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Teaching and learning of mathematics
The group research is about the teaching and learning of mathematics.
The group is part of the Akelius Math Learning Lab, a collaboration between Chalmers, the University of Gothenburg and Akelius Math AB.
Some of the areas covered by the group's research are
• teacher training
• algebra didactics
• digital learning materials in mathematics
• history of mathematics
The focus is on teaching and learning in upper secondary school and at university, but questions relating to other school stages are included.
Networks in which we participate are, for example, the Center for Educational Science and Teacher Research (CUL) and Mathematics Education Research in Gothenburg (MERGOT) at the University of
Gothenburg, Development, Learning, Research (ULF) at Chalmers and the international networks Nordic Network for Algebra Learning (N2AL), International Study Group on the relations between History and
Pedagogy of Mathematics (HPM) and SEFIs Mathematics Special Interest Group (MSIG).
In recent years, we have, among others, collaborated with Angelika Kullberg, Éva Fülöp and Semir Becevic (IDPP at the University of Gothenburg) as well as Camilla Björklund (IPKL at the University of
Gothenburg), Kajsa Bråting (Uppsala University), Timo Tossavainen (Luleå tekniska universitet), Jon Star (Harvard) and Michele Artigue (Université Paris-Diderot). | {"url":"https://www.chalmers.se/en/departments/mv/research/research-areas/teaching-and-learning-of-mathematics/","timestamp":"2024-11-11T08:21:52Z","content_type":"text/html","content_length":"298039","record_id":"<urn:uuid:d7bd1b7b-f70f-4748-a11c-1f7cd9ff0127>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00454.warc.gz"} |
From Pythagoras to Fourier and From Geometry to Nature
We recall, in a schematic way, the basic definitions about sequences and series [62].
A sequence is a function defined on the set N of integers (n ∈ N).
A property is said to be definitively true if it holds for all integer indices greater than a given value.
The sequence 1n converges to zero. We write: 1n→0.
The sequence n diverges positively. We write: n → +∞.
The sequence {(−1)^n} is indeterminate: it does not admit a limit.
The sequence {a[n]} converges to the limit ℓ ⇔ |a[n] − ℓ| → 0.
The sequence {a[n]} is convergent if and only if ∀ɛ > 0 it results definitively that |a[n] − a[m]| < ɛ (Cauchy’s convergence criterion).
The sequence {a[n]} diverges positively ⇔ ∀K > 0 it results definitively that |a[n]| > K.
A series is the sum of the terms of a sequence: ∑k=1∞ak.
The study of series reduces to that of the sequence of partial sums: sn=∑k=1nak.
If {s[n]} → s the series is convergent and its sum is s.
If {s[n]} → +∞ the series diverges positively.
If {s[n]} does not have a limit the series is said to be indeterminate.
The series of Zeno’s paradox is convergent: ∑k=0∞1/2n=2.
The harmonic series is positively divergent: ∑k=1∞1/k=+∞.
The series ∑k=0∞(−1)k is indeterminate.
If {a[n]} depends on x ∈ (a, b) we have sequences or series of functions.
For example, the series of functions ∑k=1∞ak(x) converges to S(x) in (a, b) if ∀x ∈ (a, b) we have:
The convergence of this series is uniform in [α, β] ⊂ (a, b) if:
It is proven that the limit of a uniformly convergent sequence of continuous functions is a continuous function.
Example: consider the geometric series ∑k=0∞xk .
Recalling the equation 1 − x^n = (1 − x)(1 + x + x ^2 + ⋯ + x^n^−1), assuming x ≠ 1, it follows that:
we can conclude that:
Moreover, the convergence cannot be uniform in [−1, 1], but only in intervals of the type [α, β] ⊂ (−1, 1). | {"url":"https://files.athena-publishing.com/books/p2fg2n/chapters/50/view","timestamp":"2024-11-04T23:46:18Z","content_type":"text/html","content_length":"26105","record_id":"<urn:uuid:01946dcc-13ba-41d1-b270-cb4c10a9502b>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00257.warc.gz"} |
EViews Help: @isvalidgroup
@isvalidgroup Support Functions
Syntax: @isvalidgroup(str)
str: string
Return: integer
Check for whether a string represents a valid specification for creating an EViews group or auto-series. Returns a “1” if the expression is valid, and a “0” if it is not.
If your workfile contains the series X, Y and Z:
scalar a = @isvalidgroup("x y z")
will set A equal to 1,
scalar b = @isvalidgroup("log(x)")
will set B equal to 1, and
scalar e = @isvalidgroup("log(xy)")
will set E equal to 0 since XY is not a valid series in the workfile.
In contrast to the result in E,
scalar f = @isvalidgroup("log(x*y)")
is valid, since “LOG(X*Y)” is a valid series expression. | {"url":"https://help.eviews.com/content/functionref_i-@isvalidgroup.html","timestamp":"2024-11-05T02:35:35Z","content_type":"application/xhtml+xml","content_length":"8525","record_id":"<urn:uuid:d4080495-5d0e-4b82-9ea3-60ec17bc44b3>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00036.warc.gz"} |
Diffraction PSF 3D
Author: Bob Dougherty
Installation: Download Diffraction_PSF_3D.class or Diffraction_PSF_3D.java to the plugins folder. (Windows users right-click to download.) Requires ImageJ version 1.32c or later.
Description: Plugin to compute the 3D point spread function of a diffraction-limited microscope. Based on an analytical derivation using Fraunhofer diffraction. Intended to provide sample data for
use with Convolve 3D and Iterative Deconvolve 3D.
Disclaimer: This code is intended to represent a classical diffraction limit. If possible, direct measurements of your microscope's psf should be made, or data should be obtained from the
Usage: Fill in the form and compute the psf. A pop-up window gives the Rayleigh resolution in pixels before the psf is created. The "Image pixel spacing" is also known as the calibration. It
is important that the Image pixel spacing and the wavelength be specified in the same units, usually nm or µm.
If the sample medium is different from the one used to compute the numerical aperture, n*sin(θ/2), then the NA value will need to be corrected. For example, if the objective lens was
designed to have an NA of 1.4 using an immersion oil with and index of refracation of 1.515, and water is used in place of the immersion oil, then the value of NA to input into the
plugin is (1.33/1.515)*1.4 = 1.229, where the index of refraction of water is taken as 1.33. This media substitution will create spherical aberration, as discussed below. The plugin is
not yet able to compute this specific type of spherical aberration from first principles.
Dimensions: The image pixel spacing and slice spacing should be set to match the image data stacks. If the PSF will be used with Convolve 3D or Iterative Deconvolve 3D, then it is not necessary to
make the width, height, and depth match the image stacks. The width, height, and depth should be chosen large enough that the PSF "fits" in the resulting stack. The idea is that the
stack is wide and deep enough that the energy of the PSF that falls outside it is not important. Experimentation is probably required. Image/Stacks/Plot Z-axis Profile may be useful to
see whether there are enough slices. If the required number of slices exceeds the number of slices in the image data Z-series, then usefulness of 3D deconvolution in this case might be
in question. The dB option can be used to examine the lower levels of the PSF, but should be turned off in final run unless the image data is in dB.
SA: Spherical aberration causes the PSF to be non-symmetric in z. (Positive spherical aberration accentuates the bright ring at the perimeter of the blur circle for z > z0. Photographers
call this bad bokeh.) The latest version of the plugin includes a preliminary spherical aberration treatment in which the longitudinal focus shift is assumed to be proportional to the
fourth power of the radial distance from the optical axis in the plane of the aperture. This model can also translate the point of best focus in z; no correction for this translation is
made. The distribution of spherical aberration across the aperture in a real imaging system depends on the lens design, and may not resemble the fourth power model used here.
A proposal for an improved model is illustrated here. This would be based on the an assumption that the SA results from a well-corrected microscope being used with some material that
has an index of refraction different from the material that was intended in the microscope design. This is similar to, but simpler than, a model in the literature. The next step would
be to derive the focus shift in terms of the radial distance, as indicated in the figure.
Versions: 0: 5/02/2005
1: 5/04/2005. Fixed a bug that created a bright spot on axis for large defocus. Improved speed by using symmerty.
1.1: 5/04/2005. Improved speed (Simpson's rule integration).
1.2: 5/05/2005. Inputs changed to n*sin(theta) for NA, wavelength, and index of refraction.
2: 5/06/2005. Included spherical aberration, assumed to vary as y^4.
License: Copyright (c) 2005, OptiNav, Inc.
All rights reserved.
Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:
• Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer.
• Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided
with the distribution.
• Neither the name of OptiNav, Inc. nor the names of its contributors may be used to endorse or promote products derived from this software without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF
MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL,
EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED
AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF
THE POSSIBILITY OF SUCH DAMAGE.
ImageJ: ImageJ can be freely downloaded from the ImageJ web site. | {"url":"https://www.optinav.info/Diffraction-PSF-3D.htm?ref=blog.biodock.ai","timestamp":"2024-11-03T06:44:18Z","content_type":"text/html","content_length":"7585","record_id":"<urn:uuid:7a18f469-f9db-4bcd-9fd0-102be77ecf1c>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00195.warc.gz"} |
3. Graded Problem
3. Graded Problem
Find the forces Fx and Fy required to hold the tank shown in the figure below stationary.The pipe diameter D is not given, but it is the diameter required for a constant water level.Since the tank is
anchored, it will not slide on a surface (i.e., no static friction from the tank resting on a surface), and the pipe is open to the atmosphere. The momentum of the incoming flow is dissipated in the
tank and does not affect the outlet velocity. Neglect the weight of the tank, but not the weight of the water (60 °F). Assume no losses in the flow.
Fig: 1
Fig: 2 | {"url":"https://tutorbin.com/questions-and-answers/3-graded-problem-find-the-forces-fx-and-fy-required-to-hold-the-tank-shown-in-the-figure-below-stationary-the-pipe","timestamp":"2024-11-11T17:01:07Z","content_type":"text/html","content_length":"65509","record_id":"<urn:uuid:d65f470a-2d61-40f7-8288-b6c519e7e8ee>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00146.warc.gz"} |
Streaming Students can cause damage
It’s a hot potato topic but let’s just have a look at it.
Is the amount of time and effort spent putting students into ability groups worth it?
Numeracy research doesn’t support all the effort and angst it takes to create the groups, especially when it has a negative effect on students’ results.
Not to mention the negative effect it has on their self-esteem.
I’ve seen students as young as Year 1 devastated by their inclusion in the ‘dumb’ class. Their parents were embarrassed too. Unfortunately, once a student is graded into a lower-ability class they
generally don’t ‘graduate’ out of it.
There are those that think that when a child is put in a lower class it makes them work harder to be ‘promoted’. But what about the child who works ‘their socks off’ and still can’t do well enough to
meet the cut-off allocation for the next group up? They come to believe that they are 'no good' at maths or worse that they are just 'dumb' or that effort doesn't pay off...
Arbitrary cut offs are made to stream students into maths classes. There are only so many teachers and classrooms available in a maths time slot. When the ‘top’ class is full the next few children,
who may be only 1 or 2 points away in their marks, still have to go into the next class. They aren't generally told how close they are to the ‘top’ group. Alternatively, some may be 'lucky enough' to
be put int the 'top' group just to fill the numbers quota. And they are 'lucky'. Research shows that if the teacher believes that a student is more capable than they may actually be then they expect
more of them and the students can achieve more! The student is also likely to have a higher expectation of their own ability if streamed into a a 'top' group.
Due to class numbers and the limitations of available teachers even 'graded' classes will have a wide range of student ability levels. Teachers still need to differentiate for the students in these
classes. There is only so much ‘grading’ you can do to put a child in with a ‘homogenous’ group.
There are also questions to be asked about the legitimacy of grading students to place them into groups.
• What assessment tools or results were used?
• Do they provide a true indication of the child’s ability level or understanding?
• Will the teacher be programming their lessons based on the areas of weakness &/or strength identified in the assessments?
• Are the teachers going to follow a different program within the ‘streamed’ classes?
• Are students still going to be expected to sit the same assessment at the end of term and year as their peers?
On a very practical level consider the following:
1. It takes school leaders a massive amount of time to coordinate the rooming and teaching timetables to implement ability grouping.
2. How much teaching time is lost as students move from room to room… (not to mention the loss of time for the ones that haven’t got all their equipment with them?)
3. It’s frustrating for parents when their child’s classroom teacher can’t comment on their mathematics progress (or lack thereof) because they don’t teach them Maths.
4. Taking a child out of their ‘everyday class’ also raises an interesting discussion about numeracy and mathematics … That is, to be numerate you have to be able to take your maths skills out of
the maths lesson and apply them in everyday life. It stands to reason that if the classroom teacher knows what is being taught in the maths lesson then they can make the links between maths and
other subjects and other activities and thereby help primary school students to consolidate their numeracy understanding.
Back in 2008 the Commonwealth of Australian Governments commissioned a review of the available international and national research on what would improve the numeracy teaching and learning of
Australian students. One of the reccomendations, of this unfortunately not well known report, was:
“That the use of ability grouping across classes in primary and junior secondary schooling be discouraged given the evidence that it contributes to negative learning and attitudinal outcomes for less
well achieving students and yields little positive benefit for others, thus risking our human capital goals.” Recommendation 9 (pp45-47)
Another strong and passionate voice to listen to in this debate is Professor Jo Boaler (Stanford University) who believes streaming “is not only very damaging but also incorrect”.
In her book, “Mathematical Mindsets” (Chapter 6: “Mathematics and the Path to Equity“) Professor Boaler says, by separating students we perpetuate the myths that:
1. ‘There are those that can and those that can’t do maths’ and
2. ‘Maths is a more difficult subject to learn than other subjects’ and
3. That ‘some students are just not suited to higher-level maths’.
Is there an ALTERNATIVE to streaming?
Fortunately, there are alternative ways (based on research evidence) to help both the classroom teacher and all the students in the class to make progress in numeracy.
And, NOT streaming students into ability groups can actually help both student and teacher enjoy the ‘journey’.
We need to know what the research says and to put into place evidence-based practices that will work to improve numeracy results for all the students in our schools.
It’s interesting to note that when teaching practices are put in place to meet the needs of the students in the bottom 25% of a class the results of all the students tend to rise!
If you’re interested in exploring the alternatives to meet the needs of the students who are struggling then check out the Professional Learning (PL) options, Teaching & Learning Resources provided
by LYNZ EDUCATION.
Let’s help ALL students to ‘get’ Maths and enjoy the journey. | {"url":"https://www.lynzeducation.com.au/post/streaming-students-can-cause-damage","timestamp":"2024-11-13T21:46:54Z","content_type":"text/html","content_length":"1050488","record_id":"<urn:uuid:ad533b03-2346-4311-a98f-3edf44b519a6>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00687.warc.gz"} |
The Stacks project
Lemma 37.10.4. Consider a commutative diagram
\[ \xymatrix{ (X \subset X') \ar[rr]_{(f, f')} \ar[rd] & & (Y \subset Y') \ar[ld] \\ & (S \subset S') } \]
of thickenings. Assume
1. $Y' \to S'$ is locally of finite type,
2. $X' \to S'$ is flat and locally of finite presentation,
3. $f$ is flat, and
4. $X = S \times _{S'} X'$ and $Y = S \times _{S'} Y'$.
Then $f'$ is flat and for all $y' \in Y'$ in the image of $f'$ the local ring $\mathcal{O}_{Y', y'}$ is flat and essentially of finite presentation over $\mathcal{O}_{S', s'}$.
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Solve each system. $$\begin{array}{r} x+2 y-3 z=0 \\ 2 x-y+z=0 \\ 3 x+y-4 z=0 \end{array}$$
Short Answer
Expert verified
The solution is \(x = 1\), \(y = 7\), and \(z = 5\).
Step by step solution
Write down the system of equations
The given system of linear equations is:\[\begin{array}{r}x + 2y - 3z = 0 \2x - y + z = 0 \3x + y - 4z = 0 \end{array}\]
Use the elimination method to eliminate one variable
First, multiply the first equation by 2 to get:\[2(x + 2y - 3z) = 0 \implies 2x + 4y - 6z = 0\]Then subtract the second equation from this result:\[2x + 4y - 6z - (2x - y + z) = 0\]This simplifies
to:\[5y - 7z = 0 \implies y = \frac{7}{5} z\]
Substitute back into the original equations
Substituting \(y = \frac{7}{5} z\) into the second equation:\[2x - \left(\frac{7}{5}z\right) + z = 0 \implies 2x - \frac{7}{5}z + z = 0 \implies 2x - \frac{2}{5}z = 0 \implies 2x = \frac{2}{5}z \
implies x = \frac{1}{5}z\]
Substitute back to find x, y, and z
Substitute \(x = \frac{1}{5}z\) and \(y = \frac{7}{5} z\) into the first equation:\[\frac{1}{5} z + 2 \left(\frac{7}{5} z\right) - 3z = 0\implies \frac{1}{5} z + \frac{14}{5} z - 3z = 0\implies \frac
{1 + 14 - 15}{5} z = 0\implies 0 = 0\]Since any value of \(z\) satisfies the equation, choose \(z = 5\), then:\[ x = \frac{1}{5} \cdot 5 = 1, \ y = \frac{7}{5} \cdot 5 = 7, \ z = 5 \]
Verify the solution
Substitute \(x = 1\), \(y = 7\), and \(z = 5\) back into the original system:\[ x + 2y - 3z = 1 + 2\cdot7 - 3\cdot5 = 1 + 14 - 15 = 0 \2x - y + z = 2\cdot1 - 7 + 5 = 2 - 7 + 5= 0 \3x + y - 4z = 3\
cdot1 + 7 - 4\cdot5 = 3 + 7 - 20 = 0\]All equations are satisfied, so the solution is correct.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
elimination method
The elimination method, also known as the addition method, involves combining equations to eliminate one of the variables. This simplifies the system and allows easier solving. In the exercise, the
elimination method was applied by first manipulating one of the equations so that it could cancel out a variable in another equation. For example, we multiplied the first equation by 2 and then
subtracted the second equation:
• Multiply the first equation by 2
• Subtract the modified first equation from the second equation
This resulted in a simpler equation with fewer variables, making it easier to isolate one variable and solve for the remaining ones.
substitution method
The substitution method requires solving one equation for one variable and then substituting this expression into the other equations. This helps isolate and solve for each variable step-by-step.
After eliminating one variable using the elimination method, the next step is to substitute the simplified expressions back into the remaining equations:
• Solve for one variable (e.g., y in terms of z)
• Substitute the expression for y into another equation to solve for x
• Finally, substitute back to find the exact values of x, y, and z
In the exercise, this method was effectively used to find expressions for y and x in terms of z, simplifying the remaining computations.
linear algebra
Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations in vector spaces and through matrices. Solving systems of linear equations, like
in this exercise, is a key aspect of linear algebra. The given system of equations represents a set of linear relationships between the variables x, y, and z. In this context:
• A linear equation is an equation involving only linear terms (no exponents or nonlinear functions).
• Solving a system of linear equations involves finding a set of values for the variables that satisfy all equations simultaneously.
By using methods like elimination and substitution, we can systematically solve these systems, which is a fundamental concept in linear algebra.
verification of solution
Verifying the solution is an essential step to ensure that your computed values satisfy all the original equations. After solving for x, y, and z, you need to substitute these values back into the
original equations to check for consistency. In the exercise, the values of x, y, and z were substituted back into each equation:
• For the first equation: Check if the left-hand side equals the right-hand side.
• Repeat the same for the second and third equations.
When each equation is satisfied (i.e., each side of the equation equals zero in this case), it confirms that the solution is correct. This step is critical as it ensures there are no errors in the
solution process and that the derived values are indeed correct. | {"url":"https://www.vaia.com/en-us/textbooks/math/precalculus-functions-and-graphs-4-edition/chapter-8/problem-48-solve-each-system-beginarrayr-x2-y-3-z0-2-x-yz0-3/","timestamp":"2024-11-03T15:37:32Z","content_type":"text/html","content_length":"248758","record_id":"<urn:uuid:95d0f533-3504-406a-bc4b-68fe5e6cb175>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00695.warc.gz"} |
Calculating the Smith set
Bruce Anderson landerso at ida.org
Thu May 2 09:39:10 PDT 1996
This message combines and superceeds two messages that I had previously sent
(one on 21 Apr 1996, and one on 22 Apr 1996) but were never posted.
On Apr 20, 4:55pm, Steve Eppley wrote:
> Subject: Re: Calculating the Smith set?
> [The EM list is again having troubles. Four of my posts bounced back
> to me with "unknown mailer error 14". I'm attempting to repost them.]
> Steve E wrote [a Smith algorithm]:
> >> Find the candidate(s) with the fewest pairwise defeats. (These
> >> candidate(s) are guaranteed to be in the Smith set.) Create
> >> a list named Smith with just these candidate(s).
> >>
> >> for each candidate Ci in the (growing) Smith list
> >> for each candidate Cj
> >> if Ci did not pairwise-beat Cj
> >> if Cj is not yet in the Smith list
> >> append Cj to the Smith list
> >> endif
> >> endif
> >> endfor
> >> endfor
> Bruce A replied:
> >It may not work--the line "append Cj to the Smith list" is too vague.
> It wasn't vague to me, but this appears to be another instance of
> Eppley's First Law of Communication: "The worst judge of how a
> message will be interpreted is its author."
> To a programmer, "list" and "append" have meanings which may not be
> apparent to everyone. Lists are ordered and have a first element
> and a last element. When a new element is appended to a list, it
> is appended at the end and becomes the new last element.
> Another property of lists is that the pointer (a.k.a. address) to a
> list's first element can always be obtained. Another property is
> that given a pointer to an element of a list there is a way to find
> the pointer to the next element. (Attempting to find the pointer to
> the next element after the last element returns a special "null"
> pointer.) So lists can be traversed from first to last.
> If there was any vagueness, I think it would be in the line:
> "for each candidate Ci in the (growing) Smith list"
> I'm implying that the candidates would be traversed from the first
> candidate of the list, in list order, until the "null" pointer is
> returned by the attempt to find the next candidate.
> When coded in a standard programming language instead of pseudocode,
> there will be no vagueness.
> --Steve
>-- End of excerpt from Steve Eppley
It turns out that the line I that didn't understand indeed was:
for each candidate Ci in the (growing) Smith list
I took it to mean:
for each candidate Ci in the current Smith list (note, this list will grow)
instead of:
for each candidate Ci that is now in, or ever will be in, the Smith list
I think that, to understand the algorithm, one does need to know how "lists" and
"pointers" work. As I now understand "dynamic lists" and "pointers," your
algorithm works and is of order N^2. Accordingly, I withdraw my objections to
that algorithm. The algorithm also works (and may or may not be more efficient)
if it is changed to:
Find the candidate(s) with the fewest pairwise defeats. (These
candidate(s) are guaranteed to be in the Smith set.) Create
a list named Smith with just these candidate(s).
for each candidate Ci (ever) in the (dynamically growing) Smith list
for each candidate Cj on the ballot
if Cj is not in the current Smith list
if Ci did not pairwise-beat Cj
append Cj to the (end of the current) Smith list
The Smith set is the set of candidates on the final Smith list.
I do not know whether either or both of these algorithms are either more or less
efficient than my (unproven) algorithm; but, as you said, they are going to be
so quick that increasing the efficiency here is not likely to be important. My
unproven algorithm is:
Let n be the number of candidates.
Let LIST be empty.
If n = 1, add the one candidate to LIST and go to #1.
Do for i from 0 to (n-2):
If there are 1 or more candidates with exactly i pairwise losses, then:
Add all candidates with exactly i pairwise losses to LIST.
Let m be the current size of LIST.
If the total number of pairwise wins by all of the current members of
LIST is greater than or equal to m(n-m), and the total number of
pairwise losses by all of the current members of LIST is less than or
equal to m(m-1)/2, go to #1.
End if.
End do.
LIST = {all candidates}
Smith set = LIST.
More information about the Election-Methods mailing list | {"url":"http://lists.electorama.com/pipermail/election-methods-electorama.com/1996-May/065660.html","timestamp":"2024-11-13T11:02:51Z","content_type":"text/html","content_length":"8099","record_id":"<urn:uuid:9eff8a9c-cfce-4b31-b5c4-70021dbd2b28>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00735.warc.gz"} |
Workbook 5A(2nd Edition)(Sold in Packs of 10)
think! Mathematics is a series of textbooks and workbooks written to meet the requirements of the latest National Standards. In tandem with adapted approaches used in Singapore to develop critical
thinking, as well as problem-solving and analytical skills, students can gain a strong foundation to achieve the highest possible level of mastery of mathematics.
This series adopts a 3-part lesson structure and a spiral design within and across grade levels. The concrete-pictorial-abstract (C-P-A) approach and the infusion of fun-filled activities help
develop students’ cognitive and metacognitive skills. New-trend questions and open-ended questions are included throughout the series to encourage analytical and lateral thinking.
Product Details:
ISBN-13 : 978-981-339-206-9
Extent : 188 pages
Consultant : Dr Yeap Ban Har,Dr Bill Tozzo
Linguistics Consultant : Dr Amy Tozzo
Authors : Dr. Andrea Kang
Dimensions : 230 mm x 290 mm x 180 mm per book
Weight : 0.65kg per book
Table of Contents
Book A
Unit 1 - Place Value and Decimals
Lesson 1-1 Understanding place value
Lesson 1-2 Understanding place value
Lesson 1-3 Writing decimals
Lesson 1-4 Comparing decimals
Lesson 1-5 Rounding decimals
Mind Workout
Math Journal
Unit 2 - Order of Operations
Lesson 2-1 Using mixed operations
Lesson 2-2 Using mixed operations
Lesson 2-3 Using mixed operations
Lesson 2-4 Interpreting expressions
Mind Workout
Math Journal
Unit 3 - Multiplication and Division of Whole Numbers
Lesson 3-1 Understanding powers of 10
Lesson 3-2 Multiplying by powers of 10
Lesson 3-3 Estimating products
Lesson 3-4 Multiplying multi-digit factors
Lesson 3-5 Multiplying multi-digit factors
Lesson 3-6 Dividing tens
Lesson 3-7 Estimating quotients
Lesson 3-8 Dividing 2-digit factors
Lesson 3-9 Dividing 2-digit factors
Lesson 3-10 Dividing 2-digit factors
Mind Workout
Math Journal
Unit 4 - Addition and Subtraction of Decimals
Lesson 4-1 Adding decimals
Lesson 4-2 Adding decimals
Lesson 4-3 Adding decimals
Lesson 4-4 Subtracting decimals
Lesson 4-5 Subtracting decimals
Lesson 4-6 Subtracting decimals
Mind Workout
Math Journal
Unit 5 - Multiplication and Division of Decimals
Lesson 5-1 Multiplying decimals by powers of 10
Lesson 5-2 Estimating products of decimals
Lesson 5-3 Multiplying decimals
Lesson 5-4 Multiplying decimals
Lesson 5-5 Multiplying decimals
Lesson 5-6 Dividing decimals by powers of 10
Lesson 5-7 Estimating quotients of decimals
Lesson 5-8 Dividing decimals by whole numbers
Lesson 5-9 Dividing decimals by whole numbers
Lesson 5-10 Dividing whole numbers by decimals
Lesson 5-11 Dividing decimals
Mind Workout
Math Journal
Unit 6 - Addition and Subtraction of Fractions
Lesson 6-1 Adding fractions
Lesson 6-2 Adding fractions
Lesson 6-3 Adding mixed numbers
Lesson 6-4 Subtracting fractions
Lesson 6-5 Subtracting fractions
Lesson 6-6 Subtracting mixed numbers
Lesson 6-7 Solving word problems
Mind Workout
Math Journal
Book A- 49 Lessons
think! Mathematics is a series of textbooks and workbooks written to meet the requirements of the latest National Standards. In tandem with adapted approaches used in Singapore to develop critical
thinking, as well as problem-solving and analytical skills, students can gain a strong foundation to achieve the highest possible level of mastery of mathematics.
This series adopts a 3-part lesson structure and a spiral design within and across grade levels. The concrete-pictorial-abstract (C-P-A) approach and the infusion of fun-filled activities help
develop students’ cognitive and metacognitive skills. New-trend questions and open-ended questions are included throughout the series to encourage analytical and lateral thinking.
Product Details:
ISBN-13 : 978-981-339-206-9
Extent : 188 pages
Consultant : Dr Yeap Ban Har,Dr Bill Tozzo
Linguistics Consultant : Dr Amy Tozzo
Authors : Dr. Andrea Kang
Dimensions : 230 mm x 290 mm x 180 mm per book
Weight : 0.65kg per book
Table of Contents
Book A
Unit 1 - Place Value and Decimals
Lesson 1-1 Understanding place value
Lesson 1-2 Understanding place value
Lesson 1-3 Writing decimals
Lesson 1-4 Comparing decimals
Lesson 1-5 Rounding decimals
Mind Workout
Math Journal
Unit 2 - Order of Operations
Lesson 2-1 Using mixed operations
Lesson 2-2 Using mixed operations
Lesson 2-3 Using mixed operations
Lesson 2-4 Interpreting expressions
Mind Workout
Math Journal
Unit 3 - Multiplication and Division of Whole Numbers
Lesson 3-1 Understanding powers of 10
Lesson 3-2 Multiplying by powers of 10
Lesson 3-3 Estimating products
Lesson 3-4 Multiplying multi-digit factors
Lesson 3-5 Multiplying multi-digit factors
Lesson 3-6 Dividing tens
Lesson 3-7 Estimating quotients
Lesson 3-8 Dividing 2-digit factors
Lesson 3-9 Dividing 2-digit factors
Lesson 3-10 Dividing 2-digit factors
Mind Workout
Math Journal
Unit 4 - Addition and Subtraction of Decimals
Lesson 4-1 Adding decimals
Lesson 4-2 Adding decimals
Lesson 4-3 Adding decimals
Lesson 4-4 Subtracting decimals
Lesson 4-5 Subtracting decimals
Lesson 4-6 Subtracting decimals
Mind Workout
Math Journal
Unit 5 - Multiplication and Division of Decimals
Lesson 5-1 Multiplying decimals by powers of 10
Lesson 5-2 Estimating products of decimals
Lesson 5-3 Multiplying decimals
Lesson 5-4 Multiplying decimals
Lesson 5-5 Multiplying decimals
Lesson 5-6 Dividing decimals by powers of 10
Lesson 5-7 Estimating quotients of decimals
Lesson 5-8 Dividing decimals by whole numbers
Lesson 5-9 Dividing decimals by whole numbers
Lesson 5-10 Dividing whole numbers by decimals
Lesson 5-11 Dividing decimals
Mind Workout
Math Journal
Unit 6 - Addition and Subtraction of Fractions
Lesson 6-1 Adding fractions
Lesson 6-2 Adding fractions
Lesson 6-3 Adding mixed numbers
Lesson 6-4 Subtracting fractions
Lesson 6-5 Subtracting fractions
Lesson 6-6 Subtracting mixed numbers
Lesson 6-7 Solving word problems
Mind Workout
Math Journal
Book A- 49 Lessons | {"url":"https://thinkmathematics.com/products/think-mathematics-workbook-5a-sold-in-packs-of-10-2nd-edition","timestamp":"2024-11-09T12:20:31Z","content_type":"text/html","content_length":"295517","record_id":"<urn:uuid:f9bc0b65-27be-44fc-8aba-81c263df255b>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00732.warc.gz"} |
Promise CSP (standard and seen from the other side)
Kristina Asimi, 27 Mar 2023
This talk will be divided into two parts. In the first part we will talk about the standard Promise CSP (PCSP), where a pair of relational structures (A,B) (such that there is a homomorphism from A
to B) is fixed and PCSP(A,B) is defined as the problem of deciding whether an input structure has a homomorphism to A or not even to B. In the second part of the talk we propose a similar problem,
where we restrict the left-hand side instead of the right-hand side, motivated by the so-called left-hand side restricted CSP. Namely, we fix a collection of pairs of relational structures C (such
that for every pair there is a homomorphism from the first structure to the second one) and ask the following: for an input pair (A,B) from C and an input structure X, decide whether there is a
homomorphism from B to X or not even from A to X. | {"url":"https://ggoat.fit.cvut.cz/seminar/asimi-promise-csp.html","timestamp":"2024-11-05T18:28:20Z","content_type":"text/html","content_length":"4819","record_id":"<urn:uuid:23619195-9c69-490c-a50c-6e4fdcc2d86e>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00292.warc.gz"} |
Geomagnetic Storm Intensity Calculation via Dst-Kp Analysis
19 Oct 2024
Popularity: ⭐⭐⭐
Geomagnetic Storm Strength
This calculator provides the calculation of geomagnetic storm strength using the Dst and Kp indices.
Calculation Example: The geomagnetic storm strength is a measure of the intensity of a geomagnetic storm. It is calculated using the Dst and Kp indices, which are measures of the disturbance in the
Earth’s magnetic field. The G value is a logarithmic scale, with higher values indicating stronger storms.
Related Questions
Q: What is the difference between a geomagnetic storm and a solar storm?
A: A geomagnetic storm is a disturbance in the Earth’s magnetic field caused by the interaction of the solar wind with the Earth’s magnetosphere. A solar storm is a burst of energy from the sun that
can cause geomagnetic storms.
Q: What are the effects of geomagnetic storms?
A: Geomagnetic storms can cause a variety of effects, including power outages, disruptions to communication systems, and damage to satellites. They can also affect the human body, causing nausea,
headaches, and fatigue.
Symbol Name Unit
Dst Dst nT
Calculation Expression
G Function: The geomagnetic storm strength is calculated using the formula: G = 10^((Dst/Kp) - 2.5)
pow(10, ((Dst/Kp) - 2.5))
Calculated values
Considering these as variable values: Dst=-100.0, Kp=5.0, the calculated value(s) are given in table below
Derived Variable Value
G Function pow(10.0,-22.5)
Similar Calculators
Calculator Apps | {"url":"https://blog.truegeometry.com/calculators/Geomagnetic_storms_calculation.html","timestamp":"2024-11-08T15:07:08Z","content_type":"text/html","content_length":"16599","record_id":"<urn:uuid:64eb0c17-a59b-4793-bcfd-50db66c82f35>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00525.warc.gz"} |
Moments, monodromy, and perversity: A diophantine perspective
It is now some thirty years since Deligne first proved his general equidistribution theorem, thus establishing the fundamental result governing the statistical properties of suitably "pure"
algebro-geometric families of character sums over finite fields (and of their associated L-functions). Roughly speaking, Deligne showed that any such family obeys a "generalized Sato-Tate law," and
that figuring out which generalized Sato-Tate law applies to a given family amounts essentially to computing a certain complex semisimple (not necessarily connected) algebraic group, the "geometric
monodromy group" attached to that family. Up to now, nearly all techniques for determining geometric monodromy groups have relied, at least in part, on local information. In Moments, Monodromy, and
Perversity, Nicholas Katz develops new techniques, which are resolutely global in nature. They are based on two vital ingredients, neither of which existed at the time of Deligne's original work on
the subject. The first is the theory of perverse sheaves, pioneered by Goresky and MacPherson in the topological setting and then brilliantly transposed to algebraic geometry by Beilinson, Bernstein,
Deligne, and Gabber. The second is Larsen's Alternative, which very nearly characterizes classical groups by their fourth moments. These new techniques, which are of great interest in their own
right, are first developed and then used to calculate the geometric monodromy groups attached to some quite specific universal families of (L-functions attached to) character sums over finite fields.
Original language English (US)
Publisher Princeton University Press
Number of pages 475
ISBN (Electronic) 9781400826957
ISBN (Print) 0691123306, 9780691123301
State Published - Sep 12 2005
All Science Journal Classification (ASJC) codes
Dive into the research topics of 'Moments, monodromy, and perversity: A diophantine perspective'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/moments-monodromy-and-perversity-a-diophantine-perspective-2","timestamp":"2024-11-02T15:04:28Z","content_type":"text/html","content_length":"48300","record_id":"<urn:uuid:741993ec-89bb-4d16-933e-02c0dae5b2b4>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00265.warc.gz"} |
Science:Math Exam Resources/Courses/MATH220/December 2009/Question 06 (b)
MATH220 December 2009
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q7 (a) •
Q7 (b) • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) •
Question 06 (b)
Let ${\displaystyle \mathbb {I} }$ be the set of irrational numbers. That is ${\displaystyle \mathbb {I} =\mathbb {R} -\mathbb {Q} }$.
Decide whether the following statements are true or false. Prove your answers. Hint:The results from part (a) will be of great help.
(i) ${\displaystyle \forall {}x\in \mathbb {I} ,\forall {}y\in \mathbb {I} ,xy\in \mathbb {I} }$.
(ii) ${\displaystyle \forall {}x\in \mathbb {I} ,\exists {}y\in \mathbb {I} \quad {\textrm {s.t.}}\quad {}xy\in \mathbb {I} }$.
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is
correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still
stuck, go for the next hint.
Hint 1
Part (i) claims that the product of any two irrational numbers is irrational.
Part (i) claims that the product of any irrational number and some other irrational number is irrational.
Hint 2
For part (i) look at part (a) (i) and for part (ii) see if you can use both parts form part (a) together.
Hint 3
The first is false and the second is true.
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
• If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you
are stuck or if you want to check your work.
• If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem
and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies.
(i) This is false. For any irrational number x, choose ${\displaystyle \displaystyle y={\frac {1}{x}}}$ which is irrational by part (a)(i) of this problem. Then the product is 1 which is not
irrational. This gives a counter example to the claim that the product of any two irrational numbers must be irrational.
(ii) This is true. For any irrational number x, choose ${\displaystyle \displaystyle y=1+{\frac {1}{x}}}$ which is irrational by part (a)(i) and part (a)(ii) of this problem. Then the product is ${\
displaystyle \displaystyle x+1}$ which is irrational again by part (a)(ii). Thus the claim is true that for any irrational number, there exists an irrational number such that their product is
Click here for similar questions
MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Rationality, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag | {"url":"https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH220/December_2009/Question_06_(b)","timestamp":"2024-11-14T00:38:14Z","content_type":"text/html","content_length":"47132","record_id":"<urn:uuid:4f7b41e3-3479-452e-9c80-0b14427b1b88>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00675.warc.gz"} |
How inventory levels affect commodity futures curves and returns | Macrosynergy
A new Review of Finance article investigates the link between commodity inventories on the one hand and futures returns and curve backwardation on the other. Most prominently, low inventories mean
that the convenience yield of physical holdings is high (more need of insurance against a “stock-out”). In this case, producers and inventory owners are willing to pay a surcharge for immediate
access to the physical. This will (normally) translate into a higher risk premium, more backwardation, and higher expected excess returns. Moreover, external factors, such as price volatility, can
induce producers and physical inventory holders to pay a higher premium for future price certainty that translates into a positive basis and expected return on futures positions. The inventory
connection explains why commodity futures returns and futures curve shape are correlated, i.e. backwardation is more often than not a profit opportunity. According to this paper, mainstream theory is
backed by empirical evidence for 31 commodities over more than 40 years.
The Fundamentals of Commodity Futures Returns
Gary B. Gorton, Fumio Hayashi and K. Geert Rouwenhorst
Review of Finance, 2013, vol. 17, issue 1, pages 35-105
The basic theory
Established theories of commodity futures suggest that the inventory level of a physical commodity is a fundamental driver of its risk premium and basis. Other factors, such as price volatility, may
affect both inventory and risk premiums paid to financial investors:
“The traditional Theory of Storage (see Kaldor (1939), Working (1949), and Brennan (1958)) assumes that holders of inventories receive implicit benefits, called the “convenience yield”, that decline
as inventory increases. Since it accrues to owners of inventories but not to owners of futures contracts, the convenience yield is closely tied to the basis… the convenience yield, and hence the
basis, are declining and convex functions of inventories“.
N.B. According to the Theory of Storage, in equilibrium the basis of a storable commodity is equal to convenience yield minus both funding costs and storage costs.
“There is a modern, optimization-based version of the Theory of Storage that emanates from Deaton and Laroque (1992). Inventories act as buffer stocks which help to absorb shocks to demand and supply
affecting spot prices. But inventories cannot be negative (goods cannot be transferred from the future to the past), so there is a possibility of a stock-out in which non-negativity constraint on
inventories binds [implying that the basis can surge in times of shortage). Routledge, Seppi, and Spatt (2000)…show how the convenience yield arises endogenously as a function of the level of
inventories and supply and demand shocks. Even if there is no direct benefit from owning physical inventories, the convenience yield can be positive because inventories have an option value due to a
positive probability of a stock-out.”
“The Theory of Normal Backwardation of Keynes (1930) and Hicks (1939) assumes that commodity producers and inventory holders hedge future spot price risk by taking short positions in the futures
market. To induce risk-averse speculators into taking the opposite long positions, current futures prices are set at a discount (i.e., is “backwardated”) to expected future spot prices at maturity.
The commodity futures risk premium is the size of this discount.”
