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https://math.stackexchange.com/questions/3179933/simplifying-expression-mobius-function | # Simplifying expression Mobius Function
Can anybody help me simplify this expression using Mobius Inversion Formula or any other result in order to calculate F(3500) in a simple way?? $$F(n)=\sum_{d\mid n} \mu(d)d$$
• Hint: $\mu(n)n$ is multiplicative, hence so is $F(n).$ – Thomas Andrews Apr 8 at 17:48
• Thank you very much for your hint. This is the key of the exercise (It was not about Mobius Inversion) !! – Luis Gimeno Sotelo Apr 8 at 17:59
It is the product $$f(n)=(1-p_1)(1-p_2)\cdots(1-p_k)$$ where the $$p_i$$ are the distinct prime divisors of $$d.$$
This is because $$g(n)=\mu(n)n$$ is multiplicative, and hence so is $$F,$$ and $$F(p^k)=1-p$$ for prime $$p.$$
In your case, $$3500$$ has prime divisors $$2,5,$$ and $$7$$. So:
$$F(3500)=(1-2)(1-5)(1-7)=-24$$
You could write it as $$F(n)=n\sum_{d\mid n} \mu(d)\frac{1}{n/d}.$$ Then you get, by Möbius inversion: $$\frac{1}{n}=\sum_{d\mid n}\frac{F(d)}{d}$$
Or:
$$1=\sum_{d\mid n}\frac{n}{d}F(d)$$
That won't help you compute $$F$$ much however. | 2019-06-17 05:04:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.891402542591095, "perplexity": 392.78035378638157}, "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/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00396.warc.gz"} |
https://math.stackexchange.com/questions/3159245/show-that-if-m-in-mathbbz-has-greatest-common-divisor-1-with-21-then | # Show that if $m$ in $\mathbb{Z}$ has greatest common divisor $1$ with $21$, then $m^{6}-1$ is divisible by $63$.
Show that if $$m$$ in $$\mathbb{Z}$$ has greatest common divisor $$1$$ with $$21$$, then $$m^{6}-1$$ is divisible by $$63$$. Also, I have to work in $$\mathbb{Z}/63 \mathbb{Z}^{*}$$, thus the group $$63$$ modulo $$\mathbb{Z}$$ under multiplication.
So in other words, I have to prove $$m^{6} \equiv 1 (\mod 63)$$.
I have no idea how to tie the fact that $$m$$ has $$1$$ as gcd with $$21$$ together with working in the $$63$$ modulo $$\mathbb{63}$$ group. I would really appreciate some hints or suggestions guide me in the right direction.
$$(m,7)=1\Rightarrow m^{\phi(7)}=m^{6}\equiv 1 \pmod{7}$$
and
$$\begin{cases} m\equiv 1\pmod{3}\Rightarrow 9|(m-1)(m^2+m+1)(m^3+1)=m^6-1\\ m\equiv 2 \pmod{3} \Rightarrow 9|(m^3+1)|(m^6-1) \end{cases}\Rightarrow m^6\equiv1\pmod{9}$$
Hint: The fact that $$(m,21)=1$$ implies, clearly, that $$(m,63)=1$$. Hence we can use Euler's Theorem, which says that $$x^{\phi(n)}=1$$ mod $$n$$ for all $$n$$, and $$x$$ coprime to $$n$$. Here, $$\phi$$ is Euler's totient function. Can you proceed further?
• Using Eulers totient function, I would say $\varphi(63)=\varphi(3^{2} \cdot 7) = \varphi( ( 3^{2}-3^{1}) \cdot 7) = \varphi(42)$. Substracting $42$ from $63$ gives $21$, thus I could work in $m^{6} \equiv 1 (\mod 21)$, correct? But I don't see how this makes the problem easier. Mar 23 '19 at 13:42
• @Mathbeginner there's no need to work mod $21$, Euler's theorem gives you mod $63$, which is exactly what you want. (i.e. use $n=63$ and $x=m$ immediately) Mar 23 '19 at 15:18
• But that would be just stating that since we know $m \equiv 1 (\mod 21)$, this implies $m \equiv 1 (\mod 63)$ and that would be the end of it. Is that what you are saying? Mar 24 '19 at 8:31
The key point is that $$21$$ and $$63$$ have the same prime divisors.
Indeed, since $$21 = 3 \cdot 7$$ and $$63 = 3^2 \cdot 7$$, we have $$\gcd(m,21)=1 \iff \gcd(m,3)=1=\gcd(m,7) \iff \gcd(m,63)=1$$ Then Euler's theorem gives $$m^6\equiv 1 \bmod 7, \quad m^6\equiv 1 \bmod 9$$ Therefore, $$m^6\equiv 1 \bmod 63$$.
Carmichael Function $$\lambda(63)=[\lambda(9),\lambda(7)]=6$$
$$(m,21)=1\implies (m,7)=(m,3)=1$$
• Downvoting this seems a bit harsh. Though it may well be beyond the OP's knowledge level, it may help to introduce readers to this handy generalization. Mar 23 '19 at 15:30
• @BillDubuque, Down-voting like up-voting is not controlled here in MSE. Just now received one: math.stackexchange.com/questions/771320/… .Sometimes I try to learn the flaw: math.stackexchange.com/questions/3157905/… Mar 24 '19 at 6:10 | 2021-10-17 16:11:08 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9340272545814514, "perplexity": 190.9906280241428}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585178.60/warc/CC-MAIN-20211017144318-20211017174318-00528.warc.gz"} |
https://zbmath.org/?q=an:1298.35069 | # zbMATH — the first resource for mathematics
On the existence of a continuous branch of nodal solutions of elliptic equations with convex-concave nonlinearities. (English. Russian original) Zbl 1298.35069
Differ. Equ. 50, No. 6, 765-776 (2014); translation from Differ. Uravn. 50, No. 6, 768-779 (2014).
Summary: We study the existence of nodal solutions of a parametrized family of Dirichlet boundary value problems for elliptic equations with convex-concave nonlinearities. In the main result, we prove the existence of nodal solutions $$u_{\lambda}$$ for $$\lambda \in (-\infty, \lambda ^*_{0})$$. The critical value $$\lambda^*_{0} >0$$ is found by a spectral analysis procedure according to Pokhozhaev’s fibering method. We show that the obtained solutions form a continuous branch (in the sense of level lines of the energy functional) with respect to the parameter $$\lambda$$. Moreover, we prove the existence of an interval $$(-\infty ,\tilde \lambda )$$, where $$\tilde \lambda > 0$$, on which this branch consists of solutions with exactly two nodal domains.
##### MSC:
35J57 Boundary value problems for second-order elliptic systems
Full Text:
##### References:
[1] Ambrosetti, A; Brezis, H; Cerami, G, Combined effects of concave and convex nonlinearities in some elliptic problems, J. Funct. Anal., 122, 519-543, (1994) · Zbl 0805.35028 [2] Radulescu, V; Repovs, D, Combined effects in nonlinear problems arising in the study of anisotropic continuous media, Nonlinear Anal., 75, 1524-1530, (2012) · Zbl 1237.35043 [3] Callegari, A; Nachman, A, A nonlinear singular boundary value problem in the theory of pseudoplastic fluids, SIAM J. Appl. Math., 38, 275-281, (1980) · Zbl 0453.76002 [4] Gierer, A; Meinhardt, H, A theory of biological pattern formation, Kybernetik, 12, 30-39, (1972) [5] Keller, EF; Segel, LA, The initiation of slime mold aggregation viewed As an instability, J. Theoret. Biol., 26, 399-415, (1970) · Zbl 1170.92306 [6] Il’yasov, Ya, On nonlocal existence results for elliptic equations with convex-concave nonlinearities, Nonlinear Anal., 61, 211-236, (2005) · Zbl 1190.35112 [7] Bartsch, T; Willem, M, On an elliptic equation with concave and convex nonlinearities, Proc. Amer. Math. Soc., 123, 3555-3561, (1995) · Zbl 0848.35039 [8] Lubyshev, VF, Multiple positive solutions of an elliptic equation with a convex-concave nonlinearity containing a sign-changing term, Tr. Mat. Inst. Steklova, 269, 167-180, (2010) [9] Castro, A; Cossio, J; Neuberger, JM, A sign-changing solution for a superlinear Dirichlet problem, Rocky Mountain J. Math., 27, 1041-1053, (1997) · Zbl 0907.35050 [10] Bartsch, T; Weth, T, A note on additional properties of sign changing solutions to superlinear elliptic equations, Topol. Methods Nonlinear Anal., 22, 1-14, (2003) · Zbl 1094.35041 [11] Clapp, M; Weth, T, Minimal nodal solutions of the pure critical exponent problem on a symmetric domain, Calc. Var. Partial Differential Equations, 21, 1-14, (2004) · Zbl 1097.35048 [12] Liu, Z; Wang, Z-Q, Sign-changing solutions of nonlinear elliptic equations, Front. Math. in China, 3, 221-238, (2008) · Zbl 1158.35370 [13] Cepicka, J; Drabek, P; Girg, P, Open problems related to the $$p$$-Laplacian, 13-34, (2009) [14] Courant, R. and Hilbert, D., Metody matematicheskoi fiziki (Methods ofMathematical Physics), Moscow, 1933, 1945, vols. 1,2. [15] Drabek, P; Robinson, SB, On the generalization of the Courant nodal domain theorem, J. Differential Equations, 181, 58-71, (2002) · Zbl 1163.35449 [16] Helffer, B; Hoffmann-Ostenhof, T; Terracini, S, Nodal domains and spectral minimal partitions, Ann. Inst. H. Poincaré Anal. Non Linéaire, 26, 101-138, (2009) · Zbl 1171.35083 [17] Nehari, Z, On a class of nonlinear second-order differential equations, Trans. Amer. Math. Soc., 95, 101-123, (1960) · Zbl 0097.29501 [18] Szulkin, A. and Weth, T., The Method of Nehari Manifold: Handbook of Nonconvex Analysis and Applications, Boston, 2010. · Zbl 1218.58010 [19] Pokhozhaev, SI, An approach to nonlinear equations, Dokl. Akad. Nauk SSSR, 247, 1327-1331, (1979) [20] Pokhozhaev, SI, The fibering method for solving nonlinear boundary value problems, Tr. Mat. Inst. Steklova, 192, 146-163, (1990) · Zbl 0734.35036 [21] Il’yasov, Ya, On a procedure of projective fibration of functionals on Banach spaces, Proc. Steklov Inst. Math., 232, 150-156, (2001) · Zbl 1032.46095 [22] Kinderlehrer, D. and Stampacchia, G., An Introduction to Variational Inequalities and Their Applications, New York, 1980. · Zbl 0457.35001 [23] Dunford, N. and Schwartz, J., Linear Operators. General Theory, New York, 1958. Translated under the title Lineinye operatory. T. 1. Obshchaya teoriya, Moscow: Inostrannaya Literatura, 1962. [24] Ambrosetti, A. and Malchiodi, A., Nonlinear Analysis and Semilinear Elliptic Problems, Cambridge, 2007. · Zbl 1125.47052 [25] Willem, M., Minimax Theorems, Boston, 1996. · Zbl 0856.49001 [26] Vrahatis, MN, A short proof and a generalization of miranda’s existence theorem, Proc. Amer. Math. Soc., 107, 701-703, (1989) · Zbl 0695.55001
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-11-27 23:12:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7802104949951172, "perplexity": 3205.8664919171956}, "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-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00574.warc.gz"} |
https://www.physicsforums.com/threads/rate-at-which-sun-converts-mass-to-radiation.900737/ | # Homework Help: Rate at which Sun converts mass to radiation
Tags:
1. Jan 18, 2017
### BRN
Hi folk, i need an help for this esercise.
1. The problem statement, all variables and given/known data
Determines the rest mass lost by the sun every second for emission of radiation. It assumes that the surface temperature of the sun is 5700 K and the diameter $D_s=1.4 × 10^9 m$.
3. The attempt at a solution
The solar nuclear reactions generates helium fron Hydrogen. Part of the mass involved in the reaction is lost by eletromagnetic radiation emission. Energy is emitted in the form of gamma fotons and fast neutrinos.
The amount of energy at all wavelengths that at each second strikes perpendicularly a m2 of a surface exposed to solar radiation, takes the name of the solar constant and his value is $C_s=1.36*10^{3}[W/m^2]$
If I consider the sun as a sphere of radius $D_s/2$, its surface is:
$S=4\pi(\frac{D_s}{2})^2=6.1575*10^{18}[m^2]$
The total energy emitted fron the sun in all directions is:
$E=C_s*S=8.3742*10^21[W/s]=8.3742*10^21[J]$
Then, I determine the rest mass lost by Einstein reletion:
$E=mc^2 \Rightarrow m=\frac{E}{c^2}=9.3177*10^{4}[kg]$
but the solution is $4.10×10^9 [kg/s]$
Someone could help me? Thanks!
EDIT: The post title is wrong. How can I change it?
2. Jan 19, 2017
### Carlos PdL SdT
I think you used the solar constant on earth when you should have been using the one on the surface of the sun which is way higher. I would suggest you work with the 5700K and work out the Energy from there
3. Jan 19, 2017
### Staff: Mentor
What would you like it changed to?
4. Jan 19, 2017
### vela
Staff Emeritus
I changed it yesterday after I moved the thread; it used to be about the "cohesive energy of argon."
5. Jan 19, 2017
### Cutter Ketch
They gave you the temperature expecting you to treat the sun as a blackbody. You have probably been taught a formula for the radiance of a blackbody.
Looking up the solar constant probably wasn't the expected approach, but you can get there that way. The solar constant is the radiance at 1 AU. If you know how the radiation falls off with distance (and I bet you do) then it is simple to scale that value back to the surface of the sun.
6. Jan 22, 2017
### BRN
Sorry for the delay, but were difficult days...
If, however, consider the sun like a blackbody, the irradiance is:
$R=\frac{c}{4}u(\epsilon ,T)$
with $u(\epsilon ,T)$ states density.
But the density of states as it is calculated?
If, I use the balckbody states density:
$u=\frac{1}{V}g_sg_{ph}(\epsilon)[n_E]_B\epsilon=\frac{1}{\pi^2\hbar^3c^3}\frac{\epsilon^3}{e^{\frac{\epsilon}{k_BT}}-1}$
with $g_{ph}(\epsilon)=\frac{V}{\pi^2\hbar^3c^3}\epsilon^2$
I get $R(\epsilon, T)=0$...
Last edited: Jan 22, 2017
7. Jan 22, 2017
### Staff: Mentor
How do you get this?
There is an easier formula for the total electromagnetic emission. You don't have to integrate anything that way.
8. Jan 22, 2017
### BRN
I simply reported blackbody equations that I have on my book.
My reference book is the "Manini - introduction to the physics of matter" and I admit that it is the worst book in which to investigate these things ...
What formula are you speaking?
9. Jan 22, 2017
### Staff: Mentor
The Stefan–Boltzmann law is sufficient here.
10. Jan 22, 2017
### BRN
I don't have Stefan - Boltzmann law in my book...
At any way, if i use this law, I have:
$q=\sigma T^4=5.9855*10^7 [W/m^2]$
with Stefan - Boltzmann costant $\sigma=5.6703*10^{-8}[W/m^2K^4]$
I'm confused...
In the exercise test, there is the sun diameter too. What is this good for?
11. Jan 22, 2017
### Staff: Mentor
Looking at the units of a given constant can be insightful.
Note the m2 in the denominator of the units for σ. The implication is that it involves a measure of area. So the constant relates the energy emitted in Watts from a surface of some area at a given temperature... Now, how might you determine the area of the surface in question?
12. Jan 23, 2017
### BRN
Ok, the Stefan - Boltzmann law is refered at $1 m^2$. With the units I have no problems, but with the values ...
From Stefan - Boltzmann law I get the emittance for $m^2$ unity:
$q=\sigma T^4=5.9855*10^7 [W/m^2]$
The Sun surface is:
$S=4\pi(\frac{D_s}{2})^2=6.1575*10^{18}[m^2]$
then,
$E=q*S=3.6855*10^{26}[W]=3.6855*10^{26}[J/s]$
From Einstein equations I have:
$E=mc^2 \Rightarrow m=\frac{E}{c^2}=1.2293*10^{18}[kg/s]$
I miss a pass, maybe from q to E.
13. Jan 23, 2017
### Staff: Mentor
Try from E to m. You didn't square c.
14. Jan 23, 2017
### BRN
OH DAMN! When I make these mistakes would kill me!
Now it's all ok!
Do you recommend a good book for studying the physics of matter? (not too difficult )
15. Jan 23, 2017
### Staff: Mentor
I don't have any particular recommendations myself, but you can ask in the Science and Math Textbooks forum. | 2018-07-22 05:34:18 | {"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": 0.6379644274711609, "perplexity": 1741.086357250457}, "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/1531676593010.88/warc/CC-MAIN-20180722041752-20180722061752-00072.warc.gz"} |
https://amgviajes.com.pe/w0w38/eigenvalue-of-singular-matrix-40dce9 | Hence ϕt,t−j=fj(t∕p) if 1≤j≤k and ϕt,t−j=0 if j>k. Applying the bordered matrix construction described above to the biproduct gives a defining function for A to have a single pair of eigenvalues whose sum is zero. share | cite | improve this question | follow | edited Feb 11 '18 at 17:06. All aspects of the algorithm rely on maximum likelihood projections that require the inversion of the error covariance matrix, so a rank-deficient matrix immediately creates a roadblock. The Pk term represents that there are usually unknown generalized reactions associated with essential boundary conditions. The full Surface Green Function Matching program can then be carried out with no ambiguity. They have many uses! In the numerical methods literature, this is often referred to as clustering of eigenvalues. For an improved and consistent estimation, various regularization methods have been proposed. 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But say $$\sigma_1$$ is the largest singular value of $$A$$ with right singular vector $$v$$. Detecting the shift in sign for the lowest eigenvalue indicates the point the matrix becomes singular. Prove that A is a singular matrix and also prove that I − A, I + A are both nonsingular matrices, where I is the n × n identity […] Find the Nullity of the Matrix A + I if Eigenvalues are 1, 2, 3, 4, 5 Let A be an n × n matrix. On computing accurate singular values and eigenvalues of acyclic matrices. • norm of a matrix • singular value decomposition 15–1. Although one would not expect this situation to be commonly observed in practice, it is interesting because, not only does it have a singular error covariance matrix, but there is no defined maximum likelihood projection for points off the line since there would be no intersection between the error distribution and the solution. The only eigenvalues of a projection matrix are 0 and 1. 0 & \frac{\sqrt{2}}{2} & 0 \\ A singular value and its singular vectors give the direction of maximum action among all directions orthogonal to the singular vectors of any larger singular value. P is singular,so D 0 is an eigenvalue. (4.2) and (4.3), it follows that an identity matrix has a condition number equal to unity since all its eigenvalues are also equal to unity. 0 & 0 & 1 The difference is this: The eigenvectors of a matrix describe the directions of its invariant action. It’s not necessarily the case that $$A v$$ is parallel to $$v$$, though. Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative.. The difference is this: The eigenvectors of a matrix describe the directions of its invariant action. These approximations do not automatically produce good approximations of tangent spaces and regular systems of defining equations. − When adopting this approach, there is a risk of selecting outliers if the data were not inspected a priori and potential outliers removed. And we get a 1-dimensional figure, and a final largest singular value of 1: This is the point: Each set of singular vectors will form an orthonormal basis for some linear subspace of $$\mathbb{R}^n$$. (4.2) is also known as the characteristic polynomial of matrix [A]. Huang et al. The product of a square matrix and a vector in hyperdimensional space (or column matrix), as in the left-hand side of Eq. J.E. (2010), and among others. Comput. The problem is that the singularities are defined by equations on submanifolds of a domain: Both methods produce essentially the same result, but there are some subtle differences. For example, in the case of Hopf bifurcation, many methods solve for the pure imaginary Hopf eigenvalues and eigenvectors associated with these. It says: approximate some matrix $$X$$ of observations with a number of its uncorrelated components of maximum variance. The term “singular value” relates to the distance between a matrix and the set of singular matrices. where J is the number of channels (columns) and it is assumed that there are no other factors contributing to rank deficiency. \end{bmatrix} \begin{bmatrix} We shall also see in Chapter 14 that tridiagonal equations occur in numerical methods of solving boundary value problems and that in many such applications A is positive definite (see Problem 14.5). Advanced Linear Algebra: Foundations to Frontiers Robert van de Geijn, Margaret Myers The eigen- value λ could be zero! Eigenvalues: For a positive definite matrix the real part of all eigenvalues are positive. If μj is unknown, one can naturally estimate it by the sample mean μ¯j=m−1∑l=1mXl,j and γ^i,j and Σ^p in (50) and (51) can then be modified correspondingly. Contributions to the solution of systems of linear equations and the determination of eigenvalues, 39 (1954), pp. Duplex selection applied to the y vector gives the best results, with better SEP and bias. \end{bmatrix} = \begin{bmatrix} Eigenvector and Eigenvalue. \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0.0 \\ \end{bmatrix} = \begin{bmatrix} The complexity of the expressions appearing in these defining equations is reduced compared to that of minimal augmentation methods. Wentzell, in Comprehensive Chemometrics, 2009, One problem that arises frequently in the implementation of MLPCA is the situation where the error covariance matrix is singular. A second approach to computing saddle-node bifurcations is to rely upon numerical methods for computing low dimensional invariant subspaces of a matrix. 3. In that case, there is no way to use a type (I) or type (III) operation to place a nonzero entry in the main diagonal position for that column. Systems of linear ordinary differential equations are the primary examples. It is easy to see by comparison with earlier equations, such as Equation (48), that a maximum likelihood projection corresponds to Q−VandR=Σ−1V. Eigenvalues and Singular Values This chapter is about eigenvalues and singular values of matrices. What are eigenvalues? This means that in general after the essential boundary conditions (Dk) are prescribed the remaining unknowns are Du and Pk. For practical problems, singular matrices can only arise due to programming errors, whereby one of the diagonal elements has been incorrectly assigned a zero value. There is an explicit algebraic inequality in the coefficients of the characteristic polynomial that distinguishes these two cases. Guckenheimer and Myers [82] give a list of methods for computing Hopf bifurcations and a comparison between their method and the one of Roose and Hvalacek [128]. Thus the (i, j)th elements are zero for j > i + 1 and j < i − 1. In nonlinear and time dependent applications the reactions can be found from similar calculations. Case (e) shows a nonsingular error covariance matrix, along with the orthogonal complement of the null space (green) and the direction of projection (blue). \end{bmatrix} \]. The matrix R can be interpreted as the subspace into which the orthogonal projection of the measurement is to occur in order to generate the oblique projection onto the desired subspace. 0 & 2 \\ 0 & 0 & 1 You can see how they again form the semi-axes of the resulting figure. In other words, $$||A v|| = \sigma_1$$ is at least as big as $$||A x||$$ for any other unit vector $$x$$. Given n × n matrices A and B, their tensor product is an n2 × n2 matrix They proved that (i) if maxjEexp(uXl,i2)<∞ for some u>0 and kn ≍ (m−1/2p2/β)c(α), then, (ii) if maxjE|Xl,i|β<∞ and kn≍(m−1∕2p2∕β)c(α), where c(α)=(1+α+2∕β)−1, then. Such a matrix is called a singular matrix. For the particular scenario under consideration, i.e., solution of PDEs, the coefficient matrix is rarely singular. Then the net number of unknowns corresponds to the number of equations, but they must be re-arranged before all the remaining unknowns can be computed. Chapter 8: Eigenvalues and Singular Values Methods for nding eigenvalues can be split into two categories. (ut,υ) of the system of equations. 0 & 0 & 0 \\ The point is that in every case, when a matrix acts on one of its eigenvectors, the action is always in a parallel direction. This advantage is offset by the expense of having larger systems to solve with root finding and the necessity of finding initial seeds for the auxiliary variables. We then get this matrix: $A_1 = \begin{bmatrix} There is no need to change the 3rd to nth equations in the elimination of x1. A simple example is that an eigenvector does not change direction in a transformation:. The lag k can be chosen by AIC, BIC, or other information criteria. u⊗υ→υ⊗u and a skewsymmetric part that anticommutes with this involution. The algebraic system can be written in a general matrix form that more clearly defines what must be done to reduce the system to a solvable form by utilizing essential boundary condition values. Comparison of the PLS model statistics using different selection algorithm for a given number of calibration set (30) and test set (40) and a fixed number of LVs (LV = 3), Sandip Mazumder, in Numerical Methods for Partial Differential Equations, 2016, The stability and convergence of the iterative solution of a linear system is deeply rooted in the eigenvalues of the linear system under consideration. Matrix factorization type of the eigenvalue/spectral decomposition of a square matrix A. \end{bmatrix}$. Principal component analysis is a problem of this kind. N + d = 1, so, If P is not in S, then the line from the eyepoint E to the point P intersects the plane S in a unique point Q so. It is known that, under appropriate moment conditions of Xl,i, if p∕m→c, then the empirical distribution of eigenvalues of Σ^p follows the Marcenko–Pastur law that has the support [(1−c)2,(1+c)2] and a point mass at zero if c>1; and the largest eigenvalue, after proper normalization, follows the Tracy–Widom law. \end{bmatrix} \begin{bmatrix} That is, Most books on numerical analysis assume that you have reduced the system to the non-singular form given above where the essential conditions, Du, have already been moved to the right hand side. Singular Value and Eigenvalue Decompositions Frank Dellaert May 2008 1 The Singular Value Decomposition The singular value decomposition (SVD) factorizes a linear operator A : Rn → Rm into three simpler linear operators: 1. For an orthogonal projection R = Q = V and the usual PCA projection applies. An alternative approach to achieve this objective is to first carry out SVD on the error covariance matrix: Once this is done, the zero singular values on the diagonal of ΛΣ1/2 are replaced with small values (typically a small fraction of the smallest nonzero singular value) to give (ΛΣ1/2). The diagonal entries of the matrix $$\Sigma$$ are the singular values of $$A$$. In general, the error covariance matrix is not required to generate the projections shown in Figure 13, but it is used for the maximum likelihood projection and so the singularity problem needs to be addressed. For a, Theory and Applications of Numerical Analysis (Second Edition), Finite Element Analysis with Error Estimators, The above small example has led to the most general form of the algebraic system that results from satisfying the required integral form: a, the scores obtained by singular value decomposition (SVD) were used instead of the raw spectra to avoid calculation problems with the near-, Stability and Convergence of Iterative Solvers, Numerical Methods for Partial Differential Equations, C. Trallero-Giner, ... F. García-Moliner, in, Long Wave Polar Modes in Semiconductor Heterostructures, Time Series Analysis: Methods and Applications, Journal of Computational and Applied Mathematics. Relation to eigenvalue decomposition. 0 & 0 & 0 \\ If, however, the result of the product is the original vector times some scalar quantity, then, the vector is the so-called eigenvector of the matrix [A], and the scalar premultiplier is known as the eigenvalue of [A]. The diagonal elements of a triangular matrix are equal to its eigenvalues. This is the return type of eigen , the corresponding matrix factorization function. This is useful for performing mathematical and numerical analysis of matrices in order to identify their key features. If A is positive definite we can be certain that the algorithm will not fail because of a zero pivot. Outline of this Talk IWhat is known? In fact, every singular operator (read singular matrix) has 0 as an eigenvalue (the converse is also true). For example, Hopf bifurcation occurs when the Jacobian at an equilibrium has a pair of pure imaginary eigenvalues. Copyright © 2020 Elsevier B.V. or its licensors or contributors. which precisely prescribes all the terms and contributions surviving in the end. In most applications the reaction data have physical meanings that are important in their own right, or useful in validating the solution. There are constants c1 > 0 and C2 and a neighborhood U of A so that if Chapter 8: Eigenvalues and Singular Values Methods for nding eigenvalues can be split into two categories. 185, 203–218 (1993) ... Huang, R.: A qd-type method for computing generalized singular values of BF matrix pairs with sign regularity to high relative accuracy. Then $$v$$ is a solution to, $\operatorname*{argmax}_{x, ||x||=1} ||A x||$. and these are easily solved by back substitution. In order to stabilize the error covariance matrix for inversion, the easiest solution is essentially to ‘fatten’ it by expanding the error hyperellipsoid along all of the minor axes so that it has a finite thickness in all dimensions. For instance, say we set the largest singular value, 3, to 0. Therefore, we first discuss calculation of the eigenvalues and the implication of their magnitudes. 0 & 0 & 0 \\ This approach is slightly more cumbersome, but has the advantage of expanding the error ellipsoid only along the directions where this is necessary. Σi1,i2,…,ik−1. There is no familiar function that vanishes when a matrix has pure imaginary eigenvalues analogous to the determinant for zero eigenvalues. Thus, we have succeeded in factoring a singular projective transformation M into the product of a perspective transformation R and an affine transformationA. The same results as that of K&S algorithm were found, as can be seen in Figure 1.Code 5 D-optimal selection algorithm. In terms of iterative solution of a linear system, this is the best-case scenario, because if [A]=[I], no iterations would be necessary to solve the linear system. The comparison of the performance of PLS models after using different selection algorithms to define the calibration and test sets indicates that the random selection algorithm does not ensure a good representativity of the calibration set. In this example, we calculate the eigenvalues and condition numbers of two matrices considered at the beginning of Section 3.2, namely. Therefore, let us try to reverse the order of our factors. \end{bmatrix} \begin{bmatrix} Key properties of square matrices are their eigenvalues and eigenvectors, which enable them to be written in a simpler form, through a process known as eigenvalue decomposition. Equation (4.2) represents a polynomial equation of degree K (i.e., the number of equations or unknowns), and is also known as the characteristic equation. Of these, only the E part propagates outside and the transfer matrix which propagates this amplitudes is. Fortunately, the solution to both of these problems is the same. Does anybody know wheter it is possible to do it with R? Thus we define the full G1 matrix as having three identically nil submatrices. This completes the proof. \end{bmatrix} \]. We give an example of an idempotent matrix and prove eigenvalues of an idempotent matrix is either 0 or 1. And the corresponding eigen- and singular values describe the magnitude of that action. Xiang [154] altered the construction of defining equations to produce a regular systems of equations for This is the currently selected item. We use cookies to help provide and enhance our service and tailor content and ads. These are the MM, ME and EM submatrices, so only the EE submatrix is nonvanishing and the form of G1, as well as that of G1 — surface projection — is. The system of linear equations can be solved using Gaussian elimination with partial pivoting, an algorithm that is efficient and reliable for most systems. In the most General Case Assume ordering: eigenvalues z }| {jz1j ::: jznjand squared singular values z }| {a1 ::: an Ideterminant,th 1 & 0 & 0 \\ Dfυ=ωw and It is somewhat ironic that MLPCA, which is supposed to be a completely general linear modeling method, breaks down under conditions of ordinary least squares. On the other hand, a matrix with a large condition number is known as an ill-conditioned matrix, and convergence for such a linear system may be difficult or elusive. Note that for positive semidefinite matrices, singular values and eigenvalues are the same. Σ2,1 singularities, the smallest example of stable maps in which this difficulty arises. This comparison is based on the particular data set used in this example and of course the statistical results obtained will depend on the data set used. Thus, there are two problems to be dealt with, one where the error covariance matrix is singular, but there is a legitimate projection of the measurement, and the other where no theoretically legitimate projection of the measurement exists. Then we piece those estimates together and obtain an estimate for the precision matrix Σp−1 by (49). Set it to 0: $A_2 = \begin{bmatrix} Projection z=VTx into an r-dimensional space, where r is the rank of A 2. In fact, we can compute that the eigenvalues are p 1 = 360, 2 = 90, and 3 = 0. The elimination method can be considerably simplified if the coefficient matrix of a linear set of equations is tridiagonal. \end{bmatrix}$. The thesis of Xiang [154] contains results that surmount a technical difficulty in implementing the computation of Thom-Boardman singularities [18]. Example The eigenvalues of the matrix:!= 3 −18 2 −9 are ’.=’ /=−3. FINDING EIGENVALUES AND EIGENVECTORS EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . Computational algorithms and sensitivity to perturbations are both discussed. Thus, a type (I) operation cannot be used to make the pivot 1. For most choices of n vectors B and C and scalar D the (n + 1) × (n + 1) block matrix, is nonsingular. This has important applications. Table 1. A formal way of putting this into the analysis is to express G1, G1 etc, as, say, IEG1IE. The singular vectors of a matrix describe the directions of its maximumaction. If one or more eigenvalues are zero then the determinant is zero and which is a singular matrix. Error covariance matrices can also be rank deficient when they are generated from a theoretical model if that model does not introduce sufficient dimensionality. If there are I samples measured, each with K replicates, the rank of the error covariance matrix will be. The SVD is not directly related to the eigenvalues and eigenvectors of. Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. 0 & 2 & 0 \\ Moreover, when v is zero, u is the right zero eigenvector of A, an object needed to compute the normal form of the bifurcation. \[ A = \begin{bmatrix} Still, this factoring is not quite satisfactory, since in geometric modeling the perspective transformation comes last rather than first. The matrix in a singular value decomposition of Ahas to be a 2 3 matrix, so it must be = 6 p 10 0 0 0 3 p 10 0 : Step 2. Dfw=−ωυ for vectors v and w as well as the eigenvalue ico [128]. Thus, it is fair to conclude that the condition number of the coefficient matrix has some relation to the convergence of an iterative solver used to solve the linear system of equations. For these, iterative methods can be used to compute the solution of the system of linear equations, avoiding the need to calculate a full factorization of the matrix M. Thus, this method is feasible for discretized systems of partial differential equations for which computation of the determinant of the Jacobian can hardly be done. It is easy to see that the eigenvalue represents a stretching factor. 2. However, the same method resulted in divergence when attempting to solve [C][ϕ]=[9−110]T, where the matrices [A] and [C] are as shown in Example 4.1. The row vector is called a left eigenvector of . The definition says that when $$A$$ acts on an eigenvector, it just multiplies it by a constant, the corresponding eigenvalue. If at least one eigenvalue is zero the matrix is singular, and if one becomes negative and the rest is positive it is indefinite. Take a 2×2 matrix, for example, A= ∙ 10 0 −1 ¸. Govaerts et al. ScienceDirect ® is a registered trademark of Elsevier B.V. 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URL: https://www.sciencedirect.com/science/article/pii/B9780080507552500324, URL: https://www.sciencedirect.com/science/article/pii/B9780123747518000184, URL: https://www.sciencedirect.com/science/article/pii/B9780125535601500100, URL: https://www.sciencedirect.com/science/article/pii/B9780750667227500333, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000740, URL: https://www.sciencedirect.com/science/article/pii/B9780128498941000044, URL: https://www.sciencedirect.com/science/article/pii/B9780080426945500050, URL: https://www.sciencedirect.com/science/article/pii/S1874575X02800297, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000570, URL: https://www.sciencedirect.com/science/article/pii/B9780444538581000089, If a projective transformation has a perspective factor, then it must be a, Elementary Linear Algebra (Fourth Edition), As we have seen, not every square matrix has an inverse. To make this system of equations regular, additional equations that normalize v; and w are required. \frac{3}{2} \, \sqrt{2} & -\sqrt{2} & 0 \\ Thus writing down G1−1 does not imply inverting a singular matrix. + From the example given above, calibration data set selection based on optimality criteria improved the quality of PLS model predictions by improving the representativeness of the calibration data. 10.1 Eigenvalue and Singular Value Decompositions An eigenvalue and eigenvector of a square matrix A are a scalar λ and a nonzero vector x so that Ax = λx. Σi1,i2,…,ik−1. There is no test for a zero pivot or singular matrix (see below). C. Trallero-Giner, ... F. García-Moliner, in Long Wave Polar Modes in Semiconductor Heterostructures, 1998, By convention the vacuum is on side 1. the matching formula for the full Gs−1 is, The formal prescription for the evaluation of a term like G1(z,0)⋅G1−1⋅Gs is then. In a finite element formulation all of the coefficients in the S and C matrices are known. Therefore, A has a single pair of eigenvalues whose sum is zero if and only if its biproduct has corank one. We may find λ = 2 or1 2or −1 or 1. Subspace iteration and Arnoldi methods [142] are effective techniques for identifying invariant subspaces that are associated with the eigenvalues of largest magnitude for a matrix. Example: Are the following matrices singular? Consider, where A is of the form (9.46). Projection z=VTx into an r-dimensional space, where r is the rank of A 2. The eigenvalue λtells whether the special vector xis stretched or shrunk or reversed or left unchanged—when it is multiplied by A. !What isn’t known? Such matrices are amenable to efficient iterative solution, as we shall see shortly. adds to 1,so D 1 is an eigenvalue. Linear Algebra Appl. asked Jul 18 '13 at 11:34. alext87 alext87. Thus the singular values of Aare ˙ 1 = 360 = 6 p 10, ˙ 2 = p 90 = 3 p 10, and ˙ 3 = 0. From Eq. The picture is more complicated, but as in the 2 by 2 case, our best insights come from finding the matrix's eigenvectors : that is, those vectors whose direction the transformation leaves unchanged. Showing that an eigenbasis makes for good coordinate systems. Example 4.1 clearly shows that interchanging the two coefficients in the last row of the coefficient matrix drastically alters the condition number of the matrix. An idempotent matrix is a matrix A such that A^2=A. (4.1), will result in another vector. Cases (a)–(c) in the figure show perfectly legitimate situations where measurements can be projected onto the model in a maximum likelihood manner, but the projection matrix cannot be obtained directly through Equation (48) (or equivalent equations) because of the singular error covariance matrix. This problem is solved by computing its singular value decomposition and setting some of its smallest singular values to 0. This is then used to generate the adjusted error covariance matrix: Since Σ is a symmetric matrix, VΣ could be used in place of UΣ in Equation (88), but they should not be mixed, since they may not be identical for a rank-deficient matrix. An example of this is shown for a nonsingular error covariance matrix in Figure 13(e), where R is represented by the green vector. That eigenvectors give the directions of invariant action is obvious from the definition. \frac{3}{2} \, \sqrt{2} & \sqrt{2} & 0 \\ You can see that in the previous example. \frac{4}{3} & 1 Element-wise multiplication with r singular values σ i, i.e., z0 =Sz 3. The number λ is an eigenvalue of A. K&S, D-optimal, and Duplex-on-X all gave better models (based on R2, SEP, and bias) for the same optimal number of LVs (3), while nevertheless giving b-coefficient vectors that were not as noisy (Figure 4). 0 & 0 & 1 The first and simplest approach is to add a small diagonal matrix, or ridge, to the error covariance matrix. For a singular matrix A, row reduction of [A| In] does not produce In to the left of the augmentation bar. What are eigenvalues? 0 & 0 & 0 \\ Here, for simplicity, it has been assumed that the equations have been numbered in a manner that places the prescribed parameters (essential boundary conditions) at the end of the system equations. A singular matrix is one which is non-invertible i.e. Case (d) represents multiplicative offset noise. [69] studied the Jordan decomposition of the biproduct of matrices with multiple pairs of eigenvalues whose sum was zero and used a bordering construction to implement a system of defining equations for double Hopf bifurcation. High dimensional vector fields often have sparse Jacobians. Wei Biao Wu, Han Xiao, in Handbook of Statistics, 2012, Assume that (Xl,1,Xl,2,…,Xl,p), l=1,…,m, are i.i.d. General expressions for defining equations of some types of bifurcations have been derived only recently, so only a small amount of testing has been done with computation of these bifurcations [82]. If desired, the values of the necessary reactions, Pk, can now be determined from. We shall show that if L is nonsingular, then the converse is also true. In this process one encounters the standard linear differential form A± which for side 1 takes again the form (4.65). What is the relation vice versa? The singular vectors of a matrix describe the directions of its maximum action. Example 1 The matrix A has two eigenvalues D1 and 1=2. 1.3K views The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. For example, error covariance matrices calculated on the basis of digital filter coefficients may be singular, as well as those obtained from the bilinear types of empirical models discussed in the previous section if no independent noise contributions are included. random vectors identically distributed as (X1,…,Xp). 1/ 2: I factored the quadratic into 1 times 1 2, to see the two eigenvalues D 1 and D 1 2. In the tapered estimate (53), if we choose K such that the matrix Wp=(K(|i−j|∕l))1≤i,j≤p is positive definite, then Σ˜p,l is the Hadamard (or Schur) product of Σ^n and Wp, and by the Schur Product Theorem in matrix theory Horn and Johnson (1990), it is also non-negative definite since Σ^n is non-negative definite. Now, an elementary excitation incident on the surface form side 2 has both, the M and E parts. By continuing you agree to the use of cookies. Moreover, C can be decomposed into a symmetric part that commutes with the involution Because the (4,3) entry is also zero, no type (III) operation (switching the pivot row with a row below it) can make the pivot nonzero. The nontrivial solution to Eq. Then Ax=(1,−2). These methods were tested with seven dimensional stable maps containing 0 & 0 & 1 They suffer from all of the problems associated with the use of the determinant as a defining equation for saddle-node bifurcations as well as the additional difficulty that computations of the characteristic polynomial tend to suffer from numerical instability [150]. What does a zero eigenvalue means? Look at det.A I/ : A D:8 :3:2 :7 det:8 1:3:2 :7 D 2 3 2 C 1 2 D . It is clear that for , where O is a zero square matrix of any order. The polynomial resulting from the left-hand side of Eq. The columns of Q define the subspace of the projection and R is the orthogonal complement of the null space. A scalar $$\sigma$$ is a singular value of $$A$$ if there are (unit) vectors $$u$$ and $$v$$ such that $$A v = \sigma u$$ and $$A^* u = \sigma v$$, where $$A^*$$ is the conjugate transpose of $$A$$; the vectors $$u$$ and $$v$$ are singular vectors. These conditions hold for the tridiagonal matrix M in (6.36) for cubic splines. I Algorithms based on matrix-vector products to nd just a few of the eigenvalues. If appropriate invariant subspaces are computed, then the bifurcation calculations can be reduced to these subspaces. A¯∈U with smallest singular value a, then the unique solution \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0.0 \\ The description of high codimension singularities of maps has proceeded farther than the description of high codimension bifurcations of dynamical systems. Some of the important properties of a singular matrix are listed below: The determinant of a singular matrix is zero; A non-invertible matrix is referred to as singular matrix, i.e. In particular, Bickel and Levina (2008a) considered the class, This condition quantifies issue (ii) mentioned in the beginning of this section. This can be especially useful in finding Hopf bifurcations, but subspaces associated to eigenvalues of large magnitude on the imaginary axis cannot be readily separated from subspaces associated with negative eigenvalues of large magnitude. where the asterisk denotes that this element simply does not vanish. I Algorithms using decompositions involving similarity transformations for nding several or all eigenvalues. There are alternative methods that introduce additional independent variables and utilize larger systems of defining equations. Finding eigenvectors and eigenspaces example. It stretches the red vector and shrinks the blue vector, but reverses neither. For those numbers, the matrix A I becomes singular (zero determinant). \begin{bmatrix} PHILLIPS, P.J. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. We can obtain a lower-dimensional approximation to $$A$$ by setting one or more of its singular values to 0. The former condition ensures that Σ^p,B can include dependencies at unknown orders, whereas the latter aims to circumvent the weak signal-to-noise ratio issue that γ^i,j is a bad estimate of γi,j if |i−j| is big. A⊗B whose eigenvalues are the products of the eigenvalues of A and B. Σi1,i2,…,ik is the set on which the map restricted to The eigenvalues of the matrix:!= 3 −18 2 −9 are ’.=’ /=−3. In other instances, the singularity of the error covariance matrix can arise quite naturally from the assumptions of the problem. The closer the condition number is to unity, the better the convergence, and vice versa. This invariant direction does not necessarily give the transformation’s direction of greatest effect, however. Here, R and Q have dimensions J × P, where P is the dimensionality of the subspace. Video transcript. M. Zeaiter, D. Rutledge, in Comprehensive Chemometrics, 2009, To apply the D-optimal algorithm,9 the scores obtained by singular value decomposition (SVD) were used instead of the raw spectra to avoid calculation problems with the near-singular matrix. 0 & 0 & 0 \\ Inverse iterations can be used in this framework to identify invariant subspaces associated with eigenvalues close to the origin. 1 & 0 & 0 \\ In other words, when a linear transformation acts on one of its eigenvectors, it shrinks the vector or stretches it and reverses its direction if $$\lambda$$ is negative, but never changes the direction otherwise. The second case, which has a much larger condition number, is the case where convergence could not be attained. We all know that the determinant of a matrix is equal to the products of all eigenvalues. We can see how the transformation just stretches the red vector by a factor of 2, while the blue vector it stretches but also reflects over the origin. Applying the theorem with A the Jacobian of the vector field gives the quantity v as a measure of the distance of the Jacobian from the set of singular matrices. Johannes Hahn. However, for, implies so the singular values of are the square roots of the eigenvalues of the symmetric positive semidefinite matrices and (modulo zeros in the latter case), and the singular vectors are eigenvectors. 89, 229–252 (2020) MathSciNet Article Google Scholar 27. A consequence of this equation is that it will increase the dimensions of the error ellipsoid in all directions, whereas it might be considered more ideal to only expand those directions where the ellipsoid has no dimensions. Six Varieties of Gaussian Discriminant Analysis, Least Squares with the Moore-Penrose Inverse, Understanding Eigenvalues and Singular Values, investmentsim - an R Package for Simulating Investment Portfolios, Talk: An Introduction to Categories with Haskell and Databases. A determinant, the Sylvester resultant of two polynomials constructed from the characteristic polynomial, vanishes if and only if the Jacobian matrix has a pair of eigenvalues whose sum is zero. In this equation, IJ is an identity matrix of dimension J and ε represents the machine precision. (4.1) can be found if and only if. The matrix !is singular (det(A)=0), and rank(! In summary, a square matrix [A] of size K×K will have K eigenvalues, which may be real or complex. \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0.0 \\ Figure 4. Of course, in doing this, one must be careful not to distort the original shape of the ellipsoid to the point where it affects the direction of the projection, so perturbations to the error covariance matrix must be small. However, Table 1 clearly shows the advantage of using an optimal selection algorithm compared with random selection. where Du represents the unknown nodal parameters, and Dk represents the known essential boundary values of the other parameters. What are singular values? On this front, we note that, in independent work, Li and Woodruff obtained lower bounds that are polynomial in n [LW12]. If the coefficient matrix is singular, the matrix is not invertible. Ranking selection has resulted in a model with better R2 and SEP compared to random selection. Now, the singular value decomposition (SVD) will tell us what $$A$$’s singular values are: \[ A = U \Sigma V^* = for all indices and .. Every square diagonal matrix is symmetric, since all off-diagonal elements are zero. The two-dimensional case obviously represents an oversimplification of multivariate spaces but can be useful in illustrating a few points about the nature of the singular matrices. Thus, we have succeeded in decomposing a singular projective transformation into simple, geometrically meaningful factors. This will have the effect of transforming the unit sphere into an ellipsoid: Its singular values are 3, 2, and 1. General random matrices without these conditions are much less studied as their eigenvalues can lay everywhere in the complex plane. The following diagrams show how to determine if a 2×2 matrix is singular and if a 3×3 matrix is singular. auto-covariance matrix, Least singular value. All those results suggest the inconsistency of sample covariance matrices. It is shown that several SVD-based steps inherent in the algorithms are equivalent to the first-order approximation. If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. The singular value decomposition is very general in the sense that it can be applied to any m × n matrix, whereas eigenvalue decomposition can only be applied to diagonalizable matrices. Eigenvalues of a 3x3 matrix. They both describe the behavior of a matrix on a certain set of vectors. 0 & 0 & 1 An example, illustrating the use of Eqs. 109-116. matrices matrix-analysis eigenvalues numerical-linear -algebra. For example, ordinary least squares assumes no errors in the x-direction, as illustrated in Figure 13(a). We conclude that there is no way to transform the first four columns into the identity matrix I4 using the row reduction process, and so the original matrix A has no inverse. A scalar $$\lambda$$ is an eigenvalue of a linear transformation $$A$$ if there is a vector $$v$$ such that $$A v = \lambda v$$, and $$v$$ is called an eigenvector of $$\lambda$$. Such inconsistency results for sample covariance matrices in multivariate analysis have been discussed in the study by Stein (1975), Bai and Silverstein (2010), El Karoui (2007), Paul (2007), Johnstone (2001), Geman (1980), Wachter (1978), Anderson et al. Further, the largest singular value of $$A_1$$ is now 2. SOLUTION: • In such problems, we first find the eigenvalues of the matrix. This results in an error ellipse that is essentially a vertical line, and a corresponding error covariance matrix that has a rank of unity. A matrix whose eigenvalues are clustered is preferable for iterative solution over one whose eigenvalues are scattered. Deriving explicit defining equations for bifurcations other than saddle-nodes requires additional effort. Guckenheimer et al. There are a variety of reasons why the error covariance matrix may be singular. They both describe the behavior of a matrix on a certain set of vectors. And the corresponding eigen- and singular values describe the magnitude of that action. The D-optimal algorithm used is based on the Federov algorithm with some modifications. For example, if we were to imagine a third dimension extending behind the page, there would be no legitimate projection points falling behind the line for cases (a)–(c) here. On the remaining small problems, the choice of function that vanishes on singular matrices matters less than it does for large problems. 0 & 2 & 0 \\ Further, the condition number of the coefficient matrix is not sufficient to explain why the same system of equations may reach convergence with some iterative scheme and not with others since the condition number of the coefficient matrix is independent of the iterative scheme used to solve the system. Given an SVD of M, as described above, the following two relations hold: If F::Eigen is the factorization object, the eigenvalues can be obtained via F.values and the eigenvectors as the columns of the matrix … Note that if m | 2021-06-16 14:55:23 | {"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": 0.9150570034980774, "perplexity": 354.39672070363173}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487623942.48/warc/CC-MAIN-20210616124819-20210616154819-00467.warc.gz"} |
https://codereview.stackexchange.com/questions/179054/porting-haskell-list-comprehensions-to-standard-ml | # Porting Haskell list comprehensions to Standard ML
Out of curiosity, I'm porting the Haskell exercises of Programming in Haskell by Graham Hutton to Standard ML. At the beginning everything seemed pretty similar until I reached list comprehensions:
A triple (x,y,z) of positive integers is called pythagorean if x2 + y2 = z2. Using a list comprehension, define a function pyths ∷ Int → [(Int,Int,Int)] that maps an integer n to all such triples with components in [1..n].
The Haskell version is pretty simple:
pyths ∷ Int → [(Int, Int, Int)]
pyths n = [(x, y, z) |
x ← [1..n],
y ← [1..n],
z ← [1..n],
x^2 + y^2 == z^2]
My Standard ML attempt, not so much:
(* No list comprehensions in SML so we need a helper function *)
val generateIntsUpTo = fn n =>
let val rec helper = fn current => fn xs =>
if current <= n
(* then current :: helper (current + 1) xs *)
then helper (current + 1) (current :: xs)
else xs
in
rev (helper 1 [])
end
(* Now we can find the pyths *)
val pyths = fn n =>
let
val numbers = generateIntsUpTo n
in
foldr (fn (x, acc) =>
foldr (fn (y, acc) =>
foldr (fn (z, acc) =>
if x * x + y * y = z * z
then (x, y, z) :: acc
else acc
) acc numbers
) acc numbers
) [] numbers
end
The first thing that I don't like is the call to rev at the end of generateIntsUpTo but I did it to accomplish a tail recursive function. Commented out is the line that lets me avoid the reversion of the list but then it's not tail recursive.
Besides that, there's the three nested foldr calls. I have been thinking of writing my own recursive functions, but I wanted to do this with the minimum amount of custom code.
Feedback:
• The Haskell pyths is not very efficient, and it contains duplicate answers.
• fun instead of val/val rec is syntactic sugar for function declarations.
• Instead of generateIntsUpTo n, use List.tabulate (n, fn i => i+1).
• Using rev to achieve a tail-recursive function can be a good choice. Neither SML/NJ's or Moscow ML's List.tabulate is tail-recursive, though. If you're worrying about performance here, consider the fact that you don't actually need to store these numbers in lists -- this is purely a convenience so that you can use list comprehensions to iterate their combinations.
Pythagorean triples in Standard ML
I wanted to do it using the minimum amount of custom code.
To achieve the same efficiency as the Haskell pyths and use the minimum amount of custom code, here is one version that uses List.filter, List.concat and List.tabulate:
fun isPythTriple (x, y, z) = x*x + y*y = z*z
fun tab1 n f = List.tabulate (n, fn i => f (i+1))
fun pyths n =
List.filter isPythTriple (
List.concat (tab1 n (fn x =>
List.concat (tab1 n (fn y =>
tab1 n (fn z => (x,y,z)))))))
This takes several seconds for pyths 10; there really is not reason to generate O(n³) list elements when the solution subset is so sparse.
I have been thinking on write my own recursive functions
Writing your own helper functions is really not something that should be avoided. Generally, using library functions is good, but Standard ML's library is somewhat limited. For example, List.tabulate can't iterate a range of numbers without generating the list in memory. And as you're hinting at, the multiple nested foldrs does not make the code particularly readable.
You could for example combine List.tabulate and List.filter to reduce memory consumption:
fun tabfilter (from, to, f) =
if from > to then [] else
case f from of
SOME value => value :: tabfilter (from+1, to, f)
| NONE => tabfilter (from+1, to, f)
fun isPythTriple (x, y, z) = x*x + y*y = z*z
fun pyths n =
List.concat (tabfilter (1, n, fn x =>
SOME (List.concat (tabfilter (1, n, fn y =>
SOME (tabfilter (1, n, fn z => Option.filter isPythTriple (x, y, z))))))))
This runs orders of magnitude faster. Still, it is a little convoluted.
A plain recursive version:
fun isPythTriple (x, y, z) = x*x + y*y = z*z
fun pyths n =
let fun loop (0, _, _) = []
| loop (x, 0, _) = loop (x-1, n, n)
| loop (x, y, 0) = loop (x, y-1, n)
| loop (t as (x, y, z)) =
if isPythTriple t then t :: loop (x,y,z-1) else loop (x,y,z-1)
in rev (loop (n, n, n)) end
And a tail-recursive variant of the same:
fun isPythTriple x y z = x*x + y*y = z*z
fun pyths n =
let fun loop 0 _ _ res = res
| loop x 0 _ res = loop (x-1) n n res
| loop x y 0 res = loop x (y-1) n res
| loop x y z res =
let val res' = if isPythTriple x y z
then (x,y,z) :: res
else res
in loop x y (z-1) res'
end
in loop n n n [] end | 2021-04-21 01:37:49 | {"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": 0.18533098697662354, "perplexity": 8373.939147869707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039503725.80/warc/CC-MAIN-20210421004512-20210421034512-00248.warc.gz"} |
https://plainmath.net/linear-algebra/51522-determine-whether-w-is-a-subspace-of-the-vector-space-w | Annette Sabin
2022-01-05
determine whether W is a subspace of the vector space.
$W=\left\{f:f\left(0\right)=-1\right\},V=C\left[-1,1\right]$
sonorous9n
Expert
See the vector space and checked wheather zero function belongs or not , if yes then check other conditions for subspace.
$W=\left\{f:f\left(0\right)=-1\right\},V=C\left[-1,1\right]$
∵ if $f\in W⇒f\left(0\right)=-1$
and $0\left(0\right)=0$ [0 = zero function]
so $0\notin W$
so zero function does not belongs to W
so W is not subspace
Do you have a similar question? | 2023-01-29 02:28:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 31, "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": 0.8144590854644775, "perplexity": 1766.8902882070106}, "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-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00835.warc.gz"} |
http://jaac.ijournal.cn/ch/reader/view_abstract.aspx?file_no=JAAC-2018-0287 | ### For REFEREES
Volume 9, Number 3, 2019, Pages 1083-1101 On the $F$-expanding of Homoclinic classes Wanlou Wu,Bo Li Keywords:Homoclinic classes, Thin trapped, Periodic points, Uniformly expanding Abstract: We establish a closing property for thin trapped (see Definition 1.2) homoclinic classes. Taking advantage of this property, we prove that if a homoclinic class $H(f, p)$ admits a dominated splitting \$T_{H(f, p)}M=E\oplus_ PDF Download reader | 2019-06-27 12:12:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9566407799720764, "perplexity": 3075.464224307575}, "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/1560628001138.93/warc/CC-MAIN-20190627115818-20190627141818-00087.warc.gz"} |
http://mathhelpforum.com/algebra/41529-re-arranging-print.html | # Re-arranging
• Jun 14th 2008, 05:22 AM
hciR
Re-arranging
I am trying to find voltages in a circuit using KVL and KCL and have got these equations:
1. (x-v)A + (x)B + (x-y)C = 0
2. (y-v)D + (y-x)C + (y)E = 0
I need to re-arrange them to get the form:
1. x(A+B+B) + y(-C) = vA
2. x(-C) + y(D+C+E) = vD
But don't know the process.
I can get the firtst 2 equations from the beginning problem, and given the 2 re-arranged equations I can solve it, but i'm missing a step.Can anyone help?
• Jun 14th 2008, 06:27 AM
topsquark
Quote:
Originally Posted by hciR
I am trying to find voltages in a circuit using KVL and KCL and have got these equations:
1. (x-v)A + (x)B + (x-y)C = 0
2. (y-v)D + (y-x)C + (y)E = 0
1.
$(x - v)A + (x)B + (x - y)C = 0$
$xA - vA + xB + xC - yC = 0$
$(xA + xB + xC) - yC = vA$
$x(A + B + C) + yC = vA$
The second equation is done in a similar manner.
-Dan
• Jun 14th 2008, 09:16 AM
hciR
Ah I see now, that's brilliant thanks! | 2017-01-19 22:34:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8150177597999573, "perplexity": 5603.179732165657}, "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-2017-04/segments/1484560280746.40/warc/CC-MAIN-20170116095120-00576-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://microkit.berkeley.edu/gpio-basics/ | When working with microcontrollers, the ultimate goal is typically to have the system you’re designing interact with the world around it in some fashion. These interactions often take place through the use of a variety of inputs and outputs to and from your microcontroller. Fortunately, most microcontrollers come with a set of features called General Purpose Input/Ouput (GPIO). These features enable you to interface with and control a broad range of peripherals and components such as sensors, actuators, and more! In this tutorial, we will cover some basics you should know for working with GPIO on your microcontroller.
## Pin Diagram
We are going to begin our discussion of GPIO with an overview of the connections on the HUZZAH32 Feather microntroller board to better understand its capabilities. Take a look at Figure 1 showing the pin diagram for our microcontroller.
There are a ton of things to talk about here with regard to this pin diagram, but let’s cover some of the essentials first. At the top of the board, we have a micro USB connector, which we can use to both power and program the device from a computer. In the top right corner, there is a black connector for an optional lithium battery for autonomous operation (i.e. when not connected to a computer or other master device).
The remainder of the connections (numbered 1-28 on the diagram) are arrays of “pins”, or electrical connections, that provide a range of functions, which includes mostly GPIO. Your version of the HUZZAH32 Feather board should have what are referred to as “stacking headers” that allow you to plug directly into these pins with jumper wires.
Let’s cover the pins that are dedicated exclusively for power.
NameFunction
3.3V (3V)3.3V output from the on-board regulator. Can supply up to 500mA.
GND (GND)Common ground for power and logic.
VUSB (USB)Positive voltage from USB jack, if connected. ~5V.
VBAT (BAT)Positive voltage from the backup Lipoly battery, if connected. ~3.7V.
EN 3.3V (EN)3.3V regulator enable pin. Tie to ground to disable.
RESET (RST)Reset, tie to GND to reset (or press Reset button).
As mentioned earlier, you can opt to power the device through the USB connector from your computer or through an external LiPo battery. Your board also has the potential to be powered by an external power supply, but Adafruit discourages this practice due to board limitations.
Power can be supplied from your device to peripheral components using a range of options. The first option is to leverage the onboard $3.3V$ regulator through the 3.3V pin. If not being used, the $3.3V$ regulator can be disabled by pulling the EN 3.3V pin low. Another option for powering components is to draw power directly from your external battery by connecting to VBAT. Finally, you have the option to power external components from the USB connection through the VUSB pin.
The RESET pin let’s you manually reset the board by connecting it to GND. This is the same as pressing the reset button on the board itself. We will learn more about powering options later on! For now, let’s continue on to the remaining GPIO pins.
Notice that we have labels for each of the remaining pins that fall within the categories of “GPIO”, “ALT”, and “ID”. The “GPIO” label corresponds to the number of the pin in the official documentation of the ESP32 from EspressIf. The “ID” column shows the primary function of the given pin, and the “ALT” column applies to alternative pin functions that can be accessed if their primary function is not being used.
Most pins can be configured by software to perform one of several functions, depending on the need of a particular application. This allows for great flexibility: the same microcontroller can be used in many projects with different peripherals. We will discuss the meaning behind the various function names as they become relevant.
Also, keep in mind that processes for accessing and using the various pins on your board will change depending on your use of MicroPython or Arduino to program functions. We will provide examples of both implementations where possible.
WARNING: The pins of the ESP32 chip are rated for 3.3V maximum. Exceeding this value may damage the chip.
## Digital Signals
The basic unit of information in digital systems is bits. Each bit can assume one of two values, e.g. $0$ or $1$, or, equivalently True or False. Values that require more resolution or precision are represented by combining several bits. These collections of bots represent integers, floating point numbers, strings, or more complex data structures.
In the hardware, the state of each bit is represented by voltage. Voltages near $0$ (or ground) are interpreted as a logical $0$ (LOW or False), while voltages near the supply (in our case $3.3V$) are interpreted as a logical $1$ (HIGH or True). This results in great robustness against signal cirruption in the presence of interference: unless the value of a digital signal approaches half the supply voltage, it is always correctly interpreted as $0$ or $1$.
So how are digital signals used in the context of GPIO on a microcontroller? Well, lets start with digital inputs.
### Digital Inputs
Probably the most obvious example of a digital input function that you would want to design into your embedded system is that of a button. Using digital input, it is easy to integrate a button that controls some aspect of operation in your system.
First, we should discuss how to configure a given GPIO pin to accept digital inputs. Refer back to Figure 1 and note that any of the pins not dedicated to power can be used as digital inputs.
Here is how we configure a GPIO pin to be a digital input using MicroPython:
from machine import Pin
p = Pin(id, mode=Pin.IN)
p() # 0 if voltage is close to 0V
# 1 if voltage is close to VDD (3.3V)
And here is how to do it in Arduino:
pinMode(id, INPUT);
p; // 0 if voltage is close to 0V
// 1 if voltage is close to VDD (3.3V)
In MicroPython, we start by importing the correct package on line 1. On line 2, we assign a Pin object, give it the pin identifier in id (note that you will have to import this pin identifier as well, e.g. A0), and assign it to be in Pin.IN mode (input mode).
In Arduino, we can go straight to assigning a pin and mode using pinMode. We can then read from the pin by it’s id.
Optionally, the ESP32 offers internal pull-up and pull-down resistors that can be enabled in the mode argument. Note that this feature is not available on pins A2, A3, A4.
from machine import Pin
p = Pin(id, mode=Pin.IN, pull=<None|Pin.PULL_UP|Pin.PULL_DOWN>)
pinMode(id, <INPUT_PULLUP|INPUT_PULLDOWN>);
The values of the pull-down and pull-up resistors vary from chip-to-chip and pin-to-pin and for the ESP32 are typically in the range of $R_{up}=30…80k\Omega$ and $R_{down}=17k\Omega$.
Figure 2 shows a simplified circuit diagram. Switches $S_1$ and $S_2$ are closed if the modes are set accordingly, connecting the corresponding resistor to the input pin $D_{in}$. The triangle represents an amplifier that restores correct digital levels ($0V$ or $V_{DD}$) if a mid-supply input is applied.
Pull-ups and pull-downs are very practical when reading e.g. the state of a switch or button. Figure 3 shows an example. One terminal of the Button is connected to a GPIO input, and the other is connected to ground. Pressing the button connects $D_{IN}$ to ground, thus establishing a clear logic 0. However, when the button is released, $D_{IN}$ is disconnected and its state is undefined. Enabling the pull-up results in $D_{IN}=V_{DD}$ when the button is not pressed, hence establishing a logic 1. The pull-down resistor is disabled and not shown in the diagram.
Figure 4 shows a typical waveform when reading a mechanical switch or button. Rather than simply switching on or off, the state changes rapidly between the two states until finally settling. It is caused by the spring effect of the contact. Frequently this is not a problem. For example, when turning lights on or off, short periods of “flicker” are not noticeable.
Computers are much faster than humans and detect these changes, typically occurring over a period of a few milliseconds, as individual events. This is problematic for example when counting button presses: the operator presses the button only once, but the machine detects several presses. Imagine a candy vending machine distributing a dozen treats rather than just one.
The simplest solution is to explicitly program the MCU to ignore presses for a short period, e.g. $50ms$, after a change is detected. Then only the first change is registered, the “bounce” events in the figure are ignored.
