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http://mathhelpforum.com/advanced-statistics/204383-solving-probabilities-involves-percentage.html | # Math Help - Solving probabilities that involves percentage.
1. ## Solving probabilities that involves percentage.
I'm weak when it comes to solving probability questions that has to do with percentage. please help me.
1. Muscular dystrophy is a genetically controlled disease. The probability of having this disease in any new born of a specific parents is 25%. If the couple has 5 children, find the probabilities that:
(a) 5 will have a muscular dystrophy
(b) Fewer than 2 will have a muscular dystrophy
2. If 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts:
(a) 1 is defective
(b) None of the bolts is defective
At most 2 bolts will be defective
tq
2. ## Re: Solving probabilities that involves percentage.
Originally Posted by fara
I'm weak when it comes to solving probability questions that has to do with percentage. please help me.
1. Muscular dystrophy is a genetically controlled disease. The probability of having this disease in any new born of a specific parents is 25%. If the couple has 5 children, find the probabilities that:
(a) 5 will have a muscular dystrophy
(b) Fewer than 2 will have a muscular dystrophy
2. If 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts:
(a) 1 is defective
(b) None of the bolts is defective
At most 2 bolts will be defective
tq
a) Each child has a 1/4 chance of getting the disease, so the probability of all of them having the disease is 1/4 x 1/4 x 1/4 x 1/4 x 1/4 = (1/4)^5.
3. ## Re: Solving probabilities that involves percentage.
Originally Posted by fara
1. Muscular dystrophy is a genetically controlled disease. The probability of having this disease in any new born of a specific parents is 25%. If the couple has 5 children, find the probabilities that:
(a) 5 will have a muscular dystrophy
(b) Fewer than 2 will have a muscular dystrophy
2. If 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts:
(a) 1 is defective
(b) None of the bolts is defective
At most 2 bolts will be defective
tq
HINTS:
1b) 0.63281
2a) 0.4096
2b) 0.4096
2c) 0.9728
4. ## Re: Solving probabilities that involves percentage.
Originally Posted by MaxJasper
HINTS:
1b) 0.63281
2a) 0.4096
2b) 0.4096
2c) 0.9728
Those aren't hints, those are answers. Answers which the OP most likely already has if he/she looks in the back of the book for.
A hint is something which gives the OP some direction to solve the problem him/herself. Such as for 1.b) you will need to use the Binomial Distribution. | {"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": 0.8279657959938049, "perplexity": 1582.3563285333723}, "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-2015-27/segments/1435375097512.42/warc/CC-MAIN-20150627031817-00256-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://hal.inria.fr/inria-00509219 | # On the decoding of binary cyclic codes with the Newton's identities
3 SALSA - Solvers for Algebraic Systems and Applications
LIP6 - Laboratoire d'Informatique de Paris 6, Inria Paris-Rocquencourt
Abstract : We revisit in this paper the concept of decoding binary cyclic codes with Gröbner bases. These ideas were first introduced by Cooper, then Chen, Reed, Helleseth and Truong, and eventually by Orsini and Sala. We discuss here another way of putting the decoding problem into equations: the Newton's identities. Although these identities have been extensively used for decoding, the work was done manually, to provide formulas for the coefficients of the locator polynomial. This was achieved by Reed, Chen, Truong and others in a long series of papers, for decoding quadratic residue codes, on a case-by-case basis. It is tempting to automate these computations, using elimination theory and Gröbner bases. Thus, we study in this paper the properties of the system defined by the Newton's identities, for decoding binary cyclic codes. This is done in two steps, first we prove some facts about the variety associated to this system, then we prove that the ideal itself contains relevant equations for decoding, which lead to formulas. Then we consider the so-called online Gröbner bases decoding, where the work of computing a Gröbner basis is done for each received word. It is much more efficient for practical purposes than preprocessing and substituting into the formulas. Finally, we conclude with some computational results, for codes of interesting length (about one hundred).
Document type :
Journal articles
Domain :
Cited literature [43 references]
https://hal.inria.fr/inria-00509219
Contributor : Daniel Augot <>
Submitted on : Tuesday, August 10, 2010 - 6:27:09 PM
Last modification on : Tuesday, May 14, 2019 - 10:53:26 AM
Long-term archiving on: Friday, November 12, 2010 - 11:46:08 AM
### Files
gbdecode-revised.pdf
Files produced by the author(s)
### Citation
Daniel Augot, Magali Bardet, Jean-Charles Faugère. On the decoding of binary cyclic codes with the Newton's identities. Journal of Symbolic Computation, Elsevier, 2009, Gröbner Bases in Cryptography, Coding Theory, and Algebraic Combinatorics, 44 (12), pp.1608-1625. ⟨10.1016/j.jsc.2008.02.006⟩. ⟨inria-00509219⟩
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http://mathoverflow.net/questions/35140/interesting-applications-in-pure-mathematics-of-first-year-calculus/35157 | # Interesting applications (in pure mathematics) of first-year calculus
What interesting applications are there for theorems or other results studied in first-year calculus courses?
A good example for such an application would be using a calculus theorem to prove a result in group theory. On the other hand, the importance of calculus in applied mathematics or in physics is well known, therefore is not a good example.
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I presume applications in pdes and differential geometry do not qualify as interesting as well? – Deane Yang Aug 10 '10 at 16:57
@ Deane Yang: You presume right :) – Pandora Aug 10 '10 at 17:04
Right, surely she means "trivial to demonstrate", which seems to be true. – Pete L. Clark Aug 10 '10 at 21:55
I think there's a fundamental question that needs to be addressed before this question can be answered:What do we mean by "first year calculus"? This varies a lot-from highly theoretical honors courses like Spivak to plug and chug courses like Stewart's.What's meant by first year calculus in general? – Andrew L Aug 12 '10 at 2:56
@Irene Ok,but I think this has to come with a disclaimer in that case. Spivak is NOT your typical first year calculus course for typical undergraduates in the U.S. – Andrew L Aug 12 '10 at 22:54
You're looking for something fun for a calculus course? If a rectangle $R$ is tiled by rectangles, each of which has a side with integer length, then $R$ has a side with integer length. This is from
Wagon, Stan. Fourteen proofs of a result about tiling a rectangle. Amer. Math. Monthly 94 (1987), no. 7, 601--617. MR935845
and one of those fourteen proofs goes by a double integral.
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Following on from the Galois theory example of Johannes, one straightforward way to produce an explicit polynomial with non-soluble Galois group over ${\mathbb Q}$ is to use an irreducible quintic with exactly three real roots, which necessarily has Galois group $S_5$. To check that an explicit polynomial (such as $x^5-4x+2$ if I am not mistaken, I am typing from memory) has this latter property reduces to standard calculus arguments such as "differentiate, find turning points, estimate values, use intermediate value theorem". I always find this calculus interlude at the end of half a semester of algebra quite amusing.
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The interesting application in Spivak's Calculus is the proof of the irrationality of pi. I guess this is the proof due to Niven.
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The mean-value theorem (of differential calculus) can be used to prove that Liouville numbers are transcendental. The proof is quite simple, taking only a couple of lines. See Theorem 191 of Hardy and Wright's "An Introduction to the Theory of Numbers" on Google books.
I believe, historically, that these were the first known examples of transcendental numbers.
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An interesting application of calculus is the elementary polynomial case of Mason's ABC theorem. This yields, for instance, a completely trivial proof of the polynomial case of FLT (Fermat's Last Theorem). That this works so effectively for polynomials (functions) vs. numbers is due to the fact that for functions we have available the derivative, which implies that we can exploit Wronskians as a measure of algebraic independence. Such Wronskian estimates serve as fundamental tools in diophantine approximation. See my post [1] for further details and references.
[1] sci.math.research, 1996/07/17
poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1]
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The intermediate value theorem is a basic ingredient in a Galois theory-based proof of the fundamental theorem of algebra. It is used as "Every real polynomial of odd degree has a real zero".
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That's not calculus! (Although it is taught in many "Calculus" courses, so this is an acceptable answer). – Zen Harper Aug 11 '10 at 6:48
That proof is valid over any real-closed field. – Greg Marks Aug 11 '10 at 20:35
The irrationality of $e$ !!
First use the Taylor expansion of $e^x$ to show that $|e-S_n|<\frac{3}{(n+1)!}$ where $S_n= 1+\frac{1}{1!}+\frac{1}{2!}+\cdots + \frac{1}{n!}$. Then, deduce the irrationally of $e$.
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This is really a subject within calculus. – Ryan Reich Aug 10 '10 at 21:32
@Menny: that argument establishes the irrationality of $e$, not of $e!!$, which lies somewhere in between the irrationality of $e$ and its transcendence. [:)] – Pete L. Clark Aug 10 '10 at 21:59
An example that I like is the proof that $e^{A+B}=e^A e^B$ for commuting matrices $A,B$. Since the matrix exponential is defined by the usual exponential series, we have to prove that
$\sum \frac{(A+B)^n}{n!}=\sum\frac{A^n}{n!}\sum\frac{B^n}{n!}$
This follows, without actually computing the two sides, by observing that it is the same computation as for real numbers $A,B$ (because $A$ and $B$ commute). And for real numbers we know the result is correct by first-year calculus.
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This proof is slick, but I am not convinced of its merits: are you proposing to prove an $\textit{algebraic}$ identity using a functional interpretation? In principle (although not in this particular instance), it could happen that different algebraic expressions evaluate to the same function (if this had happened for numbers, but not for matrices, it wouldn't have worked). Perhaps, I am misunderstanding what you mean when you say "This follows, without actually computing the two sides, by the same computation as for real numbers". – Victor Protsak Aug 11 '10 at 0:34
John, by "same computation" do you mean the following: we would like to interpret $e^A e^B = e^{A + B}$ as an identity in formal power series with commuting variables $A, B$. To do so, we observe the map, given by Taylor series, from two-variable analytic real functions to $R[[A,B]]$, contains both sides of the equation, and that the equation holds of the actual functions by means of some calculus argument. – Ryan Reich Aug 11 '10 at 1:45
Ryan, thanks; I think that's the formal argument behind my rough idea. – John Stillwell Aug 11 '10 at 3:50
The combinatorial proof is more conceptual and almost as short. The left-hand side is the data type for $(A \cup B)$-multisets. The right-hand side is the data type for pairs of $A$-multisets and $B$-multisets. Obviously they are isomorphic. (You can now take images to analytic power series.) This is essentially the same as the combinatorial proof of the binomial formula but without the messy detour through binomial coefficients. – Per Vognsen Aug 11 '10 at 16:22
Nice proof, Per. I also like to see applications of combinatorics to first-year calculus :) – John Stillwell Aug 11 '10 at 23:03
Going in a completely different direction: a surprising application of calculus is the use of the Leibniz and chain rules to differentiate data types to create new types that represent structures with 'holes' in them. See here for an elementary exposition.
(This is closely related to the differentiation of generating functions and combinatorial species.)
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I think you could probably show a smart group of first year calculus students how to get an exact formula for the Fibonnaci numbers using generating functions, which basically just boils down to knowing partial fraction decomposition and a few standard power series. You could then point them to Wilf's book if this makes them curious about generating functions in combinatorics.
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Do approximations of $\pi$ count? If so, see http://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106 and the references there.
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The notion of a formal derivative of a polynomial over some ring comes from the ordinary derivative of a polynomial over the real and complex numbers. Furthermore, results true over the real numbers, such as that $(fg)'=f'g+g'f$ and $(f \circ g)' = (f' \circ g) g'$, continue to hold over arbitrary rings. However, these results are much easier to prove over the real numbers using analytic techniques, and one might legitimately argue that mathematicians were only led to the corresponding formal results by the inspiration of the results in calculus.
Furthermore, using something along the lines of the Lefschetz principle, one can probably derive the identities for formal derivatives from the corresponding facts for derivatives of polynomials over the complex numbers.
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Two comments. First, I don't agree that it is "much easier" to use the limit defn. of the real derivative to prove differentiation rules than it is to use induction for proving the same rules on polynomials over any (comm.) ring. Second, it is not legitimate to argue that the formal derivative was introduced only after calculus was developed. The formal derivative on real polynomials came first! See J. Grabiner, "The Changing Concept of Change: the Derivative from Fermat to Weierstrass," Math. Magazine 56 (1983), 195-206. On JSTOR this is at jstor.org/stable/2689807 – KConrad Aug 10 '10 at 21:09
To avoid any misunderstanding, I am not suggesting the formal defn. of derivative should come first in teaching, since of course the real derivative in calculus provides extra geometric intuition. – KConrad Aug 10 '10 at 21:11
Can you give some references for derivatives of polynomials over general rings? (The stranger, the better). Thanks! – Jose Brox Aug 11 '10 at 11:50
A few years ago I gave a departmental colloquium talk, aimed at beginning M.A. students, on "An application of calculus to ring theory." A slightly facetious little abstract can be found here. The example establishing the main results—very well known to workers in commutative ring theory—was the ring of germs at $0$ of class $C^{\infty}$ functions on $\mathbb{R}$. A bit of calculus is needed in verifying the requisite properties.
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A nice application of calculus that leads to a surprising and far reaching result, first obtained by the great Gauss himself, is the computation of the Arithmetic-Geometric Mean of two numbers $a > b > 0$. A comparatively short way to this end is presented on the first pages of J. and P. Borweins "Pi and the AGM".
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See Robert M. Young's lovely book titled "Excursions In Calculus" for some examples.
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A cool example: The intermediate value theorem may be used to prove the following theorem about continued fractions:
Let $\alpha>1$, and suppose that $$\left|\alpha-\frac{p}{q}\right|<\frac{1}{2q^2}.$$ Then, $\dfrac{p}{q}$ is one of the convergents (truncated continued fractions) of $\alpha$.
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In number theory, here are four applications of techniques or results in first-year calculus.
(1) Finding equations of tangent lines by first-semester calculus methods lets us add points on elliptic curves using the Weierstrass equation for the curve. This is more algebraic geometry than number theory, so I'll add that the methods show if the Weierstrass equation has rational coefficients then the sum of two rational points is again a rational point.
(2) The recursion in Newton's method from differential calculus is the basic idea behind Hensel's lemma in $p$-adic analysis (or, more simply, lifting solutions of congruences from modulus $p$ to modulus $p^k$ for all $k \geq 1$).
(3) The infinitude of the primes can be derived from the divergence of the harmonic series (the zeta-function at 1), which is based on a bound involving the definition of the natural logarithm as an integral.
(4) Unique factorization in the Gaussian integers can be derived from the Leibniz formula $$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots = \sum_{n \geq 0} \frac{(-1)^n}{2n+1}$$ by interpreting it as a case of Dirichlet's class number formula $2\pi h/(w\sqrt{|D|}) = L(1,\chi_D)$ for $\chi_D$ the primitive quadratic character associated to ${\mathbf Q}(\sqrt{D})$ where $D$ is a negative fundamental discriminant, $h$ is the class number of ${\mathbf Q}(\sqrt{D})$ and $w$ is the number of roots of unity in ${\mathbf Q}(\sqrt{D})$. Taking $D = -4$ turns the left side into $2\pi h/(4\sqrt{4}) = (\pi/4)h$, so the Leibniz formula is equivalent to $h = 1$, which is another way of saying $\mathbf Z[i]$ is a PID or equivalently (for Dedekind domains) a UFD.
Here are two more applications, not in number theory directly.
(5) Gerry Edgar mentions in his answer Niven's proof of the irrationality of $\pi$, which is available in Spivak's calculus book. The same ideas imply irrationality of $e^a$ for every positive integer $a$, which in turns easily implies irrationality of $e^r$ for nonzero rational $r$ and thus also irrationality of $\log r$ for positive rational $r \not= 1$. The calculus fact in the proof of irrationality of the numbers $e^a$ is that for all positive integers $n$ the polynomial $$\frac{x^n(1-x)^n}{n!}$$ and all of its higher derivatives take integer values at $0$ and $1$. That implies a certain expression involving a definite integral is a positive integer, and then with the fundamental theorem of calculus that same expression turns out to be less than 1 for large $n$ (where "large" depends on the hypothetical denominator of a rational formula for $e^a$), and that is a contradiction.
(6) Prove that if $f$ is a smooth function (= infinitely differentiable) on the real line and $f(0) = 0$ then $f(x) = xg(x)$ where $g$ is a smooth function on the real line. There is no difficulty in defining what $g(x)$ has to be if it exists at all, namely $$g(x) = \begin{cases} f(x)/x, & \text{ if } x \not= 0, \\ f'(0), & \text{ if } x = 0. \end{cases}$$ And easily the function defined this way is continuous on the real line and satisfies $f(x) = xg(x)$. But why is this function smooth at $x = 0$ (smoothness away from $x = 0$ is easy)? You can try to do it using progressively messier formulas for higher derivatives of $g$ at 0 by taking limits, but a much slicker technique is to use the fundamental theorem of calculus to write $$f(x) = f(x) - f(0) = \int_0^x f(t)\,dt = x\int_0^1 f(xu)\,du,$$ which leads to a different formula for $g(x)$ that doesn't involve cases: $$g(x) = \int_0^1 f(xu)\,du.$$ If you're willing to accept differentiation under the integral sign (maybe that's not in the first-year calculus curriculum, but we used first-year calculus to get the slick formula for $g(x)$) then the right side is easily checked to be a smooth function of $x$ from $f$ being smooth.
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Shanks' simplest cubic $x^3-ax^2-(a+3)x-1$ has discriminant $D=(a^2+3a+9)^2$ and hence 3 real roots. A calculus way to see this is to rewrite it as inverting $f(x)=a$ where $$f(x)=\frac{x^3-3x-1}{x^2+x}=x-1 -\frac{1}{x}-\frac{1}{x+1}$$ whose graph has three monotone branches which clearly intersects $y=a$ at three real points for any real $a$. In particular, we can pick $a$ to be any of the (real) roots which means we can iterate the construction and get a cubic tower of totally real fields. Also (though this is not relevant to the question), it's nice to see $f$ is a trace $$f(x)=x+\rho(x)+\rho^2(x),$$ where $\rho(x)=-1/(x+1)$ is of order 3 in $\in PSL_2(Z)$ which show that if $\alpha$ is a root of the cubic , then so is $\rho(\alpha)$.
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What I like(d) most is defining an analytic function that describes some number theoretic phenomena. One thing I remember is from Winfried Kohnen's postech lecture http://www.mathi.uni-heidelberg.de/~winfried/siegel2.pdf , see pages 1-3 for more details. He starts with the standard inner product on $\mathbb{R}^m$ viewed as a quadratic form $$Q(x):=x^t x.$$ We are interested in the number $r_Q(t)$ of tuples of squares of inetegers that add up to a natural number $t$, i.e.
$$r_Q(t):= \# \left\{ g \in \mathbb{Z}^m : Q(g)=(g_1)^2+ \dots + (g_4)^2=t \right\} .$$
They can be computed via this power series
$$\theta_Q(z) = 1+ \sum_{t\geq 1} r_Q(t)\ \exp(2\pi i tz)$$
that is in fact $\theta_Q$ is a modular form of weight 2 w.r.t. $\Gamma_0$. Therefore (ok here is some kind of black box for the students), its Fourier coefficients can be given by $$r_Q(t)= 8 \left( \sigma_1(t)-4\cdot \sigma_1\left(\frac{t}{4}\right) \right)$$
where $\sigma_k(t)$ denotes the divisor function $$\sigma_k(t):=\sum_{d|t} d^k.$$
While writing this I was wondering whether the prime number theorem and elegant proofs of the fundamental theorem of algebra are too well known.
p.s. sorry for messing up the formulas again.
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https://www.physicsforums.com/threads/force-on-a-wire-carrying-current.612705/ | # Force on a wire carrying current
1. Jun 9, 2012
### ehabmozart
It is given in my book. hat F=B I l sin theta where theta is the angle between B and the current I... I guess this is no right.. I mean, how would the current make an angle with B, it should b the length of wire itself... Secondly F=Bqv is the reason behind a charged particle which enter perpendicularly to a magnetic field moves in a circle.. My doubt is that at point, this point charge will be inline with the field line.. From where will it get force at that point.. Talking about the quarter circle... I need more clarification and it would be more than amazing if there are some illustrative diagrams!
2. Jun 9, 2012
### MalachiK
The angle in your first equation IS the angle between the current and the field lines. If you imagine a uniform B field you can also imagine a wire in this field pointing in various different directions.
Current is a vector, and only the component of the current vector that cuts across the field lines causes the force. When the current is at right angles to the field, θ = 90 and sin θ =1. This means that all of the current contributes to the force so you get the maximum force = BIL.
On the other hand, when the current is parallel to the field, θ = 0 and sin θ = 0. This makes your force = BIL × 0 = 0 as there is no component of the current cutting across the field lines. | {"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": 0.8780502676963806, "perplexity": 420.3857088743446}, "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-2018-47/segments/1542039746205.96/warc/CC-MAIN-20181122101520-20181122123520-00175.warc.gz"} |
https://astarmathsandphysics.com/gcse-physics-notes/4444-the-pot-of-gold-at-the-centre-of-the-earth.html?tmpl=component&print=1&page= | ## The Pot of Gold at the Centre of the Earth
The density of the Earth is not constant. The lightest elements, or the elements that form gases, form the Earth's atmosphere. The density of the Earth's atmosphere at sea level is about
$1.29 kg/m^3$
. At the top of the Earth's atmosphere the density is very low indeed. The next density part of the Earth is the oceans. The density of pure water is
$1000 kg/m^3$
. The Earth's ocean contain dissolved substances - mostly salt and some gases. The density of the oceans is about
$1030 kg/m^3$
.
The solid parts is the Earth - the crust - is much denser than water. The crust has a density of about
$2,700 kg/m^3$
, and below this is the Earth's upper mantle with a density of
$3,400 kg/m^3$
and the Earth's lower mantle with a density of
$4,400kg/m^3$
. Then there is the Earth's outer core with a density of
$9,800kg/m^3$
and the inner core with a density of
$12,100kg/^3$
.
The density of gold is
$19,300 kg/m^3$
. Solid gold is the eighth densest element, and all the elements denser than gold are quite uncommon. It follows that gold should sink to the centre of the Earth and the Earth should have a gold rich core. Unfortunately, we cannot mine to gold and become rich We would have to drill through 100km of rock, then thousands of km of molten rock. If we can reach the core, where the temperature is about 6,300 Degrees Kelvin, are the drilling tools would be melted | {"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": 0.9489923119544983, "perplexity": 652.6100276951489}, "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-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00631.warc.gz"} |
https://link.springer.com/chapter/10.1007%2F978-3-030-27661-4_13 | # Kinetic Theory of Gases
• Gregor Skačej
• Primož Ziherl
Chapter
## Abstract
Calculate the average speed of particles in an ideal gas! The velocity is characterized by the Maxwell–Boltzmann distribution. | {"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": 0.97931307554245, "perplexity": 1158.0180646915312}, "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-2019-47/segments/1573496665575.34/warc/CC-MAIN-20191112151954-20191112175954-00111.warc.gz"} |
http://behindtheguesses.blogspot.com/2009/06/derivative-and-integral-of-heaviside.html | ## Tuesday, June 30, 2009
### Derivative and Integral of the Heaviside Step Function
The Setup
(a) Large horizontal scale
(b) ``Zoomed in''
Figure 1: The Heaviside step function. Note how it doesn't matter how close we get to
$x=0$ the function looks exactly the same.
The Heaviside step function $H(x)$, sometimes called the Heaviside theta function, appears in many places in physics, see [1] for a brief discussion. Simply put, it is a function whose value is zero for $x<0$ and one for $x>0$. Explicitly,
$H(x)=\begin{cases}0& x<0,\\1& x>0\end{cases}.$
(1)
We won't worry about precisely what its value is at zero for now, since it won't effect our discussion, see [2] for a lengthier discussion. Fig. 1 plots $H(x)$. The key point is that crossing zero flips the function from 0 to 1.
Derivative -- The Dirac Delta Function
(a) Dirac delta function
(b) Ramp function
Figure 2: The derivative (a), and the integral (b) of the Heaviside step function.
Say we wanted to take the derivative of $H$. Recall that a derivative is the slope of the curve at at point. One way of formulating this is
$\frac{dH}{dx}=\lim_{\Delta x \rightarrow 0} \frac{\Delta H}{\Delta x}.$
(2)
Now, for any points $x<0$ or $x>0$, graphically, the derivative is very clear: $H$ is a flat line in those regions, and the slope of a flat line is zero. In terms of (2), $H$ does not change, so $\Delta H=0$ and $dH/dx=0$. But if we pick two points, equally spaced on opposite sides of $x=0$, say $x_-=-a/2$ and $x_+=a/2$, then $\Delta H=1$ and $\Delta x=a$. It doesn't matter how small we make $a$, $\Delta H$ stays the same. Thus, the fraction in (2) is
$\frac{dH}{dx}=\lim_{a \rightarrow 0} \frac{1}{a}$
$=\infty.$
(3)
Graphically, again, this is very clear: $H$ jumps from 0 to 1 at zero, so it's slope is essentially vertical, i.e. infinite. So basically, we have
$\delta(x)\equiv\frac{dH}{dx}=\begin{cases}0& x<0\\\infty& x=0\\0& x>0\end{cases}.$
(4)
This function is, loosely speaking, a ``Dirac Delta'' function, usually written as $\delta(x)$, which has seemingly endless uses in physics.
We'll note a few properties of the delta function that we can derive from (4). First, integrating it from $-\infty$ to any $x_-<0$:
$\int_{-\infty}^{x_-}\delta(x)dx=\int_{-\infty}^{x_-}\left(\frac{dH}{dx}\right)dx$
$=H(x_-)-H(-\infty)$
$=0$
(5)
since $H(x_-)=H(-\infty)=0$. On the other hand, integrating the delta function to any point greater than $x=0$:
$\int_{-\infty}^{x_+}\delta(x)dx=\int_{-\infty}^{x_+}\left(\frac{dH}{dx}\right)dx$
$=H(x_+)-H(-\infty)$
$=1$
(6)
since $H(x_+)=1$.
At this point, I should point out that although the delta function blows up to infinity at $x=0$, it still has a finite integral. An easy way of seeing how this is possible is shown in Fig. 2(a). If the width of the box is $1/a$ and the height is $a$, the area of the box (i.e. its integral) is $1$, no matter how large $a$ is. By letting $a$ go to infinity we have a box with infinite height, yet, when integrated, has finite area.
Integral -- The Ramp Function
Now that we know about the derivative, it's time to evaluate the integral. I have two methods of doing this. The most straightforward way, which I first saw from Prof. T.H. Boyer, is to integrate $H$ piece by piece. The integral of a function is the area under the curve,1 and when $x<0$ there is no area, so the integral from $-\infty$ to any point less than zero is zero. On the right side, the integral to a point $x$ is the area of a rectangle of height 1 and length $x$, see Fig. 1(a). So, we have
$\int_{-\infty}^x H\,dx=\begin{cases}0& x<0,\\x& x>0\end{cases}.$
(7)
We'll call this function a ``ramp function,'' $R(x)$. We can actually make use of the definition of $H$ and simplify the notation:
$R(x)\equiv\int H\,dx=xH(x)$
(8)
since $0\times x=0$ and $1\times x=x$. See Fig. 2(b) for a graph -- and the reason for calling this a ``ramp'' function.
But I have another way of doing this which makes use of a trick that's often used by physicists: We can always add zero for free, since ${anything}+0={anything}$. Often we do this by adding and subtracting the same thing,
$A=(A+B)-B,$
(9)
for example. But we can use the delta function (4) to add zero in the form
$0=x\,\delta(x).$
(10)
Since $\delta(x)$ is zero for $x\neq 0$, the $x$ part doesn't do anything in those regions and this expression is zero. And, although $\delta(x)=\infty$ at $x=0$, $x=0$ at $x=0$, so the expression is still zero.
So we'll add this on to $H$:
$H=H+0$
$=H+x\,\delta(x)$
$=H+x\frac{dH}{dx} \quad \quad \quad \text{by (4)}$
$=\frac{dx}{dx}H+x\frac{dH}{dx}$
$=\frac{d}{dx}\left[xH(x)\right],$
(11)
where the last step follows from the ``product rule'' for differentiation. At this point, to take the integral of a full differential is trivial, and we get (8).
References
[1] M. Springer. Sunday function [online]. February 2009. Available from: http://scienceblogs.com/builtonfacts/2009/02/sunday_function_22.php [cited 30 June 2009].
[2] E.W. Weisstein. Heaviside step function [online]. Available from: http://mathworld.wolfram.com/HeavisideStepFunction.html [cited 30 June 2009].
1 To be completely precise, it's the (signed) area between the curve and the line $x=0$.
1. Not to nit pick, but unless I am misunderstanding the entire post (not to mention verious math, physics, and engineering classes I've taken) I believe the line after (6) should read "since H(x+)=1" not "since H(x-)=1"
2. Yup, typo. It's fixed. Thanks for catching it.
3. Thanks a lot for this, very useful for figuring out my assignment.
4. hi,thanks for this,just one question: how did u set x*delta(x)=0 at x=0 as delta(x) in zero would be infinity so x*delta(x)=0*infinity which is NaN?
5. Mehrtash,
It's a bit unrigorous. Basically, the delta function tends towards infinity at x=0, but the area under it is finite, so it's not ``really'' infinity, it's only sort of ``tends to infinity,'' whatever that means. So, 0*(something not really infinite) is 0.
6. This explanation of the integral of the delta function is extremely helpful. In our EE classes they don't give us textbooks with this derivation, and other internet sources just state the result. Thank you for clearing up my misunderstanding.
7. Anonymous - glad I was able to help!
8. Hi, your explanation is very good. But I am looking for the derivative and integration of functions like f(x)H(x) ,e.g., {exp(ax)}H(b-x). If I am not very wrong D{f(x)H(x)}= f'(x)H(x)+f(x){delta(X)} and then by integrating this result I should get back the original function, at this point I am confused how to handle the delta function.
['D' stands for differentiation]
9. Tanay -- that's right. To integrate backwards, first integrate f'(x)H(x) by parts:
Integral[f'(x)H(x)dx] = f(x)H(x) -Integral[f(x)H'(x)dx] = f(x)H(x) - Integral[f(x)delta(x)dx]
This second integral cancels with the Integral[f(x)delta(x)dx], leaving you with f(x)H(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": 71, "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": 0.955184817314148, "perplexity": 331.37395911655364}, "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-2013-20/segments/1368701638778/warc/CC-MAIN-20130516105358-00054-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/520367/linearly-independent-vectors-and-matrix | Linearly independent vectors and matrix
If $\{v_{1},v_{2},\cdots,v_{n}\}$ is $n$ linearly independent vectors in $\mathbb{R}^{n}$, what would be necessary and sufficient condition of $A$ ($n\times n$ matrix) $A$ so that the vectors $Av_{1}$, $Av_{2}$, $\cdots$, $Av_{n}$ are linearly independent?
-
The vectors $Av_1, \ldots ,Av_n$ are linearly independent if, and only if, $\det \left([Av_1 | \ldots |Av_n]_{n\times n}\right)\neq 0$.
Now note that $[Av_1 | \ldots |Av_n]_{n\times n}=A[v_1|\ldots |v_n]_{n\times n}$.
@Mark It's a notation which suggests that I'm 'writing the matrix by columns'. The $i^{\text{th}}$ column of the matrix $[Av_1 | \ldots |Av_n]_{n\times n}$ is $Av_i$. Is that clear now? – Git Gud Oct 9 '13 at 16:52
So, the matrix product $\left([Av_1 | \ldots |Av_n]_{n\times n}\right)$ has to be invertible here – Mark Oct 9 '13 at 17:05 | {"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": 0.9580143690109253, "perplexity": 202.80178053350687}, "config": {"markdown_headings": false, "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-2016-07/segments/1454701157443.43/warc/CC-MAIN-20160205193917-00231-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/behaviour-of-limits-and-their-effect-on-equations.698552/ | # Behaviour of limits and their effect on equations
1. Jun 24, 2013
### titowakoru
Hi,
I was wondering if some one could check my understanding of limits please.
If a limit is presetned as say x << y or x >> y am I right in thinking that x or y can be ignored as they are small enough to be insignificant? So, for example, if I had equation which was
sin(x/y)
in the situation where x << y only values of y are significant and where x >> y then only values of x are significant.
Or do I have this completely wrong??
2. Jun 24, 2013
### Office_Shredder
Staff Emeritus
You have it wrong.
If you have an addition, for example 2x+3y and you are told x<<y then the x is so small that it is insignificant and you can approximate this by 3y and be OK.
But if you have a ratio like x/y, then being told x<<y doesn't let you make any simplification. If you try throwing away the x and writing it as 1/y then you're probably off by orders of magnitude which isn't a very good thing.
Sample calculations: If x<<y approximate the following:
1) $$\frac{2x + 3y}{4x - y}$$
In this case the numerator is approximately 3y and the denominator is approximately -y, so this is going to be approximately -3.
2)
$$\frac{ 2xy}{y}$$
The answer is obviously 2x. If you throw away the x because it's a lot smaller than y, you'll end up getting 2, which is not close to 2x unless x happens to be close to 1.
3. Jun 24, 2013
### Staff: Mentor
From what you've written, I don't think you're asking about limits, but instead are asking about how to approximate expressions in two variables, and it's given that x << y or y << x. Note that sin(x/y) is NOT an equation. An equation has an = symbol in it.
The following is an example of a limit:
$$\lim_{x \to 0}\frac{\sin (x)}{x}$$
It's possible to have limits in which a point (x, y) is approaching some fixed point (x0, y0), but it's very unusual for it to be given that x << y or the other way around.
Similar Discussions: Behaviour of limits and their effect on equations | {"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": 2, "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": 0.900115430355072, "perplexity": 592.5527277264912}, "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-2017-43/segments/1508187823605.33/warc/CC-MAIN-20171020010834-20171020030834-00433.warc.gz"} |
https://blog.zenggyu.com/en/post/2018-03-11/understanding-the-bias-variance-decomposition-with-a-simulated-experiment/ | # Understanding the Bias-Variance Decomposition with A Simulated Experiment
## Forewords
The bias-variance tradeoff is an essential factor to consider when choosing a model to minimize test error. To truly understand the underlying concepts, it is helpful to learn how exactly the test error can be decomposed into bias and variance. However, for simplicity, most textbooks do not offer a precise derivation of the process, which may lead to confusion1.
To address the problem, this post provides two ways to demonstrate the bias-variance decomposition. In the first part, a precise mathematical derivation of the decomposition process is provided, along with some illustrative text. In the second part, a simulated experiment in R is presented to demonstrate the theory in a more realistic background.
## Some theoretical background
Consider a regression problem where the observed outcome is denoted by $Y$ and the predictor is denoted by $X$. The relationship between the observed outcome and the predictor can be defined as:
$Y = f(X) + \epsilon$
where $f()$ denotes the function that maps the predictor (i.e., $X$) to the true outcome (i.e., $f(X)$), and $\epsilon$ denotes an error term.
In real-life scenarios, $f()$ and $\epsilon$ are unknown. But if we have a training set $\tau$, we can obtain an estimate of $f()$, which is denoted by $\hat{f_{\tau}}()$. We can then use $\hat{f_{\tau}}(X)$ to predict the outcome. Using squared-error loss, the expected prediction error based on a test set (i.e., test error) can be defined as2:
$E[(Y - \hat{f}(X)) ^ 2] = E[(f(X) + \epsilon - \hat{f_{\tau}}(X)) ^ 2]$
Here are some things to keep in mind before performing the bias-variance decomposition. First, the values of $X$ and $\epsilon$ are those from the test set. Second, the value of $Y$ (i.e., $f(X) + \epsilon$) is dependent on both $\epsilon$ and $X$. Third, the realization of $\hat{f_{\tau}}()$ is dependent on the training set $\tau$ that is used to obtain the estimate, and therefore the value of $\hat{f_{\tau}}(X)$ is dependent on both $X$ and $\tau$. Additionally, it should also be noted that during the derivation process, $\epsilon$ is assumed to have a mean of zero.
Keeping the dependencies in mind, the expression can be rewritten as3:
$E_{\tau}[E_{X, \epsilon}[(Y - \hat{f}) ^ 2]] = E_{\tau}[E_{X, \epsilon}[(f + \epsilon - \hat{f}) ^ 2]]$
Since $X$ and $\epsilon$ are independent, $E_{X, \epsilon}[...]$ can be expanded to $E_{X}[E_{\epsilon}[...]]$. Also, we can use the Fubini’s theorem to rearrange the order in which $E_{\tau}[...]$, $E_{X}[...]$, $E_{\epsilon}[...]$ is evaluated. Then, the above expression becomes:
$E_{\tau}[E_{X, \epsilon}[(Y - \hat{f}) ^ 2]] = E_{X}[E_{\tau}[E_{\epsilon}[(f + \epsilon - \hat{f}) ^ 2]]$
Now, let’s solve the inner-most expectation, i.e., $E_{\epsilon}[...]$ (recall that $\epsilon$ has a mean of zero):
\begin{align} E_{\epsilon}[(Y - \hat{f}) ^ 2] &= E_{\epsilon}[(f + \epsilon - \hat{f}) ^ 2]\\ &= E_{\epsilon}[(f - \hat{f}) ^ 2 + \epsilon ^ 2 + 2\epsilon(f - \hat{f})]\\ &= (f - \hat{f}) ^ 2 + E_{\epsilon}[\epsilon ^ 2] + 2(f - \hat{f})E_{\epsilon}[\epsilon]\\ &= (f - \hat{f}) ^ 2 + (E_{\epsilon}[\epsilon ^ 2] - E_{\epsilon}[\epsilon] ^ 2) + E_{\epsilon}[\epsilon] ^ 2 + 2(f - \hat{f})E_{\epsilon}[\epsilon]\\ &= (f - \hat{f}) ^ 2 + Var_{\epsilon}[\epsilon] + 0 + 0\\ &= (f - \hat{f}) ^ 2 + Var_{\epsilon}[\epsilon] \end{align}
Then the second inner-most expectation, i.e., $E_{\tau}[...]$:
\begin{align} E_{\tau}[E_{\epsilon}[(Y - \hat{f}) ^ 2]] &= E_{\tau}[(f - \hat{f}) ^ 2 + Var_{\epsilon}[\epsilon]]\\ &= E_{\tau}[(f - \hat{f}) ^ 2] + Var_{\epsilon}[\epsilon]\\ &= E_{\tau}[f ^ 2 + \hat{f} ^ 2 - 2f\hat{f}] + Var_{\epsilon}[\epsilon]\\ &= f ^ 2 + E_{\tau}[\hat{f} ^ 2] - 2fE_{\tau}[\hat{f}] + Var_{\epsilon}[\epsilon]\\ &= f ^ 2 + E_{\tau}[\hat{f} ^ 2] - 2fE_{\tau}[\hat{f}] + Var_{\epsilon}[\epsilon] + E_{\tau}[\hat{f}] ^ 2 - E_{\tau}[\hat{f}] ^ 2\\ &= (f ^ 2 + E_{\tau}[\hat{f}] ^ 2 - 2fE_{\tau}[\hat{f}]) + (E_{\tau}[\hat{f} ^ 2] - E_{\tau}[\hat{f}] ^ 2) + Var_{\epsilon}[\epsilon]\\ &= (f - E_{\tau}[\hat{f}]) ^ 2 + Var_{\tau}[\hat{f}] + Var_{\epsilon}[\epsilon] \end{align}
And finally, the outer-most expectation, i.e., $E_{X}[...]$:
\begin{align} E_{X}[E_{\tau}[E_{\epsilon}[(Y - \hat{f}) ^ 2]]] &= E_{X}[(f - E_{\tau}[\hat{f}]) ^ 2 + Var_{\tau}[\hat{f}] + Var_{\epsilon}[\epsilon]]\\ &= E_{X}[(f - E_{\tau}[\hat{f}]) ^ 2] + E_{X}[Var_{\tau}[\hat{f}]] + Var_{\epsilon}[\epsilon] \end{align}
The above equation shows that the expected test error can be decomposed into:
• $Var_{\epsilon}[\epsilon]$, i.e., irreducible error. This error is the amount by which the observed outcome differs from the true outcome. It could be caused by many limitations of data (e.g., random noise, measurement error, unmeasured predictors, unmeasurable variation), and cannot be avoided unless $\epsilon$ has a variance of zero.
• Reducible error. This error is caused by the algorithm we choose to model the relationship, and it can be minimized using appropriate modeling techniques. The reducible error can be further decomposed into:
• $E_{X}[(f - E_{\tau}[\hat{f}]) ^ 2]$, i.e., bias. This error represents the amount by which the mean of $\hat{f_{\tau}}(X)$ differs from $f(X)$. It is caused by erroneous assumptions in the learning algorithm, e.g., approximating an extremely complicated real-life problem by a much simpler algorithm.
• $E_{X}[Var_{\tau}[\hat{f}]]$, i.e., variance. This error represents the amount by which $\hat{f_{\tau}}(X)$ differs from its own mean. It is caused by the sensitivity of the learning algorithm to small fluctuations in the training set.
The decomposition tells us that, in order to minimize the expected test error, we need to select a learning algorithm with both low variance and low bias. However, typically, the more complicated the algorithm, the lower the bias but the higher the variance, and hence the tradeoff.
## A simulated experiment
The objective of this experiment was to use an example to verify the equation:
$E_{\tau}[E_{X, \epsilon}[(Y - \hat{f}) ^ 2]] = E_{X}[(f - E_{\tau}[\hat{f}]) ^ 2] + E_{X}[Var_{\tau}[\hat{f}]] + Var_{\epsilon}[\epsilon]$
Unlike real-life scenarios, where $f()$ and the distribution of the irreducible error $\epsilon$ are unknown, this experiment provided explicit definitions to perform the simulation:
1. $f() = sin()$.
2. $\epsilon$ had a mean of zero and a standard deviation of 0.1 (hence a variance of 0.01).
3. Additionally, $X \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
A simple linear regression algorithm was used to estimate $f()$, and the results were then used to verify the equation.
Step 1:
Some setups.
library(tidyverse)
set.seed(1) # for reproducible results
Step 2:
Generated 101 data sets, each containing 100 observations. The first data set serves as a test set, while the remaining 100 data sets serve as training sets.
data_sets <- lapply(1:101, function(...) {
tibble(
X = runif(100, min = -pi / 2, max = pi / 2), # the predictor
fX = sin(X), # the true outcome, i.e., f(X)
e = rnorm(100, mean = 0, sd = 0.1), # irreducible error
Y = fX + e # the observed outcome
)
}) %>% {
list(test_set = .[[1]],
training_sets = .[-1])
}
Step 3:
Train the linear model on each of the 100 training sets, and get 100 realizations of $\hat{f_{\tau}}()$, which is stored in the variable f_estimates in the code below.
f_estimates <- data_sets$training_sets %>% lapply(function(set_i) { lm(Y ~ X, data = set_i) }) Step 4: Use each of the 100 $\hat{f_{\tau}}()$ to predict the outcome of the 100 observations in the test set. This process yields 10000 predictions (i.e., $\hat{f_{\tau}}(X)$, which is stored in a column named fX_estimates in the data frame results. Note that each prediction can be uniquely identified by the combination of an observation_id and a training_set_id4, which are also assigned to the data frame. results <- lapply(1:100, function(i) { data_sets$test_set %>%
mutate(fX_estimates = predict(f_estimates[[i]], newdata = data_sets$test_set), observation_id = 1:100, training_set_id = i) }) %>% bind_rows() Step 5: Compute the value for each of the terms in the equation: 1. expected_error = $E_{\tau}[E_{X, \epsilon}[(Y - \hat{f}) ^ 2]]$ 2. bias = $E_{X}[(f - E_{\tau}[\hat{f}]) ^ 2]$ 3. variance = $E_{X}[Var_{\tau}[\hat{f}]]$ 4. irreducible_error = $Var_{\epsilon}[\epsilon]$ Note that observation_id determines the value of $X$ and $\epsilon$, therefore $E_{X}[...]$, $E_{\epsilon}[...]$ and $E_{X, \epsilon}[...]$ can all be considered as $E_{observation\_id}[...]$. Similarly, $E_{\tau}[...]$ can be considered as $E_{training\_set\_id}[...]$. These notes help to understand how the calculations are performed in the code below. However, you may need to run the code yourself and examine the results to really relate the code to the math expressions5. expected_error <- results %>% with({ (Y - fX_estimates) ^ 2 }) %>% mean() bias <- results %>% split(.$observation_id) %>%
sapply(function(ith_observation) {
(ith_observation$fX[1] - mean(ith_observation$fX_estimates)) ^ 2 # ith_observation$fX is a vector with 100 identical values, but only one is needed for the calculation }) %>% mean() variance <- results %>% split(.$observation_id) %>%
sapply(function(ith_observation) {
var(ith_observation$fX_estimates) }) %>% mean() irreducible_error <- var(data_sets$test_set$e) Finally, show the results: expected_error #> [1] 0.0150791 bias + variance + irreducible_error #> [1] 0.01549488 The expected error and the sum of its components are very close. However, you may wonder why there is a slight inconsistency. This is because in the derivation process, it is assumed that the irreducible error $\epsilon$ has a mean of zero, and therefore terms that include $E_{\epsilon}[\epsilon]$ are dropped; but this assumption is not strictly satisfied in the experiment: mean(data_sets$test_set\$e)
#> [1] -0.001757949
## Conclusions
The experiment simulated some data with a sinusoidal function, and used a simple linear regression algorithm to estimate the true function and to predict the outcome in the test set. The results showed that the expected test error was almost identical to the sum of the bias, the variance and the irreducible error. The slight inconsistency could be explained by the fact that one of the assumptions underlying the derivation process, i.e., $E_{\epsilon}[\epsilon] = 0$, was not strictly satisfied in the experiment.
1. A common source of confusion is that textbooks like to simplify math expressions. For example, in section 7.3 of The Elements of Statistical Learning, the text seems to suggest that only one input point ($X = x_{0}$) is needed to compute the test error. This confusion led to the failure of my first attempt to verify the decomposition.
2. Some textbooks would drop the error term $\epsilon$ and just derive $E[(f(X) - \hat{f_{\tau}}(X)) ^ 2]$.
3. To simplify the notations while still keeping things clear, $f(X)$ and $\hat{f_{\tau}}(X)$ are respectively abbreviated as $f$ and $\hat{f}$; the dependencies are presented as subscripts of $E[...]$.
4. observation_id identifies the observation in the test set to be predicted; training_set_id identifies the training set to train $\hat{f_{\tau}}()$, which is then used to make the prediction.
5. The R code is rather simplistic (e.g., training_set_id is not used for the calculation), this is because many operations are vectorized, which is a typical feature of R. | {"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": 3, "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": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9931227564811707, "perplexity": 731.7399871535096}, "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-2023-06/segments/1674764500273.30/warc/CC-MAIN-20230205161658-20230205191658-00654.warc.gz"} |
https://www.physicsforums.com/threads/equation-for-flux-in-reflected-slab-reactor.817789/ | # Equation for flux in reflected slab reactor
1. Jun 6, 2015
### StudioSaturn
So, I've been at this for a while now. Im using my textbook (Introduction to NE, Lamarsh) and I cannot find a solution to calculate the thermal and fast fluxes in the reflector of a critical slab reactor. I have only been able to find solutions for the spherical shape, even online. As I understand it, the general solution is
And for spherical geometry, the fast flux can be resolved in a finite reflector to be
If anyone could point me to a source online, or if you could offer any help here it would be appreciated. I have been able to calculate the thermal and fast fluxes in the core, but the reflector has me stumped. Thanks in advance!
2. Jun 6, 2015
### StudioSaturn
These are the equations I used for the core.
For thermal flux in the core:
And to get the fast flux from the thermal flux:
Also, I know that at the reflector-core interface, the fluxes must be continuous, and there is no neutron source term in the reflector. 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": 0.9072031378746033, "perplexity": 546.5675071703682}, "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-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00352.warc.gz"} |
http://mathoverflow.net/questions/118475/vanishing-of-tor | Vanishing of Tor
Let $(R, \mathfrak{m})$ be a commutative Noetherian local ring and $M$ a finitely generated $R$-module. Let $x_1,...,x_t$ be an $M$-regular sequence and $I = (x_1,...,x_t)$. Is it true that $$\mathrm{Tor}_1^R(R/I^n, M) = 0$$ for all $n \geq 1$?
-
It's true for $n=1$ if the sequence is also $R$-regular, since then $\mathrm{Tor}_{1}^{R}(R/I,M) \simeq \mathrm{Tor}_{1}^{R/I}(R/I,M/I)$. (see Lemma 18.2.iii in Matsumura's CRT.) – David Hansen Jan 9 '13 at 19:11
Lemma 18.2 in Matsumura need $x$ is both $R$-regular and $M$-regular. – Pham Hung Quy Jan 10 '13 at 2:41
So Lemma 18.2 applies at least when $M$ has finite projective dimension. – Mahdi Majidi-Zolbanin Jan 10 '13 at 4:22
$I^{n-1}/I^n$ is a free $R/I$-module, so the statement follows from a simple induction. – Angelo Jul 25 '13 at 5:41
I've posted a proof here for the special case when $M$ is cyclic. Furthermore, I've mentioned that the result holds for finitely generated modules when the sequence is $R$-regular and $M$-regular.
Very nice proof. Thanks you very much YACP. The case $M$ is cyclic is an exercise in the book about tight closure of C. Huneke. Your proof even better than its solution (in my opinion). – Pham Hung Quy Jan 12 '13 at 15:43 | {"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": 0.9665011763572693, "perplexity": 316.8519907635178}, "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-2016-07/segments/1454701168065.93/warc/CC-MAIN-20160205193928-00167-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.arxiv-vanity.com/papers/math/0701727/ | # The determinant of the Dirichlet-to-Neumann map for surfaces with boundary
Colin Guillarmou Laboratoire J. Dieudonné
CNRS, Université de Nice
Parc Valrose
06100 Nice
France
and Laurent Guillopé Laboratoire J. Leray
CNRS, Université de Nantes
2, rue de la Houssinière
BP 92208, 44322 Nantes cedex 03
France
###### Abstract.
For any orientable compact surface with boundary, we compute the regularized determinant of the Dirichlet-to-Neumann (DN) map in terms of particular values of dynamical zeta functions by using natural uniformizations, one due to Mazzeo-Taylor, the other to Osgood-Phillips-Sarnak. We also relate in any dimension the DN map for the Yamabe operator to the scattering operator for a conformally compact related problem by using uniformization.
## 1. Introduction
Let be a connected compact Riemannian manifold with boundary, then the Dirichlet-to-Neumann (DN) map is the map
N:C∞(∂¯X)→C∞(∂¯X)
defined by the following problem: let and let be the solution of
Δ¯gu=0,u|∂¯X=f,
then if is the interior pointing vector field which is normal to , we set
Nf:=−∂nu|∂¯X.
It is well-known that is an elliptic self-adjoint pseudo-differential operator on with principal symbol , if (see [36, 7.11] for example). It is then possible to define its determinant by the Ray-Singer method [33]. Indeed, if is an elliptic self-adjoint pseudo-differential operator of order with positive principal symbol, we can set, following [35, 33, 24],
det(A)=e−∂sζA(0),ζA(s)=Tr(A−s)
where is a priori defined for but has a meromorphic extension to with no pole at . If we apply this to , we obtain since , indeed by Green’s identity one has thus is the space of constant functions on . We then have to modify the definition of : if is the orthogonal projection in onto the kernel , we take defined by
det′(N)=e−∂sζ∗N(0),ζ∗N(s)=Tr(N−s(1−Π))
which is well defined as before. To compute , we first show the
###### Theorem 1.1.
If is a Riemannian surface with boundary and if is the length of the boundary for the metric , the value is a conformal invariant of the conformal manifold with boundary .
Note that this was proved in the case where is a topological disc by Edward-Wu [7].
Consequently, it is sufficient to study the case of particular conformal representative in the conformal class, that is to use uniformization.
The first natural uniformization we will use has been proved by Mazzeo and Taylor [26], it picks a complete constant negative curvature metric in the conformal class: indeed, they show that there exists a unique conformally compact metric on the interior of such that has curvature and is conformal to . The manifold is then isometric to an infinite volume quotient of the hyperbolic plane by a convex co-compact group of isometries. We use this uniformization to compute , although the DN map in this case does not really make sense, but instead we have the scattering operator.
Before stating the result, we need to recall a few definitions about Riemann surfaces and their Selberg (resp. Ruelle) zeta function. Let be a Fuchsian subgroup with only hyperbolic elements (i.e. fixing points at the boundary of ), the quotient is a geometrically finite complete hyperbolic manifold. We recall that any is conjugated to the dilation , with translation length in the hyperbolic half-plane model , note that the set of primitive conjugacy classes of is in one-to-one correspondence with the set of primitive closed oriented geodesics , the length of the closed geodesic corresponding to being equal to . There is a dynamical Ruelle type zeta function defined by the formula111The original zeta function of Ruelle was actually defined by the inverse of this one, we prefer to use the convention of Fried [9].
(1.1) RΓ(λ):=∏[γ]∈[Γ](1−e−λℓ(γ))
and the Selberg zeta function
(1.2) ZΓ(λ):=∏k∈N0RΓ(λ+k).
These products converge for where is the exponent of convergence of the Poincaré series of , equal to only if is cocompact. Moreover they admit an analytic extension222For the compact case, this is a consequence of Selberg trace formula, here this follows from Fried [9] and Patterson-Perry [31] for instance. to and verify the identity
(1.3) RΓ(λ)=ZΓ(λ)/ZΓ(λ+1).
We are able to compute the determinant of the DN map using the uniformization of [26]:
###### Theorem 1.2.
Let be a smooth compact orientable connected Riemannian surface of Euler characteristic , with boundary of length , and let be the Dirichlet-to-Neumann operator of on . Let be the unique up to isometry, infinite volume, complete hyperbolic metric on conformal to and let the geometrically finite Fuchsian group such that is isometric to the space form . If we denote by the Ruelle zeta function of , we have
det′(N)ℓ(∂¯X)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1% if χ(¯X)=1,ℓ(γ)/πif χ(¯X)=0,[(2πλ)χ(¯X)−1RΓ(λ)]|λ=0/χ(¯X)if χ(¯X)<0.
In the second case , the group is cyclic elementary, generated by the hyperbolic isometry with translation length , length of the unique closed geodesic of the cylinder .
The proof of this theorem is based on a functional equation for Selberg zeta function for convex co-compact groups obtained in previous work [17] and the observation that the DN map for is, modulo constant, the scattering operator of the uniformized non-compact manifold at the parameter value : this is discussed in more generality at the end of the introduction. We emphasize that the Theorem holds even when the boundary has more than one connected component, an important fact that we need in the proof being that is always equal to the space of constants and not the locally constant functions.
Remark : Note that for odd dimensional closed hyperbolic manifolds , the value for some acyclic representation of the of the unit tangent bundle is the Reidemeister torsion of by a result of Fried [10].
Remark : It is also worth to say that the proof shows that is always a resonance of multiplicity with resonant state for the Laplacian on any convex co-compact surface except when it’s a cylinder where it is then of multiplicity ; as a byproduct it also gives the exact order of vanishing of at , which was not apparently known in that case.
The next natural uniformizations for oriented compact surfaces with boundary are given by Osgood-Phillips-Sarnak [29] (see also Brendle [3]), they are of two types: each conformal class of a metric on an oriented compact surface with boundary has a unique
• metric with constant Gauss curvature and with totally geodesic boundary.
• flat metric with constant geodesic curvature boundary.
The Gauss curvature on and the geodesic curvature of the boundary are linked through the Gauss-Bonnet formula
∫¯XKdvolg+∫∂¯Xkdℓg=2πχ(¯X).
The flat uniformization has been used by Edward-Wu [7] to show that for a topological disc (i.e. ), their explicit computation is possible thanks to the circular symmetry of the uniformized flat disc with constant geodesic curvature. In a similar way, we give in the appendix the explicit computation for the planar annulus whose boundary is the union of two concentric circles and show that it fits with the value found in Theorem 1.2 for the hyperbolic cylinder conformal to this annulus. In the case , the computation of in terms of geometric quantities by using the flat uniformization does not seem apparent at all. As for the constant curvature with geodesic boundary uniformization, the topological disc (i.e. ) is uniformized by a half-sphere of curvature , the topological cylinder (i.e. ) by a flat cylinder and in both cases, the value can be easily computed using decomposition in spherical harmonics of the Laplacian, essentially like for the flat uniformization. However, when , the constant curvature uniformization with totally geodesic boundary appears to be more useful to compute . Indeed it yields a metric on which is isometric to for some discrete group of isometries of (containing symmetries of order ) and the double of along the boundary is the closed hyperbolic surface where is the index subgroup of orientation preserving isometries of . The Mayer-Vietoris formula for determinants by Burghelea-Friedlander-Kappeler [4] reads in this case
det′(N)ℓ(∂¯X)=−12πχ(¯X)det′(ΔG∖H2)(det(ΔG0∖H2))2
where is the Dirichlet realization of the Laplacian on . We are thus interested in the value of the regularized determinants of these Laplacians. The determinant has been computed by Sarnak and Voros [34, 37] in terms of the derivative at of the Selberg zeta function defined by (1.2) with . Using a trace formula of [19], we prove a similar formula for in terms of a Selberg zeta function at , where the natural Ruelle and Selberg zeta functions for this case with boundary are defined as follows (see [19, Section 5]): let be the lengths of the geodesic boundary components of and let be the set of primitive oriented closed geodesics of length and with geometric reflections (according to the geometric optic law) on , then the zeta functions333The function we use is actually the square root of that of [19]. are defined by the following products:
(1.4) R∂¯X(λ):=N∏j=1(1−e−λℓj)2,RG0(λ):=∏c∈[C](1−(−1)nce−λℓc)(1−e−(λ+1)ℓc),ZG0(λ):=∏k∈N0R∂¯X(λ+2k)RG0(λ+2k).
We show the
###### Theorem 1.3.
Let be a compact oriented surface with boundary, with negative Euler characteristic . Let be the unique, up to isometry, constant negative curvature metric with totally geodesic boundary on and let be the discrete group such that is isometric to . Let be the subgroup of of orientation preserving isometries and be the associated Selberg zeta function of and , then
det′(N)ℓ(∂¯X)=−Z′G(1)(ZG0(1))2eℓ(∂¯X)/42πχ(¯X)
Thus, if is the uniformization of given by Theorem 1.2, then
[λχ(¯X)−1RΓ(λ)]|λ=0=−Z′G(1)ZG0(1)2eℓ(∂¯X)4(2π)−χ(¯X).
Although the products defining do not converge, we can view the last identity of Theorem 1.3 as a relation between length spectrum of and , which does not appear obvious at all. Let us also remark that the determinant of the Laplacian on an hyperbolic compact surface has different expressions with Selberg zeta values, the Sarnak-Voros [34, 37] one related to the Fuchsian uniformization and the McIntyre-Takhtajan [27] related to the Schottky uniformization.
