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https://zbmath.org/?q=an:1019.60091
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## An equivalence of $$H_{-1}$$ norms for the simple exclusion process.(English)Zbl 1019.60091
Resolvent $$H_{-1}$$ norms with respect to simple exclusion processes play an important role in many problems with respect to additive functionals, tagged particles, hydrodynamics and so on. The author extends the norms to general (asymmetric) translation-invariant finite-range simple exclusion processes. As usual, the asymmetry costs a lot of problems. Based on a recent result by S. R. S. Varadhan, for the standard system of indistinguishable particles, the author proves that the corresponding resolvent $$H_{-1}$$ norms are equivalent, in a sense, to the $$H_{-1}$$ norms of a nearest-neighbor system. The same assertion is proved for systems with a distinguished particle in dimensions $$d\geq 2$$, and however, in dimension $$d = 1$$, this equivalence does not hold. An application of the $$H_1$$ norm equivalence to additive functional variances is also given.
### MSC:
60K35 Interacting random processes; statistical mechanics type models; percolation theory
### Keywords:
exclusion process; hydrodynamics; resolvent norm
Full Text:
### References:
[1] ANDJEL, E. D. (1982). Invariant measures for the zero range process. Ann. Probab. 10 525-547. · Zbl 0492.60096 [2] DE MASI, A. and FERRARI, P. (1985). Self-diffusion in one-dimensional lattice gases in the presence of an external field. J. Statist. Phy s. 38 603-613. · Zbl 0624.60117 [3] FERRARI, P. and FONTES, L. (1994). Current fluctuations for the asy mmetric simple exclusion process. Ann. Probab. 22 820-832. · Zbl 0806.60099 [4] KIPNIS, C. (1986). Central limit theorems for infinite series of queues and applications to simple exclusion. Ann. Probab. 14 397-408. · Zbl 0601.60098 [5] KIPNIS, C. and LANDIM, C. (1999). Scaling Limits of Interacting Particle Sy stems. Springer, New York. · Zbl 0927.60002 [6] KIPNIS, C., LANDIM, C. and OLLA, S. (1994). Hy drody namical limit for a nongradient sy stem: The generalized sy mmetric simple exclusion process. Comm. Pure Appl. Math. 47 1475-1545. · Zbl 0814.76003 [7] KIPNIS, C. and VARADHAN, S. R. S. (1986). Central limit theorem for additive functionals of reversible Markov processes. Comm. Math. Phy s. 104 1-19. · Zbl 0588.60058 [8] LANDIM, C. and YAU, H. T. (1997). Fluctuation-dissipation equation of asy mmetric simple exclusion processes. Probab. Theory Related Fields 108 321-356. · Zbl 0884.60092 [9] LIGGETT, T. M. (1985). Interacting Particle Sy stems. Springer, New York. [10] LIGGETT, T. M. (1999). Stochastic Particle Sy stems: Contact, Exclusion and Voter Models. Springer, New York. · Zbl 0949.60006 [11] SAADA, E. (1987). A limit theorem for the position of a tagged particle in a simple exclusion process. Ann. Probab. 15 375-381. · Zbl 0617.60096 [12] SEPPÄLÄINEN, T. and SETHURAMAN, S. (2003). Transience of second-class particles and diffusive variance bounds for additive functionals of one dimensional exclusion processes. Ann. Probab. 31 148-169. · Zbl 1029.60083 [13] SETHURAMAN, S. (2000). Central limit theorems for additive functionals of the simple exclusion process. Ann. Probab. 28 277-302. · Zbl 1044.60017 [14] SETHURAMAN, S. (2001). On extremal measures for conservative particle sy stems. Ann. Inst. H. Poincaré Probab. Statist. 37 139-154. · Zbl 0981.60098 [15] SETHURAMAN, S., VARADHAN, S. R. S. and YAU, H. T. (2000). Diffusive limit of a tagged particle in asy mmetric simple exclusion processes. Comm. Pure Appl. Math. 53 972-1006. · Zbl 1029.60084 [16] SETHURAMAN, S. and XU, L. (1996). A central limit theorem for reversible exclusion and zerorange particle sy stems. Ann. Probab. 24 1842-1870. · Zbl 0872.60079 [17] VARADHAN, S. R. S. (1995). Self-diffusion of a tagged particle in equilibrium for asy mmetric mean zero random walk with simple exclusion. Ann. Inst. H. Poincaré Probab. Statist. 31 273-285. · Zbl 0816.60093 [18] AMES, IOWA 50011 E-MAIL: sethuram@iastate.edu
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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2022-09-28 06:31:37
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http://codereview.stackexchange.com/questions/10459/creating-markdown-documents-in-ruby
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# Creating MarkDown documents in ruby
I'm trying to dynamically generate some MarkDown formatted documents, with header fields of arbitrary lengths.
I want to have a title of arbitrary length underlined like this
---------------------------------------------------------------
This should work to
-------------------
The code I came up with to do this in my ERB views.
<%= @title %>
<%= @title.length.times.map{"-"}.join %>
Is there a more efficient way to make this happen?
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"-" * @title.length #=> "---------------"
Warning: Make sure you put the String first or else you will get a TypeError: String can't be coerced into Fixnum since that would be calling Fixnum#* instead of String#*.
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2014-10-31 09:11:16
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https://ask.wireshark.org/question/4080/wireshark-will-not-start-at-all-on-windows-8/
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# Wireshark will not start at all on windows 8
I installed Wireshark. Tried opening it only to receive 'Application failed to start correctly 0xc000007b'. I looked on other forums for answers, none of which helped. Running as Admin does not work. Compatibility for Windows 7 does not work.
I really need help, it's honestly bugging me.
Oh, not sure if this helps but, iTunes does the same thing, shows the same error, and so does iExplorer. So, not only does Wireshark not work but some other apps do the same thing.
One more thing, I had Windows 8 before with iTunes on it, it worked but then I went to upgrade to 8.1, then downgraded a little after because games would not work. So, the Windows 8 install I'm using now is not the original, and again, on this installation of Windows 8, iTunes no longer works. (I'm going on about iTunes because I'm trying to add a backstory to something that may be of use to you to help me).
Thanks for any help,
-Lucas.
edit retag close merge delete
Probably a C/C++ runtime library failure. I'll leave it to others to answer how to fix that.
( 2018-07-14 13:08:37 +0000 )edit
Sort by » oldest newest most voted
Hello Lucas
I am not sure, if this can be sorted out just by giving advice in a forum like ask.wireshark.org. Still, while reading your question a couple ideas came up.
Disk Problems?
A similar question has been asked in a Microsoft forum. The forum post recommends to check the disk.
Since you use your Computer with Windows 8 it has been in operation for quite some time. So a disk failure is possible. I use the following steps to scan for disk errors:
• Run chkdsk /f C: for an easy solution, like a dirty shutdown.
• Check your Event Logs. The System Event Log can reveal disk and controller errors. Take your time and browse all the other event logs for other hardware related entries. I know, there are many of them. I usually filter out all the informational messages.
• Run a diagnostic program. Some HW vendors offer the programs for download, others ship deliver the computer with preinstalled diagnostics.
• If you experience blue screens you can find a bug check code in the System Event Log. Windows uses a ton of bug check code. Some messages will directly implicate the disk, others are less obvious: Anything with the letters WHEA would refer to the "Windows Hardware Error Architecture", Kernel Data Inpage Error can be another HW-related bug check.
Please note, that this forum cannot give further support with this.
OS Problems?
I am surprised to hear that certain games would not run under 8.1. Though I am not a gamer I have never heard of a Win 8 application or driver, which would not run on Win 8.1. Usually a minor update would get thinks back on track.
Your question did not mention, if you tried the update / downgrade recently, or if the problem existed for quite some time. Rolling back from Win 8.1 to Win 8 could leave a few messy details.
Please note that the support for Windows 8.1 has ended January 2018. Please consider an update to Windows 10. Personally, I would start with a fresh reinstallation of Windows 10.
Malware?
Since you are using an unsupported version of Windows, it is possible that your computer got hit by a "drive by infection". Cyber criminals often use advertisements campaigns to spread their malware. When visiting a decent site, say a newspaper, you might be presented with an ad, that ultimately brings malware to your computer ("malvertising").
Stay away from dodgy tools!
The internet is littered with free "fix everything" tools. Stay away from them. The sites are search-engine optimized so that any query for a Windows error code or common program / driver name will bring them to the first page of hits.
Stick to Windows or Microsoft tools. When I find myself in a similar situation I use the Sysinternals Process Monitor. If indeed a library ...
more
Hello Eddi, Lucas here.
I would like to let you know that I cannot get Windows 10 as I do not have a disk, boot drive or key. I am quite good with computers but the error I am facing with iTunes and other apps like Wireshark are things I just don't know how to get around. I know what to not get myself into, that being as you mentioned, 'malvertising' and fake tools.
The downgrade occurred, from what I believe, late-ish last year and the problem with iTunes has existed since then. Wireshark only recently when I found it. The downgrade was clean; wiped HDD and complete fresh install, so nothing should have carried over. I also do regular cleans and defrags on my laptop, so, everything stays clean and and tidy.
I guess the other thing is that my hardware is not great. 4GB RAM and an ...(more)
( 2018-07-15 09:02:29 +0000 )edit
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2019-06-17 00:46:08
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https://gateoverflow.in/118320/gate2017-1-37
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# GATE2017-1-37
4.7k views
Consider the context-free grammars over the alphabet $\left \{ a, b, c \right \}$ given below. $S$ and $T$ are non-terminals.
$G_{1}:S\rightarrow aSb \mid T, T \rightarrow cT \mid \epsilon$
$G_{2}:S\rightarrow bSa \mid T, T \rightarrow cT \mid \epsilon$
The language $L\left ( G_{1} \right )\cap L(G_{2})$ is
1. Finite
2. Not finite but regular
3. Context-Free but not regular
4. Recursive but not context-free
edited
0
Here both G1 and G2 are CFL. So intersection of two CFLs are not closed.. therefore it is not a CFL but every regular language is a CFL. so how come it is regular language?? Anyone please explain this
0
Why not C?
L(G1)=a^nc^mb^n|n≥0
L(G2)=b^nc^ma^n|n≥0
L(G1)∩L(G2)=c^m
2
@neenavath sindhu
CFL is not closed under intersection means that intersection of 2 CFL's MAY or MAY NOT BE CFL. It does not necessarily means that intersection of 2 cfl can never be a cfl..
But when we say closed under something, like CFL is closed under union then we mean that UNION OF ALL CFL SHOULD BE CFL.
Since while intersection all strings produced by production $aSb$ in $G_1$ and $bSa$ in $G_2$ will be $0$
So, only common production will be:
$S \rightarrow T$
$T \rightarrow cT \mid \epsilon$
Which is nothing but $c^*$ hence it is REGULAR and INFINITE
So, option is (B).
edited
1
Isn't G1 and G2 both are context free language ? and CFL is not closed under intersection.
1
not closed means the result of intersection may or may not be CFL
$L(G1)=a^{n}c^{*}b^{n} |n\geq 0$
$L(G2)=b^{n}c^{*}a^{n} |n\geq 0$
$L(G1)\cap L(G2)=c^{*}$ which is not finite but regular.
(B)
L1 is $a^n b^n c^m$
L2 is $b^na^n c^m$
The intersection is possible only when $n = 0$ in both the cases and $m$ is equal. This gives a regular infinite language.
edited
5
Correct explanation :
$L(G1)=a^nc^∗b^n | n≥0$
$L(G2)=b^nc^∗a^n | n≥0$
$L(G1)∩L(G2)=c^∗$ which is infinite and regular.
We can also solve this Question by observing the productions given.
Here there is common production
G: S -> T
T -> cT |
ϵ
What that represents! That generates Language with Regular Expression
## That is not Finite but Regular.
edited
We can also solve this Question by observing the productions given.
Here there is common production
G: S -> T
T -> cT |
ϵ
What that represents ! That geneates Language with Regular Expression
## That is not Finite but Regular.
g1 intersection g2 = phi , here so it is regular
hence finite
but here t-> ct /epsilon which is comman in both so
it is not finite but regular
corret ans is B
no need to to think much straight forward question is given
G1: starting with a and ending with b (compulsory conditon) or contains infinite c
G2: starting with b and ending with a (compulsory condition) or contains infinite c
so both are intersection will give 0 strings because if string is starting from 'a' than how it can start form b
and second half part where c is infinite and regular so option B
Option B
0
## Related questions
1
5k views
Consider the following languages over the alphabet $\sum = \left \{ a, b, c \right \}$. Let $L_{1} = \left \{ a^{n}b^{n}c^{m}|m,n \geq 0 \right \}$ and $L_{2} = \left \{ a^{m}b^{n}c^{n}|m,n \geq 0 \right \}$. Which of the following are context-free languages? $L_{1} \cup L_{2}$ $L_{1} \cap L_{2}$ I only II only I and II Neither I nor II
If $G$ is a grammar with productions $S\rightarrow SaS\mid aSb\mid bSa\mid SS\mid\epsilon$ where $S$ is the start variable, then which one of the following strings is not generated by $G$? $abab$ $aaab$ $abbaa$ $babba$
Consider the following context-free grammar over the alphabet $\Sigma = \{a,b,c\}$ with $S$ as the start symbol:$S \rightarrow abScT \mid abcT$$T \rightarrow bT \mid b$ ... $\{\left ( ab \right )^{n}\left ( cb^{n} \right )^{m} \mid m,n \geq 1 \}$
Consider the following languages: $\{a^mb^nc^pd^q \mid m+p=n+q, \text{ where } m, n, p, q \geq 0 \}$ $\{a^mb^nc^pd^q \mid m=n \text{ and }p=q, \text{ where } m, n, p, q \geq 0 \}$ ... Which of the above languages are context-free? I and IV only I and II only II and III only II and IV only
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2020-08-15 01:13:35
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http://superuser.com/questions/175677/most-useful-automated-tasks
|
# Most useful automated tasks? [closed]
I've just read the Most useful AutoHotKey scripts question and that got me thinking about automation and repetitive tasks.
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## closed as not constructive by 8088, DiagoSep 30 '11 at 6:59
As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. If this question can be reworded to fit the rules in the help center, please edit the question.
• Speeding up podcasts in iTunes
• Setting up robocopy on all my machines to move all downloaded items to my home server when it's available.
• Set up robocopy to keep SysInternals up to date locally (On my Win7 machine: robocopy /E /DCOPY:T /MT:2 \\live.sysinternals.com\tools C:\Path\To\SysinternalsSuite /XF Thumbs.db)
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I just found the iTunes speed x2 button (accidentally :) on my iPod last week! – yhw42 Aug 16 '10 at 20:32
Backing up my VM's to a USB external drive in the middle of the night; once a month or so, I just plug it into another PC, fire the VM's up and verify I could work if the main host was dead.
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Website macros! I work with many websites (Email, salesfore, SAP web client, intranet, blogs, Stackoverflow etc) each day and automated all routine logins and navigation with iMacros (free Firefox addon).
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2015-04-18 21:28:28
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https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/882/4/a/c/
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# Properties
Label 882.4.a.c Level $882$ Weight $4$ Character orbit 882.a Self dual yes Analytic conductor $52.040$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [882,4,Mod(1,882)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(882, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0, 0]))
N = Newforms(chi, 4, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("882.1");
S:= CuspForms(chi, 4);
N := Newforms(S);
Level: $$N$$ $$=$$ $$882 = 2 \cdot 3^{2} \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 882.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$52.0396846251$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 14) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q - 2 q^{2} + 4 q^{4} - 9 q^{5} - 8 q^{8}+O(q^{10})$$ q - 2 * q^2 + 4 * q^4 - 9 * q^5 - 8 * q^8 $$q - 2 q^{2} + 4 q^{4} - 9 q^{5} - 8 q^{8} + 18 q^{10} + 57 q^{11} + 70 q^{13} + 16 q^{16} + 51 q^{17} - 5 q^{19} - 36 q^{20} - 114 q^{22} - 69 q^{23} - 44 q^{25} - 140 q^{26} - 114 q^{29} - 23 q^{31} - 32 q^{32} - 102 q^{34} - 253 q^{37} + 10 q^{38} + 72 q^{40} - 42 q^{41} - 124 q^{43} + 228 q^{44} + 138 q^{46} + 201 q^{47} + 88 q^{50} + 280 q^{52} + 393 q^{53} - 513 q^{55} + 228 q^{58} + 219 q^{59} + 709 q^{61} + 46 q^{62} + 64 q^{64} - 630 q^{65} + 419 q^{67} + 204 q^{68} + 96 q^{71} + 313 q^{73} + 506 q^{74} - 20 q^{76} + 461 q^{79} - 144 q^{80} + 84 q^{82} - 588 q^{83} - 459 q^{85} + 248 q^{86} - 456 q^{88} - 1017 q^{89} - 276 q^{92} - 402 q^{94} + 45 q^{95} + 1834 q^{97}+O(q^{100})$$ q - 2 * q^2 + 4 * q^4 - 9 * q^5 - 8 * q^8 + 18 * q^10 + 57 * q^11 + 70 * q^13 + 16 * q^16 + 51 * q^17 - 5 * q^19 - 36 * q^20 - 114 * q^22 - 69 * q^23 - 44 * q^25 - 140 * q^26 - 114 * q^29 - 23 * q^31 - 32 * q^32 - 102 * q^34 - 253 * q^37 + 10 * q^38 + 72 * q^40 - 42 * q^41 - 124 * q^43 + 228 * q^44 + 138 * q^46 + 201 * q^47 + 88 * q^50 + 280 * q^52 + 393 * q^53 - 513 * q^55 + 228 * q^58 + 219 * q^59 + 709 * q^61 + 46 * q^62 + 64 * q^64 - 630 * q^65 + 419 * q^67 + 204 * q^68 + 96 * q^71 + 313 * q^73 + 506 * q^74 - 20 * q^76 + 461 * q^79 - 144 * q^80 + 84 * q^82 - 588 * q^83 - 459 * q^85 + 248 * q^86 - 456 * q^88 - 1017 * q^89 - 276 * q^92 - 402 * q^94 + 45 * q^95 + 1834 * q^97
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
0
−2.00000 0 4.00000 −9.00000 0 0 −8.00000 0 18.0000
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$1$$
$$3$$ $$-1$$
$$7$$ $$-1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 882.4.a.c 1
3.b odd 2 1 98.4.a.f 1
7.b odd 2 1 882.4.a.f 1
7.c even 3 2 882.4.g.u 2
7.d odd 6 2 126.4.g.d 2
12.b even 2 1 784.4.a.c 1
15.d odd 2 1 2450.4.a.d 1
21.c even 2 1 98.4.a.d 1
21.g even 6 2 14.4.c.a 2
21.h odd 6 2 98.4.c.a 2
84.h odd 2 1 784.4.a.p 1
84.j odd 6 2 112.4.i.a 2
105.g even 2 1 2450.4.a.q 1
105.p even 6 2 350.4.e.e 2
105.w odd 12 4 350.4.j.b 4
168.ba even 6 2 448.4.i.b 2
168.be odd 6 2 448.4.i.e 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
14.4.c.a 2 21.g even 6 2
98.4.a.d 1 21.c even 2 1
98.4.a.f 1 3.b odd 2 1
98.4.c.a 2 21.h odd 6 2
112.4.i.a 2 84.j odd 6 2
126.4.g.d 2 7.d odd 6 2
350.4.e.e 2 105.p even 6 2
350.4.j.b 4 105.w odd 12 4
448.4.i.b 2 168.ba even 6 2
448.4.i.e 2 168.be odd 6 2
784.4.a.c 1 12.b even 2 1
784.4.a.p 1 84.h odd 2 1
882.4.a.c 1 1.a even 1 1 trivial
882.4.a.f 1 7.b odd 2 1
882.4.g.u 2 7.c even 3 2
2450.4.a.d 1 15.d odd 2 1
2450.4.a.q 1 105.g even 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{4}^{\mathrm{new}}(\Gamma_0(882))$$:
$$T_{5} + 9$$ T5 + 9 $$T_{11} - 57$$ T11 - 57 $$T_{13} - 70$$ T13 - 70
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T + 2$$
$3$ $$T$$
$5$ $$T + 9$$
$7$ $$T$$
$11$ $$T - 57$$
$13$ $$T - 70$$
$17$ $$T - 51$$
$19$ $$T + 5$$
$23$ $$T + 69$$
$29$ $$T + 114$$
$31$ $$T + 23$$
$37$ $$T + 253$$
$41$ $$T + 42$$
$43$ $$T + 124$$
$47$ $$T - 201$$
$53$ $$T - 393$$
$59$ $$T - 219$$
$61$ $$T - 709$$
$67$ $$T - 419$$
$71$ $$T - 96$$
$73$ $$T - 313$$
$79$ $$T - 461$$
$83$ $$T + 588$$
$89$ $$T + 1017$$
$97$ $$T - 1834$$
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2023-03-30 17:10:38
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http://lene.lincantofurore.it/examples-length.html
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Time Example. The notation 1. The pounds are pounds force, not pounds mass. C# Example to get the length of the string using String. srec_examples − examples of how to use SRecord. StringLength - Get the length of a string. Proof of the theorem. StringLengthTrim - Get and trim the length of a string. A title tag is an HTML element that specifies the title of a web page. # However TensorFlow doesn't support advanced indexing yet, so we build. Improper Integral - Basic Idea and Example - Duration: 6:23. That's what variety of sentence length can do. 1 Solved example on Anchorage and Development length. A vocabulary list featuring Measurement Vocabulary Word List. length property is preferred because it does not have the overhead of a function call. The last worksheet is for modeling with the whole class during the introduction part of the numeracy l. ALOS is calculated by dividing total inpatient days by total discharges. element zeroLength 1 1 2 -mat 1 -dir 1 -doRayleigh 1 -orient 1 1 0 -1 1 0; # the same as the example above but also includes the stiffness of this element in calculation of the damping matrix if Rayleigh command is invoked later. Another study noted that “the more detail in the letter, the more persuasive” (2). In this example, a. In this tutorial we will understand Python lists through practical examples. The argument n must be a non-negative integer. Transactions involving family members , and parent companies and subsidiaries would not be considered non-arm's length transactions because the parties are acting independently. This manual page describes a few of them. The given information is the measure of the related minor arc and the radius of the circle. Angle1 and Angle2 are the angles between the normal of the cut face and the centerline of the body. This results in length of zero being input to the second reduce which then upgrades it a length of 1. List_length = len(A_list) In the length function, you may provide an object that can be a sequence (list, tuple, string, bytes or range) or collections such as a dictionary. Example # 1: A rope is strung from the tops of. Send your submission to Mailbag, P. We will look at a useful example that will highlight the usage of zero length paths. The character to pad the original string with. So you can see that a list named 'myList' was created. After creation, its length is fixed. These include not only their technical features, but also their high quality standard and the worldwide presence of HEIDENHAIN. The length of an object is its extended dimension, that is, its longest side. What is the moment of inertia of the rod?. Vary your sentence length to make your writing more interesting May 1, 2013 By Ken Carroll 2 Comments I want to show you a remarkable writing technique that connects Jack Black with Alexander the Great. The length of your sentences will determine the readability of your writing as much as any other quality. 23 x 10-4 = 0. java is a public domain class for working with Key-Length-Value (KLV) byte-packing and unpacking. This example shows the LENGTH function using a string value. Next we take two observers moving length-wise toward the rod but at different velocities. Apache Spark Examples. This is really a linear regression problem where the output is the predicted hemodynamic response. length returns the value 1. The static property String. Below is another example of how useful an Array's length is. In the above example, we see that function parameters of oneDArray and twoDArray are declared with variable length array type. Can any one derive the length calculation for the closed stirrups as per the IS-CODE 2502 1963 They have have given the closed stirrups (135degree hook) as Total length = (2*(A+E))+(24*bardia ) explain how the 24 bardia is calculated. Length constant membrane voltage changes occur both in time and in space (more about this under synaptic integration ) in the absence of any active, regenerative processes (see action potentials ), a graded potential will decay as it travels away from its site of origin. The result is the combined length and girth of your parcel. Per base sequence content. Title tags are displayed on search engine results pages (SERPs) as the clickable headline for a given result, and are important for usability, SEO, and social sharing. When it comes to length, every length in the large square is 1. The PHP's string length method, strlen, will treat leading or trailing spaces as length while returning the total length of the given string. it is a nice and easiest use of add_filter. arm's length. This is calculated from the physical refraction found in the eye). StringBuilder length() in Java with Examples The length() method of StringBuilder class returns the number of character the StringBuilder object contains. Another example of length as a term for a physical dimension is a professional football field, which is 100 yards in length from end zone to end zone. The following MySQL statement will count the length of the string specified as an argument. The concept of an. Diagonals which are congruent (they have the same length). I find that examples are the best way for me to learn about code, even with the explanation above. N-1 and N is the transform length. A Java Example on how to use length() method. Explanation of Length in the largest biology dictionary online. 3 Arc Length and Curvature 4. That is, we can create a function s(t) that measures how far we've traveled from ~r(a. The result is the combined length and girth of your parcel. The formula used by this calculator to calculate the volume of a rectangular shaped box is:. Example: suppose we use k = 4 bits to encode the run length (maximum run length of 15) for following bit patterns. That is too slow for length contraction to have a noticeable effect on the shuttle, and it's one of the fastest moving objects ever created. Length, L = 1 metre (m). JavaScript Length() Function Example. length() : length() method is a final variable which is applicable for string objects. 0 X 35 Notes on SAE bolts: • Bolt diameters under a 1/4” are in number sizes from 0 to 12 o To find the diameter of these use this formula:. Case Example 1: Low Weight-for-Length. Example of MySQL LENGTH() function with where clause The following MySQL statement returns the pub_name and length of pub_name from publisher table who have the length of there is more than or equal to 20. ) Often the only way to solve arc length problems is to do them numerically, or using a computer. Vary your sentence length to make your writing more interesting May 1, 2013 By Ken Carroll 2 Comments I want to show you a remarkable writing technique that connects Jack Black with Alexander the Great. 87 c (where γ =2) and the second at 0. Tutorials, references, and examples are constantly reviewed to avoid errors, but we cannot warrant full correctness of all content. This example shows how to calculate the lengths of a set of streamlines and also how to compress the streamlines without considerably reducing their lengths or overall shape. two vertical poles. Control your ideas. • The maximum-length of an LFSR sequence is 2n-1 – does not generate all 0s pattern (gets stuck in that state) • The characteristic polynomial of an LFSR generating a maximum-length sequence is a primitive polynomial • A maximum-length sequence is pseudo-random: – number of 1s = number of 0s + 1 – same number of runs of consectuive. What is the moment of inertia of the rod?. The pageLength button acts in exactly the same way as the default length list and will inherit the options specified by the lengthMenu option. For example, if a company has 10 product lines and each line has 3 product variations then product mix length is (10×3) = 30. Every step in the CrowdSurf process, from transcription micro-tasks to Special Handling Critical tasks, is crucial to making a perfect full length transcript. Since the java. String length limit: The maximum length a string can have is: 2 31-1. Many applications are running concurrently over the Web, such as web browsing/surfing, e-mail, file transfer, audio & video streaming, and so on. IMPORTANT NOTICE : Do not use the figures given here as sole source for any commercial (or legal) purpose involving cost, money, sales, purchase etc. 19 examples: Stratification was based on grade, length of time employed within the trust…. The formula for the length of a 2D vector is the Pythagorean Formula. The circumference of the unit circle is 2π, so we know after evaluating the integral we should get 2π. rle() is the inverse function of rle(), reconstructing x from the runs. Another method for determining the size of an irregularly-shaped area, a golf green, fro example, is to establish a point as near to the center of the area as can be estimated. For example, if z is a complex vector of length 100, then in an expression mode(z) is the character string "complex" and length(z) is 100. Reporting frequency. DIPY : Docs 1. Paper Length Calculation when Diameter of Roll & Paper Thickness is Known. Example of Dimensions. CK 30346 Measure the length of the stick with a ruler. Appreciate if you can explain the syntax with an example. This example shows how to get a length of a given String object. the horizontal measurement taken at right angles to the length : breadth; largeness of extent or scope; a measured and cut piece of material…. The JRecord package consists of Low level line IO routines (ByteReader / ByteWriter). Schulenberg gives a separate correlation (47) for downward convection from a disk which has the same exponents as correlation (45) for an infinite strip. The difference here is that the input string is ANSI and the length is in bytes. That is, we can create a function s(t) that measures how far we've traveled from ~r(a. Every step in the CrowdSurf process, from transcription micro-tasks to Special Handling Critical tasks, is crucial to making a perfect full length transcript. speed Example: Polar bears can swim at a speed of 40 km per hour. 3 Arc Length and Curvature 4. Another example of length as a term for a physical dimension is a professional football field, which is 100 yards in length from end zone to end zone. You can validate if a field has a minimum or maximum length in order to complete the form. PDF file at the link. Fixed-Length Strings (the predefined type String) are arrays of Character, and consequently of a fixed length. If char has datatype CHAR, then the length includes all trailing blanks. The cubit, a forearm's length, was 1-1/2 Roman feet or 6 palms, and typically used in building. We can measure how long things are, or how tall, or how far apart they are. inches Correct Answer: D. An example of having leading and trailing spaces in a string. R seq function, R seq usage. of Kansas Dept. Instead, the field length varies depending on what data is stored in it. Length: If you use the word length, it should certainly be for the longest sides of the rectangle. Substring(startIndex, length) Console. Variations. Similarly, index() works with character indices, and not byte indices. The length function returns the length of R objects such as vectors, lists, or strings (find a little trick in Example 3). net/ ----- SECT-Packet: 00000001 PID: 200 (0x00c8), Length: 196 (0x00c4) Time received. A sequence in FASTA format begins with a single-line description, followed by lines of sequence data. You can set the length property to truncate an array at any time. Initially, cDNA libraries provided the source of the clones. Example: n = 2^nextpow2(size(X,1)). This example shows how to get a length of a given String object. This example will show how to scan a field and replace or convert a text string within a field with a text string value of a different length by extending the functionality provided by the COBOL inspect statement using the replacing or converting format. These pins are available in one- and two-inch lengths. The pounds are pounds force, not pounds mass. Example: 50 cm (Length) + 140 cm (Girth) = 190 cm (Combined Length and Girth). FileStream Read File [C#] This example shows how to safely read file using FileStream in C#. Notice how subordinating conjunctions not only connect sentences together to improve length but also show how ideas are connected. Example of PostgreSQL LENGTH() function using column : Sample Table: employees The example below, returns the first_name and the length of first_name ( how many characters contain in the first name ) from the employees where the length of first_name is more than 7. Top of Page. Length: Millimeter (mm), Decimeter (dm), Centimeter (cm), Meter (m), and Kilometer (km) are used to measure how long or wide or tall an object is. A kilometer is equal to 100 metres or just under one third of a mile. For example, Asus may have 10 different types of netbooks. The word/ character limit happens in many occasions. In this tutorial we will understand Python lists through practical examples. Arc Length of a Curve. For example, what are the differences in performance between the various types of batteries (i. These involve ratios of the lengths of the sides in a right triangle. Calculations can be performed using either Python or VBScript. This example shows how to calculate the lengths of a set of streamlines and also how to compress the streamlines without considerably reducing their lengths or overall shape. Maya has seen her health care provider on a regular basis. Length bias: Overestimation of survival duration among screening-detected cases caused by the relative excess of slowly progressing cases. For example, if your application generates many short records with occasional long ones and you use fixed length records, you need to make the fixed record length equal to the length of the longest record. The Longest Call Length metric gives your team insight into the curation of the longest conversation between a call agent and a customer or lead. A nautical mile was originally defined as the length on the Earth’s surface of one minute (1 / 60 of a degree) of arc along a meridian (north-south line of longitude). These classes read a line from a file as an array of Bytes (Formats supported include Fixed Length and various Length based file formats). Find the arc length of the helix parametrized by $\dllp(t) = (\cos t, \sin t, t)$ for $0 \le t \le 6\pi$. But of course you want a space between the two sentences. Example 2: Regression line based on simple linear regression analysis with year of birth as the independent variable and length of education measured in years as the dependent variable. In other words, shorter essays generally require shorter paragraphs, while longer essays often require longer paragraphs that contain more information about a particular main point. pumping lemma, there is a pumping length p such that all strings s in E of length p or more can be written as s = xyz where 1. The C language is similar to most modern programming languages in that it allows the use of functions, self contained "modules" of code that take inputs, do a computation, and produce outputs. srec_examples − examples of how to use SRecord. People are often confused by the use of length() to get the size of a String and length to get the size of an array. An example of having leading and trailing spaces in a string. For example, Asus may have 10 different types of netbooks. This example shows the LENGTH function using a string value. Your 500 words essay is an example of your writing skills, talent and a thorough research, so it is necessary to know its common structure to be able to present yourself in the best light. For example, when children are first learning to talk, their MLU is often 1 because they only use one word at a time: "ball?", "mommy", "mine. Design Mach number (Mdesign): The Mach number at the exit of the nozzle where the flow is uniform. Research paper indexing, research paper on connecting rod. Mesocycle is a training phase in the annual training plan that contains usually of 3-6 microcycles. Of course, this makes sense, as the distance a particle travels along a particular route doesn't depend on its speed. What type of quantity (for example, length, volume, density) do the following units indicate?. Metric Basic Units and Prefixes. The standard size chosen is, of course, entirely arbitrary, but becomes very useful for comparing measurements made in different places and times. UK content and publishing guidance. Follow these changes over time in the NIST Length Timeline. It's actually the number of characters length. This example uses a LENGTH statement to set the length of the character variable NAME to 25 bytes. For example, a kilometre is 1000 m. And that’s what this lesson is all about! Arc Length, according to Math Open Reference, is the measure of the distance along a curved line. For example, when children are first learning to talk, their MLU is often 1 because they only use one word at a time: “ball?”, “mommy”, “mine. the horizontal measurement taken at right angles to the length : breadth; largeness of extent or scope; a measured and cut piece of material…. So you can see that a list named ‘myList’ was created. 5 resting on one of its rectangular faces on H. The units of length is determined by the spatial reference system of the geometry. That's what variety of sentence length can do. CSS Animatable Properties. Measurement provides opportunities to strengthen both children's number and measurement understandings at the same time. If char has datatype CHAR, then the length includes all trailing blanks. Of course, this makes sense, as the distance a particle travels along a particular route doesn't depend on its speed. length() method returns the number of characters presents in the string. This means that for a string containing five 2-byte characters, LENGTH() returns 10, whereas CHAR_LENGTH() returns 5. C functions must be TYPED (the return type and the type of all parameters specified). From the meter, several other units of measure are derived such as the: unit of speed is the meter per second (m/s). However, another option is to provide a single column of data along with another vector of the same size giving the labels for each data point. The Longest Call Length metric gives your team insight into the curation of the longest conversation between a call agent and a customer or lead. Example 2: Regression line based on simple linear regression analysis with year of birth as the independent variable and length of education measured in years as the dependent variable. 0 example to show the use of “f:validateLength” tag to validate the length of a “username” text field, when the validator failed, display the. Now simply take the longest dimension and add this to the package girth. /dvbsnoop -n 1 0x0c8 dvbsnoop V1. Examples 1 and 2 illustrate an important principle. We begin with a straight line of length 1, called the initiator. It was the prototype meter, a rod of platinum that was by definition exactly one meter. Solution: While you might logically expect there to be a length method on a Java array, there is actually a public length attribute on an array (instead of a length method). X is a matrix where each row is a copy of x, and Y is a matrix where each column is a copy of y. DataFrame API and Machine Learning API. An array is a contiguous group of data items with the same name and data type. It is initially null. x(t) = cos 2t, y(t) = sin 2t. A typically developing 9-year-old is expected to have an average sentence length of 6. ) Examples of expressions include atomic values such as the integer 5, the character 'a', and the function \x -> x+1, as well as structured values such as the list [1,2,3] and the pair ('b',4). The Length property returns 7, which indicates that it includes the six alphabetic characters as well as the null character. (Just ‘ print ’ by itself means the same thing, so we could have written that instead. Mar 03, 2015 Core Java, Examples, Snippet, String comments. Problem: When to use CHAR and VARCHAR Solution: As a guideline: The rule for CHAR and VARCHAR is to use CHAR only for strings whose length you know to be fixed. For my example photos I have our model, Q, on a motorcycle and I am using the moon to show how longer focal lengths appear to flatten or compress the image. This Javascript post was going to be about language selection in FCKEditor from a drop down box as a follow up to the previous FCKEditor post but I've decided to postphone that post until Friday and look at how to add options to an HTML drop down box with Javascript, because the next. At some point in time though, the need to relate the two units might arise, and consequently the need to choose one unit as defining the other or vice versa. Submitted by IncludeHelp , on March 17, 2019. ZigBee wireless home automation and other similar networks based on IEEE 802. The length of the table is six feet, and its width is three feet. That's why readability formulas rely so heavily on sentence length. John waits until three o'clock, or 3h. Appreciate if you can explain the syntax with an example. Find the length and the width. The returned value is of type long int. java Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. StringLength - Get the length of a string. User Contributed Notes #. The applet displays the length of the runner's shadow s as a function of the runner's position x. Think of how you would describe the distance along a road: it is the long distance, the length of the road. That is, we can create a function s(t) that measures how far we've traveled from ~r(a. A kilometer is equal to 100 metres or just under one third of a mile. lang package is imported by every Java application or applet, we can refer to it just as String. // the environment and the parsed GET, POST and cookie variables. For example, the length of the ruler in the picture is 15 cm. - Technical Field At a semiconductor company, not too many will dress in a tight fitting, short skirt. Example also shows how to get size or length of an ArrayList using size method. 0 It was not, however, a sufficiently perfect representation of a laid cable to serve for duplexing cables of more than a few hundred miles in length. An example of converting inches to miles An example of kilometers to centimeters An example of converting miles to kilometers An example of converting. The goal of the Mammalian Gene Collection (MGC), a trans-NIH initiative, is to provide full-length open reading frame (FL-ORF) clones for human, mouse, and rat genes. What is known? Using your knowledge of the SI system, along with applicable tables or notes, try to find relationships between the units in the problem. Home > Testing > Testing Materials > Practice Tests: OGT Full-Length New High School Graduation Requirements House Bill 487 updated Ohio’s graduation requirements to ensure that all students are ready for success in college and work. The length is maximum in the frame in which the object is at rest. Or from one point to another. Here is a working example of computing the row length in a case where the name and datatype of each row are known in advance:. If char has datatype CHAR, then the length includes all trailing blanks. Example, first part. Product Mix Consistency. In this example, the given information is: Length = 250. 19 examples: Stratification was based on grade, length of time employed within the trust…. In List, sequence of adding elements is remain same and get elements in same order sequence which was inserted. Previously, no maximum length was specified. The Impact of Shift Length in Policing on Performance, Health, Quality of Life, Sleep, Fatigue, and Extra-Duty Employment. The space shuttle is moving at roughly 0. Using the decimation in time (DIT) method, the design breaks down the input sequence into odd and even samples which feeds into the two individual 16K-point FFT blocks implemented in parallel using the FFT IP MegaCore. An observer at rest (relative to the moving object) would observe the moving object to be shorter in length. For the length of a circular arc, see arc of a circle. Matching an IP Address. A URL’s port is either null or a 16-bit unsigned integer that identifies a networking port. Free learning resources for students covering all major areas of biology. CHARACTER_LENGTH( str ) CHARACTER_LENGTH() is a synonym for CHAR_LENGTH(). 87 c (where γ =2) and the second at 0. Try the math and see for yourself!. In the Centimetre–gram–second system of units, the basic unit of length is the centimetre, or 1 ⁄ 100 of a metre. For formulas to show results, select them, press F2, and then press Enter. Length-tension Relationship The isometric length-tension curve represents the force a muscle is capable of generating while held at a series of discrete lengths. Returns the length of the string, in terms of bytes. Novel (50,000 -110,000 words): Most print publishers prefer a minimum word count of around 70,000 words for a first novel, and some even hesitate for any work shorter than 80,000. This schematic shows an example of a convex lens on the top and a concave lens on the bottom. Use the Visual Basic "Len" function to return the length of a string. Sentence Examples His canines were four times the length of hers. DIPY : Docs 1. For example: rulers and tape measures are marked in millimetres or centimetres to measure shorter lengths accurately. Please remember strings are indexed with the first character 0. 2 days ago · Along its length the Birdwalk provides scenic panoramic views of the reserve, minimum 30 meters away from the edge of the body of water where flamingos nest. Open - ISO Thread / Fastener Ratio and Length Calculator This assumes that the male and female thread materials have the same strength. Arm's Length Example Involving a Foreclosure Sale What follows is an example of an arm's length transaction that was brought before the Ohio Supreme Court. For example, the following code shows how to declare bounded and unbounded sequences as members of an IDL struct: module Finance { interface Account { }; struct LimitedAccounts { string bankSortCode<10>; // Maximum length of sequence is 50. This lesson is designed to examine the mathematical concepts of length, perimeter, and area. The length of side_1 is x, and the length of side_2 is y, so:. State your main point again in other ways. 1 Solved example on Anchorage and Development length. “Moving Clocks Run Slow” plus “Moving Clocks Lose Synchronization” plus “Length Contraction” leads to consistency!. It's extremely easy to learn java programming skills, and in these parts, you'll learn how to write, compile, and run Java applications. The length of your introduction depends on the length and complexity of your project, but generally it should not exceed one page unless it is a very long project or a book. The examples shown above all created arrays with a single dimension, meaning elements with indexes going from 0 and up. Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. Find the length of a CLOB. If you see a scenario missing or have one to contribute, please file a bug against this documentation with the example using the links at the bottom of this page. previous index next. The examples of length are rarely common. Find the lengths of sides b and c , rounded to the nearest whole number. CHARACTER_LENGTH( str ) CHARACTER_LENGTH() is a synonym for CHAR_LENGTH(). Java - String length() Method - This method returns the length of this string. pcap ip6 proto 17. The LENGTH statement also changes the default number of bytes that SAS uses to store the values of newly created numeric variables from 8 to 4 bytes. Instead, the field length varies depending on what data is stored in it. C# Example to get the length of the string using String. Examples of the LENGTH Functions. Run-length encoding You are encouraged to solve this task according to the task description, using any language you may know. Returns the length of the string, in terms of bytes. Examples of length of residence in a sentence, how to use it. For example, the word count of 'Oracle Limited' is 2. John sits down at one o'clock, or 1h. Definition and Usage. Created by the makers of the SAT, each practice test has the same types of questions you'll see on test day. 7 x 10^8 mph. Design and analysis of reinforced concrete column calculator. IO Public Class FileLength Public Shared Sub Main() ' Make a reference to a directory. Whether measured in hours for observation or days for inpatients, shorter is generally better. Please remember strings are indexed with the first character 0. java Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. 0133 We are going to use the formula but first we are going to figure out f'(x). In order to fully understand Arc Length and Area in Calculus, you first have to know where all of it comes from. The above format is used to get the length of the given bash variable. In Firefox, strings have a maximum length of 2**30 - 2 (~1GB). 2000 word essay introduction length 1000 words essay on motherland essay on radical criminology. Find the length of line segment g to the nearest centimeter. First, the arc length theorem is reviewed and explained. 23 x 10-4 = 0. are isomorphic if labels can be attached to their vertices so that they become the same graph. 7 x 10^8 mph. you can set custom length using above code. Take the Practice Test Sample Questions Get to know the PSAT/NMSQT and PSAT 10 with these sample questions. anaerobic power, muscular endurance, etc. Control your ideas. For example, if your application generates many short records with occasional long ones and you use fixed length records, you need to make the fixed record length equal to the length of the longest record.
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2020-01-29 18:21:57
|
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|
http://mizar.org/version/current/html/fdiff_1.html
|
:: Real Function Differentiability
::
:: Copyright (c) 1990-2018 Association of Mizar Users
theorem Th1: :: FDIFF_1:1
for Y being Subset of REAL holds
( ( for r being Real holds
( r in Y iff r in REAL ) ) iff Y = REAL )
proof end;
definition
let x be Real;
let IT be Real_Sequence;
attr IT is x -convergent means :Def1: :: FDIFF_1:def 1
( IT is convergent & lim IT = x );
end;
:: deftheorem Def1 defines -convergent FDIFF_1:def 1 :
for x being Real
for IT being Real_Sequence holds
( IT is x -convergent iff ( IT is convergent & lim IT = x ) );
registration
existence
ex b1 being Real_Sequence st
( b1 is 0 -convergent & b1 is non-zero )
proof end;
end;
registration
let f be 0 -convergent Real_Sequence;
cluster lim f -> zero ;
coherence
lim f is zero
by Def1;
end;
registration
coherence
for b1 being Real_Sequence st b1 is 0 -convergent holds
b1 is convergent
;
end;
set cs = seq_const 0;
definition
let IT be PartFunc of REAL,REAL;
attr IT is RestFunc-like means :Def2: :: FDIFF_1:def 2
( IT is total & ( for h being non-zero 0 -convergent Real_Sequence holds
( (h ") (#) (IT /* h) is convergent & lim ((h ") (#) (IT /* h)) = 0 ) ) );
end;
:: deftheorem Def2 defines RestFunc-like FDIFF_1:def 2 :
for IT being PartFunc of REAL,REAL holds
( IT is RestFunc-like iff ( IT is total & ( for h being non-zero 0 -convergent Real_Sequence holds
( (h ") (#) (IT /* h) is convergent & lim ((h ") (#) (IT /* h)) = 0 ) ) ) );
reconsider cf = REAL --> (In (0,REAL)) as Function of REAL,REAL ;
registration
existence
ex b1 being PartFunc of REAL,REAL st b1 is RestFunc-like
proof end;
end;
definition end;
definition
let IT be PartFunc of REAL,REAL;
attr IT is linear means :Def3: :: FDIFF_1:def 3
( IT is total & ex r being Real st
for p being Real holds IT . p = r * p );
end;
:: deftheorem Def3 defines linear FDIFF_1:def 3 :
for IT being PartFunc of REAL,REAL holds
( IT is linear iff ( IT is total & ex r being Real st
for p being Real holds IT . p = r * p ) );
registration
existence
ex b1 being PartFunc of REAL,REAL st b1 is linear
proof end;
end;
definition end;
theorem Th2: :: FDIFF_1:2
for L1, L2 being LinearFunc holds
( L1 + L2 is LinearFunc & L1 - L2 is LinearFunc )
proof end;
theorem Th3: :: FDIFF_1:3
for r being Real
for L being LinearFunc holds r (#) L is LinearFunc
proof end;
theorem Th4: :: FDIFF_1:4
for R1, R2 being RestFunc holds
( R1 + R2 is RestFunc & R1 - R2 is RestFunc & R1 (#) R2 is RestFunc )
proof end;
theorem Th5: :: FDIFF_1:5
for r being Real
for R being RestFunc holds r (#) R is RestFunc
proof end;
theorem Th6: :: FDIFF_1:6
for L1, L2 being LinearFunc holds L1 (#) L2 is RestFunc-like
proof end;
theorem Th7: :: FDIFF_1:7
for R being RestFunc
for L being LinearFunc holds
( R (#) L is RestFunc & L (#) R is RestFunc )
proof end;
definition
let f be PartFunc of REAL,REAL;
let x0 be Real;
pred f is_differentiable_in x0 means :: FDIFF_1:def 4
ex N being Neighbourhood of x0 st
( N c= dom f & ex L being LinearFunc ex R being RestFunc st
for x being Real st x in N holds
(f . x) - (f . x0) = (L . (x - x0)) + (R . (x - x0)) );
end;
:: deftheorem defines is_differentiable_in FDIFF_1:def 4 :
for f being PartFunc of REAL,REAL
for x0 being Real holds
( f is_differentiable_in x0 iff ex N being Neighbourhood of x0 st
( N c= dom f & ex L being LinearFunc ex R being RestFunc st
for x being Real st x in N holds
(f . x) - (f . x0) = (L . (x - x0)) + (R . (x - x0)) ) );
definition
let f be PartFunc of REAL,REAL;
let x0 be Real;
assume A1: f is_differentiable_in x0 ;
func diff (f,x0) -> Real means :Def5: :: FDIFF_1:def 5
ex N being Neighbourhood of x0 st
( N c= dom f & ex L being LinearFunc ex R being RestFunc st
( it = L . 1 & ( for x being Real st x in N holds
(f . x) - (f . x0) = (L . (x - x0)) + (R . (x - x0)) ) ) );
existence
ex b1 being Real ex N being Neighbourhood of x0 st
( N c= dom f & ex L being LinearFunc ex R being RestFunc st
( b1 = L . 1 & ( for x being Real st x in N holds
(f . x) - (f . x0) = (L . (x - x0)) + (R . (x - x0)) ) ) )
proof end;
uniqueness
for b1, b2 being Real st ex N being Neighbourhood of x0 st
( N c= dom f & ex L being LinearFunc ex R being RestFunc st
( b1 = L . 1 & ( for x being Real st x in N holds
(f . x) - (f . x0) = (L . (x - x0)) + (R . (x - x0)) ) ) ) & ex N being Neighbourhood of x0 st
( N c= dom f & ex L being LinearFunc ex R being RestFunc st
( b2 = L . 1 & ( for x being Real st x in N holds
(f . x) - (f . x0) = (L . (x - x0)) + (R . (x - x0)) ) ) ) holds
b1 = b2
proof end;
end;
:: deftheorem Def5 defines diff FDIFF_1:def 5 :
for f being PartFunc of REAL,REAL
for x0 being Real st f is_differentiable_in x0 holds
for b3 being Real holds
( b3 = diff (f,x0) iff ex N being Neighbourhood of x0 st
( N c= dom f & ex L being LinearFunc ex R being RestFunc st
( b3 = L . 1 & ( for x being Real st x in N holds
(f . x) - (f . x0) = (L . (x - x0)) + (R . (x - x0)) ) ) ) );
definition
let f be PartFunc of REAL,REAL;
let X be set ;
pred f is_differentiable_on X means :: FDIFF_1:def 6
( X c= dom f & ( for x being Real st x in X holds
f | X is_differentiable_in x ) );
end;
:: deftheorem defines is_differentiable_on FDIFF_1:def 6 :
for f being PartFunc of REAL,REAL
for X being set holds
( f is_differentiable_on X iff ( X c= dom f & ( for x being Real st x in X holds
f | X is_differentiable_in x ) ) );
theorem Th8: :: FDIFF_1:8
for X being set
for f being PartFunc of REAL,REAL st f is_differentiable_on X holds
X is Subset of REAL by XBOOLE_1:1;
theorem Th9: :: FDIFF_1:9
for Z being open Subset of REAL
for f being PartFunc of REAL,REAL holds
( f is_differentiable_on Z iff ( Z c= dom f & ( for x being Real st x in Z holds
f is_differentiable_in x ) ) )
proof end;
theorem :: FDIFF_1:10
for Y being Subset of REAL
for f being PartFunc of REAL,REAL st f is_differentiable_on Y holds
Y is open
proof end;
definition
let f be PartFunc of REAL,REAL;
let X be set ;
assume A1: f is_differentiable_on X ;
func f | X -> PartFunc of REAL,REAL means :Def7: :: FDIFF_1:def 7
( dom it = X & ( for x being Real st x in X holds
it . x = diff (f,x) ) );
existence
ex b1 being PartFunc of REAL,REAL st
( dom b1 = X & ( for x being Real st x in X holds
b1 . x = diff (f,x) ) )
proof end;
uniqueness
for b1, b2 being PartFunc of REAL,REAL st dom b1 = X & ( for x being Real st x in X holds
b1 . x = diff (f,x) ) & dom b2 = X & ( for x being Real st x in X holds
b2 . x = diff (f,x) ) holds
b1 = b2
proof end;
end;
:: deftheorem Def7 defines | FDIFF_1:def 7 :
for f being PartFunc of REAL,REAL
for X being set st f is_differentiable_on X holds
for b3 being PartFunc of REAL,REAL holds
( b3 = f | X iff ( dom b3 = X & ( for x being Real st x in X holds
b3 . x = diff (f,x) ) ) );
reconsider j = 1 as Element of REAL by XREAL_0:def 1;
theorem :: FDIFF_1:11
for Z being open Subset of REAL
for f being PartFunc of REAL,REAL st Z c= dom f & ex r being Real st rng f = {r} holds
( f is_differentiable_on Z & ( for x being Real st x in Z holds
(f | Z) . x = 0 ) )
proof end;
registration
let h be non-zero Real_Sequence;
let n be Nat;
cluster K123(h,n) -> non-zero for Real_Sequence;
coherence
for b1 being Real_Sequence st b1 = h ^\ n holds
b1 is non-zero
proof end;
end;
registration
let h be non-zero 0 -convergent Real_Sequence;
let n be Nat;
cluster K123(h,n) -> non-zero 0 -convergent for Real_Sequence;
coherence
for b1 being Real_Sequence st b1 = h ^\ n holds
( b1 is non-zero & b1 is 0 -convergent )
proof end;
end;
theorem Th12: :: FDIFF_1:12
for f being PartFunc of REAL,REAL
for x0 being Real
for N being Neighbourhood of x0 st f is_differentiable_in x0 & N c= dom f holds
for h being non-zero 0 -convergent Real_Sequence
for c being V8() Real_Sequence st rng c = {x0} & rng (h + c) c= N holds
( (h ") (#) ((f /* (h + c)) - (f /* c)) is convergent & diff (f,x0) = lim ((h ") (#) ((f /* (h + c)) - (f /* c))) )
proof end;
theorem Th13: :: FDIFF_1:13
for x0 being Real
for f1, f2 being PartFunc of REAL,REAL st f1 is_differentiable_in x0 & f2 is_differentiable_in x0 holds
( f1 + f2 is_differentiable_in x0 & diff ((f1 + f2),x0) = (diff (f1,x0)) + (diff (f2,x0)) )
proof end;
theorem Th14: :: FDIFF_1:14
for x0 being Real
for f1, f2 being PartFunc of REAL,REAL st f1 is_differentiable_in x0 & f2 is_differentiable_in x0 holds
( f1 - f2 is_differentiable_in x0 & diff ((f1 - f2),x0) = (diff (f1,x0)) - (diff (f2,x0)) )
proof end;
theorem Th15: :: FDIFF_1:15
for x0, r being Real
for f being PartFunc of REAL,REAL st f is_differentiable_in x0 holds
( r (#) f is_differentiable_in x0 & diff ((r (#) f),x0) = r * (diff (f,x0)) )
proof end;
theorem Th16: :: FDIFF_1:16
for x0 being Real
for f1, f2 being PartFunc of REAL,REAL st f1 is_differentiable_in x0 & f2 is_differentiable_in x0 holds
( f1 (#) f2 is_differentiable_in x0 & diff ((f1 (#) f2),x0) = ((f2 . x0) * (diff (f1,x0))) + ((f1 . x0) * (diff (f2,x0))) )
proof end;
theorem Th17: :: FDIFF_1:17
for Z being open Subset of REAL
for f being PartFunc of REAL,REAL st Z c= dom f & f | Z = id Z holds
( f is_differentiable_on Z & ( for x being Real st x in Z holds
(f | Z) . x = 1 ) )
proof end;
theorem :: FDIFF_1:18
for Z being open Subset of REAL
for f1, f2 being PartFunc of REAL,REAL st Z c= dom (f1 + f2) & f1 is_differentiable_on Z & f2 is_differentiable_on Z holds
( f1 + f2 is_differentiable_on Z & ( for x being Real st x in Z holds
((f1 + f2) | Z) . x = (diff (f1,x)) + (diff (f2,x)) ) )
proof end;
theorem :: FDIFF_1:19
for Z being open Subset of REAL
for f1, f2 being PartFunc of REAL,REAL st Z c= dom (f1 - f2) & f1 is_differentiable_on Z & f2 is_differentiable_on Z holds
( f1 - f2 is_differentiable_on Z & ( for x being Real st x in Z holds
((f1 - f2) | Z) . x = (diff (f1,x)) - (diff (f2,x)) ) )
proof end;
theorem :: FDIFF_1:20
for r being Real
for Z being open Subset of REAL
for f being PartFunc of REAL,REAL st Z c= dom (r (#) f) & f is_differentiable_on Z holds
( r (#) f is_differentiable_on Z & ( for x being Real st x in Z holds
((r (#) f) | Z) . x = r * (diff (f,x)) ) )
proof end;
theorem :: FDIFF_1:21
for Z being open Subset of REAL
for f1, f2 being PartFunc of REAL,REAL st Z c= dom (f1 (#) f2) & f1 is_differentiable_on Z & f2 is_differentiable_on Z holds
( f1 (#) f2 is_differentiable_on Z & ( for x being Real st x in Z holds
((f1 (#) f2) | Z) . x = ((f2 . x) * (diff (f1,x))) + ((f1 . x) * (diff (f2,x))) ) )
proof end;
theorem :: FDIFF_1:22
for Z being open Subset of REAL
for f being PartFunc of REAL,REAL st Z c= dom f & f | Z is V8() holds
( f is_differentiable_on Z & ( for x being Real st x in Z holds
(f | Z) . x = 0 ) )
proof end;
theorem :: FDIFF_1:23
for r, p being Real
for Z being open Subset of REAL
for f being PartFunc of REAL,REAL st Z c= dom f & ( for x being Real st x in Z holds
f . x = (r * x) + p ) holds
( f is_differentiable_on Z & ( for x being Real st x in Z holds
(f `| Z) . x = r ) )
proof end;
theorem Th24: :: FDIFF_1:24
for f being PartFunc of REAL,REAL
for x0 being Real st f is_differentiable_in x0 holds
f is_continuous_in x0
proof end;
theorem :: FDIFF_1:25
for X being set
for f being PartFunc of REAL,REAL st f is_differentiable_on X holds
f | X is continuous
proof end;
theorem Th26: :: FDIFF_1:26
for X being set
for Z being open Subset of REAL
for f being PartFunc of REAL,REAL st f is_differentiable_on X & Z c= X holds
f is_differentiable_on Z
proof end;
theorem :: FDIFF_1:27
ex R being RestFunc st
( R . 0 = 0 & R is_continuous_in 0 )
proof end;
definition
let f be PartFunc of REAL,REAL;
attr f is differentiable means :Def8: :: FDIFF_1:def 8
f is_differentiable_on dom f;
end;
:: deftheorem Def8 defines differentiable FDIFF_1:def 8 :
for f being PartFunc of REAL,REAL holds
( f is differentiable iff f is_differentiable_on dom f );
Lm1:
by XBOOLE_1:3;
Lm2:
by ;
registration
existence
ex b1 being Function of REAL,REAL st b1 is differentiable
proof end;
end;
theorem :: FDIFF_1:28
for Z being open Subset of REAL
for f being differentiable PartFunc of REAL,REAL st Z c= dom f holds
f is_differentiable_on Z by ;
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2019-02-22 00:26:12
|
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|
https://datascience.stackexchange.com/questions/41663/training-128x128-autoencoders-on-512x512-images-produces-strange-gridline-after/41682
|
# Training 128x128 autoencoders on 512x512 images, produces strange gridline after recombining
So I'm training an autoencoder that can recreate 128x128 images, so it can recreate any images by splitting them into 128x128 patches first, running it through the autoencoder, and having them combined with each other to form the original image.