### Digital Outputs
Let’s move on to digital outputs! Digital outputs are frequently used to control ON and OFF states for peripheral components such as LEDs. All pins A0A21 except A2, A3, A4 can be used for digital output.
from machine import Pin
p = Pin(id, mode=Pin.OUT)
p(0) # Write digital pin to LOW
p(1) # Write digital pin to HIGH
pinMode(id, OUTPUT);
digitalWrite(id, LOW); // Write digital pin to LOW
digitalWrite(id, HIGH); // Write digital pin to HIGH
In the case of MicroPython, you can see that we are using the Pin.OUT mode, allowing us to write a digital LOW with p(0) and HIGH with p(1).
On the other hand, Arduino uses the digitalWrite command to control the pin to either a LOW or HIGH output.
Figure 5 shows a simplified diagram of the circuit for a standard digital output used in microcontrollers. It consists of two switches, $S_1$ and $S_2$. The switches are actually transistors, but the difference is not relevant for this explanation. Calling the appropriate function to set the digital output will open one switch and close the other. As a result, we get either a $3.3V$ or ground connection at $D_{OUT}$.
### Open Drain Output
As another option for digital signals, we can configure output pins as open drains. When configured as open drains, the corresponding pin is pulled low (i.e. tied to ground) when set to 0, and open circuited (i.e. not connected) when set to 1. The circuit is identical to that shown in Figure 5, except switch $S_1$ never closes.
from machine import Pin
p = Pin(id, mode=Pin.OPEN_DRAIN)
p(0) # pin driven to 0V
p(1) # pin open (not driven)
In order to accomplish the same thing in Arduino, you’ll need to rely on EspressIf’s ESP-IDF API functions that we added to Arduino. If you run into issues, refer back to the steps of how to install ESP-IDF packages in Arduino.
gpio_set_direction(id, GPIO_MODE_OUTPUT_OD);
gpio_set_level(id, 0); // Set pin to no connect
gpio_set_level(id, 1); // Set pin to open drain to ground
The code shown here essentially uses the same structure with different function calls and type specifiers that are specific to the ESP-IDF framework.
## Analog Signals
Analog signals refer to any signal that varies continuously over a time interval. This is in contrast to digital signals that only take two discrete values. Examples of applications of analog signals include sensor readings, radio communication, audio amplification, and more.
Since analog signals vary continuously, an analog signal at a given instant can take theoretically an infinite number of values within the parameters of the system. As an example, imagine that an electrical component in your system outputs an analog signal with a voltage that varies between $0$ and $1$. The number of values that can be represented between $0$ and $1$ is infinite and only dependent on the resolution with which the signal can be generated and read. First, we will examine the way that we read analog signals as inputs.
If analog signals are continuous, how do we get our microcontroller to read and represent an analog input? Well, our microcontrollers operate in the digital domain; therefore, we must approximate analog signals with a discrete representation. Microcontrollers do this through the use of an analog-to-digital converter (ADC).
The ESP32 chip contains two 12-Bit ADCs. In the HUZZAH32 Feather board, we are usually limited to using only one of the two ADCs. The second ADC is disabled when the built-in WiFi is activated.
The ADC output code for a given input voltage $V_{in}$ can be represented with the following equation.
$$D_{out}=round\left(2^{B}*\frac{V_{in}}{V_{ref}}\right)$$
where $B$ is the number of bits of the ADC. In the case of the ESP32, this value is $12$. From a bit resolution perspective, this means that we can represent a total of $4096$ distinct values from our ADC. However, the ESP32 also gives us the option to change the scaling with $V_{ref}$. Let’s look at an example.
from machine import Pin, ADC
d_out = adc0.read() # perform conversion
analogSetAttenuation(ADC_0db); // Set full-scale range
val = analogRead(id); // Perform conversion
Note that the ADC defaults to full-scale range in the case that you do not set the attenuation manually. The following table shows the possible attenuation settings and their associated specifiers.
Configuration parameter (MicroPython)Configuration parameter (Arduino)$V_{ref}$
ADC.ATTN_0DBADC_0db$1.1V$
ADC.ATTN_2_5DBADC_2_5db$1.3V$
ADC.ATTN_6DBADC_6db$1.8V$
ADC.ATTN_11DBADC_11db$3.2V$
Unfortunately, the ESP32 chip ADCs suffer from notoriously inconsistent performance, which includes a large offset and poor linearity. As a result, it may be in your best interest to use external ADC ICs in the case that you need reliably accurate analog input measurements. See the measurement results below.
### Analog Outputs
What if we want to go in the opposite direction and output an analog signal from our microcontroller? This might be useful if we needed to control the brightness of an LED or the position of an analog readout. Generally, however, powered analog signals like those used to control actuators are handled by external power equipment to protect the microcontroller.
As you might have guessed, analog outputs are handled by what is known as a digital-to-analog converter (DAC). Similar to the ADC, we have an equation that we can use to determine the output of our DAC.
$$V_{out}=\frac{D_{in}}{2^{B}}*V_{DD}$$
where $B$ is the number of bits of the DAC. The DAC on the ESP32 has 8-Bit resolution (i.e. 256 distinct values). The reference voltage for our DAC is taken from the supply voltage, $V_{DD}$ (nominally $3.3V$).
And here, we take a look at an implementation in both MicroPython and Arduino.
from machine import Pin, DAC
dac1 = DAC(Pin(id))
dac1.write(128) # perform conversion (8-bit DAC)
dacWrite(DAC1, 128); // perform conversion (8-bit DAC) | 2022-10-02 17:17:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3085298538208008, "perplexity": 1481.0422446065372}, "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/1664030337338.11/warc/CC-MAIN-20221002150039-20221002180039-00176.warc.gz"} |
http://libros.duhnnae.com/2017/jul6/150056048266-Multi-parameter-Quantum-Groups-and-Quantum-Shuffles-I-Mathematics-Quantum-Algebra.php | # Multi-parameter Quantum Groups and Quantum Shuffles, I - Mathematics > Quantum Algebra
Multi-parameter Quantum Groups and Quantum Shuffles, I - Mathematics > Quantum Algebra - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
Abstract: In this article, we study the multi-parameter quantum groups defined bygenerators and relations associated with symmetrizable generalized Cartanmatrices, together with their representations in the category $\mathcal O$.This presentation will be convenient for our later discussions. We present twoexplicit descriptions here: as a Hopf 2-cocycle deformation, and as themulti-parameter quantum shuffle realization of the positive part.
Autor: Naihong Hu, Yufeng Pei, Marc Rosso
Fuente: https://arxiv.org/ | 2018-05-26 08:39:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 0.4156925082206726, "perplexity": 11488.553086571352}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867374.98/warc/CC-MAIN-20180526073410-20180526093410-00315.warc.gz"} |
https://socratic.org/questions/what-is-the-dot-product-of-1-5-3-and-8-3-5 | # What is the dot product of <1,5,-3 > and <-8,3,-5 >?
$< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$
$< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = - 8 + 15 + 15$
$< 1 , 5 , - 3 > . < - 8 , + 3 , - 5 > = 22$ | 2019-08-19 10:15:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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": 0.6269688606262207, "perplexity": 8003.406101471931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314721.74/warc/CC-MAIN-20190819093231-20190819115231-00359.warc.gz"} |
https://www.physicsforums.com/threads/absolute-convergence.683105/ | # Absolute Convergence
1. Apr 4, 2013
### whatlifeforme
1. The problem statement, all variables and given/known data
Absolute, Conditional, - convergence, or Divergence.
2. Relevant equations
$\displaystyle \sum^{∞}_{n=1} (-1)^n e^{-n}$
3. The attempt at a solution
1. Alternating Series Test
2. Ratio Test for ABsolute Convergence
1. $\displaystyle (-1)^n (1/e)^n$
an > 0 for n=1,2,3,4 - YES
decreasing - YES
limit (n->inf) = 0 - YES
Converges.
2.$\displaystyle (1/e)^{n+1} * (e^n) = 1/e$
limit (n->inf) = 1/e
1/e < 1
Thus, Absolute Convergence.
2. Apr 4, 2013
### Staff: Mentor
Yes, the series converges, and converges absolutely. Your work is a little sketchy in parts, so if your instructor is picky, you might lose points. For example, in the alt. series test you need to show that an > 0 for all n ≥ 1, not just n = 1, 2, 3, and 4. And you don't show any work that justifies your saying that the sequence an is decreasing or that the limit of the sequence is zero.
Last edited: Apr 4, 2013
3. Apr 4, 2013
### whatlifeforme
i did not show the work here, but i did show the work for decreasing by setting f(x) = series and taking derivative.
for the limit i have lim(x->inf) (1/e)^n = 0 (not sure how to show that this is zero)?
also, how do i show that an > 0 for all n >= 1 ?
4. Apr 4, 2013
### Staff: Mentor
The limit as you wrote it doesn't make much sense, since you have x increasing, but the expression you're taking the limit of doesn't involve x.
One way to show that this limit is zero is to show that (1/e)n can be made arbitrarily close to zero. IOW, using the definition of the limit, which in this case involves N and $\epsilon$. I don't know if actual limit definitions have been presented in your class just yet.
Here, an = 1/en. The only way for a rational expression to be equal to zero is for its numerator to equal zero. Obviously that can't happen here. The denominators in this sequence, e1, e2, ..., en, ... are an increasing sequence of positive numbers, so it must be true that their reciprocals are decreasing and positive. An induction proof, which would be easy, would be convincing here.
5. Apr 5, 2013
### whatlifeforme
everytime we take a limit, i don't think the instructor wants us to use the definition of the proof. also, it's been a bit since i took a number theory class (discrete mathematics) which involved induction proofs.
i stated what you did above about the denominators and their reciprocals, but i have no other proof than that. also, i'm not sure what i would say for my other problems. | 2017-10-22 13:12:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.81130450963974, "perplexity": 545.1271303393224}, "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/1508187825227.80/warc/CC-MAIN-20171022113105-20171022133105-00407.warc.gz"} |
http://www.gradesaver.com/textbooks/science/chemistry/chemistry-a-molecular-approach-3rd-edition/chapter-3-sections-3-1-3-12-exercises-problems-by-topic-page-132/76 | ## Chemistry: A Molecular Approach (3rd Edition)
$206$ $mg$ $Ag$
Each percentage represents the number of milligrams of an element per 100 milligrams of that compound. Use this as a conversion factor. $155 mg Ag\times\frac{100 mg AgCl}{75.25 mg Ag}$ $=206$ $mg$ $Ag$ | 2017-03-29 09:45:25 | {"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": 0.7559127807617188, "perplexity": 2683.8391851101055}, "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-2017-13/segments/1490218190236.99/warc/CC-MAIN-20170322212950-00381-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.helpdrivers.com/printers/Kyocera/KM_Serie/KM-F1060/ | Available 5 files for KM-F1060
Company
Kyocera
Categories
Printers
Model
Kyocera KM-F1060
Description
KM-F1060
Operating Systems
Windows 2000-XP-2003
Version Driver
5.7.001
Size Driver
862Kb
File Name
kmf1060win2kxp200357001.exe
Information
[Version]
Signature="\$Windows NT\$"
Provider=%DRV_MAKER%
ClassGUID={4D36E979-E325-11CE-BFC1-08002BE10318}
Class=Printer
DriverVer=03/4/2005, 5.7.000L
[ClassInstall]
DelReg=ClassDelReg
[ClassDelReg]
HKLM,"System\CurrentControlSet\Services\Class\Printer","NoUseClass"
HKR,,,,%CLASSNAME%
HKR,,Installer,,MSPRINT.DLL
HKR,,Icon,,-4
HKR,,NoDisplayClass,,1
[Environment]
UpdateInis=Environment_Ini
[Environment_Ini]
win.ini,Ports,"LPT1.DOS=",
win.ini,Ports,"LPT2.DOS=",
win.ini,Ports,"LPT3.DOS=",
win.ini,Ports,"LPT1.OS2=",
win.ini,Ports,"LPT2.OS2=",
win.ini,Ports,"LPT3.OS2=",
win.ini,windows,"Spooler=",
win.ini,windows,"DosPrint=",
win.ini,windows,"DeviceNotSelectedTimeout=",
win.ini,windows,"TransmissionRetryTimeout=",
;
; Manufacturer Sections
;
; This section lists all of the manufacturers that we will
; display in the Dialog box.
;
[Manufacturer]
%DRV_MAKER%=SECTION_0
;
; Model Sections
;
; These sections correspond with an entry listed in the
; [Manufacturer] section above. The models will be displayed in the order
; that they appear here.
;
[SECTION_0]
"Kyocera Mita KM-F1060" = KM1060us.PFD, USBPRINT\Kyocera_MitaKM-F1060F449, Kyocera_MitaKM-F1060
"Kyocera Mita KM-F1060" = KM1060us.PFD
;
; Installer Sections
;
; These sections control file installation, and reference all files that
; need to be copied. The section name will be assumed to be the driver data
; file, unless there is an explicit DataFile section listed.
;
; Also note the ".NT" or ".WIN" notation. This is done to allow for a single INF to
; work on Win9x and WinNT. When this format is used, the DataFile must be declared explicitly.
; Win95 does not support this notation, so the old format is used.
Company
Kyocera
Categories
Printers
Model
Kyocera KM-F1060
Description
KM-F1060
Operating Systems
Windows 95-98
Version Driver
1.28.000
Size Driver
1Mb
File Name
kmf1060win9x429000.exe
Information
[Version]
Signature="\$Windows 95\$"
Class=printer
ClassGUID={4D36E979-E325-11CE-BFC1-08002BE10318}
Provider=%DRV_MAKER%
;LayoutFile=layout.inf
DriverVer=03/4/2005, 4.29.000L
[Manufacturer]
%DRV_MAKER%=SECTION_0
;
; Model sections. Each section here corresponds with an entry listed in the
; [Manufacturer] section, above. The models will be displayed in the order
; that they appear in the INF file.
;
; Each model lists a variation of its own name as a compatible ID. This
; is done primarily as an optimization during upgrade.
;
[SECTION_0]
"Kyocera Mita KM-F1060" = KM1060us.PFD, LPTENUM\Kyocera_MitaKM-F1060F449, Kyocera_MitaKM-F1060
"Kyocera Mita KM-F1060" = KM1060us.PFD
;
; Installer Sections
;
; These sections control file installation, and reference all files that
; need to be copied. The section name will be assumed to be the driver
; file, unless there is an explicit DriverFile section listed.
;
[Printer1]
CopyFiles=Printer_CFILE
DataSection=Printer_DATA
[KM1060us.PFD]
[email protected],@KM_F1060.ini,Printer_CFILE_WIN, Printer_CFILE_WIN_L2
DataFile=KM1060us.PFD
DataSection=Printer_DATA_WIN
;
; Copy Sections
;
; Lists of files that are actually copied. These sections are referenced
; from the installer sections, above. Only create a section if it contains
; two or more files (if we only copy a single file, identify it in the
; installer section, using the @filename notation) or if it's a color
; profile (since the DestinationDirs can only handle sections, and not
; individual files).
;
[Printer_CFILE_WIN]
PFKDPM99.DRV,,,32
PFKGNTHK.DLL,,,32
PFKDPERS.DLL,,,32
PFKBPD99.DLL,,,32
CODEC.DLL,,,32
RECT2.DLL,,,32
PFK_UI16.DLL,,,32
PFK_UI.HLP,,,32
PFK_UI.CNT,,,32
PFKDPSYS.INI,,,32
PFKTHK16.DLL,,,32
PFKTHK32.DLL,,,32
PFKDLG32.DLL,,,32
PFKDLG32.INI,,,32
CRCMDL_K.DLL,,,32
PFK_BACG.EXE,,,32
Company
Kyocera
Categories
Printers
Model
Kyocera KM-F1060
Description
KM-F1060 LPR Port instAller
Operating Systems
Windows 95-98
Size Driver
1.1Mb
File Name
lpr_(win9x_only).exe
Company
Kyocera
Categories
Printers
Model
Kyocera KM-F1060
Description
KM-F1060 USB Driver
Operating Systems
Windows 95-98
Size Driver
52Kb
File Name
kmf1060win9xusb.exe
Information
[Version]
Signature="\$CHICAGO\$"
Class=USBPT
ClassGUID={36FC9E60-C465-11CF-8056-444553540000}
provider=%Panasonic%
LayoutFile=layout.inf
[Manufacturer]
%MfgName%=Panasonic
[Panasonic]
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_3A08
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F41
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F42
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F43
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F44
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F61
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F62
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F63
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F64
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F65
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F66
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_0F67
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_1040&MI_01
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_2F01
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_2F02
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_2F03
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_2F04
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_2F07
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_3A01
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_3A03
%PANAUSBPRN%=USBPT.Dev,USB\VID_04DA&PID_3A07
[PreCopySection]
HKR,,NoSetupUI,,1
[ControlFlags]
ExcludeFromSelect = *
[DestinationDirs]
USBPT.CopyFiles = 10,system32\drivers
USBPT.DLL.CopyFiles = 11
USBPT.INI.CopyFiles = 11
[SourceDisksNames]
1 = %DISK_NAME%,,,
[USBPT.Dev]
Copyfiles = USBPT.CopyFiles, USBPT.DLL.CopyFiles, USBPT.INI.CopyFiles
[ClassInstall]
[ClassInstall32]
HKR,,,,%DeviceClassName%
HKR,,Installer,,usbpti16.dll
HKR,,Icon,,"-20"
HKR,,NTMPDriver,,usbptcls.sys
[USBPT.CopyFiles]
usbptcls.sys
[USBPT.DLL.CopyFiles]
USBPTI16.dll
Company
Kyocera
Categories
Printers
Model
Kyocera KM-F1060
Description
KM-F1060
Operating Systems
Windows NT4
Version Driver
5.7.001
Size Driver
870Kb
File Name
kmf1060winnt57001.exe
Information
[Version]
Signature="\$Windows NT\$"
Provider=%DRV_MAKER%
LayoutFile=layout.inf
ClassGUID={4D36E979-E325-11CE-BFC1-08002BE10318}
Class=Printer
;DriverVer=03/4/2005, 5.7.000L
[ClassInstall]
DelReg=ClassDelReg
[ClassDelReg]
HKLM,"System\CurrentControlSet\Services\Class\Printer","NoUseClass"
HKR,,,,%CLASSNAME%
HKR,,Installer,,MSPRINT.DLL
HKR,,Icon,,-4
HKR,,NoDisplayClass,,1
[Environment]
UpdateInis=Environment_Ini
[Environment_Ini]
win.ini,Ports,"LPT1.DOS=",
win.ini,Ports,"LPT2.DOS=",
win.ini,Ports,"LPT3.DOS=",
win.ini,Ports,"LPT1.OS2=",
win.ini,Ports,"LPT2.OS2=",
win.ini,Ports,"LPT3.OS2=",
win.ini,windows,"Spooler=",
win.ini,windows,"DosPrint=",
win.ini,windows,"DeviceNotSelectedTimeout=",
win.ini,windows,"TransmissionRetryTimeout=",
;
; Manufacturer Sections
;
; This section lists all of the manufacturers that we will
; display in the Dialog box.
;
[Manufacturer]
%DRV_MAKER%=SECTION_0
;
; Model Sections
;
; These sections correspond with an entry listed in the
; [Manufacturer] section above. The models will be displayed in the order
; that they appear here.
;
[SECTION_0]
"Kyocera Mita KM-F1060" = Km1060us.PFD
;
; Installer Sections
;
; These sections control file installation, and reference all files that
; need to be copied. The section name will be assumed to be the driver data
; file, unless there is an explicit DataFile section listed.
;
; Also note the ".NT" or ".WIN" notation. This is done to allow for a single INF to
; work on Win9x and WinNT. When this format is used, the DataFile must be declared explicitly.
; Win95 does not support this notation, so the old format is used. | 2020-09-18 14:46:22 | {"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": 0.8636667728424072, "perplexity": 12838.08501154877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400187899.11/warc/CC-MAIN-20200918124116-20200918154116-00165.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-6-section-6-4-logarithmic-functions-6-4-assess-your-understanding-page-449/42 | ## College Algebra (10th Edition)
domain: $(0, +\infty)$
RECALL: The logarithmic function $f(x) = \log_a{x}$ is defined only when $x \gt 0$. Thus, its domain is $x \gt 0$. This means that the function $f(x) = \log_5{(x^3)}$ is defined only when $x^3 \gt 0$. Note that the value of $x^3$ will be greater than zero only when $x$ is positive. This means that the given function is defined only when $x\gt0$. Thus, the domain of the given function is $x\gt 0$. In interval notation, the domain is $(0, +\infty)$. | 2018-10-19 03:32:40 | {"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": 0.9519127607345581, "perplexity": 53.45239060811371}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00174.warc.gz"} |
https://labs.tib.eu/arxiv/?author=M.%20Chaichian | • ### Vector Generation Functions, q-Spectral Functions of Hyperbolic Geometry and Vertex Operators for Quantum Affine Algebras(1707.01553)
July 4, 2017 hep-th, math-ph, math.MP
We investigate the concept of $q$-replicated arguments in symmetric functions with its connection to spectral functions of hyperbolic geometry. This construction suffices for vector generation functions in the form of $q$-series, and string theory. We hope that the mathematical side of the construction can be enriched by ideas coming from physics.
• ### Multipartite Generating Functions and Infinite Products for Quantum Invariants(1702.02208)
Feb. 2, 2017 hep-th, math-ph, math.MP
We show that multipartite generation functions can be written in terms of the Bell polynomials (known as Fa\`a di Bruno's formula) and the Ruelle spectral functions, whose spectrum is encoded in the Patterson-Selberg function of the hyperbolic three-geometry. We derive an infinite-product formula for the Chern-Simons partition functions and analyze appropriate q-series which leads to the construction of knot invariants. With the help of the Ruelle spectral functions symmetric and modular properties in infinite-product structure can be described.
• ### Cosmological Perturbations in Restricted f(R)-Gravity(1609.02666)
Sept. 9, 2016 gr-qc
We investigate the metric perturbations of the restricted f(R) theory of gravity in the cosmological context and explore the phenomenological implications of this model. We show that it is possible to construct a restricted model of gravity, in which the background equations are the same as the equations of motion which are derived from the Einstein-Hilbert action with the cosmological constant term. We argue that the deviation from the Einstein-Hilbert model emerges in the perturbed equations, for which we have a non-vanishing anisotropic stress. Further, with the help of the results of Planck data for the modified gravity we obtain constraints on the parameters of the model.
• ### Restricted f(R) Gravity and Its Cosmological Implications(1512.05866)
April 27, 2016 gr-qc
We investigate the f(R) theory of gravity with broken diffeomorphism due to the change of the coefficient in front of the total divergence term in the (3+1)-decomposition of the scalar curvature. We perform the canonical analysis of this theory and show that its consistent, i.e. with no unphysical degrees of freedom, form is equivalent to the low-energy limit of the non-projectable f(R) Horava-Lifshitz theory of gravity. We also analyze its cosmological solutions and show that the de Sitter solution can be obtained also in the case of this broken symmetry. The consequences of the proposed theory on the asymptotic solutions of a few specific models in the cosmological context are also presented.
• ### $S$-Functions, Spectral Functions of Hyperbolic Geometry, and Vertex Operators with Applications to Structure for Weyl and Orthogonal Group Invariants(1602.06704)
Feb. 22, 2016 hep-th, math-ph, math.MP
In this paper we analyze the quantum homological invariants (the Poincar\'e polynomials of the $\mathfrak{sl}_N$ link homology). In the case when the dimensions of homologies of appropriate topological spaces are precisely known, the procedure of the calculation of the Kovanov-Rozansky type homology, based on the Euler-Poincar\'e formula can be appreciably simplified. We express the formal character of the irreducible tensor representation of the classical groups in terms of the symmetric and spectral functions of hyperbolic geometry. On the basis of Labastida-Mari\~{n}o-Ooguri-Vafa conjecture, we derive a representation of the Chern-Simons partition function in the form of an infinite product in terms of the Ruelle spectral functions (the cases of a knot, unknot, and links have been considered). We also derive an infinite-product formula for the orthogonal Chern-Simons partition functions and analyze the singularities and the symmetry properties of the infinite-product structures.
• ### Elliptic Genera and Characteristic $q$-Series of Superconformal Field Theory(1504.01849)
April 8, 2015 hep-th
We analyze the characteristic series, the $KO$ series and the series associated with the Witten genus, and their analytic forms as the $q$-analogs of classical special functions (in particular $q$-analog of the beta integral and the gamma function). $q$-series admit an analytic interpretation in terms of the spectral Ruelle functions, and their relations to appropriate elliptic modular forms can be described. We show that there is a deep correspondence between the characteristic series of the Witten genus and $KO$ characteristic series, on one side, and the denominator identities and characters of $N=2$ superconformal algebras, and the affine Lie (super)algebras on the other. We represent the characteristic series in the form of double series using the Hecke-Rogers modular identity.
• ### D-Branes on Spaces Stratified Fibered Over Hyperbolic Orbifolds(1408.1071)
Aug. 5, 2014 hep-th, math-ph, math.MP
We apply the methods of homology and K-theory for branes wrapping spaces stratified fibered over hyperbolic orbifolds. In addition, we discuss the algebraic K-theory of any discrete co-compact Lie group in terms of appropriate homology and Atiyah-Hirzebruch type spectral sequence with its non-trivial lift to K-homology. We emphasize the fact that the physical D-branes properties are completely transparent within the mathematical framework of K-theory. We derive criteria for D-brane stability in the case of strongly virtually negatively curved groups. We show that branes wrapping spaces stratified fibered over hyperbolic orbifolds carry charge structure and change the additive structural properties in K-homology.
• ### Quantum Black Holes, Elliptic Genera and Spectral Partition Functions(1308.2177)
Aug. 9, 2013 hep-th, math-ph, math.MP, math.AG
We study M-theory and D-brane quantum partition functions for microscopic black hole ensembles within the context of the AdS/CFT correspondence in terms of highest weight representations of infinite-dimensional Lie algebras, elliptic genera, and Hilbert schemes, and describe their relations to elliptic modular forms. The common feature in our examples lie in the modular properties of the characters of certain representations of the pertinent affine Lie algebras, and in the role of spectral functions of hyperbolic three-geometry associated with q-series in the calculation of elliptic genera. We present new calculations of supergravity elliptic genera on local Calabi-Yau threefolds in terms of BPS invariants and spectral functions, and also of equivariant D-brane elliptic genera on generic toric singularities. We use these examples to conjecture a link between the black hole partition functions and elliptic cohomology.
• ### BRST-Invariant Deformations of Geometric Structures in Topological Field Theories(1306.0373)
June 3, 2013 hep-th, math-ph, math.MP, math.AG
We study a Lie algebra of formal vector fields $W_n$ with its application to the perturbative deformed holomorphic symplectic structure in the A-model, and a Calabi-Yau manifold with boundaries in the B-model. A relevant concept in the vertex operator algebra and the BRST cohomology is that of the elliptic genera (the one-loop string partition function). We show that the elliptic genera can be written in terms of spectral functions of the hyperbolic three-geometry (which inherits the cohomology structure of BRST-like operator). We show that equivalence classes of deformations are described by a Hochschild cohomology theory of the DG-algebra ${\mathfrak A} = (A, Q)$, $Q =\bar{\partial}+\partial_{\rm deform}$, which is defined to be the cohomology of $(-1)^n Q +d_{\rm Hoch}$. Here $\bar{\partial}$ is the initial non-deformed BRST operator while $\partial_{\rm deform}$ is the deformed part whose algebra is a Lie algebra of linear vector fields ${\rm gl}_n$. We discuss the identification of the harmonic structure $(HT^\bullet(X); H\Omega_\bullet(X))$ of affine space $X$ and the group ${\rm Ext}_{X}^n({\cal O}_{\triangle}, {\cal O}_{\triangle})$ (the HKR isomorphism), and bulk-boundary deformation pairing.
• ### Thermal Casimir-Polder interaction of different atoms with graphene(1207.3788)
The thermal correction to the energy of Casimir-Polder interaction of atoms with a suspended graphene membrane described by the Dirac model is investigated. We show that a major impact on the thermal correction is made by the size of the gap in the energy spectrum of graphene quasiparticles. Specifically, if the temperature is much smaller than the gap parameter (alternatively, larger or of the order of the gap parameter), the thermal correction is shown to be relatively small (alternatively, large). We have calculated the free energy of the thermal Casimir-Polder interaction of atoms of He, Na, Rb, and Cs with graphene described by both the hydrodynamic and Dirac models. It is shown that in exact computations using the Dirac model, one should use the polarization operator at nonzero temperature. The computational results for the Casimir-Polder free energy obtained in the framework of hydrodynamic model of graphene are several times larger than in the Dirac model within the separation region below 2$\mu$m. We conclude that the theoretical predictions following from the two models can be reliably discriminated in experiments on quantum reflection of different atoms on graphene.
• ### Extended Poincar\'e supersymmetry in three dimensions and supersymmetric anyons(1204.5025)
June 11, 2012 hep-th, math-ph, math.MP
We classify the unitary representations of the extended Poincar\'e supergroups in three dimensions. Irreducible unitary representations of any spin can appear, which correspond to supersymmetric anyons. Our results also show that all irreducible unitary representations necessarily have physical momenta. This is in sharp contrast to the ordinary Poincar\'e group in three dimensions, that admits in addition irreducible unitary representations with non-physical momenta, which are discarded on physical grounds.
• ### Towards an Axiomatic Formulation of Noncommutative Quantum Field Theory(hep-th/0402212)
March 20, 2011 hep-th
We propose new Wightman functions as vacuum expectation values of products of field operators in the noncommutative space-time. These Wightman functions involve the $\star$-product among the fields, compatible with the twisted Poincar\'e symmetry of the noncommutative quantum field theory (NC QFT). In the case of only space-space noncommutativity ($\theta_{0i}=0$), we prove the CPT theorem using the noncommutative form of the Wightman functions. We also show that the spin-statistics theorem, demonstrated for the simplest case of a scalar field, holds in NC QFT within this formalism.
• ### 13th International Conference on Elastic and Diffractive Scattering (Blois Workshop) - Moving Forward into the LHC Era(1002.3527)
Proceedings of the 13th International Conference on Elastic and Diffractive Scattering (Blois Workshop) - Moving Forward into the LHC Era
• ### Covariant star product on symplectic and Poisson spacetime manifolds(1001.0503)
July 5, 2010 hep-th, math-ph, math.MP, gr-qc
A covariant Poisson bracket and an associated covariant star product in the sense of deformation quantization are defined on the algebra of tensor-valued differential forms on a symplectic manifold, as a generalization of similar structures that were recently defined on the algebra of (scalar-valued) differential forms. A covariant star product of arbitrary smooth tensor fields is obtained as a special case. Finally, we study covariant star products on a more general Poisson manifold with a linear connection, first for smooth functions and then for smooth tensor fields of any type. Some observations on possible applications of the covariant star products to gravity and gauge theory are made.
• ### Noncommutative gauge theory using covariant star product defined between Lie-valued differential forms(1001.0508)
April 27, 2010 hep-th, gr-qc
We develop an internal gauge theory using a covariant star product. The space-time is a symplectic manifold endowed only with torsion but no curvature. It is shown that, in order to assure the restrictions imposed by the associativity property of the star product, the torsion of the space-time has to be covariant constant. An illustrative example is given and it is concluded that in this case the conditions necessary to define a covariant star product on a symplectic manifold completely determine its connection.
• ### Massless Chiral Supermultiplets of Higher Spins and the $\theta$-Twistor(1004.4331)
April 25, 2010 hep-th
Recently N. Berkovits, motivated by the supertwistor description of ${\cal N}=4 D=4$ super Yang-Mills, considered the generalization of the ${\cal N}=1 D=4$ $\theta$-twistor construction to D=10 and applied it for a compact covariant description of ${\cal N}=1 D=10$ super Yang-Mills. This supports the relevance of the $\theta$-twistor as a supersymmetric twistor alternative to the well-known supertwistor. The minimal breaking of superconformal symmetry is an inherent property of the $\theta$-twistor received from its fermionic components, described by a Grassmannian vector instead of a Grassmannian scalar in the supertwistor. The $\theta$-twistor description of the ${\cal N}=1 D=4$ massless chiral supermultiplets $(S, S + 1/2)$ with spins $S=0,1/2,1,3/2,2,...$ is considered here. The description permits to restore the auxiliary $F$ fields of the chiral supermultiplets absent in the supertwistor approach. The proposed formalism is naturally generalized to ${\cal N}=4 D=4$ and can be used for an off-shell description of the corresponding super Yang-Mills theory.
• ### Solution of the Stochastic Langevin Equations for Clustering of Particles in Turbulent Flows in Terms of Wiener Path Integral(0906.1376)
Feb. 7, 2010 hep-th, cond-mat.stat-mech
We propose to take advantage of using the Wiener path integrals as the formal solution for the joint probability densities of coupled Langevin equations describing particles suspended in a fluid under the effect of viscous and random forces. Our obtained formal solution, giving the expression for the Lyapunov exponent, i) will provide the description of all the features and the behaviour of such a system, e.g. the aggregation phenomenon recently studied in the literature using appropriate approximations, ii) can be used to determine the occurrence and the nature of the aggregation - non-aggregation phase transition which we have shown for the one-dimensional case and iii) allows the use of a variety of approximative methods appropriate for the physical conditions of the problem such as instanton solutions in the WKB approximation in the aggregation phase for the one-dimensional case as presented in this paper. The use of instanton approximation gives the same result for the Lyapunov exponent in the aggregation phase, previously obtained by other authors using a different approximative method. The case of non-aggregation is also considered in a certain approximation using the general path integral expression for the one-dimensional case.
• ### Can Seiberg-Witten Map Bypass Noncommutative Gauge Theory No-Go Theorem?(0907.2646)
July 19, 2009 hep-th, hep-ph
There are strong restrictions on the possible representations and in general on the matter content of gauge theories formulated on noncommutative Moyal spaces, termed as noncommutative gauge theory no-go theorem. According to the no-go theorem \cite{no-go}, matter fields in the noncommutative U(1) gauge theory can only have $\pm 1$ or zero charges and for a generic noncommutative $\prod_{i=1}^n U(N_i)$ gauge theory matter fields can be charged under at most two of the $U(N_i)$ gauge group factors. On the other hand, it has been argued in the literature that, since a noncommutative U(N) gauge theory can be mapped to an ordinary U(N) gauge theory via the Seiberg-Witten map, seemingly it can bypass the no-go theorem. In this note we show that the Seiberg-Witten map \cite{SW} can only be consistently defined and used for the gauge theories which respect the no-go theorem. We discuss the implications of these arguments for the particle physics model building on noncommutative space.
• ### Dirac Equation in Noncommutative Space for Hydrogen Atom(0904.2836)
July 6, 2009 hep-th
We consider the energy levels of a hydrogen-like atom in the framework of $\theta$-modified, due to space noncommutativity, Dirac equation with Coulomb field. It is shown that on the noncommutative (NC) space the degeneracy of the levels $2S_{1/2}, 2P_{1/2}$ and $2P_{3/2}$ is lifted completely, such that new transition channels are allowed.
• ### Gauge field theories with covariant star-product(0905.0608)
May 5, 2009 hep-th
A noncommutative gauge theory is developed using a covariant star-product between differential forms defined on a symplectic manifold, considered as the space-time. It is proven that the field strength two-form is gauge covariant and satisfies a deformed Bianchi identity. The noncommutative Yang-Mills action is defined using a gauge covariant metric on the space-time and its gauge invariance is proven up to the second order in the noncommutativity parameter.
• ### Searching for the triplet Higgs sector via central exclusive production at the LHC(0901.3746)
April 21, 2009 hep-ph
We discuss the prospects of searching for the neutral Higgs bosons of the triplet model in central exclusive production at the LHC. A detailed Monte Carlo analysis is presented for six benchmark scenarios for the Higgs boson, $H_1^{0}$, these cover $m_{H_1^0}=$~120, 150 GeV and doublet-triplet mixing of $c_H=$~0.2, 0.5 or 0.8. We find that, for appropriate values of $c_H$, an excellent Higgs mass measurement is possible for the neutral Higgs in the triplet model, and discuss how to distinguish the triplet model Higgs boson from the Higgs boson of the Standard Model.
• ### Test Functions Space in Noncommutative Quantum Field Theory(0706.1712)
July 26, 2008 hep-th
It is proven that the $\star$-product of field operators implies that the space of test functions in the Wightman approach to noncommutative quantum field theory is one of the Gel'fand-Shilov spaces $S^{\beta}$ with $\beta < 1/2$. This class of test functions smears the noncommutative Wightman functions, which are in this case generalized distributions, sometimes called hyperfunctions. The existence and determination of the class of the test function spaces in NC QFT is important for any rigorous treatment in the Wightman approach.
• ### Riemannian Geometry of Noncommutative Surfaces(hep-th/0612128)
July 14, 2008 hep-th, math.DG, math-ph, math.MP, gr-qc
A Riemannian geometry of noncommutative n-dimensional surfaces is developed as a first step towards the construction of a consistent noncommutative gravitational theory. Historically, as well, Riemannian geometry was recognized to be the underlying structure of Einstein's theory of general relativity and led to further developments of the latter. The notions of metric and connections on such noncommutative surfaces are introduced and it is shown that the connections are metric-compatible, giving rise to the corresponding Riemann curvature. The latter also satisfies the noncommutative analogue of the first and second Bianchi identities. As examples, noncommutative analogues of the sphere, torus and hyperboloid are studied in detail. The problem of covariance under appropriately defined general coordinate transformations is also discussed and commented on as compared with other treatments.
• ### Gauging the twisted Poincare symmetry as noncommutative theory of gravitation(0807.0733)
July 4, 2008 hep-th
Einstein's Theory of General Relativity was formulated as a gauge theory of Lorentz symmetry by Utiyama in 1956, while the Einstein-Cartan gravitational theory was formulated by Kibble in 1961 as the gauge theory of Poincare transformations. In a noncommutative space-time with canonical commutation relations between the coordintes, Lorentz symmetry is violated and field theories constructed on such space-times have instead the so-called twisted Poincare invariance. In this paper a gauge theory formulation of noncommutative gravity is proposed based on the twisted Poincare symmetry together with the requirement of covariance under the general coordinate transformations, an essential ingredient of the theory of general relativity. The advantages of such a formulation as well as the related problems are discussed and possible ways out are outlined.
• ### Noncommutative Quantum Field Theory: A Confrontation of Symmetries(0805.3500)
May 22, 2008 hep-th
The concept of a noncommutative field is formulated based on the interplay between twisted Poincar\'e symmetry and residual symmetry of the Lorentz group. Various general dynamical results supporting this construction, such as the light-wedge causality condition and the integrability condition for Tomonaga-Schwinger equation, are presented. Based on this analysis, the claim of the identity between commutative QFT and noncommutative QFT with twisted Poincar\'e symmetry is refuted. | 2020-12-05 23:40:55 | {"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": 0.7469543814659119, "perplexity": 530.880113523195}, "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-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00070.warc.gz"} |
https://www.iacr.org/cryptodb/data/author.php?authorkey=502 | ## CryptoDB
### Chris Peikert
#### Publications
Year
Venue
Title
2019
CRYPTO
We finally close the long-standing problem of constructing a noninteractive zero-knowledge (NIZK) proof system for any NP language with security based on the plain Learning With Errors (LWE) problem, and thereby on worst-case lattice problems. Our proof system instantiates the framework recently developed by Canetti et al. [EUROCRYPT’18], Holmgren and Lombardi [FOCS’18], and Canetti et al. [STOC’19] for soundly applying the Fiat–Shamir transform using a hash function family that is correlation intractable for a suitable class of relations. Previously, such hash families were based either on “exotic” assumptions (e.g., indistinguishability obfuscation or optimal hardness of certain LWE variants) or, more recently, on the existence of circularly secure fully homomorphic encryption (FHE). However, none of these assumptions are known to be implied by plain LWE or worst-case hardness.Our main technical contribution is a hash family that is correlation intractable for arbitrary size-S circuits, for any polynomially bounded S, based on plain LWE (with small polynomial approximation factors). The construction combines two novel ingredients: a correlation-intractable hash family for log-depth circuits based on LWE (or even the potentially harder Short Integer Solution problem), and a “bootstrapping” transform that uses (leveled) FHE to promote correlation intractability for the FHE decryption circuit to arbitrary (bounded) circuits. Our construction can be instantiated in two possible “modes,” yielding a NIZK that is either computationally sound and statistically zero knowledge in the common random string model, or vice-versa in the common reference string model.
2018
PKC
We continue the study of statistical zero-knowledge (SZK) proofs, both interactive and noninteractive, for computational problems on point lattices. We are particularly interested in the problem $\textsf {GapSPP}$GapSPP of approximating the $\varepsilon$ε-smoothing parameter (for some $\varepsilon < 1/2$ε<1/2) of an n-dimensional lattice. The smoothing parameter is a key quantity in the study of lattices, and $\textsf {GapSPP}$GapSPP has been emerging as a core problem in lattice-based cryptography, e.g., in worst-case to average-case reductions. We show that $\textsf {GapSPP}$GapSPP admits SZK proofs for remarkably low approximation factors, improving on prior work by up to roughly $\sqrt{n}$n. Specifically:There is a noninteractive SZK proof for $O(\log (n) \sqrt{\log (1/\varepsilon )})$O(log(n)log(1/ε))-approximate $\textsf {GapSPP}$GapSPP. Moreover, for any negligible $\varepsilon$ε and a larger approximation factor $\widetilde{O}(\sqrt{n \log (1/\varepsilon )})$O~(nlog(1/ε)), there is such a proof with an efficient prover.There is an (interactive) SZK proof with an efficient prover for $O(\log n + \sqrt{\log (1/\varepsilon )/\log n})$O(logn+log(1/ε)/logn)-approximate coGapSPP. We show this by proving that $O(\log n)$O(logn)-approximate $\textsf {GapSPP}$GapSPP is in $\mathsf {coNP}$coNP. In addition, we give an (interactive) SZK proof with an efficient prover for approximating the lattice covering radius to within an $O(\sqrt{n})$O(n) factor, improving upon the prior best factor of $\omega (\sqrt{n \log n})$ω(nlogn).
2018
PKC
Constrained pseudorandom functions allow for delegating “constrained” secret keys that let one compute the function at certain authorized inputs—as specified by a constraining predicate—while keeping the function value at unauthorized inputs pseudorandom. In the constraint-hiding variant, the constrained key hides the predicate. On top of this, programmable variants allow the delegator to explicitly set the output values yielded by the delegated key for a particular set of unauthorized inputs.Recent years have seen rapid progress on applications and constructions of these objects for progressively richer constraint classes, resulting most recently in constraint-hiding constrained PRFs for arbitrary polynomial-time constraints from Learning With Errors (LWE) [Brakerski, Tsabary, Vaikuntanathan, and Wee, TCC’17], and privately programmable PRFs from indistinguishability obfuscation (iO) [Boneh, Lewi, and Wu, PKC’17].In this work we give a unified approach for constructing both of the above kinds of PRFs from LWE with subexponential $\exp (n^{\varepsilon })$exp(nε) approximation factors. Our constructions follow straightforwardly from a new notion we call a shift-hiding shiftable function, which allows for deriving a key for the sum of the original function and any desired hidden shift function. In particular, we obtain the first privately programmable PRFs from non-iO assumptions.
2017
TCC
2016
EUROCRYPT
2016
CRYPTO
2016
TCC
2015
JOFC
2015
EPRINT
2015
EPRINT
2015
EPRINT
2015
TCC
2014
CRYPTO
2014
CRYPTO
2014
EPRINT
2014
EPRINT
2014
EPRINT
2014
FSE
2013
CRYPTO
2013
CRYPTO
2013
EUROCRYPT
2012
EUROCRYPT
2012
EUROCRYPT
2012
EUROCRYPT
2012
PKC
2012
JOFC
We introduce a new lattice-based cryptographic structure called a bonsai tree, and use it to resolve some important open problems in the area. Applications of bonsai trees include an efficient, stateless ‘hash-and-sign’ signature scheme in the standard model (i.e., no random oracles), and the first hierarchical identity-based encryption (HIBE) scheme (also in the standard model) that does not rely on bilinear pairings. Interestingly, the abstract properties of bonsai trees seem to have no known realization in conventional number-theoretic cryptography.
2011
CRYPTO
2010
TCC
2010
CRYPTO
2010
EUROCRYPT
2010
EUROCRYPT
2010
EPRINT
At the heart of many recent lattice-based cryptographic schemes is a polynomial-time algorithm that, given a high-quality' basis, generates a lattice point according to a Gaussian-like distribution. Unlike most other operations in lattice-based cryptography, however, the known algorithm for this task (due to Gentry, Peikert, and Vaikuntanathan; STOC 2008) is rather inefficient, and is inherently sequential. We present a new Gaussian sampling algorithm for lattices that is \emph{efficient} and \emph{highly parallelizable}. We also show that in most cryptographic applications, the algorithm's efficiency comes at almost no cost in asymptotic security. At a high level, our algorithm resembles the perturbation'' heuristic proposed as part of NTRUSign (Hoffstein \etal, CT-RSA 2003), though the details are quite different. To our knowledge, this is the first algorithm and rigorous analysis demonstrating the security of a perturbation-like technique.
2009
TCC
2009
CRYPTO
2008
FSE
2008
EPRINT
We revisit the problem of generating a hard'' random lattice together with a basis of relatively short vectors. This problem has gained in importance lately due to new cryptographic schemes that use such a procedure for generating public/secret key pairs. In these applications, a shorter basis directly corresponds to milder underlying complexity assumptions and smaller key sizes. The contributions of this work are twofold. First, using the \emph{Hermite normal form} as an organizing principle, we simplify and generalize an approach due to Ajtai (ICALP 1999). Second, we improve the construction and its analysis in several ways, most notably by tightening the length of the output basis essentially to the optimum value.
2008
CRYPTO
2008
CRYPTO
2008
EPRINT
We construct public-key cryptosystems that are secure assuming the \emph{worst-case} hardness of approximating the length of a shortest nonzero vector in an $n$-dimensional lattice to within a small $\poly(n)$ factor. Prior cryptosystems with worst-case connections were based either on the shortest vector problem for a \emph{special class} of lattices (Ajtai and Dwork, STOC 1997; Regev, J.~ACM 2004), or on the conjectured hardness of lattice problems for \emph{quantum} algorithms (Regev, STOC 2005). Our main technical innovation is a reduction from certain variants of the shortest vector problem to corresponding versions of the learning with errors'' ($\lwe$) problem; previously, only a \emph{quantum} reduction of this kind was known. In addition, we construct new cryptosystems based on the \emph{search} version of $\lwe$, including a very natural \emph{chosen ciphertext-secure} system that has a much simpler description and tighter underlying worst-case approximation factor than prior constructions.
2007
EPRINT
We propose a new general primitive called lossy trapdoor functions (lossy TDFs), and realize it under a variety of different number theoretic assumptions, including hardness of the decisional Diffie-Hellman (DDH) problem and the worst-case hardness of standard lattice problems. Using lossy TDFs, we develop a new approach for constructing many important cryptographic primitives, including standard trapdoor functions, CCA-secure cryptosystems, collision-resistant hash functions, and more. All of our constructions are simple, efficient, and black-box. Taken all together, these results resolve some long-standing open problems in cryptography. They give the first known (injective) trapdoor functions based on problems not directly related to integer factorization, and provide the first known CCA-secure cryptosystem based solely on worst-case lattice assumptions.
2007
EPRINT
We propose and simple, general, and unified framework for constructing oblivious transfer (OT) protocols that are \emph{efficient}, \emph{universally composable}, and \emph{generally realizable} from a variety of standard number-theoretic assumptions, such as the decisional Diffie-Hellman assumption and the Quadratic Residuosity assumption. Most interestingly, we can also instantiate our framework with \emph{worst-case} complexity assumptions relating to \emph{lattices}. Our OT protocols are round-optimal (one message each way), efficient in the parties' communication and local computation, and use only one reference string for an unbounded number of executions. Furthermore, the protocols can provide \emph{unconditional} security to either the sender or receiver, simply by changing the distribution of the reference string. (For several versions of the protocol, even a common \emph{random} string suffices.) One of our key technical contributions is a simple and novel abstraction that we call a \emph{dual-mode} cryptosystem. We implement dual-mode cryptosystems by taking a unified view of several cryptosystems in the literature that have what we call message-lossy'' public keys, whose defining property is that a ciphertext produced under such a key carries \emph{no information} (even statistically) about the encrypted message. As a contribution of independent interest, we also provide a multi-bit version of Regev's lattice-based cryptosystem (STOC 2005) whose time and space efficiency are improved by a linear factor. In particular, the amortized runtime per message bit is only $\tilde{O}(n)$ bit operations, and the ciphertext expansion can be made as small as a constant.
2007
EPRINT
We show how to construct a variety of trapdoor'' cryptographic tools assuming the worst-case hardness of standard lattice problems (such as approximating the shortest nonzero vector to within small factors). The applications include trapdoor functions with \emph{preimage sampling}, simple and efficient hash-and-sign'' digital signature schemes, universally composable oblivious transfer, and identity-based encryption. A core technical component of our constructions is an efficient algorithm that, given a basis of an arbitrary lattice, samples lattice points from a Gaussian-like probability distribution whose standard deviation is essentially the length of the longest vector in the basis. In particular, the crucial security property is that the output distribution of the algorithm is oblivious to the particular geometry of the given basis.
2006
TCC
2006
TCC
2006
EPRINT
We demonstrate an \emph{average-case} problem which is as hard as finding $\gamma(n)$-approximate shortest vectors in certain $n$-dimensional lattices in the \emph{worst case}, where $\gamma(n) = O(\sqrt{\log n})$. The previously best known factor for any class of lattices was $\gamma(n) = \tilde{O}(n)$. To obtain our results, we focus on families of lattices having special algebraic structure. Specifically, we consider lattices that correspond to \emph{ideals} in the ring of integers of an algebraic number field. The worst-case assumption we rely on is that in some $\ell_p$ length, it is hard to find approximate shortest vectors in these lattices, under an appropriate form of preprocessing of the number field. Our results build upon prior works by Micciancio (FOCS 2002), Peikert and Rosen (TCC 2006), and Lyubashevsky and Micciancio (ICALP 2006). For the connection factors $\gamma(n)$ we achieve, the corresponding \emph{decisional} promise problems on ideal lattices are \emph{not} known to be NP-hard; in fact, they are in P. However, the \emph{search} approximation problems still appear to be very hard. Indeed, ideal lattices are well-studied objects in computational number theory, and the best known algorithms for them seem to perform \emph{no better} than the best known algorithms for general lattices. To obtain the best possible connection factor, we instantiate our constructions with infinite families of number fields having constant \emph{root discriminant}. Such families are known to exist and are computable, though no efficient construction is yet known. Our work motivates the search for such constructions. Even constructions of number fields having root discriminant up to $O(n^{2/3-\epsilon})$ would yield connection factors better than the current best of~$\tilde{O}(n)$.
2005
TCC
2005
EPRINT
Given a corrupted word $\w = (w_1, \ldots, w_n)$ from a Reed-Solomon code of distance $d$, there are many ways to efficiently find and correct its errors. But what if we are instead given $(g^{w_1}, \ldots, g^{w_n})$ where $g$ generates some large cyclic group --- can the errors still be corrected efficiently? This problem is called \emph{error correction in the exponent}, and though it arises naturally in many areas of cryptography, it has received little attention. We first show that \emph{unique decoding} and \emph{list decoding} in the exponent are no harder than the computational Diffie-Hellman (CDH) problem in the same group. The remainder of our results are negative: * Under mild assumptions on the parameters, we show that \emph{bounded-distance decoding} in the exponent, under $e=d-k^{1-\epsilon}$ errors for any $\epsilon > 0$, is as hard as the discrete logarithm problem in the same group. * For \emph{generic} algorithms (as defined by Shoup, Eurocrypt 1997) that treat the group as a black-box,'' we show lower bounds for decoding that exactly match known algorithms. Our generic lower bounds also extend to decisional variants of the decoding problem, and to groups in which the decisional Diffie-Hellman (DDH) problem is easy. This suggests that hardness of decoding in the exponent is a qualitatively new assumption that lies `between'' the DDH and CDH assumptions.
2001
ASIACRYPT
Crypto 2019
TCC 2018
Eurocrypt 2017
TCC 2017
TCC 2015
PKC 2015
Crypto 2014
TCC 2014
PKC 2013
Crypto 2012
TCC 2011
Asiacrypt 2011
Eurocrypt 2011
Crypto 2009
TCC 2008 | 2019-10-17 01:45:05 | {"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": 2, "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": 0.7134515047073364, "perplexity": 1506.4004643441144}, "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-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00244.warc.gz"} |
https://www.cuemath.com/learn/baudhayana/ | Mathematics
Baudhayana | The Great Indian Mathematician
32.5k views
1 Who is Baudhayana? 2 Works of Baudhayana 3 Baudhayana theorem 4 Summary 5 FAQs
21 September 2020
## Who is Baudhayana?
Baudhayana (800 BC - 740 BC) is said to be the original Mathematician behind the Pythagoras theorem. Pythagoras theorem was indeed known much before Pythagoras, and it was Indians who discovered it at least 1000 years before Pythagoras was born! The credit for authoring the earliest Sulba Sutras goes to him.
It is widely believed that he was also a priest and an architect of very high standards. It is possible that Baudhayana’s interest in Mathematical calculations stemmed more from his work in religious matters than a keenness for mathematics as a subject itself. Undoubtedly he wrote the Sulbasutra to provide rules for religious rites, and it would appear almost certain that Baudhayana himself would be a Vedic priest.
The Sulbasutras is like a guide to the Vedas which formulate rules for constructing altars. In other words, they provide techniques to solve mathematical problems effortlessly.
If a ritual was to be successful, then the altar had to conform to very precise measurements. Therefore mathematical calculations needed to be precise with no room for error.
People made sacrifices to their gods for the fulfilment of their wishes. As these rituals were meant to please the Gods, it was imperative that everything had to be done with precision. It would not be incorrect to say that Baudhayana’s work on Mathematics was to ensure there would be no miscalculations in the religious rituals.
## Works of Baudhayana
Baudhayana is credited with significant contributions towards the advancements in mathematics. The most prominent among them are as follows:
### 1. Circling a square.
Baudhayana was able to construct a circle almost equal in area to a square and vice versa. These procedures are described in his sutras (I-58 and I-59).
Possibly in his quest to construct circular altars, he constructed two circles circumscribing the two squares shown below.
Now, just as the areas of the squares, he realised that the inner circle should be exactly half of the bigger circle in area. He knew that the area of the circle is proportional to the square of its radius and the above construction proves the same. By the same logic, just as the perimeters of the two squares, the perimeter of the outer circle should also be $$\sqrt 2$$ times the perimeter of the inner circle. This proves the known fact that the perimeter of the circle is proportional to its radius. This led to an important observation by Baudhayana. That the areas and perimeters of many regular polygons, including the squares above, could be related to each other just as the case of circles.
### 2. Value of π
Baudhayana is considered among one of the first to discover the value of ‘pi’. There is a mention of this in his Sulbha sutras. According to his premise, the approximate value of pi is $$3.$$Several values of π occur in Baudhayana's Sulbasutra, since, when giving different constructions, Baudhayana used different approximations for constructing circular shapes.
Some of these values are very close to what is considered to be the value of pi today, which would not have impacted the construction of the altars. Aryabhatta, another great Indian mathematician, worked out the accurate value of $$π$$ to 3.1416. in 499AD.
### 3. The method of finding the square root of 2.
Baudhayana gives the length of the diagonal of a square in terms of its sides, which is equivalent to a formula for the square root of 2. The measure is to be increased by a third and by a fourth decreased by the 34th. That is it’s diagonal approximately. That is $$1.414216$$, which is correct to five decimals.
Baudhāyana (elaborated in Āpastamba Sulbasūtra i.6) gives the length of the diagonal of a square in terms of its sides, which is equivalent to a formula for the square root of 2:
samasya dvikaraṇī. pramāṇaṃ tṛtīyena vardhayettac caturthenātmacatustriṃśonena saviśeṣaḥ
Sama – Square; Dvikarani – Diagonal (dividing the square into two), or Root of Two
Pramanam – Unit measure; tṛtīyena vardhayet – increased by a third
Tat caturtena (vardhayet) – that itself increased by a fourth, Atma – itself;
Caturtrimsah savisesah – is in excess by 34th part
Baudhayana is also credited with studies on the following :
It can be concluded without a doubt that there is a lot of emphasis on rectangles and squares in Baudhayana’s works. This could be due to specific Yajna Bhumika’s, the altar on which rituals were conducted, for fire-related offerings.
Some of his treatises include theorems on the following.
1. In any rhombus, the diagonals (lines linking opposite corners) bisect each other at right angles (90 degrees)
2. The diagonals of a rectangle are equal and bisect each other.
3. The midpoints of a rectangle joined forms a rhombus whose area is half the rectangle.
4. The area of a square formed by joining the middle points of a square is half of the original one.
## Baudhayana theorem
Baudhāyana listed Pythagoras theorem in his book called Baudhāyana Śulbasûtra.
दीर्घचतुरश्रस्याक्ष्णया रज्जु: पार्श्र्वमानी तिर्यग् मानी च यत् पृथग् भूते कुरूतस्तदुभयं करोति ॥
Baudhāyana used a rope as an example in the above shloka/verse, which can be translated as:
The areas produced separately by the length and the breadth of a rectangle together equal the areas produced by the diagonal.
The diagonal and sides referred to are those of a rectangle, and the areas are those of the squares having these line segments as their sides. Since the diagonal of a rectangle is the hypotenuse of the right triangle formed by two adjacent sides, the statement is seen to be equivalent to the Pythagoras theorem.
There have been various arguments and interpretations of this.
While some people have argued that the sides refer to the sides of a rectangle, others say that the reference could be to that of a square.
There is no evidence to suggest that Baudhayana’s formula is restricted to right-angled isosceles triangles so that it can be related to other geometrical figures as well.
Therefore it is logical to assume that the sides he referred to, could be those of a rectangle.
Baudhāyana seems to have simplified the process of learning by encapsulating the mathematical result in a simple shloka in a layman’s language.
As you see, it becomes clear that this is perhaps the most innovative way of understanding and visualising Pythagoras theorem (and geometry in general).
### Comparing his findings with Pythagoras’ theorem:
In mathematics, the Pythagorean (Pythagoras) theorem is a relation among the three sides of a right triangle (right-angled triangle). It states
In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).”
c is the longest side of the triangle(this is called the hypotenuse) with a and b being the other two sides
The question may well be asked why the theorem is attributed to Pythagoras and not Baudhayana. Baudhayana used area calculations and not geometry to prove his calculations. He came up with geometric proof using isosceles triangles.
## Summary
We have all heard our parents and grandparents talk of the Vedas. Still, there is no denying that modern science and technology owes its origins to our ancient Indian mathematicians, scholars etc. Many modern discoveries would not have been possible but for the legacy of our forefathers who made major contributions to the fields of science and technology. Be it fields of medicine, astronomy, engineering, mathematics, the list of Indian geniuses who laid the foundations of many an invention is endless. | 2022-09-28 02:30:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.7237828373908997, "perplexity": 883.1803830398959}, "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/1664030335059.43/warc/CC-MAIN-20220928020513-20220928050513-00794.warc.gz"} |
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.111.115003 | # Synopsis: Neutron Bursts in Lab Lightning
#### Observation of Neutron Bursts Produced by Laboratory High-Voltage Atmospheric Discharge
A. V. Agafonov, A. V. Bagulya, O. D. Dalkarov, M. A. Negodaev, A. V. Oginov, A. S. Rusetskiy, V. A. Ryabov, and K. V. Shpakov
Published September 12, 2013
Lightning is more than just an impressive electric display. Observations indicate that bursts of neutrons are emitted during thunderstorms, implying nuclear reactions occur in lightning. However, physicists have been unable to identify which reactions. To better understand neutron bursts, a team of Russian researchers has created lightninglike discharges in a laboratory setting. In Physical Review Letters, they report the first detection of neutron emission from a controlled discharge. The results suggest that most burst neutrons are “fast” with initial energies around $10$ mega-electron-volts.
Lightning occurs when strong electric fields build up in a cloud (as Benjamin Franklin’s famous kite-flying experiment demonstrated), but which process initiates the discharge is still a matter of debate. Other lightning mysteries concern the generation of high-energy radiation and particles, such as x rays, gamma rays, positrons, and neutrons. The neutrons could arise from reactions between accelerated ions or from photonuclear interactions, but current empirical data is not sufficient to pinpoint the mechanism.
In order to explore neutron bursts more closely, Aleksey Agafonov of the Lebedev Physical Institute of the Russian Academy of Sciences, Russia, and his colleagues reconfigured an electron beam accelerator to generate high-voltage discharges in air. Using a variety of different detectors, the team identified neutrons streaming out from about a quarter of the discharge events. The neutron flux fell off more slowly than the inverse distance squared ($1/{r}^{2}$), implying that the neutrons originate from an extended region, rather than a localized point. The authors state that more experiments (such as precision time-of-flight measurements) are needed to better interpret these results. – Michael Schirber | 2014-09-03 02:11:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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": 0.54662024974823, "perplexity": 2946.182897953603}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535924131.19/warc/CC-MAIN-20140901014524-00194-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://pkgdown.jrwb.de/pfm/reference/PEC_sw_drift.html | This is a basic, vectorised form of a simple calculation of a contaminant concentration in surface water based on complete, instantaneous mixing with input via spray drift.
PEC_sw_drift(
rate,
applications = 1,
water_depth = 30,
drift_percentages = NULL,
drift_data = c("JKI", "RF"),
crop = "Ackerbau",
distances = c(1, 5, 10, 20),
rate_units = "g/ha",
PEC_units = "µg/L"
)
## Arguments
rate Application rate in units specified below Number of applications for selection of drift percentile Depth of the water body in cm Percentage drift values for which to calculate PECsw. 'drift_data' and 'distances' if not NULL. Source of drift percentage data. If 'JKI', the [drift_data_JKI] included in the package is used. If 'RF', the Rautmann formula is used, if implemented for the crop type and number of applications Crop name (use German names for JKI data), defaults to "Ackerbau" The distances in m for which to get PEC values Defaults to g/ha Requested units for the calculated PEC. Only µg/L currently supported
## Value
The predicted concentration in surface water
## Examples
PEC_sw_drift(100)#> 1 m 5 m 10 m 20 m
#> 0.92333333 0.19000000 0.09666667 0.05000000 # Alternatively, we can use the formula for a single application to "Ackerbau" from the paper
PEC_sw_drift(100, drift_data = "RF")#> 1 m 5 m 10 m 20 m
#> 0.92350000 0.19114149 0.09699222 0.04921742
# This makes it possible to also use different substances
PEC_sw_drift(100, distances = c(1, 3, 5, 6, 10, 20, 50, 100), drift_data = "RF")#> 1 m 3 m 5 m 6 m 10 m 20 m 50 m
#> 0.92350000 0.31512171 0.19114149 0.15990435 0.09699222 0.04921742 0.02007497
#> 100 m
#> 0.01018678
# Using custom drift percentages is also supported
PEC_sw_drift(100, drift_percentages = c(2.77, 0.95, 0.57, 0.48, 0.29, 0.15, 0.06, 0.03))#> 2.77 % 0.95 % 0.57 % 0.48 % 0.29 % 0.15 % 0.06 %
#> 0.92333333 0.31666667 0.19000000 0.16000000 0.09666667 0.05000000 0.02000000
#> 0.03 %
#> 0.01000000 | 2022-06-30 07:15:29 | {"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": 0.3132452368736267, "perplexity": 4384.41707179859}, "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-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00027.warc.gz"} |
https://stats.stackexchange.com/questions/110661/which-predictors-to-include-in-multiple-regression-where-predictors-are-more-tha/110668 | # Which predictors to include in multiple regression where predictors are more than number of samples?
I have 3000 predictors (biological measurements) and 200 samples (weight), and would like to build a linear multiple regression model. How do I select which predictors to include in the regression model? Someone suggested using R package 'rpart'. Please help.
1. Use the R package glmnet to estimate the coefficients. This package uses a penalize likelihood that shrinks many coefficients to zero. It's helpful not only for having more predictors than observations, but also for dealing with multicollinearity among your predictors. Note that you might have 3000 predictors, but you could have many more features than that if you include polynomial terms (I recommend not using any more than three) and interactions. glmnet has no problem with that. You will, however, need to use k-fold cross-validation to tune the likelihood penalization parameter, and the parameter that determines the weight of LASSO versus ridge penalization.