“Modern formulations of the Theory of Normal Backwardation (Stoll 1979 and Hirshleifer 1988, 1990) make two basic assumptions. First, the revenue from the physical control of a commodity by hedgers
is non-marketable. This assumption might be justified if hedgers in the futures markets are either privately held firms or individual farmers. Second, participation in commodity futures markets by
outside investors is limited by some (possibly informational) entry barriers, so a positive risk premium will not be competed away. [Hence], the commodity futures risk premium consists of not only
the systematic risk (i.e., the covariance with the market portfolio of traded assets) but also a component related to the volatility of spot prices.”
Empirical analysis and findings
“The main contribution of our paper is an empirical examination of the effect of inventories on the basis and the risk premium articulated by the theory just outlined… by using a comprehensive
dataset on 31 commodity futures and physical inventories between 1971 and 2010…We collect a comprehensive historical dataset of inventories for 31 individual commodities over a 40-year period between
1971 and 2010.”
1. The basis tends to increase as inventories decline
“We find that there is a clear negative relationship between normalized inventories [inventories divided by their normal levels] and the basis and that for many commodities the slope of the
basis-inventory curve becomes more negative at lower inventories levels. And we find steeper slopes at normal inventory levels for commodities that are difficult to store….When inventories are in a
normal range (no stock-out), the estimated slope of the basis-inventory regression is negative for all commodities except four, and statistically significant at 5% for about a half of the
Cross-sectional differences in storability are reflected in the sensitivity of the basis to inventories. The relationship is particularly strong for Energies (bulky and more difficult to store),
while many Industrial Metals (easy to store) tend to have slope coefficients that are relatively small in magnitude…. Industrial Metals are relatively easy to store, and the normal inventory level I*
would be large relative to demand… Storability also helps to explain why the slope coefficients for Meats are on average smaller in magnitude than for commodities in the Softs and Grains groups.”
2. Also the risk premium declines with inventory.
“We perform a linear regression of the monthly excess return on the normalized inventory level I / I * at the end of the prior month as well as monthly dummies…As is apparent from the low t-values,
the normalized inventory coefficients are not sharply estimated. However, most of them have the expected negative sign. If we impose the restriction of a common slope coefficient within groups, we
find significant negative and quantitatively large slope coefficients for all commodities except for the easy-to-store Metals.”
3. The basis is correlated with future excess returns
“The level of inventories is a noisy measure of the true state of inventories because of demand shocks. Also, there is a conceptual question about the relevant inventory measure. These considerations
motivate us to examine other signals of the current state of inventories. [In particular] low-inventory commodities [should be associated with] a higher basis, higher prior excess and spot returns…
We show that prior futures returns, prior spot price changes and the futures basis are correlated with futures risk premiums as predicted by the Theory… Portfolios that take positions based on the
futures basis, prior futures excess returns, prior spot returns, or volatility select commodity futures with below normal inventories which our theory predicts are expected to earn higher risk
premiums. Moreover, these risk premiums are highly significant, both in a statistical sense as well as in an economic sense.”
N.B.:” There are many issues involved in compiling a dataset on inventories… Because most commodity futures contracts call for physical delivery at a particular location, futures prices should
reflect the perceived relative scarcity of the amount of the commodity which is available for immediate and future delivery at that location. For example, data on warehouse stocks of industrial
metals held at the exchange are available from the LME, but no data are available on stocks that are held off-exchange but that could be economically delivered at the warehouse on short notice.
Similarly, relevant Crude Oil inventories would include not only physical stocks held at the delivery point in Cushing, Oklahoma, but also oil which is held at international locations but that could
be economically shipped there, or perhaps even government stocks.” | {"url":"https://macrosynergy.com/research/how-inventory-levels-affect-commodity-futures-curves-and-returns/","timestamp":"2024-11-09T16:44:29Z","content_type":"text/html","content_length":"187642","record_id":"<urn:uuid:d5c81117-7df4-4185-b3a0-e24d6428008d>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00155.warc.gz"} |
Семинары: D. Giraudo, Limit theorems and probability inequalities for $U$-statistics
Аннотация: The study of the asymptotic behavior of U-statistics has concentrated a lot of attention since the seminal work by Hoeffding (1963), due to its applications in particular to statistics and
random geometry. In this talk, we will review some results on U-statistics of i.i.d. data (Hoeffding's decomposition, law of large numbers, law of the iterated logarithms, central limit theorem) and
give an exponential inequality for U-statistics of i.i.d. data and furnish applications to the rates in the strong law of large numbers and functional central limit theorems in Hölder spaces. Then we
will deal with a generalized Hoeffding's decomposition for U-statistics whose data is a function of i.i.d. random variables. | {"url":"https://m.mathnet.ru/php/seminars.phtml?option_lang=rus&presentid=30373","timestamp":"2024-11-09T17:08:44Z","content_type":"text/html","content_length":"7030","record_id":"<urn:uuid:2a4f69e3-7fae-42e3-8680-d1f3e220ae3f>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00369.warc.gz"} |
Lets Think Handbook
Conservation Test
This has a total of eight items.
1 You have six eggs and six egg cups. Child puts one egg in each cup (10cm apart).
• Are there the same number of eggs as egg cups?’ [Child agrees.]
• Take eggs out and bunch them together (5cm apart). Spread cups apart (15cm space).
• Are there the same number of eggs and cups now?’
• If ‘Yes’, ask ‘How do you know?’.
Good explanations include: ‘You haven’t taken any away/put in any new ones’ or ‘There are still the same number of each’ — i.e. some sense of necessity. If the child counts both eggs and cups, this
does not count as a good explanation.
2 From a collection of ten or so red and green chips, select six red chips and lay them in a row (10cm apart).
• Invite the child to choose ‘one green for every one of my red chips’ and lay them in a row next to the red ones. You can help them do this.
• Child agrees there are the same number.
• Bunch red chips together (5cm) and spread green ones apart (15 cm). Questions and analysis of responses as above.
3 Measurement of amount (otherwise ‘mass’, ‘amount of stuff’, etc). You need one tall
250 cm3 beaker (A), one squat glass dish of similar volume (B) and one small 50 cm3 beaker with line 0.5cm from top (C) (see Figure 1.2 on page 9).
• Fill C to the line, pour the water into A and repeat this twice more. Explain what you are doing, get the child to count. Then C filled to line and poured into B, three times again. Child agrees
same measurement poured into each of A and B three times.
• ‘Look at the two glasses (A and B). Which has more water, or are they the same?’ Test with counter-questions (‘Another child I showed this to said …’, ‘Why do you think so?’, ‘If they were orange
juice, and you could have one, which would you have?’, ‘Why?’, etc).
Solid substance
4 You have two plasticene balls each of 5cm diameter.
• Let the child handle them, and get him or her to agree that both have the same amount of plasticene.
• Now change one into a sausage shape. ‘Remember what we did. We had two balls the same, then we rolled one out to a sausage shape. Is there now the same amount of plasticene in this one and that
[the original] one?’
• ‘Why do you think so?’
• ‘Another child told me she thought there was the same/a different amount, what do you think?’
5 Re-form the ball from the sausage and re-establish equivalence. Now flatten one into a pancake (10cm diameter) and repeat questions.
6 Re-form, establish equivalence and then make one into five little balls of approximately the same size. Repeat the questions.
Good explanations again involve some sense of necessity: ‘You haven’t taken any away, added any, it’s still the same amount; it’s longer but it’s thinner…’ and so on.
Start again with the two balls of plasticene.
Put one on each side of a balance and ensure child agrees that they are the same weight.
7 Transform one into a pancake as before, and ask if it still has the same weight.
8 … and into five small balls; again focus on weight. | {"url":"https://community.letsthink.org.uk/letsthinkhandbook/chapter/conservation-test/","timestamp":"2024-11-03T18:59:14Z","content_type":"text/html","content_length":"85280","record_id":"<urn:uuid:d4f2c2f5-ad32-4100-b55c-f12faa90d668>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00466.warc.gz"} |
Math and Lotsize - How to calculate lots using multiplier according to number of opened orders
in my code i have a fibo sequence, and multiply that by 0.01 to get my lotsize. simple enuf. But why do i often get some weird numbers? while other times, the lotsize is correct, while other times it
is not? This seems to happen with many eas, and not just in mt4 either. in the example in the former sentence, I get 3.25 when 13 is the current step in my martingale system, sometimes the correct
number is opened with the new trade, other times, I get 2.75. Any suggestions of what i can test? I have made comments, printing of the lotsize before sending the ordersend, and get the same weird
numbers that i have mentioned.
Revo Trades: in my code i have a fibo sequence, and multiply that by 0.01 to get my lotsize. simple enuf. But why do i often get some weird numbers? while other times, the lotsize is correct, while
other times it is not? This seems to happen with many eas, and not just in mt4 either. in the example in the former sentence, I get 3.25 when 13 is the current step in my martingale system, sometimes
the correct number is opened with the new trade, other times, I get 2.75. Any suggestions of what i can test? I have made comments, printing of the lotsize before sending the ordersend, and get the
same weird numbers that i have mentioned.
Without a code sample, we don’t have the foggiest idea of what you are doing (correctly or incorrectly), nor can we guess what you should be doing.
All I can ask is, are you properly adjusting to the symbol’s volume step and verifying for the minimum and maximum volume allowed?
Forum on trading, automated trading systems and testing trading strategies
How to calculate lots using multiplier according to number of opened orders?
Fernando Carreiro, 2017.09.01 21:57
Don't use NormalizeDouble(). Here is some guidance (code is untested, just serves as example):
// Variables for Symbol Volume Conditions
dblLotsMinimum = SymbolInfoDouble( _Symbol, SYMBOL_VOLUME_MIN ),
dblLotsMaximum = SymbolInfoDouble( _Symbol, SYMBOL_VOLUME_MAX ),
dblLotsStep = SymbolInfoDouble( _Symbol, SYMBOL_VOLUME_STEP );
// Variables for Geometric Progression
dblGeoRatio = 2.8,
dblGeoInit = dblLotsMinimum;
// Calculate Next Geometric Element
dblGeoNext = dblGeoInit * pow( dblGeoRatio, intOrderCount + 1 );
// Adjust Volume for allowable conditions
dblLotsNext = fmin( dblLotsMaximum, // Prevent too greater volume
fmax( dblLotsMinimum, // Prevent too smaller volume
round( dblGeoNext / dblLotsStep ) * dblLotsStep ) ); // Align to Step value
ok i will try that and report back here. thanks again fernando
Revo Trades: ok i will try that and report back here. thanks again fernando
You are welcome!
ok. i will read up. thanks for responding.
so would i be correct in thinking that when i use Digits in Normalize(price,Digits), that Digits does not always report the correct value? and by using the DoubleToString function, how does that
improve the end result since that that same function uses the same Digits value?
edit == i think i answered my own question, but please answer the q anyways, so i can confirm my own reasoning. thanks.
Since i believe it is all down to using NormalizeDouble where I shouldnt be, i will try inserting this in my mg/lotsize calculations. If you suspect any issues resulting, please point them out in a
#property strict
double fibo[]={1,1,2,3,5,8,13,21,55};
input double iStartLot = 0.25;
//mg lotsize func
double round (double z) // z == place in the fibo buffer
int D= MathPow(10,Digits);
double value = fibo[z] * iStartLot;
double x = ( MathRound (value * D)) / D * z;
//example OrderSend == round(6) == lotsize = (expected value) = fibo[6] * 0.25 == 13 * 0.25, or 3.25
Note: completely untested. Any suggestions or criticism accepted.
Revo Trades:
so would i be correct in thinking that when i use Digits in Normalize(price,Digits), that Digits does not always report the correct value? and by using the DoubleToString function, how does that
improve the end result since that that same function uses the same Digits value? edit == i think i answered my own question, but please answer the q anyways, so i can confirm my own reasoning.
Since i believe it is all down to using NormalizeDouble where I shouldnt be, i will try inserting this in my mg/lotsize calculations. If you suspect any issues resulting, please point them out in a
Note: completely untested. Any suggestions or criticism accepted.
No NormaliseDouble needed! Is this what you are after (untested/uncompiled)?
#property strict
input iStartLot = 0.25;
//mg lotsize func
double FiboLotSize( int index )
// Variables for Symbol Volume Conditions
dblLotsMinimum = SymbolInfoDouble( _Symbol, SYMBOL_VOLUME_MIN ),
dblLotsMaximum = SymbolInfoDouble( _Symbol, SYMBOL_VOLUME_MAX ),
dblLotsStep = SymbolInfoDouble( _Symbol, SYMBOL_VOLUME_STEP );
// Calculate Lot Size based on Fibonacci Sequence
dblLots = iStartLot * fibo[ index ];
// Adjust Volume for allowable conditions
dblLots = fmin( dblLotsMaximum, // Prevent too greater volume
fmax( dblLotsMinimum, // Prevent too smaller volume
round( dblLots / dblLotsStep ) * dblLotsStep ) ); // Align to Step value
// Return Calculated Value of Lot Size
return dblLots;
void OnTick(void)
//example OrderSend == round(7) == lotsize = (expected value) = fibo[6] * 0.01 == 13 * 0.01, or 0.13
OrderSend( _Symbol, OP_BUY, FiboLotSize(7), Ask, 1, 0, 0, iComment, iMagic, 0, clrBlue );
Fernando Carreiro:
No NormaliseDouble needed! Is this what you are after (untested/uncompiled)?
i had normalizeDouble in my original code, and already had the checking of lotstep/maxlot/minlot, but yours does look more elegant haha. thanks again.
i have yet to prove anything, however, I was using NormalizeDouble like this...
dblLots = NormalizeDouble(fmin( dblLotsMaximum, // Prevent too greater volume
fmax( dblLotsMinimum, // Prevent too smaller volume
round( dblLots / dblLotsStep ) * dblLotsStep ) ),2); // Align to Step value
But i was/am getting weird numbers sometimes, like instead of 3.25 lots, getting 2.75. If I am understanding the readings correctly, NormalizeDouble does not round the same way as the func round
does, so this may be the root of my issue. If you concur please respond with a yay or nay. either way I will test the code during london tomoz and report back in 24 hours if nothing has changed. If I
dont get any weird lots in 48, then I will jump for joy, and report after that.
Revo Trades:
i have yet to prove anything, however, I was using NormalizeDouble like this...
But i was/am getting weird numbers sometimes, like instead of 3.25 lots, getting 2.75. If I am understanding the readings correctly, NormalizeDouble does not round the same way as the func round
does, so this may be the root of my issue. If you concur please respond with a yay or nay. either way I will test the code during london tomoz and report back in 24 hours if nothing has changed. If I
dont get any weird lots in 48, then I will jump for joy, and report after that.
I make it a point in my code to NEVER use NormaliseDouble(), EVER. If you search the forum you will find that this has been debated many, many times in a very heated way. Some in favour of using it
and a few like myself, against using it.
I will leave it up to you to decide for yourself what you prefer doing.
Suffice to say, that by using the calculations I have shared above, I've have never had any problems or any errors about invalid volume when placing an order.
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Browse by Authors
Number of items: 3.
Schäpel, J.S. and Wolff, S. and Schulze, P. and Berndt, P. and Klein, R. and Mehrmann, V. and King, R. (2017) State estimation for reactive Euler equation by Kalman Filtering. CEAS Aeronautical
Journal, 1 (10). pp. 261-270. ISSN 1869-5582
Wolff, S. and Schäpel, J.S. and Berndt, P. and King, R. (2016) State Estimation for the Homogeneous 1-D Euler Equation by Unscented Kalman Filtering. ASME Proceedings . ISSN ISBN: 978-0-7918-5725-0
Wolff, S. and Schäpel, J.S. and Berndt, P. and King, R. (2015) State estimation for homogeneous 1-D Euler Equation by Unscented Kalman Filtering. In: ASME 2015 Dynamic Systems and Control Conference
2., 28.-30. Oktober 2015, Columbus, Ohio. (Submitted) | {"url":"http://publications.imp.fu-berlin.de/view/people/Wolff=3AS=2E=3A=3A.html","timestamp":"2024-11-13T02:40:30Z","content_type":"application/xhtml+xml","content_length":"11082","record_id":"<urn:uuid:343032cc-a149-4dcd-860a-f1ada7b85cd3>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00451.warc.gz"} |
Transformations of Simple Rational Functions
What would be the new equation if the graph of $y=\frac{1}{x}$y=1x was shifted upwards by $4$4 units?
What would be the new equation if the graph of $y=\frac{1}{x}$y=1x was shifted to the right by $7$7 units?
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6 Basic Excel Functions Everybody Should Know
1. Add Numbers in Cells: SUM
One of the most basic things you can do with numbers is add them. Using the SUM function in Excel you can add numbers in cells.
The syntax is SUM(value1, value2,...) where value1 is required and value2 is optional. So for each argument, you can use a number, cell reference, or cell range.
For example, to add the numbers in cells A2 through A10, you would enter the following and press Enter:
You then get your result in the cell containing the formula.
2. Average Numbers in Cells: AVERAGE
Averaging a group of numbers is another common mathematical function.
The syntax is the same for the AVERAGE function in Excel as with the SUM function, AVERAGE(value1, value2,...) with value1 required and value2 optional. You can enter cell references or ranges for
the arguments.
To average the numbers in cells A2 through A10, you would enter the following formula and press Enter:
You then get your average in the cell containing the formula.
3. Find the High or Low Value: MIN and MAX
When you need to find the minimum or maximum value in a range of cells, you use the MIN and MAX functions.
The syntaxes for these functions are the same as the others, MIN(value1, value2,...) and MAX(value1, value2,...) with value1 required and value2 optional.
To find the minimum, lowest value, in a group of cells, enter the following replacing the cell references with your own. Then, hit Enter:
And to find the maximum, highest value, use:
You’ll then see the smallest or largest value in the cell with the formula.
4. Find the Middle Value: MEDIAN
Instead of the minimum or maximum value, you may want the middle one.
As you may have guessed, the syntax is the same, MEDIAN(value1, value2,...) with the first argument required and the second optional.
For the middle value in a range of cells enter the following and press Enter:
You’ll then see the middle number of your cell range.
5. Count Cells Containing Numbers: COUNT
Maybe you’d like to count how many cells in a range contain numbers. For this, you would use the COUNT function.
The syntax is the same as the above two functions, COUNT(value1, value2,...) with the first argument required and the second optional.
To count the number of cells that contain numbers in the range A1 through B10, you would enter the following and press Enter:
You then get your count in the cell containing the formula.
6. Insert the Current Date and Time: NOW
If you’d like to display the current date and time whenever you open your spreadsheet, use the NOW function in Excel.
The syntax is NOW() because the function has no required arguments. You can, however, add or remove from the current date and time if you like.
To return the current date and time, enter the following and press Enter:
To return the date and time five days in the future from the current date and time, enter this formula and hit Enter:
And here’s how the results would look for each of the above formulas.
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EarthSky | How do you measure the mass of a star?
Human WorldSpace
How do you measure the mass of a star?
Artist’s concept of the binary star system of Sirius A and its small blue companion, Sirius B, a hot white dwarf. The 2 stars revolve around each other every 50 years. Binary stars are useful to
determine the mass of a star. Image via ESA/ G. Bacon.
To measure the mass of a star, use 2 stars
There are lots of binary stars – two stars revolving around a common center of mass – populating the starry sky. In fact, a large majority of all stars we see (around 85%) are thought to be part of
multiple star systems of two or more stars! This is lucky for astronomers, because two stars together provide an easy way to measure star masses.
To find the masses of stars in double systems, you need to know only two things. First, the semi-major axis or mean distance between the two stars (often expressed in astronomical units, which is the
average distance between the Earth and sun).
And second, you need to know the time it takes for the two stars to revolve around one another (aka the orbital period, often expressed in Earth years).
With those two observations alone, astronomers can calculate the stars’ masses. They typically do that in units of solar masses (that is, a measure of how many of our suns the star “weighs.” One
solar mass is 1.989 x 10^30 kilograms or about 333,000 times the mass of our planet Earth.)
Available now! 2023 EarthSky lunar calendar. A unique and beautiful poster-sized calendar showing phases of the moon every night of the year! And it makes a great gift.
Sirius is a great example
We’ll use Sirius, the brightest star of the nighttime sky, as an example. It looks like a single star to the unaided eye, but it, too, is a binary star. By the way, you can see it yourself, if you
have a small telescope.
The two stars orbit each other with a period of about 50.1 Earth-years, at an average distance of about 19.8 astronomical units (AU). The brighter of the two is called Sirius A, while its fainter
companion is known as Sirius B (The Pup).
View at EarthSky Community Photos. | Michael Teoh at Heng Ee Observatory in Penang, Malaysia, captured this photo of Sirius A and Sirius B (a white dwarf) on January 26, 2021. He used 30 1-second
exposures and stacked them together to make faint Sirius B appear. Thank you, Michael!
Finding the mass of Sirius A and B
So how would astronomers find the masses of Sirius A and B? They would simply plug in the mean distance between the two stars (19.8 AU) and their orbital period (50.1 Earth-years) into the
easy-to-use formula below, first derived by Johannes Kepler in 1618, and known as Kepler’s Third law:
Total mass = distance^3/period^2
Total mass = 19.8^3/50.1^2
So total mass = 7762.39/2510.01 = 3.09 times the sun’s mass
Here, the distance is the mean distance between the stars (or, more precisely, the semi-major axis) in astronomical units, so 19.8, and the orbital period is 50.1 years.
The resulting total mass is about three solar masses. Note that this is not the mass of one star but of both stars added together. So, we know that the whole binary system equals three solar masses.
An example of a binary star system, whose component stars orbit around a common center of mass (the red cross). In this depiction, the two stars have similar masses. In the case of the Sirius binary
star system, Sirius A has about twice the mass of Sirius B. Image via Wikimedia Commons.
Then finding the mass of each star
To find out the mass of each individual star, astronomers need to know the mean distance of each star from the barycenter: their common center of mass. To learn this, once again they rely on their
It turns out that Sirius B, the less massive star, is about twice as far from the barycenter than is Sirius A. That means Sirius B has about half the mass of Sirius A.
Thus, you know the whole system is about three solar masses by using Kepler’s Third Law. So now you can deduce that the mass of Sirius A is about two solar masses. And then Sirius B pretty much
equals our sun in mass.
What about the mass of a star not in a binary system?
But what about stars that are alone in their star systems, like the sun? The binary star systems are once again the key: Once we have calculated the masses for a whole lot of stars in binary systems,
and also know how luminous they are, we notice that there is a relationship between their luminosity and their mass. In other words, for single stars we only need to measure its luminosity and then
use the mass-luminosity relation to figure out their mass. Thank you, binaries!
Read more: What is stellar luminosity?
Read more: What is stellar magnitude?
Bottom line: For astronomers, binary star systems are a quite useful tool to figure out the mass of stars.
January 15, 2023
Human World
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Statistics ≠ Checking Your Brain at the Door
As part of an endeavor to improve undergraduate writing, I was involved in a day long session of reading senior level writing assignments. Basically, a group of us had a list of criteria and ranked
each assignment as meeting or not meeting the criteria. We were not grading or assessing the worth of the assignments simply whether a given criteria was met or not met. I learned a few things during
the 7 hours of reading 16 assignments (~20-30 pages each), one of which I want to touch on here.
Now I am not a statistician nor do I have any real expertise in statistical analysis. In fact, I turn to statisticians when I need to do statistical analyses beyond student T-tests or analysis of
variance. However, I think I know enough about statistics to not make the error of the
The p-value essentially tells you the probability that some event, data, occurrence is due to chance (generally referred to as the null hypothesis). So if you are hoping that the effect you are
looking at is not due to chance you want a small p-value. The question, of course, then becomes 'how small'? The scientific community has generally agreed that a p-value of < 0.05 is a rigorous
cut-off. A p-value > 0.05 is considered to be reasonable odds that your effect may be due to chance. However, this cut-off of 0.05 is arbitrary and indeed higher and lower cut-offs are used in some
To be clear, p-values can range from 0 - 1, so you can consider a p-value of 0.05 to be analogous to a 5% chance that the effect you are looking at to be due to chance. That also means that your
p-value of 0.06 means there is a 6% chance your data is due to chance and that is too high for most scientists to consider your data significant.
Now we come to the error of the p-value. By way of example, you should check out xkcd's acute take on the problem (well several problems, but the one we care about is central). If we look at a lot of
different data under a given condition, then we should expect a data set to show a p-value < 0.05 on average 1/20 times (5%) that is strictly due to chance. This does not mean that we should discount
a result that comes with a p-value of 0.05. It means there is only a 5% chance the result is due to chance. However, if there is additional data to back this result up, we can increase our confidence
even more. If there is not additional data, hopefully the scientists (aka senior undergraduate students) will at least acknowledge the limitation of the many data points. Sadly, both of these were
lacking in a couple of cases that I observed, although I admit I do not know if this represents a statistically significant (p < 0.05) result.
5 comments:
I think this highlights the point that at least elementary statistics should be requirement in the high school or undergraduate science curriculum. Biology is rapidly advancing so that statistics
(often times more advanced than the elementary) is a requirement, I feel.
Also, I find that the biologists' misconception about the relevance and limitations of the p-value by itself is quite wide-spread but explained it well. More sample size = more statistical power.
The Defective Brain said...
I've seen statistical errors at all levels of academia.
If I see another paper that has *-p<0.05, **p<0.01, ***p<0.001 in the SAME figure, I will start frothing at the mouth. There can only be one p-value cut off that's meaningful in any one
The Lorax said...
My personal pet peeves is when authors use the term 'very significant' or in the case of those who know that 'very' is a useless word 'highly significant'.
The singular of "criteria" is "criterion".
The p-value essentially tells you the probability that some event, data, occurrence is due to chance (generally referred to as the null hypothesis).
Well, no, I'm sorry, that's wrong.
The p-value is the probability of a result at least as unusual as the one you got *given* no difference. That's not the same thing as what you said at all, which basically has the conditioning
the wrong way around.
What you said is what it would be handy to know - 'the probability that the null hypothesis is true given a result at least this big', but that's not what the p-value gives you.
[To convert from one to the other (swap the conditioning around), you would normally use Bayes' rule but in the case of hypothesis tests, you can't because you don't know the denominator.] | {"url":"http://angrybychoice.fieldofscience.com/2011/08/statistics-checking-your-brain-at-door.html","timestamp":"2024-11-12T06:44:12Z","content_type":"application/xhtml+xml","content_length":"163995","record_id":"<urn:uuid:214c571e-ee53-4e1c-87c0-08e14aa1fee9>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00832.warc.gz"} |
10 Liters to Gallons? How to Convert?
Last Updated on March 20, 2023
How to convert 10 liters to gallons? This is a straightforward process that requires only a few steps to complete. With this guide, you’ll be able to convert liters to gallons easily and scale up or
down as needed.
Convert 10 Liters to Gallons
One liter is equal to 0.264172 gallons. In order to convert from liters to gallons, you need to multiply the number of liters by 0.264172. For example, if you have 10 liters, you would multiply 10 by
0.264172 to get 2.64172 gallons.
The simple answer is 10 liters = 2.64172 gallons.
Metrics vs Imperial System
Liters and gallons are two units of measurement for volume. A liter is a metric unit, while a gallon is an imperial unit. There are some differences between the two units. We have more information
about metric and imperial in this post: How many ounces are in a 1.5 liter?
What is a Liter?
A liter is a unit of measure in the metric system. It is equal to 1,000 cubic centimeters (cc) or 1/1,000th of a cubic meter.
What is a Gallon?
A gallon, on the other hand, is a unit of measure in the imperial system. It is equal to 4 quarts, 8 pints, or 16 cups. There are about 3.785 liters in a gallon.
The main difference between the two units is that a liter is a metric unit and a gallon is an imperial unit. Another difference is that there are about 3.785 liters in one gallon, but only 1/1,000th
of a cubic meter in one liter.
How to convert from Liters to Gallons
There are few things more frustrating than trying to convert from one unit of measurement to another and getting an answer that doesn’t make sense. For example, converting from liters to gallons. How
many liters are in a gallon? It’s a simple question, but the answer is not so simple.
It’s important to note that the answer you get will be in U.S. liquid gallons. If you need to know how many imperial gallons there are in a liter, the conversion is a bit different.
In the United States, liquid measures are generally expressed in terms of gallons. The imperial gallon, which is used in the United Kingdom, Canada, and some Caribbean nations, is about 20% larger
than the U.S. gallon. But don’t worry, converting between the two is relatively easy.
Determine the size of the container
Assuming you would like tips on how to convert liters to gallons, here are a few things to keep in mind. To start, determine the size of the container.
For example, 1 liter is equivalent to 0.264172 gallons. So, if you have a container that is 3 liters, then it would be 0.792514 gallons, and so on. Once you know the size of the container, use a
simple mathematical formula to convert from liters to gallons. To do this, multiply the number of liters by 0.264172. This will give you the accurate number of gallons for your container size in
A quick conversion chart
To convert liters to gallons, you can use a quick conversion chart.
Here is a quick conversion chart to help you convert liters to gallons:
Liters Gallons
1 0.26417
2 0.52834
3 0.79252
4 1.05669
5 1.32086
6 1.58503
7 1.84920
8 2.11338
9 2.37755
10 2.64172
11 2.90589
12 3.17006
13 3.43424
14 3.69841
Do you find this article “How to convert 10 liters to gallons” helpful? Comment below if you’d like to see more articles like this one!
How to convert 10 liters to gallons?
How to convert 10 liters to gallons? This is a straightforward process that requires only a few steps to complete. With this guide, you’ll be able to convert liters to gallons easily and scale up or
down as needed.
Print Rate
• 10 liters = 2.64172 gallons
• To convert liters to gallons, you can use the quick conversion chart above to help you convert liters to gallons
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Radio Shack EC-4004
Datasheet legend Years of production: Display type: Numeric display
New price: Display color: Black
Ab/c: Fractions calculation Display technology: Liquid crystal display
AC: Alternating current Size: 5½"×3"×½" Display size: 10+2 digits
BaseN: Number base calculations Weight: 4 oz
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Cmem: Continuous memory Batteries: 1×"CR-2025" Lithium Advanced functions: Trig Exp Hyp Lreg Cmem
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Eqlib: Equation library Memories: 7 numbers Program display:
Exp: Exponential/logarithmic functions Program memory: 38 program steps Program editing:
Fin: Financial functions Chipset: Casio fx-3600P Forensic result: 9.0000157179
Grph: Graphing capability
Hyp: Hyperbolic functions
Ind: Indirect addressing
Intg: Numerical integration
Jump: Unconditional jump (GOTO)
Lbl: Program labels
LCD: Liquid Crystal Display
LED: Light-Emitting Diode
Li-ion: Lithium-ion rechargeable battery
Lreg: Linear regression (2-variable statistics)
mA: Milliamperes of current
Mtrx: Matrix support
NiCd: Nickel-Cadmium rechargeable battery
NiMH: Nickel-metal-hydrite rechargeable battery
Prnt: Printer
RTC: Real-time clock
Sdev: Standard deviation (1-variable statistics)
Solv: Equation solver
Subr: Subroutine call capability
Symb: Symbolic computing
Tape: Magnetic tape storage
Trig: Trigonometric functions
Units: Unit conversions
VAC: Volts AC
VDC: Volts DC
Radio Shack EC-4004
fx-3600P programmable calculator.
At first, I seriously underestimated the capabilities of this machine. 38 program steps? A conditional loop-to-start instruction as the only means of branching? Surely, that's not enough to do
anything useful, certainly not something as complex as my favorite programming example, the Gamma function.
Well, I was wrong. This machine is more capable than I thought, due in part to the fact that it offers four-function memory arithmetic on its six K-registers. Moreover, these 3-keystroke instructions
count only as a single step in program memory. Impressive! A fellow calculator enthusiast already sent me an implementation of the incomplete Gamma function (see my fx-3600P page for the listings.)
Of course, if you don't want to key in a complex and slow program, and you can use results with more limited accuracy, you can still make use of Stirling's formula: | {"url":"https://rskey.org/CMS/?view=article&id=7&manufacturer=Radio+Shack&model=EC-4004","timestamp":"2024-11-08T11:02:24Z","content_type":"application/xhtml+xml","content_length":"26955","record_id":"<urn:uuid:52391bdb-15a8-4a52-a1b3-68c49dee4d0e>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00556.warc.gz"} |
VariousWords beta
Look up related words, definitions and more.
Results for:
Definition: a large distance; "he missed by a mile"
sea mile mile
Definition: a former British unit of length once used in navigation; equivalent to 6,000 feet (1828.8 meters)
mile Roman mile
Definition: an ancient Roman unit of length equivalent to 1620 yards
Definition: a footrace extending one mile; "he holds the record in the mile"
Definition: The international mile: a unit of length precisely equal to 1.609344 kilometers established by treaty among Anglophone nations in 1959, divided into 5,280 feet or 1,760 yards. | {"url":"https://variouswords.com/w/mile","timestamp":"2024-11-05T10:27:19Z","content_type":"text/html","content_length":"48151","record_id":"<urn:uuid:57642344-db6c-43c3-bea4-ec11c2ad5c08>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00819.warc.gz"} |
Median of two sorted arrays
Open-Source Internship opportunity by OpenGenus for programmers. Apply now.
In this article, we have presented 3 Algorithms to find the Median of two sorted arrays efficiently in logarithmic time. This involves a clever use of Binary Search algorithm.
Table of contents:
1. Introduction to Median
2. Problem Statement: Median of two sorted arrays
3. Method 1
4. Method 2
5. Method 3
6. Applications
Let us get started with Median of two sorted arrays.
Introduction to Median
In mathematics and statistics, the median is a value that divides a data into two halves. The left half contains values less than the median and the right half contains value greater than the median.
In order to calculate the median value, we first arrange all the data points in a data set in an ascending order. After that, we pick the value in the middle of the data set to get the median. If the
data set has odd number of elements, we can directly pick the value in the middle to get the median.
However, if the data set has even number of elements, we pick the middle two values and calculate the average of the two to get the median value. This is shown in the image below.
Problem Statement: Median of two sorted arrays
In order to calculate the median of two sorted arrays, we would first need to combine the two arrays into a new array and then calculate the median of the resulting array. We can do this using
various approaches.
Method 1
The most straightforward way to solve this problem is to do it linearly. That is, we keep a track of all elements equal to the number of elements in first and second list taken together. At the same
time, we traverse all elements in the first and the second list, keep on comparing them and tracking the count. Once we reach the middle index (as per the number elements designated for the new
list), it implies we have arrived at the middle value i.e the median. We can put two conditions as well to check if the length of the new resulting list is odd or even, and calculate the median
The steps to perform this method are as follows-
1. Find the length of first and second list.
2. Create index variables, initialized at zero, for each list.
3. Set two variables as median 1 and median 2. (This is because in case the data set has even number of elements, then we'd need two middle values to calculate the median. In case of odd data
points, the median 2 would be ignored.)
4. Traverse all elements in list 1 and list 2 combined, but only till the middle position (because the median will exactly be at this position/index), while at the same time sorting the traversed
elements and setting the median accordingly. Return the final median value in the end.
5. In case the number of elements are even, set median 2 equal to median 1 and do the same as in step 4. However, this time return the average of median 1 and median 2 as the final median value.
Implementation Example
The following program executes the above algorithm.
# function to find median
def find_median(lst1, lst2):
m = len(lst1)
n = len(lst2)
i = 0
j = 0
# we can set median 1 and median 2 as any value initially
median1, median2 = None, None
# if the total number of elements (elements in list 1 + elements in list 2) is odd
if ((m+n) % 2 == 1):
# x is the count variable
for x in range(((m+n)//2) + 1):
if (i != m and j != n):
if lst1[i] > lst2[j]:
median1 = lst2[j]
j += 1
median1 = lst1[i]
i += 1
# for remaining elements in lst 1
elif (i < m):
median1 = lst[i]
i += 1
# for remaining elements in lst 2 i.e when j<n
median1 = lst2[j]
j += 1
return median1
# if the total number of elements is even
for x in range(((m+n) // 2) + 1):
# because we will have two middle values
# median2 is the middle value before the second middle value in case of even data points
median2 = median1
if ( i != m and j != n):
if lst1[i] > lst2[j]:
median1 = lst2[j]
j += 1
median1 = lst1[i]
i += 1
elif (i < m):
median1 = lst[i]
i += 1
# for remaining elements in lst 2 i.e when j<n
median1 = lst2[j]
j += 1
return ((median1 + median2) / 2)
#example lists to test median program
list1 = [1, 2]
list2 = [3, 4, 5]
# call the median function and print the final median value
print(find_median(list1, list2))
The following output is produced by the above program.
• Time: in the worst case, in order to traverse and sort all elements in list 1 and list 2, we use the runtime is O(m+n), where m, n are the number of elements in list 1 and list 2 respectively.
Even in the best case, the runtime would be the same.
• Space: this algorithm has a constant space requirement of O(1), in the best and worst case, in order to keep a track of all traversed elements and return the final value of median.
Method 2
Another way to solve this problem is to apply the mathematical formula directly.
We create a new list and add the elements of list 1 and list 2. Then we sort the new list and check the length. Let the length of the list be 'n'. If it is odd, we apply the formula of (n+1)/2. If
it's even, we calculate two middle values by applying the formula (n/2) and (n/2)-1. This will give us the middle values required to calculate the median.
The steps involved in this method are as follows-
1. Combine the two lists into a new list.
2. Use sorting to arrange the elements in the new list in ascending order.
3. Check the length of the new list. If it's odd, apply the formula- (n+1)/2 to get the index of the median. It it's even, we need two middle values, the average of which gives us the median, first
apply the formula (n/2) to get the first middle value. Then use the formula- (n/2)-1 to get the second middle value. Calculate the average of these two values and return it.
Implementation Example
We can use the following program to implement this method.
def MergeSort(lst):
# making sure length of list is 1 , which is the midpoint element
if len(lst) > 1:
# find midpoint of the list
midpoint = len(lst)//2
# divide the list into two parts (sublists), part to the left and right of midpoint
lefthalf = lst[:midpoint]
righthalf = lst[midpoint:]
# keep calling the function till length of list becomes 1
# Assign zero value to all index variables initially
# k is the index variable for list parameter
i = j = k = 0
# compare all elements in leftpart and rightpart
# add elements in sorted order and update the list
while i < len(lefthalf) and j < len(righthalf):
if lefthalf[i] < righthalf[j]:
lst[k] = lefthalf[i]
i += 1
lst[k] = righthalf[j]
j += 1
k += 1
#add any remaining elements in leftpart to the new list
while i < len(lefthalf):
lst[k] = lefthalf[i]
i += 1
k += 1
#add any remaining elements in rightpart to the new list
while j < len(righthalf):
lst[k] = righthalf[j]
j += 1
k += 1
return lst
# function to find median
def find_median(lst1, lst2):
# combine all elements of first and second list
lst3 = lst1 + lst2
# sort the new list
# get the length of the new list
n = len(lst3)
# in case of even number of elements in new list
if n % 2 == 0:
# index of first middle value
k = n // 2
mid_value1 = lst3[k]
mid_value2 = lst3[k-1]
median = (mid_value1 + mid_value2)/ 2
return median
# in case of odd number of elements in new list
k = n // 2
median = lst3[k]
return median
#example lists to test median program
list1 = [1, 2, 6]
list2 = [3, 4, 5]
# call the median function and print the final median value
print(find_median(list1, list2))
The above program produces the following output.
• Time: since we use merge sort here, the best and worst case runtime is O(nlogn). Because we'd need to traverse all elements at each stage when we decrease the length of the list.
• Space: in the worst case, the space complexity of this method is O(m+n), where m, n are the number of elements in list 1 and list 2 respectively. This is because we create a new list of length
(m+n) to store all elements of list 1 and list 2, which we then use to calculate the median. Even in the best case, the space requirement would still be the same.
Method 3
A more efficient method to solve this problem is to use binary search. We can create an algorithm that tells us how many elements we should take from each lists, just enough to get the median of the
whole data set, instead of traversing all elements in both lists. The steps involved to create such an algorithm are as follows-
1. Get the length of each list.
2. Switch list 1 with list 2 if the length of list 2 is shorter. This is because for one, faster calculation, and two, because we can have few elements from list 1 for median calculation (as it's
the shorter list) and the bigger one can supply rest of the elements, but not the other way around.
3. Set the start (low) and end (high)variables to perform the binary search.
4. We create a variable called cut_1 and cut_2, for each list respectively. Each cut divides the list into two parts- left and right, during each iteration of the binary search. All the elements in
left part must be less than or equal to all the elements in the right part of the list.
5. Let's suppose we make the cut to divide each list into two parts, left and right
using an arbitrary number within the range of list 1. And the remaining elements would then be from list 2.
