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Complexity Issues And Computational Phenomena
Complexity Issues And Computational Phenomena
These concern the theoretical behaviour of analytic algorithms and the exhibition of unusual related computational phenomena.
□ J.M. Borwein and P.B. Borwein, "On the complexity of familiar functions and numbers," SIAM Review, 30(1988),589-601.
□ J.M. Borwein and P.B. Borwein,"Ramanujan and Pi," Scientific American, February, 1988,112-117.
□ J.M. Borwein and P.B. Borwein, "Pi and the AGM: Topics in Analytic Number Theory and Computational Complexity", ( Wiley, New York, 1987).
□ J.M. Borwein and P.B. Borwein, "Strange series evaluations and high precision fraud," MAA Monthly, 99(1992),622-640.
□ J.M. Borwein and P.B. Borwein, "Some observations on computer assisted analysis," Notices Amer. Math. Soc, 39(1992),825-829.
□ J.M. Borwein and P.B. Borwein, and K. Dilcher, "Euler numbers, asymptotic expansions and pi," MAA Monthly, 96(1989),681-687. | {"url":"http://wayback.cecm.sfu.ca/projects/complexity_issues_and_computational_phenomena.html","timestamp":"2024-11-04T13:52:33Z","content_type":"text/html","content_length":"2643","record_id":"<urn:uuid:6095f3a3-bb59-4aec-9417-6be6f4d44e78>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00698.warc.gz"} |
Real Reason to object to Big Bang "Time Line"
<span class='quoteauthor'>RussT
And in Post # 73 Wayne correctly says...[/quote said:</span>
Originally Posted by Wayne
The fact that you think there is a speed of time shows you that you need to learn more about physics. The "speed" of something is the distance it travels / time so show me what that means. I'll
give you a clue the times will cancel out and leave you with distance. This doesn't mean 2 things separated by distance will appear to each other as having the same time. This would only be true
if light propagated instantly which it doesn't.
[/B.BUT, here is the Real key to finally understanding Time, and Wayne doesn't even really know what he actually did/showed here...in Post # 75
Originally Posted by Wayne
Agreed. All that says is that to the photon there is no time or distance. Whoopy. For the rest of us that have rest mass Relative's claims have already been proven to be wrong not only in thought
experiments but actual experiments. Flipping from our reference frame to a photon's frame, which to us is not a valid reference frame anyway, makes no sense and does not support Relative's idea.
All it does is bring the equations to a singularity and thus you can back out and get any meaningful answer. You might as well claim 1 / 0 = 2 + 5....you can't get past the 1/0.
Originally Posted by Bob Angstom
A double constant c makes no sense to me either.
This is absolutely correct...A "Constant" CANNOT have 2 different definitions, and the ONLY one that is Valid is Light speed at Constant 186,282.397mps with Time as a Constant
The definitions at the division by 0 of 1. 0 Time from Point A to Point B to ANY distance to Infinity 2. "Instantaneous" from point A to Point B at ANY distance to Infinity 3. distance contracted
to 0 from Infinity are ALL Singularity definitions and DO NOT EXIST AT ALL!!!
And here is the "REAL" kicker, and I have absolutely determined this to be 100% correct...and I do not believe that this arguement/pure logic has ever been presented to ANY Relativity Guru, so
just hand waving it aside, which seems to be the first thing that BAUTians are prone to do, would actually show that 'real consideration' and proper analysis of things being presented in honest
attempts at actually coming to terms with what is "REAL" or not is sorely lacking...
Anyway....here it is...
It has to do with supposedly *Approaching lightspeed "c"* or 'taking the limit'...and the 'real' problem with having 2 defintions of 1 Constant both called "c"
As a observer/ship gets closer and closer to supposedly going the speed of light, once they are almost at the Max....ie'as close to light speed as possible without readhing "c" (supposedly
because nothing with mass can OR it would take Infinite energy to do so...but both of those are actually meaningless)....then you get this...
Originally Posted by Grey
A more precise wording might be that if you travel from here to Alpha Centauri, moving at arbitrarily close to the speed of light, the trip will take roughly four years for outside observers, but
will be arbitrarily short for the person traveling. Depending on how fast you go, the trip could take a year as measured by the traveler, or a day, or a second, or a nanosecond. Essentially no
time as measured by an outside observer, if the traveler is moving quickly enough. And it still scales. If you're traveling fast enough that time dilation means you measure a nanosecond to go
four light years, then it will take you a whole second to travel four billion light years.
Now, I included Grey's whole paragraph, BUT the bold is the key...
IFFFFFFFFFFFFF you went just that smidgeon faster, to actually get to the supposed "c", you would go to Infinity Infintely Fast!
SO, you are NOT approaching light speed at all, you are approaching going Infinitely fast to Infinity, and all this time, for the last 100 years, this has NOT even been realized!
OR, you are "Shrinking"/contracting 4 Billion Light Years worth of Space/Distance if you got there in 1 second down to 186, 282.397 miles, but here too you are still approaching infintely fast,
NOT light speed.
Sorry, BUT there is NO frame where photons are or ever could travel from Point A to Point B in 0 time...All photons that we see/dectect and measure are traveling at Constant "c" of 186,282.397
Time dilation does not exist and swapping two observers motions is absolutely meaningless.
SO, as I have stated before....the ONLY thing that is "Real", is what is currently being called "Our Perspective" or "Earth Rest Frame" where time and light speed "c" are both Constant, and that has
absolutely Nothing to do with "Relativity".
The SR 'rest frame observer' CANNOT be equal to 'the earth rest frame observer', SO Both of the SR observers are NOT seeing any reality at all.
all throughout this thread Wayne Francis was correctly telling
that they could NOT use "Instant" photons to see any 'reality' IE: distant things as they are NOW, not even realizing that he was correctly "Falsifying" SR as well!!!
BUT, to finally justify his position he would, as most do, switch from the "Linear Singularity" of SR to the "Spherical Singularity"...the Naked Infinite Singularity of Big Bang fame... the one that
goes from a 'Point to Infinity' and includes the so called 'States/fates of the Universe'...
Einsteins orginal Lambda or "Static Universe" (where there is a "Force" holding the Universe back from contracting in on itself in perfect balance....which has been falsified) the "Closed Universe"
(where gravity wins and the universe collapses...which has now been falsified) or the "Open Universe" (where the accelerated Expansion goes on forever...(But, guess what....WIMPS don't even
exist....so now what???) and where Lambda was added to the CDM concordance model...surprise-surprise-surprise...they "FOUND" the 74% of the Universe that was "Missing")
BUT...when they consider the "Static Universe", "Closed Universe", "Open Universe"...they are sure they are covering "Every" potentiality of "How the Universe can be working...when in
"Reality"...NONE of those even exist, and NEVER even had a chance of existing.
Since I started, 5 years ago, from the Premise of "When Do SMBH's become part of a galaxies life", I have been dealing with 'singularities' and everything from "First Principles" and finally I do
understand the whole thing, and have step by step been able to correctly 'Falsify', from first principles, Relativity.
All I have been doing in reality is "Fixing" the singularity problems, BUT unfortunately that means that Relativity NEVER even had a chance of being "Real"...there are NO "Instant" photons anywhere
in our Universe(s) and without those (Singularity definitions) it is "Impossible" that Relativity could even have ever existed at all....and yet they are defending it to the death.,.:<((( | {"url":"http://forums.nimblebrain.net/cgi-bin/topic_show.pl?tid=200","timestamp":"2024-11-12T11:37:27Z","content_type":"text/html","content_length":"39272","record_id":"<urn:uuid:fc33220e-07cf-407b-84c5-41ce0ab76576>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00828.warc.gz"} |
Logan, Author at Only Code
Unlike simple algorithms, efficient sorting algorithms typically have an average time complexity of O(n logn).
Sorting algorithms are methods used to rearrange elements in a list or array into a particular order, typically either ascending or descending.
Timsort with its time complexity of O(nlogn) in the worst case and O(n) in the best case, has become the default sorting algorithm for Python, Java, and other programming languages.
Treap is an efficient and simple data structure that combines the best of both binary search trees and heaps. It provides the benefits of efficient search, insert, and delete operations.
Bayesian inference is central to probabilistic models and algorithms like Naive Bayes classifiers, which are widely used in machine learning.
Typically, login forms display input characters as asterisks or bullets to prevent shoulder surfing and other forms of eavesdropping.
Regex is efficient because most programming languages implement optimized regex engines that handle pattern matching using sophisticated algorithms like finite automata, ensuring fast performance.
Scientific notation is a method of expressing very large or very small numbers in a compact and readable form.
NLTK (Natural Language Toolkit) is a powerful library in Python that provides easy-to-use interfaces to over 50 corpora and lexical resources, including WordNet.
Caesar Cipher is one of the simplest and oldest encryption techniques, named after Julius Caesar, who used it to protect his military communications. | {"url":"https://www.onlycode.in/author/onlycode_oz40jd/","timestamp":"2024-11-11T08:07:09Z","content_type":"text/html","content_length":"187069","record_id":"<urn:uuid:7699e4e0-211d-4e94-ad1b-e956b68d278c>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00480.warc.gz"} |
Introducting the Software Engineering Working Group and {mmrm}
Mixed Models for Repeated Measures - Why is it a Problem?
Primary Goal: Collaborate to engineer R packages that implement important statistical methods to fill in critical gaps
Secondary Goal: Develop and disseminate best practices for engineering high-quality open-source statistical software
To run an MMRM model in SAS it is recommended to use either the PROC MIXED or PROC GLM procedures.
Less model assumptions are applied in PROC MIXED, primarily how one treats missingness.
We will compares the PROC MIXED procedure to the {mmrm} package in the following attributes:
Both languages have online documentation of the technical details of the estimation and degrees of freedom methods and the different covariance structures available.
One major advantage of the {mmrm} over PROC MIXED is that the unit testing in {mmrm} is transparent. It uses the {testthat} framework with {covr} to communicate the testing coverage. Unit tests can
be found in the GitHub repository under ./tests.
The integration tests in {mmrm} are set to a tolerance of 10e^-3 when compared to SAS outputs.
Method {mmrm} PROC MIXED
ML X X
REML X X
For users that need more structure {mmrm} is easily extensible via feature requests in the GitHub repository.
Covariance structures {mmrm} PROC MIXED
Unstructured (Unweighted/Weighted) X/X X/X
Toeplitz (hetero/homo) X/X X/X
Compound symmetry (hetero/homo) X/X X/X
Auto-regressive (hetero/homo) X/X X/X
Ante-dependence (hetero/homo) X/X X
Spatial exponential X X
Method {mmrm} PROC MIXED
Contain X* X
Between/Within X* X
Residual X* X
Satterthwaite X X
Kenward-Roger X X
Kenward-Roger (Linear)** X X
**This is not equivalent to the KR2 setting in PROC MIXED
Have an interest in working on these topics? Come work with us, information on the SWE WG can be found here: ASA BIOP SWE WG | {"url":"https://www.openstatsware.org/slides/r-adoption-jan2023.html","timestamp":"2024-11-05T20:24:51Z","content_type":"text/html","content_length":"51518","record_id":"<urn:uuid:6926de82-788d-498a-b7bc-77fb9bf9d7f0>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00734.warc.gz"} |
Dynamics of Processes On & Off Networks
Bring together different research areas (and their corresponding communities): network science, mathematics, computer science, signal processing and epidemiology.
When Jun 20, 2016 09:00 to
Jun 22, 2016 05:00
Where Lyon
Add event to calendar vCal
Motivation and Aims
Large-scale networks with complex interaction patterns between elements are found in abundance in both nature and man-made systems (e.g., genetic pathways, ecological networks, social networks,
networks of scientific collaboration, WWW, peer-to-peer networks, power grid etc.). The main aim of this workshop is to explore the statistical dynamics on and of such networks. “Dynamics on
networks” refers to the different types of so-called processes (e.g. proliferation, diffusion etc.) that take place on networks. The functionality/efficiency of such processes is strongly affected by
the topology as well as the dynamic behavior of the network. On the other hand, “Dynamics of networks” mainly refers to various phenomena (for instance self-organization) that go on in order to bring
about certain changes in the topology of the network. It has become clear that the stability and robustness of highly dynamical networks such as networks of mobile agents, and the study of dynamical
processes on networks are among the hottest new theoretical challenges for Complex Network research. Accordingly, Dynamics On and Of Complex Networks will focus on these topics.
One way to address this challenge of data representation is to consider not only individual entities, but also relationships between them. Considering relationships between entities leads to dealing
with graphs. Graphs are a convenient data model for representing massive digital data sets in many applications. With such a unified representation, many information processing tasks become graph
analysis problems. However, graphs are not only of interest for representing the data to facilitate its analysis, but also for defining graph-theoretical algorithms that enable the processing of data
associated with both graph edges and vertices. Indeed, more generally, massive datasets represented as graphs can be seen as a set of data samples, with one sample at each vertex in the graph. In
such a scenario, the high-dimensional data associated to vertices can be viewed as graph signals. There is emerging interest in the development of algorithms that enable the processing of graph
signals: one might be interested in filtering, clustering, or compressing this structured type of signals. The analysis of graph signals faces several open challenges, mainly because of the
combinatorial nature of the involved signals, that are not necessarily embedded in Euclidean spaces. This is leading to the emergence of a new research field on “Graph Signal Processing” at the
intersection of computer science, mathematics, and signal processing. This new research field aims at developing novel approaches enlarging the scope of traditional signal processing methods and
applications, so that they can be applied to arbitrary graph signals.
The workshop expects burgeoning multi-disciplinary research contributions that combine methods from computer science, statistical physics, graph signal processing, nonlinear dynamics, epidemiology,
econometrics and social network theory, to study common problems in systems exhibiting a complex network structure (e.g., biological systems, linguistic systems, social systems and various other
man-made systems like the Internet, WWW, peer-to-peer systems etc.). The workshop will particularly promote research contributions on dynamical networks and dynamics on networks. Accordingly, one of
the major goals of the workshop is to bring together different research areas (and their corresponding communities): network science, mathematics, computer science, signal processing and | {"url":"https://www.ixxi.fr/agenda/seminaires/seminaires-2015/dynamics-of-processes-on-off-networks","timestamp":"2024-11-11T04:13:09Z","content_type":"text/html","content_length":"26431","record_id":"<urn:uuid:dca5771c-809e-41ee-ad4b-6d6e4850b3c8>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00192.warc.gz"} |
UV light is responsible for sun tanning. How do you find the wavelength (in nm) of an ultraviolet photon whose energy is 6.4 x 10^-19? | Socratic
UV light is responsible for sun tanning. How do you find the wavelength (in nm) of an ultraviolet photon whose energy is #6.4 x 10^-19#?
1 Answer
The photon will have a wavelength of about $310$ nm.
The energy of a photon is proportional to its wavelength and is given by the equation:
$E = \frac{h c}{\lambda}$
Where $h$ is Planck's constant ($\cong 6.626 \times {10}^{-} 34$ Js)
And $c$ is the speed of light in vacuum ($\cong 2.9979 \times {10}^{8}$ m/s)
Thus $h c \cong 1.9864 \times {10}^{-} 25$ Jm
We can manipulate the previous equation to solve for wavelength:
$\lambda \cong \frac{h c}{E} = \frac{1.9864 \times {10}^{-} 25}{6.4 \times {10}^{-} 19}$
$\lambda \cong 0.310 \times {10}^{-} 6 = 310 \times {10}^{-} 9$ meters
Thus $\lambda \cong 310$ nm. One can confirm from other sources that this places the photon in the UV B wavelength range.
Impact of this question
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Fractals – Science4All
Can you glue opposite edges of a square? Relativity 7
Chaos Theory, Meteorology, Navier-Stokes, Wolfram (Hiking in Modern Math 5/7)
Fractals, Mandelbrot, Pixar (Trek through Math 4/8)
The Tortuous Geometry of the Flat Torus
Take a square sheet of paper. Can you glue opposite sides without ever folding the paper? This is a conundrum that many of the greatest modern mathematicians, like Gauss, Riemann, and Mandelbrot,
couldn't figure out. While John Nash did answer yes, he couldn't say how. After 160 years of research, Vincent Borrelli and his collaborators have finally provided a revolutionary and breathtaking
example of a bending of a square sheet of paper! And it is spectacularly beautiful!
Self-Reference, Math Foundations and Gödel’s Incompleteness
Although highly appreciated by artists, self-reference has been a nightmare for mathematicians. It took one of the greatest, Kurt Gödel, to provide a better understanding of it. This article deals
with paradoxes, recursion, fractals and Gödel's incompleteness theorems.
From Britain’s coast to Julia set: an introduction to fractals
Introduction I could have started this article by showing you beautiful examples of fractals, found in nature or invented by mathematicians, in two or three dimensions, constructed deterministically
or not. But I won’t. In fact, you wouldn’t need to read this article to find such examples. I’d rather start by telling you a story. It […] | {"url":"https://www.science4all.org/tag/fractals/","timestamp":"2024-11-03T23:08:58Z","content_type":"text/html","content_length":"32288","record_id":"<urn:uuid:4a8b773e-e148-4874-b500-06d20bc673b5>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00805.warc.gz"} |
Taryn Nefdt, Author at ExcelDemy
Hi Agnius
Thank you for your comment. Point taken. However, these are topics tested in many Excel exams and there is often confusion since workbook level is actually protecting the worksheet structure, so it
was in that context that the tutorial was made and made in order to address those issues.
This topic (worksheet level and workbook level protection) is covered in the detailed syllabus for the Microsoft Excel Expert Exam for 2013 that is exams (77-427 and 77-428), in the official book
released by Microsoft Press – MOS 2013 Study Guide Microsoft Excel Expert by Mark Dodge.
So it is relevant to go over, for people who are studying for the MOS Excel Expert Exam and any other Excel exam. Also while I agree with you that encryption and VBA also play a role in more advanced
level protection (I will do another tutorial on the more advanced level options :-)) .
My personal opinion is sometimes depending on one’s organization, one’s needs may not be that complex so its worthwhile knowing what Excel has available in terms of simpler options. It is one layer
of protection and something is better than nothing at all, in my opinion. A workbook that has sensitive information that has no protection at all, is less secure than a workbook that has worksheet
level and workbook level protection. Also from the psychological perspective there is now some form of inhibiting barrier, if one uses worksheet level or workbook level standard protection options.
However, you have opened up a very interesting debate with your points – is something in terms of protecting one’s data at a simpler level really better than nothing at all or should one address this
issue from the VBA and encryption level only.
I will in the next tutorial, address some of the points you’ve raised and give you credit for the interesting questions/points you’ve posed in the debate section of the next tutorial on Encryption
and VBA protection. | {"url":"https://www.exceldemy.com/author/taryn-nefdt/page/3/","timestamp":"2024-11-13T16:23:54Z","content_type":"text/html","content_length":"226549","record_id":"<urn:uuid:1387136a-f352-4331-a359-ce938f8440a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00820.warc.gz"} |
Soon purchased an office chair for $167.87. She received a $19.95 rebate from the manufacturer an $8.95 rebate from the store. The sales tax in her state is 55%. What is the final price? | HIX Tutor
Soon purchased an office chair for $167.87. She received a $19.95 rebate from the manufacturer an $8.95 rebate from the store. The sales tax in her state is 55%. What is the final price?
Answer 1
The final price is #$215.40#.
A rebate is a partial refund, so start by subtracting the two rebates from the original price.
#$167.87 - $19.95 = $147.92#
#$147.92 - $8.95 = $138.97#
Next, the sales tax is 55%. In decimal terms, this is 0.55. However, she still has to pay full price for the chair AND the sales tax, so we add 1. Now multiply the decimal by the current chair price,
and find your final answer.
#$138.97 * 1.55 = $215.4035#
You end up with a longer decimal, but since we only go to the hundreds place (pennies) when paying for things, you have to round to the hundreds place. So your final answer is #$215.40#.
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Answer 2
To find the final price, subtract the rebates from the original price, then add the sales tax. So, $167.87 -$19.95 - $8.95 =$138.97. Then, calculate the sales tax: $138.97 * 0.55 =$76.43. Finally,
add the sales tax to the discounted price: $138.97 +$76.43 = $215.40. Therefore, the final price is$215.40.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Soumava's Algorithm | Brilliant Math & Science Wiki
This algorithm was conceived by Soumava Pal. This is his original note.
Soumava develops a numerical method to compute pairs of \((x, y)\) such that \(x^y = y^x = k,\) where \(k > 1.\)
Given the parameters \(x_0\), a seed, and \(h\), the tolerance limit, Soumava's algorithm is as follows:
1. [Initialize] \(x \leftarrow x_0\)
2. [Compute] \(y \leftarrow k^{\frac{1}{x}}\)
3. [Iterate] While \( (x^y \sim k) > h \): \[(x,y) \leftarrow \left(k^{k^{\frac{-1}{x}}}, k^{k^{\frac{-1}{y}}}\right) \]
4. [Output] Return \((x,y)\)
The proof is cluttered, we could clean it up.
Let \(<a>\) be a sequence defined as follows: \[a_0 = x_0, \quad a_{i+1} = k^{\frac{1}{a_i}}. \] Furthermore, let \(<x>\) be the subsequence formed by the even terms of \(<a>\): \[x_i = a_{2i}.\]
Similarly, let \(<y>\) be the subsequence formed by the odd terms of \(<a>\): \[y_i = a_{2i+1}.\] We claim that \[\lim_{i\to \infty} x_i^{y_i} = y_i^{x_i} = k.\ _\square\]
We split the proof into two parts. First we prove that the sequences converge, and then we show that the limits satisfy the equation.
\(P\) = \(a_{1}\) = \(k^{1/m}\), \(b_{j+1}\) = \(k^{1/b_{j-1}}\) for all \(j \in N,\) where \(N\) is the set of all positive integers, and \(m\) is any positive real number (domain of \(m\)
is \((0, \infty)\)).
Let us take the two sub-sequences \(\{ a_{1}, a_{3}, a_{5}, \ldots \}\) and \(\{a_{2}, a_{4}, a_{6}, \ldots\}\) that are the sub-sequences formed by the odd numbered terms and by the even
numbered terms, respectively.
Let the sequence formed by the odd numbered terms be \(M\) and the sequence formed by the even numbered terms be \(T.\)
Now we express \(a_{n+2}\) in terms of \(a_{n}\). By definition,
\[a_{n+2} = k^{1/a_{n+1}} =k^{1/{k^{1/a_{n}}}}. \qquad (1)\]
Our first observation is that the two sub-sequences that I have considered are both bounded below by \(0\) and above by \(k\) itself.
The lower bound is easy to observe because we know that the exponential function \(a^x\) is positive for all real \(x\) and positive \(a\). And our sequence is of the exponential form only. So \
(0\) is indeed a lower bound.
Now we can establish the validity of \(k\) as an upper bound by the method of contradiction.
Let us assume that the sequence \(P\) contains a number which is greater than \(k\) itself, i.e. \(a_{h} = v>k\) for some \(h\) in \(N\).
But \(a_{h}= k^{1/{a_{h-1}}}= v\) necessarily implies that \(a_{h-1}\) is less than \(1\) and greater than \(0,\) i.e. \(1/{a_{h-1}}\) is greater than \(1\), and \(k^{1/{a_{h-1}}}\) is
greater than \(k\). All this holds if \(a_{h-1}\) is less than \(1\). But again \(a_{h-1}\) = \(k^{1/{a_{h-2}}}\).
Now if we take logarithm to the base \(k\) on both sides, then on LHS we get something negative as \(a_{h-1}\) belongs to \((0,1)\) by assumption and \(k\) is greater than \(1\) (domain of \
(k\) is \((1,\infty)\)). But on RHS we have \(1/{a_{h-2}}\) which is positive because \(a_{h-2}\) is positive as the lower bound of the sequence \(\{a_{1}, a_{2}, a_{3}, \ldots\}\) is 0. Thus
a negative number on the LHS is equal to a positive number on the RHS by assumption that \(a_{h}>k\) for some positive integer \(h\).
So we have a direct contradiction if \(a_{h}>k\) for some positive integer \(h\), and therefore this assumption must be false. So the negation of this statement must be true. That is ‘\(a_{h}
<k\) for all positive integers \(h'\). So we have established that \(k\) itself is an upper bound.
Our second aim is to prove that these two sub-sequences are monotonic.
Let us consider the sub-sequence \(M=\{a_{1}, a_{3}, a_{5}, \ldots \}\) and we take \(M_{1} = a_{1}, M_{2} = a_{3}, M_{3} = a_{5}\), and in general \(M_{k} = a_{2k-1}\) for all \(k\in N\).
We know by order property (law of trichotomy) that given two reals \(a\) and \(b\), either \(a=b\), or \(a>b\), or \(a<b\). Here we observe that \(M_{n}\) is not equal to \(M_{n+1}\) for any
positive integer \(n\), because of the relation between two consecutive terms of the sequence \(\{a_{1}, a_{3}, a_{5}, \ldots\}\) derived in \((1)\) above.
So \(M_{1}>M_{2}\) or \(M_{2}>M_{1}\).
Case I:
\(M_{1}>M_{2}\), i.e. \(a_{1}>a_{3}\)
Next we proceed by the principle of mathematical induction to show that \(M_{n}>M_{n+1}\) for all positive integers \(n\). In fact we have already checked the first step for \(n=1\).
Next we assume the validity of the statement to hold true for some \(n=k\), i.e. \(M_{k}>M_{k+1}\) or \(a_{m}>a_{m+2}\) (where obviously \(m={2k-1}\)).
\[ M_{k}&>M_{k+1}\\ a_{m}&>a_{m+2}\\ 1/{a_{m}} &< 1/{a_{m+2}} \qquad (\text{reciprocating the positive numbers reverses the sign of inequality})\\ k^{1/{a_{m}}} &< k^{1/{a_{m+2}}}\\ 1/{k^{1/
{a_{m}}}} &> 1/{k^{1/{a_{m+2}}}} \qquad (\text{reciprocating the positive numbers reverses the sign of inequality})\\ k^{1/{k^{1/{a_{m}}}}} &> k^{1/{k^{1/{a_{m}}}}}\\ a_{m+2} &> a_{m+4}, \
qquad (\text{by the relation derived in (1)}) \]
\[M_{k+1} > M_{k+2}.\]
So we have \(M_{1}\) > \(M_{2}\) and \(M_{k}\) > \(M_{k+1}\), which implies \(M_{k+1}\) > \(M_{k+2}\) for all positive integers \(k\).
So \(M\) is a strictly decreasing monotone sequence.
Having established that, we also see that \(M_{1}\) > \(M_{2}\) i.e. \(a_{1}\) > \(a_{3}\) implies that
\[ 1/{a_{1}} &< 1/{a_{3}}\\ k^{1/{a_{1}}} &< k^{1/{a_{3}}}\\ a_{2} &< a_{4}. \]
Therefore \(a_{1}\)>\(a_{3}\) implies that \(a_{2}\) < \(a_{4}\). If we consider the sub-sequence \(T\) consisting of \(\{a_{2}, a_{4}, a_{6}, \ldots \},\) then we have \(a_{2} < a_{4}\), i.e. \
(T_{1} < T_{2}\). By a similar inductive argument as above, we can see that
\(T\) is a strictly increasing monotone sequence.
It is to be noted that the conclusions drawn above are under the heading of Case I.
Case II:
\(M_{1} < M_{2}\)
Under this consideration, we proceed exactly as in Case I and prove that \(M\) is a strictly increasing monotone sequence.
And on the same lines as in Case I, we have that \(T\) is a strictly decreasing monotone sequence.
1. In both the cases which are mutually exhaustive, we see that \(M\) and \(T\) are monotone sequences of the opposite nature, i.e. if one is increasing the other is decreasing.
2. Also \(M\) and \(T\) are sub-sequences of a sequence \(P\) which is bounded as a whole with \(0\) as a lower bound and \(k\) as an upper bound.
3. So \(M\) and \(T\) themselves are also bounded with \(0\) as a lower bound for both and \(k\) as an upper bound for both.
So from 1, 2 and 3 we see that M and T are both monotonic and bounded.
By the Bolzano-Weierstrauss theorem, any monotonic bounded sequence has a finite limit.
Since the sequences \(M\) and \(T\) meet the requirements of the theorem, they are convergent and have a finite limit. Let the limit of \(M_{n}\) as \(n\) tends to \(∞\) be \(z\) and the
limit of \(T_{n}\) as \(n\) tends to \(∞\) be \(g\).
Now, we set out to prove that the limits indeed satisfy the equation:
Because the sequence converges, as \(i \to \infty\), \[(x_i,y_i) \to (x_{i+1}, y_{i+1}) \to \lim_{ r \to \infty}(x_r,y_r) = (x', y'). \qquad \text{(say)} \] By construction, \[ y_i &= k^{\frac{1}
{x_i}} \\ y' &= k^{\frac{1}{x'}} \\ {(y')}^{x'} &= k. \] Similarly, \[ x_{i+1} &= k^{\frac{1}{y_i}} \\ x' &= k^{\frac{1}{y'}} \\ {(x')}^{y'} &= k. \]
Therefore, \[(x')^{y'} = (y')^{x'} = k. \ _\square \]
1 def soumava(x0,h,k):
2 x = x0
3 y = k**(1.0/x)
4 while abs(x**y - k) > h:
5 x, y = k**k**(-1.0/x), k**k**(-1.0/y)
6 return x, y
This section requires expansion
The convergence rate is most likely exponential. | {"url":"https://brilliant.org/wiki/soumavas-algorithm/","timestamp":"2024-11-10T17:31:13Z","content_type":"text/html","content_length":"52741","record_id":"<urn:uuid:ecd7471d-2c54-4011-ac1c-90c5dd08cc15>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00265.warc.gz"} |
MAT 232 - Differential Equations
Course Objectives
1.To introduce the basic concepts required to understand, construct, solve and interpret differential equations. 2.To teach methods to solve differential equations of various types. 3.To give an
ability to apply knowledge of mathematics on engineering problem
Course Description
Classification of Differential Equations, First order equations;linear equations, separable equations, change of variableand integrating factor, existence and uniqueness theorems, applications.
Second order linear equations; linear equationswith constant coeficients, Homogeneous equations, the method of reduction of order, Nonhomogeneous Equations, themethod of undetermined coefficients,
the method of variation of parameters, Higher order linear equations. Euler-Cauchy equation. Power series method; solution around ordinary and regular-singular points. Laplace transformation;basic
definition and theorems, solution of initial value problems, convolution, delta function, transfer function. Systemsof linear differential equations; fundamental theories, solutions of homogeneous
and nonhomogeneous system ofdifferantial equations, solutions using Laplace transformation
Course Coordinator
Ayşe Peker
Course Language | {"url":"https://ninova.itu.edu.tr/en/courses/faculty-of-science-and-letters/11981/mat-232/","timestamp":"2024-11-05T06:06:22Z","content_type":"application/xhtml+xml","content_length":"8218","record_id":"<urn:uuid:750a7b27-2d9b-4c5d-a0b6-d69b07c1584f>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00287.warc.gz"} |
Partial Dependency Plots and GBM
My favorite "go-to-first" modeling algorithm for a classification or regression task is Gradient Boosted Regression Trees (Friedman, 2001 and 2002), especially as coded in the GBM package of R.
Gradient boosted regression is one of the focuses of my masters thesis (along with random forests) and has gotten more and more attention as more and more data mining contests have been won, at least
in part, by employing this method. Some of the reasons I pick Gradient Boosted Regression Trees as the best off the shelf predictive modeling algorithm available are:
• High predictive accuracy across many domains and target data types
□ Ability to specify various loss functions (Gaussian, Bernoulli, Poisson etc.) as well as run survival analysis via Cox proportional hazards, quantile regression etc.
• Handles mixed data types (continuous and nominal)
• Seamlessly deals with missing values
• Contains out-of-bag (OOB) estimates of error
• Contains variable importance measures
• Contains variable interaction measures / detection
• Allows estimates of marginal effects of a predictor (s) via Partial Dependency Plots.
This latter point is a nice feature coded into the GBM package that gives the analyst the ability to produce univariate and bivariate partial dependency plots. These plots enable the researcher to
understand the effect of a predictor variable (or interaction between two predictors) on the target outcome, given the other predictors (partialling them out - after accounting for their average
The technique is not unique to Gradient Boosted Regression Trees and in fact is a general method for understanding any black box modeling algorithm. When we use ensembles for instance, this tends to
be the only way to understand how changing the value of a predictor, say a binary predictor from 0 to 1 or a continuous predictor from within it's observed range, effects the outcome, given the model
(i.e. accounting for the impact of other predictors in the model).
The idea of these “marginal” plots is to displays the modeled outcome, over the range of a given predictor. Hastie (2009) describes this general technique as considering the full model function $f$,
depending on a small subset of predictors we are interested in, $X_{s}$, where this subset is typically one or two predictors in practice, and the other predictors in the model , $X_{c}$. This full
model function is then written as $f(X)=f(X_{s},X_{c})$ where $X=X_{s} \cup X_{c}$ . A partial dependence plot displays the marginal expected value of $f(X_{s})$, by averaging over the values of $X_
{c}$. Practically, this would be given by $f(X_{s}=x_{s})=Avg(f(X_{s}=x_{s}, X_{ci}))$ where the expected value of the function for a given value (vector) of the subset of predictors is the average
value of the model output setting the subset predictors to the value of interest and averaging the result with the values of $X_{c}$ as they exist in the data set.
A brute force method would be to create a new data set from the training data, but repeating it for every pattern in $X_{s}$ of interest plugged in and then using the model to output the average
value for each distinct value of $X_{s}$. | {"url":"http://adventuresindm.blogspot.com/2013/01/partial-dependency-plots-and-gbm.html","timestamp":"2024-11-09T17:09:54Z","content_type":"application/xhtml+xml","content_length":"59445","record_id":"<urn:uuid:0f10a378-445d-4bac-803d-717abb163a21>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00055.warc.gz"} |
Formula using Sum and Countif not returning expected outcomes
Here's a version of the formula that works. However I need to include a second value and get an error when I try. I'm sure it's the formula syntax. Not sure if there is a better way to get the right
outcome. I basically am trying to determine % Complete OR Not Applicable
Working with just Pass as a value
=SUM((COUNTIF(CHILDREN(Status22), ="Pass")) / SUM(COUNT(CHILDREN(Status22))))
When a Not Applicable is selected the above formula results in a reduction to the % Complete and the desired outcome would be for Pass and Not Applicable to both count so if a pass or not applicable
are used the % complete would be the same.
=SUM((COUNTIF(CHILDREN(Status22), ="Pass") + (COUNTIF(CHILDREN(Status22), ="Not Applicable")) / SUM(COUNT(CHILDREN(Status22))))
When I use the formula above I get a number much higher than 100% which is the max I should be getting
Not sure the right way to structure the formula to get a % complete based on the sum of both Pass and Not Applicable versus overall number of children. Which is how I think the formula should be
structured. Any suggestions?
• Without taking a hard look at it... try this revision?
=COUNTIF(CHILDREN(Status22), ="Pass") + COUNTIF(CHILDREN(Status22), ="Not Applicable") / COUNT(CHILDREN(Status22))
You don't need all the SUM's cause the countif will automatically sum it for you and you're doing simple math to add and then divide each part.
• Th formula you gave me did not work. It returned 700%.
However the following formula did work
=SUM(COUNTIF(CHILDREN(Status22), ="Pass") + COUNTIF(CHILDREN(Status22), ="Not Applicable")) / COUNT(CHILDREN(Status22))
Thank you!
• Hmmm. Cool. Glad I helped you get on the right track!
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/22361/formula-using-sum-and-countif-not-returning-expected-outcomes","timestamp":"2024-11-07T19:03:34Z","content_type":"text/html","content_length":"415448","record_id":"<urn:uuid:755140af-e374-4df5-947d-606811548b49>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00453.warc.gz"} |
Applied Statistics Discipline
An Applied Statistics Discipline is an applied mathematics discipline for the applied statistics domain (that studies applied statistics tasks).
• Example(s):
• Counter-Example(s):
• See: Biostatistics, Data Mining Discipline, Entropy, Statistics, Actuarial Science, Insurance, Finance, Astrostatistics, Biostatistics, Biology, Medical Statistics, Business Analytics,
Chemometrics, Chemistry.
• (Wikipedia, 2016) ⇒ http://wikipedia.org/wiki/List_of_fields_of_application_of_statistics Retrieved:2016-3-22.
□ Statistics is the mathematical science involving the collection, analysis and interpretation of data. A number of specialties have evolved to apply statistical theory and methods to various
disciplines. Certain topics have "statistical" in their name but relate to manipulations of probability distributions rather than to statistical analysis.
☆ Actuarial science is the discipline that applies mathematical and statistical methods to assess risk in the insurance and finance industries.
☆ Astrostatistics is the discipline that applies statistical analysis to the understanding of astronomical data.
☆ Biostatistics is a branch of biology that studies biological phenomena and observations by means of statistical analysis, and includes medical statistics.
☆ Business analytics is a rapidly developing business process that applies statistical methods to data sets (often very large) to develop new insights and understanding of business
performance & opportunities
☆ Chemometrics is the science of relating measurements made on a chemical system or process to the state of the system via application of mathematical or statistical methods.
☆ Demography is the statistical study of all populations. It can be a very general science that can be applied to any kind of dynamic population, that is, one that changes over time or
☆ Econometrics is a branch of economics that applies statistical methods to the empirical study of economic theories and relationships.
☆ Environmental statistics is the application of statistical methods to environmental science. Weather, climate, air and water quality are included, as are studies of plant and animal
☆ Epidemiology is the study of factors affecting the health and illness of populations, and serves as the foundation and logic of interventions made in the interest of public health and
preventive medicine.
☆ Geostatistics is a branch of geography that deals with the analysis of data from disciplines such as petroleum geology, hydrogeology, hydrology, meteorology, oceanography, geochemistry,
☆ Operations research (or Operational Research) is an interdisciplinary branch of applied mathematics and formal science that uses methods such as mathematical modeling, statistics, and
algorithms to arrive at optimal or near optimal solutions to complex problems.
☆ Population ecology is a sub-field of ecology that deals with the dynamics of species populations and how these populations interact with the environment.
☆ Psychometric is the theory and technique of educational and psychological measurement of knowledge, abilities, attitudes, and personality traits.
☆ Quality control reviews the factors involved in manufacturing and production; it can make use of statistical sampling of product items to aid decisions in process control or in accepting
☆ Quantitative psychology is the science of statistically explaining and changing mental processes and behaviors in humans.
☆ Reliability Engineering is the study of the ability of a system or component to perform its required functions under stated conditions for a specified period of time
☆ Statistical finance, an area of econophysics, is an empirical attempt to shift finance from its normative roots to a positivist framework using exemplars from statistical physics with an
emphasis on emergent or collective properties of financial markets.
☆ Statistical mechanics is the application of probability theory, which includes mathematical tools for dealing with large populations, to the field of mechanics, which is concerned with
the motion of particles or objects when subjected to a force.
☆ Statistical physics is one of the fundamental theories of physics, and uses methods of probability theory in solving physical problems.
☆ Statistical thermodynamics is the study of the microscopic behaviors of thermodynamic systems using probability theory and provides a molecular level interpretation of thermodynamic
quantities such as work, heat, free energy, and entropy.
• http://conferences.nib.si/AS2009/ Applied Statistics 2009 International Conference/
□ Papers from diverse areas of statistics and methodology are appreciated:
☆ Biostatistics
☆ Bioinformatics
☆ Data Collection
☆ Data Mining
☆ Design of experiments
☆ Econometrics
☆ Mathematical Statistics
☆ Measurement
☆ Modeling and Simulation
☆ Network Analysis
☆ Sampling Techniques
☆ Social Science Methodology
☆ Statistical Applications
☆ Statistical Education
☆ Other Areas of Statistics
□ Besides new or improved statistical methods, cross-discipline and applied paper submissions are especially welcome.
• http://www.wiley.com/bw/journal.asp?ref=0035-9254 Journal of the Royal Statistical Society: Series C (Applied Statistics)
□ QUOTE: The Journal of the Royal Statistical Society, Series C (Applied Statistics) is a journal of international repute for statisticians both inside and outside the academic world. The
journal is concerned with papers which deal with novel solutions to real life statistical problems by adapting or developing methodology, or by demonstrating the proper application of new or
existing statistical methods to them. At their heart therefore the papers in the journal are motivated by examples and statistical data of all kinds. The subject-matter covers the whole range
of inter-disciplinary fields, e.g. applications in agriculture, genetics, industry, medicine and the physical sciences, and papers on design issues (e.g. in relation to experiments, surveys
or observational studies).
A deep understanding of statistical methodology is not necessary to appreciate the content. Although papers describing developments in statistical computing driven by practical examples are
within its scope, the journal is not concerned with simply numerical illustrations or simulation studies. The emphasis of Series C is on case-studies of statistical analyses in practice.
• http://www.tandf.co.uk/journals/titles/02664763.asp Journal of Applied Statistics
□ Subjects: Business Mathematics; Mathematical Statistics; Medical Statistics; Statistical Theory & Methods; Statistics; Statistics for Social Sciences; Statistics for the Biological Sciences;
□ Journal of Applied Statistics provides a forum for communication between both applied statisticians and users of applied statistical techniques across a wide range of disciplines. These areas
include business, computing, economics, ecology, education, management, medicine, operational research and sociology, but papers from other areas are also considered. The editorial policy is
to publish rigorous but clear and accessible papers on applied techniques. Purely theoretical papers are avoided but those on theoretical developments which clearly demonstrate significant
applied potential are welcomed. Each paper is submitted to at least two independent referees. Each issue aims for a balance of methodological innovation, thorough evaluation of existing
techniques, case studies, speculative articles, book reviews and letters. | {"url":"https://www.gabormelli.com/RKB/Applied_Statistics","timestamp":"2024-11-13T21:44:01Z","content_type":"text/html","content_length":"56226","record_id":"<urn:uuid:46778a79-5e9e-4160-a676-5e16576ca64b>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00830.warc.gz"} |
What is the relationship between
What is the relationship between balanced forces and motion?
When the motion of an object changes, the forces are unbalanced. Balanced forces are equal in size and opposite in direction. When forces are balanced, there is no change in motion.
What is the relationship between unbalanced forces and second law of motion?
Newton’s Second Law of Motion is concerned with the effect that unbalanced forces have on motion. An unbalanced force acting on an object causes it to accelerate. The more mass the object has, the
more inclined it is to resist any change to its motion. For example, if you apply the same unbalanced force to a mass of.
What is the relationship between net force and Newton’s first law?
Newton’s first law says that if the net force on an object is zero ( Σ F = 0 \Sigma F=0 ΣF=0\Sigma, F, equals, 0), then that object will have zero acceleration. That doesn’t necessarily mean the
object is at rest, but it means that the velocity is constant.
Does Newton’s first law deal with balanced forces?
This means that for an inertial reference frame, Newton’s first law is valid. Equilibrium is achieved when the forces on a system are balanced. A net force of zero means that an object is either at
rest or moving with constant velocity; that is, it is not accelerating.
What is balanced force give example?
When two forces are the same strength but act in opposite direction, they are called balanced forces. Again, tug-of-war is a perfect example. If the people on each side of the rope are pulling with
the same strength, but in the opposite direction, the forces are balanced. The result is no motion.
What kind of relationship is Newton’s second law?
Newton’s second law says that the acceleration and net external force are directly proportional, and there is an inversely proportional relationship between acceleration and mass. Also, force and
acceleration are in the same direction.
What are some examples of the second law of motion?
Newton’s Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is
the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.
What is inertia depend on?
Inertia is that quantity which depends solely upon mass. The more mass, the more inertia. Momentum is another quantity in Physics which depends on both mass and speed.
What happens if there is no inertia?
If there is no inertia the body will continuously keep moving or if it is at rest it will continue to be at rest. Inertia is the measure of mass. If there is no inertia then there will be no mass.
The particles will have infinite acceleration.
Newton’s laws of motion imply the relationship between an object’s motion and the forces acting on it. In the first law, we come to understand that an object will not change its motion unless a force
acts on it. The second law states that the force on an object is equal to its mass times its acceleration. And, finally, the third law states that
When does an object in motion have a balanced force?
Once the box hits the water, the forces are balanced (50 N down and 50 N up). However, an object in motion (such as the box) will continue in motion at the same speed and in the same direction.
How are unbalanced forces different from balanced forces?
Balanced and Unbalanced Forces. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
How does Newton’s first law affect the motion of a system?
Again looking at Figure 1 (a), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion
of a system, according to Newton’s first law. (The internal forces actually cancel, as we shall see in the next section.) | {"url":"https://witty-question.com/what-is-the-relationship-between-balanced-forces-and-motion/","timestamp":"2024-11-11T08:30:28Z","content_type":"text/html","content_length":"71088","record_id":"<urn:uuid:1577bf7d-bf34-4a4f-ba3e-22d4bc4bfe78>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00615.warc.gz"} |
The zen of gradient descent
Ben Recht spoke about optimization a few days ago at the Simons Institute. His talk was a highly entertaining tour de force through about a semester of convex optimization. You should go watch it.
It’s easy to spend a semester of convex optimization on various guises of gradient descent alone. Simply pick one of the following variants and work through the specifics of the analysis: conjugate,
accelerated, projected, conditional, mirrored, stochastic, coordinate, online. This is to name a few. You may also choose various pairs of attributes such as “accelerated coordinate” descent. Many
triples are also valid such as “online stochastic mirror” descent. An expert unlike me would know exactly which triples are admissible. You get extra credit when you use “subgradient” instead of
“gradient”. This is really only the beginning of optimization and it might already seem confusing.
Thankfully, Ben kept things simple. There are indeed simple common patterns underlying many (if not all) variants of gradient descent. Ben did a fantastic job focusing on the basic template without
getting bogged down in the details. He also made a high-level point that I strongly agree with. Much research in optimization focuses on convergence rates. That is, how many update steps do we need
to minimize the function up to an epsilon error? Often fairly subtle differences in convergence rates are what motivates one particular variant of gradient descent over another. But there are
properties of the algorithm that can affect the running time more powerfully than the exact convergence rate. A prime example is robustness. basic gradient descent is robust to noise in several
important ways. Accelerated gradient descent is much more brittle. Showing that it is even polynomial time (and under what assumptions) is a rather non-trivial exercise depending on the machine
model. I’ve been saying for a while now that small improvements in running time don’t trump major losses in robustness. The situation in optimization is an important place where the trade-off between
robustness and efficiency deserves attention. Generally speaking, the question “which algorithm is better” is rarely answered by looking at a single proxy such as “convergence rate”.
With that said, let me discuss gradient descent first. Then I will try to motivate why it makes sense to expect an accelerated method and how one might have discovered it. My exposition is not
particularly close to Ben’s lecture. In particular, mistakes are mine. So, you should still go and watch that lecture. If you already know gradient descent, you can skip/skim the first section.
The basic gradient descent method
The goal is to minimize a convex function \(f\colon\mathbb{R}^n\rightarrow\mathbb{R}\) without any constraints. We'll assume that \(f\) is twice differentiable and strongly convex. This means that we
can squeeze a parabola between the tangent plane at \(x\) given by the gradient and the function itself. Formally, for some \(\ell>0\) and all \(x,z\in\mathbb{R}^n:\)
\[ f(z) \ge f(x) + \nabla f(x)^T(z-x) + \frac \ell 2 \|z-x\|^2.\]
At the same time, we don’t want the function to be “too convex”. So, we’ll require the condition, often called smoothness:
\[ f(z) \le f(x) + \nabla f(x)^T(z-x) + \frac L 2 \|z-x\|^2.\]
This is a Lipschitz condition on the gradient map in disguise as it is equivalent to:
\[\|\nabla f(x) - \nabla f(z)\|\le L\|x-z\|.\]
Let's be a bit more concrete and consider from here on the important example of a convex function \(f(x) = \frac 12 x^T A x - b^T x,\) where \(A\) is an \(n\times n\) positive definite matrix and \(b
\) is a vector. We have \(\nabla f(x) = Ax - b.\) It's an exercise to check that the above conditions boil down to the spectral condition: \(\ell I \preceq A \preceq LI.\) Clearly this problem has a
unique minimizer given by \(x^*=A^{-1}b.\) In other words, if we can minimize this function, we'll know how to solve linear systems.
Now, all that gradient descent does is to compute the sequence of points
\[x_{k+1} = x_k - t_k \nabla f(x_k)\]
for some choice of the step parameter \(t_k.\) Our hope is that for some positive \(\alpha < 1\),
\[\|x^* - x_{k+1}\| \le \alpha\|x_k-x^*\|,\]
If this happens in every step, gradient descent converges exponentially fast towards the optimum. This is soberly called linear convergence in optimization. Since the function is smooth, this also
guarantees convergence in objective value.
Choosing the right step size \(t\) is an important task. If we choose it to small, our progress will be unnecessarily slow. If we choose it too large, we will overshoot. A calculation shows that if
we put \(t= 2/(\ell + L)\) we get \(\alpha = (L-\ell)/(L+\ell).\) Remember that \(\kappa = L/\ell\) is condition number of the matrix. More generally, you could define the condition number of \(f\)
in this way. We have shown that
\[\|x_k-x^*\| \le \left(\frac{\kappa -1}{\kappa + 1}\right)^k\|x_0-x^*\|.\]
So the potential function (or Lyapunov function) drops by a factor of roughly \(1-1/\kappa\) in every step. This is the convergence rate of gradient descent.
Deriving acceleration through Chebyshev magic
What Nesterov showed in 1983 is that we can improve the convergence rate of gradient descent without using anything more than gradient information at various points of the domain. This is usually
when people say something confusing about physics. It's probably helpful to others, but physics metaphors are not my thing. Let me try a different approach. Let's think about why what we were doing
above wasn't optimal. Consider the simple example \(f(x)= \frac12 x^T A x- b^T x.\) Recall, the function is minimized at \(A^{-1}b\) and the gradient satisfies \(\nabla f(x) = Ax-b.\) Let's start
gradient descent at \(x_0=tb.\) We can then check that
\[x_k = \Big(\sum_{j=0}^k (I-A’)^k\Big)b’\]
where \(A'=tA\) and \(b'=tb.\) Why does this converge to \(A^{-1}b\)? The reason is that what gradient descent is computing is a degree \(k\) polynomial approximation of the inverse function. To see
this, recall that for all scalars \(|x|<1,\)
\[\frac{1}{x} = \sum_{k=0}^\infty (1-x)^k.\]
Since the eigenvalues of \(A'\) lie within \((0,1),\) this scalar function extends to the matrix case. Moreover, the approximation error when truncating the series at degree \(k\) is \(O( (1-x)^k)).
\) In the matrix case this translates to error \(O(\| (I-A')^k \|) = O( (1-\ell/L)^k).\) This is exactly the convergence rate of gradient descent that we determined earlier.
Why did we go through this exercise? The reason is that now we see that to improve on gradient descent it suffices to find a better low-degree approximation to the scalar function \(1/x.\) What we'll
be able to show is that we can save a square root in the degree while achieving the same error! Anybody familiar with polynomial approximation should have one guess when hearing "quadratic savings in
the degree":
Chebyshev polynomials
Let's be clear. Our goal is to find a degree \(k\) polynomial \(q_k(A)\) which minimizes the residual
\[r_k = \|(I-Aq_k(A))b\|.\]
Put differently we are looking for a polynomial of the form \(p_k(z)=1-zq(z).\) What we want is that the polynomial is as small as possible on the location of the eigenvalues of \(A\) which lie in
the interval \([\ell,L].\) At the same time, the polynomial must satisfy \(p_k(0)=1.\) This is exactly the property that Chebyshev polynomials achieve with the least possible degree! Quantitatively,
we have the following lemma that I learned from Rocco Servedio. As Rocco said in that context:
There's only one bullet in the gun. It's called the Chebyshev polynomial.
Lemma. There is a polynomial \(p_k\) of degree \(O(\sqrt{(L/\ell)\log(1/\epsilon)})\) such that \(p_k(0)=1\) and \(p_k(x)\le\epsilon\) for all \(x\in[\ell,L].\)
The lemma implies that we get a quadratic savings in degree. Since we can build \(p_k\) from gradient information alone, we now know how to improve the convergence rate of gradient descent. It gets
better. The Chebyshev polynomials satisfy a simple recursive definition that defines the \(k\)-th degree polynomial in terms of the previous two polynomials. This means that accelerated gradient
descent only needs the previous two gradients with suitable coefficients:
\[x_k = x_{k-1} - \alpha_k \nabla f(x_{k-1}) + \beta_k \nabla f(x_{k-2}).\]
Figuring out the best possible coefficients \(\alpha_k,\beta_k\) leads to the above convergence rate. What's amazing is that this trick works for any convex function satisfying our assumptions and
not just the special case we dealt with here! In fact, this is what Nesterov showed. I should say that the interpretation in terms of polynomial approximations is lost (as far as I know).The
polynomial approximation method I described was known much earlier in the context of eigenvalue computations. This is another fascinating connection I'll describe in the next section.
Let me add that it can be shown that this convergence rate is optimal for any first-order (gradient only method) by taking \(A\) to be the Laplacian of a path of length \(n\). This is true even in
our special case. It's optimal though in a weak sense: There is a function and a starting point such that the method needs this many steps. I would be interesting to understand how robust this lower
bound is.
The connection to eigenvalue methods
Our discussion above was essentially about eigenvalue location. What does polynomial approximation have to do with eigenvalues? Recall, that the most basic way of computing the top eigenvalue of a
matrix is the power method. The power method corresponds to a very basic polynomial, namely \(p_k(x) = x^k.\) This polynomial has the effect that it maps \(1\) to \(1\) and moves every number \(|x|<1
\) closer to \(0\) at the rate \(|x|^k.\) Hence, if the top eigenvalue is \(\lambda\) and the second eigenvalue is \((1-\epsilon)\lambda,\) then we need about \(k\approx 1/\epsilon\) iterations to
approximately find \(\lambda.\) Using exactly the same Chebyshev idea, we can improve this to \(k=O(\sqrt{1/\epsilon})\) iterations! This method is often called Lanczos method. So, we have the
precise correspondence:
The power method is to Lanczos as basic gradient descent is to Accelerated gradient descent!
I find this quite amazing. In a future post I will return to the power method in greater detail in the context of noise-tolerant eigenvalue computation.
Why don’t we teach gradient descent in theory classes?
I’m embarrassed to admit that the first time I saw gradient descent in full generality was in grad school. I had seen the Perceptron algorithm in my last year as an undergraduate. At the time, I was
unaware that like so many algorithms it is just a special case of gradient descent. Looking at the typical undergraduate curriculum, it seems like we spend a whole lot of time iterating through
dozens of combinatorial algorithms for various problems. So much so that we often don’t get around to teaching something as fundamental as gradient descent. It wouldn’t take more than two lectures to
teach the contents of this blog post (or one lecture if you’re Ben Recht). Knowing gradient descent seems quite powerful. It’s not only simple and elegant. It’s also the algorithmic paradigm behind
many algorithms in machine learning, optimization and numerical computation. Teaching it to undergraduates seems like a must. I just now realize that I haven’t been an undergraduate in a while. Time
flies. So perhaps this is already happening.
More pointers
• Trefethen-Bau, “Numerical Linear Algebra”. My favorite book on the topic of classical numerical methods by far.
• Ben Recht’s lecture notes here and here, his Simons talk.
• Sebastien Bubeck’s course notes are great!
• For lack of a better reference, the lemma I stated above appears as Claim 5.4 in this paper that I may have co-authored.
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Whats Bigger a Tablespoon or Dessert Spoon? ✅ | Meal Delivery Reviews
Whats Bigger a Tablespoon or Dessert Spoon
Whats Bigger A Tablespoon Or Dessert Spoon?
A dessertspoon is a bit larger than a teaspoon and smaller than a tablespoon. It is used for eating dessert. A teaspoon is used for mixing the content of a cup or glass. A dessert spoon is equivalent
to 2 teaspoons, whereas a teaspoon’s unit of measure is worth one-third of a tablespoon.
Does 2 dessert spoons equal 1 tablespoon?
A tablespoon is 15 mls, a dessertspoon 10 and tsp 5. 2 dessert spoons is one table spoon.
Is a desert spoon bigger than a teaspoon?
A dessert spoon is a spoon designed specifically for eating dessert and sometimes used for soup or cereals. Similar in size to a soup spoon (intermediate between a teaspoon and a tablespoon) but with
an oval rather than round bowl, it typically has a capacity around twice that of a teaspoon.
What size is a dessert spoon?
On occasion, you will see the term “dessert spoon” used as a unit of measurement, because the standard capacity is two teaspoons, and two dessert spoons makes up a tablespoon. Incidentally, for those
who prefer their measurements in milliliters, the capacity of a dessert spoon is approximately 12 milliliters.
What is bigger than a tablespoon?
In every cutlery set or flatware, there are forks and knives, bigger spoons and smaller spoons. The bigger ones are called tablespoons while the smaller ones are called teaspoons.
How can I measure a tablespoon without a tablespoon?
Use a measuring cup. One tablespoon is equal to about one-sixteenth of a dry measuring cup. If the set comes with a one-eighth dry measuring cup, use half of that to approximate one tablespoon.
How big is a tbsp?
A tablespoon is a unit of measure equal to 1/16 cup, 3 teaspoons, or 1/2 fluid ounce in the USA. It is either approximately or (in some countries) exactly equal to 15 mL. “Tablespoon” may be
abbreviated as T (note: uppercase letter), tbl, tbs or tbsp.
How many desert spoons are in a table spoon?
In the UK a teaspoon is 5ml, a dessertspoon is 10ml and a tablespoon is 15ml but as Delia started developing recipes before measuring spoons were common place she always uses a kitchen spoon as a
tablespoon so in Delia’s recipes, where she says tablespoon it is the equivalent to 20mls (or 2 dessertspoons).
What are the different size spoons?
Spoons vary in length (11″, 13″, 15″, 18″, 21″) for ease of use in cooking or serving. Spoons can have plastic handles that are heat-resistant. Level scoops, ladles, and portion servers provide more
accurate portion control than serving spoons that are not volume-standardized measure.
Is it dessert spoon and tablespoon same thing?
One level dessertspoon (Also known as dessert Spoon or abbreviated as dstspn) is equal to two teaspoons (tsp), 10 milliliters (mLs). A US tablespoon (tbls) is three teaspoons (15mL). In Britain and
Australia, for dry ingredients, a 2 rounded or heaped teaspoonful is often specified instead.
Is a dessert spoon the same as a tablespoon?
A dessertspoon is a bit larger than a teaspoon and smaller than a tablespoon. It is used for eating dessert. A teaspoon is used for mixing the content of a cup or glass. A dessert spoon is equivalent
to 2 teaspoons, whereas a teaspoon’s unit of measure is worth one-third of a tablespoon.
What are the 3 different types of spoons?
Based on the purpose of use we can be divided into 3 main types of spoons: spoons for eating and spoons for cooking, in addition, there will be some types of spoons used in other special things that
we will learn together today.
Is dessert spoon same as tablespoon?
A dessertspoon is a bit larger than a teaspoon and smaller than a tablespoon. It is used for eating dessert. A teaspoon is used for mixing the content of a cup or glass. A dessert spoon is equivalent
to 2 teaspoons, whereas a teaspoon’s unit of measure is worth one-third of a tablespoon.
What can I use to measure 1 tablespoon?
Use a measuring cup. One tablespoon is equal to about one-sixteenth of a dry measuring cup. If the set comes with a one-eighth dry measuring cup, use half of that to approximate one tablespoon.
Is a regular spoon a tablespoon?
A typical large dinner spoon measures about one tablespoon. This is not common, but some consider tablespoons to be used in regular soup and cereal bowls.
How large is a tablespoon?
Traditional definitions. In nutrition labeling in the US and the UK, a tablespoon is defined as 15 ml (0.51 US fl oz). A metric tablespoon is exactly equal to 15 ml (0.51 US fl oz).
What are the different sizes of spoons?