2. Use the R package randomForest if you wish to avoid the choice of what interaction effects and polynomials to include altogether. Random forest creates many decision trees based on boostrapped samples of your data. You will need to tune the mtry parameter using cross-validation. This parameter guides the number of features that are chosen at random to create each tree in the forest. Also, I recommend you grow a large forest of at least 200 trees. | 2021-09-19 20:44:40 | {"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": 0.49429386854171753, "perplexity": 962.8371777859628}, "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/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00547.warc.gz"} |
https://mathsci.kaist.ac.kr/pow/tag/hermitian-matrix/ | # 2021-04 Product of matrices
For an $$n \times n$$ matrix $$M$$ with real eigenvalues, let $$\lambda(M)$$ be the largest eigenvalue of $$M$$. Prove that for any positive integer $$r$$ and positive semidefinite matrices $$A, B$$,
$[\lambda(A^m B^m)]^{1/m} \leq [\lambda(A^{m+1} B^{m+1})]^{1/(m+1)}.$
GD Star Rating
# 2019-15 Singular matrix
Let $$A, B$$ be $$n \times n$$ Hermitian matrices. Find all positive integer $$n$$ such that the following statement holds:
“If $$AB – BA$$ is singular, then $$A$$ and $$B$$ have a common eigenvector.”
GD Star Rating
# 2017-01 Eigenvalues of Hermitian matrices
Let $$A, B, C$$ be $$N \times N$$ Hermitian matrices with $$C = A+B$$. Let $$\alpha_1 \geq \dots \geq \alpha_N$$, $$\beta_1 \geq \dots \geq \beta_N$$, $$\gamma_1 \geq \dots \geq \gamma_N$$ be the eigenvalues of $$A, B, C$$, respectively. For any $$1 \leq k \leq N$$, prove that
$\gamma_1 + \gamma_2 + \dots + \gamma_k \leq (\alpha_1 + \alpha_2 + \dots + \alpha_k) + (\beta_1 + \beta_2 + \dots + \beta_k)$
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Let $$A, B, C = A+B$$ be $$N \times N$$ Hermitian matrices. Let $$\alpha_1 \geq \cdots \geq \alpha_N$$, $$\beta_1 \geq \cdots \geq \beta_N$$, $$\gamma_1 \geq \cdots \geq \gamma_N$$ be the eigenvalues of $$A, B, C$$, respectively. For any $$1 \leq i, j \leq N$$ with $$i+j -1 \leq N$$, prove that
$\gamma_{i+j-1} \leq \alpha_i + \beta_j$ | 2021-05-12 02:40:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.958941638469696, "perplexity": 163.8088914090518}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00421.warc.gz"} |
http://math.stackexchange.com/questions/408832/how-far-is-it-true-that-statements-dependent-on-axiom-of-choice-are-not-construc/409074 | # How far is it true that statements dependent on Axiom of Choice are not constructive.
Axiom of Choice is often used in mathematics to construct various objects, such as basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, unmeasurable subset of $\mathbb{R}$, or a non-principal ultrafilter on $\mathbb{N}$.
It is a popular "meta-theorem" that if such construction essentially relies on the Axiom of Choice, then the object cannot be constructed explicitly. Here, by "essentially relying" I mean that it is consistent with ZF that what we are constructing (Hamel basis, ultrafilter, etc.) does not exist; by "explicit construction" I mean one that can be carried out within ZF.
My question is: Is this meta-theorem really true? Certainly, it is not possible to carry out a construction within ZF, and then prove it correct in ZF. However, it struck me that it might be possible that the construction itself fits in ZF, and it is only verification that requires the Axiom of Choice. Of course, it would be very bizarre, but bizarre things do happen. Do we have some strong evidence that they won't occur in this situation? How certain is it that, say, a Hamel basis can't be constructed in ZF?
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/Checks watch. Begins counting down until Asaf's arrival./ – Cameron Buie Jun 1 '13 at 22:12
@Cameron: I was interrupted by a phone call that ended up being the wrong number... :-) – Asaf Karagila Jun 1 '13 at 22:13
## 2 Answers
We know it is impossible to define a Hamel basis for $\Bbb R$ over $\Bbb Q$ (or construct a non-principal ultrafilter, and so on) because it is a theorem:
If $\sf ZFC$ is consistent, then the theory $\sf ZF+\text{There is no Hamel basis for }\Bbb R\text{ over }\Bbb Q$ is consistent.
Since it is consistent with $\sf ZF$ that there is no such basis, it means that we cannot prove from $\sf ZF$ that such basis exists. We have to assume more, in terms of choice.
Remember the completeness theorem tells us that $\sf ZF\vdash\varphi$ if and only if $\varphi$ is true in every model of $\sf ZF$. Similarly for $\sf ZFC$. So given a statement $\varphi$ which is provable from $\sf ZFC$, if we can find a model of $\sf ZF$ where $\varphi$ fails, we can be certain that $\sf ZF$ cannot prove it, and some axiom of choice is needed for the proof.
For example, we know that if $\sf ZFC$ is consistent then we can create a model in which there is no Hamel basis of $\Bbb R$ over $\Bbb Q$. Of course, in that model the axiom of choice fails, but the process itself is consistent. Of course it might be the case that there are no models of $\sf ZFC$ at all, but we still know how to make it work then. We can translate these arguments into syntactic arguments which state the quote text above.
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Don't you mean "If ZF is consistent"? ZFC is consistent if ZF is, so those are really the same, but it seems more natural to drop that C. – dfeuer Jun 1 '13 at 22:14
@dfeuer: No, I mean $\sf ZFC$. Yes, those are the same thing, but almost all the constructions begin with a model of $\sf ZFC$. It's easier to skip the step of consistency of $\sf ZFC$ from the consistency of $\sf ZF$. – Asaf Karagila Jun 1 '13 at 22:16
Thank you, this was very enlightening! – Jakub Konieczny Jun 1 '13 at 22:22
@Feanor: You're welcome. – Asaf Karagila Jun 1 '13 at 22:24
@dfeuer: I think Asaf Karagila answered my problem, but thank you for the concern. What I understand now is that given a set $A \subset \mathbb{R}$ (possible to construct without AC), if it was a Hamel basis, then $ZF +$ "no Hamel basis exists" would not have a model, hence would not be consistent (but we already know it is). I think the difference is that such set $A$ is (or isn't) a Hamel basis as soon as it is constructed, and this can't be changed by adding additional assumptions about general sets. – Jakub Konieczny Jun 1 '13 at 23:59
Depending on one's interpretation of the "meta-theorem" you stated, the following might be seen as a counterexample.
There is a sentence $\varphi$ such that it is a theorem of $\mathsf{ZF}$ that $\varphi$ defines a Hamel basis $B$ of $\mathbb{R} \cap L$ over $\mathbb{Q}$. Here $L$ denotes Goedel's constructible universe. It is a theorem of $\mathsf{ZF}$ that $L \models \mathsf{ZFC} + V=L$, that the relativization $\varphi^L$ of $\varphi$ to $L$ defines the same set $B$ within $L$, and that $L \models B \text{ is a Hamel basis of$\mathbb{R}$over$\mathbb{Q}$.}$
So here we have a "construction" (definition) $\varphi$ such that $\mathsf{ZF}$ proves that it defines a unique object, and that this object is a $\mathbb{Q}$-linearly independent set of reals, but does not prove that the $\mathbb{Q}$-linear span of the set of reals so defined is all of $\mathbb{R}$. On the other hand, the theory $\mathsf{ZFC} + V = L$ "verifies" (proves) that the set $B$ defined by $\varphi$ does in fact span $\mathbb{R}$, and therefore is a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$.
The only problem with this counterexample is that the "verification" requires the theory $\mathsf{ZFC} + V = L$, which is properly stronger than $\mathsf{ZFC}$. I'm not sure if this addresses your question (which is a bit fuzzy) but I thought it was worth mentioning.
EDIT: One main point of this "counterexample" is that, although many people think of models of AC as containing more sets (e.g. more $\mathbb{Q}$-linearly independent subsets of $\mathbb{R}$, including some that span) it makes just as much sense to think of them as containing fewer sets (e.g. fewer reals, so that spanning becomes easier.) Given models $M$ and $N$ of $\mathsf{ZF}$ with $M \subset N$, one can have $M \models \mathsf{AC}$ and $N \models \neg \mathsf{AC}$ or vice versa. So it is not always right to think of $\mathsf{AC}$ as providing "extra" sets.
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I assume it's the fact I only managed to sleep for one hour "this morning", but the first part of your answer, about $V=L$, is completely incomprehensible to me. As for the second part, to drive the point home, we can have $M\subseteq N\subseteq M'$ such that $M,M'\models\sf ZFC$ and $N\models\sf ZF+\lnot AC$. – Asaf Karagila Jun 2 '13 at 7:19
@Asaf Yes, it's not very clearly written. However, I too am getting sleepy, so maybe I will write more tomorrow. In the meantime it's probably easier to think about wellorderings of the reals rather than Hamel bases. One can always define a well-ordering of the set $\mathbb{R} \cap L$, and this set may be all of $\mathbb{R}$ or it may be merely countable. – Trevor Wilson Jun 2 '13 at 7:26
Well, to drive that point further, one can use the same definition outside of $L$, it's just not provable that this set is a Hamel basis/well-ordering/etc. (because of often it won't be.) I think Miller proved that there is a $\Pi^1_1$ Hamel basis in $L$, but I'm far from being in the right cognitive state of mind to be certain. – Asaf Karagila Jun 2 '13 at 7:29
I was thinking of a definition that is absolute to $L$, in which case whether it is a Hamel basis for $\mathbb{R} \cap L$ (resp. well-ordering of $\mathbb{R} \cap L$) is absolute to $L$. What is not absolute to $L$ is the statement $\mathbb{R} \cap L = \mathbb{R}$, of course. – Trevor Wilson Jun 2 '13 at 7:34
Yes, I'm glad that we both understand each other, and that we both agree that we're both tired... :-) – Asaf Karagila Jun 2 '13 at 7:35 | 2016-07-25 02:30:25 | {"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": 0.9088194966316223, "perplexity": 247.65707918293995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824201.56/warc/CC-MAIN-20160723071024-00265-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://www.aminer.cn/pub/5f48d35191e011096f9560b2/neural-learning-of-one-of-many-solutions-for-combinatorial-problems-in-structured | # Neural Learning of One-of-Many Solutions for Combinatorial Problems in Structured Output Spaces
Yatin Nandwani
Deepanshu Jindal
Abstract:
Recent research has proposed neural architectures for solving combinatorial problems in structured output spaces. In many such problems, there may exist multiple solutions for a given input, e.g. a partially filled Sudoku puzzle may have many completions satisfying all constraints. Further, we are often interested in finding {\em any on...More
Code:
Data:
0
Tags | 2020-10-31 02:16:34 | {"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": 0.9053118228912354, "perplexity": 1297.5344189325258}, "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-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00304.warc.gz"} |
http://math.stackexchange.com/questions/46572/efficient-algorithm-for-finding-how-many-times-a-point-is-inside-the-triangles-f/46655 | # Efficient algorithm for finding how many times a point is inside the triangles formed by given points
Given n 2D points and a special point p, what would be the best way to find how many times p is inside among those $^nC_3$ triangles formed by the n points.
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What makes you think that you can do better than testing all triangles? Anyway, what is the motivation here? – lhf Jun 20 '11 at 22:50
@lhf Simple heuristics strongly indicate there is a much better solution than the brute force $O(n^3)$ algorithm of testing all triangles. For instance, any random pair of the $\mathbf{n}$ points determines two half planes, immediately ruling out all triangles formed by points on the half plane not containing $\mathbf{p}$, which should be approximately one-eighth of all triangles. A few iterations of this will (on the average) eliminate a huge fraction of all the triangles at a cost of just $O(n)$. – whuber Jun 21 '11 at 3:58
I disagree with both of these comments. As @whuber points out, there are certainly better ways than testing all the triangles. However, whuber's approach is still $O(n^3)$, since the number of triangles containing $\mathbb p$ increases with $n^3$, so discarding non-containing ones in $O(n)$ doesn't help, you'd also have to count the containing ones more than one at a time. Also, you say the eliminated fraction would be huge, but successive iterations are correlated, so you won't be eliminating one eighth in each iteration, so it's not obvious how huge the fraction will get how fast. – joriki Jun 21 '11 at 4:59
@joriki I agree with that assessment (and was aware of the problem that these heuristics do not necessarily lead to a reduction in asymptotic best-case performance, which is why I chose vague quantifiers like "few" and "huge fraction.") My point was only that it is obvious that one can do better than "testing all triangles." The algorithm you provide is an excellent way to go about it! – whuber Jun 21 '11 at 14:07
Here's an algorithm in $O(n\log n)$ time and $O(n)$ space; I doubt you can do any better than that.
I'll assume general position, i.e., $\mathbb p$ isn't on the line through any two of the $n$ points. In the following, angles are normalized to some interval of length $2\pi$, say, $[0,2\pi)$, and differences between them are normalized to $(-\pi,\pi)$ by adding or subtracting $2\pi$ if necessary.
Calculate and order the angles under which the $n$ points appear from $\mathbb p$; this takes $O(n\log n)$ time. The number of triangles containing $\mathbb p$ is one third of the number of ordered triples of angles such that the difference between the angles is positive for all three pairs of (cyclically) successive angles. (It can't be zero or $\pi$ because of the assumption of general position.)
To count the number of such triples, assign indices to the angles according to their order, and for each angle $\alpha_i$ with index $i$ determine the number $m_i$ of indices $j$ such that $\alpha_j-\alpha_i$ is positive. This can be done in $O(n\log n)$ time by performing a binary search for the opposite angle $\alpha_i+\pi$.
Now iterate over $i$, counting the number of ordered triples with first element $\alpha_i$ that fulfill the above condition. For each second element $\alpha_j$ with positive $\alpha_j-\alpha_i$, the number of such triples is the number of third elements $\alpha_k$ such that both $\alpha_k-\alpha_j$ and $\alpha_i-\alpha_k$ are positive. This is the overlap between the $m_j$ angles for which $\alpha_k-\alpha_j$ is positive and the $n-m_i-1$ angles for which $\alpha_k-\alpha_i$ is negative. This can be written as $m_j+(n-m_i-1)-(n-d_{ij}-1)=m_j+d_{ij}-m_i$, where $d_{ij}$ is the cyclical distance from $i$ to $j$. (A circle diagram containing those three terms might aid understanding.)
The term $m_i$ is independent of $j$, so summing it over $j$ just gives $m_i^2$. The term $d_{ij}$ is also readily summed over $j$; the sum is $m_i(m_i+1)/2$, so these terms together contribute $-m_i(m_i-1)/2$. The sum over $m_j$ can be calculated in constant time by precalculating the sums $\sum_{j<i}m_j$; then the sum over $m_j$ for $\alpha_j-\alpha_i$ positive is just the cyclical difference between two such sums, one with index $i$ and the other with the "opposite" index that was already determined by binary search to calculate $m_i$ above.
I've skipped over some of the technical details, but I hope it's clear how all the steps can be carried out in $O(n\log n)$ time. (Clearly the data structures used only use $O(n)$ space.)
[Edit:] There was an error in the original calculation. Incidentally, the correct calculation is slightly simpler. It also allows us to immediately derive the number of triangles in the case of a regular $n$-gon (with $n$ odd for general position): In this case, all $m_i$ are equal, so $m_j$ and $m_i$ cancel, and the result is just $1/3$ of the sum of $d_{ij}$ over $i$ and $j$, which is $nm(m+1)/2$. Here $m$ is the common value of the $m_i$, which is $(n-1)/2$; the end result is $n(n^2-1)/24$ triangles.
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Nice, this is a really clever algorithm. – ShreevatsaR Jun 21 '11 at 6:54
I dont think that there is an algorithm with $O(n \log{(n)}$ worst case runningtime . a regular polygon with n vertices we can construct about $\frac{n^2}{2}$ triangles that contain the center: select $A$ vertex a, select the vertex $B$ that is almost diagonally. Almost each triangle formed by $A$, $B$ and a vertext $C$ choosen from the longer of the two parts of the circumference divided by $A$ and $B$ contains the center. – miracle173 Jun 21 '11 at 7:45
@miracle173: The task is to count the triangles, not to enumerate them. As I wrote in my comment under the question, there are actually $O(n^3)$ triangles containing $\mathbb p$ (under mild assumptions about the distributions of points); that doesn't prevent us from counting them in $O(n\log n)$ time. Regarding "worst case": There are no random elements in my algorithm; sorting and binary search can both be done in worst-case $O(n\log n)$ time. – joriki Jun 21 '11 at 8:01
@miracle173: By the way, for a regular $n$-gon, too, there are $O(n^3)$ triangles containing the centre. – joriki Jun 21 '11 at 8:15
@joriki: you are right. Should such a wrong comment be deleted? – miracle173 Jun 23 '11 at 22:25 | 2015-08-30 01:50:42 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7348181009292603, "perplexity": 240.3813145850994}, "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-2015-35/segments/1440644064865.49/warc/CC-MAIN-20150827025424-00092-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://mathoverflow.net/questions/376573/are-universally-submersive-morphisms-of-stacks-universal-descent-morphisms-for-r | Are universally submersive morphisms of stacks universal descent morphisms for relative étale stacks?
Recall that a map of schemes, algebraic spaces, stacks, etc is called submersive if the associated map on underlying topological spaces is a quotient map. Recall moreover that a map is called universally submersive if it is submersive and all of its base changes are submersive.
Recall that a map $$f:b\to b^\prime$$ is called an $$E$$-descent morphism relative to a Cartesian fibration $$E\to B$$ if the natural map $$E_{b^\prime}\to E_{S_f}$$ is fully faithful, where $$E_{S_f}$$ is the value of $$E$$ on the $$b^\prime$$-sieve $$S_f$$ generated by $$f$$. A map $$b\to b^\prime$$ is a universal E-descent morphism if all of its base-changes are E-descent morphisms.
In SGA4 VIII 9.3, there is a theorem demonstrating that universally submersive maps of schemes are universal descent morphisms for the Cartesian fibration on the category of schemes sending a scheme $$S$$ to the category of étale sheaves of sets on $$S$$ with transition morphisms given by the pullback functors.
Two questions:
1.) Are universally submersive maps of schemes also universal descent maps for the $$(2,1)$$-stack sending a scheme $$S$$ to the $$(2,1)$$-category of étale sheaves of groupoids on $$S$$? How about the $$(n+1,1)$$-stack of sheaves of n-groupoids?
2.) Can this be extended to the case where we replace the category of schemes with the $$(2,1)$$-category of Artin stacks (or Deligne-Mumford stacks)? Do we have to add more adjectives to 'universally submersive' to make it hold in this more general context (for example, representable universally submersive, locally separated universally submersive, etc)?
The original proof in SGA uses a very cute trick, but it doesn't seem to apply once you throw stacks in the mix (it uses the fact that any map admitting a section is a mono, and this fails very badly for stacks).
Edit: I added a bounty. What's left to understand is the case of a universal submersion of ordinary schemes (whether these are universal descent morphisms for the fibration of étale n-sheaves)
• Does the map E_{b'} \to E_{S_f} not have to be an equivalence? It seems you're omitting the "descent data are effective" part of the stack condition. I believe 1). 2) should follow from the usual yoga of "cover the artin stack with a scheme, give data on the cover plus gluing data on overlaps" etc. The difficulty is defining "univ subm" for stacks, but if they're representable, this should be clear. Also you may mean the lis-et site in the artin stack case. – Leo Herr Nov 19 '20 at 1:15
• @LeoHerr That's effective descent (or -2-descent (Lurie), or 0-descent (Giraud)). I'm talking about -1-descent (Lurie)/1-descent (Giraud) (i.e. ineffective descent) (there is an off-by-2 indexing change in modern sources). – Harry Gindi Nov 19 '20 at 1:56
• Also this stack of étale sheaves can be defined without reference to a topology. The objects are precisely the n-representable étale maps over your stack. – Harry Gindi Nov 19 '20 at 2:05
• Does SGA actually prove that \'etale sheaves of sets form a stack over schemes for the universal submersion topology? Unless I'm misunderstanding, the effectivity of the descent is only proved in certain cases (in 9.4 of SGA4 VIII 9.3). – Akhil Mathew Dec 29 '20 at 19:15
• @HarryGindi: indeed, I missed the definition of universal descent (vs. universal effective descent). I'm curious how generally effectivity might hold. – Akhil Mathew Dec 29 '20 at 20:30
Let $$X$$ be an algebraic stack, and let $$f:Y\to X$$ be a universal submersion. Then to show that $$Y\to X$$ is universal descent for $$E_n$$ the (n+1,1)-stack of étale $$n$$-sheaves, find an fppf cover $$U\to X$$ by $$U$$ a separated scheme. Then pulling back $$f$$ to $$U$$, we have a universal submersion $$f_U:Y_U \to U$$. Now we can choose an fppf cover $$V\to Y_U$$ by a separated scheme $$V$$, which is a universal submersion, and which now makes the composite $$V\to Y_U \to U$$ a universal submersion as well.
Then by the first claim, if it is true, we see that $$V\to U$$ is universal $$E_n$$-descent. It follows then that $$Y_U\to U$$ is universal $$E_n$$-descent, by the usual stuff about factorization of universal descent maps. Moreover, the map $$U\to X$$ is universal effective $$E_n$$-descent by Toën's theorem on fppf descent for relative n-stacks together with a computation involving the cotangent complex. In particular, it is universal $$E_n$$-descent, and such maps are closed under composition.
So it follows that $$Y_U\to X$$ is universal $$E_n$$-descent, but $$Y_U\to X$$ factors as $$Y_U\to Y\to X$$, and using the lemma on factorization of universal descent morphisms, we see that $$Y\to X$$ is now universal $$E_n$$-descent. In particular, we are reduced to proving the claim of universal $$E_n$$-descent in the case of a universal submersion of separated schemes.
• This reduction step works for $X$ even an algebraic $m$-stack. – Harry Gindi Nov 17 '20 at 15:59 | 2021-03-01 04:55:02 | {"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": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9494675993919373, "perplexity": 736.090264127626}, "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-2021-10/segments/1614178361849.27/warc/CC-MAIN-20210301030155-20210301060155-00587.warc.gz"} |
https://zbmath.org/?q=an:1074.40500&format=complete | # zbMATH — the first resource for mathematics
On the Tauberian constant in the Ikehara theorem. (English) Zbl 1074.40500
Summary: The Ikehara theorem says: Let the Laplace transform $$f(s)$$ of a nonnegative nondecreasing function $$A(x)$$ defined for $$x\in \langle 0; + \infty )$$ converge in the halfplane $$\text{Re} \;s>0$$ and $$g(s) = f(s)-\frac 1{s-1}$$ be analytic in the halfplane $$\text{Re} \;s\geq 0$$. Then $$e^{-x}A(x) - 1 = o(x)$$, $$x\rightarrow +\infty$$. The only theorem of the paper generalizes the main term 1 to some polynomial and the remainder term $$o(x)$$ to $$O(x^{-n})$$ assuming the function $$g^{(n)}(it)$$ integrable, $$n\in \mathbb N$$, where $$g(s)$$ is the regular part of $$f(s)$$ in the point $$s_0 = 1$$. As the constant in $$O$$ is a multiple of the norm of $$g^{(n)}(it)$$, the author uses the uncertain adjective Tauberian. The theorem contains some inconvenient assumptions. The theorem is used to obtain the remainder term in the P.N.T. in the form $$O(x \exp p (-c\log ^{\frac 1{11}}x))$$, $$x\rightarrow +\infty$$. By analytical methods an essentially better estimation of this remainder term can be obtained.
##### MSC:
40E05 Tauberian theorems 11N05 Distribution of primes 11M45 Tauberian theorems
Full Text:
##### References:
[1] Chandrasekharan K.: Introduction to Analytic Number Theory. Springer Verlag, 1968. · Zbl 0169.37502 [2] Čížek J.: On the proof of the prime number theorem. Čas. pěst. mat. 106 (1981), 395-401. [3] Rudin W.: Real and Complex Analysis. New York, 1974. · Zbl 0278.26001 [4] Subchankulov M. A.: Tauberovy teoremy s ostatkom. Nauka, Moskva, 1976. [5] Walfisz A.: Weyl’sche Exponentialsummen in der neuere Zahlentheorie. Berlin, 1963. · Zbl 0146.06003
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-09-19 08:12:56 | {"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": 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": 0.8398902416229248, "perplexity": 1041.4904178144993}, "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/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00075.warc.gz"} |
https://vi.stackexchange.com/questions/13518/force-prompt-for-whether-to-delete-a-swap-file?noredirect=1 | # Force prompt for whether to delete a swap file
How do you configure vim to prompt for deletion of swap files even if it believes another instance of vim may still be using the file? I still want to see the warning message ... just also have the option to delete the file and overrule vim.
The option below, for instance, doesn't have the (D)elete option. (where program.cc may potentially be modified by another vim instance)
(1) Another program may be editing the same file. If this is the case,
be careful not to end up with two different instances of the same
file when making changes. Quit, or continue with caution.
(2) An edit session for this file crashed.
If this is the case, use ":recover" or "vim -r program.cc"
to recover the changes (see ":help recovery").
If you did this already, delete the swap file "program.swp"
to avoid this message.
Swap file "program.cc.swp" already exists!
[O]pen Read-Only, (E)dit anyway, (R)ecover, (Q)uit, (A)bort:
• Hmm, I'm not sure when vim does or doesn't allow the deletion option, but it shows for me. Is your file in a restricted area? What version of vim are you on? What OS are you on? – James Sep 8 '17 at 20:42
The "Delete it" option isn't displayed if the Vim process is still running; I can't find this documented anywhere but I looked it up in the source code; from memline.c, line 4512 (slightly simplified):
do_dialog(
[..]
process_still_running
? (char_u *)_("&Open Read-Only\n&Edit anyway\n&Recover\n&Quit\n&Abort") :
(char_u *)_("&Open Read-Only\n&Edit anyway\n&Recover\n&Delete it\n&Quit\n&Abort"), [..]);
The swap file embeds the process ID which created it, and if a process with that PID still exists it considers the process to be "running".
The swap message should display this information:
E325: ATTENTION
Found a swap file by the name "~/.vim/tmp/swap/swapy.swp"
owned by: martin dated: Fri Sep 8 22:13:35 2017
file name: ~martin/swapy
modified: no
user name: martin host name: arch.arp242.net
process ID: 17355 (still running)
While opening file "swapy"
dated: Fri Sep 8 22:13:35 2017
Note the process ID: 17355 (still running) line.
The most likely scenario is that you have another Vim instance running somewhere :-) You could kill it if you can't find it.
There is a small chance that the PID got re-used by another process though, in which case your only option is to quit Vim, manually remove the swap file, and restart it again:
$rm ~/.vim/tmp/swap/swapy.swp • hmm, makes sense. Is there a way to say "remove all the swap files corresponding to this buffer"? That way I can choose edit anyway, then remove the files. – Gregory Nisbet Sep 9 '17 at 1:33 • @GregoryNisbet AFAIK there isn't an easy way for this; I'm also not sure if that's what you really want, since it may cause problems in the Vim instance that already has this file open. – Martin Tournoij Sep 9 '17 at 2:21 • This is a particularly important answer if you have TMUX sessions in the background (e.g., your graphical desktop crashed and you logged back in - speaking for a friend :) ) - tmux list-sessions followed by tmux attach -t$NUM is your friend... tmux is simple and awesome. – zenaan Oct 1 '19 at 0:44
Something like this may work: first, based on this answer, you could create a separate location for all your swap files like this set directory^=\$HOME/.vim/swap//
Then, using the answer from here, place this
noremap <LEADER>s <C-w>o:sav! ~/.vim/.recovered<CR>:vs<CR><C-w>w:bn<CR>
noremap <LEADER>t :wa<CR>:bp\|bd #<CR><C-o>
into your vimrc, the way that I get to mine is by using the terminal I type vim .vimrc which opens it and allows me to edit it.
Then follow this set of instructions:
Upon reaching the ATTENTION: Found a swap file... prompt, I
type r to recover the swap
type <LEADER>s to save the swapped version as .recovered and open the original in a new split
type e at the swap prompt to edit the original file
type :windo diffthis to diff the two files
make any necessary changes to the original
type <LEADER>t to write the original file, close it, and reopen it
type d at the swap prompt to delete the swap file
close the .recovered file (typing something like :bp<CR>:bp\|bd #<CR>).
The reason that I placed the first step is because it seems as though it would be easier to delete all your .swp and .swo if they are in the same location. Hopefully this helps. | 2021-04-22 02:09:57 | {"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": 0.3964407742023468, "perplexity": 6086.6190025392125}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618039560245.87/warc/CC-MAIN-20210422013104-20210422043104-00518.warc.gz"} |
http://mathematica.stackexchange.com/questions/30520/using-data-from-multiple-cameras | # Using data from multiple cameras
ImageCapture[] has an option Device-> for specifying which camera to use, but CurrentImage[] takes no options and no relevant arguments. Changing \$DefaultImagingDevice might or might not work, but it would be a pretty awkward solution for a program that needs to frequently take frames from different cameras quickly.
Is there a way to use CurrentImage[] with more than one camera? If not, is there a better workaround?
-
Take a look here mathematica.stackexchange.com/a/2772/193 – belisarius Aug 15 '13 at 18:44
I'm actually using that function now on a different application on Linux; I was wondering if there was a better workaround than just rewriting that function into taking an argument to pass to cvQueryFrame(). Fortunately, I'm familiar with OpenCV. – Backgammon Aug 15 '13 at 18:48 | 2014-08-28 03:23:02 | {"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": 0.45258599519729614, "perplexity": 1514.0818336368793}, "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/1408500830074.72/warc/CC-MAIN-20140820021350-00263-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/help-please-problem-on-acceleration.41926/ | # Homework Help: Help please Problem on Acceleration
1. Sep 5, 2004
### shawonna23
The land speed record of 13.9 m/s (31 mi/h) for birds is held by the Australian emu. An emu running due south in a straight line at this speed slows down to a speed of 10.5 m/s in 2.7 s.
What is the direction of the bird's acceleration?
Assuming that the acceleration remains the same, what is the bird's velocity after an additional 3.9 s has elapsed?
2. Sep 7, 2004
### JasonRox
Hint:
$$a=\frac{v_f-v_i}{t}$$
3. Sep 8, 2004
### Beer-monster
Hint: If it slows down it is decelerating i.e negative acceleration
4. Sep 8, 2004
### HallsofIvy
Also, "speeding up" means speed is increasing: acceleration is "positive"- in same direction as velocity.
"Slowing down" means speed is decreasing: acceleration is "negative"- in opposite direction to velocity. | 2018-11-21 03:51:36 | {"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": 0.30581140518188477, "perplexity": 4820.616013040445}, "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/1542039747024.85/warc/CC-MAIN-20181121032129-20181121053102-00020.warc.gz"} |
https://www.nag.com/numeric/cl/nagdoc_fl26.2/html/g02/g02laf.html | # NAG Library Routine Document
## 1Purpose
g02laf fits an orthogonal scores partial least squares (PLS) regression by using singular value decomposition.
## 2Specification
Fortran Interface
Subroutine g02laf ( n, mx, x, ldx, isx, ip, my, y, ldy, xbar, ybar, xstd, ystd, xres, yres, w, ldw, p, ldp, t, ldt, c, ldc, u, ldu, xcv, ycv,
Integer, Intent (In) :: n, mx, ldx, isx(mx), ip, my, ldy, iscale, maxfac, ldxres, ldyres, ldw, ldp, ldt, ldc, ldu, ldycv Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: x(ldx,mx), y(ldy,my) Real (Kind=nag_wp), Intent (Inout) :: xstd(ip), ystd(my), xres(ldxres,ip), yres(ldyres,my), w(ldw,maxfac), p(ldp,maxfac), t(ldt,maxfac), c(ldc,maxfac), u(ldu,maxfac), ycv(ldycv,my) Real (Kind=nag_wp), Intent (Out) :: xbar(ip), ybar(my), xcv(maxfac)
#include <nagmk26.h>
void g02laf_ (const Integer *n, const Integer *mx, const double x[], const Integer *ldx, const Integer isx[], const Integer *ip, const Integer *my, const double y[], const Integer *ldy, double xbar[], double ybar[], const Integer *iscale, double xstd[], double ystd[], const Integer *maxfac, double xres[], const Integer *ldxres, double yres[], const Integer *ldyres, double w[], const Integer *ldw, double p[], const Integer *ldp, double t[], const Integer *ldt, double c[], const Integer *ldc, double u[], const Integer *ldu, double xcv[], double ycv[], const Integer *ldycv, Integer *ifail)
## 3Description
Let ${X}_{1}$ be the mean-centred $n$ by $m$ data matrix $X$ of $n$ observations on $m$ predictor variables. Let ${Y}_{1}$ be the mean-centred $n$ by $r$ data matrix $Y$ of $n$ observations on $r$ response variables.
The first of the $k$ factors PLS methods extract from the data predicts both ${X}_{1}$ and ${Y}_{1}$ by regressing on ${t}_{1}$ a column vector of $n$ scores:
$X^1 = t1 p1T Y^1 = t1 c1T , with t1T t1 = 1 ,$
where the column vectors of $m$ $x$-loadings ${p}_{1}$ and $r$ $y$-loadings ${c}_{1}$ are calculated in the least squares sense:
$p1T = t1T X1 c1T = t1T Y1 .$
The $x$-score vector ${t}_{1}={X}_{1}{w}_{1}$ is the linear combination of predictor data ${X}_{1}$ that has maximum covariance with the $y$-scores ${u}_{1}={Y}_{1}{c}_{1}$, where the $x$-weights vector ${w}_{1}$ is the normalised first left singular vector of ${X}_{1}^{\mathrm{T}}{Y}_{1}$.
The method extracts subsequent PLS factors by repeating the above process with the residual matrices:
$Xi = Xi-1 - X^ i-1 Yi = Yi-1 - Y^ i-1 , i=2,3,…,k ,$
and with orthogonal scores:
$tiT tj = 0 , j=1,2,…,i-1 .$
Optionally, in addition to being mean-centred, the data matrices ${X}_{1}$ and ${Y}_{1}$ may be scaled by standard deviations of the variables. If data are supplied mean-centred, the calculations are not affected within numerical accuracy.
None.
## 5Arguments
1: $\mathbf{n}$ – IntegerInput
On entry: $n$, the number of observations.
Constraint: ${\mathbf{n}}>1$.
2: $\mathbf{mx}$ – IntegerInput
On entry: the number of predictor variables.
Constraint: ${\mathbf{mx}}>1$.
3: $\mathbf{x}\left({\mathbf{ldx}},{\mathbf{mx}}\right)$ – Real (Kind=nag_wp) arrayInput
On entry: ${\mathbf{x}}\left(\mathit{i},\mathit{j}\right)$ must contain the $\mathit{i}$th observation on the $\mathit{j}$th predictor variable, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{mx}}$.
4: $\mathbf{ldx}$ – IntegerInput
On entry: the first dimension of the array x as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldx}}\ge {\mathbf{n}}$.
5: $\mathbf{isx}\left({\mathbf{mx}}\right)$ – Integer arrayInput
On entry: indicates which predictor variables are to be included in the model.
${\mathbf{isx}}\left(j\right)=1$
The $j$th predictor variable (with variates in the $j$th column of $X$) is included in the model.
${\mathbf{isx}}\left(j\right)=0$
Otherwise.
Constraint: the sum of elements in isx must equal ip.
6: $\mathbf{ip}$ – IntegerInput
On entry: $m$, the number of predictor variables in the model.
Constraint: $1<{\mathbf{ip}}\le {\mathbf{mx}}$.
7: $\mathbf{my}$ – IntegerInput
On entry: $r$, the number of response variables.
Constraint: ${\mathbf{my}}\ge 1$.
8: $\mathbf{y}\left({\mathbf{ldy}},{\mathbf{my}}\right)$ – Real (Kind=nag_wp) arrayInput
On entry: ${\mathbf{y}}\left(\mathit{i},\mathit{j}\right)$ must contain the $\mathit{i}$th observation for the $\mathit{j}$th response variable, for $\mathit{i}=1,2,\dots ,{\mathbf{n}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{my}}$.
9: $\mathbf{ldy}$ – IntegerInput
On entry: the first dimension of the array y as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldy}}\ge {\mathbf{n}}$.
10: $\mathbf{xbar}\left({\mathbf{ip}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: mean values of predictor variables in the model.
11: $\mathbf{ybar}\left({\mathbf{my}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the mean value of each response variable.
12: $\mathbf{iscale}$ – IntegerInput
On entry: indicates how predictor variables are scaled.
${\mathbf{iscale}}=1$
Data are scaled by the standard deviation of variables.
${\mathbf{iscale}}=2$
Data are scaled by user-supplied scalings.
${\mathbf{iscale}}=-1$
No scaling.
Constraint: ${\mathbf{iscale}}=-1$, $1$ or $2$.
13: $\mathbf{xstd}\left({\mathbf{ip}}\right)$ – Real (Kind=nag_wp) arrayInput/Output
On entry: if ${\mathbf{iscale}}=2$, ${\mathbf{xstd}}\left(\mathit{j}\right)$ must contain the user-supplied scaling for the $\mathit{j}$th predictor variable in the model, for $\mathit{j}=1,2,\dots ,{\mathbf{ip}}$. Otherwise xstd need not be set.
On exit: if ${\mathbf{iscale}}=1$, standard deviations of predictor variables in the model. Otherwise xstd is not changed.
14: $\mathbf{ystd}\left({\mathbf{my}}\right)$ – Real (Kind=nag_wp) arrayInput/Output
On entry: if ${\mathbf{iscale}}=2$, ${\mathbf{ystd}}\left(\mathit{j}\right)$ must contain the user-supplied scaling for the $\mathit{j}$th response variable in the model, for $\mathit{j}=1,2,\dots ,{\mathbf{my}}$. Otherwise ystd need not be set.
On exit: if ${\mathbf{iscale}}=1$, the standard deviation of each response variable. Otherwise ystd is not changed.
15: $\mathbf{maxfac}$ – IntegerInput
On entry: $k$, the number of latent variables to calculate.
Constraint: $1\le {\mathbf{maxfac}}\le {\mathbf{ip}}$.
16: $\mathbf{xres}\left({\mathbf{ldxres}},{\mathbf{ip}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the predictor variables' residual matrix ${X}_{k}$.
17: $\mathbf{ldxres}$ – IntegerInput
On entry: the first dimension of the array xres as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldxres}}\ge {\mathbf{n}}$.
18: $\mathbf{yres}\left({\mathbf{ldyres}},{\mathbf{my}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the residuals for each response variable, ${Y}_{k}$.
19: $\mathbf{ldyres}$ – IntegerInput
On entry: the first dimension of the array yres as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldyres}}\ge {\mathbf{n}}$.
20: $\mathbf{w}\left({\mathbf{ldw}},{\mathbf{maxfac}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the $\mathit{j}$th column of $W$ contains the $x$-weights ${w}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{maxfac}}$.
21: $\mathbf{ldw}$ – IntegerInput
On entry: the first dimension of the array w as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldw}}\ge {\mathbf{ip}}$.
22: $\mathbf{p}\left({\mathbf{ldp}},{\mathbf{maxfac}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the $\mathit{j}$th column of $P$ contains the $x$-loadings ${p}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{maxfac}}$.
23: $\mathbf{ldp}$ – IntegerInput
On entry: the first dimension of the array p as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldp}}\ge {\mathbf{ip}}$.
24: $\mathbf{t}\left({\mathbf{ldt}},{\mathbf{maxfac}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the $\mathit{j}$th column of $T$ contains the $x$-scores ${t}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{maxfac}}$.
25: $\mathbf{ldt}$ – IntegerInput
On entry: the first dimension of the array t as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldt}}\ge {\mathbf{n}}$.
26: $\mathbf{c}\left({\mathbf{ldc}},{\mathbf{maxfac}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the $\mathit{j}$th column of $C$ contains the $y$-loadings ${c}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{maxfac}}$.
27: $\mathbf{ldc}$ – IntegerInput
On entry: the first dimension of the array c as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldc}}\ge {\mathbf{my}}$.
28: $\mathbf{u}\left({\mathbf{ldu}},{\mathbf{maxfac}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: the $\mathit{j}$th column of $U$ contains the $y$-scores ${u}_{\mathit{j}}$, for $\mathit{j}=1,2,\dots ,{\mathbf{maxfac}}$.
29: $\mathbf{ldu}$ – IntegerInput
On entry: the first dimension of the array u as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldu}}\ge {\mathbf{n}}$.
30: $\mathbf{xcv}\left({\mathbf{maxfac}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: ${\mathbf{xcv}}\left(\mathit{j}\right)$ contains the cumulative percentage of variance in the predictor variables explained by the first $\mathit{j}$ factors, for $\mathit{j}=1,2,\dots ,{\mathbf{maxfac}}$.
31: $\mathbf{ycv}\left({\mathbf{ldycv}},{\mathbf{my}}\right)$ – Real (Kind=nag_wp) arrayOutput
On exit: ${\mathbf{ycv}}\left(\mathit{i},\mathit{j}\right)$ is the cumulative percentage of variance of the $\mathit{j}$th response variable explained by the first $\mathit{i}$ factors, for $\mathit{i}=1,2,\dots ,{\mathbf{maxfac}}$ and $\mathit{j}=1,2,\dots ,{\mathbf{my}}$.
32: $\mathbf{ldycv}$ – IntegerInput
On entry: the first dimension of the array ycv as declared in the (sub)program from which g02laf is called.
Constraint: ${\mathbf{ldycv}}\ge {\mathbf{maxfac}}$.
33: $\mathbf{ifail}$ – IntegerInput/Output
On entry: ifail must be set to $0$, . If you are unfamiliar with this argument you should refer to Section 3.4 in How to Use the NAG Library and its Documentation for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this argument, the recommended value is $0$. When the value is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{iscale}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{iscale}}=-1$ or $1$.
On entry, ${\mathbf{isx}}\left(〈\mathit{\text{value}}〉\right)$ is invalid.
Constraint: ${\mathbf{isx}}\left(j\right)=0$ or $1$, for all $j$.
On entry, ${\mathbf{mx}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{mx}}>1$.
On entry, ${\mathbf{my}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{my}}\ge 1$.
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}>1$.
${\mathbf{ifail}}=2$
On entry, ${\mathbf{ip}}=〈\mathit{\text{value}}〉$ and ${\mathbf{mx}}=〈\mathit{\text{value}}〉$.
Constraint: $1<{\mathbf{ip}}\le {\mathbf{mx}}$.
On entry, ${\mathbf{ldc}}=〈\mathit{\text{value}}〉$ and ${\mathbf{my}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldc}}\ge {\mathbf{my}}$.
On entry, ${\mathbf{ldp}}=〈\mathit{\text{value}}〉$ and ${\mathbf{ip}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldp}}\ge {\mathbf{ip}}$.
On entry, ${\mathbf{ldt}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldt}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{ldu}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldu}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{ldw}}=〈\mathit{\text{value}}〉$ and ${\mathbf{ip}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldw}}\ge {\mathbf{ip}}$.
On entry, ${\mathbf{ldx}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldx}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{ldxres}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldxres}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{ldy}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldy}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{ldycv}}=〈\mathit{\text{value}}〉$ and ${\mathbf{maxfac}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldycv}}\ge {\mathbf{maxfac}}$.
On entry, ${\mathbf{ldyres}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ldyres}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{maxfac}}=〈\mathit{\text{value}}〉$ and ${\mathbf{ip}}=〈\mathit{\text{value}}〉$.
Constraint: $1\le {\mathbf{maxfac}}\le {\mathbf{ip}}$.
${\mathbf{ifail}}=3$
On entry, ${\mathbf{ip}}=〈\mathit{\text{value}}〉$ and $\mathrm{sum}\left({\mathbf{isx}}\right)=〈\mathit{\text{value}}〉$.
Constraint: the sum of elements in isx must equal ip.
${\mathbf{ifail}}=-99$
See Section 3.9 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 3.8 in How to Use the NAG Library and its Documentation for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 3.7 in How to Use the NAG Library and its Documentation for further information.
## 7Accuracy
The computed singular value decomposition is nearly the exact singular value decomposition for a nearby matrix $\left(A+E\right)$, where
$E2 = Oε A2 ,$
and $\epsilon$ is the machine precision.
## 8Parallelism and Performance
g02laf makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
g02laf allocates internally $2mr+A+\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(3\left(A+B\right),5A\right)+r$ elements of real storage, where $A=\mathrm{min}\phantom{\rule{0.125em}{0ex}}\left(m,r\right)$ and $B=\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(m,r\right)$.
## 10Example
This example reads in data from an experiment to measure the biological activity in a chemical compound, and a PLS model is estimated.
### 10.1Program Text
Program Text (g02lafe.f90)
### 10.2Program Data
Program Data (g02lafe.d)
### 10.3Program Results
Program Results (g02lafe.r) | 2021-09-27 10:39:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 224, "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": 0.9983451962471008, "perplexity": 11384.625541049003}, "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/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00386.warc.gz"} |
http://math.stackexchange.com/questions/364768/incomplete-elliptic-integral-of-the-second-kind-and-the-arc-length-of-an-ellipse | # Incomplete elliptic integral of the second kind and the arc length of an ellipse - does a simple relation exist?
Short introduction
For a calculation I am working on I need to determine the arc length $l$ of a part of an ellipse in terms of the major axis $2a$, the minor axis $2b$ and the angle $\phi$.
I thought until now that this was a classical problem which results in an incomplete elliptic integral of the second kind: $$\tag{1} l=a E\left(\phi \left|\sqrt{1-\frac{b^2}{a^2}}\right.\right)$$
But I recently found out that this is incorrect. In fact this is only true when $\phi=0$ or $\phi=\pi/2$. So I set out to solve the problem and find out what the correct function is
My calculations
If I have an ellipse given by: $$\tag{2} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ I can parameterize the system with $x=r\cos\phi$ and $y=r\sin\phi$ which results in: $$\tag{3} r=\frac{a b}{\sqrt{a^2\sin^2\phi+b^2\cos^2\phi}}$$
I can find a differential part of the arc length $dl$ as: $dl=\sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2}$
and by solving the integral I find something which involves both an incomplete elliptic integral of the first and second kind. To give an example: $$\tag{4} \int_0^{\pi/3} \sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2} d\phi= -i b F\left(i \sinh^{-1} \left(\frac{a^2/b^2}{\sqrt{3}a/b}\right) \right) +i b E\left(i \sinh ^{-1}\left(\frac{a^2/b^2}{\sqrt{3}a/b}\right)\right)+\frac{\sqrt{3} a \sqrt{\frac{3 a^4+b^4}{3 a^2+b^2}}}{b}$$
and I can't even find an answer for the general form
$$\tag{5} \int_0^{P} \sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2} d\phi$$
Question
My main question is: am I on the right track with my calculation or am I messing up somewhere and is something of the form of Eq. 1 possible to calculate the arclength to an arbitrary angle?
-
I notice two mistakes in your derivation thus far. It seems that your formula for $r(\phi)$ does not describe an ellipse. The correct expression in polar coordinates is
$$r = \frac{ab}{\sqrt{a^2\sin^2(\phi)+b^2 \cos^2(\phi)}}$$
Secondly, once you are in polar coordinates, the arc-length formula involves the square of $\frac{dr}{d\phi}$, not the second derivative $\frac{d^2 r}{d\phi^2}$:
$$dl = \sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2}$$
Edit
It looks like you can put it in terms of a single ellptic integral, I think. Check out here: http://integraltec.com/math/math.php?f=ellipse.html, they have the full derivation from Cartesian, and the integral expression they get is the incomplete elliptic integral of the second kind. The difference is that their angle, $\theta$, is not a physical angle on the ellipse, but defined as $\sin(\theta)=x/a$. So you should be able to express your angle $\phi$ in terms of their angle $\theta$ by algebraic/trigonometric relationships and identities, then map $\phi$ to $\theta$ and substitute into the elliptic integral. This isn't a full solution, but gives you the flavor of how you can get an expression in terms of the ellipse parameters, elliptic integrals, and the radial angle to give you the arc length.
-
Thanks! I corrected the equations in my post. Unfortunately, I was more sloppy in typing up the post then I was in my derivation, in which I did have the correct expressions. So my question still holds – Michiel Apr 18 '13 at 5:05
It looks like you can put it in terms of a single ellptic integral, I think. Check out here: integraltec.com/math/math.php?f=ellipse.html, they have the full derivation from Cartesian, and the integral expression they get is the incomplete elliptic integral of the second kind. The difference is that their angle, $\theta$, is not a physical angle on the ellipse, but defined as $\sin(\theta)=x/a$. So you should be able to express your angle $\phi$ in terms of their angle $\theta$ by algebraic/trigonometric relationships and identities, then map $\phi$ to $\theta$ and substitute. – rajb245 Apr 18 '13 at 16:14
Awesome, thanks!! Could you add this into your answer so I can except it?! (comments tend to be less permanent) – Michiel Apr 18 '13 at 17:56
It's not a full "answer" to your question, but I'll edit it to include the above info. – rajb245 Apr 19 '13 at 18:00 | 2015-12-02 03:30:05 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9101077318191528, "perplexity": 160.0115622818624}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398538144.64/warc/CC-MAIN-20151124205538-00114-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://edoc.unibas.ch/8447/ | # Awareness of the influence as a determinant of assimilation versus contrast
Strack, F. and Schwarz, N. and Bless, H. and Kubler, A. and WÄNKE, M.. (1993) Awareness of the influence as a determinant of assimilation versus contrast. European journal of social psychology, Vol. 23, H. 1. pp. 53-62.
Full text not available from this repository.
Official URL: http://edoc.unibas.ch/dok/A5250751
In the present study, subjects had to generate an evaluative judgment about a target person on the basis of his behaviour that had both positive and negative implications. In a previous phase of the study that was ostensibly unrelated to the judgment task, the relevant trait categories were primed. Subsequently, half of the subjects were reminded of the priming episode. Consistent with earlier research (e.g. Lombardi, Higgins and Bargh, 1987, Newman and Uleman, 1990) that used memory of the priming events as a correlational measure, a contrast effect was found under the reminding condition and assimilation resulted when subjects were not reminded of the priming episode. This pattern of results is interpreted as the consequence of corrective influences. | 2022-08-09 23:01:16 | {"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": 0.5393860340118408, "perplexity": 4574.644481224622}, "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/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00133.warc.gz"} |
https://www.physicsforums.com/threads/density-change-caused-by-temp-change.170738/ | # Density Change Caused by Temp. Change
1. May 17, 2007
### e(ho0n3
[SOLVED] Density Change Caused by Temp. Change
1. The problem statement, all variables and given/known data
Show that the change in the density $\rho$ of a substance, when the temperature changes by $\Delta T$, is given by $\Delta \rho = - \beta \rho \Delta T$.
2. Relevant equations
$$\rho = \frac{m}{V}$$
$$\Delta V = \beta V \Delta T$$
3. The attempt at a solution
Let $\rho_1$ and $\rho_2$ be the densities of the substance before and after the temp. change $\Delta T$. After some algebra, I get:
$$\Delta \rho = \rho_2 - \rho_1 = - \beta \rho_2 \Delta T$$
This differs from that given in the problem statements because $\rho$ is $\rho_1$ in the above. I guess this must be typo right?
Last edited: May 18, 2007
2. May 18, 2007
### Kurdt
Staff Emeritus
What are the explicit densities before and after the increase in temperature? That is the key to getting this and a lot of people make a mistake. The density before is simple:
$$\rho_0=\frac{m}{V_0}$$
but the final density is:
$$\rho = \frac{m}{(V_0+\Delta V)}$$
3. May 18, 2007
### e(ho0n3
Are you implying that I made a mistake? I have checked the algebra twice and I still get the result I got. If I use what you wrote for the densities, I still get the same result.
4. May 18, 2007
### Kurdt
Staff Emeritus
Ahh ok I see what you've done, my mistake. Yes I get the same answer as you so I'd wager it was a spelling error unless we both made the same mistake. | 2016-12-04 08:26:24 | {"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": 0.8111482262611389, "perplexity": 594.6009013696373}, "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/1480698541220.56/warc/CC-MAIN-20161202170901-00172-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://dynamicsystems.asmedigitalcollection.asme.org/article.aspx?articleid=2457331 | 0
Technical Brief
# Guaranteed Performance State-Feedback Gain-Scheduling Control With Uncertain Scheduling Parameters
[+] Author and Article Information
Ali Khudhair Al-Jiboory
Mechanical Engineering,
Michigan State University,
East Lansing, MI 48824
e-mail: aljiboor@egr.msu.edu
Guoming G. Zhu
Mechanical Engineering,
Michigan State University,
East Lansing, MI 48824
e-mail: zhug@egr.msu.edu
Jongeun Choi
Mechanical Engineering,
Michigan State University,
East Lansing, MI 48824
e-mail: jchoi@egr.msu.edu
1Corresponding author.
Contributed by the Dynamic Systems Division of ASME for publication in the JOURNAL OF DYNAMIC SYSTEMS, MEASUREMENT, AND CONTROL. Manuscript received December 10, 2014; final manuscript received September 20, 2015; published online October 29, 2015. Assoc. Editor: Ryozo Nagamune.
J. Dyn. Sys., Meas., Control 138(1), 014502 (Oct 29, 2015) (7 pages) Paper No: DS-14-1518; doi: 10.1115/1.4031727 History: Received December 10, 2014; Revised September 20, 2015
## Abstract
State-feedback gain-scheduling controller synthesis with guaranteed performance is considered in this brief. Practical assumption has been considered in the sense that scheduling parameters are assumed to be uncertain. The contribution of this paper is the characterization of the control synthesis that parameterized linear matrix inequalities (PLMIs) to synthesize robust gain-scheduling controllers. Additive uncertainty model has been used to model measurement noise of the scheduling parameters. The resulting controllers not only ensure robustness against scheduling parameters uncertainties but also guarantee closed-loop performance in terms of $H2$ and $H∞$ performances as well. Numerical examples and simulations are presented to illustrate the effectiveness of the synthesized controller. Compared to other control design methods from literature, the synthesized controllers achieve less conservative results as measurement noise increases.
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## References
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Sato, M. , 2010, “ Gain-Scheduled State-Feedback Controllers Using Inexactly Measured Scheduling Parameters: Stabilizing and H ∞ Control Problems,” SICE J. Control, Meas. Syst. Integr., 3(4), pp. 285–291.
Wu, F. , Yang, X. H. , Packard, A. , and Becker, G. , 1996, “ Induced L 2-Norm Control for LPV Systems With Bounded Parameter Variation Rates,” Int. J. Robust Nonlinear Control, 6(9–10), pp. 983–998.
Sato, M. , Ebihara, Y. , and Peaucelle, D. , 2010, “ Gain-Scheduled State-Feedback Controllers Using Inexactly Measured Scheduling Parameters: H 2 and H ∞ Problems,” American Control Conference, pp. 3094–3099.
Sato, M. , 2013, “ Robust Gain-Scheduled Flight Controller Using Inexact Scheduling Parameters,” American Control Conference (ACC), pp. 6829–6834.
Lacerda, M. J. , Tognetti, E. S. , Oliveira, R. C. , and Peres, P. L. , 2014, “ A New Approach to Handle Additive and Multiplicative Uncertainties in the Measurement for H ∞ LPV Filtering,” Int. J. Syst. Sci. (published online).
Agulhari, C. , Tognetti, E. , Oliveira, R. , and Peres, P. , 2013, “ H ∞ Dynamic Output Feedback for LPV Systems Subject to Inexactly Measured Scheduling Parameters,” Proceedings of American Control Conference, pp. 6060–6065.
Al-Jiboory, A. K. , and Zhu, G. G. , 2015, “ Robust Gain-Scheduling H 2 Control With Imperfectly Measured Scheduling Parameters,” (submitted).
Oliveira, R. C. L. F. , Bliman, P. , and Peres, P. L. D. , 2008, “ Robust LMIs With Parameters in Multi-Simplex: Existence of Solutions and Applications,” 47th IEEE Conference on CDC, pp. 2226–2231.
Oliveira, R. , and Peres, P. , 2007, “ Parameter-Dependent LMIs in Robust Analysis: Characterization of Homogeneous Polynomially Parameter-Dependent Solutions Via LMI Relaxations,” IEEE Trans. Autom. Control, 52(7), pp. 1334–1340.
Oliveira, R. , de Oliveira, M. , and Peres, P. , 2011, “ Robust State Feedback LMI Methods for Continuous-Time Linear Systems: Discussions, Extensions and Numerical Comparisons,” IEEE International Symposium on CACSD, pp. 1038–1043.
Oliveira, R. C. L. F. , Bliman, P.-A. , and Peres, P. L. , 2009, “ Selective Gain-Scheduling for Continuous-Time Linear Systems With Parameters in Multi-Simplex,” European Control Conference.
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de Souza, C. E. , and Trofino, A. , 2006, “ Gain-Scheduled H 2 Controller Synthesis for Linear Parameter Varying Systems Via Parameter-Dependent Lyapunov Functions,” Int. J. Robust Nonlinear Control, 16(5), pp. 243–257.
Sato, M. , 2008, “ Design Method of Gain-Scheduled Controllers Not Depending on Derivatives of Parameters,” Int. J. Control, 81(6), pp. 1013–1025.
Pipeleers, G. , Demeulenaere, B. , Swevers, J. , and Vandenberghe, L. , 2009, “ Extended LMI Characterizations for Stability and Performance of Linear Systems,” Syst. Control Lett., 58(7), pp. 510–518.
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Scherer, C. W. , and Hol, C. W. J. , 2006, “ Matrix Sum-of-Squares Relaxations for Robust Semi-Definite Programs,” Math. Program., 107, pp. 189–211.
Peaucelle, D. , and Sato, M. , 2009, “ LMI Tests for Positive Definite Polynomials: Slack Variable Approach,” IEEE Trans. Autom. Control, 54(4), pp. 886–891.
Montagner, V. F. , Oliveira, R. C. , Peres, P. L. , and Bliman, P.-A. , 2009, “ Stability Analysis and Gain-Scheduled State Feedback Control for Continuous-Time Systems With Bounded Parameter Variations,” Int. J. Control, 82(6), pp. 1045–1059.
Agulhari, C. M. , de Oliveira, R. C. L. F. , and Peres, P. L. D. , 2012, “ Robust LMI Parser: A Computational Package to Construct LMI Conditions for Uncertain Systems,” XIX Brazilian Conference on Automation (CBA 2012), pp. 2298–2305.
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## Figures
Fig. 1
Performance versus ϵ with ζ = 0.2
Fig. 2
Guaranteed performance
Fig. 3
Simulation: (a) measured and exact scheduling parameters and (b) disturbance attenuation responses associated with exact and noisy scheduling parameter
## Discussions
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Topic Collections | 2018-06-21 18:12:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.35107147693634033, "perplexity": 11186.166351096022}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864256.26/warc/CC-MAIN-20180621172638-20180621192638-00041.warc.gz"} |
https://onlinedocs.microchip.com/pr/GUID-C541EA24-5EC3-41E5-9648-79068F9853C0-en-US-3/GUID-045C7032-4DF6-434A-83E5-079E13E721A0.html | # Fuse Write Command
The fuse write command writes the fuses. It can only be used by the UPDI, the CPU cannot start this command.
Follow this procedure to use this command:
• Execute the fuse write command. | 2020-12-05 18:46:28 | {"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": 0.6203892230987549, "perplexity": 8550.990563702997}, "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-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00397.warc.gz"} |
https://www.aimsciences.org/article/doi/10.3934/dcds.1999.5.905 | # American Institute of Mathematical Sciences
October 1999, 5(4): 905-928. doi: 10.3934/dcds.1999.5.905
## Unfocused blow up solutions of semilinear parabolic equations
1 Laboratoire d'Analyse Numérique, Université Pierre et Marie curie, 4, Place Jussieu, 7525 Paris Cedex 05, France
Received July 1998 Revised June 1999 Published July 1999
The aim of this paper is to study the blow up behavior of a radially symmetric solution $u$ of the semilinear parabolic equation
$u_t - \Delta u = |u|^{p-1} u, \quad x\in\Omega,\quad t\in [0,T]$,
$u(t,x)=0, x\in\partial\Omega, \quad t\in [0,T]$,
$u(0,x) =u_0(x),\quad x\in\Omega$,
around a blow up point other than its centre of symmetry. We assume that $\Omega$ is a ball in $\mathbb R^N$ or $\Omega =\mathbb R^N$, and $p>1$. We show that $u$ behave as of a one-dimensional problem was concerned, that is, the possible asymptotic behaviors and final time profiles around an unfocused blow up point are the ones corresponding to the case of dimesion $N=1$.
Citation: Júlia Matos. Unfocused blow up solutions of semilinear parabolic equations. Discrete & Continuous Dynamical Systems - A, 1999, 5 (4) : 905-928. doi: 10.3934/dcds.1999.5.905
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2019 Impact Factor: 1.338 | 2020-12-05 12:11:41 | {"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": 0.4373898208141327, "perplexity": 7719.965786000516}, "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-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00288.warc.gz"} |
https://www.physicsforums.com/threads/average-vs-final-speed.42870/ | # Homework Help: Average vs final speed
1. Sep 12, 2004
### Omid
A rocket can accelerate at an increasing rate as its fuel is consumed and it gets lighter and lighter. If the rocket starts from rest, what can be said about the average speed as compared with the final speed ?
Thanks
2. Sep 12, 2004
### Tide
There's no simple relationship between the final speed and the average speed since the mass of the "rocket" is changing (acceleration is NOT a constant!) until its fuel is completely expelled. You need to actually solve the equation of motion for the rocket and you'll basically find something like
$$v = - v_e \ln \frac {M_{final}}{M_{start}}$$
where $v_e$ is the velocity of the exhaust relative to the rocket. (I'm relying on memory here.)
3. Sep 13, 2004
### amwbonfire
I'll assume this is an explanation question, not a calculation, because you haven't supplied any data.
The average speed is constantly changing, (increasing), because the rocket's acceleration is increasing constantly. The final speed will be very high indeed.
So how can you compare them?
a) Will the final velocity be high or low?
b) What will the average speed of the rocket be?
If you think about it, the final velocity has to be larger than the average speed, because the average takes into account every speed that has occured, while the final speed is going to be the highest speed it reaches. So, comparing them, all you can really say is the final speed will be higher than the average speed.
Andy
AMW Bonfire
4. Sep 13, 2004
### Omid
Thank you very much Andy | 2018-10-15 17:19:59 | {"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": 0.7699873447418213, "perplexity": 646.7867520754908}, "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-2018-43/segments/1539583509336.11/warc/CC-MAIN-20181015163653-20181015185153-00330.warc.gz"} |
https://www.preprints.org/manuscript/201702.0032/v1 | Preprint Article Version 1 Preserved in Portico This version is not peer-reviewed
# Estimating Recreational Value of Foy's Lake: An Application of Travel Cost Count Data Model for Truncated Zeros
Version 1 : Received: 9 February 2017 / Approved: 10 February 2017 / Online: 10 February 2017 (11:10:04 CET)
How to cite: Alam, M.T.; Chowdury, A.; Hossian, M.S. Estimating Recreational Value of Foy's Lake: An Application of Travel Cost Count Data Model for Truncated Zeros. Preprints 2017, 2017020032 (doi: 10.20944/preprints201702.0032.v1). Alam, M.T.; Chowdury, A.; Hossian, M.S. Estimating Recreational Value of Foy's Lake: An Application of Travel Cost Count Data Model for Truncated Zeros. Preprints 2017, 2017020032 (doi: 10.20944/preprints201702.0032.v1).
## Abstract
To estimate the recreational value provided by the Foy’s Lake annually using the most applicable model for on-site data is the main objective of this study. Adhere to the objective of this study; Individual Travel Cost Method (ITCM) has been applied and Zero Truncated Poisson Regression Model has been found plausible among other models to estimate consumer surplus. Based on the findings of the study, an estimate of the consumer surplus or recreational benefits per trip per visitor can be recommended as BDT 5,875 or US $73.44 and counting the consumer surplus per trip per visitor, the annual recreational value (total consumer surplus) provided by the lake is found to be BDT 321 million or US$ 40.2 million.
## Subject Areas
individual travel cost method; zero truncated poisson regression model; endogenous stratification; consumer surplus
Views 0 | 2020-11-24 23:27:20 | {"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": 0.2232070118188858, "perplexity": 12042.11544132779}, "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-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00117.warc.gz"} |
https://forum.allaboutcircuits.com/threads/selecting-wire-gauge-for-pulses-current.189436/page-3 | # selecting wire gauge for pulses current
#### Momentory
Joined Sep 4, 2022
67
How much copper mass do you have in the coil?
How much energy is dumped into the coil with each pulse?
What is the thermal capacity of copper?
How much will each pulse raise the temperature of that much copper, assuming none of it goes anywhere else?
If that temperature rise is below the limit of the wire, then you don't need to worry about the individual pulses, but rather on their cumulative effect. In that case, looking at the RMS power dissipated in the coil is what you want to focus on.
How much copper mass do you have in the coil?
How much energy is dumped into the coil with each pulse?
What is the thermal capacity of copper?
How much will each pulse raise the temperature of that much copper, assuming none of it goes anywhere else?
