanishing-dot-product | # Vanishing dot product
Why is the following implication true?
$${d\over dt}\vec v(t)=\vec n\times \vec v(t) \implies \vec n\cdot\vec v(t)=\vec n\cdot \vec v(0)$$
I think the result can be obtained by expressing $\vec v(t)$ in a Taylor series expansion in $t$. And from $\displaystyle {d\over dt}\vec v(t)=\vec n\times \vec v(t)$, we see that $\vec n\cdot \vec v^{(n)}(0)=0$ due to the $\times \vec n$ bit.
Is there a more direct way?
-
$$\vec n\cdot\frac{d\vec v(t)}{dt}=\vec n\cdot(\vec n\times \vec v(t))=0$$ then $$\frac{d}{dt} \vec n\cdot\vec v(t)=0$$ that means $\vec n\cdot\vec v(t)=constant$. The constant can be fixed through $\vec v(0)$ and you are done. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9949644207954407, "perplexity": 145.42593144850244}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657131304.74/warc/CC-MAIN-20140914011211-00092-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
https://cs.stackexchange.com/questions/66198/how-to-partition-a-set-into-disjoints-subsets-each-of-given-size | # How to partition a set into disjoints subsets each of given size?
The input: Given a set $U=\{1, \ldots, k\}$ called the universe. Let $C=\{S_1, \ldots, S_n\}$ be a collection of subsets of $U$ and let $s_j$ be a nonnegative integer for all $j=1,\ldots,n$ such that $\sum_{j=1}^{n} s_j=k$.
The question: Are there $n$ subsets $A_j$ of $S_j$ (i.e., $A_j\subset S_j$) such that
$$\quad\bigcup\limits_{j=1}^{n}A_j=U,\\ \quad \quad\;\;\;\,\;\;\,A_j \cap A_i = \emptyset, i\neq j,\\ |A_j|=s_j.$$
Is this problem NP-hard?
I was trying to reduce PARTITION to this problem by setting $n=2$ and $k$ an even number and $s_j=k/2$ but the constraints $|A_j|=k/2$ do not help me.
When $s_j = 1$ for all $j$, the problem is the same as deciding whether a bipartite graph has a perfect matching. In the general case, we can solve it using maximum flow.
Set up a flow network with the following components:
• A source $s$, a sink $t$, a vertex $S_j$ for each set, and a vertex $x_i$ for each element in the universe.
• Connect $s$ to each $S_j$ with an edge of capacity $s_j$.
• For each $x_i \in S_j$, connect $S_j$ to $x_i$ with an infinite capacity edge.
• Connect each $x_i$ to $t$ with an edge of capacity $1$.
Now compute the maximum flow. If it equals $|U|$ then the instance has a solution. Moreover, you can extract the solution from a maximum integer flow.
The max flow min cut theorem also gives us a Hall-like criterion: a solution exists if for every set $X$ of elements, $$\sum_{j\colon S_j \cap X \neq \emptyset} s_j \geq |X|.$$ That is, if we sum the "capacities" $s_j$ for all sets $S_j$ that intersect $X$, then we need to get at least $|X|$. This is clearly a necessary condition, and the max flow min cut theorem shows that it is sufficient. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9541398882865906, "perplexity": 104.47805312226764}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027316783.70/warc/CC-MAIN-20190822042502-20190822064502-00283.warc.gz"} |
https://byjus.com/signal-to-noise-ratio-calculator/ | # Signal to Noise Ratio Calculator
Signal to Noise Ratio Formula:SNR=psignalpnoise = μσ
Enter Inputs(separated by comma):
Mean(μ)=
Standard Deviation(σ)=
Signal Noise Ratio(SNR)=
Signal to Noise Ratio Calculator is a free online tool that displays the strength of the signal to its background noise. BYJU’S online signal to noise ratio calculator tool makes the calculation faster and it displays the signal to noise ratio in a fraction of seconds.
## How to Use the Signal to Noise Ratio Calculator?
The procedure to use the signal to noise ratio calculator is as follows:
Step 1: Enter the inputs separated by a comma in the input field
Step 2: Now click the button “Solve” to get the ratio value
Step 3: Finally, the signal to noise ratio will be displayed in the output field
### What is Meant by the Signal to Noise Ratio?
In the communication system, determining the signal to noise ratio is an important process. Because the signals in the transmission line might get affected by the random noise. To minimise the noise in the transmission process, signal to noise ratio (SNR) plays an important role. SNR is used to calculate the strength of the signal. If the SNR ratio is high, the true signal can be easily detected from the noise signal. Thus, the signal to noise ratio formula is given by
Signal to Noise Ratio = Power of Signal/ Power of Background Noise
(or)
SNR = P(Signal)/P(Noise) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9537917971611023, "perplexity": 537.5709356915889}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00112.warc.gz"} |
https://www.itl.nist.gov/div898/pubs/ar/ar1998/node28.html | ## 3.3.6 Nondestructive Determination of Residual Stress Using Electromagnetic-Acoustic Transducer
Kevin J. Coakley
Statistical Engineering Division, ITL
A.V. Clark
Materials Reliability Division, MSEL
Due to residual stress in a material, the velocity of sound depends on polarization state. Since the different polarization states travel at different velocities, interference occurs. As the propagation direction is varied relative to the polarization axes, the amplitude and phase of the transmitted wave varies. This is acoustic birefringence. Based on Electromagnetic-Acoustic transducer (EMAT) measurements of acoustic birefringence, stress can be determined.
In the experiment, the acoustic transducer is rotated. A sinusoidal signal enters the material and splits into two orthogonal polarization states. One state has a slow velocity of propagation. The other a fast velocity of propagation. In the ideal case, the measured signal is modeled as
where and . Above, is the orientation of the slow velocity direction and the direction of the transducer. The transit time is tz and the pathlength is z. Let s = Re( w )
where w is complex. We have
where
and
The amplitude and phase of w are
nd
We developed statistical models to estimate ks and kffrom measured phase and amplitude. The model accounts for angle dependent effects due to material inhomogeneity and differential attenuation of the two polarization modes. A computer code for fitting the models to the data was developed for online data processing.
SED assisted in planning a study to compare EMAT and strain gauge measurements of stress. In the comparison study, ultrasound measurements will be made at many locations. However, strain gauge measurements will be made at just a few of these locations. We plan to compare ultrasound estimates with interpolated values of the strain gauge estimates. We developed a preliminary statistical model to predict the variance of the interpolated stress (from strain gauge measurements). Associated standard errors are also predicted.
Figure 19: Amplitude and phase of acoustic echo is measured as the transducer rotates with respect to polarization axes. Predicted amplitude and phase is based on a model which accounts for material inhomogeneity and differential attenuation in steel.
Date created: 7/20/2001
Last updated: 7/20/2001 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9474233388900757, "perplexity": 1665.123484386847}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676592309.94/warc/CC-MAIN-20180721032019-20180721052019-00137.warc.gz"} |
http://physics.stackexchange.com/questions/27023/unitarity-of-s-matrix-in-qft | # Unitarity of S-matrix in QFT
I am a beginner in QFT, and my question is probably very basic.
As far as I understand, usually in QFT, in particular in QED, one postulates existence of IN and OUT states. Unitarity of the S-matrix is also essentially postulated. On the other hand, in more classical and better understood non-relativistic scattering theory unitarity of S-matrix is a non-trivial theorem which is proved under some assumptions on the scattering potential, which are not satisfied automatically in general. For example, unitarity of the S-matrix may be violated if the potential is too strongly attractive at small distances: in that case a particle (or two interacting with each other particles) may approach each other from infinity and form a bound state. (However the Coulomb potential is not enough attractive for this phenomenon.)