In the last section we discuss in more generality (in higher dimension) the relation between Dirichlet-to-Neumann map and scattering operator. An -dimensional asymptotically hyperbolic manifold is a complete Riemannian non-compact manifold, which is the interior of a smooth compact manifold with boundary such that for any boundary defining function of (i.e. and ), then is a smooth metric on such that on . The metric induced on depends on and another choice of yields a metric on conformal to , we thus define the conformal infinity of as the conformal class of on . There is a natural meromorphic family of operators (defined in Section 2) (for ) called scattering operator, acting on , these are elliptic conformally covariant pseudo-differential operators of order with principal symbol where is a conformal representative of the conformal infinity of . When is Einstein, Graham and Zworski [15] showed that for are conformal powers of the Laplacian on the boundary , initially defined in [13]. Since has order when and the same principal symbol than a Dirichlet-to-Neumann map on the compact manifold , we may expect that it is realized as a DN map for an elliptic compact problem with boundary. We observe that when has constant scalar curvature (for instance if is Einstein), then is the Dirichlet-to-Neumann map of the conformal Laplacian on a whole class of smooth metric on , conformal to , with and with minimal boundary . Conversely it is clear that there is no constant curvature uniformization when , but instead there is a solution of a singular Yamabe problem, that is, for a given , there exists an asymptotically hyperbolic metric with constant scalar curvature on the interior in the conformal class of . The existence and regularity of such a solution of this singular Yamabe problem is due to Aviles-Mac Owen [2], Mazzeo [25] and Andersson-Chruściel-Friedrich [1]. If is the mean curvature of for and is the DN map for the conformal Laplacian , we show that is the value for a complete manifold with constant negative scalar curvature, conformal to on the interior of . Note that is known to be the natural conformally covariant operator on the boundary associated to , see [5].
Acknowledgement. We thank E. Aubry, P. Delanoë, M. Harmer, A. Hassell, R. Mazzeo and P. Perry for useful discussions and for pointing out the right references. C.G. acknowledges support of NSF grant DMS0500788, and french ANR grants JC05-52556 and JC0546063.
## 2. Computation of det′(N) using Mazzeo-Taylor uniformization
We now recall the definition of the scattering operator on an asymptotically hyperbolic manifold of dimension . From Graham-Lee [14], for any choice in the conformal infinity , such a metric can be written uniquely in a collar neighbourhood of the boundary under the form
(2.1) g=dx2+h(x)x2,h(0)=h0
for some smooth -parameter family of metric on ( is a boundary defining function of ). If has a Taylor expansion at with only even powers of , then is called even (see [16]). If is even and , and , then where444We changed the convention since in the literature, would be the scattering operator acting on . is the normalisation constant and are defined by solving the Poisson problem [15]
(2.2) (Δg−λ(n−λ))uλ=0,uλ=xn−λu−λ+xλu+λ,u±λ∈C∞(¯X),u−λ|∂¯X=f.
We see that depends on and on the choice of or equivalently on the choice of conformal representative of the conformal infinity of . Changing into with induces the scattering operator
(2.3) ^S(λ)=e−λω0S(λ)e(n−λ)ω0.
From [22, 15], is holomorphic in the half plane , moreover it is a pseudodifferential operator of order with principal symbol (thus elliptic) and it is self-adjoint when , which makes its zeta regularized determinant well defined by [24]. If the dimension is even, one shows easily that if is conformal to , the conformal relation (2.3) between the associated operators and implies that , see [17, Sec. 4] for instance.
We are back to our case of surfaces (here ), thus let be a smooth Riemannian surface with boundary. We first relate to the scattering operator of an associated non-compact hyperbolic surface. Let be a function that defines and such that for some metric on , so the normal vector field to the boundary is on . Let be the unique complete hyperbolic metric on the interior of , obtained by Mazzeo-Taylor [26], where is some smooth function on , then is an asymptotically hyperbolic manifold in the sense stated in the introduction. Then is even since the metric outside some compact is the metric on a hyperbolic funnel, that is on for some (it suffices to set to have a model form (2.1)). Therefore the geodesic function such that is like (2.1) implies and . By studying the Poisson problem at energy close to for , for any , there exists a unique such that
(2.4) (Δg−λ(1−λ))uλ=0,uλ∼x→0x1−λ(f+∞∑j=1x2jf−2j(λ))+xλ(c(λ)S(λ)f+∞∑j=1x2jf+2j(λ))
for some (we used evenness of the metric so that odd powers of are zeros, see [15]). In particular at we have and but thus
Δ¯gu=0,u∈C∞(¯X),u=f−xS(1)f+O(ρ2),
but since on we automatically get
###### Lemma 2.1.
The Dirichlet-to-Neumann map for is given by the scattering operator at energy for the Laplacian on the asymptotically hyperbolic surface conformal to , where is defined using the boundary defining function associated to the representative of the conformal infinity of .
Taking a conformal metric on gives a Laplacian and the normal vector field to the boundary becomes where . We deduce that the associated Dirichlet-to-Neumann map satisfies .
###### Theorem 2.2.
Let and be two conformally related metrics on a surface with boundary , and let be the respective Dirichlet-to-Neumann operators. Then where is the length of the boundary for the metric , .
(2.5) det′(Ni)=exp(TR(log(Ni)(1−Πi))),i=0,1
where is the Kontsevich-Vishik canonical trace defined in [24], is defined by a contour integral (see [24, 32] for details), and the orthogonal projection onto with respect to the volume density on induced by , i.e. the projection onto the constants for the volume density . It is important to note that this formula holds (i.e. Guillemin-Wodzicki residue trace does not show-up in the formula) since the DN maps have regular parity in the sense of [17, Sect. 2] and thus as well: indeed, take the Mazzeo-Taylor uniformization of (which is the same than that of ), then Proposition 3.6 of [17] shows that the scattering operator associated to has regular parity in the sense of [17, Sect. 2] since the hyperbolic metric is even; this implies by using Lemma 2.1 that has regular parity for . If is a conformal change and where , then the DN map for the metric is unitarily equivalent to the self-adjoint operator on , with kernel projector
(2.6) Πt=(ℓht(∂¯X))−1etω2⊗etω2
where is the length of for the metric . First we have that
∂t(log(Nt)(1−Πt))=(∂tNt)N−1t(1−Πt)−log(Nt)∂tΠt
where is the unique operator (modulo ) satisfying , that is : indeed, multiplying on the left by gives
Nt(1−Πt)etω02N−10etω02(1−Πt) = e−tω02(1−Π0)etω02(1−Πt) = (1−ℓht(∂¯X)e−tω0Πt)(1−Πt)=1−Πt
and the same holds by multiplying on the right by . Thus taking the log derivative of with respect to gives (by the same arguments than [17, Sec. 4]) that
∂tlog(det′(Nt))=−TR(ω0(1−Πt))−Tr((log(Nt)∂tΠt)
where Tr is the usual trace. Using (2.6), we compute but since and the trace is cyclic, we have
Tr((log(Nt)∂tΠt)=Tr(log(Nt)ω0Πt)=Tr(Πtlog(Nt))=0.
Now since the Kontsevich-Vishik trace of a differential operator is in odd dimension (see [24]), but the trace of a smoothing operator is the integral on the diagonal of its Schwartz kernel, therefore
TR(ω0Πt)=Tr(ω0Πt)=∫∂¯Xω0etω0dvolh0∫∂¯Xetω0dvolh0=∂tlog(ℓht(∂¯X))
Then integrating in we get the right law for the determinant ∎
We now prove Theorem 1.2.
Proof of Theorem 1.2: Since , we have to compute . It is clear that the kernel of is one dimensional, composed of the constants, since it is the case for . According to the main formula of Paycha-Scott [32] we have for
where is the projection onto the constants. To compute , we shall analyze in the neighborhood of .
From [17, Th. 1.3] and the fact that has no zero-eigenvalue, has no pole in a neighbourhood of and is holomorphic near with a zero of order where the multiplicity for is defined by
(2.7) νλ0:=−Tr(Resλ=λ0(∂λS(λ)S−1(λ)))
Let us now compute . We consider the largest integer such that there exists a holomorphic (in ) family of functions in with , and such that . This maximum is achieved for some , is positive and is exactly by Gohberg-Sigal theory (see [11] or [18]). Thus there exists a family of functions on , holomorphic in , with such that for some function . Then setting we get the equation
S(λ)uλ=(λ−1)(S(1)v+S′(1)u1)+O((λ−1)2)
which we multiply with , integrate and use self adjointness of with to deduce
(2.8) ⟨S(λ)uλ,u1⟩=(λ−1)∫∂¯Xu1S′(1)u1 dvolh0+O((λ−1)2).
Recall that is constant since in the kernel of , but following the notation of Fefferman-Graham [8, Th. 4.3], we have a kind of curvature defined by555They actually define where , so clearly since and . and they prove the identity666We emphasize that their proof is only based on Green’s identity and evenness of the metric expansion at the boundary. In particular it includes the case of hyperbolic surfaces.
∫∂¯XQdvolh0=−\rm 0-vol(X)
where is the renormalized volume (also called -volume) of , i.e. the constant in the expansion
Vol(x>ϵ)=c0ϵ−1+V+O(ϵ), as ϵ→0.
It is however proved, from Gauss-Bonnet formula, by Guillopé-Zworski [21] and Epstein [31, Appendix] that
(2.9) \rm 0-vol(X)=−2πχ(¯X)=2π(2g+N−2)
where is the Euler characteristic of , is its genus and the number of boundary components. It follows that the coefficient of in (2.8) does not vanish if and then .
Recall (see [15]) that is self-adjoint for real, Fredholm, analytic in near and invertible in a small pointed disc (of radius ) centered at , moreover has as isolated eigenvalue of multiplicity , then one can use Kato perturbation theory [23, VII,3] to deduce that for near , the spectrum of the operator near is an isolated eigenvalue of multiplicity , that we denote ; moreover it is holomorphic in near and there is a holomorphic normalized associated eigenvector . We have where , being the length of the curve for the metric , and we get the equation
(2.10) S(λ)wλ=α(λ)wλ,wλ=ℓ(∂¯X)−1/2+(λ−1)v+O((λ−1)2),α(λ)=(λ−1)β+O((λ−1)2)
for some and . Taking a Taylor expansion of (2.10) yields
S(1)v+S′(1)w1=βw1
where we used the notation for . Pairing as before with and using that is self adjoint and previous arguments with gives
(2.11) β=−2πχ(¯X)ℓ(∂¯X).
Now let be the orthogonal projection onto , and we define the function
h(λ):=exp(TR(log(S(λ))(1−Πλ))).
It is analytic near and thus the limit of at is by (2.5), the value we search to compute. For close to but , we have , thus using the first identity in (2.5) and the fact that is the usual trace on finite rank operators, we obtain
h(λ)=exp(TR(logS(λ))exp(−logα(λ))=det(S(λ))/α(λ).
But by (2.10), which proves that
det′(S(1))=limλ→1detS(λ)β(λ−1).
In [17], we proved the functional equation
which, following Voros [37, Eq 7.24, 7.25], can be written under the form
(2.12) det(S(λ))=ZΓ(1−λ)ZΓ(λ)((2π)1−2λΓ(λ)G(λ)2Γ(1−λ)G(1−λ)2)−χ(¯X)
where is the Barnes function (see [37, Appendix]) which satisfies in particular and . Writing in (2.12) and using that is holomorphic at implies that
det′(S(1)) = −β−1[limλ→1RΓ(1−λ)(1−λ)1−χ(¯X)]limλ→1[(1−λ)(2π)1−2λΓ(1−λ)Γ(λ)G(λ)2G(2−λ)2]−χ(¯X) = (2π)χ(¯X)−1ℓ(∂¯X)χ(¯X)limλ→1RΓ(1−λ)(1−λ)1−χ(¯X).
and we are done when .
If , is a topological disc and the uniformization that puts a complete hyperbolic metric on is the usual hyperbolic disc. The proof [17] of the formula (2.12) remains true by setting and we can proceed as before where now is the projection on the constants if is like above. We finally obtain
det′(S(1))=−ℓ(∂¯X)2πlimλ→1det(S(λ))λ−1=ℓ(∂¯X)
and we are done for this case. Notice that it matches with the result of Edward-Wu [7].
The last case corresponds to the cylinder, whose interior is uniformized by the cyclic elementary group , with a unique closed geodesic of length , the translation length of the generator . In other words, is conformal to the hyperbolic cylinder | {"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": 0.9953912496566772, "perplexity": 392.90201430140763}, "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/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00139.warc.gz"} |
https://testbook.com/objective-questions/mcq-on-moment-of-forces--5eea6a1539140f30f369f461 | # Condition of static equilibrium of a planar force system is written as
1. ΣF = 0
2. ΣM = 0
3. ΣF = 0 and ΣM = 0
4. None of these
Option 3 : ΣF = 0 and ΣM = 0
## Detailed Solution
Explanation:
Equilibrium of planar forces:
If the forces acting on the free-body are non-concurrent then the equivalent resultant will be a single force acting at a common point and a moment about the same point. The effect of such a force system will be to translate the body as well as to rotate it. Hence, for equilibrium to exist, both the force and moment must be null vectors.
i.e ΣF = 0, ΣM = 0
When the forces acting on a body lie in a plane (X - Y plane) but are non-concurrent, the body will have rotational motion perpendicular to the plane in addition to the translational motion along the plane, Hence necessary and sufficient condition for static equilibrium are,
ΣFx = 0, ΣFy = 0, ΣMz = 0
# A 1 m long uniform beam of 2 kg mass is being lifted vertically up by a force F at the 100 cm mark. What is the minimum force required to do so?
1. 1 N
2. 2 N
3. 10 N
4. 20 N
Option 3 : 10 N
## Detailed Solution
Concept:
Conditions for the system to be in equlibrium
ΣFx = 0, ΣFy = 0, ΣM = 0
Calculation:
Given:
m = 2 kg, Assume g = 10 m / s2
Lager will be the moment, smaller will be the force required to lift the rod. Hence, Applying Moment about 0 cm point we get.
w × 50 = F × 100
m × g × 50 = F × 100
2 × 10 × 50 = F × 100
F = 10 N
# A force F is acting on a bent bar which is clamped at one end as shown in the figure.The CORRECT free body diagram is
Option 1 :
## Detailed Solution
Explanation:
When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.
Following is the conclusion:
Option (B) is wrong because the ground support is shown which we never show in FBD.
Option (C) is wrong because the X component is not shown.
Option (D) is wrong because the moment produced due to the eccentric force is not shown.
Hence only option (A) is correct.
# A force f = (10î + 8ĵ - 5k̂) acts a point P(2, 5, 6). What will be the moment of the force about the point Q(3, 1, 4)?
1. – 36î + 15ĵ - 48k̂
2. 18î + 19ĵ - 37k̂
3. 32î - 31ĵ - 4k̂
4. Zero
Option 1 : – 36î + 15ĵ - 48k̂
## Detailed Solution
Solution
Concept:
Moment of a force about any point is given by
$$\overrightarrow {{m_0}} = \vec F \times \vec r$$
where $$\vec r = \overrightarrow {QP} = \;\vec P - \vec Q$$
Calculation:
Given:
$${\rm{\vec F}} = \left( {10\hat i + 8{\rm{\hat j}} - 5{\rm{\hat k}}} \right)$$P( 2, 5, 6), Q(3, 1, 4).
$$\vec r = \overrightarrow {QP} = \;\vec P - \vec Q$$
$$\vec r = \overrightarrow {PQ} = \;\vec Q - \vec P$$
$$\vec r=\left( {2\hat i + 5\hat j + 6\hat k} \right)- \left( {3\hat i + 1\hat j + 4\hat k} \right)$$
∴ $$\vec r = \left( {-1\hat i +4\hat j + 2\hat k} \right)$$
$$\therefore \;{\vec m_0} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ -1&{ 4}&{ 2}\\ {10}&8&{ - 5} \end{array}} \right|$$
$${\vec m_0} = \hat i\left( {-20 - 16} \right) - \hat j\left( { 5 - 20} \right) + \hat k\left( {-8 - 40} \right)$$
$${\vec m_0} = - 36\hat i + 15\hat j - 48\hat k$$
# Two like parallel forces are acting at a distance of 30 mm apart and their resultant is 60 N, if the line of action of the resultant is 10 mm from one of the forces, the two forces are
1. 80 N and -20 N
2. 40 N and 20 N
3. 30 N and -90 N
4. 35 N and 25 N
Option 2 : 40 N and 20 N
## Detailed Solution
Concept:
The moment about the resultant is zero.
Calculation:
Let the two forces be F1 and F2
Given:
F1 + F2 = 60 N (1) (They are like forces and parallel)
Taking the moment about the resultant
$$\begin{array}{l} \mathop F\nolimits_1 \times 10 - \mathop F\nolimits_2 \times 20 = 0\\ \mathop F\nolimits_1 = 2\mathop F\nolimits_2 \end{array}$$
By solving, we get
F2 = 20 N
F1 = 40 N
# The moment of a force about any point is equal to the algebraic sum of moments of its components about that point is stated by:
1. Lufkin’s principle
2. Varignon’s principle
3. Henry’s principle
Option 2 : Varignon’s principle
## Detailed Solution
Explanation:
Varignon’s Principle of moments (or the law of moments)
• It states, “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”
MO' = R × r = F1 × r1 + F2 × r2
Lami's theorem
• Lami's theorem states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Consider three forces FA, FB, FC acting on a particle or rigid body making angles α, β, and γ with each other.
Therefore, $$\frac{{{F_A}}}{{sin\alpha }} = \frac{{{F_B}}}{{sin\beta }} = \frac{{{F_C}}}{{sin\gamma }}$$
D’ Alembert's principle:
• It states that a moving body can be brought to equilibrium by adding inertia force to the system whose magnitude is equal to the product of mass and acceleration.
• This inertia force is in the opposite direction to that of acceleration, i.e. F = ma
Polygon Law of Forces
• It is an extension of Triangle Law of Forces for more than two forces, which states, “If a number of forces acting simultaneously on a particle, be represented in magnitude and direction, by the sides of a polygon taken in order; then the resultant of all these forces may be represented, in magnitude and direction, by the closing side of the polygon, taken in the opposite order.”
# A 16 N force produce a moment of 64 Nm. The moment arm is
1. 2 m
2. $$\sqrt 2 \;m$$
3. 8 m
4. 4 m
Option 4 : 4 m
## Detailed Solution
Concept:
Moment is the product of force multiplied by the moment arm length.
Moment = Force × (moment arm length)
Calculation:
Given:
Force = 16 N, moment = 64 Nm
Moment arm length = 64/16 m
Moment arm length = 4 m
Hence, the length of moment arm is 4 m.
# Two unlike parallel forces, each of magnitude 50 kN are 200 mm apart from each other. What will be the magnitude of the moment of the couple formed by these two forces?
1. 5 kN m
2. 10 kN m
3. 20 kN m
4. 0
Option 2 : 10 kN m
## Detailed Solution
Concept:
A pair of two equal and unlike parallel forces (i.e. forces equal in magnitude, with lines of action, parallel to each other and acting in opposite directions) is known as a couple.
A couple is unable to produce any translatory motion (i.e., motion in a straight line). But it produces a motion of rotation in the body, on which it acts.
Moment of a couple = P × a
Calculation:
Magnitude of two unlike forces = P = 50 kN
Apart distance = a = 200 mm = 0.2 m
The magnitude of the moment of the couple,
M = P × a = 50 × 0.2 = 10 kN m
# According to the law of moments, if a number of coplanar forces acting on a particle are in equilibrium, then-
1. The algebraic sum of their moments about any point in their plane is zero
2. Their lines of action are at equal distances
3. The algebraic sum of their moments about any point is equal to the moment of their resultant force about the same point
4. Their algebraic sum is zero
Option 3 : The algebraic sum of their moments about any point is equal to the moment of their resultant force about the same point
## Detailed Solution
Concept:
Varignon’s Principle of moments (or the law of moments)
It states, “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”
MO' = R × r = F1 × r1 + F2 × r2
Important Points
Lami's theorem
Lami's theorem states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Consider three forces FA, FB, FC acting on a particle or rigid body making angles α, β, and γ with each other.
Therefore, $$\frac{{{F_A}}}{{sin\alpha }} = \frac{{{F_B}}}{{sin\beta }} = \frac{{{F_C}}}{{sin\gamma }}$$
# If the line of action of all the forces are along the same line, then the forces are said to be-
1. Collinear forces
2. Coplanar concurrent forces
3. Coplanar parallel forces
4. Non-coplanar non-concurrent forces
Option 1 : Collinear forces
## Detailed Solution
Concept:
• Collinear forces: The forces, whose lines of action lie on the same line, are known as collinear forces.
• When two or more forces act on a body, they are called to form a system of forces.
• Coplanar forces: The forces, whose lines of action lie on the same plane, are known as coplanar forces.
• Concurrent forces: The forces, which meet at one point, are known as concurrent forces. The concurrent forces may or may not be collinear.
• Coplanar concurrent forces: The forces, which meet at one point and their lines of action also lie on the same plane, are known as coplanar concurrent forces.
• Coplanar non-concurrent forces: The forces, which do not meet at one point, but their lines of action lie on the same plane, are known as coplanar non-concurrent forces.
• Non-coplanar concurrent forces: The forces, which meet at one point, but their lines of action do not lie on the same plane, are known as non-coplanar concurrent forces.
• Non-coplanar non-concurrent forces: The forces, which do not meet at one point and their lines of action do not lie on the same plane, are called non-coplanar non-concurrent forces.
# A block weighing W = 20 kN is resting on an inclined plane which makes an angle of 30° to the horizontal. The component of gravity force parallel to the inclined plane is
1. 5 kN
2. 17.32 kN
3. 10 kN
4. 14.14 kN
Option 3 : 10 kN
## Detailed Solution
Concept:
w = weight of the body, θ = angle of inclination of the plane with the horizontal
Calculation:
Given:
W = 20 kN, θ = 30°
Gravity force parallel to the inclined plane is
$${F_g} = Wsin~\theta = 20 \times \sin \left( {30} \right) = 20 \times \frac{1}{2} = 10\;kN$$
The effort (P) required moving the body up an inclined plane is given by
$$P = \frac{{W.{\rm{sin}}\left( {\alpha + \phi } \right)}}{{{\rm{sin}}\left[ {\theta - \left( {\alpha + \phi } \right)} \right]}}$$
The effort (P) required moving the body down an inclined plane is given by
$$P = \frac{{W.{\rm{sin}}\left( {\alpha - \phi } \right)}}{{{\rm{sin}}\left[ {\theta - \left( {\alpha - \phi } \right)} \right]}}$$
# Two forces P and Q are acting at a particle. The angle between the forces is θ and the resultant of the forces is R. If the resolved part of R in the direction of P is 2P, then
1. P = Q cosec θ
2. P = Q sin θ
3. P = Q tan θ
4. P = Q cos θ
Option 4 : P = Q cos θ
## Detailed Solution
Concept:
Parallelogram Law of Forces:
If two forces, acting at a point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.
Resultant force Fr, of any two forces F1 and F2 with an angle θ between them, can be given by vector addition as
$$F_r^2 = F_1^2 + F_2^2 + 2{F_1}{F_2}\cos θ$$
$$F_r = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos θ}$$'
Calculation:
Given:
Two forces P and Q are acting at a particle. The angle between the forces is θ and the resultant of the forces is R:
If the resolved part of R in the direction of P is 2P:
Resolved part of R = Q cosθ + P
i.e. 2P = Q cosθ + P
P = Q cosθ
# The resultant of forces P, Q, R acting along the sides BC, CA, AB of triangle ABC passes through its circumcenter, then
1. P sin A + Q sin B + R sin C = 0
2. P cos A + Q cos B + R cos C = 0
3. P sec A + Q sec B + R sec C = 0
4. P tan A + Q tan B + R tan C = 0
Option 2 : P cos A + Q cos B + R cos C = 0
## Detailed Solution
Calculation:
Let 'O' be the circumcentre ⇒ OA = OB = OC = r
consider ΔBOC
⇒ OD = r.cos A
Similarly OE = r.cosB, OF = r.cosC
Let the moment of forces about be N0;
→ Moment due to resultant about O, will be equal to sum of moment of all force about O.
M0 = MR = MP + MQ + MR
MP + MQ + MR = P(OD) + Q(OE) + R(OF)
= (P cosA + P cosB + P cosC)r → (1)
Given, resultant passes through 'O',
⇒ MR = 0 → (2)
From (1) & (2),
r(P. cosA + Q. cos B + R.cosC) = 0
since r ≠ 0
⇒ P. cosA + Q. cos B + R.cosC = 0
Important Points
The forces P, Q, R act along the sides BC, CA, AB of Δ ABC. Their resultant passes through.
1. Incentre, if P + Q + R = 0
2. Orthocentre, if P sec A + Q sec B + R sec C = 0
3. Centroid, if P cosec A + Q cosec B + R cosec C = 0
# Stationary structure acted upon by a system of forces include external loads, reactions and body forces caused by which of the following elements
1. External tension
2. Weight
3. External moment
4. Pressure
Option 2 : Weight
## Detailed Solution
Explanation:
Effect of weight on the reaction force developed in the body:
• The weight of an object is the force due to gravity acting between the earth and the object.
• As the scale of the earth is large compared with many everyday objects, the weight force acting on an object F can be related to its mass, through the expression F = mg, where g is the acceleration due to gravity at the surface of the earth.
• When a force acts on a body, it will accelerate in the direction of the force unless there is a balancing force to oppose it. For an object sitting on a plane or in a stationary condition, this balancing force is called a normal reaction or just reaction, which acts normally and upwards (perpendicular) to the plane. If there is friction between the body and the plane which acts against any sliding tendency, a frictional force acts upwards along the plane.
• Higher the weight, the higher will the resistance against the movement, and the body will develop corresponding reactions.
• So, from the above discussion, it can be concluded that reaction is caused by the weight of the body.
# Three forces acting on a rigid body are represented in magnitude, direction and line of action by the three sides of a triangle taken in order. The forces are equivalent to a couple whose moment is equal to___
1. Thrice the area of the triangle
2. Twice the area of the triangle
3. The area of the triangle
4. Half the area of the triangle
Option 2 : Twice the area of the triangle
## Detailed Solution
Concept:
$$Moment = P × OC$$
And
$$Area\;of\;triangle = \frac{1}{2} × AB × OC$$
$$= \frac{1}{2} × P × OC$$
= $$\frac{1}{2}$$× moment
∴ Moment = Twice the area of a triangle
# A system of forces with a resultant moment but NO resultant force is called :
1. a couple
2. a resultant
3. bending
4. friction
Option 1 : a couple
## Detailed Solution
Explanation:
(i) In mechanics, a couple is a system of forces with a resultant (also known as net or sum) moment but no resultant force.
(ii) A better term is force couple or pure moment. Its effect is to create rotation without translation, or more generally without any acceleration of the center of mass. In rigid body mechanics, force couples are free vectors, meaning their effects on a body are independent of the point of application.
(iii) The resultant moment of a couple is called a torque. This is not to be confused with the term torque as it is used in physics, where it is merely a synonym of the moment.
# The dynamic effect of the resultant acting on a rigid body in equilibrium condition will be ________ effect of the given system of forces.
1. double the
2. zero the
3. the same
4. half the
Option 3 : the same
## Detailed Solution
Explanation:
Varignon’s Principle of moments (or the law of moments):
It states, “If a number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.”
i.e. R × r = F1 × r1 + F2 × r2
Therefore from the above principle, we can conclude that the dynamic effect of the resultant acting on a rigid body in equilibrium will be the same effect when acted by a given system of forces. | {"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": 0.8816826343536377, "perplexity": 816.7999396608458}, "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-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00550.warc.gz"} |
https://arxiv.org/abs/1601.07346 | # Title:Orientations of Morse flow trees in Legendrian contact homology
Abstract: Let $\Lambda$ be a spin Legendrian submanifold of the 1-jet space of a smooth manifold. We prove that the Legendrian contact homology of $\Lambda$ with integer coefficients can be computed using Morse flow trees.
Comments: 91 pages, lots of figures. More detailed explanations of how the formulas in Section 6.4 are derived, together with some corrections. Corrected a mistake in how to interpret the orientation of the space of conformal automorphisms (Section 4, eqn 4.6 and Section 6.4.F, case f0). Some changes in Section 7, due to the already mentioned changes Subjects: Symplectic Geometry (math.SG) MSC classes: 57R17, 53D42 Cite as: arXiv:1601.07346 [math.SG] (or arXiv:1601.07346v3 [math.SG] for this version)
## Submission history
From: Cecilia Karlsson [view email]
[v1] Wed, 27 Jan 2016 12:44:45 UTC (838 KB)
[v2] Mon, 17 Apr 2017 18:15:51 UTC (755 KB)
[v3] Mon, 12 Aug 2019 14:21:51 UTC (149 KB) | {"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": 0.8543408513069153, "perplexity": 2374.1376507515874}, "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-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00040.warc.gz"} |
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# Exponential and Radical Expressions Problem Set
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While the radical symbol is widely used, converting to rational exponents has advantages. Explain an advantage of rational exponents over the radical sign. Include in your answer an example of an equation easier to solve as a rational exponent rather then a radical sign.
1) Solve the following equations.
a) Show work in this space.
Show work in this space.
Show work in this space.
2) Is an identity (true for all values of x)?
3) x -2 -1 0 1 2
y .25 .5 1 2 4
Given the table above, graph the function, identify the graph of the function (line, parabola, hyperbola, or exponential), explain your choice, and give the domain and range as shown in the graph, and also the domain and range of the entire function.
Graph
Graph Type
Explanation
Domain
Range
4) For the function, y = 1/(x -1)
a) Give the y values for x = -2, -1, 0, 1, 2, 3
X
Y
Show work in this space.
b) Using these points, draw a curve.
Show graph here.
5) For the equation, perform the following:
a) Solve for all values of x that satisfies the equation.
Show work in this space
b) Graph the functions on the same graph (by plotting points if necessary). Show the points of intersection of these two graphs.
c) How does the graph relate to part a?
6) A right triangle is a triangle with one angle measuring 90°. In a right triangle, the sides are related by Pythagorean Theorem, where c is the hypotenuse (the side opposite the 90° angle). Find the hypotenuse when the other 2 sides' measurements are 3 feet and 4 feet.
Show work in this space
7) Suppose you travel north for 35 kilometers then travel east 65 kilometers. How far are you from your starting point? North and east can be considered the directions of the y- and x-axis respectively.
Show work in this space.
8) The volume of a cube is given by V = s^3. Find the length of a side of a cube if the Volume is 729 cm3.
Show work in this space.
#### Solution Preview
Please see the detailed solution with graphs in the attached files. | {"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": 0.8152986764907837, "perplexity": 785.9375621246058}, "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-2021-21/segments/1620243991943.36/warc/CC-MAIN-20210513173321-20210513203321-00471.warc.gz"} |
http://crypto.stackexchange.com/questions/6313/is-aes-reducible-to-an-np-complete-problem | # Is AES reducible to an NP-complete problem?
Is breaking AES NP-hard? Can the security of AES be reduced to a NP-complete problem?
If it is reducible, what does the reduction look like? If it is not reducible, why do we assume it is secure?
-
We know no such reductions for any cryptosystem we use in practice. Not AES, nor RSA. Such reduction isn't too useful in the first place, since complexity classes talk about asymptotic cost, whereas we want to know if a certain concrete key-size is secure. – CodesInChaos Feb 10 at 19:34
How does it "seems clear"? It is clear that some problems relating to some attacks lie in NP (assuming some kind of generalization which allows growing key sizes, simply guessing the right key and then verifying them solves it – the same goes for secret-key crypto), I don't see how you get that they are NP-hard. – Paŭlo Ebermann Feb 10 at 20:20
"Being broken when $P = NP$" doesn't mean "being unbroken when $P \neq NP$". RSA is broken when factoring is easy, but neither is factoring known to be NP-complete nor is it known that RSA can only be broken by factoring the modulus (this is just the best known way nowadays). It is similar for most other public-key algorithms. (And even less for symmetric "bit-shuffling" crypto.) – Paŭlo Ebermann Feb 10 at 22:18
Actually, it's not at all true that someone showing $P=NP$ implies that all cryptography is broken. For one, an existence proof (that is, a proof that $P=NP$ that does not give an explicit reduction) doesn't affect the practical security at all. In addition, an explicit reduction of an NP-hard problem of size $n$ that takes either $2^{1000} n$ or $n^{1000}$ operations would suffice to show $P=NP$, but would be of no practical concern. To be of concern, we would have to show that large NP-hard problems can be practically solved. – poncho Feb 11 at 18:06
@CodesInChaos why not make that first remark an answer? – owlstead Feb 12 at 0:44
There are some complexity-theoretic reasons to believe that cryptography can't be based on NP-completeness. For one example, see this paper by Akavia et al. Basically it all boils down to the mismatch between average-case hardness required for cryptography, and worst-case hardness required for NP-hardness. Moreover, many (most? all?) hard algebraic problems which serve as the basis for cryptography are in NP $\cap$ co-NP (for one such example, consider factoring). Such problems cannot be NP-complete unless NP = co-NP. You can see more discussion in this stackoverflow thread. | {"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": 0.8001295924186707, "perplexity": 1055.4139448817425}, "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-2013-48/segments/1386163806278/warc/CC-MAIN-20131204133006-00080-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://socratic.org/questions/59ee6b7111ef6b09d35af574 | Chemistry
Topics
# Question af574
Oct 23, 2017
3.17%
#### Explanation:
The idea here is that you need to use the known composition of the solution to figure out the mass of glucose, the solute, present in exactly $\text{100 g}$ of the solution $\to$ this is the solution's percent concentration by mass.
To do that, you can use the fact that solutions are homogeneous mixtures, which implies that they have the same composition throughout.
So if you dissolve $\text{7.79 g}$ of glucose in $\text{237.7 g}$ of water, you can say that the mass of the solution will be equal to
overbrace("7.79 g")^(color(blue)("mass of solute")) + overbrace("237.7 g")^(color(blue)("mass of solvent")) = overbrace("245.5 g")^(color(blue)("mass of solution"))
So, you know that this solution contains $\text{7.79 g}$ of solute in $\text{245.5 g}$ of solution, so you can say that $\text{100 g}$ of this solution will contain
100 color(red)(cancel(color(black)("g solution"))) * "7.79 g glucose"/(245.5color(red)(cancel(color(black)("g solution")))) = "3.17 g glucose"#
This means that the solution's percent concentration by mass will be equal to
$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% m/m = 3.17% glucose}}}}$
The answer is rounded to three sig figs, the number of sig figs you have for the mass of glucose.
##### Impact of this question
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http://mathhelpforum.com/math-topics/174093-newtons-law-cooling.html | # Math Help - Newtons Law of Cooling
1. ## Newtons Law of Cooling
Hi all,
Ive been trying to work this question out on my own for over a week now and cant seem to crack it. I hope its in the right Sub forum and im really sorry if its not.
A hot metal body is allowed to cool in still air and the difference in temperature, θ (Celsius) between the body and its surroundings is recorded at, t (seconds) from the beginning of the cooling period.
its gives me the following:
(t) θ
40.5 60
96.0 50
137.0 40
210.5 30
327.0 20
488.5 10
The question asks me to verify graphically the relationship between θ and t is of the form: θ=θoe-kt (-kt is to the power of θoe) Hence estimate the values of the constants θo and k. Then using this relationship compile a table of values of θ for values of t between 0 and 600 at intervals of 60.
I have taken the natural log (ln) of each of the temperatures in the table above and plotted them in the graph. I drew a line of best fit and found the y intercept, and took the natural exponential of it which is:
e4.25 = 70.11
I think 70.11 is the initial temperature at time 0. so θo= 70.11?
Is k the gradient of the line? if so how do i calculate the values of θ?
All help is greatly appreciated, it would be amazing to get this out of the way!
2. Newton's law says that $\theta = \theta_0e^{-kt}$. Take natural logarithms of that equation: $\ln\theta = \ln\theta_0 - kt$. So what you should do is to graph the logarithm of theta against t. The graph should then look like a straight line, with gradient –k and intercept $\ln\theta_0$. I haven't tried to check the numbers, but it looks as though that is what you have done. If so, you're right!
Once you have the values for $\theta_0$ and k, you can plug them into the original formula $\theta = \theta_0e^{-kt}$, and use it to find $\theta$ for any given value of t.
3. Ok i think im right so far. All i need now is to estimate the values of the constants θo and k.
I have worked out
θo = 70.11 and k = -0.004
Now to compile a table of values of θ for values of t between 0 and 600 at intervals of 60.
(t) θ
0 70.11
60 ?
120 ?
180 ?
240 ?
300 ?
And so on...
do i still use ?
I cant figure out how to do this?!
4. Thanks Opalg! But, plugging them straight into the formula is the bit im having trouble with! Would you be able to explain or show and example please?
5. Once you've done as Opalg suggested, you'll need to fit a straight line to the data using linear regression (least squares fit) or some such estimator. What do you get when you do that?
6. Sorry i dont know what that means! Im just not so confident about changing the subject of the formula to t.
7. Originally Posted by skyhook2
Ok i think im right so far. All i need now is to estimate the values of the constants θo and k.
I have worked out
θo = 70.11 and k = -0.004
Now to compile a table of values of θ for values of t between 0 and 600 at intervals of 60.
(t) θ
0 70.11
60 ?
120 ?
180 ?
240 ?
300 ?
And so on...
do i still use ?
I cant figure out how to do this?!
If $\theta_0 = 70.11$ and $k=-0.004$, then your formula is $\theta = 70.11e^{-0.004t}$. For example, when t = 120, $\theta$ is given by
$\theta = 70.11e^{-0.004\times120} = 70.11e^{-0.48} \approx 70.11\times0.62\approx43.4.$
But there's something wrong with this! It says that after 120 seconds, the temperature difference between the hot object and its surroundings (43.4 degrees) is greater than it was at t=60 seconds, namely 40.5 degrees according to the given data. In other words, the object has started to heat up again!
Once you see that this can't be correct, you should look back to check the values that you got for $\theta_0$ and k. The value of $\theta_0$ is the temperature difference at time 0, which should obviously be greater than its value at any later time. But you are told that $\theta = 488.5$ when t=10. So the value $\theta_0 = 70.11$ is very wrong. I should think it would be about ten times that.
8. Here's what I got - I think it looks pretty reasonable.
9. opalg, if you look back to the first part it gives me the following (you have theta and t the wrong way round :
(t) ----------- θ
40.5 ----------- 60
96.0 ----------- 50
137.0 ----------- 40
210.5 ----------- 30
327.0 ----------- 20
488.5 ----------- 10
I have worked out the gradient wrong!
I think it is -0.2.
10. Originally Posted by skyhook2
you have theta and t the wrong way round
That would explain it! | {"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": 15, "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": 0.8321773409843445, "perplexity": 510.39843368462857}, "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/1404776426734.39/warc/CC-MAIN-20140707234026-00098-ip-10-180-212-248.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/two-true-false-questions-i-dont-understand.560369/ | # Two true/false questions I don't understand
1. Dec 14, 2011
### IntegrateMe
1. If a is a positive, then the function h(x) = (ln(ax^2)+x)/x is an antiderivative of j(x) = (2-ln(ax^2))/x^2
So, I used Wolfram and took the integral of j(x) with different values for a and always got ln(ax^2)/x, so I put false. However, the answer is true, and I can't figure out why!
2. If x = a is a critical point of a function m(x), then m'(a) = 0.
For this one I put true, and the answer is false. Is it because m'(a) can also be undefined?
Thank you!
2. Dec 14, 2011
### Dick
The difference between (ln(ax^2)+x)/x and ln(ax^2)/x is a constant. What constant? So they are both antiderivatives of (2-ln(ax^2))/x^2. And sure, m'(a) might be undefined at a critical point.
3. Dec 14, 2011
### IntegrateMe
If f"(a) = 0, then f has an inflection point at x = a.
The answer is false, but I thought inflection points were where the second derivative is equal to 0? Any clarification on this one?
4. Dec 14, 2011
### Dick
Same deal. An inflection point is where the concavity changes from concave up to concave down or vice versa. Define f(x)=x^2 for x>=0 and -x^2 for x<0. The derivative exists and is continuous and x=0 is an inflection point, but the second derivative doesn't exist there.
5. Dec 14, 2011
### IntegrateMe
But this problem is explicitly saying that the second derivative is equal to 0 at a. So shouldn't a on the original graph be an inflection point?
6. Dec 14, 2011
### Dick
Yeah, I was going backwards. Think about f(x)=x^4. Is x=0 an inflection point?
7. Dec 14, 2011
### IntegrateMe
No, it's not! I can see what you mean graphically, but can you explain it more clearly? I'm sorry I didn't follow the first time.
8. Dec 14, 2011
### Dick
I mean that if f(x)=x^4 then f''(x)=0. But f(x) is concave up everywhere.
9. Dec 14, 2011
### IntegrateMe
So if we're given a function and asked to find inflection points, how can we verify that it's an inflection point besides checking that the second derivative is equal to 0? Do we have to check the first derivative and see if it changes from +/- or -/+ at that point as well?
10. Dec 14, 2011
### Dick
No, I think you have to check if the second derivative changes sign.
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http://www.physicsforums.com/showthread.php?p=4188078 | # Does there exist a pair of interesecting lines that arent coplanar
by Bipolarity
Tags: coplanar, exist, interesecting, lines, pair
P: 783 Suppose you have two lines that intersect at the point O. Let's say the lines are OP and OQ. Could you construct a plane that contains both these lines? I can visually imagine this happening in the 3-space, but I may be wrong. I was just wondering, because this allows one to define an angle between two lines in 3-space if I remember correctly. What about in an n-space? Must two lines form a plane in an n-space? I imagine that in the same way two points define a line in n-space, two lines can define a plane in n-space, but because of my limited knowledge of linear algebra, I can't prove this. But I'm hoping to learn, perhaps someone can offer me their insights. BiP
Mentor P: 16,545 Yes, this is indeed true. Let's prove it. But to prove that, we need to know what a line is. What would you think the definition of a line is in n-dimensional space? Let's work with lines through the origin since it's easier.
P: 5,462 Have you given up on this? For n dimensions a straight line is represented by a set of (n-1) simultaneous equations. It is only represented by a single equation in 2D. In general $$\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}} = ...........$$ Which is complete for n=3 (2 equations) and continued by the dots.
Math
Emeritus
Thanks
PF Gold
P: 38,879
## Does there exist a pair of interesecting lines that arent coplanar
Another way of looking at it, in three space, is this: given two intersecting lines we can define a vector in the direction of each line. The cross product of the two vectors is perpendicular to both and the plane defined by that perpendicular vector and the point of intersection of the two lines contains both lines.
For example, we can always set up a coordinate system so that the origin is at the point of intersection and the x-axis extends along one of the lines. Parametric coordinates for that line, in that coordinate system, are x= t, y= 0, z= 0. We can take x= as, y= bs, z= ds as parametric equations for the other line. <1, 0, 0> is a vector in the direction of the first vector and <a, b, c> is a vector in the direction of the other line. Their cross product, <0, -c, b> is perpendicular to both and the plane -cy+ bz= 0 contains both lines.
As for defining an angle in n-space, most commonly we define it by exending the formula for dot product in two and three dimensional spaces:
$u\cdot v= |u||v|cos(\theta)$ where |u| and |v| are the lengths of vectors u and v and $\theta$ is the angle between them. As I say, we can prove that is true in two and three dimensions and then use it as definition of "angle" between two lines in higher dimensions.
P: 783
Quote by HallsofIvy Another way of looking at it, in three space, is this: given two intersecting lines we can define a vector in the direction of each line. The cross product of the two vectors is perpendicular to both and the plane defined by that perpendicular vector and the point of intersection of the two lines contains both lines. For example, we can always set up a coordinate system so that the origin is at the point of intersection and the x-axis extends along one of the lines. Parametric coordinates for that line, in that coordinate system, are x= t, y= 0, z= 0. We can take x= as, y= bs, z= ds as parametric equations for the other line. <1, 0, 0> is a vector in the direction of the first vector and is a vector in the direction of the other line. Their cross product, <0, -c, b> is perpendicular to both and the plane -cy+ bz= 0 contains both lines. As for defining an angle in n-space, most commonly we define it by exending the formula for dot product in two and three dimensional spaces: $u\cdot v= |u||v|cos(\theta)$ where |u| and |v| are the lengths of vectors u and v and $\theta$ is the angle between them. As I say, we can prove that is true in two and three dimensions and then use it as definition of "angle" between two lines in higher dimensions.
Thank you for that comprehensive reply! Therefore, am I correct in concluding that for any two vectors there exsits a plane on which they both lie?
If one or more of the vectors is a zero vector, then any plane containing the other vector must necessarily contain the first vector.
The vectors, whether parallel or not, must intersect because they can translated to have their terminal points at the origin. If the vectors intersect, the terminal points of each vector along with the point of intersection form a triangle. Since a triangle lies on a plane, the vectors must be coplanar.
If the vectors are identical, then any plane containing a line which in turn contains either vector must contain both vectors. We know that there exists a plane containing any line in n-space and that there exists a line containing any vector in n-space. (is this true?)
Is this reasoning correct? In n-space also?
BiP
P: 5,462 I wish HOI hadn't introduced vectors, I find they confuse things. Any single line in space sweeps out a plane. A plane divides n (n>2) space into two halves. Consequently if you have two lines the planes they sweep out either intersect or are parallel. Two planes intersect in a line. Two lines may neither intersect nor be parallel in any space of dimension greater than or equal to 3. These are called skew lines.
P: 4,570 Following on from what Studiot said, you can find these intersections using normal matrix row-reduction and you can determine skew-lines by seeing if you get an inconsistent solution to your set of equations.
Math Emeritus Sci Advisor Thanks PF Gold P: 38,879 It is a fundamental theorem of three dimensional Euclidean geometry that two intersecting lines lie in a unique plane. It is also true that two parallel lines lie in a unique plane. Two lines that are not parallel and do not intersect, "skew" lines, do not lie in a plane.
P: 18
Quote by Studiot A plane divides n (n>2) space into two halves.
Is this really true? So am I correct in guessing one could prove that the amount of space on one side of a plane is equally as infinite as on the other side??
P: 5,462
Is this really true? So am I correct in guessing one could prove that the amount of space on one side of a plane is equally as infinite as on the other side??
A point divides a line into two parts.
Each part goes from the point to 'infinity'
Formally we call these parts intervals.
The 'infinite' end is open.
We can include the whole line by including the point in one interval or other, but not both.
P: 18 right, that is pretty intuitive isn't it. Could a line segment divide a number line into uneven parts? Considering the line segment is finite
Sci Advisor P: 1,716 As I remember it is actually a postulate of Euclidean geometry that two intersecting lines determine a plane, not a theorem. The Cartesian coordinate system is set up to be a model of Euclid's postulates. One can see this from distance measurement along Cartesian straight lines because it obeys the Pythagorean theorem. Once you know this all else follows.
Math
Emeritus
Thanks
PF Gold
P: 38,879
Quote by Studiot A plane divides n (n>2) space into two halves.
Quote by boings Is this really true? So am I correct in guessing one could prove that the amount of space on one side of a plane is equally as infinite as on the other side??
That depends upon what you mean by "plane". Some texts use the term "plane" to mean a "flat" subset of $R^n$ of dimension 2 no matter what n is and use the term "hyper-plane" to mean a "flat" subset of dimension n-1. Other texts use the term "plane" to mean a "flat" subspace of dimension n-1.
P: 5,462 Thank you Halls of Ivy for tightening up on my terminology I really should have said hyperplane, dimension (n-1), and reserved 'plane' for dimension n=2.
Related Discussions Differential Geometry 9 Precalculus Mathematics Homework 9 Precalculus Mathematics Homework 1 Atomic, Solid State, Comp. Physics 20 Calculus & Beyond Homework 5 | {"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": 0.8678972125053406, "perplexity": 348.3075628743584}, "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-2014-15/segments/1397609533308.11/warc/CC-MAIN-20140416005213-00116-ip-10-147-4-33.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/117100/maclaurin-expansion-log-left-frac1x1-x-right-show-equality-of-two-su | # Maclaurin expansion $\log\left( \frac{1+x}{1-x}\right)$, show equality of two sums
I am supposed to find the Maclaurin expantion of
$\log\left( \frac{1+x}{1-x} \right)$
So I noticed the obvious that $\log(1+x) - \log(1-x)$
Then Maclaurin polynomial of $\log (1+x)$ equals $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$
So by doing a quick quick change of variables (-x) I obtained the maclaurin expansion of $\log(1-x)$
In total, I obtained that the Maclaurin expansion of $\log\left( \frac{1+x}{1-x} \right)$ equals.
$\displaystyle P(x) = \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n}$
But in my book the answer says $2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$, I can see that this is correct, by writing out a few terms. But how do I show this by algebra?
My question is therefore, how do I show that
$\displaystyle \sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n} - \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(-x)^n}{n} = 2 \sum_{n=0}^{\infty} \frac{1}{1+2n} x^{2n+1}$
?
-
Use $\log(1+x) = \log(1-(-x))$ and expand around $x=0$ just as you did for $\log(1-x)$. – Sasha Mar 6 '12 at 14:54
But I have already done this note the $x-2$ – N3buchadnezzar Mar 6 '12 at 14:55
$1+x$ becomes $1+x-2 = x-1$ after replacing $x$ with $x-2$, not $1-x$ as required. To make $1+x$ become $1-x$, replace $x$ everywhere with $-x.$ – Ragib Zaman Mar 6 '12 at 14:58
Where did you get $x-2$ from? – Ilya Mar 6 '12 at 14:58
Silly mistake! I will fix it now – N3buchadnezzar Mar 6 '12 at 15:01
The expansion for $\log (1+x)$ is not $\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}$, but $$\begin{equation*} \log (1+x)=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}% x^{n+1}=\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n+1}}{n}x^{n+1} \end{equation*}.$$
Consequently
$$\begin{eqnarray*} \log (1-x) &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}\left( -x\right) ^{n+1}=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left( -1\right) ^{n+1}}{n+1}x^{n+1} \\ &=&-\sum_{n=0}^{\infty }\frac{1}{n+1}x^{n+1}, \end{eqnarray*}$$
and $$\begin{eqnarray*} \log \left( \frac{1+x}{1-x}\right) &=&\log (1+x)-\log (1-x) \\ &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n+1}x^{n+1}+\sum_{n=0}^{ \infty }\frac{1}{n+1}x^{n+1} \\ &=&\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}+1}{n+1}x^{n+1} \\ &=&\sum_{n=0}^{\infty }\frac{2}{2n+1}x^{2n+1}, \end{eqnarray*}$$
because $$\begin{equation*} \left( -1\right) ^{n}+1=\left\{ \begin{array}{c} 2\quad \text{if }n\text{ even} \\ 0\quad \text{if }n\text{ odd}. \end{array} \right. \end{equation*}$$
-
It makes :
${\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}}-(-1)^{n+1+n}\dfrac{x^{n}}{n}={\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}}+\dfrac{x^{n}}{n}={\displaystyle \sum_{n=0}^{\infty}\dfrac{2x^{n}}{2n+1}}+{\displaystyle \sum_{n=1}^{\infty}-\dfrac{x^{n}}{2n}+\dfrac{x^{n}}{2n}}={\displaystyle \sum_{n=0}^{\infty}\dfrac{2x^{n}}{2n+1}}$
-
What did you do from 2 to 3 ? – N3buchadnezzar Mar 6 '12 at 15:28
I'm french, so I hope that it will be "understandable". I separated the even and odd terms. – matovitch Mar 6 '12 at 15:32
In fact you have $(-1)^{n+1} (-x)^n = -(x)^n$. Thus if you look the term of the two sum you have: $x^n\frac{1 - (-1)^{n}}{n}$ and for the complete sum : $2\sum_{k .odd} \frac{x^k}{k}$ it's exactly that you want.
- | {"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": 0.8960088491439819, "perplexity": 433.0082999582622}, "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-2016-07/segments/1454701159654.65/warc/CC-MAIN-20160205193919-00096-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.jobilize.com/online/course/4-2-graphs-of-exponential-functions-by-openstax?qcr=www.quizover.com | 4.2 Graphs of exponential functions
Page 1 / 6
• Graph exponential functions.
• Graph exponential functions using transformations.
As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a real-world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events.
Graphing exponential functions
Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ whose base is greater than one. We’ll use the function $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.\text{\hspace{0.17em}}$ Observe how the output values in [link] change as the input increases by $\text{\hspace{0.17em}}1.$
$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $f\left(x\right)={2}^{x}$ $\frac{1}{8}$ $\frac{1}{4}$ $\frac{1}{2}$ $1$ $2$ $4$ $8$
Each output value is the product of the previous output and the base, $\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$ We call the base $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ the constant ratio . In fact, for any exponential function with the form $\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x},$ $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of $\text{\hspace{0.17em}}a.$
Notice from the table that
• the output values are positive for all values of $x;$
• as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases, the output values increase without bound; and
• as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ decreases, the output values grow smaller, approaching zero.
[link] shows the exponential growth function $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.$
The domain of $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$ is all real numbers, the range is $\text{\hspace{0.17em}}\left(0,\infty \right),$ and the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$
To get a sense of the behavior of exponential decay , we can create a table of values for a function of the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ whose base is between zero and one. We’ll use the function $\text{\hspace{0.17em}}g\left(x\right)={\left(\frac{1}{2}\right)}^{x}.\text{\hspace{0.17em}}$ Observe how the output values in [link] change as the input increases by $\text{\hspace{0.17em}}1.$
$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $g\left(x\right)=\left(\frac{1}{2}{\right)}^{x}$ $8$ $4$ $2$ $1$ $\frac{1}{2}$ $\frac{1}{4}$ $\frac{1}{8}$
Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio $\text{\hspace{0.17em}}\frac{1}{2}.$
Notice from the table that
• the output values are positive for all values of $\text{\hspace{0.17em}}x;$
• as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases, the output values grow smaller, approaching zero; and
• as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ decreases, the output values grow without bound.
[link] shows the exponential decay function, $\text{\hspace{0.17em}}g\left(x\right)={\left(\frac{1}{2}\right)}^{x}.$
The domain of $\text{\hspace{0.17em}}g\left(x\right)={\left(\frac{1}{2}\right)}^{x}\text{\hspace{0.17em}}$ is all real numbers, the range is $\text{\hspace{0.17em}}\left(0,\infty \right),$ and the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$
Characteristics of the graph of the parent function f ( x ) = b x
An exponential function with the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ $\text{\hspace{0.17em}}b>0,$ $\text{\hspace{0.17em}}b\ne 1,$ has these characteristics:
• one-to-one function
• horizontal asymptote: $\text{\hspace{0.17em}}y=0$
• domain:
• range: $\text{\hspace{0.17em}}\left(0,\infty \right)$
• x- intercept: none
• y- intercept: $\text{\hspace{0.17em}}\left(0,1\right)\text{\hspace{0.17em}}$
• increasing if $\text{\hspace{0.17em}}b>1$
• decreasing if $\text{\hspace{0.17em}}b<1$
[link] compares the graphs of exponential growth and decay functions.
Given an exponential function of the form $\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ graph the function.
1. Create a table of points.
2. Plot at least $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ point from the table, including the y -intercept $\text{\hspace{0.17em}}\left(0,1\right).$
3. Draw a smooth curve through the points.
4. State the domain, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ the range, $\text{\hspace{0.17em}}\left(0,\infty \right),$ and the horizontal asymptote, $\text{\hspace{0.17em}}y=0.$
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 72, "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": 0.828159749507904, "perplexity": 599.9823384697337}, "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-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00092.warc.gz"} |
https://www.cbs.nl/nl-nl/achtergrond/2012/50/hot-deck-imputation-of-numerical-data-under-edit-restrictions | # Hot deck imputation of numerical data under edit restrictions
A common problem faced by statistical institutes is that data may be missing from collected data sets. The typical way to overcome this problem is to impute the missing data. The problem of imputing missing data is complicated by the fact that statistical data collected by statistical institutes often have to satisfy certain edit rules, which for numerical data usually take the form of linear restrictions. Standard imputation methods for numerical data as described in the literature generally do not take such linear edit restrictions on the data into account. Hot-deck imputation techniques form a well-known class and relatively simple to apply class of imputation methods. In this paper we extend this class of imputation methods so that linear edit restrictions are satisfied. | {"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": 0.8946496248245239, "perplexity": 564.2631914626019}, "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-2020-34/segments/1596439740838.3/warc/CC-MAIN-20200815094903-20200815124903-00288.warc.gz"} |
https://plainmath.net/1653/find-parametric-equations-lines-line-through-origin-parallel-vector-plus | # Find parametric equations for the lines in The line through the origin parallel to the vector 2j + k
Question
Find parametric equations for the lines in The line through the origin parallel to the vector $$2j + k$$
2021-02-13
Step 1 Given The vector equation is $$0i + 2j + k$$ The line passing through origin , origin is ( 0 ,0 ,0) The parametric equations for the line passing through the point $$p0 = ( x_0 ,y_0 ,z_0 )$$ and parallel to the vector is $$v =(ai + bj + ck)$$
$$x = x_0 + at , y = y_0 + bt , z = z_0 + ct$$ Step 2 To find parametric equations for the lines Here $$a = 0 , b = 2 , c = 1 , x_0 = 0 ,y_0 = 0 and z_0 = 0$$ Now plug this values in parametric equation $$x = 0 + 0t , y = 0 + 2t , z = 0 + 1t$$
$$x = 0 , y = 2t , z = t$$ Therefore parametric equations for the line is $$x = 0 , y = 2t and z = t$$ .
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Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$x=e^{-8t}\ \cos(8t),\ y=e^{-8t}\ \sin(8t),\ z=e^{-8t},\ (1,\ 0,\ 1)$$
Curves to parametric equations Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique. The line segment starting at $$P(-1,\ -3)\ \text{and ending at}\ Q(6,\ -16)$$
Consider the helix represented investigation by the vector-valued function $$r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >$$. Solve for t in the relationship derived in part (a), and substitute the result into the original set of parametric equations. This yields a parametrization of the curve in terms of the arc length parameter s. | {"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": 0.87465500831604, "perplexity": 183.83204675654082}, "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-2021-25/segments/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00047.warc.gz"} |
https://www.physicsforums.com/threads/coefficient-of-friction-problem.139498/ | # Coefficient of Friction problem!
1. Oct 22, 2006
### demode
The question states: "What is the coefficient of friction between a sled and a plane inclined at 30 degrees from the horizontal if the sled just slides without accelerating when given an initial push?"
I have drawn a free-body diagram and labeled all of the forces (I think i labeled all of them) but now I don't know what to do.. What equations will I need to write and what numbers will I need to use? If somebody could guide me through this it would be most appreciated..
2. Oct 22, 2006
### Hootenanny
Staff Emeritus
Welcome to the forums,
Could you describe your FBD, what forces do you have acting in which directions?
3. Oct 22, 2006
### demode
Last edited by a moderator: May 2, 2017
4. Oct 22, 2006
### Hootenanny
Staff Emeritus
There is one error there, would the weight really be acting in that direction?
5. Oct 22, 2006
### demode
Whoops, shouldn't it be going from the sled STRAIGHT down, perpendicular to the bottom of the triangle?
6. Oct 22, 2006
### Hootenanny
Staff Emeritus
Indeed it should.