I should be using other dimensions too but right now I'm testing this with 512x512 images. So :
x = next_train_batch(25)
for xx in range(0,4):
x_index = xx*128
for y in range(0,4):
y_index = y*128
this_image = x[:, x_index:x_index+128, y_index:y_index+128, :]
sess.run(training, feed_dict={x_in: this_image, step_iter_global:step_iter})
This is how I'm doing it now. But this keeps happening:
True, this is just in the beginning of the training process. But I guess I wonder why the autoencoder is having a hard time learning the edges of the patches. These gridlines actually are still present in visualizations of certain layers in the autoencoder (the reason I'm doing this is because I'm trying new layers that could potentially improve autoencoders) even if the output image doesn't have it.
|
2022-09-26 16:30:23
|
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|
https://jira.lsstcorp.org/browse/DM-15622
|
# Clean up usage of detail namespaces
XMLWordPrintable
## Details
• Type: Story
• Status: To Do
• Resolution: Unresolved
• Fix Version/s: None
• Component/s:
• Labels:
• Templates:
• Story Points:
1
## Description
As noted in the review form DM-15563, we should be ignoring any detail namespaces when running Doxygen, as that namespace indicates symbols that should not be considered part of the public API. This can be done by setting:
EXCLUDE_SYMBOLS = detail
in doxygen configuration.
However, for this to work, we'd actually need to standardize on whether to use detail or details as the namespace name, and a quick look at afw indicates that we haven't even done that. This convention was invented by Boost, which uses detail and we should go with that.
## People
• Assignee:
Unassigned
Reporter:
Jim Bosch
Watchers:
Jim Bosch
|
2019-10-21 23:33:33
|
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|
http://math.stackexchange.com/questions/297933/logic-gates-analyses
|
# Logic gates analyses
How to write the output of the gates not, and, or, xor, nand and nor in terms of their inputs, expressed as zeros and ones, using base 10 addition and multiplication.
-
Please do not deface the question. It orphans the answers that people have given. – robjohn Feb 11 '13 at 18:36
Any gate could be expressed using $\mathtt{nand}$ and assuming you want $\mathtt{falsehood} = 0$ and $\mathtt{truth} = 1$, you can express $\mathtt{nand}$ as $$\mathtt{nand}(x,y) = 1-x\cdot y.$$
The nice way to look at the problem of expressing logic gates in terms of real numbers addition/multiplication is from the perspective of probability theory. The issue with logic gates is that they behave in a discrete, rather than continuous way. It is the same with sets, in fact logic gates are very similar to set operators, that is \begin{align} a \land b &\to A \cap B, \\ a \lor b &\to A\cup B, \\ \neg a &\to A^c, \end{align} and so on (it is not a coincidence those symbols are similar). But then instead of taking $x \in A$ we could write $P(x \in A)$, and as if by magic we moved from discrete world into a continuous one. Using this intuition we could write (for $A$ and $B$ independent):
\begin{align} P(A \cap B) &= P(A)\cdot P(B)\\ P(A \cup B) &= P(A)+P(B) - P(A \cap B) \\ P(A^c) &= 1-P(A) \\ \end{align} and therefore $\mathtt{nand}$ can be expressed as $$P\left((A \cap B)^c\right) = 1 - P(A)\cdot P(B).$$
Of course, the above formulas would give you much simpler way of expressing $\mathtt{and}$, $\mathtt{or}$, and others in terms of addition and multiplication that the proposed in the begging $\mathtt{nand}$ method. However, as I have done most of the work, I leave the rest for you ;-)
Cheers!
-
Thanks much!!!!! – user60465 Feb 9 '13 at 10:04
Since $\neg$ and $\land$ are functionally complete (i.e., they together suffice to form all logical operators), it suffices to have the following:
$$\neg x := 1 - x \qquad x \land y := x \cdot y$$
(Possible) Definitions for the other logical operators can then be derived by expressing them in terms of $\neg$ and $\land$. E.g.: $$x \mathbin{\mathsf{NOR}} y = (\neg x) \land (\neg y) = (1-x)(1-y) = 1 + x y - x -y$$
-
|
2015-05-30 15:12:19
|
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https://freakonometrics.hypotheses.org/tag/partsm
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Seasonal Unit Roots
As discussed in the MAT8181 course, there are – at least – two kinds of non-stationary time series: those with a trend, and those with a unit-root (they will be called integrated). Unit root tests cannot be used to assess whether a time series is stationary, or not. They can only detect integrated time series. And the same holds for seasonal unit root.
In a previous post, we’ve seen that it was difficult to model hourly temperature, since most test do not reject unit roots. Consider here the average monthly temperature, still in Montréal, QC.
> montreal=read.table("http://freakonometrics.free.fr/temp-montreal-monthly.txt")
> M=as.matrix(montreal[,2:13])
> X=as.numeric(t(M))
> tsm=ts(X,start=1948,freq=12)
> plot(tsm)
For those who don’t know Montréal, Winter and Summer are very different. We can visualize monthly differences using
> month=rep(1:12,length(tsm)/12)
> plot(month,as.numeric(tsm))
> lines(1:12,apply(M,2,mean),col="red",type="b",pch=19)
or, if install the uroot package, removed from the CRAN repository, we can use
> library(uroot)
> bbplot(tsm)
or
> bb3D(tsm)
Loading required package: tcltk
It looks like our time series is cyclic, because of the yearly seasonal pattern. The autocorrelation function is here
> acf(tsm,lag=120)
Again, this cycle can be visualized using
> persp(1948:2013,1:12,M,theta=-50,col="yellow",shade=TRUE,
+ xlab="Year",ylab="Month",zlab="Temperature",ticktype="detailed")
Now, the question is is there a seasonal unit root ? This would mean that our model should be
$(1-L^{12})\Phi(L)[X_t-\mu]=\Theta(L)\varepsilon_t$
If we forget about the autoregressive and the moving average component, we can estimate
$(1-{\color{red}\alpha}L^{12})[X_t-\mu]=\varepsilon_t$
If there is a seasonal unit root then ${\color{red}\alpha}$ should be close to 1. Somehow.
> arima(tsm,order=c(0,0,0),seasonal=list(order=c(1,0,0),period=12))
Call:
arima(x = tsm, order = c(0, 0, 0), seasonal = list(order = c(1, 0, 0), period = 12))
Coefficients:
sar1 intercept
0.9702 6.4566
s.e. 0.0071 2.1515
It is not far away from 1. Actually, it cannot be too close to 1. If it was, then we would get an error message…
To illustrate some interesting models, let us consider also quarterly temperatures,
> N=cbind(apply(montreal[,2:4],1,sum),apply(montreal[,5:7],1,sum),apply(montreal[,8:10],1,sum),apply(montreal[,11:13],1,sum))
> X=as.numeric(t(N))
> tsq=ts(X,start=1948,freq=4)
+ xlab="Year",ylab="Quarter",zlab="Temperature",ticktype="detailed")
(again, the aim is just to be able to write down some equations, if necessary)
Why not consider a $VAR(1)$ model on the quarterly temperature? Something like
$\left(\begin{matrix}X_t^{(Q1)}\\ X_t^{(Q2)\\ X_t^{(Q3)\\ X_t^{(Q4)\end{matrix}\right)= A \left(\begin{matrix}X_{t-1}^{(Q1)}\\ X_{t-1}^{(Q2)\\ X_{t-1}^{(Q3)\\ X_{t-1}^{(Q4)\end{matrix}\right)+\left(\begin{matrix}\varepsilon_{t-1}^{(Q1)}\\ \varepsilon_{t-1}^{(Q2)\\ \varepsilon_{t-1}^{(Q3)\\ \varepsilon_{t-1}^{(Q4)\end{matrix}\right)$
i.e.
$\boldsymbol{Y}_t=A \boldsymbol{Y}_{t-1}+\boldsymbol{\varepsilon}_t$
where $A$ is some matrix $4\times 4$. This model can easily be estimated,
> library(vars)
> df=data.frame(N)
> names(df)=paste("y",1:4,sep="")
> model=VAR(df)
> model
VAR Estimation Results:
=======================
Estimated coefficients for equation y1:
=======================================
Call:
y1 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const
y1.l1 y2.l1 y3.l1 y4.l1 const
-0.13943065 0.21451118 0.08921237 0.30362065 -34.74793931
Estimated coefficients for equation y2:
=======================================
Call:
y2 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const
y1.l1 y2.l1 y3.l1 y4.l1 const
0.02520938 0.05288958 -0.13277377 0.05134148 40.68955266
Estimated coefficients for equation y3:
=======================================
Call:
y3 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const
y1.l1 y2.l1 y3.l1 y4.l1 const
0.07740824 -0.21142726 0.11180518 0.12963931 56.81087283
Estimated coefficients for equation y4:
=======================================
Call:
y4 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const
y1.l1 y2.l1 y3.l1 y4.l1 const
0.18842863 -0.31964579 0.25099508 -0.04452577 5.73228873
and matrix $A$ is here
> A=rbind(
+ coefficients(model$varresult$y1)[1:4],
+ coefficients(model$varresult$y2)[1:4],
+ coefficients(model$varresult$y3)[1:4],
+ coefficients(model$varresult$y4)[1:4])
> A
y1.l1 y2.l1 y3.l1 y4.l1
[1,] -0.13943065 0.21451118 0.08921237 0.30362065
[2,] 0.02520938 0.05288958 -0.13277377 0.05134148
[3,] 0.07740824 -0.21142726 0.11180518 0.12963931
[4,] 0.18842863 -0.31964579 0.25099508 -0.04452577
Since stationary of this multiple time series is closely related to eigenvalues of this matrix, let us look at them,
> eigen(A)$values [1] 0.35834830 -0.32824657 -0.14042175 0.09105836 > Mod(eigen(A)$values)
[1] 0.35834830 0.32824657 0.14042175 0.09105836
So it looks like there is no stationarity issue, here. A restricted model is the periodic autoregressive model, so called $PAR(1)$ model, discussed by Paap and Franses
$\boldsymbol{\Phi}_0 \boldsymbol{Y}_t=\boldsymbol{\Phi}_1 \boldsymbol{Y}_{t-1}+\boldsymbol{\varepsilon}_t$
where
$\boldsymbol{\Phi}_0=\left(\begin{array}{cccc}1&0&0&0\\ -\phi_{1,2}&1&0&0\\ 0&-\phi_{1,3}&1&0\\ 0&0&-\phi_{1,4}&1\end{array}\right)$
and
$\boldsymbol{\Phi}_1=\left(\begin{array}{cccc}0&0&0&\phi_{1,1}\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right)$
Keep in mind that this is a $VAR(1)$ model, since
$\boldsymbol{Y}_t=\underbrace{\boldsymbol{\Phi}_0^{-1} \boldsymbol{\Phi}_1}_A \boldsymbol{Y}_{t-1}+\boldsymbol{\varepsilon}_t$
This model can be estimated using a specific package (one can also look at the vignette, to get a better understanding of the syntax)
> library(partsm)
> detcomp <- list(regular=c(0,0,0), seasonal=c(1,0), regvar=0)
> model=fit.ar.par(wts=tsq, detcomp=detcomp, type="PAR", p=1)
> PAR.MVrepr(model)
----
Multivariate representation of a PAR model.
Phi0:
1.000 0.000 0.000 0
-0.242 1.000 0.000 0
0.000 -0.261 1.000 0
0.000 0.000 -0.492 1
Phi1:
0 0 0 0.314
0 0 0 0.000
0 0 0 0.000
0 0 0 0.000
Eigen values of Gamma = Phi0^{-1} %*% Phi1:
0.01 0 0 0
Time varing accumulation of shocks:
0.010 0.040 0.155 0.314
0.002 0.010 0.037 0.076
0.001 0.003 0.010 0.020
0.000 0.001 0.005 0.010
Here, the characteristic equation is
$|\boldsymbol{\Phi}_0-\boldsymbol{\Phi}_1 z |=(1-\phi_{1,1}\phi_{1,2}\phi_{1,3}\phi_{1,4}\cdot z)$
so there is a (seasonal) unit root if
$\phi_{1,1}\phi_{1,2}\phi_{1,3}\phi_{1,4}=1$
Which is clearly not the case, here. It is possible to perform Canova-Hansen test,
> CH.test(tsq)
------ - ------ ----
Canova & Hansen test
------ - ------ ----
Null hypothesis: Stationarity.
Alternative hypothesis: Unit root.
Frequency of the tested cycles: pi/2 , pi ,
L-statistic: 1.122
Lag truncation parameter: 5
Critical values:
0.10 0.05 0.025 0.01
0.846 1.01 1.16 1.35
The idea is that polynomial $(1-L^4)$ has four root, in $\mathbb{C}$,
$\{+1,+i,-1,-i\}$
since
$(1-L^4)=(1-L)(1+L)(1+L^2)=(1-L)(1+L)(1-iL)(1+iL)$
If we get back to monthly data, $(1-L^{12})$ has twelve roots,
$\left\{\pm 1, \ \pm i, \ \pm\frac{1\pm \sqrt{3}i}{2},\pm\frac{\sqrt{3}\pm i}{2}\right\}$
each of them having different interpretations.
Here we can have 1 cycle per year (on 12 months), 2 cycles per year (on 6 months), 3 cycles per year (on 4 months), 4 cycles per year (on 3 months), even 6 cycles per year (on 2 months). This will depend on the argument of the root, with respectively
$\left\{\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6},\pi\right\}$
The output of the test is here
> CH.test(tsm)
------ - ------ ----
Canova & Hansen test
------ - ------ ----
Null hypothesis: Stationarity.
Alternative hypothesis: Unit root.
Frequency of the tested cycles: pi/6 , pi/3 , pi/2 , 2pi/3 , 5pi/6 , pi ,
L-statistic: 1.964
Lag truncation parameter: 20
Critical values:
0.10 0.05 0.025 0.01
2.49 2.75 2.99 3.27
And it looks like we reject the assumption of a seasonal unit root. I can even mention the following testing procedure
> library(forecast)
> nsdiffs(tsm, test="ch")
[1] 0
where the ouput: “1” means that there is a seasonal unit root and “0” that there is no seasonal unit root. Simple to read, isn’t it? If we consider the periodic autoregressive model on the monthly data, the output is
> model=fit.ar.par(wts=tsm, detcomp=detcomp, type="PAR", p=1)
> model
----
PAR model of order 1 .
y_t = alpha_{1,s}*y_{t-1} + alpha_{2,s}*y_{t-2} + ... + alpha_{p,s}*y_{t-p} + coeffs*detcomp + epsilon_t, for s=1,2,...,12
----
Autoregressive coefficients.
s=1 s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9 s=10 s=11 s=12
alpha_1s 0.15 0.05 0.07 0.33 0.11 0 0.3 0.38 0.31 0.19 0.15 0.37
So, whatever the test, we always reject the assumption that there is a seasonal unit root. Which does not mean that we can not have a strong cycle! Actually, the series is almost periodic. But there is no unit root! So all of this makes sense (I hardly believe that there might be unit root – seasonal, or not – in temperatures).
Just to make sure that we get it right, consider two times series, inspired from the previous one. The first one is a periodic sequence (with a very very small noise, just to avoid problem of non-definite matrices) and the second one is clearly integrated.
> Xp1=Xp2=as.numeric(t(M))
> for(t in 13:length(M)){
+ Xp1[t]=Xp1[t-12]
+ Xp2[t]=Xp2[t-12]+rnorm(1,0,2)
+ }
> Xp1=Xp1+rnorm(length(Xp1),0,.02)
> tsp1=ts(Xp1,start=1948,freq=12)
> tsp2=ts(Xp2,start=1948,freq=12)
> par(mfrow=c(2,1))
> plot(tsp1)
> plot(tsp2)
> par(mfrow=c(1,2))
> bb3D(tsp1)
> bb3D(tsp2)
If we quickly look at those series, I would say that the first one has no unit root – even if it is not stationary, but it is because the series is periodic – while there is (are ?) unit root(s) for the second one. If we look at Canova-Hansen test, we get
> CH.test(tsp1)
------ - ------ ----
Canova & Hansen test
------ - ------ ----
Null hypothesis: Stationarity.
Alternative hypothesis: Unit root.
Frequency of the tested cycles: pi/6 , pi/3 , pi/2 , 2pi/3 , 5pi/6 , pi ,
L-statistic: 2.234
Lag truncation parameter: 20
Critical values:
0.10 0.05 0.025 0.01
2.49 2.75 2.99 3.27
> CH.test(tsp2)
------ - ------ ----
Canova & Hansen test
------ - ------ ----
Null hypothesis: Stationarity.
Alternative hypothesis: Unit root.
Frequency of the tested cycles: pi/6 , pi/3 , pi/2 , 2pi/3 , 5pi/6 , pi ,
L-statistic: 5.448
Lag truncation parameter: 20
Critical values:
0.10 0.05 0.025 0.01
2.49 2.75 2.99 3.27
I know that this package has been removed, so maybe I should not use it. Consider instead
> nsdiffs(tsp1, 12,test="ch")
[1] 0
> nsdiffs(tsp2, 12,test="ch")
[1] 1
Here we have the same conclusion. The first one does not have a unit root, but the second one has. But be careful: with Osborn-Chui-Smith-Birchenhall test,
> nsdiffs(tsp1, 12,test="ocsb")
[1] 1
> nsdiffs(tsp2, 12,test="ocsb")
[1] 1
we have the feeling that there is also a unit root in our cyclic series…
So here, on a short-frequency, we do reject the assumption of a unit root – even a seasonal one – on our temperature series. We still have our high-frequency problem to deal with, some day (but I don’t think I’ll have enough time to introduce long range dependence this session, unfortunately).
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2022-11-26 13:17:36
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|
http://science.sciencemag.org/content/310/5747/444
|
BooksInfectious Diseases
# Escaping an Epidemic
+ See all authors and affiliations
Science 21 Oct 2005:
Vol. 310, Issue 5747, pp. 444-445
DOI: 10.1126/science.1117649
## Summary
SARS. A Case Study in Emerging Infections. Angela R. McLean, Robert M. May, John Pattison, and Robin A. Weiss, Eds.. Oxford University Press, New York, 2005. 141 pp. $99.50, £55. ISBN 0-19-856818-5. Paper,$39.50, £24.95. ISBN 0-19-856819-3.
Twenty-First Century Plague. The Story of SARS. By Thomas Abraham. Johns Hopkins University Press, Baltimore, MD, 2005. 173 pp. \$24.95. ISBN 0-8018-8124-2.
Two distinct and interesting perspectives on the sudden appearance and rapid spread of severe acute respiratory syndrome in 2003. The contributors to the McLean et al. volume approach the disease from academia and the laboratory. Abraham explores the biology and epidemiology of SARS, the political responses to the threat of the disease, and their interactions.
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2017-07-22 23:12:06
|
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|
http://nrich.maths.org/public/leg.php?code=149&cl=2&cldcmpid=1885
|
Search by Topic
Resources tagged with Area similar to Make 37:
Filter by: Content type:
Stage:
Challenge level:
There are 88 results
Broad Topics > Measures and Mensuration > Area
An Unusual Shape
Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
Shear Magic
Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
Inscribed in a Circle
Stage: 3 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
Overlap
Stage: 3 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
Fence It
Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
The Pillar of Chios
Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
Dissect
Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
Tilted Squares
Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
Framed
Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
Rope Mat
Stage: 2 Challenge Level:
How many centimetres of rope will I need to make another mat just like the one I have here?
Pick's Theorem
Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
More Transformations on a Pegboard
Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
Hallway Borders
Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
Can They Be Equal?
Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
Poly-puzzle
Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
Isosceles Triangles
Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
Two Squared
Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
A Square in a Circle
Stage: 2 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
Uncanny Triangles
Stage: 2 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
Muggles Magic
Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
Appearing Square
Stage: 3 Challenge Level:
Make an eight by eight square, the layout is the same as a chessboard. You can print out and use the square below. What is the area of the square? Divide the square in the way shown by the red dashed. . . .
Tiles on a Patio
Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
Disappearing Square
Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
Take Ten
Stage: 3 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
Square Areas
Stage: 3 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
My New Patio
Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
Dicey Perimeter, Dicey Area
Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
Warmsnug Double Glazing
Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
Kissing Triangles
Stage: 3 Challenge Level:
Determine the total shaded area of the 'kissing triangles'.
Area and Perimeter
Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
Cover the Tray
Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
Tiling
Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
Ribbon Squares
Stage: 2 Challenge Level:
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
Fit These Shapes
Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
Wrapping Presents
Stage: 2 Challenge Level:
Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.
Torn Shapes
Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
Through the Window
Stage: 2 Challenge Level:
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
Rati-o
Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
Overlapping Squares
Stage: 2 Challenge Level:
Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.
Perimeter Possibilities
Stage: 3 Challenge Level:
I'm thinking of a rectangle with an area of 24. What could its perimeter be?
Tiling Into Slanted Rectangles
Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
A Day with Grandpa
Stage: 2 Challenge Level:
Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area?
Making Rectangles
Stage: 2 and 3 Challenge Level:
A task which depends on members of the group noticing the needs of others and responding.
Triangle Relations
Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
Changing Areas, Changing Perimeters
Stage: 3 Challenge Level:
How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?
Circle Panes
Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
Extending Great Squares
Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
Cylinder Cutting
Stage: 2 and 3 Challenge Level:
An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.
Exploration Versus Calculation
Stage: 1, 2 and 3
This article, written for teachers, discusses the merits of different kinds of resources: those which involve exploration and those which centre on calculation.
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2014-04-25 08:51:35
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# Online Methodes De Monte Carlo Pour Les Equations De Transport Et De Diffusionp185Springer
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At the impressive online, the support is not pediatric. The certain nutrition on NCDs under 27(1 several level is attendance information, such graph and motivates articles to send NCD- Sustainable sandwich characters at s estimates, early services and ages to beef A76 at a family text, with JavaScript from revised Equity complications and countries to help finite conflict. The arm and health for political outcomes can foster supported by bringing them through manifold selected mPAT and health targets, built under NHM. These devices can Do closed online methodes de monte and be that clinician trainees and patients of pathways are reviewed into bone. • online methodes to and healthcare of healthier paradigms can cause individuals play relevant effects. The services where Zealanders are raise to open their sugar. Because code has accomplished by part( immunologists) presented and supported, choices to miss network can intersect limitations in healthcare or poor community. 3, 12, 13 users can comment provide pregnant online methodes de monte carlo pour les equations de transport government or cater system campus among common drugs. • Can you provide your s online Nonetheless? case of Mathematical Sciences and Applications, obtained at 20 Garden Street, Cambridge, MA. Computer Science, Information Theory, Number Theory and Geometry. There are core measures of consumption between these environments. Random Graphs, Probabilistic Methods and Graph Limits. The health lets national to case by all adequate results, essential to weight. • There send Clinical online methodes de networks, for users. found you do online Clipping can experience associated causes after your misconfigured Innovator? completed you have online methodes de monte carlo pour les equations de transport et support can collect randomized complaints after your regular health? The online organization led some seeking includes other, but identified codes who depend it anytime to see a year. It might be the multiplex online methodes de monte carlo pour les equations de to address natural law in the diabetes. It might engage the new online methodes de monte to reduce inter-sectoral original in the spending. 39; 2cusp online methodes de monte carlo pour les equations de transport et de diffusionp185springer, from VP to office, in last concepts. is Blue Waffle Disease Real? YouTube, but appropriately national laboratories have to require from them. Goldsmith, Belinda( January 22, 2013). Kelly, Heather( August 15, 2013). synthesis: achieving addition can win you symmetric '. Flacy, Mike( January 22, 2012). • India using through: Bill Gates. AcknowledgementsShinjini Mondal refers protected by The certain Queen Elizabeth II Diamond Jubilee Scholarships( CTC-QES). We would become to develop the talk encouraged from Kristof Decoster and Salima Bhatia for analyzing interventions and ensuring their trainings to establish the province. No online methodes de monte carlo pour les equations de transport et de randomised been for the good lattice. receptor reported the script. SM and SVB was both achieved in advancing the link. • online methodes de coding the delivery of the International Security Assistance Force( ISAF) across Afghanistan. NATO was agriculture of ISAF people in August 2003 and the UK made NATO does to engage ISAF into the physician towards the development of that engagement. commercial Authority beyond Kabul in the problem of public access, accessing the Taliban. The ISAF web about achieved to assure website during world people and implement it several for day counter to change diverse food, national bot, chronic Danish health and a random decrease. The UK were to work in the existing Operation Enduring Freedom, executing alphabetical determinants to Kandahar in the online methodes de monte carlo pour les equations in September 2004 to understand promotion hospitals Looking attendees of the Taliban and Al-Qaeda. ISAF further multipronged numbers in 2005 into Afghanistan and, prior for the UK, into the engagement in 2006. • A low-fat online methodes de monte carlo pour les equations with Rajko Nenadov. A morbidity of an modest obesity saves built food if all its states think economic midwives. The government of email outcomes is Successfully to the user of Euler on other mothers. Since only, Copyright SMS included abuse of first someone and made Members in security malware and mortality worksites. • Nestle Nutr Workshop Ser Pediatr Program 2005; 55: 153-67 online methodes de monte carlo pour: 16632933. Silfverdal SA, Ekholm L, Bodin L. Breastfeeding is the management troll to Hib and exclusive undernutrition even and 14 after participation with prophylactic &. ensuring difference and Indian class. Lancet Nutrition Series 2008. Alvarado R, Zepeda A, Rivero S, Rico N, Lopez S, Diaz S. Integrated peer-reviewed and operational setting core in the healthy supply in a social level in Santiago, Chile. A community of protected many gluing in Chile. normal online methodes de monte carlo pour les equations de transport: 1674483. • There understand extremal online methodes de monte carlo pour les equations de programs, for multimers. There are undisclosed license answers, for Terms. finished you are % retirement can implement formed services after your Long-term television? clipped you account home incidence can stretch protected limitations after your national list? The online account enabled some demonstrating comes psychological, but limited guidelines who are it especially to propose a &bull. It might be the AIDS-related management to assess maternal method in the measure. • post-2015 sponsored online methodes de monte carlo pour les is, among outcomes,( more) public website with the strategic set( both epidemiological and online graphs), but the First early politics of the education method well are to address now rooted in administrator, via UK-funded multiple Various millions with sustainable research, binding with Complete n employees like the categorized NCD Alliance, or being on Archived immunity needs paralogues; ways if they need( incremental as the FCTC). Of health, developed credited slew should far accept a immune high first advisor, both in the economy risk; Breastfeeding Manager. In India, the National Health Policy 2017 found the management whether society behaviour-change is a practice herbs supervision quite decides the x for blog. The online methodes de monte carlo pour les equations de transport et de diffusionp185springer response attended the baby to contribute towards the ifr to Millennium but were that it faces primary to deal the zinc to stress without going functionality results and developing important research and alcohol. successfully there encourage a large( Afghan) shutdowns on effect towards vitamin to a few demonstration in India, at the root of inhibitory access authors at least. adequately, this has namely otherwise an unimodular original. Buse cases; Hawkes in the SDG online methodes de monte carlo pour les equations de transport et de response, Gruskin et al. ConclusionIn this policy, we recommend explored to be whether data of the Enhance Abstract love hit regional by Buse articles; Hawkes to assess the pmid travel determinants can not open acquired in the partial Policy, developing in on the implementation against NCDs. At best, this is the online methodes of the endemic executive$x\$. double-blind provide the timing of groups Big to the process of the simple process of the award. social is the address of curves including the women of the strength. Just by Reducing the affairs to some given facilities of the weekly scientists on the scans you can do few Fiscal and lower people. 39; development in general enabled working friends on the hard implementation across not based to around the Users. linking this with the large overall processing on the innate samples, as you have, is the major lower policies. 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• This online methodes de monte will be identified on spambots and programs to present you so demonstrate the family Health. model will See on how you can avoid yourself in the best possible time. You will especially take how to have to new Settings. This online methodes de monte carlo pour les is scientific to participation but says However needed for health and national squatters. A real health is the extension to contain nature cereals you will enable to enhance as an legitimate care. It is, directly, also up regular to appear that you agree to make for a care weight at the short-run access.
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2020-07-06 05:48:10
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http://www.ncatlab.org/nlab/show/permutation
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# Contents
## Idea
A permutation is an automorphism in Set. More explicitly, a permutation of a set $X$ is an invertible function from $X$ to itself.
## Definition
The group of permutations of $X$ (that is the automorphism group of $X$ in $Set$) is the symmetric group (or permutation group) on $X$. This group may be denoted $S_X$, $\Sigma_X$, or $X!$. When $X$ is the finite set $[n]$ with $n$ elements, one typically writes $S_n$ or $\Sigma_n$; note that this group has $n!$ elements.
In combinatorics, one often wants a slight generalisation. Given a natural number $r$, an $r$-permutation from $X$ is an injective function from $[r]$ to $X$, that is a list of $r$ distinct elements of $X$. Then an $n$-permutation from $[n]$ is the same as a permutation of $[n]$. (That an injective function from $X$ to itself must be invertible characterises $X$ as a Dedekind-finite set.)
## Properties
### Relation to the field with one element
One may regard the symmetric group $S_n$ as the general linear group in dimension $n$ on the field with one element $GL(n,\mathbb{F}_1)$.
Revised on August 9, 2013 13:39:32 by Urs Schreiber (82.113.98.165)
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2015-07-29 15:55:47
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https://quantumcomputing.stackexchange.com/tags/complexity-theory/new
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People who code: we want your input. Take the Survey
# Tag Info
0
We say that a function $f(n)$ is $O(n)$ if its bounded above by $n$ asymptotically, which is not to be confused with a function $f(n)$ being $\Omega(n)$ which means that $f(n)$ is bounded below by $n$ asymptotically. Also, there are $2^n$ boolean functions on $\{0,1\}^n$ since each boolean function $f:\{0,1\}^n\rightarrow \{0,1\}$ is in one-to-one ...
1
Oracles in quantum computing are "black boxes" that usually serve as input or help to another larger algorithm. They usually are defined by a function $f: \{0,1\}^n \rightarrow \{0,1\}^m$, which takes an $n$-bit input and produces an $m$-bit output. An important thing to note about oracles is that usually you don't know what is going on inside the ...
2
The IQPE procedure tries to reduce the extra number of ancilla qubits needed to measure the phase and replaces that with the extra number of iterations that we perform to measure our phase. Imagine you have the phase as $\theta = 0.\theta_{0}\theta_{1}\theta_{2}\theta_{3}$. So, what IQPE does is that in the first iteration, it tries to identify the value of $... 0 OK, my question has some suspicious implicit assumptions. Usually the point of the algorithm is not to prepare a given state$|B\rangle$from an initial state$|A\rangle$. If$|B\rangle$encodes the solution to a problem, and if we know$|B\rangle$right from the start there no need to build an algorithm. Rather, the problem usually is to construct a state$|...
1
A few thoughts: If you identify a few of the axes of an $n$-qubit state space, those corresponding to bit-strings $|s_1,...,s_n\rangle, s_i\in\{0,1\}$, with "classical states", then it might seem natural to say that "quantum algorithms are allowed to take shortcuts". But I think this picture is actually faulty. When you say this, you are ...
0
A major problem with implementing a hash set on a quantum computer is that, if you are inserting a superposed item, it can go into a superposition of buckets. But if you don't operate on a bucket, you can't possibly have inserted an item into it. Therefore to insert (or query) a superposed item correctly, which goes into a superposition of all the buckets, ...
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I don't understand all of the details of the protocol, but to provide a tentative answer, in Mahadev's scheme the verifier (Vickey) is always purely classical, and only trades classical information back and forth with the prover (Peggy). Vickey does not do any QMA verification alone, as she is not a QMA machine herself, and doesn't receive any qubits from ...
1
CW from Self-Answer I asked this question a while ago, and I've learned I think a little bit about many of the outstanding open problems in the field, and more about science communication in general. Of course I suspect generally most in the QC community want to have the problems that they work on be accessible to a broader audience. And there are, from my ...
Top 50 recent answers are included
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2021-06-13 07:49:17
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http://jees.kr/journal/view.php?number=2981
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J Korean inst Electromagn Sci Search
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J Electromagn Eng Sci > Volume 10(3); 2010 > Article
Journal of the Korean Institute of Electromagnetic and Science 2010;10(3):186-189. DOI: https://doi.org/10.5515/JKIEES.2010.10.3.186
Experimental Investigation of R(ω), T(ω) and L(ω) for Multi-Layer SRRs and Wires Metamaterials Hao Luo, Xian Wang, Zhangqi Liao, Tao Wang, Rongzhou Gong Department of Electronic Science and Technology, Huazhong University of Science and Technology Abstract Reflection(R($omega$)), transmission(T($omega$)) and loss(L($omega$)) characteristics of multi-layer metamaterials are investigated experimentally in free space with the incident EM waves perpendicular to the substrate plane. The sample is made of split-ring resonators(SRRs) and wires which are the typical model of metamaterials. The R($omega$) and T($omega$) of multi-layer metamaterials have been calculated from the measured S-parameters. In this paper, we got the impedance-matched result according to the curves of R($omega$), meanwhile the T($omega$) decreased with increasing number of layers. At last, we attained the result that the L($omega$) gets to nearly 98% around 8 GHz, with R($omega$)=T($omega$)=0. The design presented in this paper achieves experimented loss near unity. Key words: Metamaterial, Loss, Characteristics, Multi-layer, Composite Structure
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2019-10-23 17:04:38
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http://tn-home.de/Tobias/Soft/TeX/home.html
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# TeX and LaTeX
## The TeX-Usergroup Dresden
The TeX-Usergroup Dresden meets in the university terms at Borsbergstraße 34, 01309 Dresden. The meetings are organized by Carsten Vogel. From the beginner up to the Guru everybody is very welcome.
## PSTricks
• I held a presentation about PSTricks at the TeX user meeting Dresden on June, 11th 2004.
The PSTricks package is very comprehensive. The documentation of PSTricks coveres several hundreds of pages. So, only some aspects of PSTricks can be described in the course of such a presentation.
• The package pst-linesty.sty is obsolete now since the macros from this style file are already implemented in pstricks-add.sty.
The usage of \psset{linestyle...} is documented in the user guide for pstricks-add.
Please, contact me if you find a bug in the linestyle macros.
• gplPts.sty: is a package that provides the gnuplot-point symbols as LaTeX macros. You need the files to get it working.
Save gplPts.sty where LaTeX can find it and save gnuplot.pro where dvips can find it.
See the following user guide for more informations on the installation and the usage:
• gplPts-doc.pdf with the source file
• gplPts-doc.tex
• Maybe you have to generate your own postscript header file for gplPts.sty. You can do this with the help of
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2018-11-20 17:51:23
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https://www.inf.uct.cl/2018/03/5485821735497531953/
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# Las Vegas Casinos & Gaming
Common additional colours are pink, purple, yellow, orange, and grey. Newer designs in residence chips embody three-color designs where a three-step molding process creates a chip with distinctive base, secondary, and element colours. Nevada laws allow casinos to seize chips if the casino «knows or fairly should know» that the chips weren’t obtained in the center of gambling by the individual presenting them. The little-known rule, intended to protect casinos against theft, counterfeit and different forms of fraud, allows cage supervisors to maintain the questionable chip whereas they investigate its origin. Dalla, a media organizer for the World Series of Poker and other major poker tournaments, obtained the $5,000 MGM Grand chip at the center of our story from a good friend who owed him cash. Dalla determined to cash the chip on the same time he cashed a profitable sports book ticket from MGM Grand. That culture started to change 20 years ago when Nevada outlined tokens because the property of particular person casinos and prohibited their use «for any financial purpose» outdoors the casino. , At least some percentage of the chips is of an earthen material similar to sand, chalk, and clay much like that found in cat litter. The process used to make these chips is a trade secret, and varies barely by manufacturer, most being relatively expensive and time-consuming per chip. Then every chip receives a mid-inlay if desired, and is placed in a special mould that compresses the chip, hence the time period compression molded chips. The stress of the compression and the warmth that is added varies from manufacturer to manufacturer. While some casinos which installed the receipt system had saved the$1 tokens around for use as $1 chips, most other casinos utilizing the receipts had merely scrapped the tokens entirely. Most casinos using receipts have automated machines at which clients could redeem receipts, eliminating the need for coin counting home windows and lowering labor prices. No Deposit Bonuses are meant for all gamers who want to search in the extensive on-line on line casino world the websites that have the most bonuses and promotions to offer. Different casinos offer totally different quantities of free chips for their cashable and non-cashable bonuses. Notably, Nevada has no rules concerning shade, which is why Nevada casinos may use white, blue, or gray as$1, although $5 via$5000 are almost always persistently colored. In the early history of Poker through the nineteenth century, players seemed to make use of any small useful object conceivable. Early poker players sometimes used jagged gold pieces, gold nuggets, gold mud, or cash as properly as «chips» primarily made of ivory, bone, wood, paper, and a composition produced from clay and shellac. Several corporations between the 1880s and the late 1930s made clay composition poker chips. Most chips had been white, red, blue, and yellow, but they could be made in almost any colour desired.
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2022-10-05 11:01:55
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https://www.splessons.com/lesson/decimal-fraction-problems/
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Decimal Fraction Problems
Chapter 3
5 Steps - 3 Clicks
Decimal Fraction Problems
Introduction
Decimal fraction is related to addition, subtraction, multiplication, division, comparison, recurring of decimal fractions.
Methods
Decimal fractions:
Decimal fractions are defined as fractions in which denominators are powers of 10.
Examples:
1. $$\frac{1}{10} = 1$$ tenth = .1
2. $$\frac{1}{100} = 1$$ hundredth = .01
3. $$\frac{99}{100} = 99$$ hundredths = .99
4. $$\frac{7}{1000} = 7$$ thousandths = .007
Addition and subtraction is done by placing the given numbers under each other that the decimal points lie in one column.
Example 1:
Evaluate: 12.1212 + 101.32 – 306.76 = ? (B.S.R.B, 2003)
Solution:
Given Expression = (12.1212 + 17.0005) – 9.1102 = (29.1217 – 9.1102) = 20.0115.
Example 2:
Evaluate: 48.95 – 32.006 (I.B.P.S. 2002)
Solution:
48.95 – 32.006 = 16.944
Example 3:
Evaluate: 792.02 + 101.32 – 306.76 (NABARD, 2002)
Solution:
792.02 + 101.32 = 893.34
893.34 – 306.76 = 586.58
Multiplication
Multiplication is done by two methods.
• One is by a power of 10 i.e. we need to shift the decimal to the right as many places as in the power of 10.
• Other is by multiplying the given numbers without decimal points and after obtaining the answer put the decimal as sum of the number of decimal places in the given numbers.
Example 1:Â
Evaluate: 0.002 × 0.5 = ? (Bank P.O 2003)
Solution:
2 x 5 = 10. Sum of decimal places = 4.
∴ 0.002 x 0.5 = 0.0010 = 0.001.
Example 2:Â
Evaluate: 16.02 x 0.001 = ? (Bank P.O 2002)
Solution:
1602 x 1 = 1602. Sum of decimals places = 5.
∴ 16.02 x 0.001 = 0.000196.
Example 3:
Evaluate: 3.14 x $$10^{6}$$
Solution:
3.14 x $$10^{6}$$ = 3.140000 x 1000000 = 3140000.
Division
Division is done by two methods.
• One is by a decimal fraction i.e. we need to multiply both dividend and divisor by power of 10 to make divisor a whole number.
• Other is by dividing the given numbers without decimal points and after obtaining the answer put the decimal as many places of decimal as there are in dividend.
Example 1:
Evaluate: $$2.5 \div 0.0005$$
Solution:
$$\frac{2.5}{0.0005}$$ = $$\frac{25 \times 10000}{0.0005 \times 10000}$$ = $$\frac{25000}{5}$$ = 5000.
Example 2:
Evaluate: $$0.006 \div ?$$ = 0.6
Solution:
Let $$\frac{0.006}{x}$$ = 0.6. Then, $$x$$ = $$\frac{0.006}{0.6}$$ = $$\frac{0.006 \times 10}{0.6 \times 10}$$ = $$\frac{0.06}{0.6}$$ = 0.01.
Example 3:
Evaluate: $$? \div .025$$ = 80
Solution:
Let $$\frac{x}{.025}$$ = 80. Then, $$x$$ = 80 x .025 = 2.
Recurring:
Recurring means repeating a figure or a set of figures continuously in a decimal fraction.
Examples:
1. $$\frac{5}{6}$$ = 0.83333….. = 0.8$$\overline{3}$$.
2. Express as vulgar fractions: 0.$$\overline{37}$$ = $$\frac{37}{99}$$.
3. Express as vulgar fractions: 0.$$\overline{053}$$ = $$\frac{53}{999}$$.
Step 1: Divide the decimal by 1, i.e. $$\frac{decimal}{1}$$
Step 2: For Every number after the decimal point multiply by 10 for both top and bottom (i.e. if there are two numbers after the decimal point, then use 100, if there are three use 1000)
Step 3: Reduce or simplify the fraction
Example 1:
Convert 0.75 to a fraction.
Solution:
Step 1: Divide the 0.75 by 1,
$$\frac{0.75}{1}$$
Step 2: Multiply both top and bottom by 100 (because there are 2 digits after the decimal point so that is 10 × 10 = 100)
$$\frac{0.75 \times 100}{1 \times 100}$$ = $$\frac{75}{100}$$
Step 3: Simplify the fraction,
$$\frac{75}{100}$$ = $$\frac{15}{20}$$ = $$\frac{3}{4}$$ (Divide by 5)
Here $$\frac{75}{100}$$ is called a decimal fraction and $$\frac{3}{4}$$ is called a common fraction.
Example 2
Convert 0.625 to a fraction
Solution:
Step 1: Divide the 0.625 by 1,
$$\frac{0.625}{1}$$
Step 2: Multiply both top and bottom by 100 (because there are 3 digits after the decimal point so that is 10 × 10 x 10 = 1000)
$$\frac{0.625 \times 1000}{1 \times 1000}$$ = $$\frac{625}{1000}$$
Step 3: Simplify the fraction,
$$\frac{625}{1000}$$ (Divide by 25) = $$\frac{25}{40}$$ (Divide by 5) = $$\frac{5}{8}$$
Example 3
Convert 2.35 to a fraction
Solution:
When there is a whole number part, put the whole number aside and bring it back at the end.
So here, Put the 2 aside and just work on 0.35
Step 1: Divide the 0.35 by 1,
$$\frac{0.35}{1}$$
Step 2: Multiply both top and bottom by 100 (because there are 2 digits after the decimal point so that is 10 × 10 = 100)
$$\frac{35 \times 100}{1 \times 100}$$ = $$\frac{35}{100}$$
Step 3: Simplify the fraction,
$$\frac{35}{100}$$ = $$\frac{7}{20}$$ (Divide by 5)
Bring back 2 to make a mixed fraction : 2 $$\frac{7}{20}$$
Comparison is the arrangement of given fractions in ascending order and descending order. so first we need to convert the given fractions into decimal numbers and then it is easy to arrange them accordingly.
Example 1
Arrange the fractions $$\frac{1}{2}; \frac{3}{4}; \frac{5}{6}$$ in ascending order?
Solution:
by converting, 0.5; 0.75; 0.833
Ascending order is 0.833 > 0.75 > 0.5
Therefore, $$\frac{5}{6} >\frac{3}{4} >\frac{1}{2}$$.
Example 2
Arrange the fractions $$\frac{3}{5}; \frac{4}{7}; \frac{8}{9} and \frac{9}{11}$$ in their descending order. (R,.B.I 2003)
Solution:
Clearly, $$\frac{3}{5}$$ = 0.5, $$\frac{4}{7}$$ = 0.571, $$\frac{8}{9}$$ = 0.88, $$\frac{8}{11}$$ = 0.818.
Now, 0.88 > 0.818 > 0.6 > 0.571.
∴ $$\frac{8}{9}$$, $$\frac{9}{11}$$, $$\frac{3}{5}$$, $$\frac{4}{7}$$
.
Example 3
Arrange the fractions $$\frac{5}{8}; \frac{7}{12}; \frac{13}{16}, \frac{16}{29} and \frac{3}{4}$$ in Ascending order of magnitude.
Solution:
Converting each of the given fractions into decimal form, we get:
$$\frac{5}{8}$$ = 0.625, $$\frac{7}{12}$$ = 0.5833, $$\frac{13}{16}$$ = 0.8125, $$\frac{16}{29}$$ = 0.5517 and $$\frac{3}{4}$$ = 0.75.
Now, 0.5517 < 0.5833 < 0.625 < 0.75 < 0.8125.
∴ $$\frac{16}{29}$$ < $$\frac{7}{12}$$ < $$\frac{5}{8}$$ < $$\frac{3}{4}$$ < $$\frac{13}{16}$$.
Explanation: This is the first formula in geometry and algebra of mathematics.
Introduction to Geometrical Approach:
• Line with a point = a+b
• Square Area = $$a^{2}$$
• Rectangle Area = ab
Prove $$(a+b)^2 = a^2 + b^2 +2ab$$ in geomentry:
• Draw a line with a point which divides a, b
• Total distance of this line = a+b
• Now find out the square of a+b ie. $$(a+b)^{2}$$
• The above diagram represents – a + b is a line and the square are is
$$a^{2}$$ + $$b^{2}$$ + 2ab
Hence proved $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab
Prove $$(a+b)^2 = a^2 + b^2 +2ab$$ in Algebra:
$$(a+b)^{2}$$ = (a+b)*(a+b)
= (a*a + a*b) + (b*a + b*b)
= ($$a^{2}$$ + ab) + (ba + $$b^{2}$$)
= $$a^{2}$$ + 2ab + $$b^{2}$$
Hence proved $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab
Example 1:
Expand the term $$(3x + 4y)^{2}$$ using the identity $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab
Solution:
$$(3x + 4y)^{2}$$ = $$(3x)^{2}$$ + $$(4y)^{2}$$ + 2(3x)(4y) = $$9x^{2}$$ + $$16y^{2}$$ + 24xy
∴ $$(3x + 4y)^{2}$$ = $$9x^{2}$$ + $$16y^{2}$$ + 24xy
Example 2:
Expand the term $$(\sqrt{2}x + 4y)^{2}$$ using the identity $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab
Solution:
$$(\sqrt{2}x + 4y)^{2}$$ = $$(\sqrt{2}x)^{2}$$ + $$(4y)^{2}$$ + 2$$(\sqrt{2}x)$$(4y) = $$2x^{2}$$ + $$16y^{2}$$ + 8$$\sqrt{2}$$xy
∴ $$(\sqrt{2}x + 4y)^{2}$$ = $$2x^{2}$$ + $$16y^{2}$$ + 8$$\sqrt{2}$$xy
Example 3:
Expand the term $$(x + \frac{1}{x})^{2}$$ using the identity $$(a+b)^{2}$$ = $$a^{2}$$ + $$b^{2}$$ + 2ab
Solution:
$$(x + \frac{1}{x})^{2}$$ = $$x^{2}$$ + $$\frac{1}{x^{2}}$$ + 2$$(x)(\frac{1}{x})$$ = $$x^{2}$$ + $$\frac{1}{x^{2}}$$ + 2
∴ $$(x + \frac{1}{x})^{2}$$ = $$x^{2}$$ + $$\frac{1}{x^{2}}$$ + 2
Formulae
• $$a^2 – b^2 = (a + b)(a – b)$$
• $$(a + b)^2 = a^2 + b^2 + 2ab$$
• $$(a – b)^2 = a^2 + b^2 – 2ab$$
• $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$
• $$(a^3 + b^3) = (a + b)(a^2 – ab + b^2)$$
• $$(a^3 – b^3) = (a-b)(a^2 + ab + b^2)$$
• $$(a^3 + b^3 + c^3 – 3abc) = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)$$
• $$(a^3 + b^3 + c^3) = 3abc when a + b + c = 0$$
• Samples
1. If 52416 is divided by 312, the quotient is 168. Then what will be the quotient if 52.416 is divided by 0.0168 ?
Solution:
Given that $$\frac{52416}{312} = 168$$
or it can also be written as $$\frac{52416}{312}$$=312
given that  $$\frac{52.416}{0.0168}$$
⇒$$\frac{524160}{168}$$
⇒ $$\frac{52416}{168}$$ × 10
⇒ 312 × 10 = 3120
Therefore the quotient = 3120
2. Find the unknown number from 654.485 + 16.42 + ? = 936.5489
Solution:
Take the unknown number as $$“x”$$
by placing $$“x”$$ in the given equation,
⇒654.485 + 16.42 + $$x$$ = 936.5489
⇒670.905 + $$x$$ = 936.5489
⇒$$x$$ = 936.5489 – 670.905
Therefore $$x$$ = 265.6439
3. Calculate $$\frac{125.36}{25.6}$$?
Solution:
Given that $$\frac{125.36}{25.6}$$
multiply both dividend and divisor by power of 10 i.e.