The first question is why this cannot happen in the relativistic situation, say in QED. Why electron and positron (or better anti-muon) cannot approach each other from infinity and form a bound state?
As far as I understand, this would contradict the unitarity of S-matrix. On the other hand, in principle S-matrix can be computed, using Feynmann rules, to any order of approximation in the coupling constants. Thus in principle unitarity of S-matrix could be probably checked in this sense to any order.
The second question is whether such a proof, for QED or any other theory, was done anywhere? Is it written somewhere?
-
Why do you say that two particles can't form a bound state in QFT? I'm pretty sure there are two-dimensional integrable field theories with scattering $A+B \to C$ and where $A$, $B$ and $C$ are perfectly stable particle states. – Sidious Lord Feb 26 '12 at 15:36
@Sidious Lord: Can I read somewhere about such examples? Can it happen in QED? (As far as I heard, the 2d case is somewhat exceptional in QED: in the Schwinger model polarization of vacuum has an effect of creation of a bound state of electron-positron pair which is a free boson. But I might be wrong about this, I do not really know this.) – MKO Feb 26 '12 at 18:56
Hi @Dilaton: Concerning the tag edit(v3) I would suggest the unitarity tag and the s-matrix-theory tag instead of the qed tag (because OP is really asking about qft) and the research-level tag (because the question is textbook material). – Qmechanic Jan 1 '13 at 16:27
Thanks @Qmechanic, it never hurts when you hava a look at it too when I retag, since you are much much much more knowledgable. I change the tags as you suggest. And happy new year to you :-) – Dilaton Jan 1 '13 at 17:33
In principle, bound states are possible in a QFT. In this case, their states must be part of the S-matrix in- and out- state space in order that the S-matrix is unitary. (Weinberg, QFT I, p.110)
However, for QED proper (i.e., without any other species of particles apart from photon, electron, and positron) it happens that there are no bound states; electron and positron only form positronium, which is unstable, and decays quickly into two photons. http://en.wikipedia.org/wiki/Positronium
[Edit: Positronium is unstable: http://arxiv.org/abs/hep-ph/0310099 - muonium is stable electromagnetically (i.e., in QED + muon without weak force), but decays via the weak interaction, hence is unstable, too: http://arxiv.org/abs/nucl-ex/0404013. About how to make muonium, see page 3 of this article, or the paper discovering muonium, Phys. Rev. Lett. 5, 63–65 (1960). There is no obstacle in forming the bound state; due to the attraction of unlike charges, an electron is easily captured by an antimuon.]
Note that the current techniques for relativistic QFT do not handle bound states well. Bound states of two particles are (in the simplest approximation) described by Bethe-Salpeter equations. The situation is technically difficult because such bound states always have multiparticle contributions.
-
Unitarity of the S-matrix can be checked perturbatively. Bound states tend to be non-perturbative effects, so may not show up naive perturbative calculations. Unfortunately, the datailed proof is not discussed in many places. One book that has it is Scharf's book on QED. When looking through other books you should look for keywords like optical theorem and Cutkosky rules. Bound states are usefully discussed in the last chapter of vol.1 of Weinberg's tretease on QFT.
-
Problems with proofs in QED and other QFTs are due to wrong coupling term like $jA$ which is not correct alone and is corrected with counterterms. In addition, these counterterms cannot be treated exactly but only perturbatively so the true interaction of true constituents is not seen. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9023042917251587, "perplexity": 781.4120681680848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00541-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.woodlandsciviccentre.com/everything-about-the-photoelectric-effect-paradox/ | Albert Einstein may be famous for his theory of relativity, but it was his research on the photoelectric effect that won him a Nobel Prize for Physics. The photoelectric effect is a core concept of quantum physics, and it has brought about giant leaps in humanity’s understanding of the quantum nature of light and electrons. However, when the effect was first observed and subsequently studied, many physicists at the time referred to this phenomenon as a paradox.
To understand how Einstein resolved this paradox, we must first understand the photoelectric effect and the resulting observations that perplexed many physicists at the time.
What is the photoelectric effect?
The photoelectric effect is the discharge of electrons when electromagnetic radiation hits a material. Heinrich Hertz first discovered this phenomenon when he observed electric currents being produced when ultraviolet light shone on a piece of metal.
After Hertz’s discovery, studies were conducted by physicists on this phenomenon, and they confirmed two observations:
• The energy of the individual photoelectrons grows when the frequency or colour of the light changes. However, no change is observed when the intensity or brightness of the light changes.
• The photoelectric current is determined by the light’s intensity. An increase in light intensity leads to a proportional increase in the number of emitted electrons.
These two observations baffled physicists due to their misunderstanding of light as a wave phenomenon. The energy of a wave phenomenon is dependent on the amplitude of the wave and not its frequency, but this concept does not correspond with the observations made.
Max Planck played an essential role in Einstein resolving this paradox. Planck’s theory that the atoms’ energy can only take on discrete values led Einstein to realise that electromagnetic waves have a particle nature.
Einstein theorised that light is comprised of photons, and the energy of the photons in each quantum of light matches the frequency multiplied by a constant h, which is referred to as Planck’s constant. The photons’ energy calculation is represented by the formula E = hf, where E is the photons’ energy, h is Planck’s constant and f is the light’s frequency.
He also deduced that shining a light source with sufficient energy on metal will cause the discharge of electrons. A photon above the threshold frequency will have sufficient energy to eject an electron, producing the photoelectric effect observed by Hertz. As such, it is the frequency, and not the intensity, that determines the emission of electrons. The kinetic energy of the displaced electron is determined by the formula Kmax = hf – Φ, where h is Planck’s constant and f is the photons’ frequency. Φ represents the work function, which is the minimum energy necessary to displace the electron.
Conclusion
Einstein’s revelation is critical to the foundation of quantum mechanics, which is fundamental to quantum physics. It has changed the way generations of scientists view the properties of electromagnetic radiation.
His discovery has also fueled further research into the area and resulted in another physicist, Louis De Broglie, uncovering that light contains both wave-like and particle-like characteristics. The resolution of the photoelectric effect paradox demonstrated that one discovery has the potential to lead to another. This drive to uncover new things gives us a better understanding of physics and how the universe works.
If you find yourself struggling with this topic, you can consider taking up a tuition class. We offer H2 physics tuitionand the photoelectric phenomenon is part of the core syllabus that we will be covering in our classes. We also offer O level physics tuition as well. If you are interested, you can sign up for our physics tuition classes today!
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https://zbmath.org/?q=ai%3Aburity.ricardo+se%3A00000057 | ×
# zbMATH — the first resource for mathematics
The depth of the Rees algebra of three general binary forms. (English) Zbl 1423.13007
In the paper under review, the authors prove that the Rees algebra of an ideal generated by three general binary forms of same degree $$d \geq 5$$ has depth one. This result is in sharp contrast to recent akin statements regarding the depth of the Rees algebra of an ideal $$I$$ when $$I$$ is an almost complete intersection, (see [J. Hong et al., J. Symb. Comput. 43, No. 4, 275–292 (2008; Zbl 1139.13013); J. Commut. Algebra 5, No. 2, 231–267 (2013; Zbl 1274.13015); M. E. Rossi and I. Swanson Contemp. Math. 331, 313–328 (2003; Zbl 1089.13501); A. Simis and S. O. Tohǎneanu, Collect. Math. 66, No. 1, 1–31 (2015; Zbl 1329.13008)], where it has been proved that the Rees algebra is almost Cohen-Macaulay.
The main tools that are used in the proof are: the Ratliff-Rush filtration and the Huckaba-Marley test. The authors also give a conjecture that implies the main theorem of the paper.