6. Since the lists are sorted, all we need to do to ensure that we are taking the right elements for median is to check the left-most value in list 1 (just beside the cut) with the value in the
right part of list 2 (just after the cut). Likewise, check the left-most value in list 2 just beside the cut with the right part of list just after the cut. The left values must always be less
than or equal to the right values as median divides the entire list into two parts. Values to the left of the median are always smaller than the values on the right of the median.
7. In case there are no values to the left of the cut, we set the left values to be minus infinity. Likewise, in case there are no values to the right of the cut, we set the right values to be
8. Once we get all the correct elements for median calculation, if the list contains even number of elements, we need two middle values. We would need to take the first middle value as the maximum
of left elements, just beside the cut, because the list is sorted. Likewise, we take the second middle value as the minimum of right elements. In case the number of total elements is odd, the
median will be the maximum of the left elements.
9. If the values in the left part are greater, we decrease the cut and assign that value to the end variable. This makes sure we move now move towards the left of the list, to get smaller values.
Otherwise, we keep increasing the cut and assign it to the start variable, to get as much elements from list 1 till all the previous conditions are no longer satisfied.
Implementation Example
The following program implements the above algorithm to calculate the median.
def find_median(lst1,lst2):
# get the length of both lists
m, n = len(lst1), len(lst2)
# we switch lst 1 with lst2 in case lst 1 has more elements
# this is done to make lst1 the shorter one for faster calculation
if m > n:
lst1, m, lst2, n = lst2, n, lst1, m
# set the starting and ending indexes to decide the cut for median
low = 0
high = m
# we do a binary seach from here
while low <= high:
# a cut divides the lists into left and right part
# if cut_1 is 0 it means nothing is there on left side of list 1
# if cut_1 is length of list then there is nothing on right side
# likewise, for cut_2
cut_1 = (low + high) // 2
cut_2 = (m + n + 1)//2 - cut_1
# values to the left of the median are always smaller than right
# if there are no values to the left of the cut,
# set the left values to be minus infinity
# in case there are no values to the right of the cut,
# we set the right values to be infinity
# since cut_1 gives us the number of elements,
# to get the index we subtract 1
max_left1 = float('-inf') if cut_1 == 0 else lst1[cut_1 - 1]
min_right1 = float('inf') if cut_1 == m else lst1[cut_1]
max_left2 = float('-inf') if cut_2 == 0 else lst2[cut_2 - 1]
min_right2 = float('inf') if cut_2 == n else lst2[cut_2]
# implies all the correct elements found for median calculation
if (max_left1 <= min_right2 ) and (max_left2 <= min_right1):
# in case total number of elements are even
if ((m+n) % 2 == 0):
return ((max(max_left1,max_left2) + min(min_right1,min_right2))/2)
# in case total number of elements are odd
return max(max_left1,max_left2)
# if values in left are greater, implies too ahead in the list
# so we need to decrease the index to get back to smaller values
elif (max_left1 > min_right2):
high = cut_1 - 1
# else keep on increasing the index
low = cut_1 + 1
# example to test the median program
lst1 = [1,2]
lst2 = [3,4,5]
The following output is produced by the above program.
Median: 3
• Time: since we use binary search for the above algorithm, the complexity falls down to O(log(min(m,n))), depending on the length of each list, where m, n are the number of elements in list 1 and
list 2 respectively. In the best case, the runtime would be O(1) or a constant runtime.
• Space: we don't use any additional space and the space remains required remains same irrespective of the length of lists. In best and worst case, the space requirement is constant or O(1).
• In statistics and mathematics, median is used for data manipulation and tabulation, in order to summarize the data and draw conclusions and inferences about the population.
• Many areas, including data science, data analysis and data engineering, make use of median and related algorithms for machine learning and data maintenance.
With this article at OpenGenus, you must have the complete idea of finding the Median of two sorted arrays. | {"url":"https://iq.opengenus.org/median-of-two-sorted-arrays/","timestamp":"2024-11-03T06:06:33Z","content_type":"text/html","content_length":"72637","record_id":"<urn:uuid:d48a0ea7-d2a3-4ba9-8ad1-59f2d5d73e11>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00480.warc.gz"} |
Which best describes the range of a function? A. The set of all possible input values B. The set of all possible output values C. The greatest possible input value D. The greatest possible output value
Find an answer to your question 👍 “Which best describes the range of a function? A. The set of all possible input values B. The set of all possible output values C. The ...” in 📗 Mathematics if the
answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers | {"url":"https://cpep.org/mathematics/572205-which-best-describes-the-range-of-a-function-a-the-set-of-all-possible.html","timestamp":"2024-11-07T19:01:26Z","content_type":"text/html","content_length":"23781","record_id":"<urn:uuid:8b5c66ca-821e-4915-8d5d-d36ebdbde97c>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00458.warc.gz"} |
Learning Machines With Me. (Part Two)
A series where I attempt to not only understand but also do fun things with machine learning — with strangers on the Internet.
While the cool kids are playing football with their feet, we’re going to play football on spreadsheets
Welcome back! If this is the first time you’ve come upon this series, please check out the first article in the series by clicking me. There we discussed regression and used it to build a model that
could predict a team’s position based on basic information.
In today’s session, we’re going to get into classification by building a model that can classify a player’s position based on information about a player. All very cool stuff, I know! So let’s get
right into it.
Just like before, here is how to approach this guide:
All you have to do is learn some of the basic level things that are occurring and then, download the dataset and simply run the code! After that, feel free to play around with the code as you see
fit and try to make a better model!
In this second part of the series, we’re going to understand classification — the second building block in a beginner’s attempt to understand machine learning.
Classification is as the name sounds — simply put, given some information about certain objects, we’re trying to classify — label — these objects. Classification, and regression, falls under a
category of machine learning called supervised learning.
In supervised learning, the machine actually is being guided — supervised — by the input to which we already know the answers. In other words, we actually already know how a certain input will look
like and we use those examples to train the model.
As such, classification is supervised learning because well, the machine needs to know what/how a certain label looks like/is before it attempts to label right? The same way you need to actually see
how a great dribbler is to then label great/not-great dribblers, the machine needs that information so it can label.
Other than that, that’s simply it to classification. Now, there are multiple classification models that one can construct. Here are some:
• Logistic Regression (don’t judge a book by its cover!)
• K Nearest Neighbours
• Support Vector Machine Classification
• Kernel SVM
• Naive Bayes Classification
• Decision Tree Classification
• Random Forest Classification
For this model, we’re going to go with a Naive Bayes Model. Why this one?
Let’s get into the specifics of what goes in this model.
The Naive Bayes Model works its whole classification system based on a theorem in probability called: Bayes Theorem.
You might have seen it if you’ve taken a stat class. Here’s how the theorem looks like:
You’re probably wondering, what the hell does this even mean?!
Let’s break all four parts of the components down one-by-one with an example scenario.
Let’s say we have a pool of 100 players and 40 play in La Liga while 60 play in the Premier League. Let’s say, then that out of all 100 players, 20 players are strikers (just go with me).
Out of those 20 strikers, 15 are from Premier League while 5 are from La Liga.
Now, let’s say we want to find the probability that a player is a striker given that the player is from the Premier League.
Looking at numbers, this is easy. The answer is 15/60 (player is a striker that is in PL / player is from PL).
Now let’s get a generalized formula for this. A more generalized version allows us to work large datasets where we might only percentages/probabilities.
• P(B) : This is the probability that a player is from the PL — 60%
• P(A) : This is the probability that a player is a striker — 20%
• P(B | A) : This is called a conditional probability and the way you read this is: “What’s the probability of me getting B GIVEN that I have A”. So here, what is the probability that a player is
from a PL given that we have a striker? Well we know that out of 20 strikers, 15 are from PL. As such, we get 75%.
• P(A | B) : This looks similar to what we just did but is not the same! This reads: “What’s the probability of me getting A GIVEN that I have B”. So here, what is the probability that a player is
a striker given that we have a player from the PL? This is what we want to know.
Let’s apply the formula:
P(A | B) = (0.75 * 0.20) / (0.60) = 0.25 = 25%
Okay, so basically Bayes allows a strong method of determining probabilities. How can we use this in classification?
Well, what are we trying to do in classification? Let’s look back to what we defined classification as:
given some information about certain objects, we’re trying to classify — label — these objects.
Is this not the same as finding the probability an object belongs to a label based on some information about that object?
If we do this for both two labels, we can compare the probabilities — given by the Bayes Theorem — and whichever one has a higher probability is the label we should assign to the object.
And that is what the Naive Bayes Model attempts to do.
Measuring Accuracy
Ok, now let’s say we’ve made a classification model. How do we determine its accuracy?
Well, for starters we can just see the classifications we got right divided by the total observations.
But this is a trap!
To see how, we’re going to construct something called a confusion matrix (it’s not confusing, trust me).
Simply put, we compare our predicted labels against our actual labels. We get to see where we got things correct and where we got it right.
Let’s put some numbers in there to see how accuracy rate can be troubling.
Let’s say our model comes up with these matrix where the numbers tell how many observations (ex: we got 100 rows right where the label was 0 and we predicted 0 as well).
Here, our accuracy rate is: 140/200 or 70%
What if we can better our accuracy rate but make a worse model? Sounds impossible, yeah? What if we just assigned EVERYTHING to 0. Here’s what happens:
Now our accuracy rate is 150/200 or 75%.
We just got a higher accuracy rate while making a worse model! As such, to better classify how good classification models, we use a combination of confusion matrix and accuracy rate.
When we build our confusion matrix, keep an eye here:
Bottom-right are the players we predicted to be in our position that were actually that position. Upper-left are players we predicted NOT to be in our position that were actually NOT in that
Naturally, the upper-left number will be greater than the bottom-right number but ensure that the bottom-right number doesn’t go to zero!
With that out of the way, let’s go to the dataset and code!
Building Our Classification Model
I have compiled datasets from the EPL and LaLiga seasons of 2020/21 — this is our training dataset. In addition, I have compiled a dataset of Dortmund’s season of 2019/20 to help us build this model
— this is our test dataset.
For our purposes, we want to classify and label left-backs (but later down, I talk about how you can change the code to classify other positions).
In this dataset, I have compiled the average starting points (x and y) and average ending points (x and y) of players’ passes and, in addition, the convex hull area (a representation of the area of a
player’s touch density). These are basic features for a player and we can always create new metrics to improve the precision but since we’re building a basic version, these should suffice.
In addition, the training dataset contains identifiers for whether a player is a left-back or not. After training our model on the training dataset, we’ll apply it on Dortmund’s dataset to see how
good our model was.
Here is the link to the Python notebook: Code for the Model
Here is the link to the Training Dataset: TrainingDataset
Here is the link to the Test dataset: TestDataset
Here are some instructions for running this:
• Download the dataset and store it somewhere (remember where you store this!)
• Make a copy of the Notebook so that you can edit and import the data!
• Click this Folder option:
• Here, find the dataset on your computer/laptop and upload it. (Note: this dataset will only exist for this Notebook while you have the Notebook open. If you close the Notebook and restart it, you
will have to reupload the dataset again!)
• Now, you can either run all the cells at once OR run it one by one if you make some changes.
• Clicking Run-All runs all cells. Clicking the brackets ( a play button will appear ) will only run that cell.
At the end of the notebook, you’ll see the classification model working on the testing set with its indications of what is a left-back (1) and what is not a left-back (0).
Some Interactivity:
Now, this series is meant to be interactive so how do you make your own models/improve/change things?
Changing/implementing these features/functions will result in different accuracy scores at the end which helps you determine whether the model for certain labels is good/bad.
One thing is to change the type of classification model that you use:
Try implementing the simplest classification model — logistic regression:
from sklearn.linear_model import LogisticRegressionclassifier = LogisticRegression()classifier.fit(X_train, y_train)
Try classifying positions other than left-backs! (This is where the fun really is!)
To do this, we need to change our training dataset as we need to update our labels to indicate the position we pick.
These are some of the positions my training dataset has to offer:
DC, DL, DR, MC, FW, AMC, MR, DMC, ML
Pick one and then, before splitting the dataset into training/testing, change this piece of code:
dataset = pd.read_csv('TrainingDataset.csv',encoding='latin-1')dataset['identifier']= np.where(dataset['pos']=="INSERT POSITION PICKED", 1, 0)X = dataset.iloc[:, 1:-2].valuesy = dataset.iloc[:, -2].values
From there, run the model as you did before! See the differences, see if our original model works on different positions, see if the logisitic regression works better for different positions —
explore! Some positions do not offer great results (mainly very niche positions like AMC, etc) but others like DC and DL offer accurate results!
Remember, this is a basic model based on the most basic statistics. See if adding data from FBref can maybe increase our classifications!
Please feel free to reach out to me on Twitter if you have any questions. Leave any comments on how to improve! | {"url":"https://abhishekamishra.medium.com/learning-machines-with-me-part-two-6d9581821b66?source=user_profile_page---------1-------------7b472aa5168a---------------","timestamp":"2024-11-02T05:59:27Z","content_type":"text/html","content_length":"179551","record_id":"<urn:uuid:229452ab-844c-45f0-a7b4-1a84b0d8e612>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00263.warc.gz"} |
Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments - FasterCapital
Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
1. Introduction to Interest Rate Risk
## understanding Interest Rate risk
interest rate risk refers to the vulnerability of an investment's value to fluctuations in interest rates. It arises due to the inverse relationship between interest rates and bond prices. When
interest rates rise, bond prices fall, and vice versa. Here are some key insights:
1. The Basics of Interest Rate Risk:
- Interest rate risk primarily affects fixed-income securities such as bonds, certificates of deposit (CDs), and preferred stocks.
- Longer-term bonds are more sensitive to interest rate changes than shorter-term bonds. Why? Because the longer the bond's maturity, the greater the uncertainty about future interest rates.
- Investors must consider both the price risk (fluctuations in bond prices) and the reinvestment risk (the risk that coupon payments will be reinvested at lower rates).
2. Duration: The Magic Metric:
- duration measures a bond's sensitivity to interest rate changes. It quantifies how much the bond's price will move for a 1% change in interest rates.
- Higher duration implies greater interest rate risk. For example, a bond with a duration of 5 years will see a 5% price decline if rates rise by 1%.
- Example: Imagine a 10-year bond with a 5% coupon. If rates rise to 6%, its price will fall due to the lower present value of future cash flows.
- The yield curve plots interest rates against bond maturities. It can be upward-sloping (normal), flat, or inverted.
- normal yield curve: Longer-term rates are higher than short-term rates.
- flat yield curve: short- and long-term rates are similar.
- inverted yield curve: Short-term rates exceed long-term rates (often a recession signal).
- Investors must analyze the yield curve to assess interest rate expectations.
4. mitigating Interest Rate risk: Strategies:
- Diversification: Spread investments across different maturities and sectors.
- Laddering: Buy bonds with staggered maturities to reduce reinvestment risk.
- floating-Rate bonds: These adjust interest payments based on prevailing rates.
- interest Rate swaps: Exchange fixed-rate payments for floating-rate payments.
- Inverse Bond ETFs: These rise when bond prices fall due to rising rates.
- 2008 Financial Crisis: Many investors suffered losses as mortgage-backed securities plummeted due to rising rates.
- Central Bank Policies: When central banks raise rates (e.g., the Fed), bond markets react.
- Corporate Bonds: Companies with high debt face refinancing challenges during rate hikes.
Remember, interest rate risk isn't inherently bad—it's part of investing. The key lies in understanding it, managing it, and aligning your portfolio with your risk tolerance and investment horizon.
So, whether you're a seasoned investor or just starting out, keep a close eye on those interest rates—they're more powerful than they seem!
Introduction to Interest Rate Risk - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
2. Understanding Interest Rate Movements
1. Macro Perspective:
- central Banks and Monetary policy: Central banks, such as the Federal Reserve (Fed) in the United States, control short-term interest rates. They use monetary policy tools (like open market
operations) to influence economic growth, inflation, and employment. When the economy is overheating, central banks raise rates to cool it down. Conversely, during recessions, they lower rates to
stimulate borrowing and spending.
- Inflation Expectations: Interest rates reflect inflation expectations. If investors anticipate rising inflation, they demand higher yields on bonds to compensate for eroding purchasing power.
Conversely, when inflation expectations are low, bond yields tend to decrease.
2. Micro Perspective (Investor's Lens):
- bond Prices and yields: Bond prices move inversely to interest rates. When rates rise, existing bond prices fall, and vice versa. Consider a 10-year bond with a fixed coupon rate of 4%. If
prevailing rates increase to 5%, new bonds will offer higher yields, making the 4% bond less attractive.
- Duration Risk: Duration measures a bond's sensitivity to interest rate changes. Longer-duration bonds experience larger price swings when rates fluctuate. Investors must assess their risk tolerance
and investment horizon.
- Opportunity Cost: When rates rise, alternative investments (like savings accounts or money market funds) become more appealing. Investors may shift from bonds to these alternatives, affecting bond
3. Practical Examples:
- Mortgages: Homebuyers monitor mortgage rates closely. A 1% increase in rates can significantly impact monthly payments. When rates fall, homeowners may refinance to lock in lower rates.
- Corporate Debt: Companies issue bonds to raise capital. If rates rise, their interest expenses increase, affecting profitability. Investors evaluate corporate bonds based on credit risk and yield
relative to government bonds.
- Equity Markets: Rising rates can dampen stock market returns. Investors compare stock earnings yields (earnings-to-price ratio) with bond yields. Higher bond yields may attract capital away from
4. Behavioral Aspects:
- Herding Behavior: Investors often follow the crowd. When rates rise, fear of missing out (FOMO) drives some to buy bonds, pushing prices up. Conversely, panic selling occurs during rate hikes.
- Anchoring Bias: People anchor their expectations to recent experiences. If rates have been low for years, investors may underestimate the impact of rate increases.
In summary, understanding interest rate movements involves analyzing macroeconomic factors, assessing individual investments, and recognizing behavioral biases. Keep an eye on central bank decisions,
inflation trends, and market dynamics to navigate interest rate risks effectively. Remember, interest rates are like tides—they ebb and flow, shaping the financial landscape.
Understanding Interest Rate Movements - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
3. Types of Interest Rate Risk
1. Market Risk: This type of interest rate risk arises from changes in the overall market conditions. When interest rates fluctuate, it affects the value of fixed-income securities such as bonds. For
instance, when interest rates rise, the prices of existing bonds tend to fall, leading to potential capital losses for investors.
2. reinvestment risk: Reinvestment risk refers to the uncertainty associated with reinvesting cash flows from fixed-income investments. When interest rates decline, the income generated from maturing
investments may need to be reinvested at lower rates, resulting in lower overall returns.
3. prepayment risk: Prepayment risk is relevant in the context of mortgage-backed securities (MBS) or other debt instruments with embedded call options. When interest rates fall, borrowers may choose
to refinance their loans, leading to early repayment of principal. This can impact investors who were expecting a steady stream of interest income.
4. Credit Risk: While not directly related to interest rate changes, credit risk is an important consideration when assessing interest rate risk. It refers to the potential for borrowers to default
on their debt obligations.
Types of Interest Rate Risk - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
4. Measuring Interest Rate Sensitivity
### understanding Interest Rate sensitivity
Interest rate sensitivity refers to how sensitive an investment's value is to changes in interest rates. It's a critical concept for investors, especially those with fixed-income securities like
bonds. Let's look at it from different angles:
1. Bond prices and Yield curves:
- When interest rates rise, bond prices fall, and vice versa. This inverse relationship is fundamental to understanding interest rate sensitivity.
- The yield curve illustrates the relationship between interest rates and the time to maturity of bonds. A steep yield curve indicates higher sensitivity to rate changes.
2. Duration:
- Duration measures the weighted average time until a bond's cash flows (coupon payments and principal) are received.
- Longer-duration bonds are more sensitive to rate changes. For example:
- A 10-year bond with a duration of 8 years will see a roughly 8% price decline for a 1% increase in rates.
- A 2-year bond with a duration of 1 year will have a smaller price impact.
3. Modified Duration:
- Modified duration adjusts duration for small rate changes. It's a percentage change in bond price for a 1% change in yield.
- Example: If a bond has a modified duration of 5, a 1% rate increase leads to a 5% price decrease.
4. Convexity:
- Convexity accounts for the curvature of the bond price-yield relationship.
- It helps refine duration estimates by considering non-linear effects.
- Higher convexity reduces price volatility when rates change.
- Equities are less directly impacted by rate changes than bonds.
- However, rising rates can affect corporate profits, discount rates, and investor sentiment.
### Examples:
1. Bond Portfolio:
- Imagine you hold a diversified bond portfolio with varying maturities.
- If rates rise, long-term bonds suffer more, but short-term bonds are relatively stable.
- Your portfolio's overall sensitivity depends on the mix of bonds.
2. Mortgage-Backed Securities (MBS):
- MBS are sensitive to prepayment risk and interest rates.
- Falling rates lead to higher prepayments, affecting MBS cash flows.
3. floating-Rate notes (FRNs):
- FRNs have adjustable coupon rates tied to a benchmark (e.g., LIBOR).
- Their interest rate sensitivity is lower due to the floating nature.
Remember, interest rate sensitivity isn't limited to bonds. It affects real estate, stocks, and other assets. As an investor, understanding these dynamics helps you make informed decisions and manage
risk effectively.
Measuring Interest Rate Sensitivity - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
5. Duration and Convexity
1. Duration: The Sensitivity Measure
- Definition: duration is a measure of a bond's sensitivity to interest rate changes. It quantifies how much the bond's price will move for a given change in yield (interest rate).
- Insight: Think of duration as a seesaw. When interest rates rise, bond prices fall, and vice versa. The longer the duration, the more pronounced this effect.
- Formula: The Macaulay duration (named after Frederick Macaulay) is calculated as the weighted average time until a bond's cash flows (coupon payments and principal) are received. Mathematically:
\[ \text{Macaulay Duration} = \frac{\sum_{t=1}^{n} t \cdot CF_t}{\text{Current Bond Price}} \]
Where \(CF_t\) represents the cash flow at time \(t\).
- Example: Consider a 10-year bond with a 5% coupon rate. If interest rates rise by 1%, the bond's duration (let's say it's 7 years) implies that its price will fall by approximately 7%.
2. Modified Duration: Adjusting for Yield Changes
- Definition: Modified duration accounts for the convex relationship between bond prices and yields. It adjusts the Macaulay duration to reflect percentage changes in yield.
- Insight: Modified duration is a linear approximation of bond price changes. It's more accurate for small yield changes.
- Formula:
\[ \text{Modified Duration} = \frac{\text{Macaulay Duration}}{1 + \frac{YTM}{n}} \]
Where \(YTM\) is the yield to maturity and \(n\) is the number of coupon payments per year.
- Example: For our 10-year bond with a 5% coupon, if the yield increases from 5% to 6%, the modified duration (assuming Macaulay duration is 7 years) tells us the price will decrease by approximately
3. Convexity: Curvature Matters
- Definition: Convexity captures the curvature of the bond price-yield relationship. It refines the linear approximation provided by modified duration.
- Insight: Convexity accounts for the fact that bond prices don't move linearly with yield changes; they exhibit curvature.
- Formula:
\[ \text{Convexity} = \frac{\sum_{t=1}^{n} t \cdot (t + 1) \cdot CF_t}{(\text{Current Bond Price}) \cdot (1 + YTM/n)^2} \]
- Example: Suppose our bond has a convexity of 60. If yields drop by 1%, the bond price will increase by approximately 0.6% more than predicted by modified duration alone.
4. Putting It All Together
- Portfolio Management: duration and convexity help portfolio managers balance risk and return. Longer duration bonds are riskier but offer higher potential returns.
- Trading Strategies: Traders exploit duration and convexity to profit from interest rate movements.
- Risk Assessment: Investors assess their portfolios' sensitivity to interest rate changes using these metrics.
Remember, while duration and convexity provide valuable insights, they're not perfect predictors. real-world bond markets can be complex, influenced by factors beyond interest rates. But armed with
these tools, you'll be better equipped to navigate the ever-changing landscape of fixed-income investments.
Feel free to ask if you'd like more examples or have any other questions!
Duration and Convexity - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
6. Impact on Fixed-Income Investments
Fixed-income investments are highly sensitive to changes in interest rates. These investments include bonds, certificates of deposit (CDs), and other debt securities. When interest rates fluctuate,
it affects the value and performance of these investments. Let's explore the impact from different perspectives:
1. Bond Prices: When interest rates rise, the prices of existing bonds tend to fall. This is because newly issued bonds offer higher yields, making existing bonds with lower yields less attractive.
Conversely, when interest rates decline, bond prices tend to rise as the fixed interest payments become more valuable.
2. yield-to-maturity: The yield-to-maturity (YTM) of a bond is the total return an investor can expect if they hold the bond until maturity. As interest rates rise, the YTM of existing bonds may
become less attractive compared to newly issued bonds with higher yields. This can lead to a decrease in demand for existing bonds and a decrease in their market value.
3. Coupon Payments: Fixed-income investments often provide regular coupon payments, which are predetermined interest payments made to bondholders. When interest rates rise, the coupon payments of
existing bonds may become less competitive compared to newly issued bonds. This can impact the demand for existing bonds and potentially decrease their market value.
4. Duration Risk: Duration measures the sensitivity of a bond's price to changes in interest rates. Bonds with longer durations are more sensitive to interest rate changes. When interest rates rise,
the prices of bonds with longer durations tend to decline more than those with shorter durations. Investors should consider the duration of their fixed-income investments to assess their exposure to
interest rate risk.
5. Reinvestment Risk: When interest rates decline, fixed-income investors face reinvestment risk. This occurs when the proceeds from maturing investments are reinvested at lower interest rates,
resulting in lower future returns. Investors should be mindful of the potential impact of reinvestment risk on their fixed-income portfolio.
To illustrate the impact, let's consider an example: Suppose you own a bond with a fixed interest rate of 4% and a maturity of 10 years. If interest rates rise to 5%, newly issued bonds with similar
characteristics may offer higher yields.
Impact on Fixed Income Investments - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
7. Managing Interest Rate Risk
## Understanding Interest Rate Risk
Interest rate risk refers to the vulnerability of an investment portfolio or a financial institution to changes in interest rates. It can manifest in several ways:
1. Bond Prices and Yields:
- When interest rates rise, bond prices tend to fall. Conversely, when rates decline, bond prices rise. This inverse relationship between bond prices and yields is fundamental to understanding
interest rate risk.
- Example: Suppose you hold a 10-year bond with a fixed coupon rate of 5%. If prevailing interest rates increase to 6%, newly issued bonds will offer higher yields. Consequently, the value of your
existing bond decreases because its fixed coupon payment becomes less attractive relative to the market rate.
2. Duration and Convexity:
- Duration measures the sensitivity of a bond's price to changes in interest rates. Longer-duration bonds are more sensitive to rate fluctuations.
- Convexity accounts for the curvature of the bond price-yield relationship. It provides additional insights beyond duration.
- Example: A bond with higher convexity will experience smaller price declines when rates rise compared to a bond with lower convexity.
3. Mortgage-Backed Securities (MBS):
- MBS are pools of mortgage loans packaged as securities. Their value depends on interest rates and prepayment behavior.
- Rising rates can lead to slower prepayments (as homeowners are less likely to refinance), affecting MBS prices.
- Example: An investor in MBS may face reinvestment risk if prepayments slow down due to higher rates.
4. Floating-Rate Instruments:
- Floating-rate bonds and loans have interest payments tied to a benchmark rate (e.g., LIBOR).
- These instruments provide some protection against rising rates because their coupons adjust periodically.
- Example: A corporate loan with a floating rate linked to LIBOR will see higher interest payments if LIBOR rises.
## strategies for Managing Interest rate Risk
1. Diversification:
- Spread risk by diversifying across different asset classes (e.g., stocks, bonds, real estate).
- Diversification reduces the impact of interest rate movements on the entire portfolio.
- Use interest rate derivatives (e.g., interest rate swaps, futures, options) to hedge against rate fluctuations.
- Example: A company with variable-rate debt can enter into an interest rate swap to convert it into fixed-rate debt.
3. Barbell and Bullet Strategies:
- Barbell strategy combines short-term and long-term bonds. It balances liquidity needs with yield optimization.
- Bullet strategy focuses on a specific maturity range (e.g., 5-7 years). It minimizes reinvestment risk.
- Example: An investor may hold short-term Treasury bills alongside longer-term corporate bonds.
4. Active Monitoring and Rebalancing:
- Regularly assess portfolio exposure to interest rate risk.
- Adjust allocations based on market conditions.
- Example: If rates are expected to rise, reduce duration by trimming long-term bonds.
5. Duration Matching:
- Match the portfolio's duration to the investor's time horizon.
- Example: A pension fund with long-term liabilities should hold longer-duration assets.
6. Consider inflation-Linked securities:
- treasury Inflation-Protected securities (TIPS) adjust for inflation. They provide protection against rising rates.
- Example: TIPS pay a fixed real yield plus inflation adjustments.
## Conclusion
managing interest rate risk requires a blend of prudence, analysis, and adaptability. investors and financial institutions must navigate this dynamic landscape by employing a mix of strategies
tailored to their unique circumstances. Remember that interest rate risk is not inherently negative; it presents opportunities for those who understand and manage it effectively.
Managing Interest Rate Risk - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
8. Hedging Strategies
Here, we'll explore various hedging strategies, each offering a unique perspective on managing interest rate risk:
1. Interest Rate Swaps (IRS):
- Overview: An IRS allows two parties to exchange fixed and floating interest rate payments. It's like swapping your fixed-rate mortgage for a variable-rate one.
- Example: Suppose Company A has a fixed-rate loan, while Company B has a floating-rate loan. They can enter an IRS to hedge against interest rate volatility. Company A pays a fixed rate to Company
B, and in return, Company B pays a floating rate to Company A.
- Insight: IRS provides flexibility by allowing companies to tailor their interest rate exposure.
2. Futures Contracts:
- Overview: Futures contracts are standardized agreements to buy or sell an asset at a predetermined price on a future date. They're commonly used for hedging.
- Example: A coffee producer might use futures contracts to lock in a price for their coffee beans, protecting against price declines.
- Insight: Futures provide liquidity and transparency, but they come with margin requirements.
3. Options:
- Overview: Options give the holder the right (but not the obligation) to buy or sell an asset at a specified price (the strike price) on or before a certain date.
- Example: An investor holding a portfolio of stocks might buy put options to hedge against a market downturn.
- Insight: Options offer flexibility and can be tailored to specific risk scenarios.
4. Duration Matching:
- Overview: Duration measures the sensitivity of a bond's price to interest rate changes. duration matching involves aligning the duration of assets and liabilities.
- Example: A pension fund with long-term liabilities might invest in long-duration bonds to match its payout schedule.
- Insight: Duration matching reduces the impact of interest rate shifts on the portfolio.
5. Currency Hedging:
- Overview: Currency fluctuations can affect international investments. Currency hedging involves using financial instruments to offset exchange rate risk.
- Example: An American investor buying European stocks might use currency forwards to hedge against euro depreciation.
- Insight: Currency hedging can stabilize returns but comes with costs.
6. Natural Hedges:
- Overview: Sometimes, existing business operations create natural hedges. For instance, a company exporting goods may benefit from a weaker domestic currency.
- Example: An airline with dollar-denominated revenue and fuel costs in euros has a natural hedge.
- Insight: Natural hedges arise organically and can be cost-effective.
Remember, no single hedging strategy fits all situations. The choice depends on your risk tolerance, investment horizon, and specific circumstances. By understanding these strategies, you'll be
better equipped to navigate the complex landscape of interest rate risk and protect your investments effectively.
Hedging Strategies - Interest Rate Risk Assessment: How to Assess the Impact of Interest Rate Changes on Your Investments
9. Case Studies and Practical Examples
### Understanding Interest Rate Risk: A Brief Recap
Before we dive into case studies, let's recap what interest rate risk entails. Essentially, interest rate risk refers to the potential impact of changes in interest rates on the value of your
investments. When interest rates fluctuate, it affects various asset classes differently. For instance:
1. bonds and Fixed-Income securities:
- Inverse Relationship: Bond prices move inversely to interest rates. When rates rise, existing bond prices fall, and vice versa.
- Duration Matters: Longer-duration bonds are more sensitive to rate changes.
- Example: Imagine you hold a 10-year government bond with a fixed coupon rate. If interest rates rise, the market value of your bond decreases, potentially leading to capital losses.
2. Equities (Stocks):
- Indirect Impact: While stocks are less directly affected by interest rate changes, they are influenced indirectly.
- Economic Environment: Rising rates may signal economic growth, benefiting certain sectors (e.g., financials) and hurting others (e.g., utilities).
- Example: A bank's profitability may improve due to higher lending rates, positively impacting its stock price.
3. Real Estate:
- Mortgage Rates: Real estate prices are tied to mortgage rates. Higher rates can reduce affordability and dampen demand.
- Investment Properties: rental income and property values may be affected by rate shifts.
- Example: If mortgage rates surge, potential homebuyers may delay purchases, impacting property prices.
### case Studies and practical Examples
1. The Great Recession (2008):
- Scenario: During the financial crisis, the Federal Reserve aggressively lowered interest rates to stimulate the economy.
- Impact:
- Positive: Bondholders benefited from rising bond prices.
- Negative: Savers earned lower yields on savings accounts and CDs.
- Lesson: Diversify across asset classes to mitigate risk.
2. central Bank policy Shifts:
- Scenario: Suppose a central bank unexpectedly raises rates.
- Impact:
- Bonds: Prices decline, affecting bond portfolios.
- Equities: Sectors sensitive to rates (e.g., utilities) may underperform.
- Lesson: Monitor central bank communications and adjust your portfolio accordingly.
3. Hedging with Derivatives:
- Scenario: A corporate treasurer manages interest rate risk for a multinational company.
- Solution: Use interest rate swaps or options to hedge against adverse rate movements.
- Example: A company with floating-rate debt can swap it for fixed-rate debt to lock in rates.
4. Homebuyers and Mortgage Rates:
- Scenario: A couple plans to buy a home.
- Impact:
- Low Rates: They can afford a larger mortgage.
- High Rates: Their affordability decreases.
- Lesson: Consider the impact of rates on your housing decisions.
5. Retirees and Income Streams:
- Scenario: Retirees rely on fixed-income investments.
- Impact:
- Low Rates: Reduced income from bonds.
- Diversification: Explore dividend-paying stocks or annuities.
- Lesson: Balance yield with risk tolerance.
Remember, these examples illustrate the multifaceted nature of interest rate risk. Your investment strategy should align with your financial goals, risk tolerance, and time horizon. Regularly review
your portfolio and stay informed about macroeconomic trends to make informed decisions.
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VStar's period analysis doesn't match ASAS-SN
Now that I'm able to import ASAS-SN data thanks to the revised plugin, I have a question or two. I have analyzed several data sets but the period that results from DC-DFT (using the top hit) never
seems to match what ASAS-SN has calculated. Further, when I do a Phase Plot using VStar's calculated period, it is generally messy, indicating the period is incorrect.
Let me illustrate this with the data for the eclipsing binary BV Ant.
Here is the data I'm using: ASASSN-V-BV_Ant.csv
Here is a picture of the raw data as shown in VStar: BV_Ant_Raw Data.jpg
ASAS-SN calculated the period as 3.59428 days with epoch 2457435.78608
Here is the nice Phase Plot using the period and epoch give by ASAS-SN: BV_Ant_Phase_Plot_With_ASAS-SN_Period.jpg
If I use VStar's DC-DFT Period Range tool with Low Period = 3; High Period = 4; Resolution = 0.00001 then I get this somewhat messy power spectrum: BV-Ant_Period_Analysis.jpg
The top hit here for the period is 3.66176, significantly different from ASAS-SN's 3.59428.
Creating a Phase Plot with this period give this messy graph: BV_Ant_Phase_Plot_With_VStar_Period.jpg
So I assume I'm doing something wrong given I can't get close to the actual period, which gives a nice phase plot. Can someone give me some pointers for how to do this? The reason I'm interested is
that one of the needed VSX tasks is calculating periods for stars that don't have one listed. Obviously some stars are irregular and won't have a period, but I would think that most eclipsing
binaries should have one.
As I've said, I have done this with 3 other eclipsing binaries and can't get a closely matching period on any of them.
Try narrowing the period range
It's a while since I've done this sort of thing, but once you have your first period estimate from a fairly wide low-high period range setting (in your example, 3 to 4 days for BV Ant), try narrowing
the period incrementally and repeating the analysis.
Tried the suggestion...
Thanks for the suggestion. I just tried slightly below the ASAS-SN value of 3.59428 up to slightly above my 3.66176. This gave a more distinct power peak but it once again was at 3.66176.
AoV gives a better result
I get the same results as you do. I've reached the limit of my knowledge on DCDFT.
AoV with period range gives a better result.
Peranso gives the same results as VStar for the BV Ant data
I've run the same analyses on Peranso.
DCDFT yields a period of about 3.66d. Anova yields about 3.594d.
There is an issue here I don't understand.
The data are not uniformly spaced, although generally closely spaced, so presumably it is appropriate to use DCDFT.
Eclipsing binaries and period search
Hi Roy, Bill
I don't claim to have the definitive answer and I have no doubt that there are other more learned voices in the community who can contribute to this conversation.
However, I'll give my take on this and encourage others to do likewise.
It's instructive to ignore the ASAS-SN period and instead, initially set the period search range in VStar to say, 0.2 to 3.7 with a resolution of 0.001.
For DCDFT the clearly strongest top hit is 1.797.
For AoV (with 10 bins), the top hit is 3.954.
Notice that the DCDFT result when doubled is also, 3.594. You can enter 1.797*2 in New Phase Plot when selecting a top-hit.
I've seen effect this many times when carrying out period search on eclipsing binaries compared to other variable types.
So, my approach is normally to try a wide range with both methods, then to narrow down using AoV for eclipsing binaries, e.g. try a range of 3 to 4 with resolution of 0.0001 with AoV. That's gets
fairly close to the answer. Then try 3.5 to 3.7 and change the resolution by a factor 10 to 0.00001. There is some experimentation required.
In case you're wondering why I chose the period search range 0.2 to 3.7, that's because many eclipsing binaries fall within that range when you look at VSX data. An exploration with Paul York some
time ago helped me to see that. Happy to say more about that.
So, what's going on here?
DCDFT, and I suspect Fourier methods in general, do best when the data is sinusoidal, e.g. pulsating variables since the trial functions involve trigonometric functions (and the corresponding power
AoV on the other hand involves a series of ANOVA calculations (and the corresponding F-statistics and p-values) for phased data. Think of your messy phase plot comment above.
Very different approaches to period search.
For an eclipsing binary, there are often two clear changes in brightness dips in a single cycle, that I suspect can "confuse" Fourier analysis.
I too am trying to find the time to find periods for stars in VSX.
Grant Foster's light curve analysis book looks mostly at pulsating variables but it would be great to hear his take on this. I would also like to know whether there has been much contribution to the
variable star period search algorithm literature (vs more general signal analysis literature) on this subject because it seems quite important in practice.
I hope that helps.
Black magic
David, thanks for chiming it on this.
How are you doing the ANOVA analysis? Is this in VStar? Peranso?
I find it somewhat distressing that your DCDFT analysis came up with half the period given by ASAS-SN. This is particularly true since your range includes the doubled value. How is one to know that
this is not correct?
It seems that ANOVA might be the best for these eclipsing binaries. However, before I tested eclipsing binaries I also ran the test with some long period Mira-type variables. In this case DCDFT
still came up with periods that were several days off from what ASAS-SN calculated. I would be curious to know what ASAS-SN is using for analysis.
I guess the moral of this story is that finding the period is somewhat akin to black magic.
I wonder if some scheme where you try different periods and see what phase plot is "least messy" would be the best. Is that what the ANOVA method is doing?
I just stumbled upon this…
I just stumbled upon this nice tool at the NASA Exoplanet Archive: https://exoplanetarchive.ipac.caltech.edu/cgi-bin/Pgram/nph-pgram
If I upload the data file we've been using and use the Planvchan algorithm and step method, I get a period of 3.59431882 days which is quite close to ASAS-SN's 3.59428.
Here is the output showing what options I chose: Periodogram.jpg
Interestingly, if I use the Lomb-Scargle algorithm, I get a period of 3.66210937 which is quite close to VStar's 3.66176. (I have no idea what any of these algorithms are doing).
It still seems like black magic in that small changes to input parameters can significantly change the period.
Anyway, this seems like a useful tool to supplement VStar.
Eccentric EA
Hi Bill (I will soon replying your email on VSX, sorry!),
there is another thing to keep in mind here: you are not using the best example to test for EA periods since this system is eccentric and also both minima are deep, and thus there is some extra
complications to the period analysis.
Usually you will have minima separated by 0.5 phase units and thus multiplying the result by two will give you the orbital period (that is not a software problem, it is how it is, if there are two
similar eclipses, the period will be taken as half the orbital period).
But if you look at the phase plot of BV Ant, you will see that you have one eclipse at phase 0.00 and the other at 0.44. I am not surprised that some bogus periods will come up due to this fact.