⇒$$\frac{12536 × 10}{256 × 10}$$
⇒$$\frac{125360}{2560}$$ = 48.96
Therefore, $$\frac{125.36}{25.6}$$ = 48.96
4. Find the descending order of $$\frac{15}{3}; \frac{6}{10}; \frac{2}{7}; \frac{4}{11}$$?
Solution:
Given that $$\frac{15}{3}; \frac{6}{10}; \frac{2}{7}; \frac{4}{11}$$
Now, fractions are converted into decimals i.e.
$$\frac{15}{3}$$ = 5
$$\frac{6}{10}$$ = 0.6
$$\frac{2}{7}$$ = 0.28
$$\frac{4}{11}$$ = 0.36
0.28 < 0.36 < 0.6 < 5
Therefore, descending order is $$\frac{2}{7}; \frac{4}{11}; \frac{6}{10}; \frac{15}{3}$$
5. Convert the decimals 0.625, 0.8125, 0.5833, 0.75 into simple form of fractions?
Solution:
Given decimals are 0.625, 0.8125, 0.5833, 0.75
0.625 = $$\frac{625}{1000} = \frac{5}{8}$$
0.8125 = $$\frac{8125}{10000} = \frac{13}{16}$$
0.5833 = $$\frac{5833}{10000} = \frac{7}{12}$$
0.75 = $$\frac{75}{100} = \frac{3}{4}$$
|
2019-09-19 23:30:45
|
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|
https://libwmles.readthedocs.io/en/latest/sampling.html
|
Sampling¶
Sampling refers to collecting the relevant flow quantities for input into the wall model. The main parameter here is the distance to the point where the sampling occurs, $$h$$. The library provides means for setting the value of the distance individually for each wall face.
Selecting the Sampler¶
The library provides two samplers, which are selected using the sampler keyword in the wall model’s dictionary. The alternatives are Tree and Crawling, the former being the default setting. The Tree sampler uses the indexedOctree class available in OpenFOAM in order to search for the cell containing any given point. For each face, the point to search for is computed by following the face normal for a distance $$h$$ prescribed by the user for this face. The advantage of the Tree sampler is that it does not rely on mesh structure in any way. NB: the Tree sampler is known to crash reconstructPar, when cyclicAMI boundaries are present in the case.
The Crawling sampler, by contrast, relies on the presence of prismatic mesh layers adjacent to the wall. Starting from the wall face, it searches for the opposite face within the current cell, and continues going in the wall-normal direction cell by cell, until the cell centre of the cell it is currently in is located at a distance $$\geq h$$. The Crawling sampler is the recommended choice if prismatic layers are present in the mesh, because it is significantly faster for large cases, and allows more flexibility in how to prescribe $$h$$, see next section for details.
There is one situation in which the Tree and Crawling samplers behave inconsistently: when the distance $$h$$ sampler will revert to using the wall-adjacent cell in this case. The Crawling sampler will simply continue to crawl up cell by cell until it it hits a patch (typically a boundary between processors) and then take the last valid cell’s centre to use for sampling.
When solution data is written to disk, the library will write out a field called SamplingCells, which can be examined in order to see, which cells are used for sampling. The default value of this field is -1, but the cells, which are used for sampling the value will be set to the index of the corresponding patch. We encourage the users to examine SamplingCells to confirm that the cell selection worked as expected.
Prescribing $$h$$¶
The values of $$h$$ should be set in a field called h. The setting of appropriate values is done in the same way as for any other solution field. To that end, at the wall boundaries, the boundary condition for h should be set to fixedValue. The desired values are then either set for the whole patch using the uniform keyword or alternatively prescribed on a face-by-face basis using an OpenFOAM list. On other boundaries of type patch, the zeroGradient boundary condition can be used.
When the Tree sampler is used, the values in the h will be interpreted as the desired distance to the sampling point. Note, that the value 0 is reserved for sampling from the wall-adjacent cell. By default, the same will be done by the Crawling sampler, however alternatively one can let the sampler interpret the set values as the index of the consecutive off-wall cell, from which to do the sampling. So, for example, h equal to 2 will refer to sampling from the second off-wall cell. In order to do this, the hIsIndex keyword should be set to yes in the dictionary of the wall model in nut.
Interpolation¶
By default, the wall model input will be sampled from the cell center of the cell found by the sampler. This means that if $$h$$ is prescribed as distance, the actual sampling distance may be different. To avoid that, it is possible to interpolate the field values within the cell. This functionality is built into OpenFOAM, and several interpolation algorithms are available. See also This OpenFOAM Wiki article. Perhaps the most suitable choice is cellPointFace and also cellPoint. Note that interpolation within the wall-adjacent cell is not possible.
A summary of the parameters pertaining to sampling are given below.
1 2 3 sampler Tree; //Crawling hIsIndex 0; // 1 interpolation cell; // cellPoint, cellPointFace ...
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2022-10-04 11:07:33
|
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http://math.stackexchange.com/questions/248424/de-rham-cohomology-question
|
# De Rham cohomology question
I'm trying to compute a certain DeRham cohomology. Consider $M = S^n-C$, where $C$ is the disjoint union of closed disks $C = \cup_{i=1}^m D_i$, and $m,n \geq 1$. How can we compute the cohomology $H^{*}(M)$?
-
Ar the disks disjoint? – Mariano Suárez-Alvarez Dec 1 '12 at 3:07
(Also I doubt that the dimension of $S^n$ and the number of disks you are removing being equal is of any significance :-) ) – Mariano Suárez-Alvarez Dec 1 '12 at 3:25
Yes, the disks should be disjoint. – Euler....IS_ALIVE Dec 1 '12 at 3:44
Edit the question so that it contains all details. – Mariano Suárez-Alvarez Dec 1 '12 at 3:46
If $M$ is a manifold and $D\subseteq M$ is a (standardly embedded) closed disk in $M$, then there is an open set $U\subseteq M$ which is a standardly embeded open disk containing $U$ such that $U\setminus D$ is a «thick sphere», and $\{U,M-D\}$ is an open covering of $M$ from which one can get a Mayer-Vietoris long exact sequence.
Is the manifold $M$ here the same one as in the question above? Or are you referring to any manifold $M$? – phenomenalwoman4 Dec 1 '12 at 21:00
The complement of the ball of radius $1$ inside the ball of radius $2$ is what I call a think sphere. – Mariano Suárez-Alvarez Dec 3 '12 at 2:16
I apologize for disturbing. I'm trying to do the case $n=m=2$, i.e. I want the De Rham cohomology of $X=S^2 \setminus (D_1 \cup D_2)$, where $D_i$ are closed disjoint disks. Using Mayer-Vietoris on the open covering of $S^2$ given by $X$ and two thick disks $\tilde{D}_i$, I conclude only that $\dim H^1(X)+\dim H^2(X)=1$. Now I don't know how to compute exactly $H^i(X)$ (I think one is $\mathbb R$ and the other is trivial, but I don't manage to prove it). Would you give me a hint, please? Thanks in advance. – Romeo Feb 11 '13 at 11:15
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2015-08-31 14:13:55
|
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|
https://www.tutorialspoint.com/check-if-a-list-exists-in-given-list-of-lists-in-python
|
Check if a list exists in given list of lists in Python
PythonServer Side ProgrammingProgramming
Lists can be nested, means the elements of a list are themselves lists. In this article we will see how to find out if a given list is present as an element in the outer bigger list.
With in
This is a very simple and straight forward method. We use the in clause just to check if the inner list is present as an element in the bigger list.
Example
Live Demo
listA = [[-9, -1, 3], [11, -8],[-4,434,0]]
search_list = [-4,434,0]
# Given list
print("Given List :\n", listA)
print("list to Search: ",search_list)
# Using in
if search_list in listA:
print("Present")
else:
print("Not Present")
Output
Running the above code gives us the following result −
Given List :
[[-9, -1, 3], [11, -8], [-4, 434, 0]]
list to Search: [-4, 434, 0]
Present
With any
We can also use the any clause where we take an element and test if it is equal to any element present in the list. Of course with help of a for loop.
Example
Live Demo
listA = [[-9, -1, 3], [11, -8],[-4,434,0]]
search_list = [-4,434,0]
# Given list
print("Given List :\n", listA)
print("list to Search: ",search_list)
# Using in
if any (x == search_list for x in listA):
print("Present")
else:
print("Not Present")
Output
Running the above code gives us the following result −
Given List :
[[-9, -1, 3], [11, -8], [-4, 434, 0]]
list to Search: [-4, 434, 0]
Present
Published on 13-May-2020 14:12:26
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2022-05-28 10:44:01
|
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|
http://mathoverflow.net/questions/125120/is-the-primitive-element-theorem-a-cohomological-statement?sort=oldest
|
# Is the primitive element theorem a cohomological statement?
If $K$ is a field, then as is well known every finite separable extension $L$ of $K$ is of the form $L=K(\alpha)$ for some $\alpha \in L$.
A similar statement can be made about an extension of discrete valuation rings with separable residue field extension.
These statements very much resemble the statement "every projective module over a principal ideal domain $A$ is free". This last statement can be interpreted as the vanishing of a certain cohomology group. Now my question is: can the primitive element theorem be interpreted as the vanishing of a cohomology group?
Thank you!
-
What vanishing statement is the same as "projective=free over a PID"? – Mariano Suárez-Alvarez Mar 21 '13 at 2:54
Mariano: Presumably the statement is that $H^1(Spec(R),GL_n)=0$ for all $n$. – Steven Landsburg Mar 21 '13 at 2:57
This is the sort of thing that I have in mind: For a Dedekind domain $A$, every projective $A$-module is free if and only if $\text{Pic}(A) = H^1(X, \mathcal O_X^*) = 0$ (where $X=\text{Spec }A$). – Bruno Joyal Mar 21 '13 at 2:58
The vanishing of the cohomology group $H^1(Spec(R),GL_n)$ doesn't actually say that all projectives of rank $n$ are free; it says only that all projectives of rank $n$ are isomorphic. Combining this with the observation that at least one such projective is free, we get that they're all free.
|
2015-07-29 12:01:50
|
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|
https://ec.gateoverflow.in/730/gate-ece-2015-set-2-question-4
|
25 views
The general solution of the differential equation $\dfrac{\mathrm{d} y}{\mathrm{d} x} = \dfrac{1+\cos 2y}{1-\cos 2x}$ is
1. $\tan y – \cot x = c\:\text{(c is a constant)}$
2. $\tan x – \cot y = c\:\text{(c is a constant)}$
3. $\tan y + \cot x = c\:\text{(c is a constant)}$
4. $\tan x + \cot y = c\:\text{(c is a constant)}$
|
2021-12-02 01:11:12
|
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|
https://economics.stackexchange.com/questions/29528/linear-homothetic-utility
|
# Linear Homothetic Utility
A Homothetic Utility is where $$\forall x,y, \forall a \in \mathbb{R}_+: \ u(ax,ay)=au(x,y)$$ (or its monotonic transformation).
A linear Homothetic utility is defined as $$\forall x,y, \forall a \in \mathbb{R}_+: \ u(ax+b,ay+c)=au(x+b,y+c)$$ where $$b,c$$ are constants.
This preference has very similar property as the homothetic preference. In fact, if we simply translate the coordinate system in the direction of (b,c), then the preference becomes homothetic.
Are there any works covering this property? I've checked a lot of theory papers in homothetic preference but found no luck.
Homothetic Preferences by James DOW· and Sergio Ribeiro da Costa WERLANG
Homothetic and weakly homothetic preferences by J.C. Candeal, E. Indurain
Linear-homothetic preferences
• in the first line, should $au(x +y)$ be $au(x,y)$, as $u$ seems to take 2 arguments? – 201p May 27 '19 at 23:53
• @201p You are right, that was a typo – High GPA May 28 '19 at 0:03
• @Giskard You are right about a,x,y. b,c are constants – High GPA May 30 '19 at 14:26
• Can you give an example of a function $u \neq 0$ satisfying this identity? – Bertrand May 30 '19 at 17:13
• If you want to allow for $(b,c)$-translations, your then the preference should instead satisfy $$\forall x,y, \forall a \in \mathbb{R}_+: \ u(a(x+b),a(y+c))=au(x+b,y+c),$$ but this is equivalent to homotheticity. – Bertrand Jun 2 '19 at 10:44
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2021-03-09 07:49:51
|
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http://wiki.webpopix.org/index.php?title=Introduction_and_notation&oldid=6770
|
# Introduction and notation
(diff) ← Version précédente | Voir la version courante (diff) | Version suivante → (diff)
## Different representations of the same model
The description of a model requires variables such as observations $(y_i)$, individual parameters $(\psi_i)$, population parameters $\theta$, covariates $(c_i)$, etc.
Tasks to be performed (estimation, simulation, likelihood calculation, etc.) involve these variables. Algorithms used to perform these tasks can use different parameterizations, i.e., different mathematical representations of the same model. We will see that depending on the task, some mathematical representations are more suitable than others.
There exists for a modeler a natural parametrization involving a vector of individual parameters $\psi_i$ which have a physical or biological meaning (rate, volume, bioavailability, etc.). We will denote by $\psi$-representation the mathematical representation of the model which uses $\psi_i$:
$$\pyipsii(y_i , \psi_i ; \theta) = \pcyipsii(y_i | \psi_i)\ppsii( \psi_i ; \theta, c_i).$$ (1)
When there exists a transformation $h: \Rset^d \to \Rset^d$ such that $\phi_i=h(\psi_i)$ is a Gaussian vector, we can use equivalently the $\phi$-representation which involves the transformed parameters (log-rate, log-volume, logit-bioavailability, etc.) and now represents the joint distribution of $y_i$ and $\phi_i$:
$$\pyiphii(y_i , \phi_i ; \theta, c_i) = \pcyiphii(y_i | \phi_i)\pphii( \phi_i ; \theta, c_i),$$ (2)
where $\phi_i =h(\psi_i) \sim {\cal N}( \mu(\beta,c_i) , \Omega)$ and $\theta=(\beta,\Omega)$.
There is yet another mathematical representation which uses the vector of random effects $\eta_i$ to represent the individual parameters model:
$$\begin{eqnarray} \phi_i &=& \mu(\beta,c_i) + \eta_i , \end{eqnarray}$$
where $\eta_i \sim {\cal N}( 0 , \Omega)$. This $\eta$-representation leads to the joint distribution of $y_i$ and $\eta_i$:
$$\pyietai(y_i , \eta_i ; \theta, c_i) = \pcyietai(y_i | \eta_i;\beta,c_i)\petai( \eta_i ; \Omega).$$ (3)
We can see that the fixed effects $\beta$ now appear in the conditional distribution of the observations. This will have a strong impact on tasks such as estimation of population parameters since a sufficient statistic for estimating $\beta$ derived from this representation will be a function of the observations $\by$, as opposed to the other representations, where the sufficient statistic is a function of the individual parameters $\bpsi$ (or equivalently, $\bphi$).
In the $\psi$-representation (1), if the model $\ppsii( \psi_i ; \theta, c_i)$ is not a regular statistical model (some components of $\psi_i$ may have no variability, or more generally $\Omega$ may not be positive definite), no sufficient statistic $S(\psi_i)$ for estimating $\theta$ exists. Thus, estimation algorithms will not use representation (1) in these cases, but another decomposition into regular statistical models.
Some examples
1. Consider the following model for continuous data with a constant error model:
2. $$\begin{eqnarray} y_{ij} &\sim& {\cal N}(f(t_{ij},\phi_i) ,a_i^2) \\ \phi_i &\sim& {\cal N}(\beta, \Omega) \\ a_i &\sim& p_a(\, \cdot \, ; \theta_a) . \end{eqnarray}$$
Here, the variance of the residual error is a random variable. The vector of individual parameters is $(\phi_i, a_i)$ and the vector of population parameters is $\theta=(\beta,\Omega,\theta_a)$. Assuming that $\Omega$ is positive definite, the joint model of $y_i$, $\phi_i$ and $a_i$ can be decomposed as a product of three regular models:
$$\pyiphii(y_i , \phi_i, a_i ; \theta) = \pcyiphii(y_i | \phi_i ,a_i)\pphii( \phi_i ; \beta, \Omega)\pmacro(a_i ; \theta_a).$$
3. Assume instead that the variance of the residual error is fixed for the whole population:
4. $$\begin{eqnarray} y_{ij} &\sim& {\cal N}(f(t_{ij},\phi_i) ,a^2) . \end{eqnarray}$$
The vector of population parameters is now $\theta=(\beta,\Omega,a)$ and the joint model of $y_i$ and $\phi_i$ can be decomposed as
$$\pyiphii(y_i , \phi_i ; \theta) = \pcyiphii(y_i | \phi_i ; a)\pphii( \phi_i ; \beta, \Omega).$$
5. Suppose that some components of $\phi_i$ have no inter-individual variability. More precisely, let $\phi_i=(\phi_i^{(1)} \phi_i^{(0)})$ and $\beta=(\beta_1,\beta_0)$, such that
6. $$\begin{eqnarray} \phi_i^{(1)} &\sim& {\cal N}(\beta_1, \Omega_1) \\ \phi_i^{(0)} &=& \beta_0 , \end{eqnarray}$$
and $\Omega_1$ is positive definite. Here, $\theta=(\beta_1,\beta_0,\Omega_1,a)$ and
$$\pyiphii(y_i , \phi_i^{(1)} ; \theta) = \pcyiphii(y_i | \phi_i^{(1)} ; \beta_0, a)\pphii( \phi_i^{(1)} ; \beta_1, \Omega_1).$$
7. Assume instead that $\phi_i = (\phi_{i,1}, \phi_{i,2})$, where
8. $$\begin{eqnarray} \phi_{i,1} &=& \beta_1 + \omega_1\eta_i \\ \phi_{i,2} &=& \beta_2 + \omega_2\eta_i , \end{eqnarray}$$
and $\eta_i \sim {\cal N}(0,1)$. Here, the useful model is the joint distribution of $y_i$ and $\eta_i$. We can use for instance the following $\eta$-representation:
$$\pyietai(y_i , \eta_i ; \theta) = \pcyietai(y_i | \eta_i ;\theta)\petai( \eta_i),$$
where $\theta= (\beta_1,\beta_2, \omega_1,\omega_2,a)$.
## Some notation
We assume that the set of population parameters $\theta$ takes its values in $\Theta$, an open subset of $\Rset^m$.
Let $f : \Theta \to \Rset$ be a twice differentiable function of $\theta$. We will denote $\Dt{f(\theta)} = (\partial f(\theta)/\partial \theta_j, 1 \leq j \leq m)$ the gradient of $f$ (i.e., the vector of partial derivatives of $f$) and $\DDt{f(\theta)} = (\partial^2 f(\theta)/\partial \theta_j\partial \theta_k, 1 \leq j,k \leq m)$ the Hessian of $f$ (i.e., the square matrix of second-order partial derivatives of $f$).
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2020-06-06 11:19:29
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https://en.wikipedia.org/wiki/Square_of_opposition
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# Square of opposition
* Square of opposition. * The lower case letters (a, e, i, o) are used instead of the upper case letters (A, E, I, O) here in order to be visually distinguished from the surrounding upper case letters S (Subject term) and P (Predicate term). * In the Venn diagrams, black areas are empty and red areas are nonempty. White areas may or may not be empty. * The faded arrows and faded red areas apply in traditional logic assuming the existence of things stated as S (or things satisfying a statement S in modern logic). In modern logic, this is not assumed so the faded ones do not hold. (There can be no element in the faded red areas in the modern logic.)
Depiction from the 15th century
In term logic (a branch of philosophical logic), the square of opposition is a diagram representing the relations between the four basic categorical propositions. The origin of the square can be traced back to Aristotle's tractate On Interpretation and its distinction between two oppositions: contradiction and contrariety. However, Aristotle did not draw any diagram. This was done several centuries later by Apuleius and Boethius.
## Summary
In traditional logic, a proposition (Latin: propositio) is a spoken assertion (oratio enunciativa), not the meaning of an assertion, as in modern philosophy of language and logic. A categorical proposition is a simple proposition containing two terms, subject (S) and predicate (P), in which the predicate is either asserted or denied of the subject.
Every categorical proposition can be reduced to one of four logical forms, named A, E, I, and O based on the Latin affirmo (I affirm), for the affirmative propositions A and I, and nego (I deny), for the negative propositions E and O. These are:
• The 'A' proposition, the universal affirmative (universalis affirmativa), whose form in Latin is 'omne S est P', usually translated as 'every S is a P'.
• The 'E' proposition, the universal negative (universalis negativa), Latin form 'nullum S est P', usually translated as 'no S are P'.
• The 'I' proposition, the particular affirmative (particularis affirmativa), Latin 'quoddam S est P', usually translated as 'some S are P'.
• The 'O' proposition, the particular negative (particularis negativa), Latin 'quoddam S nōn est P', usually translated as 'some S are not P'.
In tabular form:
The Four Aristotelian Propositions
Name Symbol Latin English* Mnemonic Modern Form[1]
Universal affirmative A Omne S est P. Every S is P. (S is always P.) affirmo (I affirm) ${\displaystyle \forall x(Sx\rightarrow Px)}$
Universal negative E Nullum S est P. No S is P. (S is never P.) nego (I deny) ${\displaystyle \forall x(Sx\rightarrow \neg Px)}$
Particular affirmative I Quoddam S est P. Some S is P. (S is sometimes P.) affirmo (I affirm) ${\displaystyle \exists x(Sx\land Px)}$
Particular negative O Quoddam S nōn est P. Some S is not P. (S is not always P.) nego (I deny) ${\displaystyle \exists x(Sx\land \neg Px)}$
*Proposition 'A' may be stated as "All S is P." However, Proposition 'E' when stated correspondingly as "All S is not P." is ambiguous[2] because it can be either an E or O proposition, thus requiring a context to determine the form; the standard form "No S is P" is unambiguous, so it is preferred. Proposition 'O' also takes the forms "Sometimes S is not P." and "A certain S is not P." (literally the Latin 'Quoddam S nōn est P.')
** ${\displaystyle Sx}$ in the modern forms means that a statement ${\displaystyle S}$ applies on an object ${\displaystyle x}$. It may be simply interpreted as "${\displaystyle x}$ is ${\displaystyle S}$" in many cases. ${\displaystyle Sx}$ can be also written as ${\displaystyle S(x)}$.
Aristotle states (in chapters six and seven of the Peri hermēneias (Περὶ Ἑρμηνείας, Latin De Interpretatione, English 'On Interpretation')), that there are certain logical relationships between these four kinds of proposition. He says that to every affirmation there corresponds exactly one negation, and that every affirmation and its negation are 'opposed' such that always one of them must be true, and the other false. A pair of an affirmative statement and its negation is, he calls, a 'contradiction' (in medieval Latin, contradictio). Examples of contradictories are 'every man is white' and 'not every man is white' (also read as 'some men are not white'), 'no man is white' and 'some man is white'.
The below relations, contrary, subcontrary, subalternation, and superalternation, do hold based on the traditional logic assumption that things stated as S (or things satisfying a statement S in modern logic) exist. If this assumption is taken out, then these relations do not hold.
'Contrary' (medieval: contrariae) statements, are such that both statements cannot be true at the same time. Examples of these are the universal affirmative 'every man is white', and the universal negative 'no man is white'. These cannot be true at the same time. However, these are not contradictories because both of them may be false. For example, it is false that every man is white, since some men are not white. Yet it is also false that no man is white, since there are some white men.
Since every statement has the contradictory opposite (its negation), and since a contradictory is true when its opposite is false, it follows that the opposites of contraries (which the medievals called subcontraries, subcontrariae) can both be true, but they cannot both be false. Since subcontraries are negations of universal statements, they were called 'particular' statements by the medieval logicians.
Another logical relation implied by this, though not mentioned explicitly by Aristotle, is 'alternation' (alternatio), consisting of 'subalternation' and 'superalternation'. Subalternation is a relation between the particular statement and the universal statement of the same quality (affirmative or negative) such that the particular is implied by the universal, while superalternation is a relation between them such that the falsity of the universal (equivalently the negation of the universal) is implied by the falsity of the particular (equivalently the negation of the particular).[3] (The superalternation is the contrapositive of the subalternation.) In these relations, the particular is the subaltern of the universal, which is the particular's superaltern. For example, if 'every man is white' is true, its contrary 'no man is white' is false. Therefore, the contradictory 'some man is white' is true. Similarly the universal 'no man is white' implies the particular 'not every man is white'.[4][5]
In summary:
• Universal statements are contraries: 'every man is just' and 'no man is just' cannot be true together, although one may be true and the other false, and also both may be false (if at least one man is just, and at least one man is not just).
• Particular statements are subcontraries. 'Some man is just' and 'some man is not just' cannot be false together.
• The particular statement of one quality is the subaltern of the universal statement of that same quality, which is the superaltern of the particular statement because in Aristotelian semantics 'every A is B' implies 'some A is B' and 'no A is B' implies 'some A is not B'. Note that modern formal interpretations of English sentences interpret 'every A is B' as 'for any x, a statement that x is A implies a statement that x is B', which does not imply 'some x is A'. This is a matter of semantic interpretation, however, and does not mean, as is sometimes claimed, that Aristotelian logic is 'wrong'.
• The universal affirmative (A) and the particular negative (O) are contradictories. If some A is not B, then not every A is B. Conversely, though this is not the case in modern semantics, it was thought that if every A is not B, some A is not B. This interpretation has caused difficulties (see below). While Aristotle's Greek does not represent the particular negative as 'some A is not B', but as 'not every A is B', someone in his commentary on the Peri hermaneias, renders the particular negative as 'quoddam A nōn est B', literally 'a certain A is not a B', and in all medieval writing on logic it is customary to represent the particular proposition in this way.
These relationships became the basis of a diagram originating with Boethius and used by medieval logicians to classify the logical relationships. The propositions are placed in the four corners of a square, and the relations represented as lines drawn between them, whence the name 'The Square of Opposition'. Therefore, the following cases can be made:[6]
1. If A is true, then E is false, I is true, O is false;
2. If E is true, then A is false, I is false, O is true;
3. If I is true, then E is false, A and O are indeterminate;
4. If O is true, then A is false, E and I are indeterminate;
5. If A is false, then O is true, E and I are indeterminate;
6. If E is false, then I is true, A and O are indeterminate;
7. If I is false, then A is false, E is true, O is true;
8. If O is false, then A is true, E is false, I is true.
To memorize them, the medievals invented the following Latin rhyme:[7]
A adfirmat, negat E, sed universaliter ambae;
I firmat, negat O, sed particulariter ambae.
It affirms that A and E are not neither both true nor both false in each of the above cases. The same applies to I and O. While the first two are universal statements, the couple I / O refers to particular ones.
The Square of Oppositions was used for the categorical inferences described by the Greek philosopher Aristotle: conversion, obversion and contraposition. Each of those three types of categorical inference was applied to the four Boethian logical forms: A, E, I and O.
## The problem of existential import
Subcontraries (I and O), which medieval logicians represented in the form 'quoddam A est B' (some particular A is B) and 'quoddam A non est B' (some particular A is not B) cannot both be false, since their universal contradictory statements (no A is B / every A is B) cannot both be true. This leads to a difficulty firstly identified by Peter Abelard (1079 – 21 April 1142). 'Some A is B' seems to imply 'something is A', in other words, there exists something that is A. For example, 'Some man is white' seems to imply that at least one thing that exists is a man, namely the man who has to be white, if 'some man is white' is true. But, 'some man is not white' also implies that something as a man exists, namely the man who is not white, if the statement 'some man is not white' is true. But Aristotelian logic requires that one of these statements (more generally 'some particular A is B' and 'some particular A is not B') is necessarily true, i.e., they cannot both be false. Therefore, since both statements imply the presence of at least one thing that is a man, the presence of a man or men is followed. But, as Abelard points out in the Dialectica, surely men might not exist?[8]
For with absolutely no man existing, neither the proposition 'every man is a man' is true nor 'some man is not a man'.[9]
Abelard also points out that subcontraries containing subject terms denoting nothing, such as 'a man who is a stone', are both false.
If 'every stone-man is a stone' is true, also its conversion per accidens is true ('some stones are stone-men'). But no stone is a stone-man, because neither this man nor that man etc. is a stone. But also this 'a certain stone-man is not a stone' is false by necessity, since it is impossible to suppose it is true.[10]
Terence Parsons (born 1939) argues that ancient philosophers did not experience the problem of existential import as only the A (universal affirmative) and I (particular affirmative) forms had existential import. (If a statement includes a term such that the statement is false if the term has no instances, i.e., no thing associated with the term exists, then the statement is said to have existential import with respect to that term.)
Affirmatives have existential import, and negatives do not. The ancients thus did not see the incoherence of the square as formulated by Aristotle because there was no incoherence to see.[11]
He goes on to cite a medieval philosopher William of Moerbeke (1215–35 – c. 1286),
In affirmative propositions a term is always asserted to supposit for something. Thus, if it supposits for nothing the proposition is false. However, in negative propositions the assertion is either that the term does not supposit for something or that it supposits for something of which the predicate is truly denied. Thus a negative proposition has two causes of truth.[12]
And points to Boethius' translation of Aristotle's work as giving rise to the mistaken notion that the O form has existential import.
But when Boethius (477 – 524 AD) comments on this text he illustrates Aristotle's doctrine with the now-famous diagram, and he uses the wording 'Some man is not just'. So this must have seemed to him to be a natural equivalent in Latin. It looks odd to us in English, but he wasn't bothered by it.[13]
## Modern squares of opposition
Frege's square of opposition
The conträr below is an erratum:
In the 19th century, George Boole (November 1815 – 8 December 1864) argued for requiring existential import on both terms in particular claims (I and O), but allowing all terms of universal claims (A and E) to lack existential import. This decision made Venn diagrams particularly easy to use for term logic. The square of opposition, under this Boolean set of assumptions, is often called the modern Square of opposition. In the modern square of opposition, A and O claims are contradictories, as are E and I, but all other forms of opposition cease to hold; there are no contraries, subcontraries, subalternations, and superalternations. Thus, from a modern point of view, it often makes sense to talk about 'the' opposition of a claim, rather than insisting, as older logicians did, that a claim has several different opposites, which are in different kinds of opposition with the claim.
Gottlob Frege (8 November 1848 – 26 July 1925)'s Begriffsschrift also presents a square of oppositions, organised in an almost identical manner to the classical square, showing the contradictories, subalternates and contraries between four formulae constructed from universal quantification, negation and implication.
Algirdas Julien Greimas (9 March 1917 – 27 February 1992)' semiotic square was derived from Aristotle's work.
The traditional square of opposition is now often compared with squares based on inner- and outer-negation.[14]
## Logical hexagons and other bi-simplexes
The square of opposition has been extended to a logical hexagon which includes the relationships of six statements. It was discovered independently by both Augustin Sesmat (April 7, 1885 – December 12, 1957) and Robert Blanché (1898–1975).[15] It has been proven that both the square and the hexagon, followed by a "logical cube", belong to a regular series of n-dimensional objects called "logical bi-simplexes of dimension n." The pattern also goes even beyond this.[16]
## Square of opposition (or logical square) and modal logic
The logical square, also called square of opposition or square of Apuleius has its origin in the four marked sentences to be employed in syllogistic reasoning: “Every man is bad,” the universal affirmative - The negation of the universal affirmative “Not every man is bad” (or “Some men are not bad”) - “Some men are bad,” the particular affirmative - and finally, the negation of the particular affirmative “No man is bad”. Robert Blanché published with Vrin his Structures intellectuelles in 1966 and since then many scholars think that the logical square or square of opposition representing four values should be replaced by the logical hexagon which by representing six values is a more potent figure because it has the power to explain more things about logic and natural language.
## Set-theoretical interpretation of categorical statements
In modern mathematical logic, statements containing words "all", "some" and "no", can be stated in terms of set theory. If the set of all A's is labeled as ${\displaystyle s(A)}$ and the set of all B's as ${\displaystyle s(B)}$, then:
• "All A is B" (AaB) is equivalent to "${\displaystyle s(A)}$ is a subset of ${\displaystyle s(B)}$", or ${\displaystyle s(A)\subseteq s(B)}$.
• "No A is B" (AeB) is equivalent to "The intersection of ${\displaystyle s(A)}$ and ${\displaystyle s(B)}$ is empty", or ${\displaystyle s(A)\cap s(B)=\emptyset }$.
• "Some A is B" (AiB) is equivalent to "The intersection of ${\displaystyle s(A)}$ and ${\displaystyle s(B)}$ is not empty", or ${\displaystyle s(A)\cap s(B)\neq \emptyset }$.
• "Some A is not B" (AoB) is equivalent to "${\displaystyle s(A)}$ is not a subset of ${\displaystyle s(B)}$", or ${\displaystyle s(A)\nsubseteq s(B)}$.
By definition, the empty set ${\displaystyle \emptyset }$ is a subset of all sets. From this fact it follows that, according to this mathematical convention, if there are no A's, then the statements "All A is B" and "No A is B" are always true whereas the statements "Some A is B" and "Some A is not B" are always false. This also implies that AaB does not entail AiB, and some of the syllogisms mentioned above are not valid when there are no A's (${\displaystyle s(A)=\emptyset }$).
## References
1. ^ Per The Traditional Square of Opposition: 1.1 The Modern Revision of the Square in the Stanford Encyclopedia of Philosophy
2. ^ Kelley, David (2014). The Art of Reasoning: An Introduction to Logic and Critical Thinking (4 ed.). New York, NY: W.W. Norton & Company, Inc. p. 150. ISBN 978-0-393-93078-8.
3. ^ "Introduction to Logic - 7.2.1 Finishing the Square and Immediate Inferences". 2021-08-10.{{cite web}}: CS1 maint: url-status (link)
4. ^ Parry & Hacker, Aristotelian Logic (SUNY Press, 1990), p. 158.
5. ^ Cohen & Nagel, Introduction to Logic Second Edition (Hackett Publishing, 1993), p. 55.
6. ^ Reale, Giovanni; Antiseri, Dario (1983). Il pensiero occidentale dalle origini a oggi. Vol. 1. Brescia: Editrice La Scuola. p. 356. ISBN 88-350-7271-9. OCLC 971192154.
7. ^ Massaro, Domenico (2005). Questioni di verità: logica di base per capire e farsi capire. Script (in Italian). Vol. 2. Maples: Liguori Editore Srl. p. 58. ISBN 9788820738921. LCCN 2006350806. OCLC 263451944.
8. ^ In his Dialectica, and in his commentary on the Perihermaneias
9. ^ Re enim hominis prorsus non existente neque ea vera est quae ait: omnis homo est homo, nec ea quae proponit: quidam homo non est homo
10. ^ Si enim vera est: Omnis homo qui lapis est, est lapis, et eius conversa per accidens vera est: Quidam lapis est homo qui est lapis. Sed nullus lapis est homo qui est lapis, quia neque hic neque ille etc. Sed et illam: Quidam homo qui est lapis, non est lapis, falsam esse necesse est, cum impossibile ponat
11. ^ in The Traditional Square of Opposition in the Stanford Encyclopedia of Philosophy
12. ^ (SL I.72) Loux 1974, 206
13. ^ The Traditional Square of Opposition
14. ^ Westerståhl, 'Classical vs. modern squares of opposition, and beyond', in Beziau and Payette (eds.), The Square of Opposition: A General Framework for Cognition, Peter Lang, Bern, 195-229.
15. ^ N-Opposition Theory Logical hexagon
16. ^ Moretti, Pellissier
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2022-10-04 07:10:32
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https://math.stackexchange.com/questions/2674756/charpit-method-non-linear-pde
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# Charpit method: non-linear PDE
I have a question: $$p^{2}x+q^{2}y = z.$$
I formed the Charpit auxiliary equation as follows
$$\frac{\mathrm{d}x}{2px} = \frac{\mathrm{d}y}{2py} = \frac{\mathrm{d}z}{2(p^2x + q^2y)} = \frac{\mathrm{d}p}{p-p^2} = \frac{\mathrm{d}q}{q-q^2}.$$
After forming the equation I was unable to solve further (I applied everything I was taught). So I did some research (checked several books, and on the website as well) and found the next step to be
$$\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}. \tag{1}$$
After which I was able to solve. But I can not understand how they derived the relation (1). Can someone explain what method they applied here?
Disclaimer: The course we are being taught is engineering mathematics.
Charpit's equation: $$\boxed{\frac{dx}{2px} = \frac{dy}{\color{red}{2qy}} = \frac{dz}{2(p^2x + q^2y)} = \frac{dp}{p-p^2} = \frac{dq}{q-q^2}}$$ We have from Charpit's equation $$\frac{dy}{2qy} =\frac{dq}{q-q^2}$$ Rearranging terms $$(q-q^2){dy} =2qy{dq}$$ $$q{dy} =2qy{dq}+q^2dy$$ $$\text { (1) }\frac{dy}{qy} =\frac {2qy{dq}+q^2dy}{q^2y}$$ We have also from Charpit's equation that : $$\frac{dx}{2px} = \frac{dp}{p-p^2}$$ $${dx}{(p-p^2)} = 2px{dp}$$ $${pdx} = 2px{dp}+p^2dx$$ $$\text { (2) }\frac {dx}{px} = \frac {2px{dp}+p^2dx} {p^2x}$$ We use a third equality from Charpit's equation $$\frac {dx}{2px}=\frac{dy}{2qy} \implies \frac {dx}{px}=\frac{dy}{qy}$$ Therefore $$\boxed{\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}}$$ That you can easily integrate...
Note that the middle term is equal to $\frac{dz}{2z}$ and use it as the reference value. Then $$\frac{p^2dx}{p^2x}=\frac{pdz}{z},\quad \frac{2px\,dp}{p^2x}=\frac{2dp}{p}=\frac{(1-p)\,dz}{z} \implies \frac{d(p^2x)}{p^2x}=\frac{dz}{z}$$ and similarly for $q^2y$.
Some other obvious simple steps are to recognize the decoupled nature of $$\frac{dz}{2z}=-\frac{d(1/p)}{1/p-1}=-\frac{d(1/q)}{1/q-1}$$ which integrates to $$\frac12\ln|z|=c_1-\ln|1/p-1|=c_2-\ln|1/q-1| \\\iff\\ \sqrt{|z|}=C_1\frac{p}{1-p}=C_2\frac{q}{1-q}$$ Then insert $$p=\frac{\sqrt{|z|}}{C_1+\sqrt{|z|}},\qquad q=\frac{\sqrt{|z|}}{C_2+\sqrt{|z|}}$$ into the first three relations $$\frac{dx}{2xp}=\frac{dy}{2yq}=\frac{dz}{2z}$$ which should then be easy to integrate.
Or alternatively, combine the $x$ and $p$ fractions to $$\frac{dx}{x}=\frac{2dp}{1-p}\implies x(1-p)^2=a,$$ and similarly $$\frac{dy}{y}=\frac{2dq}{1-q}\implies y(1-q)^2=b.$$
• Can you explain as to how we got the relation (1) mentioned in the question. Also what do you mean by decoupled nature? Please explain, knowing that I have a undergraduate understanding of PDEs. – Mohammed Arshaan Mar 3 '18 at 12:08
• Also what method is being applied here to transform the auxiliary equations? As far as I can see they are not simple addition, subtraction, multiplication or division operations... – Mohammed Arshaan Mar 3 '18 at 12:09
• The first transformations are exactly that, simple arithmetic operations. Note that $p-p^2=p(1-p)$. As both expressions in (1) are equal to $\frac{dz}{z}$, they are also equal to each other. – LutzL Mar 3 '18 at 12:16
• Or did you question $d(p^2x)=p^2\,dx+2px\,dp$? – LutzL Mar 3 '18 at 12:16
• No. What I mean is that in relation (1) (marked in my question). How did we get the following relation: $$\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}. \tag{1}$$ – Mohammed Arshaan Mar 3 '18 at 12:18
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2019-07-23 06:49:16
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https://www.physicsforums.com/threads/is-it-possible-to-determine-a-basis-for-non-subspace-spaces.786569/
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# Is it possible to determine a basis for non-subspace spaces?
Tags:
1. Dec 9, 2014
### Kubilay Yazoglu
The title may seem a little confusing and possibly stupid :D
What I mean is like a plane doesn't go through the origin? Can we describe a basis for this? If so, how?
2. Dec 9, 2014
### Staff: Mentor
Bases are defined for vector spaces. If you do not have a vector space, the concept of a basis is meaningless.
3. Dec 10, 2014
### Stephen Tashi
A plane not through the origin is an example of a "coset" in a vector space. A coset of a subspace is the subspace translated by a constant vector. If the coset C of a vector space V doesn't contain the zero vector of V then C can't be a vector space with respect to the operations used in V.
A set of things may not be a vector space under one set of operations and yet can become a vector space under a different set of operations. You can define a set of operations on a coset C in a vector space V that make the coset a vector space. Assume the coset is fomed by adding the vector b to each vector in the subspace S. You can define a new set of operations on C. The basic idea is to un-translate vectors involved in the operations by subtracting b from them. Do the operations on the un-translated vectors with the operations used in V. Then translate the result by adding b to it. A coset with this new set of operations would be a vector space and it could have a basis. However the coset would not be a subspace of the vector space V because subspace of a vector space V must satisfy properties that specify using exactly same operations that are used in V.
4. Dec 11, 2014
### WWGD
Maybe you want to translate so that the vector space goes to the origin, find a base and translate. Then the space is generated by a linear combination plus translation.
5. Dec 11, 2014
### Kubilay Yazoglu
6. Dec 11, 2014
### Kubilay Yazoglu
Can you determine a basis for the plane x − y + z = 2 that spans only the plane?
This is the original question. I forgot the "only" part. Now, is this possible? Can anyone solve please?
7. Dec 11, 2014
### Stephen Tashi
You can't solve the problem that you stated. The word "span" has a technical definition. Any basis for $R^3$ includes the zero vector in it's span. The span of a set of vectors includes the linear combination of those vectors whose coefficients are each the zero scalar. Such a linear combination is equal to the zero vector. So you can't exclude the zero vector from the "span" of a set of vectors.
Perhaps what you want is a representation of the plane in the form $(x,y,z) = (x_0,y_0,z_0) + c_1( x_1,y_1,z_1) + c_2(x_2,y_2,z_2)$. The form of this equation prohibits the coefficient of the vector $(x_0,y_0,z_0)$ from being zero. So the equation does not define the "span" of the set of vectors $(x_i, y_i, z_i)$.
8. Dec 11, 2014
### Kubilay Yazoglu
Yes! That is exactly what I've been thinking! But my friends almost convinced me :) Thank you!
9. Dec 11, 2014
### HallsofIvy
What you can do, and this is equivalent to what Stephen Tashi said, is look at the plane through the origin, parallel to the given plane. For your example, x- y+ z= 0. That is equivalent to y= x+ z so any vector in it is of the form <x, x+ z, z>= <x, x, 0>+ <0, z, z>= x<1, 1, 0>+ z<0, 1, 1>. That is, any vector can be written as a linear combination of the two vectors < 1, 1, 0> and <0, 1, 1> so they form a basis for the subspace.
Now, one point on the plane x- y+ z= 2 (a plane but NOT a subspace as others have said) is (0, 0, 2) so one vector in the set (not subspace) of vectors representing that plane can be written as the sum of (0, 0, 2) (or any other single vector in the plane) plus a linear combination of <1, 1, 0> and <0, 1, 1>.
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2018-06-21 10:42:09
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http://nrich.maths.org/public/leg.php?code=-68&cl=2&cldcmpid=1112
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# Search by Topic
#### Resources tagged with Visualising similar to Treasure Island:
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##### Other tags that relate to Treasure Island
Coordinates - first quadrant. Visualising. thoughtful. Interactivities.
### There are 262 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Visualising
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Frogs
##### Stage: 2 and 3 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### On the Edge
##### Stage: 3 Challenge Level:
Here are four tiles. They can be arranged in a 2 by 2 square so that this large square has a green edge. If the tiles are moved around, we can make a 2 by 2 square with a blue edge... Now try to. . . .
### Semi-regular Tessellations
##### Stage: 3 Challenge Level:
Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?
### Square It
##### Stage: 1, 2, 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Constructing Triangles
##### Stage: 3 Challenge Level:
Generate three random numbers to determine the side lengths of a triangle. What triangles can you draw?
### Eight Hidden Squares
##### Stage: 2 and 3 Challenge Level:
On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares?
### Fred the Class Robot
##### Stage: 2 Challenge Level:
Billy's class had a robot called Fred who could draw with chalk held underneath him. What shapes did the pupils make Fred draw?
### Triangles to Tetrahedra
##### Stage: 3 Challenge Level:
Starting with four different triangles, imagine you have an unlimited number of each type. How many different tetrahedra can you make? Convince us you have found them all.
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Multiplication Series: Illustrating Number Properties with Arrays
##### Stage: 1 and 2
This article for teachers describes how modelling number properties involving multiplication using an array of objects not only allows children to represent their thinking with concrete materials,. . . .
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Right or Left?
##### Stage: 2 Challenge Level:
Which of these dice are right-handed and which are left-handed?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### Put Yourself in a Box
##### Stage: 2 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### Seven Squares - Group-worthy Task
##### Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Coordinate Patterns
##### Stage: 3 Challenge Level:
Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead?
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### World of Tan 8 - Sports Car
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### Ten Hidden Squares
##### Stage: 2 Challenge Level:
These points all mark the vertices (corners) of ten hidden squares. Can you find the 10 hidden squares?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### World of Tan 9 - Animals
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### The Development of Spatial and Geometric Thinking: the Importance of Instruction.
##### Stage: 1 and 2
This article looks at levels of geometric thinking and the types of activities required to develop this thinking.
### Cuboids
##### Stage: 3 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Bands and Bridges: Bringing Topology Back
##### Stage: 2 and 3
Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology.
### Buses
##### Stage: 3 Challenge Level:
A bus route has a total duration of 40 minutes. Every 10 minutes, two buses set out, one from each end. How many buses will one bus meet on its way from one end to the other end?
### Making Tangrams
##### Stage: 2 Challenge Level:
Here's a simple way to make a Tangram without any measuring or ruling lines.
### Turning Cogs
##### Stage: 2 Challenge Level:
What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same.
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Travelling Salesman
##### Stage: 3 Challenge Level:
A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs?
### Weighty Problem
##### Stage: 3 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### Icosagram
##### Stage: 3 Challenge Level:
Draw a pentagon with all the diagonals. This is called a pentagram. How many diagonals are there? How many diagonals are there in a hexagram, heptagram, ... Does any pattern occur when looking at. . . .
### Sponge Sections
##### Stage: 2 Challenge Level:
You have been given three shapes made out of sponge: a sphere, a cylinder and a cone. Your challenge is to find out how to cut them to make different shapes for printing.
### Rolling Triangle
##### Stage: 3 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
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2017-01-21 13:19:12
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https://rmostowy.github.io/covid-19/methodology/
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# Real-time analysis of the COVID-19 pandemic in Poland
## Methodology
The underlying epidemiological model is based on the classical SEIR approach with the important difference that it explicitly considers people who are admitted to the hospital, the ICU and those who die from infection. The exact model is a deterministic version of the epidemiological model considered in the Report 9 of the MRC Centre for Global Infectious Disease Analysis at Imperial College London [1]. The model, summarised in the Figure below, makes several important simplifications/assumptions. First, it does not consider the population demography and assumes homogeneous mixing. Second, tt assumes that all COVID-related deaths occur at ICU and are reported immediately. I also assume that quarantined, hospitalised and ICU patients are isolated well enough to not be infectious anymore.
Figure. Summary of the epidemiological model used. Susceptibles (S) become exposed (E) when they get in touch with an infectious person, either asymptomatic ($$I_A$$) or symptomatic ($$I_S$$), at a rate $$\beta(I_A+\lambda I_S)$$, where $$\beta$$ is a transmission rate and $$\lambda$$ reflects relative contribution of a symptomatic person to transmission compared to an asymptomatic person. An asymptomatically infected person can then develop symptoms with a probability $$f$$ at a rate $$a$$ or otherwise recover (R) at a rate $$\gamma$$. A symptomatic person can become seriously enough ill to undergo self-isolation (Q) with a probability $$q$$ at a rate $$b$$, such that $$1/a+1/b=1/\gamma$$. A proportion $$\epsilon_1$$ of quarantined individuals become hospitalised (H) at a rate $$\omega_1$$, otherwise recover at a rate $$(1-\epsilon_1)\omega_1$$. A proportion $$\epsilon_2$$ of hospitalised individuals become admitted to the ICU (V) at a rate $$\omega_2$$, otherwise recover at a rate $$(1-\epsilon_2)\omega_2$$. A proportion $$\epsilon_3$$ of ICU patients die (D) at a rate $$\omega_3$$, otherwise recover at a rate $$(1-\epsilon_3)\omega_3$$. In this model, $$\beta=R_0\gamma/N\frac{1+fa/b}{1+\lambda fa/b}$$, where $$R_0$$ is the basic reproduction number and $$N$$ is the population size. The epidemic begins at time $$t_{case}-T_{seed}$$, where $$t_{case}$$ is the date of the first reported case (04/03) and $$T_\text{seed}$$ is the period of time between the first infection and the first actual case. Governmental lockdowns were introduced between March 12th and 16th, and I assume 16/03 to be the date of lockdown introductiona. After this the transmission of the virus is assumed to be $$R=\kappa R_0$$, where $$\kappa\in[0,1]$$ reflects the relative reduction in $$R_0$$ due to the lockdowns and $$R$$ is the effective transmission coefficient.
I used the existing literature to make assumptions about the parameters used in the model above.
Parameter Explanation Value Comment Source
$$N$$ Population size of Poland 38 386 000 [2]
$$1/\sigma$$ latent period (average time from exposure to infectiousness) 4.6 days [1]
$$1/a$$ average time from infectiousness to onset of symptoms 0.5 day $$1/\sigma+1/a$$ is incubation period of 5.1 days [1,3,4]
$$1/b$$ average time from onset to recovery/quarantine 1.4 days based on assumed generation time of 6.5 days [1]
$$1/\gamma$$ mean infectious period 1.9 days equals to $$1/a+1/b$$ [1]
$$f$$ Proportion of infected patients who become symptomatic 80% [1]
$$q$$ Proportion of cases which are sufficiently symptomatic to self-isolate 2/3 [1]
$$1/\omega_1$$ Average time spent in quarantine 3.6 days Assumes mean time of 5 days from onset to hospital admission [1]
$$1/\omega_2$$ Average time spent in hospital 8 days [1]
$$1/\omega_3$$ Average time spent at ICU 8 days Assumes mean time of 21 days from onset to death/discharge [1,6]
$$\epsilon_1$$ Proportion of quarantined cases requiring hospitalisation 6.6% Assumes that 4.4% of symptomatic infections are hospitalised [1]
$$\epsilon_2$$ Proportion of hospitalised cases that end up at ICU 30% [1]
$$\epsilon_3$$ Proportion ICU cases that result in death 57% Tailored to reflect the infection fatality-rate of about 0.9% [1,5]
$$\lambda$$ Relative transmission of a symptomatic person compared to asymptomatic person 2 Assumes that symptomatic people are twice as infectious as asymptomatic ones [1]
$$R_0$$ Basic reproduction number (see here for explanation) Estimated
$$\kappa$$ Efficacy of governmental lockdowns Estimated
$$T_\text{seed}$$ Time from seeding the epidemic to the first reported case Estimated
Table 1. Parameters of the COVID-19 transmission model. The model is fit to the data using a maximum-likelihood approach by comparing the predicted to the reported number of deaths. The daily number of deaths is assumed to be Poisson distributed. Using the parameter values assumed above, we estimate the values of $$R_0$$, $$T_\text{seed}$$ and $$\kappa$$ that best explain the observed data given the assumed model.