##### MSC:
13A02 Graded rings 13A30 Associated graded rings of ideals (Rees ring, form ring), analytic spread and related topics 13C15 Dimension theory, depth, related commutative rings (catenary, etc.) 13D02 Syzygies, resolutions, complexes and commutative rings 13D40 Hilbert-Samuel and Hilbert-Kunz functions; Poincaré series
Full Text:
##### References:
[1] 1994 [2] Heinzer, W.; Lantz, D.; Shah, K., The Ratliff-Rush ideals in a noetherian ring, Comm. Algebra, 20, 2, 591-622, (1992) · Zbl 0747.13002 [3] Hong, J.; Simis, A.; Vasconcelos, W. V., On the homology of two-dimensional elimination, J. Symb. Comp, 43, 4, 275-292, (2008) · Zbl 1139.13013 [4] Hong, J.; Simis, A.; Vasconcelos, W. V., The equations of almost complete intersections, Bull. Braz. Math. Soc, 43, 2, 171-199, (2012) · Zbl 1260.13021 [5] Hong, J.; Simis, A.; Vasconcelos, W. V., Extremal Rees algebras, J. Comm. Algebra, 5, 2, 231-267, (2013) · Zbl 1274.13015 [6] Huckaba, S.; Marley, T., Hilbert coefficients and the depths of associated graded rings, J. London Math. Soc, 56, 1, 64-76, (1997) · Zbl 0910.13008 [7] Kustin, A.; Polini, C.; Ulrich, B., Rational normal scrolls and the defining equations of Rees algebras, J. Reine Angew. Math, 650, 23-65, (2011) · Zbl 1211.13005 [8] Rossi, M. E.; Swanson, I., Contemporary Mathematics, 331, Notes on the behavior of the Ratliff-Rush filtration, (2003), Providence, RI: American Mathematical Society, Providence, RI [9] Simis, A.; Tohǎneanu, S., The ubiquity of Sylvester forms in almost complete intersections, Collect. Math, 66, 1, 1-31, (2015) · Zbl 1329.13008
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8452163934707642, "perplexity": 2730.318194304777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703513144.48/warc/CC-MAIN-20210117174558-20210117204558-00602.warc.gz"} |
http://talkstats.com/threads/ranking-question.71080/ | # Ranking question
#### leejones15
##### New Member
I have data from a student survey in which they ranked a list of potential difficulties. There were given 13 choices, and had to rank them 1-13, with an option to not rank something if it didn't affect them. The data I have is how many times a phrase got each rank (e.g. "Trouble communicating" got ranked #1 three times, #2 once...) and the percentage each phrase got each rank.
What is the proper way to compile the data into an overall ranking, and what is the proper way to report this on a table? Some non-statistics sources said to get a raw score for each rank by multiplying the weighted rank by the number of times it was given that rank (e.g. #1 three times means 13*3) and adding up the values.
What is the "proper" research way of doing this? Is there a "proper" way?
I'm going to be presenting these results at a poster session at a research conference, so I'd like to make sure I'm on point.
P.S.- If it helps, I use Stata and SPSS
Thanks
#### Karabiner
##### TS Contributor
Since this is ordinal data I'd use the median ranks. Perhaps with
interquartile range, in order to indicate degree of agreement.
The multiplication algorithm makes the heroic assumption that
the differences between ranks are evenly spaced.
With kind regards
Karabiner
Last edited: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8557657599449158, "perplexity": 1677.204288731843}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00530.warc.gz"} |
https://www.bionicturtle.com/forum/threads/merton-model-valuation.23646/ | What's new
# Merton model valuation
#### evelyn.peng
##### Active Member
Hi,
I have a practice question (non-BT) related to the use of Merton's debt valuation model.
An investor has a large position of bonds issued by XYZ Limited. He has hedged these bonds with equity using Merton’s debt valuation model. Suppose the value takes an unprecedented tumble, but the value of equity remains stable: the investor would make a loss.
Consider the following statements:
I. A liquidity crisis, similar to the one experienced in 2008, increased the liquidity component of credit spreads
II. Risk-free rate of interest fell
III. Risk-free rate of interest increased
IV. Volatility fell
V. Volatility increased
Which of the statements above would explain why the investor’s hedge strategy failed?
A. I and V
B. II and IV
C. I, II, and IV
D. III and V ----- Correct answer according to the answer key
My solution:
the investor is long debt, and short equity.
Equity = call option
Debt value = Face Value of Debt - Put on the Asset
if risk free rate increase, it would slightly increase the value of the call/equity. It would decrease the value of the Debt.
If the volatility increased, it would increase the value of the call/equity, and increase the value of the put option, so decrease the value of the debt.
Conversely if risk free rate decreased, it would slightly decrease the value of the call/equity. It would increase the value of the Debt - which contradicts case facts.
Same thing for the volatility decrease, it would decrease the value of the call/equity, and decrease the value of the put option, and increase the value of the debt - again contracting case facts.
Statement #I doesn't factor into anything.
So the only logical answer is III and V. However, I feel like the wording of the question is flawed. The hedge did not fail per se, as the equity and debt valuation were somewhat offsetting each other. Just wondered if anyone else have any thoughts on this question?
Last edited by a moderator: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8158506751060486, "perplexity": 2473.1052653286993}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00176.warc.gz"} |
http://mathhelpforum.com/differential-equations/86169-how-do-i-solve-laplace-transform-print.html | # How do I solve this Laplace Transform?
• April 28th 2009, 04:01 AM
LooNiE
How do I solve this Laplace Transform?
The equation I have is 3(dy/dx) + 4y = 4 +6x + 4x^2
I am given that y = 1 when x = 0.
Basically, I can do the initial transform, but afterwards I get completely cnofused. Any help is appreciated.
• April 28th 2009, 04:11 AM
mr fantastic
Quote:
Originally Posted by LooNiE
The equation I have is 3(dy/dx) + 4y = 4 +6x + 4x^2
I am given that y = 1 when x = 0.
Basically, I can do the initial transform, but afterwards I get completely cnofused. Any help is appreciated.
Please post what you've done and where you got stuck.
• April 28th 2009, 04:46 AM
LooNiE
what i've done
Ok what I have done is take L.T's, which gives me this:
dont know what the symbol here is but its always Y bar
3(sy - 1) +4y = 4/s + 6/s^2 + 8/s^3
So then I get:
y(3s + 4) = 4/s + 6/s^2 + 8/s^3 + 3
I dont know how to proceed from here. I could obviously divide the rand hand side by (3s+4) but im not completely certain if I need to do anything else first.
• April 28th 2009, 04:50 AM
mr fantastic
Quote:
Originally Posted by LooNiE
Ok what I have done is take L.T's, which gives me this:
dont know what the symbol here is but its always Y bar
3(sy - 1) +4y = 4/s + 6/s^2 + 8/s^3
So then I get:
y(3s + 4) = 4/s + 6/s^2 + 8/s^3 + 3
I dont know how to proceed from here. I could obviously divide the rand hand side by (3s+4) but im not completely certain if I need to do anything else first.
Do divide. Then express each term on the right hand side in partial fraction form. Then look up your tables to get the corresponding inverse Laplace transforms.
Check your answer by confirming that it satisfies the DE and confirming that y(0) = 1.
• April 28th 2009, 05:04 AM
LooNiE
If I divide on the right hand side, they are all already in partial fracion form. From my teacher, I know that the answer is y = 1 + x^2, but I dont see how I can get the answer from what I have.
• April 28th 2009, 05:30 AM
mr fantastic
Quote:
Originally Posted by LooNiE
If I divide on the right hand side, they are all already in partial fracion form. From my teacher, I know that the answer is y = 1 + x^2, but I dont see how I can get the answer from what I have.
