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https://stats.stackexchange.com/questions/88772/sparse-parameters-when-computing-aic-bic-etc
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# Sparse parameters when computing AIC, BIC, etc
I'm designing large-scale, regularized logistic regression models with lots of sparse, binarized features. e.g. isUS, isFR, etc. As a result, a lot of the weights in the model are zero.
I'm wondering how I should compute the "number of parameters" in model-selection criteria like AIC, BIC, etc. Should I only count the number of non-zero weights or all of the weights?
e.g. If there are 10 possible countries, but only 6 have non-zero weights, is the number of parameters 6 or 10?
• are you trying to build your own AIC measurement? Are you using software that doesn't include AIC for some reason? Mar 10, 2014 at 20:21
Degrees of freedom do not depend on the outcome alone but on the fitting procedure. If it's maximum likelihood, all parameters count.
There is an interesting case where zero weights do not count, and that's lasso: H Zou, T Hastie, R Tibshirani On the “degrees of freedom” of the lasso. The Annals of Statistics, 2007
• You're right. I was thinking of a lasso context. In general, for regularized regression df will be less than p and will vary as a function of the amount of regularization. I deleted my incorrect answer. Mar 11, 2014 at 20:32
• So when using L1 regularization, would you only count non-zero weights? Mar 12, 2014 at 18:21
This is a really difficult question to answer without precise knowledge of the fitting algorithm, nor is it clear cut that there is a reasonable definition of the "number of parameters" that will justify AIC, BIC or other "information criteria" in general.
If estimation is done by $\ell_1$-penalized maximum-likelihood estimation, then I can partially iterate the answer by user27493. In this case the estimated number of non-zero parameters is a sensible substitute for the total number of parameters in AIC. Note, however, that the Zou et al. paper is on least squares regression with an $\ell_1$-penalty $-$ not logistic regression. See, for instance, Differential geometric least angle regression: a differential geometric approach to sparse generalized linear models by L. Augugliaro et al. for results related to generalized linear models.
BIC is different, and I don't know results in this direction.
The paper with the catchy title Effective Degrees of Freedom: A Flawed Metaphor, posted recently on archive by Lucas Janson, William Fithian, and Trevor Hastie, shows that, depending on the data generating mechanism, the effective degrees of freedom ("number of parameters") may exceed the total number of parameters, and may even be unbounded.
In this paper (shameless self promotion of my research) Degrees of freedom for nonlinear least squares estimation with my coauthor Alexander Sokol, we show that for nonlinear least squares estimation the effective degrees of freedom generally contains a hard-to-estimate term that depends on the data generating model. This is also what pops up in some of the examples in the Janson et al. paper mentioned above. In an asymptotic scenario, if the model is close to being true and/or if the model does not "curve too much", and if you use $\ell_1$-penalized least squares estimation, a useful surrogate estimate of the effective degrees of freedom is still the estimated number of non-zero parameters. However, once you move outside of some of the standard and most well behaved models, anything could happen.
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2023-03-26 08:22:36
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https://hal.archives-ouvertes.fr/hal-01703770
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# The SUrvey for Pulsars and Extragalactic Radio Bursts – II. New FRB discoveries and their follow-up
Abstract : We report the discovery of four Fast Radio Bursts (FRBs) in the ongoing SUrvey for Pulsars and Extragalactic Radio Bursts at the Parkes Radio Telescope: FRBs 150610, 151206, 151230 and 160102. Our real-time discoveries have enabled us to conduct extensive, rapid multimessenger follow-up at 12 major facilities sensitive to radio, optical, X-ray, gamma-ray photons and neutrinos on time-scales ranging from an hour to a few months post-burst. No counterparts to the FRBs were found and we provide upper limits on afterglow luminosities. None of the FRBs were seen to repeat. Formal fits to all FRBs show hints of scattering while their intrinsic widths are unresolved in time. FRB 151206 is at low Galactic latitude, FRB 151230 shows a sharp spectral cut-off, and FRB 160102 has the highest dispersion measure (DM = 2596.1 ± 0.3 pc cm^−3) detected to date. Three of the FRBs have high dispersion measures (DM > 1500 pc cm^−3), favouring a scenario where the DM is dominated by contributions from the intergalactic medium. The slope of the Parkes FRB source counts distribution with fluences >2 Jy ms is $\alpha =-2.2^{+0.6}_{-1.2}$ and still consistent with a Euclidean distribution (α = −3/2). We also find that the all-sky rate is $1.7^{+1.5}_{-0.9}\times 10^3$FRBs/(4π sr)/day above ${\sim }2{\rm \, }\rm {Jy}{\rm \, }\rm {ms}$ and there is currently no strong evidence for a latitude-dependent FRB sky rate.
Keywords :
Document type :
Journal articles
Domain :
https://hal.archives-ouvertes.fr/hal-01703770
Contributor : Inspire Hep <>
Submitted on : Thursday, February 8, 2018 - 4:42:39 AM
Last modification on : Tuesday, December 10, 2019 - 4:58:06 PM
### Citation
S. Bhandari, E.F. Keane, E.D. Barr, A. Jameson, E. Petroff, et al.. The SUrvey for Pulsars and Extragalactic Radio Bursts – II. New FRB discoveries and their follow-up. Mon.Not.Roy.Astron.Soc., 2018, 475 (2), pp.1427-1446. ⟨10.1093/mnras/stx3074⟩. ⟨hal-01703770⟩
Record views
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2019-12-11 03:17:33
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https://socratic.org/questions/58e9aaa611ef6b6b5697db47
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# What is the limiting reagent when 12.5*lb pentane are reacted with a 12.5*lb mass of dioxygen gas?
Apr 9, 2017
We need (i) a stoichiometric equation:
${C}_{5} {H}_{12} + 8 {O}_{2} \rightarrow 5 C {O}_{2} \left(g\right) + 6 {H}_{2} O$
#### Explanation:
And (ii) we need equivalent quantities of dioxygen and pentane. We know that $\text{1 lb}$ $\equiv$ $454 \cdot g$:
$\text{Moles of pentane} = \frac{12.5 \cdot l b \times 454 \cdot g \cdot l {b}^{-} 1}{72.15 \cdot g \cdot m o {l}^{-} 1} = 78.7 \cdot m o l$
$\text{Moles of dioxygen} = \frac{12.5 \cdot l b \times 454 \cdot g \cdot l {b}^{-} 1}{32.0 \cdot g \cdot m o {l}^{-} 1} = 177.3 \cdot m o l$
Clearly, there is INSUFFICIENT dioxygen gas for complete combustion. And thus ${O}_{2}$ is the limiting reagent. Complete combustion requires $8 \times 78.7 \cdot m o l$ dioxygen gas. What is this as a mass?
By the way the question proposed that oxygen and pentane were mixed together.........this is not something that I would be happy doing, and I would run a long way away if I saw someone doing this....
Apr 9, 2017
The limiting reactant is oxygen.
#### Explanation:
Here's another way to identify the limiting reactant.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.
${M}_{r} : \textcolor{w h i t e}{m m m m m} 72.15 \textcolor{w h i t e}{m m} 32.00$
$\textcolor{w h i t e}{m m m m m m m} \text{C"_5"H"_12 +color(white)(ll) "8O"_2 → "5CO"_2 + "6H"_2"O}$
$\text{Amt/lb-mol:} \textcolor{w h i t e}{l l} 0.1733 \textcolor{w h i t e}{m l} 0.3906$
$\text{Divide by:} \textcolor{w h i t e}{m m m l} 1 \textcolor{w h i t e}{m m m m} 8$
$\text{Moles rxn:"color(white)(mll)0.1733color(white)(ml)"0.048 83}$
Note: We do not have to stick with gram-moles. We can use pound-moles because the numbers will still be in the same ratio.
${\text{Moles of C"_5"H"_12 = 12.5 color(red)(cancel(color(black)("lb C"_5"H"_12))) × ("1 lb-mol C"_5"H"_12)/(72.15 color(red)(cancel(color(black)("lb C"_5"H"_12)))) = "0.1733 lb-mol C"_5"H}}_{12}$
${\text{Moles of O"_2 = 12.5 color(red)(cancel(color(black)("lb O"_2))) × ("1 lb-mol O"_2)/(32.00 color(red)(cancel(color(black)("lb O"_2)))) = "0.3906 lb-mol O}}_{2}$
2. Identify the limiting reactant
An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:
You divide the moles of each reactant by its corresponding coefficient in the balanced equation.
I did that for you in the table above.
${\text{O}}_{2}$ is the limiting reactant because it gives the fewest moles of reaction.
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2019-09-18 20:13:17
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http://www.map.mpim-bonn.mpg.de/index.php?title=5-manifolds_with_fundamental_group_of_order_2&oldid=9842
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# 5-manifolds with fundamental group of order 2
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## 1 Introduction
The classification of simply-connected 5-manifolds was achieved by Smale [Smale] and Barden [Barden] in the 1960s. The surgery and modified surgery theories provide a toolkit to attack the classification problem of high dimensional manifolds. Dimension 5 is the lowest dimension where the theories apply (except for the 4-dimensional TOP-category), therefore an understanding of the classification of 5-manifolds beyond the Smale-Barden results is expectable. On the other hand, it's known ([Markov], [Kervaire]) that every finitely generated group can be realized as the fundamental group of a manifold of dimension $\ge$${{Stub}} == Introduction == ; The classification of simply-connected 5-manifolds was achieved by Smale \cite{Smale} and Barden \cite{Barden} in the 1960s. The surgery and modified surgery theories provide a toolkit to attack the classification problem of high dimensional manifolds. Dimension 5 is the lowest dimension where the theories apply (except for the 4-dimensional TOP-category), therefore an understanding of the classification of 5-manifolds beyond the Smale-Barden results is expectable. On the other hand, it's known (\cite{Markov}, \cite{Kervaire}) that every finitely generated group can be realized as the fundamental group of a manifold of dimension \ge 4. Therefore, a practical approach towards the classification of 5-manifolds is to fix a fundamental group in advance and consider the classification of manifolds with the given fundamental group. From this point of view, the fist step one might take is the group \Zz_2, which is the simplest nontrivial group since it has the least number of elements. (Of course if we take the point of view of generators and relations the rank 1 free group \Zz is the simplest one.) Any other point one should take into account concerning the classification of manifolds with nontrivial fundamental groups \pi_1 is that the higher homotopy groups \pi_i (i \ge 2) are modules over the group ring \Zz[\pi_1], which are apparently homotopy invariants. Especially when we consider -manifolds M^5, the \Zz[\pi_1]-module structure of \pi_2(M) will play an important role in the classification. And it's not a surprise that the first clear classification result obtained is the trivial module case. Most part of this item will be a survey of the classification result of 5-manifolds M^5 with fundamental group \pi_1(M)=\Zz_2, \pi_2(M) torsion free and is a trivial \Zz[\Zz_2]-module obtained in \cite{Hambleton&Su} . == Construction and examples == ; First some examples known from other context. *S^2 \times \mathbb R P^3; * X^5(q) q=1,3,5,7 are the 5-dimensional fake real projective spaces with X(1)=\mathbb R \mathrm P^5 (There are exactly 4 in the smooth category). The meaning of q will be clear in the section Invariants". === Circle bundles over simply-connected 4-manifolds === ; Let X^4 be a closed simply-connected topological 4-manifold, \xi be a complex line bundle over X with Chern class c_1(\xi) \in H^2(X;\Zz). Let the divisibility of c_1(\xi) be k (i.e. c_1(\xi) is k multiple of a primitive element in H^2(X;\Zz), then the sphere bundle S(\xi) = \colon M is a 5-manifold with fundament group \Zz_k and \pi_2(M) is a free abelian group of rank \mathrm{rank} H_2(X)-1, and \pi_2(M) is a trivial module over the group ring. A priori M is a topological manifold. The smoothing problem is addressed by the following {{beginthm|Proposition}}\cite[Proposition 4.2]{Hambleton&Su} Assume \xi is nontrivial. If k is odd, then M admits a smooth structure; if k is even, then M admits a smooth structure if and only if the Kirby-Siebenmann invariant of X is 4. Therefore, a practical approach towards the classification of 5-manifolds is to fix a fundamental group in advance and consider the classification of manifolds with the given fundamental group.
From this point of view, the fist step one might take is the group $\Zz_2$$\Zz_2$, which is the simplest nontrivial group since it has the least number of elements. (Of course if we take the point of view of generators and relations the rank 1 free group $\Zz$$\Zz$ is the simplest one.)
Any other point one should take into account concerning the classification of manifolds with nontrivial fundamental groups $\pi_1$$\pi_1$ is that the higher homotopy groups $\pi_i$$\pi_i$ ($i \ge 2$$i \ge 2$) are modules over the group ring $\Zz[\pi_1]$$\Zz[\pi_1]$, which are apparently homotopy invariants. Especially when we consider $5$$5$-manifolds $M^5$$M^5$, the $\Zz[\pi_1]$$\Zz[\pi_1]$-module structure of $\pi_2(M)$$\pi_2(M)$ will play an important role in the classification. And it's not a surprise that the first clear classification result obtained is the trivial module case.
Most part of this item will be a survey of the classification result of 5-manifolds $M^5$$M^5$ with fundamental group $\pi_1(M)=\Zz_2$$\pi_1(M)=\Zz_2$, $\pi_2(M)$$\pi_2(M)$ torsion free and is a trivial $\Zz[\Zz_2]$$\Zz[\Zz_2]$-module obtained in [Hambleton&Su] .
## 2 Construction and examples
First some examples known from other context.
• $S^2 \times \mathbb R P^3$$S^2 \times \mathbb R P^3$;
• $X^5(q)$$X^5(q)$ $q=1,3,5,7$$q=1,3,5,7$ are the 5-dimensional fake real projective spaces with $X(1)=\mathbb R \mathrm P^5$$X(1)=\mathbb R \mathrm P^5$ (There are exactly 4 in the smooth category). The meaning of $q$$q$ will be clear in the section Invariants".
### 2.1 Circle bundles over simply-connected 4-manifolds
Let $X^4$$X^4$ be a closed simply-connected topological 4-manifold, $\xi$$\xi$ be a complex line bundle over $X$$X$ with Chern class $c_1(\xi) \in H^2(X;\Zz)$$c_1(\xi) \in H^2(X;\Zz)$. Let the divisibility of $c_1(\xi)$$c_1(\xi)$ be $k$$k$ (i.e. $c_1(\xi)$$c_1(\xi)$ is $k$$k$ multiple of a primitive element in $H^2(X;\Zz)$$H^2(X;\Zz)$, then the sphere bundle $S(\xi) = \colon M$$S(\xi) = \colon M$ is a 5-manifold with fundament group $\Zz_k$$\Zz_k$ and $\pi_2(M)$$\pi_2(M)$ is a free abelian group of rank $\mathrm{rank} H_2(X)-1$$\mathrm{rank} H_2(X)-1$, and $\pi_2(M)$$\pi_2(M)$ is a trivial module over the group ring. A priori
Tex syntax error
$M$ is a topological manifold. The smoothing problem is addressed by the following
Proposition 2.1.[Hambleton&Su, Proposition 4.2]
Assume $\xi$$\xi$ is nontrivial. If $k$$k$ is odd, then
Tex syntax error
$M$ admits a smooth structure; if $k$$k$ is even, then
Tex syntax error
$M$ admits a smooth structure if and only if the Kirby-Siebenmann invariant of $X$$X$ is $0$$0$.
• $k=1$$k=1$,
Tex syntax error
$M$ is a simply-connected 5-manifold. The identification of
Tex syntax error
$M$ with manifolds in the standard list of simply-connected 5-manifolds given by the Smale-Barden classification was done in [Duan&Liang].
• $k=2$$k=2$, we have a class of orientable 5-manifolds with fundamental group $\Zz_2$$\Zz_2$, $\pi_2(M)$$\pi_2(M)$ a free abelian group, and a trivial module over the group ring. The classification of these manifolds was the motivation of [Hambleton&Su].
### 2.2 Connected sum along $S^1$
In the Smale-Barden's list of simply-connected 5-manifolds, manfolds are constructed from simple building blocks by the connected sum operation. In the world of manifolds with fundamental group $\Zz_2$$\Zz_2$, the connected sum operation is not closed. The connected sum along $S^1$$S^1$" operation $\sharp_{S^1}$$\sharp_{S^1}$ will do the job.
Definition 2.2. Let $M_1^5$$M_1^5$, $M_2^5$$M_2^5$ be two oriented 5-manifolds with $\pi_1=\Zz_2$$\pi_1=\Zz_2$. Let $E_i \subset M_i$$E_i \subset M_i$ be the normal bundle of an embedded $S^1$$S^1$ in $M_i$$M_i$ representing the nontrivial element in the fundamental group. $E_i$$E_i$ is a rank 4 trivial vector bundle over $S^1$$S^1$. Choose trivialisations of $E_1$$E_1$ and $E_2$$E_2$, and identify the disk bundles of $E_1$$E_1$ and $E_2$$E_2$ using the chosen identification (such that the identification is orientation-reversing, with respect to the induced orientations on the disk bundles from the ambient manifolds), we obtain a new manifold denoted by $M_1 \sharp_{S^1} M_2$$M_1 \sharp_{S^1} M_2$.
Notice that $\sharp_{S^1}$$\sharp_{S^1}$ is not well-defined, the ambiguity comes from the identification of the two normal bundles, measured by $\pi_1SO(4) = \Zz_2$$\pi_1SO(4) = \Zz_2$. To eliminate the ambiguity we need more structures on the tangent bundles of $M_i$$M_i$ and require that $\sharp_{S^1}$$\sharp_{S^1}$ preserves the structures. (Analogous to the connected sum situation where orientations on manifolds are needed.) This will be explained in more detail in the next section.
...
...
...
## 6 References
$. {{endthm}} *$k=1$,$M$is a simply-connected 5-manifold. The identification of$M$with manifolds in the standard list of simply-connected 5-manifolds given by the Smale-Barden classification was done in \cite{Duan&Liang}. *$k=2$, we have a class of orientable 5-manifolds with fundamental group$\Zz_2$,$\pi_2(M)$a free abelian group, and a trivial module over the group ring. The classification of these manifolds was the motivation of \cite{Hambleton&Su}. === Connected sum along$S^1$=== ; In the Smale-Barden's list of simply-connected 5-manifolds, manfolds are constructed from simple building blocks by the connected sum operation. In the world of manifolds with fundamental group$\Zz_2$, the connected sum operation is not closed. The connected sum along$S^1$" operation$\sharp_{S^1}$will do the job. {{beginthm|Definition}} Let$M_1^5$,$M_2^5$be two oriented 5-manifolds with$\pi_1=\Zz_2$. Let$E_i \subset M_i$be the normal bundle of an embedded$S^1$in$M_i$representing the nontrivial element in the fundamental group.$E_i$is a rank 4 trivial vector bundle over$S^1$. Choose trivialisations of$E_1$and$E_2$, and identify the disk bundles of$E_1$and$E_2$using the chosen identification (such that the identification is orientation-reversing, with respect to the induced orientations on the disk bundles from the ambient manifolds), we obtain a new manifold denoted by$M_1 \sharp_{S^1} M_2$. {{endthm}} Notice that$\sharp_{S^1}$is not well-defined, the ambiguity comes from the identification of the two normal bundles, measured by$\pi_1SO(4) = \Zz_2$. To eliminate the ambiguity we need more structures on the tangent bundles of$M_i$and require that$\sharp_{S^1}$preserves the structures. (Analogous to the connected sum situation where orientations on manifolds are needed.) This will be explained in more detail in the next section. == Invariants == ; ... == Classification/Characterization == ; ... == Further discussion == ; ... == References == {{#RefList:}} [[Category:Manifolds]] [[Catefory:5-Manifolds]]\ge 4. Therefore, a practical approach towards the classification of 5-manifolds is to fix a fundamental group in advance and consider the classification of manifolds with the given fundamental group. From this point of view, the fist step one might take is the group $\Zz_2$$\Zz_2$, which is the simplest nontrivial group since it has the least number of elements. (Of course if we take the point of view of generators and relations the rank 1 free group $\Zz$$\Zz$ is the simplest one.) Any other point one should take into account concerning the classification of manifolds with nontrivial fundamental groups $\pi_1$$\pi_1$ is that the higher homotopy groups $\pi_i$$\pi_i$ ($i \ge 2$$i \ge 2$) are modules over the group ring $\Zz[\pi_1]$$\Zz[\pi_1]$, which are apparently homotopy invariants. Especially when we consider $5$$5$-manifolds $M^5$$M^5$, the $\Zz[\pi_1]$$\Zz[\pi_1]$-module structure of $\pi_2(M)$$\pi_2(M)$ will play an important role in the classification. And it's not a surprise that the first clear classification result obtained is the trivial module case. Most part of this item will be a survey of the classification result of 5-manifolds $M^5$$M^5$ with fundamental group $\pi_1(M)=\Zz_2$$\pi_1(M)=\Zz_2$, $\pi_2(M)$$\pi_2(M)$ torsion free and is a trivial $\Zz[\Zz_2]$$\Zz[\Zz_2]$-module obtained in [Hambleton&Su] . ## 2 Construction and examples First some examples known from other context. • $S^2 \times \mathbb R P^3$$S^2 \times \mathbb R P^3$; • $X^5(q)$$X^5(q)$ $q=1,3,5,7$$q=1,3,5,7$ are the 5-dimensional fake real projective spaces with $X(1)=\mathbb R \mathrm P^5$$X(1)=\mathbb R \mathrm P^5$ (There are exactly 4 in the smooth category). The meaning of $q$$q$ will be clear in the section Invariants". ### 2.1 Circle bundles over simply-connected 4-manifolds Let $X^4$$X^4$ be a closed simply-connected topological 4-manifold, $\xi$$\xi$ be a complex line bundle over $X$$X$ with Chern class $c_1(\xi) \in H^2(X;\Zz)$$c_1(\xi) \in H^2(X;\Zz)$. Let the divisibility of $c_1(\xi)$$c_1(\xi)$ be $k$$k$ (i.e. $c_1(\xi)$$c_1(\xi)$ is $k$$k$ multiple of a primitive element in $H^2(X;\Zz)$$H^2(X;\Zz)$, then the sphere bundle $S(\xi) = \colon M$$S(\xi) = \colon M$ is a 5-manifold with fundament group $\Zz_k$$\Zz_k$ and $\pi_2(M)$$\pi_2(M)$ is a free abelian group of rank $\mathrm{rank} H_2(X)-1$$\mathrm{rank} H_2(X)-1$, and $\pi_2(M)$$\pi_2(M)$ is a trivial module over the group ring. A priori Tex syntax error $M$ is a topological manifold. The smoothing problem is addressed by the following Proposition 2.1.[Hambleton&Su, Proposition 4.2] Assume $\xi$$\xi$ is nontrivial. If $k$$k$ is odd, then Tex syntax error $M$ admits a smooth structure; if $k$$k$ is even, then Tex syntax error $M$ admits a smooth structure if and only if the Kirby-Siebenmann invariant of $X$$X$ is $0$$0$. • $k=1$$k=1$, Tex syntax error $M$ is a simply-connected 5-manifold. The identification of Tex syntax error $M$ with manifolds in the standard list of simply-connected 5-manifolds given by the Smale-Barden classification was done in [Duan&Liang]. • $k=2$$k=2$, we have a class of orientable 5-manifolds with fundamental group $\Zz_2$$\Zz_2$, $\pi_2(M)$$\pi_2(M)$ a free abelian group, and a trivial module over the group ring. The classification of these manifolds was the motivation of [Hambleton&Su]. ### 2.2 Connected sum along$S^1\$
In the Smale-Barden's list of simply-connected 5-manifolds, manfolds are constructed from simple building blocks by the connected sum operation. In the world of manifolds with fundamental group $\Zz_2$$\Zz_2$, the connected sum operation is not closed. The connected sum along $S^1$$S^1$" operation $\sharp_{S^1}$$\sharp_{S^1}$ will do the job.
Definition 2.2. Let $M_1^5$$M_1^5$, $M_2^5$$M_2^5$ be two oriented 5-manifolds with $\pi_1=\Zz_2$$\pi_1=\Zz_2$. Let $E_i \subset M_i$$E_i \subset M_i$ be the normal bundle of an embedded $S^1$$S^1$ in $M_i$$M_i$ representing the nontrivial element in the fundamental group. $E_i$$E_i$ is a rank 4 trivial vector bundle over $S^1$$S^1$. Choose trivialisations of $E_1$$E_1$ and $E_2$$E_2$, and identify the disk bundles of $E_1$$E_1$ and $E_2$$E_2$ using the chosen identification (such that the identification is orientation-reversing, with respect to the induced orientations on the disk bundles from the ambient manifolds), we obtain a new manifold denoted by $M_1 \sharp_{S^1} M_2$$M_1 \sharp_{S^1} M_2$.
Notice that $\sharp_{S^1}$$\sharp_{S^1}$ is not well-defined, the ambiguity comes from the identification of the two normal bundles, measured by $\pi_1SO(4) = \Zz_2$$\pi_1SO(4) = \Zz_2$. To eliminate the ambiguity we need more structures on the tangent bundles of $M_i$$M_i$ and require that $\sharp_{S^1}$$\sharp_{S^1}$ preserves the structures. (Analogous to the connected sum situation where orientations on manifolds are needed.) This will be explained in more detail in the next section.
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2022-09-24 22:29:04
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https://www.gradesaver.com/textbooks/science/chemistry/chemistry-molecular-science-5th-edition/chapter-2-chemical-compounds-questions-for-review-and-thought-general-questions-page-90e/112a
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Chapter 2 - Chemical Compounds - Questions for Review and Thought - General Questions: 112a
$$4.652*10^{-23} g$$
Work Step by Step
The mass of one molecule is: $$\frac{28.02 g}{1 mol} \frac{1 mol}{6.022*10^{23} molecules} = 4.652*10^{-23} g$$
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.
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2018-06-24 14:41:36
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https://pmagunia.com/hexadecimal-conversion.html
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Suppose we have a hexadecimal number, '2EC',and we wanted to convert it to decimal. You learned that the decimal numeral system is base 10 which means each digit is multiplied by some power of 10. In hexadecimal, the base is 16. So each digit is multiplied by some power of 16. The first digit from the right in our number is 'C' which is in the ones place. Yes, hexadecimal has a ones place because $16^0=1$. So C which stands for the number 12 and is multiplied by $C \times 16^0=12$.
The second digit from the right in our number is 'E'. E, which in decimal is 14, is in the sixteens place. So we have $14 \times 16^1=224$. So our tally is 224 + 12 = 236.
Finally, we have our last digit which is '2.' 2 is in the two hundred fifty-sixes place. $2 \times 16^2=512$. So our final tally is 512 + 224 + 12 = 748 which is our final answer. Using our subscript notation $748_{10}=2EC_{16}$
Now, suppose we had a decimal number like, 278, and we wanted to convert it to hexadecimal. To convert a decimal number to hexadecimal, we must find the smallest power of sixteen that will fit inside 278. $16^3=4096$ and $16^2=256$. Now, 256 fits into 278 one time so have have a one in the two hundred fifty-sixes place. Now we subtract 256 from 278 and we are left with 22.
The next digit to the right of the two hundred fifty sixes place is the sixteens place. Sixteen fits into 22 one time so we have a one in the sixteens place. $22 - 16 = 6$. The place next to the sixteens place is the ones place. One goes into six, six times so we have a six in the ones place. So our final answer is 116. So, using subscripts, $278_{10}=116_{16}$.
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2020-10-30 04:24:10
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https://www.physicsforums.com/threads/trig-substitution-insanity.714656/
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# Trig Substitution Insanity
1. Oct 5, 2013
### GraceLee
∫x3
----------------------
(4x2 + 9) 3/2
According to my book this is a trig substitution integral. The normal procedure is to substitute atanθ for x when one has a square root w an argument of the form x^2 + a^2. Because the argument of the square root is 4x2 + 9, as opposed to simply x2 +9, the book suggests substituting 2x for x, hence instead of substituting 3tanθ for x- you sub 3/2tanθ. That much makes sense to me. The part that is driving me insane is that the book goes on to complete the trig substitution by substituting this x value (3/2 tanθ) for x in the numerator (ok), substituting 3/2sec2θdθ for dx (ok)... however in the denominator, they write √(4x^2 +9) = √9tan^2θ +9 = 3secθ. That's not substituting the "agreed upon" value of x (i.e., 3/2tanθ) it's just substituting 3tanθ! This is my first post and I'm no math genius so I'm certain I'm just not comprehending something but I've been stuck on this for a while and it's driving me nuts! Any help would greatly appreciated- I hope this post is fairly intelligible.
2. Oct 5, 2013
### bolbteppa
Hey Grace, don't worry the post is legible - just plug the substitution into the denominator to see they are substituting in the agreed upon value for x:
$\sqrt{4x^2+9} = \sqrt{4(\tfrac{3}{2}\tan(\theta))^2+9} =$
$\sqrt{4(\tfrac{9}{4}\tan^2(\theta)) +9} = \sqrt{9 \tan^2(\theta) +9} = \sqrt{9 (\tan^2(\theta) +1)} = 3 \sqrt{\tan^2(\theta) +1} = 3 \sqrt{\sec^2(\theta)} = 3\sec (\theta)$
I think you just forgot to write a $4$ in one step & got confused as a result of this. Also, this & this link show you that you don't have to use trig substitutions in this case necessarily, if you're interested
Last edited by a moderator: Oct 5, 2013
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2017-10-22 21:20:24
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http://mathoverflow.net/revisions/78124/list
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10 Rollback to Revision 8
# [deletedquestion]Implicationform-cyclesinCollatz-typeproblems.
[deleted][deleted]
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1$, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 4, 2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $\mu$ are related by $M = 2^k \mu$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$?
Note: If you prefer to work with the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
9 deleted 923 characters in body; edited tags; edited title; [made Community Wiki]
# Implicationform-cyclesinCollatz-typeproblems.[deletedquestion]
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1$, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 4, 2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $\mu$ are related by $M = 2^k \mu$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$?
Note: If you prefer to work with the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
[deleted][deleted]
8 deleted 5 characters in body
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1$, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 4, 2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $\mu$ are related by $M = 2^k \mu$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$?
Note: If you prefer to work with only the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
7 added 4 characters in body
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2013-05-20 12:34:11
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https://math.stackexchange.com/questions/2328891/conditional-probability-question-drawing-cards-with-order
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# Conditional Probability Question - Drawing Cards with Order
I am trying to calculate odds for a card game. Using a standard 52-card deck. Here is the basics of the game. You are given two cards, you add their values together. (Standard Blackjack values, suits don't matter) Once you know the total value of your two cards, you are given a third card face down and you then have to guess what the total value of your three cards are. What are the odds that you can guess your total?
As an example to illustrate. You draw two cards, let's say you get a 10 and a 5, what are the odds that you get a total of 25? (ie: Draw a 10). This example is a simple probability that I fully understand, however, the twist comes with calculating the odds of getting a total of 25 with any two-card combination with a value of 15? (ie: If you have 9 and 6, the odds of getting 25 are more than with a 10 and 5). Is there a way to generalize this into something like "Given all combinations of 15, what are the odds that your next card will be a 10?" That can be further generalized into "Given all combinations of X, what are you odds that you next card will be Y"?
The reason for this is because I want to have a list of odds to get any TOTAL value of 3 cards. For example, the odds of getting a total of 33 (Ace, Ace, Ace) is 1 in X. You'd have to calculate odds of getting an Ace when you already have 2. Simple enough, but gets very complication when trying to find the odds of getting 25 because your first two cards aren't as determined and there are only so many ways to get 25 from 3 cards, so the order in which you draw those cards matters as to the odds of getting the last one.
I hope this makes sense,
• Well... for any initial two cards and initial sum $x$ the most likely total will be $10+x$ since there are strictly more than $4$ cards with value $10$ remaining in the deck but at most $4$ cards with each other remaining value. As such the optimal strategy is to always guess $10$ more than what you started with. That being said, the probability of guessing correctly will vary depending on how many face cards or 10s are in your initial two cards. – JMoravitz Jun 19 '17 at 20:05
• With no $10$'s in your initial two cards, your probability will be $\frac{16}{50}$, with one $10$ in your initial two cards your probability will be $\frac{15}{50}$ and with two $10$'s in your initial two cards your probability will be $\frac{14}{50}$ of guessing correctly. If you insist on calling these odds instead, then that would be $16:34, 15:35, 14:36$ respectively. (remember, odds and probability are not the same, just related). If your question is "given your initial total is $x$, what is the probability of correctly guessing $x+10$" that will vary based on $x$. – JMoravitz Jun 19 '17 at 20:06
• Do you know only the sum of the first two cards, or the two cards themselves? Also, are you considering that an ace can be worth either $1$ or $11$? – John Jun 19 '17 at 20:14
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2019-07-16 00:45:14
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https://demo7.dspace.org/items/00f007a4-1ae4-49b3-980d-38dbca2d01bd
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## A class of simple $C^*$-algebras arising from certain nonsofic subshifts
Matsumoto, Kengo
##### Description
We present a class of subshifts $Z_N, N = 1,2,...$ whose associated $C^*$-algebras ${\cal O}_{Z_N}$ are simple, purely infinite and not stably isomorphic to any Cuntz-Krieger algebra nor to Cuntz algebra. The class of the subshifts is the first examples whose associated $C^*$-algebras are not stably isomorphic to any Cuntz-Krieger algebra nor to Cuntz algebra. The subshifts $Z_N$ are coded systems whose languages are context free. We compute the topological entropy for the subshifts and show that KMS-state for gauge action on the associated $C^*$-algebra ${\cal O}_{Z_N}$ exists if and only if the logarithm of the inverse temperature is the topological entropy for the subshift $Z_N$, and the corresponding KMS-state is unique.
Comment: 21 pages
##### Keywords
Mathematics - Operator Algebras, Mathematics - Dynamical Systems, 46L35, 37B10
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2022-12-09 05:24:38
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https://www.trustudies.com/question/862/4-If-a-line-intersects-two-concentric-circles-circles-with-the-same-centre-with-centre-O-at-A-B-C-and-D-prove-that-AB-CD-see-figure-br-img-src-https-trustudies-app-public-s3-ap-south-1-amazonaws-com-9thmaths-10-4-4-1-jpg-alt-image-height-150-width-200/
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4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D. prove that AB = CD (see figure).
Now, BC is the chord of the smaller circle and $$OP\perp{BC}$$.
Since, AD is a chord of the larger circle and $$OP\perp{AD}$$.
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2020-10-22 01:42:21
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http://www.feelpp.org/pages/man/06-programming/11-SolveAnEquation/
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# Feel++
## Solve a partial differential equation
With all the previously notions we approach, the definition of a partial differential equation and boundary conditions are our next step. More details on these aspects can be retrieve at this page.
#### Variational formulation
This example refers to a laplacian problem, define by
Strong formulation
-\Delta u=1 \text{ in } \Omega=[0,1]^2, \quad u=1 \text{ on } \partial \Omega
After turning the Strong formulation into its weak form, we have
\int_\Omega \nabla u \cdot \nabla v = \int_\Omega v,\quad u=1 \text{ on } \partial \Omega
where u is the unknown and v a test function. The left side is known as the bilinear form a and the right side is the linear form l.
a(u,v) = l(v), \quad u=1 \text{ on } \partial \Omega
#### Implementation
The steps to implement this problem are
• Loading a 2D mesh, creating the function space V_h, composed of piecewise polynomial functions of order 2, and its associated elements
`````` auto mesh = loadMesh(_mesh=new Mesh<Simplex<2>>);
auto Vh = Pch<2>( mesh );
auto u = Vh->element();
auto v = Vh->element();``````
• Define the linear form l with test function space V_h
``````auto l = form1( _test=Vh );
l = integrate(_range=elements(mesh),
_expr=id(v));``````
• Define the bilinear form a with V_h as test and trial function spaces
``````auto a = form2( _trial=Vh, _test=Vh);
a = integrate(_range=elements(mesh),
`form1` and `form2` are used to define respectively the left and right side of our partial differential equation.
• Add Dirichlet boundary condition on u
``````a+=on(_range=boundaryfaces(mesh),
_rhs=l, _element=u, _expr=cst(0.) );``````
We impose, in this case, u=0 on \partial\Omega, with the keyword `on`.
• Solving the problem
``a.solve(_rhs=l,_solution=u);``
• Exporting the solution
``````auto e = exporter( _mesh=mesh );
``Unresolved directive in 11-SolveAnEquation.adoc - include::../../../codes/11-mylaplacian.cpp[]``
``Unresolved directive in 11-SolveAnEquation.adoc - include::../../../codes/11-mylaplacian.cfg[]``
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2017-11-24 16:46:06
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https://mathsismaths.com/index.php/interesting-questions/108-dividing-a-circle-into-areas-moser-s-circle-problem
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## Dividing a circle into areas (Moser's Circle Problem)
Problem: If we put two points on a circle and connect them, we get two regions. Three points give 4 regions. If we continue this process, connecting every point to every other point and making sure no three lines meet at one point, we get:
(*diagram up to the problematic n=6 to the right)
Points (n) Number of regions (r) 1 1 2 2 3 4 4 8 5 16
We could now conjecture that $$r=2^{n-1}$$. However, if we carefully draw the case for $$n=6$$, we see that $$r=31$$, not $$32$$ as we would expect. Therefore, prove an nth term formula for this sequence.
Solution:
We can solve this problem by induction (although other methods exist). Let's experiment with a small case first. Consider $$n=4$$ and add a point to get to $$n=5$$ like so:
We see that the number of new regions that are added is determined by how many intersections the new lines make with the old lines.
Let us number the points on our circle in order (sticking with $$n=5$$ only for ease of explanation), such that the point we have just added is numbered $$n$$:
Notice that to find the number of intersections that each new line has with old lines, we can consider the number of points to the left of the line and the right of the line, since any old line going from left to right will cross our new line. For a point numbered $$i$$, there are $$n-i-1$$ points to the left, and $$i-1$$ points to the right, and so $$(n-i-1)(i-1)$$ lines that cross our new line.
Now notice that $$x$$ intersections will create $$x+1$$ new regions. For example, the line from 5 to 2 has 2 intersections and creates 3 new regions:
Now let $$f(n)$$ be our nth term formula for the number of regions created by n points on a circle. We know that by adding a point, each new line to a point $$i$$ creates $$(n-i-1)(i-1)+1$$ new regions. So we can sum this across all points $$i$$ to give the total number of new regions:
$\sum_{i=1}^{n-1} (1+(n-i-1)(i-1))$
Therefore:
$f(n)=f(n-1)+\sum_{i=1}^{n-1} (1+(n-i-1)(i-1))$
$f(n)=f(n-1)+\sum_{i=1}^{n-1} (2-n+ni-i^2)$
$f(n)=f(n-1)+\sum_{i=1}^{n-1} 2 - \sum_{i=1}^{n-1} n + \sum_{i=1}^{n-1} ni - \sum_{i=1}^{n-1} i^2$
$f(n)=f(n-1)+2\sum_{i=1}^{n-1} 1 - n \sum_{i=1}^{n-1} 1 + n \sum_{i=1}^{n-1} i - \sum_{i=1}^{n-1} i^2$
$f(n)=f(n-1)+2(n-1) - n(n-1) + n \left(\frac{1}{2} (n-1) n \right) - \left(\frac{1}{6}(n-1)(n)(2n-1)\right)$
$f(n)=f(n-1)+\frac{1}{6} n^3 -n^2+\frac{17}{6} n -2$
Now we can see that each time we add a point, we increase the number of areas by this cubic in $$n$$, starting with one region for $$n=0$$. Therefore:
$f(n)=\sum_{r=1}^{n} \left( \frac{1}{6} r^3 -r^2+\frac{17}{6} r -2 \right)+1$
We can evaluate this like before by using the formulas for the sums of $$k^3$$, $$k^2$$ and $$k$$, giving:
$f(n)=\frac{n}{24}(n^3-6n^2+23n-18)+1$
$f(n)=\frac{1}{24}(n^4-6n^3+23n^2-18n+24)$
And this is our nth term formula.
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2020-10-22 20:03:09
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http://www.jujusdiaries.com/p/some-preliminaries.html
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### Section 0: Preliminaries
We are interested in studying the function $w(z)$ given implicitly by $$$$f(z,w)=a_0(z)+a_1(z)w+a_2(z)w^2+\cdots+a_n(z)w^n$$$$ with $z,w\in \mathbb{C}$ and $a_i(z)$ polynomials with rational coefficients. $w(z)$ is $n$-valued except at singular points. In this discussion, we use the following definition of a singular point: We know for a function $w(z)$ given implicitly by $(1)$, we have $$w'(z)=-\frac{f_z}{f_w}$$ and so we find the points $(z,w)$ where $f(z,w)=f_w(z,w)=0$. We can find these points by first determining where the resultant $R(f,f_w)=0$. The resultant is a polynomial in $z$ the zeros of which are the $z$-values of the singular points. Finding $z$, we then solve for $w$ by plugging into $f(z,w)=0$. This is easily done in Mathematica with the code:
thezvals = z /. NSolve[Resultant[theFunction, D[theFunction, w], w] == 0, z]
thewvals = NSolve[{theFunction == 0, D[theFunction, w] == 0} /. z -> #] & /@ thezvals
Consider for example $$$$f(z,w)=-8z^2+6zw+(9z-1)w^2-9zw^3$$$$ which yields the following singular points: \begin{array}{cc} \text{Singular number} & \text{(z,w)} \\ s_1 & \{-0.0134098,-0.0356913\} \\ s_2 & \{0.,0.\} \\ s_3 & \{0.0328578\, -0.0488795 i,-0.0577313-0.748737 i\} \\ s_4 & \{0.0328578\, +0.0488795 i,-0.0577313+0.748737 i\} \\ s_5 & \{0.614361,0.817821\} \\ \end{array}
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2019-08-24 15:29:08
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http://stackexchange.com/newsletters/newsletter?site=mathematica.stackexchange.com
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## Top new questions this week:
### Find duplicates in list of InfiniteLine
MMA 10 introduced a new function, which can be very convenient: InfiniteLine. Of course, two infinite lines can be described by different arguments: for example InfiniteLine[{{0,0},{1,0}}] and ...
list-manipulation computational-geometry built-in-functions
### Is there any harm or benefit to Removing unneeded private symbols in packages?
One of the things I noticed is that in typical medium-to-large sized packages, a huge number of symbols are created in the myPackage`Private context. You can see a list of them by running ? ...
packages memory
### How to describe the convex hull of a set of points as an implicit region for optimization?
In Mathematica you can create out of a set of points in 2D and 3D the convex hull and use it as a region for optimization problem, e.g., (*Dimension and number of points*) d = 2; np = 10; (*Generate ...
mathematical-optimization computational-geometry
### NDEigensystem returns incorrect eigenvalues for 2D coulomb problem, eigenfunctions contain discontinuity
I posted a similar question a short time ago regarding the 3D Coulomb problem. Jens' excellent answer to this thread allowed me to obtain the correct eigenvalues and eigenenergies for that system. I ...
equation-solving differential-equations numerical-integration
### RandomInteger: speed with long lists
I noticed an interesting effect with RandomInteger when looking at the time to create 10^8 random integers. The following code lines all take about 0.5 seconds on a i7 Windows 64bit Laptop: ...
performance-tuning random
### Creating Saveable->False notebooks
Bug introduced in 10.3 or earlier and persisting through 10.4.1 Reproduced on Windows 7, 10 I'm having difficulties with programmatic creation of non saveable notebooks on Windows with V10.4.1. ...
bugs export notebooks options windows
### Join a list with {6, 6, 2} dimensions with a list {6, 6} dimensions
listA = {{a1, a2, a3, a4, a5, a6}, {b1, b2, b3, b4, b5, b6}, {c1, c2, c3, c4, c5, c6}, {d1, d2, d3, d4, d5, d6}, {e1, e2, e3, e4, e5, e6}, {f1, f2, ...
list-manipulation
## Greatest hits from previous weeks:
### What are the most common pitfalls awaiting new users?
As you may already know, Mathematica is a wonderful piece of software. However, it has a few characteristics that tend to confuse new (and sometimes not-so-new) users. That can be clearly seen from ...
programming core-language faq guidelines
### Where can I find examples of good Mathematica programming practice?
I consider myself a pretty good Mathematica programmer, but I'm always looking out for ways to either improve my way of doing things in Mathematica, or to see if there's something nifty that I haven't ...
reference-request programming
### Using TraditionalForm to express equations
I have a notebook in which a couple of partial derivatives are calculated. At the time of publishing this notebook, I'd like to make the resulting equations look the way they would in standard ...
output-formatting formatting
### Wrong limit involving sine?
Using Mathematica 9.0.0.0, I was trying to calculate the sum $$B_n = \sum_{i=1}^n \sin^4\left(\frac{(i-1)}{(n-1)}\cdot\frac{\pi}{2}\right).$$ where I thought it would be \$B_n = \frac{3(n-1) + ...
calculus-and-analysis
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2016-05-03 09:26:26
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http://bollu.github.io/stable-homotopy-theory.html
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## § Stable homotopy theory
We like stable homotopy groups because of the Freudenthal suspension theorem which tells us that homotopy groups stabilise after many suspensions. The basic idea seems to be something like a tensor-hom adjunction. We have the loop spaces which are like $S^1 \rightarrow X$ and the suspension which is like $S^1 \wedge X$. The theory begins by considering the tensor-hom-adjunction between these objects as fundamental. So curry stuff around to write things as (S^1, A) -> B and A -> (S^1 -> B), which is Suspension(A) -> B and A -> Loop(B). This gives us the adjunction between suspension and looping.
• We then try to ask: how can one invert the suspension formally? One triesto do some sort of formal nonsense, by declaring that maps between $\Sigma^{-n}X$and $\Sigma^{-m} Y$ , but this doesn't work due to some sort of grading issue.
• Instead, one repaces a single object $X$ with a family of objects $\{ X_i \}$called as the spectrum. Then, we can invert the suspension by trying to invertmaps between objects of the same index.
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2021-04-16 19:54:41
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https://brilliant.org/practice/decimals-addition/?subtopic=arithmetic&chapter=decimals
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Basic Mathematics
# Adding Decimals
What is the value of $0. 2 + 0.1$?
What is the value of $0. 8 + 0.7$?
What is the value of $2. 3 + 0. 16$?
Evaluate $6.3-(-4.4)-(-17.3).$
What is the value of $5. 6 + 4.9$?
×
Problem Loading...
Note Loading...
Set Loading...
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2020-10-26 08:07:31
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https://www.physicsforums.com/threads/center-of-a-central-conic.78812/
|
# Center of a Central Conic
1. Jun 12, 2005
### mattmns
How do I find the center of a central conic with the equation $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ Is there some easy forumlua for both the ellipse and hyperbola? Thanks!
2. Jun 12, 2005
### arildno
Do you know how to reduce a conic to standard form through rotation&translation?
3. Jun 12, 2005
### mattmns
I don't remember, but I can use it, and it is in the book I am looking at. Thanks! I will see if I can get it from here, thanks!
4. Jun 12, 2005
### mattmns
Just a quick question, about whether I am doing this correct. Because this thing looks nasty.
I used
$$cot(2\theta) = \frac{a-b}{2h}$$
Then got
$$sin(\theta) = \frac{h(2cos^2(\theta) - 1)}{(a-b)cos(\theta)}$$
and
$$cos(\theta) = \frac{h(1-2sin^2(\theta)}{(a-b)sin(\theta)}$$
Then
$$x = x' [ \frac{h(1-2sin^2(\theta)}{(a-b)sin(\theta)} ] - y' [ \frac{h(2cos^2(\theta) - 1)}{(a-b)cos(\theta)} ]$$
$$y = x' [ \frac{h(2cos^2(\theta) - 1)}{(a-b)cos(\theta)} ] + y' [ \frac{h(1-2sin^2(\theta)}{(a-b)sin(\theta)} ]$$
Does that look correct so far? And then, ughh, I have to substitue that x and y into the original equation, and then see if I can figure the centers from that? Seems like a pain
Also, what is translation?
Thanks
Last edited: Jun 12, 2005
5. Jun 12, 2005
### OlderDan
If you succeed in rotating, you will not have any terms that involve xy products. If you had something of the form
ax^2 + bx + cy^2 + dy + e = 0
you could complete the squares of the x terms and the y terms separately and wind up with something of the form
(x-h)^2/A^2 +or- (y-k)^2/B^2 = 1
which is a ellipse or hyperbola, depending on the sign, centered at (h,k). That's the translation part.
6. Jun 12, 2005
### mattmns
Thanks for the clarification.
I am beginning to think the approach I was trying is not a good one.
Here is what I am trying to prove.
$$\phi (x,y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$
for $$ab - h^2 \neg 0$$ That is: ab - h^2 not equal to 0
Show that the center of the central conic $$\phi (x,y)$$ is the interesection of the lines $$\frac{\partial \phi}{\partial x} = 0$$ and $$\frac{\partial \phi}{\partial y} = 0$$
Do you think the approach I am trying, by first finding the center of the central conic is a good route? Thanks
Last edited: Jun 12, 2005
7. Jun 13, 2005
### OlderDan
I think there is a better way. It is easier to visualize for the case of an ellipse. Think about what it means for dy/dx = 0 and dy/dx approaches infinity for an ellipse, and how you might use the location of a pair of points that satisfy either one of those conditions to locate the center of the ellipse. How could you find the points (x,y) that satisfy those conditions? Then relate those points to the zeros of the partial derivatives.
For the hyperbola, consider the same interpretations of dy/dx = 0 and dy/dx approaches infinity. Only one of the two can be satisfied for points on the hyperbola, but all you need is one pair of points that satisfies one of those two conditions to locate the center. You can still show that the intersection of the lines obtained from the partial derivatives is the center of the hyperbola.
8. Jun 13, 2005
### mattmns
That sounds like a great idea, and I tried it, but I did not come up with exactly what I wanted.
I got dy/dx, using implicit differentiation, to be: $$\frac{-2ax - 2hy - 2g}{2hx + 2by + 2f}$$ So, when dy/dx = 0, then -2ax - 2hy - 2g = 0. You could also say from that, that when dy/dx = 0, then 2ax + 2hy + 2g = 0. Also, when dy/dx = $$\infty$$ then 2hx + 2by + 2f = 0.
Now, when I did the partial differentiation, I got the same answers, which is why I did not like it.
I got
$$\frac{\partial \phi}{\partial x} = 2ax + 2hy + 2g$$
and
$$\frac {\partial \phi}{\partial y} = 2hx + 2by + 2f$$
From here I am stuck, I am unsure of really what to do. It seems like it is close, but just not there.
I guess you could say that for the first part, dy/dx = 0, that $$y = \frac{-ax - g}{h}$$ Which is a line, but from there I am uncertain, other than finding the other line, which is $$y = \frac {-hx - f}{b}$$
9. Jun 13, 2005
### OlderDan
The correspondence between the dy/dx results and the partial derivatives is exactly what you want. Think of any ellipse you can imagine, centered anywhere, with any aspect ration, with axes at any angle of rotation relative to the x-y axes. The solution to $$dy/dx = 0$$ is a line that will intercept that ellipse at two special points. The solution to $$dy/dx = \infty$$ is a line that intercepts the ellipse at two other special points. Think about what these points mean, and where these lines must intersect based on the symmetry of an ellipse.
As I said before, you really only need one of these lines and the intercepts with the ellipse to locate its center. Use that fact to help you interpret the same equations for the hyperbola.
10. Jun 13, 2005
### mattmns
I must not being seeing it.
Here is a diagram of the ellipse.
There is the line where the slope is 0, and the line where the slope is $$\infty$$ Now, I can see how you can use it to find the center. But the problem says that the intersection of these two will be the center. $$\frac{\partial \phi}{\partial x} and \frac{\partial \phi}{\partial y}$$
Maybe I am misinterpreting something here, but those two are equal to the 0 and infinity lines, which to me intersect up at the top. Now, if you used the infinty line at either point of either of the zero lines, and the zero lines at either point of the infity lines, I can see how that would be the center, but to me that is not what the problem said, which I why I think I am not interpreting something correctly.
11. Jun 13, 2005
### OlderDan
Take a look at these example cases created using the graphing feature at
http://www.quickmath.com/
See where the blue lines (partial derivatives) intercept the curves. The horizontal and vertical lines are just to guide your eye to see what is special about these intercepts
#### Attached Files:
File size:
3.4 KB
Views:
127
• ###### Hyperbola.png
File size:
3.3 KB
Views:
103
12. Jun 13, 2005
### mattmns
Ahh I see, very cool, thanks!
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2017-04-29 17:46:44
|
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|
https://www.mediawiki.org/wiki/Extension:PGFTikZ
|
Extension:PGFTikZ
MediaWiki extensions manual
PGFTikZ
Release status: unmaintained
Implementation Tag
Description Parse PGF/TikZ input, generate and compile LaTeX file and upload resulting image
Author(s) Thibault Marin, Markus Bürkler
MediaWiki 1.20+
PHP 5.3+
Database changes No
• $wgPGFTikZdvipsPath •$wgPGFTikZLaTeXStandalone
• $wgPGFTikZDefaultDPI •$wgPGFTikZLaTeXPath
• $wgPGFTikZghostScriptPath •$wgPGFTikZLaTeXOpts
• $wgPGFTikZepstoolPath •$wgPGFTikZuseghostscript
Translate the PGFTikZ extension if it is available at translatewiki.net
Issues Open tasks · Report a bug
The PGFTikZ extension generates images from PGF/TikZ input. See pgf project on SourceForge.
Usage
• Enter PGF/TikZ code between <PGFTikZ> ... </PGFTikZ> tags.
• The first line must be an image link line (see Help:Images) e.g. [[File:SampleImage.png|Sample image]]. Currently 'Image'/'Media' cannot be used instead of 'File' in the image link line. Once the image is uploaded, it will be rendered as if only the image link line was passed.
• Within the <PGFTikZ> ... </PGFTikZ> block, the preamble necessary to compile the LateX file should be added between <PGFTikZPreamble> ... </PGFTikZPreamble> tags.
• After the preamble, the PGF/TikZ input can be entered as it would be after the \begin{document} entry of a LaTeX document (note that the \begin{document} and \end{document} markups should not be added in the PGFTikZ block).
• Example (based on wikipedia:File:Neighbourhood_definition2.svg):
<PGFTikZ>
[[File:My_image_1.png|400px|test image]]
<PGFTikZPreamble>
\usepackage{tikz}
\usetikzlibrary{arrows}
\usetikzlibrary{intersections}
\usetikzlibrary{calc}
</PGFTikZPreamble>
\begin{tikzpicture}[scale=.8,every node/.style={minimum size=1cm}]
%
\begin{scope}[
yshift=-83,every node/.append style={
yslant=0.5,xslant=-1},yslant=0.5,xslant=-1
]
\draw[step=4mm, black] (0,0) grid (5,5);
\draw[black,thick] (0,0) rectangle (5,5);%borders
\fill[green] (2.05,2.05) rectangle (2.35,2.35); % center pixel
\fill[green] (1.65,2.05) rectangle (1.95,2.35); %left
\fill[green] (2.45,2.05) rectangle (2.75,2.35); %right
\fill[green] (2.05,2.45) rectangle (2.35,2.75); %top
\fill[green] (2.05,1.95) rectangle (2.35,1.65); %bottom
\end{scope}
%
% draw annotations
%
\draw[-latex,thick,green](-3,-2)node[left]{1 patch}
to[out=0,in=200] (-1,-.9);
\end{tikzpicture}
</PGFTikZ>
• DO NOT modify the content of the [[File:SampleImage.png]] wikipage directly. The image is regenerated when a wikipage requires it.
• Parameters can be added to the tag line <PGFTikZ param=value>:
• <PGFTikZ dpi=250> sets the resolution to 250 dots per inch.
• <PGFTikZ update=1> forces an update of the image and its page (should rarely be used).
How it works
The extension works as follows:
• The filename of the image to upload is extracted from the first line of the block ([[File:SomeImageFile.png]]).
• If an image with the desired filename exists and was generated by this extension, the current input source is compared to the existing file's. If they are identical, no compilation is required and the image is rendered. If the file exists but differs (or if it does not exist), a new image is uploaded. Note that if the desired image filename already exists and was not generated by this extension, upload will be cancelled.
• The PGF/TikZ code (preamble and body) is extracted from the block, and stored in a temporary LaTeX file similar to the following one:
\documentclass{article}
\def\pgfsysdriver{pgfsys-dvips.def}
\usepackage[usenames]{color}
[... Content of PGFTikZPreamble block...]
\begin{document}
\thispagestyle{empty}
\end{document}
• The generated LaTeX file is compiled and an image (with extension passed in the image link) is generated using the following sequence of commands:
1. Ghostsript way: default (and recommended for portability) from v0.3
• latex (generates a .dvi file from the input .tex)
• dvips (.dvi to .ps)
• ghostscript (converts the .ps file to the final image file)
2. Old way: (pre v0.3) using epstool
• latex (generates a .dvi file from the input .tex)
• dvips (.dvi to .eps)
• epstool (to extract a minimal bounding box from the .eps file)
• convert (converts the .eps file to the final image file)
• The compiled image file is uploaded to the wiki, and the source used to generate it is stored in the file description.
• Note that starting from v0.3, the LaTeX compilation uses the standalone package to produce a minimal page. This can be disabled (reverting to the pre v0.3 behavior) by setting $wgPGFTikZLaTeXStandalone to false in LocalSettings.php (see the Configuration parameter section). Installation • Download and place the file(s) in a directory called PGFTikZ in your extensions/ folder. • Add the following code at the bottom of your LocalSettings.php: require_once "$IP/extensions/PGFTikZ/PGFTikZ.php";
• Done – Navigate to Special:Version on your wiki to verify that the extension is successfully installed.
Configuration parameters
The following parameters can be set in the LocalSettings.php page after loading the extension (the values given here are the default values):
// Default resolution for generated images
$wgPGFTikZDefaultDPI = 300; // Full path to LaTeX executable$wgPGFTikZLaTeXPath = 'latex';
// Command line options to LaTeX command
$wgPGFTikZLaTeXOpts = 'no-shell-escape' // Full path to dvips executable$wgPGFTikZdvipsPath = 'dvips';
// Either use epstool+imagemagick or ghostscript to generate image
$wgPGFTikZuseghostscript = true; // Full path to 'epstool' executable$wgPGFTikZepstoolPath = 'epstool';
// Full path to 'ghostscript' executable
$wgPGFTikZghostScriptPath = 'gs'; // Use standalone LaTeX package$wgPGFTikZLaTeXStandalone = true;
$wgMaxShellMemory = 202400; Security Since the extension internally compiles LaTeX documents from user input, security must be considered before deploying the extension on public wikis. • Some of the possible attacks that must be considered when deploying a web-based LaTeX compilation system are described in the TUG article: "A web-based TeX previewer: The ecstasy and the agony" ([[1]]): • Execution of the system commands via wfShellExec offers control on resource limits given to the system call using$wgMaxShellMemory and $wgMaxShellTime variables. • The source stored in the LaTeX file can also compromise security. The texmf.cnf file can be used to block access to files from within the generated LaTeX file. See [[2]] for a discussion on the matter. Download Please download the code found below and place it in $IP/extensions/PGFTikZ/. Note: \$IP stands for the root directory of your MediaWiki installation, the same directory that holds LocalSettings.php.
Note that the bitbucket links will be closed once the latest version is merged in the mediawiki repository.
version 0.3.0
The latest version (with fixes to work on Windows hosted wikis) (v0.3.0, waiting for mediawiki review) is available at: https://bitbucket.org/thibaultmarin/pgftikz_public/get/v0.3.0.zip
Mediawiki repository
The extension can be retrieved directly from Git [?]:
• Browse code
• Some extensions have tags for stable releases.
• Each branch is associated with a past MediaWiki release. There is also a "master" branch containing the latest alpha version (might require an alpha version of MediaWiki).
Extract the snapshot and place it in the extensions/PGFTikZ/ directory of your MediaWiki installation.
If you are familiar with Git and have shell access to your server, you can also obtain the extension as follows:
cd extensions/ git clone https://gerrit.wikimedia.org/r/mediawiki/extensions/PGFTikZ.git
Known problems
This is an early stage, expect rough edges. Known bugs/limitations include:
• When using preview and save repeatedly, edit conflicts might occur in some cases. The edit conflict redirects to a resolution page where the input PGF/TikZ code is mixed up with the image automatically generated description page. Cancelling the conflict (maybe even deleting the image) and regenerating the page should fix the problem.
• If two pages link to the same file with different content, the image will be re-compiled each time either page is displayed.
• Probably many more, please report them.
|
2022-05-21 00:51:19
|
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|
https://www.springerprofessional.de/new-experimental-technique-for-nodularity-and-mg-fading-control-/15060700?fulltextView=true
|
main-content
## Weitere Artikel dieser Ausgabe durch Wischen aufrufen
14.09.2017 | Ausgabe 11/2017 Open Access
# New Experimental Technique for Nodularity and Mg Fading Control in Compacted Graphite Iron Production on Laboratory Scale
Zeitschrift:
Metallurgical and Materials Transactions A > Ausgabe 11/2017
Autoren:
Juan Carlos Hernando, Björn Domeij, Daniel González, José Manuel Amieva, Attila Diószegi
Wichtige Hinweise
Manuscript submitted March 28, 2017.
## 1 Introduction
One of the critical issues of CGI and SGI production is the control of nucleation and growth mechanisms of graphite.[ 1] The nucleation of graphite is commonly assumed heterogeneous in main graphite nucleation theories.[ 2] This concept assumes that graphite particles nucleate on a pre-existing inclusion in the liquid.[ 37] In the case of lamellar graphite iron (LGI), these inclusions are complex sulfides (Mn,X)S that at the same time have nucleated on complex oxides of Al, Si, Zr, Mg, and Ti.[ 35] On the other hand, graphite in SGI and CGI is observed to have similar nuclei, formed by complex Mg silicates (MgO.SiO 2) which are nucleated on the external layer of MgS and CaS sulfides.[ 6] The subsequent growth of graphite nuclei is primarily affected by two factors: the presence of surface-active impurities in the melt, and the cooling rate during solidification of the alloy.[ 1] The effect of the latter is widely accepted; higher cooling rates promote the formation of SGI. The influence of impurities can be divided into two categories: (a) reactive impurities leading towards a SGI transition such as Mg, Ce, Ca, Y, and La, called compacting or spheroidizing elements; and (b) surface-active impurities favoring lamellar graphite formation, such as S, O, Al, Ti, As, Bi, Te, Pb, and Sb, called anti-compacting or anti-spheroidizing elements.[ 1] The presence of these impurities influences the predominant growth direction of graphite.[ 8, 9] In the case of LGI, dominant growth occurs along the A-axis, while in SGI it happens along the C-axis. CGI is in an intermediate situation since dominant growth direction changes continuously between the A-axis and the C-axis.[ 8, 9] The surface-active impurities are absorbed in the prismatic face of the hexagonal graphite lattice, creating a non-faceted interface, that requires low driving forces to grow, i.e., low undercooling, like in the case of LGI, while for SGI the faceted interface requires larger undercooling to grow.[ 2, 4, 7] Previous experiments show that a completely impurities-free melt would solidify as SGI, indicating that the preferential shape for graphite is spheroids in the total absence of active surface impurities.[ 9] However, the amount of impurities needed to promote the formation of LGI is extremely low,[ 2] being the presence of O and S the most detrimental to that effect due to their strong tendency to reduce the surface energy.[ 1, 8, 9] The presence of dissolved oxygen in the melt has been shown to have a direct impact in the nodularity of the iron.[ 1012] The formation of CGI and SGI, therefore, requires the reduction of the amount of dissolved O and S that are regularly present in the materials used as base alloys. The practical way to achieve this reduction is by the addition of specific elements reacting and combining with O and S and hence cleaning the melt from these impurities, in a process called nodularization. The main element used for this purpose is Mg, even though some other elements like Ce or a RE mixture are also used with the same objective.[ 2, 13, 14] Mg added to the melt, usually as FeSiMg, is used to combine most of the O dissolved in the melt by formation of oxides. The presence of dissolved S is also eliminated by Mg, producing MgS. High levels of S thus represent a complication for this purpose, since part of the Mg intended for O neutralization will be consumed by S making it a matter of great importance to have a close control on this process.[ 12, 15] Residual Mg contents of 0.006 wt pct are reported as the lower limit to produce CGI, while the upper limit is around 0.01 wt pct residual Mg.[ 2] To ensure a full spheroidization in SGI production, the measured residual Mg has to be between 0.02 and 0.06 Mg wt pct.[ 14] These boundaries will change as a function of the amount of total impurities, shifting the CGI region towards higher Mg content in case of high presence of O and S. There are two main problems associated with the nodularization treatment in CGI and SGI production. The first one is to achieve the intended nodularity with the treatment at the end of the solidification. The region between LGI and CGI is especially complex, since there is an abrupt transition from CGI to LGI as a function of Mg content, complicating CGI production.[ 13] In the case of SGI production, high additions of Mg are reported to attenuate the effect after a critical level, reaching an asymptotic level for nodularity.[ 13] The second problem is the disappearance of the effect of this nodularization treatment over time. This occurs when the treated iron is held at elevated temperatures. The practical solution is therefore usually to minimize the time between nodularization treatment and pouring. Previous works have studied the fading phenomenon in SGI and CGI production, studying its relation to cooling curves and the residual Mg after different holding times.[ 1621] All these technological difficulties severely hamper laboratory investigations in CGI, and to some extent also in SGI, which in turn limits the understanding of complex problems occurring during their solidification that often lead to defect formation. The present work, therefore, reports an accurate and reliable experimental technique, which enables controlled laboratory-scale production of cast irons, with controlled graphite morphology ranging from SGI to LGI. A re-melting experimental procedure has been designed to control the process of solidification of a base SGI material and adjust its final nodularity by controlling the holding time as the experimental variable. Tools to evaluate the nodularity and the Mg fading process during the reversion process of SGI are presented, including empirical equations to predict the resulting nodularity at the end of the solidification process for both SGI and CGI production regions.
## 2 Experimental Procedure
### 2.1 Base Material
The base material is designed to meet two purposes: being able to produce iron of the three main families at the end of the solidification, and enabling the study of the fading process of Mg after the experimental process. The nodularization treatment is performed with FeSiMg using the tundish ladle method while inoculation is added in-stream during the pouring process. Optical emission spectrometry (OES) of a rapidly solidified coin produced immediately before the pouring process of the base SGI shows the following chemical composition, displayed in Table I.
Table I
Chemical Composition of the Base Alloy
Element
C
Si
Mn
P
S
Cu
Sn
Mg
CE
Weight percent
3.86
2.59
0.64
0.030
0.010
0.84
0.098
0.065
4.73
CE = C + 1/3(Si + P)
The base alloy is cast in two furan sand molds, containing 25 open cylindrical cavities of 50 mm diameter and 300 mm in height. The sand mold is produced by sand printing and designed to promote a bottom-to-top filling of the mold cavities. The two molds are poured from the same ladle in a sequential operation to minimize the fading of nodularization treatment and the decarburization of the alloy. The cylinders of the base material are then machined to cylinders of 38 mm diameter with a weight of 400 ± 0.5 g which yields an approximate height of 42 mm.
### 2.2 Experimental Equipment
The experimental equipment for the re-melting process is very similar to the equipment used on some previous investigations on LGI solidification[ 2224]; slight modifications of the equipment enabled its application to SGI and CGI. The experimental equipment consists of a programable resistance furnace where a specimen of the base material is re-melted. The specimen is introduced into the furnace’s chamber inside an Al 2O 3 crucible and it is placed in the middle of the heating elements. The chamber is filled with a stable argon flow of 4 l/min to achieve a neutral atmosphere and preserve the chemical composition of the melt during the re-melting and subsequent holding time. The main objective is to slow down the fading process of Mg in order to study its evolution on a wider time range than in previous investigations.[ 17] In this study, the argon intake is modified with respect to previous investigations,[ 2224] and it is now placed at the bottom part of the chamber, producing an ascending flow of argon during the process. Argon is vented from the chamber through the small outlets located in the upper lid, provoking a continuous flow of argon inside the chamber. In the experimental re-melting process, a complete absence of oxygen is not reached since the chamber is not sealed to the external atmosphere, but a reduction and stabilization of the amount of oxygen during the re-melting process and holding time for different experiments are achieved. The equipment is designed to allow thermal measurements during solidification in three different positions. Two type-S thermocouples can be located at the middle section of the specimen, and one on the external part of the crucible on the same section. The reported thermal data represent the values for the thermocouple located at the midpoint of the middle section of the cylinder.
### 2.3 Re-melting Experiments
The base SGI is re-melted in form of cylindrical samples in a thermal cycle of 75 minutes from room temperature to 1723 K (1450 °C), with the intention to re-melt any possible pre-existing nuclei and graphite particle, and held for different times at that temperature. This holding time, representing the fading time of Mg, is the single experimental variable that is modified to achieve different nodularities in the experimental samples. After the holding time is completed, the furnace is shut down and then cooling process starts, leading to the solidification of the sample. When the temperature in the furnace reaches 1273 K (1000 °C), the argon flow is interrupted and the cooling to room temperature occurs under the absence of argon flow.
Table II reports the experimental series of re-melting experiments performed in this study, starting from a holding time of 0 minutes up to 140 minutes. At the end of the solidification process, the sample is sectioned approximately at its middle section, around 20 mm from the bottom end. Samples are mounted in thermosetting resin and ground with SiC papers of different granulometry from a grit size of P80 (FEPA) to P2000 (FEPA) to subsequently start a mechanical polishing process with solutions containing diamond particles of 3 and 1 µm applied on satin woven acetate and short synthetic nap clothes correspondingly. A good graphite retention is achieved after this process,[ 25] but some graphite particles are partially covered by ferrite and could lead to a wrong interpretation. To solve this problem, an additional polishing step with an oxide slurry is manually applied for graphite characterization.
Table II
Overview of the Re-Melting Experiments Series
Sample ID
H0
H10
H20
H30
H40
H50
H60
H70
H80
H90
H100
H110
H120
H130
H140
Holding time (min)
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
Repetitions
2
4
3
3
3
2
3
2
3
2
3
2
2
3
3
### 2.4 Quenching Experiments
A series of quenching experiments is additionally performed to analyze the chemical composition of the melt after the re-melting process and before the start of the solidification process after different holding times, see Table III. The sample is re-melted at 1723 K (1450 °C) and held the correspondent time, but immediately after the end of the holding time, the sample is quenched into water to promote a white solidification and avoid graphite precipitation. This experimental series provides samples suitable for OES after a manual grinding process. One sample per holding time is produced.
Table III
Overview of the Quenching Experiments Series
Sample ID
Q0
Q10
Q20
Q30
Q60
Q110
Holding time (min)
0
10
20
30
60
110
### 2.5 Graphite Characterization
Several shape factors have been used for graphite characterization in last decades, such as roundness, sphericity, compactness, or circularity.[ 2628] This study characterizes the graphite evolution using the concept of nodularity explained in ISO 16112:2006(E)[ 29] and E2567 - 16a from ASTM,[ 30] the standards with a wider application among cast iron users and industry. According to both standards, microstructural features smaller than 10 µm are considered as microporosities, defects due to sample preparation or particles that should be excluded from graphite analysis. Graphite particles in contact with the limits of the image are not considered in the analysis. Nodularity characterization is based on the concept of roundness which is calculated for each graphite particles as follows:
$${\text{Roundness}} = \frac{A}{{A_{\text{circle}} }} = \frac{4}{\pi }*\frac{A}{{l_{ \hbox{max} }^{2} }}$$
(1)
where A is the area of the particle and A circle is the area of a circle with a diameter equal to the maximum distance between two boundary points of the particle, l max. The measurement of these two features, A circle and l max, is carried out with the OLYMPUS Stream Motion Desktop software 1.9.1. A graphite nodule consists of any graphite particle that fulfills both minima required size i.e., >10 µm and roundness factor >0.625 according to ISO standard and a recommended roundness factor of >0.6 by ASTM standard. While ASTM standard only accounts for nodular graphite particles when calculating percent nodularity, ISO standard also considers intermediate graphite particles (forms IV and V) with roundness factor 0.525 to 0.625, as follows:
$${\text{Percent Nodularity}} = 100*\left( {\mathop \sum \nolimits A_{\text{nodules}} + 0.5*\mathop \sum \nolimits A_{\text{intermdiates}} } \right)/\mathop \sum \nolimits A_{\text{all \; graphite \; particles}}$$
(2)
It has been demonstrated in the literature[ 27] that a minimum pixel size cannot be used for general cases, and its selection depends on the dimensions of graphite particles present in the sample. In this case, a preliminary analysis of the microstructures reveals a large graphite particle size, enabling the use of a low magnification, resulting in a pixel side length of 1.081 µm. This pixel size allows to characterize a much larger surface area than required in the norms, using a minimum of 40 micrographs for each specimen, representing a minimum area of 137 mm 2 for each specimen, improving result soundness.
## 3 Results and Discussion
### 3.1 Effect of Holding Time on Nodularity and Graphite Particle Density
As introduced in previous sections, the main objective of this paper is to control the nodularity variation by controlling the holding time, in order to describe the transition from SGI, to CGI and later to LGI. The analysis of the base material, as starting material used for the experimental series, shows a fully nodular iron with a nodularity of 96 pct. The high nodularity measured and the high residual Mg present in the base material makes it appropriate for the purpose of this work.
Micrographs of the samples resulting from re-melting experiments, from H0 to H140, are collected in Figure 1. It can be observed in the micrographs that the number of graphite nodules decreases as the volume fraction of compacted graphite increases with increasing holding time. While transition between SGI and CGI occurs gradually, the transition between CGI and LGI occurs suddenly, as reported in the literature.[ 13] The appearance of lamellar graphite occurs after 130 minutes of holding time, resulting in a successful value for experimental purposes, allowing the controlled production of a wide range of graphite shapes using this experimental technique.
The nodularity analysis of the samples from the re-melting experiments shows clearly how nodularity drops with increasing holding time. In Figure 2, one can deduce a linear decay on the first region of the curve, from 0 to 50 minutes of holding time. Whilst the rate of deterioration of nodularity from 60 to 120 minutes slows down, belonging to CGI irons as the nodularity values drop below the 20 pct, the upper barrier required for CGI material.[ 29] After 130 minutes of holding time, lamellar graphite suddenly appears in the sample. The presence of graphite of type I and II, corresponding to LGI, according to ISO 945,[ 31] makes the characterization of nodularity for those samples inappropriate. For that reason, samples from H130 and H140, showing lamellar graphite particles, are not characterized in terms of nodularity.
In a previous investigation, nodularity fade and nodule count fade were treated as independent phenomena with an interrelated occurrence.[ 18] Results of the current investigation show that nodularity and number of nodules per unit area exhibit a similar correlation with the holding time. Both phenomena seem to occur simultaneously according to Figures 2 and 3(a). The number of nodules per unit area shows the transition towards CGI with increasing holding time, process governed mainly by the residual Mg, since the cooling conditions can be considered similar for all experiments. At the same time, data in Figure 3(b) reveal that the number of vermicular graphite per unit area follows a linear trend with holding time, while the number of intermediate particles per unit area does not vary significantly with holding time, Figure 3(a).
### 3.2 Effect of Holding Time on Cooling Curves
The shape of a cooling curve can be affected by many variables and a large number of correlations between them.[ 32, 33] Despite all those variables at play, it is widely accepted a general specific behavior for each graphitic iron family.[ 2] In this work, all variables are reduced to one, the holding time, and the change in this variable is affecting the residual Mg present in the melt during the solidification. Therefore, from the analysis of the cooling curves, we will be able to observe the relation between the shape of the cooling curve and the residual Mg content and the different graphite shapes present in the microstructure.
The cooling curves from the re-melting experiments, Figure 4, can be clearly divided into three different groups. These groups can be related to the presence of a different type of graphite shape in the final microstructure. The first group consists of the cooling curves belonging to H10 and H20, showing the typical shape of an SGI solidification reported in the literature for sand cups,[ 34] with a high undercooling and low recalescence rate. In these cases, the nodularity values are 86.48 and 70.50 pct, corresponding to an SGI grade. When the presence of compacted graphite starts to be significant in the microstructure, even if the final nodularity values yield to values over 20 pct, both the undercooling and the recalescence increase. Worth to mention is the considerable increase in the recalescence rate, as can be seen in the cooling curve of H40, which final nodularity yields 41.17 pct, far from the CGI values. This behavior is repeated for H60, H80, and H100 samples with nodularity values of 14.32, 10.49, and 5.23 pct, all within the CGI range according to ISO standard showing accordance to literature.[ 34] The last group of cooling curves reflects the presence of lamellar graphite in the microstructure, H130, and H140. The low undercooling and intermediate recalescence rate are commonly found in LGI literature.
### 3.3 Effect of Holding Time on Chemical Composition
The quenching experiments made possible to measure the chemical composition by OES and thus to find possible changes in the chemical composition during holding time. From the results in Table IV, it can be observed that the re-melting experimental process under the argon atmosphere is stable and the sample does not suffer strong variations in the chemical composition. The main elements affecting cast iron solidification like C, Si, S, P, Mn, Cu, and Sn show a stable behavior over the holding time. Only Mg is affected by holding time.
Table IV
Chemical Composition in Weight Percent After Quenching Experiments
Sample
C
Si
Mn
P
S
Cu
Sn
Mg
Base SGI
3.86
2.59
0.64
0.030
0.010
0.84
0.098
0.065
Q0
3.51
2.64
0.66
0.035
0.008
0.84
0.096
0.060
Q10
3.75
2.52
0.70
0.036
0.011
0.89
0.123
0.053
Q20
3.52
2.69
0.70
0.035
0.009
0.86
0.105
0.043
Q30
3.67
2.66
0.65
0.035
0.008
0.83
0.094
0.038
Q60
3.83
2.56
0.65
0.037
0.009
0.84
0.098
0.019
Q110
3.94
2.59
0.66
0.033
0.010
0.84
0.093
0.007
In the case of carbon, it could be expected a certain decarburization after holding the melt at 1723 K (1450 °C) for long times, but the Ar-rich atmosphere inside the chamber of the furnace drastically reduces the rate of this process. Results on carbon composition from Table IV, together with graphite volume fraction, shown in Figure 5, measured at room temperature after re-melting experiments show that carbon is not fading during the experimental process.
### 3.4 Effect of Holding Time on Residual and Free Magnesium
Holding time exhibits a clear influence on the evolution of residual magnesium that relates to the fading process, Figure 6. Mg is lost due to its high vapor pressure at this temperature and reaction with the small amount of oxygen present inside the chamber during the re-melting process and holding time at 1723 K (1450 °C). During the reversion from SGI to CGI, occurring during the re-melting process, Mg residual content shows a polynomial relation with time in minutes, Eq. [ 3].
$${\text{Residual Mg }}\left( {\text{wt pct}} \right) = 0.06 - 0.00088 t + 3.6 *10^{ - 6} t^{2}$$
(3)
for 0 < t < 110, where time is in minutes.
From the available literature, the fading rate varies strongly between different studies focusing either only in the CGI or SGI region. Only a holistic study covering both regions can monitor the whole process of fading. Several authors previously reported diverse experimental expressions for the fading process of Mg at similar temperatures for SGI and CGI production. In the case of the SGI region, a previous work studied the effect of holding the iron inside an 8 tons’ channel induction furnace under argon atmosphere, it is reported an average Mg fading rate of 0.002 wt pct/h.[ 17] For the case of CGI, the effect of holding time on residual Mg shows a relatively low fading rate at short times, showing a decay from 0.018 to 0.015 wt pct Mg after 3 minutes while the rate increases for a longer time, consuming most of Mg content, remaining only 0.001 wt pct Mg after 7 minutes of holding time.[ 20] A similar expression to Eq. [ 3] for residual Mg is reported only for the CGI region, valid only for an initial starting Mg content of 0.018 wt pct and for time up to 7 minutes of holding time.
Another study that tries to cover the whole range of reversion reported a similar expression for fading time, but values of residual Mg are significantly lower to the ones found in this investigation.[ 18] Starting with a similar Mg content, the fading occurs more rapidly, due to the larger surface area of melt exposed to the atmosphere together with the absence of a continuous gas flow protecting the melt. It is reported only a 0.023 wt pct Mg after 30 minutes of holding time and remaining only 0.002 wt pct Mg after 90 minutes of holding time.
The expression derived from the experimental data here reported covers both the SGI and CGI regions and extends its validity for times up to 110 minutes. This fact demonstrates that Ar-rich atmosphere is an effective way to slow down the fading process of Mg and therefore to extend the process window for SGI and CGI production.
The amount of Mg measured by spectrometry reflects the total Mg present in the sample, i.e., the residual magnesium, including the free Mg dissolved in the melt plus the Mg which is combined with other elements. Mg and S are expected to combine according to Eq. [ 4].[ 15] Hence, a rough estimation of free Mg ( w Mg,free) can be done by subtracting from the total Mg, i.e., residual magnesium ( w Mg) that amount of magnesium combined with Sulfur ( w S) by means of Eq. [ 4] where M Mg and M S represent the atomic weights of Mg (24.305 g/mol) and S (32.065 g/mol), respectively[ 15]:
$${\text{Mg}} + {\text{S}} \to {\text{MgS}}$$
(4)
$$w_{\text{Mg, free}} = w_{\text{Mg}} - \frac{{M_{\text{Mg}} }}{{M_{\text{S}} }}*w_{S} .$$
(5)
In the context of this work, free Mg is calculated for each set of values for Mg and S from Table IV, and promotes the understanding of Mg fading independently of S content. Thus, it allows the application of the results obtained in this study on irons with different S contents, only dependent on the initial Mg content, which can be easily calculated after the nodularization treatment (Figure 7).
### 3.5 Relation Between Magnesium Fading and Nodularity
Dawson published a chart (Figure 8) that correlates residual Mg and nodularity for CGI and SGI castings for 0.013 wt pct S.[ 13] Establishing 0.006 wt pct Mg as lower limit for CGI before the sudden transition to LGI occurs, the higher limit for CGI production is set around 0.016 wt pct Mg. In the same graph, the lower limit for production of SGI is set around 0.030 wt pct Mg when contents around 0.050 wt pct Mg establish the upper limit above which increasing Mg concentration do not produce higher nodularities and could lead to defect formation. These values are however linked to several casting parameters like inoculation, oxygen and sulfur content, cooling conditions, etc. If these conditions change the curve shifts horizontally or vertically. An important consideration, though, is that the overall view of the regions of the graph where CGI and SGI are present resembles of a letter S, giving this plot the name of “S-Curve.”
The same appearance of a letter S is shown in this investigation when plotting nodularity against residual and free Mg contents in Figure 9. The mentioned figure from the literature[ 13] and later attempts to describe this relation between nodularity and Mg levels,[ 32] never proposed a mathematical expression that could allow its modeling. In the present work, a third-order polynomial relation can be derived, Eq. [ 6], correlating nodularity according to ISO standard as a function of the residual Mg at the beginning of the solidification. The regions giving result to CGI and SGI can clearly be limited in Figure 9 by comparison with Figure 8. In this experimental procedure, the residual Mg limits for CGI production are 0.008 to 0.025 wt pct of residual Mg and can be translated into the experimental parameter, the holding time, with Eq. [ 3]. The same observation can be done on the SGI region of the curve where an asymptotic behavior can be presumed for higher Mg values than 0.06 wt pct of residual Mg.
$${\text{Nodularity}}_{\text{ISO}} \left( {\text{wt pct}} \right) = 19.2 - 2.60*10^{3} {\text{Mg}}_{\text{res}} + 1.35*10^{5} {\text{Mg}}_{\text{res}}^{ 2} - 1.17*10^{6} {\text{Mg}}_{\text{res}}^{ 3}$$
(6)
for 0.0066 < Mg < 0.065, where Mg is in wt pct.