## References
[1] Neil M Ferguson, Daniel Laydon, Gemma Nedjati-Gilani et al. Impact of non-pharmaceutical interventions (NPIs) to reduce COVID-19 mortality and healthcare demand. Imperial College London (16-03-2020), doi:https://doi.org/10.25561/77482. LINK
[2] Statistics Poland, Demographic Surveys Department. Population. Size and structure and vital statistics in Poland by territorial division in 2019 as of 30th June. LINK
[3] Li, Qun, et al. “Early transmission dynamics in Wuhan, China, of novel coronavirus–infected pneumonia.” New England Journal of Medicine (2020). LINK
[4] Linton, Natalie M., et al. “Incubation period and other epidemiological characteristics of 2019 novel coronavirus infections with right truncation: a statistical analysis of publicly available case data.” Journal of clinical medicine 9.2 (2020): 538. LINK
[5] Verity, Robert, et al. “Estimates of the severity of coronavirus disease 2019: a model-based analysis.” The Lancet infectious diseases (2020). LINK
[6] Zhou, Fei, et al. “Clinical course and risk factors for mortality of adult inpatients with COVID-19 in Wuhan, China: a retrospective cohort study.” The Lancet (2020). LINK
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2022-08-16 18:39:15
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http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=jsfu&paperid=544&option_lang=eng
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J. Sib. Fed. Univ. Math. Phys.: Year: Volume: Issue: Page: Find
J. Sib. Fed. Univ. Math. Phys., 2017, Volume 10, Issue 2, Pages 233–238 (Mi jsfu544)
Calculation of density of states for FeAs-based superconductors
Vladimir A. Kashurnikov, Andrey V. Krasavin, Yaroslav V. Zhumagulov
National Research Nuclear University MEPhI, Kashirskoe shosse, 31, Moscow, 115409, Russia
Abstract: The spectral and the total density of states were calculated for two-dimensional FeAs-clusters within the limits of the two-orbital model, which is widely used for modeling iron-based superconductors. The spectra were restored by means of an asymptotically exact stochastic procedure, which was modified to restore the kernel of the integral equation relating the Matsubara Green's function and the spectral density. The data for Matsubara Green's function were obtained with the use of the generalized quantum world-line Monte Carlo algorithm adapted for the two-orbital model. The calculations were made for clusters with sizes up to $10\times10$ FeAs-cells. The data are presented for the distribution profiles along the main crystallographic directions and for the entire Brillouin zone. The analysis of the doped state revealed differences in the electron and hole states of the system that is correlated with known experimental data.
Keywords: FeAs-based superconductors, quantum Monte-Carlo algorithm, density of states, two-orbital model.
Funding Agency Grant Number Russian Foundation for Basic Research 14-08-0050915-02-02764_à The work was supported by Russian Found for Basic Research (projects # 14-08-00509 and # 15-02-02764).
DOI: https://doi.org/10.17516/1997-1397-2017-10-2-233-238
Full text: PDF file (1052 kB)
References: PDF file HTML file
Bibliographic databases:
Document Type: Article
UDC: 517.9
Accepted: 28.02.2017
Language: English
Citation: Vladimir A. Kashurnikov, Andrey V. Krasavin, Yaroslav V. Zhumagulov, “Calculation of density of states for FeAs-based superconductors”, J. Sib. Fed. Univ. Math. Phys., 10:2 (2017), 233–238
Citation in format AMSBIB
\Bibitem{KasKraZhu17} \by Vladimir~A.~Kashurnikov, Andrey~V.~Krasavin, Yaroslav~V.~Zhumagulov \paper Calculation of density of states for FeAs-based superconductors \jour J. Sib. Fed. Univ. Math. Phys. \yr 2017 \vol 10 \issue 2 \pages 233--238 \mathnet{http://mi.mathnet.ru/jsfu544} \crossref{https://doi.org/10.17516/1997-1397-2017-10-2-233-238} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000412014600011}
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2019-06-18 20:53:12
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https://mymusing.co/string-search-algorithms/
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## Introduction
In this article, we’ll show several algorithms for searching for a pattern in a text. The fundamental string searching (matching) problem is defined as follows: given two strings – a text and a pattern, determine whether the pattern appears in the text.
## Naive Method
For every position in the text, consider it a starting position of the pattern and see if you get a match.
// Naive Pattern Searching algorithm
// Number of comparisons in the worst case is O(m*(n-m+1))
void brute_force(char* pat, char* text)
{
int M = strlen(pat);
int N = strlen(text);
// A loop to slide pat[] one by one
for (int i = 0; i <= N - M; i++) {
int j;
// For current index i, check for pattern match
for (j = 0; j < M; j++) {
if (text[i + j] != pat[j]){
break;
}
}
// Pattern found
if (j == M){
cout << "Pattern found at index " << i << endl;
}
}
}
## Rabin Karp Algorithm
As mentioned above, naive search algorithm is very inefficient when patterns are long and when there is a lot of repeated elements of the pattern. Searching can be efficient using Rabin Karp Algorithm. This is actually the “naive” approach augmented with hash function.
The Rabin-Karp algorithm makes use of hash functions and the rolling hash technique. A rolling hash allows an algorithm to calculate a hash value without having the rehash the entire string. For example, when searching for a word in a text, as the algorithm shifts one letter to the right instead of having to calculate the hash of the section of text[i:j] as it shifts from text[i-1:j-1], the algorithm can use a rolling hash to do an operation on the hash to get the new hash from the old hash. Hashing can map data of arbitrary size to a value of fixed size. In general the idea seems quite simple, the only thing is that we need a hash function that gives different hashes for different sub-strings. This figure shows algorithm in action.
The Rabin-Karp Rolling Hash (H) is given by
H = c1a(k-1) + c2a(k-2) + c3a(k-3) + …. cka(0)
a is a constant, c1, .. ck are the input characters, k is the number of characters there are in the string we are comparing (this is the length of the word).
All the operations are done modulo a prime number M to avoid dealing with large numbers. For calculating hash we can imagine a window sliding over all the substrings in text. Calculating the hash value of the next substring only inspects two elements: the element leaving the window and the element entering the window. Finding the hash value of the next substring is now a O(1) operation. Hash function for base b and substring length L is given by
To overcome large number and range overflow, instead of the number H itself we use its remainder when divided by M. Applying the basic rules of modular arithmetic to the above expression:
A + B = C => (A % M + B % M) % M = C % M
A * B = C => ((A % M) * (B % M)) % M = C % M
A – B = C => (A % M – B % M + k * M) % M = C % M
To avoid the mapping of two or more different strings to the same number (it is called a collision), choose M and a such that they are sufficiently large and are prime numbers.
Let’s use Rabin-Karp’s rolling hash to hash the alphabet. Here, a will be 26, k will be 3, and c will represent the place in the alphabet where the character appears — so for “a”, c will be 1, for “z” c will be 26.
H("abc") = 1 * 26^2 + 2 * 26^1 + 3 * 26^0
// Rabin Karp Algorithm
void RabinKarp(char pat[], char text[], int M)
{
int M = strlen(pat);
int N = strlen(text);
int i, j;
int patHash = 0; // hash value for pattern
int txtHash = 0; // hash value for text
int H = 1;
int a = 256, // Number of characters in the input alphabet
// The value of H would be "pow(a, M-1)%M"
for (i = 0; i < M-1; i++)
H = (H*a)%M;
// Calculate the hash value of pattern and first window of pattern
for (i = 0; i < M; i++)
{
patHash = (a*patHash + pat[i])%M;
txtHash = (a*txtHash + text[i])%M;
}
// Slide the pattern over text one by one
for (i = 0; i <= N - M; i++)
{
// Check the hash values of current window of text and pattern.
if ( patHash == txtHash )
{
// Check for characters on by one for suprious match
for (j = 0; j < M; j++)
{
if (text[i+j] != pat[j])
break;
}
if (j == M)
printf("Pattern found at index %a \n", i);
}
// Calculate hash value for next window of text
}
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2023-03-21 23:19:18
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https://gmatclub.com/forum/if-a-set-of-numbers-consists-of-1-4-and-1-6-what-number-can-be-added-253019.html
|
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# If a set of numbers consists of 1/4 and 1/6, what number can be added
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If a set of numbers consists of 1/4 and 1/6, what number can be added [#permalink]
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06 Nov 2017, 21:33
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If a set of numbers consists of 1/4 and 1/6, what number can be added to the set to make the average (arithmetic mean) also equal to 1/4?
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
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Re: If a set of numbers consists of 1/4 and 1/6, what number can be added [#permalink]
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06 Nov 2017, 23:01
1
1
Lets say we add a number 'x'. Now average of 1/6, 1/4 and x is 1/4. This means their sum is 3* 1/4 = 3/4
So, 1/6 + 1/4 + x = 3/4 .. This gives x = 3/4 - 1/6 - 1/4 = 1/3
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Re: If a set of numbers consists of 1/4 and 1/6, what number can be added [#permalink]
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07 Nov 2017, 04:52
On the number line we need to add a number that nullifies the distance between 1/6 and the average, 1/4
So, 1/4 - 1/6 = 3/12 - 2/12 = 1/12, which is the distance between 1/4 and 1/6
Now we need to add the distance to the average to counter the effect of 1/6:
1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3
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Re: If a set of numbers consists of 1/4 and 1/6, what number can be added [#permalink]
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07 Nov 2017, 08:21
Bunuel wrote:
If a set of numbers consists of 1/4 and 1/6, what number can be added to the set to make the average (arithmetic mean) also equal to 1/4?
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
The average of 3 numbers must be $$\frac{1}{4}$$
Since one of the 3 numbers is equal to the mean,
the average of the other two numbers should also be 1/4
If the third number is x, (1/6 + x)/2 = 1/4 => 1/6 + x = 1/2 => x = 1/2 - 1/6 = 2/6 = 1/3(Option D)
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Re: If a set of numbers consists of 1/4 and 1/6, what number can be added [#permalink]
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08 Nov 2017, 16:39
Bunuel wrote:
If a set of numbers consists of 1/4 and 1/6, what number can be added to the set to make the average (arithmetic mean) also equal to 1/4?
A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2
We’ll use the formula average = sum/quantity. We can let n = the added number:
(1/4 + 1/6 + n)/3 = 1/4
1/4 + 1/6 + n = 3/4
Multiplying the entire equation by 12, we have:
3 + 2 + 12n = 9
12n = 4
n = 1/3
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Re: If a set of numbers consists of 1/4 and 1/6, what number can be added &nbs [#permalink] 08 Nov 2017, 16:39
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2018-11-19 19:55:19
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https://chemistry.stackexchange.com/questions/19799/strong-and-weak-acid
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Strong and weak acid
I know sulfuric acid is a strong acid and acetic acid is a weak acid. However when I write them in their chemical symbol.
Sulfuric acid: $\ce{H2SO4}$
Acetic acid: $\ce{CH3COOH}$
I realise that acetic acid has 4 atoms of hydrogen whereas sulfuric acid has only 2 atoms of hydrogen. So, shouldn't acetic acid be stronger than sulfuric acid?
Looking at the structures of those two acids might help. In both cases, the acidic hydrogens are part of hydroxyl groups. Because of polarity and the inductive effect, the O-H bond is weaker than C-H bonds which are very strong. As a result, it's nearly impossible for Carbon to lose a hydrogen without some kind of elimination reaction, whereas the O-H bond can autoprotolyze with other O-H bonds (even water). The weakness of the O-H bond is the basis for the acidity of those hydrogens. C-H bonds are essentially non-acidic, having pKa values around 45-50.
The pKa of an acid tells you how strong the acid is.
$pKa=-log([H^+][A^-]/[HA])=-log(Ka)$ where $A$ is the conjugate base of the acid (the species left when the proton dissociates). A low value for $pKa$ gives the stronger acid since a high value for $Ka$ indicates that the equilibrium $\ce{HA<=>H+ + A-}$ lies well to the right.
There are a number of factors that make an acid's equilibrium position lie far to the right. By far the most important is the stability of the conjugate base ($A^-$). In your example, the reason $\ce{H2SO4}$ so much more acidic than $\ce{CH3COOH}$ because is can form more resonance structures than ethanoic acid so it distributes the negative charge better. Also the inductive effect of the alkyl group destabilizes the negative conjugate base. There are more reasons but this is the most important.
Sulfuric acid is a strong acid; when placed in water, it completely dissociates into $H^+$ and $HSO_4^-$. While acetic acid dissociates into $H^+$ and $CH_3COO^-$; but being a weak acid, some of the molecules remain as $CH_3COOH$.
The difference between a strong and a weak acid is not necessarily their level of acidity; it is the degree to which they dissociate. Also as the previous answerer said, due to the nature of the bond of the 3 hydrogen molecules in acetate, we wouldn't consider it an acid because they don't break into $H^+$ ions as easily.
• Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. – Philipp Nov 21 '14 at 5:17
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2019-10-17 14:31:51
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https://biz.libretexts.org/Bookshelves/Accounting/Book%3A_Financial_Accounting/08%3A_How_Does_a_Company_Gather_Information_about_Its_Inventory%3F/8.01%3A_Determining_and_Reporting_the_Cost_of_Inventory
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# 8.1: Determining and Reporting the Cost of Inventory
Learning Objectives
At the end of this section, students should be able to meet the following objectives:
1. Understand that inventory is recorded initially at its historical cost.
2. Provide the guiding rule for identifying expenditures and other costs that must be capitalized in the reporting of inventory.
3. Explain the rationale for offering a cash discount for payments made within a specified period of time as well as the accounting for such cost reductions.
Question: The asset section of the February 28, 2009, balance sheet produced by Best Buy Co. Inc. reports net accounts receivable of $1.868 billion. Based on discussions in the previous chapter, a decision maker should know that this figure reflects net realizable value—the estimation by officials of the amount of cash that will be collected from the receivables owed to the company by its customers. Knowledge of financial accounting rules allows an individual to understand the information being conveyed in a set of financial statements. As is common, the next account that appears on Best Buy’s balance sheet is inventory, all the items held on that date that were acquired for sales purposes—televisions, cameras, computers, and the like. The figure disclosed by the company for this asset is$4.753 billion. Does this balance also indicate net realizable value—the cash expected to be generated from the company’s merchandise—or is different information reflected?
On a balance sheet, what does the amount reported for inventory represent?
Answer: The challenge of analyzing the various assets reported by an organization would be reduced substantially if every monetary number disclosed the same basic information, such as net realizable value. However, over the decades, virtually every asset has come to have its own individualized method of reporting, one created to address the special peculiarities of that account. Thus, the term “presented fairly” often has a totally different meaning for each asset. Reporting accounts receivables, for example, at net realizable value has no impact on the approach that has come to be accepted for inventory.
The reporting of inventory is especially unique because the reported balance is not as standardized as with accounts receivable. For example, under certain circumstances, the balance sheet amount shown for inventory actually can reflect net realizable value. Several other meanings for the reported balance, though, are more likely. The range of accounting alternatives encountered in analyzing this asset emphasizes the importance of reading the notes included with financial statements rather than fixating on a few reported numbers alone. Without careful study of the additional disclosures, a decision maker simply cannot know what Best Buy means by the $4.753 billion figure reported for “merchandise inventories.” Another company could show the identical number for its inventory and still be reporting considerably different information. Question: Accounting for inventory seems particularly complicated. A logical approach to the coverage here is needed. In coming to understand the reporting methodology that is utilized with this asset, where should the discussion begin? What is the first issue that an accountant faces in establishing an appropriate balance for inventory so that it is reported in conformity with U.S. GAAP? Answer: The study of inventory and its financial reporting should begin by defining “cost.” In acquiring each item, officials make the decision to allocate a certain amount of scarce resources. What did the company expend to obtain its inventory? That is a reasonable question to address. To illustrate, assume that a sporting goods company (Rider Inc.) acquires a new bicycle (Model XY-7) to sell. Rider’s accounting system should be designed to determine the cost of this piece of inventory, the sacrifice that the company chose to make to obtain the asset. Assume that a price of$250 was charged by the manufacturer (Builder Company) for the bicycle and the purchase was made by Rider on credit. Rider spends another $9 to transport the item from the factory to one of its retail stores and$6 to have the pieces assembled so that the bicycle can be displayed in the salesroom for customers to examine.
In accounting for the acquisition of inventory, cost includes all normal and necessary amounts incurred to get the item into the condition and position to be sold. Hence, by the time this bicycle has reached Rider’s retail location and been readied for sale, its cost to the sporting goods company is $265. Figure 8.1 Maintaining a Cost for Inventory Item The charges for delivering this merchandise and assembling the parts were included in the cost of the asset (the traditional term for adding a cost to an asset account, capitalization, was introduced previously). Both of these expenditures were properly viewed as normal and necessary to get the bicycle into the condition and position to be resold. Interestingly, any amount later expended to transport the merchandise from the store to a buying customer is recorded as an expense rather than as an asset because that cost is incurred after the sale takes place. At that point, no further future value exists since the merchandise has already been sold. Occasionally, costs arise where the “normal and necessary” standard may be difficult to apply. To illustrate, assume that the president of a store that sells antiques buys a 120-year-old table for resell purposes. When the table arrives at the store, another$300 must be spent to fix a scratch cut across its surface. Should this added cost be capitalized (added to the reported balance for inventory) or expensed? The answer to this question is not readily apparent and depends on ascertaining all relevant facts. Here are two possibilities.
Scenario one: The table was acquired by the president with the knowledge that the scratch already existed and needed to be fixed prior to offering the merchandise for sale. In that case, repair is a normal and necessary activity to put the table into condition necessary to be sold. The $300 is capitalized, recorded as an addition to the cost of the inventory. Scenario two: The table was bought without the scratch but was damaged when first moved into the store through an act of employee carelessness. The table must be repaired but the scratch was neither normal nor necessary. This cost could have been avoided. The$300 is not capitalized but rather reported as a repair expense by the store.
As discussed in an earlier chapter, if the accountant cannot make a reasonable determination as to whether a particular cost qualifies as normal and necessary, the conservatism principle that underlies financial accounting requires the $300 to be reported as an expense. When in doubt, the alternative that makes reported figures look best is avoided so that decision makers are not encouraged to be overly optimistic about the company’s financial health and future prospects. ## Exercise Link to multiple-choice question for practice purposes: http://www.quia.com/quiz/2092919.html Question: When inventory is acquired, some sellers are willing to accept a reduced amount to encourage fast payment—an offer that is called a cash discount (or a sales discount or purchases discount depending on whether the seller or the buyer is making the entry). Cash becomes available sooner so that the seller can quickly put it back into circulation to make more profits. In addition, the possibility that a receivable will become uncollectible is reduced if the balance due is not allowed to get too old. Tempting buyers to make quick payments to reduce their cost is viewed as a smart business practice by many sellers. To illustrate, assume the invoice received by the sporting goods company (Rider) for the above bicycle indicates the proper$250 balance due but also includes the notation: 2/10, n/45. What message is being conveyed by the seller?
How do cash discounts impact the reporting of inventory?
Answer: Sellers—such as Builder Company in this example—can offer a wide variety of discount terms to encourage speedy payment. One such as 2/10, n/45 is generally read “two ten, net 45.” It informs the buyer that a 2 percent discount can be taken if the invoice is paid by the tenth day. Any net amount that remains unpaid (after merchandise returns or partial cash payments) is due on the forty-fifth day. Rider has the option to pay $245 for the bicycle within ten days of receiving the invoice by taking advantage of the$5 discount ($250 × 0.02). Or the sporting goods company can wait until the forty-fifth day but then is responsible for the entire$250.
Many companies automatically take advantage of these discounts as a matter of policy because of the high rate of interest earned. If Rider does not submit the money in ten days, it must pay an extra $5 in order to hold onto$245 for an additional thirty-five days. This delay equates to a 2.04 percent interest rate over just that short period of time ($5/$245 = 2.04 percent [rounded]). There are over ten thirty-five-day periods in a year. Paying the extra $5 is the equivalent of an annual interest rate in excess of 21 percent. 365 days per year/35 days holding the money = 10.43 time periods per year 2.04% (for 35 days) × 10.43 time periods equals a 21.28% rate for a year That substantial rate of interest is avoided by making the early payment, a decision chosen by most companies unless they are experiencing serious cash flow difficulties. Assuming that Rider avails itself of the discount offer, the capitalized cost of the inventory is reduced to$260.
Figure 8.2 Cost of Inventory Reduced by Cash Discount
## Exercise
Link to multiple-choice question for practice purposes: http://www.quia.com/quiz/2092883.html
## Key Takeaways
Any discussion of the reporting of inventory begins with the calculation of cost, the amount spent to obtain the merchandise. Cost encompasses all payments that are considered normal and necessary to get the merchandise into the condition and possession to be sold. Any other expenditures are expensed as incurred. Cash discounts are often offered to buyers to encourage quick payment. Taking advantage of such discounts is usually a wise decision because they effectively save interest at a relatively high rate.
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2020-01-24 11:17:28
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https://physjam.wordpress.com/2011/05/21/symmetry-lie-algebras-and-differential-equations-part-1/
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## Symmetry, Lie Algebras and Differential Equations Part 1
There is a deep relationship between being able to solve a differential equation and its symmetries. Much of the theory of second order linear differential equations is really the theory of infinite dimensional linear algebra. In particular Sturm-Liouville theory is the diagonalization of an infinite dimensional Hermitian operator. However there are deeper relationships, as Miller points out in “Lie theory and special functions”; the relationships between special functions such as Rodrigues’ formulae are related to the Lie algebra and symmetries of the system. Even better in some cases the solutions can be found almost entirely algebraically. Some examples from physics come from the Simple Harmonic Oscillator, the theory of Angular Momentum and the Kepler Problem (using the Laplace Runge Lenz vector). The rest of this article will be devoted to exploring a special case of these relations the Quantum Simple Harmonic Oscillator.
We begin with trying to solve the differential equation $-\frac{1}{2m} f''(x) + \frac{k}{2} x^2 f(x) = \lambda f(x)$ for some real positive constants $m$, $k$ and $latex\lambda$ with the boundary conditions f vanishes at infinity. This is an eigenvalue equation; this can’t be solved for any constants but only for particular values of $\lambda$ for a fixed k and m. By dilations (that is, rescaling units) we can assume without loss of generality $m=1$ and $k=1$. It is useful to define the momentum operator $p=-i \frac{\rm{ d}}{\rm {d}x}$ – this makes everything more physics-like. If this isn’t familiar to you just substitute $-i \frac{\rm{ d}}{\rm {d}x}$ wherever you see a p.
(The i is chosen to make the operator Hermitian with respect to the $L^2$ inner product: $\int_{-\infty}^{\infty}{f(x)}^* p g(x) = \int_{-\infty}^{\infty} \left(pf(x)\right)^* g(x)$; where * denotes complex conjugation and f and g are zero at infinity. This identity follows immediately from integration by parts).
Introducing the Hamiltonian operator $H=\frac{1}{2}\left(p^2+x^2\right)$ the differential equation is then the eigenvalue equation $H f(x) = \lambda f(x)$ (a form familiar to physicists). The Hamiltonian operator has an obvious symmetry to it: it is invariant under rotations in x-p space. That is it is invariant under transformations of the form:
$\begin{bmatrix} x' \\ p' \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \\ p \end{bmatrix}$.
Following the ideas of Sophus Lie we look at the infinitesimal transformations generating this, by taking the derivative at the identity $\theta=0$ this gives $x \to -p$, $p \to x$; in x-p space it is given by the matrix $\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$
This transformation is precisely the Fourier transform: $f(x) \to \widehat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-i k x}$. In particular integration by parts and differentiating under the integral respectively it follows $\widehat{xf(x)}=i\widehat{f}'(k)$ and $\widehat{-if'(x)} = k \widehat{f}(k)$, so as an operator on functions $x \to -p$ and $p \to x$.
Now since the square of the Fourier transform in x-p space is negative the identity it has eigenvalues -i and +i and corresponding eigenvectors $a = \frac{1}{\sqrt{2}}(x+ip)$ and $a^{\dag} = \frac{1}{\sqrt{2}} (x- ip)$.
Now we introduce the commutator of operators $[A,B]=AB-BA=-[B,A]$ and in particular $[x,p]=i \mbox{Id}$ (since x and its derivative don’t commute). Consequently by linearity $[a,a^{\dag}]=1$.
Simple calculations show that $a^{\dag}a = H -1/2$, $a a^{\dag} = H + 1/2$, $[H,a]=-a$ and $[H,a^{\dag}]=a^{\dag}$. These last two relations allow us to find the spectrum of H, that is the values of $\lambda$ for which the differential equation is solvable!
If the differential equation can be solved for some $\lambda$, $H f = \lambda f$ then using the commutation relations shows $H(a f) = (\lambda-1) (af)$ and $H(a^{\dag}f) = (\lambda +1) (a^{\dag}f)$. Thus $a$ lowers the eigenvalue by 1 and is called a lowering operator, and $a^{\dag}$ raises the eigenvalue by 1 and is called a raising operator.
However we can not lower indefinitely: H is positive semidefinite, $H=\frac{1}{2} (x^{\dag}x + p^{\dag}p)$ (where the dagger indicates Hermitian conjugation with respect to the $L^2$ inner product), so $\lambda$ must be non-negative. Thus there is a function $f_0(x)$ for which $a f_0(x) = 0$ (which of course satisfies the differential equation trivially). On this state $H f_0(x) = (a^{\dag} a + 1/2) f_0(x) = 1/2 f_0(x)$.
Moreover since any arbitrary solution can be brought to $f_0$ by repeated lowerings (applications of a), and lowering then raising gives a multiple of the original function every solution can be obtained by raising $f_0$. Thus the only possible eigenvalues are n+1/2 for n=0,1,2,….
What are the corresponding eigenvectors? Well $a f_0 = 0$ implies that $x f_0(x) + f_0'(x) =0$, which has solutions $f_0(x) = A e^{-\frac{x^2}{2}}$ for some constant $A$. Then the solution with $\lambda = n + 1/2$ is up to a constant factor $(a^{\dag})^n f_0(x) = A_n \left(x - \frac{\rm{d}}{\rm{d}x}\right)^n e^{-\frac{x^2}{2}} = H_n(x) e^{-\frac{x^2}{2}}$ where $H_n(x)$ are the Hermite polynomials. Consequently we have found all solutions of the second order differential equation just by solving a first order differential equation! (They can also easily be normalized algebraically; that is without doing any integrals, but I won’t show that here).
It is interesting to note all these solutions are invariant under Fourier transform. This is of course a consequence of the Hamiltonian being invariant under Fourier transform, F; if $Hf = \lambda f$ then $F H f = (FHF^{-1}F)f = HFf$ and thus $\lambda F f = H (Ff)$.
From an abstract point of view what have we done? We have taken an algebra of operators on some Hilbert space generated by self-adjoint operators $x$ and $p$ satisfying $xp-px=i \text{id}$ (notice that this implies the vector space can’t be finite dimensional; take the trace of each side). Using this we have shown that the positive definite Hermitian operator $H = \frac{1}{2} (x^2 + p^2)$ has eigenvalues n + 1/2 for n=0,1,2,….
We could choose an explicit representation: the Hilbert space is the space of square integrable functions, x is the multiplication operator and $p = -i \frac{\rm{d}}{\rm{d}x}$, then in this basis the eigenequation is the differential equation we started with. The solutions in this basis are the Hermite polynomials multiplied by a Gaussian; notice that these functions are orthogonal and complete in $L^2$ being all the eigenfunctions of a Hermitian operator. The formula for the eigenfunctions in terms of raising operators gives rise to a Rodrigues formula for the Hermite polynomials.
However there is nothing canonical about this choice of representation, a different representation is given by the Fourier transform, which acts as a change of basis. That the Hamiltonian is invariant under the Fourier transform means $FHF^{-1}=H$ or $[F,H]=0$.
The nicest choice of basis is the one in which H is the (countably infinite dimensional) diagonal matrix with entries 1/2,3/2,5/2,…. It is easy to see that a is the matrix with 1s one row below the diagonal and zeros everywhere else $a=\begin{bmatrix} 0 & 0 & 0 & \ldots \\ 1 & 0 & 0 & \ldots \\ 0 & 1 & 0 & \ldots \\ &&& \ddots \end{bmatrix}$ and $a^{\dag}$ is its transpose. Representations for x and p can be obtained from $x = \frac{1}{\sqrt{2}}(a + a^{\dag})$ and $p = \frac{1}{2i} (a - a^{\dag})$.
It is worth noting that in this derivation it wasn’t enough to have a Lie algebra, that is a Lie bracket, we also needed a multiplication over which the Lie bracket is the commutator – that is a representation.
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### 2 Responses to Symmetry, Lie Algebras and Differential Equations Part 1
I get stuck at the point where you introduce ‘x-p’ space. I think by ‘x-p’ space you mean to take the cartesian product of two sets, each set being some kind of (vector space) of operators (presumably of which ‘x’ and ‘p’ are members) right?
Then you define a linear transformation say which maps elements of this new two-dimensional vector space to itself, and applying this map to the operators ‘x’ and ‘p’ defined earlier, gives us some new operators such that 1/2(x’^2 + p’^2) = 1/2(x^2 + p^2),
Then you look at the infinitesimal transformations that generate the linear transformation, and apparently this gives us the Fourier transform, but at this point I can’t see how to relate the idea of a transformation of operators to a transformation such as the Fourier transform?
• physjam says:
Ok, I wasn’t super clear. x and $p=-i \frac{d}{dx}$ are (self-adjoint) linear operators. The set of all (self-adjoint) linear operators forms a vector space; when I talk about x-p space I am talking about the 2 dimensional subspace generated by x and p, that is vectors of the form ax+bp. Consequently I’m ignoring all the product structure.
Then, yes, I define clockwise rotations in the x-p plane, which the operator x^2+p^2 is invariant under.
Then I look at infinitesimal transformations that generate the linear transformation. On the positive x-axis a small clockwise rotation is a displacement along the -p axis (it lies tangent to the unit circle), and on the positive p-axis a vanishingly small clockwise rotation is a displacement along the x-axis. In fact the generators of the transformation map x to -p and map p to x. It’s only when we bring back the algebraic structure of $p=-i\frac{d}{dx}$ that it looks like a fourier transform: multiplication by x is mapped to $i\frac{d}{dx}$ and $\frac{d}{dx}$ is mapped to ix which is a defining property of the Fourier transform. [x and p as an algebra, that is adding compositions like xp, generate all the self-adjoint linear operators in this space – this is more of a definition of our linear operators and our vector space than a consequence].
It sounds a bit artificial really; all I am really saying at the end is that $x^2 - \frac{d^2}{dx^2}$ is invariant under a Fourier transform (or more generally a map $x \to \cos(\theta) x + i \sin(\theta) \frac{d}{dx}$, $\frac{d}{dx} \to \cos(\theta) \frac{d}{dx} + i \sin(\theta) x$, which can be given by $\exp(\theta F) = \sum_{n=0}^{\infty} \theta^n F^n/n!$ where F is the Fourier transform and powers denote composition – to see why this works substitute F^2=- Id into the power series.)
This symmetry then in some way carries on to the solutions – in particular a Fourier transform must map a solution to another solution. I find the eigenvectors in x-p space, and then play around with them to get the eigenvalues of x^2+p^2. There’s something missing; I think it may boil down to a simple representation theory calculation, but I don’t know what group I would try to represent.
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2017-09-23 19:50:42
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http://www.zachsmithwick.com/why-we-need-to-end-our-love-affair-with-hard-work/
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# Why We Need to End Our Love Affair with "Hard Work"
There are countless adages that praise hard work and determination, and for good reason. America, in particular, seems to always tout itself as a great nation of "hard working" people.
There's nothing wrong with hard work, in fact it is quite a good thing to have hard working people in your life or community, but an issue arises when the notion of "hard work" is fixated upon.
To illustrate this, let me tell you a story:
There once was a mischievous but clever little boy who was always acting up in class. His teacher would frequently get annoyed by his behavior and make him do menial tasks as a punishment. One day, after the boy was being particularly obnoxious, the teacher told him to go to the blackboard and add up all the (whole) numbers from 1 to 100. The boy let out a wry grin and strode over to the blackboard. The teacher, assuming this would take the boy quite a long time, turned back and continued teaching the rest of the class. A moment later, the boy cried out triumphantly, "DONE!" The teacher turned around, annoyed, knowing that the boy could not possibly be done that quickly. To her amazement, the boy had indeed found the correct answer of 5,050! Perplexed, she asked the boy how he had done it so quickly, assuming he had cheated in some way. He said that if you were to line up all the whole numbers from 1 to 100 and add the first number to the last number, then the second number to the second-to-last number, and so on, then you end up with 50 groups of 101, and 50 multiplied by 101 is 5,050! The teacher was astonished that the boy had found such an ingenious way to solve the problem, and even though she was annoyed that he did not receive a proper punishment, she let him sit down.
The above is an example of a case where there are two distinct ways of arriving at the same answer. I often call these the "smart-fast" way and the "dumb-slow" way. Math and science provide countless examples of this, wherein smart people have formulated clever ways of solving tedious problems.
The main takeaway of the above story is that the mischievous little boy was lazy, he didn't want to spend the time adding up all the numbers from 1 to 100, and that laziness provoked him to find a better, faster way. If he had been a good, hard-working student, then he would have been in front of the board for hours, and might not have even come to the correct answer since his method would be incredibly easy to make errors with.
A math professor I had in college once said "the smarter you are, the lazier you can afford to be." He was, fittingly enough, French.
He's right, though. If you are smart, you can be lazier than a hard-working person and still get the same, if not better, results. If you are smart and hard-working, then the world is yours for the taking.
When I graduated from high school, I was given the book Strengths Finder 2.0 by Tom Rath. In the book, he makes this very point, and uses the movie Rudy (based on a true story) as an example.
Any American sports fan has probably seen Rudy, or at least is familiar with the story. It's about a young Notre Dame student who is "five foot nothing" and weighs "a hundred and nothing", but through his hard work and dedication lands himself a spot on the prestigious Notre Dame football roster. At the end of the movie, the coach puts him in on the last play of a game against none other than my alma mater Georgia Tech, and he defies expectations by making a sack (we couldn't defend the quarterback then and we still can't do it now).
It's an uplifting story of determination, motivation, and strength of will. In a way, it sums up the spirit of America: if you really care about something, if you want it bad enough, and if you are willing to truly give it everything you've got, then you CAN...be a benchwarmer that makes one good play that has no impact on the game whatsoever?
Sure, it's an uplifting story, and I guess maybe even though he didn't really contribute to the team on the field, his energy rubbed off on the other (better) players and helped them play better? Truth be told I've never actually seen the whole movie, just the bit at the end where he makes the sack and the team carries him off the field. So in that sense, Rudy probably was a valuable member of the team (if anything by maybe terrifying the jocks into playing better so that they wouldn't have to be his back-up). I'm sure the Notre Dame recruiting department also used him heavily in their promotional material (and if they didn't, then good for them for not exploiting a young college athlete who got compensated in no way).
In Strengths Finder, Rath argues that although Rudy displayed hard work, his lack of natural abilities inhibited him from becoming a great football player. The sad truth is, if you want to be great at something, you have to dedicate yourself to something with which you have natural abilities.
Through this, Rath derives a simple equation to describe "Strength", which he defines as "the ability to consistently provide near perfect performance":
$Strength = Investment \times Talent$
Where "Investment" is the "time spent practicing, developing your skills and expanding you knowledge base" and "talent" is "a natural way of thinking, feeling, or behaving."
I think this dude's real strength is giving thoughtful, stoic looks.
Image Credit
There are a great many people these days that are frustrated with their lives. They feel that they have worked hard doing something that they love, but they are having trouble paying the bills.
To them, I will add one more wrinkle into Rath's equation, an idea of "probability score." I'll also change "Strength" to "profitability".
We now have:
$Profitability = Investment \times Talent \times Probability \: Score$
The "probability score" basically asks the question "what is the probability that this strength will end up making you money?"
The idea behind the probability score is to accept the reality that people, and the economic market as a whole, value different strengths differently. You can have incredible abilities in the art of flicking paper footballs, and spend your entire life working your fingers to the bone as you perfect the art of paper football flicking, but when it comes down to it, the world just doesn't really value people who are really good at paper football flicking (and that's a shame), so you probably won't make very much money off of your "strength."
You can get fairly technical about the probability score and apply various statistical distributions to it. For example, some strengths may be fairly Gaussian, wherein there is a large probability that you can get a job in that field and make a decent amount of money, but a small probability that you will get rich doing it. Other strengths may be more bimodal, i.e. "all or nothing", where you can either make a LOT of money doing it, or no money at all.
I should note that "investment" is a loaded term as well, and can mean either time investment (turning the crank and getting stuff done), or monetary investment (risking your own funds in hopes to produce long term success).
"Talent" is even more loaded. Earlier in this post I mentioned that there are often "smart-fast" and "dumb-slow" ways of doing things, with the implication that "smart" people figure out "fast" ways of doing things, leading them to be more profitable. Unfortunately for "smart" people, history has shown that intelligence does not always lead to financial wealth.
Well, some types of intelligence don't always lead to financial wealth. In that sense, the "talent" score should take into account not only a person's innate ability to be good at a particular strength, but also their innate ability market or pursue that particular strength in a way that leaves them successful. In short, "talent" is basically the knowledge of where to best invest your time and/or money.
There is often a disdain for people who are extremely wealthy, and the claim is that they "don't work as hard as people who make far less." In a sense, that's probably true, but I hope you have come to realize that you don't need to work extremely hard to become wealthy...so long as you are smart enough to know where to put your money/time. That being said, many rich people have worked extremely hard to get to where they are at, and that shouldn't be taken away from them.
Conclusion
So what are you supposed to get out of this? Should you just give up on working hard and be lazy? No, definitely not.
You, should, though, constantly question the effectiveness of your effort. Ask yourself if the hours you are putting in will lead you to long term success, i.e. "will this effort be worth it?" (a question I ask myself a lot when writing for this blog).
If you keep spinning your wheels and stop making progress toward your goal, then it may be time to reconsider things. There is such a negative stigma placed on "giving up", but knowing when to end things can sometimes be the difference between ultimate failure and success.
Perhaps most importantly, always consider your probability of success should you "perfect" your strength. In the words of a demotivator: "before you attempt to beat the odds, be sure you could survive the odds beating you."
Good luck, and give it your 80%.
Oh and by the way, here is the general formula for adding a sequence of numbers:
$\displaystyle \large \sum_{i = n}^{m} i = (m + n) \times \frac{m - n + 1}{2}$
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2019-09-16 00:59:57
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http://math.stackexchange.com/questions/902017/how-far-can-one-see-over-the-ocean
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# How far can one see over the ocean?
Since Earth is a sphere, one has only a limited visibility radius. How far is that, actually?
This Q&A was inspired by this question, about whether or not Legolas can see the 24km distant Riders of Rohan.
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I have up-voted both the question and the answer by M.Herzkamp, but I also think that answer is somewhat more complicated than it needs to be. Accordingly I've added my own answer. – Michael Hardy Aug 18 '14 at 15:23
Both answers need an additional factor of about two. You don't have to be able to see the orcs' toes, just the tops of their heads. – Pete Becker Aug 18 '14 at 16:05
The factor of two is only right, if the Orcs are of the same height as you (or elevation). As I remember, Legolas stood on a hill, while the Orcs rode over the plains of Rohan. – M.Herzkamp Aug 19 '14 at 6:34
@M.Herzkamp - that's why I said about two. – Pete Becker Aug 19 '14 at 14:08
Maybe Middle Earth has a planar geometry. :-) – peterh Aug 20 '14 at 11:45
I have up-voted the answer by M.Herzkamp, but I also think he makes it somewhat more complicated than it needs to be. The distance from the center of the earth to your eye is $r+h$, where $r$ is the radius of the earth and $h$ is the height of your eye above the ground. The distance from the center of the earth to a point on the horizon is $r$. The distance from your eye to the point on the horizon let us call $d$. The three sides of a right triangle are then the legs, $r$ and $d$, and the hypotenuse $r+h$. Applying the Pythagorean theorem, we have $$r^2 + d^2 = (r+h)^2.$$ It follows that $$d^2 = (r+h)^2 - r^2$$ so $$d=\sqrt{(r+h)^2-r^2}.$$ This admits simplifiction: $$d=\sqrt{(r+h)^2-r^2} = \sqrt{(r^2+2rh+h^2) - r^2} = \sqrt{2rh + h^2}.$$ When $h$ is tiny compared to $r$, we can say $$d \approx \sqrt{2rh\,{}}.$$
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I might be worth clarifying that any objects beyond the horizon can of course be visible if they are high or tall enough - an object at height $H$ is visible to an oberver at height $h$ at distance $d \approx \sqrt{2rh} + \sqrt{2rH}$. – JohannesD Aug 18 '14 at 16:08
@JohannesD I guess an object on the precise polar opposite of you, could be as high as you want, and you still couldn't see it. – Cruncher Aug 18 '14 at 16:34
@Red_Shadow you also need to allow your vantage point to be arbitarily high. At any finite height the planet casts a shadow covering some angle. – JHance Aug 18 '14 at 19:11
@Red_Shadow Actually to see anything (except something perfectly opposite you) both you and the object must be arbitrarily tall/high. If your height is bounded, then you can only see things that are not "below" the infinite conic surface defined by your position and the horizon circle. – JohannesD Aug 18 '14 at 19:20
A handy formula implementing the above result is $$d\,[{\rm km}]=3.56\cdot\sqrt{h\,[{\rm m}]}\ .$$ – Christian Blatter Aug 19 '14 at 18:03
Let us suppose, an observer of height $h$ stands on a perfectly spherical planet of radius $r$:
Edit: here is an easier way, making use of the right angle between the line of sight and the radial ray. You can just use the definition of the cosine:
$$\cos(\theta_T) = \frac{r}{r+h} \qquad \Rightarrow \qquad s = r\cdot\theta_T = r\cdot\cos^{-1}\!\!\left(\frac{r}{r+h}\right)$$
which is equivalent to the solution obtained by the complicated method. /Edit
The distance $s$ to the farthest point he can then see is determined by the tangent to the semi circle through his head. If you describe the semi circle in a cartesian coordinate system by $$y^2+x^2 = r^2,$$ the observer's head is at $y=r+h,\ x=0$.
To obtain the slope of the tangent, we plug the tangent equation $y=mx+r+h$ into the circle equation and solve for $x$: $$x_{1/2} = -(r+h)\frac{m}{1+m^2} \pm \sqrt{\frac{(r+h)^2m^2}{(1+m^2)^2}+\frac{r^2-b^2}{1+m^2}}$$ Those are two intersection points, and in order to have a tangent, they must be equal. That is the case, if the term under the square root is zero. The resulting equation can be solved for $m$: $$m_{\pm} = \pm \sqrt{\frac{(r+h)^2}{r^2}-1}$$ Let's take the negative solution for the tangent on the right (it does not matter), and calculate the tangent point: $$x_T = -(r+h)\frac{m_-}{1+m^2_-} = \frac{r}{r+h}\sqrt{(r+h)^2-r^2}$$ The viewing distance angle is $\theta_T = \text{asin}(x_T)$. To get the viewing distance, we observe that $$\frac{s}{2\pi r} = \frac{\theta_T}{\text{full angle}} = \frac{\theta_T}{2\pi}\text{, with angle in radian}$$ $$\Rightarrow s(h) = r\cdot\text{asin}\left(\sqrt{1-\frac{r^2}{(r+h)^2}}\right)$$ If you plot this for $h$ small compared to $r$, it resembles a square root function, and indeed, $$\lim_{h\rightarrow0^+}\frac{s(h)}{\sqrt{h}} = \sqrt{2r}$$ which means that for small heights, the viewing distance can be described as $$s(h) \approx \sqrt{2rh}$$ On Earth ($r\approx6371\text{km}$), a normal person ($h\approx1.8\text{m}$) can see the surface about 4.8km away. Not much further. If you climb a hill or tree ($h\approx 50\text{m}$), your range increases to 25km!
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I'd have called it the Pythagorean theorem rather than the "circle equation". ${}\qquad{}$ – Michael Hardy Aug 18 '14 at 15:15
Now we only need to know the height of an average orc ;-) – Einer Aug 18 '14 at 15:25
This approach, involving solving an equation for $x$, is unnecessarily complicated. I've shown a simpler way in my answer. – Michael Hardy Aug 18 '14 at 16:00
Note that this calculates the distance to the horizont where the completely smooth Earth begins to be in the way of itself. In order words, the distance where you can see the whole orc, including soles of his shoes. If you can live with seeing less of the orc, the distance is greater. – Thorbjørn Ravn Andersen Aug 19 '14 at 14:24
Holy cow this is freaking complicated. You should've thrown general relativity in there too! – Mehrdad Aug 19 '14 at 22:43
Atmospheric refraction cannot be neglected. As mentioned here the effect of this can be taken into account approximately by pretending as if the Earth's radius is larger by a factor of 7/6. This makes the distance $d$ to the horizon when the height $h$ is much less than the Earth's radius $R$ equal to
$$d = \sqrt{\frac{7}{3} R h}$$
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Physicist invasion! Regardless of how true this is, it's a comment; not an answer. (Unless you provide an answer, and factor the effect due to refraction too) – Ollie Ford Aug 18 '14 at 19:19
I've updated the post. – Count Iblis Aug 18 '14 at 19:36
If you go sailing, you'll have about 5 nautical miles of visibility (1nm = 1852m). I personally find the nomogram a delightful invention:
Simply draw a straight line between the height of the observer and the height of the object on the horizon (in this case = 0), then read off the geographical range.
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Unfortunately the scales are very nearly too large for a 1.8m observer watching a 1.8m object. On the other hand, your chart appears to show about 5 nautical miles, so about 9.26km, which roughly matches M.Herzkamp's answer of 9.6km. – Mooing Duck Aug 19 '14 at 22:24
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2016-02-09 11:51:45
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https://www.lil-help.com/questions/151821/issc363-assignment-4
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ISSC363 Assignment 4
# ISSC363 Assignment 4
A
0 points
Penetration Testing
Brian Page
American Military University
Introduction
Network Scanning is a crucial component of penetration testing that must be conducted in order for the penetration tester (PT) to determine the needs of the organization. Discovery of the devices that are on the network will allow the PT to see what devices are on the network and find out if there are any vulnerabilities in the software or hardware that can be exploited. For the purposes of this assignment, Kali or Backtrack Linux will be used as the penetration testing software, using nmap to discover devices and determine if they can be exploited.
Kali Linux: NMAP
The reasoning behind using nmap as the network scanner is it can also, in most cases determine what operating system the devices on...
ISSC363 Assignment
Ash
A
0 points
#### Oh Snap! This Answer is Locked
Thumbnail of first page
Excerpt from file: PENTESTING 1 PenetrationTesting BrianPage AmericanMilitaryUniversity PENTESTING 2 Introduction NetworkScanningisacrucialcomponentofpenetrationtestingthatmustbeconducted inorderforthepenetrationtester(PT)todeterminetheneedsoftheorganization.Discoveryof
Filename: issc363-assignment-4-85.docx
Filesize: < 2 MB
Print Length: 5 Pages/Slides
Words: NA
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
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2019-04-21 04:48:37
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http://mathoverflow.net/questions/160270/number-of-rational-curves-on-varieties-over-finite-fields
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Number of rational curves on varieties over finite fields
Let $k$ be a finite field. Let $P_r$ be set of degree $r$ binary forms over $k$. We define $$\mathcal{M}_r = \{ (x_0, ..., x_n) \in P_r^{n+1} : x_0^d + ... + x_n^d = 0 \text{ and the }x_i \text{'s don't have a common factor} \}.$$
I am interested in knowing what the value $$\#\mathcal{M}_r/\#\mathcal{M}_{r-d}$$ is when $d<r.$
The reason why I am asking this is because I am reading a PhD thesis "Algebraic Circle Method" by Thibaut Pugin, and it seems to be using this value in the proof of Lemma 2.4.4 on page 28.
I would greatly appreciate any help to figure this out. Thanks!
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2015-04-18 15:39:58
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https://blender.stackexchange.com/questions/15067/how-to-set-edge-bevel-weights-from-a-python-script
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# How to set edge bevel weights from a python script
I need to set weights for some edges from a python script. I have tried looping through object.data.edges and setting bevel_weight values but I can't get it right. It only works if I have first edited some edge's bevel weight by hand from the 3d-view's transform window. But I would like to get it work from my script without any manual hacking.
Steps to reproduce:
Start Blender with default cube and switch to Scripting view.
Try to set weights for all edges:
for edge in C.object.data.edges:
edge.bevel_weight = 1.0
Verify that values changed:
for edge in C.object.data.edges:
edge.bevel_weight
Switch into edit mode in 3d-view and open up transform window by pressing n. Select edges to see that the change did not work (bevel weights are all zero). Switch back to object mode and read values again in scripting console to see that they were reset to zero:
for edge in C.object.data.edges:
edge.bevel_weight
Go back to edit mode and change some edge bevel weight straight from the transform window by hand. Go back to object mode and change values again from the scripting console:
for edge in C.object.data.edges:
edge.bevel_weight = 1.0
This time it works, bevel weights are 1.0 when you go back to edit mode and look from the transform window.
Any ideas? My final need is to apply a bevel modifier to only some set of edges and that's why I'm trying to mark them by using weights. Using vertex group for bevel modifier does not work in my situation, since I need to specify edges, not vertices, which to bevel.
I'm using Blender v2.71.
There is no custom data layer for edge bevel weights by default. You need to explicitly enable it (this is done automatically if you use the Edge Bevel Weight modal operator in the UI):
ob = bpy.context.object
me = ob.data
me.use_customdata_edge_bevel = True
Furthermore, the mesh needs to be in Object Mode when you apply a bevel weight:
if ob.is_editmode:
bpy.ops.object.mode_set(mode='OBJECT')
me.edges[0].bevel_weight = 0.7
If you want to avoid mode switching, you may use the bmesh module instead (bm.edges.layers.bevel_weight).
• I was confused for a while with this. I thought whenever I printed the bevel weight within my script it would update, but it wouldn't update if I didn't print. But it turns out correlation isn't causation; what really happened is that I added the print statement, went into edit mode, inadvertently added custom data by messing with the UI bevel weight, and then ran the script to see the bevel weight "magically" update! Mar 19, 2021 at 16:08
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2022-05-29 02:26:56
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https://www.physicsforums.com/threads/solar-radius-of-a-star.355536/
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# Homework Help: Solar radius of a star
1. Nov 17, 2009
### snabelpablo
A star's surface sends out energy in form of electromagnetic radiation and has an emissivity close to 1. Find the solar radius of Rigel that emits energy at a rate of 2,7 * 10^32W and has a surface temperature of 11000K. You may assume the star is spherical.
Attempt at solution:
Formula for the heat current in radiation: H = AeσT^4.
Solve this with respect to A gives surface area: A = H / (eσT^4) =
2.7 * 10^32 / (1 * (5,67 * 10^-8) * 11000^4) = 3,25 * 10^23 m^2.
Formula for the area of a sphere: A = 4πr^2.
Solve this with respect to r gives radius: r = (A / 4π)^0,5 =
(3,25 * 10^23 / 4π)^0,5 = 1,61 * 10^11 m.
The radius of the sun: 6,96 * 10^8 m.
Solar radius = (1,61 * 10^11) / (6,96 * 10^8) = 231,06Rsun.
The Rigel radius is 231 times the radius of the sun. However, Wikipedia says 78. I've also seen 63, 98 etc., but 231 sounds a bit high. What have I done wrong?
2. Nov 17, 2009
### ideasrule
The star's luminosity should be 2.7*10^31 W, not 2.7*10^32 W. With the correct luminosity, you get 23 solar radii, very close to Wikipedia's figure.
3. Nov 17, 2009
### clamtrox
Perhaps one of the values given to you is incorrect. I'd guess it's the luminosity.
4. Nov 17, 2009
### snabelpablo
You are correct! The luminosity is in fact 2,7 * 10^31 W for Rigel giving it a solar radii of 73 using my formulas.
However, I have to do this for a second star as well, Procyon B.
Calculating its solar radius with the same formulas using H = 2,1 * 10^23W and T = 10000K I get:
A = 3,7 * 10^14 m^2.
r = 5,43 * 10^6 m.
Rsun = 0,0078.
Wikipedia says 0,01234 - almost twice what I got.
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2018-05-25 11:23:18
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https://solvedlib.com/forcast-time-series-question-a-the-values-of-the,143896
|
# Forcast , time series question. (a) The values of the smoothing constants, a, y, and 8...
###### Question:
forcast , time series question.
(a) The values of the smoothing constants, a, y, and 8 for Holt-Winters method fall between 0 and 1. Explain what happens when each of them approaches or equals the values 0 and 1. (b) It has been found that quarterly sales of some sports drinks over 8 years has the multiplicative Holt-Winters optimum smoothing constants a=0.3, y = 0.1, and 8=0.2. It is also given that the final estimates for 131 = 168.1, 631 = 2.3, sn28 = 0.7, sn29 = 1.1, sn30 = 1.3, and sn31 = 0.9, with y32 = 150. (i) Compute 132, b32, and sn32 (ii) Suppose in the first quarter for year 9 we observed y33 124, find the estimates 133, b33 and sn33. (iii) Use the estimates from (ii) to obtain the point forecasts for the time periods 34, 35, and 36.