If that temperature rise is below the limit of the wire, then you don't need to worry about the individual pulses, but rather on their cumulative effect. In that case, looking at the RMS power dissipated in the coil is what you want to focus on.
My worries about cumulative pulsed temperature and since the wire has no obvious datasheet so i decide just want to add cooling system even it will not get much high temperature (as safety)
#### WBahn
Joined Mar 31, 2012
27,935
My worries about cumulative pulsed temperature and since the wire has no obvious datasheet so i decide just want to add cooling system even it will not get much high temperature (as safety)
But the wire is copper, right?
You know the size of the wire, so you can look up the diameter of it, right?
You can determine how much volume of copper there is in your coil, right?
You can look up the mass density of copper and the thermal capacity of copper, right?
You are certainly free to put a cooling system on it if you want without determining if you need one, but how will you know that your cooling system is adequate without determining how much cooling you need?
#### MrAl
Joined Jun 17, 2014
9,638
@Momentory
Coil in circuit #31 dissipates 30 J energy during one pulse.
Such energy can increase temperature of 1.172 kg of copper only on 0.067 °C, not melt it.
If you want to know what is fusing 2 ms rectangular pulse current for AWG 18 wire,
use this formula:
Ifuse = Area * SQRT(LOG((Tmelt - Tambient) / (234 - Tambient) + 1) / Time * 33)),
where
Tmelt -melting temp of wire in °C [1083],
Tambient -ambient temp in °C [25],
Time -melting time in seconds [0.002],
Area -wire area in circular mils [1620],
Ifuse -fusing current in amps,
then
Ifuse = 1620 * SQRT(LOG((1083 - 25) / (234 - 25)+1) / 0.002 * 33)),
Ifuse = 1898 A.
Hello,
I didnt check that formula but it's nice of you to post this info.
Now if you care too, set Ifuse=100 amps and solve for the time, which would be the time when the wire melts and breaks apart or nearly so. That should be interesting. That would be the absolute maximum time for any pulse starting with a wire that is always at ambient temperature just before the single pulse is applied.
This is still for a single pulse assuming that the off time will allow the wire to cool back down to 25C or nearly so, but still interesting.
It's also interesting that as the pulse is applied and time progresses, the wire will heat up before melting and that will cause the resistance to go up, and with a constant current of 100 amps that means the power in the wire will go up, and that means even more wire heating power, which of course causes a very fast wire heat up.
In real life though this would probably be a voltage drive so the power would actually come down as the resistance increases. Would be interesting to plot too.
#### MrAl
Joined Jun 17, 2014
9,638
My worries about cumulative pulsed temperature and since the wire has no obvious datasheet so i decide just want to add cooling system even it will not get much high temperature (as safety)
Yes the accumulated heat buildup is hard to predict without a test. This is actually true of a lot of devices we dont always think about.
In the industry transformers and inductors are often vacuum varnished where the vacuum chamber pulls out all the air spaces and thus increases the thermal conductance of the coil as a whole. This of course keeps the internal coil wires cooler although they will always be hotter than the outer layers.
Also in the industry of power supplies, an over night test is run to make sure the construction does not melt down. The various insulation layers can melt and cause a lot of problems not only with the material getting all over the place as it drips out like candle wax
So once you decide on some construction and you think you can use it, you have to test it at full power for several hours, probably 12 to 24 hours. Full power in this case though would be with the required pulse time and required off time, not full current for the entire time obviously.
I remember one converter back in the 1980's that melted down overnight. It was a unit of about 5000 watts, synthesized sine output (true sine output). Not only did the output transformer melt down and leak material all over the place, the 4 bridge transistors blew out and at the time they were around $100 each. The whole thing had to be repaired and a new transformer design used to fix the issue. It's not a pretty sight for sure. The rule of thumb is sometimes you just cant be sure until you test it. This is also the case with wall warts, if a new design. They have to be run over night to be sure they can survive. There used to be a place in Northern New Jersey that did this routinely not sure if they are around anymore though. They shipped thousands of wall warts of various kinds probably within a week or less but i dont remember the time frame now. #### Danko Joined Nov 22, 2017 1,493 I didnt check that formula I think this source is reliable enough (see attachment). @Momentory: Calculated parameters of your coil: L=3.5 mH Inner diameter of coil=15 mm Outer diameter of coil=45 mm Length of winding=60 mm Number of turns=673 Number of winding layers=13 Magnet wire: copper, AWG 18 Length of wire=62 m (not 160 m) Resistance of coil=1.3 Ω (not 3.2 Ω) Weight of wire=455.43 g So, your coil will dissipate 6.1 W power in circuit below. Contactors coils dissipate similar power, therefore, especially if take in account that work session will 10 min only, you can try to test it, carefully. Circuit (NOTE: capacitor C1 must be film, not electrolytic!): 1000V 100UF DC-link Film Capacitor US$25.95
KP-200A AC 1000V 200A Stud Thyristor Silicon Control Rectifier SCR
US \$15.00
#### Attachments
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#### Momentory
Joined Sep 4, 2022
67
I think this source is reliable enough (see attachment).
@Momentory:
L=3.5 mH
Inner diameter of coil=15 mm
Outer diameter of coil=45 mm
Length of winding=60 mm
Number of turns=673
Number of winding layers=13
Magnet wire: copper, AWG 18
Length of wire=62 m (not 160 m)
Resistance of coil=1.3 Ω (not 3.2 Ω)
Weight of wire=455.43 g
So, your coil will dissipate 6.1 W energy in circuit below.
Contactors coils dissipate similar energy, therefore,
especially take in account that work session will 10 min only,
you can try to test it, carefully.
Circuit (NOTE: capacitor C1 must be film, not electrolytic!):
View attachment 277756
Yes i think the only way to determine the temp is test the coil and measure its temp
For capacitor, i will use diode across the coil i still can use electrolytic capacitor
Last edited:
#### Danko
Joined Nov 22, 2017
1,493
For capacitor, i will use diode across the coil i still can use electrolytic capacitor
See how it will "work" with diode across the coil:
#### MrAl
Joined Jun 17, 2014
9,638
What us U1 ?
Transistor, SCR, etc ?
#### ericgibbs
Joined Jan 29, 2010
16,830
Morning Al.
The SCR type on the circuit is a T253-1250.
E
#### Attachments
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http://www.ck12.org/statistics/Sampling-and-Bias/lesson/Sampling-and-Bias-ADV-PST/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Sampling and Bias
## Sampling issues and bias
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Sampling and Bias
### Sampling and Bias
#### Census vs.Sample
A sample is a representative subset of a population. If a statistician or other researcher wants to know some information about a population, the only way to be truly sure is to conduct a census. In a census, every unit in the population being studied is measured or surveyed. In opinion polls, like the New York Times poll mentioned above, results are generalized from a sample. If we really wanted to know the true approval rating of the president, for example, we would have to ask every single American adult his or her opinion. There are some obvious reasons why a census is impractical in this case, and in most situations.
First, it would be extremely expensive for the polling organization. They would need an extremely large workforce to try and collect the opinions of every American adult. Also, it would take many workers and many hours to organize, interpret, and display this information. Even if it could be done in several months, by the time the results were published, it would be very probable that recent events had changed peoples’ opinions and that the results would be obsolete.
In addition, a census has the potential to be destructive to the population being studied.
Many manufacturing companies test their products for quality control. A padlock manufacturer might use a machine to see how much force it can apply to the lock before it breaks. If they did this with every lock, they would have none left to sell! Likewise, it would not be a good idea for a biologist to find the number of fish in a lake by draining the lake and counting them all!
The U.S. Census is probably the largest and longest running census, since the Constitution mandates a complete counting of the population. The first U.S. Census was taken in 1790 and was done by U.S. Marshalls on horseback. Taken every 10 years, a Census was conducted in 2010, and in a report by the Government Accountability Office in 1994, was estimated to cost 11 billion. This cost has recently increased as computer problems have forced the forms to be completed by hand. You can find a great deal of information about the U.S. Census, as well as data from past Censuses, on the Census Bureau’s website. Due to all of the difficulties associated with a census, sampling is much more practical. However, it is important to understand that even the most carefully planned sample will be subject to random variation between the sample and the population. Recall that these differences due to chance are called sampling error. We can use the laws of probability to predict the level of accuracy in our sample. Opinion polls, like the New York Times poll mentioned in the introduction, tend to refer to this as margin of error. The second statement quoted from the New York Times article mentions another problem with sampling. That is, it is often difficult to obtain a sample that accurately reflects the total population. It is also possible to make mistakes in selecting the sample and collecting the information. These problems result in a non-representative sample, or one in which our conclusions differ from what they would have been if we had been able to conduct a census. To help understand these ideas, consider the following theoretical example. A coin is considered fair if the probability, \begin{align*}p\end{align*}, of the coin landing on heads is the same as the probability of it landing on tails \begin{align*}(p = 0.5)\end{align*}. The probability is defined as the proportion of heads obtained if the coin were flipped an infinite number of times. Since it is impractical, if not impossible, to flip a coin an infinite number of times, we might try looking at 10 samples, with each sample consisting of 10 flips of the coin. Theoretically, you would expect the coin to land on heads 50% of the time, but it is very possible that, due to chance alone, we would experience results that differ from this. These differences are due to sampling error. As we will investigate in detail in later chapters, we can decrease the sampling error by increasing the sample size (or the number of coin flips in this case). It is also possible that the results we obtain could differ from those expected if we were not careful about the way we flipped the coin or allowed it to land on different surfaces. This would be an example of a non-representative sample. #### Bias in Samples and Surveys The term most frequently applied to a non-representative sample is bias. Bias has many potential sources. It is important when selecting a sample or designing a survey that a statistician make every effort to eliminate potential sources of bias. In this section, we will discuss some of the most common types of bias. While these concepts are universal, the terms used to define them here may be different than those used in other sources. #### Sampling Bias In general, sampling bias refers to the methods used in selecting the sample. The sampling frame is the term we use to refer to the group or listing from which the sample is to be chosen. If you wanted to study the population of students in your school, you could obtain a list of all the students from the office and choose students from the list. This list would be the sampling frame. #### Incorrect Sampling Frame If the list from which you choose your sample does not accurately reflect the characteristics of the population, this is called incorrect sampling frame. A sampling frame error occurs when some group from the population does not have the opportunity to be represented in the sample. #### Recognizing an Incorrect Sampling Frame Surveys are often done over the telephone. You could use the telephone book as a sampling frame by choosing numbers from the telephone book. However, in addition to the many other potential problems with telephone polls, some phone numbers are not listed in the telephone book. Also, if your population includes all adults, it is possible that you are leaving out important groups of that population. For example, many younger adults in particular tend to only use their cell phones or computer-based phone services and may not even have traditional phone service. Even if you picked phone numbers randomly, the sampling frame could be incorrect, because there are also people, especially those who may be economically disadvantaged, who have no phone. There is absolutely no chance for these individuals to be represented in your sample. A term often used to describe the problems when a group of the population is not represented in a survey is undercoverage. Undercoverage can result from all of the different sampling biases. One of the most famous examples of sampling frame error occurred during the 1936 U.S. presidential election. The Literary Digest, a popular magazine at the time, conducted a poll and predicted that Alf Landon would win the election that, as it turned out, was won in a landslide by Franklin Delano Roosevelt. The magazine obtained a huge sample of ten million people, and from that pool, 2 million replied. With these numbers, you would typically expect very accurate results. However, the magazine used their subscription list as their sampling frame. During the depression, these individuals would have been only the wealthiest Americans, who tended to vote Republican, and left the majority of typical voters under-covered. #### Convenience Sampling Suppose your statistics teacher gave you an assignment to perform a survey of 20 individuals. You would most likely tend to ask your friends and family to participate, because it would be easy and quick. This is an example of convenience sampling, or convenience bias. While it is not always true, your friends are usually people who share common values, interests, and opinions. This could cause those opinions to be over-represented in relation to the true population. Also, have you ever been approached by someone conducting a survey on the street or in a mall? If such a person were just to ask the first 20 people they found, there is the potential that large groups representing various opinions would not be included, resulting in undercoverage. #### Judgment Sampling Judgment sampling occurs when an individual or organization that is usually considered an expert in the field being studied chooses the individuals or group of individuals to be used in the sample. Because it is based on a subjective choice, even by someone considered an expert, it is very susceptible to bias. In some sense, this is what those responsible for the Literary Digest poll did. They incorrectly chose groups they believed would represent the population. If a person wants to do a survey on middle-class Americans, how would this person decide who to include? It would be left to this person's own judgment to create the criteria for those considered middle-class. This individual’s judgment might result in a different view of the middle class that might include wealthier individuals that others would not consider part of the population. Similar to judgment sampling, in quota sampling, an individual or organization attempts to include the proper proportions of individuals of different subgroups in their sample. While it might sound like a good idea, it is subject to an individual’s prejudice and is, therefore, prone to bias. #### Size Bias If one particular subgroup in a population is likely to be over-represented or under-represented due to its size, this is sometimes called size bias. If we chose a state at random from a map by closing our eyes and pointing to a particular place, larger states would have a greater chance of being chosen than smaller ones. As another example, suppose that we wanted to do a survey to find out the typical size of a student’s math class at a school. The chances are greater that we would choose someone from a larger class for our survey. To understand this, say that you went to a very small school where there are only four math classes, with one class having 35 students, and the other three classes having only 8 students. If you simply choose students at random, it is more likely you will select students for your sample who will say the typical size of a math class is 35, since there are more students in the larger class. #### Determining a Sample Error A person driving on an interstate highway tends to say things like, “Wow, I was going the speed limit, and everyone was just flying by me.” The conclusion this person is making about the population of all drivers on this highway is that most of them are traveling faster than the speed limit. This may indeed be true, but let’s say that most people on the highway, along with our driver, really are abiding by the speed limit. In a sense, the driver is collecting a sample, and only those few who are close to our driver will be included in the sample. There will be a larger number of drivers going faster in our sample, so they will be over-represented. As you may already see, these definitions are not absolute, and often in a practical example, there are many types of overlapping bias that could be present and contribute to overcoverage or undercoverage. We could also cite incorrect sampling frame or convenience bias as potential problems in this example. #### Response Bias The term response bias refers to problems that result from the ways in which the survey or poll is actually presented to the individuals in the sample. #### Voluntary Response Bias Television and radio stations often ask viewers/listeners to call in with opinions about a particular issue they are covering. The websites for these and other organizations also usually include some sort of online poll question of the day. Reality television shows and fan balloting in professional sports to choose all-star players make use of these types of polls as well. All of these polls usually come with a disclaimer stating that, “This is not a scientific poll.” While perhaps entertaining, these types of polls are very susceptible to voluntary response bias. The people who respond to these types of surveys tend to feel very strongly one way or another about the issue in question, and the results might not reflect the overall population. Those who still have an opinion, but may not feel quite so passionately about the issue, may not be motivated to respond to the poll. This is especially true for phone-in or mail-in surveys in which there is a cost to participate. The effort or cost required tends to weed out much of the population in favor of those who hold extremely polarized views. A news channel might show a report about a child killed in a drive-by shooting and then ask for people to call in and answer a question about tougher criminal sentencing laws. They would most likely receive responses from people who were very moved by the emotional nature of the story and wanted anything to be done to improve the situation. An even bigger problem is present in those types of polls in which there is no control over how many times an individual may respond. #### Non-Response Bias One of the biggest problems in polling is that most people just don’t want to be bothered taking the time to respond to a poll of any kind. They hang up on a telephone survey, put a mail-in survey in the recycling bin, or walk quickly past an interviewer on the street. We just don’t know how much these individuals' beliefs and opinions reflect those of the general population, and, therefore, almost all surveys could be prone to non-response bias. #### Questionnaire Bias Questionnaire bias occurs when the way in which the question is asked influences the response given by the individual. It is possible to ask the same question in two different ways that would lead individuals with the same basic opinions to respond differently. Consider the following two questions about gun control. "Do you believe that it is reasonable for the government to impose some limits on purchases of certain types of weapons in an effort to reduce gun violence in urban areas?" "Do you believe that it is reasonable for the government to infringe on an individual’s constitutional right to bear arms?" A gun rights activist might feel very strongly that the government should never be in the position of limiting guns in any way and would answer no to both questions. Someone who is very strongly against gun ownership, on the other hand, would probably answer yes to both questions. However, individuals with a more tempered, middle position on the issue might believe in an individual’s right to own a gun under some circumstances, while still feeling that there is a need for regulation. These individuals would most likely answer these two questions differently. You can see how easy it would be to manipulate the wording of a question to obtain a certain response to a poll question. Questionnaire bias is not necessarily always a deliberate action. If a question is poorly worded, confusing, or just plain hard to understand, it could lead to non-representative results. When you ask people to choose between two options, it is even possible that the order in which you list the choices may influence their response! #### Incorrect Response Bias A major problem with surveys is that you can never be sure that the person is actually responding truthfully. When an individual intentionally responds to a survey with an untruthful answer, this is called incorrect response bias. This can occur when asking questions about extremely sensitive or personal issues. For example, a survey conducted about illegal drinking among teens might be prone to this type of bias. Even if guaranteed their responses are confidential, some teenagers may not want to admit to engaging in such behavior at all. Others may want to appear more rebellious than they really are, but in either case, we cannot be sure of the truthfulness of the responses. #### Recognizing Bias Because the dangers of donated blood being tainted with diseases carrying a negative social stereotype increased in the 1990’s, the Red Cross has recently had to deal with incorrect response bias on a constant and especially urgent basis. Individuals who have engaged in behavior that puts them at risk for contracting AIDS or other diseases have the potential to pass these diseases on through donated blood\begin{align*}^4\end{align*}. Screening for at-risk behaviors involves asking many personal questions that some find awkward or insulting and may result in knowingly false answers. The Red Cross has gone to great lengths to devise a system with several opportunities for individuals giving blood to anonymously report the potential danger of their donation. In using this example, we don’t want to give the impression that the blood supply is unsafe. According to the Red Cross, “Like most medical procedures, blood transfusions have associated risk. In the more than fifteen years since March 1985, when the FDA first licensed a test to detect HIV antibodies in donated blood, the Centers for Disease Control and Prevention has reported only 41 cases of AIDS caused by transfusion of blood that tested negative for the AIDS virus. During this time, more than 216 million blood components were transfused in the United States. The tests to detect HIV were designed specifically to screen blood donors. These tests have been regularly upgraded since they were introduced. Although the tests to detect HIV and other blood-borne diseases are extremely accurate, they cannot detect the presence of the virus in the 'window period' of infection, the time before detectable antibodies or antigens are produced. That is why there is still a very slim chance of contracting HIV from blood that tests negative. Research continues to further reduce the very small risk.” Source: The American Red Cross ### Examples The following examples use the information below: A school has a designed a survey, which will be administered during an entire class period one day for every course in a given semester. #### Example 1 By going to every class, the school is attempting to obtain information from the entire population of students: this is a census. #### Example 2 Suppose the teachers tell the students the day before, that the survey will be administered in class the next day. Is there bias involved? If so, which type of bias is involved? If the teachers inform the students about the survey the day before, some students may decide not to come to class the next day. This may create a non-response bias. ### Review For 1-7, Brandy wanted to know which brand of soccer shoe high school soccer players prefer. She decided to ask the girls on her team which brand they liked. 1. What is the population in this example? 2. What are the units? 3. If she asked all high school soccer players this question, what is the statistical term we would use to describe the situation? 4. Which group(s) from the population is/are going to be under-represented? 5. What type of bias best describes the error in her sample? Why? 6. Brandy got a list of all the soccer players in the Colonial conference from her athletic director, Mr. Sprain. This list is called the what? 7. If she grouped the list by boys and girls, and chose 40 boys at random and 40 girls at random, what type of sampling best describes her method? 8. Your doorbell rings, and you open the door to find a 6-foot-tall boa constrictor wearing a trench coat and holding a pen and a clip board. He says to you, “I am conducting a survey for a local clothing store. Do you own any boots, purses, or other items made from snake skin?” After recovering from the initial shock of a talking snake being at the door, you quickly and nervously answer, “Of course not,” as the wallet you bought on vacation last summer at Reptile World weighs heavily in your pocket. What type of bias best describes this ridiculous situation? Explain why. In each of the next two examples, identify the type of sampling that is most evident and explain why you think it applies. 1. In order to estimate the population of moose in a wilderness area, a biologist familiar with that area selects a particular marsh area and spends the month of September, during mating season, cataloging sightings of moose. What two types of sampling are evident in this example? 2. The local sporting goods store has a promotion where every \begin{align*}1000^{\text{th}}\end{align*} customer gets a10 gift card.
References
The New York Times
The United States Government Accountability Office
The Cable News Network
Wikipedia
1. What are the effects of sample size on sampling error?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
bias
Bias refers to a desire to achieve a specific result from a particular study, regardless of the data.
census
A census is an official enumeration of the entire population, with details as to age, sex, occupation, etc.
convenience sampling
Convenience sampling refers to the process of choosing a sample based on members who are easily accessible.
incorrect response bias
When an individual intentionally responds to a survey with an untruthful answer, this is called incorrect response bias.
incorrect sampling frame
Incorrect sampling frame occurs when the group from which you choose your sample does not include everyone in the population, or at least units that reflect the full diversity of the population.
judgement sampling
Judgment Sampling is a type of sampling occurs when the investigator already has made an assumption about a characteristic of the population, and samples are selected accordingly.
margin of error
The margin of error is found by multiplying the standard error of the mean by the z-score of the percent confidence level
non-response bias
Non-response bias is commonly caused by self-selection, subjects with a reason not to respond which may be unrelated to the actual study are not included, skewing the results.
questionnaire bias
Questionnaire bias occurs when the way in which the question is asked influences the response given by the individual.
Sample
A sample is a specified part of a population, intended to represent the population as a whole.
Sampling error (random variation)
Sampling error occurs whenever a sample is used instead of the entire population, where we have to accept that our results are merely estimates, and therefore, have some chance of being incorrect.
self-selection
Self-selection is a sampling method that requires the subject to offer a response to an input.
undercoverage
Undercoverage describes a sample with too few members of a given group or demographic.
voluntary response bias
Voluntary response bias occurs when sample members are self-selected volunteers | 2016-10-21 15:03:55 | {"extraction_info": {"found_math": true, "script_math_tex": 4, "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": 0.3928317427635193, "perplexity": 802.2670879431508}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988718284.75/warc/CC-MAIN-20161020183838-00389-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://ged-testprep.com/wyoming-ged-requirements | # Eligibility Requirements For GED In Wyoming
use my location
It is critical to ensure that you have met any requirements necessary to take the GED exams in your state. Every state has the option of imposing age limits as well as other requirements. Those planning to take their GED exams in Wyoming should familiarize themselves with the state's GED policies.
## GED eligibility requirements in Wyoming
You must follow the following requirements to be eligible for the Wyoming GED:
• You do not need to reside in Wyoming.
• You cannot be a high school graduate or enrolled in an accredited high school.
• You must be at least 18 years old. You can take the test at the age of 16 or 17 if you satisfy some conditions.
• Cost: $80 (all subjects) or$20 (per subject)
• The minimum passing score for each subject is 145
• Accommodations: You must request disability accommodations via the GED® Testing Service website.
• The State of Wyoming will provide you with a GED certificate if you pass the exam, showing that you have attained the same level of knowledge as a high school graduate.
You may learn more about how to obtain a GED in Wyoming by reading the answers to the questions we receive the most below. Besides, to help you easily prepare for your coming Wyoming GED exam, we offer hundreds of GED practice test questions on this website. Let's give it a try!
## In Wyoming, what high school equivalency tests are available?
GED and HiSET are the two high school equivalency tests offered in Wyoming. Both are accepted at 99% of colleges and both are equivalent to a high school diploma. With a Wyoming HiSET or Wyoming GED, you can find a fantastic job, earn a degree, or join the military.
## What subjects are tested on the GED in Wyoming?
You will be tested in math, social studies, science, and language arts on the GED in Wyoming.
## How long is the GED exam in Wyoming?
SUBJECT TIME Math 115 minutes Social Studies 90 minutes Science 90 minutes Language Arts 150 minutes Total 7 hours, 41 minutes
## How many questions are there on the GED test in Wyoming?
SUBJECT # OF QUESTIONS Math 46 Social Studies 35 Science 34 Language Arts 46
## If I don't pass a subject on the Wyoming GED exam, can I retake it?
For each subject, you can retake it up to 2 times without having to wait. For the fourth attempt, you have to wait 60 days. You have an unlimited number of retakes per year.
## Can I take the Wyoming GED exam online?
You can take the Wyoming GED test online or at an official Wyoming GED testing center. You can use our list of Wyoming testing centers to locate the one that is most convenient for on-site testing.
There will be an Online Proctored (OP) GED test option starting on September 1st, 2021, so you can take your GED test remotely (online).
## What age must I be to take the Wyoming GED exam?
In Wyoming, you must be at least 18 years old to take the GED exam.
If you are 17 years old, you must submit a completed Age Waiver Application to request a waiver from the minimum age requirement.
If you are 16 years old, you must submit a completed Age Waiver Application to request a waiver from the minimum age requirement. You must also provide a written statement of support from a parent, legal guardian, an ABE instructor, or a GED examiner, indicating the extenuating circumstances.
## How can I obtain a copy of my Wyoming GED test transcript?
You can ask for a copy of your Wyoming GED transcript by contacting the GED testing service. | 2023-02-06 00:55:30 | {"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": 0.19118906557559967, "perplexity": 2954.2112371045914}, "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/1674764500294.64/warc/CC-MAIN-20230205224620-20230206014620-00600.warc.gz"} |
https://root-forum.cern.ch/t/creating-th2f-and-th3f-using-macros-in-ttree-draw-and-ttree-project/21534 | # Creating TH2F and TH3F using macros in TTree::Draw and TTree::Project?
I just discovered the feature where TTree::Draw uses MakeProxy to read macros, e.g.
I can already think of a few handy uses of this. One thing that doesn’t seem possible, however, is drawing to a multidimensional histogram with this feature. Could it be possible for a macro with a signature
to be used to draw a TH2F? Additionally, could this feature be used with TTree::Project? (which appears to just produce a string and passes it to TTree:Draw).
This is an interesting idea (where one challenge is to properly handle the case where the user had defined both the expression and expression2d function). Would you be interested in helping to implement it?
Thanks,
Philippe. | 2022-07-02 03:20:46 | {"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": 0.8157796263694763, "perplexity": 1820.950460630449}, "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-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00015.warc.gz"} |
https://plainmath.net/11644/sketch-region-bounded-curves-equal-region-volume-generated-revolving | Sketch the region bounded by the curves: y=lnx,y=0y=lnx,y=0 and x=ex=e, then find, the area of this region, and find the volume of the solid generated by revolving this area about the line x=−2?x=−2?
Question
Integrals
Sketch the region bounded by the curves: $$\displaystyle{y}={\ln{{x}}},{y}={0}{y}={\ln{{x}}},{y}={0}$$ and $$\displaystyle{x}={e}{x}={e}$$, then find, the area of this region, and find the volume of the solid generated by revolving this area about the line $$\displaystyle{x}=−{2}?{x}=−{2}$$?
2021-01-29
The sketch is easily obtainable through a graphing software/website, it’s also very easy to sketch it from basic principles. As for the area of the figure,
You should notice that we are essentially finding the area of the curve $$\displaystyle{\ln{{\left({x}\right)}}}$$ from 11 to e.
$$\displaystyle{\int_{{{1}}}^{{{e}}}}{\ln{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left[{x}{\ln{{x}}}−{x}\right]}{e}{1}={1}$$
As for the solid of revolution, we have to use the washer method, which has a ring of constant outer radius (e+2)(e+2) and a ring of inner radius ey+2
Therefore, we can just compute:
$$\displaystyle\pi{\int_{{{0}}}^{{{1}}}}{\left({e}+{2}\right)}^{{{2}}}-{\left({e}^{{{y}}}+{2}\right)}^{{2}}{\left.{d}{y}\right.}=π{\left[{\left({e}+{2}\right)}^{{2}}-{\frac{{{e}^{{{2}}}}}{{{2}}}}-{4}{e}+{\frac{{{1}}}{{{2}}}}\right]}={\frac{{π{\left({e}^{{2}}+{9}\right)}}}{{{2}}}}$$
Relevant Questions
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify your results using the integration capabilities of a graphing utility.
$$\displaystyle{y}={\cos{{2}}}{x}$$
y=0
x=0
$$\displaystyle{x}=\frac{\pi}{{4}}$$
Find the volume of the solid generated by revolving the revolving the region bounded by the graphs of the equations and inequalities about the y-axis
$$\displaystyle{x}^{{2}}-{y}^{{2}}={25},{x}\geq{0},{y}=-{5},{y}={5}$$
Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. $$\displaystyle{y}={5}\frac{{\left({16}−{x}^{{2}}\right)}^{{1}}}{{2}}$$ , y = 0, x = 2, x = 3, about the x-axis
The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.
(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
(e) What is the moment generating function for X?
Random variables X and Y are uniformly distributed on the region bounded by the x and y axes, and the curve $$y=1-x^{2}$$.
Calculate E(XY)
A medical technician is trying to determine what percentage of apatient's artery is blocked by plaque. To do this, she measures theblood pressure just before the region of blockage and finds that itis $$\displaystyle{1.20}\times{10}^{{{4}}}{P}{a}$$, while in the region of blockage it is $$\displaystyle{1.15}\times{10}^{{{4}}}{P}{a}$$. Furthermore, she knows that blood flowingthrough the normal artery just before the point of blockage istraveling at 30.0 cm/s, and the specific gravity of this patient'sblood is 1.06. What percentage of the cross-sectional area of thepatient's artery is blocked by the plaque?
The top string of a guitar has a fundamental frequency of 33O Hz when it is allowed to vibrate as a whole, along all its 64.0-cm length from the neck to the bridge. A fret is provided for limiting vibration to just the lower two thirds of the string, If the string is pressed down at this fret and plucked, what is the new fundamental frequency? The guitarist can play a "natural harmonic" by gently touching the string at the location of this fret and plucking the string at about one sixth of the way along its length from the bridge. What frequency will be heard then?
The crane shown in the drawing is lifting a 182-kg crate upward with an acceleration of $$\displaystyle{1.5}\frac{{m}}{{s}^{{2}}}$$. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 130 kg. The cable is then wound ontoa hollow cylindrical drum that is mounted on the deck of the crane.The mass of the drum is 150 kg, and its radius is 0.76 m. The engine applies a counter clockwise torque to the drum in order towind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
This problem is about the equation
dP/dt = kP-H , P(0) = Po,
where k > 0 and H > 0 are constants.
If H = 0, you have dP/dt = kP , which models expontialgrowth. Think of H as a harvesting term, tending to reducethe rate of growth; then there ought to be a value of H big enoughto prevent exponential growth.
Problem: find acondition on H, involving $$\displaystyle{P}_{{0}}$$ and k, that will prevent solutions from growing exponentially.
... | 2021-06-18 18:05:08 | {"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": 0.6993043422698975, "perplexity": 582.4472356869047}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00369.warc.gz"} |
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## anonymous 4 years ago Expand and simplify the first 3 terms of the binomal power: (3y+5)^9 PLEASE HELP!!
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1. mathteacher1729
Do you know the binomial theorem? It lets you find the nth term of insane expressions like this. :)
2. anonymous
I know usually how to solve using the pascal's triangle, but I don't know how to do it for just the first 3 terms
3. mathteacher1729
http://regentsprep.org/Regents/math/algtrig/ATP4/bintheorem.htm might be useful.
4. mathteacher1729
Check out the examples. Then substitute your numbers in and see what happens. NOTE -- you will get BIG NUMBERS.
5. anonymous
Okay, that was useful, but we're not supposed to be using the formula for this question. Do you know if there's any way to solve it just using pascal's triangle?
6. mathteacher1729
You'd have to expand out to the 9th row! Have fun with that! Fortunately you can do it in excel. :) http://jwilson.coe.uga.edu/EMAT6680/Parsons/MVP6690/Essay1/excel.html
7. Zarkon
use the binomial coefficients ${n\choose r}=\frac{n!}{r!(n-r)!}$
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https://croteaurmt.ca/inspirational-movies-ltjvp/acidic-oxides-list-362aab | Aluminum oxide shows acid and basic properties of an oxide, it is amphoteric. \underset{\large{Amphoteric}}{\underbrace{Al_2O_3,\: SiO_2}}\hspace{20px} It may also help to examine the physical properties of oxides, but it is not necessary. The oxide that gives a base in water is known as a basic oxide. The basicity of these oxides increase down each group. The Acidic Environment > 2. Basic Oxides. Na2O and MgO). Acidic oxides form acids with water: eg SO2 + H2O --> H2SO3 Basic oxides form basic salts (alkaline) with water: eg. With the oyxgen exhibiting an oxidation number of -2. By reaction of oxygen with compounds at higher temperatures: At higher temperatures, oxygen also reacts with many compounds forming oxides, e.g., $2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2$, $2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2$, $C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O$, $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$, $2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2$, $Cu(OH)_2 \xrightarrow{\Delta} CuO + H_2O$, By oxidation of some metals with nitric acid, $2Cu + 8HNO_3 \xrightarrow{Heat} 2CuO + 8NO_2 + 4H_2O + O_2$, $Sn + 4HNO_3 \xrightarrow{Heat} SnO_2 + 4NO_2 + 2H_2O$, By oxidation of some non-metals with nitric acid, $C + 4HNO_3 \rightarrow CO_2 + 4NO_2 + 2H_2O$. Since oxygen is highly reactive in nature, it reacts with metallic and non-metallic elements and forms oxides of those elements. Sometimes acidic oxides are formed when metals (with higher oxidation states) react … On treatment with an acid, compound oxides give a mixture of salts. $\underset{\text{Ferro-ferric oxide}}{Fe_3O_4} + 8HCl \rightarrow \underset{\text{ferric chloride}}{2FeCl_3} + \underset{\text{ferrous chloride}}{FeCl_2} + 4H_2O \label{28}$. The oxides of elements in a period become progressively more acidic as one goes from left to right in a period of the periodic table. An acidic oxide is an oxide which when combined with water gives off an acid. Alkali metal and alkaline earth oxides are basic (dissolve in acid). When a substance reacts chemically, both as a base or acid it termed as an amphoteric solution. \underset{\large{Acidic}}{\underbrace{P_4O_{10},\: SO_3,\:Cl_2O_7}}\hspace{20px}\). A basic oxide is an oxide which when combined with water gives off a base. Basic oxides are oxides of metals, which is also known as metallic oxides. It is difficult to break the covalent O-H bonds easily. Oxygen can thus be obtained from acidified water by its electrolysis. Oxides: Group 1 metals react rapidly with oxygen to produce several different ionic oxides, usually in the form of $$M_2O$$. Based on their acid-base characteristics oxides are classified as acidic, basic, amphoteric or neutral: There are different properties which help distinguish between the three types of oxides. Superoxides: Often Potassium, Rubidium, and Cesium react with excess oxygen to produce the superoxide, $$MO_2$$. $Cs + O_2 \rightarrow CsO_2 \label{21}$. While reacting with alkalies they form salt and water showing acidic properties. Al2O3. Aluminium oxide (Al2O3) is an amphoteric oxide; it can act as a base or acid. Petrucci, Ralph, William Harwood, Jeffry Madura, and Geoffrey Herring. Metal oxides. CO doesn’t produce a salt when reacted with an acid or a base. Oxides as Acid and Basic Anhydrides. Which of the following element will form acidic oxides of type E_2O_3? Due to high electronegativity, oxygen can react with almost all the elements except noble gases. Hence, electrical energy through the electrolysis process is applied to separate dioxygen from water. Sulfur dioxides, carbon dioxide, carbon monoxide, nitrates and organic particles emitted from manufacturing industries, automobile engines and farming chemicals accumulate in the atmosphere. Legal. most acidic oxide is- MgO CaO Na2O Al2O3. When these oxides are dissolved in stoichiometric water, and acid can be produced: SO3 + H 2O → H 2SO4 and nitrous acid, nitric acid, carbonic acid etc..........I For example, with base different aluminate salts will be formed: $$Rb + O_2 \; (excess) \rightarrow RbO_2$$. For example, when HSO4- reacts with water it will make both hydroxide and hydronium ions: $HSO_4^- + H_2O \rightarrow SO_4^{2^-} + H_3O^+ \label{11}$, $HSO_4^- + H_2O \rightarrow H_2SO_4 + OH^- \label{12}$. CaO + H2O --> Ca(OH)2 Amfotheric oxides have both acidic and base properties: eg. Ammonium Boron Oxide. 2O ––> Ca2++ 2OH–, moderately strong base [O2–] + H. 2O ––> 2OH–K > 1022. It is important to remember that the trend only applies for oxides in their highest oxidation states. Two of the most toxicologically significant compounds are nitric oxide (NO) and nitrogen dioxide (NO 2).Other gases belonging to this group are nitrogen monoxide (or nitrous oxide, N 2 O), and nitrogen pentoxide (NO 5).. Nitrogen dioxide is produced for the manufacture of nitric acid. Oxide - Oxide - Nonmetal oxides: All nonmetals form covalent oxides with oxygen, which react with water to form acids or with bases to form salts. Acidic oxides are oxygen-containing compounds that dissolve in water to form acidic solutions. New Jersey, NJ: Prentice Hall, 2007. These oxides, however, do not give hydrogen peroxide by action with dilute acids. The major types of o… Yes, an example is carbon monoxide (CO). Oxides of nitrogen are a mixture of gases that are composed of nitrogen and oxygen. This oxygen comes from either air or water. Question 3: Ans: C since Magnesium Oxide is the only Basic Oxide in the list and it will gives a pH higher than 7. Can an oxide be neither acidic nor basic? Arsenic pentoxide, white solid, highly toxic, oxidizing agent. Ammonium Antimony Tungsten Oxide. Notice how the amphoteric oxides (shown in blue) of each period signify the change from basic to acidic oxides, The figure above show oxides of the s- and p-block elements. Examples include CO2,N O,N O2,SO2,SO3........ Another name for these oxides is acid anhydride. Keep Up The Good Work! Amphoterism depends on the oxidation states available to a chemical species. The trend in acid-base behavior can be summarized as follows: Acidity increases from left to right, ranging from strongly basic oxides on the left to strongly acidic ones on the right, with an amphoteric oxide (aluminum oxide) in the middle. Watch the recordings here on Youtube! asked Apr 7, 2018 in Chemistry by shabnam praween (137k points) p - block element +1 vote. For example $$ZnO$$ exhibits basic behavior with $$HCl$$, Similarly, $$Al_2O_3$$ exhibits basic behavior with $$H_2SO_4$$. Li2O). When a small amount of acid is added to water ionization is initiated which helps in electrochemical reactions as follows. [ "article:topic", "shrestha", "Acidic Oxides", "basic oxides", "Neutral Oxides", "showtoc:no" ], http://www.wou.edu/las/physci/ch412/oxides.html, information contact us at info@libretexts.org, status page at https://status.libretexts.org. Dioxides on reaction with concentrated HCl yield Cl2 and on reacting with concentrated H2SO4 yield O2. Metal oxides on the left side of the periodic table produce basic solutions in water (e.g. Why is it difficult to obtain oxygen directly from water? If soluble in water, they react with water to produce hydroxides (alkalies) e.g., $\ce{ CaO + H_2O \rightarrow Ca(OH)_2} \label{6}$, $\ce{ MgO + H_2O \rightarrow Mg(OH)_2} \label{7}$, $\ce{ Na_2O + H_2O \rightarrow 2NaOH } \label{8}$. Selected data for the Oxides of Nitrogen. Oxides are binary compounds of oxygen with another element, e.g., CO2, SO2, CaO, CO, ZnO, BaO2, H2O, etc. 877. 9th Edition. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Category: Oxides. They react with acids to produce salts, e.g., $\ce{ MgO + 2HCl \rightarrow MgCl_2 + H_2O } \label{9}$, $\ce{ Na_2O + H_2SO_4 \rightarrow Na_2SO_4 + H_2O} \label{10}$. Antimony pentoxide, yellow solid, oxidizing agent. sulfides are usually oxidized when heated with oxygen. A peroxide is a metallic oxide which gives hydrogen peroxide by the action of dilute acids. with the oxidation number of the oxygen equal to -1/2. An amphoteric solution is a substance that can chemically react as either acid or base. For example, carbon dioxide reacts with water to produce carbonic acid: This is the reason why indicator paper must be moist in order to detect whether a non-metal oxide gas is acidic. Aluminum Oxide. There are a few oxides, such as NO 2 and ClO 2 in which the central atom's oxidation state does not correspond to the oxidation state of that element in a stable or known oxo acid. When heated with oxygen, compounds containing carbon and hydrogen are oxidized. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. An oxide that combines with water to give an acid is termed as an acidic oxide. Ammonium Oxide. If we take a closer look at a specific period, we may better understand the acid-base properties of oxides. Aluminium oxide, very hard, used as abrasive. Have questions or comments? Antimony trioxide, white solid. (NH4)2B4O7. Cl2O, SO2, P4O10). Oxides are chemical compounds with one or more oxygen atoms combined with another element (e.g. $H_2 + O_2 \rightarrow H_2O_2 \label{20}$. Arsenic trioxide, white solid, highly toxic. $PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O \label{24}$, $2PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O + O_2 \label{25}$. Acidic oxides are compounds that can form an acidic solution when dissolved in water. N 2 O and NO are commonly called nitrous oxide and nitric oxide, respectively. NH4OSbW. A typical example is sulfur trioxide, which dissolves to form sulfuric acid. $4 Li + O_2 \rightarrow 2Li_2O \label{19}$. There are two general statements that describe the behaviour of acidic oxides. ; Hence, basic oxides neutralize acids. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Below are a few. Amphoteric oxides exhibit both basic as well as acidic properties. 2 NO 2 + 2 OH - NO 2- + NO 3- + H 2 O. (ii) Potassium oxide is basic whereas nitrous oxides is acidic (iii) Aluminium and zinc oxides are amphoteric (iv) Sulphur trioxides is acidic whereas phosphorus pentoxide is basic (v) Carbon dioxide is neutral whereas sulphur dioxide is amphoteric Generally Group 1 and Group 2 elements form bases called base anhydrides or basic oxides e.g., $\ce{K_2O \; (s) + H_2O \; (l) \rightarrow 2KOH \; (aq) } \label{5}$. Thus Al2O3 entails the marking point at which a change over from a basic oxide to acidic oxide occurs. Reason. $ZnO + 2HCl \rightarrow \underset{\large{zinc\:chloride}}{ZnCl_2}+H_2O\,(basic\: nature) \label{13}$, $ZnO + 2NaOH \rightarrow \underset{\large{sodium\:zincate}}{Na_2ZnO_2}+H_2O\,(acidic\: nature) \label{14}$, $Al_2O_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2O\,(basic\: nature) \label{15}$, $Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2+H_2O\,(acidic\: nature) \label{16}$. Ans: C cos Acidic Oxide are formed when a non-metal combines with oxygen. An amphoteric oxide is an oxide that can act as either an acid or base in a reaction to produce a salt and water. dichloride oxide; arsenic oxide (III) Hyposulfuric anhydride; phosphorus oxide (III) carbon dioxide; titanium oxide; silicic anhydride; nitrogen oxide (V) perchloric oxide; manganese oxide (VI) Etymology "Basic oxides" is a compounds word of "Basic" and "oxides". Let me know if this chemistry blogsite has been helpful in any ways. The basic trend in oxide acidity across the rows of the periodic table is: The basic trend in oxide activity down the period groups of the periodic table is: Non-metals are found mostly in the top-right corner of the periodic table, and thus have acidic oxides. An amphoteric solution is a substance that can chemically react as either acid or base. Acidic oxides are the oxides of non-metals (Groups 14-17) and these acid anhydrides form acids with water: $\ce{SO_2 + H_2O \rightarrow H_2SO_3} \label{1}$, $\ce{ SO_3 + H_2O \rightarrow H_2SO_4} \label{2}$, $\ce{CO_2 + H_2O \rightarrow H_2CO_3} \label{3}$. purple: basic oxides blue: amphoteric oxides pink: acidic oxides. Acidic oxides are the oxides of non-metals (Groups 14-17) and these acid anhydrides form acids with water: Sulfurous Acid $\ce{SO_2 + H_2O \rightarrow H_2SO_3} \label{1}$ Sulfuric Acid $\ce{ SO_3 + H_2O \rightarrow H_2SO_4} \label{2}$ Carbonic Acid $\ce{CO_2 + H_2O \rightarrow H_2CO_3} \label{3}$ Water as such is a neutral stable molecule. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Missed the LibreFest? ; Basic oxides that are soluble in water are called alkalis.. Recall: Solutions of these alkalis in water have a pH value of more than 7. $[H_2O\:(acidulated)\rightleftharpoons H^+\,(aq)+OH]^-\times4$, $[H^+\,(aq)+e^-\rightarrow\dfrac{1}{2}H_2(g)]\times4$, $4OH^-\,(aq)\rightarrow O_2+2H_2O + 4e^-$, $2H_2O \xrightarrow{\large{electrolysis}} 2H_2\,(g) + O_2\,(g)$. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Non-metal oxides on the right side of the periodic table produce acidic solutions (e.g. Basic Oxides(usually “ionic”) CaO + 2H. Pages in category "Acidic oxides" The following 18 pages are in this category, out of 18 total. However, it is also possible for an oxide to be neither acidic nor basic, but is a neutral oxide. Table 18.3. By direct heating of an element with oxygen: Many metals and non-metals burn rapidly when heated in oxygen or air, producing their oxides, e.g., $P_4 + 5O_2 \xrightarrow{Heat} 2P_2O_5$. All the oxides of nitrogen (except NO and N20) and phosphorus are strongly acidic. (NH4)2O. Acidic Oxides. Retrieved from " https://simple.wikipedia.org/w/index.php?title=Neutral_oxide&oldid=7178542 ". An amphoteric solution is a substance that can chemically react as either acid or base. The word oxides referred to the chemical compounds that one or more oxygen atoms combined with another element such as H 2 O or CO 2.Based on their acid-base characteristics oxides can be classified into four categories: acidic oxides, basic oxides, and amphoteric oxides and neutral oxides. Basic oxides are the oxides of metals. BaO(s) + H2O(l) -> Ba+2 + 2OH- Most non-metals on the RHS of the PT form acidic oxides, excluding noble gases, which do not form any oxides, and a few elements, such as Al, Sn, Zn, Pb, which form amphoteric oxides. The basic trend in oxide activity down the period groups of the periodic table is: Acidic → Amphoteric → Basic. By thermal decomposition of certain compounds like hydroxides. Dioxides like PbO2 and MnO2 also contain higher percentage of oxygen like peroxides and have similar molecular formulae. 1) Nitrous oxide (N 2 O) 2) Nitric oxide (NO) 3) Carbon monoxide (CO) 4) Water (H 2 O) 5) Manganese (IV) oxide (MnO 2 ) There are 7 neutral oxides discovered till date. Acidic oxides are formed when a non-metal reacts with oxygen. $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 \label{22}$, $Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2 \label{23}$. Basic oxides react with acid to produce salt and water only.. Notice that this is a neutralization reaction. Most nonmetal oxides are acidic and form oxyacids, which is also possible for an,! Off a base or acid more oxygen than the corresponding basic oxide acid ) acidic solution when in... Oldid=7178542 oxygen atom attached to another element and have similar molecular.. Chemical species then mix with water and other materials before falling to the ground produce the superoxide \..., 1525057, and 1413739 oxide ; it can acidic oxides list as a base or acid it termed as because... 2 OH - NO 2- + NO 3- + H 2 O and NO are commonly called oxide..... Notice that this is a substance that can chemically react as either acid or a base or acid termed! Metallic oxide which acidic oxides list combined with water gives off an acid, compound oxides give mixture! Over from a basic oxide, oxidizing agent of oxo acids or anions by disproportionation hard, used abrasive. Neither acidic nor basic, but is a metallic oxide which gives hydrogen peroxide by action! Due to high electronegativity, oxygen can thus be obtained from acidified water by its electrolysis except., we may better understand the acid-base properties of an oxide, is. Nor basic, but is a substance that can chemically react as acidic oxides list acid or base, e.g. Sodium. From a basic oxide, e.g., Sodium, calcium and barium peroxides is termed as oxides because,. Change over from a basic oxide is an amphoteric solution electronegativity, oxygen can be! Shows both acidic and base properties: eg not reflect recent changes )... Groups of the periodic table is: acidic → amphoteric → basic separate dioxygen water! The elements except noble gases acid ), they form amphoteric oxides both... These then mix with water to form acidic solutions yield O2 or a base or acid groups of oxygen. One oxygen atom attached to another acidic oxides list 20 } \ ] before falling to the ground non-metal with..., Sodium, calcium and barium peroxides Madura, and thus have acidic oxides covalent bonds! Oxides give a mixture of gases that are composed of nitrogen blogsite has been helpful any. Neither acidic nor basic, but it is not necessary oxide ; it can act either. Pentoxide, white solid, highly toxic, oxidizing agent as acids when they in... Mixture of oxo acids or anions by disproportionation in oxide activity down the groups! With an acid, compound oxides give a mixture of oxo acids or anions by disproportionation oxides. Multiple oxidation states elements except noble gases one which shows both acidic and form oxyacids, in. Metallic oxide which when combined with water to give salts a neutralization reaction applies for oxides in their highest states... When dissolved in water that this is a substance that can chemically react as either acid or base only. They react with almost all the oxides of nitrogen oxygen, compounds containing carbon and hydrogen are oxidized SO3. Give hydrogen peroxide by action with dilute acids ) is an oxide combines. Is sulfur trioxide, which in turn yield hydronium ions ( H3O+ ) in aqueous solution acid basic! That this is a substance that can chemically react as either an acid is termed as because... In aqueous solution oyxgen exhibiting an oxidation number of -2 can thus be obtained from water. Higher percentage of oxygen like acidic oxides list and have similar molecular formulae Lithium Sodium. Peroxides and have similar molecular formulae also contain higher percentage of oxygen like peroxides and have molecular... Separate dioxygen from water typical example is carbon monoxide ( CO ) found mostly in top-right... Notice that this is a substance that can chemically react as acid. Like PbO2 and MnO2 also contain higher percentage of oxygen like peroxides and have similar formulae! This is a neutralization reaction they form salt and water showing acidic properties act... Aqueous solution ] + H. 2o –– > Ca2++ 2OH–, moderately strong base O2–... Oxides act as acids when they react with almost all the oxides of nitrogen ( NO. Oxide, very hard, used as abrasive is it difficult to obtain oxygen directly from water is! Or base in nature, it is not necessary which when combined with water gives off a base of oxygen... Produce a salt and water a typical example is carbon monoxide ( CO.! Nitric oxide, respectively 20 } \ ] SO3........ another name for these oxides, but is! Is amphoteric must be in its highest possible oxidation state because the only... Acidic solutions with an acid or a base or acid compounds that dissolve in water is as! Entails the marking point at which a change over from a basic oxide produce a and. Of acid is added to water ionization is initiated which helps in electrochemical reactions as follows attached to element... Element will form acidic oxides of those elements HCl yield Cl2 and on with... Left side of the periodic table produce basic solutions in water is known as a in. Oxides on the oxidation number of the periodic table produce basic solutions in water to acidic! + 2H when dissolved in water in wate our status page at https: //status.libretexts.org water, showing properties... + O_2 \rightarrow 2Li_2O \label { 20 } \ ] basicity of these oxides is acid anhydride period we! Solutions in water O_2 \ ; ( excess ) \rightarrow RbO_2 \.! Recent changes acidic oxides list ) Potassium, Rubidium, and thus have acidic are! Understand the acid-base properties of amphoteric oxides pink: acidic → amphoteric → basic information us! - block element +1 vote trioxide, which is also possible for an oxide, respectively ( 137k points p! Helps in electrochemical reactions as follows acid, compound oxides give a mixture of oxo acids or by. Chemical compounds with one or more oxygen than the corresponding basic oxide, respectively also possible for an which! Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 reacting with alkalies acidic oxides list form oxides. Numbers 1246120, 1525057, and Cesium react with acid to produce salt water... Section above on properties of oxides, but it is not necessary carbon and hydrogen are oxidized for! Then mix with water to form sulfuric acid possible oxidation state because the trend applies. Pbo2 and MnO2 also contain higher percentage of oxygen like peroxides and have similar molecular formulae like and! Acidic aqueous solutions and can react directly with bases to give salts........ another name for these oxides increase each!, and 1413739 acidic oxide is an amphoteric solution is difficult to obtain oxygen directly water! As a base or acid it termed as oxides because here, oxygen is in combination only. Has been helpful in any ways the top-right corner of the oxygen to... Or anions by disproportionation generate a mixture of gases that are composed of nitrogen ( NO! Chemical species oxides generate a mixture of oxo acids or anions by.... Gives off an acid, compound oxides give a mixture of gases that are composed of nitrogen and.. | 2021-10-27 23:42:43 | {"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": 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": 0.7363619208335876, "perplexity": 6727.010156687106}, "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-2021-43/segments/1634323588244.55/warc/CC-MAIN-20211027212831-20211028002831-00511.warc.gz"} |
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# Module Overview template
## Overview
Description and summary of module.
### Learning Outcomes
Upon completion of this module, you will be able to:
### Why This is Important
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http://accessemergencymedicine.mhmedical.com/content.aspx?bookid=434§ionid=41825434 | Chapter II-8
A 50-year-old man presented to the ED with abdominal pain. He had end-stage renal disease that was being managed with peritoneal dialysis.
His abdominal radiograph is shown in Figure 1.
###### Figure 1
To better localize the radiopaque object, a lateral view of the abdomen was added by the radiologist.
• What is the radiopaque object seen in the mid-abdomen?
There are several factors that contribute to the radiopacity of an object. We usually think that the intrinsic radiopacity of the material that makes up an object is the major determinant of its radiographic appearance. However, the shape of the object can play a major role in determining its radiographic appearance. In this case, the apparently radiopaque object is actually composed of relatively radiolucent material.
The intrinsic radiopacity of a substance depends, in part, on the atomic numbers of its constituent atoms. A substance is more radiopaque if it contains atoms of high atomic number such as calcium, iodine, barium, or lead.
X-ray radiation is absorbed when it ejects an electron from one of the inner orbitals of the atom—the photoelectric effect (Figure 2). More x-ray energy is absorbed when the electron is tightly bound to its orbital. The force binding an electron to the orbital is determined by the electrical charge of the nucleus, i.e., the number of protons in the nucleus, which is equivalent to the atomic number of the element. This is why bone, which contains calcium (atomic number 20), is more radiopaque than soft tissue, which is made up mostly of carbon (atomic number 6), hydrogen (atomic number 1), and oxygen (atomic number 8). Iodine (atomic number 53) is the key constituent of radiocontrast material and lead (atomic number 82) is an effective barrier to x-rays.
###### Figure 2
The photoelectric effect.
In this patient, however, the object’s shape (i.e., the patients abdomen) rather than its intrinsic radiopacity is more important in determining its radiographic appearance. The round radiopaque mass seen in the middle of the abdomen on the AP view cannot represent an object within the abdominal cavity, e.g., a solid ball that was swallowed or an intra-abdominal tumor, because such an object would be visible within the abdomen on the lateral view (Figure 1). Alternatively, a thin flat object can be difficult to see when viewed from the side. However, a thin flat disc of sufficient radiopacity to have this appearance on the AP view would appear as a very radiopaque line when viewed from the side on a lateral view.
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### AccessEmergency Medicine Full Site: One-Year Subscription
Connect to the full suite of AccessEmergency Medicine content and resources including advanced 8th edition chapters of Tintinalli’s, high-quality procedural videos and images, interactive board review, an integrated drug database, and more. | 2017-03-24 10:13:43 | {"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": 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": 0.19616927206516266, "perplexity": 1478.7649861770244}, "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-2017-13/segments/1490218187792.74/warc/CC-MAIN-20170322212947-00606-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://collegephysicsanswers.com/openstax-solutions/calculate-angular-momentum-ice-skater-spinning-600-revs-given-his-moment-inertia | Question
(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is $0.400 \textrm{ kg} \cdot \textrm{m}^2$. (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s?
a) $15.1 \textrm{ kg} \cdot \textrm{m}^2 \textrm{s}$
b) $1.92 \textrm{ kg}\cdot \textrm{m}^2$
c) $-0.503 \textrm{ kg} \cdot \textrm{m}^2 \textrm{/s}^2$
Solution Video | 2019-02-22 05:59:44 | {"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": 0.8979484438896179, "perplexity": 1182.9909758911663}, "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-2019-09/segments/1550247513661.77/warc/CC-MAIN-20190222054002-20190222080002-00240.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-6-systems-of-equations-and-inequalities-6-3-solving-systems-using-elimination-practice-and-problem-solving-exercises-page-378/25 | ## Algebra 1
Multiply the first equation by 2. Add the equations. $\ \ \ 4x-\ \ 7y=\ \ 15\ \ \ \$multiply by 2$\ \ \ \ 8x-14y=\ \ 30$ $-8x+14y=-30\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{-8x+14y=-30}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0=0$ Since the equation is always true, the system has infinitely many solutions. | 2020-02-19 07:28:47 | {"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": 0.7541052103042603, "perplexity": 349.977288846155}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00309.warc.gz"} |
https://electronics.stackexchange.com/questions/442303/free-piston-stirling-engine-tuning-capacitor-series-vs-parallel | # Free Piston Stirling Engine Tuning Capacitor - Series Vs Parallel
Why is a tuning capacitor placed in series with a free piston stirling alternator rather than parallel with it?
I am accustomed to seeing power factor correction capacitors placed in parallel in school rather than in series, why would this application be different?
To be honest I have never seen an example of power factor correction being used on an AC source, only on AC loads so maybe that has something to do with it.
Also, it's called a tuning capacitor, so maybe it has something to do with controlling the resonance of the engine piston (RLC circuit)? I know that both the piston and displacer must have the correct resonant frequency in order for the engine to operate correctly.
EDIT: I just found some more info about about the multiple purposes of a tuning capacitor in "Tradeoff Between Magnet Volume and Tuning Capacitor in a Free Piston Stirling Engine Power Generation System" (https://doi.org/10.4271/929262) This paper doesn't really answer my question, though.
• The paper you cite seems to explain it pretty well. What's not clear? – Spehro Pefhany Jun 6 at 18:35
• The paper explains the purposes of the tuning capacitor, but I don't think it explained why it is placed in series rather than parallel with the alternator output. – D. Goodell Jun 6 at 18:39
Series-tuned circuits have their minimum impedance at resonance, while parallel-tuned circuits have their maximum impedance at resonance.
It would seem that the idea here is to nullify the effects of the leakage inductance of the generator coil on power delivery, so minimum impedance is what you want.
Fun Stuff!
I'm guessing you have a linear alternator, which can be roughly modeled just as shown in your diagram: a sinusoidal voltage source from the back emf of the alternator with an output impedance consisting of the inductance and resistance of the coil. The sinusoidal frequency will be the oscillation of the piston. The series inductance means that the output voltage would vary depending on load and frequency if the capacitor were not present.
At your operating frequency, the capacitor and inductance, if the capacitance is properly chosen, would make the output impedance purely resistive and equal to the coil's resistance and the capacitor's ESR. By designing a control system to supply the proper heat to maintain the frequency at a constant value, the system's output looks like an AC source with a resistive output impedance. You must maintain this osillation frequency; if the frequency changes from your resonant point the output impedance will become inductive or capacitive.
Good Luck!
The total impedance of a series RLC circuit is R + j(Xl - Xc), which has a magnitude of R = $$\\sqrt{R^2 + (X_L - X_C)^2}\$$, so the reactance of the inductance can be cancelled out by resonating it with the capacitor (pick C such that $$\\omega L = \frac{1}{\omega C} \$$ so C= $$\\frac{1}{\omega^2L}\$$). That maximizes the magnitude of the current output.
However there are other considerations (potential demagnetization) that are covered in your linked paper. | 2019-11-20 19:44:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5961623191833496, "perplexity": 865.5618850208167}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670601.75/warc/CC-MAIN-20191120185646-20191120213646-00514.warc.gz"} |
https://justrakudoit.wordpress.com/page/2/ | Set Operations
June 21, 2013
A month or so ago, while working on ABC, I tried to use [∪] (reduce on set union) in my grammar actions. Unfortunately, it blew up when I tried it in Niecza. I assumed the problem was that I hadn’t implemented zero or one arguments forms of set union, and made a quick effort to add them. It didn’t work, so I worked around the problem by using the set constructor directly and added it to my to-do list.
I finally got around to tackling the problem again this week, and my earlier difficulties were quickly explained when I looked at the stack trace for the error message. ∪ turns out to be list associative. That means if you write $a ∪$b ∪ $c, the actual call generated is infix:<∪> ($a, $b,$c), and the reduce meta-op generates that call internally. So I needed to change infix:<∪> from a binary sub to an N-ary sub.
Previously the code looked like this
proto sub infix:<∪>($,$ --> Set) is equiv(&infix:<|>) {*}
multi sub infix:<∪>(Any $a, Any$b --> Set) { $a.Set ∪$b.Set }
multi sub infix:<∪>(Set $a, Set$b --> Set) { Set.new: $a.keys,$b.keys }
multi sub infix:<∪>(Baggy $a, Any$b --> Bag) { $a ∪$b.Bag }
multi sub infix:<∪>(Any $a, Baggy$b --> Bag) { $a.Bag ∪$b }
multi sub infix:<∪>(Baggy $a, Baggy$b --> Bag) {
Bag.new-from-pairs(($a.Set ∪$b.Set).map({ ; $_ =>$a{$_} max$b{$_} })) } So the rules were, if you had a Baggy (ie Bag or KeyBag), you promoted both arguments to Bag. Otherwise both arguments were promoted to Set. Also note that, because of this, the proto signature was wrong — it could generate a Set or a Bag. I replaced them with a single sub: only sub infix:<∪>(\|$p) is equiv(&infix:<|>) {
my $set = Set.new:$p.map(*.Set.keys);
if $p.grep(Baggy) { my @bags =$p.map(*.Bag);
Bag.new-from-pairs($set.map({ ;$_ => [max] @bags>>.{$_} })); } else {$set;
}
}
Line 2 creates the Set version of the union. Then we check to see if any of the arguments are Baggy using $p.grep(Baggy). (That filters out all the non-Baggy from the argument list, and then (because it is used in a boolean context) returns true if the resulting list has any elements in it.) If there is a Baggy, then we convert all the arguments to Bag. Line 5 is a spiffy bit of code that creates a new Bag using the maximum count for each element found in $set. Finally, if there were no Baggyarguments, just return $set. Once I had that working, I did the same for infix:<∩>. Then I started looking at infix:<(-)>. I’ve only ever seen set difference used as a binary operation, so my first step was to think about how to generalize it to N arguments. My thought was $a (-) $b (-)$c should be ($a (-)$b) (-) $c. If someone has a good reason this shouldn’t be the case, please let me know! Then the next question was, what should Baggy objects do here? Previously set difference always converted its arguments to Set. But it seemed to me there was a pretty obvious way to do it. After consideration, I concluded it should only use the Bag form if the first argument (the “set” things are being subtracted from) was a Baggy. Here’s my current code for infix:<∖> (note the new name, which is the ISO symbol for set difference rather than the common backslash): only sub infix:<∖>(\|$p) is equiv(&infix:<^>) {
return ∅ unless $p; if$p[0] ~~ Baggy {
my @bags = $p.map(*.Bag); my$base = @bags.shift;
Bag.new-from-pairs($base.keys.map({ ;$_ => $base{$_} - [+] @bags>>.{$_} })); } else { my @sets =$p.map(*.Set);
my $base = @sets.shift; Set.new:$base.keys.grep(* ∉ @sets.any );
}
}
This works (I think — I haven’t really tested the Baggy functionality, I’ve only run the existing Set tests). But it breaks some spectests (which assume set difference always returns a Set). And it involves the unspecified details mentioned above, so I’m stepping out on a bit of a limb. So I’m posting here looking for feedback before I commit these changes.
(Edited to add: I’ve also added infix:<⊖> for symmetric difference; that’s one of the standard symbols used, but it’s not ISO standard.)
Philosophical Issues with Rakudo’s .parse
March 6, 2013
For a little bit now I’ve had the ABC module working on both Niecza and Rakudo. I got a new Linux box yesterday; this morning I was playing around with my ABC tools there, trying to see if I could get them to work. I installed them with panda, so I was using ABC with Niecza by default on my MBP and with Rakudo by default on my new machine. I tried it with a file of my own compositions. Apparently I hadn’t tried this in a while, because it had several issues. And the two systems handled them completely differently.
On Niecza, I got this message:
Unhandled exception: Did not match ABC grammar: last tune understood:
A Hat for the Whales
d|:e~A3 BGEF|G2BG DG (3ABd|eBdB ~A3G|Bdef gfed|
e~A3 BGEF|G2BG DG (3ABd|eBdB AAGA|[1Beed e3d:|
On Rakudo, there were no error messages, but “A Hat for the Whales” was the last tune processed, even though there were a dozen more tunes in the ABC file after it.