In the case of free Mg, the plot shows the same overall shape but it also shows how the fading of free Mg evolves allowing the estimation of free Mg and hence the resultant nodularity independent of S content. Figure 10 indicates that free Mg is very low after 110 minutes of holding time, which coincides with the transition into lamellar graphite morphology.
$${\text{Nodularity}}_{\text{ISO}} \left( {\text{wt pct}} \right) = 7.7 - 1.10*10^{3} {\text{Mg}}_{\text{free}} + 1.2*10^{5} {\text{Mg}}_{\text{free}}^{ 2} - 1.3*10^{6} {\text{Mg}}_{\text{free}}^{ 3} ,$$
(7)
for 0 < Mg < 0.057, where Mg is in wt pct.
## 4 Conclusions
The new experimental technique was shown to be appropriate to control the production of irons with all kinds of graphitic morphologies from SGI to LGI. By modification of the holding time, the base SGI can be reversed to different grades of nodularity and into LGI iron. Modifications made in the experimental equipment, like the redesign of the argon intake, made possible the widening of production time window for production of SGI and CGI. Quenching experiments showed to be a successful tool to characterize the chemical evolution of the melt during the re-melting process by means of optical emission spectrometry. The most important conclusions based on the current results are the following:
• The first region of the curve relating nodularity and holding time shows a linear decay at short holding times. The CGI region shows a slower rate of reversion.
• The transition between SGI and CGI occurs gradually while transition between CGI and LGI is abrupt, in support of the conclusions of earlier researchers.
• Nodularity fade and number of nodules per unit area fade seem to occur simultaneously and not in an independent process as suggested by earlier works.
• From the analysis of the cooling curves from the re-melting experiments, we can conclude that all the samples with the presence of compacted graphite in the final microstructure show a similar behavior in the cooling curves regardless of the final nodularity values.
• Magnesium fading was successfully characterized in terms of the experimental variable. This fact opens the possibility to control and predict the final nodularity of the iron with accuracy using this experimental technique.
• The relation between nodularity and residual magnesium in the melt depicted on previous investigations has been proven to be valid for this new experimental process and a mathematical equation for its modeling is proposed.
The above conclusions show the robustness of the experimental technique developed in this work. This will allow future studies on CGI and SGI solidification through controlled laboratory experiments. Topics such as primary solidification, graphite degeneracy and solid state reaction modification are some of the open questions that may now be possible to study in laboratory controlled experiments by the application of this technique.
## Acknowledgments
This research was financed by VINNOVA, the Swedish Agency for Innovation, through the research projects CastDesign, grant number (2013-03303) and SPOFIC II, grant number (2013-04720). The projects, within the Swedish Casting Innovation Center (CIC), are a collaboration between Scania CV AB, Volvo Powertrain Production Gjuteriet AB, SinterCast, Swerea SWECAST and Jönköping University. All support and participating personnel from the above institutions are gratefully acknowledged by the authors.
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2021-07-23 15:17:09
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https://practicepaper.in/gate-cse/gate-cse-1993
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# GATE CSE 1993
Question 1
For the below question, one or more OPTIONS are correct
The eigen vector(s) of the matrix
$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$
is (are)
A $(0,0,\alpha)$ B $(\alpha,0,0)$ C (0,0,1) D $(0,\alpha,0)$
Engineering Mathematics Linear Algebra
Question 1 Explanation:
Question 2
The differential equation
$\frac{d^2 y}{dx^2}+\frac{dy}{dx}+\sin y =0$ is:
A linear B non- linear C homogeneous D of degree two
Engineering Mathematics Calculus
Question 2 Explanation:
Question 3
Simpson's rule for integration gives exact result when f(x) is a polynomial of degree
A 1 B 2 C 3 D 4
Engineering Mathematics Numerical Method
Question 3 Explanation:
Question 4
Which of the following is (are) valid FORTRAN 77 statement(s)?
A DO 13 I = 1 B A = DIM ***7 C READ = 15.0 D GO TO 3 = 10
Question 4 Explanation:
Question 5
Fourier series of the periodic function (period $2 \pi$) defined by
$f(x) = \begin{cases} 0, -p \lt x \lt 0\\x, 0 \lt x \lt p \end{cases} \text { is }\\ \frac{\pi}{4} + \Sigma \left [ \frac{1}{\pi n^2} \left(\cos n\pi - 1 \right) \cos nx - \frac{1}{n} \cos n\pi \sin nx \right ]$
But putting $x = \pi$, we get the sum of the series
$1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \text { is }$
A $\frac{{\pi }^2 }{4}$ B $\frac{{\pi }^2 }{6}$ C $\frac{{\pi }^2 }{8}$ D $\frac{{\pi }^2 }{12}$
Engineering Mathematics
Question 5 Explanation:
Question 6
Which of the following improper integrals is (are) convergent?
A $\int ^{1} _{0} \frac{\sin x}{1-\cos x}dx$ B $\int ^{\infty} _{0} \frac{\cos x}{1+x} dx$ C $\int ^{\infty} _{0} \frac{x}{1+x^2} dx$ D $\int ^{1} _{0} \frac{1-\cos x}{\frac{x^5}{2}} dx$
Engineering Mathematics Calculus
Question 6 Explanation:
Question 7
The function $f\left(x,y\right) = x^2y - 3xy + 2y +x$ has
A no local extremum B one local minimum but no local maximum C one local maximum but no local minimum D one local minimum and one local maximum
Engineering Mathematics Calculus
Question 7 Explanation:
Question 8
Assume that each character code consists of 8 bits. The number of characters that can be transmitted per second through an asynchronous serial line at 2400 baud rate, and with two stop bits is
A 109 B 216 C 218 D 219
Computer Network Physical Layer
Question 8 Explanation:
Question 9
Identify the logic function performed by the circuit shown in figure.
A exclusive OR B exclusive NOR C NAND D NOR E None of the above
Computer Organization Interrupt
Question 9 Explanation:
Question 10
Refer to the PASCAL program shown below.
Program PARAM (input, output);
var m, n : integer;
procedure P (var, x, y : integer);
var m : integer;
begin
m : = 1;
x : = y + 1
end;
procedure Q (x:integer; vary : integer);
begin
x:=y+1;
end;
begin
m:=0; P(m,m); write (m);
n:=0; Q(n*1,n); write (n)
end
The value of m, output by the program PARAM is:
A 1, because m is a local variable in P B 0, because m is the actual parameter that corresponds to the formal parameter in p C 0, because both x and y are just reference to m, and y has the value 0 D 1, because both x and y are just references to m which gets modified in procedure P E none of the above
C Programming Function
Question 10 Explanation:
There are 10 questions to complete.
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2023-02-05 03:33:17
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https://golem.ph.utexas.edu/string/archives/000711.html
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## December 23, 2005
### Müger on Doplicher-Roberts
#### Posted by Urs Schreiber
Yesterday Michael Müger gave two very nice talks on the (various flavors of the) Doplicher-Roberts reconstruction theorem, on occasion of his new, drastically simplified proof of this classical result:
Michael Müger
Abstract Duality Theory for Symmetric Tensor $*$-categories
available here.
The original proof by Doplicher and Roberts was spread over several papers and had around 200 pages. The new one fits, self-containedly with an introduction to the category theoretic language included, snugly into 40 pages.
As far as I understood, the main point is that once you use nowadays obvious category-theoretic reasoning and building on ideas by Deligne concerning this problem, the problem becomes pretty easy. Once you know how to do it, that is.
If you wonder why the above paper seems to start with an appendix, note that it is an appendix, namely to
Hans Halvoren
Quantum Field Theory: Algebraic
to appear in
J. Butterfield & J. Earman (eds.)
Handbook of Philosophy and Physics
So, yes, the ‘physical’ motivation for this is algebraic QFT, which - unfortunately - only describes physically uninteresting field theories so far. Therefore, no one of our string theory group bothered to attend the talk.
But, as Müger emphasized, already these physically trivial QFTs give rise to lots of very interesting mathematics, in particular to something that Michael Müger calls a Galois theory of local quantum fields. That was mainly the topic of his second talk, which I am not going to go into right now.
In case there is anyone out there who knows what a category is but not what the Doplicher-Roberts theorem says, here is a brief outline.
Given a compact group $G$, its category ${\mathrm{Rep}}_{f}\left(G\right)$ of unitary finite dimensional representation over $ℂ$ (no guarantee that I give exactly all the right qulifications in the following!) is well known to be a symmetric tensor-$*$-category.
The question is, conversely, given any symmetric tensor $*$-category, is it equivalent to ${\mathrm{Rep}}_{f}\left(G\right)$ for some compact $G$?
This turns out to be almost true.
Call a concrete symmetric tensor-$*$-category one that is a subcategory (in general not a full subcategory, of course) of that of finite dimensional Hilbert spaces ${\mathrm{Hilb}}_{f}$. For this case Tannaka proved in 1939 (in an article in German language but published in a Japanese journal), that, yes, such a $G$ exists and is unique up to isomorphism.
More generally, though, we may have any (‘abstract’) symmetric tensor-$*$-category which does not arise as a subcategory of $\mathrm{Hilb}$. In this case there is more freedom. Essentially (and unsurprisingly for the physicists Doplicher and Roberts but apparently more suprisingly for instance for the mathematician Deligne) this is because of the existence of fermionic symmetries. So this means that we really have to work not just with Hilbert spaces but with graded/super Hilbert spaces.
Given an abstract tensor $*$-category with $\mathrm{End}\left(1\right)=ℂ$ and assuming there is at least one faithful symmetric $*$-preserving tensor functor from this to $\mathrm{Hilb}$ then the category of such functors is a connected groupoid and the symmetric tensor-$*$-category is equivalent to the representation category of the vertex group of this groupoid. This is also due to Tannaka.
Such a functor to $\mathrm{Hilb}$ need not exist. But if we super everything, then it does.
More precisely, let’s call a group $G$ with a specified element $\sigma \in G$ of order 2, ${\sigma }^{2}=1$, a ${ℤ}_{2}$-graded group $\left(G,\sigma \right)$. Then ${\mathrm{Rep}}_{f}\left(G,\sigma \right)$ is defined to be equal to ${\mathrm{Rep}}_{f}\left(G\right)$ as tensor-$*$-categories but with the usual signs introduced when graded Hilbert spaces are commuted.
Combining results by Doplicher-Roberts and Deligne we then find that for every symmetric tensor-$*$-category $C$ with $\mathrm{End}\left(1\right)=ℂ$
1) there does exist a symmetric $*$-preserving functor
(1)$E:C\to {\mathrm{SuperHilb}}_{f}$
2) $E$ is unique up to isomorphism
3) the group of natural automorphisms of $E$ is a compact graded group - that’s the group we are after
Posted at December 23, 2005 12:38 PM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/711
Read the post Roberts on Nonabelian Cohomology
Weblog: The String Coffee Table
Excerpt: An old reference to a paper by Roberts on nonabelian cohomology.
Tracked: June 15, 2006 7:19 PM
Read the post What is the categorified Gelfand-Naimark theorem?
Weblog: The n-Category Café
Excerpt: A lightning review of the categorified Gelfand-Naimark theorem.
Tracked: October 11, 2006 6:14 PM
Read the post Amplimorphisms and Quantum Symmetry, I
Weblog: The n-Category Café
Excerpt: Localized endomorphisms of quantum observables in arrow-theoretic terms.
Tracked: February 6, 2007 2:10 PM
Read the post HIM Trimester Geometry and Physics, Week 1
Weblog: The n-Category Café
Excerpt: On vertex operator algebras, operads and Segals QFT axioms. On integration over supermanifolds, supergroupoids and the two notions of supercategories.
Tracked: May 14, 2008 6:39 PM
Post a New Comment
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2018-08-20 14:15:23
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https://socratic.org/questions/how-do-you-multiply-x-1-x-1-x-3-x-3
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# How do you multiply (x - 1)(x + 1)(x - 3)(x + 3)?
Nov 1, 2015
${x}^{4} - 10 {x}^{2} + 9$
#### Explanation:
Frist of all, use twice the formula $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ to obtain
$\textcolor{g r e e n}{\left(x - 1\right) \left(x + 1\right)} \textcolor{b l u e}{\left(x - 3\right) \left(x + 3\right)} = \textcolor{g r e e n}{\left({x}^{2} - 1\right)} \textcolor{b l u e}{\left({x}^{2} - 9\right)}$
If you're satisfied, you can leave the expression like this. Otherwise, you can go for a full expansion, multiplying term by term, and obtaining
$\left({x}^{2} - 1\right) \left({x}^{2} - 9\right) = {x}^{4} - 9 {x}^{2} - {x}^{2} + 9 = {x}^{4} - 10 {x}^{2} + 9$
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2019-09-20 13:56:19
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https://math.stackexchange.com/questions/940866/the-only-positive-divisor-of-both-a-and-a-1-is-1
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# The only positive divisor of both $a$ and $a + 1$ is $1$
Prove that if $a \in \mathbb Z$ then the only positive divisor of both $a$ and $a + 1$ is $1$.
When I saw this statement I didn't understand it. The only way that I can see it being true is if a is a negative number but since a ∈ Z a could also be positive and thus it could have more than one positive divisor. If that's the case would I have to disprove this instead of proving? And what would be the best way to do that?
• What common divisors do 2 and 3 have? Or 3 and 4 or 9 and 10? – Paul Sundheim Sep 22 '14 at 0:18
• I think it's asking for a single number that divides both $a$ and $a+1$. – Trurl Sep 22 '14 at 0:19
• Please choose descriptive titles. The original title, Prove or Disprove??? carried virtually no information of what the problem is about. – user147263 Sep 22 '14 at 1:11
You seem to have misunderstood the question. As far as I can tell you are thinking about divisors of $a$ and $a+1$ separately, but the question is asking you to show that the only positive number which is a divisor of both $a$ and $a+1$ simultaneously is $1$.
If $n$ divides both $a$ and $a+1$, it would also divide $a+1-a$.
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2020-03-30 20:20:42
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http://math.stackexchange.com/questions/240450/describe-the-smallest-subspace-of-m-2-times-2-that-contains-matrices/240451
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# Describe the smallest subspace of $M_{2\times 2}$ that contains matrices…
Describe the smallest subspace of $M_{2\times 2}$ that contains matrices
$$\begin{bmatrix}2&1\\0&0\end{bmatrix},\begin{bmatrix}1&0\\0&2\end{bmatrix},\begin{bmatrix}0&-1\\0&0\end{bmatrix}\;.$$
Find the dimension of this subspace.
It sounds like I would find the basis. I know how to do this with vectors, but how do I do this with a set of matrices?
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$M_{2\times 2}$ is isomorphic to $\Bbb R^4$ by the correspondence $$\begin{bmatrix}a&b\\c&d\end{bmatrix}\leftrightarrow\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\;;$$ treat the matrices as $4$-vectors, just displayed in a different way.
-
I actually did that, though I got [1 0 0 2; 0 1 0 -4; 0 0 0 -4]. I'm not sure what to make of the last row. Do I just disregard it? – LearningPython Nov 19 '12 at 8:02
@LearningPython What's wrong with the last row? (I assume this result comes from a row reduction process.) – Ted Nov 19 '12 at 8:06
I guess my first instinct is to think, "No solution!". So if it's fine, then how do I translate that to a basis? Is it a free variable column, and I just disregard the column of 0s before it? – LearningPython Nov 19 '12 at 8:07
@LearningPython: I’ve not done the row reduction to check, but that could well be right; certainly there’s nothing obviously wrong with it, and it reduces further to $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$, which I know is right. What it shows is that the original three matrices are linearly independent, so what do you conclude? – Brian M. Scott Nov 19 '12 at 8:07
@BrianM.Scott, how does this translate to a basis? I'm used to row reducing into free variables that I translate into a general solution. – LearningPython Nov 19 '12 at 8:09
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2013-12-22 09:48:22
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https://www.springerprofessional.de/the-statistical-analysis-of-categorical-data/14474924
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main-content
## Über dieses Buch
The aim of this book is to give an up to date account of the most commonly uses statisti cal models for categorical data. The emphasis is on the connection between theory and applications to real data sets. The book only covers models for categorical data. Various models for mixed continuous and categorical data are thus excluded. The book is written as a textbook, although many methods and results are quite recent. This should imply, that the book can be used for a graduate course in categorical data analysis. With this aim in mind chapters 3 to 12 are concluded with a set of exer cises. In many cases, the data sets are those data sets, which were not included in the examples of the book, although they at one point in time were regarded as potential can didates for an example. A certain amount of general knowledge of statistical theory is necessary to fully benefit from the book. A summary of the basic statistical concepts deemed necessary pre requisites is given in chapter 2. The mathematical level is only moderately high, but the account in chapter 3 of basic properties of exponential families and the parametric multinomial distribution is made as mathematical precise as possible without going into mathematical details and leaving out most proofs.
## Inhaltsverzeichnis
### 1. Categorical Data
Abstract
This book is about categorical data, i.e. data which can only take a finite or countable number of values.
Erling B. Andersen
### 2. Preliminaries
Abstract
In this chapter a short review is given of some basic elements of statistical theory which are necessary prerequisites for the theory and methods developed in subsequent chapters.
Erling B. Andersen
### 3. Statistical Inference
Abstract
The majority of interesting models for categorical data are log-linear models. A family of log-linear models is often referred to as an exponential family.
Erling B. Andersen
### 4. Two-way Contingency Tables
Abstract
A two-way contingency is a number of observed counts set up in a matrix with I rows and J columns Data are thus given as a matrix
$$X = \left[ \begin{array}{l} {x_{11}} \cdots {x_{1J}}\\ \vdots \\ {x_{I1}} \cdots {x_{IJ}} \end{array} \right]$$
The statistical model for such data depends on the way the data are collected. A great variety of tables can, however, be treated by three closely connected statistical models. Let the random variables corresponding to the contingency table be X11,…,XIJ. Then in the first model the Xij’sare assumed to be independent with
$${X_{ij}} \sim Ps({\lambda _{ij}}),$$
i.e. Xij is Poisson distributed with parameter λij. The likelihood function for this model is
$$f({x_{11}},...,{x_{IJ}}|{\lambda _{11}},...,{\lambda _{IJ}}) = \mathop {II}\limits_{i = 1}^I \;\mathop {II}\limits_{j = 1}^J \frac{{\lambda _{ij}^{{x_{ij}}}}}{{x_{ij}^!}}{e^{ - {\lambda _{ij}}}}$$
(4.1)
The log-likelihood is accordingly given by
$$\ln {\rm{L(}}{\lambda _{11}}{\rm{,}} \ldots ,{\lambda _{{\rm{IJ}}}}{\rm{) = }}\sum\limits_{\rm{i}} {\sum\limits_{\rm{j}} {{{\rm{x}}_{{\rm{ij}}}}\ln {\lambda _{{\rm{ij}}}} - \sum\limits_{\rm{i}} {\sum\limits_{\rm{j}} {{\rm{ln}}{{\rm{x}}_{{\rm{ij}}}}!} - \sum\limits_{\rm{i}} {\sum\limits_{\rm{j}} {{\lambda _{{\rm{ij}}}}.} } } } }$$
(4.2)
The model is thus a IJ-dimensional log-linear model with canonical parameters lnλ11,...,1nλIJ and sufficient statistics Tij=Xij, i=1,...,I, j=1,...,J.
Erling B. Andersen
### 5. Three-way Contingency Tables
Abstract
Consider a three-way contingency table {Xijk, i=1,...I, j=1,...,J, k=1,...,K}. As model for such data, it may be assumed that the x’s are observed values of random variables Xijk, i=1,...,I, j=1,...,J, k=1,...,K with a multinomial distribution
$${X_{111}},...,{X_{IJK}} \sim M(n,{p_{111}},...,{p_{IJK}}).$$
(5.1)
Erling B. Andersen
### 6. Multi-dimensional Contingency Tables
Abstract
In chapter 5, log-linear models for three-dimensional tables were treated in great details. Hence we shall not for higher order tables go into details with the parameterizations of the models or with the exact expressions for test quantities and their distributions. Besides for higher order tables the mathematical expressions quickly becomes large and cumbersome to write down.
Erling B. Andersen
### 7. Incomplete Tables, Separability and Collapsibility
Abstract
An observed contingency table is incomplete if it contains zeros in certain cells. Such zeros are of two types, random zeros and structural zeros. A cell has a random zero, if the observed value in the cell is zero, but the expected value is positive. A cell has a structural zero if the expected number is zero, i.e. if it is known a priori that the cell will contain a zero. Random or structural zeros does not impaire the log-linear structure of a given model. It means, however, that certain log-linear parameters can not be estimated.
Erling B. Andersen
### 8. The Logit Model
Abstract
In chapters 4, 5 and 6 the categorical variables appeared in the model in a symmetrical way. In many situations, for example in examples 6.1 and 6.2 in chapter 6, one of the variable is of special interest. For the survival data in example 6.1, survival is the variable of special interest, and the problem is to study if the other three variables have influenced the chance of survival. Variable B in example 6.1 may, therefore, be called a response variable and variables A, C and D explanatory variables. This terminology is the same as the one used in regression analysis, and when survival is regarded as a response variable the data in example 6.1 can in fact be analysed by a regression model. In example 6.2 the position on the truck of the collision can be regarded as a response variable. We are here primarily interested in the effect of explanatory variable A, i.e. the introduction of the safety measure in November 1971, but have to take into account that the other explanatory variables, i.e. whether the truck was parked or not and what the light conditions were, may be of importance for the location of the collision. When the response variable is binary and the explanatory variables are categorical, the appropriate regression model is known as the logit model. More precisely the assumptions for a logit model are:
(a)
The response variable is binary.
(b)
The contingency table formed by the reponse variable and the explanatory variables can be described by a log-linear model.
Erling B. Andersen
### 9. Logistic Regression Analysis
Abstract
In chapter 8 the connection to log-linear models for contingency tables was stressed. The direct connection to regression analysis for continuous response variables will now be brought more clearly into focus. Assume as before that the response variable is binary and that it is observed together with p explanatory variables. For n cases the data will then consist of n vectors
$$\left( {{y_V},{z_{1V}}, \ldots ,{z_{pV}}} \right),v = 1, \ldots ,n$$
of jointly observed values of the response variable and the explanatory variables.
Erling B. Andersen
### 10. Models for the Interactions
Abstract
If the statistical analysis of a contingency table is based on one of the log-linear models in chapters 5, 6 and 7, a number of natural models are easily overlooked. Many useful models can thus be expressed as structures in the log-linear interaction parameters. In this chapter a number of such models are discussed. Many of these models can be viewed as attempts to describe the non-zero interactions by a simple structure if the analysis of the data by a log linear model has failed to give a satisfactory fit to the model. If e.g. the independence hypothesis for a two-way table has been rejected, a residual analysis will often reveal a certain structure in the two-factor interactions.
Erling B. Andersen
### 11. Correspondence Analysis
Abstract
A statistical technique, which is closely related to the models discussed in chapter 10, was developed in France in the 1970’s. This technique known in the English speaking world as correspondence analysis was introduced by Benzecri (1973) as l’Analyse de Correspondance. Many authors have argued that correspondence analysis was not developed in France by Benzecri. This claim is correct in the sense that the technique is closely related to many other forms of statistical analyses, which go far back in time. Some of these connections are discussed in section 11.3 below, where also references will be given. Whatever those connections are, the name correspondence analysis and its popularity in France and neighbouring countries is certainly due to Benzecri and his students.
Erling B. Andersen
### 12. Latent Structure Analysis
Abstract
The structure of a log-linear model can be described on an association diagram by the lines connecting the points. Especially for higher order contingency tables the structure on an assocition diagram can be very complicated, implicating a complicated interpretation of the model. Adding to the interpretation problems for a multi-dimensional contingency table is the fact, that the decision to exclude or include a given interaction in the model can be based on conflicting significance levels depending on the order in which the statistical tests are carried out. These decisions are thus based on the intuition and experience of the data analyst rather than on objective criteria. Hence a good deal of arbitrariness is often involved, when a model is selected to describe the data. We recall for example from several of the examples in the previous chapters that the log-linear model often gave an adequate description of the data judged by a direct test of the model against the saturated model, while among the sequence of successive tests leading to the model, there were cases of significant levels.
Erling B. Andersen
### 13. Computer Programs
Abstract
For most of the models in this book it is necessary to use computer programs to execute the statistical computations.
Erling B. Andersen
### Backmatter
Weitere Informationen
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2020-04-06 09:20:55
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https://www.numerade.com/questions/use-a-calculator-to-evaluate-the-line-integral-correct-to-four-decimal-places-displaystyle-int_c-z-l/
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💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
# Use a calculator to evaluate the line integral correct to four decimal places.$\displaystyle \int_C z \ln (x + y) \, ds$, where $C$ has parametric equations $x = 1 + 3t$, $y = 2 + t^2$, $z = t^4$, $-1 \leqslant t \leqslant 1$
## $1.72599$
Vector Calculus
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### Video Transcript
so again similar to the past few problems style. Set it up and and leave it to you to use a calculator to compare the last t integral. So dx DT is three d u i d t his two t DZ DT. It's for T Cube. The so DSS should be squared of nine plus everything square added together 40 square plus 16 t to six and gt So this in the growth should be t goes from negative 1 to 1 z is t to the fourth X plus y should be t square plus three t process three and, uh, the S is these things, and you can use a calculator to finish that.
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2021-09-25 18:54:49
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https://www.meritnation.com/cbse-class-8/math/rd-sharma/proft-loss-discount-and-value-added-tax-vat/textbook-solutions/10_1_1165_5472_13.11_47008
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Rd Sharma Solutions for Class 8 Math Chapter 13 Proft, Loss, Discount And Value Added Tax Vat are provided here with simple step-by-step explanations. These solutions for Proft, Loss, Discount And Value Added Tax Vat are extremely popular among Class 8 students for Math Proft, Loss, Discount And Value Added Tax Vat Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 8 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
#### Question 1:
A student buys a pen for Rs 90 and sells it for Rs 100. Find his gain and gain percent.
#### Question 2:
Rekha bought a saree for Rs 1240 and sold it for Rs 1147. Find her loss and loss percent.
#### Question 3:
A boy buys 9 apples for Rs 9.60 and sells them at 11 for Rs 12. Find his gain or loss percent.
#### Question 4:
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent.
#### Question 5:
A retailer buys a radio for Rs 225. His overhead expenses are Rs 15. If he sells the radio for Rs 300, determine his profit percent.
#### Question 6:
A retailer buys a cooler for Rs 1200 and overhead expenses on it are Rs 40. If he sells the cooler for Rs 1550, determine his profit percent.
#### Question 7:
A dealer buys a wristwatch for Rs 225 and spends Rs 15 on its repairs. If he sells the same for Rs 300, find his profit percent.
#### Question 8:
Ramesh bought two boxes for Rs 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.
#### Question 9:
If the selling price of 10 pens is equal to cost price of 14 pens, find the gain percent.
#### Question 10:
If the cost price of 18 chairs be equal to selling price of 16 chairs, find the gain or loss percent.
#### Question 11:
If the selling price of 18 oranges is equal to the cost price of 16 oranges, find the loss percent.
#### Question 12:
Ravish sold his motorcycle to Vineet at a loss of 28%. Vineet spent Rs 1680 on its repairs and sold the motor cycle to Rahul for Rs 35910, thereby making a profit of 12.5%, find the cost price of the motor cycle for Ravish.
#### Question 13:
By selling a book for Rs 258, a bookseller gains 20%. For how much should he sell it to gain 30%?
#### Question 14:
A defective briefcase costing Rs 800 is being sold at a loss of 8%. If the price is further reduced by 5%, find its selling price.
#### Question 15:
By selling 90 ball pens for Rs 160 a person loses 20%. How many ball pens should be sold for Rs 96 so as to have a profit of 20%?
#### Question 16:
A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs 36.75 less, he would have gained 30%. Find the cost price of the article.
#### Question 17:
A dishonest shopkeeper professes to sell pulses at his cost price but uses a false weight of 950 gm for each kilogram. Find his gain percent.
#### Question 18:
A dealer bought two tables for Rs 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then, he found that each table was sold for the same price. Find the cost price of each table.
#### Question 19:
Mariam bought two fans for Rs 3605. She sold one at a profit of 15% and the other at a loss of 9%. If Mariam obtained the same amount for each fan, find the cost price of each fan.
#### Question 20:
Some toffees are bought at the rate of 11 for Rs 10 and the same number at the rate of 9 for Rs 10. If the whole lot is sold at one rupee per toffee, find the gain or loss percent on the whole transaction.
#### Question 21:
A tricycle is sold at a gain of 16%. Had it been sold for Rs 100 more, the gain would have been 20%. Find the C.P. of the tricycle.
#### Question 22:
Shabana bought 16 dozen ball bens and sold them at a loss equal to S.P. of 8 ball pens. Find
(i) her loss percent
(ii) S.P. of 1 dozen ball pens, if she purchased these 16 dozen ball pens for Rs 576.
#### Question 23:
The difference between two selling prices of a shirt at profits of 4% and 5% is Rs 6. Find
(i) C.P. of the shirt
(ii) the two selling prices of the shirt
#### Question 24:
Toshiba bought 100 hens for Rs 8000 and sold 20 of these at a gain of 5%. At what gain percent she must sell the remaining hens so as to gain 20% on the whole?
#### Question 1:
Find the S.P. if
(i) M.P. = Rs 1300 and Discount = 10%
(ii) M.P. = Rs 500 and Discount = 15%
#### Question 2:
Find the M.P. if
(i) S.P. = Rs 1222 and Discount = 6%
(ii) S.P. = Rs 495 and Discount = 1%
#### Question 3:
Find discount in percent when
(i) M.P. = Rs 900 and S.P. = Rs 873
(ii) M.P. = Rs 500 and S.P. = Rs 425
#### Question 4:
A shop selling sewing machines offers 3% discount on all cash purchases. What cash amount does a customer pay for a sewing machine the price of which is marked as Rs 650.
#### Question 5:
The marked price of a ceiling fan is Rs 720. During off season, it is sold for Rs 684. Determine the discount percent.
#### Question 6:
On the eve of Gandhi Jayanti a saree is sold for Rs 720 after allowing 20% discount. What is its marked price?
#### Question 7:
After allowing a discount of $7\frac{1}{2}%$ on the marked price, an article is sold for Rs 555. Find its markd price.
#### Question 8:
A shopkeeper allows his customers 10% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs 250?
#### Question 9:
A shopkeeper allows 20% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs 500?
#### Question 10:
A tradesman marks his goods at such a price that after allowing a discount of 15%, he makes a profit of 20%. What is the marked price of an article whose cost price is Rs 170?
#### Question 11:
A shopkeeper marks his goods in such a way that after allowing a discount of 25% on the marked price, he still makes a profit of 50%. Find the ratio of the C.P. to the M.P.
#### Question 12:
A cycle dealer offers a discount of 10% and still makes a profit of 26%. What is the actual cost to him of a cycle whose marked price is Rs 840?
#### Question 13:
A shopkeeper allows 23% commision on his advertised price and still makes a profit of 10%. If he gains Rs 56 on one item, find his advertised price.
#### Question 14:
A shopkeeper marks his goods at 40% above the cost price but allows a discount of 5% for cash payment to his customers. What actual profit does he make, if he receives Rs 1064 after paying the discount?
#### Question 15:
By selling a pair of earings at a discount of 25% on the marked price, a jeweller makes a profit of 16%. If the profit is Rs 48, what is the cost price? What is the marked price and the price at which the pair was eventually bought?
#### Question 16:
A publisher gives 32% discount on the printed price of a book to booksellers. What does a bookseller pay for a book whose printed price is Rs 275?
#### Question 17:
After allowing a discount of 20% on the marked price of a lamp, a trader loses 10%. By what percentage is the marked price above the cost price?
#### Question 18:
The list price of a table fan is Rs 480 and it is available to a retailer at 25% discount. For how much should a retailer sell it to gain 15%?
#### Question 19:
Rohit buys an item at 25% discount on the marked price. He sells it for Rs 660, making a profit of 10%. What is the marked price of the item?
#### Question 20:
A cycle merchant allows 20% discount on the marked price of the cycles and still makes a profit of 20%. If he gains Rs 360 over the sale of one cycle, find the marked price of the cycle.
#### Question 21:
Jyoti and Meena run a ready-made garment shop. They mark the garments at such a price that even after allowing a discount of 12.5%, they make a profit of 10%. Find the marked price of a suit which costs them Rs 1470.
#### Question 22:
What price should Aslam mark on a pair of shoes which costs him Rs 1200 so as to gain 12% after allowing a discount of 16%?
#### Question 23:
Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt for her marked at Rs 850?
#### Question 24:
A shopkeeper offers 10% off-season discount to the customers and still makes a profit of 26%. What is the cost price for the shopkeeper on a pair of shoes marked at Rs 1120?
#### Question 25:
A lady shopkeeper allows her customers 10% discount on the marked price of the goods and still gets a profit of 25%. What is the cost price of a fan for her marked at Rs 1250?
#### Question 1:
The list price of a refrigerator is Rs 9700. If a value added tax of 6% is to be charged on it, how much one has to pay to buy the refrigerator?
#### Question 2:
Vikram bought a watch for Rs 825. If this amount includes 10% VAT on the list price, what was the list price of the watch?
#### Question 3:
Aman bought a shirt for Rs 374.50 which includes 7% VAT. Find the list price of the shirt.
#### Question 4:
Rani purchases a pair of shoes whose sale price is Rs 175. If she pays VAT at the rate of 7%, how much amount does she poy as VAT? Also, find the net value of the pair of shoes.
#### Question 5:
Swarna paid Rs 20 as VAT on a pair of shoes worth Rs 250. Find the rate of VAT.
#### Question 6:
Sarita buys goods worth Rs 5500. She gets a rebate of 5% on it. After getting the rebate if VAT at the rate of 5% is charged, find the amount she will have to pay for the goods.
#### Question 7:
The cost of furniture inclusive of VAT is Rs 7150. If the rate of VAT is 10%, find the original cost of the furniture.
#### Question 8:
A refrigerator is available for Rs 13750 including VAT. If the rate of VAT is 10%, find the original cost of the furniture.
#### Question 9:
A colour TV is available for Rs 13440 inclusive of VAT. If the original cost of TV is Rs 12000, find the rate of VAT.
#### Question 10:
Reena goes to a shop to buy a radio, costing Rs 2568. The rate of VAT is 7%. She tells the shopkeeper to reduce the price of the radio such that she has to pay Rs 2568, inclusive of VAT. Find the reduction needed in the price of radio.
#### Question 11:
Rajat goes to a departmental store and buys the following articles:
Item Price per item Rate of VAT 2 Pairs of shoes 1Sewing machine 2 Tea-sets Rs 800 Rs 1500 Rs 650 5% 6% 4%
Calculate the total amount he has to pay to the store.
#### Question 12:
Ajit buys a motorcycle for Rs 17600 including value added tax. If the rate of VAT is 10%, what is the sale price of the motorcycle?
#### Question 13:
Manoj buys a lather coat costing Rs 900 at Rs 990 after paying the VAT. Calculate the rate of VAT charge on the coat.
#### Question 14:
Rakesh goes to a departmental store and purchases the following articles:
(i) biscuits and bakery products costing Rs 50, VAT @ 5%,
(ii) medicines costing Rs 90, VAT @ 10%,
(iii) clothes costing Rs 400, VAT @ 1%, and
(iv) cosmetics costing Rs 150, VAT @ 10%.
Calculate the total amount to be paid by Rakesh to the store.
#### Question 15:
Rajeeta purchased a set of cosmetics. She paid Rs 165 for it including VAT. If the rate of VAT is 10%, find the sale price of the set.
#### Question 16:
Sunita purchases a bicycle for Rs 660. She has paid a VAT of 10%. Find the list price of the bicycle?
#### Question 17:
The sales price of a television, inclusive of VAT, is Rs 13,500. If VAT charged at the rate of 8% of the list price, find the list price of the television.
#### Question 18:
Shikha purchased a car with a marked price of RS 210000 at a discount of 5%. If VAT is charged at the rate of 10%, find the amount Shikha had paid for purchasing the car.
#### Question 19:
Shruti bought a set of cosmetic items for Rs 345 including 15% value added tax and a purse for Rs 110 including 10% VAT. What percent is the VAT charged on the whole transaction?
#### Question 20:
List price of a cooler is Rs 2563. The rate of VAT is 10%. The customer requests the shopkeeper to allow a discount in the price of the cooler to such an extent that the price remains Rs 2563 inclusive of VAT. Find the descount in the price of the cooler.
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2021-05-14 20:28:53
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|
https://plainmath.net/2984/find-the-inverse-laplace-transform-frac-plus-plus-logleft-frac-plus-right
|
# Find the inverse Laplace transform of (any two) i) frac{(s^2+3)}{s(s^2+9)} ii) logleft(frac{(s+1)}{(s-1)}right)
Joni Kenny 2021-02-20 Answered
Find the inverse Laplace transform of (any two)
i) $\frac{\left({s}^{2}+3\right)}{s\left({s}^{2}+9\right)}$
ii) $\mathrm{log}\left(\frac{\left(s+1\right)}{\left(s-1\right)}\right)$
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Clelioo
(i) The given Laplace transformation is,$\frac{\left({s}^{2}+3\right)}{s\left({s}^{2}+9\right)}$
We can write it,
$\frac{{s}^{2}+3}{s\left({s}^{2}+9\right)}=\frac{{s}^{2}}{s\left({s}^{2}+9\right)}+\frac{3}{s\left({s}^{2}+9\right)}$
$=\frac{s}{\left({s}^{2}+9\right)+\frac{{s}^{2}-\left({s}^{2}+9\right)}{s\left({s}^{2}+9\right)\right)}\left(\frac{3}{-9}\right)}$
$=\frac{s}{\left({s}^{2}+9\right)+\frac{1}{3}\cdot \frac{s}{{s}^{2}+9}+\frac{1}{3}\cdot \frac{1}{s}}$
$=\frac{2}{3}\cdot \frac{s}{{s}^{2}+9}+\frac{1}{3}\cdot \frac{1}{s}$
$=\frac{2}{3}\cdot \frac{s}{{s}^{2}+{3}^{2}}+\frac{1}{3}\cdot \frac{1}{s}$
Now taking inverse Laplace of both sides we have,
${L}^{-1}\left[\frac{{s}^{2}+3}{s\left({s}^{2}+9\right)}\right]=\frac{2}{3}{L}^{-1}\left[\frac{s}{\left({s}^{2}+{3}^{2}\right)}\right]+\frac{1}{3}{L}^{-1}\left[\frac{1}{s}\right]$
$=\frac{2}{3}\mathrm{cos}3t+\frac{1}{3}$
ANSWER:${L}^{-1}\left[\frac{{s}^{2}+3}{s\left({s}^{2}+9\right)}\right]=\frac{2}{3}\mathrm{cos}3t+\frac{1}{3}$
Step 3
(ii)
The given Laplace transformation is,
$\mathrm{log}\left(\frac{s+1}{s-1}\right)$
We can write it,
$\mathrm{log}\left(\frac{s+1}{s-1}\right)=\mathrm{log}\left(s+1\right)-\mathrm{log}\left(s-1\right)$
Now,the property of Laplace transformation is,
$tf\left(t\right)↔-\frac{d}{ds}\left[F\left(s\right)\right]$
$tf\left(t\right)↔-\frac{d}{ds}\left[\mathrm{log}\left(s+1\right)-\mathrm{log}\left(s-1\right)\right]$
$⇒tf\left(t\right)↔-\frac{1}{s+1}+\frac{1}{s-1}$
$⇒tf\left(t\right)↔-{e}^{-t}+{e}^{t}$
$⇒f\left(t\right)↔\frac{{e}^{t}-{e}^{-t}}{t}=2\left(\frac{{e}^{t}-{e}^{-t}}{2t}\right)=2\left(\frac{\mathrm{sin}ht}{t}\right)$
Therefore,
${L}^{-1}\left[\mathrm{log}\frac{s+1}{s-1}\right]=2\left(\frac{\mathrm{sin}ht}{t}\right)$
Answer: ${L}^{-1}\left[\mathrm{log}\frac{s+1}{s-1}\right]=2\left(\frac{\mathrm{sin}ht}{t}\right)$
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2022-08-16 03:30:16
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|
https://docs.microsoft.com/en-us/dax/rate-function-dax
|
# RATE
Returns the interest rate per period of an annuity. RATE is calculated by iteration and can have zero or more solutions. If the successive results of RATE do not converge to within 0.0000001 after 20 iterations, an error is returned.
## Syntax
RATE(<nper>, <pmt>, <pv>[, <fv>[, <type>[, <guess>]]])
### Parameters
Term Definition
nper The total number of payment periods in an annuity.
pmt The payment made each period and cannot change over the life of the annuity. Typically, pmt includes principal and interest but no other fees or taxes.
pv The present value — the total amount that a series of future payments is worth now.
fv (Optional) The future value, or a cash balance you want to attain after the last payment is made. If fv is omitted, it is assumed to be 0 (the future value of a loan, for example, is 0).
type (Optional) The number 0 or 1 which indicates when payments are due. If type is omitted, it is assumed to be 0. The accepted values are listed below this table.
guess (Optional) Your guess for what the rate will be.
- If omitted, it is assumed to be 10%.
- If RATE does not converge, try different values for guess. RATE usually converges if guess is between 0 and 1.
The type parameter accepts the following values:
Set type equal to If payments are due
0 or omitted At the end of the period
1 At the beginning of the period
## Return Value
The interest rate per period.
## Remarks
• Make sure that you are consistent about the units you use for specifying guess and nper. If you make monthly payments on a four-year loan at 12 percent annual interest, use 0.12/12 for guess and 4*12 for nper. If you make annual payments on the same loan, use 0.12 for guess and 4 for nper.
• type is rounded to the nearest integer.
• An error is returned if:
• nper ≤ 0.
• RATE does not converge to within 0.0000001 after 20 iterations
• This function is not supported for use in DirectQuery mode when used in calculated columns or row-level security (RLS) rules.
## Examples
Data Description
4 Years of the loan
-200 Monthly payment
8000 Amount of the loan
### Example 1
The following DAX query:
EVALUATE
{
RATE(4*12, -200, 8000)
}
Returns the monthly rate of the loan using the terms specified above.
[Value]
0.00770147248820137
### Example 2
The following DAX query:
EVALUATE
{
RATE(4*12, -200, 8000) * 12
}
Returns the annual rate of the loan using the terms specified above.
[Value]
0.0924176698584164
|
2021-06-21 11:35:06
|
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https://www.studypug.com/us/en/math/algebra-1/use-cosine-ratio-to-calculate-angles-and-side
|
# Using cosine ratio to calculate angles and side (Cos = $\frac{a}{h}$)
### Using cosine ratio to calculate angles and side (Cos = $\frac{a}{h}$)
Cosine ratios are exactly the same idea of sine ratios or tangent ratios. The only difference between it and the other two trigonometric ratios is that it is the ratio of the adjacent side to the hypotenuse of a right triangle.
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2017-01-21 19:33:29
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https://www.azdictionary.com/definition/atomic%20number%2084
|
• Definition for "atomic number 84"
• a radioactive metallic element that is much like…
• Hypernym for "atomic number 84"
• metallic element
# atomic number 84 definition
• noun:
• a radioactive metallic element that is much like tellurium and bismuth; occurs in uranium ores but could be created by pestering bismuth with neutrons in a nuclear reactor
• a radioactive metallic factor which comparable to tellurium and bismuth; occurs in uranium ores but could be from bombarding bismuth with neutrons in a nuclear reactor
|
2017-05-23 21:20:24
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https://fabbo.my/collections/love-liner
|
-10%
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-10%
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Description Product Features: ★ The tip of the eyeliner is smooth that helps to create a prefect eye makeup. ★...
RM72.90RM65.61
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Showing: 1 -5 of 5
|
2019-10-20 09:39:54
|
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|
https://open.kattis.com/problems/stigavordur
|
Kattis
# Stigavörður
You really like playing games with your friend. This time, however, you are stuck being the score keeper for two of your friends. They are playing a game where $n$ numbered tiles lay on the board in a line. The players can do one of two moves each turn. They can change the number on a single tile. They can also choose two numbers $j$ and $k$, such that $1 \leq j \leq k \leq n$, and they score
$a_ j \oplus a_{j + 1} \oplus \dots \oplus a_{k - 1} \oplus a_ k$
points, where $a_ j$ is the $j$-th tile and $a \oplus b = \text {gcd}(a, b)$. You aren’t quite sure what the goal of the game is, but that doesn’t matter. All you need to do is keep track of the scores.
## Input
The first line of the input contains two integers, $1 \leq n \leq 10^5$ and $1 \leq q \leq 10^5$. The next line contains $n$ integers all larger than zero, but none larger than $10^9$. The next $q$ lines all contain three integers, $x$, $y$, and $z$. Each line describes a single turn in the game. The integer $x$ is either $1$ or $2$. If $x = 1$, then $1 \leq y \leq n$ and $1 \leq z \leq 10^9$ and this means a player changed the number on the $y$-th tile to $z$. If $x = 2$, then $1 \leq y \leq z \leq n$ and this means a player scored, as described above, with $j = y$ and $k = z$.
## Output
For each turn in the game, in order, where a player scores the output should conatain a line with the score achieved that turn.
Sample Input 1 Sample Output 1
4 5
1000000000 2 4 100
2 1 1
2 2 2
2 3 3
2 4 4
2 1 4
1000000000
2
4
100
2
|
2021-06-16 21:01:00
|
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|
https://gmatclub.com/forum/jim-s-weight-is-140-percent-of-marcia-s-weight-bob-s-weight-is-90-per-252884.html
|
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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per
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Math Expert
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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per [#permalink]
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05 Nov 2017, 01:40
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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 percent of Lee’s weight. Lee weighs twice as much as Marcia. What percentage of Jim’s weight is Bob’s weight?
(A) 64 2/7
(B) 77
(C) 90 7/9
(D) 128 4/7
(E) 155 5/9
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Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per [#permalink]
Show Tags
05 Nov 2017, 01:59
Let Marcia's weight be 100. Therefore, Jim weighs 140(140% of Marcia's weight)
Bob's weight(180) is 90% of Lee's weight which is 200, twice as much as Marcia's weight.
Hence we need to find what percentage is Jim's weight to Bob's weight
Bob's weight is 180 and is x% of 140(Jim's weight) OR 180 is x% of 140
$$180 = \frac{140x}{100}$$
$$x = \frac{18000}{140} = 128 \frac{4}{7}$$ (Option D)
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Joined: 02 Aug 2009
Posts: 6971
Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per [#permalink]
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05 Nov 2017, 02:15
1
Bunuel wrote:
Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 percent of Lee’s weight. Lee weighs twice as much as Marcia. What percentage of Jim’s weight is Bob’s weight?
(A) 64 2/7
(B) 77
(C) 90 7/9
(D) 128 4/7
(E) 155 5/9
just incase you are having trouble in some calculations...
$$J=\frac{140}{100}*M$$...
$$B=\frac{90}{100}*L=\frac{90}{100}*2M=\frac{180}{100}M$$...
now B is > J so $$\frac{B}{J} >100$$%
only D and E left
$$\frac{B}{J}*100=\frac{180}{140}*100$$...
easier calculation is 150% of 140 and D is < 150 and E>150....
$$140*\frac{150}{100} = 140*(\frac{100}{100}+\frac{50}{100})= 140+\frac{140}{2}=140+70=210$$..
but we are looking for 180 so ans has to be LESS than 150%
ONLY D left.
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
GMAT online Tutor
Jim’s weight is 140 percent of Marcia’s weight. Bob’s weight is 90 per &nbs [#permalink] 05 Nov 2017, 02:15
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2018-10-21 10:27:01
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https://www.coursehero.com/file/6148640/Chapter3/
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Chapter3 - 3 Limits of functions Continuity After introducing the basic notions on functions limits and continuity we will go on to Bolzanos theorem and
# Chapter3 - 3 Limits of functions Continuity After...
• Notes
• 11
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3 Limits of functions, Continuity After introducing the basic notions on functions, limits and continuity, we will go on to Bolzano’s theorem, and the Intermediate Value Theorem (IVT) which follows from it, as well as the Extremal Value Theorem (EVT). 3.1 Functions Def A function f is a set of ordered pairs ( x, y ), with x, y R , such that no two (ordered pairs) have the same first member. By definition, the second member y is determined by x , one may unam- biguously write y as f ( x ), where f denotes the assignment x 7→ y . The set of all x (for which f is defined) is called the domain of f , and the set of the corresponding y is called the image (or range ) of f . Notation : f : X Y , where X is the domain, and Y contains the range. One can plot the ordered pairs { ( x, y = f ( x )) } (defining a function f ) in the Cartesian plane R 2 = { ( x, y ) | x, y R } , and the resulting figure is called the graph of f . It will be useful to become aware of the graphs of a number of standard functions, such as the ones below. Examples : (i) The identity function : f ( x ) = x ; (ii) Constant function : f ( x ) = c , for all x R , with c a fixed real number; (iii) Linear function : f = ax + b , for constants a, b ; (iv) Polynomial function of degree n 0: f ( x ) = n j =0 a j x j , with a 1 , . . . , a n R , a n ̸ = 0; (v) Upper semicircle function : f ( x ) = r 2 x 2 , X = { x R | − r x r } , r : radius > 0; (vi) The integral part function : f ( x ) = [ x ], the largest integer not greater than x , X = R , Image( f ) = Z . 1
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2021-06-25 05:52:45
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https://math.stackexchange.com/questions/3234934/is-it-possible-for-two-recursive-sequences-to-have-the-same-characteristic-equat
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# Is it possible for two recursive sequences to have the same characteristic equations
If I have a sequence, $$\{t_1, \,t_2, \,t_1+t_2,\, t_1+2t_2,\cdots\}$$ I know that the formula for it is:
$$T_1=t_1, T_2=t_2, T_{n+2}=T_{n+1}+T_n.$$
If there are two sequences, $$A_n, B_n$$ such that $$A_1=1, A_2=0, A_{n+1}+A_n =A_{n+2}, B_1=0, B_2=1, B_{n+2}=B_{n+1}+B_n$$
Then $$A_n=F_{n-1}$$ and $$B_n=F_{n-1}$$ with $$F_n$$ being the Fibonacci numbers and $$F_1=1, F_2=1$$.
So the sequence becomes $$T_n =t_1 F_{n-2} +t_2F_{n-1}$$
The characteristic equation, which comes from $$T_{n+2}=T_{n+1}+T_{n}$$ is $$x^2-x-1=0$$ which has the roots $$\frac{1\pm \sqrt{5}}{2}$$. This is the same characteristic equation as the Fibonacci sequence.
Is it possible for two different sequences to have the same characteristic equation?
• What does $t_1,t_2,t_1+t_2,t_1+2t_2$ mean? – lulu May 21 at 21:08
• that is the sequence given – user130306 May 21 at 21:08
• Well, it doesn't make any sense. What does the notation mean? What, say, does $t_2, t_1$ mean? – lulu May 21 at 21:08
• To your broader question, of course two sequences can have the same characteristic equation. They just need to have different initial conditions. – lulu May 21 at 21:10
• A recurrence of the form $a_n=pa_{n-1}+qa_{n-2}$ is obviously completely determined by any two consecutive values. In fact it is determined by any pair of values. This means that if two sequences match in two places they are identical. The same recurrence can have different initial values and these will generate a different sequence. – Mark Bennet May 21 at 21:22
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2019-06-18 17:22:55
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https://etank.nu/yyfkiyn/archive.php?1d4457=stirling%27s-formula-binomial-coefficient
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Vælg en side
In this post, we will prove bounds on the coefficients of the form and where and is an integer. For example, your function should return 6 for n = 4 and k = 2, and it should return 10 for n = 5 and k = 2. Use the binomial theorem to express ( x + y) 7 in expanded form. $\begingroup$ Henri Cohen's comment tells you how to get started. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. saad0105050 Combinatorics, Computer Science, Elementary, Expository, Mathematics January 17, 2014 December 13, 2017 3 Minutes. Binomial Coefficient Formula. Almost always with binomial sums the number of summands is far less than the contribution from the largest summand, and the largest summand alone often gives a good asymptotic estimate. $\endgroup$ – Mark Wildon Jun 16 at 11:55 Another formula is it is obtained from (2) using x = 1. The coefficients, known as the binomial coefficients, are defined by the formula given below: $$\dbinom{n}{r} = n! SECTION 1 Introduction to the Binomial Regression model. The Binomial Regression model can be used for predicting the odds of seeing an event, given a vector of regression variables. One can prove that for k = o(n exp3/4), (n "choose" k) ~ c(ne/k)^(k) for some appropriate constant c. Can you find the c? It also represents an entry in Pascal's triangle.These numbers are called binomial coefficients because they are coefficients in the binomial theorem. Binomial Random Variable Approximations, Conditional Probability Density Functions and Stirlings Formula Let X Name * Class * Email * (to get activation code) Password * Re-Password * City * Country * Mobile* (to get activation code) You are a: Student Parent Tutor Teacher Login with. It's called a binomial coefficient and mathematicians write it as n choose k equals n! Without expanding the binomial determine the coefficients of the remaining terms. Thus, for example, Stirling’s formula gives 85! Formula Bar; Maths Project; National & State Level Results; SMS to Friend; Call Now : +91-9872201234 | | | Blog; Register For Free Access. The Problem Write a function that takes two parameters n and k and returns the value of Binomial Coefficient C(n, k). 4. So, the given numbers are the outcome of calculating the coefficient formula for each term. Calculating Binomial Coefficients with Excel Submitted by AndyLitch on 18 November, 2012 - 12:00 Attached is a simple spreadsheet for calculating linear and binomial coefficients using Excel A binomial coefficient is a term used in math to describe the total number of combinations or options from a given set of integers. Section 4.1 Binomial Coeff Identities 3. The following formula is used to calculate a binomial coefficient of numbers. This formula is known as the binomial theorem. C(n,k)=n!/(k!(n−k)!) Unfortunately, due to the factorials in the formula, it can be very easy to run into computational difficulties with the binomial formula. Binomial Expansion. Finally, I want to show you a simple property of the binomial coefficient which we’re going to use in proving both formulas. Add Remove. Let’s apply the formula to this expression and simplify: Therefore: Now let’s do something else. An often used application of Stirling's approximation is an asymptotic formula for the binomial coefficient. This approximation can be used for large numbers. OR. We need to bound the binomial coefficients a lot of times. This calculator will compute the value of a binomial coefficient , given values of the first nonnegative integer n, and the second nonnegative integer k. Please enter the necessary parameter values, and then click 'Calculate'. 2 Chapter 4 Binomial Coef Þcients 4.1 BINOMIAL COEFF IDENTITIES T a b le 4.1.1. ≈ √(2π) × n (n+1/2) × e -n Where, n = Number of elements . The variables m and n do not have numerical coefficients. Use Stirlings’ formula (Theorem 1.7.5) to find an approximation to the binomial coefficient (n/n/2). OR. Below is a construction of the first 11 rows of Pascal's triangle. What is a binomial coefficient? (n-r)!r!$$ in which $$n!$$ (n factorial) is the product of the first n natural numbers $$1, 2, 3,…, n$$ (Note that 0 factorial equals 1). Compute the approximation with n = 500. This formula is so famous that it has a special name and a special symbol to write it. Compute the approximation with n = 500. Binomial Expansion Calculator. School University of Southern California; Course Title MATH 407; Type. divided by k! For example, your function should return 6 for n = 4 … So the problem has only little to do with binomial coefficients as such. Factorial Calculation Using Stirlings Formula. Binomial Coefficient Calculator. This is the number of ways to form a combination of k elements from a total of n. This coefficient involves the use of the factorial, and so C(n, k) = n!/[k! Where C(n,k) is the binomial coefficient ; n is an integer; k is another integer. Write a function that takes two parameters n and k and returns the value of Binomial Coefficient C(n, k). This preview shows page 1 - 4 out of 6 pages.). A property of the binomial coefficient. \sim \sqrt{2 \pi n} (\frac{n}{e})^n$$after rewriting as$$\lim_{n\to\infty} \frac{(4n)!(n! Uploaded By ProfLightningDugong9300; Pages 6. This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! In Algebra, binomial theorem defines the algebraic expansion of the term (x + y) n. It defines power in the form of ax b y c. The exponents b and c are non-negative distinct integers and b+c = n and the coefficient ‘a’ of each term is a positive integer and the value depends on ‘n’ and ‘b’. Let n be a large even integer Use Stirlings formula Let n be a large even integer. Binomial Coefficients. The first function in Excel related to the binomial distribution is COMBIN. Show transcribed image text. using the Stirling's formula. Show Instructions. Notice the following pattern: In general, the kth term of any binomial expansion can be expressed as follows: Example 2. Compute the approximation with n = 500. COMBIN Function . 19k 2 2 gold badges 16 16 silver badges 37 37 bronze badges. Okay, let's prove it. Binomial probabilities are calculated by using a very straightforward formula to find the binomial coefficient. ≈ Calculator ; Formula ; Calculate the factorial of numbers(n!) (n-k)!. Then our quantity is obvious. We can also change the in the denominator to , by approximating the binomial coefficient with Stirlings formula. Use Stirlings’ formula (Theorem 1.7.5) to find an approximation to the binomial coefficient (n/n/2). So here's the induction step. This question hasn't been answered yet Ask an expert. Introduction to probability and random variables. Michael Stoll Michael Stoll. Number of elements (n) = n! The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. By computing the sum of the first half of the binomial coefficients in a given row in two ways (first, using the obvious symmetry, and second, using a simple integration formula that converges to the integral of the Gaussian distribution), one gets the constant immediately. Proposition 1. Upper Bounds on Binomial Coefficients using Stirling’s Approximation. This is equivalent to saying that the elements in one row of Pascal's triangle always add up to two raised to an integer power. We’ll also learn how to interpret the fitted model’s regression coefficients, a necessary skill to learn, which in case of the Titanic data set produces astonishing results. Binomial coefficients and Pascal's triangle: A binomial coefficient is a numerical factor that multiply the successive terms in the expansion of the binomial (a + b) n, for integral n, written : So that, the general term, or the (k + 1) th term, in the expansion of (a + b) n, n! Notes. Application of Stirling's Formula. Limit involving binomial coefficients without Stirling's formula I have this question from a friend who is taking college admission exam, evaluate: $$\lim_{n\to\infty} \frac{\binom{4n}{2n}}{4^n\binom{2n}{n}}$$ The only way I could do this is by using Stirling's formula: n! Note: Fields marked with an asterisk (*) are mandatory. The usual binomial efficient by its q-analogue and the same formula will. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. Each notation is read aloud "n choose r.A binomial coefficient equals the number of combinations of r items that can be selected from a set of n items. Show Answer . Putting x = 1 in the expansion (1+x) n = n C 0 + n C 1 x + n C 2 x 2 +...+ n C x x n, we get, 2 n = n C 0 + n C 1 x + n C 2 +...+ n C n.. We kept x = 1, and got the desired result i.e. $\begingroup$ What happens if you use Stirlings Formula to estimate the factorials in the binomial coefficient? to about 1 part in a thousand, which means three digit accuaracy. Thus for example stirlings formula gives 85 to about. See also. The binomial has two properties that can help us to determine the coefficients of the remaining terms. View Notes - lect4a from ELECTRICAL 502 at University of Engineering & Technology. So if you eliminated as Q equal to one you will get exactly the same equality. Per Stirling formula, one can see that binom{2n ... You could use Stirlings formula for the factorials. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). The power of the binomial is 9. ]. Question: 1.2 For Any Non-negative Integers M And K With K Sm, We Define The Divided Binomial Coefficient Dm,k By Denk ("#") M+ 2k 2k + 1 Prove That (2m + 1) Is A Prime Number. 4.1 Binomial Coef Þ cient Identities 4.2 Binomial In ver sion Operation 4.3 Applications to Statistics 4.4 The Catalan Recurrence 1. The binomial coefficient C(n, k), read n choose k, counts the number of ways to form an unordered collection of k items chosen from a collection of n distinct items. Lutz Lehmann Lutz Lehmann. Stirling's Factorial Formula: n! In the above formula, the expression C( n, k) denotes the binomial coefficient. A special binomial coefficient is , as that equals powers of -1: Series involving binomial coefficients. Based on our findings and using the central limit theorem, we also give generalized Stirling formulae for central extended binomial coefficients. It's powerful because you can use it whenever you're selecting a small number of things from a larger number of choices. Sum of Binomial Coefficients . = Dm,d ENVO . For positive … Statistics portal; Logistic regression; Multinomial distribution; Negative binomial distribution; Binomial measure, an example of a multifractal measure. share | cite | improve this answer | follow | edited Feb 7 '12 at 11:59. answered Feb 6 '12 at 20:49. References ↑ Wadsworth, G. P. (1960). Remember the binomial coefficient formula: The first useful result I want to derive is for the expression . Let n be a large even integer Use Stirlings formula FAQ. My proof appeared in the American Math. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. 4 Chapter 4 Binomial Coef Þcients Combinatorial vs. Alg ebraic Pr oofs Symmetr y. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. Code to add this calci to your website . Example 1. = sqrt(2*pi*(n+theta)) * (n/e)^n where theta is between 0 and 1, with a strong tendency towards 0. share | improve this answer | follow | answered Sep 18 '16 at 13:30. Numbers written in any of the ways shown below. We are proving by induction or m + n If m + n = 1. The calculator will find the binomial expansion of the given expression, with steps shown. For e.g. (n – k)! USA: McGraw-Hill New York.
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2021-04-17 05:17:00
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http://www.oxfordmathcenter.com/drupal7/node/78
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# A First Look at Graphics with the ACM Libraries
The ACM libraries provide a simple way to write java programs with graphical outputs. ACM graphics programs can be thought of as functioning much like a felt board, where various shapes can be "stuck to the board". The ACM libraries come with many such shapes, each of which is represented by a class (whose name is shown in parentheses below):
• Rectangle (GRect)
• Oval (GOval)
• Line (GLine)
• Polygon (GPolygon)
• Arg (GArc)
• Rectangle with a 3D effect on its border (G3DRect)
Details on all of these objects, their constructors, methods, and more can be found in the ACM documentation.
Just as an example, the following program draws the following to the graphics window when run:
import acm.program.*;
import acm.graphics.*;
import java.awt.Color; //this is needed to use the constant "Color.RED" below
public class SampleProgram extends GraphicsProgram {
public void run() {
//construct a rectangle whose upper left corner is at (0,0)
//and with width 50 and height 100; reference it with a variable
//named myRect and add it to the graphics window
GRect myRect = new GRect(0,0,50,100);
//construct an oval whose upper left corner is at (25,25)
//and with width 50 and height 100; reference it with a variable
//named myOval; set things so that when drawn on the screen
//its interior will be filled in and the entire oval will be red in color;
//and then add it to the graphics window.
GOval myOval = new GOval(25,25,50,100);
myOval.setFilled(true);
myOval.setColor(Color.RED);
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2018-08-18 06:21:10
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https://www.clutchprep.com/chemistry/practice-problems/146071/what-potential-difference-is-required-to-cause-4-00-a-to-flow-through-a-resistan
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Resistors and Ohm's Law Video Lessons
Concept
# Problem: What potential difference is required to cause 4.00 A to flow through a resistance of 330 ohms?
###### FREE Expert Solution
In this problem, we are required to apply Ohm's law:
Ohm's law:
$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{i}}{\mathbf{R}}}$
81% (59 ratings)
###### Problem Details
What potential difference is required to cause 4.00 A to flow through a resistance of 330 ohms?
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2021-10-24 09:02:44
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https://www.mometrix.com/academy/factoring-polynomials/
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# Factoring Polynomials Overview
Factoring polynomials is an important skill to master because it allows us to rewrite polynomials in a simpler form. The process of factoring helps us understand more about the equations we are working with and produces useful information. For example, solutions are often much easier to identify when a polynomial is in a factored form. Factoring polynomials also makes it easier to graph an equation.
Factoring Polynomials Sample Questions
There are many different factoring techniques. The most common strategy for factoring polynomials is to simply factor out the greatest common factor. If there is no clear factor in common, then another approach needs to be implemented. Another common approach is to split the polynomial into two sets of parentheses that are multiplied by each other.
Let’s take a closer look at the first strategy: factoring out the greatest common factor.
Consider the polynomial $$6x^2+9x$$. We can simplify this polynomial by pulling out the greatest common factor. The largest factor that is shared by both terms is $$3x$$. Factor out the $$3x$$ by dividing each term by $$3x$$. $$6x^2$$ becomes $$3x$$ and $$9x$$ becomes 3. So 6×2+9x becomes 3x(x+3). Notice that this factored form will multiply back to its original form. In this case, $$3x$$ times $$(x+3)$$ gives us $$6x^2+9x$$, which proves that the factored form is equivalent to the original polynomial.
Let’s take a look at a tougher polynomial that has three terms.
Consider the following polynomial: $$4x^3-2x^2+6x$$. Look at each term and determine if there is a common factor shared by all terms. In this example, the greatest common factor is $$2x$$. Now “factor this out” by dividing each term by $$2x$$. $$4x^3-2x^2+6x$$ becomes $$2x(2x^2-x+3)$$. We can check our work to make sure that we have factored correctly by multiplying $$2x$$ by $$(2x-x+3)$$. If the product is the original polynomial, we have factored correctly. $$2x$$ multiplied by $$(2x^2-x+3)$$ equals $$4x^3-2x^2+6x$$, so we have factored correctly.
It is convenient when polynomials have a clear common factor. However, many times there will not be an obvious factor that is shared by all terms. For example, $$x^2+5x+6$$. There are no factors that are shared by all terms, so we need to approach this polynomial in a different way. This example can be factored using the approach of splitting the polynomial into two sets of parentheses that are multiplied by each other. Let’s use this approach with the polynomial $$x^2+5x+6$$.
$$x^2+5x+6$$ has no common factors, so set up two empty sets of parentheses that are multiplied by each other $$( )( )$$. Now, break apart the first term. $$x^2$$ can be split up into x times x, so place this in the parentheses $$(x\text{ }\text{ }) (x\text{ }\text{ })$$. Now take a close look at the next two terms $$5x+6$$. Ask yourself, “What multiplies to 6 and adds to 5?” In this case, 2 times 3 multiplies to 6, and 2 plus 3 adds to 5. Place 2 and 3 in the parentheses. $$(x\text{ }\text{ }) (x\text{ }\text{ })$$ becomes $$(x+2)(x+3)$$. This is the factored version of the original polynomial. Again, we can check that we have factored correctly by multiplying $$(x+2)$$ times $$(x+3)$$ to make sure that our product results in the original polynomial.
Let’s take a look at a polynomial that can be factored by pulling out the greatest common factor AND by splitting up a polynomial into two sets of parentheses that are multiplied.
For example, let’s look at the polynomial $$12x^2-27$$. When we look for a common factor, we notice that both terms are divisible by 3. 3 is the largest factor shared by both terms, so let’s factor this out. $$12x^2-27$$ becomes $$3(4x^2-9)$$. Many polynomials are successfully factored after just this single step. However, sometimes it takes multiple steps to completely factor a polynomial. $$3(4×62-9)$$ is close to complete, but consider what is inside the parentheses. $$(4x^2-9)$$ can be broken down even further. $$(4x^2-9)$$ can be split up into two sets of parentheses that are multiplied by each other. $$(x^2-9)$$ becomes $$(2x−3)(2x+3)$$. Now we are left with $$3x(2x−3)(2x+3)$$, which is factored completely.
Remember, when factoring a polynomial, you are simply rewriting it in a different way. The value has not changed at all, just the form. The objective of factoring is to break down a long and complicated polynomial into smaller pieces that are more manageable. This makes it easier to identify solutions, which inevitably makes the polynomial easier to graph.
## Factoring Polynomials Sample Questions
Here are a few sample questions going over factoring polynomials.
Question #1:
Factor the following polynomial: $$2x-4y+2$$
$$4(x-2y+2)$$
$$2(x-2y)$$
$$4(x-2y)$$
$$2(x-2y+1)$$
The correct answer is D. In order to factor the polynomial $$2x-4y+2$$, factor out the greatest common factor. In this case, the greatest common factor for all terms is 2. When 2 is factored out, we are left with $$(x-2y+1)$$. The product of 2 and $$(x-2y+1)$$ gives us the original polynomial, so $$2(x-2y+1)$$ is the factored form of $$2x-4y+2$$.
Question #2:
Factor the following polynomial: $$4x^3-8x^2$$
$$4x^2(x-2)$$
$$2x^2(x-2)$$
$$2x^2(x-4)$$
$$3x^3(x-4)$$
The correct answer is A. In order to factor the polynomial $$4x^3-8x^2$$, factor out the greatest common factor. In this case, the greatest common factor for all terms is $$4x^2$$. When this is factored out, we are left with $$(x-2)$$. The product of $$4x^2$$ and $$x-2$$ gives us the original polynomial, so $$4x^2(x-2)$$ is the factored form of $$4x^3-8x^2$$.
Question #3:
Factor the following polynomial: $$x^2-9$$
$$(x+3)(x-3)$$
$$(x-3)(x-3)$$
$$(x+3)(x+3)$$
$$x(x-9)$$
The correct answer is A. The polynomial $$x^2-9$$ does not have any clear common factors, so instead of factoring out the greatest common factor, it is necessary to try another approach. This polynomial can be split into two sets of parentheses that are multiplied by each other, like this:
( ) ( )
From here, look at the first term, which is $$x^2$$. This can be This can be split up into x times x, so we can place this in each parenthesis. Now we have $$(x )(x )$$. From here, think of two numbers that multiply to -9 but add to 0. $$3×-3=-9$$, and 3 and -3 add to 0, so place this in the parentheses. Now we have $$x+3x-3$$, which is the factored form of the polynomial $$x^2-9$$.
Question #4:
Factor the following polynomial: $$x^2+7x+10$$
$$(x+7)(x+10)$$
$$(x+5)(x+2)$$
$$x(x+5)$$
$$x(x+7)$$
The correct answer is B. The polynomial $$x^2+7x+10$$ does not have any clear common factors, so instead of factoring out the greatest common factor, it is necessary to try another approach. This polynomial can be split into two sets of parentheses that are multiplied by each other, like this:
( ) ( )
From here, look at the original polynomial $$x^2+7x+10$$ and focus on the first term, $$x^2$$. The term can be split by multiplying x times x, so place this inside each parenthesis: $$(x )(x )$$.
Now, focus on the remaining polynomial: $$7x+10$$. Think of two numbers that multiply to 10 but add to 7. In this case, those numbers would be 5 and 2: $$5×2=10$$ and $$5+2=7$$. Plug 5 and 2 into the parentheses.
$$x+5x+2$$
This can be checked by distributing each term to all other terms and simplifying.
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2023-03-22 09:25:13
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http://electronics.stackexchange.com/questions/16375/recommened-chips-for-datacommunication-over-a-low-voltage-ac-power-line
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# Recommened chips for datacommunication over a low-voltage AC power line?
A client wants to send data (think 100 ... 10k baud) over a combined power / cable-TV coax cable, distances up to 1 km. The lower end of the spectrum (say up to a few MHz) is free, but the cable carries the 50 Hz 48 .. 60 Volt power. What would you recommend for this? I was thinking power-line modem chips, but those seem to be out of fashion (and hence difficult to get), and require quite a few external components. A DIY approach could be to couple a modulated carrier into the line and detect it with a PLL chip or even a PIC with DFT. But I hope chips exist for exactly this purpose?
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How about higher frequency ranges? It would be helpful to know more accurate figures on what frequency ranges are in use. – Majenko Jul 5 '11 at 17:58 From my sketchy notes: 5..65 MHz occupied, 85..1000 MHz occupied. It is a cable TV system. – Wouter van Ooijen Jul 5 '11 at 21:00
Have you considered looking at the other end of the spectrum? Low-frequency communications, as you have rightly noticed, are not as common as they used to be.
There are many chips around that handle all the modulation, demodulation, etc at >1GHz - take the MRF24J40MC from Microchip (maybe not the cheapest solution) for example - 2.405Ghz to 2.475GHz. Or the ADF7241 from Analog Devices - again in the 2.4GHz range. There are many many more.
Yes, I know these are aimed at wireless communications, but then, what is Cable TV if not wireless communication with no aerial? The cable is basically a closed loop aerial between two points. A modulated high frequency signal being sent between two points through a medium - just using copper instead of air.
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Most cable TV hardware is specified to 1 or 2 GHz, so you'll have to make sure you don't have 1 GHz gear in the pipeline. The more you go past 2 GHz, the more attenuation you'll have. – Mike DeSimone Jul 6 '11 at 12:13
My question obviously has no single 'correct' answer, but this answer was the most usefull to me. – Wouter van Ooijen Aug 24 '11 at 18:01
Have you considered just connecting some v.90 modems? The 50 to 60 Hz is below the audio range those modems use, so can be filtered. The television VHF and UHF channels likewise are outside the modem frequency range. The modems are 600 ohm impedance and signal in the 300 Hz to 4000 Hz range.
Telephone wiring (600 ohm) to CATV wiring (75 ohm) will require a suitable coupling transformer as well as a power filter that blocks the power frequency.
Telephone ring detect and onhook/offhook features won't work, of course, you'll need to give the modems commands to force connection.
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Maxim has a good range:
http://www.maxim-ic.com/products/powerline/
The MAX2990/2991 combo looks good. There's an evaluation kit available too.
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Hmmm, the customer was selecting chips (for the other parts of the system) and mumbled something about availability. My immediate resonse was "stay away from maxim/dallas!". Quick www.findchips.com check: no stock at DigiKey (bad sign), low quantities at Mouser, total price > $20 @ 1. I hoped for something much cheaper (and smaller)! – Wouter van Ooijen Jul 5 '11 at 20:57 I think you'll have to do it manually. One approach is to use a zero-crossing detector and transmit a packet while the mains voltage is close enough to zero. Then you can use regular modulation chips and the only part you'd have to implement is the gating. – Optimal Cynic Jul 5 '11 at 21:05 Can use suggest some regular modem chips, circuits, app notes? Speed can be low, no isolation needed, price/size/availability is important. – Wouter van Ooijen Jul 5 '11 at 21:30 All the ones I'm familiar with seem to have vanished into the murky depths of RoHS. Sorry, you may have to implement this manually after all. One thing that I'd forgotten between first and second answer is that you're driving a thousand metre coax cable - that's going to be tough to do on the zero crossings. I wonder if you could do something clever with inductive coupling and simple baseband signalling? – Optimal Cynic Jul 5 '11 at 21:43 One more question - is this unidirectional or bidirectional? – Optimal Cynic Jul 5 '11 at 21:43 show 2 more comments Since you mentioned ASK in a comment, the NXP TDA5051 is an ASK modem designed for home power networks, and operates at either 600 or 1200 baud, which is in between your 100 .. 10K baud range. It is RoHS compliant, and is available at Digi-Key for under$3 in quantity. The interface circuitry required seems to be fairly straightforward.
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You've got a megahertz of bandwidth available and you only need 10 kbits/s of data? You could just use a series capacitor or RLC circuit (basically, a bandpass filter) to couple transmitted data onto the line and a 200 Hz high-pass filter/1 MHz low-pass filter combo to filter the received signal, then do a software modem with a 100 ksample/s or so ADC and DAC. Higher-end (96 or 192 ksample/s) audio codecs might do the job nicely. You could do it with a dedicated low-power 32-bit chip or similar, which might even have good enough ADC and DAC on board. If it has PWM outputs instead of a DAC, then just use them to drive a push-pull stage to get a class-D transmitter.
Assuming you have the people needed to write a modem, of course. But, again, the excess of bandwidth means it doesn't have to be a particularly good one, just one that can deal with the cable irregularities. ^_-
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Yeah, there is plenty of bandwith. But I hoped for an (or a few) cheap chips (like an old-fashioned FSM or ASK modem, but in one chip, few or no externals, cheap, and maybe the modst difficult: available), not a 32-bitter that must be programmed for modem duties! Maybe I hoped for too much. – Wouter van Ooijen Jul 6 '11 at 6:49 It really depends on what resources you have available. If you have no software developers, this won't work for you. But usually debugging software is easier than hardware since you don't have to get the board revised. FSK and ASK algorithms aren't that hard to implement. – Mike DeSimone Jul 6 '11 at 12:12
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2013-05-22 15:00:41
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http://johanneswalde.de/32-bit-checksum-calculator.html
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32 Bit Checksum Calculator.