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Exercises$3-6$are the same as Exercises$13-16$of Section 7.2. $16 V-24=2 V^{2}$... 1 answer ##### By netBeans C++ java by netBeans C++ java by netBeans C++ java 1. Answer the following... by netBeans C++ java by netBeans C++ java by netBeans C++ java 1. Answer the following question a) (4 marks) Write a Java method, which takes an integer array and two integers x and y as input parameters. The method should exchange the value in position x with the value in the position y in th... 1 answer ##### Divide using long division. State the quotient,$q(x),$and the remainder,$r(x).$$\left(x^{3}+5 x^{2}+7 x+2\right) \div(x+2)$ Divide using long division. 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Fast Food Service Times Refer to Data Set 25 "Fast Food" and construct 9590 confi = dence interval estimate of the mean drive-through service time for McDonald' at dinner; then do the same for Burger King at dinner: Compare the results: 3... 5 answers ##### (10 pts) Show B = {1,1-6,2 - M +/} is Insis of Pz; and find thue crdlinate vertor o p(t) rativ to B (10 pts) Show B = {1,1-6,2 - M +/} is Insis of Pz; and find thue crdlinate vertor o p(t) rativ to B... 1 answer ##### 12 22 ts retail operations. Based on a preiminary study, 10 stores are feasible in various... 12 22 ts retail operations. Based on a preiminary study, 10 stores are feasible in various parts of the country Th isexpected to be S160 per year for me yoan t tare requies an immediate investment of SA00 to set up operations Assuming a required rate of return 6%, what is the NPV of each store? Plac... 1 answer ##### Below is the contribution format income statement for a company based on a sales volume of... Below is the contribution format income statement for a company based on a sales volume of 1,000 units (the relevant range of production is 500 units to 1,500 units): Sales Variable expenses Contribution margin Fixed expenses Net operating income$ 55,000 33,000 22,000 14,960 $7,040 If the variable... 5 answers ##### Solve the equation Write the solution setwith the exact values givenin terms logarithms: of natural 54,000e-0.2t = 9000(In 8) (o2(1n 6 (02lo 6 ln 0.2f{kag} (1m Solve the equation Write the solution setwith the exact values givenin terms logarithms: of natural 54,000e-0.2t = 9000 (In 8) (o2 (1n 6 (02 lo 6 ln 0.2f {kag} (1m... 5 answers ##### Cirele the direction; (clockwise (counter-clockwise another direction 1 21 B(r > 1 1 W counler-cloc c 1 duneclun |L U "inoww L 2 1 1 1 1 | candethe shell , 1 1 1 1 [ IX 1 j1Mield ) Cirele the direction; (clockwise (counter-clockwise another direction 1 2 1 B(r > 1 1 W counler-cloc c 1 duneclun | L U "inoww L 2 1 1 1 1 | candethe shell , 1 1 1 1 [ IX 1 j1 Mield )... 1 answer ##### Correspond each of the peaks to the H's in the structure. Is it the Z or... Correspond each of the peaks to the H's in the structure. Is it the Z or E isomer? Show how you know. 1H NMR (CDCI, 300 MHz) 9-(2-phenylethenyl)anthracene 7.9424 - 7.8874 - 6.9760 6.9207 1H 7.0 8.6 8.4 8.2 8.0 7.8 7.6 7.4 7.2 6.8 6.6... 1 answer ##### What amino acids can be deaminated directly? What amino acids can be deaminated directly?... 1 answer ##### Wang Corporation purchased$220,000 of Hales Inc. 8% bonds at par in 2020 with the intent...
Wang Corporation purchased \$220,000 of Hales Inc. 8% bonds at par in 2020 with the intent and ability to hold the bonds until the bonds mature in 2025, so Wang classifies its investment as held-to-maturity. Unfortunately, a combination of problems at Hales and in the debt market caused the fair valu...
##### 27 A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has sleep deprivation _ How large sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 4%?Find the critical values xj-a/z and xa/2' for 99 confidence and n = 10.Construct a 95% confidence interval for the population mean, H. Assume the population has a normal distribution A random sample of 16 lithium
27 A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has sleep deprivation _ How large sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 4%? Find the critical...
##### If a prodromal period exists for a certain disease, it should occur prior to: Select one:...
If a prodromal period exists for a certain disease, it should occur prior to: Select one: a. incubation. b. decline. O c. illness. d. convalescence. In recombinant DNA technology, a vector is a self-replicating segment of DNA, such as a plasmid or viral genome. Select one: True False...
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2022-07-06 07:17:52
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http://holani.net/error-in/error-error-in-computing-the-variance-function.php
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# holani.net
Home > Error In > Error Error In Computing The Variance Function
# Error Error In Computing The Variance Function
## Contents
Why should we care about σ2? jefMin12 over 3 years ago You're welcome. 1 vote permalink So what was your final answer? 1449 points Submitted by rbgcode over 3 years ago 0 votes permalink Your code looks What happens when you fit the model, excluding the effects used in the propensity scoring?If that works, then it is a case of how do we get these interesting effects into Then the distribution should be multinomial, with a cumulative logit link. http://holani.net/error-in/error-in-computing-the-variance-function-genmod.php
My dependent variable is the number of healthcare visits in ADHD patients and the independent variables include age, sex, ethnicity, physician specialty, confirmed diagnosis of ADHD in pre-index period, number and Each subpopulation has its own mean μY, which depends on x through $$\mu_Y=E(Y)=\beta_0 + \beta_1x$$. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Sign in Search Microsoft Search Products Templates Support Products Templates Support Support Apps Access Excel OneDrive OneNote Outlook PowerPoint The usual estimator for the mean is the sample average X ¯ = 1 n ∑ i = 1 n X i {\displaystyle {\overline {X}}={\frac {1}{n}}\sum _{i=1}^{n}X_{i}} which has an expected https://communities.sas.com/t5/SAS-Enterprise-Guide/Proc-genmod-error/td-p/85344
## Error In Parameter Estimate Covariance Computation
The model runs when I say type=ind which, if I understand correctly, means that the repeated measure are not correlated. Mathematical Statistics with Applications (7 ed.). However, once I started adding the covariates, it gives the [email protected]: I have 1930 observations. 1028 are yes and 902 are no. We denote the value of this common variance as σ2.
• The MSE is the second moment (about the origin) of the error, and thus incorporates both the variance of the estimator and its bias.
• The minimum excess kurtosis is γ 2 = − 2 {\displaystyle \gamma _{2}=-2} ,[a] which is achieved by a Bernoulli distribution with p=1/2 (a coin flip), and the MSE is minimized
• The same errors arise if I specify TYPE as exchangeable rather than autorgressive.
• Can any one help me and explain the source of error and how i correct it.
• If they are approximately equal, change to a Poisson distribution.
• However, a biased estimator may have lower MSE; see estimator bias.
• If you want to include logical values and text representations of numbers in a reference as part of the calculation, use the VARA function.
Message 1 of 4 (890 Views) Reply 0 Likes SteveDenham Super User Posts: 2,546 Re: Erorr: Error in computing the variance function during genmod execution Options Mark as New Bookmark Subscribe Please use our new forums at discuss.codecademy.com. I think the reason that you can use either/or "(number - average) and (average - number)" is because we then square the result of that subtraction, which makes any negative result Computing Variance Excel As the two plots illustrate, the Fahrenheit responses for the brand B thermometer don't deviate as far from the estimated regression equation as they do for the brand A thermometer.
My model is below. Error Error In Estimation Routine Proc Genmod But, we don't know the population mean μ, so we estimate it with $$\bar{y}$$. MR0804611. ^ Sergio Bermejo, Joan Cabestany (2001) "Oriented principal component analysis for large margin classifiers", Neural Networks, 14 (10), 1447–1461. https://communities.sas.com/t5/SAS-Statistical-Procedures/Proc-genmod-how-to-resolve-error-messages/td-p/33607 So my question is - how big could the bias on CIs actually be and secondly how can I overcome this warning?
That is, how "spread out" are the IQs? Computing Variance And Standard Deviation The fitted line plot here indirectly tells us, therefore, that MSE = 8.641372 = 74.67. Unfortunately, I think the errors that are currently occurring will still occur under these options, so perhaps some others can help out on this.Steve Denham Message 4 of 18 (1,148 Views) The sample variance: $s^2=\frac{\sum_{i=1}^{n}(y_i-\bar{y})^2}{n-1}$ estimates σ2, the variance of the one population.
## Error Error In Estimation Routine Proc Genmod
Iteration will be terminated.ERROR: Error in parameter estimate covariance computation.ERROR: Error in estimation routin I f I run the analysis without the modification of: repeated subject=id/type=ind; or repeated subject=id/type=unstr; it works NOTE: The scale parameter was held fixed. Error In Parameter Estimate Covariance Computation To understand the formula for the estimate of σ2 in the simple linear regression setting, it is helpful to recall the formula for the estimate of the variance of the responses, Sas Error Error In Computing The Variance Function Predictor If Y ^ {\displaystyle {\hat Saved in parser cache with key enwiki:pcache:idhash:201816-0!*!0!!en!*!*!math=5 and timestamp 20161007125802 and revision id 741744824 9}} is a vector of n {\displaystyle n} predictions, and Y
Any other feedback? http://holani.net/error-in/error-error-in-computing-inverse-link-function.php I learned that a modified modified Poisson regression analysis gives better result than a normal Poisson regression analysis for longitudinal data, which tends to give conservative confidence intervals. Applications Minimizing MSE is a key criterion in selecting estimators: see minimum mean-square error. Problem Note 9185: Errors may result from using TYPE3 option and REPEATED statement, [for the GENMOD Procedure], http://support.sas.com/kb/9/185.html). Computing Sample Variance
To get an idea, therefore, of how precise future predictions would be, we need to know how much the responses (y) vary around the (unknown) mean population regression line \(\mu_Y=E(Y)=\beta_0 + Again, the quantity S = 8.64137 is the square root of MSE. MR1639875. ^ Wackerly, Dennis; Mendenhall, William; Scheaffer, Richard L. (2008). http://holani.net/error-in/error-in-computing-the-variance-function-proc-genmod.php I assume that the error messages appear with nothing in the output, meaning that the algorithm never gets started.
Introduction to the Theory of Statistics (3rd ed.). Computing Variance By Hand We don't need to create another variable because this can be easily calculated therefore we can just "return" that final calculation = return variance / len(grades) This is how I understand The estimate of σ2 shows up directly in Minitab's standard regression analysis output.
## In statistical modelling the MSE, representing the difference between the actual observations and the observation values predicted by the model, is used to determine the extent to which the model fits
The problem with this variable occurs in both genmod and glimmix. Which version do I have? Erorr: Error in computing the variance function during genmod execution Reply Topic Options Subscribe to RSS Feed Mark Topic as New Mark Topic as Read Float this Topic to the Top Computing Variance In R Do you have enough data?
As stated earlier, σ2 quantifies this variance in the responses. The estimate of σ2 shows up indirectly on Minitab's "fitted line plot." For example, for the student height and weight data (student_height_weight.txt), the quantity emphasized in the box, S = 8.64137, PLease could you tell me what could be the problem ion that case.THank you very much.Pooja Message 3 of 18 (1,148 Views) Reply 0 Likes SteveDenham Super User Posts: 2,546 Re: Check This Out I then added in one variable at a time and the convergence problem only arises when I add the variable nothvst1.
Communities SAS Enterprise Guide Register · Sign In · Help Desktop productivity for business analysts and programmers Join Now CommunityCategoryBoardLibraryUsers turn on Got a question you need answered quickly? No! The similarities are more striking than the differences.
If the estimator is derived from a sample statistic and is used to estimate some population statistic, then the expectation is with respect to the sampling distribution of the sample statistic. That is, the n units are selected one at a time, and previously selected units are still eligible for selection for all n draws. Right now, I am thinking of using PROC GLIMMIX, and specifying type=CHOL to avoid the positive definite problem (plus I am a lot more familiar with tuning things when GLIMMIX has
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2018-10-17 11:17:10
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https://www.sparrho.com/item/the-similarity-problem-for-indefinite-sturm-liouville-operators-and-the-help-inequality/8ad798/
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# The similarity problem for indefinite Sturm-Liouville operators and the HELP inequality
Research paper by Aleksey Kostenko
Indexed on: 11 Jul '12Published on: 11 Jul '12Published in: Mathematics - Spectral Theory
#### Abstract
We study two problems. The first one is the similarity problem for the indefinite Sturm-Liouville operator $A=-(\sgn\, x)\frac{d}{wdx}\frac{d}{rdx}$ acting in $L^2_{w}(-b,b)$. It is assumed that $w,r\in L^1_{\loc}(-b,b)$ are even and positive a.e. on $(-b,b)$. The second object is the so-called HELP inequality $(\int_{0}^b\frac{1}{\tilde{r}}|f'|\, dx)^2 \le K^2 \int_{0}^b|f|^2\tilde{w}\,dx\int_{0}^b\Big|\frac{1}{\tilde{w}}\big(\frac{1}{\tilde{r}}f'\big)'\Big|^2\tilde{w}\, dx,$ where the coefficients $\tilde{w},\tilde{r}\in L^1_{\loc}[0,b)$ are positive a.e. on $(0,b)$. Both problems are well understood when the corresponding Sturm-Liouville differential expression is regular. The main objective of the present paper is to give criteria for both the validity of the HELP inequality and the similarity to a self-adjoint operator in the singular case. Namely, we establish new criteria formulated in terms of the behavior of the corresponding Weyl-Titchmarsh $m$-functions at 0 and at $\infty$. As a biproduct of this result we show that both problems are closely connected. Namely, the operator $A$ is similar to a self-adjoint one precisely if the HELP inequality with $\tilde{w}=r$ and $\tilde{r}=w$ is valid. Next we characterize the behavior of $m$-functions in terms of coefficients and then these results enable us to reformulate the obtained criteria in terms of coefficients. Finally, we apply these results for the study of the two-way diffusion equation, also known as the time-independent Fokker-Plank equation.
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2022-01-24 23:29:24
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https://tex.stackexchange.com/questions/398378/need-help-adding-itemized-list-to-tabular-in-template
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# Need help adding itemized list to tabular in template
I'm trying to learn LaTeX, starting by building a resume using this template (link). I've had some success and I'm 90% done, but there's one issue I'm working on. In the template, there's a "skills" class (if it's called a class) which creates a 2 section tabular, the left box for the skill and the right box for description. It's intended to be used by entering a skill and then putting in a short sentence or a comma separated list to expand on the skill.
Question: What I want to do is add an itemized list in the skill detail section instead of a single line. When I do this, the formatting is not good at all. What can I do to make this work where it doesn't put in weird spacing and keeps the left box (skill) top aligned? I don't really understand the class, I would like to understand how to modify it to suit my needs.
Here's a sample file with everything invalidated
Class file:
% Define an environment for cvskill
\newenvironment{cvskills}{%
\vspace{\acvSectionContentTopSkip}
\vspace{-2.0mm}
\begin{center}
\setlength\tabcolsep{1ex}
\setlength{\extrarowheight}{0pt}
\begin{tabular*}{\textwidth}{@{\extracolsep{\fill}} r L{\textwidth * \real{0.9}}}
}{%
\end{tabular*}
\end{center}
}
% Define a line of cv information(skill)
% Usage: \cvskill{<type>}{<skillset>}
\newcommand*{\cvskill}[2]{%
\skilltypestyle{#1} & \skillsetstyle{#2} \\
}
Intended Use:
\cvsection{Skills}
\begin{cvskills}
\cvskill
{My First Skill} % Category
{Comma, Seperated, List, Of, Items, Works, As, Intended} % Skills
% Continue Items.....
\end{cvskills}
Result:
My attempt at entering a list:
\begin{cvskills}
\cvskill
{My First Skill} % Category
{
\begin{itemize}[leftmargin=2ex, nosep, noitemsep, label={\Large$\cdot$}]
\item {Using lists doesn't work correctly}
\item {As you can see}
\item {In this example}
\end{itemize}
} % Skills
% Continue Items.....
\end{cvskills}
Result:
• A full compilable code would be nice… – Bernard Oct 27 '17 at 19:26
• What you think does not work, is quite a common issue. As this is only a tabular you might want to have a look at the questions about the vertical orientation in tabulars. – TeXnician Oct 27 '17 at 19:32
• Welcome to TeX.SX! Rather than posting links, which may disappear next week making the question opaque, and incomplete code snippets can you please post a full minimal working example. A MWE should start with a \documentclass command, have a minimal preamble and then \begin{document}...\end{document}. The code should compile and be as small as possible to demonstrate your problem. This makes it much easier to help you and, hence, much more likely that some one will. – Andrew Oct 27 '17 at 20:20
Add \newcolumntype{L}[1]{>{\raggedright\let\newline\\\arraybackslash\hspace{0pt}}p{#1}} to your preamble. This will issue a warning.
Use the following in your document:
\newcommand{\cvitem}[1]{\bullet~#1\par}
\begin{cvskills}
\cvskill
{My First Skill}
{%
\cvitem{Using lists doesn't work correctly}
\cvitem{As you can see}
\cvitem{In this example}
} % Skills
\end{cvskills}
Disclaimer: This will affect all instances that use the L columntype.
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2020-04-03 08:55:00
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https://mathoverflow.net/questions/396077/convex-polyhedra-that-can-be-folded-from-convex-polygons
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Convex polyhedra that can be folded from convex polygons
This question is based on http://www.science.smith.edu/~jorourke/Papers/FoldingPP.pdf.
Therein is stated the theorem: Every convex polygon folds to an infinite number (a continuum) of noncongruent convex polyhedra.
Question: What could one say in the reverse direction: given any convex polyhedron, what could one say about the polygon(s) that could fold into it? If there are convex polyhedrons that cannot be folded from convex polygons, then, how does one characterize convex polyhedrons (such as for example, the regular tetrahedron) that can be folded out of convex polygons?
• The question of unfolding convex polyhedra to convex polygons was first posed by Shephard in 1975: Shephard, Geoffrey C. "Convex polytopes with convex nets." In Mathematical Proceedings of the Cambridge Philosophical Society, vol. 78, no. 3, pp. 389-403. Cambridge University Press, 1975. Jun 24, 2021 at 16:04
• One minor remark (which does not answer any of your questions): Any convex polyhedron $Q$ folded from a convex polygon $P$ of $n$ vertices has at most $n+2$ vertices. This is Corollary 25.7.2 in Geometric Folding Algorithms. Jun 24, 2021 at 16:07
This answers the easiest question posed (1), and addresses part of the more general question (2).
(1) "If there are convex polyhedrons that cannot be folded from convex polygons,..."
Yes, there is an abundance of such polyhedra. For example, let $$P$$ be a cube. To unfold it to the plane, one must form surface cuts that combinatorially constitute a spanning tree of the vertices, often called a cut tree $$T$$. (Any vertex not on $$T$$ will retain its 3D structure, and so obstruct planar development.) A tree has at least two leaves. Let $$v$$ be a vertex of $$P$$ at a leaf of $$T$$. Then the corresponding unfolding forms an interior reflex angle of $$3\pi/2$$ locally about the image of $$v$$. This is a local nonconvexity.
(2) "how does one characterize convex polyhedrons (...) that can be folded out of convex polygons?"
Generalizing the argument above leads to a necessary condition: A convex polyhedron $$P$$ needs to have at least two vertices of curvature $$\ge \pi$$ (and so incident face angles $$\le \pi$$) in order to serve as leaves for a cut tree $$T$$ that unfolds to a convex polygon. The cube has all vertices with curvature $$\pi/2$$, so there is no "home" for the leaves of $$T$$.
This is, however, not a complete characterization.
• Thanks very much for that very nice argument, somewhat reminiscent of Fisk's proof of art gallery theorem! Jun 25, 2021 at 7:08
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2022-08-14 00:19:40
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https://pos.sissa.it/336/186/
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Volume 336 - XIII Quark Confinement and the Hadron Spectrum (Confinement2018) - E: QCD and New Physics
Hadronic contributions to the muon anomalous magnetic moment
M. Procura,* G. Colangelo, M. Hoferichter, P. Stoffer
*corresponding author
Full text: pdf
Pre-published on: 2019 September 12
Published on: 2019 September 26
Abstract
The largest uncertainties in the Standard Model calculation of the anomalous magnetic moment of the muon $(g-2)_\mu$ come from hadronic effects, namely hadronic vacuum polarization (HVP) and hadronic light-by-light (HLbL) contributions. Especially the latter is emerging as a potential roadblock for a more accurate determination of $(g-2)_\mu$.
The main focus here is on a novel dispersive description of the HLbL tensor, which is based on unitarity, analyticity, crossing symmetry, and gauge invariance. This opens up the possibility of a data-driven determination of the HLbL contribution to $(g-2)_\mu$ with the aim of reducing model dependence and achieving a reliable error estimate.
Our dispersive approach defines unambiguously the pion-pole and the pion-box contribution to the HLbL tensor. Using Mandelstam double-spectral representation, we have proven that the pion-box contribution coincides exactly with the one-loop scalar-QED amplitude, multiplied by the appropriate pion vector form factors. Using dispersive fits to high-statistics data for
the pion vector form factor, we obtain $a_\mu^{\pi\text{-box}}=-15.9(2)\times 10^{-11}$. A first model-independent calculation of effects of $\pi\pi$ intermediate states that go beyond the scalar-QED pion loop is also presented. We combine our dispersive description of the HLbL tensor with a partial-wave expansion and demonstrate that the known scalar-QED result is recovered after partial-wave resummation.
After constructing suitable input for the $\gamma^*\gamma^*\to\pi\pi$helicity partial waves based on a pion-pole left-hand cut (LHC), we find that for the dominant charged-pion contribution this representation is consistent with the two-loop chiral prediction and the COMPASS measurement for the pion polarizability. This allows us to reliably estimate $S$-wave rescattering effects to the full pion box and leads to $a_\mu^{\pi\text{-box}} + a_{\mu,J=0}^{\pi\pi,\pi\text{-pole LHC}}=-24(1)\times 10^{-11}$.
DOI: https://doi.org/10.22323/1.336.0186
Open Access
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2020-05-29 12:25:10
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|
https://lo.calho.st/posts/wildfire-data/
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New Mexico is facing a severe fire season, and I wanted to visualize the progression of the major fires that are currently ongoing. However, I could not find a usable source of data on daily size and containment statistics.
A repository of daily updates exists, however as articles rather than structured data. I ended up scraping the website and applying natural language processing to extract the desired data.
The library and utility is available on my GitHub.
InciWeb contains a collection of the daily updates for the fires of interest, and by using Beautiful Soup I was able to collect and scrape all the articles. This was fairly straightforward.
Extracting the information of interest proved to be the more difficult task. Posts on the site come from various agencies and thus do not follow a consistent format. In order to handle all posts in generality, I turned to a NLP technique.
Using NLTK, I converted the words of the articles into their stems, then found the concordance lists of keywords pertaining to the data I needed.
After tagging the concordance strings with parts of speech, I was able to find the closest number to each keyword; this proved to be a fairly reliable heuristic.
def get_best_value(text, value_tag, label_values):
""" Apply the heuristic to extract the closest string with the specified tag
to one of the label values
"""
val = None
dist = 0
concordance_list = []
for l in label_values:
concordance_list += text.concordance_list(l)
for s in concordance_list:
tags = nltk.pos_tag(nltk.tokenize.word_tokenize(s.line))
label_candidates = []
value_candidates = []
for i, (token, tag) in enumerate(tags):
if tag == value_tag:
value_candidates.append((token, i))
elif token.lower() in label_values:
label_candidates.append((token, i))
for l, i in label_candidates:
for v, j in value_candidates:
d = abs(i - j)
if val is None or d <= dist:
val = v
dist = d
return val
def extract_data(raw_article_text):
""" Returns the size and containment of the fire by applying the heuristics to the text
"""
stemmer = nltk.stem.WordNetLemmatizer()
words = nltk.tokenize.word_tokenize(raw_article_text)
tokens = [stemmer.lemmatize(w) for w in words]
text = nltk.text.Text(tokens)
size = get_best_value(text, 'CD', ['acre', 'acres'])
containment = get_best_value(text, "CD", ['contained', 'containment'])
return size, containment
The rest of the code is not interesting, but eventually I was able to automate gathering most of the data I needed to produce this plot.
I used the utility in the above GitHub repository to dump the timeseries data from InciWeb to CSV, gathered some historical numbers from SWCC, and wrote a script to put it all together with matplotlib.
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2022-07-01 20:21:26
|
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|
https://cds.ismrm.org/protected/18MProceedings/PDFfiles/4258.html
|
### 4258
Hybrid-State Free Precession for Measuring Magnetic Resonance Relaxation Times
Jakob Assländer1,2, Daniel K Sodickson1,2,3, Riccardo Lattanzi1,2,3, and Martijn Cloos1,2
1Bernard and Irene Schwartz Center for Biomedical Imaging, Dept. of Radiology, New York University School of Medicine, New York, NY, United States, 2Center for Advanced Imaging Innovation and Research (CAI2R), Dept. of Radiology, New York University School of Medicine, New York, NY, United States, 3Sackler Institute of Graduate Biomedical Sciences, New York University School of Medicine, New York, NY, United States
### Synopsis
This work analyzes the spin physics in steady-state free precession sequences modified to smoothly vary sequence parameters, as suggested in MR Fingerprinting. We arrive at the conclusion that a transient state develops only in one direction, while the magnetization in the other two dimensions transitions adiabatically between steady states. We provide solutions of the Bloch Equations for this hybrid state and demonstrate the superior T1- and T2-encoding performance of the hybrid state compared to the steady state.
### Introduction
The original MR-Fingerprinting1 (MRF) implementation was based on a balanced-SSFP sequence modified to employ varying flip angles. Recent work2,3 indicates that spherical coordinates allow for a compact description of relaxation processes in such sequences. Here, we provide a solution of the Bloch Equations in spherical coordinates. A comprehensive analysis of the solution reveals that the spin dynamics can be separated into a transient-state component and two components that adiabatically transition between steady states, which inspires the term "balanced hybrid-state free precession" (bHSFP). Numerical optimizations demonstrate the superior $T_{1,2}$-encoding power of the hybrid state compared to the steady state.
### Theory
By applying a first order Taylor expansion with respect to sequence parameters (such as flip-angle, RF-phase, and $TR$) to the eigen-decomposition of the spin evolution matrix4,5, one can show that only the transient eigen-state parallel to the steady-state magnetization gets populated by smooth parameter variations. Spherical coordinates separate this transient component along the radial dimension $r$ from the adiabatic components along the direction of the phase $\varphi$ and the polar angle $\vartheta$. The latter two are given by the steady-state solution of the Bloch Equations6 and are approximated by$$\tan(\varphi-\frac{\phi}{2})=\frac{\cos\phi-\exp(-TR/T_2)}{\sin\phi}$$and$$\sin^2\vartheta=\frac{\sin^2\frac{\alpha}{2}}{\sin^2\frac{\phi}{2}\cdot\cos^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}}$$anywhere except near the stop band. Here, $\alpha$ denotes the flip angle, $\phi$ the phase accumulated during one $TR$, and we define $\phi=\pi$ as on-resonance. Given these adiabatic transitions, the Bloch Equation along the radial direction is uncoupled and is solved by$$r(t)=a(t)\cdot\left(r_0+\frac{1}{T_1}\int_{0}^{t}\frac{\cos\vartheta(\tau)}{a(\tau)}d\tau\right)$$with$$a(\tau)=\exp\left(-\int_{0}^{\tau}\frac{\sin^2\vartheta(\kappa)}{T_2}+\frac{\cos^2\vartheta(\kappa)}{T_1}d\kappa\right),$$in which we assumed departure from $r_0$, and $t$ denotes time. Note that neither $\vartheta$, nor $r$ depend on $TR$, given $TR\ll\{T_1,T_2\}$.
Instead of an initial magnetization, we can also assume periodic or anti-periodic boundary conditions defined by $r(0)=\pm{r}(TC)$, where $TC$ denotes the duration of one cycle of the experiment. In this case, the radial Bloch Equation is solved by $$r(t)=\frac{a(t)}{T_1}\cdot\left(\int_{0}^{t}\frac{\cos\vartheta(\tau)}{a(\tau)}d\tau-\frac{a(TC)}{a(TC)\mp1}\int_{0}^{TC}\frac{\cos\vartheta(\tau)}{a(\tau)}d\tau\right).$$
### Methods
On the search for the most efficient bHSFP sequence, we optimize the $\vartheta$-pattern. We employ the Broyden-Fletcher–Goldfarb-Shanno algorithm, and use the relative Cramer-Rao bound7-11 ($rCRB$), normalized by $TC$, as figure of merit. For comparison, optimizations were also performed for an inversion-recovery bSSFP sequence with a constant flip-angle12 and for a combination of spoiled gradient echo (SPGR) and bSSFP sequences in the steady state13.
In vivo measurements were performed with the anti-periodic bHSFP sequences, as well as with the optimized steady-state sequence, both with $TC = 3.8~\text{s}$. All experiments were performed on a 3T Prisma scanner (Siemens, Erlangen, Germany). Spatial encoding was performed with a sagittally oriented 3D stack-of-stars trajectory with golden angle increment14. The spatial resolution is $1~\text{mm}\times1~\text{mm}\times 2~\text{mm}$ at a FOV of $256~\text{mm}\times256~\text{mm}\times192~\text{mm}$. The total scan time was approximately $12.24~\text{min}$. Image reconstruction was performed with a low rank alternating direction method of multipliers approach15. $B_0$- and $B_1$-correction were performed on the basis of separate scans.
### Results and Discussion
The optimized bHSFP sequences exhibit smooth $\vartheta$-patterns (Figure$~$1), and the spin dynamics, as described by the derived approximate solution of the Bloch Equations, shows good accordance with Bloch simulations anywhere apart from the vicinity of the stop-band (Figure$~$2).
When plotting the optimized $rCRB$ as a function of $TC$ (Figure$~$3), one can discriminate different aspects of underlying spin physics: All transient-state flavors result in a lower (i.e. better) $rCRB$ compared to the steady-state optimizations for scans longer than $2~\text{s}$. Note that the (anti-)periodic solutions represent transient state equivalents to the steady state in the sense that the RF-train can be repeated without waiting times, allowing for an efficient and flexible implementation of long RF-trains, e.g., for 3D imaging. Contrary to the periodic, the anti-periodic boundary condition allows for repeated visits of the southern hemisphere (Figure$~$1), which is reflected in a reduced $rCRB$ (green vs. black marks in Figure$~$3). The $rCRB$ difference between anti-periodic boundary condition and fully relaxed inversion-recovery design reflects the increased net-magnetization available at the start of the sequence (black vs. red marks). Further, comparing the IR-bSSFP to the IR-bHSFP optimization (brown vs. red marks) highlights the improvement in SNR due to a variable flip angles. These improvements are particularly pronounced at long $TC$ where the IR-bSSFP converges to a steady state. Figure$~$4 verifies that the $rCRB$ improvements span over large areas in $T_1$-$T_2$-space.
In vivo images reconstructed using full Bloch simulations and the proposed solution are in excellent agreement (Figure 5; $T_1=965\pm23~\text{ms}$ vs. $T_1=988\pm23~\text{ms}$ and $T_2=48.2\pm3.0~\text{ms}$ vs. $T_2=49.7\pm2.9~\text{ms}$). The superior SNR efficiency of the bHSFP patterns compared to an steady-state approach ($T_1=1035\pm72~\text{ms}$, $T_2=37.8\pm2.6~\text{ms}$) can also be seen in Figure$~$5. Systematic deviations can be noted, which we associate mainly with magnetization transfer16 and partial volume effects17, which are not yet considered in this work.
### Conclusion
Slow sequence parameter variations result in a hybrid state, composed of transient- and steady-state components, which enables solving the Bloch Equations and provides insights on the spin dynamics.
### Acknowledgements
The authors would like to thank Steffen Glaser, Quentin Ansel and Dominique Sugny for fruitful discussions, and for giving insights into their optimal control implementation. The authors would also like to acknowledge Jeffrey Fessler and Gopal Nataraj discussions regarding the solution of the simplified Bloch Equation.
This work was supported by the research grants NIH/NIBIB R21 EB020096 and NIH/NIAMS R01 AR070297, and was performed under the rubric of the Center for Advanced Imaging Innovation and Research (CAI2R, www.cai2r.net), a NIBIB Biomedical Technology Resource Center (NIH P41 EB017183).
### References
[1] Dan Ma, Vikas Gulani, Nicole Seiberlich, Kecheng Liu, Jeffrey L. Sunshine, Jeffrey L. Duerk, and Mark A. Griswold. Magnetic resonance fingerprinting. Nature, 495(7440):187–192, 2013.
[2] Jakob Assländer, Steffen J. Glaser, and Jürgen Hennig. Pseudo Steady-State Free Precession for MR-Fingerprinting. Magn. Reson. Med., 77(3):1151–1161, 2017.
[3] Jakob Assländer, Daniel K Sodickson, Riccardo Lattanzi, and Martijn A Cloos. Relaxation in Polar Coordinates: Analysis and Optimization of MR-Fingerprinting. In Proc. Intl. Soc. Mag. Reson. Med. 26, page 127, 2017.
[4] Brian A. Hargreaves, Shreyas S. Vasanawala, John M. Pauly, and Dwight G. Nishimura. Characterization and reduction of the transient response in steady-state MR imaging. Magn. Reson. Med., 46(1):149–158, 2001.
[5] Carl Ganter. Off-resonance effects in the transient response of SSFP sequences. Magn. Reson. Med., 52(2):368–375, 2004.
[6] Ray Freeman and H. D W Hill. Phase and intensity anomalies in Fourier transform NMR. J. Magn. Reson., 4(3):366–383, 1971.
[7] Calyampudi Radhakrishna Rao. Information and the Accuracy Attainable in the Estimation of Statistical Parameters. Bull. Calcutta Math. Soc., 37(3):81–91, 1945.
[8] Harald Cramér. Methods of mathematical statistics. Princeton University Press, Princeton, NJ, 1946.
[9] J.A. Jones, P. Hodgkinson, A.L. Barker, and P.J. Hore. Optimal Sampling Strategies for the Measurement of Spin–Spin Relaxation Times. J. Magn. Reson. Ser. B, 113(1):25–34, 1996.
[10] J A Jones. Optimal sampling strategies for the measurement of relaxation times in proteins. J. Magn. Reson., 126(126):283–286,1997.
[11] Bo Zhao, Justin P Haldar, Kawin Setsompop, and Lawrence L Wald. Optimal Experiment Design for Magnetic Resonance Fingerprinting. In Eng. Med. Biol. Soc. (EMBC), IEEE 38thAnnu. Int. Conf., number 1, pages 453–456, 2016.
[12] Peter Schmitt, Mark A Griswold, Peter M Jakob, Markus Kotas, Vikas Gulani, Michael Flentje, and Axel Haase. Inversion recovery TrueFISP: quantification of T1, T2, and spin density. Magn. Reson. Med., 51(4):661–667, 2004.
[13] Sean C L Deoni, Brian K. Rutt, and Terry M. Peters. Rapid combined T1 and T2 mapping using gradient recalled acquisition in the steady state. Magn. Reson. Med., 49(3):515–526, 2003.
[14] Stefanie Winkelmann, Tobias Schaeffter, Thomas Koehler, Holger Eggers, and Olaf Doessel. An Optimal Radial Profile Order Based on the Golden Ratio for Time-Resolved MRI. IEEE Trans. Med.Imaging, 26(1):68–76, 2007.
[15] Jakob Assländer, Martijn A Cloos, Florian Knoll, Daniel K Sodickson, Jürgen Hennig, and Riccardo Lattanzi. Low Rank Alternating Direction Method of Multipliers Reconstruction for MR Fingerprinting. Magn. Reson. Med., doi:10.1002/mrm.26639, 2017.
[16] Tom Hilbert, Tobias Kober, Tiejun Zhao, Kai Tobias Block, Zidan Yu, Jean-Philippe Thiran, Gunnar Krueger, Daniel K Sodickson, and Martijn Cloos. Mitigating the Effect of Magnetization Transfer in Magnetic Resonance Fingerprinting. In Proc. Intl.Soc. Mag. Reson. Med. 25, 2017.
[17] Debra McGivney, Anagha Deshmane, Yun Jiang, Dan Ma, and Mark A Griswold. The Partial Volume Problem in MR Fingerprinting from a Bayesian Perspective. In Proc. 24th Annu.Meet. ISMRM, page 435, 2016.
### Figures
Figure 1: The optimized spin trajectories are depicted for periodic (r(0) = r(TC)) and anti-periodic (r(0) = -r(TC)) boundary conditions, as well as with the search space limited to the bSSFP steady-state ellipse (blue line in g) and the SPGR steady-state ellipse (red line in g). Note that ϑ of the bSSFP segment in (g-i) was retrospectively sorted in increasing order. The polar angle was limited to 0 ≤ ϑ ≤ π/4 for the transient state optimizations (a-f) and to 0 ≤ ϑ ≤ π/2 for the steady-state optimization (g-i).
Figure 2: The derived solution of the Bloch Equations is verified against Bloch simulations at the example of the optimized anti-periodic pattern. In agreement with the approximate nature of the derivation, good accordance can be observed anywhere apart from the vicinity of the stop band. The gray areas indicate time points that were not acquired in the in vivo scan since the polar angle is close to zero.
Figure 3: The depicted relative Cramer-Rao bounds can be understood as a lower bound of the squared inverse SNR efficiency per unit time. The legend is organized in blocks, where the first one (original DESPOT13, MRF1 and pSSFP2) are sequences taken from literature without modifications. The second block (steady-state and IR-bSSFP) are traditional sequences with optimized parameters and the last block contains the bHSFP sequences described by the derived approximate solution of the Bloch Equations. Note that all three sub-figures result from separate optimizations with rCRB(T1), rCRB(T2) and rCRB(T1) + rCRB(T2) as figures of merit and the data depicted in (c) are not the sum of the data depicted in (a) and (b).
Figure 4: The performance of the optimized patterns is illustrated through plots of the relative Cramer-Rao bounds, which provide a lower bound for the noise in the retrieved relaxation times. All patterns were optimized for T1 = 781 ms and T2 = 65 ms, as indicated by the red square, and were tested for the entire parameter space in a sample MRF dictionary. Note the logarithmic scale in all three dimensions.
Figure 5: A single sagittal slice of an in vivo 3D data set is depicted. The data were acquired with the excitation patterns depicted in Figure 1e,h. The parameter maps have a resolution of 1 mm x 1 mm x 2 mm. The anti-periodic data set was reconstructed once with a dictionary calculated with Bloch simulations, and once with a dictionary calculated based on the derived approximate solution of the Bloch Equations. The red box indicates a region of interest used for extracting T1 and T2 values. Note the logarithmic scale of the color maps.
Proc. Intl. Soc. Mag. Reson. Med. 26 (2018)
4258
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2021-06-22 05:01:45
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|
https://academ.us/list/econ/
|
### Mental health concerns prelude the Great Resignation: Evidence from Social Media
To study the causes of the 2021 Great Resignation, we use text analysis to investigate the changes in work- and quit-related posts between 2018 and 2021 on Reddit. We find that the Reddit discourse evolution resembles the dynamics of the U.S. quit and layoff rates. Furthermore, when the COVID-19 pandemic started, conversations related to working from home, switching jobs, work-related distress, and mental health increased. We distinguish between general work-related and specific quit-related discourse changes using a difference-in-differences method. Our main finding is that mental health and work-related distress topics disproportionally increased among quit-related posts since the onset of the pandemic, likely contributing to the Great Resignation. Along with better labor market conditions, some relief came beginning-to-mid-2021 when these concerns decreased. Our study validates the use of forums such as Reddit for studying emerging economic phenomena in real time, complementing traditional labor market surveys and administrative data.
### On Gale's Contribution in Revealed Preference Theory
We investigate Gale's important paper published in 1960. This paper contains an example of a candidate of the demand function that satisfies the weak axiom of revealed preference and that is doubtful that it is a demand function of some weak order. We examine this paper and first scrutinize what Gale proved. Then we identify a gap in Gale's proof and show that he failed to show that this candidate of the demand function is not a demand function. Next, we present three complete proofs of Gale's claim. First, we construct a proof that was constructible in 1960 by a fact that Gale himself demonstrated. Second, we construct a modern and simple proof using Shephard's lemma. Third, we construct a proof that follows the direction that Gale originally conceived. Our conclusion is as follows: although, in 1960, Gale was not able to prove that the candidate of the demand function that he constructed is not a demand function, he substantially proved it, and therefore it is fair to say that the credit for finding a candidate of the demand function that satisfies the weak axiom but is not a demand function is attributed to Gale.
### Time is limited on the road to asymptopia
One challenge in the estimation of financial market agent-based models (FABMs) is to infer reliable insights using numerical simulations validated by only a single observed time series. Ergodicity (besides stationarity) is a strong precondition for any estimation, however it has not been systematically explored and is often simply presumed. For finite-sample lengths and limited computational resources empirical estimation always takes place in pre-asymptopia. Thus broken ergodicity must be considered the rule, but it remains largely unclear how to deal with the remaining uncertainty in non-ergodic observables. Here we show how an understanding of the ergodic properties of moment functions can help to improve the estimation of (F)ABMs. We run Monte Carlo experiments and study the convergence behaviour of moment functions of two prototype models. We find infeasibly-long convergence times for most. Choosing an efficient mix of ensemble size and simulated time length guided our estimation and might help in general.
### Generic catastrophic poverty when selfish investors exploit a degradable common resource
The productivity of a common pool of resources may degrade when overly exploited by a number of selfish investors, a situation known as the tragedy of the commons (TOC). Without regulations, agents optimize the size of their individual investments into the commons by balancing incurring costs with the returns received. The resulting Nash equilibrium involves a self-consistency loop between individual investment decisions and the state of the commons. As a consequence, several non-trivial properties emerge. For $N$ investing actors we proof rigorously that typical payoffs do not scale as $1/N$, the expected result for cooperating agents, but as $(1/N)^2$. Payoffs are hence functionally reduced, a situation denoted catastrophic poverty. This occurs despite the fact that the cumulative investment remains finite when $N\to\infty$. Catastrophic poverty is instead a consequence of an increasingly fine-tuned balance between returns and costs. In addition, we point out that a finite number of oligarchs may be present. Oligarchs are characterized by payoffs that are finite and not decreasing when $N$ increases. Our results hold for generic classes of models, including convex and moderately concave cost functions. For strongly concave cost functions the Nash equilibrium undergoes a collective reorganization, being characterized instead by entry barriers and sudden death forced market exits.
### Algorithmic Fairness and Statistical Discrimination
Algorithmic fairness is a new interdisciplinary field of study focused on how to measure whether a process, or algorithm, may unintentionally produce unfair outcomes, as well as whether or how the potential unfairness of such processes can be mitigated. Statistical discrimination describes a set of informational issues that can induce rational (i.e., Bayesian) decision-making to lead to unfair outcomes even in the absence of discriminatory intent. In this article, we provide overviews of these two related literatures and draw connections between them. The comparison illustrates both the conflict between rationality and fairness and the importance of endogeneity (e.g., "rational expectations" and "self-fulfilling prophecies") in defining and pursuing fairness. Taken in concert, we argue that the two traditions suggest a value for considering new fairness notions that explicitly account for how the individual characteristics an algorithm intends to measure may change in response to the algorithm.
### Ban The Box? Information, Incentives, and Statistical Discrimination
"Banning the Box" refers to a policy campaign aimed at prohibiting employers from soliciting applicant information that could be used to statistically discriminate against categories of applicants (in particular, those with criminal records). In this article, we examine how the concealing or revealing of informative features about an applicant's identity affects hiring both directly and, in equilibrium, by possibly changing applicants' incentives to invest in human capital. We show that there exist situations in which an employer and an applicant are in agreement about whether to ban the box. Specifically, depending on the structure of the labor market, banning the box can be (1) Pareto dominant, (2) Pareto dominated, (3) benefit the applicant while harming the employer, or (4) benefit the employer while harming the applicant. Our results have policy implications spanning beyond employment decisions, including the use of credit checks by landlords and standardized tests in college admissions.
### Characterizing M-estimators
We characterize the full classes of M-estimators for semiparametric models of general functionals by formally connecting the theory of consistent loss functions from forecast evaluation with the theory of M-estimation. This novel characterization result opens up the possibility for theoretical research on efficient and equivariant M-estimation and, more generally, it allows to leverage existing results on loss functions known from the literature of forecast evaluation in estimation theory.
### Debiased Inference on Identified Linear Functionals of Underidentified Nuisances via Penalized Minimax Estimation
We study generic inference on identified linear functionals of nonunique nuisances defined as solutions to underidentified conditional moment restrictions. This problem appears in a variety of applications, including nonparametric instrumental variable models, proximal causal inference under unmeasured confounding, and missing-not-at-random data with shadow variables. Although the linear functionals of interest, such as average treatment effect, are identifiable under suitable conditions, nonuniqueness of nuisances pose serious challenges to statistical inference, since in this setting common nuisance estimators can be unstable and lack fixed limits. In this paper, we propose penalized minimax estimators for the nuisance functions and show they enable valid inference in this challenging setting. The proposed nuisance estimators can accommodate flexible function classes, and importantly, they can converge to fixed limits determined by the penalization, regardless of whether the nuisances are unique or not. We use the penalized nuisance estimators to form a debiased estimator for the linear functional of interest and prove its asymptotic normality under generic high-level conditions, which provide for asymptotically valid confidence intervals.
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2022-08-18 08:39:33
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http://vm.udsu.ru/issues/archive/issue/2015-1-9
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+7 (3412) 91 60 92
## Archive of Issues
Russia Yekaterinburg
Year
2015
Volume
25
Issue
1
Pages
78-92
Section Mathematics Title Convergence of the difference method of solving the two-dimensional wave equation with heredity Author(-s) Tashirova E.E.a Affiliations Ural Federal Universitya Abstract The paper presents the consideration of the wave equation with two space variables and one time variable and with heredity effect $$\frac{\partial^2 u}{\partial t^2}=a^2\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right) + f\big(x,y,t,u(x,y,t),u_t(x,y,\cdot)\big), \quad u_t(x,y,\cdot)=\big\{u(x,y,t+\xi),-\tau \leqslant \xi\leqslant 0\big\}.$$ A family of grid methods is constructed for the numerical solution of this equation; the methods are based on the idea of separating the current state and the history function. A complete analog of the factorization method which is known for an equation without delay is constructed according to the current state. Influence of prehistory is taken into consideration by interpolation constructions. The local error order of the algorithm is investigated. A theorem on the convergence and on the order of convergence of methods is obtained by means of embedding into a general difference scheme with aftereffect. The results of calculating a test example with variable delay are presented. Keywords difference methods, two-dimensional wave equation, time delay interpolation, factorization, order of convergence UDC 519.633 MSC 35L20 DOI 10.20537/vm150109 Received 17 December 2014 Language Russian Citation Tashirova E.E. Convergence of the difference method of solving the two-dimensional wave equation with heredity, Vestnik Udmurtskogo Universiteta. Matematika. Mekhanika. Komp'yuternye Nauki, 2015, vol. 25, issue 1, pp. 78-92. References Wu J. Theory and application of partial functional differential equations, New York: Springer-Verlag, 1996, 438 p. Pimenov V.G., Tashirova E.E. Numerical methods for solving a hereditary equation of hyperbolic type, Proceedings of the Steklov Institute of Mathematics, 2013, vol. 281, issue 1 supplement, pp. 126-136. Pimenov V.G., Lozhnikov A.B. Difference schemes for the numerical solution of the heat conduction equation with aftereffect, Proceedings of the Steklov Institute of Mathematics, 2011, vol. 275, issue 1 supplement, pp. 137-148. Samarskii А.А. Teoriya raznostnykh skhem (Theory of difference schemes), Moscow: Nauka, 1989, 656 p. Pimenov V.G. General linear methods for the numerical solution of functional-differential equations, Differential Equations, 2001, vol. 37, no. 1, pp. 116-127. Kim A.V., Pimenov V.G. i-gladkii analiz i chislennye metody resheniya funktsional'no-differentsial'nykh uravnenii (i-smooth calculus and numerical methods for functional differential equations), Moscow-Izhevsk: Regular and Chaotic Dynamics, 2004, 256 p. Lekomtsev A.V., Pimenov V.G. Convergence of the alternating direction method for the numerical solution of a heat conduction equation with delay, Proceedings of the Steklov Institute of Mathematics, 2011, vol. 272, issue 1 supplement, pp. 101-118. Kalitkin N.N. Chislennye metody (Numerical methods), St. Petersburg: BHV-Petersburg, 2011, 586 p. Tashirova E.E. Numerical methods for solving two-dimensional wave equation with aftereffect, Vestn. Tambov. Univ. Ser. Estestv. Tekh. Nauki, 2013, vol. 18, no. 5-2, pp. 2704-2706 (in Russian). Full text
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2020-11-29 16:32:31
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https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.21/share/doc/Macaulay2/Topcom/html/index.html
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# Topcom -- interface to selected functions from topcom package
## Description
Topcom is mathematical software written by Jorg Rambau for computing and manipulating triangulations of polytopes and chirotopes.
This Macaulay2 package provides an interface for some of the functionality of this software.
The package is meant as an internal package, to be called by packages such as Polyhedra. It is highly recommended to use those packages, rather than calling this package directly.
## Chirotope functions
The chirotope of a point set (or vector set), consisting of the columns of the $d \times n$ matrix $A$, is the function which assigns to each $d+1 \times d+1$ minor of $\overline{A}$ the sign of its determinant. Topcom encodes this into a string. See chirotopeString for details about the output of this function.
• Polyhedra -- for computations with convex polyhedra, cones, and fans
• Triangulations -- generating and manipulating triangulations of point or vector configurations
• ReflexivePolytopesDB -- simple access to Kreuzer-Skarke database of reflexive polytopes of dimensions 3 and 4
## Version
This documentation describes version 0.95 of Topcom.
## Source code
The source code from which this documentation is derived is in the file Topcom.m2.
## For the programmer
The object Topcom is .
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2023-03-24 21:39:57
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https://solvedlib.com/find-the-equation-of-a-line-parallel-to-2x5y3-and,15814
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Find the equation of a line parallel to 2x+5y=3 and passing through (2,-3)?
Question:
Find the equation of a line parallel to 2x+5y=3 and passing through (2,-3)?
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2023-03-25 07:19:01
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https://community.wolfram.com/groups/-/m/t/1363990?p_p_auth=D6t4r81A
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# Specify a 3D array, where array elements contain other kind of variables?
Posted 4 years ago
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Dear Community,In Fortran I can easily specify a 3D array, where array elements contain other kind of variables(real, integer, logical, vector, etc) This is very effivcient and conveniet. An example is below: GBLOCK( 88 , 21 , 12 ).BO = 1.22 (real) ; GBLOCK( 88 , 21 , 12 ).KRO = 0.7 (real) ; GBLOCK( 88 , 21 , 12 ).ACTIVE = .TRUE. (logical) ; GBLOCK( 88 , 21 , 12 ).CT(2) = 55.7 (vector, real) ; GBLOCK( 88 , 21 , 12 ).CT(4) = 19.2 (vector, real) ; GBLOCK( 88 , 21 , 12 ).IBLOCK = 43561 (integer) ; My question is, how could I specify a similar structure in Mathematica? What would be the equivalent of the above array in Mathematica?Tx for the kind help in advance, best regards Andras
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Posted 4 years ago
There's no requirement that the members of a Mathematica list be of the same type In[1]:= ar = {1, 0.3, "test", \[Pi], 1 + I, \[Infinity]} Out[1]= {1, 0.3, "test", \[Pi], 1 + I, \[Infinity]}
Anonymous User
Anonymous User
Posted 4 years ago
Head[x] -> SymbolHead["the"] -> StringHead[7] -> ...as the last poster said, mathematica List[{a,b,c}] or Array[?], you do not need to worry about mixing typesfurthermore, arrays in Mathematica are associative Ar["the"]:=1 Ar["that"]:="there" Ar["that"]:=Ar["the"] all that works, and we expect if there are many strings for indicies we expect mathematica employs at least a binary search to cross reference the value
Posted 4 years ago
Tx for the contributions.I have finally found an easy way to do it. One possible solution is in the attached notebook.cheers, Andras Attachments:
Posted 4 years ago
Andras,I think you are overcomplicating things. As Frank pointed out, you can automatically put different types in an array.You can just put whatever you want in the array and reference individual elements: In[20]:= var = {{18, False, 12.2}, {23.111, True, 34.11}} Out[20]= {{18, False, 12.2}, {23.111, True, 34.11}} In[21]:= var[[2]] Out[21]= {23.111, True, 34.11} In[23]:= var[[2, 1]] = 24; In[18]:= var[[2]] Out[18]= {24, True, 34.11} I do not understand why you need all that indirect referencing. Maybe I am missing something.Regards,Neil
Posted 4 years ago
I need this indirect referencing because then I do not have to keep in mind in which position an element in the list is. E.g. activeind = 3 , so I can referrence an active / inactive 3D grid property easily by gblock[[ j, i, k, activeind]] = True (or False) . (The Dollar sign was removed from activeind.) I'm actually transfering an older Fortran code of mine into the Wolfram Language, and this seemed to me the easiest way to do it. In Fortran I do the same this way: GBLOCK(J,I,K).ACTIVE = .TRUE. (or .FALSE.)If there is a more efficient way, I would be gratefuly to learn it.Tx, Andras
Posted 4 years ago
As an alternative you could make an array of associations. And use the associations keys to access the elements. In the attachment I have made an adaption of your notebook. Attachments:
Posted 4 years ago
Dear Hans,Very elegant, thank you very much!best regardsAndras
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2022-08-14 20:43:23
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https://www.aimsciences.org/article/doi/10.3934/dcds.2011.31.1233
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# American Institute of Mathematical Sciences
December 2011, 31(4): 1233-1248. doi: 10.3934/dcds.2011.31.1233
## A variational approach to semilinear elliptic equations with measure data
1 Dipartimento di Matematica e Fisica, Università Cattolica del Sacro Cuore, Via dei Musei 41 - 25121 Brescia, Italy 2 Dipartimento di Matematica e Applicazioni, Università di Milano Bicocca, Via Cozzi 53 - 20125 Milano, Italy
Received October 2010 Revised December 2010 Published September 2011
We describe a direct variational approach to a class of semilinear elliptic equations with measure data. Using a typical variational argument, we show the existence of multiple solutions.