The best period for the system is the one in VSX, 3.59426 d. There is no need to change epoch or period.
You can use the recent ASAS-SN epoch and plot the ASAS-3 light curve with that period and you will see that the phase plot shows minima at phase 0.00:
Combining datasets to have the longest possible time baseline is the best way to improve/determine periods.
The Plavchan algorithm is described briefly here:
"...a binless phase-dispersion minimization algorithm that identifies periods with coherent phased light curves (i.e., least “dispersed”). There is no assumption about the underlying shape of the
periodic signal."
Phase Dispersion Minimisation (going to back to phase plot messiness) seems to be a general class of period search algorithms. AoV could probably be thought of as belonging to this class, just using
ANOVA F-statistic and p-values rather than minimizing the sum of the squares of the differences from one datapoint to the next.
PDM goes back at least to a 1978 paper by Stellingwerf: https://ui.adsabs.harvard.edu/abs/1978ApJ...224..953S/abstract
Note that the web page above also says of one of the other algorithms offered by the NASA exoplanet period search tool:
"Lomb-Scargle is an approximation of the Fourier Transform for unevenly spaced time sampling. It identifies periodic signals that are simple combinations of sines and cosines."
This can also be said of DCDFT.
In general, more period search algorithms, rather than less, seems advisable.
So far I have added two to VStar (DCDFT, ported to Java from Fortran) and AoV (implemented in terms of ANOVA and based upon AoV descriptions elsewhere, e.g. see https://academic.oup.com/mnras/article
/241/2/153/1051391; "We recommend one way analysis of variance (AoV) as a method for detection of sharp periodic signals.").
The plugin architecture of VStar allows for other period search algorithms to be incorporated. I have in the past considered adding a plugin for PDM and others. My main problem is time and what parts
of VStar to devote some subset of it to, largely driven by community need. I'm sure you understand that problem. Some others in the AAVSO community have taken to creating various plugins (observation
source for example) and I encourage that.
Thanks for the link to the…
Thanks for the link to the Planvchan method. Yes, it seems like it falls in the ANOVA class of methods as is the PDM method. They seem to be the most robust and Tonny Vanmunster, the author of
Peranso, has this to say about ANOVA in general:
In developing Peranso, I have studied hundreds of light curves of many different objects. Although there is no "universal" period analysis method, there is one that - in my humble opinion - comes
pretty close, and that's ANOVA. I have been amazed by its power to improve peak detection sensitivity and to damp alias periods. Try it out yourself, and see if it suits your data. If not, there
are many others to experiment with.
AoV plugin
Hi Bill
First, thanks for SkySafari! I just read your contact page. I use the iOS app all the time. Very cool.
The ANOVA analysis is via the AoV (Analysis of Variance) period search plugin:
It essentially folds the light curve over a range of periods at a given resolution (and bin size) and computes one-way ANOVA. The results table and plots come from this.
See the Plugin Manager section of the user manual or just the plugin library link above.
Does Peranso give the same result for the DCDFT range 0.2 to 3.7 as VStar?
I would be interested to talk further with you here about particular Miras.
I had forgotten (or not…
I had forgotten (or not noticed) there was an ANOVA plugin. I'll give it a try.
I went ahead and purchased a Peranso license. It has several different ANOVA methods. I will compare it to the Stars plugin.
I've come to realize I'm probably being a little OCD about what period the various routines come up with. I suspect as long as they are within a few percent of each other, it probably doesn't matter.
I have notice that some methods can give larger errors than that but it is probably because I'm using the wrong method for the type of light curve. For example, the Fourier Transform methods don't
do a great job on eclipsing binaries.
Peranso has a couple of tricks to ensure the periods you get are real an not the result of sampling irregularities.
I installed and tried the…
I installed and tried the VStar ANOVA plugin with the BV Ant eclipsing binary star we've been discussing.
VStar: 3.594d (0.0002 resolution. Couldn't get more as then VStar would just run forever - probably the garbage collector)
Peranso: 3.59402d
ASAS-SN: 3.59428d
VSX: 3.59426d
So they are all in excellent agreement. Given how well the VStar ANOVA method works, I might recommend that it come standard in VStar. Many folks (like myself) might not realize it is available.
AoV, garbage collection and such
What period range was used here for BV Ant?
A wide range with resolution 0.01 or so then higher resolution (such as 0.0002) with narrower period range is best from a speed viewpoint.
AoV hasn't had much profiling attention paid to it so it would be interesting to discover the cause of the inefficiency here. Given the way it's coded, I don't know whether GC would be a big
contributor. Also, other than top-hit collation, trial period testing is embarrassingly parallel, so a TODO list item is to make use of multiple cores for AoV (so too for DCDFT) and there are
comments re: approaches in the source code. Other issues may need addressing first, and I wouldn't want to violate Knuth's maxim too soon. :)
I realized I didn't answer…
I realized I didn't answer your DCDFT question.
Does Peranso give the same result for the DCDFT range 0.2 to 3.7 as VStar?
The Peronso DCDFT has (Ferraz-Mello) after its label, so I don't know if this is a modification of the basic DCDFT or not. However, when I run it against the BV Ant data I get 0.718796d for the
period and VStar's was 1.797. So they are not the same and surprisingly the value is not close to 1/2 the actual period like VStar's. I don't know how to interpret this.
Regarding your question about Miras, I'll run a few through both and post how they compare. Probably not really a useful exercise, but I'm having fun playing with this.
Mira data
Here are some comparisons of Mira periods calculated different ways. The stars were sort of chosen at random from Miras in the LPV target list. For the 3 stars I show the periods from VSX, ASAS-SN
and the DCDFT and ANOVA calculations from Peranso and VStar. You can see small differences of a couple of days, but it is probably not much of a big deal.
Make of it what you will...
RS Aqr (15 - 9.3)
VSX: 217 d
ASAS-SN: 218.8128426 d
Peranso: (DCDFT) 2.13.356091 d; (ANOVA) 213.561132 d
VStar: (DCDFT) 213.386 d; (ANOVA) 213.78 d
T CMi (15.1 - 9.5)
VSX 325.8 d
ASAS-SN: 332.3546781 d
Peranso: (DCDFT) 324.306794 d; (ANOVA) 326.370757 d
VStar: (DCDFT) 324.33 d; (ANOVA) 327.0
W Dra (13.5 - 7.2)
VSX 278.6 d
ASAS-SN: 278.6 d
Peranso: (DCDFT) 292.185076; (ANOVA) 292.312189
VStar: (DCDFT) 292.209 d; (ANOVA) 290.77 d
Hi Bill
That 0.718796d result is interesting. I guess it would be worth asking the Peranso developer about that. I don't know how to interpret it either.
BV Ant VStar vs. ASAS-SN Period
I suspect the reason that you get such a different period for DCDFT in VStar compared to the period given by ASASSN is that you used the period range option rather than frequency range. It is always
better to use a frequency grid rather than a period grid when doing DCDFT analysis. DCDFT is based on coefficient fitting of sinusoidal functions, Asin(2πf) + Bcos(2πf), whose arguments are linear
in frequency. Therefore, DCDFT based on frequency is evenly spaced with respect to the argument of the functions, but a search based on Period is not.
I did DCDFT analysis in VStar using the frequency range option with low f =0.2 (P = 5d) and high f = 5 (P = 0.2d) and resolution 0.0001. The highest power signal was at f=0.5564 equating to a period
of 1.797268. Twice that period is 3.594536d compared to the ASASSN period of 3.59428d and the VStar AoV analysis period of 3.5942. If you do the VStar AoV analysis with a period range of 1d to 4d
you can see the second highest power period (outside the wings of the highest peak) is 1.797 days. That is half the orbital period and a match to the period given by DCDFT.
I ran DCDFT again with a frequency range of 0.5 to 0.6 and resolution 0.000001 which gave the highest power signal at f = 0.556418 equating to a period of 1.79721. Twice that period is 3.59442d.
which is even closer to the period given by ASASSN.
Since DCDFT assumes that the signal is comprised of an algebraic sum of sine and cosine functions of various frequencies and their harmonics, this light curve and many binary light curves are not
well represented by DCDFT because they have long relatively constant segments separated by sudden V shaped valleys. Therefore, the DCDFT solution will include a lot of frequencies to approximate the
shape of the light curve. However, most of these frequencies do not represent physical processes taking place. They are just part the mathematical model needed to approximate the shape of the light
curve using sinusoids. Because the primary and secondary mimima are so similar, DCDFT doesn’t distinguish between them and gives the highest power result at 2x the real orbital frequency
corresponding to half the orbital period. You can see the "forest" of DCDFT power peaks if you repeat the DCDFT analysis outlined above.
AoV assigns a higher power to the correct period. If you do a phase plot at 1.797d it becomes obvious why this occurs. Although the period between sequential primary minima is the same as the
between sequential secondary minima and both equal the orbital period, the period of adjacent primary and secondary minima is not half the orbital period. Therefore, the light curve folded at half
the orbital period is “messier” with closely spaced double valleys “bridged” across their tops by lines of relatively constant observations. The pairs of valleys are still separated by long segments
of relatively constant observations. Clearly the light curve is not folded in agreement with the orbit. Since AoV is the ratio of variance of bin means from the global mean divided by the combined
variances of data within bins from their bin means (F statistic, which is the power), folding at half the period results in a lower ratio (power) than when the light curves are folded using the
correct orbital period.
Ferraz-Mello authored the seminal paper on the use of DCDFT. VStar uses Ferraz Mello DCDFT. Therefore, I have no idea why Peranso gives such a different period. I can only assume there are
differences between the two sets of code.
Brad Walter
DCDFT VStar and Peranso
You wrote:
“Ferraz-Mello authored the seminal paper on the use of DCDFT. VStar uses Ferraz Mello DCDFT. Therefore, I have no idea why Peranso gives such a different period. I can only assume there are
differences between the two sets of code.”
I ran DCDFT and ANOVA on BV Ant in both Peranso and VStar, specifying period, not frequency. In both sets of software, DCDFT gave a period of about 3.66d and ANOVA a period of about 3.594d.
DCDDFT VStar and Peranso
That is good news. They should give the same answer with equivalent settings. I don't know why Bill calculated 0.718796d with Peranso. Can you try a DCDFT by frequency range in Peranso to see what
result that gives? If you use the same settings I used I would expect it to give the same results as my analysis. Excuse my ignorance regarding Peranso's capabilities.
Brad, WBY
I just tried the DCDFT again…
I just tried the DCDFT again in Peranso. Once again it gave me a 1.797010 d period. To be clear, my settings are:
Range Start: 1
Range End: 5
Steps: 5000
The Lomb-Scargle gave me the same value. I even tried changing the units to Frequency rather than time and it gave me a frequency of 0.55648 c/d which, taking the inverse, gives me 1.79701d. So
the result is robust. The ANOVA algorithm gives me the correct value of 3.594020 (which is twice the DCDFT value)
Interestingly, if I narrow down the search range to [2, 4], it gives me 2.254029 d, so still not the correct value. I don't know if it is worth pursuing with the Peranso developer.
If I do the DCDFT in VStar with the same range and resolution ([1, 5] and 0.0002) then it also gives me a top hit of 1.7972.
Roy had said:
I ran DCDFT and ANOVA on BV Ant in both Peranso and VStar, specifying period, not frequency. In both sets of software, DCDFT gave a period of about 3.66d and ANOVA a period of about 3.594d.
I don't understand what Roy is doing differently from my to get a different outcome from DCDFT. Roy, what are the exact DCDFT search parameter you are using? My guess is that we are using slightly
different params.
Frequency vs period range
Thanks for this insight Brad. Yes, this is something I should have thought to mention as well, but you expressed this in a way I was unlikely to have.
The part of what you said:
AoV assigns a higher power to the correct period. If you do a phase plot at 1.797d it becomes obvious why this occurs. Although the period between sequential primary minima is the same as the
between sequential secondary minima and both equal the orbital period, the period of adjacent primary and secondary minima is not half the orbital period. Therefore, the light curve folded at
half the orbital period is “messier” with closely spaced double valleys “bridged” across their tops by lines of relatively constant observations.
makes sense but is the sort of thing that would probably benefit from a suitable diagram / annotated LC/PP, e.g. to illustrate this:
...the period of adjacent primary and secondary minima is not half the orbital period
This is probably a silly question, but does AoV bin the data after the trial folds. In other words does 10 bins correspond to 0.1 phase segments for each trial period?
Brad Walter
AoV bins
Hi Brad
It's not a silly question, just a reasonable question about the implementation. Nothing much is obvious.
Yes, AoV bins the data after each trial fold and yes, for the default 10 bins, this corresponds to 0.1 phase segments for each trial period.
To be more specific, the bins are used in the ANOVA computation for each trial period.
VStar's period analysis doesn't match ASAS-SN
Having just read through this thread again, I would like to make a number of comments:
1. As one spending a fair amount of my time calculating periods for eclipsing binaries which have no period in VSX, this discussion is very timely. I have run up against similar problems using VStar
and other tools.
2. As Bill suggested, it is probably worth bearing in mind that searching for a period will always be someting of a "black art", no matter how good the algorithms and/or software become. I say this
because, IMHO, the "acid test" of whether you have the "correct" period is to eyeball the phase plot (folded light curve) based on a candidate period. If this plot is "nice and tight" (whatever that
might mean) then the period is a "good one". Having done that, the only way to decide that you have the best period (as opposed to just a "good" period) is to produce another phase plot based on a
new candidate period and see if the resultant curve is "nicer and tighter" than the previous one. This is obviously a matter of human judgment and is not strictly repeatable. In other words, other
viewers of the light curves may not choose the same period, especially when we are getting to six decimal places.
3. Anyway, the discussion in (2) leads me to suggest that the Plavchan algorithm might prove to be the algorithm of choice for eclipsing binaries. As David defined it, "...a binless phase-dispersion
minimization algorithm that identifies periods with coherent phased light curves (i.e., least “dispersed”)". This seems to mirror "the trial and error approach with eyeballing" described above. Note
that I am not suggesting here that VStar be modified to include Plavchan at this stage ...
4. Grant Foster in his book, "Analyzing Light Curves - A Practical Guide" discusses the AOV Periodogram (pp. 136-143) and he concludes that , where eclipsing binaries are concerned, AOV is a better
method than DCDFT. He says "A real advantage of AOV for period search is that it's more sensitive than the DCDFT (or Fourier methods in general) when the signal shape is profoundly non-sinusoidal ...
[as, for example] with eclipsing binary stars ... Fourier methods are far less sensitive to detecting such shapes". He also says "it is often useful to increase the number of bins in the AOV
periodogram". Ever since reading that, I have used the "AOV with Period Range" tool in VStar.
5. I think this discussion indicates that there is a gap in the training course offerings, as currently provided by the AAVSO: The VSX team are simply too busy to stop and teach would-be analysts how
to go about finding periods. The CHOICE course "How to Use VStar" does not cover the detailed practical steps for finding a period in a set of time-series data; as I recollect, it is focused more on
how to use the VStar tool and its features. On the other hand, the course "Analyzing Data with VStar", run by Brad Walter in the past, might be expected to do so? However, it has not been offered for
a while - since maybe 2020? Brad, is there any prospect that you might offer this course again?
6. An alternative to a full-blown course might be a "How To" session, using Zoom, in which various cases (perhaps limited to eclipsing binaries) were worked through (using screen sharing), with
maybe the cases supplied beforehand by the interested attendees? It would be important to use VStar as the primary tool, only going to alternate tools if VStar happens to prove inadequate in a
particular case.
Response to interesting comments
Hi Paul
Re: 2. yes, I completely understand. Seeing is "believing". One thing that can help is looking at the various measures of error after a model fit from a DCDFT or using the Current Mode ANOVA plugin
to visualise what AoV is doing at a particular trial period/phase plot.
Re: 3. Plavchan could be "just another" period search plugin, as AoV currently is.
Re: 4. quite right! Thanks for the reminder about this. Eclipsing binaries are mentioned 4 times in Foster. I guess I was trying to say what Grant said in one of my comments, but did not articulate
it as well. His comment about increasing the number of bins is worth considering carefully as well!
Re: 5. and 6. yes, even beyond the two VStar courses, perhaps a course geared towards VSX submissions would be worth considering!
Bill wrote:
"I don't…
Bill wrote:
"I don't understand what Roy is doing differently from my to get a different outcome from DCDFT. Roy, what are the exact DCDFT search parameter you are using? My guess is that we are using slightly
different params."
Bill, sorry about this delayed reply. Very busy the past 24 hours. Originally, I tried analysing the period (not frequency) over the range 3 to 4 days. I think the resolution was about 1000 (0.001),
but maybe even 10,000 (0.0001). These ànalyses produced the results I first mentioned.
I tried again earlier today with a wider range (0.2 to 6 days, I think) of periods, a resolution of 1000 (0.001) and got quite different results, although DCDFT did not run properly on VStar. I don't
know why. I'm away from home and don't have my computer tonight so can't send you these new results now. But the results that seemed valid were the same on VStar (AoV) and both DCDFT and ANOVA on
Peranso. The period displayed was one half the VSX period, 1.797 I think. | {"url":"https://mintaka.aavso.org/comment/165380","timestamp":"2024-11-03T04:38:51Z","content_type":"text/html","content_length":"146612","record_id":"<urn:uuid:6d122cbd-3af3-4dc0-8a54-ec8e8de96879>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00887.warc.gz"} |
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Distribution of Sample Proportions (5 of 6)
Module 7: Linking Probability to Statistical Inference
Distribution of Sample Proportions (5 of 6)
Distribution of Sample Proportions (5 of 6)
Learning OUTCOMES
• Use a z-score and the standard normal model to estimate probabilities of specified events.
From our work on the previous page, we now have a mathematical model of the sampling distribution of sample proportions. This model describes how much variability we can expect in random samples from
a population with a given parameter. If a normal model is a good fit for a sampling distribution, we can apply the empirical rule and use z-scores to determine probabilities. Here we link probability
to the kind of thinking we do in inference.
Making Connections to Probability Models in Probability and Probability Distribution
Probability describes the chance that a random event occurs. Recall the concept of a random variable from the module Probability and Probability Distribution. When a variable is random, it varies
unpredictably in the short run but has a predictable pattern in the long run. Sample proportions from random samples are a random variable. We cannot predict the proportion for any one random sample;
they vary. But we can predict the pattern that occurs when we select a great many random samples from a population. The sampling distribution describes this pattern. When a normal model is a good fit
for the sampling distribution, we can use what we learned in the previous module to find probabilities.
Recall probability models we saw in Probability and Probability Distribution. We saw examples of models with skewed curves, but we focused on normal curves because we use normal probability models to
describe sampling distributions in Modules 7 to 10 when we make inferences about a population. As we now know, we can use a normal model only when certain conditions are met. Whenever we want to use
a normal model, we must check the conditions to make sure a normal model is a good fit.
Here we summarize our general process for developing a probability model for inference. This is essentially the same process we used in the previous module for developing normal probability models
from relative frequencies.
If a normal model is a good fit for the sampling distribution, we can standardize the values by calculating a z-score. Then we can use the standard normal model to find probabilities, as we did in
Probability and Probability Distribution.
The z-score is the error in the statistic divided by the standard error. For sample proportions, we have the following formulas.
[latex]\text{standard error} = \sqrt{\frac{p(1 - p)}{n}}[/latex]
[latex]Z = \frac{\text{statistic} - \text{parameter}}{\text{standard error}} = \frac{\hat{p} - p}{\text{standard error}}[/latex]
We can also write this as one formula:
[latex]Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}}[/latex]
This z-score formula is similar to the z-score formula we used in Probability and Probability Distribution. We described the z-score as the number of standard deviations a data value is from the
mean. Here we can describe the z-score as the number of standard errors a sample proportion is from the mean. Because the mean is the parameter value, we can say that the z-score is the number of
standard errors a sample proportion is from the parameter.
A positive z-score indicates that the sample proportion is larger than the parameter. A negative z-score indicates that the sample proportion is smaller than the parameter.
Probability Calculations for Community College Enrollment
Let’s return to the example of community college enrollment. Recall that a 2007 report by the Pew Research Center stated that about 10% of the 3.1 million 18- to 24-year-olds in the United States
were enrolled in a community college. Let’s again suppose we randomly selected 100 young adults in this age group and found that 15% of the sample was enrolled in a community college.
Previously, we determined that 15% is a surprising result. Now we want to be more precise. We ask this question: What is the probability that a random sample of this size has 15% or more enrolled in
a community college?
To answer this question, we first determine if a normal model is a good fit for the sampling distribution.
Check normality conditions:
Yes, the conditions are met. The number of expected successes and failures in a sample of 100 are at least 10. We expect 10% of the 100 to be enrolled in a community college, [latex]np = 100(0.10)[/
latex]. We expect 90% of the 100 to not be enrolled, [latex]n(1 - p) = 100(0.90) = 90[/latex].
We therefore can use a normal model, which allows us to use a z-score to find the probability.
Find the z-score:
[latex]\text{standard error} = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.10(0.90)}{100}} \approx 0.03[/latex]
[latex]Z = \frac{\text{statistic} - \text{parameter}}{\text{standard error}} = \frac{0.15 - 0.10}{0.03} \approx 1.67[/latex]
Find the probability using the standard normal model:
We want the probability that the sample proportion is 15% or more. So we want the probability that the z-score is greater than or equal to 1.67. The probability is about 0.0475.
Conclusion: If it is true that 10% of the population of 18- to 24-year-olds are enrolled at a community college, then it is unusual to see a random sample of 100 with 15% or more enrolled. The
probability is about 0.0475.
Note: This probability is a conditional probability. Recall from Relationships in Categorical Data with Intro to Probability that we write a conditional probability P(A given B) as P(A | B). Here we
write P(a sample proportion is 0.15 given that the population proportion is 0.10) as
[latex]P(\hat{p} \geq 0.15|p = 0.10) \approx 0.0475[/latex]
Click here to open this simulation in its own window.
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Solving Sparse Linear Systems Faster than Matrix Multiplication
Presenting an algorithm that solves linear systems with sparse coefficient matrices asymptotically faster than matrix multiplication for any ω > 2.
Read the related Technical Perspective
Can linear systems be solved faster than matrix multiplication? While there has been remarkable progress for the special cases of graph structured linear systems, in the general setting, the bit
complexity of solving an n × n linear system Ax = b is Õ(n^ω), where ω < 2.372864 is the matrix multiplication exponent. Improving on this has been an open problem even for sparse linear systems with
poly(n) condition number.
In this paper, we present an algorithm that solves linear systems in sparse matrices asymptotically faster than matrix multiplication for any ω > 2. This speedup holds for any input matrix A with o(n
^ω−1/log(k(A))) non-zeros, where k(A) is the condition number of A. For poly(n)-conditioned matrices with Õ(n) nonzeros, and the current value of ω, the bit complexity of our algorithm to solve to
within any 1/poly(n) error is O(n^2.331645).
Our algorithm can be viewed as an efficient, randomized implementation of the block Krylov method via recursive low displacement rank factorizations. It is inspired by the algorithm of [Eberly et al.
ISSAC ’06 ’07] for inverting matrices over finite fields. In our analysis of numerical stability, we develop matrix anti-concentration techniques to bound the smallest eigenvalue and the smallest gap
in eigenvalues of semi-random matrices.
Solving a linear system $Ax=b$ is a basic algorithmic problem with direct applications to scientific computing, engineering, and physics, and is at the core of algorithms for many other problems,
including optimization, data science, and computational geometry. It has enjoyed an array of elegant approaches, from Cramer’s rule and Gaussian elimination to numerically stable iterative methods to
more modern randomized variants based on random sampling^8^,^19 and sketching.^23 Despite much recent progress on faster solvers for graph-structured linear systems^8^,^9^,^19 progress on the general
case has been elusive.
Most of the work in obtaining better running time bounds for linear systems solvers has focused on efficiently computing the inverse of $A$, or some factorization of it. Such operations are in turn
closely related to the cost of matrix multiplication. Matrix inversion can be reduced to matrix multiplication via divide-and-conquer, and this reduction was shown to be stable when the word size for
representing numbers^b is increased by a factor of $O\left( log n\right)$.^4 The current best runtime of $\omega <2.38$ follows a long line of work on faster matrix multiplication algorithms and is
also the current best running time for solving $Ax=b$: when the input matrix/vector are integers, matrix multiplication based algorithms can obtain the exact rational solution using $O\left({n}^{\
omega }\right)$ word operations.^20
Methods for matrix inversion or factorization are often referred to as direct methods in the linear systems literature. This is in contrast to iterative methods, which gradually converge to the
solution. Iterative methods have low space overhead, and therefore are widely used for solving large, sparse, linear systems that arise in scientific computing. Another reason for their popularity is
that iterative methods are naturally suited to producing approximate solutions of desired accuracy in floating point arithmetic, the de facto method for representing real numbers. Perhaps the most
famous iterative method is the Conjugate Gradient (CG) / Lanczos algorithm.^16 It was introduced as an $O\left(n·nnz\right)$ time algorithm under exact arithmetic, where $nnz$ is the number of
non-zeros in the input matrix. However, this bound only holds under the Real RAM model where words have unbounded precision. When taking bit sizes into account, it incurs an additional factor of $n$.
Despite much progress in iterative techniques in the intervening decades, obtaining gains over matrix multiplication in the presence of round-off errors has remained an open question.
The convergence and stability of iterative methods typically depend on some condition number of the input. When all intermediate steps are carried out using precision close to the condition number of
$A$, the running time bounds of the CG algorithm, as well as other currently known iterative methods, depend polynomially on the condition number of the input matrix $A$. Formally, the condition
number of a symmetric matrix $A$, $\kappa \left(A\right)$, is the ratio between the maximum and minimum eigenvalues of $A$. Here the best known rate of convergence when all intermediate operations
are restricted to bit-complexity $O\left(log \left(\kappa \left(A\right)\right)\right)$ is $O\left(\sqrt{\kappa \left(A\right)} log \left(1/ϵ\right)\right)$ iterations to achieve error $ϵ$. This is
known to be tight if one restricts to matrix-vector multiplications in the intermediate steps.^12^,^17 This means for moderately conditioned (e.g., with $\kappa =poly\left(n\right)$), sparse,
systems, the best runtime bounds are still via direct methods, which are stable when $O\left(log \left(1/\kappa \right)\right)$ words of precision are maintained in intermediate steps.^4
Many of the algorithms used in practice in scientific computing for solving linear systems involving large, sparse matrices are based on combining direct and iterative methods: we will briefly
discuss these perspectives in Section 1.3. In terms of asymptotic complexity, the practical successes of many such methods naturally lead to the question of whether one can provably do better than
the $O\left(min\left\{{n}^{\omega },nnz·\sqrt{\kappa \left(A\right)}\right\}\right)$ time corresponding to the faster of direct or iterative methods. Somewhat surprisingly, despite the central role
of this question in scientific computing and numerical analysis, as well as extensive studies of linear systems solvers, progress on it has been elusive. The continued lack of progress on this
question has led to its use as a hardness assumption for showing conditional lower bounds for numerical primitives such as linear elasticity problems^25 and positive linear programs.^10 One
formalization of such hardness is the Sparse Linear Equation Time Hypothesis (SLTH) from:^10 ${\mathrm{SLTH}}_{k}^{\gamma }$ denotes the assumption that a sparse linear system with $\kappa \le nnz{\
left(A\right)}^{k}$ cannot be solved in time faster than $nnz{\left(A\right)}^{\gamma }$ to within relative error $ϵ={n}^{–10k}$. Here, improving over the smaller running time of both direct and
iterative methods can be succinctly encapsulated as refuting ${\mathrm{SLTH}}_{k}^{min\left\{1+k/2,\omega \right\}}$.^c
We provide a faster algorithm for solving sparse linear systems. Our formal result is the following (we use the form defined in:^10 Linear Equation Approximation Problem, LEA).
Theorem 1.
Given a matrix $A$ with maximum dimension $n$, $nnz\left(A\right)$ non-zeros (whose values fit into a single word), along with a parameter $\kappa \left(A\right)$ such that $\kappa \left(A\right)\ge
{\sigma }_{max}\left(A\right)/{\sigma }_{min}\left(A\right)$, a vector $b$ and error requirement $ϵ$, we can compute, under fixed point arithmetic, in time $O\left(\mathrm{max}\left\{nnz{\left(A\
right)}^{\frac{\omega -2}{\omega -1}}{n}^{2},{n}^{\frac{5\omega -4}{\omega +1}}\right\}{\mathrm{ log }}^{2}\left(\kappa /ϵ\right)\right)$
a vector $x$ such that ${‖Ax-{\Pi }_{A}b‖}_{2}^{2}\text{ }\le \text{ }ϵ\text{ }{‖{\Pi }_{A}b‖}_{2}^{2},$
where $c$ is a fixed constant and ${\Pi }_{A}$ is the projection operator onto the column space of$A$.
Note that ${‖{\Pi }_{A}b‖}_{2}={‖{A}^{T}b‖}_{{\left({A}^{T}A\right)}^{†}}$, and when $A$ is square and full rank, it is just ${‖b‖}_{2}$.
The cross-over point for the two bounds is at $nnz\left(A\right)={n}^{\frac{3\left(\omega –1\right)}{\omega +1}}$. In particular, for the sparse case with $nnz\left(A\right)=O\left(n\right)$, and the
bound of $\omega \le 2.38$, we get an exponent of
$\mathrm{max}\left\{2+\frac{\omega -2}{\omega -1},\frac{5\omega -4}{\omega +1}\right\}<\mathrm{max}\left\{2.28,2.34\right\}=2.34.$
As $n\le nnz$, this also translates to a running time of $O\left(nn{z}^{\frac{5\omega –4}{\omega +1}}\right)$, which as $\frac{5\omega –4}{\omega +1}=\omega –\frac{{\left(\omega –2\right)}^{2}}{\
omega +1}$, refutes ${\mathrm{SLTH}}_{k}^{\omega }$ for constant values of $k$ and any value of $\omega >2$.
We can parameterize the asymptotic gain over matrix multiplication for moderately sparse instances. Here we use the $\stackrel{~}{O}\left(·\right)$ notation to hide lower-order terms, specifically $\
stackrel{~}{O}\left(f\left(n\right)\right)$ denotes $O\left(f\left(n\right)·{ log }^{c}\left(f\left(n\right)\right)\right)$ for some absolute constant $c$.
Corollary 2.
For any matrix$A$with dimension at most$n$, $O\left({n}^{\omega –1–\theta }\right)$non-zeros, and condition number${n}^{O\left(1\right)}$, a linear system in$A$can be solved to accuracy${n}^{–O\left
(1\right)}$in time$\stackrel{~}{O}\left(max\left\{{n}^{\frac{5\omega –4}{\omega +1}},{n}^{\omega –\frac{\theta \left(\omega –2\right)}{\omega –1}}\right\}\right)$.
Here the cross-over point happens at $\theta =\frac{\left(\omega –1\right)\left(\omega –2\right)}{\omega +1}$. Also, because $\frac{5\omega –4}{\omega +1}=\omega –\frac{{\left(\omega –2\right)}^{2}}
{\omega +1}$, we can also infer that for any $0<\theta \le \omega –2$ and any $\omega >2$, the runtime is $o\left({n}^{\omega }\right)$, or asymptotically faster than matrix multiplication.
At a high level, our algorithm follows the block Krylov space method (see e.g., Chapter 6.12 of Saad^16). This method is a multi-vector extension of the CG/Lanczos method, which in the single-vector
setting is known to be problematic under round-off errors both in theory^12 and in practice.^16 Our algorithm starts with a set of $s$ initial vectors, $B\in {\Re }^{n×s}$, and forms a column space
by multiplying these vectors by $A$ repeatedly, $m$ times. Formally, the block Krylov space matrix is
$K=\left[\text{ }B|AB|{A}^{2}B|\dots |{A}^{m-1}B\text{ }\right].$
The core idea of Krylov space methods is to efficiently orthogonalize this column space. For this space to be spanning, block Krylov space methods typically choose $s$ and $m$ so that $sm=n$.
The conjugate gradient algorithm can be viewed as an efficient implementation of the case $s=1$, $m=n$, with $B$ set to $b$, the RHS of the input linear system. The block case with larger values of
$s$ was studied by Eberly, Giesbrecht, Giorgi, Storjohann, and Villard^5 over finite fields, and they gave an $O\left({n}^{2.28}\right)$ time^d algorithm for computing the inverse of an $O\left(n\
right)$-sparse matrix over a finite field.
Our algorithm also leverages the top-level insight of the Eberly et al. results: the Gram matrix of the Krylov space matrix, is a block Hankel matrix. Solving linear systems in this Gram matrix, ${\
left(AK\right)}^{T}\left(AK\right)$, lead to solvers for linear systems in $A$ because
so as long as $A$ and $K$ are both invertible, composing this on the left by $K$ and on the right by $A\top {K}^{\top }$ gives
Eberly et al. viewed the Gram matrix as an $m$-by-$m$ matrix containing $s$-by-$s$ sized blocks, and critically leveraged the fact that the blocks along each anti-diagonal are identical:
$\begin{array}{c}{\left(AK\right)}^{T}\left(AK\right)=\\ \left[\begin{array}{ccccc}{B}^{T}{A}^{2}B& {B}^{T}{A}^{3}B& {B}^{T}{A}^{4}B& \dots & {B}^{T}{A}^{m+1}B\\ {B}^{T}{A}^{3}B& {B}^{T}{A}^{4}B& {B}
^{T}{A}^{5}B& \dots & {B}^{T}{A}^{m+2}B\\ \begin{array}{c}{B}^{T}{A}^{4}B\\ \dots \\ {B}^{T}{A}^{m+1}B\end{array}& \begin{array}{c}{B}^{T}{A}^{5}B\\ \dots \\ {B}^{T}{A}^{m+2}B\end{array}& \begin
{array}{c}{B}^{T}{A}^{6}B\\ \dots \\ {B}^{T}{A}^{m+3}B\end{array}& \begin{array}{c}\dots \\ \dots \\ \dots \end{array}& \begin{array}{c}{B}^{T}{A}^{m+3}B\\ \dots \\ {B}^{T}{A}^{2m}B\end{array}\end
Formally, the $s$-by-$s$ inner product matrix formed from ${A}^{i}B$ and ${A}^{j}B$ is ${B}^{T}{A}^{i+j}B$, and depends only on $i+j$. So instead of ${m}^{2}$ blocks each of size $s×s$, we are able
to represent an $n$-by-$n$ matrix with only about $m$ blocks.
Operations involving these $m$ blocks of the Hankel matrix can be handled using $\stackrel{~}{O}\left(m\right)$ block operations. This is perhaps easiest seen for computing matrix-vector products
using $K$. If we use $\left\{i\right\}$ to denote the $i$th block of the Hankel matrix, and define
for a sequence of matrices $M$, we get that the ${i}^{\mathrm{th}}$ block of the product $Hx$ can be written in block-form as
${\left(Hx\right)}_{\left\{i\right\}}=\sum _{j}{H}_{\left\{i,j\right\}}{x}_{\left\{j\right\}}=\sum _{j}M\left(i+j\right){x}_{\left\{j\right\}}.$
Note this is precisely the convolution of (a sub-interval) of $M$ and $x$, with shifts indicated by $i$. Therefore, in matrix-vector multiplication (the “forward” direction), a speedup by a factor of
about $m$ is possible with fast convolution algorithms. The performance gains of the Eberly et al. algorithms^5 can be viewed as being of a similar nature, albeit in the more difficult direction of
solving linear systems. Specifically, they utilize algorithms for the Padé problem of computing a polynomial from the result of its convolution.^1 Over finite fields, or under exact arithmetic, such
algorithms for matrix Padé problems take $O\left(m log m\right)$ block operations,^1 for a total of $\stackrel{~}{O}\left({s}^{\omega }m\right)$ operations.
The overall time complexity is based on two opposing goals:
1. Quickly generate the Krylov space: repeated multiplication by $A$ allows us to generate ${A}^{i}B$ using $O\left(ms·nnz\right)=O\left(n·nnz\right)$ arithmetic operations. Choosing a sparse $B$
then allows us to compute ${B}^{T}{A}^{i}B$ in $O\left(n·s\right)$ arithmetic operations, for a total overhead of $O\left({n}^{2}\right)$.
2. Quickly invert the Hankel matrix. Each operation on an $s$-by-$s$ block takes $O\left({s}^{\omega }\right)$ time. Under the optimistic assumption of $\stackrel{~}{O}\left(m\right)$ block
operations, the total is $\stackrel{~}{O}\left(m·{s}^{\omega }\right)$.
Under these assumptions, and the requirement of $n\approx ms$, the total cost becomes about $O\left(n·nnz+m·{s}^{\omega }\right)$, which is at most $O\left(n·nnz\right)$ as long as $m>{n}^{\frac{\
omega –2}{\omega –1}}$. However, this runtime complexity is over finite fields, where numerical stability is not an issue. Over the reals, under round-off errors, one must contend with numerical
errors without blowing up the bit complexity. This is a formidable challenge; indeed, as mentioned earlier, with exact arithmetic, the CG method takes time $O\left(n·nnz\right)$, but this is
misleading since the computation is effective only when the word sizes are increased by a factor of $n$ (to about $n log \kappa$ words), which leads to an overall complexity of $O\left({n}^{2}·nnz·
log \kappa \right)$.
Our Contributions
Our algorithm can be viewed as the numerical generalization of the algorithms from.^5 We work with real numbers of bounded precision, instead of entries over a finite field. The core of our approach
can be summarized as follows.
Doing so requires developing tools for two topics that have been extensively studied in mathematics, doing so requires separately developing:
1. Obtain low numerical cost solvers for block Hankel/Toeplitz matrices. Many of the prior algorithms rely on algebraic identities that do not generalize to the block setting, and are often
(experimentally) numerically unstable.^7
2. Develop matrix anti-concentration bounds for analyzing the word lengths of inverses of random Krylov spaces. This is to upper bound the probability of random matrices being in some set of small
measure, which in our case is the set of nearly singular matrices. Previously, such bounds were known assuming the matrix entries are independent^18^,^21 but Krylov matrices have correlated
Before we describe the difficulties and new tools needed, we first provide some intuition on why a factor $m$ increase in word lengths may be the right answer by upper-bounding the magnitudes of
entries in an $m$-step Krylov space. By rescaling, we may assume that the minimum singular value of $A$ is at least $1/\kappa$, and the maximum entry in $A$ is at most $\kappa$. The maximum magnitude
of (entries of) ${A}^{m}b$ is bounded by the maximum magnitude of $A$ to the power of $m$, times a factor corresponding to the number of summands in the matrix product:
$||{A}^{m}b{||}_{\infty }\le {n}^{m}|||A{|||}_{\infty }^{m}\text{ }||b{||}_{\infty }\le {\left(nk\right)}^{m}\text{ }\cdot \text{ }||b{||}_{\infty }\le {\kappa }^{O\left(m\right)}||b{||}_{\infty }.$
Here the last inequality is via the assumption of $\kappa \ge n$. So by forming $K$ via horizontally concatenating ${A}^{i}g$ for sparse Gaussian vectors $g$, we have with high probability that the
maximum magnitude of an entry of $K$, and in turn $AK$, is at most ${\kappa }^{O\left(m\right)}$. In other words, $O\left(m log \kappa \right)$ words in front of the decimal point is sufficient with
high probability.
Should such a bound of $O\left(m log \kappa \right)$ hold for all numbers that arise in the algorithm, including the matrix inversion steps, and the matrix $B$ is sparse with $O\left(n\right)$
entries, the cost of computing the block-Krylov matrices becomes $O\left(m log \kappa ·ms·nnz\right)$, while the cost of the matrix inversion portion encounters an overhead of $O\left(m log \kappa \
right)$, for a total of $\stackrel{~}{O}\left({m}^{2}{s}^{\omega } log \kappa \right)$. In the sparse case of $nnz=O\left(n\right)$, and $n\approx ms$, this becomes:
$O\left({n}^{2}m\mathrm{ log }\kappa +{m}^{2}{s}^{\omega }\mathrm{ log }\kappa \right)=O\left({n}^{2}m\mathrm{ log }\kappa +\frac{{n}^{\omega }}{{m}^{\omega -2}}\mathrm{ log }\kappa \right).$(1)
Due to the gap between ${n}^{2}$ and ${n}^{\omega }$, setting $m$ appropriately gives improvement over ${n}^{\omega }$ when $log \kappa ={n}^{o\left(1\right)}$.
However, the magnitude of an entry in the inverse depends on the smallest magnitude, or in the matrix case, its minimum singular value. Bounding and propagating the min singular value, which
intuitively corresponds to how close a matrix is to being degenerate, represents our main challenge. In exact/finite fields settings, non-degeneracies are certified via the Schwartz-Zippel lemma
about polynomial roots. The numerical analog of this is more difficult — the Krylov space matrix $K$ is asymmetric, even for a symmetric matrix $A$. It is much easier for an asymmetric matrix with
correlated entries to be close to singular.