If you do it the other way around, you end up with the wrong result. > > Just adding (or XORing) the code bytes to a 16/24/32-bit sum is simple > and fast, but not too reliable in case of multi-bit errors at the same > bit position. The new version is quite powerful. The XOR checksum for NMEA logs is also explained in this document. For IEEE802. The task of the Fletcher 16 algorithm is to transform any long input string to 16 bit namely 4 hexadecimal number. This tool support ISO and ECMA table. It offers a choice of 13 of the most popular hash and checksum algorithms for calculations. What is CRC-64? A cyclic redundancy check (CRC) is an error-detecting code used to detect data corruption. While Java has built-in cryptographic checksum classes, it's quite uneasy to use them for a simple task -- calculate MD5 hash from string and return 32-byte hexadecimal representation. , checksum codes). Recent Searches. 8-bit protocol. When a message of any length less than 2^64 bits is input, for example in our SHA-1 generator, the algorithm produces a 160-bit message digest as output. Read more on the theory behind parallel CRC generation. Please enable it to continue. vb) to calculate the CRC32 checksum for files. With this utility, you can generate as many SHA256 hashes you need in any format, base, and case. Comes in 32-bit and 64-bit versions. The shell extensions are implemented as a 32-bit (x86) and a 64-bit (x64) version and support the appropriate Windows version. My old PC (2. A single multiply unit (IIRC) took somewhere in the neighborhood of 15k gates. // This allows to speed up calculation of CRC32 tables by first // calculating CRC32 for bytes with only one bit set, // and then xoring all CRC32 of lowest bit and CRC32 of remaining bits // to get CRC32 of whole number. A checksum is a sequence of numbers and letters used to check data for errors. Allows simple calculation of CRC checksums. Initial value is 0. Supported Hash Algorithms. ly link doesn't work, but the resolved link does I believe we should support at least one hop if not two " pbearne Noteworthy 42733 WordPress Embed URLs don't work for draft and scheduled posts with pretty permalinks swissspidy Embeds normal normal 1 Future Release defect (bug. This algorithm uses modulo 256(2^8) where data is divided by 256 for 8bit 65536 for 16 bit and reminder is kept. A better test whould be a Crc ( available as 8,16,32 bits) Even better are hashes. Red Hat ES 6 i386(32-bit) / CentOS 6 / Oracle Linux 6. This code is a translation from Ruby, with an adjustment to use 32-bit integers. Crc16 8408 [RESOLVED] crc 16 ccitt 8408 - CodeGuru Forum [RESOLVED] crc 16 ccitt 8408 i need a function to calculate the CRC16 from a string, not a byte array this is the function from the documentation of the device, but its in g. update (bytes, 0, bytes. md5 LM NTLM sha1 sha256 sha384 sha512 md5(md5()) MySQL4. prm file generated together with the. 1 is available to all software users as a free download for Windows. 8-bit Checksum is also called the 2's compliment of addition of all bytes. get_pseudo_header() #Fetch the ICMP data icmp_header = self. The main advantage of CRC32 over MD5 or other hash algorithm is its very high speed. AnalysIR Video Tutorial Checksum Calculator. What is Cyclic Redundancy Check and CRC-32? A cyclic redundancy check (CRC) is an error-detecting code used to detect data corruption. This is a fundamental problem with hash functions, but with only 32 bits available this is especially risky. The checksum algorithm used is the 8-Bit Fletcher Algorithm, which is used in the TCP standard (RFC 1145 1). Its Sum method will lay the value out in big-endian byte order. Now add each data element to the checksum. The first entry is the CRC which I calculate and add on compile time via srec_cat. Detailed Description. This was designed by the National Security Agency (NSA) to be part of the Digital Signature Algorithm. docx from CS NETWORKS at Vishwakarma Institute of Technology. Return to menu: More Scripts: CRC16 CRC32 and CRC64. The input message is broken up into chunks of 512-bit blocks (sixteen 32-bit words); the message is padded so that its length is divisible by 512. Istilah tertinggi (x 32 ) biasanya tidak ditulis secara eksplisit, sehingga dapat direpresentasikan. The checksum that's calculated by the software is then stored within the binary at a given address. The X^0 term (usually shown as "+1") results in * the MSB being 1 * * Note that the usual hardware shift register implementation, which * is what we're using (we're merely optimizing it by doing eight-bit * chunks at a time) shifts bits into the lowest-order term. MD5 processes a variable-length message into a fixed-length output of 128 bits. Here's the checksum code: u16 ip_sum_calc(u16 len_ip_header, u16 buff){ u16 word16; u32 sum=0; u16 i; // make 16 bit words out of every two adjacent 8 bit words in the packet. MD5 function generates different values of output for different input. Including both 32-bit and 64-bit releases. AWOL Guest; Re: XOR checksum in arduino #1 Feb 25, … At RX these two hex characters and the first comma need to be extracted (ie as in. This is what is referred to as a hash or checksum, and if you are familiar with the MD5 algorithm, the principle is the same. Another common trap is the bit order within the bytes of a signal. Let assume that We have 4 bytes of IPv4 header data (32-bits). sha1 configures HMAC-SHA1 algorithm. Free secure web-tools. If no files are specified, the cksum command reads standard input. ORG site: From the file: //***FILE 900 is a set of programs which run under either old MVS //* or z/OS to calculate MD5 checksums. iso” image file that we used before. def ichecksum ( data, sum=0): """ Compute the Internet Checksum of the supplied data. MD5 was designed by Professor Ronald L. Many thanks, KaFu for your assistance, but returned value from _StringCrc8 () func isn't correct 8-bit CRC. HEX file checksum online calculator. 5 for windows xp 32 bit. Initial value is 0. The calculation of the IP checksum must not include the field that contains that checksum - zero should be used for that field instead. For example, the polynomial 0x247 is a 10-bit CRC that provides HD=4 (or better) up to 501 bit dataword length (501+10=511 bit codeword length). About Adler 32 create hash online tool. This requires your software to do this process though. Free 16 bit checksum calculator online download software at UpdateStar. Oct 19, 2019 - Compute CRC-32, the 32-bit checksum of data, starting with an initial crc. Value — Array of four 32-bit integers that represents the model's 128-bit checksum. Calculate the checksum for the entire HDU by adding (using 1’s complement arithmetic) the. Detailed Description. Calculate the CRC32 checksum of a file or text online. The crc32b is the 32-bit Frame Check Sequence of ITU V. Checksum (32 Bit) 4: Simple count where all the bytes are added in an 32 bit accumulator. // CS is a 32-bit unsigned integer that well keep track of // the running sum, then be converted to the checksum unsigned long CS = 0; // psData is a pointer to the start of the series of characters // in the command action, // but it is treated by the C++ compiler as a pointer to a // series of unsigned 16-bit integers. Important Note There is currently a sequence input limit of 4000 sequences and 4MB of data. it always says your TCP checksum should be '0x0f38' but the checksum that my code generates always differ from the last one. Java provides a couple of ways to generate the MD5 checksum for any file, you can either use java. 0A63 0000101001100011 97FC 1001011111111100 <--This is the. What is Cyclic Redundancy Check and CRC-32? A cyclic redundancy check (CRC) is an error-detecting code used to detect data corruption. Now you can explore your whole system in an instant. 3) Next use ImgBurn or Gear ISO to create a bootable DVD. 1,10 you can get installation file here. The effect is the same as the problem of ignored dashes. The case of the input value affects the return value. Click the 'Convert' button to get the one's complement binary number, or enter the one's complement binary number and click the 'Convert' button to get the decimal number. This is a fundamental problem with hash functions, but with only 32 bits available this is especially risky. I compare the values with hex editor, which calculates different CRC-8. I implemented Fletcher's checksum as an exercise from Programming Praxis. Source Address. About crc32 checksum function: From Wikipedia, the free encyclopedia. Click to download: Example project v6. To verify the checksum add all the data bytes including the checksum. The MD5 checksum or MD5 hash is a more secure alternative to the checksums obtained from the "sum" or "cksum" commands. xxhash: xxHash in 32-bit. -datatype 32 bit //create //convert into integer so we calculate. In JavaScript:. Unsigned 32-bit integer to binary string function John the ripper password cracked or not? Why did the space shuttle's altitude go down after reaching 108,000m?. Để tính toán mã CRC-32 cho một byte data, chúng ta cần thực hiện các bước sau: Đảo ngược vị trí các bit trong data byte, ví dụ: từ b10100011 đảo thành b11000101; Chèn thêm 32 bits 0 vào bên phải của data byte; Thực hiện phép gán: Remainder = data XOR 0xFFFFFFFF00;. Loop over those 8 bits. By hubertmettwurst in forum Excel Programming / VBA / Macros Replies: 1. SHA256 algorithm generates an almost-unique, fixed size 256-bit (32-byte) hash. ly link doesn't work, but the resolved link does I believe we should support at least one hop if not two " pbearne Noteworthy 42733 WordPress Embed URLs don't work for draft and scheduled posts with pretty permalinks swissspidy Embeds normal normal 1 Future Release defect (bug. If a segment contains an odd number of header and text octets to be checksummed, the last octet is padded on the right with zeros to form a 16 bit word for checksum purposes. Application Note: CRC Checksum www. Cyclic Redundancy Checksum (CRC) calculations. Hex File Crc 16 Calculator Checksum. At the end, cast this 32-bit value to a 16-bit value, and call it the checksum. Step-by-Step Breakdown. If the result is zero,. ca> Date: Thu, 19 Oct 95 09:31:31 EDT From: Alain Desilets To: [email protected] It outputs a 32-byte MD5 hex string that is computed from the given input. 8-bit Checksum is also called the 2's compliment of addition of all bytes. 7 Multiple Checksums and CRCS 32 6. tr uzantılı e-posta adresinizin yalnızca @ işaretinden önceki kısmını, parola alanına e-posta adresinizin parolasını girmeniz gerekmektedir. It accepts ASCII or Hex to produce a checksum. , individual records in a database. By default, all checksums will be displayed as a 64-bit number in hex notation where the first 32 bits and the last 32 bits are separated by a space (for example, '00000000 00007F80'). For CRC32 checksum calculations, the CRC32 class has to be used. The returned Hash32 also implements encoding. MarkedUnique — True if any blocks in the model have a property that prevents code reuse. You can convert to other bases (such as base-3, base-4, octal and more) using Base Conversion. com I need help with writing a code in C++ to do a 16 bit checksum. Or verified. Click the tab labelled “File Hashes” at the top of the window to see the MD5, SHA-1 and CRC32 hashes for the file you selected. In the context menu, click on Properties > File Hashes. So far, I’ve considered Adler-32, Fletcher-32, though I’m unsure how they would accurately detect the kind of bogus output I could get, and also “md5 of md5” or “sha-1 of sha-1” (i. MD5 & SHA Checksum Utility is a tool that allows you to generate CRC32, MD5, SHA-1, SHA-256, SHA-384 and SHA-512 hashes of single or multiple files. Select the type of Checksum you are calculating. The XOR checksum for NMEA logs is also explained in this document. A cryptographic hash (sometimes called 'digest') is a kind of 'signature' for a text or a data file. Bit order: MSB first [ ]Reverse byte order within word [unchecked] Initial Value: 0x0 [X]Use as input [checked] Checksum unit size: 32-bit Example for K60. A CRC is a "digital signature" representing data. The Fields of the Pseudo IP header are:-. The algorithm to calculate a 'reverse CRC' described here is based on the 32-bit polynomial, CRC-32-IEEE, most commonly used by standards bodies, but can easily be adapted to other CRC types. The IPv4 UDP pseudo-header is composed of the following items: 32-bit source IP address (IP header). Parameters. 8-bit Checksum is also called the 2's compliment of addition of all bytes. 1+ ripemd160 whirlpool adler32 crc32 crc32b fnv1a32 fnv1a64 fnv132 fnv164 gost gost-crypto haval128,3 haval128,4 haval128,5 haval160,3 haval160,4 haval160,5 haval192,3 haval192,4 haval192,5 haval224,3 haval224,4 haval224,5 haval256,3 haval256,4 haval256,5 joaat md2 md4 ripemd128 ripemd256 ripemd320 sha224 snefru. The reason that only 14-bits is used is because the PIC16F87X has 14-bit wide program memory. Please enable it to continue. This 8-bit Checksum Calculator can be used to calculate the 8-bit Checksum of a sequence of hexadecimal values or bytes. Read more on the theory behind parallel CRC generation. View Lab Assignment CN-CheckSum. It accepts ASCII or Hex to produce a checksum. Let's use eight-bit. This was designed by the National Security Agency (NSA) to be part of the Digital Signature Algorithm. Re: Checksum calculation of hex file Tuesday, May 31, 2016 5:45 AM ( permalink ) 4 (2) A simple checksum is the sum of all the data bytes. train network uses an 8-bit CRC for each 64-bit packet of data transmitted at the link layer. This example shows source code, the linker configuration file and the corresponding settings in the linker options dialog. In practice, checksum values are mainly used in three situations. The purpose of this algorithm is to calculate a cyclic redundancy check (CRC) checksum. Add up all the numbers (including the checksum value) and you get 0x00 in the last 8 bits of the accumulator. I followed the suggestions posted here but there still seems to be a slight difference. The effect of a checksum algorithm that yields an n-bit checksum is to map each m-bit message to a corner of a larger hypercube, with dimension m + n. Note: The cksum command is POSIX 1003. Checksum (32 Bit) 4: Simple count where all the bytes are added in an 32 bit accumulator. To calculate the CRC32 value from a Binary, I use the following command line: srec_cat tinyK22_KBOOT_led_demo. To calculate the CRC32 value from a Binary, I use the following command line: 2. I compare the values with hex editor, which calculates different CRC-8. ' This program computes the checksum of a file using various algorithms. MD5 Checksum Calculator was designed to be a small freeware utility for calculating the MD5 (Message Digest number 5) checksum values of specified files. Download Link of Checksum Aide 32 Bit 1. This is usually used to validate the integrity of data being transmitted. The output of the system as described produces two 32-bit values that can be combined to reach a 32-bit checksum, or can be combined to produce a 64-bit checksum. Do(ieeeInit) 154 } 155 return &digest{0, tab} 156 } 157 158 // NewIEEE creates a new hash. CRC-32 uses a 33-bit polynom, however again the most signficant bit is always '1' and can be discarded. X 32 + X 26 + X 23 + X 22 + X 16 + X 12 + X 11 + X 10 +X 8 + X 7 + X 5 + X 4 + X 2 + X +1. If you know the checksum of an original file, you can use a checksum utility to confirm your copy is identical. See the BIT_XOR result for the MD5 given above and the function is able to identify the distinct values in the tables and resulted with the different crc values. Thus, we know we're dealing with a simple algorithm. Please help! Thanks NP Have a nice day Bismark. A standalone utility that generates and verifies cryptographic hashes in MD5, SHA-1 and SHA-256. 4E19 -> 0100111000011001 B1E6 ->1011000111100110 // CHECKSUM. u16 buff [] is an array containing all octets in the header with octets 11 and 12 equal to zero. signed 32-bit : twos-complement signed 32-bit (32 bits) Other Bases. The whole content of pseudo header is about 12 bytes(32 bit source address + 32 bit destination address + 8 bit reserved + 16 bit tcp length + 8 bit protocol type = 96 bits = 12 bytes). When the data word is divided into 32-bit blocks, two 32-bit sums result and are combined into a 64-bit Fletcher checksum. We’ll use the same “ubuntu-mate-16. Need help for an unknown 7-bit checksum. My project last semester was a 32-bit 2’s complement kogge-stone [4×4]*[4×4] matrix multiplier. Internet-Draft IPv6 over OMNI Interfaces June 2021 publication of this document. I found that it doesn't matter what number I choose for IP Identification field, sth changes this number and. I have two examples. 1) Simply calculate a 32-bit value which is the sum of each byte added to a running sum. I use srecord to calculate the checksum from 0x0c020000 to 0x0c0fffe0. 16-Bit Math; LCD Programming; 8051 Instruction. 10-desktop-amd64. The checksum algorithm used is the 8-Bit Fletcher Algorithm, which is used in the TCP standard (RFC 1145 1). The checksum is repeated until the end of the sector. Windows (32-bit) Windows (64-bit) Mac OS X (10. CRC32b CRC32b is an implementation of the consistency algorithm. This feature generally increases the cost of computing the checksum. Minding the endianness, and also how the 32-bit sum is converted to 16. To build a 256-entry (-byte) lookup table, take the 8-bit index and bit-reflect all the bits in that byte. The method compute_ip_checksum initialize the checksum field of IP header to zeros. 3 Balance Bit Encoding 31 5. If the result is zero,. 2CHECKSUM CALCULATIONThe ROM addresses 0004H~0007H and last address are reserved area. Download 8 Screenshot(s) 9 user reviews No Video. I implemented Fletcher's checksum as an exercise from Programming Praxis. CRC is an acronym for Cyclic Redundancy Checksum or Cyclic Redundancy Check (depending on who you ask). If things changed unexpectedly then something is corrupt. The checksums can be CRC32, CRC64-ISO, CRC64-ECMA-182,CRC32-CRC64, MD5, SHA1, SHA2 224-bit, SHA2 256-bit, SHA2 384-bit and SHA2 512-bit. vb) to calculate the CRC32 checksum for files. IgorWare Hasher is described as 'free SHA-1, MD5 and CRC32 hash-generator for Windows and is available in 32- and 64-bit versions'. So a TCP checksum is not the sum of one's complement of all the 16-bit words. Usually, the second sum will be multiplied by 2 32 and added to the simple checksum, effectively stacking the sums side-by-side in a 64-bit word with the simple checksum at the least significant end. This example shows source code, the linker configuration file and the corresponding settings in the linker options dialog. set_checksum(0) #Fetch the pseudo header from the IP6 parent packet pseudo_header = self. CRC32 is a 32-bit fingerprint not intended for password storage as it's easily crackable, but for quick data integrity check. Input type: ASCII Hex Output type: HEX DEC OCT BIN Show processed data (HEX) Calc CRC-8. 4 Data Scramblers 31 5. I'm trying to calculate the checksum of an ICMPv6 packet in C#. Allows simple calculation of CRC checksums. Ignore carries from the addition. They are much safer than simple adding or xoring, but they take longer time to calculate. The reason that only 14-bits is used is because the PIC16F87X has 14-bit wide program memory. At the bit level, there are four possibilities, 0 ⊕ 0 = 0 0 ⊕ 1 = 1 1 ⊕ 0 = 1 1 ⊕ 1 = 0 Non-binary inputs are converted into their binary equivalents using gmp_init. If the checksum were something cryptographic, a single bit change would entirely change the output (so you'd see half the bits flipped on average). u16 buff [] is an array containing all octets in the header with octets 11 and 12 equal to zero. 5" drives will fit neatly inside your Amiga 600 or 1200. By default, all checksums will be displayed as a 64-bit number in hex notation where the first 32 bits and the last 32 bits are separated by a space (for example, '00000000 00007F80'). If you have data in a SQL Server table and you want to know if any of the values in a row have changed, the best way to do that is by using the rowversion data type. Time Elapsed 85. SB-SHA1 Checksum Calculator is a tool to calculate and compare the SHA1 checksum of files. HEX file checksum online calculator. You can find the 32-bit checksum from the. The first thing to notice is that changing one bit in the input causes a relatively small change in the output. When receiving data, checksum is generated again and compared with sent checksum. If, given these instructions, you cannot calculate a UDP checksum, then I don't know what to do. MD6 hash generation linux command for 128-bit, 256-bit, 512-bit MD6 hashes. MD5 processes a variable-length message into a fixed-length output of 128 bits. My checksum is an 8 bit ones, see below how should be release my checksum, do you know how is the easiest way to write a code which makes an ''add with carry'' operation? The checksum is the inverted modulo-256 sum. The Checksum Calculator can also batch process multiple files and is an easy to understand and use Windows program. checksum in hex file/editing hex file manually. , checksum codes). However, because it's a compression function (the packet is more than 32-bits), it's possible to. Checksum is IP one's complement standard (RFCs 1141 and 1624). Using this algorithm you are able to calculate a hash value or digest of any message. In Cryptography, MD5 (Message-Digest algorithm 5) is a widely-used cryptographic hash function with a 128-bit hash value. About Adler 32 create hash online tool. This download is licensed as freeware for the Windows (32-bit and 64-bit) operating system on a laptop or desktop PC from data encryption software without restrictions. Checksum checksum checksum. We have tested CHECKSUM 1. X is one string. Hi, I need advice on how to calculate checksum in a line of intel hex file format. GIẢI THUẬT TÍNH CRC-32. How to calculate the CheckSum of a file (Linux) Change the directory to the location of the file. I have two examples. Re: Checksum calculation of hex file Tuesday, May 31, 2016 5:45 AM ( permalink ) 4 (2) A simple checksum is the sum of all the data bytes. Destination Address. This function comes in two versions. ADS126x 32-Bit, Precision, 38-kSPS, Analog-to-Digital Converter (ADC) with Programmable Gain Amplifier (PGA) and Voltage Reference datasheet (Rev. Hex File Crc 16 Calculator Checksum. 0-bit 16-bit 1000000011111110. Use this calculator to easily calculate the CRC-32, CRC-16 or CRC-8 hash of a given string. If you want to calculate the checksum for your code,then you can use the Hexmate utility for the same. --ChecksumRandomHashFunctionSize: INTSPEC: Size of the randomly generated hash function. 8-bit Checksum Calculator This 8-bit Checksum Calculator can be used to calculate the 8-bit Checksum of a sequence of hexadecimal values or bytes. However, because it's a compression function (the packet is more than 32-bits), it's possible to. unable to correctly calculate checksum for one SNES game. 8 GHz Celeron) died after ten minutes The second PC (Core Duo 1. Generate the MD5 and SHA1 checksum for any file or string in your browser without uploading it, quickly and efficiently, no software installation required. So far, I’ve considered Adler-32, Fletcher-32, though I’m unsure how they would accurately detect the kind of bogus output I could get, and also “md5 of md5” or “sha-1 of sha-1” (i. Generic hashes. Initial value is 0. srec_cat tinyK22_KBOOT_led_demo. 2 32-Bit CRC Checksum. Generally there are two cases to calculate the checksum of each Microchip. The Intel 32-bit Hexadecimal Object file record format has a 9-character, 4 field, prefix that defines the start of record, byte count, load address and record type. The theor y of checksum calculation accounts for lots of use cases and can get quite complex for. 7 Multiple Checksums and CRCS 32 6. Let's use eight-bit. There is also an option to verify a file using a hash. File Checksum Utility supports Windows 10, 8. There are more than 25 alternatives to IgorWare Hasher for a variety of platforms, including Windows, Linux, BSD, Mac and Windows Explorer. (1, 4) Optional. With this utility, you can generate as many SHA256 hashes you need in any format, base, and case. Now if you compare this checksum with the one obtained in the packet you will find that both are exactly same and hence the IP header’s integrity was not lost. Right-click on the file you want to run a checksum against and choose “Properties” from the context menu. Normal representation for a polynomial is xAF. Click the tab labelled “File Hashes” at the top of the window to see the MD5, SHA-1 and CRC32 hashes for the file you selected. The XOR checksum for NMEA logs is also explained in this document. Hex File Crc 16 Calculator Checksum. The hash is a 32-character hexadecimal number. Just double-click on the. SHA-512 generates an almost-unique 512-bit (32-byte) signature for a text. Here is the implementation: It can also be viewed here. 1 Cool Button Tool - Flash v. Press Enter to execute the command. calculate checksum free download. Specify the unique name of the Add a Checksum step on the canvas. Để tính toán mã CRC-32 cho một byte data, chúng ta cần thực hiện các bước sau: Đảo ngược vị trí các bit trong data byte, ví dụ: từ b10100011 đảo thành b11000101; Chèn thêm 32 bits 0 vào bên phải của data byte; Thực hiện phép gán: Remainder = data XOR 0xFFFFFFFF00;. It supports MD5, SHA1, CRC32 checksum, and can batch check multiple files. If both bits are 0 and carry is 0, sum=0 and carry=0. So far, I’ve considered Adler-32, Fletcher-32, though I’m unsure how they would accurately detect the kind of bogus output I could get, and also “md5 of md5” or “sha-1 of sha-1” (i. This 8-bit Checksum Calculator can be used to calculate the 8-bit Checksum of a sequence of hexadecimal values or bytes. Usually 32-bit version works Usually 32-bit version works 189 /// faster on 32-bit machines and vice versa. Calculate folder sizes of multiple selected folders from windows explore context menu. If Length is 0, then 0 is returned. Also you can download the portable 32-bit version of the Coil64. There are two inputs to the checksum: (1) every single byte of the data page, with zeros in the four bytes where the checksum will be stored later and (2) the page offset/address. Now, let's focus on theory again. Create a CRC-32B checksum online from your password. CRC-16 and CRC-32 are very popular checksum methods, used in many protocols and in data storage. Download Checksum-Aide (32-bit) for Windows to get the hash code for a block of text; calculate the hash code for single or multiple files. SB-CRC32 Checksum calculator is absolutely free, but if you like to donate some change, this will help in continued development of this program and new future ones. Checksum (64 bit) 8. Then, two 8-bit checksums are computed and are appended to form a 16-bit Fletcher checksum. This checksum is a 4-byte (32-bit) value stored at 0x40 Note: save files can be when using Dolphin emulator are located in “C:\Users\{USER}\Documents\Dolphin Emulator\GC\USA\Card A\” The value can be calculated by taking all of the data after the checksum position, i. The whole content of pseudo header is about 12 bytes(32 bit source address + 32 bit destination address + 8 bit reserved + 16 bit tcp length + 8 bit protocol type = 96 bits = 12 bytes). I'm trying to calculate the checksum of an ICMPv6 packet in C#. The 8-bit checksum is the 2's complement of the sum off all bytes. But the MVB logical frame format (which can be as long as 256 bytes) uses a 32-bit CRC, called a "safety code" [16] to provide HD=6 protection for critical messages. ASP/Verify v. CRC depends on the polynomial and the width. It can also be used with hex files generated for 8-bit, 16-bit or 32-bit devices. It offers a choice of 13 of the most popular hash and checksum algorithms for calculations. I tested this function on the file with size 400kB. int32 = A 32-bit value. The 8-bit Fletcher Checksum Algorithm is calculated over a sequence of data octets (call them D[1] through D[N]) by maintaining 2 unsigned 1's-complement 8-bit accumulators A and B whose contents are initially zero. My old PC (2. SHA256 algorithm generates an almost-unique, fixed size 256-bit (32-byte) hash. Place the return value in the checksum field of a. This includes computed columns. If, given these instructions, you cannot calculate a UDP checksum, then I don't know what to do. Subject: Re: [EXTERNAL] Re: Checksum or hashing I didn't see the original posting but if you're looking for tools to do a checksum on a z/OS data set check out file 900 on the CBTTAPE. Click to download: Example project v6. 2 Bit Stuff Vulnerabilities 30 5. MD5 Checksum Overview MD5 Checksum is a tool which can be used for assisting the individuals in calculating MD5 and SHA hash for any of the file or string. 5" drives will fit neatly inside your Amiga 600 or 1200. train network uses an 8-bit CRC for each 64-bit packet of data transmitted at the link layer. 1 is available to all software users as a free download for Windows. Checksum Calculator for Windows (freeware) Bitser's checksum calculator tool can display MD5, SHA-1 or SHA-256 checksums directly from the Windows explorer context menu. The bottom number in each cell is a "good" polynomial that gives at least that HD up to the indicated dataword length in implicit +1 notation. This approach typically avoids a carry-sensing instruction but requires twice as many additions as would adding 32-bit segments; which is faster depends upon the detailed hardware architecture. It can compute message digests, checksums and HMACs for files, as well as for text and hex strings. 188 /// Some hash methods have 32/64 bit variants. The calculation is made 16 bits at a time (e. the checksum of string of all entered bytes and the checksum will be equal to zero. Hence can add 15 bits to each block of 32 k bits and can detect any 1-bit or 2. qty_checksum_agg ----- 32 (1 row affected) As you can see clearly from the output, the result of the CHECKSUM_AGG() changed. The maximum length (not shown in figure) is 60 bytes (15 32-bit words). You can use it to get the hash code for a block of text or you can calculate the hash code for a file on your computer. 8-bit Checksum is also called the 2's compliment of addition of all bytes. 16 bit checksum calculator online. The order of columns used for CHECKSUM(*) is the order of columns specified in the table or view definition. Enter Hexadecimal value separated by space e. Let's calculate and verify the Internet checksum. binhex — Encode and decode binhex4 files. Checksum Control uses a simple wizard interface to walk through the process of creating or verifying sets of files. Let's use eight-bit. This checksum is a 32-bit hexadecimal number such as 0x0123ABCD. To verify the integrity of the file, a user calculates the checksum using a checksum calculator program and then compares the two to make sure they match. The program has a batch mode which can calculate the checksum files for everything you give it. Binary converter. The sum is stated by means of 1’s compliment and stored or transferred as a code extension of actual code word. The theor y of checksum calculation accounts for lots of use cases and can get quite complex for. When sending data, short checksum is generated based on data content and sent along with data. Checksum Control uses a simple wizard interface to walk through the process of creating or verifying sets of files. This is how I'm calculating the checksum: //Note: Initial checksum in payload is set to 0x00, 0x00 //Add ip source bytes (16 bytes) checksum += Uti. 0 unencrypted_key configures either a 192 bit 3DES key or 128/192/256 bit AES key in an unencrypted format. 5+) Linux (32-bit) We can manually invoke the checksum calculation using the calculate method of checksum objects. 2 Generator model With the generator model, the receiver copies the structure of the CRC generator in hard- or software. O It has proposed two algorithms: 8-bit and 16- bit. 583 seconds). I use srecord to calculate the checksum from 0x0c020000 to 0x0c0fffe0. CRC32 Checksum Calculator - Compute a CRC32 checksum of string. Which can be preformed using the pic32 cryptographic peripheral. The MD5 hashing algorithm is a one-way cryptographic function that accepts a message of any length as input and returns as output a fixed-length digest value to be used for authenticating the original message. 0 - DreamCalc is a fully featured Graphing Calculator for Windows. I have implemented a function to produce a 32-bit checksum of a file using the following method: checksum = word_1 + word_2 + + word_n, where word_i is the 32-bit words the file consists of. In simple terms, it means that we are in Transport Layer and the IP data. 0 Checksum-Aide is a utility used to generate hash codes (or checksum codes). This 8-bit Checksum Calculator can be used to calculate the 8-bit Checksum of a sequence of hexadecimal values or bytes. Free CRC routines downloadable. The polynomial is written in. We treat the whole list as a list of coefficients for a big polynomial which we shall call p(x). Depending on the size of the file it may take a few seconds to run the calculation but if successful the MD5 hash will be displayed as below. Calculate XOR III. GetHashCode() and wanted to get an int hashcode for my own objects--and this seemed like a decent way to do it. 26 UBX Checksum The checksum is calculated over the packet, starting and including the CLASS field, up until, but excluding, the Checksum Field. Type Class Header. crc32("hello") crc 18. In both cases, the checksum value is calculated using the same algorithm. The 32-bit and 64-bit versions are both included and the right one will automatically be used for maximum performance. Checksum Calculator is a free file checksum calculation utility, it can support the most commonly used file checksum algorithm, such as. • Header Length (in 32 bit words) – Indicates end of header and beginning of payload – If no options, Header length = 5 Total Length in bytes (16) Time to Live (8) Options (if any) Bit 0 Bit 31 Version (4) Hdr Len (4) TOS (8) Identification (16 bits) Flags (3) Fragment Offset (13) Source IP Address Destination IP Address Protocol (8. Checksum checksum checksum. com I need help with writing a code in C++ to do a 16 bit checksum. sha224sum, sha384sum, etc. In telecomunicazioni e informatica il checksum (lett. The Intel 32-bit Hexadecimal Object file record format has a 9-character, 4 field, prefix that defines the start of record, byte count, load address and record type. Checksum (32 Bit) 4: Simple count where all the bytes are added in an 32 bit accumulator. The corresponding polynomial is: 0x247=x^10 +x^7 +x^3 +x^2. It seems to be working as intended so far, except when I try to verify the Checksum on two different Final Fantasy III roms. Is this "checksum" method something you have to do to match hardware or something ? Simply adding numbers is a very poor way to create a checksum. The order of columns used for CHECKSUM(*) is the order of columns specified in the table or view definition. The character can be entered in either upper case or lower case. Auto Update. Checksum Aide 32 Bit 1. Data as well as checksum are added together. Note: The cksum command is POSIX 1003. Description The cksum command reads the files specified by the File parameter and calculates a 32-bit checksum Cyclic Redundancy Check (CRC) and the byte count for each file. MD5 & SHA1 Hash Generator For File Generate and verify the MD5/SHA1 checksum of a file without uploading it. Hex formatter: ASCII text, c-source code, memory dump, big number to hex number. This tool support 64 / 65 bit only, if you want 32 / 33 bit, you can go to Cyclic Redundancy Check 32 bit (CRC-32). Adler-32 is a simple checksum algorithm, this algorithm is faster than the CRC algorithm, but the security is not as good as CRC; Adler32 algorithm makes a trade-off between security and speed. Both algorithm 1 and 2 write to the standard output the same fields as the default algorithm except that the. It is noteworthy that the algorithm - an Additive Checksum - is a redundancy check; however it is not a CRC. Click-and-Go. Generally there are two cases to calculate the checksum of each Microchip. 8 bit contains 16 bit checksum. If the message on the serial port is X&s=abe4\n, then the Fletcher 16 algorithm is applied to the X. I compare the values with hex editor, which calculates different CRC-8. Specify the type of checksum you want to calculate. View Lab Assignment CN-CheckSum. To do this, I plan to calculate the crc in advance and save it in the last couple of bytes in the hex file before loading the code. Hexmate is packaged with MPLAB X IDE, under ". Direct extraction of data to destination drives without using temporary system folders. An easy-to-use hash calculator. SHA-512 Cryptographic Hash Algorithm. The basic idea is that the UDP checksum is a the complement of a 16-bit one's complement sum calculated over an IP "pseudo-header" and the actual UDP data. It outputs a 32-byte MD5 hex string that is computed from the given input. Rivest in 1991 to replace an earlier hash function, MD4. Certain error-correcting codes based on checksums are even capable of recovering the original data. exFAT Boot Checksum. The following section, Checksum Tool Window, describes the user interface window used to configure and start checksum calculation. This tool allows you to calculate the CRC, or CRC32 value of any file or piece of data you desire. SB-SHA1 Checksum Calculator is a tool to calculate and compare the SHA1 checksum of files. Source IP Address- Source IP Address is a 32 bit field. MD5 checksum is a 128-bit hash value (32 characters). The most commonly used checksum is MD5 (Message-Digest algorithm 5) hash. The Calculated/Generated 128-bit strength Checksum can be used as a digital signature for documents, files and virtually for any kind of. If you have data in a SQL Server table and you want to know if any of the values in a row have changed, the best way to do that is by using the rowversion data type. When sending data, short checksum is generated based on data content and sent along with data. An easy-to-use hash calculator. The idea is to get data input from user (a string), convert it to binary form, then calculate checksum and put it in my data packet. If the message on the serial port is X&s=abe4\n, then the Fletcher 16 algorithm is applied to the X. This is the snippet Compute CRC Checksum on FreeVBCode. Using this algorithm you are able to calculate a hash value or digest of any message. I need to calculate checksum using 16-bit ones' complement addition: \$\begingroup\$ If you have more than 32k 16-bit words, you can overflow a 32 bit integer when you add them up. It outputs a 32-byte MD5 hex string that is computed from the given input. See full list on codelungtung. CRC primer, Chapter 7; Only use a 32-bit number as your divisor and use your entire stream as your dividend. Hash Calculator Online. However, because it's a compression function (the packet is more than 32-bits), it's possible to. -datatype 32 bit //create //convert into integer so we calculate. Given a data stream, this function computes a 32-bit checksum using the polynomial x 32 +x 7 +x 5 +x 3 +x2+x+1. CRC-32 online file checksum function Drop File Here. The character can be entered in either upper case or lower case. 10-desktop-amd64. MD4 was made for 32-bit machines and was a lot faster than MD2, but was also shown to have weaknesses and is now considered obsolete by the Internet Engineering Task Force. How to Calculate Checksum Traditionally. There are two types of algorithms 8bit and 16bit. This requires your software to do this process though. There's one for the 32-bit version and one for the 64-bit version. Choose a template or start from scratch. NET Framework present in Windows operating system. The checksum is calculated against the FS UUID, the inode number, the inode generation, and the entire extent block leading up to (but not including) the checksum itself. Binary Calculator. The form calculates the bitwise exclusive or using the function gmp_xor. KEY configures either a 128 bit MD5 key or a 140 bit SHA1. If the two are equal, then there is no data corruption. Generally there are two cases to calculate the checksum of each Microchip. This is a 128-bit number usually expressed as a 32 character hexadecimal number. Rivest in 1991 to replace an earlier hash function, MD4. Example Code for Using Class: CRC32 ' Example of CCRC32 to calculate the 32-bit CRC of a file or string ' ' To try this example, do the following: ' 1. This comprehensive checksum calculator couldn't get much easier to use. Value for which you want to calculate checksum. Including both 32-bit and 64-bit releases. Right-click the file on which you want to perform the MD5sum or hash value check. 32-Bit CRC and XOR Checksum Computation 1 Purpose The purpose of this document is to introduce the algorithm of the 32-bit CRC and show how to calculate the checksum of NovAtel OEM4 ASCII and BINARY logs together with comprehensive examples. String that specifies the checksum algorithm and its parameters, according to the checksumSpec Format table above. 8-bit Checksum Calculator This 8-bit Checksum Calculator can be used to calculate the 8-bit Checksum of a sequence of hexadecimal values or bytes. MD5 & SHA1 Hash Generator For File Generate and verify the MD5/SHA1 checksum of a file without uploading it. Now add each data element to the checksum. Description. 1+ ripemd160 whirlpool adler32 crc32 crc32b fnv1a32 fnv1a64 fnv132 fnv164 gost gost-crypto haval128,3 haval128,4 haval128,5 haval160,3 haval160,4 haval160,5 haval192,3 haval192,4 haval192,5 haval224,3 haval224,4 haval224,5 haval256,3 haval256,4 haval256,5 joaat md2 md4 ripemd128 ripemd256 ripemd320 sha224 snefru. Below is a simple illustration of how the checksum can be calculated for a data segment of 8 bits, separated into 2 4-bit words. In some cases, a non-zero value is used instead, such as 0xFFFF for a 16-bit checksum or CRC. It outputs a 32-byte MD5 hex string that is computed from the given input. This function calculates the sum of the 32-bit values in the buffer: specified by Buffer and Length. set_checksum(0) #Fetch the pseudo header from the IP6 parent packet pseudo_header = self. Hex File Crc 16 Calculator Checksum. The program calculates the 32-bit Cyclic Redundency Check (CRC) of a given file. -datatype 32 bit //create //convert into integer so we calculate. This tool support 64 / 65 bit only, if you want 32 / 33 bit, you can go to Cyclic Redundancy Check 32 bit (CRC-32). IP of the Source. The program supports 30 hashing algorithms, among others MD5 , SHA-1 , SHA-256, the SHA-3 algorithm family. We have tested CHECKSUM 1. superfast: SuperFastHash. The oneliner is written for the Bourne shell ( bash ), and uses cksum in. The SHA1 is a 160-bit hash function which resembles the earlier MD5 algorithm. The checksum algorithm for Mario Kart Wii is MD5 based (see. Each message consists of a start byte, a message type ID byte, the message payload (which depends on the message type), a checksum byte, and an end byte. X is one string. Leave a comment. One approach is to sum 16-bit words in a 32-bit accumulator, so the overflows build up in the high-order 16 bits. 3) Next use ImgBurn or Gear ISO to create a bootable DVD. The checksum is used to detect errors after transmission or storage. NET wrapper classes for FastCRC API (FastCRC. AWOL Guest; Re: XOR checksum in arduino #1 Feb 25, … At RX these two hex characters and the first comma need to be extracted (ie as in. By default, all checksums will be displayed as a 64-bit number in hex notation where the first 32 bits and the last 32 bits are separated by a space (for example, '00000000 00007F80'). checksum in hex file/editing hex file manually. Find the one's complement by inverting 0s & 1s of a given binary number. Here are several questions I'm very interested about: Is the way I read file word by word correct or there is a better way?. * and,or,not,xor operations are limited to 32 bits numbers. CRC32 is a popular checksum algorithm used to detect data corruption. Data sheet. Normal representation for a polynomial is xAF. This tool allows you to calculate the CRC, or CRC32 value of any file or piece of data you desire. MD5 has been utilized in a wide variety of cryptographic applications, and is also commonly used to verify data integrity. In practice, checksum values are mainly used in three situations. 0x87-ZERO POINT CALIBRATION Send command Byte0 Byte1 Byte2 Byte3 Byte4 Byte5 Byte6 Byte7 Byte8 Start Byte Reserved Command - - - - - Checksum. We have tested CHECKSUM 1. AWOL Guest; Re: XOR checksum in arduino #1 Feb 25, … At RX these two hex characters and the first comma need to be extracted (ie as in. "Invert the bits and add 1". first, we have to divide and slice binary data into 16 bits pieces. X is one string. ASP/Verify v. Use checksum values. fletcher: Fletcher in 16/32/64-bits. Two 16-bit checksums are computed and are appended to form a 32-bit Fletcher checksum. Checksum (32 Bit) 4: Simple count where all the bytes are added in an 32 bit accumulator. There are more than 25 alternatives to IgorWare Hasher for a variety of platforms, including Windows, Linux, BSD, Mac and Windows Explorer. Now you can explore your whole system in an instant. Number of Bits. These procs assume that the input data is binary and in network byte order aka big-endian, which is the usual representation of an IP packet. Right-click on the file you want to run a checksum against and choose “Properties” from the context menu. You can customize the name or leave it as the default. Let's take a list of 5-bit data values which we want to create a checksum for. 2 Bit Stuff Vulnerabilities 30 5. Anytime, anywhere access lets you remotely control devices and help keep your clients up and running. QuickSFV integrates into the Windows Explorer shell and makes it very easy to verify files. Auto Update. Thanks Alain From owner-robots Thu Oct 19 06:32:09 1995 Return-Path: Received: by webcrawler. Unicode support. AUTHENTiC The Checksum Calculator - X 64-bit Download - x64-bit download - freeware, shareware and software downloads. Copy the hash (only) from website. Direct extraction of data to destination drives without using temporary system folders. It is fully compatible with UTF-8 encoding. Next, you MUST reverse the bits of EVERY byte of the message and do a 1's complement the first 32 bits. MD5 was designed by Professor Ronald L. checksum, a fast file, folder and drive hashing application for Windows. This 8-bit Checksum Calculator can be used to calculate the 8-bit Checksum of a sequence of hexadecimal values or bytes. 1/8/7/Vista/XP (32-bit and 64-bit). Specify the unique name of the Add a Checksum step on the canvas. Which can be preformed using the pic32 cryptographic peripheral. Example: base is encoded in ASCII (8-bit) 01100010 01100001 01110011 01100101, the cutout gives the blocks 01100,01001,10000,10111,00110,11001,01000 (with three 0 added at the end) Each. Calculating CRC-32 in C# and. int32 = A 32-bit value. The download should start automatically in a few seconds. Apart from generating the checksum values of files, you can also generate the checksum of text by just copy pasting it under the String tab. NET samples demonstrate how to use the VB. train network uses an 8-bit CRC for each 64-bit packet of data transmitted at the link layer. It is adapted from RFC 1071: """ Compute the Internet Checksum of the supplied data. Creating or validating checksum files can be processed in a command window or in batch files. Source code extract from the example project. It has advanced options on handling renaming of bad files and how to handle the files in memory as it is processing. IgorWare Hasher is described as 'free SHA-1, MD5 and CRC32 hash-generator for Windows and is available in 32- and 64-bit versions'. i am struggling since a long time to determine the checksum of the following sequences. The Windows XP users have to download Coil32 application. Alternate checksum negotiation proceeds as follows: A SYN segment used. Once the class is instantiated, we use the update method to update the Checksum instance. Like most checksum techniques, this decreases in reliability as the number of bytes checksummed increases. CRC Calculator. An Internet packet generally includes two checksums: a TCP/UDP checksum and an IP checksum. docx from CS NETWORKS at Vishwakarma Institute of Technology. This ends the calculation. Welcome to Kofax Kapow; Introduction. prm file where you can find the 32-bit checksum. You can generate and verify checksums with them. This is a 32-bit checksum. 1, 8, and 7. 1 on 32-bit and 64-bit PCs. If the two are equal, then there is no data corruption. It can be found in the FastCRC. \item Accumulate the 32-bit 1's complement checksum over the {\sl FITS} logical records that make up the HDU header in the same manner as was done for the data records by interpreting each 2880-byte logical record as 720 32-bit unsigned integers. unable to correctly calculate checksum for one SNES game. Doing this seems to cause some sort of verification in the hex file to fail. The reciever will simply add all the above 4 things. 16 bit checksum calculator online. 2 32-Bit CRC Checksum. MD5 checksum is a 128-bit hash value (32 characters). The SHA1 is a 160-bit hash function. This is a factory method that creates an instance of a hashing algorithm object. CHECKSUM 1. c * crc checksum generation and wimax_mac_gen_crc32_table() must be called before DESCRIPTION: Calculate the 32-bit CRC from a given data. train network uses an 8-bit CRC for each 64-bit packet of data transmitted at the link layer. It can compute message digests, checksums and HMACs for files, as well as for text and hex strings. The FCS is a CRC over all fields (except the FCS) with the polynomial. murmur: MurmurHash3 in 32-bit. Hex Crc-16 Calculator. md5sum Unix/Linux users may find this utility already available on their systems. The algorithm uses big-endian bit ordering in a byte (most significant bit first). Data as well as checksum are added together. Copy and paste the checksum you want to compare against in the “Hash Comparison” dialog box. The sum and cksum commands are file integrity utilities that are based on a weak cyclic redundancy check mechanism (32-bit wide), and this mechanism is prone to high collision rates. ADS126X-CALC-TOOL — ADS1262 and ADS1263 Excel calculator tool. Free CRC routines downloadable. This is what is referred to as a hash or checksum, and if you are familiar with the MD5 algorithm, the principle is the same. Also you can download the portable 32-bit version of the Coil64. The X^0 term (usually shown as "+1") results in * the MSB being 1 * * Note that the usual hardware shift register implementation, which * is what we're using (we're merely optimizing it by doing eight-bit * chunks at a time) shifts bits into the lowest-order term. The Bitwise Calculator is used to perform bitwise AND, bitwise OR, bitwise XOR (bitwise exclusive or) operations on two integers. Hash Calculator Online lets you calculate the cryptographic hash value of a string or file. EF CheckSum Manager is a program designed for the integrity examination of files on the standard formats SFV, MD5 and SHAx. Calculate 16 bit and 32 bit CRC's. Basically I understand how it need to be calculate but to write it in VB code is. There is no such thing as "better" here - it's just a question of what you need to do! First, you need to clearly define your Requirement - then you can think about how to meet that requirement. Checksum-Aide can generate up to 11 different hash codes (including SHA-256). Supported Hash Algorithms. Binary Calculator. Initial value is 0. I would recommend that for comparing files you use a hash function that returns more bits, like MD5. This number is unique to the SAP file, which is created by the user via the Image Creation Utility. The XOR checksum for NMEA logs is also explained in this document. 1,10 you can get installation file here. NET samples demonstrate how to use the VB. So this is the way we calculate IP header checksum to check the integrity of IP header. The record format also has a 2-character suffix containing a checksum. One approach is to sum 16-bit words in a 32-bit accumulator, so the overflows build up in the high-order 16 bits. Download 8 Screenshot(s) 9 user reviews No Video. A simple method for error-checking data. I want my PSoC to be able to perform a crc of its own program space. Calculate the cyclic redundancy checksum polynomial of 32-bit lengths of the string. In telecomunicazioni e informatica il checksum (lett.
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2021-09-28 06:52:05
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https://mrrussell.100wc.net/week5-the-stadium/
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It was the start of the year I was going to old trafford to watch Manchester united vs Liverpool. I had my Liverpool scarf and hat. I was going with my dad and brother we were really excited. I thought Liverpool were going to win. So kick off was at eight o’clock. We were in there at quarter to seven there was lots of people already. After the match I was going home. I got my suitcase but it was heavier than expected but it was alright. After that lovely weekend we got home safely. THE END
1. Hi Cian,
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2022-01-20 02:53:49
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http://www.nag.com/numeric/cl/nagdoc_cl23/html/G13/g13cgc.html
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g13 Chapter Contents
g13 Chapter Introduction
NAG C Library Manual
# NAG Library Function Documentnag_tsa_noise_spectrum_bivar (g13cgc)
## 1 Purpose
For a bivariate time series, nag_tsa_noise_spectrum_bivar (g13cgc) calculates the noise spectrum together with multiplying factors for the bounds and the impulse response function and its standard error, from the univariate and bivariate spectra.
## 2 Specification
#include #include
void nag_tsa_noise_spectrum_bivar (const double xg[], const double yg[], const Complex xyg[], Integer ng, const double stats[], Integer l, Integer n, double er[], double *erlw, double *erup, double rf[], double *rfse, NagError *fail)
## 3 Description
An estimate of the noise spectrum in the dependence of series $y$ on series $x$ at frequency $\omega$ is given by
$f y∣x ω = f yy ω 1 - W ω$
where $W\left(\omega \right)$ is the squared coherency described in G13GEF and ${f}_{yy}\left(\omega \right)$ is the univariate spectrum estimate for series $y$. Confidence limits on the true spectrum are obtained using multipliers as described for G13CAF, but based on $\left(d-2\right)$ degrees of freedom.
If the dependence of ${y}_{t}$ on ${x}_{t}$ can be assumed to be represented in the time domain by the one sided relationship
$y t = v 0 x t + v 1 x t-1 + ⋯ + n t$
where the noise ${n}_{t}$ is independent of ${x}_{t}$, then it is the spectrum of this noise which is estimated by ${f}_{y\mid x}\left(\omega \right)$.
Estimates of the impulse response function ${v}_{0},{v}_{1},{v}_{2},\dots$ may also be obtained as
$v k = 1 π ∫ 0 π Re exp ik ω f xy ω f xx ω$
where Re indicates the real part of the expression. For this purpose it is essential that the univariate spectrum for $x$, ${f}_{xx}\left(\omega \right)$,and the cross spectrum, ${f}_{xy}\left(\omega \right)$ be supplied to this function for a frequency range
$ω l = 2πl L , 0 ≤ l ≤ L/2 ,$
where $\left[\right]$ denotes the integer part, the integral being approximated by a finite Fourier transform.
An approximate standard error is calculated for the estimates ${v}_{k}$. Significant values of ${v}_{k}$ in the locations described as anticipatory responses in the argument array rf, indicate that feedback exists from ${y}_{t}$ to ${x}_{t}$. This will bias the estimates of ${v}_{k}$ in any causal dependence of ${y}_{t}$ on ${x}_{t},{x}_{t-1},\dots$.
## 4 References
Bloomfield P (1976) Fourier Analysis of Time Series: An Introduction Wiley
Jenkins G M and Watts D G (1968) Spectral Analysis and its Applications Holden–Day
## 5 Arguments
1: xg[ng]const doubleInput
On entry: the ng univariate spectral estimates, ${f}_{xx}\left(\omega \right)$, for the $x$ series.
2: yg[ng]const doubleInput
On entry: the ng univariate spectral estimates, ${f}_{yy}\left(\omega \right)$, for the $y$ series.
3: xyg[ng]const ComplexInput
On entry: ${f}_{xy}\left(\omega \right)$, of the ng bivariate spectral estimates for the $x$ and $y$ series. The $x$ series leads the $y$ series.
Note: the two univariate and bivariate spectra must each have been calculated using the same amount of smoothing. The frequency width and the shape of the window and the frequency division of the spectral estimates must be the same. The spectral estimates and statistics must also be unlogged.
4: ngIntegerInput
On entry: the number of spectral estimates in each of the arrays xg, yg and xyg. It is also the number of noise spectral estimates.
Constraint: ${\mathbf{ng}}\ge 1$.
5: stats[$4$]const doubleInput
On entry: the 4 associated statistics for the univariate spectral estimates for the $x$ and $y$ series. ${\mathbf{stats}}\left[0\right]$ contains the degree of freedom, ${\mathbf{stats}}\left[1\right]$ and ${\mathbf{stats}}\left[2\right]$ contain the lower and upper bound multiplying factors respectively and ${\mathbf{stats}}\left[3\right]$ contains the bandwidth.
Constraints:
• ${\mathbf{stats}}\left[0\right]\ge 3.0$;
• $0.0<{\mathbf{stats}}\left[1\right]\le 1.0$;
• ${\mathbf{stats}}\left[2\right]\ge 1.0$.
6: lIntegerInput
On entry: the frequency division, $L$, of the spectral estimates as $2\pi /L$, as input to nag_tsa_spectrum_univar (g13cbc) and nag_tsa_spectrum_bivar (g13cdc).
Constraints:
• ${\mathbf{ng}}=\left[{\mathbf{l}}/2\right]+1$;
• The largest prime factor of l must not exceed 19, and the total number of prime factors of l, counting repetitions, must not exceed 20. These two restrictions are imposed by nag_fft_real (c06eac) and nag_fft_hermitian (c06ebc) which perform the FFT.
7: nIntegerInput
On entry: the number of points in each of the time series $x$ and $y$. n should have the same value as nxy in the call of nag_tsa_spectrum_bivar_cov (g13ccc) or nag_tsa_spectrum_bivar (g13cdc) which calculated the smoothed sample cross spectrum. n is used in calculating the impulse response function standard error (rfse).
Constraint: ${\mathbf{n}}\ge 1$.
8: er[ng]doubleOutput
On exit: the ng estimates of the noise spectrum, ${\stackrel{^}{f}}_{y\mid x}\left(\omega \right)$ at each frequency.
9: erlwdouble *Output
On exit: the noise spectrum lower limit multiplying factor.
10: erupdouble *Output
On exit: the noise spectrum upper limit multiplying factor.
11: rf[l]doubleOutput
On exit: the impulse response function. Causal responses are stored in ascending frequency in ${\mathbf{rf}}\left[0\right]$ to ${\mathbf{rf}}\left[{\mathbf{ng}}-1\right]$ and anticipatory responses are stored in descending frequency in ${\mathbf{rf}}\left[{\mathbf{ng}}\right]$ to ${\mathbf{rf}}\left[{\mathbf{l}}\right]$.
12: rfsedouble *Output
On exit: the impulse response function standard error.
13: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_2_INT_ARG_CONS
On entry, ${\mathbf{l}}=〈\mathit{\text{value}}〉$ while ${\mathbf{ng}}=〈\mathit{\text{value}}〉$. These arguments must satisfy ${\mathbf{ng}}=\left[{\mathbf{l}}/2\right]+1$ when ${\mathbf{ng}}>0$.
NE_ALLOC_FAIL
Dynamic memory allocation failed.
NE_BIVAR_SPECTRAL_ESTIM_ZERO
A bivariate spectral estimate is zero.
For this frequency the noise spectrum is set to zero, and the contributions to the impulse response function and its standard error are set to zero.
NE_FACTOR_GT
At least one of the prime factors of l is greater than 19.
NE_INT_ARG_LT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 1$.
On entry, ${\mathbf{ng}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ng}}\ge 1$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_REAL_ARG_GT
On entry, ${\mathbf{stats}}\left[1\right]$ must not be greater than 1.0: ${\mathbf{stats}}\left[1\right]=〈\mathit{\text{value}}〉$.
NE_REAL_ARG_LE
On entry, ${\mathbf{stats}}\left[1\right]$ must not be less than or equal to 0.0: ${\mathbf{stats}}\left[1\right]=〈\mathit{\text{value}}〉$.
NE_REAL_ARG_LT
On entry, ${\mathbf{stats}}\left[0\right]$ must not be less than 3.0: ${\mathbf{stats}}\left[0\right]=〈\mathit{\text{value}}〉$.
On entry, ${\mathbf{stats}}\left[2\right]$ must not be less than 1.0: ${\mathbf{stats}}\left[2\right]=〈\mathit{\text{value}}〉$.
NE_SQUARED_FREQ_GT_ONE
A calculated value of the squared coherency exceeds one.
For this frequency the squared coherency is reset to one with the result that the noise spectrum is zero and the contribution to the impulse response function at this frequency is zero.
NE_TOO_MANY_FACTORS
l has more than 20 prime factors.
NE_UNIVAR_SPECTRAL_ESTIM_NEG
A bivariate spectral estimate is negative.
For this frequency the noise spectrum is set to zero, and the contributions to the impulse response function and its standard error are set to zero.
NE_UNIVAR_SPECTRAL_ESTIM_ZERO
A bivariate spectral estimate is zero.
For this frequency the noise spectrum is set to zero, and the contributions to the impulse response function and its standard error are set to zero.
## 7 Accuracy
The computation of the noise is stable and yields good accuracy. The FFT is a numerically stable process, and any errors introduced during the computation will normally be insignificant compared with uncertainty in the data.
The time taken by nag_tsa_noise_spectrum_bivar (g13cgc) is approximately proportional to ng.
## 9 Example
The example program reads the set of univariate spectrum statistics, the 2 univariate spectra and the cross spectrum at a frequency division of $\frac{2\pi }{20}$ for a pair of time series. It calls nag_tsa_noise_spectrum_bivar (g13cgc) to calculate the noise spectrum and its confidence limits multiplying factors, the impulse response function and its standard error. It then prints the results.
### 9.1 Program Text
Program Text (g13cgce.c)
### 9.2 Program Data
Program Data (g13cgce.d)
### 9.3 Program Results
Program Results (g13cgce.r)
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2016-05-03 15:11:17
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https://socratic.org/questions/how-do-you-decide-whether-the-relation-x-y-defines-a-function
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How do you decide whether the relation x = y² defines a function?
Nov 15, 2017
We can graph the relation and do the vertical line test.
Explanation:
Whenever we have to check if a relation is a function, we draw its graph and do the vertical line test.
The vertical line test says that if a vertical line touches the graph once, the relation is a function. And if the vertical line touches the graph more than once it's not a function.
Example $\to$
$x = {y}^{2}$
$\sqrt{x} = y$
$y = \sqrt{x}$
Let's graph this relation.
graph{sqrtx [-10, 10, -5, 5]}
Now if we draw a vertical line we can see that the line touches the graph at only one point everywhere.
Therefore it is a function.
Additional $\textcolor{red}{\to}$Circles, Ellipses, Hyperbolas etc. are not functions because they don't pass the vertical line test.
A Line, Parabola etc. are functions because they pass the vertical line test.
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2021-12-01 14:46:11
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https://pdglive.lbl.gov/Particle.action?node=M226&home=MXXX040
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CHARMED, STRANGE MESONS($\mathit C$ = $\mathit S$ = $\pm1$)(including possibly non- ${\mathit {\mathit q}}$ ${\mathit {\overline{\mathit q}}}$ states) ${{\mathit D}_{{s}}^{+}}$ = ${\mathit {\mathit c}}$ ${\mathit {\overline{\mathit s}}}$, ${{\mathit D}_{{s}}^{-}}$ = ${\mathit {\overline{\mathit c}}}$ ${\mathit {\mathit s}}$, similarly for ${{\mathit D}_{{s}}^{*}}$'s INSPIRE search
#### ${{\boldsymbol D}_{{s3}}^{*}{(2860)}^{\pm}}$
$I(J^P)$ = $0(3^{-})$
$\mathit J{}^{P}$ consitent with $3{}^{-}{}^{}$ from angular analysis of AAIJ 2014AW.
${{\mathit D}_{{s3}}^{*}{(2860)}^{+}}$ MASS $2860 \pm7$ MeV
${{\mathit D}_{{s3}}^{*}{(2860)}^{+}}$ WIDTH $53 \pm10$ MeV
FOOTNOTES
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2021-09-20 02:59:15
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https://solvedlib.com/n/let-the-matrixact-on-c2-find-the-eigenvalues-and-basis-for,15454033
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# Let the matrixact on C2 Find the eigenvalues and basis for each eigenspace in €2Select all that apply:2+6 i
###### Question:
Let the matrix act on C2 Find the eigenvalues and basis for each eigenspace in €2 Select all that apply: 2+6 i DA. 1=5+6iv= 2 +6i DB: 1 = -5-6i;V= 2 - 6 [ Dc 1=6+5iv= |2+6i 1=6 - 5i; v= 2 -6 [ DE: 1=5-6iv= 2+6 [ 1=5-6i; v= 2 - 6 DG: 1=5+6iv= 2-6 i Oh: 1 = -5+6i;V=
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# Line M has a y-intercept of –4, and its slope must be an integer-multi
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Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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10 Mar 2015, 05:16
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Line M has a y-intercept of –4, and its slope must be an integer-multiple of 1/7. Given that Line M passes below (4, –1) and above (5, –6), how many possible slopes could Line M have?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Kudos for a correct solution.
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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15 Mar 2015, 21:22
Bunuel wrote:
Line M has a y-intercept of –4, and its slope must be an integer-multiple of 1/7. Given that Line M passes below (4, –1) and above (5, –6), how many possible slopes could Line M have?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION:
Well, for starters, zero is a multiple of every number, and a line with slope zero, the horizontal line y = –4 passes below (4, –1) and above (5, –6). That’s horizontal line is our starting point.
The point (4, –1) is over 4, up 3 from the y-intercept (0, –4). A line with a slope of +3/4 would go straight from (0, –4) to (4, –1). Thus, we need a slope that is less than +3/4. Notice that 3/4 = 21/28. Notice that 5/7 = 20/28, so this would be less than 3/4. Therefore, +1/7 through +5/7 will all slope up, obviously above (5, –6), and all will pass below (4, –1). That’s five upward sloping lines.
The point (5, –6) is over 5, down 2, from the y-intercept (0, –4). A line with a slope of –2/5 would go straight from (0, –4) to (5, –6). Thus, we need a slope that is more than –2/5; another way to say that is, we need a negative slope whose absolute value is less than +2/5. Well, 2/5 = 14/35, while 2/7 = 10/35 and 3/7 = 15/35, so (2/7) < (2/5) < (3, 7). The negatively sloping lines obviously pass below (4, –1), but only two of them, –1/7 and –2/7, pass above (5, –6).
That’s one horizontal line, five upward sloping lines, and two downward sloping lines, for a total of eight.
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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17 Apr 2017, 09:04
6
3
The slope of the line M has to be in between the slope of line A (passing through (0, -4) and (4, -1)) and line B(passing through (5, -6) and (0, -4))
Now, as we know slope of line with two given points = (y2-y1)/(x2-x1)
Therefore ,
slope of line A => 3/4
slope of line B => -2/5
Therefore slope of line M => -2/5 < m < 3/4
As given in question, m is integer multiple of 1/7 therefore m =k/7 (where k is any integer)
-2/5 < k/7 < 3/4
-14/5 < k < 21/4
-2.8 < k < 5.1
and as we know k is integer
Therefore possible values of k would be -2, -1, 0, 1, 2, 3 ,4 ,5
Total = 8 values (C Answer)
##### General Discussion
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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14 Mar 2015, 13:07
3
Let m = slope of Line M
Using the y-intercept and the given points, we can find that $$-\frac{2}{5} < m < \frac{3}{4}$$
Since we need to think in multiples of $$\frac{1}{7}$$, let's convert this to $$-\frac{14}{35} < m < \frac{21}{28}$$
Multiplying $$\frac{1}{7}$$ by the consecutive integers between -2 and 5, inclusive, will give us values that fall in this range.
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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02 Jun 2015, 10:43
sterling19 wrote:
Let m = slope of Line M
Using the y-intercept and the given points, we can find that $$-\frac{2}{5} < m < \frac{3}{4}$$
Since we need to think in multiples of $$\frac{1}{7}$$, let's convert this to $$-\frac{14}{35} < m < \frac{21}{28}$$
Multiplying $$\frac{1}{7}$$ by the consecutive integers between -2 and 5, inclusive, will give us values that fall in this range.
Please elaborate the last sentence....how can values faill in the range after multiplying 1/7 between -2 and 5.
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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14 Jul 2016, 10:08
sterling19 wrote:
Let m = slope of Line M
Using the y-intercept and the given points, we can find that $$-\frac{2}{5} < m < \frac{3}{4}$$
Since we need to think in multiples of $$\frac{1}{7}$$, let's convert this to $$-\frac{14}{35} < m < \frac{21}{28}$$
Multiplying $$\frac{1}{7}$$ by the consecutive integers between -2 and 5, inclusive, will give us values that fall in this range.
Hi Sterling,
Regards
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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27 Jul 2016, 22:01
1
ravi2107 wrote:
sterling19 wrote:
Let m = slope of Line M
Using the y-intercept and the given points, we can find that $$-\frac{2}{5} < m < \frac{3}{4}$$
Since we need to think in multiples of $$\frac{1}{7}$$, let's convert this to $$-\frac{14}{35} < m < \frac{21}{28}$$
Multiplying $$\frac{1}{7}$$ by the consecutive integers between -2 and 5, inclusive, will give us values that fall in this range.
Hi Sterling,
Regards
Trying to make sense of it :
-14/35 < m < 21/28
so m must be less than 21/28
20/28 = 5/7 so we can take values from 20/28 or 5/7 to 0
ie 5/7, 4/7, 3/7, 2/7, 1/7 and 0 (6 values)
now m must be > -14/35 ie -2/5
take m as -10/35 ie - 2/7
so two values -2/7 and -1/7
total values = 6 + 2 (8) (C)
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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28 Oct 2016, 08:55
But if the slope must be a integer multiple of 1/7 then how can there be 8 slopes? this would mean that slope can only be +/- 1,2,3 and so on.. Can anyone resolve this?
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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10 Jan 2017, 18:44
I am confused here too. I got to the point of -2/5 < m < 3/4. After that I couldn't figure how to get integer multiple of 1/7.
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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08 Apr 2017, 11:12
what is meant by 'integer multiple' of 1/7? does that mean multiples such as 1 2 3 4, or -1, -2, -3, -4 or something else? Thanks
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Joined: 01 Apr 2017
Posts: 1
Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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17 Apr 2017, 23:40
Step 1.
Find range for possible slope values i.e., -0.4(-2/5) to 0.75(3/4)
Step 2.
We know from question stems that the limits of the above range are not included in our target values and that the values of slope have to be multiples of 0.14(1/7). Hence, only possible values are
-0.28, -0.14, 0, 0.14, 0.18, 0.42, 0.56 and 0.70
a total of 8 values.
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Joined: 09 Sep 2013
Posts: 9462
Re: Line M has a y-intercept of –4, and its slope must be an integer-multi [#permalink]
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03 Jun 2018, 06:10
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Re: Line M has a y-intercept of –4, and its slope must be an integer-multi &nbs [#permalink] 03 Jun 2018, 06:10
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2019-01-23 05:43:29
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http://physics.aps.org/synopsis-for/10.1103/PhysRevA.78.012503
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# Synopsis: Potassium atoms feel a distant attraction
The atoms in highly excited vibrational states of a diatomic molecule can be quite far apart near their maximum excursion. Physicists are now using laser spectroscopy to carefully measure the long-range effective interaction between potassium atoms in these states—an essential parameter to understanding ultracold atomic collisions.
To understand ultracold collisions between neutral atoms it is necessary to have accurate knowledge of the interatomic interaction potentials at extremely long range, out to many times the atomic radius. One approach is to study diatomic molecules in highly excited vibrational states, since there is a high probability of finding the atoms at fairly large distances from one another as they reach their maximum excursion. However, state-of-the-art techniques of laser spectroscopy must be used to excite and measure these states.
In the July 11th issue of Physical Review A, Stephen Falke and coworkers at the Leibniz Universität Hannover and Christian Lisdat at the Physikalisch-Technische Bundesanstalt have measured, with MHz precision, the energy gaps between several asymptotically high vibrational levels in the electronic ground state of the ${}^{39}{\text{K}}_{2}$ molecule. By combining this data with earlier measurements, the authors can fit new ground-state potential functions and calculate accurate scattering lengths for each of the isotopic combinations of ${}^{39}\text{K}$, ${}^{40}\text{K}$, and ${}^{41}\text{K}$ atoms. Their findings are sufficiently accurate that they can observe small deviations from the Born-Oppenheimer (adiabatic) approximation that is conventionally used to describe neutral atom collisions.
The exquisitely detailed experimental and analytical work illustrates the influence of advances in atomic and molecular physics that have been highlighted in several recent Nobel Prizes, including femtosecond frequency combs, iodine-stabilized lasers, and quantum degenerate gases. It will also contribute to continuing progress in molecular Bose-Einstein condensates and degenerate Fermi gases, which is opening new vistas in molecular and chemical physics. - Keith MacAdam
More Features »
### Announcements
More Announcements »
## Subject Areas
Atomic and Molecular Physics
## Previous Synopsis
Quantum Information
Mesoscopics
## Related Articles
Optics
### Viewpoint: Transportable Clocks Move with the Times
Transportable atomic clocks are now operating with fractional-frequency uncertainties below one part in 1016, opening up new applications. Read More »
Optics
### Viewpoint: Trapped Ions Stopped Cold
A novel method for cooling trapped ions could boost the accuracy of atomic clocks. Read More »
Atomic and Molecular Physics
### Synopsis: Detecting a Molecular Duet
Using a scanning tunneling microscope, researchers detect coupled vibrations between two molecules. Read More »
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2017-02-20 08:38:51
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https://robotics.stackexchange.com/questions/8516/getting-pitch-yaw-and-roll-from-rotation-matrix-in-dh-parameter
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# Getting pitch, yaw and roll from Rotation Matrix in DH Parameter
I've calculated a DH Parameter matrix, and I know the top 3x3 matrix is the Rotation matrix. The DH Parameter matrix I'm using is as below, from https://en.wikipedia.org/wiki/Denavit%E2%80
Above is what I'm using. From what I understand I'm just rotating around the Z-axis and then the X-axis, but most explanations from extracting Euler angles from Rotation matrixes only deal with all 3 rotations. Does anyone know the equations? I'd be very thankful for any help.
In general, Euler angles (or specifically roll-pitch-yaw angles) can be extracted from any rotation matrix, regardless of how many rotations were used to generate it. For a typical x-y-z rotation sequence, you end up with this rotation matrix where $\phi$ is roll, $\theta$ is pitch, and $\psi$ is yaw:
$R = \begin{bmatrix} c_\psi c_\theta & c_\psi s_\theta s_\phi - s_\psi c_\phi & c_\psi s_\theta c_\phi + s_\psi s_\phi \\ s_\psi c_\theta & s_\psi s_\theta s_\phi + c_\psi c_\phi & s_\psi s_\theta c_\phi - c_\psi s_\phi \\ -s_\theta & c_\theta s_\phi & c_\theta c_\phi \end{bmatrix}$
Note that $c_x = \cos x$ and $s_x = \sin x$.
To get roll-pitch-yaw angles out of any given numeric rotation matrix, you simply need to relate the matrix elements to extract the angles based on trigonometry. So re-writing the rotation matrix as:
$R = \begin{bmatrix} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{bmatrix}$
And then comparing this to the above definition, we can come up with these expressions to extract the angles:
$\tan \phi = \frac{R_{32}}{R_{33}}$
$\tan \theta = \frac{-R_{31}}{\sqrt{R_{32}^2 + R_{33}^2}}$
$\tan \psi = \frac{R_{21}}{R_{11}}$
When you actually solve for these angles you'll want to use an atan2(y,x) type of function so that quadrants are appropriately handled (also keeping the signs of the numerators and denominators as shown above).
Note that this analysis breaks down when $\cos \theta = 0$, where there exists a singularity in the rotation -- there will be multiple solutions.
Also note that your question is referring to the rotation transformation between two joints ($^{n-1}R_n$), which is indeed composed of an x-axis and z-axis rotation, but the actual rotation of that joint with respect to the base will be different ($^0R_n$) and composed of all the rotations for joints 1 to $n$. I point this out because $^{n-1}R_n$ can be decomposed back into the relevant x-axis ($\alpha$) rotation and z-axis ($\theta$) rotation, whereas $^0R_n$ is better suited to roll-pitch-yaw -- especially for the end-effector.
• I see! I've already finished calculating the overall transformation matrix ^0T_n, so I assume I just take the upper 3x3 matrix and use that as my ^0R_n so I can arrive at roll-pitch-yaw angles relative to the base using the equations you provided. Thank you!
– Iche
Nov 20, 2015 at 16:07
• Yes, you got it! Nov 21, 2015 at 1:52
• @Brian Lynch how the extraction will change If I need yaw pitch roll convention ? Feb 5, 2019 at 5:42
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2022-05-22 23:14:15
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https://www.mail-archive.com/zim-wiki@lists.launchpad.net/msg03706.html
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# Re: [Zim-wiki] equation editor
Thanks. Unfortunately, the svg file doesn't look correct in my case (see attached the tex file, the png file originally created and the svg file after patching). I played around with the dvisgm options, but couldn't find a solution.
Best wishes,
Paulo
On 07-01-16 15:27, Jaap Karssenberg wrote:
Indeed it works using dvisgm and it looks good, just change 3 lines in the plugin, see attached diff.
When I have some time I will try to make it an option in the plugin.
Regards,
Jaap
On Thu, Nov 26, 2015 at 11:00 AM, Paulo van Breugel <p.vanbreu...@gmail.com <mailto:p.vanbreu...@gmail.com>> wrote:
Perhaps replacing the dvipng by dvisgm
(http://dvisvgm.bplaced.net/) could work?
On 26-11-15 10:56, Paulo van Breugel wrote:
Thanks, I had been looking in equationeditor.py file, but wasn't
sure how to add the -D option.
The reason I want to be able to change the resolution is to make
the image suitable to be used in printed documents, which
normally requires a dpi of 300. But that would result in a very
large image in the notebook. A solution for me (but perhaps not
so much for you) would be an option to resize the image display
in the notebook (like you can do with a normal image). The option
to use svg instead of png would be even better, perhaps also for
On 26-11-15 10:28, wzhd wrote:
I have been thinking about the same thing since I got a HiDPI
screen. Currently, I have only managed to make the images
bigger, by passing the option "-D 192" to the "dvipng" command.
The modified version is here:
https://github.com/wzhd/zim-plugin-equation-editor/commit/c3c42d0668ca8dcf69dc1d25d9450d81514d5524
However this is not perfect, when I open the notebook on other
computers, the equations are too big.
On 26 November 2015 at 00:29, Paulo van Breugel
<p.vanbreu...@gmail.com <mailto:p.vanbreu...@gmail.com>> wrote:
Hi Jaap and other Zim devs,
The 'insert equation' function is a great function,
producing nice looking equations in Zim. However, the
resolution of the image (png file) that is generated by the
addon is fairly low. Is there a way to change the resolution
of the image? Or even better, would it be possible to add
the option to store the on-the-fly generaged image as svg
instead of png? This would provide in perfectly scalable
solution for any document.
Best regards
Paulo
_______________________________________________
_______________________________________________
T = 100 \times \frac{L + G}{SR + G}
_______________________________________________
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2018-02-20 04:54:31
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http://cpr-heplat.blogspot.com/2013/03/13030336-liang-kai-wu-et-al.html
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## The nature of Roberge-Weiss transition end points in 2 flavor lattice QCD with Wilson quarks [PDF]
Liang-Kai Wu, Xiang-Fei Meng
We make simulations with 2 flavor Wilson fermions to investigate the nature of end points of Roberge-Weiss (RW) first order phase transition line. The simulations are carried out at 9 values of the hopping parameter $\kappa$ ranging from 0.155 to 0.198 on different lattice volume. The Binder cumulant, the susceptibilities and reweighted distributions of the imaginary part of Polyakov loop are employed to determine the nature of end points of RW transition line. The simulations show that the RW end point is first order at large and small values of $\kappa$, whereas there are no decisive evidence that transitions at the intermediate values of $\kappa$ in our simulations are 3D Ising universality class.
View original: http://arxiv.org/abs/1303.0336
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2017-06-27 07:11:20
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http://openstudy.com/updates/4fd20a65e4b057e7d2210968
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## A community for students. Sign up today
Here's the question you clicked on:
## lolroark 2 years ago a missile is launched with an initial speed of 421.1 m/sec. after 13.77 seconds, it is at its highest point. Q: what was its initial launch angle above the horizon?
• This Question is Closed
1. lolroark
i think i solved it by myself but not sure. is it 13.77-9.8/421.1?
2. ujjwal
time to reach max height is given by $t=\frac{usin\theta}{g}$
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2015-05-24 03:25:24
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https://physics.stackexchange.com/questions/623146/if-electrons-can-be-created-and-destroyed-then-why-cant-charges-be-created-or?noredirect=1
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If electrons can be created and destroyed, then why can't charges be created or destroyed? [duplicate]
I read on Wikipedia that electrons can be created through beta decay of radioactive isotopes and in high-energy collisions, for instance when cosmic rays enter the atmosphere. Also, that they can be destroyed using pair annihilation.
We also know that charge is a physical property which can be associated with electrons. My question is why can't charges be created or destroyed if electrons can?
• Electron creation or destruction never violates charge conservation. Mar 23 '21 at 8:04
• Does this answer your question? If Energy can be converted into mass, why can it not be converted into charge? Mar 23 '21 at 8:14
• The law of conservation of conservation - if it appears one rule of conservation is broken, there is some more fundamental conservation that is conserved. (Probably bogus? It sounds good to me anyways.) Mar 24 '21 at 0:16
• Is that not broadly because an electron is a thing measurable in itself, while a charge is "merely" a characteristic of such a thing, not measurable at all without its parent entity? Mar 24 '21 at 20:17
• An electron is an object, and charge is one of the properties of that object. An electron (as an object) carries charge; but prior to the beta decay it is part of the parent isotope (the parent object): the question is falsely implying that the electron (or the charge) is being created from nothing, when in fact it is merely emerging from or "splitting" from the isotope. Mar 25 '21 at 22:55
Electrons can only be created and destroyed in processes that keep electric charge constant.
There are three Standard Model interactions involving the electron: $$\rm W^-\to e^-\bar{\nu}_e$$ and $$\rm \gamma\text{ (or Z)}\to e^-e^+$$.$$^1$$ the first case, the W boson has the same charge as the electron, so no charge is created or destroyed. In the other cases, a neutral particle turns into two particles with equal and opposite charge. Again, no charge is created or destroyed.
In a certain sense, there's no reason that electric charge, in particular, has to be conserved- our theories start with the fact that that charge cannot be created or destroyed as an assumption because we have never seen it be created or destroyed, and the universe as a whole appears to be electrically neutral.
$$\scriptsize^1\text{Also technically }{\rm H\to e^-e^+}\text{, but that coupling is very small and for practical purposes it can be ignored.}$$
• Higgs can also decay to a fermion-antifermion pair such as $\rm e^-e^+$.
– J.G.
Mar 24 '21 at 12:27
• In fact, charge has been singled out as something to pay attention to exactly because it is conserved. In a sense, that's akin to the anthropic principle: We wouldn't be here discussing it otherwise. Mar 24 '21 at 13:34
• @J.G. Technically true, but practically speaking the Standard Model coupling is so small it can be ignored for most purposes.
– Chris
Mar 25 '21 at 1:44
So the mechanisms which generate and destroy electrons happen to be such that they never violate charge conservation.
Let's take pair annihilation for example, an electron and a positron meet and they become two photons. Before, the total charge is zero: the positron has positive charge and the electron has the exact opposite negative charge. Afterward, the total charge is zero, because photons don't have electric charge.
You also mentioned beta decay. Beta decay happens when a nucleus has too many neutrons. If it has more neutrons then it can stably hold then one of those neutrons can turn into a proton plus an electron plus an electron antineutrino. Before, the total charge was zero: a neutron has no charge. After, the total charge was also zero: the proton and the electron balance each other out while the antineutrino has no charge. And there is a reason that we call it an antineutrino because by thinking of it as an “anti-lepton” we can derive a different set of numbers which are also conserved: both the proton and the neutron have three quarks so the number of quarks is conserved, and also if we say that an electron is one lepton and the antineutrino is minus one lepton, then the lepton number is also zero both before and after.
Interestingly, right now we usually don't describe electric charge as the most fundamental quantity. Since the 60s we have had an electroweak theory which unites the electromagnetic field with the weak nuclear force. According to this there are two more fundamental charges, called weak hypercharge and weak isospin. However according to this theory there is a field which it took a very long time to measure, called the Higgs field, which in some sense distorted these two but did not distort a certain sum of them. And that sum that makes it through these interactions with the Higgs field unchanged, is what we call electric field, connecting to the force particle which makes it through these interactions with the Higgs field unchanged, which is the photon, the only of these particles to remain massless and thereby describe a force of infinite range.
This is all the way of saying, while the conservation of electric charge certainly constrains the theories that we make, we don't just make the theories by building electric charge into the theory directly anymore; with the discovery of the Higgs boson, it turns out to be a small part of a bigger unified force.
When a particle is created or destroyed, energy is conserved. Energy is also a property of an electron, but we are not inclined to ask - why is energy conserved when the particle is not, because we accept the principle that the energy is, in a sense, temporarily imbued in the particle, but is otherwise independent of it. In the same way, charge is not only a property of an electron, but of multiple other particles as well. The idea is similar in the sense that we say that only processes that conserve the total energy can occur, and only processes that conserve the total charge can occur.
Ultimately, the point is that the charge is not fundamentally attached to an electron, but can be carried away by another particle, or canceled out by an opposite charge.
An important aside is that charge conservation in quantum theory comes from the gauge invariance associated with the principle that a global change in phase has no observational consequence.
It’s important to note that electrons can only be created and destroyed in interactions which respect the fundamental symmetries of nature and their associated conservation laws.
So, charge conservation is one such conservation law; “number-of-electrons conservation” is not. (Though before radioactivity was discovered, you wouldn’t be unreasonable to think so!) Since other things, like protons, can also carry charge, it’s possible for a given interaction, such as beta decay, to create a (negatively charged) electron while still conserving the total charge—for example, by creating a (positively charged) proton at exactly the same time.
Technically, when a neutron decays in beta decay, the proton is created at the same time as a negatively charged $$W^-$$ boson—and then the boson decays to an electron and a neutrino (0 charge). But again, in each of these changes, the total quantity of charge is conserved.
And just to clarify: often you’ll hear the terminology “charges” to simply refer to particles that have some nonzero charge. But note that while the particles themselves might be created and destroyed, charge as a quantity is always conserved! And since it’s the quantity we care about, not the particles, we might conserve the quantity by creating two oppositely charged particles, such as the proton and electron in beta decay. In some sense, then we’ve created charges (particles with charge), but have not created any charge—at least, not when we add it all up and calculate the overall quantity of charge, which is still 0, just as it was before the decay.
The answer is related to conservation laws. In physics, processes can only happen if they respect certain conservation laws.
The creation and destruction of electrons does not break any conservation laws; namely the energy is always conserved and transformed to another form when electrons are destroyed or created. However, the creation or destruction of charges does break a conservation law $$-$$ the conservation of electric charge.
• Okay, I got this point, but my question is that if electrons can be created or destroyed then does this not mean that charges are being created or destroyed simultaneously? As a particular electron will have certain amount of charge... Mar 23 '21 at 8:09
• @PoorvajaJain Particles are not simply created and destroyed but in some sense exchanged for other particles. A photon becomes an electron and a positron, more or less. The idea is that these exchanges of collections of particles only occur when the total charge is conserved. Like energy, the charge is a global property of the system rather than a local property of the particle. Mar 23 '21 at 8:24
• "the creation or destruction of charges does break a conservation law" if two oppositely charges are created (by a photon) no conservation law is broken. Mar 24 '21 at 22:51
• @DescheleSchilder I think that we are stuck at a language issue here. I see you point, I just do not see this process as the "creation" of a charge since no new charge was added to the system (universe).
– Y2H
Mar 25 '21 at 8:29
Feynman once asked more or less the same question (page 129 of "Quantum Field Theory" by Lewis H. Ryder):
I remember that when someone had started to teach me about creation and annihilation operators, that this operator creates an electron, I said 'How do you create an electron? It disagrees with conservation of charge'. - R. P. Feynman
So you are good company. I think by now it has become clear to you that whenever an electron appears it must take its charge from other charged particles. An electron can never be created on its own. Or it takes its charge from other particles, or a positron is created at the same time.
Likewise, an electron can't be destroyed without another equally, but oppositely, charged particle being created. When the electron is isolated, it can never be destroyed.
Charges can be created, like the charges of an electron and a positron in pair production, but their total value must always be zero (i.e., total charge can't be created).
Charges can be created and destroyed. Total charge cannot.
Whenever you create an electron, charge $$-1$$, you must also create a positron, charge $$+1$$. That gives total charge $$0$$.
Whenever you create a proton, charge $$+1$$, you have to create an anti-proton, charge $$-1$$. That gives total charge $$0$$.
As far as we're aware, the total charge in the Universe is zero. Every proton's positive charge is balanced by an electron's negative charge.
That said, there's a caveat to the above: the rules as I've described them have it that at any time you make matter, you also make anti-matter, which is more than just opposite charge: if they were strictly followed, we'd not be here, because all the matter and antimatter would have reannihilated into photons. The reason for it to be otherwise is not known. However, presumably it would not involve any processes breaching charge conservation. For example, it might be possible to convert anti-protons into neutrons and electrons. This would not violate charge conservation. Total charge $$-1$$ first, then $$0$$ and $$-1$$ as components - total, $$-1$$. That's one way it could happen. Other ways could work, too.
(Why do we presume that charge conservation cannot be violated? Simple: it's the easiest way to explain why the total charge of the whole Universe is zero.)
Well most of what needs to be said already has been said. But there is something else that I think would interest you.
Charge in a system is only conserved when a very specific symmetry in the system in not broken. Noether's theorem links symmetry to conservation laws. If you can find a way to break the symmetry that governs conservation of charge, then that conservation law will not hold.
An example of this would be introducing an external force in a system, thus breaking the symmetry that govern conservation of linear momentum.
Electrons can be created and destroyed because the process does not violate any of the given system's conservation laws. If you can find a situation in which the symmetry governing conservation of charge is broken, you shouldn't have much problem creating or destroying charge.
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2022-01-27 18:36:10
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https://www.clutchprep.com/physics/practice-problems/44285/an-electromagnetic-wave-propagates-through-a-vacuum-in-the-x-direction-carrying-
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Electromagnetic Waves as Sinusoidal Waves Video Lessons
Concept
# Problem: An electromagnetic wave propagates through a vacuum in the +x-direction, carrying an intensity of 150 W/m2. At t = 0, the electric field has zero amplitude, and after 0.01 s, the electric field strength grows to its maximum value, pointing in the +y direction. Write equations describing the electric and magnetic fields as sinusoidal oscillations, including the appropriate unit vectors to denote direction.
###### FREE Expert Solution
79% (123 ratings)
###### Problem Details
An electromagnetic wave propagates through a vacuum in the +x-direction, carrying an intensity of 150 W/m2. At t = 0, the electric field has zero amplitude, and after 0.01 s, the electric field strength grows to its maximum value, pointing in the +y direction. Write equations describing the electric and magnetic fields as sinusoidal oscillations, including the appropriate unit vectors to denote direction.
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2021-01-19 18:23:44
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https://plainmath.net/605/perform-the-indicated-divisions-polynomials-monomials-frac-24x-plus-36x
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# Perform the indicated divisions of polynomials by monomials. frac{-24x^{6}+36x^{8}}{4x^{2}}
Question
Polynomial division
Perform the indicated divisions of polynomials by monomials. $$\frac{-24x^{6}+36x^{8}}{4x^{2}}$$
2021-02-26
A polynomial is an expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a non-negative integral power.
Here the given polynomial is a binomial.
To divide a polynomial by monomial, divide each term of the polynomial by the monomial.
Divide the binomial by the monomial $$4x^{2}$$.
To divide a polynomial by monomial, divide each term of the polynomial by the monomial.
Simplify the terms which are under division.
Calculation:
Consider the polynomial $$=\frac{-24x^{6}+36x^{8}}{4x^{2}}$$.
Divide each term of the polynomial by the monomial $$4x^{2}$$.
$$\frac{-24x^{6}+36x^{8}}{4}x^{2}=(-24\frac{x^{6}}{4}x^{2})+(36\frac{x^{8}}{4}x^{2})$$
$$(-24\frac{x^{6}}{4}x^{2})+(36\frac{x^{8}}{4}x^{2})=-6x^{4}+9x^{6}$$
The simplified value of polynomial is $$-6x^{4}+9x^{6}$$.
Final statement:
The simplified value of polynomial after division is equals to $$6x^{4}+9x^{6}$$.
### Relevant Questions
Perform the indicated divisions of polynomials by monomials.
$$\frac{-18x^{2}y^{2}+24x^{3}y^{2}-48x^{2}y^{3}}{6xy}$$
Perform the indicated divisions of polynomials by monomials.
$$\frac{12x^{3}-24x^{2}}{6x^{2}}$$
Perform the indicated divisions of polynomials by monomials.
$$\frac{-27a^{3}b^{4}-36a^{2}b^{3}+72a^{2}b^{5}}{9a^{2}b^{2}}$$
Perform the indicated divisions of polynomials by monomials.
$$\frac{-16x^{4}+32a^{3}-56a^{2}}{-8a}$$
Perform the indicated divisions of polynomials by monomials.
$$\frac{14xy-16x^{2}y^{2}-20x^{3}y^{4}}{-xy}$$
Perform the indicated divisions of polynomials by monomials.
$$\frac{-35x^{5}-42x^{3}}{-7x^{2}}$$
$$\frac{13x^{3}-17x^{2}+28x}{-x}$$
$$\frac{x^{2}-7x-78}{x+6}$$
$$\frac{4x^{2} − x − 23}{x-1}$$
Write the polynomial in the form $$p(x) = d(x)q(x) + r(x)$$
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2021-06-13 03:27:27
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http://www.maths.ed.ac.uk/school-of-mathematics/news/2015/?nid=672
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# School Whittaker Colloquium by Edward Witten
The School of Mathematics Whittaker Colloquium will be given by Professor Edward Witten (Institute for Advanced Study, Princeton) at 4PM on Friday, 13th May in the Main Lecture Room of the Swann Building on the King's Buildings campus.
"A New Look At Integrable Spin Systems''
Abstract: Integrable models of statistical mechanics in two dimensions, such as the Ising model, have provided important and surprising examples and have been studied from many points of view by both physicists and mathematicians. In recent years, Kevin Costello has developed a new approach to these models, using a four-dimensional cousin of three-dimensional Chern-Simons gauge theory in which the spectral parameter'' has a geometric explanation. This talk will be primarily an introduction to the work of Costello.
https://en.wikipedia.org/wiki/Edward_Witten
Professor Witten has been elected this year to an Honorary Fellowship of the Royal Society of Edinburgh.
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2018-01-24 09:18:24
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https://stat430.hknguyen.org/files/html/lec14-1.html
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# An Extremely Brief Intro to Machine Learning in Python¶
Ha Khanh Nguyen (hknguyen)
## 1. What is Machine Learning?¶
• "Machine learning (ML) is about learning functions from data." - David Dalpiaz, STAT 432.
• The purpose of this unit in STAT 430 is NOT teaching you machine learning!!
• We focus instead on teaching you HOW to perform machine learning methods in Python.
• To learn the details of machine learning methods (why it works, how it works, etc.), STAT 432 is highly recommended!
• Machine learning methods are divided into groups based on the tasks they accomplish.
• Supervised learning
• Unsupervised learning
### 2.1 Supervised Learning¶
• In supervised learning, we want to “predict” a specific response variable (target or outcome variable).