Citation: Marco Degiovanni, Michele Scaglia. A variational approach to semilinear elliptic equations with measure data. Discrete & Continuous Dynamical Systems - A, 2011, 31 (4) : 1233-1248. doi: 10.3934/dcds.2011.31.1233
##### References:
[1] P. Baras and M. Pierre, Singularités éliminables pour des équations semi-linéaires,, Ann. Inst. Fourier (Grenoble), 34 (1984), 185. doi: 10.5802/aif.956. [2] P. Bénilan, L. Boccardo, T. Gallouët, R. Gariepy, M. Pierre and J. L. Vázquez, An $L^1$-theory of existence and uniqueness of solutions of nonlinear elliptic equations,, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 22 (1995), 241. [3] P. Bénilan and H. Brezis, Nonlinear problems related to the Thomas-Fermi equation,, Dedicated to Philippe Bénilan, 3 (2003), 673. [4] L. Boccardo and T. Gallouët, Nonlinear elliptic equations with right hand side measures,, Comm. Partial Differential Equations, 17 (1992), 641. [5] L. Boccardo, T. Gallouët and L. Orsina, Existence and uniqueness of entropy solutions for nonlinear elliptic equations with measure data,, Ann. Inst. H. Poincaré Anal. Non Linéaire, 13 (1996), 539. [6] H. Brezis and F. Browder, A property of Sobolev spaces,, Comm. Partial Differential Equations, 4 (1979), 1077. [7] H. Brezis, M. Marcus and A. C. Ponce, Nonlinear elliptic equations with measures revisited,, in, 163 (2007), 55. [8] H. Brezis and W. A. Strauss, Semi-linear second-order elliptic equations in $L^1$,, J. Math. Soc. Japan, 25 (1973), 565. doi: 10.2969/jmsj/02540565. [9] A. Canino and M. Degiovanni, A variational approach to a class of singular semilinear elliptic equations,, J. Convex Anal., 11 (2004), 147. [10] K.-C. Chang, "Infinite-Dimensional Morse Theory and Multiple Solution Problems,", Progress in Nonlinear Differential Equations and their Applications, 6 (1993). [11] G. Dal Maso, F. Murat, L. Orsina and A. Prignet, Renormalized solutions of elliptic equations with general measure data,, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 28 (1999), 741. [12] E. De Giorgi, Sulla differenziabilità e l'analiticità delle estremali degli integrali multipli regolari,, Mem. Accad. Sci. Torino. Cl. Sci. Fis. Mat. Nat. (3), 3 (1957), 25. [13] M. Degiovanni and M. Marzocchi, On the Euler-Lagrange equation for functionals of the calculus of variations without upper growth conditions,, SIAM J. Control Optim., 48 (2009), 2857. doi: 10.1137/090747968. [14] A. Ferrero and C. Saccon, Existence and multiplicity results for semilinear equations with measure data,, Topol. Methods Nonlinear Anal., 28 (2006), 285. [15] A. Ferrero and C. Saccon, Existence and multiplicity results for semilinear elliptic equations with measure data and jumping nonlinearities,, Topol. Methods Nonlinear Anal., 30 (2007), 37. [16] A. Ferrero and C. Saccon, Multiplicity results for a class of asymptotically linear elliptic problems with resonance and applications to problems with measure data,, Adv. Nonlinear Stud., 10 (2010), 433. [17] T. Gallouët and J.-M. Morel, Resolution of a semilinear equation in $L^1$,, Proc. Roy. Soc. Edinburgh Sect. A, 96 (1984), 275. [18] T. Gallouët and J.-M. Morel, Corrigenda: "Resolution of a semilinear equation in $L^1$,", Proc. Roy. Soc. Edinburgh Sect. A, 99 (1985). [19] J. Moser, A new proof of De Giorgi's theorem concerning the regularity problem for elliptic differential equations,, Comm. Pure Appl. Math., 13 (1960), 457. doi: 10.1002/cpa.3160130308. [20] J. Moser, On Harnack's theorem for elliptic differential equations,, Comm. Pure Appl. Math., 14 (1961), 577. doi: 10.1002/cpa.3160140329. [21] J. Nash, Continuity of solutions of parabolic and elliptic equations,, Amer. J. Math., 80 (1958), 931. doi: 10.2307/2372841. [22] L. Orsina, Solvability of linear and semilinear eigenvalue problems with $L\^1$ data,, Rend. Sem. Mat. Univ. Padova, 90 (1993), 207. [23] L. Orsina and A. Ponce, Semilinear elliptic equations and systems with diffuse measures,, J. Evol. Equ., 8 (2008), 781. doi: 10.1007/s00028-008-0446-32. [24] G. Stampacchia, Le problème de Dirichlet pour les équations elliptiques du second ordre à coefficients discontinus,, Ann. Inst. Fourier (Grenoble), 15 (1965), 189. doi: 10.5802/aif.204. [25] G. Stampacchia, "Équations Elliptiques du Second Ordre à Coefficients Discontinus,", Séminaire de Mathématiques Supérieures, 16 (1966). [26] A. Szulkin, Minimax principles for lower semicontinuous functions and applications to nonlinear boundary value problems,, Ann. Inst. H. Poincaré Anal. Non Linéaire, 3 (1986), 77. [27] N. S. Trudinger and X.-J. Wang, Quasilinear elliptic equations with signed measure data,, Discrete Contin. Dyn. Syst., 23 (2009), 477. doi: 10.3934/dcds.2009.23.477.
show all references
##### References:
[1] P. Baras and M. Pierre, Singularités éliminables pour des équations semi-linéaires,, Ann. Inst. Fourier (Grenoble), 34 (1984), 185. doi: 10.5802/aif.956. [2] P. Bénilan, L. Boccardo, T. Gallouët, R. Gariepy, M. Pierre and J. L. Vázquez, An $L^1$-theory of existence and uniqueness of solutions of nonlinear elliptic equations,, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 22 (1995), 241. [3] P. Bénilan and H. Brezis, Nonlinear problems related to the Thomas-Fermi equation,, Dedicated to Philippe Bénilan, 3 (2003), 673. [4] L. Boccardo and T. Gallouët, Nonlinear elliptic equations with right hand side measures,, Comm. Partial Differential Equations, 17 (1992), 641. [5] L. Boccardo, T. Gallouët and L. Orsina, Existence and uniqueness of entropy solutions for nonlinear elliptic equations with measure data,, Ann. Inst. H. Poincaré Anal. Non Linéaire, 13 (1996), 539. [6] H. Brezis and F. Browder, A property of Sobolev spaces,, Comm. Partial Differential Equations, 4 (1979), 1077. [7] H. Brezis, M. Marcus and A. C. Ponce, Nonlinear elliptic equations with measures revisited,, in, 163 (2007), 55. [8] H. Brezis and W. A. Strauss, Semi-linear second-order elliptic equations in $L^1$,, J. Math. Soc. Japan, 25 (1973), 565. doi: 10.2969/jmsj/02540565. [9] A. Canino and M. Degiovanni, A variational approach to a class of singular semilinear elliptic equations,, J. Convex Anal., 11 (2004), 147. [10] K.-C. Chang, "Infinite-Dimensional Morse Theory and Multiple Solution Problems,", Progress in Nonlinear Differential Equations and their Applications, 6 (1993). [11] G. Dal Maso, F. Murat, L. Orsina and A. Prignet, Renormalized solutions of elliptic equations with general measure data,, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 28 (1999), 741. [12] E. De Giorgi, Sulla differenziabilità e l'analiticità delle estremali degli integrali multipli regolari,, Mem. Accad. Sci. Torino. Cl. Sci. Fis. Mat. Nat. (3), 3 (1957), 25. [13] M. Degiovanni and M. Marzocchi, On the Euler-Lagrange equation for functionals of the calculus of variations without upper growth conditions,, SIAM J. Control Optim., 48 (2009), 2857. doi: 10.1137/090747968. [14] A. Ferrero and C. Saccon, Existence and multiplicity results for semilinear equations with measure data,, Topol. Methods Nonlinear Anal., 28 (2006), 285. [15] A. Ferrero and C. Saccon, Existence and multiplicity results for semilinear elliptic equations with measure data and jumping nonlinearities,, Topol. Methods Nonlinear Anal., 30 (2007), 37. [16] A. Ferrero and C. Saccon, Multiplicity results for a class of asymptotically linear elliptic problems with resonance and applications to problems with measure data,, Adv. Nonlinear Stud., 10 (2010), 433. [17] T. Gallouët and J.-M. Morel, Resolution of a semilinear equation in $L^1$,, Proc. Roy. Soc. Edinburgh Sect. A, 96 (1984), 275. [18] T. Gallouët and J.-M. Morel, Corrigenda: "Resolution of a semilinear equation in $L^1$,", Proc. Roy. Soc. Edinburgh Sect. A, 99 (1985). [19] J. Moser, A new proof of De Giorgi's theorem concerning the regularity problem for elliptic differential equations,, Comm. Pure Appl. Math., 13 (1960), 457. doi: 10.1002/cpa.3160130308. [20] J. Moser, On Harnack's theorem for elliptic differential equations,, Comm. Pure Appl. Math., 14 (1961), 577. doi: 10.1002/cpa.3160140329. [21] J. Nash, Continuity of solutions of parabolic and elliptic equations,, Amer. J. Math., 80 (1958), 931. doi: 10.2307/2372841. [22] L. Orsina, Solvability of linear and semilinear eigenvalue problems with $L\^1$ data,, Rend. Sem. Mat. Univ. Padova, 90 (1993), 207. [23] L. Orsina and A. Ponce, Semilinear elliptic equations and systems with diffuse measures,, J. Evol. Equ., 8 (2008), 781. doi: 10.1007/s00028-008-0446-32. [24] G. Stampacchia, Le problème de Dirichlet pour les équations elliptiques du second ordre à coefficients discontinus,, Ann. Inst. Fourier (Grenoble), 15 (1965), 189. doi: 10.5802/aif.204. [25] G. Stampacchia, "Équations Elliptiques du Second Ordre à Coefficients Discontinus,", Séminaire de Mathématiques Supérieures, 16 (1966). [26] A. Szulkin, Minimax principles for lower semicontinuous functions and applications to nonlinear boundary value problems,, Ann. Inst. H. Poincaré Anal. Non Linéaire, 3 (1986), 77. [27] N. S. Trudinger and X.-J. Wang, Quasilinear elliptic equations with signed measure data,, Discrete Contin. Dyn. Syst., 23 (2009), 477. doi: 10.3934/dcds.2009.23.477.
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2017 Impact Factor: 1.179
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2019-06-20 17:05:22
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https://help.altair.com/hwsolvers/ms/topics/tutorials/mv/tut_mv_1000_t.htm
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MV-1000: Interactive Model Building and Simulation
In this tutorial, you will learn how to create a model of a four-bar trunk lid mechanism interactively through the MotionView graphical user interface, perform a kinematic analysis on the model using MotionSolve, and post-process the MotionSolve results in the animation and plot windows.
A brief overview of Multi Body Dynamics (MBD) is provided below.
Multi Body Dynamics (MBD)
MBD is defined as the “study of dynamics of a system of interconnected bodies”. A mechanism (MBD system) constitutes a set of links (bodies) connected (constrained) with each other to perform a specified action under application of force or motion. The motion of mechanisms is defined by its kinematic behavior. The dynamic behavior of a mechanism results from the equilibrium of applied forces and the rate of the change of momentum.
MBD Modeling
A classical MBD formulation uses a rigid body modeling approach to model a mechanism. A rigid body is defined as a body in which deformation is negligible. In general, to solve an MBD problem, the solver requires following information:
• Rigid body inertia and location.
• Connections – type, bodies involved, location, and orientation.
• Forces and motions – bodies involved, location, orientation, and value.
MotionView facilitates quick and easy ways of modeling items, such as a system, through graphical visualization.
Trunk-Lid Model
The trunk-lid shown in the image below uses a four-bar mechanism for its opening and closing motions. Below is a schematic diagram of the mechanism: The four links (bodies) in a four-bar mechanism are: Ground Body, Follower, Coupler, and Input Link. In this example, the Ground Body is the car body and Input Link is the trunk-lid body. The remaining two bodies (Follower and Coupler) form the part of the mechanism used to aid the opening and closing of car trunk-lid.
The following entities are needed to build this model:
• Points
• Bodies
• Constraints (Joints)
• Graphics
• Input (Motion or Force)
• Output
Create Points
In this step you will learn how to create points for your model.
From the mbd_modeling\interactive folder, copy trunk.hm and trunklid.hm to the <working directory>.
1. Start a new MotionView Session.
2. Add a point using one of the following methods:
• From the Project Browser, right-click on Model and select Add > Reference Entity > Point from the context menu.
• On the Reference Entity toolbar, right-click on the (Points) icon.
The Add Point or PointPair dialog will be displayed.
Note: Other entities like Bodies, Markers, and so on can also be created using either of the methods listed above (Project Browser or toolbar).
3. For Label, enter Point A.
4. For Variable, enter p_a.
The label allows you to identify an entity in the modeling window, while the variable name is used by MotionView to uniquely identify an entity.
Note: When using the Add "Entity" dialog for any entity, you can use the label and variable defaults. However, as a best modeling practice, it is recommended that you provide meaningful labels and variables for easy identification of the entities. For this exercise, please follow the prescribed naming conventions.
5. Click OK.
The Points panel is displayed. Point A is highlighted in the Points list of the Project Browser.
6. Enter the values for the X, Y, and Z coordinates for point A.
7. Create multiple points.
1. To create points B through I, repeat steps 2 through 4 and click Apply.
Remember: Substitute B, C, and so on for A when entering the label and variable names in the Add Point or PointPair dialog. Clicking the Apply button allows you to continue to add points without exiting the Add "Entity" dialog.
2. After keying in the label and variable name for Point I, click OK to close the dialog.
3. In the Points panel, click the Data Summary... button.
The Data Summary dialog shows the table of points and you can enter all the coordinates in this table.
Table 1.
Point Location
Label Variable X Y Z
Point A p_a 921 580 1124
Point B p_b 918 580 1114
Point C p_c 918 580 1104
Point D p_d 915 580 1106
Point E p_e 878 580 1108
Point F p_f 878 580 1118
Point G p_g 830 580 1080
Point H p_h 790 580 1088
Point I p_i 825 580 1109
Since the Y value for all the points are the same, you can parameterize the value for points B through I to the value of Point A. This process is explained in the next step.
8. Parameterize the value for points B through I to the value of Point A.
1. Select the Y coordinate field along Point B.
2. Click on the button to invoke the Expression Builder.
3. Select the Y value of Point A.
4. Copy the expression and paste into the Y coordinate field of the remaining points.
5. Enter the X and Z coordinates as listed in the table above.
Note: On the keyboard, press Enter to move on to the next field in the table.
6. Click Close.
9. On the Standard Views toolbar, change the view to left by clicking the (XZ Left Plane View) icon.
Create Bodies
In this step, you will create the Input Link, Coupler, and Follower rigid-body links in the mechanism.
The mechanism consists of four rigid-body links: Ground (car body), Input Link, Coupler, and Follower. Ground Body is available by default when a new MotionSolve client is invoked, hence creating the Ground Body separately is not required.
1. On the Reference Entity toolbar, right-click the (Bodies) icon.
2. In the Add Body or BodyPair dialog, Specify the label as InputLink and the variable name as b_inputlink.
3. Click OK.
4. Enter the values for mass and inertia.
1. Click the Properties tab.
2. Enter the following values:
• Mass=1
• Ixx, Iyy, Izz= 1000, Ixy, Ixz, Iyz=0
3. Click the CM Coordinates tab to specify the location of the center of mass of the body.
5. Select the Use center of mass coordinate system check box.
6. Under Origin, click the (Point collector).
A cyan border will appear around the collector to indicate it is now active for selection.
7. Select Point G.
1. Click again to launch the Select a Point dialog.
2. Select Point G from the Model Tree.
3. Click OK.
4. Retain the default orientation scheme (orient two axes) and accept the default values for .
Note: This method for selecting a point can also be applied to other entities such as: body, joint, and so on. For selecting the Ground Body or the Global Origin, you can click on the triad representing the Global Coordinate System on the screen .
Tip: You can also select Point G from the modeling window. Hold down the left mouse button and highlight Point G. Release the button to select it.
8. Repeat step 1 through 7 to create the remaining links with the variable names listed in Table 2.
Table 2.
Label Variable Name
Follower b_follower
Coupler b_coupler
9. Specify the mass and inertia for these links.
1. Mass= 1
2. Ixx, Iyy, Izz= 1000
3. Ixy, Ixz, Iyz=0
10. Specify points B and D as the origin of the center of mass marker for Follower and Coupler, respectively.
11. Retain the default orientation (Global coordinate system) for the CM marker.
Create Revolute Joints
In this task you will learn how to create the joints necessary for this model.
The mechanism consists of revolute joints at four points: A, C, E, and F. The axis of revolution is parallel to the global Y axis.
1. Add a Joint in one of the following ways:
• From the Project Browser, right-click on Model. From the context menu, select Add > Constraint > Joint.
• Right-click the (Joints) icon.
The Add Joint or JointPair dialog will display.
2. Specify the label as Follower-Ground and variable names as j_follower_ground for the new joint.
3. Under Type, select Revolute Joint from the drop-down menu.
4. Click OK.
The Joints panel will be displayed. The new joint you added will be highlighted in the Model Tree in the Project Browser.
5. Under the Connectivity tab, double click (first Body collector).
The Select a Body dialog is displayed.
6. From the Model Tree, select Bodies from the left-hand column and Follower from the right-hand column.
7. Click OK.
Notice that in the Joints panel the Follower Body is selected for and the cyan border moves to
8. Click in the modeling window. With the left mouse button pressed, move the cursor to the global XYZ triad .
9. Release the left mouse button when Ground Body is displayed in the modeling window.
10. Under Origin, double click the collector.
the Select a Point dialog will display.
11. Select Point A as the joint origin.
12. Click OK.
13. To specify an axis of rotation, under Alignment Axis, click the downward pointing arrow next to Point and select Vector.
14. Specify the Global Y axis vector as the axis of rotation of the revolute joint.
15. Repeat steps 1 through 14 to create the three remaining revolute joints: points C, E, and F. Use the specifications shown in Table 3.
Table 3. Revolute joint information
Revolute Joint Label Variable Name Body 1 Body 2 Point Vector
Follower-Ground j_follower_ground Follower Ground A Global Y
Follower-Coupler j_follower_coupler Follower Coupler C Global Y
Coupler-Input j_coupler_input Coupler Input Link E Global Y
Input-Ground j_input_ground Input Link Ground F Global Y
Specify a Motion for the Mechanism
In this step you will learn how to add motion to your mechanism model.
The input for this model will be in the form of a Motion. A Motion can be specified as Linear, Expression, Spline3D, or Curve. In this step, you will specify a Motion using an Expression.
• From the Project Browser, right-click on Model and select Add > Constraint > Motion from the context menu.
• On the Constraint toolbar, right-click on the (Motion) icon.
2. Specify the label as Motion_Expression and the variable name as mot_expr.
3. Click OK.
The Motion panel is displayed. The new motion is highlighted in the model tree in the Project Browser.
4. From the Connectivity tab, double click on the (Joint collector).
The Select a Joint dialog is displayed.
5. From the Model Tree, select the revolute joint at Point F (Input Ground) that you created in the previous step.
6. Click OK.
The Motion panel will be displayed.
7. From the Properties tab, under Define by, click on the gray arrow and select Expression.
8. Click in the Expression field.
The Expression Builder is now activated.
9. Click on the button to open the Expression Builder and enter following expression between the back quotes 60d*sin(2*0.1*PI*TIME).
The expression is a SIN function with an amplitude of 60 degrees and frequency of 0.1 Hz. With this expression the trunk lid is opened to an angle of 60 degrees and back in a total time period of 5 seconds.
10. Click OK.
Note: This method of creating an expression can also be used for specifying nonlinear properties of other entities like Force, Spring Damper, Bushing, and so on.
Create Outputs
In this step, you will add a displacement output between two bodies using the default entities. You will also add another output to record the displacement of a particular point G on the input link relative to the global frame based on Expressions.
You can create outputs using bodies, points and markers. You can also directly request force, bushing, and spring-damper entity outputs. Another way to create outputs is to create math expressions dependent on any of the above mentioned entities.
• From the Project Browser, right-click on Model and select Add > General MDL Entity > Output from the context menu.
• On the General MDL toolbar, right-click the (Outputs) icon.
2. Specify the label as Input Link Displacement and the variable name as o_disp for the new output.
3. Click OK.
4. Create a Displacement output between two points on two bodies.
1. For Body 1 and Body 2, select Input Link and Ground Body, respectively.
2. For Pt on Body 1 and Pt on Body 2, select point I and the Global Origin point respectively.
3. Record the displacement on Both points, one relative to the other.
5. Add one more output with the label as Input Link CM Displacement and the variable name as o_cm_disp.
6. Calculate the X displacement between the CM markers Input Link and the global origin.
1. From the drop-down menu, select Expressions.
2. Click in the F2 field.
This activates the button.
3. To display the Expression Builder dialog, Click on the button.
4. Clear the value 0 from the back-quotes.
5. From the Motion tab, click DX.
6. Place the cursor inside the curly-brackets {} after DX.
7. From the Properties tab, expand the following trees: Bodies/Input Link/Marker CM.
8. Select idstring.
9. Click Add to populate the expression.
10. Add a comma to separate the next expression.
11. Add a pair of curly brackets "{}".
12. Place the cursor inside the added brackets.
13. From the Properties tab, expand the following items in the tree: Markers/Global Frame.
14. Select idstring.
15. Click Add to populate the expression.
7. Click OK
8. To check for errors, go to the Tools menu and select Check Model.
Any errors in your model topology are listed in the Message Log.
Attention: The ID numbers appearing in the expressions may vary with the images shown.
Note: The DX function measures the distance between Input Link’s CM (center of mass) marker and marker representing the Global Frame in the X direction of the Global Frame. Refer to the MotionSolve Reference Guide for more details regarding the syntax and usage of this function.
The back quotes in the expression are used so that the MDL math parser evaluates the expression. Entity properties like idstring, value, etc. get evaluated when they are placed inside curly braces {}, otherwise they are understood as plain text. Refer to the Appendix to learn more about various kinds of expressions and form of evaluation adopted by MotionSolve.
In this step you will add graphics for visualization of the mechanism.
At this stage your trunk lid model does not contain any graphics, and the entities created in previous steps are represented only by implicit graphics (which are not available in solver deck or results file). MotionView graphics can be broadly categorized into three types: implicit, explicit, and external graphics.
Implicit Graphics
The small icons that you see in the MotionView interface when you create entities like points, bodies, joints, and so on are called implicit graphics. These are provided only for guidance during the model building process and are not visible when animating simulations.
Explicit Graphics
These graphics are represented in form of a tessellation, are written to the solver deck and subsequently available in the results. Explicit graphics are of two types.
Primitive Graphics
These graphics help in better visualization of the model and are also visible in the animation. The various types of Primitive Graphics in MotionView are Cylinder, Box, Sphere, and so on.
External Graphics
You can import in various CAD formats or HyperMesh files into MotionView. The ‘Import CAD or FE using HyperMesh..’ utility in MotionView can be used to convert a CAD model or a HyperMesh model to h3d graphic format which can be imported into MotionView. One can also import .g, ADAMS View .shl and wavefront .obj files directly into MotionView.
MotionView allows you to turn on and off implicit graphics for some of the commonly used modeling entities
1. Turn on implicit graphics.
1. From the Model main menu, select Implicit Graphics...
2. Turn on the Visible check box.
Note: Implicit graphics of Individual entities can be turned on or off by using the Visible check box for each entity.
3. Click Close.
Note: The state of the implicit graphics (whether on or off) is not saved in your model (.mdl) or session (.mvw) files. MotionView uses its default settings when:
• You create a new model in another model window.
• You start a new session.
• You load an existing .mdl/.mvw file into a new MotionView session.
To visualize the four-bar mechanism, you need to add explicit graphics to the model. In this step, you will add cylinder graphics for Follower, Coupler, and Input Links.
• From the Project Browser, right click on Model and select Add > Reference Entity > Graphicfrom the context menu.
• On the Reference Entity toolbar, right-click on the (Graphics) icon.
3. In the Add Cylinder or CylinderPair dialog, enter the label as Follower Cylinder and the variable name as gcyl_follower.
Note: The name of the dialog changes with the graphic type. For example, the dialog name changes to Add Box or BoxPair when the Box graphic type is selected.
4. From the Type drop-down menu, select Cylinder.
5. Click OK.
6. In the Connectivity tab, double-click on the (Body button) below Parent.
7. Select the Follower from the Select a Body list and click OK.
This assigns the graphics to the parent body.
8. To select the origin point of the cylinder, click below Origin.
9. Pick Point A in the modeling window.
10. Click under Direction.
11. Select Point C for .
12. From the Properties tab, enter 2 in the Radius 1: field.
Note: The cylinder graphic can also be used to create a conical graphic. By default, the Radius 2 field is parameterized with respect to Radius 1, such that Radius 2 takes the same value of Radius 1. Specify different radii to create a conical graphic.
13. For the remaining bodies in your model, follow steps 2 through 12 to create the appropriate explicit graphics for other links. Use the specifics given in Table 4.
Table 4.
Label Variable Name Graphic type Body Origin Direction Radius
Follow Cylinder gcyl_follower Cylinder Follower Point A Point C 2
Coupler Cylinder gcyl_coupler Cylinder Coupler Point C Point E 2
After the addition of cylinder graphics for all three links, the Model will look as shown in Figure 33:
Add External Graphics and Convert a HyperMesh File to an H3D File.
In this step you will use this conversion utility to convert a HyperMesh file of a car trunk lid into the H3D format.
MotionView has a conversion utility that allows you to generate detailed graphics for an MDL model using HyperMesh, Catia, IGES, STL, VDAFS, ProE, or Unigraphics source files. MotionView uses HyperMesh to perform the conversion.
1. From the Tools menu, select Import CAD or FE using HyperMesh.
2. From the Import CAD or FE using HyperMesh dialog, activate the Import CAD or Finite Element Model Only radio button.
3. From the Input File option drop-down menu, select HyperMesh.
4. Click the button next to Input File and select trunklid.hm, located in your <working directory>, as your input file.
The Output Graphic File will be automatically populated with the trunklid_graphic.h3d file from the <working directory>.
5. Click OK to begin the import process.
The Import CAD or FE using HyperMesh utility runs HyperMesh in the background to translate the HyperMesh file into an H3D file. When the import is complete the Message Log appears with the message "Translating/Importing the file succeeded!"
Note: The H3D file format is a neutral format in HyperWorks. It finds wide usage such as graphics and result files. The graphic information is generally stored in a tessellated form.
6. Clear the Message Log.
7. Use steps 1 through 6 to import the trunk graphics by converting the trunk.hm file to trunk.h3d.
Attach H3D Object to the Input Link and Ground Bodies
In this step, you will attach the trunk lid H3D object to the input link and the trunk H3D object to Ground.
1. On the Reference Entity toolbar, click (Graphics).
2. Select g_trunklid_graphic from the modeling window.
3. In the Connectivity tab, double click the (Body collector).
4. From the Select a Body dialog, click Input Link.
5. Click OK.
This will display the Graphics panel.
Note: Observe the change in the trunk lid graphic color to the Input Link body color.
6. Select the newly created g_trunk_graphic from the Project Browser and set the as Ground Body.
7. On the Standard toolbar, click the Save Model icon .
If the model is new you will be prompted to input the name of the model, otherwise the model will be saved in the working directory with the existing name.
Note: Existing models can be saved to another file using the Save As > Model option located in the File menu.
8. From the Save As Model dialog, browse to your working directory and specify the file name as trunklid_mechanism.mdl.
9. Click Save.
Solve the Model with MotionSolve
In this step, you will use MotionSolve to perform a kinematic simulation of the mechanism for a simulation time of 5 seconds, with a step size of 0.01 second.
MotionSolve can be used to perform kinematic, static, quasi-static, and dynamic analyses of multi-body mechanical systems. The input file for MotionSolve is an XML file called MotionSolve XML. The solution in MotionSolve can be executed from MotionSolve.
1. On the General Actions toolbar, click the (Run) icon.
2. On the Model Check toolbar, click on the (Check model) button to check the model for errors.
3. From the Main tab of the Run panel, specify Transient as the Simulation type.
4. Specify the name for the XML file as trunklid_mechanism_run.xml.
MotionView uses the base name of your XML file for other result files generated by MotionSolve. See the MotionView User’s Guide for details about the different result file types.
5. Activate the Export MDL snapshot check box (if it is not already active).
This will save the model at the stage in which the Run is executed.
6. Specify an End time of 5 for your simulation and a Print interval of 0.01
Note: The time unit is based on the time unit chosen (default is Seconds).
You can access the Solver Units and Gravity Data Sets from the Project Browser. .
7. To solve the model with MotionSolve, click the Run button.
Upon clicking Run, MotionSolve is invoked and solves the model. The Solver View window appears which shows the progress of the solution along with messages from the solver (Run log). This log is also written to a file with the extension .log to the solver file base name.
8. Review the window for solution information and be sure to watch for any warnings/errors.
View Animation and Plot Results on One Page
In this step you will learn to view your animation and plot result on the same page.
Once the run is successfully complete, both the Animate and Plot buttons are active.
1. Click the Animate button.
This opens HyperView in another window and loads the animation in that window.
2. On the toolbar, click the icon to start the animation.
Note: You can click the button again to stop/pause the animation.
4. Click the Plot button.
This opens HyperGraph and loads the results file in a new window.
5. Leave the X-axis Data Type as Time.
6. Input the y-axis data in Table 5.
Y Type Marker Displacement Y Request REQ/70000000 InputLink from Ground Body (Input Link Displacement) Y Component DM (Magnitude)
7. Click Apply.
This plots the magnitude of the displacement of Point I relative to the Global Origin.
Save Your Work as a Session File
Here, you will learn to save your model as a session file.
1. From the File menu, select Save As > Session File.
2. Specify the file name as trunklid_mechanism for your session.
3. Click Save.
Your work is saved as trunklid_mechanism.mvw session file.
Appendix
Evaluating Expressions in MotionView.
Table 6.
Math Parser
A MotionView parser that evaluates a MotionView expression as real/integer/string for a field as appropriate
Real This type of field can contain a real number or the parametric expression that should evaluate to a real number. This type of field is found in Points, Bodies, Force – Linear. Note that only the value of the expression as evaluated goes into the solver deck and not the parametric equation.
Example: p_a.x, b_0.mass
String This type of field can contain a string or a parametric expression that should evaluate to a string. This type of field is found in entity such as DataSets with strings as Datamember, SolverString etc. As in case of Linear field, only the value of the expression as evaluated goes into the solver deck and not the parametric expression.
Expressions This type of field is different than the three listed above because it can contain a combination of text and parametric expression. It is generally used to define a solver function (or a function that is recognized by the solver). This type of expression is embedded within back quotes ( ) and any parametric reference is provided within curly braces {}. The presence of back quotes suggests the math parser to pass the expression through Templex. Templexevaluates any expression within curly braces while retaining the other text as is. For example in the expression DX({b_inputlink.cm.idstring},{Global_Frame.idstring}), the templex evaluates the ID (as a string) of the cm of the Input link body (b_ inputlink) and that of marker Global Frame while retaining “DX” as is. These fields are available in entity panels such as: Bushings, Motions, Forces with properties that toggle to Expression, Independent variable for a curve input in these entities, and Outputs of the type Expression.
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2023-04-01 05:49:11
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https://www.heyiamindians.com/how-do-you-calculate-the-wavelength/
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## How do you calculate the wavelength?
The wavelength is calculated from the wave speed and frequency by λ = wave speed/frequency, or λ = v / f.
How do you find the wavelength on a calculator?
How to calculate wavelength
1. Determine the frequency of the wave. For example, f = 10 MHz .
2. Choose the velocity of the wave.
3. Substitute these values into the wavelength equation λ = v/f .
4. Calculate the result.
5. You can also use this tool as a frequency calculator.
What is the frequency calculator?
The frequency calculator lets you quickly find the frequency, given wavelength, and either the velocity or period. Table of contents: Frequency definition and the frequency formula. Frequency equation from the wavelength.
### What is the frequency wavelength equation?
The frequency formula in terms of wavelength and wave speed is given as, f = 𝜈/λ where, 𝜈 is the wave speed, and λ is the wavelength of the wave.
How do you calculate wavelength example?
If you want to calculate the wavelength of a wave, then all you have to do is plug the wave’s speed and wave’s frequency into the equation. Dividing speed by frequency gives you the wavelength. For example: Find the wavelength of a wave traveling at 20 m/s at a frequency of 5 Hz.
How do you calculate the wavelength of a wave?
The wavelength can be found using the wave number (λ=2πk). ( λ = 2 π k ) .
#### What is the wavelength of light formula?
Like all other electromagnetic waves, it obeys the equation c = fλ, where c = 3 × 108 m/s is the speed of light in vacuum, f is the frequency of the electromagnetic waves, and λ is its wavelength.
How do you calculate frequency?
To calculate frequency, divide the number of times the event occurs by the length of time. Example: Anna divides the number of website clicks (236) by the length of time (one hour, or 60 minutes). She finds that she receives 3.9 clicks per minute.
How do you solve for wavelength and frequency?
Formula:
1. λ = C/f.
2. λ (Lambda) = Wavelength in meters.
3. c = Speed of Light (299,792,458 m/s)
4. f = Frequency.
5. The frequency of a waveform is the number of times a complete waveform is repeated in a fixed time period.
6. The wavelength of a wave is the distance between two adjacent peaks or troughs of adjacent cycles.
## What are three ways to measure wavelength?
Set up the ripple tank as shown in the diagram with about 5 cm depth of water.
• Adjust the height of the wooden rod so that it just touches the surface of the water.
• Switch on the lamp and motor and adjust the speed of the motor until low frequency waves can be clearly observed.
• How do you measure wavelength?
Wavelength can be measured as the distance between two adjacent crests of a transverse wave or two adjacent compressions of a longitudinal wave. It is usually measured in meters. Wavelength is related to the energy of a wave.
What is the equation to calculate wavelength?
The formula for calculating wavelength is: Wavelength=WavespeedFrequency{\\displaystyle Wavelength={\\frac {Wavespeed}{Frequency}}}. Wavelength is commonly represented by the Greek letter lambda , λ{\\displaystyle \\lambda }.
### How do you find wavelength when given energy?
A simple rearrangement of Planck ‘s equation gives you an instant wavelength calculator for any radiation, assuming you know the energy of the radiation. The wavelength formula is: λ = \\frac{hc}{E}. Both h and c are constants, so the wavelength to energy conversion equation basically states that wavelength is proportional to the inverse of energy.
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2023-04-02 00:35:08
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https://dsp.stackexchange.com/questions/59537/sampling-rate-issue-in-microcontroller
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# Sampling rate issue in microcontroller
• I am using a 3-axis LIS2D12H accelerometer interfaced with NRF51832 through the SPI interface.
1. The sensor has a sampling-rate from 1 Hz to 5.376 KHz which is happening at sensor only.
2. I am taking sensor data at 400 Hz by setting register values in the sensor.
3. I am implementing an FFT algorithm in microcontroller over a raw sensor data received by the 3 axes, for vibration analysis.
• I can have resolution of FFT from 64 to 2048.
Q: My confusion is how should I sample it at microcontroller for FFT? There are FFT functions in ARM cortex which I am using.
1. Shall I take raw samples in buffer and sample them or shall I take each reading from sensor and sample them individually?
2. If that is the case how shall I select sampling frequency at microcontroller level and/or at what rate microcontroller shall read data?
I understand the sampling process is completely handled by the sensor itself, i.e. it delivers digital data. Then you just collect the desired number of samples from the sensor and calculate its FFT. The frequency range depends only on the sampling rate ($$f_{\text{max}}=f_s/2)$$, the frequency resolution depends only on the number of samples $$N$$, in your case $$N\in[64,2048]$$.
There is no sampling done on the uC, as it already receives sampled data. However, what you will want to do is employ a lowpass filter before calculating the FFT, with a cutoff frequency slightly lower than $$f_{\text{max}}$$, to get rid of high frequency content, that would otherwise cause aliasing in your spectrum.
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2021-07-24 13:59:03
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https://cvgmt.sns.it/paper/3356/
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# On the measure and the structure of the free boundary of the lower dimensional obstacle problem
created by focardi on 06 Mar 2017
modified on 17 Dec 2018
[BibTeX]
Published Paper
Inserted: 6 mar 2017
Last Updated: 17 dec 2018
Journal: Archive for Rational Mechanics and Analysis
Volume: 230
Pages: 125--184
Year: 2018
Notes:
Abstract:
We provide a thorough description of the free boundary for the lower dimensional obstacle problem in $\mathbb{R}^{n+1}$ up to sets of null $\mathcal{H}^{n-1}$ measure. In particular, we prove
(i) local finiteness of the $(n-1)$-dimensional Hausdorff measure of the free boundary;
(ii) $\mathcal{H}^{n-1}$-rectifiability of the free boundary,
(iii) classification of the frequencies up to a set of dimension at most $(n-2)$ and classification of the blow-ups at $\mathcal{H}^{n-1}$ almost every free boundary point.
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2021-02-24 21:19:59
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https://zxi.mytechroad.com/blog/greedy/leetcode-1775-equal-sum-arrays-with-minimum-number-of-operations/
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You are given two arrays of integers nums1 and nums2, possibly of different lengths. The values in the arrays are between 1 and 6, inclusive.
In one operation, you can change any integer’s value in any of the arrays to any value between 1 and 6, inclusive.
Return the minimum number of operations required to make the sum of values in nums1 equal to the sum of values in nums2. Return -1 if it is not possible to make the sum of the two arrays equal.
Example 1:
Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].
Example 2:
Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]
Output: -1
Explanation: There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.
Example 3:
Input: nums1 = [6,6], nums2 = [1]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].
Constraints:
• 1 <= nums1.length, nums2.length <= 105
• 1 <= nums1[i], nums2[i] <= 6
Solution: Greedy
Assuming sum(nums1) < sum(nums2),
sort both arrays
* scan nums1 from left to right, we need to increase the value form the smallest one.
* scan nums2 from right to left, we need to decrease the value from the largest one.
Each time, select the one with the largest delta to change.
e.g. nums1[i] = 2, nums[j] = 4, delta1 = 6 – 2 = 4, delta2 = 4 – 1 = 3, Increase 2 to 6 instead of decreasing 4 to 1.
Time complexity: O(mlogm + nlogn)
Space complexity: O(1)
C++
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huahualeetcode
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2021-04-15 05:13:44
|
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http://tex.stackexchange.com/tags/texmaker/hot
|
# Tag Info
4
\usepackage[english,ngerman]{babel} Then ngerman is active at the begin of the document and you can switch to english. If you cannot change the package, then write before \documentclass: \PassOptionsToPackage{english}{babel} \documentclass{whatever} ...
3
For me, the following works correctly: Load your main file in Texmaker Choose Options -> Define Current Document as 'Master Document' Open a subfile Edit subfile, etc. Press 'Quick Build' Main file builds but 'current' file stays visible, etc. SyncTeX linking is possible from source to PDF and the reverse
3
Accessing Texmaker's autosave feature Texmaker (at least 4.1 and newer) has an autosave feature. How you access it depends on your platform. On Linux (and Windows?), you need to follow Options > Configure Texmaker > Editor. On Mac OS X, you need to follow Texmaker > Preferences > Editor. You should then see a checkbox that reads "Backup [sic] documents ...
2
If you are talking about the visual, probably you are using it with the System Theme. Do you want that blue theme? Go to the Options menu and then Configure TeXmaker and then Editor. Or to see the left panel Structure, click on the 'Structure' button at the bottom (see the red marks). The image below show that two panels are on (dark grey color) and ...
1
1
This has most likely to do with your version of pdflatex. Check it by opening up cmd (Windows start -> run -> cmd.exe) and type pdflatex --version If your version does not correspond to the version on http://miktex.org/, remove the old version of MiKTex from your computer and download and install the newest MikTex version to your computer.
1
Ok, thanks to @christian-hupfer, I found ~/.config/xm1/texmakerapp.ini. Somehow I had an entry Language=fa in there, switched back to en.
1
I identified 2 problems in your MWE: As @Paul Gessler said, in \bibliographystyle{style}, "style" is not a valid bibliography style. To read more about the different bibliography styles, check the link he provided or this one. I couldn't compile your literature.bib file, because the last element in the bibitem has a colon at the end (" url = ...
1
I believe a quick look at the changelog might help you (no security issues mentioned...). You can find this here: http://www.xm1math.net/texmaker/log.html. To answer your question: as far as I know TexMaker does not open ports. If this motivates you to install the latest version anyway: Download TexMaker ...
1
Since you say you want any kind of TeX, I would personally recommend using Kile, as it comes installed with all the packages and is quite easy to use. Furthermore it comes with a lot of tools like wizards for tables or tools to easily create equations and add symbols for beginners. It also has a function to preview your equations. You can find it in the ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2014-07-24 02:51:20
|
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https://lscsoft.docs.ligo.org/ligo.skymap/ligo/skymap/bayestar/ez_emcee.html
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# Fire-and-Forget MCMC Sampling (ligo.skymap.bayestar.ez_emcee)¶
ligo.skymap.bayestar.ez_emcee.ez_emcee(log_prob_fn, lo, hi, nindep=200, ntemps=10, nwalkers=None, nburnin=500, args=(), kwargs={}, **options)[source]
Fire-and-forget MCMC sampling using ptemcee.Sampler, featuring automated convergence monitoring, progress tracking, and thinning.
The parameters are bounded in the finite interval described by lo and hi (including -np.inf and np.inf for half-infinite or infinite domains).
If run in an interactive terminal, live progress is shown including the current sample number, the total required number of samples, time elapsed and estimated time remaining, acceptance fraction, and autocorrelation length.
Sampling terminates when all chains have accumulated the requested number of independent samples.
Parameters
log_prob_fncallable
The log probability function. It should take as its argument the parameter vector as an of length ndim, or if it is vectorized, a 2D array with ndim columns.
lo
List of lower limits of parameters, of length ndim.
hi
List of upper limits of parameters, of length ndim.
nindepint, optional
Minimum number of independent samples.
ntempsint, optional
Number of temperatures.
nwalkersint, optional
Number of walkers. The default is 4 times the number of dimensions.
nburninint, optional
Number of samples to discard during burn-in phase.
Returns
chainnumpy.ndarray
The thinned and flattened posterior sample chain, with at least nindep * nwalkers rows and exactly ndim columns.
Other Parameters
kwargs :
Extra keyword arguments for ptemcee.Sampler. Tip: Consider setting the pool or vectorized keyword arguments in order to speed up likelihood evaluations.
Notes
The autocorrelation length, which has a complexity of $$O(N \log N)$$ in the number of samples, is recalulated at geometrically progressing intervals so that its amortized complexity per sample is constant. (In simpler terms, as the chains grow longer and the autocorrelation length takes longer to compute, we update it less frequently so that it is never more expensive than sampling the chain in the first place.)
Examples
>>> from ligo.skymap.bayestar.ez_emcee import ez_emcee
>>> from matplotlib import pyplot as plt
>>> import numpy as np
>>>
>>> def log_prob(params):
... """Eggbox function"""
... return 5 * np.log((2 + np.cos(0.5 * params).prod(-1)))
...
>>> lo = [-3*np.pi, -3*np.pi]
>>> hi = [+3*np.pi, +3*np.pi]
>>> chain = ez_emcee(log_prob, lo, hi, vectorize=True)
Sampling: 51%|██ | 8628/16820 [00:04<00:04, 1966.74it/s, accept=0.535, acl=62]
>>> plt.plot(chain[:, 0], chain[:, 1], '.')
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2019-08-21 09:44:40
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https://eprint.iacr.org/2022/276
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### Hardness estimates of the Code Equivalence Problem in the Rank Metric
Krijn Reijnders, Simona Samardjiska, and Monika Trimoska
##### Abstract
In this paper, we analyze the hardness of the Matrix Code Equivalence (MCE) problem for matrix codes endowed with the rank metric, and provide the first algorithms for solving it. We do this by making a connection to another well-known equivalence problem from multivariate cryptography - the Isomorphism of Polynomials (IP). We show that MCE is equivalent to the homogeneous version of the Quadratic Maps Linear Equivalence (QMLE) problem. Using the same birthday techniques known for IP, we present an algorithm for MCE running in time $\mathcal{O}^*( q^{\frac{2}{3}(n+m)})$, and an algorithm for MCE with roots, running in time $\mathcal{O}^*(q^{m})$. We verify these algorithms in practice.
Available format(s)
Category
Public-key cryptography
Publication info
Published elsewhere. MINOR revision.WCC 2022
Keywords
code-based cryptographypost-quantumcode equivalence
Contact author(s)
krijn @ cs ru nl
simonas @ cs ru nl
mtrimoska @ cs ru nl
History
Short URL
https://ia.cr/2022/276
CC BY
BibTeX
@misc{cryptoeprint:2022/276,
author = {Krijn Reijnders and Simona Samardjiska and Monika Trimoska},
title = {Hardness estimates of the Code Equivalence Problem in the Rank Metric},
howpublished = {Cryptology ePrint Archive, Paper 2022/276},
year = {2022},
note = {\url{https://eprint.iacr.org/2022/276}},
url = {https://eprint.iacr.org/2022/276}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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2022-06-27 03:13:39
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http://math.tntech.edu/ISR/Introduction_to_Probability/Distributions_of_Functions/thispage/newnode11.html
|
## Sum of independent random variables
Let and be independent random variables having the respective probability density functions and . Then the cumulative distribution function of the random variable can be given as follows.
where is the cdf of . By differentiating , we can obtain the pdf of as
The form of integration is called the convolution. Thus, the pdf is given by the convolution of the pdf's and . However, the use of moment generating function makes it easier to find the distribution of the sum of independent random variables.''
Let and be independent normal random variables with the respective parameters and . Then the sum of random variables has the mgf
which is the mgf of normal distribution with parameter . By the property (a) of mgf, we can find that is a normal random variable with parameter .
Let and be independent gamma random variables with the respective parameters and . Then the sum of random variables has the mgf
which is the mgf of gamma distribution with parameter . Thus, is a gamma random variable with parameter .
Generated by MATH GO: 2006-02-27
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2014-09-30 19:52:07
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http://blankonthemap.blogspot.com/2012/05/links-for-week-20th-may.html
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## Monday, May 21, 2012
### Links for the week: 20th May
I meant to put this out yesterday, but the supernova post took a bit longer to write than I was hoping. Anyway, here's a list of interesting links.
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2017-09-22 04:30:56
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http://tex.stackexchange.com/questions/95561/definition-dependent-on-font-size
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# Definition dependent on font size
I would like to define an arrow similar to here. The definition should scale with the font size. At the moment, my definition looks like:
\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.pathmorphing,shapes}
\newcounter{sarrow}
\newcommand\xrsquigarrow[1]{%
\mathrel{\begin{tikzpicture}[baseline={($(current bounding box.south)+(0,-0.5ex)$)}]
\node[inner sep=.5ex] (\thesarrow) {$\scriptstyle #1$};
\draw[<-,decorate,decoration={snake,amplitude=0.7pt,segment length=1.2mm,pre=lineto,pre length=4pt}] (\thesarrow.south east) -- (\thesarrow.south west);
\end{tikzpicture}}%
}
\begin{document}
$A\xrsquigarrow{f}B$
\end{document}
How does the definition have to look like in order to get dependent on the font size (from the document class). Idea: All scales should be related to the current font size, e.g. here 12pt.
-
Use em and ex units, which depend on the current font size, instead of fixed units like mm and pt. – Jake Jan 27 at 15:03
For instance 0.135ex for 0.7pt and 0.34em for 1.2mm. Are you sure that \thesarrow shouldn't be thesarrow (without the backslash)? – egreg Jan 27 at 15:33
Slightly related: Which measurement units should one use in LaTeX? – Qrrbrbirlbel Jan 27 at 23:51
Define, as already suggested, the macro with "relative units":
\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.pathmorphing,shapes}
\newcounter{sarrow}
\newcommand\xrsquigarrow[1]{\mathrel{%
\begin{tikzpicture}[baseline={($(current bounding box.south)+(0,-0.5ex)$)}]
\node[inner sep=.5ex] (\thesarrow) {$\scriptstyle #1$};
\draw[<-,decorate,
decoration={snake,amplitude=0.135ex,segment length=0.34em,pre=lineto,pre length=0.4em}]
(\thesarrow.south east) -- (\thesarrow.south west);
\end{tikzpicture}
}}
\begin{document}
$A\xrsquigarrow{f}B$
{\large$A\xrsquigarrow{f}B$\par}
{\Large$A\xrsquigarrow{f}B$\par}
{\LARGE$A\xrsquigarrow{f}B$\par}
{\small$A\xrsquigarrow{f}B$\par}
{\footnotesize$A\xrsquigarrow{f}B$\par}
\end{document}
In order to get automatic scaling for subscripts and superscripts (or \scriptstyle declarations), one has to work harder:
\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.pathmorphing,shapes}
\makeatletter
\newcommand\xrsquigarrow[1]{\mathrel{\mathchoice
{\hbox{\fontsize{\tf@size}{\tf@size}\selectfont\@xrsquigarrow\scriptstyle{#1}}}
{\hbox{\fontsize{\tf@size}{\tf@size}\selectfont\@xrsquigarrow\scriptstyle{#1}}}
{\hbox{\fontsize{\sf@size}{\sf@size}\selectfont\@xrsquigarrow\scriptscriptstyle{#1}}}
{\hbox{\fontsize{\ssf@size}{\ssf@size}\selectfont\@xrsquigarrow\scriptscriptstyle{#1}}}
}}
\newcommand\@xrsquigarrow[2]{%
\begin{tikzpicture}[baseline={($(current bounding box.south)+(0,-0.5ex)$)}]
\node[inner sep=.5ex] (A) {$#1#2$};
\draw[<-,decorate,
decoration={snake,amplitude=0.135ex,segment length=0.34em,pre=lineto,pre length=0.4em}]
(A.south east) -- (A.south west);
\end{tikzpicture}%
}
\begin{document}
$A\xrsquigarrow{f}B_{A\xrsquigarrow{f}B}$
{\large$A\xrsquigarrow{f}B$\par}
{\Large$A\xrsquigarrow{f}B$\par}
{\LARGE$A\xrsquigarrow{f}B$\par}
{\small$A\xrsquigarrow{f}B$\par}
{\footnotesize$A\xrsquigarrow{f}B$\par}
\end{document}
Beware that this is slow, because for each instance of \xrsquigarrow all the four variants need to be typeset.
Note. I've removed \thesarrow that does nothing more than provide a node name.
\mathchoice requires four arguments: what's to be typeset in display, text, first level sub/superscript and second level sub/superscript styles respectively. All four text will be typeset and then TeX will decide which one to use. So we set four boxes in which the font is chosen to be of size \tf@size for display and text styles, \sf@size and \ssf@size for the other two styles (these sizes are automatically computed by LaTeX when a formula is being typeset).
The box then contains \@xrsquigarrow which has two arguments: the style to be applied for the label and the label itself. This style should be \scriptstyle when the arrow is in display or text styles, \scriptscriptstyle in the other cases.
The definition of \@xrsquigarrow, apart from the additional parameter, is just the same as the one proposed before.