$\frac{4}{s (3s + 4)}$ etc. are not in partial fraction form. eg. The partial fraction form of $\frac{4}{s (3s + 4)}$ is $\frac{A}{s} + \frac{B}{3s + 4}$ where you have to find the value of A and B. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.841660737991333, "perplexity": 784.6715092614904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398474527.16/warc/CC-MAIN-20151124205434-00102-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://stats.stackexchange.com/questions/515082/ratio-of-two-weighted-sum-of-bernoulli-random-variables | # Ratio of two weighted sum of Bernoulli random variables
Given a vector $$X=(x_1,\dots x_N) \in R_+^N$$ and $$N$$ i.i.d Bernoulli random variable $$A_i$$ with $$P(A_i= 1)=p = 1 -P(A_i = 0)$$
First we define the random variable $$T$$ as $$T = \frac{\sum_{i}x_i * A_i}{\sum_i A_i}$$
If we let $$0/0 = 0$$, then $$T$$ should be an unbiased estimator of the sample mean of $$X$$, that is $$E[T] = \sum_{i} x_i/n$$. I am wondering if there is any closed form formula to express the variance of $$T$$? Any upper bound on the variance in terms of $$N$$ is also helpful here.
Thanks.
### In Short
A variation of this problem has been described by Grab and Savage (1954) 'Tables of the Expected Value of 1/X for Positive Bernoulli and Poisson Variables' JASA Vol. 49, No. 265
The following situation often arises in sampling problems. An observation $$y_i \, (i=1, \cdots x)$$ is made of $$x$$ individuals and the average, $$Y = (y_1 + y_2 + \dots + y_x)/x,$$ is computed.
If the $$y_i$$'s are independent observations on a random variable with mean value $$\mu$$ and variance $$\sigma^2$$, we find that the mean value of $$Y$$ is $$\mu$$. However, if $$x$$, the sample size, is a random variable, then the variance of $$Y$$ is not $$\sigma^2/x$$ but it is $$\sigma^2E(1/x)$$.
### Long
With the definition $$0/0 = 0$$ it becomes a bit difficult because this will give some probability that $$T=0$$. Below I will consider the problem as truncated such that you never have $$\sum A_i =0$$ or $$T=0$$.
Your problem is equivalent to sampling from $$X$$ without repetition, and taking the mean of the sample, where the sample size $$n$$ is determined by a binomial distribution.
If $$X$$ is very large in comparison to the sample size then you can estimate this by sampling with repetition.
In that approximation you can compute the variance by considering the moments conditional on the sample size $$n = \sum A_n$$. The variance scales like $$\frac{Var(X)}{n}$$ and the total variance will be a sum of those variances for different $$n$$ * $$Var(T) \approx Var(X) \sum_{n=1}^N n^{-1}{Pr}\left[\sum A_i = n\right] = Var(T)E(n^{-1})$$
For this expectation of $$n^{-1}$$, I have not an easy solution. Of course, you can simply compute it as a sum, but when $$N$$ becomes large you might want to have some expression to estimate it. I have created a separate question for this: Expectation of negative moment $E[k^{-1}]$ for zero-truncated Poisson distribution
### Simulation
Below is a comparison using the integral expression for $$E(n^{-1})$$ from the other question. I used $$X = \lbrace{1,2,\dots,1000\rbrace}$$. It illustrates that it works for smaller $$p$$. the discrepancy between the computation and simulation at higher $$p$$ is due to the approximation of sampling without repetition by sampling with repetition.
x = 1:1000
n = length(x)
p = 0.01
lambda = n*p
#### compute estimate
compVar <- function() {
var_x = mean(x^2)-mean(x)^2
E_inv_n = exp(-lambda)/(1-exp(-lambda))*(expint::expint_Ei(lambda)-log(lambda)-0.57721)
### result = 94.17509
E_inv_n*var_x
}
### Simulate
sim <- function() {
a <- rbinom(n,1,p) ## Bernoulli variable
result <- sum(a*x)/sum(a)
return(result)
}
simVar <- function() {
n_sim <- 10^4
T_sim <- replicate(n_sim,sim())
return(mean(T_sim^2, na.rm = TRUE)-mean(T_sim, na.rm = TRUE)^2)
}
### compare for different values of p
sim_v <- c()
comp_v <- c()
set.seed(1)
prng <- 10^seq(-2,-0.2,0.2)
for (ps in prng) {
p = ps
lambda = n*p
sim_v <- c(sim_v,simVar())
comp_v <- c(comp_v,compVar())
}
plot(prng,comp_v, log = 'xy',
xlab = expression(P*(A[i] == 0)),
ylab = "variance", type = "l", ylim = c(3*10^1,10^4), xlim = c(0.01,1),
xaxt = "n", yaxt = "n")
points(prng,sim_v, pch = 21 , col = 1, bg = 0)
logaxis <- function(pos,order,tck = -0.005, las = 1) {
axis(pos, at = 10^order, las = las)
for (subord in order[-1]) {
axis(pos, at = 0.1*10^subord*c(1:9),labels=rep("",9), tck = tck)
}
}
logaxis(pos = 1, order = c(-3:0))
logaxis(pos = 2, order = c(1:4), las = 2)
legend(0.01,100,c("simulation","computed"), lty = c(0,1), pch = c(21,NA))
* (note that this simple computation of the variance of the compound distribution works because the mean is the same I every condition) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 41, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9748675227165222, "perplexity": 716.4659498916338}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00093.warc.gz"} |
https://math.stackexchange.com/questions/392249/how-to-understand-cx-bounded-borel-measurable-functions | # How to understand C(X)'' = bounded Borel measurable functions?
Let $X$ be a compact metric space and $C(X)=\{ f:X\rightarrow \mathbb{R} \ | \ \ f \ continuous\}$ with the uniform norm. It is a separable Banach space.
1) I'm aware of the fact that $C(X)^*$, the space of continuous linear functionals $C(X)\rightarrow \mathbb{R}$ coincide with $M(X)$, the space of signed regular borel measures on $X$.
This intuitively makes sense because $\mu\in M(X)$ can be naturally be seen as an evaluation map $f\mapsto \mathbb{R}$ by means of integration.
2) I'm also aware that $(C(X)^*)^*$, the dual of the dual, coincide with the set of bounded Borel measurable functions $F:X\rightarrow\mathbb{R}$ (source: P. Lax, Functional Analysis).
UPDATE Landscape provided a counterexample in the comments. My source was: P. Lax "Functional Analysis" 2002, Page 82, Theorem 14(ii). I guess this might be a mistake, perhaps fixed in some errata somewhere.
New question: C.f. Landscape's example, let $g_A$ be the extension (given by Hahn-Banach) of $f_A$ to $C([0,1])^{**}$. Looking at $g_A$ as a function $X\rightarrow\mathbb{R}$, $g_A$ must satisfy by construction the equation $\int_{[0,1]}g_A \ d \ \delta_x= 1$ if $x\in A$ and $0$ otherwise. This seems to imply that $g_A$ is the characteristic function of $A$. Hence not Borel if $A$ is not Borel.
Thus, $g_A$ as a function $(X\rightarrow\mathbb{R})$ is uniquely determined by the construction starting from $f_A$. But is it $g_A$ as an element of $C([0,1])^{**}$ uniquely determined? If so, it looks to me we would have a reasonably well defined notion of integration for non-measurable sets.
Thanks!
Screen shot of Lax's theorem 14 on page 82:
• Things in $C(X)^{**}$ are things that morally can be integrated with respect to every Borel measure. Does that give you intuition about Q1? May 15 '13 at 8:48
• Mariano thanks for the answer. This is helpful. But also, e.g., analytic functions can be integrated and are not borel. But beside this observation... Why is this the right way to look at C(X)**? After all, we do not look at $C(X)^*$ as things that can be "integrated" with respect to every continuous $f\in C(X)$, don't we? Please bear with my ignorance here. May 15 '13 at 8:56
• Let me also ask, just to be sure. it correct to write that the action of $F\in C(X)^{**}$ on measures is expressed as $F(\mu)=\int F d \mu$, right? May 15 '13 at 8:58
• The assertion in your question 1 is incorrect. Please the comment given by commenter here.