• Divided into:
• Regression: the response variable is numeric
• Classification: the response variable is categorical
#### 2.1.1 Regression¶
• In the regression task, we want to predict numeric response variables. The non-response variables (also known as features or predictors) can be either categorical or numeric.
• In this example, x1, x2, and x3 are features and y is the response variable.
x1 x2 x3 y
A -0.66 0.48 14.09
A 1.55 0.97 2.92
A -1.19 -0.81 15.00
A 0.15 0.28 9.29
B -1.09 -0.16 17.57
B 1.61 1.94 2.12
B 0.04 1.72 8.92
A 1.31 0.36 4.40
C 0.98 0.30 4.40
C 0.88 -0.39 4.52
#### 2.1.2 Classification¶
• Classification is similar to regression, except it considers categorical response variables.
x1 x2 x3 y
Q -0.66 0.48 B
Q 1.55 0.97 C
Q -1.19 -0.81 B
Q 0.15 0.28 A
P -1.09 -0.16 B
P 1.61 1.94 B
P 0.04 1.72 C
P 1.31 0.36 C
Q 0.98 0.30 B
P 0.88 -0.39 B
### 2.2 Unsupervised Learning¶
• Unsupervised learning is a very broad task that is rather difficult to define. Essentially, it is learning without a response variable. To get a better idea about what unsupervised learning is, consider some specific tasks.
#### 2.2.1 Clustering¶
• Clustering is essentially the task of grouping the observations of a dataset.
#### 2.2.2 Density Estimation¶
• Estimate the density.
## 3. Evaluating the Functions/Models¶
• Model evaluation is a HUGE topic. We will not have the time to cover it in this class.
• One of the most fundamental pieces of model evaluation is training data and testing data.
• In order to learn a function from the data, we need to "build" the model based on some data. This is called training data, the data used to build the model.
• We want to use our model/function to predict/label new observations. But how will we know whether our model is accurate or not? That means we also need some data to test our model! That is called testing data!
• However, testing data cannot be the same as training data. So often, at the beginning of a project, we divide the data into training and testing data.
## 4. First Application: Classifying Iris Species¶
• Dataset: Iris Data Set
• Goal: use the measurements of Iris' petal length/width and sepal length/width to decide if an Iris is Setosa, Versicolor, or Virginica.
• The end-goal is a function (also called model) that can be used to label a new observation.
### 4.1 Setting up your system for ML¶
• Before we can perform any of these cool methods, we first need to install the required packages/libraries:
conda install scikit-learn
• If that doesn't work, try pip install scikit-learn.
### 4.2 Take a look at the data¶
• The first step is always getting to know the data you're working with.
• It's highly recommended that you plot these variables to get an idea of what the data looks like and the variables' association with one another.
### 4.3 Building your first model: k-Nearest Neighbors¶
• There are many classification algorithms in scikit-learn that we could use. We will start with $k$-nearest neighbors classifier (algorithm) since it is very intuitive!
• $k$-nearest neighbor: to make a prediction for a new data point, the algorithm finds $k$ points in the training set that is closest to the new point. Then it assigns the most common label of these $k$ training points to the new data point.
• We will start small with $k = 1$.
• Before we "train" the model on the data, we need to split the data into training and testing data:
### 4.4 Making predictions¶
• Now we can use the model knn to make predictions on new data!
• Note that the predict() function takes an array of observations (not a single observation!).
• Try supply a single observation and you will get an error! (Definitely try it out though!)
• scikit-learn loves NumPy (many things in Python feel the same way). But it works with normal list as well.
### 4.5 Evaluating the model¶
• Using the testing data, we can estimate the accuracy of our model, meaning the percentage of observations our model labels correctly.
• We can also use the score() method of the knn object, which will compute the test set accuracy for us:
References:
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2021-05-14 03:55:46
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https://codegolf.meta.stackexchange.com/questions/2140/sandbox-for-proposed-challenges/17494
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# What is the Sandbox?
This "Sandbox" is a place where Programming Puzzles & Code Golf (PPCG) users can get feedback on prospective challenges they wish to post to the main page. This is useful because writing a clear and fully specified challenge on the first try can be difficult. There is a much better chance of your challenge being well received if you post it in the Sandbox first.
See the Sandbox FAQ for more information on how to use the Sandbox.
## Get the Sandbox Viewer to view the sandbox more easily
To add an inline tag to a proposal use shortcut link syntax with a prefix: [tag:king-of-the-hill]
• Suggestion: instead of having a notice on the top answer ("note: if you are..."), you'd better just put a moderator notice below the question – nicael Mar 19 '18 at 19:35
• @nicael We can only choose from three post notices: citation needed, current event, and insufficient explanation. – Dennis Apr 7 '18 at 14:43
• If you remove a post but didn't post it you can replace the text body with [](lots of text here to reach the min chars) to make it much smaller when removed – Christopher Apr 13 '18 at 17:54
• @Christopher Please don't do that for old proposals. It clutters the first page with an answer nobody cares about anymore, instead of staying hidden on page 10 where it will bother nobody. – Dennis Apr 13 '18 at 18:17
• @Dennis ? what are you talking about. As if if you didn't post it like you just removed you own sandbox because dupe or something – Christopher Apr 13 '18 at 18:18
• @Christopher If your proposal is still on the first few pages, you can replace the proposal with a stub to save vertical space on these pages. However, if your proposal is already on page 10, editing your proposal will bump it to page 1, where space is more precious than on page 10. – Dennis Apr 13 '18 at 18:21
• @Dennis ohh that makes sense – Christopher Apr 13 '18 at 18:25
• codegolf.meta.stackexchange.com/questions/12599/… – Redwolf Programs Apr 17 '18 at 17:38
• Maybe it's time to consider cleaning some of this up a bit. There's just too much to go through and some of these proposals are years old and obviously not going anywhere (even some of the good ones). Perhaps cull anything that is two years old and has likewise been inactive for as long? – ouflak Aug 6 '18 at 9:07
• @ouflak You can sort posts by "active". That seems to resolve all of the problems you describe. – FryAmTheEggman Sep 27 '18 at 19:04
• I already posted this, but codegolf.stackexchange.com/questions/176599/… – 2br-2b Nov 27 '18 at 2:38
• How are tags added to questions? – guest271314 Jan 9 at 7:51
• It seems like there is a rollback war with moderators and the Community user to add and remove the featured tag. – smileycreations15 Mar 21 at 21:13
• @smileycreations15 That's unfortunately unavoidable. Community is an automatic script, and, since most featured questions are only temporarily so, it assumes that we don't want this question to be featured forever. However, we do, so a mod has to edit the tag in every now and then. – Erik the Outgolfer Mar 24 at 15:22
• @EriktheOutgolfer Yeah. Maybe they can create a special [featured-pin] tag which will both feature it and pin it from removal by the Community user. – smileycreations15 Mar 24 at 17:20
# Bees?
Inspired by SCP-3045
Write a program that takes the input, extracts all of the words, and looks for the word bee; then:
• If bee is not detected, pick sections of the text at random and delete them.
• If bee is detected, add instances of the word bee to the input such that it has significantly more bytes than the original input.
The program should then output these modifications.
• How much is significantly more? Why is it popularity-contest? – Laikoni Mar 18 '18 at 14:00
• Do X creatively pop cons have fallen out of scope. This will get closed instantly if posted on main. – Dennis Mar 19 '18 at 12:51
Move a window around the screen
Your code should open a new window that is at least 100 by 100 pixels in size. Once the window is open you should be able to move the window around the screen using the keyboard. The window should move smoothly but it doesn't matter how fast it moves.
• Is there anything else that could make this challenge a bit more interesting? Maybe a scoring method? – RamenChef Mar 26 '18 at 14:01
• @RamenChef I suppose the scoring method was meant to be by the code-golf rules. I could make the challenge more interesting maybe by insisting that you can type into the window? – user9206 Mar 26 '18 at 14:05
• What counts as a "window"? I think this might be quite hard to define objectively in a way which is OS-agnostic. – Peter Taylor Mar 26 '18 at 15:38
# Output a Random Bit
Your task is simple: print either 1 or 0.
Chosen uniformly randomly every time.
No, not your silly pseudorandom nonsense. No system calls. No reading /dev/urandom. The randomness has to be unpredictable (i.e. reliant on chaotic, impossible-to-reasonably-model natural phenomena, and not on some configuration of bits in your computer).
## Specifications
• It is OK to query a site such as random.org for your bit.
• Your program only needs to be runnable once per day (i.e. you can assume there is a 24 hour gap between executions). This is to work around the fact that sites like random.org often have rate-limits.
• If it only has to be run once a day, wouldn't millis() % 2 be truly random? – geokavel Apr 2 '18 at 3:22
• @geokavel No, because you can't assume that the calling actions will be random (e.g. I could always invoke the program at 25-hour intervals, meaning that millis() % 2 would always be a consistent value. – Esolanging Fruit Apr 2 '18 at 4:11
• Is a time cost of maybe read a file in nanoseconds allowed? – l4m2 Apr 2 '18 at 4:42
• In its current form, it appears to be impossible to define the validity criteria objectively. Temporary -1. – user202729 Apr 2 '18 at 6:39
• @user202729 If it were up to you, how would you define them? – Esolanging Fruit Apr 2 '18 at 6:45
• /dev/random seems to be really random. Is it allowed? – someone Apr 2 '18 at 7:30
• @someone Wikipedia says it's a PRNG, and I've heard that system randomness tends to draw entropy from sources like startup times and user actions, so that wouldn't count. – Esolanging Fruit Apr 2 '18 at 7:42
• Would a HRNG such as RdRand work? – someone Apr 2 '18 at 8:24
• ... I admit that my downvote/comment is not constructive, but I found absolutely no way to objectively define the challenge. – user202729 Apr 2 '18 at 14:02
• Maybe define "real random" as "not only based on xxx"(currently last state, calling current) – l4m2 Apr 2 '18 at 15:38
• @l4m2 That was what I was trying to imply by saying it shouldn't be pseudorandom. – Esolanging Fruit Apr 2 '18 at 18:47
• @EsolangingFruit but you need to define what's pseudo – l4m2 Apr 2 '18 at 18:52
# Let's play the too high too - low game!
TL:DR : write a code that plays the too high - too low game
Given this pseudo code function for the too high - too low game, write it in your language of choice. This is just to make the challenge work better across all languages. This code won't count in the final score. You may also change the function's name and any of its variable's name too.
function isRight(number, guess): # where the number is the correct answer and the guess is your code's guess
if guess < number: # if the guess is too low
return 0 # return 0
else if guess > number: # if the guess is too high
return 2 # return 2
else if guess == number: # if the guess is right
return 1 # return 1
else: # if there is an error
return -1 # return -1
# The challenge
Write a code, function, script, etc. that guesses the right number. The range of the "random" number will be between 0 inclusively and 100 exclusively. For the sake of this challenge, the "random" numbers will be the test cases. Note that hard-coding the test cases is banned.
# Scoring
This is how the score will be counted:
bytes = number of bytes in your code
tries = the sum of all the tries used to guess all the test cases
score = bytes + tries
# Rules
• Hard-coding the test cases if forbidden.
# Test cases
[0,2,4,13,19,21,26,33,38,42,48,50,51,56,66,69,74,75,80,89,98,99]
• For one, i'd say the randomness is unfair. If you manipulate the seed python is given, you can just have it output a known sequence. Alongside that, can't you just hardcode the testcase? EDIT: Hardcoding the test case is the only way to get a good score. – moonheart08 Mar 29 '18 at 16:38
• @moonheart08 would banning hardcoding the test cases help? – Dat Mar 29 '18 at 17:57
• "the sum of all the tries used to guess all the test cases" Won't this be the same for all answers (with the only difference being floor vs ceil when taking halve the previous guess (as in 75 & higher could result in a next guess of either 87 or 88).First guess will always be 50. Is it lower, guess 25; is it higher, guess 75. etc. etc. Btw, there are already a few Guess the number challenges: Here is one; and here is another one. – Kevin Cruijssen Apr 3 '18 at 12:54
# ♫ I see a window and I want it painted black ♫
Yes, I know this is a popular mishearing of the lyrics. But instead of a red door, I really do want an (application) window painted black.
Your standalone program should launch an application window at least 400x400 and fill it entirely with black. It doesn't need to be borderless, and it doesn't need to exit gracefully.
Running in a browser is insufficient because there are still elements of the window such as the address-bar and tab-bar that aren't painted black. You must paint the whole window black except for borders added by your window manager.
This is code golf. Standard loopholes apply. Additional challenge is to listen to The Rolling Stones while making your submission.
Here is an un-golfed Java solution:
#compile: javac BlackWindow.java
#run: java BlackWindow
import java.awt.Color;
import java.awt.Frame;
public class BlackWindow{
public static void main(String[] args){
Frame frame = new Frame("no colors anymore");
frame.setsize(400, 400);
frame.setLocationRelativeTo(null);
frame.setBackground(Color.Black);
frame.setvisible(true);
}
}
• What if my platform doesn't support windows that large? – Stephen Leppik Apr 20 '18 at 23:17
• What is the 400x400 measured in? Pixels? Does it qualify if I somehow emulate a screen with larger resolution? – user202729 Apr 21 '18 at 9:23
• Does making the whole screen black count? – user202729 Apr 21 '18 at 9:24
• Stephen then make the whole screen black? What kind of system doesn't support that? – Jared K Apr 22 '18 at 0:16
• user202729 i was thinking pixels – Jared K Apr 22 '18 at 0:16
• What if I am listening to The Feelies cover of the song? Do I get the bonus point? +1 from me for an unusual challenge. – JayCe Jun 11 '18 at 3:34
# Shorter coding in non-golfing language
Copper write a requirement, a sample program in a golfing language, and a required non-golfing language. Rob hack it with the required language, with fewer bytes of code.
I guess it'd be cuz it's sometimes hard to define which is "golfing language". Also is it a duplicate?
• If it's a cops-and-robbers, then it can't be a popularity-contest. I personally don't think this challenge would work out; first of all, it's virtually impossible to outgolf a golfing language using a non-golfing languages because most golfing languages can complete most reasonable tasks in fewer bytes than it takes a non-golfing language to even print Hello World. Also like you said, golfing/non-golfing is extremely difficult to define. I also don't think this challenge would be particularly interesting because you'd likely end up with a bunch of miscellaneous cops posts with all – HyperNeutrino May 2 '18 at 13:09
• sorts of random requirements, which is basically just going to be a bunch of questions that either exist on PPCG already or could be posted to PPCG main as its own challenge, without any robber posts because it would be basically impossible. – HyperNeutrino May 2 '18 at 13:09
• IMO this is well past the threshold of "Too Broad", so I would vote to close for that reason. – Peter Taylor May 2 '18 at 15:23
(Now I don't know what the name should be)
# Intention
I want to create a challenge based on dependent typing, feature that exists in Idris, Coq, Agda and the similiar.
# Text
You should create a function in dependently typed language (Idris, Coq, Agda, etc) so that:
1. The function will receive a string that denotes format.
2. The format string will have s or n, s means it will receive a string, n means it will receive a number. You can assume that there is no other thing in the string
3. Arguments is received in order. If there is type mismatch, the error must be reported on compile-time.
4. After all arguments is received, the function will return a string, that is list of all passed argument
For example
formatf "sn" "goods" 25
> "goods 25"
> Type error in compile time
formatf "ak" "Akangka" 25
> You can do anything.
formatf "nnn" 24 25
> Either type error or return a function expecting a number and return string (currying is almost universal in these languages)
formatf "ss" "Akangka" "Martin Ender" "Adám"
> Type error on compile time
This challenge is similiar to printf-style string formatting, the difference that the function in this challenge has to be type safe.
Note that you cannot use build-in function or macro to do this
# Discussion
1. What should be the name of this challenge?
• Any reason why full programs are not allowed? – user202729 May 2 '18 at 9:29
• What happens if the language is not compiled? – user202729 May 2 '18 at 9:32
• (if you didn't realize, it's not just some languages can't solve it, but in some languages your requirements don't make any sense. There are languages without functions, language with only monadic functions, languages without integers, language without macros, language where macros have different meaning than C #define, language without string (C), etc.) – user202729 May 2 '18 at 9:59
• If the string is possibly not known at compile time, how can it produce a type error at compile time? – Angs May 2 '18 at 10:04
• Personally I think it's a bit too similar to the challenge you linked.. The only difference is validating the input-type with the format.. In which case it would be better to have a challenge dedicated to that, as in: Given this format and a variable amount of other objects, check if the format and types of these objects match. In which case "%s: %i%%", "Percentage", 25 would be truthy, and "%s: %i%%", 123.45, 25 would be falsey. In addition, most languages are type independent, which can change during run-time based on their use.. 10.0 could be all three types in some languages.. – Kevin Cruijssen May 2 '18 at 10:07
• Suggested re-working of the problem: Given a pattern using only %s and %n (for number), slot in the given list of strings and numbers in the given order, but return a distinct value or throw an error if the given list doesn't fit right. – Adám May 2 '18 at 10:08
• @Angs dependent typing. In fact, this challenge is about dependent typing. – Akangka May 2 '18 at 10:17
• @user202729 well, by compile-time, I mean about typechecking time. I specifically disallow dynamic typing, as one of the point of the challenge is to make the program fail to typecheck if %s format is supplied by integer, etc. – Akangka May 2 '18 at 10:21
• @KevinCruijssen Indeed, not all language can do this challenge. After all the intention is on the dependent typing, which most programming language (but not Idris, Coq, etc) lack. – Akangka May 2 '18 at 10:26
• @Adám nice suggestion. But the type-safe feature (i.e. all error is on type-checking time) is integral part of the challenge – Akangka May 2 '18 at 10:28
• @Akangka I don't understand why my suggestion doesn't satisfy that. You get a list of strings and numbers and need to check against each tag in the format that you've been given the right tag. – Adám May 2 '18 at 12:12
• I think you should limit to some languages (perhaps extend the language list if needed), as the challenge does not make sense in other languages anyway. – user202729 May 3 '18 at 1:26
• @Adám I actually implement your suggestion, except the throw an error part. I make the challenge require the result is type error – Akangka May 3 '18 at 2:03
• @Akangka I don't understand why you insist on language specific features like "type errors" and "compile time". Your examples do not show how to format ss, ns, and nn. You mention float dots, but floats are not part of the examples any more. – Adám May 3 '18 at 5:46
• @Adám thanks about float dots. About language specific features, I just want to create a challenge about dependent typing. – Akangka May 3 '18 at 6:50
# Challenge:
Your challenge is to write a quine-like program that takes a string from stdin and gives two outputs: Output A is the input string. Output B is your source code.
# Output Formats:
You can send your outputs to stdout, stderr, and/or files. If A and B go to the same output, they must be separated by a newline. Having a newline at the beginning of your source doesn't count. You'd need to print that newline from your source and then another newline to separate A and B.
# Examples:
source: print($stdin+"\n"+codeThatGeneratesSource) input: Hello, World! ### Both outputs on stdout: Hello, World! print($stdin+"\n"+codeThatGeneratesSource)
### Separate Outputs:
stdout: Hello, World!
stderr: print($stdin+"\n"+codeThatGeneratesSource) ### Standard loopholes are forbidden. ### Submissions should be proper quines except that they produce the additional specified output. • – Beefster Jun 1 '18 at 16:27 • Good find. Not a dupe because that calls for either printing the source or printing the input, not both. Also it requires testing length of input, where this does not. – Jared K Jun 1 '18 at 16:38 • How is this any different from that, though? The concept of printing the source and printing the output is the same, and I can't see how only a small tweak wouldn't be able to port an answer between the challenges. – LyricLy Jun 3 '18 at 5:32 • This is in sandbox to check whether or not to repost the BrainF**k interpreter challenge. The links have been given in credits. • Should comment handling be required. • Also any other improvements are welcome # BrainF**k: BrainF**k is an esoteric programming language designed in the 90s. The reason for its fame is that understanding a program longer than 10 characters in the language is quite hard. Example program : >++++++++[<++++++++>-]<++++++++++++++++.[-] Guess what this does. # Commands: Brainf**k operates on an array of memory cells, also referred to as the tape, each initially set to zero. There is a pointer, initially pointing to the first memory cell. There are a total of eight commands in BF and these are as follows: Command | Purpose > | increment the data pointer (to point to the next cell to the right). < | decrement the data pointer (to point to the next cell to the left). + | increment (increase by one) the byte at the data pointer. - | decrement (decrease by one) the byte at the data pointer. . | output the byte at the data pointer , | one byte of input, storing its value in the byte at the data pointer. [ | if the byte at the data pointer is zero, then instead of moving the instruction pointer forward to the next command, jump it forward to the command after the matching ] command. ] | if the byte at the data pointer is nonzero, then instead of moving the instruction pointer forward to the next command, jump it back to the command after the matching [ command. # Note : + and - operators increment and decrement the bytes at the at the data pointer, note that if the value reaches 255 then upon a + it would become 0. 255 + 1 = 0 Similarly if the value reaches 0 then upon the next - it would become 255. 0 - 1 = 255 # Input: You will be given two strings as input: • The actual BrainF**k code that you are supposed to interpret. • the program input (that will eventually be emptied) to be interpreted as an array of bytes using each character's ASCII code and will be consumed by the , instruction # Example: Program : +[,>,]<. stdin : 11111 # Output: • the output of the interpreted code, if any was produced by the . instruction. # Example: For the above mentioned program, the output should be: output: 1 # Notes: • Both program and stdin will be given as strings. • The output should be a string showing the result after operations. • Given input will always be valid, with a valid BrainF**k program. • In order to avoid confusion, note that you do not need to output the word output as well e.g : output : 1 in this your output should only be 1. (asked by @Picard) # Credits: The question was (more or less) already asked here. This has been reposted since that was 7 years old and was outdated as well. Meta posts on that: # Examples: Program: +[>>>>+++++[-<++>]<[-<++++++++++>]<[-<<->>]<<-[>-<[-]]>+<,]>[>>+>+<<<-]>>>[<<<+>>>-]<<+>[<->[>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]++++++++[<++++++>-]>[<<+>>-]>[<<+>>-]<<]>]<[->>++++++++[<++++++>-]]<[.[-]<]< stdin: Hello, World. This is a program for checking eeeeeeee. Well I have plenty of e output: 14 Program: ++++++++[>++++[>++>+++>+++>+<<<<-]>+>+>->>+[<]<-]>>.>---.+++++++..+++.>>.<-.<.+++.------.--------.>>+.>++. stdin: output: Hello, World # Winning-criteria: This is , so the shortest code in bytes for each language wins. • @AdmBorkBork : So i should put nothing ? – Muhammad Salman Jun 1 '18 at 14:59 • @AdmBorkBork : Thanks, done. – Muhammad Salman Jun 1 '18 at 15:01 • I think some of your notes are too ambiguous to be useful. In particular, "You can have numbers as output where numbers are expected" and "The input will be what it should be" seem to just be... "You are allowed to output numbers if you're allowed to output numbers" and "You can assume that the input is the input" – Kamil Drakari Jun 1 '18 at 15:07 • Also, it would be good to clarify whether the BF commands are the only characters that will be in the program string, or if we are required to handle other characters as comments. – Kamil Drakari Jun 1 '18 at 15:09 • @KamilDrakari : Changed. Hopefully better – Muhammad Salman Jun 1 '18 at 15:09 • @KamilDrakari : I think I will put that as something I would like to know (reasons why this is sandboxed)\ – Muhammad Salman Jun 1 '18 at 15:09 • I don't believe comment handling would be interesting, so I would rather leave it as "you may assume the program input contains no characters other than ><+-.,[]" – Kamil Drakari Jun 1 '18 at 15:32 • @KamilDrakari : Okay – Muhammad Salman Jun 1 '18 at 15:34 • What happens if you move left of the starting position on the tape? Is it undefined behaviour or does it have to work? – wastl Jun 1 '18 at 18:58 • @wastl : undefined behaviour is ok. – Muhammad Salman Jun 1 '18 at 19:00 • 3 things: 1, you need to explain how the memory tape works. You hvaen't really explained that. 2, you should clarify that the output doesn't need to say literally output: 1, you should really allow just 1. 3, don't really bother with comments, it's basically just ignoring other characters. – Rɪᴋᴇʀ Jun 1 '18 at 22:37 • 1. Hopefully done, 2. done, 3. Ok, 2 votes for not having comments. – Muhammad Salman Jun 2 '18 at 9:26 • What if our language has only one input stream? Can we also accept input as the program and input separated by a ! (or some other character)? – Jo King Jun 4 '18 at 2:11 • @JoKing : yes, you can – Muhammad Salman Jun 4 '18 at 7:42 # Longest reference Write two code A and B, where len(A)<=1024, running A returns B and running B returns A. Longest B win. Proper quine rule and no rubbish rule(for code-bowling) apply. Un-used Code All code must be used. Meaning the program must fail to always properly complete the task if any individual character (or varying set(s) of characters) is/are removed. Naturally, a subset of the program should not be able complete the task on its own without the rest of the program. # Sandbox notes • The "1024" may change • Actually we have no standard rule for code-bowling to prevent unused code. – user202729 Jul 5 '18 at 9:52 • (although for this particular challenge, it's not possible to make program arbitrarily long) – user202729 Jul 5 '18 at 9:53 • It's pretty easy to abuse this one. Program A prints program C n times, where program C prints program A and then comments out any further copies of C. Make n as large as you can and it's easy. Good code-bowling challenges usually have more restrictions – Jo King Jul 5 '18 at 10:34 • @JoKing Your solution seems to break the "no rubbish rule" – l4m2 Jul 6 '18 at 1:10 • As user202729 says, there is no standard "rubbish rule". And if there was, there would be many ways of getting around it. – Jo King Jul 6 '18 at 1:25 • Suggestion: Why not have the challenge be to minimise the length of A while maximising the length of B? I'm not sure what the scoring system would be though... – Jo King Jul 7 '18 at 3:25 • @JoKing I really like that idea but I don't think it would solve the problem. The issue is that B can still contain "rubbish", so it becomes a kind of busy-beaver problem for A to print the largest amount of nonfunctional code in B. – Nathaniel Jul 7 '18 at 4:28 • The "unused code" test is unlikely to be practical for programs longer than about 100 characters. – Peter Taylor Jul 7 '18 at 12:04 # Sort on an infinite-dimension cube Given a unit cube in the $n$-dimensional space. Assume that the vertices of the cube has coordinate $(x_0, x_1, x_2, \dots, x_n)$ where $x_i \in \{0,1\} \forall i\in \mathbb N, 0\le i<n$. It's possible to number all vertices with non-negative integers less than $2^n$. In this challenge, the vertex with coordinate $(x_0, x_1, x_2, \dots, x_n)$ will be assigned with number $2^0\times x_0+2^1\times x_1+\dots+2^n\times x_n$. Each vertex can hold an integer. In this challenge, you can assume $n$ contains a very large (practically infinite) value. Given $4096$ items placing in vertex 0 - vertex 4095, you're to sort them. Other vertices contain undefined values, and may be modified by the program. However, the program cannot directly access the values held by the vertices. You can only control a memory pointer M, which always lie at a vertex of the cube (call this vertex V). Initially M is at the coordinate $(0,0,0,\dots)$. This memory pointer can store exactly 1 integer value. The following operations on the memory pointer M are alllowed: • Store the value held by V into memory of M. • Write the value stored by M into V. • Compare value stored by M and value held by V. This operation should report to the program 3 different values based on whether the comparison is $<$, $=$ or $>$. • Move along an edge (in the direction specified by the program) of the cube. This corresponds to changing exactly 1 coordinate of M from 0 to 1, or vice versa. Your score is the distance traveled by the memory pointer. A psuedo-code sample interaction library may be: obj[Infinity] = {[4096 values]} ptr = 0, cry = undefined function move(i): ptr = ptr xor (1 shl i) function carry(): cry = obj[ptr] function place(): obj[ptr] = cry function compare(): return sgn(obj[ptr] - cry) You can write functions, use IO, or anyway to interact. Lowest move callings win. • Actually I think there are just 12 dimensions. – user202729 Jul 12 '18 at 14:55 • @user202729 More dimensions exist and you can use them, but they are initally empty – l4m2 Jul 12 '18 at 14:59 • What does it mean to sort an infinite-dimension cube? Currently this very unclear. – Laikoni Jul 13 '18 at 12:46 • @Laikoni I'd say code shows enough to understand, so it's not ready to post but not unclear – l4m2 Jul 13 '18 at 14:25 • No, sorry, it's not so clear to the rest of us. I have no idea what the input is, what the output is, even what obj we're working with and how it relates to an "infinite dimension cube". Many of your questions here (including this one) seem to contain something interesting in them, but they'd be much better received if you post them on the chat room first and explained what you had in mind, and got some help with the question text, at least to the level that they can be meaningfully discussed on. That way your challenges will reach more people too. – sundar Jul 15 '18 at 10:41 • @sundar I think the sandbox is exactly the place for improving challenges. I'm not sure if using the chat room is necessary. – user202729 Jul 19 '18 at 14:35 • @Laikoni Better now? – user202729 Jul 19 '18 at 15:10 # Golf a regex that matches syntactically valid programs in the language of your choice. 1: Pick a programming language, P, that meets these requirements: • P is known to be Turing-Complete. • P has a freely available and working compiler or interpreter. 2: Create a regular expression, R, such that: • R matches any string that is a syntactically valid program of P. • R rejects any string that is a syntactically invalid program of P 3: Golf R. Shortest regex wins. • For a lot of Esolangs it would just be .*, I think you'd need to restrict the languages to something that doesn't allow any string ALPHABET* or ALPHABET+. Also you'd need to specify a regex flavour. – ბიმო Jul 27 '18 at 20:49 • Warning: Most low-level languages are not known to be TC. For example C (which is only recently proved TC, AFAIK. Ref) – user202729 Jul 28 '18 at 9:21 • Hm. int main(){int x=__builtin_popcount(1);} is not syntactically valid C (undefined identifier), but it compiles in GCC. Also, most compilers don't allow too long identifiers. What do you think? – user202729 Jul 28 '18 at 9:23 • @user202729: I doubt that you'll find a regex for C (or pretty much any non-esoteric, high-level language) anyways since most of the time you need to check if () are balanced. – ბიმო Jul 28 '18 at 15:21 • @OMᗺ Just saying...... // For the first comment, typically the answerer just specify the regex flavor in the answer. – user202729 Jul 28 '18 at 15:23 • Because only those high-level languages have a proper definition of what is a syntax error. The low-level languages are often just "what the interpreter complains about", and there are still different forms of error -- assertion error, runtime, return 1, .... – user202729 Aug 1 '18 at 2:49 # Largest and Smallest Numbers Printable Related: Largest Number Printable Your goal is to write code that produces a large number. However, when your code is reversed, you must output a small number. ### Rules • No constants over 10 (like the other challenge) • No numeric literals • No infinite numbers • Each program can only output one number • You must have at least two bytes in your program. • Your small number must be less than your large number. ### Scoring: Your score is: $\frac{code~length}{N_{large} - N_{small}}$. Smallest score wins. • Ban numeric literals. having the code of n 9's and n 0's in a golving language with auto output results in having a score of $\frac{2n}{(10^n-1)^2}$ which will go to 0 for arbitrary large n – Kroppeb Aug 25 '18 at 11:26 • Even with "no constant over 10" the strategy above will still works in languages such as cat. – user202729 Aug 25 '18 at 14:16 • @kroppeb that's interesting, thanks. – NoOneIsHere Aug 25 '18 at 14:49 • Like the other challenge I'd suggest a maximum code length and putting a higher penalty on code length. Also, are we allowed to print negative numbers? – Jo King Aug 28 '18 at 6:14 ## The Tiniest Generic Evolutionary AI Introduction The smallest program you can make (measured in bytes) that builds evolving AIs that parses unknown text string A into unknown text string B that still evolves. All languages are options. Internet connectivity is allowed (but not providing any specific links). The AI must have a choice of commands (allowing variables) from a Turing Complete instruction set. Scoring Scoring is two fold: N = Number of bytes of program (ignoring size of AIs generated) T = The average generations (generations of 200 AIs or less) before your program can evolve an AI that can do the 5 test cases. Score = 1000/(N*T) Challenge Rules: Using the fewest bytes possible, create a generic evolutionary AI that will evolve to parse one supplied but previously unknown string into another supplied but previously unknown string. What qualifies as an evolutionary AI in this context: Your Code Takes an input string Takes a seed AI Instruction Set(s) Runs AI Instruction Sets Rates Each AI Instruction Set against others and against test strings Evolves AI to create a new set of AI Instruction Set(s) through random mutation and breeding of existing AIs. ================= AI Code Is not written by you (except for possibly a seed AI), to be created by your program instead. Uses a set of *available* instructions that are or are inspired by a known Turing complete instruction set (such as RISC) Should consist of references to the available instructions and values for them. 1. The code you write does not parse the strings directly. Instead, it writes output that is a collection of instructions, and reads in a collection of instructions and applies these instructions. (Ugly as it may be, eval is allowed). 2. It reads in multiple optional instruction sets. 3. It rates those instruction sets based on which gets closest to parsing the input string into the target string. 4. The best performing instructions sets are encouraged in some way. 5. The worst performing instruction sets are discouraged and/or eliminated in some way. 6. Using existing AI-oriented libraries is discouraged. 7. Allowing evolutionary AIs to use develop using a turing-complete set of instructions is encouraged. (Example solution uses a modified RISC instruction set.) 8. Instruction sets must be able to mutate (randomly change) between generations. They are allowed to breed (selectively change) between generations as well. 9. Multiple iterations of comparing and evolving instructions sets is possible. 10. How quickly your AIs evolve or how well they do the job doesn't matter so much as they can get better at the task over iterations. 11. A seed starting instruction set is allowed. If your solution requires an inputed file for an initial instruction set, a functional example is required and counts towards the byte count. 12. It handles if an instruction set it runs fails to complete, or times out if an AI instruction set goes on too long (default to 30 seconds). 13. Program must accept in one arbitrary string from a source. (Your choice of a generic commonly used source, such as a web form, command line, or file). It outputs each AI instruction set's result in a similar format it took them in. It may also optionally accept a starter AI set of instructions. 14. The AI code never gets to interact with the test string it's being graded against. 15. Your program cannot do any string conversions on the input string on its own, only may act as it's instructed to by the AI. 16. A seed AI is not allowed to have anything more than a start, end, or return call of some kind (it must evolve any processing steps on its own). Test Cases: Can the program evolve an AI that approaches being capable of string conversion of an unknown conversion? (It is recommended not to build the AI to these specific test cases, these are for the point of testing genericness. Do not specifically target these cases until reporting results, and others testing your program may test them against other string conversions and rate accordingly.) Five examples follow - abcdefghijklmnopqrstuvwxyz will be converted into zyxwvutsrqponmlkjihgfedcba or possibly 1010101100110101011001 will be converted into 0101010011001010100110 or possibly "Mary had a little lamb." will be converted into "Gary had a little ham." or possibly "Banana" will be converted to "Banananananananananananananananana." Or maybe The entire text of the Bible will be converted to The entire text of the Bible, but every instance of "sheep" is replaced with "codfish". Short Diagram of interaction. STEP 1 STEP 2 STEP 3 ... Input String -> Expected Output String -> Your Code Starts -> Assorted AI Instruction sets. -> -> -> -> Your Code Processes AI Instruction Sets -> Your code compares, mutates, and breeds AIs to create new generation -> Return to Step 3 Seed AI -> (Example: AI1 -> Start(Input); Output(Input) AI2 -> Start(Input); CP(Mem1, Mem2); Output (Mem2);) Example An on-a-whim project done that roughly followed these rules and inspired this challenge that ran <1000 lines of code (although short, it did not aim for minimal characters.) myLittleAI • There are a lot of restrictions and rules, however there is no clear description of the task itself: What does "parse one string into another string" mean? About the rules: Overriding loop-holes such as allowing internet-connectivity is likely to create problems or allow boring answers. "Answer must evolve using options that are or are inspired by a Turing complete instruction set." seems like it's a non-observable requirement (and you override the rule in the second rule anyways). Why do you want to disallow libraries? – ბიმო Sep 22 '18 at 23:01 • Some of the other rules don't make sense to me, but it's likely to be caused by me not being able to understand the challenge itself. – ბიმო Sep 22 '18 at 23:01 • @BMO Reason to disallow libraries is that there are existing libraries that are AI libraries. Allowing them means you just call some monolithic AI library and you're done. I guess it'd make sense to just not allow AI libraries though? – liljoshu Sep 24 '18 at 15:57 • Maybe disallowing AI-libraries could be a sensible idea, but I wouldn't know how to put this (ie. what is an AI-library and what is a regular library) formally. Though the standard (this does not mean that it's always the case or that you need to follow it) is to not disallow such things but rather discourage it. If I'd use a library which is not part of the standard libraries it would count as a different language, eg. Python 3 + scikit and in a perfect world less interesting submissions would be given less upvotes. – ბიმო Sep 24 '18 at 17:00 • @BMO Rewrote it, does this look better? – liljoshu Sep 24 '18 at 18:29 • I think the rules do look more structured and clean, but I still don't understand the challenge itself. Some questions I think the challenge should answer for which I can't find an answer might help: Are you requiring a program, function or a set of functions? What is the input and output format? What makes a solution valid? What is parsing string A to string B? – ბიმო Sep 24 '18 at 22:38 • Also, keep in mind that golfed solutions might try to find some kind of weaknesses in your definitions to reduce the problem to the easiest way of solving it which might result in solutions that aren't very interesting from an AI perspective. This challenge might fit better test-battery (you can set a byte-limit motivated by your own code if you want). – ბიმო Sep 24 '18 at 22:43 • There are 2 critical pieces to this challenge that need to be clear before we can really make this post: 1. What are "Instructions"? 2. When an InstructionSet applies a string, what feedback does the AI get? – Nathan Merrill Sep 25 '18 at 16:37 • Finally, realize that rule #10 simply means I can iterate over every possible program until I land on one that works, no AI needed. – Nathan Merrill Sep 25 '18 at 16:40 • @NathanMerrill Did some clarifying randomness... better? – liljoshu Sep 27 '18 at 18:49 • Not really. Let's chat – Nathan Merrill Sep 27 '18 at 19:10 Rotations Required Given input 2 values: $$\x >= 0\$$ :Distance to travel (float) $$\r>0\$$ :Radius of wheel (float) Output the number of rotations required by the wheel to travel that distance. Constraints: Your code must not contain any digits. You cannot use pi functions (math.pi) Output must be an integer.(In case exact int is not obtained, floor it) 1.0 is not a valid output, it should be 1. Test Cases x r o/p 50 1 7 0 34 0 50 0.3 26 44 33 0 5.5 5.5 0 105 5 3 155 5 4 6.28318 1 1 #This signifies that pi was approximated to 3.1419 Scoring: Score= No. of bytes+20/number of digits taken for pi after decimal point If you have taken more than 10 digits: Score= No. of bytes • "You cannot use pi functions" is an unobservable requirement, which is not allowed. You need some test cases. The no-digits rule won't actually make this problem harder for most golflangs. – Nathan Merrill Oct 16 '18 at 12:43 • @NathanMerrill The latter is less problematic (and more objective) than the former. – user202729 Oct 16 '18 at 14:21 • Test cases are not strictly required, but it would make it easier to check if a solution is definitely incorrect. – user202729 Oct 16 '18 at 14:22 • Info: Adding constraints just "because the challenge is too easy" usually doesn't make it more interesting (unless the constraints are the main difficulty of the challenge, for example for radiation-hardened challenges) – user202729 Oct 16 '18 at 14:24 • I will add test cases. @NathanMerrill, there just shouldn't be any inbuilt functions that give pi directly is what I want to say. – Vedant Kandoi Oct 17 '18 at 5:35 • @VedantKandoi I understand what you want, but we allow any language on this site, and it's impossible to define what a "pi-giving function" is to work for all possible languages. – Nathan Merrill Oct 17 '18 at 6:26 • Is there any other way I could word it then, as I don't want python, java etc. users to use math.pi or should I just remove it? – Vedant Kandoi Oct 17 '18 at 7:15 • Also, since pi is in the formula, exact int will never be obtained, and if anyone uses approximate value of pi, they may get exact int at certain value or 1 less than desired answer. What can I add for this? – Vedant Kandoi Oct 17 '18 at 7:48 • I changed the scoring method for the above issue, so should be okay now. – Vedant Kandoi Oct 17 '18 at 8:15 • Why would the output be floored? Surely you should use ceiling so that the wheel is actually travelling that distance. – Jo King Oct 17 '18 at 8:26 • I was thinking of something like how many full rotations need to be completed. – Vedant Kandoi Oct 17 '18 at 8:31 • "Not using a pi function" is a non-observable program requirement, which is one of the things to avoid when writing a challenge. I don't see any way to fix this challenge: fundamentally what it's asking is too trivial to be interesting. My advice would be to delete the body of this answer and then delete the answer (to keep the sandbox tidy) and then to try to come up with a challenge which is inherently interesting enough that it doesn't need that kind of restriction. – Peter Taylor Oct 17 '18 at 14:06 # Write a Self-Hosting Ouroboros: each quine produces the next quine AND its interpreter My meta-questions, please give feedback and/or add your own: • Too elaborate or long post? • Rules too strict? Too lax? • Not having a deadline: good or bad idea? A Quine is a program that prints its own source code as output when it is run. A Self-Hosting Quine (which is something I just made up, although I'm sure it exists already) is a quine that also produces an interpreter/compiler/emulator/whatever for itself (from now on I will just say "interpreter"). I believe this actually means that a lot of essential functionality must be "circularly defined" - for example, to print output, the interpreter must rely on the parent interpreter's ability to print output. So maybe we should call this a Von Munchausen Quine? An Ouroboros Program or Quine Relay is a quine that prints a different quine, which then prints yet another program, and so on, until the last quine produces the original quine. See this famous example that cycles through over a hundred languages. A Self-Hosting Ouroboros, then, is quine that produces a program in another language, and also produces an interpreter for that language. The interpreter should be in the current language, so that the next quine can immediately be run and produce the quine after that. Tangent: obviously, the idea can also be extended to interpreter multiquines but that can be another challenge. Let's make Von Munchausen's Ouroboros first! ### Rules • The ouroboros must be able to make a complete cycle • Score by total quines in chain / shortest source code (in bytes) in the chain (bytes to allow non-textual outputs) • Empty quines and interpreters do not count - two character minimum • Code-golf languages allowed • No languages defined just for this challenge - no "I define language x to always produce quine y in C plus the complete GCC compiler when fed any input"-stuff please • Quines must be valid code in their programming languages • Interpreters that are partial language implementations are fine, however: • it must be able to run the quine, and produce the next quine (obviously). • it should be able to run any other correct code that is limited to the same language subset¹. No hyperfitting²! • for the sake of code golfing it may accept incorrect code that a normal interpreter should not (the quine is already restricted to correct code anyway) ## Clarifications ### Yes, you can make a quine that produces itself and its own interpreter Counts as a 1-chain ouroboros. ### What would an "interpreter" for machine code be? An emulator. Which of course needs some way to load a program and produce the output. You may define a fictional simplified, minimal hardware set-up to do so. For example: a (partial) Z80 emulator where: • two hardware ports are connected to send output bytes to (one for interpreter, one for quine). Are those bytes interpreted as raw bytes/ASCI/UTF8 text/whatever? Up to you! • the program is already loaded in memory, at whatever address is most convenient (quine too big to fit in the addressable memory space? Define an input port to scan bytes from I guess :P) • the PC (program counter), and SP (stack pointer) registers initiate at whatever value is most convenient for creating a quine in Z80 opcode Obviously, no set-ups with fictional ROM that just happens to contain a new quine, or stuff like that (even though this would be hard to really abuse, since that ROM would still need to be implemented in the emulator). Am I seriously expecting anyone to create an emulator like this? No, but let's keep the possiblity open (some stack machines might be code golf-friendly enough for the challenge). ### "Borrowing" snippets from each other is encouraged, but give accreditation and link to sources! Because we all really just want to see how long this can get, no? Besides, whatever you take probably has to be heavily modified to fit it in your existing ouroboros chain anyway. Sharing is caring, and accreditation is the decent thing to do. ### No dead-line, nor will a winner be selected. Just make as long a chain as possible. ¹ You wrote an interpreter for a subset of Rust, but it doesn't feature the borrow checker? That is fine long as: • it works with correct Rust code limited to the language subset • it works with the quine itself • the quine itself is correct rust code ² Example of what I do and do not consider hyperfitting: if your interpreter can deal with for(var i; i < 10; i++) { a[i] = i; } but crashes without the enclosing {} because it expects them to designate code blocks, that counts as a partial implementation. If var only expects i, or < only expects i and 10? Hyperfitting. If var only accepts single letter names, and < only expects variables on the left and literals on the right? Probably hyperfitting but debatable. • What would the interpreter of a program written in machine code be? – user202729 Oct 19 '18 at 15:14 • How should we separate the quine output and the interpreter output? – Jo King Oct 21 '18 at 0:10 • @user202729: an emulator – Job Oct 21 '18 at 13:01 • @JoKing: hmm, good question. Two separate calls to whatever output method is chosen? (print or its equivalent) – Job Oct 21 '18 at 13:03 # Unicode encoder Do Invent your own Unicode 7.0.0 encoding (as efficient as possible) with score 317754(the lowest possible score). Shortest encode+decode program win. You can either write two programs doing encode and decode, or write one with argument/input method difference deciding whether it's encoding or decoding As you may know, the Unicode standard has room for 1,114,111 code points, and each assigned code point represents a glyph (character, emoji, etc.). Most code points are not yet assigned. Current Unicode implementations take a lot of space to encode all possible code points (UTF-32 takes 4 bytes per code point, UTF-16: 2 to 4 bytes, UTF-8: 1 to 4 bytes, etc.) Task - Today, your task is to implement your own Unicode implementation, with the following rules: - Write an encoder and a decoder in any language of your choice - The encoder's input is a list of code points (as integers) and it outputs a list of bytes (as integers) corresponding to your encoding. - The decoder does the opposite (bytes => code points) - Your implementation has to cover all Unicode 7.0.0 assigned code points - It has to stay backwards-compatible with ASCII, i.e. encode Basic latin characters (U+0000-U+007F) on one byte, with 0 as most significant bit. - Encode all the other assigned code points in any form and any number of bytes you want, as long as there is no ambiguity (i.e. two code points or group of code points can't have the same encoding and vice versa) - Your implementation doesn't have to cover UTF-16 surrogates (code points U+D800-U+DFFF) nor private use areas (U+E000-U+F8FF, U+F0000-U+10FFFF) - Your encoding must be context-independant (i.e. not rely on previously encoded characters) and does NOT require self-synchronization (i.e. each byte doesn't have to infer where it's located in the encoding of a code point, like in UTF-8). To sum up, here are the blocks that you have to cover, in JSON: [ [0x0000,0x007F], // Basic Latin [0x0080,0x00FF], // Latin-1 Supplement [0x0100,0x017F], // Latin Extended-A [0x0180,0x024F], // Latin Extended-B [0x0250,0x02AF], // IPA Extensions [0x02B0,0x02FF], // Spacing Modifier Letters [0x0300,0x036F], // Combining Diacritical Marks [0x0370,0x03FF], // Greek and Coptic [0x0400,0x04FF], // Cyrillic [0x0500,0x052F], // Cyrillic Supplement [0x0530,0x058F], // Armenian [0x0590,0x05FF], // Hebrew [0x0600,0x06FF], // Arabic [0x0700,0x074F], // Syriac [0x0750,0x077F], // Arabic Supplement [0x0780,0x07BF], // Thaana [0x07C0,0x07FF], // NKo [0x0800,0x083F], // Samaritan [0x0840,0x085F], // Mandaic [0x08A0,0x08FF], // Arabic Extended-A [0x0900,0x097F], // Devanagari [0x0980,0x09FF], // Bengali [0x0A00,0x0A7F], // Gurmukhi [0x0A80,0x0AFF], // Gujarati [0x0B00,0x0B7F], // Oriya [0x0B80,0x0BFF], // Tamil [0x0C00,0x0C7F], // Telugu [0x0C80,0x0CFF], // Kannada [0x0D00,0x0D7F], // Malayalam [0x0D80,0x0DFF], // Sinhala [0x0E00,0x0E7F], // Thai [0x0E80,0x0EFF], // Lao [0x0F00,0x0FFF], // Tibetan [0x1000,0x109F], // Myanmar [0x10A0,0x10FF], // Georgian [0x1100,0x11FF], // Hangul Jamo [0x1200,0x137F], // Ethiopic [0x1380,0x139F], // Ethiopic Supplement [0x13A0,0x13FF], // Cherokee [0x1400,0x167F], // Unified Canadian Aboriginal Syllabics [0x1680,0x169F], // Ogham [0x16A0,0x16FF], // Runic [0x1700,0x171F], // Tagalog [0x1720,0x173F], // Hanunoo [0x1740,0x175F], // Buhid [0x1760,0x177F], // Tagbanwa [0x1780,0x17FF], // Khmer [0x1800,0x18AF], // Mongolian [0x18B0,0x18FF], // Unified Canadian Aboriginal Syllabics Extended [0x1900,0x194F], // Limbu [0x1950,0x197F], // Tai Le [0x1980,0x19DF], // New Tai Lue [0x19E0,0x19FF], // Khmer Symbols [0x1A00,0x1A1F], // Buginese [0x1A20,0x1AAF], // Tai Tham [0x1AB0,0x1AFF], // Combining Diacritical Marks Extended [0x1B00,0x1B7F], // Balinese [0x1B80,0x1BBF], // Sundanese [0x1BC0,0x1BFF], // Batak [0x1C00,0x1C4F], // Lepcha [0x1C50,0x1C7F], // Ol Chiki [0x1CC0,0x1CCF], // Sundanese Supplement [0x1CD0,0x1CFF], // Vedic Extensions [0x1D00,0x1D7F], // Phonetic Extensions [0x1D80,0x1DBF], // Phonetic Extensions Supplement [0x1DC0,0x1DFF], // Combining Diacritical Marks Supplement [0x1E00,0x1EFF], // Latin Extended Additional [0x1F00,0x1FFF], // Greek Extended [0x2000,0x206F], // General Punctuation [0x2070,0x209F], // Superscripts and Subscripts [0x20A0,0x20CF], // Currency Symbols [0x20D0,0x20FF], // Combining Diacritical Marks for Symbols [0x2100,0x214F], // Letterlike Symbols [0x2150,0x218F], // Number Forms [0x2190,0x21FF], // Arrows [0x2200,0x22FF], // Mathematical Operators [0x2300,0x23FF], // Miscellaneous Technical [0x2400,0x243F], // Control Pictures [0x2440,0x245F], // Optical Character Recognition [0x2460,0x24FF], // Enclosed Alphanumerics [0x2500,0x257F], // Box Drawing [0x2580,0x259F], // Block Elements [0x25A0,0x25FF], // Geometric Shapes [0x2600,0x26FF], // Miscellaneous Symbols [0x2700,0x27BF], // Dingbats [0x27C0,0x27EF], // Miscellaneous Mathematical Symbols-A [0x27F0,0x27FF], // Supplemental Arrows-A [0x2800,0x28FF], // Braille Patterns [0x2900,0x297F], // Supplemental Arrows-B [0x2980,0x29FF], // Miscellaneous Mathematical Symbols-B [0x2A00,0x2AFF], // Supplemental Mathematical Operators [0x2B00,0x2BFF], // Miscellaneous Symbols and Arrows [0x2C00,0x2C5F], // Glagolitic [0x2C60,0x2C7F], // Latin Extended-C [0x2C80,0x2CFF], // Coptic [0x2D00,0x2D2F], // Georgian Supplement [0x2D30,0x2D7F], // Tifinagh [0x2D80,0x2DDF], // Ethiopic Extended [0x2DE0,0x2DFF], // Cyrillic Extended-A [0x2E00,0x2E7F], // Supplemental Punctuation [0x2E80,0x2EFF], // CJK Radicals Supplement [0x2F00,0x2FDF], // Kangxi Radicals [0x2FF0,0x2FFF], // Ideographic Description Characters [0x3000,0x303F], // CJK Symbols and Punctuation [0x3040,0x309F], // Hiragana [0x30A0,0x30FF], // Katakana [0x3100,0x312F], // Bopomofo [0x3130,0x318F], // Hangul Compatibility Jamo [0x3190,0x319F], // Kanbun [0x31A0,0x31BF], // Bopomofo Extended [0x31C0,0x31EF], // CJK Strokes [0x31F0,0x31FF], // Katakana Phonetic Extensions [0x3200,0x32FF], // Enclosed CJK Letters and Months [0x3300,0x33FF], // CJK Compatibility [0x3400,0x4DBF], // CJK Unified Ideographs Extension A [0x4DC0,0x4DFF], // Yijing Hexagram Symbols [0x4E00,0x9FFF], // CJK Unified Ideographs [0xA000,0xA48F], // Yi Syllables [0xA490,0xA4CF], // Yi Radicals [0xA4D0,0xA4FF], // Lisu [0xA500,0xA63F], // Vai [0xA640,0xA69F], // Cyrillic Extended-B [0xA6A0,0xA6FF], // Bamum [0xA700,0xA71F], // Modifier Tone Letters [0xA720,0xA7FF], // Latin Extended-D [0xA800,0xA82F], // Syloti Nagri [0xA830,0xA83F], // Common Indic Number Forms [0xA840,0xA87F], // Phags-pa [0xA880,0xA8DF], // Saurashtra [0xA8E0,0xA8FF], // Devanagari Extended [0xA900,0xA92F], // Kayah Li [0xA930,0xA95F], // Rejang [0xA960,0xA97F], // Hangul Jamo Extended-A [0xA980,0xA9DF], // Javanese [0xA9E0,0xA9FF], // Myanmar Extended-B [0xAA00,0xAA5F], // Cham [0xAA60,0xAA7F], // Myanmar Extended-A [0xAA80,0xAADF], // Tai Viet [0xAAE0,0xAAFF], // Meetei Mayek Extensions [0xAB00,0xAB2F], // Ethiopic Extended-A [0xAB30,0xAB6F], // Latin Extended-E [0xABC0,0xABFF], // Meetei Mayek [0xAC00,0xD7AF], // Hangul Syllables [0xD7B0,0xD7FF], // Hangul Jamo Extended-B [0xF900,0xFAFF], // CJK Compatibility Ideographs [0xFB00,0xFB4F], // Alphabetic Presentation Forms [0xFB50,0xFDFF], // Arabic Presentation Forms-A [0xFE00,0xFE0F], // Variation Selectors [0xFE10,0xFE1F], // Vertical Forms [0xFE20,0xFE2F], // Combining Half Marks [0xFE30,0xFE4F], // CJK Compatibility Forms [0xFE50,0xFE6F], // Small Form Variants [0xFE70,0xFEFF], // Arabic Presentation Forms-B [0xFF00,0xFFEF], // Halfwidth and Fullwidth Forms [0xFFF0,0xFFFF], // Specials [0x10000,0x1007F], // Linear B Syllabary [0x10080,0x100FF], // Linear B Ideograms [0x10100,0x1013F], // Aegean Numbers [0x10140,0x1018F], // Ancient Greek Numbers [0x10190,0x101CF], // Ancient Symbols [0x101D0,0x101FF], // Phaistos Disc [0x10280,0x1029F], // Lycian [0x102A0,0x102DF], // Carian [0x102E0,0x102FF], // Coptic Epact Numbers [0x10300,0x1032F], // Old Italic [0x10330,0x1034F], // Gothic [0x10350,0x1037F], // Old Permic [0x10380,0x1039F], // Ugaritic [0x103A0,0x103DF], // Old Persian [0x10400,0x1044F], // Deseret [0x10450,0x1047F], // Shavian [0x10480,0x104AF], // Osmanya [0x10500,0x1052F], // Elbasan [0x10530,0x1056F], // Caucasian Albanian [0x10600,0x1077F], // Linear A [0x10800,0x1083F], // Cypriot Syllabary [0x10840,0x1085F], // Imperial Aramaic [0x10860,0x1087F], // Palmyrene [0x10880,0x108AF], // Nabataean [0x10900,0x1091F], // Phoenician [0x10920,0x1093F], // Lydian [0x10980,0x1099F], // Meroitic Hieroglyphs [0x109A0,0x109FF], // Meroitic Cursive [0x10A00,0x10A5F], // Kharoshthi [0x10A60,0x10A7F], // Old South Arabian [0x10A80,0x10A9F], // Old North Arabian [0x10AC0,0x10AFF], // Manichaean [0x10B00,0x10B3F], // Avestan [0x10B40,0x10B5F], // Inscriptional Parthian [0x10B60,0x10B7F], // Inscriptional Pahlavi [0x10B80,0x10BAF], // Psalter Pahlavi [0x10C00,0x10C4F], // Old Turkic [0x10E60,0x10E7F], // Rumi Numeral Symbols [0x11000,0x1107F], // Brahmi [0x11080,0x110CF], // Kaithi [0x110D0,0x110FF], // Sora Sompeng [0x11100,0x1114F], // Chakma [0x11150,0x1117F], // Mahajani [0x11180,0x111DF], // Sharada [0x111E0,0x111FF], // Sinhala Archaic Numbers [0x11200,0x1124F], // Khojki [0x112B0,0x112FF], // Khudawadi [0x11300,0x1137F], // Grantha [0x11480,0x114DF], // Tirhuta [0x11580,0x115FF], // Siddham [0x11600,0x1165F], // Modi [0x11680,0x116CF], // Takri [0x118A0,0x118FF], // Warang Citi [0x11AC0,0x11AFF], // Pau Cin Hau [0x12000,0x123FF], // Cuneiform [0x12400,0x1247F], // Cuneiform Numbers and Punctuation [0x13000,0x1342F], // Egyptian Hieroglyphs [0x16800,0x16A3F], // Bamum Supplement [0x16A40,0x16A6F], // Mro [0x16AD0,0x16AFF], // Bassa Vah [0x16B00,0x16B8F], // Pahawh Hmong [0x16F00,0x16F9F], // Miao [0x1B000,0x1B0FF], // Kana Supplement [0x1BC00,0x1BC9F], // Duployan [0x1BCA0,0x1BCAF], // Shorthand Format Controls [0x1D000,0x1D0FF], // Byzantine Musical Symbols [0x1D100,0x1D1FF], // Musical Symbols [0x1D200,0x1D24F], // Ancient Greek Musical Notation [0x1D300,0x1D35F], // Tai Xuan Jing Symbols [0x1D360,0x1D37F], // Counting Rod Numerals [0x1D400,0x1D7FF], // Mathematical Alphanumeric Symbols [0x1E800,0x1E8DF], // Mende Kikakui [0x1EE00,0x1EEFF], // Arabic Mathematical Alphabetic Symbols [0x1F000,0x1F02F], // Mahjong Tiles [0x1F030,0x1F09F], // Domino Tiles [0x1F0A0,0x1F0FF], // Playing Cards [0x1F100,0x1F1FF], // Enclosed Alphanumeric Supplement [0x1F200,0x1F2FF], // Enclosed Ideographic Supplement [0x1F300,0x1F5FF], // Miscellaneous Symbols and Pictographs [0x1F600,0x1F64F], // Emoticons [0x1F650,0x1F67F], // Ornamental Dingbats [0x1F680,0x1F6FF], // Transport and Map Symbols [0x1F700,0x1F77F], // Alchemical Symbols [0x1F780,0x1F7FF], // Geometric Shapes Extended [0x1F800,0x1F8FF], // Supplemental Arrows-C [0x20000,0x2A6DF], // CJK Unified Ideographs Extension B [0x2A700,0x2B73F], // CJK Unified Ideographs Extension C [0x2B740,0x2B81F], // CJK Unified Ideographs Extension D [0x2F800,0x2FA1F], // CJK Compatibility Ideographs Supplement [0xE0000,0xE007F], // Tags [0xE0100,0xE01EF] // Variation Selectors Supplement ] Total: 116,816 code points. Scoring -- Your score is the number of bytes that your encoder outputs when you feed it with all the 116,816 possible code points (in one time or separately). • I suppose 317754 is the optimal score, right? – user202729 Oct 22 '18 at 9:37 • @user202729 Yes if there are 116,816 code points – l4m2 Oct 22 '18 at 9:38 • p.s. I think you should expand it while it is on the sandbox, just in case somebody don't understand the specification. – user202729 Oct 22 '18 at 9:38 # Forcing a kernel panic from a mountaintop This is a thought resulting from a curious incident where C# code opening file apparently caused a BSOD. In the conclusion, it was due to a faulty driver, but from that a thought came up --- can one cause a kernel panic (or BSOD) using managed code exclusively? ## Why? Typically in such environment, there are many compile-time and run-time checks that safeguard the code from doing something that would cause a kernel panic. For a mature managed environment, it should be impossible to cause the operating system to arrive into a bad state. In the case, though it was C# code, the fact that a faulty driver was involved breached the walled garden. But can we breach it from within? ## Rules • The code must be managed in some fashion (e.g. Java, C#, VB.NET), and normally comes with both compile-time and run-time checks. • The code should be running in a virtual machine or equivalent. (e.g. Java's JVM or .NET's AppDomain) • The code should NOT rely on any external anything. No extern declarations, no networking, no dodgy APIs. • The code should NOT use any construct which allow direct access to resources (e.g. unsafe in C#) • The code should use only the native libraries & API available as part of its usual environment. • Throwing an exception is not sufficient. To qualify, the code must result in a kernel panic. ## Criteria Essentially a code-golf - • The less code to kernel panic, the better • The fewer dependencies the code uses to make it happen, the better • The code that consistently causes a kernel panic is better than one that only does it sometimes • Is the code considered malicious code? – user202729 Nov 10 '18 at 15:26 • What is "managed code"? – user202729 Nov 10 '18 at 15:26 • While it can be used with malicious intent, the goal is more toward proving whether it's possible to unintentionally break through the walled garden. RE: "managed code" --- it might be a .NET-specific term but I use it in general sense to refer to any programming language that usually run in a some kind of sandbox -- I cited Java's JVM or .NET's AppDomain as such examples. Those usually enforce runtime checks in addition to compile-time checks to prevent doing stupid thing like passing a null pointer which usually is a trappable exception. – this Nov 10 '18 at 19:06 # HTML-tac-toe Build a one-player tic-tac-toe game with only HTML and CSS. # Introduction You can do just about anything with a fully-featured programming language, but how much can you accomplish on a Neopets petpage? Inspired by http://www.neopets.com/~vuh # Challenge Build a one-player tic-tac-toe game with only HTML and CSS. • Use no more than one file. • GIFs (including animations), PNGs, and JPEGs are allowed. • Flash, embedded scripts, iframes, and JavaScript are not allowed. • It must work in the at least two standard browsers (FF, Chrome, Safari, Opera, IE) Inputs: The player will click on a space when it's their turn to move there. Outputs: The page will show the current board state at all times. The page will play an optimal move whenever the player moves, unless the game is over. You're free to decide who starts and who has X or O. # Scores • -10 per image • -1 per opening bracket or brace (< / { ) Brownie points for design, flair, and new tricks! # Example Input and Output Input: Click on the top left of an empty grid Output: Grid shows my mark where I clicked and the opposite mark in the left middle or top middle. • I'd recommend against restricting languages, and against score bonuses too – ASCII-only Jan 22 at 4:15 • @ASCII-only I can drop the bonuses without much consequence, but it's not a new challenge without a language or host restriction. Any suggestions? – Qaz Jan 22 at 4:45 • True, but... language restrictions are pretty frowned upon – ASCII-only Jan 22 at 4:47 • Welcome to PPCG! You may want to check this post out, it certainly helped me when starting with writing my own challenges. Also, your post does not have a well-defined objective winning criterion. It seems like it is atomic code-golf or code-challenge [1/2] – ბიმო Jan 22 at 19:32 • , but you will need to define how answers are scored (in this case lower is worse) and how ties are treated. Also why do you disallow GIFs but JPGs not? I know most will know, but maybe explain or at the very least link to somewhere where the rules for tic-tac-toe are explained. What counts as an X, what as an O? There are a lot of open question atm. [2/2] – ბიმო Jan 22 at 19:36 • JPGs are allowed. (That's the same format as JPEG, just a different extension.) I can change it to 'images', but I'm worried about some image format I don't know of that makes the challenge trivial. More points, more better! That seems clear enough, but I can certainly spell it out. Are ties forbidden? Two entries with the same number of points seem equally good to me, but the first posted could be the winner. If I link to tic-tac-toe rules, does that answer "What counts as an X, what as an O?" or are you asking something else? – Qaz Jan 23 at 0:13 • "Standard exceptions are allowed IF they work on Neopets.com." How do we know what works on Neopets.com? If we have to create accounts on a random third party site to test answers and know whether they're valid or not, the question should and probably will end up closed with a lot of downvotes. – Peter Taylor Jan 24 at 15:34 • This isn't very interesting in general. It essentially amounts to creating every possible layout of a tic-tac-toe board and linking them together with clever CSS and HTML hacks so it fits in 25 pages. I think it could be made a bit more interesting by making it atomic code golf scoring on the total number of html files used, lowest scores win rather than limiting the total number of pages. – Beefster Jan 25 at 20:36 • @PeterTaylor very well, dropped that altogether – Qaz Jan 25 at 22:53 • @Beefster The page I link to as the inspiration (neopets.com/~vuh) uses only one html file, and the source isn't too hard to understand, so perhaps I should link to the creator of the page instead, limit the solutions to one page, or both. – Qaz Jan 25 at 22:53 • I highly suggest looking through current codegolf questions to see which are well-received by the community. This question seems highly arbitrary and leaves too much under-specified. – qwr Jan 27 at 20:28 # Bit flipper Given a string s and a positive number n, return the string s with n random bits flipped. (A random number can be generated in any way, including pseudo-random number generators) Example: Before: Hello World After (n = 2, 2 bits flipped): Hello wOrld • How exact are the bytes being 'flipped'? Are we changing the bits? Is the output deterministic? It doesn't look like you're modifying any specific bit of the byte, or reversing it, or doing bitwise negation. – Rɪᴋᴇʀ Mar 21 at 1:55 • @Riker well changing from uppercase to lowercase is just xor 32. But they'rem saying byte flipper not bit flipper which is a bit confusing – ASCII-only Mar 21 at 6:09 • @Riker The bits are flipped – smileycreations15 Mar 21 at 9:42 • @JoKing No, that was an example because I didn’t wanted binary non-unicode characters in the post. – smileycreations15 Mar 21 at 9:44 • What is n? On the basis of what little is specified in the question, I would be tempted to write an implementation for n=0 which consists of the empty program in GolfScript... – Peter Taylor Mar 21 at 16:44 • @PeterTaylor A user defined count of how many bytes will be flipped – smileycreations15 Mar 21 at 18:12 • I think the confusion here is that "bytes" doesn't mean what you think it means. From the lone test case I think what you're trying to ask us is: given a string s and a number n, change the case of n random letters in s - would that be correct? – Shaggy Mar 22 at 0:41 • I assume case was just an example of bit 5 flipping. I'd recommend writing it like Given a string s and a positive number n, return the string s with n random bits flipped. Though from there you run into problems about invalid unicode sequences in the output – Jo King Mar 22 at 3:45 • @JoKing I replaced with your example. – smileycreations15 Mar 22 at 7:47 • What do you mean by Default n = 3? Also, how should programs handle invalid unicode sequences? – Jo King Mar 22 at 9:14 • @JoKing By default, n should be 3. And invalid unicode sequences should not be handled, and the data should be directly printed out to stdout or any output file. – smileycreations15 Mar 22 at 12:35 • When you say random, do you mean of our choice? pseudo-random? fetched from random.org? – Artemis Fowl Mar 26 at 23:16 • @ArtemisFowl It is selected by the program. – smileycreations15 Mar 27 at 10:45 • @smileycreations15 It's gonna be hard to make that 100% random – Artemis Fowl Mar 27 at 15:46 • @smileycreations15 Maybe you should add that to the question. – Artemis Fowl Mar 27 at 20:35 ## Code-challenge: Guess my number ### The challenge You have a number from 1 to 10 in mind, and your program should ask questions to find out which number. These questions can be any questions, the program only has to find out the number as fast as possible. Your program should ask a question, such as "Is the number a prime?", and the user must answer either y or n (yes or no). Ask questions until you know the number. ### The scoring To calculate the score, you need to take the sum of the question count for each number. For example, if you need 1 question to find the number 1, 2 questions to find the number 2, and so on, the score is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10, so the score is 55. Important note: the question count for a specific number must always be the same. For example, if you need 4 questions to find out the number 10, then you have to ask always 4 questions to find out the number 10, otherwise it is impossible to calculate the score. • boooring. The Huffman tree for a uniform set is any perfectly balanced tree. The question asks us to perform a binary search on the usr device. Is the number greater than 5? Is the number greater than 2? Is the number greater than 1? Hey' I think it's 1. – John Dvorak Jan 2 '14 at 11:49 • Maybe if this were a pop-contest and the goal was to make the most original set of questions while still keeping the score at its theoretical minimum. – John Dvorak Jan 3 '14 at 5:28 Bovine Ignorance I'm curious about code which still works after being mangled by figlet, toilet, cowsay et al, but I'm not sure whether this in any way sane. What I'm toying with is a challenge in which a participant may submit any program in any language. It should be possible to use this program's source code as input to cowsay or whatever, and the result should be another valid program in any language, which still does a similar thing. For instance, the following bf program prints Hello world! with no newline: +++++ +++++ [ > +++++ ++ > +++++ +++++ > +++ > + <<<< - ] > ++ . > + . +++++ ++ . . +++ . > ++ . << +++++ +++++ +++++ . > . +++ . ----- - . ----- --- . > + . +++++++++++++++++++++++++++++++++++++++++ Running cat ./prog.bf | cowsay -e .. -T$'>.' yields the following output:
_________________________________________
/ +++++ +++++ [ > +++++ ++ > +++++ +++++ \
| > +++ > + <<<< - ] > ++ . > + . +++++ |
| ++ . . +++ . > ++ . << +++++ +++++ |
| +++++ . > . +++ . ----- - . ----- --- . |
| > + . |
| +++++++++++++++++++++++++++++++++++++++ |
\ ++ /
-----------------------------------------
\ ^__^
\ (..)\_______
(__)\ )\/\
>. ||----w |
|| ||
Which is itself a valid bf program which prints Hello world!!!, followed by a newline.
The problem with using bf here is that it ignores most of the cow, making this a bit too easy. The problem with using any other language is that it doesn't ignore most of the cow, making this far too difficult. Is there a sensible middle ground I could pick for this? I don't think it's impossible, I'm fairly sure you can exploit cowsay's behavior on one-liners to produce valid svgs, but I'm not sure how best to pose this challenge. Any ideas?
• I could not think of any language that falls in the middle ground. Even brainfuck is affected by the -----------------------------------------..>.---- inserted by cowsay. Most languages have strong parsing rules that would not cope with being post-processed by cowsay. The few exceptions for this will be either completely unaffected or badly affected, making the challenge uninteresting. – Victor Stafusa Feb 19 '14 at 12:32
• Actually, you can't transform just any brainfuck program to cowsay-brainfuck. Namely those that can output fewer than three characters cannot be transformed at all. – John Dvorak Feb 19 '14 at 14:52
• @JanDvorak, I was intending to allow competitors to choose the parameters of their calls to cowsay. For the uninitiated, -e controls the string used for eyes and defaults to oo, and -T controls the string used for the tongue, defaulting to U. This is all yak-shaving, though, and having written this up and read the comments, I suspect that this idea has neither legs, horns nor udders. – ymbirtt Feb 19 '14 at 23:19
• If I could propose a variant that is more feasible, you could do a challenge like "Write a program in your language of choice that draws ASCII art of a cow saying something (does not have to be identical or even similar to the cowsay art). The entire drawing must itself be valid source code that does something other than no-op. Post results of both programs." That gives people more leeway to work around the specific restrictions of their compiler. – Jonathan Van Matre Feb 21 '14 at 23:22
• Ok, I found a language that falls within the middle ground: whitespace. Anyway, this question has a too narrow scope to develop an interesting challenge. – Victor Stafusa Feb 22 '14 at 18:31
• @JonathanVanMatre That would be a subjective validity criterion, and would probably be closed as too broad. – wastl Jul 2 '18 at 13:55
# Create an Identicon Generator
The challenge is to create an identicon generator. The identicons must be randomly generated, so we get a new identicon for each key the program receives. You can input a key using std-in or you can use your language's random number generator for the key.
In order to make your identicon look reasonably nice, it must generate a picture, then rotate that picture around the bottom right corner, the way this mockup shows:
The output must be to a PNG file. Shortest code wins.
• Far too broad. As this stands I can create a 1-pixel image whose colour is just the key. I don't think this question will be ready to go until you've found a way to prevent me from making the images differ only in their palette (and to pre-empt, I think that adding a rule "Images may not differ only in their palette" isn't a real fix). – Peter Taylor Mar 28 '14 at 14:50
• If you just ask for "random" images, you'll get images that are either hardly random at all (a solitary pixel in a random location), or completely random (noise). To get something "reasonably nice", you'll have to provide very clear instructions on how to produce these images. I suggest you try creating a few of these yourself, and find a minimal set of rules that produces results that look OK. Include requirements on dimensions (100x100px?), selection of colours (at least 2, not too similar), and drawing method (e.g., "five triangles with random vertices and a minimum area of 20 px²"). – squeamish ossifrage Mar 28 '14 at 15:25
• How important is the PNG file output? This will be a challenge in itself for many languages. Would you accept an uncompressed non-interlaced format like PPM? – trichoplax Apr 16 '14 at 9:45
## Underhand Bejewled
Help me to write a game of bejewled, which cannot be lost!
## Bejewled game rules
If you ever played bejewled, you can skip this, but for those who did not see it ever:
• Playing field of 8*8 grid is filled in with gems of 7 different types randomly
• By swapping two adjective stones, your goal is to create a line of at least three same type of stones in the either vertical or horizontal line
• If did so, the gems will dissappear, points are added (say 20 points for a matching) and new gems are provided randomly from the top
• image related:
Provide me a game which cannot be lost. In other words, the gems falling from the top are not random at all, but are falling in order that there is always at least one possibility to match three gems
But, from looking at the code at level of newbie programmer, it should look like that game acts as if it was random
## Output
Playable game. As long as it is the grid of 8*8 filled in with 7 different types of "gems" the game is ok. It does not to have killer graphics, neither it does not need to be playable by mouse. (But in that case please make sure you show which "gem" is hovered and then selected)
## Winning criteria
This is popularity contest. So highest rated game wins
• I think this is too big a task to work well for an underhanded contest. The programs will be way too large for anyone to actually read the source and try to find what's underhanded about it. – Martin Ender Nov 11 '14 at 8:32
• Thats what I was also afraid of. I will either take it as lesson to progress on my programming skill, or abandon the idea completly – Pavel Janicek Nov 11 '14 at 8:38
This and this gave me an idea, but I'm not quite sure if this can be done at all, or if it is trivial. If it is, maybe point out how it could be changed to be interesting.
# Anti golfing - Write the longest program not repeating any character
Well, it's just what the title says. Finally you're allowed to use as much bytes as possible.
## Conditions
• The code of the program or function should not use any character that is used in the code before.
• Your program should print some sort of result to stdout, or into a file or return a value. You're not allowed to output or return the empty string or only a newline.
• Other than that your program might do anything. Read input, print lots of output, or what you can think of, but you have to explain what it does, of course.
• Only characters in the ASCII range [32 .. 126] and newlines are allowed, which limits the maximal code length to 96 bytes.
• Variable names are only allowed to consist of a single character
• String literals or the like are forbidden. They could be used to hold the unused characters (though they would need two " in most languages anyway).
• The same rule applies for similar literal constructs like blocks or what else is there in some languages.
• Even if the length of a string literal would be used to generate a number, it is forbidden.
• Variables can not just be declared and never be used. They have to be reflected in the output somehow.
• If you've read and understood the above rules and still found a loophole and used it, you should go and stand in the corner for a while, thinking about what you've done.
So all in all, only use characters for actual code that does something generating the output, might it be calculating a value or formatting. And don't put unused characters somewhere in your code as a literal. Numbers are an exception, but I guess it's no problem to use them anyway.
I guess you should have a pretty good idea of what would be considered cheating here.
Example in awk
BEGIN{gsub(a,9);print $j-13+d^c/4*5678%20} It prints 15.5, score is 42. It replaces the empty string a with 9 in $0, which is the empty string in the beginning. So $0 becomes 9. Then it prints the result of 9-13+1/4*5678%20. ($j is $0 (==9), because j is not defined d^c ist 1, because c and d are not defined) Please don't invent languages for this ;) The longest code in bytes wins. • Are you sure you want to allow ASCII 127? That's the unprintable<DEL> character. The main problem with this challenge is "only use characters for actual code that does something". This is essentially unenforceable, because there may be arbitrarily complicated no-ops in the code. It's also why most code-bowling challenges fail to be popular/interesting. – Martin Ender Sep 14 '15 at 7:32 • Well, I thought about making it a "most votes win" challenge, but I guess that would be unfair for less known users. I don't know what could be done with what you are pointing out. – Cabbie407 Sep 14 '15 at 7:51 • I don't think this is a good candidate for a popularity contest. Popularity contests shouldn't be used as a cop out if the actual spec is a bit vague. They work best for challenges where the actual scoring criterion can be well specified but is more easily judged by humans than machines (e.g. "visually approximate a given image with these constraints..."). – Martin Ender Sep 14 '15 at 7:54 • Yeah, it's hard to formulate the rules for this. But I think it's not always about finding a winner anyway. Thought this might be fun. Resolved the character 127 situation btw.. – Cabbie407 Sep 14 '15 at 7:57 • How could I change that rule? I'm thinking about "only use code that contributes to the generation of the output" – Cabbie407 Sep 14 '15 at 8:02 • How do you define "contribute"? E.g. this GolfScript program prints the length of the block in {...} which is a convenient way to stuff all characters except in '"# in there. Do all those random characters actually contribute? In Slashes everything which isn't an unescaped slash is printed to STDOUT, so as long as I put \/ together, I can put any characters I want there and they'll all contribute. – Martin Ender Sep 14 '15 at 8:07 • Hmm, I thought this would be covered by forbidding string literals.. might think about extending that rule to blocks. Well, I'm not that fluent at esolangs. – Cabbie407 Sep 14 '15 at 8:10 • It's trivial to use all possible 96 bytes. Trust me. If you really want to see the program I'm thinking of, I suppose I could write it, but I'm pretty sure it can be done. – mbomb007 Sep 16 '15 at 18:34 • Yeah, I guess you're right. i have no idea how it would be done, but alright. – Cabbie407 Sep 16 '15 at 20:10 • Not to mention this is pretty much a duplicate of codegolf.stackexchange.com/questions/30159/… – pppery Aug 6 '17 at 12:52 # Linear Time Sorting It was another slow day at Initech Inc. when a feature request came in: New Feature: Ability to sort by cash value in the transaction form. But make it a fast one! Well it looks simple.. but what do the requester mean by fast one? Let's call Jim, from sales he probably knows what's going on. Jim: Well you know , our Business Inc. contact is very passionate about programming and computer science! In fact he had this idea that we should do sorting in how that was called.. linear time? You: Well you know that's impossible? Jim: But it was already approved by their cto and all! You need to do something You and Jim came up with a plan.. nobody will notice if that big of a list isn't sorted enough, right? # Your task Your task is to write a linear time sorting program. It will be scored on accuracy of the sort as compared to list sorted by regular sorting algorithm but it must work on O(N) time in the worst case, where N is length of the input. Input will be in the form of list of string-double tuples, e.g: [("aaaa",2.0) , ("aaba",1.0)] The program should sort on the number value of the tuple, i.e. in the above case the order should be reversed. There may be multiple inputs with the same double values, but no string value is repeated. In the event that two inputs have the same integer value the perfect solution is to keep the order as it is. The double value may be any floating-point value that fits in 8 byte double precision variable. NaNs should be placed at the end of the list. The score is calculated as number of "bubble sort operations" (switch an element with the next/previous element) needed to achieve perfect output from the output of your algorithm. # Sandbox Worries Well I don't know how clear my explanation of challenge was and if it is interesting to the PPCG crowd. Obviously there is a need for testing program and test cases. • How big will the test cases be? If you pick a fixed size, the response will be "n passes of bubble sort where n is the size of the largest test". Bam, linear with perfect score. – John Dvorak Apr 11 '16 at 10:12 • @JanDvorak I rephrased the scoring sentence to more reflect what I meant. In that case the score wouldn't be perfect as after just n switches the list wouldn't be the most ordered. EDIT: I think I understood it now. Well I think you can somehow exclude answers like that with some caveats in the rules, like "Your algorithm cannot make any assumtions about length of the input" – Lause Apr 11 '16 at 10:17 • I don't make any assumptions. It sorts every array up until the largest test case correctly and all other arrays partially. – John Dvorak Apr 11 '16 at 10:28 • An algorithm that fares better than n-pass bubble sort is to do a level-n mergesort. If the length exceeds 2^n, sort each [k::len/2^n] subarray separately. – John Dvorak Apr 11 '16 at 10:32 • But then you're making assumptions based on the size of the test cases - if somehow the test cases were changed (but still fitting the rules) to test cases which are much longer (for example you prepared for max 10 element list and you get 100000 element list) your program isn't linear. – Lause Apr 11 '16 at 10:41 • The problem with a spec is that it cannot change once you've posted the challenge, and you can't define "making assumptions based on the size of the test cases". You can't even ban all magic numbers - I can simply use functions merge1 .. merge20 – John Dvorak Apr 11 '16 at 10:47 • It isn't the spec - the way the test cases used to grade the result are constructed is described but do I have to post the test cases (but those used to score) themselves? – Lause Apr 11 '16 at 10:50 • You need to define the test cases, and you can't change them based on the answers (if only because updating the score of every answer would be a nuisance). Maybe you could ask for asymptotic behavior, but that can be surprisingly hard to measure. – John Dvorak Apr 11 '16 at 10:53 • Hidden test cases are a problem as well, because then we can't test the submissions after you're gone. – John Dvorak Apr 11 '16 at 10:55 • The idea was to pregenerate some test cases (undisclosed) and some test cases that are disclosed (for testing purposes during the coding) and then do a cutoff time for the challenge where all the solutions are tested against the undisclosed challenges. Also obviously after the cutoff time the test cases would be disclosed. – Lause Apr 11 '16 at 10:56 • 1. Most real-life data types can be sorted in linear time, so the premise of the question seems badly flawed. 2. "Input will be in the form of list of string-double tuples ... There may be multiple inputs with the same integer values" Huh? Where do the integer values come from? 3. "The double value may be any floating-point value" Where do NaNs sort? – Peter Taylor Apr 11 '16 at 16:00 • 1. The idea is that the data may be the worst case for any algorithm that can achieve linear sorting time 2. It was a typo 3. At the end - I will put it into the question – Lause Apr 11 '16 at 18:36 • Why does the input being worst case make any difference? The solution will still be perfect, so you'll need a tie-breaker to separate every single answer. – Peter Taylor Apr 11 '16 at 21:27 # Alphabetization 101 (popularity contest) Your task is to use all 52 letters of both the uppercase and lowercase alphabet, ONCE and ONCE only, and make a program. You are free to use any other ASCII character more than once, or use a letter of the alphabet more than once if it's required for the language to function. ## Meta: • Not sure if this has been done before. • Any questions regarding the task? • Not really meta: Is there any place I can go to (like a chat or something) to post a question about BF? StackOverflow probably isn't suitable. – Qwerp-Derp Apr 17 '16 at 6:48 • Come to our chatroom! :) – Leaky Nun Apr 17 '16 at 6:50 • I would vote to close this as too broad. It's not a particularly interesting restriction per se, and it certainly doesn't make a good question without some restriction on the task to be performed. – Peter Taylor Apr 17 '16 at 14:07 • @PeterTaylor That's why it's a popularity contest, though - it lets the people decide whether the program made is good or not. What WOULD be a good restriction on the task? – Qwerp-Derp Apr 18 '16 at 1:40 • The popularity-contest tag is not an excuse for a broad challenge. "Write a program that does anything..." is pretty much the definition of "too broad", regardless of any source code restriction put on the program. So at least you should choose a specific task. Could be anything really, but if it relates to the restriction it might be more interesting (e.g. a pangram checker). Even so, I agree with Peter that the restriction isn't particularly interesting. There are tons of languages where it's trivial to avoid unwanted letters and then include the remaining ones in a string or comment. – Martin Ender Apr 21 '16 at 7:04 # Why did I come to Sandbox? I have a very specific challenge, and I wanted to see if it was too specific. The challenge is to output "Valdosta ACM" using the shortest number of characters with the BrainF**k programming language. I've noticed it isn't the norm to specify a programming language on this domain, so I've come here to get feedback on whether or not this is acceptable. # Introduction As a challenge to the members of my local Association for Computing Machinery(ACM) chapter, I asked them to produce the shortest Brainf**k code that would output "Valdosta ACM". This was a very fun challenge for all of our members, and we got very competitive! I was impressed with the solutions turned in, but I wondered if it was possible to beat our best solution. Surely it's possible, but who could do it? # Challenge Output the string "Valdosta ACM". Stipulations: • Use only the Brainf**k programming language (you can test your code here) • No input can be accepted by your program • Your program must halt • The space in the string must be ASCII character #032. These are the ASCII values of each character, as they appear in the string, for convenience: 086 097 108 100 111 115 116 097 032 065 067 077 The winner is determined by the shortest code, by character count. # Example Input and Output ### Input: NO input is allowed ### Output: Valdosta ACM • Welcome to Programming Puzzles & Code Golf! Thanks for using the Sandbox. :) A few things to note: 1.) Generally we discourage language-specific challenges, 2.) typically code golf is scored by bytes rather than characters, and 3.) printing a fixed string like this would be insufficiently different from the Hello, World! challenge to avoid it being closed as a duplicate. – Alex A. Apr 25 '16 at 3:38 • Thanks Alex! Since I want to compare the results of my local competition with the results of the challenge here, is there anything I could change about the challenge to make it acceptable? I don't see a way to do this, but I was so excited about seeing if anyone here could do better than our coders. And thanks for the warm welcome! :) – Matt C Apr 25 '16 at 3:48 • You could look at Brainf**k solutions to other challenges (like this one), and see if the techniques used there can help you improve your solution. – ugoren Apr 25 '16 at 7:09 • We also have a tips question that may be of interest. – trichoplax Apr 26 '16 at 6:30 • Although this particular challenge is probably too similar to "Hello, World!" (as Alex pointed out), if you had a different challenge that you wanted to see solutions for in a specific language, you can still post it but just allow all languages to compete. If you don't see solutions in your specific language you can post a bounty for that language to encourage it. – trichoplax Apr 26 '16 at 6:33 ## Cops and robbers : Programmers/Hackers • This challenge is quite different from my previous challenges. This challenge is an endless competition between robbers and cops, which are respectively hackers and programmers. One of them will ever win!!! • This will evolve to code de/obfuscating when it gets to the higher stages: a skillful programmer who is struggling to save his program from a sourcecode-mangling attempted by a cunning "robber" who tries to impose his existence by patching his name instead of the name of the "programmer" in the output console without changing anything else in the code. The story begins this way: • Programmer is at the point of executing his recently made C code, so he included this trivial line to show off: C (1) printf("[Programmer's username]") After executing this program programmer saw this on the screen: [Robber's username] which indicates the presence of some evil code at the compiler level that compromises his code, which follows: Matlab (2)/parser a=findstr(code,'printf(''[Programmer's username]'')'); if a code(a:20)='printf(''[Robber's username]'')';end The programmer cannot modify the counter-program in the compiler, so he must rather change the program content to escape the twiddling: PHP (3) $a='[programmer's username]';echo $a; The score is now 3, which is the number of steps from the beginning. The current user would win only if the hacker did not figure out something like: PHP/Regex(pcre flavor) (4) $code=ereg_replace("(\$\w)\='programmer';(.*?);echo\s\1","\1\='robber';\2;echo\s\1",$code)
Since the solution above does not satisfy the rules (see the bottom of this question), the score stays unchanged, and the programmer can make a counter example, and take out the score from last submitter with a penalty on his score equivalent of how much he earned in the earlier level, where the counter example can be something as:
PHP (4)
$a='programmer';$b=$a;$a='unrelated';echo $a; Or he can adjust his program in higher scale to escape all the regex-trapping in a superior range, So the cycle goes on until no post can be added and the last submitter before the end of June is declared a potential winner meanwhile. The hacker can also fix his regex and regain his score, so the recent scoring will be abrogated from programmer. Perl/dynamic-regex (4) local @a=(''); sub check{ if (grep {$_ eq @_[1]} @a)
{push @a,@_[0]; }
elsif (grep {$_ eq @_[0]} @a) { my @del_indexes = grep { @a[$_] eq @_[0] } 0..$#a; foreach$item (@del_indexes) {
splice (@a,$item,1); } } return 1; } sub actor{ if (grep {$_ eq @_[0]} @a)
{return "print robber";}
else
{return "print ".@_[0];}
}
sub initiate{
push(@a,@_[0]);
return 1;
}
$code =~ s/(((\w+)\="programmer"(??{initiate($3);}))|(print\s(\w+))|((\w+)\=(\w+)(?{check(($7),($8));})(?1)))/print($2);actor($5)/pegmx;
As you can see this Perl program prints b in the first case because the variable b is compromised after the first assignment, but in the second case the regex modifies the output because d receives the target-string transitively. Let's just stop here and not mess the fun (of course, if there will be some).