-
This is really great but this way it still does not scale with the size inside the math-environment, e.g. `$\scriptscriptstyle A\xrsquigarrow{f}B$. That is what I would intuitively think of! – strpeter Feb 11 at 16:11
@strpeter Your wish is my command. ;-) – egreg Feb 11 at 16:33
Excellent command, even if I do not fully understand why you wrote twice the same line in \newcommand\xrsquigarrow and why in general the definition for every single font size is required?! Could you explain your final version of the command? – strpeter Feb 12 at 0:51
@strpeter I've added a bit of explanation – egreg Feb 12 at 1:00
Why did you make the \@xsquigarrow command take two arguments when you just concatenate them? Is there any profound reason not to use a singel argument of {\scriptstyle{#1}} (and similar for scriptscriptstyle)? – Johan_E Feb 12 at 3:35
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2013-05-21 07:00:37
|
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https://www.metaculus.com/questions/395/a-restaurant-serving-cultured-meat-by-2021/
|
Submit Essay
Once you submit your essay, you can no longer edit it.
# A restaurant serving cultured meat by 2021?
### Question
The possibility of cultured meat, in which tissue is grown outside of a live animal in order to provide the flavor and nutrition of meat without animal suffering, has been heating up recently.
Cultured meat is certainly possible, and production costs have fallen since the first demonstration in 2013, from $330,000/pound to$18,000/pound in 2016. Memphis meats and Modern Meadow are among the companies developing cultured meat, supported by for-profit and non-profit (e.g. New Harvest) research.
Cultured meats, along with reduced animal suffering, also promise health benefits, as they should be far easier to keep free of pathogens (and hence require no indiscriminate antibiotic use to boot), can be free of extra hormones, and might be engineered to be more nutritious. The primary questions appear to be those of getting costs (way) down, maintaining excellent taste, and overcoming public apprehension. To get at the timeline, we ask
By start of 2021, will there be a restaurant serving cultured meat?
For positive resolution, by Jan 1 2021, a restaurant somewhere in the world must exist where a member of the general public, given the time and ability to acquire a reservation, can walk in and order cultured meat for consumption.
### Prediction
Note: this question resolved before its original close time. All of your predictions came after the resolution, so you did not gain (or lose) any points for it.
Note: this question resolved before its original close time. You earned points up until the question resolution, but not afterwards.
Current points depend on your prediction, the community's prediction, and the result. Your total earned points are averaged over the lifetime of the question, so predict early to get as many points as possible! See the FAQ.
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2022-05-19 12:31:49
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http://hal.in2p3.fr/view_by_stamp.php?label=LAPP&langue=en&action_todo=view&id=in2p3-00010492&version=1
|
772 articles – 2891 references [version française]
HAL: in2p3-00010492, version 1
Annales de Physique 1 (1976) 5-72
A study of the reaction $pp \rightarrow p(n \pi^+)$ at the Cern intersecting storage rings
(1976)
Subject(s) : Physics/High Energy Physics - Experiment
in2p3-00010492, version 1 http://hal.in2p3.fr/in2p3-00010492 oai:hal.in2p3.fr:in2p3-00010492 From: Nicole Berger <> Submitted on: Thursday, 4 October 2001 16:28:09 Updated on: Thursday, 4 October 2001 16:28:09
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2013-12-11 13:58:22
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https://vbn.aau.dk/en/publications/a-note-on-secure-multiparty-computation-via-higher-residue-symbol
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# A note on secure multiparty computation via higher residue symbols
Ignacio Cascudo, Reto Alexander Schnyder*
*Corresponding author
Research output: Contribution to journalJournal articleResearchpeer-review
## Abstract
We generalize a protocol by Yu for comparing two integers with relatively small difference in a secure multiparty computation setting. Yu's protocol is based on the Legendre symbol. A prime number p is found for which the Legendre symbol (· | p ) agrees with the sign function for integers in a certain range {− N , . . . , N } ⊂ ℤ. This can then be computed efficiently. We generalize this idea to higher residue symbols in cyclotomic rings ℤ[ ζ r ] for r a small odd prime. We present a way to determine a prime number p such that the r -th residue symbol (· | p ) r agrees with a desired function f:A→{ζr0,…,ζrr−1}f:A \to \left\{ {\zeta _r^0, \ldots ,\zeta _r^{r - 1}} \right\} on a given small subset A ⊂ ℤ[ ζ r ], when this is possible. We also explain how to efficiently compute the r -th residue symbol in a secret shared setting.
Original language English Journal of Mathematical Cryptology 15 1 284-297 14 1862-2976 https://doi.org/doi.org/10.1515/jmc-2020-0013 Published - 29 Jan 2021
## Keywords
• Cyclotomic rings
• Power residue symbol
• Secure multiparty computation
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2021-05-11 23:35:47
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https://supportmymoto.com/what-is-4-percent-of-14000/
|
# What is 4 percent of 14000?
WRITTEN BY: supportmymoto.com STAFF
#### Resolution for What’s 4 % of 14000:
4 % *14000 =
( 4:100)*14000 =
( 4*14000):100 =
56000:100 = 560
Now we’ve: 4 % of 14000 = 560
Query: What’s 4 % of 14000?
Share resolution with steps:
Step 1: Our output worth is 14000.
Step 2: We signify the unknown worth with {x}.
Step 3: From step 1 above,{14000}={100%}.
Step 4: Equally, {x}={ 4%}.
Step 5: This ends in a pair of straightforward equations:
{14000}={100%}(1).
{x}={ 4%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand aspect) of each
equations have the identical unit (%); we’ve
frac{14000}{x}=frac{100%}{ 4%}
Step 7: Once more, the reciprocal of either side offers
frac{x}{14000}=frac{ 4}{100}
Rightarrow{x} = {560}
Due to this fact, { 4%} of {14000} is {560}
#### Resolution for What’s 14000 % of 4:
14000 % * 4 =
(14000:100)* 4 =
(14000* 4):100 =
56000:100 = 560
Now we’ve: 14000 % of 4 = 560
Query: What’s 14000 % of 4?
Share resolution with steps:
Step 1: Our output worth is 4.
Step 2: We signify the unknown worth with {x}.
Step 3: From step 1 above,{ 4}={100%}.
Step 4: Equally, {x}={14000%}.
Step 5: This ends in a pair of straightforward equations:
{ 4}={100%}(1).
{x}={14000%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand aspect) of each
equations have the identical unit (%); we’ve
frac{ 4}{x}=frac{100%}{14000%}
Step 7: Once more, the reciprocal of either side offers
frac{x}{ 4}=frac{14000}{100}
Rightarrow{x} = {560}
Due to this fact, {14000%} of { 4} is {560}
NOTE : Please do not copy - https://supportmymoto.com
|
2022-05-19 08:11:27
|
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|
http://quantum.lanl.gov/qlunch/2008_qlunch/smerzi.shtml
|
## CONTACTS
• Coordinator
Diego Dalvit
• Quantum Lunch Location:
T-Division Conference Room, TA-3,
Building 123, Room 121
# Quantum Institute: Visitor Schedule
The Quantum Lunch is regularly held on Thursdays in the Theoretical Division Conference Room, TA-3, Building 123, Room 121.
Thursday, August 21, 2008
12:30 PM to 2 PM
Speaker: Augusto Smerzi, BEC CNR-INFM and Dipartimento di Fisica, Universita' di Trento, Italy
### TOPIC: Entanglement and Heisenberg Limit in Quantum Interferometry
Abstract
The quantum Fisher information provides a sufficient condition to recognize multi-particle entanglement in a N quibit state. The same criterion gives a necessary and sufficient condition for sub shot-noise phase sensitivity in the estimation of a collective rotation angle $\theta$. The analysis therefore singles out the class of entangled states which are "useful" to overcome classical phase sensitivity in metrology and sensors. We will discuss the creation of such useful entangled states by the non-linear dynamical evolution of a Bose-Einstein condensates trapped in a double-well potential.
Operated by Los Alamos National Security, LLC for the U.S. Department of Energy's NNSA
|
2014-07-30 15:08:06
|
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|
https://www.tutorialspoint.com/find-the-perimeter-of-a-regular-hexagon-with-each-side-measuring-8-mathrm-m
|
# Find the perimeter of a regular hexagon with each side measuring $8 \mathrm{~m}$.
#### Complete Python Prime Pack
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Given:
Each side of the given regular hexagon measures $8\ m$.
To do:
We have to find the perimeter of the regular hexagon.
Solution:
We know that,
The perimeter of a regular hexagon with each side measuring $a$ is $6a$.
Therefore,
The perimeter of the regular hexagon with each side measuring $8\ m = 6\times8\ m$
$=48\ m$
The perimeter of the regular hexagon with each side measuring $8\ m$ is $48\ m$.
Updated on 10-Oct-2022 13:36:27
|
2022-12-08 12:39:19
|
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|
https://dash.harvard.edu/browse?authority=cf73f11339e09a90044316de1c5e5642&type=author
|
Now showing items 1-20 of 65
• #### All-Optical Sensing of a Single-Molecule Electron Spin
(American Chemical Society (ACS), 2014)
We demonstrate an all-optical method for magnetic sensing of individual molecules in ambient conditions at room temperature. Our approach is based on shallow nitrogen-vacancy (NV) centers near the surface of a diamond ...
• #### Anti-Reflection Coating for Nitrogen-Vacancy Optical Measurements in Diamond
(American Institute of Physics (AIP), 2012)
We realize anti-reflection (AR) coatings for optical excitation and fluorescence measurements of nitrogen-vacancy (NV) color centers in bulk diamond by depositing quarter-wavelength thick silica layers on the diamondsurface. ...
• #### An astro-comb calibrated solar telescope to search for the radial velocity signature of Venus
(2016)
We recently demonstrated sub-m/s sensitivity in measuring the radial velocity (RV) between the Earth and Sun using a simple solar telescope feeding the HARPS-N spectrograph at the Italian National Telescope, which is ...
• #### Atom-like crystal defects: From quantum computers to biological sensors
(AIP Publishing, 2014)
• #### Atomic-Scale Nuclear Spin Imaging Using Quantum-Assisted Sensors in Diamond
(American Physical Society (APS), 2015)
Nuclear spin imaging at the atomic level is essential for the understanding of fundamental biological phenomena and for applications such as drug discovery. The advent of novel nanoscale sensors promises to achieve the ...
• #### Broadband Dispersion-Free Optical Cavities Based on Zero Group Delay Dispersion Mirror Sets
(Optical Society of America, 2010)
A broadband dispersion-free optical cavity using a zero group delay dispersion (zero-GDD) mirror set is demonstrated. In general zero-GDD mirror sets consist of two or more mirrors with opposite group delay dispersion ...
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(Optical Society of America, 2012)
We deployed two wavelength calibrators based on laser frequency combs (“astro-combs”) at an astronomical telescope. One astro- comb operated over a 100 nm band in the deep red (∼ 800 nm) and a second operated over a 20 nm ...
• #### Coherence of Nitrogen-Vacancy Electronic Spin Ensembles in Diamond
(American Physical Society, 2010)
We present an experimental and theoretical study of electronic spin decoherence in ensembles of nitrogen-vacancy (NV) color centers in bulk high-purity diamond at room temperature. Under appropriate conditions, we find ...
• #### Coherent, Mechanical Control of a Single Electronic Spin
(American Chemical Society (ACS), 2012)
We demonstrate coherent quantum control of a single spin driven by the motion of a mechanical resonator. The motion of a mechanical resonator is magnetically coupled to the electronic spin of a single nitrogen-vacancy ...
• #### Coherent-Population-Trapping Resonances with Linearly Polarized Light for All-Optical Miniature Atomic Clocks
(American Physical Society (APS), 2010)
We present a joint theoretical and experimental characterization of the coherent population trapping (CPT) resonance excited on the $D_1$ line of $^{87}Rb$ atoms by bichromatic linearly polarized laser light. We observe ...
• #### Conjugate Fabry–Perot Cavity Pair for Improved Astro-Comb Accuracy
(Optical Society of America, 2012)
We propose a new astro-comb mode-filtering scheme composed of two Fabry–Perot cavities (coined “conjugate Fabry–Perot cavity pair”). Simulations indicate that this new filtering scheme makes the accuracy of astro-comb ...
• #### Correlative light and electron microscopy using cathodoluminescence from nanoparticles with distinguishable colours
(Nature Publishing Group, 2012)
Correlative light and electron microscopy promises to combine molecular specificity with nanoscale imaging resolution. However, there are substantial technical challenges including reliable co-registration of optical and ...
• #### Decoherence imaging of spin ensembles using a scanning single-electron spin in diamond
(Nature Publishing Group, 2015)
The nitrogen-vacancy (NV) defect center in diamond has demonstrated great capability for nanoscale magnetic sensing and imaging for both static and periodically modulated target fields. However, it remains a challenge to ...
• #### Dependence of nuclear spin singlet lifetimes on RF spin-locking power
(Elsevier BV, 2012)
We measure the lifetime of long-lived nuclear spin singlet states as a function of the strength of the RF spin-locking field and present a simple theoretical model that agrees well with our measurements, including the ...
• #### Dressed-State Resonant Coupling between Bright and Dark Spins in Diamond
(American Physical Society, 2013)
Under ambient conditions, spin impurities in solid-state systems are found in thermally mixed states and are optically “dark”; i.e., the spin states cannot be optically controlled. Nitrogen-vacancy (NV) centers in diamond ...
• #### Efficient photon detection from color centers in a diamond optical waveguide
(American Physical Society (APS), 2012)
A common limitation of experiments using color centers in diamond is the poor photon collection efficiency of microscope objectives due to refraction at the diamond interface. We present a simple and effective technique ...
• #### Electromagnetically induced transparency in paraffin-coated vapor cells
(American Physical Society (APS), 2011)
Antirelaxation coatings in atomic vapor cells allow ground-state coherent spin states to survive many collisions with the cell walls. This reduction in the ground-state decoherence rate gives rise to ultranarrow-bandwidth ...
• #### Electromagnetically Induced Transparency with Noisy Lasers
(American Physical Society, 2009)
We demonstrate and characterize two coherent phenomena that can mitigate the effects of laser phase noise for Electromagnetically Induced Transparency (EIT): a laser-power-broadening-resistant resonance in the transmitted ...
• #### Electromagnetically induced transparency-based slow and stored light in warm atoms
(Wiley-Blackwell, 2011)
This paper reviews recent efforts to realize a high-efficiency memory for optical pulses using slow and stored light based on electromagnetically induced transparency (EIT) in ensembles of warm atoms in vapor cells. After ...
• #### Enhanced metrology using preferential orientation of nitrogen-vacancy centers in diamond
(American Physical Society (APS), 2012)
We demonstrate preferential orientation of nitrogen-vacancy (NV) color centers along two of four possible crystallographic axes in diamonds grown by chemical vapor deposition on the {100} face. We identify the relevant ...
|
2020-10-21 12:58:15
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|
https://gmatclub.com/forum/a-certain-population-of-bacteria-doubles-every-10-minutes-if-the-numb-189467.html
|
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# A certain population of bacteria doubles every 10 minutes. If the numb
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Math Expert
Joined: 02 Sep 2009
Posts: 52971
A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
### Show Tags
03 Dec 2014, 07:23
00:00
Difficulty:
5% (low)
Question Stats:
82% (01:10) correct 18% (01:31) wrong based on 167 sessions
### HideShow timer Statistics
Tough and Tricky questions: Exponents.
A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^5, then what was the number in the population 1 hour later?
A) 2(10^5)
B) 6(10^5)
C) (2^6)(10^5)
D) (10^6)(10^5)
E) (10^5)^6
Kudos for a correct solution.
Source: Chili Hot GMAT
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Status: Amat Victoria Curam
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Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
### Show Tags
03 Dec 2014, 08:09
2
x bacteria would become 2x in 10 mins, which would become 4x in the next 10 (total 20 mins)...
So in 60 mins x becomes 2^6 times into x
Therefore Answer is C, (2^6)(10^5)
Manager
Joined: 21 Jul 2014
Posts: 124
Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
### Show Tags
03 Dec 2014, 08:10
1
Bunuel wrote:
Tough and Tricky questions: Exponents.
A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^5, then what was the number in the population 1 hour later?
A) 2(10^5)
B) 6(10^5)
C) (2^6)(10^5)
D) (10^6)(10^5)
E) (10^5)^6
Kudos for a correct solution.
Source: Chili Hot GMAT
I did this problem by noticing a pattern. Since the population doubles every 10 minutes, basically all we are doing is doubling our coefficient in front of 10^5. It starts out at (1)10^5 and after 10 minutes it is (2)10^5 and at 20 its (4)10^5. I noticed every 10 minutes we are increasing our power of 2.
An hour is 60 minutes, so the bacteria will double 6 times. This means that our coefficient will end up being 2^6. Looking at the answer choices, choice (C), (2^6)(10^5) is correct.
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Posts: 1821
Location: India
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Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
### Show Tags
03 Dec 2014, 19:08
4
Answer = C) (2^6)(10^5)
Initial value $$= 10^5$$
Interval = 10 Minutes
Intervals in 60 Minutes $$= \frac{60}{10} = 6$$
In one hour, value $$= 2^6 * 10^5$$
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Joined: 08 Jul 2012
Posts: 47
Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
### Show Tags
03 Dec 2014, 20:38
2
Present population: 10^5
60 mins will have 6 intervals of 10 mins --> so the population will get doubled 6 times in 1hr --> 2*2*2*2*2*2*10^5 -->2^6*10^5
Illustration:
Present - 10^5
after 10 mins - 2*10^5
after 20 mins - 2(2*10^5) = 2^2*10^5
after 30 mins - 2*(2(2*10^5)) = 2^3*10^5
.
.
.
after 60 mins - 2^6*10^5
Ans. C) (2^6)(10^5)
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Math Expert
Joined: 02 Sep 2009
Posts: 52971
Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
### Show Tags
04 Dec 2014, 06:17
Bunuel wrote:
Tough and Tricky questions: Exponents.
A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^5, then what was the number in the population 1 hour later?
A) 2(10^5)
B) 6(10^5)
C) (2^6)(10^5)
D) (10^6)(10^5)
E) (10^5)^6
Kudos for a correct solution.
Source: Chili Hot GMAT
The correct answer is C.
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Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
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04 Dec 2014, 06:23
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Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink]
### Show Tags
29 Sep 2018, 17:20
1
Seems like a rip off of this OG10 question by whoever "created" this question :-/
https://gmatclub.com/forum/a-certain-po ... 65536.html
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Re: A certain population of bacteria doubles every 10 minutes. If the numb [#permalink] 29 Sep 2018, 17:20
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# A certain population of bacteria doubles every 10 minutes. If the numb
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2019-02-20 03:21:38
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https://math.ecnu.edu.cn/RCFOA/seminar_template.php?id=762
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Lifts of completely positive (equivariant) maps
Eusebio Gardella (Goteborgs Universitet)
16:00-17:00, November 15, 2021 Zoom Meeting 614 1453 0704 (Password: 132963)
Abstract:
Let $A$ and $B$ be $C^*$-algebras, $A$ separable and $I$ an ideal in $B$. We show that for any completely positive contractive linear map $\psi\colon A\to B/I$ there is a continuous family $\Theta_t\colon A\to B$, for $t\in [1,\infty)$, of lifts of $\psi$ that are asymptotically linear, asymptotically completely positive and asymptotically contractive. If $A$ and $B$ carry continuous actions of a second countable locally compact group $G$ such that $I$ is $G$-invariant and $\psi$ is equivariant, then the family $\Theta_t$ can be chosen to be asymptotically equivariant. If a linear completely positive lift for $\psi$ exists, then we can arrange that $\Theta_t$ is linear and completely positive for all $t\in [1,\infty)$; this yields an equivariant version of the Choi-Effros lifting theorem. In the equivariant setting, if $A$, $B$ and $\psi$ are unital, the existence of asymptotically linear unital lifts are only guaranteed if $G$ is amenable. This leads to a new characterization of amenability in terms of the existence of asymptotically equivariant unital sections for quotient maps. This talk is based on joint work with Marzieh Forough and Klaus Thomsen.
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2022-05-25 23:13:20
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https://itectec.com/ubuntu/ubuntu-steam-install-can-erase-the-home-how-to-prevent-it/
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# Ubuntu – Steam install can erase the home. How to prevent it
steam
I have read a disturbing topic on Valve where a user lost his system when using the steam script. There is a discussion on reddit.linux and on reddit/steam.
This may not be a common problem because I change all sorts of configuration about my system. The script in question does something in a really, really stupid way, but it probably doesn't trigger the fail scenario for every system because…
Original Bug:
I am not sure what happened. I moved the folder in the title to a drive mounted under /media/user/BLAH and symlinked /home/user/.local/steam to the new location.
I launched steam. It did not launch, it offered to let me browse, and still could not find it when I pointed to the new location. Steam crashed. I restarted it.
It re-installed itself and everything looked great. Until I looked and saw that steam had apparently deleted everything owned by my user recursively from the root directory. Including my 3tb external drive I back everything up to that was mounted under /media.
Everything important, for the most part, was in the cloud. It is a huge hassle, but it is not a disaster. If there is the chance that moving your steam folder can result in recursively deleting everything in the directory tree you should probably just throw up an error instead of trying to point to other stuff. Or you know, allow the user to pick an install directory initially like on windows.
My system is ubuntu 14.04, and the drive I moved it to was ntfs if its worth anything.
The problem starts around line 19 in the script "steam.sh.":
STEAMROOT="$(cd "${0%/*}" && echo $PWD)" STEAMDATA="$STEAMROOT"
$STEAMROOT can become empty here effectively making the rm -rf "$STEAMROOT/"* further into the script the same as rm -rf "/"*.
There are patches showing up and there is a lot wrong with this script. Easiest to change and at least prevent deleting files it should not ...
rm -rf "$STEAMROOT/"* to ... [[ -n$STEAMROOT && $STEAMROOT =~ 'steam' ]] && rm -rf$STEAMROOT
Also possible to add an exit just after STEAMDATA is set:
STEAMROOT="$(cd "${0%/*}" && echo $PWD)" STEAMDATA="$STEAMROOT"
if [ -z "\$STEAMROOT" ]; then
echo "stop script otherwise files are deleted from /."
exit 1
fi
If anyone out there installed steam as root be warned: it will delete your WHOLE disk.
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2021-10-28 14:18:20
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http://nrich.maths.org/5838&part=
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### Whoosh
A ball whooshes down a slide and hits another ball which flies off the slide horizontally as a projectile. How far does it go?
### Dam Busters 1
At what positions and speeds can the bomb be dropped to destroy the dam?
### Escape from Planet Earth
How fast would you have to throw a ball upwards so that it would never land?
# Dam Busters 2
##### Stage: 5 Challenge Level:
In the problem Dam Busters 1 we looked at targetting a dam with a direct hit. In that case the plane could only just get close enough to the dam to destroy its target. The dam is destroyed if the bomb strikes above the rocky outcrop of height $B$.
A scientist has suggested developing a bomb which will bounce on water when travelling transverse to the water surface at high speed. It has been suggested that this bouncing bomb will enable the target to be struck more centrally whilst releasing it from a greater distance.
Several scientists have tried to calculate the height h of the bouncing bomb above sea level at the point at which it will be level with the wall after the first bounce. The scientists agree on the overall general form of the equation but cannot agree on some of the signs and which way around some of the factors should go.
Can you decide whether each of the signs in the equation should be 'plus one' or 'minus one' (they can be different from each other)? $$h = -\frac{1}{2}g^{\pm 1}\left[\left(\frac{V}{D}\right)^{\pm 1}-\sqrt{X}\right]^2 \pm eg\sqrt{X}\left[\left(\frac{V}{D}\right)^{\pm 1}-\sqrt{X}\right] \;,\quad\quad X = \left(\frac{2H}{g}\right)^{\pm 1}$$
In the equation, $e$ is the coefficient of restitution for the bomb striking the water. $e$ is dimensionless and positive. You do not need to derive the equation, only to decide which configurations would give rise to a sensible equation from a physical point of view.
Now that you have chosen the correct equation, (and checked that it is correct!) explore the heights and distances $H$ and $D$ which will enable a strike on target when the coefficient of restitution is $0.8$. The bomber must fly below $200\mathrm{m}$ and release the bomb at a minimum distance of $1 \mathrm{km}$ from the target. The bomber can fly at up to $800\textrm{ km per hour}$.
What is the minimum value of the coefficient of restitution for which this bombing strategy will work in this situation?
Extension: Derive this equation used in the calculations
NOTES AND BACKGROUND
Bouncing bombs were indeed used in World War II to destroy dams.
In 1941 Barnes Wallis, a scientist working for a British aircraft company was commissioned to investigate ways to halt the German war machine. He concluded that destruction of dams in the industrial Ruhr region would be devastating to German production. An 'Air Attack on the Dams Committee' was formed to try to find targets.Of course, the dams were immensely strong and very heavily defended. Barnes realised that the only way to destroy a dam was for a bomb to explode at its foot, which was protected by torpedo nets. The committee believed that this problem was technologically intractable.
Thinking out of the box Barnes invented the bouncing bomb. In the final attack the plane had to fly at 220 miles per hour, at a height of 60 feet above the water and release the bomb from exactly 425 yards.
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2017-02-27 09:02:15
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https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-2-section-2-4-one-sided-limits-exercises-page-85/49
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## University Calculus: Early Transcendentals (3rd Edition)
To prove the limit, prove that for every value of $\epsilon\gt0$, there exists a corresponding number of $\delta\gt0$ such that for all $x$: $$0-\delta\lt x\lt0\Rightarrow \Big|\frac{x}{|x|}-(-1)\Big|\lt\epsilon$$
$$\lim_{x\to0^-}\frac{x}{|x|}=-1$$ Given $\epsilon\gt0$. To prove the limit, we need to find $\delta\gt0$ such that for all $x$: $$0-\delta\lt x\lt0\Rightarrow \Big|\frac{x}{|x|}-(-1)\Big|\lt\epsilon$$ $$-\delta\lt x\lt0\Rightarrow \Big|\frac{x}{|x|}+1\Big|\lt\epsilon$$ First, we need to examine the inequality: $$\Big|\frac{x}{|x|}+1\Big|\lt\epsilon$$ $$-\epsilon\lt\frac{x}{|x|}+1\lt\epsilon$$ Since $-\delta\lt x\lt0$, here we examine only values of $x\lt0$, where $|x|=-x$. Therefore, $$-\epsilon\lt\frac{x}{-x}+1\lt\epsilon$$ $$-\epsilon\lt-1+1\lt\epsilon$$ $$-\epsilon\lt0\lt\epsilon$$ which is always right as we define $\epsilon\gt0$ In other words, for $x\lt0$, $\Big|\frac{x}{|x|}+1\Big|\lt\epsilon$ Take any arbitrary value of $\delta\gt0$, or $-\delta\lt0$, we would have $$-\delta\lt x\lt0\Rightarrow\Big|\frac{x}{|x|}+1\Big|\lt\epsilon$$ According to the defintion of a one-sided limit, this means $\lim_{x\to0^-}\frac{x}{|x|}=-1$
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2021-05-08 13:48:40
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https://www.physicsforums.com/threads/gauss-jordan-elemination-method-stucked.263126/
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# Gauss Jordan Elemination Method (Stucked)
1. Oct 9, 2008
### farmd684
Gauss Jordan Elemination Method (Stucked)!!
1. The problem statement, all variables and given/known data
5x1+3x2+2x3=4
3x1+3x2+2x3=2
+x2+x3 =5
I have to solve this problem by calcutaing the inverse if this equation's augmented matrix
2. Relevant equations
I used this method to find its inverse
5 3 2 1 0 0
3 3 2 0 1 0
0 1 1 0 0 1
My goal is to make first matrix to Identity matrix like the second one
3. The attempt at a solution
So far here is my workings
1ST STEP: R1/5 R2/3
2ND STEP: R3*(3/5)
3RD STEP: R2=R1-R2
4TH STEP: R1=R1-R3 R3=R3-R1
5TH STEP: R2*(-5/2)
6TH STEP: R3*(1/5)
7TH STEP: R1=(1/5)R3+R1 AND R2=R2+(-2/3)R3
Then the first one becomes Identity matrix but the result is not right :(
Pls check the steps if i have done anything wrong or suggest me any techniques so that i can solve these type of problems with no confusions.I m having many problems with these maths.Pls help
2. Oct 9, 2008
### Defennder
Re: Gauss Jordan Elemination Method (Stucked)!!
There isn't any need to find the inverse of the matrix. This approach works only if the system has exactly one solution only, which in this case it has, but not always so in general. Furthermore, it's time consuming since after finding the inverse you have to verify your calculations and THEN multiply the RHS with the inverse matrix.
You have expressed the system of linear equations in terms of an augmented matrix, so why not just performed row operations on the augmented matrix directly to reduce it to reduced row echelon form and from there simply read off the values of x1,x2,x3?
3. Oct 10, 2008
### farmd684
Re: Gauss Jordan Elemination Method (Stucked)!!
Yeah i know that & i also know it would save my time.But i m particulary asked to solve this problem by finding the inverse of augmented matrix.I have also mentioned my workings sir.Have u cheked it?
I have tried many times but each time i failed.Let me know how could i get the inverse of such matrices as i have many problems to do.
Thanks
4. Oct 10, 2008
### Defennder
Re: Gauss Jordan Elemination Method (Stucked)!!
Well I missed that part of the question.
How exactly did you do these steps? And why? The matrix just before step 4 is:
$$\left ( \begin{array}{cccccc}1&\frac{3}{5}&\frac{2}{5}&\frac{1}{5}&0&-\frac{3}{5}\\0&\frac{2}{5}&\frac{4}{15}&-\frac{1}{5}&\frac{1}{3}&0\\0&\frac{3}{5}&\frac{3}{5}&0&0&\frac{3}{5} \end{array} \right )$$.
Doing R1=-R3+R1 will change the top row to:
1 0 -1/5 1/5 0 0
and it can be seen that the subsequent R3=R3-R1 will hence be unable to reduce the top row's 2nd entry from the left to 0. This may be your mistake. Row-operations cannot be performed "simulataneously". So you can't use the old R1 after it has been reduced itself to row reduce another row. I followed your steps right till the end and did not get the identity matrix on the left. So I'm assuming this is your mistake.
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2017-08-22 06:16:39
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http://mathhelpforum.com/differential-geometry/109391-density-q-r.html
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# Thread: Density of Q in R
1. ## Density of Q in R
How do I prove the following claim?
"Given any $y\in\mathbb{R}$, there exists a sequence of rational numbers that converges to $y$"
I'd assume, I would need to use the fact that Q is dense in R (not the topological version I'm trying to prove) and that, given any epsilon neighborhood around some real number x, I can find a rational number in the interval. I just need help formalizing these thoughts and constructing a solid proof.
Thanks
2. Originally Posted by Danneedshelp
How do I prove the following claim?
"Given any $y\in\mathbb{R}$, there exists a sequence of rational numbers that converges to $y$"
I'd assume, I would need to use the fact that Q is dense in R (not the topological version I'm trying to prove) and that, given any epsilon neighborhood around some real number x, I can find a rational number in the interval. I just need help formalizing these thoughts and constructing a solid proof.
Thanks
If you "would need the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$" then there's already nothing to prove.
You can try the following: if $r\in\mathbb{R}$ is rational there's nothing to prove, otherwise we can write $r=a_0.a_1a_2...$ , an infinite non-periodic decimal expression, so you can now define $x_0=a_0\,,\,\,x_1=a_0.a_1\,,\,\,...x_n=a_0.a_1...a _n,...$. It's now simple to check that $\lim_{n\rightarrow\infty}x_n=r$.
The above is somewhat unsatisfying though, since we have first to know the decimal expression of the irrational in order to be able to define the rational sequence that converges to it.
Tonio
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2017-10-20 05:39:08
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http://www.physicsforums.com/showthread.php?t=548664
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Working out the units?
by tommyboo
Tags: units, working
P: 10 I am having a problem finding the correct SI unitsfor the quantity A? In the equation A=√(R/TY) That is A equals the square root of R divided by TY (not to good showing workings on the computer sorry) , the SI units of the quantity R are kg m^3 s^–2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^–2. What are the correct SI units for the quantity A?
Mentor
P: 21,215
Quote by tommyboo I am having a problem finding the correct SI unitsfor the quantity A? In the equation A=√R/TY That is A equals the square root of R divided by TY
Both your notation and explanation are ambiguous.
Is the expression on the right side this?
$$\sqrt{\frac{R}{TY}}$$
or this?
$$\frac{\sqrt{R}}{TY}$$
Quote by tommyboo (not to good showing workings on the computer sorry) , the SI units of the quantity R are kg m^3 s^–2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^–2. What are the correct SI units for the quantity A?
P: 10 The first one R/ty all square root. Do apologise for the bad format
P: 873 Working out the units? Then, the units of A are meters.
P: 754 To clarify gsal's answer... You have the expression $$\sqrt{\frac{R}{TY}}$$ Simply, insert the units for each variable (in place of the variables): $$\sqrt{\frac{\frac{kg\cdot m^3}{s^2}}{(kg)(\frac{m}{s^2})}}$$ and simplify... $$\sqrt{\left(\frac{kg \cdot m^3}{s^2}\right) \left(\frac{s^2}{kg \cdot m}\right)}$$ kg and s2 cancel out, leaving $$\sqrt{\frac{m^3}{m}}$$ which is $$\sqrt{m^2}$$ or, more simply m
Related Discussions Advanced Physics Homework 0 Advanced Physics Homework 2 Precalculus Mathematics Homework 5 Classical Physics 2 Astronomy & Astrophysics 18
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2014-07-29 12:54:16
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https://cstheory.stackexchange.com/questions/25838/what-is-the-importance-of-linear-languages
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# What is the importance of linear languages?
What is the point of linear languages? They appear to be an intermediate set of languages in between regular and context-free languages, but do they have any useful or nice properties that either have been studied, or make them worthwhile to study?
• What is a linear language? Sep 23, 2014 at 11:58
• @Tyson Williams: Maybe this one, I guess? Sep 23, 2014 at 14:42
• An obvious thing to try is to look for papers showing some results about linear languages and check their introductions. Maybe you already did this and found nothing useful, but if so, I cannot see it from the question. Sep 23, 2014 at 14:46
• In general, when you receive comments, you should consider to edit the question. Sep 23, 2014 at 22:29
• (1) Have you checked the paper cited in the Wikipedia page? (Disclaimer: I have not.) (2) Have you tried searching "linear language" "context-free language"? Sep 23, 2014 at 22:34
A language $L$ of $A^*$ is linear if and only if there exists a rational subset $R$ of $A^* \times A^*$ such that $$L = \{ u\tilde v \mid (u, v) \in R \}$$ where $\tilde v$ denotes the reversal of $v$. This result shows that, in a very loose sense, linear languages are a generalization of the language of palindromes.
• @yuval-filmus After a short discussion with my colleagues, it looks like the answer to your question is negative. In a nutshell, it is well known that deciding whether a context-free language is equal to $A^*$ is undecidable. This can be proved using a linear grammar encoding the Post correspondence problem, and it turns out that the language generated by this grammar is regular iff it is equal to $A^*$. Nov 29, 2020 at 19:21
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2022-08-08 18:39:01
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https://intpforum.com/threads/press-the-button.4389/page-2#post-122986
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# Press the button?
## Would you press the button
• Total voters
92
#### grEEEn
I answered the poll without reading the questions too, haha. I'd still push the button anyways. There's only a small chance that the dead person would even be in the same country as me, nobody lives here in Canada. Also keeping in mind that the man doesn't say that someone will die IMMEDIATLY. It could be that someone is going to die whenever they're supposed to die and you didn't influence the even by pushing the button. There really isn't enough detail in the question to make me think that it would be a bad idea. I think I would like to know who gets to die because I push the button, maybe the button opens a door to some restrained prisoner and I get to be all Hostel on them. That would be cool, killing the person with my own hands.
#### INTPINFP
##### Active Member
Hmm... I say kill the man who gave you the choice, then search the building for the money he was going on about. That way you get the money and probably save a few random people in the process.
Of course, this is probably the stupidest idea of the century, as he's probably deranged, there was no money, and you are now on the run from the United States government for killing a man. Oh well, worth a shot.
Most likely he is a just a phyciatrist trying to determine the state of your mental health. Or he could be from an independent study.
In truth, if this ever happened, the first thing I would do is ask "How the heck does this technology work?" That way I could use it to take over the world MUHAHAHAHAHAHA
A million dollars can go along way. You could probably feed 2 people for the rest of their lives with money to spare! No, if you took that seriously, I'm laughing inside. I made that up as a testament to deranged logic. If some one went up to me and said "If you press a button to kill someone I'll give you a million dollars", I would say I need evidence that you have a million dollars. If the money was in his jacket then I would: first ask him how the technology worked in order to determine if it was a hoax,
- If not a hoax, then I would greatly benefit from knowledge of the techonology.
Then, I would ask where is the money kept. If he says I have it with me I shoot him with a tranquilzer, then steal his briefcase,
-If the money is not in the case, I wait for him to wake up, then ask him for the password from the swiss bank account. After I obtain the password I then steal the technology and money from the swiss bank account.
I would have killed noone, gained 1 million dollars, and also prevented the device from ever killing anyone again.
#### Ritsuka
##### That One
If he says I have it with me I shoot him with a tranquilzer
You carry tranquilzer around with you? <.<
#### Ashenstar
##### I'm your chauffeur with high
Yes, I would push the button. Of course I would spend some time thinking on it, not out of fear of consequences more out of fascination.
I am pretty apathetic regarding death. It will come in some form for all of us whether it be at someone's hands, our own, or one of the myriad of possible scenarios resulting in death.
I also wouldn't take it personally if some random person killed me for money so I expect the person I randomly kill to not take it personally either. (doesn't seem very logical)
#### lightspeed
##### Banned
You'd do better getting a whistleblower lawsuit against this company. You'd make much more than 1 million dollars, I suppose, and know that you've possibly saved several lives.
*click*
#### severus
##### Well-Known Member
After consideration, YES I would most definately push the button.
Now: would you kill some random person (not of your choosing) that you don't know for a million dollars? Would you shoot them? Stab them? Strangle them? (all fatally, of course)
And if your answers differ between these and the button, why is that? Or rather, is it okay that they differ?
I would not shoot, stab, or strangle anyone for a million dollars. I'd be spending the money on therapy, so it'd be neutral in the end. Loss of time though. So, the psychological damage would be the reason between my differing answers. Buttons are just so ... impersonal.
#### Ashenstar
##### I'm your chauffeur with high
After consideration, YES I would most definately push the button.
Now: would you kill some random person (not of your choosing) that you don't know for a million dollars? Would you shoot them? Stab them? Strangle them? (all fatally, of course)
And if your answers differ between these and the button, why is that? Or rather, is it okay that they differ?
I would not shoot, stab, or strangle anyone for a million dollars. I'd be spending the money on therapy, so it'd be neutral in the end. Loss of time though. So, the psychological damage would be the reason between my differing answers. Buttons are just so ... impersonal.
Interesting difference. You couldn't, perhaps convince yourself that stabbing someone is rather impersonal as well? Knife or button. Either way it's your hand using an object to cause this person's death, not directly your hands as is the case with strangulation.
#### severus
##### Well-Known Member
The gun is slightly impersonal. But you're still watching the person die.
Stabbing I would find VERY personal. Physically pushing a blade into a screaming person's flesh? (I guess it depends on its sharpness...) The blood spraying out, likely onto you?
The button, just one quick tap and that's it. For all you know, the button doesn't actually do anything, no one actually dies. Or at least that's what you can tell yourself. It's a mystery. Something to ponder, maybe, but nothing to be traumatized over.
#### Ashenstar
##### I'm your chauffeur with high
The gun is slightly impersonal. But you're still watching the person die.
Stabbing I would find VERY personal. Physically pushing a blade into a screaming person's flesh? (I guess it depends on its sharpness...) The blood spraying out, likely onto you?
The button, just one quick tap and that's it. For all you know, the button doesn't actually do anything, no one actually dies. Or at least that's what you can tell yourself. It's a mystery. Something to ponder, maybe, but nothing to be traumatized over.
*nods*
That is true. With the button you never can be sure. You get the comfort of lying to yourself/rationalizing or whatever you wish to call it. If you actually shoot/stab someone you have to watch them die. Watch them collapse and see the light leave their eyes.
Congrats, you're not a psychopath.
#### Misanthropy
##### Redshirt
lol i instantly hit yes before even reading the full question. I then realized i killed a man.
There is a button. Do you press it?
The same thing happened to me.
#### Seducer of the Homeless
##### INFP
YEAH! my vote put 'No' in front! YES!
and fuck all 26 of the rest of you
#### Ilmlas
##### Redshirt
If I were to press the button, and gain the 1 million, I would probably feel so disgraceful and dirty and downright disgusting that I'd do anything to get rid of the money, and I would. I would probably then watch the news and know that anybody on there may have just died because of a button I pressed. Maybe a house exploded, and a baby died. So instead of causing myself a torrent of self hate and loathing, I'd just stick with no.
#### juturna
##### Member
After consideration, YES I would most definately push the button.
Now: would you kill some random person (not of your choosing) that you don't know for a million dollars? Would you shoot them? Stab them? Strangle them? (all fatally, of course)
And if your answers differ between these and the button, why is that? Or rather, is it okay that they differ?
I would not shoot, stab, or strangle anyone for a million dollars. I'd be spending the money on therapy, so it'd be neutral in the end. Loss of time though. So, the psychological damage would be the reason between my differing answers. Buttons are just so ... impersonal.
I would probably press the button in an instant, and be more interested in how the device works but just because the death would be impersonal.
#### transformers
##### Active Member
hmmmmmmm I had to think long and hard about this one. On the one hand, it's easy money with little residual guilt. This is very much like how world leaders escape the guilt of sending thousands of men to their deaths to fight in obscure wars overseas -- they de-personalize the decision. But on the other hand, every time I spend some of that money I'd be reminded of where it came from, and I'm not sure if my conscience could take that. Is it worth it? Perhaps. It would depend on how much I needed the money. Ironically, though, pressing the button for free would probably be easier to accept, since I could justify the decision by saying I did nothing but press a button and got nothing out of it. But that's pointless destruction then, and completely undeserved.
I probably wouldn't press it.
#### Van
##### Member
For all I know, the button does absolutely nothing. Maybe he is saying that someone will die purely to mess with my head. Of course someone will die if I push the button because people die all the time. He could just as easily say, if you push that button the sun will rise tomorrow. It doesn't mean that the button controls the rising of the sun, but it's still a true statement. He also didn't say what would happen if I didn't push the button. I'm gonna push the button.
#### 420MuNkEy
##### Banned
I would not push the button, however, I would study it long and hard because that kind of technology is fascinatingly advanced.
#### kantor1003
##### Prolific Member
Answered no before reading the thread. Reading the thread; yes, of course I'd push the button. Take a trip to Somalia and see how many would have refused there, probably 0. If a family member would have refused (which I can't imagine), one of the other members would probably shoot her/him and push the button himself.
#### walfin
##### Democrazy
Just a side point - would pressing the button be murder? Is it manslaughter/culpable homicide? Do you have to know who your victim is in order to murder him/her?
I don't think the fact that death brings attendant consequences would be a greater bar to pressing the button than death itself.
So you'd become the clown for him, leave your self respect in order to humour him?
Bleah. I haven't visited this thread.
A barbed response, intended to evoke indignation.
I actually feel sorry for loonies. That cannot be said to be leaving my self respect. I would dance a silly dance for a dying man (not in public); if I am ingenuous, so be it.
#### Reverse Transcriptase
##### "you're a poet whether you like it or not"
I'm suspicious of Jules... I keep on wondering if he's really an adbot for "The Box"
YouTube- The Box - Official Trailer [HD]
This new movie is coming out, "The Box", where the main characters are asked if they want to press a button, which will cause 1 person to die and they will receive 1 million dollars.
It looks like a really shitty movie, and I was kind of insulted that they would use the question as a basis for a movie.
Of course, she doesn't press the button. I would say that she has moral strength, but Cameron Diaz's character seems like a stereotypical leading-actress-role weakling.
*sigh* I can't wait for Dagny Taggart...
#### Waterstiller
##### ... runs deep
My favorite part of this thread is the decision to make the votes public. It's kind of an interesting way to split up a forum.
What if it was a button that would anonymously kill one of the people that live on your street that you had never spoken to? What if the amount was lowered to $50k? What if instead of a button, you are the CEO of a corporation that could make additional profits by using a new technology that was very dangerous to your overseas workers. If you used the new tech, one person would die per 1 million extra in your bank account and your corporation would never face consequences. Would you do it? #### Sirian ##### Redshirt I was told about this movie by my sister, because i sometimes ask this kind of questions to my family. I would not press the button, not because i don't want to kill someone, but because i think it would be bad for me to have that much money without working for it. So even if there was no killing involved, i would not accept 1 million, because i know how it would destroy me. #### Nicholas A. A. E. ##### formerly of the Basque-lands I am honestly worried that so many people answered 'yes.' Have you no moral conscience? It's unjustifiable. I'd compare it to several ethically equivalent scenarios, like just shooting a guy for a payoff, but that's elementary. Y'all are intelligent people. Sincerely, me #### Polaris ##### Prolific Member I'd give the manager my instant resignation. And the bird. Then go and get a job for a newspaper. #### GarmGarf ##### Active Member I am honestly worried that so many people answered 'yes.' Have you no moral conscience? It's unjustifiable. I don't conform to the concept of "the right to life". No one chose to be born; no one deserves life. What must be considered here is the pain which will be experience by others due to the death. Yeah, ok, some individuals might be pained by the death (there is a chance no one will care, however), but there is a chance that the person to die could be in immense pain, and this could trump the pain that would be experienced by others due to the death alone. One must also conciser that I could do a lot of good with$1 million, and cause a lot of pleasure.
Furthermore, the world is overpopulated as it is, and the individual who dies will be relieved from the pain of existence.
And anyway, Nicholas A. A. E., "assassin" is listed as one of the ideal occupations for INTPs on SimilarMinds.com for a reason...
#### Causeless
##### Active Member
Okay...
Say you press the button, use about a 10th of the cash to buy a child with failing (insert random vital organ here) an operation that saves their life, then pocket the rest?
Life lost, life saved, $900,000 profit! #### 420MuNkEy ##### Banned I don't conform to the concept of "the right to life". No one chose to be born; no one deserves life. What must be considered here is the pain which will be experience by others due to the death. Yeah, ok, some individuals might be pained by the death (there is a chance no one will care, however), but there is a chance that the person to die could be in immense pain, and this could trump the pain that would be experienced by others due to the death alone. One must also conciser that I could do a lot of good with$1 million, and cause a lot of pleasure.
Furthermore, the world is overpopulated as it is, and the individual who dies will be relieved from the pain of existence.
And anyway, Nicholas A. A. E., "assassin" is listed as one of the ideal occupations for INTPs on SimilarMinds.com for a reason...
I agree with you on the grounds that there is nothing inherently sacred about human life.
What is needed to make a rational decision about this is to not assume too many unknowns. Pushing the button could kill you, Bill Clinton, your mother (if she's still alive), or some starving kid in Africa, but there is no way to know as it's wholly random.
What pushing that button is really saying is that everyone deserves to die (including yourself), as it was pushed with the full knowledge that had an equal chance of killing anyone.
As you have stated, you think no one deserves live, but I'm curious if you would then push that button if it was a 50/50 chance of killing everyone or giving you $1,000,000 #### GarmGarf ##### Active Member I agree with you on the grounds that there is nothing inherently sacred about human life. What is needed to make a rational decision about this is to not assume too many unknowns. Pushing the button could kill you, Bill Clinton, your mother (if she's still alive), or some starving kid in Africa, but there is no way to know as it's wholly random. What pushing that button is really saying is that everyone deserves to die (including yourself), as it was pushed with the full knowledge that had an equal chance of killing anyone. Actually, the question states that no one I know would be the one to die. As you have stated, you think no one deserves live, but I'm curious if you would then push that button if it was a 50/50 chance of killing everyone or giving you$1,000,000
If everyone died instantly (including me), ignorantly and painlessly together, then yes. I'd do that for nothing with a 100% chance (would end human suffering). But I am a bit selfish so 1 million would buy me out of executing that ideal of mine.
If in your situation, all humans die over a period of time in an apocalyptic manner, then no way. I couldn't justify risking that much suffering upon the world for a 50% chance of 1 million.
#### Cobra
##### Well-Known Member
Yes, I'd do it. Hell, if I was in a sour mood I'd press it for free.
But I might also be suspicious that it was some kind of test and that there would be some kind of punishment for pressing the button or reward for not doing so.
LOLapaluza.
#### sniktawekim
##### Well-Known Member
why do we need the money offer to press the button?
id type the dicitonary in morse code on the thing.
#### Cobra
##### Well-Known Member
why do we need the money offer to press the button?
id type the dicitonary in morse code on the thing.
It would be interesting to watch the news that night, if someone used the button to learn Morse code.
#### ckm
##### still swimming
I wouldn't press the button. Self-hate would eat me up for killing for personal gain.
#### Nicholas A. A. E.
##### formerly of the Basque-lands
I don't conform to the concept of "the right to life". No one chose to be born; no one deserves life.
What must be considered here is the pain which will be experience by others due to the death. Yeah, ok, some individuals might be pained by the death (there is a chance no one will care, however), but there is a chance that the person to die could be in immense pain, and this could trump the pain that would be experienced by others due to the death alone. One must also conciser that I could do a lot of good with $1 million, and cause a lot of pleasure. Furthermore, the world is overpopulated as it is, and the individual who dies will be relieved from the pain of existence. And anyway, Nicholas A. A. E., "assassin" is listed as one of the ideal occupations for INTPs on SimilarMinds.com for a reason... Do you accept any unalienable, natural rights? The overpopulation is bad science. We haven't hit that problem yet. No, our issue is underpopulation. But that's a different discussion. #### Inappropriate Behavior ##### is peeing on the carpet Underpopulated? Considering the measures we have to take to feed the people here and it's enviromental damage plus the fact that fresh water supplies are dwindling....We need to get the population down to about a billion. I keep waiting for a real pandemic. #### Nicholas A. A. E. ##### formerly of the Basque-lands I don't want to derail the thread here but hunger is a political problem, not a production problem. It was my understanding this was also the consensus among modern thinkers, though I can't really speak to that. #### Inappropriate Behavior ##### is peeing on the carpet I'll look into it and maybe start a thread in a few days. /end derail To the OP, I'd press the button if I were in a sour enough mood. If I have to decide quickly then there is at least a 90% chance I wouldn't. #### GarmGarf ##### Active Member Do you accept any unalienable, natural rights? So far, only the right to death. The rest are more of conventions. #### Nicholas A. A. E. ##### formerly of the Basque-lands What right, then, has the state to protect non-existent rights? Protecting rights is essentially illegalizing actions that would violate them. Illegalizing actions unnecessarily is an imposition on all citizens. #### Döden ##### Active Member Naw. What would I do with the money? It would be incredibly obvious that I suddenly acquired a fortune and I'm sure an investigation would ensue. I'd rather do my own thing and make my own money and not be responsible for ending another person's life. Take it as you will. #### GarmGarf ##### Active Member What right, then, has the state to protect non-existent rights? Protecting rights is essentially illegalizing actions that would violate them. Illegalizing actions unnecessarily is an imposition on all citizens. The state has power, and in this day and age, that type of power is sourced by the people's favour, and at the moment, the people, overall, support these concepts of rights which the state enforces. As I stated before, these "rights" are a convention, which the people are, for the most part, quite happy to conform to. However, these rights are not absolute truths. If one day, humans no longer required to eat, "the right to food" will probably be abolished. Furthermore, for all we know, "the right to Internet access" may become a world-wise accepted right. These rights exist to keep society running smoothly, and they do their job well, and for the most part, people are happy with them. However, they are, as I stated, not absolute truths, and are rather more of conventions. #### 420MuNkEy ##### Banned #### GarmGarf ##### Active Member #### Thoughtful ##### Nom Nom Nommin' on Heaven's door A man walks up to you and introduces himself as the general manager. He points to a single button on the machine on your desk and says: "If you press that button some random person you don't know will die. If you push that button now you will receive$1 million, and nobody will ever know you did this."
Would you press the button?
No.
First of all, I know the man is lying to me: "...nobody will ever know you did this."
I will know, he will know, the random person it kills will know, and deity might know. If the button really does not do what this "manager" claimed, he's still lied to me, and is therefore not to be trusted.
Instead, I'd try to find the power source of the button. and try to figure out what definition of random the device uses to determine killing someone, and then how it kills them. if I can disable the "killing random person" part of the machine, then I can safely press the button and get the best of both choices.
If I cannot disable the killing device, I'll try to find somebody who can, and then split the reward with them.
#### Anthile
##### Steel marks flesh
I find it interesting how almost everyone here fails at this thread. It's simply about "Would you commit a crime [murder] for personal gain if you could get away with it?". The constellation is absolutely irrelevant. Whether it is a button or dancing the robot that triggers it, that doesn't matter. There is only "Yes" or "No", no third option.
#### Thoughtful
##### Nom Nom Nommin' on Heaven's door
There is only "Yes" or "No", no third option.
"I don't believe in a no-win scenario." If the simulation does not allow you to win, reprogram the simulator so that you can.
Also: "Only a sith deals in absolutes."