7. Oct 22, 2006
### demode
With that being said, I am going to need to write my equations.. I believe that frictional force can be modeled with the equation (Ff = μ Fn) and Fn in this context is equal to mgcos30. Also, mgsin30 minus the frictional force would be equal to ma. However, if we don't know M how can we solve for the coefficient of friction in the first equation?
8. Oct 22, 2006
### arildno
I assume you with M means the mass of the sled.
Why don't you set up Newton's 2.law and see what can be done with that pernicious M?
9. Oct 22, 2006
### demode
I still don't understand what to substitute into Newton's 2.Law; Acceleration must be zero if the sled slides without accelerating with an initial push, and we don't know the value for m, as well as the value for the net force.. By playing around with the forces I labeled, I got the equation (mgsin30 - μk*mgcos30) = ma) but this may not be correct.
Ahhh, I can't seem to understand this =(
10. Oct 22, 2006
### arildno
"Also, mgsin30 minus the frictional force would be equal to ma."
Correct!
Hence, with a=0, do you agree that we have:
$$mg\sin(30)=F_{fric}$$
where $F_{fric}$ is the frictional force?
11. Oct 22, 2006
### arildno
This IS correct.
So, what does that equation say when you know that a=0?
12. Oct 22, 2006
### demode
It will say after simplifying:
4.9m - μk * 8.48m = 0
13. Oct 22, 2006
### arildno
To set in numbers is NOT simplification!!
Do as follows:
$$mg\sin(30)=\mu{m}g\cos(30)$$
Now, divide with $mg\cos(30)$:
$$\frac{mg\sin(30)}{mg\cos(30)}=\frac{\mu{mg}\cos(30)}{mg\cos(30)}$$
Now, what factors cancels in each of the fractions?
14. Oct 22, 2006
### demode
ahh i got it now.. Thanks so much for your help everyone! | {"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": 0.9325551390647888, "perplexity": 1597.65871973408}, "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-2019-04/segments/1547583728901.52/warc/CC-MAIN-20190120163942-20190120185942-00447.warc.gz"} |
https://www.physicsforums.com/threads/first-order-linear-de.127556/ | First order Linear DE
• Start date
name
6
0
Hello all,
I'm trying to prove to myself that the following solution to the DE shown works. I can't start using it until i prove to my self it works (it's this psycological thing i have were i can't use anything unless i know where it comes from). :rofl:
Here is the Equation and it's solution
http://img142.imageshack.us/img142/3437/defa7.png [Broken]
and here is me trying to prove to my self it works...
http://img137.imageshack.us/img137/1831/desolhv6.png [Broken]
Am I doing anything wrong? Or can anyone please show me a proof which shows that this is a solution to the Differential Equation?
edit: I don't think this is a homework question, as you know, im just trying to prove it to my self.
Last edited by a moderator:
Related Differential Equations News on Phys.org
HallsofIvy
Homework Helper
41,713
876
Let me see if I can put into Latex what you have so others won't have to wait for those to load:
Your first reference asserts that the solution to the first order, linear, differential equation
$$\frac{dx}{dt}= ax(t)+ f(t)$$
with x(0) and f(t) given is
$$x(t)= e^{at}x(0)+ \int_0^t e^{a(t-s)}ds[/itex] 2) You method of solution is: an integrating factor for the problem is e-at so [tex]e^{-at}\frac{dx}{dt}= ae^{-at}x(t)+ e^{-at}f(t)$$
$$e^{-at}\frac{dx}{dt}- ae^{-at}x(t)= e^{-at}f(t)$$
$$\frac{de^{-at}x}{dt}= e^{-at}f(t)$$
Yes, so far this is completely correct. You then integrate to get
$$e^{-at}x(t)= \int e^{-at}f(t)dt+ C$$
so
$$x(t)= e^{at}\int e^{-at}f(t)dt+ Ce^{at}$$
and try to determine C by setting t= 0
$$x(0)= e^{a0}\int e^{-a(0)}f(0)dt+ Ce^{a0}$$
That's your mistake! You are treating the "t" inside the integral as if it were the same as the "t" outside. It's not- it's a "dummy" variable.
Remember that $\int_0^1 t^2dt$= 3. You can't "set" t equal to 0 and declare that $\int_0^1 0^2 dt= 3$!
Go back and use a different variable in your integral:
$$e^{-at}x(t)= \int^t e^{-as}f(s)ds$$
Notice the single "t" as a limit on the integral. That tells people we mean for the final result of the integral to be in the variable t. Also notice there is no "C". Strictly speaking, that is included in the indefinite integral. A better technique, which you should learn, is to write that indefinite integral as a definite integral with a variable limit:
$$e^{-at}x(t)= \int_{0}^t e^{-as}f(s)ds+ C$$
I now have "+ C" because choosing a lower limit is the same as choosing a specific constant for the indefinite integral which we don't want to do yet.
I took the lower limit as 0 because we know x(0). The upper limit is the variable t. Of course, if t= 0, that integral is from 0 to 0 and so is 0 no matter what is being integrated:
$$e^{-a(0)}x(0)= x(0)= \int_0^0 e^{-as}f(s)ds+ C= C$$
so
$$e^{-at}x(t)= \int_0^t e^{-as}f(s)ds+ x(0)$$
Now multiply by eat to get
$$x(t)= e^{at}\int_0^t e^{-as}f(s)ds+ x(0)e^{at}$$
$$x(t)= \int_0^t e^{a(t-s)}f(s)ds+ x(0)e^{at}$$
as claimed. (Of course we can take that eat inside the integral as if it were a constant because it does not depend on the variable of integration, s.)
(You don't think this is a homework question? Don't you know for sure?:rofl:
name
6
0
HallsofIvy said:
Let me see if I can put into Latex what you have so others won't have to wait for those to load:
...
(You don't think this is a homework question? Don't you know for sure?:rofl:
HallsofIvy,
Thank you very much for that. What a disgrace, this is even a fundemental part of first year calculas!. I kind of knew something was wrong in that line - hence those red question marks.
As for the homework part, :tongue2: What i meant to say was that i don't think this should be in the homework section (I wasn't sure what constitutes as "homework" in this forum). And Latex looks powerful, I think i'd better learn it.
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• Solo and co-op problem solving | {"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": 0.9244908690452576, "perplexity": 822.2131598714855}, "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-2019-43/segments/1570987813307.73/warc/CC-MAIN-20191022081307-20191022104807-00238.warc.gz"} |
https://pugliablog.info/the-relationship-between/the-relationship-between-voltage-and-current-in-a-transformer.php | # The relationship between voltage and current in a transformer
### RELATIONSHIP BETWEEN PRIMARY AND SECONDARY WINDINGS
As just explained, the turns ratio of a transformer affects current as well as voltage . If voltage is doubled in the secondary, current is halved in the secondary. A transformer needs an alternating current that will create a changing magnetic field. The ratio between the voltages in the coils is the same as the ratio of the. Assuming that the transformer is % efficient (no energy is lost between its primary coil and Ip is the current in the primary coil in amperes (amps), A.
Copper loss can be minimized by using the proper diameter wire. Large diameter wire is required for high-current windings, whereas small diameter wire can be used for low-current windings. Eddy-Current Loss The core of a transformer is usually constructed of some type of ferromagnetic material because it is a good conductor of magnetic lines of flux.
Whenever the primary of an iron-core transformer is energized by an alternating-current source, a fluctuating magnetic field is produced. This magnetic field cuts the conducting core material and induces a voltage into it. The induced voltage causes random currents to flow through the core which dissipates power in the form of heat. Since the thin, insulated laminations do not provide an easy path for current, eddy-current losses are greatly reduced.
Hysteresis Loss When a magnetic field is passed through a core, the core material becomes magnetized. To become magnetized, the domains within the core must align themselves with the external field.
If the direction of the field is reversed, the domains must turn so that their poles are aligned with the new direction of the external field. Power transformers normally operate from either 50 Hz, or Hz alternating current. Each tiny domain must realign itself twice during each cycle, or a total of times a second when 50 Hz alternating current is used.
The energy used to turn each domain is dissipated as heat within the iron core. Hysteresis loss can be held to a small value by proper choice of core materials. The input power is equal to the product of the voltage applied to the primary and the current in the primary. The output power is equal to the product of the voltage across the secondary and the current in the secondary. The difference between the input power and the output power represents a power loss.
You can calculate the percentage of efficiency of a transformer by using the standard efficiency formula shown below: If the input power to a transformer is watts and the output power is watts, what is the efficiency? Hence, the efficiency is approximately The voltage, current, and power-handling capabilities of the primary and secondary windings must also be considered.
The maximum voltage that can safely be applied to any winding is determined by the type and thickness of the insulation used. When a better and thicker insulation is used between the windings, a higher maximum voltage can be applied to the windings.
The maximum current that can be carried by a transformer winding is determined by the diameter of the wire used for the winding. If current is excessive in a winding, a higher than ordinary amount of power will be dissipated by the winding in the form of heat.
This heat may be sufficiently high to cause the insulation around the wire to break down. If this happens, the transformer may be permanently damaged. The power-handling capacity of a transformer is dependent upon its ability to dissipate heat.
If the heat can safely be removed, the power-handling capacity of the transformer can be increased.
### BBC - GCSE Bitesize Science - Transformers : Revision, Page 4
This is sometimes accomplished by immersing the transformer in oil, or by the use of cooling fins. The power-handling capacity of a transformer is measured in either the volt-ampere unit or the watt unit.
Two common power generator frequencies 50 hertz and hertz have been mentioned, but the effect of varying frequency has not been discussed.
If the frequency applied to a transformer is increased, the inductive reactance of the windings is increased, causing a greater ac voltage drop across the windings and a lesser voltage drop across the load. However, an increase in the frequency applied to a transformer should not damage it. But, if the frequency applied to the transformer is decreased, the reactance of the windings is decreased and the current through the transformer winding is increased. If the decrease in frequency is enough, the resulting increase in current will damage the transformer.
For this reason a transformer may be used at frequencies above its normal operating frequency, but not below that frequency. A brief discussion of some of these applications will help you recognize the importance of the transformer in electricity and electronics. These transformers have two or more windings wound on a laminated iron core.
The number of windings and the turns per winding depend upon the voltages that the transformer is to supply. Their coefficient of coupling is 0. You can usually distinguish between the high-voltage and low-voltage windings in a power transformer by measuring the resistance. The low-voltage winding usually carries the higher current and therefore has the larger diameter wire.
This means that its resistance is less than the resistance of the high-voltage winding, which normally carries less current and therefore may be constructed of smaller diameter wire.
## Transformers - Higher tier
So far you have learned about transformers that have but one secondary winding. The typical power transformer has several secondary windings, each providing a different voltage. The schematic symbol for a typical power-supply transformer is shown in figure This proportion also shows the relationship between the number of turns in each winding and the voltage across each winding. This proportion is expressed by the equation: Notice the equation shows that the ratio of secondary voltage to primary voltage is equal to the ratio of secondary turns to primary turns.
The equation can be written as: The following formulas are derived from the above equation: If any three of the quantities in the above formulas are known, the fourth quantity can be calculated.
A transformer has turns in the primary, 50 turns in the secondary, and volts applied to the primary Ep. What is the voltage across the secondary E s? There are turns of wire in an iron-core coil. If this coil is to be used as the primary of a transformer, how many turns must be wound on the coil to form the secondary winding of the transformer to have a secondary voltage of one volt if the primary voltage is five volts?
The ratio of the voltage 5: Sometimes, instead of specific values, you are given a turns or voltage ratio. In this case, you may assume any value for one of the voltages or turns and compute the other value from the ratio. For example, if a turn ratio is given as 6: The transformer in each of the above problems has fewer turns in the secondary than in the primary.
As a result, there is less voltage across the secondary than across the primary. The ratio of a four-to-one step-down transformer is written as 4: A transformer that has fewer turns in the primary than in the secondary will produce a greater voltage across the secondary than the voltage applied to the primary.
### BBC - GCSE Bitesize: Transformers - Higher tier
A transformer in which the voltage across the secondary is greater than the voltage applied to the primary is called a STEP-UP transformer. The ratio of a one-to-four step-up transformer should be written as 1: Notice in the two ratios that the value of the primary winding is always stated first.
The magnetic field produced by the current in the secondary interacts with the magnetic field produced by the current in the primary. This interaction results from the mutual inductance between the primary and secondary windings.
• Conservation of energy in transformers
It is also the means by which energy is transferred from the primary winding to the secondary winding. The inductance which produces this flux is also common to both windings and is called mutual inductance.
Figure 11 shows the flux produced by the currents in the primary and secondary windings of a transformer when source current is flowing in the primary winding. When a load resistance is connected to the secondary winding, the voltage induced into the secondary winding causes current to flow in the secondary winding.
This current produces a flux field about the secondary shown as broken lines which is in opposition to the flux field about the primary Lenz's law. Thus, the flux about the secondary cancels some of the flux about the primary. | {"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": 0.9152534604072571, "perplexity": 490.77676041145463}, "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-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00557.warc.gz"} |
https://www.physicsforums.com/threads/physics-pressure-density-question.258941/ | # Physics pressure/density question
1. Sep 24, 2008
### mint911
Physics "pressure/density" question
1. The problem statement, all variables and given/known data
A cube of wood of side 4.0 cm floats in fresh water with one quarter showing above the surface.
a.i.) Calculate the water pressure on the base of the cube
a.ii.) Calculate the density of the wood
b.) What fraction of the volume of the cube would be immersed if it was floated in olive oil?
2. Relevant equations
Olive oil = density/kg m -3 of 920.
2. Sep 24, 2008
### ||spoon||
Re: Physics "pressure/density" question
You are supposed to show an attempt at a solution first, people will not simply solve the question for you...
If you are having trouble finding where to start think about the total amount of water being displaced. If it hadn't been displaced what could you have been able to say about the forces acting on it??
-Spoon
Similar Discussions: Physics pressure/density question | {"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": 0.8115964531898499, "perplexity": 1506.747198465051}, "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-2017-17/segments/1492917118851.8/warc/CC-MAIN-20170423031158-00159-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://svn.geocomp.uq.edu.au/escript/trunk/doc/cookbook/example07.tex?view=markup&sortby=date&sortdir=down&pathrev=6651 | # Contents of /trunk/doc/cookbook/example07.tex
Revision 6651 - (show annotations)
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1 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 3 % Copyright (c) 2003-2018 by The University of Queensland 4 5 % 6 % Primary Business: Queensland, Australia 7 % Licensed under the Apache License, version 2.0 8 9 % 10 % Development until 2012 by Earth Systems Science Computational Center (ESSCC) 11 % Development 2012-2013 by School of Earth Sciences 12 % Development from 2014 by Centre for Geoscience Computing (GeoComp) 13 % 14 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 15 16 The acoustic wave equation governs the propagation of pressure waves. Wave 17 types that obey this law tend to travel in liquids or gases where shear waves 18 or longitudinal style wave motion is not possible. An obvious example is sound 19 waves. 20 21 The acoustic wave equation is defined as; 22 \begin{equation} 23 \nabla ^2 p - \frac{1}{c^2} \frac{\partial ^2 p}{\partial t^2} = 0 24 \label{eqn:acswave} 25 \end{equation} 26 where $p$ is the pressure, $t$ is the time and $c$ is the wave velocity. In this 27 chapter the acoustic wave equation is demonstrated. Important steps include the 28 translation of the Laplacian $\nabla^2$ to the \esc general form, the stiff 29 equation stability criterion and solving for the displacement or acceleration solution. 30 31 \section{The Laplacian in \esc} 32 The Laplacian operator which can be written as $\Delta$ or $\nabla^2$, is 33 calculated via the divergence of the gradient of the object, which in this 34 example is the scalar $p$. Thus we can write; 35 \begin{equation} 36 \nabla^2 p = \nabla \cdot \nabla p = 37 \sum_{i}^n 38 \frac{\partial^2 p}{\partial x^2_{i}} 39 \label{eqn:laplacian} 40 \end{equation} 41 For the two dimensional case in Cartesian coordinates \autoref{eqn:laplacian} 42 becomes; 43 \begin{equation} 44 \nabla^2 p = \frac{\partial^2 p}{\partial x^2} 45 + \frac{\partial^2 p}{\partial y^2} 46 \end{equation} 47 48 In \esc the Laplacian is calculated using the divergence representation and the 49 intrinsic functions \textit{grad()} and \textit{trace()}. The function 50 \textit{grad{}} will return the spatial gradients of an object. 51 For a rank 0 solution, this is of the form; 52 \begin{equation} 53 \nabla p = \left[ 54 \frac{\partial p}{\partial x _{0}}, 55 \frac{\partial p}{\partial x _{1}} 56 \right] 57 \label{eqn:grad} 58 \end{equation} 59 Larger ranked solution objects will return gradient tensors. For example, a 60 pressure field which acts in the directions $p _{0}$ and $p 61 _{1}$ would return; 62 \begin{equation} 63 \nabla p = \begin{bmatrix} 64 \frac{\partial p _{0}}{\partial x _{0}} & 65 \frac{\partial p _{1}}{\partial x _{0}} \\ 66 \frac{\partial p _{0}}{\partial x _{1}} & 67 \frac{\partial p _{1}}{\partial x _{1}} 68 \end{bmatrix} 69 \label{eqn:gradrank1} 70 \end{equation} 71 72 \autoref{eqn:grad} corresponds to the Linear PDE general form value 73 $X$. Notice however, that the general form contains the term $X 74 _{i,j}$\footnote{This is the first derivative in the $j^{th}$ 75 direction for the $i^{th}$ component of the solution.}, 76 hence for a rank 0 object there is no need to do more then calculate the 77 gradient and submit it to the solver. In the case of the rank 1 or greater 78 object, it is also necessary to calculate the trace. This is the sum of the 79 diagonal in \autoref{eqn:gradrank1}. 80 81 Thus when solving for equations containing the Laplacian one of two things must 82 be completed. If the object \verb!p! is less than rank 1 the gradient is 83 calculated via; 84 \begin{python} 85 gradient=grad(p) 86 \end{python} 87 and if the object is greater then or equal to a rank 1 tensor, the trace of 88 the gradient is calculated. 89 \begin{python} 90 gradient=trace(grad(p)) 91 \end{python} 92 These values can then be submitted to the PDE solver via the general form term 93 $X$. The Laplacian is then computed in the solution process by taking the 94 divergence of $X$. 95 96 Note, if you are unsure about the rank of your tensor, the \textit{getRank} 97 command will return the rank of the PDE object. 98 \begin{python} 99 rank = p.getRank() 100 \end{python} 101 102 103 \section{Numerical Solution Stability} \label{sec:nsstab} 104 Unfortunately, the wave equation belongs to a class of equations called 105 \textbf{stiff} PDEs. These types of equations can be difficult to solve 106 numerically as they tend to oscillate about the exact solution, which can 107 eventually lead to a catastrophic failure. To counter this problem, explicitly 108 stable schemes like the backwards Euler method, and correct parameterisation of 109 the problem are required. 110 111 There are two variables which must be considered for 112 stability when numerically trying to solve the wave equation. For linear media, 113 the two variables are related via; 114 \begin{equation} \label{eqn:freqvel} 115 f=\frac{v}{\lambda} 116 \end{equation} 117 The velocity $v$ that a wave travels in a medium is an important variable. For 118 stability the analytical wave must not propagate faster then the numerical wave 119 is able to, and in general, needs to be much slower then the numerical wave. 120 For example, a line 100m long is discretised into 1m intervals or 101 nodes. If 121 a wave enters with a propagation velocity of 100m/s then the travel time for 122 the wave between each node will be 0.01 seconds. The time step, must therefore 123 be significantly less than this. Of the order $10E-4$ would be appropriate. 124 This stability criterion is known as the Courant\textendash 125 Friedrichs\textendash Lewy condition given by 126 \begin{equation} 127 dt=f\cdot \frac{dx}{v} 128 \end{equation} 129 where $dx$ is the mesh size and $f$ is a safety factor. To obtain a time step of 130 $10E-4$, a safety factor of $f=0.1$ was used. 131 132 The wave frequency content also plays a part in numerical stability. The 133 Nyquist-sampling theorem states that a signals bandwidth content will be 134 accurately represented when an equispaced sampling rate $f _{n}$ is 135 equal to or greater then twice the maximum frequency of the signal 136 $f_{s}$, or; 137 \begin{equation} \label{eqn:samptheorem} 138 f_{n} \geqslant f_{s} 139 \end{equation} 140 For example, a 50Hz signal will require a sampling rate greater then 100Hz or 141 one sample every 0.01 seconds. The wave equation relies on a spatial frequency, 142 thus the sampling theorem in this case applies to the solution mesh spacing. 143 This relationship confirms that the frequency content of the input signal 144 directly affects the time discretisation of the problem. 145 146 To accurately model the wave equation with high resolutions and velocities 147 means that very fine spatial and time discretisation is necessary for most 148 problems. This requirement makes the wave equation arduous to 149 solve numerically due to the large number of time iterations required in each 150 solution. Models with very high velocities and frequencies will be the worst 151 affected by this problem. 152 153 \section{Displacement Solution} 154 \sslist{example07a.py} 155 156 We begin the solution to this PDE with the centred difference formula for the 157 second derivative; 158 \begin{equation} 159 f''(x) \approx \frac{f(x+h - 2f(x) + f(x-h)}{h^2} 160 \label{eqn:centdiff} 161 \end{equation} 162 substituting \autoref{eqn:centdiff} for $\frac{\partial ^2 p }{\partial t ^2}$ 163 in \autoref{eqn:acswave}; 164 \begin{equation} 165 \nabla ^2 p - \frac{1}{c^2h^2} \left[p_{(t+1)} - 2p_{(t)} + 166 p_{(t-1)} \right] 167 = 0 168 \label{eqn:waveu} 169 \end{equation} 170 Rearranging for $p_{(t+1)}$; 171 \begin{equation} 172 p_{(t+1)} = c^2 h^2 \nabla ^2 p_{(t)} +2p_{(t)} - 173 p_{(t-1)} 174 \end{equation} 175 this can be compared with the general form of the \modLPDE module and it 176 becomes clear that $D=1$, $X_{i,j}=-c^2 h^2 \nabla ^2 p_{(t)}$ and 177 $Y=2p_{(t)} - p_{(t-1)}$. 178 179 The solution script is similar to others that we have created in previous 180 chapters. The general steps are; 181 \begin{enumerate} 182 \item The necessary libraries must be imported. 183 \item The domain needs to be defined. 184 \item The time iteration and control parameters need to be defined. 185 \item The PDE is initialised with source and boundary conditions. 186 \item The time loop is started and the PDE is solved at consecutive time steps. 187 \item All or select solutions are saved to file for visualisation later on. 188 \end{enumerate} 189 190 Parts of the script which warrant more attention are the definition of the 191 source, visualising the source, the solution time loop and the VTK data export. 192 193 \subsection{Pressure Sources} 194 As the pressure is a scalar, one need only define the pressure for two 195 time steps prior to the start of the solution loop. Two known solutions are 196 required because the wave equation contains a double partial derivative with 197 respect to time. This is often a good opportunity to introduce a source to the 198 solution. This model has the source located at it's centre. The source should 199 be smooth and cover a number of samples to satisfy the frequency stability 200 criterion. Small sources will generate high frequency signals. Here, when using 201 a rectangular domain, the source is defined by a cosine function. 202 \begin{python} 203 U0=0.01 # amplitude of point source 204 xc=[500,500] #location of point source 205 # define small radius around point xc 206 src_radius = 30 207 # for first two time steps 208 u=U0*(cos(length(x-xc)*3.1415/src_radius)+1)*\ 209 whereNegative(length(x-xc)-src_radius) 210 u_m1=u 211 \end{python} 212 213 \subsection{Visualising the Source} 214 There are two options for visualising the source. The first is to export the 215 initial conditions of the model to VTK, which can be interpreted as a scalar 216 surface in \mayavi. The second is to take a cross section of the model which 217 will require the \textit{Locator} function. 218 First \verb!Locator! must be imported; 219 \begin{python} 220 from esys.escript.pdetools import Locator 221 \end{python} 222 The function can then be used on the domain to locate the nearest domain node 223 to the point or points of interest. 224 225 It is now necessary to build a list of $(x,y)$ locations that specify where are 226 model slice will go. This is easily implemented with a loop; 227 \begin{python} 228 cut_loc=[] 229 src_cut=[] 230 for i in range(ndx/2-ndx/10,ndx/2+ndx/10): 231 cut_loc.append(xstep*i) 232 src_cut.append([xstep*i,xc[1]]) 233 \end{python} 234 We then submit the output to \verb!Locator! and finally return the appropriate 235 values using the \verb!getValue! function. 236 \begin{python} 237 src=Locator(mydomain,src_cut) 238 src_cut=src.getValue(u) 239 \end{python} 240 It is then a trivial task to plot and save the output using \mpl 241 (\autoref{fig:cxsource}). 242 \begin{python} 243 pl.plot(cut_loc,src_cut) 244 pl.axis([xc[0]-src_radius*3,xc[0]+src_radius*3,0.,2*U0]) 245 pl.savefig(os.path.join(savepath,"source_line.png")) 246 \end{python} 247 \begin{figure}[h] 248 \centering 249 \includegraphics[width=6in]{figures/sourceline.png} 250 \caption{Cross section of the source function.} 251 \label{fig:cxsource} 252 \end{figure} 253 254 255 \subsection{Point Monitoring} 256 In the more general case where the solution mesh is irregular or specific 257 locations need to be monitored, it is simple enough to use the \textit{Locator} 258 function. 259 \begin{python} 260 rec=Locator(mydomain,[250.,250.]) 261 \end{python} 262 When the solution \verb u is updated we can extract the value at that point 263 via; 264 \begin{python} 265 u_rec=rec.getValue(u) 266 \end{python} 267 For consecutive time steps one can record the values from \verb!u_rec! in an 268 array initialised as \verb!u_rec0=[]! with; 269 \begin{python} 270 u_rec0.append(rec.getValue(u)) 271 \end{python} 272 273 It can be useful to monitor the value at a single or multiple individual points 274 in the model during the modelling process. This is done using 275 the \verb!Locator! function. 276 277 278 \section{Acceleration Solution} 279 \sslist{example07b.py} 280 281 An alternative method to the displacement solution, is to solve for the 282 acceleration $\frac{\partial ^2 p}{\partial t^2}$ directly. The displacement can 283 then be derived from the acceleration after a solution has been calculated 284 The acceleration is given by a modified form of \autoref{eqn:waveu}; 285 \begin{equation} 286 \nabla ^2 p - \frac{1}{c^2} a = 0 287 \label{eqn:wavea} 288 \end{equation} 289 and can be solved directly with $Y=0$ and $X=-c^2 \nabla ^2 p_{(t)}$. 290 After each iteration the displacement is re-evaluated via; 291 \begin{equation} 292 p_{(t+1)}=2p_{(t)} - p_{(t-1)} + h^2a 293 \end{equation} 294 295 \subsection{Lumping} 296 For \esc, the acceleration solution is preferred as it allows the use of matrix 297 lumping. Lumping or mass lumping as it is sometimes known, is the process of 298 aggressively approximating the density elements of a mass matrix into the main 299 diagonal. The use of Lumping is motivated by the simplicity of diagonal matrix 300 inversion. As a result, Lumping can significantly reduce the computational 301 requirements of a problem. Care should be taken however, as this 302 function can only be used when the $A$, $B$ and $C$ coefficients of the 303 general form are zero. 304 305 More information about the lumping implementation used in \esc and its accuracy 306 can be found in the user guide. 307 308 To turn lumping on in \esc one can use the command; 309 \begin{python} 310 mypde.getSolverOptions().setSolverMethod(SolverOptions.HRZ_LUMPING) 311 \end{python} 312 It is also possible to check if lumping is set using; 313 \begin{python} 314 print(mypde.isUsingLumping()) 315 \end{python} 316 317 \section{Stability Investigation} 318 It is now prudent to investigate the stability limitations of this problem. 319 First, we let the frequency content of the source be very small. If we define 320 the source as a cosine input, then the wavlength of the input is equal to the 321 radius of the source. Let this value be 5 meters. Now, if the maximum velocity 322 of the model is $c=380.0ms^{-1}$, then the source 323 frequency is $f_{r} = \frac{380.0}{5} = 76.0 Hz$. This is a worst case 324 scenario with a small source and the models maximum velocity. 325 326 Furthermore, we know from \autoref{sec:nsstab}, that the spatial sampling 327 frequency must be at least twice this value to ensure stability. If we assume 328 the model mesh is a square equispaced grid, 329 then the sampling interval is the side length divided by the number of samples, 330 given by $\Delta x = \frac{1000.0m}{400} = 2.5m$ and the maximum sampling 331 frequency capable at this interval is 332 $f_{s}=\frac{380.0ms^{-1}}{2.5m}=152Hz$ this is just equal to the 333 required rate satisfying \autoref{eqn:samptheorem}. 334 335 \autoref{fig:ex07sampth} depicts three examples where the grid has been 336 undersampled, sampled correctly, and over sampled. The grids used had 337 200, 400 and 800 nodes per side respectively. Obviously, the oversampled grid 338 retains the best resolution of the modelled wave. 339 340 The time step required for each of these examples is simply calculated from 341 the propagation requirement. For a maximum velocity of $380.0ms^{-1}$, 342 \begin{subequations} 343 \begin{equation} 344 \Delta t \leq \frac{1000.0m}{200} \frac{1}{380.0} = 0.013s 345 \end{equation} 346 \begin{equation} 347 \Delta t \leq \frac{1000.0m}{400} \frac{1}{380.0} = 0.0065s 348 \end{equation} 349 \begin{equation} 350 \Delta t \leq \frac{1000.0m}{800} \frac{1}{380.0} = 0.0032s 351 \end{equation} 352 \end{subequations} 353 Observe that for each doubling of the number of nodes in the mesh, we halve 354 the time step. To illustrate the impact this has, consider our model. If the 355 source is placed at the center, it is $500m$ from the nearest boundary. With a 356 velocity of $380.0ms^{-1}$ it will take $\approx1.3s$ for the wavefront to 357 reach that boundary. In each case, this equates to $100$, $200$ and $400$ time 358 steps. This is again, only a best case scenario, for true stability these time 359 values may need to be halved and possibly halved again. 360 361 \begin{figure}[ht] 362 \centering 363 \subfigure[Undersampled Example]{ 364 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm,clip]{figures/ex07usamp.png} 365 \label{fig:ex07usamp} 366 } 367 \subfigure[Just sampled Example]{ 368 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm,clip]{figures/ex07jsamp.png} 369 \label{fig:ex07jsamp} 370 } 371 \subfigure[Over sampled Example]{ 372 \includegraphics[width=0.45\textwidth,trim=0cm 6cm 5cm 6cm,clip]{figures/ex07nsamp.png} 373 \label{fig:ex07nsamp} 374 } 375 \caption{Sampling Theorem example for stability investigation} 376 \label{fig:ex07sampth} 377 \end{figure} 378 | {"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": 0.9134993553161621, "perplexity": 1437.591080043045}, "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/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00115.warc.gz"} |
https://blogs2.datall-analyse.nl/2016/02/18/error_propagation_r_monte_carlo_copulas/ | # Error propagation in R: Monte Carlo simulations using copulas
Physicists and engineers often have to calculate the uncertainty in a derived quantity.
For instance, a test engineer repeatedly measures two separate angles. The uncertainty (or error) in the measurements of each of these angles appears to be +/- 1 degree. Subsequently, the engineer calculates the sum of these two angles. This sum is a derived quantity. But note that this derived quantity is composed of two measured quantities each having their own uncertainties, so what is the uncertainty (or error) in this derived quantity? In other words, how propagates the uncertainty from the measured quantities (the two angles) to a derived quantity (sum of two angles).
The following R code calculates the uncertainty in a derived quantity using a Monte Carlo simulation. Moreover, the Monte Carlo simulation employs copulas that fall into the Archimedean class of copulas. This class consists of members such as the Clayton and Gumbel copula. These Archimedean copulas make it possible to calculate the uncertainty in a derived quantity when 1) the measured quantities are dependent (correlated), and 2) the measured quantities follow a normal or some asymmetric distribution (e.g., an exponential, lognormal, or Weibull distribution).
R code that uses a Gaussian copula for calculating the uncertainty in a derived quantity can be found in this blog post.
A comprehensive introduction to Archimedean copulas is given by Salvadori et al. in their 2007 book Extremes in nature: an approach using copulas.
See this page for an overview of all of Stefan’s R code blog posts.
R code (last update: 2016-10-20)
```library(copula)
library(survival)
##generate artificial data
#- denote the measured quantities by l and b
#- the derived quantity is equal to l*b
#- assume that l follows a normal and b an exponential distribution
#- the measured quantities l and b are dependent
#- use a Clayton copula for simulating the dependence between l and b
#Clayton copula
myCop <- claytonCopula(param=.7, dim=2)
#margins: normal and exponential distributions
myMvdc <- mvdc(copula=myCop, c("norm", "exp"),
list(list(mean=25, sd=.02), list(rate=.05)))
#draw sample
ns <- 50 #sample size
lb <- rMvdc(ns, myMvdc)
#inspect data
plot(lb[,1], lb[,2], xlab="l", ylab="b")
##fit Archimedean copula to data
#fitted are a Clayton and Gumbel copula
#for other copulas see Salvadori et al., Appendix C, pp. 233-264
#obtain start values for fitting the copulas
expRate <- exp(-coef(survreg(Surv(lb[,2])~1, dist="exponential")))
normMean <- mean(lb[,1])
nNorm <- length(lb[,1])
sdNorm <- sqrt((nNorm-1)/nNorm)*sd(lb[,1])
#fit Clayton copula to data
myClayton <- mvdc(copula=claytonCopula(param=1, dim=2), c("norm", "exp"),
list(list(mean=0, sd=1), list(rate=1)))
tauEst <- cor(lb[,1], lb[,2], method="kendall")
thetaClayton <- (2*tauEst)/(1-tauEst)
start <- c(normMean, sdNorm, expRate, thetaClayton)
fitClayton <- fitMvdc(lb, myClayton, start=start)
fitClayton
#estimate of derived quantity (which is the equal to l*b)
fitClayton@estimate[1]*(1/fitClayton@estimate[3])
#fit Gumbel copula to data
myGumbel <- mvdc(copula=gumbelCopula(param=1.5, dim=2), c("norm", "exp"),
list(list(mean=0, sd=1), list(rate=1)))
tauEst <- cor(lb[,1], lb[,2], method="kendall")
thetaGumbel <- 1 / (1-tauEst)
start <- c(normMean, sdNorm, expRate, thetaGumbel)
fitGumbel <- fitMvdc(lb, myGumbel, start=start)
fitGumbel
#estimate of derived quantity (which is the equal to l*b)
fitGumbel@estimate[1]*(1/fitGumbel@estimate[3])
#compare log-likelihoods of the fitted Clayton and Gumbel copulas
fitClayton@loglik
fitGumbel@loglik
##diagnostics
#Kendall plot and level curves are used as diagnostic tools
#generate sequence of cdf values for u and v
u <- seq(0, 1, length.out=100)
v <- seq(0, 1, length.out=100)
uv <- as.matrix(expand.grid(u,v))
#observations
uobs <- lb[,1]
vobs <- lb[,2]
nobs <- length(uobs) #sample size
#1) Kendall plot
#Kendall's measure (non-parametric)
kn <- sapply(1:nobs, function(i) {
sum(as.numeric(uobs < uobs[i] & vobs < vobs[i])) / (nobs-1)})
#Kendall's measure (parametric)
#(see Salvadori et al., pp. 147-148)
#Clayton copula
myCopClayton <- copClayton
myCopClayton@theta <- fitClayton@mvdc@copula@parameters
KuClayton <- pK(u, cop=myCopClayton, d=2)
#Gumbel copula
myCopGumbel <- copGumbel
myCopGumbel@theta <- fitGumbel@mvdc@copula@parameters
KuGumbel <- pK(u, cop=myCopGumbel, d=2)
#Kendall plot
plot(sort(kn), (1:nobs)/nobs, type="p", xlim=c(0,1), log="y",
ylab="Kc", xlab="t", col=gray(.5))
lines(u, KuClayton, col="blue")
lines(u, KuGumbel, col="darkgreen")
legend("bottomright", lty=1, col=c("blue", "darkgreen"), legend=c("Clayton", "Gumbel"))
#2) compare the level curves of the theoretical and empirical copula
#see Salvadori et al., pp. 140-142
#empirical copula (Salvadori et al., p.140, eq. (3.31))
ec <- C.n(u=uv, X=pobs(cbind(uobs,vobs)))
ec <- matrix(ec, nrow=length(u))
cL <- contourLines(u, v, z=ec, levels=seq(.1,.9,.1))
#Clayton copula
contour(claytonCopula(param=fitClayton@mvdc@copula@parameters), pCopula)
invisible(lapply(cL, lines, lwd=1, lty=2, col="red"))
#Gumbel copula
contour(claytonCopula(param=fitGumbel@mvdc@copula@parameters), pCopula)
invisible(lapply(cL, lines, lwd=1, lty=2, col="red"))
##Monte Carlo simulation
#proceed with the fitted Clayton copula for running the Monte Carlo simulation
#set up a Clayton copula with Student and exponential margins
corBLfit <- fitClayton@mvdc@copula@parameters
meanBfit <- fitClayton@estimate[1]
nb <- length(lb[,1])
sdBfit <- sqrt(nb/(nb-1))*fitClayton@estimate[2] #unbiased MLE of the sd
rateLfit <- fitClayton@estimate[3]
mcClayton <- mvdc(copula=claytonCopula(param=corBLfit, dim=2),
margins=c("t", "exp"),
paramMargins=list(list(df=nb-1), list(rate=rateLfit)))
#Monte Carlo simulation
nsim <- 100000
blSim <- rMvdc(nsim, mcClayton)
blSim[,1] <- meanBfit + blSim[,1]*sdBfit
prodSim <- blSim[,1]*blSim[,2]
#results Monte Carlo simulation
hist(prodSim) #skewed/asymmetric distribution
mean(prodSim)
sd(prodSim)
#percentile confidence intervals
alpha <- .05
quantile(prodSim, probs=c(alpha/2, 1-alpha/2))
#function for calculating the shortest coverage interval
sci <- function (values, alpha=.05){
sortedSim <- sort(values)
nsim <- length(values)
covInt <- sapply(1:(nsim-round((1-alpha)*nsim)), function(i) {
sortedSim[1+round((1-alpha)*nsim)+(i-1)]-sortedSim[1+(i-1)]})
lcl <- sortedSim[which(covInt==min(covInt))]
ucl <- sortedSim[1+round((1-alpha)*nsim)+(which(covInt==min(covInt))-1)]
c(lcl, ucl)
}
#shortest 95% coverage interval
(simInt <- round(sci(prodSim, alpha=.05), 4))
#draw limits
hist(prodSim)
abline(v=simInt, col="blue", lty=2, lwd=2)
##confidence interval estimation when taking the sample size into account
#the above procedure for calculating the confidence interval does not take the
#sample size (i.e., n=50) into account, whereas the following procedure does
n <- nrow(lb) #original sample size
Nmc <- 50000 #number of samples
meanNmc <- numeric(Nmc)
#sample with replacement from skewed distribution of l*b
meanNmc <- replicate(Nmc, mean(sample(prodSim, n, replace=TRUE)))
#Central Limit Theory states that the sampling distribution from any
#distribution approaches a normal distribution as long as the sample
#size is sufficiently large
#thus, taking repeatedly a sample from the above skewed distribution (of l*b)
#should yield a sampling distribution that is approximately normal, assuming the
#orginal sample size (of n=50) was sufficiently large
hist(meanNmc) #approaches a normal distribution
mean(meanNmc)
sd(meanNmc)
#percentile confidence intervals
alpha <- .05
quantile(meanNmc, probs=c(alpha/2, 1-alpha/2))
#shortest 95% coverage interval
(simInt <- round(sci(meanNmc, alpha=.05), 4))
#draw limits
hist(meanNmc)
abline(v=simInt, col="blue", lty=2, lwd=2)
##parametric bootstrap method
#compare the latter Monte Carlo simulation confidence intervals (i.e., when the sample size
#of n=50 was taken into account) with those obtained by a parametric bootstrap method
#initialize bootstrap
B <- 5000 #number of bootstrap samples
estimateB <- numeric(B)
n <- nrow(lb)
#perform bootstrap
for (i in 1:B) {
xyb <- rMvdc(n, mcClayton)
xyb[,1] <- meanBfit + xyb[,1]*sdBfit
mxb <- mean(xyb[,1])
myb <- 1/exp(-coef(survreg(Surv(xyb[,2])~1, dist="exponential")))
estimateB[i] <- mxb*myb
}
#check results bootstrap
hist(estimateB)
mean(estimateB)
sd(estimateB)
#percentile confidence intervals
alpha <- .05
quantile(estimateB, probs=c(alpha/2, 1-alpha/2))
#shortest 95% coverage interval
(simIntB <- round(sci(estimateB, alpha=.05), 4))
#draw limits
hist(estimateB)
abline(v=simIntB, col="blue", lty=2, lwd=2)
##coverage probability of the confidence intervals when taking sample size into account
#coverage probability estimation of the Monte Carlo confidence intervals
#when repeatedly sampling from the skewed distribution of l*b
#the following code demonstrates that the coverage probability of these latter
#confidence intervals is close to the nominal level
#note: running this code may take some time
#true (population) values
#assume that l and b follow a normal and exponential distribution, respectively
meanLT <- 25 #mean of l
sdLT <- .2 #standard deviation of l
rateBT <- .05 #mean of b is 20, thus rate is 1/20=.05
XYt <- meanLT*(1/rateBT) #best estimate of derived quantity (i.e., l*b)
#initialize Monte Carlo simulation
nsim <- 500 #number of simulations
alpha <- .05 #nominal coverage probability is equal to 1-alpha
m <- 50 #sample size
#population Clayton copula
myMvdc <- mvdc(copula=claytonCopula(param=.7, dim=2), c("norm", "exp"),
list(list(mean=25, sd=.02), list(rate=.05)))
#fitted Clayton
myClayton <- mvdc(copula=claytonCopula(param=1, dim=2), c("norm", "exp"),
list(list(mean=0, sd=1), list(rate=1)))
#estimate the coverage probability of the Monte Carlo simulation
nMC <- 100000
nSS <- 50000
uclB <- numeric(nsim)
lclB <- numeric(nsim)
meanNmcXY <- numeric(nMC)
for (j in 1:nsim) {
#sample from population copula
xyb <- rMvdc(m, myMvdc)
#fit Clayton copula to sample
expRate <- exp(-coef(survreg(Surv(xyb[,2])~1, dist="exponential")))
normMean <- mean(xyb[,1])
sdNorm <- sqrt((m-1)/m)*sd(xyb[,1])
tauEst <- cor(xyb[,1], xyb[,2], method="kendall")
thetaClayton <- (2*tauEst)/(1-tauEst)
fitClayton <- fitMvdc(xyb, myClayton, start=c(normMean, sdNorm, expRate, thetaClayton))
#Monte Carlo simulation using fitted Clayton copula
corXYfit <- fitClayton@mvdc@copula@parameters
meanXfit <- fitClayton@estimate[1]
sdXfit <- sqrt(m/(m-1))*fitClayton@estimate[2]
rateYfit <- fitClayton@estimate[3]
mcClayton <- mvdc(copula=claytonCopula(param=corXYfit, dim=2),
margins=c("t", "exp"),
paramMargins=list(list(df=m-1), list(rateYfit)))
xySim <- rMvdc(nMC, mcClayton)
xySim[,1] <- meanXfit + xySim[,1]*sdXfit
hatXYsim <- xySim[,1]*xySim[,2]
#sample with replacement from skewed distribution
meanNmcXY <- replicate(nSS, mean(sample(hatXYsim, m, replace=TRUE)))
#shortest coverage interval
sortedSim <- sort(meanNmcXY)
covInt <- sapply(1:(nSS-round((1-alpha)*nSS)), function(i) {
sortedSim[1+round((1-alpha)*nSS)+(i-1)]-sortedSim[1+(i-1)]})
lclB[j] <- sortedSim[which(covInt==min(covInt))]
uclB[j] <- sortedSim[1+round((1-alpha)*nSS)+(which(covInt==min(covInt))-1)]
}
#coverage probability
mean(lclB < XYt & uclB > XYt) #result will be 94% (approximately)
#average width of confidence intervals
mean(uclB-lclB) #result will be 275 (approximately)``` | {"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": 0.9616755843162537, "perplexity": 752.4946294520405}, "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-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00730.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/163880-joint-pmf-print.html | # joint pmf
• November 20th 2010, 01:09 PM
chutiya
joint pmf
if X and Y are discrete RVs with pmf $p(x,y) = a\; \dfrac{k!}{x!y!(k-x-y)!}$ where x and y are $non \; negative \; integers \; and\; x+y\leq k$.
I want to find the value of a.
I know that I need to show that the pmf sums up to 1.
$\sum_x \sum_y p(x,y)=1$
since x and y are non negatve: 0<=x+y<=k
here i am getting confused in where x and y range from.
can anyone help me how to solve this?
• November 21st 2010, 04:42 AM
Robb
$\sum_{x=0}^{k} \sum_{y=0}^{k-x} p(x,y)=1$
so,
$a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}=a\cdot 3^k =1$
Hence $a=\left(\frac{1}{3}\right)^k$
• November 21st 2010, 06:26 AM
chutiya
Thank you so much.
But how do we get $\sum_{y=o}^{k-x}\dfrac{1}{y!(k-x-y)!}= \dfrac{2^{k-x}}{(k-x)!}$
is this a gamma function? Could you explain please.
• November 22nd 2010, 04:54 AM
Robb
Refer to the section on series of binomial coefficients
Binomial coefficient - Wikipedia, the free encyclopedia
$a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!}{x!y!(k-x-y)!} = a \cdot \sum_{x=0}^{k} \sum_{y=0}^{k-x} \dfrac{k!\cdot \left(k-x\right)!}{x!y!(k-x-y)!(k-x)!} =a \cdot \sum_{x=0}^{k} \dfrac{k!}{x!(k-x)!}\sum_{y=0}^{k-x} \dfrac{\left(k-x\right)!}{y!(k-x-y)!}$
$(Y+X)^n = \sum_{k=0}^n {n \choose k} X^{n-k}Y^K$
$a \cdot \sum_{x=0}^{k} \dfrac{k!\cdot 2^{k-x}}{x!(k-x)!}= a \cdot \sum_{x=0}^\infty {k \choose x} 2^{k-x}1^k=a \cdot \left(2+1\right)^k$ | {"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": 10, "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": 0.9376780390739441, "perplexity": 1202.0318736547522}, "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-2016-22/segments/1464049276537.37/warc/CC-MAIN-20160524002116-00131-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.gamedev.net/forums/topic/216542-am-i-on-the-right-track/ | • Announcements
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am i on the right track?
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coke_baby 122
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Guest Anonymous Poster
If you have a two rigid bodies with a bar between them that links them together (a pin joint) than your physics should still be simple. If a force is applied to one of the objects, then the mass that is accelerated along the vector of the bar is the sum of the masses of the 2 objects (lets assume a weightless bar for now) however... the mass that is accelerated in the direction perpendicular to the bar''s vector is just the mass upon which the force is applied. Try it out on paper... connect two balls (in outter space perhaps) with this "pin joint bar" & apply a force on one of the balls at an angle that is 45 degrees to the bar''s vector. For every iteration following, you must take into account each of the ball''s velocity vectors (really their momentum vectors are of interest) & again... the momentum of each object along the vector of the bar must be calculated using the total mass of the two objects, however the momentum of each object along the vector perpendicular to the bar vector should be calculated with each of the respective object''s mass only, not both.
I hope that was clear as mud.
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coke_baby 122
i kind of see what you're getting at with the bar.
i'm trying to find a way that i can keep my existing physics for the rigid bodies as they are, and add in forces or mass to compensate. I'm coming back to the idea below, that i should bring in forces that compensate for the extra mass:
say i have a pin joint and am applying the force directly along the vector of the bar, to the first mass (actually a rigid body, not a point mass) as you describe.
in order to get the second mass to move as if it were connected by the bar, i can either i can apply the same force to the second body and change the masses of both objects OR
apply a force to both the second body in the direction of the initial force as well as another force in the opposite direction to the first body, leaving both rigid bodies with the same mass.
ie: if both bodies were of the same mass, i'd end up with this:
----->F(B1)<-----F/2 XXXXBARXXX F/2------->(B2)
i think this would keep motion correct for forces applied perpindicular to the bar, as a dot product will reduce this to zero (edit: for the second mass).
does that make sense? would it work?
[edited by - coke_baby on March 30, 2004 2:35:41 PM] | {"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": 0.8646542429924011, "perplexity": 533.0944900592335}, "config": {"markdown_headings": false, "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-2017-34/segments/1502886120573.75/warc/CC-MAIN-20170823152006-20170823172006-00274.warc.gz"} |
https://link.springer.com/article/10.1007%2Fs00211-015-0734-5 | Numerische Mathematik
, Volume 132, Issue 4, pp 767–806
# Multi-index Monte Carlo: when sparsity meets sampling
• Abdul-Lateef Haji-Ali
• Fabio Nobile
• Raúl Tempone
Article
## Abstract
We propose and analyze a novel multi-index Monte Carlo (MIMC) method for weak approximation of stochastic models that are described in terms of differential equations either driven by random measures or with random coefficients. The MIMC method is both a stochastic version of the combination technique introduced by Zenger, Griebel and collaborators and an extension of the multilevel Monte Carlo (MLMC) method first described by Heinrich and Giles. Inspired by Giles’s seminal work, we use in MIMC high-order mixed differences instead of using first-order differences as in MLMC to reduce the variance of the hierarchical differences dramatically. This in turn yields new and improved complexity results, which are natural generalizations of Giles’s MLMC analysis and which increase the domain of the problem parameters for which we achieve the optimal convergence, $${\mathcal {O}}(\mathrm {TOL}^{-2}).$$ Moreover, in MIMC, the rate of increase of required memory with respect to $$\mathrm {TOL}$$ is independent of the number of directions up to a logarithmic term which allows far more accurate solutions to be calculated for higher dimensions than what is possible when using MLMC. We motivate the setting of MIMC by first focusing on a simple full tensor index set. We then propose a systematic construction of optimal sets of indices for MIMC based on properly defined profits that in turn depend on the average cost per sample and the corresponding weak error and variance. Under standard assumptions on the convergence rates of the weak error, variance and work per sample, the optimal index set turns out to be the total degree type. In some cases, using optimal index sets, MIMC achieves a better rate for the computational complexity than the corresponding rate when using full tensor index sets. We also show the asymptotic normality of the statistical error in the resulting MIMC estimator and justify in this way our error estimate, which allows both the required accuracy and the confidence level in our computational results to be prescribed. Finally, we include numerical experiments involving a partial differential equation posed in three spatial dimensions and with random coefficients to substantiate the analysis and illustrate the corresponding computational savings of MIMC.
## Mathematics Subject Classification
65C05 65N30 65N22
## Notes
### Acknowledgments
Raúl Tempone is a member of the Special Research Initiative on Uncertainty Quantification (SRI-UQ), Division of Computer, Electrical and Mathematical Sciences and Engineering (CEMSE) at King Abdullah University of Science and Technology (KAUST). The authors would like to recognize the support of KAUST AEA project “Predictability and Uncertainty Quantification for Models of Porous Media” and University of Texas at Austin AEA Round 3 “Uncertainty quantification for predictive modeling of the dissolution of porous and fractured media”. F. Nobile acknowledges the support of the Swiss National Science Foundation under the Project No. 140574 “Efficient numerical methods for flow and transport phenomena in heterogeneous random porous media”. The authors would also like to thank Prof. Mike Giles for his valuable comments on this work.
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## Authors and Affiliations
• Abdul-Lateef Haji-Ali
• 1
Email author
• Fabio Nobile
• 2
• Raúl Tempone
• 1
1. 1.Applied Mathematics and Computational SciencesKing Abdullah University of Science and Technology (KAUST)ThuwalSaudi Arabia
2. 2.MATHICSE-CSQI, Ecole Polytechnique Fédérale de LausanneLausanneSwitzerland | {"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": 0.8485870957374573, "perplexity": 3707.2533760300175}, "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-2019-26/segments/1560627999779.19/warc/CC-MAIN-20190624231501-20190625013501-00220.warc.gz"} |
https://www.physicsforums.com/threads/electric-field-caused-by-a-uniform-linear-charge.276550/ | # Electric field caused by a uniform linear charge
1. Dec 2, 2008
### gotem3303
A uniform linear charge density of $$\lambda$$ = 2.0 nC/m is distributed along the x-axis from x = 0 to x =3m. Which integral is correct for the magnitude of the electric field at x = -4 m on the x axis?
I'm going over a old test and he didn't circle the correct answer but I'm trying to understand the denominator.
I'm pretty sure I have everything but the denominator, would it be $$\int$$$$\frac{18dx}{(4+x)^{2}}$$ from 0 to 3?
2. Dec 2, 2008
### Redbelly98
Staff Emeritus
Looks okay, except that you did not write any of the units.
Similar Discussions: Electric field caused by a uniform linear charge | {"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": 0.9630283117294312, "perplexity": 730.2162711682337}, "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/1502886103316.46/warc/CC-MAIN-20170817131910-20170817151910-00050.warc.gz"} |
http://mathhelpforum.com/calculus/133921-q2iia8-integration.html | # Math Help - Q2iiA8-integration
1. ## Q2iiA8-integration
Hi:
I have this differentiation problem which am completely not sure about. I have done the question part way, but would like to know if am heading in the right direction before I take the next step.
thank you
Attached Thumbnails
2. That looks difficult - better substitute u = 7 + e^(2x)
3. You mean?
Attached Thumbnails
4. Just in case a picture helps...
... where
... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
The general drift being...
Spoiler:
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
5. So this leaves me with a quotient. Should i now use integration by parts?
Attached Thumbnails
6. Originally Posted by stealthmaths
Hi:
I have this differentiation problem which am completely not sure about. I have done the question part way, but would like to know if am heading in the right direction before I take the next step.
thank you
You could just observe that:
$\frac{d}{dx} \left[\frac{1}{3}(7+e^{2x})^{3/2}\right]=e^{2x}(7+e^{2x})^{1/2}$
then use the fundamental theorem of calculus to find the integral.
CB
7. hi:
so I have got 143.0 (3s.f.) | {"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": 1, "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": 0.9598013758659363, "perplexity": 1288.804585923505}, "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/1406510264270.11/warc/CC-MAIN-20140728011744-00254-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://slideplayer.com/slide/3431885/ | # AP CALCULUS 1005: Secants and Tangents. Objectives SWBAT determine the tangent line by finding the limit of the secant lines of a function. SW use both.
## Presentation on theme: "AP CALCULUS 1005: Secants and Tangents. Objectives SWBAT determine the tangent line by finding the limit of the secant lines of a function. SW use both."— Presentation transcript:
AP CALCULUS 1005: Secants and Tangents
Objectives SWBAT determine the tangent line by finding the limit of the secant lines of a function. SW use both informal and precise mathematical language to describe the tangent line of a function
Average Rates of Change The AVERAGE SPEED (average rate of change) of a quantity over a period of time is the amount of change divided by the time it takes. In general, the average rate of change of a function over an interval is the amount of change divided by the length of the interval. Therefore, the average rate of change can be thought of as the slope of a secant line to a curve.
Average Rate of Change Known Formula Average Rate of Change : (in function notation) ax Slope of a secant
Slope of a Tangent Calculus I -The study of ___________________________________ Slope : (in function notation) ax Move x closer to a Rates of change
A. THE DERIVATIVE (AT A POINT) WORDS: Layman’s description: The DERIVATIVE is a __________________ function. Built on the ________________ formula. The derivative (slope of a tangent line) is the limit of the slopes of the secants as the two points are brought infinitely close together Rate of change slope
Let T(t) be the temperature in Dallas( in o F) t hours after midnight on June 2, 2001. The graph and table shows values of this function recorded every two hours,. What is the meaning of the secant line (units)? Estimate the value of the rate of change at t = 10. t T 073 2 73 470 669 8 72 10 81 Estimate the rate of change tangent line As the points get closer together the ROC approach ROC at 10
EX: THE DERIVATIVE AT A POINT EX 1: at a = 4 Notation: Words: f(a) = 8
Equation of the Tangent To write the equation of a line you need: a) b) Point- Slope Form: point slope (4,8) m = 6
There is local linearity When the secant lines get very near a point it acts like a tangent line
Normal to a Curve The normal line to a curve at a point is the line perpendicular to the tangent at the point. Therefore, the slope of the normal line is the negative reciprocal of the slope of the tangent line. EX 1(cont): at a = 4 Find the equation of the NORMAL to the curve
EX: THE DERIVATIVE AT A POINT EX 2: at x = -2 Method: y = 1 Mantra:
At a Joint Point SKIP Piece Wise Defined Functions: The function must be CONTINUOUS Derivative from the LEFT and RIGHT must be equal. The existence of a derivative indicates a smooth curve; therefore,
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Download ppt "AP CALCULUS 1005: Secants and Tangents. Objectives SWBAT determine the tangent line by finding the limit of the secant lines of a function. SW use both."
Similar presentations | {"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": 0.8870858550071716, "perplexity": 532.9493688651319}, "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-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00082.warc.gz"} |
http://alicematters.web.cern.ch/?q=comment/88092 | by Virginia Greco. Published: 05 November 2016
The period of the year when ALICE plays the leading role at LHC has finally arrived: after smashing protons for months, the machine will start accelerating lead ions on November 7th and continue until the winter shut-down. After ramp up and adjustment, the accelerator will deliver stable beams and proton-lead collisions, first at 5 TeV and then at 8 TeV of energy per colliding nucleon pair.
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This year’s run is divided in two phases. In the first, proton and lead beams will be collided at 5 TeV. This energy is equivalent to that of the Pb-Pb collisions in 2015, and of a special pp reference data sample which is being collected through Run 2, thus researchers will be able to make direct comparisons between all three at the same energy. In this year’s 5 TeV p-Pb run, ALICE scientists aim at collecting much larger statistics of events than in 2013, in view of a campaign of high-precision measurements on the p-Pb system.
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An exciting period for ALICE’s researchers lies ahead.
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https://www.gerad.ca/fr/papers/G-2005-56 | Groupe d’études et de recherche en analyse des décisions
# Extension of the Weiszfeld Procedure to a Single Facility Minisum Location Model with Mixed Norms
This paper presents a general mixed-norm minisum problem for locating a single facility in continuous space. It is assumed that several transportation modes exist between the new facility and a given set of fixed points (the customers), each mode being represented by a different $\ell_p$ norm. A simple extension of Weiszfeld’s well-known iterative procedure is proposed to solve the model. Convergence properties and opti- mality criteria are derived, and computational results are given. | {"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": 1, "/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": 0.9288347363471985, "perplexity": 1495.8464716654667}, "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-2018-05/segments/1516084891316.80/warc/CC-MAIN-20180122113633-20180122133633-00301.warc.gz"} |
http://theinfolist.com/php/SummaryGet.php?FindGo=Talk:Crest_factor | TheInfoList
1. waveforms in table should be normalized to RMS, not peak. this is the normative normalization method.