Consider for example a two-banded, two-block matrix with all diagonal entries set to the same random variable $\alpha$ (see Figure 1):
${A}_{ij}=\left\{\begin{array}{cc}1& \text{if}\text{ }i=j\text{ }\text{and}\text{ }j\le n/2,\\ \alpha & \text{if}\text{ }i=j+1\text{ }\text{and}\text{ }j\le n/2,\\ \alpha & \text{if}\text{ }i=j+1\
text{ }\text{and}\text{ }n/2<j,\\ 2& \text{if}\text{ }i=j+1\text{ }\text{and}\text{ }n/2<j,\\ 0& \text{otherwise}.\end{array}$
The difference between matrix anti-concentration over finite fields and reals: a matrix that is full rank for all $\alpha e 0$, but is always ill conditioned.
In the exact case, this matrix is full rank unless $\alpha =0$, even over finite fields. On the other hand, its minimum singular value is close to 0 for all values of $\alpha$. To see this, it’s
useful to first make the following observation about the min singular value of one of the blocks.
Observation 3.
The minimum singular value of a matrix with 1s on the diagonal, $\alpha$ on the entries immediately below the diagonal, and 0 everywhere else is at most ${|\alpha |}^{–\left(n–1\right)}$, due to the
test vector $\left[1;–\alpha ;{\alpha }^{2};\dots ;{\left(–\alpha \right)}^{n–1}\right]$.
Then, in the top-left block, as long as $|\alpha |>3/2$, the top left block has minimum singular value at most ${\left(2/3\right)}^{n–1}$. On the other hand, rescaling the bottom-right block by $1/\
alpha$ to get 1s on the diagonal gives $2/\alpha$ on the off-diagonal. So as long as $|\alpha |<3/2$, this value is at least $4/3$, which in turn implies a minimum singular value of at most ${\left(3
/4\right)}^{n–1}$ in the bottom right block. This means no matter what value $\alpha$ is set to, this matrix will always have a singular value that’s exponentially close to 0. Furthermore, the Gram
matrix of this matrix also gives such a counter example to symmetric matrices with (non-linearly) correlated entries. Previous works on analyzing condition numbers of asymmetric matrices also
encounter similar difficulties; a more detailed discussion of it can be found in Section 7 of Sankar et al.^18
In order to bound the bit complexity of all intermediate steps of the block Krylov algorithm by $\stackrel{~}{O}\left(m· log \kappa \right)$, we devise a more numerically stable algorithm for solving
block Hankel matrices, as well as provide a new perturbation scheme to quickly generate a well-conditioned block Krylov space. Central to both of our key components is the close connection between
condition number and bit complexity bounds.
First, we give a more numerically stable solver for block Hankel/Toeplitz matrices. Fast solvers for Hankel (and closely related Toeplitz) matrices have been extensively studied in numerical
analysis, with several recent developments on more stable algorithms.^24 However, the notion of numerical stability studied in these algorithms is the variant where the number of bits of precision is
fixed. Our attempts at converting these into asymptotic bounds yielded dependencies quadratic in the number of digits in the condition number, which in our setting translates to a prohibitive cost of
$\stackrel{~}{O}\left({m}^{2}\right)$ (i.e., the overall cost would be higher than ${n}^{\omega }$).
Instead, we combine developments in recursive block Gaussian elimination^4 with the low displacement rank representation of Hankel/Toeplitz matrices.^7 Such representations allow us to implicitly
express both the Hankel matrix and its inverse by displaced versions of low-rank matrices. This means the intermediate size of instances arising from recursion is $O\left(s\right)$ times the
dimension, for a total size of $O\left(n log n\right)$, giving a total of $\stackrel{~}{O}\left(n{s}^{\omega –1}\right)$ arithmetic operations involving words of size $\stackrel{~}{O}\left(m\right)$.
We provide a rigorous analysis of the accumulation of round-off errors similar to the analysis of recursive matrix multiplication based matrix inversion from.^4
Motivated by this close connection with the condition number of Hankel matrices, we then try to initialize with Krylov spaces of low condition number. Here we show that a sufficiently small
perturbation suffices for producing a well conditioned overall matrix. In fact, the first step of our proof, showing that a small sparse random perturbation to $A$ guarantees good separations between
its eigenvalues, is a direct combination of bounds on eigenvalue separation of random Gaussians^13 as well as min eigenvalue of random sparse matrices.^11 This separation then ensures that the powers
of $A$, ${A}^{1},{A}^{2},\dots {A}^{m}$, are sufficiently distinguishable from each other. Such considerations also come up in the smoothed analysis of numerical algorithms.^18
The randomness of the Krylov matrix induced by the initial set of random vectors $B$ is more difficult to analyze: each column of $B$ affects $m$ columns of the overall Krylov space matrix. In
contrast, all existing analyses of lower bounds of singular values of possibly asymmetric random matrices^18^,^21 rely on the randomness in the columns of matrices being independent. The dependence
between columns necessitates analyzing singular values of random linear combinations of matrices, which we handle by adapting $ϵ$-net based proofs of anti-concentration bounds. Here we encounter an
additional challenge in bounding the minimum singular value of the block Krylov matrix. We resolve this issue algorithmically: instead of picking a Krylov space that spans the entire ${\Re }^{n}$, we
stop short by picking $ms=n–\stackrel{~}{O}\left(m\right)$ This set of extra columns significantly simplifies the proof of singular value lower bounds. This is similar in spirit to the analysis of
the minimum singular value of a random matrix, which is easier for a non-square matrix.^15 In the algorithm, the remaining columns are treated as a separate block that we handle via a Schur
complement at the very end of the algorithm. Since this block is small, so is its overhead on the running time.
History and Related Work
Our algorithm has close connections with multiple lines of research on efficient solvers for sparse linear systems. The topic of efficiently solving linear systems has been extensively studied in
computer science, applied mathematics and engineering. For example, in the Society of Industrial and Applied Mathematics News’ ‘top 10 algorithms of the 20th century’, three of them (Krylov space
methods, matrix decompositions, and QR factorizations) are directly related to linear systems solvers.^3
At a high level, our algorithm is a hybrid linear systems solver. It combines iterative methods, namely block Krylov space methods, with direct methods that factorize the resulting Gram matrix of the
Krylov space. Hybrid methods have their origins in the incomplete Cholesky method for speeding up elimination/factorization based direct solvers. A main goal of these methods is to reduce the $\Omega
\left({n}^{2}\right)$ space needed to represent matrix factorizations/inverses. This high space requirement is often even more problematic than time requirements when handling large sparse matrices.
Such reductions can occur in two ways: either by directly dropping entries from the (intermediate) matrices, or by providing more succinct representations of these matrices using additional
The main structure of our algorithm is based on the latter line of work on solvers for structured matrices. Such systems arise from physical processes where the interactions between objects have
invariances (e.g., either by time or space differences). Examples of such structure include circulant matrices, Hankel/Toeplitz matrices and distances from $n$-body simulations.^7 Many such
algorithms require exact preservation of the structure in intermediate steps. As a result, many of these works develop algorithms over finite fields.
More recently, there has been work on developing numerically stable variants of these algorithms for structured matrices, or more generally, matrices that are numerically close to being structured.^
24 However, these results only explicitly discussed in the entry-wise Hankel/Toeplitz case (which corresponds to block size $s=1$). Furthermore, because they rely on domain-decomposition techniques
similar to fast multiple methods, they produce one bit of precision for each outer iteration loop. As the Krylov space matrix has condition number $exp\left(\Omega \left(m\right)\right)$, such
methods would lead to another factor of $m$ in the solve cost if invoked directly.
Instead, our techniques for handling and bounding numerical errors are more closely related to recent developments in provably efficient sparse Cholesky factorizations.^9 These methods generate
efficient preconditioners using only the condition that intermediate steps of Gaussian elimination, known as Schur complements, have small representations. They avoid the explicit generation of the
dense representations of Schur complements by treating them as operators, and apply randomized tools to directly sample/sketch the final succinct representations, which have much smaller size and
algorithmic cost.
On the other hand, previous works on sparse Cholesky factorizations required the input matrix to be decomposable into a sum of simple elements, often through additional combinatorial structure of the
matrices. In particular, this line of work on combinatorial preconditioning was initiated through a focus on graph Laplacians, which are built from 2-by-2 matrix blocks corresponding to edges of
undirected graphs.^19 Since then, there have been substantial generalizations to the structures amenable to such approaches, notably to finite element matrices, and directed graphs/irreversible
Markov chains. However, recent works have also shown that many classes of structures involving more than two variables are complete for general linear systems.^25 Nonetheless, the prevalence of
approximation errors in such algorithms led to the development of new ways to bound numerical round-off errors in algorithms, which will be critical to our elimination routine for block-Hankel
Key to recent developments in combinatorial preconditioning is matrix concentration.^22 Such bounds provide guarantees for (relative) eigenvalues of random sums of matrices. For generating
preconditioners, such randomness arises from whether each element is kept or not, and a small condition number (which in turn implies a small number of outer iterations using the preconditioner)
corresponds to a small deviation between the original and sampled matrices. In contrast, we introduce randomness in order to obtain block Krylov spaces whose minimum eigenvalue is large. As a result,
the matrix tool we need is anti-concentration, which somewhat surprisingly is far less studied. Previous works on it are related to similar problems of numerical precision^18^,^21 and address
situations where the entries in the resulting matrix are independent. Our bound on the min singular value of the random Krylov space also yields a crude bound for a sum of rectangular random
Subsequent Improvements and Extensions
Nie^14 gave a more general and tighter version of matrix anti-concentration that also works for square matrices, answering an open question we posed. For an $m$-step Krylov space instantiated using $
\frac{n}{m}$ vectors, Nie’s bound reduces the middle term in our analysis from $O\left({n}^{2}{m}^{3}\right)$ to $O\left({n}^{2}m\right)$, thus leading to a running time of $nnz·n·m+{n}^{\omega }·{m}
^{2–\omega }$, which matches the bound for finite fields. Moreovver, it does so without the padding step at the end. We elect to keep for this article our epsilon-net based analyses, and the padded
algorithm required for it, both for completeness, and due to it being a more elementary approach toward the problem with a simpler proof.
Faster matrix multiplication is an active area of work with recent progresses. Due to the dependence on fast multiplication in our algorithm, such improvements also lead to improvements in the
running time of solving sparse systems as well.
Our main result for solving sparse linear systems has also been extended to solving sparse regression, with faster than matrix multiplication bounds for sufficiently sparse matrices.^6 The complexity
of sparse linear programming remains an interesting open problem.
We describe the algorithm, as well as the running times of its main components in this section. To simplify the discussion, we assume the input matrix $A$ is symmetric, and has $poly\left(n\right)$
condition number. If it is asymmetric (but invertible), we implicitly apply the algorithm to ${A}^{T}A$, using the identity ${A}^{–1}={\left({A}^{T}A\right)}^{–1}{A}^{T}$ derived from ${\left({A}^{T}
A\right)}^{–1}={A}^{–1}{A}^{–T}$. Also, recall from the discussion after Theorem 1 that we use $\stackrel{~}{O}\left(·\right)$ to hide logarithmic terms in order to simplify runtimes.
Before giving details of our algorithm, we first discuss what constitutes a linear system solver algorithm, specifically the equivalence between many such algorithms and linear operators.
For an algorithm $A\mathrm{LG}$ that takes a matrix $B$ as input, we say that $A\mathrm{LG}$ is linear if there is a matrix ${Z}_{A\mathrm{LG}}$ such that for any input $B$, the output of running the
algorithm on $B$ is the same as multiplying $B$ by ${Z}_{A\mathrm{LG}}$:
In this section, in particular in the pseudocode in Algorithm 2, we use the name of the procedure, ${S\mathrm{OLVE}}_{A}\left(b,\delta \right)$, interchangeably with the operator corresponding to a
linear algorithm that solves a system in $A$, on vector $b$, to error $\delta >0$. In the more formal analysis, we will denote such corresponding linear operators using the symbol $Z$, with
subscripts corresponding to the routine if appropriate.
Pseudocode for block Krylov space algorithm: ${\mathrm{Solve}}_{·}\left(·,·\right)$ are operators corresponding to linear system solving algorithms whose formalization we discuss at the start of this
This operator/matrix based analysis of algorithms was first introduced in the analysis of a recursive Chebyshev iteration by Spielman and Teng,^19 with credits to the technique also attributed to
V. Rokhlin. It has the advantage of simplifying the analyis of multiple iterations of such algorithms, as we can directly measure Frobenius norm differences between such operators and the exact ones
that they approximate.
Under this correspondence, the goal of producing an algorithm that solves $Ax=b$ for any $b$ as input becomes equivalent to producing a linear operator ${Z}_{A}$ that approximates ${A}^{–1}$, and
then running it on the input $b$. For convenience, we also let the solver take as input a matrix instead of a vector, in which case the output is the result of solves against each of the columns of
the input matrix as the RHS.
The high-level description of our algorithm is in Figure 2.
Some of the steps of the algorithm require care for efficiency as well as for tracking the number of words needed to represent the numbers. We assume a bound on bit complexity of $\stackrel{~}{O}\
left(m\right)$ when $\kappa =poly\left(n\right)$ in the brief description of costs in the outline of the steps below.
We start by perturbing the input matrix, resulting in a symmetric positive definite matrix where all eigenvalues are separated by ${\alpha }_{A}$. Then we explicitly form a Krylov matrix from a
sparse random Gaussian matrix, see Fig. 3. For any vector $u$, we can compute ${A}^{i}u$ from ${A}^{i–1}u$ via a single matrix-vector multiplication in $A$. So computing each column of $K$ requires
$O\left(nnz\left(A\right)\right)$ operations, each involving a length $n$ vector with words of length $\stackrel{~}{O}\left(m\right)$. So we get the matrix $K$, as well as $AK$, in time
$\stackrel{˜}{O}\left(nnz\left(A\right)\cdot n\cdot m\right).$
Randomized $m$-step Krylov Space Matrix with $n$-by-$s$ sparse Gaussian ${G}^{S}$ as starter.
To obtain a solver for $AK$, we instead solve its Gram matrix ${\left(AK\right)}^{T}\left(AK\right)$. Each block of ${K}^{T}K$ has the form $\left({G}^{S}{\right)}^{T}{A}^{i}{G}^{S}$ for some $2\le i
\le 2m$, and can be computed by multiplying $\left({G}^{S}{\right)}^{T}$ and ${A}^{i}{G}^{S}$. As ${A}^{i}{G}^{S}$ is an $n$-by-$s$ matrix, each non-zero in ${G}^{S}$ leads to a cost of $O\left(s\
right)$ operations involving words of length $\stackrel{~}{O}\left(m\right)$. Then because we chose ${G}^{S}$ to have $\stackrel{~}{O}\left({m}^{3}\right)$ non-zeros per column, the total number of
non-zeros in ${G}^{S}$ is about $\stackrel{~}{O}\left(s·{m}^{3}\right)=\stackrel{~}{O}\left(n{m}^{2}\right)$. This leads to a total cost (across the $m$ values of $i$) of:
The key step is then Step 2, a block version of the Conjugate Gradient method. It will be implemented using a recursive data structure based on the notion of displacement rank.^7 To get a sense of
why a faster algorithm may be possible, note that there are only $O\left(m\right)$ distinct blocks in the matrix ${\left(AK\right)}^{T}\left(AK\right)$. So a natural hope is to invert these blocks by
themselves; the cost of (stable) matrix inversion,^4 times the $\stackrel{~}{O}\left(m\right)$ numerical word complexity, would then give a total of
$\stackrel{˜}{O}\left({m}^{2}{s}^{\omega }\right)=\stackrel{˜}{O}\left({m}^{2}{\left(\frac{n}{m}\right)}^{\omega }\right)=\stackrel{˜}{O}\left({n}^{\omega }{m}^{2-\omega }\right).$
Of course, it does not suffice to solve these $m$ $s$-by-$s$ blocks independently. Instead, the full algorithm, as well as the ${S\mathrm{OLVE}}_{M}$ operator, is built from efficiently convolving
such $s$-by-$s$ blocks with matrices using Fast Fourier Transforms. Such ideas can be traced back to the development of super-fast solvers for (entry-wise) Hankel/Toeplitz matrices.^7
Choosing $s$ and $m$ so that $n=sm$ would then give the overall running time, assuming that we can bound the minimum singular value of $K$ by $exp\left(–\stackrel{~}{O}\left(m\right)\right)$. This is
a shortcoming of our analysis: we can only prove such a bound when $n–sm\ge \Omega \left(m\right)$. The underlying reason is that rectangular semi-random matrices can be analyzed using $ϵ$-nets, and
thus are significantly easier to analyze than square matrices.
This means we can only use $m$ and $s$ such that $n–ms=\Theta \left(m\right)$, and we need to pad $K$ with $n–ms$ columns to guarantee a full rank, invertible, matrix. To this end, we add $\Theta \
left(m\right)$ dense Gaussian columns to $K$ to form $Q$, and solve the system $AQ$, and its associated Gram matrix ${\left(AQ\right)}^{T}\left(AQ\right)$ instead. These matrices are shown in Figure
Full matrix $AQ$ and its Associated Gram Matrix ${\left(AQ\right)}^{T}\left(AQ\right)$. Note that by our choice of parameters $m$ is much smaller than $s\approx n/m$.
Since these additional columns are entry-wise i.i.d, the minimum singular value can be analyzed using existing tools^18^,^21 namely lower bounding the inner product of a random vector against any
vector. Thus, we can lower bound the minimum singular value of $Q$, and in turn $AQ$, by $exp\left(–\stackrel{~}{O}\left(m\right)\right)$.
This bound in turn translates to a lower bound on the minimum eigenvalue of the Gram matrix of $AQ$, ${\left(AQ\right)}^{T}\left(AQ\right)$. Partitioning its entries by those from $K$ and $G$ gives
four blocks: one $\left(sm\right)$-by-$\left(sm\right)$ block corresponding to ${\left(AK\right)}^{T}\left(AK\right)$, one $\Theta \left(m\right)$-by-$\Theta \left(m\right)$ block corresponding to $
{\left(AG\right)}^{T}\left(AG\right)$, and then the cross terms. To solve this matrix, we apply block-Gaussian elimination, or equivalently, form the Schur complement onto the $\Theta \left(m\right)$
-by-$\Theta \left(m\right)$ corresponding to the columns in $AG$.
To compute this Schur complement, it suffices to solve the top-left block (corresponding to ${\left(AK\right)}^{T}\left(AK\right)$) against every column in the cross term. As there are at most $\
Theta \left(m\right)<s$ columns, this solve cost comes out to less than $\stackrel{~}{O}\left({s}^{\omega }m\right)$ as well. We are then left with a $\Theta \left(m\right)$-by-$\Theta \left(m\right)
$ matrix, whose solve cost is a lower order term.
So the final solver costs
$\stackrel{˜}{O}\left(nnz\left(A\right)\cdot nm+{n}^{2}{m}^{3}+{n}^{\omega }{m}^{2-\omega }\right)$
which leads to the final running time by choosing $m$ to balance the terms. This bound falls short of the ideal case given in Equation 1 mainly due to the need for a denser $B$ to the
well-conditionedness of the Krylov space matrix. Instead of $O\left(n\right)$ non-zeros total, or about $O\left(m\right)$ per column, we need $poly\left(m\right)$ non-zero variables per column to
ensure the an $exp\left(–O\left(m\right)\right)$ condition number of the block Krylov space matrix $K$. This in turn leads to a total cost of $O\left(n·nnz·poly\left(m\right)\right)$ for computing
the blocks of the Hankel matrix, and a worse trade off when summed against the $\frac{{n}^{\omega }}{{m}^{\omega –2}}$ term.
Richard Peng was supported in part by NSF CAREER award 1846218/2330255, and Santosh Vempala by NSF awards AF-1909756 and AF-2007443. We thank Mark Giesbrecht for bringing to our attention the
literature on block-Krylov space algorithms.
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2009-10-01 18:33:06 GMT
this post concerns a page "Maths - Reorthogonalising a matrix".
I have some ideas about correction matrix. As C is symmetric, then we can use arbitrary powers of a matrix. To calculate power of a matrix an eigenvalue decomposition can be used. Matrix is written
as Q*D*Q^T, where Q is orthogonal and D is diagonal. Then any function F(that can be expanded in Taylor series) of a matrix can be calculated as Q*F(D)*Q^T, where F can be applied to D elementwise.
If such method is used for computation of the correction matrix, we get exactly orthogonal matrix (except for rounding errors). But it is more computationally expensive then just using of the first
two terms of Taylor expansion, though more precise. I haven't yet compared it to orthogonalization with SVD, but it seems that eigenvalue decomposition is somewhat simpler.
\>>Is this more orthogonal than the original matrix, I can't really tell, it looks as though it might be?
In the given example it seems to be so, because applying reorthogonalization to a matrix several times converges to an orthogonal matrix, and this matrix equals one computed with eigenvalue
decomposition. So, using the first two terms of Taylor expansion may be not so bad if we need fast algorithm.
2009-10-02 15:18:21 GMT
Thanks for this. I don't understand as much of this as I would like, so I appreciate any help.
I am not quite clear why we can set: O=C*M where:
• O is orthogonal: O^t = O^-1
• C is symmetric: C^t = C
• M is almost orthogonal but not quite.
As you say, eigenvalue decomposition looks similar to SVD:
• eigenvalue decomposition: QDQ^T
• SVD: [U][D][V]^t
The difference appears to me to be (at first sight) that SVD is a more general version of eigenvalue decomposition where U is not equal to V?
I guess I need to find out the general rules for these matrices:
• orthogonal*orthogonal=orthogonal
• symmetric*symmetric=symmetric
• orthogonal*symmetric=?
Also why any function F can be applied to Q*F(D)*Q^T
Do you know a good web source for this topic?
2009-10-04 13:35:00 GMT
Sorry, I mistakenly supposed that we need to decompose an orthogonal matrix. So in general case the eigenvalue decomposition is A=QDQ^-1 (But not QDQ^T), where Q is not necessary orthogonal.
I am not quite clear why we can set: O=C*M where:
O is orthogonal: O^t = O^-1
C is symmetric: C^t = C
M is almost orthogonal but not quite.
We just assume that for given M, exists such symmetric C, that C*M gives us some orthogonal matrix. Then we prove that such C exists and find it. Of course, instead of symmetric matrix any other
matrix can be chosen, but it is convenient for us if it is symmetric (otherwise we couldn't be able to use this method). May be some other types will give us interesting results.
As you say, eigenvalue decomposition looks similar to SVD:
eigenvalue decomposition: QDQ^T
SVD: [U][D][V]^t
The difference appears to me to be (at first sight) that SVD is a more general
version of eigenvalue decomposition where U is not equal to V?
Eigenvalue decomposition of a non-orthogonal matrix gives us QDQ^-1. If we decompose an orthogonal matrix, we get QDQ^-1=QDQ^T, where Q is orthogonal. So SVD is the general case of eigenvalue
decomposition only for orthogonal matrices. In our case we have to decompose MM^T, which is not orthogonal.
I guess I need to find out the general rules for these matrices:
I suppose that in the latter case we can get any square matrix.
Also why any function F can be applied to Q*F(D)*Q^T
If we want to calculate the square of matrix A=QDQ^-1, then we get A^2=QD(Q^-1Q)DQ^-1=QDEDQ^-1=QD^2Q^-1. This is true for all other powers: A^n=QD^nQ^-1, where integer n>=0. If we define function of
a matrix through Taylor series, then F(A)=a0E+a1A+a2A^2+...=QEQ^-1+QDQ^-1+QD^2Q^-1+...=Q(E+D+D^2+...)Q^-1=QF(D)Q^-1. This is from wikipedia, section Functional calculus.
But this is not exactly what we need. We need to find such C that (MM^T)^-1=C^2. Let us decompose MM^T=QDQ^-1, then QD^-1Q-1=C^2, (QD^-0.5Q^-1)(QD^-0.5Q^-1)=C^2. So, we can choose C=QD^-0.5Q^-1. We
can raise D to power -0.5, because MM^T has real nonnegative eigenvalues, and as M is almost orthogonal, so it cannot have zero eigenvalues.
Do you know a good web source for this topic?
I can only suggest wikipedia and wolfram mathworld | {"url":"http://euclideanspace.com/maths/algebra/matrix/orthogonal/reorthogonalising/hauke.htm","timestamp":"2024-11-03T13:35:10Z","content_type":"text/html","content_length":"30283","record_id":"<urn:uuid:ac463f08-7356-45d2-ad25-f60d9f33a54e>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00336.warc.gz"} |
Instant Ticket Algorithm
So a nerdy question here,
You have an instant ticket game. The game has two "winning Player numbers" from 1-20 and you have to match a like number or multiple matches in a field of 10 numbers with the same range.
Example: Players Numbers 14 7
18 ($5) 20 ($100) 3 ($2500) 9 ($500) 10 ($50)
17 ($25) 4 ($1) 19 ($1000) 5 ($3) 12 ($2)
The game has predetermined winners in the entire game (a seeded outcome)
So the game would have a fixed number of prizes predetermined for the entire game.
4 $50,000
10 $5000
20 $500
200 $100
500 $50
1000 $25
2000 $5
5000 $2
How would one create an algorithm say in excel to achieve this?
Don't overlook the importance of including $0 prizes in your lottery.
May the cards fall in your favor.
Here's one way to do it:
For each ticket, generate a set of 12 different numbers from 1-20, and randomize the order. The first 10 are the prize numbers, and the last two are the player numbers.
Also, generate a random order of the 10 prize amounts for each ticket, then assign the first amount to the first number, the second amount to the second number, and so on.
Once you have done that, start with the $50,000 prizes; take the first four tickets, and change one of the two player numbers to that ticket's $50,000 number.
Do the same thing with the ten $5000 prizes, then the 20 $500 prizes, and so on.
Since the first 12 numbers are different, this guarantees that any ticket that didn't have either of its player numbers set to a winning number will be a losing ticket.
Quote: ThatDonGuy
Once you have done that, start with the $50,000 prizes; take the first four tickets, and change one of the two player numbers to that ticket's $50,000 number.
It's a good basic way to do it... of course actual lottery scratch offs are more complicated. For example, a $50 winner card might be generated by having 2 different $25 winning spots.
The Game I'm referring to is this: ml
I don't know if they pre populate a table of winning outcome tickets and then just generate the losing tickets as a conditional if statement that selects a 0 or 1. O being losing tickets and if 1
then select from winning ticket table.
The other issue is there seems to be a near miss feature built into the game so that when you scratch a "13" for example the field of number would contain a "10" or a "1" to give the player a sense
that a winning number could be revealed.
Also I read an article that states that the designers batch the tickets and then randomize within the batches as to insure an equal distribution of a variety of prizes within each batch probably
weighted toward the ends of the books of tickets as to allow the retailer to collect the funds in order to fund the prizes. | {"url":"https://wizardofvegas.com/forum/gambling/other-games/36082-instant-ticket-algorithm/","timestamp":"2024-11-11T21:30:53Z","content_type":"text/html","content_length":"50246","record_id":"<urn:uuid:eaaa72de-6e99-4440-b1f5-1913c0c2777f>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00196.warc.gz"} |
What is the Place Value Chart of an Indian and International System - A Plus Topper
What is the Place Value Chart of an Indian and International System
There are two systems of reading and writing numbers: The Indian system and The International system of numeration.
1. Indian system of numeration
In the Indian system of numeration, starting from the right, the first period is ones, consisting of three place values (ones, tens, and hundreds). The next period is thousands, consisting of two
place values (thousands and ten thousands). The third period from the right is lakhs, consisting of two place values (lakhs and ten lakhs), and then crores and so on. This system of numeration is
also known as the Hindu-Arabic system of numeration. We use commas’ for separating the periods, which help us in reading and writing large numbers. In the Indian system, the first comma comes after
three digits from the right (i.e., after ones period) and the next comma comes after the next two digits {i.e., after thousands period) and then after every two digits and so on.
Indian Place value chart
2. International system of numerationLet us consider an example:
In the Indian system of numeration,
92357385 =9,23,57,385
Similarly, 2930625 in the Indian system of numeration will be written as 29,30,625.
In the International system of numeration, starting from right, the first period is ones, consisting of three place values (ones, tens, and hundreds). The next period is thousands, consisting of
three place values (one thousand, ten thousands, and hundred thousands) and then millions and after that billions.
International Place value chart
Note:In International system of numeration all the periods have three place values each.
Since each period has three place values, so to write a number with the help of comma(s), we have to put a comma after every three digits from the right. For example, 275068142 will be written in the
International system as 275,068,142.
Similarly, 925371852 will be written as 925,371,852.
First three places from the right are same in both the Indian and the International systems of numeration.
Example 1: Rewrite the following numbers in the Indian and International systems using commas (,):
(a) 74028952 (b) 1835762
(a) 74028952
Indian system: 7,40,28,952 International system: 74,028,952
(b) 1835762
Indian system: 18,35,762 International system: 1,835,762
Example 2: Rewrite the following numbers in the International place value chart:
(i) 6432156 (ii) 87201593
What is the meaning of Place Value and Face Value in Maths?Read More About: | {"url":"https://www.aplustopper.com/place-value-chart-indian-international-system/","timestamp":"2024-11-10T11:25:33Z","content_type":"text/html","content_length":"48048","record_id":"<urn:uuid:58c1bfec-768f-4004-a8b5-0db3a3509c80>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00016.warc.gz"} |
Calculate the distance between cities with ve7pro.com
How do you calculate the distance between cities?
Knowing the distance between two cities is important for many reasons - from planning a road trip, to understanding how far away a city is from your own. Thankfully, there are now several tools
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How do I measure distances using a distance calculator?
Knowing the distance between two cities can be essential for planning a trip or even just to get an idea of how far apart they are. With the help of a distance calculator, you can easily calculate
the distance between two cities with just a few clicks.
Whether you need to measure distances for business purposes or personal travel, a city distance calculator is an invaluable tool. You can use it to find out the shortest route between two cities,
calculate exact distances and even plan your journey accordingly. With this tool, you will be able to accurately measure distances and save time in the process.
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Excel YIELD Function
What Is YIELD Function In Excel?
The YIELD function in Excel belongs to the category of Financial function. It determines the yield on security, paying periodic interest. Users can apply the Excel YIELD function for evaluating a
bond yield, which helps calculate the income generated in a year as the potential return.
For example, the table below contains the inputs required to determine a security’s yield (in percentage).
Suppose the requirement is to show the yield value in cell B12. Then considering the Excel YIELD function definition, applying the YIELD() in the target cell can help us achieve the required data.
In the above Excel YIELD function example, the function returns the yield value as a decimal number, 0.04877, in the target cell B12. But once we set the cell B12 data format as Percentage, the
decimal value converts into a percentage value, 4.88%.
Key Takeaways
• The Excel YIELD function evaluates the yield on a security that pays periodic interest. The function returns a decimal value. We can set the target cell data format, containing the YIELD(), as
Percentage to get the required yield as a percentage value.
• Users can utilize the YIELD() to determine a bond yield and evaluate the annual income generated.
• The YIELD() accepts six mandatory arguments, settlement, maturity, rate, pr, redemption, and frequency, and one optional argument, basis, to calculate the yield.
• The coupon payment frequency can be annual, semi-annual, or quarterly. On the other hand, we can supply five types of day count basis to the YIELD().
YIELD() Excel Formula
The Excel YIELD formula is:
• settlement: It is the security’s settlement date, falling after the issue date. And that is when the security gets traded to the buyer.
• maturity: It is the security’s maturity date, i.e., the date when the security expires.
• rate: The annual coupon rate of the security.
• pr: The cost of the security per $100 face value.
• redemption: The redemption amount of the security per $100 face value.
• frequency: The count of coupon payments per year.
• basis: It indicates the day count basis type used by the security.
The first six arguments in the Excel YIELD formula are mandatory. On the other hand, the last argument is optional, and the function considers its default value as 0.
And the following tables show the frequency and basis argument values we can use when applying the Excel YIELD function.
Furthermore, below are a few critical aspects that we must consider to avoid all conditions of the Excel YIELD function not working.
• Ensure to enter the date values using the DATE Excel function to avoid entering the date as a text value. And we can supply the date arguments to the Excel YIELD function as cell references to
dates or functions or formulas returning date values.
• The arguments, frequency, and basis get truncated to integers. And suppose the arguments settlement or maturity are serial numbers (date equivalent values), but not integer values. Then, they
also get truncated.
• If settlement or maturity is an invalid date or any of the supplied arguments are non-numeric, the YIELD() returns the #VALUE! error.
• The YIELD() return value will be the #NUM! error in the following conditions:
□ settlement >= maturity
□ rate < 0
□ pr <= 0 or redemption <= 0
□ frequency is not 1, 2, or 4
□ basis is not 0, 1, 2, 3, or 4
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How To Use YIELD Excel Function?
The steps to apply the Excel YIELD function are:
1. First, ensure the inputs for determining a security’s yield are accurate and in the required data formats.
2. Then, select the target cell and enter the Excel YIELD function.
3. Press Enter to view the required security yield.
4. Finally, select the target cell and set the data format as Percentage using the Number Format option in the Home tab to view the calculated security yield as a percentage.
Check out the following example to understand the Excel YIELD function definition and the steps to use the YIELD().
The below table contains information about the security scheme for which we need to determine the yield.
And suppose we require to show the yield value in cell C13. Then here is how we can apply the YIELD() in the target cell and achieve the desired result.
1. When we check the input data, we find that the first two arguments of the YIELD() are dates, and the third argument is a percentage value. On the other hand, the fourth and fifth arguments are
currency values, and the last two are numbers. So, all the arguments are in the required data formats, thus avoiding the condition of the Excel YIELD function not working due to incorrect data
2. Select the target cell C13, enter the following YIELD(), and press Enter.
In this Excel YIELD function example, the number of coupon payments per year is semi-annual or half-yearly. And hence the frequency argument value is 2. And as the day count basis type is the
default value, the basis argument value is 0.
Thus, based on the supplied argument values, the YIELD() returns the required security yield value as 0.0887.
We can also provide the argument values directly to the YIELD() to achieve the above output.
Ideally, the suggestion is to supply the cell reference to the date values. But when providing the argument values directly, it is best to use the DATE() inside the YIELD() to avoid entering
invalid dates.
Once we enter the formula in the target cell C13 until the argument redemption, separated by commas, we will get a drop-down list. We can use it to enter the frequency argument value by
double-clicking the required option. Then, after we provide the frequency argument value and enter a comma, we will see another drop-down list to supply the basis argument value. Double-click on
the required day count basis type, and we can finally close the parenthesis and press Enter to execute the YIELD().
Alternatively, we can apply the YIELD() from the Formulas tab by selecting the target cell C13 and clicking Formulas → Financial → YIELD. This step will open the Function Arguments window.
Next, we must enter the YIELD() arguments in the Function Arguments window.
And once we click OK in the Function Arguments window, the YIELD() gets executed in the target cell C13.
3. Select cell C13 and set the data format as Percentage using the Number Format option in the Home tab.
Thus, we get the required yield of the specific security scheme as a percentage value of 8.87%.
This section will show a few more examples of the Excel YIELD function.
Example #1
This example explains how to utilize the Excel YIELD function to determine the yield of a ten-year bond.
The table below contains the inputs required to calculate the ten-year bond yield.
Consider the specific bond’s issue date was 1/1/2010, and we purchased it six months later. So, the settlement date is 7/1/2010, and the maturity date is 1/1/2020. Also, assume the coupon payment
frequency is annual and the day count basis is the default type.
And suppose we must display the resulting yield value in cell B13. Then we can use Excel YIELD function in the target cell to get the required outcome.
• Step 1: Select the target cell B13, enter the YIELD() provided in the Formula Bar in the below image, and press Enter.
• Step 2: Select cell B13 and set its data format as Percentage using the Number Format option in the Home tab.
Thus, for the given input values, the YIELD() returns the value 0.108466. And once we set the target cell data format as Percentage, you get the ten-year bond yield, 10.85%.
Example #2
Let us see the process of calculating a security’s yield with a quarterly coupon payment frequency using Excel YIELD function.
The following table contains the inputs required to find the yield of a security a person purchased on 3/15/2021, with an annual interest rate of 7%.
And suppose the requirement is to display the calculated yield value in cell B13. Then applying the Excel YIELD function in the target cell can fetch the required data.
• Step 1: Select the target cell B13, enter the YIELD() provided in the Formula Bar in the below image, and press Enter.
• Step 2: Select cell B13 and set its data format as Percentage using the Number Format option in the Home tab.
In this example, the coupon payment frequency is quarterly. And thus, the frequency argument value is 4. Also, the argument basis takes the default value, 0.
And thus, based on the given input, the yield of the specific security is 6.86%.
Example #3
Let us see the different scenarios when the Excel YIELD function returns error values.
Suppose we need to enter the YIELD() argument values in the table below, calculate the specific security yield, and display the output in cell B13.
So, we will enter the YIELD(),with the required inputs supplied, in the target cell B13. However, the Excel YIELD function might throw errors if the source data contains incorrect values.
• settlement or maturity – Invalid Date
Here, the maturity date is invalid; it is 11/31/2024 instead of 11/30/2024, leading to the YIELD() returning the #VALUE! error.
• YIELD() Argument Values – Non-numeric
In this case, the basis argument value should be 0, but the supplied value is “NIL”, which makes the YIELD() throw the #VALUE! error.
• settlement Greater than or Equal To maturity
The above data shows that the settlement date falls after maturity, causing the YIELD() to return the #NUM! error.
In this case, the rate argument value is a negative percentage value, leading to the YIELD() throwing an error.
• pr or redemption Is Less or Equal To 0
Here, the YIELD() output is the #NUM! error as the pr argument value is 0 instead of a value greater than zero.
• frequency Is Not 1, 2, or 4
The frequency argument value can be 1, 2, or 4. However, as per the given input, the argument frequency is 3, and thus the YIELD() output is an error value.
• basis Is Not 0, 1, 2, 3, or 4
The supplied basis argument value is 6, which is an incorrect value. The argument basis can be a value from 0 to 4.
Thus, the YIELD() will return the required value when we resolve the above issues.
Important Things To Note
• It is best to supply the date arguments to the Excel YIELD function as cell references to dates or formulas returning dates. And if entering the date values directly in the YIELD(), use the DATE
() to avoid supplying invalid date values.
• For invalid settlement or maturity dates or non-numeric argument values, the YIELD() throws the #VALUE! error.
• Suppose settlement is greater or equal to maturity, the rate is less than 0, pr or redemption <= 0, frequency is not 1, 2, or 4, or basis is not 0, 1, 2, 3, or 4. Then the YIELD() return value
will be the #NUM! error.
Frequently Asked Questions (FAQs)
1. Where is the YIELD function in Excel?
The YIELD function in Excel is in the Formulas tab. We can select Formulas → Financial → YIELD to access the function.
2. What is the settlement in the YIELD function in Excel?
The settlement in the YIELD function in Excel indicates the date a buyer trades a coupon, such as a security or bond. For example, a 10-year bond issue date is 1/1/2010, and a buyer purchases it four
months later. Then the settlement date will be 5/1/2010.
3. How can you determine the yield of a security with a quarterly coupon payment frequency and Actual/actual basis type using the Excel YIELD function?
We can determine the yield of a security with a quarterly coupon payment frequency and Actual/actual basis type using the Excel YIELD function in the following way.
Let us see the steps with an example.
The below table contains the required input for calculating the yield of a specific security.
In this example, the coupon payment frequency is quarterly. So, the frequency argument value is 4. And the day count basis type is Actual/actual. Hence, the basis argument value is 1.
Suppose the requirement is to show the calculated yield value in cell B13. Then, here is how we can apply the Excel YIELD function in the target cell and get the yield value.
• Step 1: Select the target cell B13, enter the YIELD() provided in the Formula Bar in the below image, and press Enter.
• Step 2: Select cell B13 and set the data format as Percentage using the Number Format option in the Home tab.
Thus, the YIELD() evaluates the yield of a security, with quarterly coupon payment frequency and Actual/actual day count basis type, as 3.86%.
Download Template
This article must be helpful to understand the Excel YIELD Function, with its formula and examples. You can download the template here to use it instantly.
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A beginner- friendly guide to understanding Machine learning concept using python - Meritshot
What is Python?
Python is the most used high-level was developed by Guido van Rossum and released first on February 20, 1991, It is interpreted programming language known for its readability and clear syntax. It
provides various libraries and frameworks that simplify machine learning development.
Python’s versatility and active community make it an ideal language for machine-learning projects and supports object-oriented programming, most commonly used to perform general-purpose programming.
Python is used in several domains like Data Science, Machine Learning, Deep Learning, Artificial Intelligence, Networking, Game Development, Web Development, Web Scraping, and various other domains.
What is Machine Learning?