Without thinking about it too much I corrected this batch of errors. Now on Niecza the entire file parsed, but I got the error
Unhandled exception: Illegal key signature B minor
Under Rakudo, the same ABC file just appeared to work.
What’s going on here? Well, key signature processing looks something like this (simplified):
token key-def { <basenote> <chord_accidental>? <mode>? }
token mode { <minor> | <major> }
token minor { "m" ["in" ["o" ["r"]?]?]? }
token major { "maj" ["o" ["r"]?]? }
No space allowed between basenote and mode. So when you invoke parse with that rule on “B minor”, in Niecza it fails, because the entire string was not successfully parsed. In Rakudo it succeeds, because “B” is a valid key signature, and it doesn’t care that the entire string was not parsed. In practice, this means that three of the tunes in the collection got the wrong key signature: B major instead of the correct B minor.
To my mind, Niecza’s behavior is more intuitive. If you’re trying to parse a file, parsing half of it and then silently ignoring the rest is wildly unhelpful behavior. In every use I’ve had for parse so far, I’ve wanted it to parse everything in the string I sent to the function.
But hey, I understand that some people might have different needs. Can we at least add a flag that will give you a parsing failure if the entire string is not parsed? :all, perhaps?
STEP Actions
October 16, 2012
I’ve started adding actions to the STEP code, targeted (at least at first) at collecting up the entity instance names in a parameter list to make it easier to analyze what’s in a STEP file.
I need to do this more often, because I spent about an hour fumbling about making stupid mistakes. And even now that I’ve got it working, I’m haunted by the notion there should be a simpler approach. Anyway, here’s what I’ve got at the moment:
class ISO_10303_21::Actions {
method entity_instance_name($/) { make [~$/] }
method parameter($/) { for <typed_parameter untyped_parameter omitted_parameter> ->$s {
return make $/{$s}.ast if $/{$s}.defined;
}
}
method omitted_parameter($/) { make [] } method untyped_parameter($/) {
for <entity_instance_name list_of_parameters> -> $s { return make$/{$s}.ast if$/{$s}.defined; } make []; } method typed_parameter($/) { make $<parameter>.ast } method list_of_parameters($/) { make merge-arrays(@($<parameter>)».ast); } method parameter_list($/) { make merge-arrays(@($<parameter>)».ast); } } What’s bugging me are the parameter and untyped_parameter definitions — those seem like they should be easier to express somehow. At least, that basic functionality is going to be pretty common. Oh, and I just found (and fixed) a bug in the code looking at this blog post. Feel like I had something more to say, but I’m tired and heading to bed. Parsing STEP October 10, 2012 A month ago I started perl6-ISO_10303-21. In plain English, it’s the beginnings of a toolset for working with CAD files in the STEP format. All it does is parse so far. The STEP (that’s ISO-10303) Part 21 standard provides a BNF (I think) grammar for ASCII STEP files. It was downright trivial to translate it to a Perl 6 grammar — looking at the commits, it took me about two days to get it to read the first file. THen it sat there for a long time until I starting throwing files at it. Since then I’ve made a couple of simple changes. First, I replace C-style comments in the file with spaces before parsing; the spec’s grammar doesn’t mention them at all. Second, I discovered three of the basic sample files I was using were actually illegal, with newlines embedded in strings. I fretted about what to do for a while: should I expand the grammar to accept illegal but “standard” files, or keep the grammar strict so that I could use it to test files my own code exports for correctness? Then I realized I could have my cake and eat it too: grammar ISO_10303_21::LooseGrammar is ISO_10303_21::Grammar { token non_q_char { <special> | <digit> | <space> | <lower> | <upper> | \v } } I just defined a new subclass, changing the non_q_char token to also accept vertical whitespaces, too. Now ISO_10303_21::Grammar is strict, and ISO_10303_21::LooseGrammar is loose. The next obstacle I’ve run into was speed. The 1.3 meg sample file took about five minutes to prase in Rakudo, and 10+ hours (!) in Niecza. Luckily jnthn++ was on the job, and sent me a patch about fifteen minutes after I sent him the profile of the parse today. (Rakudo’s got --profile built in.) That shaved off 1/6th of the running time. He also pointed out that tokens like token standard_keyword { <upper> [ <upper> | <digit> ]* } were unnecessarily creating a Match object for each character in the keyword. Since all we want is the full text of the keyword, this could be changed to token standard_keyword { <.upper> [ <.upper> | <.digit> ]* } and it would suppress those Match objects. That reduced the running time by another 10%. Another few days like this and it will be fast enough to do some useful work for me… Sums of Fourth Powers October 5, 2012 Another fun John Cook post has sent me off into playing with numbers and Perl 6. He gives a footnote saying that Euler discovered 635318657 = 158^4 + 59^4 = 134^4 + 133^4 and that this was the smallest number known to be the sum of two fourth powers in two ways. It seems odd now to think of such questions being unresolved. Today we’d ask Hardy “What do you mean 635318657 is the smallest known example? Why didn’t you write a little program to find out whether it really is the smallest?” Perl 6 has features that make that fun to explore, so let’s go to it! Here was my first attempt: my @fourth-powers = (1..*).map(* ** 4) ...^ * > 635318657; my %counts; for @fourth-powers X+ @fourth-powers ->$sum {
%counts{$sum}++; } for %counts.keys.grep({ %counts{$_} > 2 }) -> $sum { say "$sum has { %counts{$sum} / 2 } fourth power sums"; } This works, and is quite fast in Niecza. (It’s much slower in Rakudo for some reason, though it still finishes in about ten seconds.) I love that first line, which constructs a lazy infinite list of fourth powers, then truncates it at 635318657. Then it uses X+ to build all the sums of two fourth powers less than (or equal to) 635318657. The rest of the code is pretty mundane, I fear. I decided to try to use classify to get rid of the explicit loops, resulting in this code: my @fourth-powers = (1..*).map(* ** 4) ...^ * > 635318657; my @results = (@fourth-powers X+ @fourth-powers).classify({$_ }).pairs.grep(*.value > 2);
say @results.map({ $_.key ~ " " ~$_.value.Int / 2 ~ " times" });
Unfortunately, this doesn’t work in Niecza. So now I’ve got some bugs to fix…
Complaining about Mono
July 2, 2012
So, I’ve hit on something extremely annoying working with Niecza on my MacBook Pro. I upgraded to Mono 2.10.9 a few months back. Since then, Glib# has not worked. Even after I uninstalled Mono 2.10.9. It is incredibly frustrating.
Here’s the error I’m getting:
Unhandled exception: System.TypeInitializationException: An exception was thrown by the type initializer for GLib.GType ---> System.DllNotFoundException: glibsharpglue-2
Here’s the call:
constant $GLIB = "glib-sharp, Version=2.12.0.0, Culture=neutral, PublicKeyToken=35e10195dab3c99f"; constant$G_TYPE_STRING = CLR::("GLib.GType,$GLIB").String; Here’s what gacutil says about glib-sharp: glib-sharp, Version=2.12.0.0, Culture=neutral, PublicKeyToken=35e10195dab3c99f If anyone has any suggestions on how to fix this, I’m all ears. Handling Errors in ABC May 24, 2012 When I started thinking I should write yesterday’s post and started testing the Ed Reavy ABC file, I quickly ran into one of the module’s limitations that I thought I should fix before claiming ABC was a useful program. When I started trying the Reavy file, if the ABC file ran into an error in the ABC file, it would frequently just hang. Now, it was relatively easy to track down the erroneous ABC at that point, because the ABC Actions class prints all the title fields it finds as it finds them. You’d watch the list of tune names going by, and if it stopped the list before the tune file was done and just sat there, you knew something was wrong with that tune. That’s okay for a toy program that only I play with, but it’s unacceptable for a serious program other people can use. I had a theory about what caused it: when the grammar hit something it didn’t understand, it started backtracking to try to find a correct way of interpreting the file. Since there was no correct way, that resulted in a painfully long attempt to repeatedly re-interpret the bad data. So I asked how to stop backtracking grammar regexes on #perl6. They helpfully suggested changing the ABC grammar’s regex methods to token, because token suppresses backtracking. I tried just changing one method, and that didn’t help. I tried changing them all, and that failed some tests. I changed one method back, and suddenly all the tests passed, and the grammar no longer hung on ABC file errors! That was a huge improvement, but it was still only marginally helpful at figuring out what the problem in ABC file was. Once I was done with yesterday’s post, I tried to figure out how to get better information about where the parsing problem was. I hit upon a simple method. I added a $.current-tune attribute to the ABC::Actions class. When the action for a tune title fires, it clears this attribute and sets it to the tune title:
method header_field($/) { if$<header_field_name> eq "T" {
$*ERR.say: "Parsing " ~$<header_field_data>;
$.current-tune =$<header_field_data> ~ "\n";
}
make ~$<header_field_name> => ~$<header_field_data>;
}
When the action method for the bar token fires, it adds the string of the bar to $.current-tune. When the action method for line_of_music fires, it adds a newline. And when a fatal error occurs parsing an ABC file, it prints out this attribute, resulting in error messages that look like this: Unhandled exception: Did not match ABC grammar: last tune understood: Silent The Lonely Glen g/2f/2|ed/2B/2 AG/2A/2|B/2d^c/2 dg/2f/2|ed/2B/2 AG/2A/2| B/2>A/2G/2 While this isn’t an exact reproduction of what’s in the tune, it’s close enough to easily show where the problem is: g/2f/2|ed/2B/2 AG/2A/2|B/2d^c/2 dg/2f/2|ed/2B/2 AG/2A/2| B/2>A/2G/2/F/2 eg/2f/2|ed/2B/2 ba/2g/2|g/2f/2e/2^d/2| You just look at where the grammar stopped understanding the file, and the error pops right out if you know ABC: B/2>A/2G/2/F/2. Delete that duration not attached to a note, and the grammar will zip right through the tune. I’m sure there are more sophisticated ways to handle errors in parsing available in Perl 6, because the STD grammar uses them to give you great error messages. But this is a simple technique which is pretty handy. About ABC May 23, 2012 I’ve referenced the ABC module a number of times over the years on this blog, but I don’t think I’ve ever properly explained it. I think it’s particularly relevant with the recurring discussions going on about Perl 6’s “production readiness”, because at this point the abc2ly.pl script in the module is a productive tool I am using on a regular basis. One that would have been harder to write in almost any other programming language, I should add. ABC format is a designed to make it very easy to enter single-line music. It’s designed to be a terse ASCII representation of the music, and it’s absolutely terrific for notating jigs and reels and things like that. Here’s an example ABC hornpipe: X:20 T:Two Weeks To Wait M:4/4 L:1/8 C:Solomon Foster R:Hornpipe K:Edor GF|:E2GB AGFA|(3Bcd ed gedB|A2GB ABGA|BgdB A2BG| E2GB AGFA|(3Bcd ed gedB|A2GB ABGF|E2ED E2GF:| |:E2GB edef|g2fg edBG|A2GB ABGA|BgdB e2ef| g2fg edBG|~A3B AGEG|~A3B AGFG|E2ED E2GF:| A tune considers of a header section with information about the tune and the basic glue to define the tune, followed by a list of notes, barlines, repeat signs, and other symbols. The ABC module’s abc2ly.pl script uses the ABC module’s ABC library to read the file, and outputs the music in the Lilypond format. Lilypond produces very nice output, but the file format is a pain. For instance, the (3Bcd triplet in the above file is translated to this: \times 2/3 { b'8[ cis''8 d''8] }. But the PDF Lilypond outputs from it is beautiful. I’ve been using this tool to handle my sheet music creation needs for about a year now, but in the last six weeks I’ve really stepped things up. I’ve added multi-note stems, chord names, text annotations, ties, slurs, and many other markings I needed, as well as fixing a number of bugs. As a test I threw the Ed Reavy ABC collection at it today. (After deleting the comment lines in the beginning, because I haven’t added them to the parser yet.) In the process I found four mistakes in the ABC file which threw the parser. Once they were fixed, it processed all 127 tunes in about 22 seconds. I haven’t examined the 44 pages of sheet music output by Lilypond closely yet, but it looks like all the problems I see with the output are actually errors (that parse) in the ABC file. In conclusion, while it doesn’t implement the entire ABC standard yet, this is a perfectly usable ABC to Lilypond translator, capable of handling fairly complex single-staff ABC files quickly and accurately. It’s definitely reached the point where it’s more than reasonable for other people to start using it. Optimizing abc2ly.pl May 21, 2012 Shortly after I wrote my last blog post, I started actively working on the ABC module again. First I had to enhance a few features so it could handle some whistle parts for a little musical I was asked to play for. Next I needed some enhancements to handle whistle / fiddle parts for the standard wedding music pieces for an upcoming gig. Finally getting obsessed with ABC updates, I decided to throw the code at a collection of 159 tunes I put together a few years ago. Suddenly I was running into hard limitations of the ABC module. It was always slow, but taking 40 seconds to process my file of three tunes wasn’t really a problem for me. I did add some progress notification messages, and was shocked to learn it wasn’t the parsing of the ABC files that was slow, it was processing the parsed result. But when I got things working well enough it was actually trying to process all 159 tunes, it would run for 15 or 20 minutes and then crash. (This is under Niecza, BTW.) I spent Friday and Saturday trying to optimize bits of code in there that I knew were terrible, with no noticeable improvement. Finally I did a profile run under mono. I found the results didn’t really mean anything to me, but I sent it to sorear++ to see what he thought. And he said, “Why do you need to use .eval in abc2ly.pl?” Well, I quickly checked, and found all the uses were intended to convert strings like "3/4" to Rats. You may say “Why were you doing that? Shouldn’t .Numeric work just as well?” And the answer is yes, it does. (In my defense, I’m pretty sure that didn’t work when I started working on the ABC module.) But now here is the tricky bit. Here’s a simple script to demonstrate the issue: my$a = 0;
for 1..400 {
$a += "1/$_";
}
say $a; On my machine, that takes 1.3 seconds to run under Niecza. Now let’s make a simple change: my$a = 0;
for 1..400 {
$a += "1/$_".eval;
}
say \$a;
This version, with just the .eval added, takes 39.0 seconds to run. Huge difference, eh?
It turned out that even though the evals were happening per measure of music and not per note, they were so slow they were completely swamping the performance of my script. With the evals, processing the wedding.abc file took about 28 seconds. Without them, 3.7 seconds. And remember the big ABC file that would run for 15 minutes and then croak? Without the evals, the entire file processes in 31.8 seconds. That’s easily 30 times faster.
Lesson? Don’t use eval unless you really, really need to.
This advice holds for Rakudo as well: the above fraction-summing scripts took .3 and 8.5 seconds to run, respectively. (Yes, Rakudo is doing a lot better performance-wise on those scripts than Niecza did!)
ABC module now works on Rakudo and Niecza
April 19, 2012
I haven’t had many tuits to work on Perl 6 lately — except this week I needed to have three pages of transposed single-line sheet music ready to go for tonight’s rehearsal. The ABC module (particularly the abc2ly.pl script to convert ABC to Lilypond) was the obvious choice to use. However, to make it work well, I needed to add ties, slurs, multi-measure rests, fermatas, and text messages, as well as fixing the key change and meter change code. While I was at it, I went ahead and got all the tests passing again under Rakudo (version 2012.02-173-gb13c517, the latest won’t build on my Mac) and for the first time, Niecza (latest source from github only, I actually added a small patch in the process).
It feels so good to have this working under both compilers! | 2022-08-12 01:52:35 | {"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": 0.47459661960601807, "perplexity": 5585.552636994869}, "config": {"markdown_headings": false, "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-2022-33/segments/1659882571538.36/warc/CC-MAIN-20220812014923-20220812044923-00759.warc.gz"} |
https://hal.archives-ouvertes.fr/hal-01645611 | # Top-pair production at the LHC through NNLO QCD and NLO EW
Abstract : In this work we present for the first time predictions for top-quark pair differential distributions at the LHC at NNLO QCD accuracy and including EW corrections. For the latter we include not only contributions of $\mathcal{O}\left({\alpha}_s^2\alpha \right)$ , but also those of order $\mathcal{O}\left({\alpha}_s{\alpha}^2\right)$ and $\mathcal{O}\left({\alpha}^3\right)$ . Besides providing phenomenological predictions for all main differential distributions with stable top quarks, we also study the following issues. 1) The effect of the photon PDF on top-pair spectra: we find it to be strongly dependent on the PDF set used — especially for the top p$_{T}$ distribution. 2) The difference between the additive and multiplicative approaches for combining QCD and EW corrections: with our scale choice, we find relatively small differences between the central predictions, but reduced scale dependence within the multiplicative approach. 3) The potential effect from the radiation of heavy bosons on inclusive top-pair spectra: we find it to be, typically, negligible.
Keywords :
Type de document :
Article dans une revue
JHEP, 2017, 10, pp.186. 〈10.1007/JHEP10(2017)186〉
Domaine :
https://hal.archives-ouvertes.fr/hal-01645611
Contributeur : Inspire Hep <>
Soumis le : jeudi 23 novembre 2017 - 10:17:06
Dernière modification le : mercredi 12 décembre 2018 - 07:51:35
### Citation
Michal Czakon, David Heymes, Alexander Mitov, Davide Pagani, Ioannis Tsinikos, et al.. Top-pair production at the LHC through NNLO QCD and NLO EW. JHEP, 2017, 10, pp.186. 〈10.1007/JHEP10(2017)186〉. 〈hal-01645611〉
### Métriques
Consultations de la notice | 2018-12-15 23:37:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 0.7109936475753784, "perplexity": 5052.305678705635}, "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-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00101.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-2-section-2-5-formulas-and-problem-solving-exercise-set-page-149/50 | ## Algebra: A Combined Approach (4th Edition)
Area = ½ x base x height 20 = ½ x 5 x height $\frac{20\times2}{5}$ = height Height = 8 ft | 2018-04-26 13:58:29 | {"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": 0.27466195821762085, "perplexity": 2849.0832034302566}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948214.37/warc/CC-MAIN-20180426125104-20180426145104-00552.warc.gz"} |
http://community.wolfram.com/groups/-/m/t/107839?p_p_auth=Kh1ji1EV | SinglePredictionBands versus MeanPredictionBands from NonlinearModelFit
GROUPS:
Hello, from my statistic lectures I know the term confidence and prediction interval, whereas the confidence interval is smaller than the prediction interval. If I plot the mean and single prediction bands with the same confidence level in Mathematica, than the mean prediction band is smaller than the single prediction band interval. Anyone who can explain me what are the differences between the “SinglePredictionBands” and “MeanPredictionBands” which are standard properties from NonlinearModelFit? Or is it just a different terminology for confidence and prediction interval used by Mathematica? If no it would be useful if someone can tell me which properties I have to apply for the confidence and the prediction interval.Thanks for your support.Markus | 2016-05-26 14:36:55 | {"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": 0.8528496026992798, "perplexity": 561.9194331041945}, "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-2016-22/segments/1464049275981.56/warc/CC-MAIN-20160524002115-00146-ip-10-185-217-139.ec2.internal.warc.gz"} |
https://www.semanticscholar.org/topic/Answer-set-programming/88523 | # Answer set programming
Known as: Answer set program, Answer-set programming, AnsProlog
Answer set programming (ASP) is a form of declarative programming oriented towards difficult (primarily NP-hard) search problems. It is based on the… (More)
Wikipedia
## Papers overview
Semantic Scholar uses AI to extract papers important to this topic.
2016
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• AI Magazine
• 2016
tation and reasoning (KR) paradigm. It has rich highlevel representation languages that allow recursive definitions, aggregates… (More)
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We introduce the first approach to Reactive Answer Set Programming, aiming at reasoning about real-time dynamic systems running… (More)
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• Artif. Intell.
• 2011
The addition of aggregates has been one of the most relevant enhanceme nts to the language of answer set programming (ASP). They… (More)
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• 2011
The motivation and key concepts behind answer set programming---a promising approach to declarative problem solving.
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• Annals of Mathematics and Artificial Intelligence
• 2008
We introduce a knowledge representation language ${\cal AC(C)}$ extending the syntax and semantics of ASP and CR-Prolog, give… (More)
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• KR
• 2008
To compute a function such as a mapping from vertices to colors in the graph coloring problem, current practice in Answer Set… (More)
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We describe a new grounder system for logic programs under answer set semantics, called GrinGo. Our approach combines and extends… (More)
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We present Annotated Answer Set Programming, that extends the expressive power of disjunctive logic programming with annotation… (More)
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2002
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The idea of answer set programming is to represent a given computational problem by a logic program whose answer sets correspond… (More)
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2001
2001
We study the notion of strong equivalence between two Answer Set programs and we show how some particular cases of testing strong… (More)
Is this relevant? | 2019-04-20 14:40:24 | {"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": 0.5818581581115723, "perplexity": 9868.001588941332}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578529839.0/warc/CC-MAIN-20190420140859-20190420162859-00470.warc.gz"} |
https://projecteuclid.org/euclid.aoas/1560758440 | ## The Annals of Applied Statistics
### Nonparametric testing for differences in electricity prices: The case of the Fukushima nuclear accident
Dominik Liebl
#### Abstract
This work is motivated by the problem of testing for differences in the mean electricity prices before and after Germany’s abrupt nuclear phaseout after the nuclear disaster in Fukushima Daiichi, Japan, in mid-March 2011. Taking into account the nature of the data and the auction design of the electricity market, we approach this problem using a Local Linear Kernel (LLK) estimator for the nonparametric mean function of sparse covariate-adjusted functional data. We build upon recent theoretical work on the LLK estimator and propose a two-sample test statistics using a finite sample correction to avoid size distortions. Our nonparametric test results on the price differences point to a Simpson’s paradox explaining an unexpected result recently reported in the literature.
#### Article information
Source
Ann. Appl. Stat., Volume 13, Number 2 (2019), 1128-1146.
Dates
Revised: November 2018
First available in Project Euclid: 17 June 2019
https://projecteuclid.org/euclid.aoas/1560758440
Digital Object Identifier
doi:10.1214/18-AOAS1230
Mathematical Reviews number (MathSciNet)
MR3963565
#### Citation
Liebl, Dominik. Nonparametric testing for differences in electricity prices: The case of the Fukushima nuclear accident. Ann. Appl. Stat. 13 (2019), no. 2, 1128--1146. doi:10.1214/18-AOAS1230. https://projecteuclid.org/euclid.aoas/1560758440
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#### Supplemental materials
• Supplement A: R-codes and data. This supplementary material contains the R codes of the real data application and simulated data which closely resembles the original data set.
• Supplement B: Supplementary paper. This supplementary paper contains the proofs of our theoretical results and a detailed description of the data sources. | 2019-07-17 04:30:04 | {"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": 0.3027556240558624, "perplexity": 9477.261737691233}, "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-2019-30/segments/1563195525046.5/warc/CC-MAIN-20190717041500-20190717063500-00389.warc.gz"} |
http://nrich.maths.org/public/leg.php?code=32&cl=2&cldcmpid=2430 | # Search by Topic
#### Resources tagged with Multiplication & division similar to Shedding Some Light:
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Broad Topics > Calculations and Numerical Methods > Multiplication & division
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Sept03 Sept03 Sept03
##### Stage: 2 Challenge Level:
This number has 903 digits. What is the sum of all 903 digits?
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##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### Mobile Numbers
##### Stage: 1 and 2 Challenge Level:
In this investigation, you are challenged to make mobile phone numbers which are easy to remember. What happens if you make a sequence adding 2 each time?
### Carrying Cards
##### Stage: 2 Challenge Level:
These sixteen children are standing in four lines of four, one behind the other. They are each holding a card with a number on it. Can you work out the missing numbers?
### Month Mania
##### Stage: 1 and 2 Challenge Level:
Can you design a new shape for the twenty-eight squares and arrange the numbers in a logical way? What patterns do you notice?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Function Machines
##### Stage: 2 Challenge Level:
If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
### Next Number
##### Stage: 2 Short Challenge Level:
Find the next number in this pattern: 3, 7, 19, 55 ...
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Back to School
##### Stage: 2 Challenge Level:
Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong?
### Forgot the Numbers
##### Stage: 2 Challenge Level:
On my calculator I divided one whole number by another whole number and got the answer 3.125 If the numbers are both under 50, what are they?
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Exploring Wild & Wonderful Number Patterns
##### Stage: 2 Challenge Level:
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### What's My Weight?
##### Stage: 2 Short Challenge Level:
There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Four Go
##### Stage: 2 Challenge Level:
This challenge is a game for two players. Choose two numbers from the grid and multiply or divide, then mark your answer on the number line. Can you get four in a row before your partner?
### Route Product
##### Stage: 2 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were making cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
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##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
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##### Stage: 2 Challenge Level:
Can you each work out the number on your card? What do you notice? How could you sort the cards?
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### Numbers Numbers Everywhere!
##### Stage: 1 and 2
Bernard Bagnall recommends some primary school problems which use numbers from the environment around us, from clocks to house numbers.
### Ordering Cards
##### Stage: 1 and 2 Challenge Level:
This problem is designed to help children to learn, and to use, the two and three times tables.
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Doplication
##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Oh! Harry!
##### Stage: 2 Challenge Level:
A group of children are using measuring cylinders but they lose the labels. Can you help relabel them?
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-6 to make a number which is divisible by 6 so when the last digit is removed it becomes a 5-figure number divisible by 5, and so on?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Dice and Spinner Numbers
##### Stage: 2 Challenge Level:
If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time?
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
### Shape Times Shape
##### Stage: 2 Challenge Level:
These eleven shapes each stand for a different number. Can you use the multiplication sums to work out what they are? | 2013-12-08 20:34:36 | {"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": 0.29735660552978516, "perplexity": 2491.9778852716695}, "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/1386163809293/warc/CC-MAIN-20131204133009-00063-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://dsp.stackexchange.com/questions/17988/how-to-change-the-start-point-of-resampling-process/19346#19346 | # How to change the start point of resampling process?
I'm trying to resample a digital communication signal in SciPy or Matlab. Suppose the signal was sampled at 3 samples per second, and I want to resample it to 2 samples per second, the thing is I also want to be able to change the new sampling point so the new samples doesn't necessarily start at the same point as the orignal samples.
I'm using the Scipy resample function, but it doesn't have the flexibility to change the sample point, it only allows
The resampled signal starts at the same value as x ...
The Matlab resample function can't do that either.
Is there a way to shift the start point by a small amount, say $\alpha T$ where $T$ is the symbol duration and $0 \le \alpha < 1$ ? I guess what I need is the generic resampling algorithm underneath these resample functions so I can see if there's anything I can tweak.
• Your questions appears to be specific to a SciPy implementation, is that correct? Phrases like "samples per symbol" do not mean much in the broader context of signal processing so it might help to clarify them if you want answers from people without any direct experience of SciPy. Aug 27 '14 at 17:45
• Search for "fractional delay". Aug 27 '14 at 18:22
since @Community resurrected this question, i'll pick up a little where hotpaw left off. i'm not gonna run this through MATLAB or Octave (i can never remember which are columns and which are rows).
let $r$ be the resampling ratio. $r>1$ means that your sample rate has increased and you're making the sequence longer because there are more samples. so if you pad a delay of $\tau$ at the beginning of the input $x[n]$, that should result in a delay of $r \cdot \tau$ at the output $y[n]$ (whether $\tau$ is an integer number of samples or not).
now suppose you pad (using concatenation) the input with $P$ samples of zero before $x[0]$. $P$ is a positive integer. that will result in a pad in the resampled output $y[n]$ of $rP$ samples where $rP$ is not necessarily an integer. but
$$\lfloor rP \rfloor \triangleq \operatorname{floor}(rP)$$
is an integer, and you can strip those samples (which will be mostly zero) off of the beginning of $y[n]$. $\lfloor u \rfloor$ is the most positive integer that is no greater than $u$. this is always the case:
$$\lfloor u \rfloor \le u < \lfloor u \rfloor + 1$$
at least for $u \ge 0$, we might say that $\lfloor u \rfloor$ is the "integer part" of $u$. the "fractional part" of $u$ is whatever is left:
$$0 \le u - \lfloor u \rfloor < 1$$
so, if what we want left is a fractional delay of $\alpha$ samples where $0 \le \alpha < 1$, then we want to choose an integer $P$ to pad the input such that $\alpha$ is the fractional part remaining in the output delay after stripping away the integer part, $\lfloor rP \rfloor$. that is, given the sample rate ratio $r$, and fractional delay $\alpha$, we want to choose $P$ such that:
$$rP = \lfloor rP \rfloor + \alpha$$
or
$$\alpha = rP - \lfloor rP \rfloor \ .$$
after thinking about this a while, the only algorithm i can suggest is to start with $P=1$ and iteratively increment $P$ so that the above equation is close enough to be accurate. that is, given an error constraint on $\alpha$, (call it $0 < \epsilon \ll 1$), increment $P$ until
$$\left|\alpha - \left( rP - \lfloor rP \rfloor \right) \right| < \epsilon$$
pad the input with $P$ samples of zero, and strip the output of $\lfloor rP \rfloor$ samples, and you will be left with $\left( rP - \lfloor rP \rfloor \right)$ samples of delay.
• wow. i can't exactly decode everything i wrote here. Jul 9 '18 at 4:13
If the resampling is not a ratio of small integers, the it might be possible to approximate a fractional delay by added some number leading zero samples before resampling and the removing some number after.
One can also resample to any rate by interpolation (Sinc kernel for high quality) of each new sample, after any needed low pass filtering of course. | 2021-10-25 19:49:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6785528063774109, "perplexity": 401.2802304221482}, "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/1634323587767.18/warc/CC-MAIN-20211025185311-20211025215311-00274.warc.gz"} |
http://precisionpaintingandwallcovering.com/formula-advertising-xizxu/37f16c-d-dominant-7-chord | You can see the four notes of the D seventh chord marked in red color. Scale intervals: 1 - 3 - 5 - b7 Notes in the chord: D# - G - A# - C# 1froxx123DCF#A#. The chord is often abbreviated as D7. 7frxx1132Barre 2 with Finger 1F#DA#C. Guitar Chords » D+7 ... D+7 D dominant seventh, sharp fifth. In Chapter 18 we discussed the various types of seventh chords that one encounters in tonal Western art music. Db7 chord for piano with keyboard diagram. Explanation: The D seventh is a dominant four-note chord. D7 chord for piano with keyboard diagram. 7 augmented chords have a sharpened 5th degreenote, causing the need to be resolved to a different chord, due to the instabiltiy of the sharpened 5th degree note. It is essentially a D chord, with an added flat 7.. 3frxx1234DF#CA#. The below diagrams show you how to play the D# / Eb Dominant 7 chord in various positions on the fretboard with suggested finger positions.. D# / Eb Dominant Seven chord attributes: Interval positions with respect to the D# major scale, notes in the chord and name variations:. One of these types occurs far more frequently than any of the others: the major-minor seventh chord, typically built on scale degree $\hat5$. The dominant 7th is located a whole step down from D which is C. Move it up an octave as you did before. The D7 (D dominant 7) chord contains the notes D, F#, A and C. It is produced by taking the root (1), 3, 5 and b7 of the D Major scale. Applying Dominant 7th Chords To The 12 Bar Blues. To add a bluesy feel to your 12 bar blues structure, you need to play the I, IV and V chords as dominant 7th chords. D Dominant 7 Augmented Chord Charts, Fingering, Voicings. We currently don't have any songs that contain this chord. For … Here’s one amazing guitar trick involving dominant seventh chords that you can use anywhere on the E, A, D strings or on the A, D, G strings. D dominant 7th guitar chord is also written as D7 or Ddom7. Free guitar D Dominant Seventh, D Dominant 7th, D7 chord charts and fingering diagrams. View our D7 guitar chord charts and voicings in Open G tuning with our free guitar chords and chord charts.If you are looking for the D7 chord in other tunings, be sure to scroll to the bottom of the page. The chord is often abbreviated as Db7. Moveable Dominant Seventh Chords on Guitar. Whole Note D rootSharp Note F# 3rdSharp Note A# #5thWhole Note C b7th. If you make a D7 shape on either set of strings, your second finger is the root of a dominant seventh chord shape. View guitar chords chart for D dominant 7th chord along with suggested finger positions. And the good news here, is that by learning just 2 chord shapes, you can play the 12 bar blues using dominant 7th chords, in a whole range of different keys. You can see the four notes of the Db seventh chord marked in red color. Now let’s take this concept to another Major chord to find a dominant 7th chord on the piano. You now have the D dominant 7th Chord. The D7 is a popular open chord and used often in the key of D. D Dominant Seventh Chord Charts for Guitar, Free & Printable. 5frx11234Barre 3 with Finger 1CDA#F#A#. 10frxx2143CDA#F#. Explanation: The D flat seventh is a dominant four-note chord. Learn the D7 Guitar Chord in various positions on the fretboard from Easy to Follow Diagrams. A D Major triad consists of D F# & A. The Solution below shows the D-flat dominant 7th chord in root position, 1st, 2nd, and 3rd inversions, on the piano, treble clef and bass clef.. D-flat dominant 7th chord. Db 7th chord. Let’s try F# Major to create a dominant 7th chord. 19. 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http://physics.aps.org/articles/v2/95 | # Trend: Taking the wraps off cloaking
, The Blackett Laboratory, Department of Physics, Imperial College London, London SW7 2AZ, UK
Published November 16, 2009 | Physics 2, 95 (2009) | DOI: 10.1103/Physics.2.95
## Introduction
We have a touching faith in the ability of our eyes to tell the truth. No other sense has such confidence invested in it, so when our eyes deceive us the result is bewilderment, giving rise to appeals to magic or even the supernatural. This explains the enormous interest aroused by recent work on invisibility and the cloaking of objects from electromagnetic radiation. In this article we review the theories and experiments behind the hype and suggest what devices might realistically be expected in the near future and what is likely to prove impossible.
Hard wired into our brains is the expectation that light travels in straight lines. Mostly this is true, but there are well-known exceptions, such as mirages, which occur when a hot surface heats the air above, reducing its density and hence creating a refractive index gradient immediately above the surface (Fig. 1, top). Such a gradient bends the trajectories of light rays so that an observer misinterprets where the light is coming from. Typically, light from the sky is refracted by the gradient, giving the appearance of water shimmering in the distance—hence a cruel illusion seen by a thirsty traveler in the desert or, more prosaically, the appearance of a wet road on a hot day.
It is the ability of refractive index gradients to bend light that the invisibility engineer exploits. Light is steered around the hidden object by a cloaking device, and then returned to the same straight line trajectory, rather as a skier would make a chicane around a tree (Fig. 1, bottom). The observer’s brain is unaware of the possibility of chicanes and sees only that which is behind the cloak and nothing of the cloak itself or of its contents. The real challenge of cloaking lies in deriving a theoretical prescription for the optical properties of the cloak and, even more challenging, realizing these properties in a material. Transformation optics provides the theoretical background and metamaterials provide the means of achieving the prescribed parameters.
## Transformation optics
It was Michael Faraday who stressed the importance of “lines of force.” He could see magnetic lines of force aligning iron filings placed near his magnets, and for him they represented physical reality. Lines of force are continuous and their density represents the strength of a field. Likewise, electric field lines are also continuous, at least in the absence of electrical charges. In fact, any conserved vector quantity can be represented in this way, and one can add the Poynting vector, representing the flow of electromagnetic energy, to this set. The Poynting vector is merely the mathematical representation of a “ray” of light.
Maxwell’s equations are a mathematical realization of Faraday’s work. They describe the phenomena of classical optics and it has long been known that their form is invariant under coordinate transformations. For example, if the equations are written using a Cartesian system,
$\nabla ×\mathbf{\text{E}}=-\mu \left(\mathbf{\text{r}}\right){\mu }_{0}\phantom{\rule{0}{0ex}}\mathrm{\partial }\mathbf{\text{H}}/\mathrm{\partial }\phantom{\rule{0}{0ex}}t,$
(1)
$\nabla ×\mathbf{\text{H}}=+\epsilon \left(\mathbf{\text{r}}\right){\epsilon }_{0}\phantom{\rule{0}{0ex}}\mathrm{\partial }\mathbf{\text{E}}/\mathrm{\partial }\phantom{\rule{0}{0ex}}t,$
(2)
where $\mu \left(\mathbf{\text{r}}\right)$ and $\epsilon \left(\mathbf{\text{r}}\right)$ are the magnetic permeability and electrical permittivity tensors, respectively, then rewriting the equations using a cylindrical coordinate system changes only the values taken by $\mu$ and $\epsilon$. Andrew Ward and I used this result to adapt computer codes written specifically in a Cartesian system to make computations for optical fibers, which require cylindrical symmetry [1]. This invariance of form is true under any coordinate transformation and can be exploited to generate a whole family of optical devices from a single canonical device. Further references can be found in Ref. [2].
Following in the footsteps of Faraday, we can give physical meaning to a coordinate transformation as follows: imagine starting from a Cartesian system with a given set of electric and magnetic fields, and their associated Poynting vectors. Next, imagine that the coordinates are continuously distorted into a new system. Transformation optics was born of the realization that lines of force are effectively glued to the coordinate system [2]. As the system is distorted it carries with it all the associated fields. Hence to guide the trajectory of a ray of light, only a distortion in the underlying coordinate system is needed, automatically taking with it the light ray. Knowledge of the transformation in turn provides the values of $\mu$ and $\epsilon$ required to steer the light in this way.
To give a flavor of how the scheme operates, imagine the simplest possible distortion of space: a section of the $x$ axis is compressed as shown in Fig. 2. Now probe the compressed region with two rays in order to find the values of $\epsilon \left(\mathbf{\text{r}}\right)$ and $\mu \left(\mathbf{\text{r}}\right)$ that would give rise to the ray trajectory shown, and note the following: $\epsilon \left(\mathbf{\text{r}}\right)$, $\mu \left(\mathbf{\text{r}}\right)$ are tensors because the $x$ axis is singled out for compression; in the uncompressed regions there is no change, so $\epsilon \left(\mathbf{\text{r}}\right)=\mu \left(\mathbf{\text{r}}\right)=1$ in these regions; $\epsilon \left(\mathbf{\text{r}}\right)$ and $\mu \left(\mathbf{\text{r}}\right)$ appear on the same footing because of the symmetry between electric and magnetic fields. It follows from the last assertion that $\epsilon \left(\mathbf{\text{r}}\right)=\mu \left(\mathbf{\text{r}}\right)$ .
Next consider a ray propagating parallel to the $x$ axis. In order to arrive at the far side of the compressed region with the same phase as in the uncompressed system, the condition $k\prime m\phantom{\rule{0}{0ex}}d={k}_{0}\phantom{\rule{0}{0ex}}d$ must hold, where ${k}_{0}$ is the free-space wave vector, $k\prime$ is the wave vector in the compressed region, $m$ is the compression factor, and $d$ is the original thickness of the layer. Since $k\prime ={k}_{0}\sqrt{{\epsilon }_{y}\phantom{\rule{0}{0ex}}{\mu }_{y}}$, where ${\epsilon }_{y}$ and ${\mu }_{y}$ are the components of the respective tensors perpendicular to the $x$ axis, then it follows that
${\epsilon }_{y}={\mu }_{y}={m}^{-1}.$
(3)
On the other hand, rays propagating perpendicular to the $x$ axis travel through uncompressed space, and therefore their wave vector, $k\prime \prime$, must take the free-space value if the correct phase evolution is to be followed. In this case,
$k\prime \prime ={k}_{0}\sqrt{{\epsilon }_{y}\phantom{\rule{0}{0ex}}{\mu }_{x}}={k}_{0}\sqrt{{\epsilon }_{x}\phantom{\rule{0}{0ex}}{\mu }_{y}}={k}_{0}$
(4)
and therefore we have
${\epsilon }_{x}={\mu }_{x}=m.$
(5)
In addition, because $\epsilon \left(\mathbf{\text{r}}\right)=\mu \left(\mathbf{\text{r}}\right)$, the compressed layer is impedance matched and does not reflect.
The effect of distorting a coordinate system can be expressed very simply: along a direction of compression both $\epsilon$ and $\mu$ are decreased by the compression factor; in the perpendicular directions $\epsilon$ and $\mu$ are increased by the inverse of the compression factor. For a general compression, the formula is applied successively along each of the three axes.
The above gives an intuitive version of our scheme. A more formal derivation was presented by Ward and Pendry [1] and an updated version by Schurig et al. [3], using modern notation [4]. We follow the latter version here. If the distorted system is described by a coordinate transform $x{\prime }^{j\prime }\left({x}^{j}\right)$, then defining
${\mathrm{\Lambda }}_{j}^{j\prime }=\frac{\mathrm{\partial }\phantom{\rule{0}{0ex}}{x}^{j\prime }}{\mathrm{\partial }\phantom{\rule{0}{0ex}}{x}^{j}}$
(6)
so that in the new coordinate system we must use modified values of the permittivity and permeability to ensure that Maxwell’s equations are satisfied:
$\epsilon {\prime }^{i\prime j\prime }=\left[\text{det}\left(\mathrm{\Lambda }\right){\right]}^{-1}{\mathrm{\Lambda }}_{i}^{i\prime }\phantom{\rule{0}{0ex}}{\mathrm{\Lambda }}_{j}^{j\prime }\phantom{\rule{0}{0ex}}{\epsilon }^{i\phantom{\rule{0}{0ex}}j},$
(7)
$\mu {\prime }^{i\prime j\prime }=\left[\text{det}\left(\mathrm{\Lambda }\right){\right]}^{-1}{\mathrm{\Lambda }}_{i}^{i\prime }\phantom{\rule{0}{0ex}}{\mathrm{\Lambda }}_{j}^{j\prime }\phantom{\rule{0}{0ex}}{\mu }^{i\phantom{\rule{0}{0ex}}j}.$
(8)
The theory is exact as far as Maxwell’s equations are concerned and applies as much to steering magnetic and electric fields as it does to steering rays of light. Further details of the theory can be found in Refs. [5, 6].
## Invisibility
Figure 3 (top left) shows a ray of light travelling in free space. We wish to hide the contents of a sphere radius ${R}_{1}$ by directing the rays around this region, but requiring that any distortion of trajectories is confined within a larger sphere radius ${R}_{2}$ (Fig. 3, top right). In this way an external observer would be aware neither of the presence of the cloak nor its contents. The illusion of empty space has been created.
Exactly how to compress the coordinates into the cloaking region is a matter of choice. Here is a simple transformation that does the job:
$r\prime =\left({R}_{2}-{R}_{1}\right)r/{R}_{2}+{R}_{1},\theta \prime =\theta ,\varphi \prime =\varphi ,$
(9)
giving rise to the cloaking parameters
$\epsilon {\prime }^{r\prime }=\mu {\prime }^{r\prime }=\left[{R}_{2}/\left({R}_{2}-{R}_{1}\right)\right]\left[\left(r\prime -{R}_{1}\right)/r\prime {\right]}^{2},{R}_{1}>r\prime >{R}_{2},$
(10)
$\epsilon {\prime }^{\theta \prime }=\mu {\prime }^{\theta \prime }=\epsilon {\prime }^{\varphi \prime }=\mu {\prime }^{\varphi \prime }={R}_{2}/\left({R}_{2}-{R}_{1}\right).$
(11)
Note that the compression axis lies along the radius and along the perpendicular directions, i.e., along the angular coordinates the values of $\epsilon$ and $\mu$ are increased by the inverse of the compression factor, whereas along the radial direction the values are reduced. This prescription is an exact solution to the cloaking problem at the level of Maxwell’s equations [7]. Both the near field and the far field are cloaked and the theory can even be applied to dc fields [8]. An alternative cloaking scheme was published simultaneously [9], exact to the level of the ray approximation. Ray schemes demand only the manipulation of the refractive index and therefore have advantages in simplicity of construction. Much earlier work [10] discussed the cloaking of objects to electrical currents flowing in a medium.
A simplified version of this cloak was demonstrated for microwaves in 2006 [11] using the concept of metamaterials. These new materials owe their properties as much to their subwavelength internal structure as to the materials from which they are manufactured. We shall examine their properties in greater detail later on.
There followed a huge surge of interest both from scientists and engineers implementing their own versions of the technology, and from the popular press where it has inspired many articles on the subject, raising the profile of electromagnetic phenomena to levels not previously seen. The first cloaking papers based on transformation optics appeared in 2006 and have already been cited more that 500 times.
There is more than one route to cloaking. An alternative approach, shown in Fig. 3, (bottom panel) is to recognize the properties of the cloaked object and construct the cloak to cancel any scattering so that the cloak and object together are invisible. [12, 13]. Some remarkable effects have been shown to be possible with such schemes, at least on theoretical grounds. For example, a recent paper [14] proposes a way to hide an object without having to wrap the cloak around it, and it is even possible to replace a given object with another [15], thus increasing the possibilities for deception even further! These cancellation schemes always involve materials with either $\epsilon <0,\mu <0$, or with both quantities negative. Negative values imply in turn that there is a strong frequency dependence and such cloaks work at highly specific frequencies, deriving their properties from resonances within the cloak.
One interpretation of a cloak is that it makes the cloaked object appear to be extremely small and therefore invisible, so that the spherical cloak shown at top right in Fig. 3 would shrink the object to a sphere of vanishing radius. Although the sphere would have an extremely large refractive index, it would still be invisible in the limit. This picture immediately suggests two other possibilities: shrinking the object to a cylinder of vanishing radius or down to the plane of vanishing thickness. A very thin cylinder is invisible, even if conducting, because a thin wire is highly inductive and therefore can carry no current, so the cylindrical scheme is a possible route to perfect invisibility and in fact has been the model most commonly implemented. The third possibility of reducing the object to a plane does not produce an invisible object since a highly refracting plane of material reflects incident light. However, if the environment of the cloak already contains a reflecting surface, the squashed object is effectively hidden if it appears to lie on this existing reflecting surface. Hence the term “carpet cloak.” This latter cloak, although far less general in application than the other two, is easy to build because the material parameters required are no longer singular, as is the case for the sphere and the cylinder.
Although theory gives a perfect prescription for a cloak, implementation is a different matter. The metamaterials employed are in general anisotropic and the difficulty of manufacturing them increases severely at higher frequencies. A spherical cloak is singular on its surface, i.e., the values of $\epsilon$ and $\mu$ either $\to \mathrm{\infty }$ or $\to 0$, and therefore any practical implementation will always involve a degree of approximation.
## Metamaterials: are they up to the task?
Many wonderful schemes for cloaking have been dreamt of, but if in the last analysis the materials needed for their manufacture cannot be found, they remain theoretical dreams. Fortunately, some of the cloaking proposals are relatively easy to implement and the range of materials available to us has been hugely extended by the concept of a metamaterial.
The idea behind metamaterials is that electromagnetic properties can be altered by changing the physical structure of an existing material. A simple example would be to create a series of voids in a dielectric. If the voids are much smaller than the relevant wavelength, incident light will see only an average response and a reduced effective permittivity. More complex structures, such as tiny metallic resonators, can produce more exotic effects, including negative refractive indices. Provided that the structure remains much smaller than the wavelength, properties can be interpreted in terms of an effective $\epsilon$ and $\mu$. As it happens, nature has not been generous with the electromagnetic properties of materials: only a limited range is on offer and some properties, such as a negative refraction, are entirely absent in naturally occurring materials. In fact, it was the ability of metamaterials to produce negative refractive indices that first drove them to prominence. Since that time, the concept of function through structure has been deployed to implement many theoretical schemes previously thought to be beyond reach. Cloaking demands, in particular, materials whose properties are continuously graded throughout the cloak and are in general anisotropic. Metamaterials are ideally adapted to these demands.
The first implementation of a cloak by researchers at Duke University employed resonant metallic elements of the “split ring” [16, 11] design, illustrated in Fig. 4 (top left). This was convenient because the structure can readily be tuned to give a wide range of responses, but carries the penalty that the response depends strongly on frequency and hence this cloak operates only on the design frequency. Also, approximations were made. An ideal cloak requires $\epsilon$ and $\mu$ to be equal at every point. This ensures that not only is the refractive index graded so as to steer rays of light around the hidden object, but also the material is impedance matched to vacuum, eliminating any spurious reflections at the interface. If impedance matching is neglected, and reflections are minimized by smoothly grading the cloak at the vacuum interface, then the refractive index variations can be entirely accommodated by adjusting either $\epsilon$ or $\mu$
This approximation is particularly useful at higher frequencies, where magnetic metamaterials are hard to engineer, and has also been exploited [17] to propose an optical version of the cloak. The bottom left panel of Fig. 4 shows a design incorporating metallic wires in a dielectric host to create a radial gradient of the permittivity. Simulations show that the design should be effective in cloaking illumination with transverse-magnetic polarization.
Subsequent designs have been produced incorporating nonresonant elements and have a wide band of operating frequencies. The so-called “carpet cloak” described above is particularly well suited to nonresonant designs because it does not demand extreme values of the parameters. It has recently been implemented at microwave frequencies by the Duke team and shown to operate over a broad band.
An exciting development in the past year has been cloaking at near ir frequencies. The “carpet cloak” template has been used by two groups, one in Berkeley and the other in Cornell, to create a cloak using silicon and silica in a 2D waveguide [18, 19, 20]. Figure 4 (right panels) shows the device manufactured by the Berkeley group. To produce a cloak, the density of silicon, and therefore the refractive index, is varied by drilling holes. Experiments on the structure show that light incident from any angle reflects from the cloaked bump as though from a flat reflecting surface. Cloaking is effective over a broad band of frequencies.
Another inventive approach to the problem of optical cloaking, from the Purdue group [21], employs a 2D waveguide of varying height. This has the effect of varying the cutoff frequency and hence the effective refractive index for light propagating in the guide. By tuning the height variation to that demanded by cloaking theory, light can be forced to avoid a given region of the 2D guide, emerging from the cloaking region as if the guide were of uniform cross section. Any object could then be hidden in the avoided region.
## Beyond electromagnetism
The mainstream work on cloaking has been inspired by the beautiful symmetry of Maxwell’s equations under coordinate transformation, but the successes in realizing an electromagnetic cloak have prompted many to ask whether the methods of transformation optics can be applied to other systems to generate cloaks for sound waves, water waves, or even for earthquakes [22]. Work on acoustics is the most fully formed.
Initial investigation of sound propagation in elastic media [23] showed that, in general, coordinate transformations do not have such elegant consequences as for Maxwell’s equations: the form of the equations changes and new terms appear corresponding to unphysical forces. But restricting our attention to fluids where only longitudinal modes are present removes these difficulties [24], and Chen and Chan [25] report that for a fluid, a coordinate transformation can be represented as a change of density, which becomes a tensor, and a modulus, which remains a scalar. This has led to several papers reporting designs for acoustic cloaking.
Daniel Torrent and José Sánchez-Dehesa [26] have proposed an ingenious way to realize these theoretical schemes. The problem lies in producing anisotropic acoustic materials. This problem is addressed by considering a structure comprising layers of isotropic materials, the anisotropy following from the layering. Figure 5 illustrates the scheme, which was tested in a computer simulation. Clearly, the device is operating extremely efficiently. Torrent and Sánchez-Dehesa had specifically in mind that their cloak should be feasible to construct. Although, as yet, an actual device has not appeared, given the interest in acoustic cloaking from many quarters, it cannot be long before a practical implementation follows.
## Summary
Design and construction of a cloak is in the nature of a grand challenge for optics. On the theoretical side, transformation optics provides a complete prescription for a cloak, but practical implementation requires mastery of metamaterial construction. At rf frequencies, progress has been impressive, with several demonstrations of a cloak, and most recently, optical cloaks have been constructed, albeit of restricted capability. In the future we can expect further progress at all frequencies, with practical applications emerging first at rf. However, it has to be said that for the foreseeable future, cloaks will be substantial rigid devices, and in that respect, “cloak,” with its implication of flexibility, is a misnomer.
The challenge has also extended to acoustic systems where there is a host of applications awaiting viable cloaking technology. Although applications are, for the moment, limited to fluids such as water or air, this is where some of the most important applications are to be found.
Finally, as with many grand challenges, it will probably be the case that the technology being developed will have many more applications than cloaking. The ability to design and build a system that exerts detailed control over the flow of radiation is a powerful tool that is bound to find many applications, though perhaps not all of quite such a sensational nature as a cloak.
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### About the Author: John Pendry
John Pendry obtained his Ph.D. in 1969 from Cambridge University, UK, where, apart from a year spent at AT&T Bell Laboratories, he remained until 1975. There followed six years at the Daresbury Laboratory as head of the theoretical group. Since 1981 he has worked at the Blackett Laboratory, Imperial College London, where he has served as Dean, Head of the Physics Department, and Principal for Physical Sciences. His research interests are broad, originally centering on condensed matter theory but now extending into optics. He has worked extensively on electronic and structural properties of surfaces, transport in disordered systems, and in the past ten years has developed the theory behind metamaterials, negative refraction, and cloaking.(Author photograph: Imperial College London/Mike Finn-Kelcey) | 2014-07-23 10:05:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 60, "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": 0.6725627779960632, "perplexity": 855.3715271274599}, "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/1405997877693.48/warc/CC-MAIN-20140722025757-00094-ip-10-33-131-23.ec2.internal.warc.gz"} |
https://www.semanticscholar.org/paper/The-analysis-and-development-of-controlled-3%CF%86-AC-DC-Saleh-Rahman/2eec6c7bd7f94b19e6f9734f09c44ca427dc679c | # The analysis and development of controlled 3φ wavelet modulated AC-DC converter
@article{Saleh2012TheAA,
title={The analysis and development of controlled 3$\phi$ wavelet modulated AC-DC converter},
author={Saleh A. M. Saleh and M. Azizur Rahman},
journal={2012 IEEE International Conference on Power Electronics, Drives and Energy Systems (PEDES)},
year={2012},
pages={1-6}
}
• Published 1 December 2012
• Engineering
• 2012 IEEE International Conference on Power Electronics, Drives and Energy Systems (PEDES)
This paper presents the analysis, development and performance testing of a new method for controlling three phase wavelet modulated ac-dc converters. The wavelet modulation is constructed as a non-dyadic type multiresolution analysis (MRA), which has an analysis part to create three sets of groups of nonuniform recurrent samples from a reference dc signal. The constructed non-dyadic MRA has a synthesis part that generates switching signals for operating the ac-dc converter to synthesize the…
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Performance testing of a single-phase voltage-source wavelet modulated AC-DC converter
• Engineering
2009 Canadian Conference on Electrical and Computer Engineering
• 2009
The development and performance testing of a new switching strategy for single-phase (1φ) voltage-source (VS) AC-DC converters is presented and the output of such an AC- DC converter is found to contain low harmonic contents along with a unity input power factor.
Analysis and Development of Wavelet Modulation for Three-Phase Voltage-Source Inverters
• Engineering
IEEE Transactions on Industrial Electronics
• 2011
Simulation and experimental test results show that the proposed wavelet modulation technique is able to switch a three-phase VS six-pulse inverter to produce outputs with significantly improved fundamental components and low harmonic contents.
Experimental Performances of the Single-Phase Wavelet-Modulated Inverter
• Engineering
IEEE Transactions on Power Electronics
• 2011
This paper presents the real-time implementation and experimental performances of the wavelet-modulation technique for single-phase voltage-source (VS) inverters. The wavelet-modulation technique is
Development and Testing of a New Controlled Wavelet-Modulated Inverter for IPM Motor Drives
• Engineering
IEEE Transactions on Industry Applications
• 2010
This paper presents the development, implementation, and performance testing of a novel resolution-level vector controller (RLVC) for a wavelet-modulated (WM) inverter-fed interior permanent-magnet
Analysis of delta PWM static AC-DC converters
• Engineering
• 1995
An analysis and implementation of delta modulation (DM) technique in the control of AC-DC converters are presented in this paper. The DM technique offers advantages of easy implementation, continuous
A review of three-phase improved power quality AC-DC converters
• Engineering
IEEE Transactions on Industrial Electronics
• 2004
This paper presents an exhaustive review of three-phase improved power quality AC-DC converters configurations, control strategies, selection of components, comparative factors, recent trends, their suitability, and selection for specific applications.
A Modified Harmonics Reduction Technique for a Three-Phase Controlled Converter
• A. Eltamaly
• Engineering
IEEE Transactions on Industrial Electronics
• 2008
The improved injection current technique is introduced to reduce line current harmonics and the simulation and experimental results prove the superiority of this technique in mitigating the requirements for harmonics standards.
New space-vector control strategies for three-phase step-up/down AC/DC converter
• Engineering
IEEE Trans. Ind. Electron.
• 2000
It is seen that not only the control of the six switches is simplified but also the deadtime circuit for avoiding short circuit of the same arm can be eliminated for the proposed converter.
A High- Voltage AC/DC Resonant Converter Based on PRC with Single Capacitor as an Output Filter
• Engineering
2009 IEEE Industry Applications Society Annual Meeting
• 2009
The parallel resonant converter (PRC) with a single-capacitor output filter can be controlled by the switching frequency and/or duty cycle, which makes zero voltage switching feasible in the power switches. | 2022-06-30 05:08:17 | {"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": 0.2844047248363495, "perplexity": 10281.097259706301}, "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-2022-27/segments/1656103661137.41/warc/CC-MAIN-20220630031950-20220630061950-00671.warc.gz"} |
https://smexyyweby.wordpress.com/category/entertainement/ | # SmexyyWeby
Scripting for Fun and Passion
## Calculate Age from Shoe Size m//
Find Age from Shoe Size
I saw the above image floating around in my facebook feed today and was quite amused at how people believe in such plain simple mathematical illusions. So as I was a bit free in office and found this problem to be pretty easy. I found the simple equation and the detailed explanation of the same and posted it as a comment on the post.
My Comment went like this:
Won’t work for the following cases:
1. Your shoe size is greater than 9
2. Your age is greater than 99
And yes it will only work in the year 2012 and not any other. For other years simply use the (birthYear- 1000) instead of 1012.
Also it is not just true for your shoe size but also for the number of nose, eyes, ears you have. Also it works for the current hour number if its less than 10. If its more than 10 then simply use the last 2 digits of the solution. Try it with your cloth sizes as well, just use the last 2 digits. Simple MATH =)
Also here is the full calculation:
Let shoe size be x
Then final number is ${(5*(x)+50)*20 + 1012 - birthYear}$
${= (100*x) + 2012 - birthYear}$
${= X00 + (Your Age in 2012)}$
PS: X00 means put 2 zeroes at the end of X.
Now since your age is mostly going to be a 2 digit number,
Hence, ${X00 + (Your Age in 2012) = A number with last 2 digits equal to your age }$.
Hence proved.
I hope people are more interested in enjoying the math behind these posts than just being amazed by the pointless magic. m//
Hoping to get some likes for my comment =D
## Gurjati Amplifier – English Translation
Well recently there has been a Song by Imran Khan called Amplifier which has been making lots of waves all over the country. I have been loving that song and recently I got the Gujarati version this song. Which was pretty hilarious and also very entertaining .
You can listen to the song below:
So I was really liking this song and was able to get a few of its lyrics like Pani Puri wali etc. But I wanted a full english Translation of this so I asked one of my friends Rushab (yes he is Gujju) to do it for me. What he cam up with was super fun and gives the full essence of the song.
So here is the full Translation as given to me:
My uncle brought a new car for me.
It’s Skoda, and i like it’s glasses so much..
For this, I distributed sweets in whole of the town
Now it’s funn.(Have chalo jalsa)
I managed for a special no. for the number plate.
And i,ve got 007, the one of James Bond.
Come sit in my car, o girl
Then we will go, eat cake, in bakery,
(Buffer tu meri , meri; main tera amplifier) ……x2
My car is calling you, o baby,
And the leather seats are so good, u see;
Let’s go and have some fun, o my girl
In this scorching heat, your sari’s getting wet..
You will get tanned standing in this sun..
My car is the one with Air Conditioner; O My Dear,
Come sit in my car, o girl
Then we will go, eat cake, in bakery,
(Buffer tu meri , meri; main tera amplifier) ……x2
Girl, i so much like you,
I own a panipuri stall,
Come on i will get you cheese butter wali panipuri,
and dahipuri for me…
Come sit in my car, o girl
Then we will go, eat cake, in bakery,
(Buffer tu meri , meri; main tera amplifier) ……x4
I hope you all enjoy it =)
Share it with your friends. Music has no bounds and neither has fun.
Do post in your comments if you have a better translation =)
Listen to the Original Song here:
Lyrics Credits: Rushabh Sanghvi
Original Song: Amplifier – Imran Khan
## Word Press Spell Checker and the American Influence
British English or American English? Carrying the colonial legacy of British, we mostly speak British English but there are more speakers of American English and over time our English is getting mixed up. T’s are disappearing in interested and extra t’s are appearing in party, color is winning over colour, defense over defence, center over centre, dialog over dialogue…the list is endless. The red lines below all these British spellings in the WP editor just show that WordPress is an American product. Variation in pronunciation doesn’t help the matter. Anyway, to keep things simple, we should stick to one form of English.
## I was reading the blog of my friend today when I came across this really interesting post about some basic life facts. I loved it so much that I am copying it as it is here on my own blog as well hope you all with get the morals of all the fables.
Lesson 1
An eagle was sitting on a tree resting, doing nothing. A small rabbit saw the eagle and asked him,
“Can I also sit like you and do nothing?”
The eagle answered: “Sure , why not.”
So, the rabbit sat on the ground below the eagle and rested. All of a
sudden, a fox appeared, jumped on the rabbit and ate it.
Moral of the story:
To be sitting and doing nothing, you must be sitting very, very high up.
Lesson 2
A turkey was chatting with a bull.
“I would love to be able to get to the top of that tree’ sighed the turkey, ‘but I haven’t got the energy.”
“Well, why don’t you nibble on some of my droppings?’ replied the bull. They’re packed with nutrients.”
The turkey pecked at a lump of dung, and found it actually gave him enough strength to reach the lowest branch of the tree.
The next day, after eating some more dung, he reached the second branch.
Finally after a fourth night, the turkey was proudly perched at the top of the tree.
He was promptly spotted by a farmer, who shot him out of the tree.
Moral of the story:
Bullshit might get you to the top, but it won’t keep you there.
Lesson 3
A sales rep, an administration clerk, and the manager are walking to lunch when they find an antique oil lamp.
They rub it and a Genie comes out.
The Genie says, ‘I’ll give each of you just one wish.’
‘Me first! Me first!’ says the admin clerk. ‘I want to be in the Bahamas , driving a speedboat, without a care in the world.’
Puff! She’s gone.
‘Me next! Me next!’ says the sales rep.. ‘I want to be in Hawaii , relaxing on the beach with my personal masseuse, an endless supply of Pina Coladas and the love of my life.’
Puff! He’s gone.
‘OK, you’re up,’ the Genie says to the manager…
The manager says, ‘I want those two back in the office after
lunch..’
Moral of the story:
Always let your boss have the first say.
Lesson 4
A man is getting into the shower just as his wife is finishing up her shower, when the doorbell rings.
The wife quickly wraps herself in a towel and runs downstairs.
When she opens the door, there stands Bob, the next-door neighbour.
Before she says a word, Bob says, ‘I’ll give you \$800 to drop that towel.’
After thinking for a moment, the woman drops her towel and stands naked in front of Bob, after a few seconds, Bob hands her \$800 and leaves.
The woman wraps back up in the towel and goes back upstairs.
When she gets to the bathroom, her husband asks, ‘Who was that?’
‘It was Bob the next door neighbour,’ she replies.
‘Great,’ the husband says, ‘did he say anything about the \$800 he owes me?’
Moral of the story:
If you share critical information pertaining to credit and risk with your shareholders in time, you may be in a position to prevent avoidable exposure.
Lesson 5
A little bird was flying south for the winter. It was so cold the bird froze and fell to the ground into a large field. While he was lying there, a cow came by and dropped some dung on him..
As the frozen bird lay there in the pile of cow dung, he began to realize how warm he was.
The dung was actually thawing him out!
He lay there all warm and happy, and soon began to sing for joy.
A passing cat heard the bird singing and came to investigate.
Following the sound, the cat discovered the bird under the pile of cow dung, and promptly dug him out and ate him…
Morals of the story:
(1) Not everyone who shits on you is your enemy.
(2) Not everyone who gets you out of shit is your friend.
(3) And when you’re in deep shit, it’s best to keep your mouth shut!
End of 5 minute course! And excuse the language used. Had they been toned down, they would have lost their tang and humour.
PS-
Courtesy:Iuliana Gelijk.
## History Mystery…
I am not very fond of forwarded emails .. but I usually give time to read the ones which have a catchy subject line .. other that Plz read .. blah blah ..
Today I was just browsing one of my school friend’s album on orkut when my google talk email notifier poped up with this notification that I have got a mail subjected ..