As per the grid lines, the waveforms in the table are normalized to a 'peak' level of one. I don't know exactly what you mean by 'normative normalization method' but normalizing to peak value of 1 makes the most sense to me visually. --Kvng (talk) 19:26, 8 March 2010 (UTC)
waveform type, waveform plot, peak value, crest factor (linear), crest factor (dB)
RMS value of 1 is implied for all waveforms so it doesn't need to be a column.
non-relevant columns should be pushed to the end, or removed. right now they are at the center of the table, even though this is not an article about mean-rectifed magnitude, and waveform factor. (and waveform factor is not even related to crest factor/PAPR..)
3. all these mathematical derivations (square roots and ) are obscuring the main data. consider hiding them / moving them.
217.132.27.200 (talk) 21:19, 4 February 2008 (UTC)
let me correct myself re #2 - when waveforms are normalized to 1 RMS, the "peak value" column is equal to the "crest factor (linear)" column. so both columns should be merged into 1 column. this makes the table even simpler and more understandable.
I'm ready to do the corrections myself, but i would like some OK first so that some whimsical admin won't revert my work.
217.132.27.200 (talk) 21:23, 4 February 2008 (UTC)
Table
The table of the article is wrong.
The mean values of sine, triangle and square wave signals are all 0.
The values given are for (perfectly) rectified signals.
KjellElec
Yes. Also, the meaning of waveform factor is not explained. Omegatron 04:33, 21 October 2006 (UTC)
It appears to be the RMS divided by the mean, though I don't know how it is used. — Omegatron 07:52, 9 March 2007 (UTC)
I've made some bold edits to the table. Waveform factor seemed only to complicate things it is gone. As a reference point, peak level in all cases is 1. So perhaps we don't need the Peak magnitude column. The stragglers at the end of the table might want to be moved to their own table. --Kvng (talk) 19:26, 8 March 2010 (UTC)
You are correct and the article is still wrong. Sine triangle, and square have PAR=1.Skeptonomicon (talk) 22:36, 12 April 2016 (UTC)
Also: QPSK (which has zero-crossings in the constellation diagram) is incorrectly shown to have lower PAPR than OQPSK (which has no zero crossings and would ideally be constant modulus, except for overshoot); this shows the table is incorrect. — Preceding unsigned comment added by 2610:148:610:3C0D:B464:91F3:2A41:BDEE (talk) 17:22, 3 January 2020 (UTC)
Reformat
Please break up the table into individual images and editable math formulas in an HTML table. I'll do it if no one else does, but I don't have time right now. — Omegatron 19:38, 20 October 2006 (UTC)
The table should also be expanded to include Gaussian white noise, pink noise, compressed mastered CD audio, raw audio from a microphone, etc. etc. — Omegatron 19:59, 20 October 2006 (UTC)
Starting:
Wave type Waveform Mean value (rectified) Waveform factor RMS value Crest factor
Sine wave ${\displaystyle {2 \over \pi }\approx 0.637}$ ${\displaystyle {\pi \over 2{\sqrt {2}}}\approx 1.11}$ ${\displaystyle {1 \over {\sqrt {2}}}\approx 0.707}$ ${\displaystyle {\sqrt {2}}\approx 1.414}$
Half-wave rectified sine ${\displaystyle {1 \over \pi }\approx 0.318}$ ${\displaystyle {\pi \over 2}\approx 1.571}$ ${\displaystyle {1 \over 2}=0.5}$ ${\displaystyle 2}$
Full-wave rectified sine ${\displaystyle {2 \over \pi }\approx 0.637}$ ${\displaystyle {\pi \over 2{\sqrt {2}}}\approx 1.11}$ ${\displaystyle {1 \over {\sqrt {2}}}\approx 0.707}$ ${\displaystyle {\sqrt {2}}\approx 1.414}$
Triangle wave ${\displaystyle {1 \over 2}=0.5}$ ${\displaystyle {2 \over {\sqrt {3}}}\approx 1.155}$ ${\displaystyle {1 \over {\sqrt {3}}}\approx 0.577}$ ${\displaystyle {\sqrt {3}}\approx 1.732}$
Sawtooth wave ${\displaystyle {1 \over 2}=0.5}$ ${\displaystyle ?}$ ${\displaystyle ?}$ ${\displaystyle ?}$
Square wave ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle 1}$ ${\displaystyle 1}$
Obviously we should create dedicated images. These are just placeholders. Omegatron 04:12, 21 October 2006 (UTC)
I created better images and they are now in the article and here. — Omegatron 05:05, 9 March 2007 (UTC)
I think we should more clearly discuss the mean values are obtained by rectification. At a quick-glance, these values seem incorrect, though they are just rectified.
What do you mean? — Omegatron 05:05, 9 March 2007 (UTC)
Waveform factor is the ratio of DC average to RMS and is used to scale resistors for measurements with DC or AC meters. The waveform factor for the half wave rectified sine wave should be 2.22 as the DC average is VP/Pi. —Preceding unsigned comment added by 212.77.217.195 (talk) 11:32, 26 September 2007 (UTC)
Peak Value?
Suppose we have a waveform whose peak values are asymmetric. That is, the magnitude of the negative peak is different from the magnitude of the positive peak. Which peak is used when calculating crest factor: the higher or lower magnitude? Or should the average of the peak magnitudes be used?
65.161.52.184 21:22, 18 December 2006 (UTC) Scott L.
The higher value would be used. — Omegatron 05:05, 9 March 2007 (UTC)
Whole premise of article is inconsistent, Term is ill defined
The term peak to average power ratio or crest factor, as used in electronics, is the peak of the power envelope divided by the average of the power envelope, and is not typically applied to signals directly. PAPR and PAR are almost exclusively used to talk about envelopes, though I am less sure that crest factor is limited to envelopes.
As an example of the inconsistency, a QPSK signal with all zeros (no modulation) is a sine wave, yet the table has different values for sine wave and QPSK. QPSK is correctly shown as the ratio of the peak power envelope to the average power envelop, but the sine is ignoring the envelope and incorrectly listing a PAR greater than one. The PAR of a sine wave is one. Skeptonomicon (talk) 22:31, 12 April 2016 (UTC)
Also, in the table, QPSK (which has zero-crossings like a sine wave) is shown to have lower PAPR than OQPSK (which has no zero crossings) and would ideally be constant modulus (except for overshoot); this shows the table is incorrect. — Preceding unsigned comment added by 2610:148:610:3C0D:B464:91F3:2A41:BDEE (talk) 17:15, 3 January 2020 (UTC)
Crest factor vs PAR
The source I am reading says that PAR is "identical to the traditional crest factor", implying that crest factor is an older term. — Omegatron 04:19, 26 July 2007 (UTC)
I still hear the term crest factor used all the time. Much more, in fact, than PAR
Me too, in my field it's much more common. It probably depends what field you're in --mcld (talk) 13:01, 29 January 2009 (UTC)
"Crest factor" is in IEE Std. 100. "Peak to average ratio" is not. A quick look on Google initially shows PAR meters are tropical fish tank light meters measuring Photosynthetically active radiation. We need a reference for PAR meter being a recognized term in some industry, else I'd be inclined to cut it out. --Wtshymanski (talk) 16:38, 29 January 2009 (UTC)
Yes, "crest factor" is defined in IEEE Std. 100[1] as the ratio of a peak quantity to the RMS value of the quantity. Examples provided are: (1) & (2) voltage, (3) any periodic function, etc.
I'm sorry, but "Average" or "Mean"?. And the crest factor is the "peak to rms" or "peak to average (or mean)"? It's very confused for me.--Vmsa (talk) 01:57, 31 May 2013 (UTC)
References
crest factor for stochastic signals
The crest factor as given here does not to seem very practical for stochastic signals. The reason is that it can always happen that there occurs an rare, but extremely high value. This high value would determine the crest factor for the whole measurement. I think that practically, some kind of "decaying maximum" is used in real measuring devices. In other words, the maximum is "forgotten" after a while. The article should describe how this is typically done in practice. --62.159.14.3 (talk) 12:32, 12 June 2009 (UTC)
For real-time measurements there's also a practical problem with using a true average. In practice, there are time constants associated with both peak and average readings. These are commonly first-order low-pass filters. --Kvng (talk) 19:07, 8 March 2010 (UTC)
Isn't MS an indication of average power - not RMS? —Preceding unsigned comment added by 116.193.184.242 (talk) 08:59, 21 October 2010 (UTC)
No: Root_mean_square#Average_electrical_power 71.167.72.23 (talk) 14:58, 14 April 2014 (UTC)
PAR vs PAPR
I'm confused. PAR interpreted literally cannot be a synonym of crest factor, since average voltage != RMS voltage. For any waveform without DC, the average voltage is 0, so PAR is always infinite? I thought PAR was implicitly always a power ratio, since average power = RMS voltage and they then produce the same number. Now the article says PAR and PAPR are separate things? — Omegatron (talk) 16:31, 18 September 2013 (UTC)
PAPR of a sine wave
I've got a CCDF measurement up in my Keysight VSA software. I'm inputting a 2GHz sine wave, and the CCDF tells me the PAPR is roughly 0.2dB. told me 3dB, and now I'm spiraling into a world of confusion.
Your measurement for 2GHz sine wave aka CW is correct. PAPR for such a signal is 0dB. and also some authors have fooled themselves to believe that sine wave has PAPR of 3dB. It is true if the sinewave is an envelope of a RF signal.
In case a RF signal has an envelope of sinewave, in other words is an AM-modulated RF signal, it will have PAPR of 3dB if AM modulation index is 100%. In case a RF signal has an envelope of DC, you could call it a sine wave if you want, it will have PAPR of 0dB.
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Don't write 'Effective value'; write 'RMS'
In first paragraph, don't write 'Effective value'; write 'RMS'. If you write 'effective value' that phrase should be defined and/or hyperlinked. It's just another layer for the reader to try to figure out. Just use 'RMS' then everyone knows what the crest (peak) is being compared to. 12.33.223.210 (talk) 18:57, 4 October 2018 (UTC)
I just asked my wife and she doesn't know what "RMS" means. "Effective value" is at least English and not acronymese. --Wtshymanski (talk) 16:23, 9 October 2018 (UTC) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "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": 0.8412171006202698, "perplexity": 2673.7545920056054}, "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-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00071.warc.gz"} |
https://arxiv.org/abs/math/0606088 | math
(what is this?)
(what is this?)
# Title: Linear Equations in Primes
Abstract: Consider a system \Psi of non-constant affine-linear forms \psi_1,...,\psi_t: Z^d -> Z, no two of which are linearly dependent. Let N be a large integer, and let K be a convex subset of [-N,N]^d. A famous and difficult open conjecture of Hardy and Littlewood predicts an asymptotic, as N -> \infty, for the number of integer points n in K for which the integers \psi_1(n),...,\psi_t(n) are simultaneously prime. This implies many other well-known conjectures, such as the Hardy-Littlewood prime tuples conjecture, the twin prime conjecture, and the (weak) Goldbach conjecture. <p>
In this paper we (conditionally) verify this asymptotic under the assumption that no two of the affine-linear forms \psi_1,...,\psi_t are affinely related; this excludes the important binary'' cases such as the twin prime or Goldbach conjectures, but does allow one to count non-degenerate'' configurations such as arithmetic progressions. Our result assumes two families of conjectures, which we term the Inverse Gowers-norm conjecture GI(s) and the Mobius and Nilsequences Conjecture MN(s), where s \in {1,2,...} is the complexity of the system and measures the extent to which the forms \psi_i depend on each other. For s = 1 these are essentially classical, and the authors recently resolved the cases s = 2.<p>
Our results are therefore unconditional in the case s = 2, and in particular we can obtain the expected asymptotics for the number of 4-term progressions p_1 < p_2 < p_3 < p_4 <= N of primes, and more generally for any (non-degenerate) problem involving two linear equations in four prime unknowns.
Comments: 84 pages, numerous small changes made in the light of comments from the referees Subjects: Number Theory (math.NT); Dynamical Systems (math.DS) Cite as: arXiv:math/0606088 [math.NT] (or arXiv:math/0606088v2 [math.NT] for this version)
## Submission history
From: Ben Green [view email]
[v1] Sun, 4 Jun 2006 11:02:25 GMT (79kb)
[v2] Tue, 22 Apr 2008 17:00:46 GMT (79kb) | {"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": 0.911682665348053, "perplexity": 1244.2151302494706}, "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-2016-40/segments/1474738659865.46/warc/CC-MAIN-20160924173739-00158-ip-10-143-35-109.ec2.internal.warc.gz"} |
https://research.brighton.ac.uk/en/publications/understanding-changes-in-uptake-and-release-of-serotonin-from-gas | # Understanding changes in uptake and release of serotonin from gastrointestinal tissue using a novel electroanalytical approach
G. Marcelli, Bhavik Patel
Research output: Contribution to journalArticlepeer-review
## Abstract
Serotonin (5-HT) is well known to be a key neurotransmitter within the gastrointestinal (GI) tract, where it is responsible for influencing motility. Obtaining dynamic information about the neurotransmission process (specifically the release and reuptake of 5-HT) requires the development of new approaches to measure the extracellular 5-HT concentration profile. In this work constant-potential amperometry has been utilised at +650 mV vs. Ag|AgCl to measure in vitro the overflow of 5-HT. Steady-state levels of 5-HT have been observed, due to continuous mechanical stimulation of the tissue from the experimental protocol. Measurements are conducted at varying tissue–electrode distances in the range of 5 to 1100 µm. The difference in the current from the bulk media and that from each tissue–electrode distance is obtained, and the natural log of this current is plotted versus the tissue–electrode distance. The linear fit to the log of the current is derived, and its intercept, I0, with the vertical axis and its slope are calculated. The reciprocal of the slope, indicated as slope−1, is used as a marker of reuptake. The ratio between intercept, I0, and the reciprocal of the slope, I0/slope−1, is a measure of the flux at the tissue surface and it can be used as a marker for the 5-HT release rate. Current measurements for ileum and colon tissue indicated a significantly higher reuptake rate in the colon, showed by a lower slope−1. In addition, the ratio, I0/slope−1, indicated that the colon has a higher 5-HT flux compared to the ileum. Following the application of the serotonin selective reuptake inhibitor (SSRI), fluoxetine, both tissues showed a higher value of slope−1, as the reuptake process is blocked preventing clearance of 5-HT. No differences were observed in the ratio, I0/slope−1, in the ileum, but a decrease was observed in the colon. These results indicate that ileum and colon are characterised by different reuptake and release processes. The new approach we propose provides pivotal information on the variations in the signalling mechanism, where steady state levels are observed and can be a vital tool to study differences between normal and diseased tissue and also the efficacy of pharmacological agents
Original language English 2340-2347 8 Analyst 135 9 https://doi.org/10.1039/C0AN00260G Published - 2 Jul 2010
### Bibliographical note
© 2010 The Royal Society of Chemistry | {"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": 0.848637044429779, "perplexity": 3051.646863346508}, "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/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00253.warc.gz"} |
https://stats.stackexchange.com/questions/212813/help-me-understand-the-quantile-inverse-cdf-function/212828 | # Help me understand the quantile (inverse CDF) function
I am reading about the quantile function, but it is not clear to me. Could you provide a more intuitive explanation than the one provided below?
Since the cdf $F$ is a monotonically increasing function, it has an inverse; let us denote this by $F^{−1}$. If $F$ is the cdf of $X$, then $F^{−1}(\alpha)$ is the value of $x_\alpha$ such that $P(X \le x_\alpha) = \alpha$; this is called the $\alpha$ quantile of $F$. The value $F^{−1}(0.5)$ is the median of the distribution, with half of the probability mass on the left, and half on the right. The values $F^{−1}(0.25)$ and $F^{−1}(0.75)$ are the lower and upper quartiles.
• You should learn to use math markup, see my edits! May 16, 2016 at 12:07
• This is a model of concise explanation at a certain level and contains an example already. It's unclear what level of explanation you seek. An answer could be 10 times longer than this depending on what you don't know. E.g. do you know a cdf is? do you know what 'monotonically increasing' means? do you know what an inverse function is? We're only part way through the first sentence. Your question is equivalent to a statement that you don't understand (all) this and although we have no reason to doubt you, that is not at all a precise question. May 16, 2016 at 12:57
All this may sound complicated at first, but it is essentially about something very simple.
By cumulative distribution function we denote the function that returns probabilities of $$X$$ being smaller than or equal to some value $$x$$,
$$\Pr(X \le x) = F(x).$$
This function takes as input $$x$$ and returns values from the $$[0, 1]$$ interval (probabilities)—let's denote them as $$p$$. The inverse of the cumulative distribution function (or quantile function) tells you what $$x$$ would make $$F(x)$$ return some value $$p$$,
$$F^{-1}(p) = x.$$
This is illustrated in the diagram below which uses the normal cumulative distribution function (and its inverse) as an example.
## Example
As an simple example, you can take a standard Gumbel distribution. Its cumulative distribution function is
$$F(x) = e^{-e^{-x}}$$
and it can be easily inverted: recall natural logarithm function is an inverse of exponential function, so it is instantly obvious that quantile function for Gumbel distribution is
$$F^{-1}(p) = -\ln(-\ln(p))$$
As you can see, the quantile function, according to its alternative name, "inverts" the behaviour of cumulative distribution function.
## Generalized inverse distribution function
Not every function has an inverse. That is why the quotation you refer to says "monotonically increasing function". Recall that from the definition of the function, it has to assign for each input value exactly one output. Cumulative distribution functions for continuous random variables satisfy this property since they are monotonically increasing. For discrete random variables cumulative distribution functions are not continuous and increasing, so we use generalized inverse distribution functions which need to be non-decreasing. More formally, the generalized inverse distribution function is defined as
$$F^{-1}(p) = \inf \big\{x \in \mathbb{R}: F(x) \ge p \big\}.$$
The definition, translated to plain English, says that for given probability value $$p$$, we are looking for some $$x$$, that results in $$F(x)$$ returning value greater or equal then $$p$$, but since there could be multiple values of $$x$$ that meet this condition (e.g. $$F(x) \ge 0$$ is true for any $$x$$), so we take the smallest $$x$$ of those.
## Functions with no inverses
In general, there are no inverses for functions that can return same value for different inputs, for example density functions (e.g., the standard normal density function is symmetric, so it returns the same values for $$-2$$ and $$2$$ etc.). The normal distribution is an interesting example for one more reason—it is one of the examples of cumulative distribution functions that do not have a closed-form inverse. Not every cumulative distribution function has to have a closed-form inverse! Hopefully in such cases the inverses can be found using numerical methods.
## Use-case
The quantile function can be used for random generation as described in How does the inverse transform method work?
• This answer works well up until the penultimate paragraph. By the time you get there, you have asserted that every continuous CDF has an inverse but then you appear to have offered the Normal distribution as a counterexample to that very statement. That is potentially very confusing.
– whuber
May 5, 2017 at 13:23
• @whuber you are right, added one sentence to make it more clear.
– Tim
May 5, 2017 at 13:26
• Tim, and I added one more word to make it even clearer :) May 5, 2017 at 13:27
• @Tim great answer but could you shed some light on the definition of the inverse cdf $F^{-1}(u)=\inf\{x:F(x) \ge u\}$? As you mentioned we ask what $x$ would make $F(x)=p$. I understand the $\inf$ part as follows. Since the cdf is monotone increasing there are many values all satisfying $F(x) \ge u$ but the $\inf$ would give the greatest lower bound, i.e. fix a unique point and by doing so define generalized inverse. Does this make sense ? Oct 16, 2018 at 19:29
• @AlexanderCska Yes, basically, multiple F(x) values are greater then u, so we take the lower bound, "the smallest value that meets this condition".
– Tim
Oct 17, 2018 at 6:41
I'd like to add one more remark. Not every monotonically increasing function has an inverse function. Actually only strictly monotonically increasing/decreasing functions have inverse functions.
For monotonically increasing cdf which are not strictly monotonically increasing, we have a quantile function which is also called the inverse cumulative distribution function. You can find more details here.
Both inverse functions (for those strictly increasing cdf) and quantile functions (for those monotonically increasing but not strictly monotonically increasing cdfs) can be denoted as $$F^{-1}$$, which can be confusing sometimes.
Chapter 2 of the book "Statistical Distributions" by Forbes, Evans, Hastings, and Peacock has a concise summary with consistent notation.
A quantile is any possible value (e.g. in context of a random draw) of a variable, that is, a variate. The authors give an example of a sample space of tossing 2 coins as the set {HH, HT, TH, TT}. The number of heads in that sample is a quantile of the ordered set {0, 1, 2}.
For a probability distribution or mass function, you are plotting the variate on the x-axis and the probability on the y-axis.
If you knew the probability and the function and wanted to deduce the variate on the x-axis from it, you would invert the function or approximate an inversion of it to get x, knowing y.
The discrete or continuous values along the y-axis for the discrete or continuous pdf might not be increasing and there may be multiple x's which would result in the same y.
The CDF (cumulative distribution function) is more convenient as the function plotted is increasing along the x-axis and the y-axis. Extracting the quantile, that is, the variate from CDF is usually easier math.
There are a few diagrams in the book demonstrating properties of the discrete probability distribution, and the CDF in chapter 2 and those are shown in answers posted to your question above this one also (though I can't see them while I'm typing this answer).
Table 2.1 has a concise summary of many terms and item 4 is for the inverse distribution function or quantile function (of probability alpha) and refers to determining x from the inverse function which takes the probability as an argument.
The book is a practical handbook on the subject with examples, though implementing the inverse functions requires other resources (like pre-computed tables findable at NIST or published approximation algorithms etc. https://www.itl.nist.gov/div898/handbook/eda/section3/eda367.htm).
(NOTE: everything past the 1st sentence was added in response to the comment from gung.)
• Can you give a precis of their summary? Otherwise, this should be a comment. Sep 1, 2020 at 20:42 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.893828809261322, "perplexity": 276.15516797394315}, "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-2022-40/segments/1664030337853.66/warc/CC-MAIN-20221006155805-20221006185805-00379.warc.gz"} |
https://sciencenotes.org/calculate-percent-error/?shared=email&msg=fail | # Calculate Percent Error 9
Percent error is the percent difference between a measured and expected value. (image: Sherman Geronimo-Tan)
Percent Error Definition
Percent error, sometimes referred to as percentage error, is an expression of the difference between a measured value and the known or accepted value. It is often used in science to report the difference between experimental values and expected values.
The formula for calculating percent error is:
Note: occasionally, it is useful to know if the error is positive or negative. If you need to know the positive or negative error, this is done by dropping the absolute value brackets in the formula. In most cases, absolute error is fine. For example, in experiments involving yields in chemical reactions, it is unlikely you will obtain more product than theoretically possible.
### Steps to Calculate the Percent Error
1. Subtract the accepted value from the experimental value.
2. Take the absolute value of step 1
3. Divide that answer by the accepted value.
4. Multiply that answer by 100 and add the % symbol to express the answer as a percentage.
Now let’s try an example problem.
You are given a cube of pure copper. You measure the sides of the cube to find the volume and weigh it to find its mass. When you calculate the density using your measurements, you get 8.78 grams/cm3. Copper’s accepted density is 8.96 g/cm3. What is your percent error?
Solution:
experimental value = 8.78 g/cm3
accepted value = 8.96 g/cm3
Step 1: Subtract the accepted value from the experimental value.
8.78 g/cm3 – 8.96 g/cm3 = -0.18 g/cm3
Step 2: Take the absolute value of step 1
|-0.18 g/cm3| = 0.18 g/cm3
Step 3: Divide that answer by the accepted value.
Step 4: Multiply that answer by 100 and add the % symbol to express the answer as a percentage.
0.02 x 100 = 2
2%
The percent error of your density calculation was 2%.
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## 9 thoughts on “Calculate Percent Error”
• Mark
Steps 1 and 3 use the wrong values. Since the experimental value is smaller than the accepted value it should be a negative error.
• Mary Andrews
Percent error is always represented as a positive value. The difference between the actual and experimental value is always the absolute value of the difference. |Experimental-Actual|/Actualx100 so it doesn’t matter how you subtract. The result of the difference is positive and therefore the percent error is positive.
• David
Percent error is always positive, but step one still contains the error initially flagged by Mark. The answer in that step should be negative:
experimental-accepted=error
8.78 – 8.96 = -0.18
In the article, the answer was edited to be correct (negative), but the values on the left are still not in the right order and don’t yield a negative answer as presented.
• mytrialbuisness
Say if you wanted to find acceleration caused by gravity, the accepted value would be the acceleration caused by gravity on earth (9.81…), and the experimental value would be what you calculated gravity as 🙂
• jessica bedford
okay say you have two different percent errors, because of a word problem were there the end result id two groups of percent errors, how do you know which one is the accurate 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": 0.8597095608711243, "perplexity": 883.1192145247583}, "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-2019-09/segments/1550247514804.76/warc/CC-MAIN-20190222074154-20190222100154-00639.warc.gz"} |
https://www.physicsforums.com/threads/centripetal-acceleration-part-ii.219491/ | # Centripetal Acceleration Part II
1. Mar 3, 2008
### BitterSuites
[SOLVED] Centripetal Acceleration Part II
1. The problem statement, all variables and given/known data
(From previous question) In order for a satellite to move in a stable circular orbit of radius 6689 km at a constant speed, its centripetal acceleration must be inversely proportional to the square of the radius of the orbit.
What is the speed of the satellite? The G is 6.67259e-11 and the mass of the earth is 5.98e24kg.
The answer to this was 7723.55 m/s.
Part II:
Find the time required to complete one orbit. Answer in units of h.
2. Relevant equations
The book only gives one equation for T referencing orbits, which is v=2pir/T
3. The attempt at a solution
v=2pi/T
7723.55 = 2pi/T
7723.55T = 2pi
T = 2pi/7723.55
T = .000814
I think the homework system even laughed at me when I entered that answer :)
2. Mar 3, 2008
### Staff: Mentor
Right.
Wrong.
Compare the two equations. (You left off the r.)
3. Mar 3, 2008
### BitterSuites
*sigh* I keep making really silly mistakes.
So I input the correct equation but it came out correct.
v=2pir/t
7723.55 m/s = 2pi * 6689000m/T
7723.55T = 4.20282e7
T = 5221.57 seconds
Convert to hours
5221.57/60 = 90.6928 minutes
90.6928/60 = 1.51155 hours
Thank you so much.
Last edited: Mar 3, 2008
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https://www.physicsforums.com/threads/accuracy-of-experimental-values.135027/ | # Accuracy of experimental values
1. Oct 5, 2006
### mbrmbrg
We did an experiment with abysmal accuracy. I'm now writing up my report and would like to meaningfully express the error between calculated and measured values.
EXAMPLE
calculated T=0.284
measured T = 0.201
I want to express the difference between them as follows:
% difference between the two = (1 - 0.201/0.284)(100%) = 29%
But as ridiculous as the errors are throughout the experiment, they're not that bad!
Take the difference between the two values and then divide by one of the two values? But which one?
I know how to take percent error for something like $107.45\pm\0.08$ but this seems to be a different cup of tea.
Last edited: Oct 5, 2006
2. Oct 6, 2006
### andrevdh
Your 29% error value is correct. I assume that the calculation was also done with the measurements? It is impossible to comment on experiments, even if you tell us exactly what you did. Experiments are a very personal thing. We can only offer suggestions but even then may be missing something very obvious. So it is best to fool around with an experiment and try and push it in all directions in order to gain confidence in the result.
3. Oct 6, 2006
### mbrmbrg
The experiment was a basic one about Newton's second law:
a dynamics cart was placed on a track, a mass was attatched to a string that ran from the cart over a pulley toward the floor.
T is the tension in the string connecting the cart and the suspended mass.
The calculation was $T = \frac{(m_1)(m_2)}{m_1+m_2}g$
and the measured value of T was obtained using a force sensor hooked up to a Pasco interface.
So assuming the mass was decently accurate (and this balance hasn't created crazy abberations in other experiments), the calculated T should be OK.
In the brief pre-lab, the professor asked us if we all believed that F=ma, and when we said yes, he replied that we wouldn't so much after this experiment. But 30%?!?! Yow.
4. Oct 6, 2006
### andrevdh
Are you sure that the measured tension is the smaller one and not the other way round?
Last edited: Oct 6, 2006
5. Oct 9, 2006
### mbrmbrg
In 6 out of 7 trials, the measured tension was indeed smaller than the calculated tension.
Not only that, but measurments say that if there's a heavier cart and a lighter cart with the same suspended mass, the string attatched to the heavier cart will have a considerably smaller tension than the lighter cart--the exact opposite of what the calculations predict!
I'm hoping for a really good post-lab to explain why we spent two hours on an experiment that the professor knew would give such screwy results.
6. Oct 9, 2006
### andrevdh
Well according to my analysis the tension in the string should increase from a case where no rolling friction, f, is acting on the cart:
$$T_o = \frac{m_1 m_2 g}{m_1 + m_2}$$
to a case where the cart do experience rolling friction:
$$T_f = f(1 - \frac{m_1}{m_1 + m_2}) + T_o$$
which is not what you are reporting.
The heavier cart will experience a larger rolling frictional force than the lighter one.
Did you perhaps notice if the maximum force that the force sensor could measure was 50 N?
7. Oct 9, 2006
### andrevdh
Your tensions look very small to me. They are approximately the weight of only 20 grams!
8. Oct 9, 2006
### mbrmbrg
Could you explain when you're talking about what a little more?
49 N, actually . Why would that make a difference, if the largest measured tension was 0.267 N and the largest calculated tension was 0.4729 N?
And you're on the money: 0.201 N was the measured tension when the suspended mass was 30g.
Sorry, I really should draw up a table and attatch it... Do you want me to do that? At this point, I don't think I need enough help to make it worth it, but if you're interested in a screwy experiment, I'll be happy to share.
Last edited: Oct 9, 2006
9. Oct 9, 2006
### mbrmbrg
And andrevdh, please pardon my manners (or lack thereof).
Thank you very much for your help.
10. Oct 11, 2006
### andrevdh
Both unexpected observations can be explained if the pulley is generating a bit of friction (taken that you did the experiment with quite small masses).
To come back to the force sensor. If its maximum capability is 50 N (5 kg) one can expect that it will not be that accurate at 20 gram (0.020 kg). So any measurements at such small tensions may have quite significant errors (uncertainties). This means that the error in the measured value may be so large that it includes the calculated value as a possible outcome for the experiment.
As far as friction on the pulley goes consider this: Friction in the pulley will increase the tension on the "input side" (masshanger pulling on the pulley) and decrease it on the "output side" (force sensor) of the pulley. This can account for the difference between the measured and calculated tension.
A decrease in tension with a heavier cart can be explained as follows. The pulley is most likely to generate larger friction as lower rotational speeds - that is a heavier cart will experience a lower acceleration and therefore measure a lower tension than that of a lighter cart.
You can test this hypothesis by using two force sensors simultaneously. Suspend another force sensor from the string and run the cart (without the mass hanger since the mass of the force sensor will be sufficient). Then compare the two tensions to see whether the input tension is more than the output tension.
Last edited: Oct 11, 2006 | {"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": 0.8024595975875854, "perplexity": 1197.5917296547639}, "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/1521257647530.92/warc/CC-MAIN-20180320185657-20180320205657-00436.warc.gz"} |
http://math.stackexchange.com/questions/131402/if-f-is-lipschitz-and-g-is-differentiable-then-g-circ-f-is-differentiable-w | # If $f$ is Lipschitz and g is differentiable, then $g\circ f$ is differentiable where $g'(b)=0$.
I'm having problems showing this result:
Let $f:U\rightarrow\mathbb{R}^m$ be a Lipschitz function with $a\in U$ and $g:V\rightarrow\mathbb{R}^p$, $V$ open, with $f(U)\subset V$ and $b=f(a)$. If $g'(b)=0$ then $g\circ f:U\rightarrow \mathbb{R}^p$ is differentiable in $a$, with $(g\circ f)'(a)=0$.
I have tried to use the definition of $g$ being differentiable in $b=f(a)$ and using $f(v)$ as the increment $h$ so:
$g(f(a)-f(v))-g(f(a))=g'(f(a))\cdot f(v)+r(f(a),f(v))$
Since $g'(b)=g'(f(a))=0$, we have:
$g(f(a)-f(v))-g(f(a))=r(f(a),f(v))$
I have tried to do through the inverse way but couldn't do it either. Can someone help me?
-
You may start by using that $f$ is Lipschitz, to replace the first term in you last equation. – Andrés Caicedo Apr 13 '12 at 18:34
You need to show that $||f\circ g(a+h)-f\circ g(a)||=||h||R(h)$ where $R(h)\to 0$ as $h\to 0$. Now, you have $||f\circ g(a+h)-f\circ g(a)|| \le ||g(a+h)-g(a)||=||h||R(h)$ since $g'(a)=0$. – Ivan Apr 13 '12 at 18:36
@AndresCaicedo, the Lipschitz property is only valid under absolute value. – Marra Apr 13 '12 at 19:28
@Ivan, I think you confused f with g when doing the composition. – Marra Apr 13 '12 at 19:28
@GustavoMarra Yes, so: Do you see what to do about that? Where are the absolute values going to come from? – Andrés Caicedo Apr 13 '12 at 20:19
The definition of differentiability of $g$ is more commonly written $$g(y)=g(x)+g'(x)\cdot (y-x)+o(|y-x|)\qquad\text{as } y\to x.$$ Substitute in $x=f(a)$ and $w=f(v)$ and let $v\to w$.
@GustavoMarra: You need it in order to see that $o(|f(v)-f(a)|)=o(|v-a|)$. – Harald Hanche-Olsen Apr 14 '12 at 7:21 | {"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": 0.9435655474662781, "perplexity": 166.03472061364778}, "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-2016-07/segments/1454701174607.44/warc/CC-MAIN-20160205193934-00208-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://philpapers.org/rec/WEHFOH | # Fragments of HA based on Sigma_1 induction
Archive for Mathematical Logic 37 (1):37-49 (1997)
Abstract In the first part of this paper we investigate the intuitionistic version \$iI\!\Sigma_1\$ of \$I\!\Sigma_1\$ (in the language of \$PRA\$ ), using Kleene's recursive realizability techniques. Our treatment closely parallels the usual one for \$HA\$ and establishes a number of nice properties for \$iI\!\Sigma_1\$ , e.g. existence of primitive recursive choice functions (this is established by different means also in [D94]). We then sharpen an unpublished theorem of Visser's to the effect that quantifier alternation alone is much less powerful intuitionistically than classically: \$iI\!\Sigma_1\$ together with induction over arbitrary prenex formulas is \$\Pi_2\$ -conservative over \$iI\!\Pi_2\$ . In the second part of the article we study the relation of \$iI\!\Sigma_1\$ to \$iI\!\Pi_1\$ (in the usual arithmetical language). The situation here is markedly different from the classical case in that \$iI\!\Pi_1\$ and \$iI\!\Sigma_1\$ are mutually incomparable, while \$iI\!\Sigma_1\$ is significantly stronger than \$iI\!\Pi_1\$ as far as provably recursive functions are concerned: All primitive recursive functions can be proved total in \$iI\!\Sigma_1\$ whereas the provably recursive functions of \$iI\!\Pi_1\$ are all majorized by polynomials over \${\Bbb N}\$ . 0 \$iI\!\Pi_1\$ is unusual also in that it lacks closure under Markov's Rule \$\mbox{MR}_{PR}\$
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2 ( #245,904 of 722,826 ) | {"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": 0.8225614428520203, "perplexity": 4546.740934859755}, "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-2013-48/segments/1386163052727/warc/CC-MAIN-20131204131732-00058-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/toe-e-n-f-1-n-3-c-2-n-3.33368/ | # TOE? -> E(n) = f^(1+(n-3)) * c^(2+(n-3))
1. Jun 30, 2004
### dperez3894
TOE? --> E(n) = f^(1+(n-3)) * c^(2+(n-3))
TOE --> E(n) = f^(1+(n-3)) * c^(2+(n-3))
Here's a possible TOE(Theory Of Everything). In order to show how the equation was derived, lets look at mass and how it relates to time in Einstein's Relativity equations;
Mass In Relativity
m = 1/(1-(v^2/c^2))^.5
Time in Relativity
t = (1-(v^2/c^2))^.5
Mass is a result of the inversion of Time at any speed below "c" if
(1-(v^2/c^2))^.5
in the equation;
m = 1/(1-(v^2/c^2))^.5
is replaced with "t"
Mass = 1/t
The answer for Mass derived from the Relativity equations is similar to the answer of another equation;
Frequency = 1/t
Therefore;
f = m = 1/t
Freqency = Mass
Which shows mass is a frequency; i.e. a vibration. The concept of wavelength is also connected to Frequency = Mass;
Frequency = Wavelength/Time
Frequency = Mass
Mass = Wavelength/Time
f = m = wavelength/time
Both of the Frequency and Wavelength connections to Mass reveals the superstring theory where all subatomic particles in the Universe are made up of superstrings vibrating at different frequencies with different wavelengths.
Now that we know Frequency = Mass, let's put it into E=mc^2;
E = fc^2
All forms of energy, including light energy are basically vibrations at different frequencies. Charles Hinton, as well as Bernard Reimann, said light is a vibration of an unseen 4th Dimension.
Since light photons are a form of electromagnetic energy, it's logical to conclude the entire electromagnetic spectrum from heat to gamma rays are vibrations in 3-Dimensional space caused by the 4th Dimension.
Since all subatomic particles are made up of superstrings of energy that vibrate at particular frequencies, the vibration of the superstrings would logically be vibrations in 3-Dimensional space caused by the 4th Dimension.
By logically extending this thought to other dimensions, it would mean that energy and matter in each dimension manifests from vibrations at particular frequencies which are caused by the next higher dimension.
If the equation;
E = fc^2
works for the 3rd Dimension, then we'd have to modify the equation so it works for all dimensions.
By placing (n-3) in the exponents of the variables and use (n) to represent a dimension, we could make an equation that could work in any dimension;
E(n) = f^(1+(n-3)) * c^(2+(n-3))
Dimension Zero
E0 = f^-2 * c^-1
1st Dimension
E1 = f^-1 * c^0
2nd Dimension
E2 = f^0 * c^1
3rd Dimension
E3 = f^1 * c^2
4th Dimension
E4 = f^2 * c^3
5th Dimension
E5 = f^3 * c^4
6th Dimension
E6 = f^4 * c^5
7th Dimension
E7 = f^5 * c^6
8th Dimension
E8 = f^6 * c^7
9th Dimension
E9 = f^7 * c^8
10th Dimension
E10 = f^8 * c^9
The equation;
E(n) = f^(1+(n-3)) * c^(1+(n-3))
Could be a possible candidate for TOE.
2. Jul 1, 2004
Staff Emeritus
Is not "time" but the conversion factor for time in one frame as a function of time in the other frame, where the two frames have a relative velocity of v. The conversion factor goes by the designation gamma ($$\gamma$$} and the square root function in its denominator is called beta ($$\beta$$). So it's not true that $$t = \gamma$$ and all your reasoning that follows that fails. | {"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": 0.8795446157455444, "perplexity": 2223.854193955643}, "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/1495463610342.84/warc/CC-MAIN-20170528162023-20170528182023-00271.warc.gz"} |
http://mathhelpforum.com/trigonometry/38427-solving-trigonometry-equation.html | # Math Help - Solving trigonometry equation
1. ## Solving trigonometry equation
Solve the equation cos2 x-sin2x=cosx 0≤x<2π
2. Hello,
Originally Posted by kelsey3
Solve the equation cos2 x-sin2x=cosx 0≤x<2π
Is it $\cos^2x$ or $\cos 2x$ ?
Is it $\sin^2x$ or $\sin 2x$ ?
3. Originally Posted by Moo
Hello,
Is it $\cos^2x$ or $\cos 2x$ ?
Is it $\sin^2x$ or $\sin 2x$ ?
I was thinking it was $cos2x$ and $sin2x$.... then i got confused with what i was doing....
4. Its squared
5. Originally Posted by kelsey3
Its squared
For the two ones ?
6. Originally Posted by kelsey3
Its squared
that makes a difference! thank you! i'll see if i can do anything with that.... i bet Moo can figure it out long before me though!
7. ya they're both squared
8. Originally Posted by kelsey3
Solve the equation cos2 x-sin2x=cosx 0≤x<2π
$\cos^2(x)-(1-\cos^2(x))=\cos(x)$
$2\cos(x)-\cos(x)-1=0 \iff (2\cos(x)+1)(\cos(x)-1)=0$
Now we just need to solve
$2\cos(x)+1=0$ and
$\cos(x)-1=0$
Good luck.
9. Ok.
$\cos^2x-\sin^2x=\cos(2x)$
~~~~~~~~
Why ?
We know that $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$\cos(a+a)=\cos(2a)=\cos^2(a)-\sin^2(a)$
~~~~~~~~
The equation to solve is now :
$\cos(2x)=\cos(x)$
Now, you should know that if $\cos(a)=\cos(b)$, then $a=b+2k \pi$ or $a=-b+2k \pi$
Try to find the solutions, it has nothing difficult
----------------------------------
Other method, maybe longer, but I prefer it :
Remember that $\sin^2 x+\cos^2x=1 \implies \sin^2x=1-\cos^2x$
The equation is now :
$\cos^2x-1+\cos^2x=\cos x$
$2\cos^2x-\cos x-1=0$
$(\cos x-1)(2 \cos x+1)=0$
-> $\cos x-1=0$
or $2 \cos x+1=0$
Edit : too slow
10. aw moo! you were too slow! I bet it's my fault! hehe....
Wow, i never would have gotten that.... But I get it now! Thank you Kelsey3 for askin'!
11. $
\cos^2{x}-\sin^2{x}=\cos{x}
$
$
\cos^2{x}+\sin^2{x}=1
$
So,
$
\sin^2{x}=1-\cos^2{x}
$
Substitute in the original equation:
$
\cos^2{x}-(1-\cos^2{x})=\cos{x}
$
$
\cos^2{x}-1+\cos^2{x}=\cos{x}
$
$
2\cos^2{x}-1=\cos{x}
$
$
2\cos^2{x}-\cos{x}-1=0
$
$
(2\cos{x}+1)(\cos{x}-1)=0
$
$
\cos{x}=-0.5,\;1
$
First case:
$
\cos{x}=-0.5
$
$
x=\pm\frac{2\pi}{3}+2n\pi
$
$
x=\frac{2\pi}{3},\;\frac{4\pi}{3}
$
Second case:
$
\cos{x}=1
$
$
x=\frac{\pi}{2}+2n\pi
$
$
x=\frac{\pi}{2}
$
Final solution:
$
x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac {\pi}{2}
$ | {"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": 43, "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": 0.945955216884613, "perplexity": 3897.9190006940244}, "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-2015-18/segments/1430453759945.83/warc/CC-MAIN-20150501041559-00068-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/483136/real-and-rational-numbers | # Real and Rational Numbers
Intuitively, we often think of real numbers as existing in one-to-one correspondence with the points on a continuously drawn line, the real number line. One way of expressing the completeness of the real numbers is to say that the real line has no holes.
That is, the values that are “missing” from the rationals—such as, for example, 2 —are present in the real numbers. What do you see as the possible limitations of using this intuitive idea to prove the existence of certain limits in the real numbers? Do you think it is sufficient for the purposes of most students who study calculus to simply accept the existence of these limits without proof?
-
For a first calculus course that is taken by "everybody," certainly. Though the students need to be exposed to enough nasty examples that they will know that blind manipulation can lead to incorrect conclusions. – André Nicolas Sep 3 '13 at 17:01
Did you mean to write $\sqrt 2$ instead of $2$ for your example? – Omnomnomnom Sep 3 '13 at 17:03
Well, in higher up math, you don't "prove" these exist, you construct them to exist. So it's not clear how you can prove to a calculus student that they exist. Instead, for calculus, you assume it as an axiom of the real numbers. – Thomas Andrews Sep 3 '13 at 17:06 | {"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": 0.9032726287841797, "perplexity": 311.72987797045613}, "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/1422122691893.34/warc/CC-MAIN-20150124180451-00052-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://iacr.org/cryptodb/data/author.php?authorkey=12258 | ## CryptoDB
### Stella Wohnig
#### Publications
Year
Venue
Title
2022
ASIACRYPT
Ring signatures allow a user to sign messages on behalf of an \emph{ad hoc} set of users - a ring - while hiding her identity. The original motivation for ring signatures was whistleblowing [Rivest et al. ASIACRYPT'01]: a high government employee can anonymously leak sensitive information while certifying that it comes from a reliable source, namely by signing the leak. However, essentially all known ring signature schemes require the members of the ring to publish a structured verification key that is compatible with the scheme. This creates somewhat of a paradox since, if a user does not want to be framed for whistleblowing, they will stay clear of signature schemes that support ring signatures. In this work, we formalize the concept of universal ring signatures (URS). A URS enables a user to issue a ring signature with respect to a ring of users, independently of the signature schemes they are using. In particular, none of the verification keys in the ring need to come from the same scheme. Thus, in principle, URS presents an effective solution for whistleblowing. The main goal of this work is to study the feasibility of URS, especially in the standard model (i.e. no random oracles or common reference strings). We present several constructions of URS, offering different trade-offs between assumptions required, the level of security achieved, and the size of signatures: \begin{itemize} \item Our first construction is based on superpolynomial hardness assumptions of standard primitives. It achieves compact signatures. That means the size of a signature depends only logarithmically on the size of the ring and on the number of signature schemes involved. \item We then proceed to study the feasibility of constructing URS from standard polynomially-hard assumptions only. We construct a non-compact URS from witness encryption and additional standard assumptions. \item Finally, we show how to modify the non-compact construction into a compact one by relying on indistinguishability obfuscation. \end{itemize}
#### Coauthors
Pedro Branco (1)
Nico Döttling (1) | {"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": 0.8239393830299377, "perplexity": 1191.2910489098015}, "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-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00328.warc.gz"} |
http://mathhelpforum.com/calculus/153113-area-triangle-print.html | # Area of Triangle
• Aug 8th 2010, 08:48 PM
ilovemymath
Area of Triangle
The tangent and normal to the curve $y=4(sqroot(x+2))$ at the point A(7,12) cut the x-axis at points B and C respectively. Calculate the area of the triangle ABC.
• Aug 8th 2010, 09:01 PM
eumyang
$y = 4\sqrt{x + 2}$
$y' = \dfrac{2}{\sqrt{x +2}}$
At point A(7, 12), the slope of the tangent line is $m = \dfrac{2}{\sqrt{7 +2}} = \dfrac{2}{3}$. The slope of the normal line would be $\dfrac{-1}{m} = -\dfrac{3}{2}$.
The equation of the tangent line would be
$y = m(x - x_1) + y_1$
Plug in m, x1, y1, and also y = 0 to solve for x to get your point B:
$0 = \frac{2}{3}(x - 7) + 12$
...
The equation of the normal line would be
$y = \frac{-1}{m}(x - x_1) + y_1$
Plug in -1/m, x1, y1, and also y = 0 to solve for x to get your point C:
$0 = -\frac{3}{2}(x - 7) + 12$
...
ABC would be a right triangle so let AB be your base and AC be your height (it doesn't matter which is which). Find the distances AB and AC using the distance formula, then use the area of the triangle formula $A = \frac{1}{2}bh$ to find the area. | {"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": 10, "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": 0.8663588762283325, "perplexity": 385.5072317540502}, "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/1512948514051.18/warc/CC-MAIN-20171211203107-20171211223107-00074.warc.gz"} |
http://cms.math.ca/cmb/kw/automorphisms | location: Publications → journals
Search results
Search: All articles in the CMB digital archive with keyword automorphisms
Expand all Collapse all Results 1 - 4 of 4
1. CMB 2010 (vol 54 pp. 141)
Kim, Sang Og; Park, Choonkil
Linear Maps on $C^*$-Algebras Preserving the Set of Operators that are Invertible in $\mathcal{A}/\mathcal{I}$ For $C^*$-algebras $\mathcal{A}$ of real rank zero, we describe linear maps $\phi$ on $\mathcal{A}$ that are surjective up to ideals $\mathcal{I}$, and $\pi(A)$ is invertible in $\mathcal{A}/\mathcal{I}$ if and only if $\pi(\phi(A))$ is invertible in $\mathcal{A}/\mathcal{I}$, where $A\in\mathcal{A}$ and $\pi:\mathcal{A}\to\mathcal{A}/\mathcal{I}$ is the quotient map. We also consider similar linear maps preserving zero products on the Calkin algebra. Keywords:preservers, Jordan automorphisms, invertible operators, zero productsCategories:47B48, 47A10, 46H10
2. CMB 2009 (vol 52 pp. 535)
Daigle, Daniel; Kaliman, Shulim
A Note on Locally Nilpotent Derivations\\ and Variables of $k[X,Y,Z]$ We strengthen certain results concerning actions of $(\Comp,+)$ on $\Comp^{3}$ and embeddings of $\Comp^{2}$ in $\Comp^{3}$, and show that these results are in fact valid over any field of characteristic zero. Keywords:locally nilpotent derivations, group actions, polynomial automorphisms, variable, affine spaceCategories:14R10, 14R20, 14R25, 13N15
3. CMB 2008 (vol 51 pp. 481)
Bayart, Frédéric
Universal Inner Functions on the Ball It is shown that given any sequence of automorphisms $(\phi_k)_k$ of the unit ball $\bn$ of $\cn$ such that $\|\phi_k(0)\|$ tends to $1$, there exists an inner function $I$ such that the family of non-Euclidean translates" $(I\circ\phi_k)_k$ is locally uniformly dense in the unit ball of $H^\infty(\bn)$. Keywords:inner functions, automorphisms of the ball, universalityCategories:32A35, 30D50, 47B38
4. CMB 2008 (vol 51 pp. 261)
Neeb, Karl-Hermann
On the Classification of Rational Quantum Tori and the Structure of Their Automorphism Groups An $n$-dimensional quantum torus is a twisted group algebra of the group $\Z^n$. It is called rational if all invertible commutators are roots of unity. In the present note we describe a normal form for rational $n$-dimensional quantum tori over any field. Moreover, we show that for $n = 2$ the natural exact sequence describing the automorphism group of the quantum torus splits over any field. Keywords:quantum torus, normal form, automorphisms of quantum toriCategory:16S35 | {"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": 0.9604371786117554, "perplexity": 673.9680531138669}, "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/1406510272940.33/warc/CC-MAIN-20140728011752-00100-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/93720/lhospitals-rule-for-infty-infty | # L'Hospital's Rule for $\infty/\infty$
Arthur Mattuck in his Introduction to Analysis book, pg. 220 says, in order to prove L'Hospital's Rule for $\infty/\infty$ case,
Let $L=\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ and choose $a$ so that
$\frac{f'(x)}{g'(x)} \stackrel{\approx}{\epsilon} L$ for $x>a$.
And prove two approximations below (valid for $x\gg 1$)
$$\frac{f(x)}{g(x)} \stackrel{\approx}{\epsilon} \frac{f(x)-f(a)}{g(x)-g(a)} \stackrel{\approx}{\epsilon} L$$
His hint is, for the first approximation, that we write
$$f(x) - f(a) = f(x) [ 1 - f(a)/f(x)]$$
and use limit theorem, for the second, use the Cauchy Mean-value Theorem.
Anyone know how to pursue this proof? Thanks,
-
Could you please explain the notation $\stackrel{\approx}{\epsilon}$? – Brad Dec 23 '11 at 17:27
My guess is that it means $|f'(x)/g(x) - L| < \epsilon$. I agree that it isn't very standard and should be explained. – Dylan Moreland Dec 23 '11 at 17:29
To the OP: Do you mean to write $f'(x)/g'(x)$? – Dylan Moreland Dec 23 '11 at 17:29
Brad, if $a$ and $b$ are within $\epsilon$ of eachother, then Mattuck uses $a \stackrel{\approx}{\epsilon} b$. – BB_ML Dec 23 '11 at 18:08
Dylan, correction, yes the first ratio is between derivatives. – BB_ML Dec 23 '11 at 18:19
The first approximation should be clear since we can, for any $\epsilon$, choose $x_0$ sufficiently large such that $$1 - \epsilon < \frac{1 - f(a)/f(x)}{1-g(a)/g(x)} < 1 + \epsilon$$ whenever $x > x_0$.
For the second approximation: by the Cauchy MVT, there exists $c, a < c < x$ such that $(f(x) - f(a))g'(c) = (g(x)-g(a))f'(c)$, that is: $$\frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c)}{g'(c)}$$ which is $\epsilon$-close to $L$. | {"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": 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": 0.9821859002113342, "perplexity": 440.1705846515695}, "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/1422122152935.2/warc/CC-MAIN-20150124175552-00018-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/400169/why-is-the-dimension-of-sl2-mathbbh-equal-to-15 | # Why is the dimension of $SL(2,\mathbb{H})$ equal to $15$?
Let me ask a very basic question which is inspired by reading M. Atiyah's "Geometry and physics of knots". Could you explain me (or give a reference to) the definition of the special linear group $SL(2,\mathbb{H})$?
What I don't understand is how to compute the determinant of a matrix with noncommutative entries. Further, even if there was some way to define it, the naive counting gives $12$ for the dimension of $SL(2,\mathbb{H})$, whereas it is clear from the context that it should be equal to $15$.
To solve the contradiction I thought that, probably, $SL(2,\mathbb{H})$ is not defined as a matrix group with determinant equal to $1$, but rather as a group of fractional linear transformations of $\mathbb{HP}^1$. Could someone please confirm if this is correct?
-
There is a notion of a determinant for matrices with coefficients in a noncommutative ring due to Dieudonné. Also the quaternions can be seen as a subring of $M_{2\times 2}(\Bbb C)$, so that $M_{2\times 2}(\Bbb H)\subset M_{4\times 4}(\Bbb C)$ as a subring. That gives you at least one possible definition of a $\Bbb C$-valued determinant. You could search "quaternionic determinant" on google, and "Dieudonné determinant". As for dimension, I would expect that the above determinant is a submersion onto $\Bbb C^*$ so that $SL(2,\Bbb H)$ should have complex dimension 15. – Olivier Bégassat May 23 '13 at 17:38
The dimesion statement in my comment is false. – Olivier Bégassat May 23 '13 at 17:45
1. Lie algebra $sl_2(\mathbb H)$ is easier to define: it's the Lie algebra generated by traceless matrices. In non-commutative case the space of traceless matrices is not closed under Lie bracket — that's why dim>12. In fact, $sl_2(\mathbb H)$ is the Lie algebra of matrices with pure imaginary trace.
2. $SL(2,\mathbb R)\cong Spin(2,1)$, $SL(2,\mathbb C)\cong Spin(3,1)$. Now $Spin(5,1)$ is connected and 1-connected, has Lie algebra $sl_2(\mathbb H)$ (and acts by conformal transformations on $\mathbb HP^1$, btw) — so $SL(2,\mathbb H)\cong Spin(5,1)$ (and is 15-dimensional). (All this is explained somewhere in Baez's Octonions, I believe.)
3. It seems (after googling) that $SL_2(\mathbb H)$ can also be defined as the commutator subgroup of $GL_2(\mathbb H)$ and this is exactly the subgroup of matrices with Dieudonné determinant equal to 1. | {"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": 0.9711840748786926, "perplexity": 212.99668014427067}, "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-2015-40/segments/1443736678818.13/warc/CC-MAIN-20151001215758-00068-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://www.math.ac.cn/xshd/xsbg/201804/t20180416_403539.html | Endoscopy and cohomology of U(n-1,1)
2018.05.09（星期三），16:30-17:30
点：数学院南楼N913
要：We apply the endoscopic classification of automorphic representations for inner forms of unitary groups to bound the growth of cohomology in congruence towers of locally symmetric spaces associated with U(n-1,1). Our bound is sharper than the bound predicted by Sarnak-Xue for general locally symmetric spaces. This is joint work with Simon Marshall. | {"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": 0.9840741157531738, "perplexity": 722.1584485325147}, "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/1531676593223.90/warc/CC-MAIN-20180722120017-20180722140017-00607.warc.gz"} |
https://tex.stackexchange.com/questions/229543/double-tilde-symbol-under-letter?noredirect=1 | # double tilde symbol under letter
Currently, I'm representing second and fourth order tensor with this (taken from someone, and i don't really get how it works):
% tensor 2:
\newcommand{\tend}[1]{\oalign{\mbox{\boldmath$#1$}\crcr\hidewidth$\scriptscriptstyle\sim$\hidewidth}}
%tensor 4:
\newcommand{\tenq}[1]{\tend{\tend #1}\vphantom{#1}}
Which looks like this:
I very much don't like the result: boldface is strange, with scales badly, there is too much spacing for the fourth order tensor... I found the untertilde package, with looks more robust for second order, but i cannot see how to make 4th order from it. I also found this topic, bold widetilde, which doe not adress symbol stacking for 4th order.
Is there a simple elegant way to do? My main concern here is consistant boldface for the tilde and the letter, and proper vertical spacing of the 4th order.
Thanks
• Can you clarify what you want: everything bold, nothing bold,...? \oalign is a plain TeX command for building minature tables, used for stacking symbols. \crcr indicates a newline. \hidewdith...\hidewidth disguises the width of the intervening material. – Andrew Swann Feb 22 '15 at 15:48
Here I use stacks to recursively place \sim under the argument, based on the value of the optional argument. In the definition, the [1pt] is the under-gap from the argument and the [0pt] is the vertical separation between \sim characters in a multi-stack. These values can be altered to suit (including being made negative).
I have not used any bold font for the tensor itself, though it could be added, if desired, to the definition, or at time of invocation.
The specification of \def\useanchorwidth{T} says to ignore the width of the \sim underset in setting the horizontal spacing. The only time that could be an issue is if you, for example, used adjacent invocations on narrow arguments, e.g., \tenq[2]{i}\tenq[3]{j}. While the \useanchorwidth line could be removed, in which case \tenq with narrow arguments would always take up at least the width of a \scriptscriptstyle\sim, my first step would instead be to manually add \, space on those very rare occasions when needed.
\documentclass{article}
\usepackage{stackengine}
\stackMath
\newcommand\tenq[2][1]{%
\def\useanchorwidth{T}%
\ifnum#1>1%
\stackunder[0pt]{\tenq[\numexpr#1-1\relax]{#2}}{\scriptscriptstyle\sim}%
\else%
\stackunder[1pt]{#2}{\scriptscriptstyle\sim}%
\fi%
}
\begin{document}
$\tenq{\sigma}\neq\tenq[2]{\Lambda}\neq\tenq[3]{\Delta}\neq\tenq[4]{\psi}$
\end{document}
If you like the first one, you can use the same definition for \tenq but using \approx instead of \sim.
Also use \bm from the same package instead of \boldmath.
