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Re: [ontolog-forum] Types of Formal (logical) Definitions inontology
To: andrei rodin "henson" Wed, 9 Jul 2014 19:31:23 -0500
```Indeed category theory is a mathematical formalism which is being used for ontology, even in cases where the users are unaware of it. The reasons are straight forward. A category is really a directed graph with a path composition operation and possibly other language constructions. Many applications in science and engineering make heavy use of directed graphs, including John’s bridge example in more ways than one. (01) Benzene is a good example test case for an ontology language. The idea is to give an axiom set in a logic for the class of benzene molecules. Then use automated reasoning answer questions about the structure of benzene molecules. For example, one would like to ask in the ontology language the question “whether any benzene molecule has a carbon ring? “ Automated reasoning can give the correct answer only when all of the benzene molecules in any valid interpretation have the structure such as is shown in the diagram : image: http://media1.picsearch.com/is?8Pu5G3VX2yKV0e-u-4aq4q12Q8DGv_UKZslfRR_RSLM&height=254 Reasoning on DL axiom sets cannot answer this question correctly, as the DL axioms will always have a tree structured model which does not conform to this picture. Remember, the diagram and the axiom set are description templates, rather than descriptions of a specific molecule. (02) However, one can give first order axioms for Benzene which constrain all the models to consist of molecules which only have the correct shape. This is not a trivial result. I repeat the comment that approach separates the logical language used to express the axioms from the logic’s term language which is the ontology language. The logical langauge is classical first order, the term language may or may not be classical (03) A category theory can be used as an ontology language by giving FOL axioms for the language constructions such as path composition. One needs a bit more ontology language constructions than a directed graph for this example, but not a lot. The axioms use two sorts, nodes and arrows. I will call them types and maps. A category theory person would call them objects and maps. Some of this was noted in some of the earlier exchanges with Alex. The axioms for composition are of course partial, as has been commented on before. But as the conditions for composition to be well defined are decidable, there really is no problem with proof theory or model theory. (04) The axioms for a category or directed graph with path composition can be found in the Lambek and Scott book in the early pages. Of course a category can be very large, but in engineering examples generally have finite models. For example the benzene description with the right constructions generates a category which has a canonical minimal finite model which looks like the graphs one sees in Wikipedia. (05) With regard to whether predicates are viewed as a form of classification, I would say yes. But there is an important caveat. There are two logics involved. One is the logic in which the axiom set is expressed which is called the external logic. The other is that in a category with additional structure such as a Cartesian close category or elementary topos, there is an internal logic. These frameworks have a type constant symbol Omega. An internal formula is a map q:X → Omega. If Omega reduces to {true,false} then the internal logic looks like the external one. However, one can add truth value constants in addition to true:X → Omega, and false:X → Omega such as maybe:X → Omega. Then one gets a non-traditional logic. Type theory, and rule systems such as Hilog, use this device of having an internal representation of higher order logic as terms within a first order presentation logic. The internal logic admits various generalization of classical logic. The external predicates classify the terms, i.e., syntactic structures. The internal predicates classify things in the interpretation. As an example of this idea, the external predicates allows one to deduce that “the gun shot hit the target = maybe”. The statement is an internal formulae and the equation is external. (06) I would modify John’s comment slightly. If you have to ask what it is from the viewpoint of ontology, the worst place to find out are standard text books on category theory. The reason for this is it is usually presented as a theory within set theory, rather than as a first order theory. Presenting topos theory as a theory within set theory obviates ability to serve as an ontology language. (07) By the way, the signature of the Benzene axiom set, as I would do it, has 3 type symbol constants, Benzene, Hydrogen and Carbon. It also has 12 part maps, and 12 type constructions, and 12 connection maps. I have not, of course, indicated what the axioms are which are needed for the benzene axiom set. (08) Henson Graves (09) -----Original Message----- From: John F Sowa Sent: Wednesday, July 09, 2014 12:01 PM To: ontolog-forum@xxxxxxxxxxxxxxxx Subject: Re: [ontolog-forum] Types of Formal (logical) Definitions inontology (010) Frank, Henson, and Ed, (011) FG > Given the discussions on Category Theory, I was curious if the > community views Predicates as a form of categorization/classification. (012) Two points: (013) 1. Category theory is a mathematical system that might be used for ontology. But if you have to ask what it is, don't use it. Brief summary: http://en.wikipedia.org/wiki/Category_theory (014) 2. Every class, type, or category in any ontology can be associated with a predicate that is true of everything in the class, type, or category and false of everything not in it. (015) HG > I believe the problem is that OWL axiom sets always have a tree > model whether you want it or not. OWL axiom sets may have other > models which are not trees. (016) EJB > Not quite 'trees' -- classification lattices, with 'property links' > that produce more or less general semantic networks, including cycles. (017) I agree with Henson. The OWL *classes* are organized in lattices. But the *models* (in Tarski's sense) are limited to trees. (018) For example, a benzene ring has a cycle. OWL cannot state the critical properties that require a model of benzene to have a cycle. (019) Just look at any bridge truss, and you'll find interconnected triangles that create huge numbers of cycles. It's impossible to find any complex artifact or biological system without cycles. OWL cannot specify, analyze, or reason about those cycles. (020) HG > Where they have problems is in constructing OWL axiom sets for > a molecule such as H2O which constrains the models to only consist > of molecules with exactly three atoms. There are a lot of papers > by Horrocks and Motik and students on this topic. (021) I know Horrocks and Motik. They're intelligent, but they're hopelessly *arrogant* to claim that they are qualified to tell experts in every branch of science, engineering, and business what models they're allowed to use, represent, and reason with. (022) HG > This has led them to try graph extensions of DL and logic > programming extensions. (023) Yes. And every one of those extensions adds more and more complexities to an overly complex system. By comparison, the LP systems are easier to learn and use and far more efficient for equivalent applications. (024) Programmers and computer scientists have had far better methods for controlling complexity and avoiding issues of undecidability. They use various terms: structured programming, structured analysis, design patterns, and knowledge compilers. (025) When I wrote my article "Fads and Fallacies in Logic", the primary fad I had in mind was OWL. And the primary fallacy was decidability as the bogeyman: http://www.jfsowa.com/pubs/fflogic.pdf (026) By the way, Jim Hendler was the editor of the journal in which it was published. And he liked that article very much. (027) John (028) _________________________________________________________________ Message Archives: http://ontolog.cim3.net/forum/ontolog-forum/ Config Subscr: http://ontolog.cim3.net/mailman/listinfo/ontolog-forum/ Unsubscribe: mailto:ontolog-forum-leave@xxxxxxxxxxxxxxxx Shared Files: http://ontolog.cim3.net/file/ Community Wiki: http://ontolog.cim3.net/wiki/ To join: http://ontolog.cim3.net/cgi-bin/wiki.pl?WikiHomePage#nid1J (029) _________________________________________________________________ Message Archives: http://ontolog.cim3.net/forum/ontolog-forum/ Config Subscr: http://ontolog.cim3.net/mailman/listinfo/ontolog-forum/ Unsubscribe: mailto:ontolog-forum-leave@xxxxxxxxxxxxxxxx Shared Files: http://ontolog.cim3.net/file/ Community Wiki: http://ontolog.cim3.net/wiki/ To join: http://ontolog.cim3.net/cgi-bin/wiki.pl?WikiHomePage#nid1J (030) ```
Current Thread Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, (continued) Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, henson Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, Alex Shkotin Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, John F Sowa Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, henson Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, Alex Shkotin Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, henson Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, Alex Shkotin Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, John F Sowa Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, henson Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, John F Sowa Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, henson <= Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, Alex Shkotin [ontolog-forum] Predicate Collections for Semantic Relationships, Frank Guerino Re: [ontolog-forum] Predicate Collections for Semantic Relationships, Ravi Sharma Re: [ontolog-forum] Predicate Collections for Semantic Relationships, John F Sowa Re: [ontolog-forum] Predicate Collections for Semantic Relationships, Frank Guerino Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, Barkmeyer, Edward J Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, Alex Shkotin Re: [ontolog-forum] Types of Formal (logical) Definitions inontology, henson [ontolog-forum] Semantic Markers in "THE STRUCTURE OF A SEMANTIC THEORY", 1963, John Bottoms | 2,482 | 10,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.912697 |
prodisi.wordpress.com | 1,591,251,368,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439019.86/warc/CC-MAIN-20200604032435-20200604062435-00530.warc.gz | 491,044,234 | 22,558 | ## Tentukanlah nilai akhir dari kode berikut:
A = TRUE
B= FALSE
C = TRUE
D = A AND B AND C
E = C OR A OR B
F = NOT (C AND E)
G = A IMP B EQV C
H = B XOR A XOR C AND C AND B
Print D, E, F, G, H | 83 | 203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-24 | latest | en | 0.287611 |
https://www.usefullinks.org/cat/Quadratic_forms.html | 1,718,467,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00335.warc.gz | 939,639,534 | 7,747 | Perfect lattice
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https://www.thespringstore.com/part-number-system-for-conical-springs-in-metric.html | 1,685,389,146,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00628.warc.gz | 1,149,743,328 | 42,744 | # Part Number System For Conical Compression Springs in Metric
## The Spring Store Part Number System For Conical Compression Springs in Metric Millimeter
Part Number
#### CC
Conical Compression Spring
Wire Diameter
#### -16256
Small Outer Diameter
#### -25400
Large Outer Diameter
Total Coils
Material Type
Free Length
End Type
Finish
Unit of Measure
### The Spring Store By Acxess Spring Part Number System In Metric For Conical Compression Springs
We have developed an easy stock spring part number system using the actual dimensions of the spring. This part number system for springs was developed to facilitate the creation and understanding of part numbers for springs. By using the actual dimensions of the spring in the spring part number we have made our part numbers for springs simple, but yet descriptive of the actual spring measurements in real time. Our part numbers consist of a series of numbers to describe each individual spring. Each part number is separated by hyphens to indicate different dimensional characteristics of a spring. Each series describes a different dimension. Below you will find a breakdown explanation of how The Spring Store part number system works for conical compression spring in millimeters.
• The letter P stands for Part Number
• The letter CC stands for Conical Compression Spring
• The first series of numbers indicates Wire Diameter=WD
• The second series, following the first hyphen, indicates Small Outer Diameter=SOD
• The second series, following the first hyphen, indicates Large Outer Diameter=LOD
• The third series, following the second hyphen, indicates Total Coils=TC
• The fourth series, following the third hyphen, indicates Material Type
• The fifth series, following the fourth hyphen, indicates Free Length=FL
• The sixth series, following the fifth hyphen, indicates End Type
• The seventh series, following the sixth hyphen, indicates Finish
• The eighth series, following the last hyphen, indicates Unit of Measure Millimeters=MM
Please look at the part number example below for a further explanation.
#### PCC2413-16256-25400-7000-MW-38100-CG-N-MM
• P=Part Number
• CC=Conical Compression Spring
• 2413=2.413 Wire Diameter=WD
• 16256=16.256 Small Outer Diameter=SOD
• 25400=25.400 Large Outer Diameter=LOD
• 7000=7.000 Total Coils=TC
• MW=Material Type
• 38100=38.100 Free Length=FL
• CG=End Type
• N=Finish
• MM=Unit Of Measure Millimeters
#### Nomenclature:
Material Type: BC=Beryllium Copper, HD=Hard Drawn, MW=Music Wire, OT=Oil Tempered, PB=Phosphor Bronze, SST=Stainless Steel
End Type: C=Closed Ends, CG=Closed and Ground Ends, O=Open Ends
FinishBO=Black Oxide, GI=Gold Irridite, N=None, Z=Zinc
Unit of Measure: Millimeter=MM | 650 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | longest | en | 0.820433 |
http://bnash.wikidot.com/symplectic-integrator | 1,534,461,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211316.39/warc/CC-MAIN-20180816230727-20180817010727-00399.warc.gz | 60,654,775 | 7,573 | Symplectic Integrator
If we use position along a curve as our our coordinate, the Hamiltonian is given by p_z.
(1)
\begin{align} H=1+\delta -(1+hx)\frac{A_s}{B\rho}-(1+hx)\sqrt{(1+\delta)^2-p_x^2-p_y^2} \end{align}
Expanding and splitting, we get $H=H_1+H_2$
(2)
\begin{align} H_1 = (1+hx)\frac{p_x^2+p_y^2}{2(1+\delta)} \end{align}
(3)
\begin{align} H_2 = -(1+hx)\frac{A_s}{B\rho}-(1+\delta)hx \end{align}
# 4th order Symplectic integrator
Now, suppose we can solve H_1 and H_2 independently.
Define
(4)
\begin{align} d_1 = d_4 = \frac{1}{2-2^{1/3}} L \end{align}
(5)
\begin{align} d_2 = d_3 = \frac{1-2^{1/3}}{2(2-2^{1/3})} L \end{align}
(6)
\begin{align} k_1 =k_3 = \frac{1}{2-2^{1/3}} L \end{align}
(7)
\begin{align} k_2 = -\frac{2^{1/3}}{2-2^{1/3}} L \end{align}
Now, suppose we can solve $H_1$ (drift) and $H_2$ (kick) independently, and we notate
(8)
$$e^{:H_1 d:} = D(d)$$
(9)
$$e^{:H_2:k} = K(k)$$
Then the 4th order integrator is
(10)
$$D(d_1) K(k_1) D(d_2) K(k_2) D(d_2) K(k_1) D(d_1)$$
Idea for a fast symplectic integrator
we want to define a grid of initial conditions, z1 … zn. We then define the one turn map via the mapped valus of these initial conditions: ringpass(ring,[z1 z2 … zn]) | 529 | 1,215 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 7, "equation": 3, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-34 | longest | en | 0.646802 |
http://stackoverflow.com/questions/9044834/an-algorithm-on-mathematica-to-calculate-the-determinant-of-a-nn-matrix | 1,405,207,805,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776435471.2/warc/CC-MAIN-20140707234035-00054-ip-10-180-212-248.ec2.internal.warc.gz | 133,482,760 | 15,946 | # An algorithm on mathematica to calculate the determinant of a n*n matrix:
I am working on an algorithm which calculates the determinant of any n*n matrix, here is my code:
`````` Laplace[matrix_List] := Module[{a = matrix, newmatrix, result = 0},
If [Length[a] == 1, result = Total[Total[a]],
For [i = 1, i <= Length[a], i++,
newmatrix = Drop[a, {i}, {1}];
result = result + (-1)^(i + 1) *
Total[Total[Take[a, {i}, {1}]]]*
Laplace[newmatrix];
]
]; result]
``````
It works recursively, it works for a 2*2 matrix(I have checked with Det[]), but it doesn't work for any matrix of higher degree than 2!
I would like to solve this solution myself - I want to implement this myself, rather than simply using `Det` - but I would appreciate it if someone could explain what is wrong with the recursion here?
-
Localize 'i'. Else it messes up because it changes in recursive calls. Also it does not bode well for a matrix such as {{i, j}, {k, l}}. Could also try this variant: Laplace[mat : {{a_}} /; MatrixQ[mat]] := a Laplace[mat_?MatrixQ] /; Length[mat] == Length[mat[[1]]] := Laplace[mat] = Sum[(-1)^j*mat[[j, 1]]*Laplace[Drop[mat, {j}, {1}]], {j, Length[mat]}] – Daniel Lichtblau Jan 28 '12 at 21:19
Take a look at this answer : stackoverflow.com/questions/8507654/… – Artes Jan 29 '12 at 13:00
Why not use `Det[]`? – Jack Maney Jun 12 '12 at 20:40
@Jack there's a deleted answer on which the OP comments that they are deliberately trying to implement the determinant themselves, rather than use `Det`. I'll update the question to include. – AakashM Jun 12 '12 at 21:53
@John: You may wish to use something other than `Laplace` for your function name. That's the name of a built in symbol. Standard practice is to typically use lower case names. – Mike Bantegui Jun 12 '12 at 22:01 | 542 | 1,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2014-23 | latest | en | 0.848472 |
https://www.businessinsider.com.au/trump-abolish-alternative-minimum-tax-2005-return-2017-3 | 1,606,243,879,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176922.14/warc/CC-MAIN-20201124170142-20201124200142-00605.warc.gz | 601,746,889 | 18,298 | The tax Trump paid in 2005 is the tax Trump wants to abolish
President Donald Trump wants to abolish the “Alternative Minimum Tax” (AMT) that, according to a copy of his federal tax return, cost him \$US31 million in 2005.
That \$US31 million was a majority of all the \$US38.5 million in taxes he paid on his \$US153 million in total income for that year. The White House confirmed part of the president’s 2005 tax returns ahead of a report from MSNBC host Rachel Maddow.
Trump said he wanted to abolish the AMT in 2016 when he was on the campaign trail:
“All … Americans will get a simpler tax code with four brackets — 0%, 10%, 20% and 25% — instead of the current seven. This new tax code eliminates the marriage penalty and the Alternative Minimum Tax while providing the lowest tax rate since before World War II.”
If the AMT didn’t exist, Trump would have paid only a fraction of his 2005 bill. Trump’s return shows that alongside the AMT adjustment the other taxes he paid total only about \$US7 million, or 4.5% of his \$US153 million in total income. As it is, Trump paid an effective tax rate of about 25% on his income.
Here is a basic summary of Trump’s income and tax data:
• Total income: \$US153 million
• Writedown on losses: -\$US103 million
• Income after writedown: \$US50 million
• AMT: -\$US31 million (20% rate)
• Regular income tax: -\$US5.3 million (4.5%)
• Other taxes: -\$US1.9 million (1%)
• Total taxes Trump paid in 2005: -\$US38.5 million (25%)
• Total taxes Trump would have paid absent the AMT: \$US7 million
• *Numbers may not add due to rounding and various tax provisions
Under Trump’s proposal, he would have paid about 5.5% in tax rather than the 25% he actually paid. That would have lowered his tax rate to below that for people who earn less than \$US100,000, according to the New York Times. It would also have saved him about \$US31 million.
You can debate the numbers because the tax rules generate them according to the structure in place in 2005, not the structure that would be in place if the AMT were abolished, which might change how Trump’s income and taxes would be calculated. Despite that, the numbers do give a rough guide to how much Trump would personally save if his AMT proposal were enacted.
The intent of the AMT is to ensure that wealthy and self-employed Americans pay tax rates that are comparable to those paid by ordinary workers on a salary or a simple wage. The AMT was introduced in 1970 to solve the problem of wealthy people paying lower tax rates that those poorer than them.
The US tax code contains many provisions allowing wealthy and self-employed people to take “deductions” — business expenses they paid throughout the year — against the taxes they owe. These deductions often lead to a situation in which a wealthy person with multiple streams of income might pay a lower tax rate, or even less tax overall, than someone on an average annual wage.
The intent of the AMT is that if your income is above a certain level (\$US109,475 in 2005) you pay a flat minimum tax rate regardless of the deductions you’re trying to write off.
Trump’s proposal to abolish the AMT would replace it with a simpler set of bands in which the very rich pay a top tax rate of 25%. Trump’s campaign said many deductions would be abolished to pay for it:
“Reducing or eliminating most deductions and loopholes available to the very rich.”
But the rich would keep some valuable (but unspecified) deductions:
“Those within the 25% bracket will keep fewer deductions. Charitable giving and mortgage interest deductions will remain unchanged for all taxpayers.”
NOW WATCH: A body-language expert analyses Trump’s unique handshakes
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You are here: Home » Content » Mathematics Grade 6 » To recognise the place values of digits
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# To recognise the place values of digits
Module by: Siyavula Uploaders. E-mail the author
## Memorandum
HD TD D H T E 1.1 6 0 9 8 0 1.2 5 0 8 1 3 6 1.3 5 0 4 6 0 0 1.4 8 2 6 0 4 0 1.5 4 0 7 0 0 5
3.1 576 826 = 500 000 + 70 000 + 6 000 + 800 + 20 + 6
3.2 894 392 = 800 000 + 90 000 + 4 000 + 300 + 90 + 2
## LEANER SECTION
### To use a series of techniques to perform calculations [LO 1.10.3]
1. Let’s see if you are able to apply your knowledge about hundred thousands. Write the following numbers correctly into the table:
HT TT T H T U 1.1 60 980 ___ ___ ___ ___ ___ ___ 1.2 508 136 ___ ___ ___ ___ ___ ___ 1.3 Five hundred and four thousand six hundred ___ ___ ___ ___ ___ ___ 1.4 Eight hundred and twenty six thousand and forty ___ ___ ___ ___ ___ ___ 1.5 Four hundred and seven thousand and five ___ ___ ___ ___ ___ ___
2. If you are able to “read” a number line correctly, you will be able to determine exactly where any number fits in. Work with a partner and use arrows to indicate the approximate positions of the following numbers in the number line.
2.1 318 500
2.2 347 899
2.3 372 000
2.4 398 799
2.5 403 000
3. In Activity 1.3 you have had the opportunity to write numbers in expanded notation.
See if you can also do it correctly with the following numbers:
e.g. 26 113 = 20 000 + 6 000 + 100 + 10 + 3
3.1 576 826 =
3.2 894 392 =
#### DO YOU STILL REMEMBER?
= means it is equal to
¹ means it is not equal to
< means it is smaller than
Ø means it is bigger than
## Assessment
Learning Outcome 1: The learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills.
Assessment Standard 1.4: We know this when the learner recognises the place value of digits in:
1.4.1 whole numbers to at least 9-digit numbers
Learning Outcome 1: The learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills.
Assessment Standard 1.10: We know this when the learner uses a range of techniques to perform written and mental calculations with whole numbers including:
1.10.3: building up and breaking down numbers
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| External bookmarks | 1,457 | 5,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2014-23 | latest | en | 0.790852 |
http://hipem.com.br/black-cane-ogdkxx/absolute-value-function-graph-4d5b38 | 1,719,250,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00312.warc.gz | 13,893,531 | 6,275 | To gain access to our editable content Join the Algebra 2 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. Here is the graph of f (x) = | x|: f (x) = | x| The graph looks like a "V", with its vertex at (0 Given an absolute value function, the student will analyze the effect on the graph when f(x) is replaced by af(x), f(bx), f(x – c), and f(x) + d for specific positive and negative real values. Next lesson. No, they do not always intersect the horizontal axis. The process of graphing absolute value equations can be reduced to some specific steps which help develop any kind of absolute value graph. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. View Graphing_Absolute_Value_Functions_1-12.pdf from MATH 5310 at East Tennessee State University. Which absolute value function has a graph that is wider than the parent function, f(x) = |x|, and is translated to the right 2 units? No, they do not always intersect the horizontal axis. Piecewise functions. Practice: Graph absolute value functions. Type in any equation to get the solution, steps and graph This website uses cookies to … Absolute Value Functions and Graphs - Word Docs & PowerPoints. y - k = |x - h| Step 3 : To get the vertex of the absolute value function above, equate (x - h) and (y - k) to zero, The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. Scaling & reflecting absolute value functions: graph. Graphing the Absolute Value Function The graph of the absolute value function f (x) = | x| is similar to the graph of f (x) = x except that the "negative" half of the graph is reflected over the x-axis. Free absolute value equation calculator - solve absolute value equations with all the steps. Graphing absolute value functions. The absolute value function is commonly used to measure distances between points. The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. This is the currently selected item. (-∞, ∞) (-∞, 0) (-6, 0) (0, ∞) Absolute value graphs review. Algebra 1 Name_ ID: 1 Absolute Value Functions Date_ Period_ Identify the … When we look at the above graph, clearly the vertex is (0, 0) Step 2 : Write the given absolute value function as . The graph of an absolute value function will intersect the vertical axis when the input is zero. 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Input is zero of absolute value graph Functions and Graphs - Word Docs & PowerPoints MATH 5310 at Tennessee... | 2,424 | 12,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-26 | latest | en | 0.875813 |
https://ibrahimhasnat.com/leetcode-226-invert-binary-tree-python-solution/ | 1,722,980,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00853.warc.gz | 256,565,766 | 22,425 | # LeetCode 226 | Invert Binary Tree | Python | Solution
Invert Binary Tree is one of the most popular programming problems. It’s an easy problem. I assume that you have a basic understanding of Binary Tree and Recursion.
LeetCode 226
The problem description is straightforward. Simply put, we have to invert a binary tree and then return the root.
More precisely, to solve this, we need to swap the left subtree with the right subtree of a node. And we have to do this swapping in every step. We will solve this problem recursively. The algorithm we will use is DFS.
Let’s see the solution in Python.
```# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if not root:
return None
left = self.invertTree(root.left)
right = self.invertTree(root.right)
root.left = right
root.right = left
return root```
The time complexity of this problem is O(n), where n is the number of nodes. And space complexity is also O(n). This n is the number stack size in the memory for recursive function calls.
Scroll to Top | 287 | 1,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-33 | latest | en | 0.783825 |
http://mclements.net/blogWP/index.php/2019/02/17/fractional-octaves/ | 1,627,731,150,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00499.warc.gz | 32,467,033 | 8,105 | # Fractional Octaves
I’ve been working with parametric EQ settings lately; here’s a quick cheat sheet.
Overview
We perceive the frequencies of sounds logarithmically. Each doubling of frequency is an octave. Thus, the difference between 40 and 80 Hz sounds the same as the difference between 4000 and 8000 Hz. Even though the latter difference is 10 times greater, it sounds the same to us. This gives a range of audible frequencies between 9 to 10 octaves, which is much wider than the range of frequencies of light that we can see.
Ratios
Two frequencies 1 octave apart have a frequency ratio of 2:1; one has twice the frequency of the other. A half octave is halfway between them on a logarithmic scale. That is, some ratio R such that f1 * R * R = f2. Since f2 = 2 * f1, R is the square root of 2, or about 1.414. Sanity check: 40 * 1.414 = 56.6, and 56.6 * 1.414 = 80. Thus 56.6 Hz is a half-octave above 40, and a half-octave below 80. Even though 60 Hz is the arithmetic half-way point between 40 and 80 Hz, to our ears 56.6 sounds like the half-way point between them.
More generally, the ratio for the fractional octave 1/N, is 2^(1/N). Above, N=2 so the half-octave ratio is 1.414. If N=3 we have 1/3 octave ratio which is 2^(1/3) = 1.260. Here is a sequence taken to 4 significant figures:
• 1 octave = 2.000
• 3/4 octave = 1.682
• 1/2 octave = 1.414
• 1/3 octave = 1.260
• 1/4 octave = 1.189
• 1/5 octave = 1.149
• 1/6 octave = 1.122
• 1/7 octave = 1.104
• 1/8 octave = 1.091
• 1/9 octave = 1.080
• 1/10 octave = 1.072
• 1/11 octave = 1.065
• 1/12 octave = 1.059
The last is special because in western music there are 12 notes in an octave. With equal temperament tuning, every note has equally spaced frequency ratios. Thus the frequency ratio between any 2 notes is the 12th root of 2, which is 1.059:1. Every note is about 5.9% higher in frequency than the prior note.
Bandwidth with Q
Another way to express the frequency range or bandwidth of a parametric filter is Q. Narrow filters have big Q values, wide filters have small Q values. A filter 2 octaves wide (1 octave on each side of the center frequency) has Q = 2/3 = 0.667.
For a total bandwidth of N octaves (N/2 on each side of center frequency), the formula is:
`Q = sqrt(2^N) / (2^N - 1)`
Here are some example values. You can check them by plugging into the formula.
• N=2, Q=0.667
• N=1.5, Q=0.920
• N=1, Q=1.414
• N=2/3, Q=2.145
• N=1/2, Q=2.871
Note that these N octave fractions are total width, which is twice the above table which shows octave on each side of the center frequency.
Gotchas
Whatever tool you’re using for this, make sure you know whether it expects total bandwidth around the center frequency, or bandwidth on each side. And make sure you know whether it expects frequency ranges as raw ratios, fractions of an octave, or Q.
Real-World Correction
The above formula comes straight from any textbook. But these Q factors may give wider ranges than expected, due to an assumption it makes. This assumption is that the range of the filter is where the peak amplitude (at its center) drops to half its value. So the filter is still taking effect at these edges. If you want the filter to taper to zero at the edges, you need to use a bigger Q value to get a narrower filter. Roughly speaking, this means multiply the Q value by 2.0.
For example consider a filter that is -4 dB at 3,000 Hz, 3/4 octave wide on each side. That is a ratio of 1.682:1, so this filter tapers to zero at 3,000 / 1.682 = 1,784 and 3,000 * 1.682 = 5,045 Hz. Total width is 1.5 octaves (5,045 / 1,784 = 2.83 = 2^1.5). The above formula says this is Q=0.92. But that will be a wider filter. It will reduce to half (roughly +2 dB) at 1,784 and 5,045 Hz. If you want it to taper to zero at these edged then use Q = 0.92 * 2.0 = 1.84.
Note: this is an approximate / rough guide.
Example
Suppose you are analyzing frequency response and see a peak between frequencies f1 and f2. You want to apply a parametric EQ at the center point that tapers to zero by f1 and f2.
First, find the logarithmic midpoint. Compute the ratio f2 / f1 and take its square root to get R. Multiple f1 by R, or divide f2 by R and you’ll have the logarithmic midpoint.
For example if f1 is 600 Hz and f2 is 1700 Hz, the ratio is 2.83:1, so R = sqrt(2.83) = 1.683. Double check our work: 600 * 1.683 = 1010 and 1010 * 1.683 = 1699. Close enough.
So 1,010 Hz is the logarithmic midpoint between 600 and 1700 Hz. We center our frequency here and we want it to taper to zero by 600, and 1700. That range is a ratio of 1.683 on each side, which in the above list is 3/4 octave, or Q=0.920. Multiply Q by 2.0 to get Q=1.84 since you want this filter to have no effect (taper to zero) at these 2 endpoint frequencies. So now we know the center frequency and width of our parametric EQ. | 1,468 | 4,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-31 | latest | en | 0.919771 |
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A239426 21*n^4 - 36*n^3 + 25*n^2 - 8*n + 1. 4
1, 3, 133, 931, 3441, 9211, 20293, 39243, 69121, 113491, 176421, 262483, 376753, 524811, 712741, 947131, 1235073, 1584163, 2002501, 2498691, 3081841, 3761563, 4547973, 5451691, 6483841, 7656051, 8980453, 10469683, 12136881, 13995691, 16060261, 18345243 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS For n > 0: a(n) = A219069(2*n-1,n). LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1). FORMULA a(n) = (1-3*n+3*n^2) * (1-5*n+7*n^2) = A003215(n-1) * A239449(n). G.f.: ( -1+2*x-128*x^2-286*x^3-91*x^4 ) / (x-1)^5 . - R. J. Mathar, Mar 31 2014 MATHEMATICA Table[(21 n^4 - 36 n^3 + 25 n^2 - 8 n + 1), {n, 0, 40}] (* Vincenzo Librandi, Mar 21 2014 *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 3, 133, 931, 3441}, 40] (* Harvey P. Dale, May 04 2016 *) PROG (Haskell) a239426 n = (((21 * n - 36) * n + 25) * n - 8) * n + 1 (Magma) [21*n^4-36*n^3+25*n^2-8*n+1: n in [0..31]]; // Vincenzo Librandi, Mar 21 2014 CROSSREFS Sequence in context: A213203 A199141 A152435 * A157086 A366306 A051376 Adjacent sequences: A239423 A239424 A239425 * A239427 A239428 A239429 KEYWORD nonn,easy AUTHOR Reinhard Zumkeller, Mar 19 2014 STATUS approved
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Last modified May 25 17:00 EDT 2024. Contains 372804 sequences. (Running on oeis4.) | 762 | 1,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.573677 |
http://www.learningaboutelectronics.com/Articles/How-to-build-a-metal-detector-circuit.php | 1,701,318,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00528.warc.gz | 75,081,756 | 5,026 | How to Build a Metal Detector Circuit
# How to Build a Metal Detector Circuit
In this project, we will demonstrate how to build a simple metal detector circuit.
The device we then build will function as a metal detector that can scout out metal objects, such as coins, nails, keys such as car keys you may not be able to find, and even gold if you're looking for in a beach (though this one may not have industrial strength). This metal detector can detect certain kinds of metal- especially iron-containing metals, which are called ferrous metals, even if under a half-inch of drywall or sand.
How this metal detector works is that it uses an IC that generates an AC signal that goes through a coil. Metal objects are objects which conduct electricity, so a current can be induced in these metal objects. When the coil in the metal detector comes near a metal object, the electromagnetic field in the coil induces currents in the metal object. The electromagnetic field generated by the metal changes the current in the coil. When the signal changes, the IC turns on an LED< alerting the user to the presence of a metal.
Components Needed
• TDA0161 Proximity Detector (IC1)
• 2 1KΩ Resistors (R1, R4)
• 10KΩ Potentiometer (R2)
• 330Ω Resistor (R3)
• 120Ω Resistor (R5)
• 2N3904 Transistor (Q1)
• 2 4.7nF ceramic capacitors (C1, C2)
• 680 picohenry bobbin-type Inductor
• Battery Holder for 4 AA batteries (6V)
• LED
• SPST Switch
### Metal Detector Circuit Schematic Diagram
Below is the complete electronic circuit for the metal detector that we are building:
• Inductor L1- The inductor L1 forms a parallel circuit with the capacitor C1 to form an LC parallel circuit. When a signal that oscillates at several KHz passes through this circuit, the signal creates an electric field around the coil. When you bring the coil near a metallic object, that electric field induces an oscillating signal in the object. So when the oscillating signal has been induced in the metallic object, the signal in the object creates an electric field that induces current in the coil. This current changes the oscillating signal running through the LC parallel circuit.
• TDA0161 Proximity Detector IC- This IC is a proximity detector. This IC suplies the oscillating signal that is sent through the LC parallel circuit. The IC also responds to any changes in the signal. The IC has an output of 1 milliamperes (mA) or less if the coil is far from a metallic object and an output of 0mA or higher if the coil is near a metallic object.
Thus, this IC is at the heart of this circuit. When the object is far from a metallic object, the current which the IC produces is insufficient to drive the LED. Thus, the LED does not turn on. When the coil is near a metallic object, the IC produces sufficient power to drive the LED and it turns on.
• Resistor R1 and Potentiometer R2- These resistors are used to calibrate the TDA0161 IC to the LC circuit. You calibrate it by adjusting the potentiometer to change the current output it creates in accordance with the proximity of metal to the coil. You can adjust so that it can detect metals at the distances which you want it to. By increasing the potentiometer resistance, the IC will create less current output. Therefore, a metal must be placed closer to the coil in order for the LED to light. By decreasing the potentiometer resistance, the IC produces less current output, so the metal doesn't have to be placed as close to the coil. It's up to you to set the adjustment.
• 2N3904 Transistor (Q1)- The 2N3904 transistor provides amplification, so that there is sufficient current to power the LED. Without this transistor, there would not be enough power to turn on the LED.
• LED- The LED in the circuit serves as an indicator to when there is a presence of a metal. When a metal is in close proximity to our electronic circuitry, the LED turns on. This shows we have found metal. When the LED is off (not lit), then our metal detector has not detected metal and indicates no metal is in close proximity.
• 6 volts- The 6 volts is the supply power to the entire circuit. This 6 volts is supplied through 4 AA batteries in series. Being that each battery supplies 1.5 volts, 4 AA batteries (1.5V * 4) provides 6V. This 6 volts gives power to every component in the circuit.
• SPST Switch- The SPST switch allows us to shut off power to the circuit, if we want the metal detector power shut off, just like any electronic device would have. This just serves asn an on/off switch.
• This circuit detects metal through the coil, L1. Once metal is placed near the inductor L1, it will trigger current production from the proximity detector IC, which in turn lights the LED. So to test this circuit, just place a metallic object near this inductor. When done, the LED should turn on. When the metallic object is moved away, the LED should shut off.
Some may like a metal detector that has an LED that lights up in the presence of a metal, but there are other alternatives you can do. It doesn't have to be an LED. It can be a buzzer that sounds off when metal is detected. It can be a piezo buzzer, a siren, an alarm, etc. Any alternative can be possible if power is allocated properly. So keep this in mind if you don't want an LED to light but a buzzer to sound.
Related Resources
How to Build an AC Fan Circuit | 1,242 | 5,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.880549 |
https://studylibid.com/doc/4280146/poster-id-69-icced | 1,679,385,828,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00398.warc.gz | 613,879,959 | 13,008 | # Poster ID 69 ICCED
```Regression Model Program Bill Of Quantity
Simple Home Construction
Departement Of Civil Engineerinng
Nusa Putra University
Abstract
- Every construction of home, made a Bill of Quantity is very important. Due to the basis for outlining the budget plan as well as the basis for calculating
human resource needs, money resources, material resources and other resources are depend on the Bill Of Quantity. But, without supported by the completed
data, the bill of quantity can’t be compiled. How to compile Bill of Quantity if the information data is known only building area without any other data.
The solution of the problem is “model”, by the existing model, the program can be formulate to compile the Bill of Quantity in the house construction project. To get the model and arrange the program, the research is done by using quantitative method, the research is also supported by primary data. That is by
doing the project of house construction directly. And secondary data by survey and interview to the construction contractor. Data is processed then do the regression analysis to produce the model as a formula to arrange the Bill of Quantity.
This research is finally yield a program to develop Bill of Quantity with building area as the data input. The Bill of Quantity is automatically arranged by
entering the variable of building area.
introduction
The house is a basic human need so that the construction of houses in big cities, towns and villages is
always done. Even the Indonesian government has set it in Act No. 1 of 2011 on housing and residential
areas. Each undertaking the house construction, needs to be prepared design drawings and resources
that include, material, human, money, machinery and other tools, as well as implementation of construction method, as well as implementation of construction method.
This research is intend to find the Bill of Quantity (BOQ) of house construction with data input only
building area, for example having a plan to build a house with 3 bedrooms, each of them is 9m 2, living
room is 12m2, kitchen is 9m2 and terrace is 6m2, so the total of building area is 74m2. So, how about the
BOQ? This study is aims to produce a simple method to find the BOQ with building area as data input
OBJECTIVE
This research uses quantitative method, to complete this
research then required data, the required data is Bill of Quantity in budget
source consist of project document and contract document. Then the data
to get the model. Furthermore the model into a Bill of Quantity formula
plan of home construction project, the data
is processed and done the regression analysis
compiled on the excel program.
Secondary data as many as 24, are obtained by interview and survey
to construction contractor. Secondary sampling data obtained from The Spring housing as many as 250 units taken
5 samples, from The View housing as many
as 250 units taken 5 samples, from Melati Loka housing as many as 275 units taken 5 sample, from Cibeureum Permai housing as many
as 350 units taken 5 samples and individual house taken 4 samples. The number of primary and secondary data is 38 samples.
To perform regression analysis,
firstly processed the data into measured variables.
It can be assumed that the dependent variable (Y)
is the volume and extent of the item of the building component,
expressed y1 (the volume of concrete structure), y2 (wall area),
y3 (number of sills, door),
y4 (floor area),
y5(sanitary number) ,
y6 (the number of ME),
y7 (foundation volume),
y8 (roof area),
y9 (paint area),
y10(ceiling area
while the independent variable (X) is the building area
CONCLUSION
By finding the regression model as a formula to identify the volume and extent of building components of
2
2
minimalist house with minimum building area of 21 m and maximal 290 m . Then by doing the 12 steps, Bill of
Quantity program of house development project has been arranged, just only enter the variable x (building area)
into the program, automatically the volume and extent of building components on Bill of Quantity of home construction project will be identified. The resulting model as the formula is:
Model y1 (struktur beton) = 3.927 + 0.171X.
Model y2(dinding) = 100.968 + 2.126X.
Model y3 (kusen pintu, jendela) = 10.518 + 0.090X.
Model y4 (lantai) = 0.449 + 1.074X.
Model y5 (sanitary) = 9.879 + 0.072X.
Model y6 (ME) = 12.863 + 0.323X.
Model y7 (fondasi) = 2.416 + 0.125X.
Model y8 (atap) = 37.322 + 0.463X.
Model y9 (pengecatan) = 126.447 + 5.334X.
Model y10(plafond) = 13.388 + 0.953X.
The validation model has been tested and applied to Bill of Quantity program with accurate results.
``` | 1,116 | 4,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-14 | latest | en | 0.936899 |
https://www.glocktalk.com/threads/basic-electrical-exam.117968/ | 1,511,185,249,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00727.warc.gz | 808,747,135 | 20,071 | close
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# Basic Electrical Exam
Discussion in 'The Lighter Side' started by lethal tupperwa, Nov 26, 2002.
1. ### lethal tupperwa
Messages:
9,495
1,660
Joined:
Aug 20, 2002
Location:
Virginia
When taking some special classes in prep for my electrical exam sample test
were given to help prepare us for the exam. The following was one of the
ones that was given to us and I thought that you might want to see how hard
these things really are. There is no wonder that when I took my exam only
17% of those taking the test passed and most of those 17% were retakes.
The exam was a 100 question, multiple choice, 8 hour, two part test.
Rick
BASIC ELECTRICAL EXAM
1. What color is a green ground screw? ________________
2. When hooking up a 200 volt heater, you must use 220 volt wire.
true____ false____
3. Electricity will leak out of pipes if they are not connected with rain
tight fittings.
true____ false____
4. To trip a breaker, you must stick your foot out at it walks by.
true____ false____
5. When dealing with conduit, the O.D. must exceed the I.D. or the hole
will be on the outside.
true____ false____
6. A keyless fixture cannot be unlocked.
true____ false____
7. A circuit breaker reads "20" on the handle. This means it can only trip
20 times before it is worn out.
true____ false____
8. If you plug something marked 110 volts into a 120 volt outlet, 10 volts
will leak out and make a mess.
true____ false____
9. The gauge of wire tells you how many plugs you can hook into it.
true____ false____
10. When pulling two 4/0 wires into a 1/2" PVC conduit the "PVC" stands for
"Pipe Very Crowded".
true____ false____
11. An OHM is a Hindu measurement of voltage.
true____ false____
12. A flush mounted device may only be hooked up to a toilet.
true____ false____
13. Electrical Inspectors are also known as _________________________.
14. Service entrance conductors purchased from an electrical parts house
are very dangerous to install.
true____ false____
15. High voltage wire is used in the upper levels of tall buildings, where
as low voltage wire is usually
found in basements or underground.
true____ false____
16. If you have a molded case circuit breaker, the mold can be washed off
with warm soapy water.
true____ false____
2. ### larry_minnSilver MemberMillennium Member
Messages:
12,550
3,545
Joined:
Dec 16, 1999
Location:
Minnesota
Number 13 is the one I have trouble with. I can think of a dozen answers and all are correct. Most I can't print here.
Messages:
15,908 | 660 | 2,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-47 | longest | en | 0.885583 |
https://assignmentgrade.com/question/639850/ | 1,611,619,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704792131.69/warc/CC-MAIN-20210125220722-20210126010722-00529.warc.gz | 220,911,452 | 6,515 | What are three things all circuits need? - AssignmentGrade.com
# What are three things all circuits need?
QUESTION POSTED AT 16/04/2020 - 06:27 PM
### Answered by admin AT 16/04/2020 - 06:27 PM
I would say an energy source, a load, and conductors.
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QUESTION POSTED AT 27/05/2020 - 11:25 PM
### What is meant when it is said that stars have a life cycle? Stars produce living things. Stars travel in circular orbits during their lives Stars are born, mature, and eventually die. Stars move about in their galaxies
QUESTION POSTED AT 26/05/2020 - 11:20 AM
### A circuit has a 9.0 v power supply and two resistors connected in parallel. one resistor is 10 , and the other is 15 . what is the current across the 15 resistor?
QUESTION POSTED AT 26/05/2020 - 07:28 AM
### An 8.0-v power supply delivers a 1.75-a current to a circuit. calculate the power provided to the circuit.
QUESTION POSTED AT 26/05/2020 - 07:19 AM
### The potential difference in a simple circuit is 12 V and the resistance is 23 Ω . What current flows in the circuit? Answer in units of A
QUESTION POSTED AT 26/05/2020 - 01:10 AM
### What factor determines the primary differences among living things? A)whether or not their bodies contain nucleic acids B)the type of complex carbohydrates in their diet C)the order of nucleotides in their DNA D)the variety of proteins in their diet
QUESTION POSTED AT 24/05/2020 - 05:52 PM
### When choosing your driving speed, the most important thing to consider is:?
QUESTION POSTED AT 23/05/2020 - 04:33 PM
### Living things all share which of the following characteristic?
QUESTION POSTED AT 11/05/2020 - 06:15 AM
### A 100.0 resistor, a 50.0 resistor, and a 40.0 resistor are connected in parallel and placed across a 12.0 V battery. ( a. What is the equivalent resistance of the parallel circuit? Incorrect: Your answer is incorrect. ( b. What is the current delivered by the battery? A ( c. What is the current through each branch of the circuit? A (through 100.0 resistor) A (through 50.0 resistor) A (through 40.0 resistor)
QUESTION POSTED AT 09/05/2020 - 12:17 AM
### According to Ohm's law, a circuit with a high resistance _____. - will have a low electric current - will have a high electric current - will have an equal electric current - will have an electric current but the relative amount cannot be determined
QUESTION POSTED AT 08/05/2020 - 07:37 PM
### The purpose of a battery in a circuit is to _____. - provide a conductive path - increase the potential of charges in the circuit - decrease the potential of charges in the circuit - add charges to the circuit
QUESTION POSTED AT 08/05/2020 - 07:28 PM
### The smallest level of organization in living things is the atom the cell the small tissue the nervous system
QUESTION POSTED AT 28/04/2020 - 09:21 AM
### What is the best term to describe information sent as patterns and codes in the controlled flow of electrons through a circuit? A. Electron circuit B. Digital signal C. Analog signal D. Integrated signal
QUESTION POSTED AT 28/04/2020 - 08:50 AM
### Select all that apply. Which of the following do not illustrate a complete circuit? A B C D
QUESTION POSTED AT 28/04/2020 - 08:41 AM
### Terrence connects several lightbulbs in a parallel circuit. Electricity flows through the circuit, and all the bulbs light up. What will happen if Terrence removes lightbulb number 2
QUESTION POSTED AT 28/04/2020 - 08:09 AM
### Locate the element calcium (Ca) on the periodic table and click on the square. Read about the properties of calcium. Why might calcium be important in the diet of many living things?
QUESTION POSTED AT 28/04/2020 - 06:58 AM
### In what type of circuit will the equivalent resistance be less than the resistors in the circuit?
QUESTION POSTED AT 28/04/2020 - 06:41 AM
### Power source that provides a current of 1.6 A to two 5 Ω resistors connected in series is moved to a parallel circuit that consists of three identical resistors. In the parallel circuit, the overall current is 2.0 A. The value of a resistor used in the parallel circuit is? A)5.3Ω B)8.0Ω C)12Ω D)24ΩIt's D :P
QUESTION POSTED AT 28/04/2020 - 05:12 AM
### What is a path through which electric charges travel? A) electric circuit B) resistance C) voltage D) electric current
QUESTION POSTED AT 28/04/2020 - 05:08 AM
### Which item(s) would be sufficient to make a circuit?
QUESTION POSTED AT 28/04/2020 - 03:44 AM | 1,733 | 6,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-04 | latest | en | 0.872154 |
https://answers.yahoo.com/question/index?qid=20190523224803AABi1p3 | 1,560,854,366,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998716.67/warc/CC-MAIN-20190618103358-20190618125358-00072.warc.gz | 348,543,494 | 23,470 | # HELP WITH VECTORS !!!?