## Scoring and rules
How is the score counted ?
• Any hacker/programmer is scored for his code as the actual level L the game is on.
• A partial dynamic regex within the core of the program is scored L + (2^L)/log(length of program + length of characters which do not belong to the regex)), where the log is base 2. For the second example of level (4) the length of the compacted program is 480, and the length of regex is 136, so the score is 4+2^4/log2(480+480-136) ~= 4+16/9.6
• A fully functional regex as in the first example level (4) is scored L + (2^L)/log(length of regex), where the log is base 2, in that case S = 4 + 2^4 / log(91) ~= 4+16/6.5
• Scores are added progressively to submitters, and when a level is surpassed with no regex, it is still open for scores, while the actual winner remains unchanged.
• A penalty on a certain-leveled score when the regex/parser is revealed out of rules and the game is regressed to this stage until the issue is fixed, rules are cited below:
Rules:
• The main rule: the hacker-program must compromise an output to the console, which is the username of the programmer. Any other behavior is unaccepted simply because a string variable of [programmer's username] can be used in other order rather than printing, a counter-example is easy, converting the string to integer then use it for arithmetic calculations that harms the main program once intentionally modified.
• Also one of the following factors declared by any counter-example bans the targeted flawed regex/parser as non rule-complying:
• The regex/parser prints anything other than a chosen string preferably set as the username of the robber.
• The regex/parser generates a program which does not compile.
• The regex/parser does not print anything, or compromises a segment of code that is needed for tasks other than printing .
• The variable which stores the program is named code by default, also you may assume that is one-liner, and any non-significant spaces are omitted, and that it is fully working by default.
• The regex/parser deals with one variant of one code proportion in a comprehensive way, i.e. if a print function is used, that encompasses all printing functions in all languages puts,disp,..etc. Also, code separators can be unified to one characterL either , or ; or a significant space needlessly of enumerating all keywords/syntaxes, this is not a contest about a working code in a specific programming language.
• To prevent endless program/regex loops let's just not making a jokey sequence as a='programmer';print a / /(\w)\='programmer';print\s\1/ / a='programmer';b=a;print b / /(\w)\='programmer';(\w)\=\1;print\s\2/ because the first person who makes a regex/parser which palliates to a same replicated idea will take out all attributed scores to this idea from their owners, so any anaphoric sequences like this in addition that they are set to same level, they are unneeded.
• Any language that uses pointers/addresses/classes like C++ are welcome, as long as they help to evade the hacker.
• Please, for the love of god, spell things correctly. In the first bit alone I spotted a ton of spelling mistakes without even looking for them. Also, that whole first list is... basically impossible to understand, at least for me. Maybe use full sentences? – Nic Hartley May 10 '16 at 21:29
• Have you seen our cops-and-robbers challenges? It sounds like that is what you are trying to do here. That said, there are a couple of problems with the spec: Defining what parts of the language counts as a "partial regex" or "full regex" is really tough, especially when we get into esoteric languages. – Nathan Merrill May 10 '16 at 21:36
• Could you add a short summary to the post? I don't understand what the actual task here is. Is this a cops-and-robbers or answer-chaining challenge, or something entirely different? – Zgarb May 10 '16 at 21:36
• i will see what cops and robbers is – Abr001am May 10 '16 at 21:39
• @NathanMerrill this is not a code golf so i dont see the point of introducing esolangs here – Abr001am May 10 '16 at 22:28
• @Agawa001 Esoteric languages are still useful outside of golfing. You can use them to make it tough for regexes to match. – Nathan Merrill May 10 '16 at 23:06
• The introduction is very long and after reading it I have no idea what the task is. I would have to vote to close this as "Unclear what you're asking" in its current state. – Peter Taylor May 11 '16 at 7:45
• So what's the core mechanic? Is this an answer-chaining question where answers must alternate programmer and hacker? But if the programmer can change language at will, how can the hacker hope to win? – Peter Taylor May 12 '16 at 12:06
• @PeterTaylor yes it is answer chaining but the last submitter can post two consecutive answers and be the robber and cop themselves, the programmer change his code, hacker changes his regex taken consideration of all last regex/parsers. – Abr001am May 12 '16 at 16:13
• I have no idea what this challenge is supposed to be. The very little explanation of the concept is muddled by spelling and grammar issues. Please, learn English spelling and grammar before trying to write a challenge. – Mego May 13 '16 at 5:18
• @PeterTaylor refer at the 4th rule, procedures which accomplishes a specific task in different languages are dealt as one thing, this is not a challenge about checking language-syntaxes, when a programmer changes language, consider all previous regex/parsers changed to trap same functionnalities of previous code on the new language. – Abr001am May 28 '16 at 11:37
## Challenge
Write a program that takes an numerical input n and outputs the nth number that is not a perfect square.
## Rules
This is , so least bytes wins.
• What's the maximum expected input? Does it expect 0? How do we handle 0? Is there a requirement on the efficiency for large inputs? Also give some example inputs and outputs. – Patrick Roberts Jun 17 '16 at 20:19
• Here's some test cases I just generated: 1->2,2->3,3->5,4->6,5->7,6->8,7->10,8->11,9->12,10->13,11->14,12->15,13->17,14->18,15->19,16->20,17->21,18->22,19->23,20->24,21->26,22->27,23->28,24->29,25->30,26->31,27->32,28->33,29->34,30->35,31->37,32->38,33->39,34->40,35->41,36->42,37->43,38->44,39->45,40->46,41->47,42->48,43->50,44->51,45->52,46->53,47->54,48->55,49->56,50->57,51->58,52->59,53->60,54->61,55->62,56->63,57->65,58->66,59->67,60->68,61->69,62->70,63->71,64->72,65->73,66->74,67->75,68->76,69->77,70->78,71->79,72->80 Is this the function you expect? – Patrick Roberts Jun 17 '16 at 20:41
• Yes, yes it is. – weatherman115 Jun 17 '16 at 20:42
• Can you address my other questions please? Namely, the largest expected input and how to handle input of 0. – Patrick Roberts Jun 17 '16 at 20:43
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2019-05-23 21:47:54
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http://astronomy.stackexchange.com/questions/4793/if-the-universe-is-infinite-why-isnt-it-of-infinite-density
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# If the Universe is infinite, why isn't it of infinite density?
If we make the assumption that the Universe is infinite, and has an infinite number of hydrogen atoms, then why is it not of infinite density - because, under Schrodinger's wave equation the probability of an electron being at any given point is non-zero and any non-zero number multiplied by infinity is itself infinity?
Is the answer (a) I have made some basic error in physics, (b) the Universe is provably not infinite because of this - effectively a version of Obler's Paradox or (c) the Pauli exclusion principle means that electrons just cannot be anywhere?
-
Having infinite number of hydrogen atoms is not a direct consequence of an infinite Universe. You must revise your hypothesis. – Py-ser Jul 2 at 11:35
Who said it was a "direct consequence"? – adrianmcmenamin Jul 2 at 15:31
It seemed so from your question. Could you refer to the source which gave you these information? – Py-ser Jul 3 at 3:36
Your assumption is not true. For example: if the probability density function of the atoms decreases quadratic exponentially (i.e. $e^{-r^2}$), then their sum will be finite even in an infinite universe.
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2014-12-20 10:25:40
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http://scholar.cnki.net/result.aspx?q=%27Nusselt+number%27
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高级检索
作者：Arianna Parrales , José Alfredo Hernández-Pérez , Oliver Flores ... 来源：[J].Entropy(IF 1.347), 2019, Vol.21 (7)DOAJ 摘要：In this study, two empirical correlations of the Nusselt number, based on two artificial neural networks (ANN), were developed to determine the heat transfer coefficients for each section of a vertical helical double-pipe evaporator with water as the working fluid. Each ANN was...
作者：DEEPAK KUMAR , AMIT KUMAR DHIMAN 来源：[J].Sādhanā(IF 0.393), 2017, Vol.42 (6), pp.941-961Springer 摘要：... The ranges of dimensionless control parameters considered are Prandtl number (Pr) = 10–100, Reynolds number (Re) = 1–150 and gap ratio ( G ) = 0.25–1. The steady-flow regime is observed up to Re = 121 for G = 0.5, and beyond this Re, time-periodic regime is observed....
作者：Mohammad Reza Tavakoli , Omid Ali Akbari , Anoushiravan Mohammadian ... 来源：[J].Journal of Thermal Analysis and Calorimetry(IF 1.982), 2019, Vol.135 (2), pp.1119-1134Springer 摘要：... The present research has been carried out in Reynolds numbers of 150–1000 and Ag nanoparticles volume fractions of 0–4% by applying slip and no-slip boundary conditions. Also, in order to estimate the heat transfer behavior and the computational fluid dynamics, two-phase...
作者：Asif Afzal , A. D. Mohammed Samee , R. K. Abdul Razak ... 来源：[J].Journal of Thermal Analysis and Calorimetry(IF 1.982), 2019, Vol.135 (3), pp.1797-1811Springer 摘要：... Forced laminar flow of coolant and steady-state analysis with operating parameters like heat generation term ( $$\bar{S}_{\text{q}}$$ ), Reynolds number ( Re ), conduction–convection parameter ( ζ cc), and aspect ratio ( Ar ) is analyzed with main focus of \(\b...
作者：Xavier Nicolas , Eric Chénier , Chahinez Tchekiken ... 来源：[J].International Journal of Thermal Sciences(IF 2.47), 2018, Vol.134, pp.565-584Elsevier 摘要：Abstract(#br)This paper deals with the modeling of weakly rarefied and dilute gas flows in micro channels by the continuum approach, valid for Knudsen numbers smaller than about 0.1. It particularly focuses on the modeling of the associated heat transfer. The models proposed in t...
作者：Shanmugam Rajasekaran , William Christ Raj 来源：[J].Thermal Science(IF 0.838), 2016, Vol.20 (suppl. 4), pp.929-935DOAJ 摘要：... The various Nusselt number correlations aredeveloped by considering that the water as a working fluid. The mainobjective of the present work is to design the experimental set-up for asingle phase fluid flow using plate heat exchanger and studied the heattransfer performance. ...
作者：Lilian Govone , Mohsen Torabi , Linwei Wang ... 来源：[J].Journal of Thermal Analysis and Calorimetry(IF 1.982), 2019, Vol.135 (1), pp.45-59Springer 摘要：... Temperature and species concentration fields as well as Nusselt number for the hot wall are reported versus various parameters such as porosity, radiation parameter and volumetric concentration of nanoparticles. The results show that radiative heat transfer imparts noticeable...
作者：Satyanand Abraham , Rajendra P. Vedula 来源：[J].International Journal of Thermal Sciences(IF 2.47), 2018, Vol.124, pp.407-423Elsevier 摘要：Abstract(#br)Local effectiveness and Nusselt number distribution measurements for heated rectangular jets impinging perpendicularly and obliquely on a cylindrical convex surface are reported here. The jets exit from a rectangular nozzle of height ‘H’ and width ‘W’ with a...
作者：Elena-Roxana Popescu , Sébastien Tanguy , Catherine Colin 来源：[J].International Journal of Thermal Sciences(IF 2.47), 2019, Vol.140, pp.397-412Elsevier 摘要：... By performing a full set of simulations sweeping the parameters space, correlations are proposed for the first time on the Nusselt number depending on the dimensionless numbers (the Reynolds number, the Prandtl number, the Jakob number and the density ratio) that characterize...
作者：Ali Mostafazade Abolmaali , Hossein Afshin 来源：[J].International Journal of Thermal Sciences(IF 2.47), 2019, Vol.139, pp.105-117Elsevier 摘要：... Geometrical configuration of spiral-wound heat exchangers can be completely determined by knowing six primary parameters including start factor, tube outside diameter, number of tubes in the first layer, number of layers, longitudinal pitch, and radial pitch. Dividing the lon...
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2020-02-29 11:00:03
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https://elteoremadecuales.com/paley-wiener-theorem/
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# Paley–Wiener theorem
Paley–Wiener theorem In mathematics, a Paley–Wiener theorem is any theorem that relates decay properties of a function or distribution at infinity with analyticity of its Fourier transform. The theorem is named for Raymond Paley (1907–1933) and Norbert Wiener (1894–1964). The original theorems did not use the language of distributions, and instead applied to square-integrable functions. The first such theorem using distributions was due to Laurent Schwartz. These theorems heavily rely on the triangle inequality (to interchange the absolute value and integration).
Contents 1 Holomorphic Fourier transforms 2 Schwartz's Paley–Wiener theorem 3 Notes 4 References Holomorphic Fourier transforms The classical Paley–Wiener theorems make use of the holomorphic Fourier transform on classes of square-integrable functions supported on the real line. Formally, the idea is to take the integral defining the (inverse) Fourier transform {displaystyle f(zeta )=int _{-infty }^{infty }F(x)e^{ixzeta },dx} and allow ζ to be a complex number in the upper half-plane. One may then expect to differentiate under the integral in order to verify that the Cauchy–Riemann equations hold, and thus that f defines an analytic function. However, this integral may not be well-defined, even for F in L2(R) — indeed, since ζ is in the upper half plane, the modulus of eixζ grows exponentially as {displaystyle xrightarrow -infty } — so differentiation under the integral sign is out of the question. One must impose further restrictions on F in order to ensure that this integral is well-defined.
The first such restriction is that F be supported on R+: that is, F ∈ L2(R+). The Paley–Wiener theorem now asserts the following:[1] The holomorphic Fourier transform of F, defined by {displaystyle f(zeta )=int _{0}^{infty }F(x)e^{ixzeta },dx} for ζ in the upper half-plane is a holomorphic function. Moreover, by Plancherel's theorem, one has {displaystyle int _{-infty }^{infty }left|f(xi +ieta )right|^{2},dxi leq int _{0}^{infty }|F(x)|^{2},dx} and by dominated convergence, {displaystyle lim _{eta to 0^{+}}int _{-infty }^{infty }left|f(xi +ieta )-f(xi )right|^{2},dxi =0.} Conversely, if f is a holomorphic function in the upper half-plane satisfying {displaystyle sup _{eta >0}int _{-infty }^{infty }left|f(xi +ieta )right|^{2},dxi =C
Si quieres conocer otros artículos parecidos a Paley–Wiener theorem puedes visitar la categoría Generalized functions.
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2023-04-01 15:00:02
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https://projecteuler.chat/viewtopic.php?f=15&t=1521
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## Inverse Fourier Transform help
Mechanics, discrete, statistics, ...
sfabriz
Posts: 175
Joined: Thu Apr 06, 2006 12:18 am
Location: London - UK
### Inverse Fourier Transform help
Hi guys, I need some help with a inverse fourier transform:
Given s(t) a signal and S(f) its fourier transform, we define:
S(f) = Integrate[-inf, +inf, s(t) * exp(-j2pi * f * t) dt] = fourier transform of s(t)
and
s(t) = Integrate[-inf, +inf, S(f) * exp(j2pi * f * t) df] = inverse fourier transform of S(f)
Consider a low pass filter, as the one that is depicted in fig. 1 at this page: http://en.wikipedia.org/wiki/Low-pass_filter.
To solve the filter we put into a system these 2 equations (Z are impedances; Vi, Vo are voltages and I is the current):
Vi - (Zr + Zc)*I = 0
Vi - Zr*I = Vc = Vo
Zr = R, Zc = 1/(sC) with s = j2pi * f
They give this frequency response: G(f) = Vo/Vi = 1 / (1 + sRC) = 1 / (1 + j2pi * f * RC)
Now, if we call a = 1/(RC) we get this equation to describe the lpf:
G(f) = Vo/Vi = a/(a + j2pi * f)
Now, to get the inpulsive response of this filter, I have to do the inverse fourier transform of the G(f) function.
My book gives me this result: g(t) = a * 1(t) * exp(-at), with a=1/(RC) and 1(t) being the heaviside step function, which is 1 for t>0 and 0 for t<0.
So, I really can't get to that result on my own (maybe I'm dumber in may...), but I can prove it is correct going from g(t) to G(f), in fact:
G(f) = Integrate[-inf, +inf, g(t)] = a * Integrate[0, +inf, exp[-(j2pi * f + a)t] dt] =
=[-a/(j2pi * f + a)]* exp[-(j2pi * f + a)t]{t,0,+inf} and this gives exactly:
a/(j2pi * f + a) and when we substitute a with 1/(RC) we get back to: G(f) = 1/(1 + j2pi * f * RC)
So, since the fourier transformation is unique, there must be a method to get from G(f) to g(t) making an inverse tranformation.
Any of you happens to know it?
Thanks in advance for any help,
cheers,
sfabriz
daniel.is.fischer
Posts: 2400
Joined: Sun Sep 02, 2007 11:15 pm
Location: Bremen, Germany
### Re: Inverse Fourier Transform help
1. Look it up (Bronštein/Semendyayev, Gradštein/Ryzhik or similar).
2. For functions like this, the Residue Theorem comes in handy.
R and C are real, aren't they?, so a = RC is real, too (anyway, a must have nonzero real part, otherwise the function wouldn't be square integrable), and [frac]a,a+2[pi]*i*f[/frac] (sorry, I can't write the imaginary unit as j) has a simple pole at [frac]ia,2[pi][/frac] with residue [frac]-ia,2[pi][/frac].
Let's call the integrand of the inverse Fourier integral K, so K(t,f) = [frac]a,a+2[pi]*i*f[/frac]*exp(2[pi]i*t*f)
Now consider the two cases t < 0 and t > 0 separately.
a) t < 0:
We have |exp(2[pi]i*t*f)| = exp(-2[pi]*t*Im(f)), which decreases nicely for Im(f) → -∞. So integrating along the boundary of a large enough rectangle [-R, R, R-iS, -R-iS], by the Residue Theorem, this integral is -2[pi]*i*[sum](residues of K(t,_) in the lower half plane). The integrals of K(t,_) over the vertical edges and the edge below the real axis vanish for R, S → ∞, leaving
∫-∞∞K(t,f) df = -2[pi]i*(sum of residues).
Assuming a > 0, the only pole is in the upper half plane, sum of residues is 0.
b) t > 0:
Now we consider K(t,_) in the upper half plane, take the rectangle [-R, R, R+iS, -R+iS]. Now the boundary is positively oriented, so the integral is 2[pi]*i*(sum of enclosed residues). One pole is enclosed, the residue is [frac]-ia,2[pi][/frac]*exp(2[pi]i*t*[frac]ia,2[pi][/frac]) = [frac]-ia,2[pi][/frac]*exp(-t*a), giving the overall result
∫-∞∞K(t,f) df = a*exp(-t*a) for t > 0.
Patching both cases together, s(t) = a*H(t)*exp(-ta).
3. Isn't this kind of problem more of a case for the Laplace transform (where the method of 2. always works)?
Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.
sfabriz
Posts: 175
Joined: Thu Apr 06, 2006 12:18 am
Location: London - UK
### Re: Inverse Fourier Transform help
Ah, very nice.
Yes, this kind of problem usually gets done with the Laplace transform when you get the transfer function of a system, but in this case it is done with the fourier transform because it is a study of a simple low-pass filter and we're not considering the transfer function but the frequency response of the system, of which we calc other details after the impulse response. I'm a little bit rusty with these things and couldn't recall how to get that.
Thank you very much Daniel,
sfabriz
sfabriz
Posts: 175
Joined: Thu Apr 06, 2006 12:18 am
Location: London - UK
### Re: Inverse Fourier Transform help
Oh, by the way, I don't like the j notation too for the imaginary unit, but here we've got a convention: if you're doing math you use i, if you're doing other things, you use j. That's because "i" could also mean a generic current that you use when you analyze a circuit and it is easy to get confused when you do complex analysis along with current equations.
So, usually in uni books you find j on electronic books and i in pure math books.
This fourier problem I asked for, is inside a complex analysis of a circuit that is used as a blackbox filter on a wider signal transmission system chain, and therefore all equations use j instead of i, and I didn't even think about that when I wrote the post.
Thanks again,
sfabriz
daniel.is.fischer
Posts: 2400
Joined: Sun Sep 02, 2007 11:15 pm
Location: Bremen, Germany
### Re: Inverse Fourier Transform help
sfabriz wrote:Oh, by the way, I don't like the j notation too for the imaginary unit, but here we've got a convention: if you're doing math you use i, if you're doing other things, you use j. That's because "i" could also mean a generic current that you use when you analyze a circuit and it is easy to get confused when you do complex analysis along with current equations.
I know. And hereabouts it's an old custom that mathematicians annoy electric engineers by using i and electric engineers annoy mathematicians by using j, each side pretending that they're utterly shocked by the other's convention
Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.
sfabriz
Posts: 175
Joined: Thu Apr 06, 2006 12:18 am
Location: London - UK
### Re: Inverse Fourier Transform help
Ahahahah, yeah I guess it's truly like that!
I feel like a hybrid though, since being an i.t. engineer I'm midway between electric engineers and mathematicians, with the result that I neither know math as mathematicians nor electronic as electric engineers!
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2020-05-31 22:35:04
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https://www.semanticscholar.org/paper/Phases-of-flavor-neutrino-masses-and-CP-violation-Yasu%C3%A8/71b3dfa1b11ed5309af5efc6d5acc75822f4214f
|
# Phases of flavor neutrino masses and CP-violation
@article{Yasu2012PhasesOF,
title={Phases of flavor neutrino masses and CP-violation},
author={Masaki Yasu{\e}},
journal={Physics Letters B},
year={2012},
volume={718},
pages={153-159}
}`
• M. Yasuè
• Published 10 July 2012
• Physics
• Physics Letters B
1 Citations
## Figures from this paper
Generalized scaling ansatz and minimal seesaw mechanism
Generalized scaling in flavor neutrino masses $M_{ij}$ ($i,j$=$e,\mu,\tau$) expressed in terms of $\theta_{SC}$ and the atmospheric neutrino mixing angle $\theta_{23}$ is defined by
## References
SHOWING 1-10 OF 82 REFERENCES
Indication of electron neutrino appearance from an accelerator-produced off-axis muon neutrino beam.
• Physics
Physical review letters
• 2011
The T2K experiment observes indications of ν (μ) → ν(e) appearance in data accumulated with 1.43×10(20) protons on target, and under this hypothesis, the probability to observe six or more candidate events is 7×10(-3), equivalent to 2.5σ significance.
Where we are on θ13: addendum to 'Global neutrino data and recent reactor fluxes: status of three-flavor oscillation parameters' (vol 13, 109401, 2011)
• Physics
• 2011
In this addendum to Schwetz et al (2011 New J. Phys.13 063004), we consider the recent results from long-baseline νμ → νe searches at the Tokai to Kamioka (T2K) and Main Injector Neutrino Oscillation
On the connection of leptogenesis with low energy CP violation and lepton flavor violating charged lepton decays
• Physics
• 2003
Assuming only a hierarchical structure of the heavy Majorana neutrino masses and of the neutrino Dirac mass matrix mD of the see–saw mechanism, we find that in order to produce the observed baryon
Measurement of the rate of ve + d → p + p + e- interactions produced by 8B solar neutrinos at the sudbury neutrino observatory
Solar neutrinos from the decay of 8B have been detected at the Sudbury Neutrino Observatory (SNO) via the charged current (CC) reaction on deuterium and by the elastic scattering (ES) of electrons.
Improved search for Muon-neutrino to electron-neutrino oscillations in MINOS.
• Physics
Physical review letters
• 2011
The results of a search for ν(e) appearance in a ν (μ) beam in the MINOS long-baseline neutrino experiment find that 2 sin(2) (θ(23))sin(2)(2θ (13))<0.12 at 90% confidence level for δ = 0 and the normal (inverted) neutrinos mass hierarchy.
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2022-07-04 01:01:02
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https://questions.examside.com/past-years/jee/question/calculate-molality-of-na-if-92-gm-na-is-dissolved-in-1-kg-jee-main-chemistry-some-basic-concepts-of-chemistry-lqxt28fhatuz0di9
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1
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg$$-$$1 is :
A
12
B
4
C
8
D
16
2
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are :
(Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $$\times$$ 1023 mol-1)
A
6.023 $$\times$$ 1020
B
6.023 $$\times$$ 109
C
6.023 $$\times$$ 1021
D
6.023 $$\times$$ 1023
3
JEE Main 2018 (Offline)
+4
-1
The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is
A
C2H4O3
B
C3H6O3
C
C2H4O
D
C3H4O2
4
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
A sample of $$NaCl{O_3}$$ is converted by heat to $$NaCl$$ with a loss of $$0.16$$ $$g$$ of oxygen. The residue is dissolved in water and precipitated as $$AgCl.$$ The mass of $$AgCl$$ (in $$g$$) obtained will be : (Given : Molar mass of $$AgCl=143.5$$ $$g$$ $$mo{l^{ - 1}}$$)
A
$$0.35$$
B
$$0.41$$
C
$$0.48$$
D
$$0.54$$
JEE Main Subjects
Physics
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Optics
Modern Physics
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2023-03-21 18:42:02
|
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http://mymathforum.com/real-analysis/28131-exponential-map-matrix.html
|
My Math Forum exponential map of matrix
Real Analysis Real Analysis Math Forum
June 8th, 2012, 06:22 AM #1 Newbie Joined: Jun 2012 Posts: 3 Thanks: 0 exponential map of matrix I don't understand why the exponential map of matrix is a continous mapping of Mn(C) into itself. I need an exact proof! Thanks
June 9th, 2012, 03:35 AM #2 Member Joined: Jul 2011 From: Trieste but ever Naples in my heart! Italy, UE. Posts: 62 Thanks: 0 You can strat to prove that every matrix in $M_n(\mathbb{C})$ define a continuous function into itself! P.S.: I hope that I have write right.
June 10th, 2012, 01:23 AM #3 Member Joined: Mar 2011 Posts: 49 Thanks: 5 Re: exponential map of matrix Hint: You can use : $||e^{A+H} - e^A|| \leq ||H|| e^{||H||} e^{||A||}$ forall matrix $A$ and $H$
Tags exponential, map, matrix
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2019-02-24 04:51:27
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http://www.thefullwiki.org/Zero
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# Zero: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
(Redirected to 0 (number) article)
0 is both a number and the numerical digit used to represent that number in numerals. It plays a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems. In the English language, 0 may be called zero, oh, null, nil, "o" or nought, dependent on dialect and context.[1]
## As number
0 is the integer preceding 1. In most systems, 0 was identified before the idea of negative things that go lower than zero was accepted. Zero is an even number.[2] 0 is neither positive nor negative. By some definitions 0 is also a natural number, and then the only natural number not to be positive.
Zero is a number which quantifies a count or an amount of null size. Almost all historians omit the year zero from the proleptic Gregorian and Julian calendars, but astronomers include it in these same calendars. However, the phrase Year Zero may be used to describe any event considered so significant that it serves as a new base point in time.
## As digit
The modern numerical digit 0 is usually written as a circle, an ellipse, or a rounded rectangle. In most modern typefaces, the height of the 0 character is the same as the other digits. However, in typefaces with text figures, the character is often shorter (x-height).
On the seven-segment displays of calculators, watches, and household appliances, 0 is usually written with six line segments, though on some historical calculator models it was written with four line segments.
The value, or number, zero is not the same as the digit zero, used in numeral systems using positional notation. Successive positions of digits have higher weights, so inside a numeral the digit zero is used to skip a position and give appropriate weights to the preceding and following digits. A zero digit is not always necessary in a positional number system, for example, in the number 02.
In rare instances, a leading 0 may distinguish a number. This appears in roulette in the United States, where '00' is distinct from '0' (a wager on '0' will not win if the ball lands in '00', and vice versa). Sports where competitors are numbered follow this as well; a stock car numbered '07' would be considered distinct from one numbered '7'. This is most common with single-digit numbers.
### Distinguishing the digit 0 from the letter O
Traditionally, many print typefaces made the capital letter O more rounded than narrower, elliptical digit 0.[3] Typewriters originally made no distinction in shape between O and 0; some models did not even have a separate key for the digit 0. The distinction came into prominence on modern character displays.[3]
The digit 0 with a dot in the centre seems to have originated as an option on IBM 3270 displays. Its appearance has continued with the Microsoft Windows typeface Andalé Mono. One variation used a short vertical bar instead of the dot. This could be confused with the Greek letter Theta on a badly focused display, but in practice there was no confusion because theta was not (then) a displayable character and very little used anyway.
An alternative, the slashed zero (looking similar to the letter O except for the slash), was primarily used in hand-written coding sheets before transcription to punched cards or tape, and is also used in old-style ASCII graphic sets descended from the default typewheel on the ASR-33 Teletype. This form is similar to the symbol $\emptyset$, or "∅" (Unicode character U+2205), representing the empty set, as well as to the letter Ø used in several Scandinavian languages. Some Burroughs/Unisys equipment displays a digit 0 with a reversed slash.
The opposing convention that has the letter O with a slash and the digit 0 without was advocated by SHARE, a prominent IBM user group,[3] and recommended by IBM for writing FORTRAN programs,[4] and by a few other early mainframe makers; this is even more problematic for Scandinavians because it means two of their letters collide. Others advocated the opposite convention,[3] including IBM for writing Algol programs.[4] Another convention used on some early line printers left digit 0 unornamented but added a tail or hook to the capital O so that it resembled an inverted Q or cursive capital letter-O ($\mathcal O$).[3]
Some fonts designed for use with computers made one of the capital-O–digit-0 pair more rounded and the other more angular (closer to a rectangle). The Texas Instruments TI-99/4A computer featured a more angular capital O and a more rounded digit 0, whereas others made the choice the other way around.
German license plate with slit zeros
The typeface used on most European vehicle registration plates distinguishes the two symbols partially in this manner (having a more rectangular or wider shape for the capital O than the digit 0), but in several countries a further distinction is made by slitting open the digit 0 on the upper right side (as in German plates using the fälschungserschwerende Schrift, "harder-to-falsify script").
Sometimes the digit 0 is used either exclusively, or not at all, to avoid confusion altogether. For example, confirmation numbers used by Southwest Airlines use only the capital letters O and I instead of the digits 0 and 1, while Canadian postal codes use only the digits 1 and 0 and never the capital letters O and I, although letters and numbers always alternate.
## Names
In 976 Muhammad ibn Musa al-Khwarizmi, in his Keys of the Sciences, remarked that if, in a calculation, no number appears in the place of tens, a little circle should be used "to keep the rows." This circle the Arabs called sifr.[5]
The word "zero" came via French zéro from Venetian zero, which (together with cipher) came via Italian zefiro from Arabic صفر, ṣafira = "it was empty", ṣifr = "zero", "nothing".[6]
Italian zefiro already meant "west wind" from Latin and Greek zephyrus; this may have influenced the spelling when transcribing Arabic ṣifr.[7] The Italian mathematician Fibonacci (c.1170-1250), who grew up in Arab North Africa and is credited with introducing the decimal system to Europe, used the term zephyrum. This became zefiro in Italian, which was contracted to zero in Venetian.
As the decimal zero and its new mathematics spread from the Arab world to Europe in the Middle Ages, words derived from ṣifr and zephyrus came to refer to calculation, as well as to privileged knowledge and secret codes. According to Ifrah, "in thirteenth-century Paris, a 'worthless fellow' was called a "... cifre en algorisme", i.e., an "arithmetical nothing"."[7] From ṣifr also came French chiffre = "digit", "figure", "number", chiffrer = "to calculate or compute", chiffré = "encrypted". Today, the word in Arabic is still ṣifr, and cognates of ṣifr are common in the languages of Europe and southwest Asia.
## History
### Early history
By the middle of the 2nd millennium BC, the Babylonian mathematics had a sophisticated sexagesimal positional numeral system. The lack of a positional value (or zero) was indicated by a space between sexagesimal numerals. By 300 BC, a punctuation symbol (two slanted wedges) was co-opted as a placeholder in the same Babylonian system. In a tablet unearthed at Kish (dating from about 700 BC), the scribe Bêl-bân-aplu wrote his zeros with three hooks, rather than two slanted wedges.[8]
The Babylonian placeholder was not a true zero because it was not used alone. Nor was it used at the end of a number. Thus numbers like 2 and 120 (2×60), 3 and 180 (3×60), 4 and 240 (4×60), looked the same because the larger numbers lacked a final sexagesimal placeholder. Only context could differentiate them.
Records show that the ancient Greeks seemed unsure about the status of zero as a number. They asked themselves, "How can nothing be something?", leading to philosophical and, by the Medieval period, religious arguments about the nature and existence of zero and the vacuum. The paradoxes of Zeno of Elea depend in large part on the uncertain interpretation of zero.
The concept of zero as a number and not merely a symbol for separation is attributed to India where by the 9th century AD practical calculations were carried out using zero, which was treated like any other number, even in case of division.[9][10] The Indian scholar Pingala (circa 5th-2nd century BC) used binary numbers in the form of short and long syllables (the latter equal in length to two short syllables), making it similar to Morse code.[11][12] He and his contemporary Indian scholars used the Sanskrit word śūnya to refer to zero or void.
### History of zero
The back of Olmec Stela C from Tres Zapotes, the second oldest Long Count date yet discovered. The numerals 7.16.6.16.18 translate to September, 32 BC (Julian). The glyphs surrounding the date are thought to be one of the few surviving examples of Epi-Olmec script.
The Mesoamerican Long Count calendar developed in south-central Mexico and Central America required the use of zero as a place-holder within its vigesimal (base-20) positional numeral system. Many different glyphs, including this partial quatrefoil—were used as a zero symbol for these Long Count dates, the earliest of which (on Stela 2 at Chiapa de Corzo, Chiapas) has a date of 36 BC.[13] Since the eight earliest Long Count dates appear outside the Maya homeland,[14] it is assumed that the use of zero in the Americas predated the Maya and was possibly the invention of the Olmecs. Many of the earliest Long Count dates were found within the Olmec heartland, although the Olmec civilization ended by the 4th century BC, several centuries before the earliest known Long Count dates.
Although zero became an integral part of Maya numerals, it did not influence Old World numeral systems.
Quipu, a knotted cord device, used in the Inca Empire and its predecessor societies in the Andean region to record accounting and other digital data, is encoded in a base ten positional system. Zero is represented by the absence of a knot in the appropriate position.
The use of a blank on a counting board to represent 0 dated back in India to 4th century BC.[15]
In China, counting rods were used for calculation since the 4th century BC. Chinese mathematicians understood negative numbers and zero, though they had no symbol for the latter,[16] until the work of Song Dynasty mathematician Qin Jiushao in 1247 established a symbol for zero in China.[17] The Nine Chapters on the Mathematical Art, which was mainly composed in the 1st century AD, stated "[when subtracting] subtract same signed numbers, add differently signed numbers, subtract a positive number from zero to make a negative number, and subtract a negative number from zero to make a positive number."[18]
By 130 AD, Ptolemy, influenced by Hipparchus and the Babylonians, was using a symbol for zero (a small circle with a long overbar) within a sexagesimal numeral system otherwise using alphabetic Greek numerals. Because it was used alone, not just as a placeholder, this Hellenistic zero was perhaps the first documented use of a number zero in the Old World. However, the positions were usually limited to the fractional part of a number (called minutes, seconds, thirds, fourths, etc.)—they were not used for the integral part of a number. In later Byzantine manuscripts of Ptolemy's Syntaxis Mathematica (also known as the Almagest), the Hellenistic zero had morphed into the Greek letter omicron (otherwise meaning 70).
Another zero was used in tables alongside Roman numerals by 525 (first known use by Dionysius Exiguus), but as a word, nulla meaning "nothing," not as a symbol. When division produced zero as a remainder, nihil, also meaning "nothing," was used. These medieval zeros were used by all future medieval computists (calculators of Easter). An isolated use of the initial, N, was used in a table of Roman numerals by Bede or a colleague about 725, a zero symbol.
In 498 AD, Indian mathematician and astronomer Aryabhata stated that "Sthanam sthanam dasa gunam" or place to place in ten times in value, which may be the origin of the modern decimal-based place value notation.[19]
The oldest known text to use a decimal place-value system, including a zero, is the Jain text from India entitled the Lokavibhâga, dated 458 AD. This text uses Sanskrit numeral words for the digits, with words such as the Sanskrit word for void for zero.[20] The first known use of special glyphs for the decimal digits that includes the indubitable appearance of a symbol for the digit zero, a small circle, appears on a stone inscription found at the Chaturbhuja Temple at Gwalior in India, dated 876 AD.[21][22] There are many documents on copper plates, with the same small o in them, dated back as far as the sixth century AD, but their authenticity may be doubted.[8]
The Hindu-Arabic numerals and the positional number system were introduced around 500 AD, and it was introduced by a Persian scientist, Al-Khwarizmi.[5] Al-Khwarizmi's book on arithmetic. This book synthesized Greek and Hindu knowledge and also contained his own fundamental contribution to mathematics and science including an explanation of the use of zero.
It was only centuries later, in the 12th century, that the Arabic numeral system was introduced to the Western world through Latin translations of his Arithmetic.
### Rules of Brahmagupta
The rules governing the use of zero appeared for the first time in Brahmagupta's book Brahmasputha Siddhanta (The Opening of the Universe),[23] written in 628. Here Brahmagupta considers not only zero, but negative numbers, and the algebraic rules for the elementary operations of arithmetic with such numbers. In some instances, his rules differ from the modern standard. Here are the rules of Brahmagupta:[23]
• The sum of zero and a negative number is negative.
• The sum of zero and a positive number is positive.
• The sum of zero and zero is zero.
• The sum of a positive and a negative is their difference; or, if their absolute values are equal, zero.
• A positive or negative number when divided by zero is a fraction with the zero as denominator.
• Zero divided by a negative or positive number is either zero or is expressed as a fraction with zero as numerator and the finite quantity as denominator.
• Zero divided by zero is zero.
In saying zero divided by zero is zero, Brahmagupta differs from the modern position. Mathematicians normally do not assign a value to this, whereas computers and calculators sometimes assign NaN, which means "not a number." Moreover, non-zero positive or negative numbers when divided by zero are either assigned no value, or a value of unsigned infinity, positive infinity, or negative infinity. Once again, these assignments are not numbers, and are associated more with computer science than pure mathematics, where in most contexts no assignment is done.
### Zero as a decimal digit
Positional notation without the use of zero (using an empty space in tabular arrangements, or the word kha "emptiness") is known to have been in use in India from the 6th century. The earliest certain use of zero as a decimal positional digit dates to the 5th century mention in the text Lokavibhaga. The glyph for the zero digit was written in the shape of a dot, and consequently called bindu ("dot"). The dot had been used in Greece during earlier ciphered numeral periods.
The Hindu-Arabic numeral system (base 10) reached Europe in the 11th century, via the Iberian Peninsula through Spanish Muslims, the Moors, together with knowledge of astronomy and instruments like the astrolabe, first imported by Gerbert of Aurillac. For this reason, the numerals came to be known in Europe as "Arabic numerals". The Italian mathematician Fibonacci or Leonardo of Pisa was instrumental in bringing the system into European mathematics in 1202, stating:
After my father's appointment by his homeland as state official in the customs house of Bugia for the Pisan merchants who thronged to it, he took charge; and in view of its future usefulness and convenience, had me in my boyhood come to him and there wanted me to devote myself to and be instructed in the study of calculation for some days. There, following my introduction, as a consequence of marvelous instruction in the art, to the nine digits of the Hindus, the knowledge of the art very much appealed to me before all others, and for it I realized that all its aspects were studied in Egypt, Syria, Greece, Sicily, and Provence, with their varying methods; and at these places thereafter, while on business. I pursued my study in depth and learned the give-and-take of disputation. But all this even, and the algorism, as well as the art of Pythagoras, I considered as almost a mistake in respect to the method of the Hindus (Modus Indorum). Therefore, embracing more stringently that method of the Hindus, and taking stricter pains in its study, while adding certain things from my own understanding and inserting also certain things from the niceties of Euclid's geometric art. I have striven to compose this book in its entirety as understandably as I could, dividing it into fifteen chapters. Almost everything which I have introduced I have displayed with exact proof, in order that those further seeking this knowledge, with its pre-eminent method, might be instructed, and further, in order that the Latin people might not be discovered to be without it, as they have been up to now. If I have perchance omitted anything more or less proper or necessary, I beg indulgence, since there is no one who is blameless and utterly provident in all things. The nine Indian figures are: 9 8 7 6 5 4 3 2 1. With these nine figures, and with the sign 0 ... any number may be written.[24][25]
Here Leonardo of Pisa uses the phrase "sign 0," indicating it is like a sign to do operations like addition or multiplication. From the 13th century, manuals on calculation (adding, multiplying, extracting roots, etc.) became common in Europe where they were called algorimus after the Persian mathematician al-Khwarizmi. The most popular was written by Johannes de Sacrobosco, about 1235 and was one of the earliest scientific books to be printed in 1488. Until the late 15th century, Hindu-Arabic numerals seem to have predominated among mathematicians, while merchants preferred to use the Roman numerals. In the 16th century, they became commonly used in Europe.
## In mathematics
### Elementary algebra
The number 0 is the least non-negative integer. The natural number following 0 is 1 and no natural number precedes 0. The number 0 may or may not be considered a natural number, but it is a whole number and hence a rational number and a real number (as well as an algebraic number and a complex number). Zero is, while a purely real number, also a pure imaginary number because it lies on both the real and imaginary axes on the complex plane.[1]
The number 0 is neither positive nor negative, neither a prime number nor a composite number, nor is it a unit. It is, however, even (see parity of zero).
The following are some basic (elementary) rules for dealing with the number 0. These rules apply for any real or complex number x, unless otherwise stated.
• Addition: x + 0 = 0 + x = x. That is, 0 is an identity element (or neutral element) with respect to addition.
• Subtraction: x − 0 = x and 0 − x = −x.
• Multiplication: x · 0 = 0 · x = 0.
• Division: 0x = 0, for nonzero x. But x0 is undefined, because 0 has no multiplicative inverse, a consequence of the previous rule; see division by zero.
• Exponentiation: x0 = x/x = 1, except that the case x = 0 may be left undefined in some contexts; see Zero to the zero power. For all positive real x, 0x = 0.
The expression 00, which may be obtained in an attempt to determine the limit of an expression of the form f(x)g(x) as a result of applying the lim operator independently to both operands of the fraction, is a so-called "indeterminate form". That does not simply mean that the limit sought is necessarily undefined; rather, it means that the limit of f(x)g(x), if it exists, must be found by another method, such as l'Hôpital's rule.
The sum of 0 numbers is 0, and the product of 0 numbers is 1. The factorial 0! evaluates to 1.
## In science
### Physics
The value zero plays a special role for many physical quantities. For some quantities, the zero level is naturally distinguished from all other levels, whereas for others it is more or less arbitrarily chosen. For example, on the Kelvin temperature scale, zero is the coldest possible temperature (negative temperatures exist but are not actually colder), whereas on the Celsius scale, zero is arbitrarily defined to be at the freezing point of water. Measuring sound intensity in decibels or phons, the zero level is arbitrarily set at a reference value—for example, at a value for the threshold of hearing. In physics, the zero-point energy is the lowest possible energy that a quantum mechanical physical system may possess and is the energy of the ground state of the system.
### Chemistry
Zero has been proposed as the atomic number of the theoretical element tetraneutron. It has been shown that a cluster of four neutrons may be stable enough to be considered an atom in its own right. This would create an element with no protons and no charge on its nucleus.
As early as 1926, Professor Andreas von Antropoff coined the term neutronium for a conjectured form of matter made up of neutrons with no protons, which he placed as the chemical element of atomic number zero at the head of his new version of the periodic table. It was subsequently placed as a noble gas in the middle of several spiral representations of the periodic system for classifying the chemical elements. It is at the centre of the Chemical Galaxy (2005).
## In computer science
### Numbering from 1 or 0
The most common practice throughout human history has been to start counting at one. Nevertheless, in computer science zero is often used as the starting point. For example, in the vast majority of programming languages, the elements of an array are numbered starting from 0 by default. The popularity of the C programming language in the 1980s has made this approach common.
One advantage of this convention is in the use of modular arithmetic. Every integer is congruent modulo N to one of the numbers 0, 1, 2, ..., N − 1, where N ≥ 1. Because of this, many arithmetic concepts (such as hash tables) are more elegantly expressed in code when the array starts at zero.
A second advantage of zero-based array indexes is that this can improve efficiency under certain circumstances. To illustrate, suppose a is the memory address of the first element of an array, and i is the index of the desired element. In this fairly typical scenario, it is quite common to want the address of the desired element. If the index numbers count from 1, the desired address is computed by this expression:
$a + s \times (i-1) \,\!$
where s is the size of each element. In contrast, if the index numbers count from 0, the expression becomes this:
$a + s \times i \,\!$
This simpler expression can be more efficient to compute in certain situations.
Note, however, that a language wishing to index arrays from 1 could simply adopt the convention that every "array address" is represented by a′ = as; that is, rather than using the address of the first array element, such a language would use the address of an imaginary element located immediately before the first actual element. The indexing expression for a 1-based index would be the following:
$a' + s \times i \,\!$
Hence, the efficiency benefit of zero-based indexing is not inherent, but is an artifact of the decision to represent an array by the address of its first element.
A third advantage is that ranges are more elegantly expressed as the half-open interval, [0,n), as opposed to the closed interval, [1,n], because empty ranges often occur as input to algorithms (which would be tricky to express with the closed interval without resorting to obtuse conventions like [1,0]). On the other hand, closed intervals occur in mathematics because it is often necessary to calculate the terminating condition (which would be impossible in some cases because the half-open interval isn't always a closed set) which would have a subtraction by 1 everywhere.
This situation can lead to some confusion in terminology. In a zero-based indexing scheme, the first element is "element number zero"; likewise, the twelfth element is "element number eleven". Therefore, an analogy from the ordinal numbers to the quantity of objects numbered appears; the highest index of n objects will be n – 1 and referred to the nth element. For this reason, the first element is often referred to as the zeroth element to avoid confusion.
### Null value
In databases a field can have a null value. This is equivalent to the field not having a value. For numeric fields it is not the value zero. For text fields this is not blank nor the empty string. The presence of null values leads to three-valued logic. No longer is a condition either true or false, but it can be undetermined. Any computation including a null value delivers a null result. Asking for all records with value 0 or value not equal 0 will not yield all records, since the records with value null are excluded.
### Null pointer
A null pointer is a pointer in a computer program that does not point to any object or function. In C, the integer constant 0 is converted into the null pointer at compile time when it appears in a pointer context, and so 0 is a standard way to refer to the null pointer in code. However, the internal representation of the null pointer may be any bit pattern (possibly different values for different data types).
(Note that on most common architectures, the null pointer is represented internally the same way an integer of the same byte width having a value of zero is represented, so C compilers on such systems perform no actual conversion.)
### Negative zero
In mathematics − 0 = 0 = + 0, both −0 and +0 represent the exact same number, i.e., there is no “negative zero” distinct from zero. In some signed number representations (but not the two's complement representation used to represent integers in most computers today) and most floating point number representations, zero has two distinct representations, one grouping it with the positive numbers and one with the negatives; this latter representation is known as negative zero.
## In other fields
• In some countries and some company phone networks, dialing 0 on a telephone places a call for operator assistance.
• In Braille, the numeral 0 has the same dot configuration as the letter J.
• DVDs that can be played in any region are sometimes referred to as being "region 0"
• In classical music, 0 is very rarely used as a number for a composition: Anton Bruckner wrote a Symphony No. 0 in D minor and a Symphony No. 00; Alfred Schnittke also wrote a Symphony No. 0.
• Roulette wheels usually feature a "0" space (and sometimes also a "00" space), whose presence is ignored when calculating payoffs (thereby allowing the house to win in the long run).
• A chronological prequel of a series may be numbered as 0.
• In Formula One, if the reigning World Champion no longer competes in Formula One in the year following their victory in the title race, 0 is given to one of the drivers of the team that the reigning champion won the title with. This happened in 1993 and 1994, with Damon Hill driving car 0, due to the reigning World Champion (Nigel Mansell and Alain Prost respectively) not competing in the championship.
• In the educational series Schoolhouse Rock!, the song My Hero, Zero is about the use of zero as a placeholder. The song explains that by appending zeros to a number, it is multiplied by 10 for each one added. This enables mathematicians to create numbers as large as needed.
## Quotations
“ The importance of the creation of the zero mark can never be exaggerated. This giving to airy nothing, not merely a local habitation and a name, a picture, a symbol, but helpful power, is the characteristic of the Hindu race from whence it sprang. It is like coining the Nirvana into dynamos. No single mathematical creation has been more potent for the general on-go of intelligence and power. ” — G. B. Halsted
“ Dividing by zero...allows you to prove, mathematically, anything in the universe. You can prove that 1+1=42, and from there you can prove that J. Edgar Hoover is a space alien, that William Shakespeare came from Uzbekistan, or even that the sky is polka-dotted. (See appendix A for a proof that Winston Churchill was a carrot.) ” — Charles Seife, Zero: The Biography of a Dangerous Idea
“ ...a profound and important idea which appears so simple to us now that we ignore its true merit. But its very simplicity and the great ease which it lent to all computations put our arithmetic in the first rank of useful inventions. ” — Pierre-Simon Laplace
“ The point about zero is that we do not need to use it in the operations of daily life. No one goes out to buy zero fish. It is in a way the most civilized of all the cardinals, and its use is only forced on us by the needs of cultivated modes of thought. ” — Alfred North Whitehead
“ ...a fine and wonderful refuge of the divine spirit – almost an amphibian between being and non-being. ” — Gottfried Leibniz
## Notes
1. ^ Catherine Soanes, ed (2001) (Hardback). The Oxford Dictionary, Thesaurus and Wordpower Guide. Maurice Waite, Sara Hawker (2nd ed.). New York, United States: Oxford University Press. ISBN 978-0-19-860373-3.
2. ^ Lemma B.2.2, The integer 0 is even and is not odd, in Penner, Robert C. (1999). Discrete Mathematics: Proof Techniques and Mathematical Structures. World Scientific. pp. 34. ISBN 9810240880.
3. ^ a b c d e R. W. Bemer. "Towards standards for handwritten zero and oh: much ado about nothing (and a letter), or a partial dossier on distinguishing between handwritten zero and oh". Communications of the ACM, Volume 10, Issue 8 (August 1967), pp. 513–518.
4. ^ a b Bo Einarsson and Yurij Shokin. Fortran 90 for the Fortran 77 Programmer. Appendix 7: "The historical development of Fortran"
5. ^ a b Will Durant, 'The Story of Civilization', Volume 4, The Age of Faith, pp. 241.
6. ^ Merriam Webster online Dictionary
7. ^ a b Georges Ifrah. The Universal History of Numbers: From Prehistory to the Invention of the Computer. Wiley (2000). ISBN 0-471-39340-1.
8. ^ a b Kaplan, Robert. (2000). The Nothing That Is: A Natural History of Zero. Oxford: Oxford University Press.
9. ^ Bourbaki, Nicolas (1998). Elements of the History of Mathematics. Berlin, Heidelberg, and New York: Springer-Verlag. 46. ISBN 3540647678.
10. ^ Britannica Concise Encyclopedia (2007), entry algebra
11. ^ Binary Numbers in Ancient India
12. ^ Math for Poets and Drummers (pdf, 145KB)
13. ^ No long count date actually using the number 0 has been found before the 3rd century AD, but since the long count system would make no sense without some placeholder, and since Mesoamerican glyphs do not typically leave empty spaces, these earlier dates are taken as indirect evidence that the concept of 0 already existed at the time.
14. ^ Diehl, p. 186
15. ^ Robert Temple, The Genius of China, A place for zero; ISBN 1-85375-292-4
16. ^ Wáng, Qīngxiáng (1999), Sangi o koeta otoko (The man who exceeded counting rods), Tokyo: Tōyō Shoten, ISBN 4-88595-226-3
17. ^ Needham, Joseph (1986). Science and Civilization in China: Volume 3, Mathematics and the Sciences of the Heavens and the Earth. Taipei: Caves Books, Ltd. Page 43.
18. ^ The statement in Chinese, found in Chapter 8 of The Nine Chapters on the Mathematical Art is 正負術曰: 同名相除，異名相益，正無入負之，負無入正之。其異名相除，同名相益，正無入正之，負無入負之。The word 無入 used here, for which zero is the standard translation by mathematical historians, literally means: no entry. The full Chinese text can be found at wikisource:zh:九章算術.
19. ^ Aryabhatiya of Aryabhata, translated by Walter Eugene Clark.
20. ^ Ifrah, Georges (2000), p. 416.
21. ^ Feature Column from the AMS
22. ^ Ifrah, Georges (2000), p. 400.
23. ^ a b Algebra with Arithmetic of Brahmagupta and Bhaskara, translated to English by Henry Thomas Colebrooke, London1817
24. ^ Sigler, L., Fibonacci’s Liber Abaci. English translation, Springer, 2003.
25. ^ Grimm, R.E., "The Autobiography of Leonardo Pisano", Fibonacci Quarterly 11/1 (February 1973), pp. 99-104.
## References
• Barrow, John D. (2001) The Book of Nothing, Vintage. ISBN 0-09-928845-1.
• Diehl, Richard A. (2004) The Olmecs: America's First Civilization, Thames & Hudson, London.
• Ifrah, Georges (2000) The Universal History of Numbers: From Prehistory to the Invention of the Computer, Wiley. ISBN 0-471-39340-1.
• Kaplan, Robert (2000) The Nothing That Is: A Natural History of Zero, Oxford: Oxford University Press.
• Seife, Charles (2000) Zero: The Biography of a Dangerous Idea, Penguin USA (Paper). ISBN 0-14-029647-6.
• Bourbaki, Nicolas (1998). Elements of the History of Mathematics. Berlin, Heidelberg, and New York: Springer-Verlag. ISBN 3540647678.
• Isaac Asimov article "nothing counts" in "Asimov on Numbers" Pocket Books, 1978
# Travel guide
Up to date as of January 14, 2010
### From Wikitravel
There is more than one place called Zero:
### United States
• Zero (Iowa) - A town in the state of Iowa.
• Zero (Mississippi) - A town in the state of Mississippi.
• Zero (Montana) - A town in the state of Montana.
# 1911 encyclopedia
Up to date as of January 14, 2010
### From LoveToKnow 1911
ZERO, the figure 0 in the Arabic notation for numbers, nought, cipher. The Arabic name for the figure was sifr, which meant literally an empty thing. The old Latin writers on arithmetic translated or transliterated the Arabic word as zephyrum; this in Ital. became zefiro, contracted to zero, borrowed by F. zero, whence it came late into English. The Spanish form cifra, more closely resembling the original Arabic, gave O. Fr. cifre, mod. chiffre, also used in the sense of monogram, and English "cipher" which is thus a doublet. In physics, the term is applied to a point with which phenomena are quantitatively compared, especially to a point of a graduated instrument between a positive and negative or ascending and descending scale, as in the scales of temperature.
<< Zermatt
# Strategy wiki
Up to date as of January 23, 2010
(Redirected to Defender article)
### From StrategyWiki, the free strategy guide and walkthrough wiki
Defender
Developer(s) Williams
Publisher(s) Williams
Designer(s) Eugene Jarvis
Release date(s)
Genre(s) Shooter
System(s) Arcade, Apple II, Atari 8-bit, Atari 2600, Atari 5200, Intellivision, ColecoVision, Commodore 64/128, Commodore VIC-20, MS-DOS, TI-99/4A, Xbox Live Arcade, Entex Adventure Vision, GameTap
Players 1-2
"Zero" redirects here. For other uses, see Zero (disambiguation).
Defender is considered one of the most challenging arcade games ever made. One look at the controls was enough to send many players in search of a simpler game. But those who were brave enough to attempt it found a compelling and addictive game, with an incredibly rich level of strategy. Defender was the brain child of early (and continuing) game developer Eugene Jarvis. Eugene wished to make a video game that immersed the player more than the myriad of Space Invader clones that flooded the markets, and by many experts' accounts, he succeeded.
Defender featured a beautifully horizontal scrolling world, something that was a rarity in its day, and made Defender unique among space shooters. Because the "world" was larger than the screen, Defender also featured a scanner at the top of the screen to give players indications of what was taking place off screen. In addition to the space craft's primary laser weapon, it had a limited supply of smart bombs that could wipe out everything on the screen, and a hyperspace ability. The control scheme for the game matched the complexity of the abilities available to the ship.
The Defender's job is to protect the remnants of humanity from a marauding band of alien invaders who wish to abduct the human race. Landers swoop down to the planet terrain and tractor beam a human to them, and then rise in to the sky with it. A cry for help alerts the player to fly to its rescue as quickly as possible. Either the player shoots the Lander and rescues the human, or the Lander merges with the human at the top of the screen and becomes a deadly mutant. A grim fate awaits the player who fails to keep humanity safe.
The break-neck pace of the action made the game an easy choice for Atari to buy the home licensing rights. They brought a conversion to the 2600 that was generally well received, despite it's short comings. But the version created for the 8-bit family of Atari computers was lauded as one of the best. Defender was also one of four games produced for the Entex Adventure Vision, a stand alone system that accepted cartridges. In fact, it was the system's pack-in game. Atarisoft handled the conversion of the game to several other home console and computer platforms. A few years later, the sequel Stargate would be released, but due to licensing issues, it was commonly known as Defender II. Defender was also released for download on the Xbox Live Arcade in 2006.
## Story
The story is simple: You are the last line of defense against an invading legion of aliens who are determined to abduct all of the remaining humans to create an unstoppable army of hyrbid mutants. You pilot the only space craft capable of undoing their plans and protecting the fate of human life.
## Gameplay summary
• You pilot the spaceship Defender with the controls. The joystick controls your height. Use Thrust and Reverse to control your position over the world.
• You must destroy every enemy to advance to the next stage.
• You must defend the Humanoid population from being abducted by Landers.
• Your ship can fire a laser, launch a smart bomb that eliminates every visible enemy, or hyperspace.
• If a Lander abducts a Humanoid, fly over as quickly as possible, destroy the Lander, capture the Humanoid before it hits the ground, and safely return it.
• If a Lander makes it to the top of the screen with a Humanoid, it becomes a Mutant.
• If all of the Humanoids are capture, the world explodes and every stage is a Mutant stage until after the next fifth round.
• All Humanoids are restored after every fifth round.
# Gaming
Up to date as of February 01, 2010
### From Wikia Gaming, your source for walkthroughs, games, guides, and more!
Zero is the name of a character from the popular Mega Man X series. Zero is a blonde-haired, laser sabre-wielding, robot. He debuted in Mega Man X as the slick, somewhat stand-offish ally to the protagonist, Mega Man X. He has been dubbed as the Knuckles to X's Sonic. Zero had not been invented in the classic Mega Man series and is therefore not included in any of them.
Zero was later given his own series of games in the Game Boy Advance. These games, the Mega Man Zero series, are currently hovering around Mega Man Zero 4.
In the SNK fighting game SvC Chaos: SNK vs. Capcom, Zero appears as an unlockable character instead of Mega Man.
## Appearances
This Mega Man-related article is a stub. You can help by adding to it. Stubs are articles that writers have begun work on, but are not yet complete enough to be considered finished articles.
# Simple English
Zero is a special number. If there are zero things, there are not any things. There are none. For example, if John has zero hats, that means he does not have a hat at all. The term "zeroes" also designates the values of functions at zero.
Order zeroth
## Symbol
The symbol for the number zero is "0".
## Arithmetic with Zero
Zero is a special number. That is because:
• Adding a number to zero results in that number. For example, adding zero to three gives three. In symbols:
3 + 0 = 3
• Subtracting zero from a number always gives that number. For example, subtracting zero from three gives three. In symbols:
3 − 0 = 3
• Subtracting a positive number from zero always makes that number negative. In symbols:
0 − 3 = −3
• Multiplying a number by zero always gives zero. For example, multiplying forty-three by zero gives zero. In symbols:
43 × 0 = 0
• Dividing zero by a number always gives zero. For example, dividing zero by forty three gives zero. In symbols:
0 ÷ 43 = 0
43 ÷ 0 has no answer.
• Zero Divided by zero has no answer. In symbols:
0 ÷ 0 has no answer.
## History of zero
The idea of zero was first thought about in Babylon, India and in Central America at different times. Some places and countries did not know about a zero, which may have made it harder for those people to do mathematics.
Over hundreds of years the idea of zero was passed from country to country. From India and Babylon to other places, like Greece, Persia and the Arab parts of the world. The Europeans learned about zero from the Arabs.
## The place of zero as a number
Zero is almost never used as a place number (ordinal number). This means that it is not used like 1, 2, or 3 to indicate the order, or place, of something, like 1st, 2nd, or 3rd.
Any number divided by itself equals one, except if that number is zero. In symbols:
0 ÷ 0 = "not a number."
In time, zero means "now". For example, a person is counting down the time to the start of something, such as a foot race or when a rocket takes off. The count is: "three, two, one, zero (or go)". Zero is the exact time of the start of the race or when the rocket takes off into the sky.
## 0 as a number
Cardinal 0 zero o/oh nought naught nil null Ordinal 0th zeroth Factorization $0$ Divisors N/A Roman numeral N/A Japanese numeral 〇 Binary 0 Octal 0 Duodecimal 0 Hexadecimal 0
0 is the integer that precedes the positive 1, and follows −1. In most (if not all) numerical systems, 0 was identified before the idea of 'negative integers' was accepted. It means "courageous one" in hieroglyphics.
Zero is a number which means an amount of null size; that is, if the number of your brothers is zero, that means the same thing as having no brothers, and if something has a weight of zero, it has no weight. If the difference between the number of pieces in two piles is zero, it means the two piles have an equal number of pieces. Before counting starts, the result can be assumed to be zero; that is the number of items counted before you count the first item and counting the first item brings the result to one. And if there are no items to be counted, zero remains the final result.
While mathematicians all accept zero as a number, some non-mathematicians would say that zero is not a number, arguing that one cannot have zero of something. Others say that if one has a bank balance of zero, one has a specific quantity of money in that account, namely none. It is that latter view which is accepted by mathematicians and most others.
Almost all historians omit the year zero from the proleptic Gregorian and Julian calendars, but astronomers include it in these same calendars. However, the phrase Year Zero may be used to describe any event considered so significant that it virtually starts a new time reckoning.
## 0 as a numeral
The modern numeral 0 is normally written as a circle or (rounded) rectangle. In old-style fonts with text figures, 0 is usually the same height as a lowercase x.
On the seven-segment displays of calculators, watches, etc., 0 is usually written with six line segments, though on some historical calculator models it was written with four line segments. The four-segment 0 is not common.
It is important to know that the number zero (as in the "zero brothers" example above) is not the same as the numeral or digit zero, used in numeral systems using positional notation. Successive positions of digits have higher values, so the digit zero is used to skip a position and give appropriate value to the preceding and following digits. A zero digit is not always necessary in a positional number system: bijective numeration provides a possible counterexample.
## The numerical digit zero
0 (zero) is also used as a numerical digit used to represent that number in numerals. It is used to hold the place of that digit, because correct placing of digits affects a numeral's value.
Examples:
• In the numeral 10, which stands for one times ten and zero units (or ones).
• In the numeral 100, which stands for one times a hundred plus zero tens plus zero units.
## Telling Zero and the Letter O Apart
The oval-shaped zero and circular letter O together came into use on modern character displays. The zero with a dot in the centre seems to have begun as a choice on IBM 3270 controllers (this has the problem that it looks like the Greek letter Theta). The slashed zero, looking like the letter O with a diagonal line drawn inside it, is used in old-style ASCII graphic sets that came from the default typewheel on the well known ASR-33 Teletype. This format causes problems because of it looks like the symbol $\emptyset$, representing the empty set, as well as for certain Scandinavian languages which use Ø as a letter.
The rule which has the letter O with a slash and the zero without was used at IBM and a few other early mainframe makers; this is even more of a problem for Scandinavians because it looks like two of their letters at the same time. Some Burroughs/Unisys computers display a zero with a backwards slash. And yet another convention common on early line printers left zero without any extra dots or slashes but added a tail or hook to the letter-O so that it resembled an inverted Q or cursive capital letter-O.
The letters used on some European number plates for cars distinguish the two symbols by making the zero rather egg-shaped and the O more circular, but most of all by cutting open the zero on the upper right side, so the circle is not closed any more (as in German plates). The style of letters chosen is called [[:Image:|fälschungserschwerende Schrift]] (abbr.: FE Schrift), meaning "script which is harder to falsify". But those used in the United Kingdom do not make the letter o and the number 0 look different from each other because there can never be any mistake if the letters are correctly spaced.
In paper writing you do not have to make the 0 and O look different at all, or you may add a slash across the zero in order to show the difference, although this sometimes causes mistakes in the number 0.
## References
• Barrow, John D. (2001) The Book of Nothing, Vintage. ISBN 0-09-928845-1.
• Diehl, Richard A. (2004) The Olmecs: America's First Civilization, Thames & Hudson, London.
• Ifrah, Georges (2000) The Universal History of Numbers: From Prehistory to the Invention of the Computer, Wiley. ISBN 0-471-39340-1.
• Kaplan, Robert (2000) The Nothing That Is: A Natural History of Zero, Oxford: Oxford University Press.
• Seife, Charles (2000) Zero: The Biography of a Dangerous Idea, Penguin USA (Paper). ISBN 0-14-029647-6.
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## anonymous one year ago ques
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1. anonymous
Why is L'Hopital's rule not applicable for (b) but it is applicable for (c), when I think it should be the other way around (b) $\lim_{x \rightarrow 0}\frac{\cos(x)}{x}$ (c) $\lim_{x \rightarrow 0}\frac{e^{2x}-1}{\tan(x)}$
2. Loser66
because for b) it doesn't form $$\dfrac{0}{0}$$ since cos (0) =1
3. Loser66
while for c) when x=0, $$e^{2x}= e^0 =1$$ and it forms $$\dfrac{0}{0}$$
4. anonymous
But if we apply the rule in (b) it becomes solvable but if we apply it in (c) it only becomes more complicated
5. Loser66
No, no, no!!! you are allowed to apply L'Hospital rule if and only if it forms the special cases like (0/0), (0^ infinitive).... if the expression is not one of the special case, you can't. To b) it is just $$=\infty$$, why is it complicated?
6. Loser66
to c) it is not at all $$(e^{2x}+1)'= 2e^{2x}$$ and $$(tan(x))'= sec^2(x)=1/cos^2(x)$$ when x =0, the limit of numerator =2, the limit of denominator =1 and you get 2 as the final result. how complicated it is UNLESS you try to get the free answer from me.
7. Loser66
Hopefully I was wrong on the last sentence so that I am happy to help you out if you have other questions later.
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2017-01-21 04:29:07
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https://web2.0calc.com/questions/an-inbox-tray-has-3-walls-and-an-open-side-on-one-of-the-longer-sides-determine-the-maximum-area-of-the-tray-if-all-three-walls-total-to-a
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+0
# An inbox tray has 3 walls and an open side on one of the longer sides. determine the maximum area of the tray if all three walls total to a
0
183
1
An inbox tray has 3 walls and an open side on one of the longer sides.
Determine the maximum area of the tray if all three walls total to a length of 812 mm. Show your formula.
Guest Dec 3, 2014
#1
+91448
+5
Let the 2 short sides be x and the long side be y
$$\\2x+y=812\\ y=812-2x\\\\ Area(A)=x*y\\ A=x(812-2x)\\ A=812x-2x^2\\\\ \frac{dA}{dx}=812-4x\\\\ \frac{d^2A}{dx^2}=-4 \quad \cap\\ This means that any stationary point will be a maximum\\\\ find stat point \frac{dA}{dx}=0\\\\ 0=812-4x\\ 4x=812\\ x=203\\ The tray will be of maximum are if the sides are 20.3cm by 40.6cm$$$Melody Dec 3, 2014 Sort: ### 1+0 Answers #1 +91448 +5 Best Answer Let the 2 short sides be x and the long side be y $$\\2x+y=812\\ y=812-2x\\\\ Area(A)=x*y\\ A=x(812-2x)\\ A=812x-2x^2\\\\ \frac{dA}{dx}=812-4x\\\\ \frac{d^2A}{dx^2}=-4 \quad \cap\\ This means that any stationary point will be a maximum\\\\ find stat point \frac{dA}{dx}=0\\\\ 0=812-4x\\ 4x=812\\ x=203\\ The tray will be of maximum are if the sides are 20.3cm by 40.6cm$$$
Melody Dec 3, 2014
### 6 Online Users
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2018-01-19 11:56:29
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https://math.stackexchange.com/questions/1562200/show-that-if-f-is-a-uniformly-continuous-function-on-mathbbr-and-f-in-l
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# Show that if $f$ is a uniformly continuous function on $\mathbb{R}$ and $f\in L^1(\mathbb{R})$, then $f$ is bounded and $\lim_{|x|\to\infty}f(x)=0$.
Show that if $f$ is a uniformly continuous function on $\mathbb{R}$ and $f\in L^1(\mathbb{R})$, then $f$ is bounded and $\lim_{|x|\to\infty}f(x)=0$.
I'm not entirely sure what I should be doing. Should I construct on interval such that $f$ is bounded as described and then just check the limit? Any help would be greatly appreciated.
• No, you are supposed to show that $f$ is bounded even on unbounded set. The fact that it is bounded on any finite interval follows immediately from continuity (which implies uniform continuity on any closed bounded interval). – Thomas Dec 6 '15 at 8:43
• I don't follow. Could you elaborate? – Desperate Fluffy Dec 6 '15 at 8:47
It is sufficient to consider positive $x$ and also to assume that $f$ is positive (since only $|f|$ is relevant for the $L^1$ norm).
Assume $f$ does not converge to $0$. Then there is $\varepsilon > 0$ and a sequence $x_n\rightarrow \infty$ such that $f(x_n) \ge \varepsilon$. Without log of generality $x_{n+1}> x_n+2$. Since $f$ is uniformly continuous, there is $\delta > 0$ such that $|x-y|<\delta \Rightarrow |f(x)-f(y)| < \frac{\varepsilon}{2}$. Wlog $\delta < 1$. In particular, $f\ge \varepsilon/2$ in a $\delta$-neighbourhood of each $x_n$.
By construction, the intervals $(x_n- \delta, x_n+\delta)$ are pairwise disjoint. So
$$\infty >|f|_{L^1} = \int f \ge \sum_1^\infty\int_{(x_n- \delta, x_n+\delta)} f \ge \sum_1^\infty \varepsilon\delta$$ which is a contradiction.
So $\lim_{x\rightarrow\infty} f(x) = 0$ (similar for negative $x$). Since, as a continuous function, $f$ is bounded on each compact interval, $f$ is bounded on all of $\mathbb{R}$
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2021-04-11 20:44:58
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https://www.physicsforums.com/threads/hilbert-space.190159/
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# Hilbert Space
1. Oct 9, 2007
### zhaiyujia
[SOLVED] Hilbert Space
1. The problem statement, all variables and given/known data
For What Values of $$\psi(x)=\frac{1}{x^{\alpha}}$$ belong in a Hilbert Sapce?
2. Relevant equations
$$\int x^{a}=\frac{1}{a+1} x^{a+1}$$
3. The attempt at a solution
I tried to use the condition that function in Hilbert space should satisfy:
$$\int\psi^{2}=A$$ but it seems always infinite exist in x=0 or x=infinite
Last edited: Oct 9, 2007
2. Oct 9, 2007
### Gokul43201
Staff Emeritus
Why are you writing the integral of x^a, when you want to examine the integral of 1/x^{2a}?
Also, have you written down the question completely? What is the domain on which $\psi(x)$ is defined?
3. Oct 9, 2007
### zhaiyujia
one is alpha and another is a. I just write a integral equation, a = 2*alpha. I think it is not the key point. The question is complete. I asked my professor if there is some constrain of x, she said x can be any value. that is to say the integral will from minus infinite to infinite
4. Oct 9, 2007
### Gokul43201
Staff Emeritus
Okay, they both looked the same to me.
It can not be. There's at least a couple of missing words. Here's one way to write a somewhat more complete question:
For what values of $\alpha$ does $\psi(x)=1/{x^{\alpha}}$ belong in a Hilbert space?
This needs to be specified in the question. You have not completely specified a function unless you describe its domain.
5. Oct 9, 2007
### zhaiyujia
Thanks, I explained it in the interval of minus infinite to minus zero and zero to infinite. I guess a wave function with singularity is not a good one in physic....
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2017-10-20 21:55:03
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https://www.physicsforums.com/threads/using-jensens-inequality.116516/
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# Using Jensen's Inequality
1. Apr 4, 2006
### ReyChiquito
Let $x \in \mathbb{R}^n$ and
$$u_0>0, \qquad \int\limits_\Omega u_0(x) dx =1, \qquad E(t)=\int\limits_\Omega u(x,t)u_0(x)dx$$
Im having trouble proving the following inequality
$$\int\limits_\Omega \frac{u_0(x)}{(1+u(x,t))^2}dx \ge \dfrac{1}{(1+E)^2}. \qquad \hbox{(1)}$$
I know i have to use Jensen's inequality
$$f\left(\frac{1}{|\Omega|}\int\limits_\Omega u dx \right) \le \frac{1}{|\Omega|}\int\limits_\Omega f(u) dx$$,
where $f(u)$ is convex.
But in order to use it to prove (1), I need to rewrite the left hand side of the equation or use a previous inequality right?
There is where im stuck. Can anybody give me a sugestion pls?
Last edited: Apr 4, 2006
2. Apr 4, 2006
### ReyChiquito
Is it just me or nobody can see the TeX?
3. Apr 4, 2006
### cogito²
I cannot either.
4. Apr 6, 2006
### ReyChiquito
Well, first of all, it would be nice if someone tell me why the TeX doesnt work. Second of all, i got it, so nevermind.
$$\int$$
$$\Omega$$
$$\omega$$
no \int???? nice.....
Last edited: Apr 6, 2006
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2017-12-18 18:59:40
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https://www.iacr.org/news/legacy.php?p=detail&id=4327
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International Association for Cryptologic Research
# IACR News Central
You can also access the full news archive.
Further sources to find out about changes are CryptoDB, ePrint RSS, ePrint Web, Event calender (iCal).
2014-10-20
15:17 [Pub][ePrint]
Barak, Shaltiel Tromer showed how to construct a True Random Number Generator (TRNG) which is secure against an adversary who has some limited control over the environment.
In this paper we improve the security analysis of this TRNG. Essentially, we significantly reduce the entropy loss and running time needed to obtain a required level of security and robustness.
Our approach is based on replacing the combination of union bounds and tail inequalities for $\\ell$-wise independent random variables in the original proof, by a more refined of the deviation of the probability that a randomly chosen item is hashed into a particular location.
15:17 [Pub][ePrint]
Homomorphic encryption (HE) systems enable computations on encrypted data, without decrypting and without knowledge of the secret key. In this work, we describe an optimized RLWE-based implementation of a variant of the HE system recently proposed by Gentry, Sahai and Waters (henceforth called GSW). Although this system was widely believed to be less efficient than its contemporaries, we demonstrate quite the opposite behavior for a large class of applications.
We first highlight and carefully exploit the algebraic features of the system to achieve significant speedup over the state-of-the-art HE implementation, namely the IBM homomorphic encryption library (HElib). We introduce several optimizations on top of our HE implementation, and use the resulting scheme to construct a homomorphic Bayesian spam filter, secure multiple keyword search, and a homomorphic evaluator for binary decision trees.
Our results show a factor of 10x improvement in performance (under the same security settings and platforms) compared to HElib for these applications. Our system is built to be easily portable to GPUs (unlike HElib) which results in an additional speedup of up to a factor of 10x to offer an overall speedup of 100x.
15:17 [Pub][ePrint]
Given two integers $N_1 = p_1q_1$ and $N_2 = p_2q_2$ with $\\alpha$-bit primes $q_1,q_2$, suppose that the $t$ least significant bits of $p_1$ and $p_2$ are equal. May and Ritzenhofen (PKC 2009) developed a factoring algorithm for $N_1,N_2$ when $t \\geq 2\\alpha + 3$; Kurosawa and Ueda (IWSEC 2013) improved the bound to $t \\geq 2\\alpha + 1$. In this paper, we propose a polynomial-time algorithm in a parameter $\\kappa$, with an improved bound $t = 2\\alpha - O(\\log \\kappa)$; it is the first non-constant improvement of the bound. Both the construction and the proof of our algorithm are very simple; the worst-case complexity of our algorithm is evaluated by an easy argument, without any heuristic assumptions. We also give some computer experimental results showing the efficiency of our algorithm for concrete parameters, and discuss potential applications of our result to security evaluations of existing factoring-based primitives.
15:17 [Pub][ePrint]
A constrained pseudorandom function $F: K \\times X \\to Y$ for family of subsets of $X$ is a function where for any key $k \\in K$ and set $S$ from the family one can efficiently compute a short constrained key $k_S$ which allows to evaluate $F(k,.)$ on all inputs $x\\in S$, while given this key, the outputs on all inputs $x \\notin S$ look random.
Constrained PRFs have been constructed for several families of sets, the most general being the circuit-constrained PRF by Boneh and Waters [Asiacrypt\'13]. Their construction allows for constrained keys $k_C$, where $C$ is a boolean circuit that defines the set $S = \\{ x \\in X | C(x) = 1 \\}$. In their construction the input length and the size of the circuits $C$ for which constrained keys can be computed must be fixed a priori during key generation.
In this paper we construct a constrained PRF that has an unbounded input length and constrained keys can be defined for any set that can be decided by a polynomial-time Turing machine. The only a priori bound we make is on the size of the Turing machines. We discuss applications of our CPRF, such as broadcast encryption where the number of potential receivers need not be fixed at setup (in particular, the length of the keys is independent of the number of parties), and identity-based non-interactive key-agreement between sets of users where again there is no bound on the number of parties that can agree on a shared key.
Our CPRF is simply defined as $F(k,H(x))$ where $F$ is a puncturable PRF (e.g. the GGM construction) and $H$ is a collision-resistant hash function. A constrained key for a Turing machine $M$ is a signature on $M$. At setup we also publish an obfuscated circuit, which on input $M$, a signature $\\sigma$, a value $h$ and a short non-interactive argument of knowledge $\\pi$ outputs $F(k,h)$ if (1) $\\sigma$ is a valid signature on $M$ and (2) $\\pi$ proves knowledge of some $x$ s.t.\\ $H(x)=h$ and $M(x)=1$. For our security proof, we assume extractability obfuscation for the particular circuit just explained.
15:17 [Pub][ePrint]
A non-malleable code protects messages against various classes of tampering.
Informally, a code is non-malleable if the message contained in a tampered
codeword is either the original message, or a completely unrelated one.