MWE:
\documentclass{article}
\usepackage{amsmath,bm}
% tensor 2:
\newcommand{\tend}[1]{\hbox{\oalign{$\bm{#1}$\crcr\hidewidth$\scriptscriptstyle\bm{\sim}$\hidewidth}}}
%tensor 4:
\newcommand{\tenq}[1]{\hbox{\oalign{$\bm{#1}$\crcr\hidewidth$\scriptscriptstyle\bm{\approx}$\hidewidth}}}
\begin{document}
$\tend{\sigma}\neq\tenq{\Lambda}$
\end{document}
• I'd put \m@th into the $...$ to remove any non-zero \mathsurround. – Henri Menke Feb 22 '15 at 20:10
• thanks for this answer! Yours and Steven's are very much ok, but I'll accept Steven's for consistancy of the tilde notation and modularity. – Napseis Feb 23 '15 at 9:04
Just for fun.
\documentclass{standalone}
\usepackage{amsmath,bm}
\newlength{\fillwidth}
% creates a tilde which slightly overlaps what's above and below, centered in a space \fillwidth wide
\newcommand{\flatsim}{\hbox to \fillwidth{\hfil\raisebox{0pt}[.05ex][.05ex]{$\scriptscriptstyle\bm{\sim}$}\hfil}}
\newcommand{\triple}[1]{\settowidth{\fillwidth}{$\bm{#1}$}%
\vtop{\baselineskip=0pt\hbox{$\bm{#1}$}\hbox{\rule{0pt}{.2ex}}\flatsim\flatsim\flatsim}}
\begin{document}
\triple{\sigma}
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https://statalasso.github.io/docs/regularized_reg/ | # Regularized regression
lasso2 solves the elastic net problem
where
• $(y_i - x_i'\beta)^2$ is the residual sum of squares (RSS),
• $\beta$ is a $p$-dimensional parameter vector,
• $\lambda$ is the overall penalty level, which controls the general degree of penalization,
• $\alpha$ is the elastic net parameter, which determines the relative contribution of $\ell_1$ (lasso-type) to $\ell_2$ (ridge-type) penalization. $\alpha=1$ corresponds to the lasso; $\alpha=0$ is ridge regression.
• $\Psi$ is a $p$ by $p$ diagonal matrix of predictor-specific penalty loadings.
• $N$ is the number of observations
In addition, lasso2 estimates the square-root lasso (sqrt-lasso) estimator, which is defined as the solution to the following objective function:
lasso2 implements the elastic net and sqrt-lasso using coordinate descent algorithms. The algorithm (then referred to as “shooting”) was first proposed by Fu (1998) for the lasso, and by Van der Kooij (2007) for the elastic net. Belloni et al. (2011) implement the coordinate descent for the sqrt-lasso, and have kindly provided Matlab code.
Penalized regression methods, such as the elastic net and the sqrt-lasso, rely on tuning parameters that control the degree and type of penalization. The estimation methods implemented in lasso2 use two tuning parameters: $\lambda$ and $\alpha$.
## How to select the tuning parameters
lassopack offers three approaches for selecting the “optimal” $\lambda$ and $\alpha$ value, which are implemented in lasso2, cvlasso and rlasso, respectively.
1. Cross-validation: The penalty level $\lambda$ may be chosen by cross-validation in order to optimize out-of-sample prediction performance. $K$-fold cross-validation and rolling cross-validation (for panel and time-series data) are implemented in cvlasso. cvlasso also supports cross-validation across $\alpha$.
2. Theory-driven: Theoretically justified and feasible penalty levels and loadings are available for the lasso and sqrt-lasso via the separate command rlasso. The penalization is chosen to dominate the noise of the data-generating process (represented by the score vector), which allows derivation of theoretical results with regard to consistent prediction and parameter estimation. Since the error variance is in practice unknown, Belloni et al. (2012) introduce the rigorous (or feasible) lasso that relies on an iterative algorithm for estimating the optimal penalization and is valid in the presence of non-Gaussian and heteroskedastic errors. Belloni et al. (2016) extend the framework to the panel data setting. In the case of the sqrt-lasso under homoskedasticity, the optimal penalty level is independent of the unknown error variance, leading to a practical advantage and better performance in finite samples (see Belloni et al., 2011, 2014).
3. Information criteria: $\lambda$ can also be selected using information criteria. lasso2 calculates four information criteria: Akaike Information Criterion (AIC; Akaike, 1974), Bayesian Information Criterion (BIC; Schwarz, 1978), Extended Bayesian information criterion (EBIC; Chen & Chen, 2008) and the corrected AIC (AICc; Sugiura, 1978, and Hurvich, 1989). | {"extraction_info": {"found_math": true, "script_math_tex": 21, "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": 0.9211451411247253, "perplexity": 1647.7836405525215}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "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-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00451.warc.gz"} |
http://winnerscience.com/2012/01/09/difference-between-displacement-current-and-conduction-current/ | DIFFERENCE BETWEEN DISPLACEMENT CURRENT AND CONDUCTION CURRENT
CONCEPT OF DISPLACEMENT CURRENT (DIFFERENCE BETWEEN DISPLACEMENT CURRENT AND CONDUCTION CURRENT)
Let there be a parallel R-C network with a voltage V as shown in fig .Let the current through resistor R is Ic and by Ohm’s law it is given by
Ic=V/R
And current through capacitor C is Id and is given by
Id=dQ/dt
Id=Cdv/dt ( dQ=Cdv)(1)
In practice ,the current does not flow through the capacitor . But ,the current that flows out of one electrode of capacitor equals the current that flows in to the other electrode. The net effect is as if there is a current flowing through the path containing the capacitor. But current, Icactually flows through the resistor.
Hence ,from the above result ,current flowing through the resistor is known as conduction current and it obeys Ohm’s law,while the current flowing through the capacitor is commonly known as Displacement current.
Mathematical Proof.As the electric field inside each element equals the voltage V across the element divided by its length d
That is E=V/d or V=ED (2)
Now the current density in resistor is given by
Jc=Ic/A=σE (3)
Where A= cross-sectional area
σ=conductivity of resistance element
Also capacitance of a parallel plate capacitor is given by
C=ε0A/d (4)
Now rewrite equation(1)
Id=C dV/dt
By substituting the values of V and C from equations (2 and 4) in above equation,we get
Id= ε 0A/d( E/t)
Id= ε0A E/t (5)
Therefore current density Jd inside capacitor is
Jd=Id/A
Substititing value of Id from equation (5) in above equation,we get
Jd= ε 0/A E/t
Or Jd= ε0 E/t (6a)
Or Jd= D/t (6b)
Where D= ε 0E=Electric displacement vector
And Jd=Displacement current density
Equation (6a) proves that displacement current density arises whenever there will be change in electric field E that is (E/t≠0)
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"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": 0.9491568207740784, "perplexity": 3052.8468518162294}, "config": {"markdown_headings": false, "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-2018-51/segments/1544376829568.86/warc/CC-MAIN-20181218184418-20181218210418-00404.warc.gz"} |
http://www.cfd-online.com/Forums/main/81190-boundary-conditions-using-simpler-algorithm-print.html | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- Main CFD Forum (http://www.cfd-online.com/Forums/main/)
- - Boundary conditions using SIMPLER algorithm (http://www.cfd-online.com/Forums/main/81190-boundary-conditions-using-simpler-algorithm.html)
lost.identity October 19, 2010 09:20
Boundary conditions using SIMPLER algorithm
I have seen a number of forum posts on this topic but I'm still struggling with my boundary conditions.
I'm running a 1-D spherical calculation (only radial dependence considered). I decided to have velocity boundary conditions at the inlet and outlet. So the BCs I have are:
inlet:
outlet: which is zero gradient BC.
I understand that for subsonic simulations you either have to specify velocity or pressure BCs. I have a staggered grid where the velocities are staggered forward.
My discretised equation for the pressure looks of the form
I used the discretised equation to calculate the pressure values at the boundary. At the inlet since I have therefore, I do not have in the discretised pressure equation and for the outlet I do not have . Is this correct?
johnhelt October 19, 2010 11:01
How can you have 0 velocity in and non-zero velocity out, unless of course you are generating matter inside your volume?
lost.identity October 20, 2010 05:54
I'm studying combustion where the zero-upstream velocity is a far-field velocity and there is a flame in the middle of the domain and the boundary condition at the outlet is a zero-gradient boundary condition but the flow reaches about 60 m/s at this outlet.
I was at first baffled by your question but I think the combustion, creates density difference which creates mass.
johnhelt October 20, 2010 07:54
Could you maybe sketch your geometry?
I understood it as:
Flame (r=0)-----inlet (r=r_i)-----------outlet (r=r_o)----> radial coordinate
Please bear with me if that is not how it looked :-)
lost.identity October 20, 2010 08:46
It's actually,
Outlet (r=0) ----------------------------------------Flame------------------Inet (r=R)
I use a normalised temperature to define the flame location. I study spherical explosions where the flame starts at the centre and moves radially outward. Therefore at the centre (r=0) is the highest temperature (burnt side) where the normalised temperature C=1. I end simulations before the flame reaches R (end of the domain). Since there is no flame here, the normalised temperature C=0 at the outlet. So the profile of normalised temperature (C) looks like a backward ramp function.
So the flame actually moves from left to right during my calculations. However, relative to the flame is the flow velocity, so for a particle fixed on the flame the flow direction will be from right-to-left. That's why I defined my inlet to be at r=R, and outlet to be at r=0.
johnhelt October 20, 2010 09:36
Ok, I think I get it finally. These are suggestions, I am by no means a CFD expert ...
With regards to the pressure: for symmetry you should have a zero gradient at r=0, so then P(N)=P(N-1), where N is the outlet node and N-1 is the node right next to it (here to the EAST). You could do that by setting , , and in the pressure-equation written above.
At the inlet boundary (r=R), I would specify that the pressure on the EAST point would be equal to the ambient pressure P0. You could do by lumping into the source term (b). In b, I think the velocities at the faces are also given, so there you would have to specify that the velocity at east face is 0.
Let me hear what you think.
Regards
lost.identity October 20, 2010 13:00
Hi thanks very much for the quick reply.
I'm still confused. This is a lengthy reply but I'm trying to understand certain things between what I have now and what you've just told me.
So I'm using SIMPLER as I said before and I use a staggered grid.
What I have now is a zero-gradient BC at the outlet (r=0) for velocity which says . Am I correct in thinking that if I have this velocity BC then I can't have a pressure BC at the same location for incompressible flow?
At this boundary I discretise the pressure equation to obtain (just neglecting a_w term):
and according to SIMPLER method the mass source term b looks like
------------- (1)
where r_P, r_w and r_e are the radii at the relevant points along the grid (these terms only come about because of the spherical discretisaiton). Also my grid is staggered and I store the velocities at the velocity-cell faces. So with my current boundary conditions I set , where the pseduo-velocities are only calculated for the internal nodes. The problem with this is that the term doesn't go to zero at this location for spherical coordinates. This is ecause even though the velocities and are the same, the change in radii and creates a source. So even if I have neglected the in the discretised equation for this boundary . Therefore, this violates symmetry. So am I right in thinking that rather than having zero-gradient with respect to velocity I should have zero-gradient with respect to pressure?
With regards to the inlet (r=R) BC, I have the discretised equation
by just neglecting a_E. A problem I have now is that I don't know what the cell face velocities should be at this location for a staggered grid to evaluate according to Eq.(1). At the end I do not have since it's that point is outside the grid, but I set . However, my boundary condition for . So despite the fact that there is a radius change the mass term b goes to zero here since the velocities are zero anyway. I'm really not sure about this bit.
lost.identity October 20, 2010 16:10
Actually I realise that I don't include the centre point of the sphere, r=0 because it creates a singularity. Instead I start it at r=a, some finite value. So I'm not sure if I need to have symmetry BCs in this case.
johnhelt October 21, 2010 10:18
If your centre point for the pressure discretization is at 0, then wouldn't the control volume at that point would be a sphere, rather than a spherical shell? Then you only have flow out of there, right?. I think this should change your discretized equations there.
With regards to the inlet, you could move the grid for the pressure so it does not end at r=R, but at . Then as you are staggering the velocity grid forward your last velocity node will be at r=R, right? In 'b' in the pressure-equation you could set u_e = 0, while keeping u_w from your solution to the momentum equation. Also you set a_E = 0 here, as you already wrote.
Again, I'm by no means sure of this.. I don't remember seeing an example in the textbooks I've been reading except for Patankar, regarding discretization in spherical coordinates and that wasn't for the NS equations).
Hopefully someone more qualified will answer you soon :confused:
ahazbavi October 25, 2010 18:27
Hello, I am a mechanical engineering.I need to use a subroutine ables to solve (in unsteady condition) 3-D fluid flow (the pressure and velocity and temperature) fields in cylindrical coordinate .Who Can send me the source code for SIMPLE / SIMPLER scheme (preferably on fortran) so that I can get & proceed from it.
please send me through my email, [email protected].
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https://complementarytraining.net/optimal-force-velocity-profile-for-sprinting-is-it-all-bollocks-part-6/ | Optimal Force-Velocity Profile for Sprinting: Is It All Bollocks? – Part 6
Optimal Force-Velocity Profile for Sprinting: Is It All Bollocks? – Part 6
Previous Part:
In this sixth and final installment of this article series, I will explore whether the “causal” interpretation of the Force-Velocity Profiling (FVP; $F_0$, $V_0$ or their other combination variant $P_{max}$ and $S_{fv}$) “holds water” using N=1 testing. Ideally, this should be done with more athlete, which we might actually do in the near future, but for now a simple common sense and one athlete sprinting is more than enough to simulate the debate as well as to question FVP parameters as determinants of performance. You can read my initial beef with this model on Twitter here and here.
Purpose of the FVP (from the realist perspective) is to uncover mechanical determinants of sprinting, and further to consider them as causal traits of the individual responsible for sprint performance (Figure 1).
Figure 1: Realist viewpoint of assuming $F_0$ and $V_0$ to be causal determinants of performance and represent individual traits/qualities
This realist viewpoint sees FVP ($F_0$, $V_0$ and particularly their other combination variant $P_{max}$ and $S_{fv}$) as causal, responsible for sprint performance given the environmental conditions and constraints (e.g., wind speed and temperature, additional load or resistance and so forth).
We have already explored sensitivity of this model to changes in athlete body weight in terms of its effect of sprint performance (summarized by MSS and MAC parameters), assuming $F_0$ and $V_0$ stay the same. Here we will expand this theoretical model simulation by estimating changes in sprint performance under different types of environmental conditions.
These conditions involve (1) extra horizontal resistance, in Newtons (force), but for the sake of simplicity expressed as kilograms (i.e., $100\;N = \frac{100}{9.81} = 10.19\;kg$), (2) additional inertia in a form of a weight vest, which only have vertical resistance but increase the athlete mass, and (3) a sled which is a combination of the two: providing extra inertia and horizontal resistance. For a sled, horizontal resistance (i.e., force) is assumed to be constant and equal to 0.4 of the sled weight (e.g., if the sled weights 20kg, then the horizontal resistance is equal to $20 \times 9.81 \times 0.4 = 78.5 \; N$).
If we assume that the $F_0$ and $V_0$, which are estimated during the normal sprinting, stay the same across conditions (and according to the realist model from Figure 1 they should since they represent individual traits or mechanical determinants), then we can estimate the sprint performance parameters changes. Here we assume we have athlete that weights 85 kg, height of 185 cm, with MSS and MAC equal to 9 $m^{-1}$ and 9 $m^{-2}$ respectively. Estimated $F_0$ and $V_0$ during normal sprint are equal to 680 N and 9.34 $m^{-1}$ respectively. These are considered to be mechanical determinants of this individual.
Then we increase the load used for resistance, vest, and sled conditions, from 0 to 25kg and estimate sprint performance parameters. Since I am bad at math, I will use optimization to do so.
Results are shown in Figure 2.
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regular price $350 NOW$245/ first year | {"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": 0.8148535490036011, "perplexity": 1707.8230511208894}, "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/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00310.warc.gz"} |
http://scholar.cu.edu.eg/?q=hassanibrahim/publications | ## Publications
Export 50 results:
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2017
Ibrahim, H., and others, Phys. Rev., vol. C95, no. 3, pp. 035209, 2017. Abstract
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2016
Ibrahim, H., and others, Phys. Rev., vol. D94, no. 5, pp. 052003, 2016. Abstract
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2015
Ibrahim, H., and others, Phys. Rev., vol. C92, no. 5, pp. 055202, 2015. Abstract
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Ibrahim, H., and others, Phys. Rev., vol. C92, no. 1, pp. 015208, 2015. Abstract
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Ibrahim, H., and others, Phys. Lett., vol. B744, pp. 309-314, 2015. Abstract
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Ibrahim, H., and others, Phys. Rev., vol. C91, pp. 034308, 2015. Abstract
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2014
Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 113, no. 2, pp. 022002, 2014. Abstract
Double-spin asymmetries and absolute cross sections were measured at large Bjorken $x$ (0.25 $\le x \le$ 0.90), in both the deep-inelastic and resonance regions, by scattering longitudinally polarized electrons at beam energies of 4.7 and 5.9 GeV from a transversely and longitudinally polarized $^3$He target. In this dedicated experiment, the spin structure function $g_2$ on $^3$He was determined with precision at large $x$, and the neutron twist-three matrix element $d_2^n$ was measured at $\left< Q^2\right>$ of 3.21 and 4.32 GeV$^2$/$c^2$, with an absolute precision of about $10^{-5}$. Our results are found to be in agreement with lattice QCD calculations and resolve the disagreement found with previous data at $\left< Q^2\right> =$ 5 GeV$^2$/$c^2$. Combining $d_2^n$ and a newly extracted twist-four matrix element, $f_2^n$, the average neutron color electric and magnetic forces were extracted and found to be of opposite sign and about 30 MeV/fm in magnitude.
Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 113, no. 23, pp. 232505, 2014. Abstract
We present a precise measurement of double-polarization asymmetries in the He→3(e→,e′d) reaction. This particular process is a uniquely sensitive probe of hadron dynamics in He3 and the structure of the underlying electromagnetic currents. The measurements have been performed in and around quasielastic kinematics at Q2=0.25(GeV/c)2 for missing momenta up to 270 MeV/c. The asymmetries are in fair agreement with the state-of-the-art calculations in terms of their functional dependencies on pm and ω, but are systematically offset. Beyond the region of the quasielastic peak, the discrepancies become even more pronounced. Thus, our measurements have been able to reveal deficiencies in the most sophisticated calculations of the three-body nuclear system, and indicate that further refinement in the treatment of their two-and/or three-body dynamics is required.
Collaboration, J. L. H. A., Phys.Rev., vol. C90, no. 5, pp. 055209, 2014. Abstract
An experiment to measure single-spin asymmetries of semi-inclusive production of charged pions in deep-inelastic scattering on a transversely polarized He3 target was performed at Jefferson Laboratory in the kinematic region of 0.16<x<0.35 and 1.4<Q2<2.7GeV2. Pretzelosity asymmetries on He3, which are expressed as the convolution of the h1T⊥ transverse-momentum-dependent distribution functions and the Collins fragmentation functions in the leading order, were measured for the first time. Under the effective polarization approximation, we extracted the corresponding neutron asymmetries from the measured He3 asymmetries and cross-section ratios between the proton and He3. Our results show that both π± on He3 and on neutron pretzelosity asymmetries are consistent with zero within experimental uncertainties.
Collaboration, J. L. H. A., J.Phys., vol. G41, pp. 105109, 2014. Abstract
The five-fold differential cross section for the (12)C(e,e′p)11B reaction was determined over a missing momentum range of 200–400 MeVc−1, in a kinematics regime with xB>1 and Q2=2.0 (GeVc−1)2. A comparison of the results with previous lower missing momentum data and with theoretical models are presented. The extracted distorted momentum distribution is shown to be consistent with previous data and extends the range of available data up to 400 MeVc−1. The theoretical calculations are from two very different approaches, one mean field and the other short range correlated, yet for this system the two approaches show striking agreement with the data and each other up to a missing momentum value of 325 MeVc−1. For larger momenta, the calculations diverge which is likely due to the factorization approximation used in the short range approach.
Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 113, no. 2, pp. 022502, 2014. Abstract
We report the first measurement of the target-normal single-spin asymmetry in deep-inelastic scattering from the inclusive reaction $^3$He$^{\uparrow}\left(e,e' \right)X$ on a polarized $^3$He gas target. Assuming time-reversal invariance, this asymmetry is strictly zero in the Born approximation but can be non-zero if two-photon-exchange contributions are included. The experiment, conducted at Jefferson Lab using a 5.89 GeV electron beam, covers a range of $1.7 < W < 2.9$ GeV, $1.02$ GeV, which is non-zero at the $2.89\sigma$ level. Our measured asymmetry agrees both in sign and magnitude with a two-photon-exchange model prediction that uses input from the Sivers transverse momentum distribution obtained from semi-inclusive deep-inelastic scattering.
Collaboration, J. L. H. A., Phys.Rev., vol. C90, no. 5, pp. 055201, 2014. Abstract
We report the first measurement of target single spin asymmetries of charged kaons produced in semi-inclusive deep inelastic scattering of electrons off a transversely polarized $^3{\rm{He}}$ target. Both the Collins and Sivers moments, which are related to the nucleon transversity and Sivers distributions, respectively, are extracted over the kinematic range of 0.1$<$$x_{bj}$$<$0.4 for $K^{+}$ and $K^{-}$ production. While the Collins and Sivers moments for $K^{+}$ are consistent with zero within the experimental uncertainties, both moments for $K^{-}$ favor negative values. The Sivers moments are compared to the theoretical prediction from a phenomenological fit to the world data. While the $K^{+}$ Sivers moments are consistent with the prediction, the $K^{-}$ results differ from the prediction at the 2-sigma level.
Collaboration, J. L. H. A., Phys.Rev., vol. C89, no. 4, pp. 042201, 2014. Abstract
We report the first measurement of target single-spin asymmetries (A$_N$) in the inclusive hadron production reaction, $e~$+$~^3\text{He}^{\uparrow}\rightarrow h+X$, using a transversely polarized $^3$He target. The experiment was conducted at Jefferson Lab in Hall A using a 5.9-GeV electron beam. Three types of hadrons ($\pi^{\pm}$, $\text{K}^{\pm}$ and proton) were detected in the transverse hadron momentum range 0.54 \$
Allada, K., and others, Phys. Rev., vol. C89, no. 4, pp. 042201, 2014. Abstract
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2012
Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 108, pp. 052001, 2012. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 109, pp. 192501, 2012. Abstract
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Abrahamyan, S., and others, Phys. Rev. Lett., vol. 109, pp. 192501, 2012. Abstract
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2011
Collaboration, J. L. H. A., Phys.Rev., vol. C83, pp. 025201, 2011. Abstract
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Collaboration, J. L. H. A., Phys.Rev., vol. C83, pp. 025201, 2011. Abstract
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Collaboration, J. L. H. A., Phys.Rev., vol. C84, pp. 055204, 2011. Abstract
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Collaboration, J. L. H. A., Phys.Rev., vol. C84, pp. 055204, 2011. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 107, pp. 262501, 2011. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 107, pp. 072003, 2011. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 107, pp. 072003, 2011. Abstract
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2009
Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 103, pp. 202501, 2009. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 103, pp. 202501, 2009. Abstract
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2008
Collaboration, J. L. H. A., Science, vol. 320, pp. 1476-1478, 2008. Abstract
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Collaboration, J. L. H. A., Science, vol. 320, pp. 1476-1478, 2008. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 101, pp. 182502, 2008. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 101, pp. 182502, 2008. Abstract
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2007
Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 99, pp. 242501, 2007. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 99, pp. 242501, 2007. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 99, pp. 052501, 2007. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 99, pp. 052501, 2007. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 99, pp. 072501, 2007. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 98, pp. 032301, 2007. Abstract
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Collaboration, J. L. H. A., Phys.Rev.Lett., vol. 98, pp. 032301, 2007. Abstract
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Acha, A., and others, Phys. Rev. Lett., vol. 98, pp. 032301, 2007. Abstract
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Collaboration, J. L. H. A., Phys.Rev., vol. C75, pp. 025201, 2007. Abstract
n/a | {"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": 0.8834983706474304, "perplexity": 3553.280978045624}, "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/1500549423774.37/warc/CC-MAIN-20170721122327-20170721142327-00097.warc.gz"} |
http://camiloarcaya.info/2015/10/ | ## Hemeroteca del mes octubre, 2015
#### A toy quantum analog of enzymes
Por • 31 oct, 2015 • Category: Crítica
We present a quantum system incorporating qualitative aspects of enzyme action in which the possibility of quantum superposition of several conformations of the enzyme-substrate complex is investigated. We present numerical results showing quantum effects that transcend the case of a statistical mixture of conformations.
#### Topological Ramsey numbers and countable ordinals
Por • 30 oct, 2015 • Category: Educacion
We study the topological version of the partition calculus in the setting of countable ordinals. Let α and β be ordinals and let k be a positive integer. We write β→top(α,k)2 to mean that, for every red-blue coloring of the collection of 2-sized subsets of β , there is either a red-homogeneous set homeomorphic to α or a blue-homogeneous set of size k . The least such β is the topological Ramsey number Rtop(α,k) . We prove a topological version of the Erdos-Milner theorem, namely that Rtop(α,k) is countable whenever α is countable. More precisely, we prove that Rtop(ωωβ,k+1)≤ωωβ⋅k for all countable ordinals β and finite k .
#### Topological dynamics and the complexity of strong types
Por • 30 oct, 2015 • Category: Ambiente
We develop topological dynamics for the group of automorphisms of a monster model of any given theory. In particular, we find strong relationships between objects from topological dynamics (such as the generalized Bohr compactification introduced by Glasner) and various Galois groups of the theory in question, obtaining essentially new information about them, e.g. we present the closure of the identity in the Lascar Galois group of the theory as the quotient of a compact, Hausdorff group by a dense subgroup. We apply this to describe the complexity of bounded, invariant (not necessarily Borel) equivalence relations, obtaining comprehensive results, subsuming and extending the existing results and answering some open questions from earlier papers. We show that, in a countable theory, any such relation restricted to the set of realizations of a complete type over ∅ is type-definable if and only if it is smooth. Then we show a counterpart of this result for theories in an arbitrary (not necessarily countable) language, obtaining also new information involving relative definability of the relation in question.
#### Continuum as a primitive type
Por • 30 oct, 2015 • Category: Opinion
The paper is the revision and extended version of the Section 6 of the paper {\em Types and operations} arXiv:1501.03043. Here, primitive types (corresponding to the intuitive concept of Continuum) are introduced along with primitive operations, constructors, and relations.
#### The Insecure Future of the World Economic Growth
Por • 30 oct, 2015 • Category: Economía
Growth rate of the world Growth Domestic Product (GDP) is analysed to determine possible pathways of the future economic growth. The analysis is based on using the latest data of the World Bank and it reveals that the growth rate between 1960 and 2014 was following a trajectory approaching asymptotically a constant value. The most likely prediction is that the world economic growth will continue to increase exponentially, which over a sufficiently long time is unsustainable. A more optimistic but less realistic prediction is based on the assumption that the growth rate will start to decrease linearly. In this case, the world economic growth is predicted to reach a maximum, if the growth rate is going to decrease linearly with time, or to follow a logistic trajectory, if the growth rate is going to decrease linearly with the size of the world GDP.
#### Can we have another light (~ 145 GeV) Higgs boson?
Por • 30 oct, 2015 • Category: Leyes
A second light Higgs boson, with mass of approximately 145 GeV, is predicted by non-minimal Supersymmetric models. This new particle can account for an apparent \sim 3 \sigma excess recorded by the CMS experiment at the Large Hadron Collider (LHC) during Run 1. We show how this can be explained in a particular realisation of these scenarios, the (B-L) Supersymmetric Model (BLSSM), which also has other captivating features, like offering an explanation for neutrino masses and relieving the small hierarchy problem of the Minimal Supersymmetric Standard Model (MSSM).
#### Cuba and US Foreign Policy
Por • 28 oct, 2015 • Category: sociologia
It’s worth to note that STRATFOR experts use the example of Venezuela, the Cuban nearest and closest ally in the hemisphere, in their assessments of prospects for US-Cuba relationship. They believe that ultimately the Venezuela’s future will rely on global oil prices and Venezuelan President Nicolas Maduro’s ability to simultaneously manage unrest on the streets and counter challengers within the government. In fact, several hours after the U.S.-Cuba prisoner swap was announced, Maduro publicly said Venezuela would be willing to improve its stagnant political ties with the United States.
#### Strategic Engineered Migration as Weapon of War
Por • 28 oct, 2015 • Category: Política
Who is providing guarantees that encourage hundreds of thousands of people to rush from other continents to Europe and why? What we are seeing now is the practical implementation of theoretical calculations of a strategic nature. Such strategies have been under development for a long time. One of them is a study by the Belfer Center for Science and International Affairs at Harvard University that bears the name «Strategic Engineered Migration as a Weapon of War», which the author also uses for the title of this article. The study was first published in 2008 in the Civil Wars journal. Using a combination of statistical data and case study analysis, the author of the work, Kelly Greenhill, provides answers to the following questions: can refugees be a specific type of weapon, can this weapon only be used in wartime or in peacetime as well, and just how successful can its exploitation be? On the whole, Greenhill answers these questions in the affirmative.
#### Policías en peligro
Por • 28 oct, 2015 • Category: Nacionales
Los venezolanos tienen miedo y restringen sus actividades normales por temor a ser víctimas
#### Heideggers Dinge
Por • 28 oct, 2015 • Category: Filosofía
This paper discusses the notion of a thing (Ding) in Heidegger. Its aim is to explain the systematic place of that notion in Heidegger’s thought in relation to his ontological discourse: as what is explained through different understandings of being, things allow for a simultaneous differentiation and discussion of the different epochs in the so-called history of being. Thus a phenomenology of things and thingness serves as frame of reference for all explications of ‘what there is.’ If Heidegger is a realist, it is not because he attributes reality to all that is, but rather because all explanations of being refer back to how things are discovered as meaningful. | {"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": 0.8517570495605469, "perplexity": 1543.968742728751}, "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-2017-43/segments/1508187828411.81/warc/CC-MAIN-20171024105736-20171024125736-00687.warc.gz"} |
https://en.wikipedia.org/wiki/Isospin | Isospin
In nuclear physics and particle physics, isospin is a quantum number related to the strong interaction. More specifically, isospin symmetry is a subset of the flavour symmetry seen more broadly in the interactions of baryons and mesons.
The name of the concept contains the term spin because its quantum mechanical description is mathematically similar to that of angular momentum (in particular, in the way it couples; for example, a proton-neutron pair can be coupled either in a state of total isospin 1 or in one of 0.[1]). Unlike angular momentum, however, it is a dimensionless quantity, and is not actually any type of spin.
Etymologically, the term was derived from isotopic spin, a confusing term to which nuclear physicists prefer isobaric spin, which is more precise in meaning. Before the concept of quarks were introduced, particles that are affected equally by the strong force but had different charges (e.g. protons and neutrons) were treated as being different states of the same particle, but having isospin values related to the number of charge states.[2]. A close examination of isospin symmetry ultimately led directly to the discovery and understanding of quarks, and of the development of Yang–Mills theory. Isospin symmetry remains an important concept in particle physics.
Quark content and isospin
In the modern formulation, isospin (I) is defined as a vector quantity in which up and down quarks have a value of I = 12, with the 3rd-component (I3) being 12 for up quarks, and −12 for down quarks, while all other quarks have I = 0. In general, for hadrons[3], therefore
${\displaystyle I_{3}={\frac {1}{2}}(n_{u}-n_{d})\ }$
where nu and nd are the numbers of up and down quarks respectively.
In any combination of quarks, the 3rd component of the isospin vector (I3) could either be aligned between a pair of quarks, or face the opposite direction, giving different possible values for total isospin for any combination of quark flavours. Hadrons with the same quark content but different total isospin can be distinguished experimentally, verifying that flavour is actually a vector quantity, not a scalar (up vs down simply being a projection in the quantum mechanical z-axis of flavour-space).
For example, a strange quark can be combined with an up and a down quark to form a baryon, but there are two different ways the isospin values can combine - either adding (due to being flavour-aligned) or cancelling out (due to being in opposite flavour-directions). The isospin 1 state (the Sigma 0) and the isospin 0 state (the Lambda 0) have different experimentally detected masses and half-lives.
Isospin and symmetry
Isospin is regarded as a symmetry of the strong interaction under the action of the Lie group SU(2), the two states being the up flavour and down flavour. In quantum mechanics, when a Hamiltonian has a symmetry, that symmetry manifests itself through a set of states that have the same energy (the states are described as being degenerate). In simple terms, that the energy operator for the strong interaction gives the same result when an up quark and an otherwise identical down quark are swapped around.
Like the case for regular spin, the isospin operator I is vector-valued: it has three components Ix, Iy, Iz which are coordinates in the same 3-dimensional vector space where the 3 representation acts. Note that it has nothing to do with the physical space, except similar mathematical formalism. Isospin is described by two quantum numbers: I, the total isospin, and I3, an eigenvalue of the Iz projection for which flavor states are eigenstates, not an arbitrary projection as in the case of spin. In other words, each I3 state specifies certain flavor state of a multiplet. The third coordinate (z), to which the "3" subscript refers, is chosen due to notational conventions which relate bases in 2 and 3 representation spaces. Namely, for the spin-12 case, components of I are equal to Pauli matrices divided by 2 and Iz = 12 τ3, where
${\displaystyle \tau _{3}={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}}$.
While the forms of these matrices are isomorphic to those of spin, these Pauli matrices only act within the Hilbert space of isospin, not that of spin, and therefore is common to denote them with τ rather than σ to avoid confusion.
Although isospin symmetry is actually very slightly broken, SU(3) symmetry is more badly broken, due to the much higher mass of the strange quark compared to the up and down. The discovery of charm, bottomness and topness could lead to further expansions up to SU(6) flavour symmetry, which would hold if all six quarks were identical. However, the very much larger masses of the charm, bottom and top quarks means that SU(6) flavour symmetry is very badly broken in nature (at least at low energies) and assuming this symmetry leads to qualitatively and quantitatively incorrect predictions. In modern applications, such as lattice QCD, isospin symmetry is often treated as exact while the heavier quarks must be treated separately.
Hadron nomenclature is based on isospin.[4]
• Particles of total isospin 32 are named Delta baryons and can only be made by a combination of any three up or down quarks.
• Particles of total isospin 1 can be made from two up quarks, two down quarks, or one of each:
• Particles of total isospin 12 can be made from:
• a single up or down quark together with an additional quark of higher flavour - strange (kaons), charm (D meson), or bottom (B meson)
• a single up or down quark together with two additional quarks of higher flavour - Xi baryon
• an up quark, a down quark, and either an up or a down quark - (nucleons. Note that three identical quarks would be forbidden by the Pauli exclusion principle
• Particles of total isospin 0 can be made from
• one up quark and one down quark - eta mesons
• one up quark and one down quark, with an additional quark of higher flavour - Lambda baryons
• anything not involving any up or down quarks
History
Original motivation for isospin
Isospin was introduced as a concept in 1932, well before the 1960s development of the quark model. The man who introduced it, Werner Heisenberg[5], did so to explain symmetries of the then newly discovered neutron (symbol n):
• The mass of the neutron and the proton (symbol p) are almost identical: they are nearly degenerate, and both are thus often called nucleons. Although the proton has a positive electric charge, and the neutron is neutral, they are almost identical in all other aspects.
• The strength of the strong interaction between any pair of nucleons is the same, independent of whether they are interacting as protons or as neutrons.
This behavior is not unlike the electron, where there are two possible states based on their spin. Other properties of the particle are conserved in this case. Heisenberg introduced the concept of another conserved quantity that would cause the proton to turn into a neutron and vice versa. In 1937, Eugene Wigner introduced the term "isospin" to indicate how the new quantity is similar to spin in behavior, but otherwise unrelated.[6].
Protons and neutrons were then grouped together as nucleons because they both have nearly the same mass and interact in nearly the same way, if the (much weaker) electromagnetic interaction is neglected. In particle physics, the near mass-degeneracy of the neutron and proton points to an approximate symmetry of the Hamiltonian describing the strong interactions. It was thus convenient to treat them as being different states of the same particle.
Heisenberg's particular contribution was to note that the mathematical formulation of this symmetry was in certain respects similar to the mathematical formulation of spin, whence the name "isospin" derives. The neutron and the proton are assigned to the doublet (the spin-12, 2, or fundamental representation) of SU(2). The pions are assigned to the triplet (the spin-1, 3, or adjoint representation) of SU(2). Though, there is a difference from the theory of spin: the group action does not preserve flavor (specifically, the group action is an exchange of flavour).
Similar to a spin 12 particle, which has two states, protons and neutrons were said to be of isospin 12. The proton and neutron were then associated with different isospin projections I3 = +12 and −12 respectively.
Although the neutron does in fact have a slightly higher mass due to isospin breaking (this is now understood to be due to the difference in the masses of the up and down quarks and the effects of the electromagnetic interaction), the appearance of an approximate symmetry is useful even if it doesn't exactly hold; the small symmetry breakings can be described by a perturbation theory, which gives rise to slight differences between the near-degenerate states
When constructing a physical theory of nuclear forces, one could simply assume that it does not depend on isospin, although the total isospin should be conserved.
The particle zoo
These considerations would also prove useful in the analysis of meson-nucleon interactions after the discovery of the pions in 1947. The three pions (
π+
,
π0
,
π
) could be assigned to an isospin triplet with I = 1 and I3 = +1, 0 or −1. By assuming that isospin was conserved by nuclear interactions, the new mesons were more easily accommodated by nuclear theory.
As further particles were discovered, they were assigned into isospin multiplets according to the number of different charge states seen: 2 doublets I = 12 of K mesons (
K
,
K0
),(
K+
,
K0
), a triplet I = 1 of Sigma baryons (
Σ+
,
Σ0
,
Σ
), a singlet I = 0 Lambda baryon (
Λ0
), a quartet I = 32 Delta baryons (
Δ++
,
Δ+
,
Δ0
,
Δ
), and so on.
The power of isospin symmetry and related methods comes from the observation that families of particles with similar masses tend to correspond to the invariant subspaces associated with the irreducible representations of the Lie algebra SU(2). In this context, an invariant subspace is spanned by basis vectors which correspond to particles in a family. Under the action of the Lie algebra SU(2), which generates rotations in isospin space, elements corresponding to definite particle states or superpositions of states can be rotated into each other, but can never leave the space (since the subspace is in fact invariant). This is reflective of the symmetry present. The fact that unitary matrices will commute with the Hamiltonian means that the physical quantities calculated do not change even under unitary transformation. In the case of isospin, this machinery is used to reflect the fact that the mathematics of the strong force behaves the same if a proton and neutron are swapped around (in the modern formulation, the up and down quark).
An example : Delta baryons
For example, the particles known as the Delta baryons—baryons of spin32 were grouped together because they all have nearly the same mass (approximately 1232 MeV/c2), and interact in nearly the same way.
They could be treated as the same particle, with the difference in charge being due to the particle being in different states. Isospin was introduced in order to be the variable that defined this difference of state. In an analogue to spin, an isospin projection (denoted I3) is associated to each charged state; since there were four Deltas, four projections were needed. Like spin, isospin projections were made to vary in increments of 1. Hence, in order to have four increments of 1, an isospin value of 32 is required (giving the projections I3 = 32, 12, −12, −32). Thus, all the Deltas were said to have isospin I = 32 and each individual charge had different I3 (e.g. the
Δ++
was associated with I3 = +32).
In the isospin picture, the four Deltas and the two nucleons were thought to simply be the different states of two particles. The Delta baryons are now understood to be made of a mix of three up and down quarks - uuu ((
Δ++
), uud (
Δ+
), udd (
Δ0
), and ddd (
Δ
); the difference in charge being difference in the charges of up and down quarks ((+23e and −13e respectively); yet, they can also be thought of as the excited states of the nucleons.
Gauged isospin symmetry
Attempts have been made to promote isospin from a global to a local symmetry. In 1954, Chen Ning Yang and Robert Mills suggested that the notion of protons and neutrons, which are continuously rotated into each other by isospin, should be allowed to vary from point to point. To describe this, the proton and neutron direction in isospin space must be defined at every point, giving local basis for isospin. A gauge connection would then describe how to transform isospin along a path between two points.
This Yang–Mills theory describes interacting vector bosons, like the photon of electromagnetism. Unlike the photon, the SU(2) gauge theory would contain self-interacting gauge bosons. The condition of gauge invariance suggests that they have zero mass, just as in electromagnetism.
Ignoring the massless problem, as Yang and Mills did, the theory makes a firm prediction: the vector particle should couple to all particles of a given isospin universally. The coupling to the nucleon would be the same as the coupling to the kaons. The coupling to the pions would be the same as the self-coupling of the vector bosons to themselves.
When Yang and Mills proposed the theory, there was no candidate vector boson. J. J. Sakurai in 1960 predicted that there should be a massive vector boson which is coupled to isospin, and predicted that it would show universal couplings. The rho mesons were discovered a short time later, and were quickly identified as Sakurai's vector bosons. The couplings of the rho to the nucleons and to each other were verified to be universal, as best as experiment could measure. The fact that the diagonal isospin current contains part of the electromagnetic current led to the prediction of rho-photon mixing and the concept of vector meson dominance, ideas which led to successful theoretical pictures of GeV-scale photon-nucleus scattering.
The introduction of quarks
Combinations of three u, d or s-quarks forming baryons with spin-32 form the baryon decuplet.
Combinations of three u, d or s-quarks forming baryons with spin-12 form the baryon octet
The discovery and subsequent analysis of additional particles, both mesons and baryons, made it clear that the concept of isospin symmetry could be broadened to an even larger symmetry group, now called flavor symmetry. Once the kaons and their property of strangeness became better understood, it started to become clear that these, too, seemed to be a part of an enlarged symmetry that contained isospin as a subgroup. The larger symmetry was named the Eightfold Way by Murray Gell-Mann, and was promptly recognized to correspond to the adjoint representation of SU(3). To better understand the origin of this symmetry, Gell-Mann proposed the existence of up, down and strange quarks which would belong to the fundamental representation of the SU(3) flavor symmetry.
In the quark model, the isospin projection (I3) followed from the up and down quark content of particles; uud for the proton and udd for the neutron. Technically, the nucleon doublet states are seen to be linear combinations of products of 3-particle isospin doublet states and spin doublet states. That is, the (spin-up) proton wave function, in terms of quark-flavour eigenstates, is described by[2]
${\displaystyle \vert p\uparrow \rangle ={\frac {1}{3{\sqrt {2}}}}\left({\begin{array}{ccc}\vert duu\rangle &\vert udu\rangle &\vert uud\rangle \end{array}}\right)\left({\begin{array}{ccc}2&-1&-1\\-1&2&-1\\-1&-1&2\end{array}}\right)\left({\begin{array}{c}\vert \downarrow \uparrow \uparrow \rangle \\\vert \uparrow \downarrow \uparrow \rangle \\\vert \uparrow \uparrow \downarrow \rangle \end{array}}\right)}$
and the (spin-up) neutron by
${\displaystyle \vert n\uparrow \rangle ={\frac {1}{3{\sqrt {2}}}}\left({\begin{array}{ccc}\vert udd\rangle &\vert dud\rangle &\vert ddu\rangle \end{array}}\right)\left({\begin{array}{ccc}2&-1&-1\\-1&2&-1\\-1&-1&2\end{array}}\right)\left({\begin{array}{c}\vert \downarrow \uparrow \uparrow \rangle \\\vert \uparrow \downarrow \uparrow \rangle \\\vert \uparrow \uparrow \downarrow \rangle \end{array}}\right)}$
Here, ${\displaystyle \vert u\rangle }$ is the up quark flavour eigenstate, and ${\displaystyle \vert d\rangle }$ is the down quark flavour eigenstate, while ${\displaystyle \vert \uparrow \rangle }$ and ${\displaystyle \vert \downarrow \rangle }$ are the eigenstates of ${\displaystyle S_{z}}$. Although these superpositions are the technically correct way of denoting a proton and neutron in terms of quark flavour and spin eigenstates, for brevity, they are often simply referred to as "uud" and "udd". Note also that the derivation above assumes exact isospin symmetry and is modified by SU(2)-breaking terms.
Similarly, the isospin symmetry of the pions are given by:
{\displaystyle {\begin{aligned}\vert \pi ^{+}\rangle &=\vert u{\overline {d}}\rangle \\\vert \pi ^{0}\rangle &={\frac {1}{\sqrt {2}}}\left(\vert u{\overline {u}}\rangle -\vert d{\overline {d}}\rangle \right)\\\vert \pi ^{-}\rangle &=-\vert d{\overline {u}}\rangle \end{aligned}}}
Although the discovery of the quarks led to reinterpretation of mesons as a vector bound state of a quark and an antiquark, it is sometimes still useful to think of them as being the gauge bosons of a hidden local symmetry.[7]
Weak isospin
Isospin is similar to, but should not be confused with weak isospin. Briefly, weak isospin is the gauge symmetry of the weak interaction which connects quark and lepton doublets of left-handed particles in all generations; for example, up and down quarks, top and bottom quarks, electrons and electron neutrinos. By contrast (strong) isospin connects only up and down quarks, acts on both chiralities (left and right) and is a global (not a gauge) symmetry. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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": 0.9561570882797241, "perplexity": 704.924409666378}, "config": {"markdown_headings": false, "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-2018-39/segments/1537267156224.9/warc/CC-MAIN-20180919122227-20180919142227-00415.warc.gz"} |
http://chasethedevil.blogspot.com/2013/07/bessel-and-harmonic-kinks-in-forward.html | ## Tuesday, July 02, 2013
### Bessel and Harmonic Kinks in the Forward
As Bessel (sometimes called Hermite) spline interpolation is only C1, like the Harmonic spline from Fritsch-Butland, the forward presents small kinks compared to a standard cubic spline. Hyman filtering also creates a kink where it fixes the monotonicity. Those are especially visible with a log scale in time. Here is how it looks on the Hagan-West difficult curve. | {"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": 0.845329761505127, "perplexity": 3675.3898825610654}, "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/1495463607846.35/warc/CC-MAIN-20170524131951-20170524151951-00470.warc.gz"} |
https://www.isa-afp.org/entries/Floyd_Warshall.html | # The Floyd-Warshall Algorithm for Shortest Paths
Title: The Floyd-Warshall Algorithm for Shortest Paths Authors: Simon Wimmer and Peter Lammich Submission date: 2017-05-08 Abstract: The Floyd-Warshall algorithm [Flo62, Roy59, War62] is a classic dynamic programming algorithm to compute the length of all shortest paths between any two vertices in a graph (i.e. to solve the all-pairs shortest path problem, or APSP for short). Given a representation of the graph as a matrix of weights M, it computes another matrix M' which represents a graph with the same path lengths and contains the length of the shortest path between any two vertices i and j. This is only possible if the graph does not contain any negative cycles. However, in this case the Floyd-Warshall algorithm will detect the situation by calculating a negative diagonal entry. This entry includes a formalization of the algorithm and of these key properties. The algorithm is refined to an efficient imperative version using the Imperative Refinement Framework. BibTeX: ```@article{Floyd_Warshall-AFP, author = {Simon Wimmer and Peter Lammich}, title = {The Floyd-Warshall Algorithm for Shortest Paths}, journal = {Archive of Formal Proofs}, month = may, year = 2017, note = {\url{http://isa-afp.org/entries/Floyd_Warshall.html}, Formal proof development}, ISSN = {2150-914x}, }``` License: BSD License Depends on: Refine_Imperative_HOL | {"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": 0.8801962733268738, "perplexity": 389.58430532998693}, "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-2018-34/segments/1534221210413.14/warc/CC-MAIN-20180816034902-20180816054902-00608.warc.gz"} |
http://math.stackexchange.com/questions/3444/intuition-for-the-definition-of-the-gamma-function | # Intuition for the definition of the Gamma function?
In these notes by Terence Tao is a proof of Stirling's formula. I really like most of it, but at a crucial step he uses the integral identity
$$n! = \int_{0}^{\infty} t^n e^{-t} dt$$
coming from the Gamma function. I have a mathematical confession to make: I have never "grokked" this identity. Why should I expect the integral on the right to give me the number of elements in the symmetric group on $n$ letters?
(It's not that I don't know how to prove it. It's quite fun to prove; my favorite proof observes that it is equivalent to the integral identity $\int_{0}^{\infty} e^{(x-1)t} dt = \frac{1}{1 - x}$. But if someone were to ask me, "yes, but why, really?" I would have no idea what to say.)
So what are more intuitive ways of thinking about this identity? Is there a probabilistic interpretation? What kind of random variable has probability density function $\frac{t^n}{n!} e^{-t}$? (What does this all have to do with Tate's thesis?)
As a rough measure of what I'm looking for, your answer should make it obvious that $t^n e^{-t}$ attains its maximum at $t = n$.
Edit: The kind of explanation I'm looking for, as I described in the comments, is similar to this explanation of the beta integral.
-
I can answer exactly one sentence of your question: $\frac{t^n}{n!}e^{-t}$ is the probability density function of the sum of n+1 i.i.d. exponential random variables of unit rate. – Rahul Narain Aug 27 '10 at 5:47
I would add the "special-functions" tag, created by myself yesterday, to include gamma, zeta,and other functions generally called Special Functions. – Américo Tavares Aug 27 '10 at 6:16
+1 for the comprehensive explanation. However I ask if you mean that Calculus is not intuitive? – Américo Tavares Aug 27 '10 at 6:40
@Americo: that is my point! It is very frustrating. I believe that in mathematics nothing is a trick if seen from a sufficiently high level and I am hoping someone can tell me what that level is. – Qiaochu Yuan Aug 27 '10 at 15:37
@Qiaochu: The method of stationary phase (Tao's notes), btw, is motivated by the central limit theorem. The gamma distribution, as you now know from other comments and another answer, is a sum of independent exponentials. The CLT implies it should be close to Gaussian for large n. This means the integrand should be well approximated by a parabolic fit to its logarithm. A good choice is the Taylor series around the maximum. This lets us approximate the integral by the Normal integral (equivalently, erf) with parameters depending on n, whence we obtain asymptotic formulae. – whuber Aug 29 '10 at 16:39
show 4 more comments
I haven't quite got this straight yet, but I think one way to go is to think about choosing points at random from the positive reals. This answer is going to rather longer than it really needs to be, because I'm thinking about this in a few (closely related) ways, which probably aren't all necessary, and you can decide to reject the uninteresting parts and keep anything of value. Very roughly, the idea is that if you "randomly" choose points from the positive reals and arrange them in increasing order, then the probability that the (n+1)th point is in a small interval (t,t+dt) is a product of probabilities of independent events, n factors of t for choosing n points in the interval [0,t], one factor of $e^{-t}$ as all the other points are in $[t,\infty)$, one factor of dt for choosing the point in $(t,t+dt)$, and a denominator of $n!$ coming from the reordering. At least, as an exercise in making a simple problem much harder, here goes...
I'll start with a bit of theory before trying to describe intuitively why the probability density $\frac{t^n}{n!}e^{-t}$ pops out.
We can look at the homogeneous Poisson process (with rate parameter 1). One way to think of this is to take a sequence on independent exponentially distributed random variables with rate parameter 1, $S_1,S_2,\ldots$, and set $T_n=S_1+\cdots+S_n$. As has been commented on already, $T_{n+1}$ has the probability density function $\frac{t^n}{n!}e^{-t}$. I'm going to avoid proving this immediately though, as It would just reduce to manipulating some integrals. Then, the Poisson process $X(t)$ counts the number of times $T_i$ lying in the interval [0,t].
We can also look at Poisson point processes (aka, Poisson random measures, but that Wikipedia page is very poor). This is just makes rigorous the idea of randomly choosing unordered sets of points from a sigma-finite measure space $(E,\mathcal{E},\mu)$. Technically, it can be defined as a set of nonnegative integer-valued random variables $\{N(A)\colon A\in\mathcal{E}\}$ counting the number of points chosen from each subset A, such that $N(A)$ has the Poisson distribution of rate $\mu(A)$ and $N(A_1),N(A_2),\ldots$ are independent for pairwise disjoint sets $A_1,A_2,\ldots$. By definition, this satisfies $$\begin{array}{}\mathbb{P}(N(A)=n)=\frac{\mu(A)^n}{n!}e^{-\mu(A)}.&&(1)\end{array}$$ The points $T_1,T_2,\ldots$ above defining the homogeneous Poisson process also define a Poisson random measure with respect to the Lebesgue measure $(\mathbb{R}_+,{\cal B},\lambda)$. Once you forget about the order in which they were defined and just regard them as a random set that is, which I think is the source of the $n!$. If you think about the probability of $T_{n+1}$ being in a small interval $(t,t+\delta t)$ then this is just the same as having $N([0,t])=n$ and $N((t,t+\delta t))=1$, which has probability $\frac{t^n}{n!}e^{-t}\delta t$.
So, how can we choose points at random so that each small set $\delta A$ has probability $\mu(\delta A)$ of containing a point, and why does (1) pop out? I'm imagining a hopeless darts player randomly throwing darts about and, purely by luck, hitting the board with some of them. Consider throwing a very large number $N\gg1$ of darts, independently, so that each one only has probability $\mu(A)/N$ of hitting the set, and is distributed according to the probability distribution $\mu/\mu(A)$. This is consistent, at least, if you think about the probability of hitting a subset $B\subseteq A$. The probability of missing with all of them is $(1-\mu(A)/N)^N=e^{-\mu(A)}$. This is a multiplicative function due to independence of the number hitting disjoint sets. To get the probability of one dart hitting the set, multiply by $\mu(A)$ (one factor of $\mu(A)/N$ for each individual dart, multiplied by N because there's N of them). For n darts, we multiply by $\mu(A)$ n times, for picking n darts to hit, then divide by n! because we have over-counted the subsets of size n by this factor (due to counting all n! ways of ordering them). This gives (1). I think this argument can probably be cleaned up a bit.
Getting back to choosing points randomly on the positive reals, this gives a probability of $\frac{t^n}{n!}e^{-t}dt$ of picking n in the interval $[0,t]$ and one in $(t,t+dt)$. If we sort them in order as $T_1\lt T_2\lt\cdots$ then $\mathbb{P}(T_1\gt t)=e^{-t}$, so it is exponentially distributed. Conditional on this, $T_2,T_3,\ldots$ are chosen randomly from $[T_1,\infty)$, so we see that the differences $T_{i+1}-T_{i}$ are independent and identically distributed.
Why is $\frac{t^n}{n!}e^{-t}$ maximized at t=n? I'm not sure why the mode should be a simple property of a distribution. It doesn't even exist except for unimodal distributions. As $T_{n+1}$ is the sum of n+1 IID random variables of mean one, the law of large numbers suggests that it should be peaked approximately around n. The central limit theorem goes further, and gives $\frac{t^n}{n!}e^{-t}\approx\frac{1}{\sqrt{2\pi n}}e^{-(t-n)^2/{2n}}$. Stirling's formula is just this evaluated at $t=n$.
What's this to do with Tate's thesis? I don't know, and I haven't read it (but intend to), but have a vague idea of what it's about. If there is anything to do with it, maybe it is something to do with the fact that we are relating the sums of independent random variables $S_1+\cdots+S_n$ distributed with respect to the Haar measure on the multiplicative group $\mathbb{R}_+$ (edit: oops, that's not true, the multiplicative Haar measure has cumulative distribution given by log, not exp) with randomly chosen sets according to the Haar measure on the additive group $\mathbb{R}$.
-
thanks for this very interesting answer! As you say, the story is not quite straight yet, but I think these are the ideas I wanted to hear about. – Qiaochu Yuan Aug 28 '10 at 2:19
Thanks for writing this up. I have been musing over the same ideas, so I think I can make the final connection with the symmetric group: recall how the Poisson distribution is computed as a limit of Binomials. In tracing through the derivation, you see that the 1/k! terms come from Binomial coefficients; those, originally, count ways of selecting subsets from a set. The denominator (k!) counts the ways of permuting elements in a k-element subset. This completes the train of thought: counting subsets --> symmetric group --> Binomial distribution --> Poisson --> Gamma. – whuber Aug 28 '10 at 16:42
The geometric approach works.
Let’s compute the volume of the $2n$ dimensional ball, $D^{2n}$, in two ways. One way is extremely clever but has been known for centuries and provides interesting insights: it’s based on Liouville’s trick. Specifically, we will compute two integrals in polar coordinates, one of which is the volume of the ball and the other of which reduces to a product of one-dimensional integrals. Both integrands will depend (at most) on the radial coordinate $r$, which lets us separate out the surface area of the boundary of the ball as a common factor. Write this surface area as $S_{2n-1}$.
There’s essentially just one way to do this trick: integrate $\exp(-r^2)$. Its integral over $\mathbb{R}^{2n}$ equals
$$S_{2n-1} \int_0^\infty {\exp\left(- r^2 \right) r^{2n-1} dr}.$$
However, because $r^2 = x_1^2 + x_2^2 + \ldots + x_{2n}^2$, the integrand (in Cartesian coordinates $\left( x_1, x_2, \ldots, x_{2n} \right)$) factors as $\exp\left(-r^2 \right) = \exp\left(-x_1^2 \right) \cdots \exp\left(-x_{2n}^2 \right)$, each of which must be integrated from $-\infty$ to $+\infty$. Whence
$$S_{2n-1} \int_0^\infty {\exp \left(- r^2 \right) r^{2n-1} dr} = \left( \int_{- \infty}^ \infty {\exp \left( -x^2 \right) dx} \right) ^{2n}.$$
I will call the left hand integral $\tfrac{1}{2} \Gamma \left(n \right)$, because that is what it is (as a simple change of variables shows). In the same notation, $\Gamma \left(1/2 \right) = \int_{-\infty}^\infty {\exp\left(-x^2 \right) dx}$. Algebraic re-arrangement of the foregoing yields the volume of $D^{2n}$ as
$$|D^{2n} | = S_{2n - 1} \int_0^1 {r^{2n - 1} dr} = \frac{{S_{2n - 1} }} {{2n}} = \frac {\Gamma \left(1/2 \right)^{2n}} { n \Gamma \left(n \right) }.$$
That was the first method: the result is a familiar one, but has been left expressed in a way that better reveals its origins in polar and Cartesian integration.
The next way to compute the ball's volume is, I believe, new. It is inspired by Cavalieri’s Principle: the idea that you can shift slices of a solid around without changing the volume of the solid. The generalization is to move two-dimensional slices around and to change their shapes while you do so, but in a way that does not change their areas. It follows that the new solid has the same (hypervolume) as the original, although it might have a completely different shape.
We will compute the volume of a region $Q_n$ in $\mathbb{R}^{2n}$. It is conveniently described by identifying $\mathbb{R}^{2n}$ with $\mathbb{C}^{n}$, using coordinates $z_i = \left( x_{2i - 1}, x_{2i} \right)$, in terms of which
$$Q_n = \{ \mathbf{z} \in \mathbb{C}^n :0 \leq \left| {z_1 } \right| \leq \left| {z_2 } \right| \leq \cdots \leq \left| {z_n } \right| \leq 1 \}.$$
If these were real variables, we could make the volume-preserving transformation $w_1 = z_1, w_2 = z_2 – z_1, \ldots , w_i = z_i – z_{i-1}, \ldots, w_n = z_n – z_{n-1}$, with the sole restriction that the sum of the $w_i$ (all of which are nonnegative) not exceed 1. Because they are complex variables, though, we have to consider the area of an annulus bounded by $z_{i-1}$ and $z_i$: it is proportional to $z_i^2 – z_{i – 1}^2$. The circle of the same area has radius $w_i$ for which $w_i^2 = z_i^2 – z_{i – 1}^2$. Therefore, if we define new variables $w_i$ according to this formula, we obtain a new region- - one of substantially different shape- - having the same volume. This region is defined by $\left| {w_1 }^2 \right| + \cdots + \left| {w_n }^2 \right| \le 1$: that is, it’s our old friend $D^{2n}$. Therefore, the volume of $Q_n$ equals the volume of $D^{2n}$ .