Nikki is driving a motorboat across a river that flows at 5 m/s. The motorboat has a speed of 12 m/s in still water. She heads out from one bank in a direction perpendicular to the current. How far does she travel in 3 minutes. | 66 | 253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-26 | longest | en | 0.956914 |
https://www.lmfdb.org/EllipticCurve/6.6.371293.1/79.3/d/7 | 1,713,487,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00334.warc.gz | 765,914,510 | 121,776 | Properties
Label 6.6.371293.1-79.3-d7 Base field $$\Q(\zeta_{13})^+$$ Conductor norm $$79$$ CM no Base change no Q-curve no Torsion order $$2$$ Rank $$0$$
Related objects
Show commands: Magma / PariGP / SageMath
Base field$$\Q(\zeta_{13})^+$$
Generator $$a$$, with minimal polynomial $$x^{6} - x^{5} - 5 x^{4} + 4 x^{3} + 6 x^{2} - 3 x - 1$$; class number $$1$$.
sage: R.<x> = PolynomialRing(QQ); K.<a> = NumberField(R([-1, -3, 6, 4, -5, -1, 1]))
gp: K = nfinit(Polrev([-1, -3, 6, 4, -5, -1, 1]));
magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![-1, -3, 6, 4, -5, -1, 1]);
Weierstrass equation
$${y}^2+\left(a^{3}-2a+1\right){x}{y}+\left(a^{4}+a^{3}-3a^{2}-3a+1\right){y}={x}^{3}+\left(-a^{4}+a^{3}+4a^{2}-4a-3\right){x}^{2}+\left(-56a^{5}+144a^{4}+90a^{3}-446a^{2}+276a-42\right){x}-1161a^{5}+2386a^{4}+3200a^{3}-7763a^{2}+1278a+1453$$
sage: E = EllipticCurve([K([1,-2,0,1,0,0]),K([-3,-4,4,1,-1,0]),K([1,-3,-3,1,1,0]),K([-42,276,-446,90,144,-56]),K([1453,1278,-7763,3200,2386,-1161])])
gp: E = ellinit([Polrev([1,-2,0,1,0,0]),Polrev([-3,-4,4,1,-1,0]),Polrev([1,-3,-3,1,1,0]),Polrev([-42,276,-446,90,144,-56]),Polrev([1453,1278,-7763,3200,2386,-1161])], K);
magma: E := EllipticCurve([K![1,-2,0,1,0,0],K![-3,-4,4,1,-1,0],K![1,-3,-3,1,1,0],K![-42,276,-446,90,144,-56],K![1453,1278,-7763,3200,2386,-1161]]);
This is a global minimal model.
sage: E.is_global_minimal_model()
Invariants
Conductor: $$(a^5-4a^3+3a-2)$$ = $$(a^5-4a^3+3a-2)$$ sage: E.conductor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor norm: $$79$$ = $$79$$ sage: E.conductor().norm() gp: idealnorm(ellglobalred(E)[1]) magma: Norm(Conductor(E)); Discriminant: $$(-a^5+4a^3-3a+2)$$ = $$(a^5-4a^3+3a-2)$$ sage: E.discriminant() gp: E.disc magma: Discriminant(E); Discriminant norm: $$-79$$ = $$-79$$ sage: E.discriminant().norm() gp: norm(E.disc) magma: Norm(Discriminant(E)); j-invariant: $$-\frac{41494566893660714555288080}{79} a^{5} + \frac{51497801567134241086634054}{79} a^{4} + \frac{195058086408065742590919756}{79} a^{3} - \frac{213001575932755291713651877}{79} a^{2} - \frac{197618395730591263439326644}{79} a + \frac{172124231458098209062968630}{79}$$ sage: E.j_invariant() gp: E.j magma: jInvariant(E); Endomorphism ring: $$\Z$$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) sage: E.has_cm(), E.cm_discriminant() magma: HasComplexMultiplication(E); Sato-Tate group: $\mathrm{SU}(2)$
Mordell-Weil group
Rank: $$0$$ Torsion structure: $$\Z/2\Z$$ sage: T = E.torsion_subgroup(); T.invariants() gp: T = elltors(E); T[2] magma: T,piT := TorsionSubgroup(E); Invariants(T); Torsion generator: $\left(-\frac{17}{4} a^{5} + \frac{19}{4} a^{4} + \frac{35}{2} a^{3} - \frac{35}{2} a^{2} - \frac{31}{4} a + \frac{13}{2} : \frac{13}{4} a^{5} - \frac{51}{8} a^{4} - \frac{93}{8} a^{3} + \frac{157}{8} a^{2} + \frac{15}{4} a - \frac{51}{8} : 1\right)$ sage: T.gens() gp: T[3] magma: [piT(P) : P in Generators(T)];
BSD invariants
Analytic rank: $$0$$ sage: E.rank() magma: Rank(E); Mordell-Weil rank: $$0$$ Regulator: $$1$$ Period: $$202.53939857027286036588199073950547562$$ Tamagawa product: $$1$$ Torsion order: $$2$$ Leading coefficient: $$1.32957$$ Analytic order of Ш: $$16$$ (rounded)
Local data at primes of bad reduction
sage: E.local_data()
magma: LocalInformation(E);
prime Norm Tamagawa number Kodaira symbol Reduction type Root number ord($$\mathfrak{N}$$) ord($$\mathfrak{D}$$) ord$$(j)_{-}$$
$$(a^5-4a^3+3a-2)$$ $$79$$ $$1$$ $$I_{1}$$ Non-split multiplicative $$1$$ $$1$$ $$1$$ $$1$$
Galois Representations
The mod $$p$$ Galois Representation has maximal image for all primes $$p < 1000$$ except those listed.
prime Image of Galois Representation
$$2$$ 2B
$$3$$ 3B
Isogenies and isogeny class
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2, 3, 4, 6 and 12.
Its isogeny class 79.3-d consists of curves linked by isogenies of degrees dividing 12.
Base change
This elliptic curve is not a $$\Q$$-curve.
It is not the base change of an elliptic curve defined over any subfield. | 1,715 | 4,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.32726 |
https://la.mathworks.com/matlabcentral/cody/problems/1793-02-vector-variables-3/solutions/2120537 | 1,606,294,825,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00456.warc.gz | 389,415,308 | 17,013 | Cody
# Problem 1793. 02 - Vector Variables 3
Solution 2120537
Submitted on 7 Feb 2020 by Takeyoshi Terui
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
cVec = 5:-.2:-5; c = VectorFunc(); assert(isequal(c,cVec))
cVec = Columns 1 through 17 5.0000 4.8000 4.6000 4.4000 4.2000 4.0000 3.8000 3.6000 3.4000 3.2000 3.0000 2.8000 2.6000 2.4000 2.2000 2.0000 1.8000 Columns 18 through 34 1.6000 1.4000 1.2000 1.0000 0.8000 0.6000 0.4000 0.2000 0 -0.2000 -0.4000 -0.6000 -0.8000 -1.0000 -1.2000 -1.4000 -1.6000 Columns 35 through 51 -1.8000 -2.0000 -2.2000 -2.4000 -2.6000 -2.8000 -3.0000 -3.2000 -3.4000 -3.6000 -3.8000 -4.0000 -4.2000 -4.4000 -4.6000 -4.8000 -5.0000
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 398 | 912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.379755 |
http://www.forexsure.net/advanced-forex/trading-strategies/185-the-average-true-range.html | 1,579,591,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601615.66/warc/CC-MAIN-20200121044233-20200121073233-00341.warc.gz | 221,504,739 | 6,998 | # The Average True Range
Developed by J. Welles Wilder, the Average True Range (ATR) measures volatility of the forex market. Originally Wilder derived the True Range index for commodities and daily prices. Later Wilder realized that a simple range calculation was not always efficient as applied to market volatility trends. Consequently, he smoothed the True Range using the moving average and obtained the Average True Range (ATR). In other words, ATR is the moving average of the TR for a prescribed time period (usually 2 weeks).
Recall that the TR is given by TR = H – L , for the “normal” days, TR = H – Cl, for days that open with an upward gap, TR = Cl – L, for days which opened with a downward gap, where H is the today's high, L the today's low and Cl the yesterday's close. The Figure below illustrates how the ATR works. During more volatile markets ATR moves up whereas for less volatile market ATR moves down. When the candlestick price bars are short the Forex traders see the ATR indicator moving lower. If price bars begin to grow the ATR indicator line will rise.
Let us show how to avoid the whipsaws. Recall that the whipsaw is a point in the market where the price moves sharply up or down. In the early stages the whipsaw looks similar to the start of a new trend. However, instead of continuing the trend or leveling off it will suddenly dive back down or up to a price close to where it started.
Suppose that the trader has a breakout system that tells where to enter. Consider a breakout system that triggers an entry buy order once market breaks above its previous day high. Let’s say this high was at 1.3000 for EURUSD. Without any filters we would buy at 1.3002, but are we risking to be whipsawed.
With ATR filter the forex trader measures the ATR first. Suppose that he has found that for EURUSD the 14 day ATR stands at 110 pips. He may choose to enter at breakout + 20% ATR (110 x 20% = 22 pips). Therefore, instead of rushing in on a breakout and risking to be whipsawed, he enters at 1.3000 + 22 pips = 1.3022. Finally, although the trader gave up some initial pips on a breakout, an additional measure has been taken to avoid being whipsawed.
Top Forex Brokers Reviews here | 527 | 2,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-05 | latest | en | 0.936482 |
http://mywallpapers.mobi/mv-kini-lawyer-determinant-of-a-matrix-math-is-fun/ | 1,558,525,128,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256797.20/warc/CC-MAIN-20190522103253-20190522125253-00035.warc.gz | 145,095,370 | 27,751 | mv kini lawyer Determinant of a Matrix – Math is Fun - MywallpapersMobi
# mv kini lawyer Determinant of a Matrix – Math is Fun
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Math Algebra (all content) Matrices Determinants & inverses of large matrices
## Determinants & inverses of large matrices
• Determinant of a 3×3 matrix: standard method (1 of 2)
This is the currently selected item.
• Determinant of a 3×3 matrix: shortcut method (2 of 2)
• Practice: Determinant of a 3×3 matrix
• Inverting a 3×3 matrix using Gaussian elimination
• Inverting a 3×3 matrix using determinants Part 1: Matrix of minors and cofactor matrix
• Inverting a 3×3 matrix using determinants Part 2: Adjugate matrix
• Practice: Inverse of a 3×3 matrix
Next tutorial
Solving equations with inverse matrices
Math Algebra (all content) Matrices Determinants & inverses of large matrices
# Determinant of a 3×3 matrix: standard method (1 of 2)
## Determinants & inverses of large matrices
• Determinant of a 3×3 matrix: standard method (1 of 2)
This is the currently selected item.
• Determinant of a 3×3 matrix: shortcut method (2 of 2)
• Practice: Determinant of a 3×3 matrix
• Inverting a 3×3 matrix using Gaussian elimination
• Inverting a 3×3 matrix using determinants Part 1: Matrix of minors and cofactor matrix
• Inverting a 3×3 matrix using determinants Part 2: Adjugate matrix
• Practice: Inverse of a 3×3 matrix
Next tutorial
Solving equations with inverse matrices
Tags
Determinant of a matrix
## Video transcript
As a hint, I will take the
determinant of another 3 by 3 matrix. But it's the exact same
process for the 3 by 3 matrix that you're trying to
find the determinant of. So here is matrix A.
Here, it's these digits. This is a 3 by 3 matrix. And now let's evaluate
its determinant. So what we have to remember
is a checkerboard pattern when we think of
3 by 3 matrices: positive, negative, positive. So first we're going to
take positive 1 times 4. So we could just
write plus 4 times 4, the determinant of 4 submatrix. And when you say,
what's the submatrix? Well, get rid of the column
for that digit, and the row, and then the submatrix
is what's left over. So we'll take the
determinant of its submatrix. So it's 5, 3, 0, 0. Then we move on to the
second item in this row, in this top row. But the checkerboard
pattern says we're going to take
the negative of it. So it's going to be
negative of negative 1– let me do that in a slightly
different color– of negative 1 times the determinant
of its submatrix. You get rid of this
row, and this column. You're left with 4,
3, negative 2, 0. And then finally, you
have positive again. Positive times 1. This 1 right over here. Let me put the positive
in that same blue color. So positive 1, or plus
1 or positive 1 times 1. Really the negative is where
it got a little confusing on this middle term. But positive 1 times 1 times the
determinant of its submatrix. So it's submatrix is
this right over here. You get rid of the row,
get rid of the column 4, 5, negative 2, 0. So now we just have to evaluate
these 2 by 2 determinants. So the determinant
right over here is going to be 5 times
0 minus 3 times 0. And all of that is going
to be multiplied times 4. Well this is going
to be 0 minus 0. So this is all just a 0. So 4 times 0 is just a 0. So this all simplifies to 0. Now let's do this term. We get negative negative 1. So that's positive 1. So let me just make
these positive. Positive 1, or we
could just write plus. Let me just write it here. So positive 1 times
4 times 0 is 0. So 4 times 0 minus
3 times negative 2. 3 times negative
2 is negative 6. So you have 4– oh, sorry,
you have 0 minus negative 6, which is positive 6. Positive 6 times 1 is just 6. So you have plus 6. And then finally you have
this last determinant. You have– so it's going
to be plus 1 times 4 times 0 minus 5 times negative 2. So this is going to be
equal to– it's just going to be equal with–
1 times anything is just the same thing. 4 times 0 is 0. And then 5 times negative
2 is negative 10. But we're going to
subtract a negative 10. So you get positive 10. So this just simplifies
to 10, positive 10. So you're left with,
let me be clear. This is 0, all of this
simplifies to plus 6, and all of this
simplifies to plus 10. And so you are left with, if
you add these up, 6 plus 10 is equal to 16. So the trick here
is to just make sure you remember the
checkerboard pattern, and you don't mess up with all
the negative numbers and all of the multiplying.
Determinant of a 3×3 matrix: shortcut method (2 of 2)
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Determinant of a 3×3 matrix: shortcut method (2 of 2)
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How to Find the Determinant of a 4×4 MatrixNext Lesson
# Finding the Determinant of a 3×3 Matrix
Chapter 16
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Lesson 13
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What is a determinant and how do you find it? This lesson explains what a determinant is and shows you a step-by-step process for finding the determinant of a 3 x 3 matrix.
## Setting Up the Problem
We are about to go over how to find the determinant for a 3 x 3 matrix, but first we’ll need to know what a determinant is. A determinant is a single specific number associated with a specific square matrix. We should note that determinants are only defined for square matrices. Let’s take a look at the process used to find the determinant for a specific matrix.
• Step 1 – Write the matrix
We have to know what we’re working on, right? Well, here is the 3 x 3 matrix we are going to be using for this exercise.
The 3 x 3 refers to the number of rows and columns in our matrix. Since it has three rows and three columns, we call it a 3 x 3 matrix. Since the number of columns and rows are equal, this is a square matrix – which means that it will have a determinant.
• Step 2 – Write the matrix with determinant symbols
There is only a small difference in this image and the last one: the brackets have turned into straight lines. Mathematically speaking, however, this indicates a very large difference. The matrix represents a whole series of relationships between numbers while the determinant is just a single number.
• Step 3 – Write the matrix without brackets or determinant symbols
Now that we know the matrix we are working on, what a determinant is and how it’s written – we can start the process of finding the determinant. This step involves just writing the columns and numbers without any other symbols.
Simple, right? Now let’s keep going to the next step.
• Step 4 – Add the first two columns to the right
Now, to the right of our 3 columns we are going to add two more columns. Not just any columns though – we are simply going to repeat the first two columns from our matrix.
The dotted line in this picture is just for demonstration purposes – it’s not necessary to put this in when you are working on other determinant problems. Although if it helps you keep track of where you are in the process, you can certainly keep it.
• Step 5 – Add multiplications of first down diagonal
Start with the number in the first row and first column and multiply together the three numbers in the diagonal going down and to the right. In the image above these three numbers are circled. We are going to add the numbers from the down diagonals together.
• Step 6 – Add multiplications of second and third down diagonals
Repeat Step 5 for the second and third down diagonals. Again, we are choosing the three numbers from our extended matrix that are on a diagonal that goes down and to the right. Once we have these numbers, we are going to multiply them together and add them to our growing expression. Don’t multiply the numbers at this point – we’ll do that later on.
• Step 7 – Subtract multiplications of up diagonals
Now we are going to do a similar process with the up diagonals. For each diagonal, we are still going to be choosing three numbers to multiply, but for the up diagonals, we are going to be subtracting the multiplication terms instead of adding.
Do you notice how we have a large subtraction sign before each of the up diagonal terms? It’s very easy to get the wrong sign in this step, so make sure you know why it’s there.
• Step 8 – Compute results
Now that we have all of the terms for computing our determinant, we can start doing the operations. Remember, a negative times a negative is a positive, and if any of the multipliers are 0, then that term is going to be equal to 0.
What we have after all of the above steps is:
Determinant of A
= +(1)(3)(2) + (-4)(-1)(2) + (0)(0)(0) – (2)(3)(0) – (0)(-1)(1) – (2)(0)(-4)
When we evaluate each of the multiplication terms we get:
= 6 + 8 + 0 – 0 – 0 – 0
## Solution
Completing the last step of our process will give an answer of 14. This means that the determinant of our matrix turns out to be 14.
## How are Determinants Used
Getting this determinant may seem anticlimactic because we don’t yet know how to use it. Determinants are very useful for more advanced math: eigenvalues and eigenvector problems, for example. However, those concepts are beyond what we can cover in this lesson.
You may have noticed that the symbol for the determinant of a matrix looks a lot like the absolute value symbol and we ended up with a positive determinant. Does this mean that determinants can’t be negative?
This seems like a reasonable assumption, but it turns out to be incorrect. To prove it, all we have to do is take our existing matrix and change the sign of every term in it. When we then go through and evaluate the determinant, we will get negative 14 instead positive 14. So, determinants can be both positive and negative.
What about zero determinants? Determinants can be equal to zero, but this only happens when the lines of the square matrix are dependent on each other. Let’s say that our matrix represented a series of 3 equations for 3 unknown variables and we wanted to find out what those variable are. A determinant of 14, because it is not equal to zero, indicates that the three equations are independent of each other – which in turn means that our system of equations has a singular solution. If our determinant was zero, it would essentially mean that we have less than three independent equations, and our system of equations would not have a solution other than all variables equal to zero.
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• What is a Matrix?
5:39
• How to Write an Augmented Matrix for a Linear System
4:21
• How to Perform Matrix Row Operations
5:08
• Matrix Notation, Equal Matrices & Math Operations with Matrices
6:52
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6:29
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• How to Find the Inverse of a 4×4 Matrix
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6:10
• How to Solve Linear Systems Using Gauss-Jordan Elimination
5:00
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4:31
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7:02
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tell me more | 6,868 | 28,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-22 | latest | en | 0.822253 |
k83q9p3iem.talkiforum.com | 1,591,200,323,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00559.warc.gz | 60,688,673 | 3,869 | All Forums > General Discussion
# Voltage Drop explanations...
posted Sep 16, 2012 16:41:48 by SarahElizabethRaymer
Voltage drop has been something I have struggled with, and still I know I don't understand it completely- but I will, eventually. The thing is there is so much more to it than just this basic equation…
When calculating voltage drop in AC conductors, it is standard practice to use the nominal voltage on the AC side. For example, if you were calculating the voltage drop in a conductor in a 240V residential system, you would use the 240V. However, when it comes to PV there are more variables to consider and even knowing the nominal voltage or array voltage wouldn't really be completely all inclusive to determine the drop in voltage. A PV system designer may decide to use the short circuit current [Isc] as a conservative measure in the voltage drop calculation to consider the potential voltage drop under extreme conditions imposed by high currents from fluctuations in irradiance. Also, the designer may chose to consider temperature fluctuations and the possible drop in system voltage due to high ambient heat, using the adjusted high- temperature system voltage on the DC side.
For example, when using a voltage drop formula to determine the PV output circuit conductor size in a grid-tied PV System, you would use the array voltage [module] and the maximum power current [Imp] of the module to determine the voltage drop potential under standard operation conditions.
In the case of #'s 47 and 49 of the 2009 NABCEP PVI SG, we are calculating a stand alone system. It is stated in these questions what we must consider "that Im is flowing" [that] but not that the voltage is that of the modules- therefor we use the nominal voltage of the batteries.
When calculating the voltage drop in question #50 of the 2009 PVI Study Guide, we are asked to calculate the voltage drop "under maximum power conditions at STC". This makes the question different in stating that we are designing for a stand alone system under maximum power conditions which would only be achieved with a MPPT tracker- in which case we could consider the voltage of the array.
Confusing? Yes.
The first thing I would do if I was taking an exam and I saw this would be to say – “wait just one gosh darned second…”. I would discern that because of the similarity in these questions and this little differentiation in wording, that something was being asked of me that was ‘specific’ and that there was something fishy. Something is different in these 2 questions, but you may not realize what was really being asked in the previous questions [47] until you get to the last one [50].
I hope this clarifies! | 568 | 2,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-24 | latest | en | 0.932098 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=eulermac | 1,502,926,233,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102757.45/warc/CC-MAIN-20170816231829-20170817011829-00095.warc.gz | 614,554,572 | 24,173 | eulermac - Maple Programming Help
Home : Support : Online Help : Mathematics : Power Series : Expansions : eulermac
eulermac
Euler-Maclaurin summation
Calling Sequence eulermac(expr, x) or eulermac(expr, x = a..b) eulermac(expr, x, n) or eulermac(expr, x = a..b, n)
Parameters
expr - expression in x x - independent variable a, b - interval over which the approximation to the sum is computed n - (optional) integer (degree of summation)
Description
• The forms eulermac(expr, x) and eulermac(expr, x, n) compute asymptotic approximations to sum(expr, x). If F(x) = eulermac(f(x), x), then $F\left(x+1\right)-F\left(x\right)$ is asymptotically equivalent to f(x). The order of the approximation is specified by n, and defaults to Order - 1.
• The forms eulermac(expr, x=a..b) and eulermac(expr, x=a..b, n) compute nth degree Euler-Maclaurin summation formulas for expr (thus n terms of the expansion are given). If n is not specified, it is assumed to be Order - 1.
Examples
> $\mathrm{eulermac}\left(\frac{1}{x},x\right)$
${\mathrm{ln}}{}\left({x}\right){-}\frac{{1}}{{2}{}{x}}{-}\frac{{1}}{{12}{}{{x}}^{{2}}}{+}\frac{{1}}{{120}{}{{x}}^{{4}}}{-}\frac{{1}}{{252}{}{{x}}^{{6}}}{+}{\mathrm{O}}{}\left(\frac{{1}}{{{x}}^{{8}}}\right)$ (1)
> $\mathrm{eulermac}\left(\frac{1}{k},k=1..x\right)$
${{∫}}_{{1}}^{{x}}\frac{{1}}{{k}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{k}{+}{\mathrm{γ}}{+}\frac{{1}}{{2}{}{x}}{-}\frac{{1}}{{12}{}{{x}}^{{2}}}{+}\frac{{1}}{{120}{}{{x}}^{{4}}}{-}\frac{{1}}{{252}{}{{x}}^{{6}}}{+}{\mathrm{O}}{}\left(\frac{{1}}{{{x}}^{{8}}}\right)$ (2)
> $\mathrm{eulermac}\left(\frac{1}{{x}^{2}},x\right)$
${-}\frac{{1}}{{x}}{-}\frac{{1}}{{2}{}{{x}}^{{2}}}{-}\frac{{1}}{{6}{}{{x}}^{{3}}}{+}\frac{{1}}{{30}{}{{x}}^{{5}}}{-}\frac{{1}}{{42}{}{{x}}^{{7}}}{+}{\mathrm{O}}{}\left(\frac{{1}}{{{x}}^{{9}}}\right)$ (3)
> $\mathrm{eulermac}\left(\frac{1}{{k}^{2}},k=1..x,4\right)$
${{∫}}_{{1}}^{{x}}\frac{{1}}{{{k}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{k}{+}\frac{{1}}{{6}}{}{{\mathrm{π}}}^{{2}}{-}{1}{+}\frac{{1}}{{2}{}{{x}}^{{2}}}{-}\frac{{1}}{{6}{}{{x}}^{{3}}}{+}\frac{{1}}{{30}{}{{x}}^{{5}}}{+}{\mathrm{O}}{}\left(\frac{{1}}{{{x}}^{{7}}}\right)$ (4) | 919 | 2,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-34 | latest | en | 0.252117 |
https://medium.com/@zrjzhuruijie/facebook-%E8%BF%9B%E5%86%9B%E4%B9%8B%E8%B7%AF-%E9%99%84%E9%9D%A2%E7%BB%8F-995dbb681615 | 1,516,426,933,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00383.warc.gz | 737,230,505 | 24,349 | # Facebook 进军之路(附面经)
FB的面试流程很规范，包括电面和线下面试。
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solution link：http://www.jiuzhang.com/solution/valid-palindrome/
**Q2：大字符串相乘**
Lintcode原题link：http://www.lintcode.com/en/problem/big-integer-multiplication/
solution link：http://www.jiuzhang.com/solution/big-integer-multiplication/
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Lintcode原题link：http://www.lintcode.com/en/problem/paint-house-ii/
solution link：http://www.jiuzhang.com/solutions/paint-house-ii/
**Q2：合并k个排序链表**
Lintcode原题link：http://www.lintcode.com/en/problem/merge-k-sorted-lists/
solution link：http://www.jiuzhang.com/solutions/merge-k-sorted-lists/
**2nd Round:**
**Q1：搜索旋转排序数组**
Lintcode原题链接：http://www.lintcode.com/en/problem/search-in-rotated-sorted-array/
solution link：http://www.jiuzhang.com/solutions/search-in-rotated-sorted-array/
**Q2：正则表达式匹配**
Lintcode原题链接：http://www.lintcode.com/en/problem/regular-expression-matching/
solution link：http://www.jiuzhang.com/solutions/regular-expression-matching/
**3rd Round: **
**4th Round:**
One clap, two clap, three clap, forty?
By clapping more or less, you can signal to us which stories really stand out. | 469 | 1,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-05 | longest | en | 0.255843 |
http://www.netlib.org/lapack/lawn41/node35.html | 1,484,733,464,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280266.9/warc/CC-MAIN-20170116095120-00407-ip-10-171-10-70.ec2.internal.warc.gz | 603,746,528 | 3,058 | Next: Tests for the Orthogonal Up: The Linear Equation Test Previous: Tests for General and Contents
Tests for Triangular Matrices
The triangular test paths, xTR, xTP, and xTB, include a number of pathological test matrices for testing the auxiliary routines xLATRS, xLATPS, and xLATBS, which are robust triangular solves used in condition estimation. The triangular test matrices are summarized in Table 3. To generate unit triangular matrices of predetermined condition number, we choose a special unit triangular matrix and use plane rotations to fill in the zeros without destroying the ones on the diagonal. For the xTB path, all combinations of the values 0, 1, 17#17, 18#18, and 19#19 are used for the number of offdiagonals 22#22, so the diagonal type is not necessary.
Types 11-18 for the xTR and xTP paths, and types 10-17 for xTB, are used only to test the scaling options in xLATRS, xLATPS, and xLATBS. These subroutines solve a scaled triangular system 50#50 or 51#51, where 52#52 is allowed to underflow to 53#53 in order to prevent overflow in 54#54. A growth factor is computed using the norms of the columns of 16#16, and if the solution can not overflow, the Level 2 BLAS routine is called. Types 11 and 18 test the scaling of 55#55 when 55#55 is initially large, types 12-13 and 15-16 test scaling when the diagonal of 16#16 is small or zero, and type 17 tests the scaling if overflow occurs when adding multiples of the columns to the right hand side. In type 14, no scaling is done, but the growth factor is too large to call the equivalent BLAS routine.
Table 3: Test matrices for triangular linear systems
Test matrix type TR, TP TB Diagonal 1 Random, 23#23 2 1 Random, 24#24 3 2 Random, 25#25 4 3 Scaled near underflow 5 4 Scaled near overflow 6 5 Identity 7 6 Unit triangular, 23#23 8 7 Unit triangular, 24#24 9 8 Unit triangular, 25#25 10 9 Matrix elements are O(1), large right hand side 11 10 First diagonal causes overflow, offdiagonal column norms 56#56 12 11 First diagonal causes overflow, offdiagonal column norms 57#57 13 12 Growth factor underflows, solution does not overflow 14 13 Small diagonal causes gradual overflow 15 14 One zero diagonal element 16 15 Large offdiagonals cause overflow when adding a column 17 16 Unit triangular with large right hand side 18 17
The tests performed for the triangular routines are similar to those for the general and symmetric routines, including tests of the inverse, solve, iterative refinement, and condition estimation routines. One additional test ratio is computed for the robust triangular solves:
• 58#58
Table 4 shows the test ratios computed for the triangular test paths.
Table 4: Tests performed for triangular linear systems
Test ratio TR, TP TB 33#33 1 35#35 2 1 36#36 3 2 36#36, refined 4 3 (backward error)38#38 5 4 48#48 6 5 49#49 7 6 59#59 8 7
Next: Tests for the Orthogonal Up: The Linear Equation Test Previous: Tests for General and Contents
Susan Blackford 2001-08-13 | 799 | 2,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-04 | latest | en | 0.815102 |
https://www.emathhelp.net/en/calculators/calculus-1/online-graphing-calculator/?y=asinh%28x%29 | 1,716,534,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058684.66/warc/CC-MAIN-20240524060414-20240524090414-00053.warc.gz | 667,827,274 | 9,584 | Graph of $y = f{\left(x \right)} = \operatorname{asinh}{\left(x \right)}$
The calculator will draw the graph of the function $y = f{\left(x \right)} = \operatorname{asinh}{\left(x \right)}$.
To draw a parabola, circle, ellipse or hyperbola, choose the "Implicit" option.
You can pass a function as a parameter: https://www.emathhelp.net/en/calculators/calculus-1/online-graphing-calculator/?y=sin(x)&y=cos(x) will draw y=sin(x) and y=cos(x).
Related calculator: 3D Graphing Calculator
Pan the graph (move it) by holding the Shift key and dragging the graph with the mouse.
Zoom the graph in and out by holding the Shift key and using the mouse wheel.
The above operations can be very slow for more than 2 graphs. To avoid it, delete all graphs, pan and zoom, and then plot the graphs again.
Save as... | 222 | 809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.842908 |
https://developer.zebra.com/comment/44301 | 1,606,882,559,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00410.warc.gz | 266,329,366 | 21,841 | # How to center text on Card?
I am having trouble center aligning text on a card for a Zebra ZXP Series 3 using the Card Printing SDK.
I found in one of your examples codes how to center a barcode for the Card. But how can I center text for a card?
For example, for a barcode its:
// Barcode Drawing---------------------------------------------------
int rotation = 0; // origin lower left and no rotation
int barcodeType = 0; // Code 39
int barcodeWidthRatio = 2; // narrow bar = 2 dots, wide bar = 5 dots
int barcodeMultiplier = 2; // {2..9}
int barcodeHeight = 75; // 75 dots
int textUnder = 1; // true
string barcodeData = "123456789";
//To calculate the full length of a Code 39 bar code:
//L = [(C+2) (3R + 7) - 1] X Where
//L = Length of bar code
//C = Number of characters
//R = Ratio of wide-to-narrow bars
//X = Number of dots times 0.0033 inches per dot (0.08847 mm per dot); for the 5:2 ratio, X = Dots times 2
//See ZXP3 SDK Manual for the forumulas used to calcuate the length of other barcode types.
// Calculate the length of the barcode
int C = barcodeData.Length;
double R = 5.0 / 2.0;
int X = 2;
int length = (int)((C + 2) * (3 * R + 7) - 1) * X;
int startX = (int)Math.Floor((CARD_WIDTH - length) / 2.0); //Center barcode horizontally
int startY = (int)Math.Floor(((CARD_HEIGHT - barcodeHeight) / 2.0) + barcodeHeight); //Center barcode vertically
// Sends Barcode data to the Monochrome Buffer
if (DrawBarcode(startX, startY, rotation, barcodeType, barcodeWidthRatio, barcodeMultiplier, barcodeHeight, textUnder, _asciiEncoder.GetBytes(barcodeData), out errValue) == 0)
{
msg = "Printing DrawBarcode Error: " + errValue.ToString();
return;
}
But how and where can i find the formula for centering a label / text on a card??? It says it's in the manual but it isn't can someone please provide this formula?
Joshua Preston
As of right now, I am having
As of right now, I am having to manually pad the text left and right in C# to try and get it to appear “centered” on the card. I have different font sizes on the card so It’s really hard when I don’t know the formula on how to center the text on the card. If I could just get the formula, the generic formula to center any text on the card – that would fix my issue! I am only printing on the front side of the card and it is just text.
Right now I am using this to pad my text with spaces left and right. It's sketchy and it appears to look centered in some cases but there may be cases where it isn't. I of course can't sit here and test the 3 billion different name combinations I may be putting on the card.
public string SetTextAlignmentToCenter(string textToCenterAlign, int lengthOfLine)
{
// # of Characters you can put per line on the Badge. Badge is Landscape Orientation
string centeredText = textToCenterAlign.PadLeft(((lengthOfLine - textToCenterAlign.Length) / 2) + textToCenterAlign.Length).PadRight(lengthOfLine);
return centeredText;
}
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Stephen Troup
I stopped using the graphics
I stopped using the graphics methods used in the sdk for this reason. I extracted the .net graphics object that the zebra graphics sdk uses to print to a card. I use the following routine to get access to it.
internal static System.Drawing.Graphics GetGraphicsObject(ref ZMTGraphics.ZMotifGraphics g)
{
System.Reflection.FieldInfo dynField = g.GetType().GetField("helper", System.Reflection.BindingFlags.NonPublic | System.Reflection.BindingFlags.Static);
object gg = dynField.GetValue(null);
return (System.Drawing.Graphics)gg.GetType().GetField("graphics").GetValue(gg);
}
The reason I came up with this methods was because I needed to measure a string to be able to shrink to fit and the sdk doesn't provide a public method to do it.
If I were printing on the landscape on a black panel then I'd do
int datalen;
byte[] bmpFrontMonoKPanels;
ZMTGraphics.ZMotifGraphics g = new ZMTGraphics.ZMotifGraphics();
System.Drawing.Drawing2D.GraphicsState gfxstate;
g.ClearGraphics();
g.InitGraphics(0, 0, ZMTGraphics.ZMotifGraphics.ImageOrientationEnum.Landscape, ZMTGraphics.ZMotifGraphics.RibbonTypeEnum.MonoK);
gfx = GetGraphicsObject(ref g);
gfxstate = gfx.Save();
<use the .net system.drawing.graphics gfx object to do my printing>
gfx.Restore(gfxstate);
bmpFrontMonoKPanels = g.CreateBitmap(out dataLen);
g.ClearGraphics();
You just need to use the standard methods for drawing objects i.e. exactly the same as when printing to the old printer object. So finally to draw centred text, create a rectangle that's as big as the region you want the text to appear in, then use the gfx.drawstring overload that has the system.drawing.stringformat and use that to centre the text. No need to pad the string or calculate anything, unless it's too big to fit.
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Joshua Preston
Unfortunately I do not have
Unfortunately I do not have the ZMTGraphics. I am using the ZXP 3 not the ZXP 7
I only have:
ZBRGraphics.cs
ZBRPrinter.cs
There still has to be a formula to calculate how to horizontally center the text on a card. They have a formula to calculate centering a bar code above in my OP - so I just need them to provide me the one for the text
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Stephen Troup
Yep, sorry though you'd
Yep, sorry though you'd switched to the driverless sdk not the older one as the driverless zxp3 uses the same graphics library as the 7.
Have a look at the ZBRGDIDrawTextRectEx function, as it provides an alignment parameter. Similar to what I said for the 7, create a rectangle that covers the area you want the text to appear and then pass it's position, size and text alignment along with the text and font. Notes on the routine at on page 148 of the sdk manual.
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Joshua Preston
After hours of fighting with
After hours of fighting with the text alignment I finally got a response from Zebra. You have to use the DrawTextEx() method and pass it in the alignment parameter
[DllImport("ZBRGraphics.dll", EntryPoint = "ZBRGDIDrawTextEx", CharSet = CharSet.Auto,
SetLastError = true)]
static extern int ZBRGDIDrawTextEx(int x, int y, int angle, int alignment, byte[] text, byte[] font, int fontSize, int fontStyle, int color, out int err);
public int DrawTextEx(int x, int y, int angle, int alignment, byte[] text, byte[] font, int fontSize, int fontStyle, int color, out int err)
{
return ZBRGDIDrawTextEx(x, y, angle, alignment, text, font, fontSize, fontStyle, color, out err);
}
How to use it (from the "SDK Manual"):
int x = 0;
int y = 0;
int angle = 0; //0 degrees rotation (no rotation)
int alignment = 4; //center justified
string TextToPrint = "Printed Text";
byte[] text = null;
string FontToUse = "Arial";
byte[] font = null;
int fontSise = 12;
int fontStyle = 1; //bold
int color = 0x0FF0000; //black
int err = 0;
int result = 0;
//use the function:
System.Text.ASCIIEncoding ascii = new System.Text.ASCIIEncoding();
text = ascii.GetBytes(TextToPrint);
font = ascii.GetBytes(FontToUse);
result = ZBRGDIDrawTextEx(x, y, angle, alignment, text, font, fontSize,fontStyle, color, out err);
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Steven Moore
This is a total hack, and I
This is a total hack, and I am not sure it is accurate, but printing is basically centered on a vertical card using this calculation:
printVal is the text to print. The Y coordinate is fixed. X is variable.
// center name
// 2.11667 is mm/letter for Ariel 6 pt.
// 12 is dots per mm
// 2 splits it in half for the offset
//15.66 is estimated dots per letter at Ariel 8 pt
if (param == "name")
{
int m_length = (int)Math.Round((((printVal.Length) * 15.66) / 2), MidpointRounding.AwayFromZero);
int m_xcor = paramVal.XCor - m_length;
//Draw a Text
//retValue = graphics.DrawText(m_xcor, paramVal.YCor, ASCIIEncoding.ASCII.GetBytes(printVal),
// ASCIIEncoding.ASCII.GetBytes(paramVal.Font), paramVal.FontSize, lookup[paramVal.FontStyle].Single(),
// Convert.ToInt32(paramVal.FontColor, 16), out errValue);
retValue = graphics.DrawText(m_xcor, paramVal.YCor, ASCIIEncoding.Default.GetBytes(printVal),
ASCIIEncoding.ASCII.GetBytes(paramVal.Font), paramVal.FontSize, lookup[paramVal.FontStyle].Single(),
Convert.ToInt32(paramVal.FontColor, 16), out errValue);
}
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You voted ‘up’ | 2,195 | 8,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-50 | latest | en | 0.638226 |
https://math.stackexchange.com/questions/3792242/the-existence-of-the-power-set-of-an-infinite-set | 1,618,090,017,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038059348.9/warc/CC-MAIN-20210410210053-20210411000053-00620.warc.gz | 504,196,126 | 43,155 | # The existence of the power set of an infinite set
A large part of the set theory is devoted to infinities of various kinds, and this has been built on Cantor's groundbreaking work on uncountable sets. However, even Cantor's proof is based on the assumption that certain sets exist, namely that the power set of a countably infinite set exists. Sure, after assuming that the set of natural numbers has a power set, Cantor's proof shows that this power set is uncountable. However, he does not address why such a power set should exist in the first place. Is the existence of the power set of an infinite set assumed as an axiom of set theory?
Intuitively, I found the existence of such a power set troubling. By definition, most of the elements of this set consist of purely random infinite sets. For example, the power set of natural numbers can be partitioned into:
1. Those elements that can be constructed by an algorithm (i.e. a Turing Machine), such as all the finite sets, or the infinite sets with a constructive algorithm (e.g., set of odd numbers, set of natural numbers that build the digits of $$\pi$$ in some form such as {3, 14, 159, ...}, etc.)
2. All the other elements that no Turing Machine can generate them, e.g., an infinite set of entirely random natural numbers.
These purely random sets have (1) an infinite amount of information packed into them, and (2) we cannot even construct them. Both these aspects are unsettling to me.
Setting the intuition aside, is there a concrete reason why mathematicians accept the existence of the power set of an infinite set? Can we prove that such a set must exist? Has there been any inquiry on whether the existence of such a set would result in inconsistencies or not?
• Well, yes, one of the axioms of ZF(C) is the existence of the power set of any set. – Berci Aug 15 '20 at 23:07
• The real numbers have the same size as the power set of the natural numbers, so if you accept that the points on a line are in 1-1 correspondence with the reals, you accept the existence of at least one power set of an infinite set. – Rivers McForge Aug 15 '20 at 23:07
• People struggled really hard with this kind of stuff in the early 20th Century. People wore themselves out, and now they put this on the proverbial back burner. – Stephen Montgomery-Smith Aug 15 '20 at 23:50
• @RiversMcForge: Are you claiming that ZFC-power set+existence of the reals implies that the power set of the natural numbers exists? I don't think that's obvious. – tomasz Aug 15 '20 at 23:50
• Prove using what? This sort of fundamental question can be very sensitive to the foundations you use, including the axioms and the logic. – tomasz Aug 15 '20 at 23:59
This should perhaps be a comment, but it's too long. The question emphasizes a dichotomy among the subsets of $$\mathbb N$$: computable sets versus all the others. But there's actually a much richer spectrum here: computable sets, computably enumerable sets (like the set of code numbers of Turing machines that eventually halt when started on an empty tape), arithmetically definable sets, predicative sets, constructible sets (in the sense of Gödel), etc.
Early in the 20th century, there was considerable discussion (or dispute) as to whether mathematics should work with completely arbitrary sets or should require some degree of definability. This question played a central role in discussions about the axiom of choice. Eventually, when mathematical logic had been developed to the point where one could talk precisely about the many levels of definability, it became clear that the natural way to proceed is to let "set" mean "completely arbitrary set"; if one wants to work instead with definable sets, then one should say so and one should specify exactly what sorts of definitions are to be allowed. This decision in favor of arbitrary sets is the basis for most set theorists' acceptance of the axiom of choice.
It is entirely possible for someone (like me) who accepts arbitrary sets (and the ZFC axioms that are intended to describe them) to also consider more restricted universes in which only the computable sets are guaranteed to exist. One important axiom system often used to describe such a universe is called $$\text{RCA}_0$$ (abbreviating "recursive comprehension axiom"). Another axiom system, $$\text{ACA}_0$$, describes a universe with arithmetical sets; KP+Inf provides for the existence of hyperarithmetical sets; etc.
From the point of view of mathematics in general, it is interesting to ask whether various well-known theorems (like the Bolzano-Weierstrass theorem or the existence of prime ideals in non-degenerate rings, or the dominated convergence theorem) are provable with such limited supplies of sets. The detailed study of such questions is called "reverse mathematics", because to verify that one has just the right axiom system A for proving some theorem T, the usual method is to deduce A from T --- the reverse of the usual business of mathematics, deducing theorems from axioms. The standard reference for this topic is Stephen Simpson's book "Systems of Second-Order Arithmetic".
I should also mention that sets at the other end of the spectrum, random sets that have very high information content, have also been studied extensively by computability theorists. As with definability, randomness also has a spectrum; there are different levels of randomness. And the two spectra are (somewhat) related: A subset of $$\mathbb N$$ is random (to some degree) if it lies in all definable (to some degree) sets of probability 1 (in the standard probability space of subsets of $$\mathbb N$$).
The "undefinable" end of the spectrum also contains non-random sets of various sorts, for example the so-called generic sets, which lie in all definable sets that are comeager (i.e., their complement is of first Baire category). As with randomness, there are levels of genericity, related to levels of definability.
• Thank you for your thorough response. Regarding the existing (though non-mainstream) schools of thought such as finitists, or predicativists, is the reason for these mathematicians to not accept "completely arbitrary sets" as real, just a matter of taste? or are there more concrete reasons for them not to accept such sets? e.g. do they think accepting $P(\mathbb{N})$ can lead to inconsistencies? – nebeleh Aug 18 '20 at 7:28
• My impression is that there are different opinions among those who don't accept arbitrary sets (of natural numbers). Some don't think the notion of "arbitrary set" is clear enough to be used in mathematics. Or at least not clear enough to be used in foundations of mathematics. Others expect the notion to lead to inconsistency. Others expect inconsistency just from the notion of natural numbers, without even worrying about sets. – Andreas Blass Aug 18 '20 at 16:34
• Besides Stephen Simpson's book that you mentioned, can I ask what books/articles do you suggest for further reading on these topics? Most of the set theory books that I have read do not discuss these foundational matters in much depth and mostly focus on the set theory as it is practiced by most mathematicians. On the other hand, the philosophy of math books that I have seen do not go much into the technical details of these topics. Where do you think is a good starting point to get a deeper understanding of these viewpoints? Thank you. – nebeleh Aug 19 '20 at 0:01
Firstly, let's discuss the situation in classical logic (ordinary logic where $$P \lor \neg P$$ is a tautology). Suppose the existence of a set $$S$$, and suppose that the set $$\{f : S \to 2\}$$ exists, where $$2$$ is some fixed 2-element set $$\{\top, \bot\}$$. In other words, suppose that the collection of all functions $$S \to 2$$ forms a set. In this case, we can plainly see that the power set of $$S$$ exists by the Axiom of Replacement, since every subset $$J \subseteq S$$ defines a characteristic function $$f : S \to 2$$ by $$f(j) = \top$$ if $$j \in J$$, and $$f(x) = \bot$$ if $$x \notin J$$, and every function $$f : S \to 2$$ defines a subset $$\{x : f(x) = \top\}$$. Thus, if we want the collection of functions between two fixed sets to form a set, we must accept the existence of power sets. On the other hand, accepting the existence of power sets (along with the existence of cartesian products, which follows from ZF's replacement axiom schema) allows one to define the set of functions between two sets. Basically, if we want to consider sets of functions with a given domain and codomain, we also must accept the existence of powersets.
In general, this concept generalises to intuitionist logic (logic which doesn't necessarily accept $$P \lor \neg P$$ as a tautology) as follows: let $$1$$ be the one-element set, and suppose the set $$\Omega = \{S : S \subseteq 1\}$$ exists. Then the powerset of an arbitrary set $$S$$ exists iff the collection $$\{f : S \to \Omega\}$$ is a set. For if we have the set $$\{f : S \to \Omega\}$$, then just as before we can define for every $$f : S \to \Omega$$ a subset $$\{x \in S : f(x) = 1\}$$; and conversely, for every $$J \subseteq S$$, we can define a function $$f(x) = \{y \in 1 : x \in J\}$$, $$f : S \to \Omega$$, so we can again apply Replacement.
If we want the power set of $$1$$ to exist and we want function sets to exist, we must accept the existence of power sets as a necessary consequence. In classical logic, we can prove that the powerset of $$1$$ exists since it is equal to $$\{\emptyset, 1\}$$ which exists by the axiom of pairing.
There are weaker theories of set theory in which the law of excluded middle ($$P \lor \neg P$$) is not considered a tautology and in which function sets $$\{f : A \to B\}$$ exist but it cannot be proven that powersets exist (even $$\Omega = P(1)$$ may not exist). An example is CZF. Other examples of such theories (moving beyond the realm of "set theory") include Homotopy Type Theory and, in general, the theory of $$\Pi W$$ pretoposes.
If we eliminate the power set axiom from $$ZFC$$, then power sets can't be proven to exist. This is why the power set axiom is, in fact, an axiom. But then we'd also be forced to give up the existence of sets of functions, which would make set theory a completely miserable place for most mathematics.
Indeed, it seems difficult to even discuss point-set topology at all without the axiom of the power set. We'd essentially have to write that branch of mathematics off.
Of course, the axiom of the powerset does introduce one complication; it's possible that $$ZF$$ including powerset is inconsistent but $$ZF$$ without powerset is consistent. In fact, because we can prove that $$ZFC -$$powerset is consistent within $$ZFC$$, we cannot possibly prove that $$ZFC$$ is consistent from $$ZFC -$$ powerset unless both theories are inconsistent. But at the end of the day, constant fear of foundational inconsistencies is no way to live mathematical life.
• I am curious: how do you show that ZFC$^-$ has a model from ZFC? – tomasz Aug 16 '20 at 0:02
• @tomasz Just take $\kappa$ an uncountable regular cardinal; then let $H_\kappa$ be the set of all sets with hereditary size $< \kappa$. Then $(H_\kappa, \in_{H_\kappa})$ is a model of $ZFC$ without powerset. In particular, it can be shown in ZFC that every successor cardinal is regular. Thus, we can take the set of all sets with hereditary size at most countably infinite. This will form a model of $ZFC$ without powerset. In particular, in this model, we would have only one infinite cardinal. – Doctor Who Aug 16 '20 at 2:20
• @tomasz: You can find the details in Ken Kunen’s Set Theory: An Introduction to Independence Proofs, Ch. IV, Theorem $6.5$. – Brian M. Scott Aug 16 '20 at 3:26
• Thank you. Let me rephrase what I've understood from your answer. You accept the Cartesian product of two sets exists. Now, if we're mapping set $S$ to set $\{\top, \bot\}$, every function defines one of the subsets of $S$, hence $P(S)$ should exist. That is fine, however, my question is why should the Cartesian product of two sets exist in the first place? For example, if $S$ is infinite, some of these functions can be constructed (e.g. map odd members to $\top$ and evens to $\bot$), however, most cannot be constructed or written in any form. Why do we accept their existence? – nebeleh Aug 16 '20 at 3:49 | 2,974 | 12,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 46, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-17 | latest | en | 0.93872 |
https://www.physicsforums.com/threads/mixing-ice-with-water.227995/ | 1,597,426,145,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739347.81/warc/CC-MAIN-20200814160701-20200814190701-00484.warc.gz | 779,281,944 | 14,430 | # Mixing ice with water
## Homework Statement
If you mixed 200g if ice that is at -5°C with 20g of water that is at 15°C, what will be the temperature and condition of the final state once equilibrium is achieved?