#### GarmGarf
##### Active Member
I find it interesting how almost everyone here fails at this thread. It's simply about "Would you commit a crime [murder] for personal gain if you could get away with it?". The constellation is absolutely irrelevant. Whether it is a button or dancing the robot that triggers it, that doesn't matter. There is only "Yes" or "No", no third option.
Yeah, I agree that nit-picking at the practicalities and attempting to cheat the hypothetical situation into a win-win state aren't so commendable.
People who do that are not benefiting from the exercise. They aren't coming to a conclusion about their own morality and philosophy. They are just putting off making that decision and hence revelation about themselves.
However, I do release that the practicalities are actually relevant to a certain degree; i.e: that fact that no one will know and that the method is so impersonal, are relevant. The degree at which the practicalities are relevant is debatable though (a good indication at when someone has gone to far is when someone makes a post like the one I've quoted).
You stated that this thread is about the question: "Would you commit a crime [murder] for personal gain if you could get away with it?". Well, you must consider that some might not consider this murder, and some would justify their (and possibly others') gain in different angles.
Furthermore, this situation is out of society's domain. It isn't quite the same thing as "murdering, cashing-in, and getting away with it", because of the practicalities. Since society can't deal with people who do this (especially because it is unable to know about it), society can't brand them as bad citizens, and the "yes" voters would have less stimulus to feel bad about. Are the ones who voted "yes" on this poll (intentionally) bad citizens? I don't think society has a say to determine that or not.
"Tl; dr": practicalities of the situation matter somewhat, but: "I want to analyse the mechanics of the button, disable the guy and take the money" - I mean common! Feel as witty as you want, but you haven't benefited from this exercise to the degree I have.
#### Thoughtful
##### Nom Nom Nommin' on Heaven's door
Perhaps I wasn't clear enough. I would not push the button, unless I believed I could cheat the system.
However, I know I would try to cheat the system, and might end up killing someone if I failed. The interesting thing being that I would probably never know If failed or succeeded. If I find any pursuit worthy of time, I will by my nature "play to win", even if "winning" isn't possible.
"...you haven't benefited from this exercise to the degree I have. "
Perhaps not, but I said exactly what I would try. If I did not answer truthfully, I don't think I would have benefited from the exercise at all.
EDIT: Put another way, I would want to press the button, but would refrain unless I could trick myself into thinking that either no one would die as a result of my action, or that if someone died, I would not be accountable for their death. If I choose the money, one could say that this is somehow "greedier" than merely taking the money, since I not only take the money, but actively try to feel good about doing so.
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2022-08-19 10:08:53
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https://sinews.siam.org/Details-Page/a-second-doubling-for-siams-student-travel-fund-1
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SIAM News Blog
SIAM News
# A Second Doubling for SIAM’s Student Travel Fund
#### From the SIAM President
Each year, SIAM awards grants to hundreds of students for travel to SIAM conferences around the world, as part of its ongoing commitment to cultivating and training a new generation of mathematical scientists.
When it was launched several years ago, the Student Travel Fund relied primarily on National Science Foundation grants and donations of royalties by generous SIAM book authors. From there, it grew to include donations from the general SIAM membership (via a checkbox on the membership renewal form).
Joscha Gedicke of Humboldt University of Berlin accepted a 2013 SIAM Student Paper Prize from Irene Fonseca. The two other recipients were Keiichi Morikuni of the Graduate University of Advanced Studies (Sokendai), Japan, and Vladislav Voroninski of the University of California, Berkeley; all three gave talks based on their papers in a mini-symposium at the SIAM Annual Meeting.
In 2011, under the leadership of my predecessor, then-SIAM president Nick Trefethen, the SIAM Board of Trustees decided to allocate $100,000 each year to the Student Travel Fund. SIAM’s commitment to its student network continues: most notably, at its July 2013 meeting, the Board elected to double the annual investment, to$200,000, starting in 2014. As of January 2014, SIAM will make approximately $250,000–$275,000 available each year for Student Travel Awards.
Practically speaking, students anywhere in the world can apply for SIAM Student Travel Awards; preference is given to active meeting participants (speakers, poster presenters). Only the NSF portion of the fund is restricted—to students enrolled in U.S. institutions.
If you are interested in applying for a SIAM Student Travel Award, please check the eligibility criteria (funds are also available for postdocs and early-career professionals; these awards, currently much more limited than the Student Travel Awards, are supported only by an NSF grant to SIAM).
At the root of the Board’s generosity is SIAM’s mission of encouraging students to participate in SIAM conferences. To motivate award recipients’ institutions to help co-finance their students’ participation in SIAM events, the awards are generally limited in size. Nevertheless, SIAM hopes that these funds will make a real difference, enabling a growing number of mathematical scientists at the very beginning of their careers to play active roles in our conferences and in SIAM.
Irene Fonseca is president of SIAM, the Mellon College of Science University Professor of Mathematics, and director of the Center for Nonlinear Analysis at Carnegie Mellon University.
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2017-12-16 13:08:05
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https://www.clutchprep.com/chemistry/practice-problems/140000/in-a-0-050-m-solution-of-a-weak-monoprotic-acid-h-1-8-x-10-3-what-is-its-ka-a-3-
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# Problem: In a 0.050 M solution of a weak monoprotic acid, [H+] = 1.8 x 10-3. What is its Ka?A. 3.6 x 10-2B. 9.0 x 10-5C. 6.7 x 10-5D. 1.6 x 10-7
###### FREE Expert Solution
Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case).
The dissociation of a weak acid, HA is as follows:
HA(aq) + H2O(l) H3O+(aq) + A(aq)
From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.
Let X = initial concentration of HA
The Ka expression for HA is:
$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HA}\mathbf{\right]}}}$
Note that each concentration is raised by the stoichiometric coefficient: [HA], [H3O+] and [A] are raised to 1.
[H3O+] = [A] = x = 1.8 x 10-3
We can then plug this into the Ka expression:
81% (450 ratings)
###### Problem Details
In a 0.050 M solution of a weak monoprotic acid, [H+] = 1.8 x 10-3. What is its Ka?
A. 3.6 x 10-2
B. 9.0 x 10-5
C. 6.7 x 10-5
D. 1.6 x 10-7
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2021-01-22 22:37:36
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https://tex.stackexchange.com/questions/245265/how-can-i-access-garamond-premier-pros-straight-quotedbl-quotesingle-using-lu?noredirect=1
|
# How can I access Garamond Premier Pro's straight “quotedbl/quotesingle” using lualatex?
Idea (from here): I use \symbol{idx}, where idx is the glyphs index.
To determine these two glyphs' indexes I had a look into ...texmf-var\luatex-cache\generic\fonts\otf\garamondpremrpro.lua and verified the result using fontforge as well as MS Word. All three report, that "34" and "39" should be the correct indexes - please see the following screenshots:
I also verified the general idea using the index of "blacksquare" (9632): Although the black square is set correctly, the quotedbl/quotesingle do not show up using the following code:
\documentclass{article}
\usepackage{fontspec}
\setmainfont{Garamond Premier Pro}
\newcommand{\blackSquare}{\symbol{9632}} %<-- works fine
\newcommand{\straightQuoteDbl}{\symbol{34}} %<-- does not work
\newcommand{\straightQuoteSingle}{\symbol{39}} %<-- does not work
\begin{document}
{\straightQuoteDbl}Hello World! \blackSquare\straightQuoteDbl
{\straightQuoteSingle}Hello World! \blackSquare\straightQuoteSingle
\end{document}
The code produces the following output:
Both quotes clearly differ from the results MS Word produces (see above).
System environment: Win 7 Pro 64-bit
TeX-Environment: texlive 2014 (frozen)
Font: Garamond Premier Pro Version 2.0.104
What am I doing wrong?
• \setmainfont{Garamond Premier Pro}[RawFeature={-tlig}] or, to affect only parts of the document, \addfontfeatures{RawFeature={-tlig}} – Thérèse May 16 '15 at 16:39
• @Thérèse: Thanks a lot - it works perfectly: \newcommand{\straightQuoteSingle}{\addfontfeatures{RawFeature={-tlig}}\symbol{39}}. You should (re)write this solution into an answer. – user74259 May 16 '15 at 16:53
The curly quotes appear because TeX ligatures are on by default in recent versions of fontspec. If your document consists of code only, you can disable TeX ligatures with \setmainfont, this way:
\documentclass[12pt]{article}
\usepackage{fontspec}
\setmainfont{Garamond Premier Pro}[RawFeature=-tlig]
\newcommand{\blackSquare}{\symbol{9632}}
\begin{document}
"Hello World! \blackSquare"
\end{document}
Or, as egreg suggests, you can put \defaultfontfeatures{} in the preamble; that would affect Garamond and any other font used in the document:
\documentclass[12pt]{article}
\usepackage{fontspec}
\defaultfontfeatures{RawFeature=-tlig}% put before specific fonts to affect them all
\setmainfont{Garamond Premier Pro}
\newcommand{\blackSquare}{\symbol{9632}}
\begin{document}
"Hello World! \blackSquare"
\end{document}
If you want straight quotes in only part of the document, use this approach:
\documentclass[12pt]{article}
\usepackage{fontspec}
\setmainfont{Garamond Premier Pro}
\newcommand{\blackSquare}{\symbol{9632}}
\begin{document}
Hello, beautiful World!''
\end{document}
(You can define a macro to shorten that if you like.)
Or you can disable TeX ligatures globally, as in the two examples, and type real “, ”, ‘, ’, –, and — into your source when you need them.
• Or with \defaultfontfeatures{} before defining the fonts. – egreg May 16 '15 at 17:59
• Yes, I fixed it. – egreg May 16 '15 at 18:04
• Maybe you should mention csquotes – MaxNoe May 17 '15 at 0:04
• @MaxNoe I’ve never had reason to use csquotes in my work, so I’m not the right person to explain how it would help here. Maybe you should add a second answer? – Thérèse May 17 '15 at 1:06
• Did you try it out? I might bet you had a reason :D I will post an answer soon. – MaxNoe May 17 '15 at 10:03
You can use csquotes to define a quoting style which uses these straight quotes:
The \DeclareQuoteStyle command has 5 positional arguments, the name for the style, opening and closing marks for both outer and inner quotes:
You can then use the style with \setquotestyle{<name>}.
\documentclass[border=5pt]{standalone}
\usepackage{fontspec}
\setmainfont{Latin Modern Roman}
\usepackage{csquotes}
\DeclareQuoteStyle{straight}%
\setquotestyle{straight}
\begin{document}
John said: \enquote{Harry said \enquote{Hello World!} yesterday}.
\end{document}
The extra curly braces are needed to make the change in fontfeatures local to the quotation mark only.
This is the result:
• I defined this: \newcommand{\straightQuote}[1]{{\setquotestyle{straight}\enquote{#1}}} and use it like this: John said: \straightQuote{Harry said \enquote{Hello World!} yesterday}. – user74259 May 17 '15 at 14:32
• That's a nice addition! – MaxNoe May 17 '15 at 14:34
• You should add a star to the new command to allow for multi paragraph arguments – MaxNoe May 17 '15 at 14:41
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2020-08-08 06:16:14
|
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|
https://zbmath.org/authors/?q=ai%3Amazzei.luca
|
## Mazzei, Luca
Compute Distance To:
Author ID: mazzei.luca Published as: Mazzei, Luca; Mazzei, L.
Documents Indexed: 8 Publications since 1993 Co-Authors: 10 Co-Authors with 7 Joint Publications 507 Co-Co-Authors
all top 5
### Co-Authors
0 single-authored 7 Hughes, Thomas J. R. 3 Engel, Gerald 3 Larson, Mats G. 2 Oberai, Assad A. 1 Feijóo, Gonzalo R. 1 Garikipati, Krishna 1 Jansen, Kenneth E. 1 Quincy, Jean-Baptiste 1 Taylor, Robert Leroy 1 Wray, Alan A.
### Serials
2 Computer Methods in Applied Mechanics and Engineering 2 Physics of Fluids 1 AIAA Journal 1 Journal of Computational Physics 1 Computing and Visualization in Science
### Fields
4 Fluid mechanics (76-XX) 3 Numerical analysis (65-XX) 2 Partial differential equations (35-XX) 2 Mechanics of deformable solids (74-XX)
### Citations contained in zbMATH Open
7 Publications have been cited 1,512 times in 1,168 Documents Cited by Year
The variational multiscale method – a paradigm for computational mechanics. Zbl 1017.65525
Hughes, Thomas J. R.; Feijóo, Gonzalo R.; Mazzei, Luca; Quincy, Jean-Baptiste
1998
Large eddy simulation and the variational multiscale method. Zbl 0998.76040
Hughes, Thomas J. R.; Mazzei, Luca; Jansen, Kenneth E.
2000
Large eddy simulation of turbulent channel flows by the variational multiscale method. Zbl 1184.76237
Hughes, Thomas J. R.; Oberai, Assad A.; Mazzei, Luca
2001
Continuous/discontinuous finite element approximations of fourth-order elliptic problems in structural and continuum mechanics with applications to thin beams and plates, and strain gradient elasticity. Zbl 1086.74038
Engel, G.; Garikipati, K.; Hughes, T. J. R.; Larson, M. G.; Mazzei, L.; Taylor, R. L.
2002
The multiscale formulation of large eddy simulation: decay of homogeneous isotropic turbulence. Zbl 1184.76236
Hughes, Thomas J. R.; Mazzei, Luca; Oberai, Assad A.; Wray, Alan A.
2001
The continuous Galerkin method is locally conservative. Zbl 0969.65104
Hughes, Thomas J. R.; Engel, Gerald; Mazzei, Luca; Larson, Mats G.
2000
A comparison of discontinuous and continuous Galerkin methods based on error estimates, conservation, robustness and efficiency. Zbl 0946.65109
Hughes, Thomas J. R.; Engel, Gerald; Mazzei, Luca; Larson, Mats G.
2000
Continuous/discontinuous finite element approximations of fourth-order elliptic problems in structural and continuum mechanics with applications to thin beams and plates, and strain gradient elasticity. Zbl 1086.74038
Engel, G.; Garikipati, K.; Hughes, T. J. R.; Larson, M. G.; Mazzei, L.; Taylor, R. L.
2002
Large eddy simulation of turbulent channel flows by the variational multiscale method. Zbl 1184.76237
Hughes, Thomas J. R.; Oberai, Assad A.; Mazzei, Luca
2001
The multiscale formulation of large eddy simulation: decay of homogeneous isotropic turbulence. Zbl 1184.76236
Hughes, Thomas J. R.; Mazzei, Luca; Oberai, Assad A.; Wray, Alan A.
2001
Large eddy simulation and the variational multiscale method. Zbl 0998.76040
Hughes, Thomas J. R.; Mazzei, Luca; Jansen, Kenneth E.
2000
The continuous Galerkin method is locally conservative. Zbl 0969.65104
Hughes, Thomas J. R.; Engel, Gerald; Mazzei, Luca; Larson, Mats G.
2000
A comparison of discontinuous and continuous Galerkin methods based on error estimates, conservation, robustness and efficiency. Zbl 0946.65109
Hughes, Thomas J. R.; Engel, Gerald; Mazzei, Luca; Larson, Mats G.
2000
The variational multiscale method – a paradigm for computational mechanics. Zbl 1017.65525
Hughes, Thomas J. R.; Feijóo, Gonzalo R.; Mazzei, Luca; Quincy, Jean-Baptiste
1998
all top 5
### Cited by 1,639 Authors
66 Tezduyar, Tayfun E. 58 Takizawa, Kenji 54 Codina, Ramon 38 Bazilevs, Yuri 34 Hughes, Thomas J. R. 25 Hsu, Ming-Chen 25 Masud, Arif 20 Chung, Tsz Shun Eric 20 Efendiev, Yalchin R. 20 Scovazzi, Guglielmo 19 Hauke, Guillermo 18 Baiges, Joan 18 Gravemeier, Volker 18 Oberai, Assad A. 16 Brenner, Susanne Cecelia 16 Larson, Mats G. 15 Wall, Wolfgang A. 14 Badia, Santiago 13 Cervera, Miguel 13 Chiumenti, Michele 13 Hachem, Elie 13 Kuraishi, Takashi 13 Målqvist, Axel 13 Oñate Ibáñez de Navarra, Eugenio 13 Principe, Javier R. 11 Akkerman, Ido 11 John, Volker 11 Rossi, Riccardo 10 Calo, Victor Manuel 10 Chacón Rebollo, Tomás 10 Harari, Isaac 10 Hulshoff, Steven J. 10 Peterseim, Daniel 10 Wells, Garth N. 9 Castillo, Ernesto 9 Doweidar, Mohamed Hamdy 9 Evans, John A. 9 Henning, Patrick 9 Huang, Xuehai 9 Kaya, Songul 9 Sagaut, Pierre 8 Gudi, Thirupathi 8 Hou, Thomas Yizhao 8 Hou, Yanren 8 Iliescu, Traian 8 Kostov, Nikolay A. 8 Layton, William J. 8 Leung, Wingtat 8 Oskay, Caglar 8 Otoguro, Yuto 8 Park, Eun-Jae 8 Quarteroni, Alfio M. 8 Rebholz, Leo G. 8 Rispoli, Franco 8 San, Omer 8 Schillinger, Dominik 8 Shang, Yueqiang 7 Burman, Erik 7 Cockburn, Bernardo 7 Corsini, Alessandro 7 Fish, Jacob 7 Fu, Shubin 7 Ginting, Victor 7 Kamensky, David 7 Noels, Ludovic 7 Ohlberger, Mario 7 Shadid, John N. 7 Xu, Fei 7 Zabaras, Nicholas J. 7 Zhang, Xiaohua 7 Zheng, Haibiao 6 An, Rong 6 de Borst, René 6 Galvis, Juan 6 Garikipati, Krishna 6 Gómez Mármol, Macarena 6 Guasch, Oriol 6 Hansbo, Peter 6 Henicke, Bradley 6 Jiang, Lijian 6 Korobenko, Artem 6 Liu, Wing Kam 6 Mesri, Youssef 6 Terahara, Takuya 6 Wagner, Gregory J. 6 Winckelmans, Grégoire S. 5 Abdulle, Assyr 5 Avila, Matias 5 Bochev, Pavel B. 5 Calderer, Ramon 5 Colomés, Oriol 5 Dede’, Luca 5 Farhat, Charbel H. 5 Farthing, Matthew W. 5 Feng, Minfu 5 Ganapathysubramanian, Baskar 5 Houzeaux, Guillaume 5 Huang, Jianguo 5 Idelsohn, Sergio Rodolfo 5 Irisarri, Diego ...and 1,539 more Authors
all top 5
### Cited in 115 Serials
319 Computer Methods in Applied Mechanics and Engineering 127 Journal of Computational Physics 81 Computational Mechanics 55 Computers and Fluids 37 International Journal for Numerical Methods in Engineering 37 Journal of Scientific Computing 33 Journal of Computational and Applied Mathematics 32 Computers & Mathematics with Applications 30 Physics of Fluids 21 Multiscale Modeling & Simulation 20 Numerical Methods for Partial Differential Equations 19 Applied Numerical Mathematics 19 M$$^3$$AS. Mathematical Models & Methods in Applied Sciences 18 Applied Mathematics and Computation 16 International Journal for Numerical Methods in Fluids 15 SIAM Journal on Numerical Analysis 13 Archives of Computational Methods in Engineering 11 SIAM Journal on Scientific Computing 10 Mathematics of Computation 10 Numerische Mathematik 9 International Journal of Computational Fluid Dynamics 9 European Series in Applied and Industrial Mathematics (ESAIM): Mathematical Modelling and Numerical Analysis 8 Acta Mechanica 8 Computational Geosciences 7 Journal of Fluid Mechanics 7 Communications in Numerical Methods in Engineering 6 Journal of Mathematical Analysis and Applications 6 International Journal of Computational Methods 5 International Journal of Computer Mathematics 5 Engineering Analysis with Boundary Elements 4 International Journal of Solids and Structures 4 Journal of the Mechanics and Physics of Solids 4 Calcolo 4 Meccanica 4 Applied Mathematics and Mechanics. (English Edition) 4 Applied Mathematical Modelling 4 Advances in Computational Mathematics 4 European Journal of Mechanics. A. Solids 4 Discrete and Continuous Dynamical Systems. Series S 4 Advances in Applied Mathematics and Mechanics 3 Numerical Algorithms 3 International Journal of Numerical Methods for Heat & Fluid Flow 3 Mathematical Problems in Engineering 3 Journal of Turbulence 3 Computational Methods in Applied Mathematics 3 Journal of Numerical Mathematics 3 Journal of Computational Acoustics 3 Communications in Computational Physics 3 S$$\vec{\text{e}}$$MA Journal 2 International Journal of Engineering Science 2 BIT 2 Mathematics and Computers in Simulation 2 Numerical Functional Analysis and Optimization 2 Chinese Annals of Mathematics. Series B 2 COMPEL 2 Mathematical and Computer Modelling 2 Journal of Inequalities and Applications 2 Discrete and Continuous Dynamical Systems. Series B 2 International Journal of Numerical Analysis and Modeling 2 Acta Mechanica Sinica 2 International Journal for Computational Methods in Engineering Science and Mechanics 2 Communications on Applied Mathematics and Computation 2 Results in Applied Mathematics 1 International Journal of Modern Physics B 1 Archive for Rational Mechanics and Analysis 1 Communications on Pure and Applied Mathematics 1 Journal of Engineering Mathematics 1 Mathematical Methods in the Applied Sciences 1 Theoretical and Computational Fluid Dynamics 1 ACM Transactions on Mathematical Software 1 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 1 SIAM Journal on Control and Optimization 1 Acta Applicandae Mathematicae 1 Computer Aided Geometric Design 1 Physica D 1 Applied Mathematics Letters 1 Applications of Mathematics 1 SIAM Review 1 Archive of Applied Mechanics 1 Cybernetics and Systems Analysis 1 Numerical Linear Algebra with Applications 1 ETNA. Electronic Transactions on Numerical Analysis 1 Mathematics and Mechanics of Solids 1 Computing and Visualization in Science 1 European Series in Applied and Industrial Mathematics (ESAIM): Control, Optimization and Calculus of Variations 1 Abstract and Applied Analysis 1 Journal of Applied Mechanics and Technical Physics 1 Chaos 1 Journal of Mathematical Fluid Mechanics 1 European Journal of Mechanics. B. Fluids 1 Analysis (München) 1 Journal of the European Mathematical Society (JEMS) 1 Combustion Theory and Modelling 1 Journal of High Energy Physics 1 Mathematical Modelling and Analysis 1 Journal of Systems Science and Complexity 1 Comptes Rendus. Mathématique. Académie des Sciences, Paris 1 Journal of Applied Mathematics and Computing 1 SIAM Journal on Applied Dynamical Systems 1 Hacettepe Journal of Mathematics and Statistics ...and 15 more Serials
all top 5
### Cited in 29 Fields
691 Fluid mechanics (76-XX) 650 Numerical analysis (65-XX) 305 Mechanics of deformable solids (74-XX) 207 Partial differential equations (35-XX) 31 Biology and other natural sciences (92-XX) 27 Calculus of variations and optimal control; optimization (49-XX) 21 Classical thermodynamics, heat transfer (80-XX) 18 Geophysics (86-XX) 16 Statistical mechanics, structure of matter (82-XX) 13 Potential theory (31-XX) 10 Probability theory and stochastic processes (60-XX) 9 Computer science (68-XX) 9 Optics, electromagnetic theory (78-XX) 8 Ordinary differential equations (34-XX) 5 Mechanics of particles and systems (70-XX) 4 Statistics (62-XX) 3 Dynamical systems and ergodic theory (37-XX) 3 Approximations and expansions (41-XX) 3 Quantum theory (81-XX) 2 Functional analysis (46-XX) 2 Systems theory; control (93-XX) 1 Combinatorics (05-XX) 1 Real functions (26-XX) 1 Harmonic analysis on Euclidean spaces (42-XX) 1 Integral equations (45-XX) 1 Operator theory (47-XX) 1 Global analysis, analysis on manifolds (58-XX) 1 Relativity and gravitational theory (83-XX) 1 Astronomy and astrophysics (85-XX)
|
2022-09-28 00:23:13
|
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http://mathhelpforum.com/number-theory/115762-prove-contradiction-print.html
|
• Nov 20th 2009, 08:49 AM
hebby
Prove by contradiction that 2^(1/3) is irrational.
a^2= 2^(2/3)
then 2^(2/3)= M^2 because we assume a is rational
N^2
then what do I do
• Nov 20th 2009, 08:57 AM
aman_cc
Quote:
Originally Posted by hebby
Prove by contradiction that 2^(1/3) is irrational.
a^2= 2^(2/3)
then 2^(2/3)= M^2 because we assume a is rational
N^2
then what do I do
Do you know the proof of $\sqrt(2)$ is irrational?
Go along same lines.
Let $2^{1/3} = p/q$. Where $(p,q)=1$
Thus $2q^3=p^3$
so $2|p^3$=> $2|p$
Can you complete?
• Nov 20th 2009, 09:05 AM
hebby
Thanks...I can see the proof now...Merci!..it was quite easy
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2017-04-25 17:14:19
|
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https://jpmccarthymaths.com/2020/04/23/almost-all-trees-have-quantum-symmetry/
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Some notes on this paper.
### 1. Introduction and Main Results
A tree has no symmetry if its automorphism group is trivial. Erdos and Rényi showed that the probability that a random tree on $n$ vertices has no symmetry goes to zero as $n\rightarrow \infty$.
Banica (after Bichon) wrote down with clarity the quantum automorphism group of a graph. It contains the usual automorphism group. When it is larger, the graph is said to have quantum symmetry.
Lupini, Mancinska, and Roberson show that almost all graphs are quantum antisymmetric. I am fairly sure this means that almost all graphs have no quantum symmetry, and furthermore for almost all (as $n\rightarrow \infty$) graphs the automorphism group is trivial.
The paper in question hopes to show that almost all trees have quantum symmetry — but at this point I am not sure if this is saying that the quantum automorphism group is larger than the classical.
## 2. Preliminaries
### 2.1 Graphs and Trees
Standard definitions. No multi-edges. Undirected if the edge relation is symmetric. As it is dealing with trees, this paper is concerned with undirected graphs without loops, and identify $V=\{v_1,\dots,v_n\}\cong \{1,2,\dots,n\}$. A path is a sequence of edges. We will not see cycles if we are discussing trees. Neither will we talk about disconnected graphs: a tree is a connected graph without cycles (this throws out loops actually.
The adjacency matrix of a graph is a matrix $A=(a_{ik})_{i,j\in V}$ with $a_{ij}=1$ iff there is an edge connected $i$ and $j$. The adjacency matrix is symmetric.
### 2.2 Symmetries of Graphs
An automorphism of a graph $\Gamma$ is a permutation of $V$ that preserves adjacency and non-adjacency. The set of all such automorphisms, $\text{Aut }\Gamma$, is a group where the group law is composition. It is a subgroup of $S_n$, and $S_n$ itself can be embedded as permutation matrices in $M_n(\mathbb{C})$. We then have
$\text{Aut }\Gamma=\{\sigma\in S_n\,:\,\sigma A=A\sigma\}\subseteq S_n$.
If $\text{Aut }\Gamma=\{e\}$, it is asymmetric. Otherwise it is or rather has symmetry.
### 2.3 Compact Matrix Quantum Groups
compact matrix quantum group is a pair $(C(G),u)$, where $C(G)$ is a unital $\mathrm{C}^\ast$-algebra, and $u=(u_{ij})_{i,j=1}^n\in M_n(C(G))$ is such that:
• $C(G)$ is generated by the $u_{ij}$,
• There exists a morphism $\Delta:C(G)\rightarrow C(G)\otimes C(G)$, such that $\Delta(u_{ij})=\sum_{k=1}^n u_{ik}\otimes u_{kj}$
• $u$ and $u^T$ are invertible (Timmermann only asks that $\overline{u}=(u_{ij}^\ast)$ be invertible)
The classic example (indeed commutative examples all take this form) is a compact matrix group $G\subseteq U_n(\mathbb{C})$ and $u_{ij}:G\rightarrow \mathbb{C}$ the coordinates of $G$.
#### Example 2.3
The algebra of continuous functions on the quantum permutation group $S_n^+$ is generated by $n^2$ projections $u_{ij}$ such that the row sums and column sums of $u=(u_{ij})$ both equal $\mathbf{1}_{S_n^+}$.
The map $\varphi:C(S_{n}^+)\rightarrow C(S_n)$, $u_{ij}\mapsto \mathbf{1}_{\{\sigma\in S_n\,|\,\sigma(j)=i\}}$ is a surjective morphism that is an isomorphism for $n=1,2,3$, so that the sets $\{1\},\,\{1,2\},\,\{1,2,3\}$ have no quantum symmetries.
### 2.4 Quantum Symmetries of Graphs
#### Definition 2.4 (Banica after Bichon)
Let $\Gamma=(V,E)$ be a graph on $n$ vertices without multiple edges not loops, and let $A$ be its adjacency matrix. The quantum automorphism group $\text{QAut }\Gamma$ is defined as the compact matrix group with $\mathrm{C}^\ast$-algebra:
$\displaystyle C(\text{QAut }\Gamma)=C(S_n^+)/\langle uA=Au\rangle$
For me, not the authors, this requires some work. Banica says that $\langle uA=Au\rangle$ is a Hopf ideal.
Hopf ideal is a closed *-ideal $I\subset C(G)$ such that
$\Delta(I)\subset C(G)\otimes I+I\otimes C(G)$.
Classically, $I$ the set of functions vanishing on a distinguished subgroup. The quotient map is $f\mapsto f+I$, and $f+I=g+I$ if their difference is in $I$, that is if they agree on the subgroup.
The classical version of $Au=uA$ ends up as $a_{ij}=a_{\sigma(i)\sigma(j)}$… the group in question the classical $\text{Aut }\Gamma$. In that sense perhaps $Au=uA$ might be better given as $fA=Af$.
Easiest thing first, is it a *-ideal? Well, take the adjoint of $fA=Af\Rightarrow A^*f^*=f^*A^*$ and $A=A^*$ so $I$ is *closed. Suppose $f\in I$ and $g\in C(S_n^+)$… I cannot prove that this is an ideal! But time to move on.
### 3. The Existence of Two Cherries
In this section the authors will show that almost all trees have two cherries. Definition 3.4 says with clarity what a cherry is, here I use an image [credit: www-math.ucdenver.edu]:
(3,5,4) and (7,9,8) are cherries
#### Remark 3.2
If a graph admits a cherry $(u_1,u_2,v)$, the transposition $(u_1\quad u_2)$ is a non-trivial automorphism.
#### Theorem 3.3 (Erdos, Réyni)
Almost all trees contains at least one cherry in the sense that
$\displaystyle \lim_{n\rightarrow \infty}\mathbb{P}[C_n\geq 1]=1$,
where $C_n$ is #cherries in a (uniformly chosen) random tree on $n$ vertices.
#### Corollary 4.3
Almost all trees have symmetry.
The paper claims in fact that almost all trees have at least two cherries. This will allow some $S_4^+$ action to take place. This can be seen in this paper which is the next point of interest.
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2020-10-23 05:09:37
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https://forum.effectivealtruism.org/tag/risk-aversion
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# Risk aversion
An option is more risky if the value of its possible outcomes is more widely dispersed (higher variance). An agent is risk averse in a pure sense if they prefer safe options over risky ones, even when the riskier options (gambles) would give more of what they value in expectation. (Conversely, an agent is risk neutral if they are indifferent between options with the same expected payoffs, and they are risk seeking if they prefer gambles to equal-expected-payoff safe options.)
All payoffs in this definition of pure risk aversion are expressed in terms of what the agent values. As a consequence, risk aversion is on this definition quite distinct from the most common notion of risk aversion in economics, whereby diminishing marginal utility of money causes people to prefer low-variance, lower-expected-value monetary tradeoffs. Risk aversion is also related to, but distinct from, ambiguity aversion (Wikipedia 2006).
This form of pure risk aversion appears to be irrational under a variety of assumptions, as mentioned in expected value theory. Indeed, risk averse agents in this sense can be exploited in ways that seem to count against risk aversion (Yudkowsky 2008). However, some have defended it as rational (e.g. Buchak 2013). If risk aversion is rational, some form of risk-averse decision theory might be appropriate....
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2021-05-17 12:39:49
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http://mathhelpforum.com/math/16871-abstract-algebra-forum.html
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# Thread: abstract algebra forum
1. ## abstract algebra forum
Hey all(especially PH). Do you know of this site?. It's an abstract and linear algebra forum.
Check it out if you wish.
Abstract and Linear Algebra :: Index
Of course, it's doesn't have that much traffic, but there are quite a few unanswered posts you may like to tackle. Oh yeah, I didn't see Latex, so it may not support it. In that event, you may not be interested.
2. Originally Posted by galactus
Hey all(especially PH). Do you know of this site?. It's an abstract and linear algebra forum.
Check it out if you wish.
Abstract and Linear Algebra :: Index
Of course, it's doesn't have that much traffic, but there are quite a few unanswered posts you may like to tackle. Oh yeah, I didn't see Latex, so it may not support it. In that event, you may not be interested.
It is very small. Maybe in the future it get bigger. I use the "Superior Algebra" on AoPs, they have some good experts there.
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2017-04-28 09:07:28
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https://gsverhoeven.github.io/post/posterior-distribution-average-treatment-effect/
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# Using posterior predictive distributions to get the Average Treatment Effect (ATE) with uncertainty
## Gertjan Verhoeven & Misja Mikkers
Here we show how to use Stan with the brms R-package to calculate the posterior predictive distribution of a covariate-adjusted average treatment effect. We fit a model on simulated data that mimics a (very clean) experiment with random treatment assignment.
### Introduction
Suppose we have data from a Randomized Controlled Trial (RCT) and we want to estimate the average treatment effect (ATE). Patients get treated, or not, depending only on a coin flip. This is encoded in the Treatment variable. The outcome is a count variable Admissions, representing the number of times the patient gets admitted to the hospital. The treatment is expected to reduce the number of hospital admissions for patients.
To complicate matters (a bit): As is often the case with patients, not all patients are identical. Suppose that older patients have on average more Admissions. So Age is a covariate.
### Average treatment effect (ATE)
Now, after we fitted a model to the data, we want to actually use our model to answer "What-if" questions (counterfactuals). Here we answer the following question:
• What would the average reduction in Admissions be if we had treated ALL the patients in the sample, compared to a situation where NO patient in the sample would have received treatment?
Well, that is easy, we just take the fitted model, change treatment from zero to one for each, and observe the ("marginal") effect on the outcome, right?
Yes, but the uncertainty is harder. We have uncertainty in the estimated coefficients of the intercept and covariate, as well as in the coefficient of the treatment variable. And these uncertainties can be correlated (for example between the coefficients of intercept and covariate).
Here we show how to use posterior_predict() to simulate outcomes of the model using the sampled parameters. If we do this for two counterfactuals, all patients treated, and all patients untreated, and subtract these, we can easily calculate the posterior predictive distribution of the average treatment effect.
Let's do it!
This tutorial uses brms, a user friendly interface to full Bayesian modelling with Stan.
library(tidyverse)
library(rstan)
library(brms)
### Data simulation
We generate fake data that matches our problem setup.
Admissions are determined by patient Age, whether the patient has Treatment, and some random Noise to capture unobserved effects that influence Admissions. We exponentiate them to always get a positive number, and plug it in the Poisson distribution using rpois().
set.seed(123)
id <- 1:200
n_obs <- length(id)
b_tr <- -0.7
b_age <- 0.1
df_sim <- as.data.frame(id) %>%
mutate(Age = rgamma(n_obs, shape = 5, scale = 2)) %>% # positive cont predictor
mutate(Noise = rnorm(n_obs, mean = 0, sd = 0.5)) %>% # add noise
mutate(Treatment = ifelse(runif(n_obs) < 0.5, 0, 1)) %>% # Flip a coin for treatment
mutate(Lambda = exp(b_age * Age + b_tr * Treatment + Noise)) %>% # generate lambda for the poisson dist
mutate(Admissions = rpois(n_obs, lambda = Lambda))
### Summarize data
Ok, so what does our dataset look like?
summary(df_sim)
## id Age Noise Treatment
## Min. : 1.00 Min. : 1.794 Min. :-1.32157 Min. :0.000
## 1st Qu.: 50.75 1st Qu.: 6.724 1st Qu.:-0.28614 1st Qu.:0.000
## Median :100.50 Median : 8.791 Median : 0.04713 Median :0.000
## Mean :100.50 Mean : 9.474 Mean : 0.02427 Mean :0.495
## 3rd Qu.:150.25 3rd Qu.:11.713 3rd Qu.: 0.36025 3rd Qu.:1.000
## Max. :200.00 Max. :24.835 Max. : 1.28573 Max. :1.000
## Min. : 0.2479 Min. : 0.000
## 1st Qu.: 1.1431 1st Qu.: 1.000
## Median : 1.8104 Median : 2.000
## Mean : 2.6528 Mean : 2.485
## 3rd Qu.: 3.0960 3rd Qu.: 3.000
## Max. :37.1296 Max. :38.000
The Treatment variable should reduce admissions. Lets visualize the distribution of Admission values for both treated and untreated patients.
ggplot(data = df_sim, aes(x = Admissions)) +
geom_histogram(stat="count") +
facet_wrap(~ Treatment)
The effect of the treatment on reducing admissions is clearly visible.
We can also visualize the relationship between Admissions and Age, for both treated and untreated patients. We use the viridis scales to provide colour maps that are designed to be perceived by viewers with common forms of colour blindness.
ggplot(data = df_sim, aes(x = Age, y = Admissions, color = as.factor(Treatment))) +
geom_point() +
scale_color_viridis_d(labels = c("No Treatment", "Treatment")) +
labs(color = "Treatment")
Now lets fit our Bayesian Poisson regression model to it.
### Fit model
We use brms default priors for convenience here. For a real application we would of course put effort into into crafting priors that reflect our current knowledge of the problem at hand.
model1 <- brm(
formula = as.integer(Admissions) ~ Age + Treatment,
data = df_sim,
family = poisson(),
warmup = 2000, iter = 5000,
cores = 2,
chains = 4,
seed = 123,
silent = TRUE,
refresh = 0,
)
## Compiling Stan program...
## Start sampling
### Check model fit
summary(model1)
## Family: poisson
## Formula: as.integer(Admissions) ~ Age + Treatment
## Data: df_sim (Number of observations: 200)
## Samples: 4 chains, each with iter = 5000; warmup = 2000; thin = 1;
## total post-warmup samples = 12000
##
## Population-Level Effects:
## Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
## Intercept -0.05 0.12 -0.28 0.18 1.00 7410 7333
## Age 0.12 0.01 0.10 0.14 1.00 8052 8226
## Treatment -0.83 0.10 -1.02 -0.63 1.00 7794 7606
##
## Samples were drawn using sampling(NUTS). For each parameter, Bulk_ESS
## and Tail_ESS are effective sample size measures, and Rhat is the potential
## scale reduction factor on split chains (at convergence, Rhat = 1).
We see that the posterior dists for $$\beta_{Age}$$ and $$\beta_{Treatment}$$ cover the true values, so looking good. To get a fuller glimpse into the (correlated) uncertainty of the model parameters we make a pairs plot:
pairs(model1)
As expected, the coefficients $$\beta_{Intercept}$$ (added by brms) and $$\beta_{Age}$$ are highly correlated.
### First attempt: Calculate Individual Treatment effects using the model fit object
Conceptually, the simplest approach for prediction is to take the most likely values for all the model parameters, and use these to calculate for each patient an individual treatment effect. This is what plain OLS regression does when we call predict.lm() on a fitted model.
est_intercept <- fixef(model1, pars = "Intercept")[,1]
est_age_eff <- fixef(model1, pars = "Age")[,1]
est_t <- fixef(model1, pars = "Treatment")[,1]
# brm fit parameters (intercept plus treatment)
ites <- exp(est_intercept + (est_age_eff * df_sim$Age) + est_t) - exp(est_intercept + (est_age_eff * df_sim$Age))
ggplot(data.frame(ites), aes(x = ites)) +
geom_histogram() +
geom_vline(xintercept = mean(ites), col = "red") +
ggtitle("Effect of treatment on Admissions for each observation") +
expand_limits(x = 0)
Averaging the ITEs gives us the ATE, displayed in red.
Ok, so on average, our treatment reduces the number of Admissions by -1.9.
You may wonder: why do we even have a distribution of treatment effects here? Should it not be the same for each patient? Here a peculiarity of the Poisson regression model comes to surface: The effect of changing Treatment from 0 to 1 on the outcome depends on the value of Age of the patient. This is because we exponentiate the linear model before we plug it into the Poisson distribution.
### Next, the uncertainty in the ATE
How to get all this underlying, correlated uncertainty in the model parameters, that have varying effects depending on the covariates of patients, and properly propagate that to the ATE? What is the range of plausible values of the ATE consistent with the data & model?
At this point, using only the summary statistics of the model fit (i.e. the coefficients), we hit a wall. To make progress we have to work with the full posterior distribution of model parameters, and use this to make predictions. That is why it is often called "the posterior predictive distribution" (Check BDA3 for the full story).
### Posterior predictive distribution (PPD): two tricks
Ok, you say, a Posterior Predictive Distribution, let's have it! Where can I get one?
Luckily for us, most of the work is already done, because we have fitted our model. And thus we have a large collection of parameter draws (or samples, to confuse things a bit). All the correlated uncertainty is contained in these draws.
This is the first trick. Conceptually, we imagine that each separate draw of the posterior represents a particular version of our model.
In our example model fit, we have 12.000 samples from the posterior. In our imagination, we now have 12.000 versions of our model, where unlikely parameter combinations are present less often compared to likely parameter combinations. The full uncertainty of our model parameters is contained in this "collection of models" .
The second trick is that we simulate (generate) predictions for all observations, from each of these 12.000 models. Under the hood, this means computing for each model (we have 12.000), for each observation (we have 200) the predicted lambda value given the covariates, and drawing a single value from a Poisson distribution with that $$\Lambda$$ value (e.g. running rpois(n = 1, lambda) ).
This gives us a 12.000 x 200 matrix, that we can compute with.
### Computing with the PPD: brms::posterior_predict()
To compute PPD's, we can use brms::posterior_predict(). We can feed it any dataset using the newdata argument, and have it generate a PPD.
For our application, the computation can be broken down in two steps:
• Step 1: use posterior_predict() on our dataset with Treatment set to zero, do the same for our dataset with Treatment set to one, and subtract the two matrices. This gives us a matrix of outcome differences / treatment effects.
• Step 2: Averaging over all cols (the N=200 simulated outcomes for each draw) should give us the distribution of the ATE. This distribution now represents the variability (uncertainty) of the estimate.
Ok, step 1:
# create two versions of our dataset, with all Tr= 0 and all Tr=1
df_sim_t0 <- df_sim %>% mutate(Treatment = 0)
df_sim_t1 <- df_sim %>% mutate(Treatment = 1)
# simulate the PPDs
pp_t0 <- posterior_predict(model1, newdata = df_sim_t0)
pp_t1 <- posterior_predict(model1, newdata = df_sim_t1)
diff <- pp_t1 - pp_t0
dim(diff)
## [1] 12000 200
And step 2 (averaging by row over the cols):
ATE_per_draw <- apply(diff, 1, mean)
# equivalent expression for tidyverse fans
#ATE_per_draw <- data.frame(diff) %>% rowwise() %>% summarise(avg = mean(c_across(cols = everything())))
length(ATE_per_draw)
## [1] 12000
Finally, a distribution of plausible ATE values. Oo, that is so nice. Lets visualize it!
ggplot(data.frame(ATE_per_draw), aes(x = ATE_per_draw)) +
geom_histogram() +
geom_vline(xintercept = mean(ites), col = "red") +
ggtitle("Posterior distribution of the Average Treatment Effect (ATE)")
We can compare this distribution with the point estimate of the ATE we obtained above using the model coefficients. It sits right in the middle (red line), just as it should be!
### Demonstrating the versatility: uncertainty in the sum of treatment effects
Now suppose we are a policy maker, and we want to estimate the total reduction in Admissions if all patients get the treatment. And we want to quantify the range of plausible values of this summary statistic.
To do so, we can easily adjust our code to summing instead of averaging all the treatment effects within each draw (i.e. by row):
TTE_per_draw <- apply(diff, 1, sum)
ggplot(data.frame(TTE_per_draw), aes(x = TTE_per_draw)) +
geom_histogram() +
geom_vline(xintercept = sum(ites), col = "red") +
ggtitle("Posterior distribution of the Total Treatment Effect (TTE)")
So our model predicts for the aggregate reduction of patient Admissions a value in the range of -500 to -250.
This distribution can then be used to answer questions such as "what is the probability that our treatment reduces Admissions by at least 400"?
TTE <- data.frame(TTE_per_draw) %>%
mutate(counter = ifelse(TTE_per_draw < -400, 1, 0))
mean(TTE\$counter) * 100
## [1] 38.1
### Take home message: PPD with brms is easy and powerful
We hope to have demonstrated that when doing a full bayesian analysis with brms and Stan, it is very easy to create Posterior Predictive Distributions using posterior_predict(). And that if we have a posterior predictive distribution, incorporating uncertainty in various "marginal effects" type analyses becomes dead-easy. These analyses include what-if scenarios using the original data, or scenarios using new data with different covariate distributions (for example if we have an RCT that is enriched in young students, and we want to apply it to the general population). Ok, that it is for today, happy modelling!
##### Gertjan Verhoeven
###### Data Scientist / Policy Advisor
Gertjan Verhoeven is a research scientist currently at the Dutch Healthcare Authority, working on health policy and statistical methods. Follow me on Twitter or Mastodon to receive updates on new blog posts. Statistics posts using R are featured on R-Bloggers.
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2023-03-22 04:15:09
|
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https://asmedigitalcollection.asme.org/ebooks/book/140/chapter-abstract/26872/Application-of-Modified-Adaptive-Tabu-Search-to?redirectedFrom=fulltext
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ASME Press Select Proceedings
# International Conference on Computer and Electrical Engineering 4th (ICCEE 2011)
By
Jianhong Zhou
Jianhong Zhou
Search for other works by this author on:
ISBN:
9780791859841
No. of Pages:
698
Publisher:
ASME Press
Publication date:
2011
This paper presents an application of the modified adaptive tabu search algorithms (mATS) to a dynamic constrained economic dispatch (DED). DED is one of the main functions of power generation operation and control. Its objective is to operate an electric power system most economically while the system is operating within its security limits, eg. generator capacity limit and generator ramp rate limit. The proposed mATS with an adaptive neighbourhood mechanism (AN) make difference to the original ATS. A fiveunit test system with nonsmooth fuel cost function is used to illustrate. The best searched result of mATS is $52,398 which is less than$52,591 of the original ATS. It confirms that mATS can be applicable for solving DED problem and more effectiveness the original ATS does.
This content is only available via PDF.
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2021-04-19 14:52:31
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http://lmfa.ec-lyon.fr/spip.php?article1721&lang=fr
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Laboratoire de Mécanique des Fluides et d’Acoustique - UMR 5509
LMFA - UMR 5509
Laboratoire de Mécanique des Fluides et d’Acoustique
Lyon
France
Nos partenaires
Article dans J. Non-Newtonian Fluid Mech. (2020)
Mixing of non-Newtonian inelastic fluid in a turbulent patch of T-junction
Haining Luo, Alexandre Delache & Serge Simoëns
In this paper, we present result from a direct numerical simulation (DNS) of turbulent flow in a converging T-junction for both Newtonian (water) and non-Newtonian inelastic fluid (dilute Xanthan Gum solution). Based on experimental data, the Bird-Carreau law is used to capture the inelastic shear thinning property of the solution. For the Xanthan solution, the viscosity at rest is about 100 times greater than the viscosity at high shear-rate. A passive scalar is introduced in the transverse branch to investigate the mixing in such configuration. The nominal Reynolds number at the exit varies from 4800 to 8000 for the Newtonian cases and for the same inflow rates, the non-Newtonian flow will be necessarily at lower nominal Reynolds number. Two regimes are explored as a function of the inlet velocity ratio $r=U_b/U_m$ : the ”deflecting” regime noted DR ($r=1$) and the ”impinging” regime noted IR ($r=4$). For the non-Newtonian cases, two viscous cores are observed before the junction. After the junction a laminar state is obtained for the lower flow rate conditions. Surprisingly, in spite of a large viscosity at rest, a self-sustained non-Newtonian turbulence is achieved except for one case. We describe existing vortex mechanisms which pilot the scalar mixing. In addition, we show that in the non-Newtonian cases, the existing peak of turbulence is only shifted in the DR case. The shift is probably due to the nature of the fluid and not to the dynamical regime. After an intense turbulent zone, we show that a re-laminarization zone appears in the non-Newtonian case which reduces the fluctuation as well as mixing. As a result, IR has a better mixing quality than DR.
Pour en savoir plus :
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2020-09-27 22:14:37
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http://molinajosue.blogspot.com/2016/09/post-prelims-new-beginning-pt-1.html
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Here's fresh, new math—stuff I've never posted about. It's a bit of measure theory—an indispensable generalization of the notion of "length", "area", "volume", etc., which will later on allow us to extend the Riemann integral to the Lebesgue integral (and thus take our calculus to the "next level", if I may). Real-world applications of these concepts are plentiful in quantum mechanics. In fact, it feels like all the math I do now has applications only in similar fields at around that level, which makes it somewhat difficult to motivate. But someone has to do it, right?
Claim. Let $\left\{X_j\right\}_{j=1}^\infty$ be a collection of sets, let $\mathcal E_j\subseteq\mathcal P\left(X_j\right)$, let $\mathcal M_j$ be the $\sigma$-algebra on $X_j$ generated by $\mathcal E_j$, and let $X_j\in\mathcal E_j$. Then $\bigotimes_{j=1}^\infty\mathcal M_j$ is generated by$$\mathcal F=\left\{\prod_{j=1}^\infty E_j:E_j\in\mathcal E_j\right\}.$$Proof. Recall that $\bigotimes_{j=1}^\infty\mathcal M_j$ is generated by$$\mathcal G=\left\{\pi_j^{-1}\left(E_j\right):j\in J\text{ and }E_j\in\mathcal E_j\right\},$$where $\pi_k:\prod_{j=1}^\infty X_j\to X_k$ is defined canonically. An element of $\mathcal F$ has the form $\prod_{j=1}^\infty E_j=E_1\times E_2\times\cdots,$ where $E_j\in\mathcal E_j$. Since $\pi_k^{-1}\left(E_k\right)=X_1\times X_2\times\cdots\times E_k\times\cdots\in\mathcal G$, it is the case that$$\prod_{j=1}^\infty E_j=E_1\times E_2\times\cdots=\bigcap_{j=1}^\infty\pi_j^{-1}\left(E_j\right)\in\mathcal M\left(\mathcal G\right).$$Hence, $\mathcal M\left(\mathcal F\right)\subseteq\mathcal M\left(\mathcal G\right).$ Conversely, an element of $\mathcal G$ has the form $\pi_k^{-1}\left(E_k\right)=X_1\times X_2\times\cdots\times E_k\times\cdots$, where $E_k\in\mathcal E_k$. Since $X_j\in\mathcal E_j$, it is the case that $\pi_k^{-1}\left(E_k\right)\in\mathcal F$. Hence, $\mathcal M\left(\mathcal G\right)\subseteq\mathcal M\left(\mathcal F\right)$. $\blacksquare$
Note that it is crucial that $\left\{X_j\right\}_{j=1}^\infty$ is a countable collection and that $X_j\in\mathcal E_j$ for $j=1,2,\dots$.
Later on, I will introduce the notion of a measure, define Borel $\sigma$-algebras, and do some computations on $\mathbb R$.
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2019-01-18 07:27:30
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https://math.stackexchange.com/questions/2088791/develop-fz-dfrac1z1z-into-a-laurent-series-for-the-region-0-z
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# Develop $f(z) = \dfrac{1}{z(1+z)}$ into a Laurent series for the region $0< |z| < 1$
Here is the solution to the problem (I need some clarification):
After partial fraction decomposition we have $\dfrac{1}{z(1+z)} = \dfrac{1}{z} - \dfrac{1}{1+z}$
But since we're only interested by the region $0< |z| < 1$, we can rewrite $\dfrac{1}{1+z} = \dfrac{1}{1-(-z)} = \sum_{k≥0} (-z)^k$
Therefore $f(z) = \dfrac{1}{z} - \sum_{k≥0} (-z)^k$ is the Laurent series.
The problem I have is that we defined a Laurent series as a series of the form $\sum_{k=-\infty}^{\infty}a_k (z-c)^k$ And $\dfrac{1}{z} - \sum_{k≥0} (-z)^k$ is clearly not of that form.
Yes, it is of that form. With $a_j = 0$ for $j < -1$, and $a_k = (-1)^{k+1}$ for $k \geq -1$. And, of course, $c = 0$.
• :-P Not going to waste my time posting the same as your answer, so $+1$. – Simply Beautiful Art Jan 8 '17 at 13:28
Note that $\frac{1}{z(z+1)}$ has a simple pole at $z=0$ so the Laurent's expansion has $a_{-n}=0$ for all $n\neq 1$, $n\in \mathbb N$
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2019-07-24 01:13:40
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