Now for the punch line: $Q_n$ is a fundamental domain for the action of $S[n]$, the symmetric group, on the product of $n$ disks $T^{2n} = \left( D^2 \right) ^n$; $S[n]$ acts by permuting the Complex coordinates $z_1, \ldots, z_n$. The volume of $T^{2n}$ equals $|D^2|^n = \pi ^n$. Writing $|S[n]|$ for the number of permutations and equating our two completely different calculations of the volume of the $2n$ ball gives
$$\pi ^ n / |S[n]| = \frac {\Gamma \left(1/2 \right)^{2n}} { n \Gamma \left(n \right) },$$
whence
$$|S[n]| = \frac{{\pi ^n n\Gamma \left( n \right)}}{{\Gamma \left( {1/2} \right)^{2n} }}.$$
This simplifies: the volume formula for $n = 2$ must give the area of the unit circle, equal to $\pi$, whence $\Gamma \left( 1/2 \right)^2 = \pi$. Finally, then,
$$|S[n]| = n\Gamma \left( n \right).$$
I will finish by remarking that Liouville's method is a perfectly natural thing to encounter when working with the multivariate Normal distribution, so it's not really an isolated trick, but is rather a pretty basic result expressing a defining property of Normal (Gaussian) variates. There are, of course, many other ways to compute the volume of $D^{2n}$, but this one gives us the Gamma function directly.
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thanks for your answer. The second method is very interesting; I would appreciate a little more elaboration on the first method. In particular I would appreciate an explanation which does not have to use the word "trick" :) – Qiaochu Yuan Aug 29 '10 at 6:46
The first method is motivated by the hope of finding integrable real functions f and g for which f(sqrt(x^2 + y^2 + ... + z^2)) = g(x)g(y)...g(z): the lhs will be the integrand in polar coordinates and the rhs is the factorization that makes everything work. Assuming f differentiable, equating partial derivatives on both sides shows that f'(r)/(r f(r)) must be a constant, whence f(r) = A*exp(b*r^2). We need b < 0 for integrability. This determines the form of g; a simple choice is g(x)=exp(-x^2). (I didn't go into this originally because the first method is now standard and well documented.) – whuber Aug 29 '10 at 16:29
@Qiaochu: The final paragraph was put there precisely to avoid being accused of using a trick! ;-) The reference to statistics is that the study of variances leads one to look at the distributions of sums of squares of Gaussian variables (called chi-squared distributions). Obtaining volumes of hyperspheres drops out almost by accident as a byproduct of this calculation. – whuber Aug 29 '10 at 16:44
This is another answer in terms of Poisson processes and the Gamma distribution, and it still uses a bit of calculus which you might call a trick, but I think at least it does add another bit of intuition:
Consider the homogeneous Poisson process with rate parameter 1; this means we are counting the number of occurrences of an event that happens with rate 1. Let's calculate the probabilities $p_k(t)$ that we are in state $k$ at time $t$, i.e., that the event occurs $k$ times in the interval $[0,t]$.
Since the event happens with rate 1, probability mass flows from $p_k(t)$ to $p_{k+1}(t)$ with rate 1. This means that ${p'}_0(t) = -p_0(t)$ and ${p'}_{k+1}(t) = p_k(t) - p_{k+1}(t)$. Also, $p_0(0) = 1$ and $p_{k+1}(0) = 0$. Here comes the bit of calculus: these equations have the solution $p_k(t) = \frac{t^k}{k!} e^{-t}$. (A bit fuzzily, we can read this in two parts: $\frac{t^k}{k!}$ is 1 integrated $k$ times, and $e^{-t}$ represents probability mass being lost at rate 1 to states further down the line. See below for yet another fuzzy explanation.)
Now consider the waiting time $T_k$ until the $k$'th occurrence. Clearly, $T_k = t$ means that the transition from state $k-1$ to state $k$ happens at time $t$, so the probability of $T_k \le t$ is the probability that the transition happens before time $t$, and the density is the derivative of this, i.e., the rate at which probability mass flows from state $k-1$ to $k$. This equals the occurrence rate (i.e., 1) times $p_{k-1}(t)$.
So the probability density of the random variable $T_k$ is $1 \cdot p_{k-1}(t) = \frac{t^{k-1}}{(k-1)!} e^{-t}$ (for $t \ge 0$). Since the probability that there is no occurrence ever is obviously zero, $\int_{0}^{\infty} \frac{t^{k-1}}{(k-1)!} e^{-t} dt = 1$.
Incidentally, this is related to a way of thinking about why $\sum_{k=0}^\infty \frac{t^k}{k!} = \lim_{n\to\infty} (1 + \frac{t}{n})^n$. Suppose you start with one unit of money in an account and get 100% interest, continuously compounded. However, the interest from the original account (number 0) is paid not the original account, but to account #1; interest from account #1 is paid to account #2, and so on.
Then the money $m_k(t)$ in account $k$ equals $1$ integrated $k$ times, and the total money $m(t)$ is $\sum_{k=0}^\infty m_k(t) = \sum_{k=0}^\infty \frac{t^k}{k!}$. But on the other hand, $m(t)$ is continuously compounded at 100% interest, so $m(t) = \lim_{n\to\infty} (1 + \frac{t}{n})^n$ by the usual reasoning.
This gives another fuzzy argument why we should have $p_k(t) = \frac{t^k}{k!} e^{-t}$. The change in the $p_k(t)$ over time consists of two parts: on the other hand, each $p_{k+1}(t)$ increases at rate $p_k(t)$; on the other hand, each $p_k(t)$ decreases at rate $p_k(t)$. If we view the $p_k(t)$ as accounts, since we are taking money out of every account at the constant rate 1, the effect is to decrease the total amount at the constant rate 1, i.e. by a factor of $e^{-t}$, which cancels out the increase of $e^t$ due to the accruing interest. It makes some intuitive sense that we can model this effect by simply rescaling the amount of money in each of the accounts by $e^{-t}$.
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You are really asking for a direct connection between some property of $n!$ and the integral. This can be done from the recursion
$S[n] = nS[n-1], S[0] = 1$
where $S[n]$ is the order of the symmetric group on $n$ elements. The exponential generating function for this series equals $1/(1-t)$. As in your favorite proof, replace that by $\int_{0}^{\infty} e^{(x-1)t} dt$, expand $e^{xt}$ as a series, reverse the order of summation and integration, and you recover a power series in $t$ whose terms are precisely $\int_{0}^{\infty} t^n e^{-t} dt/n!$. The result follows upon term-by-term comparison.
I realize that's not a whole lot more insight, but it does show explicitly a connection between a defining property of $n!$ and the integral for the $\Gamma$ function.
As far as your subsequent musings go: the integrand is the probability density of a Gamma variate, of course! One useful relation is that the Gamma distribution with parameter $n$ is the sum (i.e., convolution) of $n$ independent exponential variates (i.e., Gamma variates with $n=1$). The expectation of an exponential variate is 1 (this is easy), whence the expectation of a Gamma variate must be $n$ (because expectations add), strongly suggesting the mode of its pdf should be near $n$ (as justified for large $n$ by the Central Limit Theorem).
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The recursion is precisely what I did not want to hear about; I have no intuition for why this sequence of integrals should satisfy this recursion. But the last paragraph is helpful; thanks. – Qiaochu Yuan Aug 27 '10 at 15:36
It would help to understand better what kind of "intuition" you are looking for. For example, it's a standard Calculus exercise in integration by parts to show that this sequence of integrals satisfies the recursion, so for some people that would be good enough. I take it you're looking for a deeper connection with the Symmetric groups, but (i) why should there be one that goes beyond the mere counting of their elements and (ii) what would constitute a deeper connection for you? – whuber Aug 27 '10 at 16:08
i) Experience has taught me that this is a valuable question to ask. If I put on my John Baez hat, I might say that I don't understand this identity well enough to categorify it yet. ii) Ideally, something deep enough to categorify. For example, many simple identities involving binomial coefficients can be proven by mindlessly manipulating series, but they can also be proven bijectively, and these bijective proofs are often (at least to my mind) much more interesting. The analogue of a bijective proof of an integral is a probabilistic one; for example, the evaluation of the... – Qiaochu Yuan Aug 27 '10 at 16:25
(cont.) beta function with integer parameters has a very nice probabilistic proof. I am wondering if something similar is true here. For example, a direct proof that the gamma distribution is the sum of n independent exponential variables. – Qiaochu Yuan Aug 27 '10 at 16:28
I understand. Another train of thought: the gamma function shows up naturally in computing volumes of hyperspheres, whence one obtains that the volume of the 2n-dimensional unit ball is Pi^n/n!. Interestingly, Pi^n is the volume of the product of n unit disks (a solid torus), on which the symmetric group naturally acts by permuting coordinates. So: is there some natural geometric way to show that the volume of a fundamental domain of this action must equal the volume of the unit ball? – whuber Aug 27 '10 at 19:55
Sorry to "revive," but here's something I noticed while writing 2012 Fall OMO Problem 25. IMO it gives a kind of neat perspective on the gamma function (as a particular case of a "continuous" generating function), so hopefully this is not too off-topic. It may be somewhat related to George Lowther's answer above, but I don't have the background to fully understand/appreciate his post. Also, there might be a bit of seemingly irrelevant setup here.
Anyways, first consider the following "discrete" problem:
Find the number of integer solutions to $a+b+c+d=18$ with $0\le a,b,c,d\le 6$.
This is fairly standard via PIE or generating functions: for a decent discussion, see this AoPS thread.
Now consider a close "continuous" variant:
Let $N$ be a positive integer and $M$ be a positive real number. Find the probability that $y_1,y_2,\ldots,y_{N-1}\le M$ for a random solution in nonnegative reals to $y_0+y_1+\cdots+y_N=1$.
The direct generalization would have $y_0,y_N\le M$ as well, but I'll keep it like this since the OMO problem uses this version (and both versions allow essentially the same solution).
It's easy to generalize the PIE solution to the discrete version, but here's what I got when I tried to generalize the generating functions solution (unfortunately it's not really rigorous, but I feel like it should all be correct).
To extend the discrete generating functions solution, suppose we work with formal power "integrals" indexed by nonnegative reals instead of formal power series (indexed by nonnegative integers). As a basic example, the integral $\int_0^\infty x^t\; dt$ has a $1$ coefficient for each $x^t$ term, and thus corresponds to $y_0,y_N$---the discrete analog would be something like $\sum_{t=0}^{\infty}x^t$. For $y_1,\ldots,y_{N_1}$, which are bounded above by $M$, we instead have $\int_0^M x^t\; dt$. By "convolution" the desired probability is then $$\frac{[x^1](\int_0^M x^t\; dt)^{N-1}(\int_0^\infty x^t\; dt)^2}{[x^1](\int_0^\infty x^t\; dt)^{N+1}}.$$ But $\int_0^M x^t\; dt = \frac{x^M - 1}{\ln{x}}$ and for positive integers $L$, $(-\ln{x})^{-L} = \frac{1}{(L-1)!}\int_0^\infty x^t t^{L-1}\; dt$. Note that this is essentially the gamma function when $x=e^{-1}$, and for $L=1$ we have $\int_0^\infty x^t\; dt = (-\ln{x})^{-1}$. This is not hard to prove by differentiating (w.r.t. $x$) under the integral sign, but for integer $L$ there's a simple combinatorial proof of $$\frac{1}{(L-1)!}\int_0^\infty x^t t^{L-1}\; dt = (\int_0^\infty x^t\; dt)^L$$ which may provide a bit of intuition for the gamma function. Indeed, suppose we choose $L-1$ points at random from the interval $[0,t]$; the resulting sequence is nondecreasing with probability $\frac{1}{(L-1)!}$. These $L-1$ "dividers" split the interval $[0,t]$ into $L$ nonnegative reals adding up to $t$, which corresponds to the convolution on the RHS. It's not hard to check that this yields a bijection, which explains the $[x^t]$ coefficient of the LHS ($\frac{t^{L-1}}{(L-1)!}$). (Compare with the classical "balls and urns" or "stars and bars" argument for the discrete analog. In fact, it shouldn't be hard to interpret the $[x^t]$ coefficient as a limiting case of the discrete version.)
Now back to the problem: by "convolution" again (but used in a substantive way this time) we find \begin{align*} [x^1]\left(\int_0^M x^t\; dt\right)^{N-1}\left(\int_0^\infty x^t\; dt\right)^2 &=[x^1](1-x^M)^{N-1}(-\ln{x})^{-(N-1)}(-\ln{x})^{-2} \\ &=[x^1](1-x^M)^{N-1}\frac{1}{N!}\int_0^\infty x^t t^N\; dt \\ &=\frac{1}{N!}\sum_{0\le k\le N-1,\frac{1}{M}}(-1)^k\binom{N-1}{k}(1-kM)^N \end{align*} and $$[x^1](\int_0^\infty x^t\; dt)^{N+1} =[x^1](-\ln{x})^{-(N+1)} =[x^1]\frac{1}{N!}\int_0^\infty x^t t^N\; dt =\frac{1}{N!}.$$ The desired probability thus comes out to $\sum_{0\le k\le N-1,\frac{1}{M}}(-1)^k\binom{N-1}{k}(1-kM)^N$.
You can see how this explicitly applies to the OMO problem in my post here. Everything I wrote above is based off of the "Official 'Continuous' Generating Function Solution" there.
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https://www.aimsciences.org/journal/1078-0947/2020/40/10 | # American Institute of Mathematical Sciences
ISSN:
1078-0947
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## Discrete & Continuous Dynamical Systems - A
October 2020 , Volume 40 , Issue 10
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2020, 40(10): 5617-5638 doi: 10.3934/dcds.2020240 +[Abstract](172) +[HTML](58) +[PDF](407.58KB)
Abstract:
This paper studies the very weak solution to the steady MHD type system in a bounded domain. We prove the existence of very weak solutions to the MHD type system for arbitrary large external forces \begin{document}$({\bf f},{\bf{g}})$\end{document} in \begin{document}$L^r({\Omega})\times [X_{\theta',q'}({\Omega})]'$\end{document} and suitable boundary data \begin{document}$({\mathcal B}^0,{\mathcal U}^0)$\end{document} in \begin{document}$W^{-1/p,p}({\partial}{\Omega})\times W^{-1/q,q}({\partial}{\Omega})$\end{document}, under certain assumptions on \begin{document}$p,q,r,\theta$\end{document}. The uniqueness of very weak solution for small data \begin{document}$({\bf f},{\bf{g}},{\mathcal B}^0,{\mathcal U}^0)$\end{document} is also studied.
2020, 40(10): 5639-5710 doi: 10.3934/dcds.2020241 +[Abstract](61) +[HTML](22) +[PDF](922.2KB)
Abstract:
This article is a review of Lévy processes, stochastic integration and existence results for stochastic differential equations and stochastic partial differential equations driven by Lévy noise. An abstract PDE of the typical type encountered in fluid mechanics is considered in a stochastic setting driven by a general Lévy noise. Existence and uniqueness of a local pathwise solution is established as a demonstration of general techniques in the area.
2020, 40(10): 5711-5728 doi: 10.3934/dcds.2020242 +[Abstract](69) +[HTML](16) +[PDF](412.68KB)
Abstract:
The goal of this paper is to study the two-dimensional inviscid Boussinesq equations with temperature-dependent thermal diffusivity. Firstly we establish the global existence theory and regularity estimates for this system with Yudovich's type initial data. Then we investigate the vortex patch problem, and proving that the patch remains in Hölder class \begin{document}$C^{1+s}\; (0<s<1)$\end{document} for all the time.
2020, 40(10): 5729-5754 doi: 10.3934/dcds.2020243 +[Abstract](72) +[HTML](20) +[PDF](517.03KB)
Abstract:
We provide an estimation of the dissipation of the Wasserstein 2 distance between the law of some interacting \begin{document}$N$\end{document}-particle system, and the \begin{document}$N$\end{document} times tensorized product of the solution to the corresponding limit nonlinear conservation law. It then enables to recover classical propagation of chaos results [20] in the case of Lipschitz coefficients, uniform in time propagation of chaos in [17] in the case of strictly convex coefficients. And some recent results [7] as the case of particle in a double well potential.
2020, 40(10): 5755-5764 doi: 10.3934/dcds.2020244 +[Abstract](52) +[HTML](20) +[PDF](319.65KB)
Abstract:
We provide the simple proof of the Adams type inequalities in whole space \begin{document}${\mathbb R}^{2m}$\end{document}. The main tools are the Fourier rearrangement technique introduced by Lenzmann and Sok [16], a Hardy–Rellich type inequality for radial functions, and the sharp Moser–Trudinger type inequalities in \begin{document}${\mathbb R}^2$\end{document}.
2020, 40(10): 5765-5793 doi: 10.3934/dcds.2020245 +[Abstract](82) +[HTML](26) +[PDF](697.12KB)
Abstract:
A boolean network is a map \begin{document}$F:\{0,1\}^n \to \{0,1\}^n$\end{document} that defines a discrete dynamical system by the subsequent iterations of \begin{document}$F$\end{document}. Nevertheless, it is thought that this definition is not always reliable in the context of applications, especially in biology. Concerning this issue, models based in the concept of adding asynchronicity to the dynamics were propose. Particularly, we are interested in a approach based in the concept of delay. We focus in a specific type of delay called firing memory and it effects in the dynamics of symmetric (non-directed) conjunctive networks. We find, in the caseis in which the implementation of the delay is not uniform, that all the complexity of the dynamics is somehow encapsulated in the component in which the delay has effect. Thus, we show, in the homogeneous case, that it is possible to exhibit attractors of non-polynomial period. In addition, we study the prediction problem consisting in, given an initial condition, determinate if a fixed coordinate will eventually change its state. We find again that in the non-homogeneous case all the complexity is determined by the component that is affected by the delay and we conclude in the homogeneous case that this problem is PSPACE-complete.
2020, 40(10): 5795-5814 doi: 10.3934/dcds.2020246 +[Abstract](67) +[HTML](19) +[PDF](427.19KB)
Abstract:
By establishing Multiplicative Ergodic Theorem for commutative transformations on a separable infinite dimensional Hilbert space, in this paper, we investigate Pesin's entropy formula and SRB measures of a finitely generated random transformations on such space via its commuting generators. Moreover, as an application, we give a formula of Friedland's entropy for certain \begin{document}$C^{2}$\end{document} \begin{document}$\mathbb{N}^2$\end{document}-actions.
Li Ma and
2020, 40(10): 5815-5830 doi: 10.3934/dcds.2020247 +[Abstract](66) +[HTML](18) +[PDF](316.0KB)
Abstract:
The effects of weak and strong advection on the dynamics of reaction-diffusion models have long been investigated. In contrast, the role of intermediate advection still remains poorly understood. This paper is devoted to studying a two-species competition model in a one-dimensional advective homogeneous environment, where the two species are identical except their diffusion rates and advection rates. Zhou (P. Zhou, On a Lotka-Volterra competition system: diffusion vs advection, Calc. Var. Partial Differential Equations, 55 (2016), Art. 137, 29 pp) considered the system under the no-flux boundary conditions. It is pointed that, in this paper, we focus on the case where the upstream end has the Neumann boundary condition and the downstream end has the hostile condition. By employing a new approach, we firstly determine necessary and sufficient conditions for the persistence of the corresponding single species model, in forms of the critical diffusion rate and critical advection rate. Furthermore, for the two-species model, we find that (i) the strategy of slower diffusion together with faster advection is always favorable; (ii) two species will also coexist when the faster advection with appropriate faster diffusion.
2020, 40(10): 5831-5843 doi: 10.3934/dcds.2020248 +[Abstract](81) +[HTML](12) +[PDF](330.67KB)
Abstract:
We prove the existence of at least one positive solution for a Schrödinger equation in \begin{document}$\mathbb{R}^N$\end{document} of type
with a vanishing potential at infinity and subcritical nonlinearity \begin{document}$f$\end{document}. Our hypotheses allow us to consider examples of nonlinearities which do not verify the Ambrosetti-Rabinowitz condition, neither monotonicity conditions for the function \begin{document}$\frac{f(x, s)}{s}$\end{document}. Our argument requires new estimates in order to prove the boundedness of a Cerami sequence.
2020, 40(10): 5845-5868 doi: 10.3934/dcds.2020249 +[Abstract](90) +[HTML](23) +[PDF](4683.93KB)
Abstract:
In this paper, we propose and investigate a memory-based reaction-diffusion equation with nonlocal maturation delay and homogeneous Dirichlet boundary condition. We first study the existence of the spatially inhomogeneous steady state. By analyzing the associated characteristic equation, we obtain sufficient conditions for local stability and Hopf bifurcation of this inhomogeneous steady state, respectively. For the Hopf bifurcation analysis, a geometric method and prior estimation techniques are combined to find all bifurcation values because the characteristic equation includes a non-self-adjoint operator and two time delays. In addition, we provide an explicit formula to determine the crossing direction of the purely imaginary eigenvalues. The bifurcation analysis reveals that the diffusion with memory effect could induce spatiotemporal patterns which were never possessed by an equation without memory-based diffusion. Furthermore, these patterns are different from the ones of a spatial memory equation with Neumann boundary condition.
2020, 40(10): 5869-5895 doi: 10.3934/dcds.2020250 +[Abstract](106) +[HTML](24) +[PDF](925.6KB)
Abstract:
The centralizer (automorphism group) and normalizer (extended symmetry group) of the shift action inside the group of self-homeomorphisms are studied, in the context of certain \begin{document}$\mathbb{Z}^d$\end{document} subshifts with a hierarchical supertile structure, such as bijective substitutive subshifts and the Robinson tiling. Restrictions on these groups via geometrical considerations are used to characterize explicitly their structure: nontrivial extended symmetries can always be described via relabeling maps and rigid transformations of the Euclidean plane permuting the coordinate axes. The techniques used also carry over to the well-known Robinson tiling, both in its minimal and non-minimal versions.
2020, 40(10): 5897-5909 doi: 10.3934/dcds.2020251 +[Abstract](82) +[HTML](11) +[PDF](360.25KB)
Abstract:
We study KMS states for gauge actions with potential functions on Cuntz–Krieger algebras whose underlying one-sided topological Markov shifts are continuously orbit equivalent. As a result, we have a certain relationship between topological entropy of continuously orbit equivalent one-sided topological Markov shifts.
2020, 40(10): 5911-5928 doi: 10.3934/dcds.2020252 +[Abstract](48) +[HTML](15) +[PDF](388.36KB)
Abstract:
In this paper, an infinite dimensional KAM theorem with double normal frequencies is established under qualitative non-degenerate conditions. This is an extension of the degenerate KAM theorem with simple normal frequencies in [3] by Bambusi, Berti and Magistrelli. As applications, for nonlinear wave equation and nonlinear Schr\begin{document}$\ddot{\mbox{o}}$\end{document}dinger equation with periodic boundary conditions, quasi-periodic solutions of small amplitude and quasi-periodic solutions around plane wave are obtained respectively.
2020, 40(10): 5929-5954 doi: 10.3934/dcds.2020253 +[Abstract](82) +[HTML](21) +[PDF](598.66KB)
Abstract:
This paper is mainly concerned with the classification of the general two-component \begin{document}$\mu$\end{document}-Camassa-Holm systems with quadratic nonlinearities. As a conclusion of such classification, a two-component \begin{document}$\mu$\end{document}-Camassa-Holm system admitting multi-peaked solutions and \begin{document}$H^1$\end{document}-norm conservation law is found, which is a \begin{document}$\mu$\end{document}-version of the two-component modified Camassa-Holm system and can be derived from the semidirect-product Euler-Poincaré equations corresponding to a Lagrangian. The local well-posedness for solutions to the initial value problem associated with the two-component \begin{document}$\mu$\end{document}-Camassa-Holm system is established. And the precise blow-up scenario, wave breaking phenomena and blow-up rate for solutions of this problem are also investigated.
2020, 40(10): 5955-5972 doi: 10.3934/dcds.2020254 +[Abstract](80) +[HTML](11) +[PDF](368.8KB)
Abstract:
In this paper we consider viscosity solutions of a class of non-homogeneous singular parabolic equations
where \begin{document}$-1<\gamma<0$\end{document}, \begin{document}$1<p<\infty$\end{document}, and \begin{document}$f$\end{document} is a given bounded function. We establish interior Hölder regularity of the gradient by studying two alternatives: The first alternative uses an iteration which is based on an approximation lemma. In the second alternative we use a small perturbation argument.
2020, 40(10): 5973-5990 doi: 10.3934/dcds.2020255 +[Abstract](96) +[HTML](23) +[PDF](419.24KB)
Abstract:
We study the influence of Szegő projector on the \begin{document}$L^2-$\end{document}critical non linear focusing Schrödinger equation, leading to the quintic focusing NLS–Szegő equation on the line
It has no Galilean invariance but the momentum \begin{document}$P(u) = \langle -i\partial_x u, u\rangle_{L^2}$\end{document} becomes the \begin{document}$\dot{H}^{\frac{1}{2}}-$\end{document}norm. Thus this equation is globally well-posed in \begin{document}$H^1_+ = \Pi(H^1(\mathbb{R}))$\end{document}, for every initial datum \begin{document}$u_0$\end{document}. The solution \begin{document}$L^2-$\end{document}scatters both forward and backward in time if \begin{document}$u_0$\end{document} has sufficiently small mass. By using the concentration–compactness principle, we prove the orbital stability of some weak type of the traveling wave : \begin{document}$u_{\omega, c}(t, x) = e^{i\omega t}Q(x+ct)$\end{document}, for some \begin{document}$\omega, c>0$\end{document}, where \begin{document}$Q$\end{document} is a ground state associated to Gagliardo–Nirenberg type functional
for some \begin{document}$\gamma\geq 0$\end{document}. Its Euler–Lagrange equation is a non local elliptic equation. The ground states are completely classified in the case \begin{document}$\gamma = 2$\end{document}, leading to the actual orbital stability for appropriate traveling waves. As a consequence, the scattering mass threshold of the focusing quintic NLS–Szegő equation is strictly below the mass of ground state associated to the functional \begin{document}$I^{(0)}$\end{document}, unlike the recent result by Dodson [8] on the usual quintic focusing non linear Schrödinger equation.
2020, 40(10): 5991-6014 doi: 10.3934/dcds.2020256 +[Abstract](190) +[HTML](27) +[PDF](423.8KB)
Abstract:
The paper is devoted to investigating a semilinear parabolic equation with a nonlinear gradient source term:
where \begin{document}$p>m+2$\end{document}, \begin{document}$m\geq0$\end{document}. Zhang and Hu [Discrete Contin. Dyn. Syst. 26 (2010) 767-779] showed that finite time gradient blowup occurs at the boundary and the accurate blowup rate is also obtained for super-critical boundary value. Throughout this paper, we present a complete large time behavior of a classical solution \begin{document}$u$\end{document}: \begin{document}$u$\end{document} is global and converges to the unique stationary solution in \begin{document}$C^1$\end{document} norm for subcritical boundary value, and \begin{document}$u_x$\end{document} blows up in infinite time for critical boundary value. Gradient growup rate is also established by the method of matched asymptotic expansions. In addition, gradient estimate of solutions is obtained by the Bernstein-type arguments.
2020, 40(10): 6015-6041 doi: 10.3934/dcds.2020257 +[Abstract](67) +[HTML](16) +[PDF](913.98KB)
Abstract:
We investigate dynamical systems obtained by coupling two maps, one of which is chaotic and is exemplified by an Anosov diffeomorphism, and the other is of gradient type and is exemplified by a N-pole-to-S-pole map of the circle. Leveraging techniques from the geometric and ergodic theories of hyperbolic systems, we analyze three different ways of coupling together the two maps above. For weak coupling, we offer an addendum to existing theory showing that almost always the attractor has fractal-like geometry when it is not normally hyperbolic. Our main results are for stronger couplings in which the action of the Anosov diffeomorphism on the circle map has certain monotonicity properties. Under these conditions, we show that the coupled systems have invariant cones and possess SRB measures even though there are genuine obstructions to uniform hyperbolicity.
Xiao Wen and
2020, 40(10): 6043-6059 doi: 10.3934/dcds.2020258 +[Abstract](71) +[HTML](13) +[PDF](341.76KB)
Abstract:
We prove that every singular hyperbolic chain transitive set with a singularity does not admit the shadowing property. Using this result we show that if a star flow has the shadowing property on its chain recurrent set then it satisfies Axiom A and the no-cycle conditions; and that if a multisingular hyperbolic set has the shadowing property then it is hyperbolic.
2020, 40(10): 6061-6088 doi: 10.3934/dcds.2020259 +[Abstract](75) +[HTML](19) +[PDF](449.9KB)
Abstract:
We study the regularity of exceptional actions of groups by \begin{document}$C^{1, \alpha}$\end{document} diffeomorphisms on the circle, i.e. ones which admit exceptional minimal sets, and whose elements have first derivatives that are continuous with concave modulus of continuity \begin{document}$\alpha$\end{document}. Let \begin{document}$G$\end{document} be a finitely generated group admitting a \begin{document}$C^{1, \alpha}$\end{document} action \begin{document}$\rho$\end{document} with a free orbit on the circle, and such that the logarithms of derivatives of group elements are uniformly bounded at some point of the circle. We prove that if \begin{document}$G$\end{document} has spherical growth bounded by \begin{document}$c n^{d-1}$\end{document} and if the function \begin{document}$1/\alpha^d$\end{document} is integrable near zero, then under some mild technical assumptions on \begin{document}$\alpha$\end{document}, there is a sequence of exceptional \begin{document}$C^{1, \alpha}$\end{document} actions of \begin{document}$G$\end{document} which converge to \begin{document}$\rho$\end{document} in the \begin{document}$C^1$\end{document} topology. As a consequence for a single diffeomorphism, we obtain that if the function \begin{document}$1/\alpha$\end{document} is integrable near zero, then there exists a \begin{document}$C^{1, \alpha}$\end{document} exceptional diffeomorphism of the circle. This corollary accounts for all previously known moduli of continuity for derivatives of exceptional diffeomorphisms. We also obtain a partial converse to our main result. For finitely generated free abelian groups, the existence of an exceptional action, together with some natural hypotheses on the derivatives of group elements, puts integrability restrictions on the modulus \begin{document}$\alpha$\end{document}. These results are related to a long-standing question of D. McDuff concerning the length spectrum of exceptional \begin{document}$C^1$\end{document} diffeomorphisms of the circle.
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https://groupprops.subwiki.org/wiki/Balanced_subgroup_property_(function_restriction_formalism) | # Balanced subgroup property (function restriction formalism)
This article defines a subgroup metaproperty: a property that can be evaluated to true/false for any subgroup property
View a complete list of subgroup metaproperties
View subgroup properties satisfying this metaproperty| View subgroup properties dissatisfying this metaproperty
VIEW RELATED: subgroup metaproperty satisfactions| subgroup metaproperty dissatisfactions
BEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]
## Definition
### Symbol-free definition
A subgroup property is said to be a balanced subgroup property if it can be expressed via a function restriction expression with both the left side and the right side being equal.
### Definition with symbols
A subgroup property is said to be a balanced subgroup property if it can be expressed as $a \to a$ where $a$ is a function property. In other words, a subgroup $H$ satisfies the property in a group $G$ if and only if every function on $G$ satisfying property $a$ in $G$ restricts to a function satisfying property $a$ on $H$.
## Examples
### Characteristic subgroup
The property of a subgroup being characteristic is expressible as a balanaced subgroup property in the function restriction formalism as follows:
Automorphism $\to$ Automorphism
## Relation with other metaproperties
### T.i. subgroup properties
Clearly, any balanced subgroup property with respect to the function restriction formalism is both transitive and identity-true. Hence, it is a t.i. subgroup property.
Interestingly, a partial converse holds by the balance theorem: every t.i. subgroup property that can be expressed using the function restriction formalism, is actually a balanced subgroup property. In fact, more strongly, a balanced expression for the property can be obtained by using either the right tightening operator or the left tightening operator to any starting expression.
### Intersection-closedness
In general, a balanced subgroup property need not be intersection-closed.
### Join-closedness
In general, a balanced subgroup property need not be join-closed. | {"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": 16, "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": 0.9078460335731506, "perplexity": 1324.2715968457399}, "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-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00551.warc.gz"} |
http://www.maa.org/press/periodicals/convergence/connecting-greek-ladders-and-continued-fractions-matching-continued-fraction-convergents-to-greek | # Connecting Greek Ladders and Continued Fractions - Matching Continued Fraction Convergents to Greek Ladder Approximations
Author(s):
Kurt Herzinger (United States Air Force Academy) and Robert Wisner (New Mexico State University)
As we noted earlier, the simple continued fraction converging to $\sqrt{2}$ is $[1,2,2,2 \dots]$. It is easy to check that the $n$th convergent of this continued fraction is equal to the value of $\frac{y_n}{x_n}$ produced by the classic Greek ladder for $\sqrt{2}$ by hand (or computer) up to any $n$ we wish. A general proof for all $n$ is left to the reader but we should note that the proof is a special case of the induction proof we will demonstrate later in this section. Similarly, the simple continued fraction converging to $\sqrt{3}$ is $[1,1,2,1,2,1,2,\dots]$ and the sequence of convergents of this continued fraction is equal to the sequence of rational numbers produced by the classic Greek ladder for $\sqrt{3}$. We might expect this relationship to continue for $\sqrt{k}$ for any positive integer $k,$ but when $k = 5$ we see the simple continued fraction that converges to $\sqrt{5}$ is $[2,4,4,4,4\dots],$ which produces convergents $S_1 = 2,\ S_2 = \frac{9}{4}=2.25,\ S_3=\frac{38}{17}=2.23529,\ S_4=\frac{161}{72}=2.23611,\ \dots.$ However, the classic Greek ladder approximating $\sqrt{5}$ is:
$x_n$ $y_n$ ${y_n}/{x_n}$ $n=1$ $1$ $1$ $1$ $n=2$ $2$ $6$ $3$ $n=3$ $8$ $16$ $2$ $n=4$ $24$ $56$ $2.3333$ $\vdots$ $\vdots$ $\vdots$ $\vdots$
For $k \ge 5$ we know that the first convergent of the simple continued fraction for $\sqrt{k}$ is at least $2,$ whereas the classic Greek ladder always start with $1$. The reader is encouraged to investigate and attempt to prove that even adjusting the starting values of the Greek ladder so that $y_1/x_1 = S_1$ will not make the two sequences equal.
The question we pose is, "can we find a continued fraction converging to $\sqrt{k}$ such that the sequence of convergents is identical to the sequence of approximations that comes from the classic Greek ladder for all $k \ge 2$?" The answer to our question is, "yes." However, the continued fraction in question will not be a simple continued fraction. For example, $[1;4,2;4,2;4,2;\dots]$ is a continued fraction that converges to $\sqrt{5}$ and the sequence of convergents of this continued fraction is equal to the sequence $\frac{y_n}{x_n}$ created by the classic Greek ladder for $\sqrt{5}$.
Claim: We claim that the sequence of approximations created by the classic Greek ladder for $\sqrt{k}$ is identical to the convergents of the continued fraction
$[1; k-1,2;k-1,2;k-1,2;\dots].$
The proof will proceed by induction; however, the induction argument is somewhat more sophisticated than standard induction proofs that establish common identities such as
$1+2+\cdots+n = \frac{(n)(n+1)}{2}$
for all positive integers $n.$
We believe that students who have had some experience with basic induction arguments will find it enlightening to examine this proof. Such students might attempt this proof on their own before reading what follows. We believe this exercise will be useful as a project in an introductory proofs course or an introductory Combinatorics class studying recursion. For a more challenging project, instructors might require that students discover the continued fraction on their own. In this case, students will need to create several examples, look for patterns, and test conjectures before attempting a proof.
Proof of the claim: We start with the classic Greek ladder for $\sqrt{k}$:
$x_n$ $y_n$ ${y_n}/{x_n}$ $n=1$ $1$ $1$ $1$ $n=2$ $2$ $k+1$ $\frac{k+1}{2}$ $n=3$ $k+3$ $3k+1$ $\frac{3k+1}{k+3}$ $n=4$ $4k+4$ $k^2 +6k+1$ $\frac{k^2 +6k+1}{4k+4}$ $\vdots$ $\vdots$ $\vdots$ $\vdots$
Recall that $x_n = x_{n-1} + y_{n-1}$ and $y_n = kx_{n-1} + y_{n-1}$ for $n \ge 2.$
Now, consider the continued fraction:
$1+\frac{k-1}{2+\frac{k-1}{2 + \frac{k-1}{2+\frac{k-1}{\vdots}}}}$
For $n = 1$, we have $S_1 = 1 = \frac{y_1}{x_1}$.
For $n=2$, we have $S_2 = 1+\frac{k-1}{2} = \frac{k+1}{2} = \frac{y_2}{x_2}$.
This establishes the truth of our statement for $n=1$ and $n=2$ and hence the base cases for our induction argument. Before proceeding with the inductive step we take a moment to note something about the convergents of the continued fraction:
$S_3-1 = \frac{k-1}{2+\frac{k-1}{2}}=\frac{k-1}{{S_2}+1}$
$S_4-1 = \frac{k-1}{2+\frac{k-1}{2 + \frac{k-1}{2}}}=\frac{k-1}{{S_3}+1}$
$\vdots$
Following the discussion that led to Equation (1), we see that for $n\ge3$ we have,
$S_n-1 = \frac{k-1}{S_{n-1}+1}\quad{\rm{(Equation}}\,\,2).$
Now, our induction hypothesis states that for $n\ge3$ we assume that the $(n-1)$th convergent of the continued fraction is equal to $\frac{y_{n-1}}{x_{n-1}}$. That is, we assume that
$S_{n-1} = \frac{y_{n-1}}{x_{n-1}}.$
Then, from Equation (2) we have
$S_n-1 = \frac{k-1}{S_{n-1}+1}$
$= \frac{k-1}{\frac{y_{n-1}}{x_{n-1}}+1}$
$= \frac{k-1}{\frac{x_{n-1}+y_{n-1}}{x_{n-1}}}$
$= \frac{(k-1)x_{n-1}}{x_{n-1}+y_{n-1}}.$
Adding $1$ to both sides of the equation yields
$S_n = \frac{(k-1)x_{n-1}}{x_{n-1}+y_{n-1}}+1$
$= \frac{kx_{n-1}+y_{n-1}}{x_{n-1}+y_{n-1}}.$
Recalling the recursive definitions of $x_n$ and $y_n$ we have
$S_n = \frac{y_n}{x_n}.$
We have shown for $n\ge 3$ that if ${S_{n-1}}={\frac{y_{n-1}}{x_{n-1}}},$ then $S_n=\frac{y_n}{x_n}$. This completes the proof by induction that $S_n=\frac{y_n}{x_n}$ for all $n\ge1.$
The key step in the preceding proof was establishing the recursive relation in Equation (2). This relation is very similar to Equation (1). If students have seen the proof that $[1,2,2,\dots]$ converges to $\sqrt{2}$, then they may be on the look-out for a similar sort of relation. The instructor will need to be ready to help students realize Equation (2) in order to keep them moving forward on the proof.
Kurt Herzinger (United States Air Force Academy) and Robert Wisner (New Mexico State University), "Connecting Greek Ladders and Continued Fractions - Matching Continued Fraction Convergents to Greek Ladder Approximations," Convergence (January 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": 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": 0.9766071438789368, "perplexity": 164.77179599949602}, "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-2015-35/segments/1440645340161.79/warc/CC-MAIN-20150827031540-00009-ip-10-171-96-226.ec2.internal.warc.gz"} |
http://ima.umn.edu/2008-2009/T9.26-27.08/abstracts.html | HOME » SCIENTIFIC RESOURCES » Volumes
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Abstracts and Talk Materials
Mathematical and Computational Approaches to Quantum Chemistry
September 26 - 27, 2008
Quantum Chemistry aims at understanding the properties of matter through the modelling of its behaviour at a subatomic scale, where matter is described as an assembly of nuclei and electrons.
At this scale, the equation that rules the interactions between these constitutive elements is the Schrdinger equation. It can be considered (except in few special cases notably those involving relativistic phenomena or nuclear reactions) as a universal model for at least three reasons. First it contains all the physical information of the system under consideration so that any of the properties of this system can be deduced in theory from the Schrdinger equation associated to it. Second, the Schrdinger equation does not involve any empirical parameter, except some fundamental constants of Physics (the Planck constant, the mass and charge of the electron, ...); it can thus be written for any kind of molecular system provided its chemical composition, in terms of natures of nuclei and number of electrons, is known. Third, this model enjoys remarkable predictive capabilities, as confirmed by comparisons with a large amount of experimental data of various types.
Unfortunately, the Schrödinger equation cannot be directly simulated, except for very small chemical systems. It indeed reads as a time-dependent 3(M+N)-dimensional partial differential equation, where M is the number of nuclei and N the number of the electrons in the system under consideration. On the basis of asymptotic and semiclassical limit arguments, it is however often possible to approximate the Schrdinger dynamics by the so-called Born-Oppenheimer dynamics, in which nuclei behave as classical point-like particles. The internuclei (or interatomic) potential can be computed ab initio, by solving the time-independent electronic Schrödinger equation.
The latter equation is a 3N-dimensional partial differential equation (it is in fact a spectral problem), for which several approximation methods are available. The main of them are the wavefunction methods and the Density Functional Theory (DFT).
In my first lecture (Mathematical modelling of electronic structures), I will present the mathematical properties of the time-independent electronic Schrödinger equation, and show how to construct variational approximations of this equation, in the framework of wavefunction methods. I will mainly deal with the Hartree-Fock approximation; more advanced wavefunction methods will then be presented in the lectures by L. Slipchenko and A. Krylov.
In my second lecture (Mathematical aspects of density functional theory), I will examine the mathematical foundations of DFT. I will compare the constrained-search approach proposed by Levy and involving pure states, with the one proposed by Lieb and involving mixed states. These two approaches lead to Kohn-Sham and extended Kohn-Sham models respectively. I will then review the mathematical properties of the Kohn-Sham LDA and GGA models (corresponding to the first two rungs of the ladder of approximations previously presented by J. Perdew). Lastly, I will introduce the concept of bulk (or thermodynamic) limit, which allows one to rigorously derive DFT models for the condensed phase from molecular DFT models by letting the number of nuclei and electrons go to infinity in an appropriate way.
Coupled-cluster (CC) and equation-of-motion coupled-cluster (EOM-CC) methods are the most reliable and versatile tools of electronic structure theory. The exponential CC ansatz ensures size-extensivity. By increasing the excitation level, systematic approximations approaching the exact many-body solution are possible. EOM extends the CC methodology (applicable to the wave functions dominated by a single Slater determinant) to the open-shell and electronically excited species with multi-configurational wave-functions. The lecture will present an overview of CC and EOM-CC methods and highlight their important formal properties.
1. T. Helgaker, P. Jorgensen, and J. Olsen, Molecular electronic structure theory; Wiley & Sons, 2000.
2. A. I. Krylov, Equation-of-motion coupled-cluster methods for open-shell and electronically excited species: The hitchhiker's guide to Fock space Ann. Rev. Phys. Chem. v. 59, 433 (2008).
3. D. Mukherjee and S. Pal, Use of cluster expansion methods in the open-shell correlation problem, Adv. Quantum Chem. v. 20, 291 (1989).
4. R.J. Bartlett and J.F. Stanton, Applications of post-Hartree-Fock methods: A tutorial, Rev. Comp. Chem. v. 5, 65 (1994).
Density Functional Theory is one of the most successful approaches for computing the electronic structure of materials and is currently used to study thousand-atom systems today. The goals of this tutorial are two-fold. First, I will present the basic equations and ideas behind the solution of the many-body electronic Schrodinger equation through the Density Functional approach that leads to the Kohn-Sham equations. I will then discuss the most commonly used approximations and how they translate into the two main types of numerical algorithms used to solve the Kohn-Sham equations. In the second part of this talk, I will outline the major computational components of plane wave DFT codes, which represent the dominant approach for the simulation of materials science problems. Finally, I will discuss some of the computational challenges, including parallelization, of the resulting large-scale simulations.
Combined with 9/26 abstract.
Introduction to quantum mechanics
September 26, 2008
My task is to discuss the basic principles of Quantum Mechanics which are crucial for the electronic structure theory. The following topics will be covered: the correspondence principle which connects Classical Mechanics and Quantum Mechanics; the uncertainty principle and related questions; the superposition principle. We shall discuss the Hilbert space of wavefunctions, and the operators associated with the observables. We shall illustrate the theory by considering the properties of angular momentum (orbital and spin). The theory of hydrogen atom will constitute the important part of the lecture.
Electronic structure theory predicts what atoms, molecules and solids can exist, and with what properties. The density functional theory (DFT) of Hohenberg, Kohn and Sham 1964-65 is now the most widely used method of electronic structure calculation in both condensed matter physics and quantum chemistry. Walter Kohn and John Pople shared the 1998 Nobel Prize in Chemistry for this theory and for its computational implementation. This tutorial will begin by explaining why the computational demands are far smaller in DFT than in many-electron wavefunction theory, especially for large molecules and for solids. Two fundamental theorems of DFT will be presented and proven by the transparent constrained-search approach of Levy: (1) The ground-state electron density n of a system of N electrons in the presence of an external scalar multiplicative potential v determines v and hence all properties of the system. (2) There exists a universal density functional Q[n] such that minimization of Q[n] + for given N and v yields the ground-state density n and energy E. For accurate approximation of Q[n], one needs to introduce single-electron Kohn-Sham orbitals that yield the non-interacting kinetic energy part of Q[n] exactly, reducing the ground-state problem to a problem of noninteracting electrons moving in a selfconsistent density-dependent effective potential. Then the only part of Q[n] that must be approximated is the many-body exchange-correlation energy E_xc[n], which will be defined precisely. From the definition, many exact properties of E_xc[n] can be derived and employed as constraints to construct nonempirical or empirical approximations to E_xc[n]. Some important constraints will be reviewed. A ladder of approximations, from the simplest local density approximation to the most elaborate nonlocal approximations, will be reviewed, the surprising success of even the simplest one will be explained, and error estimates for each level will be presented. It will become apparent that local or semilocal approximations can suffice for most properties of most systems at or near the equilibrium nuclear positions, but that full nonlocality is needed to describe situations in which electrons are shared between separated subsystems, with noninteger average electron number in each subsystem. The exact density functional theory of such systems, as presented by Perdew, Parr, Levy and Balduz 1982, will be presented. The time-dependent density functional theory, which can describe time-dependent and excited states, will be briefly reviewed. Some outstanding physical and mathematical problems of DFT will be summarized.
Wave function methods in chemistry
September 26, 2008
After separating the electronic end nuclear coordinates through the Born-Oppenheimer approximation, one may attempt to solve the electronic Schrodinger equation by a hierarchy of wave function techniques. The lowest level in this hierarchy and the core method of the wave function quantum chemistry is the Hartree-Fock (HF) model, in which each electron moves in a mean field created by all other electrons. In order to get a chemical accuracy, one needs to correlate the motion of the electrons. This may be achieved by either the perturbation theory or by the configuration interaction procedure, employed on the top of the HF wave function. Basic ideas and approximations used in these wave function methods, as well as numerical approaches, challenges, and limitations will be discussed.
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http://www.gigahedron.de/wpgh/the-fractional-symmetric-rigid-rotor/ | # The fractional symmetric rigid rotor
author:
R. Herrmann
abstract:
Based on the Riemann fractional derivative the Casimir operators and multipletts for the fractional extension of the rotation group SO(n) are calculated algebraically. The spectrum of the corresponding fractional symmetric rigid rotor is discussed. It is shown, that the rotational, vibrational and $\gamma$-unstable limits of the standard geometric collective models are particular limits of this spectrum. A comparison with the ground state band spectra of nuclei shows an agreement with experimental data better than 2%. The derived results indicate, that the fractional symmetric rigid rotor is an appropriate tool for a description of low energy nuclear excitations. | {"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": 0.9562006592750549, "perplexity": 753.8423332550335}, "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-2022-40/segments/1664030337339.70/warc/CC-MAIN-20221002181356-20221002211356-00330.warc.gz"} |
https://keio.pure.elsevier.com/en/publications/experimental-study-and-prediction-by-computational-fluid-dynamics | Experimental study and prediction by computational fluid dynamics on self-induced sloshing due to bubble flow in a rectangular vessel
Research output: Contribution to journalArticlepeer-review
1 Citation (Scopus)
Abstract
Self-induced sloshing is an oscillatory phenomenon of a free liquid surface due to the flow of a fluid. This phenomenon has been reported in some gas–liquid reactors, and it is important to predict and prevent its occurrence for the safe operation of the reactors. However, the fundamental knowledge on the self-induced sloshing by bubble flow, such as the frequency and amplitude, is insufficient, and the occurrence condition has not been clarified. The purpose of this study is to investigate the characteristics of self-induced sloshing by bubble flow experimentally. We attempt to reproduce self-induced sloshing by using computational fluid dynamics (CFD) and establish a CFD model for the prediction of the occurrence of self-induced sloshing. In the experiments, air bubbles were dispersed into a liquid from the bottom of a rectangular vessel. The effects of the air flow rate and static liquid height on the characteristics of self-induced sloshing were investigated experimentally by image analysis. The occurrence of self-induced sloshing was confirmed by increasing the airflow rate at a specific static liquid height. The amplitude reached a maximum at the static liquid height, where self-induced sloshing was most likely to occur, and the frequency decreased with increasing static liquid height. Next, in order to reproduce the self-induced sloshing through CFD, an appropriate drag model of the bubbles was selected. Although the amplitude was overestimated due to the absence of the foam layer, the predicted frequency agreed well with the experimental value. Finally, the movement of the circulation flow was analyzed, and its correlation with the self-induced sloshing was clarified.
Original language English 51-57 7 JOURNAL OF CHEMICAL ENGINEERING OF JAPAN 54 2 https://doi.org/10.1252/JCEJ.20WE007 Published - 2021
Keywords
• Bubble Flow
• CFD
• Self-Induced Sloshing
ASJC Scopus subject areas
• Chemistry(all)
• Chemical Engineering(all)
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http://aas.org/archives/BAAS/v26n4/aas185/abs/S5205.html | The Globular Cluster System of the dE,N galaxy N3115 DW1
Session 52 -- Elliptical Galaxies
Display presentation, Tuesday, 10, 1995, 9:20am - 6:30pm
## [52.05] The Globular Cluster System of the dE,N galaxy N3115 DW1
Patrick R. Durrell, Dean E. McLaughlin, William E. Harris (McMaster U.), David A. Hanes (Queens U.)
A bright ($M_V\sim-17$) dE,N galaxy near the isolated S0 galaxy NGC 3115 has a surprisingly populous system of globular clusters. We have studied this system using $BV$ CCD photometry from CTIO. A population of $\sim$40 clusters is clearly seen above the background out to a distance of $\sim50^{\prime \prime}$ ($\sim 2$ kpc for $d=8$ Mpc) from the nucleus. The mean $(B-V)_o$ of the clusters is 0.75$\pm0.17$, indicating that most (if not all) of the clusters are genuinely $old$ globulars, and not a population of clusters of intermediate or young age. The observed colors then yield a mean [Fe/H] of $\sim-1.1$, which is more metal-poor (in the mean) than the dwarf itself, implying that the clusters formed during or before the bulk of the galaxy's stellar population. Nevertheless, this is a surprisingly metal-rich cluster system for a dwarf elliptical. | {"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": 0.9177101850509644, "perplexity": 4585.312133760849}, "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/1408500835699.86/warc/CC-MAIN-20140820021355-00151-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/151087-mann-whitney-rank-sum.html | ## Mann Whitney Rank Sum
Hello,
I'm using the 'Ranksum' formula in Matlab, which I believe is their equivalent of the Mann Whitney rank sum test. I am using frequency distribution data. Say I had two datasets as below:
(a)
x 1 2 3 4 5
a 2 3 4 0 0
b 3 4 2 1 1
(b)
x 1 2 3 4 5
a 2 3 4
b 3 4 2 1 1
where column x is an incremental value, a is a frequency distribution from sample 1 and b is a frequency distribution from sample 2. This is just junk data btw I haven't tried it on this. Basically the difference between the two datasets is that in (a) I have shown that the frequency distribution at 4 and 5 is 0, whereas in (b) I have shown it as NaN. As it is a rank sum test, this changes the result of the test, and consequently whether my actual data is significant or not! My question is, which is the better way to represent frequency distribution data in this instance? I suspect it is probably (b) because you are essentially distributing a set amount of points throughout x, and when there are no points in 4 and 5, this should be no data, but I just want to clarify if this is correct! The ranksum allows two datasets of different lengths, incidentally. | {"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": 0.8549990057945251, "perplexity": 305.9272890914725}, "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-2017-39/segments/1505818687820.59/warc/CC-MAIN-20170921134614-20170921154614-00628.warc.gz"} |
https://www.colorado.edu/physics/thomas-degrand | Professor
Physics
Office: DUAN F319
## Research Interests:
I study the properties of strongly-interacting systems, most of which appear in the context of elementary particle physics, with a combination of analytic and numerical techniques.
I am interested in the physics of strongly interacting quantum fields. The prototype of such system is Quantum Chromodynamics, the theory of quarks and gluons which describes the strong nuclear force. Succesful calculations in this area of science predict (or postdict) the properties of strongly interacting particles like the proton in terms of the properties of its constituents and of their interactions. Lately I have begun to apply these techniques to conjectured models of new physics beyond the Standard Model, which might be observed at the Large Hadron Collider. I use a mix of analytic techniques and numerical simulation on small and large computers in my research. In addition, I have an amateur interest in all areas of theoretical physics.
## Selected Publications:
1. T. DeGrand, "Lattice baryons in the 1/N expansion,'' Phys. Rev. D 86, 034508 (2012)[arXiv:1205.0235 [hep-lat]].
2. S. Catterall, P. H. Damgaard, T. DeGrand, R. Galvez and D. Mehta, "Phase Structure of Lattice N=4 Super Yang-Mills,'' JHEP 1211, 072 (2012) [arXiv:1209.5285 [hep-lat]].
3. T. DeGrand, Y. Shamir and B. Svetitsky, "Infrared fixed point in SU(2) gauge theory with adjoint fermions,'' Phys. Rev. D 83 (2011) 074507 [arXiv:1102.2843 [hep-lat]]. | {"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": 0.8926529288291931, "perplexity": 1244.966784190318}, "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-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00049.warc.gz"} |
http://3cq.org/standard-error/when-to-use-standard-deviation-or-standard-error.php | Home > Standard Error > When To Use Standard Deviation Or Standard Error
# When To Use Standard Deviation Or Standard Error
## Contents
Sep 16, 2013 All Answers (9) Eik Vettorazzi · University Medical Center Hamburg - Eppendorf Hi Jasmine, this is already discussed in https://www.researchgate.net/post/Which_of_the_following_measures_is_better_to_show_the_differences cheers Sep 16, 2013 Gregory Verleysen · University Subscribe to R-bloggers to receive e-mails with the latest R posts. (You will not see this message again.) Submit Click here to close (This popup will not appear again) Warning: The Contrary to popular misconception, the standard deviation is a valid measure of variability regardless of the distribution. Note: The Student's probability distribution is a good approximation of the Gaussian when the sample size is over 100. this content
The mean of these 20,000 samples from the age at first marriage population is 23.44, and the standard deviation of the 20,000 sample means is 1.18. Two data sets will be helpful to illustrate the concept of a sampling distribution and its use to calculate the standard error. The goal should be to have one standard method to describe the distribution of a study sample, thereby reducing confusion among the readers of biomedical research papers. It seems from your question that was what you were thinking about. http://bja.oxfordjournals.org/content/90/4/514.long
## Standard Error And Standard Deviation Difference
Anaesthesia journals could easily avoid this statistical error by requiring authors to adhere to statistical recommendations, for instance through a more stringent statistical review process. If it is large, it means that you could have obtained a totally different estimate if you had drawn another sample. Review authors should look for evidence of which one, and might use a t distribution if in doubt.
How are they different and why do you need to measure the standard error? We observe the SD of $n$ iid samples of, say, a Normal distribution. The following expressions can be used to calculate the upper and lower 95% confidence limits, where x ¯ {\displaystyle {\bar {x}}} is equal to the sample mean, S E {\displaystyle SE} Error And Deviation In Chemistry Basic concepts of statistical reasoning: standard errors and confidence intervals.
Learn R R jobs Submit a new job (it's free) Browse latest jobs (also free) Contact us Welcome! Standard Error Vs Standard Deviation Example This use is not only statistically inappropriate, it also makes the reader assume a much smaller variability of the sample. For any random sample from a population, the sample mean will usually be less than or greater than the population mean. https://www.r-bloggers.com/standard-deviation-vs-standard-error/ share|improve this answer edited Jun 10 at 14:30 Weiwei 48228 answered Jul 15 '12 at 13:39 Michael Chernick 25.8k23182 2 Re: "...consistent which means their standard error decreases to 0"
The ages in one such sample are 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. Standard Error Matlab It is useful to compare the standard error of the mean for the age of the runners versus the age at first marriage, as in the graph. Cryptic message How to defeat the elven insects using modern technology? The aim of this study was to evaluate the frequency of inappropriate use of the sem in four leading anaesthesia journals in 2001.
## Standard Error Vs Standard Deviation Example
As the sem is always less than the sd, it misleads the reader into underestimating the variability between individuals within the study sample. http://www.graphpad.com/guides/prism/6/statistics/stat_standard_deviation_and_standar.htm Excluded from this analysis were articles using median and range, and articles that solely used inferential statistics such as confidence intervals (CI). Standard Error And Standard Deviation Difference Common mistakes in interpretation Students often use the standard error when they should use the standard deviation, and vice versa. Standard Error In R The standard deviation (sd) describes the variability between individuals in a sample; the standard error of the mean (sem) describes the uncertainty of how the sample mean represents the population mean.
The age data are in the data set run10 from the R package openintro that accompanies the textbook by Dietz [4] The graph shows the distribution of ages for the runners. http://3cq.org/standard-error/when-to-use-standard-error-versus-standard-deviation.php BMJ 1994;309: 996. [PMC free article] [PubMed]4. The term may also be used to refer to an estimate of that standard deviation, derived from a particular sample used to compute the estimate. Another way of considering the standard error is as a measure of the precision of the sample mean.The standard error of the sample mean depends on both the standard deviation and Standard Error In Excel
The two can get confused when blurring the distinction between the universe and your sample. –Francesco Jul 15 '12 at 16:57 Possibly of interest: stats.stackexchange.com/questions/15505/… –Macro Jul 16 '12 Philadelphia: Churchill Livingston, 2000; 753–92 ↵ Streiner DL. The standard error is used to construct confidence intervals. have a peek at these guys J Paediatr Child Health 2000; 36: 502–5 CrossRefMedlineWeb of Science ↵ Webster CS, Merry AF.