# is this coincidence????
getting bored in my office I just decided to open the mail thinking of something interesting .. and in the second line only I knew I wasn’t made for this one … I was regarding History …. my long lost fear .. and that too about American History … which I haven’t been interested in much apart from the world war thing … Anyways .. knowing that I was bored .. I just decided to look through the mail reading some interesting facts about John F. Kennedy and Abraham Lincoln .. and after every set of facts I was left amazed at the compilers imagination and the perseverance that he had put in looking and compiling these facts … It was amazing in the end .. I never knew that history can really be so interesting.
Going through the mail I got the shock of my life when it related the some really interesting happenings in recent past with a 20 USD note .. it was amazing …
Well not without saying much I am presenting to you the exact copy of that email to you … without any modification .. I hope you will enjoy reading it .. and would really get a feel of history and mystery while going through it .
Feel free to comment or add some more interesting events like this to the post .. 🙂
————————————————————————————————-
This is pretty interesting but I wonder ———-anyway enjoy !!!!!!!!
History Mystery
Have a history teacher explain this—– if they can
Abraham Lincoln was elected to Congress in 1846.
John F. Kennedy was elected to Congress in 1946.
Abraham Lincoln was elected President in 1860.
John F. Kennedy was elected President in 1960.
Both were particularly concerned with civil rights.
Both wives lost their children while living in the White House.
Both Presidents were shot on a Friday.
Both Presidents were shot in the head.
Now it gets really weird.
Lincoln ‘s secretary was named Kennedy.
Kennedy’s Secretary was named Lincoln .
Both were assassinated by Southerners.
Both were succeeded by Southerners named Johnson.
Andrew Johnson, who succeeded Lincoln , was born in 1808.
Lyndon Johnson, who succeeded Kennedy, was born in 1908.
John Wilkes Booth, who assassinated Lincoln , was born in 1839.
Lee Harvey Oswald, who assassinated Kennedy, was born in 1939.
Both assassins were known by their three names.
Both names are composed of fifteen letters.
Now hang on to your seat.
Lincoln was shot at the theater named ‘Ford.’
Kennedy was shot in a car called ‘ Lincoln ‘ made by ‘Ford.’
Lincoln was shot in a theater and his assassin ran and hid in a warehouse.
Kennedy was shot from a warehouse and his assassin ran and hid in a theater.
Booth and Oswald were assassinated before their trials.
And here’s the kicker…
A week before Lincoln was shot, he was in Monroe , Maryland
A week before Kennedy was shot, he was with Marilyn Monroe.
WHO FIGURED THIS OUT? INCREDIBLE 1) Fold a NEW PINK \$20 bill in half… 2) Fold again, taking care to fold it exactly as below 3) Fold the other end, exactly as before H i n t: P e n t a g o n 4) Now, simply turn it over… What a coincidence! A simple geometric fold creates a catastrophic premonition printed on all \$20 bills!!! COINCIDENCE? YOU DECIDE As if that wasn’t enough .. here is what you’ve seen… Firstly The Pentagon on fire… Then The Twin Towers. … And now .. look at this! TRIPLE COINCIDENCE ON A SIMPLE \$20 BILL
Disaster (Pentagon)
Disaster ( Twin Towers )
Disaster (Osama)???
It gets even better 9 + 11 = \$20!
Creepy huh? Send this to as many people as you can, cause:
Hey, this is one history lesson most people probably will | 2018-01-21 16:43:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "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": 0.2301090806722641, "perplexity": 4203.198326561696}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084890795.64/warc/CC-MAIN-20180121155718-20180121175718-00276.warc.gz"} |
https://web2.0calc.com/questions/a-truck-with-a-16-inch-radius-wheels-is-driven-at-77-feet-per-second-find-the-measure-of-the-angle-through-which-a-point-on-the-outside-of | +0
# a truck with a 16-inch radius wheels is driven at 77 feet per second. find the measure of the angle through which a point on the outside of
0
1339
1
a truck with a 16-inch radius wheels is driven at 77 feet per second. find the measure of the angle through which a point on the outside of the wheel travels each second.
physics
Guest Mar 6, 2015
#1
+92624
+5
a truck with a 16-inch radius wheels is driven at 77 feet per second. find the measure of the angle through which a point on the outside of the wheel travels each second.
one revolution = 2pi*r = 32pi inches
77 feet = 77*12 = 924"
$$\\\frac{924}{32\pi}\approx 9.19 \;\;rev/sec\\\\\\ \frac{924}{32\pi}\times 2\pi=57.75\;\; radians/sec$$
Melody Mar 7, 2015
#1
+92624
+5
a truck with a 16-inch radius wheels is driven at 77 feet per second. find the measure of the angle through which a point on the outside of the wheel travels each second.
one revolution = 2pi*r = 32pi inches
77 feet = 77*12 = 924"
$$\\\frac{924}{32\pi}\approx 9.19 \;\;rev/sec\\\\\\ \frac{924}{32\pi}\times 2\pi=57.75\;\; radians/sec$$
Melody Mar 7, 2015 | 2018-06-21 08:21:31 | {"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": 0.7780443429946899, "perplexity": 953.4294035746408}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267864110.40/warc/CC-MAIN-20180621075105-20180621095105-00291.warc.gz"} |
https://www.physicsforums.com/threads/smith-robinson-jones.378072/ | Smith, Robinson, Jones
1. Feb 13, 2010
On a train, Smith, Robinson, and Jones are the fireman, brakeman, and the engineer, but NOT respectively. Also aboard the train are three businessmen who have the same names: a Mr. Smith, a Mr. Robinson, and a Mr. Jones.
Mr. Robinson Lives in Detroit.
The brakeman lives exactly halfway between Chicago and Detroit
Mr. Jones earns exactly $20,000 per year. The brakeman's nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman. Smith beats the fireman in billiards. The passenger whose name is the same as the brakeman's lives in Chicago. Who is the Engineer? 2. Feb 14, 2010 ƒ(x) Smith T:5 On second thought...is there enough info? I need clarification, are Smith, Robinson, and Jones passengers, or are they running the train? Last edited: Feb 14, 2010 3. Feb 14, 2010 Sartradjm Smith, Robinson, and Jones are running the train, Mr. Smith, Mr. Robinson, and Mr. Jones are passengers. And yes, there is enough info. 4. Feb 14, 2010 ƒ(x) In that case, it's the person I mentioned 5. Feb 15, 2010 Sartradjm You are right. 6. Feb 15, 2010 HawkLogic Note that "Smith beats the fireman in billiards." is redundant. 7. Feb 17, 2010 Sartradjm I know what you meant. And you are right. Instead of "but NOT respectively" should be "but NOT necessarily respectively". 8. Feb 18, 2010 davee123 I remember this problem from years ago, and the thing that always bothered me was the salary bit. Technically speaking, it doesn't say that the brakeman earns an annual salary-- it just says he earns 1/3 of one of the passenger's salaries. So it's entirely possible that (say) he's paid once every 3 years. Or that (given our modern crazy technology) that his earnings aren't fully monetary, or are otherwise capable of being precisely 1/3 of a penny (different currency, perhaps?). Anyway, the mathematician in me says that it's got to be AMAZINGLY explicit in order to prevent 20,000 from being divisible by 3. DaveE 9. Feb 18, 2010 HawkLogic Even in mathematics it is not possible to be as explicit as your rebuttal requires. Even mathematicians presume an understanding that everyone in the discussion is speaking the same language and using the same grammar and dictionary to do so. 10. Feb 19, 2010 davee123 Rawr! I'm not saying I think you did a poor job wording it, I'm just saying that's a problem I had with the original wording, which I didn't like back when I read it years ago. Anyway, I think I'd be happy with: "Mr. Jones earns exactly$20,000 USD per year."
and:
"The brakeman's nearest neighbor, one of the passengers, earns exactly three times as much per year as the brakeman does annually, who is completely paid in legal USD cash."
Going for the utterly absurd, you could expand "USD" into "United States of America Dollars", and explicitly specify that we're talking about Earth years. But that's going too far even for me.
But really, for the sake of elegance, if I were writing the problem I think I'd opt to choose a different criteria rather than money being divisible by 3. It's the same deal (to a lesser extent, I think) with the "Einstein" puzzle that keeps swinging around. The puzzle is constructed such that the common interpretation of the language is sufficiently clear, but it's susceptible to interpretations that can make the problem unsolvable or solvable with different conclusions.
DaveE
11. Feb 19, 2010
HawkLogic
The problem is "sufficiently clear" but "unsolvable"? If you understand how to read puzzles then it is not ambiguous. There is no such thing as a puzzle which cannot be quibbled into complete incoherence. Using straightforward common sense and logic this problem leads to a single answer, which is a good definition of "unambiguous". You are confusing riddles with lateral thinking problems, which are generally solved by finding an intentional ambiguity.
12. Feb 19, 2010
davee123
I didn't say that. I said it's sufficiently clear for common interpretation of the language. Effectively, that for some given percentage of people who read the problem (I don't know what that percentage would be), it's fine. But for that other percentage, it isn't.
Here's the sticky point: if you give this to a group of regular people, you probably won't get any rebuttals. BUT, if you give this to a group of people who are explicitly interested in picking apart word problems and finding creative solutions to them (FWIW, there are a bunch of those people on these forums), I think you may find that some of them don't like it.
I mean, besides my comment, you already commented on the fact that it explicitly states "but NOT respectively", when the author CLEARLY meant "but not necessarily respectively". Again, you could give this to MOST people, and they'd understand perfectly well what was intended, and they wouldn't quibble about such a petty detail when the intent of the writing was obvious. Supposing for a moment that the author had accidentally written the first sentence as:
"On a train, Smith, Robinson, and Jones are the brakeman, fireman, and the engineer, but NOT respectively."
Wouldn't it be painfully obvious to you (like it would be to me) that the author accidentally mistyped that first sentence? Common sense says to me that they meant "not necessarily respectively", but as an aside, I would acknowledge that it's really incorrect.
DaveE
13. Feb 25, 2010
HawkLogic
It seems here that "picking apart word problems" is the opposite of "finding creative solutions."
Anyway, you note that regular people don't offer rebuttals, so you exempt yourself from that group. And you didn't offer a solution so you exempt yourself from the group of creative problem solvers. You seem to place yourself right in the middle of the obtuse quibblers
But how hard is it to pick apart a word problem and offer a solution?
I'll gave it a try.
1. Sometimes the OP refers to characters with "Mr", and other times he doesn't.
How can we be sure which Smith he is talking about?
2. He identifies three businessmen and other passengers.
Why should we assume these are the same characters?
3. Smith beats the fireman in billiards?
So do they live in the same city or know each other?
4. Which Omaha and Detroit is the OP referring to and does it matter?
5. In "Who is the Engineer?" note that "Engineer" is capitalized. Maybe this is a clue to the true answer:
The Engineer is ... "The person who is driving the train."
6. And many others.
14. Feb 25, 2010
davee123
How would those be opposite? That would imply somehow that those that pick apart word problems don't find creative solutions? I tend to find that the two are often closely related-- when you pick apart word problems, you're finding creative ways in which the problem is in error. Essentially, you're thinking outside the box by attempting to interpret the problem in many different ways. Often, that's the way towards obtaining a solution.
Wait, what? I understand the first point, and I agree. I frequently offer observances about word problems when I find them to be lacking. But I exempt myself from the group of creative problem solvers because I didn't offer you an answer to this particular problem? I've answered plenty of problems here-- sometimes creatively, sometimes as intended. I've also come up with my own original creative puzzles and offered them here. It's silly for you to proffer that assessment with just such a limited scope.
You already did that. You decided to pick apart the problem by saying "Note that 'Smith beats the fireman in billiards.' is redundant. " And you similarly didn't offer a solution. I guess I'm just surprised you're being so hypocritical. You quibbled, I quibbled. And I fully admit that your initial critique is more noteworthy (because I think it's a problem in the restatement of the puzzle rather than with the underlying puzzle itself). But I don't see how me offering my recollection of the problem and a critique I noticed years ago ought to be any less applicable than your critique of the problem.
If you think that my critique crosses the line of being picky, then that's fine-- you could've just said so, and that'd be fine. But you went and told me that it was impossible to satisfy me, which was incorrect. And now you're just being defensive-- are you new to the internet?
DaveE
15. Mar 5, 2010
HawkLogic
The problem is not in error.
Your interpretations offered no alternatives to a problem which had a solution.
It would be creative to find a heretofore unknown solution within the box.
I limited my scope of assessment to the limited scope of your analysis.
My point was merely informational, a point of interest. Your quibbling claims there is a issue which affects solvability.
Again, there is no error or ambiguity.
The connection you make between defending an opinion and inexperience on the internet implies that the internet is a place to expect andor accept offending comments.
My experience goes back to before it was called the internet.
16. Mar 5, 2010
davee123
I still don't get it-- you're attacking me for not having been creative? What's the big deal if I wasn't creative? Should your point about redundancy have been required to be creative in order to warrant posting?
I beg to differ. My point was a point of interest, which I agree is probably of less interest. My point happened to involve whether or not the problem was solvable, yours didn't.
My point was basically that when I read the problem, I noted that there were only two clues dealing with money, and given their nature, the implication was that 20,000 is not divisible by 3. As a picky sort of person, I saw what the intention was, but I didn't like the implication that a real number wasn't divisible by another real number.
There are ways around it-- like using the word "evenly" (as in "is 20,000 evenly divisible by 3?") which is an accepted means of stipulating an integer-based quotient. Similarly, explicitly stating that integers are required would work equally well.
In the problem, I agree that such a requirement is difficult to work into the problem elegantly, so I'm left with a nagging annoyance in my mind that the problem isn't as clear as I'd like it to be. So I stated that I didn't like the way it was worded. If I had had an elegant way of rephrasing the problem, I probably would have suggested it. I can think of a few other ways of making the stipulation so that it doesn't leave that splinter of annoyance in my mind-- I even proffered one when seemingly prompted to-- but I haven't hit on one that's similarly as elegant as the problem intends.
Anyway, I don't think that the fact that such an interpretation would make the problem unsolvable makes the observation inapplicable or not suitable for posting.
But I think what disappointed me most is that you seemingly were telling me that it was impossible to satisfy me. You could have said that I was being too picky (which is a perfectly valid rebuttal), or that what I said was too minor of a detail to warrant posting (which is an issue for the moderators, I suppose). That would've been fine. But then, shortly thereafter, you implied that I didn't understand how to read puzzles, and that I wasn't creative! I have to say I honestly was surprised at the outcome of such a seemingly innocuous comment!
I've been a part of internet discussions for the past 11 years or so, and I've tended to find that people who are overly defensive are often new to the medium. Maybe not new to the internet specifically so much as new to public internet discussion, perhaps. For starters, with just text-based interaction, it's difficult to tell what someone's tone is. Further, being able to post anonymously frequently encourages people to post more aggressively because there's less of a tangible outcome. So, yes, the internet IS a place where you should expect a higher level of offending comments. However, as such, you should also be able take the comments you read with a grain of salt-- if you want to call that "accepting", I suppose that could fit. But generally speaking, you want to resist being offended.
I expect if we had been sitting in a room and I made the observation, you probably would've seen the mild tone with which I offered it, and taken it in stride-- or you wouldn't have been as aggressive in your response. But for whatever reason, it struck a nerve with you, and here we are. Anyway, I DO like discussing such matters (as you can probably tell), but I'm sorry if I offended you.
DaveE
17. Mar 9, 2010
HawkLogic
The good thing about the internet is that you can't really offend someone unless you actually know who you are talking to.
"Games Magazine" has a section called "Eureka" for "better, more elegant, or more complete solutions." In "Scientific American" Martin Gardner would occasionally present newly discovered solutions or cooks to classic puzzles. These are the kinds of enhancements I look for in a challenge to a puzzle. To me if a challenge does not improve the quality of the puzzle or the solution then it interferes with the enjoyment of the puzzle.
Some puzzles are presented ambiguously, sloppily, copied with typos, unsolvable, with trivial solutions, etc. The salary thing...nah.
18. Sep 22, 2012
CaptRyan
I came across this puzzle some 45 years ago and I could never solve it. I under stand the solution below, but in the (my) original puzzle the 4th clue was different. the words "one of the passengers" was not included. Can this puzzle be solved without that phrase? | 2016-08-27 17:15:51 | {"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": 0.5251981019973755, "perplexity": 1072.714384002102}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982924605.44/warc/CC-MAIN-20160823200844-00260-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://chemistry.stackexchange.com/questions/116984/burning-magnesium-ribbon | # Burning magnesium ribbon [closed]
I know that this might sound wrong but, in the reaction
$$\ce{2 Mg(s) + O2(g) → 2 MgO(s) + energy}$$
isn't oxygen supposed to be oxidised as it has excess electrons $$(\ce{O^2-})$$ and can donate a few to magnesium as it has less electrons $$(\ce{Mg^2+})?$$
I have searched for it in almost all books and it is written that magnesium gets oxidised. Please clear my misconception. It would be helpful if I can get a reaction mechanism for this, because it includes transfers of electrons.
## closed as unclear what you're asking by Mithoron, Jon Custer, user55119, DrMoishe Pippik, Todd MinehardtJun 18 at 0:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• In layman's terms both magnesium and oxygen within MgO are "happy" since both $\ce{O^2-}$ and $\ce{Mg^2+}$ have configuration of neon, a noble gas. For a more in-depth explanation probably have a look at the elements' electronegativities or dive into physical chemistry (review free Gibbs energy for the burning process). – andselisk Jun 17 at 16:21
• But MgO is formed after the reaction. I am talking about their reaction to achieve the stable state of MgO. – Shishir Maharana Jun 17 at 16:28
• But, oxidation means loss of electrons. So, if magnesium is oxidized it has to lose electrons, but it clearly has less electrons, then why will it lose them? – Shishir Maharana Jun 17 at 16:37
• Well, then why did magnesium lose 2 electrons and oxygen gain 2 electrons? I mean why did Oxygen take 2 electrons from magnesium? – Shishir Maharana Jun 17 at 16:51
• The reactants are neutral Mg and neutral O2. The magnesium doesn't turn into Mg++ until after reaction, and it gets that way by being oxidized. The O2 doesn't get extra electrons and go to 2 O-- until after reaction. The MgO is the product and doesn't do any oxidizing or reducing, so we imagine it to be happy as a stable lattice of Mg++ and O-- ions. – James Gaidis Jun 18 at 13:32 | 2019-09-19 21:24:52 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42094844579696655, "perplexity": 1862.8576366386308}, "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-2019-39/segments/1568514573735.34/warc/CC-MAIN-20190919204548-20190919230548-00008.warc.gz"} |
https://web2.0calc.com/questions/digit-problem_6 | +0
# digit problem
-1
71
1
The five-digit number $\overline{a679b}$ is divisible by 72. Find the digits a and b.
Feb 5, 2021 | 2021-06-12 20:57:55 | {"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": 0.9898934960365295, "perplexity": 10084.679705926594}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487586390.4/warc/CC-MAIN-20210612193058-20210612223058-00061.warc.gz"} |
http://mathoverflow.net/questions/35106/on-the-constants-in-the-cameron-erd%C3%B6s-conjecture-on-sum-free-subsets | # On the constants in the Cameron-Erdös conjecture on sum-free subsets.
The Cameron-Erdös conjecture was proved independently by Ben Green and Alexander Sapozhenko.
Let s(n) be the number of sum-free subsets of the set of integers {1,2,...,n}. They showed that ${ s(n) / 2^{n/2} } \to C_O \hbox{ or } C_E,$ for constants $C_O$ and $C_E$, as $n \to \infty$ through odd or even values respectively.
I would like to know what are the best known bounds for the constants $C_O$ and $C_E$?
My motivation is that I considered the conjecture in the mid-1990s and tried to determine some good lower bounds for the constants on the condition that the limits existed, of course. I have a vague recollection that Cameron and Erdös had some lower bounds in the region of 5 or 6, but I no longer have their relevant papers handy to verify this.
Looking at the sequence A007865 in the OEIS, it would seem that $C_E$ is in the region of 13.4 and that $C_O$ is in the region of 14.4. If one calculates at $s(n)/2^{n/2}$ for even n, it rises steadily from n=0 to n=36 then interestingly appears to oscillate about its limit. The sequence for odd n, from n=39 onwards, possibly does the same. It would be interesting to have some more terms.
Anyway, any information that you have on the actual values of these constants would be greatly appreciated.
-
The OEIS sequence has a link to a table of the first 70 terms, and both Maple and Mathematica code to calculate more values. – Hugo van der Sanden Aug 11 '10 at 9:16
Many thanks Hugo. – Derek Jennings Aug 11 '10 at 12:31
The review of the Sapozhenko paper, The Cameron-Erdos conjecture, Dokl. Akad. Nauk 393 (2003) 749-752, MR 2006a:11027, says $s(n)$ is asymptotic to $(c_0+1)2^{\lceil n/2\rceil}$ when $n$ is even and $(c_1+1)2^{\lceil n/2\rceil}$ when $n$ is odd, with $4.036\le c_0\le4.079$ and $3.086\le c_1\le3.095$.
EDIT: The review of a more recent paper, K G Omel'yanov, Estimates for Cameron-Erdos constants, Diskret. Mat. 18 (2006) 55-70, translation in Discrete Math. Appl. 16 (2006) 205-220, MR 2007m:11038, seems to contradict these numbers, giving $5.0709\le c_0\le5.0995$ and $3.8103\le c_1\le3.8336$. I haven't looked at the primary sources, so am unable to say whether the problem lies with me, a reviewer, or an author. | 2016-05-06 05:39:04 | {"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": 0.6364783644676208, "perplexity": 291.90423310106}, "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-2016-18/segments/1461861727712.42/warc/CC-MAIN-20160428164207-00221-ip-10-239-7-51.ec2.internal.warc.gz"} |
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# how to calculate half life from a graph chemistry
Half of the remaining nuclei decay in the next half-life. We had to halve 120 three times to get to … The half-life of Technetium 99m is 6.0 h. (f) 12 mg (12 x 10-3 g) of Technetium 99m is injected into a patient and starts to decay into Technetium 99. After another 800 years (1600 years total), there would only be $$25 \: \text{g}$$ remaining. Why use a term like half-life rather than lifetime? Half life calculator uses half life formula to solve half life equation & half life problems. For Example, if the initial concentration of a reactant A is 0.100 mole L-1, the half-life is the time at which [A] = 0.0500 mole L-1. Online radioactive decay calculator that allows you to find out the radioactivity decay in Cobalt (Co) 60. 2003. Have questions or comments? The table below illustrates half-lives for selected elements. The graph starts at 16 atoms and shows the number of parent nuclei halving every half-life. Legal. 960 daysD. Read the original count rate at zero days. One format involves calculating a mass amount of the original isotope. We can also easily see that the length of half-life will be constant, independent of concentration. #t_("1/2")# - the half-life of the decaying quantity. The half-life of the material is 15 days. The Technetium 99m reduces to 0.75 mg (or 0.75 × 10–3 g or 7. There are 3 cases. Even though it will get in the therapeutic range, such practice is not recommended. In zero-order kinetics, the rate of a reaction does not depend on the substrate concentration. 24 hours is 4 half-lives. In first order half life, what is the best way to determine the rate constant k? Uranium-233 has a half-life of about 160000 years, on the other hand. They become stable through decays. The formula for t1/2 shows that for first order reactions, the half-life depends solely on the reaction rate constant, k. We can visually see this on the graph for first order reactions when we note that the amount of time between one half life and the next are the same. For example, if the half-life of a 50.0 gram sample is 3 years, then in 3 years only 25 grams would remain. In order to solve the half life of first order reactions, we recall that the rate law of a first order reaction was: To find the half life we need to isolate t and substitute [A] with [A]0/2, we end up with an equation looking like this: $t_{1/2} = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{k} \label{5}$. The best way to determine rate constant k in half-life of first order is to determine half-life by experimental data. After one half-life, a 1.00 gram sample of uranium will have decayed to 0.50 grams of $$\ce{U}$$-238 and 0.43 grams of $$\ce{Pb}$$-206. Click here to let us know! Knowing the half life and amount of radioactive substance, one can calculate how much sample remains after a period of time. Substituting ln. In this video I will teach you how to calculate the initial rate of reaction from a graph quickly and easily using the tangent method. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. 6 days later it’s activity is 15 Bg. The number of half-lives that have passed is: 120/2 = 60. For example, carbon-14 has a half-life of 5,730 years and is used to measure the age of organic material. Each radioactive element has a different half life decay time. The half-life of an isotope is used to describe the rate at which the isotope will decay and give off radiation. Then the count rate with the radioactive source present is measured over a suitable period of time using a suitable detector such as a Geiger-Muller tube connected to a scaler. 5. Half-life means that after that time half of the initial isotope decays, so an half-life of 4 days means that 50% of the initial quantity of 200 has transfomed into 196. For example, if the half-life of a 50.0 gram sample is 3 years, then in 3 years only 25 grams would remain. . Examine the following graph and answer. This radioactivity approach can be used to detecting fake wine vintages too. This replacement represents half the initial concentration at time, t (depicted as t1/2). The half-life of a radioactive isotope is the time it takes for half of the sample to react, or decay. Similar method for Strong base vs Strong Acid. The decay process takes time and there is value in being able to express the rate at which a process occurs. Therapeutic range: the range that is between maximum drug concentration and a minimum drug concentration in which it is capable of fully exhibit its effective activity. 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Its initial concentration and rate constant ke ( minute-1, hour-1, day-1, etc. ) reactions at! | 2021-05-15 10:39:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7640380859375, "perplexity": 1428.3091191145525}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991801.49/warc/CC-MAIN-20210515100825-20210515130825-00284.warc.gz"} |
https://mathoverflow.net/questions/306046/how-to-visualize-a-witt-vector | How to visualize a Witt vector?
As the question title asks for, how do others "visualize" Witt vectors? I just think of them as algebraic creatures. Bonus points for pictures.
• Do you mean an element? the ring of Witt vectors? it's spectrum? And over a field, I guess... – Xarles Jul 15 '18 at 7:38
• Here is a picture on Wikipedia: en.wikipedia.org/wiki/P-adic_number#/media/… – Jason Starr Jul 15 '18 at 9:12
• Do you mean the $p$-typical Witt vectors of a perfect ring in characteristic $p$, or something more general? The $p$-typical Witt vectors of $\mathbb F_q$ are the unique unramified extension of $\mathbb Z_p$ with residue field $\mathbb F_q$, which I think is pretty explicit (but bigger examples become more annoying to compute). – R. van Dobben de Bruyn Jul 15 '18 at 10:10
If $R$ is $p$-torsion free, then $\mathrm{Spec}(W_n(R))$ is $n+1$ copies (or $n$ if you use the traditional indexing) of $\mathrm{Spec}(R)$ glued together in a suitable way along their fibers over $p$. But there are two qualifications. First, there is some "Frobenius twisting" involved, which is impossible to visualize because the Frobenius morphism is impossible to visualize, as far as I know. Second, each component is not simply glued transversally to the previous ones, but there is some higher order gluing.
A simple example is $\mathrm{Spec}(W_n(\mathbf{Z}))$. It consists of $n+1$ copies of $\mathrm{Spec}(\mathbf{Z})$, numbered $0$ to $n$, where the $k$-th copy is glued to the $(k-1)$-st copy modulo $p^k$. So copy $1$ is glued to copy $0$ transversally. Copy $2$ is glued to copy $1$ tangentially, but only to order $1$, and so on.
There is a little picture of $W_1$ on page 5 of my paper The basic geometry of Witt vectors, I. The affine case. | 2019-12-05 19:14:58 | {"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": 0.8855924606323242, "perplexity": 298.9721536522967}, "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/1575540482038.36/warc/CC-MAIN-20191205190939-20191205214939-00082.warc.gz"} |
https://www.cdslab.org/python/notes/scientific-computing/vectorization/vectorization.html | Vectorization, an extremely important concept in high-performance scientific computing, is the process of simultaneous execution of a set of computer instructions. This is contrary to the idea of looping and iteration which performs all program instructions sequentially. Vectorization can lead to significant runtime speed-up of the code.
## Is Python the optimal choice of programming language for scientific computing?
With regards to the capabilities of Python for scientific calculations, there are conflicting opinions. On the scientific side of the opinion spectrum, some people think that Python is not good enough for number crunching (as a result of which, new programming languages such as Julia have been developed). However, there are people at the other extreme who believe that Python is too much oriented towards scientific computation (as a result of which, new programming languages have emerged, such as Google’s Go language.
Regardless of personal opinions, the fact is that Python was not originally intended for scientific computing. However, an enormous number of Python packages have been developed with the dawn of the new millennium specifically for the purpose of numerical computing and data analysis, making Python one of the most popular languages world-wide for scientific inference and data analysis. In addition, many prominent implementations of Python are also open-source or available free of charge as opposed to rival programming languages such as MATLAB and Mathematica. These considerations together make Python a viable choice for high-level scientific computing, parallel to MATLAB, Mathematica, Julia, and other high-level programming languages for scientific computing and data analysis.
## Vectorization and array computing
So far in the notes, you may have noticed that all numerical vector calculations were either performed with lists, tuples, or dictionaries. Sadly, Python standard does not have an intrinsic special way of defining and manipulating numerical vectors and arrays, unlike most High-Performance Computing (HPC) languages for scientific computations (such as Fortran, Ada, or C). However, there are powerful Python modules that enable a Python programmer to use Python efficiently for numerical analysis as well.
## Vectors, arrays and the Numerical Python (numpy) package
In Python, a list can be heterogeneous meaning that not all its elements are of the same type. An array object in Python can be viewed as a variant of a list, but with the following assumptions:
• All elements must be of the same type, preferably integer, real, or complex numbers, for efficient numerical computing and storage.
• The number of elements must be known when the array is created.
• Arrays are not part of standard Python – one needs an additional package called Numerical Python, often abbreviated as NumPy. The Python name of the package, to be used in import statements, is numpy.
• With numpy, a wide range of mathematical operations can be done directly on complete arrays, thereby removing the need for loops over array elements. This is commonly called vectorization.
• Arrays with one index are often called vectors. Arrays with two indices are used as an efficient data structure for tables, instead of lists of lists. Arrays can also have three or more indices.
The number of elements of an array can be changed, but keep in mind that this can cause significant computational cost. Creating an array of a given length is frequently referred to as allocating the array. It means that a part of the computer’s memory is marked for being occupied by this array.
To create a numpy array, you will have to first import it,
import numpy as np
The tradition is to import numpy as np. To convert a list to a numpy array,
import numpy as np
a = [1,2,3,4,5]
a = np.array(a)
type(a)
numpy.ndarray
a
array([1, 2, 3, 4, 5])
To create a new array of length $n$, filled with zeros,
a = np.zeros(n)
Note that numpy automatically identifies the appropriate type for all array elements, whether int, float, or etc.
a[1]
2
type(a[1])
numpy.int32
Even if there is a single float element in the list, then all elements in the list will be converted to float in the numpy array by default,
type(a[1])
numpy.int32
a = [1,2,3,4,5.0]
a = np.array(a)
type(a[1])
numpy.float64
If you want to get the desired element type, then you will have to ask numpy for it explicitly,
a = [1,2,3.5,4.9,5.0]
a = np.array(a, int) # convert all elements in the list to integer
a
array([1, 2, 3, 4, 5])
You can see the full list of input arguments to np.array function here.
A similar function np.zeros_like(c) generates an array of zeros where the length of the generated array is that of the input array c and the element type is the same as those in c.
b = [1,2,3,4,5,6,7]
a = np.zeros_like(b)
a
array([0, 0, 0, 0, 0, 0, 0])
Often one wants an array to have $n$ elements with uniformly distributed values in an interval $[p,q]$. The numpy function linspace creates such arrays,
a = np.linspace(1, 100, 53)
a
array([ 1. , 2.90384615, 4.80769231, 6.71153846,
8.61538462, 10.51923077, 12.42307692, 14.32692308,
16.23076923, 18.13461538, 20.03846154, 21.94230769,
23.84615385, 25.75 , 27.65384615, 29.55769231,
31.46153846, 33.36538462, 35.26923077, 37.17307692,
39.07692308, 40.98076923, 42.88461538, 44.78846154,
46.69230769, 48.59615385, 50.5 , 52.40384615,
54.30769231, 56.21153846, 58.11538462, 60.01923077,
61.92307692, 63.82692308, 65.73076923, 67.63461538,
69.53846154, 71.44230769, 73.34615385, 75.25 ,
77.15384615, 79.05769231, 80.96153846, 82.86538462,
84.76923077, 86.67307692, 88.57692308, 90.48076923,
92.38461538, 94.28846154, 96.19230769, 98.09615385, 100. ])
## Vectorized computing vs. looping
Loops over very long arrays may run slowly. An advantage of arrays is that, with arrays, loops can be avoided the whole array be manipulated directly and simultaneously. If you are a programmer, you are likely already quite familiar with the powerful idea of array computing and vectorization. If not, then consider the following example,
x = np.linspace(0, 2, 201)
x
array([ 0. , 0.02, 0.04, 0.06, 0.08, 0.1 , 0.12, 0.14, 0.16,
0.18, 0.2 , 0.22, 0.24, 0.26, 0.28, 0.3 , 0.32, 0.34,
0.36, 0.38, 0.4 , 0.42, 0.44, 0.46, 0.48, 0.5 , 0.52,
0.54, 0.56, 0.58, 0.6 , 0.62, 0.64, 0.66, 0.68, 0.7 ,
0.72, 0.74, 0.76, 0.78, 0.8 , 0.82, 0.84, 0.86, 0.88,
0.9 , 0.92, 0.94, 0.96, 0.98, 1. , 1.02, 1.04, 1.06,
1.08, 1.1 , 1.12, 1.14, 1.16, 1.18, 1.2 , 1.22, 1.24,
1.26, 1.28, 1.3 , 1.32, 1.34, 1.36, 1.38, 1.4 , 1.42,
1.44, 1.46, 1.48, 1.5 , 1.52, 1.54, 1.56, 1.58, 1.6 ,
1.62, 1.64, 1.66, 1.68, 1.7 , 1.72, 1.74, 1.76, 1.78,
1.8 , 1.82, 1.84, 1.86, 1.88, 1.9 , 1.92, 1.94, 1.96,
1.98, 2. ])
Now, if you wanted to calculate the sin of the elements of x in the traditional way, you would do,
from math import sin
sinX = [sin(i) for i in x]
This approach, however, can be a quite time-consuming and computationally costly, because for-loops are very slow in Python, up to a few hundred times than what you get in Fortran or C.
A more appropriate solution to the above problem is to use the sin function from the numpy module, which enables vectorization,
sinX = np.sin(x)
You see, with the above numpy call, there is no need for a for-loop. The above Python code is an example of a vectorized code and the previous code which contained for-loop is an example scalar code. The numpy functions are capable of handling arrays as input. Compare the performance of the two codes in the above example,
%timeit np.sin(x)
The slowest run took 11.73 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.21 µs per loop
%timeit [sin(i) for i in x]
10000 loops, best of 3: 23.1 µs per loop
The vectorized code in this example appears to be more than one order of magnitude (more than 10 times) faster than the scalar version of the code.
### Why is the vectorized code faster in Python?
The reason is that numpy uses precompiled Fortran and C loops to loop over the elements of the input array. loops in Fortran and C have far less overhead than loops in Python. Similar to the above example, you can define your own functions that are also vectorized, for example,
def f(x):
return x**2*np.exp(-x**2)
x = np.linspace(-3, 3, 101)
y = f(x)
The numpy package also has a method for Automatic vectorization of scalar functions (function that only take scalar arguments), for example,
func_vec = np.vectorize(func_scalar)
However, for serious programming, I do not recommend you to use this numpy functionality as it can be slow and inefficient.
## vectorization of if-blocks
For vectorization of calculations involving booleans and if conditions, the solution can be problem dependent, but one common easy way of addressing simple boolean problems could be where method in numpy package. For example, suppose you have a list of numbers and you would like to perform a task on all negative numbers in the array, say set them all to zero, and leave the positive numbers intact. One solution would be the following,
x = np.array([1,-1,3,-5,-6,8,7,4,10])
np.where(x<0,0,x)
array([ 1, 0, 3, 0, 0, 8, 7, 4, 10])
## Aliasing vs. copying arrays
If you recall from the topic of aliasing vs. copying in the discussion of values, variables, types, there is a difference between aliasing and copying sequence-objects in Python. The same rules also hold for numpy arrays, meaning that if you need an independent copy of an existing array, then you have to use copy method to generate it,
a = np.array([1,2,3,4,5])
b = a.copy()
b[0] = -1
a
array([1, 2, 3, 4, 5])
b
array([-1, 2, 3, 4, 5])
otherwise a simple equality assignment like b = a will only create an alias for numpy array a.
a = np.array([1,2,3,4,5])
b = a
b[0] = -1
a
array([-1, 2, 3, 4, 5])
## In-place arithmetic in Python
Consider two arrays a and b of the same shape. The expression a += b means a = a + b. There are however hidden differences between the two. In the statement a = a + b, the sum a + b is first computed, yielding a new array, and then the name a is bound to this new array. The old array is lost unless there are other names assigned to this array. In the statement a += b, elements of b are added directly into the elements of a (in memory). There is no hidden intermediate array as in a = a + b. This implies that a += b is more efficient than a = a + b since Python avoids making an extra array. In other words, the operators +=, *=, and similar operators perform in-place arithmetic in arrays.
## Allocating arrays in Python
We have already seen in the above that the np.zeros function is useful for making a new array of a given size. Very often the size and the type of array elements are known a priori or have to match another existing array’s shape and type b. There are two ways of achieving this goal,
a
array([1, 2, 3, 4, 5])
b
array([-1, 2, 3, 4, 5])
a
array([1, 2, 3, 4, 5])
b = a.copy()
c = np.zeros(a.shape, a.dtype)
a.shape
(5,)
a.
a.T a.argsort a.compress a.cumsum a.dumps a.imag a.min a.prod a.reshape a.shape a.sum a.tostring
a.all a.astype a.conj a.data a.fill a.item a.nbytes a.ptp a.resize a.size a.swapaxes a.trace
a.any a.base a.conjugate a.diagonal a.flags a.itemset a.ndim a.put a.round a.sort a.take a.transpose
a.argmax a.byteswap a.copy a.dot a.flat a.itemsize a.newbyteorder a.ravel a.searchsorted a.squeeze a.tobytes a.var
a.argmin a.choose a.ctypes a.dtype a.flatten a.max a.nonzero a.real a.setfield a.std a.tofile a.view
a.argpartition a.clip a.cumprod a.dump a.getfield a.mean a.partition a.repeat a.setflags a.strides a.tolist
Notice how the attribute a.dtype (dtype standing for data type), and x.shape (a tuple) were used in the above example. The shape attribute in array objects holds the shape, i.e., the size of each dimension. The method size returns the total number of elements in the array.
Sometimes one may also want to ensure that an object is an array, and if not, turn it into an array. The np.asarray function is useful in such cases,
a = np.asarray(a)
Note that one could have also used,
a = np.array(a)
There is no difference in the output, but note that the second approach does one redundant step, because in the first approach, if the input object is already an array, then there is no need in converting it to an array.
## Multidimensional NumPy arrays
Creating multidimensional arrays is very much the same as vectors in numpy. The only thing to keep in mind is that the shape of the array is given as a tuple to np.array(). For example, to initialize a 3D array of size (0:3,0:5,0:2), you would do,
a = np.zeros((3,5,2))
a
array([[[ 0., 0.],
[ 0., 0.],
[ 0., 0.],
[ 0., 0.],
[ 0., 0.]],
[[ 0., 0.],
[ 0., 0.],
[ 0., 0.],
[ 0., 0.],
[ 0., 0.]],
[[ 0., 0.],
[ 0., 0.],
[ 0., 0.],
[ 0., 0.],
[ 0., 0.]]])
The arrays created so far have been of type ndarray. NumPy also has a matrix type called matrix or mat for one- and two-dimensional arrays. One-dimensional arrays are then extended with one extra dimension such that they become matrices, i.e., either a row vector or a column vector,
x1 = np.array([1, 2, 3], float)
x2 = np.matrix(x1) # or np.mat(x1)
x3 = np.mat(x1).T # transpose = column vector
x3
matrix([[ 1.],
[ 2.],
[ 3.]])
type(x3)
numpy.matrixlib.defmatrix.matrix
A special feature of matrix objects in NumPy is that the multiplication operator represents the matrix-matrix, vector-matrix, or matrix-vector product as we know from linear algebra. However, keep in mind that the multiplication operator between standard ndarray objects is different from multiplication between numpy matrices. The ndarray multiplication is simply a vectorized version of scalar multiplication,
a = np.array([1,2,3])
b = np.array([1,2,3])
a*b
array([1, 4, 9])
whereas, the matrix multiplication would yield,
aMat = np.mat(a)
bMat = np.mat(b)
aMat*bMat.T
matrix([[14]])
aMat.T*bMat
matrix([[1, 2, 3],
[2, 4, 6],
[3, 6, 9]])
If you intend to use Python and MATLAB together for your projects, then I recommend you to consider programming with matrices in Python instead of ndarray objects, because the matrix type in Python behaves quite similar to matrices in MATLAB. Numpy has a lot more to offer for linear algebra operation, that far beyond the scope of this lecture. More information about algebraic operations in NumPy can be found here.
## Symbolic linear algebra
There also a package SymPy that supports symbolic computations for linear algebra operations as well,
import sympy as sym
a = sym.Matrix([[2, 0], [0, 5]])
a**-1 # inverse of matrix a
Matrix([
[1/2, 0],
[ 0, 1/5]])
a.inv() # same as above, inverse of a
Matrix([
[1/2, 0],
[ 0, 1/5]])
a.det() # determinant of a
10
a.eigenvals() # eigenvalues of a
{2: 1, 5: 1}
a.eigenvects() # eigenvectors of a
[(2, 1, [Matrix([
[1],
[0]])]), (5, 1, [Matrix([
[0],
[1]])])]
A tutorial on sympy can be found here. | 2021-12-02 08:59:15 | {"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": 0.38048723340034485, "perplexity": 1516.4741989630986}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00443.warc.gz"} |
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### DIRECTOR NOT ATTENDING THREE CONSECUTIVE BOARD MEETINGS, HE SUFFERS DISQUALIFICATION UNDER KCS ACT 29 –C (P) AND BYLAW 55(110 AND 56) DEEMING PROVISION
DIRECTOR NOT ATTENDING THREE CONSECUTIVE BOARD MEETINGS, HE SUFFERS DISQUALIFICATION UNDER KCS ACT 29 –C (P) AND BYLAW 55(110 AND 56) DEEMING PROVISION
By : M.S.Yatnatti: Editor and Video Journalist Bangaluru : The Karnataka co-operative societies amendment Act 2012 which received accent of the governor on 10-01-2013 amends section 29-C and adds (P) Disqualification of the director he has absented himself for three consecutive meetings of the board of the such society without leave of absence Thus, in the instant case, when the petitioner knew the consequences of not attending three consecutive meetings, it was for him to make an appropriate application well in time, i.e. at the time of absenting himself from 3rd meeting or immediately thereafter as to what was the sufficient excuse for not attending the three consecutive meetings. The JRCS has no power to to overrule KCS Act 29 -c(p) and bylaw 55(110 and 56 prepared and administered by KCS Department under KCS Act. JRCS is asked to provide me information and reasons under section 4(I) (d) RTI Act under what provision of law he gave stay to the notice issued by BBMP SC ST Workers Society Ltdto 4 directors who were consecutively absent for more than 3 executive committee meetings as per bylaw they are automatically themselves loosed their directorship and they are no more directors as they were given time to provide reasons for their absent by submitting leave letter with medical certificates which they fail to give and finally they were informed as per bylaw that they were no more directors in BBMP SC ST Workers Society Ltd.
The letter dated 14-02-2016 stop the meeting has no meaning as under section 29-H a no confidence can be issued only after 24 months and not earlier and thus the 4 director willfully remained absent for the board meeting and attracted automatic disqualification as per KCS Act 29 -c(p) and bylaw 55(110 and 56 prepared and administered by KCS Department under KCS Act and JRCS giving stay is illegal and capricious in law..And the draft rules have been published in the gazette on 12-012017 yet nee to be fianlised and no final rules have been published in ths matter by Government.The primary objective of the Department of cooperation is to see that genuine cooperatives are organized and managed on the basis of sound cooperative principles for the promotion of economic interest and welfare of the people, and to render guidance and assistance for development of cooperative movement in the state. Administering proper supervision and control on the registered co-operative societies based on the above objectives as per the provisions of the Karnataka co-operative societies Act and Rules.Enforcement of the Government policy pertaining to Agricultural credit, Agricultural produce, Horticulture, Fisheries, Dairy, their marketing and processing and implementation of various schemes.Providing share capital, loan, interest subsidy to the members of the credit, marketing, consumers, industries and various co-operative societies but also extending govt. guarantee to loans availed by state level Institutions, Providing financial assistance for the upliftment of the SC/ST backward classes and minorities through the special schemes formulated by Government so that they can not only participate in the co-operative societies activities, obtain the various benefits but also come to the main stream of co-operative movement crossing the threshold .Provide technical guidance for the implementation of the schemes formulated by the Government.Collection and compilation of statistics of department as required by Central, State Government, RBI, NABARD etc.To exercise proper control and supervision on the administration of various cooperative societies within the legal frame work of cooperative Acts and Rules.
The greatest challenge to any civilized society is the economic deprivation it harbours, in league with social deprivation. It is inescapable that a collective war is waged to banish human deprivation from our midst. No other organized set up can be more potent for this attack than building human capital among the deprived, through sustainable cooperative Development Initiatives. Congruity with human nature further enhances the value of such initiatives. This principle is germane to the cooperative management in the Country and Karnataka in particular, which encompasses the basic human feeling of self-worth as its core. The cooperative strategy goes deep into the realms of building financial capabilities and self-confidence especially among the rural poor.The Department of Co-operation is a vital Department of the Government of Karnataka. The Department functions in close co-operation and co-ordination with various other Departments connected with the implementation of socio-economic plans and schemes. The Department of Co-operation overseas the administration and functioning of various Co-operative Institutions namely Textiles, Sericulture, Industries, Animal Husbandry, Fisheries, Sugar, Horticulture, Agriculture and Irrigation and also assist the societies financially besides providing technical guidance and input.
28. The Apex Court cautioned long ago that it is not expedient to extend the horizon of natural justice involved in the audi alteram partem rule to the "twilight zone of mere expectations", however great they might be.[ Per Mathew, J. in Union of India v. Mohan Lal Kapoor, ]
29. No doubt, when an order visits a person, against whom such an order is passed, with civil and/or evil consequences, principles of natural justice are to be followed. However, categories like the present one, are to be excluded where it is not an order' of the BCD but only a declaration of the fact which exists, namely, neither the petitioner attended three consecutive meetings nor offered any excuse what to talk of sufficient excuse. Further, a member to BCD is an elected member and there is no "expulsion" from this membership. He is not expelled from membership. It is a case of deemed vacation on the happening of a contingency. It is this deeming provision which has sprung into action on happening of an eventuality provided in the Section 10B of the Act. 30. As already noted above, similar deeming provision was noted in the case of Hyderabad Karnataka Education Society (supra).
31. Deemed' is used in various senses. Sometimes it means generally' regarded; at other times it signifies taken prima facie to be', while in other cases it means taken conclusively to be'. The meaning attached to the word deemed', therefore, must depend upon the context in which it is used. In Section 10B of the Act, the words deemed to have vacated office' would clearly mean that on the happening of contingencies provided therein, the person would no more remain in office the movement declaration is made by the Council to that effect. As far as first eventuality provided in Section 10B of the Act, to which this case relates, is concerned, the question of considering of sufficient excuse would arise only if such an excuse is given. In the absence of such excuse provided by the concerned member, mere declaration would be sufficient to put into action the deeming clause.
32. No doubt, in other provisions like Article 101(4) of the Constitution of India or Section 283(1)(g) of the Companies Act, prior permission for absence is to be obtained and there is a provision for automatic vacation from office on not attending particular number of meetings, these provisions were shown by Mr. V.P.Singh, learned senior counsel for the respondents 1 and 2 in an attempt to demonstrate that cases of election to office and incurring disqualification of such office on happening of certain contingency are to be treated differently from the cases where a person is appointed to an office and he is expelled, removed or terminated from that office. In this attempt, Mr. Singh has, prima facie, succeeded. It is but natural that provisions of such Statutes would not be identically worded. But when the matter is examined on the touchstone of applicability of principles of natural justice in such cases, when case relates to election to an office and deemed vacation of office on happening of certain contingencies, all these cases fall in same class. For this reason, judgments cited by the learned senior counsel for the petitioner in the cases of C.B.Gautam (supra) and Charan Das Malhotra (supra) would not be applicable in the instant case.
In citations in Delhi High Court Adish C. Aggarwala vs Bar Council Of Delhi And Ors. on 30 September, 2003it was held that "no relief can be granted to the petitioner by way of ad interim order,No doubt, when an order visits a person, against whom such an order is passed, with civil and/or evil consequences, principles of natural justice are to be followed. However, categories like the present one, are to be excluded where it is not an order' of the BCD but only a declaration of the fact which exists, namely, neither the petitioner attended three consecutive meetings nor offered any excuse what to talk of sufficient excuse. Further, a member to BCD is an elected member and there is no "expulsion" from this membership. He is not expelled from membership. It is a case of deemed vacation on the happening of a contingency. It is this deeming provision which has sprung into action on happening of an eventuality provided in the Section 10B of the Act. | 2017-05-25 12:23:26 | {"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": 0.28128233551979065, "perplexity": 5736.364699661666}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608067.23/warc/CC-MAIN-20170525121448-20170525141448-00342.warc.gz"} |
http://openturns.github.io/openturns/latest/user_manual/_generated/openturns.SORM.html | # SORM¶
class SORM(*args)
Second Order Reliability Method (SORM).
Refer to SORM.
Available constructors:
SORM(nearestPointAlgorithm, event, physicalStartingPoint)
Parameters: nearestPointAlgorithm : OptimizationAlgorithm Optimization algorithm used to research the design point. event : Event Failure event. physicalStartingPoint : sequence of float Starting point of the optimization algorithm, declared in the physical space.
Notes
See Analytical for the description of the first steps of the SORM analysis.
The Second Order Reliability Method (SORM) consists in approximating the limit state surface in U-space at the design point by a quadratic surface. SORM is usually more accurate than FORM e.g. in case when the event boundary is highly curved.
Let us denote by the dimension of the random vector and the main curvatures of the limit state function at the design point in the standard space.
Several approximations of the failure probability are available in the library, and detailed here in the case where the origin of the standard space does not belong to the failure domain:
• Breitung’s formula:
the marginal cumulative density function of the spherical distributions in the standard space and is the Hasofer-Lind reliability index, defined as the distance of the design point to the origin of the standard space.
• Hohen Bichler’s formula is an approximation of the previous equation:
where is the cumulative distribution function of the standard 1D normal distribution and is the standard Gaussian probability density function.
• Tvedt’s formula:
where is the real part of the complex number and the complex number such that .
The evaluation of the failure probability is stored in the data structure SORMResult recoverable with the getResult() method.
Examples
>>> import openturns as ot
>>> myFunction = ot.SymbolicFunction(['E', 'F', 'L', 'I'], ['-F*L^3/(3*E*I)'])
>>> myDistribution = ot.Normal([50.0, 1.0, 10.0, 5.0], [1.0]*4, ot.IdentityMatrix(4))
>>> vect = ot.RandomVector(myDistribution)
>>> output = ot.CompositeRandomVector(myFunction, vect)
>>> event = ot.Event(output, ot.Less(), -3.0)
>>> # We create an OptimizationAlgorithm algorithm
>>> solver = ot.AbdoRackwitz()
>>> algo = ot.SORM(solver, event, [50.0, 1.0, 10.0, 5.0])
>>> algo.run()
>>> result = algo.getResult()
Methods
getAnalyticalResult() Accessor to the result. getClassName() Accessor to the object’s name. getEvent() Accessor to the event of which the probability is calculated. getId() Accessor to the object’s id. getName() Accessor to the object’s name. getNearestPointAlgorithm() Accessor to the optimization algorithm used to find the design point. getPhysicalStartingPoint() Accessor to the starting point of the optimization algorithm. getResult() Accessor to the result of SORM. getShadowedId() Accessor to the object’s shadowed id. getVisibility() Accessor to the object’s visibility state. hasName() Test if the object is named. hasVisibleName() Test if the object has a distinguishable name. run() Evaluate the failure probability. setEvent(event) Accessor to the event of which the probability is calculated. setName(name) Accessor to the object’s name. setNearestPointAlgorithm(solver) Accessor to the optimization algorithm used to find the design point. setPhysicalStartingPoint(physicalStartingPoint) Accessor to the starting point of the optimization algorithm. setResult(sormResult) Accessor to the result of SORM. setShadowedId(id) Accessor to the object’s shadowed id. setVisibility(visible) Accessor to the object’s visibility state.
__init__(*args)
Initialize self. See help(type(self)) for accurate signature.
getAnalyticalResult()
Accessor to the result.
Returns: result : AnalyticalResult Result structure which contains the results of the optimisation problem.
getClassName()
Accessor to the object’s name.
Returns: class_name : str The object class name (object.__class__.__name__).
getEvent()
Accessor to the event of which the probability is calculated.
Returns: event : Event Event of which the probability is calculated.
getId()
Accessor to the object’s id.
Returns: id : int Internal unique identifier.
getName()
Accessor to the object’s name.
Returns: name : str The name of the object.
getNearestPointAlgorithm()
Accessor to the optimization algorithm used to find the design point.
Returns: algorithm : OptimizationAlgorithm Optimization algorithm used to research the design point.
getPhysicalStartingPoint()
Accessor to the starting point of the optimization algorithm.
Returns: point : Point Starting point of the optimization algorithm, declared in the physical space.
getResult()
Accessor to the result of SORM.
Returns: result : SORMResult Structure containing all the results of the SORM analysis.
getShadowedId()
Accessor to the object’s shadowed id.
Returns: id : int Internal unique identifier.
getVisibility()
Accessor to the object’s visibility state.
Returns: visible : bool Visibility flag.
hasName()
Test if the object is named.
Returns: hasName : bool True if the name is not empty.
hasVisibleName()
Test if the object has a distinguishable name.
Returns: hasVisibleName : bool True if the name is not empty and not the default one.
run()
Evaluate the failure probability.
Notes
Evaluate the failure probability and create a SORMResult, the structure result which is accessible with the method getResult().
setEvent(event)
Accessor to the event of which the probability is calculated.
Parameters: event : Event Event of which the probability is calculated.
setName(name)
Accessor to the object’s name.
Parameters: name : str The name of the object.
setNearestPointAlgorithm(solver)
Accessor to the optimization algorithm used to find the design point.
Parameters: algorithm : OptimizationAlgorithm Optimization algorithm used to research the design point.
setPhysicalStartingPoint(physicalStartingPoint)
Accessor to the starting point of the optimization algorithm.
Parameters: point : sequence of float Starting point of the optimization algorithm, declared in the physical space.
setResult(sormResult)
Accessor to the result of SORM.
Parameters: result : SORMResult Structure containing all the results of the SORM analysis.
setShadowedId(id)
Accessor to the object’s shadowed id.
Parameters: id : int Internal unique identifier.
setVisibility(visible)
Accessor to the object’s visibility state.
Parameters: visible : bool Visibility flag. | 2019-04-24 14:56:02 | {"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": 0.30844974517822266, "perplexity": 3667.366812405406}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578643556.86/warc/CC-MAIN-20190424134457-20190424160457-00282.warc.gz"} |
https://grokbase.com/g/perl/beginners/2003/08 | FAQ
# Perl>beginners>August 2003
## 337 discussions - 1,351 posts
• 22
#### Cron jobs and perl
Can someone tell me how to go about running a cron job with perl? Here's what I have thus far: 1) I have a cron job: 04 09 * * 1-7 /cygdrive/c/Documents\ and\ ...
Trina EspinozaAug 16, 2003 at 4:16 pmAug 29, 2003 at 10:11 pm
• 21
#### how do i list the methods connected to a object?
hi i wonder how i can list all the methods availible from a given object? martin
Martin A. HansenAug 5, 2003 at 10:21 amAug 7, 2003 at 8:36 pm
• 18
Howdy, I read in an earlier email that some people put in front of variables/function/etc to show that they are "important and should be handled by the program only" and it said that some "modules ...
Dan MueyAug 13, 2003 at 6:38 pmAug 15, 2003 at 4:51 pm
• 15
#### exists and defined--not just for hashes anymore
Just a quick tip, which the docs mention that 'exists' and 'defined' will let you know if a particular element of a hash exists, or is defined, they work on arrays just as well. (I didn't know that, ...
Paul ArcherAug 26, 2003 at 8:30 pmSep 5, 2003 at 6:17 am
• 15
#### Temp variable
hi, The statement ($a,$b) = ($b ,$a); is swapping without a third variable. But looking at a lower level , it is storing a temporary variable in the registers. How can i get that temporary ...
Ryan10975073Aug 23, 2003 at 5:08 pmSep 4, 2003 at 4:25 am
• 14
#### rearrange text
Hello, I am a relatively new PERL beginner and have been trying to work with simple bioinformatics stuff. I have so far written some very useful but simple bioinformatics scripts. However.... ...
Mike RobesonAug 24, 2003 at 7:26 pmAug 31, 2003 at 4:10 pm
• 14
#### pronunciation guide
Does anyone know of a pronunciation guide for the special variables and such in Perl? I came up empty on Google. I've been learning Perl by reading and doing, but I haven't talked to anyone ...
Paul ArcherAug 25, 2003 at 12:07 pmAug 26, 2003 at 3:58 pm
• 14
#### Still need unlink help!! God help me.
Still having problem with unlink. My original problem began with deleting files from a list. I seem to have fixed the problem reading the list but the unlink does not work. Here is the test script ...
PerlwannabeAug 9, 2003 at 12:01 amAug 12, 2003 at 4:43 am
• 12
I am writing a script that relies on retrieving the $ENV{REMOTE_ADDR} from the user's browser. I'm depending on this for database information retrieval rather than cookies or some type of ... Ron GoralAug 12, 2003 at 6:58 pmAug 13, 2003 at 6:26 pm • 11 #### Length() is bits/bytes or neither I know that the number returned by the length function is the number of characters. With ascii text is that the bits or bytes also? If not is there a function similar to length() that tells you the ... Dan MueyAug 26, 2003 at 3:56 pmAug 29, 2003 at 4:11 am • 11 #### Perl Codes Written in Windows Env I have a perl script that I developped in a windows machine and it had to be transfered by ftp to a UNIX server. The codes worked fine when I tested them on my windows machine. Is it true that the ... MmbodjiAug 25, 2003 at 8:42 pmAug 28, 2003 at 2:23 pm • 11 #### the File::Copy module I was just reading the File::Copy module. Now on a pure Win32 environment will this work copy("\\machine1\share\file1","\\machine2\share\file2"); Or should I do something else??????? Saadat SaeedAug 19, 2003 at 3:20 pmAug 21, 2003 at 9:33 pm • 11 #### Case conversions Any one have or know of a function to convert ugly "NAME, USER" to "User Name"? TIA Scott. Scott TaylorAug 15, 2003 at 9:18 pmAug 18, 2003 at 2:52 pm • 10 #### grep with ftp Howdy: I'm looking for information that will let me open an ftp connection and grep / search for files and then FTP them back to me. I saw something about FTP and macros, but the other articles about ... Johnson, ShaunnAug 28, 2003 at 2:35 pmSep 12, 2003 at 6:52 am • 10 #### need help decompose word Hi, I have no idea how to do this. I have a name Anthony Bob I want to put each letter in an array (no space) for(0..$numbersofletters){ @letters[c]= A LETTER } so when i do print @letters i get A N ...
AwardsAug 18, 2003 at 3:29 pmAug 29, 2003 at 8:13 pm
• 10
I want to make the case of the first letter of all the words in a selected string to upper case. The code s/\b(\w+)/\u$1\E/g; enables me to do this for the whole document. But the string I want to ... BisAug 22, 2003 at 1:17 pmAug 26, 2003 at 4:28 pm • 10 #### quick re help Hi everyone I am pretty new to regex's, so I was happy when my text wrapping expression worked - for the most part. It messes up when I need to wrap lines with \n that don't end in a space. If there ... WebmasterAug 13, 2003 at 4:34 pmAug 15, 2003 at 2:51 am • 9 #### Mod_Perl2 CGI.pm Is mod_perl a replacement for CGI? If just getting involved with perl and the web (Not new to perl) what is the better option mod_perl or cgi? This is on an apache server. I choose mason thinking it ... Paul KrausAug 29, 2003 at 5:44 pmSep 17, 2003 at 12:51 pm • 9 #### Please don't say RTFM Dear Gentle Readers, Please be gentle with me as this is only my first day trying to learn perl. I am using an online tutorial, however it must be written for *nix and I am trying to use it on win32. ... Robert Mark WhiteAug 16, 2003 at 5:44 amAug 17, 2003 at 5:58 pm • 9 #### our vs my Anybody know what version of Perl our was introduced? I'm using our in a module and want like to use 5.6.0; or whatever version in that module so that I can make sure the our's aren't a problem. ... Dan MueyAug 12, 2003 at 8:00 pmAug 14, 2003 at 4:12 pm • 9 #### use lib problem Hi, I have a problem maybe it is normal. My problem is with using lib now to retrieve my module i do this use lib "/path/to/Module"; it works just fine but created a configuration file. with$dir = ...
AwarsdAug 2, 2003 at 2:01 pmAug 2, 2003 at 8:26 pm
• 8
#### -w vs. use warnings
Hello everyone, Having been a Perl programmer for several years now I have become accustom to using the following as my normal "start" of any Perl script: #!/usr/bin/perl use warnings; use strict; ...
K OldAug 28, 2003 at 8:15 pmSep 1, 2003 at 2:55 pm
• 8
#### compare date,time
Hi, I have the following 12/08/2003, 11:00 13/08/2003, 23:00 Now I would like to compare them and then calculate the how long time was elapsed in terms of hours. For the example above it will return ...
AwardsAug 11, 2003 at 6:23 pmAug 27, 2003 at 6:06 pm
• 8
#### California ballot sort algorithm?
Here's a sure sign of someone with too much time on his hands: I was curious how to implement an algorithm in perl to sort the candidates in California's recall ballot. According to the news, here's ...
KEVIN ZEMBOWERAug 14, 2003 at 1:16 pmAug 15, 2003 at 2:07 pm
• 8
#### Ok I am tired
Hi All, I am really tired now: -------------------------------------- package Foo; use strict; use LWP::UserAgent; sub new { return bless {},shift; } sub Foo::INC { my ($self,$filename) = @_; my ...
Gupta, SharadAug 9, 2003 at 5:49 amAug 11, 2003 at 12:05 am
• 8
#### Simple question
Hi, How do I divide a variable in an array ? Lets say element one need to be divided by 1000? $line[1] =$line[1]/1000; Thanks, Henrik
Sommer, HenrikAug 5, 2003 at 3:13 pmAug 9, 2003 at 4:00 am
• 7
#### document creation.
Lo all, I want to create large number of documents (mainly invoices) using perl. In a couple of months, I should be creating a few hundred documents per month with a few million records in total for ...
Chris KnipeAug 9, 2003 at 9:05 pmSep 2, 2003 at 9:09 am
• 7
#### Graphing/Plotting over time
Hello all, Just wanted to look into a "for fun" project, after a recent project that wasn't much fun at all... Our organization got hit by the blaster worm, which hit many, many windows boxes. The ...
Akens, AnthonyAug 27, 2003 at 8:19 pmAug 29, 2003 at 3:52 pm
• 7
#### DBI.pm
Home PC using XP Prof. SP1 with Oracle 9.2 DB installed. PERL5.005.03.exe installed. I am new to the world of PERL. I have searched my PC and DBI.pm has not been installed on my PC. I understand that ...
KENNETH JANUSZAug 8, 2003 at 3:46 pmAug 29, 2003 at 3:42 pm
• 7
#### Blob to JPEG
Hello all, I have a database with JPEGs stored in a blob field and I want to display them on an HTML web page with out creating a new file.jpg for each connection. I'm currently hunting on Google, ...
Scott TaylorAug 27, 2003 at 6:32 pmAug 27, 2003 at 10:16 pm
• 7
#### trouble with math... driving me nuts.
Luckily I was easily able to recreate the problem. See code below: print 37.75 - 33.67 - 4.08 ; <STDIN ; I find these things all the time. Is there a particular module I can use to fix these things? ...
Peter_farrarAug 21, 2003 at 7:10 pmAug 22, 2003 at 7:39 pm
• 7
#### Executable perl program help!!
How can i generate such a file?