Spoons vary in length (11″, 13″, 15″, 18″, 21″) for ease of use in cooking or serving. Spoons can have plastic handles that are heat-resistant. Level scoops, ladles, and portion servers provide more
accurate portion control than serving spoons that are not volume-standardized measure.
What is the use of dessert spoon?
A dessert spoon is a spoon which is midway between the size of a teaspoon and a tablespoon. You use it to eat desserts. Place the dessert spoon near the dessert bowls. Bring out the dessert bowls and
dessert spoons after the entrée.
Is a regular spoon 1 tablespoon?
A typical large dinner spoon measures about one tablespoon. This is not common, but some consider tablespoons to be used in regular soup and cereal bowls.
How can I measure 1 tbsp without measuring spoon?
If you don’t have a measuring spoon, you can use measurement equivalents. For example, 3 level teaspoons makes 1 tablespoon. 1/16 of a cup is also equivalent to 1 tablespoon, and 15 ml of any liquid
is equal to 1 tablespoon.
What can I use if I don’t have a tablespoon?
If you are missing a tablespoon, simply measure out three level teaspoons instead. Measure 1/16 of a cup. A tablespoon is equivalent to 1/16 of a cup, which will allow you to easily measure out that
amount without a measuring spoon. | {"url":"https://ageekoutside.com/whats-bigger-a-tablespoon-or-dessert-spoon/","timestamp":"2024-11-03T07:27:41Z","content_type":"text/html","content_length":"54137","record_id":"<urn:uuid:9ea03d1b-745a-4fa3-add1-e11deeae8575>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00480.warc.gz"} |
Planet sides separation by
Planet sides separation by “day” and “night”
Each planet in “from space” view has two sides: “day” – illuminated by Sun, and the opposite – “night”, with no Sun lighting.
Earth sides separation by “day” and “night”
While modeling a scene with planet, it is not difficult to achieve this effect by lighting setup. Leave only one lighting source in scene and place it in the right place. However, the result is not
so good as real space photographs. To achieve a greater similarity, the proper texture with a darkened face and bright lights – night lights of big cities, mast be mapped to the “night” side of
modelling planet.
Prepare two textures: “day” and “night” – for each side of our modelling planet:
Earth “day” surface map (image from nasa.gov)
Earth “night” surface map (image from nasa.gov)
Create Earth:
1. Create a sphere
1. shift+a – Mesh – UV Sphere
2. In the T-bar
1. Transform
1. set Shading to Smooth
3. Add modifier Subdivision Surface
Create a light source – the “Sun”:
2. Create plane and refer its flat side to sphere-Earth:
1. shift+a – Mesh – Plane
2. Move and scale it a bit
1. g – x – 5 – enter
1. Light plane coordinates after moving are X = 5, Y = 0, Z = 0. They will be needed to us in the future, in customization nodes tree.
2. s – 2 – enter
3. Bind it:
1. Select plane
2. Select shpere with pressed shift
3. ctrl+t – Track To Constrait
4. Rotate plane always looking to sphere by the side
1. Select plane
2. Enter Edit mode (tab)
3. r – y – 90 – enter
4. Off Edit mode (tab)
5. Create light source material for plane
1. In Node Editor
1. Create new material for the plane
2. Replace Diffuse node with Emission node
1. Select Diffuse node
2. shift+s – Shaider – Emission
1. Strength = 5
Setting light
Correctly map prepared texture to sphere-planet. Make a separate node branches for the “day” and “night” patterns:
Setting “day” and “night” textures
It remains to properly combine created node branches by the terminator – line that separates the illuminated side of the planet from the darken side, to use “day” texture on the illuminated side of
the Earth, and “night” – on the darken.
Unfortunately Blender has no input node parameter, to get luminance value at current shaider point. However, to solve this problem, we can use the fact that at any point of the spherical surface of
our planet normal vector is known. Lets remember a bit of vector algebra and take a look at sphere cut. We may notice that the angle between the sphere surface normal vectors (blue) and the vectors
of incoming to the sphere surface light (yellow) evenly changes from 0 to 90 degrees just like the sphere surface illunination changes.
Normals and lighting vectors scheme
Lets admit the angle of 0 degrees for full illumination. Then any increasing the value of this angle corresponds to decreasing in surface luminance at current point. It remains to get this angle
value and use it as mapping factor of “day” and “night” Earth textures.
To get the desired angle we have normal vectors of the planet surface, but there is no vector of the incoming to the surface light. However, knowing the position of the light source, we can get the
desired vector. There are two vectors: the first – from the point of origin to the location point of the light source (red) and the second – from the point of origin to the lication point of the
sphere-planet (green). The required vector (yellow) will be the result of subtraction of these two vectors.
Getting lighting vector
Transfer the resulting vector in nodes system:
3. In the Node Editor in Earth’s material create three value nodes – to define the light source position.
1. shift + a – Input – Value
2. shift + d – move
3. shift + d – move
4. Combine this three nodes to Vector
1. shift+a – Converter – CombineXYZ
2. Value nodes outputs connect to the node CombineXYZ input
5. Set light source X-Y-Z coordinates to the Value nodes (values from st. 2.2.1.1.).
Setting lighting position in nodes tree
4. Add Object Info node – to get our sphere-planet location
1. shift+a – Input – Object Info
To get incoming light vector it is necessary to subtract location vector of the planet from the location vector of the “Sun”.
5. Add vector operations node:
1. shift+a – Converter – Vector Math
1. Choose subtraction: Subtract
2. Connect Vector output (CombineXYZ) to the upper Vector input (Vector Math)
3. Connect Location output (Object Info) with the lower Vector input (Vector Math)
6. Normalize the resulting vector – make its length equal to 1.
1. Add Vector Math node
1. shift+a – Converter – Vector Math
1. Choose a normalization operation: Normalize
2. The upper Vector input (Vector Math) connect with Vector output (Subtract)
3. Lower input Vector values (Vector Math) set to 1,1,1
Getting lighting vector in nodes tree
The result is a normalized incoming light vector.
Now we need to get the angle between this vector and the normal vectors of the sphere surface. The dot product of two vectors – value that characterizes the ratio of the lengths of two vectors and
angle between them. The normal vectors of sphere surface are initially normalized and we normalized calculated incoming light vector manually so the lengths of all the vectors are equal to 1, and the
ratio of their lengths is equal to 1 too. The scalar product of vectors in this case depends only on the angle between vectors and is equal to the cosine of this angle. Cosine values varies from 0 to
1, which is more easy to node system than the angular values.
Make the scalar product of the desired angle in node tree:
7. Add node, which will give us the value of the normal vector at the current shaider point:
1. shift+a – Input – Geometry
8. Add vector operations node:
1. shift+a – Converter – Vector Math
2. Set the operation of the scalar product (Dot Product)
3. Connect the upper Vector input (Dot Product) with Vector output (Normalize)
4. Connect the lower Vector input (Dot Product) wit Normal output (Geometry)
Getting angle between normal vector and lighting vector
The result is a value representing the angle between the planet’s surface normals and incoming to the planet’s surface light. Use this value to mix two prepared textures of the day and the night.
9. Add Mix Shader node
1. shift+a – Shaider – Mix Shaider
1. Connect Shaider output (Mix Shaide) with Surface input (Material Output)
2. Connect the upper Shaider input (Mix Shaider) with BSDF output (Diffuse) node with the “daytime” texture
3. Connect the lower Shaider input (Mix Shaider) with BSDF output (Diffuse) node with the “night” texture
4. Connect Factor input (Mix Shaider) with Value output (Dot Product)
Now the “daily” texture is shown on the illuminated side of the planet , and “night” texture – on its dark side. However, the angle between planet surface normals and incoming light, depending on
which we put a texture mixture factor, changes uniformly from the brightest point of the surface to the darkest, so textures start to mix almost immediately, the terminator of the planet is too
blurred, and the most part of the “day” side of the planet overlaps with “night” texture. Make simple correction and add Color Ramp node to our node tree:
10. Add Color Ramp node
1. shift+a – Converter – Color Ramp
2. insert it into connection between the Vector output (Dot Product) and Factor input (Mix Shaider)
3. Move right (white) Color Ramp node slider to the left, to correct terminator borders.
Now there is clear “day” texture on the illuminated side of our planet, “night” on the dark side, and throug the terminator we can see a rapid transition from one state to another.
Finished node tree looks like this:
Finished nodes tree
Configure final scene and send it to render.
While configurating your scene, remember – if you move the lighting source, also enter new coordinates of the light source to the Earth node tree to separate the “day” and the “night” correctly.
Textures of the Earth’s day surface and Earth’s night surface taken from nasa.gov site for educational purposes only.
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Analogous Estimating | Definition, Examples, Pros & Cons - Project-Management.info
Analogous Estimating | Definition, Examples, Pros & Cons
Analogous estimating is a top-down estimation technique for estimating the cost, resources and durations of projects (according to PMBOK®, 6^th edition, ch. 6.4.2, 7.2.2, 9.2.2). While it is less
accurate than other methods, it can be used to produce an order of magnitude or an initial estimate. Therefore, it is a common technique during the selection or initiation of projects.
In this article, we will give an overview of this technique, its definition, typical uses as well as examples of analogous estimating in project management.
What Is Analogous Estimating?
Analogous estimating is an estimation technique is also referred to as top-down estimating. It involves leveraging the estimators’ experience or historical data from previous projects by adopting
observed cost, duration or resource needs to a current project or portions of a project. Analogous estimating does not require data manipulation or statistical adjustments.
This technique is useful if you need to produce estimates without having plenty of information available. This may be the case during project selection or initiation phases, when overseeing a bunch
of projects at the portfolio-level (source: PMI Practice Standard for Project Estimating), or in the early stages of a project. Estimations can relate to a whole project or parts of a project, such
as work packages or activities.
The PMI project management framework lists analogous estimating under the techniques of the processes estimate costs, estimate activity durations and estimate activity resources (PMBOK® Guide, 6^th
edition, ch. 6.4.2, 7.2.2, 9.2.2).
Analogous estimating is typically used to get 4 types of estimates:
• a single-point or absolute value estimate,
• a ratio estimate,
• an estimate range, and
• a three-point estimate (often defined as a subcategory of range estimates).
What Is an Absolute or One-Point Estimate?
This term refers to an estimation result that consists of a single absolute value. For instance, if the cost of a previous project used to be $100,000 and it is estimated that a new, similar project
requires a similar budget, the analogous estimate would be $100,000, an absolute value.
Determination of a one-point estimate using analogous estimating based on historical data (click to enlarge the chart).
What Is a Ratio Estimate?
A ratio estimate describes the relative application of historical data or experience to a current project. One form is estimating by applying a factor to observed historical values. Estimators might
expect, for instance, that the current project will require 125% of the time of the previous project.
Another use is the estimation of a breakdown or parts of the full project cost. Based on historical data, a company may conclude that the expenses for user acceptance tests typically amount to 25% of
the total cost of an IT project, for instance.
This approach assumes a linear relationship between different aspects of a project. It is not dissimilar from a basic implementation of a parametric estimation. Yet it tends to be rather expert
judgment-based and lack the statistical evidence.
What Is an Estimate Range?
A range estimate comprises of a range of possible values, rather than a single number. It is however often accompanied by a most likely estimate. A common form of range estimates is the three-point
estimation (which is sometimes referred to as a type of estimate on its own).
Chart showing a range estimate determined through analogous estimating based on historical data (click to enlarge the image).
What Is a Three-Point Estimate?
Three-point estimating requires the project manager or the team to come up with three different estimates:
• optimistic estimate,
• pessimistic estimate, and
• most likely estimate.
These values are then transformed into a final estimate using the triangular or the PERT distribution. Read our detailed guide for more details.
What Is the Difference between Parametric and Analogous Estimating?
Parametric estimating uses historical data in a different way than analogous estimating. It requires calculations and adjustments to account for the characteristics of the current project.
This is typically done using a statistical approach. It involves identifying parameters in historical data that correlate with cost, duration or resource-related values of a project. Inserting the
parameters of the current project will then lead to the estimates for the current endeavor. The implementation of this technique varies greatly among organizations – it generally ranges from a simple
‘rule of three’ calculation to complex statistical models and algorithms.
Analogous estimating, on the other hand, does not usually involve an adjustment of data. It also does not require statistical evidence of assumed relationships. Instead, it relies more on expert
All in all, parametric estimating tends to produce more accurate results thanks to its statistical and data-driven approach. Analogous estimating, on the other hand, requires fewer data and fewer
resources and can therefore be used when only a little information is known.
How Do You Apply Analogous Estimating?
Analogous estimating usually involves the following steps:
1. Creating a list of similar projects (you can start with a longlist first and refine it later).
2. Getting the historical cost, durations and/or resource requirements and additional details on the characteristics of past projects, e.g. scope, activities, complexity, environmental factors, etc.
3. Refining the longlist by removing previous projects that are not deemed relevant anymore. The result is a shortlist of projects, that are similar to the current one.
4. Deciding which type of estimates is needed, based on the stakeholders’ requirements, the availability of data and the confidence of the estimators. Refer to the first section of this article for
the definitions of the different result types of analogous estimating.
5. Select or calculate the estimate from the historical data.
Read this step-by-step instruction for more details and guidance.
Advantages and Disadvantages of Analogous Estimating
Like any other approach to estimating cost, schedule or resources, the analogous estimation technique comes with a number of advantages and disadvantages.
• Analogous estimating typically does not require a lot of resources or time.
• This type of estimating can be performed with very limited available data.
• It is therefore ideal in the project initiation phase and for activities for which not much information and historical data are available.
• These estimates can be ideal for high-level assessments and strategic considerations, as the accuracy is often sufficient for working on the ‘big picture’. It can then be used in program
management or for early stakeholder communications, for instance.
• An analogous estimate is often an initial estimate for a project or parts of a project at a time when not much information is available. It will then be refined over time (similar to the concept
of progressive elaboration).
• Estimates tend to be rough and they are often not very accurate.
• The underlying assumption is that historical data or experience of the estimators would be applicable to the current project. If it turns out that this assumption was incorrect, the estimate will
be useless.
• In practice, top-down estimates can sometimes be driven by political considerations or even pressure rather than based on the project-specific characteristics or the expertise of the subject
matter experts.
• The high level and the potential inaccuracy of analogous estimates put certain limitations on their use for decision-making or project planning and controlling.
What Are the Typical Uses of Analogous Estimating in Project Management?
Analogous estimates are used by project managers of almost all kinds of projects. Its application is often dependent on the project’s phase and the availability of data rather than the subject matter
of a project or activity.
Within the lifecycle of a project, analogous estimates are particularly common in the project selection or initiation phases (source). However, the assumptions for the cost-benefit analysis of change
processes or less significant parts of a project could also be estimated using this method.
This technique can also be applied to any level of granularity within the work breakdown structure. However, it is particularly common to estimate entire projects or larger portions of a project in
an early planning stage.
You may consider applying the analogous estimation technique in cases where
• resources for estimating are limited,
• not many details about the project (and/or comparable projects) are known, or
• a rough estimate fits for the purpose.
In practice, project managers tend to rely on analogous estimating mainly in situations where only limited resources or little input information are available. Producing rough results that are ‘good
enough’ for a phase or a part of a project is another typical use case.
Another example is program management (source: PMI Practice Standard for Project Estimating): For the high-level management of a portfolio of projects, rough estimates are often sufficient to produce
data as a basis for strategic decisions.
When projects have been running for some time, such estimates are usually refined or replaced based on more accurate types of estimates (e.g. bottom-up or parametric estimates).
Read on for an example of how the different types of analogous estimating can be used in practice.
Example of Analogous Estimating in Projects
In this example, the different types of the analogous estimation technique are applied to the following situation:
An IT vendor is asked by a prospective customer to estimate the implementation cost of off-the-shelf software. The vendor has done similar types of jobs a couple of times before and stored the key
indicators of past projects in a dedicated database. The database shows the following data for a long list of comparable projects:
Historical project data Cost (in $1,000) Duration (in days)
Project A 100 40
Project B 200 70
Project C 80 50
Project D 160 50
Project E 120 60
In order to determine an analogous estimate, the estimators compare the characteristics of the upcoming project with those of the 6 previous projects for which they obtained the historical cost and
duration values.
One-Point Estimate
The team applies some expert judgment and concludes that the characteristics of the current project are similar to Project E. There is actually an overlap in terms of scope, complexity and
availability of resources. Subsequently, they are using the observed cost and duration of that project as their analogous estimate (E):
E_cost = $120,000
E_duration = 60 days
Range of Estimates and three-point estimate
If the estimators are not able to find an exact match in their historical data, they tend to prefer estimating a range instead of a single value.
In this case, they exclude project C as they consider it an outlier in terms of scope (narrower than the current scope) and cost (low). Their estimates are:
E_cost_min = $100,000
E_cost_max = $200,000
E_duration_min = 40 days
E_duration_min = 70 days
The estimators communicate this range. However, as this range is quite broad, the stakeholders ask the estimators to come up with a ‘most likely’ estimate. The team then uses the above-mentioned
one-point estimate (based on identical considerations) as the ‘most likely’ estimate:
E_cost_mostlikely = $120,000
E_duration_mostlikely = 60
For a three-point estimate that can be further processed in a triangular or PERT distribution, the team uses the minimum and maximum estimates as the optimistic and pessimistic points, respectively.
Ratio Estimate
The estimators could also determine a ratio estimate. For instance, if they expected the current project to incur 30% higher cost and take 20% more time than Project A, their estimates would be:
E_cost = $100,000 x 1.3 = $130,000
E_duration = 40 days x 1.2 = 48 days
For some projects, a breakdown into different parts of a project may be required. According to the team’s experience, project efforts and time are usually distributed as follows:
• Project management: 10%
• Installation: 10%
• Customization: 50%
• Documentation: 10%
• Testing and quality assurance: 20%
Applied to their previous estimate, their numbers are as follows:
Typical share Cost estimate Time estimate
Total estimate – 120 48
Project management 10% 12 4.8
Installation 10% 12 4.8
Customization 50% 60 24
Documentation 10% 12 4.8
Testing and quality assurance 20% 24 9.6
Uses of these Analogous Estimates
Depending on the confidence of the estimators, these estimates may or may not be deemed good enough to quote a price for their customer’s project.
In any case, the team will probably share the range (cost of 100 to 200 and a duration of 40 to 70 days) as an order of magnitude. At the same time, they might want to request further details from
their prospective client, such as a list of requirements or areas that need to be customized. This feedback can then be used to produce more accurate estimates.
For their internal communication, e.g. with their account management team, the estimators may want to use the ‘most likely’ or the ratio estimates. These numbers can help obtain a decision of whether
the vendor is willing and able to pursue this opportunity. They may also rely on the breakdown (ratio estimate) to check the availability of resources, e.g. whether the customizing experts required
for this project would be available for 24 days.
If analogous estimating is used in internal projects of an organization, the ratio estimate can also be useful to calculate resource requirements in the early stages of a project and to determine to
which extent departments or entities are affected by a project.
In any case, these analogous estimates are still quite rough, so an organization will likely want to replace them with more accurate figures over the course of a project.
Although the analogous or top-down estimation technique is often rough and high-level, it is very relevant in practice. Where not much information is available (yet) or an order of magnitude is
needed rather than a definitive estimate, analogous estimating can be the method of choice.
Nevertheless, if you need more accurate numbers, you should consider using other estimation techniques. You will find an overview of the different methods in this article on cost estimation and this
article on duration estimation. | {"url":"https://project-management.info/analogous-estimating/","timestamp":"2024-11-08T04:50:25Z","content_type":"text/html","content_length":"113213","record_id":"<urn:uuid:0236d7b1-423b-4b9c-b8fe-88e0f46c5619>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00004.warc.gz"} |
Chapter-Wise MCQ Questions for Class 8 Maths Quizzes with Answers
MCQ Questions for CBSE Class 8 Maths with Answers
Are you searching for CBSE Class 8 Maths MCQ Quiz Questions with Answers to practice? Then, you have come to the right spot. Quiz Questions is the best way to secure max marks in the final
examinations & it also improves your subject knowledge. So, practicing from Class 8 Maths Multiple Type Quiz Questions with Answers will let you know your level of preparation along with strengths &
weak points. Grab chapter-wise Practice Tests from here & secure more in CBSE MCQ Questions for Class 8 Maths.
Chapter-wise CBSE Maths MCQ Practice Test Questions for Class 8 with Answers
We at MCQMojo.com assists students with effective preparation resources for tough Class 8 Maths MCQ Questions. Find here a list of chapter-wise practice test CBSE Board MCQ Quiz Questions for Class 8
Advantages of Taking MCQ Quiz Test for CBSE Class 8 Maths Exam
Practice Tests help students in preparing these MCQ Questions in an effective manner & make them get familiar with the MCQ section in Board Exam Paper. Well, here are some of the main advantages of
attempting Objective Type Questions Practice tests for Class 8 Maths below:
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Frequently Asked Questions on MCQ Quizzes for CBSE Class 8 Maths
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Seven Core TI-Nspire Applications - dummies
The TI-Nspire handheld device has seven applications from which to choose — Calculator application, Graphs application, Geometry application, Lists & Spreadsheets application, Data & Statistics
application, Notes application, and Vernier DataQuest.
• Calculator application: In this application, you perform calculations. You also enter and view expressions, equations, and formulas, all of which are displayed in a format similar to what you see
in a textbook. A variety of built-in templates is also available to give you the power to represent just about any mathematical concept symbolically.
• Graphs application: In this application, you graph equations, expressions, and a variety of functions. Variables and sliders allow you to investigate the effect of certain parameters dynamically.
Analyze the graph to find critical points and the values of local extrema.
• Geometry application: In this application, you can explore synthetic geometry concepts, that is, geometry not associated with the coordinate plane. Also, the Geometry application allows you to
integrate coordinate geometry and synthetic geometry. Watch as connections between these two areas are made dynamically, in real time.
• Lists & Spreadsheet application: In this application, you investigate numeric data, some of which is captured from the Graphs application and some of which resides entirely within the Lists &
Spreadsheet application. Like a computer spreadsheet program, this application allows you to label columns, insert formulas into cells, and perform a wide range of statistical analyses.
• Data & Statistics application: Used in conjunction with Lists & Spreadsheet, this application allows you to visualize one-variable and two-variable data sets. Data & Statistics allows you to
create a variety of statistical graphs, including scatter plots, histograms, box-and-whisker plots, dot plots, regression equations, and normal distributions. You can also manipulate a data set
(either numerically or graphically) and watch the corresponding change in the other representation.
• Notes application: The Notes application enables you to put math into writing. Three templates make the Notes application a robust and integral part of any TI-Nspire document. With the Notes
application, you can pose questions, review or write geometric proofs, and provide directions for an activity. Interactive math boxes link to all the other applications.
• Vernier DataQuest application: This application can be used along with probes (like the CBR2 motion detector) to collect real data. There are three views available within the DataQuest
application that allow for multiple representations of the data. You can even discard the parts of the data that you do not want to include.
About This Article
This article is from the book:
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How to calculate how much stock you want to buy
To calculate first the percentage of your funds you want to spend on a given stock, take your amount multiplied by the percent (We have $90000 and want to spend 20% of that, so we type in 90000 x 0.2
= ). We now know we want to spend not more than $18000.
Now subtract the commission for the stock purchase ($9.99). Then divide the remaining number (17990) by the ask price of the stock ($32.77), like this: 17990 / 32.77 = . Round the number you get down
to a whole number (remove the numbers after the decimal). Now you know you want to buy 549 stocks at the price of $32.77 each, which won’t cost more than $17,990.
You can check your math by multiplying the number of stocks (549) by the price (32.77). 549 x 32.77 = 17990. | {"url":"https://coolstuffinterestingstuffnews.com/how-to-calculate-how-much-stock-you-want-to-buy/","timestamp":"2024-11-11T01:53:03Z","content_type":"text/html","content_length":"45147","record_id":"<urn:uuid:16d6c768-0e74-4fde-a6cc-4d90e83f9f2c>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00128.warc.gz"} |
matrix in computer science pdf
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Let R is relation on set of real numbers such that aRb=a<b then... | Filo
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a. reflexive
b. symmetric
c. transitive
d. none
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Question Text Let is relation on set of real numbers such that then is
Updated On Nov 14, 2022
Topic Algebra
Subject Mathematics
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Transforming Trajectories | FTCLib Docs
Trajectories can be transformed from one coordinate system to another and moved within a coordinate system using the relativeTo and the transformBy methods. These methods are useful for moving
trajectories within space, or redefining an already existing trajectory in another frame of reference.
Neither of these methods changes the shape of the original trajectory.
The relativeTo Method
The relativeTo method is used to redefine an already existing trajectory in another frame of reference. This method takes one argument: a pose, (via a Pose2d object) that is defined with respect to
the current coordinate system, that represents the origin of the new coordinate system.
For example, a trajectory defined in coordinate system A can be redefined in coordinate system B, whose origin is at (2, 2, 30 degrees) in coordinate system A, using the relativeTo method.
Pose2d bOrigin = new Pose2d(2, 2, Rotation2d.fromDegrees(30));
Trajectory bTrajectory = aTrajectory.relativeTo(bOrigin);
In the diagram above, the original trajectory (aTrajectory in the code above) has been defined in coordinate system A, represented by the black axes. The red axes, located at (2, 2) and 30° with
respect to the original coordinate system, represent coordinate system B. Calling relativeTo on aTrajectory will redefine all poses in the trajectory to be relative to coordinate system B (red axes).
The transformBy Method
The transformBy method can be used to move (i.e. translate and rotate) a trajectory within a coordinate system. This method takes one argument: a transform (via a Transform2d object) that maps the
current initial position of the trajectory to a desired initial position of the same trajectory.
For example, one may want to transform a trajectory that begins at (2, 2, 30 degrees) to make it begin at (4, 4, 50 degrees) using the transformBy method.
Transform2d transform = new Pose2d(4, 4, Rotation2d.fromDegrees(50)).minus(trajectory.getInitialPose());
Trajectory newTrajectory = trajectory.transformBy(transform);
In the diagram above, the original trajectory, which starts at (2, 2) and at 30° is visible in blue. After applying the transform above, the resultant trajectory’s starting location is changed to (4,
4) at 50°. The resultant trajectory is visible in orange. | {"url":"https://docs.ftclib.org/ftclib/v1.2.0/pathing/trajectory/transforming-trajectories?fallback=true","timestamp":"2024-11-06T12:28:38Z","content_type":"text/html","content_length":"246725","record_id":"<urn:uuid:555c8b1d-4d6d-4140-b326-25d8ff896958>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00641.warc.gz"} |
Post shut-in arrest and recession solutions for a deflating hydraulic fracture in a permeable elastic medium
Published: 31 May 2022| Version 1 | DOI: 10.17632/4nxghz7cmr.1
Anthony Peirce
This data set provides the numerical solutions generated using an Implicit Moving Mesh Algorithm (IMMA) that has been adapted to include the k-g, r-g multiscale and k and r-vertex asymptotes to model
the post-shut-in arrest and recession of a radial hydraulic fracture in a permeable elastic medium. The theory behind this work is described in the paper "The arrest and recession dynamics of a
deflating radial hydraulic fracture in a permeable elastic medium" published in the JMPS (https://doi.org/10.1016/j.jmps.2022.104926). For the simulations I set the following parameters to unity: Ep
= 1 mup = 1 Cp = 1 Q0 = 1 Since V0 = Q0*ts, from eq (39) in the paper you can vary omega = ts/tmm~ = ts (since all the material parameters in t_mm~ are 1) and choose the value of phi^V that you want
by setting Kp as follows: Kp = (TS'*(phi).^(-65/9)).^(1/26); Recall: tmmt = (mup^4*Q0^6/Cp^18/Ep^4)^(1/7); omega = Ts/tmmt; phiV = (Ep^21*mup^5*Cp^10*Q0*Ts/Kp^26)^(9/65); Issue the command load
(Extract_Radial_MKC1_Ts_1em3_phi_50); to access the structures Results, which contains the structure Input. The file names embed the two dimensionless parameters: omega=Ts/tmm~=Ts (since tmm~=1) and
phiV. To avoid decimal points in the file name I have multiplied the phiV value by 100. So the above data file is for the case omega=10^{-3} and phiV=0.5. Input = struct('Ep',Ep,... % Pa plane strain
modulus 'mup',mup,... % Pa*s, alternate fluid viscosity 'Cp',Cp,... % m/s^1/2, alternate Carter's leak-off coefficient 'Kp',Kp,... % Pa*m^1/2, alternate fracture toughness 'Q0',Q0,... % m/s,
injection rate 'Ts',Ts,... % shut-in time 'omega',omega,... % dimensionless shut-in time 'phiV',phiV,... % arrest regime parameter 'Nr',Nr,... % number of grid points in r direction 'Nt',itcol); %
number of time steps till collapse Results=struct('pt',P(1,1:itcol),... % wellbore pressure versus t 'Rt',R(1:itcol),... % fracture radius versus t 'wt',W(1,1:itcol),... % wellbore aperture versus t
'eta',eta(1:itcol),... % efficiency versus t 'pr',P(:,1:itcol),... % fluid pressure versus r at all times Nt 'wr',W(:,1:itcol),... % fracture width versus r at all times Nt (because of the moving
mesh to plot in real space use plot(rho*R(it),wr(:,it))) 'rho',rho,... % lateral spatial coordinate 't',time(1:itcol),... % time 'keyindx',[its ita itd itcol],... % key indices keyindx(1)=its
(shut-in index), keyindx(2)=ita (arrest), keyindx(3)=itr (recession), keyindx(4)=itc (collapse) 'Input',Input); % Input Structure
Steps to reproduce
See the following article in the JMPS for a detailed description: https://doi.org/10.1016/j.jmps.2022.104926
The University of British Columbia
Hydraulic Fracturing
Related Links | {"url":"https://data.mendeley.com/datasets/4nxghz7cmr/1","timestamp":"2024-11-02T12:26:33Z","content_type":"text/html","content_length":"114013","record_id":"<urn:uuid:230027c9-69b6-43a0-8841-50479f51d221>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00391.warc.gz"} |
wrapping M5-branes on a Riemann surface
4158 views
AdS/CFT gives us a way to use geometry to study field theory! I am trying to wrap M5-branes on a Riemann surface $\Sigma_{g}$. In my problem, for a Riemann surface in 11d, the normal bundle is max
$SO(5)$. Here is my question: How do we put $SO(2)$ in $SO(5)$?
Urs Schreiber suggests the following mathematically precise interpretation of the question, which probably addresses the concerns of those who commented or who had voted to put the OP on hold:
There is a famous construction of (N=2)-supersymmetric 4-dimensional Yang-Mills field theories and their Seiberg-Witten theory from the N=(2,0)-superconformal 6-dimensional field theory on the
worldvolume of M5-branes: by Kaluza-Klein-compactifying the latter on a Riemann surface. This construction was revived more recently in 2009 by the influential article
On page 22 of this article, around the displayed formula (2.27), the author mentions that the Kaluza-Klein compactification of the 6d theory on a Riemann surface involves a “well known twisting
procedure” of the holonomy of the Riemann surface by choosing an SO(2)-subgroup of the SO(5) group that is the “R-symmetry” group of the 6-dimensional supersymmetric field theory (the group under
which its supercharges transform).
Question: What is this “well known twisting procedure” exactly, and how does it work? Of course I know how to find $SO(2)$-subgroups of $SO(5)$, but what does such a choice amount to in the
context of the construction of an N=2, D=4 SYM from the 6d-field theory on the 5-brane? Where is this twisting procedure discussed in the literature?
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user Irina
Most voted comments show all comments
In my limited experience, I tried to get guided by mathematical naturalness and elegance. All answers were more generally helpful. Thank you for heads up)
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user Irina
Urs Schreiber, Thanks very much, I am glad for the references.
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user Irina
@UrsSchreiber, this is from an American television comedy called "Green Acres," pre-internet. One farmhand is cutting bread with a saw. Another says, "Don't go that way, you're cutting against the
grain." The cosmopolitan New Yorker says "There's no grain in bread!" One of the farmhands says "Wanna bet?" So, branes in mathematics and all.
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user Will Jagy
The twist can be summarized as follows: the (2,0) theory has 5 scalar fields (which can be seen as arising from the 5 transverse directions to an M5 brane), acted on by the R-symmetry group SO(5). In
order to define the theory on a general Riemann surface (times flat spacetime), you need to specify how these fields should transform: three will continue to be scalars, but two transform as a
section of the cotangent bundle to the Riemann surface. This is why you need an embedding of SO(2) into SO(5) (just rotating the first two variables).
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user David Ben-Zvi
For references try Witten's long paper on knot homology - the section on fivebranes discusses a twist of the (2,0) theory which is holomorphic in two dimensions and topological in four. Or try the
fundamental series of papers of Gaiotto-Moore-Neitzke, this construction is essential to their work (and they're my source).
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user David Ben-Zvi
Most recent comments show all comments
@Urs Schreiber, for my part at least, the question was put on hold because no one understands exactly what the OP wants to know about embeddings of SO(2) into SO(5). Branes are neither here nor
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user HJRW
Now enroute nevertheless, thanks to the efficiency and generosity of your answers, and remarkably, I learnt a lot. I am beginning to feel like I live in a Seinfeld episode. Though it is a bit
embarrassing to have units appearing so prominently in the answers. If the questions fails to be relevant here, I have an excuse.
This post imported from StackExchange MathOverflow at 2014-12-24 12:03 (UTC), posted by SE-user Irina
Now that the question is open again (now in my paraphrasing), maybe I'll repost my reply from the nForum with some brief comments thrown in:
That the 6-dimensional (2,0)-superconformal QFT on the worldvolume of the M5-brane yields N=2 D=4 super Yang-Mills theory under Kaluza-Klein compactification on a 2d Riemann surface was known since
about the mid 90s. Edward Witten had famously advertized this in the Proceedings to Graeme Segal's 60th birthday conference that by this construction the remaining invariance under Moebius
transformations of that compactification manifold geometrically explains the "Montonen-Olive"/"electric-magnetic" S-duality invariance of (super) Yang-Mills theory.
Later he realized further compactification of this down to 2-dimensions as a geometric realization of geometric Langlands duality. In the course of this the N=2 D=4 super Yang-Mills theory is
"topologically twisted" in a way analogous to the well-known twisting of N=4 SYM that goes back to the work that won him the Fields medal. The twisting of the N=2 theory then also showed up in the
more recent work by Gaiotto et al. that lead to the AGT correspondence.
While the details for the topological twisting of the N=2 supersymmetric field theory are a tad more involved than those of the N=4 theory, the basic idea is the same: one picks an embedding of the
spacetime rotation symmetry into the R-symmetry group (the one under which the supercharges transform) and then asks for a linear combination $Q$ of the supercharges that is held fixed by the
resulting external+internal symmetry transformations. The cohomology of this $Q$ is then seen to pick inside the quantum observables of the origional super gauge field theory those of a topological
field theory. That is the topologically twisted theory.
Pointers to more details on this topological twisting that the above questing is after are collected here:
Pointers specifically concerning the twistied Kaluza-Klein compactification of the M5-brane on a Riemann surface to a topologically twisted N=2 super Yang-Mills theory are here:
This post imported from StackExchange MathOverflow at 2014-12-24 12:04 (UTC), posted by SE-user Urs Schreiber | {"url":"https://www.physicsoverflow.org/25571/wrapping-m5-branes-on-a-riemann-surface","timestamp":"2024-11-07T21:46:30Z","content_type":"text/html","content_length":"148912","record_id":"<urn:uuid:cef5a199-c4ba-4622-8349-33982798ca60>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00124.warc.gz"} |
How to Make Math Accessible Online - KnowledgeOneHow to Make Math Accessible Online - KnowledgeOne
Have you ever struggled to display algebra or complex math online? Do you want to design your online STEM course to make it more accessible to people who are blind? Are you curious how to get popular
screen readers like NVDA or JAWS to read equations properly?
If so, then this post may be for you.
STEM courses are increasingly being offered online as a cost-effective way to educate large numbers of students (Chirikov et al., 2020; Xu & Xu, 2019). However, online environments impose unique
constraints on our ability to design these environments for the needs of our learners.
The Social Model of Disability & The Failures of Modern Media
Gravel et al. argued in 2015 that it is “our learning environments, first and foremost, that are disabled” (as cited in Nieminen & Pesonen, 2020, p.5). To better understand the accessibility of the
internet, I reviewed Jung et al. to unearth some hard facts about the media landscape today.
They report an astounding finding: “All 30 visualizations we sampled from major news outlets did not contain alt text associated with the visualizations…Similarly, the IEEE VIS & TVCG collection did
not include any text alternatives to the figures” (2022, p. 1098).
The researchers cite the Washington Post, New York Times, and fivethirtyeight.com as exemplars of modern media. The findings are even more alarming once you realize that the IEEE Visualization
Conference brings together industry and academic experts in the field of visualization, and the IEEE Transactions on Visualization and Computer Graphics (TVCG) publishes peer-reviewed articles on
this subject bimonthly.
Given this snapshot, it’s safe to say that neither academia nor major media are rising to the challenge of making data, visualizations, or even images accessible to the blind and visually impaired
Accessibility Online
Accessibility online is often done using major screen readers. What here is the problem exactly? Simply put, the internet was not designed to display mathematical notation in the way that we commonly
interact with it in the classroom (usually on a chalkboard or a whiteboard, which has zero spatial restrictions).
If we consider PDFs, one of the major ways of publishing literature, the software doesn’t display mathematical equations correctly because it lacks the notational flexibility to divide the same blank
page into the correct set of layers and symbolic relationships.
Thankfully, Microsoft Word’s Equation Editor closes this gap substantially. Professors familiar with mathematical notation will find it easy to write equations correctly. But NVDA, the most commonly
used screen reader that is also free, struggles to work well with it.
You might imagine that Microsoft Word’s built-in screen reader, called ReadAloud, would read equations written in the editor just fine. However, this is not the case, and strange bugs are
The other major competitor to NVDA is JAWS, but it is not free. While universities might retain licenses for JAWS to service the needs of students with disabilities, this is not true for all
institutions. Since the software is expensive, there is a substantive financial barrier for learners.
This encouraged us to design a solution that worked well with NVDA and JAWS.
Enter LaTeX
LaTeX (pronounced ‘lay-tech’) is the art of converting mathematical notation into code and vice versa. It is designed for the production of technical and scientific documentation, and it’s free!
So how does it work? Suppose we imagine a math professor has written an equation using Word’s Equation Editor. In that case, we can convert it into a script by selecting the option to read the
equation as ‘linear’ instead of ‘professional.’
The output can be read in any HTML document so long as the LaTeX is bookended with characters to identify the LaTeX to the web browser. This is ( and ), respectively.
If you have worked with Moodle or any other content management system, there’s often a text editor equipped with the function to write in HTML, aka the ‘source code.’
Then, when you publish your web doc, the LaTeX gets converted back into the original equation. This is how you make math accessible online.
There are two major benefits to this.
1. The primary benefit is that blind or visually impaired learners can read the equations with the correct level of specificity via their screen reader of choice. Since the math has been hard-coded,
fewer bugs, confusing expressions, and errant terms occur.
2. It’s worth mentioning that sighted users benefit from this accessibility as equations are now interactive and can be enlarged.
To understand how wrong this can go, Microsoft’s screen reader, ReadAloud, may switch languages part-way when reading equations written using Microsoft’s own Equation Editor. This is a jarring
experience and renders the equation illegible. The omission of a single word can change the _______ of the whole sentence.
The drawback is that LaTeX is a language and as such, requires a learning experience designer or programmer to familiarize themselves with its terms and syntax in order to resolve equations that
don’t complete the conversion into LaTeX script successfully. Given that a basic understanding of the math is also required, unfamiliarity with the scripting language could result in broken equations
and interminable bugs.
The Importance of Alt Text
The other important piece to an accessible math course is the importance of alt text, which is the invisible ink of the internet. Chart data, graphs, spreadsheets, pie charts – data visualizations
can be central to a message but without any alt text describing the trend, data, or message, the learner relying on a screen reader is effectively shut out of the conversation.
How can we improve this practice? By using a good standard provided by Jung et al. (2022): For a detailed walkthrough, please consult the video included as part of this post.
1. Describe the chart type (i.e. line, pie, graph, etc.)
2. Provide a brief description.
3. Prime the reader for the detailed description (i.e. ‘Detailed Description:’
4. Describe the axes and the general trend(s)
5. Use plain language and,
6. Avoid double quotation marks (“) as this ends the alt text. Substitute single quotes (‘) instead.
By following these guidelines, you too can be better than the Washington Post, the New York Times, and FiveThirtyEight. I believe in you!
Out of curiosity, I wanted to verify Jung et al.’s findings since they published their paper two years ago.
A review of the first images I could discover on the websites of the Washington Post, 538.com, and the New York Times revealed that only the New York Times had since made the images for their stories
accessible. There’s no explanation for the Washington Post however 538 has a more difficult task at hand since the amount of data contained in any one of their visualizations is more than you could
listen to in an afternoon.
For researchers and practitioners interested in further research on this topic, this is a good place to start.
As we learned, the internet was not designed with people who are blind in mind. Educators are encouraged to craft their online learning spaces to be as open and accessible as possible to all
For STEM courses, we learned that Microsoft’s Equation Editor is quite helpful for producing somewhat legible equations for screen readers. It works quite well with JAWS, but for NVDA, the commonly
used free screen reader, you will have to convert the equations into LaTeX before you publish them in a web document (e.g., HTML).
If you are brave and want to learn how to code, LaTeX is free, and you can certainly pick it up. It will reduce the bugs you encounter when publishing accessible equations because it gives you more
control over how screen readers will interpret the math.
Despite these advances in accessibility, there are still two key issues that make higher-level STEM courses difficult to produce. For one, you may struggle to conduct an expert review. While it is
possible for you to produce accessible math online, the Venn diagram of experts in a particular STEM subject who are also blind or visually impaired is vanishingly small.
It’s a bit of a Catch-22. We lack expert review for online STEM courses, especially for accessibility, because those types of courses have historically not accommodated the needs of blind learners.
The second issue involves the linguistic nature of math. When a professor reads out an equation, certain relationships or nuances in the notation are omitted since it is assumed that the learner can
read and verify the equation for themselves at the same time.
A classic example of this is the molecular formula for water, H[2]O. Generally, we do not say out loud that the 2 is a subscript of the H as that might be perceived as pedantic. However, for learners
with visual impairments, complex equations with numbers that describe both the number of atoms as well as the general composition of the molecule, the output from the screen reader might be
Simply put, as an equation’s complexity increases, the subtle omissions required to describe it with brevity can distort its overall meaning.
More work is needed to help standardize how STEM professors read out equations. This will help educators and learning experience designers craft more accessible online learning environments. However,
this is easier said than done.
Complex equations can take quite a long time to read out, and this is especially true if you are trying to explicate all of the relationships involved. This may lead to cognitive overload as the
beginning of the equation fades from short-term memory before it can be converted into meaning by the learner.
If you’re doing substitutions, it can be confusing when you consider transcripts, particularly if you’re trying to transcribe something very verbose, which can become almost unintelligible if the
equation is long enough.
This is to say that there are very real constraints on the professor’s ability to produce legible content and on learners’ capacity to sit through and digest that content.
I hope this article and the attached video will give you a lot to think about when designing an online learning environment that is accessible to people who are blind. While this is a niche topic, it
is important to push accessibility forward to reduce ostracization and improve our potential for stimulating rich minds.
For privacy reasons YouTube needs your permission to be loaded. For more details, please see our
Privacy Policy
Chirikov, I., Semenova, T., Maloshnok, N., Bettinger, E. & Kizilcec, R. F. (2020). Online education platforms scale college STEM instruction with equivalent learning outcomes at lower cost. Science
Advances, 6(15). DOI: 10.1126/sciadv.aay5324.
LaTeX Project. (2023). LaTeX – A document preparation system. https://www.latex-project.org/.
Nieminen, J. H. & Pesonen, H. V. (2020). Taking universal design back to its roots: Perspectives on accessibility and identity in undergraduate mathematics. Education Sciences, 10(1), 12. DOI: https:
Xu, D. & Xu, Y. (2019). The promises and limits of online higher education: Understanding how distance education affects access, cost, and quality. American Enterprise Institute (AEI). https://
Learning Experience Designer @KnowledgeOne. Poet. Musician. Producer. Community Organizer. Web Developer. | {"url":"https://knowledgeone.ca/how-to-make-math-accessible-online/","timestamp":"2024-11-07T05:35:05Z","content_type":"text/html","content_length":"99645","record_id":"<urn:uuid:ac849b83-70aa-4194-acf8-6c103322821b>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00617.warc.gz"} |
Video - Towards Trustworthy Scientific Machine Learning: Theory, Algorithms, and Applications
Machine learning (ML) has achieved unprecedented empirical success in diverse applications. It now has been applied to solve scientific problems, which has become an emerging field, Scientific
Machine Learning (SciML). Many ML techniques, however, are very complex and sophisticated, commonly requiring many trial-and-error and tricks. These result in a lack of robustness and
interpretability, which are critical factors for scientific applications. This talk centers around mathematical approaches for SciML, promoting trustworthiness. The first part is about how to embed
physics into neural networks (NNs). I will present a general framework for designing NNs that obey the first and second laws of thermodynamics. The framework not only provides flexible ways of
leveraging available physics information but also results in expressive NN architectures. The second part is about the training of NNs, one of the biggest challenges in ML. I will present an
efficient training method for NNs - Active Neuron Least Squares (ANLS). ANLS is developed from the insight gained from the analysis of gradient descent training. | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=video&page=2&listStyle=gallery&sort_index=Dept&order_type=asc&document_srl=996744","timestamp":"2024-11-10T14:13:42Z","content_type":"text/html","content_length":"52726","record_id":"<urn:uuid:71cebba1-31cb-48b5-8d3d-719c901b1632>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00116.warc.gz"} |
A “better” class of puzzle
Due to the last few weeks being the only time of the year when I actually do some real work, as opposed to pretending (hope my employers don’t read this blog!), I haven’t done many puzzles in the
last 2-3 weeks.
At first, I thought the lack of practice was going to handicap me with this puzzle, as I got down to 16ac without putting in an answer. No need to worry though, as the top half came very quickly
The bottom half didn’t take long either, and with the help of a fair smattering of general knowledge, I ended up solving the puzzle in about 8 minutes. Considering the grid was still blank after
about 90 seconds, I am actually quite proud of that time.
This was an excellent puzzle – I always enjoy crosswords that test your knowledge as much as your ability to work out wordplay. I also noticed a “themelet” around gambling
1 ZOOM-LENS – Lens being a city in France, of course, as any football fan would instantly know.
9 UNDERMILK-WOOD – brilliant wordplay leading to Dylan Thomas’s famous radio play
10 BET-RAY – as an ex-bookie, I am disappointed I didn’t get this clue quicker. A Yankee is a type of bet where you make four selections, and have a combination of 11 bets, guaranteeing a return if
you have at least two winners
11 O(<=tar)O-RIO – I guessed that Samson was the name of an oratorio, and upon checking, discovered that Handel wrote it.
13 STATE-CR.-A-F.T.
15 DIGS – good one
18 GO FOR-BROKER(r)
21 CONSPIRE – (prison)* in CE
22 PI-RATE – as in Captain Hook
23 STRAIGHT FLUSH
25 SE(X)T ON
2 OPULENT – (up l note)*
3 MODERN TIMES – fantastic Chaplin movie
4 (p)EARLY
5 (<=nips)-OZ.-A – Baruch Spinoza, a Dutch 17th century philosopher
7 (d)R(a)I(n)-O
8 (g)UNDO(IN)G
12 ORDER AROUND – my employees would never have got this clue!
14 COGNITION – O in (gin tonic)*
17 LO(OK)SE-(grime)E – “butcher’s” is Cockney rhyming slang for LOOK (butcher’s hook)
20 KITCHEN – (the nick)*
16 comments on “A “better” class of puzzle”
1. 6:39 for this, profiting from spotting 9A quickly.
1. Off hand I can’t think of any other radio play famous enough to make it to the Times crossword so, like you, PB, I wrote it in early.
It was useful as I had pencilled in DAN for 7D tentatively hoping that the old city of that name (in Caanan) may have been a port, so I was able to correct that to RIO before it messed up my
thinking in the NE corner.
1. War of the Worlds?
1. Yes, maybe, but I think UMW was originally devised as a radio play whereas WotW was a book first and later adapted for radio/film.
I agree with you about STRAIGHT FLUSH.
2. 6:07 for me, although I have to admit I put a couple in without worrying about working out the wordplay. There are a lot of FRESHERs here in Cambridge this week, which may have helped. I also
have a good friend doing a PhD on Spinoza. 26A I think falls into the “Old chestnut” category. Jason J
3. Took the plunge this morning and joined the Times club, thus had the luxury of solving while at home wrapped in five duvets. OK, slight exaggeration.
20 minutes or so, which I’m very happy with.
Very best of luck to all at Cheltenham this weekend. Have fun!
4. After a mediocre week I was hoping for a confidence boost today before Sunday. I didn’t get it, after an utter disaster in the top left, only resolved by an adjacent amateur photographer who
volunteered the ‘zoom’ bit of ZOOM LENS, whereafter I finally got OPULENT, MODERN (TIMES) and BETRAY. Maybe ‘Yankee’ = BET is questionable without a ‘perhaps’ or similar, but that’s really no
5. 9:36 for me, but I was flagging after doing Wednesday’s and Thursday’s. I enjoyed this puzzle very much, but, like Neil, I’d have liked some sort of indication that “Yankee” is an example of a
BET; similarly for “Good group of clubs” (STRAIGHT FLUSH), where “perhaps” would have been easy to slip in.
6. 15-16 minutes here. I too had an unsolved grid until 16ac. From there, the bottom half flew in, but some of the top half had me scratching my head. For the second day running, it took far too
long to spot a simple anagram indication (this time at 2dn). The ZOOM bit of 1ac also took far longer than it should. Perhaps its a good job I decided not to try to qualify for Cheltenham. Good
luck to all those who did qualify. On that note, what chance is there that the regional finals will be rekindled in the future?
1. I can’t see them doing regional finals unless a generous sponsor can be found. It’s a pity, as many solvers who would be quite happy to go to Glasgow or York are unlikely to trek to
Cheltenham. Now that the link with the literary festival is pretty nominal, the one-day final could quite reasonably be held in different places each year, like the Open golf – there must be
lots of other universities with similar facilities that could be used. I may try suggesting this, though the organizers have probably had enough interfering suggestions from me …
7. Am I the only one who would have liked a ‘maybe’ or similar for the clubs? Unless I’ve misunderstood this, a straight flush can be in any suit.
8. “Yankee” and “straight flush” have been mentioned as needing a “perhaps” or some such. So also, I think, does “deal” in 9A. Yet things are OK in 15A, 22A, 8D. Are rules about all this changing?
9. Three puzzles in a week where I couldn’t get a glue, gak! Had ?e?ray from the fish, couldn’t think of anything that meant either yankee or shop, so I took a misguided stab at DEFRAY. Rest of
crossword took very little time. Oh, well… you learn something every month.
10. Another one over an hour. I, too, missed the anagram in 2d and bogged down for easily 15 minutes!
Didn’t know ‘bet’ for yankee so had to go searching for proof!
11. I, who lag fainting behind all of you most of the time, found this maybe the easiest puzzle ever. Filled most of it in nonstop, then even after “stuckness” found that the rest filled in steadily.
Too bad I never time myself.
I was mystified by “Yankee,” even though I am one, but “ray” and “shop” worked so I wrote it in anyway. “Under Milk Wood” was my first entry, and “Zoom Lens” my last — I had no idea Lens was a
French city.
12. Magnificent 7 “easies”:
6a Show surprise (4-2)
16a Mollusc left in river (4)
C L AM. Detritus from a picnic basket in a punt?
26a Local leader ignored by head of state (8)
(P) RESIDENT
6d Look like the next scene in film (4,5)
TAKE AFTER
19d More familiar person up for the first time? (7)
22d Adverts for pants (5)
24d Grass, speed not ecstasy (3)
RAT (E) | {"url":"https://timesforthetimes.co.uk/a-better-class-of-puzzle","timestamp":"2024-11-05T16:10:57Z","content_type":"text/html","content_length":"187234","record_id":"<urn:uuid:9927b3fb-a7cc-494a-a4f7-302842e085dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00699.warc.gz"} |
Create a new time geometry, which is oriented with the given plane orientation.
The function uses the currently selected time point to retrieve the current base geometry from the given time geometry. A new SlicedGeometry3D is created and initialized with the given parameters and
the extracted base geometry. This new SlicedGeometry3D is then used to replace the geometries of each time step of a cloned version of the given TimeGeometry. This allows to only replace the
contained geometries for each time step, keeping the time steps (time point + time duration) from the original time geometry.
timeGeometry The TimeGeometry which should be used for cloning.
orientation The AnatomicalPlane that specifies the new orientation of the time geometry.
top If true, create the plane at top, otherwise at bottom.
frontside Defines the side of the plane.
rotated Rotates the plane by 180 degree around its normal. s.a. PlaneGeometry::InitializeStandardPlane
The given TimeGeometry must cover the currently selected time point. If not, an exception is thrown. If the given TimeGeomety is a nullptr, a nullptr is returned immediately.
The created geometry as TimeGeometry. | {"url":"https://docs.mitk.org/nightly/namespacemitk_1_1SliceNavigationHelper.html","timestamp":"2024-11-14T14:25:25Z","content_type":"application/xhtml+xml","content_length":"18323","record_id":"<urn:uuid:b081f91e-0e07-4bd8-8be5-b660d23ba50b>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00618.warc.gz"} |
Math 5: Spring 2018
Early textbook sections
Useful links
Class Notes
Practice Tests, Tests
Test 1 Practice, Solutions
Test 1 Solutions
Test 2 Practice, Solutions
Test 2 Solutions
Test 3 Practice, Solutions
Test 3 Solutions
To study for the Final Exam, use Tests 1-3 and the practice tests. On the Final Exam, you will be asked to solve an LP using the simplex method.
Quiz Solutions
Quiz 1
Quiz 2 - In-class activity
Quiz 3
Quiz 4
Quiz 5 - In-class test review | {"url":"https://myweb.liu.edu/~dredden/OldCourses/5s18/Handouts.html","timestamp":"2024-11-06T09:35:30Z","content_type":"text/html","content_length":"2134","record_id":"<urn:uuid:0a473307-04b1-4b58-8708-61ec18c4bc0e>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00441.warc.gz"} |
Complex Number
Use the Complex Number System
Learning Objectives
By the end of this section, you will be able to:
• Evaluate the square root of a negative number
• Add and subtract complex numbers
• Multiply complex numbers
• Divide complex numbers
• Simplify powers of
Before you get started, take this readiness quiz.
1. Given the numbers ⓐ rational numbers, ⓑ irrational numbers, ⓒ real numbers.
If you missed this problem, review (Figure).
2. Multiply:
If you missed this problem, review (Figure).
3. Rationalize the denominator:
If you missed this problem, review (Figure).
Evaluate the Square Root of a Negative Number
Whenever we have a situation where we have a square root of a negative number we say there is no real number that equals that square root. For example, to simplify x so that x^2 = –1. Since all real
numbers squared are positive numbers, there is no real number that equals –1 when squared.
Mathematicians have often expanded their numbers systems as needed. They added 0 to the counting numbers to get the whole numbers. When they needed negative balances, they added negative numbers to
get the integers. When they needed the idea of parts of a whole they added fractions and got the rational numbers. Adding the irrational numbers allowed numbers like
But now we will expand the real numbers to include the square roots of negative numbers. We start by defining the imaginary unit
Imaginary Unit
The imaginary unit i is the number whose square is –1.
We will use the imaginary unit to simplify the square roots of negative numbers.
Square Root of a Negative Number
If b is a positive real number, then
We will use this definition in the next example. Be careful that it is clear that the i is not under the radical. Sometimes you will see this written as i is not under the radical. But the
Write each expression in terms of i and simplify if possible:
Write each expression in terms of i and simplify if possible:
Write each expression in terms of i and simplify if possible:
Now that we are familiar with the imaginary number i, we can expand the real numbers to include imaginary numbers. The complex number system includes the real numbers and the imaginary numbers. A
complex number is of the form a + bi, where a, b are real numbers. We call a the real part and b the imaginary part.
Complex Number
A complex number is of the form a + bi, where a and b are real numbers.
A complex number is in standard form when written as a and b are real numbers.
We summarize this here.
Real number
Imaginary number
Pure imaginary number
The standard form of a complex number is
The diagram helps us visualize the complex number system. It is made up of both the real numbers and the imaginary numbers.
Add or Subtract Complex Numbers
We are now ready to perform the operations of addition, subtraction, multiplication and division on the complex numbers—just as we did with the real numbers.
Adding and subtracting complex numbers is much like adding or subtracting like terms. We add or subtract the real parts and then add or subtract the imaginary parts. Our final result should be in
standard form.
Remember to add both the real parts and the imaginary parts in this next example.
Multiply Complex Numbers
Multiplying complex numbers is also much like multiplying expressions with coefficients and variables. There is only one special case we need to consider. We will look at that after we practice in
the next two examples.
In the next example, we multiply the binomials using the Distributive Property or FOIL.
In the next example, we could use FOIL or the Product of Binomial Squares Pattern.
Use the Product of Binomial Squares Pattern,
Multiply using the Binomial Squares pattern:
Multiply using the Binomial Squares pattern:
Since the square root of a negative number is not a real number, we cannot use the Product Property for Radicals. In order to multiply square roots of negative numbers we should first write them as
complex numbers, using
To multiply square roots of negative numbers, we first write them as complex numbers.
In the next example, each binomial has a square root of a negative number. Before multiplying, each square root of a negative number must be written as a complex number.
To multiply square roots of negative numbers, we first write them as complex numbers.
We first looked at conjugate pairs when we studied polynomials. We said that a pair of binomials that each have the same first term and the same last term, but one is a sum and one is a difference is
called a conjugate pair and is of the form
A complex conjugate pair is very similar. For a complex number of the form
Complex Conjugate Pair
A complex conjugate pair is of the form
We will multiply a complex conjugate pair in the next example.
From our study of polynomials, we know the product of conjugates is always of the form difference of squares. We can multiply a complex conjugate pair using this pattern.
The last example we used FOIL. Now we will use the Product of Conjugates Pattern.
Notice this is the same result we found in (Figure).
When we multiply complex conjugates, the product of the last terms will always have an
This leads us to the Product of Complex Conjugates Pattern:
Product of Complex Conjugates
If a and b are real numbers, then
Multiply using the Product of Complex Conjugates Pattern:
Use the Product of Complex Conjugates Pattern,
Simplify the squares.
Multiply using the Product of Complex Conjugates Pattern:
Multiply using the Product of Complex Conjugates Pattern:
Divide Complex Numbers
Dividing complex numbers is much like rationalizing a denominator. We want our result to be in standard form with no imaginary numbers in the denominator.
How to Divide Complex Numbers
We summarize the steps here.
How to divide complex numbers.
1. Write both the numerator and denominator in standard form.
2. Multiply the numerator and denominator by the complex conjugate of the denominator.
3. Simplify and write the result in standard form.
Divide, writing the answer in standard form:
Divide, writing the answer in standard form:
Divide, writing the answer in standard form:
Be careful as you find the conjugate of the denominator.
Simplify Powers of i
The powers of i. Let’s evaluate the powers of
We summarize this now.
If we continued, the pattern would keep repeating in blocks of four. We can use this pattern to help us simplify powers of i. Since i^4 = 1, we rewrite each power, i^n, as a product using i^4 to a
power and another power of i.
We rewrite it in the form q, is the quotient of n divided by 4 and the exponent, r, is the remainder from this division. For example, to simplify i^57, we divide 57 by 4 and we get 14 with a
remainder of 1. In other words,
Access these online resources for additional instruction and practice with the complex number system.