In general, the use of the sem should be limited to inferential statistics where the author explicitly wants to inform the reader about the precision of the study, and how well Standard Error Vs Standard Deviation Excel Case reports and review articles were not considered, except for case series comprising several cases and using descriptive statistics to describe the study sample. Two sample variances are 80 or 120 (symmetrical).
## Sokal and Rohlf (1981)[7] give an equation of the correction factor for small samples ofn<20.
A total of 257 articles fulfilled the search criteria in Anesthesiology, 405 articles in Anesthesia & Analgesia, 137 in the British Journal of Anaesthesia, and 61 in the European Journal of All Rights Reserved. You pay me a dollar if I'm correct, otherwise I pay you a dollar. (With correct play--which I invite you to figure out!--the expectation of this game is positive for me, Standard Deviation Of The Mean The graph shows the ages for the 16 runners in the sample, plotted on the distribution of ages for all 9,732 runners.
The mean of all possible sample means is equal to the population mean. Related Content Load related web page information Share Email this article CiteULike Delicious Facebook Google+ Mendeley Twitter What's this? The distribution of these 20,000 sample means indicate how far the mean of a sample may be from the true population mean. http://3cq.org/standard-error/why-is-standard-error-smaller-than-standard-deviation.php This gives 9.27/sqrt(16) = 2.32.
J. Philadelphia: American College of Physicians, 1997 ↵ Carlin JB, Doyle LW. This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called BMJ 1995;310: 298. [PMC free article] [PubMed]3.
Methods. The standard deviation (sd) describes the variability between individuals in a sample; the standard error of the mean (sem) describes the uncertainty of how the sample mean represents the population mean. Most confidence intervals are 95% confidence intervals. Standard error of the mean (SEM) This section will focus on the standard error of the mean.
It makes them farther apart. Mahajan View full editorial board For Authors Instructions to authors Online submission Submit Now! | {"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": 0.804413914680481, "perplexity": 1104.2363414860636}, "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-2018-05/segments/1516084890928.82/warc/CC-MAIN-20180121234728-20180122014728-00771.warc.gz"} |
http://mathoverflow.net/questions/54975/when-is-a-finitely-generated-group-finitely-presented/54978 | # When is a finitely generated group finitely presented?
I think the question is very general and hard to answer. However I've seen a paper by Baumslag ("Wreath products and finitely presented groups", 1961) showing, as a particular case, that the lamplighter group is not finitely presented. To prove this, he gives conditions to say if a wreath product of groups is finitely presented. The question is: which ways (techniques, invariants, etc) are available to determine whether a finitely generated group is also finitely presented? For instance, is there another way to show that fact about the lamplighter group?
Thanks in advance for references and comments.
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This question seems like it's asking for a list. If it has no single correct answer, it should probably be Community Wiki. – HJRW Feb 10 '11 at 2:41
Maybe this is too basic, but algebraic (ex. nilpotent) and geometric (ex. hyperbolic) properties can give f.p. automatically. – Steve D Feb 10 '11 at 4:48
## 2 Answers
One general method is to consider an infinite presentation of the group, and then show that every finite subset of the set of relations defines a group with clearly different property. for example, the lamplighter group has the presentation $\langle \ldots a_{-n}, \ldots, a_1, a_2, \ldots, a_n,\ldots,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$. Every finite subpresentation defines a group that has as a quotient one of the following groups $H_n=\langle a_{-n}, \ldots, a_1, a_2, \ldots, a_n,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$ for some $n$. The group $H_n$ is an HNN extension of a finite Abelian group $\langle a_{-n},\ldots, a_n\rangle$ with the free letter $t$. Hence $H_n$ is a virtually free group, in particular, $H_n$ contains a non-Abelian free subgroup. Therefore every finite subpresentation defines a group containing a free non-Abelian subgroup, while the Lamplighter group is solvable and thus cannot contain a free non-Abelian subgroup. Similarly lacunary hyperbolic but not hyperbolic groups given by presentations satisfying small cancelation conditions or their generalizations are infinitely presented since every finite subpresentation of their presentation defines a hyperbolic group.
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Can this type of argument be used to show that the group has no finite presentation? (Or does it only show what it seems to show, namely that the group has no finite sub-presentation of a given infinite presentation.) – aaron Feb 10 '11 at 13:26
If a finitely generated group $G$ is finitely presented with finite set of relators $Q$, then every infinite presentation $R$ has a finite subset that is also a presentation. Indeed, consider any proof of $Q$ using $R$. It involves only finite subset $R'$ of $R$. The relations from $R'$ define the group $G$. Indeed, let $G'$ be the group defined by $R'$. Then the identity map on the generating set induces a hom. $\phi: G'\to G$. All relations of $Q$ hold in $G'$, so the kernel of $\phi$ is trivial, and $G'=G$. – Mark Sapir Feb 10 '11 at 13:36
... of course, you need to use the fact that if $G$ has finite presentation with one generating set, then for any other finite generating set, it also has finite presentation (just rewrite all relations in the new generating set). – Mark Sapir Feb 10 '11 at 14:27
@Mark: Very nice. Thanks. – aaron Feb 10 '11 at 21:27
An often-used method is to compute $H_2$. If the group is finitely presentable then $H_2$ is of finite rank with any coefficients.
For instance, you can use this technique to show that if $q:F\to\mathbb{Z}$ is the map from the free group of rank two that sends both generators to one then the fibre product $H\subseteq F\times F$, ie $(q\times q)^{-1}$ of the diagonal, is infinitely presented.
A famous theorem of Bestvina and Brady shows that this doesn't always work: they give a similar example which is infinitely presented but has finite-rank $H_2$.
A related technique shows that this question is indeed `very hard'. Grunewald showed that the fibre product coming from a surjection $f:F\to Q$ is finitely presented if and only $Q$ is finite. It follows that you cannot in general tell if a recursively presented group is (in)finitely presented.
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...with any small choice of coefficients. – Mariano Suárez-Alvarez Feb 10 '11 at 3:40
Mariano - right. I suppose I want the coefficient module itself to be finitely generated. – HJRW Feb 10 '11 at 14:56 | {"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": 0.9526816606521606, "perplexity": 263.4656176615448}, "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-2015-32/segments/1438042992543.60/warc/CC-MAIN-20150728002312-00099-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://karatoyama.iinaa.net/s-tuji.htm | # o
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https://simple.m.wikipedia.org/wiki/Radioactive_decay | emission of subatomic particles by the decay of an atomic nucleus
Radioactive decay happens to some chemical elements. Most chemical elements are stable. Stable elements are made up of atoms that stay the same. Even in a chemical reaction, the atoms themselves do not change ever.
The trefoil symbol is used to indicate radioactive material.
In the 19th century, Henri Becquerel discovered that some chemical elements have atoms that change over time. In 1898, Marie and Pierre Curie called this phenomenon radioactive decay.[1] Becquerel and the Curies were awarded the Nobel Prize in Physics for this discovery, in 1903.
## Example
Most carbon atoms have six protons and six neutrons in their nucleus. This carbon is called carbon-12s (six protons + six neutrons = 12). Its atomic weight is 12. If a carbon atom has two more neutrons it is carbon-14. Carbon-14 acts chemically like other carbon, because the six protons and six electrons are what govern its chemical properties. In fact, carbon-14 exists in all living things; all plants and animals contain carbon-14. However, carbon-14 is radioactive. It decays by beta decay to become nitrogen-14. Carbon-14, in the small amounts found about us in nature, is harmless. In archeology, this kind of carbon is used to determine the age of wood and other formerly living things. The method is called radiocarbon dating.
## Different kinds of decay
Ernest Rutherford found that there are different ways in which these particles penetrate matter. He found two different kinds, which he called alpha decay and beta decay.[2] Paul Villard discovered a third kind in 1900. Rutherford called it gamma decay, in 1903.[3]
The change from radioactive carbon-14 to stable nitrogen-14 is a radioactive decay. It happens when the atom emits an alpha particle. An alpha particle is a pulse of energy as an electron or positron leaves the nucleus.
Other kinds of decay were discovered later. The types of decay are different from each other because different types of decay produce different kinds of particles. The starting radioactive nucleus is called the parent nucleus and the nucleus that it changes into is called the daughter nucleus. The high-energy particles produced by radioactive materials are called radiation.
These various kinds of decay can happen sequentially in a "decay chain". One kind of nucleus decays to another kind, which decays again to another and so on until it becomes a stable isotope and the chain comes to an end.
## Speed of decay
The speed at which this change happens, is different for each element. Radioactive decay is governed by chance: The time it takes, on average for half the atoms of a substance to change is named the half-life. The rate is given by an exponential function. As an example, iodine (131I) has a half-life of about 8 days. That of plutonium ranges between 4 hours (243Pu) and 80 million years[4] (244Pu)
## Nuclear transformations and energy
Radioactive decay changes an atom from one that has higher energy inside its nucleus into one with lower energy. The change of energy of the nucleus is given to the particles that are created. The energy released by radioactive decay may either be carried away by a gamma ray electromagnetic radiation (a type of light), a beta particle or an alpha particle. In all those cases, the change of energy of the nucleus is carried away. And in all those cases, the total number of positive and negative charges of the atom's protons and electrons sum to zero before and after the change.
## Alpha decay
During alpha decay, the atomic nucleus releases an alpha particle. Alpha decay causes the nucleus to lose two protons and two neutrons. Alpha decay causes the atom to change into another element, because the atom loses two protons (and two electrons). For example, if Americium were to go through alpha decay it would change into Neptunium because Neptunium is defined by having two protons fewer than Americium. Alpha decay usually happens in the most heavy elements, such as uranium, thorium, plutonium, and radium.
Alpha particles cannot even go through a few centimeters of air. Alpha radiation cannot hurt humans when the alpha radiation source is outside the human body, because human skin does not let the alpha particles go through. Alpha radiation can be very harmful if the source is inside the body, such as when people breathe dust or gas containing materials which decay by emitting alpha particles (radiation).
## Beta decay
There are two kinds of beta decay, beta-plus and beta-minus.
In beta-minus decay, the nucleus gives out a negatively charged electron and a neutron changes into a proton:
${\displaystyle n^{0}\rightarrow p^{+}+e^{-}+{\bar {\nu }}_{e}}$ .
where
${\displaystyle n^{0}}$ is the neutron
${\displaystyle \ p^{+}}$ is the proton
${\displaystyle e^{-}}$ is the electron
${\displaystyle {\bar {\nu }}_{e}}$ is the anti-neutrino
Beta-minus decay happens in nuclear reactors.
In beta-plus decay, the nucleus releases a positron, which is like an electron but it is positively charged, and a proton changes into a neutron:
${\displaystyle \ p^{+}\rightarrow n^{0}+e^{+}+{\nu }_{e}}$ .
where
${\displaystyle \ p^{+}}$ is the proton
${\displaystyle n^{0}}$ is the neutron
${\displaystyle e^{+}}$ is the positron
${\displaystyle {\nu }_{e}}$ is the neutrino
Beta-plus decay happens inside the sun and in some types of particle accelerators.
## Gamma decay
Gamma decay happens when a nucleus produces a high-energy packet of energy called a gamma ray. Gamma rays do not have electrical charge, but they do have angular momentum. Gamma rays are usually emitted from nuclei just after other types of decay. Gamma rays can be used to see through material, to kill bacteria in food, to find some types of disease, and to treat some kinds of cancer. Gamma rays have the highest energy of any electromagnetic wave, and gamma ray bursts from space are the most energetic releases of energy known.
## References
1. Pierre Curie, Marie Curie, G. Bémont (1898). "Sur une nouvelle substance fortement radio-active contenue dans la pechblende". Comptes rendus hebdomadaires des séances de l'Académie des sciences (in French) (127): 1215–1217.CS1 maint: multiple names: authors list (link)
2. Ernest Rutherford (1899). "Uranium radiation and the electrical conduction produced by It". Philosophical Magazine. 47 (5): 116. More than one of |number= and |issue= specified (help)
3. Ernest Rutherford (1903). "The magnetic and electric deviation of the easily absorbed rays from Radium" (PDF). Philosophical Magazine. 5 (6): 177. doi:10.1080/14786440309462912. More than one of |number= and |issue= specified (help)
4. D. C. Hoffman, F. O. Lawrence, J. L. Mewherter, F. M. Rourke (1971). "Detection of Plutonium-244 in Nature". Nature (234): 132–134. doi:10.1038/234132a0.CS1 maint: multiple names: authors list (link). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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": 0.892636775970459, "perplexity": 1283.670875633952}, "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/1634323585203.61/warc/CC-MAIN-20211018124412-20211018154412-00252.warc.gz"} |
http://mathhelpforum.com/calculus/88323-lagrange-multiplier-problem.html | # Math Help - Lagrange Multiplier problem
1. ## Lagrange Multiplier problem
Can someone explain me step by step how to solve this problem
Minimize
with constraint
Find Partial Derivatives
Fx
Fy
FLambda
F acheives its minimum at
x=
y=
lambda=
f=
2. Originally Posted by v3ndetta
Can someone explain me step by step how to solve this problem
Minimize
with constraint
Find Partial Derivatives
Fx
Fy
FLambda
F acheives its minimum at
x=
y=
lambda=
f=
$F(x,y,\lambda)=-9x^2-2xy+y^2+1-\lambda(6x-y-12)$ | {"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": 1, "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": 0.9810800552368164, "perplexity": 3099.287039361547}, "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-2014-10/segments/1394678704362/warc/CC-MAIN-20140313024504-00061-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://physics.stackexchange.com/tags/wave-particle-duality/new | Tag Info
-1
There is only one logical explanation: the diffraction rings appear, one way or other, because of the positive ions inevitably present in a metal lattice strafed with energetic particles. Firstly this variant should be verified by diffractograms through earthing thin metal foils (or even through metal foils connected to a weak source of free electrons), ...
2
Bragg appears to to try to explain wave-particle duality by waving his hands rapidly. The moment of observation (the present) is now. Extrapolating backwards in time from that moment (when I observe a particle) I can deduce that this thing I observed was a particle in the past. But I cannot deduce much about its future - the further out you try to predict, ...
1
What Bragg meant by now, I cannot know. There is a huge amount of writers who meant all sort of things, and it's impossible to know all these views. The question whether the quantum system (quantum particle) is a wave or is a particle, preoccupies the physicists even today, 91 years after de Broglie formulated his $\lambda = h/p$ formula. You see, both ...
0
First at all light is electromagnetic radiation. To produce EM radiation you need excited electrons (or protons or some of this stuff), they emit photons. The light you see is thermal radiation from sun, laser, LED, light bulb. Then ever one detect light in detail he will end with the detection of photons. The first who claimed that light consists of quanta ...
0
I refer to your main questions, 1) the function $ξ$ I have written above is the wave function $\psi$ from the quantum mechanics with $s$ acting as $x$ (in $\psi$)? If not, then what is the relation between them? If you would have written the function $ξ$ correctly as in the article about the 2slit experiment, yes, the amplitude on the screen were given ...
1
The functions you write down are solutions of Maxwell's equations (if you think of them as lone, Cartesian components) and, as such, have an exact relationship the one-photon quantum state of the quantum photon field. Now, whether this is a photon wave function depends on your definitions. If you want to write down the quantum state of a one-photon, so ...
-6
The experiment is meant to show the wave-particle duality of light. When light goes through a slit, it is attracted gravitationally by the side-walls of the slit and seems to curve around the exit corners. Since light is radially projected, some light smears along the sidewall of the slit also and is refracted, though this amount of light is negligible when ...
1
She means to say that it would behave as if both the slits were open and fall on the screen according to the double slit pattern. The entire double slit pattern comes from interference.
1
The radiative transfer equation is a simplified model for describing light transfer. Of course it is possible to derive the radiative transfer equation by the Boltzmann equation for a photon density function $f(x,t)$: $$\partial_t f(x,t) + v_x \partial_x f(x,t) = (\partial_t f(x,t))_{coll}.$$ Here, the term $(\partial_t f(x,t))_{coll}$ is the gain and ...
Top 50 recent answers are included | {"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": 0.8799808621406555, "perplexity": 432.65667932472167}, "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-2015-11/segments/1424936464840.47/warc/CC-MAIN-20150226074104-00141-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://itprospt.com/num/3465432/sketch-each-of-the-images-under-the-specified-transformation | 5
# Sketch each of the images under the specified transformation_Tis the expansion and contraction represented by Tlx; y) = (ix2y)...
## Question
###### Sketch each of the images under the specified transformation_Tis the expansion and contraction represented by Tlx; y) = (ix2y)
Sketch each of the images under the specified transformation_ Tis the expansion and contraction represented by Tlx; y) = (ix2y)
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Hypothesis Tests About Proportion/Mean Define the population parameter or distribution about which hypotheses are t0 be tested: The Null Hypothesis (Ho): 0 = The Alternative Hypothesis (Ha): M < M > p # P > P = p = Significance level: 0.05 The Test Statistic b: z = Pal_Pa X - Lo a. t = s/Vn... | {"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": 0.9216862916946411, "perplexity": 1641.8233764495028}, "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-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00513.warc.gz"} |
https://socratic.org/questions/what-is-the-name-of-the-compound-cr-2-co-3-3 | Chemistry
Topics
# What is the name of the compound Cr_2(CO_3)_3?
Apr 24, 2016
Chromium(III) carbonate
#### Explanation:
In order to name an ionic compound, you must identify the cation, which is the positively charged ion, and the anion, which is the negatively charged ion.
Cations are always written first in the chemical formula of an ionic compound, followed by the anions.
"Cr"_ color(red)(2) ("CO"_ 3)_color(blue)(3)
It's important to remember that ionic formulas are written using the crisscross rule, which states that when a cation and an anion form an ionic compound, the charge on the cation becomes the subscript of the anion and the charge of the anion becomes the subscript of the cation in the chemical formula of the compound.
In your case, the compound contains chromium, $\text{Cr}$, as its cation. Since it has a subscript of $\textcolor{red}{2}$, it follows that the charge of the anion must be equal to $\textcolor{red}{2 -}$.
Now, the anion is actually a polyatomic ion called the carbonate anion. Notice that it's written between parentheses, which tells you that it contains one atom of carbon and three atoms of oxygen, and that it has a subscript of $\textcolor{b l u e}{3}$.
This means that the charge on the cation will be equal to $\textcolor{b l u e}{3 +}$.
This means that the ions that make up this compound are
$\textcolor{red}{2} \times \text{Cr"^(color(blue)(3+)" }$ and " " color(blue)(3) xx "CO"_3^(color(red)(2-)
Because chromium is a transition metal, which means that it can have multiple oxidation states, you must use Roman numerals to express its oxidation state in the compound.
Since it carries a $\textcolor{b l u e}{3 +}$ charge, it follows that you must use the (III) Roman numeral.
The cation will thus be chromium(III) and the anion will be the carbonate anion, which means that the compound's name will be
chromium(III) carbonate -> "Cr"_2("CO"_3)_2
##### Impact of this question
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http://mathhelpforum.com/trigonometry/135177-smallest-p-value-sin.html | # Thread: smallest "p" value for sin
1. ## smallest "p" value for sin
If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees?
How can I show that it really is 360?
2. Hello donnagirl
Originally Posted by donnagirl
If p is a positive real number and 2 sin x = 2 sin(x + p) for every real value of x, what is the smallest possible value for p, in degrees?
How can I show that it really is 360?
Divide by $2$; then expand $\sin (x+p)$, and say:
$\sin x \equiv \sin(x+p)$
$\Rightarrow \sin x \equiv \sin x \cos p + \cos x \sin p$
Then compare the coefficients of $\sin x$ and $\cos x$:
$\Rightarrow \left\{ \begin{array}{l} \cos p =1\\ \sin p = 0\end{array}\right .$
And the smallest positive value of $p$ that satisfies these equations is $p = 360^o$
3. Hmmm, I still don't see it Grandad--how did you obtain the coefficients values?
4. Hello donnagirl
Originally Posted by donnagirl
Hmmm, I still don't see it Grandad--how did you obtain the coefficients values?
You'll see that I've used a $\equiv$ sign, rather than a $=$ sign, to show that the equation is true for all values of $x$. This means that we can say that the coefficients of $\sin x$ and $\cos x$ on both sides will be the same. So:
$\sin x \equiv \sin x \cos p + \cos x \sin p$
$\Rightarrow \color{red}1\color{black}\sin x+\color{red}0\color{black}\cos x \equiv \color{red}\cos p\color{black}\sin x + \color{red}\sin p\color{black}\cos x$
$\Rightarrow \left\{\begin{array}{l}\cos p = 1\\ \sin p = 0\end{array}\right .$
OK now? | {"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": 17, "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": 0.8996433615684509, "perplexity": 389.7770866042896}, "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-2013-48/segments/1386163047545/warc/CC-MAIN-20131204131727-00056-ip-10-33-133-15.ec2.internal.warc.gz"} |
http://16pi2.com/maxwell-s-equations/46-on-electromagnetic-induction | ## Electromagnetic Momentum of a Current
(22) We may begin by considering the state of the field in the neighbourhood of an electric current. We know that magnetic forces are excited in the field, their direction and magnitude depending according to known laws upon the form of the conductor carrying the current. When the strength of the current is increased, all the magnetic effects are increased in the same proportion. Now if the magnetic state of the field depends on motions of the meditum, a certain force must be exerted in order to increase or diminish these motions, and when the motions are excted they continue, so that the effect of the connexion between the current and the electromagnetic field surrounding it, is to endow the current with a kind of momentum, just as the connexion between the driving-point of a machine and a fly-wheel endows the driving-point with an additional momentum, which may be called the momentum of the fly-wheel reduced to the driving-point. The unbalanced force acting on the driving-point increases this momentum, and is measured by the rate of its increase.
In the case of electric currents, the resistance to sudden increase or diminution of strength produces effects exactly like those of momentum, but the amount of this momentum depends on the shape of the conductor and the relative position of its different parts.
The momentum described here is provided by the primary angular momentum of the electron and photon. It is seen in the APM that what we consider at the macro level of existence (that angular momentum is a property of something), at the quantum level angular momentum is a thing itself. It is because electrons and photons are primary angular momentum that they can appear as either waves or particles. The structures of the electron and photon are explained in Secrets of the Aether.
Primary angular momentum is a more primary level of existence than are atoms. The property of solidity of solid particles only begins at the level of bound particles (such as atoms). More primary existences such as electrons and photons appear less solid, yet they still possess (and are) angular momentum. The Aether is even more primary than electrons and protons, and there are yet more primary levels of existence than Aether, such as the Gforce and dark matter.
Maxwell's point is that current has momentum and there is a delay in its effect upon the magnetic field when the current is changed in magnitude.
### Mutual Action of two Currents
(23) If there are two electric currents in the field, the magnetic force at any point is that compounded of the forces due to each currently separately, and since the two currents are in connextion with every point of the field, they will be in connexion with each other, so that any increase or diminution of the one will produce a force acting with or contrary to the other.
### Dynamical Illustration of Reduced Momentum.
(24) As a dynamical illustration, let us suppose a body C so connected with two independent driving-points A and B that its velocity is $$p$$ times that of A together with $$q$$ times that of B. Let $$u$$ be the velocity of A, $$v$$ that of B, and $$w$$ that of C, and let $$\delta x$$, $$\delta y$$, $$\delta z$$ be their simultaneous displacements, then by the general equation of dynamics[1],
$$C\frac{dw}{dt}\delta z=X\delta x+ Y\delta y$$
where X and Y are the forces acting at A and B.
But
$$\frac{dw}{dt}=p\frac{du}{dt}+q\frac{dv}{dt}$$
and
$$\delta z=p\delta x + q\delta y$$
Substituting, and remembering that $$\delta x$$ and $$\delta y$$ are independent,
$$\begin{array}\\X =\frac{d}{dt}(Cp^{2}u+Cpqv)\\ Y=\frac{d}{dt}(Cpqu+Cq^{2}v) \end{array}$$
We may call Cp2u+Cpqv the momentum of C referred to A, and Cpqu+Cq2v its momentum referred to B; then we may say that the effect of the force X is to increase the momentum of C referred to A, and that of Y to increase its momentum referred to B.
If there are many bodies connected with A and B in a similar way but with different values of p and q, we may treat the question in the same way by assuming
$$L=\sum(Cp^{2}), M=\sum(Cpq), and N=\sum(Cq^{2}),$$
where the summation is extended to all the bodies with their proper values of C, p, and q. Then the momentum of the system referred to A is
and referred to B,
and we shall have
where X and Y are the external forces on A and B.
The above is a mechanical analogy that Maxwell provides for quantifying reduced momentum. By Maxwell's definition, A and B are already velocities because p times A and q times B are velocities and p and q are dimensionless. His initial logic is flawed in that p and q are magnitudes of "driving- points" A and B. There is no scientific basis for separating a magnitude from its unit. The variables p and q cannot be used for defining L, M, N based upon the logic presented. Regardless, Maxwell clearly intended L, M, and N to be units of mass.
The "driving points" Maxwell are talking about would be the points of connection between matter and Aether. Matter is inherently encapsulated by Aether. The two meet at a boundary, which quantifies as the unit of conductance. Conductance is a property of all surfaces, whether the surfaces are of matter or space. Actually, conductance is strictly a property of the surface of space. The material counterpart to conductance is magnetic flux. Magnetic flux is reciprocal to conductance in the Aether Physics Model.
If the velocity of A be increased at the rate du/dt, then in order to prevent B from moving a force, ?=d/dt(Mu) must be applied to it.
This effect on B, due to an increase of the velocity of A, corresponds to the electromotive force on one circuit arising from an increase in the strength of a neighbouring circuit.
This dynamical illustration is to be considered merely as assisting the reader to understand what is meant in mechanics by Reduced Momentum. The facts of the induction of currents as depending on the variations of the quantity called Electromagnetic Momentum, or Electrotonic State, rest on the experiments of Faraday[2], Felici16, &c.
One cannot simply discuss electrical processes, which are based upon dimensions of charge, in terms of mechanical processes, which are based upon units of angular momentum. In order to equate mechanical processes with electrical processes, one must directly correlate magnetic charge (not electrostatic charge) with mass, as there is an absolute mass to magnetic charge ratio in the Universe. Magnetic charge was not recognized during Maxwell's time, as it was quantified as a moving electrostatic charge. The ratio of mass to electrostatic charge is not constant, as it changes due to varying quantities of electrons, protons, and neutron in matter.
## Coefficients of Induction for Two Circuits
(26) In the electromagnetic field the values of L, M, N depend on the distribution of the magnetic effects due to the two circuits, and this distribution depends only on the form and relative position of the circuits. Hence, L, M, N are quantities depending on the form and relative positions of the circuits, and are subject to variation with the motion of the conductors. It will be presently seen that L, M, N are geometrical quantities of the nature of lines, that is, of one dimension in space; L depends on the form of the first conductor, which we shall call A, N on that of the second, which we shall call B, and M on the relative position of A and B.
(27) Let ? be the electromotive force acting on A, x the strength of the current, and R the resistance, the $$Rx$$ will be the resisting force. In steady currents the electromotive force just balances the resisting force, but in variable currents the resultant force $$\xi=Rx$$ is expended in increasing the "electromagnetic momentum," using the word momentum merely to express that which is generated by a force acting during a time, that is, a velocity existing in a body.
Maxwell has just defined what is today known as Ohm's law. Since Maxwell was on the cutting edge of technology at his time, we can forgive the misuse of the phrase "electromotive force" in describing the potential. Force and potential are different units of physical behavior. In today's detailed and extended body of scientific knowledge, we must be explicit with our terminology in physics.
In the case of electric currents, the force in action is not ordinary mechanical force, at least we are not as yet able to measure it as common force, but we call it electromotive force, and the body moved is not merely the electricity in the conductor, but something outside the conductor, and capable of being affected by other conductuctors in the neighbourhood carrying currents. In this it resembles rather the reduced momentum of a driving-point of a machine as influenced by its mechanical connexions, than that of a simple moving body like a cannon ball, or water in a tube.
The "something" outside the conductor is the Aether, and it structurally quantified in the Aether Physics Model.
### Electromagnetic Relations of two Conducting Circuits.
(28) In the case of two conducting circuits, A and B, we shall assume that the electromagnetic momentum belonging to A is
$$Lx + My$$
and that belonging to B,
$$Mx + Ny$$
where L, M, and N correspond to the same quantities in the dynamical illustration, except that they are supposed to be capable of variation when the conductors A or B are moved.
If L, M, and N are the same quantities as the dynamical illustration, then they are in units of mass. However, the equations following only work if L,M, and N are in units of inductance. There is no physical basis for this arbitrary change from mass to inductance.
Then the equation of the current x in A will be
$$\xi=Rx+\frac{d}{dt}(Lx+My)$$
and that of $$y$$ in B
$$\eta=Sy+\frac{d}{dt}(Mx+Ny)$$
where $$\xi$$ and $$\eta$$ are the electromotive forces, x and y are the currents, and R and S the resistances in A and B respectively.
The units do not match in the above equations. Maxwell apparently forgot to account for $$dt$$.
### Induction of one Current by another.
(29) Case 1st. Let there be no electromotive force on B, except that which arises from the action of A, and let the current of A increase from 0 to the value $$x$$, then
$$Sy+\frac{d}{dt}(Mx+Ny)=0$$
whence
$$Y=\int_{0}^{t}ydt=-\frac{M}{S}x$$
that is, a quantity of electricity Y, being the total induced current, will flow through B when $$x$$ rises from 0 to $$x$$. This is induction by variation of the current in the primary conductor. When M is positive, the induced current due to increase of the primary current is negative.
The above equation is false since the units don't match. The simplified integral must also be false. Also, since L, M, and N are simply the masses of something (as defined in section (24), it makes no sense to describe M in terms of being positive or negative.
## Induction by Motion of Conductor
(30) Case 2nd. Let $$x$$ remain constant, and let M change from M to M', then
$$Y=-\frac{M'-M}{S}x$$
so that if M is increased, which it will be by the primary and secondary circuits approaching each other, there will be a negative induced current, the total quantity of electricity passed through B being Y.
This is induction by the relative motion of the primary and secondary conductors.
Whether by using Standard Model units or APM quantum measurements units, the "quantity of electricity" passing through B and labeled Y has no meaning. There is no empircally measured unit having three dimensions of charge, let alone over a two dimensional area.
The bad habit of Maxwell to constantly reuse the same variables for different units tend to hide his errors from most readers. However, by systematically going through the paper in a math program, such as MathCAD, it is easier to keep track of the units and catch errors.
There can be no doubt that Y has been redefined in this case to be a "quantity of electricity," but there is no basis for a unit with cubed charge over a surface.
Based upon Maxwell's subsequent theory following below, the best that can be guessed is that Maxwell changed the units of L, M, and N from mass to inductance without giving any reason for it. Even still, the equation in section (30) produces a unit of charge squared (coul^2 in MKS units, coul^4 in QMU units), which has not been empirically defined.
## Equation of Work and Energy
(31) To form the equation between work done and energy produced, multiply (1) by $$x$$ and (2) by $$y$$, and add
$$\xi x+\eta y=Rx^{2}+Sy^{2}+x\frac{d}{dt}(Lx+My)+y\frac{d}{dt}(Mx+Ny)$$
Here $$\xi x$$ is the work done in unit of time by the electromotive force $$\xi$$ acting on the current $$x$$ and maintaining it, and $$\eta y$$ is the work done by the electromotive force $$\eta$$. Hence the left-hand side of the equation represents the work done by the electromotive forces in unit of time.
We begin to see Maxwell's errors incorporated into his theory. He is squaring the current, but not the resistance. No physical basis is given for squaring the current. Also, the units do not match in the above equation (in any system of units).
From the perspective of the APM, it is apparent Maxwell misinterpreted the unit of Charge Resonance (current squared in MKS units) due to the incorrect notation of charge dimensions in the units. The APM quantum measurement unit of Charge Resonance is:
$$chrs={e_{emax}}^{2}\cdot {F_{q}}^{2}$$
Charge resonance is equal to current times frequency.
$$chrs=curr \cdot freq$$
## Heat produced by the Current
(32) On the other side of the equation we have, first,
$$Rx^{2}+Sy^{2}=H$$
Another error in judgment by Maxwell is in assuming that power is temperature. In the APM, temperature is specifically the unit of velocity squared.
$$temp=velc^{2}$$
Which represents the work done in overcoming the resistance of the circuits in unit of time. This is converted into heat. The remaining terms represent work not converted into heat. They may be written
$$\frac{1}{2}\frac{d}{dt}(Lx^{2}+2Mxy+Ny^{2})+\frac{1}{2}\frac{dL}{dt}x^{2}+\frac{dM}{dt}xy+\frac{1}{2}\frac{dN}{dt}y^{2}$$
The above terms do not produce the unit of energy, or any other known physical unit. This, again, is a clear example of a mistake in Maxwell's work.
## Intrinsic Energy of the Currents
(33) If L, M, N are constant, the whole work of the electromotive forces which is not spent against resistance will be devoted to the development of the currents. The whole instrinsic energy of the currents is therefore
$$\frac{1}{2}Lx^{2}+Mxy+\frac{1}{2}Ny^{2}=E$$
In order for the above equation to produce the unit of energy, L, M, and N would have to be in units of inductance. Maxwell gives no clue that L,M, and N have been changed to inductance. Even after switching to inductance, the equation could only work when the charge dimension in the unit of current is single dimension. Since a strong case can be made that all charge should always be distributed, this equation of Maxwell's (even after being corrected for inductance) has no physical basis.
This energy exists in a form imperceptible to our senses, probably as actual motion, the seat of this motion being not merely the conducting circuits, but the space surrounding them.
The energy as he quantified it is simply non-existent since the equation has no foundation.
(34) The remaining terms,
$$\frac{1}{2}\frac{dL}{dt}x^{2}+\frac{dM}{dt}xy+\frac{1}{2}\frac{dN}{dt}y^{2}=W$$
Represent the work done in unit of time arising from the variations of L, M, and N, or, what is the same thing, alterations in the form and position of the conducting circuits A and B.
Now if work is done when a body is moved, it must arise from ordinary mechanical force acting on the body while it is moved. Hence this part of the expression shows that there is a mechanical force urging every part of the conductors themselves in that direction in which L, M, and N will be most increased.
The existence of the electromagnetic force between conductors carrying currents is therefore a direct consequence of the joint and independent action of each current on the electromagnetic field. If A and B are allowed to approach a distance $$ds$$, so as to increase M from M to M' while the currents are $$x$$ and $$y$$, then the work done will be
$$(M'-M)xy$$
Apparently, Maxwell intends to change L, M, and N to inductance
We will also change the units to single dimension charge for the moment
and the force in the direction of ds will be
and this will be an attraction if x and y are of the same sign, and if M is increased as A and B approach.
It appears, therefore, that if we admit that the unresisted part of electromotive force goes on as long as it acts, generating a self-persistent state of the current, which we may call (from mechanical analogy) its electromagnetic momentum, and that this momentum depends on circumstances external to the conductor, then both induction of currents and electromagnetic attractions may be proved by mechanical reasoning.
Clearly, Maxwell is wrong. He has not provided a mechanical analogy linked to an electromagnetic analogy. He switched the definitions of the L,M, and N variables without giving the proper logic, which if intentional, would be unethical in science. Regardless of the switch, the charge dimensions of the units are wrong due to an error that preceded Maxwell in converting cgs units to MKS units. And even when using Maxwell's own units his equations do not always balance.
What I have called electromagnetic momentum is the same quantity which is called by Faraday the electrotonic state of the circuit, every change of which involves the action of an electromotive force, just as change of momentum involves the action of mechanical force.
Maxwell is clearly wrong. Potential is not force. In the Aether Physics Model, potential is equal to pressure times volume per strong charge.
Potential in the APM may also view as force across a length per strong charge:
Since potential is a volumetric pressure, potential may only exist as a force as long is the force is longitudinal. That is, the force inherent to potential is applied across a radius and in all directions.
If, therefore, the phenomena described by Faraday in the Ninth Series of his experimental Researches were the only known facts about electric currents, the laws of Ampere relating to the attraction of conductors carrying currents, as well as those of Faraday about the mutual induction of currents, might be deduced by mechanical reasoning.
In order to bring these results within the range of experimental verification, I shall next investigate the case of a single current, of two currents, and of the six currents of the electric balance, so as to enable the experimenter to determine the values of L, M, N.
## Case of a single Circuit
(35) The equation of the current x in a circuit whose resistance is R, and whose coefficient of self-induction is L, acted on by an external electromotive force ?, is
Whenever the distributed charge based units (inductance for example) are equated with single dimension charge based units, the equations are wrong according to the APM. For the time being, we will use Maxwell's units.
When ? is constant, the solution is of the form
where a is the value of the current at the commencement, and b is the final value. the total quantity of electricity which passes in time t, where t is great, is
The value of the integral of x2 with respect to the time is
The actual current changes gradually from the initial value of a to the final value b, the values of the integrals of x and x2 are the same as if a steady current of intensity ½(a+b) were to flow for a time 2 L/R, and were then succeeded by the steady current b. The time 2 L/R is generally so minute a fraction of a second, that the effects on the galvanometer and dynamometer may be calculated as if the impulse were instantaneous.
Just because humans cannot easily discern a small effect does not mean one can arbitrarily assume instantenous or null results. This assumption only introduces errors (or covers them up, as the case may be.)
If the circuit consists of a battery and a coil, then, when the circuit is first completed, the effects are the same as if the current had only half its final strength during the time 2 L/R. This diminution of the current, due to induction, is sometimes called the counter-current.
(36) If an additional resistance r is suddenly thrown into the circuit, as by the breaking contact, so as to force the current to pass through a thin wire of resotance r, then the original current is a=?/R, and the final current is b=?/R+r.
The current of induction is then
and continues for a time 2L/R+r. This current is greater than that which the battery can maintain in the two wires R and r, and may be sufficient to ignite the thin wire r.
When contact is broken by separating the wires in air, this additional resistance is given by the interposed air, and since the electromotive force across the new resistance is very great, a spark will be forced across.
If the electromotive force is of the form E sin pt, as in the case of a coil revolving in a magnetic field, then
where ?2 = R2 + L2p2, and tan ? = Lp/R.
## Case of Two Circuits
(37) Let R be the primary circuit and S the secondary circuit, then we have a case similar to that of the induction coil.
The equations of currents are those marked A and B, and we may here assume L, M, N as constant because there is no motion of the conductors. The equations then become
To find the total quantity of electricity which passes, we have only to integrate these equations with respect to t; then if x0, y0 be the strengths of the currents at time 0, and x1, y1, at time t, and if X, Y be the quantities of electricity passed through each circuit during time t,
1. Lagrange, Méc. Anal. ii. 2. § 5. | {"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": 21, "x-ck12": 0, "texerror": 0, "math_score": 0.9511624574661255, "perplexity": 513.190921050021}, "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-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00635.warc.gz"} |
https://research.tudelft.nl/en/publications/flight-path-planning-in-a-turbulent-wind-environment | # Flight path planning in a turbulent wind environment
Uwe Fechner, Roland Schmehl
Research output: Chapter in Book/Conference proceedings/Edited volumeChapterScientificpeer-review
5 Citations (Scopus)
## Abstract
To achieve a high conversion efficiency and at the same time robust control of a pumping kite power system it is crucial to optimize the three-dimensional flight path of the tethered wing. This chapter extends a dynamic system model to account for a realistic, turbulent wind environment and adds a flight path planner using a sequence of attractor points and turn actions. Path coordinates are calculated with explicit geometric formulas. To optimize the power output the path is adapted to the average wind speed and the vertical wind profile, using a small set of parameters. The planner employs a finite state machine with switch conditions that are highly robust towards sensor errors. The results indicate, that the decline of the average power output of pumping kite power systems at high wind speeds can be mitigated. In addition it is shown, that reeling out towards the zenith after flying figure eight flight maneuvers significantly reduces the traction forces during reel-in and thus increases the total efficiency.
Original language English Airborne Wind Energy Advances in Technology Development Roland Schmehl Springer 361-390 30 978-981-10-1947-0 9789811019463 https://doi.org/10.1007/978-981-10-1947-0_15 Published - 2018
### Publication series
Name Green Energy and Technology 9789811019463 18653529 18653537
### Bibliographical note
Green Open Access added to TU Delft Institutional Repository ‘You share, we take care!’ – Taverne project https://www.openaccess.nl/en/you-share-we-take-care Otherwise as indicated in the copyright section: the publisher is the copyright holder of this work and the author uses the Dutch legislation to make this work public.
## Fingerprint
Dive into the research topics of 'Flight path planning in a turbulent wind environment'. Together they form a unique fingerprint.
• ### REACH: Resource Efficient Automatic Conversion of High-Altitude Wind
Schmehl, R., Peschel, J. O. & Schelbergen, M.
1/12/1531/08/19
Project: Research
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=16&t=6315&p=15861 | ## Question 10 on Quiz 1 Preparation Fall 2013 in the Workbook
Katie Clark 3B
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am
### Question 10 on Quiz 1 Preparation Fall 2013 in the Workbook
I have reworked question 10 and still cannot get the correct answer. The question reads, "If molybdenum is irradiated with light of wavelength of 294 nm, what is the maximum possible kinetic energy of the emitted electrons?". I got the correct answer for the first part of the question, "what is the minimum energy needed to produce this effect?". I then said set the kinetic energy equal to the energy of the photon minus the work function. Can someone explain what I'm doing wrong? Thanks!
Serena Zhang 3D
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am
### Re: Question 10 on Quiz 1 Preparation Fall 2013 in the Workb
I think what you are doing is correct, although in my workbook the wavelength provided is 194nm. I'm not sure if 294nm was a typo or if you miswrote the number in your calculations, in which case that should fix 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": 0.8555127382278442, "perplexity": 683.9625417896046}, "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/1585370497042.33/warc/CC-MAIN-20200330120036-20200330150036-00039.warc.gz"} |
https://wetalkuav.com/journal/43d364-electromagnetic-induction-examples | About Us | Lenzâs law states that when an EMF is generated by a change in magnetic flux according to Faradayâs Law, the polarity of the induced EMF is such, that it produces an induced current whose magnetic field opposes the initial changing magnetic field which produced it. Induction Experiments (Faraday / Henry) - If the magnetic flux through a circuit changes, an emf and a current are induced. It is produced by subjecting a metal to a changing magnetic field. Some of the common uses of the electromagnetism are in: Motors and Generators: In small toy motors we use permanent magnets as the sources of the magnetic field, but in large industrial motors we use field coils which act as an electromagnet when a current is provided. 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And when Faraday presented his discovery, one person asked him, so what? using askIItians. One of the most widely known uses is in electrical generators (such as ⦠A1 nad A2 be their respective areas of cross-section. What is the emf in the coil when the current in the solenoid changes at the rate of 10 Amp/s? This video shows how Faraday's Law is used to calculate the magnitude of the induced voltage in a coil of wire. physics 112N 3 magnetic ï¬ux! Students can hear the music through the speaker even though there is no direct connection. From the above observation we conclude that, the magnitude of torque required on the loop to keep it moving with constant ω would be [(BAωN)2/R] |sin2ωt|. If you're behind a web filter, please make sure that the domains ⦠Introduction to Electromagnetic Induction, AC Circuits, and Electrical Technologies. The diagram below shows a segment of a wire as it moves through a region of uniform magnetic field B.Show the direction of the induced current flow through the wire. Example Problems for algebra-based physics (from College Physics 2nd Edition by Knight, Jones, and Field): Example Problems (Magnetic Induction and Lenz's Law) Solutions to Example Problems (Magnetic Induction and Lenz's Law) Example Problems for calculus-based physics (from Fundamentals of Physics 9th Edition by Halliday, Resnick, and Walker): Use Coupon: CART20 and get 20% off on all online Study Material, Complete Your Registration (Step 2 of 2 ), Live 1-1 coding classes to unleash the creator in your Child. This type of radiation derives from what our eyes perceive as a clear, observable field of view. From the above observation we conclude that, the virtual current in the circuit would be 60.67 amp. Write something. The strength of the magnetic field is 0.5 T and the side of the loop is 0.2 m. What is the magnetic flux in the loop? Today, electromagnetic induction is used to power many electrical devices. Terms & Conditions | Register yourself for the free demo class from Generally, Michael Faraday is recognized with the innovation of induction in the year 1831. Due to this flux, an emf is induced in the coil. For example, an electric generator produces a current because of electromagnetic induction. Due to this flux, an emf is induced in the coil. Magnetic flux and Faraday's law. When the coils are stationary, no current is induced. Electromagnetism is a branch of Physics, that deals with the electromagnetic force that occurs between electrically charged particles. A copper disc 20 cm in diameter rotates with an angular velocity of 60 rev s-1 about its axis. Write something. _________________________________________________________________________________________. Transformers are critical in electrical transmission because they can step voltage up or down as needed during its journey to consumers. Read here to know about the electromagnetism and its uses in everyday life. Electromagnetic Induction. Faraday's experiment showing induction between coils of wire: The liquid battery (right) provides a current which flows through the small coil (A), creating a magnetic field. Faradayâs law of electromagnetic induction (referred to as Faradayâs law) is a basic law of electromagnetism predicting how a magnetic field will interact with an electric circuit to produce an electromotive force (EMF). So, this phenomenon of generating induced emf or current because of changing flux is called the Electromagnetic Induction. When changing current is passed through a closed coil, varying magnetic flux develops in it. This induced emf generates an induced current in it. This either happens when a conductor is placed in a moving magnetic field (when using AC power source) or when a conductor is constantly moving in a stationary magnetic field. FAQ's | conceptual. Faradayâs law of induction is one of the four equations in Maxwellâs equations, governing all electromagnetic phenomena. One of the major applications is the production of electricity. Example Definitions Formulaes. Faradayâs law of induction, in physics, a quantitative relationship between a changing magnetic field and the electric field created by the change, developed on the basis of experimental observations made in 1831 by the English scientist Michael Faraday.The phenomenon called electromagnetic induction was first noticed and investigated by Faraday; the⦠Faradayâs laws of electromagnetic induction Based on his studies on the phenomenon of electromagnetic induction, Faraday proposed the following two laws. number, Please choose the valid Generally, Michael Faraday is recognized with the innovation of induction in the year 1831. seems we can induce a current in a loop with a changing magnetic ï¬eld. Solved Examples on Electromagnetic Induction and Alternating Current Question 1:- A copper disc 20 cm in diameter rotates with an angular velocity of 60 rev s-1 about its axis. One is attached to a music source, such as a small radio or iPod, and the other is attached to an external speaker. 1. Magnitude of induced current = [BAωN/R] |sin ωt|. According to Faraday’s law, magnitude of induced e.m.f is. Electromagnetic Induction was first discovered way back in the 1830âs by Michael Faraday. Do you know that a kind of electromagnetic radiation have the potential to kill cancer cells? for a plane surface of area A = = physics 112N 5 Remember from Year 12 The relationship between electricity and magnetism (Motor effect) ⢠When current flows through a wire a magnetic field is created around the wire ⢠And if the wire is coiled (to form a solenoid) Just like; Lessons. If you're seeing this message, it means we're having trouble loading external resources on our website. Unit: Electromagnetic induction. There are many applications of Faradayâs Law of induction, as we will explore in this tutorial and others. Advanced Knowledge of Electromagnetic Induction. Electromagnetic induction is one of the major portion of many subjects like physics, basic electrical and others. The disc is placed in a magnetic field of induction 0.2 T acting parallel to the axis of rotation of the disc. Electromagnetic induction A changing magnetic flux induces a current into a coil. The definition of electromagnetic induction is the creation of voltage or an electromotive force across a conductor within a varying magnetic field. The phenomenon of electromagnetism is broadly used in many electrical devices and in machines. First law Whenever the amount of magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. To test his hypothesis he made ⦠And this phenomena is called electromagnetic induction. Wind pushes the blades of the turbine, spinning a shaft attached to magnets. Enroll For Free. Solution: Using Flemingâs right-hand rule or the right-hand slap rule, the direction of motion of the wire is D. Figure shows a bar magnet falling towards a solenoid. A coil replaced with another coil that has loops 2 times the initial loops and the rate of change of magnetic flux is constant. This wind turbine in the Thames Estuary in the UK is an example of induction at work. physics 112N 2 experimental basis of induction! Calculate the magnitude of the e.m.f. These field coils in large motors then becomes the source of the magnetic field. induced between the axis of rotation and the rim of the disc. Electromagnetic induction is the use of the movement of magnets around a coil of wire to create an electrical current through the wire. The area that combines magnetics and fluid flow is known as magnetohydrodynamics. A ⦠Please refer to the examination notes which you can use for preparing and revising for exams. The induction cooker uses a magnetic field to produce eddy currents in the metal frying pan by a process known as electromagnetic induction. - A time-varying magnetic field can act as source of electric field. grade, Please choose the valid Known : Initial loops (N) = 1. In which direction did the wire move to produce the induced current? Michael Faraday is generally credited with the discovery of induction in 1831, and James Clerk Maxwell mathematically described it as Faraday's law of induction. An electric generator rotates a coil in a magnetic field, inducing an EMF given as a function of time by $$\mathrm{ε=NABw \sin Ït}$$. Electromagnetic Induction. Sitemap | Compute the mutual inductance of the two circuits. Lenzâs Law of Electromagnetic Induction. EMI is the interference from one electrical or electronic system to another caused by the electromagnetic fields generated by its operation. Related Concepts. Assuming that at t = 0, the plane of the loop is normal to the lines of force, find an expression for the peak value of the emf and current Induced in the loop. These notes will help you to revise the concepts quickly and get good marks. Electromagnetic forces are not only of importance in solid materials. So, power input = heat dissipation per second. ?§:¢0ÂFBx\$ !«¤i@Ú¤¹H§È[EE1PLÊ
⢡V¡6£ªQP¨>ÔUÔ(j Electromagnetic induction (also known as Faraday's law of electromagnetic induction or just induction, but not to be confused with inductive reasoning), is a process where a conductor placed in a changing magnetic field (or a conductor moving through a stationary magnetic field) causes the production of a voltage across the conductor. Franchisee | Flux ϕ2 through coil crated by current i1 in solenoid is ϕ2 = N2(B1A2), So magnitude of induced emf = E2 = M|di1/dt|, Signing up with Facebook allows you to connect with friends and classmates already induced between the axis of rotation and the rim of the disc would be 0.377 V. _____________________________________________________________________________________. Figure 1. Àp|îOÃàX The electromagnetic force is one of the four fundamental forces and exhibits electromagnetic fields such as magnetic fields, electric fields, and light.It is the basic reason electrons bound to the nucleus and responsible for the complete structure of the nucleus. Ex.1 A square loop ACDE of area 20 cm2 and resistance 5 W is rotated in a magnetic field B = 2T through 180° This process of electromagnetic induction, in turn, ⦠Email, Please Enter the valid mobile It is a system that converts mechanical energy into electrical energy. Letâs understand the Lenzâs law by below given two definitions and then we shall see few examples to understand the concept of Electromagnetic Induction in more simple way so that it is easy for new students to absorb the complexity of the concept. This phenomenon is known as electromagnetic induction. The faster the magnet the higher the induced current. A rectangular loop of N turns of area A and resistance R rotates at a uniform angular velocity ω about Y-axis. A long solenoid of length 1 m, cross sectional area 10 cm2, having 1000 turns has wound about its centre a small coil of 20 turns. Slide 2 / 47 Multiple Choice. So electromagnetic induction is a phenomena in which when you change the magnetic field through a coil it induces a voltage or a current. Area swept by radius vector during one revolution, A = (area swept in one revolution) × (Number of revolutions per second). Privacy Policy | A conducting sheet lies in a plane perpendicular to a magnetic field $$\displaystyle \vec{B}$$ that is below the sheet. “Relax, we won’t flood your facebook Lenz's law describes the direction of the induced field. Careers | Today, electromagnetic induction is used to power many electrical devices. When a wire that's near to a changing magnetic field produces electric current, this phenomenon is called induction. For applications and consequences of the law, see Electromagnetic induction. The plane of the area of antenna is inclined at 47º with the direction of Earthâs magnetic field. RD Sharma Solutions | Tutor log in | Electric motors, power generators and transformers all work because of induction. Many of our electrical home appliances use electromagnetism as a basic principle of working. Demo 1: As the magnet is moved in, the magnetic flux through the solenoid changes and an induced current appears (Faraday's law). Learn. Slide 3 / 47 1 A square loop of wire is placed in a uniform magnetic field perpendicular to the magnetic lines. Electromagnetic induction is the complementary phenomenon to electromagnetism. Gill offers full end-to-end development services of intrinsically safe products. Generator: Generate electricity with a bar magnet! Write something completely different. Write something. Electromagnetic Induction: This simple animation shows the current induced in a coil when a magnet is moved near the coil. Electromagnetic induction is a phenomenon that explains how EMF and current is or can be induced in a coil when a coil and a magnetic field interact. In the above figure, A rectangular conductor width sides are placed in between a magnetic field. electromagnetic induction. An induction motor is an asynchronous AC motor where power is transferred to the rotor by electromagnetic induction, much like transformer action. Discover the physics behind the phenomena by exploring magnets and how you can use them to make a bulb light. Electrical bell: In electrical bells electromagnetic coils are us⦠{{{;}#âtp¶8_\. _______________________________________________________________________________________________. Electromagnetic induction 1. Examples of Faradayâs Law: Rotating Loop in a Field Now consider a circular loop with area 1 m 2 and three turns of wire (N = 3) rotating in a magnetic field with a ⦠13.6 Eddy Currents. The disc is placed in a magnetic field of induction 0.2 T acting parallel to the axis of rotation of the disc. Visible Light Waves. , An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1W and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit. Electromagnetic Induction: Solved Example Problems EXAMPLE 4.1 A circular antenna of area 3 m2 is installed at a place in Madurai. õMFk¢ÍÑÎè t,:.FW Ðè³èô8ú¡c1L&³³³Ó9
Æa¦±X¬:Öë År°bl1¶ So, this phenomenon of generating induced emf or current because of changing flux is called the Electromagnetic Induction. Flux and magnetic flux (Opens a modal) What is magnetic flux? The induced emf lasts so long as the change in magnetic flux continues. Figure shows the direction of the induced current in a wire. Electromagnetic induction is the use of the movement of magnets around a coil of wire to create an electrical current through the wire. A ⦠(Opens a modal) Faraday's Law Introduction (Opens a modal) Lenz's Law (Opens a modal) Faraday's Law example (Opens a modal) What is Faraday's law? Determine the ratio of initial and final induced emf. Examples from Classical Literature This video shows how Faraday's Law is used to calculate the magnitude of the induced voltage in a coil of wire. askiitians. Electromagnetic Induction is a current produced because of voltage production (electromotive force) due to a changing magnetic field. This process of electromagnetic induction, in turn, ⦠When Michael Faraday made his discovery of electromagnetic induction in 1831, he hypothesized that a changing magnetic field is necessary to induce a current in a nearby circuit. Thus from the above observation we conclude that, the magnitude of the e.m.f. Faraday's Magnetic Field Induction Experiment. Electromagnetic induction (also known as Faraday's law of electromagnetic induction or just induction, but not to be confused with inductive reasoning), is a process where a conductor placed in a changing magnetic field (or a conductor moving through a stationary magnetic field) causes the production of a voltage across the conductor. Faradayâs law of induction, in physics, a quantitative relationship between a changing magnetic field and the electric field created by the change, developed on the basis of experimental observations made in 1831 by the English scientist Michael Faraday.The phenomenon called electromagnetic induction was first noticed and investigated by Faraday; the⦠Demo 1: As the magnet is moved in, the magnetic flux through the solenoid changes and an induced current appears (Faraday's law). Pay Now | To obtain the magnitude of torque required on the loop to keep it moving with constant ω, we have to equate power input is equal to heat dissipation per second. Electromagnetic or magnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field. Here are 10 examples of electromagnetic radiation which we come across daily and the harmful effects that result from it: 1. name, Please Enter the valid Electromagnetic induction occurs when a circuit with an alternating current flowing through it generates current in another circuit simply by being placed nearby. A very important application has to do with audio and video recording tapes. If $$\displaystyle \vec{B}$$ oscillates at a high frequency and the conductor is made of a material of low resistivity, the region above the sheet is effectively shielded from $$\displaystyle \vec{B}$$. Media Coverage | This produces a Voltage or EMF (Electromotive Force) across the electrical conductor. Problems practice. Register online for Physics tuition on Vedantu.com to score ⦠Electromagnetic induction A changing magnetic flux induces a current into a coil. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Electromagnetic Induction - definition The property due to which a changing magnetic field within a closed conducting coil induces electric current in the coil is called electromagnetic induction. The definition of electromagnetic induction is the creation of voltage or an electromotive force across a conductor within a varying magnetic field. Electromagnetic induction, induced EMF â problems and solutions. Electromagnetic Induction. Many questions are asked in academic and competitive exam and even in interviews of many high profile companies. In case of an ac, the voltage leads the current in phase by angle. So it is required to have an in-depth knowledge of the same. Electromagnetic Induction Practice Problems. Blog | As the disc rotates, any of its radii cuts the lines of force of magnetic field. Dear Falling Behind in Studies? If we take an example of an electric fan, the motor works on the principle of electromagnetic induction. 1. Other uses for electromagnetic induction include electric motors used in anything from washing machines to trains, electric hobs and cookers, transformers, welding and ⦠The electromagnetic induction of membrane sodium potassium ATPase inhibition mediated ATP synthesis may also be a similar primitive source of cellular energy. Electromagnetic Induction or Induction is a process in which a conductor is put in a particular position and magnetic field keeps varying or magnetic field is stationary and a conductor is moving. Electric and hybrid vehicles also take advantage of electromagnetic induction. Preparing for entrance exams? Electromagnetic Induction Examples. Applications of Electromagnetic Induction. Explanation with an example DC Generator works on the principle of Faradayâs Law of Electromagnetic Induction. CBSE Class 12 Physics Electromagnetic Induction Solved Examples. Refund Policy, Register and Get Free access of Online Study Material, Please choose a valid One limiting factor that inhibits widespread acceptance of 100% electric vehicles is that the lifetime of the battery is not as long as the time you get to drive on a full tank of gas. The loop lies in a uniform magnetic field B in the direction of X-axis. AP Physics Chapter 20 Electromagnetic Induction - Examples: pacemakers stopping, migrating birds get lost, GPS won t work, etc. Faraday noticed that when he moved a permanent magnet in and out of a coil or a single loop of wire it induced an E lectro M otive F orce or emf, in other words a Voltage, and therefore a current was produced. For example, in metals processing using induction furnaces, electromagnetic forces are important to understand, since molten metals are typically highly conductive. 21. Solved Examples on Electromagnetic Induction and Alternating Current Question 1:-A copper disc 20 cm in diameter rotates with an angular velocity of 60 rev s-1 about its axis. useful to deï¬ne a quantity called magnetic ï¬ux! Examples of electromagnetic induction include: moving a magnet inside a wire coil; generating the high voltage necessary to ionise the vapour in a fluorescent tube and cause the spark needed to ignite the explosive mixture in a petrol engine; changing the voltage of an alternating current, using a transformer. Electricity,is something that we take for granted,which is sad,because it's a fascinating phenomenon. Electromagnetic interference (or EMI) is a disruption that affects an electrical circuit because of either electromagnetic induction or externally emitted electromagnetic radiation. The disc is placed in a magnetic field of induction 0.2 T acting parallel to the axis of rotation of the disc. The flow of eddy currents in the frying pan produces heat. - A time-varying electric field can act as source of magnetic field. news feed!”. This induced emf generates an induced current in it. Join Our Performance Improvement Batch. Contact Us | When changing current is passed through a closed coil, varying magnetic flux develops in it. What is the magnitude of torque required on the loop to keep it moving with constant ω? subject, Electromagnetic Induction and Alternating Current, Solved Examples on Electromagnetic Induction and Alternating Current, A copper disc 20 cm in diameter rotates with an angular velocity of 60 rev s, A rectangular loop of N turns of area A and resistance R rotates at a uniform angular velocity, From the above observation we conclude that, the magnitude of torque required on the loop to keep it moving with constant ω would be, A long solenoid of length 1 m, cross sectional area 10 cm, Structural Organisation in Plants and Animals, French Southern and Antarctic Lands (+262), United state Miscellaneous Pacific Islands (+1), Complete JEE Main/Advanced Course and Test Series, Complete AIPMT/AIIMS Course and Test Series. Gill Research & Development Limited (Gill R&D), part of the Gill Group of product engineering companies, promotes its ability to provide full end-to-end design to certified manufacturing of intrinsically safe (IS) products. School Tie-up | Also find the virtual current in the circuit. Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 6 - Electromagnetic Induction prepared by expert Physics teachers from latest edition of CBSE(NCERT) books. Two coils of wire are held close to each other, but not touching. This phenomenon âelectromagnetic inductionâ is explained by Faradayâs laws of electromagnetic induction. The faster the magnet the higher the induced current. Maxwell - An induced current (and emf ) is generated when: (a) we move a magnet Apart from these major examples, there are numerous applications that use electromagnetic induction principle for their functioning such as electric power transmission, induction cookers, industrial furnaces, medical equipments, electromagnetic flow sensors, musical instruments (like electric violin and electric guitar) and so on. | {"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": 0.8455770611763, "perplexity": 637.7935817317815}, "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/1610703500028.5/warc/CC-MAIN-20210116044418-20210116074418-00511.warc.gz"} |
http://www-old.newton.ac.uk/programmes/AGA/seminars/2007040216301.html | # AGA
## Seminar
### Convergence of resonances on thin branched quantum wave guides
Post, O (Humboldt)
Monday 02 April 2007, 16:30-17:00
Seminar Room 1, Newton Institute
#### Abstract
We consider convergence results of a family of noncompact, thin branched quantum waveguides (QWG) to the associated quantum graph. The branched quantum waveguide can either be a thin neighbourhood of the (embedded) quantum graph or be defined as a manifold without boundary (like the surface of a pipeline network approaching the metric graph). On the QWG has boundary, we consider the (Neumann) Laplacian; on the metric graph we consider the Laplacian with free boundary conditions. Our main result is a convergence result for the spectrum and resonances under some natural uniformity conditions on the spaces.