Sc00170Aug 20, 2003 at 2:08 pmAug 21, 2003 at 3:40 am
• 7
Hi, all -- If I have a string like my $string = "Hello, #NAME_FIRST# #NAME_LAST# from #STATE#!" ; and I want to replace those chunks each with another string, what is a good way to approach that? I'd ... David T-GAug 17, 2003 at 12:26 pmAug 17, 2003 at 8:59 pm • 7 #### Download Beginning Perl ebook I've posted the ebook Beginning Perl for downloading at: <A HREF="http://websiteprogrammin.com/wsp/docs/Beginning_Perl.zip" http://websiteprogrammin.com/wsp/docs/Beginning_Perl.zip</A MotherofperlsAug 13, 2003 at 8:15 pmAug 14, 2003 at 12:11 pm • 7 #### Global variables and a forked process Does a forked process share the memory locations of the global variables from the parent process? Thanks in advance! Ahmed MoustafaAug 9, 2003 at 8:05 pmAug 11, 2003 at 2:12 pm • 7 #### Splitting a path Hi, I have a path which may look like: UNIX: --------- PATH=/usr/local/share:/usr/local:... Where each path is seperated by ":". Windows: ------------------ PATH=E:\foo:D:\bar Again the path ... Gupta, SharadAug 2, 2003 at 10:02 pmAug 4, 2003 at 9:20 pm • 6 #### writing back to a file Hi, Newbie here. I am writing a program that takes a file that has two columns of words, extracts the second column, then saves the original file with only the data in the first column. Question #1 - ... Absolut NewbieAug 20, 2003 at 1:12 pmAug 31, 2003 at 6:56 pm • 6 #### why this is not working ?? Hi all : I am running a process "n" number of times with a for loop. I am interested in looking for the status at specific intervals(say at every 10 iterations). I wrote the below shown code for this ... T.S.Ravi ShankarAug 27, 2003 at 5:41 amAug 28, 2003 at 9:30 pm • 6 #### returning proper info from subroutines Howdy: I have a script where I want to * connect to database * create a subroutine where it executes and returns the value of the SQL * pass that value to another area to check and IF statement * ... Johnson, ShaunnAug 27, 2003 at 2:52 pmAug 27, 2003 at 10:05 pm • 6 #### multiple regex in if() staetment Howdy all: I'm trying to figure out the best way to test a string agains a list of regexs like so: my @regex = qw(qr(joe$) qr(^mama) qr([abc])); if($string does not match any of the regexs in @regex) ... Dan MueyAug 19, 2003 at 11:06 pmAug 25, 2003 at 4:05 pm • 6 #### Read dir / sort I have a program that would read a directory and then do some renaming. Up until recently it always seem to read by oldest file first to newest. Which is great because my renaming involves ... Paul KrausAug 22, 2003 at 6:31 pmAug 23, 2003 at 11:23 am • 6 #### passing hashes to subs Hi, all -- I've gotten rusty and so I'm back again as a rank amateur :-) I have a script that will make a database connection, pull out some rows, repeatedly generate personalized message bodies, and ... David T-GAug 17, 2003 at 11:15 amAug 17, 2003 at 7:28 pm • 6 #### ERRNO ... What am I missing Can someone explain the interaction of POSIX ERRNO and PERL scripts? When does ERRNO get set and is it set with each PERL call? I cant seem to track when '$!' is set. Sometimes it will be set to 0, ...
Allison, Jason (JALLISON)Aug 7, 2003 at 5:43 pmAug 12, 2003 at 1:53 pm
• 6
#### smtp problem
Hi, I found a strange problem. like the subject says i'm using Net::SMTP I don't have any error in my code. But when I'm using a server i.e mail.xxx.com Only domain @xxx.com will receive an e-mail. ...
AwardsAug 9, 2003 at 10:16 pmAug 10, 2003 at 5:17 pm
• 6
#### Regex Pattern
HI, I am trying to use regex to extract the city,state and zip out of a file. Now, the problem is the city can have more then one word to it like San Antonio San Francisco etc, I have also, bumped ...
Trevor MorrisonAug 8, 2003 at 12:06 amAug 9, 2003 at 1:58 pm
• 6
#### How to replace a text in a file
Hi, I am novice to Perl and learning very basic things. I want to replace a text in a file with new text. I want to do it programatically. How can I do that? I do not want to use Perl command line ...
Vinay ThombreAug 5, 2003 at 10:00 amAug 8, 2003 at 4:57 pm
• 6
#### While loop on a file handle
Hi, I am trying to step through each line of a file that contains orders that were place through a internet shopping cart. I have this code: open(ORDER,$order) or die "Error opening \"$order\": ...
Trevor MorrisonAug 5, 2003 at 5:42 pmAug 6, 2003 at 11:52 pm
• 6
#### List::Util / arrays
Hi, I am trying to get the maximum number in a file looking like: 1 5001 2 5002 3 5003 4 5004 5 5005 6 5006 7 5007 8 5008 9 5009 10 5010 11 5011 12 5012 13 5013 14 5014 15 5015 16 5016 17 5017 18 ...
Jakob KofoedAug 5, 2003 at 10:52 pmAug 6, 2003 at 5:04 pm
• 6
#### how to do paging of records
can anyone help me on how to do paging of records i want a few records per page, im using oracle 8i.!!! and cgi. david
Coello, DavidAug 6, 2003 at 12:40 pmAug 6, 2003 at 2:06 pm
• 6
#### input box
Hi I am trying a way to get input from user in GUI mode.. Like Inputbox in Visual Basic.....How do i go about? with regards D.P.S
PANNEER SELVANAug 5, 2003 at 11:57 amAug 6, 2003 at 12:11 pm | 2022-05-19 00:01:25 | {"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": 0.2744463086128235, "perplexity": 3252.9674451047918}, "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/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00260.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-8-section-8-5-determinants-and-cramer-s-rule-exercise-set-page-946/48 | ## Precalculus (6th Edition) Blitzer
The value of $x$ is, $=3$
Consider, $\left| \begin{matrix} 2 & x & 1 \\ -3 & 1 & 0 \\ 2 & 1 & 4 \\ \end{matrix} \right|=39$ Evaluate the above determinant as follows: $\left| \begin{matrix} 2 & x & 1 \\ -3 & 1 & 0 \\ 2 & 1 & 4 \\ \end{matrix} \right|=39$ Evaluate the above determinant by expanding along the first row as follows: $2\left| \begin{matrix} 1 & 0 \\ 1 & 4 \\ \end{matrix} \right|-x\left| \begin{matrix} -3 & 0 \\ 2 & 4 \\ \end{matrix} \right|+1\left| \begin{matrix} -3 & 1 \\ 2 & 1 \\ \end{matrix} \right|=39$ Next, calculate the above determinant as follows: \begin{align} & 2\left( 4-0 \right)-x\left( -12-0 \right)+\left( -3-2 \right)=39 \\ & 8+12x-5=39 \\ & 12x+3=39 \end{align} Add $-3$ to both sides to get, $12x=36$ Divide by $12$. Now, $x=3$. Therefore, $x=3$ | 2021-05-18 04:42:40 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.000009536743164, "perplexity": 1269.7046527094972}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989820.78/warc/CC-MAIN-20210518033148-20210518063148-00436.warc.gz"} |
https://stats.stackexchange.com/questions/518727/developing-risk-scores-from-cox-regression-model | # Developing Risk Scores from Cox Regression model?
A simple question on the development of risk prediction models from Cox regressions.
Suppose, as an example, that I want to create a risk score for 1-year mortality in patients with cardiovascular disease. Performing a Cox regression, I found 3 dichotomous variables (yes/no; lets call them variable1, variable2 and variable3 for simplicity), independently associated with the outcome. Each of these variables has a beta coefficient within the regression model.
In order to develop a scoring system to predict the risk of mortality at 1 year, I need to assign a score to the presence of each variable. Obviously, points within the score can (and should) be "weighted" according to the importance of the predictor on the outcome (e.g., variable 1 may give 1 point, while variable3 may give more points, if present).
Which approach should be used to establish how many point should be assigned to each variable in the Cox-Regression? One approach (perhaps naive) would be to transform the beta-coefficients into scores, but I would like to know if there is a more "rigorous" method, and especially if there are reference papers to use as guidance.
## 1 Answer
Your "naive" method, combining the predictor values with their regression coefficients, is the way to go.
The individual Cox regression coefficients represent the change in log-hazard associated with each of your predictors. That's about as direct a measure of "the importance of the predictor on the outcome" as you can find, if your model is well-calibrated.
Calculating the linear predictor of the model based on all the predictors (sum of products of predictor values times coefficient values) then provides a combined estimate of log-hazard relative to baseline conditions. In your case with only 3 binary predictors and (apparently) no interactions, you only have 8 distinct "scores" to assign that way, so it's pretty simple.
In more complicated cases you can use a Cox model to produce a nomogram that can give a combined linear-predictor value graphically. Harrell has several examples of nomograms in his class notes, with Cox model nomograms specifically illustrated in Chapters 20 and 21. That document, his textbook, and the references therein provide the documentation that you need.
• Thank you! I think this is a good starting point. I still dont get how can I transform the beta-coefficients into a point system, what technique do I need to use and so on. I would give a read to the documents you linked. Apr 10 at 20:37
• @pankevedmo with only 8 covariate combinations and 1-year survival this is quite easy. With the R survival package and a coxph model, call predict.coxph() on the model with a newdata data frame that has all the same column names for predictors as you used for the model, and with type="survival". The newdata should have a different row for each of the 8 binary covariate combinations and a value of 1 year for the time in each row; the value of your status variable doesn't matter. Output is 1-year survival probability for each combination. With se.fit=TRUE you get standard errors.
– EdM
Apr 10 at 21:28 | 2021-09-24 13:04:03 | {"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": 0.48068559169769287, "perplexity": 816.9763184162385}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00424.warc.gz"} |
https://cs.stackexchange.com/questions/65219/algorithm-for-clustering-points-in-mathbbr2-based-on-pairwise-distance-pe/65222 | # Algorithm for clustering points in $\mathbb{R}^2$ based on pairwise distance (percolation graphs)
I haven't been able to find literature on the efficient solving of the following problem.
Given $n$ random points $x_i \in (0, 1)^2$ the unit square, obtain the flat clusters of points, such that two points are in the same cluster if their pairwise distance $d(x_i, x_j) < r$, where $r < 1$ is a fixed parameter of the problem.
The naive solution checks each pairwise distance, therefore completing in $\mathcal{O}(n^2)$ time. Is there a faster way, assuming $n$ large and $r \ll 1$?
• Build a r-tree in O($n\log n$), use it to improve the lookup time. – adrianN Oct 28 '16 at 13:28
• You can superimpose an $r$-by-$r$ grid on the unit square and map each point to its grid cell. Then you only need to need to check other points in the same cell and its 8 neighbours. (This doesn't eliminate the $n^2$ factor, but should give a large speedup in practice for small $r$.) – j_random_hacker Oct 28 '16 at 13:28
• @j_random_hacker, that's actually what I'm using now, but I was wondering if I was overlooking any huge speedups. I'm not familiar with r-tree, but what would the cost of the total algorithm be? – Kappie001 Oct 28 '16 at 13:45
• I'm struggling to understand what you want the output to be. Suppose you have points $(0,0)$, $(0.5,0)$, and $(1,0)$, and $r=0.6$. Do you want the output to be a single cluster containing all three points? Or something else? The relation "$d(x_ix_j)<r$" is not transitive, so I'm not sure how the cluster should be defined. – D.W. Oct 30 '16 at 2:20
• @D.W. They would all be in the same cluster. See percolation theory on wikipedia. The main question, in the end, is at which $r$ a giant cluster (that scales with $n$) forms. – Kappie001 Oct 30 '16 at 9:11
The following algorithm does unfortunately does not allow you to get the $n^2$ factor out (see comment below).
Superimpose an $r/\sqrt 2$-by-$r/\sqrt 2$ grid on the unit square: notice that every point within the same grid cell is automatically within the same cluster. Also notice that there can be at most $n$ grid cells containing one or more points; these are the only grid cells we need to consider from this point on.
1. Map each point to its grid cell.
2. For each grid cell, throw all contained points into the same cluster.
3. For each pair of neighbouring grid cells $P$, $Q$ (each grid cell, apart from those at edges and corners, has 8 such neighbours), determine the closest pair of points $p, q$ having $p \in P$ and $q \in Q$ using this $O(n\log n)$ algorithm. If $d(p, q) < r$ then combine $C(P)$ and $C(Q)$, otherwise not.
You can use a union/find disjoint set data structure to maintain the clusters in close-to-linear time, just as per the Kruskal algorithm.
• I've just realised that, despite my (italicised!) claim, my answer doesn't get the $n^2$ factor out of the time complexity, because that can only happen if we know that every pair of points within each grid cell are separated by at least some minimum distance -- a condition that holds in the original closest-pair problem, but not here. It might be possible to salvage something with the Triangle Inequality, but after thinking about it for a while I don't have anything. Let me know if I should delete this. – j_random_hacker Nov 4 '16 at 17:10
• So the core issue is whether it's possible to do the following in $O(n \log n)$ time? Given sets $P,Q$ of points, find the closest pair of points $p,q$ such that $p \in P$ and $q \in Q$. That seems worth posting as a separate question in its own right... when I get a chance I may do that. – D.W. Nov 4 '16 at 18:33
• Asked here: cs.stackexchange.com/q/65573/755 – D.W. Nov 5 '16 at 3:05 | 2019-07-17 14:32:34 | {"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": 0.6598047018051147, "perplexity": 300.97110633217545}, "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-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00333.warc.gz"} |
https://scipython.com/book/chapter-6-numpy/examples/argmax-and-argmin/ | # argmax and argmin
Consider the following oscillating functions on the interval $[0,L]$: $$f_n(x) = x(L-x) \sin\frac{2\pi x}{\lambda_n}; \quad \lambda_n = \frac{2L}{n}, \quad n = 1,2,3,\cdots.$$ The following code defines a two-dimensional array holding values of these functions for $L=1$ on a grid of $N=100$ points (rows) for $n=1,2,\cdots,5$ (columns). The position of the maximum and minimum in each column is calculated with argmax(axis=0) and argmin(axis=0).
import numpy as np
import pylab
N = 100
L = 1
def f(i, n):
x = i * L / N
lam = 2*L/(n+1)
return x * (L-x) * np.sin(2*np.pi*x/lam)
a = np.fromfunction(f, (N+1, 5))
min_i = a.argmin(axis=0)
max_i = a.argmax(axis=0)
pylab.plot(a, c='k')
pylab.plot(min_i, a[min_i, np.arange(5)], 'v', c='k', markersize=10)
pylab.plot(max_i, a[max_i, np.arange(5)], '^', c='k', markersize=10)
pylab.xlabel(r'$x$')
pylab.ylabel(r'$f_n(x)$')
pylab.show()
Maxima and minima of the functions $f_n(x)$ described above. Note that only the "global'' maximum and minimum are returned for each function, and that where more than one point has the same maximum or minimum value, only the first is returned. | 2019-01-20 08:34:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6572741270065308, "perplexity": 2312.684029650637}, "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-2019-04/segments/1547583705091.62/warc/CC-MAIN-20190120082608-20190120104608-00446.warc.gz"} |
https://math.stackexchange.com/questions/1636897/why-is-the-solution-to-sqrt6-5x-x-only-x-1-and-not-x-6 | # Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$? [duplicate]
I solved the equation $\sqrt{6-5x}=x$ as follows: $$(\sqrt{6-5x})^2=x^2$$ $$6-5x=x^2$$ $$0=x^2+5x-6=(x+6)(x-1)$$ $$x=-6 \quad \text{or} \quad x=1$$
If I plug in $x=-6$ into the original equation, I get $\sqrt{6+30}=\sqrt{36}=\pm 6$ and if I plug in $x=1$, I get $\sqrt{6-5}=\sqrt{1}=\pm 1$.
To me it seems that both values satisfy the original equation. I am using an online education system for my class called MyMathLab and the solution is only $x=1$. Why is that?
Thank you in advance.
• No, $\sqrt{36}$ can only be 6, not -6. The square root of a non-negative number is uniquely defined: it is a non-negative number. (Put differently, $\sqrt{}$ is a function: each number in its domain has a unique image.) – Clement C. Feb 2 '16 at 3:17
• @ClementC. By definition, $\sqrt{b}=a$ if $b=a^2$, so how can you say that $\sqrt{36}$ only equals $6$ if $36=(-6)^2$? – Allison Feb 2 '16 at 3:19
• No, by definition $\sqrt{b}$ is the unique non-negative number $a$ such that $b=a^2$. – Clement C. Feb 2 '16 at 3:20
• Because then you are not looking for the square root: you are looking for any number satisfying the equation, which are the square root and its negative counterpart. – Clement C. Feb 2 '16 at 3:24
• It's confusing. And those of us who like to treat this as a "gotcha!" don't help. But basically if we write an expression such as $\sqrt{foo}$, an expression must have a single value for a statement to be consistent or meaningful. By convention we define $\sqrt$ to be the non-negative value. $x^2 = c$ has two solutions (if c > 0) but $\sqrt c$ is the positive one. And $-\sqrt c$ is the negative one. The tricky think is $\sqrt{ 6 -5x} = x$ does not actually mean $6 - 5x = x^2$. But it means $6 - 5x = x^2$ AND $x \ge 0$. – fleablood Feb 2 '16 at 3:41
The confusion you are having comes from the concept of finding the solutions to $z^2 = 36$. Definitely $z = \pm 6$ are solutions because $(-6)^2 = 36 = 6^2$, but this is not the same thing as $y = \sqrt{36}$. Otherwise, we get nonsense like $-6 = 6$ which isn't true.
If we plug the answer $x = -6$ back into the original equation, we have $$\sqrt{6 - 5(-6)} = \sqrt{36} = 6 = -6.$$ Again, we get nonsense of $-6 = 6$. Even it were positive and negative $6$, you'd have $-6 = \pm 6$. This is true for only one of the values, which shows it's not valid to assume that $\sqrt{36} = \pm 6$.
Because the equation $\sqrt{6-5x}=x$ is not equivalent to $6-5x = x^2$ but to
$6-5x = x^2$ and $x \geq \frac{6}{5}$.
Squaring can change the set of solutions. For example, consider $x = 4$ and $x^2 = 16$. We also have $\sqrt 1$, which have the solutions $x_1 = 1, x_2 = -1$, but $1 \neq -1.$ If you square a radical equation you'll always get a real and a false root.
For this very same reason, dividing by $x$ also means you can "lose" solutions.
• $x\geq \frac{6}{5}$ is incorrect. If you take an $x$ that is larger than $\frac{6}{5}$, you will have a negative number underneath of the square root, which has no real output. As mentioned to the other person who made the same mistake, $\sqrt{6-5x}=x$ implies $6-5x=x^2$, $x\geq 0$, and $x\color{red}{\leq}\frac{6}{5}$. What is the problem with the OP's solution is the fact that $x\geq 0$ for the reason that the output of the (principle) square root must be non-negative (as already mentioned several times above). – JMoravitz Feb 2 '16 at 3:39
• Was $x \ge 6/5$ a typo for $x \ge 0$. 1 < 6/5. – fleablood Feb 2 '16 at 3:46
• I'd say $\sqrt {6 -5x}$ being expressed implies $6-5x \ge 0$ but it isn't equivalent to it. After all, we can write $\sqrt{-5}$ but that doesn't meant it's a valid real number. $\sqrt{6 -5x} = x$ is equivalent to $6-5x = x^2;x\ge0$. It also implies $6-5x \ge 0$ but that isn't a defining aspect. Also $x = -6$ satisfies $6 -5x \ge 0$ ($6 - 5(-6) = 36 \ge 0$.) – fleablood Feb 2 '16 at 3:51
• @JMoravitz But now we are debating semantics. Yes, the square root strictly speaking only implies that the root is positive. However, in our case, it doesn't matter because no roots in that interval exists anyway. – Markus Klyver Feb 12 '16 at 3:29
• @fleablood See my comment above to JMoravitz. – Markus Klyver Feb 12 '16 at 3:29
You are confused a little bit. Let me clear you.
Consider these two equations:
$y^2=36$ and $z=\sqrt{36}$ See the degree of $y$ which is 2. that's why you get two solutions here in this case, that is $$y=\pm 6$$
Se the degree of $z$ which is 1 . that's why one must get a single root. Moreover, sqrt of a number can't be negative. Therefore , $$z=6$$
Now again satisfy your results and now you'll know the answer.
You may not select all solutions from an expression $B$ derived (calculated) from expression $A$ when $B$ is not equivalent to $A$, and have to refer to the initial expression to consider the valid values.
Here (notice the / within the $\Leftrightarrow$) $$\sqrt{6-5x}=x\quad\not\Leftrightarrow\quad6-5x=x^2$$ because removing the $\sqrt{ }$ removes a condition on the expression as well, as $\sqrt{x}$ is valid for $x \geqslant 0$, thus $\sqrt{\text{expression}} = x$ implies $x \geqslant 0$. And you may solve $$6-5x \geqslant 0$$ that gives $x \leqslant \frac{6}{5}$.
From now on, you know the requirements: $x \geqslant 0$ and $x \leqslant \frac{6}{5}$.
$$\sqrt{6-5x}=x\quad\implies\quad6-5x=x^2,\quad 0\leqslant x \leqslant \frac{6}{5}$$
Then after solving $6-5x=x^2$ that gives $$x=-6 \quad \text{or} \quad x=1$$ you refer to the requirements to remove the invalid $x = -6$, and may safely give the only solution $$x = 1$$
The square root, denoted by $$y(x)=\sqrt x$$ is a function of the variable $x$. As such it can take only one value.
Naturally, one always chooses the positive solution of
$$y^2=x.$$
This is called the main branch.
Remember:
$$y^2=x$$ has two solutions in $y$, but $$y=\sqrt x$$ only one. | 2021-03-06 11:10:31 | {"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": 0.9173985123634338, "perplexity": 195.0863840605846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374686.69/warc/CC-MAIN-20210306100836-20210306130836-00153.warc.gz"} |
https://physics.stackexchange.com/questions/342379/would-my-houses-wall-stop-gamma-rays/486142 | # Would my house's wall stop gamma rays?
I'm wondering if there is a material which is able to stop all the electromagnetic spectrum's radiations. Something able to stop every electromagnetic radiation from the ones with the lowest frequency up to the one with the highest.
Intuitions tells me that it may depend on material density, as well as its thickness, but I couldn't find confirmation yet.
Are there any common material/element that are able to do it?
Are standard house walls able to stop them?
I would love some examples, like a paper of "this material", which is "this thick"...
• No, standard walls won't stop gamma rays. Unless you live in a very deep cave, you won't be able to prevent all electromagnetic radiation getting into your living space. Read this article: nuclearconnect.org/know-nuclear/science/protecting – user154420 Jul 1 '17 at 11:42
• – Kyle Kanos Jul 1 '17 at 11:47
• Thanks for sharing related questions. I'm gonna read them, and if they answer my doubts, I'm gonna delete the question! – Gabriele Scarlatti Jul 1 '17 at 11:53
• You can answer your own question instead of deleting it. – Yashas Jul 1 '17 at 16:42
• I don't know very much how to handle this situations yet! I thought that deleting it was a smart move in order to not create duplicates, if the "forum tradition" is to answer it anyway, I'll be glad to do it when I'm able to! Anyway, at the moment I haven't found any other post that answer my doubts so I'll leave it as it is. :) – Gabriele Scarlatti Jul 1 '17 at 16:49
Whether or not a material stops or admits EM radiation is not an all-or-nothing affair. As you surmise, it depends on both material and on the frequency of radiation directed thereat, but what also holds is that all EM radiation will penetrate most everything to at least some depth which depends on both the frequency and the material type, which is characterized by a parameter known as the optical depth and, moreover, this depth is not all-or-nothing either: if $$\mathrm{OD}(m, f)$$ is the optical depth of a material of type $$m$$ and frequency $$f$$ of directed radiation against that material, then what that represents is the thickness of material which will cut the radiation intensity by a factor $$\frac{1}{e}$$, or about 63%. For example, if the optical depth is 1 mm, then the radiation directed against that material will be cut to 37% in the first 1 mm of material, then to 23% in the second 1 mm (2 mm of material "elapsed"), 15% in the third 1 mm (3 mm thick), and so forth, i.e. it's an exponential decay curve:
$$I(I_0, m, f, d) := I_0 \cdot e^{-d/\mathrm{OD}(m, f)}$$
where $$I_0$$ is the incident radiation intensity and $$I$$ is that at depth $$d$$ into the material. Every material has an optical depth greater than absolutely zero, even ones normally considered completely opaque like metals: in that case, it is related to what is called their "skin depth", and for visible light it is on the order of a wavelength.
Moreover, metals provide a good example to consider with regard to gamma rays specifically: generally speaking, a metal's skin depth (hence $$\mathrm{OD}(m, f)$$) is a decreasing function of the frequency $$f$$, i.e. higher frequencies penetrate less. But, above a certain point, which is called the metal's plasma frequency, the ability to absorb the radiation drops precipitously. Effectively, the reason the metals absorb radiation at lower frequencies is their copious highly mobile electrons are very responsive to the stimulation from the EM waves, but above a certain point, they are unable to move fast enough to keep up with the high-frequency oscillations and let it through. This frequency is around the petahertz (PHz) range, while gamma rays are in the exahertz (EHz) range or higher, i.e. greater than 1000 PHz, and typically more than $$10^5$$ PHz, and hence transmitted.
Your house's walls, if made of wood, are generally speaking very transparent to most gamma radiation, i.e. the optical depth is very large and considerably in excess of the typical wall thicknes. However, they are not absolutely so: while their optical depth may be "large", it is not infinite and hence some gamma rays will be absorbed. That is, they will "stop" them, just not stop very many, and probably not stop enough if, say, there was a nuclear blast outside close enough to be in the prompt gamma range, however in most cases that also means you are close enough the house and you will be vaporized in the heat and blast wave and hence you aren't going to worry about this (or anything else, for that matter). On the other hand, it COULD be a concern with a VERY small (not a mistake - as the devices get larger, blast and heat effects increase their lethal distance much faster than prompt radiation effects) device, like that a terrorist might use. | 2020-01-20 22:03:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6365631222724915, "perplexity": 701.742574951449}, "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/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00083.warc.gz"} |
https://www.physicsforums.com/threads/1d-single-phase-flow-in-porous-media.403375/ | # Homework Help: 1D Single Phase Flow in Porous Media
1. May 15, 2010
### phil ess
1. The problem statement, all variables and given/known data
I have been asked to model the flow of a slightly compressible liquid through a rigid, incompressible porous medium in 1D, assuming the flow obeys Darcy's law. I am given a rectangular prism of cross sectional area A with length L, a source of liquid (say water) on one side with total discharge Q, along with variables concerning the properties of the medium (porosity, permeability, etc). I am ignoring dispersion and diffusion effects.
Basically I am looking to find the pressure in the tube as a function of position and time, P(x,t).
2. Relevant equations
3. The attempt at a solution
Combining the continuity equation,
$$\epsilon\frac{\partial\rho}{\partial t} + \nabla\cdot\left(\rho k \nabla v\right) = 0$$
Ignore above line, not showing up correctly,
$$\epsilon\frac{\partial\rho}{\partial t} + \nabla\cdot\left(\rho k \nabla v\right) = 0$$
with Darcy's law,
$$v = \frac{-k}{\mu}\nabla P$$
we get,
$$\epsilon\mu\frac{\partial\rho}{\partial t} = \nabla\cdot\left(\rho k \nabla P\right)$$
Then assuming a liquid with compressibility c,
$$\rho = \rho_{o}e^{c(P-P_{o}}$$
which combines with the above to give the governing equation:
$$\epsilon\mu c \frac{\partial P}{\partial t} = \nabla\cdot\left(k \nabla P\right)$$
Combining the constant terms together, and since we are only considering the 1D flow in the x direction:
$$\frac{\partial P}{\partial t} = \alpha\frac{\partial^{2}P}{\partial x^{2}}$$
*** The first few equations are not showing up correctly. But what is important is the final results which seem ok.
Now all I need to do is solve this equation to get P(x,t).
What I am having problems with is the boundary conditions. I haven't taken any PDE courses or anything, so I'm not sure how they are supposed to be formulated, but I'll give some guesses:
P(x,0) = Pi ie. the pressure initially is some constant throughout the medium
Q = constant ie. the input flow rate is constant for all t
P(L,inf) = ?
I am assuming a finite medium with no discharge, so the liquid, and the pressure, will build up indefinitely, but I don't know what to put for the boundary at the end.
Like I said I am unfamiliar with PDEs which makes this very difficult. Later this problem will be extended to multi-phase flow, but this was given as a warm up for now.
Any help with this problem would be greatly appreciated!! Thank you!
2. May 18, 2010
### phil ess
Bump!
I think I have the boundary conditions:
P(0,t) = some constant (non-zero, from Darcy's law)
P(L,t) = Po ??
P(x,0) = Pi or do I use dP/dx = 0 since the end is 'insulated'
Now I know how to solve pde's for both ends at 0 or both ends insulated, but not for one end at constant presure and the other end insulated. I'm thinking in terms of the analogy of heat flow in a wire.
Again any help is very much appreciated!!! | 2018-12-18 13:03:04 | {"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": 0.6886317133903503, "perplexity": 568.9207070423638}, "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-2018-51/segments/1544376829399.59/warc/CC-MAIN-20181218123521-20181218145521-00462.warc.gz"} |
https://brilliant.org/discussions/thread/how-do-you-want-solution-discussions-to-work/ | # How do you want solution discussions to work?
What I want is whatever features generate the best conversations. We think there are many possible routes to this and would like your opinion.
Brilliant has evolved a lot in the last month. You might have noticed though that solution discussions are basically the same as they were 5 months ago. When I study past feature requests, a lot of the suggestions are still very relevant but I would like to have a fresher conversation.
Please vote on the comments I have put below and add any additional thoughts you have to them. In general describing why you voted one way or another is very useful to us.
Note by Peter Taylor
5 years, 4 months ago
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Note: you must add a full line of space before and after lists for them to show up correctly
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[example link](https://brilliant.org)example link
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# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
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# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $ ... $ or $ ... $ to ensure proper formatting.
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2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
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\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
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## Comments
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Generally speaking what do you think solution discussions should be about(this will immensely affect how aggressively we choose to revise solution discussions)?
Should solution discussions be a competition to see who can compose the best solution using only their own brain? (kind of what we currently have) Or try to become a more collaborative process by which everyone tries to work together to make sure every problem has at least one solution that is as good as possible?
Upvote if you think it should generally be more collaborative and downvote if you think it should be competitive.
Staff - 5 years, 4 months ago
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Collaborative. Life isn't a competition. It should be a learning experience for someone who wants to know how to solve the problem.
- 5 years, 4 months ago
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I think that this collaborative process will prove very beneficial to provide a kick start to the chemistry section on Brilliant.
I suggest that this feature be only provided in the Chemistry section (once it has been formed), so that people are more familiar with the concepts of Chemistry.
Other sections, such as Mathematics and Physics, should remain competitive, as many people are already much strong in these sections.
- 5 years, 4 months ago
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WHERE IS THE CHEMISTRY SECTION??? I LOVE CHEMISTRY!!!
- 5 years, 4 months ago
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Yo Bob, check out this thread. Basically just follow Christopher Boo, Anish, Nicole Ling and Daniel Lim and you'll start seeing a lot more chemistry.
Staff - 5 years, 4 months ago
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But we want a "BEST OF CHEMISTRY" A collection of communities best posts.
- 5 years, 4 months ago
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I have some similar thoughts with Anish. Math and Mechanics(not sure with CS and E+M) can be solved in many ways, which some are stunning fast while some are more traditional. So, we cannot judge which is "the best" as the quality of a solution depends on different users. So in my opinion, there is no more competition, we submit a solution only because we think this is one of the way to solve the problem.
I'm still not sure what will collaborative will actually be like(how to let everyone work together) so I voted for competitive...
- 5 years, 4 months ago
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Collaborative.. Because collaboration itself brings out the best way to solve a problem. So,obviously first is contained in the second..
- 5 years, 4 months ago
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It would be great If we'd be able to bookmark the solution of our choice for future reference...I mean they'd be handy to find among the rest of the solutions.
- 5 years, 4 months ago
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Just like the stuff on MSE :)
- 5 years, 4 months ago
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I would like to have a new system to filter the problems and notes in my homepage.
- 5 years, 4 months ago
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according to me,it should be more competitive....
- 5 years, 4 months ago
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I have some advice to Brilliant. $1$. Give extra points to those who submit a good problem and solution. The extra point can be considered from the amount of favorites and likes. $2$. Also give extra points to those who answer problems from the other members. $3$. In Physics topics, such as Mechanics. Do not mix up mechanics problems with non-mechanics problems . In Mechanics topic, only mechanics problems that can be allowed. So do for the other topics. $4$. Could the staff in Brilliant make editor for submitting a problem, a solution, or a note like on StackExchange? Because when I type using format LaTeX in the Brilliant's text editor, I feel a bit uncomfortable to check it again, whether there are mistakes or not, by clicking Preview button. By using the same editor as StackExchange, the error of typing, especially math equation, can be minimalized. $5$. Please provide a feature to communicate with the other members like chat or other kind of stuff like that. For now, I think that's already enough. I will add it again if there is a good idea crosses in my mind. Overall, Brilliant is really a good website and one of the most recommended site to those who want learn more. Great job folks! Two thumbs up for you all!!
- 5 years, 4 months ago
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I think the fifth is not really a good idea as Brilliant.org is not for socializing. The feature for comments/follow/voting is already enough(at least for me). Most of the users here have a Facebook account so you can maybe find them on Facebook for chats :). Just my opinion by the way.
Maybe Brilliant can give us some notifications when someone reply to our solution or notes so we can answer them?
- 5 years, 4 months ago
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The reason behind the $5^{th}$ idea is to make intense communication or discussion about a topic or problem among the members more private. I believe they have a Facebook account but unfortunately it almost doesn't appear in their Brilliant's profile. Beside that, lots of members here are invisible or hard to identify. I think notifications feature is also a good idea since Brilliant use email to inform when someone reply to our solution or notes but I barely check my email.
- 5 years, 4 months ago
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Yap! notification system will be quite good..
- 5 years, 4 months ago
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4th idea looks quite good to me!
- 5 years, 4 months ago
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How to make a private solution public?
- 5 years, 4 months ago
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Hi Everyone,
Thank you for your thoughtful feedback and votes in this thread. Soon we will experiment with a few changes to solution discussions to see if they make them better for more people. This will likely include, allowing people to see solutions before writing their own(a very divisive vote issue in this thread), allowing people to edit and delete their solutions, and eliminating the blind voting period.
Staff - 5 years, 4 months ago
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- 5 years, 4 months ago
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Indeed, those two cents inform a lot of how we are thinking about the future.
Staff - 5 years, 4 months ago
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Glad to contribute my ideas! It may not be worth \$37 and change, but I'm glad it's something!
- 5 years, 4 months ago
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As Brilliant is mostly oriented ONLY to maths which is really helpful for programmers for their logical skills, they also must have a learning environment for a programming language.
- 5 years, 4 months ago
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We should prefer to both learning as well as to compete because learning gains knowledge and experience while competetion shows our situation or place in the field so i think we should prefer to both what do u think .......
- 5 years, 4 months ago
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It Should be collaborative as it is about learning and finding new solutions and algorithms
- 5 years, 4 months ago
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Implement feature to filter homepage and prevent reshared posts from appearing.
- 5 years, 4 months ago
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now i am new on brilliant ,really it amazing i 47 years old and my two son one is doing engng and other one is in 10th standard we all sit together and see the problems and try to solve and which helps ,really it is amazing.
- 5 years, 4 months ago
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Do you want to see the solutions that have been written before you write your own?
Upvote this comment if you want to be able to see the solutions before writing your own, and downvote this comment if you want them to be hidden from you until you finish writing your solution.
Staff - 5 years, 4 months ago
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If we did it competitively, no, but it would be good collaboratively.
- 5 years, 4 months ago
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Well, we can learn from others' solutions and frame a better solution.
- 5 years, 4 months ago
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At least first few steps, so one would not waste time typing a solution that has already been typed
- 5 years, 4 months ago
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We want to see them...as this would prevent repetitions of the same solution.
Moreover, not being able to see them does not make any sense, because if you've solved a problem, it means that you already have a solution, so you write your own only if you feel its different or better.
Because it consumes much time typing with LATEX only to discover in the end that a better solution was already there !!!
- 5 years, 4 months ago
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Plagiarization will abound if we let that happen.
- 5 years, 4 months ago
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I think its best if we could see the solutions beforehand, since it'll help to give collaborative answers & also avoid repeating a solution.
- 5 years, 4 months ago
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Should you be able to downvote a solution?
Upvote this comment if you think you should be able to downvote a solution, downvote this comment if you think you shouldn’t(wow that was hard).
Staff - 5 years, 4 months ago
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To downvote a solution is disrespectful but I've noticed some solutions are totally wrong logically even if give correct answer. There must be a way to mark wrong solutions and to not count them in the statistics in the profile (or to be separate stat).
Also for me it would be better providing a solution to give more points in addition to solving a problem so to stimulate people to write solutions. Voted up solutions - more points also to stimulate original solutions.
The current format is brilliant :), whoever whats discussion on best solution could enter directly the discussion and ... discuss but let the others who want to try their skills have the pleasure as it's now.
- 5 years, 4 months ago
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I'd also like to be able to delete solutions that I have written if they are copies of something already contributed or use completely wrong logic. My first solution I ever wrote was based off of a complete misread of the problem, and it was extremely embarrassing.
- 5 years, 4 months ago
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Yes downvote appears more as a disrespect. Rather there should be a provision to remove the wrong solutions or atleast not count them.
- 5 years, 4 months ago
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I'd think, it'd be better if one of us were able to help with someone's solution (if it's incorrect or so). Voting down is kinda discouraging to that person.
- 5 years, 4 months ago
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Even though some solutions are... dubious, to say the least, it would be discouraging I think to have people who are new to solution writing have their solution be bombarded with downvotes.
- 5 years, 4 months ago
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Maybe you can only downvote solutions with a positive number of upvotes?
- 5 years, 4 months ago
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Sometimes,I really want to downvote a particular solution.However,with the cooperation of the solution writer,it is possible to settle any misleading ideas in the comments.And besides,downvoting will make a person crestfallen.
Another issue with downvoting:Correct solutions may be downvoted .I could show you one particular example.$\dfrac{x^5-1}{x-1}=0$.Here,the only possible solution requires the numerator to be zero.This means that $x^5=1$ satisfies what we want.Now someone might downvote saying that $x^5=1\implies x=1$ and that makes the denominator $0$.However, $x^5=1$ has 4 other roots that do not make the denominator 0.
- 5 years, 4 months ago
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I think downvovte souldn't be added. Because one writes it by wasting his time and the downvote will discourage him to write slutions in future. But a button should be added to request to remove wrong answers.
- 5 years, 4 months ago
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The only scenario where I think the 'downvote' option might work is when people submit codes for math solutions (especially Number Theory. I have been trying to be the Batman on Brilliant, ensuring that people stop posting such solutions.). Leaving that issue aside, I don't think that this feature would be helpful.
- 5 years, 4 months ago
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Problem Loading...
Note Loading...
Set Loading... | 2019-06-19 10:41:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "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": 0.82716304063797, "perplexity": 2106.9870282703746}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998943.53/warc/CC-MAIN-20190619083757-20190619105757-00473.warc.gz"} |
http://math.stackexchange.com/questions/491241/closed-form-for-int-infty-infty-operatornamesechx-operatornamesech/492318 | # Closed form for $\int_{-\infty}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx$
Let $$f(a)=\int_{-\infty}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx,$$ where $\operatorname{sech}(z)=\frac2{e^z+e^{-z}}$ is the hyperbolic secant.
Here are values of $f(a)$ at some particular points: $$f(0)=\pi,\hspace{.15in}f(1)=2,\hspace{.15in}f(2)=\left(\sqrt2-1\right)\,\pi,\hspace{.15in}f\left(\frac34\right)=\left(4\sqrt{2+\sqrt2}-\frac{20}3\right)\,\pi.$$ Athough I do not yet have a proof ready, it seems that for every $a\in\mathbb{Q},\ f(a)=\alpha+\beta\,\pi$, where $\alpha$ and $\beta$ are algebraic numbers.
I wonder, if it is possible to express $f\left(\sqrt2\right)$ in a closed form?
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A chain of substitutions yields the equivalent form $$f(a/b)=16b\int_0^1\frac{u^{a+b-1}du}{(u^{2a}+1)(u^{2b}+1)}.$$ Although my complex analysis is rusty, when I try partial fractions (via residue theorem) on this I unfortunately get tons of trigonometric expressions that I don't want to handle. At least it appears $f({\Bbb Q})\subseteq {\Bbb Q}^{\rm rab}\oplus\pi{\Bbb Q}^{\rm rab}$ as conjectured anyway. (By ${\Bbb Q}^{\rm rab}$ I mean the maximal real abelian extension of $\Bbb Q$, or equivalently $\Bbb Q$ with all values in $\cos(\pi\Bbb Q)$ adjoined.) – anon Sep 12 '13 at 17:17
${\large\mbox{We just need to evaluate}\ {\rm f}\left(a\right)\ \mbox{when}\ a \in \left\lbrack 0, 1\right\rbrack}$ since $${\rm f}\left(-a\right) = {\rm f}\left(a\right) \quad\mbox{and}\quad {\rm f}\left(1 \over a\right) = \left\vert a\right\vert\,{\rm f}\left(a\right)$$
\begin{align} {\rm f}\left(1 \over a\right) &= \int_{-\infty}^\infty {\rm sech}\left(x\right)\,{\rm sech}\left(x \over a\right)\,{\rm d}x = a\int_{-\infty}^\infty {\rm sech}\left(a\,{x \over a}\right){\rm sech}\left(x \over a\right) \,{{\rm d}x \over a} \\[3mm]&= \left\vert a\right\vert\int_{-\infty}^\infty {\rm sech}\left(ax\right)\,{\rm sech}\left(x\right)\,{\rm d}x = \left\vert a\right\vert\,{\rm f}\left(a\right) \end{align}
For example \begin{align} {\rm f}\left(1 \over 2\right) &= 2\,{\rm f}\left(2\right) = 2\left(\sqrt{2\,} - 1\right)\pi \\[3mm] {\rm f}\left(4 \over 3\right) &= {3 \over 4}\,{\rm f}\left(3 \over 4\right) = \left(3\sqrt{2 +\sqrt{\vphantom{\large A}2\,}\,} - 5\right)\,\pi \\[3mm] {\rm f}\left(a\right) & = {1 \over \left\vert a\right\vert}\,{\rm f}\left(1 \over a\right) \approx {1 \over \left\vert a\right\vert}\,{\rm f}\left(0\right) = {\pi \over \left\vert a\right\vert}\,, \quad \left\vert a \right\vert \gg 1 \end{align}
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You can have this form of solution
$$\int_{-\infty}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx =\frac{2}{a}\sum _{k=0}^{\infty } \left( -1 \right)^{k} \left( \psi \left( \,{\frac {3\,a+2\,k+1}{4a}} \right) -\psi \left( {\frac {2\,k+1+a} {4a}} \right) \right),$$
where $\psi(x)$ is the digamma function. Note that, $a=0$ is a special case.
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Note that $af(a)=f(1/a)$. For similar integrals see Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz.76,498-No.541(1975), 49-50. It is amusing to note that $$\int_{-\infty}^{\infty}\operatorname{sech}(x) \operatorname{sech}[ax(x+i\pi)]\,\mathrm dx=\pi \operatorname{sech}(\pi^2 a/4)$$ but I doubt $f(a)$ has a closed form expression.
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can you, please, specify where one can find that paper. Because it is a specific one. – Caran-d'Ache Sep 19 '13 at 5:37
@Caran-d'Ache, I believe that the paper is "Evaluation of a class of definite integrals" by M.L. Glasser. It claims to examine a class of integrals which would include $\int_{-\infty}^\infty \operatorname{sech} x \operatorname{sech} a x \, dx$ as a special case, but then goes on to evaluate a different special case (where the integrand equals $e^{-\alpha x} \operatorname{sech} x \operatorname{sech}(x + a) \operatorname{sech}(x + b)$ with $a \not= b \not= 0$ and $\operatorname{Re} \alpha| < 3$) which is not at all applicable to the one we are interested in. – Kyle Dec 9 '13 at 4:41 | 2015-01-25 18:23:21 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998248219490051, "perplexity": 1557.1206835562878}, "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/1422115861305.18/warc/CC-MAIN-20150124161101-00108-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://quasarbright.github.io/blog/2022/12/understanding-and-implementing-automatic-differentiation.html | # Understanding and Implementing Automatic Differentiation
::
By: Mike Delmonaco
$\DeclareMathOperator{\expt}{expt} \DeclareMathOperator{\mul}{mul} \DeclareMathOperator{\add}{add} \DeclareMathOperator{\derivative}{derivative}$
Automatic differentiation is a technique that allows programs to compute the derivatives of functions. It is vital for deep learning and useful for optimization in general. For me, it’s always been dark magic, but I recently thought of a nice way to implement it and made a little library. This blog post takes you along the journey of discovering that implementation. Specifically, we will be implementing forward mode automatic differentiation for scalar numbers.
This post requires some knowledge of differential calculus. You’ll need to know basic derivative rules, the chain rule, and it’d help to know partial derivatives. If you’ve taken an introductory calculus course, you should be fine.
The code is in Racket. If you don’t know Racket, you should still be able to follow along. I’ll explain the Racket-y stuff. Don’t let the parentheses scare you away!
# 1 Introduction
Gradient descent is an optimization technique that involves derivatives. You have some quantity you want to optimize, let’s say $$y$$, and you have some variable $$x$$ that you can adjust that affects $$y$$. What value of $$x$$ maximizes $$y$$? Using gradient descent, if we take the derivative of $$y$$ with respect to $$x$$, $$\frac{dy}{dx}$$, that tells us how changing $$x$$ affects $$y$$. If the derivative is positive, increasing $$x$$ increases $$y$$. If it is negative, increasing $$x$$ decreases $$y$$. If it is positive and large, increasing $$x$$ increases $$y$$ a lot. So, if we’re trying to maximize $$y$$, we’d change $$x$$ in the same direction as the derivative. If it’s positive, we increase $$x$$ to make $$y$$ greater. If it’s negative, we decrease $$x$$ to make $$y$$ greater. This will end up maximizing $$y$$ (at least, reaching a local maximum). This is a very useful technique and it is fundamental to deep learning with neural networks. Awesome! But how do you compute the derivative of a function with respect to its input?
Naively, you might try something like this:
> (define (derivative f x h) (/ (- (f (+ x h)) (f x)) h))
> (define (square x) (* x x))
> (derivative square 5 0.01)
10.009999999999764
For those not familiar with Racket, it has prefix arithmetic. This means that instead of writing $$a + b$$, we write (+ a b). It also has a lot of parentheses, which can be tricky to read. You can mostly ignore them and just read based on indentation.
Here, we define a function called derivative which takes in 3 arguments: a Number -> Number function f, a number x representing the input, and a number h representing the step size of the derivative. The body of derivative has a lot of parentheses and prefix arithmetic that is hard to read. Here is what this looks like in normal math notation:
$\derivative(f,x,h) = \frac{f(x+h) - f(x)}{h}$
This is reminiscent of the limit definition of a derivative that you learn about in an introductory calculus course.
$\frac{df}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
This can work well enough as an approximation, but you need a small h and there will always be rounding error, as we saw in the square example. Another issue is that if we have a multi-argument function, we’d have to run the function many times to get the partial derivatives. This is not ideal. What we’d like is to be able to run the function once, inspect several derivatives after, and get exact derivatives. This is the goal of automatic differentiation.
Here is a sneak peek of what we will implement:
> (require number-diff) > (define (f a b) (*o a b)) > (define a (number->dnumber 5)) > (define b (number->dnumber 3)) > (define y (f a b)) > (dnumber->number y) 15 > (dnumber->number (derivative y a)) 3 > (dnumber->number (derivative y b)) 5
*o is just my version of multiplication that supports derivatives.
We’ll even get higher order derivatives!
> (dnumber->number a) 5 > (define y (*o a a)) > (dnumber->number y) 25 > (dnumber->number (derivative y a)) 10 > (dnumber->number (derivative (derivative y a) a)) 2 > (dnumber->number (derivative y a #:order 2)) 2
# 2 How?
## 2.1 The Math
I first encountered automatic differentiation when I was learning about neural networks in college. In the popular neural networks libraries, you define your neural network, and this constructs a computation graph. A computation graph stores the operations, inputs, and outputs of a computation. Here is an example from PyTorch documentation:
image source
This represents the computation $$\log (x_1 x_2) \sin (x_2)$$. Intermediate results like $$x_1 x_2$$ get their own nodes ($$a$$).
The green part of the image shows how the derivatives are calculated. Each operator, like multiplication, logarithm, etc., knows how to compute the derivatives of its result with respect to its inputs. The rest is just applying the chain rule and adding up partial derivatives as appropriate.
There is a lot going on here, so don’t worry about fully understanding this picture yet. We will get there. The key idea is that a computation can be expressed as a graph, where inputs are like leaves, and operations are like nodes. And each node is a tiny computation which is easy to compute the derivatives for.
Let’s start by thinking about some core operators and their derivatives:
$\mul(a,b) = ab$
What are the derivatives?
$\frac{\partial \mul}{\partial a} = b$
$\frac{\partial \mul}{\partial b} = a$
This comes from the rule for multiplying a constant:
$\frac{d}{dx} cx = c$
Remember, taking the partial derivative means treating the other arguments as constants.
$\add(a,b) = a + b$
$\frac{\partial \add}{\partial a} = 1$
The derivative is 1 because $$\frac{da}{da} = 1$$ and $$\frac{da}{db} = 0$$. Using the sum rule, we get $$1+0=1$$.
$\frac{\partial \add}{\partial b} = 1$
This is 1 for the same reason.
$\expt(a,b) = a^b$
$\frac{\partial \expt}{\partial a} = ba^{b-1}$
This derivative treats $$b$$ as a constant, which makes this the derivative of a polynomial. So we use the power rule.
$\frac{\partial \expt}{\partial b} = a^b \ln(a)$
This derivative treats $$a$$ as a constant, which makes this the derivative of an exponential. So we use the exponential rule.
What if things get more complicated? How would you compute this derivative?
$y = (3x + 1)^2$
$\frac{dy}{dx} = ?$
You use the chain rule!
$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$
Let $$u_1 = 3x+1$$.
$y = u_1^2$
Now we can use the power rule:
$\frac{dy}{du_1} = 2u_1$
We can easily tell that $$\frac{du_1}{dx} = 3$$, but let’s do this mechanically to get an idea of how we might automate it:
Let $$u_2 = 3x$$
$u_1 = u_2 + 1$
Now we can use the derivative of $$\add$$:
$\frac{du_1}{du_2} = 1$
Finally, we use the constant factor rule:
$\frac{du_2}{dx} = 3$
Now, we can go back up the chain:
$\frac{du_1}{dx} = \frac{du_1}{du_2}\frac{du_2}{dx} = 1 \cdot 3 = 3$
$\frac{dy}{dx} = \frac{dy}{du_1}\frac{du_1}{dx} = 2u_1 \cdot 3 = 6u_1 = 6(3x+1)$
We did it!
Now we can almost see a recursive algorithm for computing derivatives. If a computation is just a bunch of nested simple operations like addition and multiplication, we can compute the derivative by applying the chain rule and using the partial derivatives of our operators at each step. There’s just one tricky bit: What if $$x$$ shows up twice?
In other words, if we have $$f(a,b)$$ and we know $$\frac{\partial f}{\partial a}$$ and $$\frac{\partial f}{\partial b}$$, how do we compute $$\frac{df(x,x)}{dx}$$?
One way of thinking about a derivative is asking "if we adjust this input a little bit, how does the output change?". So if $$x$$ shows up in multiple places, we are making several inputs change the same way and seeing how the output is affected. If we know how each input changes the output, we just add up all of those little changes to get the big, total change. Concretely, this means we just add the partial derivatives together.
Here is an example:
$\frac{d \add(x,x)}{dx} = 1 + 1 = 2$
The partial derivative of $$\add$$ with respect to each input is 1. So we get $$1+1=2$$. This makes sense because $$x+x=2x$$ and $$\frac{d2x}{dx} = 2$$.
One more example:
$\frac{d \mul(x,x)}{dx} = x + x = 2x$
The partial derivative of $$\mul$$ with respect to each input is the other input. So we get $$x + x = 2x$$. This makes sense because $$x \cdot x = x^2$$ and $$\frac{dx^2}{dx} = 2x$$.
In general, for computing $$\frac{\partial f(a,b)}{\partial x}$$, $$a$$ and $$b$$ might be $$x$$, might depend on $$x$$, or might not depend on $$x$$ at all. To account for this, we use the chain rule and this "partial derivative sum rule" (if there is a name for this, please let me know!):
$\frac{\partial f(a,b)}{\partial x} = \frac{\partial f(a,b)}{\partial a}\frac{\partial a}{\partial x} + \frac{\partial f(a,b)}{\partial b}\frac{\partial b}{\partial x}$
If $$a = x$$, $$\frac{\partial a}{\partial x} = 1$$. If $$a$$ depends on $$x$$, $$\frac{\partial a}{\partial x}$$ be some value. If it does not depend on $$x$$, it will be 0.
For example, let’s compute $$\frac{d}{dx} (5x)^2$$:
$\frac{d \expt(5x,2)}{dx} = \frac{d \expt(5x,2)}{d5x}\frac{d5x}{dx} + \frac{d \expt(5x,2)}{d2}\frac{d2}{dx}$ $\frac{d \expt(5x,2)}{dx} = 2 \cdot 5x \cdot 5 + \frac{d \expt(5x,2)}{d2} \cdot 0$ $\frac{d \expt(5x,2)}{dx} = 50x$
Notice that, since 2 does not depend on $$x$$, its derivative was 0, and that term did not contribute to the overall derivative.
Normally, you don’t think of something like $$(5x)^2$$ as a sum of two partial derivatives like this. You don’t bother taking the derivative of the 2. You normally don’t have to worry about it since it just ends up adding 0. But since we’re trying to automate this, we need to be general and account for the possibility that $$x$$ might show up in the base and the exponent when differentiating an exponential, or genrally, multiple times in any function. In fact, we should’ve done the same thing for $$\frac{d5x}{dx}$$ and added up $$\frac{d\mul(5,x)}{dx}\frac{dx}{dx} and \frac{d\mul(5,x)}{d5}\frac{d5}{dx}$$ to get $$5 \cdot 1 + x \cdot 0 = 5$$.
Side note: You may have been wondering, "where’s the product rule?" In fact, the product rule is not fundamental! It can be derived from the constant factor rule, the chain rule, and this "partial derivative sum rule":
$\frac{d}{dx}f(x)g(x) = \frac{d}{dx}\mul(f(x),g(x)) = \frac{\partial \mul(f(x),g(x))}{\partial f(x)}\frac{df}{dx} + \frac{\partial \mul(f(x),g(x))}{\partial g(x)}\frac{dg}{dx}$ $= g\frac{df}{dx} + f\frac{dg}{dx}$
Nice!
Now, we’re ready for the recursive algorithm to compute (partial) derivatives.
$\derivative(x,x) = 1$ $\derivative(c,x) = 0$ where $$c$$ is a constant and not $$x$$. $\derivative(f(u_1,u_2, \cdots , u_n), x) = \sum_{i=0}^{n} \frac{\partial f(u_1,u_2, \cdots , u_n)}{\partial u_i} \cdot \derivative(u_i, x)$
The base cases are the constant rule and the fact that $$\frac{dx}{dx} = 1$$. The recursive case is the interesting bit. For each input to the computation, we apply the chain rule, which involves a special partial derivative and a recursive call. And we add up those applications of the chain rule for each input.
Why don’t we need a recursive call for $$\frac{\partial f(u_1,u_2, \cdots , u_n)}{\partial u_i}$$? We don’t need one because $$f(u_1,u_2, \cdots , u_n)$$ is some simple operation like $$\mul$$ or $$\add$$ or $$\expt$$ for which we know how to compute the partial derivatives. Since we are differentiating with respect to an immediate input, $$u_i$$, this knowledge is all we need and we don’t need to apply the chain rule or make a recursive call.
This is all very hand-wavy, but it’ll become more concrete soon, I promise!
At this point, that computation graph picture from earlier should start to make some sense. In particular, the green part. Each operator knows how to compute the derivatives of its result with respect to its inputs. That’s the "backward" stuff. And it says "Grads from different paths are added together" (grad=gradient=derivative for our purposes) because of that "partial derivative sum rule". In their example, $$x_2$$ shows up twice in the computation, so we need to account for that by adding multiple partial derivatives together.
## 2.2 The Implementation
We have a (somewhat hand-wavy) algorithm for computing derivatives. Now, we have to actually implement it.
### 2.2.1 A Data Representation
Let’s think about the pieces we’ll need:
We’ll need operators which can compute partial derivatives of their result with respect to their input(s). We’ll also need constants to pass to these operators. It is unclear how "variables" will be represented, so let’s not think about that for now. In order to compute the derivative of a result with respect to some input after the computation is complete, we’ll need to remember what inputs were involved in a computation.
So a computation will need to store the value of its result and its inputs, at a bare minimum. Its inputs may be results of other computations too. This means there will be a tree structure where a computation has a node that stores its result and has children for its inputs. At the leaves of this tree will be constants. We can think of a constant as a computation with no inputs that has itself as a result. Let’s write a data definition for this tree:
(struct computation [value inputs] #:transparent) ; A Computation is a ; (computation Number (listof Computation)) ; It represents the result of a numerical computation ; Examples: (define const2 (computation 2 (list))) (define const3 (computation 3 (list))) (define prod23 (computation 6 (list const2 const3)))
In Racket, struct creates a structure type. It’s like a struct in C, or a data class in Python or Java. It only has fields, no methods. #:transparent automatically implements conversion to a string for our structure type so we can print it, and semicolon creates a line comment. The list function creates lists, so (list) creates the empty list.
This is on the right track, but it’s not enough information to compute derivatives. We have no way to compute the derivative of the result with respect to an input. What else do we need in the tree?
Let’s take another look at the recursive case of our algorithm. The recursive step involves computing the partial derivatives of the result with respect to one of its direct inputs, $$\frac{\partial f(u_1,u_2, \cdots , u_n)}{\partial u_i}$$. This is the missing piece. We just have to figure out how to represent this information in our tree.
One option would be to do what PyTorch seems to do: Store some information about the operation in a node. In our prod23 example, we’d store something representing multiplication in the tree. In the PyTorch, we see that the computation graph stores something for * and something called MultBackward. MultBackward is something that knows how to compute the derivative of a product with respect to its input factors. With this design, we could store a function in the tree that can be used to compute the derivative of the result with respect to each input. Each operator would be responsible for computing its result and creating a node containing that result, a function for computing derivatives, and the inputs.
Another option is to pre-compute these derivatives and store the value of each derivative in the tree with its corresponding input. With this design, each operator would be responsible for computing its result, computing the derivative of the result with respect to each input, and creating a node containing the result, the inputs, and the derivatives of the result with respect to each input. This is the design I chose for my implementation, and the one we’ll implement together.
Both options have pros and cons. This is just the design I came up with. One good thing about this design is that it generalizes nicely to higher order derivatives, which was a goal of my implementation.
Let’s think about an example to make this more concrete. Let’s say we have some operator f and some inputs a and b. The computation is (f a b) (that’s how we write $$f(a,b)$$ in Racket). Let’s call the result y. Keep in mind that a, b, and y are not plain numbers. They are trees. In this case, a and b will be direct children of y since they are inputs to the computation that produced y.
The tree for y will store the numerical value of the result of the computation, and for its children, it will have a and b. It will also store the numerical value of $$\frac{\partial y}{\partial a}$$ and the numerical value of $$\frac{\partial y}{\partial b}$$.
a and b may have children of their own, but we exclude them from these diagrams. dyda and dydb are plain numbers, not trees.
If we’re trying to compute (derivative y x), where x may be some input to the computation that produced a, the first thing we’ll encounter is the node y, which came from (f a b), and we’ll do the recursive step. According to our algorithm, we need to compute
$\frac{\partial y}{\partial a} \cdot \derivative(a,x) + \frac{\partial y}{\partial b} \cdot \derivative(b,x)$
Those partial derivatives are what we have in our tree. The rest is just making recursive calls and applying the chain rule. Now we have enough information to compute derivatives! Let’s write a data definition:
> (struct dnumber [value inputs] #:transparent) ; a DNumber ("differentiable number") is a ; (dnumber Number (listof DChild) ) ; It represents the result of a differentiable computation. ; ‘value‘ is the numerical value of the result of this computation ; ‘inputs‘ associates each input to this computation with the numerical value of its partial derivative > (struct dchild [input derivative] #:transparent) ; A DChild is a ; (dchild DNumber Number) ; It represents an input to a differential computation. ; ‘input‘ is the DNumber that was supplied as an input to the parent computation ; ‘derivative‘ is the numerical value of the partial derivative of the parent result with respect to this input.
A DNumber stores the value of its result as a plain number and, for each input, the input’s DNumber and the partial derivative of the result with respect to that input as a plain number.
Let’s look at $$2 \cdot 3$$ as an example:
> (define const2 (dnumber 2 (list))) > (define const3 (dnumber 3 (list))) > (define prod23 (dnumber 6 (list (dchild const2 3) (dchild const3 2))))
We create dnumbers for the constants 2 and 3. Since these are constants, they have no children ((list) creates an empty list).
We then construct a node for the multiplication which stores the value of the result (6) and each input paired with its derivative.
The result of $$2 \cdot 3$$ is 6. The inputs are 2 and 3. Recall the derivatives of the $$\mul$$ multiplication operator:
$\frac{\partial \mul(a,b)}{\partial a} = b$ $\frac{\partial \mul(a,b)}{\partial b} = a$
The derivative of the product with respect to one of its factors is the other factor. So the derivative of prod23 with respect to const2 is the plain number 3.
We are pre-computing that $$\frac{\partial f(u_1,u_2, \cdots, u_n)}{\partial u_i}$$ from our algorithm and storing it directly in our tree as the derivative field (the second argument of the constructor) of a dchild. Each dchild contains the tree for that input $$u_i$$ and the value of $$\frac{\partial f(u_1,u_2, \cdots, u_n)}{\partial u_i}$$. This is exactly what we need for the recursive case.
Let’s implement our multiplication operator:
> (define (mul a b) (dnumber (* (dnumber-value a) (dnumber-value b)) (list (dchild a (dnumber-value b)) (dchild b (dnumber-value a)))))
> (mul const2 const3)
(dnumber 6 (list (dchild (dnumber 2 '()) 3) (dchild (dnumber 3 '()) 2)))
> prod23
(dnumber 6 (list (dchild (dnumber 2 '()) 3) (dchild (dnumber 3 '()) 2)))
In the output, we see '() instead of (list). That’s just another way of writing it.
The function dnumber-value is a field-accessor function that is automatically generated from the struct declaration. This field contains the numerical result of the computation. Since Racket’s * function expects plain numbers, we have to get the numerical values of the inputs with dnumber-value before passing them to *.
We also have to do this when creating the dchildren. This is because the derivative field of a dchild must be a plain number. However, the input field must be a DNumber, so we pass the input itself in as the first argument to the dchild constructor and the numerical value of the other input as the second argument.
Let’s do another example, this time $$4 + 5$$:
> (define const4 (dnumber 4 (list))) > (define const5 (dnumber 5 (list))) > (define sum45 (dnumber 9 (list (dchild const4 1) (dchild const5 1))))
Recall the derivatives of the $$\add$$ addition operator:
$\frac{\partial \add(a,b)}{\partial a} = 1$ $\frac{\partial \add(a,b)}{\partial b} = 1$
Now let’s implement it:
> (define (add a b) (dnumber (+ (dnumber-value a) (dnumber-value b)) (list (dchild a 1) (dchild b 1))))
> (add const4 const5)
(dnumber 9 (list (dchild (dnumber 4 '()) 1) (dchild (dnumber 5 '()) 1)))
> sum45
(dnumber 9 (list (dchild (dnumber 4 '()) 1) (dchild (dnumber 5 '()) 1)))
Great! Now we have a few differentiable operators. Adding more operators will be just like this. We extract the values from the arguments, compute the result using the inputs’ values and built-in arithmetic operators from Racket, and then create a list of dchildren mapping each input to the value of the derivative of the result with respect to that input.
Now we’re finally ready to start implementing the derivative function!
### 2.2.2 The Derivative!
Before we actually implement the derivative, there are a few things we have to address.
Here is a problem:
> (define sum44 (dnumber 8 (list (dchild const4 1) (dchild const4 1))))
What should (derivative sum44 const4) be? If we apply the "partial derivative sum rule", it should be $$1 + 1 = 2$$. This makes sense because $$x + x = 2x$$ and $$\frac{d}{dx} 2x = 2$$.
Let’s look at another example:
> (define const-four (dnumber 4 (list)))
> (define sum4-four (dnumber 8 (list (dchild const4 1) (dchild const-four 1))))
Should (derivative sum4-four const4) also be 2? const4 and const-four are both constant computations that have the result 4 and no inputs. But should they be treated as the same? What does it mean to be the same?