Key Concepts
• Square Root of a Negative Number
□ If b is a positive real number, then
Real number
Imaginary number
Pure imaginary number
□ A complex number is in standard form when written as a + bi, where a, b are real numbers.
• Product of Complex Conjugates
□ If a, b are real numbers, then
• How to Divide Complex Numbers
1. Write both the numerator and denominator in standard form.
2. Multiply the numerator and denominator by the complex conjugate of the denominator.
3. Simplify and write the result in standard form.
Section Exercises
Practice Makes Perfect
Evaluate the Square Root of a Negative Number
In the following exercises, write each expression in terms of i and simplify if possible.
Add or Subtract Complex Numbers In the following exercises, add or subtract.
Multiply Complex Numbers
In the following exercises, multiply.
In the following exercises, multiply using the Product of Binomial Squares Pattern.
In the following exercises, multiply.
In the following exercises, multiply using the Product of Complex Conjugates Pattern.
Divide Complex Numbers
In the following exercises, divide.
Simplify Powers of i
In the following exercises, simplify.
Writing Exercises
Explain the relationship between real numbers and complex numbers.
Aniket multiplied as follows and he got the wrong answer. What is wrong with his reasoning?
Explain how dividing complex numbers is similar to rationalizing a denominator.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ On a scale of
Chapter Review Exercises
Simplify Expressions with Roots
In the following exercises, simplify.
Estimate and Approximate Roots
In the following exercises, estimate each root between two consecutive whole numbers.
In the following exercises, approximate each root and round to two decimal places.
Simplify Variable Expressions with Roots
In the following exercises, simplify using absolute values as necessary.
Use the Product Property to Simplify Radical Expressions
In the following exercises, use the Product Property to simplify radical expressions.
In the following exercises, simplify using absolute value signs as needed.
Use the Quotient Property to Simplify Radical Expressions
In the following exercises, use the Quotient Property to simplify square roots.
Simplify expressions with
In the following exercises, write as a radical expression.
In the following exercises, write with a rational exponent.
In the following exercises, simplify.
Simplify Expressions with
In the following exercises, write with a rational exponent.
In the following exercises, simplify.
Use the Laws of Exponents to Simplify Expressions with Rational Exponents
In the following exercises, simplify.
Add and Subtract Radical Expressions
In the following exercises, simplify.
Multiply Radical Expressions
In the following exercises, simplify.
Use Polynomial Multiplication to Multiply Radical Expressions
In the following exercises, multiply.
Divide Square Roots
In the following exercises, simplify.
Rationalize a One Term Denominator
In the following exercises, rationalize the denominator.
Rationalize a Two Term Denominator
In the following exercises, simplify.
Solve Radical Equations
In the following exercises, solve.
Solve Radical Equations with Two Radicals
In the following exercises, solve.
Use Radicals in Applications
In the following exercises, solve. Round approximations to one decimal place.
Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula
Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula
Evaluate a Radical Function
In the following exercises, evaluate each function.
Find the Domain of a Radical Function
In the following exercises, find the domain of the function and write the domain in interval notation.
Graph Radical Functions
In the following exercises, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.
Evaluate the Square Root of a Negative Number
In the following exercises, write each expression in terms of i and simplify if possible.
Add or Subtract Complex Numbers
In the following exercises, add or subtract.
Multiply Complex Numbers
In the following exercises, multiply.
In the following exercises, multiply using the Product of Binomial Squares Pattern.
In the following exercises, multiply using the Product of Complex Conjugates Pattern.
Divide Complex Numbers
In the following exercises, divide.
Simplify Powers of i
In the following exercises, simplify.
Practice Test
In the following exercises, simplify using absolute values as necessary.
In the following exercises, simplify. Assume all variables are positive.
In the following exercises, solve.
In the following exercise, ⓐ find the domain of the function ⓑ graph the function ⓒ use the graph to determine the range.
complex conjugate pair
A complex conjugate pair is of the form a + bi, a – bi.
complex number
A complex number is of the form a + bi, where a and b are real numbers. We call a the real part and b the imaginary part.
complex number system
The complex number system is made up of both the real numbers and the imaginary numbers.
imaginary unit
The imaginary unit i^2 = –1 or
standard form
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Practical AI in banking
Everyone’s excited about artificial intelligence. But most people, in most jobs, struggle to see the how AI can be used in the day-to-day work they do. This post, and others to come, are all about
practical AI. We’ll dial the coolness factor down a notch, but we explore some real gains to be made with AI technology in solving business problems in different industries.
This post demonstrates a practical use of AI in banking. We’ll use machine learning, specifically neural networks, to enable on-demand portfolio valuation, stress testing, and risk metrics.
I spend a lot of time talking with bankers about AI. It’s fun, but the conversation inevitably turns to concerns around leveraging AI models, which can have some transparency issues, in a
highly-regulated and highly-scrutinized industry. It’s a valid concern. However, there are a lot of ways the technology can be used to help banks –even in regulated areas like risk –without
disrupting production models and processes.
Banks often need to compute the value of their portfolios. This could be a trading portfolio or a loan portfolio. They compute the value of the portfolio based on the current market conditions, but
also under stressed conditions or under a range of simulated market conditions. These valuations give an indication of the portfolio’s risk and can inform investment decisions. Bankers need to do
these valuations quickly on-demand or in real-time so that they have this information at the time they need to make decisions.
However, this isn’t always a fast process. Banks have a lot of instruments (trades, loans) in their portfolios and the functions used to revalue the instruments under the various market conditions
can be complex. To address this, many banks will approximate the true value with a simpler function that runs very quickly. This is often done with first- or second-order Taylor series approximation
(also called quadratic approximation or delta-gamma approximation) or via interpolation in a matrix of pre-computed values. Approximation is a great idea, but first- and second-order approximations
can be terrible substitutes of the true function, especially in stress conditions. Interpolation can suffer the same draw-back in stress.
An American put option is shown for simplicity. The put option value is non-linear with respect to the underlying asset price. Traditional approximation methods, including this common second-order
approximation, can fail to fit well, particularly when we stress asset prices.
Improving approximation with machine learning
Machine learning is technology commonly used in AI. Machine learning is what enables computers to find relationships and patterns among data. Technically, traditional first- order and second-order
approximation is a form of classical machine learning, such as linear regression. But in this post we’ll leverage more modern machine learning, like neural networks, to get a better fit with ease.
Neural networks can fit functions with remarkable accuracy. You can read about the universal approximation theorem for more about this. We won’t get into why this is true or how neural networks work,
but the motivation for this exercise is to use this extra good-fitting neural network to improve our approximation.
Each instrument type in the portfolio will get its own neural network. For example, in a trading portfolio, our American options will have their own network and interest rate swaps, their own
The fitted neural networks have a small computational footprint so they’ll run very quickly, much faster than computing the true value of the instruments. Also, we should see accuracy comparable to
having run the actual valuation methods.
The data, and lots of it
Neural networks require a lot of data to train the models well. The good thing is we have a lot of data in this case, and we can generate any data we need. We’ll train the network with values of the
instruments for many different combinations of the market factors. For example, if we just look at the American put option, we’ll need values of that put option for various levels of moneyness,
volatility, interest rate, and time to maturity.
Most banks already have their own pricing libraries to generate this data and they may already have much of it generated from risk simulations. If you don’t have a pricing library, you may work
through this example using the Quantlib open source pricing library. That’s what I’ve done here.
Now, start small so you don’t waste time generating tons of data up front. Use relatively sparse data points on each of the market factors but be sure to cover the full range of values so that the
model holds up under stress testing. If the model was only trained with interest rates of 3 -5 percent, it’s not going to do well if you stress interest rates to 10 percent. Value the instruments
under each combination of values.
Here is my input table for an American put option. It’s about 800k rows. I’ve normalized my strike price, so I can use the same model on options of varying strike prices. I’ve added moneyness in
addition to underlying.
This is the input table to the model. It contains the true option prices as well as the pricing inputs. I used around 800K observations to get coverage across a wide range of values for the various
pricing inputs. I did this so that my model will hold up well to stress testing.
The model
I use SAS Visual Data Mining and Machine Learning to fit the neural network to my pricing data. I can use either the visual interface or a programmatic interface. I’ll use SAS Studio and its
programmatic interface to fit the model. The pre-defined neural network task in SAS Studio is a great place to start.
Before running the model, I do standardize my inputs further. Neural networks do best if you’ve adjusted the inputs to a similar range. I enable hyper-parameter auto-tuning so that SAS will select
the best model parameters for me. I ask SAS to output the SAS code to run the fitted model so that I can later test and use the model.
The SAS Studio Neural Network task provides a wizard to specify the data and model hyper parameters. The task wizard generates the SAS code on the right. I’ve allowed auto-tuning so that SAS will
find the best model configuration for me.
I train the model. It only takes a few seconds. I try the model on some new test data and it looks really good. The picture below compares the neural network approximation with the true value.
The neural network (the solid red line) fits very well to the actual option prices (solid blue line). This holds up even when asset prices are far from their base values. The base value for the
underlying asset price is 1.
If your model’s done well at this point, then you can stop. If it’s not doing well, you may need to try a deeper model, or different model, or add more data. SAS offers model interpretability tools
like partial dependency to help you gauge how the model fits for different variables.
Deploying the model
If you like the way this model is approximating your trade or other financial instrument values, you can deploy the model so that it can be used to run on-demand stress tests or to speed up intra-day
risk estimations. There are many ways to do this in SAS. The neural network can be published to run in SAS, in data-base, in Hadoop, or in-stream with a single click. I can also access my model via
REST API, which gives me lots of deployment options. What I’ll do, though, is use these models in SAS High-Performance Risk (HPRisk) so that I can leverage the risk environment for stress testing and
simulation and use its nice GUI.
HPRisk lets you specify any function, or method, to value an instrument. Given the mapping of the functions to the instruments, it coordinates a massively parallel run of the portfolio valuation for
stress testing or simulation.
Remember the SAS file we generated when we trained the neural network. I can throw that code into HPRisk’s method and now HPRisk will run the neural network I just trained.
I can specify a scenario through the HPRisk UI and instantly get the results of my approximation.
I introduced this as a practical example of AI, specifically machine learning in banking, so let’s make sure we keep it practical, by considering the following:
• Only approximate instruments that need it. For example, if it's a European option, don’t approximate. The function to calculate its true price, the Black-Scholes equation, already runs really fast.
The whole point is that you’re trying to speed up the estimation.
• Keep in mind that this is still an approximation, so only use this when you’re willing to accept some inaccuracy.
• In practice, you could be training hundreds of networks depending on the types of instruments you have. You’ll want to optimize the training time of the networks by training multiple networks at
once. You can do this with SAS.
• The good news is that if you train the networks on a wide range of data, you probably won’t have to retrain often. They should be pretty resilient. This is a nice perk of the neural networks over
the second-order approximation whereby parameters need to be recomputed often.
• I’ve chosen neural networks for this example but be open to other algorithms. Note that different instruments may benefit from different algorithms. Gradient boosting and others may offer simpler,
more intuitive models, that get similar accuracy.
When it comes to AI in business, you’re most likely to succeed when you have a well-defined problem, like our stress testing that takes too long or isn’t accurate. You also need good data to work
with. This example had both, which made it a good candidate for to demonstrate practical AI.
More resources
Interested in other machine learning algorithms or AI technologies in general? Here are a few resources to keep learning.
Article: A guide to machine learning algorithms and their applications
Blog post: Which machine learning algorithm should I use?
Video: Supervised vs. Unsupervised Learning
Article: Five AI technologies that you need to know
Leave A Reply
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How Does Buoyancy Work on a Boat? - SwimPoolHub
How Does Buoyancy Work on a Boat?
Buoyancy works on a boat by creating an upward force that opposes the weight of the boat in water. This is due to the density of the boat being less than the density of the water it displaces.
As a result, the boat floats on the surface of the water rather than sinking. Buoyancy is a fascinating phenomenon that allows boats to float on water despite their weight. Boats, whether they are
large ships or small vessels, rely on the concept of buoyancy to stay afloat.
But how exactly does buoyancy work on a boat? We will explore the mechanics behind buoyancy and how it enables boats to float effortlessly on water. We will delve into the principles of density and
displacement, and understand why the density of the boat’s material and the water it displaces play a crucial role in determining its buoyant force. So, let’s dive deeper into the world of buoyancy
and unravel the mysteries of how it works on a boat.
The Archimedes Principle
Boats float on water because of the principle of buoyancy, which can be explained by the Archimedes Principle. This principle states that when an object is placed in a fluid, it experiences an upward
force equal to the weight of the fluid it displaces.
This is why boats, which are hollow and filled with air, can float. The air inside the boat is less dense than water, so it provides the buoyant force that counteracts the weight of the boat. As long
as the total density of the boat, including everything inside it, is less than the density of the water it displaces, the boat will float.
This is why even heavy ships can float on water, as long as their density is lower than that of the water. Understanding the relationship between buoyancy and weight is crucial for comprehending how
boats stay afloat.
Determining Fluid Displacement
Buoyancy is the term used to describe the upward force that counteracts the weight of an object when placed in water. This principle applies to boats as well. Determining fluid displacement is
crucial in understanding how buoyancy works on a boat.
By measuring the amount of fluid displaced, we can calculate the volume of water or other fluids that the boat displaces. Converting volume to weight is another important step. This helps us
understand how the weight of the boat is supported by the buoyant force.
The principle of flotation is at play here, as the boat will float when the weight of the boat is less than the weight of the fluid it displaces. By following these principles and understanding how
ships use buoyancy, we can grasp the concept of how boats float on water.
The Principle Of Flotation
Buoyancy is the force that allows boats to float on water. The principle of flotation, also known as Archimedes’ principle, explains how objects can either float or sink. Objects float when the
density of the total volume, including the air inside the boat, is less than the density of the water it displaces.
This buoyant force pushes the boat upward, counteracting the weight of the boat. On the other hand, if the object’s density is greater than the density of water, it sinks. To demonstrate this
principle, you can conduct a simple experiment by placing a brick in water.
Due to its high density, the brick sinks. Understanding buoyancy is crucial for ship and boat design, ensuring they stay afloat even when loaded with heavy cargo.
Buoyancy On Ships
Buoyancy on ships is a fascinating concept that plays a crucial role in their ability to float and stay afloat. How ships utilize buoyancy is closely tied to the displacement of water by the ship
itself. When a ship is placed in water, it displaces a certain volume of water equal to its own weight.
This displaced water exerts an upward force on the ship, known as buoyant force, which counteracts the downward force of the ship’s weight. As a result, the ship floats on the surface of the water.
The key to a ship’s buoyancy lies in ensuring that its overall density is less than that of the water it is floating in.
By carefully managing the weight distribution and design of a ship, naval architects and engineers are able to achieve this delicate balance and create vessels that are buoyant and seaworthy.
The Plimsoll Line
Buoyancy is the upward force that counters the weight of an object in water. This force allows boats to float. The air inside a ship is less dense than water, which contributes to its buoyancy. For a
boat to float, the average density of the ship and everything inside it must be less than the same volume of water.
If the object’s density is less than that of water, it will float. On the other hand, if the density is greater, the object will sink. The concept of buoyancy is vital in ensuring ship safety and
stability. Understanding the significance of the Plimsoll Line, which indicates the maximum safe load, is crucial for maintaining the balance of a boat and preventing it from sinking.
Density And Buoyancy
Density and buoyancy are closely intertwined concepts that explain how objects, including boats, float on water. The key to understanding buoyancy lies in the concept of density. Density refers to
the amount of matter packed into a given volume. When an object is submerged in a fluid, such as water, it displaces a volume of that fluid equal to its own volume.
If the object’s density is less than the density of the fluid, it will experience an upward force called buoyancy, which allows it to float. This is because the fluid exerts a greater pressure on the
bottom of the object than on the top, resulting in a net upward force.
On the other hand, if the object’s density is greater than the fluid’s density, it will sink. In the case of a boat, the density of the boat’s hull and other components must be carefully designed to
ensure that it is less than the density of water, allowing it to float effortlessly.
Buoyant Force And Floating
Buoyancy is a fascinating concept that plays a crucial role in the floating of boats. The buoyant force, which is exerted by a fluid on an immersed object, helps a ship stay afloat. This force pushes
the object upwards, allowing it to overcome its weight and float on the water’s surface.
The key factor in determining whether an object will float or sink is its density. If the object’s density is less than that of the fluid it is immersed in, it will float. Conversely, if the object’s
density is greater than the fluid’s density, it will sink.
Additionally, the shape and size of the object also influence the buoyant force. By understanding the principles of buoyancy, engineers, and designers can create ships that can effortlessly navigate
through water bodies.
The Role Of Weight In Floating
Buoyancy allows boats to float on water because of the relationship between weight and volume. The concept of buoyancy is governed by Archimedes’ principle, which states that an object immersed in a
fluid experiences an upward force equal to the weight of the fluid displaced.
This means that the weight of the boat is counteracted by the buoyant force exerted by the water. Even heavy ships can float as long as the weight of the ship, including its contents and the air
inside, is less dense than the water it displaces.
The buoyant force pushes the ship upwards, preventing it from sinking. If the ship’s density is greater than that of the water, it will sink. Thus, a ship’s ability to float is a result of the
balance between its weight and the buoyant force acting upon it.
The Science Behind Floating
Buoyancy is the upward force that counteracts the weight of an object in water. For a boat to float, it must displace a volume of water equal to its weight. The key factor in determining whether an
object will float is its average density compared to that of water.
If the average density of the boat, including the air inside it, is less than the density of the water, it will float. The buoyant force pushes the boat upwards, keeping it afloat. On the other hand,
if the object’s density is greater than that of water, it will sink.
Understanding the principles of buoyancy is essential in boat design and ensures safe navigation on water.
Frequently Asked Questions Of How Does Buoyancy Work On A Boat
How Does Buoyancy Make A Boat Float?
Buoyancy allows a boat to float by creating an upward force that opposes the weight of the boat. The density of the boat and everything inside it, including the air, must be less than the same volume
of water for it to float.
How Do Boats Achieve Buoyancy?
Boats achieve buoyancy because the air inside the boat is less dense than water, making it float.
How Do Boats Float On Water Density?
Boats float on water because the air inside them is less dense than water, making the total density of the boat and its contents lower than that of the same volume of water. This buoyant force keeps
the boat afloat.
How Does Ship Float In Water Although It Is Heavy?
Ships float in water because of buoyancy. The buoyant force pushes the ship upwards, allowing it to float even though it is heavy. The ship’s density must be less than that of water for it to float.
How Does Buoyancy Work On A Boat?
Boats float due to the principle of buoyancy, which states that the buoyant force exerted on an object in a fluid is equal to the weight of the fluid displaced by the object.
To understand how buoyancy works on a boat, it is important to recognize that buoyancy is an upward force that opposes the weight of an object placed in water. This force allows the boat to float
instead of sinking. The concept of buoyancy is based on Archimedes’ principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object.
When a boat is placed in water, it displaces an amount of water equal to its weight. This displacement creates an upward force that counteracts the weight of the boat, allowing it to stay afloat. The
boat floats because the average density of the boat and everything inside, including the air, is less than the density of the water.
Understanding buoyancy is crucial for boat designers and engineers as it helps them ensure the stability and safety of the boat. By applying the principles of buoyancy, they can design boats that can
support their intended load and navigate through various water conditions.
Buoyancy plays a vital role in how boats float. The upward force it creates allows boats to overcome the downward force of their weight and stay afloat. By understanding how buoyancy works, we can
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Matchsticks Number Sequence Puzzle | Best Riddles and Brain Teasers
What is the next number in the matchsticks number sequence below?
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How to use the FIND function in ExcelHow to use the FIND function in Excel
How to use the FIND function in Excel
In this article, we will learn How to use the FIND function in Excel.
Find any text in Excel
In Excel, you need to find the position of a text. Either you can count manually watching closely each cell value. Excel doesn't want you to do that and neither do we. For example finding the
position of space character to divide values wherever space exists like separating first name and last name from the full name. So let's see the FIND function syntax and an example to illustrate its
In the cell Jimmy Kimmel, Space character comes right after the y and before K. 6th is the occurrence of Space char in word starting from J.
FIND Function in Excel
FIND function just needs two arguments (third optional). Given a partial text or single character to find within a text value is done using the FIND function.
FIND function Syntax:
=FIND(find_text, within_text, [start_num])
Find_text : The character or set of characters you want to find in string.
Within_text : The text or cell reference in which you want to find the text
[start_num] : optional. The starting point of search.
Example :
All of these might be confusing to understand. Let's understand how to use the function using an example. Here Just to explain FIND function better, I have prepared this data.
In Cell D2 to D4, I want to find the location of “Hero” in cell B2, “@” in cell B3 and “a” in B4, respectively.
I write this FIND formula in Cell D2 and drag it down to cell D4.
Use the Formula in D4 cell
And now i have the location of the given text:
Find the Second, Third and Nth Occurrence of Given Characters in the Strings.
Here we have some strings in range A2:A4. In cell C2, C3, and C4 we have mentioned the characters that we want to search in the strings. In D2, D3, and D4 we have mentioned the occurrence of the
character. In the adjacent cell I want to get the position of these occurrences of the characters.
Write this formula in the cell E2 and drag down.
This returns the exact positions (19) of the mentioned occurrence (4) of the space character in the string.
How does it work?
The technique is quite simple. As we know the SUBSTITUTE function of Excel replaces the given occurrence of a text in string with the given text. We use this property.
So the formula works from inside.
SUBSTITUTE(A2,C2,"~",D2): This part resolves to SUBSTITUTE("My name is anthony gonsalvis."
," ","~",4). Which ultimately gives us the string "My name is anthony~gonsalvis."
P.S. : the fourth occurrence of space is replaced with "~". I replaced space with "~" because I am sure that this character will not appear in the string by default. You can use any character that
you are sure will not appear in the string. You can use the CHAR function to insert symbols.
Search substring in Excel
Here we have two columns. Substring in Column B and Given string in Column A.
Write the formula in the C2 cell.
Find function takes the substring from the B2 cell of Column B and it then matches it with the given string in the A2 cell of Column A.
ISNUMBER checks if the string matches, it returns True else it returns False.
Copy the formula in other cells, select the cells taking the first cell where the formula is already applied, use shortcut key Ctrl + D.
As you can see the output in column C shows True and False representing whether substring is there or not.
Here are all the observational notes using the FIND function in Excel
Notes :
FIND returns the first found location by default.
If you want a second found location of text then provide start_num and it should be greater than first location of the text.
If the given text is not found, FIND formula will return #VALUE error.
FIND is case sensitive. If you need an case insentive function, use the SEARCH function.
FIND does not support wildcard characters. Use the SEARCH function if you need to use wildcards to find text in a string.
Hope this article about How to use the FIND function in Excel is explanatory. Find more articles on searching partial text and related Excel formulas here. If you liked our blogs, share it with your
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Related Articles :
How to use the SEARCH FUNCTION : It works same as the FIND function with only one difference of finding case insensitive ( like a or A are considered the same) in Excel.
Searching a String for a Specific Substring in Excel : Find cells if cell contains given word in Excel using the FIND or SEARCH function.
Highlight cells that contain specific text : Highlight cells if cell contains given word in Excel using the formula under Conditional formatting
How to Check if a string contains one of many texts in Excel : lookup cells if cell contains from given multiple words in Excel using the FIND or SEARCH function.
Count Cells that contain specific text : Count number of cells if cell contains given text using one formula in Excel.
How to lookup cells having certain text and returns the Certain Text in Excel : find cells if cell contains certain text and returns required results using the IF function in Excel.
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How to use the IF Function in Excel : The IF statement in Excel checks the condition and returns a specific value if the condition is TRUE or returns another specific value if FALSE.
How to use the VLOOKUP Function in Excel : This is one of the most used and popular functions of excel that is used to lookup value from different ranges and sheets.
How to use the SUMIF Function in Excel : This is another dashboard essential function. This helps you sum up values on specific conditions.
How to use the COUNTIF Function in Excel : Count values with conditions using this amazing function. You don't need to filter your data to count specific values. Countif function is essential to
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A Logical Basis for Cumulative Defense Strategy and the Mathematical Analysis of Defense Strategy & Countermeasures (MADSC)
• Post by: David Wallace
• November 11, 2021
• Comments off
David Wallace, Chair, ASCE/EWRI WISE SC (Water Infrastructure Security Enhancements Standards Committee), Lloyd Foster, Actuary, Computational Mathematician, Foster Colley Enterprises
View PDF
The Water Sector has experienced an inadvertent gap in physical security by using risk models not suited for the specific needs of community water systems. Assessment results founded on the
probability of adversarial threats have referenced historical frequency, likelihood, or available intelligence which has often skewed security recommendations, causing an unintended reduction in
security countermeasures. This approach has also unintentionally promoted a lax culture of security in the face of emerging threats. Resiliency must be improved.
The increase of foreign and domestic threats mandates a serious evaluation of existing security methodologies, guidelines, and vulnerability assessments. A more specific and yet comprehensive defense
strategy with quantitative and qualitative measurements will be presented in this informational brief to show how the Water Sector can optimize security countermeasures and achieve an objective cost
to benefit ratio based on the analysis.
Design Basis Threat (DBT)
After 9/11 and the birth of the U.S. Department of Homeland Security (DHS), the fight was on to secure America’s critical infrastructure against subsequent attacks. If the U.S. could suffer an aerial
attack from adversaries using our own airplanes against us, there was no telling what was possible. In the race for solutions, Design Basis Threat (DBT) was quickly adopted from the U.S. Nuclear
Regulatory Commission’s 1979 DBT Rule as a methodology for protecting water, wastewater, and storm water utilities. It was not recognized at the time that the DBT methodology would contain criteria
for the Water Sector that were unattainable in the way that it was for the Atomic and Nuclear Energy sectors. The result has been a vast underinvestment in security countermeasures, loss of key
stakeholder trust in the accuracy of the guidelines, and an operational culture that has been disconnected from physical security priorities.
The ANSI/ASCE/EWRI 56-10, Guidelines for the Physical Security of Water Utilities, incorporated the Design Basis Threat (DBT) methodology as a guideline for securing the Water Sector, along with ASCE
(2010b) which incorporated the DBT methodology for wastewater/stormwater utilities. DBT assigns a security approach to defend against a hypothetical attack by estimating the objectives and motives of
a potential assailant according to the threat classifications of Vandal, Criminal, Saboteur, or Insider.^1,2 The persistent problem has been the near impossibility of determining who the adversary
might be, assessing their motives and objectives, and then selecting which countermeasures to use accordingly.
Risk Analysis and Management for Critical Asset Protection (RAMCAP)
The RAMCAP methodology was first introduced to Nuclear Power Plants (NPP) in 2005. In 2010 RAMCAP was published by the American Water Works Association (AWWA) with the modified name of RAM-W, adopted
from Sandia Laboratories. According to the RAM-W model, risk is defined as:
Risk = Likelihood (Specific Attack) x Vulnerability (Specific Attack) x Consequence (of the Attack)^3
RAM-W further states that identifying these threats is only possible with the use of “available intelligence.”^3 Once again, historical frequency (likelihood) and available intelligence are critical
to the success of these models.
Frequentist Probability Method and Malevolent Threats
The examination and scoring of threats, vulnerabilities, and consequences would seem reasonable, but the idea was to base threat likelihood on historical data. This worked for natural hazards, but
not for malevolent attacks against the Water Sector. A statistically valid set of data simply does not exist to make this relevant. The study of historical data for determining probability is known
as Frequentist Probability and has been defined in the Department of Homeland Security Risk Lexicon-2010 Edition and states in the annotation that:
“1) Within the frequentist probability interpretation, precise estimation of new or rarely occurring events, such as the probability of a catastrophic terrorist attack, is generally not possible.
2) Frequentist probabilities generally do not incorporate “degree of belief” information, such as certain types of intelligence information.” ^4
Recognizing that determining the likelihood of a specific attack based on available intelligence is not possible, an innovative approach must be taken to effectively characterize adversarial risk.
Significant Developments to the Risk Equation
In 1981, the document “On the Quantitative Definition of Risk,” written by Kaplan and Garrick, proposed using “triplets” to describe risk as a set of probabilities, scenarios, and consequences.^5
Kaplan and Garrick defined the risk as having three components, a scenario (si), the probability of the scenario (pi) and the consequence of the scenario (ci). With the knowledge of this criteria,
the Probability of Success of a Given Threat (PS|T) could be determined.
Here, the probability of the scenario could be applied to natural disasters or even man-made accidents, but estimating adversarial threats was still plagued since determining the probability of the
scenario was based on historical frequency, which could not be determined.
In 2010, a document called “A Risk Informed Method for Enterprise Security (RIMES)”^ 6 was introduced by Sandi National Laboratories where Wyss, et al. resurfaced the Kaplan and Garrick triplets with
a modification that leveraged the approach by replacing p[i], the probability of the scenario, with d[i], the degree of difficulty. This improved the equation in determining the probability of
success of a given threat for an identified scenario, degree of difficulty, and the consequence. With probability based on historical frequency removed, this puts the focus on the degree of
difficulty, measuring what is known instead of trying to measure what is not known.
For Defense in Depth terms, if an adversary on foot were to choose a path of jumping over a fence to initiate the compromise of a targeted asset (scenario), with the intent of a catastrophic failure
of a water treatment plant (consequence), then the security countermeasures required to be defeated would represent the degree of difficulty. If the triplets for security risk <s[i], d[i], c[i]> are
known, then they become a function of the conditional probability of success of a given threat, P[S|T].
Now the question becomes how to determine what the degree of difficulty is. This measurement cannot stem from evaluating a dataset of adversarial capability as this data does not exist. And even if
it did, there are an infinite number of variables that could be introduced into the equation that would significantly complicate the formula. The degree of difficulty must be measured across the
security countermeasures that comprise the defensive steps an adversary must cross to reach the target. Once this is determined, the degree of vulnerability directly corresponds to the probability of
success of a given threat, P[S|T].
Cumulative Defense Strategy© (CDS)
With a few small adjustments, non-offensive defense strategies can be cleaned up and made more effective. The Defense in Depth methodology uses diverse protective measures along each potential
adversarial path^7, but this can be further defined by requiring the increase of the quality and quantity of security countermeasures throughout the scale of each defense layer. We have termed this
Cumulative Defense Strategy^©. Incrementally adding quality and quantity of security countermeasures throughout the defense layer increases the required resources for an adversary’s success and
consequently addresses the variance in adversarial capability. This incremental increase establishes the Minimum Difficulty Threshold (MDT) level required at each step. The Cumulative Defense
Strategy^© method also enables a consistency of approach on which mathematical analysis can be performed to determine the probability of success of a given threat and cost/benefit analysis.
Mathematical Analysis of Defense Strategy & Countermeasures© (MADSC)
The key to unlocking the new risk definition was discovered in 2021 by Lloyd Foster and David Wallace by mathematically calculating the degree of difficulty to achieve the probability of success of a
given threat for a scenario and consequence identified. The new math model analyzes the current defensive countermeasures and determines the optimized placements mathematically and objectively for
improvements. The process of evaluation is called the Mathematical Analysis of Defense Strategy & Countermeasures^© (MADSC), and it requires the sequential increase of quantitative and qualitative
countermeasures. The MADSC analysis is performed by analyzing the defensive layers that comprise the entire defense strategy. Each defense layer addresses an attack vector. Path analysis of each
attack vector is then evaluated for the number of ordinal steps and the subset of countermeasures within each of these ordinal steps. For instance, an adversary on foot may have to breach six ordinal
steps to reach a critical asset, and each step consists of multiple countermeasures that collectively bolster each ordinal step. The steps are ordinal in nature because they must be sequentially
crossed to reach the critical asset. At each ordinal step, the quantity of countermeasures must increase to maintain a growing Minimum Difficulty Threshold (MDT), which in turn increases the quality
of each ordinal step. As referenced above, this additive nature is known as the Cumulative Defense Strategy^©.
The MADSC analysis then calculates the effective difficulty of defensive countermeasures through the coupling of two different math models, which in Latin is called a “copula.” Copulas were invented
in 1959 by Abe Sklar, and about a dozen formulas have been invented since then to solve unique needs. This latest version is called the Foster-Wallace Formula^©. The Foster-Wallace Formula^© is
unique in that it couples the Probability Density Function (PDF) of a Geometric Distribution across the ordinal steps which are required to be defeated by an adversary, with the Cumulative
Distribution Function (CDF) of a Gamma Distribution across the total defensive countermeasures within the subsets of the ordinal steps. This coupling provides the ability to generate the Joint
Probability of compromise at each ordinal step and at each defensive countermeasure, mathematically expressed as:
It is not necessary to understand how the formula works to understand the results of the MADSC methodology. The results are truly clear in revealing locations that are well protected and other areas
that need specific improvement. In either case, a clear optimization of countermeasures is provided. Joint Probability conveys what is the probability that the entirety of countermeasures could be
compromised at a given point in the breach process, specifically at a given countermeasure point within an ordinal step. It can also be used to express effective difficulty. Conversely, the
percentage remaining can be used to represent the remaining difficulty required for a complete compromise of countermeasures. Therefore, the placement for the best use of investment becomes clear for
guarding against compromise and achieving the highest levels of intruder delay to allow for adequate response time.
Unfortunately, most risk assessments currently written for the Water Sector are based on a faulty foundation of risk understanding and lead to a flawed conclusion. The risk assessment inaccuracies,
due to the insufficient risk formulas, have resulted in an underinvestment of defensive countermeasures, or a reactionary over-investment from those who are aware of the problem. The more skilled
risk teams who serve in the Water Sector have been aware of this problem for years but have struggled to find solutions without an adequate risk equation and formulated solution.
Utilizing the incremental increase of quantity and quality of security countermeasures and ordinal steps, evaluated with the Mathematical Analysis of Defense Strategy & Countermeasures^©, and
measured with the Foster-Wallace Formula^©, a robust Cumulative Defense Strategy^© can be implemented for optimal security countermeasure placement in the Water Sector. This new methodology will
yield the most effective way of screening the various capabilities of potential adversaries with increasing difficulty levels and provides fiscally sound security practices for cost-to-benefit
analysis and budgeting. The implementation of these methodologies in the Water Sector could help lead the way in physically securing other DHS critical security sectors.
1. American Society of Civil Engineers (ASCE) (2010a). Guidelines for the Physical Security of Water Utilities, ANSI/ASCE/EWRI 56-10. pp. 3-6. ASCE, Reston, VA.
2. ASCE. (2010b). Guidelines for the Physical Security of Wastewater/Stormwater Utilities, ANSI/ASCE/EWRI 57-10. pp. 67-72. ASCE, Reston, VA.
3. American Water Works Association. (2013). Risk and Resilience Management of Water and Wastewater Systems, J100-10 (R13). pp. 5, 13. AWWA, Denver, CO.
4. McNamara, P.A., and R. Beers. (2010). DHS Risk Lexicon 2010 Edition. pp. 16-17, 23-25. Department of Homeland Security (DHS). Washington, DC.
5. Kaplan, S., and J.B. Garrick. (1981). On The Quantitative Definition of Risk, Risk Analysis, vol. 1, pp. 11-27. https://doi.org/10.1111/j.1539-6924.1981.tb01350.x
6. Wyss, G.D., J.F. Clem, J.L. Darby, K. Dunphy-Guzman, J.P. Hinton, and K.W. Mitchiner. (2011). A Risk Informed Method for Enterprise Security. p. 2., Sandia National Laboratories, Albuquerque, NM.
7. Sandia National Laboratories. (2016). International Training Course on the Physical Protection of Nuclear Facilities and Materials. slide 15. Albuquerque, NM. https://shareng.sandia.gov/itc/
Authors’ Bios
David Wallace has over two decades of field experience in critical security vulnerability assessments, with deployments in over 40 states including Alaska, Hawaii, the U.S. Territory of San Juan, and
Canada. His education includes graduate work in Homeland Security at Penn State University and comparative homeland security studies in Israel through Homeland Security International. His site
deployments have included federal law enforcement, numerous military bases, a joint operations headquarters (JOC), state and local government entities, water utilities, Department of Transportation,
health care, commercial and distribution facilities, and sites of national interests. David has a strong interest in national security and advanced methodologies for securing critical infrastructure
sites to protect the American way of life. His current efforts include compiling government documentation and academic research to gain a global picture of influencing factors, trends, and threats to
help inform and safeguard key infrastructure.
Lloyd Foster has over 20 years of professional experience building and validating complex risk models and carries over 3 decades as a fully qualified actuary. His formal training and considerable
experience include stochastic calculus, advanced statistical techniques, object- oriented programming using C++, and advanced applications in Mathematica in mathematical and statistical modeling.
Lloyd also specializes in Cumulative Distribution Function, Probability Density Function, Copulas, Ordinary Differential Equations and Partial Differential Equations, with applications to security | {"url":"https://csrac.us/journal_article/a-logical-basis-for-cumulative-defense-strategy-and-the-mathematical-analysis-of-defense-strategy-countermeasures-madsc/","timestamp":"2024-11-10T08:41:34Z","content_type":"text/html","content_length":"113584","record_id":"<urn:uuid:51e88847-74c3-4380-849f-c619eccda20f>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00206.warc.gz"} |
Avoid These Mistakes While Preparing For Class 8 Maths Olympiad
International Mathematics Olympiad or IMO is conducted by SFO and students from all around the world apply for the exam. These competitive exams need a good amount of preparation and practice to
crack them. The International Maths Olympiad is for the students of classes 1-12. The Examination pattern is multiple choice questions and the students from class 1 to class 4 have to answer 35
questions while students from classes 5-13 have to answer 50 questions in just one hour. Competitive exams like IMO interest a lot of students, hence a lot of students prepare for the same. While
preparing for exams like the International Maths Olympiad, it is very common to make mistakes from the preparation level only. A student starts making mistakes months before the actual exam starts. A
high profile exam like IMO needs proper systematic and correct preparation. Here are a few mistakes that students make while preparing for IMO, International Maths Olympiad that they should avoid:
1) Don’t prepare without knowing the syllabus
A lot of students every year prepare and appear for IMO without even knowing the proper syllabus. A lot of students just follow NCERT for the Olympiad exam which is not at all-sufficient. Although
one section of the examination comes from the syllabus of class 8 the difficulty level of those questions are higher. Without proper coaching and study material, the preparation becomes a joke.
2) Don’t follow one book for preparation
Looking at the number of pages, a student thinks that one book is enough for them to qualify for the Math Olympiad but in reality, there is not a single book that can explain all the concepts and
fundamentals that’s there in the syllabus. One book can be good for the number system and another book can be good for algebra. However, preparing one book for the Olympiad will just give half the
knowledge. There are many other books, previous year question papers and sample papers to refer to while preparing.
3) Don’t get Chained
Many students while preparing for the Olympiad try to solve a particular question without even thinking about the other possibilities of solving the same question. They get stuck wasting too much
time solving one single question. In competitive exams like IMO every second matters. Wasting time like this while preparing makes it even worse.
4) Don’t get demotivated
A lot of students get excited by the examination as they get to represent their zone and in the second round even their country. The Olympiad unfortunately is not subjected to questions only from the
NCERT books. The question paper of the Maths Olympiad is divided into 4 sections, 1) Logical Reasoning 2) Mathematical Reasoning 3) Everyday Mathematics 4) Achievers Section. As the difficulty level
of the question increases, the mistakes also increase and the morale decreases. After not being able to solve questions with a slightly higher difficulty level they just give up whereas the right way
of answering the question is just by analysing the question.
5) Start Learning from the mistakes
Students after solving the questions don’t pay a lot of attention to the mistakes they have made. When a student encounters a tough question, they directly go for the answer key, without even trying
to solve the question in the first place. The wrong questions are just left behind unnoticed. To crack IMO, students should always learn from their mistakes. They should solve similar questions that
give them trouble.
6) Don’t ignore the Previous Year Question Papers
The most important mistake a student does is they completely ignore the IMO Maths Olympiad previous year question paper class 8 2011. Students often take it as a task to get all the minimum 5 years
of the previous year question paper and then solve it. Solving the previous year question paper will give you a rough idea of the whole examination pattern. There have been a lot of instances where a
few questions have also been the same from previous years. Students ignoring previous year question papers will be deprived of freebies.
7) Manage your Time Properly
Another mistake students do is they don’t manage their time properly while preparing for the International Maths Olympiad because of which in the exam hall they’re not able to complete or attempt all
the questions on time. Solving a problem just for the sake of solving it will not do any good. A student appearing for these exams has to be fast and smart. If a student doesn’t utilise and manage
their time properly during the early days of preparations, solving 50 questions in 60 mins will look like a humongous task
8) Follow your study routine properly
A lot of students don’t follow their timetable properly and many don’t even make or have one. Many students don’t even know how to make a timetable. Students even after knowing their strengths and
weaknesses are still not able to categorise which topic needs more time and which topic needs less. A timetable should be made in such a way that it compliments your day to day life.
9) Please make notes
A lot of students don’t even make notes for the last minute preparation, this leads to a massive panic situation before the exam. Preparing notes also keeps the students updated with their progress.
A good note is neither lengthy nor too short, although it varies from student to student, the last minute before the exam notes should have all the important points but not in detail.
10) Get yourself a good Mentor
Try to get yourself a good mentor, because, without a good mentor, there will be no proper guidance for your Olympiad. It is important to get someone that understands and helps you whenever you need
It is very common to make these kinds of mistakes. However, ignorance is dangerous and if you don’t follow the correct steps while preparing for the International Maths Olympiad, your chances of
qualifying for the exam decreases. Hence keep a note of the following mistakes that a student makes while preparing, try to avoid all the above-mentioned problems. Trust the process and have
confidence in yourself, having a little faith in yourself is good. | {"url":"https://www.articlemarketingnews.com/avoid-these-mistakes-while-preparing-for-class-8-maths-olympiad/","timestamp":"2024-11-07T16:56:15Z","content_type":"text/html","content_length":"87394","record_id":"<urn:uuid:090fc856-782c-45f6-9372-cc879778670b>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00673.warc.gz"} |
How many bacteria would be present 15 hours after the experiment began if a set of bacteria begins with 20 and doubles every 2 hours? | Socratic
How many bacteria would be present 15 hours after the experiment began if a set of bacteria begins with 20 and doubles every 2 hours?
1 Answer
The answer is $3811$ bacteria.
Using a function to describe exponential growth, we can say that
$A = {A}_{\circ} \cdot {e}^{k \cdot t}$, where
$A$ - is the amount we need to find out (in this case, the number of bacteria after 15 hours of growth);
${A}_{\circ}$ - the initial number of bacteria;
$k$ - growth rate;
$t$ - time;
We were given $t$=$15$ hours and ${A}_{\circ}$=$20$ bacteria; however, both $k$ and $A$ need to be determined.
We will determine $k$ by using the fact that the number of bacteria doubles every two hours - this means that after the first 2 hours, we will have 40 bacteria. So,
$40 = 20 \cdot {e}^{k \cdot 2}$, which gives us a $k$ = $0.35$.
Therefore, $A = 20 \cdot {e}^{0.35 \cdot 15} = 3811$ bacteria.
Impact of this question
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On the demand generated by a smooth and concavifiable preference ordering
It is shown that if a consumer's preference ordering is strictly convex and is representable by means of a concave, twice continuously differentiable utility function, then the partial derivative of
a demanded commodity with respect to its price is bounded from above in a neighborhood of a price vector at which the demand fails to be differentiable. In the case of two commodities, if the demand
does not possess finite derivatives with respect to prices at a certain point, then the partial 'derivative' of a commodity with respect to its price is equal to minus infinity. The same result holds
for n commodities under 'almost every' choice of coordinates in the commodity space. If preferences are weakly convex but the same representation assumption holds, demand may not be single-valued but
own-price difference quotients are still bounded from above.
All Science Journal Classification (ASJC) codes
• Economics and Econometrics
• Applied Mathematics
Dive into the research topics of 'On the demand generated by a smooth and concavifiable preference ordering'. Together they form a unique fingerprint. | {"url":"https://pure.psu.edu/en/publications/on-the-demand-generated-by-a-smooth-and-concavifiable-preference-","timestamp":"2024-11-14T09:09:39Z","content_type":"text/html","content_length":"48183","record_id":"<urn:uuid:1d40cb49-1d95-4d6d-bfbc-39d5f680b88f>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00327.warc.gz"} |
The 10x programmer - James Hiebert's Blog
The 10x programmer - James Hiebert's Blog
Ever since the popularization of the software engineering bible, “The Mythical Man Month”, software engineers have loved to debate the merits of one particular concept in the book: The 10x
programmer. I can’t remember exactly how it is stated, but something along the lines of the fact that “good” programmers are generally 5-10 times more productive than mediocre programmers.
This claim seems to elicit a wide range of visceral reactions and opinions from all over the spectrum (opinions on the Internet?! gasp). People get really into debating whether this is a thing or
not. Just google The Myth of the 10x programmer and you’re sure to get loads of opinions.
Personally, I have never had any trouble believing this to be the case. I consider myself to be a fairly decent developer, but I’ve also had the privilege of working with a few folks that just blew
me completely out of the water. Most of these folks now work at the big 3 or a few equally prestigious companies. I have to say that it’s a humbling experience to personally watch (while holding a
stopwatch) someone spend 15 minutes to write a program that you were struggling to write for several days! Let’s say that I spent 2 x 8 hour days working on this program. 0.25 hour / 16 hours is a
factor of 64 (way more than 10x!) in productivity.
Likewise, I have worked with some folks where I was the one on the 10x side of the equation. I have had situations where I could trivially envision the problem space, knew a wide range of the
possible solutions and hammered out a working prototype in the amount of time where the other person was writing up a fundamental question on StackExchange. Knowledge, experience, practice and
judgment come together to make some people orders of magnitude more productive programmers than others. | {"url":"https://vuink.com/post/wnzrf-d-duvroreg-d-danzr/blog/work/2016/11/20/The-10x-Developer-d-dhtml","timestamp":"2024-11-02T08:15:35Z","content_type":"text/html","content_length":"235126","record_id":"<urn:uuid:8eeb2a1a-baa1-4d37-8c4b-01dad306e2d3>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00855.warc.gz"} |
Lesson 4
Quadrilaterals in Circles
Lesson Narrative
In this lesson, students explore an outcome of the Inscribed Angle Theorem in an analysis of cyclic quadrilaterals. These are quadrilaterals that have a circumscribed circle, or a circle that passes
through each vertex of the quadrilateral.
First, students try to draw circumscribed circles for several quadrilaterals. They observe that it is possible to circumscribe some but not all quadrilaterals. Then, they use inscribed angles to
prove that those quadrilaterals that are cyclic have supplementary pairs of opposite angles. They construct the circumscribed circle for a cyclic quadrilateral with a 90 degree angle, and they
explore the idea that the center of the circumscribed circle is equidistant from each vertex of the figure.
As students draw a conclusion from repeated calculations of the measures of angles in cyclic quadrilaterals, they are looking for regularity in repeated reasoning (MP8).
Learning Goals
Teacher Facing
• Prove (using words and other representations) properties of angles for a quadrilateral inscribed in a circle.
Student Facing
• Let’s investigate quadrilaterals that fit in a circle.
Required Preparation
If students will do the digital activity for the activity Construction Ahead, prepare class access to internet-enabled devices, ideally 1 for every 1-2 students. If students will use the activity
from the printed materials, prepare access to geometry toolkits.
The scientific calculators are for the extension to the activity Inscribed Angles and Circumscribed Circles.
Student Facing
• I can prove a theorem about opposite angles in quadrilaterals inscribed in circles.
CCSS Standards
Building On
Building Towards
Glossary Entries
• circumscribed
We say a polygon is circumscribed by a circle if it fits inside the circle and every vertex of the polygon is on the circle.
• cyclic quadrilateral
A quadrilateral whose vertices all lie on the same circle.
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A valid CSS unit. (like "100\ which will be coerced to a string and have "px" appended.
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a directory to place the visualization. If NULL, a temporary directory is used when the offline object is printed.
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: You are given two 32-bit numbers, N
: You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to insert Minto N such that M starts at bit j and ends at bit i. You can assume that the bits j through i
have enough space to fit all of M. That is, if M = 10011, you can assume that there are at least 5
: You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to insert Minto N such that M starts at bit j and ends at bit i. You can assume that the bits j through i
have enough space to fit all of M. That is, if M = 10011, you can assume that there are at least 5
Database System Concepts
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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Section: Chapter Questions
: You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method
to insert Minto N such that M starts at bit j and ends at bit i. You can assume that the bits j through
i have enough space to fit all of M. That is, if M = 10011, you can assume that there are at least 5
bits between j and i. You would not, for example, have j = 3 and i = 2, because M could not fully
fit between bit 3 and bit 2
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How to perform a pipe stress analysis | Consulting - Specifying Engineer
How to perform a pipe stress analysis
Understanding the various types of pipe stresses, the process, and best practices are necessary to perform effective pipe stress analyses.
Learning Objectives
• Define and evaluate the pipe stress analysis process.
• Understand pipe stress analysis.
• Learn how to model a piping system and pressure design basics.
Pipe stress analysis is an analytical method to determine how a piping system behaves based on its material, pressure, temperature, fluid, and support. Pipe stress analysis is not an accurate
depiction of the piping behavior, but it is a good approximation.
The analytical method can be by inspection, simple to complex hand calculations, or a computer model. The computer models can vary from 1-D beam elements to complex, finite element models. For
instance, if it is a water system with no outside forces applied to the piping system, inspection or hand calculations are usually sufficient. If it is a high-pressure, high-temperature,
hazardous-fluids system, and/or large outside forces are applied to the piping system, a computer-aided model may be required.
Understanding pipe stress analysis software does not make for a solid foundation of pipe stress analysis. It’s important to understand the various types of pipe stresses, the process, and other items
related to pipe stress analysis for best practices in performing a pipe stress analysis.
There are many piping codes and standards that could be used during a pipe stress analysis depending on the application (power, process chemical, gas distribution) and location (country or local
jurisdiction). However, to keep things simple, this discussion is based on American Society of Mechanical Engineers (ASME) B31.1 Power Piping. The physics of pipe stress analysis does not change with
piping code.
Pipe stress analysis should be done primarily to provide safety to the public, whether you are designing a building heating system or a high-pressure gas line in a refinery. Public safety is
paramount. The National Society of Professional Engineers (NSPE) Code of Ethics’ first cannon is: “Hold paramount the safety, health, and welfare of the public.”
On a good day, a pipe failure is only a broken support that the owner does not call the designer/engineer about. On a bad day, the owner requires the designer/engineer to pay for the damage and the
engineer to provide a solution for free. On a horrible day, someone is killed.
Another reason a pipe stress analysis is performed is to increase the life of piping. Most engineers won’t consider a piece of pipe to be equipment, but it is no different than a pump. Both have
moving parts and must be designed and maintained properly to ensure a proper life. Pipe stress analysis also is used to protect equipment, because a pipe is nothing more than a big lever arm
connected to a delicate piece of equipment. If not properly supported and designed, it can have devastating effects on that equipment.
There are several common reasons that could warrant a pipe stress analysis, in addition to those above. They include:
• Elevated temperatures (>250°F).
• Pressure mandated (300 psig).
• Sensitive equipment connections.
• Large D/t ratio (>50).
• Piping subject to external pressures.
• Critical services.
The key when performing a pipe stress analysis is determining the required level of detail.
How to model the piping system
Pipe stress analysis computer models are a series of 3-D beam elements that create a depiction of the piping geometry. Three-dimensional beam elements are the most efficient way to model the piping
system, but not necessarily the most accurate; and without complex finite element models, it is nearly impossible to account for everything. However, it is known from historical empirical testing
that these methods and 3-D beam computer models demonstrate enough behavior that they are a good approximation. In addition, piping codes, such as ASME B31, have safety margins that allow for
approximation. That being said, there are some pitfalls with modeling piping systems that one should avoid:
• The computer models are only as good as the information entered into them. It is important when developing a pipe stress analysis, as with any finite element analysis (FEA) model, to also
understand the physics and boundary conditions of the model.
• Elements used to model the piping system have their limitations. One-dimensional beam elements are great for straight pieces of piping, but not so good with pipe fittings (elbows, tees, reducers,
etc.). Therefore, ASME has developed stress-intensification factors (SIFs) for piping fittings through empirical testing. They allow for greater approximation without using complex FEA models
with shells, plates, and brick elements.
It is important to make sure these limitations are considered when developing a pipe stress analysis. Most pipe stress analyses do not perform like a high-powered FEA software package.
Three-dimensional beam element
The 3-D beam element behaviors are dominated by bending moments. As mentioned above, it is efficient for most analyses and sufficient for system analysis. However, there are downsides to using a 3-D
beam element:
• No localized effects will be seen on the pipe wall.
• No second-order effects.
• No large rotation.
• No accounting for a large shear load.
□ Wall deflection occurs before bending failure.
□ Short, fat cantilever versus long and skinny.
• No shell/wall effects can be seen.
The main types of piping stresses
There are five primary piping stresses that can cause failure in a piping system: hoop stress, axial stress, bending stress, torsional stress, and fatigue stress.
Hoop stress is the result of pressure being applied to the pipe either internally or externally. Because pressure is uniformly applied to the piping system, hoop stress also is considered to be
uniform over a given length of pipe. Note that hoop stress will change with diameter and wall thickness throughout the piping system. Hoop stress is most commonly represented by the following
Axial stress results from the restrained axial growth of the pipe. Axial growth is caused by thermal expansion, pressure expansion, and applied forces. If a pipe run can grow freely in one direction,
there is no axial present—at least in theory. When comparing axial growth caused by pressure, steel-pipe growth is minimal at over 100 ft and can be ignored. Composite piping such as fiber
re-enforced pipe (FRP) or plastic pipe will exhibit noticeable growth, as much as 2 to 3 in. over 100 ft under the right conditions (200 to 300 psi). The primary reason for the difference in growth
rates under pressure is related to the modulus of elasticity. Steel has a modulus of elasticity of approximately 30 x 10^6 psi, whereas composites will be 2 to 3 orders of magnitude or less. Axial
stress is represented by the axial force over the pipes cross-sectional area:
Bending stress is the stress caused by body forces being applied to the piping. Body forces are the pipe and medium weight, concentrated masses (valves, flanges), occasional forces (seismic, wind,
thrust loads), and forced displacements caused by growth from adjacent piping and equipment connections. Body forces create a resultant moment about the pipe, for which the stress can be represented
by the moment divided by the section modulus:
Torsional stress is the resultant stress caused by the rotational moment around the pipe axis and is caused by body forces. However, because a piping system most likely will fail in bending before
torsion, most piping codes ignore the effects of torsion.
Fatigue stress is created by continuous cycling of the stresses that are present in the piping. For example, turning a water faucet on and off all day will create a fatigue stress, albeit low,
because of the pressure being released and then built up. In power-cycle applications, the cycling of a steam turbine from low to high pressure/temperature creates a fatigue stress. Fatigue stress
results in a reduction of allowable strength in the piping system and is commonly caused by cycling of:
• Pressure.
• Temperature.
• Vibration, flow induced or cause by rotating equipment.
• Occasional loads (a gentle breeze caused the Tacoma Narrows Bridge in Washington State to collapse from fatigue).
Allowable code stresses
Piping codes, such as those published by ASME, provide an allowable code stress, which is the maximum stress a piping system can withstand before code failure. A code failure is not necessarily a
piping failure. This is because of safety factors built into piping codes. ASME codes consider three distinct types of stress: sustained stress, displacement (thermal or expansion) stress, and
occasional stress.
Sustained or longitudinal stress is developed by imposing loads necessary to satisfy the laws of equilibrium between external and internal forces. Sustained stresses are not self-limiting. If the
sustained stress exceeds the yield strength of the piping material through the entire thickness, the prevention of failure is entirely dependent on the strain-hardening properties of the material.
Displacement stress is developed by the self-constraint of the piping structure. It must satisfy an imposed strain pattern rather than being in equilibrium with an external load. Displacement
stresses are most often associated with the effects of temperature; however, external displacements, such as building settlements, are considered a displacement stress.
Occasional stress is “The sum of longitudinal stresses produced by internal pressure, live and dead loads, and those produced by occasional loads,” according to ASME B31.1, paragraph 102.3.3(A).
Occasional stresses can exceed the allowable code stress by a given percentage depending on frequency and duration of the load; for ASME piping codes, this is typically 15% or 20%. For example, wind
loads can only exceed the allowable code stress by 15% due to their frequency, but seismic loads can exceed by 20% due to the relative infrequency of the loads.
Pressure design basics
As a pipe stress analyst, it is critical to understand how wall thickness is determined. If the pipe wall is too thin, it will not matter how the pipe is supported; it will fail. Typically, the
engineer designing the system also will determine the wall thickness; however, the wall thickness is also verified during the pipe stress analysis. Most engineers are more concerned with mass flow
and pressure drop, therefore the effects of pipe size and wall thickness may be lost on them. Going to a thicker pipe wall or a larger pipe size may be worth the material costs, versus facing design
issues and added pipe-support costs in labor and materials.
Hoop stress (simplified) is . ASME codes apply a safety factor of two when determining wall thickness based on hoop stress, yielding:
The safety factor is to account for the additional stresses caused by bending and axial stresses to be applied later. Through basic algebraic manipulation, the code equation for wall thickness is:
A is the additional thickness added to the pipe corrosion, erosion, and wear during normal operation. The value of A is left up to the designer by ASME. However, most people consider 0.0625 in. to be
an acceptable value.
The minimum will thickness (actual) shown above is based on the internal diameter (ID) of the piping. The main difference in the two wall thickness equations is the simplified version is more
conservative, quicker, and easier to calculate for scheduled pipe. The actual version is closer to the measured hoop stress. Most stress analysis programs default to calculating hoop stress based ID.
Lastly, ASME codes require that minimum thickness account for the 12.5% mill tolerance:
Please note that when factoring in the 12.5% mill tolerance, multiplying by 1.125 is not the same as dividing by 0.875.
Sustained stresses
For someone who is new to pipe stress analysis—there is no reason sustained stresses in the pipe should be greater than 55% of the standard allowable stress. There are a couple of reasons why. First,
recommended pipe support spans are governed by deflection, and not by allowable stress, to ensure proper flow and drainage. The second is from the discussion above, the wall thickness is based on a
safety factor of two, which is removed from the sustained-stress equation.
Manufacturers Standardization Society (MSS) SP-58: Pipe Hangers and Supports—Materials, Design, Manufacture, Selection, Application, and Installation recommends support spans to be based on
deflection criteria of approximately 0.125 in. or less between supports. The deflection criteria assume a simply supported beam. However, a supported piping system is a continuously supported beam
that reduces reaction and moments at each support, further reducing the deflection between supports. This negates the bending moments between supports and reduces the bending moment term of sustained
Below is the sustained equation from ASME B31.1:
The simplified hoop-stress term is in the equation above, is based on minimum wall thickness, and is approximately at 50% of allowable stress, based on the wall thickness safety factor. However, in
the equation above, hoop stress is based on nominal wall thickness, which is at least 1/0.875 times greater than minimum wall thickness. Conversely, if hoop stress as a function of minimum wall
thickness is 50% of allowable code stress, then hoop stress as a function of nominal wall thickness is 50% x 0.875 = 43.75%.
As mentioned above, the sustained-stress equation is based on nominal wall thickness, with extra wall thickness for milling and corrosion. Because there is extra wall thickness, the pipe has extra
strength available to resist deflection. Furthermore, to achieve pipe failure from deflection, the supported pipe spans would be at least three to four times greater in length than the recommended
MSS SP-58 spans. The moment due to dead weight contributes approximately 10% code stress to the equation above when using MSS SP-58 recommended pipe-support spans.
Looking back at the sustained-stress equation above, if you assume 10% code stress from the deadweight moments and 44% code stress from hoop stress, the sustained stress should be approximately 54%
or less. If this is not the case, there are usually excessive deflections at a bend and/or concentrated mass in the piping, creating a higher-than-expected bending moment from an unbalanced system
(see Table 1).