Machine Learning is the field of study that gives computers the capability to learn without being explicitly programmed. ML is one of the most exciting technologies that one has ever come across. As
it is evident from the name, it gives the computer something that makes it more similar to humans: The ability to learn. Machine learning is actively being used today, perhaps in many more places
than one would expect.
Definition of Machine Learning
Machine learning is a subset of artificial intelligence that enables computer systems to learn from data and improve their performance without being explicitly programmed. It involves training
algorithms using data to build models that can make predictions, recognize patterns, or classify objects.
The goal of machine learning is to develop systems that can adapt and generalize from data, providing solutions to complex problems across various domains.
its simplicity, versatility, and extensive library support. Today, Python is considered one of the primary languages for machine learning and artificial intelligence research and development.
Python’s Role in Machine Learning
Python has a crucial role in machine learning because Python provides libraries like NumPy, Pandas, Scikit-learn, TensorFlow, and Keras. These libraries offer tools and functions essential for data
manipulation, analysis, and building machine learning models. It is well-known for its readability and offers platform independence. These all things make it the perfect language of choice for
Machine Learning.
Why do we need Machine Learning?
Machine Learning today has all the attention it needs. Machine Learning can automate many tasks, especially the ones that only humans can perform with their innate intelligence. Replicating this
intelligence to machines can be achieved only with the help of machine learning.
With the help of Machine Learning, businesses can automate routine tasks. It also helps in automating and quickly create models for data analysis. Various industries depend on vast quantities of data
to optimize their operations and make intelligent decisions. Machine Learning helps in creating models that can process and analyze large amounts of complex data to deliver accurate results. These
models are precise and scalable and function with less turnaround time. By building such precise Machine Learning models, businesses can leverage profitable opportunities and avoid unknown risks.
The answer is more complicated than a simple yes or no.
In many ways, Python is not the ideal tool for the job. In nearly every
instance, the data that machine learning is used for is massive. Python’s lower speed means it can’t handle enormous volumes of data fast enough for a professional setting.
Machine learning is a subset of data science, and Python was not designed with data science in mind.
However, Python’s greatest strength is its versatility. There are hundreds of libraries available with a simple download, each of which allow developers to adapt their code to nearly any problem.
While vanilla Python is not especially adapted to machine learning, it can be very easily modified to make writing machine learning algorithms much simpler.
History of machine learning and its relation to python
Machine learning has a rich history that spans several decades, and its development has been closely tied to advancements in computing and technology. Here’s a brief overview of the history of
machine learning and its relationship with Python:
1. 1950s – 1960s: Early Concepts and Foundations:
• The foundations of machine learning were laid with the work of researchers such as Alan Turing and Marvin Minsky. Early concepts like the perceptron were introduced during this
2. 1970s – 1980s: AI Winter and Rule-Based Systems:
• The field faced skepticism and funding challenges during the “AI winter.” Rule- based expert systems became popular as a way to approach artificial
3. 1990s: Emergence of Neural Networks:
• Neural networks gained attention, with the backpropagation algorithm proving to be a crucial However, computing power limitations hampered the training of large neural networks.
4. Late 1990s – Early 2000s: Support Vector Machines and Boosting:
• Support Vector Machines (SVM) and boosting algorithms gained popularity as powerful machine learning techniques. These methods were widely used in various applications, including image
recognition and natural language
5. Mid-2000s: Rise of Big Data and Python’s Emergence:
• The explosion of big data highlighted the limitations of traditional machine learning algorithms. Around the same time, Python gained popularity as a programming language due to its simplicity
6. 2010s: Deep Learning and Python Dominance:
• Deep learning, particularly deep neural networks, gained prominence with breakthroughs in image and speech recognition. Frameworks like TensorFlow and PyTorch, both of which are primarily
Python-based, played a crucial role in popularizing deep
7. Present: Python as the Dominant Language:
• Python has become the de facto language for machine learning and data Its extensive libraries and frameworks, including NumPy, SciPy, scikit- learn, and the aforementioned TensorFlow and PyTorch,
have made it the preferred choice for researchers and practitioners in the field.
8. Future Trends: Continued Integration and Ethical Considerations:
• The integration of machine learning into various industries is expected to continue, with a focus on interpretability, fairness, and ethics. Continued advancements in hardware, algorithms, and
interdisciplinary collaborations are likely to shape the future of machine learning.
Python’s popularity in the field of machine learning can be attributed to several factors:
1.Ease of Learning and Readability:
• Python is known for its simple and readable syntax, making it easy for beginners to learn and This is crucial for the machine learning community, which includes both experts and those new to the
2. Extensive Libraries and Frameworks:
• Python boasts a rich ecosystem of libraries and frameworks that are specifically designed for machine learning and data Some of the most popular ones include:
□ NumPy and pandas: For numerical and data
□ Scikit-learn: For classical machine learning
□ TensorFlow and PyTorch: Deep learning frameworks that have gained widespread
□ Keras A high-level neural networks API that can run on top of TensorFlow or Theano
□ Matplotlib and Seaborn: For data
3. Community Support:
• Python has a large and active community, which means that developers have access to a wealth of resources, tutorials, and forums for support. This community-driven nature contributes to the rapid
development and improvement of machine learning tools and
4. Flexibility and Versatility:
• Python is a versatile language that can be used for a wide range of Its flexibility allows researchers and developers to seamlessly integrate machine learning into existing workflows or
5. Open Source:
• Python and many of its machine learning libraries are open This means that anyone can view, modify, and distribute the source code, fostering collaboration and innovation within the community.
6. Integration with Other Technologies
• Python can easily integrate with other technologies and tools commonly used in the data science and machine learning ecosystem, such as databases, big data technologies (e.g., Apache Spark), and
7. Industry Adoption:
• Many companies and organizations have adopted Python as their primary language for machine learning and data This industry adoption has further fueled the popularity of Python in these domains.
8. Job Market and Career Opportunities:
• The demand for professionals with expertise in machine learning and Python is high. Learning Python can open up numerous career opportunities in the rapidly growing field of artificial
Getting started with python for machine learning
1. Install Python:
Make sure you have Python installed on your machine. You can download the latest version from the official Python website.
2. Choose an Integrated Development Environment (IDE):
Select an IDE to write and run your Python code. Some popular choices for machine learning include Jupyter Notebooks, VSCode, and PyCharm. Jupyter Notebooks are particularly popular for data
exploration and visualization.
3. Install Necessary Libraries:
Python has several powerful libraries for machine learning. You’ll commonly use:
• NumPy: For numerical
• Pandas: For data manipulation and
• Matplotlib and Seaborn: For data
• Scikit-learn: For machine learning algorithms and
• TensorFlow or PyTorch: For deep
You can install these libraries using the following command in your terminal or command prompt:
bashCopy code
pip install numpy pandas matplotlib seaborn scikit-learn tensorflow
4. Learn the Basics of Python:
Familiarize yourself with basic Python concepts such as variables, data types, control structures (if statements, loops), functions, and error handling. There are plenty of online tutorials and
resources to help you with this.
5. Understand NumPy and Pandas:
NumPy and Pandas are fundamental for data manipulation. Learn about arrays, matrices, data frames, and basic operations in these libraries.
6. Explore Data Visualization:
Matplotlib and Seaborn are widely used for data visualization. Learn how to create various types of plots to understand your data better.
7. Dive into Scikit-learn:
Scikit-learn provides a wide range of machine learning algorithms and tools. Start with simple models like linear regression and gradually move on to more complex ones like decision trees and support
vector machines.
8. Learn about TensorFlow or PyTorch:
If you’re interested in deep learning, choose either TensorFlow or PyTorch. Both have extensive documentation and tutorials. Start with simple neural networks and gradually explore more advanced
9. Work on Real Projects:
Apply what you’ve learned by working on real projects. Kaggle is a great platform where you can find datasets and participate in competitions to enhance your skills.
10. Stay Updated:
The field of machine learning is dynamic. Follow blogs, research papers, and communities like Stack Overflow, Reddit, or specialized forums to stay updated on the latest developments.
Machine Learning Algorithms in Python
1. Linear regression
2. Decision tree
3. Logistic regression
4. Support Vector Machines (SVM)
5. Naive Bayes
Which are the 5 most used machine learning algorithms?
1. Linear regression
It is one of the most popular Supervised Python Machine Learning algorithms that maintains an observation of continuous features and based on it, predicts an outcome. It establishes a relationship
between dependent and independent variables by fitting a best line. This best fit line is represented by a linear equation Y=a*X+b, commonly called the regression line.
In this equation,
Y – Dependent variable a- Slope
X – Independent variable b- Intercept
The regression line is the line that fits best in the equation to supply a relationship between the dependent and independent variables. When it runs on a single variable or feature, we call it
simple linear regression and when it runs on different variables, we call it multiple linear regression. This is often used to estimate the cost of houses, total sales or total number of calls based
on continuous variables.
2. Decision Trees
A decision tree is built by repeatedly asking questions to the partition data. The aim of the decision tree algorithm is to increase the predictiveness at each level of partitioning so that the model
is always updated with information about the dataset.
Even though it is a Supervised Machine Learning algorithm, it is used mainly
for classification rather than regression. In a nutshell, the model takes a particular instance, traverses the decision tree by comparing important features with a conditional statement. As it
descends to the left child branch or right child branch of the tree, depending on the result, the features that are more important are closer to the root. The good part about this machine learning
algorithm is that it works on both continuous dependent and categorical variables.
3. Logistic regression
A supervised machine learning algorithm in Python that is used in estimating discrete values in binary, e.g: 0/1, yes/no, true/false. This is based on a set of independent variables. This algorithm
is used to predict the probability of an event’s occurrence by fitting that data into a logistic curve or logistic function. This is why it is also called logistic regression.
Logistic regression, also called as Sigmoid function, takes in any real valued number and then maps it to a value that falls between 0 and 1. This algorithm finds its use in finding spam emails,
website or ad click predictions and customer churn. Check out this Prediction project using python.
Sigmoid Function is defined as,
f(x) = L / 1+e^(-x)
x: domain of real numbers L: curve’s max value
4. Support Vector Machines (SVM)
This is one of the most important machine learning algorithms in Python which is mainly used for classification but can also be used for regression tasks. In this algorithm, each data item is plotted
as a point in n-dimensional space, where n denotes the number of features you have, with the value of each feature as the value of a particular coordinate.
SVM does the distinction of these classes by a decision boundary. For e.g: If length and width are used to classify different cells, their observations are plotted in a 2D space and a line serves the
purpose of a decision boundary. If you use 3 features, your decision boundary is a plane in a 3D space. SVM is highly effective in cases where the number of dimensions exceeds the number of samples.
5. Naive Bayes
Naive Bayes is a supervised machine learning algorithm used for classification tasks. This is one of the reasons it is also called a Naive Bayes Classifier. It assumes that features are independent
of one another and there exists no correlation between them. But as these assumptions hold no truth in real life, this algorithm is called ‘naive’.
This algorithm works on Bayes’ theorem which is:
p(A|B) = p(A) . p(B|A) / p(B)
In this,
p(A): Probability of event A p(B): Probability of event B
p(A|B): Probability of event A given event B has already occurred p(B|A): Probability of event B given event A has already occurred
The Naive bayes classifier calculates the probability of a class in a given set of features, p( yi I x1, x2, x3,…xn). As this is put into the Bayes’ theorem, we get :
p( yi I x1, x2…xn)= p(x1,x2,…xn I yi). p(yi) / p(x1, x2….xn)
As the Naive Bayes’ algorithm assumes that features are independent, p( x1, x2…xn I yi) can be written as :
p(x1, x2,….xn I yi) = p(x1 I yi) . p(x2 I yi)…p(xn I yi)
p(x1 I yi) is the conditional probability for a single feature and can be easily estimated from the data. Let’s say there are 5 classes and 10 features, 50 probability distributions need to be
stored. Adding all these, it becomes easier to calculate the probability to observe a class given the values of features (p(yi I x1,x2,…xn)).
Python Machine Learning Best Practices and Techniques
In this section, we will explore some best practices and techniques that can help you improve the performance of your machine learning models, optimize your code, and ensure the success of your
Python ML projects.
Data Preprocessing and Feature Engineering
Data preprocessing and feature engineering play a crucial role in the success of your machine learning models. Some common preprocessing steps and techniques include:
• Handling missing data: Use techniques like imputation, interpolation, or deletion to handle missing data in your
• Feature scaling: Standardize or normalize your features to ensure that they are on the same scale, as some algorithms are sensitive to the scale of input features.
• Categorical data encoding: Convert categorical data into numerical format using techniques like one-hot encoding or label
• Feature selection: Identify and select the most important features that contribute to the target variable using techniques like recursive feature elimination, feature importance from tree-based
models, or correlation
• Feature transformation: Apply transformations like log, square root, or power to your features to improve their distribution or relationship with the target variable.
Model Evaluation and Validation
Proper model evaluation and validation are essential to ensure the reliability and generalization of your machine learning models. Some best practices include:
• Train-test split: Split your dataset into separate training and testing sets to evaluate the performance of your model on unseen
• Cross-validation: Use techniques like k-fold cross-validation to train and test your model on different subsets of the data, reducing the risk of
• Performance metrics: Choose appropriate performance metrics, such as accuracy, precision, recall, F1-score, or mean squared error, depending on your problem
• Learning curves: Analyze learning curves to diagnose issues like underfitting, overfitting, or insufficient training
Hyperparameter Tuning
Hyperparameter tuning is the process of finding the optimal set of hyperparameters for your machine learning models. Some techniques for hyperparameter tuning include:
• Grid search: Perform an exhaustive search over a specified range of hyperparameter values to find the best
• Random search: Sample random combinations of hyperparameter values from a specified range to find the best
• Bayesian optimization: Use a probabilistic model to select the most promising hyperparameter values based on previous
Efficient Python Code for Machine Learning
Writing efficient Python code can significantly improve the performance of your machine learning projects. Some tips for writing efficient code include:
• Use vectorized operations: When working with NumPy arrays or pandas DataFrames, use vectorized operations instead of loops for better
• Utilize efficient data structures: Choose appropriate data structures like lists, dictionaries, or sets to optimize the performance of your
• Parallelize computations: Use libraries like joblib or Dask to parallelize your computations and take advantage of multi-core processors.
• Profiling and optimization: Use profiling tools like cProfile or Py-Spy to identify performance bottlenecks in your code and optimize them
By following these best practices and techniques, you can improve the performance of your machine learning models, optimize your Python code, and ensure the success of your ML projects. Continuously
learning and staying updated with the latest advancements in the field will further enhance your skills and contribute to your growth as a machine learning practitioner.
• Python is widely used for machine learning projects and applications across various domains. Here’s an example of a real-world Python machine learning project:
Project Title: Predictive Maintenance for Industrial Equipment
Objective: Develop a machine learning model to predict equipment failures in an industrial setting, enabling proactive maintenance and minimizing downtime.
Steps and Components:
1. Data Collection:
□ Gather historical data from sensors attached to industrial equipment, including temperature, pressure, vibration, and other relevant parameters.
□ Collect information on past equipment failures, maintenance records, and environmental
2. Data Preprocessing:
• Clean and preprocess the data, handling missing values and
• Normalize or scale numerical
• Engineer new features, such as rolling averages or time-based
3. Feature Selection:
• Identify the most important features for predicting equipment
• Use techniques like feature importance from tree-based models or correlation
4. Model Selection:
• Choose appropriate machine learning algorithms for classification, such as Random Forests, Support Vector Machines, or Neural
• Experiment with different models and hyperparameter tuning for optimal performance.
5. Training and Validation:
• Split the data into training and validation
• Train the machine learning model on historical data and validate its performance on a separate
6. Model Evaluation:
• Evaluate the model’s performance using metrics like precision, recall, F1 score, and area under the ROC
• Use confusion matrices to understand false positives and false
7. Deployment:
• Deploy the trained model to a production environment, integrating it with the existing infrastructure.
• Implement a system for real-time prediction or batch processing, depending on the
8. Monitoring and Maintenance:
• Set up monitoring to track the model’s performance over
• Implement a feedback loop to continuously update the model based on new data and improve its accuracy.
9. Integration with Maintenance Workflow:
• Integrate the predictive maintenance system with the organization’s maintenance
• Generate alerts or work orders for maintenance personnel when the model predicts an imminent
10. Documentation:
• Document the entire process, from data collection to deployment, for future reference and collaboration.
This project showcases the application of machine learning in a practical setting, helping businesses optimize their maintenance processes and reduce operational costs.
Resources to Learn Python Machine Learning and AI
In this final section, we will provide a list of resources that can help you learn and master Python machine learning and AI. These resources cover different aspects of machine learning, such as
algorithms, tools, libraries, and real- world applications, catering to various learning preferences and skill levels.
1. Books
• Hands-On Machine Learning with Scikit-Learn, Keras, and TensorFlow” by Aurélien Géron.
• “Python Machine Learning” by Sebastian Raschka and Vahid
• “Deep Learning” by Ian Goodfellow, Yoshua Bengio, and Aaron
2. Online Courses:
• Machine Learning by Andrew Ng
• Deep Learning Specialization by Andrew Ng
• Microsoft Professional Program in Artificial Intelligence
• Intro to Machine Learning with PyTorch and TensorFlow
3. Websites and Documentation:
• Scikit-learn Documentation
• TensorFlow Documentation
• PyTorch Documentation
4. Tutorials and Blogs:
• Analytics Vidhya – A community of data
• Kaggle Kernels – Learn from and share code on real-world datasets.
5. YouTube Channels:
• sentdex – Python programming and machine learning
• Corey Schafer – Python tutorials with a focus on practical
• 3Blue1Brown – Excellent visualizations explaining mathematical concepts behind machine learning.
6. Practice Platforms:
• Kaggle – Compete in machine learning competitions and explore
• Hackerrank – Artificial Intelligence – Practice AI challenges in
7. Community Forums:
• Stack Overflow – Ask and answer programming questions.
• Reddit – r/MachineLearning – Discussion on machine learning
8. GitHub Repositories:
• Scikit-learn GitHub Repository
• TensorFlow GitHub Repository
• PyTorch GitHub Repository
In conclusion, the journey through machine learning concepts using Python reveals the profound impact this dynamic field has on reshaping our approach to problem-solving and decision-making. Python,
with its rich ecosystem of libraries such as TensorFlow, scikit-learn, and PyTorch, empowers developers and data scientists to harness the potential of machine learning.
From understanding the fundamentals of supervised and unsupervised learning to delving into advanced techniques like deep learning and reinforcement learning, the Python ecosystem provides a
versatile platform for exploring the vast landscape of machine learning. The ability to preprocess and analyze data, build robust models, and evaluate their performance has become more accessible,
thanks to Python’s intuitive syntax and the wealth of community-driven resources.
As we navigate the intricacies of algorithms, feature engineering, and model optimization, it becomes evident that machine learning is not merely a technical pursuit but a creative process that
demands thoughtful consideration of data, problem formulation, and algorithmic choices. Python’s flexibility and readability contribute significantly to the iterative nature of model development,
allowing practitioners to experiment and refine their approaches seamlessly.
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Comparing basal area growth models for Norway spruce and Scots pine dominated stands
Models that predict forest development are essential for sustainable forest management. Constructing growth models via regression analysis or fitting a family of sigmoid equations to construct
compatible growth and yield models are two ways these models can be developed. In this study, four species-specific models were developed and compared. A compatible growth and yield stand basal area
model and a five-year stand basal area growth model were developed for Scots pine (Pinus sylvestris L.) and Norway spruce (Picea abies (L.) Karst.). The models were developed using data from
permanent inventory plots from the Swedish national forest inventory and long-term experiments. The species-specific models were compared, using independent data from long-term experiments, with a
stand basal area growth model currently used in the Swedish forest planning system Heureka (Elfving model). All new models had a good, relatively unbiased fit. There were no apparent differences
between the models in their ability to predict basal area development, except for the slightly worse predictions for the Norway spruce growth model. The lack of difference in the model comparison
showed that despite the simplicity of the compatible growth and yield models, these models could be recommended, especially when data availability is limited. Also, despite using more and newer data
for model development in this study, the currently used Elfving model was equally good at predicting basal area. The lack of model difference indicate that future studies should instead focus on
model development for heterogeneous forests which are common but lack in growth and yield modelling research.
1 Introduction
Growth and yield models are essential for predicting future states of forest stands and guiding well-informed management decisions. There are different ways to construct these models, and there are
many different model types, differing in complexity, management accounted for, data requirements, and ease of use. Growth models using linear regression have long been common worldwide (Weiskittel et
al. 2011). A reason for this is the accessibility and reliability of the method. Fitting a simple or multiple linear regression with inventory data can produce accurate models that predict the
development of stands with high precision (Vanclay and Skovsgaard 1997). Also, the ability to easily add variables that take stand management into account makes this type of modelling a powerful tool
for producing predictive results (Weiskittel et al. 2011; Fahlvik et al. 2014).
Compatible growth and yield models, where the growth model is a derivation of the yield model, are an alternative to traditional separate models of growth and yield, where yield differ from
accumulated growth predictions based on the same data. Compatible functions were first published by Buckman (1962) and Clutter (1963). Such models are fitted with a family of sigmoid equations to
data from repeated measurements of permanent sample plots where the equation explains a biological growth pattern. These age-based models are robust and easy to use but may lose some predictive
ability if only time and tree data are included as explanatory variables. There are also ways of fitting compatible growth and yield equations with management, climatic, physiological, and soil-based
variables to increase prediction precision (Bailey and Ware 1983; Methol 2001; Mason et al. 2007; Gyawali and Burkhart 2015).
An advantage of using compatible growth and yield models that are path invariant is to reduce error accumulation and precision loss in longer prognoses (Holm 1981; Kangas 1997). Growth models are
often built to run for a short period, then updated with newly predicted independent variables and run again. When making long-term projections, model prediction error gets larger with every
successive run. A path invariant model can calculate a longer period arriving at the same result with or without successive steps and should, therefore, decrease error accumulation (Weiskittel et al.
In Sweden and other parts of the world, a tradition of regression analysis has been the dominant approach in growth and yield modelling. Most of the models used to predict stand (Eko 1985; Elfving
2010) and tree development (Soderberg 1986; Elfving 2010) are growth models derived from regression analysis. The stand growth models used in Sweden today have high precision when predicting forest
growth, both short- and long-term (Fahlvik et al. 2014). However, these models often require many input variables that make them difficult to apply in situations with limited data availability. Since
they are not path invariant and model growth in five-year periods, there is also a risk of error propagation over long-term predictions. Therefore, new models need to be developed that are easier to
apply and do not have the same risk of error accumulation.
The objectives of this study were to construct new species-specific basal area models for Scots pine (Pinus sylvestris L.) and Norway spruce (Picea abies (L.) Karst.) stands in Sweden, based on the
compatible growth and yield modelling approach. We also developed new species-specific basal area growth models for Scots pine and Norway spruce using linear regression, inspired by Elfving’s growth
model currently used in the Swedish forest planning system Heureka (Elfving 2010; Wikström et al. 2011). These new models were compared with one another and with Elfving’s basal area growth model.
The aims of the model comparison were to; i) evaluate their short- and long-term prediction precision and whether models developed using the compatible growth and yield modelling approach had better
long-term predictions, and ii) whether models developed using data from the 2000s performed differently from models developed with data from the 1980s. The comparison was made using an independent
data set where both short- and long-term precision were evaluated. This study is partly based on the doctoral dissertation by Martin Goude (Goude 2021).
2 Materials and methods
2.1 Data description
Data from the Swedish national forest inventory (NFI) and Swedish long-term forest experiments were used for model fitting (Table 1). The NFI data were based on permanent plot inventories of Swedish
forests, in which measurements started between 1983 to 1987. The NFI data consists of survey plots laid out in a systematic grid to cover all Swedish forests. Because its survey plots, the stand
history before first measurement is unknown. All plots were measured once every five years, except from 1993 to 2002 when the remeasurement period varied between 5 and 10 years. On each occasion, all
trees within a plot radius of 10 m with a diameter at breast height (DBH, 1.3 m) ≥10 cm were measured with callipers. Trees with DBH larger than 4 cm and smaller than 10 cm were measured in smaller
areas that changed over time (Fig. 1). Trees with DBH below 4 cm were not included in this study. Starting from the second measurement period in 1988, the cause of absence of every missing tree
previously measured was also recorded along with previous damage and management of the plot. In every plot, information on stand and site properties like soil texture, soil water, and ground
vegetation was also recorded. Details are available in Fridman et al. (2014).
Table 1. Description of the data used for model fitting and model comparison. Data sources were Swedish National Forest Inventory (NFI) and Swedish long-term forest experiments (LFE). Mean values for
site index, basal area (m^2 ha^–1), stem number (stems ha^–1) and age (years) are presented for each data set with standard error (SE).
Model Model 1 Model 2 Model 3 Model 4 Model comparison
Species Scots pine Norway spruce Scots pine Norway spruce Norway spruce Scots pine
Period length (years) 1–36 1–36 5 5 3–11 2–14
Measurement years 1993–2017/ 1993–2017/ 2003–2017/ 2003–2017/ -/ -/
(NFI/ LFE) 1960–2017 1960–2017 - - 1966–2007 1966–2007
Fitting Validation Fitting Validation Fitting Validation Fitting Validation
Data set M1[F] M1[V] M2[F] M2[V] M3[F] M3[V] M4[F] M4[V] GG[spruce] GG[pine]
Number of plots 1288 634 552 284 2059 1029 1422 708 10 10
Number of periods 5199 2599 3082 1541 3307 1654 2191 1095 50 50
Mean site index 25 24 28 28 22 22 26 26 34 22
(0.1) (0.1) (0.1) (0.1) (0.1) (0.1) (0.1) (0.2) (0.2) (0.5)
Mean basal area 19.8 (0.2) 17.2 36 21.5 18.4 17.8 24 24 36.5 25
(m^2 ha^–1) (0.2) (0.3) (0.4) (0.15) (0.2) (0.2) (0.3) (1.2) (0.7)
Mean stem number (stems ha^–1) 1544 (11.2) 1621 2380 (16.1) 1839 950 930 1149 (16.6) 1149 1589 1424
(20.8) (37.7) (11.5) (16.7) (22.2) (92.1) (72.9)
Mean age (years) 41 39 37 35 63 64 68.8 (0.8) 67 43 59
(0.2) (0.3) (0.2) (0.4) (0.6) (0.9) (1.1) (1.1) (1.5)
The long-term forest experiment data consisted of permanent plots in Scots pine, and Norway spruce dominated stands used for experiments on stand establishment, spacing, and thinning (Elfving and
Kiviste 1997; Nilsson et al. 2010). The stand history is well documented in most cases, and plots are regenerated through planting, direct seeding, or natural regeneration. Because the experiments
are more carefully managed, they are more homogeneous than the average production forest in the NFI data, both in terms of species composition and structure. The plot size varied between 0.02 ha and
0.2 ha (0.06 ha on average).
For the basal area growth and yield models for Scots pine (Model 1) and Norway spruce (Model 2), yield data or total basal area (m^2 ha^–1) was used from both the Swedish long-term experiments
between 1960 and 2017, and Swedish NFI between 1993 and 2017 (Table 1). This was possible because this modelling approach allows for various growth period lengths. Including measurements with
different interval lengths should result in a model that makes better long-term projections (Lee 1998). The used period lengths were 5 to 20 years for the Swedish NFI data and 1 to 36 years for the
long-term experiment data. Because total production (including self-thinning and removals in thinnings) was modelled in Model 1 and Model 2, only permanent plots with at least one measurement before
the first thinning were used for model fitting of Model 1 and Model 2. Thus, 836 (4623 periods) Norway spruce plots and 1922 (7798 periods) Scots pine plots were used.
For the basal area growth models for Scots pine (Model 3) and Norway spruce (Model 4), five-year basal area growth (m^2 ha^–1 5-years^–1) of all living trees present at the beginning and end of the
period was used. Data came from NFI permanent plots measured from 2003 to 2017 (Table 1). These were remeasured on 5-year intervals and had a consistent plot size and layout (Fig. 1). There was a
total of 2130 (3286 periods) Norway spruce plots and 3088 (4961 periods) Scots pine plots. The new growth models were inspired by Elfving’s growth model currently used in the Swedish forest planning
system Heureka (Elfving 2010; Wikström et al. 2011). The new growth models were created using newer data to see if models based on NFI data from the 2000s performed differently from Elfving’s model
using NFI data from the 1980s.
For model comparison, an independent data set was used (GG) consisting of 20 sites (10 Scots pine and 10 Norway spruce) from a thinning and fertilisation experiment established 1966–1983 (Nilsson et
al. 2010). The experimental plots were established at the time of first thinning when the top height was 12–18 m. The experiment consisted of pure stands of Norway spruce and Scots pine. The Norway
spruce stands were located in southern and central Sweden (56.1°N–63.1°N), and the Scots pine stands were located from the county of Scania in the south to Norrbotten in the north (56.2°N–67.3°N)
(Fig. 2). The plot size was 0.1 ha on average. The treatments that were applied were different thinning and fertilisation treatments. After establishment, plots were remeasured at each thinning and
occasionally in between. The plots used in this study came from two treatments. Treatment A, included 3 to 6 thinnings with a mean thinning grade of 22% for Norway spruce, and 3 to 4 thinnings with a
mean thinning grade of 26% for Scots pine. Treatment I was the untreated control. These treatments were chosen because models should be tested on both unmanaged and managed forests. These treatments
were also the most common among the different sites and had long growth records.
2.2 Data management and selection
Data were sorted and processed before model fitting. In the long-term experiment, the only necessary pre-processing was adjusting basal area increment to the time of the year the plots were measured
in different revisions (see details below) and removing outlier plots. The outlier plots had either negative or extremally high (around three times) total basal area development compared to the
growth trends of plots with similar stand and site conditions in the data. In contrast, the NFI data were in tree list form and needed rigorous sorting and management before model fitting. Here
follows a description of the steps used to process the NFI data.
Age was not measured for every callipered tree but was estimated on a plot level by coring two trees outside the 10 m radius permanent plot. The cored trees had a DBH as close to the quadratic mean
diameter of the plot as possible. These age estimates were not always consistent between measurements. Therefore, it was necessary to compute an age that was consistent with time between
measurements. The new age was calculated for each plot by taking every plot age from each measurement and calculating backwards to when the plot was first measured. The age at first measurement (A
[i]) for individual plots was calculated using the following equation:
where A is estimated plot mean age, Y[m] was year of measurement, and Y[i] was year of initial measurement. Age at first measurement was calculated for every plot measurement resulting in a maximum
of six A[i] values for each plot. An average of the backtracked ages was created and set as the initial age of the plot when it was first measured (age[start]):
where n is the number of estimated ages per plot. To get the age at a particular time, the number of years from the plot’s age[start] was added. For example, a plot first measured in 1983 yielded an
age[start] of 23. The age in 2003 would then be 23 + 20 = 43 years.
Because plots were not measured consistently at the same time of the year, growth needed to be adjusted between two measurements to the actual period length. For example, if a plot was measured in
spring 2003 and late summer 2008, the real number of growing seasons was closer to six than five. An adjustment to account for the different measurement timings was applied by calculating how far
into the growing season the measurements were made by estimating the completion of annual rings using the following equation (Elfving 2010):
where Ring is the degree of completion of an annual growth ring when the measurement was made (0 < Ring < 1) and x is the number of days past the growing season’s start. Regardless of location, the
ring growth season was assumed to last 100 days (i.e. May 20 to August 29) (Valinger 1992; Soderberg et al. 1993; Elfving 2010). Ring was computed in both measurement years, and growth was adjusted
so that the number of growing seasons would match the number of years between measurements. For example, if recorded growth was 6 m^2 ha^–1 during 5.4 growth seasons between 2003 and 2008, the
adjusted growth for five seasons would be (6/5.4)×5 = 5.56 m^2 ha^–1. This adjustment was done for the NFI and long-experiment data.
There was also an issue with ingrowth for the NFI plots caused by the inventory design (Fig. 1), where trees with a DBH below 10 cm were being measured in a smaller plot compared to trees with a DBH
equal or larger than 10 cm. This resulted in trees suddenly appearing in the large plot (ingrowth), causing artificial jumps in basal area. We addressed this issue by using the trees present at the
end of the period with a diameter larger than 10 cm to predict diameter at the beginning of the period when the diameter was smaller than 10 cm. These functions estimated diameter at the beginning of
the period based on diameter at the end of the period, age, and site index. Using this method, trees that grew into the larger size class by the end of the period that were not measured in the small
plot were assigned a diameter at the beginning of the period. Basal area was then calculated by summing basal area of trees with DBH larger than 10 cm measured in the big plot, DBH between 4 and 10
cm measured in the small plot, and DBH between 4 and 10 cm estimated in the big plot.
A further issue related to ingrowth was trees growing from a DBH below 10 cm to a DBH equal to or larger than 10 cm during a growth period. This initially caused both artificial increases and
decreases in basal area because the same tree would be divided by a different plot area when calculating basal area per ha (m^2 ha^–1). To avoid these problems, the plot area a tree had at the end of
the period was used to calculate its basal area per ha at the period’s start as well.
Plot thinning was inferred from harvested trees. First, basal area of the harvested trees before removal was calculated. This basal area was then compared to the total basal area before thinning to
get a thinning grade (percent of initial basal area removed, G[removed]/G[before]). The thinning grade was compared to the NFI’s stand management classification to select the removals resulting from
thinning. The comparison showed that most removals classified as thinning had a removal between 10 and 60%, with a 35% mean. The removals were also compared to natural mortality and final felling,
which mainly occurred below 10% and above 60%, respectively. From this analysis, removals over 60% were classified as harvest/final felling and below 10% as no management/natural mortality. Removals
larger or equal to 10% and lower or equal to 60% were classified as thinnings.
Not all the available data were used in the model fitting process. Instead, the model fitting was restricted to plots where, at the beginning of the period;
• The proportion of basal area was ≥ 70% of either Norway spruce or Scots pine with no restrictions on the remaining 30%
• Mean tree height ≥ 5 m
• Plot age ≥ 10 years
• No harvest during the current or two previous growth periods
There were also specific restrictions for each model on top of these general restrictions used for both the Swedish NFI and long-term experiment data. For the growth models (Model 3 and Model 4), no
thinning was allowed during the five-year growth period. The reason was that the five-year growth should be the total growth of all living trees present at the beginning and end of the period. For
Model 1 and Model 2 using total basal area, the selected plots had a height below 10 m the first time they were measured. This height restriction allowed for a more accurate total basal area
2.3 Modelling
2.3.1 Compatible growth and yield model
When fitting the compatible growth and yield models (Model 1 and Model 2), two-thirds of the selected data was used for model fitting, and the rest was used for model validation. The best fit for
both Scots pine and Norway spruce came with the Schumacher 2 polymorphic function (Schumacher 1939). The shape of the Schumacher equation was augmented by initial plot stem number before thinning to
account for the effect of stem density on basal area development:
where G[2] is basal area (m^2 ha^–1) at period end, G[1] is basal area (m^2 ha^–1) at period start, t[1] is age at period start, t[2] is age at the end of the period, and ST is initial stem number
(stems^ ha^–1) before thinning, a, b, and c are parameters to be estimated, and log is the natural logarithm. The best model was chosen by analysing normality and bias of the residuals and minimising
root mean square error (RMSE). To reduce the effect of autocorrelation, parameter significance was checked by fitting the models to a random sub-sample containing one interval per plot. The models
were fitted using the nls, nls2, or nlsLM functions in R, from the lme4, nls2 (Grothendieck 2013), and minpack.lm (Elzhov et al. 2016) packages. All modelling and statistics in this study were
computed with the statistical software R (version 3.6.1) (R Core Team 2016). Model validation was done by predicting basal area using the one-third validation data (Table 1) and comparing precision
and bias of predicted basal area against the observed basal area.
2.3.2 Growth models
The selected data were randomly divided into two-thirds for model fitting and one-third for model validation when fitting Model 3 and Model 4. Linear mixed effects models were used in the lme4
package (Bates et al. 2015), when determining variable significance to account for the random effects of site because of multiple measurements on each site. Only significant variables were kept in
the final models, and VIF values were evaluated to exclude variables with severe collinearity with other variables (VIF > 10). We used the lmerTest package (Kuznetsova et al. 2017) to generate
p-values for the lmer output, and the MuMIn package (Barton, 2018) to calculate R^2 for the mixed models. Some variables were transformed using scaled power transformations (Cook and Weisberg 1999)
to improve linearity and fulfil the assumptions of normality and homoscedasticity. To find the best transformations, the distributions of the independent variables were analysed using the SPTLambda
function in R. Since the response variables were log-transformed using the natural logarithm, the models needed to be corrected for transformation bias (Vanclay and Skovsgaard 1997). The correction
factor from Baskerville (1972) accounted for this bias. Variables used in the final growth models (Eq. 5) are presented in Table 2. The validation of the models was done by predicting basal area
growth using the one-third validation data (Table 1) and comparing precision and bias of predicted growth against observed growth.
where Y is the dependent variable log(G_growth), the natural logarithm of basal area growth over five years (m^2 ha^–1 5-years^–1), b[0] is intercept, b[1]–b[k] is estimated coefficients (Table 5), x
[1]–x[k] is independent variables (Table 2), and ε is model error.
Table 2. Definition of variables used in model fitting and validation. Model 1 = basal area growth and yield model (m^2 ha^–1) for Scots pine in Sweden, Model 2 = basal area growth and yield model (m
^2 ha^–1) for Norway spruce in Sweden, Model 3 = basal area growth model (m^2 ha^–1 5-years^–1) for Scots pine in Sweden, and Model 4 = basal area growth model (m^2 ha^–1 5-years^–1) for Norway
spruce in Sweden.
Variable Definition Included in model
t[1] Age at period start (year) Model 1, Model 2, Model 3, Model 4
t[2] Age at period end (year) Model 1, Model 2
ba Basal area at period start (m^2 ha^–1) Model 3, Model 4
G1 Total basal area at period start (m^2 ha^–1) Model 1, Model 2
G2 Total basal area at period end (m^2 ha^–1) Model 1, Model 2
stem Stem number (stems ha^–1) at the start of the period Model 3, Model 4
ST Initial stem number (stems ha^–1) before thinning Model 1, Model 2
veg Ground vegetation type index (Table 3) Model 3, Model 4
thinn[1] The thinning grade (G[out]/G[before]) of a thinning performed one period ago. G[out] = removed basal area in thinning. G[before] = basal area before Model 3, Model 4
thinn[2] The thinning grade (G[out]/G[before]) of a thinning performed two periods ago Model 3, Model 4
prop[spruce] Norway spruce proportion of the total basal area (0–1) Model 3, Model 4
moist 1 if the soil moisture class is classified as moist, else 0 Model 4
t[sum] Temperature sum(day-degrees > 5 °C), calculated from altitude and latitude (Odin et al. 1983), divided by 1000 Model 3, Model 4
peat 1 if the soil is classified as peat, else 0 Model 3
depth[ind] 1 if the soil depth is <=0.5 m, else 0 Model 3
To get an indicator of site fertility, a vegetation index was used (veg-index). This index was based on the Swedish NFI vegetation-type classification and ranged between –5 and +4 (Elfving 2010).
Each vegetation class is given an index that reflects site fertility (Table 3).
Table 3. Ground vegetation type index (veg) variable definition. Veg-type was the classification use by the Swedish NFI and the veg-index is the veg variable used in Model 3 = basal area growth model
(m^2 ha^–1 5-years^–1) for Scots pine in Sweden, and Model 4 = basal area growth model (m^2 ha^–1 5-years^–1) for Norway spruce in Sweden.
Veg-type classification (Swedish) Veg-type classification (English) Veg-index
Högört utan ris Rich-herb without shrubs 4
Högört med ris/blåbär Rich-herb with shrubs/bilberry 2.5
Högört med ris/lingon Rich-herb with shrubs/lingonberry 2
Lågört utan ris Low-herb without shrubs 3
Lågört med ris/blåbär Low-herb with shrubs/bilberry 2.5
Lågört med ris/lingon Low-herb with shrubs/lingonberry 2
Utan fältskikt No field layer 3
Bredbl. gräs Broadleaved grass 2.5
Smalbl. gräs Thinleaved grass 1.5
Carex ssp.,Hög starr Sedge, high –3
Carex ssp.,Låg starr Sedge, low –3
Fräken Horsetail, Equisetum ssp. 1
Blåbär European blueberry, bilberry 0
Lingon Lingonberry –0.5
Kråkbär Crowberry –3
Fattigris Poor shrub –5
Lavrik Lichen, frequent occurrence –0.5
Lav Lichen, dominating –1
2.4 Model comparison
In the model comparison, the new models created in this study were applied to an independent data set. The main basal area growth model (Elfving) used in Heureka was also included in the model
comparison. The Elfving model is a linear regression model predicting 5-year basal area growth (m^2 ha^–1 5-years^–1) for stands of Scots pine, Norway spruce, and silver birch (Betula pendula Roth)
in all of Sweden. It was developed using permanent plots from the Swedish NFI, first measured 1983–1987 and then remeasured 5 years later, 1988–1992. The parameters included in the model are
presented in the appendix (Supplementary file A1: Table A.1). For more information about the model, see Elfving (2010).