– 23rd
May 15 '13 at 10:20
• Hello Landscape, thanks! this is quite surprising as I think Lex's book is a very solid reference. Regarding the counterexample proposed in the page you linked. To apply Hahn-Banach it is required for $A$ not just to be subset of $[0,1]$ but to be a sub-linear space. Does this make any difference? May 15 '13 at 11:42
Let $X$ be a compact metric space. Let $\mathcal B$ be the collection of Borel sets on $X$. Banach space $C(X)$ is the set of all (necessarily bounded) continuous real-valued functions $f : X \to \mathbb R$, with norm $$\|f\|_\infty = \max\{|f(x)| : x \in X\}$$
Banach space $M(X)$ is the set of all countably-additive signed ($\mathbb R$-valued) measures on the sigma-algebra $\mathcal B$. The norm is the total variation: for $\mu \in M(X)$, write $\mu = \mu^+ - \mu^{-}$ where $\mu^+$ and $\mu^-$ are positive measures, singular to each other, and let $$\|\mu\|_1 = \mu^+(X)+\mu^-(X)$$
The pairing $M(X) \times C(X) \to \mathbb R$ defined by $$\langle\mu,f\rangle = \int_X f \;d\mu$$ identifies $C(X)^\ast$ isometrically with $M(X)$ in these norms.
Another description of $M(X)$ may be obtained as follows. Let $\{\tau_i : i \in I\}$ be a maximal (under inclusion) family of mutually singular probability measures on $(X,\mathcal B)$. (Such a family exists by Zorn's Lemma. But it is not unique.) Identify $M(X)$ isometrically with the $l^1$-direct sum of the family of all the $L^1(\tau_i)$ spaces: $$M(X) \approx \left(\bigoplus_{i\in I} L^1(X,\mathcal B,\tau_i)\right)_{1}$$
Once we describe $M(X)$ in this way, we get the corresponding description of the dual as the $l^\infty$-sum of the spaces $L^\infty(\tau_i)$. $$C(X)^{\ast\ast}\approx \left(\bigoplus_{i\in I} L^\infty(X,\mathcal B,\tau_i)\right)_{\infty}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9220346808433533, "perplexity": 194.04420844467208}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00067.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=46&t=69429&p=281806 | ## Textbook Question 9C:5
SLai_1I
Posts: 93
Joined: Wed Sep 30, 2020 9:52 pm
### Textbook Question 9C:5
I was going through the textbook problems for the coordination compounds section and had trouble with problems 5b. and 5d. How can CO3 2- sometimes be bidentate, and how is oxalate bidentate? Please help if you can!
Mahnoor_Wani_1I
Posts: 118
Joined: Wed Sep 30, 2020 9:35 pm
Been upvoted: 3 times
### Re: Textbook Question 9C:5
When you draw out the lewis structure of CO3 2- you can see that there are two lewis structures. One structure has one CO double bond and two single CO bond. In this structure the Oxygen is capable of interacting with a transitional metal at two different places so it is classified as bidentate. The other structure has two CO double bond and one single CO bond that is capable of interacting with TM so it is classified as just monodentate(not sure what it is called).
Joseph Hsing 2C
Posts: 85
Joined: Wed Sep 30, 2020 9:42 pm
### Re: Textbook Question 9C:5
To figure out if a ligand is polydentate, look at the molecular structure and look if there are atoms like carbon between the donor atoms. Spacing is the key factor. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.859981894493103, "perplexity": 3440.8811034908963}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178359082.48/warc/CC-MAIN-20210227174711-20210227204711-00299.warc.gz"} |
https://www.physicsforums.com/threads/proof-of-irrationality-of-sqrt-2.817396/ | # Proof of irrationality of sqrt(2)
1. Jun 4, 2015
### ViolentCorpse
Hi,
There's something I don't understand in the popular "proof by contradiction" of sqrt(2) after the following step:
a2/b2 = 2
a2=2b2
The above equation implies that a2 is even. Fair enough. But the way I see it, the above equation also makes it impossible for a to be even, as
a=sqrt(2)*b.
This implies that a should actually be an irrational number as it is a multiple of sqrt(2). Though it seems obvious to me that if some number n^2 is even, then n must also be even, but in the context of the equation a2=2b2, it doesn't seem possible.
Shouldn't this be the end of the proof? We started by assuming that a and b are whole numbers and a/b can be used to represent sqrt(2), but found that a, at least, is actually irrational. Doesn't that count as a contradiction?
I'm really confused. :(
2. Jun 4, 2015
### micromass
It does, but you don't know that a priori. At this stage of the proof, you don't know yet that $\sqrt{2}$ is irrational.
3. Jun 4, 2015
### ViolentCorpse
Ohhhhh! I'm an utter idiot! I can't help laughing at myself now. Literally. :p
Thank you so much, micromass! I really appreciate it!
4. Jun 4, 2015
### ViolentCorpse
One another thing I would like to ask. When we say that we are assuming an irreducible fraction, are we just saying it? I'm trying to say that our math should be aware of the assumptions we are making so when we write down the premise of the proof, we should be translating that assumption into mathematical form.
How does the math know that a/b is not supposed to be a reducible fraction?
5. Jun 4, 2015
### bhillyard
The irreducibility of a/b is assumed by the prover. At a later stage in the proof we show that a and b must have a common factor - there lies the contadiction that is the heart of the proof.
6. Jun 4, 2015
### Svein
The proof goes like this: Assume that $(\frac{a}{b})^{2} = 2$ where a and b are integers (and b≠0). We can also assume that the fraction is irreducible (a and b have no common factors). Then $a^{2}=2\cdot b^{2}$, which implies that a2 is even. But the only way a2 can be even is that a is even. Therefore you can write a = 2⋅p, where p is an integer. This gives $(\frac{2p}{b})^{2}=2$ or $4\cdot p^{2} = 2\cdot b^{2}$ which gives $b^{2}=2\cdot p^{2}$ which implies that b2 is even. But the only way b2 can be even is that b is even. Therefore you can write b = 2⋅q, where q is an integer. But this again says that $\frac{a}{b}=\frac{2p}{2q}$ which means that a and b have a common factor (2) which is contrary to the assumption.
7. Jun 4, 2015
### ViolentCorpse
Thanks I understand pretty much all of the proof now except the use of assumption which is troubling me.
The first step of the proof is:
sqrt(2) = a/b.
Then we just tell ourselves that a/b is an irreducible fraction. There should be a way of letting the math know precisely what assumptions we are making because the statement (a/b)^2 = 2 could mean anything, so if it gives us a fraction with common factors, it's not really a contradiction because we only made that assumption verbally.
I'm sorry if I sound obnoxious. I'm just a bit of a dimwit.
8. Jun 4, 2015
### MrAnchovy
No we don't, we make sure it is an irreducible fraction. How? You have given the answer yourself...
then what do you think you should do?
9. Jun 4, 2015
### ViolentCorpse
I'm sorry I don't get it. How do we make sure it is an irreducible fraction?
10. Jun 5, 2015
### Svein
Basis knowledge of fractions: If there is a common factor in the numerator and denominator, we can remove it in both places (if the common factor is c, we have $\frac{a}{b}=\frac{c\cdot p}{c\cdot q}= \frac{c}{c}\cdot\frac{p}{q}=1\cdot \frac{p}{q}$.