Standard span guidelines
Below are some general thoughts on standard pipe spans to consider:
• Fluid has a greater impact as the pipe size becomes larger. Water weight is more than pipe weight for 12 in. nominal pipe size (NPS) for standard wall thickness (STD), or greater.
• When concentrated loads, such as flanges, valves, and piping specialties, are present between pipe supports, the recommended span should be reduced to account for them.
• A pipe support should be placed within one-third the recommended span of a rotating equipment connection to minimize vertical load and moments at connection. In most cases, this support should be
a variable spring to help with adjustment and reduce translation vibration.
• When piping changes horizontal direction, the recommended span between pipe supports shall be reduced by 25%.
Displacement stresses
In most cases, if displacement or expansion stresses are perceived to be a concern (e.g., elevated temperatures), then a computerized pipe stress analysis is required. If a computerized analysis is
performed, displacement stresses should be kept at 80% to 90% of what the code allows.
Typically, this recommendation is met by ensuring the equipment connection loads are within published allowable code stresses through adding flexibility to the piping system. Flexible piping systems
typically have low displacement stresses because the piping can grow freely.
Occasional stresses
Occasional stresses in the piping system are caused by short-term events, such as seismic, wind, and relief-thrust loads. These three loads comprise most of the possible occasional load combinations.
Because occasional stresses are short-term, most piping codes allow for increased pipe stresses for a brief period. ASME codes typically allow an increase of:
• Fifteen percent if the event lasts less than 8 hours and no more than 800 hours per year
• Twenty percent if the event lasts less than 1 hour and no more than 80 hours per year.
Typically, wind loads fall under the 15% increase category, where seismic and relief thrust would be a 20% increase.
If occasional stresses are perceived to be a concern or are complex in nature, a computerized pipe stress analysis is warranted. However, in most cases. adding lateral restraints for every three or
four nominal pipe-support spans will cover most seismic or wind loadings, unless they are in a high seismic zone, such as California, or are subjected to coastal wind loading with sustained hurricane
Keep pipe analysis records
Most people believe that a computer printout is a sufficient record of a pipe stress analysis. This a big mistake that can be avoided with little effort. Creating a record of your work is about more
than keeping a hard copy or PDF of the computer-aided pipe stress analysis. It means documenting a trail of all inputs, not just the drawings used to create the piping geometry. Items that could be
included are the piping and instrument diagrams, system parameters, load cases, and any corresponding external forces applied to the piping system, pipe-support locations, and type of pipe support
used. Most pipe stress analysis records will fill a three-ring binder.
As most consulting engineers have internal quality assurance/quality control procedures, develop a standard list of the inputs commonly used and corresponding reference for the information. This
would provide the checker of a calculation a place to sign off, indicating they concur with the input and acknowledge the source of the input. In the end, your documentation should tell a complete
Monte Engelkemier is the group engineering lead for piping, mechanical, and equipment in the starches, sweeteners, and texturizers division of Cargill. Prior to that, he was a member of Stanley
Consultants for 12 years, where he authored this article before taking his current position.
Do you have experience and expertise with the topics mentioned in this content? You should consider contributing to our WTWH Media editorial team and getting the recognition you and your company
deserve. Click here to start this process. | {"url":"https://www.csemag.com/articles/how-to-perform-a-pipe-stress-analysis/","timestamp":"2024-11-03T09:33:36Z","content_type":"text/html","content_length":"252469","record_id":"<urn:uuid:dcc5ec65-fdd1-4676-a664-240b41e90635>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00759.warc.gz"} |
What Are The Factors Of 69? - Tech Stray
What Are The Factors Of 69?
So do you have to discover the elements of sixty nine? In this brief guide, we’re going to describe what the elements of sixty nine are, the way you discover them, and list the detail pairs of sixty
nine to make the calculation art work. Let’s dive!
Sixty nine Definition Factors
When we talk approximately the factors of sixty nine, we really endorse all the fantastic and horrible integers (complete numbers) that may be divided further into sixty nine. If you’re taking 69 and
divide it by way of way of actually without a doubt certainly one of its factors, the answer can be some one of a kind issue of 69.
Sixty nine. How To Discover The Elements Of
We have honestly said that a hassle is a spread of that can be divided lightly into 69. So the way you find and list all the factors of sixty nine is to undergo each quantity as plenty as and along
side 69 and take a look at which numbers bring about a very good quotient (this means that that that no decimal locations).
Doing this via hand in big numbers can be time-eating, but it is as an alternative easy for a pc software to do it. Our calculator has completed it for you. Here are all the factors of sixty nine:
sixty nine 1 = 69
sixty nine three = 23
69 23 = 3
sixty nine 69 = 1
All those elements can be used to divide 69 and get a whole range. The complete listing of terrific factors for sixty nine are:
1, 3, 23, and 69
sixty nine. Negative Elements Of
Technically, you can also have horrible elements of sixty nine in math. If you need to calculate the elements of numerous for homework or a check, often the instructor or take a look at can be mainly
looking for exceptional numbers.
However, we are able to great convert wonderful numbers to terrible and those terrible numbers additionally may be factors of sixty nine:
-1, -three, -23, and -69
How Many Elements Does sixty nine Have?
As we are capable of see from the above calculation that there are a entire of 4 high pleasant elements for sixty nine and 4 horrible factors for 69 for an entire of 8 elements for the quantity 69.
Sixty nine has 4 awesome factors and sixty nine has 4 terrible factors. What are the negative numbers that may be a factor of 69?
Sixty 9. Upload The Factors Of
A issue pair is a aggregate of factors that can be improved to equal sixty nine. For 69, all feasible component pairs are listed underneath:
1 x sixty nine = sixty nine
3 x 23 = 69
We’ve moreover written a guide that is going into a bit greater element about trouble pairs for 69 in case you’re concerned!
As earlier than, we can also listing all horrible problem pairs for 69:
-1 x -sixty nine = 69
-three x -23 = sixty nine
Note in terrible problem pairs that due to the reality we’re multiplying minus with the resource of minus, the quit result is a fantastic quantity.
So there you’ve got it. A Complete Guide to the Factors of sixty nine. You have to now have the information and talents to calculate your own factors and factors for any type of your choice.
Feel free to attempt the calculator beneath to test each one-of-a-kind variety or, if you’re feeling fancy, take a pencil and paper and try to do it via using way of hand. Just make certain to pick
out the smallest variety!
Cite, Hyperlink Or Reference This Web Page
If you discover this cloth beneficial to your research, please do us a tremendous pick out and use the device under to ensure you speak with us nicely anywhere you use it. We simply recognize your
Sixty nine. Factors Of
The factors of sixty nine and the excessive factors of 69 are one-of-a-kind because of the reality seventy-9 is a composite quantity. Also, no matter being carefully related, the high elements of
sixty nine and the pinnacle factors of sixty nine are not precisely the identical.
In any case, by way of studying you may discover the solution to the question What are the factors of 69? And the entirety else you need to understand approximately the problem.
What Are The Elements Of sixty nine?
They are: sixty nine, 23, 3, 1. These are all factors of sixty nine, and each get right of entry to in the list can divide sixty nine without punctuation (module 0). Therefore the issue and
denominator of 69 may be used interchangeably.
As in the case of any natural massive variety extra than 0, the giant range itself, right right right here sixty nine, further to at least one are the factors and divisors of sixty nine.
Sixty 9. Top Factors Of
The excessive factors of 69 are the ones excessive numbers that exactly divide 69 without a the relaxation described thru Euclidean branch. In different terms, a top factorization of sixty nine
divides sixty nine with none break, modulo 0.
For sixty nine, the pinnacle elements are: 3, 23. By definition, 1 is not a top variety.
In addition to 1, what separates the top elements and the immoderate factors of the variety 69 is the phrase “high”. The first listing consists of both composite and top numbers, while the latter
consists of fine excessive numbers.
Sixty 9. Prime Factorization Of
The pinnacle factorization of sixty nine is three × 23. This is a very particular listing of pinnacle factors with their multiples. Note that the excessive factorization of 69 does no longer
encompass the no 1, but it consists of each instance of a remarkable high factorization.
69 is a composite massive range. Unlike pinnacle numbers, which have handiest one aspect, mixed numbers like 69 have at the least factors.
To make clear the which means that that of 69, 23, 3, 1 the rightmost and rightmostose the second rightmost and the second leftmost get right of entry to to achieve the 2nd factorization which
moreover produces 69.
The top factorization or integer factorization of 69 way identifying the set of high numbers which, at the same time as extended collectively, produce the perfect quantity 69. This is likewise
referred to as pinnacle decomposition of 69.
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• BPC •
Autor: BLINDER, S. M. & FANO, GUIDO
Año: 2017
Género: FÍSICA, MATEMÁTICAS, CIENCIA Y TECNOLOGÍA
Formato: PDF
This book is designed to make accessible to nonspecialists the still evolving concepts of quantum mechanics and the terminology in which these are expressed. The opening chapters summarize elementary
concepts of twentieth century quantum mechanics and describe the mathematical methods employed in the field, with clear explanation of, for example, Hilbert space, complex variables, complex vector
spaces and Dirac notation, and the Heisenberg uncertainty principle. After detailed discussion of the Schrödinger equation, subsequent chapters focus on isotropic vectors, used to construct spinors,
and on conceptual problems associated with measurement, superposition, and decoherence in quantum systems. Here, due attention is paid to Bell's inequality and the possible existence of hidden
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secure communication will result. This book will be of interest to a wide readership seeking to understand modern quantum mechanics and its potential applications. | {"url":"http://labiblioteca.mx/1153.html","timestamp":"2024-11-06T18:03:38Z","content_type":"text/html","content_length":"3085","record_id":"<urn:uuid:10948afd-0992-42a2-bc9e-296804800c81>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00309.warc.gz"} |
The Magic Cafe Forums - Seven Queens "number to card" calculation (difficult)
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Nashville TN
960 Posts
Seven Queens "number to card" calculation (difficult).
Note that this thread is for other magician articles/threads to reference.
The easy "card to number" calculation is documented elsewhere on another thread. This thread is specifically on how to calculate "number to card" for the Seven Queens stack (this will also work on
the Jackknife stack and the King Deuce stack).
Below is the Seven Queens stack:
7S, QD, 8D, KC, 10H, 9C, AC, 3S, 6H, 5C, JS, 2H, 4D
7H, QS, 8S, KD, 10C, 9D, AD, 3H, 6C, 5D, JH, 2C, 4S
7C, QH, 8H, KS, 10D, 9S, AS, 3C, 6D, 5S, JC, 2D, 4H
7D, QC, 8C, KH, 10S, 9H, AH, 3D, 6S, 5H, JD, 2S, 4C
Note that the possible offsets that may be used are: 0 or 13 or 26 or 39.
The Suits associated number: 1=spades, 2=hearts, 3=clubs, 4=diamonds. Note that 0 is also equal to diamonds.
Below is the Harry Riser type of twin pairs:
Ace and 7 (have a sharp angle at top)
2 and Queen (2 headed Queen)
3 and 8 (the 3 looks like half of an eight)
4 and King (the K and 4 have 4 corners)
5 and 10 (five and dime store)
6 and 9 (same symbol upside down)
Jack stands alone.
Also note that we have an imaginary 8 rung ladder with rungs numbered 0, 1, 2, 3, 0, 1, 2, 3.
This ladder will come in handy below. These rung numbers do NOT directly relate to the suits, it's the number of steps that are taken between the rungs that relate to the suits (this will be
explained a little further below).
Note that the term "mod 4" just means the REMAINDER number when dividing by 4.
A spectator names any number from one to 52.
The magician (or shill) mentally calculates:
1. Subtracts the nearest OFFSET that is lower than the named number (this gives a number from one to 13).
2. The Harry Riser type pair is the value of the card.
3. To determine the suit of the card:
a. Mod 4 the Harry Riser pair (result will be 0 or 1 or 2 or 3). From the bottom of the ladder this is the STARTING RUNG.
b. The TARGET RUNG is the first rung above that is equal to THE FIRST DIGIT OF THE OFFSET.
c. The NUMBER OF STEPS to get from the starting rung to the target rung equates to the suit value ie: 1 step=spades, 2 steps=hearts, 3 steps=clubs, 4 steps=diamonds (or zero steps equals diamonds).
The above rules work on the Seven Queens stack, Jackknife stack, and King Deuce stack.
The below examples are only for the Seven Queens stack.
Example: spectator names the number 28.
Magician or shill calculates:
Nearest lower possible offset is 26.
28 - 26 is 2
The 2's twin is the Queen (12).
Mod 4 the 12 gives 0.
How many steps to get from the 0 rung to the first digit of 26 rung (rung 2)?
We must take two steps up:
(rung 1, then rung 2).
The two steps equates to hearts.
Thus the answer is the Queen of Hearts.
Example: spectator names the number 18.
Magician or shill calculates:
Nearest lower possible offset is 13.
18 - 13 is 5
The 5's twin is the 10 (the card value).
Mod 4 the 10 gives 2.
How many steps to get from the '2" rung to the first digit of 13 rung (rung 1)?
We must take three steps up:
(rung 3, then rung 0, then rung 1).
The three steps equates to clubs.
Thus the answer is the Ten of Clubs.
Example: spectator names the number 39.
Magician or shill calculates:
Nearest lower possible offset is 26.
39 - 26 is 13
The King (13) twin is the 4 (the card value).
Mod 4 the 4 gives 0.
How many steps to get from the "0" rung to the first digit of 26 rung (rung 2)?
We must take two steps up:
(rung 1, then rung 2).
The two steps equates to hearts.
Thus the answer is the Four of Hearts.
Example: spectator names the number 6.
Magician or shill calculates:
Nearest lower possible offset is 0.
6 - 0 is 6
The 6 twin is the 9 (the card value).
Mod 4 the 9 gives 1.
How many steps to get from the "1" rung to the first digit of 0 rung (next rung 0)?
We must take three steps up:
(rung 2, then rung 3, then rung 0).
The three steps equates to clubs.
Thus the answer is the Nine of Clubs.
Example: spectator names the number 20.
Magician or shill calculates:
Nearest lower possible offset is 13.
20 - 13 is 7.
The 7's twin is the Ace (1). The card value.
Mod 4 the 1 gives 1 .
How many steps to get from the "1" rung to the first digit of 13 rung (rung 1)?
We must take zero or 4 steps up:
(rung 2, then rung 3, then rung 0, then rung 1).
The 0 or 4 steps equates to diamonds.
Thus the answer is the Ace of Diamonds.
Note that "zero steps up" or "four steps up" will get to the same target rung number thus zero or four steps up means diamonds. | {"url":"https://themagiccafe.com/forums/viewtopic.php?topic=750788#1","timestamp":"2024-11-05T07:38:38Z","content_type":"application/xhtml+xml","content_length":"13648","record_id":"<urn:uuid:2c2c4001-e1be-4634-b176-5311f3e6ed7e>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00746.warc.gz"} |
Discussion with Zielinski, physical meaning of Einstein's theory of gravity
On Nov 28, 2010, at 2:37 PM, Paul Zielinski wrote:
On Sun, Nov 28, 2010 at 1:54 PM, JACK SARFATTI <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:
On Nov 28, 2010, at 1:29 PM, Paul Zielinski wrote:
Look, I'm not trying to undermine your condensate model. I'm just trying to see how it plays out physically. How it may be capable of explaining gravitational attraction in physical terms.
Gravity attraction is explained exactly the same as Einstein explained it - geodesics in curved spacetime.
My condensate - tetrad model gives emergent
Guv + kTuv = 0
as the end product.
Hence gravity attraction is explained.
Your tetrads describe vacuum Weyl curvature as observed from moving reference frames.
Right. In principle they describe Ricci curvature also where Tuv =/= 0 as well. For example, one can imagine tiny strain gauges implanted in the Earth. Or drill holes and lower detectors on cables -
those would be static LNIFs.
The E-H field equations determine Ricci curvature at the source. How does you model bridge that gap? How do you get Ricci curvature from the Goldstone phases? Doesn't your formal analogy relate the
Goldstone phases directly to the vacuum tetrads?
The vacuum is still there even when Tuv =/= 0. One does not literally have to have actual detectors in place. That is impossible. The point is the Einstein field equations
Guv + kTuv = 0 + initial/boundary/final conditions
describe the GMD field.
A particular representation for guv is a pattern of LNIF detectors that could be there in a counterfactual sense. If you put one there you would get the numbers given by the guv solution.
For example when you write the SSS solution for a black hole outside its horizon
gtt = 1 - rs/r = - 1/grr
rs/r < 1
g(r) ~ (c^2rs/r^2)(1 - rs/r)^-1/2 ---> infinity as r ---> rs from the outside ~ Unruh temperature
that only works for the static rocket detectors in Hawking's picture!
Similarly for the deSitter solution in the static LNIFs where we are at r = 0 in Tamara Davis's Fig 1.1.
gtt = 1 - / ^2 = - 1/grr
r < /^-1/2
g(r) --> 2c^2(/ )(1 - / ^2)^-1/2 ---> infinity as r ---> /^-1/2 from the inside only when / > 0.
g(0) = 0 we are on a de Sitter geodesic.
but which way is g(r) pointing? - is the question.
gtt = 1 + 2VNewton/c^2
when VNewton = -c^2rs/r
-dVNewton /dr = -c^2rs/r^2 attraction pointing toward smaller r
In contrast when VNewton = -c^2/ ^2
-dVNewton /dr = +2c^2/
therefore dS / > 0 is virtual boson dark energy repulsion away from r = 0.
In contrast, AdS / < 0 is closed loop virtual fermion dark matter attraction toward r = 0.
Note, there is no event horizon at r = /^1/2 in the AdS case.
Note that the potential / ^2 has QCD-like asymptotic freedom as r --> 0 and it also has confinement as r ---> /^-1/2
when / ~ 1/fermi^2 instead of / ~ 1/Area of future horizon
where the parallel Regge trajectories for hadronic resonances are explained i.e. hadrons as rotating black holes in Salam's f-gravity.
Or is there more to your model than that? Can you really get matter-induced Ricci curvature from the Goldstone phases of the post-inflation condensate?
Since I get Einstein's field equation - short answer is yes.
The point is once I derive the tetrads for LIFs from the gradients of the Goldstone phases then simply use standard arguments that Einstein used to introduce Tuv with his Guv.
Technically I get the curvature 2-form R^I^J from the Goldstone phases including their 3rd order "jerk" partial derivatives - hence the kind nonlocality that we see in radiation reaction in charged
particle mechanics leading to Wheeler-Feynman type picture - from Dirac's anticipatory picture using advanced potentials. Clearly the third order partial derivatives demand Aharonov's post-selection
final boundary condition.
The basic field partial differential equations of Einstein are really 3rd order not second order in the Goldstone phase Cartan 0-forms from the cohering of the false vacuum into the present vacuum at
If you impose flat space decoupled from time, i.e. Galilean group that's Newton's gravity force picture.
You can still have Newtonian gravity in Minkowski spacetime, but it would violate the maximum signal propagation speed c.
Curved spacetime eliminates Newton's gravity force and replaces it by the invariant pattern of geodesics in curved spacetime.
Of course.
The equivalence principle is a qualitative PHYSICAL gap that cannot be crossed over to the electro-weak-strong forces.
I'm not sure what this means.
You have a formal analogy between the Goldstone phases and the tetrads of GTR. Why not flesh that out with physical explanations?
I have. The physical explanation is exactly the same as that for superflow in superfluids - except now it's a 4D supersolid with plastic distortions of the point gravity monopole defects in the
condensate. You simply do not understand what I have been saying in its fullness.
As to the vacuum fine. But how do you get Ricci curvature at the source? And how do you get from that to the Weyl curvature of the vacuum? No Ricci curvature, no E-H field equations. What I don't
understand is how you get an analog of Ricci curvature from your tetrad-Goldstone phase analogy. Also why do you need diffeomorphism gauge invariance?
I don't need it. Nature needs it.
Well I disagree.
As you know I think you are mistaken here.
I've given you a clear cut mathematical argument as to why this fails, based on the geodesic equation.
I think your argument is based on a profound misunderstanding of the physical meaning of Einstein's GR. So I don't accept it. It's too easy to get lost in all the excess baggage formalism that is a
dense fog hiding the physics.
All that means is that locally coincident accelerating LNIF frames see the same objective curved spacetime invariant patterns of geodesics and their relative deviation.
I have a complete physical picture - not just formalism.
Yes of course Riemann curvature is no problem, it's a tensor quantity. The problem is that gravitational deformation of the geodesics is locally determined by the metric gradients g_uv, w(x), and not
the curvature R^u_vwl(x).
You are mistaken here. This is the key error in your attempt. You don't understand that the Levi-Civita connection from first-order partial derivatives of the metric tensor does not affect the
gravitational deformation pattern of the tangent bundle of geodesics. The Levi-Civita connection only describes the acceleration of the detectors measuring the non-accelerating geodesic test
particles. The gravity deformation is only the covariant curl piece of the Levi-Civita connection with itself. Those are second-order partial derivatives of the metric tensor, but they are, at a
deeper level of the Dirac substrate - third-order partial derivatives of the eight coherent Goldstone phases of the post-inflation vacuum superconductor whose point -ike monopole defects form the
Kleinert world crystal lattice.
Active diffeomorphism invariance implies that there is no physical distinction between coordinate artifacts
g'_u'v', w(x') =/= 0 (1)
on the one hand, and actual geometric gradients
g'_uv, w(x) =/= 0 (2)
on the other. I say this is simply wrong. The gradients (2) deform the geodesics; the gradients (1) do not. The transformations (2) accelerate free test with respect to the source (or vice versa);
while the transformations (1) do not. That is a physical distinction. I would have thought you'd be better off without it. I know this is Rovelli's hobby horse, but I think I can knock it and the
"hole" argument down in 3-4 lines. Babak and Grishchuk didn't get that far in their paper.
I don't believe you.
I showed you the argument based on the geodesic equation Jack. Ball's in your court.
OK, explain why there is no physical distinction between (1) and (2) above, with reference to the geodesic equation. FYI some top people in foundations of physics reject active diffeomorphism
invariance as distinguishing GTR from any other theory based on a spacetime manifold -- e.g., Cartan's spacetime formulation of Newtonian theory.
I don't believe them because they do not understand the physical meaning of Einstein's GR. They know how to manipulate the formal symbols, but lack the physical understanding.
You can disagree Jack, but don't try to tell me this is a "crank" position. It's not.
It is a mistaken model. It's wrong because you have made a false premise. Your starting point is mathematically and conceptually wrong. The first-order metric gradients do not change the objective
pattern of the geodesics as you assume.
On Sat, Nov 27, 2010 at 2:44 PM, Jack Sarfatti <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:
The real gravity field comes from the set of coherent vacuum phase gradients analogous to superflow. | {"url":"https://stardrive.org/index.php/all-blog-articles/2735-Discussion-with-Zielinski,-physical-meaning-of-Einstein's-theory-of-gravity","timestamp":"2024-11-12T00:45:18Z","content_type":"text/html","content_length":"26160","record_id":"<urn:uuid:ec65c405-04fc-4058-a137-8a037f3c2662>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00848.warc.gz"} |
how to maximize the value of z?
Not open for further replies.
When a distance of 1 unit is moved along x-axis, there is 1 unit height increase along z-axis. Likewise, when a distance of 1 unit is moved along y-axis, there is 2 unit height increase along z-axis.
It means that when you move a distance of √2 along diagonal, there would be a 3-unit height increase along z-axis. Please see the the
We only want to move 1-unit along the diagonal and still want the height along z-axis to be maximum. It means that we need a certain combination of x- and y-units.
I think that 0.45 units along x-axis and 0.9 units along y-axis will give us the maximum height along z which is 2.25. Please have a look
I'm not sure if the combination given above is correct, and also how the method works. It looks like an optimization problem. Not sure. Could you please guide me? Thank you.
When a distance of 1 unit is moved along x-axis, there is 1 unit height increase along z-axis. Likewise, when a distance of 1 unit is moved along y-axis, there is 2 unit height increase along
z-axis. It means that when you move a distance of √2 along diagonal, there would be a 3-unit height increase along z-axis. Please see the the
We only want to move 1-unit along the diagonal and still want the height along z-axis to be maximum. It means that we need a certain combination of x- and y-units.
I think that 0.45 units along x-axis and 0.9 units along y-axis will give us the maximum height along z which is 2.25. Please have a look
I'm not sure if the combination given above is correct, and also how the method works. It looks like an optimization problem. Not sure. Could you please guide me? Thank you.
Your answer is correct. X-distance is 1/sqrt(5 )= 0.447214, and y-distance is 2/sqrt(5) = 0.894427 . Did you solve it by incremental iteration or by differential calculus? It is a simple calculus
Thank you, Ratch.
I was discussing it with someone and I was given the number but even that person wasn't really sure about it. I don't know how he came up with these numbers. How do you do it by differential
calculus? I'd prefer to know about differential calculus method than iteration method which is more of a 'mechanical' algorithm. Could you please help? Thanks.
Thank you, Ratch.
I was discussing it with someone and I was given the number but even that person wasn't really sure about it. I don't know how he came up with these numbers. How do you do it by differential
calculus? I'd prefer to know about differential calculus method than iteration method which is more of a 'mechanical' algorithm. Could you please help? Thanks.
If the x-distance is "x", what is the y-distance?
If the x-distance is "x", what is the y-distance?
"y"! I'm sorry that I don't get your point.
"y"! I'm sorry that I don't get your point.
The diagonal is a function of the x-distance and the y-distance. If you know the x-distance, you can easily calculate the y-distance. Hint: Pythagoras's theorem
It looks like that you have misinterpreted the original query.
We don't know the x-distance.
The problem is about x,y,z plane. We are given that when a distance of 1 unit is moved along x-axis, there is 1 unit height increase along z-axis. Likewise, when a distance of 1 unit is moved along
y-axis, there is 2 unit height increase along z-axis. It means that when you move a distance of √2 along diagonal, there would be a 3-unit height increase along z-axis.
Then, using the given slope ratios, i.e. 1-z/1-x and 2-z/1-y, we are asked to find the combination of optimum x-distance and y-distance which gives maximum height for z.
Thank you.
It looks like that you have misinterpreted the original query.
We don't know the x-distance.
The problem is about x,y,z plane. We are given that when a distance of 1 unit is moved along x-axis, there is 1 unit height increase along z-axis. Likewise, when a distance of 1 unit is moved
along y-axis, there is 2 unit height increase along z-axis. It means that when you move a distance of √2 along diagonal, there would be a 3-unit height increase along z-axis.
Then, using the given slope ratios, i.e. 1-z/1-x and 2-z/1-y, we are asked to find the combination of optimum x-distance and y-distance which gives maximum height for z.
Thank you.
I fully understand the problem. You have not answered the question I asked. If I move x amount of distance along the x-axis until the diagonal is one, how much will I move along the y-axis. It is a
simple geometric relationship. Use the "hint" I gave you previously. Forget about the "z" direction for now.
Pythagoras theorem:
hypotenuse^2 = base^2 + perpendicular^2
diagonal^2 = x-distance^2 + y-distance^2
y-distance^2 = diagonal^2 - x-distance^2
y-distance = sqrt{diagonal^2 - x-distance^2}
y-distance = sqrt{1 - x-distance^2}
I hope that this is what you were asking for.
Pythagoras theorem:
hypotenuse^2 = base^2 + perpendicular^2
diagonal^2 = x-distance^2 + y-distance^2
y-distance^2 = diagonal^2 - x-distance^2
y-distance = sqrt{diagonal^2 - x-distance^2}
y-distance = sqrt{1 - x-distance^2}
I hope that this is what you were asking for.
Correct. So now you know the y-distance in terms of the x-distance. So we can write the height equation. It is:
So that equation determines height Z if the diagonal is one and x is known. Now, using your knowledge of differential calculus, can you determine the value of x such that Z is at its maximum? And,
once you know x, you can easily calculate the y-distance.
Is it correct?
y-distance = sqrt{1 - x-distance^2}=sqrt{1 - 0.45^2}=0.9
Is it correct?
y-distance = sqrt{1 - x-distance^2}=sqrt{1 - 0.45^2}=0.9
From what I can see of the calculations, it appears to be correct. Much of the image is truncated, however.
From what I can see of the calculations, it appears to be correct. Much of the image is truncated, however.
There are only two lines of calculations. First, derivative is found and then maximum value of z is found by setting the found derivative expression to zero.
It looks like that it's settled. Thank you so much!
Not open for further replies. | {"url":"https://www.electro-tech-online.com/threads/how-to-maximize-the-value-of-z.152221/","timestamp":"2024-11-08T04:43:43Z","content_type":"text/html","content_length":"154224","record_id":"<urn:uuid:93a12bbc-48ef-4f79-a560-ede6f81098b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00858.warc.gz"} |
The maximum $k$-colorable subgraph (M$k$CS) problem is to find an induced $k$-colorable subgraph with maximum cardinality in a given graph. This paper is an in-depth analysis of the M$k$CS problem
that considers various semidefinite programming relaxations including their theoretical and numerical comparisons. To simplify these relaxations we exploit the symmetry arising from permuting the
colors, … Read more | {"url":"https://optimization-online.org/author/j-c-veralizcano/","timestamp":"2024-11-13T21:27:37Z","content_type":"text/html","content_length":"103574","record_id":"<urn:uuid:723c6053-74c2-4dd1-b2ae-56b81b49417f>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00448.warc.gz"} |
Codac: constraint-programming for robotics
Codac (Catalog Of Domains And Contractors) is a C++/Python library providing tools for constraint programming over reals, trajectories and sets. It has many applications in state estimation or robot
What is constraint programming?
In this paradigm, users concentrate on the properties of a solution to be found (e.g. the pose of a robot, the location of a landmark) by stating constraints on the variables. Then, a solver performs
constraint propagation on the variables and provides a reliable set of feasible solutions corresponding to the problem. In this approach, the user concentrates on what is the problem instead of how
to solve it, thus leaving the computer dealing with the how.
What about mobile robotics?
In the field of robotics, complex problems such as non-linear state estimation, parameter estimation, delays, SLAM or kidnapped robot problems can be solved in a very few steps by using constraint
programming. Even though the Codac library is not meant to target only robotics problems, the design of its interface has been largely influenced by the needs of the above class of applications.
Codac provides solutions to deal with these problems, that are usually hardly solvable by conventional methods such as particle approaches or Kalman filters.
In a nutshell, Codac is a constraint programming framework providing tools to easily solve a wide range of problems.
• constraint-programming • dynamical-systems • state-estimation
• mobile robotics • tubes • SLAM
• interval-analysis • localization • solver
We only have to define domains for our variables and a set of contractors to implement our constraints. The core of Codac stands on a Contractor Network representing a solver.
In a few steps, a problem is solved by
1. Defining the initial domains (boxes, tubes) of our variables (vectors, trajectories)
2. Take contractors from a catalog of already existing operators, provided in the library
3. Add the contractors and domains to a Contractor Network
4. Let the Contractor Network solve the problem
5. Obtain a reliable set of feasible variables
For instance.
Let us consider the robotic problem of localization with range-only measurements. A robot is described by the state vector \(\mathbf{x}=\{x_1,x_2,\psi,\vartheta\}^\intercal\) depicting its position,
its heading and its speed. It evolves between three landmarks \(\mathbf{b}_1\), \(\mathbf{b}_2\), \(\mathbf{b}_3\) and measures distances \(y_i\) from these points.
The problem is defined by classical state equations:
\[\begin{split}\left\{ \begin{array}{l} \dot{\mathbf{x}}(t)=\mathbf{f}\big(\mathbf{x}(t),\mathbf{u}(t)\big)\\ y_i=g\big(\mathbf{x}(t_i),\mathbf{b}_i\big) \end{array}\right.\end{split}\]
where \(\mathbf{u}(t)\) is the input of the system, known with some uncertainties. \(\mathbf{f}\) and \(g\) are non-linear functions.
First step.
Defining domains for our variables.
We have three variables evolving with time: the trajectories \(\mathbf{x}(t)\), \(\mathbf{v}(t)=\dot{\mathbf{x}}(t)\), \(\mathbf{u}(t)\). We define three tubes to enclose them:
dt = 0.01 # timestep for tubes accuracy
tdomain = Interval(0, 3) # temporal limits [t_0,t_f]=[0,3]
x = TubeVector(tdomain, dt, 4) # 4d tube for state vectors
v = TubeVector(tdomain, dt, 4) # 4d tube for derivatives of the states
u = TubeVector(tdomain, dt, 2) # 2d tube for inputs of the system
float dt = 0.01; // timestep for tubes accuracy
Interval tdomain(0, 3); // temporal limits [t_0,t_f]=[0,3]
TubeVector x(tdomain, dt, 4); // 4d tube for state vectors
TubeVector v(tdomain, dt, 4); // 4d tube for derivatives of the states
TubeVector u(tdomain, dt, 2); // 2d tube for inputs of the system
We assume that we have measurements on the headings \(\psi(t)\) and the speeds \(\vartheta(t)\), with some bounded uncertainties defined by intervals \([e_\psi]=[-0.01,0.01]\), \([e_\vartheta]=
x[2] = Tube(measured_psi, dt).inflate(0.01) # measured_psi is a set of measurements
x[3] = Tube(measured_speed, dt).inflate(0.01)
x[2] = Tube(measured_psi, dt).inflate(0.01); // measured_psi is a set of measurements
x[3] = Tube(measured_speed, dt).inflate(0.01);
Finally, we define the domains for the three range-only observations \((t_i,y_i)\) and the position of the landmarks. The distances \(y_i\) are bounded by the interval \([e_y]=[-0.1,0.1]\).
e_y = Interval(-0.1,0.1)
y = [Interval(1.9+e_y), Interval(3.6+e_y), \ # set of range-only observations
b = [[8,3],[0,5],[-2,1]] # positions of the three 2d landmarks
t = [0.3, 1.5, 2.0] # times of measurements
Interval e_y(-0.1,0.1);
vector<Interval> y = {1.9+e_y, 3.6+e_y, 2.8+e_y}; // set of range-only observations
vector<Vector> b = {{8,3}, {0,5}, {-2,1}}; // positions of the three 2d landmarks
vector<double> t = {0.3, 1.5, 2.0}; // times of measurements
Second step.
Defining contractors to deal with the state equations.
The distance function \(g(\mathbf{x},\mathbf{b})\) between the robot and a landmark corresponds to the CtcDist contractor provided in the library. The evolution function \(\mathbf{f}(\mathbf{x},\
mathbf{u})=\big(x_4\cos(x_3),x_4\sin(x_3),u_1,u_2\big)\) can be handled by a custom-built contractor:
ctc_f = CtcFunction(
Function("v[4]", "x[4]", "u[2]",
"(v[0]-x[3]*cos(x[2]) ; v[1]-x[3]*sin(x[2]) ; v[2]-u[0] ; v[3]-u[1])"))
CtcFunction ctc_f(
Function("v[4]", "x[4]", "u[2]",
"(v[0]-x[3]*cos(x[2]) ; v[1]-x[3]*sin(x[2]) ; v[2]-u[0] ; v[3]-u[1])"));
Third step.
Adding the contractors to a network, together with there related domains, is as easy as:
cn = ContractorNetwork() # creating a network
cn.add(ctc_f, [v, x, u]) # adding the f constraint
for i in range (0,len(y)): # we add the observ. constraint for each range-only measurement
p = cn.create_interm_var(IntervalVector(4)) # intermed. variable (state at t_i)
# Distance constraint: relation between the state at t_i and the ith beacon position
cn.add(ctc.dist, [cn.subvector(p,0,1), b[i], y[i]])
# Eval constraint: relation between the state at t_i and all the states over [t_0,t_f]
cn.add(ctc.eval, [t[i], p, x, v])
ContractorNetwork cn; // creating a network
cn.add(ctc_f, {v, x, u}); // adding the f constraint
for(int i = 0 ; i < 3 ; i++) // we add the observ. constraint for each range-only measurement
IntervalVector& p = cn.create_interm_var(IntervalVector(4)); // intermed. variable (state at t_i)
// Distance constraint: relation between the state at t_i and the ith beacon position
cn.add(ctc::dist, {cn.subvector(p,0,1), b[i], y[i]});
// Eval constraint: relation between the state at t_i and all the states over [t_0,t_f]
cn.add(ctc::eval, {t[i], p, x, v});
Fourth step.
Solving the problem.
Fifth step.
Obtain a reliable set of feasible positions: a tube, depicted in blue. The three yellow robots illustrate the three instants of observation. The white line is the unknown truth.
You just solved a non-linear state-estimation without knowledge about initial condition.
In the tutorial and in the examples folder of this library, you will find more advanced problems such as Simultaneous Localization And Mapping (SLAM), data association problems or delayed systems.
Want to use Codac? The first thing to do is to install the library, or try it online:
Then you have two options: read the details about the features of Codac (domains, tubes, contractors, slices, and so on) or jump to the standalone tutorial about how to use Codac for mobile robotics,
with telling examples.
The following tutorial is standalone and tells about how to use Codac for mobile robotic applications, with telling examples:
We suggest the following BibTeX template to cite Codac in scientific discourse:
title={The {C}odac Library},
journal={Acta Cybernetica},
series = {Special Issue of {SWIM} 2022},
author={Rohou, Simon and Desrochers, Benoit and {Le Bars}, Fabrice}, | {"url":"https://codac.io/","timestamp":"2024-11-02T13:45:45Z","content_type":"text/html","content_length":"61923","record_id":"<urn:uuid:b3817efc-9ffe-4bd1-90b1-601864a22605>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00361.warc.gz"} |
Lifted Probabilistic Inference by Variable Elimination (Gelifte probabilistische inferentie door variabele eliminatie)
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Lifted Probabilistic Inference by Variable Elimination (Gelifte probabilistische inferentie door variabele eliminatie)
Boek - Dissertatie
Representing, learning, and reasoning about knowledge are central to artificial intelligence (AI). A long standing goal of AI is unifying logic and probability, to benefit from the strengths of both
formalisms. Probability theory allows us to represent and reason in uncertain domains, while first-order logic allows us to represent and reason about structured, relational domains. Many real-world
problems exhibit both uncertainty and structure, and thus can be more naturally represented with a combination of probabilistic and logical knowledge. This observation has led to the development of
probabilistic logical models (PLMs), which combine probabilistic models with elements of first-order logic, to succinctly capture uncertainty in structured, relational domains, e.g., social networks,
citation graphs, etc. While PLMs provide expressive representation formalisms, efficient inference is still a major challenge in these models, as they typically involve a large number of objects and
interactions among them. Among the various efforts to address this problem, a promising line of work is lifted probabilistic inference. Lifting attempts to improve the efficiency of inference by
exploiting the symmetries in the model. The basic principle of lifting is to perform an inference operation once for a whole group of interchangeable objects, instead of once per individual in the
group. Researchers have proposed lifted versions of various (propositional) probabilistic inference algorithms, and shown large speedups achieved by the lifted algorithms over their propositional
counterparts. In this dissertation, we make a number of novel contributions to lifted inference, mainly focusing on lifted variable elimination (LVE). First, we focus on constraint processing, which
is an integral part of lifted inference. Lifted inference algorithms are commonly tightly coupled to a specific constraint language. We bring more insight in LVE, by decoupling the operators from the
used constraint language. We define lifted inference operations so that they operate on the semantic level rather than on the syntactic level, making them language independent. Further, we show how
this flexibility allows us to improve the efficiency of inference, by enhancing LVE with a more powerful constraint representation. Second, we generalize the `lifting' tools used by LVE, by
introducing a number of novel lifted operators in this algorithm. We show how these operations allow LVE to exploit a broader range of symmetries, and thus expand the range of problems it can solve
in a lifted way. Third, we advance our theoretical understanding of lifted inference by providing the first completeness result for LVE. We prove that LVE is complete---always has a lifted
solution---for the fragment of 2-logvar models, a model class that can represent many useful relations in PLMs, such as (anti-)symmetry and homophily. This result also shows the importance of our
contributions to LVE, as we prove they are sufficient and necessary for LVE to achieve completeness. Fourth, we propose the structure of first-order decomposition trees (FO-dtrees), as a tool for
symbolically analyzing lifted inference solutions. We show how FO-dtrees can be used to characterize an LVE solution, in terms of a sequence of lifted operations. We further make a theoretical
analysis of the complexity of lifted inference based on a corresponding FO-dtree, which is valuable for finding and selecting among different lifted solutions. Finally, we present a pre-processing
method for speeding up (lifted) inference. Our goal with this method is to speed up inference in PLMs by restricting the computations to the requisite part of the model. For this, we build on the
Bayes-ball algorithm that identifies the requisite variables in a ground Bayesian network. We present a lifted version of Bayes-ball, which works with first-order Bayesian networks, and show how it
applies to lifted inference.
Jaar van publicatie:2013 | {"url":"https://www.researchportal.be/nl/publicatie/lifted-probabilistic-inference-variable-elimination-gelifte-probabilistische-inferentie","timestamp":"2024-11-06T02:30:42Z","content_type":"text/html","content_length":"66122","record_id":"<urn:uuid:adfd3bb4-36e3-4246-b189-f6e83ac9a92d>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00222.warc.gz"} |
1. Introduction2. Theoretical Results3. ConclusionCite this paperReferences
In order to analyse and connect the two aforementioned equations, following facts will be presented [1] :
・ Schrodinger equation is the cornerstone of quantum mechanics while Black-Scholes equation is the cornerstone of quantitative economics. The following facts are to be used:
・ If we have a quantum particle, its position is a random variable; at the same time the price of security is a random variable.
・ The Schrodinger equation demands a complex state function. At the same time, Black-Scholes is a real partial differential equation that always yields a real valued expression for the option price.
Black- Scholes equation is like a Schrodinger equation for imaginary time.
・ As Schrodinger equation demands a complex state function, the price of the option is analogous to the state function and the aforementioned state function requires a probabilistic interpretation.
On the contrary, the option price is directly observable and does not ask for probabilistic treatment.
・ At the same time, in that sense referring to probabilistic interpretation above, Schrodinger equation has to
satisfy the following condition required by the probabilistic interpretation in quantum mechanics
while at the same time the value of is arbitrary.
・ At the same time, we must refer to the Hamiltonians. All the Hamiltonians in quantum mechanics are Hermitian and therefore all eigenvalues are real. At the same time, Black-Scholes Hamiltonians
that affect option price are not Hermitian and this causes eigenvalues that are complex.
・ Complex eigenvalues of Hamiltonians that are obtained in finance lead to a more complicated analysis than one encountered in quantum mechanics, in particular according to Belal E. Baaquie there is
no well-defined procedure applicable to all Hamiltonians for choosing the set of functions that yield the completeness equation. The special cases where a similarity transformation leads to an
equivalent Hermitian Hamiltonian yields a natural choice for the set of complete eigenfunctions.
・ Schrodinger equation is time reversible and is an initial value problem and time evolution is given by while Black-Scholes process is time irreversible due to its Hamiltonian being non-Hermitian
and at the same time because pricing kernel is determined by the time-irreversible semi-group.
In order to analyse and see the similarity between Black-Scholes equation, it will be derived but in the formalism of quantum mechanics.
The Black-Scholes equation, but considering at the same time option price with constant volatility is given by:
If the change of variable is implemented:
Then, the Black-Scholes-Schrodinger equation is obtained [1] .
where the Black-Scholes Hamiltonian [2] is given by
If we view the Black-Scholes as a quantum mechanical system, it only has one degree of freedom, x, with following analogies with Schrodinger equation:
・ Volatility-inverse of mass.
・ Drift term-(velocity-dependent) potential.
・ Price of the option C-Schrodinger state function.
In order to see if Black-Scholes Hamiltonian is Hermitian or anti-Hermitian, we must define and derive the following equations.
It is known that a matrix M has a Hermitian conjugate defined by. The Hermitian conjugate on arbitrary operator O is given by [2] :
It is important to establish if Hamiltonian is Hermitian or anti-Hermitian because it is necessary to be aware of the space that an operator acts on, whether it acts on some N space or its dual
space. The difference in finance is important.
In order to analyse Black-Scholes and Schrodinger equation, the state space in quantum mechanics must be defined. It is important to explain completeness equation. Completeness equation refers to the
existence of basis vector so that any arbitrary vector can be represented as a linear combination of these basis states [1] .
For all the application that will be examined, particle moves on a “continuous” line R; each point on the continuous line is a possible state for the system and thus the aforementioned particle
requires continuously infinitely many independent basis vectors. However, the completeness equation of a two state is consequently to a N state system and then the limit of is taken.
In order to introduce completeness equation, firstly we will introduce the example from the Belal E. Baaquie [1] . He is considering an electron moving in a space, its position is denoted by x and it
can hop on the discrete points on a lattice, the points are given by.
The basis states are labeled by and can be represented by an infinite column vector with the only non-zero entry being unity in the nth position. So we have the following:
: completeness Equation (8)
where above is the infinite-dimensional unit matrix. As we want the movement on lattice not to be discrete than continuous, we have to introduce the limit of. The state vector for the particle is
given by the “ket” and its dual by the “bra” vector. This is where Hermitian of Hamiltonian plays the major role. Ket and bra vector are denoted respectively:
-“ket vector” and -“bra vector”
In terms of the, the following is defined:
where the scalar product is given by Dirac delta function:
The completeness equation is given by:
-the completeness Equation (11)
where is the identity operator on (function) state space.
The presented completeness equation is a key equation in the analysis of the state space. For the case of two quantum particles with positions on x, y the completeness equation is given by:
where. If we want to generalize the completeness equation to many particles, the following equation is obtained for three particles.
where. If we want to generalise the equation to n particles, the following result is obtained:
The two vectors represented by and can be mapped to each other. The aforementioned completeness equation gives the following:
At the same time, according to completeness equation, we must derive the Hermitian adjoint between and
its adjoint operator. The completeness equations presented above yield the following:
The differential operator is anti_Hermitian. Now we will analyse the co-ordinate operator:
The co-ordinate operator is Hermitian.
It was very important to introduce the following operators, as Hamiltonian operator, which is denoted by H, evolves the system in time and it is the most important operator in option pricing. The
operator is non-Hermitian and is defined in the following.
We have introduced the co-ordinate operator and differential operator. As Belal E. Baaquie [2] points out there are special eigenstates that are of particular importance for all operators. For the
co-ordinate operator, equation can be written as:
This vector is called an eigenstate of the co-ordinate operator with real eigenvalue x since is Hermitian.
According to Belal E. Baaquie, the eigenvalue equation for a non-Hermitian Hamiltonian H is given by a generalization of the the equation and the following equation. There exists special quantum
states which are called energy eigenstates with real energy eigenvalues that form a complete set of states and are given by:
By using the above equation, we obtain the following equations for eigenvalues and eigenfunctions:
The density of states for eigenvalue E is defined by, marked as:
The completeness equation is:
Hamiltonian operator is an operator
On the other side, if we want to write down the Schrodinger equation, we first need to specify the degrees of freedom of the system and at the same time it is necessary to specify the Hamiltonian H
of the system that describes the range of energy as well as form of energy the system can have.
The celebrated Schrodinger equation is given by
At the same time, we will be considering the quantum particle with m moving in one dimension in potential, the Schrodinger equation is given:
where the Hamiltonian operator acts on dual basis.
The Hamiltonian for the quantum particle moving in one dimension is given by:
If we now compare Schrodinger Hamiltonian and Black-Scholes Hamiltonian, we can conclude the following:
-Schrodinger Hamiltonian (26)
-Black-Scholes Hamiltonian (27)
Schrodinger potential is equal to and
Schrodinger is equal to.
After having demonstrated the similarity between Black-Scholes Hamiltonian and Schrodinger Hamiltonian, it can be proved that that the price of the option satisfies the (imaginary time) Schrodinger
equation [3] :
with the final value fixed by the payoff function as follows:
Comparing the wave function of quantum mechanics, the option price is directly observable, at the same time there is no concept of quantum measurement in option theory. The similarity of option
pricing with quantum mechanics, at this stage, is mathematical:
They can be described by an infinite-dimensional linear vector space and linear operators like H acting on vector space.
At the same time, we must assume that Hamiltonian has the following general form [2] :
where is an arbitrary function of x. It is the volatility of stock price and indicates the degree to which the evolution of stock price is random.
The famous Black-Scholes Hamiltonian is not Hermitian and it was derived above. It is important to be aware that this is the property of all the Hamiltonians in finance.
By having presented Hamiltonians, we would like to give the final Black-Scholes Schrodinger equation [4] :
In terms of the variable and time t, the Black-Scholes-Schrodinger equation for option pricing is given by [5] :
It was proved that by using Schrodinger equation and Black-Scholes Hamiltonian, the famous Black-Scholes equation can be derived. It appears in this form since variable S is the variable of choice in
the most literature in finance. This represents the basics of quantum social science [5] . | {"url":"https://www.scirp.org/xml/59468.xml","timestamp":"2024-11-05T19:10:31Z","content_type":"application/xml","content_length":"28612","record_id":"<urn:uuid:4fb094a5-1fa5-4d2f-a12e-af9f5d7f4331>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00210.warc.gz"} |
Rank-Nullity Theorem in Linear Algebra
This is a development version of this entry. It might change over time and is not stable. Please refer to release versions for citations.
In this contribution, we present some formalizations based on the HOL-Multivariate-Analysis session of Isabelle. Firstly, a generalization of several theorems of such library are presented. Secondly,
some definitions and proofs involving Linear Algebra and the four fundamental subspaces of a matrix are shown. Finally, we present a proof of the result known in Linear Algebra as the “Rank-Nullity
Theorem”, which states that, given any linear map f from a finite dimensional vector space V to a vector space W, then the dimension of V is equal to the dimension of the kernel of f (which is a
subspace of V) and the dimension of the range of f (which is a subspace of W). The proof presented here is based on the one given by Sheldon Axler in his book Linear Algebra Done Right. As a
corollary of the previous theorem, and taking advantage of the relationship between linear maps and matrices, we prove that, for every matrix A (which has associated a linear map between finite
dimensional vector spaces), the sum of its null space and its column space (which is equal to the range of the linear map) is equal to the number of columns of A.
July 14, 2014
Added some generalizations that allow us to formalize the Rank-Nullity Theorem over finite dimensional vector spaces, instead of over the more particular euclidean spaces. Updated abstract.
Session Rank_Nullity_Theorem | {"url":"https://devel.isa-afp.org/entries/Rank_Nullity_Theorem.html","timestamp":"2024-11-04T11:19:24Z","content_type":"text/html","content_length":"13998","record_id":"<urn:uuid:065da617-2580-4369-817e-796cc0530173>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00875.warc.gz"} |
reduced density matrix
Electronic structure calculations, in particular the computation of the ground state energy, lead to challenging problems in optimization. These problems are of enormous importance in quantum
chemistry for calculations of properties of solids and molecules. Minimization methods for computing the ground state energy can be developed by employing a variational approach, where the
second-order reduced … Read more
Large-scale semidefinite programs in electronic structure calculation
Employing the variational approach having the two-body reduced density matrix (RDM) as variables to compute the ground state energies of atomic-molecular systems has been a long time dream in
electronic structure theory in chemical physics/physical chemistry. Realization of the RDM approach has benefited greatly from recent developments in semidefinite programming (SDP). We present the
actual … Read more | {"url":"https://optimization-online.org/tag/reduced-density-matrix/","timestamp":"2024-11-12T07:08:28Z","content_type":"text/html","content_length":"86529","record_id":"<urn:uuid:16fd1429-9287-4417-93c8-51ee1614df5a>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00343.warc.gz"} |
Differential Equation (DE)
Definition 1: Independent and Dependent Variables and Parameters
The dependent variable(s) of a DE are the unknown functions that we want to solve for i.e. f(x), y(x, t), etc. The independent variable(s) of a DE are the variable(s) that the independent
variable(s) depend on i.e. x, t, etc. A parameter is a term that is an unknown but is not an independent or dependent variable i.e. a, b, α, β etc.
Definition 2: Order of a DE
The order of a DE is the order of the highest derivative.
Definition 3: ODEs and PDEs
A DE is an Ordinary Differential Equation (ODE) if it only contains ordinary derivatives (i.e. no partial derivatives).
A DE is a Partial Differential Equation (PDE) if it contains at least one partial derivative of a independent variable.
Definition 4: Linear and nonlinear DEs
A DE that contains no products of terms involving the dependent variable(s) is called linear.
If a DE is not linear then it is nonlinear. | {"url":"https://stevengong.co/notes/Differential-Equation","timestamp":"2024-11-09T02:58:16Z","content_type":"text/html","content_length":"14518","record_id":"<urn:uuid:7d813bf6-d524-46b6-a91a-5d932bfd2edd>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00627.warc.gz"} |
Ethyl acetate approach [CIELO] - Preparation - Welcome to the DMT-Nexus
Thanks Loveall.
I’ll do a side-by-side of std vs boiling water in near future.
Also will tinker with the Plan B approach.
HLP, when you have results and time, please post your method and comments.
Expecting to post results table tomorrow for 24-run experiment, in original post #568.
Loveall wrote:
Also, we have people reporting crystals for paste with densities anywhere from 48 to 60g per 1/4 cup, so my theory on paste consistency being a factor seems bunk.
We would need the density of paste that results in goo to know for sure whether density is a non-factor.
BTW, I wish I could play along, but I don't have my materials so not able to do any extractions right now. I'm actively following this thread though.
Is anyone actually getting crappy product for reasons other than learning curve? I’m all for fixing obvious problems, and making the tek as robust as possible, but, other than learning curve, what
exactly is the problem?
No disrespect intended, but, so far, based on my numerous, consecutive results, the standard tek has held up, in terms of yield and apparent quality, as good or better than all the variations and
deadend “fixes”.
I do believe that issues may arise from how the powder is produced (particle size, uniformity, moisture content, plant tissue composition, etc.). But, I would try to fix the root cause of these,
rather than futz with the actual extraction tek.
For now, I’ll be focused on cleaning the stuff generated by poor technique/learning curve, and on solvent re-use.
When I have enough of my own material, I’ll develop a powder prep tek for it that matches my current starting material.
Imo, we have a great standard tek! If we have a good cleanup method for problem runs, and can rehab and extend solvent life, seems to me we’ll have a simple, complete, effective, and high-purity
mescaline citrate extraction method.
But, if someone thinks these other efforts are really worthwhile, why not try to re-create the problem that is of concern. This would be my initial approach to solving it.
shroombee wrote:
Loveall wrote:
Also, we have people reporting crystals for paste with densities anywhere from 48 to 60g per 1/4 cup, so my theory on paste consistency being a factor seems bunk.
We would need the density of paste that results in goo to know for sure whether density is a non-factor.
BTW, I wish I could play along, but I don't have my materials so not able to do any extractions right now. I'm actively following this thread though.
I agree that we wound need to know the density of paste that resulted in goo to know for sure, it just seems unlikely since the large range 48 to 60g per 1/4 cup was OK.
Shroombee, what is your opinion on what causes goo for some? As an early pioneer you saw different results, interested in your current perspective. Is it paste MC%? Adherence to proper technique?
Starting material?
Reading back through the thread, not fully dried and not fully ground cactus could cause goo. Maybe the cactus powder needs to be very dry and very fine for the TEK to work, otherwise there could be
too much "free" water. Maybe a finer powder can grab onto water more efficiently than one with larger particles. Maybe the amount of surface-water contact matters along with MC%. Also, as has been
mentioned before, fully drying cactus material changes the solubility of some components (e.g. by denaturing proteins).
Or it could just be not adhering to the TEK steps. Or both could contribute. Or could be other reasons of course.
HLP, aside from your currently poor yields which sound like they are due to low alk material, how would you evaluate your recent results, how they have changed over time, and why do you think they
have changed?
Anyone else who has had yield or quality issues with product produced using this tek, please chime in, too.
Results posted for 24-run experiment,
see post #568
Loveall wrote:
Shroombee, what is your opinion on what causes goo for some? As an early pioneer you saw different results, interested in your current perspective. Is it paste MC%? Adherence to proper technique?
Starting material?
I don't have a guess. But your question inspired a thought... now that a group of us can consistently create xtals, how about we intentionally try to create goo by deviating one variable at a time
from a known good process that yields xtals?
Cheelin wrote:
Wow, what a tremendous amount of quality work 🙏
Essentially, the chilled approach and microwave approach lowered yields (repeating your previous result). Salting method had no major impact (new result). What about xtal size, any comments/
observations there?
The microwave option has been removed from the TEK, perhaps we should also remove the chilled paste option too?
9% MC on your cactus powder is interesting, definitely want to measure mine now. Perhaps we should control this variable, what if some has 20% MC? After controlling MC, yields would be more accurate
/standard, at best it could even make the process more robust or even break down chlorophyll and make EA easier to reuse (see table below from attached paper). A simple oven step at low heat (~200F)
for a few hours should do the trick. Like shroombee said, this MC% could swing the actual paste ratio as ~ 1/(1-MC).
shroombee wrote:
how about we intentionally try to create goo by deviating one variable at a time from a known good process that yields xtals?
I suggest we don't do this. There are a lot of things to change and even if we find goo, how do we know that was the same problem that an actual user had? Point is, there are lots of ways to break
the process, I think it is better to focus on making it more robust. I believe it is pretty robust already, but I would like to add simple controls if they are easy (e.g cactus powder MC) and remove
some color from EA so it is easier to reuse. Would also be nice to get product to not stick to the walls (e.g. boiling water paste).
Thanks Loveall.
I thought the first pic in the 24-run post would say it all about particle size, but my pics just don’t display here like they do on my phone. The passive method is like having a snowglobe with
small snowflakes in it; when filtered, the delicate snowflakes break into beautiful needles that sparkle in the light, The active stirring method is beautiful in it’s own way, with immediate
precipitation of very tiny particles that don’t trigger the thought of “crystals” imo, looking more like powdered sugar.
I know there are hcl bigots out there who claim that this tek precipitates citric acid, and I must admit that the active stir product’s appearance always triggers that thought in my mind. BUT, a
simple taste comparison between the the passive and active product simply shows that they are identical from a taste perspective, with absolutely zero perception of acid (wine research has
determined humans can detect flavor & perception differences down to the parts per trillion range). I challenge anyone who doubts my claim about zero perceptible acidity, to produce this tek’s
product and then taste it, followed by a taste of the most minuscule amount citric acid, and then honestly tell me that this tek, when properly done, produces citric acid precipitate of sufficient
quantity to be detected by the human tongue.
Regarding product sticking to the walls, my work shows this to be a trivial issue. It will be impossible to completely eliminate crystals sticking to the walls/bottom of the salting jar. Even the
active stirring leaves “crystals” adhered to the wall, albeit smaller and fewer of them than the passive option. A way to minimize the adhered crystals left in the jar and ending up in the evap dish
is as follows:
1. Use a Wide-Mouth jar/beaker for salting/crystallization, so that there is no shoulder in the jar to entrap loose crystals
2, The jar/beaker should have a volume 2X the volume of salting solution, so that adequate swirling can occur before pouring the solution through filter
3. Begin the filtering process with vigorous swirling of the solution, then pour approx 3/4 of solution into filter, re-swirl remaining solution vigorously, and in one quick motion dump the contents
in the filter
4. Using a pipette or syringe, take 10-20mL (for 1/2 pint of solution, more if a bigger load) of Fresh EA, from the top of the j/b drain onto & circling the side walls, swirl vigorously and dump,
repeat once and let j/b dry.
5. After j/b is dry, boil a little distilled water, follow step 4 but substituting the boiling water for the EA. After swirling, but before dumping (this time into glass evaporating dish), tilt j/b
and turn so that the warm water briefly covers all the walls of the jar,
The above process will minimize the amount of evaporation needed and residue produced . Take a look at the splits between product that ends up in the filter and evap residue, in my results tables.
The residue amounts are tiny.