Q=mcΔT, Q=mLf
## The Attempt at a Solution
I know that the amount of ice is to great to change with such little water, so I know the final condition will be a 220g block of ice at some temperature between -5°C and 0°C. Since the mass of the water is added to the mass of the ice does this imply that Qgained≠Qlost?
That's what I'm having trouble on, deriving the equation to solve for temperature if they are not equal. | 162 | 639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-34 | latest | en | 0.960447 |
https://discourse.pymc.io/t/combining-theano-variables-and-numpy-arrays-in-custom-step-function/6117 | 1,656,761,988,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104054564.59/warc/CC-MAIN-20220702101738-20220702131738-00531.warc.gz | 254,717,247 | 5,816 | # Combining Theano variables and Numpy arrays in custom step function
Hi all - I’m working on a fairly simple case of implementing a custom step method. I want to sample from a conjugate posterior Dirichlet distribution using a custom step, but I can’t quite figure out how best to combine the Theano variable from the rest of the graph with the Numpy random sampling functions which are available to sample from a Dirichlet. I am aware of the gamma-sum representation of the Dirichlet, but I don’t see any straightforward way to do this in Theano short of implementing my own routines to sample Dirichlet variates. Any ideas or advice? The code below shows my problem.
``````from pymc3.step_methods.arraystep import BlockedStep
import pymc3 as pm
import numpy as np
class ConjugateDirichletUpdate(BlockedStep):
def __init__(self, var, concentration, counts):
self.var = var
self.vars = [var]
self.conc = concentration
self.counts = counts
def step(self, point):
new = point.copy()
alpha = self.conc + self.counts
#What I want: new[self.var] = np.random.dirichlet(alpha)
# This samples, though it's with the wrong distribution:
new[self.var] = np.random.dirichlet(self.counts)
return new
J = 4
counts = np.sum(np.random.uniform(size=[10,4]) > 0.5, axis=0)
with pm.Model():
tau = pm.Exponential('tau', lam=1)
alpha = pm.Deterministic('alpha', tau*np.ones(J))
p = pm.Dirichlet('p', a=alpha)
x = pm.Multinomial('x', p=p, n=counts.sum(), observed=counts)
step = ConjugateDirichletUpdate(p, alpha, counts)
trace = pm.sample(step=[step], chains=1,cores=1)
``````
I worked out how to do this. Here’s a gist that shows an implementation for a simple case: https://gist.github.com/ckrapu/8be4da91a70763ee62f889c6cd98f700
The basic idea is that instead of manipulating the Theano symbolic variables as they come into the step method, work instead with the `point` object which contains their numerical values.
1 Like
As a follow up question, I’d like to figure out how to use a `Deterministic` variable within a step method. Currently, deterministic variables are not recorded in the `point` object passed to the step method. It’s not clear exactly how the state of those variables can be passed via 'point`.
If you’re interested, here is an example of a conjugate step method for the same underlying Categorical/Dirichlet pairing. In this case, a (transition) matrix with Dirichlet rows is updated according to a matrix of “observed” transition counts.
This step method was recently updated to handle more than just the most basic transition matrix graphs (i.e. stacked rows of Dirichlet distributions), which neatly illustrates the inevitable challenges one faces when creating such step methods. In other words, conjugate step methods face a generalizability challenge that methods like HMC don’t because they have a strong dependency on the exact form of a graph.
The example above addresses some of this but still has large deficiencies. The way to address this more broadly involves improvements to the underlying graph library (i.e. Theano-PyMC) and a framework that fits this type of work (see Symbolic PyMC).
Also, here are some more conjugate step methods in the old PyMC2. They follow roughly the same pattern as PyMC3 step methods—just without the Theano.
3 Likes
Yes, that is precisely the type of example I’m looking for. Related to the `pymc3-hmm` repository, I’ve often come across use cases where a dynamic Gaussian model / Kalman filter- centric distribution would also be of great use. Thanks for providing these links. | 849 | 3,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.733128 |
http://www.stokastik.in/leetcode-reverse-pairs/ | 1,553,379,126,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203021.14/warc/CC-MAIN-20190323201804-20190323223804-00142.warc.gz | 361,154,533 | 10,211 | # Leetcode : Reverse Pairs
Problem Statement
Solution:
The brute force approach is to consider all pairs (i, j) such that i < j and check whether they satisfy the condition that nums[i] > 2*nums[j]. The time complexity for this approach is definitely O(N2) where N is the number of elements in the array 'nums'. We can improve this time complexity by reducing the number of comparisons required for each element. For e.g. if we are on the i-th element in the array, then all values j > i such that nums[i] > 2*nums[j] are the valid candidates. Thus if we traverse the array from back and maintain a sorted array T of the values 2*nums[j]+1 for all nums[j] seen so far, then we need to check at the i-th element how many elements in the current array T has values less than or equal to nums[i]. For e.g. if the array is like this:
[2,4,3,5,1]
Then when we traverse from the back and reach the 2nd element (at i=1) i.e. 4, the sorted array T would look like [3, 7, 11], because 2*1+1=3, 2*3+1=7 and 2*5+1=11. Now only 3 is less than or equal to 4, thus for the 2nd element only possible pair is (i, j) = (1,4). But challenge is to keep the array T sorted every-time. We cannot afford to sort the array every-time after adding a new element. One possible mechanism is to keep a Balanced Binary Search Tree of the values 2*nums[j]+1 with an additional value at each node indicating how many nodes are there in its left sub-tree.
In this way adding a new node is O(logN), updating the left-subtree count value of its parent by 1 (if it is left node of its parent) up to the root node is also O(logN). Thus with this modified BST, we can compute the number of such reverse pairs in O(N*logN) complexity. But most of the time I have found that an alternative solution exists for a BST method that does not require us to create and maintain a balanced BST.
We will use a modified Merge Sort based approach for this problem instead of a BST with the same running time complexity but reduced space-time complexity.
The idea is to sort the array in decreasing order using merge sort and while doing the merge operation on the two sorted sub-arrays, we can also count the number of reverse pairs for the merged array.
For e.g. given an array nums=[a1, a2, ..., aN], in merge sort, first we divide this array into two equal parts:
x1 = [a1, a2, ..., aN/2], and x2 = [aN/2+1, aN/2+2, ..., aN]
We sort x1 and x2 recursively using merge sort to get the sorted arrays (in decreasing order) y1 and y2:
y1 = [b1, b2, ..., bN/2], and y2 = [c1, c2, ...,cN/2]
We will count the number of reverse pairs in y1 and y2 then merge them and return back. Let us say that the number of reverse pairs in the array y1 be p1 and in the array y2 be p2, then the number of reverse pairs for the merged array will also include the reverse pairs between y1 and y2 i.e. for each element in y1 which are the corresponding reverse pairs from y2.
For the 1st element in y2 i.e. c1, the possible reverse pairs from y1 will be [b1, b2, ..., bk1] such that bk1 >= 2*c1+1 and bk1+1 < 2*c1+1 because the elements are in descending order. Similarly for the next element c2, the possible reverse pairs will include all the reverse pairs from c1, plus additional elements [bk1+1, bk1+2, ..., bk2], such that bk2 >= 2*c2+1 and bk2+1 < 2*c2+1 and so on.
Let us denote the number of reverse pairs for an element cj from y2 by N(cj), then we have:
Thus, N(cj+1) = N(cj) + length([bkj+1, bkj+2, ..., bk(j+1)])
#ReversePairs([y1, y2]) = p1 + p2 + N(c1) + N(c2) + ... + N(cN/2)
Observe that we never scan each element in y1 or y2 more than once. Thus the time complexity of computing the number of reverse pairs is O(len(y1)+len(y2)). Thus the overall time complexity of sorting + counting reverse pairs is O(N*logN). The python code for the same is as follows:
```class Solution(object):
def count_reverse(self, nums, left, right):
if left < right:
mid = (left+right)/2
a, x = self.count_reverse(nums, left, mid)
b, y = self.count_reverse(nums, mid+1, right)
n, m = len(a), len(b)
i, j, cnt, last_cnt, sum_cnt = 0, 0, 0, 0, 0
while True:
if i < n and j < m and a[i] > 2*b[j]:
cnt += 1
i += 1
elif (i < n and j < m and a[i] <= 2*b[j]) or (i == n and j < m):
last_cnt += cnt
sum_cnt += last_cnt
cnt = 0
j += 1
else:
break
out, i, j = [], 0, 0
while True:
if (i < n and j < m and a[i] > b[j]) or (i < n and j == m):
out.append(a[i])
i += 1
elif (i < n and j < m and a[i] <= b[j]) or (i == n and j < m):
out.append(b[j])
j += 1
else:
break
return out, sum_cnt + x + y
return [nums[left]], 0
def reversePairs(self, nums):
if len(nums) == 0:
return 0
out, cnt = self.count_reverse(nums, 0, len(nums)-1)
return cnt```
In each recursion, we return the sorted array as well as the number of reverse pairs in the sorted array.
The BST technique is preferred if we need to build a streaming system for this problem because with each new element in the stream we can add that element efficiently to the Balanced BST, whereas merge sort needs to be called on the entire array each time a new element is added.
Categories: PROBLEM SOLVING | 1,524 | 5,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2019-13 | latest | en | 0.876855 |
https://www.cfd-online.com/Forums/openfoam-solving/72092-k-epsilon-turbulence-model.html | 1,510,949,817,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803906.12/warc/CC-MAIN-20171117185611-20171117205611-00204.warc.gz | 758,179,339 | 18,501 | # k-epsilon turbulence model
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January 25, 2010, 12:10 k-epsilon turbulence model #1 Member James Baker Join Date: Dec 2009 Posts: 35 Rep Power: 9 I have looked everywhere I know to look, but can not find what the constants in the k, e, alphat, mut files are actually for. From what I understand, the k is the kinetic energy, e is the dissipation rate, mut is the turbulent viscosity, and alphat has to do with the thermal diffusity. But those are all variable AFAIK. What do the constants represent in these files and how would they be calculated. Thank you for any light on this subject!
January 27, 2010, 20:33 #2 Member James Baker Join Date: Dec 2009 Posts: 35 Rep Power: 9 Nobody knows?
January 28, 2010, 04:14 #3 Senior Member matej forman Join Date: Mar 2009 Location: Brno, Czech Republic Posts: 133 Rep Power: 10 For example for me it is hard to answer as I do not know what constants you're talking about. Do you mean (a) turbulence model constants, (b) wall model constants (Cmu, kappa ...) or (c) some other numbers? matej
January 28, 2010, 06:41 #4 Senior Member Gijsbert Wierink Join Date: Mar 2009 Posts: 383 Rep Power: 11 Hi James, Well, k is the turbulent kinetic energy and e (or epsilon) is the dissipation rate of k. If you are wondering how to calculate basic initial values for a case, you can use §2.1.8.1 of the user guide. There is explained that you can use e.g. 5% turbulent intensity, so u' is the 0.05*U, so k = 1/2 * u' * u'. Epsilon is then C_mu^0.75 * k^1.5 / L, where C_mu is commonly 0.09 and L typically 20% of the characteristic length scale (e.g. pipe diameter). Is this of any help? Claudio87, FLZ, sbusmayer and 7 others like this. __________________ Regards, Gijs
January 28, 2010, 21:01 #5 Member James Baker Join Date: Dec 2009 Posts: 35 Rep Power: 9 Yes this information helps very much for k and epsilon. What are alphat and mut for?
January 29, 2010, 00:42
#6
Member
Jinbiao Xiong
Join Date: Oct 2009
Location: China/Japan
Posts: 50
Rep Power: 10
Quote:
Originally Posted by fijinx Yes this information helps very much for k and epsilon. What are alphat and mut for?
I am not sure what version are you using. In my case, I use nut instead of mut. I guess mut is eddy viscosity, and mut = nut*rho. But I have not idea what alphat is.
January 29, 2010, 02:23 #7 Senior Member Gijsbert Wierink Join Date: Mar 2009 Posts: 383 Rep Power: 11 Ah, yes, forgot that . "mut" is the turbulent (eddy) viscosity. Basically, turbulence makes the fluid a bit more viscous. "alphat" is, I think, the thermal diffusivity. It probably plays a role in some wall functions etc. Nikunj.R, a_habib and gallon like this. __________________ Regards, Gijs Last edited by gwierink; January 29, 2010 at 02:47. Reason: typo
January 29, 2010, 03:19 #8 Senior Member matej forman Join Date: Mar 2009 Location: Brno, Czech Republic Posts: 133 Rep Power: 10 If you're wondering what some variable means, you have several options where to look. Most convenient is to use doxygen - documentation of the source codes. For example you may use the online version at: http://foam.sourceforge.net/doc/Doxygen/html/ and look for alphat. You will get a list of links. First you'll be lost, but using it will help you to understand better how foam works (at least happened to me). for example here: http://foam.sourceforge.net/doc/Doxy...ce.html#l00246 you will find how alphat is calculated for k-eps model. You even see it in the source code, so you are pretty sure, that it really works like that in your computation. What a beauty! hope this helps matej a_habib likes this.
January 29, 2010, 03:21 #9 Member Jinbiao Xiong Join Date: Oct 2009 Location: China/Japan Posts: 50 Rep Power: 10 Right. If you are considering the heat transfer, there should be an alphat considering the turbulence heat transfer. Jinbiao yalilou likes this.
May 16, 2015, 11:26 problem chosing the RAS turbulence models for compressible fluids #10 New Member Join Date: Apr 2015 Posts: 1 Rep Power: 0 hello, i'm runnig a case on a 9 mm projectil. sould i use in RASpropreties as RAS turbulence models for compressible fluids — compressibleRASModels, the kEpsilon model?
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All times are GMT -4. The time now is 16:16. | 1,405 | 4,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-47 | longest | en | 0.90272 |
http://spmath81709.blogspot.com/2009/10/ratio-homework-scribe-for-oct-18.html | 1,508,285,875,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00282.warc.gz | 298,033,973 | 19,995 | ## Sunday, October 18, 2009
### Ratio homework scribe for oct 18
1. The ratio of the width to the length of the Canadian flag is 1:2
a.) The flag on the cover of an atlas is 12cm wide. how long is it ?
According to the ratio it is 24 cm in length because you just double the width
b.) A large flag outside a calgary school is 3m long . what is its width ?
1.5 because i just double the width to double the length | 115 | 417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-43 | longest | en | 0.918499 |
https://1-texasholdempoker.com/casinos/how-do-casinos-make-money-off-of-blackjack.html | 1,643,176,436,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00643.warc.gz | 136,284,303 | 19,874 | # How do casinos make money off of blackjack?
Contents
The basic rules of Blackjack create a house advantage. For most players and most sets of rules, the house advantage is just 1% to 5%. Let’s say an average player giving up a 2% house advantage buys in for \$100 and plays a \$10 bet for ten hands. Theoretically, he will have lost 2%, or \$2.
## How is blackjack profitable for casinos?
Casinos get more income when they find a method of increasing their house edge, the amount of time on each game and the average bet. If there are no bets or if the players are few, then the handle is less and the profit is less. … However, it is very rare to find a blackjack game that has a minimum bet of \$2 or \$5.
## How does the dealer make money in blackjack?
If a player has blackjack (and the dealer DOES NOT,) pay the player 3/2 their bet. If they bet \$5, you pay them \$7.50, for example. If the dealer also has blackjack, it’s a “push” and nobody wins. They keep their money.
## Do casinos lose money on blackjack?
Conclusion. The casinos build their blackjack games to make a profit. And if you aren’t careful, you’re going to lose your entire bankroll. The biggest mistake that blackjack gamblers make is playing 6:5 blackjack games instead of 3:2 blackjack games.
## How do casinos run blackjack?
If a player’s first two cards are an ace and a “ten-card” (a picture card or 10), giving a count of 21 in two cards, this is a natural or “blackjack.” If any player has a natural and the dealer does not, the dealer immediately pays that player one and a half times the amount of their bet.
## Do casinos lose money?
Each game you play at a casino has a statistical probability against you winning. … In 2018, commercial casino gaming revenue amounted to about \$41.7 billion; one way to think about all those profits is that they are the result of the accrual of all of the losses from casino patrons each year.
## Is owning a casino profitable?
On the games with the lowest house edge, the smallest advantage, a casino might only be generating about a 1% to 2% profit. On other games, it may make profits of up to 15 %to 25% or more. The house edge on a 00 roulette wheel is 5.26%.
## How do casino owners make money?
The way the casino makes its profit is by paying you winnings that are lower than the odds that would make a game break-even. For example, if the casino made you risk \$110 to win \$100 on a coin toss, in the long run, the casino would make a profit. 50% of the time, they’d lose \$100.
## Is Blackjack ever profitable?
No, you can’t. In online casinos, the deck is shuffled after every hand, making the card counting useless. So, even if you would follow basic blackjack strategy, you would lose money in the long run. You cannot really make money in the long run because counting cards is impossible.
## Where do casinos make the most money?
Slot machines remain the most important money-making part of casinos in the United States. In many states, casinos make between 65 and 80 percent of their gambling income from slots.
## Do dealers cheat in blackjack?
How likely is it for a blackjack dealer to cheat the players while he/she is dealing? The short answer is: It’s highly unlikely. … Blackjack is one of the easiest games to stack, because in its simplest form, the dealer has to control only two cards. About those two cards.
## Is Blackjack a skill or luck?
Purely based on statistics, some casino gamblers get lucky and win money. Blackjack, however, can be beaten based on skill—no luck involved.
## Do casinos track your winnings?
Yes, many casinos track your winnings. They continuously observe your moves and change the results in your favour. For example, if you bet in a roulette game and you won for more than three to five times, the dealer will play some buttons under the table to change results in your favour.
## Does blackjack beat 21?
A player Blackjack beats any dealer total other than a dealer’s Blackjack, including a dealer’s regular 21. Total of both dealer and player hands are the same. Offered on the dealer’s ace to insure the player’s hands against the dealer’s Blackjack.
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## What happens if the dealer busts in blackjack?
The dealer never doubles, splits, or surrenders. If the dealer busts, all remaining player hands win. If the dealer does not bust, each remaining bet wins if its hand is higher than the dealer’s and loses if it is lower.
## Does 5 cards Beat 21 in blackjack?
A hand of three or four cards worth 21 points beats everything else except a Pontoon or Five Card Trick. Hands with 20 or fewer points and fewer than five cards rank in order of their point value – the nearer to 21 the better. Hands with more than 21 points are bust and are worthless. | 1,096 | 4,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.960726 |
http://www.analyzemath.com/calculus/Problems/maximize_radius_circle.html | 1,490,278,386,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187113.46/warc/CC-MAIN-20170322212947-00446-ip-10-233-31-227.ec2.internal.warc.gz | 423,648,851 | 29,023 | # Maximum Radius of Circle - Problem with Solution
Use the derivative to find the size of an angle of a right triangle so that the radius of the circle inscribed is maximum; for a constant hypotenuse.
Problem: ABC is a right triangle and r is the radius of the inscribed circle. a) Express r in terms of angle x and the length of the hypotenuse h. b) Assume that h is constant and x varies; find x for which r is maximum. Solution to Problem: a) Let M, N and P be the points of tangency of the circle and the sides of the triangle. OM, ON and OP are perpendicular to CB, CA and AB respectively. Triangles COM and CON are rigth triangles and have two congruent sides: CO and OM and ON; the two triangles are therefore congruent. We denote the size of angle MCN by x and write tan(x / 2) = r / CM Triangles BOM and BOP are rigth triangles and have two congruent sides: BO and OM and OP; the two triangles are therefore congruent. We denote the size of angle MBP by y and write tan(y / 2) = r / BM Note that y + x = 90 which gives y / 2 = 45 - x / 2 Substitute y / 2 by 45 - x / 2 in the equation tan(y / 2) = r / BM to obtain tan(45 - x / 2) = r / BM We now solve the equation tan(x / 2) = r / CM for CM and solve equation tan(45 - x / 2) = r / BM for BM to obtain CM = r / tan(x/2) and BM = r / tan(45 - x/2) We now use the fact that h = CM + BM to write the equation h = r / tan(x/2) + r / tan(45 - x/2) = r [ 1 / tan (x/2) + 1 / tan(45 - x/2) ] We now use trigonometric identities to simplify the above equation. The first identity we will use is tan(45 - x/2) = ( tan(45) - tan(x/2) ) / ( 1 + tan(45)tan(x/2) ) = ( 1 - tan(x/2) ) / ( 1 + tan(x/2) ) The formula for h is now given by h = r [ 1 / tan (x/2) + (1 - tan(x/2)) / ( 1 + tan(x/2)) ] We now use the identity tan(x/2) = sin(x/2) / cos(x/2) and other identities to rewrite h as follows h = r [ cos(x/2) / sin(x/2) + (cos(x/2) + sin(x/2)) / (cos(x/2) - sin(x/2)) ] = r [ 1 / [ (sin(x/2) (cos(x/2) - sin(x/2)) ] ] We now solve the above equation for r to obtain r = h sin(x/2) ( cos(x/2) - sin(x/2) ) = h ( sin(x/2) cos(x/2) - sin 2(x/2)) We now use the identities sin(x/2)cos(x/2) = (1/2) sin (x) and sin 2(x/2) = 1/2 - (1/2)cos(x) to write r as follows r = (h / 2) ( sin(x) + cos(x) - 1) b) Now that we have calculated r as a function of x, and h assumed constant, let us find the first derivative of r with respect to x dr / dx = (h / 2) [ cos(x) - sin(x)] Let us find a critical points for r by solving the equation dr / dx = 0. (h / 2) [ cos(x) - sin(x)] = 0 gives cos(x) - sin(x) = 0 since h is constant and not equal to 0. cos(x) = sin(x) Square both sides cos 2(x) = sin 2(x) cos 2(x) = 1 - cos 2(x) Solve for cos x to obtain cos(x) = + 1/sqrt(2) or - 1/sqrt(2) x is a acute angle and therefore the solution to our equation is given by x = pi/4 = 45 degrees. The graph of dr/dx is shown below and dr/dx is positive for x < pi/4 and negative for x > pi/4, therefore r has a maximum at x = pi/4 = 45 degrees. x = 45 degrees is the value of angle ACB for which the radius r is maximum for a given value h of the hypotenuse. Note that in this case triangle ABC is isosceles. More references on calculus problems | 1,061 | 3,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2017-13 | longest | en | 0.875153 |
https://www.numberempire.com/269515 | 1,600,463,142,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188841.7/warc/CC-MAIN-20200918190514-20200918220514-00796.warc.gz | 938,111,956 | 6,450 | Home | Menu | Get Involved | Contact webmaster
# Number 269515
two hundred sixty nine thousand five hundred fifteen
### Properties of the number 269515
Factorization 5 * 19 * 2837 Divisors 1, 5, 19, 95, 2837, 14185, 53903, 269515 Count of divisors 8 Sum of divisors 340560 Previous integer 269514 Next integer 269516 Is prime? NO Previous prime 269513 Next prime 269519 269515th prime 3792821 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 1000001110011001011 Octal 1016313 Duodecimal 10bb77 Hexadecimal 41ccb Square 72638335225 Square root 519.14834103558 Natural logarithm 12.504379326409 Decimal logarithm 5.4305829410847 Sine -0.78817693759582 Cosine -0.61544871032611 Tangent 1.2806541379836
Number 269515 is pronounced two hundred sixty nine thousand five hundred fifteen. Number 269515 is a composite number. Factors of 269515 are 5 * 19 * 2837. Number 269515 has 8 divisors: 1, 5, 19, 95, 2837, 14185, 53903, 269515. Sum of the divisors is 340560. Number 269515 is not a Fibonacci number. It is not a Bell number. Number 269515 is not a Catalan number. Number 269515 is not a regular number (Hamming number). It is a not factorial of any number. Number 269515 is a deficient number and therefore is not a perfect number. Binary numeral for number 269515 is 1000001110011001011. Octal numeral is 1016313. Duodecimal value is 10bb77. Hexadecimal representation is 41ccb. Square of the number 269515 is 72638335225. Square root of the number 269515 is 519.14834103558. Natural logarithm of 269515 is 12.504379326409 Decimal logarithm of the number 269515 is 5.4305829410847 Sine of 269515 is -0.78817693759582. Cosine of the number 269515 is -0.61544871032611. Tangent of the number 269515 is 1.2806541379836
### Number properties
Examples: 3628800, 9876543211, 12586269025 | 599 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-40 | latest | en | 0.680059 |
http://www.digitalmars.com/d/archives/digitalmars/D/28778.html | 1,500,648,290,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423785.29/warc/CC-MAIN-20170721142410-20170721162410-00304.warc.gz | 426,601,980 | 3,832 | ## digitalmars.D - Increment/Decrement Boolean Operators
• AJG (29/29) Sep 21 2005 Hi there,
• Oskar Linde (23/55) Oct 07 2005 Isn't bit supposed to behave as a 1-bit integer? In that case, foo++ and
```Hi there,
I noticed that the Increment/Decrement operators are defined for type bool,
which is in turn, type bit. I dunno whether this is intended or not. Thinking
about it, they could be quite useful:
bool foo;
foo++; // Make foo = true.
foo--; // Make foo = false.
// etc.
However, this works only so far. In fact, it subly works but could blow up in
Internally, bit is still an integer of some sort, therefore using those
operators actually changes the quantity just like a regular integer. In other
words, it has nothing to do with on/off which is what a boolean is about.
For instance:
void main() {
bool Bar = false;
assert(!Bar); // OK.
Bar++;
assert(Bar); // OK.
Bar++;
assert(Bar); // OK.
Bar--;
assert(!Bar); // Fails.
}
Given what I said, that last assert should have passed. Since bit is an on/off
type and not a numerical type, I think these operators are not working properly.
They should be overridden for booleans appropiately.
Cheers,
--AJG.
```
Sep 21 2005
```AJG wrote:
Hi there,
Hello,
I noticed that the Increment/Decrement operators are defined for type
bool, which is in turn, type bit. I dunno whether this is intended or not.
Thinking about it, they could be quite useful:
bool foo;
foo++; // Make foo = true.
foo--; // Make foo = false.
// etc.
Isn't bit supposed to behave as a 1-bit integer? In that case, foo++ and
foo-- should overflow giving equal behavior to ++ and --: toggling the bit.
Disregarding the integer promotion rules that prevent this, foo++; should do
the same thing as foo = foo + cast(bit)1;
However, this works only so far. In fact, it subly works but could blow up
IMHO, foo++, and foo-- works as they should: modulo 2 arithmetics.
Internally, bit is still an integer of some sort, therefore using those
operators actually changes the quantity just like a regular integer. In
other words, it has nothing to do with on/off which is what a boolean is
For instance:
void main() {
bool Bar = false;
assert(!Bar); // OK.
Bar++;
assert(Bar); // OK.
Bar++;
assert(Bar); // OK.
Bar--;
assert(!Bar); // Fails.
}
I am unable to reproduce that. I get:
void main() {
bit bar = 0;
assert(!bar); // OK
bar++;
assert(bar); // OK
bar++;
assert(!bar); // OK
bar--;
assert(bar); // OK
}
(with dmd 0.133 linux)
Given what I said, that last assert should have passed. Since bit is an
on/off type and not a numerical type, I think these operators are not
working properly. They should be overridden for booleans appropiately.
In my opinion, bits/bools should (and do, as far as I can tell) follow the
rules of modulo 2 arithmetics.
Regards
Oskar Linde
```
Oct 07 2005 | 757 | 2,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-30 | longest | en | 0.908112 |
https://mystocksinvesting.com/category/intrinsic-value/page/4/ | 1,680,025,119,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00012.warc.gz | 484,365,002 | 42,213 | ## China Zaino – Fundamental & Intrinsic Value
• Post author:
Base on Q2, 2009 financial report:
• PE = 3.5
• Dividend Yield = 5.8%
• NAV = \$0.288
• Net Earning = 15.6% (18.3% in 2008)
• Current Ratio = 5.7
• ROA = 31.7% (Base on 2008 Full Year)
• ROE = 36.3% (base on 2008 Full Year)
• Cash Flow = S\$199.8 Million (27% increase with reference to 2008 Full Year)
Stock Background
• 3 years historal high = \$0.615
• Current Price = \$0.30 (only 49% of 3 years high)
Intrinsic Value Calculation
PE Model
• Fair value PE = 15, intrinsic value= \$1.23 (base on EPS \$0.08185)
• PEG = 0.24 (it looks like super under value!)
DCF Model
• Growth Rate = 15% (capped)
• Discount Rate = 5%
• Number of Shares = 945 million
• 2008A Net Operating Cash Flow = S\$55.893 million, intrinsic value = \$1.01
Discounted EPS Model
• Growth Rate = 15% (capped)
• Discount Rate = 5%
• 2009F EPS = \$0.08185
• Intrinsic Value = \$1.40
Intrinsic Value for China Zaino (the most conservative number of the three numbers) = \$1.01 (Current price is 70% discount to the intrinsic value!)
## How do I value a stock?
• Post author:
There are 3 different methods of doing a stock valuation.
• Simple Price / Earning (PE) Ratio & PEG
• Discounted Cash Flow (DCF)
• Discounted Earning Per Share (EPS)
Simple PE Ratio & PEG
As a general guideline, a company is at its fair value if the PE is about 15. If the PE ratio less than 15, it is considered under value, and vice versa. I also do a quick comparison of current PE versus the average PE historically. I will pay special attention to the stock price movement if the current PE is more higher than the historical PE average.
PEG refers PE ratio divided by company growth rate. PEG = 1 means that PE growth rate is the same as the company growth rate (measured by either EPS growth rate or net operating cash flow growth rate)
If PEG < 1, the stock price is under value.
If PEG > 1, the stock price is over value.
If PEG = 1, the stock price is at its fair value.
Discounted Cash Flow Model (DCF)
This model is to estimate the company next 10 years net operating cash flow (Future Value, FV) and re-calculate to the Present Value (PV), and add all ten years PV together. The intrinsic value can be calculated after dividing the total number of shares,. The assumption made is the company must be able to generate cash growth consistently with a CAGR (Compounded Annual Growth Rate) which computed from the past history of net operating cash flow.
Net Operating Cash Flow information can be found from the company annual reports, under the Cash Flow Statement.
Discounted Earning Per Share Model (EPS)
Similar to the DCF model, but this time Earning Per Share is being looked into. EPS information can be found at the Income Statement by getting the Net Earning number and divided by the total number of shares. By looking at the historical EPS, a CAGR for EPS growth can be calculated.
By bringing all the 10 years FV of EPS to PV, adding them together give an intrinsic value of the stock.
Valuation of a stocks need some financial background and need some practice. Two key areas to pay attention to:
(1) Where to find the information? All the financial statement can be found from the annual reports by going to the company web site. Another way is to get the summarised information from the website like Shareinvestor (for Singapore Stocks), Morningstar & MSN Money (for US Stocks)
(2) Understand the Financial Fundamental & Definitions like Present Value, Future Value, Discount Rate, CAGR and also practise how to use them. I use the financial calculator to calculate the CAGR and instrinsic value calculator to calculate the intrinsic value of the stocks. I got those simple software (formula in excel form) from my investment course.
I found that DCF model is a better model although it is a little more complicated because Cash Flow is not easily manipulated by the company accountant and cash is always easily be audited. If the company business model is solid, the net operating cash flow grows consistently every year. On the other hand, good EPS numbers do not mean the company is increasing the sales revenue and gaining competitive advantage to expand the market share. EPS can be manipulated easily as the company accountant can add whatever provision they want, using different amortization or depreciation method or using different revenue recognition method. Furthermore, the company can make the EPS more attractive by buying back shares, doing all sorts of cost cutting internally (like selling company fixed assets) to make the number looks nice.
After the intrinsic value is calculated, I compare the current stock price with the intrinsic value. If the current stock price is at least 20% discount to the intrinsic value, I will put the stocks in my watchlist and wait for the right time to buy. I will share in the next post on how I time my entry.
## China HongXing – Fundamental & Intrinsic Value
• Post author:
Base on Q2, 2009 financial report:
PE = 7.1
Dividend Yield = 1.2%
NAV = \$0.3359
Net Earning = 9.7% (15.5% in 2008)
Current Ratio = 25.02
ROA = 10.2% (Base on 2008 Full Year)
ROE = 11.6% (base on 2008 Full Year)
Cash Flow = S\$520.7Million (33.4% increase with reference to 2008 Full Year)
Stock Background
3 years historal high = \$1.42
Current Price = \$0.225 (only 16% of 3 years high)
Intrinsic Value Calculation
EPS Annual Growth = 15% (from 2005A to 2008A)
Discount Rate = 5%
Intrinsic Value = \$0.54 (58% Discount of Current Price to Intrinsic Value) | 1,408 | 5,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.833254 |
https://simplewebtool.com/converters/area/squarerodstobarns/squarerodstobarns.html | 1,716,256,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00460.warc.gz | 453,182,452 | 12,451 | Square Rods to Barns conversion
Square Rods to Barns (b) conversion calculator and how to convert.
How to convert Square Rods to Barns: 1 square rod is approximately equal to 2.53e29 barns. 1 square rod ≈ (2.53 * 1029) ≈ 2.53e29 b The areaQ(b) in barns is approximately equal to the area Q(sr) in square rods multiplied by 2.53e29.
Formula: Q(b) ≈ Q(sr) * 2.53e29
Square Rods to Barns conversion table
Square Rods Barns (b)
1 square rod 2.53e29 b
2 square rods 5.06e29 b
3 square rods 7.59e29 b
4 square rods 1.01e30 b
5 square rods 1.26e30 b
6 square rods 1.52e30 b
7 square rods 1.77e30 b
8 square rods 2.02e30 b
9 square rods 2.28e30 b
10 square rods 2.53e30 b | 246 | 670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-22 | latest | en | 0.649074 |
https://grammarlytutors.com/more-power-company-accounting-homework-help/ | 1,696,257,827,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00472.warc.gz | 316,228,047 | 11,982 | # More-power company | Accounting homework help
More-power company has projected sales of 75,000 regular sanders and 30,000 mini-sanders for the next year.The projected income statement is as follows:
Regular sanders mini-sanders Total
Sales 3,000,000 1,800,000 4,800,000
Less:variable 1,800,000 900,000 2,700,000
expenses
Contribution margin 1,200,000 900,000 2,100,000
Less:direct fixed 250,000 450,000 700,000
Product margin 950,000 450,000 1,400,000
Less:common fixed 600,000
expenses
operating income 800,000
expenses
1.set up the given income statement on a spreadsheet.Then,substitute the following sales mixes, and calculate operating income.
Regular sanders mini-sander
a.75,000 37,500
b.60,000 60,000
c.30,000 90,000
d.30,000 60,000
2.Calculate the break-even units for each product for each of the preceding sales mixes.
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https://kidsworksheetfun.com/mixed-fractions-worksheets-grade-7/ | 1,702,207,723,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101779.95/warc/CC-MAIN-20231210092457-20231210122457-00326.warc.gz | 373,478,450 | 25,747 | # Mixed Fractions Worksheets Grade 7
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Pin On Printable Blank Worksheet Template | 879 | 4,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.884475 |
https://weeklymathematics.com/tag/derived-functor/ | 1,638,501,134,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362589.37/warc/CC-MAIN-20211203030522-20211203060522-00568.warc.gz | 673,607,690 | 27,890 | ## A Road to the Grothendieck Spectral Sequence:Derived Functors II
In the previous post, I gave a little glimpse into derived functors and somewhat motivated their construction. In this post, we’ll get our hands dirty with homological algebra to continue setting up the required machinery and go through many proofs.
In the previous post, I promised to continue the proof of a lemma which establishes a long exact sequence. Before giving the proof, let me mention a few facts which will be useful.
If we’re given a short exact sequence $0 \to I \to A \to B \to 0$ in an abelian category where $I$ is an injective object. From the map $I \to I$ and the monomorphism $I \hookrightarrow A$, we can extend to a map $A \to I$ such that the composition is the identity. Using the Splitting Lemma, one obtains a non-canonical splitting $A \simeq I \oplus B$. Applying $F$,
$0 \to F(A) \to F(I) \oplus F(B) \to F(B)$.
The identity map $B \to B$ factors through the projection map $\pi:A \to B$, the same holds true after applying $F$, in particular, the last map is surjective!
$0 \to F(A) \to F(I) \oplus F(B) \to F(B) \to 0$.
### Proof of Lemma 1:
#### Step 1:A morphism between objects with injective resolution induces a chain map between the resolutions
Let $\phi:A \to B$ be a morphism between two objects with injective resolutions $A_{\bullet},B_{\bullet}$. In the figure below, the map $\phi_{0}$ is constructed from the fact that $d_{0}:A \to I_{0}$ is a monomorphism and there is a map $A \to B \to I'_{0}$ from $A$ to an injective object.
Now, there is a monomorphism $I_{0}/ker(d_{1})=I_{0}/Im(d_{0})=Coker(A \to I_{0}) \hookrightarrow I_{1}$. Next, note that by the commutativity of the square already defined, $\phi_{0}$ takes $Im(d_{0})=Ker(d_{1})$ to $Ker(d'_{1})=Im(d'_{0})$ by the fact that $d'{1}d'_{0}=0$ by exactness of the lower sequence. This means that the map $\phi_{0}$ induces a morphism $h_{0}: Coker(A \to I_{0}) \to Coker(B \to I'_{0})$ and by exactness, we can compose this with $B/Ker(d'_{1})$ to get a map $Coker(A \to I_{0}) \to I'_{1}$. Since $I'_{1}$, we get the required map $\phi_{1}$. Inductively continue this process to get the entire chain map. Note that all the maps defined from the injective object property are not unique.
#### Step 2:Proving that any two such extensions are chain-homotopic
Let $f_{n},g_{n}$ be two chain maps from $A \to I_{\bullet}$ to $B \to I'_{\bullet}$.
## A Road to the Grothendieck Spectral Sequence:Derived Functors I
This is a series of posts which will develop the required material to present the Grothendieck spectral sequence, a few applications of it in algebraic geometry(Lerray sequence, sheaf cohomology etc.) and some other categorical nonsense. This post is meant to be a light glimpse into derived functors and not a full introduction.
I’ve wanted to post this for a few months but unfortunately I overestimated how much free time I would have in my first semester of college(hurrah!). At one point, I totally forgot that I even maintain a blog! That’s why I haven’t posted for about 5 months. The title of my blog is now officially fraudulent. There were many new things(both math-related and otherwise) at UT Austin that I wanted to explore in my first semester so I could be forgiven for putting aside my blog. I think I’ve realized that it is really just a matter of consistency. If I do something regularly, I’ll stick to the routine. So maybe, the title of the blog is more aspirational than it is real.
Anyways, enough of the nonsense. Though the Grothendieck spectral sequeuce encodes a lot of data(like all other spectral sequences),it is actually surprisingly simple to motivate and feels almost ‘natural’ to construct. But I suppose that is really indicative of the Grothendieck’s spectacular reformulation of homological algebra that has seeped into modern textbooks that makes it feel so ‘natural’. It is truly a beautiful sight to witness how derived functors lead to this elegant construction.So first, let’s set up the machinery.
An object $I$ in a category $C$ is said to be an injective object if for every morphism $f:X \to I$ and every monomorphism $i:X \to Y$, there exists a morphism $h:Y \to I$ extending the map $f$ such that the diagram commutes.
In the abelian category setting, the importance lies in the fact that $I$ is an inejctive object if and only if the Hom functor is $Hom_{C}(--,I)$ is exact. If an injective object is at the beginning of a short exact sequence in $C$, the sequence splits.
A category $C$ has enough injectives if for every every object $X \in C$, there exists a monomorphism $X \hookrightarrow I$ for some injective object $I$.
An injective reslution of an object $X \in C$, an abelian category is a resolution
$0 \to X \to I_{0} \to I_{1} \to \cdots$ where $I_{k}$ are injective objects. In particular, this is a quasi isomorphism of chain complexes in $C$ given by $X \to I_{\bullet}$ where $X$ is the complex $0 \to X \to 0 \cdots$.
## Derived Functors
Let’s say $A$ is an abelian category. Consider a short exact sequence in $A$:
$0 \to L \to M \to N \to 0$
An exact functor is a functor between abelian categories which preserves such sequences. Taking the direct sum, for example, preserves preserves a short exact sequence. Accordingly, we say that functors are left and right exact if they preserve the left and right parts of the short exact sequence. It is a well known that in the case of the category of modules over a ring $R$,the covariant Hom functor is left exact. $0 \to L \to M \to N \to 0$, then
$0 \to 0 \to Hom(A,L) \to Hom(A,M) \to Hom(A,N)$ where $F_{A}(X)=Hom(A,X):A \to Ab$ is the Hom functor.
The tensor product functor $G(X)=X \otimes_{R} M$ where $M$ is an $R-module$ is known to be right exact. Many of these facts and their proofs can be found in many standard texts on commutative algebra or homological algebra. Some of the arguments in these proofs tend to be quite arduous to work through. An easier way to prove it is to notice that the functors are adjoint and show the equivalent statement that Hom preserves limits which is not too difficult. Taking the dual, yields the statement for tensoring and infact, it yields the completely general version which states that left adjoint functors preserve finite colimits using the Yoneda Lemma. See my other post on adjoint functors if you wish to learn more.
The point of derived functors(which I’ll shortly introduce) is to take these ‘incomplete’ exact sequences where we’ve ‘lost data’ to try and construct a long exact sequence. Remember that chain complex
$\cdots \xrightarrow{} C{n+1} \xrightarrow{\partial} C_{n} \xrightarrow{\partial}$
equipped with ‘boundary maps(which I’ve not labelled) allows us to compute homology $H_{n}=\frac{ker\partial}{Im\partial}$ which measures how far the sequence is from being exact. ALL the data is in the chain complex itself and the entire process of computing homology/cohomology is just a formalization which turns out to be quite handy. In the same manner, one should treat derived functors as a comfortable formalization using the data we possess. For an object $X \in A$, we know just one thing:that there is an injective resolution:
$0 \to X \to I_{0} \to I_{1} \to \cdots$.
Now, we take our left exact functor $F$(contravariant $Hom$ for example) and apply it to the injective resolution to get $F(X \to I_{\bullet})$
$0 \to F(X) \to F(I_{0}) \to F(I_{1}) \to \cdots$
Now, just ‘take homology/cohomology’ and call it the right derived functor
$R^{i}F(X)=H^{i}(F(X \to I_{\bullet}))$. But wait, did you notice something?On the left hand side, I don’t refer to the injective resolution. That is the essence of the construction, it is independent of the injective resolution of $X$ upto canonical isomorphism. A proof of this can be found in any standard textbook on algebra in the homological algebra section(Aluffi, Dummit and Foote;I think Hatcher also proves it for the dual case in the section on cohomology). Let’s take a closer look at this sequence. Since $F$ is left exact, we get the following exact sequence:
$0 \to F(A) \to F(I_{0}) \to F(I_{1})$
W get $R^{0}(F(X))=F(X)$. If we $F$ is exact, then the all $R^{i}(F(X))=0$ would be trivial for $i>0$! I guess you could think of this as a way to encode approximation just like in homology/cohomology.
The converse isn’t necessarily true. An object $X \in A$ is said to be $\bf{F-acyclic}$ for a left exact functor if $R^{i}(F(X))=0$ for $i > 0$.
The final step of this formalization is ensuring that we have the long exact sequence
## Lemma 1:
If $0 \to L \to M \to N \to 0$ is a short exact sequence in an abelian category $A$ with enough injectives and $F$ is a left exact functor, there is an LES
$0 \to L \to M \to N \to \\ \to R^{1}(F(L)) \to R^{1}(F(M)) \to R^{1}(F(N)) \to \cdots$
The proof will be deferred to the next post with discussions on its dual, other theorems and special cases such as Ext, Tor. | 2,466 | 8,976 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 84, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-49 | latest | en | 0.859633 |
https://stats.stackexchange.com/questions/594774/why-are-standard-frequentist-hypotheses-so-uninteresting | 1,726,001,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00113.warc.gz | 506,322,674 | 56,500 | # Why are standard frequentist hypotheses so uninteresting?
In almost any textbook introducing the topic of frequentist statistics, null hypotheses of the form $$H_0: \mu=\mu_0$$ or similar are presented (the coin is unbiased, two measurement devices have identical behavior, etc.). Classic statistical tests such as the $$Z$$ or $$T$$ tests are then based on rejecting these null hypotheses.
As I see it, these types of equality hypotheses are uninteresting for a number of reasons:
• In "real life" one is always only interested in some finite accuracy $$\epsilon$$, meaning the hypothesis of interest is actually of the form $$H_0: |\mu-\mu_0|<\epsilon$$.
• The equality hypothesis is a priori known to be wrong when considering continuous variables (no coin is perfectly unbiased in reality!), and as a corollary,
• The fact that the null hypothesis cannot be rejected is by definition temporary, and is an artifact of not enough data. Given enough data, any type of equality hypothesis on continuous variables will be rejected in a real world use case.
So, why are these hypotheses still used, both in text books and in applications, while it is difficult to find formulas for more interesting* "real" hypotheses?
* e.g. a question I recently asked regarding these types of hypotheses
• This is a slight tangent from the main thrust of your question, but on the topic of biased coins, you may be interested in Gelman's paper "You Can Load a Die, But You Can't Bias a Coin" Commented Nov 5, 2022 at 19:27
• I don't think what I am about to say will answer your question, however, forming an interval estimate (e.g. $(1-\alpha)100\%$ CI) basically avoids all of the shortcomings by focusing on what $\mu$ is likely to be, rather than focusing on what it isn't Commented Nov 5, 2022 at 20:03
• This of course is a loaded question, and such questions are not usually very helpful to learn something that goes beyond already existing prejudices. Commented Nov 6, 2022 at 1:02
• 1. Tests like "H0:|μ−μ0|<ϵ" are called equivalence tests; there's also noninferiority and superiority tests. en.wikipedia.org/wiki/Equivalence_test . They have not as been as widely used as they should, but I think that's in part because they require thinking (in coming up with $\varepsilon$) for more than a few seconds. 2. People often use tests in situations where their stated question of interest is clearly not a testing question. Neither of those issues is the fault of hypothesis tests, but of how they're being taught and used by people we have little influence over Commented Nov 6, 2022 at 1:37
• @Glen_b That’s been my observation, too, that people are reluctant to pick an epsilon.
– Dave
Commented Nov 6, 2022 at 4:33
You can regard this as an example of 'All models are wrong but some are useful'. Null hypothesis testing is a simplification.
Null hypothesis testing is often not the primary goal and instead it is more like a tool for some other goal, and it is used as an indicator of the quality of a certain result/measurement.
An experimenter wants to know the effect size and know whether the result is statistically significant.
For the latter, statistical significance, one can use a null hypothesis test (which answers the question whether the observation has a statistically significant deviation from zero).
The null hypothesis test and p-values are now considered as a bit of an old fashioned tool. Better expressions of experimental results are confidence intervals or intervals from Bayesian approaches.
The equality hypothesis is a priori known to be wrong
Yes, if you consider coins.
But an exception might be hardcore science like physics or chemistry where certain theories are tested. For instance the equivalence principle.
In addition if the equality hypothesis is a priori wrong, then why perform an experiment? If something is a priori wrong then the point is not to show that this something is wrong, instead the point is to show that there is an effect that can be easily measured. A casino that wants to test coins may not care about the theoretical probability that coins are not exactly p=0.5 fair and might differ by some theoretically small value, they care about finding out coins with a larger difference. And the point of the null hypothesis test is to prevent false positives.
Also note the two approaches/philosophies behind null hypothesis testing
• Fisher: You may have observed some effect, and as expected it is not zero, but if your p-value is high then it means that your test has little precision and little strength in differentiating between different effect sizes (even down to a true effect size of zero, the observed effect may have likely occured and thus statistical fluctuations govern your observation). So you better gather some more data.
The p-value and null hypothesis is a rule of thumb for indicating precision of an experiment.
• Neyman and Pearson: (from 'On the Problem of the Most Efficient Tests of Statistical Hypotheses')
Indeed, if $$x$$ is a continuous variable – as for example is the angular distance between two stars – then any value of $$x$$ is a singularity of relative probability equal to zero. We are inclined to think that as far as a particular hypothesis is concerned, no test based upon the theory of probability can by itself provide any valuable evidence of the truth or falsehood of that hypothesis.
But we may look at the purpose of tests from another view-point. Without hoping to know whether each separate hypothesis is true or false, we may search for rules to govern our behaviour with regard to them, in following which we insure that, in the long run of experience, we shall not be too often wrong.