11. Jun 5, 2015
### ViolentCorpse
I think the thing I'm having trouble grasping is how we are to make sure that a/b is already in its reduced form when we start the proof. Sure we could cancel it out later (even a and b can be reduced when we find they are both even), but I think it is important to the proof that a/b be already written in its reduced form. For example, if I write x = p/q, does that guarantee that I will get a value of x that is already reduced to it lowest form?
12. Jun 5, 2015
### Svein
No. But you know that if it is not, you can reduce it.
You have to make some assumption when it comes to fractions, otherwise you will have to deal with an infinity of fractions all representing the same number. For example 1 = n/n for all n.
13. Jun 5, 2015
### ViolentCorpse
Right. Thank you very much!
14. Jun 5, 2015
### Staff: Mentor
Well, maybe there's one number for which this isn't true... All is a very general term.
15. Jun 5, 2015
### Svein
Yes. I was a bit sloppy there. I should have said (∀n∈ℕ), but that is a bit unreadable. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9644028544425964, "perplexity": 455.64060143977946}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591578.1/warc/CC-MAIN-20180720100114-20180720120114-00439.warc.gz"} |
http://texblog.net/latex-archive/uncategorized/matrix-align-left-right/ | ## Matrices with alignment
The entries of matrix columns are centered by default. Let’s see an example:
$\begin{pmatrix} 1 & 2 & 1 \\ 0 & -2 & -3 \\ 0 & 3 & -2 \end{pmatrix}$
Output:
For right alignment you could use an array environment instead of pmatrix. I will show a possibility to get left, right or centered alignment by redefining the internal amsmath macro \env@matrix. Its original definition in version 2.13 is:
\def\env@matrix{\hskip -\arraycolsep \let\@ifnextchar\new@ifnextchar \array{*\c@MaxMatrixCols c}}
I introduce an optional parameter, its default value is c:
\makeatletter \renewcommand*\env@matrix[1][c]{\hskip -\arraycolsep \let\@ifnextchar\new@ifnextchar \array{*\c@MaxMatrixCols #1}} \makeatother
Now, if you write \begin{pmatrix}[r] … above you will get:
This topic was discussed in the Matheplanet Forum.
05. May 2008 by stefan
1. Thanks for this — I wish that it were in amslatex already!
2. You can also use the pmatrix* environment provided by the mathtools package like this:
begin{pmatrix*}[r]…
3. By a chance I found a small piece of LaTeX code that allows horizontal alignment in matrix-environments of the package “amsmath” (bmatrix, pmatrix, etc.) for each column separately. If you put the following lines in the preambule:
\usepackage{amsmath}
\makeatletter
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
\hskip -\arraycolsep
\let\@ifnextchar\new@ifnextchar
\array{#1}}
\makeatother %%% for getting a separeting vertical line in bmatrix
then you can write things like this in the LaTeX document:
$$[A|I]= \begin{bmatrix}[rrr|rrr] 1 & 5 & 1 & -1 & 0 & 0\\ 2 & 5 & 0 & 0 & 12 & 0 \\ 2 & 7 & 1 & 0 & 0 & 1 \\ \end{bmatrix}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9820745587348938, "perplexity": 3777.1664529124164}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945082.84/warc/CC-MAIN-20180421071203-20180421091203-00562.warc.gz"} |
https://physics.stackexchange.com/questions/52436/compton-scattering/52448 | # Compton Scattering
Compton Scattering essentially states that when a photon of a given wavelength hits an electron the energy level of the electron changes and the photon has its wavelength changed. This seems to be implying that it is the same photon that is reflected outwards. Do we know it is the same photon of a changed wavelength, or is it possible that the original photon merely pushed a 2nd photon out of the electron? (This would mean the original photon is not changing wavelength, that is meerly absorbed).
• I don't think the electron "absorbs" the full photon. I'm not certain, but I think the excess energy is "reflected" from the photon rather than "absorbed" and "re-emitted" by the electron. – Kenshin Jan 29 '13 at 2:58
• There is not really anything like the "same" photon. – Gunnish Jan 29 '13 at 11:54
It is my understanding that we cannot tell which of the two scenarios takes place, moreover, it does not matter, as photons having the same characteristics (such as momentum, energy, and polarization) are indistinguishable.
• This is not correct. There is a finite amount of time between absorption and re-emission of the photon---allowing the two possibilities to be distinguished. See my answer. – DilithiumMatrix Jan 29 '13 at 3:31
• Is this your personal opinion or could you give a reference? – akhmeteli Jan 29 '13 at 4:24
It is conventional to describe the out-going photon as a "different" particle. Basically because photons have only one independent property: their wavenumber ($\vec{k}$) and that is how we label them. Further as they experience no time there is not opportunity for them to change it.
This is consistent with quantum field theory where a process like this will be written with a destruction operator on the in-coming wavenumber and a creation operator on the out-going wavenumber.
A more detailed explanation of the interaction with QED The photon is absorbed by the electron, the electron is excited, the electron decays, a new photon is emitted. There is some finite, non-zero time between the absorption and emission of the photons. The process is illustrated in the following Feynman diagram from here:
• Yes this is what I thought. That absorption and re-emission would occur with a time delay. So what really happens in the scenario given by the asker? Does the photon reflect, or is it absorbed and re-emitted? – Kenshin Jan 29 '13 at 3:50
• The photon is absorbed and re-emitted – DilithiumMatrix Jan 29 '13 at 4:37
• How do you know this is the case. Do you have any references to experimental evidence confirming the time delay that we would expect from the absorption/re-emission case? – Kenshin Jan 29 '13 at 4:45
• You may be taking the Feynman diagrams too seriously. Recall that they stand in for terms in a perturbation series, and look at the labeling of the external line on those two diagrams: though they are drawn confusingly, one corresponds to the out-going photon being emitted before the in-coming photon is absorbed. The electron is not "excited" and it does not "decay", it is "off-shell" on the internal line (experimental evidence of excited lepton states would win the Nobel prize in a heartbeat). – dmckee --- ex-moderator kitten Jan 29 '13 at 16:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8576847910881042, "perplexity": 460.5760127470245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00376.warc.gz"} |
https://www.physicsforums.com/threads/simple-rlc-circuit.125132/ | # Simple RLC Circuit
1. Jul 3, 2006
Question:
A parallel RLC circuit, which is driven by a variable frequency 10-A source, has the following parameters:
$$R=500\Omega$$
$$L=0.5mH$$
$$C=20\muF$$
Find the resonant frequency, the $Q$, the average power dissipated at resonant frequency, the $BW$, and the average power dissipated at the half-power frequencies.
$Q$, $\omega_0$, and $BW$ are all straightfoward calculations.
$$Q = 100 [/itex] [tex] BW = 100 \frac{rad}{sec}$$
$$\omega_0 = 10000\frac{rad}{sec}$$
The half power frequencies are:
$$\omega_{hi} = 1005.01$$
[tex] \omega_{lo} = 995.01 [/itex]
I don't understand how to calculate the average power dissapated at resonant frequency, OR at the half power frequencies. If someone could give me a push in the right direction, that would be swell
Last edited: Jul 3, 2006
2. Jul 3, 2006
nm, figured it out. That was embarrassingly easy.
Similar Discussions: Simple RLC Circuit | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9282692074775696, "perplexity": 1594.636816594848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689615.28/warc/CC-MAIN-20170923085617-20170923105617-00031.warc.gz"} |
http://www.nanoscalereslett.com/content/6/1/183 | Nano Review
Numerical study of instability of nanofluids: the coagulation effect and sedimentation effect
Yu Ni, JianRen Fan* and YaCai Hu
Author Affiliations
State Key Laboratory of Clean Energy Utilization, Zhejiang University, Hangzhou 310027, P. R. China
For all author emails, please log on.