As regards the wiki: if it were my tek, i would drop the description of all variations from the wiki, so that the wiki shows the “standard tek”, along with 5g citric/100g material passive and active
salting/crystallization options, a problem product cleaning method, and a solvent reclaim method. I would add a comment discussing the importance of proper technique, which for most people new to
extraction involves a learning curve. I would also add a comment directing all those who can not produce the proper end product, to this thread for possible variations of the standard tek, and help.
As far as work on making the tek more robust, i’d limit future work on the wiki to 3 topics: specifying powder characteristics to some reasonable extent; developing a brown, sticky product cleaning
method; and determining the maximum number or effective solvent reuses (and the method).
All other work to improve the tek would be addressed, on this thread, and only when the problem can be recreated by someone other than the poster who has it. An appendix on the wiki would list
verified problems and their solution.
One additional issue to consider: while I understand, appreciate, and agree with this sites’s reluctance to allow easy membership, most of the issues producing the proper product from this tek will
happen to newbies who can’t post.
New results:
- Oven dried 100g of power at 175C for a few hours. After two hours the weight stopped changing at 92g. Therefore my MC% has been 8%.
- Proceeded with the standard TEK (added 308g of water to compensate).
- The extract is dark green. At this oven temp, the color of EA is not improved as far as I can see
- EA is currently xtalizing, will report on yields, I expect no change/no issues.
- Conclusion: I was able to measure my MC% and it was very close to Cheelin's. We still get different paste consistencies (mine looks looser than his). No issues with either of us getting xtals
I think the TEK is robust. Will be interested in anyone else getting goo consistently instead of xtals and their plant/prices details. As far as I know, everyone is getting xtals at this point.
Cheeling, I get the same results you do when it comes to xtalization. Dropping 5g acid gives big xtals, stirring 15g of acid gives a fine powder that tis much denser. Both are mescaline citrate with
same yields. If dropped as little as 0.8g of acid and the extract fully xtalized and was slightly acidic with pH paper. Will keep the TEK as is, but the process window is very large - which is good.
I've also squeezed the french press hard and had no issues xtalizing. After squeezing, I broke up the compacted paste and some solvent came loose which can be squeezed out. The extra drops don't
really matter, same yield after 6 pulls as the standard method.
If a lot of xtals stick to the wall hitting them with a butter knife or spoon knocks them off the wall.
We've kicked the tires of the new TEK pretty hard and it seems robust, giving similar answers. Cheelin noticed a slight drop in yield with the microwave and cold extract method so those are now gone
from the TEK. To get EA that is not extremely dark green and may be easier to reuse I added the boiling water option, that really reduces the color in the extract and does not seem to reduce yields
(would be interested in someone's check on this, maybe careful measurents do shoe a drop, IDK).
Newbies can post in the welcome area and ask about the CIELO Tek there. That seems to work (e.g. HLP).
Awesome work Cheelin.
So looking at those results the chilled EA killed yields more than anything else. Where as the microwave method without chilled EA was almost on par (within 10% yield from your best run). And same
average yield as a few other runs.
So Loveall I would ask for the purpose of solvent reuse, is it necessary to completely get rid of the microwave step from the TEK? Or does a brine wash/ hot water paste correct this?
I would argue getting rid of chilled EA is more important.
Regardless I would like to add one more variable and that is hot EA pulls, I've seen better yields doing hot pulls but crystal color suffers, however this was with 15-20mg salting. Color may be
corrected with 5mg. Thoughts?
Disclaimer: All my posts are of total fiction.
Bufotenine Benzoate
I think microwave yields are lower looking at Cheelin's data. The combined (microwave + chilled) was the lowest, making the signal seem real. Maybe we need to calculate a p value 😅
Using boiling water yielded fine for me (but also I did not notice a significant difference with the microwave, However my data was not as thorough as Cheelin's). Boiling water is a lot simpler than
the microwave. Also, not everyone has a microwave. So I replaced the microwave option with boiling water to keep the EA a light yellow/green and easier to reuse.
Your heat comment is interesting. Maybe next time I will not wait for the boiled water paste to cool and pull after 15 minutes, while heat is present. Not sure how long the chlorophyll needs to
break down, but if it is within this time frame maybe we can get the best of both worlds all at one for a quick, light colored, and high yielding extract.
I would be keen to try the boiling water instead of a microwave next run, would simplify it. I've used a hot water bath for the EA and french press in the past, this seemed to work well.
Disclaimer: All my posts are of total fiction.
Bufotenine Benzoate
Thanks _Trip_.
Loveall, I plan to do a boiling water version of the bulk wet mix std process, with a set of 3 runs similar to those in the 24-run expmt and same material, hopefully tomorrow/Friday.
If heat during the pulls leads to higher yields than the std process in someone else’s runs, i’ll consider using the materials, and doing the extra work for a 3-run set of this variation, later.
I’m very leery of doing any more runs of treatments that don’t meet or exceed std process yields, especially just to get lighter colored solvent. I have yet to see anyone demonstrate that clear
neutral yellow solvent adds any value over clear neutral green solvent.
I intend to test yellow vs green solvent reuse in multiple consecutive runs soon.
I'm pretty sure we understand the mescaline citrate salt form: (MesH)H2Cit thanks to data presented by several people.
I'm going to do a titration experiment to add to this data. Also, to make sure the salt form doesn't change under different pH conditions.
For each 1g of mescaline freebase (Mes), assumimg that to first order that is the major component being titrated,
For (MesH)H2Cit: 910mg of citric are needed to produce 1.910g of xtals
For 2(MesH)HCit: 455mg of citric are needed to produce 1.455g of xtals
For 3(MesH)Cit : 303mg of citric are needed to produce 1.303g of xtals
So by keeping track of the pH as citric acid is being added, and stopping when pH paper becomes neutral, and knowing the final yield, the salt form can be known. I typically get 1.2% yield with my
current powder, so I would need ~571mg of citric acid to neutralize the extract if the salt form is indeed (MesH)H2Cit. I can check at/above 190mg and 286mg to verify the extract is still alkaline
on pH paper.
I also wonder if adding 190mg of citric and then giving time for xtals for form would produce the 3(MesH)Cit form and neutralize the solution over time.
👍🏻 Loveall!
Tested a 45g quantity of material, bulk prep’d & wet mixed, using std process, subdivided into 3 portions. using same material used in 24-run experiment. The only difference between this treatment
and (B,0,0,0,1) is that the water used to make the paste was boiled before mixing with the pickling lime.
During the 8-minute mix, the paste was noticeably more liquidy than my normal paste, looking more like pudding than playdough. After the 10 minute rest, the paste was much more rubbery than usual,
looking more like silly putty that the usual rubbery version of playdough.
When pulling, the paste remained rubbery and clumpy throughout all 6 pulls, rather than becoming applesauce appearance by the end of the final pull. Unlike pulling the paste made from ambient temp
water, the 1st pull hardly absorbed any solvent, and although not measured, the volume of solution from each pull looked roughly equal. The color of the 1st pull was slightly darker than fresh
solvent, with each subsequent pull taking on a slightly darker shade of yellowish-green, combined pulls look the ligjtest color of any treatment yet.
The 3 runs were salted/crystalized the same way as each 3-some in the 24-run exprmt, results are in the attached table.
Additionally, three other sets of treatments were run: a 3-run test without the 1-hour settling period after the final pull; a 4-run, 2-way test of the active stirring option, using 5-minute or
10-minute magnetic stir with 24-hr rest, at the equivalent of 5 or 15g citric acid per 100g powdered material; and an 8-run, 2-way test of 20-minute magnetic stir with 4 different rest periods
before filtering (1hr, 3hr, 6hr, 12hr), at the equivalent of 5 or 15g citric acid per 100g powdered material.
All of these runs were done with the same material used in the 24-run experiment, using wet mix standard paste. The 3-run test used 45g of powder, bulk prepared, split into three portions; the 4-run
test used 60g of powder, bulk prepared, split into 4 portions; the 8-run test used two batches x 60g of powder, bulk prepared, split into 4 portions each.
Results of these runs are in the attached table.
Main Points:
1. All passively crystallized runs produced white needles and minimal, white, jar wash residue; all magnetic stir runs, regardless of citric acid option, produced white powder and minimal, white,
jar wash residue; all runs tasted identical, moderately bitter, no acidity.
2. Using boiled water, produces a paste that is more difficult to extract and reduces yield (roughly equivalent to the microwave treatment (B,0,1,0,1) yield). The mixing and/or pulling processes can
likely be adjusted to improve yield, but will require more effort than just using ambient temp water.
3. Eliminating the 1-hour settling period after the final pull, slightly reduces yield.
4. Magnetic stir period slightly affects yields: high to low yield ranking is 10-min, 5-min, 20-min.
5. Both citric acid levels produce similar yield for a given stir method, indicating that there is no need to use more than 5g citric acid per 100g of powdered material.
6. When using magnetic stirring, yield roughly peaks 3 hours after stirring.
Cheelin attached the following image(s):
I noticed the boiling water paste congeal also so I added some boiling water before pulling to loosen it up (25ml for a paste with 100g of cactus). I wanted EA to penetrate well that's why I broke
it up with fresh boiling water right before pulling. I don't think it was enough to raise the temp significantly, just enough to make a loser paste so it would be easier for EA to get in there.
I've started to go by paste feel. Looser paste with a little bit more water seems easier to pull. More compact paste is harder to pull, and yields can be lower (that is what I got with the failed
"min water" paste). No issues with goo or xtalization when I add a little more water to get the paste I want.
I worry your boiling water run may yield less because the paste was so compact. We will see. I think that the xtals that form may be larger than the standard TEK. They may form faster too. curious
to see what you get.
I've also started squeezing with the french press. After squeezing, I break up the compressed mass with a knife and a little more EA comes out, so I squeeze that out again, and repeat of needed.
Only takes a few seconds each time at the end of a pull. Sakkadelic would be proud. I think I'm getting a tiny bit more yield this way (1.3% instead of 1.2%). No crystalization issues and I've
squeezed pretty hard. We were being cautious about this when people where getting goo to avoid water and recommending only lightly squeezing or no squeezing at all, but it doesn't seem that was the
cause of some people seeing goo.
After not seeing water/particles for a few consecutive runs, I've stopped waiting for an hour. I would still recommend it for newcomers. Xtalization happens the same as when waiting (yields, etc).
Overall, the TEK is being extremely robust for me, even when I push the parameters around. Over time a feel for it develops, especially for what a good water paste consistency is.
Looking forward to your results Cheelin 🙂
Like I have said, Loveall, you nailed it, this tek would make Shulgin proud!
I think the paste making and pulling/filtering parts of the tek, as well as passive salting/crystallization are good enuff for my purposes, with my efforts hitting the point of diminishing returns.
Looking now to minimize time on the active salting/crystallization method for those who want the fastest process without yield loss. I got a little carried away and did a few more runs today than
planned, lol. I’ll show results in the earlier post, when done. | {"url":"https://www.dmt-nexus.me/forum/default.aspx?g=posts&m=1131045","timestamp":"2024-11-03T22:23:45Z","content_type":"application/xhtml+xml","content_length":"220193","record_id":"<urn:uuid:3ff526af-ebb1-450e-a1e0-00572bbcc724>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00343.warc.gz"} |
What is Schrödinger's Equation?
November 2021
In classical mechanics, arguably the most important equation is Newton's famous $$F=ma$$ It's so simple, it almost feels silly putting it on a separate line in the middle of the screen. But it
deserves the spot, because it's so central. With this equation, you can take information about a classical mechanics system and figure out how it will change for all time.
In quantum mechanics, there is a similar equation used to predict the behavior of quantum systems. It is Schrödinger's famous
$$i \hbar \frac{\partial}{\partial t} \Psi(\mathbf{r}, t) = - \frac{\hbar^2}{2m} \nabla^2 \Psi(\mathbf{r}, t) + V(\mathbf{r}, t) \Psi(\mathbf{r}, t)$$
Putting this on a separate line doesn't feel silly at all. Surprisingly, however, this equation isn't too far off from the principles of classical mechanics and \(F=ma\). Today I am going to show you
how to understand this equation a little better using classical physics and lots of math. You will need classical mechanics and calculus, but no quantum background. There are just a few facts about
quantum physics you need to accept first:
• Quantum entities can have energy from their frequencies according to the equation \(E=hf\), where \(E\) is energy, \(f\) is frequency, and \(h\) is a constant called Planck's constant.
• The wavelength \(\lambda\) of a quantum entity depends on its momentum \(p\) according to the de Broglie equation, \(\lambda = h/p\). You can actually derive this quite easily from \(E=hf\) and \
• With quantum mechanics, we don't look at the exact positions or velocities of particles. Instead, we use something called a wavefunction. With wavefunctions and some calculus, we can find the
probability distributions of position, momentum, and more, but not certain values.
It's important to note that this does not represent a true derivation: it would be easier to just accept Schrödinger's equation directly rather than accept the above facts and go through all this
math. However, hopefully this will give some intuition into Schrödinger's equation and the math behind it.
Classical Basis
You might say “Schrödinger's equation looks nothing like Newton's equation, how can they be analogous?” This is a fair point. In fact, Schrödinger's equation is a little more analogous to
conservation of energy. (More precisely, it's based on a Hamiltonian, not a true conservation of energy equation.) The equation for classical conservation of energy, where we'll start, is $$E=K+V$$
Where \(K\) is kinetic energy, \(V\) is potential energy, and \(E\) is total energy. Doesn't this already look a little like Schrödinger's equation? We have one term on the left, and it's the sum of
two terms on the right.
From here, we're going to modify this equation step by step until we end up with Schrödinger's equation. First of all, we could write \(K\) in terms of mass \(m\) and velocity \(v\). $$E= \frac{1}{2}
mv^2 +V$$ It turns out that in quantum mechanics, the momentum \(p=mv\) will be more helpful to us than the velocity. Luckily, we can write \(K\) in terms of mass and momentum as well.
$$K = \frac{mv^2}{2} = \frac{m^2v^2}{2m} = \frac{(mv)^2}{2m} = \frac{p^2}{2m}$$
$$E= \frac{p^2}{2m} + V$$ Now we are going to transition from classical mechanics to quantum mechanics. To do this, we are going to multiply by the quantum wavefunction \(\Psi\) on both sides, just
to get it into our equation. $$\boxed{E \Psi= \frac{p^2}{2m} \Psi + V \Psi}$$ But at this point, our equation is a bad mix of classical and quantum mechanics that doesn't really make sense. We said
that we don't deal with exact values of momentum in quantum mechanics, only probability distributions with our wavefunction \(\Psi\). But here we have \(p\) and \(\Psi\) in the same equation, as if
we knew exactly what the momentum \(p\) was.
We might not know what \(p\) is, but it turns out we can change \(\frac{p^2}{2m} \Psi\) to something in terms of \(\Psi\) and things we do know, so we can deal with probabilities like we're supposed
We don't really know what \(\Psi\) is, since we are keeping it general, but we can write it in general terms. How about this: $$\boxed{\Psi = Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t}}$$ Why is
only time negative?
This is a tough question, and I couldn't find a satisfactory answer online, but here's one way I found to think about it. From multivariable calculus, we have $$\frac{dx}{dt} = - \frac{\partial \Psi/
\partial t}{\partial \Psi/\partial x}$$ If we had \(e^{i \omega t}\) instead of \(e^{-i \omega t}\), this would become $$\frac{dx}{dt} = - \frac{i \omega}{i k_x} = - \frac{\omega}{k_x}$$ Later, we'll
find that \(\omega = 2 \pi f\) and \(k_x = 2 \pi p_x/h\). This gives $$\frac{dx}{dt} = \frac{p_x}{m} = - \frac{2 \pi f}{2 \pi p_x/h} = - \frac{hf}{p_x}$$ $$\frac{p_x^2}{m} = -hf$$ But this means
frequency is negative, or momentum is imaginary, obviously both making no sense. Therefore, we must have opposite signs for space and time.
This equation doesn't really tell us much about \(\Psi\). We have no idea what the values of these variables are. That's actually a good thing, because we don't have information about \(\Psi\), so we
don't want to pretend like we do and make stuff up.
All we are saying is that \(\Psi\) is some number \(A\) times something like \(e^{i \theta}\) a bunch of times. Each \(e^{i \theta}\) term shows that the wave function depends on something in some
way: \(x\), \(y\), and \(z\) for position in three dimensions, and \(t\) for time. In other words, our equation translated to English is just saying “the wavefunction depends on space and time in
some way.”
There is one more thing the equation is saying. \(e^{i \theta}\) is an oscillating function, so our wavefunction will oscillate like a wave.
How do we know it's a wave of this form? Well, actually we don't. But it turns out that if you have some solutions to the Schrödinger equation, their sum (more precisely, their linear combination)
will also be a solution. Also, there is something called the Fourier transform which says that you can write any function as a sum of sine and cosine functions (waves). Putting these two ideas
together, if we can derive the Schrödinger equation for a general wave, we can add waves together to make whatever other function we want. This sum will also be a solution since it's the sum of
individual solutions.
Kinetic Energy
So we have a wavefunction, and it's a wave. We might be interested in the wavelength (in space) and frequency (in time) of the wave. If we have \(e^{i (a) \theta}\), the “wavelength” would be \(2 \pi
/a\), since we make a full circle back to \(e^{0i}=e^{2\pi i}\) once \(\theta\) reaches \(2\pi/a\). That means for \(e^{i (k_x) x}\), our wavelength \(\lambda_x\) is \(2 \pi/k_x\). The same idea
applies to \(k_y\) and \(k_z\), for the wavelengths in the \(y\) and \(z\) directions.
But wait, remember the de Broglie equation? $$\lambda = \frac{h}{p}$$ This means that if we have the wavelength for each direction, we can easily find the momentum in that direction. $$p = \frac{h}{\
lambda}$$ $$p_x = \frac{hk_x}{2 \pi}, \ p_y = \frac{hk_y}{2 \pi}, \ p_z = \frac{hk_z}{2 \pi}$$ Let's define a new constant, \(\hbar = h/2\pi\), just to clean things up a little. $$\boxed{p_x = \hbar
k_x, \ p_y = \hbar k_y, \ p_z = \hbar k_z}$$ There's one problem: we have no idea what all these \(k\) values are. I made them up when we wrote a general equation for \(\Psi\). But something
interesting happens if we take the second derivative of \(\Psi\). We find $$\boxed{p_x^2 \Psi = - \hbar^2 \frac{\partial^2 \Psi}{\partial x^2}}$$ Proof $$\Psi = Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \
omega t}$$
$$\frac{\partial^2 \Psi}{\partial x^2} = (ik_x)^2 Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t} = (ik_x)^2 \Psi = -k_x^2 \Psi$$
$$\Psi = - \frac{1}{k_x^2} \frac{\partial^2 \Psi}{\partial x^2}$$ $$p_x = \hbar k_x \implies \frac{1}{k_x} = \frac{\hbar}{p_x}$$ $$\Psi = - \frac{1}{k_x^2} \frac{\partial^2 \Psi}{\partial x^2} = - \
frac{\hbar^2}{p_x^2} \frac{\partial^2 \Psi}{\partial x^2}$$ $$\boxed{p_x^2 \Psi = - \hbar^2 \frac{\partial^2 \Psi}{\partial x^2}}$$
For the \(y\) and \(z\) components of momentum, we'll have almost the same equation, just replace \(x\) with the new letter. For the total momentum, we have to add the squares of each component: $$p^
2 = p_x^2 + p_y^2 + p_z^2$$ Now we can divide by \(2m\) and multiply by \(\Psi\) on both sides, then plug in the equation for momentum in each component with the second derivatives.
$$\frac{p^2}{2m} \Psi = - \frac{\hbar^2}{2m} \bigg( \frac{\partial^2 \Psi}{\partial x^2} + \frac{\partial^2 \Psi}{\partial y^2} + \frac{\partial^2 \Psi}{\partial z^2} \bigg)$$
If you know multivariable calculus, you might recognize the Laplacian operator, \(\nabla^2\), in here. If not, just consider \(\nabla^2\) to be a special abbreviation for the sum of all these second
derivatives. $$\frac{p^2}{2m} \Psi = - \frac{\hbar^2}{2m} \nabla^2 \Psi$$ Remember earlier when we were working with classical mechanics? We said that \(p^2/2m\) was kinetic energy. In our new
quantum formula, we are using \(- \frac{\hbar^2}{2m} \nabla^2\) on our wavefunction to get the term corresponding to kinetic energy. Therefore, we say that \(- \frac{\hbar^2}{2m} \nabla^2\) is the
for kinetic energy. Now let's look back at our old equation, where we just took a classical equation and multiplied by \(\Psi\). $$E \Psi= \frac{p^2}{2m} \Psi + V \Psi$$ Now with our operator, we are
one step closer to the Schrödinger equation. $$\boxed{E \Psi= - \frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi}$$ In fact, what we have now is already a valid form of the Schrödinger equation if we know
our value for the energy \(E\). But we can get even more information with more math.
Total Energy
Remember I said it might be interesting to know the wavelength and frequency of the wavefunction? We tried finding the wavelength, and ended up coming much closer to the Schrödinger equation. But
let's not forget the frequency! Now let's find the frequency and hope we come even closer. Let's go back to the general wavefunction. $$\Psi = Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t}$$ The
frequency in time will be based on the \(e^{-i \omega t}\) term. Specifically, the frequency will be \(f=\omega/2\pi\), for similar reasons as the wavelength \(2\pi/k\). Let's see what happens if we
take the first derivative with respect to time.
$$\frac{\partial \Psi}{\partial t} = (-i \omega) Ae^{ik_xx} e^{ik_yy} e^{ik_zz} e^{-i \omega t} = (-i \omega) \Psi$$
$$f=\omega/2\pi \implies \frac{\partial \Psi}{\partial t} = (-i \cdot 2\pi f) \Psi$$
But wait, remember \(E=hf\)? That means if we have the frequency, we can easily find the energy. If we plug this into our earlier Schrödinger equation, this gives us $$\boxed{i \hbar \frac{\partial \
Psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi}$$ There, that's the full Schrödinger equation!
Show steps We just had $$\frac{\partial \Psi}{\partial t} = (-i \cdot 2\pi f) \Psi$$ $$f = \frac{E}{h} \implies \frac{\partial \Psi}{\partial t} = (-i \cdot 2\pi E/h) \Psi$$ Solve for \(E \Psi\). $$E
\Psi = -\frac{h}{2\pi i} \frac{\partial \Psi}{\partial t} = i \frac{h}{2 \pi} \frac{\partial \Psi}{\partial t}$$ Remember we defined \(\hbar = h/2\pi\). $$E \Psi = i \hbar \frac{\partial \Psi}{\
partial t}$$ Let's plug this back into our equation. $$E \Psi= - \frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi$$ $$\boxed{i \hbar \frac{\partial \Psi}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2 \Psi + V
When I presented the Schrödinger equation at the beginning of this post, I did one more step to make it a little extra scary. We can consider position, which we can write as a vector \(\mathbf{r}\).
The wavefunction and potential energy can then be functions of both position \(\mathbf{r}\) and time \(t\). That gives
$$i \hbar \frac{\partial}{\partial t} \Psi(\mathbf{r}, t) = - \frac{\hbar^2}{2m} \nabla^2 \Psi(\mathbf{r}, t) + V(\mathbf{r}, t) \Psi(\mathbf{r}, t)$$
What is Schrödinger's Equation?
So far, we've shown how to find Schrödinger's equation with classical mechanics, but I haven't really explained what the equation is like I promised in the title. Here are some thoughts about what it
all means.
As a summary of what Schrödinger's equation is, you can think of it as a statement of conservation of energy in quantum mechanics. One big difference is that it is a probabilistic equation, since it
tells you about the wavefunction. The wavefunction can help you predict what a particle is doing, but you can never be completely sure. In classical mechanics, you can be completely sure, at least if
your model is right.
It's interesting to see that we played a lot with the total energy \(E\) and the kinetic energy \(K\) in the transition to quantum mechanics, but the potential energy \(V\) is still just written as \
(V\). This makes some sense, since \(V\) really depends on the situation, while \(E\) and \(K\) are properties of the particle itself.
You must be wondering, why is the Schrödinger equation so much more complicated than \(E=K+V\)? Well, it doesn't have to be. It just is that way because it's more explicit about the quantities we
need. If you want the simple version, you can write $$E \Psi = \hat{H} \Psi$$ Where \(E\) is energy, and \(\hat{H}\) is called the Hamiltonian operator and equals \(- \frac{\hbar^2}{2m} \nabla^2 + V
\). These forms are equivalent, since we earlier proved that \(E \Psi = i \hbar \ \partial \Psi/\partial t\).
It seems like conservation of energy has a simple version in both classical physics and quantum physics. We just did a lot of work with a complicated analog of conservation of energy in quantum
phyiscs. Is there an analog to this in classical physics?
In general, for a conservative system, the Hamiltonian represents the sum of kinetic and potential energy. So in classical physics, we have $$E = K+V = H$$ $$E = \frac{p^2}{2m}+V$$ From here, we
can't go further without more information, so we could say that this is the classical analog of the complicated Schrödinger equation.
But if we know more about the situation, we can make this more complicated. Maybe we know the potential energy is from gravity, and maybe we know the initial energy was all gravitational potential
energy. Then we can have
$$gh_0 = \frac{1}{2}\bigg[\bigg(\frac{dx}{dt}\bigg)^2+ \bigg(\frac{dy}{dt}\bigg)^2+\bigg(\frac{dz}{dt}\bigg)^2\bigg] + gz$$
Proof $$E = \frac{p^2}{2m}+V$$ $$mgh_0 = \frac{1}{2m}p^2 + mgz$$ $$mgh_0 = \frac{1}{2m}(p_x^2+p_y^2+p_z^2) + mgz$$ $$mgh_0 = \frac{1}{2m}(m^2)(v_x^2+v_y^2+v_z^2) + mgz$$ $$mgh_0 = \frac{m}{2}(v_x^
2+v_y^2+v_z^2) + mgz$$
$$gh_0 = \frac{1}{2}\bigg[\bigg(\frac{dx}{dt}\bigg)^2+ \bigg(\frac{dy}{dt}\bigg)^2+\bigg(\frac{dz}{dt}\bigg)^2\bigg] + gz$$
This is somewhat complicated, like the more complicated version of the Schrödinger equation. It also has derivatives, which can help us find how the system will change with time. The moral of the
story is that given the basic idea of the Hamiltonian \(E=H\), and some information specific to the situation, we can plug in that information to get a more complicated but more useful equation. It's
interesting how this works for both quantum and classical physics.
The Schrödinger equation looks scary, and it is a little scary, but it's also meaningful. Just like conservation of energy in the Hamiltonian form can help us tell what will happen in classical
mechanics, the Schrödinger equation tells us what will happen to a wavefunction in quantum mechanics.
There's a lot more to this, like how exactly we use wavefunctions, and what a wavefunction is. There are many questions that still don't have answers, like what it means that all this seems
probability-based. But considering that this equation is at the heart of our universe (until someone finds a better one that explains quantum gravity or something) it's interesting to know that it's
related to classical conservation of energy, which even introductory physics students know about.
a. Quantum Physics I (B. Zwiebach, MIT)
b. Hamilton’s Equations of Motion (Jeremy Tatum, University of Victoria) | {"url":"https://www.harysdalvi.com/blog/2111","timestamp":"2024-11-13T19:24:02Z","content_type":"text/html","content_length":"19564","record_id":"<urn:uuid:026c9a2f-ed75-42f1-9853-4c3a0d352f60>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00002.warc.gz"} |
Deep Learning: Python
Python Language
1. Expand Menus ↓
Control Flow
Data Structures
Machine Learning
Exception Handling
Deep Learning in Python:
Deep learning is a subset of machine learning that utilizes artificial neural networks with multiple layers (hence "deep" learning) to learn intricate patterns and representations from data. It aims
to mimic the way the human brain works, where each layer of neurons processes information and extracts features before passing it on to the next layer. Deep learning algorithms can automatically
discover features from raw data without the need for manual feature extraction, which is a significant advantage compared to traditional machine learning techniques.
• Key components and concepts:
1. Neural Networks: Deep learning models are typically constructed using artificial neural networks, which are computational models inspired by the structure and function of biological neurons in the
human brain. Neural networks consist of interconnected layers of nodes (neurons) that process input data and produce output predictions.
2. Deep Architectures: Deep learning models contain multiple layers of neurons, allowing them to learn hierarchical representations of data. These deep architectures enable the model to extract
increasingly abstract features as information flows through successive layers.
3. Learning Representations: Deep learning algorithms learn representations of data through a process called feature learning or representation learning. By iteratively adjusting the parameters of
the neural network based on observed data (e.g., using gradient descent optimization), the model learns to automatically discover useful features and patterns from the input data.
4. Training with Backpropagation: Deep learning models are trained using an optimization algorithm called backpropagation. Backpropagation involves computing gradients of a loss function with respect
to the model's parameters, and then updating the parameters in the direction that minimizes the loss. This process allows the model to learn from its mistakes and improve its predictions over time.
5. Convolutional Neural Networks (CNNs): CNNs are a type of deep learning architecture commonly used for image recognition and computer vision tasks. They consist of multiple layers of convolutional
and pooling operations, which are specialized for extracting spatial hierarchies of features from image data.
6. Recurrent Neural Networks (RNNs): RNNs are another type of deep learning architecture designed for sequential data processing tasks, such as natural language processing and time series analysis.
RNNs have connections that form directed cycles, allowing them to maintain a memory of past inputs and make decisions based on sequential information.
7. Applications: Deep learning has been applied to a wide range of domains, including image and speech recognition, natural language processing, autonomous vehicles, medical diagnosis, and more. Its
ability to learn complex patterns from large-scale datasets has led to significant advancements in various fields.
Conclusion: Deep learning is a powerful and versatile approach to machine learning that has revolutionized the field by enabling computers to learn directly from data and solve complex tasks with
unprecedented accuracy and efficiency.
Let's create a simple feedforward neural network to classify handwritten digits from the MNIST dataset. We'll cover concepts such as model definition, data loading, training loop, loss function, and
import torch
import torch.nn as nn
import torch.optim as optim
import torchvision
import torchvision.transforms as transforms
# Define the neural network model
class NeuralNetwork(nn.Module):
def __init__(self, input_size, hidden_size, num_classes):
super(NeuralNetwork, self).__init__()
self.fc1 = nn.Linear(input_size, hidden_size)
self.relu = nn.ReLU()
self.fc2 = nn.Linear(hidden_size, num_classes)
def forward(self, x):
out = self.fc1(x)
out = self.relu(out)
out = self.fc2(out)
return out
# Hyperparameters
input_size = 784 # 28x28 pixels
hidden_size = 128
num_classes = 10
learning_rate = 0.001
batch_size = 100
num_epochs = 5
# Load the MNIST dataset
transform = transforms.Compose(
[transforms.ToTensor(), transforms.Normalize((0.5,), (0.5,))]
train_dataset = torchvision.datasets.MNIST(
root="./data", train=True, transform=transform, download=True
test_dataset = torchvision.datasets.MNIST(
root="./data", train=False, transform=transform
train_loader = torch.utils.data.DataLoader(
dataset=train_dataset, batch_size=batch_size, shuffle=True
test_loader = torch.utils.data.DataLoader(
dataset=test_dataset, batch_size=batch_size, shuffle=False
# Initialize the model
model = NeuralNetwork(input_size, hidden_size, num_classes)
# Loss and optimizer
criterion = nn.CrossEntropyLoss()
optimizer = optim.Adam(model.parameters(), lr=learning_rate)
# Training loop
total_steps = len(train_loader)
for epoch in range(num_epochs):
for i, (images, labels) in enumerate(train_loader):
# Reshape images to (batch_size, input_size)
images = images.reshape(-1, 28 * 28)
# Forward pass
outputs = model(images)
loss = criterion(outputs, labels)
# Backward pass and optimization
if (i + 1) % 100 == 0:
f"Epoch [{epoch+1}/{num_epochs}], Step [{i+1}/{total_steps}], Loss: {loss.item():.4f}"
# Test the model
with torch.no_grad():
correct = 0
total = 0
for images, labels in test_loader:
images = images.reshape(-1, 28 * 28)
outputs = model(images)
_, predicted = torch.max(outputs.data, 1)
total += labels.size(0)
correct += (predicted == labels).sum().item()
print(f"Accuracy of the network on the 10000 test images: {100 * correct / total}%")
Epoch [1/5], Step [100/600], Loss: 0.3430
Epoch [1/5], Step [200/600], Loss: 0.3055
Epoch [1/5], Step [300/600], Loss: 0.3339
Epoch [1/5], Step [400/600], Loss: 0.4905
Epoch [1/5], Step [500/600], Loss: 0.3267
Epoch [1/5], Step [600/600], Loss: 0.3499
Epoch [2/5], Step [100/600], Loss: 0.1985
Epoch [2/5], Step [200/600], Loss: 0.1345
Epoch [2/5], Step [300/600], Loss: 0.2220
Epoch [2/5], Step [400/600], Loss: 0.2771
Epoch [2/5], Step [500/600], Loss: 0.1967
Epoch [2/5], Step [600/600], Loss: 0.1771
Epoch [3/5], Step [100/600], Loss: 0.2300
Epoch [3/5], Step [200/600], Loss: 0.1877
Epoch [3/5], Step [300/600], Loss: 0.1724
Epoch [3/5], Step [400/600], Loss: 0.2479
Epoch [3/5], Step [500/600], Loss: 0.1501
Epoch [3/5], Step [600/600], Loss: 0.2179
Epoch [4/5], Step [100/600], Loss: 0.1410
Epoch [4/5], Step [200/600], Loss: 0.1556
Epoch [4/5], Step [300/600], Loss: 0.0882
Epoch [4/5], Step [400/600], Loss: 0.0866
Epoch [4/5], Step [500/600], Loss: 0.1177
Epoch [4/5], Step [600/600], Loss: 0.0655
Epoch [5/5], Step [100/600], Loss: 0.0625
Epoch [5/5], Step [200/600], Loss: 0.1334
Epoch [5/5], Step [300/600], Loss: 0.1079
Epoch [5/5], Step [400/600], Loss: 0.0611
Epoch [5/5], Step [500/600], Loss: 0.1826
Epoch [5/5], Step [600/600], Loss: 0.1376
Accuracy of the network on the 10000 test images: 96.77%
• Explanation:
1. Neural Network Model Definition: We define a simple feedforward neural network with one hidden layer using the 'nn.Module' class.
2. Hyperparameters: We define hyperparameters such as input size, hidden size, number of classes, learning rate, batch size, and number of epochs.
3. Data Loading: We use torchvision to load the MNIST dataset and create data loaders for training and testing.
4. Model Initialization: We initialize the neural network model.
5. Loss and Optimizer: We specify the loss function (cross-entropy loss) and optimizer (Adam optimizer) for training the model.
6. Training Loop: We loop through the dataset for a number of epochs, perform forward and backward passes, and update the model parameters based on the computed gradients.
7. Testing: We evaluate the trained model on the test dataset to measure its accuracy.
This example covers some fundamental concepts in deep learning using PyTorch, such as defining a neural network architecture, loading and preprocessing data, training the model, and evaluating its
performance. You can further extend this example by exploring more complex architectures, experimenting with different optimizers and learning rates, and incorporating techniques like regularization
and dropout to improve model performance.
What's Next?
We've now entered the finance section on this platform, where you can enhance your financial literacy. | {"url":"https://www.techbaz.org/courses/py-deep-learning.php","timestamp":"2024-11-13T10:59:35Z","content_type":"text/html","content_length":"36387","record_id":"<urn:uuid:1822ff00-f5cd-4c8e-be9b-5e21a436587b>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00171.warc.gz"} |
How to Combine these formulas
I posted yesterday about a grid that I have since revamped. I am super new at this and spent a lot of time trying different formulas and watching videos. The formulas are working as standalones but I
am having trouble when combining them. Standalone Formulas:
=IF(Type@row = "Medium", WORKDAY([R Date]@row, +10))
=IF(Type@row = "Large", WORKDAY([R Date]@row, +15))
=IF(Type@row = "Small", WORKDAY([R Date]@row, +5))
Combined formula (which is not working and just displays blank as you see in the red cell below):
=IF(CONTAINS(Type@row = "Small", WORKDAY([R Date]@row, +5)), IF(CONTAINS(Type@row = "Medium", WORKDAY([R Date]@row, +10)), IF(CONTAINS(Type@row = "Large", WORKDAY([R Date]@row, +15)))))
What am I doing wrong?
• The CONTAINS function doesn't work as you've written it. It's not CONTAINS ( field = something), it's CONTAINS (search_for, range)
As an individual formula it would be written like this:
=IF(CONTAINS("Small",Type@row), WORKDAY([R Date]@row, +5))
As a combined nested IF formula it would be this:
=IF(CONTAINS("Small",Type@row), WORKDAY([R Date]@row, +5) , IF(CONTAINS("Medium",Type@row), WORKDAY([R Date]@row, +10), IF(CONTAINS("Large",Type@row), WORKDAY([R Date]@row, +15) )))
BRIAN RICHARDSON | PMO TOOLS AND RESOURCES | HE|HIM
SEATTLE WA, USA
• This was extremely helpful. I tried to find the different scenarios for IF Contains and this formula worked, it just needed brackets in the type@row section. Thank you
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/122414/how-to-combine-these-formulas","timestamp":"2024-11-09T14:13:09Z","content_type":"text/html","content_length":"397484","record_id":"<urn:uuid:1ce5e081-7952-4590-b429-9c36487bb7a3>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00223.warc.gz"} |
Gabriel's horn: finite volume and infinite surface area
The bucket that can’t hold enough paint to paint itself
Gabriel’s horn is the surface created by rotating 1/x around the x-axis. It is often introduced in calculus classes as an example of a surface with finite volume and infinite surface area. If it were
a paint can, it could not hold enough paint to paint itself!
This post will do two things:
1. explain why the paradox works, and
2. explain why it’s not paradoxical after all.
Rather than working out the surface area and volume exactly as one would do in a calculus class, we’ll be a little less formal but also more general.
Original function
When you set up the integral to compute the volume of the solid bounded by rotating the graph of a function f, the integrand is proportional to the square of f. So rotating the graph of 1/x gives us
an integral whose integrand is proportional to 1/x² and the integral converges.
When you set up the integral to compute the surface area, the integrand is proportional to f itself, not its square. So the integrand is proportional to 1/x and diverges.
For the volume to be finite, all we need is that f is O(1/x), i.e. eventually bounded above by some multiple of 1/x, and in fact we could get by with less.
For the area to be infinite, it is sufficient for the function to be Ω(1/x), i.e. eventually bounded below by some multiple of 1/x. An as before, we could get by with less.
So to make another example like Gabriel’s horn, we could use any function in Θ(1/x), i.e. eventually bounded above and below by some multiple of 1/x. So we could, for example, use
f(x) = (x + cos²x) / (x² + 42)
If you’re unfamiliar with the notation here, see these notes on big-O and related notation.
Now back to the idea of filling Gabriel’s horn with paint. If we spread the paint on the outside of the can with any constant thickness, we can only cover a finite area, but the area is infinite, so
we can’t paint the whole thing.
The resolution to the paradox is that we’re requiring the paint to be more realistic than the can. We’re implicitly letting the material of our can become thinner and thinner without any limit to how
thin it could be. If we also let our paint spread thinner and thinner as well at the right rate, we could cover the can with a coat of paint.
2 thoughts on “The bucket that can’t hold enough paint to paint itself”
1. OK, I’ll bite. Since the volume is finite, we can fill the can. Doesn’t this effectively “paint” the inside of the can? So we can paint the inside, but not the outside?
2. If the paint has any minimum thickness, it won’t reach the bottom of the can.
This could even be the case with a physical can. Once the bottom is narrow enough, paint won’t keep going. | {"url":"https://www.johndcook.com/blog/2020/06/11/gabriels-horn/","timestamp":"2024-11-02T11:08:25Z","content_type":"text/html","content_length":"53485","record_id":"<urn:uuid:2d7a379f-0a60-48ac-a3c0-8a505e285ee4>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00722.warc.gz"} |
Reinforcement Learning in Central Banking
Modeling Framework¶
Reinforcement learning (RL) is a sub-discipline of machine learning concerned with solving problems of optimal control. Typical use cases are self-driving cars, robot vacuums, and game-playing AI,
but the situation faced by a central bank also falls very neatly into the RL paradigm.
RL considers an agent, usually a robot or an AI, who interacts with an external environment, comprised of states $s_t \in \mathcal{S}$. The agent does two things: takes actions $a_t \in \mathcal{A}$
and receives rewards $R: \mathcal{S} \times \mathcal{A} \mapsto \mathbb{R}$. Rewards are thus a map of state-action pairs, converting these pairs to real numbers, thereby allowing an agent to
evaluate his performance.
Feedback is introduced into the model by having agent actions determine transition probabilities between states. We consider 1st order Markov dynamics to match what we have seen in class, where the
probability of any given series of outcomes ${y_t}_0^T$ is given by $p(y_0, y_1, ... , y_T) = \Pi_{t=1}^T p(y_{t+1} | y_{t}, a_t)$
The goal of the agent is to maximize expected rewards over a planning horizon. To ensure convergence, we add a discount factor $\beta < 1$, so total expected rewards over the planning horizon is
$$\mathbb{E}\left [ \sum_{t=0}^\infty \beta^t R(s_t, a_t) \; | \; t \right ]$$
In the case of the central banker, actions $a_t$ are interest rates $i_t$, and rewards are the negative loss function (allowing for maximization rather than minimization). For a loss function, we can
use a slightly modified form of the function shown in class:
$$ \mathcal{L} = (\mathbb{E}_t\,[\pi_{t+1}^2] + \alpha i_t^2)$$
$\pi_t$ has been replace with the expected next step inflation, because we assume the usual functional form for expected inflation: $\mathbb{E}_t\,[\pi_{t+1}] = A\pi_t + Bi_t$. Note this has been
converted from a "backwards looking" AR(1) model to a "forward-looking" one, by use of the expectations operator.
Loss is the squared values of the target and the instrument, with the interest rate scaled by a "cost parameter" $\alpha > 1$, to represent a bank's risk aversion to large changes in $i_t$. To change
the problem from one of loss minimization to reward maximization, we define rewards as the negative of the one-step loss function. Plugging the value of $\mathbb{E}_t\,[\pi_{t+1}]$ and expanding, we
$$ R_{t}(\pi_{t}, i_{t}) = -(A^2\pi_{t}^2 + 2AB\pi_{t} i_{t} + B^2 i_{t}^2 + \alpha i_{t}^2)$$
The banker will chose values of $i_t$ in order to maximize rewards over the entire planning horizon. To do this, he requires a policy function $\zeta: \mathcal{S} \mapsto \mathcal{A}$. The policy
function $\zeta$ gives, for any observed value of $\pi$, a value of $i$. In class, we have imposed a linear functional form for the policy function: $i_t = \zeta(\pi_t) = F\pi_t$. Here, suppose that
the functional form of $\zeta$ is unknown. The agent can follow any policy. Whatever policy he follows, we can describe the total expected trajectory of rewards obtained from that policy as a value
function $V^\zeta_0(\pi_0, i_0 | i_0 = \zeta(\pi_0))$.
$$V^\zeta_0(\pi_0, i_0 | i_0 = \zeta(\pi_0)) = \sum_{t=0}^T \beta^t \mathbb{E}_0\,[R(\pi_t, i_t | \zeta)]$$
The optimization is solved using a Bellman equation:
$$V^\zeta_t(\pi_t, i_t) = R(\pi_t, i_t, | \zeta) + \beta \mathbb{E}_t\,[V^\zeta_{t+1} (\pi_{t+1}, i_{t+1} | \zeta)]$$
The goal is to find the optimal $\zeta^\star$, such that:
$$\zeta^\star(\pi_t) = \text{arg} \max_{\zeta} V_t^\zeta(\pi_t)$$
The optimal policy could then produce an optimal value function, $V^\star_t(\pi_t)$, for which all actions are determined ahead of time. In an RL setting, however, we would like to be able to vary
the action, in order to have the agent "learn" the optimal action by trial and error in the data, rather than solving for it analytically (as we could from this point). To do this, we define an
Action-Value Function, the celebrated "Q-function" of Q-Learning:
$$Q_t^\zeta(\pi_t, i_t) = \mathbb{E}_t\,[R_t(\pi_t, i_t | \pi, i)] + \beta \mathbb{E}_t\,[V^\zeta_{t+1}(\pi_{t+1}, | \pi)]$$
In this equation, we propose to take whatever action we like now, and then promise to follow the policy $\zeta$ later. When the policy we follow later is the optimal policy $\zeta^\star$, we get an
equation for the optimal Q-function:
$$Q_t^\star(\pi, i) = \mathbb{E}_t\,[R_t(\pi_t, i_t)] + \beta \mathbb{E}_t\,[V^\star_{t+1}(\pi_{t+1}, | \pi)]$$
And note that when we use our "free action" in the first term of the Q-function to take the optimal action, we return to the optimal value function, so:
$$V_t^\star = \max_i Q_t^\star(\pi, i)$$
By substitution we obtain the recursive Bellman optimality equation for the Q-function:
$$Q_t^\star(\pi, i) = \mathbb{E}_t [ R_t(\pi, i)] + \beta \max_{i_{t+1} \in \mathcal{A}} \mathbb{E}_t [Q_{t+1}^\star (\pi_{t+1}, i_{t+1}) | \pi, i]$$
Explicitly substituting for rewards:
$$Q_t^\star(\pi, i) = \mathbb{E}_t \left[\beta Q_{t+1}^\star - A^2\pi_{t}^2 - 2AB\pi_{t} i_{t} - B^2 i_{t}^2 - \alpha i_{t}^2 \right]$$
The Q-function is quadratic in "action" $i_t$, so we can solve for $i_t^\star$ by setting $\frac{\partial Q_t^\star(\pi, i)}{\partial i_t} = 0$ and solving for $i_t$. Doing so yields:
$$i_t^\star = -\frac{AB\pi_t}{B^2 + \alpha}$$ | {"url":"http://www.jbgrabowski.com/notebooks/central-bank/","timestamp":"2024-11-08T09:23:51Z","content_type":"text/html","content_length":"360510","record_id":"<urn:uuid:3a08612d-9198-41cb-9ab7-5c7dd9dade4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00449.warc.gz"} |
How to avoid the negative effect of via in high-speed PCB board design
1. Basic concept of via
Through hole is an important part of a multilayer PCB board, and the cost of drilling usually accounts for 30% to 40% of the cost of PCB manufacturing. In short, each hole on a PCB can be called a
through hole. From the perspective of function, vias can be divided into two categories: one is used as electrical connection between layers; Second, it is used for fixing or positioning devices. In
terms of process, these vias are generally divided into three categories, namely blind hole, buried hole and through hole. The blind hole is located on the top and bottom surfaces of the printed
circuit board, and has a certain depth. It is used to connect the surface line and the lower inner line. The depth of the hole usually does not exceed a certain ratio (aperture). Embedded hole refers
to the connection hole located in the inner layer of the printed circuit board, which will not extend to the surface of the printed circuit board. The above two types of holes are located in the
inner layer of the circuit board. Before lamination, the through hole forming process is used to complete the hole. During the hole forming process, several inner layers may be overlapped. The third
is called through hole, which passes through the whole circuit board and can be used for internal interconnection or as the installation and positioning hole of components. Because the through-hole
is easier to realize in technology and lower in cost, most printed circuit boards use it instead of the other two kinds of through-hole. The following vias, unless otherwise specified, are considered
as through-hole. From the design point of view, a via is mainly composed of two parts: one is the middle drill hole, and the other is the pad area around the drill hole. The size of these two parts
determines the size of the vias. Obviously, when designing high-speed and high-density PCB boards, designers always hope that the smaller the vias are, the better, so that more wiring space can be
left on the board. In addition, the smaller the via, the smaller its parasitic capacitance, which is more suitable for high-speed circuits. However, the reduction of hole size also brings about an
increase in cost, and the size of vias cannot be reduced without limitation. It is limited by drilling and electroplating technology: the smaller the hole, the longer the drilling time will take, and
the easier it is to deviate from the center; And when the depth of the hole exceeds 6 times of the drilling diameter, it is impossible to ensure that the hole wall can be uniformly copper plated. For
example, if the thickness (through-hole depth) of a normal 6-layer PCB is 50Mil, then under normal conditions, the diameter of the hole provided by the PCB manufacturer can only reach 8Mil. With the
development of laser drilling technology, the size of drilling can also be smaller and smaller. Generally, vias with a diameter of less than or equal to 6Mils are called micropores. Microholes are
often used in HDI (high-density interconnection structure) design. Microhole technology allows vias to be directly punched on pads, which greatly improves circuit performance and saves wiring space.
The vias on the transmission line behave as breakpoints with discontinuous impedance, which will cause signal reflection. Generally, the equivalent impedance of the vias is about 12% lower than that
of the transmission lines. For example, the impedance of a 50 ohm transmission line will decrease by 6 ohm when it passes through the vias (specifically related to the size of the vias and the
thickness of the plates, not the reduction). However, the reflection caused by the discontinuous impedance of vias is actually very small, and its reflection coefficient is only (44-50)/(44+50)=0.06.
The problems caused by vias are more concentrated on the effects of parasitic capacitance and inductance.
2. Parasitic capacitance and inductance of via
There is parasitic stray capacitance in the via itself. If it is known that the diameter of the solder mask area of the via on the floor is D2, the diameter of the via pad is D1, the thickness of the
PCB is T, and the dielectric constant of the board substrate is ε, Then the parasitic capacitance of the via is approximately C=1.41 ε The parasitic capacitance of TD1/(D2-D1) via will mainly affect
the circuit by prolonging the signal rise time and reducing the circuit speed. For example, for a PCB with a thickness of 50Mil, if the via pad diameter is 20Mil (drilling diameter is 10Mils) and the
solder mask area diameter is 40Mil, we can approximately calculate the parasitic capacitance of the via through the above formula: C=1.41x4.4 x 0.050 x 0.020/(0.040-0.020)=0.31pF The rise time change
caused by this part of capacitance is approximately T10-90=2.2C (Z0/2)=2.2x0.31x (50/2)=17.05ps. It can be seen from these values that although the effect of the rise delay caused by the parasitic
capacitance of a single via is not very obvious, if the via is used for inter layer switching for many times in the routing, multiple vias will be used, which should be carefully considered in the
design. In practical design, parasitic capacitance can be reduced by increasing the distance between via and copper clad area or reducing the diameter of pad. Parasitic capacitance and inductance
exist in vias. In the design of high-speed digital circuits, the parasitic inductance of vias often brings more harm than the parasitic capacitance. Its parasitic series inductance will weaken the
contribution of bypass capacitor and the filtering effectiveness of the whole power supply system. We can use the following empirical formula to simply calculate the approximate parasitic inductance
of a via: L=5.08h [ln (4h/d)+1] where L is the inductance of the via, h is the length of the via, and d is the diameter of the central borehole. It can be seen from the formula that the diameter of
the via has little influence on the inductance, while the length of the via has little influence on the inductance. Using the above example, it can be calculated that the inductance of the via is: L=
5.08x0.050 [ln (4x0.050/0.010)+1]=1.015nH If the rise time of the signal is 1ns, the equivalent impedance is XL=π L/T10-90=3.19 Ω. Such impedance cannot be ignored when there is high-frequency
current passing through. In particular, the bypass capacitor needs to pass through two vias when connecting the power layer and the stratum, so the parasitic inductance of the vias will be
3. How to use vias
Through the above analysis of the parasitic characteristics of vias, we can see that in high-speed PCB design, seemingly simple vias often bring great negative effects to circuit design. In order to
reduce the adverse effects caused by the parasitic effect of vias, the following measures can be taken in the design:
1) Considering the cost and signal quality, the reasonable size of vias is selected. If necessary, you can consider using vias of different sizes. For example, for power supply or ground wire vias,
you can consider using larger sizes to reduce impedance, while for signal wiring, you can use smaller vias. Of course, with the reduction of via size, the corresponding cost will increase.
2) From the two formulas discussed above, it can be concluded that the use of thinner PCB is beneficial to reduce the two parasitic parameters of vias.
3) The signal wiring on the PCB shall not change layers as much as possible, that is to say, unnecessary vias shall not be used as much as possible.
4) The pins of power supply and ground shall be punched nearby, and the lead between the via and pin shall be as short as possible. Multiple vias can be drilled in parallel to reduce the equivalent
5) Place some grounded vias near the vias for signal layer change, so as to provide near circuit for signals. You can even place some redundant grounding vias on the PCB.
6) For high-speed PCB boards with high density, micro vias can be considered. | {"url":"https://www.ipcb.com/pcb-blog/9813.html","timestamp":"2024-11-05T06:18:59Z","content_type":"application/xhtml+xml","content_length":"37582","record_id":"<urn:uuid:17fe441b-aa6e-4b05-a732-9adeae3025fa>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00568.warc.gz"} |
Space Complexity - (Computational Algebraic Geometry) - Vocab, Definition, Explanations | Fiveable
Space Complexity
from class:
Computational Algebraic Geometry
Space complexity refers to the amount of memory space required by an algorithm to run as a function of the size of the input data. It is a crucial aspect of algorithm analysis, as it provides insight
into how much memory an algorithm will need relative to its input, which is essential for understanding performance and scalability, especially in the context of computational tasks like Buchberger's
congrats on reading the definition of Space Complexity. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Buchberger's algorithm has a high space complexity because it requires storing intermediate polynomials and their relationships, which can grow rapidly with the size of the ideal being processed.
2. The space complexity of Buchberger's algorithm can be significantly affected by the number of generators in the input ideal, as more generators lead to more potential intermediate results.
3. In practice, efficient memory management strategies can help mitigate high space complexity, allowing for the processing of larger ideals without exhausting system resources.
4. Analyzing space complexity involves considering both the auxiliary space used by the algorithm and the space required for the input itself.
5. Understanding space complexity is essential for implementing Buchberger's algorithm in environments with limited memory resources, such as embedded systems or during large-scale computations.
Review Questions
• How does space complexity affect the implementation of Buchberger's algorithm in practical scenarios?
□ Space complexity significantly impacts how Buchberger's algorithm can be effectively implemented in real-world applications. Since this algorithm can require substantial memory due to its
need to store intermediate polynomials and relationships, understanding its space requirements helps developers anticipate potential issues related to memory usage. In scenarios where memory
resources are constrained, it becomes essential to optimize the implementation or seek alternative algorithms with lower space requirements.
• Evaluate how the number of generators in an ideal influences the space complexity of Buchberger's algorithm and what strategies might help manage this complexity.
□ The number of generators in an ideal directly correlates with the space complexity when using Buchberger's algorithm. More generators lead to more intermediate polynomials being created and
stored, which increases memory usage. To manage this complexity, one strategy could involve limiting the number of generators processed at one time or utilizing techniques such as modular
arithmetic to reduce the size of polynomial representations, thus optimizing memory consumption during computations.
• Assess how a thorough understanding of space complexity can enhance research efforts in computational algebraic geometry and improve algorithm design.
□ A comprehensive grasp of space complexity allows researchers in computational algebraic geometry to make informed decisions when designing and implementing algorithms like Buchberger's
algorithm. By analyzing space requirements, they can tailor algorithms to fit within available resources and optimize performance. Furthermore, this understanding fosters innovation in
creating new algorithms that balance both time and space efficiency, thereby pushing forward advancements in computational methods and applications across various fields.
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[QSMS Monthly Seminar] Yokonuma Hecke algebras and invariants of framed links
2022년 4월 QSMS Monthly Seminar
• Date: 27 May Fri PM 2:00 ~ 5:00
• Place: 27-220 (SNU)
• Speaker: Myungho Kim
Title: Yokonuma Hecke algebras and invariants of framed links
Yokonuma-Hecke algebra (of rank n) is a finite-dimensional quotient of the group algebra of the framed braid group $F_n$, which is a semidirect product of $Z^n$ with the braid group $B_n$.
Yokonuma-Hecke algebras provide a polynomial invariant for framed links along the similar line as the Iwahori-Hekce algebras of type A provide the HOMFLYPT polynomials. We will review the
construction of the invariants in this talk.
Speaker: Hanwool Bae
Title: Calabi-Yau cluster structures on Rabinowitz Fukaya categories
In this talk, I will consider the plumbing of the cotangent bundles of (d+1)-dimensional spheres along a tree T. I will discuss that its (derived) wrapped Fukaya category W, its (derived) compact
Fukaya category F, and a certain generator L of W form a (d+1)-Calabi-Yau triple, where L is given by the direct sum of cocore disks. This implies that the quotient category W/F carries a
d-Calabi-Yau cluster structure. Recently, by Ganatra-Gao-Venkatesh, the quotient category W/F was shown to be quasi-equivalent to the Rabinowitz Fukaya category if F is a Koszul dual subcategory of
W. Using this result, in the case the tree T is given by a Dynkin diagram of type A,D or E, I will discuss how to show that the Lagrangian Rabinowitz Floer homology of L and itself is isomorphic to
the path algebra of a certain quiver as a ring .
This talk is based on a joint work with Wonbo Jeong and Jongmyeong Kim. | {"url":"https://qsms.math.snu.ac.kr/index.php?mid=board_sjXR83&order_type=desc&page=4&listStyle=gallery&document_srl=2299","timestamp":"2024-11-12T03:10:15Z","content_type":"text/html","content_length":"71446","record_id":"<urn:uuid:a213c437-c5d8-4c1b-856f-55b7824c3ad2>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00514.warc.gz"} |
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Description Transforms each sample in the time series to the sum of the near samples within the given sliding window
Input Time Series – Single time series or a multiple time series
Parameters Window - Mandatory, Window size in data points
Filter Type - [Optional]
Simple Include Current - The function's value at time t is the average of N past points, including the metric value at t. t-N, t-N+1,..., t-1, t
Simple Exclude Current - The function's value at time t is the average of N previous points, excluding the metric value at t. t-N, t-N+1,...,t-1
Centered Include Current - The function's value at time t is the average of the N points around the time t (both previous and after). t-N/2, t-N/2+1,...,t, t+1, t+2,...,t+N/2. Including
the value at time t.
Centered Exclude Current - The function's value at time t is the average of N previous points, excluding the metric value at t. t-N, t-N+1,...,t-1
Output Transformed Time Series
Available Dashboards | {"url":"https://support.anodot.com/hc/en-us/articles/115002815114-MovingSum","timestamp":"2024-11-08T07:35:39Z","content_type":"text/html","content_length":"20506","record_id":"<urn:uuid:f65214e4-ef9a-431c-864f-8d33272aa7ac>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00036.warc.gz"} |
Terms II :: CC 315 Textbook
Terms II
We can describe the sizes of trees and position of nodes using different terminology, like level, depth, and height.
• Level - The level of a node characterizes the distance between the node and the root. The root of the tree is considered level 1. As you move away from the tree, the level increases by one.
□ For our family tree example, what nodes are in the following levels? Think about the answer and then click corresponding arrow. Level 1:Myra - Level 1 is always the root Level 2:Raju, Joe,
Zia - These are the nodes which are 1 edge away from the root. Level 3:Uzzi, Bert, Uma - These are the nodes which are 2 edges away from the root. Level 4:Bev, Ava, Ang - These are the nodes
which are 3 edges away from the root. Level 5:Isla - This is the only node which is 4 edges away from the root. Level 6:Eoin - This is the only node which is 5 edges away from the root.
• Depth - The depth of a node is its distance to the root. Thus, the root has depth zero. Level and depth are related in that: level = 1 + depth.
□ For our family tree example, what nodes have the following depths? Depth 0:Myra - The root will always be at depth 0. Depth 1:Raju, Joe, Zia - These are the nodes which are 1 edge away from
the root. Depth 2:Uzzi, Bert, Uma - These are the nodes which are 2 edge away from the root. Depth 3:Bev, Ava, Ang - These are the nodes which are 3 edge away from the root. Depth 4:Isla -
This is the only node which is 4 edges away from the root. Depth 5:Eoin - This is the only node which is 5 edges away from the root.