The hypothesis test is a practical device to create a decision rule. One goal is to make this rule efficient and using a likelihood ratio with a null hypothesis is one method of achieving that.
In "real life" one is always only interested in some finite accuracy $$\epsilon$$, meaning the hypothesis of interest is actually of the form $$H_0: |\mu-\mu_0|<\epsilon$$.
This is captured by hypothesis testing. An example is two one-sided t-tests for equivalence testing and can be explained with the following image and can be considered as testing three hypotheses instead of two for the absolute difference
$$\begin{array}{}H_0&:& \text{|difference|} = 0\\ H_\epsilon&:& 0 <\text{|difference|} \leq \epsilon\\ H_\text{effect}&:& \epsilon < \text{|difference|} \end{array}$$
Below is a sketch of the position of the confidence interval within these 3 regions (unlike the typical sketch of TOST, there are actually 5 situations instead of 4).
The point of observations and experiments is to find a data driven answer to questions by excluding/eliminating what is (probably) not the answer (Popper's falsification).
Null hypothesis testing does this in a somewhat crude manner and does not differentiate between the situations B, C, E. However, in many situations this is not all too much of a problem. In a lot of situations the problem is not to test tiny effects with $$H_0: |\mu-\mu_0|<\epsilon$$. The effect size is expected to be sufficiently large and above some $$\epsilon$$. In many practical cases testing $$|\text{difference}| > \epsilon$$ is nearly the same as $$|\text{difference}| > 0$$ and the null hypothesis test is a simplification. It is in the modern days of large amounts of data that effect sizes of $$\epsilon$$ play a role in results.
Before this issue was dealt with by having arbitrary cut-off values for p-values and by power analysis. If a test had a p-values below some significance level, then the conclusion is that the effect must be some effect beyond some size. These p-values are still arbitrary, also with TOST equivalence testing. A researcher has some given significance level and computes a required sample size to obtain a given power for a given effect that the researcher wants to be able to measure. The effect of replacing $$H_0$$ by some range within $$\epsilon$$ is effectively changing the power curve. For a given effect size close to $$\epsilon$$ the power is reduced and it becomes less likely to reject the null hypothesis. It is effectively just a shift in the power.
Why are standard frequentist hypotheses so uninteresting?
They are simple basic examples that allow for easy computations. It is easier to work with them. But indeed, it is more difficult to imagine the practical relevance.
You have to crawl before you walk, and simple examples like testing a coin for bias, under a null hypothesis of zero bias, make for a teachable example for complete beginners (which every single one of us was at one point).
Jumping straight to, say, equivalence testing without even discussing easier null hypothesis significance testing seems like poor pedagogy. Where the majority of statistics education seems to suffer is when it comes to teaching the numerous limitations of hypothesis testing. After all, the OP is right: we basically always know the null hypothesis to be at least a little bit false, and we probably are more interested in something along the lines of $$H_0: \vert \mu_1-\mu_2\vert<\epsilon$$. Better integration of equivalence testing like this into early statistics curricula is an interesting idea—at least after some basics are covered.
• Is it covered anywhere though? If indeed we agree that an epsilon type hypothesis should be the one practically used, you'd expect textbooks and guidance to at least mention it and state versions of standard tests for it. Commented Nov 5, 2022 at 21:58
• @nbubis Equivalence testing probably should come up more often than it does, yes, though it depends on the field where you work. If you want to show your new medication to be more effective than that of competitors, a standard test makes sense. If your big selling point is that your medication is much cheaper, then perhaps it’s enough just to show that it has efficacy equivalent to that of competitors, and then equivalence testing might be your friend.
– Dave
Commented Nov 6, 2022 at 4:43
• There is not just one but many versions of Catch-22 here. Often hypothesis testing has to be taught early because the students will meet it soon, because your colleagues assume that you will teach it, and so on. But it's uncomfortable for everybody to introduce a body of ideas together with vast reservations and criticisms of why it is limited and even wrong: uncomfortable for teachers, uncomfortable for students. Commented Nov 6, 2022 at 13:23
• An alternative that say the entire first course or text should focus on exploratory data analysis and ideas about measurement and research design without touching on statistical inference has sometimes been tried but usually abandoned. So far as I can tell, the introductory textbooks have mostly not changed much in some decades in the balance between descriptive and inferential statistics, apart from gimmickry such as photos of happy people, fun facts, silly quizzes, and even (no names here) sprinklings of off-colour jokes because supposedly that's what students will react to. Commented Nov 6, 2022 at 13:27
• Equivalence testing is in the end equivalent to constructing confidence intervals, which are intervals of non-rejected hypothesis tests. The problem is mostly the 'null' in null hypothesis testing but not 'hypothesis testing'. Commented Nov 8, 2022 at 9:06
No model is ever true. This means that not only the null hypothesis is not true, neither is the alternative, nor something like $$|\mu_1-\mu_2|<\epsilon$$. If you're interested in which model is true, you're generally lost in model-based statistics; there's nothing particularly wrong about standard null hypotheses. Whether the $$H_0$$ or any parametric model is true is simply the wrong question.
Of course you can decide yourself what you're interested in, but I find often informative whether or not data give any evidence against a simplistic random variation "nothing meaningful is going on" model. Of course we all know that non-rejection doesn't mean the null model is true, but if you can't reject it you should really really really not claim that the data show anything meaningful, and by the way then of course you can't reject $$|\mu_1-\mu_2|<\epsilon$$ either, whatever $$\epsilon$$.
Of course you should be interested in effect sizes and not only run a null hypothesis test, so that even in case your point null is rejected you can see whether data are still compatible with a ridiculously low effect (i.e., compute a confidence interval and see whether something as small as $$\epsilon$$ is in it, in case you can specify a "critical $$\epsilon$$").
Baseline: What's "interesting" is subjective, but the rationale of testing a point null is not the question whether it's true (it isn't, but it isn't alone at that), but rather whether there's any clear "signal" in the data deviating from it. In case there is, you've got to do more to learn more.
Note particularly that it is one thing to actually reject the $$H_0$$, but quite another to just claim, in case you didn't reject it, that you'd have rejected it with more data. Particularly then you won't have any idea in which direction things will go. And also, D. Mayo made the valid point that if rejecting a $$H_0$$ were so easy indeed, why is it often so hard to replicate rejections?
Another consideration is that in fact tests will not always reject a false null hypothesis with a large enough data set, because (a) many standard frequentist tests are one sided and with a large data set things may go to the wrong side, and (b) in case the nominal model is wrong (which it always is), one can in many cases even find other models that imply a low probability to reject the null hypothesis for a range of parameters and also for big data sets, for example if you use a t-test and the true underlying distribution has very heavy tails and/or produces outliers, or if you have negative correlation between observations.
• "This means that not only the null hypothesis is not true, neither is the alternative". Given that the alternative is the negation of the null, exactly one of them must be true. Arguably both are false if the mean does not exist, but that's not something that happens all the time. Commented Nov 6, 2022 at 21:24
• @BetterthanKwora The alternative in standard parametric testing is the same parametric model but with different parameters, for example observations i.i.d. Normal$(\mu,\sigma^2)$ with values $\mu$ different from the null hypothesis. Real observations are never normal as they are always discrete. Furthermore arguably nothing in the world is ever truly independent, or identically repeatable. De Finetti wrote that "probability does not exist" (meaning objective frequentist probabilities), and he had very good arguments for that. Commented Nov 6, 2022 at 21:51
The appeal of such hypotheses lies in their simplicity and the analytical tractability (or just simplicity) of testing them.*
• In "real life" one is always only interested in some finite accuracy $$\epsilon$$, meaning the hypothesis of interest is actually of the form $$H_0: |\mu-\mu_0|<\epsilon$$.
This might be true for continuous phenomena, but not for discrete ones. E.g. in genetics a gene either has an effect on something or not. (I guess there can be even better examples than that.)
The fact that the null hypothesis cannot be rejected is by definition temporary, and is an artifact of not enough data. Given enough data, any type of equality hypothesis on continuous variables will be rejected in a real world use case.
Again, this holds for continuous phenomena only.
What we can criticize is perhaps the choice of examples in textbooks. Perhaps more examples of discrete phenomena are in order.
*Your criticism could just as well be directed at statistical models. These are often quite simple (e.g. the ubiquitous linear models), and much of their appeal is also in their tractability and simplicity of interpretation. Or even models in general, as again they are simplifications of reality with all of the drawbacks (but also benefits) that come with that.
• I noted this was true only for continuous models. I think the fundamental difference between point hypotheses and say linear regression is that the latter converge to "usefulness" as we get more data, but the former becomes less useful. Commented Nov 5, 2022 at 21:57
• @nbubis Linear regression does test point hypotheses. In particular, the most common test in a regression is if some regression coefficient (or several of them) is zero or not.
– Dave
Commented Nov 6, 2022 at 4:45
• @nbubis, I think I can see your point. At the same time, sticking to a linear model of a nonlinear real-world phenomenon as the sample size grows often implies a failure to utilize the data better. (Think about learning curves in machine learning and their use in model selection and expansion of the candidate model set.) Commented Nov 6, 2022 at 8:37
• The gene example may not be a good one. All genes compete for resources to an extent, at least in theory. (In practice this effect is so small as to be unmeasurable in many cases; this is not the same as saying that it does not exist.)
– TLW
Commented Nov 6, 2022 at 20:19
• @TLW, well then, my bad. I welcome suggestions of cleaner examples! Commented Nov 6, 2022 at 20:23
(1) The more boring a null hypothesis, the more interesting it is when you are unable to reject.
E.g. after one million flips, we still cannot distinguish the coin from perfectly unbiased. (After one million patients, we still cannot distinguish the treatment from placebo.)
(2) If your question isn't a testing question, don't use a test to answer it.
E.g. instead of a yes/no question "is the coin biased", you want to estimate "how biased is the coin". (What is the impact of a disease on life expectancy.)
To calculate a p-value, you need a null hypothesis such that the test statistic has a known distribution. Simple asserting $$|\mu|<\epsilon$$ doesn't give you a distribution.
• Neither does $|\mu|=0$. Commented Mar 2, 2023 at 11:09
• @nbubis $|\mu|=0$, with other assertions, gives a distribution. For instance, with a normality assumption, this gives you a t-distribution for $\frac {\sqrt n (\bar x - \mu)}{s}$. Commented Mar 3, 2023 at 0:34
• In "real life" one is always only interested in some finite accuracy ϵ, meaning the hypothesis of interest is actually of the form H0:|μ−μ0|<ϵ.
Indeed, it is an important practical issue that a significant effect turn out to be small and therefore "insignificant" for practical purposes. As an anecdotal example: bilingual children start speaking later than their peers in monolingual families... but this discrepancy is smaller than that between boys and girls (boys start speaking later than girls.) - The effect is statistically significant, but of too little importance to deprive your child from learning a second language from their birth.
There are procedures for testing whether the effect is bigger than some pre-specified value, see Equivalence testing. However these testing procedures are based on simpler testing procedures, which therefore must be learned beforehand.
• The equality hypothesis is a priori known to be wrong when considering continuous variables (no coin is perfectly unbiased in reality!), and as a corollary,
• The fact that the null hypothesis cannot be rejected is by definition temporary, and is an artifact of not enough data. Given enough data, any type of equality hypothesis on continuous variables will be rejected in a real world use case.
The seeming paradox arises here, because we abandon some idealizations, while preserving the other: thus, while taking into account that no coin is perfect, we still assume that a) we can make an infinite number of trials, and b) the effect is important (greater than some $$\epsilon$$.) A well known adagio about models is that "All models are wrong, but some are useful" and statistical testing is a case in point.
In short story On Exactitude in Science Jorge Luis Borges has given a well-known poetic description of why any useful model/theory is necessarily approximate, whereas the one that takes into account everything is totally useless. (See here for the full text.)
I wouldn't put too much philosophical labor into contemplating the null hypothesis per se, as the OP does.
As I have discussed here, this is a device through which we can determine whether the data allow us to assert probabilistically the direction of influence.
And the direction of influence is a heavyweight, in all worlds. | 4,489 | 20,629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.942141 |
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# Solve the following equation for x : $\tan^{-1} \bigg( \large\frac{1-x}{1+x} \bigg) = \large\frac{1}{2}\tan^{-1}(x), x > 0$
Toolbox:
• $tan(x-y)=\frac{tanx-tany}{1+tanx.tany}$
• $tan\frac{\pi}{4}=1$
Let $x=tan\theta,\:\Rightarrow\:\theta=tan^{-1}x$
Substituting the value of x in the given equation, we have
$tan^{-1}\bigg(\frac{1-tan\theta}{1+tan\theta}\bigg)=\frac{1}{2}tan^{-1}tan\theta$
We know that $tan\frac{\pi}{4}=1$
$\Rightarrow\:\large\:tan^{-1}\bigg(\large\frac{tan\frac{\pi}{4}-tan\theta}{1+tan\frac{\pi}{4}.tan\theta}\bigg)=\large\frac{1}{2}.\theta$
$\Rightarrow\:\large\:tan^{-1}tan(\frac{\pi}{4}-\theta)=\large\:\frac{1}{2}\theta$
$\Rightarrow\:\large\frac{\pi}{4}-\theta=\large\frac{\theta}{2}$
$\Rightarrow\:\large\frac{3\theta}{2}=\large\frac{\pi}{4}$
$\Rightarrow\:\large\theta=\large\frac{\pi}{6}=\large\:tan^{-1}x$
$\Rightarrow\:\large\:x=\large\:tan\frac{\pi}{6}=\large\frac{1}{\sqrt3}$
edited Mar 21, 2013 | 385 | 949 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-50 | longest | en | 0.311821 |
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# econ212 practice problems1 - econ 212 Practice Problems 1...
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econ 212 Practice Problems 1. True/False Indicate whether the statement is true or false AND EXPLAIN WHY BRIEFLY. ____ 1. At the optimum, a consumer’s indifference curve must be tangent to her budget line. ____ 2. Max Gross has the utility function U ( x , y ) = max{ x , y } . If the price of x is the same as the price of y , Max will buy equal amounts of x and y . ____ 3. Clara’s utility function is U ( x , y ) = ( x + 2)( y + 1). If her consumption of both x and y are doubled, then her marginal rate of substitution between x and y remains constant. ____ 4. Charlie’s utility function is U ( x , y ) = xy 2 . His marginal rate of substitution between x and y does not change if the amount of both goods doubles. ____ 5. Ambrose’s utility function is U ( x , y ) = x + 4 y 1/2 . The price of x is \$1 and the price of y is \$2. If his income rises from \$100 to \$150, his consumption of y increases by more than 10% but less than 50%. ____ 6. Linus has utility function U ( x , y ) = x + 2 y . If the price of x is \$1 and the price of y is .50 cents then Linus must consume equal amounts of both goods in order to maximize his utility.
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Ask a homework question - tutors are online | 470 | 1,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-13 | latest | en | 0.839408 |
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# RiskChapterHWKey - Homework Solutions Introduction to Risk...
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Homework Solutions Introduction to Risk, Return, and the Opportunity Cost of Capital 17. a.Interest rates tend to fall at the outset of a recession and rise during boom periods. Because bond prices move inversely with interest rates, bonds provide higher returns during recessions when interest rates fall. b. r stock = [0.2 × ( - 5%)] + (0.6 × 15%) + (0.2 × 25%) = 13.0% r bonds = (0.2 × 14%) + (0.6 × 8%) + (0.2 × 4%) = 8.4% Variance(stocks) = [0.2 × ( - 5 - 13) 2 ] + [0.6 × (15 - 13) 2 ] + [0.2 × (25 – 13) 2 ] = 96 Standard deviation = % 80 . 9 96 = Variance(bonds) = [0.2 × (14 - 8.4) 2 ] + [0.6 × (8 - 8.4) 2 ] + [0.2 × (4 - 8.4) 2 ] = 10.24 Standard deviation = % 20 . 3 24 . 10 = c.Stocks have both higher expected return and higher volatility. More risk averse investors will choose bonds, while those who are less risk averse might choose stocks. 10-1
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20. Risk reduction is most pronounced when the stock returns vary against each other. When one firm does poorly, the other will tend to do well, thereby stabilizing the
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Ask a homework question - tutors are online | 478 | 1,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-51 | latest | en | 0.851186 |
https://betterlesson.com/lesson/reflection/16532/having-fun | 1,511,443,601,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806832.87/warc/CC-MAIN-20171123123458-20171123143458-00328.warc.gz | 591,826,959 | 18,263 | ## Reflection: Joy Unit Review Game: Trashball - Section 1: Review Game
Math class should always be a happy place but I notice more smiles on the days that we play review games. There is nothing better than having a student hit the 50 point shot during Trashball and having the class erupt in applause!
I definitely write the review game questions to prepare students for the upcoming assessment, but it is also a time to have some fun and get excited about a paper ball falling into a garbage can. I instill in students that they will get nothing from winning this game and the objective is to practice their math and have some fun as a class.
Having Fun
Joy: Having Fun
# Unit Review Game: Trashball
Unit 1: Functioning with Functions
Lesson 10 of 11
## Big Idea: Use Trashball to review the important concepts of this unit.
Print Lesson
1 teacher likes this lesson
Standards:
Subject(s):
Math, greatest integer functions, Precalculus and Calculus, combine functions, review game, piecewise-defined functions, relative extrema
55 minutes
### Tim Marley
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# Essence Collector97 (50)
### Find 3 butterflies.
• Unlocked by 337 tracked gamers (27% - TA Ratio = 1.93) 1,252
## Achievement Guide for Essence Collector
Solution
This achievement is missable
If you collect the butterflies in the same order as they appear in the game, you should unlock your achievement after finding these first 3 butterflies:
1. ER Entrance - Bathroom
Explanation: Bathrooms near the beginning of the game. Claire even mentions it being there.
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2. ER Break Room - Clock Puzzle
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Hope this helps you guys! | 274 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-51 | latest | en | 0.834282 |
https://cooking.stackexchange.com/questions/102939/how-much-caffeine-would-there-be-if-i-reuse-tea-leaves-in-a-second-brewing/102946 | 1,618,113,518,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00239.warc.gz | 305,669,005 | 39,338 | # How much caffeine would there be if I reuse tea leaves in a second brewing?
If you look for information about the amount of caffeine present in a cup of tea, the amounts vary depending on the kind of tea, if the tea is bagged or loose leaves, the brand, and other factors. Nonetheless, the values estimated range from the 25 mg in a cup of white tea to 50 mg in a cup of black tea, or maybe more.
Let's say I use 2 grams of black tea leaves and I brew them in 200 ml of almost-boiling water for the reccomended time of 5 minutes. I get a cup of tea with an estimated amount of caffeine of 50 mg. But then I reuse those same leaves and I brew them again in another 200 ml of water.
What would be the estimated amount of caffeine in that second brewing? Are there estimates for that? Can the result be at least expressed in an estimated percentage with respect to the amount obtained in the first brewing? Would that percentage be the same for other kinds of tea (green, red, oolong...)?
Related:
In general it looks like 65-75% of the total caffeine comes out in the first steeping, while 20-25% comes out in the second steeping. This was addressed in this paper which examined different types of tea. The results are summarized in this table. For more details check out this reddit thread.
• I hope they followed the international standard of making tea in the study, en.wikipedia.org/wiki/ISO_3103 – Per Alexandersson Oct 18 '19 at 19:41
• Mind quoting a few of the points from the links, in the event that the URLs become un-reachable or otherwise not-clickable? – BruceWayne Oct 18 '19 at 20:27
• @Per Alexandersson 6 mins and boiling water, it is unusable for green or oolong tea with colder water and multiple short brewing. Higher grains (loose) and green/oolong(colder) mean lower percentage in the 1st brewing, what the table shows. Additionally, brack tea is seldom used for multiple brewing. – Poutnik Oct 19 '19 at 11:57
The answer to this depends a lot on variables in the steeping process. Steeping at a higher temperature will remove more caffeine. Steeping for longer will remove more caffeine. Doing either or both of these will leave less caffeine for a second or third (or more) steeping. Using whole leaves can slow down caffeine extraction, while using fannings (as in tea bags) can speed it up.
For example, some green and white teas are steeped repeatedly for very short (~30-second) intervals. The water temperature is often quite a bit below boiling as well. In that case, the second steeping can potentially contain more caffeine than the first (as I noted in a previous answer). For a longer initial steep, the caffeine extraction for that first brewing will gradually increase, though you will get diminishing returns for very long steeps (as discussed here).
Note also that these steeping conditions significantly affect perception of caffeine content and actual extraction. As described in a previous answer, white tea is often thought to have less caffeine than black tea (mentioned in the question), but that is likely due to the fact that white and green teas are typically brewed at lower temperatures. The total amount of caffeine you might extract from a white or green tea in multiple infusions may be roughly the same as black tea. And when all types of tea are brewed under similar conditions, they will generally extract similar amounts of caffeine.
Thus, there's no simple equation to describe the relationship between the caffeine content of multiple brewings. The higher the temperature and the longer the first steep, the more caffeine will be extracted initially, leaving less caffeine available for subsequent brews made with the same leaves.
If you follow a procedure as suggested in the question (5 minutes, almost boiling water), Dugan's answer of roughly 70% caffeine extraction is a pretty reasonable estimate. In general, in those conditions, I'd say a good rule of thumb is that each steeping will extract about 2/3 of the caffeine left in the leaves. (So, first steeping is roughly 67% leaving about 33% in the leaves. Second is roughly 22%, leaving about 11% in the leaves. Third is roughly 7-8%, leaving only a tiny about of caffeine left.)
But again, that only applies to that specific scenario. Various tea studies Dugan and I have linked to can give estimates for other conditions. | 960 | 4,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-17 | latest | en | 0.914816 |
https://community.smartsheet.com/discussion/33621/sumifs-with-vlookup | 1,716,008,009,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00255.warc.gz | 162,682,155 | 106,559 | SUMIFS with VLOOKUP
Options
edited 12/09/19
On a sheet I call "Project Group Hour Totals", I am trying to add the hours from a sheet I call "Project Time Tracker", using VLOOKUP to find both my Project Category and my Project group. My Project Time Tracker Range 1 is the hours column on the project time tracker, and Range 2 includes 6 columns. The project Number being the first and the group being the last.
I've attempted this with SUMIF and AND, which returned an incorrect argument error.
=SUMIF(AND({Project Time Tracker Range 1}, VLOOKUP([Project Category]10, {Project Time Tracker Range 2}, 1, false), VLOOKUP([# of Emp / Group]10, {Project Time Tracker Range 2}, 6, false)))
I've also tried SUMIFS, which returen Unparseable
=SUMIFS({Project Time Tracker Range 1}, (VLOOKUP([Project Category]10, {Project Time Tracker Range 2},1,false),(VLOOKUP([# of Emp / Group]10,{Project Time Tracker Range 2},6,False))
• Employee
Options
Hello,
Thanks for reaching out! It sounds like you are trying to create a formula that will sum the values in the hours column from your sheet "Project Time Tracker” if all the evaluated criteria is true.
If so, then you are correct that using a SUMIFS statement would add the values in the hours range that meet the specified Project Category and Project Group criteria. Since the SUMIFS function sums numbers within a range that meet multiple criteria, you do not need to include the AND or VLOOKUP functions.
Here is an example of what the formula might look like in your sheet:
=SUMIFS({Project Time Tracker Range 1}, {Project Time Tracker Range 2}, [Project Category]@row, {Project Time Tracker Range 3}, [# of Emp / Group]@row)
In this case, the Project Time Tracker Range 1 is referring to the Hours column (the group of cells to sum), the Project Time Tracker Range 2 is referring to the Project Category column and the Project Time Tracker Range 3 is referring to the Group column in your "Project Time Tracker" sheet.
Additionally, to possibly improve performance in your sheets, I recommend substituting the row numbers with @row in your cell references. If a row is moved Smartsheet won’t need to modify the cell references, which can result in quicker sheet load and save times. You can also copy and paste this formula without having to manually change the row numbers on cell references, saving you time when you need to copy your formulas. | 564 | 2,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | latest | en | 0.833982 |
http://newton.dep.anl.gov/askasci/math99/math99223.htm | 1,411,257,825,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657134114.26/warc/CC-MAIN-20140914011214-00103-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 197,893,566 | 3,757 | Sums of Irrational Numbers ```Name: Jehanzeb Status: student Age: N/A Location: N/A Country: N/A Date: 9/25/2005 ``` Question: Is the sum of two irrational numbers equal to a rational number? Replies: The sum of two rational numbers, p/q + n/m = (pm +qn)/qm. If either p/q or n/m = X or Y, respectively, and either X or Y cannot be expressed as a ratio, then the sum cannot be expressed as a ratio. However, an irrational number can be expressed as an INFINITE sum of rational numbers. This was first noted by Newton (1676), but Euler gave the first proof (1774). So the sum of two IRRATIONAL numbers can be expressed as an INFINITE SUM of rational numbers. Vince Calder Jehanzeb, Two irrational numbers very RARELY add up to a rational number. Any integer, finite decimal, or repeating decimal is a rational number. A rational number can be represented as a fraction. An irrational number cannot. It IS true that two RATIONAL numbers add up to a rational number: two fractions always add up to a fraction. Here are two examples: (1/2)+(1/3)=(5/6), 1+3=4. The latter applies because you can represent it as (1/1)+(3/1)=(4/1). However, sqrt(2) is not rational because there is no fraction, no ratio of integers, that will equal sqrt(2). It calculates to be a decimal that never repeats and never ends. The same can be said for sqrt(3). Also, there is no way to write sqrt(2)+sqrt(3) as a fraction. In fact, the representation is already in its simplest form. To get two irrational numbers to add up to a rational number, you need to add irrational numbers such as [1+sqrt(2)] and [1-sqrt(2)]. In this case, the irrational portions just happen to cancel out, leaving: [1+sqrt(2)]+[1-sqrt(2)]=2. 2 is a rational number (i.e. 2/1). Dr. Ken Mellendorf Physics Instructor Illinois Central College Click here to return to the Mathematics Archives
NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.
For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs | 546 | 2,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2014-41 | longest | en | 0.91165 |
jaiden3n1a5.qowap.com | 1,544,840,226,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00634.warc.gz | 144,462,538 | 6,768 | # The Ultimate Guide To series
Apparently, even though the harmonic series diverges to infinity, the alternating harmonic series converges for the all-natural logarithm of 2,
Alternatively, Should the sequence of partial sums isn't going to converge into a Restrict (e.g., it oscillates or ways ), the series is said to diverge. An example of a convergent series would be the geometric series
soap opera - a serialized system commonly working with sentimentalized relatives matters which is broadcast on radio or television (usually sponsored by a firm promoting soap solutions)
Numerous strategies generally known as convergence tests can be employed to ascertain no matter whether a given series converges. Although conditions of a series can have possibly signal, convergence Attributes can frequently be computed while in the "worst case" of all phrases getting good, and afterwards applied to the particular series at hand.
Given that we fully grasp what a sequence is, we're going to consider what occurs into the phrases of the sequence at infinity (do they solution 0, a finite value, or +- infinity?).
to publish or broadcast as a serial. reeks van maak يَنْشُر قِصَّة على شَكْلٍ مُسَلْسَل издавам на части publicar em episódios vydat/vysílat na pokračování die Serie udgive som serie; sende som serie παρουσιάζω σε συνέχειες publicar/televisar por entregas järjeloona avaldama, sarjana esitama به صورت مجموعه منتشر يا پخش كردن julkaista sarjana publier/diffuser en feuilleton לְַסֵ��/לְשַׁדֵר כְּסִדרָה धरावहि�� से प्रकाशित या प्रसारित करना pretvoriti u seriju, prikazati u obliku serije folytatásokban közöl disiarkan bersambung birta sem framhaldssögu/-þátt pubblicare/trasmettere a puntate 連載する 연재하다 leisti atskiromis dalimis/tęstinį leidinį, statyti serialą izlaist sērijās/turpinājumos menerbitkan sesuatu sebagai cerita bersiri in afleveringen laten verschijnen sende ut som serie wydawać/nadawać w odcinkach په يوه مجموعي توګه خپرول publicar em episódios a publica/a difuza în foileton/în episoade издавать выпусками/сериями vydať / vysielať na pokračovanie objavljati, predvajati v nadaljevanjih objaviti u nastavcima publicera (sända) som fileöljetong (serie) พิมพ์หรือส่งกระจายเสียงเรื่องออกมาเป็นตอน ๆ dizi hâlinde yayınlamak 連載出版，連續播出 видавати випусками سلسلہ وار سیریل کے طور پر پیش یا نشر کرنا phát hành 连载或连续广播
A bunch of rock formations intently related in time of origin and distinct as a group from other formations.
Samples of this series involve works through the Chronicles of Narnia, where by the fifth guide released, The Horse and His Boy, is definitely established over the time of the very first ebook, and the sixth guide revealed, The Magician's Nephew is in fact set extended ahead of the to start with guide. This was accomplished intentionally by C. S. Lewis, a medieval literature scholar. Medieval literature didn't usually explain to a story chronologically.
At any position, the contestant in control of the wheel could spin again, ask to buy a vowel (at which issue \$250 was deducted from their rating, and only if they'd a minimum of \$250) or try to solve the puzzle; pretty early from the demonstrate's operate, a participant had to land on the Purchase a Vowel Area to be able to purchase a vowel, but this concept was scrapped right before Wheel concluded its 1st thirty day period within the air.
The planet on the Vikings is brought to everyday living in the journey of Ragnar Lothbrok, the primary Viking to arise from Norse legend and on to the web pages of background - a person on the sting of fantasy.
a. A succession of publications that present website an prolonged narrative, such as a comic e-book series, or that have very similar subjects or identical formats, like a series of cookbooks.
No matter in case you sit at the rear of the wheel or powering the driving force, your knowledge inside the 7 Series will probably be unforgettable.
Right after an eerie mist rolls into a little city, the citizens should battle the mysterious mist and its threats, preventing to maintain morality and sanity.
Set 97 yrs after a nuclear war has ruined civilization, every time a spaceship housing humanity's lone survivors sends 100 juvenile delinquents back to Earth in hopes of maybe re-populating the Earth. | 1,100 | 4,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-51 | latest | en | 0.365852 |
https://electronicsbeliever.com/led-driver-circuit-explained/ | 1,713,113,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00593.warc.gz | 203,689,065 | 34,281 | # LED Driver Circuit Explained and Available Solutions
The days of incandescent lamps are over. LED lighting takes over nowadays as they are much more energy efficient. LED lights on the other hand require a good controlling circuit to operate correctly and this is the so called LED driver circuit. LEDs are basically a form of diode that gives off light when forward biased. A diode is rated by its forward voltage of 0.3V or 0.7V for germanium and silicon respectively. For LED lights, the forward voltage is higher than the diode and may be typically as high as 2V-3.5V, per LED. Some LEDs that has higher voltage specified is already combination of several LEDs.
LEDs are DC in nature, but why LEDs lights are being used directly in replacement for incandescent and CFL lamps in AC socket? This is made possible using LED driver circuit. A LED driver circuit will convert AC into DC level that is safely usable by the LEDs. There are some available solutions out there for LED driver circuit. LED drivers are can be linear or switching. Let us familiarize these solutions.
# Linear LED Driver Circuit
Linear LED driver circuit uses a linear device to control the current to the LEDs. This circuit solution is not efficient at all and limited to small power application only. A linear LED driver is can be a simple voltage source and a current limiting resistor only; it is indeed very simple that is why still a popular solution of driving LEDs. Another advantage of a linear LED driver is that it can provide very clean light, I mean for clean is that there is no blurring or flickering effect.
### Simple Linear LED Driver Circuit
Below circuit is a very simple way of driving LEDs. Basically it is only composed of a DC voltage source and a limiting resistor, Rlimit. However, in this solution, the voltage source must be a pure DC or linear level so that the current setting to the LEDs will not change. In the event that the current to the LEDs will vary, the illumination will somewhat shows intensity change and this is not nice to see by the eyes. Another drawback of a changing LED current is that the LEDs may overheat and get damage.
In the above example circuit, the voltage source is a pure DC and the LED current set by the limiting resistor is 600mA. This gives a total LED power of 8.332W. The current limiting resistor is dissipating 3.67W. The total power applied to the circuit is 12W and the efficiency is only 69.43% which is very low.
LED Efficiency = 8.332W / 12W = 69.43%
### Linear Regulator as LED Driver
Above example is very basic and elementary approach of driving LEDs. In the event of a varying voltage source, a linear regulator can be used. A linear regulator is able to accept varying input voltage while making the output voltage constant. It is still a lossy solution of driving LEDs but better than the first approach in terms of current stability to the LEDs.
Below diagram is a typical linear regulator circuit. VOUT is the node where the load is applied and it is regulated to a voltage level set by the user. Supposing the input voltage range is 9-16V, the output voltage will remain the same; for instance 7.5V per setting. When the difference between the input and output is huge, the linear regulator dissipates huge power in order to maintain a regulated output voltage. The property of a linear regulator to maintain output voltage is making it popular to driving LEDs.
Below is a LED driver circuit using a linear regulator from Linear Technology, LT1083-12. The output of this regulator is 12V fixed. Still, a series resistor is needed to set a current level safe for the LEDs. The LED current in this circuit is 261.6mA.
LED Current = (12V – (3 X 3.128V)) / 10Ω = 261.6mA
The LED power is only 2.452W.
Power LED = 3 X 3.128V X 261.6mA = 2.45W
The power dissipated by the limit resistor is 0.684W.
Power Limit Resistor = (261.6mA)2 X 10Ω = 0.684W
The power dissipated by the linear regulator is
Power Regulator = (VIN – VOUT) X (LED Current + quiescent current) = (16V-12V) X (261.6mA +5mA)= 1.0664W.
(Quiescent current is specified in the regulator datasheet. This is only a small value and can be neglected most of the times for easier calculation.)
The circuit efficiency is
Circuit Efficiency = Power LED / (Power LED + Power Limit Resistor + Power Regulator) = 2.45W / (2.45W + 0.684W + 1.0664W) = 58.33%
The efficiency is very poor the same with the previous solution. The efficiency will further decrease when operating at higher input voltage.
### Specialized Linear LED Controller
There are specific linear ICs designed solely for LED driver application. The concept and analysis on the power section side however is
the same on with the above example.
These ICs advantage is in terms of multiple LED strings driving capability and built-in protections for short and open LEDs. Another advantage is the inclusion of a dimming function. A generic linear regulator does not have a dimming function.
One example of this solution is BD8374HFP-M from ROHM semiconductor. Below is the application circuit. It is only a single channel, with dimming capability, open and short LED protection, over voltage and over temperature protection.
For this controller, the way to set a LED current is through the RVIN_F resistor. This resistor is located in the input as opposed to the previous examples above that is located in series to the LEDs. In this solution, the LED voltage will set the output voltage of the controller IC. In using a typical voltage regulator, the output is a fixed voltage but here, the output is variable depending to the total LED forward voltage.
The total LED power is simply the sum of the LED forward voltages times the IOUT or set LED current by the resistor RVIN_F. The power dissipated by the linear IC (BD8374HFP-M) is the difference between the input voltage and total LED forward voltage drop, times the output set current. On the other hand, the power dissipation of the current setting resistor RVIN_F is just its voltage drop times the output current or the square of the output current times the resistance. The efficiency calculation is can be done the same from the above example.
# Switch Mode LED Driver Circuit
In linear mode type of LED driver, the input voltage variation is narrow as limited by the power dissipation of the linear controller. The losses are huge as well in the linear solution. These shortcomings are solved by switch mode type of LED driver. A switch mode LED driver is can be a step-down (buck), step-up (boost) or combination (buck-boost). A switch mode LED driver is can be used directly from a universal AC line; say 90-264Vrms.
### Switch Mode Principle
Switch mode means that the controlling device is operating in continuous switching state between turn on and turn off of a switching device like MOSFET or BJT. At turn on of the switch, there is ideally zero resistance so ideally zero power loss. At turn off on the other hand, there is ideally zero current thus no power loss as well. These behaviors make the switch mode solution more efficient than the linear solution. However, switch mode approach is more complicated than the linear solution and will cost higher.
### Buck Converter Derived LED Driver
Below is a generic buck converter power section circuit diagram. Buck converter is a step down switching converter. Its output is always lower than its input. The MOSFET Q1 is driven into saturation and cut-off by the PWM signal in order to generate an output voltage. The inductor L1 is serving as an energy storage which charges when the MOSFET Q1 is driven into saturation. It discharges when the MOSFET Q1 is driven into cut-off.
The capacitor C1 is serving as a reservoir also to minimize voltage fluctuations in the output rail. It charges when Q1 is driven into saturation while discharges when Q1 is driven into cut-off. The diode D1 serves as a path for the inductor current when it discharges, it is only functioning once the MOSFET Q1 is at cut-off state.
Both the MOSFET and the diode will conduct only a portion of the switching period. The relation between input and output voltage is define by the so called duty cycle. The ideal duty cycle of a buck converter is
Duty Cycle, Buck = Vout / Vin
### Example Working Buck Derived LED Driver Circuit
Below circuit is LED driver circuit based on buck topology. It is working pretty well in simulation so in actual. The controlling device is LT3474 from Linear technology.
The power path is from IN, to U1 internal switch (Q1 in the generic buck converter above), to L1 and C3 (C1 in the generic buck converter above). D1 is the inductor discharge path diode as with the D1 in the generic buck converter circuit above. The circuit allows wide variation in the input voltage in contrast to the linear solution.
The calculations of the power section of this driver circuit are the same to the generic buck converter we discussed above. This LED driver circuit has a PWM dimming capability by applying PWM signal on the PWM pin.
The simulated LED current with PWM dimming is:
As you can see on the above waveform, the LED voltage which is the output voltage of the buck is less than the input voltage which is 10V, as Buck is a step down converter. The LED current is modulated to achieve dimming.
### Boost Converter Derived LED Driver
Below is a typical boost converter power section circuit. Q1 is modulated and operates in saturation and cut-off in a fast manner. The same with the buck converter, the switching device will have ideal zero loss as during saturation there is no resistance while there is no current during cut-off ideally. When Q1 is turns on, L1 will charge and D1 is reversed biased. When Q1 turns off, L1 will reverses polarity and forward bias D1 then current will reach the output node. C1 serves as a reservoir so that there is still energy delivered to the load when the inductor is charging. Boost converter is also a duty cycle controlled, its ideal duty cycle equation is:
Duty Cycle, Boost = 1 – (VIN / VOUT)
### Example Working Boost Derived LED Driver Circuit
Below circuit is a simple LED driver derived from a boost converter.
With a boost driver, the input must always be lower compared to the total forward voltage of the LEDs. In this circuit, the input voltage is 3 while the total LED voltage is 9.64V based on simulation.
### Buck-Boost Derived LED Driver
If the application needs a very wide voltage range that cannot be accommodated by a boost or a buck alone, then consider using buck-boost derived LED driver. An example of this is below circuit from Linear Technology.
### LED Driver Circuit Derived from AC Line
What we discussed solutions above are all DCDC applications. How about if we want a LED light that we can directly plug to an AC socket, like the commercial LED lights available nowadays, what should we do? This tie we need another LED driver circuit suited for ACDC purpose. There are few things to make this possible.
### Non-Isolated Lossy ACDC LED Driver
Below circuit is a simple non-isolated ACDC LED driver. It consists only with passive devices and the Zener and diode. This is an economical solution but not efficient and safe to use. Be careful.
### Non-Isolated Non-Lossy ACDC LED Driver
Below solution is still non-isolated as there is no presence of isolation transformer. This is a solution provided by Richtek using RT8402 controller. However, this driver is efficient compared to the first circuit above. This particular solution is a buck
derived AC-DC LED driver. The bridge rectifier converts AC into DC and Q1, D1, L1 and EC1 are the buck converter power section. This is an efficient driver since Q1 is operating between saturation and cut-off. Still, be careful, this solution is non-isolated.
Another solution from Richtek using RT8487 controller:
Both solutions are commonly used in commercial low power and low cost LED lamps.
### Isolated Non-Lossy ACDC LED Driver using Flyback Derived Topology
For high power LED lights or lamps, below circuit is preferable. This is a solution from Richtek using RT7306. It is a flyback derived LED driver. The presence of the transformer introduces isolation between the AC line and the LEDs. There is no danger for electric shock if you accidentally touch the output side.
Being a flyback topology, the driver is able to handle wide input voltage range from 90-264Vac. This solution is also efficient at power less than 50W. However, at power more than 50W, the efficiency may decrease but still high enough compared to the linear solution.
1. Gavin Perry says:
Nice work,covered the range of options. I’d give other examples than LT, they are pricey. For linear look st the BCR40x series. Diodes Inc. is now sourcing them (along with Infineon who has had them for a long time) so may be more affordable.
I need learn led drivers. i am tv tecnision
sins 1995. how is aproch tutorial where is your institute please send details
narsipatnam
phone 9493567282
1. RAJAN SAKA says:
must follow books for smps circuits like I pressman
3. VIJAY SURYAWANSHI says:
I have to manufacture comercial basis all types of drives pl suggest required componant , and its source and PCB
1. electronicsbeliever says:
Hi,
There are many solutions out there available. basically when you visit controllers supplier like Analog Devices (they own now the linear technology), Texas Instruments, you can find existing designs from there datasheet. You can refer to the for a start.
4. Hakan says:
I built up Non-Isolated Lossy ACDC LED Driver circuit but not succeeded. At the led side i measured 2.7 V. But it did not lit a 5mm led also.
1. electronicsbeliever says:
Hakan, | 3,056 | 13,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-18 | longest | en | 0.948604 |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A000140 Kendall-Mann numbers: the maximal number of inversions in a permutation on n letters is floor(n(n-1)/4); a(n) is the number of permutations with this many inversions. (Formerly M1665 N0655) 13
%I M1665 N0655
%S 1,1,2,6,22,101,573,3836,29228,250749,2409581,25598186,296643390,
%T 3727542188,50626553988,738680521142,11501573822788,190418421447330,
%U 3344822488498265,62119523114983224,1214967840930909302
%N Kendall-Mann numbers: the maximal number of inversions in a permutation on n letters is floor(n(n-1)/4); a(n) is the number of permutations with this many inversions.
%C Row maxima of A008302, see example.
%C The term a(0) would be 1: the empty product is one and there is just one coefficient 1=x^0, corresponding to the 1 empty permutation (which has 0 inversions).
%C From Ryen Lapham and _Anant Godbole_, Dec 12 2006: (Start)
%C Also, the number of permutations on {1,2,...,n} for which the number A of monotone increasing subsequences of length 2 and the number D of monotone decreasing 2-subsequences are as close to each other as possible, i.e., 0 or 1. We call such permutations 2-balanced.
%C If 4|n(n-1) then (with A and D as above) the feasible values of A-D are C(n,2), C(n,2)-2,...,2,0,-2,...,-C(n,2), whereas if 4 does not divide n(n-1), A-D may equal C(n,2), C(n,2)-2,...,1,-1,...,-C(n,2). Let a_n(i) equal the number of permutations with A-D the i-th highest feasible value.
%C The sequence in question gives the number of permutations for which A-D=0 or A-D=1, i.e., it equals A_n(j) where j = floor((binomial(n,2)+2)/2). Here is the recursion: a_n(i) = a_n(i-1) + a_{n-1}(i) for 1 <= i <= n and a_n(n+k) = a_n(n+k-1) + a_{n-1}(n+k) - a_n(k) for k >= 1. (End)
%C The only two primes found < 301 are for n = 3 and 6.
%C a(n+1)/a(n) = n - 1/2 + O(1/n^{1-epsilon}) as n --> infinity (compare with A008302, A181609, A001147). - _Mikhail Gaichenkov_, Apr 11 2014
%C Define an ordered list to have n terms with terms t(k) for k=1..n. Specify that t(k) ranges from 1 to k, hence the third term t(3) can be 1, 2, or 3. Find all sums of the terms for all n! allowable arrangements to obtain a maximum sum for the greatest number of arrangements. This number is a(n). For n=4, the maximum sum 7 appears in 6 arrangements: 1114, 1123, 1213, 1222, 1231, and 1132. - _J. M. Bergot_, May 14 2015
%D F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 241.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Robert Israel, Robert G. Wilson v and N. J. A. Sloane <a href="/A000140/b000140.txt">Table of n, a(n) for n = 1..400</a> (a(1) to a(61) from Sloane, a(62) to a(350) from Wilson).
%H D. Foata, <a href="http://dx.doi.org/10.1007/978-94-010-1220-1_2">Distributions eulériennes et mahoniennes sur le groupe des permutations</a>, pp. 27-49 of M. Aigner, editor, Higher Combinatorics, Reidel, Dordrecht, Holland, 1977.
%H M. Gaichenkov, <a href="http://mathoverflow.net/questions/46368/the-property-of-kendall-mann-numbers">The property of Kendall-Mann numbers</a>, MathOverflow, 2010.
%H R. K. Guy, <a href="/A000707/a000707_2.pdf">Letter to N. J. A. Sloane with attachment, Mar 1988</a>
%H A. Waksman, <a href="http://dx.doi.org/10.1109/T-C.1970.222866">On the complexity of inversions</a>, IEEE Trans. Computers, 19 (1970), 1225-1226.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Inversion_(discrete_mathematics)">Inversion</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Q-Pochhammer_symbol">q-Pochhammer symbol</a> - _Paul Muljadi_, Jan 18 2011
%H <a href="/index/Cor#core">Index entries for "core" sequences</a>
%F Largest coefficient of (1)(x+1)(x^2+x+1)...(x^(n-1)+...+x+1). - _David W. Wilson_
%F The number of terms is given in A000124.
%F Asymptotics (Mikhail Gaichenkov, 2010): a(n) ~ 6 * n^(n-1) / exp(n). - _Vaclav Kotesovec_, May 17 2015
%e From _Joerg Arndt_, Jan 16 2011: (Start)
%e a(4) = 6 because the among the permutations of 4 elements those with 3 inversions are the most frequent and appear 6 times:
%e [inv. table] [permutation] number of inversions
%e 0: [ 0 0 0 ] [ 0 1 2 3 ] 0
%e 1: [ 1 0 0 ] [ 1 0 2 3 ] 1
%e 2: [ 0 1 0 ] [ 0 2 1 3 ] 1
%e 3: [ 1 1 0 ] [ 2 0 1 3 ] 2
%e 4: [ 0 2 0 ] [ 1 2 0 3 ] 2
%e 5: [ 1 2 0 ] [ 2 1 0 3 ] 3 (*)
%e 6: [ 0 0 1 ] [ 0 1 3 2 ] 1
%e 7: [ 1 0 1 ] [ 1 0 3 2 ] 2
%e 8: [ 0 1 1 ] [ 0 3 1 2 ] 2
%e 9: [ 1 1 1 ] [ 3 0 1 2 ] 3 (*)
%e 10: [ 0 2 1 ] [ 1 3 0 2 ] 3 (*)
%e 11: [ 1 2 1 ] [ 3 1 0 2 ] 4
%e 12: [ 0 0 2 ] [ 0 2 3 1 ] 2
%e 13: [ 1 0 2 ] [ 2 0 3 1 ] 3 (*)
%e 14: [ 0 1 2 ] [ 0 3 2 1 ] 3 (*)
%e 15: [ 1 1 2 ] [ 3 0 2 1 ] 4
%e 16: [ 0 2 2 ] [ 2 3 0 1 ] 4
%e 17: [ 1 2 2 ] [ 3 2 0 1 ] 5
%e 18: [ 0 0 3 ] [ 1 2 3 0 ] 3 (*)
%e 19: [ 1 0 3 ] [ 2 1 3 0 ] 4
%e 20: [ 0 1 3 ] [ 1 3 2 0 ] 4
%e 21: [ 1 1 3 ] [ 3 1 2 0 ] 5
%e 22: [ 0 2 3 ] [ 2 3 1 0 ] 5
%e 23: [ 1 2 3 ] [ 3 2 1 0 ] 6
%e The statistics are reflected by the coefficients of the polynomial
%e (1+x)*(1+x+x^2)*(1+x+x^2+x^3) ==
%e x^6 + 3*x^5 + 5*x^4 + 6*x^3 + 5*x^2 + 3*x^1 + 1*x^0
%e There is 1 permutation (the identity) with 0 inversions,
%e 3 permutations with 1 inversion, 5 with 2 inversions,
%e 6 with 3 inversions (the most frequent, marked with (*) ), 5 with 4 inversions, 3 with 5 inversions, and one with 6 inversions. (End)
%e G.f. = x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + 101*x^6 + 573*x^7 + 3836*x^8 + ...