#### Video
The video for this talk should appear here if JavaScript is enabled.
If it doesn't, something may have gone wrong with our embedded player.
We'll get it fixed as soon as possible. | {"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": 0.8482612371444702, "perplexity": 1991.0027066246703}, "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-2015-40/segments/1443736677918.35/warc/CC-MAIN-20151001215757-00182-ip-10-137-6-227.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/265138/the-image-of-a-map-of-separable-algebras?answertab=votes | # The image of a map of separable algebras
I'm reading Lenstra's notes on the étale fundamental group, and I've got stuck on his exercise 3.9(c). He says that if $A$ and $B$ are separable algebras over a field $K$, and $f:A \to B$ a $K$-algebra homomorphism, then the image of $f$ is precisely $\{b \in B: b \otimes 1 = 1 \otimes b\text{ in }B \otimes_A B\}$.
By his theorem 2.7, every separable $K$-algebra is a finite product of separable extension fields $K_i$ of $K$. So the morphisms of two such algebras are fairly restricted: if I'm not mistaken, if $A = \prod K_i$ and $B = \prod L_j$, and $v_i$ and $w_j$ are the respective idempotents defining these splittings, then each $f(v_i)$ is a sum of distinct $w_j$, each generating an extension field of $K_i$.
I don't know how to get from this to the claim. More generally, for any map of commutative rings $A \to B$, we have this subring $\{b\in B:b \otimes 1 = 1 \otimes b \text{ in }B \otimes_A B\}$, which I feel like I've seen in other contexts as well, though I can't remember where. Is there a high-level way of thinking about this set, or more general conditions that tell you it's equal to $f(A)$? Thank you in advance.
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One approach: you have the natural inclusion of $A$ modules $f(A) \rightarrow \{b \in B: b \otimes 1 = 1 \otimes b\}$, to prove this is an isomorphism it suffices to do it locally, which reduces you to the case that $A$ is simply a field, which makes the tensor product $B \otimes_A B$ less mysterious. – uncookedfalcon Dec 26 '12 at 4:41
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## 1 Answer
Let $A$ be a field, $f: A \rightarrow B$ a map of rings. I claim first that in this baby case $f(A) = \{b: b \otimes 1 = 1 \otimes b \in B \otimes_A B \}$. Indeed, we have the inclusion $\subseteq$, and in the other direction, if $b \notin f(A)$ we may extend $1,b$ to basis $b^i$ for $B$ over $A$, as $b^i \otimes b^j$ form a basis for $B \otimes_A B$, in particular we have $1 \otimes b \neq b \otimes 1$.
Let's reduce to this case for the question at hand: you have the natural inclusion of $A$ modules $f(A) \rightarrow \{b \in B: b \otimes 1 = 1 \otimes b\}$, to prove this is an isomorphism it suffices to do it locally (the latter is easily seen to be an $A$ module). Indeed, suppose $A = K_1 \oplus \ldots \oplus K_n$, the spectrum of $A$ is just $\mathfrak{p}_1 \ldots \mathfrak{p}_n$, where $\mathfrak{p}_i = \oplus_{j \neq i} K_j \subseteq A$.
Fix some $\mathfrak{p}_i =: p$. Now, $A_{p}$ is just $K_i$, and $f(A)_p = f_p(A_p)$, where $f_p$ is the induced map $A_p \rightarrow B_p$. We have directly that $$\{b \in B: b \otimes 1 = 1 \otimes b \in B \otimes_A B\}_p \subseteq \{b \in B_p: b \otimes 1 = 1 \otimes b \in B_p \otimes_{A_p} B_p\}$$where we use the canonical identification $(B \otimes_A B)_p \simeq B_p \otimes_{A_p} B_p$. By our earlier analysis (of $A$ just a field) then the result follows, as all of the RHS is $f_p(A_p)$.
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Thanks a lot. So the only important thing is that $A$ is a product of fields. For the more general question -- for what maps $f:A \to B$ does $f(A) = \{b \in B: b \otimes 1 = 1 \otimes b \in B \otimes_A B\}$ -- your last paragraph makes it clear that we can work locally. Do you have any idea when this happens more generally (or good examples for when it doesn't happen), other than $A$ a field? – Paul VanKoughnett Dec 26 '12 at 20:23
I really don't know man :(. This could be a great follow up question though! – uncookedfalcon Dec 27 '12 at 0:25
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http://mathhelpforum.com/calculus/44159-derivatives.html | 1. ## Derivatives
Find f: f''=20x^3 + 12x^2 + 4 f(0)=8 f(1)=5
2. Originally Posted by jkeatin
Find f: f''=20x^3 + 12x^2 + 4 f(0)=8 f(1)=5
$f''(x)=20x^3 + 12x^2 + 4$
$f'(x)=5x^4 + 4x^3 + 4x + C_1$
$f(x)=x^5 + x^4 + 2x^2 + C_1x + C_2$
now, substitute $x=0$ and $x=1$ and you will have 2 equations and two unknowns.
and then solve for $C_1$ and $C_2$. | {"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": 7, "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": 0.9448741674423218, "perplexity": 2065.68551027394}, "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-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00469-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.torahbytes.info/2018/12/rosh-chodesh-calculation.html | ### Rosh Chodesh Calculation
Q. In the Jewish calendar, months are either 29 or 30 days in length. Why is it that for a 29 day month (known as a choser), we observe only one day Rosh Chodesh, while for a 30 day month (which is referred to as a moleh) we observe 2 days Rosh Chodesh?
A. This question was posed to Rabbeinu Yeshaya ha’Rishon (b. 1180) [Teshuvos Ha’Rid 32]. He writes that one might suggest that we observe two days of Rosh Chodesh because of sfeika d’yoma (doubts in the calendar), which is the reason we observe two days of Yom Tov. However, this cannot be correct, for if so, we would observe two days Rosh Chodesh every month. Rather, the reason is as follows. The Jewish months follow the lunar cycle which is approximately 29½ days. In theory, each month on the calendar should be 29½ days. However, this cannot be done, because a month must consist of integral numbers of days, and not fractions. For this reason, we are required to alternate months between 29 and 30 days in length, so that on average, the cycle is 29½ days. On a moleh (30-day month), Rosh Chodesh should begin in the middle of the 30th day when the new moon appears. However, this cannot be done, because we may not split the holiness of a day in half. If part of the day is holy, then the entire day must be holy. Rather, in such a month we designate the 31st day as Rosh Chodesh. However, in recognition of the fact that the new moon really appeared in the middle of the 30th day, and the latter half of the 30th day was fitting to have been declared Rosh Chodesh, the Rabbis instituted that the 30th day should be observed as Rosh Chodesh as well. Rabbeinu Yeshaya points out that the custom of observing two days of Rosh Chodesh has very early sources and is alluded to in Sefer Shmuel I (20:27). More about this program and to subscribe visit https://oukosher.org/halacha-yomis-email/ | {"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": 0.8488345742225647, "perplexity": 1550.0432365671597}, "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-2021-17/segments/1618038878326.67/warc/CC-MAIN-20210419045820-20210419075820-00461.warc.gz"} |
http://mathhelpforum.com/calculus/206882-limits-equations-print.html | # limits / equations?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• November 6th 2012, 10:12 AM
pnfuller
limits / equations?
for what value of a is the following equation true?
lim ((x+a)/(x-a))^x = e
x->inf
• November 6th 2012, 10:57 AM
MarkFL
Re: limits / equations?
Take the natural log of both sides, so that you wind up with:
$\lim_{x\to\infty}\frac{\ln\left(\frac{x+a}{x-a} \right)}{\frac{1}{x}}=1$
Now use L'Hôpital's rule to find $a$.
• November 6th 2012, 12:03 PM
pnfuller
Re: limits / equations?
after applying L'Hôpital's rule i got the limit as x approaches infinity ((x-a)(-2a))/(-x^-2) = 1
and now what do i do from there?
• November 6th 2012, 12:13 PM
MarkFL
Re: limits / equations?
You have differentiated the numerator incorrectly. You want:
$\frac{d}{dx}\left(\ln\left(\frac{x+a}{x-a} \right) \right)=\frac{1}{\frac{x+a}{x-a}}\cdot\frac{-2a}{(x-a)^2}=\frac{2a}{a^2-x^2}$
Now, putting this together with the derivative of the denominator, we have:
$\lim_{x\to\infty}\frac{2ax^2}{x^2-a^2}=1$
which we can write as:
$2a\lim_{x\to\infty}\frac{1}{1-\frac{a^2}{x^2}}=1$
From here, it is a piece of cake.
• November 6th 2012, 12:20 PM
pnfuller
Re: limits / equations?
im not seeing how this is a piece of cake?
• November 6th 2012, 12:36 PM
Plato
Re: limits / equations?
Quote:
Originally Posted by pnfuller
im not seeing how this is a piece of cake?
LEARN this: $\lim _{x \to \infty } \left( {1 + \frac{a}{{x + b}}} \right)^{cx} = e^{ac}$
Yours can be rewritten an $\left( {1 + \frac{{2a}}{{x - a}}} \right)^x$.
• November 6th 2012, 12:37 PM
MarkFL
Re: limits / equations?
Sorry, bad choice of words. :)
What is $\lim_{x\to\infty}\frac{a^2}{x^2}$ ?
• November 6th 2012, 12:42 PM
pnfuller
Re: limits / equations?
the limit as a^2/x^2 approaches infinity equals 0? is that what you mean?
• November 6th 2012, 01:09 PM
MarkFL
Re: limits / equations?
Yes, so what does that tell you about the limit at the end of post #4?
• November 6th 2012, 03:06 PM
pnfuller
Re: limits / equations?
that the limit as x approaches infinity is 1/1 and it equals 1 but what does a equal?
• November 6th 2012, 03:10 PM
Prove It
Re: limits / equations?
Reread that equation carefully. You were told \displaystyle \begin{align*} 2a \cdot \lim_{x \to \infty}\frac{1}{1 - \frac{a^2}{x^2}} = 1 \end{align*}
• November 6th 2012, 03:11 PM
MarkFL
Re: limits / equations?
Okay, since the limit is 1, you simply have:
$2a=1$
Now solve for $a$.
• November 6th 2012, 03:19 PM
pnfuller
Re: limits / equations?
so a = 1/2
• November 6th 2012, 03:21 PM
Plato
Re: limits / equations?
Quote:
Originally Posted by pnfuller
that the limit as x approaches infinity is 1/1 and it equals 1 but what does a equal?
This whole thread is totally frustrating for me.
The answer is $a=0.5$ . WHY? | {"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": 12, "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": 0.9974011182785034, "perplexity": 3192.198389702727}, "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/1438043060830.93/warc/CC-MAIN-20150728002420-00028-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/vector-calc-div-identity.330215/ | # Vector Calc: Div Identity
1. Aug 11, 2009
### Payne0511
1. The problem statement, all variables and given/known data
div(grad f x grad g)=0. I need to prove this somehow.
2. Relevant equations
3. The attempt at a solution
I dont really know where to even start this at >.< any help is greatly appreciated.
2. Aug 11, 2009
### Dick
Write grad(f) and grad(g) out in components and take their cross product. Now take div of that. Use that mixed partial derivatives are equal e.g. d^2(f)/(dxdy)=d^2(f)/(dydx). That will get the job done.
3. Aug 12, 2009
### VKint
Personally, I like using index notation for this type of problem; it's a good deal more compact and, once you get used to it, very transparent. Instead of writing out all the components longhand, write $$\nabla f \times \nabla g = \tilde{\varepsilon}^{ijk} f_{,i} g_{,j} \hat{\mathbf{e}}_{k}$$ , where $$\tilde{\varepsilon}$$ is the Levi-Civita symbol (not being careful about index placement here, since we're in flat space). Then you have
\begin{align*} \nabla \cdot (\nabla f \times \nabla g) &= \tilde{\varepsilon}^{ijk} (f_{,i} g_{,j})_{,k}\\ &= \tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j} + \tilde{\varepsilon}^{ijk} f_{,i} g_{,j,k} \textrm{.} \end{align*}
Both summands in the last equation vanish. Indeed, let $$P = \tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j}$$ . Then, since partials commute, we have $$P = \tilde{\varepsilon}^{ijk} f_{,k,i} g_{,j}$$ ; by a trivial relabeling of dummy indices, this gives $$P = \tilde{\varepsilon}^{kji} f_{,i,k} g_{,j} = -\tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j} = -P$$ (since $$\tilde{\varepsilon}$$ is completely antisymmetric), so $$P = 0$$. The logic for the other summand is identical.
4. Aug 13, 2009
### gabbagabbahey
Alternatively, if you've already proven the following identities in class, you can simply combine them with $\textbf{A}=\mathbf{\nabla}f$ and $\textbf{B}=\mathbf{\nabla}g$ and the result becomes apparent.
Identities:
$$\mathbf{\nabla}\cdot(\textbf{A}\times\textbf{B})=\textbf{B}\cdot(\mathbf{\nabla}\times\textbf{A})-\textbf{B}\cdot(\mathbf{\nabla}\times\textbf{B})$$
$$\mathbf{\nabla}\times(\mthbf{\nabla}f)=0$$
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https://www.physicsforums.com/threads/why-is-acceleartion-in-denominator-2-units.69287/ | # Why is acceleartion in denominator ^2 units?
1. Mar 30, 2005
### ktpr2
Let's say i have some function S(t) where t is in seconds and S(t) gives feet per second.
As I take the derivatives of S to get the acceleration function, why is the feet unit squared? What's the best way to conceptualize what's going on?
2. Mar 30, 2005
### what
you have it a little mixed up.
S(t) would give you position not velocity
it's derivative S'(t) would give you ft/sec
an the derivative of S''(t) would give you acc. which is ft/(sec)^2
The first derivative tells you how fast your position changes with time, the second derivative is telling how fast your first rate of change is changing with respect to time, hence ft per seconds squared.
3. Mar 30, 2005
### Nylex
It's the definition of acceleration, ie. rate of change of velocity. Velocity is rate of change of position (m/s), so rate of change of velocity is m/s/s.
4. Mar 30, 2005
### dextercioby
So no feet is getting squared.On the other hand,the second does...
Daniel.
5. Mar 30, 2005
### ktpr2
That's the best explanation i've seen. Thanks.
And yeah that should be seconds squared. Also, my plural use of deriviative was in respect to s''.
6. Mar 30, 2005
### whozum
Distance is in meters
Velocity is in meters per second
Acceleration is in meters per second per second.
In each step you are dividing by seconds.
Velocity / Seconds = $$\frac{\frac{m}{s}}{s}} = \frac{m}{s} * \frac{1}{s} = \frac{m}{s^2}$$
Likewise jerk is the change in acceleration, dividing by seconds:
Acceleration / Seconds = $$\frac{\frac{m}{s^2}}{s}} = \frac{m}{s^2} * \frac{1}{s} = \frac{m}{s^3}$$
7. Mar 30, 2005
### MiniTank
i never knew there was something after acceleration :|
8. Mar 30, 2005
### whozum
Theres infinitely many, they only named the first five I think.
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http://tex.stackexchange.com/questions/191323/how-to-declare-a-macro-with-a-variable-argument-list | # How to declare a macro with a variable argument list?
I am looking for a macro with two optional arguments and tried the following code:
\documentclass{tufte-handout}
%
\usepackage{xparse}
\usepackage{lipsum}
\usepackage[demo]{graphics}
\usepackage{easyfig}
%
%% Usage \mfig[position][trim]{name-figure}{caption}{label}
%
\NewDocumentCommand{\mfig}{o o m m m}{%
\begin{marginfigure}[\IfValueTF{#1}{#1}{0}cm]
\Figure[trim={.0\width} {\IfValueTF{#2}{#2}{.05}\height}%
{.0\width} {.0\height},clip,%
width=\linewidth,keepaspectratio=true,%
caption={#4},label={#5},center,here]{#3}
\end{marginfigure}%
}
%
\begin{document}
\section{A simple test}
\lipsum[1]
\mfig{myfigure}{This is just a caption to the demo figure.}{fig:test}
Figure~\ref{fig:test} shows a test. \lipsum[3].
\end{document}
The first optional argument gives the margin figure vertical position and the second gives the trim coordinates (an inferior cut-off actually).
It is ok to to use:
\mfig[-1.5]{name}{caption}{label}
to move vertically the margin figure about 1.5cm up, for instance, and with the default trim coordinates of .05, but I am unable to use
\mfig[.55]{name}{caption}{label}
to cut-off 55% of my graphic, while using the default position of 0cm, because my "macro" is unable to recognize the optional argument #2 without to explicitly use the #1.
So, my question is: is there any way to specify only the #2 optional argument while the macro accepts the default #1 argument like the above?
-
optional arguments are positional so you can place one in a different position such as o m o m m but it's usually better to consider a variant syntax such as [a,b] or [a=x, b=y] that is the key value syntax of \includegraphics is designed to allow optional arguments without this problem, but your wrapper hides that. – David Carlisle Jul 13 '14 at 20:13
@David Yes, I can see it now ... much more clever, and simple, I believe. But I did not understood your variant syntax note. How does it changes my macro/function declaration and usage? – jotagah Jul 13 '14 at 23:03
if you have two optional arguments and put one before and one after the mandatory argument they can be distinguished. with o o m if you have just one [] argument it has to be #1 but if you have o m o you can have \foo[a]{b}[c] and distinguish \foo[a]{b} from \foo{b}[c] – David Carlisle Jul 13 '14 at 23:08
But how to use [a,b] variant? – jotagah Jul 13 '14 at 23:15
As a note, I would very much recommend using the default-value syntax for optional arguments in the argspec. Instead of oommm, use O{0cm} O{0.05} mmm. – Sean Allred Jul 14 '14 at 1:43
Why not use keys? That way you can have practically as many parameters as you would like.
Of the many ways you could do this, here's one:
In the preface you could have
\usepackage{pgfkeys}
\def\mycaption{default caption}
\def\mylabel{default label}
\def\myfigwidth{0pt}
\def\myfigheight{0.05}
\pgfkeys{/jotagah/myfig/.cd,
caption/.store in=\mycaption,
label/.store in=\mylabel,
fig width/.store in=\myfigwidth,
fig height/.store in=\myfigheight,
}
Then you could define your function as
\NewDocumentCommand{\mfig}{om}{%
\pgfkeys{/jotagah/myfig/.cd,
fig width=0cm,
fig height=0.05,
#1}%%
\begin{marginfigure}[\myfigwidth]
\Figure[trim={.0\width} {\myfgheight\height}%
{.0\width} {.0\height},clip,%
width=\linewidth,keepaspectratio=true,%
caption=\mycaption,label=\mylabel,center,here]{#2}
\end{marginfigure}%
-
@Ellett I do not know the pgfkeys package and so I have to read a bit more to understand your solution. May I ask you how to call my macro/function using only #1, only #2,both, or none of the optional arguments? – jotagah Jul 13 '14 at 23:12 | {"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": 0.8454818725585938, "perplexity": 2604.1025053132494}, "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/1440645322940.71/warc/CC-MAIN-20150827031522-00340-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://deepai.org/publication/lambda-calculus-and-probabilistic-computation | # Lambda Calculus and Probabilistic Computation
We introduce two extensions of λ-calculus with a probabilistic choice operator, modeling respectively the call-by-value and call-by-name probabilistic computation. We prove that both enjoys confluence and standardization, in an extended way: indeed these two fundamental notions have been revisited to take into account the asymptotic behaviour of terms. The common root of the two calculi is a further calculus based on Linear Logic, which allows us to develop a unified, modular approach.
## Authors
• 7 publications
• 4 publications
• ### Decomposing Probabilistic Lambda-calculi
A notion of probabilistic lambda-calculus usually comes with a prescribe...
02/19/2020 ∙ by Ugo Dal Lago, et al. ∙ 0
• ### A note on confluence in typed probabilistic lambda calculi
On the topic of probabilistic rewriting, there are several works studyin...
06/11/2021 ∙ by Rafael Romero, et al. ∙ 0
• ### Incidence Calculus: A Mechanism for Probabilistic Reasoning
Mechanisms for the automation of uncertainty are required for expert sys...
03/27/2013 ∙ by Alan Bundy, et al. ∙ 0
• ### The Bang Calculus and the Two Girard's Translations
We study the two Girard's translations of intuitionistic implication int...
04/15/2019 ∙ by Giulio Guerrieri, et al. ∙ 0
• ### A Calculus for Modular Loop Acceleration
Loop acceleration can be used to prove safety, reachability, runtime bou...
01/06/2020 ∙ by Florian Frohn, et al. ∙ 0
• ### Grid-Free Computation of Probabilistic Safety with Malliavin Calculus
We work with continuous-time, continuous-space stochastic dynamical syst...
04/29/2021 ∙ by Francesco Cosentino, et al. ∙ 0
• ### On Randomised Strategies in the λ-Calculus (Long Version)
In this work we introduce randomised reduction strategies, a notion alre...
05/10/2018 ∙ by Ugo Dal Lago, et al. ∙ 0
##### This week in AI
Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.
## I Introduction
The pervasive role of stochastic models in a variety of domains (such as machine learning, natural language, verification) has prompted a vast body of research on probabilistic programming languages; such a language supports at least discrete distributions by providing an operator which models sampling. In particular, the functional style of probabilistic programming, pioneered by
[29], attracts increasing interest because it allows for higher-order computation, and offers a level of abstraction well-suited to deal with mathematical objects. Early work [18, 25, 22, 27, 23] has evolved in a growing body of software development and theoretical research. In this context, the -calculus has often been used as a core language.
In order to model higher-order probabilistic computation, it is a natural approach to take the -calculus as general paradigm, and to enrich it with a probabilistic construct. The most simple and concrete way to do so ([24, 8, 12]) is to equip the untyped -calculus with an operator , which models flipping a fair coin. This suffices to have universality, as proved in [8], in the sense that the calculus is sound and complete with respect to computableprobability distributions. The resulting calculus is however non-confluent, as it has been observed early (see [8] for an analysis). We revise the issue in Example 1. The problem with confluence is handled in the literature by fixing a deterministic reduction strategy, typically the leftmost-outermost strategy. This is not satisfactory both for theoretical and practical reasons, as we discuss later.
In this paper, we propose a more general point of view. Our goal is a foundational calculus, which plays the same role as the -calculus does for deterministic computation. More precisely, taking the point of view propounded by Plotkin in [26], we discriminate between a calculus and a programming language. The former defines the reduction rules, independently from any reduction strategy, and enjoys confluence and standardization, the latter is specified by a deterministic strategy (an abstract machine). Standardization is what relates the two: the programming language implements the standard strategy associated to the calculus. Indeed, standardization implies the existence of a strategy (the standard strategy) which is guaranteed to reach the result, if it exists.
In this spirit, we consider a probabilistic calculus to be characterized by a specific calling mechanism; the reduction is otherwise only constrained by the need of discriminating between duplicating a function which samples from a distribution, and duplicating the result of sampling. Think of tossing a coin and duplicating the result, versus tossing the coin twice, which is indeed the issue at the core of confluence failure, as the following examples (adapted from [9, 8]) show.
###### Example 1 (Confluence).
Let us consider the untyped -calculus extended with a binary operator which models fair, binary probabilistic choice: reduces to either or with equal probability ; we write this as . Intuitively, the result of evaluating a probabilistic term is a distribution on its possible values.
1. Consider the term , where , and ; is the standard construct for exclusive , and are terms which code boolean values.
– If we first reduce , we obtain or , with equal probability . This way, evaluates to , i.e. with probability .
– If we reduce the outermost redex first, reduces to , and the term evaluates to the distribution .
The two resulting distributions are not even comparable.
2. The same phenomenon appears even if we restrict ourselves to call-by-value. Consider for example the reductions of with as in (1.), and . We obtain the same two different distributions as above.
In this paper, we define two probabilistic -calculi, respectively based on the call-by-value (CbV) and call-by-name (CbN) calling mechanism. Both enjoy confluence and standardization, in an extended way: indeed we revisit these two fundamental notions to take into account the asymptotic behaviour of terms. The common root of the two calculi is a further calculus based on Linear Logic, which is an extension of Simpson’s linear -calculus [31], and which allows us to develop a unified, modular approach.
##### Content and Contributions
In Section IV, we introduce a call-by-value calculus, denoted , as a probabilistic extension of the call-by-value -calculus of Plotkin (where the -reduction fires only in case the argument is a value, i.e. either a variable or a -abstraction.) We choose to study in detail call-by-value for two main reasons. First, it is the most relevant mechanism to probabilistic programming (most of the abstract languages we cited are call-by-value, but also real-world stochastic programs such as Church [15]). Second, call-by-value is a mechanism in which dealing with functions, and duplication of functions, is clean and intuitive, which allows us to address the issue at the core of confluence failure. The definition of value (in particular, a probabilistic choice is not a value) together with a suitable restriction of the evaluation context for the probabilistic choice, allow us to recover key results: confluence and a form of standardization (Section V). Let us remind that, in the classical -calculus, standardization means that there is a strategy which is complete for all reduction sequences, i.e., for every reduction sequence there is a standard reduction sequence from to . A standard reduction sequence with the same property exists also here. An unexpected result is that strategies which are complete in the classical case, are not so here, notably the leftmost strategy.
In Section VI we study the asymptotic behavior of terms. Our leading question is how the asymptotic behaviour of different sequences starting from the same term compare. We first analyze if and in which sense confluence implies that the result of a probabilistically terminating computation is unique. We formalize the notion of asymptotic result via limit distributions, and establish that there is a unique maximal one.
In Section VII we address the question of how to find such greatest limit distribution, a question which arises from the fact that evaluation in is non-deterministic, and different sequences may terminate with different probability. With this aim, we extend the notion of standardization to limits; this extension is non-trivial, and demands the development of new sophisticated proof methods.
We prove that the new notion of standardization supplies a family of complete reduction strategies which are guaranteed to reach the unique maximal result. Remarkably, we are able to show that, when evaluating programs, i.e., closed terms, this family does include the leftmost strategy. As we have already observed, this is the deterministic strategy which is typically adopted in the literature, in either its call-by-value ([18, 7]) or its call-by-name version ([24, 12]), but without any completeness result with respect to probabilistic computation. Our result offers an “a posteriori” justification for its use!
The study of allows us to develop a crisp approach, which we are then able to use in the study of different probabilistic calculi. Because the issue of duplication is central, it is natural to expect a benefit from the fine control over copies which is provided by Linear Logic. In Section IX we use our tools to introduce and study a probabilistic linear -calculus, . The linear calculus provides not only a finer control on duplication, but also a modular approach to confluence and standardization, which allow us to formalize a call-by-name version of our calculus, namely , in Section X. We prove that enjoys properties analogous to those of , in particular confluence and standardization.
In Section II we provide the reader with some background and motivational observations. Basic notions of discrete probability and rewriting are reviewed in Section III.
##### Related Work
The idea of extending the -calculus with a probabilistic construct is not new; without any ambition to be exhaustive, let us cite [22, 27], [24, 12, 8, 5]. In all these cases, a specific reduction strategy is fixed; they are indeed languages, not calculi, according to Plotkin’s distinction.
Confluence failure in the probabilistic case is well-analyzed in [8], which then studies the operational semantics of both a CbV and CbN language, and the translation between the two.
The issue about confluence appears every time the -calculus is extended with a choice effect: quantum, algebraic, non-deterministic. Confluence for an algebric calculus is dealt with in [1] for the call-by-value, and in [32] for the call-by-name. In the quantum case we would like to cite [7, 6, 10], which use Simpson’s calculus [31]. The ways of framing the same problem in different settings are naturally related, and we were inspired by them.
To our knowledge, the only proposal of a probabilistic -calculus in which the reduction is independent from a specific strategy is for call-by-name, namely the calculus of [19], in the line of work of differential [13] and algebric [32] -calculus. The focus in [19] is essentially semantical, as the author want to study an equational theory for the -calculus, based on an extension of Böhm trees. [19] develops results which in their essence are similar to those we obtain for call-by-name in Sec. X, in particular confluence and standardization, even if his calculus –which internalizes the probabilistic behavior– is quite different from ours, and so are the proof techniques.
Finally, we wish to mention that proposals of a probabilistic -calculus could also be extracted from semantical models, such as the one in [3], which develops an idea earlier presented in [30], and in which the notion of graph models for -calculus has been extended with a probabilistic construct.
## Ii Background and Motivational observations
In this section, we first revise -in a non-technical way- the specificities of probabilistic program, and how they differ from classical ones. We then focus on some motivational observations which are relevant to our work. First, we give an example of features which are lost if a programming language is characterized by a strategy which is not rooted in a more general calculus. Then, we illustrate some of the issues which appear when we study a general calculus, instead of a specific reduction strategy. To address these issues in the paper, will lead us to develop new notions and tools.
### Ii-a Classical vs. Probabilistic Programs
A classical program defines a deterministic input-output relation; it terminates (on a given input), or does not; if it terminates, it takes finitely many steps to do so. Instead, a probabilistic program generates a probability distribution over possible outputs; it terminates (on a given input) with a certain probability; it may take infinitely many steps even when termination has probability .
A probabilistic program is a stochastic model. The intuition is that the probabilistic program is executed, and random choices are made by sampling; this process defines a distribution over all the possible outputs of . Even if the termination probability is (almost sure termination), that degree of certitude is typically not reached in any finite number of steps, but it appears as a limit. A standard example is a term which reduces to either the normal form or itself, with equal probability . After steps, reduces to with probability . Only at the limit this computation terminates with probability .
##### Probabilistic vs. Quantitative
The notion of probabilistic termination is what sets apart probabilistic -calculus from other quantitative calculi such as those in [1, 13, 32], and from the non-deterministic -calculus [9]. For this reason, the asymptotic behaviour of terms will be the focus of this paper.
### Ii-B Confluence of the calculus is relevant to programming
Functional languages have their foundation in the -calculus and its properties, and such properties (notably, confluence and standardization) have theoretical and practical implications. A strength of classical functional languages -which is assuming growing importance- is that they are inherently parallel (we refer e.g. to [21] for discussion on deterministic parallel programming): every sub-expression can be evaluated in parallel, because of referential transparency abstracts over the execution order; still, we can perform reasoning, testing and debugging on a program using a sequential model, because the result of the calculus is independent from the evaluation order. Not to force a sequential strategy impacts the implementation of the language, but also the conception of programs. As advocated by Harper [17], the parallelism of functional languages exposes the “dependency structure of the computation by not introducing any dependencies that are not forced on us by the nature of the computation itself."
This feature of functional languages is rooted in the confluence of the -calculus, and is an example of what is lost in the probabilistic setting, if we give-up either confluence, or the possibility of non-deterministic evaluation.
### Ii-C The result of probabilistic computation
A ground for our approach is the distinction between calculus and language. Some of the issues which we will need to address do not appear when working with probabilistic languages, because they are based on a simplification of the -calculus. Programming languages only evaluate programs, i.e., closed terms (without free variables). A striking simplification appears from another crucial restriction, weak evaluation, which does not evaluate function bodies (the scope of -abstractions). In weak call-by-value (base of the ML/CAML family of probabilistic languages) values are normal forms.
What is the result of a probabilistic computation is well understood only in the case of programming languages: the result of a program is a distribution on its possible outcomes, which are normal forms w.r.t. a chosen strategy. In the literature of probabilistic -calculus, two main deterministic strategies have been studied: weak left strategy in CbV [8] and head strategy in CbN [12], whose normal forms are respectively the closed values and the head normal forms.
When considering a calculus instead of a language, the identity between normal forms and results does not hold anymore, with important consequences in the definition of limit distributions. We investigate this issue in Sec. VI. The approach we develop is general and uniform to all our calculi.
## Iii Technical Preliminaries
We review basic notions on discrete probability and rewriting which we use through the paper. We assume that the reader has some familiarity with the -calculus.
### Iii-a Basics on Discrete Probability
A discrete probability space is given by a pair , where is a countable set, and is a discrete probability distribution on , i.e. is a function from to such that . In this case, a probability measure is assigned to any subset as
. In the language of probability theory, a subset of
is called an event.
Let be as above. Any function , where is another a countable set, induces a probability distribution on by composition: i.e. . In the language of probability theory, is called a discrete random variable on .
###### Example 2 (Die).
1. Consider tossing a die once. The space of possible outcomes is the set . The probability measure of each outcome is . The event
“result is odd"
is the subset , whose probability measure is .
2. Let be a set with two elements , and the obvious function from to . induces a distribution on , with and .
### Iii-B Subdistributions and DST(Ω)
Given a countable set , a function is a probability subdistribution if . We write for the set of subdistributions on . With a slight abuse of language, we will use the term distribution also for subdistribution. Subdistributions allow us to deal with partial results and non-successful computations.
Order: is equipped with the standard order relation of functions : if for each .
Support: The support of is .
Representation: We represent a distribution by explicitly indicating the support, and (as superscript) the probability assigned to each element by . We write if and otherwise.
### Iii-C Multidistributions
To syntactically represent the global evolution of a probabilistic system, we rely on the notion of multidistribution [2].
A multiset is a (finite) list of elements, modulo reordering, i.e. ; the multiset has three elements. Let be a countable set and a multiset of pairs of the form , with , and . We call (where the index set ranges over the elements of ) a multidistribution on if . We denote by the set of all multidistributions on .
We write the multidistribution simply as . The sum of multidistributions is denoted by , and it is the concatenation of lists. The product of a scalar and a multidistribution is defined pointwise: .
Intuitively, a multidistribution is a syntactical representation of a discrete probability space where at each element of the space is associated a term of and a probability measure. To the multidistribution we associate a probability distribution as follows:
μ(M)={\footnotesizep if p=∑i∈Ipi s.t. Mi=M0otherwise;
and we call the probability distribution associated to .
###### Example 3 (Distribution vs. multidistribution).
If , then . Please observe the difference between distribution and multidistribution: if , then , but .
### Iii-D Binary relations (notations and basic definitions)
Let be a binary relation on a set . We denote its reflexive and transitive closure. We denote the reflexive, symmetric and transitive closure of . If , we write if there is no such that ; in this case, is in -normal form. Figures convention: as is standard, in the figures we depict as ; solid arrows are universally quantified, dashed arrows are existentially quantified.
##### Confluence and Commutation
Let . The relations and on commute if ( and ) imply there is such that ( and ); they diamond-commute (-commute) if ( and ) imply there is such that ( and ). The relation is confluent (resp. diamond) if it commutes (resp. -commutes) with itself. It is well known that -commutation implies commutation, and diamond implies confluence.
## Iv Call-by-Value calculus Λcbv⊕
In this section we define a CbV probabilistic -calculus, which we denote by .
### Iv-a Syntax of Λcbv⊕
#### Iv-A1 The language
Terms and values are generated respectively by the grammars:
where ranges over a countable set of variables (denoted by ). and denote respectively the set of terms and of values. Free variables are defined as usual. denotes the term obtained by capture-avoiding substitution of for each free occurrence of in .
Contexts () and surface contexts () are generated by the grammars:
where denotes the hole of the term context. Given a term context , we denote by the term obtained from by filling the hole with , allowing the capture of free variables. All surface contexts are contexts. Since the hole will be filled with a redex, surface contexts formalize the fact that the redex (the hole) is not in the scope of a -abstraction, nor of a .
denotes the set of multi-distributions on .
#### Iv-A2 Reductions
We first define reduction rules on terms (Fig. 1), and one-step reduction from terms to multidistributions (Fig. 2). We then lift the definition of reduction to a binary relation on .
Observe that, usually, a reduction step is given by the closure under context of the reduction rules. However, to define a reduction from term to term is not informative enough, because we still have to account for the probability. The meaning of the probabilistic choice is that this term reduces to either or , with equal probability . There are various way to formalize this fact; here, we use multidistributions.
##### Reduction Rules and Steps
The reduction rules on the terms of are defined in Fig. 1.
The (one-step) reduction relations are defined in Fig. 2. Observe that the probabilistic rules are closed only under surface contexts, while the reduction rule is closed under general context (hence is a conservative extension of Plotkin’s CbV -calculus, see IV-B). We denote by the union .
##### Lifting
We lift the reduction relation to a relation , as defined in Fig. 3. Observe that is a reflexive relation.
We define in the same way the lifting of any relation to a binary relation on . In particular, we lift to .
##### Reduction sequences
A -sequence (or reduction sequence) from is a sequence such that (). We write to indicate that there is a finite sequence from to , and for an infinite sequence.
##### βv equivalence
We write for the transitive, reflexive and symmetric closure of ; abusing the notation, we will write for .
##### Normal Forms
Given , a term is in normal form if there is no such that . We also write . We denote by the set of the normal forms of .
It is immediate to check that all closed -normal forms are values, however a value is not necessarily a -normal form.
#### Iv-A3 Full Lifting
The definition of lifting allows us to apply a reduction step to any number of in the multidistribution . If no is reduced, then (the relation is reflexive). Another important case is when all for which a reduction step is possible are indeed reduced. This notion of full reduction, denoted by , is defined as follows.
Obviously, . As for the case of lifting, also the notion of full lifting can be extended to any reduction. So, for any , its full lifting is denoted by .
The relation plays an important role in VII.
### Iv-B Λcbv⊕ and the λ-calculus
A comparison between and the -calculus is in order.
Let be the set of the -terms; we denote by the CbN -calculus, equipped with the reduction [4], and by the CbV -calculus, equipped with the reduction [26].
is a conservative extension of . A translation can be defined as follows, where is a fresh variables which is used by no term:
\footnotesize(x)λ=xMNλ=MλNλM⊕Nλ=zMλNλλx.Mλ=λx.Mλ
The translation is injective (if then ) and preserves values.
###### Proposition 4 (Simulation).
The translation is sound and complete. Let .
1. implies ;
2. implies there is a (unique) , with and .
### Iv-C Discussion (Surface Contexts)
The notion of surface context which we have defined is familiar in the setting of -calculus, it corresponds to weak evaluation, which we have discussed in II-C.
In , the -reduction is unrestricted. Closing the -rules under surface context expresses the fact that the -redex is not reduced under -abstraction, nor in the scope of another . The former is fundamental to confluence: it means that a function which samples from a distribution can be duplicated, but we cannot pre-evaluate the sampling. The latter is a a technical simplification, which we adopt to avoid unessential burdens with associativity. To require no reduction in the scope of is very similar to allow no reduction in the branches of an if-then-else.
## V Confluence and Standardization
### V-a Confluence
We prove that is confluent. We modularize the proof using the Hindley-Rosen lemma. The notions of commutation and -commutation which we use are reviewed in Sec. III-D.
###### Lemma (Hindley-Rosen).
Let and be binary relations on the same set . Their union is confluent if both and are confluent, and and commute.
The following criterion allows us to work pointwise in proving commutation and confluence of binary relations on multidistributions, namely and .
###### Lemma 5 (Pointwise Criterion).
Let and their lifting (as defined in IV-A2). Property (*) below implies that -commute.
(*) If and then s.t. and .
###### Proof.
We prove that (**) and imply exists such that and .
Let . By definition of Lifting, for each , we have and , with and . It is easily checked, that for each , it exists s.t. and . If either or uses reflexivity (rule ), it is immediate to obtain . Otherwise, is given by property (*). Hence satisfies (**).
We derive confluence of from the same property in the CbV -calculus [26, 28], using the simulation of Prop. 4.
###### Lemma 6.
The reduction is confluent.
###### Proof.
Assume and . We first observe that if , then and are respectively of the shape , , with and . By Prop. 4, we can project such reduction sequences on , obtaining that for each , and . Since in CbV -calculus is confluent, there are such that and . By Prop. 4.2, for each there is a unique such that , and the proof is given. ∎
We prove that the reduction is diamond, i.e., the reduction diagram closes in one step.
###### Lemma 7.
The reduction is diamond.
###### Proof.
We prove that if and , then such that and . The claim then follows by Lemma 5, by taking .
Let , and . Because of definition of surface context, the two -redexes do not overlap: is a subterm of and is a subterm of . Hence we can reduce those redexes in and , to obtain the same .
We prove commutation of and by proving a stronger property: they -commute.
###### Lemma 8.
The reductions and -commute.
###### Proof.
By using Lemma 5, we only need to prove that if and , then such that and .
The proof is by induction on . Cases and are not possible given the hypothesis.
1. Case . is the only possible -redex. Assume the -redex is inside (the other case is similar), and that , . It is immediate that satisfies the claim.
2. Case . cannot have the form because neither nor could contain a -redex.
1. Assume that the -redex is inside , and the -redex inside . We have (with ), (with ). It is immediate that satisfies the claim. The symmetric case is similar.
2. Assume that both redexes are inside . Let us write as . Assume , , therefore and . We use the inductive hypothesis on to obtain such that , , . We conclude that for , it holds that and .
###### Theorem 9.
The reduction is confluent.
###### Proof.
By Hindley-Rosen, from Lemmas 8, 6, and 7. ∎
Let us say that is a normal-forms multidistribution if all are -normal forms (i.e. ). An immediate consequence of confluence is the following:
###### Corollary 10.
The normal-forms multidistribution to which reduces, if any, is unique.
#### V-A1 Discussion
While immediate, the content of Cor. 10 is hardly useful, for two reasons. First, we know that probabilistic termination is not necessarily reached in a finite number of steps; the relevant notion is not that , but rather that of a distribution which is defined as limit by the sequence . Second, in the Plotkin CbV calculus the result of computation is formalized by the notion of value, and considering normal forms as values is unsound ([26], page 135). In Section VI-B we introduce a suitable notion of limit distribution, and study the implications of confluence on it.
### V-B A Standardization Property
In this section, we first introduce surface and left reduction as strategies for . In the setting of the CbV -calculus, the former corresponds to weak reduction, the latter to the standard strategy originally defined in [26]. We then establish a standardization result, namely that every finite -sequence can be partially ordered as a sequence in which all surface reductions are performed first. A counterexample shows that in , a standardization result using left reduction fails.
#### V-B1 Surface and Left Reduction
We remind that in the -calculus, a deterministic strategy defines a function from terms to redexes, associating to every term the next redex to be reduced. More generally, we call reduction strategy for a reduction relation such that . The notion of strategy can be easily formalized through the notion of context. With this in mind, let us consider surface and left contexts.
• Surface contexts have been defined in Sec.IV-A1.
• Left contexts are defined by the following grammar:
L::=□∣LM∣VL
Note that in particular a left contexts is a surface context.
• We call surface reduction, denoted by (with lifting ) and left reduction, denoted by (with lifting ), the closure of the reduction rules in Fig. 1 under surface contexts and left contexts, respectively. It is clear that . Observe that .
• A reduction step is deep, written , if it is not a surface step. A reduction step is internal (written ) if it is not a left step. Observe that .
###### Example 11.
• () Let , where . Then and ; instead, , .
• () Let . Then and , while and
Intuitively, left reduction chooses the leftmost of the surface redexes. More precisely, this is the case for closed terms (for example, the term has a -step, but no -step).
Surface Normal Forms: We denote by the set of -normal forms. We observe that all values are surface normal forms (but the converse does not hold): (and ). The situation is different if we restrict ourselves to close term, in fact the following result holds, which is easy to check.
###### Lemma 12.
If is a closed term, the following three are equivalent:
1. is a -normal form;
2. is a -normal form;
3. is a value.
#### V-B2 Finitary Surface Standardization
The next theorem proves a standardization result, in the sense that every finite reduction sequence can be (partially) ordered in a sequence of surface steps followed by a sequence of deep steps.
###### Theorem 13 (Finitary Surface Standardization).
In , if then exists such that and .
###### Proof.
We build on an analogous result for CbV -calculus, which is folklore and is proved explicitly in Appendix V-B . We then only need to check that deep steps commute with -steps, which is straightforward technology (the full proof is in Appendix V-B ). ∎
##### Finitary Left Standardization does not hold
The following statement is false for .
“If then exists such that and ."
###### Example 14 (Counter-example).
Let us consider the following sequence, where and . . If we anticipate the reduction of , we have , from where we cannot reach . Observe that the sequence is already surface-standard !
## Vi Asymptotic Evaluation
The specificity of probabilistic computation is to be concerned with asymptotic behavior; the focus is not what happens after a finite number of steps, but when tends to infinity. In this section, we study the asymptotic behavior of -sequences with respect to evaluation. The intuition is that a reduction sequence defines a distribution on the possible outcomes of the program. We first clarify what is the outcome of evaluating a probabilistic term, and then we formalize the idea of result “at the limit" with the notion of limit distribution (Def. 19). In Sec. VI-B we investigate how the asymptotic result of different sequences starting from the same compare.
We remind that to any multidistribution on is associated a probability distribution on , as described in Sec.III-C. We assume the following letter convention: given a multidistribution we denote the associated distribution by the corresponding greek letter If is a -sequence, then is the sequence of associated distributions.
### Vi-a Probabilistic Evaluation
We start by studying the property of being valuable (VI-A1) and by analyzing some examples (VI-A2). This motivates the more general approach we introduce in VI-A3.
#### Vi-A1 To be valuable
In the CbV -calculus, the key property of a term is to be valuable, i.e., can reduce to a value. To be valuable is a yes/no property, whose probabilistic analogous is the probability to reduce to a value. If describes the result of a computation step, the probability that such a result is a value is simply , i.e. the probability of the event . Since the set of values is closed under reduction, the following property holds:
###### Fact 15.
If and , then , with , and .
Let be a -sequence, and the sequence of associated distributions. The sequence of reals is nondecreasing and bounded, because of Fact 15. Therefore the limit exists, and is the supremum: This fact allows us the following definition.
• The sequence evaluates with probability
if , written .
• is -valuable if is the greatest probability to which a sequence from can evaluate.
###### Example 16.
Let and .
1. Consider the term where . Then . Since , is -valuable.
2. Consider the term , where . Then It is immediate that is -valuable.
3. Let , so that is a divergent term, and let . Then is -valuable.
#### Vi-A2 Result of a CbV computation
The notion of being -valuable allows for a simple definition, but it is too coarse. Consider Example 16; both points (1.) and (2.) give examples of -valuable term. However, in (1.) the probability is concentrated in the value , while in (2.) and have equal probability . Observe that and are different normal forms, and are not -equivalent. To discriminate between and , we need a finer notion of evaluation. Since the calculus is CbV, the result “at the limit" is intuitively a distribution on the possible values that the term can reach. Some care is needed though, as the following example shows.
###### Example 17.
Consider Plotkin’s CbV -calculus. Let ; the term has the following -reduction: . We obtain a reduction sequence where , . Each is a value, but there is not a "final" one in which the reduction ends. Transposing this to , let , . The -sequence is -valuable, but the distribution on values is different at every step. In other words, , the sequence has no limit. Observe that however all the values are -equivalent.
#### Vi-A3 Observations and Limit distribution
Example 17 motivates the approach that we develop now: the result of probabilistic evaluation is not a distribution on values, but a distribution on some events of interest. In the case of , the most informative events are equivalence classes of values.
We first introduce the notion of observation, and then that of limit distribution.
###### Definition 18.
A set of observations for is a set such that , and if then .
Given , has probability (similarly to the event "the result is Odd" in Example 2).
It follows immediately from the definition that, given a sequence , then for each the sequence is nondecreasing and bounded, and therefore has a limit, the . This allows us to define a distribution on .
###### Definition 19.
Let be a set of observations. The sequence defines a distribution , where , for each .
• We call such a the limit distribution of . Letter convention: greek bold letters denote limit distributions.
• The sequence converges to (or evaluates to) the limit distribution , written
.
• If has a sequence which converges to , we write
.
• Given , we denote by the set of all limit distributions of . If has a greatest element, we indicate it by .
If is clear from the context, we omit the index which specifies it, and simply write , , .
The notion of limit distribution formalizes what is the result of evaluating a probabilistic term, once we choose the set of observations which interest us. In VI-B we prove that confluence implies that has a unique maximal element.
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# compare the penetrating power of alpha, beta and gamma radiation
There is always a finite probability for a gamma to penetrate a given thickness of absorbing material and so, unlike the charged particulate radia… . Beta particles can be dangerous and any contact with the body must be avoided, though their ionization power is low. They hold the highest power of penetration. Alpha, beta and gamma radiation are all real entities in the physics world and are worth avoiding when you can manage it. Beta radiation is more penetrating than alpha radiation. Properties of Alpha, Beta and Gamma Rays, More Lecture Notes from Easy Biology Class…, BotanyZoologyBiochemistryGeneticsMolecular BiologyBiotechnologyHuman PhysiologyPlant PhysiologyMicrobiologyImmunologyEmbryologyEcologyEvolutionBiophysicsResearch Meth.BiostatisticsChemistryPhysics, Lecture NotesBiology PPTVideo TutorialsBiology MCQQuestion BankDifference betweenPractical AidsMock Tests (MCQ)Biology Exams, If you like this post, please COMMENT . A substance with such an unstable nucleus is called the radioactive substance. However, it may take a material with a greater thickness and density to stop beta particles. They are physically similar to the electrons discussed earlier in this unit, but they are not in orbit around an atom. Whereas alpha particles have the lowest penetrating ability and can be stopped by a sheet of paper. There exist three major types of radiations emitted by the radioactive particles namely: These radiations are released from the nucleus of an atom. The Gamma rays carry a large amount of energy and can also travel via thick concrete and thin lead. All types are caused by unstable atoms, which have either an excess of energy or mass (or both). They can cause serious damage if get into the body due to their high ionization power. Answer to Which radiation has the greatest penetrating power: alpha particles, beta particles, or gamma rays?. Gamma rays are highly energetic waves and are poor at ionising other atoms or molecules. Please Share with Your Friends... (Compare Alpha Particles, Beta Particles and Gamma Rays – Table), Comparison Table of Alpha Beta and Gamma Rays. Penetration power: Low: Moderate, 100 times more than alpha particles: High, 100 times more than beta particles: Effect of magnetic and electric field: Deflected towards the negative late: Deflected towards the positive plate: Not deflected: Ionizing power: Greater than beta and gamma rays: Very low: Very low: Luminescence: Produce fluorescence and phosphorescence Therefore, it is always said : "gamma rays" but for alpha and beta rays, it is sometimes said: "beta particles" and "alpha particles". Uncharged ?~0.01a, electromagnetic radiation, ~10 cm in air, can be stopped by 1mm of Aluminium, Upto a few m in air, can be stopped by a thin layer of Aluminium, Several m in air, can be stopped by a thick layer of Lead, Excited nuclei formed as a result of Gamma decay. Your email address will not be published. The following table shows the similarities and differences between alpha particles, beta particles and gamma rays. 4. This means a thin wall is sufficient for stopping alpha particles. Gamma Rays have an ionising power so low that they penetrate very deeply into matter before most of the energy has been used up. This process of decaying continues till the nucleus attains a stable stage. PropertisAlpha RaysBeta RaysGamma RaysSymbolαβγMass6.65 X 10^-27 kg5.5 x 10^-4 amuNegligibleCharge2 positive charge1 negative chargeNo chargeSpeed and NatureHigh speed helium nucleusHigh speed electronsHigh speed electromagnetic radiationsVelocity~5% of the velocity of lightNearly equal to that of lightEqual to the velocity of lightPenetration powerLowModerate, 100 times more than alpha particlesHigh, 100 times more than beta particlesEffect of magnetic and electric fieldDeflected towards the negative lateDeflected towards the positive plateNot deflectedIonizing powerGreater than beta and gamma raysVery lowVery lowLuminescenceProduce fluorescence and phosphorescenceProduce phosphorescenceProduce phosphorescenceDistance travelled2 – 4 cm2 – 3 meters500 meters, (adsbygoogle=window.adsbygoogle||[]).push({}), @. Gamma rays are the most penetrating of the radiations. Hence, the atoms eventually decay by emitting a particle that transforms when they are unstable and transforms the nucleus into a lower energy state. Predict whether each type of ionizing radiation has the ability to penetrate, or pass through, our skin and body. Beta particles ionise the gas through which they pass. 5. 3. 3. Alpha particles are produced as a result of the alpha decay of a radioactive material such as Uranium-238. The main difference between alpha beta and gamma particles is that alpha particles have the least penetration power while beta particles have a moderate penetration power and gamma particles have the highest penetration power. Their behavior differs from one another, though all the three causes some ionization and carry some penetration power. Beta rays are high energy high and speed electrons emitted from a radioactive material after the beta decay. Alpha decay is seen only in heavier elements greater than atomic number 52, tellurium. It can be stopped (or absorbed) by a sheet of paper. Ionizing radiation takes a few forms: Alpha, beta, and neutron particles, and gamma and X-rays. . Alpha radiation is the least penetrating. Beta particles can be stopped by a few millimetres of aluminium. 84 Complete the nuclear equation in your answer booklet for the beta decay of the Co-60 by writing an isotopic notation for the missing product. Gamma rays are the most difficult to stop and require concrete, lead, or … Let’s discuss the properties of beta, alpha and gamma one by one. There are many types of emmitted particles and radiation that radioisotopes produce when they decay. Students will: 1. Beta particles have a higher penetration power when compared to alpha particles and can travel through the skin with ease. During radioactivity, particles like alpha, beta & gamma rays are emitted by an atom, due to unstable atom trying to gain stability. Hence, it is the electron that is emitted by the nucleus at a rapid pace. Differentiate between radiation exposure and radiation contamination. Difference between Alpha, Beta and Gamma Radiation Key Difference: Alpha radiation can be described as the producer of high energy and fast moving helium particles. In order to reach a stable state, they must release that extra energy or mass in the form of radiation. Beta particles travel faster than alpha particles and carry less charge (one electron compared to the 2 protons of an alpha particle) and so interact less readily with the atoms and molecules of the material through which they pass. They can be stopped by … Of the three types of radiation, alpha particles are the easiest to stop. It is identical to the helium nucleus. Radium-226 is a gamma emitter. Stay tuned with Byju’s to learn more about characteristics of alpha, beta and gamma rays along with their differentiation and common features. Gamma radiations are electromagnetic radiations with high energy and high penetration capacity produced from a radioactive material after the gamma decay. Alpha rays consist of two protons and two neutrons bound tougher into particles. Difference between GM Counter and Scintillation Counter, Difference between Simple Diffusion and Facilitated Diffusion: A Comparison Table, Difference between Covalent Bond and Metallic Bond: Comparison Table, Difference between True Solution, Colloidal Solution and Suspension: A Comparison Table, Difference between Phase Contrast Microscopy and Differential Interference Contrast Microscopy: (Easy Short Notes), Difference between GM Counter and Proportional Counter, Moderate, 100 times more than alpha particles. They are also called beta particles. Being electrically neutral, the interaction of gamma rays with matter is a statistical process and depends on the nature of the absorber as well as the energy of the gamma. Beta Particle Beta-particles are given out from the nucleus of the atom while cathode rays are given out from its orbital electrons. A sheet of paper is all that is needed for the absorption of alpha rays. The penetrating power of alpha rays, beta rays, and gamma rays varies greatly. The interactions of the various radiations with matter are unique and determine their penetrability through matter and, consequently, the type and amount of shielding needed for radiation protection. Alpha-particle is highly active and energetic helium atom that contains two neutrons and protons. Express the changes in the atomic number and mass number of a radioactive nuclei when an alpha, beta, or gamma particle is emitted. The table of the summary of properties shows that alpha particles are the biggest, beta particles are very much smaller and gamma rays have no mass. 4. With all the radiation from natural and man-made sources, we should quite reasonably be concerned about how all the radiation might affect our health. The bigger the particle, the more likely it is to have a collision with the atoms of the material. Penetrating power of Beta-particles is more than that of alpha particles. The other two types of decay are seen in all of the elements. Penetrating ability depends on the size of the radioactive particle. Compare qualitatively the ionizing and penetration power of alpha particles $$\left( \alpha \right)$$, beta particles $$\left( \beta \right)$$, and gamma rays $$\left( \gamma \right)$$. The other radiations, beta and alpha, have particle charecter. Compare qualitatively the ionizing and penetration power of alpha particles $$\left( \alpha \right)$$, beta particles $$\left( \beta \right)$$, and gamma rays $$\left( \gamma \right)$$. 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