• Height of a Node - The height of a node is the longest path to a leaf descendant. The height of a leaf is zero.
□ For our family tree example, what nodes have the following heights? Height 0:Raju, Eoin, Ava, Bert, Ang - The leaves always have height 0. Height 1:Isla, Uma - `Isla -> Eoin` and `Uma -> Ang`
Height 2:Bev, Zia - `Bev -> Isla -> Eoin` and `Zia -> Uma -> Ang` Height 3:Uzzi - `Uzzi -> Bev -> Isla -> Eoin` Height 4:Joe - `Joe -> Uzzi -> Bev -> Isla -> Eoin` Height 5:Myra - `Myra ->
Joe -> Uzzi -> Bev -> Isla -> Eoin`
• Height of a Tree - The height of a tree is equal to the height of the root.
□ Our family tree would have height 5 | {"url":"https://textbooks.cs.ksu.edu/cc315/ii-trees/3-tree-traversal/4-terms-ii/","timestamp":"2024-11-14T15:41:36Z","content_type":"text/html","content_length":"52235","record_id":"<urn:uuid:88474188-0c32-4f91-a45c-69788351350b>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00218.warc.gz"} |
Introductory Chemistry, 1st Canadian Edition [Clone]
70 Dilutions and Concentrations
Learning Objective
1. Learn how to dilute and concentrate solutions.
Often, a worker will need to change the concentration of a solution by changing the amount of solvent. Dilution is the addition of solvent, which decreases the concentration of the solute in the
solution. Concentration is the removal of solvent, which increases the concentration of the solute in the solution. (Do not confuse the two uses of the word concentration here!)
In both dilution and concentration, the amount of solute stays the same. This gives us a way to calculate what the new solution volume must be for the desired concentration of solute. From the
definition of molarity,
molarity = moles of solute / liters of solution
we can solve for the number of moles of solute:
moles of solute = (molarity)(liters of solution)
A simpler way of writing this is to use M to represent molarity and V to represent volume. So the equation becomes
moles of solute = MV
Because this quantity does not change before and after the change in concentration, the product MV must be the same before and after the concentration change. Using numbers to represent the initial
and final conditions, we have
M[1]V[1] = M[2]V[2]
as the dilution equation. The volumes must be expressed in the same units. Note that this equation gives only the initial and final conditions, not the amount of the change. The amount of change is
determined by subtraction.
Example 9
If 25.0 mL of a 2.19 M solution are diluted to 72.8 mL, what is the final concentration?
It does not matter which set of conditions is labelled 1 or 2, as long as the conditions are paired together properly. Using the dilution equation, we have
(2.19 M)(25.0 mL) = M[2](72.8 mL)
Solving for the second concentration (noting that the milliliter units cancel),
M[2] = 0.752 M
The concentration of the solution has decreased. In going from 25.0 mL to 72.8 mL, 72.8 − 25.0 = 47.8 mL of solvent must be added.
Test Yourself
A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution?
135.4 mL
Concentrating solutions involves removing solvent. Usually this is done by evapourating or boiling, assuming that the heat of boiling does not affect the solute. The dilution equation is used in
these circumstances as well.
Chemistry Is Everywhere: Preparing IV Solutions
In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels). Does an aide run to a supply
cabinet and take out an IV bag containing this concentration of KCl?
Not likely. It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a stock solution, of KCl. The aide is
expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation.
Medical personnel commonly must perform dilutions for IV solutions. Source: “Infuuszakjes” by Harmid is in the public domain.
If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then it is an easy calculation to determine how much stock solution to use:
(10%)V[1] = (0.50%)(100 mL)
V[1] = 5 mL
Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough even for medical purposes.
Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the wrong dose can be
useless, harmful, or even fatal!
Key Takeaways
• Calculate the new concentration or volume for a dilution or concentration of a solution.
1. What is the difference between dilution and concentration?
2. What quantity remains constant when you dilute a solution?
3. A 1.88 M solution of NaCl has an initial volume of 34.5 mL. What is the final concentration of the solution if it is diluted to 134 mL?
4. A 0.664 M solution of NaCl has an initial volume of 2.55 L. What is the final concentration of the solution if it is diluted to 3.88 L?
5. If 1.00 mL of a 2.25 M H[2]SO[4] solution needs to be diluted to 1.00 M, what will be its final volume?
6. If 12.00 L of a 6.00 M HNO[3] solution needs to be diluted to 0.750 M, what will be its final volume?
7. If 665 mL of a 0.875 M KBr solution are boiled gently to concentrate the solute to 1.45 M, what will be its final volume?
8. If 1.00 L of an LiOH solution is boiled down to 164 mL and its initial concentration is 0.00555 M, what is its final concentration?
9. How much water must be added to 75.0 mL of 0.332 M FeCl[3](aq) to reduce its concentration to 0.250 M?
10. How much water must be added to 1.55 L of 1.65 M Sc(NO[3])[3](aq) to reduce its concentration to 1.00 M?
Dilution is a decrease in a solution’s concentration, whereas concentration is an increase in a solution’s concentration.
0.484 M
2.25 mL
401 mL
24.6 mL | {"url":"https://opentextbc.ca/introductorychemistryclone/chapter/dilutions-and-concentrations/","timestamp":"2024-11-03T17:20:04Z","content_type":"text/html","content_length":"111214","record_id":"<urn:uuid:035364ee-61b3-40e1-9d48-1d026fcb75db>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00830.warc.gz"} |
Understanding PMT in Finance: A Comprehensive Guide
In the world of finance, various formulas and concepts help individuals and businesses make informed decisions. One such key concept is the PMT function, commonly used in financial calculations. This
article will explore what PMT is, its applications, how to calculate it, and its importance in financial planning.
What is PMT?
PMT, short for "payment," refers to a financial function that calculates the payment amount for a loan based on constant payments and a constant interest rate. It is primarily used in the context of
loans and annuities, helping borrowers determine their monthly payments when they take out a loan or make regular investments.
The PMT function can be calculated using a straightforward formula or through financial calculators and software programs like Excel. The PMT function is crucial for budgeting, loan repayment
planning, and investment analysis.
The PMT Formula
The PMT formula is derived from the present value of annuities. The standard formula is as follows:
PMT = \frac{P \times r}{1 - (1 + r)^{-n}}
- \( PMT \) = Payment amount per period
- \( P \) = Principal amount (the total loan or investment amount)
- \( r \) = Interest rate per period (annual interest rate divided by the number of payment periods per year)
- \( n \) = Total number of payments (the loan term multiplied by the number of payment periods per year)
Let’s break down each component for better understanding:
1. Principal Amount (P):
This is the initial sum borrowed or invested. For instance, if you take out a mortgage for $200,000, your principal amount is $200,000.
2. Interest Rate (r):
The rate at which interest accrues on the principal. If your mortgage has an annual interest rate of 5%, and you make monthly payments, you would divide 5% by 12 months to find the monthly interest
3. Number of Payments (n):
This is the total number of payments you’ll make. For a 30-year mortgage with monthly payments, you’d have \( 30 \times 12 = 360 \) payments.
Practical Applications of PMT
Understanding how to use the PMT function is essential for various financial scenarios:
1. Loan Repayment Planning
One of the most common uses of the PMT function is for calculating monthly loan repayments. When taking out a loan, knowing your monthly payment helps you budget and manage your finances effectively.
For example, if you take out a $250,000 mortgage at a 4% annual interest rate for 30 years, you can use the PMT formula to determine your monthly payment. This knowledge allows you to understand how
much of your income will go toward loan repayments and plan accordingly.
2. Investment Planning
Investors also use the PMT function to determine the future value of regular investments. If you plan to invest a fixed amount monthly into a retirement fund, knowing the PMT helps you estimate how
much you will accumulate over time based on expected returns.
3. Comparing Financial Products
When evaluating different loan options or investment plans, the PMT function can help you compare total costs. By calculating the monthly payments for various scenarios, you can determine which
option is more financially viable.
4. Amortization Schedules
PMT is fundamental in creating amortization schedules. These schedules outline each payment made toward a loan, showing how much goes to interest and how much reduces the principal. Understanding
your amortization schedule can help you strategize additional payments to reduce total interest paid over the loan term.
Example Calculation
Let’s illustrate the PMT function with an example. Suppose you want to borrow $100,000 at a 6% annual interest rate for 15 years. We can calculate the monthly payment as follows:
1. Principal (P): $100,000
2. Annual Interest Rate: 6%
3. Monthly Interest Rate (r): \( \frac{6\%}{12} = 0.5\% = 0.005 \)
4. Total Payments (n): \( 15 \times 12 = 180 \)
Plugging these values into the PMT formula:
PMT = \frac{100,000 \times 0.005}{1 - (1 + 0.005)^{-180}}
Calculating this gives:
PMT = \frac{500}{1 - (1.005)^{-180}} \approx \frac{500}{0.413} \approx 1211.78
Thus, the monthly payment for this loan would be approximately $1,211.78.
Importance of Understanding PMT
Understanding the PMT function is crucial for anyone involved in financial planning, whether for personal use or professional purposes. Here are some key reasons why:
1. Financial Literacy
Knowledge of PMT enhances financial literacy. Being able to calculate payments helps individuals understand the implications of loans and investments, leading to better financial decisions.
2. Budgeting
Knowing your monthly payment obligations allows for better budgeting. It helps you allocate funds effectively and avoid financial pitfalls.
3. Long-Term Planning
For investors, understanding PMT can aid in long-term financial planning, ensuring that they set realistic goals based on their investment strategies.
4. Negotiation Power
When you understand how to calculate payments, you have more leverage when negotiating loan terms or investment products, as you can assess offers accurately.
In conclusion, the PMT function is a fundamental aspect of financial management. Whether you're looking to take out a loan, plan for retirement, or compare investment options, understanding how to
calculate and use PMT can significantly impact your financial decisions. By grasping this concept, you can navigate the complexities of finance more effectively, leading to informed choices and
better financial health.
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Binary Search Tree Properties | Various Binary Search Tree Properties
Updated March 10, 2023
Introduction to Binary Search Tree Properties
Binary search tree properties are defined as characteristics and traits that helps in describing a search tree to be a binary search tree. Binary search tree is defined as a sorted and ordered tree
that belongs to the class of rooted tree, a tree where one vertex is chosen as the root through which other branches are assigned a natural orientation (either towards or away from the root), type of
data structure.
This binary search tree allows developers to conduct fast lookups, insertion or removal of any root item and hence an obvious choice for any dynamic sets or lookup tables. The word binary here refers
to the split that happens when another number either needs to be inserted, or a number that is to be searched is found after traversing the tree.
Various Binary Search Tree Properties
Now we know the tree characteristics of a binary search tree it is very important to know the properties so that when we, later on in this article, look into the basic operations at a superficial
level we can easily connect on the importance of a binary tree that helps us to perform the operations in an easier way.
The following are the properties of a node-based binary tree:
1. The left subtree of the binary search tree contains those values that are lesser than the node’s key. While performing this, it is not necessary that the tree should have an equal number of left
nodes and right nodes. Though the ask is not unrealistic, the tree that is asked for having an equal number of left and right nodes or at the max one extra node either on the left or right is known
as a balanced binary search tree. With this characteristic, we can maintain that the numbers on the left side of the tree are lesser and hence needs to be traversed in case then we want to travel the
path to perform an operation that focuses on the lesser numbers.
2. On the same lines, the next characteristic of the binary tree is that the right subtree contains values that are greater than the node’s key. Likewise, to the earlier point, while performing this,
it is not necessary that the tree should have an equal number of right nodes and left nodes. With this characteristic, we can maintain that the numbers on the right side of the tree are greater and
hence needs to be traversed in case then we want to travel the path to perform an operation that focusses on the greater numbers.
3. Last but not the least property is that each of the subtrees that is formed should be a binary tree as a standalone, which essentially means that when we cut the full tree at any node, the
resulting tree should be a binary search tree in itself. This property is the most crucial one because it helps maintain the previous 2 points in the subtrees so that the path to be traversed is the
shortest one.
Let us look at an example. Let us say we have to search for a number in the tree. Now assume that only at the root node, we apply the previous 2 points. Now if we traverse, we know that we need to
traverse on the right or the left side depending on the value being greater or lesser than the root node.
Now, arises 2 conditions:
• The sub trees are not individually binary search trees: In this scenario, it is very difficult to understand the path that needs to be traversed as we don’t know which side of the tree would
contain the number that is to be searched for.
• The sub trees are not individually binary search trees: In this scenario, it becomes very easy to understand the path that needs to be traversed as we would know which side of the tree should be
traversed so that we find the appropriate path that would contain the number that is to be searched for.
With the above properties, it becomes even easier to keep the sanity of the binary search tree to be used for the use case that is built for. Let us look at the operations that a binary search tree
exploits the properties we have mentioned and helps in the easiest traversal of the tree.
• Search: The first operation that a binary tree has to look at is the search. Now, during this operation, we input the number that needs to be searched for. At first, we would compare it to the
number at the root node. Now if the number over there is found then great. Otherwise, we would see if the number is lesser than or greater than the root node. Now depending on the scenario either
the first or the second property is exploited, and that path is taken. Now since the third property says that the subtree should also be a binary search tree, hence the similar steps are followed
till we reach a node that is either the number we are searching for or the leaf node (node post which we don’t have any other nodes).
• Delete and Insert: In both these operations, we would need to use the search operation and hence the properties gets used in accordance to the need and when we reach that specific node, either
the insert or the delete operation is performed as per the need of the use case.
We have looked into the properties of the binary search tree in great detail. This article also enabled us to link how properties are linked to operation in a binary search tree and also see the
importance of properties which keeps the sanctity of the binary search tree on its use cases.
Recommended Articles
This is a guide to Binary Search Tree Properties. Here we discuss the introduction and various binary search tree properties respectively. You may also have a look at the following articles to learn
more – | {"url":"https://www.educba.com/binary-search-tree-properties/","timestamp":"2024-11-07T06:42:29Z","content_type":"text/html","content_length":"307155","record_id":"<urn:uuid:ac8545e2-a1f4-43dc-ba82-d64cf8210500>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00159.warc.gz"} |
An Atypical Metaheuristic Approach to Recognize an Optimal Architecture of a Neural Network
Abishai Ebenezer M., Arti Arya
The structural design of an Artificial Neural Network (ANN) greatly determines its classification and regression capabilities. Structural design involves both the count of hidden layers and the count
of neurons required in each of these hidden layers. Although various optimization algorithms have proven to be good at finding the best topology for a given number of hidden layers for an ANN, there
has been little work done in finding both the optimal count of hidden layers and the ideal count of neurons needed in each layer. The novelty of the proposed approach is that a bio-inspired
metaheuristic namely, the Water Cycle Algorithm (WCA) is used to effectively search space of local spaces, by using the backpropagation algorithm as the underlying algorithm for parameter
optimization, in order to find the optimal architecture of an ANN for a given dataset. Computational experiments have shown that such an implementation not only provides an optimized topology but
also shows great accuracy as compared to other advanced algorithms used for the same purpose.
Paper Citation
in Harvard Style
Ebenezer M. A. and Arya A. (2022). An Atypical Metaheuristic Approach to Recognize an Optimal Architecture of a Neural Network. In Proceedings of the 14th International Conference on Agents and
Artificial Intelligence - Volume 3: ICAART, ISBN 978-989-758-547-0, pages 917-925. DOI: 10.5220/0010951600003116
in Bibtex Style
author={Abishai Ebenezer M. and Arti Arya},
title={An Atypical Metaheuristic Approach to Recognize an Optimal Architecture of a Neural Network},
booktitle={Proceedings of the 14th International Conference on Agents and Artificial Intelligence - Volume 3: ICAART,},
in EndNote Style
TY - CONF
JO - Proceedings of the 14th International Conference on Agents and Artificial Intelligence - Volume 3: ICAART,
TI - An Atypical Metaheuristic Approach to Recognize an Optimal Architecture of a Neural Network
SN - 978-989-758-547-0
AU - Ebenezer M. A.
AU - Arya A.
PY - 2022
SP - 917
EP - 925
DO - 10.5220/0010951600003116 | {"url":"http://scitepress.net/PublishedPapers/2022/109516/","timestamp":"2024-11-10T11:52:34Z","content_type":"text/html","content_length":"6964","record_id":"<urn:uuid:98fef587-be9f-48aa-8e20-d8c4a8c6ab92>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00214.warc.gz"} |
Volume of a Cube
The sum of edges of a cube is 60, what is its volume?
STEP 1
First, let us determine the number of edges in a cube. There are 12 edges.
STEP 2
Since all edges of a cube are equal in length, we can divide the sum of the edges by the number of edges:
60/12 = 5
Thus, the length of one edge of this cube is equal to 5 units.
STEP 3
Then, to calculate the volume of a cube, its length is to be multiplied by its width, multiplied by its height,
, the length of one edge by the power of three. Thus:
5 x 5 x 5
= 125
Therefore, the volume of this cube is
125 cubic units
If you like our answer and explanation to this problem, please visit our brand new platform,
Zap Zap Math
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How do I find the values of x for which the linear approximation is accurate to within 0.1 of a function?
Hello! Could you please show me how I could find the linear approximation of a function (1 variable)? Also, could you please please show me how I could find the values of x for which the linear
approximation is accurate to within 0.1? Thank you very much! | {"url":"https://www.calculatorti.com/calculator-help/5555/values-which-linear-approximation-accurate-within-function","timestamp":"2024-11-06T18:54:55Z","content_type":"text/html","content_length":"25436","record_id":"<urn:uuid:113b719d-b902-4804-bf51-b63b5cdd05ec>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00690.warc.gz"} |
Stresses produced in gasses by temperature and concentration inhomogeneities. New types of free convection
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Reviews of topical problems
Stresses produced in gasses by temperature and concentration inhomogeneities. New types of free convection
The main results of theoretical investigation of slow $(\mathbf{Re}\sim1)$ non-isothermal (temperature drop in the gas $\theta=\Delta T/T\sim1$) are reported. These flows are described by equations
that differ from the classical Navier--Stokes equations for a compressible liquid in that the momentum equation contains besides the viscous-stress tensor, also a temperature-stress tensor of the
same order of magnitude. The question of the influence of temperature stresses on the motion of the gas are analyzed, as are the forces acting on bodies placed in the gas. This question was first
raised long ago by J. Maxwell, who used implicitly linearization in $\theta$ and reached the conclusion that the temperature stresses cause neither motion of the gas nor forces. However, when $\
theta$ is not small, a new type of convection of the gas appears in the absence of external forces (e.g., of gravitation), namely, the temperature stresses cause the gas to move near uniformly heated
(cooled) bodies; some examples of this convection are presented. In addition, for the case of small $\theta$, an electrostatic analogy is established, describing the force interaction between these
bodies as a result of the temperature stresses. The problem of the flow around a uniformly heated sphere at $\mathbf{Re}_\infty\ll1$ (the Stokes problem) is solved: the temperature stresses exert an
ever increasing influence on the resistance of the sphere with increasing sphere temperature. Analogous phenomena, produced in gas mixtures by concentration (diffusion) stresses, are indicated. | {"url":"https://ufn.ru/en/articles/1976/5/d/","timestamp":"2024-11-15T02:40:06Z","content_type":"text/html","content_length":"19011","record_id":"<urn:uuid:98a8d5d5-6c39-419d-b64d-e6ce37278991>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00418.warc.gz"} |
to Hong Kong catt
Category: main menu • concrete menu • Pounds
concrete conversion
Amount: 1 pound (lb) of mass
Equals: 0.75 Hong Kong catties (kan 斤) in mass
Converting pound to Hong Kong catties value in the concrete units scale.
TOGGLE : from Hong Kong catties into pounds in the other way around.
CONVERT : between other concrete measuring units - complete list.
Conversion calculator for webmasters.
This general purpose concrete formulation, called also concrete-aggregate (4:1 - sand/gravel aggregate : cement - mixing ratio w/ water) conversion tool is based on the concrete mass density of 2400
kg/m3 - 150 lbs/ft3 after curing (rounded). Unit mass per cubic centimeter, concrete has density 2.41g/cm3. The main concrete calculator page.
The 4:1 strength concrete mixing formula applies the measuring portions in volume sense (e.g. 4 buckets of concrete aggregate, which consists of gravel and sand, with 1 bucket of cement.) In order
not to end up with a too wet concrete, add water gradually as the mixing progresses. If mixing concrete manually by hand; mix dry matter portions first and only then add water. This concrete type is
commonly reinforced with metal rebars or mesh.
Convert concrete measuring units between pound (lb) and Hong Kong catties (kan 斤) but in the other reverse direction from Hong Kong catties into pounds.
conversion result for concrete:
From Symbol Result To Symbol
1 pound lb = 0.75 Hong Kong catties kan 斤
Converter type: concrete measurements
This online concrete from lb into kan 斤 converter is a handy tool not just for certified or experienced professionals.
First unit: pound (lb) is used for measuring mass.
Second: Hong Kong catty (kan 斤) is unit of mass.
concrete per 0.75 kan 斤 is equivalent to 1 what?
The Hong Kong catties amount 0.75 kan 斤 converts into 1 lb, one pound. It is the EQUAL concrete mass value of 1 pound but in the Hong Kong catties mass unit alternative.
How to convert 2 pounds (lb) of concrete into Hong Kong catties (kan 斤)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.75000000826734 * 2 (or divide it by / 0.5)
1 lb of concrete = ? kan 斤
1 lb = 0.75 kan 斤 of concrete
Other applications for concrete units calculator ...
With the above mentioned two-units calculating service it provides, this concrete converter proved to be useful also as an online tool for:
1. practicing pounds and Hong Kong catties of concrete ( lb vs. kan 斤 ) measuring values exchange.
2. concrete amounts conversion factors - between numerous unit pairs.
3. working with - how heavy is concrete - values and properties.
International unit symbols for these two concrete measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for pound is:
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for Hong Kong catty is:
kan 斤
One pound of concrete converted to Hong Kong catty equals to 0.75 kan 斤
How many Hong Kong catties of concrete are in 1 pound? The answer is: The change of 1 lb ( pound ) unit of concrete measure equals = to 0.75 kan 斤 ( Hong Kong catty ) as the equivalent measure for
the same concrete type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only
whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in lb - pounds for concrete
amount, the rule is that the pound number gets converted into kan 斤 - Hong Kong catties or any other concrete unit absolutely exactly. | {"url":"https://www.traditionaloven.com/building/masonry/concrete/convert-pound-lb-of-concrete-to-hong-kong-catty-concrete.html","timestamp":"2024-11-03T07:15:29Z","content_type":"text/html","content_length":"40073","record_id":"<urn:uuid:efc88fc5-189d-4338-8a7f-8b438802395c>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00038.warc.gz"} |
Exam-Style Question on Significance
Question id: 18. This question is similar to one that appeared on an IB Studies paper in 2012. The use of a calculator is allowed.
The older students from Glee High School are required to follow a two year IB Mathematics course. Data were gathered from a sample of 242 students regarding their choice of course. The following data
were recorded.
Gender Studies Standard Higher Total
Male 35 15 21 71
Female 60 30 81 171
Total 95 45 102 242
A \(\chi ^2\) test was carried out at the 5% significance level to analyse the relationship between gender and choice of mathematics course.
(a) Write down the null hypothesis, H[o], for this test.
(b) Find the expected value of female students on the Studies course.
(c) Write down the number of degrees of freedom.
(d) Use your graphic display calculator to determine the \(\chi ^2\)[calc] value.
(e) Determine whether H[o] should be accepted. Justify your answer.
One student is chosen at random from the 242 students.
(f) Find the probability that this student is male.
(g) Find the probability that the student chosen at random is on the Standard course.
Two students are chosen at random from the 242 students.
(h) Find the probability that both are on the Studies course.
(i) Find the probability that neither are on the Higher course. | {"url":"https://transum.org/Maths/Exam/Question.asp?Q=18","timestamp":"2024-11-07T14:19:24Z","content_type":"text/html","content_length":"21039","record_id":"<urn:uuid:aad95ec8-c7fd-4382-a6a2-8fd496ddb631>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00861.warc.gz"} |
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Probability Overview
Probability is the ratio of successful outcomes to the total number of outcomes. So, the probability of any event is the number of successful outcomes over the total number of outcomes. Probability
is always a number between zero and one. Zero as a probability means that the event is impossible, which means it will not happen. One means that it's certain to happen. Probabilities help us
understand commonly reported percentages, such as weather reports and election results. Now, let's that, for example,
it was reported that there's a 40% chance of rain this afternoon. What does that really mean? What that means is that 40% of the geographical area, so 40% of the geographical area contained in the TV
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Probability Real World Example
For example, let's use rolling a die to find the probability of an event. The first thing we need to do is we need to find out what the sample space is for rolling a die. So, we have to determine
what outcomes can we get when we do roll that die. So, the sample space will be the numbers one through six.
Now, let's say we want to find the probability of rolling an even. What we need to do is we need to find out the number of successful outcomes over the total number. So, what we'll do, is we'll find
the number of successful outcomes by looking at our sample space, and we are going to identify the even numbers in our set. There are three successful outcomes.
There are six total outcomes in the data set, so our probability is three over six. Now, we could reduce that number or we can divide it out, and we get our probability of rolling an even on a die is
point five. Now, what does that mean? What that means is that you should expect to roll an even number about half the time. Probability is simply the ratio of successful outcomes to the total number
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Tax forms
Has anyone else noticed that box 8TI referred to on Form2047(red),section VII no longer seems to exist on Form 2042 (blue)?
My 2042 has yet to arrive but maybe this has some bearing on it?:
I hope you can understand what the letter which Parsnips quotes is trying to convey because it's gibberish to me.[:-))]
It does on mine.
[quote user="Gardener"]It does on mine.[/quote]It does on mine too and I have had THE letter - what a fine mess we are going to be in this year!!
I think I will do my online tonight as it stands having not had the letter and leave it with them.
It seems bizarre that the forms being sent out are not uniform.I've double checked & there definitely isn't a Box 8TI on my 2042.Any suggestions where I might stick it instead-please keep it clean!
Can you do your return on line? May be the form will be the other variety if you haven't had the letter that Jay has had.
I've just got mine and it is there (back page, 6 boxes from the bottom.) Which department are you in, Malcolm? I thought all these forms were the same though.
[quote user="cooperlola"]I've just got mine and it is there (back page, 6 boxes from the bottom.) Which department are you in, Malcolm? I thought all these forms were the same though.[/quote]
Same as you here in Maine et Loire, form 2042 K préremplie arrived this morning showing Box TI and it seems almost identical to last year's form.
On the same back page at the top, please could someone explain the first line: CSG Déductible.. and in the box beside it is already printed a figure which appears to be not quite half of the sum we
paid in Prélèvements Sociaux..
At the bottom of the page in Section 8 I am going to do as I did last year and enter my teacher pension in Box TI, put a line through Box TL and ignore Box TK for the moment.- I'll take the whole lot
( in pencil!) to the Hôtel des Impôts later and see what they say.. As they've sent out the equivalent of last year's forms they probably won't say much and if that's so then I'll probably keep
schtumm too!
Hi, If you look at last years CSG demand, at the bottom it should show a figure for CSG deductible --this should tally with the figure printed on your 2042--if it doesn't , correct the 2042 figure.
Thank you, I should have thought of looking on the back of that myself! Bewitched , bothered and bewildered would describe me today...
[quote user="Gardener"]Can you do your return on line? May be the form will be the other variety if you haven't had the letter that Jay has had.[/quote]I have just done mine. It was interesting,
because I cannot remember whether or not in previous years, I had to do the Section 8 ( "T" boxes ) myself, but this year the subject didn't even come up. I declared our pensions in 1AS and 1BS as
per and then that was it - no totalling up to do at all. Anyway, very straightforward for me so that's it for another year.[:)]
Boy, totally confused with this year's declaration, or more to the point which forms I should be filling in.Last year I have to confess to using an accountant to fill in our 2042 and 2047, this year
I have decided to try and complete our return myself, but at the moment much confusion abounds.To date I have only received the 2042 (live in 79, Bressuire office) 2047 yet to appear. I have been
sent a 2042SK; last year the accountant used a straight 2042, with no added letters. Unfortunately this year I will have to declare some capital gains from share dealings in 2010, parsnips very
kindly advised me as to which box I am to declare the gains in on another thread, which would be box 3VG, but on my 2042SK form this is no section 3! The form runs from section 1, 2 then 6 and 7,
nothing in between! Even though the pages are numbered 1-4.Questions: Is the 2042SK form the correct 2042 form I should have received, or have the authorities sent me the wrong form?If in fact the
2042SK is the correct 2042 form, then where the ****** **** is box 3VG! Or more to the point where am I expected to declare my capital gains?Should I have expected to receive my 2047 form with or
without a letter by now? Or is it time to pay a visit to the tax office and request our form?Help!
Hmm. I'm looking at 2042K at the mo' and 3VG is certainly on that (2/3 of the way down P3) so maybe SK is wrong for you.
How about doing it on line? You can go back and forth and get it right before pressing the button so it really isn't that daunting, honestly.[:)]
EDIT : We usually get 2047 first so maybe a call to the office if you don't fancy doing it on line would be an idea.
[quote user="cooperlola"][quote user="Gardener"]Can you do your return on line? May be the form will be the other variety if you haven't had the letter that Jay has had.[/quote]I have just done
mine. It was interesting, because I cannot remember whether or not in previous years, I had to do the Section 8 ( "T" boxes ) myself, but this year the subject didn't even come up. I declared our
pensions in 1AS and 1BS as per and then that was it - no totalling up to do at all. Anyway, very straightforward for me so that's it for another year.[:)][/quote]Aren't the 1AS 1BS for French
pensions not overseas ones? The section 8 boxes on the blue form have always been there. The VI , VII and VIII on the pink form are all transposed into Section 8.
[quote user="Gardener"][quote user="cooperlola"][quote user="Gardener"]Can you do your return on line? May be the form will be the other variety if you haven't had the letter that Jay has had.[/
quote]I have just done mine. It was interesting, because I cannot remember whether or not in previous years, I had to do the Section 8 ( "T" boxes ) myself, but this year the subject didn't even come
up. I declared our pensions in 1AS and 1BS as per and then that was it - no totalling up to do at all. Anyway, very straightforward for me so that's it for another year.[:)][/quote] Aren't the 1AS
1BS for French pensions not overseas ones? The section 8 boxes on the blue form have always been there. The VI , VII and VIII on the pink form are all transposed into Section 8.[/quote]From the tax
FAQ's (which I follow every year without problem):
Q I have a UK non-public sector pension - how do I declare this?
A Company pensions and the UK old age state pension are entered (gross) on form 2047 section I. PENSIONS, RETRAITES, RENTES. The totals then go across to box AS/BS on the 2042.
Obviously that doesn't apply if you have a public sector pension from Britain, but we haven't!
[quote user="Gardener"][quote user="cooperlola"][quote user="Gardener"]Can you do your return on line? May be the form will be the other variety if you haven't had the letter that Jay has had.[/
quote]I have just done mine. It was interesting, because I cannot remember whether or not in previous years, I had to do the Section 8 ( "T" boxes ) myself, but this year the subject didn't even come
up. I declared our pensions in 1AS and 1BS as per and then that was it - no totalling up to do at all. Anyway, very straightforward for me so that's it for another year.[:)][/quote] Aren't the 1AS
1BS for French pensions not overseas ones? The section 8 boxes on the blue form have always been there. The VI , VII and VIII on the pink form are all transposed into Section 8.[/quote]Sorry, me
What I meant to convey was that the T boxes didn't come up in my online declaration - yes they are on the forms but I don't use those as I do it electronically.
Thanks a lot for your reply coops.If I fill in our return online how will I get on with any physical paperwork that the tax office may require? I will be claiming a tax credit for our new boiler we
had installed in 2010, and may need to present some contract notes. Is it OK to complete online, then take any paperwork that may be required into the tax office separately?Also is it possible to key
in the phase regarding being in receipt of an E121, thus stopping social charges being taken from our pensions?Lastly if I 'play around' online without hitting the submit button, and complete the
form but not submit it, is it possible to print off the forms without submitting them? At least that way I will then be in possesion of the correct forms.Sorry for all the questions.
From memory, the 2042SK is a pre-completed simplified tax declaration so it's not the one you want.
You can either visit your tax office to get a plain 2042 or download one from the impots website. Alternatively, you can make an online declaration. If you do the latter, then you don't need to
provide any supporting documentation (eg for your tax credits) but just retain them in case of a later tax audit.
Note that the 2011 2042 is not yet available for download but if you're stuck you can download the 2010 form and just alter the date on the front.....[;-)]
[quote user="Grecian"]Thanks a lot for your reply coops. If I fill in our return online how will I get on with any physical paperwork that the tax office may require? I will be claiming a tax credit
for our new boiler we had installed in 2010, and may need to present some contract notes. Is it OK to complete online, then take any paperwork that may be required into the tax office separately?
Also is it possible to key in the phase regarding being in receipt of an E121, thus stopping social charges being taken from our pensions? Lastly if I 'play around' online without hitting the submit
button, and complete the form but not submit it, is it possible to print off the forms without submitting them? At least that way I will then be in possesion of the correct forms. Sorry for all the
questions.[/quote]In his inimitable fashion, Mr Driver has answered most of this.
Can you do screen shots? If so, then you can fill in your tax form on line, take a screen shot of each page, then move away from your submission before finalising it. That way you'll have a copy of
the thing which you can print out.[:)]
EDIT : Whether this is usable as an alternative to the paper form though, I'm not sure because, iirc, the on line submission only gives you the boxes you need, it doesn't take you through the entire
Thanks a lot SD and coops for your replies, the mists are now clearing on the tax form front, well slightly.Just to get this straight in my head if I complete online then I do not supply any physical
paperwork, in my case this year the receipt for the new boiler and contract notes for any capital gains/losses made? Although obviously file everything away for any future audit. If this is the case
then I think 'having a go' completing online for the first time seems worth a shot. I think, or rather hope we have now done most of the 'firsts' in our new life here in France, and there any not
many left to confront us, but I am sure something will manifest itself to throw us into confusion in the not too distant future.Also just to confirm, I take it I can key in the sentence regarding
being in receipt of an E121 online as well.
IIRC, there is a bit at the end for notes, where you could add the E121 info, but I wouldn't swear to this in a court of law!S/D probably knows for sure! As regards the paperwork re your new boiler/
cgt etc, then I'm sure you have it correctly. Keep them on file and the chances are you will never need them.
[quote user="Grecian"]Just to get this straight in my head if I complete online then I do not supply any physical paperwork, in my case this year the receipt for the new boiler and contract notes for
any capital gains/losses made? Although obviously file everything away for any future audit. [/quote]
Correct - they'll ask for it if they want it.
Trust me.[8-|]
Slightly different question which I hope someone can answer. My friend in 61 (Orne) tells me that there is absolutely no return address on his Tax Form - this is his first and he doesn't know where
to return it to.I haven't looked at the form myself so can't vouch for the above but can you please advise where he would return the form to.Thankslaurier
As you know, the address should be printed on the top right hand corner of the front page of his form. If it isn't, he can bung is address etc into this page and the contact details for his tax
office etc will pop up:
This topic is now archived and is closed to further replies. | {"url":"https://forum.completefrance.com/topic/6811-tax-forms/","timestamp":"2024-11-12T17:12:58Z","content_type":"text/html","content_length":"354642","record_id":"<urn:uuid:d3422705-7574-49e4-837b-1a22a86b220f>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00826.warc.gz"} |
Use this applet to investigate the sum of angles in polygons. Is there a pattern? Can you predict the sum of the angles in a 20-sided polygon?
Use the applet...
Duane Habecker, Created with GeoGebra
1. Create a triangle of your choice. Measure each angle of the triangle and find their sum. Try one or two more triangles. Is their angle sum always the same? Record the angle sum for triangles in
the table below. and quadrilaterals in the table below.
2. Create a quadrilateral of your choice. Measure each angle of the quadrilateral and find their sum. Try one or two more quadrilaterals. Is their angle sum always the same? Record the angle sum for
quadrilaterals in the table below.
3. Continue to fill in the table for pentagons and hexagons. Find a pattern.
4. What is the relationship between the number of sides on the shape and the sum of the angles?
│ Polygon │Number of sides │Sum of angles│
│Triangle │ │ │
│Quadrilateral│ │ │
│Pentagon │ │ │
│Hexagon │ │ │
│Septagon │ │ │
│Octagon │ │ │
Extension: What is the sum of the angles in a 20-sided figure?
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Astronomical Units to Leagues Converter
Enter Astronomical Units
β Switch toLeagues to Astronomical Units Converter
How to use this Astronomical Units to Leagues Converter π €
Follow these steps to convert given length from the units of Astronomical Units to the units of Leagues.
1. Enter the input Astronomical Units value in the text field.
2. The calculator converts the given Astronomical Units into Leagues in realtime β using the conversion formula, and displays under the Leagues label. You do not need to click any button. If the
input changes, Leagues value is re-calculated, just like that.
3. You may copy the resulting Leagues value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Astronomical Units to Leagues?
The formula to convert given length from Astronomical Units to Leagues is:
Length[(Leagues)] = Length[(Astronomical Units)] / 3.2273405322519826e-8
Substitute the given value of length in astronomical units, i.e., Length[(Astronomical Units)] in the above formula and simplify the right-hand side value. The resulting value is the length in
leagues, i.e., Length[(Leagues)].
Calculation will be done after you enter a valid input.
Consider that the average distance from Earth to the Sun is 1 astronomical unit (AU).
Convert this distance from astronomical units to Leagues.
The length in astronomical units is:
Length[(Astronomical Units)] = 1
The formula to convert length from astronomical units to leagues is:
Length[(Leagues)] = Length[(Astronomical Units)] / 3.2273405322519826e-8
Substitute given weight Length[(Astronomical Units)] = 1 in the above formula.
Length[(Leagues)] = 1 / 3.2273405322519826e-8
Length[(Leagues)] = 30985264.4927
Final Answer:
Therefore, 1 AU is equal to 30985264.4927 lea.
The length is 30985264.4927 lea, in leagues.
Consider that the distance from Earth to Mars at its closest approach is approximately 0.5 astronomical units (AU).
Convert this distance from astronomical units to Leagues.
The length in astronomical units is:
Length[(Astronomical Units)] = 0.5
The formula to convert length from astronomical units to leagues is:
Length[(Leagues)] = Length[(Astronomical Units)] / 3.2273405322519826e-8
Substitute given weight Length[(Astronomical Units)] = 0.5 in the above formula.
Length[(Leagues)] = 0.5 / 3.2273405322519826e-8
Length[(Leagues)] = 15492632.2464
Final Answer:
Therefore, 0.5 AU is equal to 15492632.2464 lea.
The length is 15492632.2464 lea, in leagues.
Astronomical Units to Leagues Conversion Table
The following table gives some of the most used conversions from Astronomical Units to Leagues.
Astronomical Units (AU) Leagues (lea)
0 AU 0 lea
1 AU 30985264.4927 lea
2 AU 61970528.9855 lea
3 AU 92955793.4782 lea
4 AU 123941057.971 lea
5 AU 154926322.4637 lea
6 AU 185911586.9565 lea
7 AU 216896851.4492 lea
8 AU 247882115.942 lea
9 AU 278867380.4347 lea
10 AU 309852644.9275 lea
20 AU 619705289.855 lea
50 AU 1549263224.6375 lea
100 AU 3098526449.275 lea
1000 AU 30985264492.7499 lea
10000 AU 309852644927.4993 lea
100000 AU 3098526449274.9927 lea
Astronomical Units
An astronomical unit (AU) is a unit of length used in astronomy to measure distances within our solar system. One astronomical unit is equivalent to approximately 149,597,870.7 kilometers or about
92,955,807.3 miles.
The astronomical unit is defined as the mean distance between the Earth and the Sun.
Astronomical units are used to express distances between celestial bodies within the solar system, such as the distances between planets and their orbits. They provide a convenient scale for
describing and comparing distances in a way that is more manageable than using kilometers or miles.
A league is a unit of length that was traditionally used in Europe and Latin America. One league is typically defined as three miles or approximately 4.83 kilometers.
Historically, the league varied in length from one region to another. It was originally based on the distance a person could walk in an hour.
Today, the league is mostly obsolete and is no longer used in modern measurements. It remains as a reference in literature and historical texts.
Frequently Asked Questions (FAQs)
1. What is the formula for converting Astronomical Units to Leagues in Length?
The formula to convert Astronomical Units to Leagues in Length is:
Astronomical Units / 3.2273405322519826e-8
2. Is this tool free or paid?
This Length conversion tool, which converts Astronomical Units to Leagues, is completely free to use.
3. How do I convert Length from Astronomical Units to Leagues?
To convert Length from Astronomical Units to Leagues, you can use the following formula:
Astronomical Units / 3.2273405322519826e-8
For example, if you have a value in Astronomical Units, you substitute that value in place of Astronomical Units in the above formula, and solve the mathematical expression to get the equivalent
value in Leagues. | {"url":"https://convertonline.org/unit/?convert=astronomical_unit-leagues","timestamp":"2024-11-02T06:08:33Z","content_type":"text/html","content_length":"91875","record_id":"<urn:uuid:c8e6140a-0efd-4656-8ccd-6c9773ec9813>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00803.warc.gz"} |
Octblog 2024
Octblog 2024
Welcome to the upgraded site. The new ‘Hamburger’ icons in each section open up their various subsections.
The Geometry, Topology, and Chirality of the Universe.
This crude visual aid for a private lecture on the geometry of the universe.
Shows how a sphere can stereographically project onto a plane.
(Hockey ball with many drill holes for galaxies, black paint and a torch.)
This gives a lower dimensional representation of how a hypersphere can stereographically project into a sphere. Imagine that the plane into which the sphere projects represents a plane bisecting a
sphere. Although this reduced dimensional visual aid represents a hypersphere by a sphere, and a sphere by a plane, it does show how distances in a hypersphere will become exaggerated to observers
who assume that they inhabit a gravitationally flat universe and deceive them into hypothesising an expansion driven by some mysterious dark energy.
See Hypersphere Cosmology equations 16 and 17 and Septblog 2024 on this site, which make a seemingly overwhelming case for the Hypersphere model of the universe.
This month’s Apophenia concerns the Chirality (handedness) of the Universe.
We can represent a Hypersphere by a Hopf Fibration which shows it as decomposed into a fibre bundle of circles each of which passes inside all of the other circles. In Hypersphere Cosmology the
galaxies all Vorticitate around the circles of the Hopf Fibration. This gives the universe no net angular momentum although it allows it to conform to the Godel metric. The circles of the Hopf
Fibration each pass within all of the other circles.
Few if any seem to have noticed that a hypersphere submits to two possible Hopf Fibration modes as shown in the following picture.
Notice that the two Fibrations remain distinct, and we cannot superimpose them by any form of re-orientation in three dimensions, they constitute mirror images of each other. The one on the left has
a Left Handed Fibration and the one on the right has a Right Handed Fibration.
Which chirality corresponds to the universe we inhabit?
Almost certainly we inhabit a universe with a Left Chiral Vorticitation - a Left Handed Universe!
The supporting evidence seems strong but kind of weird: -
1) The weak nuclear force remains exclusively left handed. All neutrinos (the simplest matter particles) rotate anticlockwise.
2) Most terrestrial chiral biomolecules show a preference for left handedness.
3) Rotating spiral galaxies all over the universe show a preference for left handed (anticlockwise) rotation.
4) Galaxies all over the universe seem predominantly aligned with respect to their nearest three neighbouring galaxies, in left handed tetrahedral orientations.
Just how the anticlockwise (left handed) vorticitation of the entire universe could induce these mysterious microcosmic, midi-cosmic, and macrocosmic symmetry breakages remains an interesting
question, yet it seems a fair bet to predict from Hypersphere Cosmology that this universe has a Left-Chiral Vorticitation. It will however probably take a lot of long timescale observation of the
relative movements of far distant galaxies and some heavyweight computer analysis to confirm this.
Topology clarification. The keyring models of the Hopf Fibration do perhaps not make the left or right twists of the fibre bundle entirely apparent. We can easily make it more apparent with three
pieces of string. Three pieces of string can represent the spacetime world lines of three bodies in a hypersphere.
So, flip one piece of card over twice in the same direction to rotate it by 360 degrees and then free the ends from the cards and tie them together. This makes three interlinked circles, showing how
within a hypersphere all particles and bodies move along great circles which (over vast timescales) all pass within each other.
It works with any number of strings, and the result looks oddly similar to various mystical figures from the Vesica Piscis to various Celtic and Tibetan knotwork designs.
Magic News. | {"url":"https://specularium.org/blog/octblog-2024","timestamp":"2024-11-07T12:18:29Z","content_type":"text/html","content_length":"121762","record_id":"<urn:uuid:a4433d83-14c2-478f-b7f0-0cda6db2f9dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00677.warc.gz"} |
The Game Is In Theory
Game theory is promoted as a system where you can apply to any aspect of life. This area of economics has had excellent public relations for a long time. It traps people to take a simple idea and
apply it to complicated situations. Its call for imagination is muddling the gap between its usefulness and how much it is being used.
Game theory is predictive of the behavior of agents who abide by that theory, and are thus worth off due to this behavior. As how a set of fables can only give insights but never advice in real life,
game theory misses out on a lot of information and relevant details. This loss comes from formal language and abstractions that are far away from \(1 - \epsilon\) of the population.
The expected utility model proposed by von Neumann-Morgenstern was the basis for game theory. People still use this principle to model uncertainty even after Duncan and Howard^[1] criticized it for
decades and offered better alternatives. There are many ideas in game theory that have not been fully developed. And most of them may fall in the sphere of pure intellectual analysis and not always
hold practical relevance.
Game theory makes simple abstractions of strategic situations, and formalizes them into [often complex] models. It is only good at offering strategy-proof alternatives. Even so, one good real-life
example of its strategic employment is in security. Many research publications have promoted its broad usage in security. Contributions from Thomas Schelling, von Neumann and John Nash while he was
at RAND Corp., Milind Tambe^[2] and team towards better security are some examples. Discourse comprehension and formalizing argumentation is another unsolved interesting problem for game theory. The
best human argumentation model may only be achieved by a highly performant game theoretic model.
For any decision problem, the combined optimization + control theoretic approach often offers a better alternative. MIP with a class of nonlinear convex constraints using an FSM/MPC is an effective
template for this. Social and economic policies hedge on micro-economic theories with strong game theoretic considerations. Market design, however has aspects of efficiency, optimization and
non-strategic based dynamics. We should stop finding solutions in game theory to problems that theory has nothing to say about. | {"url":"https://densebit.com/posts/11.html","timestamp":"2024-11-12T05:52:48Z","content_type":"text/html","content_length":"5406","record_id":"<urn:uuid:642cff9e-0cea-47f5-82c2-4a66e943eae0>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00763.warc.gz"} |
How to Use a Concrete PSI Chart
From bridges and skyscrapers to tunnels and powerplants, we rely heavily on the stability and security of concrete. Compressive strength resilience plays an enormous role in ensuring the safety and
sturdiness of our structures, but with so many different mixtures, how do we determine which concrete blend is suitable for the job? The answer, of course, is the concrete PSI chart.
Concrete PSI charts are a simple, effective way to assess the compressive strength of different types of concrete. They allow us to make informed decisions about concrete mix ratios, empowering us to
create stronger, better concrete fit for its intended purpose.
In this article, we’ll detail how to read a concrete PSI chart, ways to use one, and how to assess the compressive strength of concrete. We’ll also show you where to find the best equipment for all
your concrete testing applications.
What is a concrete PSI Chart?
A concrete PSI chart is a table indicating the resistance of different concrete mixes during a compressive strength test. This test evaluates how well concrete can handle compression when force is
Concrete compressive strength is an essential way to determine the load-bearing capacity of a concrete mixture. An engineer can use this test to assess whether concrete is fit for its intended use.
The concrete PSI chart is an easy way to compare the PSI of different concrete types and select the right concrete mix ratio for a project. Let’s look at how you can interpret one.
Understanding the Concrete PSI Chart
To begin understanding the concrete PSI chart, let’s first examine an example of a standard concrete PSI chart.
│Concrete Grade│Concrete Mix Ratio (cement: sand : aggregate) │Compressive strength (MPa (N/mm2))│Compressive strength│
│ │ │ │(PSI) │
│M5 │1 : 5: 10 │5 MPa │725 psi │
│M7.5 │1 : 4 : 8 │7.5 MPa │1087 psi │
│M10 │1 : 3 : 6 │10 MPa │1450 psi │
│M15 │1 : 2 : 4 │15 MPa │2175 psi │
│M20 │1: 1.5 : 3 │20 MPa │2900 psi │
│M25 │1 : 1 : 2 │25 MPa │3625 psi │
│M30 │Design Mix │30 MPa │4350 psi │
│M35 │Design Mix │35 MPa │5075 psi │
│M40 │Design Mix │40 MPa │5800 psi │
│M45 │Design Mix │45 MPa │6525 psi │
│M50 │Design Mix │50 MPa │7250 psi │
│M55 │Design Mix │55 MPa │7975 psi │
│M60 │Design Mix │60 MPa │8700 psi │
│M65 │Design Mix │65 MPa │9425 psi │
│M70 │Design Mix │70 MPa │10150 psi │
This table represents a more complex concrete PSI chart than you may typically use because it illustrates concrete’s ability to withstand pressure into both metric (MPa) and imperial (PSI) units.
That said, once you understand the fundamentals behind the concrete PSI chart, it is easily readable, reliable, and beneficial when determining the best mix ratio for your concrete projects. Let’s
break it down.
Keep Reading: Concrete Curing Temperature Chart: Optimizing Strength.
Grades of Concrete and Mix Ratio
At the far left of the table, we can see the grades of concrete, denoted by ‘M’. The initial concrete grades, such as C10, C15, and C20, are used for various applications in both domestic and
commercial settings. In this instance, M stands for ‘mix’, and the number next to the M represents the compressive strength in megapascals (MPa) of a molded concrete cube or concrete cylinder
that has been cured for 28 days. The higher the ‘M’ grade, the more resistant the concrete is under compressive stress.
Both MPa and PSI measure the compressive strength of concrete. MPa is the metric version, whereas a Concrete PSI Chart, the measurement you’ll be most familiar with, is the imperial version used in
the United States.
To the right of the concrete grade, you’ll notice the concrete mix ratio. This tells us the correct proportions of cement, sand, and aggregate (in that order) to create each concrete grade. For
example, M5 concrete requires a ratio of 1:5:10. This means that 1 kg of cement, 5 kg of sand, and 10 kg of aggregates would make concrete with a compressive strength of 5 MPa.
Nominal Mix vs. Design Mix: What’s the Difference?
When reading the concrete PSI chart, you may notice that some grades are paired with a mix ratio while others are labeled design mix. Let’s explore what this means.
Nominal Mixes
Grades M5-M25 are labeled by a specific concrete mix ratio. For example, when creating M10 concrete, we know the mix proportions are one part cement, three parts sand, and six parts aggregates. We
refer to these concrete grades as ‘nominal mixes’.
A nominal mix is based on a tried-and-tested approach from years of trial and error. It requires no scientific experimentation. Anyone of any skill level can create these mixes. As long as you follow
the ratio instructions for your chosen M grade, you’ll create a low-mid strength concrete to a reasonable degree of accuracy.
That said, nominal mixes can be unreliable because they don’t consider factors such as material characteristics, curing time, and water content. For this reason, you’ll typically only use a nominal
mix to create regular concrete for use in general construction and repair projects.
Design Mixes
Grades M30 to M70 are labeled ‘design mix’. This means the concrete designer specifies the concrete mixture ratio based on scientific analysis and experimentation.
A design mix considers the water ratio in cement paste, the unique properties of the materials, and other external factors. Design mixes may also incorporate various admixtures that impact the
concrete’s PSI.
When developing a design mix, an engineer will create several batches, experimenting with different ratios to create a concrete mix with the desired PSI strength on the concrete PSI chart.
Experimentation is vital for stronger concrete mixtures because a structural engineer cannot rely on an inconsistent nominal mix ratio when constructing an important structure.
Water Ratio
A design mix considers the water ratio in cement paste, the unique properties of the materials, and other external factors. Design mixes may also incorporate various admixtures that impact the
concrete’s PSI.
When developing a design mix, an engineer will create several batches, experimenting with different ratios to create a concrete mix with the desired PSI strength on the concrete PSI chart.
Experimentation is vital for stronger concrete mixtures because a structural engineer cannot rely on an inconsistent nominal mix ratio when constructing an important structure.
Design Mix Importance – Example of Concrete PSI Chart Usage
For example, a bridge requires concrete with a high PSI strength on the concrete PSI chart. If the engineer created M40-grade concrete using a nominal ratio, there would be no way of guaranteeing the
concrete’s resistance to compressive strength. This is because the engineer didn’t consider how material properties impact the final mix. Finer aggregates may result in less resistant concrete, which
could be dangerous when used to build large-scale structures like bridges.
Creating a design mix requires comprehensive knowledge of concrete properties. This means it is only suited to those who have a lot of expertise. Design mix concrete is typical in larger, more
complex structures that will experience heavy loads and extreme wear.
More Read: Concrete Break Machine Models for Efficient Strength Testing.
Compressive Strength (MPa and PSI) on the Concrete PSI Chart
On the far right of the table, you’ll notice the different compressive strengths of the concrete mixtures. The compressive strength indicates the maximum load a concrete mix can handle before it
In the table above, compressive strength is denoted in both MPa (N/mm2) and PSI. MPa stands for Megapascal, and concrete PSI chart stands for Pounds per Square Inch.
As previously stated, MPa and PSI both measure the compressive strength of concrete, MPa being metric and PSI being imperial.
The good news is that you can translate the two easily. One Megapascal is the equivalent of 145 PSI. This means that concrete with a compressive strength resistance of 10MPa will have a PSI of 1450.
What Projects are Different Concrete Mix Ratios Useful For?
We now know that concrete with a higher PSI is more compression-resistant and a Concrete PSI Chart, but how does this apply in the real world?
Nominal concrete mixtures under 3,000 PSI shouldn’t be used for complex load-bearing structures. They are best suited to applications such as:
• Flooring, sidewalks, and driveways
• Repair work
• Temporary structures
Concrete in the range of 3,000-4,000 PSI is well-suited to basic structural components like:
• Columns, beams, slabs, and footers
• Small-scale construction projects
• Slab foundations and footings, especially in situations where heavy loads are expected to be stored or moved, such as RV pads
Design mixtures in the range of 4,000-6,000 on the concrete PSI chart are very strong. As such, an engineer may use them when building:
• Bridges and large-scale buildings
• Warehouses and factories
Higher than 7,000 PSI Concrete
Anything higher than 7,000 PSI concrete is considered ultra-high-strength concrete. These concrete mixtures are used in large-scale load-bearing structures or in situations where contamination is
possible. For example:
• Nuclear powerplants
• High-rise buildings
• Bridges and tunnels
How to Test the Compressive Strength of a Concrete Mixture?
Depending on the concrete standards you follow, you may choose to test compressive strength using either a concrete cube or concrete cylinder. Here’s a basic step-by-step guide to testing concrete
psi chart and compressive strength.
• Prepare the sample: Take a concrete sample from freshly-poured concrete. Pour the sample into your cube or cylinder mold. Use a tamper to remove excess air and compact the concrete.
• Cure the sample:Â Remove your concrete from the mold after 24 to 48 hours and cure it according to your chosen standard. This typically involves placing the concrete into a moist environment at a
specified temperature for a predetermined period (usually 28 days).
• Test the sample: Place the sample in a concrete compressor such as a compressometer or automatic compression testing machine. These machines apply load at a constant rate following recognized
• Find the PSI: Once the sample fails, calculate PSI by dividing the force applied by the cross-sectional area of the specimen. For example, a 2″ concrete cube has a cross-sectional area of 4
square inches. Let’s assume that the concrete withstood a force of 10,0000 pounds. Dividing 10000 by 4 gives us a concrete strength of 2500 PSI.
Where Can I Find the Best Material Testing Equipment?
Knowing how to interpret a concrete PSI chart is one thing. Having the right equipment for your own concrete testing is another. CertifiedMTP is your one-stop shop for all your material testing
We offer a range of compression testing equipment, molds, tampers, finishing trowels, and more, designed to make creating accurate concrete mixtures fast, easy and reliable.
Looking for something different? We also offer a range of soil testing, aggregate testing, asphalt testing, and cement testing equipment. Whatever your material testing needs, CertifiedMTP has
the equipment and expertise to help you get things done.
Can’t find what you’re looking for? Get in touch. With thousands of products from some of the world’s leading material testing brands, we’re sure to have what you need.
For more concrete testing needs, consider the popular Mini-Jaw Crusher
Related Blogs for Concrete PSI Chart | {"url":"https://blog.certifiedmtp.com/how-to-use-a-concrete-psi-chart/","timestamp":"2024-11-12T09:07:27Z","content_type":"text/html","content_length":"361653","record_id":"<urn:uuid:496095f6-d7d2-4712-9fca-853f9a516c4b>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00219.warc.gz"} |
FAQ: How do I interpret odds ratios in logistic regression?
When a binary outcome variable is modeled using logistic regression, it is assumed that the logit transformation of the outcome variable has a linear relationship with the predictor variables. This
makes the interpretation of the regression coefficients somewhat tricky. In this page, we will walk through the concept of odds ratio and try to interpret the logistic regression results using the
concept of odds ratio in a couple of examples.
From probability to odds to log of odds
Everything starts with the concept of probability. Let’s say that the probability of success of some event is .8. Then the probability of failure is 1 – .8 = .2. The odds of success are defined as
the ratio of the probability of success over the probability of failure. In our example, the odds of success are .8/.2 = 4. That is to say that the odds of success are 4 to 1. If the probability
of success is .5, i.e., 50-50 percent chance, then the odds of success is 1 to 1.
The transformation from probability to odds is a monotonic transformation, meaning the odds increase as the probability increases or vice versa. Probability ranges from 0 and 1. Odds range from 0
and positive infinity. Below is a table of the transformation from probability to odds and we have also plotted for the range of p less than or equal to .9.
p odds
.001 .001001
.01 .010101
.15 .1764706
.2 .25
.25 .3333333
.3 .4285714
.35 .5384616
.4 .6666667
.45 .8181818
.5 1
.55 1.222222
.6 1.5
.65 1.857143
.7 2.333333
.75 3
.8 4
.85 5.666667
.9 9
.999 999
.9999 9999
The transformation from odds to log of odds is the log transformation (In statistics, in general, when we use log almost always it means natural logarithm). Again this is a monotonic
transformation. That is to say, the greater the odds, the greater the log of odds and vice versa. The table below shows the relationship among the probability, odds and log of odds. We have also
shown the plot of log odds against odds.
p odds logodds
.001 .001001 -6.906755
.01 .010101 -4.59512
.15 .1764706 -1.734601
.2 .25 -1.386294
.25 .3333333 -1.098612
.3 .4285714 -.8472978
.35 .5384616 -.6190392
.4 .6666667 -.4054651
.45 .8181818 -.2006707
.5 1 0
.55 1.222222 .2006707
.6 1.5 .4054651
.65 1.857143 .6190392
.7 2.333333 .8472978
.75 3 1.098612
.8 4 1.386294
.85 5.666667 1.734601
.9 9 2.197225
.999 999 6.906755
.9999 9999 9.21024
Why do we take all the trouble doing the transformation from probability to log odds? One reason is that it is usually difficult to model a variable which has restricted range, such as probability.
This transformation is an attempt to get around the restricted range problem. It maps probability ranging between 0 and 1 to log odds ranging from negative infinity to positive infinity. Another
reason is that among all of the infinitely many choices of transformation, the log of odds is one of the easiest to understand and interpret. This transformation is called logit transformation. The
other common choice is the probit transformation, which will not be covered here.
A logistic regression model allows us to establish a relationship between a binary outcome variable and a group of predictor variables. It models the logit-transformed probability as a linear
relationship with the predictor variables. More formally, let $Y$ be the binary outcome variable indicating failure/success with $\{0,1\}$ and $p$ be the probability of $y$ to be $1$, $p = P(Y=1)$.