%p f := 1: for n from 0 to 40 do f := f*add(x^i, i=0..n): s := series(f, x, n*(n+1)/2+1): m := max(coeff(s, x, j) \$ j=0..n*(n+1)/2): printf(`%d,`,m) od: # from _James A. Sellers_, Dec 07 2000 [offset is off by 1 - _N. J. A. Sloane_, May 23 2006]
%p P:= [1]: a[1]:= 1:
%p for n from 2 to 100 do
%p a[n]:= max(eval(convert(P,list),x=1));
%p od:
%p seq(a[i],i=1..100); # _Robert Israel_, Dec 14 2014
%t f[n_] := Max@ CoefficientList[ Expand@ Product[ Sum[x^i, {i, 0, j}], {j, n-1}], x]; Array[f, 20]
%t Flatten[{1, 1, Table[Coefficient[Expand[Product[Sum[x^k, {k, 0, m-1}], {m, 1, n}]], x^Floor[n*(n-1)/4]], {n, 3, 20}]}] (* _Vaclav Kotesovec_, May 13 2016 *)
%t Table[SeriesCoefficient[QPochhammer[x, x, n]/(1-x)^n, {x, 0, Floor[n*(n-1)/4]}], {n, 1, 20}] (* _Vaclav Kotesovec_, May 13 2016 *)
%o (PARI)
%o a(n)=
%o /* return largest coefficient in product (1)(x+1)(x^2+x+1)...(x^(n-1)+...+x+1) */
%o { local(p,v);
%o p=prod(k=1,n-1,sum(j=0,k,x^j)); /* polynomial */
%o v=Vec(p); /* vector of coefficients */
%o v=vecsort(v); /* sort so largest is last element */
%o return(v[#v]); /* return last == largest */
%o }
%o vector(22,n,a(n)) /* _Joerg Arndt_, Jan 16 2011 */
%o (MAGMA) /* based on _David W. Wilson_'s formula */ PS<x>:=PowerSeriesRing(Integers()); [ Max(Coefficients(&*[&+[ x^i: i in [0..j] ]: j in [0..n-1] ])): n in [1..21] ]; // _Klaus Brockhaus_, Jan 18 2011
%o (PARI) {a(n) = if( n<0, 0, vecmax( Vec( prod(k=1, n, 1 - x^k) / (1 - x)^n)))}; /* _Michael Somos_, Apr 21 2014 */
%Y Row maxima of A008302.
%Y Odd terms are A186888.
%Y Cf. A128566.
%K nonn,easy,core,nice
%O 1,3
%A _N. J. A. Sloane_
%E Edited by _N. J. A. Sloane_, Mar 05 2011
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### How to ranging variables
Posted: Tue Mar 26, 2019 4:46 pm
Hi all,
I'm having trouble with variables that the model chooses to give them very small values but not zero, values that are not reasonable for these variables. My question is how can I give the model a range for these variables such that: on the one hand they can be zero and on the other hand they can not be for example smaller than 0.1? Something like: x=0 or x>0.1.
Currently, these variables are defined as positive variables.
Thanks,
Yehuda
### Re: How to ranging variables
Posted: Wed Mar 27, 2019 11:31 am
Hi
I'm afraid you did not explain your problem well.
would you mind making it more clear?
### Re: How to ranging variables
Posted: Wed Mar 27, 2019 5:05 pm
Thank for your response.
I'll try to make it clear.
Since small values for one of the variables in my model are causing an infeasible solution, I'm looking for a way to solve this problem.
However, a simple restriction like X>1 is inapplicable because I want to give this variable the option to be zero. Therefore, I'm trying to give this variable a specific range like:
X=0 or X>1 but I don't know how to do so.
In other words, what I'm trying to do is to create a specific range in which one of my variables can be changed. This range is not continuous. The first part is zero (only one value), and the second part is all the values above 1.
I hope my problem is clear now.
Thank you again for the willingness to help, I really appreciate it!
Yehuda
### Re: How to ranging variables
Posted: Thu Mar 28, 2019 8:37 am
Hi again.
Now I know what is your problem.
I'm afraid you can not use ".lo" or ".up" for doing this but I think there is an alternative solution:
foe example suppose you have an equation on X in your problem.
X + Y * Z >= 6 ;
firstly you define :
Code: Select all
``````positive variable X , Y , Z;
Equations
eq1
;
eq1\$(X = 0 or X > 1 ) .. X + Y * Z =e= 6;
``````
I believe it works, try it.
### Re: How to ranging variables
Posted: Thu Mar 28, 2019 2:07 pm
Thank you again.
So I tried your suggestion but it causes another problem. Now the model is declaring an endogenous problem and it prohibits the use of \$ operation.
I thought that the ranging issue will be easy to defined, but it's apparently not.
Anyway, thanks a lot!
Yehuda
### Re: How to ranging variables
Posted: Mon Apr 08, 2019 4:32 pm
Yehuda,
You might want top have a look at semi-continous variables: https://www.gams.com/latest/docs/UG_Var ... iableTypes
Best,
Fred | 680 | 2,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-34 | latest | en | 0.937019 |
https://www.greenlistlouisville.com/multiplying-fractions-worksheets-6th-grade/past-simple-regular-verbs-exercises-printable-addition-worksheets-for-first-grade/ | 1,627,477,834,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00638.warc.gz | 832,740,687 | 12,956 | Worksheets: Past Simple Regular Verbs Exercises Printable Addition Worksheets For First Grade. Theme Worksheets 4th Grade Field Trip Worksheet Analogies Worksheet 8th Grade | Greenlistlouisville
# Past Simple Regular Verbs Exercises Printable Addition Worksheets For First Grade
Published at Thursday, 26 November 2020. Worksheets. By .
By the time they are learning first grade math, kids should be ready to tackle things like the relationship between addition and subtraction, the concept of adding and subtracting two-digit numbers and learning to count beyond 100. Being able to compare numbers as larger, smaller or equal to each other is also important, as it provides the basis for recognizing whether or not the answer to a computation problem is the correct one. Children need to be allowed to master these and other essential math skills before being asked to move on to new ideas, but the modern classroom setting does not always allow for this. As focus on core curriculum begins to push complex ideas into lower grade levels, kids are expected to learn more at a younger age. First grade math still contains many fundamental concepts essential for understanding higher math, and therefore should not be rushed through. By letting a child try and re-try each new thing as it comes, online math games can give the extra time and practice that struggling students need to achieve success.
Many children are being left behind due to lack of math skills. Schools today seem to do a poor job of preparing students for math at the middle and high school level. Here are 5 tips that parents can use to help their child be successful at math. Start early. Before your child goes to preschool, they need to be familiar with small numbers, up to 10. Two is easy to teach and point out. Pair of socks, shoes, etc. Five fingers on a hand and toes on feet. Ten total fingers and toes. At the preschool level, start counting up to 20. Add small numbers, 1 plus 1 is 2. 2 plus 1 is 3. You can even begin the fraction of one half. Half a sandwich, and other food items are a great start. When finishing kindergarten, your child needs to be able to count past 20 and know what larger numbers mean as well. Not working with them, just be familiar.
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### Past Simple Regular Verbs Exercises Printable Addition Worksheets For First Grade
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Click here for a printable PDF of the student version of this exemplar Lesson: Base ten Block Challenge Grade: 3-4 Skills: addition, number sense I wanted to find out how many tens rods were in the entire school (kindergarten through eighth grade), so I checked every classroom. Soon, I found there was a pattern that helped me count. I discovered that: Each kindergarten class had 35 tens rods. Each first grade class had 70 tens rods. Each second grade class had 140 tens rods. This pattern continued all the way through eighth grade. I also know there are: 4 classes in every grade k-5 3 classes in both grades 6 & 7 2 classes in grade 8 1. How many tens rods are in the entire school? 2. Challenge: How many hundreds flats could you trade all the tens rods for? | 221 | 981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-26 | longest | en | 0.944433 |
https://www.bartleby.com/solution-answer/chapter-18-problem-35e-precalculus-mathematics-for-calculus-6th-edition-6th-edition/9780840068071/6a842323-e05f-43c6-966e-e0361bc96f68 | 1,632,242,525,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.57/warc/CC-MAIN-20210921161350-20210921191350-00612.warc.gz | 698,809,015 | 99,600 | # The closer point among P ( 3 , 1 ) and Q ( − 1 , 3 ) to the point R ( − 1 , − 1 ) .
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 1.8, Problem 35E
To determine
## To calculate: The closer point among P(3,1) and Q(−1,3) to the point R(−1,−1) .
Expert Solution
Thecloser point tothe point R(1,1) is Q(1,3) .
### Explanation of Solution
Given information:
The points P(3,1) and Q(1,3) .
Formula used:
Distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .
Calculation:
Consider the provided set of points P(3,1) and Q(1,3) . Also the point R(1,1) .
Recall that the distance d(A,B) between two points A=(x1,y1) and B=(x2,y2) in the Cartesian plane is denoted by d(A,B)=(x2x1)2+(y2y1)2 .
Evaluate the distance between P(3,1) and R(1,1) .
d(R,P)=(13)2+(11)2=16+4=20=4.47
Next evaluate the distance between Q(1,3) and R(1,1) .
d(Q,R)=(1(1))2+(13)2=16=4
Observe that d(R,P)>d(Q,P) .
That is the distance between the point Q(1,3) and the R(1,1) is less than the distance the point P(3,1) and the R(1,1) .
Therefore, the point Q(1,3) is near to R(1,1) .
Thus, the closer point to the point R(1,1) is Q(1,3) .
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! | 562 | 1,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-39 | latest | en | 0.74382 |
https://discourse.julialang.org/t/obtain-frequency-response-from-digital-filter-with-dsp-jl/44618 | 1,623,685,280,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00113.warc.gz | 209,704,769 | 5,314 | # Obtain frequency response from digital filter with DSP.jl
Hi, I have a simple and potentially very stupid question. I use `DSP.jl` to construct a filter via `digitalfilter()`, which leaves me with a filter kernel I can apply to my signal.
If I wanted to take a look at the response in the frequency domain, `freqz()` seems to be the appropriate function. However, it only accepts filter coefficient objects, which leads to the question of how to convert a filter kernel from `digitalfilter()` to such an object.
I found the documentation of what otherwise looks like an excellent and well-designed package a bit sparse (at least for people with no DSP background), but it seems like this does what I want for filters constrcuted with the `remez()` method:
``````julia> bpass = remez(35, [(0, 0.1)=>0, (0.15, 0.4)=>1, (0.45, 0.5)=>0]);
julia> b = DSP.Filters.PolynomialRatio(bpass, [1.0])
julia> f = range(0, stop=0.5, length=1000)
julia> plot(f, 20*log10.(abs.(freqz(b,f,1.0))))
``````
It seems to work just fine fĂĽr filters created by `digitalfilter()`, so am I correct that e.g. `DSP.Filters.PolynomialRatio(digitalfilter(FIRWindow(hamming(3), Highpass(10; fs=100)), [1.0])` will yield a correct frequency response?
And a bonus question: Does anyone have a good resource for learning about how these filter coefficients relate to e.g. a window-based FIR filter?
I am not an expert but some feedback:
• It seem that freqz does not compute the frequency response of FIR filters (it works alright for others).
• You can use function available in: FIRfreqz, to display your FIR filter response as follows:
``````fs = 100
f = digitalfilter(Highpass(10; fs=fs), FIRWindow(hamming(3)))
w = range(0, stop=pi, length=1024)
h = FIRfreqz(f, w)
ws = w/pi*(fs/2)
plot(ws, 20*log10.(abs.(h)), xlabel="Frequency(Hz)",ylabel="Magnitude(db)",label="10Hz high-pass" )
``````
1 Like
Thanks! That helps actually quite a lot! My understanding is that `sw += b[j]*exp(-im*w[i])^-j` is basically just a Fourier transformation, right?
Correct. Called Discrete Time Fourier Transform (DTFT) in article FIR_filter, see the FIR filter’s frequency response equation in section “Frequency response”.
As a quick hack, you could extend `freqz` to work with FIR filters.
``````freqz(h::Vector{T}, N) where T = fft([h; zeros(T, N-length(h))])
``````
Edit: In this case the corresponding normalised frequency vector (in radians) would be
``````w = range(0, step=2Ď€/N, length=N)
``````
2 Likes | 691 | 2,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-25 | latest | en | 0.891818 |
https://math6.nelson.com/studentcentre/studsurf_ch12_lesson06.html | 1,642,872,978,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303868.98/warc/CC-MAIN-20220122164421-20220122194421-00323.warc.gz | 431,089,109 | 1,966 | ## Lesson 6 - Percents as Special Ratios
Fraction Pie - Use the sliders to show the fraction as a percent, and a decimal value.
Converting a Fraction to a Percent - Follow the steps to convert a fraction to a percent.
Converting Percent to Decimals - Follow the steps to convert a percent to a decimal. | 69 | 305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-05 | latest | en | 0.782543 |
https://homework.cpm.org/category/CON_FOUND/textbook/mc1/chapter/3/lesson/3.3.3/problem/3-84 | 1,723,203,514,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00580.warc.gz | 232,678,852 | 15,112 | ### Home > MC1 > Chapter 3 > Lesson 3.3.3 > Problem3-84
3-84.
What is the distance on a ruler from $\frac { 1 } { 2 }$inch to $7\frac{3}{4}$ inches? Label your answer in inches.
Look back to 2-20 for help with estimating lengths on a ruler. To help you, here is a ruler with this length marked.
$7\frac{1}{4}\text{ inches}$ | 108 | 327 | {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-33 | latest | en | 0.840444 |
http://slideplayer.com/slide/3892172/ | 1,519,568,797,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00794.warc.gz | 339,422,543 | 27,772 | # Measures of disease occurrence and frequency
## Presentation on theme: "Measures of disease occurrence and frequency"— Presentation transcript:
Measures of disease occurrence and frequency
Epidemiology matters: a new introduction to methodological foundations Chapter 5
Epidemiology Matters – Chapter 1
Seven steps Define the population of interest Conceptualize and create measures of exposures and health indicators Take a sample of the population Estimate measures of association between exposures and health indicators of interest Rigorously evaluate whether the association observed suggests a causal association Assess the evidence for causes working together Assess the extent to which the result matters, is externally valid, to other populations Epidemiology Matters – Chapter 1
Seven measure of disease occurrence and frequency
Counts Prevalence Incidence/risk Mean/variance Median Mode Rates Epidemiology matters - Chapter 5
Tuberculosis in New York City
Tuberculosis is a reportable condition All diagnosed cases must be reported to the department of health In 2011, there were 689 new cases of tuberculosis in New York City Epidemiology matters - Chapter 5
Tuberculosis in New York City
Tuberculosis is a reportable condition All diagnosed cases must be reported to the department of health In 2011, there were 689 new cases of tuberculosis in New York City Is this information useful? Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
1. Counts Provide an absolute number of the burden of disease However counts has limited utility for two reasons The burden of disease in the population is very different if the population size is 100,000 versus 1,000,000 Some people are not at risk for developing a new onset of tuberculosis in 2011 (due to pre-existing infection), thus we need to know not only the size of the total population, but the size of the total population at risk Epidemiology matters - Chapter 5
Incidence and prevalence
Two measures overcome many of the limitations of a simple count of cases - incidence and prevalence Prevalence tells us about the proportion of cases among the total population at any given time Incidence tells us the probability of a new onset of disease among those at risk for developing the illness Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
2. Prevalence The proportion of people who have the disease (existing cases plus new cases) over the total population for a given time period Epidemiology matters - Chapter 5
Disease occurrence in a sample of Farrlandia over time
Year 1, 5 individuals developed the outcome Year 2, an additional 7 people developed the outcome Year 3, an additional 4 people developed the outcome
What is the prevalence of disease in Year 2?
What is the numerator? 5 cases in Year cases in Year 2 = 12 What is the denominator? Total sample size = 30 Prevalence = 12/30 = 0.4 The prevalence of disease in Year 2 is 40% Epidemiology matters - Chapter 5
What is the prevalence of disease in Year 3?
What is the numerator? 5 cases in Year cases in Year cases in Year 3 = 16 What is the denominator? Total sample size = 30 Prevalence = 16/30 = 0.533 The prevalence of disease in Year 2 is 53.3% Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Summary: Prevalence For prevalence, we need a numerator (number of existing cases), and denominator (total sample size), and a time period of interest The time period should be specified as much as possible For example, when we say “in Year 2” we mean over the duration of time that spanned up to Year 2 Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
3. Incidence Perhaps the most widely used tool in epidemiology Goes by many names - most common alternative name is “risk,” and less commonly, “incidence proportion” Numerator = number of new cases Denominator = population at risk of becoming a new case Specified over a specific time period Epidemiology matters - Chapter 5
What is the incidence of disease in Year 2?
What is the numerator? 7 new cases in Year 2 What is the denominator? 25 people at risk (5 people already developed the disease in Year 1 and are thus not at risk) Incidence = 7/25 = 0.28 The incidence (risk) of disease in Year 2 is 28% Epidemiology matters - Chapter 5
What is the incidence of disease in Years 2 and 3?
What is the numerator? 7 new cases in Year new cases in Year 3 = 11 What is the denominator? 25 people at risk (5 people already developed the disease in Year 1 and are thus not at risk) Incidence = 11/25 = 0.44 The incidence (risk) of disease in Years 2 and 3 is 44% Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Summary: Incidence For incidence, we need a numerator (number of new cases), and denominator (total sample size at risk), and a time period of interest The time period should again be specified as much as possible Epidemiology matters - Chapter 5
The relation between incidence and prevalence
For incidence, we need a numerator (number of new cases), and denominator (total sample size at risk), and a time period of interest The time period should again be specified as much as possible Epidemiology matters - Chapter 5
Understanding incidence and prevalence: the bathtub example
Epidemiology matters - Chapter 5
Examples of the relation between incidence and prevalence
High incidence, steady prevalence Example: highly contagious infectious disease with very short duration or a high case-fatality Low incidence, high prevalence Examples: diseases with long duration such as arthritis, diabetes, Crohn’s disease, and other chronic illnesses Epidemiology matters - Chapter 5
Examples of the relation between incidence and prevalence
Impact of a new treatment that prolongs life with the disease but does not cure it New HIV Infections People Living with HIV Epidemiology matters - Chapter 5
Summary, incidence, prevalence
Prevalence is affected by incidence and duration If a disease has short duration, Prevalence ~= incidence* If a disease has long duration, in general, Prevalence > incidence * Assumes that incidence is constant over time Epidemiology matters - Chapter 5
Mean, variance, median, mode
Health outcomes are sometimes not measured by presence or absence, but rather as a continuous measure Examples: Body Mass Index, blood pressure, cholesterol, birth weight, lung function, number of depression or anxiety symptoms In these cases, we need measures of centrality and spread to characterize occurrence and frequency Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Mean The mean is estimated by summing the outcomes for each individual and dividing that summed score by the number of individuals For example, suppose we measured BMI in a sample of 31 individuals Epidemiology matters - Chapter 5
Mean Table: Body mass index (BMI) in a random sample of 31 Farrlandians Epidemiology matters - Chapter 5
Thus, the mean BMI in our sample is 31.1
The mean is estimated by summing the outcomes for each individual and dividing that summed score by the number of individuals = 31.1 Thus, the mean BMI in our sample is 31.1 Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Variance In addition to estimating the mean of a continuous variable, it is important to estimate how close all of the individual values are to that mean For example, suppose we sampled two populations, and obtained the following histograms of their risk of disease Epidemiology matters - Chapter 5
The values of BMI in Sample 2 are closer to the mean than in Sample 1
Therefore, Sample 2 has a lower variance than Sample 1 Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Variance The spread of individual values around the mean is a measure of the variance of the data The size of the variance gives us important information about the distribution of the variable of interest within the sample A large variance tells us that while the mean may be 31.1, there is a wide range of total values across the whole sample (and, if a representative sample, underlying population) A small variance tells us that there is little variability in the sample (and, if a representative sample, underlying population) with respect to the variable of interest Epidemiology matters - Chapter 5
Mean and variance: limitations
The mean can be influenced by extremes in the data If our data had one recorded miscoded as a BMI of 550 instead of 55, the mean would be 47.1 rather than 31.1 In general, when the outcomes are not evenly distributed across a full range of potential values and instead are aggregated at the low end or the high end, the mean may not be the most informative measure of centrality For example, suppose we would like to measure the mean number of cigarettes smoked per day among a sample of adolescents Epidemiology matters - Chapter 5
Mean and variance: limitations
Table: Number of cigarettes smoked per day among a random sample of 17 adolescents Epidemiology matters - Chapter 5
Mean and variance: limitations
The mean would be 9.24 However most of the values are between 1 and 3, thus reporting an average of 9.24 cigarettes smoked in the sample is not very informative Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
5. Median The median of a variable is the numerical value that falls in the exact middle of the range of values; it is the value for which 50% of the remaining values are above and 50% are below Epidemiology matters - Chapter 5
Median 3 5 7 3 3 5 7 9 9 11 The median value is 5
The median value is 5 The median value of this variable is 7 There are six observations in this set, so that there is no single value that falls directly in the middle In this case, we take the mean of the two values most centered. Since 3 and 4 are the most centered values (2 observations fall below, and 2 observations fall above), the median of this set is the mean of 3 and 4: (3+4)/2=3.5 Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Median Considering our smoking variable, the median value would be 2 There are eight observations that fall below 2 in this string of values, and eight that fall above 2 Thus, whereas the mean number of cigarettes smoked was 9.24, the median was 2 This signals that the distribution is quite skewed by a few heavy smokers Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
6. Mode One simple measure of centrality is the most frequently observed value, which is labeled the mode Returning to our example of cigarette smoking, we can determine the following: 3 students reported smoking 1 cigarette per day 6 students reported 2 cigarettes per day 4 reported 3 cigarettes per day 1 student reported 10 per day 1 student reported 20 per day 1 student reported 40 per day 1 reported 60 per day The modal value is the value that is most frequent; given that 6 students reported 2 cigarettes per day, the modal value would be 2 Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
7. Incidence rates We have learned that “incidence” or “risk” is calculated as the number of new cases over the population at risk of becoming a new case Incidence is an accurate representation of a sample experience of health and disease when we have complete follow-up of a sample That is, each individual is observed at every measurement time point from the beginning of the study to the end Epidemiology matters - Chapter 5
Example: alcohol consumption and liver cirrhosis
Suppose we conduct a study to estimate the association between heavy alcohol consumption and liver cirrhosis We follow 20 people over time 10 are heavy alcohol consumers First, let us imagine that we had complete follow-up data on all people in the study Epidemiology matters - Chapter 5
Disease incidence over time by population exposure
four time points = 0.65 or 65% 13/20 = Let us imagine that this is the sample followed forward in time with complete follow up. Let us imagine that people in black are exposed and grey are unexposed – ignore this for now, we will return to black and grey when we learn about measures of association. Epidemiology matters - Chapter 5
Example: alcohol consumption and liver cirrhosis
Now, let us imagine that we lost some people over time Thus, we do not know whether these individuals became diseased or not Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Loss to follow up in a sample over time Epidemiology matters - Chapter 5
Incidence when there is loss to follow-up
We know that the true incidence is 65% If we only analyzed the data based on who was present at the end of the study, we would estimate incidence as 9/15 = 0.60 or 60% If we assumed that individuals who dropped out did not become diseased we would get 9/20 = 0.45 or 45% If we assumed that individuals who dropped out did become diseased we would get 14/20 = 0.70 or 70% There is one more option: a rate Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Incidence rates Incidence rates are commonly used in prospective studies in which some people are lost over time To estimate a rate over the time frame of the study, we need to know how much total time each person contributed to the study follow-up before they either developed the outcome or dropped out We term the total time that each person contributed as person-time Epidemiology matters - Chapter 5
Understanding person years
Person 2 stayed in the study all 40 years and did not develop the outcome Person 10 dropped out of the study at Year 30 Person 19 developed the outcome at Year 10 Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Understanding person years Table: Person-time and disease status among 20 subjects followed for forty years Epidemiology matters - Chapter 5
Calculating the incidence rate
The numerator is the number of cases The denominator is the total person-time In our example: 8/440 = 0.18, or a rate of 18 cases per 1,000 person-years Epidemiology matters - Chapter 5
Calculating the incidence rate
The incidence rate can be interpreted as the number of expected cases in every set of 1,000 person years That is, if we were to observe 1,000 people for 1 year, we would expect 18 cases If we were to observe 500 people for 2 years, we would still expect 18 cases The assumption underlying this is that the incidence rate is constant over time, so for every year in which 1,000 person years are observed an additional 18 cases will be expected Given this assumption, the incidence rate tells us the average number of cases per a specified set of person time Epidemiology matters - Chapter 5
Rate versus proportion: what’s the difference?
A proportion can range from 0 to 100, and the numerator is contained in the denominator A rate can range from 0 to infinity and the numerator is the number of cases whereas the denominator is the person-time at risk Incidence rates can be conceptualized as the speed at which disease is occurring in cases per person year When we have complete follow-up of a sample or a population, the rate can approximate the proportion of disease or the risk Epidemiology matters - Chapter 5
Risks and rates, an example, part 1
We have 10 people who are disease free at the start of follow-up, each followed for 1 year Three of these individuals develop the disease. All individuals are followed for the entirety of the study period The risk (incidence) of disease will be 3 out of 10, or 0.3 Assuming these individuals developed the disease just as the year was ending, and the rate would be 3 per 10 person years or 0.3 (equivalent to the risk) Epidemiology matters - Chapter 5
Rate versus proportion, an example, part 2
Now suppose that those who developed the disease did so halfway through the year 7 people were followed and did not develop the disease, i.e., 1 person year for each totaling 7 person years 3 people developed the disease, i.e., we assign each of them 0.5 person years for the midpoint of the time interval for a total of 1.5 person years Thus, the incidence rate would be 3 per 8.5 person years, or 0.35 Epidemiology matters - Chapter 5
Incidence vs. incidence rate: what’s the difference?
Because measures of incidence are so central to epidemiological investigation, the term “incidence” can be used in various contexts, and the concept that we refer to as “incidence” can go by different terms The incidence refers to the number of new cases divided by the population at risk. It is also called the incidence proportion, or the risk When we refer to “incidence”, we mean the incidence proportion, also known as the risk The incidence rate refers to the number of new cases divided by the person-time at risk contributed by members of the study When we refer to “incidence rate”, we specifically refer to a measure in which the denominator is the person-time at risk contributed by members of the study. Epidemiology matters - Chapter 5
An extra, conditional risks
We can “condition” risk estimate by other factors to begin to examine whether certain factors are associated with increased or decreased risk Let us return to our earlier example of alcohol consumption an liver cirrhosis In order to estimate whether heavy drinkers have a different incidence of cirrhosis compared with non-heavy drinkers, we can use a measure of the conditional incidence Epidemiology matters - Chapter 5
Two by two table showing exposure in each row and disease status in each column
Conditional risk of cirrhosis among heavy drinkers = 8/10 = 80% Conditional risk of cirrhosis among non-heavy drinkers = 5/10 = 50% Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Conditional risks It appears that heavy drinkers have a higher incidence of cirrhosis compared with non-heavy drinkers (Next we will learn how to quantify this) Building these 2x2 tables crossing exposure with disease and using these 2x2 tables to estimate associations will become a building block of epidemiology Epidemiology matters - Chapter 5
Epidemiology matters - Chapter 5
Summary Measures of disease occurrence and frequency in epidemiology are the cornerstone of how we build the science of population health Key measures are: incidence/risk, prevalence, mean, median, mode, incidence rates, and conditional risks Incidence rates are more appropriate than incidence when there are losses to follow-up Epidemiology matters - Chapter 5
Epidemiology Matters – Chapter 1
Seven steps Define the population of interest Conceptualize and create measures of exposures and health indicators Take a sample of the population Estimate measures of association between exposures and health indicators of interest Rigorously evaluate whether the association observed suggests a causal association Assess the evidence for causes working together Assess the extent to which the result matters, is externally valid, to other populations Epidemiology Matters – Chapter 1
Epidemiology Matters – Chapter 1
epidemiologymatters.org Epidemiology Matters – Chapter 1 | 4,171 | 19,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-09 | latest | en | 0.892999 |
https://www.lmfdb.org/EllipticCurve/Q/3360/r/ | 1,674,833,640,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494986.94/warc/CC-MAIN-20230127132641-20230127162641-00325.warc.gz | 891,902,402 | 9,785 | # Properties
Label 3360.r Number of curves $2$ Conductor $3360$ CM no Rank $1$ Graph
# Related objects
Show commands: SageMath
sage: E = EllipticCurve("r1")
sage: E.isogeny_class()
## Elliptic curves in class 3360.r
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
3360.r1 3360j1 $$[0, 1, 0, -26, -60]$$ $$31554496/525$$ $$33600$$ $$[2]$$ $$384$$ $$-0.33299$$ $$\Gamma_0(N)$$-optimal
3360.r2 3360j2 $$[0, 1, 0, -1, -145]$$ $$-64/2205$$ $$-9031680$$ $$[2]$$ $$768$$ $$0.013582$$
## Rank
sage: E.rank()
The elliptic curves in class 3360.r have rank $$1$$.
## Complex multiplication
The elliptic curves in class 3360.r do not have complex multiplication.
## Modular form3360.2.a.r
sage: E.q_eigenform(10)
$$q + q^{3} - q^{5} + q^{7} + q^{9} + 2 q^{11} - q^{15} - 6 q^{17} - 6 q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels. | 469 | 1,303 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-06 | latest | en | 0.535667 |
https://forum.math.toronto.edu/index.php?PHPSESSID=2h9fj04alq45skvg6snedf70k1&topic=1163.msg4115 | 1,653,775,707,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00145.warc.gz | 313,689,727 | 6,949 | ### Author Topic: Final exam (Read 3276 times)
#### serena li
• Newbie
• Posts: 2
• Karma: 0
##### Final exam
« on: April 04, 2018, 09:44:35 AM »
Is 2.7 and 2.8 covered in final exam? Since the announcement says that chapter 2,3,4,7,9 is covered, but 2.7 and 2.8 are not covered by term test so I am a little bit confused.
#### Victor Ivrii
• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Final exam
« Reply #1 on: April 04, 2018, 09:54:42 AM »
2.7--2.8 covered, 2.9 not
#### Meng Wu
• Elder Member
• Posts: 91
• Karma: 36
• MAT3342018F
##### Re: Final exam
« Reply #2 on: April 04, 2018, 11:35:59 AM »
2.7--2.8 covered, 2.9 not
Just want to make sure, because my Prof said there is nothing on modelling, numerical methods, exists and uniqueness theorem from chapter 2. (from Tues. B. Jacelon)
#### Victor Ivrii
• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Final exam
« Reply #3 on: April 04, 2018, 12:06:55 PM »
There is a difference "contained" and "covered". The former means that there is a directly related question, the latter that "you need to know it". In fact, existence and uniqueness are heavily used everywhere in the course
#### Meng Wu
• Elder Member
• Posts: 91
• Karma: 36
• MAT3342018F
##### Re: Final exam
« Reply #4 on: April 04, 2018, 12:52:08 PM »
Thanks for the clarification.
#### Syed Hasnain
• Full Member
• Posts: 18
• Karma: 3
• mat244h1s-winter2018
##### Re: Final exam
« Reply #5 on: April 05, 2018, 03:47:17 AM »
Just for the sake of clarity, can anyone mention the individual sections
within each chapter that is going to be covered ?
#### Victor Ivrii
• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Final exam
« Reply #6 on: April 05, 2018, 05:02:20 AM »
You may look at the forum on MAT244 @2014 Final Exam.
#### Rasagya Monga
• Newbie
• Posts: 4
• Karma: 0
##### Periodic Solutions on final?
« Reply #7 on: April 09, 2018, 11:22:46 PM »
Will periodic Solutions and Limit cycles be there on the final exam? | 666 | 1,968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.762829 |
https://www.baresmusic.com/piano/where-is-f-on-the-piano.html | 1,620,372,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00365.warc.gz | 657,256,505 | 10,399 | Where is f# on the piano
What does F# mean in piano?
F# major chord for piano (including inversions) presented by keyboard diagrams. Explanation: The regular F# chord is a triad, meaning that it consists of three notes. F# stands for F sharp .
Which key is F sharp?
Scales with sharp key signatures
Major key Number of sharps Sharp notes
G major 1 F♯
D major 2 F♯, C♯
A major 3 F♯, C♯, G♯
E major 4 F♯, C♯, G♯, D♯
Where is C# on a piano?
Let’s first of all locate C sharp on your piano keyboard . C sharp is the first black key in the set of two black keys on your keyboard .
What is M in piano chords?
F# minor chord for piano (including inversions) presented by keyboard diagrams. Explanation: The regular F# minor chord is a triad, meaning that it consists of three notes. The chord is often abbreviated as F# m . F# m stands for F sharp minor.
What is G M on piano?
G# minor chord The chord is often abbreviated as G # m . G # m stands for G sharp minor . Theory: The G# minor chord is constructed with a root, a minor thirdAn interval consisting of three semitones, the 3rd scale degree and a perfect fifthAn interval consisting of seven semitones, the 5th scale degree.
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What key is FCG am?
We can also play the four chords more than once. C-G- Am – F-C-G – Am -F etc. We have worked till now in what we call the “ key ” of C or the “ key ” of A minor. Each step of the scale will now receive a number.
How do you hold F#?
F# Chord (E Shape) Barre your first finger across all the strings on the 2nd fret. Place your 3rd finger on the 4th fret of the A string. (5th string.) Place your 4th finger on the 4th fret of the D string. (4th string.) Place your 2nd finger on the 3rd fret of the G string. (3rd string.)
What chord is C sharp?
(Learn piano chords). Notes of the C# major chord are C#-E#-G#.
What does an F sharp look like?
F sharp is the enharmonic equivalent of Gb. The G flat major scale makes use of the same keys on the piano and sounds the same as the F# major scale. The difference is the names of the notes. The notes of the G flat major scale are G♭, A♭, B♭, C♭, D♭, E♭, and F .
What note harmonizes with F#?
F# / Gb Major Scale Triads
Degrees I iii
Notes F# A#
Chords F# A#m
Notes in chord F #-A#-C# A#-C#-E#
Is E flat the same as F sharp?
Acoustically, they are exactly the same . In terms of music theory, a note would be called either D# or Eb depending on what key it appears in. Additionally, the relative minor keys of these key signatures are also “ sharp keys”: E minor, B minor, F# minor, C# minor, G# minor, D# minor, and A# minor.
You might be interested: Where can i play piano
What note harmonizes with D sharp?
The D-sharp major chord ii is the E# minor chord, and contains the notes E#, G#, and B#. This supertonic chord’s root / starting note is the 2nd note (or scale degree) of the D# major scale .
What chords are in the key of C#?
Therefore the triad chords in the key of C sharp are C#maj, D#min, E#min, F#maj, G#maj, A#maj and B#dim. The four note extended chords based on the C sharp major scale are C#maj7, D#min7, E#min7, F#maj7, G#7, A#min7 and B#m7b5. | 929 | 3,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | longest | en | 0.946499 |
https://www.plati.market/itm/k4-solution-at-53-reshebnik-termehu-targ-sm-1982/2130430?lang=en-US | 1,542,812,164,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039748901.87/warc/CC-MAIN-20181121133036-20181121155036-00255.warc.gz | 949,560,625 | 12,921 | # K4 Solution At 53, reshebnik termehu Targ SM 1982
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# Description
Meeting the challenge of K4 Version 53 of Reshebnik on theoretical mechanics to Taskbook Targ SM 1982.
Subject K4 task - complex motion of a point.
Check the condition of the task K4 (Pages 44-49, zadachnik Targ SM 1982):
A rectangular plate (Fig K4.0 -. K4.4) or round plate of radius R = 60 cm (Fig K4.5 -. K4.9) rotates about a fixed axis by law φ = f1 (t), given in the Table. K4. The positive direction of the reference angle φ shown in the figures arc arrow. Fig. 0, 1, 2, 5, 6, the axis of rotation perpendicular to the plane of the plate and passes through the point (plate rotates in its own plane); Fig. 3, 4, 7, 8, 9 OO1 rotation axis lies in the plane of the plate (plate rotates in space)
At the plate along the line BD (. Figure 0-4) or a circle of radius R (. Figure 5-9) moves the point M; .. Law its relative motion, ie the relationship s = AM = f2 (t) (s expressed in centimeters, t - in seconds) specified in the table for Fig separately. 0-4 and Fig. 5-9; ibid given dimensions b and l. Figures point M is shown in a position in which the AM s => 0 (if s <0, the point M is on the other side of the point A).
Find the absolute speed and the absolute acceleration of the point M at time t1 = 1 sec.
# Additional information
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## Transcription
2 Planck constant. The number of transitions from the energy level E to the energy level E per unit time caused by a stimulated emission of radiation may be written in the form where i = B B =σ π B hω, () 3 B = hω / ( π c )(exp( hω / kt) ) () is a blackbody emissivity (Planck s function), σ is the stimulated emission cross-section, T is the temperature, k is the Boltzmann constant, c is the speed of light, and width. is the line The number of transitions from the lower level to the upper one per unit time may be written in the form where σ is the absorption cross-section. σ π B. (3) ω = BB = h Here A, B and B are the coefficients introduced by Einstein [6], the coefficients B and B being modified to account for the line width. We denote as and the number of atoms occupying the energy levels E and E, respectively. The levels E and E are suggested for simplicity to be non-degenerated. In the equilibrium state the full number of transitions from the lower level to the upper one is equal to the number of reverse transitions : = B B = = ( A + B B ). (4) The line width is suggested to be equal to the natural line width [7]: = A + B B. (5) state. It is assumed that the lower level has a zero energy width, i.e. this level is a ground It follows from the last equation that = A / ( B B ). (6) Substituting this expression in the equation (4), we obtain / = B B = σ π B h ω. (7)
3 In the limit of high temperatures T, corresponding to the range of frequencies hω <kt, the function B ( T) is given by the Rayleigh-Jeans formula B = kt / c, (8) where is the frequency of radiation, = ω / π. Since B when T, it follows from the equation (6) that the coefficient B is depending on the temperature T in such a manner that when T. It is clear that. The ratio of frequencies of transitions caused by spontaneous and induced emission of radiation is given by the expression B B < B B s / i = A /( B B ) = ( B B)/( B B). (9) This ratio is approaching zero when T, since B B. It means that in the range of frequencies hω <kt thermal radiation is produced by the stimulated emission, whereas the contribution of a spontaneous emission may be neglected. the ratio It follows from equations (7) and (8) that in the Rayleigh-Jeans range of frequencies is given by the formula / / =σ ωkt/(πh c ). (0) An analysis of observational data on thermal radio emission from various astrophysical objects suggests that the absorption cross-section does not depend on a wavelength of radiation and has an order of magnitude of an atomic cross-section, 5 σ 0 cm [,5]. Only this assumption is consistent with the observed spectra of thermal radio emission from major planets, the spectral indices of radio emission from galactic and extragalactic sources, and the wavelength dependence of radio source size. form If the absorption cross-section is constant then the energy distribution of atoms has a / = const( E E ) kt. () For all real values of the temperature and wavelength of radiation the ratio () is much smaller than unit. For example, at λ=m the ratio () is less than, if T<0 8 K. It implies that, where is the total number of atoms. This relation was used in [5] to obtain the condition for emission. There are no reasons to expect that the Boltzmann distribution is still valid for an ensemble of atoms interacting with thermal radiation. form From the relation B B κ we obtain the stimulated emission cross-section in the 3
4 σ h ω /( πb ) = λl T /( π), () where λ is the wavelength, and l = πh c / ( kt ) T. Thus, Einstein s relation B B =, equivalent to σ = σ, is not valid in the Rayleigh-Jeans region. Consider now the Wien region hω >kt. There are strong theoretical and observational indications of the stimulated character of thermal blackbody radiation in the whole range of spectrum [8]. First, the spectral energy density of thermal blackbody radiation is described by a single Planck s function at all frequencies. Second, there are clear observational indications of the existence of thermal harmonics in stellar spectra and of laser type sources. The latter are connected with the induced origin of thermal radiation, similarly to well known maser sources. In particular, a possible non-saturated X-ray laser source emitting in the Fe J [9]. K α line at 6.49 kev was recently discovered in the radio-loud quasar MG Assuming the stimulated origin of thermal blackbody radiation, i.e. < s i, from Eq.(8) one can obtain the relation BB = σ πb /( hω), and then the stimulated emission cross-section in the form the ratio σ ( λ / π) exp( hω / kt ). (3) The absorption cross-section is likely to be constant in the whole range of spectrum, so, according to Eq.(6), is given by the formula / / ( σ ω / πc ) exp( h ω / kt), (4) which is somewhat similar to the Boltzmann law. The exact formula for the ratio can be obtained from Eq.(6) : / / =σ ω / ( πc )(exp( h ω / kt) ). (5) The function (5) has a maximum at ω =.6kT. It means that, in the field of thermal blackbody radiation, the excited levels with E E kt are the most populated. 5 umerically, ( / ) = T, so the ratio / is less than, if only T<3*0 max 7 K. For the temperatures T>4*0 7 K the population of the levels corresponding to the maximum of the function (5) is inverse. It suggests that laser type sources are more easily realized in X-ray and gamma-ray range of spectrum than in the optical region. The well known example is a possible gamma-laser in the Galactic Center emitting in the line 0.5 MeV [0]. The author is grateful to A.V.Postnikov for useful discussions. 4
5 [].M.agar, A.S.Wilson, and H.Falcke. Astrophys. J., 559, L87 (00). J.S.Ulvestad, and L.C.Ho. Astrophys. J., 56, L33 (00). [] F.V.Prigara, astro-ph/00399 (00). [3] S.R.Pottasch, Planetary ebulae (D.Reidel, Dordrecht, 984). [4].Siodmiak, and R.Tylenda. Astron. Astrophys., 373, 03 (00). [5] F.V.Prigara, astro-ph/00483 (00). [6] J.R.Singer, Masers (Wiley, ew York, 959). M.Jammer, The Conceptual Development of Quantum Mechanics (McGraw-Hill, ew York, 967). [7] L.D.Landau, and E.M.Lifshitz, Quantum Mechanics, on-relativistic Theory (Addison- Wesley, Reading, MA, 958). V.B.Berestetskii, E.M.Lifshitz, and L.P.Pitaevskii, Quantum Electrodynamics (auka, Moscow, 989). [8] F.V.Prigara, astro-ph/007 (00). [9] G.Chartas et al., astro-ph/0 (00). [0] R.E.Lingenfelter, and R.Ramaty, in AIP Proc. 83: The Galactic Center, edited by G.Riegler, and R.Blandford (AIP, ew York, 98). 5
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8 Radiative Cooling and Heating Reading: Katz et al. 1996, ApJ Supp, 105, 19, section 3 Thoul & Weinberg, 1995, ApJ, 442, 480 Optional reading: Thoul & Weinberg, 1996, ApJ, 465, 608 Weinberg et al., 1997, | 7,845 | 32,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-47 | latest | en | 0.893659 |
https://www.codespeedy.com/program-to-solve-the-knapsack-problem-in-cpp/ | 1,675,883,819,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00526.warc.gz | 731,862,498 | 16,762 | # Program to solve the knapsack problem in C++
In this tutorial, we will learn how to solve the knapsack problem using a C++ program. We can use it for good decision-making to solve real-world problems. Here we will use it to find the maximum profit that can be gained with a set of items. If you are looking for a C++ program to find the solution to the knapsack problem you are in the right place.
## Solving the knapsack problem
In the problem, there is a sack having a specific capacity and a set of items or products with their weights. So, using this information, we have to find the items so that we get the maximum profit from the products. So, to solve a given knapsack problem follow the steps given below –
• Calculate the profit-weight ratio for each item or product.
• Arrange the items on the basis of ratio in descending order.
• Take the product having the highest ratio and put it in the sack.
• Reduce the sack capacity by the weight of that product.
• Add the profit value of that product to the total profit.
• Repeat the above three steps till the capacity of sack becomes 0 i.e. until the sack is full.
## C++ program to solve the knapsack problem
We will use the structure data structure to implement the same. Firstly, we have to take the number of items or products and the capacity of the sack. Then, we have to take the profit and weight of each item from the user and calculate the profit-weight ratio. To find the maximum profit, we have to select the items with a high profit-weight ratio and put them in the sack. We get the maximum profit for the given knapsack problem when the sack is full. Let us see the C++ program to solve the same.
```#include<iostream>
#define MAX 10
using namespace std;
struct product
{
int product_num;
int profit;
int weight;
float ratio;
float take_quantity;
};
int main()
{
product P[MAX],temp;
int i,j,total_product,capacity;
float value=0;
cout<<"ENTER NUMBER OF ITEMS : ";
cin>>total_product;
cout<<"ENTER CAPACITY OF SACK : ";
cin>>capacity;
cout<<"\n";
for(i=0;i<total_product;++i)
{
P[i].product_num=i+1;
cout<<"ENTER PROFIT AND WEIGHT OF PRODUCT "<<i+1<<" : ";
cin>>P[i].profit>>P[i].weight;
P[i].ratio=(float)P[i].profit/P[i].weight;
P[i].take_quantity=0;
}
//HIGHEST RATIO BASED SORTING
for(i=0;i<total_product;++i)
{
for(j=i+1;j<total_product;++j)
{
if(P[i].ratio<P[j].ratio)
{
temp=P[i];
P[i]=P[j];
P[j]=temp;
}
}
}
for(i=0;i<total_product;++i)
{
if(capacity==0)
break;
else if(P[i].weight<capacity)
{
P[i].take_quantity=1;
capacity-=P[i].weight;
}
else if(P[i].weight>capacity)
{
P[i].take_quantity=(float)capacity/P[i].weight;
capacity=0;
}
}
cout<<"\n\nPRODUCTS TO BE TAKEN -";
for(i=0;i<total_product;++i)
{
cout<<"\nTAKE PRODUCT "<<P[i].product_num<<" : "<<P[i].take_quantity*P[i].weight<<" UNITS";
value+=P[i].profit*P[i].take_quantity;
}
cout<<"\nTHE KNAPSACK VALUE IS : "<<value;
return 0;
}```
The structure array variable ‘P’ stores the details of all items or products. After taking the input from the user, we sort the array on the basis of the profit-weight ratio. Then, we add the items profit-wise, and the weight of the item subtracts the capacity of the sack. Finally, the program displays the knapsack value with the individual items with their weights.
#### C++ program output
Finally, the above program displays the items and their weights that are to be put in the sack. And the maximum profit for the given problem which is also called the knapsack value. So, the output of the above program is as follows –
```[email protected]:~/cpp\$ g++ knapsack.cpp
[email protected]:~/cpp\$ ./a.out
ENTER NUMBER OF ITEMS : 7
ENTER CAPACITY OF SACK : 15
ENTER PROFIT AND WEIGHT OF PRODUCT 1 : 10 2
ENTER PROFIT AND WEIGHT OF PRODUCT 2 : 5 3
ENTER PROFIT AND WEIGHT OF PRODUCT 3 : 15 5
ENTER PROFIT AND WEIGHT OF PRODUCT 4 : 7 7
ENTER PROFIT AND WEIGHT OF PRODUCT 5 : 6 1
ENTER PROFIT AND WEIGHT OF PRODUCT 6 : 18 4
ENTER PROFIT AND WEIGHT OF PRODUCT 7 : 3 1
PRODUCTS TO BE TAKEN -
TAKE PRODUCT 5 : 1 UNITS
TAKE PRODUCT 1 : 2 UNITS
TAKE PRODUCT 6 : 4 UNITS
TAKE PRODUCT 3 : 5 UNITS
TAKE PRODUCT 7 : 1 UNITS
TAKE PRODUCT 2 : 2 UNITS
TAKE PRODUCT 4 : 0 UNITS
THE KNAPSACK VALUE IS : 55.3333
[email protected]:~/cpp\$```
The knapsack problem given in the output contains 7 items and the sack capacity is 15. The user enters 7 items and their respective weights. Then, the program finds the products to be taken and the knapsack value of the problem.
Also read: Calculate factorial of a number in C++ | 1,223 | 4,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-06 | latest | en | 0.874701 |
https://www.aplusphysics.com/courses/honors/waves/interference.html | 1,695,919,576,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510427.16/warc/CC-MAIN-20230928162907-20230928192907-00328.warc.gz | 691,407,013 | 8,767 | # Wave Interference
### Superposition
When more than one wave travels through the same location in the same medium at the same time, the total displacement of the medium is governed by the principle of superposition. The principle of superposition simply states that the total displacement is the sum of all the individual displacements of the waves. The combined effect of the interaction of the multiple waves is known as wave interference.