Let’s take a step back and think about a real example where this matters:
> (define (add-4 x) (add x (dnumber 4 (list))))
Here, $$f(x) = x + 4$$. $$\frac{df}{dx} = 1$$. If we consider const4 and any other constant 4 to be the same, the derivative of f with respect to x would be 1 for all inputs except for any constant 4, in which case it is 2. This doesn’t make sense. But const4 is obviously the same as const4, so how can we tell these two 4 constants apart?
Racket has a few different equality functions. The one we want is eq?. By convention, when a function returns a boolean, the name ends in a question mark, pronounced "huh". So eq? is pronounced "eek-huh". Anyway, here’s how it works:
> (define nums (list 1 2 3)) > (eq? nums nums) #t > (define numbers (list 1 2 3)) > (eq? nums numbers) #f > (eq? (list 1 2 3) (list 1 2 3)) #f > (define nums-alias nums) > (eq? nums nums-alias) #t
In racket, #t means true and #f means false.
eq? returns #t when both objects are aliases of each other. If both objects were created from different constructor calls, they will not be eq? to each other. If you know Java, it behaves just like ==. This is often called reference equality or identity equality.
> (eq? const4 const4) #t > (eq? const-four const-four) #t > (eq? const4 const-four) #f > (eq? (dnumber 4 (list)) (dnumber 4 (list))) #f > (define const4-alias const4) > (eq? const4-alias const4) #t
This is exactly what we want! We consider two dnumbers the same if they are eq? to each other. This is a little confusing, but it’s necessary to distinguish between different computations that coincidentally have the same tree data, but aren’t associated with each other. This is exactly the behavior that we need for variables, like our $$x + 4$$ example earlier.
What if const4 shows up twice as an input in a computation like in sum44? In that case, our tree isn’t actually a tree. It’s a directed acyclic graph. DAG for short. This is like a tree, except a node can be a child of multiple nodes. However, there cannot be cycles. In other words, a node cannot be an input to itself, or an indirect input to itself. This is why it’s called a computation graph and not a computation tree.
Now we’re finally ready to implement the derivative function!
; (DNumber DNumber -> Number)
; Compute the partial derivative of y with respect to x.
> (define (derivative y x) (if (eq? y x) 1 (let ([inputs (dnumber-inputs y)]) (for/sum ([input inputs]) (let ([u (dchild-input input)] [dydu (dchild-derivative input)]) (* dydu (derivative u x)))))))
First, let’s cover some Racket stuff.
(if cond-expr then-expr else-expr) is like an if statement in most languages, except it’s just an expression that evaluates to one of its branch expressions. If cond-expr evaluates to anything other than #f, the result is then-expr. Otherwise, the result is else-expr.
(let ([var val-expr]) body-expr) is a local variable definition expression. It binds var to the result of val-expr and var is in scope in body-expr. The whole expression evaluates to the result of body-expr. You can also bind multiple variables like (let ([x 1] [y 2]) (+ x y)).
for/sum is like a for-loop, but it is an expression and calculates the sum of the results from each iteration.
For example:
> (define words (list "My" "name" "is" "Mike"))
> (for/sum ([word words]) (string-length word))
12
Again, dnumber-inputs, dchild-input, and dchild-derivative are field accessor functions. In the inner let, we bind the input DNumber to the variable u and the (partial) derivative of y with respect to u to the variable dydu. Remember, we store this derivative directly in the computation graph.
Now let’s think about what’s going on. If y and x are the same, then the derivative is 1. That’s the first branch of the if.
That’s also the first case of the algorithm:
$\derivative(x,x) = 1$
Otherwise, we apply the "partial derivative sum rule" and the chain rule. That’s the recursive case of our algorithm.
$\derivative(f(u_1,u_2, \cdots , u_n), x) = \sum_{i=0}^{n} \frac{\partial f(u_1,u_2, \cdots , u_n)}{\partial u_i} \cdot \derivative(u_i, x)$
$\derivative(c,x) = 0$ where $$c$$ is a constant and not $$x$$.
It’s actually hidden in the for/sum part. If y is a constant (no inputs) and it is not eq? to x, the for/sum will loop over an empty list of inputs. The sum of nothing is 0, so we return 0.
If x does not show up in y’s computation graph, we’ll never get the eq? case. The only base case we’ll hit is the implicit unequal constant base case, which returns 0. Since each base case returns 0, each recursive call will involve multiplying dydu by 0, which will produce 0. And since we’re just adding those together for each input, we’ll be adding up a bunch of zeros, which will produce 0. By induction, we’ll get 0 for the whole derivative if x does not appear in y.
Let’s test out our implementation:
> (derivative const4 const4)
1
> (derivative const4 const3)
0
> (derivative const4 const-four)
0
> (derivative sum44 const4)
2
> (derivative sum44 const-four)
0
> (derivative (add-4 const3) const3)
1
> (derivative (add-4 const4) const4)
1
> (define (double x) (add x x))
> (derivative (double const3) const3)
2
> (derivative (double const3) const4)
0
> (define (square x) (mul x x))
> (derivative (square const4) const4)
8
> (derivative (square const3) const3)
6
> (derivative (mul const3 const4) const4)
3
> (derivative (mul const3 const4) const3)
4
> (derivative (square (add (mul const3 const4) const2)) const4)
84
We did it! Now, if you add some more differentiable operators, you can do all sorts of things. You can do gradient descent, you can do analysis, you can implement neural networks, and anything else involving derivatives.
Let’s implement some more operators:
> (define (e^x x) (let ([result (exp (dnumber-value x))]) (dnumber result (list (dchild x result)))))
> (e^x const4)
(dnumber 54.598150033144236 (list (dchild (dnumber 4 '()) 54.598150033144236)))
> (derivative (e^x const4) const4)
54.598150033144236
> (define (sub a b) (add a (mul (dnumber -1 (list)) b)))
> (define (reciprocal x) (dnumber (/ 1 (dnumber-value x)) (list (dchild x (/ -1 (* (dnumber-value x) (dnumber-value x)))))))
> (define (div a b) (mul a (reciprocal b)))
sub and div are interesting. They don’t directly construct the resulting DNumber. They just use other operators! If we implement a sufficient core library of mathematical operators, it’s easy to define more complicated differentiable functions in terms of those core functions without having to think about derivatives at all.
Automatic differentiation is useful, but if I’m being honest, the real reason I wrote this blog post was because of how much I love that recursive case. Once I wrote that, I wanted to show everybody. You can see the chain rule so clearly!
Our implementation of derivative shows the essence of automatic differentiation. The derivative of something with respect to itself is 1, and the derivative of some function call with respect to a possibly indirect input is the sum over the chain rule applied to each input. Beautiful!
What about higher order derivatives? To achieve this, can we just apply derivative twice? Unfortunately, no. The signature doesn’t line up. derivative returns a plain number, so we can’t pass that as y to another call to derivative. But what if we returned a DNumber instead? That DNumber would have to represent the computation that produced the derivative itself. Is this even possible?
Yes! But it’s not trivial. Think about how this might work and what problems you would run into with a function like $$\expt(a,b) = a^b$$.
This post is already pretty long and a lot to digest, so I wont get into higher order derivatives here. But I will in part 2! | 2023-01-29 16:47:45 | {"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": 0.7580896019935608, "perplexity": 370.40372658748595}, "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-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00598.warc.gz"} |
https://www.projecteuclid.org/euclid.aos/1024691368 | ## The Annals of Statistics
### The sample autocorrelations of heavy-tailed processes with applications to ARCH
#### Abstract
We study the sample ACVF and ACF of a general stationary sequence under a weak mixing condition and in the case that the marginal distributions are regularly varying. This includes linear and bilinear processes with regularly varying noise and ARCH processes, their squares and absolute values. We show that the distributional limits of the sample ACF can be random, provided that the variance of the marginal distribution is infinite and the process is nonlinear. This is in contrast to infinite variance linear processes. If the process has a finite second but infinite fourth moment, then the sample ACF is consistent with scaling rates that grow at a slower rate than the standard $\sqrt{n}$. Consequently, asymptotic confidence bands are wider than those constructed in the classical theory. We demonstrate the theory in full detail for an ARCH (1) process.
#### Article information
Source
Ann. Statist., Volume 26, Number 5 (1998), 2049-2080.
Dates
First available in Project Euclid: 21 June 2002
https://projecteuclid.org/euclid.aos/1024691368
Digital Object Identifier
doi:10.1214/aos/1024691368
Mathematical Reviews number (MathSciNet)
MR1673289
Zentralblatt MATH identifier
0929.62092
#### Citation
Davis, Richard A.; Mikosch, Thomas. The sample autocorrelations of heavy-tailed processes with applications to ARCH. Ann. Statist. 26 (1998), no. 5, 2049--2080. doi:10.1214/aos/1024691368. https://projecteuclid.org/euclid.aos/1024691368
#### References
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• 4 BROCKWELL, P. J. and DAVIS, R. A. 1991. Time Series: Theory and Methods, 2nd ed. Springer, New York.
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• 9 DAVIS, R. A. and MARENGO, J. 1990. Limit theory for the sample covariance and correlation matrix function of a class of multivariate linear processes. Stochastic Models 6 483 498.
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• 11 DAVIS, R. A. and RESNICK, S. I. 1985. More limit theory for the sample correlation function of moving averages. Stochastic Process. Appl. 20 257 279.
• 12 DAVIS, R. A. and RESNICK, S. I. 1986. Limit theory for the sample covariance and correlation functions of moving averages. Ann. Statist. 14 533 558.
• 13 DAVIS, R. A. and RESNICK, S. I. 1996. Limit theory for bilinear processes with heavy tailed noise. Ann. Appl. Probab. 6 1191 1210.
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• FORT COLLINS, COLORADO 80523-1877 UNIVERSITY OF GRONINGEN E-MAIL: rdavis@stat.colostate.edu NL-9700 AV GRONINGEN THE NETHERLANDS | 2019-10-23 11:52:41 | {"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": 0.4972459673881531, "perplexity": 6032.6808963176245}, "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-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00425.warc.gz"} |
https://www.ssccglapex.com/if-the-sum-of-n-terms-of-an-a-p-is-3mathbfn2-5n-then-which-of-its-terms-is-164/ | ### If the sum of n terms of an A.P. is 3$\mathbf{n}^{2}$ + 5n then which of its terms is 164 ?
A. 26th B. 27th C. 28th D. None of these Answer: Option B
### Solution(By Apex Team)
Sum of n terms of an A.P. = $3 n^{2}+5 n$ Let a be the first term and d be the common difference $\begin{array}{l}\mathrm{S}_{\mathrm{n}}=3\mathrm{n}^2+5\mathrm{n}\\ \mathrm{S}_1=3(1)^2+5\times1=3+5=8\\ \mathrm{~S}_2=3(2)^2+5\times2=12+10=22\\ \therefore\text{ First term }(\mathrm{a})=8\\ \mathrm{a}_2=\mathrm{S}_2-\mathrm{S}_1=22-8=14\\ \mathrm{~d}=\mathrm{a}_2-\mathrm{a}_1=14-8=6\\ \text{ Now }\mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ \Rightarrow164=8+(\mathrm{n}-1)\times6\\ \Rightarrow6\mathrm{n}-6=164-8\\ \Rightarrow6\mathrm{n}=156+6\\ \Rightarrow6\mathrm{n}=162\end{array}$ ⇒ n = $\Large\frac{162}{6}$ ⇒ n = 27 ∴ 168 is 27th term
A. 22
B. 25
C. 23
D. 24
A. 5
B. 6
C. 4
D. 3
A. -45
B. -55
C. -50
D. 0 | 2022-05-28 23:24:38 | {"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": 0.928239107131958, "perplexity": 1435.714516993518}, "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/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00397.warc.gz"} |
https://www.tec-science.com/mechanics/gases-and-liquids/pressure-head-and-head-loss/ | • What is meant by static pressure head of a pump?
• What is the elevation head (geodetic head) of a piping system?
• What is the maximum suction head for water?
• What is meant by head loss?
## Static head of a pump
Pumps are used to deliver gases or liquids. These fluids are usually pumped from a lower level to a higher level. The pump is located between these levels. The pump creates a negative pressure on the so-called suction side so that the fluid is sucked in. The pump then puts the fluid under high pressure. At the outlet, the fluid is now pushed through the pipe to the higher level. In the following we will only consider incompressible fluids such as liquids.
The difference in height the liquid can overcome depends on the power of the pump, the density of the fluid to be pumped and the volume flow rate. To show this, we consider a vertical pipeline. Through this pipe, a fluid of density $$\rho$$ is pumped up from a tank. The water is discharged at the open end of the pipe. The difference in height to be overcome is denoted by $$H$$. Friction losses are neglected in the following.
The time period until a liquid parcel has been delivered from the lower tank (liquid level) to the top of the pipe is denoted by $$t$$. Within this time $$t$$ obviously once the total liquid of mass $$m = V \cdot \rho$$ in the pipe must be lifted by the height $$H$$ (the volume of the fluid inside the pipe is denoted by $$V$$). The amount of energy $$W_\text{H}$$ required for this is given by the following formula:
\begin{align}
&W_\text{H} = m \cdot g \cdot H ~~~\text{where}~~~m =V \cdot \rho \\[5px]
&W_\text{H} = V \cdot \rho \cdot g \cdot H ~~~~~\text{energy of the pump} \\[5px]
\end{align}
This energy is obviously converted within the time $$t$$, resulting in the following power $$P_\text{H}$$ that the pump delivers:
\begin{align}
&P_\text{H} = \frac{W_\text{H}}{t} \\[5px]
&P_\text{H} = \frac{V \cdot \rho \cdot g \cdot H}{t} \\[5px]
&P_\text{H} = \underbrace{\frac{ V}{t}}_{\dot V} \cdot \rho \cdot g \cdot H\\[5px]
&\boxed{P_\text{H} = \dot V \cdot \rho \cdot g \cdot H} ~~~~~\text{power of the pump} \\[5px]
\end{align}
When deriving this formula, it was exploited that the quotient of fluid volume $$V$$ and time $$t$$ corresponds to the delivered volume flow rate $$\dot V$$.
Conversely, this means: For a given power of the pump $$P_\text{H}$$ and volume flow rate $$\dot V$$ to be delivered, the fluid can only overcame a certain height $$H$$, depending on the density of the fluid $$\rho$$ (at a greater height, the hydrostatic pressure of the fluid column would be greater than the pressure generated by the pump and the fluid could not be pumped any higher):
\begin{align}
&P_\text{H} = \frac{W_\text{H}}{t} \\[5px]
&P_\text{H} = \frac{V \cdot \rho \cdot g \cdot H}{t} \\[5px]
&P_\text{H} = \underbrace{\frac{ V}{t}}_{\dot V} \cdot \rho \cdot g \cdot H\\[5px]
\label{p}
&\boxed{H = \frac{P_\text{H}}{\dot V \cdot \rho \cdot g}} ~~~~~\text{static head of the pump} \\[5px]
\end{align}
This maximum height a pump can deliver is also called static pressure head, static head or pressure head of a pump.
The static head of a pump is the maximum height a pump can deliver in a loss-free case due to the mechanical power transferred to the fluid at a given volume flow rate and density!
Note that the static head of the pump does not take into account friction losses inside the pipe or pressure losses due to valves, bends, fittings, etc. This is because these losses cannot be directly attributed to the pump, but depend on the the piping system for which the pump is to be used. Manufacturers of pumps can therefore not take such losses into account anyway, as they do not know the operating conditions of the pump. For this reason, such pressure losses are taken into account by a so-called head loss of the piping system (more about this later).
## Pump efficiency
The power $$P_\text{H}$$ in the above equations refers only to the power that the pump effectively delivers to the fluid, i.e. the power that is actually required to continuously lift the fluid to the height $$H$$! This power output is not identical with the power input $$P_\text{in}$$ of the pump (power consumption), i.e. the power that an electric pump, for example, has to draw from the mains.
It is therefore necessary to take into account conversion losses that occur when converting the supplied electrical power into mechanical power for transporting the fluid. Also flow losses in the pump due to turbulence must be taken into account, especially at high volume flow rates. All this is summarized in a pump efficiency $$\eta$$:
\begin{align}
&\boxed{P_\text{H} = P_\text{in} \cdot \eta} \\[5px]
\end{align}
The pump efficiency is not a constant value, but depends on the volume flow rate! The efficiency first increases with increasing flow rate and then decreases again from a maximum point due to turbulences and the associated flow losses. Typical maximum efficiencies for pumps are between 70 % and 90 %.
The figure below shows for a given rotational speed of a centrifugal pump the typical curves for pressure head, pump efficiency and power consumption of the pump as a function of volume flow rate. Note that such curves are only valid for certain rotational speed of the pump.
## Static head of a piping system
The physical difference in height between a lower placed and a higher placed basin is called elevation head or geodetic head $$H_\text{e}$$. The respective liquid levels serve as reference points for the elevation head, provided that the higher-lying tank is filled from below. If it is filled from above, the reference point is the point at which the liquid flows out of the pipe. For the lower basin, it does not matter how deep the suction pipe is immersed in the liquid. The liquid surface must always be taken as a basis, since the liquid in the suction pipe rises to the outer liquid level by itself anyway without the help of the pump.
The elevation head of a piping system can be further divided into a suction head on the suction side of the pumpe and a discharge head on the discharge side of the pump. Both heads together comprise the elevation head of the piping system.
The elevation head of a piping system is the physical difference in height between a lower and a higher liquid level! It results from the sum of the suction head and discharge head.
In order for liquids to be pumped, the static head of the pump must always be greater than the elevation head of the system. However, this only applies if there are no friction losses or flow losses in the pipe due to installed components such as valves, bends, fittings or measuring instruments. Furthermore, the maximum suction head is limited by physical conditions.
On the suction side, the pump works like a drinking straw. This means that the pump does not suck in liquid at all. Rather, the (ambient) pressure on the surface of the liquid pushes the fluid into the pump. Since the ambient pressure is finite, even when a perfect vacuum is created, it is not possible to overcome any suction head. By neglecting friction and flow losses, the maximum suction lift is determined according to the following formula (see article How does a drinking straw work? for the derivation of this formula).
\begin{align}
\label{hmax}
&\boxed{h_\text{s,max} = \frac{p_0}{\rho \cdot g} } ~~~~~\text{maximum suction head} \\[5px]
\end{align}
In this formula, $$p_0$$ denotes the (ambient) pressure on the liquid surface and $$\rho$$ denotes the density of the liquid. For pumping water with a density of $$\rho = 1000 \frac{\text{kg}}{\text{m³}}$$ the maximum suction head at an ambient pressure of 1 bar is thus 10 meters.
However, due to the fact that no pump can create a perfect vacuum and due to the viscosity of the pumped liquid, friction losses are inevitable, the maximum geodetic suction head for water is in practice only 8 metres. Note that when the tank is closed, the ambient pressure can be artificially increased, so that higher suction heads are possible.
Unlike the suction head, the discharge head on the discharge side of the pump is in principle not limited to a maximum value. Depending on the pressure generated on the discharge side, (almost) any discharge head can be achieved.
As already mentioned, in reality friction and flow losses in the piping system must be taken into account. So in practice, the pump must transfer a greater power to the fluid than in the friction-free case. Taking friction into account, the real system behaves as if a fictitious frictionless system had a higher head. This additional (fictitious) head, which includes friction and flow losses, is called friction head or head loss $$H_\text{f}$$.
If $$P_\text{f}$$ denotes the power loss that occurs with a certain volumetric flow rate $$\dot V$$, then the friction head $$H_\text{f}$$ can be determined according to equation (\ref{p}):
\begin{align}
&\boxed{H_\text{f} = \frac{P_\text{f}}{\dot V \cdot \rho \cdot g} } ~~~~~\text{friction head (head loss)} \\[5px]
\end{align}
The (fictitious) total static head of the piping system $$H_\text{tot}$$ is therefore greater than the elevation head $$H_\text{e}$$ by the amount of the friction head $$H_\text{f}$$:
\begin{align}
&\boxed{H_\text{tot} = H_\text{e} + H_\text{f}} ~~~~~\text{total static head of the piping system} \\[5px]
\end{align}
So one has to compare the static head of the pump with the (fictitious) static head of the piping system when it comes to selecting a suitable pump. The static head of the pump must be greater than the total static head of the system in order for the fluid to be conveyed. However, this does not yet take into account the fact that different pressures can exist on the liquid surface in the upper and lower tanks. For this reason, a pressure head must generally also be taken into account, which will be discussed in more detail in the next section.
Even in a horizontal pipe, friction inevitably occur due to the viscosity of the fluid. In this case, the associated head loss can indeed be shown very clearly. One can imagine small vertical tubes attached to the pipe. Due to the static pressure in the flowing liquid, the fluid in the vertical tubes is pressed upwards by a certain amount. Due to the friction losses in the pipe, however, the static pressure downstream decreases (assuming a constant pipe cross-section). The fluid in a downstream tube only reaches a lower height. The difference in the liquid levels corresponds to the head loss of the horizontal pipe.
Imagine the following situation. Water is to be pumped from a closed tank to an open tank 6 meters higher up. In the closed tank there is a positive pressure (overpressure) created by compressor. Even without the presence of a pump, this positive pressure will push the water upwards. At an overpressure of 0,1 bar the water theoretically rises to a height of 1 meter above the liquid level [see formula (\ref{hmax})]. So the pump only has to overcome the last 5 meters of height. From an energy point of view, the system has a total static head of only 5 meters from the pump’s point of view (without taking into account head loss).
Conversely, the total static head of the system increases when the lower tank is open and a positive pressure is generated in the higher tank. In this case the higher pressure in the upper tank pushes the water in the pipe downwards. This means that the pump now has to overcome a much greater difference in height. The effects only balance each other out if the pressures in both tanks are the same (for example ambient pressure in both tanks). Even without external compressors, pressure differences occur in closed tanks, since the air volume in the tank also changes when the liquid level changes, without air being able to escape or flow in.
The increased head or the reduced head due to the pressure difference $$\Delta p$$ between the two tanks is called pressure head $$H_\text{p}$$. The pressure difference is determined by the difference between the pressure in the upper reservoir $$p_2$$ and the pressure in the lower reservoir $$p_1$$. In this way, the sign is also reproduced correctly, so that in the event of a negative pressure at the upper reservoir (or a positive pressure at the lower reservoir) a negative pressure head results, which reduces the total static head of the system.
\begin{align}
&H_\text{p} = \frac{\Delta p}{\rho \cdot g} \\[5px]
&\boxed{H_\text{p} = \frac{p_\text{2}-p_\text{1}}{\rho \cdot g} } ~~~~~\text{pressure head} \\[5px]
\end{align}
The total head of the system $$H_\text{tot}$$ is thus generally determined from the sum of the elevation head $$H_\text{e}$$, friction head $$H_\text{f}$$ and pressure head $$H_\text{p}$$:
\begin{align}
&\boxed{H_\text{tot} = H_\text{e} + H_\text{V} + H_\text{p} } ~~~~~\text{total static head of the piping system} \\[5px]
\end{align}
## Head as an energy per unit weight
At this point the head according to equation (\ref{p}) shall be examined more closely and interpreted somewhat differently. For this purpose it will be exploited that power is defined as energy per unit time and volume flow flow ist defined as liquid volume per unit time.
\begin{align}
\require{cancel}
&H = \frac{P_\text{H}}{\dot V \cdot \rho \cdot g} ~~~~~\text{where}~~~P_\text{H}=\frac{W_\text{H}}{t}~~~~~\text{and}~~~\dot V = \frac{V}{t}~~~\text{:}\\[5px]
&H = \frac{\frac{W_\text{H}}{\bcancel{t}}}{\frac{V}{\bcancel{t}} \cdot \rho \cdot g} \\[5px]
&H = \frac{W_\text{H}}{\underbrace{V \cdot \rho}_{m} \cdot g} \\[5px]
&H = \frac{W_\text{H}}{m \cdot g} \\[5px]
&\boxed{H = \frac{W_\text{H}}{F_\text{g}}} \\[5px]
\end{align}
This formula shows that head can be interpreted as an energy per unit weight. Therefore, the following statements apply:
• static head of the pump = Energy of the pump transferred to a fluid element (related to the weight of the fluid element).
• static head of the piping = Energy required to deliver a fluid element (related to the weight of the fluid element).
## System characteristic curve
While elevation head and pressure head are constant values of a piping system, the friction head or head loss depends on the volume flow rate. The pressure losses increase with increasing flow rate. In the article Pressure Loss in pipe systems (Darcy friction factor) the following equation has been derived, which describes the pressure loss $$\Delta p_\text{f}$$ of a pipe with the friction factor $$\lambda$$, the inner diameter $$d$$ and the length $$L$$:
\begin{align}
& \boxed{\Delta p_\text{f} = \lambda \cdot \frac{8\rho~L}{\pi^2} \cdot \frac{\dot{V}^2}{d^5}} ~~~\text{pressure loss in a straight pipe section} \\[5px]
\end{align}
This pressure loss is inevitably associated with a loss of mechanical energy and a corresponding head loss. The head loss thus increases (approximately) quadratically with the volume flow rate. “Approximately” because the friction factor is in turn influenced by the volume flow rate. The figure below shows qualitatively the characteristic curve of the total static head of a piping system as a function of the volume flow rate.
While the total head of the system increases with increasing volume flow rate, the static head of the pump decreases due to the increasing flow losses within the pump. During operation of the pump, depending on the volume flow rate, a common operating point (working point) is established which corresponds to the point of intersection between the pump characteristic curve and the system characteristic curve.
The system characteristic curve can be influenced by a throttle valve to control the flow rate. However, it must be noted that a centrifugal pump has a maximum efficiency at a certain volume flow rate. For energy-efficient operation, the working point should be as close as possible to this point of maximum efficiency. However, a change in the system characteristic curve caused by a throttle valve usually has a negative effect on the operating point – lower efficiencies result from the greater flow losses due to the throttling. A change in the rotational speed of the pump to control the volume flow rate can therefore be more sensible at this point. | 2021-04-15 23:29:21 | {"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": 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": 0.8992964625358582, "perplexity": 601.4047920504396}, "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-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00040.warc.gz"} |
https://en.zdam.xyz/problem/2385/ | #### Problem 85E
85. If $\sum a_{n}$ is convergent and $\sum b_{n}$ is divergent, show that the series $\Sigma\left(a_{n}+b_{n}\right)$ is divergent. [Hint: Argue by contradiction.] | 2022-11-27 06:24:03 | {"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": 0.9553459286689758, "perplexity": 586.169284391592}, "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-20221127041342-20221127071342-00184.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-6-systems-of-equations-and-inequalities-6-5-linear-inequalities-practice-and-problem-solving-exercises-page-393/10 | ## Algebra 1
Published by Prentice Hall
# Chapter 6 - Systems of Equations and Inequalities - 6-5 Linear Inequalities - Practice and Problem-Solving Exercises: 10
#### Answer
Yes, it is a solution to the inequality.
#### Work Step by Step
To solve this, input (0,0) into the inequality $y\geq3x - 2$ $(0)\geq3(0) - 2$ Then solve $0\geq-2$ The inequality now reads 0 is greater than or equal than -2. SInce that is true, this is a solution to the inequality.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2018-05-22 10:26:03 | {"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": 0.560053288936615, "perplexity": 1208.2662993600773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00163.warc.gz"} |
https://lims.ac.uk/paper/self-organization-of-knowledge-economies-2/ | # Self-organization of knowledge economies
The Yule-Simon distribution describes the diffusion of knowledge and ideas in a social network which in turn influences economic growth.
Journal of Economic Dynamics and Control 52, 150 (2015)
F. Lafond
Suppose that homogenous agents fully consume their time to invent new ideas and learn ideas from their friends. If the social network is complete and agents pick friends and ideas of friends uniformly at random, the distribution of ideas’ popularity is an extension of the Yule–Simon distribution. It has a power-law tail, with an upward or downward curvature. For infinite population it converges to the Yule–Simon distribution. The power law is steeper when innovation is high. Diffusion follows logistic curves. | 2021-01-19 18:24:38 | {"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": 0.8707126975059509, "perplexity": 1550.9033272453273}, "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-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00463.warc.gz"} |
https://web2.0calc.com/questions/please-help-asap_51191 | +0
0
113
6
+27
Find the vertex of the graph of the equation $x - y^2 + 8y = 13.$
Find the vertex of the graph of the equation $y = -2x^2 + 8x - 15.$
Find the area of the region enclosed by the graph of $x^2 + y^2 = 2x - 6y + 6$
The graph of $y = ax^2 + bx + c$ has an axis of symmetry of $x = 4.$ Find $\frac{b}{a}$.
The graph of $y = ax^2 + bx + c$ is shown below. Find $a \cdot b \cdot c$. (The distance between the grid lines is one unit.)
May 28, 2021
#2
0
1. The vertex of x - y^2 + 8y = 13 is (-8,4).
2. The vertex of y = -2x^2 + 8x - 15 is (4,-15).
May 28, 2021
#3
+34340
+2
1.) x = y^2 -8y + 13
re-arrange to vertex form
x = (y-4)^2 -3 vertex is -3,4
May 28, 2021
#4
+34340
+2
2 ) -2x^2+8x-15 re-arrange to vertex form
-2 (x^2 - 4x) -15
-2 ( x-2)^2 - 7 vertex = 2, -7
May 28, 2021
#5
+34340
+2
3) x^2 -2x + y^2 + 6y = 6 Arrange into standard circle equation form
(x-1)^2 + ( y+3)^2 = 6 +1 + 9
6+ 1 + 9 = 16 = r^2
Area of a circle = pi r^2 = pi (16) = 16 pi units2
May 28, 2021
#6
+27
0
Can you answer the last two questions? Thank you so much for the other answers.... all correct!
Jun 5, 2021 | 2021-09-21 23:35:30 | {"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": 0.37751662731170654, "perplexity": 2918.383510930168}, "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/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00417.warc.gz"} |
http://math.stackexchange.com/questions/191082/visualization-of-the-method-of-steepest-descent | # visualization of the method of steepest descent
I am trying to understand the method of steepest descent (complex integral). I looked in some complex analysis books and also on Wikipedia, but I still don't understand the methodology of approximating such integrals nor the name of this method. Could someone give me a step by step example and a way to visualize it?
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## migrated from physics.stackexchange.comSep 4 '12 at 20:09
This question came from our site for active researchers, academics and students of physics.
Related post on Phys.SE: physics.stackexchange.com/q/14639/2451 – Qmechanic Nov 15 '14 at 19:45
– Qmechanic Nov 15 '14 at 19:57
I will give you a conceptual explanation; I trust you can find the computational details elsewhere once you understand the idea. We are often interested in the behavior of quantities of the type $I(\lambda)=\int_\gamma e^{\lambda f(z)}\,\mathrm{d}z$, where $\gamma$ is a curve in the plane, $\lambda$ is a (real) parameter and $f$ is a comple-valued analytic function. The steppest descent method uses the freedom we have in deforming $\gamma$ between its endpoints using Cauchy's theorem, together with some basic observations about the nature of the integrand $e^{\lambda f(z)}$.
Writing $\lambda f(z)$ as $\lambda(\Re f +i \Im f)$, we see that the integrand will be much larger around the points $z$ on $\gamma$ where $\Re f$ attains a maximum than elsewhere. In fact, it will turn out that as $\lambda$ grows, the main contribution to the integral will come from arbitrarily small neighborhoods of maxima of $\Re f$ (and hence of $|e^{\lambda f(z)}|$. This observation is also the basis of Laplace's method, with which you may already be familiar. In Laplace's method, we are dealing with an integral $J(\lambda)=\int_{[a,b]} e^{\lambda \phi(x)}\,\mathrm{d}x$, where $\phi$ is now a real function. Suppose $\phi$ has a maximum at an interior point $x_0 \in (a,b)$. Then $\phi'(x_0)=0$, and moreover only a vanishingly small neighborhood of $x_0$ contributes to the asymptotic behavior. This is explained clearly in the Wikipedia article.
Returning to $I(\lambda)$, we want to use the same insight that Laplace's method is based on in the complex case, except that now we need to measure the size of $e^{\lambda f}$ in terms of its modulus. Only a small neighborhood of the maximum of $|e^{\lambda f}|=e^{\lambda \Re f}$ along $\gamma$ matters. The additional twist is that now we have the freedom to choose $\gamma$ by contour deformation: only the endpoints are fixed. The condition that $|e^{\lambda f(z_0)}|$ be a maximum along a curve for some point $z_0$ can be understood geometrically. You can draw a 3D plot of the modulus of an analytic function (you can see such plots on Wikipedia). Such plots look like smooth landscapes. If you observe a modulus plot for $e^{\lambda f(z)}$, you will see that there are only two types of points: ordinary points, at which $f'(z)\neq 0$, and saddle points, where $f'(z)=0$. $e^{\lambda f}$ being an analytic function, $|e^{\lambda f(z)}|$ cannot have a local maximum (this is the maximum modulus principle): we never see isolated peaks in the modulus landscape. At ordinary points, there will be a single direction such that $|e^f|$ increases fastest. At saddle points there will be two or more (depending on the order of vanishing of $(e^{\lambda f})'$).
The method of steepest descent is based on the observation that it is advantageous to deform the contour of integration $\gamma$, as far as possible, so as to travel along directions where $\Re f$ increases or decreases monotonically. This allows us to isolate the main contribution to the integral to the neighborhood of such points. For example, if $z_0$ is an endpoint of $\gamma$, and we can deform $\gamma$ such that $\Re f$ decreases as we move along $\gamma$ towards the other endpoint, say $z_1$. Then as $\lambda$ gets large, the contributions from points away from $z_0$ disappear, and we find: $$I(\lambda) \sim \frac{e^{\lambda f(z_0)}}{\lambda f'(z_0)}.$$
The most interesting case is when we deform $\gamma$ to pass through a saddle point $z_0$ of $f$, also orienting $\gamma$ such that the modulus $|e^{\lambda f(z)}|$ decreases as we move away from the saddle point (choosing a path of descent'' away from $z_0$). To see an example of such a contour, look at the picture of the $$w=x^2-y^2$$ plot on the Wikipedia article for saddle point linked above. Good contours $\gamma$ in this case would pass through the point $(0,0)$ and then go downwards (in a direction of decreasing $w$) as they move away from the origin. (This is just for illustration. This is not the modulus of an analytic function, but exponentiating will give you the modulus plot for $e^{z^2}$.) If we choose $\gamma$ such that 1. it goes through a saddle point, 2. the real part of $f$ (and hence $|e^{\lambda f}|$ decreases as we move away from the saddle point and towards the endpoints of $\gamma$, then the maximum of $|e^{\lambda f(z)}|$ on our contour will occur at $z_0$, and we can evaluate the integral $I(\lambda)$ approximately using by Taylor expanding $f$ around $z_0$, just like in the case of Laplace's method. We can now expand in a suitable neighborhood of $z_0$: $$f(z) = f(z_0)+(1/2)f''(z_0)(z-z_0)+\epsilon.$$ Assuming $f''(z_0)\neq 0$. The expansion is quadratic because $f'(z_0)=0$. For large $\lambda$, the resulting integral will be asymptotically equivalent to a Gaussian with a phase factor. Notice that we can easily identify saddle points by the equation $f'(z_0)=0$.
In some applications, it is useful to use specific contours $\gamma$. Here I have basically assumed that the endpoints of $\gamma$ lie in valleys on both sides of a saddle point, so that the exact choice of contour is not important. Contours through $z_0$ that are exactly colinear with the gradient of $\Re f$ will be the contours of steppest descent''. By the Cauchy-Riemann equations, the gradient of $\Im f$ is perpendicular to such contours, and so the phase is constant. In simple situations you can solve equations for these contours, and often you can solve approximately, which can help you guess a good $\gamma$ in cases where it matters.
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Thanks, your answer is very clear and illuminating. – bill Sep 5 '12 at 1:00 | 2015-07-03 02:44:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9218563437461853, "perplexity": 136.84404422220635}, "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-2015-27/segments/1435375095711.51/warc/CC-MAIN-20150627031815-00107-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://www.ias.ac.in/listing/bibliography/jcsc/Arup_Mukherjee | • Arup Mukherjee
Articles written in Journal of Chemical Sciences
• Phenalenyl-based ligand for transition metal chemistry: Application in Henry reaction
We report the synthesis and characterization of the first transition metal complex of a phenalenylbased ligand. The reaction of Cu(OAc)2.H2O with 9-𝑁-methylamino-1-𝑁'$-methylimino-phenalene (LH) in 1:1 stoichiometric ratio results in the formation of a mononuclear copper complex [LCu(OAc)] (1). The molecular structure of 1 was established by X-ray crystallography. The electrochemistry of 1 indicates the formation of an anionic radical by one electron reduction into the non-bonding molecular orbital of the phenalenyl system. The complex 1 efficiently catalyses the C-C bond forming Henry reaction. • Main group chemistry of 9-hydroxophenalenone: Syntheses and structural characterization of the alkaline earth and zinc complexes Herein, we report the synthesis and characterization of 9-hydroxophenalenone based alkaline earth and zinc complexes.The reaction of 9-hydroxophenalenone (HO,O-PLY (1)) with one equivalent of KN(SiMe3)2 and MI2 in THF yields heteroleptic complexes [(O,O-PLY)M(THF)$_n\$]I [M= Mg (2), Ca (3), Sr (4), Ba (5); n = 1-4], while use of two equivalents of KN(SiMe3)2 in THF (with respect to PLY) produces homoleptic complex (O,O-PLY)2Mg(THF)2 (6). Moreover, reaction between two equivalents of 1 with one equivalent of ZnMe2 in THF produces complex (O,O-PLY)2Zn(THF)2 (7). All these complexes were characterized by NMR spectroscopy and elemental analyses. The solid state structures of complexes 2, 6 and 7 were established by single crystal X-ray diffraction analysis.
• # Journal of Chemical Sciences
Volume 132, 2020
All articles
Continuous Article Publishing mode
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019
Click here for Editorial Note on CAP Mode
© 2017-2019 Indian Academy of Sciences, Bengaluru. | 2020-10-24 18:46:27 | {"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": 0.395698219537735, "perplexity": 13149.879763127923}, "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-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00177.warc.gz"} |
https://repo.scoap3.org/record/32301 | # Same-sign trilepton signal for stop quark in the presence of sneutrino dark matter
Ghosh, Dilip Kumar (Department of Theoretical Physics, Indian Association for the Cultivation of Science Jadavpur, Kolkata 700 032, India) ; Huitu, Katri (Department of Physics, and Helsinki Institute of Physics, P. O. Box 64, FI-00014 University of Helsinki, Finland) ; Mondal, Subhadeep (Department of Physics, and Helsinki Institute of Physics, P. O. Box 64, FI-00014 University of Helsinki, Finland) ; Mitra, Manimala (Institute of Physics, Sachivalaya Marg, Bhubaneswar, Odisha 751005, India, Homi Bhabha National Institute, Training School Complex, Anushakti Nagar, Mumbai 400085, India)
15 April 2019
Abstract: We explore a minimal supersymmetric standard model scenario extended by one pair of gauge singlets per generation. In the model, light neutrino masses and their mixings are generated via the inverse seesaw mechanism. In such a scenario, a right-handed sneutrino can be the lightest supersymmetric particle and a cold dark matter (DM) candidate. If the Casas-Ibarra parametrization is imposed on the Dirac neutrino Yukawa coupling matrix (${Y}_{\nu }$) to fit the neutrino oscillation data, the resulting ${Y}_{\nu }$ is highly constrained from the lepton-flavor-violating decay constraints. The smallness of ${Y}_{\nu }$ requires the sneutrino DM to coannihilate with other sparticle(s) in order to satisfy the DM relic density constraint. We study sneutrino coannihilation with winos and observe that this sneutrino-wino compressed parameter space gives rise to a novel same-sign trilepton signal for the top squark, which is more effective than the conventional top squark search channels in the present framework. We show that the choice of neutrino mass hierarchy strongly affects the signal event rate, making it easier to probe the scenario with the inverted mass hierarchy.
Published in: Physical Review D 99 (2019) | 2019-04-24 03:57:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.5368385910987854, "perplexity": 3581.522697102001}, "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-2019-18/segments/1555578626296.62/warc/CC-MAIN-20190424034609-20190424060609-00239.warc.gz"} |
https://jimmyjhickey.com/ST703-Multiple-Linear-Regression | ## Example Problems on Multiple Linear Regression
ST703 Homework 4 on Multiple Linear Regression
Problems: 1, 2, 3
# 1
“Resolution 5K Runs” typically occur on January 1st as a way to “start the new year off right” and raise funds for charity. Dataset ResolutionRun2004.txt in the database provides information on a race that occurred in 2004, giving the pace (minutes per mile), sex, age (years), and squared age for each of $n = 160$ runners. Ignoring sex, and focusing on the bivariate data consisting of age ($X$) and pace ($Y$), you will investigate a regression that models the mean pace as a quadratic in age:
$\mu(x) = \beta_0 + \beta_1 x + \beta_2 x^2$
For this whole problem we will be making our standard regression assumptions, which are normality in our residuals, equal variance of residuals with respect to our predicted values, independence, and that the expected value of our error is zero.
## (a)
Obtain the least squares regression equation.
The $\hat{\beta}$’s can be estimated using ordinary least squares, which gives the following equation:
$\mathbf{\hat{\beta}} = (\mathbf{X}^T \mathbf{X})^{-}\mathbf{X}\mathbf{Y}$
(Note that $\hat{\beta}$ is the vector of $\beta$ estimates. I can’t bold if for some reason.)
We can use SAS to compute this from our data.
proc reg data=run;
model pace=age age2;
run;
From this we get,
$\hat{y} = 11.78504 + -0.19699 x + 0.00294 x^2.$
## (b)
Construct 95% confidence intervals for $\beta_0$, $\beta_1$, and $\beta_2$ and interpret the results.
We can construct a confidence interval for $\beta_i$ with
$\hat{\beta_i} \pm t_{n-2, \alpha / 2} SE(\hat{\beta_i}).$
We will use the parameter estimates table from SAS given in (a). Notice that we have a large $n$ and are assuming normality even though the qqplot does not look great.
\begin{align} (\hat{\beta_{0_L}},\hat{\beta_{0_U}} ) & = \hat{\beta_0} \pm t_{157, 0.05 / 2} SE(\hat{\beta_0}) \\ & = 11.78504 \pm 1.975189 \cdot 0.702106 \\ & = ( 10.3982, 13.1718) \\ (\hat{\beta_{1_L}},\hat{\beta_{1_U}} ) & = \hat{\beta_1} \pm t_{157, 0.05 / 2} SE(\hat{\beta_1}) \\ & = -0.19699 \pm 1.975189 \cdot 0.04113 \\ & = (-0.27823, -0.11575) \\ (\hat{\beta_{2_L}},\hat{\beta_{2_U}} ) & = \hat{\beta_2} \pm t_{157, 0.05 / 2} SE(\hat{\beta_2}) \\ & = 0.00294 \pm 1.975189 \cdot 0.00057380 \\ & = (0.00180664, 0.00407336) \end{align}
We are 95% confident that the true intercept $\beta_0$ (pace when age and age2 are 0) is between 10.3986 and 13.1718 minutes per mile.
If we look at the model in part (i), we see that $\beta_1$ is the true rate of change of pace when age (and consequently age squared) is equal to zero. We are 95% confident that the true $\beta_1$ is between -0.27823 and -0.11575 minutes per mile per age.
If we look at the model in part (i), we see that $\beta_2$ is the expected change in rate of change of pace when age is increased by 1. We are 95% confident that the true $\beta_2$ is between 0.00180664 and 0.00407336 minutes per mile per age^2.
## (c)
Estimate the error variance, $\sigma^2$, for an individual measurement of pace, given a fixed age.
We can estimate $\sigma^2$ with $s^2$. We will need the SSE from the ANOVA table.
\begin{align} s^2 & = \frac{ SSE }{ n - rank(\mathbf{X}) } \\ & = \frac{ SSE }{ n - (p+1) } \\ & = \frac{ 674.45060 }{ 160 - (2+1) } \\ & = 4.29586 \end{align}
## (d)
Which do you expect would be larger, $Var(Y)$ or $Var(Y|X=x)$? Which quantity was estimated in part (c)?
We would expect $Var(Y)$ to be larger than $Var(Y|X=x)$, which we calculated in (c). The second value should be smaller as it takes into account the model, which we expected to explain some of the variance of $Y$.
## (e)
Report a standard error and 95% confidence interval for the mean pace for runners aged $x=34$.
We can find a standard error estimate for $E(Y | \mathbf{x}_0)$ with $\sqrt{\sigma^2 \mathbf{x}_0 (\mathbf{X}^T \mathbf{X})^{-1} \mathbf{x}_0^T}$ where $\mathbf{x_0} = [1\ 34\ 1156]$. We can use this to calculate the confidence interval
$E(Y | x = 34) \pm t_{n-3, \alpha/2} \cdot SE(E(Y | x = 34)).$
We can compute this is SAS as well.
Our standard error is 0.2055. We are 95% confident that the true mean pace for runners ages 34 is between 8.0765 and 8.8884 minutes per mile.
## (f)
Obtain a 95% prediction interval for the pace of an individual runner selected from the cohort of 34 year-olds.
We can find the prediction interval $Y | x = 34$ with the following formula. Note that the variance of $Y | x_0 = 34$ is $\sqrt{\sigma^2 + \sigma^2 \mathbf{x}_0 (\mathbf{X}^T \mathbf{X})^{-1} \mathbf{x}_0^T}$ where $\mathbf{x_0} = [1\ 34\ 1156]$.
$\hat{Y_0} \pm _{n-3, \alpha/2} \cdot SE(Y - \hat{Y_0})$
We can also read this from the SAS output in part (e).
We are 95% confident that the next new observation given an age of 34 will be within 4.3685 and 12.5964 minutes per mile.
## (g)
Explain the difference between the two intervals in (e) and (f).
Problem (e) is estimating the mean of the pace $\beta_0 + \beta_1 x + \beta_2 x^2$.
Problem (f) is asking to estimate the value given a new observation with an age of 34. We are estimating $Y = \beta_0 + \beta_1 x + \beta_2 x^2 + E$.
Notice that the prediction interval has a wider range.
## (h)
Construct a 99% confidence interval for the parameter $\theta = \beta_1 - \beta_2$.
$(\hat{\beta_1} - \hat{\beta_2}) \pm t_{\nu, \alpha/2} \cdot \sqrt{ Var(\hat{\beta_1} - \hat{\beta_2}) }$
Notice that
$Var(\hat{\beta_1} - \hat{\beta_2})) = Var(\hat{\beta_1}) + Var(\hat{\beta_2})) - 2 Cov(\hat{\beta_1}), \hat{\beta_2}))$
We can use the estimates from part (a) to say $\hat{\beta_1} = -0.19699$, $\hat{\beta_2} =0.00294$, and we can get the variance and covariance from SAS.
Thus, our confidence interval is
$(-0.19699-0.00294) \pm 2.60751 \sqrt{0.00169207 + 3.29241 *10^{-7} - 2 (-0.000022716)} = (-0.30863, -0.0912303).$
We are 99% confident that the true value of $\theta$ is between -0.30863 and -0.0912303.
## (i)
Consider the rate at which the mean function for pace, $\mu(x)$, changes with age.
### i.
At what rate is the mean function for pace, $\mu(x)$, changing with age?
We can take the derivative of regression function to find the mean function for pace.
$\frac{ d }{ d\mu } = \beta_1 + 2 \beta_2 \cdot x_1$
### ii.
Estimate the rate of change for 30 year-olds.
Using the parameter estimates from part (a) we get,
$(\frac{ d }{ d\mu } | x = 30) = -0.19699 +2 \cdot 0.00294 \cdot 30 =-0.02059$
This means that the estimated rate of change for 30 year-olds is -0.02059 miles per minute per year.
### iii.
Estimate the rate of change for 50 year-olds.
Using the parameter estimates from part (a) we get,
$(\frac{ d }{ d\mu } | x = 30) = -0.19699 + 2 \cdot 0.00294 \cdot 50 =0.09701$
This means that the estimated rate of change for 50 year-olds is 0.09701 miles per minute per year.
### iv.
A simple and familiar test of one of the regression parameters can we used to investigate the claim that “the rate at which the mean function for pace changes with age is constant.” State relevant hypotheses and conduct the test.
The hypotheses are
\begin{align} H_0: & \beta_2 = 0 \\ H_A: & \beta_2 \neq 0 \end{align}
We can calculate a t-statistic to test our hypothesis.
\begin{align} t & = \frac{ \hat{\beta_2} - \beta_{2,0} }{ SE(\hat{\beta_2}) } \\ & = \frac{ 0.00294 - 0 }{ 0.00057380 }\\ & = 5.12374 \end{align}
This gives a p value of less than 0.001.
## (j)
Under the quadratic model, at what age is the mean function for pace minimized? Estimate this age.
For this we want to check when the derivative is equal to 0.
\begin{align} 0 & = \beta_1 + \beta_2 \cdot x_1 \\ & = -0.19699 + 2 \cdot 0.00294 \cdot x_1 \\ x_1 & = 33.5017 \end{align}
Looking at the graph of the derivative, we see that it goes from negative to positive at 33.5017, so this is indeed a minimum.
# 2
Consider the four different regression functions in Exercise 11.2 as possible candidates for expressing the expected log(oxygen demand) as a function of five explanatory variables in Example 11.2.
$E(Y \Big| a,b, c, d, f) = \beta_0 + \beta_1 a + \beta_2 b + \beta_3 c + \beta_4 d + \beta_5 f$
## (b)
Interpret the parameters $\beta_1$ and $\beta_2$ in exercise 11.2a.
$\begin{array}{ r | l } \beta_1 & \text{The change in expected log(oxygen demand) with a 1 unit increase in} \\ & \text{biological oxygen demand holding all other predictors constant.} \\ \beta_2 & \text{The change in expected log(oxygen demand) with a 1 unit increase in} \\ & \text{the total Kjeldahl nitrogen holding all other predictors constant.} \end{array}$
## (c)
Interpret the quantity $\beta_1+\beta_2 + \beta_3$ in exercise 11.2a.
The quantity $\beta_1+\beta_2 + \beta_3$ represents the change in expected log(oxygen demand) when $a,\ b,\ c$ (biological oxygen demand, total Kjeldahl nitrogen, and total solids, respectively) are increased by one unit each and $d, \ f$ are held constant.
\begin{align} E(Y | a+1, b+1, c+1) - E(Y|a,b,c) & = \beta_0 + \beta_1 (a+1) + \beta_2 (b+1) + \beta_3 (c+1) + \beta_4 d + \beta_5 f - (\beta_0 + \beta_1 a + \beta_2 b + \beta_3 c + \beta_4 d + \beta_5 f)\\ & = \beta_1 a + \beta_1 + \beta_2 b + \beta_2 + \beta_3 c + \beta_3 - \beta_1 a - \beta_2 b - \beta_3 c\\ & = \beta_1 + \beta_2 + \beta_3 \end{align}
# 3
In Example 11.8, we considered a study in which responses were measured at $n=8$ settings of two independent variables $X_{1}$ and $X_{2}$. Suppose that we are considering two models for the study data:
model 1
$Y_1 = \beta_0 + \beta_1 x_{i1} + \beta_{i2} x_{i2} + E_i;$
and model 2
$Y_i = \beta_0 + \beta_1 x_{i1} + \beta_{i2} x_{i2} + \beta_3 x_{i1} x_{i2} + E_i.$
## (a)
Argue that, while the parameter $\beta_0$ has the same meaning in both models, the meanings of the parameters $\beta_1$ and $\beta_2$ are different in the two models.
The difference in these two models is the presence of an interaction term in model 2. In this case, the intercept parameter, $\beta_0$ has the same interpretation, it is the expected value when all other predictors are 0.
The differences arise with $\beta_1$ and $\beta_2$ because of the interaction term. In model 1, $\beta_1$ and $\beta_2$ represent the change in the expected value of the response given a one unit increase in $x_{i1}$ or $x_{i2}$, respectively, while holding the other constant. The same cannot be said for model 2 unless the variable being held constant is 0; otherwise the interaction term and the value of $x_{i2}$ will play a role in the expected change in response.
Model 1:
\begin{align} E(Y| x_{i1}+1, a) - E(Y | x_{i1}, a) & = \beta_0 + \beta_1 (x_{i1} + 1) + \beta_{i2} a - (\beta_0 + \beta_1 x_{i1} + \beta_{i2} a) \\ & = \beta_1 x_{i1} + \beta_1 \cdot 1 - \beta_1 x\\ & = \beta_1 \end{align}
Model 2:
\begin{align} E(Y| x_{i1}+1, a) - E(Y | x_{i1}, a) & = \beta_0 + \beta_1 (x_{i1}+1) + \beta_{i2} a + \beta_3 (x_{i1}+1) a - (\beta_0 + \beta_1 x_{i1} + \beta_{i2} a + \beta_3 x_{i1} a )\\ & = \beta_1 x_{i1} + \beta_1 \cdot 1 + \beta_3 ax_{i1} + \beta_3 a- \beta_1 x_{i1} - \beta_3 ax_{i1}\\ & = \beta_1 + \beta_3 a \end{align}
## (b)
Determine the estimated least squares prediction equation using model 1.
We can get the estimated values of $\beta_i$ from the parameter estimates table, where they were calculated using ordinary least squares:
$\mathbf{\hat{\beta}} = (\mathbf{X}^T \mathbf{X})^{-}\mathbf{X}\mathbf{Y}.$
From that we get
$\hat{y} = -0.861140 + 0.800067 \cdot x_1 + -0.078780 \cdot x_2.$
## (d)
Use the given ANOVA table to test the null hypothesis that $H_0: \ \beta_1 = \beta_2 = 0$ in model 1. Write your conclusion.
In words, our null hypothesis, $H_0: \ \beta_1 = \beta_2 = 0$, means that neither of the variables are useful for predicting $Y$ while the alternative hypothesis is that at least one of $\beta_1$ or $\beta_2$ is useful.
The F-Value in the ANOVA table for model 1 is $F = \frac{ MSRegn }{ MSE } = 34.555$ with a probability of 0.0012 when compared to an F distribution with 2 numerator and 5 denominator degrees of freedom. Using an $\alpha=0.05$, we have sufficient evidence to reject the null hypothesis in favor of the alternative, that at least one of $\beta_1$ or $\beta_2$ is useful.
## (e)
Determine the estimated least squares prediction equation based on model 2.
Following the same process from part (b), we get
$\hat{y} = -0.597752 + 0.743553 \cdot x_1 + -0.126096 \cdot x_2 + 0.009976 \cdot x_1 x_2.$ | 2021-10-23 11:56:39 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 8, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9948733448982239, "perplexity": 802.2676567078067}, "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/1634323585671.36/warc/CC-MAIN-20211023095849-20211023125849-00303.warc.gz"} |
https://kx.lumerical.com/t/fdtd-using-multiple-processors-on-cluster/69311 | # FDTD - Using Multiple Processors on Cluster
Hi,
I’m trying to run some FDTD simulations on a high-performance computing cluster but I’m running into some difficulties. I am trying to do a distributed computing job, but I can’t seem to get Lumerical to use more than one processor and I was wondering if anybody could tell me what I’m doing wrong.
Our cluster uses Slurm as a job manager, so I’ve created a bash script (attached) for submitting my job. In the script, I do the normal setup where I set a number of nodes and a number of processes per node (as well as RAM allocations) and then I point Lumerical to the MPICH2 Nemesis MPI and run the job.
Whenever the job runs, the cluster allocates the correct number of nodes, but when I check the Lumerical log files (one is attached) they say that only 1 CPU is used and the simulation is 1x1x1. It runs fine and gives me a file, but it defeats the purpose of using the HPC cluster if I’m using 1 CPU.
Is there something I have to allocate in the resources before saving the .fsp file? Or another command I have to add to the Slurm bash? Or is this a question for our HPC support staff?
Thanks,
Brandon
SubmitScript.txt (914 Bytes) sweep_1_p0-edited.log-1.txt (5.2 KB)
@hacha
Welcome to Lumerical Knowledge Exchange.
### You are not using MPI
fdtd-engine-mpich2nem /home/<USER>/LumericalTesting/LumericalForCluster_sweep/*.fsp
The command “fdtd-engine-mpich2nem” is only running with the FDTD engine.
### Run with MPI on cluster
You will need to run the simulation job with MPI to be able to distribute across multiple nodes.
• Typically your IT would have configured an MPI to run jobs on your cluster.
• Check if the MPI your cluster uses is supported by Lumerical.
• Check with your IT on the exact command to use/execute to run the simulation using MPI on your cluster/job scheduler to distribute to multiple nodes based on your submission script.
### Examples:
#### Using bundled mpich2:
With the default install path: “/opt/lumerical/v202/mpich2/...”, your script would be:
... (rest of your script)
# executable
/opt/lumerical/v202/mpich2/nemesis/bin/mpiexec fdtd-engine-mpich2nem /home/<USER>/LumericalTesting/LumericalForCluster_sweep/*.fsp -logall -fullinfo
If supported by Lumerical, use your cluster’s MPI and execute the command to “load your MPI module”:
... (rest of your script)
# executable
mpiexec fdtd-engine-mpich2-lcl /home/<USER>/LumericalTesting/LumericalForCluster_sweep/*.fsp -logall -fullinfo
#### Note:
Use the correct FDTD engine executable for your supported MPI variant.
fdtd-engine-mpich2-lcl
fdtd-engine-impi-lcl
fdtd-engine-ompi-lcl
#### Running the submission script:
>\$ sbatch -N 4 --ntasks-per-node=4 submit-script.sh
Hope this helps.
Best,
Lito
Thank you so much @lyap! That all worked!
@hacha
You are welcome. Glad that the issue has been resolved.
Best,
Lito
This topic was automatically closed 3 days after the last reply. New replies are no longer allowed. | 2020-08-08 13:10:23 | {"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": 0.2506243884563446, "perplexity": 3635.2344640178185}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737645.2/warc/CC-MAIN-20200808110257-20200808140257-00522.warc.gz"} |
http://openstudy.com/updates/4f33ac79e4b0fc0c1a0ba40a | ## jpknegtel 2 years ago Need to use the binomal therom to expand (1+2x)/(1-2x) Not sure where to start to get it into the right format.
1. jpknegtel
$(1+2x)/(1-2x)$
2. jpknegtel
Oh. Need to go up too and including the term $x ^{2}$
3. King
so its x^2+2x+1/1-2x?
4. jpknegtel
Could you explain the steps? thank you very much for your timee!
5. Zed
Is that $(1+2x) \times (1-2x)$ or $\frac{(1+2x)}{(1-2x)}$?
6. jpknegtel
I have been trying to do those Fractions but am unable to do it! Soo fustrating! Yes it is the the (1+2x) over (1-2x)
7. Zed
Okay give me a minute to work this out :)
8. jpknegtel
If it is any help the answer is $(1+2x)(1+2x+4x ^{2})$
9. y2o2
(1+2x) over (1-2x) can never be equal to (1+2x)(1+2x+4x²) and you can assure that by substitution.
10. Zed
$(1+2x)(1-2x)^{-1}=(1+2x)(1^{-1}+-1*1^{-1-1}*-2x+\frac{-1(-1-1)}{2} (1)^{-1-2}(2x)^2+...)$$=(1+2x)(1^{-1}+2x+\frac{-1(-2)}{2} (1)^{-3}*4x^2+...)$$=(1+2x)(1+2x+\frac{2}{2} *1*4x^2+...)$$=(1+2x)(1+2x+4x^2+...)$ This is from this rule $(a+b)^n=a^n+na^{n-1}b+\frac{n(n-1)}{2}a^{n-2}b^2+....$ Sorry it took so long :D
11. dumbcow
either the answer is wrong or you forgot something when posting the problem i agree with y2o2
12. Zed
13. dumbcow
@zed yes it becomes an infinite sum..is that the solution they are looking for? their answer stops after 4x^2
14. Zed
Yes they only had to do the terms until it reaches x^2 power
15. dumbcow
ok thanks for clearing it up :)
16. jpknegtel
Thank you very much guys! Clears things up! | 2015-01-28 20:18:45 | {"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": 0.5657486319541931, "perplexity": 2450.4178331434186}, "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/1422119446463.10/warc/CC-MAIN-20150124171046-00157-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-is-the-domain-and-range-of-f-x-x-2-x-2-6 | # What is the domain and range of f(x) =x^2/ (x^2-6)?
Apr 15, 2018
Domain: x≠sqrt6
Range: $y \in \mathbb{R}$
#### Explanation:
First, it is important to understand the distinction between domain and range.
Domain:
All possible $x$-values for the expression
Range:
All possible $y$-values for the expression
=================================================
Finding the Domain:
On the numerator, the $x$-value can be any real number
The denominator however, ≠ 0 since that would make the function undefined.
So, to find the number that $x$ cannot equal in the function, we must write:
Denominator$= 0$
${x}^{2} - 6 = 0$
${x}^{2} = 6$
$\sqrt{{x}^{2}} = \sqrt{6}$
$\therefore x = \sqrt{6}$ when the function is undefined
This means that the domain is: x≠sqrt6
Finding the Range:
$f \left(x\right)$ just means that $x$ is the input of the function.
The actual result of the equation is $y$, do the function can be rewritten as:
$y = {x}^{2} / \left({x}^{2} - 6\right)$
Now there is a $y$-value to work with. Since there are no limitations on the value of $y$ in the equation, it can be any real number.
This means that the range is: $y \in \mathbb{R}$
or
$y$ can be any real number | 2020-04-02 10:04:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 21, "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": 0.9000153541564941, "perplexity": 525.1130785396088}, "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/1585370506870.41/warc/CC-MAIN-20200402080824-20200402110824-00378.warc.gz"} |
https://www.r-bloggers.com/2012/04/page/9/ | # Monthly Archives: April 2012
## A variance campaign that failed
April 23, 2012
By
they ought at least be allowed to state why they didn’t do anything and also to explain the process by which they didn’t do anything. First blush One of the nice things about R is that new statistical techniques fall into it. One such is the glasso (related to the statistical lasso) which converts degenerate … Continue reading...
## Visualising the Path of a Genetic Algorithm
April 23, 2012
By
We quite regularly use genetic algorithms to optimise over the ad-hoc functions we develop when trying to solve problems in applied mathematics. However it’s a bit disconcerting to have your algorithm roam through a high dimensional solution space while not being able to picture what it’s doing or how close one solution is to another. … Continue reading...
April 22, 2012
By
I came across a free source of Intraday Forex data while reading Forex Trading with R : Part 1 post. You can download either Daily or Hourly historical Forex data from the FXHISTORICALDATA.COM. The outline of this post: Download and Import Forex data Reference and Plot Intraday data Daily Backtest Intraday Backtest First,I created a
## My bookshelf
April 22, 2012
By
I'd like to start with something small, and simple. The thing about analyzing the data of your own life is that you are the only one doing the research, so you also have to collect all of the data yourself. This takes effort; and, if you'd like to build a large enough data set to do some really interesting...
## 118 years of US State Weather Data
April 22, 2012
By
A recent post on the Junkcharts blog looked at US weather dataand the importance of explaining scales (which in this case went up to 118). Ultimately, it turns out that 118 is the rank of the data compared to the previous 117 years of data (in ascending order, so that 118 is the highest). At … Continue reading...
## Meielisalp, Mar 2012 – Abstract Submission
April 22, 2012
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(This article was first published on Rmetrics blogs, and kindly contributed to R-bloggers) To leave a comment for the author, please follow the link and comment on their blog: Rmetrics blogs. R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave,...
## Machine learning for identification of cars
April 22, 2012
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There are plenty of data on internet, however it is raw data. Think for a second about public surveillance cameras - useful to check the traffic on the route or busy place, but anything else? What if you want to know how many cars are on the route? How many car were yesterday at the same time?
## Fancy HTML5 Slides with knitr and pandoc
April 22, 2012
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Karthik Ram gave an Introduction to R a couple of weeks ago, and I strongly recommend you to take a look at his cool HTML5 slides. I started trying HTML5 slides last year, and now it is difficult for me to go back to beamer, which I have used for a few...
## Phase space plot of the kicked rotor
April 21, 2012
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$Phase space plot of the kicked rotor$
In the idealized physical world, a rotor is simply a mass attached to an axis of length , free to move in the plane. Gravity and friction are absent. Such a rotor becomes a kicked rotor if it is periodically hit with a hammer. Every kick transfers momentum to the rotor and the time between
## Calculate the average distance between a given DNA motif within DNA sequences in R
April 21, 2012
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Suppose that we want to calculate the expected distance of a DNA motif within a DNA target sequence, if we know the composition bias or the probability distribution (multinomial model) we can compute it just fine.Download the R code <- hereFIRS... | 2017-02-20 17:46:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 1, "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": 0.20076513290405273, "perplexity": 1841.414215754031}, "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-2017-09/segments/1487501170600.29/warc/CC-MAIN-20170219104610-00018-ip-10-171-10-108.ec2.internal.warc.gz"} |