Nanoscale Research Letters 2011, 6:183 doi:10.1186/1556-276X-6-183
Received: 26 September 2010 Accepted: 28 February 2011 Published: 28 February 2011
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
This study is a numerical study on the coagulation as well as the sedimentation effect of nanofluids using the Brownian dynamics method. Three cases are simulated, focusing on the effects of the sizes, volume fraction, and ζ potentials of nano-particles on the formation of coagulation and sedimentation of nanofluids. The rms fluctuation of the particle number concentration, as well as the flatness factor of it, is employed to study the formation and variation of the coagulation process. The results indicate a superposition of coagulation and sedimentation effect of small nano-particles. Moreover, it is stable of nanofluids with the volume fraction of particles below the limit of "resolution" of the fluids. In addition, the effect of ζ potentials is against the formation of coagulation and positive to the stability of nanofluids.
Introduction
The nanofluid is characterized by the fluid with nanometer-sized solid particles dispersed in solution [1], which can increase the heat transfer coefficient [2-6], enhance the critical heat flux in boiling heat transfer [7-9], reduce the wall friction force [10], improve the optical characteristics [11], etc. Nano-sized particles are utilized because of its better stability than the suspension of micro-sized particles. For a badly stable suspension, sedimentation or coagulation (agglomeration) may occur. It compromises the above-mentioned advantages of the nano-suspension.
As is well known [12], the occurrences of coagulation and sedimentation are the two main factors for the instability of nanofluid. The phenomenon of coagulation is characterized by the formation of particle clusters, i.e., particles are in contact with each other and the cohesion takes place. Then, the clusters grow up. Many researchers investigated the coagulation effect of particles by the Brownian dynamics simulation, focusing on the formation of gelation [13], coagulation rates [14], particle network [15], etc. For example, Hütter [14] identified the characteristic coagulation time scales in colloidal suspensions, and measured their dependencies on the solid content and potential interaction parameters. He also deduced different cluster-cluster bonding mechanisms in the presence of an energy barrier, etc. Besides, the sedimentation always occurs after a big particle cluster is established, i.e., the particles within the cluster sediment flow downward because of the increased effect of the gravity of the cluster over the buoyancy force of it, and reduced the effect of Brownian motion to the big cluster. Many researches were devoted to the sedimentation [16-18] using the Brownian dynamics simulation too. For example, Soppe and Jannsen [17] studied the sediment formation of colloidal particle by a process of irreversible single-particle accretion. They used the algorithm of Ermak and McCammon, incorporating the inter-particle forces and hydrodynamic interaction on the two-particle level, and analyzed the effect of two-particle hydrodynamic interactions on the sediment structure, etc. They found that the process of sediment formation by colloidal particle is the result of a delicate balance of sediment field strength, DLVO interactions, and hydrodynamic interactions.
However, there is an important issue about which few researches have been concerned: the interaction between the coagulation and sedimentation for the instability of nanofluids. For example, the processing of coagulation causes the particle clusters to grow up, and then the large clusters are more prone to sedimentation than that of small clusters because of the intensive gravity effect. In other words, the coagulation effect is able to augment the sedimentation effect. Thus, this study is intended to carry out some research on this issue, exploring the complex interaction as well as the close relation between the coagulation and the sedimentation phenomena.
Numerical model
Governing equation
In this study, the Brownian dynamics technique is employed to investigate the motion of nanoparticles. The governing equation is the so-called Langevin equation, which is formulated as follows [19]:
(1)
where the superscript 0 indicates that the variable is corresponding to the beginning of the time step Δt; ri is the ith component of the position vector of particle; Dij is the element of the diffusion tensor indexed by (i, j); Fj is the force experienced by the jth particle; kB is the Boltzmann constant; and T denotes the temperature. The displacement Rit) is a random displacement of a Gaussian distribution with a zero expectation and a () covariance (). In this study, the Rotne-Prager tensor [20] is utilized as approximations to the hydrodynamic interaction:
(2)
where η is the viscosity, a is the particle radius, δij is the Kronecker delta, is the vector from the center of particle i to the center of particle j, and is the unit tensor.
Moreover, three forces, i.e., the attractive Van der Walls force, fv, the repulsive electrostatic force by the electric double layer, fe, and the gravity force fg are considered as the forces experienced by any particle, which are formulated as follows [12]:
(3)
(4)
(5)
(6)
where A, d, ε, κ, ζ, ρf, ρ, and g are the Hamaker constant, the particle diameter, the electric permittivity of the fluid, the inverse of the double-layer thickness, the zeta potential of the suspension, the density of the fluid, the density of the particle, and the gravity acceleration, respectively.
It is noted in Equation (4) that the results for rij - d = 0 is meaningless when the contact between the two particles occur, and they will adhere to each other or rebound back. Thus, we treat the condition with as the situation when the two particles are separated, so that Equation (4) works. Otherwise, it results in coalescence between the two particles. Once the coalescence between colliding particles takes place, the clusters start growing up.
Simulation conditions
In this study, three cases with different diameters of particles (Case 1), different volume fractions (Case 2), and different zeta potentials (Case 3) are simulated, respectively (Table 1). For these cases, the parameters of the material, as well as other relevant parameters, are illustrated in Table 2.
Table 1. Three cases with different diameters of particles, volume fractions, and zeta potentials
Table 2. Parameters used in this simulation
In this simulation, the boundary conditions in the x and y directions (Table 2) in the horizontal plane are both periodic, whereas the top and bottom walls of the simulation domain in the z-direction are treated as adhesive walls to which the particles adhere immediately once they come into contact with them. It is reasonable to conclude thus, since the agglomerated particle clusters always adhere to the bottom walls or the top interfaces.
Initially, a random distribution is given to the particles. As time advances, the possible movements of particles are computed through solution of the governing equations.
Simulation results
Case 1: effect of particle sizes
This section deals with the effect of particle sizes on the coagulation and sedimentation. Figure 1 shows the simulation results at t = 0, 5, 10, and 50 μs for Case 1 due to the effects of different sizes of particles. Figure 1a, b, c, d shows the results of d = 10 nm at t = 0, 5, 10, and 50 μs, respectively. Similarly, Figure 1e, f, g, h shows the results for d = 25 nm, whereas Figure 1i, j, k, l shows them for d = 50 nm. It is seen that the coagulation takes place the most intensively and rapidly for the smallest size of particles (Figure 1a, b, c, d), moderately for the intermediate size of particles (Figure 1e, f, g, h), and weakly and slowly for the largest size of particles (Figure 1e, f, g, h, i, j, k, l). More importantly, the results of the intermediate sizes are due to coagulation but with weak sedimentation, whereas the results of the smallest sizes are due to both the effects of coagulation and sedimentation. For the largest size of the particles, it is neither due to coagulation nor sedimentation. It looks complicated. As is known, the larger particles bear the major effect of gravity, and they are the most prone to sedimentation. However, it is only true of the single particle without coagulation. With the superposition of the coagulation effect, it can amplify or augment the trend of sedimentation through coagulation. Owing to the increasing agglomeration of the particles, the gravity effect may play an important and even a dominant role, which causes possible sedimentation of the whole agglomeration (the upper part of the agglomeration in Figure 1d is due to the adhesive boundary on the upper wall). In other words, there exists a balance between the sedimentation effect of the large-sized individual particles and the sedimentation effect of small-sized aggregated particles. The former is caused solely by the gravity effect, whereas the latter is caused by the superposition of the coagulation and the gravity effects, i.e., the amplification and augmentation of the gravity effects of the aggregated particles due to coagulation.
Figure 1. Snapshots of simulation results. (Case 1 for d = 10 nm (a-d), d = 25 nm (e-h), d = 50 nm (i-l) at t = 0, 5, 10, and 50 μs respectively).