Let $x_1, \cdots, x_k$ be a set of predictor variables. Then the logistic regression of $Y$ on $x_1, \cdots, x_k$ estimates parameter values for $\beta_0, \beta_1, \cdots, \beta_k$ via maximum
likelihood method of the following equation
$$logit(p) = log(\frac{p}{1-p}) = \beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k.$$
Exponentiate and take the multiplicative inverse of both sides,
$$\frac{1-p}{p} = \frac{1}{exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k)}.$$
Partial out the fraction on the left-hand side of the equation and add one to both sides,
$$\frac{1}{p} = 1 + \frac{1}{exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k)}.$$
Change 1 to a common denominator,
$$\frac{1}{p} = \frac{exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k)+1}{exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k)}.$$
Finally, take the multiplicative inverse again to obtain the formula for the probability $P(Y=1)$,
$${p} = \frac{exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k)}{1+exp(\beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k)}.$$
We are now ready for a few examples of logistic regressions. We will use a sample dataset, https://stats.idre.ucla.edu/wp-content/uploads/2016/02/sample.csv, for the purpose of illustration. The
data set has 200 observations and the outcome variable used will be hon, indicating if a student is in an honors class or not. So our p = prob(hon=1). We will purposely ignore all the significance
tests and focus on the meaning of the regression coefficients. The output on this page was created using Stata with some editing.
Logistic regression with no predictor variables
Let’s start with the simplest logistic regression, a model without any predictor variables. In an equation, we are modeling
logit(p)= β[0 ]
Logistic regression Number of obs = 200
LR chi2(0) = 0.00
Prob > chi2 = .
Log likelihood = -111.35502 Pseudo R2 = 0.0000
hon | Coef. Std. Err. z P>|z| [95% Conf. Interval]
intercept | -1.12546 .1644101 -6.85 0.000 -1.447697 -.8032217
This means log(p/(1-p)) = -1.12546. What is p here? It turns out that p is the overall probability of being in honors class ( hon = 1). Let’s take a look at the frequency table for hon.
hon | Freq. Percent Cum.
0 | 151 75.50 75.50
1 | 49 24.50 100.00
Total | 200 100.00
So p = 49/200 = .245. The odds are .245/(1-.245) = .3245 and the log of the odds (logit) is log(.3245) = -1.12546. In other words, the intercept from the model with no predictor variables is the
estimated log odds of being in honors class for the whole population of interest. We can also transform the log of the odds back to a probability: p = exp(-1.12546)/(1+exp(-1.12546)) = .245, if we
Logistic regression with a single dichotomous predictor variables
Now let’s go one step further by adding a binary predictor variable, female, to the model. Writing it in an equation, the model describes the following linear relationship.
logit(p) = β[0 ]+ β[1]*female
Logistic regression Number of obs = 200
LR chi2(1) = 3.10
Prob > chi2 = 0.0781
Log likelihood = -109.80312 Pseudo R2 = 0.0139
hon | Coef. Std. Err. z P>|z| [95% Conf. Interval]
female | .5927822 .3414294 1.74 0.083 -.0764072 1.261972
intercept | -1.470852 .2689555 -5.47 0.000 -1.997995 -.9437087
Before trying to interpret the two parameters estimated above, let’s take a look at the crosstab of the variable hon with female.
| female
hon | male female | Total
0 | 74 77 | 151
1 | 17 32 | 49
Total | 91 109 | 200
In our dataset, what are the odds of a male being in the honors class and what are the odds of a female being in the honors class? We can manually calculate these odds from the table: for males, the
odds of being in the honors class are (17/91)/(74/91) = 17/74 = .23; and for females, the odds of being in the honors class are (32/109)/(77/109) = 32/77 = .42. The ratio of the odds for female to
the odds for male is (32/77)/(17/74) = (32*74)/(77*17) = 1.809. So the odds for males are 17 to 74, the odds for females are 32 to 77, and the odds for female are about 81% higher than the odds for
Now we can relate the odds for males and females and the output from the logistic regression. The intercept of -1.471 is the log odds for males since male is the reference group (the variable female
= 0). Using the odds we calculated above for males, we can confirm this: log(.23) = -1.47. The coefficient for female is the log of odds ratio between the female group and male group: log(1.809) =
.593. So we can get the odds ratio by exponentiating the coefficient for female. Most statistical packages display both the raw regression coefficients and the exponentiated coefficients for
logistic regression models. The table below is created by Stata.
Logistic regression Number of obs = 200
LR chi2(1) = 3.10
Prob > chi2 = 0.0781
Log likelihood = -109.80312 Pseudo R2 = 0.0139
hon | Odds Ratio Std. Err. z P>|z| [95% Conf. Interval]
female | 1.809015 .6176508 1.74 0.083 .9264389 3.532379
Logistic regression with a single continuous predictor variable
Another simple example is a model with a single continuous predictor variable such as the model below. It describes the relationship between students’ math scores and the log odds of being in an
honors class.
logit(p) = β[0 ]+ β[1]*math
Logistic regression Number of obs = 200
LR chi2(1) = 55.64
Prob > chi2 = 0.0000
Log likelihood = -83.536619 Pseudo R2 = 0.2498
hon | Coef. Std. Err. z P>|z| [95% Conf. Interval]
math | .1563404 .0256095 6.10 0.000 .1061467 .206534
intercept | -9.793942 1.481745 -6.61 0.000 -12.69811 -6.889775
In this case, the estimated coefficient for the intercept is the log odds of a student with a math score of zero being in an honors class. In other words, the odds of being in an honors class when
the math score is zero is exp(-9.793942) = .00005579. These odds are very low, but if we look at the distribution of the variable math, we will see that no one in the sample has math score lower
than 30. In fact, all the test scores in the data set were standardized around mean of 50 and standard deviation of 10. So the intercept in this model corresponds to the log odds of being in an
honors class when math is at the hypothetical value of zero.
How do we interpret the coefficient for math? The coefficient and intercept estimates give us the following equation:
log(p/(1-p)) = logit(p) = – 9.793942 + .1563404*math
Let’s fix math at some value. We will use 54. Then the conditional logit of being in an honors class when the math score is held at 54 is
log(p/(1-p))(math=54) = – 9.793942 + .1563404 *54.
We can examine the effect of a one-unit increase in math score. When the math score is held at 55, the conditional logit of being in an honors class is
log(p/(1-p))(math=55) = – 9.793942 + .1563404*55.
Taking the difference of the two equations, we have the following:
log(p/(1-p))(math=55) – log(p/(1-p))(math = 54) = .1563404.
We can say now that the coefficient for math is the difference in the log odds. In other words, for a one-unit increase in the math score, the expected change in log odds is .1563404.
Can we translate this change in log odds to the change in odds? Indeed, we can. Recall that logarithm converts multiplication and division to addition and subtraction. Its inverse, the
exponentiation converts addition and subtraction back to multiplication and division. If we exponentiate both sides of our last equation, we have the following:
exp[log(p/(1-p))(math=55) – log(p/(1-p))(math = 54)] = exp(log(p/(1-p))(math=55)) / exp(log(p/(1-p))(math = 54)) = odds(math=55)/odds(math=54) = exp(.1563404) = 1.1692241.
So we can say for a one-unit increase in math score, we expect to see about 17% increase in the odds of being in an honors class. This 17% of increase does not depend on the value that math is held
Logistic regression with multiple predictor variables and no interaction terms
In general, we can have multiple predictor variables in a logistic regression model.
logit(p) = log(p/(1-p))= β[0 ] + β[1]*x1 + … + β[k]*xk
Applying such a model to our example dataset, each estimated coefficient is the expected change in the log odds of being in an honors class for a unit increase in the corresponding predictor variable
holding the other predictor variables constant at certain value. Each exponentiated coefficient is the ratio of two odds, or the change in odds in the multiplicative scale for a unit increase in the
corresponding predictor variable holding other variables at certain value. Here is an example.
logit(p) = log(p/(1-p))= β[0 ] + β[1]*math + β[2]*female + β[3]*read
Logistic regression Number of obs = 200
LR chi2(3) = 66.54
Prob > chi2 = 0.0000
Log likelihood = -78.084776 Pseudo R2 = 0.2988
hon | Coef. Std. Err. z P>|z| [95% Conf. Interval]
math | .1229589 .0312756 3.93 0.000 .0616599 .1842578
female | .979948 .4216264 2.32 0.020 .1535755 1.80632
read | .0590632 .0265528 2.22 0.026 .0070207 .1111058
intercept | -11.77025 1.710679 -6.88 0.000 -15.12311 -8.417376
This fitted model says that, holding math and reading at a fixed value, the odds of getting into an honors class for females (female = 1)over the odds of getting into an honors class for males (
female = 0) is exp(.979948) = 2.66. In terms of percent change, we can say that the odds for females are 166% higher than the odds for males. The coefficient for math says that, holding female and
reading at a fixed value, we will see 13% increase in the odds of getting into an honors class for a one-unit increase in math score since exp(.1229589) = 1.13.
Logistic regression with an interaction term of two predictor variables
In all the previous examples, we have said that the regression coefficient of a variable corresponds to the change in log odds and its exponentiated form corresponds to the odds ratio. This is only
true when our model does not have any interaction terms. When a model has interaction term(s) of two predictor variables, it attempts to describe how the effect of a predictor variable depends on
the level/value of another predictor variable. The interpretation of the regression coefficients become more involved.
Let’s take a simple example.
logit(p) = log(p/(1-p))= β[0] + β[1]*female + β[2]*math + β[3]*female*math
Logistic regression Number of obs = 200
LR chi2(3) = 62.94
Prob > chi2 = 0.0000
Log likelihood = -79.883301 Pseudo R2 = 0.2826
hon | Coef. Std. Err. z P>|z| [95% Conf. Interval]
female | -2.899863 3.094186 -0.94 0.349 -8.964357 3.164631
math | .1293781 .0358834 3.61 0.000 .0590479 .1997082
femalexmath | .0669951 .05346 1.25 0.210 -.0377846 .1717749
intercept | -8.745841 2.12913 -4.11 0.000 -12.91886 -4.572823
In the presence of interaction term of female by math, we can no longer talk about the effect of female, holding all other variables at certain value, since it does not make sense to fix math and
femalexmath at certain value and still allow female change from 0 to 1!
In this simple example where we examine the interaction of a binary variable and a continuous variable, we can think that we actually have two equations: one for males and one for females. For males
(female=0), the equation is simply
logit(p) = log(p/(1-p))= β[0 ]+ β[2]*math.
For females, the equation is
logit(p) = log(p/(1-p))= (β[0 ]+ β[1]) + (β[2] + β[3 ])*math.
Now we can map the logistic regression output to these two equations. So we can say that the coefficient for math is the effect of math when female = 0. More explicitly, we can say that for male
students, a one-unit increase in math score yields a change in log odds of 0.13. On the other hand, for the female students, a one-unit increase in math score yields a change in log odds of (.13 +
.067) = 0.197. In terms of odds ratios, we can say that for male students, the odds ratio is exp(.13) = 1.14 for a one-unit increase in math score and the odds ratio for female students is exp
(.197) = 1.22 for a one-unit increase in math score. The ratio of these two odds ratios (female over male) turns out to be the exponentiated coefficient for the interaction term of female by math:
1.22/1.14 = exp(.067) = 1.07. | {"url":"https://stats.oarc.ucla.edu/other/mult-pkg/faq/general/faq-how-do-i-interpret-odds-ratios-in-logistic-regression/","timestamp":"2024-11-03T03:25:25Z","content_type":"text/html","content_length":"57376","record_id":"<urn:uuid:7f0c84cf-3383-484a-9410-319effadf80b>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00524.warc.gz"} |
Theory to the Mystery of the Super Massive Black Holes
1. Introduction
Edwin Hubble was the first astronomer who inferred that spiral nebulae are galaxies located at great distances from the Milky Way, publication 1929. The speeds of stars in these galaxies have been
measured based on the redshifts of the light they emit. The speed of a star results in a change in the wavelength, which is called a redshift. In the Universe, all stars at great distances shift
their wavelength towards red, and therefore, it has been inferred that the Universe is expanding. By using the redshifts, Vera Rubin measured the rotating speeds of galaxies and found that the stars
rotate in the outer part of the galaxy approximately at the same speed, Ref. [1] 1983. The speed of rotation is zero at the center of the galaxy, after which it rapidly increases until at a certain
value it remains constant. The matter of a spiral galaxy is concentrated into a great, dense bulge at the center of the galaxy, and the rest of the galactic matter forms a large rotating spiral disk.
The gravitational force of this kind of galactic matter distribution reduces approximately inversely proportional to the square of the distance from the center of the galaxy. Therefore, the constant
value speed distribution is not only resulted from the visible and dark matter in the center bulge and disk of the galaxy. There must be a great amount of some other types of dark matter affecting
the rotation of stars. The question is what type of mass distribution would achieve constant rotational speed distribution on the outer part of the galactic disk. It should probably be a ball-shaped
dark matter formation above the galactic disk. A sufficient amount of dark matter above the galactic disk was not found. The measurements of the space telescope Hubble have not given support to this
theory, Ref. [3] 1995. It seems that dark matter in the halo of the galaxy is not so massive that it could explain the constant speed distribution. After that, the search for dark matter has
primarily concentrated on fundamental particles. It must be a fundamental particle that has mass and force of gravity. The particle must not have any other forces because otherwise it would have been
observed by now. Fundamental particles like this have been searched in mine shafts that enable minimal disturbing factors. Richard Panek has written on the latest difficulties in the research of the
Universe, Ref. [4] 2020. Astronomers have calculated how fast the universe is expanding. Two different calculation methods have two different results. The cause of the discrepancy may be the fact
that physics has new unknown phenomena. In Ref. [5] 2022, Anil Ananthaswamy has written about the same theme: How fast is the Universe expanding? How much does the matter clump up? The final
conclusion is: We are missing something.
In Ref. [6] “Galaxy Rotation in the Space of Four Distance Dimensions”, the dark matter mystery has been solved in the Universe of four distance dimensions. Ref. [6] presented a new idea that dark
matter is located at the fourth distance dimension above the center of the galaxy. In the same manner as a structure of three dimensions can be drawn in the cross-sections of two dimensions, a
structure of four dimensions can also be drawn in the cross-sections of two or three dimensions. Therefore, the determination of the location of dark matter in the fourth dimension is no problem. The
study in Ref. [6] contains the solution to the dark matter mystery of spiral galaxies by using the space of four distance dimensions x,y,z,x', in which x' is the fourth distance dimension. The
four-dimensional mass M, which generates the main gravitation field of the galaxy, is located in the fourth dimension at the distance x' = X' and the other dimensions are zero x = 0, y = 0, z = 0.
The speed distribution curve of the four-dimensional mass v[M] is the effect of the mass M on the total rotational speed of the galaxy. Ref. [2] has presented the rotational speed distribution curves
of the galaxy NGC 3198. The speed distribution curve of the galactic halo in Ref. [2] corresponds to the speed distribution curve v[M] of the four-dimensional mass M in Ref. [6] . In order to find
out how well this four-dimensional model functions, the speed distribution curve v[M] was calculated, and it was compared to the halo speed distribution curve of Ref. [2] . The conclusion was that
the calculated distribution curve v[M] is a good match to the halo curve in Ref. [2] . Furthermore, four rotational speed distribution curves v[M] were calculated using different values of the
distance X', which yielded different values for the maximum radius of the galaxy. In this manner, the different galaxy models of Ref. [1] were obtained, and therefore, this solution to the dark
matter mystery was proved.
Ref. [7] “The Solution to the Dark Energy Mystery in the Universe of Four Distance Dimensions” solves the mystery of dark energy by using the structure of the four-dimensional Universe. The model of
the Universe is the surface volume of the four-dimensional spherical Universe. This type of structure of the Universe creates the same kind of an accelerating redshift increase that has been
measured. The cause of the redshift in this model of the Universe is its structure, and therefore, there is no accelerating expansion of the Universe. In order to prove this theory, the model of the
surface volume of the four-dimensional Universe was constructed, the equation of the redshift caused by this Universe was solved, and the theoretical equation was shown to be the same as the measured
redshift in the Universe. The measured redshift in the Universe was obtained from the derivative of the model of the expanding Universe in Ref. [8] . A similar model of the Universe has been
constructed by NASA [9] . The four-dimensional model of the Universe in this study yielded the Universe that was found to have decelerating expansion until this point of time and a Big Bang that was
not very big.
Black hole is a body of space which is so massive that even the speed of light is not enough to escape it. In this publication the function of the black holes is studied in our Universe of three
distance dimensions, and in the super Universe of four distance dimensions. In the for-dimensional Universe, the black holes are rotating galaxies, and if they are near enough to our
three-dimensional Universe, they can act as quasars. The quasars spit matter into space, and therefore it can be inferred that their black holes are out of our three-dimensional Universe. No matter
can escape the gravitational force of that kind of black hole. Astronomers have been searching for suitable black hole candidates, and the most famous are, Ref. [10] : Sagittarius A*, mass 7.8 × 10^
36 kg, which is the supermassive black hole at the centre of our Milky Way. It’s generally inactive, with only modest X-ray outbursts as it consumes small gas clouds. The first black hole to be
imaged right down to the event horizon, revealing the black hole’s “shadow” on the surrounding accretion disc is Messier 87 black hole, mass 1.2 × 10^40 kg. The most massive black hole is the
supermassive black hole found in the quasar TON 618, mass 12.5 × 10^40 kg.
2. Method of Calculation
In the following, a simple method has been presented to see the space of four distance dimensions, and to perform calculations in it. In Figure 1, a box is drawn in the three-dimensional space x,y,z.
On Figure 2, the same box is drawn in the two-dimensional space z,x and on Figure 3 it is drawn in the two-dimensional space z,y. There is still a third possible set of coordinates x,y, but it is not
needed to determine the shape of the box. Here, it is seen that a structure of three dimensions can be drawn in a two-dimensional coordination system. In the same manner, a structure of four
dimensions can be drawn in coordination with the three dimensions x,y,z, y,z,x', x,z,x', x,y,x'. If the structure is simple, it is possible
Figure 1. A box drawn in coordination of three-dimensional space x,y,z.
Figure 2. The box of Figure 1 drawn in coordination of two-dimensional space z,x.
Figure 3. The box of Figure 1 drawn in coordination of two-dimensional space z,y.
that only one two-dimensional co-ordination is needed to determine the form of the structure.
3. Calculation of the Rotational Speed of Galaxy
In Figure 4, the four-dimensional space x,y,z,x' is presented as a drawing of two-dimensional coordination system x,x'. The Universe is comprised of three dimensions x,y,z and the enlarged Universe
has four dimensions x,y,z,x'. The coordination system of Figure 4 is constructed from the coordination system of Figure 1 so that the galaxy rotates at the plane of x,y coordinates and the coordinate
z has been exchanged to the coordinate of the fourth-dimension x'. In the figure a star of mass m rotates round about the center of galaxy O. The rotation axis of the galaxy is the coordinate x' and
the four-dimensional mass M is at the distance X' from the center of galaxy O. The mass of the star m and four-dimensional mass M draws each other with the force of gravity. Because the
three-dimensional gravity force of the star is lacking the force of fourth dimension, the force that has an effect on the star is Fcosα.
It seems that this model fits very well for the measurements. Near the center of galaxy, rotational speeds from the redshift measurements and theoretically calculated speeds from the light
intensities of rotating stars have relatively good correspondence. In the center of galaxy, the gravity force between the star of mass m and the four-dimensional mass M is in the x' axis direction,
and it disappears completely. At the border areas of the galactic rotating disk the gravity force of the mass M increases in the x axis direction Fcosα which is according to the redshift
measurements. Even over 90% of the gravity forces acting at the border areas of galaxy cannot be explained by the masses of the known visible and dark matter. This model of the space of four
dimensions can explain the weird rotation speeds of galaxies.
The mystery of dark matter in the rotation of galaxies is a common subject matter in the study books of cosmology, Ref. [11] . This mystery has been solved in this paper by the gravitational force of
four-dimensional mass in the
Figure 4. A star with mass m rotates round about the center of galaxy O at the distance R. Fourth-dimensional mass M is located at the distance X' from the center of galaxy.
space of four dimensions. In Figure 5, the star with mass m rotates round about the center of galaxy along the circle of radius R. As the star rotates, the force of gravity which has an effect on the
star is equal to the centrifugal force of its rotation movement.
The differentia of the gravitational force is equal to the differentia of the centrifugal force.
$\frac{\gamma m\Delta M\left(x\right)}{{\left(R-x\right)}^{2}}=\frac{m}{R}\text{d}\left({v}_{n}^{2}\right)$(1)
If x < R, the differentia of the gravitational force increases the centrifugal force.
If x > R, the differentia of the gravitational force decreases the centrifugal force.
Therefore, this equations shows that the halo mass distribution structure cannot be the solution to the galaxy rotation speed mystery because Equation (1) cancels as halo structure continues about
the same mass M(x) on the both sides of the location of the mass m, and there is no rotation speed generation. The dark matter which rotates the galaxy must be on the rotation axis, or near it.
Summation yields the centrifugal force component of the visible galactic mass
$\gamma m{\sum }^{\text{}}\frac{\Delta M\left(x\right)}{{\left(R-x\right)}^{2}}=\frac{m}{R}\int \text{d}\left({v}_{n}^{2}\right)=\frac{m{v}_{n}^{2}}{R}$(2)
The star’s rotational speed round about the center of the galaxy is v. The speed distribution components of the star’s rotational speed v are as follows: the speed distribution component from the
visible mass is v[n] and from the ordinary dark matter is v[p] and the speed distribution component from the four-dimensional mass M is v[M]. The centrifugal force components are from visible mass,
from ordinary dark matter and from the four-dimensional mass M.
The following equation is obtained by adding the three centrifugal force component.
Figure 5. Calculation of the effect of the visible galactic mass ΔM(x) = M(x + dx/2)dx on the rotational speed component v[n] of the star which mass is m.
The total rotational speed curve v and the rotational speed curve of visible mass v[n] are known, the rotational speed curve of ordinary dark matter v[p] is not knows, but it can be concluded that it
is the same form as the speed curve of the visible matter v[n]. The rotational speed curve of the four-dimensional mass M v[M] is not known and it will be calculated by using the knowledge of the
rotational system. In Figure 6, a star which mass is m rotates round about the center of galaxy O at the distance R. The four-dimensional mass M is on the rotation axis x' at the distance X' from the
center of galaxy. The galaxy has two rotational axes z and x'. This can be seen as follows: The rotating star system in the four-dimensional space x,y,z,x' in Figure 6 is transformed into the system
of oscillating atom about zero point O in the three-dimensional system x,y,x'. If the oscillating atom is at the point of the mass m, and it is oscillating along the axis x about the point O, it has
two axes about which it oscillates y,x' in the three-dimensional space x,y,x'. In the same manner the star rotating round about point O has two axes z,x' round about which it rotates in
four-dimensional space x,y,z,x'. The gravity force of the three-dimensional mass decreases in the inverse value of the square of the distance, and it can be supposed that the gravity force of the
four-dimensional mass decreases in the inverse of the cube of the distance, and therefore, the gravity force of the four-dimensional mass M is
$F=\frac{\gamma mM}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{3/2}}$(6)
in which the component at the direction of x-axis is
$F\mathrm{cos}\alpha =\frac{\gamma mM}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{3/2}}\cdot \frac{R}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{1/2}}$(7)
Figure 6. A star of mass m rotates round about the center of galaxy O along a circle trajectory of the radius R. The gravity effect on the star from the four-dimensional mass M is at the direction of
x axis Fcosα.
$F\mathrm{cos}\alpha =\frac{R\gamma mM}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}$(8)
The centrifugal force component corresponding to the four-dimensional mass M is
Calculation of the speed distribution curve of the four-dimensional mass v[M]. The gravitational force is equal to the centrifugal force.
$\frac{m{\left({v}_{M}\right)}^{2}}{R}=\frac{R\gamma mM}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}$(10)
The speed distribution component of the four-dimensional mass M
${v}_{M}=\sqrt{\frac{{R}^{2}\gamma M}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}}=\frac{R\sqrt{\gamma M}}{{R}^{2}+{\left({X}^{\prime }\right)}^{2}}$(11)
The four-dimensional mass M
$M=\frac{{v}_{M}^{2}{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}{{R}^{2}\gamma }$(12)
Equation (10) at the distances R[1] and R[2] from the center of galaxy
$\frac{{m}_{1}{\left({v}_{M1}\right)}^{2}}{{R}_{1}}=\frac{{R}_{1}\gamma {m}_{1}M}{{\left({R}_{1}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}$(13)
$\frac{{m}_{2}{\left({v}_{M2}\right)}^{2}}{{R}_{2}}=\frac{{R}_{2}\gamma {m}_{2}M}{{\left({R}_{2}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}$(14)
By dividing the two equations above
$\frac{{R}_{2}^{2}{\left({v}_{M1}\right)}^{2}}{{R}_{1}^{2}{\left({v}_{M2}\right)}^{2}}=\frac{{\left({R}_{2}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}{{\left({R}_{1}^{2}+{\left({X}^{\prime }\
$a\left({R}_{1}^{2}+{\left({X}^{\prime }\right)}^{2}\right)={R}_{2}^{2}+{\left({X}^{\prime }\right)}^{2}$(17)
The distance of the four-dimensional mass M from the center of galaxy
${X}^{\prime }=\sqrt{\frac{{R}_{2}^{2}-a{R}_{1}^{2}}{a-1}}$(18)
The values of the total rotational speeds v have been calculated from the redshifts of the rotating stars of the galaxies. The rotational speed distribution component of the visible stars v[n] has
been calculated from the gravity force of their mass, which has been calculated according to their light intensity, and the result of which is like the distribution curves in Figure 7. It can be
supposed that the speed distribution curve from the ordinary dark matter v[p] is the same form as the speed distribution curve from the visible light v[n]. By using these three speed distribution
curves it is possible to calculate with Equation (6) the speed distribution curve which corresponds to the speed distribution curve v[M] of the four-dimensional mass M. In Ref. [2] has been
calculated the halo speed distribution curve, which corresponds to the curve v[M] of Equation (11). They can be compared. The system of galaxy rotation of this study is a hypothesis, and it must be
proved to be correct or not correct. With a pair of the rotational speeds v[M][1] and v[M][2] and the radii R[1] and R[2] it can be done.
The speed distribution curves of the galaxy NGC 3198 have been presented in Ref. [2] . The speed distribution curve of the galactic halo in that publication corresponds to the speed distribution
curve v[M] of the four-dimensional mass of this study. In order to find out how well this four-dimensional model functions, the speed distribution curve v[M] of the four-dimensional mass has been
calculated by using these rotational radius and speed values.
This is the best fitting to the halo component of the speed distribution curve in Ref. [2] . The fitting succeeded well enough, and it proved that the four-dimensional mass is real.
The distance of the four-dimensional mass M from the center of the galaxy, Equation (16) and (18)
Figure 7. Total speed distribution curve v and speed distribution curve of visible light v[n] which is the same shape as the speed distribution curve of ordinary dark matter v[p]. Speed distribution
curve v[M] of the four dimensional mass M has been calculated with Equation (11), and the speed curves v[p] and v[n] have been calculated from the speed curve v[M] with Equation (5) using the fact
that speed curve v is level shape curve.
${X}^{\prime }=\sqrt{\frac{{R}_{2}^{2}-a{R}_{1}^{2}}{a-1}}=\sqrt{\frac{{8}^{2}-1.64×{3}^{2}}{0.64}}×{10}^{20}\text{\hspace{0.17em}}\text{m}=8.8×{10}^{20}\text{\hspace{0.17em}}\text{m}$
The four-dimensional mass M, Equation (12)
$M=\frac{{v}_{M}^{2}{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}{{R}^{2}\gamma }$
within the accuracy of this study M = 80 × 10^61 kg.
4. Calculation of the Black Hole
The rotational speed of the super massive black hole is the speed of light c = 300,000 km/s. The rotational speed distribution curve of the four-dimensional mass M, v[M] in Figure 8, and it
corresponds to the component of halo speed distribution curve of the galaxy NGC 3198 in Ref. [2] . The four-dimensional mass which produces the speed distribution curve is M = 80 × 10^61 kg, and its
location X' = 8.8 × 10^20 m at the four-dimensional rotation axis. The dark matter
Figure 8. The rotational speed distribution curve v[M] of the four-dimensional mass M has been presented as the function of the radius R. The rotational speed distribution curve v[M] is calculated
with Equation (11) by using the values of the mass M = 80 × 10^61 kg and its distance from the center of galaxy X' = 8.8 × 10^20 m, and it is the best fitting of Equation (11) to the component of
halo speed distribution of the galaxy NGC 3198, Ref. [2] . The maximum value of the speed distribution curve is v[M] = 131 km/s at the radius R = 8.8 × 10^20 m (29.2 kpc), and it is about the same as
the maximum radius of the galaxy NGC 3198 in Ref. [2] , R = 9.0 × 10^20 m.
must be at the galaxy’s rotation axis, or near it. Otherwise, it does not rotate the galaxy. This is the only location which can generate the measured dark matter component of the galaxy’s speed
distribution curve. This can be inferred from Equation (1) and (2). Therefore, the dark matter of the galaxy NGC 3198 cannot be a halo structure. It could be a black hole, but because there is no
black hole at that position near galaxy NGC 3198, the dark matter is not in our three-dimensional Universe. In this study, a theoretical Black Hole is calculated so that the location of the
four-dimensional mass M = 80 × 10^61 kg, X' = 8.8 × 10^20 m is shifted at the rotational axis towards the center point of the galaxy’s rotation. The maximum value of the speed distribution curve of
the four-dimensional mass v[M] is at the point in which the derivative of the speed distribution curve is zero. Equation (11)
${v}_{M}=\frac{R\sqrt{\gamma M}}{{R}^{2}+{\left({X}^{\prime }\right)}^{2}}$(19)
Derivation formula
$\frac{\text{d}\left(\frac{u}{v}\right)}{\text{d}x}=\frac{\frac{\text{d}u}{\text{d}x}\cdot v-\frac{\text{d}v}{\text{d}x}\cdot u}{{v}^{2}}$(20)
Derivation of the speed distribution curve
$\frac{\text{d}{v}_{M}}{\text{d}R}=\sqrt{\gamma M}\frac{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}-2R\cdot R\right)}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}=\sqrt{\gamma M}\
frac{{\left({X}^{\prime }\right)}^{2}-{R}^{2}}{{\left({R}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}$(21)
The maximum value of the speed distribution curve v[Mx] is at the radius R[x]
$\frac{\text{d}{v}_{M}}{\text{d}R}=0⇒\sqrt{\gamma M}\frac{{\left({X}^{\prime }\right)}^{2}-{R}_{x}^{2}}{{\left({R}_{x}^{2}+{\left({X}^{\prime }\right)}^{2}\right)}^{2}}=0⇒{R}_{x}={X}^{\prime }$(22)
${v}_{Mx}=\sqrt{\gamma M}\frac{{R}_{x}}{{R}_{x}^{2}+{R}_{x}^{2}}=\sqrt{\gamma M}\frac{1}{2{R}_{x}}$(23)
${v}_{Mx}=\sqrt{\gamma M}\frac{1}{2{X}^{\prime }}$(24)
The maximum value of the speed distribution curve v[M] is at the point X' = R. It is also about the maximum radius of the galaxy NGC 3198 R = 9.0 × 10^20 m in Ref. [2] . In Figure 9 the speed
distribution curve v[M] which is generated by the four-dimensional mass M = 80 × 10^61 kg, X' = 8.8 × 10^20 m is shifted toward the center point of the galaxy, and the maximum speed of the
distribution curve increases.
The whole Black Hole is in our three-dimensional Universe when the galaxy’s fourth dimension X' is zero.
${X}^{\prime }=0$
${v}_{M0}=\sqrt{\gamma M}\frac{R}{{R}^{2}+{\left({X}^{\prime }\right)}^{2}}=\sqrt{\gamma M}\frac{1}{R}$(25)
The radius of the Black Hole R[0] corresponds to the speed of light (300 × 10^3 km/s)
Figure 9. Four rotational speed distribution curves v[M] presented as a function of radius R from the center of galaxy. The speed distribution curve v[M] which is generated by the four-dimensional
mass M = 80 × 10^61 kg, X' = 8.8 × 10^20 m is shifted toward the center point of the galaxy, and the maximum speed of the distribution curve increases. As the distance X' decreases X' = 2 × 10^20 m →
X' = 0.1 × 10^20 m (6.5 kpc → 0.32 kpc) the maximum speed of the speed distribution curve increases v[M] = 577 km/s → v[M] = 11,500 km/s. The maximum speed of 11,500 km/s is enough for a quasar.
in the speed distribution curve v[M][0] = 3 × 10^8 m/s. The point is indicated with dash lines in Figure 10.
${v}_{M0}=\sqrt{\gamma M}\frac{1}{{R}_{0}}=3×{10}^{8}\text{m}/\text{s}$(26)
The surface of the four-dimensional sphere at the radius R from the center of the sphere is a volume, and therefore the surface of a Black Hole is a volume. The volume of the center of a Black Hole
is infinite.
The maximum radius of the galaxy R[x] is at the point of the maximum value of the speed distribution curve vM. Between the maximum value of the speed distribution curve v[M] and the center point of
the galaxy there is a large hole in the gravitational field of the Black Hole, and matter falling into this hole produces the galaxy in our Universe. Therefore the maximum radius of the galaxy R[x]
is at the point of the maximum value of the speed distribution curve v[M]. Black Hole’s mass can be calculated from Equation (23).
${v}_{Mx}=\sqrt{\gamma M}\frac{1}{2{R}_{x}}$
$M=\frac{4{R}_{x}^{2}{\left({v}_{Mx}\right)}^{2}}{\gamma }$(28)
In Ref. [7] hard evidence has been presented that our three-dimensional Universe is the surface volume of four-dimensional sphere. In Table 1 the four-dimensional masses of the Black Holes at the
center of the sphere have been calculated, and it seems to be that at the longer distance there are more massive Black Holes, Equation (22) X' = R[x].
From Equation (11) the effect of four-dimensional mass M on the rotational speed distribution curve v[M] is as follows:
Figure 10. Five rotational speed distribution curves v[M] presented as a function of the radius R from the center of galaxy. The speed distribution curves in this figure have been derived by shifting
the four-dimensional mass M = 80 × 10^61 kg, distance X' = 8.8 × 10^20 m towards the center point of the galaxy. This speed distribution curve of the mass M = 80 × 10^61 kg, X' = 8.8 × 10^20 m is in
Figure 8, and it corresponds to the halo component of speed distribution curve of the galaxy NGC 3198. The four speed distribution curves indicate the effect of the fourth distance X' on the speed
distribution curves v[M]. As the distance X' decreases X' = 4 × 10^17 m → X' = 3 × 10^17 m (13 pc → 9.7 pc) the maximum rotational speed increases v[M] = 287,000 km/s → 385,000 km/s, Equation (24).
The speed of light is 300,000 km/s, and therefore the theoretical galaxy M = 80 × 10^61 kg, X' = 3 × 10^17 m is a Black Hole in our three-dimensional Universe. The radius of the theoretical Black
Hole M = 80 × 10^61 kg, X' = 0 m is according to Equation (27) is R[0] = 7.7 × 10^17 m.
Table 1. Black Hole mass M and position X' = R[x] calculated with Equation (28) from Ref. [1] .
${v}_{M}=\sqrt{\gamma M}\frac{R}{{R}^{2}+{\left({X}^{\prime }\right)}^{2}}=\sqrt{M}\frac{R\sqrt{\gamma }}{{R}^{2}+{\left({X}^{\prime }\right)}^{2}}$(29)
As it is obvious that the effects of three- and four-dimensional masses are the same, except for the effect of the distance R, it can be inferred that the coefficients of gravity are the same, except
for the quality, γ = 6.67 × 10^−11 N·m^3·kg^−2.
In Figure 11 a star is falling into the Black Hole. The transformation of the relativity measurements compatible with the invariance of the velocity of light is according to Ref. [12] (page 123), the
Lorentz transformation
${x}^{\prime }=\frac{x-vt}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$${t}^{\prime }=\frac{t-\frac{xv}{{c}^{2}}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$(30)
The distance of the observer from the Black Hole is x, the star is falling into the Black Hole at the speed of light v = c, time is t, the distance of the star from the Black Hole seen by the
observer is x – ct, the distance of the star from the Black Hole calculated with the principal of relativity is x'
${x}^{\prime }=\frac{x-ct}{\sqrt{1-\frac{{c}^{2}}{{c}^{2}}}}=\infty$${t}^{\prime }=\frac{t-\frac{xc}{{c}^{2}}}{\sqrt{1-\frac{{c}^{2}}{{c}^{2}}}}=\infty$(31)
Therefore the real distance of the star from the Black Hole is infinite, and the time has stopped, and the star cannot fall into the Black Hole at any time. The speed of the star cannot be higher
than c because otherwise $\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ would be imaginary. Most probable is that the path of motion of the star falling into the Black Hole is rotation about it. The extreme
gravitation of the Black Hole accelerates the star near the speed of light, and the falling into the Black Hole slows down.
Figure 11. A star is falling into the Black Hole. The speed of the star is v, time of the observer is t, the distance of the star from the observer is vt, and the distance of the Black Hole from the
observer is x. The distance of the Black Hole from the star as seen at the star is x', and the time at the star is t'.
5. Conclusions
In Figure 12, a star is falling into the Black Hole. In the falling movement, the change of the kinetic energy is equal to the work that is done by the force of gravity. The first calculation is in
the three-dimensional gravitational field of our
Figure 12. A star is falling into the Black Hole. The beginning of the fall is at the 0-point. The distance of the star from the 0-point is x, the distance from the 0-point to the Black Hole is x[0].
The Black Hole in the figure is the theoretical Black Hole M = 80 × 10^61 kg, X' = 0 m in Figure 10, and the radius of the Black Hole according to Equation (26) is R[0] = 7.7 × 10^17 m.
three-dimensional Universe, and the second calculation is in the four-dimensional gravitational field of the four-dimensional Universe.
The first calculation in the three-dimensional gravitation field: A star is falling into the Black Hole of the radius is R[0] = 7.7 × 10^17 m, and the starting point is x[0] = 10^20 m; the speed of
falling star is v = 0 → 3 × 10^8 m/s; the speed of light is c = 3 × 10^8 m/s.
x[0] – x 7.7 × 10^17 m
x[0] 10^20 m
c 3 × 10^8 m/s
$\frac{1}{2}m\text{d}\left({v}^{2}\right)=\frac{\gamma mM}{{\left({x}_{0}-x\right)}^{2}}\text{d}x$$\frac{1}{2}{v}^{2}={\int }_{0}^{x}\frac{\gamma M}{{\left({x}_{0}-x\right)}^{2}}\text{d}x$(32)
$\frac{1}{2}{v}^{2}=\gamma M\left(-1\right)\left(-1\right)\left[\frac{1}{{x}_{0}-x}-\frac{1}{{x}_{0}}\right]$(33)
$M=\frac{{v}^{2}}{2\gamma \left[\frac{1}{{x}_{0}-x}-\frac{1}{{x}_{0}}\right]}$
The second calculation is performed in the four-dimensional gravitational field of the four-dimensional Universe
$\frac{1}{2}m\text{d}\left({v}^{2}\right)=\frac{\gamma mM}{{\left({x}_{0}-x\right)}^{3}}\text{d}x$$\frac{1}{2}{v}^{2}={\int }_{0}^{x}\frac{\gamma M}{{\left({x}_{0}-x\right)}^{3}}\text{d}x$(35)
$\frac{1}{2}{v}^{2}=\gamma M\left(-\frac{1}{2}\right)\left(-1\right)\left[\frac{1}{{\left({x}_{0}-x\right)}^{2}}-\frac{1}{{x}_{0}^{2}}\right]$(36)
The speed distribution curves of the galaxy NGC 3198 have been presented in Ref. [2] . The speed distribution curve of the galactic halo in that publication corresponds to the speed distribution
curve v[M] of the four-dimensional mass of this study. This speed distribution curve is in Figure 8, and it is produced by the Black Hole M = 80 × 10^61 kg, X' = 8.8 × 10^20 m in the four-dimensional
Universe. In this study, this Black Hole has been shifted toward our three-dimensional Universe, and at the distance X' = 0 the radius of this theoretical Black Hole is 7.7 × 10^17 m. According to
the first calculation in the three-dimensional gravity, this theoretical Black Hole radius is resulted from the mass 5.2 × 10^44 kg. It can be compared with the real masses which have been found in
our three-dimensional Universe. The most massive Black Hole is the quasar TON 618 with a mass 66 billion times greater than our Sun, and its mass is 66 × 10^9 × 2 × 10^30 kg = 13.2 × 10^40. The first
calculation yielded the mass for the theoretical Black Hole 5.2 × 10^44 kg and therefore the Black Hole of this mass is not in our three-dimensional Universe. The theoretical Black Hole M = 80 × 10^
61 kg, X' = 0.1 × 10^20 in Figure 9 produces the gravity field with a maximum speed of rotation 11500 km/s, and it is enough to create the quasar. The fact that a quasar ejects matter along its
rotational axis is proof that the quasar is not a Black Hole in our three-dimensional universe, but rather, as suggested by its location, exists in the fourth dimension. As it is seen in the Lorentz
transformation above, the star falling into the Black Hole does not reach the Black Hole, but the observer sees that the falling stops before the Black Hole. Therefore the matter of the Black Hole is
not three-dimensional matter, but it is four-dimensional matter.
The first calculation continues. The centrifugal force is equal to the gravitational force in the three-dimensional gravitational field.
$\frac{m{v}^{2}}{R}=\frac{\gamma mM}{{R}^{2}}$$M=\frac{{v}^{2}R}{\gamma }=\frac{9×{10}^{16}×7.7×{10}^{17}}{6.67×{10}^{-11}}=10.4×{10}^{44}\text{\hspace{0.17em}}\text{kg}$(38)
The second calculation for the four-dimensional gravity yielded the mass of the Black Hole M = 80 × 10^61 kg for the direct falling into the Black Hole which is the same value that has been in the
rotational speed curves indicating the rotational falling into the Black Hole. Therefore, it can be concluded that the Black Hole of the four-dimensional matter is real. The first calculation for the
three-dimensional gravity yielded the mass of the Black Hole for the direct falling into the Black Hole M = 5.2 × 10^44 kg, and in the rotational falling into the Black Hole 10.4 × 10^44 kg.
Therefore, it can be concluded that the Black Hole of three-dimensional mass is not real.
1) The original Black Hole theory that shifting the real Black Hole of the galaxy NGC 3198 into our three-dimensional Universe was not true. The most massive Black Hole of our Universe was not
massive enough for that theory. However, there is a right kind of Black Hole for the quasar TON 618, but that Black Hole is in the four-dimensional Universe. That can explain the fact that matter can
escape from the gravitation force of quasar.
2) A common belief has been that the mysterious missing dark matter could be a halo structure around the galaxy. But it cannot be a halo structure because that kind of structure does not rotate the
galaxy. That can be inferred from Equation (1). To be able to rotate the galaxy the dark matter must be on the galaxy’s rotation axis, or near it. But the mass which is many times the mass of the
whole galaxy would be a Black Hole, and it would destroy the galaxy. Therefore it cannot be a solution to this problem. The only option is that the mysterious missing dark matter is located out of
our three-dimensional Universe.
Furthermore, the theories of Refs. [6] and [7] were proven to be true, and they also have an impact on this study. The following is presented the progress of the galaxy rotation hypothesis. This
hypothesis states that the major gravitational force of the galaxy rotation is due to the four-dimensional mass which is the Black Hole of this study.
In Figure 13, the progress of the research and testing the hypothesis is presented in the same manner as in Karl Poppers’s book [13] . The first step, the hypothesis, is that the major gravitational
field which rotates the galaxy is due to a dark fourth-dimensional matter M which is located at the galaxy’s fourth-dimensional rotation axis.
The second step is the derivation of the equations. This phase of proceeding involves the derivation of mathematics of the galaxy rotation. The speed distribution curve v[M] of the four-dimensional
mass M, Equation (11) and (12), has been calculated by using two pairs of rotational radius and speed values which are approximately the same as in Ref. [2] the speed distributions curves of the
galactic halo of the galaxy NGC 3198.
The third step is the test prediction which is that the speed distribution curve of galactic halo in Ref. [2] corresponds to the speed distribution curve vMof four-dimensional mass of this study.
Figure 13. The progress of testing the hypotheses of the fourth-dimensional mass is presented by using the method of Karl Popper, Ref. [13] .
The fourth step is the measurements. The four-dimensional mass value M = 80 × 10^61 kg was calculated with Equation (12) and it is the Black Hole of this study. Its distance above the center of
galaxy in the four-dimensional axis X' = 8.8 × 10^20 m was calculated with Equation (18). The rotational speed distribution component of the four-dimensional mass v[M] was calculated with Equation
(11), and it is presented in Figure 8. The rotational speed curve v[M] was compared to the real measurements of Ref. [2] .
The fifth step is the analysis of the result and justification.
1) By comparing the speed distribution curve of the four-dimensional mass with the speed distribution of halo in Ref. [2] , it was inferred that it corresponded approximately to the halo
distribution. The sources of inaccuracy are the redshift measurements and the evaluation of the mass distribution of ordinary dark matter. The halo speed distribution curve of Ref. [2] corresponds
quite well to the speed distribution of four-dimensional mass v[M] until the radius R = 30 kpc. In Ref. [2] the speed measurements has been done until the radius R = 30 kpc, which obviously is the
maximum radius of galaxy, but however the halo curve continue until radius R = 50 kpc. In the region 30 - 50 kpc the halo speed curve of Ref. [2] and the four-dimensional mass speed curve v[M] begins
to be separated from each other significantly. The speed curve of four-dimensional mass v[M] decreases in this region, but the halo speed curve of Ref. [2] continues to increase. It is possible that
the maximum radius of galaxy is determined according to the point in which the speed curve v[M] begins to decrease.
2) The generation process of the galactic system can be explained. The force of gravitation of the four-dimensional mass M acts on the rotational plane of galaxy. It has no component at the direction
of the fourth dimension because the three-dimensional mass of galaxy of stars, planets, gas, dust and other matter have not gravity force of four-dimensional force of gravity. The result is that the
four-dimensional mass M generates a gravitational field which has a great hole at the center of galaxy. The gravitational field of the four-dimensional mass accelerates the three-dimensional mass of
stars, planets, dust, and other matter into the speed of rotation in which it rotates round about the center of galaxy. In this manner the hole in the gravitational field of the four-dimensional mass
M fills up, and the typical constant speed outer boundary regions of the galaxies have been generated. The gravity field of the four-dimensional mass M accelerates a star to the rotational speed
somewhat above 130 km/s, in which case it retain rotating the galaxy in the border region, and if the star loses kinetic energy and speed, it begins to rotate the galaxy at the center region, and if
the star accelerates considerably more than 130 km/s, then the gravity force of the four-dimensional mass cannot hold it, and it moves out of the gravity field.
3) In Ref. [6] the effect of the distance X' on the maximum radius of galaxy has been studied. As the distance X' of the four-dimensional mass M increases, the maximum radius of the galaxy increases.
By applying Equation (29) for more massive four-dimensional masses, it can be obtained rotation curves of Ref. [1] .
4) In Ref. [7] it was proven that the structure of our three-dimensional Universe is a surface volume of four-dimensional sphere. This type of structure of the Universe creates the same kind of an
accelerating redshift increase which has been measured. This theory was proven by the same kind of Karl Poppers testing diagram as is shown above. The model of the surface volume of the
four-dimensional Universe was constructed, the equation of the redshift caused by this Universe was solved, and the theoretical equation was shown to be the same as the measured redshift in the
Universe. In this manner the four-dimensional surface volume Universe was proven to be real. This is also a proof that the mysterious dark matter in the galaxy rotation is a four-dimensional mass
which is a Black Hole.
Karl Poppers’ method of testing hypotheses has been essential in the development of the modern world. This method of progress was used for the solution to the Hill’s equation. The famous British
Nobel laureate A. V. Hill invented this equation in 1938. Within four rounds that problem was solved, Refs. [14] [15] [16] and [17] . This study is the third round of Karl Poppers’ method for solving
the structure of four-dimensional Universe.
List of Variables
Dimensions of ordinary space x,y,z
Fourth distance dimension x'
Four-dimensional mass M
Ordinary visible galactic mass M(x)
Location of the mass M on the fourth distance dimension X'
Rotational speed distribution curve v
Rotational speed distribution component of the mass M v[M]
Maximum value of the speed distribution cure v[M] v[Mx]
Radius of the maximum value of the speed distribution cure v[M] R[x]
Rotational speed distribution component of visible mass v[n]
Rotational speed distribution component of ordinary dark matter v[p]
Rotational speed distribution component of visible and dark matter v[m]
Radius of rotation of the galactic mass R
Radius of the Black hole R[0]
Rotational speed distribution curve v[M] at the distance X' = 0 v[M][0]
Mass of a star m
Black hole in the four-dimensional Universe Black Hole | {"url":"https://www.scirp.org/journal/paperinformation?paperid=125280","timestamp":"2024-11-07T19:21:57Z","content_type":"application/xhtml+xml","content_length":"231610","record_id":"<urn:uuid:4409acce-f0d2-4521-98f9-669b0ec784f1>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00354.warc.gz"} |
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M.Sc. Thesis Sharif University of Technology ; Saadat Foumani, Mahmoud
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Mikulec, I ; Rabady, D ; Rahbaran, B ; Rohringer, H ; Schofbeck, R ; Strauss, J ; Taurok, A ; Treberer Treberspurg, W ; Waltenberger, W ; Wulz, C. E ; Mossolov, V ; Shumeiko, N ; Suarez Gonzalez,
J ; Alderweireldt, S ; Bansal, M ; Bansal, S ; Cornelis, T ; De Wolf, E. A ; Janssen, X ; Knutsson, A ; Luyckx, S ; Ochesanu, S ; Roland, B ; Rougny, R ; Van De Klundert, M ; Van Haevermaet, H ;
Van Mechelen, P ; Van Remortel, N ; Van Spilbeeck, A ; Blekman, F ; Sharif University of Technology
The ratio of the top-quark branching fractions R=B(t→Wb)/B(t→Wq), where the denominator includes the sum over all down-type quarks (q = b, s, d), is measured in the tt- dilepton final state with
proton-proton collision data at s=8 TeV from an integrated luminosity of 19.7 fb-1, collected with the CMS detector. In order to quantify the purity of the signal sample, the cross section is
measured by fitting the observed jet multiplicity, thereby constraining the signal and background contributions. By counting the number of b jets per event, an unconstrained value of R=
1.014±0.003(stat.)±0.032(syst.) is measured, in a good agreement with current precision measurements in electroweak and flavour...
Article Physics Letters, Section B: Nuclear, Elementary Particle and High-Energy Physics ; Vol. 736 , 2014 , Pages 64-85 ; ISSN: 03702693 ; Sirunyan, A. M ; Tumasyan, A ; Adam, W ; Bergauer, T ;
Dragicevic, M ; Ero, J ; Fabjan, C ; Friedl, M ; Fruhwirth, R ; Ghete, V. M ; Hartl, C ; Hormann, N ; Hrubec, J ; Jeitler, M ; Kiesenhofer, W ; Knunz, V ; Krammer, M ; Kratschmer, I ; Liko, D ;
Mikulec, I ; Rabady, D ; Rahbaran, B ; Rohringer, H ; Schofbeck, R ; Strauss, J ; Taurok, A ; Treberer Treberspurg, W ; Waltenberger, W ; Wulz, C. E ; Mossolov, V ; Shumeiko, N ; Suarez Gonzalez,
J ; Alderweireldt, S ; Bansal, M ; Bansal, S ; Cornelis, T ; De Wolf, E. A ; Janssen, X ; Knutsson, A ; Luyckx, S ; Ochesanu, S ; Roland, B ; Rougny, R ; Van De Klundert ,M ; Van Haevermaet, H ;
Van Mechelen, P ; Van Remortel, N ; Van Spilbeeck, A ; Blekman, F ; Sharif University of Technology
Constraints are presented on the total width of the recently discovered Higgs boson, ΓH, using its relative on-shell and off-shell production and decay rates to a pair of Z bosons, where one Z
boson decays to an electron or muon pair, and the other to an electron, muon, or neutrino pair. The analysis is based on the data collected by the CMS experiment at the LHC in 2011 and 2012,
corresponding to integrated luminosities of 5.1fb -1 at a center-of-mass energy s=7TeV and 19.7fb -1 at s=8TeV. A simultaneous maximum likelihood fit to the measured kinematic distributions near
the resonance peak and above the Z-boson pair production threshold leads to an upper limit on the Higgs boson width of ΓH...
Article Physical Review A - Atomic, Molecular, and Optical Physics ; Vol. 89, Issue. 6 , 2014 ; ISSN: 10502947 ; Dezfouli, B. G ; Ghasemipour, F ; Rezakhani, A. T ; Saberi, H ; Sharif University
of Technology
We propose a flexible numerical framework for extracting the energy spectra and photon transfer dynamics of a unit kagome cell with disordered cavity-cavity couplings under realistic experimental
conditions. A projected-entangled pair state (PEPS) Ansatz to the many-photon wave function allows us to gain a detailed understanding of the effects of undesirable disorder in fabricating
well-controlled and scalable photonic quantum simulators. The correlation functions associated with the propagation of two-photon excitations reveal intriguing interference patterns peculiar to
the kagome geometry and promise at the same time a highly tunable quantum interferometry device with a signature for the...
Article Physica A: Statistical Mechanics and its Applications ; Vol. 404 , 2014 , Pages 200-216 ; ISSN: 03784371 ; Manzari, M. T ; Sharif University of Technology
In this paper, the Immersed Moving Boundary-Lattice Boltzmann (IMB-LB) method is compared with the single relaxation time and multiple-relaxation-time versions of the Immersed Boundary-Lattice
Boltzmann (IB-LB) method in terms of the amount of numerical velocity slip produced on solid boundaries. The comparisons are performed for both straight and curved boundaries based on the effects
of thickness of virtual domain used in the IB method for the first time, and relaxation time parameter(s) of the LB method. For the straight boundaries, a shear flow problem is studied while for
the curved boundaries, a falling circular cylinder in an infinite channel is investigated. First, sensitivities of...
Article International Journal of Production Research ; Vol. 52, issue. 10 , Nov , 2014 , pp. 2954-2982 ; ISSN: 00207543 ; Khedmati, M ; Sharif University of Technology
In this paper, a new multi-attribute T2 control chart is initially proposed to monitor multi-attribute processes based on a transformation technique. Then, the maximum likelihood estimator of a
multivariate Poisson process change point is derived for unknown changes that are assumed to belong to a family of monotonic changes. Using extensive simulation experiments, the performance of
the proposed change-point estimator is compared to the ones derived for step changes and linear-trend disturbances, when the true change types are step change, linear trends and multiple-step
changes. We show when the type of the change is not known a priori, the proposed estimator is an appropriate choice,...
Article Construction and Building Materials ; Vol. 57 , April , 2014 , pp. 69-80 ; ISSN: 09500618 ; Beygi, M. H. A ; Kazemi, M. T ; Vaseghi Amiri, J ; Rabbanifar, S ; Rahmani, E ; Rahimi, S ;
Sharif University of Technology
Self compacting concrete (SCC), as an innovative construction material in concrete industry, offers a safer and more productive construction process due to favorable rheological performance which
is caused by SCC's different mixture composition. This difference may have remarkable influence on the mechanical behavior of SCC as compared to normal vibrated concrete (NVC) in hardened state.
Therefore, it is vital to know whether the use of all assumptions and relations that have been formulated for NVC in current design codes are also valid for SCC. Furthermore, this study presents
an extensive evaluation and comparison between mechanical properties of SCC using current international codes and...
Article Applied Catalysis A: General ; Vol. 475 , April , 2014 , pp. 55-62 ; ISSN: 0926860X ; Zare, M ; Salemnoush, T ; Ozkar, S ; Akbayrak, S ; Sharif University of Technology
A novel organic-inorganic hybrid heterogeneous catalyst system was obtained from the reaction of the molybdenum(VI) complex of salicylidene 2-picoloyl hydrazone with mesoporous silica containing
3-chloropropyl groups prepared by a direct synthetic approach involving hydrolysis and co-condensation of tetraethylorthosilicate (TEOS) and 3-chloropropyltrimethoxysilane in the presence of the
triblock copolymer P123 as template under acidic conditions. Characterization of the functionalized materials by X-ray diffraction (XRD), high resolution transmission electron microscopy (HRTEM),
scanning electron microscopy (SEM), N2 adsorption/desorption, FT-IR and UV-Vis spectroscopy, and thermogravimetric...
Article Composites Part B: Engineering ; Vol. 60 , 2014 , pp. 413-422 ; ISSN: 13598368 ; Ahmadian, M. T ; Taati, E ; Sharif University of Technology
Design and development of FGMs as the heat treatable materials for high-temperature environments with thermal protection require understanding of exact temperature and thermal stress distribution
in the transient state. This information is primary tool in the design and optimization of the devices for failure prevention. Frequently FGMs are used in many applications that presumably
produce thermal energy transport via wave propagation. In this study, transient non-Fourier temperature and associated thermal stresses in a functionally graded slab symmetrically heated on both
sides are determined. Hyperbolic heat conduction equation in terms of heat flux is used for obtaining temperature... | {"url":"https://repository.sharif.edu/search/?&query=farjam--t&field=authorOther&count=20&execute=true&sort=resType&dir=1","timestamp":"2024-11-05T23:54:08Z","content_type":"text/html","content_length":"104451","record_id":"<urn:uuid:298c359a-7cc9-494e-9028-bbce085260a5>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00292.warc.gz"} |
Meaning of Remainder in Urdu
Meaning and Translation of Remainder in Urdu Script and Roman Urdu with Wikipedia Reference, Synonyms, Antonyms,
remainder bachat بچت
remainder baqiya بقيہ
remainder pass mandah پس ماندہ
remainder bacha kuchha بچا کچھا
remainder haasil حاصل
remainder wirasat وراثت
remainder guzara گزارہ
remainder baaz gasht بازگشت
In mathematics, the remainder is the amount "left over" after performing some computation. In arithmetic, the remainder is the integer "left over" after dividing one integer by another to produce an
integer quotient (integer division).
Read more at wikipedia
In mathematics, the remainder is the amount "left over" after performing some computation. In arithmetic, the remainder is the integer "left over" after dividing one integer by another to produce an
integer quotient (integer division).
butt carry-over detritus dregs excess fragment garbage hangover heel junk leavings leftover obverse oddment overplus refuse relic remains remnant residue residuum rest ruins salvage scrap stump
surplus trace vestige waste wreck wreckage origin | {"url":"https://meaning.urdu.co/remainder/","timestamp":"2024-11-03T20:13:19Z","content_type":"application/xhtml+xml","content_length":"9274","record_id":"<urn:uuid:40f6b79c-f3ee-47be-82ba-48e1e0a05df2>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00182.warc.gz"} |
Riemannian Manifold
Review of Short Phrases and Links
This Review contains major "Riemannian Manifold"- related terms, short phrases and links grouped together in the form of Encyclopedia article.
1. A Riemannian manifold is a differentiable manifold on which the tangent spaces are equipped with inner products in a differentiable fashion.
2. A Riemannian manifold (M,g) is a smooth manifold M, together with a Riemannian metric on M, i.e. (Web site)
3. A (pseudo -) Riemannian manifold is conformally flat if each point has a neighborhood that can be mapped to flat space by a conformal transformation.
4. A Riemannian manifold is called a homogeneous nilmanifold if there exist a nilpotent group of isometries acting transitively on it. (Web site)
5. A Riemannian manifold is collapsed with a lower curvature bound if the sectional curvature is at least −1 and the volume of every unit ball is small.