Question: The diagram shows two pulses approaching each other in a uniform medium. Diagram the superposition of the two pulses.
### Constructive Interference
When two or more pulses with displacements in the same direction interact, the effect is known as constructive interference. The resulting displacement is greater than the original individual pulses. Once the pulses have passed by each other, they continue along their original path in their original shape, as if they had never met. An animation of two pulses interfering constructively is shown below.
Courtesy Penn State Schuylkill
Notice the top pulse travels to the right with a positive displacement, while the middle pulse travels to the left with a positive displacement. When the two pulses meet (shown at bottom), the interfere constructively before continuing on their path as if they had never met.
### Destructive Interference
When two or more pulses with displacements in opposite directions interact, the effect is known as destructive interference. The resulting displacements negate each other. Once the pulses have passed by each other, they continue along their original path in their original shape, as if they had never met. An animation of two pulses interfering destructively is shown below.
Courtesy Penn State Schuylkill
Notice the top pulse travels to the right with a positive displacement, while the middle pulse travels to the left with a negative displacement. When the two pulses meet (shown at bottom), the interfere destructively before continuing on their path as if they had never met.
Question: Two wave sources operating in phase in the same medium produce the circular wave patterns shown in the diagram. The solid lines represent wave crests and the dashed lines represent wave troughs. Which point is at a position of maximum destructive interference?
Answer: Point B is at a position of maximum destructive interference, since point B represents the intersection of a crest and a trough.
### Standing Waves
When waves of the same frequency and amplitude traveling in opposite directions meet, a standing wave is produced. A standing wave is a wave in which certain points (nodes) appear to be standing still and other points (anti-nodes) vibrate with maximum amplitude above and below the axis.
Looking at the standing wave produced on the right, we can see a total of five nodes in the wave, and four anti-nodes. For any standing wave pattern, you will always have one more node than anti-node.
Standing waves can be observed in a variety of patterns and configurations, and are responsible for the functioning of most musical instruments. Guitar strings, for example, demonstrate a standing wave pattern. By fretting the strings, you adjust the wavelength of the string, and therefore the frequency of the standing wave pattern, creating a different pitch. Similar functionality is seen in instruments ranging from pianos and drums to flutes, harps, trombones, xylophones, and even pipe organs!
Question: The diagram shows a standing wave in a string clamped at each end. What is the total number of nodes and antinodes in the standing wave?
Answer: Five nodes and four anti-nodes. | 702 | 3,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.937419 |
https://forum.allaboutcircuits.com/threads/syncing-a-camera-and-a-pulsed-flash-light-using-a-ttl-signal.132403/page-3 | 1,639,038,472,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363689.56/warc/CC-MAIN-20211209061259-20211209091259-00378.warc.gz | 310,091,808 | 21,548 | # Syncing a camera and a pulsed flash light using a TTL signal
#### Biofluid Lab Biofluid Lab
Joined Feb 17, 2017
24
The resistor and capacitor work together to make the delay. Resistance (ohms) times capacitance (Farads) is called a "time constant". Doubling the resistance or doubling the capacitance will double tour delay. Doubling both will give you 4x the delay.
Thanks;
What about the op amp? Is it just to bring the voltage back up?
#### GopherT
Joined Nov 23, 2012
8,012
Thanks;
What about the op amp? Is it just to bring the voltage back up?
Which circuit did you end up using - I'll explain it.
#### Biofluid Lab Biofluid Lab
Joined Feb 17, 2017
24
There are no goofy questions.
1. It's Ground.
2. The shield of the cable is connected to ground, so as long as you connect everything up BNC wise, you'll always have a path to ground, like this:
View attachment 121004
3. You could, but a better idea would be to make R1 a rheostat. See 2.
4. R5 is the input resistance of the sparker and isn't part of the delay circuitry proper. Sorry for the confusion.
I used this one! Also attached to this post.
thx
#### Attachments
• 182.3 KB Views: 4
#### EM Fields
Joined Jun 8, 2016
583
Now that I was able to get what I wanted, let me ask you about more details on the delay board and what each components do.
So what is the main component causing the delay. I assume it is the capacitor, right? As far as I know, resistors cannot make delay to the signal; they only attenuate it, right? and that's why we used op amp to bring the voltage back up, right? Is it the only function of op amp in my circuit? Just to bring the voltage back up?
Thanks
Using the water analogy, let's say the capacitor is like a tank which you want to fill about half-full in a given amount of time, and that the resistor is like a pipe connecting the bottom of the tank to the water supply.
Assuming that the water pressure of the source remains constant, the tank's fill time will depend on the diameter and length of the feed pipe, the diameter of the tank's bottom, and the difference between the source pressure and the back-pressure exerted by the water column in the tank.
#### Biofluid Lab Biofluid Lab
Joined Feb 17, 2017
24
and the difference between the source pressure and the back-pressure exerted by the water column in the tank.
Thanks.
Is this section related to op amp function?
#### EM Fields
Joined Jun 8, 2016
583
Thanks.
Is this section related to op amp function?
I don't understand. | 624 | 2,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.931454 |
https://fr.slideserve.com/jgrandy/activity-8-4-powerpoint-ppt-presentation | 1,653,496,776,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00333.warc.gz | 312,217,197 | 20,114 | Activity 8 - 4
# Activity 8 - 4
Télécharger la présentation
## Activity 8 - 4
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Activity 8 - 4 Buy or Lease? 2008 Honda Accord coupe, College Park, MD 5/10/2008
2. Objectives • Determine the amortization payment on a loan using a formula • Determine the amortization payment on a loan using technology • Solve problems involving repaying a loan or liquidating a sum of money by amortization model
3. Vocabulary • Amortization – the process of repaying a loan by a series of equal payments over a specified period of time. • Leasing – “renting” a car for a specified period of time. Car returns to the dealer after that period (extra charges for damage, excess mileage may apply)
4. Activity You are interested in purchasing a new car. You have decided to by a Honda Accord for \$20,995. The credit union requires a 10% down payment and will finance the balance with an 8% interest loan for 36 months. The sales tax in your city is 7%, and the license and title charges are \$80. What is the total purchase price of the car? What is the amount of the down payment? What is the total loan amount? \$20,995(1.07) + \$80 = \$22,544.65 \$22,544.65 0.10 = \$2,254.47 Total Price – Down Payment = Loan Amount \$22,544.65 – \$2,254.47 = \$20,290.18
5. Amortization Formula The formula to determine the amount of the payment, Amt, is where Amt = amortization payment (monthly payment) PV = amount of the loan (Present Value) I = interest rate per period, and n = number of periods i Amt = PV ---------------- 1 – (1 + i)-n
6. Activity - cont The credit union will finance the balance, \$20,290.18 with an 8% interest loan for 36 months. Determine the monthly amortization payment on your car using the amortization payment formula? Use TMV Solver feature of the calculator to determine the monthly payment amount. How do they compare? i .08/12 Amt = PV ---------------- = 20,290.18 -------------------------- = \$635.82 1 – (1 + i)-n 1 – (1 + .08/12)-36 N = 36, I% = 8, PV = 20,290.18, FV = 0, P/Y = C/Y = 12, END payment = \$635.82 They are the same
7. College Funds Your parents and grandparents have been contributing to your college fund for many years. The fund currently has a total of \$16,000. The decision has been made to amortize (liquidate) that amount so you will receive equal monthly payments over the next four years of college. At the end of the four years the account will be zeroed out. If the single college account earns 6% annual interest, how much money will you receive at the beginning of each month? N = 48, I% = 6, PV = 16,000, FV = 0, P/Y = C/Y = 12, BEGIN payment = \$373.89 (book had \$375.76)
8. Leasing a Car Rather than purchasing the car, you look into leasing the car. The car dealership explains that a lease is an agreement in which you make equal monthly payments for a specific period of time. At the end of this period, you return the car to the dealer. You do not have ownership of the car and, therefore, have no equity or asset at the end of the leasing period. You have the option of purchasing the car at a predetermined price. There can be end-of-lease termination fees as well as charges for excess mileage or damage. Lease for \$249 per month for 36 months, with no security deposit, \$2500 at signing plus tax, license, and title.
9. Leasing a Car - cont Compare the costs of Leasing versus Purchasing Total Cost = Down Payment + Taxes + Monthly Payments Car Value = Selling Price – Deprecation (over 3 years)
10. Summary and Homework • Summary • Amortization is the process or repaying a loan • It can also liquidate an asset down to zero balance • The amortization payment can be determined by a formula: • Homework • pg 934–935; problems 1-3, 5 i i Amt = PV ------------------ = PV ------------------ 1 – (1 + i)-n 1 – (1/(1+i)n) | 1,064 | 3,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-21 | latest | en | 0.858739 |
https://mersenneforum.org/showthread.php?s=92fbfb1c64d72b9da6c74610c5cf2fbd&t=19829 | 1,638,332,904,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.78/warc/CC-MAIN-20211201022332-20211201052332-00099.warc.gz | 467,704,890 | 8,333 | mersenneforum.org I think I found proof that no odd perfect numbers exist!
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2014-11-16, 11:31 #1 Philly314 Nov 2014 1 Posts I think I found proof that no odd perfect numbers exist! First, we can all agree that all perfect numbers are in the form (2^(n-1))(2^n-1), right? So, if a number is odd, then it has two factors that are both odd. I can prove this by drawing a multiplication table. I'm just going to do 2 & 3, but you can check for yourself. 2 3 2 4 6 3 6 9 The only pair of factors that have an odd product are the factors that are both odd; 3 and 3. If this is the case, then if (2^(n-1))(2^n-1) equals a perfect number, then both 2^(n-1) and (2^n)-1 are odd. For (2^n)-1, any number n will be odd, except for n=0, but (2^(0-1))(2^0-1) equals (1/2)(-1), or -1/2, which is not an odd number, or a perfect number. For 2^(n-1), only n=1 makes it odd, because 2^(1-1) = 1, but that would mean that the other factor would = (2^1)-1 = 1. 1*1 = 1. Does this mean that 1 is an odd perfect number? Otherwise, there is no odd perfect number.
2014-11-16, 14:04 #2 legendarymudkip Jun 2014 23×3×5 Posts It is only even perfect numbers that have that form. As well as this, one of the factors is even, because it is a power of 2.
2014-11-16, 14:04 #3
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·23·137 Posts
Quote:
Originally Posted by Philly314 First, we can all agree that all perfect numbers are in the form (2^(n-1))(2^n-1), right?
That is a definition for an even perfect number.
Quote:
Originally Posted by Philly314 So, if a number is odd, then it has two factors that are both odd. I can prove this by drawing a multiplication table. I'm just going to do 2 & 3, but you can check for yourself. 2 3 2 4 6 3 6 9 The only pair of factors that have an odd product are the factors that are both odd; 3 and 3. If this is the case, then if (2^(n-1))(2^n-1) equals a perfect number, then both 2^(n-1) and (2^n)-1 are odd. For (2^n)-1, any number n will be odd, except for n=0, but (2^(0-1))(2^0-1) equals (1/2)(-1), or -1/2, which is not an odd number, or a perfect number. For 2^(n-1), only n=1 makes it odd, because 2^(1-1) = 1, but that would mean that the other factor would = (2^1)-1 = 1. 1*1 = 1. Does this mean that 1 is an odd perfect number? Otherwise, there is no odd perfect number.
Very clever, you just proved that an even perfect number can't be an odd perfect number.
Now there is just the small problem remaining to prove that an odd number can't be a perfect number.
BTW: Even the Wikipedia page could have saved you all this embarrassment.
2014-11-16, 14:58 #4
R.D. Silverman
Nov 2003
22·5·373 Posts
Quote:
Originally Posted by retina That is a definition for an even perfect number.Very clever, you just proved that an even perfect number can't be an odd perfect number. Now there is just the small problem remaining to prove that an odd number can't be a perfect number. BTW: Even the Wikipedia page could have saved you all this embarrassment.
You know that cranks prefer to shoot their mouth off, rather than READ,
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Wed Dec 1 04:28:24 UTC 2021 up 130 days, 22:57, 1 user, load averages: 1.19, 1.21, 1.26 | 1,123 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.931212 |
https://fr.mathworks.com/matlabcentral/answers/2016736-determine-adjacent-points-in-a-logical-matrix | 1,713,635,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00865.warc.gz | 225,356,710 | 28,078 | # Determine adjacent points in a logical matrix
2 vues (au cours des 30 derniers jours)
dormant le 4 Sep 2023
Déplacé(e) : Stephen23 le 5 Sep 2023
I have a 2-D matrix of logical values, eg
000000000
010000000
000000000
000000000
000000000
000001000
000000000
000000000
100000000
How can I create a similar sized matrix with True in all the locations adjacent to the Trues in the first matrix, eg
111000000
101000000
111000000
000000000
000011100
000010100
000011100
110000000
010000000
If a location is assigned True from two or more adjacent locations, then it should be True.
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
dormant le 4 Sep 2023
Déplacé(e) : Stephen23 le 5 Sep 2023
Thanks everyone. I knew it would be trivial, but I don't have much experience with manipulating 2D arrays.
Connectez-vous pour commenter.
### Réponse acceptée
John D'Errico le 4 Sep 2023
Modifié(e) : John D'Errico le 4 Sep 2023
Trivial. Learn to think in terms of MATLAB operations. I've added another 1 in there, just to make it clear what the problem is.
A = [0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0];
For example, what does this do? Does it get you close to what you want?
B = conv2(A,ones(3),'same')
B = 9×9
1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 2 1 0 0 0 0 0 1 2 2 1 0 0 0 0 0 1 2 2 1 0 1 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
Close, but we can fix that.
B = conv2(A,[1 1 1;1 0 1;1 1 1],'same')
B = 9×9
1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 2 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 2 2 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
Next, we need to be careful, as two elements near each other can create something bigger than 1 due to the convolution. This next will correct that.
B = conv2(A,[1 1 1;1 0 1;1 1 1],'same') ~= 0
B = 9×9 logical array
1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
The ones are adjacent to each other in the original array, and that meant that it filled in the ones in the result. We can zap them out too.
B = (conv2(A,[1 1 1;1 0 1;1 1 1],'same') ~= 0) & (~A)
B = 9×9 logical array
1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
##### 2 commentairesAfficher AucuneMasquer Aucune
dormant le 4 Sep 2023
This is perfect. Many thanks.
John D'Errico le 4 Sep 2023
Thank you. I hope the explanations made sense. The important point is to think of conv and conv2 when you have problems like this.
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### Plus de réponses (1)
DGM le 4 Sep 2023
This can also be done succinctly with image processing tools:
% a logical array
A = [0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0];
A = logical(A);
% output is a logical array
B = imdilate(A,ones(3)) & ~A
B = 9×9 logical array
1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens
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Translated by | 1,941 | 3,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.635829 |
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## Question number: 62
Appeared in Year: 2012
MCQ▾
### Question
A conducting rod of mass m and length l is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance R across the rails. In the entire space around, there is a uniform magnetic field B normal to the plane of the rod and rails. The rod is given an impulsive velocity v0.
Finally, the initial energy
### Choices
Choice (4) Response
a.
will be converted fully into magnetic energy due to induced current
b.
will enable rod to continue to move with velocity v0 since the rails are frictionless
c.
will be converted fully into heat energy in the resistor
d.
will be converted into the work done against the magnetic field
## Question number: 63
Appeared in Year: 2012
MCQ▾
### Question
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is held at a temperature of 1000C while the other one is kept at 00 C. If the two are brought into contact, then, assuming no heat loss to the environment, the final temperature that they will reach is,
### Choices
Choice (4) Response
a.
less than 500C
b.
00C
c.
more than 500C
d.
500C
f Page | 383 | 1,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-30 | longest | en | 0.883343 |
https://ptcouncil.net/how-many-ways-to-arrange-5-letters/ | 1,659,984,618,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00386.warc.gz | 448,592,755 | 5,023 | This ar covers permutations and also combinations.
You are watching: How many ways to arrange 5 letters
Arranging Objects
The number of ways of arranging n uneven objects in a line is n! (pronounced ‘n factorial’). N! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1
Example
How countless different ways have the right to the letters P, Q, R, S be arranged?
The prize is 4! = 24.
This is due to the fact that there are 4 spaces to it is in filled: _, _, _, _
The an initial space can be filled by any kind of one of the 4 letters. The second space can be to fill by any of the continuing to be 3 letters. The third room can it is in filled by any type of of the 2 staying letters and the final an are must it is in filled by the one continuing to be letter. The total variety of possible kinds is as such 4 × 3 × 2 × 1 = 4!
The variety of ways that arranging n objects, the which p of one kind are alike, q that a second kind are alike, r that a third form are alike, and so on is:
n! .p! q! r! …
Example
In how countless ways have the right to the letter in the word: STATISTICS be arranged?
There room 3 S’s, 2 I’s and also 3 T’s in this word, therefore, the variety of ways the arranging the letters are:
10!=50 4003! 2! 3!
The number of ways that arranging n unequal objects in a ring when clockwise and also anticlockwise arrangements are different is (n – 1)!
When clockwise and anti-clockwise arrangements room the same, the number of ways is ½ (n – 1)!
Example
Ten world go to a party. How numerous different ways deserve to they be seated?
Anti-clockwise and clockwise arrangements space the same. Therefore, the total number of ways is ½ (10-1)! = 181 440
Combinations
The variety of ways of choosing r objects indigenous n unlike objects is:
Example
There space 10 balls in a bag numbered indigenous 1 come 10. Three balls room selected at random. How numerous different ways are there of selecting the 3 balls?
10C3 =10!=10 × 9 × 8= 120 3! (10 – 3)!3 × 2 × 1
Permutations
A permutation is an notified arrangement.
The variety of ordered arrangements of r objects taken from n uneven objects is:
nPr = n! . (n – r)!
Example
In the match of the Day’s goal of the month competition, you had actually to choose the top 3 objectives out that 10. Since the stimulate is important, the is the permutation formula which we use.
10P3 =10! 7!
= 720
There are therefore 720 different ways of picking the top three goals.
Probability
The over facts deserve to be used to aid solve troubles in probability.
Example
In the national Lottery, 6 numbers are chosen from 49. You success if the 6 balls girlfriend pick enhance the six balls selected through the machine. What is the probability of win the nationwide Lottery?
The number of ways of selecting 6 numbers from 49 is 49C6 = 13 983 816 .
See more: What Do You Call A Baby Penguin S Called? What Are Baby Penguins Called
Therefore the probability of to win the lottery is 1/13983816 = 0.000 000 071 5 (3sf), i beg your pardon is around a 1 in 14 million chance. | 796 | 3,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-33 | latest | en | 0.911232 |
https://biggboss3.net/qa/quick-answer-how-do-you-calculate-present-value-of-interest.html | 1,607,191,398,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00400.warc.gz | 209,450,223 | 7,052 | Quick Answer: How Do You Calculate Present Value Of Interest?
What is the present value of 1?
A present value of 1 table states the present value discount rates that are used for various combinations of interest rates and time periods.
A discount rate selected from this table is then multiplied by a cash sum to be received at a future date, to arrive at its present value..
What is difference between NPV and IRR?
Net present value (NPV) is the difference between the present value of cash inflows and the present value of cash outflows over a period of time. By contrast, the internal rate of return (IRR) is a calculation used to estimate the profitability of potential investments.
How do you calculate present value?
Present value is an estimate of the current sum needed to equal some future target amount to account for various risks. Using the present value formula (or a tool like ours), you can model the value of future money….The Present Value FormulaC = Future sum.i = Interest rate (where ‘1’ is 100%)n= number of periods.
What is the formula for present value in Excel?
You would need to figure out how much is needed to invest today, or the present value. The formula for present value is PV = FV ÷ (1+r)^n; where FV is the future value, r is the interest rate and n is the number of periods.
What is the formula for calculating net present value?
It is calculated by taking the difference between the present value of cash inflows and present value of cash outflows over a period of time. As the name suggests, net present value is nothing but net off of the present value of cash inflows and outflows by discounting the flows at a specified rate.
What is the discount rate formula?
How to calculate discount rate. There are two primary discount rate formulas – the weighted average cost of capital (WACC) and adjusted present value (APV). The WACC discount formula is: WACC = E/V x Ce + D/V x Cd x (1-T), and the APV discount formula is: APV = NPV + PV of the impact of financing.
What is the formula for present value of an annuity?
The Present Value of Annuity Formula P = the present value of annuity. PMT = the amount in each annuity payment (in dollars) R= the interest or discount rate. n= the number of payments left to receive.
What is NPV example?
For example, if a security offers a series of cash flows with an NPV of \$50,000 and an investor pays exactly \$50,000 for it, then the investor’s NPV is \$0. It means they will earn whatever the discount rate is on the security.
Should present value be higher or lower?
The present value is usually less than the future value because money has interest-earning potential, a characteristic referred to as the time value of money, except during times of zero- or negative interest rates, when the present value will be equal or more than the future value.
How do I calculate accumulated present value?
The formula to find out the compute the accumulated present value of a continuous stream of income at rate R(t) , for time T years and interest rate k , compounded continuously is P(t)=∫T0R(t)e−kt dt P ( t ) = ∫ 0 T R ( t ) e − k t d t .
How do you calculate present value ratio?
Present Value Ratio (PVR) can also be used for economic assessment of project(s) and it can be determined as net present value divided by net negative cash flow at i*. | 750 | 3,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-50 | latest | en | 0.948014 |
www.zavamed.com | 1,718,735,636,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00148.warc.gz | 945,779,673 | 42,504 | We all have a good idea of how many sexual partners we’ve had. But how many partners have they had and what about the partners of those people? Even if your own number is low, the number of people you’ve been exposed to via indirect sexual contact is almost certainly far higher, including countless people you haven’t even met. The size of this network can actually be quantified and estimated based on the number of your sexual partners and their average number of partners.
In the calculator below, the number of people you’ve been exposed to within six degrees of connection is calculated. Input the number of direct partners you have had and how many partners you estimate each of those partners has had before you to see the true scale of your sexual exposures throughout six degrees.
While it might be shocking to see your number of exposures as a percentage of the population of a major city in your country or the nearest global city, indirect sexual exposure can have a far reach when a person’s sexual history is combined with that of his or her partners and partners’ partners – and so on for six degrees of connection. The implications are unnerving, as the greater this number of indirect sexual exposure, the greater one’s risk of contracting a sexually transmitted infection. Seen at this scale, the true level of risk from unprotected sex will hopefully have users think again about precarious relations.
Other recently published calculators make assumptions about the sexual history of your partners and their number of contacts. However, this can produce exaggerated results and an inflated number of indirect exposures. Our calculator allows you to select your own estimate of the average number of partners that your partners have had before you. As graphed above, the exponential behaviour of this situation is revealed. That is, your sexual exposure grows exponentially with the number of partners that your partners have had. This is further compounded by the number of people that you have slept with directly, the latter of which is revealed by the various lines on the graph.
Even if you and all of your partners have only single-digit numbers of direct sexual partners, this can quickly add up to hundreds of thousands of indirect exposures - a number that more than justifies taking precautions. Remember: Just one exposure is all it takes to contract a sexually transmitted infection, but correct usage of protection such as condoms can stop a transmissible infection in its tracks. If you aren’t using condoms, it’s all the more important to know your status with regular STI tests.
Given the sheer number of sexual exposures that many of us will face throughout our lives, taking an active role in your personal health and wellness is more important than ever. At Zavamed.com, we offer a range of treatments with medical professionals on a variety of sexual health concerns such as sexually transmitted infections (STIs). Several STI tests are available, along with a wide range of discreet treatment options, including Extended STI Test. All of this is available from the privacy of your own home. Visit Zavamed.com today to find out how you can look out for your health, both now and in the future.
## Understanding STI Risks and Transmission
### HIV/AIDS
Modes of transmission: HIV can be transmitted through unprotected sexual intercourse (vaginal, anal, or oral) with an infected partner, sharing needles or syringes contaminated with HIV-infected blood, and from mother to child during childbirth or breastfeeding.
Factors influencing transmission: Unprotected sexual activity, especially if one partner is HIV-positive, increases the risk of transmission. Other factors include sharing needles or syringes with infected individuals and receiving contaminated blood or blood products.
HIV treatment: treatment for HIV/AIDS typically involves a combination of antiretroviral therapy (ART) medications. ART helps control the virus, reduces its replication in the body, and slows down the progression of the disease. Early diagnosis and timely initiation of treatment can significantly improve outcomes for individuals living with HIV/AIDS.
HIV Testing is crucial for early detection and allows individuals to access necessary medical care and support. ZAVA offers confidential HIV testing services that can be accessed through their website: ZAVA HIV Testing.
### Chlamydia
Modes of transmission: Chlamydia is primarily spread through vaginal, anal, or oral sex with someone who has the infection. It can be transmitted even if there is no penetration, orgasm, or ejaculation. The infection can be carried in semen, pre-cum, vaginal fluids, and genital fluids. Also pregnant individuals with chlamydia can pass the infection to their baby during childbirth.
Factors influencing transmission: Unprotected sexual activity is the main factor contributing to chlamydia transmission. Having multiple sexual partners and engaging in high-risk sexual behaviors can increase the chances of acquiring or spreading the infection.
Chlamydia treatment: It is important to promptly diagnose and treat chlamydia to prevent complications and reduce the risk of transmitting the infection to others. Chlamydia testing is vital, especially for sexually active individuals or those experiencing symptoms. ZAVA provides convenient and confidential chlamydia testing services. To learn more, you can visit: ZAVA Chlamydia Testing.
### Gonorrhoea
Modes of transmission: Gonorrhoea is mainly transmitted through sexual contact, including vaginal, anal, or oral sex, with an infected individual. It can also be passed from an infected mother to her baby during childbirth.
Factors influencing transmission: Unprotected sexual activity, particularly with an infected partner, significantly increases the risk of gonorrhoea transmission. Engaging in high-risk sexual behaviors and having multiple sexual partners can also contribute to the spread of the infection.
Gonorrhoea treatment: Early diagnosis and treatment are crucial to prevent complications and curb the spread of the disease. Testing for gonorrhoea is essential, especially for individuals who are sexually active or experiencing symptoms. ZAVA offers discreet and convenient gonorrhoea testing services. To find out more, you can visit: ZAVA Gonorrhoea Testing.
### Syphilis
Modes of transmission: Syphilis is primarily transmitted through sexual contact, including vaginal, anal, or oral sex, with an infected individual. It can also be transmitted from an infected mother to her baby during pregnancy or childbirth.
Factors influencing transmission: Unprotected sexual activity, particularly with an infected partner, increases the risk of syphilis transmission. Having multiple sexual partners and engaging in high-risk sexual behaviors can further amplify the likelihood of contracting or transmitting the infection.
Syphilis treatment: Syphilis can be effectively treated with antibiotics, particularly in the early stages. It is important to diagnose and treat syphilis promptly to prevent further complications. Syphilis testing is vital for early detection and allows for timely treatment. ZAVA provides confidential and reliable syphilis testing services. To access their testing options, you can visit: ZAVA Syphilis Testing.
### Herpes
Modes of transmission: Herpes can be transmitted through direct skin-to-skin contact with an active herpes sore or lesion. This includes vaginal, anal, or oral sex, as well as kissing or touching the affected areas. Herpes can be transmitted from the mouth to the genital area or vice versa through oral-genital contact. It can also be transmitted from a mother to her baby during childbirth.
Factors influencing transmission: The risk of herpes transmission is highest when an active outbreak or visible sores are present. However, the virus can also be transmitted even in the absence of symptoms. Unprotected sexual activity and close personal contact with an infected individual increase the chances of transmission.
Herpes treatment: while there is no cure for herpes, antiviral medications can help manage symptoms, reduce the frequency and duration of outbreaks, and lower the risk of transmission. Herpes testing is important to confirm the infection and guide appropriate management and preventive measures. ZAVA offers discreet and convenient herpes testing services. To learn more about their testing options, you can visit: ZAVA Herpes Testing.
Understanding the modes of transmission and the factors influencing STI transmission is crucial for promoting safer sexual practices and implementing preventive measures to reduce the spread of STIs.
## How to Use ZAVA's Sexual Exposure Calculator
Zavamed.com encourages the embedding of this interactive and the hosting of screenshots of it as well as the graph above. To embed the (responsive) calculator, you may access an embed code in the interactive above, and as seen here. When using any of these assets, or discussing the results of this calculator, we ask that you link to this page so users and those interested may learn more about the project.
### Methodology
Sexual exposure was derived using established formulas for a finite geometric series, which, when applied to this situation, allow the following formula for sexual exposure to be derived:
## Conclusion
In conclusion, ZAVA offers a comprehensive range of STI testing services and discreet treatment options, including Extended STI Test and medication for herpes. With the convenience of accessing these services from the privacy of your own home, you can take control of your sexual health and receive the necessary care and support.
Prompt testing and timely treatment are crucial for early detection, effective management, and prevention of STI transmission. By utilizing ZAVA's services, you can prioritize your well-being and make informed decisions about your sexual health.
## ZAVA Related Research
We use cookies to provide you with a better service. Carry on browsing if you are happy with this, or find out how to manage cookies. | 1,902 | 10,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-26 | longest | en | 0.967542 |
https://topgraderesearch.com/exponential-and-logarithmic-equations-and-modeling-with-exponential-functions/ | 1,670,612,843,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711475.44/warc/CC-MAIN-20221209181231-20221209211231-00086.warc.gz | 610,826,493 | 11,748 | # Exponential-and-Logarithmic-Equations-and-Modeling-with-Exponential-Functions
The half-life of a radioactive substance is the time in which the original amount is reduced to half.
The formula for continuously compounded interest is A(t) = Pert
In the formula, A(t) = P(1 + r/n)nt, n is the number of times interest is compounded per year. | 85 | 343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.930288 |
https://www.openmiddle.com/finding-the-length-of-a-right-triangles-altitude/ | 1,685,559,534,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00745.warc.gz | 1,005,231,329 | 22,738 | # Finding the Length of a Right Triangle’s Altitude
Directions: The black triangle is a right triangle with legs 8 and 6. The vertices are at the points (0,0), (0,8), and (6,0). The red line segment is perpendicular to hypotenuse. Find the length of the red line segment.
### Hint
1) What’s the area of the triangle?
2) Would knowing an angle measurement help?
3) Are there similar triangles?
4) Could you find the equation of any of the lines?
There are many methods to finding the answer. Here they are given based on the hint question:
METHOD 1:
The area of a triangle is 0.5(b)(h). In a right triangle, we can use the legs to calculate this, so 0.5(8)(6) = 24. But the red line segment is also the height of the triangle, since it is perpendicular to the hypotenuse, which can also act as a base. So it is also possible to calculate the area by doing 0.5(hypotenuse)(red line).
Pythagoras tells us that the hypotenuse is 10 (6^2 + 8^2 = 10^2), and we already know the area of the triangle is 24, so 24 = 0.5(10)(red line) –> 24 = 5x –> x = 4.8. The red line segment is 4.8.
METHOD 2:
We know that the legs of the right triangle are 6 and 8, so we can use inverse tan to find the base angle. Base angle = arctan(8/6). Base angle = 53.13…
We see that this angle is also in a smaller right triangle formed by the red line segment. In this triangle 6 is the hypotenuse and the red line is the opposite side from the angle we found. So we can do sin(53.13…) = red line/6. Red line = 4.8
*note, you could also do this finding the top angle and using the upper triangle formed by the red line
METHOD THREE:
The red line divides the big triangle into two smaller triangles, both of which are similar to the big triangle. I will work with the bottom triangle and the big triangle. We know they are similar because they both have an angle of 90 degrees and they share the angle at the point (6,0).
The big triangle has:
short leg = 6
long leg = 8
hypotenuse = 10 (Pythagoras)
The small triangle has:
short leg = ?
long leg = red line (x)
hypotenuse = 6
So we can set up a proportion with the long legs and the hypotenuses. x/8 = 6/10 -> 10x = 48, so the red line = 4.8
METHOD FOUR:
The hypotenuse line goes through the points (6,0) and (0,8). Therefore the gradient (slope) is -8/6 = -4/3. The y-intercept is 8. Therefore the equation of the line of the hypotenuse is:
y = -4/3x + 8
Since the red line is perpendicular to the hypotenuse line, its gradient must be 3/4 (negative reciprocal or m*m = -1). Since its y-intercept is 0, the equation of the red line is:
y = 3/4x
To find the point of intersection, we can set the two equations equal to each other:
-4/3x + 8 = 3/4x (multiply both sides by 12 to eliminate fractions)
-16x + 96 = 9x
96 = 25x
x = 3.84
plug in to find y:
y = 3/4 (3.84)
y = 2.88
So the red line goes from (0,0) to (3.84, 2.88). Now we can use the distance formula to calculate the length of the line segment:
distance^2 = 3.84^2 + 2.88^2
distance^2 = 23.04
distance = 4.8
so the red line = 4.8
Source: Kate Nerdypoo | 925 | 3,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2023-23 | latest | en | 0.922328 |
https://www.teacherspayteachers.com/Product/Tulips-for-Mom-Mystery-Picture-1831632 | 1,488,030,918,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171775.73/warc/CC-MAIN-20170219104611-00336-ip-10-171-10-108.ec2.internal.warc.gz | 905,720,664 | 25,465 | Total:
\$0.00
# Tulips for Mom Mystery Picture
Subjects
Resource Types
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4.0
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0.7 MB | 9 pages
### PRODUCT DESCRIPTION
First Quadrant: Few (15) Fractional Points
This coordinate plotting and drawing activity is ideal for a Mother’s Day project while teaching, reviewing or reinforcing the use of coordinates. The large grid is suited for students that are new to this concept but this project can also be completed by students that are more advanced in coordinate plotting. The students will create a drawing of potted tulips with the word MOM. This will be created by plotting ordered pairs and then connecting them with straight lines.
This activity can be a class project or something to be worked on independently when time allows. It also works as an extra credit assignment or when there is a substitute teacher.
The graph consists of points in the first quadrant using ordered pairs with whole numbers. There are 15 points which use the fraction 1/2. If you wish, you can plot those points on the coordinate sheet for them before you run off their copy or use them to begin to teach them to plot fractional points. When finished, they can use markers, colored pencils or other medium to enhance the project.
* Coordinate graph paper for plotting (with both a light and a dark grid)
* Coordinate list
* Sample of the completed project in color
Also included is:
* Student Directions for Plotting Coordinates
* Directions for Plotting Coordinates with Fractions and Decimals
* Directions for Plotting Coordinates in Quadrant I
* Directions for Plotting Coordinates in All Quadrants
* Details for Plotting Coordinates with Fractions in Quadrant I
* Details for Plotting Coordinates with Fractions in All Quadrants
If you like this project, please let us know. There are many projects covering the entire school year and some projects which are cross curricular (Sports, Patriotism, Seasons, Holidays, History, Foreign Language and Science). Lessons begin at 2nd grade and are appropriate for the Middle School and some for High School students. These projects have been used successfully at both the Elementary and Middle School levels.
http://www.teacherspayteachers.com/Store/Anthony-And-Linda-Iorlano
Anthony & Linda Iorlano
NCTM Standards
Specify locations and describe spatial relationships using coordinate geometry and other representational systems.
Common Core State Standards
CCSS.Math.Content.5.G.A.1 Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate).
Keywords: Keywords: Coordinate Drawing, Coordinate Drafting, Coordinate Graphing, Abscissa, Ordinate, Ordered Pair, Spring, Flowers, Easter, Flower, Tulip, mom, Mother’s Day
Total Pages
9
Included
Teaching Duration
N/A
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\$3.00 | 780 | 3,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-09 | longest | en | 0.922851 |
http://www.socscistatistics.com/effectsize/Default3.aspx | 1,532,283,519,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593438.33/warc/CC-MAIN-20180722174538-20180722194538-00536.warc.gz | 539,802,509 | 4,755 | # Effect Size Calculator for T-Test
For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation.
Cohen's d = (M2 - M1) SDpooled
where:
SDpooled = √((SD12 + SD22) 2)
Glass' Delta and Hedges' G
Cohen's d is the appropriate effect size measure if two groups have similar standard deviations and are of similar size. Glass' delta, which uses only the standard deviation of the control group, is an alternative measure if each group has a different standard deviation. Hedges' g, which provides a measure of effect size weighted according to the relative size of each sample, is an alternative where there are different sample sizes.
Please enter the sample mean (M), sample standard deviation (s) and sample size (n) for each group. Two things to note: (1) if you intend to report Glass's delta, then you need to enter your control group values as Group 1; and (2) if you don't provide values for n, the calculator will still calculate Cohen's d and Glass' delta, but it won't generate a value for Hedges's g.
Group 1
Mean (M): Standard deviation (s): Sample size (n):
Group 2
Mean (M): Standard deviation (s): Sample size (n):
Please enter your values in the boxes above, and press the "Calculate" button.
Cohen's d: Not yet calculated
Glass' delta: Not yet calculated
Hedges' g: Not yet calculated
| Privacy | Donate | Contact | About | ©2018 Jeremy Stangroom | | 358 | 1,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.903472 |
https://numbas.mathcentre.ac.uk/question/47746/viet-s-copy-of-simultaneous-equations-by-elimination-2.exam | 1,582,097,080,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00106.warc.gz | 483,955,396 | 2,345 | // Numbas version: exam_results_page_options {"name": "Viet's copy of Simultaneous equations by elimination 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"ungrouped_variables": ["a", "b", "ans1", "ans2", "ans3", "yCoef", "n1", "n2", "n3", "n4"], "variables": {"b": {"templateType": "anything", "definition": "random(-10..10 except 0 except a)", "description": "", "name": "b", "group": "Ungrouped variables"}, "ans3": {"templateType": "anything", "definition": "{ans1}*{n3} - {ans2}*{n1}", "description": "", "name": "ans3", "group": "Ungrouped variables"}, "ans2": {"templateType": "anything", "definition": "{n3}*{a}+{n4}*{b}", "description": "", "name": "ans2", "group": "Ungrouped variables"}, "a": {"templateType": "anything", "definition": "random(-10..10 except 0)", "description": "", "name": "a", "group": "Ungrouped variables"}, "n1": {"templateType": "anything", "definition": "random(-10..10 except 0)", "description": "", "name": "n1", "group": "Ungrouped variables"}, "n3": {"templateType": "anything", "definition": "random(-10..10 except 0 except n1)", "description": "", "name": "n3", "group": "Ungrouped variables"}, "ans1": {"templateType": "anything", "definition": "{n1}*{a} + {n2}*{b}", "description": "", "name": "ans1", "group": "Ungrouped variables"}, "yCoef": {"templateType": "anything", "definition": "{n3}*{n2} - {n1}*{n4}", "description": "", "name": "yCoef", "group": "Ungrouped variables"}, "n4": {"templateType": "anything", "definition": "random(-10..10 except 0 except n2)", "description": "", "name": "n4", "group": "Ungrouped variables"}, "n2": {"templateType": "anything", "definition": "random(-10..10 except 0)", "description": "", "name": "n2", "group": "Ungrouped variables"}}, "statement": "", "extensions": [], "tags": [], "parts": [{"prompt": "
Solve the pair of equations
\n
\$\\begin{eqnarray} \\simplify{{n1}*x + {n2}*y} & = & \\var{ans1} &&&&&&&(1)\\\\ \\simplify{{n3}*x + {n4}*y} & = & \\var{ans2}&&&&&&&(2)\\end{eqnarray}\$
\n
\n
We are going to solve for $y$ first. To do this, we need to eliminate $x$ from the equations. The quickest way to do this is to multiply the first equation by the co-efficient of $x$ in the second equation (here $\\var{n3}$), and multiply the second equation by the co-efficient of $x$ in the first equation (here $\\var{n1}$). We then get the equations:
\n
\n
[[0]]$x +$[[1]]$y = \\simplify{{ans1}*{n3}}$
\n
[[2]]$x +$[[3]]$y = \\simplify{{ans2}*{n1}}$
\n
\n
We then subtract one new equation from the other to get:
\n
[[4]]$\\simplify{y = {ans3}}$
\n
Now we can work out $y$
\n
$y =$[[5]]
\n
and substitute this value back in to any of the previous equations to get the value for $x$.
\n
$\\simplify{{n1}*x}$ + [[6]] = $\\var{ans1}$
\n
which then solves to give $x =$[[7]].
\n
\n
Straightforward solving linear equations question
Solve the pair of equations
\n
\$\\begin{eqnarray} \\simplify{{n1}*x + {n2}*y} & = & \\var{ans1} &&&&&&&(1)\\\\ \\simplify{{n3}*x + {n4}*y} & = & \\var{ans2}&&&&&&&(2)\\end{eqnarray}\$
\n
\n
We are going to solve for $y$ first. To do this, we need to eliminate $x$ from the equations. The quickest way to do this is to multiply the first equation by the co-efficient of $x$ in the second equation (here $\\var{n3}$), and multiply the second equation by the co-efficient of $x$ in the first equation (here $\\var{n1}$). We then get the equations:
\n
\n
$\\simplify{{n3}*{n1}*x + {n3}*{n2}*y = {ans1}*{n3}}$
\n
$\\simplify{{n1}*{n3}*x + {n1}*{n4}*y = {ans2}*{n1}}$
\n
\n
We then subtract one new equation from the other to get:
\n
$\\simplify{{yCoef}y = {ans3}}$
\n
Now we can work out $y$
\n
$y = \\var{b}$
\n
and substitute this value back in to any of the previous equations to get the value for $x$.
\n
$\\simplify{{n1}*x + {n2}*{b} = {ans1}}$
\n
which then solves to give $x = \\var{a}$.
\n
\n
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https://www.numbersaplenty.com/1345333 | 1,701,912,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00060.warc.gz | 1,012,822,950 | 3,072 | Search a number
1345333 = 111941157
BaseRepresentation
bin101001000011100110101
32112100110011
411020130311
5321022313
644500221
714302153
oct5103465
92470404
101345333
11839850
1254a671
13381472
142703d3
151b893d
hex148735
1345333 has 16 divisors (see below), whose sum is σ = 1592640. Its totient is φ = 1123200.
The previous prime is 1345303. The next prime is 1345343. The reversal of 1345333 is 3335431.
It is a cyclic number.
It is not a de Polignac number, because 1345333 - 25 = 1345301 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (1345303) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 8491 + ... + 8647.
It is an arithmetic number, because the mean of its divisors is an integer number (99540).
21345333 is an apocalyptic number.
It is an amenable number.
1345333 is a deficient number, since it is larger than the sum of its proper divisors (247307).
1345333 is a wasteful number, since it uses less digits than its factorization.
1345333 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 228.
The product of its digits is 1620, while the sum is 22.
The square root of 1345333 is about 1159.8849080836. The cubic root of 1345333 is about 110.3934395825.
The spelling of 1345333 in words is "one million, three hundred forty-five thousand, three hundred thirty-three". | 453 | 1,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-50 | latest | en | 0.874119 |
https://de.scribd.com/document/192277109/Herimite-Shape-Function-for-Beam | 1,568,999,466,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574050.69/warc/CC-MAIN-20190920155311-20190920181311-00107.warc.gz | 439,510,635 | 74,065 | Sie sind auf Seite 1von 29
# Beams and frames
M = y I = / E d 2 v / dx 2 = M / EI
## Potential energy approach
Strain energy in an element of length dx is
1 M2 2 y dA dx = 2 2 EI A
2 y dA is the moment of inertia I A
## The total strain energy for the beam is given by-
1 2 2 U = EI d v / dx dx 20
## Potential energy of the beam is then given by-
1 2 2 ' = EI d v / dx dx pvdx pm vm M k vk 20 m k 0
Where-p is the distributed load per unit length -pm is the point load at point m. -Mk is the moment of couple applied at point k -vm is the deflection at point m -vk is the slope at point k.
Galerkins Approach
p M V+dV M+dM V dx Here we start from equilibrium of an elemental length. dV/dx = p dM/dx =V
d 2 v / dx 2 = M / EI
2 d2 d v 2 p=0 EI dx dx 2
## For approximate solution by Galerkins approach-
2 d d v EI 2 p dx = 0 2 dx dx
## is an arbitrary function using same basic functions as v
Integrating the first term by parts and splitting the interval 0 to L to (0 to xm), (xm to xk) and (xk to L) we getd vd d d v EI 2 dx pdx + EI 2 2 dx dx dx dx 0 0 0
L 2 2 l 2 xm
d d v d v d d v d EI EI EI 2 2 2 dx dx dx dx dx dx x 0
2 2 2
m
xk
=0
xk
Further simplifying-
d 2v d 2 ' EI dx p dx p M m m k k =0 2 2 dx dx m k 0 0
L L
and M are zero at support..at xm shear force is pm and at xk Bending moment is -Mk
FINITE ELEMENT FORMULATION Beam is divided in to elementseach node has two degrees of freedom. Degree of freedom of node j are Q2j-1 and Q2j Q2j-1 is transverse displacement and Q2j is slope or rotation. Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10
e1
e2
e3
e4
Q = [Q1, Q2 , Q3 KQ10 ]
## Q is the global displacement vector.
Local coordinatesq1 q2 e q3 q4
T
q = [q1 , q2 , q3 , q4 ] = [v1 , v , v2 , v ]
' 1 ' 2
The shape functions for interpolating v on an element are defined in terms of from 1 to 1. The shape functions for beam elements differ from those defined earlier. Therefore, we define Hermite Shape Functions
Slope=0 1
H1
Slope=0 Slope=1 Slope=0
H2
Slope=0
H3
Slope=0
Slope=1
Slope=0
H4
## Each Hermite shape function is of cubic order represented by-
H i = ai + bi + ci 2 + d i 3 K i = 1,2,3,4
The condition given in following table must be satisfied.
H1 =-1 1 =1 0
H1 H2 0 0 0 0
H2 H3 1 0 0 1
H3 H4 0 0 0 0
H4 0 1
## Finding out values of coefficients and simplifying,
1 2 H1 = (1 ) (2 + ) 4 1 2 H 2 = (1 ) ( + 1) 4 1 2 H 3 = (1 + ) (2 + ) 4 1 2 H 4 = (1 + ) ( 1) 4
## Hermite functions can be used to write v in the form-
dv v( ) = H1v1 + H 2 d
dv + H 3v3 + H 4 d 1
## dv le dv = d 2 dx Therefore, le le v( ) = H1q1 + H 2 q2 + H 3 q3 + H 4 q4 2 2 v = Hq where le le H = H1 , H 2 , H 3 , H 4 2 2
1 2 2 U e = EI d v / dx dx 2e dv 2 dv = dx le d
2
and
d v 4 d v = 2 dx le d 2
T
## substituting in above equation d v T 16 d H =q 4 2 2 dx le d
2
Where-
d 2H d 2 q
d 2H d 2
3 1 + 3 le 3 1 + 3 le , , = 2 , 2 2 2 2 2
Note that-
2 2 d = 3 1
d = 0 d = 2
1
## Where Ke is element stiffness matrix given by
12 6le 12 6le 2 2 EI 6le 4le 6le 2le ke = 3 le 12 6le 12 6le 2 2 6le 2le 6le 4le
It can be seen that it is a symmetric matrix.
ple 1 pvdx = 2 Hd q le 1
Substituting the value of H we get-
le
pvdx = f
e
eT
2 e 2 e T
## This is equivalent to the element shown belowp 1
le Ple/2 e 2
Ple/2 Ple
2/12
-Ple2/12
The point loads Pm and Mk are readily taken care of by introducing the nodes at the point of application.
1 T = Q KQ QF 2
T T
## KQ F = 0 where = admissible virtual displacement vector.
BOUNDARY CONSIDERATIONS Let Qr = a.single point BC Following Penalty approach, add 1/2C(Qr-a)2 to C represents stiffness which is large in comparison with beam stiffness terms. C is added to Krr and Ca is added to Fr to getKQ = F These equations are solved to get nodal displacements. Ca C Dof =(2i-1) C Ca Dof = 2i
## Shear Force and Bending MomentWe have,
dM d 2v M = EI 2 V = dx dx and v = Hq
V1 = R1
V2 = -R3
M1 = -R2
M 2 = R4
## Beams on elastic support
Shafts supported on ball, roller, journal bearings Large beams supported on elastic walls. Beam supported on soil (Winkler foundation). Stiffness of support contributes towards PE. Let s be the stiffness of support per unit length.