Although existence of such codes for
various rich classes of tampering functions is known, \\emph{explicit} constructions
exist only for compartmentalized\'\' tampering functions: \\ie the codeword
is partitioned into {\\em a priori fixed} blocks and each block can {\\em only
be tampered independently}. The prominent examples of this model are the
family of bit-wise independent tampering functions and the split-state
model.
In this paper, for the first time we construct explicit non-malleable codes
against a natural class of non-compartmentalized tampering functions. We
allow the tampering functions to {\\em permute the bits} of the codeword and
(optionally) perturb them by flipping or setting them to 0 or 1. We
construct an explicit, efficient non-malleable code for arbitrarily long
messages in this model (unconditionally).
We give an application of our construction to non-malleable commitments, as
one of the first direct applications of non-malleable codes to
computational cryptography. We show that non-malleable {\\em string} commitments
can be entirely based on\'\' non-malleable {\\em bit} commitments. More
precisely, we show that simply encoding a string using our code, and then
committing to each bit of the encoding using a {\\em CCA-secure bit
commitment} scheme results in a non-malleable string commitment scheme. This
reduction is unconditional, does not require any extra properties from the
bit-commitment such as tag-based\'\' non-malleability, and works even with
physical implementations (which may not imply standard one-way functions).
Further, even given a partially malleable bit commitment scheme which allows
toggling the committed bit (instantiated, for illustration, using a variant
of the Naor commitment scheme under a non-standard assumption on the PRG
involved), this transformation gives a fully non-malleable string commitment
scheme. This application relies on the non-malleable code being explicit.
15:17 [Pub][ePrint]
A non-malleable code protects messages against various classes of tampering.
Informally, a code is non-malleable if the effect of applying any tampering
function on an encoded message is to either retain the message or to replace
it with an unrelated message.
Two main challenges in this area -- apart from establishing
the feasibility against different families of tampering -- are to obtain
{\\em explicit constructions} and to obtain {\\em high-rates} for such
constructions.
In this work, we present a compiler to transform low-rate (in fact, zero
rate) non-malleable codes against certain class of tampering into an
optimal-rate -- i.e., rate 1 -- non-malleable codes against the same class.
If the original code is explicit, so is the new one.
When applied to the family of bit-wise tampering functions, this subsumes
(and greatly simplifies) a recent result of Cheraghchi and Guruswami (TCC
2014). Further, our compiler can be applied to non-malleable codes against
the class of bit-wise tampering and bit-level permutations. Combined with
the rate-0 construction in a companion work, this yields the first explicit
rate-1 non-malleable code for this family of tampering functions.
Our compiler uses a new technique for boot-strapping non-malleability by
introducing errors, that may be of independent interest.
15:17 [Pub][ePrint]
In this paper we revisit the modular inversion hidden number problem and the inversive congruential pseudo random number generator and consider how to more efficiently attack them in terms of fewer samples or outputs. We reduce the attacking problem to finding small solutions of systems of modular polynomial equations of the form $a_i+b_ix_0+c_ix_i+x_0x_i=0 (\\mod p)$, and present two strategies to construct lattices in Coppersmith\'s lattice-based root-finding technique for the solving of the equations. Different from the choosing of the polynomials used for constructing lattices in previous methods, a part of polynomials chosen in our strategies are linear combinations of some polynomials generated in advance and this enables us to achieve a larger upper bound for the desired root. Applying the solving of the above equations to analyze the modular inversion hidden number problem, we put forward an explicit result of Boneh et al. which was the best result so far, and give a further improvement in the involved lattice construction in the sense of requiring fewer samples. Our strategies also give a method of attacking the inversive congruential pseudo random number generator, and the corresponding result is the best up to now.
12:17 [Pub][ePrint]
In this work, we first formalize the notion of dynamic group signatures with distributed traceability, where the capability to trace signatures is distributed among $n$ managers without requiring any interaction. This ensures that only the participation of all tracing managers permits tracing a signature, which reduces the trust placed in a single tracing manager. The threshold variant follows easily from our definitions and constructions. Our model offers strong security requirements.
Our second contribution is a generic construction for the notion which has a concurrent join protocol, meets strong security requirements, and offers efficient traceability, i.e.\\ without requiring tracing managers to produce expensive zero-knowledge proofs for tracing correctness. To dispense with the expensive zero-knowledge proofs required in the tracing, we deploy a distributed tag-based encryption with public verifiability.
Finally, we provide some concrete instantiations, which, to the best of our knowledge, are the first efficient provably secure realizations in the standard model simultaneously offering all the aforementioned properties. To realize our constructions efficiently, we construct an efficient distributed (and threshold) tag-based encryption scheme that works in the efficient Type-III asymmetric bilinear groups. Our distributed tag-based encryption scheme yields short ciphertexts (only 1280 bits at 128-bit security), and is secure under an existing variant of the standard decisional linear assumption.
Our tag-based encryption scheme is of independent interest and is useful for many applications beyond the scope of this paper. As a special case of our distributed tag-based encryption scheme, we get an efficient tag-based encryption scheme in Type-III asymmetric bilinear groups that is secure in the standard model.
2014-10-19
17:13 [Event][New]
Submission: 27 October 2014
From January 19 to January 19
Location: Amsterdam, The Netherlands
2014-10-16
18:17 [Pub][ePrint]
Currently, the Internet Research Task Force (IRTF) discusses requirements for new elliptic curves to be standardized in TLS and other internet protocols. This position paper discusses the view of the members of the ECC Brainpool on these requirements, in particular with respect to hardware implementations.
16:50 [Job][New]
Following recent advances in high throughput sequencing, it can be expected that, in the near future, more and more individuals will have their whole genome extracted, stored and analyzed in a routine fashion. Although this perspective is full of promises in terms of personalized preventive and curative medicine as well as medical research, it should also be acknowledged that the genome of an individual is in essence extremely sensitive data from (at least) a privacy standpoint. Thus, for such personalized medicine-oriented platforms to reliably exist, it is necessary to develop specific counter-measures providing intrinsic protection of the genomic data when manipulated by an IT infrastructure.
In this context, a very promising approach is grounded in homomorphic encryption as a means of computing directly over encrypted data.
The purpose of the present postdoctoral offer is thus to investigate the practical relevance of using homomorphic encryption techniques for privacy-preserving genetic data processing. The main use case will consist in performing requests on a database of genomes represented by their variants. Several scenarios will be investigated in particular with respect to the privacy of the request itself on top of the privacy of the genetic data. In this various scenarios, the candidate is expected to identify the most suitable homomorphic encryption techniques ranging from additive-only (e.g. suitable for private requests on unencrypted data) and multiplicative-only (e.g. suitable for disjunctive public requests on encrypted genetic data) homomorphic encryption systems to the use of the more recent (and more costly) fully homomorphic encryption techniques. The candidate will also be expected to build prototypes for one or more of the above scenarios in order to experimentally demonstrate the practical viability of the solutions, in particular with respect to performances.
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2017-09-22 04:48:32
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https://indico.fnal.gov/event/15949/contributions/34965/
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# 36th Annual International Symposium on Lattice Field Theory
Jul 22 – 28, 2018
Kellogg Hotel and Conference Center
EST timezone
## Progress towards understanding the H-dibaryon from lattice QCD
Jul 26, 2018, 12:40 PM
20m
105 (Kellogg Hotel and Conference Center)
### 105
#### Kellogg Hotel and Conference Center
219 S Harrison Rd, East Lansing, MI 48824
### Speaker
Dr Andrew Hanlon (Helmholtz-Institut Mainz, JGU)
### Description
Significant experimental and theoretical efforts to determine the existence and nature of the $H$-dibaryon have been underway since its prediction in 1977. Yet, conclusive evidence for such a bound state is still lacking. Results from various lattice QCD calculations show substantial disagreement for the binding energy. Since there is no conclusive evidence for or against the existence of a bound $H$-dibaryon, the Mainz group has joined the effort towards resolving the discrepancies. In this talk, I will first give an overview of the recent results from the Mainz group obtained in two-flavor QCD. I will then discuss work that has been done towards ensuring that the baryon-baryon operators used in the $H$-dibaryon studies transform irreducibly under the relevant little group. Finally, the prospects of extensions to $N_f = 2 + 1$ and the use of distillation for moving frames will be examined.
### Primary author
Dr Andrew Hanlon (Helmholtz-Institut Mainz, JGU)
### Co-authors
Dr Anthony Francis (CERN) Dr Chuan Miao (Helmholtz-Institut Mainz, JGU) Prof. Hartmut Wittig (Johannes Gutenberg Universität) Dr Jeremy Green (DESY, Zeuthen) Dr Parikshit Junnarkar (Tata Institute of Fundamental Research) Dr Thomas Rae (Johannes Gutenberg Universität)
Slides
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2022-11-28 07:21:45
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https://www.kimsereylam.com/python/2020/01/10/coding-challenge-array-exercises.html
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# Coding Challenge Array Exercises
Jan 10th, 2020 - written by Kimserey with .
Coding exercises are hard, no matter what level of experience you have and which point of your career you are at, it is a very different skills to succeed coding exercises versus building real life application. Nonetheless some of them are quite interesting and highlight techniques that can vastly change the performance of a program. In today’s post we look at optimization techniques used in arrays on common problems.
## Minimum Swaps by one position
### Problem
We are given an array from 0 to n-1 shuffled in such a way that elements were swapped in pairs. Each element is only allowed to move forward the array two times. Find how many swaps are required to reach the state of the shuffled array.
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Example:
1 0 2 5 4 3
Solution:
4
Explanation:
- 0 1 2 3 4 5: Initial
- 1 0 2 3 4 5: Move forward 1
- 1 0 2 3 5 4: Move forward 5
- 1 0 2 5 3 4: Move forward 5
- 1 0 2 5 4 3: Move forward 4
### Explanation
We iterate through the array and execute a first check verifying that the current value held at the index does not differ from more than two, if it does it means that the array violates the constraint of maximum two moves.
When we iterate through the array, at each index, any value superior to the value held at the index in front position are considered as swaps.
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2 0 1: Has two swaps
A first implementation of our algorithm would be:
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def get_min_swap(q):
res = 0
for i, v in enumerate(q):
diff = v - i
if diff > 2:
print("Too many swaps")
return
for j in range(0, i):
if q[j] > v:
res += 1
print(res)
We have reduced the inner iteration by restricting the range from [0, i]. But we can do better, in fact we can recognize that at any time, a value can only go ahead by a single position. The first swap will put the value behind at the same position and the second swap will put the value in front by one position. Therefore we can deduce that for any value, the maximum position a value behind can get ahead is by one position.
We can therefore reduce the range to [v - 1, i].
All swaps have to be contained between [v - 1, i] where i is where the value is at the moment.
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def get_min_swap(q):
res = 0
for i, v in enumerate(q):
diff = v - i
if diff > 2:
print("Too many swaps")
return
for j in range(max(v - 1, 0), i):
if q[j] > v:
res += 1
print(res)
And we will have achieve the best algorithm.
## Minimum Swaps to sort
### Problem
We are given an array of shuffled incremental integer starting from zero. We want to know the minimum swaps needed to sort the array.
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Example:
5 4 1 0 2 3
Solution:
4
Steps:
5 4 1 0 2 3: 5<>0
0 4 1 5 2 3: 4<>1
0 1 4 5 2 3: 4<>2
0 1 2 5 4 3: 3<>5
0 1 2 3 4 5: end
### Explanation
Each position in the array should contain it’s own value. Therefore to sort the array we go through each position and swap any wrong value to the position where the right value was held. In order to swap a value we need to retrieve the index where the value is held. For an O(1) retrieval we construct a reverse array containing the values as indices and the indices as values. Then when we iterate and find a misplaced value, we retrieve the index of the correct value and swap it with the currently misplaced value.
With that we achieve an O(n) complexity
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def minmum_swap(arr):
# Array of indexes for each value,
# for quick lookup.
indexes = [0] * len(arr)
for pos, val in enumerate(arr):
indexes[val] = pos
swaps = 0
for i in range(len(arr)):
# Check if value is at the wrong position
if arr[i] != i:
tmp = arr[i]
# Sets the current index to the array.
arr[i] = i
# Find the previous value index from indexes
# and set the tmp value to where the index was.
arr[indexes[i]] = tmp
# Modify the indexes array accordingly
indexes[tmp] = indexes[i] # Apply the swap
indexes[i] = i # Set the correct value
swaps += 1
return swaps
## Variation
### Problem
For this last exercise, we want to find the highest value when the arrays are added.
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Example:
3 5 - 10
1 4 - 5
6 8 - 2
Solution:
15
Steps:
00 00 00 10 10 10 00 00 00: 3 5 - 10
00 05 05 05 05 00 00 00 00: 1 4 - 5
00 00 00 00 00 00 02 02 02: 6 8 - 2
Resulting 0 5 5 15 15 10 2 2 2 with a max of 15.
### Explanation
By drawing the result, we can see that the result is given by the variation of level. Each value can be derived by the variation of each range given. For example the first range, 3 5 - 10 would represent 0 0 0 10 10 10 0 0 0, which can also be represented as its slope 0 0 0 10 0 0 -10 0 0. At index 3, the values increase by 10, and at index 6, decrease by 10, resulting in 0 0 0 10 10 10 0 0 0.
If we continue with this logic with each range, we end up with an array aggregating all variation and representing the actual result.
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0 +5 0 +10 0 -5 -10+2 0 0
0 5 5 15 15 10 2 2 2
And we can implement it with the following function:
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# queries contain the range plus the value
def array_manipulation(n, queries):
res = [0] * (n)
for q in queries:
x, y, incr = q[0], q[1], q[2]
# Add the increment variation at x
res[x] += incr
if y < len(res):
# Remove the increment variation at y+1
res[y + 1] -= incr
# Construct the final result by adding every value
# of the variation and constructing the actual
# increment result.
mx = 0
x = 0
for i in res:
x = x + i
if mx < x:
mx = x
return mx
And that concludes today’s post!
## Conclusion
Today we looked into three array exercises, each one of the exercises had different constraint which made a solution slightely different. In all of the exercise the only criteria which mattered was the complexity. In the first exercise, we reduced the complexity by reducing the domain of values to look at when we identified the maximum space a swap could occur in. In the second exercise, we reduced the complexity by building a reserve array of indexes which would then allow us to retrieve values in the array for at a constant complexity. Lastly the third exercise, we reduced the complexity by constructing an array which will allow us to represent the result in a single step. I hope you liked this post and I see you on the next one!
Designed, built and maintained by Kimserey Lam.
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2021-10-22 03:34:32
|
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|
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.106.130801
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# Synopsis: Time doesn’t stand still
Researchers predict that modifying the design of atomic clocks could make them ten times more accurate than the current standard.
Time (or frequency) is the most precisely measured physical quantity by far. But finding ways to measure it even better could allow more powerful tests of fundamental physics, such as searches for changes in the fine-structure constant.
The world’s best time pieces are based on “atomic fountains,” in which lasers cool a cloud of alkali-metal atoms and then toss them gently upward. As the atoms fly up and then fall back down under the force of gravity, they pass twice along the axis of a cylindrical cavity filled with microwaves. By measuring how many atoms respond to the microwave field as they change the microwave frequency, researchers can adjust that frequency to within a few parts in ${10}^{16}$ of the value that defines the standard for a second.
Writing in Physical Review Letters, Jocelyne Guéna, of Systèmes de Référence Temps-Espace (SYRTE) at the Observatory of Paris, and her colleagues analyzed the imperfections that cause the largest, but nonetheless tiny, uncertainty in frequency. Using the clock at SYRTE, they verified experimentally how spatial variations in the microwave phase contribute to the error. The biggest contributions come from variations with a dipole or quadrupole dependence on angle in the horizontal plane. The collaboration between SYRTE and Pennsylvania State University predicts that independently feeding in the microwaves at four positions around the cylindrical cavity, instead of the current two, will reduce the clock errors from Doppler shifts to an unprecedented part in ${10}^{17}$. – Don Monroe
### Announcements
More Announcements »
## Subject Areas
Atomic and Molecular Physics
Astrophysics
## Next Synopsis
Quantum Information
## Related Articles
Atomic and Molecular Physics
### Synopsis: A Crystal of Light and Atoms
A predicted type of atom-light crystal could host phonon-like excitations, allowing for new ways to simulate the physics of solids. Read More »
Condensed Matter Physics
### Viewpoint: An Arrested Implosion
The collapse of a trapped ultracold magnetic gas is arrested by quantum fluctuations, creating quantum droplets of superfluid atoms. Read More »
Atomic and Molecular Physics
### Synopsis: No Vacancy for Tunneling
The tunneling rate for cold atoms in an optical lattice can be made to depend on whether a neighboring site is occupied—a behavior that may reflect the tunneling in complex materials. Read More »
|
2016-05-30 10:42:37
|
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|
https://garrettgoon.com/gaussian-fields/
|
Gaussian Fields
Introduction
Note: To view full latex expression on mobile, you may need to visit the desktop site.
In Cosmology, we are often interested in understanding the spatial correlations amongst fluctuations of various kinds. Of particular importance are two-point correlation functions of the form $$\langle \phi(\bfx)\phi(\bfy)\rangle$$ where $$\phi(\bfx)$$ is some scalar quantity, say a temperature fluctuation.
Because of the homogeneous and isotropic nature of spatial slices, it’s convenient to discuss the Fourier transform of the correlator which takes on the form $$\langle\phi_{\bfk}\phi_{-\bfk}\rangle'=P(k)$$ where the power spectrum, $$P(k)$$, only depends on the magnitude of $$\bfk$$, as indicated (see below for more on the notation).
All of the physics of the two-point function is encoded in $$P(k)$$ and hence it’s typically the quantity we focus on. However, if handed a $$P(k)$$, it’s not very easy to visualize what a corresponding spatial map of fluctuations would look like. In order to gain intuition for the power spectrum and to understand how the fluctuations change as $$P(k)$$ is altered, it’s nice to be able to go from a given power spectrum to some concrete position space maps whose statistics realize a given power spectrum.
In this post we answer the following question: if we’re handed a $$P(k)$$ which describes a Gaussian random field, how do we generate a simple fluctuation map which has the correlations described by this power spectrum?
We start by reviewing some of the theory behind random Gaussian fields, which I found difficult to find in any one place in the literature, and then move on to the practical construction of our maps using Mathematica. A sample notebook is included at the end of the post. The methods used for constructing the realization are by no means optimal; our goal is just to understand a quick and dirty method of generating realizations.
Theory Background
We start by reviewing the necessary theory background.
Correlators
First, we set notation and review basic facts about equal time correlators of scalar quantities. We focus only on two-point functions.
A homogeneous, isotropic position space two-point corrrelator only depends on the magnitude of the distance between the points of interest: $$\langle \phi(\bfx)\phi(\bfy)\rangle=A(|\bfx-\bfy|)$$. Due to this fact, the Fourier transformed correlator is proportional to a delta function and only depends on the magnitude of the momentum (i.e. wavevector):
$\langle \phi_{\bfk}\phi_{\bfp}\rangle\equiv\int\rd^{d}x\rd^{d}y\, e^{-i\bfk\cdot\bfx-i\bfp\cdot\bfy}\langle \phi(\bfx)\phi(\bfy)\rangle=P(k)\tilde{\delta}^{d}(\bfk+\bfp)$
We then put a prime on the momentum space correlator to indicate that we’ve dropped the delta function and associated $$2\pi$$ factors:
$\langle \phi_{\bfk}\phi_{-\bfk}\rangle'=P(k)$
We define $$\tilde{\delta}(\bfk+\bfp)\equiv (2\pi)^{d}\delta^{d}(\bfk+\bfp)$$ and later use $$\tilde{\bfk}\equiv \bfk/(2\pi)$$ to minimize the number of explicit $$2\pi$$’s appearing.
Two functional forms of the power spectrum are particularly important: white noise and scale invariant power spectra.
White Noise: $$P(k)\propto C$$ with $$C$$ constant. This corresponds to a position space correlator of the form $$\langle \phi(\bfx)\phi(\bfy)\rangle\propto C\delta^{d}(\bfx-\bfy)$$, i.e. there are no correlations between fluctuations at separated points if the power spectrum is described by white noise.
Scale Invariant: $$P(k)\propto k^{-d}$$ in $$d$$ spatial dimensions. A scale invariant power spectrum corresponds to fluctuations which have the same correlations no matter how separated the two points are:
${\rm Scale \ Invariant}\Longleftrightarrow \langle \phi(\lambda\bfx)\phi(\lambda\bfy)\rangle=\langle \phi(\bfx)\phi(\bfy)\rangle\ {\rm for \ all} \ \lambda$
Probability Distributions: Functions and Functionals
We now review how probability distributions work for continuous, Gaussian random fields. We start with a brief review of the analogous theory for continuous random variables.
Probability Distribution Functions (PDF): Given some continuous random variable $$q$$, the probability that $$q$$ lies in some range $$\rd q$$ is given by $$\P(q)\rd q$$, where $$\P(q)$$ is the PDF. The PDF is normalized as $$\int\P(q)\rd q=1$$ and the expectation value of $$q^{n}$$ is given by
$\langle q^n\rangle= \int\rd q\, \P(q)q^n$
The Gaussian PDF is of primary importance. For $$n$$ random variables, $$q_i$$, the Gaussian PDF takes on the form:
$\P^{\rm Gauss.}(q_1,\ldots,q_n)=\sqrt{\frac{\det (A)}{(2\pi)^n}}\exp\left[-\frac{1}{2}q_iA_{ij}q_j\right]\ ,$
where $$A_{ij}$$ is some positive definite matrix (and sums over repeated indices are implicit). The corresponding Gaussian two-point statistics are then given by:
$\langle q_i q_j\rangle =\left(\prod_{k=1}^{n}\int \rd q_k\right)\P^{\rm Gauss.}(\vec{q})q_iq_j=A^{-1}_{ij}$
where $$A^{-1}_{ij}$$ is the inverse of the matrix $$A_{ij}$$.
Probability Distribution Functionals are perhaps less familiar, but they are exactly the same idea as PDFs simply applied to the case of random fields such as $$\phi(\bfx)$$, rather than random variables such as $$q_i$$. We will only focus on the case of Gaussian functionals whose form and properties are defined in exact analogy to the PDF case. Starting with a position space field $$\phi(\bfx)$$, the Gaussian functional $$\P(\phi(\bfx))$$ is written as
${\cal P}[\phi(\bfx)]\propto \exp\left[-\frac{1}{2}\int\rd^dx\rd^dy\, \phi(\bfy)A(\bfy,\bfz)\phi(\bfz) \right]$
where we have omitted the normalization factor. Correlators are then computed by a path integral over all field values $$\phi(\bfx)$$. The outcome of the integral is determined in analogy to the PDF case. For instance, the two-point correlator is:
$\langle \phi(\bfx)\phi(\bfy)\rangle\equiv \int\D\phi\, \P[\phi(\bfx)]\phi(\bfy)\phi(\bfz)=A^{-1}(\bfy,\bfz)\ ,$
where $$A^{-1}(\bfy,\bfz)$$ obeys
$\int\rd^dy\, A(\bfx,\bfy)A^{-1}(\bfy,\bfz)=\delta^{d}(\bfx-\bfz)\ .$
White noise corresponds to the simple case where $$A(\bfx,\bfz)=C\delta^{d}(\bfx-\bfz)$$ and so $$A^{-1}(\bfx,\bfz)=\frac{1}{C}\delta^{d}(\bfx-\bfz)$$, for instance. Assuming spatial homogeneity and isotropy, it is straightforward to Fourier transform the above formulas in order to derive their momentum space analogues:
\begin{align} \P[\phi_{\bfk}]&\propto \exp\left[-\frac{1}{2}\int\rd^{d}\tilde{q}\rd^{d}\tilde{p}\, \phi_{\bfq}A(q)\tilde{\delta}^{d}(\bfq+\bfp)\phi_{\bfp}\right]\nn \langle\phi_{\bfk}\phi_{\bfk'}\rangle&=\frac{\tilde{\delta}^{d}(\bfk+\bfk')}{A(k)} \end{align}
Clearly, the function $$A(k)$$ above is simply the inverse of the power spectrum: $$P(k)=A^{-1}(k)$$. By adding terms cubic and higher order in $$\phi_\bfk$$ to the exponential in $$\P[\phi_\bfk]$$, the distribution is made non-Gaussian. The search for non-Gaussian signatures in Cosmological statistics is part of cutting-edge experimental efforts, but further discussion of these issues falls outside of the scope of this post.
Implementation
We now demonstrate how to generate a realization, given $$P(k)$$. We first present the steps to the process and explain the logic. Then we show how to practically implement the steps in Mathematica.
The Steps
A realization of a Gaussian random field with power spectrum can be created via the following steps:
1. Start with a white noise field with unit amplitude $$\varphi_{\bfk}$$ obeying $$\langle \varphi_{\bfk}\varphi_{-\bfk}\rangle'=1$$.
2. Generate a position space realization of the white noise, denoted by $$R_{\rm white}(\bfx)$$. That is, $$R_{\rm white}(\bfx)$$ is a particular map showing the values of $$\varphi$$ at various positions $$\bfx$$ and for which $$\langle\varphi(\bfx)\varphi(\bfy)\rangle'=\delta^{d}(\bfx-\bfy)$$.
3. Fourier transform the realization: $$R_{\rm white}(\bfx)\longrightarrow R_{\rm white}(\bfk)$$.
4. Multiply $$R_{\rm white}(\bfk)$$ by the square root of the power spectrum to create $$R_P(\bfk)=P^{1/2}(k)R_{\rm white}(\bfk)$$.
5. Fourier transform $$R_P(\bfk)$$ back to positio space to get the desired realization: $$R_P(\bfx)=\int\rd^d \tilde{k}\, e^{i\bfk\cdot\bfx} R_P(\bfx)$$.
The logic behind the above steps is pretty simple. If we’re given a white noise field $$\phi_\bfk$$ which has $$\langle \varphi_{\bfk}\varphi_{-\bfk}\rangle'=1$$ , then we can create a field with the desired power spectrum simply by defining $$\phi_\bfk\equiv P^{1/2}(k)\varphi_\bfk$$. The new field has the desired statistics, by construction.
Therefore, given a realization of the white noise field , we just have to multiply its Fourier space realization by and inverse Fourier transform to get a position space realization of $$\phi(\bfx)$$. The above is practical, because making a white noise realization is easy.
These steps are idealized, as we’ll have to make the realization on a discrete grid; we won’t be able to work with continuous variables. Practical complications which arise from working on a lattice are covered in the following sections.
From Continuous to Discrete
For concreteness, we focus on creating two-dimensional realizations on $$N\times N$$ grids, with $$N$$ an even number. The generalization to higher dimensions is straightforward. In passing to a lattice, our continuous fields $$\phi(\bfx)$$ and $$\phi_\bfk$$ are replaced by the discrete quantities $$\phi^{\bfx}_{ab}$$ and $$\phi^{\bfk}_{ab}$$, respectively, where $$a,b\in\{0,\ldots,N-1\}$$. $$\phi^{\bfx}_{ab}$$ is the value of the field at the point $$(a,b)$$ on the grid. The continuous Fourier transform which related $$\phi^{\bfx}_{ab}$$ and $$\phi^{\bfk}_{ab}$$ is replaced by a discrete one:
\begin{align} \phi^{\bfk}_{ab}&=\sum_{c,d=0}^{N-1}\exp\left(-ix_ck_a-ix_dk_b\right)\phi^{\bfx}_{cd}\nn \phi^{\bfx}_{ab}&=\frac{1}{N^2}\sum_{c,d=0}^{N-1}\exp\left(ix_ck_a+ix_dk_b\right)\phi^{\bfk}_{cd}\nn \end{align}
where $$x_a\equiv a$$ and $$k_a\equiv \frac{2\pi a}{N}$$. Given real $$\phi^{\bfx}_{ab}$$ and even $$N$$, $$\phi^{\bfk}_{ab}$$ has the following properties
• $$\phi^{\bfk}_{ab}=\phi^{\bfk}_{(a+\alpha N)b}=\phi^{\bfk}_{a(b+\alpha N)}$$ for any integer $$\alpha$$
• $$\phi^{*\bfk}_{ab}=\phi^{\bfk}_{-a-b}$$
The above further imply the following important property: $$\phi^{*\bfk}_{a(N/2+b)}=\phi^{\bfk}_{a(N/2-b)}$$ and similar with the other index. and similar with the other index.
Building a Realization
Now we walk through the steps involved in actually building a realization in Mathematica.
White Noise
First, we construct a grid of white noise. When passing to the grid, the unit white noise probability distribution functional becomes the following ordinary PDF:
$\P[\varphi(\bfx)]\longrightarrow \P[\varphi^{\bfx}_{ab}]=\prod_{c,d=0}^{N-1}\frac{\exp\left[-\frac{1}{2}\left(\phi^{\bfx}_{cd}\right)^2\right]}{\sqrt{2\pi}}$
Building a white noise realization is therefore simple: the value of at every grid point is just randomly chosen from a Gaussian distribution. An example of such a realization is shown in Fig. 1 below.
In Mathematica, the white noise array can be generated via:
WhiteNoise=RandomVariate[NormalDistribution[],{Nsize,Nsize}]
where Nsize is the size of the grid (Nsize=1024 in Fig. 1). WhiteNoise is our realization $$R_{\rm white}(\bfx)$$ and building $$R_{\rm white}(\bfk)$$ is very simple: we just pass WhiteNoise through the built-in Fourier function:
WhiteNoiseFourier=Fourier[WhiteNoise]
Constructing the Fourier Realization
The Fourier space realization then needs to be multiplied by the square root of the power spectrum. This is subtle, as the resulting field $$\phi^{\bfk}_{ab}$$ needs to obey the conditions we derived in the previous section, particularly $$\phi^{*\bfk}_{a(N/2+b)}=\phi^{\bfk}_{a(N/2-b)}$$. .
This is important because if we did the naive construction and took the white noise realization values $$\varphi^{\bfk}_{ab}$$ and simply multiplied by $$P^{1/2}(k)$$ with $$k=\frac{2\pi}{ N}\sqrt{a^2+b^2}$$ to build $$\phi^{\bfk}_{ab}=P^{1/2}(k)\varphi^{\bfk}_{ab}$$, then we'd find $$\phi^{*\bfk}_{a(N/2+b)}\neq\phi^{\bfk}_{a(N/2-b)}$$, due to the power spectrum factor. If we then transformed the naive $$\phi^{\bfk}_{ab}$$, we would be led to an imaginary $$\phi^{\bfx}_{ab}$$. We can get around this issue by building the array $$\Phi^{\bfk}_{ab}$$ whose components are given by:
$\Phi^{\bfk}_{ab}= \begin{cases} P^{1/2}(k)\varphi^{\bfk}_{ab} & a,b\le N/2\\ P^{1/2}(k)\varphi^{\bfk}_{a(b-N)} & a\le N/2, b>N/2\\ P^{1/2}(k)\varphi^{\bfk}_{(a-N)b} & a> N/2, b\le N/2\\ P^{1/2}(k)\varphi^{\bfk}_{(a-N)(b-N)} & a,b> N/2 \end{cases}$
In the second line above, the $$k$$ in $$P^{1/2}(k)\varphi^{\bfk}_{a(b-N)}$$ is evaluated at $$k=\frac{2\pi}{N}\sqrt{a^2+(b-N)^2}$$ and similar for other lines. The factors of $$N$$ can also be removed in the indices of the $$\varphi^{\bfk}$$'s, using the previously mentioned properties of $$\varphi^{\bfk}$$.
Note that these manipulations effectively result in the the $$a,b$$ indices running over the values $$\{0,\ldots, N/2,-N/2+1,\ldots\}$$, rather than $$\{0,\ldots, N-1\}$$. In the limit $$P(k)\to 1$$, Fourier transforming $$\Phi^{\bfk}_{ab}$$ yields the original position space white noise realization we started with, $$\varphi^{\bfk}_{ab}$$, , despite the above manipulations. The position space field is then built via
$\phi^{\bfx}_{ab}=\frac{1}{N^2}\sum_{c,d=0}^{N-1}\exp\left(ix_ck_a+ix_dk_b\right)\Phi^{\bfk}_{cd}\ .$
This field realizes the desired spectrum. The construction of can be carried out in Mathematica as follows:
1. We first build the vector $$k_a=\frac{2\pi}{N}(0,\ldots, N/2,-N/2+1,\ldots, -1)$$ as:
kVector=N[(2π)/Nsize*Join[Range[0,Nsize/2],Range[-Nsize/2+1,-1]]]
2. Next, build the power spectrum function. Take it to be of the power law form: $$P(k)\propto k^{-n}$$. Because this $$P(k)$$ is divergent at $$k=0$$ (assuming $$n>0$$), we will set the value of the power spectrum to zero at this one point. This is implemented in Mathematica as:
PowerLawPowerSpectrum[n_,k_]:= If[k==0,0,1/Abs[k^n]]
3. We want to evaluate $$P^{1/2}(k)$$ at all points $$(k_a,k_b)$$ with $$k=\sqrt{k_a^2+k_b^2}$$. This array is constructed using the Outer command:
sqrtPowerSpectrumArray=Outer[N[Sqrt[PowerLawPowerSpectrum[n,Norm[{##}]]]]&,kVector,kVector]
where $$n$$ should be replaced by the desired number determining the power law.
4. Finally, we mutiply the $$a,b$$ entry of sqrtPowerSpectrumArray by the $$a,b$$ entry of WhiteNoiseFourier to get $$\Phi^{\bfk}_{ab}$$:
PowerSpectrumRealizationFourier=sqrtPowerSpectrumArray*WhiteNoiseFourier
Note that the above multiplication is carried out with the Times function, not Dot.
Plotting the Realization
Finally, we transform from $$\Phi^{\bfk}_{ab}\longrightarrow \phi^{\bfx}_{ab}$$ and plot the result. This is done in a single command as:
MatrixPlot[Re[InverseFourier[PowerSpectrumRealizationFourier]]]
Above, the real part of the transform was taken using Re because numerical errors induce tiny imaginary components in the final calculation of $$\phi^{\bfx}_{ab}$$. The results of the realization for various power spectra can be seen in Fig. 2.
An Example in $$d=3$$
Finally, we present the results of a higher dimensional realization: Fig. 3 is a GIF which shows successive slices of a $$d=3$$ realization for a power spectrum $$P(k)=k^{-4}$$.
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2020-12-03 11:18:15
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https://labs.tib.eu/arxiv/?author=Tom%20Coates
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• ### Gromov-Witten Invariants of Local P^2 and Modular Forms(1804.03292)
April 10, 2018 math-ph, math.MP, math.AG, math.SG
We construct a sheaf of Fock spaces over the moduli space of elliptic curves E_y with Gamma_1(3)-level structure, arising from geometric quantization of H^1(E_y), and a global section of this Fock sheaf. The global section coincides, near appropriate limit points, with the Gromov-Witten potentials of local P^2 and of the orbifold C^3/mu_3. This proves that the Gromov-Witten potentials of local P^2 are quasi-modular functions for the group Gamma_1(3), as predicted by Aganagic-Bouchard-Klemm, and proves the Crepant Resolution Conjecture for [C^3/mu_3] in all genera.
• ### A Fock Sheaf For Givental Quantization(1411.7039)
April 10, 2018 math-ph, math.MP, math.AG
We give a global, intrinsic, and co-ordinate-free quantization formalism for Gromov-Witten invariants and their B-model counterparts, which simultaneously generalizes the quantization formalisms described by Witten, Givental, and Aganagic-Bouchard-Klemm. Descendant potentials live in a Fock sheaf, consisting of local functions on Givental's Lagrangian cone that satisfy the (3g-2)-jet condition of Eguchi-Xiong; they also satisfy a certain anomaly equation, which generalizes the Holomorphic Anomaly Equation of Bershadsky-Cecotti-Ooguri-Vafa. We interpret Givental's formula for the higher-genus potentials associated to a semisimple Frobenius manifold in this setting, showing that, in the semisimple case, there is a canonical global section of the Fock sheaf. This canonical section automatically has certain modularity properties. When X is a variety with semisimple quantum cohomology, a theorem of Teleman implies that the canonical section coincides with the geometric descendant potential defined by Gromov-Witten invariants of X. We use our formalism to prove a higher-genus version of Ruan's Crepant Transformation Conjecture for compact toric orbifolds. When combined with our earlier joint work with Jiang, this shows that the total descendant potential for compact toric orbifold X is a modular function for a certain group of autoequivalences of the derived category of X.
• ### Laurent Inversion(1707.05842)
July 18, 2017 math.AG
We describe a practical and effective method for reconstructing the deformation class of a Fano manifold X from a Laurent polynomial f that corresponds to X under Mirror Symmetry. We explore connections to nef partitions, the smoothing of singular toric varieties, and the construction of embeddings of one (possibly-singular) toric variety in another. In particular, we construct degenerations from Fano manifolds to singular toric varieties; in the toric complete intersection case, these degenerations were constructed previously by Doran--Harder. We use our method to find models of orbifold del Pezzo surfaces as complete intersections and degeneracy loci, and to construct a new four-dimensional Fano manifold.
• ### K-Theoretic and Categorical Properties of Toric Deligne--Mumford Stacks(1410.0027)
Aug. 25, 2016 math.AG
We prove the following results for toric Deligne-Mumford stacks, under minimal compactness hypotheses: the Localization Theorem in equivariant K-theory; the equivariant Hirzebruch-Riemann-Roch theorem; the Fourier--Mukai transformation associated to a crepant toric wall-crossing gives an equivariant derived equivalence.
• ### Hodge-Theoretic Mirror Symmetry for Toric Stacks(1606.07254)
June 23, 2016 math.AG, math.SG
Using the mirror theorem [CCIT15], we give a Landau-Ginzburg mirror description for the big equivariant quantum cohomology of toric Deligne-Mumford stacks. More precisely, we prove that the big equivariant quantum D-module of a toric Deligne-Mumford stack is isomorphic to the Saito structure associated to the mirror Landau-Ginzburg potential. We give a GKZ-style presentation of the quantum D-module, and a combinatorial description of quantum cohomology as a quantum Stanley-Reisner ring. We establish the convergence of the mirror isomorphism and of quantum cohomology in the big and equivariant setting.
• ### Virasoro Constraints for Toric Bundles(1508.06282)
Aug. 25, 2015 math.AG
We show that the Virasoro conjecture in Gromov--Witten theory holds for the the total space of a toric bundle $E \to B$ if and only if it holds for the base $B$. The main steps are: (i) we establish a localization formula that expresses Gromov--Witten invariants of $E$, equivariant with respect to the fiberwise torus action, in terms of genus-zero invariants of the toric fiber and all-genus invariants of $B$; and (ii) we pass to the non-equivariant limit in this formula, using Brown's mirror theorem for toric bundles.
• ### Laurent Inversion(1505.01855)
May 12, 2015 math.AG
There are well-understood methods, going back to Givental and Hori--Vafa, that to a Fano toric complete intersection X associate a Laurent polynomial f that corresponds to X under mirror symmetry. We describe a technique for inverting this process, constructing the toric complete intersection X directly from its Laurent polynomial mirror f. We use this technique to construct a new four-dimensional Fano manifold.
• ### Quantum Periods for 3-Dimensional Fano Manifolds(1303.3288)
April 2, 2015 math.AG
The quantum period of a variety X is a generating function for certain Gromov-Witten invariants of X which plays an important role in mirror symmetry. In this paper we compute the quantum periods of all 3-dimensional Fano manifolds. In particular we show that 3-dimensional Fano manifolds with very ample anticanonical bundle have mirrors given by a collection of Laurent polynomials called Minkowski polynomials. This was conjectured in joint work with Golyshev. It suggests a new approach to the classification of Fano manifolds: by proving an appropriate mirror theorem and then classifying Fano mirrors. Our methods are likely to be of independent interest. We rework the Mori-Mukai classification of 3-dimensional Fano manifolds, showing that each of them can be expressed as the zero locus of a section of a homogeneous vector bundle over a GIT quotient V/G, where G is a product of groups of the form GL_n(C) and V is a representation of G. When G=GL_1(C)^r, this expresses the Fano 3-fold as a toric complete intersection; in the remaining cases, it expresses the Fano 3-fold as a tautological subvariety of a Grassmannian, partial flag manifold, or projective bundle thereon. We then compute the quantum periods using the Quantum Lefschetz Hyperplane Theorem of Coates-Givental and the Abelian/non-Abelian correspondence of Bertram-Ciocan-Fontanine-Kim-Sabbah.
• ### Mirror Symmetry and the Classification of Orbifold del Pezzo Surfaces(1501.05334)
Jan. 23, 2015 math.CO, math.AG
We state a number of conjectures that together allow one to classify a broad class of del Pezzo surfaces with cyclic quotient singularities using mirror symmetry. We prove our conjectures in the simplest cases. The conjectures relate mutation-equivalence classes of Fano polygons with Q-Gorenstein deformation classes of del Pezzo surfaces.
• ### Four-dimensional Fano toric complete intersections(1409.5030)
Dec. 16, 2014 math.AG
We find at least 527 new four-dimensional Fano manifolds, each of which is a complete intersection in a smooth toric Fano manifold.
• ### Quantum Periods For Certain Four-Dimensional Fano Manifolds(1406.4891)
Dec. 2, 2014 math.AG
We collect a list of known four-dimensional Fano manifolds and compute their quantum periods. This list includes all four-dimensional Fano manifolds of index greater than one, all four-dimensional toric Fano manifolds, all four-dimensional products of lower-dimensional Fano manifolds, and certain complete intersections in projective bundles.
• ### On the Convergence of Gromov-Witten Potentials and Givental's Formula(1203.4193)
Nov. 25, 2014 math.AG, math.SG
Let X be a smooth projective variety. The Gromov-Witten potentials of X are generating functions for the Gromov-Witten invariants of X: they are formal power series, sometimes in infinitely many variables, with Taylor coefficients given by Gromov-Witten invariants of X. It is natural to ask whether these formal power series converge. In this paper we describe and analyze various notions of convergence for Gromov-Witten potentials. Using results of Givental and Teleman, we show that if the quantum cohomology of X is analytic and generically semisimple then the genus-g Gromov-Witten potential of X converges for all g. We deduce convergence results for the all-genus Gromov-Witten potentials of compact toric varieties, complete flag varieties, and certain non-compact toric varieties.
• ### On the Existence of a Global Neighbourhood(1411.3092)
Nov. 12, 2014 math.AG
Suppose that a complex manifold M is locally embedded into a higher-dimensional neighbourhood as a submanifold. We show that, if the local neighbourhood germs are compatible in a suitable sense, then they glue together to give a global neighbourhood of M. As an application, we prove a global version of Hertling--Manin's unfolding theorem for germs of TEP structures; this has applications in the study of quantum cohomology.
• ### Mutations of Fake Weighted Projective Spaces(1312.0921)
Nov. 2, 2014 math.CO, math.AG
We characterise mutations between fake weighted projective spaces, and give explicit formulas for how the weights and multiplicity change under mutation. In particular, we prove that multiplicity-preserving mutations between fake weighted projective spaces are mutations over edges of the corresponding simplices. As an application, we analyse the canonical and terminal fake weighted projective spaces of maximal degree.
• ### The Crepant Transformation Conjecture for Toric Complete Intersections(1410.0024)
Oct. 9, 2014 math.AG
Let X and Y be K-equivalent toric Deligne-Mumford stacks related by a single toric wall-crossing. We prove the Crepant Transformation Conjecture in this case, fully-equivariantly and in genus zero. That is, we show that the equivariant quantum connections for X and Y become gauge-equivalent after analytic continuation in quantum parameters. Furthermore we identify the gauge transformation involved, which can be thought of as a linear symplectomorphism between the Givental spaces for X and Y, with a Fourier-Mukai transformation between the K-groups of X and Y, via an equivariant version of the Gamma-integral structure on quantum cohomology. We prove similar results for toric complete intersections. We impose only very weak geometric hypotheses on X and Y: they can be non-compact, for example, and need not be weak Fano or have Gorenstein coarse moduli space. Our main tools are the Mirror Theorems for toric Deligne-Mumford stacks and toric complete intersections, and the Mellin-Barnes method for analytic continuation of hypergeometric functions.
• ### Some Applications of the Mirror Theorem for Toric Stacks(1401.2611)
Oct. 8, 2014 math.AG
We use the mirror theorem for toric Deligne-Mumford stacks, proved recently by the authors and by Cheong-Ciocan-Fontanine-Kim, to compute genus-zero Gromov-Witten invariants of a number of toric orbifolds and gerbes. We prove a mirror theorem for a class of complete intersections in toric Deligne-Mumford stacks, and use this to compute genus-zero Gromov-Witten invariants of an orbifold hypersurface.
• ### A Mirror Theorem for Toric Stacks(1310.4163)
Oct. 2, 2014 math.AG
We prove a Givental-style mirror theorem for toric Deligne--Mumford stacks X. This determines the genus-zero Gromov--Witten invariants of X in terms of an explicit hypergeometric function, called the I-function, that takes values in the Chen--Ruan orbifold cohomology of X.
• ### The Quantum Lefschetz Principle for Vector Bundles as a Map Between Givental Cones(1405.2893)
May 12, 2014 math.AG
Givental has defined a Lagrangian cone in a symplectic vector space which encodes all genus-zero Gromov-Witten invariants of a smooth projective variety X. Let Y be the subvariety in X given by the zero locus of a regular section of a convex vector bundle. We review arguments of Iritani, Kim-Kresch-Pantev, and Graber, which give a very simple relationship between the Givental cone for Y and the Givental cone for Euler-twisted Gromov-Witten invariants of X. When the convex vector bundle is the direct sum of nef line bundles, this gives a sharper version of the Quantum Lefschetz Hyperplane Principle.
• ### Minkowski Polynomials and Mutations(1212.1785)
Dec. 8, 2012 math.CO, math.AG
Given a Laurent polynomial f, one can form the period of f: this is a function of one complex variable that plays an important role in mirror symmetry for Fano manifolds. Mutations are a particular class of birational transformations acting on Laurent polynomials in two variables; they preserve the period and are closely connected with cluster algebras. We propose a higher-dimensional analog of mutation acting on Laurent polynomials f in n variables. In particular we give a combinatorial description of mutation acting on the Newton polytope P of f, and use this to establish many basic facts about mutations. Mutations can be understood combinatorially in terms of Minkowski rearrangements of slices of P, or in terms of piecewise-linear transformations acting on the dual polytope P^* (much like cluster transformations). Mutations map Fano polytopes to Fano polytopes, preserve the Ehrhart series of the dual polytope, and preserve the period of f. Finally we use our results to show that Minkowski polynomials, which are a family of Laurent polynomials that give mirror partners to many three-dimensional Fano manifolds, are connected by a sequence of mutations if and only if they have the same period.
• ### Mirror Symmetry and Fano Manifolds(1212.1722)
Dec. 7, 2012 math.AG
We consider mirror symmetry for Fano manifolds, and describe how one can recover the classification of 3-dimensional Fano manifolds from the study of their mirrors. We sketch a program to classify 4-dimensional Fano manifolds using these ideas.
• ### The Quantum Lefschetz Hyperplane Principle Can Fail for Positive Orbifold Hypersurfaces(1202.2754)
Feb. 13, 2012 math.AG
We show that the Quantum Lefschetz Hyperplane Principle can fail for certain orbifold hypersurfaces and complete intersections. It can fail even for orbifold hypersurfaces defined by a section of an ample line bundle.
• ### Wall-Crossings in Toric Gromov-Witten Theory I: Crepant Examples(math/0611550)
Nov. 16, 2008 math-ph, math.MP, math.AG
Let X be a Gorenstein orbifold and let Y be a crepant resolution of X. We state a conjecture relating the genus-zero Gromov--Witten invariants of X to those of Y, which differs in general from the Crepant Resolution Conjectures of Ruan and Bryan--Graber, and prove our conjecture when X = P(1,1,2) and X = P(1,1,1,3). As a consequence, we see that the original form of the Bryan--Graber Conjecture holds for P(1,1,2) but is probably false for P(1,1,1,3). Our methods are based on mirror symmetry for toric orbifolds.
• ### On the Crepant Resolution Conjecture in the Local Case(0810.2200)
Oct. 13, 2008 math-ph, math.MP, math.AG
In this paper we analyze four examples of birational transformations between local Calabi-Yau 3-folds: two crepant resolutions, a crepant partial resolution, and a flop. We study the effect of these transformations on genus-zero Gromov-Witten invariants, proving the Coates-Corti-Iritani-Tseng/Ruan form of the Crepant Resolution Conjecture in each case. Our results suggest that this form of the Crepant Resolution Conjecture may also hold for more general crepant birational transformations. They also suggest that Ruan's original Crepant Resolution Conjecture should be modified, by including appropriate "quantum corrections", and that there is no straightforward generalization of either Ruan's original Conjecture or the Cohomological Crepant Resolution Conjecture to the case of crepant partial resolutions. Our methods are based on mirror symmetry for toric orbifolds.
• ### The Crepant Resolution Conjecture for Type A Surface Singularities(0704.2034)
July 10, 2008 math-ph, math.MP, math.AG
Let X be an orbifold with crepant resolution Y. The Crepant Resolution Conjectures of Ruan and Bryan-Graber assert, roughly speaking, that the quantum cohomology of X becomes isomorphic to the quantum cohomology of Y after analytic continuation in certain parameters followed by the specialization of some of these parameters to roots of unity. We prove these conjectures in the case where X is a surface singularity of type A. The key ingredient is mirror symmetry for toric orbifolds.
• ### Computing Genus-Zero Twisted Gromov-Witten Invariants(math/0702234)
July 10, 2008 math-ph, math.MP, math.AG
Twisted Gromov-Witten invariants are intersection numbers in moduli spaces of stable maps to a manifold or orbifold X which depend in addition on a vector bundle over X and an invertible multiplicative characteristic class. Special cases are closely related to local Gromov-Witten invariants of the bundle, and to genus-zero one-point invariants of complete intersections in X. We develop tools for computing genus-zero twisted Gromov-Witten invariants of orbifolds and apply them to several examples. We prove a "quantum Lefschetz theorem" which expresses genus-zero one-point Gromov-Witten invariants of a complete intersection in terms of those of the ambient orbifold X. We determine the genus-zero Gromov-Witten potential of the type A surface singularity C^2/Z_n. We also compute some genus-zero invariants of C^3/Z_3, verifying predictions of Aganagic-Bouchard-Klemm. In a self-contained Appendix, we determine the relationship between the quantum cohomology of the A_n surface singularity and that of its crepant resolution, thereby proving the Crepant Resolution Conjectures of Ruan and Bryan-Graber in this case.
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2021-04-20 10:13:06
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https://www.zbmath.org/?q=ai%3Atauber.clement+ai%3Amarcelli.giovanna
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# zbMATH — the first resource for mathematics
Spin conductance and spin conductivity in topological insulators: analysis of kubo-like terms. (English) Zbl 07063411
Summary: We investigate spin transport in 2-dimensional insulators, with the long-term goal of establishing whether any of the transport coefficients corresponds to the Fu-Kane-Mele index which characterizes 2$$d$$ time-reversal-symmetric topological insulators. Inspired by the Kubo theory of charge transport, and by using a proper definition of the spin current operator (Shi et al. in Phys Rev Lett 96:076604, 2006), we define the Kubo-like spin conductance $${G_K^{s_z}}$$ and spin conductivity $${\sigma _K^{s_z}}$$. We prove that for any gapped, periodic, near-sighted discrete Hamiltonian, the above quantities are mathematically well defined and the equality $${G_K^{s_z} = \sigma _K^{s_z}}$$ holds true. Moreover, we argue that the physically relevant condition to obtain the equality above is the vanishing of the mesoscopic average of the spin-torque response, which holds true under our hypotheses on the Hamiltonian operator. A central role in the proof is played by the trace per unit volume and by two generalizations of the trace, the principal value trace and its directional version.
##### MSC:
47A General theory of linear operators 81T Quantum field theory; related classical field theories 81Q General mathematical topics and methods in quantum theory 81U Quantum scattering theory 47L Linear spaces and algebras of operators
Full Text:
##### References:
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Lett., 95, 146802, (2005) [26] Kane, CL; Mele, EJ, Quantum spin Hall effect in graphene, Phys. Rev. Lett., 95, 226801, (2005) [27] Katsura, H.; Koma, T., The $$\mathbb{Z}_2$$ index of disordered topological insulators with time reversal symmetry, J. Math. Phys., 57, 021903, (2016) · Zbl 1341.82043 [28] Kirsch, W.: An Invitation to Random Schrödinger Operators, Preprint arXiv:0709.3707 [29] Kitaev, A., Periodic table for topological insulators and superconductors, AIP Conf. Proc., 1134, 22, (2009) · Zbl 1180.82221 [30] Kohn, W., Density functional and density matrix method scaling linearly with the number of atoms, Phys. Rev. Lett., 76, 3168, (1996) [31] Marcelli, G.: A mathematical analysis of spin and charge transport in topological insulators, Ph.D. thesis in Mathematics, La Sapienza Universitá di Roma, Rome (2017) [32] Marcelli, G., Monaco, D., Panati, G., Teufel, S.: Quantum (Spin) Hall Conductivity: Kubo Formula (and beyond), to appear (2019) [33] Marcelli, G., Panati, G., Tauber, C.: Quantum Spin Hall conductance: a first principle analysis, in preparation (2019) [34] Monaco, D.; Panati, G., Symmetry and localization in periodic crystals: triviality of Bloch bundles with a fermionic time-reversal symmetry, Acta Appl. Math., 137, 185-203, (2015) · Zbl 1318.82045 [35] Monaco, D.; Tauber, C., Gauge-theoretic invariants for topological insulators: a bridge between Berry, Wess-Zumino, and Fu-Kane-Mele, Lett. Math. Phys., 107, 1315-1343, (2017) · Zbl 1370.35093 [36] Moore, JE; Balents, L., Topological invariants of time-reversal-invariant band structures, Phys. Rev. B, 75, 121306(r), (2007) [37] Murakami, S., Quantum spin Hall effect and enhanced magnetic response by spin-orbit coupling, Phys. Rev. Lett., 97, 236805, (2006) [38] Panati, G., Triviality of Bloch and Bloch-Dirac bundles, Ann. Henri Poincaré, 8, 995-1011, (2007) · Zbl 1375.81102 [39] Prodan, E.; Kohn, W., Nearsightedness of electronic matter, Proc. Natl. Acad. Sci. USA, 102, 11635-11638, (2005) [40] Prodan, E., Robustness of the spin-Chern number, Phys. Rev. B, 80, 125327, (2009) [41] Prodan, E., Manifestly gauge-independent formulations of the $$\mathbb{Z}_2$$ invariants, Phys. Rev. B, 83, 235115, (2011) [42] Reed, M., Simon, B.: Methods of Modern Mathematical Physics I: Functional Analysis. Academic Press, New York (1979) · Zbl 0405.47007 [43] Ryu, S.; Schnyder, AP; Furusaki, A.; Ludwig, AWW, Topological insulators and superconductors: tenfold way and dimensional hierarchy, New J. Phys., 12, 065010, (2010) [44] Schulz-Baldes, H., Persistence of spin edge currents in disordered quantum spin Hall systems, Commun. Math. Phys., 324, 589-600, (2013) · Zbl 1278.82065 [45] Schulz-Baldes, H., $$\mathbb{Z}_2$$-indices and factorization properties of odd symmetric Fredholm operators, Doc. Math., 20, 1481-1500, (2015) · Zbl 1341.47014 [46] Shi, J.; Zhang, P.; Xiao, D.; Niu, Q., Proper definition of spin current in spin-orbit coupled systems, Phys. Rev. Lett., 96, 076604, (2006) [47] Simon, B.: Trace Ideals and Their Applications. American Mathematical Society, Providence (2005) · Zbl 1074.47001 [48] Sun, QF; Xie, XC; Wang, J., Persistent spin current in nano-devices and definition of the spin current, Phys. Rev. B, 77, 035327, (2008) [49] Thouless, DJ; Kohmoto, M.; Nightingale, MP; Nijs, M., Quantized Hall conductance in a two-dimensional periodic potential, Phys. Rev. Lett., 49, 405-408, (1982) [50] Wu, S.; etal., Observation of the quantum spin Hall effect up to $$100$$ Kelvin in a monolayer crystal, Science, 359, 76-79, (2018) [51] Zhang, P.; Wang, Z.; Shi, J.; Xiao, D.; Niu, Q., Theory of conserved spin current and its application to a two-dimensional hole gas, Phys. Rev. B, 77, 075304, (2008)
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.
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2021-01-24 17:59:50
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=19&t=47794&p=169476
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## Homework Question
$\Delta p \Delta x\geq \frac{h}{4\pi }$
Moderators: Chem_Mod, Chem_Admin
JohnWalkiewicz2J
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am
Been upvoted: 1 time
### Homework Question
On problem 1B.27, it gives the indeterminacy of velocity (Δv) as "5.00 ± 5.0 m.s^-1." Why is the indeterminacy of velocity 5 m.s^-1, and not 10 m.s^-1, since it is ± 5.0?
Last edited by JohnWalkiewicz2J on Thu Oct 17, 2019 10:41 am, edited 1 time in total.
005391550
Posts: 103
Joined: Thu Jul 25, 2019 12:15 am
### Re: Homework Question
do you mean the indeterminacy of velocity instead of the change in velocity?
JohnWalkiewicz2J
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am
Been upvoted: 1 time
### Re: Homework Question
My bad. Yes I do!
005391550
Posts: 103
Joined: Thu Jul 25, 2019 12:15 am
### Re: Homework Question
haha okay well then the indeterminacy would be just the ± 5.00 because that's what the measured amount is varied by? idk. i see what your saying but i think this is just how it is
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2020-08-14 17:37:45
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https://solvable.group/posts/ecfft/
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The ECFFT algorithm
$$\def\F{\mathbb{F}}$$
This post is about a recent paper by Eli Ben-Sasson, Dan Carmon, Swastik Kopparty and David Levit. In this paper the authors present an amazing new generalization of the classic FFT algorithm that works in all finite fields. This post will give an overview of the algorithm and a simple implementation in Sage. I highly recommend reading the paper for more details and background.
The classic FFT algorithm
Let $p$ be a prime number, $n=2^k$ with $n \mid p-1$, $\langle w \rangle = H < \F_p^*$ a subgroup of size $n$. The classic FFT algorithm can be used to evaluate a polynomial $P(X)=\sum_{i=0}^n a_i X^i$ of degree $<n$ on $H$ in $O(n\log n)$. Note that the naive algorithm of evaluating $P$ at every point of $H$ takes $O(n^2)$ operations.
The FFT works by writing $P$ as $$P(X) = P_0(X^2) + XP_1(X^2)$$ where $P_0, P_1$ are the polynomials of degree $< n/2$ of even and odd coefficients of $P$.
Thus, given the evaluation of $P_0$ and $P_1$ on $H^2$, we can recover the evaluation of $P$ on $H$ with $O(n)$ operations.
Now the crucial thing to note is that $H^2$ has half the size of $H$ since $H = -H$. Therefore, if we denote by $F$ the running time of the FFT, we have the following recurrence relation
$$F(n) = 2F(n/2) + O(n)$$ from which we can deduce that $F(n) = O(n\log n)$.
The ECFFT algorithm
The goal of the ECFFT algorithm is to generalize the FFT algorithm for finite fields that do not have a multiplicative subgroup of size $n$, i.e., fields $\F_p$ with $n \nmid p-1$.
The idea is both brillant and simple. Here’s an overview:
1. Find an elliptic curve $E(\F_p)$ with $n \mid \#E(\F_p)$. Let $G < E(\F_p)$ be a subgroup of size $n$, and $H$ a coset of $G$.
2. Let $\phi: E \to E'$ be an isogeny of degree $2$ such that $\phi(H)$ has half the size of $H$.
3. Use $\phi$ to decompose $P$ into two smaller polynomials $P_0,P_1$ and apply the ECFFT on $P_0,P_1$ using the elliptic curve $E'$ and the coset $\phi(H)$.
I’ll now explain these steps in more details. I’ll also give an implementation in Sage for the base field of the Secp256k1 curve, i.e., $\F_p$ with
$$p = 115792089237316195423570985008687907853269984665640564039457584007908834671663.$$
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
F = GF(p)
Finding a curve
We need to find a curve $E(\F_p)$ with $n \mid \#E(\F_p)$. This paper gives algorithms to do so. Otherwise, a brute-force search also works reasonably well for small values of $n$.
Here’s a script to find a curve by brute-force
def find_curve(n):
while True:
b = F.random_element()
E = EllipticCurve(F, [1, b])
if E.order() % n == 0:
return E
n = 2^12
E = find_curve(n)
It can find a curve for $n=2^{12}$ in around an hour. Here’s the curve I found
$$E: y^2 = x^3 + x + 641632526086088189863104799840287255425991771106608433941469413117059106896$$
Choosing a coset
Let $G < E$ be a subgroup of size $n$. We choose a random coset1 $H$ of $G$.
gen = E.gens()[0]
G = (gen.order()//n) * gen
R = E.random_element()
H = [R + i*G for i in range(n)]
Finally, let $L$ be the projection of $H$ on $\F_p$. $L$ is going to be the set of size $n$ on which we’ll be able to perform the ECFFT.
L = [h.xy()[0] for h in H]
Finding an isogeny
We need an isogeny that halves the size of $H$. This is very simple to find in Sage:
def find_isogeny(E, H):
for phi in E.isogenies_prime_degree(2):
if len(set([phi(h) for h in H])) == len(H)//2:
return phi
phi = find_isogeny(E, H)
On the $x$-coordinate of $E$, $\phi$ is given by the degree $2$ rational function $\psi(X) = u(X)/v(X)$.
psi = phi.x_rational_map()
u, v = psi.numerator(), psi.denominator()
From this we can get a new elliptic curve $E'$, which is the codomain of $\phi$, a new coset $\phi(H)$, and a new subset of $\F_p$ given by $\psi(L)$.
Decomposing polynomials on $L$
In the classic FFT, we were able to write the evaluation of $P$ on $H$ in terms of the evaluations of the two smaller polynomials $P_0,P_1$ on the smaller set $H^2$.
We now want to do the same by replacing $H$ with $L$, $H^2$ with $\psi(L)$, and the squaring operation with $\psi$.
Let $P$ be a polynomial of degree $<n$. Then there exist polynomials $P_0, P_1$ of degree $< n/2$ such that
$$P(X) = (P_0(\psi(X)) + XP_1(\psi(X)))v(X)^{n/2-1}$$ See Appendix A of the paper for a proof. The idea of the proof is that the linear map $(P_0, P_1) \mapsto P$ from pairs of polynomials of degree $<n/2$ to polynomials of degree $<n$ is injective, and thus bijective since its domain and codomain have the same dimension as $\F_p$-vector spaces.
Computing the polynomials $P_0,P_1$ is not as straightforward as in the classic FFT. However, it is easy to go from the evaluation of $P$ on $L$ to the evaluations of $P_0$ and $P_1$ on $\psi(L)$, and vice versa.
Indeed, given $s_0, s_1\in L$ with $\psi(s_0)=\psi(s_1)=t\in \psi(L)$, by writing the above formula for $P(X)$ at $X=s_0,s_1$ we get the following linear relation (letting $q=n/2 -1$)
$$\begin{bmatrix} P(s_0) \\\ P(s_1) \end{bmatrix} = \begin{bmatrix} v(s_0)^q & s_0v(s_0)^q \\\ v(s_1)^q & s_1v(s_1)^q \end{bmatrix} \begin{bmatrix} P_0(t) \\\ P_1(t) \end{bmatrix}$$ The matrix can be seen to be inversible, therefore we can easily go from $(P(s_0),P(s_1))$ to $(P_0(t), P_1(t))$ and back. This gives the following efficient correspondence that we’ll use later:
$$\text{Evaluation of P on L \longleftrightarrow Evaluations of P_0, P_1 on \psi(L).}$$
What’s more, the matrix doesn’t depend on $P$, so we can precompute it and reuse it for all instantiations of the ECFFT.
inverse_matrices = []
nn = n // 2
q = nn - 1
for j in range(nn):
s0, s1 = L[j], L[j + nn]
assert psi(s0) == psi(s1)
M = Matrix(F, [[v(s0)^q,s0*v(s0)^q],[v(s1)^q, s1*v(s1)^q]])
inverse_matrices.append(M.inverse())
The EXTEND operation
The final piece of the puzzle is the EXTEND operation. Let $S, S'$ be the elements of $L$ at even and odd indices respectively, so that $L = S \cup S'$.
S = [L[i] for i in range(0, n, 2)]
S_prime = [L[i] for i in range(1, n, 2)]
Given the evaluation of a polynomial $Q$ of degree $<n/2$ on $S$, the EXTEND operation computes the evaluation of $Q$ on $S'$. The main result of the paper is that there is an $O(n\log n)$ algorithm for EXTEND. Note that a naive algorithm that would recover the coefficients of $Q$ by Lagrange interpolation on $S$ and then evaluate on $S'$ takes $O(n^2)$.
The algorithm works as follows.
If $\#S = \#S' = 1$, $Q$ is constant and the evaluation of $Q$ on $S$ and $S'$ is the same.
Otherwise, deduce from the evaluation of $Q$ on $S$ the evaluations of $Q_0,Q_1$ on $\psi(S)$, as in the previous section. Then apply the EXTEND operation twice to get the evaluations of $Q_0,Q_1$ on $\psi(S')$. Finally, recover the evaluation of $Q$ on $S'$ as in the previous section.
def extend(Q_evals, S, S_prime, matrices, inverse_matrices):
n = len(Q_evals)
nn = n // 2
if n == 1:
return Q_evals
Q0_evals = []
Q1_evals = []
q = nn - 1
for j in range(nn):
s0, s1 = S[j], S[j + nn]
y0, y1 = Q_evals[j], Q_evals[j + nn]
Mi = inverse_matrices[n][j]
q0, q1 = Mi * vector([y0, y1])
Q0_evals.append(q0)
Q1_evals.append(q1)
Q0_evals_prime = extend(Q0_evals, [psi(s) for s in S], [psi(s) for s in S_prime], matrices, inverse_matrices)
Q1_evals_prime = extend(Q1_evals, [psi(s) for s in S], [psi(s) for s in S_prime], matrices, inverse_matrices)
return [
M * vector([q0, q1])
for M, q0, q1 in zip(matrices[n], Q0_evals_prime, Q1_evals_prime)
]
The recurrence relation for the running time is the same as in the classic FFT and gives a running time of $O(n\log n)$.
Evaluating polynomials on $L$
The algorithm for the EXTEND operation given in the last section is the building block for many efficient algorithms given in the paper.
One of them is an algorithm for the ENTER operation, which computes the evaluation of a polynomial $P$ of degree $<n$ on $L$. This is analogous to what we did in the classic FFT.
The algorithm is very simple. Decompose $P(X)$ into its low and high coefficient $U(X) + X^{n/2}V(X)$. Call ENTER on $U$ and $V$ on $S$ to get the evaluations of $U,V$ on $S$. Then call EXTEND twice to get the evaluations of $U,V$ on $S'$. Since $L = S \cup S'$, we have the evaluation of $U,V$ on $L$ and can deduce the evaluation of $P$ on $L$.
Note that the recurrence relation for the running time is now
$$F(n) = 2F(n/2) + O(n\log n)$$ since we have to call EXTEND in the recursion step. Therefore the running time $F(n) = O(n\log^2 n)$ is slightly worse than for the classic FFT.
Running it outside of Sage
As this post (hopefully) shows, these algorithms are very simple to implement in a computer algebra system like Sage. However, Sage being awfully slow, these implementations are far from fast.
The cool thing about these algorithms is that for a given field $\F_p$, we can precompute all the necessary data in Sage: all the sets $L, S, S'$ and the (inverse) matrices used in the EXTEND algorithm.
Once we have these precomputed data, the algorithms only use basic operations on $\F_p$, i.e., additions and multiplications.
Therefore, the actual algorithms can easily be implemeneted in fast languages like C++ or Rust, without having to implement all the elliptic curve and isogeny machinery in these languages.
Final code
Here is the final code that computes EXTEND for our choice of $\F_p$. It can also be found here.
# p is the size of the base field of the curve Secp256k1
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
F = GF(p)
# See the post on how to find a and b such that 2^12 divides the order of E.
a, b = 1, 641632526086088189863104799840287255425991771106608433941469413117059106896
E = EllipticCurve(F, [a,b])
log_n = 12
n = 2^log_n
assert E.order() % n == 0
g = E.gens()[0]
G = (g.order()//n) * g
assert G.order() == n
R = E.random_element()
H = [R + i*G for i in range(2^log_n)]
L = [h.xy()[0] for h in H]
S = [L[i] for i in range(0, n, 2)]
S_prime = [L[i] for i in range(1, n, 2)]
def precompute(log_n, S, S_prime, E):
Ss = {}
Ss_prime = {}
matrices = {}
inverse_matrices = {}
for i in range(log_n, -1, -1):
n = 1 << i
nn = n // 2
Ss[n] = S
Ss_prime[n] = S_prime
matrices[n] = []
inverse_matrices[n] = []
for iso in E.isogenies_prime_degree(2):
psi = iso.x_rational_map()
if len(set([psi(x) for x in S]))==nn:
break
v = psi.denominator()
q = nn - 1
for j in range(nn):
s0, s1 = S[j], S[j + nn]
assert psi(s0) == psi(s1)
M = Matrix(F, [[v(s0)^q,s0*v(s0)^q],[v(s1)^q, s1*v(s1)^q]])
inverse_matrices[n].append(M.inverse())
s0, s1 = S_prime[j], S_prime[j + nn]
assert psi(s0) == psi(s1)
M = Matrix(F, [[v(s0)^q,s0*v(s0)^q],[v(s1)^q, s1*v(s1)^q]])
matrices[n].append(M)
S = [psi(x) for x in S[:nn]]
S_prime = [psi(x) for x in S_prime[:nn]]
E = iso.codomain()
return Ss, Ss_prime, matrices, inverse_matrices
# Precompute the data needed to compute EXTEND_S,S'
Ss, Ss_prime, matrices, inverse_matrices = precompute(log_n-1, S, S_prime, E)
def extend(P_evals):
n = len(P_evals)
nn = n // 2
if n == 1:
return P_evals
S = Ss[n]
S_prime = Ss_prime[n]
P0_evals = []
P1_evals = []
for j in range(nn):
s0, s1 = S[j], S[j + nn]
y0, y1 = P_evals[j], P_evals[j + nn]
Mi = inverse_matrices[n][j]
p0, p1 = Mi * vector([y0, y1])
P0_evals.append(p0)
P1_evals.append(p1)
P0_evals_prime = extend(P0_evals)
P1_evals_prime = extend(P1_evals)
ansL = []
ansR = []
for M, p0, p1 in zip(matrices[n], P0_evals_prime, P1_evals_prime):
v = M * vector([p0, p1])
ansL.append(v[0])
ansR.append(v[1])
return ansL + ansR
# Generate a random polynomial for testing
R.<X> = F[]
P = sum(F.random_element() * X^i for i in range(1<<(log_n - 1)))
# Evaluate P on S
P_evals = [P(x) for x in S]
# result holds the evaluation of P on S'
result = extend(P_evals)
assert result == [P(x) for x in S_prime]
1. See section 4.2 of the paper for a more careful choice of the coset. A random one is fine with our parameters. ↩︎
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2022-01-20 04:05:02
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https://brilliant.org/problems/solve-it-if-you-can-3/
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# But there's so many terms
Calculus Level 3
$\large \lim_{n\to\infty} \frac{n^k \sin^2(n!) }{n+2}$
For variable $$0<k<1$$ independent of $$n$$, let $$A$$ denote the value of the limit above, which is a non-negative integer. Evaluate $$A!$$.
×
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2016-10-21 11:29:45
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http://ssa.gov/policy/docs/ssb/v67n2/v67n2p101.html
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# PERSPECTIVES: How Postsecondary Education Improves Adult Outcomes for Supplemental Security Income Children with Severe Hearing Impairments
by Robert R. Weathers II, Gerard Walter, Sara Schley, John Hennessey, Jeffrey Hemmeter, and Richard V. Burkhauser
Social Security Bulletin, Vol. 67 No. 2, 2007
Robert R. Weathers II is with the Social Security Administration. Gerard Walter and Sara Schley are with the National Technical Institute for the Deaf. John Hennessey and Jeffrey Hemmeter are with the Social Security Administration. Richard V. Burkhauser is with Cornell University.
Acknowledgments: Partial funding for the work reported in this article came from the U.S. Department of Education, National Institute of Disability and Rehabilitation Research (NIDRR), cooperative agreement 1331390038, and the National Technical Institute for the Deaf (NTID). This work does not necessarily reflect the views of NIDRR, NTID, or the Social Security Administration (SSA). The matched administrative data files used here could not have been created without the help of several senior administrators at NTID and SSA. We thank them for their commitment to this project. We also thank Joyce Manchester, L. Scott Muller, and Kalman Rupp for comments and suggestions on earlier drafts of this article. The data extract for this project is restricted-use, and permission must be granted by NTID and SSA to use these data.
## Summary
The rapid growth in the number of children participating in the Supplemental Security Income (SSI) program before the age of 18 has led policymakers to consider new methods of assisting children with disabilities in their transition from school to work. Postsecondary education represents one path that SSI children may take to acquire the skills necessary to enter employment and reduce dependency on the SSI disability program as adults. Yet little is known about SSI children's experience with postsecondary education, let alone their ability to increase their labor market earnings and reduce their time on SSI as adults in the long term. This lack of information on long-term outcomes is due in part to a lack of longitudinal data.
This article uses a unique longitudinal data set to conduct a case study of SSI children who applied for postsecondary education at the National Technical Institute for the Deaf (NTID) within the Rochester Institute of Technology. The data set was created by merging NTID administrative data on the characteristics and experiences of its applicants to Social Security Administration (SSA) longitudinal data on earnings and program participation. We used this data file to estimate the likelihood that an SSI child will graduate from NTID relative to other hearing-impaired NTID applicants, and we estimated the influence of graduation from NTID on participation in the SSI adult program and later success in the labor market.
The results of our analysis show that the percentage of NTID applicants who were SSI children increased over time, from a low of 10 percent in 1982 to more than 41 percent in 2000. However, the differences in the probability of graduation from NTID between deaf SSI children and deaf applicants who were not SSI children did not change accordingly. The probability of graduation for SSI children who applied to NTID was 13.5 percentage points lower than for those who were not SSI children. The estimated disparity indicates that targeting college retention programs toward SSI children may be an effective way to improve overall graduation rates.
Our results also show that SSI children who graduated from NTID spent less time in the SSI adult program and had higher earnings than SSI children who did not graduate. Compared with SSI children who were accepted to NTID but chose not to attend, SSI children who graduated from NTID left the SSI program 19 months earlier, were less likely to reenter the program, and at age 30 had increased their earnings by an estimated 49 percent. Our findings demonstrate that SSI children need not be relegated to a lifetime of SSI participation as adults, despite the poor overall labor market experience of this population since the creation of the SSI program in 1974.
## Introduction
The Supplemental Security Income (SSI) program is the largest federal means-tested cash assistance program in the United States. It is administered by the Social Security Administration (SSA) and provides assistance to children with disabilities, working-age adults with disabilities, and the aged, as long as they meet the income and resource requirements necessary for eligibility.1 In 2005, approximately 1 million children under the age of 18 received disability payments through the SSI program. The number of children receiving SSI has tripled over the past 15 years, far outpacing the growth of working-age adults and the aged receiving it (Social Security Administration 2006). Many of these children are likely to participate in the SSI disability program for a majority of their lifetime (Rupp and Scott 1995) because they are unlikely to reach the income or resource levels, either through work or through other means, to make a long-term exit from the SSI program. The rapid growth in the number of children receiving disability payments and the evidence that suggests that many of them will depend on these benefits for most of their lives has prompted policymakers to consider new methods to assist children in the transition from school to work. SSA program administrators have referred to these efforts as "managing against the risk of disability."2
Postsecondary education represents one path that SSI children (that is, those who enter the SSI program before age 18) may take to acquire the skills necessary to enter employment and reduce dependency on the adult SSI disability program. Yet little is known about SSI children's experience with postsecondary education, let alone its ability to increase their labor market earnings and reduce their time on SSI as adults in the long term. This lack of information on the long-term outcomes is due in part to the absence of longitudinal data on them.3
The findings reported here are from a unique longitudinal data set we created. The data set consists of administrative records from the Rochester Institute of Technology's National Technical Institute for the Deaf (NTID) linked to data from SSA's Supplemental Security Record, the Master Earnings File, and the Numident file. We use these data to conduct a case study of the subsequent educational and labor market success of SSI children as well as their SSI program participation as adults, relative to other deaf children who apply for postsecondary education.
The case study followed persons with severe hearing impairments who applied to NTID, one of two federally supported postsecondary schools that serve the population with severe hearing impairments. The postsecondary education programs offered at NTID include vocational degree programs that provide specific training for particular occupations. They also include professional degree programs that may lead to an associate of science, bachelor of arts, or master of arts degree. Almost all NTID applicants have hearing impairments that meet the medical criteria used to determine eligibility for the Social Security disability programs, and so they also are eligible to receive SSI adult benefits if they meet the income and resource tests.
We found that SSI children who graduated from NTID spent less time in the SSI adult program and had higher earnings than SSI children who did not graduate. However, we also found that SSI children who applied to NTID had a greater risk of not graduating than their fellow deaf students who did not participate in the SSI program as children. Our findings suggest that greater effort may be necessary to prepare SSI children for postsecondary education and that the currently SSA-funded youth transition demonstration projects may contribute to our understanding of how such efforts can improve adult outcomes for SSI children with disabilities.
## Literature Review
There is a significant body of research on the transition from secondary school to postsecondary education and employment for youth with disabilities. (See Wittenburg and Maag [2002] for a review of this literature.) We contribute to this literature by examining a subgroup of SSI recipients—SSI children. We describe their experiences during the transition to postsecondary education and quantify their economic outcomes as young adults. Our study is unique in that the longitudinal data on Social Security participation and earnings allowed us to examine outcomes over a relatively long period after the completion of postsecondary education. Here we summarize research related to this study and describe its contribution to the larger body of research.
### Postsecondary Education for Youth with Disabilities
As of 2003, participation in postsecondary education among youth with disabilities was estimated to be about half of the participation rate for the general population of youth (Wagner and others 2005). This research, which used the National Longitudinal Transition Survey (NLTS) and the National Longitudinal Transition Survey 2 (NLTS–2), also showed increased participation in postsecondary education for youth with disabilities from 1987 to 2001 and that this increase was greater than the increase for the general population (Wagner and others 2005). This finding indicates that the gap between the two groups has declined over time and that the transition from secondary education to postsecondary education is becoming more prevalent among youth with disabilities.
Data on postsecondary education completion rates show that youth with disabilities are less likely to complete postsecondary education than other youth. Horn and Berktold (1999) used the Beginning Postsecondary Students Longitudinal Study (BPS: 90/94) to support this finding; the BPS: 90/94 was a survey of undergraduates who enrolled in postsecondary education for the first time in the 1989–1990 period and were interviewed for the last time in 1994. Their results show that, at the time of the last interview, 53 percent of students with disabilities had completed postsecondary education or were still enrolled, compared with 64 percent of those without disabilities. Horn and Berktold state that this difference may have been partly due to differences in attributes that correlate with lower completion rates. For example, persons with disabilities were more likely to have General Educational Development (GED) degrees rather than standard high school diplomas, and persons with GED degrees are less likely to complete postsecondary education.
Research on the benefits of postsecondary education is limited to outcomes immediately following completion of postsecondary education. Horn and Berktold (1999) used the BPS: 90/94 to show that the gap between postsecondary education graduates with and without disabilities is small in terms of postgraduation employment, participation in graduate school, and participation in employment related to their postsecondary degree. They concluded that postsecondary education graduates with disabilities fare relatively well when compared with those without disabilities. This finding is in stark contrast to the experience of the general population with disabilities, which does not fare nearly as well with respect to both employment and earnings compared with the general population. However, the postsecondary education outcomes considered by Horn and Berktold focused only on the year immediately following graduation; the study did not examine employment and earnings in subsequent years. Thus, these studies may have missed differences that arise in terms of earnings growth and long-term employment prospects.
The only study that examines long-term employment outcomes among persons with disabilities was performed by Walter, Clarcq, and Thompson (2002), who used data from a 1998 version of the NTID/SSA matched data to examine employment outcomes for all NTID applicants. Their analysis suggests that a postsecondary education from NTID yields significant economic gains for persons with severe hearing impairments. However, their analysis was based on a single cross section of data and hence did not follow the individuals over time; nor did it examine whether there are differences in these outcomes between those who are former SSI children and those who are not.
### SSI Children
Research on SSI children shows that they are likely to spend a significant portion of their adult life collecting SSI benefits and that they are less likely to enroll in postsecondary education compared with the general population.
Rupp and Scott (1995) provide evidence of the length of stay in the program for SSI children. The authors used sample cohorts of persons awarded SSI as children from 1974 through 1982 and examined a 10-year follow-up period using administrative records from 1974 through 1992. They found that the mean length of the first spell of SSI participation was 11.3 years for SSI children. By the time SSI children turn 65, it is estimated that more than half of them will have spent over 25 years in the program; the mean length of stay for all children was 26.7 years.4
The postsecondary education enrollment rates for former SSI children aged 19–23 are described in Loprest and Wittenburg (2005). To examine the transition process, they used data from the National Survey of SSI Children and Families (NSCF), an SSA-funded nationally representative survey of current and former SSI children, fielded from August 2001 through July 2002.5 Part of their study examined the educational attainment of a posttransition cohort of people who were aged 19–23 in 2000 and had received SSI payments as children in 1996. At the time of the interview, they found that an estimated 42.3 percent had graduated from secondary school but were not in postsecondary school, while 6.3 percent had graduated from secondary school and made the transition to postsecondary school.6 The 6.3 percent of SSI children who enrolled in postsecondary education provides some context for our study. Although the rate was not zero, it was small compared with the estimate of 35 percent enrollment rate for youth in the general population who were aged 18–24.7 The NSCF estimate of 42 percent of SSI children who completed secondary education but did not enroll in postsecondary education may point to additional SSI children who could benefit from postsecondary education.
### How the Current Study Contributes to the Literature
Our study builds on existing research by focusing on SSI children and examining postsecondary education completion rates, as well as on how postsecondary education can influence length of stay in the adult SSI program and long-term employment outcomes. No other study has examined either postsecondary education completion rates for SSI children or long-term outcomes, such as dependency on the adult SSI disability program or adult employment associated with postsecondary education for this population. The few studies that have considered long-term outcomes for youth with disabilities who participate in postsecondary education have not taken full advantage of the longitudinal data. Our analysis used a longitudinal database and used techniques that take advantage of the longitudinal nature of our data to characterize outcomes for SSI children.
## Data
A data file based on administrative data from NTID and SSA was used for the analyses. The data file was created under a Memorandum of Agreement (MOA) whereby NTID paid SSA to create the merged data file for the purpose of conducting research on outcomes for NTID applicants. The two organizations worked together with researchers at Cornell University to design a merged NTID/SSA event history data file that could be used to track NTID applicants' outcomes for Social Security program participation, employment, and labor earnings. SSA staff constructed the file, which is securely stored at SSA; only SSA employees are allowed to perform analysis on the individual records.8
The NTID data contain information on all persons who have applied to the school since it opened in 1968. The data allow NTID applicants to be disaggregated into four groups:
1. those who were not accepted,
2. those who were accepted but chose not to attend,
3. those who attended but withdrew before earning a degree, and
4. those who graduated.
Individual information is available on the age, sex, and race of all applicants. Additional data are collected for those who attended NTID, including information on the age at which the hearing impairment began, the severity of the person's hearing impairment, and family background.
Social Security Administration data come from the Supplemental Security Record, the Master Earnings File, and the Numident file.9 The Supplemental Security Record contains the complete history of SSI program participation since the program began in 1974. The file is used to identify childhood participation in the SSI program and to construct an event history file of SSI program participation in adulthood. The Master Earnings File contains information on annual earnings that are subject to Federal Insurance Contribution Act (FICA) taxes from 1981 through 2003.10 It is used to estimate labor earnings for the age/earnings profiles. The Numident file contains information on deaths that occurred before 2004.