It is necessary to mention that Figure 1i, j, k, l does not indicate the stability of the nanofluids. Alternatively, it indicates a relatively stable characteristic compared to Figure 1a, b, c, d. After the evolution over a long time, possible coagulation or sedimentation may also occur.
In order quantify the degree of coagulation, we need to define some functions. Let us divide the simulation domain Lx × Ly × Lz by Nx × Ny × Nz cubic meshes by the cell volume (δx × δy × δz). The mean number concentration of particles is the mean number of particles within each mesh volume (δx × δy × δz). Then, the rms value of the particle concentration R1 and the flatness factor of the number concentration R4 are formulated as follows:
(7)
(8)
The rms of concentration means the fluctuation of the number concentration of particles, and it is closely related to the formation of particle clusters due to coagulation. The flatness factor means the intensity of fluctuation of the number concentration, thereby indicating the intensity of coagulation. Thus, these two functions are helpful in enabling the quantification of particle coagulation.
Figure 2a, b shows the R1 and R4 for Case 1. It is seen from Figure 2a, b that the coagulation of the small particles is the fastest. They become almost totally coagulated immediately even at the beginning. Comparatively, the coagulation of the larger particles takes place slowly and increasing steadily. However, the final level of coagulation of the larger particles is greater than that of smaller particles. In addition, the degree as well as the rapidity of the coagulation of the intermediate particles is intermediate between that of the smaller and the larger particles.
Figure 2. The flatness factor of the concentration distribution of nanoparticles. (The R1(a) and R4(b) for Case 1).
Case 2: effects of volume fractions
In this section, the effect of volume fraction, i.e., the concentration of particles, is studied. As aforementioned, the smaller particles are more prone to coagulate than the larger particles, under the same condition of the volume fractions. However, the process of coagulation is also closely related to the number of particles contained in it.
For example, Figure 3a, b, c, d shows one of the results of Case 2 where only np = 400 particles are simulated. Compared to Figure 1a, b, c, d, it is seen that the coagulation does not takes place at t = 50 μs. It says that the coagulation is to be regarded appropriately as a process with an excess of particle content. When the particle numbers go beyond the superior limit of the resolvent, then the coagulation will certainly take place. Thus, when the np = 4200 particles are simulated, more intensive coagulations are observed correspondingly (Figure 3e, f, g, h).
Figure 3. Snapshots of simulation results. (Case 2 for np = 400 (a-d) and np = 4200 (e-h) at t = 0, 5, 10, and 50 μs respectively).
In addition, Figure 4a, b shows the R1 and R4 of Case 2. It is seen from Figure 4 that the R1 and R4 for np = 400 are always relatively of small value, indicating a stable status almost without coagulation and sedimentation, although R4 is slightly fluctuated when t < 0.06. Moreover, compared to np = 1200, it is seen that the concentration fluctuation R1 and flatness factor of concentration R4 for np = 400 are relatively of lower values. It validates the conclusions derived from the observation of Figure 3a, b, c, d, in comparison with Figure 1e, f, g, h.
Figure 4. The flatness factor of the concentration distribution of nanoparticles. (The R1(a) and R4(b) for Case 2).
With the increased number of particles, it is seen that the R1 and R4 are increased too (np = 1200 and 2100, respectively, Figure 4). However, when the particle number is extremely large, all the spaces are almost stuffed with particles, leading to a homogeneous distribution and a low fluctuation in the number concentration (np = 4200, in Figure 4).
Case 3: effects of ζ potentials
The previous sections showed the results with ζ = 0 eV. As seen from Equation (5), no repulsive effect has been considered between the particles since fe = 0. Thus, this section will focus on the effect of the repulsive effect by varying the ζ potentials.
Comparing with Figure 1e, f, g, h, it is seen that the degree of coagulation with ζ = 0.01 eV is attenuated (Figure 5a, b, c, d), and it almost disappears with ζ = 0.05 eV (Figure 5e, f, g, h). It indicates that the repulsive effect induced by the ζ potentials is beneficial to the stability of nanofluids, since it acts against the coagulation process.
Figure 5. Snapshots of simulation results. (Case 3 for ζ = 0.01 eV (a-d) and ζ = 0.05 eV (e-h) at t = 0, 5, 10, and 50 μs, respectively).
This conclusion is also validated by the variations of R1 and R4 (Figure 6a, b, respectively). It is seen that the fluctuation of the number concentration of particles with ζ = 0.05 eV increases much more slowly than the cases with smaller ζ. Although the R1 and R4 for ζ = 0.05 eV are still increasing with time, which indicates that the coagulation process may still take place after a fairly long time, their effects against the formation of coagulation are very clear. In other words, it is positively beneficial for the stability of nanofluids.
Figure 6. The flatness factor of the concentration distribution of nanoparticles. (The R1(a) and R4(b) for Case 3).
Conclusions
The findings of this study are briefly summed up as follows:
1. A complicated superposition of the coagulation and sedimentation effects for small particle is observed. The mechanisms of sedimentation for the larger and the smaller particles are different. The former is caused mainly by the great gravity effect of any individual particle, whereas the latter is mainly due to the coagulation process, and the superposition of coagulation causes the sedimentation of the whole agglomeration of particles.
2. There exists a superior limit of the fluid for particle content. When the volume fraction is below the limit, it is hard for the coagulation to occur. In contrast, the coagulation will certainly take place when the concentration of nanoparticles is beyond the capacity of "resolution" of the fluids.
3. The effect of ζ potentials is beneficial for the stability of nanofluid, since it resists the formation of coagulation. In other words, increase in the value of ζ potentials is helpful to make the nanofluid more stable.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
All authors contributed equally.
Acknowledgements
This study is supported by the National High Technology Research and Development of China 863 Program (2007AA05Z254), for which the authors are grateful.
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J Chem Phys 1969, 50:4831. Publisher Full Text | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8434062600135803, "perplexity": 1585.0981835856267}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802768982.112/warc/CC-MAIN-20141217075248-00078-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/find-the-unit-vector-perpendicular-to-a-and-b.171066/ | # Find the unit vector perpendicular to a and b
1. May 20, 2007
### anantchowdhary
1. The problem statement, all variables and given/known data
there are two vectors a and b
a=3i+2j+k
b=2i+j+2k
Find the unit vector perpendicular to a and b
2. Relevant equations
| A|^a=A
3. The attempt at a solution
I dont kno how to start
2. May 20, 2007
### neurocomp2003
go to mathworld.com and find what it means to be a perpendicular vector to
2 other vectors.
3. May 20, 2007
### cristo
Staff Emeritus
Well, calling the third vector c=(l,m,n), you should be able to find 3 equations. They will come from the facts that (i) a is perpendicular to c, (ii) b is perpendicular to c, and (iii) c is a unit vector.
4. May 20, 2007
### anantchowdhary
i tried to use the dot product and cross product but i still couldnt get the answer
5. May 20, 2007
### cristo
Staff Emeritus
Well what did you try?
6. May 20, 2007
### malawi_glenn
Do you know the properties of vector (cross) product ?
and this is precalculus math, not physics
7. May 22, 2007
### genius_me
try using crossproduct,
a x b = resultant(r)
divide a x b by the magnitude of r,
u will get ur unit vector
8. Jun 26, 2007
### krateesh
ya axb..will give us vector whose unit vector is required
9. Jun 26, 2007
### anantchowdhary
yea..thnx....gt the answer long ago! thnx nywyz
help much appreciated
Similar Discussions: Find the unit vector perpendicular to a and b | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8952285051345825, "perplexity": 3203.0727168193152}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886110485.9/warc/CC-MAIN-20170822065702-20170822085702-00174.warc.gz"} |
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