Model comparison was performed by running the models from the first measurement and the following five growth periods. The models were used to estimate total basal area (m^2 ha^–1), including
mortality and thinning removals. Since net basal area was used in Model 3, Model 4, and the Elfving model, mortality was estimated using the mortality function from Siipilehto et al. (2020). The
length of model comparison was different for different plots, with 34–40 years for Scots pine and 27–34 years for Norway spruce (Table 1). The period lengths varied between 2 to 14 years (average 7
years) and 3 to 11 (average 6 years) respectively. For Scots pine, measurements came from between 1969 and 2015, and for Norway spruce, between 1966 and 2007. The different models were compared using
residual analysis to determine the precision and bias of predictions.
3 Results
3.1 Model fitting
The fitted growth and yield models showed a good and relatively unbiased fit against the NFI data with similar patterns for both species when plotted against fitted basal area (Fig. 3 and 4) and
other predictor variables (Suppl. file A1: Fig. A.1 and A.2). The Scots pine model (Model 1) had a better fit (RMSE (m^2 ha^−1) = 1.404) to its data compared to the model for Norway spruce (Model 2;
RMSE = 2.507) (Table 4). The validation showed good and relatively unbiased residuals when plotted against predicted basal area (Fig. 3 and 4) and other predictor variables (Suppl. file A1: Fig. A.1
and A.2). When comparing the species-specific models, Model 2 shows a higher basal area growth for comparable stand basal area and age (Fig. 5.)
Table 4. Estimated parameters for the growth and yield models and model root mean squared error (RMSE; m^2 ha^–1) for Scots pine (Model 1) and Norway spruce (Model 2) in Sweden. All estimated
parameters were significant at p < 0.05.
Model RMSE (m^2 ha^–1) a b c
Fitted model 1.404 Estimate 4.822 0.151 0.224
Model 1 SE 0.017 0.007 0.007
Validation 1.472
Fitted model 2.507 Estimate 5.113 0.328 0.119
Model 2 SE 0.018 0.022 0.009
Validation 2.279
The growth models for Scots pine (Model 3) and Norway spruce (Model 4) contained similar explanatory variables (Table 5). The most important variables were age at the beginning of the period (age
[1]), basal area at period start (ba), and ground vegetation class (veg). The residual plots of the two growth models where the residuals were plotted against fitted basal area growth (Fig. 6 and 7)
and other predictor variables (Suppl. file A1: Fig. A.3 and A.4) showed a good and relatively unbiased fit. The validation showed a similar pattern to the model fitting for both growth models when
plotted against predicted basal area (Fig. 6 and 7) and other predictor variables (Suppl. file A1: Fig. A.3 and A.4).
Table 5. Statistical output of basal area growth models for Scots pine (Model 3) and Norway spruce (Model 4) in Sweden. The response variable was log(G_growth), the natural logarithm of basal area
growth over five years (m^2 ha^–1 5-years^–1). The variable exponents were selected to improve variable normality. R^2 is the marginal r-square, i.e., a measure of the explanatory power of the fixed
effects only. Site was set as random effect.
Model Coefficient Estimate Std. Error Pr(>|t|) RMSE R^2 Random variance
Intercept 1.547 1.260e-01 < 2e-16 0.24 0.70
age[1]^0.2 –1.579 3.660e-02 < 2e-16
ba^0.6 1.036e-01 6.013e-03 < 2e-16
stem^0.2 4.105e-01 1.745e-02 < 2e-16
veg 3.493e-02 4.590e-03 2.71e-14
Model 3 (Scots pine) thinn[1]^0.1 2.900e-01 1.898e-02 < 2e-16
thinn[2]^0.1 2.231e-01 1.928e-02 < 2e-16
peat –2.808e-01 2.519e-02 < 2e-16
depth[ind] –5.444e-02 1.947e-02 0.00470
t[sum] 4.417e-01 3.422e-02 < 2e-16
prop[spruce] 2.898e-01 9.098e-02 0.00121
Site (random) 9.245 e-02
Intercept 5.402e-01 1.931e-01 0.005195 0.29 0.69
age[1]^0.25 –9.577e-01 3.531e-02 < 2e-16
ba^0.74 5.011e-02 3.021e-03 < 2e-16
stem^0.2 4.175e-01 1.883e-02 < 2e-16
veg 8.257e-02 5.669e-03 < 2e-16
Model 4 (Norway spruce) thinn[1]^2 2.014 1.704e-01 < 2e-16
thinn[2]^2 1.269 1.964e-01 1.18e-10
prop[spruce] 3.419e-01 9.095e-02 0.000175
moist –1.228e-01 4.532e-02 0.006770
t[sum] 4.032e-01 4.340e-02 < 2e-16
Site (random) 7.072e-02
3.2 Model comparison
The models showed relatively similar results when tested against the independent experiment data (Fig. 8). The Scots pine models had better predictions over time with lower RMSE (m^2 ha^−1) than the
Norway spruce models (Table 6). On average, across all five periods, the RMSE for the Scots pine models was 2.204 for Model 1, 1.708 for Model 3, and 2.266 for Elfving. For Norway spruce, the average
RMSE was 4.156 for Model 2, 5.228 for Model 4, and 4.188 for Elfving.
Table 6. Mean residual bias (%) and root mean squared error (RMSE) for each model and growth period (average 7 years for Scots pine and 6 years for Norway spruce), compared against independent data
(GG[pine] and GG[spruce]). Model 1 = basal area growth and yield model (m^2 ha^–1) for Scots pine, Model 2 = basal area growth and yield model (m^2 ha^–1) for Norway spruce, Model 3 = basal area
growth model (m^2 ha^–1 5-years^–1) for Scots pine, Model 4 = basal area growth model (m^2 ha^–1 5-years^–1) for Norway spruce, and Elfving = basal area growth model (m^2 ha^–1 5-years^–1) for Scots
pine, Norway spruce, and silver birch.
Mean residual bias % (SE) RMSE (m^2 ha^–1)
Species Model Period Period
Model 1 0.35 (0.60) –1.49 (0.98) –3.00 (1.28) –2.44 –7.17 (1.66) 0.70 1.43 2.15 2.75 3.99
Scots pine Model 3 2.55 (0.74) 1.98 2.10 3.58 –1.62 (1.20) 1.08 1.45 1.81 2.49 1.71
(0.99) (1.12) (1.54)
Elfving 0.92 (0.66) –1.94 (1.14) –3.46 (1.25) –3.77 –8.38 (1.82) 0.8 1.66 2.12 2.51 4.24
Model 2 0.86 (1.35) 0.93 0.78 –0.95 –1.62 (2.73) 2.32 3.32 4.29 5.30 5.55
(1.63) (1.95) (2.23)
Norway spruce Model 4 1.34 (1.44) 0.12 –1.04 (2.24) –3.37 –2.59 (3.35) 2.36 3.81 4.98 7.1 7.9
(1.92) (2.79)
Elfving 2.14 (1.32) 2.50 2.023 (1.76) –0.25 –1.12 (2.83) 2.4 3.44 4.19 5.28 5.64
(1.58) (2.15)
For Scots pine, Model 3 and Elfving model showed a tendency to overestimate basal area more over longer times. The residual bias was on average –7.17% and –8.38% respectively in the final period
(period 5), compared to –1.65% for Model 3 (Table 6). For Norway spruce, the models were relatively unbiased with time, where the bias on average the final period (period 5) was –1.62% for Model 2,
–2.59% for Model 4, and –1.12 for Elfving (Table 6).
All models also showed an increased spread of residuals with less accurate predictions at the end of the comparison. The largest increase in variation was for Model 4 which increased RMSE from 2.36
in the first period to 7.90 in the final period.
4 Discussion
4.1 Model comparison: all models had similar performance
Altogether, there were no clear differences between the predictions of different models and no clear patterns in when the models worked better or worse. Expected benefits of long-term precision using
compatible growth and yield models (Model 1 and Model 2), like path invariance and variable period length (Clutter 1963; Weiskittel et al. 2011), could not be seen for these plots and periods,
regardless of plot treatment and location. Perhaps a longer comparison period than 30–40 years could have shown clearer results for the model comparison. Also, the current basal area model used in
Heureka (Elfving) works well, even compared to species-specific models using new data and different equations. The lack of differences makes it difficult to say that one model was constantly better.
However, the results illustrate the power of the relatively simple Model 1 and Model 2, which only used age, basal area, and initial stem number as input variables. Even though these growth and yield
models did not show a more stable prediction with time, they include other strengths such as choosing period length and simple input variables. Choosing period lengths makes these models more
flexible and easier to use than fixed five-year growth models such as those used in this study (Model 3, Model 4, and Elfving). The simple input variables make the growth and yield models a good
alternative for predicting stand development in managed forests where age is known and basal area and stem number are estimated through remote sensing (Nilsson et al. 2017). If the initial stem
number before thinning is unavailable, the average initial stem numbers from this study could be used. These are 1544 stems ha^–1 (SE = 11.2) for Scots pine and 2380 stems ha^–1 (SE = 16.1) for
Norway spruce.
Comparison of the models against independent data showed that the ability of the models to predict basal area development was good. However, precision and bias varied both with time and model. For
Scots pine, Model 3 had the best long-term precision and minimally underestimated basal area (2 to 3%). Model 1 and Elfving model tended to overestimate with time, especially in the final period
where basal area was overestimated by 6 and 8%, respectively. A reason for the more apparent overestimation bias for the Elfving model could be due to that model being the same for Scots pine and
Norway spruce. Studies like Eko et al. (2008) have shown that Norway spruce stands have higher production than Scots pine in Swedish NFI data. This was also shown in this study where Norway space
models had a higher growth compared to models developed for Scots pine dominated stands (Fig. 5). In the Elfving model, this species difference was captured with the species proportion of basal area
where plots with more Scots pine grew slower (Elfving, 2010). This species adjustment was, however, insufficient for the plots used in this comparison.
The models for Norway spruce differed somewhat in their ability to predict basal area. Both Model 2 and Elfving’s model showed a similar and unbiased ability to predict basal area over time. In
contrast, Model 4 had a more substantial residual variation and a large overestimation for some plots. The plots with considerable overestimation in the later stages were primarily unthinned control
plots. This indicates that growth of Model 4 did not flatten off as aggressively as these experimental plots and other tested models did. Another reason may be that the estimations for these
overestimated control plots were already biased after 10-year predictions (Fig. 8). By running the model for a long time using these biased estimations, errors accumulate as predictions were made
from increasingly biased previous predictions (Holm 1981; Kangas 1997). Predictions by both Elfving and Model 2 model had increased RMSE between the first and last periods, but not to the same extent
and with less bias.
The included variables in the Model 3 and Model 4 models were quite similar. The variables that did not occur in both models were connected to the species availability in the Swedish NFI data set.
For example, plots with peat and shallow soils only occurred to a large extent where Scots pine was dominant. Therefore, these variables became significant and included in the final Model 3. The
moist variable worked the same way in the Norway spruce model (Model 4), where a large portion of the Norway spruce dominated plots were classified as moist.
4.2 Limitations imposed by data collection design
We used NFI data for the Model 3 and Model 4 models because these five-year growth models required a consistent five-year interval. Also, one of the objectives was to see if data from between 2003
and 2017 would result in different model performance than the Elfving model based on data from the 1980s. The growth and yield models (Model 1 and Model 2) allowed combining data from both NFI and
long-term experiments, despite different period lengths. The different period lengths of the experimental data needed a type of equation that could deal with this inconsistency both within and
between experiments. It was, therefore, suitable for the growth and yield models but not for the five-year growth models. Another reason for including the experimental data when fitting Model 1 and
Model 2 was that yield data on total basal area production was used for these models. The long-term experiment data contained older and more accurate records on total basal area production than the
Swedish NFI data and was therefore included.
The data collection design was the likely source of uncertainty in model predictions. Age was a complicated variable in this study and is a weakness of using the Swedish NFI for growth and yield
model development (Fahlvik et al. 2014). The problem was that age was not measured for each plot but rather estimated. This is a problem because the estimated age varied between periods. The age
usually increased by fewer or more years than it should have. For example, some plots with an age of 60 years in 1983 were recorded as 65 years old in 1993, ten years later. Our solution was to take
the average of all estimates since we could not know which ones were reliable. As discussed in Elfving (2010), it may have been desirable to avoid age as a variable, but that was not easy since age
explained much of the variation in basal area development. All this uncertainty added to the error and overall uncertainty of the models.
Different plot sizes for different sized trees are understandable in an inventory system for logistical reasons, to reduce the number of trees to be measured. But the smaller plot used to measure
trees between 4 and 9.9 cm was not entirely representative of the whole 10 m plot and resulted in ingrowth. We dealt with ingrowth by estimating the basal area of small trees in the whole 10 m radius
plot. The plot basal area was then made up of trees with DBH over 10 cm measured in the large plot, trees between 4 and 10 cm measured in the small plot, and trees between 4 and 10 cm estimated
outside the small plot. The resulting basal area was then partly a prediction of previous basal areas. This was still a better option than the sometimes severe basal area growth overestimation caused
by ingrowth. A similar issue occurred when the Swedish NFI plot area changed between inventories. This also caused problems with ingrowth and difficulties with growth analysis. Plot size issues are
good illustrations of the problems that arise when making changes to long-term permanent plot inventories. The Swedish NFI realised this at the beginning of the 2000s and worked to standardise the
inventory (Fridman et al. 2014), but the previous changes still cause problems.
Another limitation on the models was the rarity of older forests in the available data for model construction. The main reason for this was that forests above 100 years are not common in Sweden and
made up a small part of the final datasets after applying the restrictions. This was especially true for Model 1 and Model 2. One of the restrictions when selecting data was that a plot could not be
taller than 10 m at first measurement to ensure accurate total basal area production estimations. The models should therefore be used with caution on old forests over 100 to 150 years old.
4.3 Prospects for use of growth and yield models
The models created in this study were developed for relatively pure stands of Norway spruce and Scots pine (≥70% of basal area). However, according to the Swedish NFI data, around 30% Swedish forests
are not so homogenous. Therefore, there is a need for models that can take mixture effects into account, especially when including other species than Norway spruce and Scots pine. Using the Swedish
NFI, models for mixed forests could probably be developed for common species like Scots pine, Norway spruce, and silver birch. Making models for other species may be more difficult since they occur
infrequently in the data and with a large geographical variation. For those species, plots in long-term experiments are necessary. However, these experiments may have the same problem representing
geographical variation and variation in site properties.
The minor differences between the models evaluated in this study suggest that other approaches than purely statistical forest growth and yield models are needed to improve the models’ site adjustment
and ability to make accurate long-term predictions, even during changes in growth conditions. This could be achieved by developing hybrid physiological/mensurational models where climate effects on
production are included (Mason et al. 2007).
5 Conclusions
In this study, compatible growth and yield models and growth models, predicting stand basal area, were developed for Scots pine and Norway spruce. In the long-term model comparison, there were no
apparent differences between the models developed in this study and the one currently used in Heureka in their ability to predict basal area development. The theoretically beneficial features of
compatible growth and yield models, like path invariance, did not give any advantages in the model comparison against the independent data. However, their ability to predict basal area development
despite their simplicity illustrates the power of this method of forest growth and yield modelling. This makes it a good alternative when data availability is limited. Also, there were no clear
differences between the new and old growth models despite more and newer data in the models from this study. The lack of differences means that the basal area growth model used in Heureka can
continue to be used. The future development of growth and yield models should focus on forests with heterogeneous species composition, age, and tree size. These forests are common and may become even
more so in the future, but there is a lack of models for this type of forests. Future studies should also investigate ways to better incorporate site properties and weather into growth and yield
models to better deal with short- and long-term changes in temperature and precipitation.
The authors gratefully acknowledge the entire team at the Swedish National Forest Inventory and Unit for Field-based Forest Research, SLU, for providing field data and support.
Funding for this work was provided by the research program Trees and Crops for the Future (TC4F) and the Swedish Research Council for Sustainable Development (FORMAS) under grants 2018-00968 and
Declaration of openness of research materials, data, and code
The datasets and the custom R-code for model development and analysis are available upon reasonable request by contacting the corresponding author.
Authors’ contributions
MG, UN, and EM developed the research idea and methodology. MG led the work with data analysis, model development, and manuscript writing, with valuable support from UN, EM, and GV.
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Simplest Form Of Fractions 10 Mind Blowing Reasons Why Simplest Form Of
Simplest Form Of Fractions 10 Mind Blowing Reasons Why Simplest Form Of
Fractions In Simplest Form Worksheets | {"url":"https://reimbursementform.com/en/40-as-a-fraction-in-simplest-form.html","timestamp":"2024-11-01T19:38:31Z","content_type":"text/html","content_length":"21362","record_id":"<urn:uuid:0d09b40a-016b-4b17-a616-0d08349bc933>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00056.warc.gz"} |
Keras documentation: PReLU layer
PReLU layer
PReLU class
Parametric Rectified Linear Unit.
It follows:
f(x) = alpha * x for x < 0
f(x) = x for x >= 0
where alpha is a learned array with the same shape as x.
Input shape
Arbitrary. Use the keyword argument input_shape (tuple of integers, does not include the samples axis) when using this layer as the first layer in a model.
Output shape
Same shape as the input.
• alpha_initializer: Initializer function for the weights.
• alpha_regularizer: Regularizer for the weights.
• alpha_constraint: Constraint for the weights.
• shared_axes: The axes along which to share learnable parameters for the activation function. For example, if the incoming feature maps are from a 2D convolution with output shape (batch, height,
width, channels), and you wish to share parameters across space so that each filter only has one set of parameters, set shared_axes=[1, 2]. | {"url":"https://keras.io/2.15/api/layers/activation_layers/prelu/","timestamp":"2024-11-02T13:56:03Z","content_type":"text/html","content_length":"14192","record_id":"<urn:uuid:716f4508-ed81-4ebc-a17b-cc0ddfc2df3a>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00553.warc.gz"} |
Polynomial Inverse - Mike
Polynomial Inverse
Mike ●November 23, 2015●2 comments
One of the important steps of computing point addition over elliptic curves is a division of two polynomials. When working in $GF(2^n)$ we don't have large enough powers to actually do a division, so
we compute the inverse of the denominator and then multiply. This is usually done using Euclid's method, but if squaring and multiplying are fast we can take advantage of these operations and compute
the multiplicative inverse in just a few steps.
The first time I ran across this method it really confused me. The process itself does not depend on normal basis math, it just looks like it. Using normal basis to compute the squares means you only
need a barrel shifter, so for the field sizes mentioned in the "One Cycle Polynomial Math" blog, this method is ideal for either FPGA or embedded implementations.
We start off with a polynomial $\alpha$ in a field $GF(2^n)$ which means $$\alpha = \sum_{i=0}^{i=n-1}c_i\beta^{2^i}=\sum_{i=0}^{i=n-1}d_i\beta^i$$ where $c_i$ and $d_i$ have values $0$ or $1$. We
want to find its inverse $\alpha^{-1}$. Using Fermat's Little Theorem we can write $\alpha^{-1} = \alpha^{2^n - 2}$. The first step is to break this down by noticing that $$\alpha^{2^n-2} = (\alpha^
{2^{n-1}-1})^2$$ From here, things get a little more complicated.
What we are going to do is build a chain of terms all of the form $\alpha^{2^k-1}$. The final term in the chain will be $\alpha^{2^{n-1}-1}$ so we just need to square this result to have our inverse.
The chain will start with $\alpha^{2^1-1}=\alpha$.
Since we are going to do an addition chain, the next form to look at is is $\alpha^{2^{k+j}-1}$. Let's expand this: $$\begin{array}{c}\alpha^{2^{k+j}-1}=\alpha^{2^{k+j}-2^j+2^j-1}\\ =\alpha^{2^{k+j}
-2^j}\cdot \alpha^{2^j-1}\\ =(\alpha^{2^k-1})^{2^j}\cdot \alpha^{2^j-1}\end{array}$$ So if we have the forms $\alpha^{2^k-1}$ and $\alpha^{2^j-1}$ we can get the form $\alpha^{2^{k+j}-1}$ by squaring
the form $\alpha^{2^k-1}$ $j$ times.
A special version of this is when $k = j$. Putting $k+k=2k$ into the above we find $\alpha^{2^{2k}-1}=(\alpha^{2^k-1})^{2^k}\cdot \alpha^{2^k-1}$ So now we have the tools to build an addition chain.
There are many ways to create a chain of sums. The simplest is to look at the binary representation of a number, use the $2k$ form for each bit position, and add the terms with the bits set. As a
concrete example, let's look at $n=106$. The first step is to subtract $1$ so we start with $105$. In binary $105 = 1101001$. The most significant bit is $2^6$ so the first 6 terms in the chain are
$1, 2, 4, 8, 16, 32, 64$. The bit position for 2^5 is set so we add the form for $32$ to get $96$, the bit for position $2^3$ is set so we add form $8$ to get $104$, and then the last bit is set so
we add the term for $1$ to get $105$. The first term is just $\alpha$. Let's call the first term $\gamma_0$. We use the doubling form 6 times and get the following
│$\gamma_0$ │$\alpha^{2^1-1}$ │
│$\gamma_1=\gamma_0^2 \cdot \gamma_0$ │$\alpha^{2^2-1}$ │
│$\gamma_2=\gamma_1^{2^2}\cdot \gamma_1$ │$\alpha^{2^4-1}$ │
│$\gamma_3=\gamma_2^{2^4}\cdot\gamma_2$ │$\alpha^{2^8-1}$ │
│$\gamma_4=\gamma_3^{2^8}\cdot\gamma_3$ │$\alpha^{2^{16}-1}$│
│$\gamma_5=\gamma_4^{2^{16}}\cdot\gamma_4$ │$\alpha^{2^{32}-1}$│
│$\gamma_6=\gamma_5^{2^{32}}\cdot\gamma_5$ │$\alpha^{2^{64}-1}$│
Now we can begin the summation part of the chain. Bit 6 and bit 5 are set in the binary representation of 105 so we start with $\gamma_7=\gamma_6^{2^{32}}\cdot \gamma_5=\alpha^{2^{96}-1}$. Bit 3 is
the next bit set so $\gamma_8=\gamma_7^{2^8}\cdot \gamma_3 = \alpha^{2^{104}-1}$. Finally the last bit is set so we have $\gamma_9=\gamma_8^2\cdot \gamma_0=\alpha^{2^{105}-1}$.
Our final step is to compute the square of $\gamma_9\colon \gamma_9^2=(\alpha^{2^{105}-1})^2=\alpha^{2^{106}-2}=\alpha^{-1}$.
The thing that confused me the most about this algorithm was the number of times to square the first term. It depends on the second term. We can see this from the derivation of the $j+k$ form when we
add and subtract the second term's $2j$ value. We could have just as easily added and subtracted $2k$, but then $k$ would be the second term.
Another thing that confused me is the arbitrariness of the addition chain. A completely valid result can be derived using the sequence $1, 2, 3, 6, 12, 13, 26, 52, 104, 105$. This method "walks" down
the bit pattern. It takes the same number of steps as the doubling first and adding terms. It might take some experimentation to determine which method (or something else entirely) is better for a
particular application.
When implemented using normal basis representation, the squaring operations can be performed in a single cycle using a barrel shifter. If implemented in a field size which allows the type 1 normal
basis, then the multiplies can be also be done very quickly. Thus, for the right field size, computing an inverse using the above algorithm can be done in very few clock cycles compared with standard
implementations which can take 100's of clocks.
[ - ]
Comment by ●November 24, 2015
I haven't found a good addition chain algorithm to minimize the # of operations; Knuth talks about it a little bit in TAOCP, and there's djb's article on Pippenger's algorithm (see one of the
references in Wikipedia:
) which I've tried to implement but it's COMPLICATED and my implementation didn't seem to work properly.
[ - ]
Comment by ●November 25, 2015
As the Wikipedia article points out, finding the minimum is an NP complete problem. That's about as hard as it gets! For most applications I'm thinking about, you only have to figure it out once, for
one field size. So spending a little time staring at possible chains is worth while. If you can see patterns in the binary representation you can reduce the chain length a little. I have no clue how
you could automate that!
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pipe bend weight calculator
Pipe Bend Weight Calculator
Calculating the weight of pipe bends is crucial in various industries, including construction and manufacturing. Accurate calculations ensure safety and efficiency in designing and installing piping
systems. To simplify this process, we present a comprehensive pipe bend weight calculator along with an informative article to guide its usage.
How to Use
1. Input Data: Enter the required parameters such as pipe diameter, bend radius, and material density into the designated fields.
2. Click Calculate: Click the “Calculate” button to obtain the weight of the pipe bend.
3. View Result: The calculated weight will be displayed below the input fields.
The formula for calculating the weight of a pipe bend is derived from the equation for calculating the volume of a cylinder:
• W = Weight of the pipe bend (in pounds or kilograms)
• π ≈ 3.14159 (mathematical constant)
• d = Diameter of the pipe (in inches or millimeters)
• R = Bend radius (in inches or millimeters)
• t = Thickness of the pipe wall (in inches or millimeters)
• ρ = Material density (in pounds per cubic inch or kilograms per cubic millimeter)
Example Solve
Let’s consider a scenario where:
• Pipe Diameter (d) = 10 inches
• Bend Radius (R) = 24 inches
• Wall Thickness (t) = 0.5 inches
• Material Density (ρ) = 0.2836 pounds per cubic inch (for carbon steel)
Plugging these values into the formula:
W≈8.81 pounds
Thus, the weight of the pipe bend is approximately 8.81 pounds.
Q: Can I use this calculator for different materials?
A: Yes, simply input the appropriate material density for accurate results.
Q: What units should I use for input data?
A: You can use either inches or millimeters for dimensions and pounds per cubic inch or kilograms per cubic millimeter for material density.
Q: Is the calculation affected by the angle of the bend?
A: No, the formula provided accounts for the angle of the bend automatically.
Efficiently determining the weight of pipe bends is crucial for various industries. By utilizing our pipe bend weight calculator and understanding the underlying formula, professionals can streamline
their design and installation processes, ensuring optimal performance and safety. | {"url":"https://calculatordoc.com/pipe-bend-weight-calculator/","timestamp":"2024-11-06T12:18:06Z","content_type":"text/html","content_length":"92472","record_id":"<urn:uuid:6e2a778e-7fa8-47c4-aa74-cc59191089b8>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00531.warc.gz"} |
References and Further Reading
4.11 References and Further Reading
Constraint satisfaction techniques are described by Mackworth [1977a], Dechter [2003], Freuder and Mackworth [2006], and Dechter [2019]. The GAC algorithm was invented by Mackworth [1977b].
Variable elimination for propositional satisfiability was proposed by Davis and Putnam [1960]. VE for optimization has been called non-serial dynamic programming and was invented by Bertelè and
Brioschi [1972]. For a more recent overview see Dechter [2019], who calls variable elimination bucket elimination (the buckets contain the constraints to be joined).
Stochastic local search is described by Spall [2003] and Hoos and Stützle [2004]. The any-conflict algorithm is based on Minton et al. [1992]. Simulated annealing was invented by Kirkpatrick et al. [
1983]. Xu et al. [2008] describe the use of algorithm portfolios to choose the best solver for each problem instance.
Genetic algorithms were pioneered by Holland [1975]. A huge literature exists on genetic algorithms; for overviews, see Mitchell [1996], Bäck [1996], Goldberg [2002], Lehman et al. [2018], Stanley
et al. [2019], and Katoch et al. [2021]
Gradient descent is due to Cauchy [1847].
Python implementations that emphasize readability over efficiency are available at AIPython (aipython.org). | {"url":"https://artint.info/3e/html/ArtInt3e.Ch4.S11.html","timestamp":"2024-11-05T22:03:53Z","content_type":"text/html","content_length":"17982","record_id":"<urn:uuid:1f86099d-426e-4502-9ca8-2a2fca184821>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00583.warc.gz"} |
Talk:Chernick's Carmichael numbers - Rosetta CodeTalk:Chernick's Carmichael numbers
does a(10) exist?
Does anyone know whether a(10) actually exists?
I've checked all values of 'm' up to 16 billion and found nothing. This is in contrast to a(9) which only required 'm' equal to 950,560.
So, if a(10) does exist, it must be very large and, given the nature of the constraints, the probability of finding 10 primes which satisfy them is beginning to look low to me. --PureFox (talk)
18:48, 3 June 2019 (UTC)
a(10) was discovered today by Amiram Eldar (the author of the A318646 sequence) for m = 3208386195840. -- Trizen (talk) 12:10, 4 June 2019 (UTC)
Yep, knowing that, I've now found a(10) to be:
So my 16 billion wasn't even in the right ballpark and I estimate it would have taken my Go program about 8.5 days to find it, albeit on slow hardware. On a fast machine, using a faster compiler and
GMP for the big integer stuff, you might be able to get this down to a few hours but it's probably best to remove it as an optional requirement as I see you've now done. Interesting task nonetheless
so thanks for creating it. --PureFox (talk) 15:03, 4 June 2019 (UTC)
Or we could keep it. The task only gets interesting at 10. See below.--Nigel Galloway (talk) 09:34, 6 June 2019 (UTC)
In C++, using 64-bit integers, it's possible to compute a(10) in just ~3.5 minutes, with GMP for primality testing, preceded by trial division over odd primes up to p=23. It's very likely
that one of the 10 factors ((6*m+1), (12*m+1), (18m+1), ...) is composite, having a small prime divisor <= p, therefore we can simply skip it, without running a primality test on it. Another
big optimization for n > 5, noticed by Thundergnat (see the Perl 6 entry), and explained bellow by Nigel Galloway, is that m is also a multiple of 5, which makes the search 5 times faster. At
the moment, the largest known term is a(11) (with m = 31023586121600) and was discovered yesterday by Amiram Eldar. -- Trizen (talk) 14:27, 6 June 2019 (UTC)
Amazing speed-up as a result of these optimizations which, of course, are always obvious after some-one else has pointed them out :)
I've now added a second Go version (keeping the first version as the zkl entry is a translation of it). Even on my modest machine, this is now reaching a(10) in about 22 minutes so I'm
very pleased. Think I'll leave a(11) to others though :) --PureFox (talk) 10:07, 7 June 2019 (UTC)
The good news is no Casting out nines. The coefficients cₙ of the polynomials for 10 are:
n cₙ
For 10 m is divisible by 64, so the sequence begins 64 128 192 256 320 384 ... it is easy to see that this continues with 4 8 2 6 0 as the last digit. Which are the same last digits as cₙ. Consider
the following table which shows the final digit of the number produced by the polynomial cₙ(m)+1:
m 2 4 6 8
Remembering that prime numbers cannot end in 5 from the above I deduce that if m is not divisible by 320 then at least 1 of the polynomials must have a non prime result. This reduces the work to be
done by 80%. Using this I obtained an answer in 6h, 31min, 16,387 ms. This could easily be halved using pseudo-primes and validating the final result. There is an obvious multi-threading strategy
which could cut the time to 1/number of threads. More interestingly programming wise is start prime testing for a given m from the 10th polynomial. Observe that if this is not prime then for m*2 the
9th polynomial can not be prime (it is the same number). Similarly for m*4 the 8th polynomial can not be prime ... down to m*128 for the 3rd polynomial and then m*384 for the first polynomial. If the
10th polynomial is prime but the Nth (down to the 3rd) is not then the same applies just shifted. Keeping track of the m that do not need to be tested should reduce the time significantly.--Nigel
Galloway (talk) 09:34, 6 June 2019 (UTC)
Promote to Task
This draft task is clearly defined and has several good implementations. Are there objections to promoting to a task?
--DavidFashion (talk) 20:36, 14 February 2020 (UTC) | {"url":"https://rosettacode.org/wiki/Talk:Chernick%27s_Carmichael_numbers?oldid=290950","timestamp":"2024-11-15T01:15:13Z","content_type":"text/html","content_length":"49930","record_id":"<urn:uuid:c4aea417-de57-4f52-afd8-7a3cbd6e9ece>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00847.warc.gz"} |
1.3 Displacement Vector in 1D | Classical Mechanics | Physics | MIT OpenCourseWare
« Previous | Next »
Distance: Is the length of the path travelled by an object between two points in space. From its definition, the distance is a scalar and it is always a positive quantity.
Displacement: Is the change in the position of an object. If at time \(t=t_1\) the object is at position \(\vec{r}(t_1)\), and at a later time \(t=t_2 > t_1\) the object is at position \(\vec{r}(t_2)
\), the displacement vector is defined as \(\Delta \vec{r} = \vec{r}(t_2) - \vec{r}(t_1)\). In one dimension, the displacement vector has one component. For example, if the motion is along the
x-axis, the displacement vector becomes \(\Delta \vec{r} = \Delta x \hat{i} = (x(t_2) - x(t_1))\hat{i}\). The component of the displacement vector can be positive, when the final position is larger
than the initial one. It can be negative, when the final position is smaller than the initial one. It can aslo be zero, if the object ends at the starting point.
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About 11152163
Algebra, Elementary (3-6) Math, Geometry, Midlevel (7-8) Math
Bachelors in Electrical and Electronics Engineering from Ryerson University
Masters in Electrical and Electronics Engineering from Oakland University
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I have 20+ years of experience as a professional engineer and have amassed a wealth of theoretical and practical knowledge in science, physics and math. I've successfully mentored dozens of engineers
during those years and have played pivotal roles in many young lives throughout my career. As a full-time teacher, I taught middle and high school math, science, physics and electronics at a
community school and continued providing grade-level tutoring on a voluntary basis for many years while working as an engineer.
I Love Tutoring Because
it fulfills a lifelong desire to share my knowledge and experience with students. My greatest satisfaction comes when students grasp the concepts I present and gain the confidence to dig deeper into
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Math - Geometry
Glenn was awesome
FP - Mid-Level Math
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Ch. 2 Key Concepts - Calculus Volume 2 | OpenStax
Key Concepts
2.1 Areas between Curves
• Just as definite integrals can be used to find the area under a curve, they can also be used to find the area between two curves.
• To find the area between two curves defined by functions, integrate the difference of the functions.
• If the graphs of the functions cross, or if the region is complex, use the absolute value of the difference of the functions. In this case, it may be necessary to evaluate two or more integrals
and add the results to find the area of the region.
• Sometimes it can be easier to integrate with respect to y to find the area. The principles are the same regardless of which variable is used as the variable of integration.
2.2 Determining Volumes by Slicing
• Definite integrals can be used to find the volumes of solids. Using the slicing method, we can find a volume by integrating the cross-sectional area.
• For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the
cross-sections are circles, and using the formula for the area of a circle.
• If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before
2.3 Volumes of Revolution: Cylindrical Shells
• The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. This method is sometimes preferable to either the method of
disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other.
• The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.
2.4 Arc Length of a Curve and Surface Area
• The arc length of a curve can be calculated using a definite integral.
• The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions
of $y.y.$
• The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
• The integrals generated by both the arc length and surface area formulas are often difficult to evaluate. It may be necessary to use a computer or calculator to approximate the values of the
2.5 Physical Applications
• Several physical applications of the definite integral are common in engineering and physics.
• Definite integrals can be used to determine the mass of an object if its density function is known.
• Work can also be calculated from integrating a force function, or when counteracting the force of gravity, as in a pumping problem.
• Definite integrals can also be used to calculate the force exerted on an object submerged in a liquid.
2.6 Moments and Centers of Mass
• Mathematically, the center of mass of a system is the point at which the total mass of the system could be concentrated without changing the moment. Loosely speaking, the center of mass can be
thought of as the balancing point of the system.
• For point masses distributed along a number line, the moment of the system with respect to the origin is $M=∑i=1nmixi.M=∑i=1nmixi.$ For point masses distributed in a plane, the moments of the
system with respect to the x- and y-axes, respectively, are $Mx=∑i=1nmiyiMx=∑i=1nmiyi$ and $My=∑i=1nmixi,My=∑i=1nmixi,$ respectively.
• For a lamina bounded above by a function $f(x),f(x),$ the moments of the system with respect to the x- and y-axes, respectively, are $Mx=ρ∫ab[f(x)]22dxMx=ρ∫ab[f(x)]22dx$ and $My=ρ∫abxf(x)dx.My=
• The x- and y-coordinates of the center of mass can be found by dividing the moments around the y-axis and around the x-axis, respectively, by the total mass. The symmetry principle says that if a
region is symmetric with respect to a line, then the centroid of the region lies on the line.
• The theorem of Pappus for volume says that if a region is revolved around an external axis, the volume of the resulting solid is equal to the area of the region multiplied by the distance
traveled by the centroid of the region.
2.7 Integrals, Exponential Functions, and Logarithms
• The earlier treatment of logarithms and exponential functions did not define the functions precisely and formally. This section develops the concepts in a mathematically rigorous way.
• The cornerstone of the development is the definition of the natural logarithm in terms of an integral.
• The function $exex$ is then defined as the inverse of the natural logarithm.
• General exponential functions are defined in terms of $ex,ex,$ and the corresponding inverse functions are general logarithms.
• Familiar properties of logarithms and exponents still hold in this more rigorous context.
2.8 Exponential Growth and Decay
• Exponential growth and exponential decay are two of the most common applications of exponential functions.
• Systems that exhibit exponential growth follow a model of the form $y=y0ekt.y=y0ekt.$
• In exponential growth, the rate of growth is proportional to the quantity present. In other words, $y′=ky.y′=ky.$
• Systems that exhibit exponential growth have a constant doubling time, which is given by $(ln2)/k.(ln2)/k.$
• Systems that exhibit exponential decay follow a model of the form $y=y0e−kt.y=y0e−kt.$
• Systems that exhibit exponential decay have a constant half-life, which is given by $(ln2)/k.(ln2)/k.$
2.9 Calculus of the Hyperbolic Functions
• Hyperbolic functions are defined in terms of exponential functions.
• Term-by-term differentiation yields differentiation formulas for the hyperbolic functions. These differentiation formulas give rise, in turn, to integration formulas.
• With appropriate range restrictions, the hyperbolic functions all have inverses.
• Implicit differentiation yields differentiation formulas for the inverse hyperbolic functions, which in turn give rise to integration formulas.
• The most common physical applications of hyperbolic functions are calculations involving catenaries. | {"url":"https://openstax.org/books/calculus-volume-2/pages/2-key-concepts","timestamp":"2024-11-06T06:07:29Z","content_type":"text/html","content_length":"320024","record_id":"<urn:uuid:dc725c2c-9c4e-4fa3-ae72-7e428d1e1351>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00024.warc.gz"} |
Sequential measurement in quantum learning
Spencer-Wood, Hector (2024) Sequential measurement in quantum learning. PhD thesis, University of Glasgow.
Full text available as:
In an increasingly quantum world with more and more quantum technologies nearing practical use, the importance of interacting directly with quantum data is becoming clear. Although doing so often
leads to advantages, it also presents us with some uniquely quantum challenges: for example, information about a quantum system cannot, in general, be extracted without disturbing the state of the
system. In this thesis, we primarily focus on how performing a learning task on quantum data disturbs it, and affects one’s ability to learn about it again in the future. In particular, we focus on
the learning task of unsupervised binary classification, and how it affects quantum data when it is performed on a subset of it. In such a binary classification task, we are given a dataset that is
made up of qubits that are each in one of two unknown pure states, and our aim is to cluster, with optimal probability of success, the data points into two groups based on their state. To investigate
how well we can perform this task sequentially, we first consider a base case of a three-qubit dataset, made up of qubits that are each in one of two unknown states, and investigate how an
intermediate classification on a two-qubit subset affects our ability to subsequently classify the whole dataset. We analytically derive and plot the tradeoff between the success rates of the two
classifications and find that, although the intermediate classification does indeed affect the subsequent one in a non-trivial way, there is a remarkably large region where the first classification
does not force the second away from its optimal probability of success. We then describe this scenario as a quantum circuit and simulate the tradeoff using Qiskit’s AerSimulator. Following on from
this, we go on to investigate whether an intermediate measurement can leave a subsequent one unaffected in the more general setting of an n-qubit dataset, again made up of qubits that are each in one
of two unknown states. We see that numerics hint that nothing about the order of the qubits in a (n − 1)-qubit dataset can be learnt without affecting a subsequent classification on the full dataset.
We make steps to prove that this is indeed the case and show that an immediate consequence of this is that, for some m > 1, a non-trivial intermediate classification on n−m qubits will always
negatively affect a subsequent one on all n qubits. We conclude this line of work by deriving two bounds to how successful an intermediate classification of n − 1 qubits can be without affecting the
following n-qubit one, hypothesising that one of these is optimal.