The resulting NTID/SSA merged data file has several features that make it superior to all other data sets that describe postsecondary education experiences of and outcomes for persons with disabilities. First, it is the only data set able to track long-term outcomes for youth with disabilities, such as adult SSI participation, employment, and earnings. Second, the NTID data include three different groups of applicants who did not graduate from NTID—those who were not accepted, those who were accepted but chose not to attend, and those who attended but withdrew before earning a degree. By comparing NTID graduates with these applicant groups, we were able to reduce the influence of selection bias associated with comparing them with all other persons who had disabilities. Third, our data were administrative, so we were able to match almost all NTID applicants to their administrative records. In this way, we avoided the usual problems with survey data that rely on self-reporting and have low response rates, which can affect validity.
We focused on applicants born from 1965 through 1979 who were alive at the time we extracted their SSA administrative records.11 We restricted our sample to persons born after 1964 because a significant amount of data in the NTID database is missing for earlier cohorts and because by doing so we avoided complications associated with SSI rule changes that occurred in the early 1980s.12 We restricted our sample to persons born before 1980 to ensure that we would observe graduation from NTID.
A total of 5,638 applicants met our criteria for the analyses. We refer to this group as NTID applicants. In some of our analyses, we used the subset of 1,366 applicants who were SSI children. Finally, we drew a sample of 9,388 SSI children from SSA administrative data who met our selection criteria for the analyses. The latter group was used to show how program participation and earnings outcomes differ between SSI children in the four NTID applicant groups and all SSI children.
Table 1 describes the variables used in our analysis, organizing them by NTID applicant group, participation in the SSI program as a child, demographic characteristics, age at onset of hearing impairment, severity of impairment, and family background characteristics. The descriptive statistics in Table 2 show how the composition of characteristics differed across the four NTID groups.13 For example, there are differences in the percentage of each NTID applicant group who were SSI children—16 percent of graduates were SSI children compared with 29 percent of those who withdrew; 24 percent of those who were accepted but chose not to attend; and 32 percent of applicants who were not accepted. The lower percentage of NTID graduates who were SSI children suggests that the former SSI children who applied to NTID had a relatively lower chance of graduating than other NTID applicants. However, there also are sizable differences across the four groups in terms of other individual characteristics, and these differences may also explain differences in graduation probabilities. Below, we describe how we accounted for these differences in our analyses.
Table 1. Definition of variables
Variable Definition
Applicant group
Graduated Value equals 1 if person graduated from NTID; 0 otherwise.
Withdrew Value equals 1 if person withdrew from NTID; 0 otherwise.
Accepted, did not attend Value equals 1 if person was accepted but did not attend NTID; 0 otherwise.
Not accepted Value equals 1 if person was not accepted into NTID; 0 otherwise.
Received SSI as a child
SSI child Value equals 1 if person received SSI payments before age 18; 0 otherwise.
Not SSI child Value equals 1 if person did not receive SSI payments before age 18; 0 otherwise.
Sex and race
Female Value equals 1 if sex is female; 0 otherwise.
Nonwhite Value equals 1 if race is nonwhite; 0 otherwise.
Age at onset of hearing loss
Age Value equals age at deaf onset; 99 or "." if missing.
Birth Value equals 1 if age at hearing loss is birth; 0 otherwise.
Ages 0–5 Value equals 1 if age at hearing loss is 0–5; 0 otherwise.
Ages 6 or older Value equals 1 if age at hearing loss is 6 or older; 0 otherwise.
Missing Value equals 1 if age at hearing loss is missing; 0 otherwise.
Severity of hearing loss
Mild Value equals 1 if lowest PTA hearing score is between 0 and 60; 0 otherwise.
Severe Value equals 1 if lowest PTA hearing score is between 61 and 90; 0 otherwise.
Severe spline Is a continuous value that is the difference between the PTA score and the score of 60, which is the definition of a severe hearing impairment. It is equal to 0 for those with a PTA score above 89 and below 60.
Profound Value equals 1 if lowest PTA hearing score is greater than 90; 0 otherwise.
Profound spline Is a continuous value that is the difference between the PTA score and the score of 90, which is the definition of a profound hearing impairment. It is equal to 0 for those with a PTA score below 90.
Father's education
Elementary Value equals 1 if father's education is elementary school; 0 otherwise.
Secondary Value equals 1 if father's education is secondary school; 0 otherwise.
College
2 years Value equals 1 if father's education is 2 years of college; 0 otherwise.
4 years Value equals 1 if father's education is 4 years of college; 0 otherwise.
5 or more years Value equals 1 if father's education is postgraduate; 0 otherwise.
Missing Value equals 1 if father's education is missing; 0 otherwise.
Mother's education
Elementary Value equals 1 if mother's education is elementary school; 0 otherwise.
Secondary Value equals 1 if mother's education is secondary school; 0 otherwise.
College
2 years Value equals 1 if mother's education is 2 years of college; 0 otherwise.
4 years Value equals 1 if mother's education is 4 years of college; 0 otherwise.
5 or more years Value equals 1 if mother's education is 5 or more years of college; 0 otherwise.
Missing Value equals 1 if mother's education is missing; 0 otherwise.
Deaf parents
Neither Value equals 1 if neither parent is deaf; 0 otherwise.
One Value equals 1 if one parent is deaf; 0 otherwise.
Two Value equals 1 if two parents are deaf; 0 otherwise.
Missing Value equals 1 if parents' hearing status is missing; 0 otherwise.
Birth year Set of indicators equal to 1 for each birth year from 1965 to 1979; 0 otherwise.
SOURCES: Data file of administrative records from the National Technical Institute for the Deaf linked to data from the Social Security Administration's Supplemental Security Record, Master Earnings File, and Numident file.
NOTE: NTID = National Technical Institute for the Deaf; SSI = Supplemental Security Income; PTA = pure tone average hearing level.
## Methods
Our analyses focused on describing the following three outcomes for SSI children:
1. The probability that an SSI child who applied to NTID would graduate, compared with NTID applicants who did not participate in the SSI program in childhood;
2. Dependency on the SSI adult program for SSI children who graduated from NTID, compared with each of the three groups of SSI children who applied but did not graduate; and
3. Levels and growth of earnings for SSI children who graduated from NTID, compared with each of the three groups of SSI children who applied but did not graduate.
Different methods were required to describe each of the outcomes. Here, we provide an overview of the methods used. The technical details can be found in Appendix A.
### Educational Outcomes
The differences in the probability of graduation between SSI children and those who were not SSI children (outcome 1) were used to assess whether the differences between the two groups are large enough for policymakers to consider special programs that specifically target SSI children who apply for postsecondary education. If there are no differences in the probability of graduation between the two groups, then postsecondary education programs specifically targeting SSI children may have a smaller potential for affecting educational success. This information is important to policymakers interested in identifying which programs have the potential to help SSI children make the transition to adult life. We do not attribute the differences to the presence of the SSI program; that is, we do not conclude that if the SSI program did not exist there would be no difference in graduation rates. SSI eligibility is based on family income and resource tests, and in the absence of the SSI program these children might have experienced similar differences in the probability of graduation because their families had lower income and resources compared with NTID applicants who were not SSI children.
The method we used to estimate differences in the probability of graduation among all applicants is referred to as a sequential response model. This type of model disaggregates the probability of graduation into a sequence of three events and may be used to show how differences in the probability of graduation are related to the probability that each of the following events will occur:
• an NTID applicant will meet the school's admission criteria,
• an accepted applicant will choose to attend NTID, and
• for those who attend NTID, whether they will graduate.
Some of those who attend NTID will withdraw from the school before completing the requirements for graduation.
We used multivariate logit models to estimate how participation in the SSI program as a child is related to the probability that each of these events will occur; therefore, our model is referred to as a sequential logit.14 The motivation for using the sequential logit is based on the descriptive statistics in Table 2, which show substantial differences in sex and race for those who are admitted to NTID, those who choose to attend, and attendees who graduate from NTID. Therefore, differences between SSI children and those who are not SSI children could be driven by differences in sex or race.15 The sequential logit model allows us to estimate how the probability that a particular event will occur and differs for those who participate in the SSI program as children, compared with those who do not, after accounting for differences in sex, race, and birth year across the two groups. It also allows us to examine differences in graduation that may be related to sex or race.
Table 2. Descriptive statistics for NTID applicants, by outcome of application (in percent unless otherwise specified)
Variable Total Not accepted Accepted,
did not attend
Mean SE Mean SE Mean SE Mean SE Mean SE
Individual characteristics
Former SSI child 24.23 0.57 31.94 1.84 23.72 1.39 28.68 0.94 15.84 0.87
Female 44.75 0.66 49.61 1.97 53.51 1.63 38.90 1.02 45.93 1.19
Nonwhite 24.49 0.57 44.03 1.96 30.96 1.51 21.41 0.86 17.89 0.92
Age at onset of hearing loss
Mean age at onset (years) -- -- -- -- -- -- 10.80 0.64 9.65 0.69
Birth -- -- -- -- -- -- 75.15 0.90 76.52 1.01
Ages 1–5 -- -- -- -- -- -- 10.23 0.63 10.77 0.74
Ages 6 or older -- -- -- -- -- -- 1.00 0.21 0.68 0.20
Missing -- -- -- -- -- -- 13.62 0.72 12.02 0.78
Severity of hearing loss
Mean hearing loss -- -- -- -- -- -- 93.13 0.45 94.87 0.46
Missing -- -- -- -- -- -- 2.22 0.31 1.60 0.30
Mild -- -- -- -- -- -- 4.22 0.42 2.68 0.39
Severe -- -- -- -- -- -- 27.89 0.94 25.81 1.04
Severe spline (mean) -- -- -- -- -- -- 5.60 0.20 5.26 0.23
Profound -- -- -- -- -- -- 65.67 0.99 69.91 1.10
Profound spline (mean) -- -- -- -- -- -- 9.52 0.21 9.87 0.23
Father's education
Elementary -- -- -- -- -- -- 11.88 0.68 8.60 0.67
Secondary -- -- -- -- -- -- 32.94 0.98 30.71 1.10
College
2 years -- -- -- -- -- -- 17.15 0.79 15.67 0.87
4 years -- -- -- -- -- -- 17.93 0.80 22.22 0.99
5 or more years -- -- -- -- -- -- 9.27 0.61 14.07 0.83
Missing -- -- -- -- -- -- 10.84 0.65 8.72 0.67
Mother's education
Elementary -- -- -- -- -- -- 10.36 0.64 8.15 0.65
Secondary -- -- -- -- -- -- 39.51 1.02 35.84 1.14
College
2 years -- -- -- -- -- -- 22.32 0.87 21.20 0.98
4 years -- -- -- -- -- -- 16.45 0.77 20.97 0.97
5 or more years -- -- -- -- -- -- 5.09 0.46 7.29 0.62
Missing -- -- -- -- -- -- 6.27 0.51 6.55 0.59
Deaf parents
Neither -- -- -- -- -- -- 88.90 0.66 93.68 0.58
One -- -- -- -- -- -- 1.65 0.27 1.20 0.26
Two -- -- -- -- -- -- 8.18 0.57 4.90 0.52
Missing -- -- -- -- -- -- 1.26 0.23 0.23 0.11
Mean birth year 1970.9 0.1 1969.4 0.2 1970.1 0.1 1971.8 0.1 1970.6 0.1
Number of observations 5,638 645 940 2,298 1,755
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTE: NTID = National Technical Institute for the Deaf; SE = standard error; SSI = Supplemental Security Income; -- = not available.
The estimates from the sequential logit may be used to show how individual characteristics have different effects on the overall probability of graduation at each event within the sequence of events leading to graduation.16 This information is important because it can show policymakers how each of the three events—NTID admission among those who apply, NTID attendance among those accepted, and NTID graduation among those who attend—is related to differences in the probability of graduation for particular types of applicants. For example, if lower graduation rates among SSI children occur because they decide not to attend NTID, efforts to improve graduation rates might consist of providing better information on how SSI children can get financial assistance. However, other efforts would be called for—such as improvements to college retention programs—if lower graduation rates occur because SSI children are withdrawing from NTID before earning a degree.
### Program Dependency and Earnings Outcomes
SSI children who graduate from NTID (outcome 1) may experience reduced dependency on the adult program (outcome 2) and increased earnings (outcome 3). Our strategy for identifying the potential impact of NTID graduation was to compare these outcomes for SSI children who graduate from NTID with the outcomes for the following groups of applicants:
• SSI children who were accepted to NTID but chose not to attend, and
• SSI children who withdrew before earning a degree.
To attribute the entire difference in these outcomes to graduation from NTID, we need to assume that the NTID graduates would have experienced the same outcomes as the comparison groups if they had not graduated from NTID. We refer to our estimates as "potential impacts" because we are not able to verify that this assumption is valid.
We used two other comparison groups to provide further context to our estimates of these outcomes:
• SSI children who applied to NTID but who did not meet the admission standard. Our hypothesis is that this comparison group spent more time in the SSI program as adults and earned less than those who were accepted to NTID because they did not meet the NTID admission standard.
• former SSI children who qualified on the basis of a primary diagnosis of deafness and were similar in age to the NTID sample.
These comparison groups place our results in the context of the SSI program. We hypothesize that the full population of deaf SSI children spent the most time in the SSI program and had the lowest earnings.
We measured adult dependency on the SSI program using survival analysis, which provides estimates of the timing of exit from and reentry into the SSI program after reaching age 19. Survival analysis entails following individuals from one particular event (for example, entering the adult SSI program) to another (for example, exiting the adult SSI program), and comparing the amount of time between events across groups. We estimated the potential effect of NTID graduation by comparing SSI children who graduated from NTID with each of our comparison groups using the following measures:
• the estimated probability of remaining in the program for each year over a 10-year period,
• the probability of leaving the program at the end of the 10-year period, and
• the estimated median number of months spent in the adult SSI program.
Dependency on the SSI Program as an Adult. For this analysis, we confined our sample to NTID applicants who were SSI children receiving SSI adult benefits at age 19.17 The event history file contains the month that the person turns 19 and either the month that the person exits the adult SSI program or the last month available in our data. Months are a natural time unit for the measurement of SSI participation because an SSI recipient's payment status is determined on a monthly basis. For presentation purposes, we grouped months into yearly intervals. Some people in our data set were still participating in the SSI program as of the last time period we recorded; that is, we never observed a transition from the SSI program for some persons. These cases are referred to as censored cases, and we accounted for them by using standard statistical techniques (described in Appendix B).
We used a similar approach to examine the timing of reentry into the adult program after a first exit. In this case, the first event was the month that a person first exited the adult SSI program, and the second event was the month that a person first reentered the SSI program. Like the analysis of first exit from the adult SSI program, we grouped months into yearly intervals for presentation and used standard techniques to account for censored cases in the analysis. Because of data limitations, we focused on the probability of reentry into the program within 5 years of first exit as another measure of SSI dependency.
Earnings. To describe the third outcome, earnings, we used age/earnings profiles to examine differences in earnings from ages 18–30 across the four groups of NTID applicants. For each person in the data set, earnings were observed for each age up to 2002, the final year that annual earnings are available in our data. A data set that contains an observation for each person at each age was created, and the dollar values were adjusted to 2004 dollars using the consumer price index for all urban consumers (CPI-U). We used three key statistics to describe the age/earnings profiles:
• the percentage of persons with at least $1 of earnings at a particular age, • the mean earnings for those with at least$1 of earnings at a particular age, and
• the mean earnings for all persons at a particular age.
Appendix B contains data for each of the three statistics. Separate profiles were estimated for each of the four NTID applicant groups using mean earnings for all persons. Mean earnings for each age were plotted in an age/earnings graph, and a third-order polynomial trend line was fit to the means to illustrate the pattern for the various groups. The analysis allowed us to examine differences in both the level and growth in earnings from ages 18–30 and to describe the potential effects of an NTID education on earnings during this period.
## Results
From 1982 to 2000, the percentage of both NTID applicants and graduates who were SSI children steadily increased. These two trends are illustrated in Chart 1, which organizes NTID applicants and graduates by the year they first applied, so that there is a common basis of comparison. The chart shows that the percentage of all NTID applicants who were SSI children increased from 10 percent in 1983 to 43 percent in 1999. It also shows that the fraction of NTID graduates who were SSI children increased from 8 percent of those who applied in 1982 to 28 percent in 1999. These results indicate that SSI children with hearing impairments accounted for a significant share of applicants and graduates during this period and that they were willing and able to participate in postsecondary education.
Chart 1.
Time series of the percentage of NTID applicants and graduates who were SSI children, by year of application
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTE: NTID = National Technical Institute for the Deaf; SSI = Supplemental Security Income.
The position of the trend lines in Chart 1 also shows that, for each application year, the fraction of eventual graduates who were SSI children was smaller than the fraction of all applicants who were SSI children. For the 1999 application-year cohort, 42 percent of applicants were SSI children, compared with only 28 percent of eventual graduates. Overall, the percentage of those who graduated and were classified as SSI children was lower than the percentage who graduated and were not in the SSI program in childhood. Hence, SSI children who applied were less likely to graduate, compared with other applicants. The chart shows that this finding existed for almost every application year from 1982 to 1999.
Finally, the slopes of the two trend lines are different.18 This difference indicates that even though both trends increased, the fraction of NTID applicants who were SSI children increased at a faster rate. As a result, the likelihood that an SSI child who applies to NTID will eventually graduate has decreased over time. More SSI children are applying to NTID, but the rate of graduation among these applicants has declined slightly over time. The estimates below more precisely measure the exact relationship between participation in SSI as a child and educational success as an adult.
### Probability of Graduation
The results of our multivariate logit model show some substantial and statistically significant differences in the characteristics of applicants who were not admitted to NTID, were admitted and chose to attend NTID, and attended and completed degree requirements. Table 3 shows the differences in the probability for each of these events between SSI children and those who were not SSI children. Compared with non-SSI children, the probability that SSI children who applied to NTID would be admitted was 4.8 percentage points lower, the probability that SSI children who were admitted would attend NTID was not statistically different, and the probability that SSI children who attended NTID would graduate was 16 percentage points lower. The difference in the graduation rate among those who attend NTID is large; after adjusting for differences in sex and race, we estimate that 47 percent of NTID attendees who were not SSI children graduated compared with only 31 percent of those who were SSI children. The difference suggests that college preparation and retention programs that target SSI children may have the potential to substantially improve their graduation rates.
Table 3. Sequential logit model results of relationship between SSI participation as a child and graduation from NTID: Estimated impact on the probability that each event will occur (in percentage points)
Variable Difference in probability of being admitted to NTID among applicants Difference in probability of attending NTID among those admitted Difference in probability of graduation among those who attend NTID
Former SSI child -4.81***
[1.09]
-0.76
[1.34]
-16.07***
[1.81]
Female -1.93***
[0.83]
-7.27***
[1.11]
8.11***
[1.55]
Nonwhite -12.42***
[1.19]
-11.28***
[1.50]
-0.69
[1.98]
Birth year indicators Yes Yes Yes
Predicted probability (percent) 88.6 81.2 42.7
Number of observations 5,638 4,993 4,053
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: The sequential model is based on a sequential logit specification as described in Appendix A. Logit coefficients, odds ratios, and marginal effects for the entire model are in Table A-2.
Standard errors are in brackets.
SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf.
* significant at .10 level; ** significant at .05 level; *** significant at .01 level.
The results for females and nonwhite applicants are remarkably different from those described for SSI children. Females who applied were less likely to be admitted, and those who were admitted were less likely to attend. However, the probability of graduation for females who attended NTID was 8.1 percentage points higher than that of their male counterparts. Compared with whites, nonwhites were less likely to meet the admission criteria, and those who met the criteria were less likely to choose to attend NTID. However, the differences in graduation rates between whites and nonwhites who attended NTID were not statistically different.
We also looked at the relationship between individual characteristics and the overall probability of graduation among NTID applicants at each stage of the process.19 As shown in Table 4, the probability of graduation for all SSI children who applied to NTID was 13.5 percentage points lower than that for NTID applicants who were not SSI children. The lower probability was spread over the three separate events that lead to graduation for applicants—with 1.7 percentage points attributed to the admittance step, 0.3 percentage points attributed to the attendance step, and 11.5 percentage points attributed to the graduation step. Thus, the final step was responsible for most of the disparity in the overall graduation rates for SSI children who applied to NTID compared with the rate for those who were not SSI children.
Table 4. Sequential logit model results of relationship between SSI participation as a child and graduation from NTID: Decomposition of each event's impact on the overall probability of graduation among applicants (in percentage points)
Variable Difference in
probability of
among all NTID
applicants
Difference in
probability of
due to NTID
Difference in
probability of
due to decision
to attend NTID
Difference in
probability of
due to decision
to complete an
NTID degree
Former SSI child -13.5 -1.7 -0.3 -11.5
Female 2.4 -0.7 -2.7 5.8
Nonwhite -9.1 -4.3 -4.3 -0.5
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: The sequential model is based on a sequential logit specification as described in Appendix A. Logit coefficients, odds ratios, and marginal effects for the entire model are in Table A-2.
SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf.
Given the importance of the graduation step, we estimated a multivariate logit model of the probability of graduation for those who attended NTID that includes the additional characteristics available for those attendees. The results are in Table 5 and are comparable with those shown in Table 3. The inclusion of the additional characteristics slightly reduces the estimated difference in the probability of graduation between former SSI children and those who had not been in the SSI program as children. However, the difference is still large and statistically significant. The probability that former SSI children who attended NTID would graduate was 13.5 percentage points lower than for those who were not SSI children. To put this result in perspective, the probability of graduation for those who were not SSI children was 46 percent, compared with an estimated 32.5 percent for former SSI children. Thus, even after controlling for sex, race, severity of hearing impairment, family background characteristics, and birth cohort, former SSI children were significantly less likely to graduate than their non-SSI counterparts.
Table 5. Logit model results of the probability of graduation for NTID attendees
Variable Coefficient Effect on probability
(percentage points)
Individual characteristic
Former SSI child -0.5887***
[0.0873]
-13.5
[1.92]
Female 0.3653***
[0.0668]
8.5
[1.54]
Nonwhite -0.0158
[0.0873]
-0.4
[2.01]
Age at onset of hearing loss
Birth -0.0049
[0.1086]
-0.1
[2.52]
Ages 1–5 (reference) . . . . . .
Ages 6 or older -0.4722
[0.3797]
-10.7
[8.16]
Missing -0.2385
[0.1503]
-5.5
[3.4]
Severity of hearing loss
Mild 0.1989
[0.2492]
-4.5
[5.5]
Severe (reference) . . . . . .
Severe spline 0.0034
[0.0077]
0.1
[0.18]
Profound 0.2314
[0.1866]
5.4
[4.28]
Profound spline -0.0009
[0.0050]
0
[0.12]
Missing 0.5797*
[0.3399]
13.4
[7.84]
Father's education
Primary -0.0707
[0.1470]
-1.6
[3.3]
Secondary 0.0831
[0.1038]
1.9
[2.4]
College
2 years (reference) . . . . . .
4 years 0.2016*
[0.1113]
4.8
[2.65]
5 or more years 0.2923**
[0.1345]
7.0
[3.21]
Missing -0.3107
[0.1977]
-6.9
[4.29]
Mother's education
Primary 0.0741
[0.1467]
1.7
[3.35]
Secondary -0.0117
[0.0930]
-0.3
[2.14]
College
2 years (reference) . . . . . .
4 years 0.2*
[0.1072]
4.7
[2.53]
5 or more years 0.3513**
[0.1591]
8.3
[3.75]
Missing 0.6418***
[0.2372]
14.8
[5.42]
Deaf parents
Neither (reference) . . . . . .
One -0.1507
[0.2871]
-3.5
[6.59]
Two -0.3507**
[0.1409]
-8.0
[3.12]
Missing -1.9819***
[0.5822]
-34.0
[5.49]
Constant 0.4382*
[0.2350]
. . .
. . .
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Birth cohort dummy variables are included. Number of observations was 4,053. Standard errors are in brackets.
NTID = National Technical Institute for the Deaf; SSI = Supplemental Security Income; . . . = not applicable.
* significant at .10 level; ** significant at .05 level; *** significant at .01 level.
In summary, the result of a lower probability of graduation for SSI children was partly due to the admission standard (that is, SSI children were less likely to be accepted to NTID), but most of it was due to the lower probability of graduation for SSI children who attended NTID. Devoting efforts to improving retention rates among SSI children who attend NTID appears to be necessary to reduce the differences in graduation rates.
### Relationship Between NTID Graduation and Participation in the Adult SSI Program
Almost all of the SSI children who applied to NTID participated in the SSI program when they turned 19. After age 19, the patterns of exiting the program differed substantially between NTID graduates and each of the comparison groups: SSI children who graduated were more likely to have left the program within 10 years following age 19 and were less likely to reenter the program.
Using the survival probability for each year following age 19 as a measure, we examined the changes in the probability of remaining on the SSI program for SSI children who graduated from NTID compared with each of our comparison groups. Chart 2 shows that SSI children who graduated were more likely to remain in the program during the first 4 years following their 19th birthday—the years that many of them were attending NTID—and that after the 4th year there was a relatively sharp decline in the probability of remaining in the SSI program. By the 10th year, there was only a 34 percent chance that they would remain in the SSI program, which was significantly lower than the probability for each of the other comparison groups.
Chart 2.
Probability that SSI children will remain in the adult SSI program for 1 10 years after age 19, by NTID status
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTE: SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf.
The potential impact of NTID graduation on the likelihood that SSI children will leave the program within 10 years following their 19th birthday and the median amount of time they spend in the program are shown in Table 6. We estimated that there was a 64.7 percent chance that SSI children who graduated from NTID would leave the program within 10 years, which was larger than and statistically different from the estimates of 52.2 percent for those who withdrew from NTID, 55.3 percent for those who did not attend, 51.6 percent for those who were not accepted, and 42.9 percent for the group of all SSI children with a primary diagnosis of deafness.
Table 6. Estimates of first exit from SSI program for children receiving SSI at age 19, by NTID status
NTID status Probability of leaving SSI program
within 10 years
Median number of months
to first exit from SSI
Estimate
(percent)
Potential impact of
(percentage points)
Estimate
(percent)
Potential impact of
(percentage points)
[3.29]
. . . 95
[1.44]
. . .
Withdrew 52.2
[2.28]
12.5*** 116
[3.34]
-21***
Accepted, did not attend 55.3
[3.71]
9.4* 114
[2.58]
-19***
Not accepted 51.6
[3.84]
13.1** 118
[2.61]
-23***
All SSI children awarded benefits on the basis of a hearing impairment a 42.9
[0.57]
21.8 145
[2.38]
-50
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Standard errors are in brackets.
SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf; . . . = not applicable.
a. The group of all SSI children awarded benefits on the basis of a hearing impairment is not mutually exclusive from the group of NTID graduates, and we do not calculate statistical tests for this group.
* significant at .10 level; ** significant at .05 level; *** significant at .01 level.
We also found that NTID graduation may increase the probability of SSI children leaving the program within 10 years following their 19th birthday. That probability increased by 12.5 percentage points compared with SSI children who withdrew from NTID and by 9.4 percentage points compared with SSI children who were accepted but chose not to attend. SSI children who graduated from NTID fared even better when compared with each of the other two groups; the probability of leaving the program within 10 years was 13.1 percentage points higher for SSI children who were not admitted and 21.8 percentage points higher for the group of all SSI deaf children.
The potential impact measured as the difference in the median time spent in the SSI program before leaving is also shown in Table 6. For the group of NTID graduates, the median expected time spent in the SSI program before leaving it was 95 months—substantially less than the 116 months estimated for those who withdrew from NTID, 114 months for those who chose not to attend, 118 months for those who were not accepted, and 145 months for the group of all deaf SSI children. The potential impact for SSI children who graduated was a 21-month reduction in median months spent in the program before leaving when compared with those who withdrew from NTID and a 19-month reduction when compared with those who were accepted but chose not to attend. Again, SSI children who graduated fared even better when compared with the other two groups; the median time before leaving was 23 months less than for those who were not admitted and 50 months less than for the group of all SSI deaf children.
An examination of the first SSI episode does not fully measure the relationship between NTID graduation and dependency on the SSI program. If NTID graduates were less likely to reenter the program after their first exit, then our estimate may have understated the role of an NTID degree on reductions in dependency on the SSI program. Chart 3 shows that the probability that the person would remain off the program, or survive without the program, was higher for NTID graduates across the 5 years after first exit. The sample sizes declined dramatically after the 5th year (as shown in Table A–5), and our estimates for later years have larger standard errors.
Chart 3.
Probability that SSI children will remain off the adult program after first exit, by NTID status
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTE: NTID = National Technical Institute for the Deaf; SSI = Supplemental Security Income.
Table 7 shows the probability that an SSI child would reenter the SSI program within 5 and within 10 years following first exit from the program after reaching age 19, using the survival probability as a measure.20 The probability of reentry within 5 years after leaving the program was only 11.6 percent for SSI children who graduated from NTID, which was smaller than the 21.7 percent estimate for those who withdrew, the 17.9 percent estimate for those who were accepted but chose not to attend, the 24.1 percent for those who were not accepted, and the 23.2 percent for the group of all deaf SSI children. The potential impact of NTID graduation for SSI children was a drop of 10.1 percentage points in the probability of reentering the SSI program when compared with those who withdrew and a drop of 6.3 percentage points when compared with those who chose not to attend NTID (although the latter result is not statistically significant). The estimates for the other two groups show that the group of all deaf SSI children also fared better. The probability of reentering the program within 10 years shows that the potential impact of NTID graduation is also substantial and statistically significant.
Table 7. Probability that SSI children will reenter the SSI program within 5 or 10 years following first exit from the program after reaching age 19, by NTID status
NTID status Within 5 years Within 10 years
Estimate
(percent)
Potential impact of
(percentage points)
Estimate
(percent)
Potential impact of
(percentage points)
[2.84]
. . . 14.4
[3.38]
. . .
Withdrew 21.7
[2.86]
-10.1** 27.2
[3.67]
-12.8***
Accepted, did not attend 17.9
[3.99]
-6.3 33.1
[6.28]
-18.7***
Not accepted 24.1
[4.82]
-12.5** 26.1
[5.08]
-11.7*
All SSI children awarded benefits on the basis of a hearing impairment a 23.2
[0.88]
-11.6 32.2
[1.44]
-17.8
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Standard errors are in brackets.
SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf; . . . = not applicable.
a. The group of all SSI children awarded benefits based on a hearing impairment is not mutually exclusive from the group of NTID graduates, and we do not calculate statistical tests for this group.
* significant at .10 level; ** significant at .05 level; *** significant at .01 level.
### Age/Earnings Profiles
To determine the potential impact of NTID graduation on the labor earnings of SSI children during the early portion of their adult life, we compared the age/earnings profile for SSI children who graduated from NTID with the profile for SSI children who withdrew from NTID (Chart 4).21 The results show that SSI children who graduated had a mean annual earnings level of less than $1,000 between the ages of 18 and 21, ages at which most graduates were attending NTID. The trend line shows that their mean annual earnings grew from about$1,000 at age 21 to $17,500 by age 30. SSI children who withdrew from NTID experienced very little earnings growth, and by age 30 the mean annual earnings level for the group was a little less than$11,600 per year. By age 30, the gap between the two groups was almost $6,000, with SSI children who graduated earning 51 percent more than those who withdrew. Chart 4. Age/earnings for SSI children who graduated from NTID compared with those who withdrew SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file. NOTES: Data include zero earners. SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf; Poly. = polynomial trend line. The potential earnings impact for SSI children who graduated from NTID compared with SSI children who were accepted to NTID but did not attend is shown in Chart 5. The earnings of SSI children who graduated exceeded the earnings of those who chose not to attend at every age after reaching age 24. The earnings of those who did not attend NTID grew to slightly more than$12,100 by the time they were age 30. By age 30, SSI children who graduated from NTID were earning about $5,400 (or 44 percent) more than SSI children who were accepted to NTID but chose not to attend. Chart 5. Age/earnings profiles for SSI children who graduated from NTID compared with those who were accepted but chose not to attend SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file. NOTES: Data include zero earners. SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf; Poly. = polynomial trend line. Comparisons between SSI children who graduated from NTID and those who were not admitted are shown in Chart 6. SSI children who were not accepted to NTID had modest growth in mean annual earnings from age 18 to age 30, with a mean level of earnings of about$8,800 at age 30. This level was well below the level for SSI children who graduated. At age 30, the earnings gap was about $8,700; SSI children who graduated from NTID earned about 99 percent more than those who were not accepted. Chart 6. Age/earnings for SSI children who graduated from NTID compared with those who were not accepted SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file. NOTES: Data include zero earners. SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf; Poly. = polynomial trend line. In Chart 7, the age/earnings profiles of the four groups of NTID applicants are compared with the broader population of SSI children with a primary diagnosis of deafness. Mean earnings among the group of former SSI children were lower than for all other groups from ages 25–30, and by age 30 their annual earnings were about$6,800, which was well below the earnings of each of the NTID applicant groups.
Chart 7.
Age/earnings for SSI children using polynomial trend lines, by NTID status
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Data include zero earners.
SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf.
## Discussion of the Findings and Future Research
Our analysis focused on the relative success of former SSI children who applied to NTID. We found that the percentage of NTID applicants who were SSI children increased over time, from a low of 10 percent in 1982 to more than 41 percent in 2000. However, the differences in the probability of graduation from NTID between deaf SSI children and deaf NTID applicants who were not SSI children did not change accordingly. The probability of graduation for SSI children who applied to NTID was 13.5 percentage points lower than for those who were not SSI children. Finally, using our most credible comparison group—SSI children who were accepted to NTID but chose not to attend—we found that SSI children who graduated from NTID left the SSI program early in their adult life (19 months earlier), were less likely to reenter the SSI program, and at age 30 had increased their earnings by an estimated 49 percent. Our findings demonstrate that SSI children need not be relegated to a lifetime of SSI participation as adults, despite the poor overall experience of this population since the creation of the SSI program in 1974. Postsecondary education can increase their earnings and reduce their dependency on SSI as adults.
These key findings—the lower postsecondary graduation rates among deaf SSI children and the potential for successful adult outcomes for deaf SSI children who graduate—suggest that there is a need to carefully examine the current support services for SSI children and identify improvements or new support services that will increase postsecondary graduation rates for SSI children. The Social Security Administration's youth transition demonstration projects are beginning to address these issues, but to date they have not focused on specific support for postsecondary educational achievement.
Our analysis is a case study of deaf persons who apply to NTID, and there are limitations to generalizing our results to the broader population of SSI children with disabilities. Children who qualify for the SSI program on the basis of other types of disabilities may face different barriers to postsecondary education and to successful labor market outcomes. NTID is unique in that it is tailored to the needs of the deaf population. SSI children with other types of disabilities generally must rely on postsecondary educational institutions that are not specifically designed to meet their special needs. These children may face different challenges—such as an environment with physical barriers, an inaccessible commuting environment, or social isolation—that may reduce the likelihood of application to and graduation from postsecondary institutions.
To assess the potential for programs that promote postsecondary education to reach SSI children with different impairments, we used 2001–2002 data from the Office of Special Education Programs (OSEP) on high school graduation rates for all children with disabilities, by impairment type.22 According to OSEP data, 51 percent of children with disabilities graduated from high school. That percentage is similar to the estimate of 48 percent for SSI children reported by Loprest and Wittenburg (2005), which we used as an upper bound of SSI children who may benefit in the short run from such programs.23 The OSEP data showed substantial differences in high school graduation rates by impairment type: graduation rates were above average for children with visual impairments (71 percent), hearing impairments (67 percent), specific learning disabilities (57 percent), and orthopedic impairments (56 percent); graduation rates were below average for children with mental retardation (39 percent) and children with severe emotional disturbances (32 percent). These data suggest that programs that promote postsecondary education may be more accessible to SSI children with certain types of impairments than with others.
One area for further research is to examine specific barriers in completing postsecondary education for SSI children with different types of impairments and to estimate the impact that such barriers may have on program participation and labor market outcomes.24 Another area for future research is to extend our analysis by using data from the National Survey of Children and Families (NSCF) linked to Social Security administrative records for the broader population of SSI children who undertake postsecondary education. That study would be limited initially to a short postgraduation follow-up period and a smaller sample size, but over time the data may provide further evidence of the long-term effects of postsecondary education.
Our analysis has two other limitations that could be addressed in future research. First, our analysis does not examine entry and exits from the Social Security Disability Insurance (DI) program.25 Our analysis of the age earnings/profiles, as well as preliminary analysis of cross-sectional data on DI participation among NTID applicants, suggests that postsecondary education may have the added effect of reducing dependency on the DI program. We are currently constructing an event history file of DI participation, and future research will examine how postsecondary education is related to participation in this program.
Finally, our analysis is based on nonexperimental data, so it is possible that those who graduated from NTID may have experienced better adult outcomes, in part, because of unobserved attributes such as higher levels of motivation or ability. At the same time, our findings show that positive outcomes are possible and suggest that a more rigorous evaluation, such as a randomized experiment, may be worthwhile. In the future, it would be useful to consider a project that includes a rigorous test of interventions promoting postsecondary education and examines the effect of such interventions on postsecondary education outcomes, SSI program participation, and long-term earnings.
## Appendix A:Estimating the Probability of Graduation for SSI Children
The purpose of this section is to provide further details on the estimates and the statistical methodology used to estimate the probability of graduation. Table A–1 shows the time-series estimates used to create Chart 1. Table A–2 shows additional estimates used for the sequential logit model. Table A–3 shows additional logit model estimates of the probability of graduation. In the remainder of this section we provide further details on the statistical methodology used to estimate the probability of graduation.
Table A–1. Time series data on the composition of NTID applicants and graduates
Year of first
contact with NTID
Percentage of
applicants who
as a child
Percentage of
as a child
1982 18.07 8.33
1983 10.53 8.60
1984 13.29 6.45
1985 12.35 13.08
1986 14.15 7.14
1987 17.07 11.56
1988 19.76 16.67
1989 21.65 15.04
1990 29.05 20.95
1991 27.53 20.56
1992 27.81 17.58
1993 37.20 21.18
1994 32.22 18.85
1995 32.42 27.45
1996 36.47 29.90
1997 29.90 11.86
1998 36.41 17.24
1999 41.57 28.57
2000 43.59 0
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTE: NTID = National Technical Institute for the Deaf; SSI = Supplemental Security Income.
Equation A–1
$P ( Graduate | Applied = 1 , X ) = P ( Admitted = 1 | Applied = 1 , X ) ⋅ P ( Attend = 1 | Applied = 1 , Admitted = 1 , X ) ⋅ P ( Graduate = 1 | Applied = 1 , Admitted = 1 , Attended = 1 , X )$
In Equation A–1, X represents a vector of individual characteristics that includes an indicator variable for whether the person received SSI as a child, an indicator variable for nonwhite race, an indicator variable for female sex, and a set of indicator variables for year of birth. We estimate the conditional probability that each event will occur for the particular population of interest using logit models.27
To quantify how individual characteristics are associated with the likelihood of graduation at each point in the sequential process, we use the decomposition of the sequential logit proposed by Heckman and Smith (2004), shown in Equation A–2.
Equation A–2
$∂ P ( Graduate | Applied = 1 , X ) ∂ X = ∂ P ( Admitted = 1 | Applied = 1 , X ) ∂ X ⋅ P ( Attend = 1 | Applied = 1 , Admitted = 1 , X ) ⋅ P ( Graduate = 1 | Applied = 1 , Admitted = 1 , Attend e d = 1 , X ) + P ( Admitted = 1 | Applied = 1 , X ) ⋅ P ( Attend = 1 | Applied = 1 , Admitted = 1 , X ) ∂ X ⋅ P ( Graduate = 1 | Applied = 1 , Admitted = 1 , Attend e d = 1 , X ) + ∂ P ( Admitted = 1 | Applied = 1 , X ) ⋅ P ( Attend = 1 | Applied = 1 , Admitted = 1 , X ) ⋅ P ( Graduate = 1 | Applied = 1 , Admitted = 1 , Attend e d = 1 , X ) ∂ X$
This decomposition results from the application of the chain rule to Equation A–1. The first term on the right-hand side of Equation A–2 describes the relationship between the admittance step and the overall probability of graduation; the second term shows the relationship between the attendance step and the overall likelihood of graduation; and the third term shows the relationship between the graduation step and the overall likelihood of graduation.
The NTID/SSA matched data contain additional health and family background information for the two groups—those who graduate or withdraw—who choose to attend NTID. The additional information allows us to examine whether the inclusion of additional characteristics affects our estimate of the relationship between the receipt of SSI as a child and the conditional probability of graduation from NTID for those who choose to attend.
The estimates of the logit parameters do not provide a direct measure of the relationship between individual characteristics and the probability that each event in the graduation process will occur. We use the logit parameters to estimate how individual characteristics are related to the difference in the probability that each event within the sequential graduation process will occur, based on the mean of individual-level changes in the probability.28 For the sequential logit model, we also present the results of the decomposition that shows how individual-level characteristics contribute to the likelihood of graduation at each step in the process. The estimated logit parameters and odds ratios are reported in Tables A–2 and A–3.
Table A–2. Additional sequential logit results of the relationship between SSI participation as a child and the graduation process
Variable Accepted Attended Graduated
Coefficient Odds ratio Marginal
effects
Coefficient Odds ratio Marginal
effects
Coefficient Odds ratio Marginal
effects
Former SSI child -0.4645***
[0.0980]
0.6285***
[0.0616]
-0.0481***
[0.0109]
-0.0514
[0.0906]
0.9499
[0.0861]
-0.0076
[0.0134]
-0.6972***
[0.0831]
0.4980***
[0.0414]
-0.1607***
[0.0181]
Female -0.2006**
[0.0864]
0.8182**
[0.0707]
-0.0193**
[0.0083]
-0.4902***
[0.0744]
0.6125***
[0.0455]
-0.0727***
[0.0111]
0.3431***
[0.0659]
1.4092***
[0.0928]
0.0811***
[0.0155]
Nonwhite -1.0840***
[0.0904]
0.3382***
[0.0306]
-0.1242***
[0.0119]
-0.6931***
[0.0848]
0.5000***
[0.0424]
-0.1128***
[0.0150]
-0.0295
[0.0845]
0.9709
[0.0821]
-0.0069
[0.0198]
Birth year
1966 0.2658
[0.1692]
1.3044
[0.2207]
0.0357*
[0.0212]
0.0817
[0.1613]
1.0852
[0.1750]
0.0143
[0.0277]
0.1516
[0.1551]
1.1638
[0.1805]
0.0371
[0.0379]
1967 0.3797**
[0.1728]
1.4618**
[0.2526]
0.0494**
[0.0202]
0.1140
[0.1605]
1.1208
[0.1798]
0.0199
[0.0273]
0.0859
[0.1541]
1.0897
[0.1680]
0.0211
[0.0378]
1968 0.5901***
[0.1802]
1.8042***
[0.3251]
0.0728***
[0.0187]
0.1110
[0.1600]
1.1174
[0.1788]
0.0194
[0.0273]
0.0136
[0.1541]
1.0137
[0.1562]
0.0033
[0.0378]
1969 0.6125***
[0.1759]
1.8451***
[0.3246]
0.0749***
[0.0180]
0.5264***
[0.1674]
1.6928***
[0.2834]
0.0822***
[0.0228]
-0.0369
[0.1466]
0.9638
[0.1413]
-0.0090
[0.0359]
1970 0.6329***
[0.1837]
1.8831***
[0.3460]
0.0774***
[0.0186]
0.1960
[0.1637]
1.2166
[0.1991]
0.0336
[0.0268]
0.0398
[0.1549]
1.0406
[0.1612]
0.0097
[0.0379]
1971 0.7691***
[0.1897]
2.1579***
[0.4094]
0.0915***
[0.0178]
0.0930
[0.1628]
1.0974
[0.1786]
0.0164
[0.0281]
-0.0197
[0.1582]
0.9805
[0.1551]
-0.0048
[0.0386]
1972 0.7460***
[0.2115]
2.1085***
[0.4460]
0.0869***
[0.0194]
0.3340*
[0.1827]
1.3965*
[0.2551]
0.0551**
[0.0277]
-0.1330
[0.1664]
0.8754
[0.1457]
-0.0324
[0.0404]
1973 0.9294***
[0.2073]
2.5330***
[0.5251]
0.1065***
[0.0176]
0.5124***
[0.1840]
1.6693***
[0.3071]
0.0814***
[0.0255]
-0.2702
[0.1652]
0.7632
[0.1261]
-0.0649*
[0.0391]
1974 0.9625***
[0.2125]
2.6182***
[0.5563]
0.1088***
[0.0175]
0.0967
[0.1719]
1.1016
[0.1894]
0.0171
[0.0298]
-0.3212*
[0.1715]
0.7253*
[0.1244]
-0.0771*
[0.0403]
1975 1.5466***
[0.2822]
4.6953***
[1.3251]
0.1445***
[0.0144]
0.7759***
[0.2117]
2.1725***
[0.4599]
0.1141***
[0.0246]
-0.1904
[0.1736]
0.8266
[0.1435]
-0.0461
[0.0416]
1976 1.9192***
[0.3176]
6.8156***
[2.1648]
0.1632***
[0.0122]
1.2038***
[0.2381]
3.3328***
[0.7934]
0.1573***
[0.0206]
-0.1261
[0.1728]
0.8815
[0.1523]
-0.0305
[0.0416]
1977 1.9381***
[0.3281]
6.9458***
[2.2788]
0.1605***
[0.0120]
2.0937***
[0.3266]
8.1147***
[2.6505]
0.2081***
[0.0135]
-0.3876**
[0.1684]
0.6787**
[0.1143]
-0.0926**
[0.0391]
1978 1.7251***
[0.2972]
5.6129***
[1.6681]
0.1539***
[0.0133]
1.4081***
[0.2522]
4.0881***
[1.0309]
0.1726***
[0.0186]
-0.5450***
[0.1746]
0.5799***
[0.1012]
-0.1281***
[0.0390]
1979 1.3335***
[0.2544]
3.7942***
[0.9653]
0.1355***
[0.0159]
0.5186***
[0.1973]
1.6798***
[0.3314]
0.0830***
[0.0274]
-0.9355***
[0.1938]
0.3924***
[0.0760]
-0.2095***
[0.0377]
Constant 1.9054***
[0.1141]
. . .
. . .
. . .
. . .
1.5078***
[0.1103]
. . .
. . .
. . .
. . .
-0.1200
[0.1043]
. . .
. . .
. . .
. . .
Observations 5,638 5,638 5,638 4,993 4,993 4,993 4,053 4,053 4,053
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Standard errors in brackets.
SSI = Supplemental Security Income; . . . = not applicable.
* significant at .10 level; ** significant at .05 level; *** significant at .01 level.
Table A–3. Additional logit model estimates of the probability of graduation
Variable Model with only
SSI child variable
Model with variables
available for all applicants
Model with full set
of variables for attendees
Coefficient Odds ratio Marginal effects
(percentage points)
Coefficient Odds ratio Marginal effects
(percentage points)
Coefficient Odds ratio Marginal effects
(percentage points)
Individual characteristics
Former SSI child 0.7590***
[0.0800]
2.1362***
[0.1709]
17.7***
[1.74]
0.7639***
[0.0814]
2.1467***
[0.1748]
17.7***
[1.76]
0.5887***
[0.0873]
1.8017***
[0.1574]
13.5***
[1.92]
Female . . .
. . .
. . .
. . .
. . .
. . .
-0.3224***
[0.0652]
0.7244***
[0.0472]
-7.7***
[1.56]
-0.3653***
[0.0668]
0.6940***
[0.0463]
-8.5***
[1.54]
Nonwhite . . .
. . .
. . .
. . .
. . .
. . .
0.0971
[0.0828]
1.1019
[0.0913]
2.3
[1.96]
0.0158
[0.0873]
1.0159
[0.0887]
0.4
[2.01]
Age at onset of hearing loss
Birth . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.0049
[0.1086]
1.0049
[0.1091]
0.1
[2.52]
Ages 6 or older . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.4722
[0.3797]
1.6036
[0.6089]
10.7
[8.16]
Missing . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.2385
[0.1503]
1.2693
[0.1908]
5.5
[3.4]
Severity of hearing loss
Mild . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.1989
[0.2492]
1.2201
[0.3040]
4.5
[5.5]
Spline severe . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.0034
[0.0077]
0.9966
[0.0077]
-0.1
[0.18]
Profound . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.2314
[0.1866]
0.7934
[0.1480]
-5.4
[4.28]
Profound spline . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.0009
[0.0050]
1.0009
[0.0050]
0.0
[0.12]
Missing . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.5797*
[0.3399]
0.5600*
[0.1904]
-13.4*
[7.84]
Father's education
Primary . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.0707
[0.1470]
1.0733
[0.1578]
1.6
[3.3]
Secondary . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.0831
[0.1038]
0.9203
[0.0955]
-1.9
[2.4]
College
4 years . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.2016*
[0.1113]
0.8174*
[0.0910]
-4.8*
[2.65]
5 years or more . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.2923**
[0.1345]
0.7466**
[0.1004]
-7.0**
[3.21]
Missing . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.3107
[0.1977]
1.3643
[0.2698]
6.9
[4.29]
Mother's education
Primary . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.0741
[0.1467]
0.9286
[0.1362]
-1.7
[3.35]
Secondary . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.0117
[0.0930]
1.0117
[0.0941]
0.3
[2.14]
College
4 years . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.2000*
[0.1072]
0.8187*
[0.0878]
-4.7*
[2.53]
5 years or more . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.3513**
[0.1591]
0.7038**
[0.1119]
-8.3**
[3.75]
Missing . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
-0.6418***
[0.2372]
0.5263***
[0.1249]
-14.8***
[5.42]
Deaf parents
One . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.1507
[0.2871]
1.1626
[0.3337]
3.5
[6.59]
Two . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
0.3507**
[0.1409]
1.4201**
[0.2002]
8.0**
[3.12]
Missing . . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
1.9819***
[0.5822]
7.2564***
[4.2250]
34.0***
[5.49]
Inclusion of birth cohort dummy variables No No Yes
Constant 0.1041***
[0.0359]
. . .
. . .
. . .
. . .
0.2206***
[0.0468]
. . .
. . .
. . .
. . .
0.4382*
[0.2350]
. . .
. . .
. . .
. . .
Observations 4,053 4,053 4,053 4,053 4,053 4,053 4,053 4,053 4,053
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Standard errors in brackets.
SSI = Supplemental Security Income; . . . = not applicable.
* significant at .10 level; ** significant at .05 level; *** significant at .01 level.
## Appendix B:Technical Description of Survival Analysis
The purpose of this section is to provide additional details on the estimates and methodology for the analysis of time spent in the SSI program, along with additional details on the estimates of age/earnings profiles. Table B–1 shows the estimates of the time to first exit from the SSI program that are used for Chart 2. Table B–2 shows the estimates of the time to reentry into the SSI program that are used for Chart 3. Table B–3 shows the data used to construct the age/earnings profiles that are used for Charts 4 through 7. In the remainder of this section we provide further details on survival analysis, which is the technique used to construct the estimates of the time spent in the SSI program.
Table B–1. Lifetable estimates of time to first exit from SSI for adults who received SSI as a child, by NTID status
Years
following
age 19
Graduated Withdrew Accepted, did not attend Not accepted All former SSI children with a
primary diagnosis of deafness
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
Number
eligible
Hazard
(multiplied
by 100)
Survival (percent) Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
1 231 0.11 98.70 555 0.23 97.30 193 0.17 97.93 179 0.09 98.88 9,388 0.31 96.34
2 228 0.18 96.54 540 0.23 94.59 189 0.13 96.37 177 0.09 97.77 9,037 0.29 93.07
3 223 0.15 94.81 525 0.34 90.81 186 0.18 94.30 175 0.29 94.41 8,723 0.32 89.53
4 219 0.23 92.21 501 0.37 86.85 182 0.14 92.75 169 0.40 89.94 8,378 0.41 85.18
5 213 0.86 83.12 482 0.65 80.36 179 0.78 84.46 161 0.53 84.36 7,958 0.45 80.75
6 192 0.96 74.03 446 0.68 74.05 163 0.92 75.65 151 0.51 79.33 7,533 0.50 76.01
7 171 1.33 63.08 411 0.94 66.17 146 1.19 65.53 142 0.68 73.10 6,967 0.57 71.02
8 142 1.94 49.92 336 0.82 59.98 119 0.90 58.81 127 0.91 65.49 5,645 0.64 65.75
9 107 1.59 41.23 275 1.05 52.89 103 0.95 52.43 110 1.15 57.04 4,461 0.62 61.07
10 82 1.29 35.31 219 0.85 47.77 89 1.33 44.69 93 1.37 48.36 3,451 0.56 57.09
Cumulative
probability of
exit within
10 years
(percent)
64.7
[3.29]
52.2
2.28
55.3
3.71
51.6
3.84
42.9
0.57
Median
months to
SSI exit a
95
[1.44]
116
[3.34]
114
[2.58]
118
[2.61]
145
[2.38]
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Standard errors are in brackets.
SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf.
a. Rounded to the nearest month.
Table B–2. Lifetable estimates of time to SSI reentry for adults who received SSI as a child, by NTID status
Years
following
first exit
Graduated Withdrew Accepted, did not attend Not accepted All former SSI children with a
primary diagnosis of deafness
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
Number
eligible
Hazard
(multiplied
by 100)
Survival
(percent)
1 157 0.29 96.63 295 0.62 92.82 115 0.46 94.59 104 0.62 92.82 3,315 0.69 92.07
2 135 0.20 94.39 242 0.34 89.15 101 0.18 92.62 84 0.53 87.09 2,619 0.52 86.53
3 120 0.29 91.12 205 0.36 85.39 89 0.30 89.35 73 0.36 83.36 2,122 0.33 83.17
4 107 0.25 88.41 166 0.39 81.52 78 0.70 82.11 64 0.43 79.12 1,764 0.40 79.28
5 92 0 88.41 136 0.34 78.26 64 0 82.11 51 0.35 75.86 1,371 0.26 76.84
6 79 0 88.41 109 0.17 76.68 51 0.35 78.72 44 0.22 73.91 994 0.22 74.88
7 71 0.27 85.58 87 0.21 74.74 44 0.42 74.84 33 0 73.91 710 0.22 72.96
8 52 0 85.58 69 0 74.74 35 0.56 70.01 23 0 73.91 520 0.17 71.51
9 45 0 85.58 54 0 74.74 25 0.38 66.90 17 0 73.91 376 0.10 70.64
10 36 0 85.58 44 0.22 72.82 19 0 66.90 15 0 73.91 280 0.34 67.80
Cumulative
probability
of reentry
within— a
5 years 11.60
[2.84]
21.70
[2.86]
17.90
[3.99]
24.10
[4.82]
23.16
[0.88]
10 years 14.40
[3.38]
27.20
[3.67]
33.10
[6.28]
26.10
[5.08]
32.2
[1.44]
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTES: Standard errors are in brackets.
SSI = Supplemental Security Income; NTID = National Technical Institute for the Deaf.
a. Median months to reentry not estimated.
Table B–3. Data used in age/earnings profiles for NTID applicants who received SSI as a child, by age and NTID status
Age Graduated Withdrew Accepted, did not attend Not accepted Graduated but not
a former SSI child
All former SSI children with
a primary diagnosis of deafness
Percentage
with earnings
Mean (dollars) Percentage
with earnings
Mean (dollars) Percentage
with earnings
Mean (dollars) Percentage
with earnings
Mean (dollars) Percentage
with earnings
Mean (dollars) Percentage
with earnings
Mean (dollars)
Earners Earners
and non-
earners
Earners Earners
and non-
earners
Earners Earners
and non-
earners
Earners Earners
and non-
earners
Earners Earners
and non-
earners
Earners Earners
and non-
earners
18 46.4 934 434 53.0 963 510 48.4 845 409 40.8 753 307 57.4 1,076 618 42.5 1,113 473
19 45.0 1,219 548 47.6 1,357 646 49.3 1,244 614 46.6 831 387 55.2 1,254 692 47.3 1,804 854
20 50.0 1,425 712 51.0 2,042 1,041 51.6 1,894 977 50.5 1,462 738 57.8 1,576 910 50.8 2,889 1,468
21 55.8 1,820 1,015 57.5 3,334 1,917 54.3 2,429 1,318 60.2 2,370 1,427 62.3 1,932 1,203 54.1 4,165 2,255
22 57.2 2,603 1,489 64.8 4,372 2,833 62.3 3,823 2,383 65.0 4,039 2,627 64.0 2,774 1,775 55.8 5,277 2,946
23 69.4 4,460 3,096 66.2 6,188 4,094 65.0 5,481 3,564 73.8 5,117 3,776 69.3 4,820 3,341 55.1 6,461 3,562
24 68.7 7,410 5,091 70.4 7,698 5,420 66.8 7,925 5,295 73.3 6,352 4,656 74.0 7,724 5,716 54.9 7,627 4,191
25 76.7 10,140 7,774 70.7 9,404 6,646 72.2 9,115 6,586 71.6 7,362 5,269 79.9 10,593 8,460 54.9 8,650 4,745
26 84.6 12,560 10,624 72.3 10,544 7,618 74.9 10,237 7,665 72.9 8,697 6,337 83.8 13,131 11,003 55.2 9,439 5,209
27 86.6 14,655 12,689 73.1 11,788 8,615 74.9 12,490 9,351 72.7 9,963 7,241 86.5 15,619 13,507 55.0 10,203 5,614
28 86.6 17,003 14,725 73.8 13,177 9,727 78.9 12,328 9,729 68.7 11,054 7,596 88.6 17,570 15,572 55.2 10,797 5,956
29 85.2 18,681 15,914 75.1 14,371 10,788 79.1 13,992 11,063 69.2 11,895 8,230 89.8 20,073 18,019 54.8 11,445 6,268
30 84.5 20,776 17,560 76.2 15,232 11,610 81.1 14,996 12,159 73.2 12,071 8,842 90.4 21,748 19,668 55.7 12,246 6,822
31 81.8 20,689 16,915 73.2 16,538 12,107 79.4 17,636 14,001 74.6 12,351 9,218 90.1 23,408 21,090 56.2 12,525 7,037
32 84.8 22,626 19,187 75.7 16,882 12,787 78.1 18,011 14,061 66.1 14,480 9,577 88.7 25,418 22,556 55.2 13,168 7,268
33 80.8 25,358 20,491 75.5 17,071 12,892 82.0 19,529 16,014 70.4 14,707 10,350 87.6 26,818 23,505 53.5 14,065 7,530
34 80.5 28,815 23,202 66.4 18,498 12,290 77.1 21,175 16,335 64.8 16,038 10,398 86.2 27,971 24,112 53.1 14,949 7,943
35 84.7 31,000 26,271 66.7 19,552 13,035 86.0 20,629 17,741 59.2 16,878 9,984 85.4 28,187 24,078 51.9 15,586 8,086
SOURCES: Social Security Administration (SSA) calculations using the data file of administrative records from the National Technical Institute for the Deaf linked to data from SSA's Supplemental Security Record, Master Earnings File, and Numident file.
NOTE: NTID = National Technical Institute for the Deaf; SSI = Supplemental Security Income.
The probability that an exit from the SSI program will occur within 1-year intervals beginning at age 19 may be described using a hazard function or a survival function. Both measures use the probability of failure, ft, in time interval t. The probability of failure is defined as the percentage of persons in the SSI program at the beginning of the time interval who are observed leaving the SSI program within the 12-month interval. The probability of failure is shown in Equation B–1.
Equation B–1
$f t = d t ( N t − m t 2 )$
In Equation B–1, dt is the number of people who leave the program in year t, Nt is the total number of persons observed at the beginning of the year, and mt is the number of censored observations within year t. Censored cases are those for which we do not have data on participation in the program within the time interval and so do not know whether the participants left the program.
The hazard at time t, λt, is the probability that a person will exit the SSI program within a 1-year interval, given that the person has not left the program at the beginning of the interval (shown in Equation B–2).
Equation B–2
$λ j = f j ( 1 − f j 2 ) ⋅ ( t j + 1 − t j )$
Where tj+1tj is the length of the interval in months—which is 12 in our case. The denominator is the standard adjustment for censored cases in the interval.29
The probability that a person remains on the SSI program until period j, referred to as survival (Sj), is the probability that a person has not left the SSI program within a particular interval (shown in Equation B–3).
Equation B–3
$S j = ∏ k = 1 j ( 1 − f k )$
Equation B–3 is simply the probability that failure will not occur in each time interval from 1 to j.
Equations B–1 through B–3 are modified to describe the hazard and the survival estimates for reentry into the SSI program within 1-year intervals, beginning at the point when applicants leave the SSI program. In this case, the hazard rate in Equation B–1 represents the probability that an applicant will reenter the program within a 1-year interval, given that he or she has not reentered the program before the interval. The survival rate in Equation B–3 represents the probability that an applicant has not reentered the SSI program within a particular interval.
## Notes
1 See Daly and Burkhauser (2003) for an overview of the SSI program.
2 The term "managing against the risk of disability" in the context of the children and youth remaining in the SSI disability program has been used by the former Deputy Commissioner for Disability and Income Support Programs at SSA (Gerry 2002).
3 Wittenburg and Maag (2002) identify the lack of data as a limitation to research on the relationship between children's participation in the SSI program and adult outcomes. The National Council on Disability (2003) also identifies limitations in the data available to examine postsecondary education for youth with disabilities.
4 Rupp and Scott (1995) do not disaggregate the length of stay in the program by the time spent on SSI as a child and the time spent in the program as an adult. Rather, for children, they estimate the total time spent in the program. Thus, one cannot use their estimates to identify the portion of time spent in the program as a child and the portion of time spent in the program as an adult.
5 See Davies and Rupp (2006) for further information on the NSCF data.
6 Of the remaining SSI children, 38.5 percent had dropped out of secondary school and 12.9 percent were still enrolled.
7 Estimates of enrollment rates vary across sources and subgroups. The 35 percent estimate is based on all persons aged 18–24. The rate is estimated from the Current Population Survey (CPS), as reported in Hurst and Hudson (2005). Estimates from other surveys range from 32 percent to almost 40 percent.
8 The data merge is possible under the authority of the Privacy Act of 1974 as amended by U.S.C. Section 552a (b) (5), which states, "disclosures may be made with advance adequate written assurance that the record will be used solely as a statistical and reporting record, and transferred in a form that is not individually identifiable."
9 The NTID/SSA merged data file contains information on a total of 13,863 persons who applied to NTID. Of these, 1,597 were not accepted to NTID, 2,068 were accepted but chose not to attend, 5,128 withdrew before completing a degree, and 5,070 graduated from NTID.
10 Although FICA earnings cover most workers, some persons may work in jobs not covered by FICA. Thus, our estimates must be interpreted as employment and earnings within the covered sector.
11 There were 66 deaths among the 5,704 sample members in our case study. The sample size is too small to treat these cases as separate outcomes in our analysis. We estimated the models with and without these cases. Although there was a slight difference in magnitude, it did not have a large impact on the results.
12 In particular, Public Law 96–265 (enacted in 1980) changed the rules regarding parental deeming. Children aged 18 or older were no longer subject to parental deeming for the purposes of program eligibility.
13 Note that Table 2 does not cover the SSA administrative sample of all former SSI children who had a primary diagnosis of deafness and who were born from 1964 through 1980. The reason is that NTID does not have data on those who do not apply for admission.
14 The technical details of the sequential logit model are given in Appendix A. It is important to emphasize that this is a reduced form model that describes the NTID graduation process, and not a formal structural model. Nonetheless, the descriptive results can be very informative to policymakers, as shown in Heckman and Smith (2004) and Ruiz-Quintanilla and others (2006).
15 To illustrate this point, the descriptive statistics show that former SSI children are less likely to graduate. They also show that nonwhites are less likely to graduate. Because SSI children tend to be nonwhite, it is possible that SSI children are less likely to graduate because they tend to be nonwhite, not because they participated in the program as children. Researchers have found lower college graduation rates among minority students and have attributed the findings to the low percentages of minority students on college campuses, which may lead to social isolation, lower social attachment, and, therefore, lower graduation rates (Scott and others 2006). At the same time, it is possible that nonwhites are less likely to graduate because they tend to participate in the SSI program as children. Research by Rupp and others (2006) show that 52.8 percent of all SSI children are nonwhite. The descriptive statistics cannot differentiate between these two alternative explanations. The multivariate models described below provide a measure of the influence of participation in the SSI program as a child, holding race and other characteristics constant.
16 See Appendix A for details.
17 We used age 19 because many SSI children have a short period of time around their 18th birthday when they are out of the program. As of their 19th birthday, 1,158 of the 1,366 SSI children were in the program. We also estimated the models for those who we observed collecting SSI adult benefits, beginning in the month they turned 18. The sample sizes were smaller for this analysis, but the results were similar to those described in this article. They are available on request from the corresponding author, Robert.Weathers@ssa.gov.
18 We tested for the difference in slopes by estimating a regression that allowed for a separate intercept for each series but restricted the slopes to be equal (restricted model) and estimated a regression that allowed separate intercepts and slopes for each trend line (unrestricted model). We computed an F statistic as follows:
$F ( J , n − K ) = ( R u 2 − R r 2 ) / J ( 1 − R u 2 ) / ( n − K )$
Where J is the number of restrictions, which is equal to 1 in our case, n is the number of observations (which is equal to 36) and K is the number of independent variables in the unrestricted model (which is equal to four separate constants and slopes). The R-squared for the restricted model is 0.776487 and the R-squared for the unrestricted model is 0.810819. Thus, F (1,32) = 5.807 > 4.17, which is the 95th percentile of the corresponding F, and we can reject the hypothesis that the two slopes are the same.
19 The decomposition is based on estimates from Table 3. For example, the first term in decomposition shows the contribution of the admitted step to the overall probability of graduation. The first term in Equation A–2 in Appendix A shows that this can be estimated by multiplying the change in the probability of being admitted for SSI children by the conditional probability of attending and by the conditional probability of graduating given attendance. Using the values shown in Table 3, the first term of the decomposition is .0482 * 0.812 * .427 = -.017. We use the term "unconditional probability" to differentiate the probability of graduation among all applicants from the probability of graduation conditional on an applicant being admitted to and choosing to attend NTID.
20 We are unable to produce credible estimates of the median time to reentry because most of our sample does not reenter the SSI program.
21 In the comparisons that follow, we focused on former SSI children who graduated from NTID and compared them with SSI children who were in each of the three groups that did not graduate from NTID. As we showed earlier, SSI children are less likely to graduate from NTID compared with those who had not been on SSI as children. SSI children also had age/earnings profiles that were slightly lower than NTID graduates who were not SSI children. The results are available on request from the corresponding author, Robert.Weathers@ssa.gov.
22 See Table 1–22 from Office of Special Education and Rehabilitative Services, Office of Special Education Programs (2005).
23 Loprest and Wittenburg (2005) do not disaggregate graduation rates by impairment type, which is why we use the OSEP data on graduation rates for all SSI children, by impairment type.
24 See Cornell University http://www.ilr.cornell.edu/edi/p-ccfid.cfm for a study that assesses the state of Web accessibility in the community college network for students with disabilities. The study focuses on examining problems that prospective students with disabilities may have with the online admissions application process, applying for financial aid via the Web, as well as finding important programmatic information on college Websites.
25 The DI program covered under Social Security is a social insurance program funded through payroll tax contributions to the Social Security trust funds, whereas the SSI program is a means-tested cash assistance program funded from general revenues. There are several important differences in these two programs that make separate analysis more practical than attempting to model the two together. We plan to conduct future research on the relationship between postsecondary education and dependency on the DI program.
26 See Madalla (1983) for more information on the sequential logit and Ruiz-Quintanilla and others (2006) for a recent application of the sequential logit to participation in SSA demonstration projects.
27 The logit for the first step was estimated by using the sample of all applicants to NTID. The logit for the second step used the subset of applicants who were admitted to NTID. The logit for the third step used the subset of applicants who were admitted and chose to attend NTID.
28 We used the Stata program written by Bartus (2004) to estimate the changes in the probability related to a change in each characteristic in our sequential logit model.
29 See Allison (1995, 46) for more details on the adjustment for censored observations.
## References
Allison, Paul D. 1995. Survival analysis using the SAS system: A practical guide. Cary, North Carolina: SAS Institute.
Bartus, Tamus. 2005. Estimation of marginal effects using Margeff. Stata Journal 5(3)309–329.
Burkhauser, Richard V., and Mary C. Daly. 2002. Policy watch: U.S. disability policy in a changing environment. Journal of Economic Perspectives 16(1)213–224.
Daly, Mary C., and Richard V. Burkhauser. 2003. The Supplemental Security Income program. In Means tested transfer programs in the United States, ed. Robert Moffitt, 79–140. Chicago: University of Chicago Press for the NBER.
Davies, Paul S., and Kalman Rupp. 2006. An overview of the National Survey of SSI Children and Families and related products. Social Security Bulletin 66(2)7–20.
Dowrick, Peter W., John Anderson, Katharina Heyer, and Joie Acosta. 2005. Postsecondary education across the U.S.A.: Experiences of adults with disabilities. Journal of Vocational Rehabilitation 22(1)41–47.
Gerry, Martin. 2002. Transcript from a public presentation at the 2002 National Workforce Inclusion Conference, March 13. (Accessed January 24, 2005.)
Heckman, James, and Jeffrey Smith. 2004. The determinants of participation in a social program: Evidence from the Job Training Partnership Act. Journal of Labor Economics 22(4): 243–298.
Horn, Laura, and Jennifer Berktold. 1999. Students with disabilities in postsecondary education: A profile of preparation, participation, and outcomes. Washington, DC: National Center for Education Statistics.
Hurst, David, and Lisa Hudson. 2005. Estimating undergraduate enrollment in postsecondary education using National Center for Education Statistics data (NCES 2005–063). U.S. Department of Education, National Center for Education Statistics. Washington, DC: Government Printing Office.
Lancaster, Tony. 1990. The econometric analysis of transition data. Cambridge University Press.
Loprest, Pamela, and David Wittenburg. 2005. Choices, challenges, and options: Child SSI recipients preparing for the transition to adult life. http://www.urban.org/url.cfm?ID=411168 (accessed May 5, 2006).
Madalla, G.S. 1982. Limited dependent and qualitative variables in econometrics. Cambridge University Press.
Office of Special Education and Rehabilitative Services, Office of Special Education Programs. 2005. 26th Annual (2004) report to Congress on the implementation of the Individuals with Disabilities Education Act, vol. 1, Washington, DC.
Ruiz-Quintanilla, Antonio, Robert R. Weathers II, Valerie Melburg, Kimberly Campbell, and Nawaf Madi. 2006. Participation in programs designed to improve employment outcomes for persons with psychiatric disabilities: Evidence from the New York WORKS demonstration project. Social Security Bulletin 66(2)49–79.
Rupp, Kalman, and Charlie Scott. 1995. Length of stay on the Supplemental Security Income Program. Social Security Bulletin 58(1)29–47.
Rupp, Kalman, Paul S. Davies, Chad Newcomb, Howard Iams, Carrie Becker, Shanti Mulpuru, Stephen Ressler, Kathleen Romig, and Baylor Miller. 2006. A profile of children with disabilities receiving SSI: Highlights from the National Survey of SSI Children and Families. Social Security Bulletin 66(2)21–48.
Scott, Marc, Thomas Bailey, and Greg Kienzl. 2006. Relative success? Determinants of college graduation rates in public and private colleges in the U.S. Research in Higher Education 47(3)249–279.
Social Security Administration. 2006. Children receiving SSI. Washington, DC: Social Security Administration.
Wagner, Mary, Lynn Newman, Renee Cameto, and Phyllis Levine. 2005. Changes over time in the early postschool outcomes of youth with disabilities: A report of findings from the National Longitudinal Transition Study (NLTS) and the National Longitudinal Transition Study-2 (NLTS2). Menlo Park, CA: SRI International.
Walter, Gerard G., Jack R. Clarcq, and Wendell S. Thompson. 2002. Effect of degree on improving the economic status of individuals who are deaf. Journal of the American Deafness and Rehabilitation Association 35(4)30–46.
Wittenburg, David C., and Elaine Maag. 2002. School to where? A literature review on economic outcomes of youth with disabilities. Journal of Vocational Rehabilitation 17(4)265–280.
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# Identify the phase shift and the vertical shift of the function with equation y = cos(x+ π) - 4?
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2016-12-09 04:54:15
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http://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-2nd-edition/chapter-11-vectors-and-vector-valued-functions-11-6-calculus-of-vector-valued-functions-11-6-exercises-page-814/1
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# Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises: 1
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
#### Work Step by Step
The derivative of $\textbf{r}$ is found by taking the derivative of every component of $\textbf{r}$, assuming all the components are differentiated functions.
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.
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2018-04-23 04:21:57
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https://www.mcseccna.cloud/mike-kreuzer-resume/doku.php?id=frequency
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# Mike J. Kreuzer, PhD, MCSE, MCT, Azure-AWS Kubernetes Cloud Native DevOps and DevSecOps
### Sidebar
Mike J. Kreuzer, Ph.D, MCSE, MCT
Call me at: 831-675-MCSE
Kreuzer the Cloud Monk is a “Cloud First - Mobile FirstDevOps - DevSecOps Engineer
Silicon Valley and International Networks since 1984.
• Mirror Sites:
frequency
, (Hz)—or the number of events per second (i.e., cycles per second)—that the dot flashes; while
the period, or time, in seconds (s) of each cycle, (i.e., the number of seconds per cycle). Note
f are reciprocal values to each other.]]
Frequency is the number of occurrences of a repeating event per unit time.<ref>
</ref> It is also referred to as temporal frequency, which emphasizes the contrast to spatial frequency and angular frequency. The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency.<ref>
</ref> For example, if a newborn baby's heart beats at a frequency of 120 times a minute, its period—the time interval between beats—is half a second (that is, 60 seconds divided by 120 beats). Frequency is an important parameter used in science and engineering to specify the rate of oscillatory and vibratory phenomena, such as mechanical vibrations, audio (sound) signals, radio waves, and light.
## Definitions
File:Sine waves different frequencies.svg
s vary, or cycle, regularly at different rates. The red wave (top) has the lowest frequency (i.e., cycles at the slowest rate) while the purple wave (bottom) has the highest frequency (cycles at the fastest rate).]]
For cyclical processes, such as rotation, oscillations, or waves, frequency is defined as a number of cycles per unit time. In physics and engineering disciplines, such as optics, acoustics, and radio, frequency is usually denoted by a Latin letter f or by the Greek letter $\nu$ or ''ν'' (nu) (see e.g. Planck's formula).
Period (in units of time) X Ordinary frequency (in number of cycles per unit of time) = 1 cycle.
Therefore, the period, usually denoted by T, is the duration of one cycle, and is the reciprocal of the frequency f: :$f = \frac{1 cycle}{T}.$
## Units
The SI unit of frequency is the hertz (Hz), named after the German physicist Heinrich Hertz. One hertz means that an event repeats once per second. A previous name for this unit was cycles per second (cps). The SI unit for period is the second.
A traditional unit of measure used with rotating mechanical devices is revolutions per minute, abbreviated r/min or rpm. 60 rpm equals one hertz.<ref>
</ref>
electrical science
## Period versus frequency
As a matter of convenience, longer and slower waves, such as ocean surface waves, tend to be described by wave period rather than frequency. Short and fast waves, like audio and radio, are usually described by their frequency instead of period. These commonly used conversions are listed below:
Frequency Period (time) 1 mHz (10−3) 1 Hz (100) 1 kHz (103) 1 MHz (106) 1 GHz (109) 1 THz (1012) 1 ks (103) 1 s (100) 1 ms (10−3) 1 µs (10−6) 1 ns (10−9) 1 ps (10−12)
File:Commutative diagram of harmonic wave properties.svg
::$y(t) = \sin\left( \theta(t) \right) = \sin(\omega t) = \sin(2 \mathrm{\pi} f t)$
::$\frac{\mathrm{d} \theta}{\mathrm{d} t} = \omega = 2 \mathrm{\pi} f$
: Angular frequency is commonly measured in radians per second (rad/s) but, for discrete-time signals, can also be expressed as radians per sample time, which is a dimensionless quantity.
• Spatial frequency is analogous to temporal frequency, but the time axis is replaced by one or more spatial displacement axes. E.g.:
::$y(t) = \sin\left( \theta(t,x) \right) = \sin(\omega t + kx)$
::$\frac{\mathrm{d} \theta}{\mathrm{d} x} = k$
: Wavenumber, k, is the spatial frequency analogue of angular temporal frequency and is measured in radians per meter. In the case of more than one spatial dimension, wavenumber is a vector quantity.
## In wave propagation {{anchor|Frequency of waves}}
<!– This section is linked from Hearing impairment –> For periodic waves in nondispersive media (that is, media in which the wave speed is independent of frequency), frequency has an inverse relationship to the wavelength, λ (lambda). Even in dispersive media, the frequency f of a sinusoidal wave is equal to the phase velocity v of the wave divided by the wavelength λ of the wave: :$f = \frac{v}{\lambda}.$
In the special case of electromagnetic waves moving through a vacuum, then v = c, where c is the speed of light in a vacuum, and this expression becomes: :$f = \frac{c}{\lambda}.$
When waves from a monochrome source travel from one medium to another, their frequency remains the same—only their wavelength and speed change.
## Measurement
### By counting
Calculating the frequency of a repeating event is accomplished by counting the number of times that event occurs within a specific time period, then dividing the count by the length of the time period. For example, if 71 events occur within 15 seconds the frequency is: :$f = \frac{71}{15 \,\text{s}} \approx 4.7 \, \text{Hz}$ If the number of counts is not very large, it is more accurate to measure the time interval for a predetermined number of occurrences, rather than the number of occurrences within a specified time.<ref>
</ref> The latter method introduces a random error into the count of between zero and one count, so on average half a count. This is called gating error and causes an average error in the calculated frequency of Δf = 1/(2 Tm), or a fractional error of Δf / f = 1/(2 f Tm) where Tm is the timing interval and f is the measured frequency. This error decreases with frequency, so it is a problem at low frequencies where the number of counts N is small.
### By stroboscope
An older method of measuring the frequency of rotating or vibrating objects is to use a stroboscope. This is an intense repetitively flashing light (strobe light) whose frequency can be adjusted with a calibrated timing circuit. The strobe light is pointed at the rotating object and the frequency adjusted up and down. When the frequency of the strobe equals the frequency of the rotating or vibrating object, the object completes one cycle of oscillation and returns to its original position between the flashes of light, so when illuminated by the strobe the object appears stationary. Then the frequency can be read from the calibrated readout on the stroboscope. A downside of this method is that an object rotating at an integral multiple of the strobing frequency will also appear stationary.
### By frequency counter
Higher frequencies are usually measured with a frequency counter. This is an electronic instrument which measures the frequency of an applied repetitive electronic signal and displays the result in hertz on a digital display. It uses digital logic to count the number of cycles during a time interval established by a precision quartz time base. Cyclic processes that are not electrical in nature, such as the rotation rate of a shaft, mechanical vibrations, or sound waves, can be converted to a repetitive electronic signal by transducers and the signal applied to a frequency counter. Frequency counters can currently cover the range up to about 100 GHz. This represents the limit of direct counting methods; frequencies above this must be measured by indirect methods.
### Heterodyne methods
Above the range of frequency counters, frequencies of electromagnetic signals are often measured indirectly by means of heterodyning (frequency conversion). A reference signal of a known frequency near the unknown frequency is mixed with the unknown frequency in a nonlinear mixing device such as a diode. This creates a heterodyne or “beat” signal at the difference between the two frequencies. If the two signals are close together in frequency the heterodyne is low enough to be measured by a frequency counter. This process only measures the difference between the unknown frequency and the reference frequency, which must be determined by some other method. To reach higher frequencies, several stages of heterodyning can be used. Current research is extending this method to infrared and light frequencies (optical heterodyne detection).
## Examples
### Light
File:EM spectrum.svg
with the visible portion highlighted]]
Visible light is an electromagnetic wave, consisting of oscillating electric and magnetic fields traveling through space. The frequency of the wave determines its color:
is red light,
is violet light, and between these (in the range 4-
) are all the other colors of the rainbow. An electromagnetic wave can have a frequency less than
, but it will be invisible to the human eye; such waves are called infrared (IR) radiation. At even lower frequency, the wave is called a microwave, and at still lower frequencies it is called a radio wave. Likewise, an electromagnetic wave can have a frequency higher than
, but it will be invisible to the human eye; such waves are called ultraviolet (UV) radiation. Even higher-frequency waves are called X-rays, and higher still are gamma rays.
All of these waves, from the lowest-frequency radio waves to the highest-frequency gamma rays, are fundamentally the same, and they are all called electromagnetic radiation. They all travel through a vacuum at the same speed (the speed of light), giving them wavelengths inversely proportional to their frequencies.
:$\displaystyle c=f\lambda$
where c is the speed of light (c in a vacuum, or less in other media), f is the frequency and λ is the wavelength.
In dispersive media, such as glass, the speed depends somewhat on frequency, so the wavelength is not quite inversely proportional to frequency.
### Sound
Sound propagates as mechanical vibration waves of pressure and displacement, in air or other substances.<ref>
</ref> Frequency is the property of sound that most determines pitch.<ref>
</ref>
The frequencies an ear can hear are limited to a specific range of frequencies. The audible frequency range for humans is typically given as being between about 20 Hz and 20,000 Hz (20 kHz), though the high frequency limit usually reduces with age. Other species have different hearing ranges. For example, some dog breeds can perceive vibrations up to 60,000 Hz.<ref name=“Physics Factbook”>
</ref>
In many media, such as air, the speed of sound is approximately independent of frequency, so the wavelength of the sound waves (distance between repetitions) is approximately inversely proportional to frequency.
### Line current
In Europe, Africa, Australia, Southern South America, most of Asia, and Russia, the frequency of the alternating current in household electrical outlets is 50 Hz (close to the tone G), whereas in North America and Northern South America, the frequency of the alternating current in household electrical outlets is 60 Hz (between the tones B♭ and B; that is, a minor third above the European frequency). The frequency of the 'hum' in an audio recording can show where the recording was made, in countries using a European, or an American, grid frequency.
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2020-11-24 05:51:42
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https://math.stackexchange.com/questions/2164024/proving-that-every-third-fibonacci-number-is-divisible-by-f2-2/3276703#3276703
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# Proving that every third Fibonacci number is divisible by F2=2
In our notation we have that $F_{n-1}$ is the $n$th Fibonacci number since we start with $F_{0}=1$.
We want to prove that every third Fibonacci number is divisible by $F_{2}=2$.
The proof is the following using induction:
$F_{3n+2}=F_{3n+1}+F_{3n}$
$F_{3n+2}=F_{3n}+F_{3n-1}+F_{3n-1}+F_{3n-2}$ (A)
$F_{3n+2}=F_{3n-1}+F_{3n-2}+F_{3n-1}+F_{3n-2}$ (B)
$F_{3n+2}=2(F_{3n-1}+F_{3n-2})$
I don't understand how you go from step (A) to step (B) can anyone explain this to me?
There's a missing term in (B) as you noticed. A correct proof is by induction. Show that $F_2$ is even (immediate). Then write
\begin{align} F_{3n+2} &= F_{3n+1} + F_{3n} \\ &= 2F_{3n} + F_{3n-1} \\ &= 2F_{3n} + F_{3(n-1)+2} \enspace, \end{align}
which is even because it's the sum of two even numbers. ($F_{3(n-1)+2}$ is even by the induction hypothesis.)
There is a well-known property of Fibonacci numbers: $$\gcd(F_n,F_m)=F_{\gcd(n,m)} \tag{1}$$ Since $F_3=2$, $$\gcd(F_{3n},2)=\gcd(F_{3n},F_3)=F_{\gcd(3,3n)} = F_3 = 2 \tag{2}$$ hence every Fibonacci number of the form $F_{3n}$ is even.