1 l 2 additional term = sv dx 2 0 v = Hq 1 = q T s H T H dx q 2 e e
1 = q T kes 2 e where kes is stiffness matrix for elastic foundation 156 22le 54 13le 2 2 sle 22le 4le 13le 3le s ke = 420 54 13le 156 22le 2 2 13 l 3 l 22 l 4 l e e e e
PLANE FRAMES
Plane structure with rigidly connected members. Similar to beams except that axial loads and deformations are present. We have 2 displacements and 1 rotation at each node. 3 dof at each node. q =[q1, q2, q3, q4, q5, q6]T
q6 (q6)
Y X
q2 q1 q1 q3 (q3)
q2
Y X
## q =[q1, q2, q3, q4, q5, q6]
l,m are the direction cosines of local coordinate system. XY l = cos() m =sin() We can see from the figure thatq3 = q3 q3 = q6 which are rotations with respect to body.
q = Lq Where-
l m 0 L= 0 0 0
m 0 0 0 0 l 0 0 0 0 0 1 0 0 0
0 0 l m 0 0 0 m l 0 0 0 0 0 1
q2, q3,q5 and q6 are beam element dof while q1andq4 are like rod element dof.
Combining two stiffness and rearranging at proper locations we get element stiffness matrix as EA l e 0 0 k 'e = EA le 0 0 EA 0 le 6 EI le2 4 EI le 0 12 EI le3 0 0 0 EA le 6 EI le2 0 12 EI 6 EI 3 le le2 6 EI 2 EI 2 le le 0 0 12 EI 6 EI 0 2 3 le le 6 EI 4 EI 2 le le 0 0
0 12 EI le3 6 EI le2
6 EI le2
2 EI le
Strain energy of an element is given byUe = 1 T 'e q' k q' 2 1 = q T LT k 'e Lq 2 by Galerkin ' s approach , = T LT k 'e Lq
Y Y X
2 e 2 e
KQ = F | 2,192 | 5,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-39 | latest | en | 0.751808 |
https://www.studypool.com/discuss/470403/physics-electric-and-current-circuit?free | 1,506,140,402,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689471.25/warc/CC-MAIN-20170923033313-20170923053313-00462.warc.gz | 872,221,700 | 14,763 | Time remaining:
physics electric and current circuit
label Physics
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5
A carbon rod with a radius of 1.4 mm is used
to make a resistor.
What length of the carbon rod should be
used to make a 1.3 Ω resistor? The resistivity
of this material is 3.9 × 10−5 Ω · m .
Apr 11th, 2015
R = ρL/A
ρ is resistivity of the material in Ω-m
L is length in meters
A is cross-sectional area in m²
A = πr² = π(0.0014)² = 4.39e-3 m²
1.3 = (3.9e-5) L / (4.39e-3)
L = [ 1.3 * (4.39e-3) ] / (3.9e-5)
L = 146.33 mm or 0.14633 m
Apr 11th, 2015
...
Apr 11th, 2015
...
Apr 11th, 2015
Sep 23rd, 2017
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Secure Information
Content will be erased after question is completed.
check_circle | 285 | 777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-39 | latest | en | 0.845662 |
https://www.albert.io/ie/gre/finding-area-of-a-rectangle-given-clues-about-the-length-and-width | 1,484,727,196,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280242.65/warc/CC-MAIN-20170116095120-00214-ip-10-171-10-70.ec2.internal.warc.gz | 889,869,322 | 20,107 | Free Version
Moderate
# Finding Area of a Rectangle Given Clues About the Length & Width
GRE-QN8ZBV
The width of a rectangle is $2x+3$. The length is five less than twice the width. What is the rectangle's area?
A
$4x+1$
B
$6x+4$
C
$8{ x }^{ 2 }+3$
D
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January 11, 2016, 11:04 Mathematics, Numerics, Derivations and OpenFOAM #1 Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51 Dear all, during the derivation of the stress equations that are implemented in OpenFOAM, I started to derive the mass, momentum and (not complete) the energy equation. Furthermore, I showed and proofed that the vector form of the equations are equal to the Cartesian notation. The content is: Derivation of mass and momentum equation General conserved equation Energy equation Reynolds-Averaging Shear-Rate Tensor Mathematics Numerics Turbulence modelling More about divDevRhoREff and divDevReff Derive shear-rate tensor for pisoFoam Different notations for the NS-Equations Matrix Algebra Cauchy - Stress tensor, shear-rate tensor and pressure http://www.holzmann-cfd.de/index.php/en/publications Feel free to comment on that and I hope that it is helpful, clear and understandable. Happy new year to everybody philippose, balkrishna, olivierG and 39 others like this. __________________ Keep foaming, Tobias Holzmann Last edited by Tobi; January 22, 2016 at 05:03. Reason: Changed the link
January 20, 2016, 13:51 #2 Senior Member Jon Elvar Wallevik Join Date: Nov 2010 Location: Reykjavik, ICELAND Posts: 103 Rep Power: 19 Nice work Tobias! I have newer seen the name “shear rate tensor” for tau (usually referred to as the extra stress tensor, as you also use). What you probably mean with “shear rate tensor” is when “tau” represents the GNM (the Generalized Newtonian Model). But the extra stress tensor “tau” can be much nastier than GNM. Regarding Chapter 6, if you want also to explain why shear rate is calculated as is done in OpenFOAM (sqrt(2 e:e)), then in my paper I show proof of that in Section 3. It also explains the GNM and it relationship with non-Newtonian fluids, with some good references (the paper is using the MRF in OpenFOAM for non-Newtonian fluids). If you use this, please make ref to my paper. Thanks. Again, nice work man. Cheers J. P.s. the reference is... J.E. Wallevik, Effect of the hydrodynamic pressure on shaft torque for a 4-blades vane rheometer, International Journal of Heat and Fluid Flow, 50 (2014) 95–102. DOI: 10.1016/j.ijheatfluidflow.2014.06.001
January 20, 2016, 15:11 #3 Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51 Hi Jon, thanks for the feedback and the hint to your paper. My colleague also uses your shear-rate calculation for visco-plastic modeling but at the moment there was no time also to include Non-Newtonian fluids. Mostly (as you already realized, I talk about Newtonian fluids). To the "extra-stress tensor" tau. In many of my literature (maybe due to the fact that most of them are German books, are using this form. Also Ferziger and Perić use this nomenclature. That the extra stress tensor could be much worser is clear, especially if we have Non-Newtonian fluids. If I have access to the paper (I think I have it at my university), I will check it out and extend it. But at the moment I have a lot of other stuff to do. By the way, there is still a problem in my reynolds-averaging that I do not know exactly. Maybe you can answer this question: I showed how we get the RANS equation and used the Buossinesq approximation where I said that the term is neglected. The question is, is it really neglected because Pope et. al. wrote that we can take this term to the pressure and get a modified pressure p': Do you know if we calculate p' in OpenFOAM or do you have any other hint to that? I am not 100% sure about it. The hint came from Gerhald Holzinger (Linz). Thanks again, Tobi __________________ Keep foaming, Tobias Holzmann
January 21, 2016, 05:30 #4 Senior Member Jon Elvar Wallevik Join Date: Nov 2010 Location: Reykjavik, ICELAND Posts: 103 Rep Power: 19 I will take a look,... pretty max out right now, but I will give feedback soon on this, if I have anything on this or not. cheers J.
January 22, 2016, 03:16
#5
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Likun
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Quote:
Originally Posted by Tobi By the way, there is still a problem in my reynolds-averaging that I do not know exactly. Maybe you can answer this question: I showed how we get the RANS equation and used the Buossinesq approximation where I said that the term is neglected. The question is, is it really neglected because Pope et. al. wrote that we can take this term to the pressure and get a modified pressure p': Do you know if we calculate p' in OpenFOAM or do you have any other hint to that? I am not 100% sure about it. The hint came from Gerhald Holzinger (Linz). Thanks again, Tobi
Hi Tobis,
I had the same question before, see this post http://www.cfd-online.com/Forums/ope...-stresses.html. And I realized that the second term is indeed negligible compared to p, unless k is very large.
Best,
Likun
January 24, 2016, 14:00 #6 Senior Member Jon Elvar Wallevik Join Date: Nov 2010 Location: Reykjavik, ICELAND Posts: 103 Rep Power: 19 Hi Tobi and Likun When going into the source code of simpleFoam, I cannot see it is possible that the solver is working with a modified pressure. If this were the case, the pressure equation would have to get the term 2/3*k*I from somewhere, but there is no hint of this term anywhere in the solver. However, I did another test. I went into ~/OpenFOAM/OpenFOAM-2.4.x/src/turbulenceModels/incompressible/RAS/kEpsilon/kEpsilon.C ...and did this (see the member function R() above the member divDevReff())... tmp kEpsilon::divDevReff(volVectorField& U) const { return ( // - fvm::laplacian(nuEff(), U) // - fvc::div(nuEff()*dev(T(fvc::grad(U)))) // -------------------------------------------------------- // Y16M01D24: - fvm::laplacian(nuEff(), U) - fvc::div( nuEff()*dev(T(fvc::grad(U))) - ((2.0/3.0)*I)*k_ ) ); } Thereafter, I tested pitzDaily with and without the modification, after running stressComponents, I saw no difference between the the two cases. So if one want, it is possible to add this, but at least in the pitsDaily case, it is not important. Note, with modification, you have to add div(((nuEff*dev(T(grad(U))))-((0.666667)*k))) Gauss linear; into ./system/fvSchemes (in divSchemes) J. Likun likes this.
January 26, 2016, 06:22 #7 Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51 Hey Jon, thanks for the clearness. As I supposed in my paper it is neglected. Thanks for checking this. @all. I found a mistake in the enthalpy equation of chapter 3. I changed it and added one more hint. The correct diffusion term is (of course): Also the temperature equation has to be changed. The new PDF is up to date. __________________ Keep foaming, Tobias Holzmann
January 30, 2016, 07:54 #8 Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51 Hi all, again I found some mistakes in my paper. I will change it the next days and let you know when it is finished and uploaded. Cheers ssss likes this. __________________ Keep foaming, Tobias Holzmann
January 31, 2016, 11:08
#9
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Sergei
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Quote:
Originally Posted by Tobi Feel free to comment on that and I hope that it is helpful, clear and understandable.
I think it would be better if your text was more in line with the books (Ferziger, Versteeg, Anderson) you reference to. Specifically for the part when you derive momentum equation (2.14). The second term in the right hand side of the equation comes with minus sign (-nabla . tau); so does components of a shear stress tensor in equations (4.1)-(4.6). Even though those minuses just cancel out in the end and the final equation (4.13) is the same as in the books, it would be more easy to follow your text if your text followed the books.
Looking at the energy equation (3.4) it is not clear where all the terms come from. I suggest that you should derive the energy equation the way you did with continuum and momentum equations, i.e. start with an infinitesimal fluid volume, formulate the equation in Cartesian scalar notation and transform it into a vector form. Here is the useful link: http://cfd.direct/openfoam/energy-equation/
Misprints:
eq. (2.21): rho was left out.
eq. (4.27): nabla was left out.
Anyway, good job!
January 31, 2016, 15:14
#10
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Tobias Holzmann
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Quote:
Originally Posted by Zeppo Very interesting! Can this article develop into anything big?
It could be, I can offer the LaTeX files and everybody can include some stuff.
Quote:
I think it would be better if your text was more in line with the books (Ferziger, Versteeg, Anderson) you reference to. Specifically for the part when you derive momentum equation (2.14). The second term in the right hand side of the equation comes with minus sign (-nabla . tau); so does components of a shear stress tensor in equations (4.1)-(4.6). Even though those minuses just cancel out in the end and the final equation (4.13) is the same as in the books, it would be more easy to follow your text if your text followed the books.
Of course the minus signs will turn to "+" at least but due to my derivation we get the -nabla . tau at the beginning. I followed the book of Bird (Transport Phenomena) here very close.
Quote:
Looking at the energy equation (3.4) it is not clear where all the terms come from. I suggest that you should derive the energy equation the way you did with continuum and momentum equations, i.e. start with an infinitesimal fluid volume, formulate the equation in Cartesian scalar notation and transform it into a vector form. Here is the useful link: http://cfd.direct/openfoam/energy-equation/
Of course it would be the best but I had no time for that and just took the term from the books. If you want, I will do the same work for energy equation to demonstrate where the terms come from (good hint). But for it was clear after the first two derivation how to do it and I was not sure if anybody is think that this could be useful. At least due to the fact, that a lot of books are out.
Quote:
Misprints: eq. (2.21): rho was left out. eq. (4.27): nabla was left out.
Thank you for the good reviewing.
Quote:
Anyway, good job!
Thank you again. Very nice critic. Now I can improve it more.
PS: I also found other mistakes (icoFoam) and derivation to POISSON equation. These are fixed. I will also fix the stuff you mentioned and maybe add Non-Newtonian + Energy-derivation to the documentation.
Good night to everybody.
Tobi
__________________
Keep foaming,
Tobias Holzmann
February 15, 2016, 10:09
#11
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I have recently written a document about the derivation of the enthalpy equation from the total energy equation and the handling of source terms. I figured it might be of some interest here so I'm posting it. Please let me know what you think and if there are errors.
Edit: Some updates to the document.
Edit 2: Added heat conduction term which arises when h=h(p,T).
Attached Files
Derivation of the enthalpy equation.pdf (107.0 KB, 130 views)
Last edited by chriss85; February 17, 2016 at 08:18.
February 17, 2016, 04:40 #12 Senior Member Join Date: Oct 2013 Posts: 397 Rep Power: 18 Has anyone checked the derivation for errors? I recently learned in a discussion with mkraposhin that there is also an additional heat conduction term when the enthalpy function depends on pressure. I have updated the document to reflect this. I have done some tests to investigate the influence of the additional mass and momentum source terms in the energy equation as well as the additional heat conduction term. In my case they provide small but visible differences so I would suggest to include them. Edit: Actually the inclusion of the E*S_M term in the enthalpy equation looks to be significant for my case at later times in the simulation. Last edited by chriss85; February 17, 2016 at 08:19.
February 17, 2016, 08:40 #13 Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51 Hey Chris, I checked your summary and its fine. Just a few hints. You never mentioned what the term stand for. Of course it is the mass source term but you should mention that it is zero. Normally we are not dealing with fusion or sth. like that. Additional, people who are not familiar with the mathematic would not know why you multiply the continuity equation by the energy E eqn. (5). Therefore (it is my opinion) the derivation is in the wrong direction. For me it would be clear to derive the conserved energy equation and use the conti to go to the non-conserved form. You did it vice versa and hence, people can get confused why you e.g. multiply the conti by E. Nontheless, nice stuff. PS: A lot of authors use the total derivative to derive conserved equations. __________________ Keep foaming, Tobias Holzmann
February 17, 2016, 08:53
#14
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Quote:
Originally Posted by Tobi You never mentioned what the term stand for. Of course it is the mass source term but you should mention that it is zero. Normally we are not dealing with fusion or sth. like that.
Agreed, some explanation of the terms might be useful. I included a mass source because it can appear in ablation or evaporation models that don't implement it through a boundary condition but as source in the first cell layer. If you can point me to a proper implementation of such models through boundary conditions I would be quite happy .
Quote:
Additional, people who are not familiar with the mathematic would not know why you multiply the continuity equation by the energy E eqn. (5). Therefore (it is my opinion) the derivation is in the wrong direction. For me it would be clear to derive the conserved energy equation and use the conti to go to the non-conserved form. You did it vice versa and hence, people can get confused why you e.g. multiply the conti by E.
In this case I started with the equation for the total energy which is present in Versteeg, "Computational Fluid Dynamics" and derived the equation which is used in the pisoCentralFoam solver by mkraposhin while considering these additional terms. I feel that this is the most generic solution because you can still just omit terms from the equations which are zero.
Quote:
PS: A lot of authors use the total derivative to derive conserved equations.
Yes, but I wanted to show how to get to the final equation that is used in the solver.
February 17, 2016, 10:30
#15
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Tobias Holzmann
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Quote:
Originally Posted by chriss85 In this case I started with the equation for the total energy which is present in Versteeg, "Computational Fluid Dynamics" and derived the equation which is used in the pisoCentralFoam solver by mkraposhin while considering these additional terms. I feel that this is the most generic solution because you can still just omit terms from the equations which are zero.
I know, also in other literature you will start with the total derivative (but here you miss always the conti equation) and you start with a non-conservative form. Of course you will end up in a conservative form because you use the continuity equation but it is not as straight forward as derive the conservative form directly and then go to the non-conservative form. If you are familiar with the guys its not a problem, but for other people its "like" a mystery. For example you use the continuity equation:
and the total derivative:
Till now everything is clear for everybody. Next step is to add the continuity equation to the total derivative like:
Now (maybe) some people are wondering why you multiply the conti by . For me it is clear why you do this and why we have to multiply by . Everybody who is familiar with the mathematic background will not complain about that but beginners or non-familiar math guys get confused or do not know why has to be added
Thats the reason why I said (in my opinion) the better way would be to derive the conservative form first and then do simplifications or not. Why I tell you this, because I got really confused about all the equations. Everywhere and everybody uses other derivations, start from integrals, total derivatives etc. It like you read this book and this and you get more confused about the equations. Of course maybe I am a stupid guy who need a strict forward line (maybe); that was the reason why I derived everything out of a volume element. Then you really get the conserved equations, which can be simplified using the conti to non-conserved equations. If you did this once, of course you can do it vice versa (:
Anyway Well done.
__________________
Keep foaming,
Tobias Holzmann
February 18, 2016, 09:57 #16 Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51 Dear all, I investigated time to extend the paper / summary as follow: Math section extended and cleared (also mistakes removed) Total-Derivatives and math operations included Derivation of kinetic energy done Derivation of total energy done Derivation of internal energy done Derivation of non-conserved equations of momentum, total energy, internal energy and kinetic energy added The document is now 45 pages big. Further improvements: speech and logical structure. I will need a bit more time to get it ready but I will keep you updated. Notice that the derivations are done using a volume element. Therefore we get the conservative forms straight forward. Xinze, JonW, chriss85 and 1 others like this. __________________ Keep foaming, Tobias Holzmann
February 19, 2016, 16:19 #17 Senior Member Matvey Kraposhin Join Date: Mar 2009 Location: Moscow, Russian Federation Posts: 355 Rep Power: 21 Dear colleagues, Do you keep in mind, that OF is intended to solve integral equations, but not differential. __________________ MDPI Fluids (Q2) special issue for OSS software: https://www.mdpi.com/journal/fluids/..._modelling_OSS GitHub: https://github.com/unicfdlab Linkedin: https://linkedin.com/in/matvey-kraposhin-413869163 RG: https://www.researchgate.net/profile/Matvey_Kraposhin
February 20, 2016, 11:34 #18 Senior Member Jon Elvar Wallevik Join Date: Nov 2010 Location: Reykjavik, ICELAND Posts: 103 Rep Power: 19 Dear Matvey I am pretty sure that most of us know that OF uses the finite volume approach. It is fairly straight forward to transform an integral equation into a differential one. See for example p. 387 in my PhD thesis @ http://ntnu.diva-portal.org/smash/re...2:124984&rvn=1 Also, when programming a solver, a differential approach is used (e.g. UEqn.H in interFoam), which OF transforms into an integral approach (in this case, by the fvm::div(), fvm::laplacian(), fvm::grad() and etc -class)), thus in the end, it is nicer to have the equations in its differential state, rather than in integral state. Hope this helps cheers J. Tobi and SHUBHAM9595 like this.
February 28, 2016, 07:21 #19 Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Tussenhausen Posts: 2,708 Blog Entries: 6 Rep Power: 51 Dear all, I am happy to announce that I have published the new paper. A lot of changes are done and can not compared to the original (: The document has 45 pages and include the following: Mathematic Notation Derivation of governing equationsContinuity Momentum Total Energy Mechanical (Kinetic) Energy Internal (Thermo) Energy Enthalpy Governing equations for engineersContinuity Momentum Enthalpy SUMMARY OF EQUATIONS Shear-Rate tensor and Navier-StokeNewtonian Fluids Relation between Cauchy, shear-rate and pressure Derivatives of momentum equations Turbulence modeling Derivations of shear-rate functions (openfoam) The equations are given in the following forms in the first part: Conserved partial Non-Conserved partial Integral Finally I added the literature I used and gave some cites. After I restarted working on the stuff, I realized that a lot of stuff is not straight forward and rearranged a lot. Furthermore the english speak was improved and checked. A few mathematic errors were found and corrected. Everything can be found on my updated page www.holzmann-cfd.de Feedback, error-report, etc. is welcomed, ramakant, JonW, Dipsomaniac and 4 others like this. __________________ Keep foaming, Tobias Holzmann
February 29, 2016, 03:41
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Originally Posted by Tobi Everything can be found on my updated page www.holzmann-cfd.de Feedback, error-report, etc. is welcomed, | 5,224 | 21,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-22 | longest | en | 0.902618 |
http://studentsmerit.com/paper-detail/?paper_id=41851 | 1,481,142,720,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542246.21/warc/CC-MAIN-20161202170902-00458-ip-10-31-129-80.ec2.internal.warc.gz | 263,124,031 | 8,282 | #### Details of this Paper
##### Determine the number of personal and dependency exemptions in each of the following
Description
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Question;33. LO.3, 4 Determine the number of personal and dependency exemptions in each of the following independent situations.a. Leo and Amanda (ages 48 and 46, respectively) are husband and wife and furnish more than 50% of the support of their two children, Elton (age 18) and Trista (age 24). During the year, Elton earns \$4,500 providing transportation for elderly persons with disabilities, and Trista receives a \$5,000 scholarship for tuition at the law school she attends.b. Audrey (age 45) is divorced this year. She maintains a household in which she, her ex-husband, Clint, and his mother, Olive, live. Audrey furnishes more than 50% of the household?s support. Olive is age 91 and blind.c. Crystal, age 45, furnishes more than 50% of the support of her married son, Andy (age 18), and his wife, Paige (age 19), who live with her. During the year, Andy earned \$8,000 from a part-time job. All parties live in Iowa (a common law state).d. Assume the same facts as in (c), except that all parties live in Washington (a community property state).34. LO.3, 4 Determine the number of personal and dependency exemptions in each of the following independent situations.a. Reginald, a U.S. citizen and resident, contributes 100% of the support of his parents, who are citizens of Canada and live there.b. Pablo, a U.S. citizen and resident, contributes 100% of the support of his parents, who are citizens of Panama. Pablo?s father is a resident of Panama, and his mother is a legal resident of the United States.c. Gretchen, a U.S. citizen and resident, contributes 100% of the support of her parents, who are U.S. citizens but are residents of Germany.35. LO.3, 4 Determine how many personal and dependency exemptions are available in each of the following independent situations. Specify whether any such exemptions would come under the qualifying child or the qualifying relative category.a. Andy maintains a household that includes a cousin (age 12), a niece (age 18), and a son (age 26). All are full-time students. Andy furnishes all of their support.b. Jackie provides all of the support of a family friend?s son (age 20) who lives with her.She also furnishes most of the support of her stepmother, who does not live with her.c. Raul, a U.S. citizen, lives in Costa Rica. Raul?s household includes his friend Helena, who is age 19 and a citizen of Costa Rica. Raul provides all of Helena?s support.d. Karen maintains a household that includes her ex-husband, her mother-in-law, and her brother-in-law (age 23 and not a full-time student). Karen provides more than half of all of their support. Karen is single and was divorced last year.36. LO.4 Jenny, age 14, lives in a household with her father, uncle, and grandmother. The household is maintained by the uncle. The parties, all of whom file separate returns, have AGI as follows: father (\$30,000), uncle (\$50,000), and grandmother (\$40,000).a. Who is eligible to claim Jenny as a dependent on a Federal income tax return?b. Which of Jenny?s relatives takes precedence in claiming the exemption? Explain.37. LO.3, 4 Sam and Elizabeth Jefferson file a joint return. They have three children, all of whom qualify as dependents. If the Jeffersons report 2013 AGI of \$322,000, what is their allowable deduction for personal and dependency exemptions?38. LO.4, 9 Wesley and Myrtle (ages 90 and 88, respectively) live in an assisted care facility and for years 2012 and 2013 received their support from the following sources.Percentage of SupportSocial Security benefits 16%Son 20Niece 29Cousin 12Brother 11Family friend (not related) 12a. Who is eligible to claim the Federal income tax dependency exemptions under a multiple support agreement?b. Must Wesley and Myrtle be claimed as dependents by the same person(s) for both2012 and 2013? Explain.c. Who, if anyone, can claim an itemized deduction for paying the medical expenses ofWesley and Myrtle?
Paper#41851 | Written in 18-Jul-2015
Price : \$22 | 988 | 4,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2016-50 | longest | en | 0.958153 |
https://aprove.informatik.rwth-aachen.de/eval/JBC-Recursion/RTA11Examples/CAppE/CAppE.jar.AProVE%202011.html | 1,725,753,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00325.warc.gz | 81,582,928 | 5,294 | ### (0) Obligation:
JBC Problem based on JBC Program:
Manifest-Version: 1.0 Created-By: 1.6.0_22 (Sun Microsystems Inc.) Main-Class: CAppE
`public class CAppE { CAppE next; public static void main(String[] args) { Random.args = args; CAppE list = createList(); cappE(Random.random()); } public static void cappE(int j) { CAppE a = new CAppE(); if (j > 0) { a.appE(j); while (a.next == null) {} } } public void appE(int i) { if (next == null) { if (i <= 0) { return; } else { next = new CAppE(); } i--; } next.appE(i); } public static CAppE createList() { CAppE result = null; int length = Random.random(); while (length > 0) { result = new CAppE(result); length--; } return result; } public CAppE() { this.next = null; } public CAppE(CAppE n) { this.next = n; }}class Random { static String[] args; static int index = 0; public static int random() { String string = args[index]; index++; return string.length(); }}`
### (1) JBC2FIG (SOUND transformation)
Constructed FIGraph.
### (2) Obligation:
FIGraph based on JBC Program:
CAppE.main([Ljava/lang/String;)V: Graph of 118 nodes with 0 SCCs.
CAppE.createList()LCAppE;: Graph of 91 nodes with 1 SCC.
CAppE.appE(I)V: Graph of 35 nodes with 0 SCCs.
### (3) FIGtoITRSProof (SOUND transformation)
Transformed FIGraph SCCs to IDPs. Logs:
Log for SCC 0:
Generated 25 rules for P and 9 rules for R.
Combined rules. Obtained 1 rules for P and 3 rules for R.
Filtered ground terms:
757_0_appE_FieldAccess(x1, x2) → 757_0_appE_FieldAccess(x2)
Cond_757_0_appE_FieldAccess(x1, x2, x3) → Cond_757_0_appE_FieldAccess(x1, x3)
874_0_appE_Return(x1) → 874_0_appE_Return
821_0_appE_Return(x1) → 821_0_appE_Return
768_0_appE_Return(x1, x2) → 768_0_appE_Return
Combined rules. Obtained 1 rules for P and 3 rules for R.
Finished conversion. Obtained 1 rules for P and 3 rules for R. System has predefined symbols.
Log for SCC 1:
Generated 17 rules for P and 3 rules for R.
Combined rules. Obtained 1 rules for P and 1 rules for R.
Filtered ground terms:
366_0_createList_LE(x1, x2, x3) → 366_0_createList_LE(x2, x3)
Cond_366_0_createList_LE(x1, x2, x3, x4) → Cond_366_0_createList_LE(x1, x3, x4)
395_0_createList_Return(x1) → 395_0_createList_Return
Filtered duplicate args:
366_0_createList_LE(x1, x2) → 366_0_createList_LE(x2)
Cond_366_0_createList_LE(x1, x2, x3) → Cond_366_0_createList_LE(x1, x3)
Combined rules. Obtained 1 rules for P and 1 rules for R.
Finished conversion. Obtained 1 rules for P and 1 rules for R. System has predefined symbols.
### (5) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
807_1_appE_InvokeMethod(768_0_appE_Return, 0) → 821_0_appE_Return
807_1_appE_InvokeMethod(821_0_appE_Return, x0) → 874_0_appE_Return
807_1_appE_InvokeMethod(874_0_appE_Return, x0) → 874_0_appE_Return
The integer pair graph contains the following rules and edges:
(0): 757_0_APPE_FIELDACCESS(x0[0]) → COND_757_0_APPE_FIELDACCESS(x0[0] > 0, x0[0])
(1): COND_757_0_APPE_FIELDACCESS(TRUE, x0[1]) → 757_0_APPE_FIELDACCESS(x0[1] + -1)
(0) -> (1), if ((x0[0] > 0* TRUE)∧(x0[0]* x0[1]))
(1) -> (0), if ((x0[1] + -1* x0[0]))
The set Q consists of the following terms:
807_1_appE_InvokeMethod(768_0_appE_Return, 0)
807_1_appE_InvokeMethod(821_0_appE_Return, x0)
807_1_appE_InvokeMethod(874_0_appE_Return, x0)
### (6) IDPNonInfProof (SOUND transformation)
The constraints were generated the following way:
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.
For Pair 757_0_APPE_FIELDACCESS(x0) → COND_757_0_APPE_FIELDACCESS(>(x0, 0), x0) the following chains were created:
• We consider the chain 757_0_APPE_FIELDACCESS(x0[0]) → COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0]), COND_757_0_APPE_FIELDACCESS(TRUE, x0[1]) → 757_0_APPE_FIELDACCESS(+(x0[1], -1)) which results in the following constraint:
(1) (>(x0[0], 0)=TRUEx0[0]=x0[1]757_0_APPE_FIELDACCESS(x0[0])≥NonInfC∧757_0_APPE_FIELDACCESS(x0[0])≥COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])∧(UIncreasing(COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])), ≥))
We simplified constraint (1) using rule (IV) which results in the following new constraint:
(2) (>(x0[0], 0)=TRUE757_0_APPE_FIELDACCESS(x0[0])≥NonInfC∧757_0_APPE_FIELDACCESS(x0[0])≥COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])∧(UIncreasing(COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])), ≥))
We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3) (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_12] + [(2)bni_12]x0[0] ≥ 0∧[(-1)bso_13] ≥ 0)
We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4) (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_12] + [(2)bni_12]x0[0] ≥ 0∧[(-1)bso_13] ≥ 0)
We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5) (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_12] + [(2)bni_12]x0[0] ≥ 0∧[(-1)bso_13] ≥ 0)
We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6) (x0[0] ≥ 0 ⇒ (UIncreasing(COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_12 + (2)bni_12] + [(2)bni_12]x0[0] ≥ 0∧[(-1)bso_13] ≥ 0)
For Pair COND_757_0_APPE_FIELDACCESS(TRUE, x0) → 757_0_APPE_FIELDACCESS(+(x0, -1)) the following chains were created:
• We consider the chain COND_757_0_APPE_FIELDACCESS(TRUE, x0[1]) → 757_0_APPE_FIELDACCESS(+(x0[1], -1)) which results in the following constraint:
(7) (COND_757_0_APPE_FIELDACCESS(TRUE, x0[1])≥NonInfC∧COND_757_0_APPE_FIELDACCESS(TRUE, x0[1])≥757_0_APPE_FIELDACCESS(+(x0[1], -1))∧(UIncreasing(757_0_APPE_FIELDACCESS(+(x0[1], -1))), ≥))
We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8) ((UIncreasing(757_0_APPE_FIELDACCESS(+(x0[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)
We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9) ((UIncreasing(757_0_APPE_FIELDACCESS(+(x0[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)
We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10) ((UIncreasing(757_0_APPE_FIELDACCESS(+(x0[1], -1))), ≥)∧[2 + (-1)bso_15] ≥ 0)
We simplified constraint (10) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(11) ((UIncreasing(757_0_APPE_FIELDACCESS(+(x0[1], -1))), ≥)∧0 = 0∧[2 + (-1)bso_15] ≥ 0)
To summarize, we get the following constraints P for the following pairs.
• 757_0_APPE_FIELDACCESS(x0) → COND_757_0_APPE_FIELDACCESS(>(x0, 0), x0)
• (x0[0] ≥ 0 ⇒ (UIncreasing(COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_12 + (2)bni_12] + [(2)bni_12]x0[0] ≥ 0∧[(-1)bso_13] ≥ 0)
• COND_757_0_APPE_FIELDACCESS(TRUE, x0) → 757_0_APPE_FIELDACCESS(+(x0, -1))
• ((UIncreasing(757_0_APPE_FIELDACCESS(+(x0[1], -1))), ≥)∧0 = 0∧[2 + (-1)bso_15] ≥ 0)
The constraints for P> respective Pbound are constructed from P where we just replace every occurence of "t ≥ s" in P by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:
POL(TRUE) = 0
POL(FALSE) = 0
POL(807_1_appE_InvokeMethod(x1, x2)) = [-1]
POL(768_0_appE_Return) = [-1]
POL(0) = 0
POL(821_0_appE_Return) = [-1]
POL(874_0_appE_Return) = [-1]
POL(757_0_APPE_FIELDACCESS(x1)) = [2]x1
POL(COND_757_0_APPE_FIELDACCESS(x1, x2)) = [2]x2
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(-1) = [-1]
The following pairs are in P>:
COND_757_0_APPE_FIELDACCESS(TRUE, x0[1]) → 757_0_APPE_FIELDACCESS(+(x0[1], -1))
The following pairs are in Pbound:
757_0_APPE_FIELDACCESS(x0[0]) → COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])
The following pairs are in P:
757_0_APPE_FIELDACCESS(x0[0]) → COND_757_0_APPE_FIELDACCESS(>(x0[0], 0), x0[0])
There are no usable rules.
### (8) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
807_1_appE_InvokeMethod(768_0_appE_Return, 0) → 821_0_appE_Return
807_1_appE_InvokeMethod(821_0_appE_Return, x0) → 874_0_appE_Return
807_1_appE_InvokeMethod(874_0_appE_Return, x0) → 874_0_appE_Return
The integer pair graph contains the following rules and edges:
(0): 757_0_APPE_FIELDACCESS(x0[0]) → COND_757_0_APPE_FIELDACCESS(x0[0] > 0, x0[0])
The set Q consists of the following terms:
807_1_appE_InvokeMethod(768_0_appE_Return, 0)
807_1_appE_InvokeMethod(821_0_appE_Return, x0)
807_1_appE_InvokeMethod(874_0_appE_Return, x0)
### (9) IDependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
### (11) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
807_1_appE_InvokeMethod(768_0_appE_Return, 0) → 821_0_appE_Return
807_1_appE_InvokeMethod(821_0_appE_Return, x0) → 874_0_appE_Return
807_1_appE_InvokeMethod(874_0_appE_Return, x0) → 874_0_appE_Return
The integer pair graph contains the following rules and edges:
(1): COND_757_0_APPE_FIELDACCESS(TRUE, x0[1]) → 757_0_APPE_FIELDACCESS(x0[1] + -1)
The set Q consists of the following terms:
807_1_appE_InvokeMethod(768_0_appE_Return, 0)
807_1_appE_InvokeMethod(821_0_appE_Return, x0)
807_1_appE_InvokeMethod(874_0_appE_Return, x0)
### (12) IDependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
### (14) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
366_0_createList_LE(0) → 395_0_createList_Return
The integer pair graph contains the following rules and edges:
(0): 366_0_CREATELIST_LE(x0[0]) → COND_366_0_CREATELIST_LE(x0[0] > 0, x0[0])
(1): COND_366_0_CREATELIST_LE(TRUE, x0[1]) → 366_0_CREATELIST_LE(x0[1] + -1)
(0) -> (1), if ((x0[0] > 0* TRUE)∧(x0[0]* x0[1]))
(1) -> (0), if ((x0[1] + -1* x0[0]))
The set Q consists of the following terms:
366_0_createList_LE(0)
### (15) IDPNonInfProof (SOUND transformation)
The constraints were generated the following way:
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.
For Pair 366_0_CREATELIST_LE(x0) → COND_366_0_CREATELIST_LE(>(x0, 0), x0) the following chains were created:
• We consider the chain 366_0_CREATELIST_LE(x0[0]) → COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0]), COND_366_0_CREATELIST_LE(TRUE, x0[1]) → 366_0_CREATELIST_LE(+(x0[1], -1)) which results in the following constraint:
(1) (>(x0[0], 0)=TRUEx0[0]=x0[1]366_0_CREATELIST_LE(x0[0])≥NonInfC∧366_0_CREATELIST_LE(x0[0])≥COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])∧(UIncreasing(COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])), ≥))
We simplified constraint (1) using rule (IV) which results in the following new constraint:
(2) (>(x0[0], 0)=TRUE366_0_CREATELIST_LE(x0[0])≥NonInfC∧366_0_CREATELIST_LE(x0[0])≥COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])∧(UIncreasing(COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])), ≥))
We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3) (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_10] + [(2)bni_10]x0[0] ≥ 0∧[(-1)bso_11] ≥ 0)
We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4) (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_10] + [(2)bni_10]x0[0] ≥ 0∧[(-1)bso_11] ≥ 0)
We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5) (x0[0] + [-1] ≥ 0 ⇒ (UIncreasing(COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_10] + [(2)bni_10]x0[0] ≥ 0∧[(-1)bso_11] ≥ 0)
We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6) (x0[0] ≥ 0 ⇒ (UIncreasing(COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_10 + (2)bni_10] + [(2)bni_10]x0[0] ≥ 0∧[(-1)bso_11] ≥ 0)
For Pair COND_366_0_CREATELIST_LE(TRUE, x0) → 366_0_CREATELIST_LE(+(x0, -1)) the following chains were created:
• We consider the chain COND_366_0_CREATELIST_LE(TRUE, x0[1]) → 366_0_CREATELIST_LE(+(x0[1], -1)) which results in the following constraint:
(7) (COND_366_0_CREATELIST_LE(TRUE, x0[1])≥NonInfC∧COND_366_0_CREATELIST_LE(TRUE, x0[1])≥366_0_CREATELIST_LE(+(x0[1], -1))∧(UIncreasing(366_0_CREATELIST_LE(+(x0[1], -1))), ≥))
We simplified constraint (7) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(8) ((UIncreasing(366_0_CREATELIST_LE(+(x0[1], -1))), ≥)∧[2 + (-1)bso_13] ≥ 0)
We simplified constraint (8) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(9) ((UIncreasing(366_0_CREATELIST_LE(+(x0[1], -1))), ≥)∧[2 + (-1)bso_13] ≥ 0)
We simplified constraint (9) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(10) ((UIncreasing(366_0_CREATELIST_LE(+(x0[1], -1))), ≥)∧[2 + (-1)bso_13] ≥ 0)
We simplified constraint (10) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(11) ((UIncreasing(366_0_CREATELIST_LE(+(x0[1], -1))), ≥)∧0 = 0∧[2 + (-1)bso_13] ≥ 0)
To summarize, we get the following constraints P for the following pairs.
• 366_0_CREATELIST_LE(x0) → COND_366_0_CREATELIST_LE(>(x0, 0), x0)
• (x0[0] ≥ 0 ⇒ (UIncreasing(COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])), ≥)∧[(-1)Bound*bni_10 + (2)bni_10] + [(2)bni_10]x0[0] ≥ 0∧[(-1)bso_11] ≥ 0)
• COND_366_0_CREATELIST_LE(TRUE, x0) → 366_0_CREATELIST_LE(+(x0, -1))
• ((UIncreasing(366_0_CREATELIST_LE(+(x0[1], -1))), ≥)∧0 = 0∧[2 + (-1)bso_13] ≥ 0)
The constraints for P> respective Pbound are constructed from P where we just replace every occurence of "t ≥ s" in P by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:
POL(TRUE) = 0
POL(FALSE) = 0
POL(366_0_createList_LE(x1)) = [-1]
POL(0) = 0
POL(395_0_createList_Return) = [-1]
POL(366_0_CREATELIST_LE(x1)) = [2]x1
POL(COND_366_0_CREATELIST_LE(x1, x2)) = [2]x2
POL(>(x1, x2)) = [-1]
POL(+(x1, x2)) = x1 + x2
POL(-1) = [-1]
The following pairs are in P>:
COND_366_0_CREATELIST_LE(TRUE, x0[1]) → 366_0_CREATELIST_LE(+(x0[1], -1))
The following pairs are in Pbound:
366_0_CREATELIST_LE(x0[0]) → COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])
The following pairs are in P:
366_0_CREATELIST_LE(x0[0]) → COND_366_0_CREATELIST_LE(>(x0[0], 0), x0[0])
There are no usable rules.
### (17) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
366_0_createList_LE(0) → 395_0_createList_Return
The integer pair graph contains the following rules and edges:
(0): 366_0_CREATELIST_LE(x0[0]) → COND_366_0_CREATELIST_LE(x0[0] > 0, x0[0])
The set Q consists of the following terms:
366_0_createList_LE(0)
### (18) IDependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
### (20) Obligation:
IDP problem:
The following function symbols are pre-defined:
!= ~ Neq: (Integer, Integer) -> Boolean * ~ Mul: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer = ~ Eq: (Integer, Integer) -> Boolean ~ Bwxor: (Integer, Integer) -> Integer || ~ Lor: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean < ~ Lt: (Integer, Integer) -> Boolean - ~ Sub: (Integer, Integer) -> Integer <= ~ Le: (Integer, Integer) -> Boolean > ~ Gt: (Integer, Integer) -> Boolean ~ ~ Bwnot: (Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer & ~ Bwand: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer && ~ Land: (Boolean, Boolean) -> Boolean
The following domains are used:
Integer
The ITRS R consists of the following rules:
366_0_createList_LE(0) → 395_0_createList_Return
The integer pair graph contains the following rules and edges:
(1): COND_366_0_CREATELIST_LE(TRUE, x0[1]) → 366_0_CREATELIST_LE(x0[1] + -1)
The set Q consists of the following terms:
366_0_createList_LE(0)
### (21) IDependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. | 7,285 | 20,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.657387 |
http://headinside.blogspot.com/2005_09_01_archive.html | 1,527,039,008,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865023.41/warc/CC-MAIN-20180523004548-20180523024548-00260.warc.gz | 128,034,126 | 270,603 | 0
## MathPath
Published on Sunday, September 25, 2005 in , , , ,
I've recently run across a site from a kindred soul called MathPath, which is available in English and French. This site is full of math tricks and discussions that regular readers of this blog will find very interesting. Much like my own Be A Genius* site, MathPath lets you pratice the tricks you learn right on the site, such as learning to square numbers up to 125 in your head! There's also a program that automatically generates Magic Squares (PC only). He offers the complete Visual Basic project files for download, so perhaps some venturesome REALbasic programmer can create a Mac version.
0
## Two Great Ideas...
Published on Sunday, September 25, 2005 in , , , , , ,
I noted earlier that one of Chris Wasshuber's own tricks was no longer available at his site, Lybrary.com. Happily, that has been rectified. Now the Ultimate Magic Square (in both King of Hearts and "That's Magic" versions) is available again. I described and reviewed this effect in my second blog post.
Lew Brooks has a great book called Stack Attack (also available in the UK) out. The basis of the book is the False False Shuffle ("FFS"), which is one of the simplest, most ingenious and convincing false shuffles I've ever seen. You'll have the basic technique mastered in only minutes, but it's very powerful. Lew then goes on to describe several tricks, which help you realize just how disarming and flexible the FFS really is. Lew is not just a great thinker, but also a master of presentation, as well.
Lew constantly get repeated requests to show "Bughouse Poker" and "Poetry Poker". Beyond just the included tricks, Lew shows you how to use the FFS with other stacks. Anyone who likes great presentations and does memorized deck work should be adding this book to their arsenal.
Just for fun: What would happen if we use ideas from both of these sources? The presentation for either version of the Ultimate Magic Square is that a spectator rolls three dice, and you're not only able to put together a magic square for that total, but you're able to lay out the cards so that they make a pre-determined design, such as the King of Hearts or the phrase "That's Magic". Mixing this idea with concepts from Stack Attack, I've come up with a great new presentation and routine for the Ultimate Magic Square that involves great apparent mental calculation and memory.
I introduce the cards (I use cards with just numbers, and no background pattern) as flash cards, and explain that I use them to train my memory and mental calculation abilities. The cards are mixed in various ways, and then memorized in their shuffled order. The spectator rolls three dice, and I then take a second, apparently to recall the order and make some mental calculations, and then lay the cards out to create a magic square!
The presentation here is that I'm challenging my own mental abilities, using mixed cards and random dice, so I can't have a pre-determined background pattern show up, as this would give away part of the method. That's why I use cards with no background pattern.
0
## New Layout for MemoryEffects.pdf
Published on Sunday, September 18, 2005 in , , , ,
Inspired by the recent redesigns at sites like Lybrary.com and MagiCentric, I decided it was time for a little redesign of my own. For those of you who are thinking, "The blog looks exactly the same!", let me explain.
What I've actually redesigned is the MemoryEffects.pdf file, which I first posted back in April.
Until this change, it was a boring list dumped directly from a word processor. With a little help from iWork Pages, however, I've made it more readable, and added an improved layout. I hope you like the new look!
As always, if you have any additions to the list, or suggestions for improving the layout, let me know in the comments!
1
## Ummm . . . WOW!
Published on Saturday, September 17, 2005 in , , , ,
I've mentioned Chris Wasshuber's Lybrary.com site before, here.
If you go there now, however, you'll see a complete redesign of the site. The redesign is so complete, that I've had to change all my links to his site in my "For Sale" links. It has a much smoother, more streamlined feel and function, and the store functions in a much more streamlined manner, as well. The site itself now has a "parchment" look, which is quite appropriate and adds to the charm of the site.
The titles themselves are better organized and easier to find, by subject or author. There are some great new offerings, such as the free ebooks and the PDF catalog. There are some items that are no longer available, which include, strangely, Chris Wasshuber's own routine, "The Ultimate Magic Square".
I must, of course, give special attention to the Memory & Mnemonics section, especially as this is the only place you can easily find Dominic O'Brien's elusive book How to Develop a Perfect Memory.
0
## More Cafe Memories
Published on Thursday, September 08, 2005 in , , , , ,
Over at the Magic Cafe, which, by the way, is celebrating its 4th year in operation this month, there are quite a few interesting items of note:
* There's a new thread about memorizing Pi.
* Another memory thread I mentioned earlier has some new updates. Dr. Wilson describes an excellent routine he recently performed, in which he remembers the phone numbers, birthdays and names of people attending the function. If you have to ask why it got the great reactions he describes, then a quick review of Dale Carnegie's principles should be in order.
Update (9/9/05): Richard Osterlind just added a great story about how memory technique got him out of an awkward situation.
* Cafe member Jean-Denis, in this Cafe post, announces his new mnemonics program (currently PC only), which helps increase the speed with which you memorize numbers. Jean-Denis mentions that he was able to increase his speed 50% in one week!
* For those who work with memorized stacks, Cafe member Nick Pudar offers his "StackView" program (also currently PC only), which allows you to see the effects of almost any manipulation of a standard deck, without having to test and re-set a real deck. For those who have Mnemonica, this is a wonderful tool for exploring "The Eight Mnemonicas".
0
## RSS Change
Published on Thursday, September 08, 2005 in ,
Due to some compatibility problems, I have changed the name of the RSS feed for this blog from "greymatters.xml" to "greymatters.rss". Please update your RSS feed readers and links accordingly. | 1,452 | 6,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-22 | latest | en | 0.964715 |
https://fenicsproject.discourse.group/t/problem-when-combining-test-and-trial-functions-from-different-spaces/7083 | 1,638,912,950,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363418.83/warc/CC-MAIN-20211207201422-20211207231422-00386.warc.gz | 314,818,572 | 7,182 | Problem when combining test and trial functions from different spaces
Hi all,
I am trying to solve a set of equations that involve velocity, pressure and temperature.
The finite element spaces that I am using for the different variables are the following:
Finite element spaces
import ufl
v_cg2 = ufl.VectorElement(“CG”, mesh.ufl_cell(), 2)
p_cg1 = ufl.FiniteElement(“CG”, mesh.ufl_cell(), 1)
T_cg2 = ufl.FiniteElement(“CG”, mesh.ufl_cell(), 2)
V = dolfinx.FunctionSpace(mesh, v_cg2) # Velocity function space
Q = dolfinx.FunctionSpace(mesh, p_cg1) # Pressure function space
W = dolfinx.FunctionSpace(mesh, T_cg2) # Temperature function space
Trial and test functions
u = ufl.TrialFunction(V) # Velocity (trial)
v = ufl.TestFunction(V) # Velocity (test)
p = ufl.TrialFunction(Q) # Pressure (trial)
q = ufl.TestFunction(Q) # Pressure (test)
T = ufl.TrialFunction(W) # Temperature (trial)
w = ufl.TestFunction(W) # Temperature (test)
The problem is that the equations are coupled, so I need to solve them simultaneously. However, I am getting errors when I try to build the bilinear form, since I need to combine test and trial functions from different element spaces.
For example, I need to compute the following term:
L1 = inner(q,(gamma*p-T))*dx
A = dolfinx.fem.assemble_matrix(L1)
A.assemble()
which gives me the following error:
Found different Arguments with same number and part.