1. In mathematics, a sub-Riemannian manifold is a certain type of generalization of a Riemannian manifold.
2. In mathematics, a Hermitian manifold is the complex analog of a Riemannian manifold. (Web site)
3. In mathematics, a Hermitian symmetric space is a Kähler manifold M which, as a Riemannian manifold, is a Riemannian symmetric space. (Web site)
1. In general, geodesics can be defined for any Riemannian manifold.
1. More generally, any embedded hypersurface in a Riemannian manifold can be cut out by an equation of the form F(P) = 0 such that dF is nowhere zero.
2. Given a control region Ω on a compact Riemannian manifold M, we consider the heat equation with a source term g localized in Ω.
1. A manifold $M$ together with a Riemannian metric tensor $g$ is called a Riemannian manifold.
1. The Riemannian metrics of tangent and cotangent bundles of a Riemannian manifold were defined by S.Sasaki [1], I.Sato [2] and N.Bhatia, N.Prakash [3]. (Web site)
1. A manifold together with a Riemannian metric tensor is called a Riemannian manifold.
1. In general, a manifold is not a linear space, but the extension of concepts and techniques from linear spaces to Riemannian manifold are natural. (Web site)
2. In this setting, the linear space has been replaced by Riemannian manifold and the line segment by a geodesic. (Web site)
1. I will discuss the Dirichlet Problem for such harmonic functions on bounded domains in a riemannian manifold. (Web site)
1. As David Henderson, Taimina's husband, has explained, a hyperbolic plane "is a simply connected Riemannian manifold with negative Gaussian curvature".
1. The Riemann sphere is only a conformal manifold, not a Riemannian manifold.
1. The converse is also true: a Riemannian manifold is hyperkähler if and only if its holonomy is contained in Sp(n).
1. Abstract: Let M be a closed Riemannian manifold of dimension d.
2. Theorem: Let be a closed Riemannian manifold.
1. So let us construct a Riemannian manifold that has vanishing curvature outside of a compact set. (Web site)
2. Conversely, we can characterize Eucldiean space as a connected, complete Riemannian manifold with vanishing curvature and trivial fundamental group.
1. A Riemannian manifold M is geodesically complete if for all, the exponential map exp p is defined for all, i.e. (Web site)
1. That makes possible to define geodesic flow on unit tangent bundle U T(M) of the Riemannian manifold M when the geodesic γ V is of unit speed.
1. Abstract: Connes has demonstrated that for a compact spin Riemannian manifold, the geodesic metric is determined by the Dirac operator.
1. If is a biwave map from a compact domain into a Riemannian manifold such that (3.19) then is a wave map. (Web site)
1. Korn and Lichtenstein proved that isothermal coordinates exist around any point on a two dimensional Riemannian manifold. (Web site)
2. Let us denote by a complete simply connected m - dimensional Riemannian manifold of constant sectional curvature k, i.e. (Web site)
1. It is known that a cut - The dimension of a cut locus on a smooth Riemannian manifold.
1. Zaved., Matematika (1987) no.5, 25-33) Grigor'yan, A., On the fundamental solution of the heat equation on an arbitrary Riemannian manifold, Math. (Web site)
1. Let (M, g) be a Riemannian manifold, and a Riemannian submanifold. (Web site)
1. Every compact, simply connected, conformally flat Riemannian manifold is conformally equivalent to the round sphere. (Web site)
2. Fourth, one can use conformal symmetry to extend harmonic functions to harmonic functions on conformally flat Riemannian manifold s. (Web site)
3. Any 2-dimensional (smooth) Riemannian manifold is conformally flat, a consequence of the existence of isothermal coordinates. (Web site)
1. The Ricci curvature is determined by the sectional curvatures of a Riemannian manifold, but contains less information. (Web site)
2. For Brownian motion on a Riemannian manifold this gives back the value of Ricci curvature of a tangent vector. (Web site)
3. However, the Ricci curvature has no analogous topological interpretation on a generic Riemannian manifold. (Web site)
1. This work proposes a novel algorithm for clustering data sampled from multiple submanifolds of a Riemannian manifold. (Web site)
1. Let (N, h) be a Riemannian manifold with LeviCivita connection ∇ and (M, g) be a submanifold with the induced metric.
1. Parallel to this discussion, the notion of a Riemannian manifold will be introduced. (Web site)
1. Abstract: The loop space of a Riemannian manifold has a family of canonical Riemannian metrics indexed by a Sobolev space parameter.
1. On a general Riemannian manifold, f need not be isometric, nor can it be extended, in general, from a neighbourhood of p to all of M.
2. It is known that the spin structure on Riemannian manifold can be extended to noncommutative geometry using the notion of spectral triple. (Web site)
1. This motivates the definition of geodesic normal coordinates on a Riemannian manifold. (Web site)
2. A space form is by definition a Riemannian manifold with constant sectional curvature.
1. Any two points of a complete simply connected Riemannian manifold with nonpositive sectional curvature are joined by a unique geodesic.
2. For example, the circle has a notion of distance between two points, the arc-length between the points; hence it is a Riemannian manifold.
1. In this paper, we discuss various concepts, definitions and properties for the functions on Riemannian manifold. (Web site)
2. In this paper, we extend the Brézis-Wainger result onto a compact Riemannian manifold. (Web site)
1. The Riemannian curvature tensor is an important pointwise invariant associated to a Riemannian manifold that measures how close it is to being flat.
1. The restriction of a Killing vector field to a geodesic is a Jacobi field in any Riemannian manifold. (Web site)
2. The restriction of a Killing field to a geodesic is a Jacobi field in any Riemannian manifold.
1. The Riemannian manifold of covariance matrices is transformed into the vector space of symmetric matrices under the matrix logarithm mapping.
2. Let denote the vector space of smooth vector fields on a smooth Riemannian manifold.
1. One can think of Ricci curvature on a Riemannian manifold, as being an operator on the tangent bundle.
2. One can think of Ricci curvature on a Riemannian manifold, as being an operator on the tangent space. (Web site)
1. On a (pseudo -) Riemannian manifold M a geodesic can be defined as a smooth curve γ(t) that parallel transports its own tangent vector.
1. We also give an example of a connection in the normal bundle of a submanifold of a Riemannian manifold and study its properties.
1. In mathematics, the volume form is a differential form that represents a unit volume of a Riemannian manifold or a pseudo-Riemannian manifold.
2. Riemannian manifold s (but not pseudo-Riemannian manifold s) are special cases of Finsler manifolds.
3. Oriented Riemannian manifold s and pseudo-Riemannian manifold s have a canonical volume form associated with them.
1. General relativity is also a local theory, but it is used to constrain the local properties of a Riemannian manifold, which itself is global.
1. A complete simply connected Riemannian manifold has non-positive sectional curvature if and only if the function f p(x) = d i s t 2(p, x) is 1- convex.
2. Let f be a smooth nondegenerate real valued function on a finite dimensional, compact and connected Riemannian manifold.
3. Distance function and cut loci on a complete Riemannian manifold. (Web site)
1. Let (M,g) be a Riemannian manifold, and S subset M a Riemannian submanifold.
1. A subset K of a Riemannian manifold M is called totally convex if for any two points in K any geodesic connecting them lies entirely in K, see also convex.
2. A function f on a Riemannian manifold is a convex if for any geodesic γ the function is convex.
3. A function f on a Riemannian manifold is a convex if for any geodesic γ the function is convex.
1. The field of velocities of a (local) one-parameter group of motions on a Riemannian manifold.
1. A case of particular interest is a metric linear connection: this is a metric connection on the tangent bundle, for a Riemannian manifold. (Web site)
1. An example of a Riemannian submersion arises when a Lie group G acts isometrically, freely and properly on a Riemannian manifold (M, g). (Web site)
1. On a Riemannian manifold one has notions of length, volume, and angle.
2. Informally, a Riemannian manifold is a manifold equipped with notions of length, angle, area, etc. (Web site)
Related Keywords
* Ambient Space * Compact * Compact Riemannian Manifold * Complete * Complete Riemannian Manifold * Complex Manifold * Complex Structure * Conformal Maps * Constant * Curvature * Diffeomorphism *
Differentiable Manifold * Differential Forms * Differential Geometry * Dimension * Elliptic Operator * Euclidean Space * Finsler Manifold * Geodesic * Geodesics * Injectivity Radius * Inner
Product * Isometric * Isometries * Manifold * Manifolds * Metric * Metric Space * Metric Tensor * Notion * Point * Riemannian Geometry * Riemannian Manifolds * Riemannian Metric * Riemann
Curvature Tensor * Scalar Curvature * Smooth Manifold * Space * Structure * Tangent Bundle * Tangent Space * Tangent Spaces * Tensor * Theorem * Topological Dimension * Vector
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What is a word that could be used to describe a Wordle guess that could possibly be the target word?
That is, the guess contains none of the letters in black squares, and all of the letters in green squares are repeated in those squares, and all of the letters in yellow squares are repeated in
different squares. The first guess satisfies this condition and so does the last guess, assuming that you solve the puzzle.
It is tempting to call such guesses “legal”, but that is not quite right. Sometimes these types of move are not even the best choice. For example, if you know the location of 4 of the letters and
have several choices for the remaining letter, the best strategy is to come up with words containing as many as possible of the possible fifth letters.
Observing members: 0
Composing members: 0
Yes, but it could be the target word, which is not true of all guesses.
”...the best strategy is to come up with words containing as many as possible of the possible fifth letters”
But this is not permitted when playing in Hard mode. In Hard mode one must play all letters that have been revealed as yellow or green.
So I could refer to what I am talking about as a hard mode guess, even if I am not playing in hard mode.
Answer this question | {"url":"https://i.fluther.com/236280/what-is-a-word-that-could-be-used-to-describe-a/","timestamp":"2024-11-10T19:19:56Z","content_type":"application/xhtml+xml","content_length":"13960","record_id":"<urn:uuid:318c5ebd-c2bb-40db-aae6-b7a41823c40b>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00814.warc.gz"} |
Transactions Online
Shigeo KAWASAKI, Harunobu SEITA, Takuo MORIMOTO, "The Moment Method Analysis as a Simulator Technique for a Dipole Antenna Using Wavelets" in IEICE TRANSACTIONS on Electronics, vol. E84-C, no. 7, pp.
914-922, July 2001, doi: .
Abstract: As a solver in a simulator, advantages of use of a wavelet function were investigated for analysis of a dipole antenna using the Moment Method. Realization of a sparse matrix due to
orthogonality and due to inherent nature of the wavelet is confirmed by observing an impedance matrix using each Daubechies' wavelet. Calculated results of the input impedance, the impedance matrix,
and the current distribution are compared in variation of the wavelet in two integral equations for a dipole antenna. Use of the Daubechies' wavelet of the high number with a small matrix and a
threshold in the Hallen's Integral Equation is suitable for the reduction of the matrix size and of the calculation cost.
URL: https://global.ieice.org/en_transactions/electronics/10.1587/e84-c_7_914/_p
author={Shigeo KAWASAKI, Harunobu SEITA, Takuo MORIMOTO, },
journal={IEICE TRANSACTIONS on Electronics},
title={The Moment Method Analysis as a Simulator Technique for a Dipole Antenna Using Wavelets},
abstract={As a solver in a simulator, advantages of use of a wavelet function were investigated for analysis of a dipole antenna using the Moment Method. Realization of a sparse matrix due to
orthogonality and due to inherent nature of the wavelet is confirmed by observing an impedance matrix using each Daubechies' wavelet. Calculated results of the input impedance, the impedance matrix,
and the current distribution are compared in variation of the wavelet in two integral equations for a dipole antenna. Use of the Daubechies' wavelet of the high number with a small matrix and a
threshold in the Hallen's Integral Equation is suitable for the reduction of the matrix size and of the calculation cost.},
TY - JOUR
TI - The Moment Method Analysis as a Simulator Technique for a Dipole Antenna Using Wavelets
T2 - IEICE TRANSACTIONS on Electronics
SP - 914
EP - 922
AU - Shigeo KAWASAKI
AU - Harunobu SEITA
AU - Takuo MORIMOTO
PY - 2001
DO -
JO - IEICE TRANSACTIONS on Electronics
SN -
VL - E84-C
IS - 7
JA - IEICE TRANSACTIONS on Electronics
Y1 - July 2001
AB - As a solver in a simulator, advantages of use of a wavelet function were investigated for analysis of a dipole antenna using the Moment Method. Realization of a sparse matrix due to
orthogonality and due to inherent nature of the wavelet is confirmed by observing an impedance matrix using each Daubechies' wavelet. Calculated results of the input impedance, the impedance matrix,
and the current distribution are compared in variation of the wavelet in two integral equations for a dipole antenna. Use of the Daubechies' wavelet of the high number with a small matrix and a
threshold in the Hallen's Integral Equation is suitable for the reduction of the matrix size and of the calculation cost.
ER - | {"url":"https://global.ieice.org/en_transactions/electronics/10.1587/e84-c_7_914/_p","timestamp":"2024-11-05T12:54:34Z","content_type":"text/html","content_length":"59799","record_id":"<urn:uuid:a817262f-f287-4be8-8a92-d4375003e8e6>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00266.warc.gz"} |
This Is the World’s Only Unbreakable Encryption
But we can’t use it to save us from the future
How strange it is to think that some part of us exists online. Who we are no longer has a purely physical answer. Our work, thoughts, relationships, and obsessions exist online, extensions of
ourselves that make it possible for people to get to know us without ever having to occupy the same rooms as us. Or even the same country. This is the transition of the human to the cyborg — we are
starting to shift more and more of our lives onto our machines.
Encrypting our messages, then, is about more than just protecting our money. It’s about protecting some aspect of our humanity as well. And up until now we thought our information was very well
protected. Cryptographers have come up with clever ways to keep our text messages and bank information safe from criminals and agencies. But all of that will someday change. As a seemingly inevitable
computer revolution looms on the horizon, it casts an uncertain shadow on the privacy of our lives. By some estimates, it may already be too late to keep our sensitive information safe.
Current methods of encryption use something known as integer factorization. The security of this method is based on a straightforward mathematical problem: given a large number, what are the factors
which multiply to give you that number? It’s simple enough in theory, but in practice it can take even the world’s most powerful supercomputers billions of years to answer. For information using the
Advanced Encryption Standard (AES) of 128, 192, or 256 bit keys, it would take the combined computational power on all of Earth trillions of years to decrypt even the smaller 128 bit key. This kind
of encryption is used for iPhone files and websites using HTTPS (like Medium).
Yet even this kind of encryption is at risk of becoming obsolete. As supercomputers advance and people have at their disposal ever greater amounts of computational power, experts understand that
current internet encryption will not remain invulnerable forever. Perhaps the biggest threat to this security method bears the face of the quantum computer. Finding the factors of large numbers is a
monumental task for a classical computer, but it’s not much of a problem at all for quantum machines.
In fact, throughout the history of ciphers and cryptography, only one method of encryption has ever been mathematically proven to offer perfect security. It is the world’s only unbreakable cipher.
Yet its strength lies not in its digital complexity, but in its real-world simplicity.
The cipher is the one-time pad (OTP), so called because a pad of numbers can only be used once and must then be disposed of afterward. While there are many variations of the OTP cipher — some using
binary, some grouping letters into sets, some using a Vigenere table, and so on — I’ll give one of the easier encryption methods below.
We begin by first creating our OTP of numbers. It’s important not to use an everyday number generator to do this. Computers rely on mathematics to come up with their “random” numbers, but patterns
are prevalent throughout math. What seems like random numbers will be vulnerable to patterns if they’re generated from a normal computer program. As this is the most important step in the OTP
encryption process, you’ll want to use Hardware Random Number Generators (RNG’s). These kind of generators are based on physical events such as electrical noise from semiconductors or the passage of
a photon through a filter. Alternatively, ten-sided dice can also be used to roll random numbers, making the entire encryption process non-digital.
An OTP of random numbers will then look something like this:
One of the most inconvenient aspects of the OTP cipher is that both the sender and recipient of a message must have an exact copy of the numbers. Anyone who has access to this can decrypt your
Now let’s imagine we want to send the word “quasar”. We first find each letter’s place in the alphabet. Q is 17 in the alphabet, u is 21, and so on. At the end we have this string of numbers to
represent the word “quasar”: 17-21-1-19-1-18. Using modular arithmetic, this string of 6 numbers will be added to the first 6 numbers in our OTP. 17 is added to 5, 21 to 19, 1 to 6, etc. The modular
arithmetic is in place so that we don’t end up with results larger than the number of letters in the alphabet, which is 26. So normally 21 + 19 = 40, but with modular addition the 40 becomes 14.
With our new set of numbers, we find the corresponding letters of the alphabet. The end result are the letters “VNGHAC” which are now the OTP encrypted version of “QUASAR”.
This encrypted word shows how an OTP cipher is more efficient than something like a Cesar cipher. With Cesar encryption “QUASAR” could become something like “XBHZHY”, in which the letters are shifted
by a certain amount. But their frequency is maintained. Because there are two letter a’s in the word quasar, there appear two letter h’s in the Cesar encryption. This is a major weakness. It means
someone trying to decrypt the message will use their knowledge of letter frequency to figure out certain letters, compromising the security of the entire message. With OTP encryption, the word quasar
may have multiple a’s yet the final encrypted word doesn’t repeat a single letter. Knowledge of letter frequency won’t be of any help.
The true strength of the OTP cipher, however, lies in the fact that there are two unknowns — the first being the encrypted text and the second being the random numbers of the pad. It is
mathematically impossible to solve this encryption no matter how much time or computational power anyone might have. This perfect security is only ensured if one follows all of the OTP rules. Do not
reuse numbers under any circumstances, destroy the OTP after use, and make sure the numbers are truly random.
OTP encryption was used during WW2 and the Cold War. Agencies like MI6 and the Russian Security Ministry still use it to this day. It’s not that a cipher like the One-time Pad is the future of
security, but there is something poetic in realizing that the world’s most impenetrable form of encryption relies not on digital and computational power, but on simple actions between two people.
Perfect secrecy can’t be found in the digital realm; but it does lie somewhere out here in the real world, taking the form of a piece of paper and a handful of dice. As the physicality of human
beings begins to bleed into the machine, what does it say about the safety and security of our information going forward? Will our future descendants belong to themselves alone — or to someone else?
Reference: [Article By: Ella Alderson – Article here; https://medium.com/swlh/this-is-the-worlds-only-unbreakable-encryption-a27691ac0890/] | {"url":"https://clctiv.com/this-is-the-worlds-only-unbreakable-encryption/?author=3","timestamp":"2024-11-07T13:00:22Z","content_type":"text/html","content_length":"74845","record_id":"<urn:uuid:5b4bd5be-4a0c-47d1-ad1f-9083391a516b>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00455.warc.gz"} |
An acute triangle has side lengths 21 cm, x cm, and 2x cm. If 21 is one of the shorter sides of the triangle, what is the greatest possible length of the longest side, rounded to the nearest tenth?18.8 cm24.2 cm42.0 cm72.7 cm
Answer:B. 24.2 cmStep-by-step explanation:We are given that,Lengths of the sides of a triangle are 21 cm, x cm and 2x cm.Using Triangle Inequality Theorem, which states that,'The sum of measure of
any two sides must be greater than the measure of the third side'.Since, 21 cm is the shorter side. We have,A) [tex]21+x>2x[/tex] i.e. [tex]21>x[/tex] or [tex]x<21[/tex].B) [tex]x+2x>21[/tex] i.e.
[tex]3x>21[/tex] or [tex]x>7[/tex]So, [tex]7<x<21[/tex]That is, [tex]14<2x<42[/tex]Thus, the length of the larger side has measure between 14 cm and 42 cm.Hence, the possible length of the longest
side is 24.2 cm. | {"url":"http://math4finance.com/general/an-acute-triangle-has-side-lengths-21-cm-x-cm-and-2x-cm-if-21-is-one-of-the-shorter-sides-of-the-triangle-what-is-the-greatest-possible-length-of-the-longest-side-rounded-to-the-nearest-tenth-18-8-cm24-2-cm42-0-cm72-7-cm","timestamp":"2024-11-08T14:13:11Z","content_type":"text/html","content_length":"30217","record_id":"<urn:uuid:bd0910cf-01ef-4685-b307-27f97f82193d>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00016.warc.gz"} |
On the relationship between the self-similarities of fractal signals and wavelet transforms
Since many natural phenomena are occasionally defined as stochastic processes and the corresponding fractal characteristics are hidden from their correlation functions or power spectra, the topic
would become very interest in signal processing. In this paper, we summarize the fractal dimensions and the relationship of the fractal in probability measure, variance, time series, time-averaging
autocorrelation, ensemble-averaging autocorrelation, time-averaging power spectrum, average power spectrum and distribution functions for stationary and nonstationary processes. We also propose that
the preservation of the one-dimensional self-similarity of a fractal signal is obtained by using the continuous wavelet transform (CWT) and the discrete wavelet transform (DWT) with the perfect
reconstruction - quadrature mirror filter structure. Moreover, we extend the results to the two-dimensional case and point out the relationship of the self-similarities between the CWT and DWT of the
fractal signals. A fractional Brownian motion process is provided as an example to show the results of this paper.
Conference Proceedings of the 1996 4th International Symposium on Signal Processing and its Applications, ISSPA'96. Part 2 (of 2)
城市 Gold Coast, Aust
期間 25/08/96 → 30/08/96
深入研究「On the relationship between the self-similarities of fractal signals and wavelet transforms」主題。共同形成了獨特的指紋。 | {"url":"https://scholar.nycu.edu.tw/zh/publications/on-the-relationship-between-the-self-similarities-of-fractal-sign","timestamp":"2024-11-14T12:23:08Z","content_type":"text/html","content_length":"54740","record_id":"<urn:uuid:71bb4695-813a-4757-a297-adc8da394879>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00818.warc.gz"} |
Why Investors Should Hold LongWhy Investors Should Hold Long
Interest rates are currently at a historical low. Since in the longer run interest rates will return to their historical average, this implies that bond prices are about to fall. Popular investment
advice therefore says that investors should shorten the maturity of their bond portfolios to minimise their losses. This argument, however, skips over the fact that longer-dated bonds pay higher
coupons as well as the fact that a substantial rise in interest rates may take quite some time to occur. We examine the combined impact of both and conclude that the current interest rate environment
in no way implies that investors should rebalance towards short-dated bonds. Extensive scenario analysis confirms that in an overall portfolio context a longer-dated bond portfolio is more efficient
than a short-dated bond portfolio, especially when long-dated liabilities are present.
A well-diversified portfolio should contain a substantial amount of fixed income. This leaves an important question, however. Should investors hold short or long(er)-dated bonds and how does this
depend on the level of interest rates? Popular advice in the current historically low interest rate environment is that, with interest rates on their way up and bond prices therefore on their way
down, investors should shorten the maturity of their bond holdings to minimise losses. However, this argument skips over the fact that longer-dated bonds pay higher coupons as well as the fact that a
substantial interest rate rise may take quite some time to occur. In this paper we integrate these arguments.
Bonds with longer maturities offer higher yields. The only exception to this rule is the case of a so-called 'inverted' yield curve when interest rates for longer maturities are lower than for short
maturities. This situation, however, only rarely occurs, and if it does, it never lasts long. Longer maturities also bear a significant market risk. If interest rates rise, investors with a
long-dated bond portfolio do not benefit from this rate rise. On the contrary, since the coupons on their bonds are fixed until maturity, these bonds will become less attractive, resulting in a drop
in value. The longer the maturity of the bonds, the greater the loss will be.
The difficulty in specifying the optimal maturity lies in the two contradicting effects described above. In the current low interest rate environment, should an investor only hold short-dated fixed
income in order to minimise his loss in case of a rate rise, or does the higher yield of long-dated bonds more than compensate for this risk? We demonstrate that in the current interest rate
environment, the optimal maturity of a fixed income portfolio will still be relatively long. Using a straightforward analysis of the total return on a fixed income portfolio, and taking into account
the expected rise in interest rates, we show that long-dated fixed income is expected to outperform short-dated fixed income. We also show what the optimal maturity would be in other interest rate
environments, such as relatively high interest rates or a flat yield curve. One might argue that long-dated fixed income is expected to perform better than short-dated fixed income at the price of
higher risk. To show that this is not the case, we perform an elaborate scenario analysis where we look at both the expected return and the risk of loss of capital over a 5-year horizon. We conclude
with the case where long-term liabilities are involved. Especially in this case the benefits of long-dated bonds are quite compelling.
Current interest rates in historical perspective
Figure 1 shows the US short (3-month) and long (10-year) interest rate over the past 15 years, as well as the difference between the two^2. The graph clearly illustrates that long rates are usually
higher than short rates. The curve was inverted only during a brief period in 1989. The fact that long-dated fixed income usually offers a higher yield than short-dated fixed income is often referred
to as the 'yield pick-up' of longer maturities. The latter stems from various sources, the most important being that investors tend to require a premium for inflation risk and lower liquidity.
Figure 1: Recent history of US short and long rates
Click on the image for an enlarged preview
Another interesting conclusion that can be drawn from Figure 1 is that over the past fifteen years the long and especially the short rate have never been as low as is currently the case. This is true
even if one considers a longer period. Market professionals generally assume that interest rates tend to return to their historical average. This effect is known as 'mean-reversion'. The short rate
mean-reversion level is assumed to be approximately 4%, whereas the long rate mean-reversion level is often thought to be around 5%^3. This means that in the current interest rate environment a rise
in interest rates is far more likely than a fall, although it should be emphasised that a further fall in interest rates is surely not impossible.
Click on the image for an enlarged preview
The so-called mean-reversion factor ^4.
Optimal maturity in the current interest rate environment
Figure 2 shows the current US yield curve and its assumed mean-reversion level. The curve is relatively steep, which means that in the current interest rate environment the spread between the long
and short rate, i.e. the yield pick-up, is quite high.
Figure 2: US yield curve of June 2004 and its mean-reversion level
Figure 3: Change in portfolio value
The expected change in value of a fixed income portfolio can be approximated by multiplying its duration^7 by the expected change of the corresponding interest rate^8. The expected change in
portfolio value due to the mean-reversion effect is shown in Figure 3, which clearly illustrates that short-dated fixed income is relatively insensitive to a change in interest rates. For longer
maturities the expected loss increases due to the increase in duration. For maturities longer than four years, however, the expected loss decreases again. On first sight this may seem strange, since
duration increases with maturity. From Figure 2, however, we see that the expected change of the interest rate is smaller the longer the time to maturity. The 3-month interest rate is approximately
3% below its mean-reversion level, but the 10-year interest rate is already quite close to its mean-reversion level.
Figure 4: Expected total return (R) of a fixed income portfolio
The expected total return (R) over a 1-year horizon is shown in Figure 4. It clearly indicates that in expected terms the yield pick-up of long-dated fixed income more then outweighs the possible
loss if interest rates rise in line with expectations.
Of course, one also has to take risk into account. If, in a portfolio context, long-dated bonds add significantly to the overall risk of a portfolio then the higher expected return is simply
compensation for the additional risk incurred. Only if long-dated bonds do not increase the risk of a typical investment portfolio we can say that such bonds are to be preferred. We will investigate
this issue further under the section "The Risk Factor" below. First, however, we check on the robustness of the above result and show how things would work out in other interest rate environments.
Robustness of results
One question is how robust the above conclusion, that in the current interest rate environment a portfolio with long-dated bonds leads to a higher expected return than a portfolio with short
maturities, is with respect to the assumed degree of mean-reversion. We therefore repeated the above analysis under different assumptions for the mean-reversion parameter, ranging from 0.1 to 0.5.
This showed that our conclusion remains valid even if we assume the degree of mean-reversion in interest rates to be much stronger. For example, Figure 5 shows the expected total return, as the sum
of the current interest rate and the expected change in market value due to the mean-reversion effect, using a mean-reversion factor equal to 0.5.
Figure 5: Expected total return and change in value of a fixed income portfolio in current interest rate environment, assuming a mean-reversion factor of 0.5
Figure 6: Expected total return and change in value of a fixed income portfolio in current interest rate environment, assuming a long rate mean-reversion level of 6%
We also repeated the analysis with higher mean-reversion levels for the long rate. Again, our conclusion remains valid. Figure 6, for example, shows the expected total return using a long rate
mean-reversion level of 6% instead of 5%. The mean-reversion factor is equal to 0.25.
Optimal maturity in other interest rate environments with a normal curve
In the analysis so far we have focused on the current interest rate environment, in which the yield curve is normally shaped, interest rates are below their historical average, and the curve is
relatively steep. But what would happen if the yield curve was less steep, or interest rates were above their historical average? Obviously, the flatter the curve, the less important the yield
pick-up argument will be. Likewise, the further interest rates are from their mean-reversion level, the more important the value argument will be.
Figure 7: Expected total return and change in value of a fixed income portfolio in case of a fictitious flat yield curve with low interest rates
Consider the case of a flat yield curve with relatively low interest rates. The yield pick-up in this case will be more or less equal to zero. However, one would make a loss if interest rates
returned to their historical average. Since short maturities are less sensitive to a change in interest rates, short maturities are in this case preferred over longer maturities. Figure 7 depicts
this conclusion graphically. Given a relatively flat curve, significantly below its long-term average, the expected drop in portfolio value increases with maturity. With hardly any yield pick-up to
compensate, this translates into a similar behavior for the expected total return. In this particular case a short-dated fixed income portfolio is optimal. It should be emphasised, however, that a
combination of low interest rates and a flat curve very rarely occurs^9.
Figure 8: Expected total return and change in value of a fixed income portfolio in case of the flat US yield curve with high interest rates of April 1989
If the yield curve is flat but rates are above their historical average, the optimal strategy is again easily figured out. From the perspective of yield pick-up one would be indifferent to either
short or long maturities. Since interest rates are expected to drop, however, one would expect to make a profit that increases with maturity. As a consequence, the fixed income portfolio should
clearly have a long maturity, as is visualised in Figure 8, which corresponds with the US yield curve of April 1989.
Finally, if the curve is steep with interest rates above their historical average, the maturity of the fixed income portfolio should of course be long. This would give the highest yield pick-up as
well as the highest profit if interest rates started to fall. It should be emphasised though, that, as is often the case with these kind of 'too-good-to-be-true' situations, the combination of a
steep curve and high interest rates hardly ever occurs.
What if the yield curve was inverted?
In the previous sections we assumed the yield curve to be normally shaped, implying a non-negative yield pick-up. It might happen, however, that the yield curve is inverted. Although this usually
happens in a high interest rate environment and does not last very long, for completeness we briefly describe the optimal maturity in the case of a steep inverted curve in both a high and low
interest rate environment.
Figure 9: Expected total return and change in value of a fixed income portfolio in case of a fictitious steep inverted yield curve with high interest rates
In the unlikely case of low interest rates the maturity should clearly be short. It gives the highest yield and no losses if interest rates rise and return to their historical average. A more
interesting case is a steep inverted curve with high interest rates. The yield pick-up in this case is negative and therefore a short maturity seems preferable. Since short maturities are more or
less insensitive to a change in interest rates, however, longer maturities have to be bought in order to take advantage of the expected fall in interest rates. Figure 9 shows that in case of a steep
inverted yield curve with relatively high interest rates, a medium-dated fixed income portfolio is optimal. It should be emphasised though that this optimal maturity strongly depends on the exact
form and level of the yield curve.
The risk factor
In the previous sections it was shown that, in terms of 1-year expected total return, in the current interest rate environment long-dated fixed income is to be preferred over short-dated fixed
income. This conclusion, however, ignores possible differences in terms of risk. Only if long-dated bonds do not add to overall portfolio risk we can truly say that they are superior to short-dated
bonds. To investigate this matter we performed a scenario analysis in which 2000 5-year scenarios were generated for interest rates and equity returns using a statistical model based on historical
data over the period 1970-2003, taking into account correlations, cross-correlations, auto-correlations, and, of course, mean-reversion in interest rates^10. Such an extensive set of scenarios not
only provides insight in the expected return but also in the risk for an investor following a certain investment strategy.
Figure 10: Risk-return profile for portfolios containing both equity and fixed income with different maturities
Consider an investor who wants to invest in both equity and fixed income. As always, his aim is to maximise return while keeping risk at an acceptable level. For practical purposes, however, this is
not specific enough. The exact definition of risk and return must depend on the specific goals of the investor. We will assume that our particular investor wants to maximise his total return after
five years, while at least keeping his capital intact. This means taking return to be the expected annual return over a 5-year period and to equate risk to the probability that after five years the
investor's asset value is less than at the outset. For different asset allocation strategies, i.e. yearly rebalanced mixtures of equity and fixed income, both expected return and risk (as defined)
are plotted in Figure 10. The upper (blue) line in Figure 10 reflects portfolios with short-dated fixed income, the middle (red) line represents portfolios with medium-dated fixed income, and the
bottom (green) line refers to portfolios with long-dated fixed income^11.
From Figure 10 we see, for example, that an allocation of 30% in equity and 70% in short-dated fixed income (the left point of the blue line) yields an expected annual return of 4.9% with a 2.7%
probability of capital after five years being less than at the outset. If one invested 40% in equity and 60% in short-dated fixed income, the latter probability would rise to around 6% but at the
same time expected return would rise to 5.4%. Figure 10 also shows that, holding on to the 40/60 allocation, if we were to invest in medium-dated bonds expected return would increase by 45 basis
points while the risk of capital after five years being below its initial value would drop as well. Investing in long-dated fixed income would yield even higher rewards. Expected return would
increase with almost 55 basis points, and the probability of a drop in capital would fall from around 6% to little over 5%. Note that maturities longer than 10 years would not significantly improve
the performance of the portfolio any further due to the fact that the current yield curve is relatively flat for maturities longer than 10 years.
As an alternative to keeping the allocation fixed at 40% equity and 60% fixed income, we could also decide to keep the risk at a fixed level. Keeping the probability of capital being less than
initial capital at 6%, Figure 10 shows that the investor should in that case invest in long-dated fixed income and at the same time enlarge his allocation in equity from 40% to 43%. Doing so would
allow him to pick up some more of the equity risk premium and thereby increase his expected annual return by more than 65 basis points, from 5.4% to little over 6%.
The above scenario analysis clearly illustrates that the current yield pick-up is large enough to compensate for the possible loss if interest rates start to rise. This is in line with the conclusion
from the previous analysis, where we only looked at expected total return.
More dynamic strategies
The scenario analysis in the previous section was based on a static strategy where every year the fixed income portfolio consists of bonds with the same maturity. It is not difficult to incorporate
dynamic strategies that take decisions based on the state of the economy. Every year the optimal maturity of the fixed income portfolio is then based on the slope of the curve and the deviation from
the mean-reversion curve. Applying these techniques would further improve the performance of the portfolio but that is beyond the scope of this paper.
Optimal maturity with long-term liabilities
Our analysis so far has been asset-only, i.e. it did not involve any liabilities. In many cases, however, medium to long-term liabilities play an important role on the balance sheet of investors.
This kind of liabilities is found, for example, in private wealth portfolios with long-term mortgages, in charities with long-term commitments to donate to social projects and in pension fund and
life insurance portfolios where obligations sometimes extend far beyond 30 years. In these cases, investing in longer-dated bonds not only leads to a more efficient asset portfolio, but also to
significant risk reduction on the total balance sheet. After investing in long-dated bonds the asset-side and the liability-side of the balance sheet both will have a long maturity. As a consequence,
both will react to a change in interest rates in more or less the same way.
Figure 11: Risk-return profile for portfolios containing both equity and fixed income with different maturities. The portfolio contains long-term liabilities in the form of a 30-year mortgage
The analysis in the previous section showed that in an asset-only context the risk-return profile in many different interest rate environments, including the current one, is optimised by investing in
long-dated bonds. The addition of long-term liabilities makes the case for long-dated investments even stronger. Figure 11 illustrates this. It shows the risk-return characteristics of our investor,
who now also carries a 30-year mortgage on his balance sheet, with a notional equal to 40% of his total assets. Investing in long-dated fixed income instead of short-dated fixed income would reduce
the risk of his capital (assets minus liabilities) after five years being less than his initial capital by more than 25% (from a probability of more than 17% to a probability of less than 13%). At
the same time this would increase his expected annual return over a 5-year horizon by almost 100 basis points.
An obvious strategy in the current historically low interest rate environment is to invest in short-dated bonds only. We showed, however, that this is far from optimal, as it does not take into
account the spread between long and short-term interest rates, which is typically positive and currently quite large. In the current environment long maturities should be preferred over short
maturities. Short maturity fixed income would only be optimal if the yield curve were substantially less steep. The following table summarises our conclusions (assuming a normally shaped yield
In a risk-return context, we showed that investing in longer-dated bonds not only adds expected return, but also reduces risk at the same time. When long-term liabilities are present, the degree of
risk reduction is much stronger because long-dated bonds form a better hedge for such liabilities. Still, even in these circumstances, it is not uncommon for managers with long-dated liability
portfolios to conclude from the, statistically correct, argument that interest rates are more likely to rise than to fall, that they should shorten the maturity of the asset portfolio. In this paper
we have (hopefully convincingly) shown that this is definitively not the case.
Fabozzi, F.J. and Fabozzi, T.D. (1995), The Handbook of Fixed Income Securities - Fourth edition, Irwin Professional Publishing.
Judge, G.G., Griffiths, W.E., Hill, R.C., Lütkepohl, H and Lee, T.C. (1985), The Theory and Practice of Econometrics - Second edition, John Wiley & Sons
Siegel, J.J. (1992), The real rate of interest from 1800-1990, Journal of Monetary Economics, Vol. 29, pp. 227-252.
Steehouwer, H. (2004), Macroeconomic Scenarios and Reality, PhD thesis.
Ziemba, W.T. and Mulvey J.M. (2001), Worldwide Asset and Liability Modelling, Cambridge University Press.
^1 Vincent van Antwerpen and Janwillem Engel are consultants and Theo Kocken CEO of Cardano Risk Management in Rotterdam, The Netherlands. Harry M. Kat is Professor of Risk Management and Director of
the Alternative Investment Research Centre, Cass Business School, City University, London
^2 Source: Bloomberg.
^3 Siegel (1992) and Steehouwer (2004), for example, present studies of interest rates over very long time periods, showing long term average rates around the levels we use.
^4 The expectation of the short rate increases with 0.25 *(4-1.3) = 0.7 from 1.3% to 2.0% in the first year. In the second year it increases with 0.25*(4-2) = 0.5 from 2.0% to 2.5%.
^5 Initially, we will restrict ourselves to expected return in determining the optimal maturity. For a complete comparison, however, the risk of different portfolios should also be taken into
account. This extra dimension will be added and discussed in section 7-9.
^6 Note that this decomposition can also be interpreted as taking the difference between the mean reversion effect and the 1-year forward curve from the current yield curve. If after one year, as a
result of the mean reversion effect, the interest rate for a certain maturity is still below the current 1-year forward rate, then the expected total return for that maturity is positive and vice
^7 Duration is defined as the weighted average of the time to receipt of the individual cash flows (coupon and notional) of the bond or portfolio. It can be used as a measure of the sensitivity of a
bond portfolio to a change of interest rates. See for example Fabozzi and Fabozzi (1995).
^8 Note that in the current interest rate environment, where interest rates are expected to rise, calculating the change in value of a fixed income portfolio using duration only, i.e. without
accounting for convexity, will somewhat overestimate this change. Comparable arguments apply for high interest rate environments.
^9 In June 2003 the US short (3-month) interest rate was just above 1%, whereas the long (10-year) interest rate was about 3.5%. In this case both effects (yield pick-up and mean reversion) are of
similar magnitude and the expected total return is close to zero for all maturities.
^10 The scenarios are constructed using a vector autoregressive model. For the theoretical introduction to VAR-models see for example Judge et al. (1985). An application of these models in an
asset-liability context can be found in part VIII of Ziemba and Mulvey (2001).
^11 The short-dated fixed income portfolio consists of bonds with an average maturity of one year. The medium-dated portfolio consists of bonds with an average maturity of three years. The long-dated
portfolio consists of bonds with an average maturity of six years. | {"url":"http://www.eurekahedge.com/Research/News/1133/_ftnref4","timestamp":"2024-11-11T00:46:28Z","content_type":"text/html","content_length":"124739","record_id":"<urn:uuid:f80912a0-a7fb-461c-addf-576e36ee7f69>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00397.warc.gz"} |
The Boost Graph Library
The Boost Graph Library (BGL)
Graphs are mathematical abstractions that are useful for solving many types of problems in computer science. Consequently, these abstractions must also be represented in computer programs. A
standardized generic interface for traversing graphs is of utmost importance to encourage reuse of graph algorithms and data structures. Part of the Boost Graph Library is a generic interface that
allows access to a graph's structure, but hides the details of the implementation. This is an “open” interface in the sense that any graph library that implements this interface will be interoperable
with the BGL generic algorithms and with other algorithms that also use this interface. The BGL provides some general purpose graph classes that conform to this interface, but they are not meant to
be the “only” graph classes; there certainly will be other graph classes that are better for certain situations. We believe that the main contribution of the The BGL is the formulation of this
The BGL graph interface and graph components are generic, in the same sense as the Standard Template Library (STL) [2]. In the following sections, we review the role that generic programming plays in
the STL and compare that to how we applied generic programming in the context of graphs.
Of course, if you are already familiar with generic programming, please dive right in! Here's the Table of Contents. For distributed-memory parallelism, you can also look at the Parallel BGL.
The source for the BGL is available as part of the Boost distribution, which you can download from here.
How to Build the BGL
DON'T! The Boost Graph Library is a header-only library and does not need to be built to be used. The only exception is the GraphViz input parser.
When compiling programs that use the BGL, be sure to compile with optimization. For instance, select “Release” mode with Microsoft Visual C++ or supply the flag -O2 or -O3 to GCC.
Genericity in STL
There are three ways in which the STL is generic.
Algorithm/Data-Structure Interoperability
First, each algorithm is written in a data-structure neutral way, allowing a single template function to operate on many different classes of containers. The concept of an iterator is the key
ingredient in this decoupling of algorithms and data-structures. The impact of this technique is a reduction in the STL's code size from O(M*N) to O(M+N), where M is the number of algorithms and N is
the number of containers. Considering a situation of 20 algorithms and 5 data-structures, this would be the difference between writing 100 functions versus only 25 functions! And the differences
continues to grow faster and faster as the number of algorithms and data-structures increase.
Extension through Function Objects
The second way that STL is generic is that its algorithms and containers are extensible. The user can adapt and customize the STL through the use of function objects. This flexibility is what makes
STL such a great tool for solving real-world problems. Each programming problem brings its own set of entities and interactions that must be modeled. Function objects provide a mechanism for
extending the STL to handle the specifics of each problem domain.
Element Type Parameterization
The third way that STL is generic is that its containers are parameterized on the element type. Though hugely important, this is perhaps the least “interesting” way in which STL is generic. Generic
programming is often summarized by a brief description of parameterized lists such as std::list<T>. This hardly scratches the surface!
Genericity in the Boost Graph Library
Like the STL, there are three ways in which the BGL is generic.
Algorithm/Data-Structure Interoperability
First, the graph algorithms of the BGL are written to an interface that abstracts away the details of the particular graph data-structure. Like the STL, the BGL uses iterators to define the interface
for data-structure traversal. There are three distinct graph traversal patterns: traversal of all vertices in the graph, through all of the edges, and along the adjacency structure of the graph (from
a vertex to each of its neighbors). There are separate iterators for each pattern of traversal.
This generic interface allows template functions such as breadth_first_search() to work on a large variety of graph data-structures, from graphs implemented with pointer-linked nodes to graphs
encoded in arrays. This flexibility is especially important in the domain of graphs. Graph data-structures are often custom-made for a particular application. Traditionally, if programmers want to
reuse an algorithm implementation they must convert/copy their graph data into the graph library's prescribed graph structure. This is the case with libraries such as LEDA, GTL, Stanford GraphBase;
it is especially true of graph algorithms written in Fortran. This severely limits the reuse of their graph algorithms.
In contrast, custom-made (or even legacy) graph structures can be used as-is with the generic graph algorithms of the BGL, using external adaptation (see Section How to Convert Existing Graphs to the
BGL). External adaptation wraps a new interface around a data-structure without copying and without placing the data inside adaptor objects. The BGL interface was carefully designed to make this
adaptation easy. To demonstrate this, we have built interfacing code for using a variety of graph dstructures (LEDA graphs, Stanford GraphBase graphs, and even Fortran-style arrays) in BGL graph
Extension through Visitors
Second, the graph algorithms of the BGL are extensible. The BGL introduces the notion of a visitor, which is just a function object with multiple methods. In graph algorithms, there are often several
key “event points” at which it is useful to insert user-defined operations. The visitor object has a different method that is invoked at each event point. The particular event points and
corresponding visitor methods depend on the particular algorithm. They often include methods like start_vertex(), discover_vertex(), examine_edge(), tree_edge(), and finish_vertex().
Vertex and Edge Property Multi-Parameterization
The third way that the BGL is generic is analogous to the parameterization of the element-type in STL containers, though again the story is a bit more complicated for graphs. We need to associate
values (called “properties”) with both the vertices and the edges of the graph. In addition, it will often be necessary to associate multiple properties with each vertex and edge; this is what we
mean by multi-parameterization. The STL std::list<T> class has a parameter T for its element type. Similarly, BGL graph classes have template parameters for vertex and edge “properties”. A property
specifies the parameterized type of the property and also assigns an identifying tag to the property. This tag is used to distinguish between the multiple properties which an edge or vertex may have.
A property value that is attached to a particular vertex or edge can be obtained via a property map. There is a separate property map for each property.
Traditional graph libraries and graph structures fall down when it comes to the parameterization of graph properties. This is one of the primary reasons that graph data-structures must be
custom-built for applications. The parameterization of properties in the BGL graph classes makes them well suited for re-use.
The BGL algorithms consist of a core set of algorithm patterns (implemented as generic algorithms) and a larger set of graph algorithms. The core algorithm patterns are
• Breadth First Search
• Depth First Search
• Uniform Cost Search
By themselves, the algorithm patterns do not compute any meaningful quantities over graphs; they are merely building blocks for constructing graph algorithms. The graph algorithms in the BGL
currently include
• Dijkstra's Shortest Paths
• Bellman-Ford Shortest Paths
• Johnson's All-Pairs Shortest Paths
• Kruskal's Minimum Spanning Tree
• Prim's Minimum Spanning Tree
• Connected Components
• Strongly Connected Components
• Dynamic Connected Components (using Disjoint Sets)
• Topological Sort
• Transpose
• Reverse Cuthill Mckee Ordering
• Smallest Last Vertex Ordering
• Sequential Vertex Coloring
Data Structures
The BGL currently provides two graph classes and an edge list adaptor:
The adjacency_list class is the general purpose “swiss army knife” of graph classes. It is highly parameterized so that it can be optimized for different situations: the graph is directed or
undirected, allow or disallow parallel edges, efficient access to just the out-edges or also to the in-edges, fast vertex insertion and removal at the cost of extra space overhead, etc.
The adjacency_matrix class stores edges in a |V| x |V| matrix (where |V| is the number of vertices). The elements of this matrix represent edges in the graph. Adjacency matrix representations are
especially suitable for very dense graphs, i.e., those where the number of edges approaches |V|^2.
The edge_list class is an adaptor that takes any kind of edge iterator and implements an Edge List Graph.
Copyright © 2000-2001 Jeremy Siek, Indiana University (jsiek@osl.iu.edu)
Lie-Quan Lee, Indiana University (llee@cs.indiana.edu)
Andrew Lumsdaine, Indiana University (lums@osl.iu.edu) | {"url":"https://www.boost.org/doc/libs/1_42_0/libs/graph/doc/index.html","timestamp":"2024-11-09T16:20:36Z","content_type":"text/html","content_length":"14622","record_id":"<urn:uuid:e2d0a652-c6da-4192-84fa-19088696712d>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00799.warc.gz"} |
Graphs of Trigonometric Functions
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Recall that when we studied Trigonometric Ratios, we looked at the Functions $\mathrm{sin}\left(x\right)$,$\mathrm{cos}\left(x\right)$ and $\mathrm{tan}\left(x\right)$. A trigonometric function is a
function that relates the size of an angle in a right-angled triangle to the lengths of its sides.
Properties of Trigonometric Graphs
There are three graphs that we are interested in when studying the graphs of trigonometric functions: the graphs of sin(x), cos(x) and tan(x). For GCSE mathematics, you kind of need to memorise what
these graphs look like. However, they do have some key properties that make them quite simple to draw. We will start with the graph of $y=\mathrm{sin}\left(x\right)$.
The graph of y=sin(x)
Graphs of Trigonometric Function- Graph of y=sin(x), Jordan Madge- StudySmarter Originals
Key properties
• We can see that the graph of y=sin(x) has a maximum value of 1 and a minimum value of -1. From this, we can conclude that the value of sin(x) can only fall between 1 and -1. Thus, if we have an
equation where $\mathrm{sin}\left(x\right)=1.4$, the equation has no solutions.
• The x values go up in 90-degree intervals and periodically repeat on a 360-degree cycle. In other words, after every 360 degrees, we notice that the graph repeats itself.
• At various points, the graph is symmetrical. For example, we have symmetry about the line $x=90°$. This will be useful to us later on when finding multiple solutions to trigonometric Equations.
Suppose $\mathrm{sin}\left(x\right)=1$. Looking at the graph, we can see that $\mathrm{sin}\left(x\right)=1$ at $x=-270°$, $x=90°$ and $x=450°$. Since the graph of $\mathrm{sin}\left(x\right)$ will
continue to oscillate infinitely, we could conclude from this that the equation $\mathrm{sin}\left(x\right)=1$ has an infinite Number of solutions. If a trigonometric equation has one solution, it
will have an infinite Number of solutions and later on we will use the symmetry property to try to find such solutions.
The official name of a graph that takes the shape of a sine graph is a sinusoidal wave. Many things naturally take the shape of a sine wave, for example, the movement of planets around the sun.
The graph of y=cos(x)
Key properties
• If you were not paying close attention in the last section, you may think that this graph is pretty much the same as the graph of $\mathrm{sin}\left(x\right)$. However, if you go back and play a
game of spotting the difference, you will notice that the graph of $\mathrm{cos}\left(x\right)$ is just the graph of $\mathrm{sin}\left(x\right)$shifted 90 degrees to the left.
• Similarly to $\mathrm{sin}\left(x\right)$, the graph of $\mathrm{cos}\left(x\right)$ also has a maximum at $1$, a minimum at $-1$, and also a symmetry property. We must just remember that the
graph of $\mathrm{cos}\left(x\right)$ starts at $1$, whereas the graph of $\mathrm{sin}\left(x\right)$ starts at $0$.
The graph of y=tan(x)
Key properties
• The graph of $\mathrm{tan}\left(x\right)$ looks quite a fair bit different to $\mathrm{cos}\left(x\right)$ and $\mathrm{sin}\left(x\right)$. However, it is similar in the sense that it is
periodic, and we can see that it repeats itself every $180$degrees.
• The graph of $\mathrm{tan}\left(x\right)$ has these things called asymptotes, which are points that the graph tends towards but never quite reaches. These are represented on the graph as dashed
lines. We can see the first positive asymptote appears at $x=90°$, and then they repeat every $180$ degrees.
• Unlike $\mathrm{cos}\left(x\right)$ and $\mathrm{sin}\left(x\right)$, the graph of $\mathrm{tan}\left(x\right)$ does not have a maximum or minimum of plus or minus $1$; it has a maximum and
minimum of plus or minus infinity. Thus, the equation $\mathrm{tan}\left(x\right)=3.8$ can be solved to obtain an infinite number of real values for x.
Graphs of Trigonometric Functions Methods
Finding solutions to trigonometric equations
In the previous section, we briefly touched upon the fact that if a trigonometric equation has one solution, it will have an infinite number of solutions. In the below section, we will be working out
how to find multiple solutions to trigonometric Equations.
Since trigonometric equations can have an infinite number of solutions, we need to specify a boundary when stating answers so that we do not spend an infinite amount of time finding every last
solution. This boundary will usually be expressed as an interval, for example, $0°\le x\le 360°$, or $-180°\le x\le 180°$. Be sure to take note of this boundary when answering questions.
Graphs of Trigonometric Functions Examples
Find the solutions to $\mathrm{sin}\left(x\right)=0.9$, for the interval $0°\le x\le 360°$.
The first step is to sketch the graph of $y=\mathrm{sin}\left(x\right)$ and $y=0.9$ on the same axis for the interval $0°\le x\le 360°$.
The points of intersection have been labelled in orange as 1 and 2, these are the solutions we are seeking to find the exact values of.
The second step is to find the exact value of the initial solution. This can be done by typing ${\mathrm{sin}}^{-1}\left(0.9\right)$ into our calculator. When we do this, we obtain $x=64.2°$. This is
clearly the first solution labelled on the diagram since it is between $0°$and $90°$.
It is important to note that your calculator should be in degree mode when calculating trigonometric values since we are working in degrees. If your calculator is in radian mode, the answer may vary
and so you may obtain the incorrect answer. You know that your calculator is in degree mode when a small D appears on the top of the screen. If you see an R or any other letter, it is in the wrong
mode and needs to be changed.
The next step is to find the other solution using the symmetrical property of the graph of sin(x). If we notice, the graph is symmetrical about id="2569857" role="math" $x=90°$. Thus, we can work out
the second solution by working out the distance between $64.2°$ and $90°$, and then adding this value to $90°$. This can be illustrated in the below diagram:
Since the distance between $64.2°$ and $90°$ is $25.8°$, the second solution is id="2569881" role="math" $90+25.8=115.8°$. Therefore, the two solutions to the equation $\mathrm{sin}\left(x\right)=
0.9$ in the interval $0°\le x\le 360°$ are id="2569884" role="math" $x=64.2°$ and id="2569885" role="math" $x=115.8°$.
Find the solutions to $\mathrm{cos}\left(x\right)=-0.2$ for the interval $-180°\le x\le 180°$.
The first step is to sketch the graphs of $y=\mathrm{cos}\left(x\right)$ and id="2569887" role="math" $y=-0.2$ on the same axes for the interval $-180°\le x\le 180°$ so that we can see the solutions
we are trying to find.
The next step is to find the initial solution by typing ${\mathrm{cos}}^{-1}\left(-0.2\right)$ into our calculator. We obtain id="2569892" role="math" $x=101.5°$. Clearly, this is the solution
labelled 2 on the diagram, since it is a little more than id="2569898" role="math" $90°$ but less than id="2569899" role="math" $180°$.
We now need to find the other solution depicted in the diagram. Since the graph of $\mathrm{cos}\left(x\right)$ is symmetrical about the line $x=0$, we can see that the other solution must be at id=
"2569894" role="math" $x=-101.5°$. Thus, the two solutions to id="2569895" role="math" $\mathrm{cos}\left(x\right)=-0.2$ in the interval $-180°\le x\le 180°$ are id="2569897" role="math" $x=101.5°$
and id="2569896" role="math" $x=-101.5°$.
Find the solutions to $\mathrm{tan}\left(x\right)=2.3$ for the interval $0°\le x\le 360°$.
The first step, as usual, is to sketch the graphs $y=\mathrm{tan}\left(x\right)$ and id="2569908" role="math" $y=2.3$ on the same axes for the interval $0°\le x\le 360°$.
We can see that there are two points of intersection and thus two solutions to $\mathrm{tan}\left(x\right)=2.3$. The first solution can be found by typing id="2569911" role="math" ${\mathrm{tan}}^
{-1}\left(2.3\right)$ into our calculator. Doing so, we obtain id="2569912" role="math" $x=66.5°$This is clearly the first solution since it is between $0$ and $90$ degrees.
The graph of $\mathrm{tan}\left(x\right)$ repeats itself periodically after $180$ degrees. Therefore, we can find the next solution by adding multiplies 180 to the initial solution. So, the second
solution is at id="2569918" role="math" $180+66.5=246.5°$. Therefore, the two solutions to $\mathrm{tan}\left(x\right)=2.3$ in the interval $0°\le x\le 360°$ are id="2569915" role="math" $x=66.5°$
and id="2569917" role="math" $x=246.5°$.
Solutions to any equations involving tan(x) can be found by adding multiples of 180 to the initial solution.
Find the solutions of $4\mathrm{tan}\left(x\right)=3$ for the interval $-180°\le x\le 180°$.
We cannot solve this equation in its current form. We first need to divide both sides by $4$ to get $\mathrm{tan}\left(x\right)$ by itself. We obtain $\mathrm{tan}\left(x\right)=\frac{3}{4}$. Now, we
can find the first solution to the equation by taking the inverse tan of both sides to get $x={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)=36.9°$.
Now, since it is tan, we know that solutions can be found by adding or subtracting multiples of $180°$ to the initial solution. Thus, the next solution will be at $36.9+180=216.9$, however, this is
out of the range. We can get another solution by subtracting$180°$ from $36.9°$ to get $-143.1°$ which is in the range. Subtracting a further ${180}^{°}$ will yield a solution out of the range,
therefore the two solutions to $4\mathrm{tan}\left(x\right)=3$ in the interval $-180°\le x\le 180°$ are $x=36.9°$ and $x=-143.1°$.
Graphs of Trigonometric Functions - Key takeaways
• There are three graphs that we are interested in when studying the graphs of trigonometric functions: the graphs of sin(x), cos(x) and tan(x).
• The graphs of sin(x) and cos(x) have a maximum value of 1 and a minimum value of -1, the graph of tan(x) has a maximum and minimum of plus or minus infinity.
• The graph of cos(x) is just the graph of sin(x) shifted to the left by 90 degrees.
• The graphs of sin(x) and cos(x) have symmetry properties that enable us to find multiple solutions when solving equations.
• For equations involving tan(x), we can get each solution by adding multiples of 180 to each solution.
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Frequently Asked Questions about Graphs of Trigonometric Functions
How to graph trigonometric functions?
The graphs of sin(x), cos(x) and tan(x) each have their own graphs with some key properties. If you learn the shape of each of the graphs, you should be able to graph any trigonometric equation.
How do you find the domain of a trigonometric functions?
Trigonometric functions have an infinite domain, so the domain you are interested in is usually specified in the question.
What are the Applications of Trigonometric Functions?
Trigonometric functions can be used to solve trigonometric equations.
How to graph the inverse of trigonometric functions?
Graph the trigonometric function and then reflect it in the line y=x.
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Perform Univariate Linear Regression Separately for Columns of X — univariate_regression
This function performs the univariate linear regression y ~ x separately for each column x of X. Each regression is implemented using .lm.fit(). The estimated effect size and stardard error for each
variable are outputted.
Z = NULL,
center = TRUE,
scale = FALSE,
return_residuals = FALSE
n by p matrix of regressors.
n-vector of response variables.
Optional n by k matrix of covariates to be included in all regresions. If Z is not NULL, the linear effects of covariates are removed from y first, and the resulting residuals are used in place
of y.
If center = TRUE, center X, y and Z.
If scale = TRUE, scale X, y and Z.
Whether or not to output the residuals if Z is not NULL.
A list with two vectors containing the least-squares estimates of the coefficients (betahat) and their standard errors (sebetahat). Optionally, and only when a matrix of covariates Z is provided, a
third vector residuals containing the residuals is returned.
n = 1000
p = 1000
beta = rep(0,p)
beta[1:4] = 1
X = matrix(rnorm(n*p),nrow = n,ncol = p)
X = scale(X,center = TRUE,scale = TRUE)
y = drop(X %*% beta + rnorm(n))
res = univariate_regression(X,y) | {"url":"https://stephenslab.github.io/susieR/reference/univariate_regression.html","timestamp":"2024-11-12T02:45:28Z","content_type":"text/html","content_length":"11563","record_id":"<urn:uuid:d1b75896-7754-4be5-aff4-10247f0ffa56>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00578.warc.gz"} |