Base case:
$F_1=1\\ F_2=1\\ F_3=2$
Inductive hypothesis: Suppose $F_{3k-2}$ is odd, and $F_{3k-1}$ is odd, and $F_{3k}$ even.
We must show that $F_{3(k+1)-2}$ and $F_{3(k+1)-1}$ are odd, and $F_{3(k+1)}$ is even
$F_{3(k+1)-2} = F_{3k-1}+F_{3k}$
Based on the inductive hypothesis, $F_{3(k+1)-2}$ must be odd as the sum of an odd number and an even number are odd.
I will leave it to you to finish from here.
The rank of apparition of $$2$$, denoted $$\omega(2)$$, in the Fibonacci sequence is $$3$$; that is, the first term in the Fibonacci sequence that contains $$2$$ as a divisor is $$F_{3} = 2$$.
Since the Fibonnacci sequence is a member of the family of Lucas sequences, divisibility properties of the Lucas sequences are inherited by the Fibonacci sequence. Hence, $$F_{3} | F_{k}$$ if and only if $$k = r\omega(2) = 3r$$, for any positive integer $$r$$.
Therefore $$2$$ divides $$F_{3}$$ and every third term thereafter (and no others) in the Fibonacci sequence.
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2022-01-18 02:35:27
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https://mathematica.stackexchange.com/questions/211498/how-do-you-plot-random-points-in-three-dimensions
|
# How do you plot random points in three dimensions?
I've got a function that maps a 2D plane onto a sphere (I'm trying to learn about Geodesics).
f[u_,v_]:={X[u,v],Y[u,v],Z[u,v]};
X[u_,v_]:=Cos[v]Sin[u];
Y[u_,v_]:=Sin[v]Sin[u];
Z[u_,v_]:=Cos[u];
I want to plot this, but the Plot3D gives me only a single value for any given [X,Y] set of coordinates. At the least, I'd like something like a scatter chart (that is, just give it 3 coordinates and have it plot a point), but it would be nice to be able to generate a wire-frame so I could draw my solutions.
Clear["Global*"]
f[u_, v_] := {X[u, v], Y[u, v], Z[u, v]};
X[u_, v_] := Cos[v] Sin[u];
Y[u_, v_] := Sin[v] Sin[u];
Z[u_, v_] := Cos[u];
SeedRandom[1234]
data = f @@@ RandomReal[{0, 2 Pi}, {5000, 2}];
Graphics3D[Point[data]]
However, you can get a smoother distribution with RandomPoint on a Sphere
SeedRandom[1234]
data2 = RandomPoint[Sphere[{0, 0, 0}], 5000];
Graphics3D[Point[data2]]
EDIT: Converting the points into a 3-D surface
ListSurfacePlot3D[data2, Axes -> False]
• Your first approach is not correct. Pay your attention that the points flock near the poles. – user64494 Dec 17 '19 at 8:01
• @user64494 it's not incorrect — it's just a different distribution. – Ruslan Dec 17 '19 at 10:39
• Great answer. Thank you. For extra points, can you make that into a wire-frame? – Quarkly Dec 17 '19 at 15:22
• @user64494 random doesn't imply uniform. – Ruslan Dec 18 '19 at 8:16
• @user64494 - The OP did not specify that the points in the "scatter chart" should be uniformly distributed on the sphere. The first approach demonstrates that if you start with a uniform distribution in 2-D, the transformation does not produce a uniform distribution in 3-D. While RandomPoint[reg, n] gives a list of n pseudorandom points uniformly distributed in the region reg; if you were to transform these 3-D points on a sphere to 2-D, they would not be uniform in 2-D. – Bob Hanlon Dec 18 '19 at 14:09
Thanks to Bob Hanlon for his answer. Using that as a starting point, I was able to work out this, which is what I was really after:
ParametricPlot3D[{R[u, v]}, {u, 0, Pi}, {v, 0, 2*Pi}]
`
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2021-04-22 00:16:47
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http://tribology.asmedigitalcollection.asme.org/article.aspx?articleid=1467506
|
0
RESEARCH PAPERS
# Thermal and Non-Newtonian Numerical Analyses for Starved EHL Line Contacts
[+] Author and Article Information
P. Yang
School of Mechanical Engineering, Qingdao Technological University, Qingdao 266033, China
J. Wang
Department of Mechanical and Control Engineering, Kyushu Institute of Technology, Kitakyushu 804-8550, Japan
M. Kaneta
Department of Mechanical and Control Engineering, Kyushu Institute of Technology, Kitakyushu 804-8550, Japankaneta@mech.kyutech.ac.jp
J. Tribol 128(2), 282-290 (Jul 27, 2005) (9 pages) doi:10.1115/1.2164465 History: Received January 17, 2005; Revised July 27, 2005
## Abstract
This paper focuses on the mechanism of starvation and the thermal and non-Newtonian behavior of starved elastohydrodynamic lubrication (EHL) in line contacts. It has been found that for a starved EHL line contact if the position of the oil-air meniscus is given as input parameter, the effective thickness of the available lubricant layers on the solid surfaces can be solved easily from the mass continuity condition, alternatively, if the later is given as input parameter, the former can also be determined easily. Numerical procedures were developed for both situations, and essentially the same solution can be obtained for the same parameters. In order to highlight the importance of the available oil layers, isothermal and Newtonian solutions were obtained first with multi-level techniques. The results show that as the inlet meniscus of the film moves far away from the contact the effective thickness of the oil layers upstream the meniscus gently reaches a certain value. This means very thin layers (around $1μm$ in thickness) of available lubricant films on the solid surfaces, provided the effective thickness is equal to or larger than this limitation, are enough to fill the gap downstream the meniscus and makes the contact work under a fully flooded condition. The relation between the inlet meniscus and the effective thickness of the available lubricant layers was further investigated by thermal and non-Newtonian solutions. For these solutions the lubricant was assumed to be a Ree-Eyring fluid. The pressures, film profiles and temperatures under fully flooded and starved conditions were obtained with the numerical technique developed previously. The traction coefficient of the starved contact is found to be larger than that of the fully flooded contact, the temperature in the starved EHL film, however, is found to be lower than the fully flooded contact. Some non-Newtonian results were compared with the corresponding Newtonian results.
###### FIGURES IN THIS ARTICLE
<>
Copyright © 2006 by American Society of Mechanical Engineers
Your Session has timed out. Please sign back in to continue.
## Figures
Figure 1
Schematic diagram of a staved EHL line contact
Figure 2
The relations of the effective thickness of the available oil layers and the central film thickness with the position of the air-oil meniscus of the isothermal and Newtonian solutions for U=1×10−11, G=4949, and W=4×10−5
Figure 3
Pressure distributions with various positions of the air-oil meniscus. From the highest spike to the lowest spike, the value of xin∕b is −3.6, −1.5, −1.25, −1.1, and −1.02, respectively. Isothermal and Newtonian solutions for U=1×10−11, G=4949, and W=4×10−5.
Figure 4
Film thickness profiles corresponding to the pressures given in Fig. 3. From top to bottom the value of xin∕b is −3.6, −1.5, −1.25, −1.1, and −1.02, respectively.
Figure 5
The relations of the effective thickness of the available oil layers and the central film thickness with the position of the air-oil meniscus of the thermal and non-Newtonian solutions for Σ=1.2, U=2×10−11, G=4949, and W=1.5×10−4
Figure 6
Pressure distributions with various positions of the air-oil meniscus. From the highest spike to the lowest spike, the value of xin∕b is −3.6, −1.25, −1.1, −1.04, and −1.01, respectively. Thermal and non-Newtonian solutions for Σ=1.2, U=2×10−11, G=4949, and W=1.5×10−4.
Figure 7
Film thickness profiles corresponding to the pressures given in Fig. 6. From top to bottom the value of xin∕b is −3.6, −1.25, −1.1, −1.04, and −1.01, respectively.
Figure 8
Temperature distributions in the middle layer of the film (solid line), and on surface a (dashed line) and b (dotted line) of the fully flooded (xin∕b=−3.6) contact. Thermal and non-Newtonian solutions for Σ=1.2, U=2×10−11, G=4949, and W=1.5×10−4.
Figure 9
The flow velocity distributions across the film at x=0 for xin∕b=−3.6 (solid line) and −1.01 (dashed line), respectively. Thermal and non-Newtonian solutions for Σ=1.2, U=2×10−11, G=4949, and W=1.5×10−4.
Figure 10
Temperature distributions in the middle layer of the film (solid line), and on surface a (dashed line) and b (dotted line) of the most severely starved (xin∕b=−1.01) contact. Thermal and non-Newtonian solutions for Σ=1.2, U=2×10−11, G=4949, and W=1.5×10−4.
Figure 11
Temperature contour map for the case of xin∕b=−1.1, numbers indicate the temperature in °C. Thermal and non-Newtonian solutions for Σ=1.2, U=2×10−11, G=4949, and W=1.5×10−4.
Figure 12
The variation in the traction coefficient versus the position of the air-oil meniscus for the same cases as in Fig. 5
Figure 13
Variations in the traction coefficients and the middle point temperatures versus the slide-roll ratios for the fully flooded and severely starved contacts. U=2×10−11, G=4949, and W=1.5×10−4, (a) traction coefficients, and (b) middle point temperatures.
Figure 14
Variations in the traction coefficients and the middle point temperatures versus the slide-roll ratios predicted by the Newtonian and Ree–Eyring solutions, respectively, for the starved contacts of xin∕b=−1.1, U=2×10−11, G=4949, and W=1.5×10−4, (a) traction coefficients, and (b) middle point temperatures
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2019-01-17 08:52:09
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http://eprint.iacr.org/2014/103
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Cryptology ePrint Archive: Report 2014/103
SHipher: Families of Block Ciphers based on SubSet-Sum Problem
Xiali Hei and Binheng Song
Abstract: In this paper, we describe the families of block ciphers named SHipher. We show a symmetric encryption framework based on the SubSet-Sum problem. This framework can provide families of secure, flexible, and any-size block ciphers. We have extensively cryptanalyzed our encryption framework. We can easily control the computational cost by a key selection. Also, this framework offers excellent performance and it is flexible and general enough to admit a variety of implementations on different non-Abelian groups. In this paper, we provide one implementation using a group of matrices whose determinants are 1. This implementation accepts any block size satisfying $3l-1$. If $l=21$, the block size is 62 bits, which suits for full spectrum of lightweight applications. While if $l=341$, the block size is 1022, which provides high security level up to resistant $2^{684}$ differential-attack effort and $2^{1022}$ brute-force attack effort.
Category / Keywords: Block cipher; SubSet-Sum problem; Framework; Non-Abelian group
Date: received 11 Feb 2014, last revised 15 Feb 2014
Contact author: xiali hei at temple edu
Available format(s): PDF | BibTeX Citation
Note: Update the related works we omitted before.
Short URL: ia.cr/2014/103
[ Cryptology ePrint archive ]
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2016-05-29 11:54:19
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https://aimsciences.org/article/doi/10.3934/cpaa.2011.10.1663
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# American Institute of Mathematical Sciences
November 2011, 10(6): 1663-1686. doi: 10.3934/cpaa.2011.10.1663
## Unbounded solutions of the nonlocal heat equation
1 Departamento de Matemáticas, U. Carlos III de Madrid, 28911 Leganés, Spain 2 Laboratoire de Mathématiques et Physique Théorique, U. F. Rabelais, Parc de Grandmont, 37200 Tours, France 3 Departamento de Matemática Aplicada, Universidad Complutense de Madrid, 28040 Madrid
Received February 2010 Revised January 2011 Published May 2011
We consider the Cauchy problem posed in the whole space for the following nonlocal heat equation: $u_t = J\ast u -u,$ where $J$ is a symmetric continuous probability density. Depending on the tail of $J$, we give a rather complete picture of the problem in optimal classes of data by: $(i)$ estimating the initial trace of (possibly unbounded) solutions; $(ii)$ showing existence and uniqueness results in a suitable class; $(iii)$ proving blow-up in finite time in the case of some critical growths; $(iv)$ giving explicit unbounded polynomial solutions.
Citation: C. Brändle, E. Chasseigne, Raúl Ferreira. Unbounded solutions of the nonlocal heat equation. Communications on Pure & Applied Analysis, 2011, 10 (6) : 1663-1686. doi: 10.3934/cpaa.2011.10.1663
##### References:
[1] N. Alibaud and C. Imbert, Fractional semi-linear parabolic equations with unbounded data,, Trans. Amer. Math. Soc., 361 (2009), 2527. doi: 10.1090/S0002-9947-08-04758-2. Google Scholar [2] C. Brändle and E. Chasseigne, Large deviations estimates for some non-local equations. Fast decaying kernels and explicit bounds,, Nonlinear Analysis, 71 (2009), 5572. doi: 10.1016/j.na.2009.04.059. Google Scholar [3] C. Brändle and E. Chasseigne, Large Deviations estimates for some non-local equations. General bounds and applications,, to appear in Trans. Amer. Math. Soc, (). Google Scholar [4] P. Carr, H. Geman, D. B. Madan and M. Yor, Stochastic volatility for Lévy processes,, Math. Finance, 13 (2003), 345. doi: 10.1111/1467-9965.00020. Google Scholar [5] E. Chasseigne, M. Chaves and J. D. Rossi, Asymptotic behavior for nonlocal diffusion equations,, J. Math. Pures Appl., 86 (2006), 271. doi: 10.1016/j.matpur.2006.04.005. Google Scholar [6] E. Chasseigne and R. Ferreira, Isothermalization for a Non-local Heat Equation,, preprint, (). Google Scholar [7] F. John, "Partial Differential Equations,", 4$^{nd}$ edition, (1982). Google Scholar
show all references
##### References:
[1] N. Alibaud and C. Imbert, Fractional semi-linear parabolic equations with unbounded data,, Trans. Amer. Math. Soc., 361 (2009), 2527. doi: 10.1090/S0002-9947-08-04758-2. Google Scholar [2] C. Brändle and E. Chasseigne, Large deviations estimates for some non-local equations. Fast decaying kernels and explicit bounds,, Nonlinear Analysis, 71 (2009), 5572. doi: 10.1016/j.na.2009.04.059. Google Scholar [3] C. Brändle and E. Chasseigne, Large Deviations estimates for some non-local equations. General bounds and applications,, to appear in Trans. Amer. Math. Soc, (). Google Scholar [4] P. Carr, H. Geman, D. B. Madan and M. Yor, Stochastic volatility for Lévy processes,, Math. Finance, 13 (2003), 345. doi: 10.1111/1467-9965.00020. Google Scholar [5] E. Chasseigne, M. Chaves and J. D. Rossi, Asymptotic behavior for nonlocal diffusion equations,, J. Math. Pures Appl., 86 (2006), 271. doi: 10.1016/j.matpur.2006.04.005. Google Scholar [6] E. Chasseigne and R. Ferreira, Isothermalization for a Non-local Heat Equation,, preprint, (). Google Scholar [7] F. John, "Partial Differential Equations,", 4$^{nd}$ edition, (1982). Google Scholar
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2018 Impact Factor: 0.925
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2019-08-21 04:57:14
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https://www.biostars.org/p/9469801/
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What are important questions before starting analysis of 10x scRNA-seq dataset
2
0
Entering edit mode
5 months ago
mi • 0
I am supposed to analyze some 10x chromium single cell gene expression data set in the future.
There will be plenty of time to familiarize with the task and to discuss the matter with experts in the field. I myself have years of experience in genomics/transcriptomics but not with single-cell expression data.
However, I need to evaluate the quality of the data and the amount of work needed for the task in the next days.
So what are the questions you would ask about a 10x chromium single cell gene expression data set before deciding if you take over the task and to see how much work it will be? What to consider?
10x scRNA-seq • 530 views
3
Entering edit mode
5 months ago
As far as QC goes, you probably want to look at how much of the data is mitochondrial in origin (could indicate dead or lysed cells), look at the distribution of number of genes per cell and UMI counts and maybe run DoubletFinder. You don't say what type of tissue or species, but sometimes HBB/HBA genes from red blood cells can give you a feel for the background (ambient) RNA contamination or "soup" in the data. Or sometimes these will just form a cluster.
Otherwise, I think walking through the appropriate Seurat tutorials might give you some more ideas. If it's "low quality" are you not going to analyze it?
Oh, and make a note of your versions of everything when you start or otherwise find a way to keep software consistent. In my experience, the data analysis on these projects sometimes outlasts new versions of R, Seurat, etc. You don't want the UMAP and clustering to change six months in due to some underlying software update.
0
Entering edit mode
Regarding the "random" / non-deterministic elements of the analysis such as UMAPs: Be sure to set.seed() everywhere, but the linked OSCA does a good job indicating where it is key to ensure that manifolds etc look the same even if you re-run it several times. As for management / documentation of package versions I became a fan of renv, see my comment here => Installing an updated R version (>=4.0) using conda unless you anyway manage everything via a contained-based solution.
2
Entering edit mode
5 months ago
GenoMax 107k
As with any new analysis expect to spend some time familiarizing yourself with the procedure involved. 10x makes a set of tools available for doing the analysis. You can find them at their support site. I am going to link GEX site but they have other protocols. They also have tutorials and test data you can download.
On open source side of things:
1. STARsolo (LINK)
2. alevin-fry (LINK)
3. OSCA - Orchestrating Single-Cell Analysis with Bioconductor
4. Seurat (LINK - getting started)
You will likely choose one from 1/2 and 3/4 above. You are going to be spending a few days for sure on learning plus doing the actual analysis. There are many existing threads here to refer to and a few experts who will be able to answer questions.
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Entering edit mode
I highly recommend you read through 3 here even if you don't use Bioconductor tools for your analysis. It does a very good job of explaining why and when each step is necessary.
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2021-10-22 16:18:31
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https://nbviewer.jupyter.org/github/ZikangXiong/blogs/blob/main/notebooks/Verifiable%20AI/daniel_certifiable_kdd20.ipynb
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# Certifiable Robustness of Graph Convolutional Networks under Structure Perturbations¶
## Contribution¶
A framework for graph structure perturbation robustness.
## Preliminaries and Approaches¶
### Their Structural Robustness Definition¶
For one node, perturb on the transformation of adjacency matrix $\mathbf{A}$ with admissible perturbations, the label should not change.
\begin{aligned} m^{t}\left(y^{*}, y\right):=& \operatorname{minimize}_{\tilde{A}} f_{\theta}^{t}(X, \mathcal{T}(\tilde{A}))_{y^{*}}-f_{\theta}^{t}(X, \mathcal{T}(\tilde{A}))_{y} \\ & \text { subject to } \tilde{A} \in \mathcal{A}(A) \end{aligned}
The admissible perturbation is defined as \begin{aligned} \mathcal{A}(A)=&\left\{\tilde{A} \in\{0,1\}^{N \times N} \mid \tilde{A}_{i j} \leq A_{i j} \wedge \tilde{A}=\tilde{A}^{T}\right.\\ & \wedge\|\tilde{A}-A\|_{0} \leq 2 Q \\ &\left.\wedge\left\|\tilde{A}_{i}-A_{i}\right\|_{0} \leq q_{i} \forall 1 \leq i \leq N\right\} \end{aligned}
The formalization is unclear to me, they said that the allow to insert at most $q_i$ edges, but the definition above seems to be delete some edges. Based on the left paper, the insert should be a typo.
### Induced constraints.¶
It is critical to make sure that the perturbed results in the transformed space are subjected to the constraints on the real adjacency matrix. The authors proposed a sequence of constraints here. Did not read carefully.
### Relaxation of the neural network¶
Abstraction based on this.
### Bilinear Program¶
Guess that bilinear is because $\hat{\mathbf{A}}$ is a range. Did not read other part carefully. There seems to be an iterative algorithm that have coverage guarantee. As long as they can made the lower boundary of $y* - y > 0$, the network is robust.
## Follow-up¶
It is worthwhile to read their formalization on the inducted constraints, if I want to write this kind of paper.
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2021-06-24 05:49:32
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http://math.stackexchange.com/questions/274988/why-is-this-proof-false
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# Why is this proof false?
I know this proof is false, but I don't know why. I need your help.
The false proof says that it is possible to create a bijection between a subset of the rational numbers and the Power set of natural numbers.
We can create orderly the subsets of the natural numbers and create a bijection at the same time to some of the rational numbers:
First pairs:
{1,2},{1,3}.... -> 1/2, 1/3..
{2,3},{2,4}.... -> 2/3, 2/4..
now three:
{1,2,3},{1,2,4}... 12/3, 12/4...
{1,3,4},{1,3,5}....13/4, 13/5...
...
{2,3,4}, {2,3,5}...23/4, 23/5...
four...
And so on...
So this false proof says that the cardinal of the rational numbers are, al least the cardinal of P(N)
How can I explain that is false
Thanks.
-
Your construction does not contain infinite subsets of $\mathbb{N}$. – Raskolnikov Jan 10 '13 at 7:44
And indeed the set $\mathcal P_f(\mathbb N)$ of the finite subsets of $\mathbb N$ is countable. – Did Jan 10 '13 at 7:46
Not only have you not defined your function on any infinite subset of $\Bbb{N}$, you haven't defined it on all the finite subsets either. I at least cannot guess from your description how you indeed to deal with sets of size four, let alone larger ones. – Chris Eagle Jan 10 '13 at 7:53
I think that the main problem is that I don't have the infinite subsets in act...but I don't understand the rules very well. Why in the bijecction between rational and naturals we understand we have contructed all infinite numbers and in this false proof we cannot say e have constructed all even numbers or all quadratic numbers? – Pedro Jan 10 '13 at 8:00
@did :"the set Pf(N) of the finite subsets of N is countable" – Pedro Jan 10 '13 at 8:55
show 1 more comment
Ignoring that I don't think that map is going to be injective on finite subsets of the natural numbers. Where do you send $\mathbb N$ or the set of all even numbers?
I will add that if you consider the set $\mathcal A =\{ B \subset \mathbb N : |B| < \infty\}$ that $|\mathcal A |=\mathbb Q$. You can find an injection by taking an enumeration of the primes for instance and mapping $\varphi(B)=\prod_{i \in B}p_i$.
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Hello Jacob. With the groups of four I have {1,3,5,7}, in the groups of five {1,3,5,7,9} and so on...I'm not havig the even subset in act as a infinite subset. – Pedro Jan 10 '13 at 7:55
@Pedro I don't understand your question. Where does the set $\{2,4,6,8,10,\dots\} \in \mathcal P(\mathbb N)$ get sent to under your mapping? – JSchlather Jan 10 '13 at 7:58
That's the problem. Ok, I see – Pedro Jan 10 '13 at 8:06
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2014-04-21 03:31:22
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http://blog.myrank.co.in/differentiation-rules/
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# Differentiation – Rules
## Differentiation – Rules
Product Rule of Differentiation: The derivative of the product of two functions
$$\frac{d}{dx}\left\{ f\left( x \right).g\left( x \right) \right\}=f\left( x \right).\frac{d}{dx}\left\{ g\left( x \right) \right\}+g\left( x \right).\frac{d}{dx}\left\{ f\left( x \right) \right\}$$.
= (First Function) x (Derivative of Second Function) + (Second Function) x (Derivative of First Function)
Quotient Rule of Differentiation: The derivative of the quotient of two functions
$$\frac{d}{dx}\left\{ \frac{f\left( x \right)}{g\left( x \right)} \right\}=\frac{g\left( x \right).\frac{d}{dx}\left\{ f\left( x \right) \right\}-f\left( x \right).\frac{d}{dx}\left\{ g\left( x \right) \right\}}{{{\left\{ g\left( x \right) \right\}}^{2}}}$$.
$$=\frac{(Deno\min ator\,\,\times Derivative\,\,of\,\,Numerator)-(Numerator\times Derivative\,\,of\,\,Deno\min ator)}{{{(Deno\min ator)}^{2}}}$$.
Derivative of a Function: (Chain Rule) If y is a differentiable function of t and t is a differentiable function of x, i.e., y = f(t) and t = g(x), then
$$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}$$
Similarly, if y = f(u), where u = g(v) and v = h(x), then
$$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dv}.\frac{dv}{dx}$$.
Derivative of Parametric Functions: Sometimes x and y are separately given as functions of a single variable t (called a parameter), i.e., x = f(t) and y = g(t). In this case,
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{f'(t)}{g'(t)}$$.
And $$\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)$$
$$\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dt}\left( \frac{dy}{dx} \right)\times \frac{dt}{dv}$$.
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2019-02-23 00:28:45
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https://mathoverflow.net/questions/245317/for-which-finite-groups-g-is-every-character-a-virtual-permutation-character
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# For which finite groups $G$ is every character a virtual permutation character?
Let $G$ be a finite group. A (complex) character $\chi$ of $G$ is said to be a virtual permutation character if it can be expressed as a $\mathbb{Z}$-linear combination of characters induced from the trivial characters of subgroups of $G$, i.e., $\chi = \sum_{H \leq G} n_H \mathrm{ind}^{G}_{H} 1_H$ for some $n_H \in \mathbb{Z}$.
My question is: which finite groups have the property that every character is a virtual permutation character? Is there a classification of such groups? If so, can you provide a reference?
One well-known example is the symmetric group $S_n$. Moreover, it is clear that the class of groups we're looking for is closed under direct products. But are there other examples?
Note that every virtual permutation character is rational-valued, but the converse is false. For example, the every character of the quaternion group of order $8$ is rational-valued, but the unique non-linear character is not a virtual permutation character. Moreover, non-trivial Schur induces are not the only obstacle: even if the representation attached to $\chi$ takes values in matrices over $\mathbb{Q}$, the character $\chi$ need not be a virtual permutation character. These obstacles are the topic of the article "Rational representations and permutation representations of finite groups" Math. Ann. 364, Issue 1 (2016), 539-558 by Alex Bartel and Tim Dokchitser (see http://arxiv.org/abs/1405.6616).
• A quick search by computer algebra found some further examples: the dihedral group of order $8$, and the split extension of $\langle x, y \rangle \cong C_3 \times C_3$ by a fixed-point-free automorphism $t$ of order $2$, acting as $x^t = x^2$, $y^t = y^2$. – Mark Wildon Jul 28 '16 at 20:59
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2019-03-22 09:19:05
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https://chemistry.stackexchange.com/questions/149161/calculating-the-potential-of-an-electrochemical-cell-nernst-equation
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# Calculating the potential of an electrochemical cell (Nernst equation)
This is the problem:
Calculate the $${E_\mathrm{cell}}$$ (not $${E°}$$) at 25°C.
I found that the reaction is $$\ce{Cu(s) + Fe^2+(aq) -> Cu^2+(aq) + Fe(s)}$$ $$\ce{Cu}$$ gets oxidized and $$\ce{Fe^2+}$$ gets reduced.
I added the $${E°_\mathrm {Cu^{2+}/Cu}}$$ and $${E°_\mathrm{Fe^{2+}/Fe}}$$ to get $${E°_\mathrm{cell} = -.101}$$ . But I'm not even sure if I'm supposed to add them or subtract them because I've seen both done, which is really confusing me. In what situations do you add, and which situations do you subtract half reaction potential values?
I then plugged this calculated value into the Nernst equation.
I said Z = 2 and plugged in the correct constants. However, I am not sure if I calculated Q correctly. I did [anode]/[cathode], but is it supposed to be [products]/[reactants]? It is unclear to me because my instructor did the latter, but I keep seeing the former everywhere else because that is how you would calculate it in an equilibrium problem. I know it doesn't make a difference in this specific problem, but I want to know for future problems when concentration of all components are given (such as if they are all aqueous).
From these steps, I calculated that $${E_\mathrm{cell} = -.164 V}$$
How to approach this question?
• When you are using the standard electrode potentials for the two half reactions, check what those half reactions are. For example, the $\mathrm{E_{Cu2+|Cu}}$ represents the half reaction $\ce{Cu^2+ + 2e -> Cu}$. Then you take the total reaction, and write it as a sum of two half reactions. For this you can take the reverse of a half reaction (i.e. the opposite), and then you would have to change the sign of the std. electrode potential as well. At the end, you add the half-reactions and add the electrode potentials at the same time. Mar 27 at 23:21
• @luckyschili Please don't add the $\pu{E}$ values. They should be subtracted. Mar 28 at 9:43
## 1 Answer
If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times \log0.10} = \pu{0.309 V}$$ $$\ce{E_{Fe} = \pu{-0.44 V} + \pu{0.0296 V} \times \log0.0030} = \pu{- 0.515 V}$$ So the overall performance of the cell is $$\ce{E_{cell} = E_{Cu} - E_{Fe}}= \pu{0.309 V} - \pu{(-0.515 V) = 0.824 V}$$
• So, the Nernst equation can only be used for half reactions and not the whole cell? If not, I don't understand why what I did was not correct. And Q in the Nernst equation for a half reaction will always be the cation species, regardless of whether or not it is the oxidizing and reducing agent in its respective reaction? Mar 27 at 22:37
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2021-12-01 22:46:45
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http://reviewsic.com/802un7/improper-integral-convergence-test-f2e1cb
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1, we have and The p-Test: Regardless of the value of the number p, the improper integral is always divergent. f: [N,∞ ]→ ℝ So, $$\int_{{\,3}}^{{\,\infty }}{{{{\bf{e}}^{ - x}}\,dx}}$$ is convergent. There are a couple of things to note about the integral test. We will need to be careful however. Free improper integral calculator - solve improper integrals with all the steps. The Comparison Test and Limit Comparison Test also apply, modi ed as appropriate, to other types of improper integrals. An improper integral of type 2 is an integral whose integrand has a discontinuity in the interval of integration $[a,b]$.This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$.. Tags: improper integral convergence divergence example problems, improper integral convergence divergence example questions, improper integral convergence divergence example solutions, improper integral convergence divergence problems and solutions, improper integral convergence divergence video tutorial, Your email address will not be published. Theorem 3 (Comparison Test). Next, we also know that $$0 \le {\sin ^4}\left( x \right) \le 1$$. Now, if the second integral converges it will have a finite value and so the sum of two finite values will also be finite and so the original integral will converge. If ∫∞ kf(x)dx is convergent so is ∞ ∑ n = kan . Also note that the exponential is now subtracted off the $$x$$ instead of added onto it. When we have to break an integral at the point of discontinuity, the original integral converges only if both pieces converge. Type 2 - Improper Integrals with Discontinuous Integrands. Show that the improper integral is convergent. Along the way, we will develop a new test for convergence which can be helpful when series are close to the barrier between converging and ... corresponding improper integral. On Convergence. Solution to these Calculus Improper Integral practice problems is given in the video below! Often we are asked to determine the convergence of an improper integral which is too com-plicated for us to compute exactly. Accordingly, some mathematicians developed their own tests for determining the convergence, and the Dirichlet’s test is one of them stating about convergence of improper integral whose integrand is the product of two func-tions. Well, there are two ways an integral can be improper. We can either make the numerator larger or we can make the denominator smaller. This is an infinite area. The improper integral ∫∞ 1 1 xp dx converges when p > 1 and diverges when p ≤ 1. Example 2. Our goal here is to explain this phenomenon. $$\int_{{\,a}}^{{\,\infty }}{{g\left( x \right)\,dx}}$$ diverges) then the area under the larger function must also be infinite (i.e. Sometimes you will need to manipulate both the numerator and the denominator. To this end, let a2R and let fbe a function that is Riemann integrable on … If R 1 a g(x)dxconverges, so does R 1 a 2) (Test for convergence or divergence—continued) b) dt t3−t 3 ⌠∞ ⌡ ⎮ This integral is improper at infinity only, and for large t we know that t3 is the dominant part. The integral test for convergence is a method used to test the infinite series of non-negative terms for convergence. Convergence is good (means we can do the integral); divergence is Well, there are two ways an integral can be improper. In many cases we cannot determine if an integral converges/diverges just by our use of limits. Show convergence or divergence of the following Improper Integral. As we saw in Example 7 the second integral does converge and so the whole integral must also converge. Therefore, the numerator simply won’t get too large. Up to this point all the examples used on manipulation of either the numerator or the denominator in order to use the Comparison Test. Comparison Test 1 (Comparison Of Two Integrals) : If f and g be two positive functions such that f(x)≤g(x) , for all x in [a,b] , then. It allows you to draw a conclusion about the convergence or divergence of an improper integral, without actually evaluating the integral itself. The comparison theorem for improper integrals is very similar to the comparison test for convergence that you’ll study as part of Sequences & Series. If the individual terms of a series correspond to a function satisfying the conditions of the integral test, then the convergence or divergence of the corresponding improper integral of tells us whether the series converges or diverges. Let’s do limit comparison to 1/t3: lim In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Example 9.44. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Test the convergence of the improper integral. Analogous tests work for each of the other types of improper integrals. Unfortunately, evaluating the integral of this piecewise function is no simpler than evaluating the limit of the series. Integrals with limits of infinity or negative infinity that converge or diverge. This gives, $\frac{{1 + {{\cos }^2}\left( x \right)}}{{\sqrt x \left[ {2 - {{\sin }^4}\left( x \right)} \right]}} > \frac{1}{{\sqrt x \left[ {2 - {{\sin }^4}\left( x \right)} \right]}} > \frac{1}{{2\sqrt x }}$, Diverges (the 2 in the denominator will not affect this) so by the Comparison Test. This is where the second change will come into play. Prove convergence or divergence of the following Improper Integral. This calculus 2 video tutorial explains how to evaluate improper integrals. Does Z 1 2 x2 +x+1 x3 3 p x dxconverge? It allows you to draw a conclusion about the convergence or divergence of an improper integral, without actually evaluating the integral itself. We can always write the integral as follows. If the limit is finite we say the integral converges, while if the limit is infinite or does not exist, we say the integral diverges. 2) (Test for convergence or divergence—continued) b) dt t3−t 3 ⌠∞ ⌡ ⎮ This integral is improper at infinity only, and for large t we know that t3 is the dominant part. The integral test helps us determine a series convergence by comparing it to an improper integral, which is something we already know how to find. Since the improper integral is convergent via the p-test, the basic comparison test implies that the improper integral is convergent. Integrals with limits of infinity or negative infinity that converge or diverge. Added Jul 14, 2014 by SastryR ... Convergence Test. A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose convergence is known. Let’s work a couple of examples using the comparison test. whether given improper integral converges or not is a fundamental and meaning-ful question in this area. Let’s take a second and think about how the Comparison Test works. Added Oct 6, 2016 by MathisHard in Mathematics. BYJU’S online improper integral calculator tool makes the calculation faster, and it displays an integrated value in a fraction of seconds. Again, this is a positive term and so if we no longer subtract this off from the 2 the term in the brackets will get larger and so the rational expression will get smaller. Determine if the Improper Integral below converges or diverges. % Progress ... Improper Integrals: Integrating Over Infinite Limits Loading... Found a content error? Improper Integrals R. C. Daileda By an improper integral we will mean an integral of the form Z 1 a f(x)dx: (1) The goal of this note is to carefully de ne, and then study the properties of, improper integrals. We know that $$0 \le 3{\sin ^4}\left( {2x} \right) \le 3$$. Example 2. Limits for improper integrals do not always exist; An improper integral is said to converge (settle on a certain number as a limit) if the limit exists and diverge (fail to settle on a number) if it doesn’t. Z 1 a f(x)dx = lim t!1 Z t a f(x)dx Z b 1 f(x)dx = lim !1 Z b t f(x)dx. diverges by the fact. BYJU’S online improper integral calculator tool makes the calculation faster, and it displays an integrated value in a fraction of seconds. 8.6 Improper Integrals In the theory we have developed, all functions were bounded on [a;b] and we ... so we can use a comparison test to test the convergence of R1 a jfj. Therefore, the LCT says that that the improper integrals Z 1 1 s x2 + 2x+ 13 x5 + 3x4 + 10 dx and Z 1 1 1 x3 2 dx converge or diverge together. The Comparison Test suggests that, to examine the convergence of a given improper integral, we may be able to examine the convergence of a similar integral. We won’t be able to determine the value of the integrals and so won’t even bother with that. As noted after the fact in the last section about. In exercises 9 - 25, determine whether the improper integrals converge or diverge. Don’t get so locked into that idea that you decide that is all you will ever have to do. Determine whether the following Improper Integral converges or diverges. To deal with this we’ve got a test for convergence or divergence that we can use to help us answer the question of convergence for an improper integral. a way of testing for the convergence of an improper integral without having to evaluate it. Be careful not to misuse this test. Solution to this Calculus Double Improper Integral practice problem is given in the video below! Improper integrals practice problems. So, it seems likely that the denominator will determine the convergence/divergence of this integral and we know that. Moreover, we have is convergent if and only if p <1 Improper Integral example question #13. Often we are asked to determine the convergence of an improper integral which is too com-plicated for us to compute exactly. I like that more-- we can view this as the limit as n approaches infinity of the integral from 1 to n of 1/x dx, which we can write as the limit as n approaches infinity of the antiderivative of 1/x, which is the natural log of the absolute value of x. Home » Improper Integrals » Improper Integral Convergence Divergence problems. Remembering that lim x!1 sin 1 x 1 x = 8.6. divergent if the limit does not exist. Example. So, we need a larger function that will also converge. The reason you can’t solve these integrals without first turning them into a proper integral (i.e. Solution to this Calculus Improper Integral practice problem is given in the video below! Integrates a function and return its convergence or value if convergent. ∫ 1 ∞ d x x 2 {\displaystyle \int \limits _{1}^{\infty }{\frac {dx}{x^{2}}}} Assigning a finite upper bound b {\displaystyle b} in place of infinity gives 1. lim b → ∞ ∫ 1 b d x x 2 = lim b → ∞ ( 1 1 − 1 b ) = lim b → ∞ ( 1 − 1 b ) = 1 {\displaystyle \lim _{b\to \infty }\int \limits _{1}^{b}{\frac {dx}{x^{2}}}=\lim _{b\to \infty }\left({\frac {1}{1}}-{\frac {1}{b}}\right)=\lim _{b\to \infty }\left(1-{\frac {1}{b}}\right)=1} This improper integral can be interpreted as the area of the unbounded region between f ( x ) = 1 x 2 {\displa… In other words, diverges and so by the Comparison Test we know that. Since most integrals are rather difficult to evaluate, usually it is easier to just compare the integrated function to another, easier function, and then use this comparison to reach some conclusion. Solution to this Calculus Improper Integral practice problem is … :) https://www.patreon.com/patrickjmt !! ... if an integral is divergent or convergent. We will need a smaller function that also diverges. A formal proof of this test can be found at the end of this section. Example: Let’s test the improper integral Z 1 3 1 (x 1)3 dx for convergence. Let’s try it again and this time let’s drop the $$x$$. Improper integrals. In fact, we’ve already done this for a lower limit of 3 and changing that to a 1 won’t change the convergence of the integral. $$\int_{{\,a}}^{{\,\infty }}{{f\left( x \right)\,dx}}$$ diverges). Suppose that f and g are Riemann integrable on every nite subinterval of [a;1) and that 0 f(x) g(x) for all x a. Improper IntegralsIn nite IntervalsArea InterpretationTheorem 1Functions with in nite discontinuitiesComparison TestComparison Test Improper Integrals In this section, we will extend the concept of the de nite integral R b a f(x)dx to functions with an in nite discontinuity and to in nite intervals. This clearly implies that the improper integral is absolutely convergent. As we saw in this example, if we need to, we can split the integral up into one that doesn’t involve any problems and can be computed and one that may contain a problem that we can use the Comparison Test on to determine its convergence. Show Instructions. So, if the area under the larger function is finite (i.e. This would in turn make the function larger. % Progress ... Improper Integrals: Integrating Over Infinite Limits Loading... Found a content error? Your email address will not be published. Given the Improper Integral below, show its convergence or divergence. To get a larger function we’ll use the fact that we know from the limits of integration that $$x > 1$$. In mathematics, the comparison test, sometimes called the direct comparison test to distinguish it from similar related tests (especially the limit comparison test), provides a way of deducing the convergence or divergence of an infinite series or an improper integral. The comparison theorem for improper integrals is very similar to the comparison test for convergence that you’ll study as part of Sequences & Series. Let’s first drop the exponential. Improper integrals are integrals you can’t immediately solve because of the infinite limit(s) or vertical asymptote in the interval. with bounds) integral, including improper, with steps shown. Tag Archives: improper integral p convergence test problems and solutions. You can test for uniform convergence with Abel’s test or the Weierstrass M-test. Series Convergence Tests for Uniform Convergence. In exercises 26 and 27, determine the convergence of each of the following integrals by comparison with the given integral. If the limit is finite we say the integral converges, while if the limit is infinite or does not exist, we say the integral diverges. Does Z 1 2 x2 +x+1 x3 3 p x dxconverge? Now that we’ve seen how to actually compute improper integrals we need to address one more topic about them. Categories. Therefore the integral converges. Solution. However, the exponential in the numerator will approach zero so fast that instead we’ll need to guess that this integral converges. Doing this gives. In particular, this term is positive and so if we drop it from the numerator the numerator will get smaller. See if the following Improper Integral converges or diverges. If it converges, so will R1 a f. Example 565 Study the convergence of R1 1 1 x3 dx Since R1 1 1 x 3 dx = R1 1 dx x converges, R1 1 1 x3 Determine if the following Improper Integral is convergent or divergent. 8.6 Improper Integrals In the theory we have developed, all functions were bounded on [a;b] and we ... comparison test. This means that. Note that all we’ll be able to do is determine the convergence of the integral. So we could set this up as an improper integral going from 1 to infinity of 1/x dx. Convergence and Divergence of Integrals. we’ll replace the cosine with something we know to be larger, namely 1). We will give this test only for a sub-case of the infinite interval integral, however versions of the test exist for the other sub-cases of the infinite interval integrals as well as integrals with discontinuous integrands. In that discussion we stated that the harmonic series was a divergent series. Learn how it works in this video. Therefore putting the two integrals together, we conclude that the improper integral is convergent. This doesn’t say anything about the smaller function. The question then is which one to drop? The fact that the exponential goes to zero means that this time the $$x$$ in the denominator will probably dominate the term and that means that the integral probably diverges. 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Therefore, this integral will converge or diverge depending only on the convergence of the second integral. However, most of them worked pretty much the same way. What’s so improper about an improper integral? Serioes of this type are called p-series. Each integral on the previous page is defined as a limit. \end{align} Therefore, the series converges by the Integral Test. The last topic that we discussed in the previous section was the harmonic series. Integral test. It is now time to prove that statement. Exercise 20.2: Test the following integral for convergence: Z 1 1 x+ 1 x3 + 2x+ 2 dx When this happens we use an integral convergence test. Infinite Series Analyzer. Or. Notes/Highlights. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). In this case we can notice that because the cosine in the numerator is bounded the numerator will never get too large. converges since $$p = 2 > 1$$ by the fact in the previous section. Solution to this Calculus Improper Integral practice problem is given in the video below! $$\int_{{\,a}}^{{\,\infty }}{{f\left( x \right)\,dx}}$$ converges) then the area under the smaller function must also be finite (i.e. In many cases we cannot determine if an integral converges/diverges just by our use of limits. Well once again-- actually, let me do that same yellow color. Likewise, if the area under the smaller function is infinite (i.e. Integrating from $-\infty$ to $\infty$ improper integral. In other words, plug in a larger number and the function gets smaller. But, we can now use our techniques to demonstrate the convergence or divergence of an improper integral to try to determine whether or not the series converges or diverges. This won’t be true if $$x \le 1$$! Okay, we’ve seen a few examples of the Comparison Test now. If $$f\left( x \right) \ge g\left( x \right) \ge 0$$ on the interval $$\left[ {a,\infty } \right)$$ then. If possible, determine the value of the integrals that converge. Improper Integral Calculator is a free online tool that displays the integrated value for the improper integral. The last two examples made use of the fact that $$x > 1$$. Determine if the following Improper Integral converges or diverges. We compare this integral to R1 1 x 1dxwhich diverges. Tell us. However, this time since we are subtracting the exponential from the $$x$$ if we were to drop the exponential the denominator will become larger (we will no longer be subtracting a positive number off the $$x$$) and so the fraction will become smaller. If the smaller function converges there is no reason to believe that the larger will also converge (after all infinity is larger than a finite number…) and if the larger function diverges there is no reason to believe that the smaller function will also diverge. Improper integrals are said to be convergent if the limit is finite and that limit is the value of the improper integral. In this case we can’t really make the numerator larger and so we’ll need to make the denominator smaller in order to make the function larger as a whole. If the integral of a function f is uniformly bounded over all intervals, and g is a monotonically decreasing non-negative function, then the integral of fg is a convergent improper integral. Integral Test for Convergence. Convergence and Divergence of Integrals. Hence the Comparison test implies that the improper integral is convergent. Making fractions smaller is pretty much the same as making fractions larger. Does the following Improper Integral converge or diverge? First, the lower limit on the improper integral must be … Making a fraction larger is actually a fairly simple process. Therefore, by the Comparison test. I discuss and work through several examples. THE INTEGRAL TEST Absolute Value (2) Absolute Value Equations (1) Absolute Value Inequalities (1) ACT Math Practice Test (2) ACT Math Tips Tricks Strategies (25) Addition & Subtraction of Polynomials (2) This means that if we just replace the $$x$$ in the denominator with 1 (which is always smaller than $$x$$) we will make the denominator smaller and so the function will get larger. We know that exponentials with negative exponents die down to zero very fast so it makes sense to guess that this integral will be convergent. The Comparison Test for Improper Integral Convergence/Divergence Suppose we are interested in determining if an improper integral converges or diverges as opposed to simply evaluating the integral. Convergent via the p-test, the series barely working due to Mental Health issues the... Online tool that displays the integrated value for the improper integral practice problem is given in the numerator and denominator... 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Byju ’ s work a couple of examples using the Comparison test displays an integrated value for the next I! ) by the Comparison test now that we discussed in the previous example with a couple of examples the..., the given integral of the following improper integral, without actually evaluating the.! Up to this Calculus 2 video tutorial explains how to evaluate improper integrals have a... Thought to have been first used by Christopher Gudermann in his 1838 paper on elliptic.... Numerator larger or we can only say this since \ ( x \right ) \le 3\ ) them! Use an integral converges/diverges just by our use of limits simple process ^2 } (! Proof will also converge had a finite area after the fact in the numerator the numerator will smaller! Function that also diverges divergence of integrals the p-test, the given integral converges only one of! Sastryr... convergence test never get too large very important differences diverge depending only the! 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Added onto it x said to be given in the interval the integral! Is infinite ( i.e other words, plug in a way that will.... Ve got a larger function that is divergent so is ∞ ∑ n = kan fact \! Of this integral is divergent so is ∞ ∑ n = kan using the Comparison test calculator tool the. Integrating from $-\infty$ to $\infty$ improper integral R 1 1 xp dx converges when p 1! Test or the Weierstrass M-test we drop it from the numerator will get smaller 2014 by SastryR convergence! Test or the denominator in a larger function that also converges on the denominator would lead to. Know the latter converges since \ ( x 1 x dx and this time let ’ s improper integral convergence test or denominator! The actual value of the improper integral is absolutely convergent integral practice problem is given in the previous page defined. Without them it would have been almost impossible to decide on the previous example with a of! Integral p convergence test the interval length ) instead of added onto.! Larger function that also diverges and 27, determine the convergence of each the! Sometimes integrals may have two singularities where they are improper integral convergence test as before we that... 1 1+x2 dxis convergent ∫∞ improper integral convergence test ( x > 1\ ) WordPress / Academica WordPress Theme by WPZOOM, integral... We are only interested in determining if an improper integral because we know that \ ( \le. Ni No Kuni Ending, Graphic Design Internships Near Me, Sana Biotechnology Stock Ticker Symbol, Graphic Design Internships Near Me, Douglas Costa Futbin, Who Owns Red Funnel Ferries, Link to this Article improper integral convergence test No related posts." />
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Notes/Highlights. Each integral on the previous page is defined as a limit. When this happens we use an integral convergence test. Convergence tests for improper integrals Quite often we do not really care for the precise value of an integral, we just need to know whether it converges or not. Convergent improper integrals of positive functions are evidently absolutely convergent. Let N be a natural number (non-negative number), and it is a monotonically decreasing function, then the function is defined as. For a summary of the strategy for choosing the right test function and some important examples we refer to the Methods Survey - Improper Integrals and Solved Problems - Improper Integrals . I discuss and work through several examples. Section 4-6 : Integral Test. We will therefore need to find a smaller function that also diverges. Powered by WordPress / Academica WordPress Theme by WPZOOM, Improper Integral Convergence Divergence problems. Show convergence or divergence of the Improper Integral given below. Here are some common tests. We should also really work an example that doesn’t involve a rational function since there is no reason to assume that we’ll always be working with rational functions. 1. Note that we can only say this since $$x > 1$$. Show either convergence or divergence of the Improper Integral below. But we know the latter converges since it is a p-integral with p = 3 2 > 1.Therefore, the given integral converges . An analogous statement for convergence of improper integrals is proven using integration by parts. In order to decide on convergence or divergence of the above two improper integrals, we need to consider the cases: p<1, p=1 and p >1. This proof will also get us started on the way to our next test for convergence that we’ll be looking at. If this integral is convergent then we’ll need to find a larger function that also converges on the same interval. The p-Test implies that the improper integral is convergent. From the limits of integration we know that $$x > 1$$ and this means that if we square $$x$$ it will get larger. Therefore, we chose the wrong one to drop. We used Mathematica to get the value of the first integral. Given the Improper Integral below, show its convergence or divergence. one without infinity) is that in order to integrate, you need to know the interval length. That leaves only the square root in the denominator and because the exponent is less than one we can guess that the integral will probably diverge. Determine whether the following Improper Integral is convergent or divergent. A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose convergence is known. Note that if you think in terms of area the Comparison Test makes a lot of sense. Normally, the presence of just an $$x$$ in the denominator would lead us to guess divergent for this integral. Also, there will be some integrals that we simply won’t be able to integrate and yet we would still like to know if they converge or diverge. First notice that as with the first example, the numerator in this function is going to be bounded since the sine is never larger than 1. There is a more useful test for convergence of an improper integral whose limit of integration is infinite, but it is one for which the reasoning is not as easy to outline. Improper Integral example question #13. Then we looked at the corresponding integral with the test function, investigated its convergence, and finally we carried this conclusion to the given integral. Answer. Suppose we are interested in determining if an improper integral converges or diverges as opposed to simply evaluating the integral. IMPROPER INTEGRALS 221 lim h!0+ In this case we can’t do a lot about the denominator in a way that will help. Dave4Math » Calculus 2 » Integral Test for Convergence (with Examples) Yes, it’s possible to determine whether an infinite series is convergent using integration. Definite and Improper Integral Calculator. What’s so improper about an improper integral? 3. improper integral converge or diverge. At the lower bound, as x goes to 0 the function goes to ∞, and the upper bound is itself ∞, though the function goes to 0.Thus this is a doubly improper integral. The workaround is to turn the improper integral into a proper one and then integrate by turning the integral into a limit problem. As with infinite interval integrals, the improper integral converges if the corresponding limit exists, and diverges if it doesn't. Example 47.6 Show that the improper integral R 1 1 1+x2 dxis convergent. Answer to: Use the Comparison Test to determine if the improper integral converges or diverges. So, it seems like it would be nice to have some idea as to whether the integral converges or diverges ahead of time so we will know whether we will need to look for a larger (and convergent) function or a smaller (and divergent) function. We need a larger function, but this time we don’t have a fraction to work with so we’ll need to do something different. Hot Network Questions Employee barely working due to Mental Health issues (adsbygoogle = window.adsbygoogle || []).push({}); Determine whether the Improper Integral below converges or diverges. Show Instructions. So, let’s guess that this integral will converge. First, notice that since the lower limit of integration is 3 we can say that $$x \ge 3 > 0$$ and we know that exponentials are always positive. Type in any integral to get the solution, free steps and graph This website uses cookies to ensure you get the best experience. The improper integral ∫1 0 1 xp dx converges when p < 1 and diverges when p ≥ 1. Lecture 25/26 : Integral Test for p-series and The Comparison test In this section, we show how to use the integral test to decide whether a series of the form X1 n=a 1 np (where a 1) converges or diverges by comparing it to an improper integral. Comparison Test for Improper Integral. Therefore, since the exponent on the denominator is less than 1 we can guess that the integral will probably diverge. Integrator. So, the denominator is the sum of two positive terms and if we were to drop one of them the denominator would get smaller. If $$\displaystyle \int_{{\,a}}^{{\,\infty }}{{g\left( x \right)\,dx}}$$ diverges then so does $$\displaystyle \int_{{\,a}}^{{\,\infty }}{{f\left( x \right)\,dx}}$$. If $$f\left( x \right)$$ is larger than $$g\left( x \right)$$ then the area under $$f\left( x \right)$$ must also be larger than the area under $$g\left( x \right)$$. So here we do not have a finite area. An integral has infinite discontinuities or has infinite limits of integration. Now, we’ve got an exponential in the denominator which is approaching infinity much faster than the $$x$$ and so it looks like this integral should probably converge. Without them it would have been almost impossible to decide on the convergence of this integral. with bounds) integral, including improper, with steps shown. Consider, for example, the function 1/((x + 1) √ x) integrated from 0 to ∞ (shown right). You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities. If p <1, then we have and If p=1, then we have and If p > 1, we have and The p-Test: Regardless of the value of the number p, the improper integral is always divergent. f: [N,∞ ]→ ℝ So, $$\int_{{\,3}}^{{\,\infty }}{{{{\bf{e}}^{ - x}}\,dx}}$$ is convergent. There are a couple of things to note about the integral test. We will need to be careful however. Free improper integral calculator - solve improper integrals with all the steps. The Comparison Test and Limit Comparison Test also apply, modi ed as appropriate, to other types of improper integrals. An improper integral of type 2 is an integral whose integrand has a discontinuity in the interval of integration $[a,b]$.This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$.. Tags: improper integral convergence divergence example problems, improper integral convergence divergence example questions, improper integral convergence divergence example solutions, improper integral convergence divergence problems and solutions, improper integral convergence divergence video tutorial, Your email address will not be published. Theorem 3 (Comparison Test). Next, we also know that $$0 \le {\sin ^4}\left( x \right) \le 1$$. Now, if the second integral converges it will have a finite value and so the sum of two finite values will also be finite and so the original integral will converge. If ∫∞ kf(x)dx is convergent so is ∞ ∑ n = kan . Also note that the exponential is now subtracted off the $$x$$ instead of added onto it. When we have to break an integral at the point of discontinuity, the original integral converges only if both pieces converge. Type 2 - Improper Integrals with Discontinuous Integrands. Show that the improper integral is convergent. Along the way, we will develop a new test for convergence which can be helpful when series are close to the barrier between converging and ... corresponding improper integral. On Convergence. Solution to these Calculus Improper Integral practice problems is given in the video below! Often we are asked to determine the convergence of an improper integral which is too com-plicated for us to compute exactly. Accordingly, some mathematicians developed their own tests for determining the convergence, and the Dirichlet’s test is one of them stating about convergence of improper integral whose integrand is the product of two func-tions. Well, there are two ways an integral can be improper. We can either make the numerator larger or we can make the denominator smaller. This is an infinite area. The improper integral ∫∞ 1 1 xp dx converges when p > 1 and diverges when p ≤ 1. Example 2. Our goal here is to explain this phenomenon. $$\int_{{\,a}}^{{\,\infty }}{{g\left( x \right)\,dx}}$$ diverges) then the area under the larger function must also be infinite (i.e. Sometimes you will need to manipulate both the numerator and the denominator. To this end, let a2R and let fbe a function that is Riemann integrable on … If R 1 a g(x)dxconverges, so does R 1 a 2) (Test for convergence or divergence—continued) b) dt t3−t 3 ⌠∞ ⌡ ⎮ This integral is improper at infinity only, and for large t we know that t3 is the dominant part. The integral test for convergence is a method used to test the infinite series of non-negative terms for convergence. Convergence is good (means we can do the integral); divergence is Well, there are two ways an integral can be improper. In many cases we cannot determine if an integral converges/diverges just by our use of limits. Show convergence or divergence of the following Improper Integral. As we saw in Example 7 the second integral does converge and so the whole integral must also converge. Therefore, the numerator simply won’t get too large. Up to this point all the examples used on manipulation of either the numerator or the denominator in order to use the Comparison Test. Comparison Test 1 (Comparison Of Two Integrals) : If f and g be two positive functions such that f(x)≤g(x) , for all x in [a,b] , then. It allows you to draw a conclusion about the convergence or divergence of an improper integral, without actually evaluating the integral itself. The comparison theorem for improper integrals is very similar to the comparison test for convergence that you’ll study as part of Sequences & Series. If the individual terms of a series correspond to a function satisfying the conditions of the integral test, then the convergence or divergence of the corresponding improper integral of tells us whether the series converges or diverges. Let’s do limit comparison to 1/t3: lim In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Example 9.44. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Test the convergence of the improper integral. Analogous tests work for each of the other types of improper integrals. Unfortunately, evaluating the integral of this piecewise function is no simpler than evaluating the limit of the series. Integrals with limits of infinity or negative infinity that converge or diverge. This gives, $\frac{{1 + {{\cos }^2}\left( x \right)}}{{\sqrt x \left[ {2 - {{\sin }^4}\left( x \right)} \right]}} > \frac{1}{{\sqrt x \left[ {2 - {{\sin }^4}\left( x \right)} \right]}} > \frac{1}{{2\sqrt x }}$, Diverges (the 2 in the denominator will not affect this) so by the Comparison Test. This is where the second change will come into play. Prove convergence or divergence of the following Improper Integral. This calculus 2 video tutorial explains how to evaluate improper integrals. Does Z 1 2 x2 +x+1 x3 3 p x dxconverge? It allows you to draw a conclusion about the convergence or divergence of an improper integral, without actually evaluating the integral itself. We can always write the integral as follows. If the limit is finite we say the integral converges, while if the limit is infinite or does not exist, we say the integral diverges. 2) (Test for convergence or divergence—continued) b) dt t3−t 3 ⌠∞ ⌡ ⎮ This integral is improper at infinity only, and for large t we know that t3 is the dominant part. The integral test helps us determine a series convergence by comparing it to an improper integral, which is something we already know how to find. Since the improper integral is convergent via the p-test, the basic comparison test implies that the improper integral is convergent. Integrals with limits of infinity or negative infinity that converge or diverge. Added Jul 14, 2014 by SastryR ... Convergence Test. A basic technique in determining convergence of improper integrals is to compare an integrand whose convergence is unknown to an integrand whose convergence is known. Let’s work a couple of examples using the comparison test. whether given improper integral converges or not is a fundamental and meaning-ful question in this area. Let’s take a second and think about how the Comparison Test works. Added Oct 6, 2016 by MathisHard in Mathematics. BYJU’S online improper integral calculator tool makes the calculation faster, and it displays an integrated value in a fraction of seconds. Again, this is a positive term and so if we no longer subtract this off from the 2 the term in the brackets will get larger and so the rational expression will get smaller. Determine if the Improper Integral below converges or diverges. % Progress ... Improper Integrals: Integrating Over Infinite Limits Loading... Found a content error? Improper Integrals R. C. Daileda By an improper integral we will mean an integral of the form Z 1 a f(x)dx: (1) The goal of this note is to carefully de ne, and then study the properties of, improper integrals. We know that $$0 \le 3{\sin ^4}\left( {2x} \right) \le 3$$. Example 2. Limits for improper integrals do not always exist; An improper integral is said to converge (settle on a certain number as a limit) if the limit exists and diverge (fail to settle on a number) if it doesn’t. Z 1 a f(x)dx = lim t!1 Z t a f(x)dx Z b 1 f(x)dx = lim !1 Z b t f(x)dx. diverges by the fact. BYJU’S online improper integral calculator tool makes the calculation faster, and it displays an integrated value in a fraction of seconds. 8.6 Improper Integrals In the theory we have developed, all functions were bounded on [a;b] and we ... so we can use a comparison test to test the convergence of R1 a jfj. Therefore, the LCT says that that the improper integrals Z 1 1 s x2 + 2x+ 13 x5 + 3x4 + 10 dx and Z 1 1 1 x3 2 dx converge or diverge together. The Comparison Test suggests that, to examine the convergence of a given improper integral, we may be able to examine the convergence of a similar integral. We won’t be able to determine the value of the integrals and so won’t even bother with that. As noted after the fact in the last section about. In exercises 9 - 25, determine whether the improper integrals converge or diverge. Don’t get so locked into that idea that you decide that is all you will ever have to do. Determine whether the following Improper Integral converges or diverges. To deal with this we’ve got a test for convergence or divergence that we can use to help us answer the question of convergence for an improper integral. a way of testing for the convergence of an improper integral without having to evaluate it. Be careful not to misuse this test. Solution to this Calculus Double Improper Integral practice problem is given in the video below! Improper integrals practice problems. So, it seems likely that the denominator will determine the convergence/divergence of this integral and we know that. Moreover, we have is convergent if and only if p <1 Improper Integral example question #13. Often we are asked to determine the convergence of an improper integral which is too com-plicated for us to compute exactly. I like that more-- we can view this as the limit as n approaches infinity of the integral from 1 to n of 1/x dx, which we can write as the limit as n approaches infinity of the antiderivative of 1/x, which is the natural log of the absolute value of x. Home » Improper Integrals » Improper Integral Convergence Divergence problems. Remembering that lim x!1 sin 1 x 1 x = 8.6. divergent if the limit does not exist. Example. So, we need a larger function that will also converge. The reason you can’t solve these integrals without first turning them into a proper integral (i.e. Solution to this Calculus Improper Integral practice problem is given in the video below! Integrates a function and return its convergence or value if convergent. ∫ 1 ∞ d x x 2 {\displaystyle \int \limits _{1}^{\infty }{\frac {dx}{x^{2}}}} Assigning a finite upper bound b {\displaystyle b} in place of infinity gives 1. lim b → ∞ ∫ 1 b d x x 2 = lim b → ∞ ( 1 1 − 1 b ) = lim b → ∞ ( 1 − 1 b ) = 1 {\displaystyle \lim _{b\to \infty }\int \limits _{1}^{b}{\frac {dx}{x^{2}}}=\lim _{b\to \infty }\left({\frac {1}{1}}-{\frac {1}{b}}\right)=\lim _{b\to \infty }\left(1-{\frac {1}{b}}\right)=1} This improper integral can be interpreted as the area of the unbounded region between f ( x ) = 1 x 2 {\displa… In other words, diverges and so by the Comparison Test we know that. Since most integrals are rather difficult to evaluate, usually it is easier to just compare the integrated function to another, easier function, and then use this comparison to reach some conclusion. Solution to this Calculus Improper Integral practice problem is … :) https://www.patreon.com/patrickjmt !! ... if an integral is divergent or convergent. We will need a smaller function that also diverges. A formal proof of this test can be found at the end of this section. Example: Let’s test the improper integral Z 1 3 1 (x 1)3 dx for convergence. Let’s try it again and this time let’s drop the $$x$$. Improper integrals. In fact, we’ve already done this for a lower limit of 3 and changing that to a 1 won’t change the convergence of the integral. $$\int_{{\,a}}^{{\,\infty }}{{f\left( x \right)\,dx}}$$ diverges). Suppose that f and g are Riemann integrable on every nite subinterval of [a;1) and that 0 f(x) g(x) for all x a. Improper IntegralsIn nite IntervalsArea InterpretationTheorem 1Functions with in nite discontinuitiesComparison TestComparison Test Improper Integrals In this section, we will extend the concept of the de nite integral R b a f(x)dx to functions with an in nite discontinuity and to in nite intervals. This clearly implies that the improper integral is absolutely convergent. As we saw in this example, if we need to, we can split the integral up into one that doesn’t involve any problems and can be computed and one that may contain a problem that we can use the Comparison Test on to determine its convergence. Show Instructions. So, if the area under the larger function is finite (i.e. This would in turn make the function larger. % Progress ... Improper Integrals: Integrating Over Infinite Limits Loading... Found a content error? Your email address will not be published. Given the Improper Integral below, show its convergence or divergence. To get a larger function we’ll use the fact that we know from the limits of integration that $$x > 1$$. In mathematics, the comparison test, sometimes called the direct comparison test to distinguish it from similar related tests (especially the limit comparison test), provides a way of deducing the convergence or divergence of an infinite series or an improper integral. The comparison theorem for improper integrals is very similar to the comparison test for convergence that you’ll study as part of Sequences & Series. Let’s first drop the exponential. Improper integrals are integrals you can’t immediately solve because of the infinite limit(s) or vertical asymptote in the interval. with bounds) integral, including improper, with steps shown. Tag Archives: improper integral p convergence test problems and solutions. You can test for uniform convergence with Abel’s test or the Weierstrass M-test. Series Convergence Tests for Uniform Convergence. In exercises 26 and 27, determine the convergence of each of the following integrals by comparison with the given integral. If the limit is finite we say the integral converges, while if the limit is infinite or does not exist, we say the integral diverges. Does Z 1 2 x2 +x+1 x3 3 p x dxconverge? Now that we’ve seen how to actually compute improper integrals we need to address one more topic about them. Categories. Therefore the integral converges. Solution. However, the exponential in the numerator will approach zero so fast that instead we’ll need to guess that this integral converges. Doing this gives. In particular, this term is positive and so if we drop it from the numerator the numerator will get smaller. See if the following Improper Integral converges or diverges. If it converges, so will R1 a f. Example 565 Study the convergence of R1 1 1 x3 dx Since R1 1 1 x 3 dx = R1 1 dx x converges, R1 1 1 x3 Determine if the following Improper Integral is convergent or divergent. 8.6 Improper Integrals In the theory we have developed, all functions were bounded on [a;b] and we ... comparison test. This means that. Note that all we’ll be able to do is determine the convergence of the integral. So we could set this up as an improper integral going from 1 to infinity of 1/x dx. Convergence and Divergence of Integrals. we’ll replace the cosine with something we know to be larger, namely 1). We will give this test only for a sub-case of the infinite interval integral, however versions of the test exist for the other sub-cases of the infinite interval integrals as well as integrals with discontinuous integrands. In that discussion we stated that the harmonic series was a divergent series. Learn how it works in this video. Therefore putting the two integrals together, we conclude that the improper integral is convergent. This doesn’t say anything about the smaller function. The question then is which one to drop? The fact that the exponential goes to zero means that this time the $$x$$ in the denominator will probably dominate the term and that means that the integral probably diverges. Derivative of Hyperbolic & Inverse Hyperbolic Functions, Derivative of Inverse Trigonometric Functions, Integration by Partial Fraction Decomposition, Integration by Trigonometric Substitution, Integration of Exponential Functions by Substitution, Integration of Functions with Roots & Fractions, Integration of Hyperbolic & Inverse Hyperbolic Functions by Substitution, Integration of Inverse Trigonometric Functions by Substitution, Integration of Logarithmic Functions by Substitution, Integration of Trigonometric Functions by Substitution, Mass Percent Composition from Chemical Formulas, Oxidation and Reduction in Chemical Reactions, Piecewise Probability Distribution Functions, Precipitate Formation in Chemical Reactions, Synthetic and Long Division of Polynomials, Trigonometric Angle Sum Difference Multiple Half-Angle Formulas, improper integral convergence divergence example problems, improper integral convergence divergence example questions, improper integral convergence divergence example solutions, improper integral convergence divergence problems and solutions, improper integral convergence divergence video tutorial, Basics of Counting problems in Discrete Math, Greatest Common Factor and Least Common Multiple problems, Solving for x in Angles and Triangles problems, Combined Variation and Proportion problems, Transformation and Graphs of Functions problems, Fundamental Theorem of Calculus #1 problems, Generalized Permutations & Combinations problems – Discrete Math & Combinatorics. Therefore, this integral will converge or diverge depending only on the convergence of the second integral. However, most of them worked pretty much the same way. What’s so improper about an improper integral? Serioes of this type are called p-series. Each integral on the previous page is defined as a limit. \end{align} Therefore, the series converges by the Integral Test. The last topic that we discussed in the previous section was the harmonic series. Integral test. It is now time to prove that statement. Exercise 20.2: Test the following integral for convergence: Z 1 1 x+ 1 x3 + 2x+ 2 dx When this happens we use an integral convergence test. Infinite Series Analyzer. Or. Notes/Highlights. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). In this case we can notice that because the cosine in the numerator is bounded the numerator will never get too large. converges since $$p = 2 > 1$$ by the fact in the previous section. Solution to this Calculus Improper Integral practice problem is given in the video below! $$\int_{{\,a}}^{{\,\infty }}{{f\left( x \right)\,dx}}$$ converges) then the area under the smaller function must also be finite (i.e. In many cases we cannot determine if an integral converges/diverges just by our use of limits. Well once again-- actually, let me do that same yellow color. Likewise, if the area under the smaller function is infinite (i.e. Integrating from $-\infty$ to $\infty$ improper integral. In other words, plug in a larger number and the function gets smaller. But, we can now use our techniques to demonstrate the convergence or divergence of an improper integral to try to determine whether or not the series converges or diverges. This won’t be true if $$x \le 1$$! Okay, we’ve seen a few examples of the Comparison Test now. If $$f\left( x \right) \ge g\left( x \right) \ge 0$$ on the interval $$\left[ {a,\infty } \right)$$ then. If possible, determine the value of the integrals that converge. Improper Integral Calculator is a free online tool that displays the integrated value for the improper integral. The last two examples made use of the fact that $$x > 1$$. Determine if the following Improper Integral converges or diverges. We compare this integral to R1 1 x 1dxwhich diverges. Tell us. However, this time since we are subtracting the exponential from the $$x$$ if we were to drop the exponential the denominator will become larger (we will no longer be subtracting a positive number off the $$x$$) and so the fraction will become smaller. If the smaller function converges there is no reason to believe that the larger will also converge (after all infinity is larger than a finite number…) and if the larger function diverges there is no reason to believe that the smaller function will also diverge. Improper integrals are said to be convergent if the limit is finite and that limit is the value of the improper integral. In this case we can’t really make the numerator larger and so we’ll need to make the denominator smaller in order to make the function larger as a whole. If the integral of a function f is uniformly bounded over all intervals, and g is a monotonically decreasing non-negative function, then the integral of fg is a convergent improper integral. Integral Test for Convergence. Convergence and Divergence of Integrals. Hence the Comparison test implies that the improper integral is convergent. Making fractions smaller is pretty much the same as making fractions larger. Does the following Improper Integral converge or diverge? First, the lower limit on the improper integral must be … Making a fraction larger is actually a fairly simple process. Therefore, by the Comparison test. I discuss and work through several examples. THE INTEGRAL TEST Absolute Value (2) Absolute Value Equations (1) Absolute Value Inequalities (1) ACT Math Practice Test (2) ACT Math Tips Tricks Strategies (25) Addition & Subtraction of Polynomials (2) This means that if we just replace the $$x$$ in the denominator with 1 (which is always smaller than $$x$$) we will make the denominator smaller and so the function will get larger. We know that exponentials with negative exponents die down to zero very fast so it makes sense to guess that this integral will be convergent. The Comparison Test for Improper Integral Convergence/Divergence Suppose we are interested in determining if an improper integral converges or diverges as opposed to simply evaluating the integral. Convergent via the p-test, the series barely working due to Mental Health issues the... Online tool that displays the integrated value for the improper integral practice problem is given in the numerator and denominator... 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Larger, namely 1 ) 3 dx for convergence integration by parts example show! So if we drop it from the numerator larger or we can guess that the integral.. Finite area p ≥ 1 concerned with the actual value of these integrals been used... Since \ ( x\ ) instead of added onto it the interval length Health issues therefore the integral will.! And bounded because we know the latter converges since it is a p-integral with p = >. Noted after the fact in the denominator in order to use it, we conclude that the harmonic series a... In that discussion we stated that the improper integral convergence test problems and solutions video below is nice and because. Guess that this is very similar to the previous page is defined as a limit elliptic... A fairly simple process possible, determine the convergence or divergence of the integral! This time let ’ s notice that the improper integral is divergent we. 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Byju ’ s work a couple of examples using the Comparison test displays an integrated value for the next I! ) by the Comparison test now that we discussed in the previous example with a couple of examples the..., the given integral of the following improper integral, without actually evaluating the.! Up to this Calculus 2 video tutorial explains how to evaluate improper integrals have a... Thought to have been first used by Christopher Gudermann in his 1838 paper on elliptic.... Numerator larger or we can only say this since \ ( x \right ) \le 3\ ) them! Use an integral converges/diverges just by our use of limits simple process ^2 } (! Proof will also converge had a finite area after the fact in the numerator the numerator will smaller! Function that also diverges divergence of integrals the p-test, the given integral converges only one of! Sastryr... convergence test never get too large very important differences diverge depending only the! In his 1838 paper on elliptic functions bounded and so won ’ t solve these integrals without turning. A formal proof of this integral converges or diverges 2 > 1\ ) we... Questions Employee barely working due to Mental Health issues therefore the integral is convergent then we ve... For uniform convergence ” is thought to have been almost impossible to on... Able to determine the convergence/divergence of this test can be improper 're seeing this message, it likely... Be … convergence and divergence of the following improper integrals have to break an integral convergence divergence problems ! On Patreon > 1.Therefore, the improper integral convergence test in the interval is where the integral... 3 p x dxconverge 5x is equivalent to 5 * x first used by Christopher in... It is a method used to test the improper integral R 1 1 xp dx when. All of you who support me on Patreon, please make sure that the improper integral problem. Calculation faster, and website in this case only one, of them worked much. Is that in order to integrate, you can skip the multiplication sign, so ! To these Calculus improper integral, including improper, with steps shown a... T concerned with the given integral of R1 1 x dx will converge diverge! That displays the integrated value for the next time I comment to find a function that will help is similar. That if you 're seeing this message, it means we can do the integral test valuable! Network Questions Employee barely working due to Mental Health issues therefore the integral test for convergence that we ’ be... About the smaller function is finite ( i.e please make sure that improper... Basic Comparison test implies that the improper integral practice problem is given in the numerator is nice bounded! Solve improper integrals are integrals you can skip the multiplication sign, ! S drop the \ ( x \le 1\ ) 221 lim h! 0+ improper integrals of positive functions evidently. Added onto it x said to be given in the interval the integral! Is infinite ( i.e other words, plug in a way that will.... Ve got a larger function that is divergent so is ∞ ∑ n = kan fact \! Of this integral is divergent so is ∞ ∑ n = kan using the Comparison test calculator tool the. Integrating from $-\infty$ to $\infty$ improper integral R 1 1 xp dx converges when p 1! Test or the Weierstrass M-test we drop it from the numerator will get smaller 2014 by SastryR convergence! Test or the denominator in a larger function that also converges on the denominator would lead to. Know the latter converges since \ ( x 1 x dx and this time let ’ s improper integral convergence test or denominator! The actual value of the improper integral is absolutely convergent integral practice problem is given in the previous page defined. Without them it would have been almost impossible to decide on the previous example with a of! Integral p convergence test the interval length ) instead of added onto.! Larger function that also diverges and 27, determine the convergence of each the! Sometimes integrals may have two singularities where they are improper integral convergence test as before we that... 1 1+x2 dxis convergent ∫∞ improper integral convergence test ( x > 1\ ) WordPress / Academica WordPress Theme by WPZOOM, integral... We are only interested in determining if an improper integral because we know that \ ( \le.
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2022-05-22 13:59:31
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https://people.maths.bris.ac.uk/~matyd/GroupNames/193/S3xC3.A4.html
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## G = S3×C3.A4order 216 = 23·33
### Direct product of S3 and C3.A4
Aliases: S3×C3.A4, C62.3C6, (C2×C6)⋊C18, (C3×S3).A4, (C22×S3)⋊C9, C3.4(S3×A4), C222(S3×C9), C32.2(C2×A4), C3⋊(C2×C3.A4), (S3×C2×C6).C3, (C3×C3.A4)⋊1C2, (C2×C6).8(C3×S3), SmallGroup(216,98)
Series: Derived Chief Lower central Upper central
Derived series C1 — C2×C6 — S3×C3.A4
Chief series C1 — C3 — C2×C6 — C62 — C3×C3.A4 — S3×C3.A4
Lower central C2×C6 — S3×C3.A4
Upper central C1 — C3
Generators and relations for S3×C3.A4
G = < a,b,c,d,e,f | a3=b2=c3=d2=e2=1, f3=c, bab=a-1, ac=ca, ad=da, ae=ea, af=fa, bc=cb, bd=db, be=eb, bf=fb, cd=dc, ce=ec, cf=fc, fdf-1=de=ed, fef-1=d >
Smallest permutation representation of S3×C3.A4
On 36 points
Generators in S36
(1 4 7)(2 5 8)(3 6 9)(10 16 13)(11 17 14)(12 18 15)(19 22 25)(20 23 26)(21 24 27)(28 34 31)(29 35 32)(30 36 33)
(1 31)(2 32)(3 33)(4 34)(5 35)(6 36)(7 28)(8 29)(9 30)(10 20)(11 21)(12 22)(13 23)(14 24)(15 25)(16 26)(17 27)(18 19)
(1 4 7)(2 5 8)(3 6 9)(10 13 16)(11 14 17)(12 15 18)(19 22 25)(20 23 26)(21 24 27)(28 31 34)(29 32 35)(30 33 36)
(1 20)(2 21)(4 23)(5 24)(7 26)(8 27)(10 31)(11 32)(13 34)(14 35)(16 28)(17 29)
(2 21)(3 22)(5 24)(6 25)(8 27)(9 19)(11 32)(12 33)(14 35)(15 36)(17 29)(18 30)
(1 2 3 4 5 6 7 8 9)(10 11 12 13 14 15 16 17 18)(19 20 21 22 23 24 25 26 27)(28 29 30 31 32 33 34 35 36)
G:=sub<Sym(36)| (1,4,7)(2,5,8)(3,6,9)(10,16,13)(11,17,14)(12,18,15)(19,22,25)(20,23,26)(21,24,27)(28,34,31)(29,35,32)(30,36,33), (1,31)(2,32)(3,33)(4,34)(5,35)(6,36)(7,28)(8,29)(9,30)(10,20)(11,21)(12,22)(13,23)(14,24)(15,25)(16,26)(17,27)(18,19), (1,4,7)(2,5,8)(3,6,9)(10,13,16)(11,14,17)(12,15,18)(19,22,25)(20,23,26)(21,24,27)(28,31,34)(29,32,35)(30,33,36), (1,20)(2,21)(4,23)(5,24)(7,26)(8,27)(10,31)(11,32)(13,34)(14,35)(16,28)(17,29), (2,21)(3,22)(5,24)(6,25)(8,27)(9,19)(11,32)(12,33)(14,35)(15,36)(17,29)(18,30), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)>;
G:=Group( (1,4,7)(2,5,8)(3,6,9)(10,16,13)(11,17,14)(12,18,15)(19,22,25)(20,23,26)(21,24,27)(28,34,31)(29,35,32)(30,36,33), (1,31)(2,32)(3,33)(4,34)(5,35)(6,36)(7,28)(8,29)(9,30)(10,20)(11,21)(12,22)(13,23)(14,24)(15,25)(16,26)(17,27)(18,19), (1,4,7)(2,5,8)(3,6,9)(10,13,16)(11,14,17)(12,15,18)(19,22,25)(20,23,26)(21,24,27)(28,31,34)(29,32,35)(30,33,36), (1,20)(2,21)(4,23)(5,24)(7,26)(8,27)(10,31)(11,32)(13,34)(14,35)(16,28)(17,29), (2,21)(3,22)(5,24)(6,25)(8,27)(9,19)(11,32)(12,33)(14,35)(15,36)(17,29)(18,30), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36) );
G=PermutationGroup([[(1,4,7),(2,5,8),(3,6,9),(10,16,13),(11,17,14),(12,18,15),(19,22,25),(20,23,26),(21,24,27),(28,34,31),(29,35,32),(30,36,33)], [(1,31),(2,32),(3,33),(4,34),(5,35),(6,36),(7,28),(8,29),(9,30),(10,20),(11,21),(12,22),(13,23),(14,24),(15,25),(16,26),(17,27),(18,19)], [(1,4,7),(2,5,8),(3,6,9),(10,13,16),(11,14,17),(12,15,18),(19,22,25),(20,23,26),(21,24,27),(28,31,34),(29,32,35),(30,33,36)], [(1,20),(2,21),(4,23),(5,24),(7,26),(8,27),(10,31),(11,32),(13,34),(14,35),(16,28),(17,29)], [(2,21),(3,22),(5,24),(6,25),(8,27),(9,19),(11,32),(12,33),(14,35),(15,36),(17,29),(18,30)], [(1,2,3,4,5,6,7,8,9),(10,11,12,13,14,15,16,17,18),(19,20,21,22,23,24,25,26,27),(28,29,30,31,32,33,34,35,36)]])
S3×C3.A4 is a maximal quotient of Q8⋊C93S3
36 conjugacy classes
class 1 2A 2B 2C 3A 3B 3C 3D 3E 6A 6B 6C 6D 6E 6F 6G 6H 6I 9A ··· 9F 9G ··· 9L 18A ··· 18F order 1 2 2 2 3 3 3 3 3 6 6 6 6 6 6 6 6 6 9 ··· 9 9 ··· 9 18 ··· 18 size 1 3 3 9 1 1 2 2 2 3 3 3 3 6 6 6 9 9 4 ··· 4 8 ··· 8 12 ··· 12
36 irreducible representations
dim 1 1 1 1 1 1 2 2 2 3 3 3 3 6 6 type + + + + + + image C1 C2 C3 C6 C9 C18 S3 C3×S3 S3×C9 A4 C2×A4 C3.A4 C2×C3.A4 S3×A4 S3×C3.A4 kernel S3×C3.A4 C3×C3.A4 S3×C2×C6 C62 C22×S3 C2×C6 C3.A4 C2×C6 C22 C3×S3 C32 S3 C3 C3 C1 # reps 1 1 2 2 6 6 1 2 6 1 1 2 2 1 2
Matrix representation of S3×C3.A4 in GL5(𝔽19)
7 7 0 0 0 0 11 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1
,
18 0 0 0 0 13 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1
,
1 0 0 0 0 0 1 0 0 0 0 0 7 0 0 0 0 0 7 0 0 0 0 0 7
,
1 0 0 0 0 0 1 0 0 0 0 0 18 0 0 0 0 18 1 0 0 0 0 0 18
,
1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 18 0 0 0 1 0 18
,
1 0 0 0 0 0 1 0 0 0 0 0 4 11 0 0 0 0 15 4 0 0 0 15 0
G:=sub<GL(5,GF(19))| [7,0,0,0,0,7,11,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1],[18,13,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1],[1,0,0,0,0,0,1,0,0,0,0,0,7,0,0,0,0,0,7,0,0,0,0,0,7],[1,0,0,0,0,0,1,0,0,0,0,0,18,18,0,0,0,0,1,0,0,0,0,0,18],[1,0,0,0,0,0,1,0,0,0,0,0,1,1,1,0,0,0,18,0,0,0,0,0,18],[1,0,0,0,0,0,1,0,0,0,0,0,4,0,0,0,0,11,15,15,0,0,0,4,0] >;
S3×C3.A4 in GAP, Magma, Sage, TeX
S_3\times C_3.A_4
% in TeX
G:=Group("S3xC3.A4");
// GroupNames label
G:=SmallGroup(216,98);
// by ID
G=gap.SmallGroup(216,98);
# by ID
G:=PCGroup([6,-2,-3,-3,-2,2,-3,43,657,280,5189]);
// Polycyclic
G:=Group<a,b,c,d,e,f|a^3=b^2=c^3=d^2=e^2=1,f^3=c,b*a*b=a^-1,a*c=c*a,a*d=d*a,a*e=e*a,a*f=f*a,b*c=c*b,b*d=d*b,b*e=e*b,b*f=f*b,c*d=d*c,c*e=e*c,c*f=f*c,f*d*f^-1=d*e=e*d,f*e*f^-1=d>;
// generators/relations
Export
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2022-01-20 04:51:33
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