Did you combine test or trial functions from different spaces?
The Arguments found are:
v_1
v_0
v_1
ERROR:UFL:Found different Arguments with same number and part.
Did you combine test or trial functions from different spaces?
The Arguments found are:
v_1
v_0
v_1
Traceback (most recent call last):
File “/root/dolfinX/Test.py”, line 110, in
A = dolfinx.fem.assemble_matrix(L1)
File “/usr/lib/python3.9/functools.py”, line 877, in wrapper
return dispatch(args[0].__class__)(*args, **kw)
File “/usr/local/dolfinx-real/lib/python3.8/dist-packages/dolfinx/fem/assemble.py”, line 283, in assemble_matrix
A = cpp.fem.create_matrix(_create_cpp_form(a))
File “/usr/local/dolfinx-real/lib/python3.8/dist-packages/dolfinx/fem/assemble.py”, line 35, in create_cpp_form
return Form(form).cpp_object
File “/usr/local/dolfinx-real/lib/python3.8/dist-packages/dolfinx/fem/form.py”, line 56, in __init
self._ufc_form, module, self._code = jit.ffcx_jit(
File “/usr/local/dolfinx-real/lib/python3.8/dist-packages/dolfinx/jit.py”, line 54, in mpi_jit
return local_jit(*args, **kwargs)
File “/usr/local/dolfinx-real/lib/python3.8/dist-packages/dolfinx/jit.py”, line 204, in ffcx_jit
r = ffcx.codegeneration.jit.compile_forms([ufl_object], parameters=p_ffcx, **p_jit)
File “/usr/local/lib/python3.9/dist-packages/ffcx/codegeneration/jit.py”, line 145, in compile_forms
ffcx.naming.compute_signature(forms, _compute_parameter_signature(p)
File “/usr/local/lib/python3.9/dist-packages/ffcx/naming.py”, line 24, in compute_signature
object_signature += ufl_object.signature()
File “/usr/local/lib/python3.9/dist-packages/ufl/form.py”, line 243, in signature
self._compute_signature()
File “/usr/local/lib/python3.9/dist-packages/ufl/form.py”, line 487, in _compute_signature
self._compute_renumbering())
File “/usr/local/lib/python3.9/dist-packages/ufl/form.py”, line 460, in _compute_renumbering
cn = self.coefficient_numbering()
File “/usr/local/lib/python3.9/dist-packages/ufl/form.py”, line 234, in coefficient_numbering
self._analyze_form_arguments()
File “/usr/local/lib/python3.9/dist-packages/ufl/form.py”, line 447, in _analyze_form_arguments
arguments, coefficients = extract_arguments_and_coefficients(self)
File “/usr/local/lib/python3.9/dist-packages/ufl/algorithms/analysis.py”, line 127, in extract_arguments_and_coefficients
error(msg)
File “/usr/local/lib/python3.9/dist-packages/ufl/log.py”, line 158, in error
raise self._exception_type(self._format_raw(*message))
ufl.log.UFLException: Found different Arguments with same number and part.
Did you combine test or trial functions from different spaces?
The Arguments found are:
v_1
v_0
v_1
What I understand then is that I cannot combine test and trial functions from different spaces. Therefore, I defined a mixed finite element space to try to fix this error:
Mixed finite element space
import ufl
v_cg2 = ufl.VectorElement(“CG”, mesh.ufl_cell(), 2)
p_cg1 = ufl.FiniteElement(“CG”, mesh.ufl_cell(), 1)
T_cg2 = ufl.FiniteElement(“CG”, mesh.ufl_cell(), 2)
mel = ufl.MixedElement([v_cg2, p_cg1, T_cg2])
Y = dolfinx.FunctionSpace(mesh, mel) # Mixed function space
and redefined the trial and test functions as follows:
Trial and test functions
u = ufl.TrialFunction(Y) # Velocity (trial)
v = ufl.TestFunction(Y) # Velocity (test)
p = ufl.TrialFunction(Y) # Pressure (trial)
q = ufl.TestFunction(Y) # Pressure (test)
T = ufl.TrialFunction(Y) # Temperature (trial)
w = ufl.TestFunction(Y) # Temperature (test)
This approach fixes the previous error, but now I am getting another one when I try to compute another term:
L2 = inner(w,div(u))*dx
A = dolfinx.fem.assemble_matrix(L2)
A.assemble()
The error now is:
Shapes do not match: <Argument id=140363992421472> and <Div id=140364133793920>.
ERROR:UFL:Shapes do not match: <Argument id=140363992421472> and <Div id=140364133793920>.
Traceback (most recent call last):
File “/root/dolfinX/Test.py”, line 105, in <module>
L2 = inner(w,div(u))*dx
File “/usr/local/lib/python3.9/dist-packages/ufl/operators.py”, line 158, in inner
return Inner(a, b)
File “/usr/local/lib/python3.9/dist-packages/ufl/tensoralgebra.py”, line 147, in __new__
error(“Shapes do not match: %s and %s.” % (ufl_err_str(a), ufl_err_str(b)))
File “/usr/local/lib/python3.9/dist-packages/ufl/log.py”, line 158, in error
raise self._exception_type(self._format_raw(*message))
ufl.log.UFLException: Shapes do not match: <Argument id=140363992421472> and <Div id=140364133793920>.
This means that the shape of w is different from the shape of div(u), which is understandable since now w and u belong to the same mixed space W (and when taking the divergence of u, the space changes).
Consequently, my question would be how can I combine test and trial functions that belong to different spaces to build the bilinear form. Does anybody have any idea?
My minimal example is:
# Import
from mpi4py import MPI
rank = MPI.COMM_WORLD.rank
import dolfinx
from ufl import Identity, div, dot, ds, dx, inner, lhs, nabla_grad, rhs, sym, Dn, grad
‘’’ Generation of geometry and mesh ‘’’
# Domain dimensions
L = 3
H = 0.5
gdim = 3
# Initialise meshing process
import gmsh
gmsh.initialize()
# Geometry
if rank == 0:
fluid = gmsh.model.occ.addBox(0, 0, 0, L, H, H)
gmsh.model.occ.synchronize()
# Generate mesh
if rank == 0:
gmsh.model.mesh.generate(gdim)
gmsh.write(“meshTVAchannel.msh”)
# Domain marker
fluid_marker = 1
if rank == 0:
volumes = gmsh.model.getEntities(dim=gdim)
gmsh.model.setPhysicalName(volumes[0][0], fluid_marker, “Fluid”)
# Boundary markers
import numpy as np
inlet_marker, outlet_marker, wall_marker = 2, 3, 4
inflow, outflow, walls = [], [], []
if rank == 0:
boundaries = gmsh.model.getBoundary(volumes)
for boundary in boundaries:
center_of_mass = gmsh.model.occ.getCenterOfMass(boundary[0], boundary[1])
if np.allclose(center_of_mass, [0, H/2, 0]):
inflow.append(boundary[1])
elif np.allclose(center_of_mass, [L, H/2, 0]):
outflow.append(boundary[1])
else:
walls.append(boundary[1])
gmsh.model.setPhysicalName(1, wall_marker, “Walls”)
gmsh.model.setPhysicalName(1, inlet_marker, “Inlet”)
gmsh.model.setPhysicalName(1, outlet_marker, “Outlet”)
# Import mesh to dolfinX
from gmsh_helpers import gmsh_model_to_mesh
mesh, facet_tags = gmsh_model_to_mesh(gmsh.model, cell_data=True, facet_data=False, gdim=3)
‘’’ Generation of finite element spaces ‘’’
# Finite element spaces
import ufl
v_cg2 = ufl.VectorElement(“CG”, mesh.ufl_cell(), 2)
p_cg1 = ufl.FiniteElement(“CG”, mesh.ufl_cell(), 1)
T_cg2 = ufl.FiniteElement(“CG”, mesh.ufl_cell(), 2)
mel = ufl.MixedElement([v_cg2, p_cg1, T_cg2])
V = dolfinx.FunctionSpace(mesh, v_cg2)
Q = dolfinx.FunctionSpace(mesh, p_cg1)
W = dolfinx.FunctionSpace(mesh, T_cg2)
Y = dolfinx.FunctionSpace(mesh, mel)
# Trial and test functions
u = ufl.TrialFunction(V)
v = ufl.TestFunction(V)
p = ufl.TrialFunction(Q)
q = ufl.TestFunction(Q)
T = ufl.TrialFunction(W)
w = ufl.TestFunction(W)
# # Trial and test functions
# u = ufl.TrialFunction(Y)
# v = ufl.TestFunction(Y)
# p = ufl.TrialFunction(Y)
# q = ufl.TestFunction(Y)
# T = ufl.TrialFunction(Y)
# w = ufl.TestFunction(Y)
# Numerical parameters
gamma = 1.017
‘’’’ Left-hand side ‘’’
# First component
L1 = inner(q,(gamma*p-T))*dx
# Second component
L2 = inner(w,div(u))*dx
L = L1 + L2
A = dolfinx.fem.assemble_matrix(L)
A.assemble()
Thank you very much in advance!
Please make sure that your code is properly formatted (indentation preserved) using 3x encapsulation, and make sure to remove all code not needed for reproducibility.
Sorry, I did not paste the code in the correct format. Please find below a minimal example for the first error and a minimal example for the second error.
Minimal example for first error
# Import
import dolfinx
from ufl import Identity, div, dot, ds, dx, inner, lhs, nabla_grad, rhs, sym, Dn, grad
# Domain dimensions
L = 3
H = 0.5
gdim = 3
# Initialise meshing process
import gmsh
gmsh.initialize()
# Geometry
fluid = gmsh.model.occ.addBox(0, 0, 0, L, H, H)
gmsh.model.occ.synchronize()
# Generate mesh
gmsh.model.mesh.generate(gdim)
# Domain marker
fluid_marker = 1
volumes = gmsh.model.getEntities(dim=gdim)
gmsh.model.setPhysicalName(volumes[0][0], fluid_marker, 'Fluid')
# Import mesh to dolfinX
from gmsh_helpers import gmsh_model_to_mesh
mesh, facet_tags = gmsh_model_to_mesh(gmsh.model, cell_data=True, facet_data=False, gdim=3)
# Finite element spaces
import ufl
v_cg2 = ufl.VectorElement('CG', mesh.ufl_cell(), 2)
p_cg1 = ufl.FiniteElement('CG', mesh.ufl_cell(), 1)
T_cg2 = ufl.FiniteElement('CG', mesh.ufl_cell(), 2)
mel = ufl.MixedElement([v_cg2, p_cg1, T_cg2])
V = dolfinx.FunctionSpace(mesh, v_cg2)
Q = dolfinx.FunctionSpace(mesh, p_cg1)
W = dolfinx.FunctionSpace(mesh, T_cg2)
Y = dolfinx.FunctionSpace(mesh, mel)
# Trial and test functions
u = ufl.TrialFunction(V)
v = ufl.TestFunction(V)
p = ufl.TrialFunction(Q)
q = ufl.TestFunction(Q)
T = ufl.TrialFunction(W)
w = ufl.TestFunction(W)
# Numerical parameters
gamma = 1.017
# First component
L1 = inner(q,(gamma*p-T))*dx
# Assemble bilinear form
A = dolfinx.fem.assemble_matrix(L1)
A.assemble()
Minimal example for the second error
# Import
import dolfinx
from ufl import Identity, div, dot, ds, dx, inner, lhs, nabla_grad, rhs, sym, Dn, grad
# Domain dimensions
L = 3
H = 0.5
gdim = 3
# Initialise meshing process
import gmsh
gmsh.initialize()
# Geometry
fluid = gmsh.model.occ.addBox(0, 0, 0, L, H, H)
gmsh.model.occ.synchronize()
# Generate mesh
gmsh.model.mesh.generate(gdim)
# Domain marker
fluid_marker = 1
volumes = gmsh.model.getEntities(dim=gdim)
gmsh.model.setPhysicalName(volumes[0][0], fluid_marker, 'Fluid')
# Import mesh to dolfinX
from gmsh_helpers import gmsh_model_to_mesh
mesh, facet_tags = gmsh_model_to_mesh(gmsh.model, cell_data=True, facet_data=False, gdim=3)
# Finite element spaces
import ufl
v_cg2 = ufl.VectorElement('CG', mesh.ufl_cell(), 2)
p_cg1 = ufl.FiniteElement('CG', mesh.ufl_cell(), 1)
T_cg2 = ufl.FiniteElement('CG', mesh.ufl_cell(), 2)
mel = ufl.MixedElement([v_cg2, p_cg1, T_cg2])
V = dolfinx.FunctionSpace(mesh, v_cg2)
Q = dolfinx.FunctionSpace(mesh, p_cg1)
W = dolfinx.FunctionSpace(mesh, T_cg2)
Y = dolfinx.FunctionSpace(mesh, mel)
# Trial and test functions
u = ufl.TrialFunction(Y)
v = ufl.TestFunction(Y)
p = ufl.TrialFunction(Y)
q = ufl.TestFunction(Y)
T = ufl.TrialFunction(Y)
w = ufl.TestFunction(Y)
# Numerical parameters
gamma = 1.017
# Second component
L2 = inner(w,div(u))*dx
# Assemble bilinear form
A = dolfinx.fem.assemble_matrix(L2)
A.assemble()
You need to split the trial function from the mixed space for each component, which can either be done by
y = ufl.TrialFunction(Y)
u, p, T = ufl.split(y)
or use ufl.TrialFunctions, as shown below:
# Import
import dolfinx
from ufl import div, dx, inner, FiniteElement, VectorElement, TrialFunctions, TestFunctions, MixedElement
from mpi4py import MPI
mesh = dolfinx.UnitCubeMesh(MPI.COMM_WORLD, 10, 10, 10)
v_cg2 = VectorElement('CG', mesh.ufl_cell(), 2)
p_cg1 = FiniteElement('CG', mesh.ufl_cell(), 1)
T_cg2 = FiniteElement('CG', mesh.ufl_cell(), 2)
mel = MixedElement([v_cg2, p_cg1, T_cg2])
V = dolfinx.FunctionSpace(mesh, v_cg2)
Q = dolfinx.FunctionSpace(mesh, p_cg1)
W = dolfinx.FunctionSpace(mesh, T_cg2)
Y = dolfinx.FunctionSpace(mesh, mel)
# Trial and test functions
u, p, T = TrialFunctions(Y)
v, q, w = TestFunctions(Y)
# Numerical parameters
gamma = 1.017
# Second component
L2 = inner(w,div(u))*dx
# Assemble bilinear form
A = dolfinx.fem.assemble_matrix(L2)
A.assemble()
`
1 Like
Thank you very much! It’s working now. | 3,873 | 12,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-49 | latest | en | 0.637694 |
http://askmetips.com/standard-error/standard-error-for-sample-mean-calculator.php | 1,529,872,147,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867055.95/warc/CC-MAIN-20180624195735-20180624215735-00087.warc.gz | 25,341,203 | 5,032 | Home > Standard Error > Standard Error For Sample Mean Calculator
# Standard Error For Sample Mean Calculator
## Contents
BurkeyAcademy 1,178 views 21:53 How to Calculate Standard Deviation - Duration: 1:29. This article is a part of the guide: Select from one of the other courses available: Scientific Method Research Design Research Basics Experimental Research Sampling Validity and Reliability Write a Paper The manual calculation can be done by using above formulas. Standard Error of Sample Means The logic and computational details of this procedure are described in Chapter 9 of Concepts and Applications. http://askmetips.com/standard-error/standard-error-of-the-sample-mean-calculator.php
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## Standard Error Calculator Excel
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All Rights Reserved. Standard Error of the Estimate A related and similar concept to standard error of the mean is the standard error of the estimate. Calculate Standard Error for the Sample Mean: Steps Example: Find the standard error for the following heights (in cm): Jim (170.5), John (161), Jack (160), Freda (170), Tai (150.5). http://www.miniwebtool.com/standard-error-calculator/ Copyright © 2016 Statistics How To Theme by: Theme Horse Powered by: WordPress Back to Top
To understand this, first we need to understand why a sampling distribution is required. Standard Error Of Measurement Calculator Find a Critical Value 7. Published on Sep 20, 2013Find more videos and articles at: http://www.statisticshowto.com Category People & Blogs License Standard YouTube License Show more Show less Loading... Working...
## Standard Error Of The Estimate Calculator
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## TweetOnline Tools and Calculators > Math > Standard Error Calculator Standard Error Calculator Enter numbers separated by comma, space or line break: About This Tool The online Standard Error Calculator is
Image: U of OklahomaThe sampling distribution of the sample mean is a probability distribution of all the sample means. statisticsfun 596,815 views 5:05 The Sampling Distribution of the Sample Mean (fast version) - Duration: 7:25. Watch QueueQueueWatch QueueQueue Remove allDisconnect Loading... Standard Error Calculator For Two Samples Standard Error of Sample Means The logic and computational details of this procedure are described in Chapter 9 of Concepts and Applications.
How to Find the Sample Mean Variance of the sampling distribution of the sample mean Calculate Standard Error for the Sample Mean Sample Mean Symbol The sample mean symbol is x̄, The formula of Mean is: The Variance of a finite population of size n is: The Standard Deviation is the square root of Variance: The Standard Error of the symbols) are just different. http://askmetips.com/standard-error/standard-error-for-the-sample-mean-calculator.php What is the Sample Mean? | 1,079 | 5,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-26 | latest | en | 0.797498 |
http://spotidoc.com/doc/277187/ | 1,579,932,060,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251669967.70/warc/CC-MAIN-20200125041318-20200125070318-00496.warc.gz | 153,133,205 | 8,848 | # Document 277187
```Sample Documents
NC Stnd Course of Study
(NC1)
EDUCAIDE SOFTWARE
c 1999 by EAS EducAide Software Inc.
or the related software is prohibited by law.
Mt. Titan Middle School
Write your answers on the answer sheets you are given. You will have 45 minutes to complete these questions.
7.
1.
If you pile apples in a pyramid with 1 in the top
layer, 4 in the next layer, 9 in the third layer, and
16 in the fourth layer, how many apples will you
need to make a pile with 6 layers?
2.
Write each of these numbers in scientific notation.
8.
a) 4,860,000,000
The formula d = r × t can be used to find the
distance d traveled when the rate r and time t are
known. Use the formula to determine how far a
bus traveling at the rate of 65 mph can travel in
4 hours?
The equation W + [ = 12 indicates that the
sum of two numbers is 12. Find four pairs of
replacements for W and [ which will make the
equation true.
b) 0.73
3.
4.
Your kitchen is 12 0 × 14 0 . Tiles come in boxes
of 100. Each tile is 4 inches square. How many
boxes of single-color tiles do you need to purchase?
Essay Questions
9.
John has the pieces of lumber pictured. Can those
four pieces be glued together to form a board with
an area of 64 sq ft?
30
30
30
50
50
30
50
50
10.
5.
6.
Boxes without lids can be formed by cutting out
corner squares from a sheet of graph paper and
folding on the lines. How many different size
boxes can be constructed from a 9 × 12 sheet?
Which box has the greatest surface area? Which
has the greatest volume? Explain.
If a giraffe stands 576 cm tall, how long a shadow
would it cast if it is compared to a man who is
180 cm tall casting a shadow 1 m long?
Mary made a series of deposits and withdrawals
from her account. Use an integer to represent the
current status of the account.
Date
Deposit
Withdrawals
2-28
3-1
3-2
3-10
\$52
\$36
\$ 8
\$16
\$ 4
\$40
√
Find a calculator value√for 2. Is the calculator
value exactly equal to 2 ? Explain.
11. Students in science class want to measure rainfall.
They plan to keep daily and monthly records.
What measurement tools do they need and to
what degree of accuracy should they try to
measure?
12. Write five expressions equivalent to 43 . In each
case explain why the expression is equal to 43 .
Copyright (c) 1999 EAS EducAide Software Inc. All rights reserved. You may distribute this document freely, provided you do not alter it in any way or remove this copyright notice.
Note: this document was produced by EducAide’s Acces program. It contains problems from the NC1 database. For more information, please call 800-669-9405 or visit www.educaide.com.
Acces format version 3.3H β
c 1997 EducAide Software
Licensed for use by (unregistered)
Mt. Titan Middle School
1/8/99
1.
91 apples
2.
a. 4.86 × 109 ; b. 7.3 × 10−1
3.
16 boxes
4.
yes
5.
3.2 m
6.
\$28
7.
260 miles
8.
(1, 11), (2, 10), (3, 9), (4, 8),
etc.
9.
4 differently sized boxes;
surface area: 10 × 7 × 1 box;
volume: 8 × 5 × 2 box
10. 1.4142 . . . ; no, [explanations
will vary]
11. [answers will vary]
12. [answers will vary]
2.
5.
8.
11.
3.
6.
9.
12.
Catalog List
1.
4.
7.
10.
NC1
NC1
NC1
NC1
BC
DD
BC
CA
2
3
7
9
NC1
NC1
NC1
NC1
CA
CB
BC
CD
7
3
4
6
NC1
NC1
NC1
NC1
BE 9
AA 16
BD 2
CA 5
Algebra 1A
Review Problems
1/8/99
1.
Which data appear linear?
a.
2.
b.
b.
5.
c.
d.
Erin earns \$24,000 per year in salary and 8% commission on her sales. If she must have a
total income of at least \$30,000 to pay her expenses, what is the least amount she must sell
in order to do this?
a. \$6,000
4.
d.
Which graph best represents the relationship between the cost of pizzas of various
diameters?
a.
3.
c.
b. \$14,000
c. \$54,000
d. \$75,000
Write a polynomial for each model. Then find the sum of the two polynomials.
a. 3x2 + 4x + 1
b. x2 + 4x + 1
c. x2 + 2x + 1
d. 2x2 + 2x − 1
Simplify:
a.
x
2
−
2x + 2 x + 1
x2 − 2x + 2
2(x + 1)(x + 2)
b.
x2 − 2x + 4
2(x + 1)(x + 2)
c.
x−4
2(x + 1)
d.
x2 − 6x + 4
2(x + 1)(x + 2)
Copyright (c) 1999 EAS EducAide Software Inc. All rights reserved. You may distribute this document freely, provided you do not alter it in any way or remove this copyright notice.
Note: this document was produced by EducAide’s Acces program. It contains problems from the NC1 database. For more information, please call 800-669-9405 or visit www.educaide.com.
Page 2
6.
There are 28 cylindrical columns in a shopping mall that need repainting. Each column
measures 14 ft tall and 4.7 ft around. If a gallon of paint covers 400 square feet, how many
gallons are needed to paint all 28 columns?
7.
The angular speed of each moving gear is inversely
proportional to the number of teeth on the gear. Suppose
the drive gear has 60 teeth and the pinion gear has
20 teeth. If the drive gear makes 10 revolutions per
minute, how many revolutions per minute will the pinion
gear make?
8.
The sum of the digits of a two-digit number is 9. If the digits are reversed the new number
is 27 more than the original number. Find the original number.
9.
Set up a proportion based on the similar figures.
3 ft
5 ft
4 ft
? ft
10.
Write an expression involving the subtraction of integers which has −8 as its answer.
11.
Sketch a reasonable graph which represents an individual’s height as he ages. Label both
axes and write a sentence or two to justify the shape or behavior of your graph.
Copyright (c) 1999 EAS EducAide Software Inc. All rights reserved. You may distribute this document freely, provided you do not alter it in any way or remove this copyright notice.
Note: this document was produced by EducAide’s Acces program. It contains problems from the NC1 database. For more information, please call 800-669-9405 or visit www.educaide.com.
Acces format version 3.3H β
c 1997 EducAide Software
Licensed for use by (unregistered)
Algebra 1A
Review Problems
1/8/99
1.
4.
c
a
2.
5.
a
c
3.
6.
d
≈ 4.6 gallons
7.
30 rpm
8.
36
9.
4
3
3.
6.
9.
NC1 FC 26
NC1 HC 11
NC1 HB 4
10. [answers will vary]
=
?
5
11. [graph]; [answers will vary]
Catalog List
1.
4.
7.
10.
NC1
NC1
NC1
NC1
EG 5
EI 3
HB 8
FA 3
2.
5.
8.
11.
NC1
NC1
NC1
NC1
FD 9
FI 10
GF 14
GD 14
``` | 1,956 | 6,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-05 | latest | en | 0.909219 |
https://oeis.org/A202109 | 1,720,921,089,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00266.warc.gz | 373,732,843 | 4,459 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A202109 a(n) = n^3*(n+1)^3*(n+2)^3/72. 1
3, 192, 3000, 24000, 128625, 526848, 1778112, 5184000, 13476375, 31944000, 70180968, 144685632, 282589125, 526848000, 943296000, 1630015488, 2729559627, 4444632000, 7057911000, 10956792000, 16663911033, 24874409472, 36501000000, 52728000000, 75075609375 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 REFERENCES Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966, p. 162. LINKS Paolo Xausa, Table of n, a(n) for n = 1..10000 Pedro A. Piza, Powers of sums and sums of powers, Math. Mag. 25 (3) (1952) 137. Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1). FORMULA a(n) = 3*(Sum_{k=1..n} k*(k+1)/2)^3. a(n) = 3*A000292(n)^3. a(n) = Sum_{k=1..n} A000217(k)^3+2*A000217(k)^4. G.f.: 3*x*(1+54*x+405*x^2+760*x^3+405*x^4+54*x^5+x^6) / (x-1)^10. - R. J. Mathar, Dec 13 2011 From Amiram Eldar, Apr 09 2024: (Start) Sum_{n>=1} 1/a(n) = 261/4 - 54*zeta(3). Sum_{n>=1} (-1)^(n+1)/a(n) = 135*zeta(3)/2 + 432*log(2) - 1521/4. (End) MATHEMATICA Array[#^3*(#+1)^3*(#+2)^3/72 &, 50] (* Paolo Xausa, Apr 07 2024 *) CROSSREFS Cf. A000217, A000292, A002117. Sequence in context: A032594 A159658 A257038 * A230171 A332957 A203749 Adjacent sequences: A202106 A202107 A202108 * A202110 A202111 A202112 KEYWORD nonn,easy AUTHOR Martin Renner, Dec 11 2011 STATUS approved
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Last modified July 13 21:32 EDT 2024. Contains 374288 sequences. (Running on oeis4.) | 742 | 1,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.525348 |
https://www.theessayworld.com/if-a-fair-6-sided-die-is-rolled-three-times-what-is-the-probability-that-exactly-one-3-is-rolled/ | 1,680,439,927,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00344.warc.gz | 1,102,310,186 | 13,964 | Question
1. # If A Fair 6-Sided Die Is Rolled Three Times, What Is The Probability That Exactly One 3 Is Rolled?
## Introduction
We can all agree that understanding the probability of something occurring is an important part of life. Whether it’s the chance of winning a game, landing a job, or even just predicting the weather, it’s always good to have a basic understanding of what could happen. That’s especially true when it comes to rolling dice. Knowing how to calculate the chances of certain numbers being rolled on a fair die is essential for any gambler or games enthusiast. But what if you want to know the odds of only one specific number being rolled out of three tries? Well, this blog post will explain exactly that — if a fair six-sided die is rolled three times, what is the probability that exactly one 3 is rolled? Read on to find out!
## The Probability Formula
If a fair-sided die is rolled three times, the probability that exactly one is rolled is 1/3. This is because there are three possible outcomes when rolling a die (1, 2, 3), and only one of those outcomes results in exactly one being rolled. Therefore, the probability formula for this scenario is 1/3.
## The Die Is Rolled Three Times
A fair-sided die is rolled three times. The probability that exactly one is rolled is 1/6 * 1/6 * 5/6 = 25/216.
## Conclusion
It is clear that when rolling a fair 6-sided die three times, the probability of exactly one 3 being rolled is 0.25 or 25%. This means that there is a 25% chance of getting one 3 if the die is rolled three times. Therefore it can be concluded that this probability falls in line with expectations and demonstrates how probabilities can be calculated when rolling dice multiple times.
2. Rolling dice can be fun, but it can also be confusing when trying to figure out the probability of what might happen. Take for example, the classic six-sided die and the probability of rolling exactly one 3 when it is rolled three times.
Before we get into the probability of rolling exactly one 3, let’s first take a look at the basics of probability. Probability is the measure of how likely something is to happen, with a probability of 1 being certain and a probability of 0 being impossible.
When it comes to rolling a six-sided die three times, there are six possible outcomes, so the probability of rolling a 3 is 1/6. That means that the probability of rolling exactly one 3 is 1/6 x 1/6 x 5/6 = 25/216.
But what if you want to know the probability of rolling exactly two 3s? Well, in this case, you can use the same calculation, but with different numbers. The probability of rolling two 3s would be 1/6 x 1/6 x 1/6 = 1/216.
Now that we’ve looked at the probability of rolling one or two 3s, let’s move on to the probability of rolling exactly one 3 when the die is rolled three times.
The probability of rolling exactly one 3 is 1/6 x 5/6 x 5/6 = 125/216. That means that there is a 57.87% chance that exactly one 3 will be rolled when the die is rolled three times.
So, if you’re ever feeling adventurous and decide to roll a six-sided die three times, there’s a good chance that you’ll get exactly one 3! | 751 | 3,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2023-14 | latest | en | 0.932874 |
https://magedkamel.com/1a-easy-introduction-to-tension-members/ | 1,723,630,289,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00050.warc.gz | 288,312,516 | 35,976 | # 1A- Easy introduction to Tension members-part-2.
Last Updated on April 27, 2024 by Maged kamel
## Tension members-part 2.
In this post, tension members-part 2, we will check the different types of steel materials with varying grades and ASTM designations. It is critical to grasp the different types of materials.
In this post, we provide a study of AISC Table 2-4.
### How do we estimate the net area of tension members?
The next topic concerns gauge distances if we have a single angle with four bolted connections arranged in two gauge lines.
Assume the applied force acts at the angle’s centroid. The first and second lines of fasteners are arranged in the same direction as the applied force.
The first and second gauge lines have two fasteners. The vertical distance between them is the gage distance, represented by the symbol g. It runs perpendicular to the direction of the force.
The inclined distance joining between the two fasteners, denoted as D1, has a horizontal distance in pitch distance S1, while the vertical distance component is gage g.
The edge distance, known as ED1, is measured from the center of a hole to the nearest edge. This value is the horizontal distance.
We can get the net width, which depends on the route of bolts; if our section is vertical, we will deduct the hole diameters from the vertical width depending on the number of bolts we face.
To get the net width while working with a zigzag line distance between the bolts, add S^2/4g, S is the longitudinal distance as a center-to-center spacing, where g is the transverse distance, and deduct the hole diameters for bolts that the section is cutting.
### Applicable ASTM specifications for structural shapes- AISC Table 2-4.
#### Carbon steel-AISC Table 2-4 for various structural shapes-CM#14.
Carbon steel is classified into five categories, beginning with ASTM A36, ASTM A53 Grade B, and ASTM A500, which includes two grades: B and C.
ASTM A501 has grades A and B, followed by ASTM A 592 C, which has grades 50 and 55.
What are the preferred material parameters based on the marked black square?
A36 with Fy=36 ksi and Fu=58-80 ksi is ideal for M, S, Mc, and L shapes.
The preferred ASTM A for pipe is ASTM A 53 grade B, with Fy=35 ksi and ultimate stress Fu=60 ksi.
For round HSS, the preferred ASTM A is ASTM A 500 grade B with Fy=42 ksi and ultimate stress Fu=58 ksi.
For rectangular HSS ASTM A 500 grade B, Fy=46 ksi and ultimate stress Fu=58 ksi.
The gray squares are for other potential grades and can be utilized if requested. This comprises ASTM A500 grade C and ASTM A501 grades A and B for HS sections.
The white square for the material specification is not applicable.
#### High strength-alloy steel-AISC Table 2-4 for Various structural shapes-CM#14.
There are four groups, including ASTM A 572, with five grades for the high-strength alloy. Grade 50 is preferred for the HP section, with yield stress Fy=50 ksi and fu=65 ksi. We have the symbol d, meaning a maximum tensile strength of 70 ksi can be specified.
Also, we have symbol e for grades 60 and 65, for which grade 60 has yield stress Fy=60 ksi and ultimate stress Fu=60 ksi.
While grade 65 has a yield stress of Fy=65 ksi and ultimate stress of Fu=65 ksi, E is for flange thicknesses of less than or equal to 2 inches only.
ASTM A 618 f, with grade I and grade ii, with yield stress Fy=50 ksi and Fu=70 ksi, the symbol f stands for it can be specified as corrosion resistant.
ASTM A-913 comes in four grades; the first grade is 50 with Fy=50 ksi with symbol g, which indicates that the minimum Fy of 50 ksi is to be applied with walls nominally ¾ inches and Fu=60 with symbol h, which states that the ultimate stress can be raised to 65 ksi and a maximum yield to tensile stress strength ratio of 0.85 can be specified.
ASTM A 992, preferred for the W section, has yield stress Fy=50 ksi and ultimate stress Fu=65 ksi. Its symbol I stands for maximum yield to tensile stress strength ratio of 0.85. The carbon equivalent formula is mandatory in ASTM A992.
.
#### Corrosion-resistant high strength low alloy for various structural shapes-CM#14.
There are three categories of corrosion-resistant high-strength low alloys: ASTM A 242, grades 42, 52, and 60, and ultimate stress Fu from 63 to 70 ksi.
We have symbols j, K, and L. The symbols J, K, and L are related to the shapes with flange thickness. For j is for flanges with a thickness greater than 2 inches only.
The specified Fy equals 42 ksi, and the ultimate stress is 63 ksi. For K for the flange, the thickness is >1.5 inches and less than or equal to 2 inches only. The specified Fy equals 46 ksi, and the ultimate stress is 67 ksi.
L is for flanges with a thickness of less than or equal to 1.50 inches. The specified Fy equals 50 ksi, and the ultimate stress is 70 ksi. The second group is for ASTM A 588, with Fy=50 Ksi and Fu=70 ksi.
ASTM A572 contains five classes, ranging from 42 to 65; hence, the required grade must be given.
#### Practice problem for ASTM A24.
ASTM A 242 is specified for a beam with W24x192; it is required to find the yield stress Fy and Fu values for the beam and the modulus of elasticity; this practice problem is only given for CM#14. As a reminder, A242 was used as corrosion resistant in tables 2-4 and later removed from CM#14 for corrosion resistance; it was available for the W section and has a grey color.
Now, it is used for plates. Based on the flange thickness, it has three yield stress grades: 42, 46, and 50 ksi, marked J,k, and l.
Based on Table 1-1, the flange thickness is 1.46 inches which is less than 1.50 inches, we will select grade 50 category L and the ultimate stress Fu is specified as 70 ksi. The modulus of elasticity is 29×10^6 psi.
## Results
-
Last Updated on April 27, 2024 by Maged kamel
Last Updated on April 27, 2024 by Maged kamel
### #3. What is preffered grade for HSS pipe section?
The next post is a Review of AISC Tables 2-5 for plates.
For a useful external Chapter 3 – Tension Members– Bartlett Quimby
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https://prime-numbers.fandom.com/wiki/Thread:6909 | 1,582,098,569,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00024.warc.gz | 512,443,087 | 76,102 | ## FANDOM
852 Pages
• Which prime numbers satisfy the expression:
666 = p(3·3·3) + p(p(3·3·3))
where p(i) means the ith prime number (that is, p(1) = 2, p(2) = 3, p(3) = 5, etc.) | 75 | 182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-10 | latest | en | 0.617801 |
https://geosci-inversion.curve.space/inversion-module/linearinversion-app | 1,685,268,768,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00529.warc.gz | 323,859,796 | 51,918 | # Exercise 3: Linear Tikhonov Inversion
Liz Maag-Capriotti
# Linear Tikhonov Inversion¶
from geoscilabs.inversion.LinearInversionDirect import LinearInversionDirectApp
from ipywidgets import interact, FloatSlider, ToggleButtons, IntSlider, FloatText, IntText
import matplotlib.pyplot as plt
import matplotlib
matplotlib.rcParams['font.size'] = 14
app = LinearInversionDirectApp()
## Background¶
This app is based upon the inversion tutorial: "INVERSION FOR APPLIED GEOPHYSICS" by Oldenburg and Li (2005).
Douglas W. Oldenburg and Yaoguo Li (2005) 5. Inversion for Applied Geophysics: A Tutorial. Near-Surface Geophysics: pp. 89-150. eISBN: 978-1-56080-171-9 print ISBN: 978-1-56080-130-6
## Purpose¶
We illustrate how a generic linear inverse problem can be solved using a Tikhonov approach. The default parameters provided for the Forward and Inverse problems below generate a reasonable example for illustrating the inversion but the learning comes when these parameters are changed and outcomes are observed.
## Outline¶
The app is divided into two sections:
### Forward Problem¶
• Mathematical Background for the Forward Problem
• Step 1: Create a model, $\mathbf{m}$.
• Step 2: Generate a sensitivity matrix $\mathbf{G}$.
• Step 3: Simulate data ($\mathbf{d} = \mathbf{G}\mathbf{m}$) and add noise.
These steps are explored individually but additional text is given in 2 Linear Tikhonov Inversion. For convenience, the widgets used to carry out all three steps are consolidated at the end of the section. A brief mathematical description is also provided.
### Inverse Problem¶
• Mathematical Background for the Inverse Problem
• Step 4: Invert the data, and explore the results
Here we provide widgets to adjust the parameters for the inverse problem. Some basic information is provided but details about the parameters are provided in the text 2 Linear Tikhonov Inversion.
## Mathematical Background for the Forward Problem¶
Let $g_j(x)$ denote the kernel function for $j$th datum. With a given model $m(x)$, the $j$th datum can be computed by solving following integral equation:
Equation 1.7 Generic representation of a linear functional for forward mapping where $d_j$ is the $j^{th}$ datum, $g_j$ the associated kernel function, and $m$ the model.
$d_j=\int^b_ag_j(x)m(x)dx$
where
Equation 2.1 Oscillatory kernel functions $g_j(x)$, each of which correspond to each datum $d_j$, that decay with depth. The rate of decay is controlled by $p_j$, and $q_j$ controls the frequency.
$g_j(x)= e^{p_jx}\cos(2\pi q_jx)$
is the $j^{th}$ kernel function. By integrating $g_j(x)$ over cells of width $\Delta x$ and using the midpoint rule cell we obtain the sensitivities
Equation 2.19 Oscillatory kernel functions, defined in Equation 2.1, written after integrating over cells of a width $\Delta x$ and using the midpoint rule for the discretized calculations in LinearTikhonovInversion_App.
$\mathbf{g}_j(\mathbf{x}) = e^{p_j\mathbf{x}} cos (2 \pi q_j \mathbf{x}) \Delta x$
where
• $\mathbf{g}_j$: $j$th row vector for the sensitivty matrix ($1 \times M$)
• $\mathbf{x}$: model location ($1 \times M$)
• $p_j$: decaying constant (<0)
• $q_j$: oscillating constant (>0)
By stacking multiple rows of $\mathbf{g}_j$, we obtain sensitivity matrix, $\mathbf{G}$:
Equation 2.20 The sensitivity matrix $\mathbf{G}$ is created by stacking multiple rows of kernel functions $\mathbf{g}_j$ (LinearTikhonovInversion_App).
\begin{aligned} \mathbf{G} = \begin{bmatrix} \mathbf{g}_1\\ \vdots\\ \mathbf{g}_{N} \end{bmatrix} \end{aligned}
Here, the size of the matrix $\mathbf{G}$ is $(N \times M)$. Finally data, $\mathbf{d}$, can be written as a linear equation:
Equation 2.3 Expression for the linear forward problem in Equation 2.2, expanded for the N-length data vector $\mathbf{d}$.
\begin{aligned} \mathbf{d} = \mathbf{G}\mathbf{m} = \begin{bmatrix} d_1\\ \vdots\\ d_{N} \end{bmatrix}\end{aligned}
where $\mathbf{m}$ is an inversion model; this is a column vector ($M \times 1$).
In real measurments, there will be various noise sources, and hence observation, $\mathbf{d}^{obs}$, can be written as
Equation 2.5 The observed data is a combination of the clean data $\mathbf{d}$ and the noise $\mathbf{n}$.
$\mathbf{d}^{obs}=\mathbf{d}+\mathbf{n}$
## Step 1: Create a model, $\mathbf{m}$¶
The model $m$ is a function defined on the interval [0,1] and discretized into $M$ equal intervals. It is the sum of a: (a) background $m_{background}$, (b) box car $m1$ and (c) Gaussian $m2$.
• m_background : background value
The box car is defined by
• m1 : amplitude
• m1_center : center
• m1_width : width
The Gaussian is defined by
• m2 : amplitude
• m2_center : center
• m2_sigma : width of Gaussian (as defined by a standard deviation $\epsilon$)
• M : number of model parameters
Q_model = app.interact_plot_model()
## Step2: Generate a sensitivity kernel (or matrix), $\mathbf{G}$¶
By using the following app, we explore each row vector of the sensitivity matrix, $\mathbf{g}_j$. Parameters of the apps are:
• M: # of model parameters
• N: # of data
• p: decaying constant (<0)
• q: oscillating constant (>0)
• ymin: maximum limit for y-axis
• ymax: minimum limit for y-axis
• show_singular: show singualr values
Q_kernel = app.interact_plot_G()
## Step 3: Simulate data, $\mathbf{d}=\mathbf{Gm}$, and add noise¶
The $j$-th datum is the inner product of the $j$-th kernel $g_j(x)$ and the model $m(x)$. In discrete form it can be written as the dot product of the vector $\mathbf{g}_j$ and the model vector $\mathbf{m}$.
Equation 2.2 The linear forward problem in Equation 1.7 evaluated for a discretized model on a 1D mesh.
\begin{aligned} d_j&=\int_0^{x_1}g_j(x)m_1dx +\int_{x_1}^{x_2}g_j(x)m_2dx+\dots \\ &=\sum^M_{i=1}\left(\int_{x_{k-1}}^{x_k}g_j\left(x\right)dx\right)m_i\\ &\\ d_j &= \mathbf g_j \mathbf m\end{aligned}
If there are $N$ data, these data can be written as a column vector, $\mathbf{d}$:
Equation 2.3 Expression for the linear forward problem in Equation 2.2, expanded for the N-length data vector $\mathbf{d}$.
\begin{aligned} \mathbf{d} = \mathbf{G}\mathbf{m} = \begin{bmatrix} d_1\\ \vdots\\ d_{N} \end{bmatrix}\end{aligned}
Observational data are always contaminated with noise. Here we add Gaussian noise $N(0,\epsilon)$ (zero mean and standard deviation $\epsilon$). Here we choose
Equation 2.6 Definition of the standard deviation as a percentage of the datum and a floor value.
$\epsilon_j = \%|d_j| + \nu_j$
Q_data = app.interact_plot_data()
## Composite Widget for Forward Modelling¶
app.reset_to_defaults()
app.interact_plot_all_three_together()
## Mathematical Background for the Inverse Problem¶
In the inverse problem we attempt to find the model $\mathbf{m}$ that gave rise to the observational data $\mathbf{d}^{obs}$. The inverse problem is formulated as an optimization problem to minimize:
Equation 2.17 Objective function for the inverse problem which combines the data misfit (Equation 2.7) and chosen definition of the model norm (e.g. Equation 2.12, Equation 2.13, Equation 2.14) with a trade-off parameter $\beta$ to balance the relative influence of these terms.
$\phi(m) = \phi_d(m)+\beta\phi_m(m)$
where
• $\phi_d$: data misfit
• $\phi_m$: model regularization
• $\beta$: trade-off (Tikhonov) parameter $0<\beta<\infty$
Data misfit is defined as
Equation 2.7 Data misfit function measures the difference between each predicted datum $d_j$ and observation $d^{obs}_j$, normalized by the estimated standard deviation $\epsilon_j$.
$\phi_d=\sum^N_{j=1}\left(\frac{d_j-d_j^{obs}}{\epsilon_j}\right)^2$
where $\epsilon_j$ is an estimate of the standard deviation of the $j$th datum, and $d_j=\mathbf{g}_j\mathbf{m}$.
The model regularization term, $\phi_m$, can be written as
Equation 2.14 Combination of the smallest (Equation 2.12) and flattest model norms (Equation 2.13), where the quantities $\alpha_s$ and $\alpha_x$ are nonnegative constants used to adjust the relative importance of each term.
$\phi_m=\alpha_s\int (m-m^{ref})^2 dx+\alpha_x\int (\frac{d(m-m^{ref})}{dx})^2 dx$
The first term is referred to as the "smallness" term. Minimizing this generates a model that is close to a reference model $\mathbf{m}_{ref}$. The second term penalizes roughness of the model. It is generically referred to as a "flattest" or "smoothness" term.
## Step 4: Invert the data, and explore the results¶
In the inverse problem we define parameters needed to evaluate the data misfit and the model regularization terms. We then deal with parameters associated with the inversion.
### Parameters¶
• mode: Run or Explore
• Run: Each click of the app, will run n_beta inversions
• Explore: Not running inversions, but explore result of the previously run inversions
#### Misfit¶
• percent: estiamte uncertainty as a percentage of the data (%)
• floor: estimate uncertainty floor
• chifact: chi factor for stopping criteria (when $\phi_d^{\ast}=N \rightarrow$ chifact=1)
#### Model norm¶
• mref: reference model
• alpha_s: $\alpha_s$ weight for smallness term
• alpha_x: $\alpha_x$ weight for smoothness term
#### Beta¶
• beta_min: minimum $\beta$
• beta_max: maximum $\beta$
• n_beta: the number of $\beta$
#### Plotting options¶
• data: obs & pred or normalized misfit
• obs & pred: show observed and predicted data
• normalized misfit: show normalized misfit
• tikhonov: phi_d & phi_m or phi_d vs phi_m
• phi_d & phi_m: show $\phi_d$ and $\phi_m$ as a function of $\beta$
• phi_d vs phi_m: show tikhonov curve
• i_beta: i-th $\beta$ value
• scale: linear or log
• linear: linear scale for plotting the third panel
• log: log scale for plotting the third panel
app.interact_plot_inversion()
References
1. Oldenburg, D. W., & Li, Y. (2005). 5. Inversion for Applied Geophysics: A Tutorial. In Near-Surface Geophysics (pp. 89–150). Society of Exploration Geophysicists. 10.1190/1.9781560801719.ch5 | 2,836 | 9,919 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 99, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-23 | longest | en | 0.668364 |
http://mathhelpforum.com/pre-calculus/209385-logs.html | 1,501,060,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426086.44/warc/CC-MAIN-20170726082204-20170726102204-00600.warc.gz | 201,923,956 | 10,972 | 1. ## Logs
OK so i have $\log_2(2x)+\log_2(x-3)=3$
I think I should rewrite it as follows $\log_2(2x(x-3))=3$ then what?
and I need help with $e^-2x+3\bullet e^-x=10$
2. ## Re: Logs
Originally Posted by M670
OK so i have $\log_2(2x)+\log_2(x-3)=3$
I think I should rewrite it as follows $\log_2(2x(x-3))=3$ then what?
So far so good. Now, since $2^{\log_2(a)} = a$ what's your next step?
-Dan
3. ## Re: Logs
Originally Posted by M670
and I need help with $e^-2x+3\bullet e^-x=10$
First the LaTeX code. Instead of using \bullet I'd use \cdot. It looks a bit nicer.
$e^{-2x} + 3 \cdot e^{-x} = 10$
For the sake of clarity, I'd multiply both sides by $e^{2x}$. This step is not vital and you can easily alter the substitution I'm going to use.
So the equation is going to be:
$1 + 3 e^x = 10e^{2x}$
$10e^{2x} - 3e^x - 1 = 0$
Now let $y = e^x$. This gives
$10y^2 - 3y - 1 = 0$
Solve this then back substitute to get x.
-Dan
4. ## Re: Logs
Originally Posted by topsquark
So far so good. Now, since $2^{\log_2(a)} = a$ what's your next step?
-Dan
I have seen in the past that you can use this $2^{\log_2(a)} = a$ to get this $(2x\cdot(x-3))=3^2$ which I believe would give me $2x^2-6x-9=0$ then I solve it using the quadratic formula. But I don't know the reason why I am able to square it and then drop the log off?
5. ## Re: Logs
Originally Posted by M670
I have seen in the past that you can use this $2^{\log_2(a)} = a$ to get this $(2x\cdot(x-3))=3^2$ which I believe would give me $2x^2-6x-9=0$ then I solve it using the quadratic formula. But I don't know the reason why I am able to square it and then drop the log off?
So I believe my answer is x= $\frac{6+\sqrt108}{4}$ | 611 | 1,691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-30 | longest | en | 0.896154 |
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