We then shift our focus to the field of indefinite causal order (ICO). Motivated by ICO’s connection to non-commutivity, we explore the idea of implementing quantum key distribution (QKD) in an
indefinite causal regime. After showing that it is possible to share a key in an ICO, we find that, unlike other QKD protocols in the literature, eavesdroppers can be detected without publicly
discussing a subset of the shared key. Indeed, we show that this is true for any individual attack in which the eavesdroppers abide by the causal structure chosen by the sharing parties. Further, we
prove the security of this protocol for a subclass of these individual attacks. We then ask whether this “private detection” is a truly consequence of ICO and show that there is a definite causal
ordered strategy that appears to yield the same phenomenon. Although we note that there are hints of some more subtle differences between the definite and indefinite causal cases, we conclude that
carrying out QKD in an ICO is unlikely to offer any advantage, at least when considered in the form that we did. Finally, we close this thesis by summarising what we have found and noting some
possible directions for future study.
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Add and Subtract Radical Expressions (Basic)
Add and Subtract Basic Radical Expressions 1
Add and Subtract Basic Radical Expressions 2
Add and Subtract Basic Radical Expressions 3
Add and Subtract Basic Radical Expressions 4
Add and Subtract Basic Radical Expressions 5
Add and Subtract Basic Radical Expressions 6
Add and Subtract Basic Radical Expressions 7
Add and Subtract Basic Radical Expressions 8
Add and Subtract Basic Radical Expressions 9
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Quantum Protocols for ML, Physics, and Finance
In this thesis, we give new protocols that offer a quantum advantage for problems in ML, Physics, and Finance.
Quantum mechanics gives predictions that are inconsistent with local realism.
The experiment proving this fact (Bell, 1964) gives a quantum protocol that is impossible to replicate using classical resources.
Such a situation, where quantum techniques outperform classical methods, is known as a quantum advantage.
Firstly, we consider the problem of adversarial robustness from ML.
Modern machine learning models have been shown to be vulnerable to small, adversarially chosen perturbations of the input.
In the standard setting, one allows perturbations that are small in the $\ell_p$ norm.
We introduce two new models and give a defense against adversarial examples in each of them.
In the first, fully classical model, we consider an adversary which is not limited by the type of perturbations he can apply but when presented with a classifier can repetitively generate random
adversarial examples.
We design a protocol in which interaction with such an adversary leads to a provable defense.
Our second model is influenced by the quantum PAC-learning setting introduced in Bshouty and Jackson, 1995.
We assume that the adversary is quantum, computationally bounded but unrestricted in other aspects.
In this model, we give an interactive protocol between the learner and the adversary that guarantees robustness.
Moreover, we show that it offers a quantum advantage as it beats a classical lower bound from Goldwasser et al., 2020.
Secondly, we consider the problem of Entanglement vs Nonlocality from Physics.
Axioms of quantum theory lead to a natural division of states into entangled and separable.
This dichotomy relies purely on a mathematical formalism.
The question of Entanglement = Nonlocality asks if it is true that for every entangled state $\rho_{AB}$ there exists an experiment between Alice and Bob and a verifier that distinguishes between two
cases: (i) Alice and Bob share $\rho_{AB}$ versus (ii) Alice and Bob share only a separable state.
It was shown (Barrett, 2002) that in the standard Bell scenario the answer is no.
We introduce a new model in which the parties are computationally bounded.
In this model, we show that if the Learning with Errors problem is quantumly hard then Entanglement = Nonlocality.
We also show a complementary result, i.e. we prove that Entanglement = Nonlocality implies $\mathtt{BQP} \neq \mathtt{\PP}$.
These two results connect this foundational problem in quantum mechanics to computational complexity.
As an implication of this connection, we show the impossibility of some protocols for the delegation of quantum computation.
Thirdly, we consider the problem of arbitrage from Finance.
We consider a situation where two parties trade securities in distant locations.
We give a quantum protocol that improves the risk-return tradeoff compared to what is possible for classical strategies.
The protocol is based on the experiment from Bell, 1964 and requires only two entangled qubits.
Due to its simplicity, this scheme might be realizable in practice in the near future.
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Mathematicians resolve the ‘not possible’ quantity puzzle | world - Bournemouthjobcentre
1 of two A mathematician solves the ‘not possible’ quantity puzzle – Picture: Pexels
Mathematicians resolve the ‘not possible’ quantity puzzle – Picture: Pixels
After three a long time of attempting, mathematicians have managed to find out the worth of a fancy quantity that was beforehand thought-about not possible to compute. Utilizing supercomputers, two
teams of researchers have revealed the ninth Dedekind quantity, or D (9) – a sequence of integers alongside the strains of recognized primes or the Fibonacci sequence.
Among the many many mysteries of arithmetic, Dedekind’s numbers, found within the nineteenth century by German mathematician Richard Dedekind, have captured the creativeness and curiosity of
researchers through the years.
Till just lately, solely Dedekind’s quantity eight was recognized, and it was solely unveiled in 1991. However now, in a shocking flip of occasions, two unbiased analysis teams from the Catholic
College of Leuven in Belgium and the College of Paderborn in Germany have achieved the unthinkable and solved the issue. sports activities.
Each research have been submitted to the arXiv preprint server: the primary on April 5 and the second on April 6. Though not but peer-reviewed, each analysis teams have come to the identical
conclusion – suggesting that Dedekind’s ninth quantity has lastly been decoded.
Dedekind’s ninth quantity, or D (9).
The worth of the ninth Dedekind quantity is calculated to be 286,386,577,668,298,411,128,469,151,667,598,498,812,366. D(9) has 42 digits in comparison with D(8) which has 23 digits.
Every Dedekind quantity represents the variety of attainable configurations of a given sort of true-false logical operation in numerous spatial dimensions. The primary quantity within the sequence, D
(0), represents the zero dimension. So D(9), which represents 9 dimensions, is the tenth quantity within the sequence.
The idea of Dedekind numbers is difficult to grasp for many who don’t like arithmetic. His calculations are very advanced, because the numbers on this sequence enhance exponentially with every new
dimension. Which means it will get more durable and more durable to quantify, in addition to it will get greater and larger – which is why the worth of D(9) has lengthy been seen as not possible to
“For 32 years, calculating D(9) was an open problem, and it was questionable whether or not it was ever attainable to calculate this quantity,” says pc scientist Lennart Van Hirtum of the College
of Paderborn, writer of one of many research.
Dedekind numbers are an growing sequence of integers. Its logic is predicated on “Montonic Boolean Capabilities” (MBFs), which choose an output primarily based on inputs that encompass solely two
attainable (binary) states, corresponding to true and false, or 0 and 1.
Boolean unary features constrain logic in such a approach that altering the quantity 0 to 1 on just one enter causes the output to alter from 0 to 1, not from 1 to 0. As an example this idea,
researchers use purple and white, as an alternative of 1 and 0 , though the essential concept is similar.
“Basically, you possibly can consider a monotonous logical operate in two, three and infinite dimensions, like a sport with a dice of n dimensions. You stability the dice on a cable after which paint
every of the remaining corners white and purple,” van Hertom explains.
“There is just one rule: you need to by no means place a white nook on prime of a purple nook. This creates a sort of vertical red-and-white cross. The thing of the sport is to see what number of
divisions there are.”
Thus, the Dedekind quantity represents the utmost variety of intersections that may happen in a dice of n dimensions that satisfies the rule. On this case, the n dimensions of the dice correspond to
the Dedekind quantity n.
For instance, Dedekind’s eighth quantity has 23 digits, which is the utmost variety of completely different divisions that may be made in an eight-dimensional dice that satisfies the rule.
In 1991, the Cray-2 supercomputer (one of the vital highly effective computer systems of the time, however much less highly effective than a contemporary smartphone) and mathematician Doug Wiedemann
took 200 hours to calculate D(8).
D(9) has virtually twice as many digits and was calculated utilizing the Noctua 2 supercomputer on the College of Paderborn. This supercomputer is able to performing a number of mathematical
operations on the identical time.
Because of the computational complexity of calculating D(9), the staff used the P coefficient formulation developed by Van Hirtum’s thesis advisor, Patrick de Causmaecker. Doing the modulus P permits
D(9) to be computed utilizing a big sum as an alternative of calculating every time period within the sequence.
In our case, profiting from the symmetries of the formulation, we have been in a position to scale back the variety of phrases to solely 5.5 * 10^18, which is a big quantity. By comparability, the
variety of grains of sand on Earth is 7.5 * 10^18, which isn’t one thing to smell out, however to a pc Nonetheless, this course of is totally manageable,” says van Hertom.
Nonetheless, the researcher believes that Dedekind’s tenth account requires a extra trendy pc than those presently in existence.
“If we calculate it now, then a processing energy equal to the total energy of the solar will probably be required,” van Hertom informed the portal. Science lives. He added that this makes
computation “virtually not possible”.
2 of two Numerical sequence was found by Richard Dedekind within the nineteenth century. – Picture: Wikimedia Commons
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Building an If/Than for a Pass Fail Based on a Cell Value
Hello Community,
I work in a shoe factory in quality control and am currently using smartsheet to enter via form our test results for bonding strength for sole adhesion tests we do in our lab. I am trying to create a
statement formula in which If the cell value is Greater than or equal to 80, it is a pass, less than it's a fail. That is an easy formula, no issue getting that to work. After I put it into place, I
realized that sometimes, i'm only getting one sample (only one right shoe, no left or vice versa), so i can't base it off of both columns, the logic would have to be based on if i have entries for
that value or not. If that makes sense....
Here's some pictures to help put it into place in a test sheet that i have:
Primary column is where I have my basic formula of =IF(AND([Column2]@row >= 80, [Column3]@row >= 80), "PASS", "FAIL").
Column 2 is an average of an array of cells where operators enter in their result for the left sample.
Column 3 is an average of an array of cells where operators enter in their result for the right sample.
Row 2 in this test has a value of zero for column 2 as we only received one sample, but the value in column 3 passes.
Is there a way to set up a formula so that when column2 equals 0, it only takes into account column3's data?
Best Answers
• Hey Walter
Did you try just copy pasting my formula in? It looks like parentheses are out of place in the second IF (your original formula).
• Hey Walter
Using my formula, I cannot replicate the error. I can replicate this error, however, when I incorrectly position the parentheses. When you converted to the column formula - were you in the cell
with the corrected formula?
Please again check your formula, perhaps again trying to copy paste my formula above into your sheet.
• Hey @Walter Mootz
=IF(AND(COUNTIFS([Column 2]@row:[Column 3]@row, @cell > 0) = 1, SUM([Column 2]@row:[Column 3]@row) >= 80), "PASS", IF(AND([Column 2]@row >= 80, [Column 3]@row >= 80), "PASS", "FAIL"))
This allows for either column to be a zero and the score comes from the completed column. If both columns are populated then both scores are taken into account.
Is this what you need?
• I ensured that i had the formula correct but got an invalid argument.
Here is a snap of my formula copied from what you posted:
Here is what I get when I hit enter:
• Hey Walter
Did you try just copy pasting my formula in? It looks like parentheses are out of place in the second IF (your original formula).
• I just suck and didn't realize i had that last parentheses in the wrong spot. Thank you Kelly for the fix. I appreciate it. I was close on what i was thinking originally, but mine involved way
too many if statements compared to yours.
Thank you so much!
• @Kelly Moore One thing i'm noticing, it works on the one where there's a zero in column 2, but not on the others.... I do have it set for a column formula, not an individual cell.
• Hmmm. I’ll have to check later after meetings. When I tested last night I tested all combination of column entries and all worked.
• Hey Walter
Using my formula, I cannot replicate the error. I can replicate this error, however, when I incorrectly position the parentheses. When you converted to the column formula - were you in the cell
with the corrected formula?
Please again check your formula, perhaps again trying to copy paste my formula above into your sheet.
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repeat unit molecular weight of polystyrene
Similarly one may ask, how do you find the number of repeat units? Students also viewed these Thermodynamics questions (b) Compute the number-average molecular weight for a polystyrene for which the
degree of polymerization is 26,000. Compute the molecular weight of a singe (1), polystyrene (PS) repeat unit. is a molecular structure, defined by the monomer, constituting part of the repeating
unit. Solution (a) The repeat unit molecular weight of polystyrene is called for in this portion of the problem. Compute the repeat unit molecular weight of polystyrene. Ionization Techniques • MALDI
– Typically ... – Ag and Cu with polystyrene polyisoprene etc. The structure of the polymer repeating unit can be represented as: The presence of the pendant phenyl (C 6 H 5) groups is key to the
properties of polystyrene. Expert Answer 100% … The molecular weight of a particular polymer molecule is a product of the degree of polymerization and the molecular weight of the repeating unit.
Polystyrene average Mw 35,000; CAS Number: 9003-53-6; Synonym: PS; Linear Formula: [CH2CH(C6H5)]n; find Sigma-Aldrich-331651 MSDS, related peer-reviewed papers, technical documents, similar products
& more at Sigma-Aldrich. Compute the repeat unit molecular weight of polystyrene. Molecular weight is an important parameter for synthetic polymers ... mass of repeat units and end-group mass
structure. For For a polystyrene solution with a bimodal molecular weight distribution, the coefficient A is smaller than for either of the monodisperse polystyrene solutions. Polyvinylidine fluoride
is usually made using emulsion polymerization. monomer. Formula to Calculate Number Average Molecular Weight. 14.4 (a) Compute the repeat unit molecular weight of polypropylene. You need to know the
number of repeat units in the chain and multiply the monomer molecular weight by that number. A . 104. b. 260. c. 401. d. 2600. e. 13. The repeat unit for polystyrene (table 14. Solution (a) The
repeat unit molecular weight of polystyrene is called for in this portion of the problem. m n= M = 350, 000 g/mol 4425 = 79.10 g/mol Thus, the molecular weight of the butadiene repeat unit is : …
When considering the molecular weight of a polymer for the above calculation, we usually take either the number average molecular weight (M n) or weight average molecular weight (M w). For a given
molecular weight the relaxation spectra at ... energy represented by this distribution was associated with an activation energy required for the motion of a chemical repeat unit. Solution (a) The
repeat unit molecular weight of polystyrene is called for in this portion of the problem. (b) Compute the number-average molecu- lar weight for a polypropylene for which the degree of polymerization
is 15,000. (b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000. You are right. 3 the number-average molecular weight of a polystyrene is
500,000 g/mol. (a) Compute the repeat unit molecular weight of polystyrene. Compute the repeat unit molecular weight of polystyrene and the number-average molecular weight of the material if the
degree of polymerisation is 25000. The effect of nanoconfinement on the glass transition temperature (Tg) of supported polystyrene (PS) films is investigated over a broad molecular weight (MW) range
of 5000−3 000 000 g/mol. We are asked to compute the degree of polymerization for polystyrene, given that the number- average molecular weight is 500,000 g/mol. Now it is possible to compute the
degree of polymerization using Equation 14.6 as 14.4 (a) Compute the repeat unit molecular weight of polystyrene. • … For polystyrene, from Table 14.3, each repeat unit … (b) Compute the
number-average molecular weight for a… Get the answers you need, now! Polystyrene MW is shown to have no significant impact on the film thickness dependence of Tg − Tg,bulk. Repeating units Polymer
molecules are very large compared with most other molecules , so the idea of a repeat unit is used when drawing a displayed formula. molecular weight of ... a function of polystyrene molecular weight
(P w) [28]. (a) Compute the repeat unit molecular weight of polystyrene. Select one: a. Problem 4.4 (a) Compute the repeat unit molecular weight of polypropylene. T able 8.7 Molecular Weights of PET
Samples . (b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000. The molecular weight is the sum of the weights of all the repeat units
plus the weight of the end groups. MW = 1,000 units x 104.15 Da/unit + 2 Da = 104,152 Da. 3) has eight carbon atoms and eight hydrogen atoms. 14. 7.7 for the mer structure of polycarbonate.) It is
defined also as monomer or monomeric unit, but not always in the correct way. The effect of nanoconfinement on the glass transition temperature (T g) of supported polystyrene (PS) films is
investigated over a broad molecular weight (MW) range of 5000-3 000 000 g/mol. Reinhard Kniewske, Werner‐Michael Kulicke, Study on the molecular weight dependence of dilute solution properties of
narrowly distributed polystyrene in toluene and in the unperturbed state, Die Makromolekulare Chemie, 10.1002/macp.1983.021841021, 184, 10, (2173-2186), (2003). A . (Use Table 4.3 in text book for
mer structure.) Its molecular weight can be as high as 5 × 10 6. 100 repeat units per polymeric chain with an average . When drawing one, you need to: Solution 4.4 (a) The repeat unit molecular
weight of polypropylene is called for in this portion of the problem. is a low molecular weight compound from which the polymer is obtained through the synthesis reaction. Calculate its degree of
polymerization. Solid polystyrene is transparent, owing to these large, ring-shaped molecular groups, which prevent the polymer chains from packing into close, crystalline arrangements. (b) Compute
the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000. Answer b: MW = 200 units x 72.06 Da/unit + 2 Da = 14,412 Da (based on 200 simple repeating
units, but this is a complicated case. If you can determine the molecular weight of the polymer chain (end group analysis, mass spectrometry (MALDI, preferably), gel permeation chromatography) then
you divide the obtained mass by the molecular weight of the repeat unit. The particular molecule contains 5.35 x 10 3 of repeat units. (a) Compute the repeat unit molecular weight of polystyrene. ...
energy extracted from the temperature shift in the relaxation spectra corresponded to the motion of a statistical unit (Kuhn's segment) in polystyrene. m = 3(AC) + 6(AH) = (3)(12.01 g/mol) + (6)
(1.008 g/mol) = 42.08 g/mol. 14.5 Below, molecular weight data for a poly- tetrafluoroethylene material are tabulated. (a) From Equation 14.6, the average repeat unit molecular weight of the
copolymer, m calculated as:, is DP From Table 14.5, the butadiene repeat unit contains 4 Carbon atoms and 6 Hydrogen atoms. General-purpose polystyrene is clear, hard, and rather brittle. M.W. (b)
Compute the number-average molecular weight for a polypropylene for which the degree of polymerization is 15,000. This can be accounted for by the molecular weight dependence of the density of repeat
units of a polymer coil in solution. In this study, synchrotron X-ray scattering and quantitative data analysis were performed on a series of 33-armed polystyrenes with four different arm… It is well
known that the glass transition temperatures, Tgs, of supported polystyrene (PS) films decrease dramatically with decreasing film thickness below 60-80 nm. 5.4 (a) Compute the repeat unit molecular
weight of polystyrene. (b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000. Compute the degree of polymerization. However, a detailed
understanding of the cause of this effect is lacking. Polystyrene 19K (5 mg/mL in THF) MALDI micro MX spec-trum. (See Sec. Polystyrene (PS) / ˌ p ɒ l i ˈ s t aɪ r iː n / is a synthetic aromatic
hydrocarbon polymer made from the monomer known as styrene. Get the detailed answer: Compute the repeat unit molecular weight of polypropylene, Compute the number-average molecular weight for a
polypropylene for whi 14.3 The number-average molecular weight of a polystyrene is 500,000 g/mol. molecular weight of polymer (g/mol) 12,000 g/mol DP molecular weight of mer (g/mol/mer) 226 g/mol/mer
===53 mers 7.4 An injectionŒmolding polycarbonate material has an average molecular weight of 25,000 g/mol. (b) Compute the number-average molecular weight for a polystyrene for which the degree of
polymerization is 25,000. It is an inexpensive resin per unit weight. About Polystyrene; 1 cubic meter of Polystyrene weighs 1 060 kilograms [kg] 1 cubic foot of Polystyrene weighs 66.17364 pounds
[lbs] Polystyrene weighs 1.06 gram per cubic centimeter or 1 060 kilogram per cubic meter, i.e. For instance a particular polythylene molecule with \(DP = 1000\) will have a molecular weight of
28,000. CHE 253 Homework #5--solution Due Date: February 26, 2014 4.4 (a) Compute the repeat unit molecular weight of polystyrene. The repeat unit molecular weight of polypropylene is just.
Polystyrene can be solid or foamed. 104.4 g/mol Compute the degree of polymerization for a polystyrene (PS) polymer with a number-average molecular weight … monomeric unit. • Molecular Weight
Distribution – Mn - Number Average – Mw - Weight Average – Pd - Polydispersity • Repeat Unit Mass – Repeat Unit Molecular Formula • End Group Mass – End Group Molecular Formula. , polystyrene ( PS )
repeat unit molecular weight of polystyrene is called for in this portion of the.! Hard, and rather brittle polythylene molecule with \ ( DP = 1000\ ) have! Plus the weight of polypropylene is just )
has eight carbon atoms and eight hydrogen.... In solution eight hydrogen atoms Use Table 4.3 in text book for mer structure ). ) will have a molecular structure, defined by the molecular weight is
the sum of weights. 28 ] eight hydrogen atoms is called for in this portion of the end groups in solution … the. Weight compound from which the degree of polymerization is 15,000 5.35 x 10 of... The
monomer, constituting part of the cause of this effect is lacking... – Ag and Cu polystyrene. Monomeric unit, but not always in the correct way ), polystyrene ( PS ) repeat unit is.. Constituting
part of the cause of this effect is lacking atoms and eight hydrogen atoms molecular. 19K ( 5 mg/mL in THF ) MALDI micro MX spec-trum find the number repeat... Material if the degree of
polymerization is 25,000 weight data for a polystyrene is called for in this of. In solution the number-average molecular weight of polystyrene the sum of the cause of repeat unit molecular weight of
polystyrene effect is lacking the you. Weight by that number the answers you need, now unit, but not always the! As high as 5 × 10 6 is defined also as monomer or monomeric unit, but not in! Smaller
than for either of the problem polymerisation is 25000 by that.! Is 26,000 the monomer, constituting part of the end groups structure, by... The repeating unit on repeat unit molecular weight of
polystyrene film thickness dependence of Tg − Tg bulk., molecular weight of polystyrene may ask, how do you find the number of repeat?. Below, molecular weight of polypropylene is called for in this
portion of monodisperse. Monomeric unit, but not always in the chain and multiply the,! 500,000 g/mol number- average molecular weight of a polystyrene for which the degree polymerization. Mass of
repeat units solution 4.4 ( a ) Compute the number-average molecular weight ( P )... • MALDI – Typically... – Ag and Cu with polystyrene polyisoprene etc Table! Particular polymer molecule is a
molecular structure, defined by the molecular weight of particular... ) [ 28 ] molecule with \ ( DP = 1000\ ) will a... Need to know the number of repeat units of a singe ( 1 ), (. Molecu- lar weight
for a… Get the answers you need, now Tg, bulk bimodal molecular weight polystyrene! A… Get the answers you need to: Compute the number-average molecular weight data for polystyrene... Ag and Cu with
polystyrene polyisoprene etc mw = 1,000 units x 104.15 repeat unit molecular weight of polystyrene + Da! Particular polythylene molecule with \ ( DP = 1000\ ) will have a molecular is... One, you
need to: Compute the repeat unit molecular weight is an important parameter for polymers. But not always in the correct way how do you find the number of repeat units of a (... As 5 × 10 6 of repeat
units particular molecule contains 5.35 x 10 3 of repeat?. Mass of repeat units plus the weight of polypropylene is just polystyrene for which the polymer obtained... Need to know the number of
repeat units is called for in this portion of the groups... Solution 4.4 ( a ) Compute the repeat unit molecular weight of a polystyrene for which repeat unit molecular weight of polystyrene of! And
rather brittle b ) Compute the number-average molecular weight of the material the... Is obtained through the synthesis reaction is the sum of the problem with bimodal! Weight for a polystyrene for
which the degree of polymerisation is 25000 unit molecular weight of polystyrene brittle! ) has eight carbon atoms and eight hydrogen atoms molecule contains 5.35 x 10 3 of repeat plus! Maldi –
Typically... – Ag and Cu with polystyrene polyisoprene etc.... In this portion of the cause of this effect is lacking weight ( P w ) 28! A polystyrene for which the degree of polymerization is 25,000
( 5 mg/mL in THF MALDI. And rather brittle poly- tetrafluoroethylene material are tabulated and multiply the monomer, constituting part of monodisperse. High as 5 × 10 6 material are tabulated in
this portion of the weights all! Cause of this effect is lacking weights of all the repeat unit molecular weight of material... ( 1 ), polystyrene ( PS ) repeat unit molecular weight a…! High as 5 ×
10 6 units x 104.15 Da/unit + 2 =. Defined by the monomer, constituting part of the degree of polymerisation repeat unit molecular weight of polystyrene 25000 to have no significant impact the... Of
28,000 ), polystyrene ( PS ) repeat unit molecular weight polystyrene. Polymerization for polystyrene, given that the number- average molecular weight of a polymer coil in solution of... 3 the
number-average molecular weight of the repeating unit for polystyrene, given that the number- average weight. You find the number of repeat units multiply the monomer, constituting part of the if.
For by the monomer, constituting part of the material if the degree of polymerization 25,000... Is 15,000 for either of the weights of all the repeat units multiply... Is defined also as monomer or
monomeric unit, but not always in the and... Problem 4.4 ( a ) the repeat unit molecular weight for a polystyrene for which the degree of and! Data for a polystyrene is 500,000 g/mol units in the
correct way general-purpose is. Function of polystyrene is called for in this portion of the repeating.. Molecule is a low molecular weight of the problem molecule contains 5.35 x 10 3 of repeat
units 5.35! – Ag and Cu with polystyrene polyisoprene etc rather brittle atoms and eight hydrogen atoms understanding of the cause this. Is 25000 is 15,000 THF ) MALDI micro MX spec-trum is called
for in portion! The polymer is obtained through the synthesis reaction in the correct way PS ) repeat unit molecular for. Thickness dependence of the monodisperse polystyrene solutions text book for
mer structure. all the repeat unit molecular weight a! With a bimodal molecular weight for a polystyrene for which the degree of is. 1000\ ) will have a molecular structure, defined by the monomer,
constituting part of repeating! This effect is lacking rather brittle can be as high as 5 10! Parameter for synthetic polymers... mass of repeat units repeat unit molecular weight of polystyrene, and
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Least Common Multiple
The Least Common Multiple (LCM) of a group of numbers is the smallest number that is a multiple of all the numbers.
You can find the LCM of two or more numbers through various methods: Division method and Multiple of a Number.
Let us help you find the LCM for your numbers. Just enter the numbers in boxes below.
To find the least common multiple of two numbers just type them in and get the solution.
Some of the "Least Common Multiple" | {"url":"https://www.geteasysolution.com/least-common-multiple","timestamp":"2024-11-06T20:06:57Z","content_type":"text/html","content_length":"29185","record_id":"<urn:uuid:29f19cdb-dd8c-48b7-af14-c3e328143374>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00484.warc.gz"} |
Automata Theory and Formal Language CS-6205 exam answers
__ is a finite set of symbols called the alphabet.
__ is the initial state from where any input is processed (q0 ∈ Q).
______ is used to derive a string using the production rules of a grammar.
“C if H” is also the same as saying _______.
“If H then ” is also the same as saying _______.
“Whenever H holds, C follows” can also be said as _______.
{Q, Σ, q0, F, δ, Q × Σ → Q} is…
* Starts from tree leaves. It proceeds upward to the root which is the starting symbol S.
* Starts with the starting symbol S. It goes down to tree leaves using productions.
A __________ of a derivation is a tree in which each internal node is labeled with a nonterminal.
A compact textual representation of a set of strings representing a language.
A conceptual design for a machine consisting of a Mill, Store, Printer, and Readers.
A DFA can remember a finite amount of information, but a can remember an infinite amount of information.
A Finite Automaton with null moves (FA-ε) does transit not only after giving input from the alphabet set but also without any input symbol. This transition without input is called a
A formal language is a set of strings, each string composed of symbols from a finite set called.
A general purpose, programmable, information processor with input and output.
A good regular expression for any valid real number using engineering notation is ______________.
A good regular expression for any valid whole number is ______________.
A good regular expression of a person's name is ______________.
A grammar G is ambiguous if there is a word w Î L(G) having are least two different parse trees.
A language accepted by Deterministic Push down automata is closed under which of the following?
A may or may not read an input symbol, but it has to read the top of the stack in every transition.
A parsing that starts from the top with the start-symbol and derives a string using a parsetree.
A PDA accepts a string when, after reading the entire string, the PDA is in a final state.
A PDA machine configuration (p, w, y) can be correctly represented as.
A PDA machine configuration (p, w, y) can be ly represented as ____________
A pushdown automaton is a way to implement a context-free grammar in a similar way to design DFA for a regular grammar.
A regular expression consisting of a finite set of grammar rules is a quadruple.
A right-most derivation of a sentential form is one in which rules transforming the are always applied.
A set has n elements, then the number of elements in its power set is?
A set of rules, P: N → (N U T)*, it does have any right context or left context.
A set of strings of a's and b's of any length including the null string. So L= { ε, a, b, aa , ab , bb , ba, aaa.......}, the regular
A set of strings of a's and b's of any length including the null string. So L= { ε, a, b, aa , ab , bb , ba, aaa.......}, the regular expression is.
A set of terminals where N ∩ T = NULL.
A string is accepted by a, iff the DFA/NDFA starting at the initial state ends in an after reading the string.
A sub-tree of a derivation tree/parse tree such that either all of its children are in the sub-tree or none of them are in the
A transition function δ : Q × (Σ ∪ {ε}) → 2Q.
A transition function: Q × (Σ∪{ε}) × S × Q × S*.
A' will contain how many elements from the original set A?
According to the 5-tuple representation i.e. FA= {Q, ∑, δ, q, F} Statement 1: q ϵ Q'; Statement 2: FϵQ
According to what concept is CFL a superset of RL?
An automaton that produces outputs based on current input and/or previous state is called.
An example string w of a language L characterized by equal number of As and Bs is _______________.
An indirect way of building a CFG is to _____________
An order rooted tree that graphically represents the semantic information a string derived from a context-free grammar. | {"url":"https://amauoed.com/courses/cs/automata-theory-and-formal-language-6205-cs","timestamp":"2024-11-05T16:15:31Z","content_type":"text/html","content_length":"105167","record_id":"<urn:uuid:c5d84a42-d733-4e70-86e4-a6142b69eb98>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00870.warc.gz"} |
Hotelling's T-Squared: Simple Definition
Hypothesis Tests
> Hotelling’s T-Squared
What Is Hotelling’s t-squared?
Hotelling’s T-Squared, sometimes written as T^2
, is the multivariate counterpart of
the T-test
[1]. “Multivariate” means that you have data for more than one parameter for each sample. For example, let’s say you wanted to compare how well two different sets of students performed in school. You
could compare
univariate data
(e.g. mean test scores) with a
. Or, you could use Hotelling’s T-squared to compare
multivariate data
(e.g. the mutivariate mean of test scores, GPA, and class grades). Hotelling’s T-Squared is based on Hotelling’s T
distribution and forms the basis for various multivariate
control charts
. It can also describe the
Mahalanobis distance
between two populations and can also be used to identify
or nonconformities in a data set. Formally, Hotelling’s T-squared distribution is defined as follows [2]: Suppose that a vector
normally distributed
with a mean of zero and unit covariance matrix N
(0, I), and M is an
matrix with a
Wishart distribution
with unit scale matrix and
n degrees of freedom
(I, m). Then, m
has a two-parameter Hotelling distribution T
(p,m). Where
• A covariance matrix shows the correlation between each pair of variables in a multivariate dataset. Covariance is a measure of how much two variables vary together.
• A unit scale matrix is a covariance matrix where all variances are equal to 1, which means that all the variables are equally spread out around their means.
Test Versions
Two versions of the test exist with the following
null hypotheses
• One sample: The multivariate vector means for a group equals a hypothetical vector of means.
• Two sample: The multivariate vector of means for two groups are equal.
For more than two samples, one option is to run a
. MANOVA is more
than Hotelling’s T-squared when there are more than two groups.
Two-sample Hotelling’s t-squared
If you know how to run a
two sample t-test
, then you know how to run a two-sample Hotelling’s T-squared.
The basic steps are the same,
although you’ll use a different formula to calculate the t-squared value and you’ll use a different table (the
) to find the
critical value
. Hotelling’s T-squared has
several advantages
over the t-test [3]:
• The Type I error rate is well controlled,
• The relationship between multiple variables is taken into account,
• It can generate an overall conclusion even if multiple (single) t-tests are inconsistent. While a t-test will tell you which variable differ between groups, Hotelling’s summarizes the
between-group differences.
test hypotheses
• Null hypothesis (H[0]): the two samples are from populations with the same multivariate mean.
• Alternate hypothesis (H[1]): the two samples are from populations with different multivariate means.
Three major assumptions
are that the samples:
Hotelling’s-T can be transformed to an F-statistic.
Like the t-test, you’ll want to find a value for T (in this case, for T-squared) and compare it to a table value; if the calculated value is greater than the table statistic, you can
reject the null hypothesis
. For ease of this calculation, Hotelling’s t
is first transformed into an
• N[1] & N[2] = sample sizes,
• p = number of variables measured,
• N[1] + N[2] – p – 1 = degrees of freedom.
Reject the null hypothesis (at a chosen
significance level
) if the calculated value is greater than the F-table critical value. Rejecting the null hypothesis means that at least one of the parameters, or a combination of one or more parameters working
together, is
significantly different
Why Is Hotelling’s T-Squared Distribution Important?
Hotelling’s T
is an important tool for identifying changes in means between multiple populations. By using
linear combinations
of variables, it allows us to compare multiple samples at once, instead of having to run separate tests for each sample. This makes it much easier and faster to identify any meaningful changes in
means between populations over time or across different groups of people. Additionally, because it uses a
chi-squared test statistic,
it helps us determine whether or not these changes are
statistically significant
– something that traditional t-tests cannot do on their own.
1. Hotelling H (1931) The generalization of Student’s ratio. Ann Math Stat. 2(3):360–378.
2. Weisstein, E. (2002). CRC Concise Encyclopedia of Mathematics. CRC Press.
3. Fang, J. (2017). Handbook of Medical Statistics. | {"url":"https://www.statisticshowto.com/hotellings-t-squared/","timestamp":"2024-11-04T03:57:05Z","content_type":"text/html","content_length":"74747","record_id":"<urn:uuid:e8ad5178-6061-484b-91a5-7509d00fdfc0>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00246.warc.gz"} |
Elder-Rule-Staircodes for Augmented Metric Spaces
An augmented metric space is a metric space (X, d_X) equipped with a function f_X: X →ℝ. This type of data arises commonly in practice, e.g, a point cloud X in ℝ^d where each point x∈ X has a density
function value f_X(x) associated to it. An augmented metric space (X, d_X, f_X) naturally gives rise to a 2-parameter filtration 𝒦. However, the resulting 2-parameter persistent homology H_∙(𝒦) could
still be of wild representation type, and may not have simple indecomposables. In this paper, motivated by the elder-rule for the zeroth homology of 1-parameter filtration, we propose a barcode-like
summary, called the elder-rule-staircode, as a way to encode H_0(𝒦). Specifically, if n = |X|, the elder-rule-staircode consists of n number of staircase-like blocks in the plane. We show that if H_0
(𝒦) is interval decomposable, then the barcode of H_0(𝒦) is equal to the elder-rule-staircode. Furthermore, regardless of the interval decomposability, the fibered barcode, the dimension function
(a.k.a. the Hilbert function), and the graded Betti numbers of H_0(𝒦) can all be efficiently computed once the elder-rule-staircode is given. Finally, we develop and implement an efficient algorithm
to compute the elder-rule-staircode in O(n^2log n) time, which can be improved to O(n^2α(n)) if X is from a fixed dimensional Euclidean space ℝ^d, where α(n) is the inverse Ackermann function. | {"url":"http://api.deepai.org/publication/elder-rule-staircodes-for-augmented-metric-spaces","timestamp":"2024-11-15T02:36:36Z","content_type":"text/html","content_length":"153803","record_id":"<urn:uuid:3368a494-b212-4b50-9e07-16b616c8cbc9>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00824.warc.gz"} |
How do you calculate total capacitance in series and parallel? | SolidWorks Assignment Help
How do you calculate total capacitance in series and parallel? Here is everything (from start up here and here). I will actually go into more detail on this but I am only posting a small percentage
because after a while the truth is that I know people already who can calculate this type of calculation, well you could probably look at the database or the package source here and see how it can be
done. It might have been my mistake by commenting from where I originally wrote that everyone in here uses these types of type of calculations or methods. Okay, let’s get started with a little bit
more info 🙂 In order to get started with my quick research around multi-pass the truth is that I want it in the following form [this is my primary intent. I want there to be a way to calculate
this type of analysis and you could also try the method (if you have input with more than one passed parameters). To see that here is the general procedure] First go to the database and try (as seen
below) for any 1-pass or other kinds of calculations. You may use built-in VBA functions like this Function(“Convert”, “{›}”).ToArray() Then go to a file called “CalculatorsData” where you can drag
/drop the files (here) Read the file with a browser (code below) Duplicate of type Double Write to the file read by the browser Write back to the file save as datame 1 of something else Done! Now you
have done with this. Here is the complete simple xl code for debugging! Re-execute on your Mac: $ for y in (1..20)/20, 20; do (printf “%d”, rand()) // read from database What can you do with these
values? There is a small convenience function(s) that can help you Generate the corresponding double values/points from somewhere (here) (because there should be as many as possible points: (1/20))
The second thing is to call $1. If you type “1”, you will generate a second pair of values. Write for each line or row of the file generated by the Get-Datasize function. Now, we need to add up the
output (here it is the I/O representation) and multiply by the maximum value you see in the file For each row in the file called “calculate” here is the List of individual column These data are then
added to Table 15 in the Table “Table 15” by selecting column “t,i,” and format it as Text in the.txt file. Now we need to create our own function to generate the sum of multiple points. Here is the
code for this function: A variable named Math = MaximiumTable() ForHow do you calculate total capacitance in series and parallel? One technique used to resolve this problem is to store the actual
capacitance value computed by the cable and then compute a total capacitance. When the cable is physically separated, it produces the total capacitance with less information and less time. Hence, the
resulting total capacitance can be used as a factor of total capacitance as measured in series. A very common starting point of electronics and electronics products is digital circuits, but I still
use capacitors as they have the same main components to be connected and decoded.
Pay For Homework
By using capacitors and by dividing the wires, you can use both to calculate the total capacitance and to store it in series. If you’re going to use any such source code, you’ll need to replace
the standard serial numbers.org with serial numbers, ASCII, and even Chinese letters like R, G, J, K, and 6 by replacing the conversion functions in.org with serial numbers and the standard numbers
by giving the.org the.org command line number of a specific serial number to call onto. We’re really going to use the above three commands.org for data on the serial numbers themselves as they
differ in complexity and cost. Below is a pair of images of the simplest code we have written. The new solution uses the conversion functions and not the standard single digit one plus or minus one.
Therefore, this technique is an extremely common means to “convert” electrical data – including serial data – to binary data and write this to the standard binary data file. This technique is
more expensive because it is easier to find a couple of to a pair of numbers for this special purpose using a single digit comparison software. Method The conversion process is simple and very
inexpensive, about 32Gbit size. First, a straight sketch of a few million wires will show you how to calculate capacitance without using a pin, and capacitance with a digital scale. You can do this
by moving a little on one digit. What is the total capacitance? Conductivity The standard scale, though small, this can’t be calculated by using a digital scale for digital circuitry hardware.
Assuming that the total capacitance isn’t close to zero, you can try.probe You’ll find the total capacitance is like a 6A – it uses just a few bits of memory – 2.35Gbit and 0.01%.
Do My College Algebra Homework
There is too much more, say, a little bit less than 2Kb, for 100000000s of bytes, just multiply it by 2 or 3 to make it 2Kb. Then, you can check capacitance by the least most significant – then
divide the excess by this 10Kb. Polarity (or FIVE) The polarity of thisHow do you calculate total capacitance in series and parallel? their explanation other parts of this tutorial, I will discuss
parallel capacitance (pC; for example the capacitance in series (pC/pB). I always prefer the horizontal, vertical, vertical, horizontal, horizontal, horizontal, horizontal, and vertical capacitance
formulas, to vertical or horizontal capacitance. For a capacitor, one may consider an open, closed circuit consisting of the full (pB) and full (pC) form of the capacitor with (u, pB) and (v, pC). If
you take the capacitance from a real example, then you need pC/pB rather than pB/pB. This is why amaxiagetransition #4 below uses this formula to describe the capacitance in series (pB, pC): If you
don’t want pB, you can use (u, pB). You can also write the formula below on line 3: for The List below: for your own table: Please find some other answers to this for more details. Example 12:
Numerical test cases I will assume two types of test cases can be produced: pC=3,5,10,18, I would like to calculate the capacitance in a series – your calculations and your formulas would be
interesting. But in this tutorial, I will show more technical tables, data not have one, and only a general explanation not explain the formulas. Example 13: Examples of capacitor figures – in this
example, you show how a capacitor is connected (pC, u) to a voltage supply (vb), then you calculate the capacitance as in 14: Example 14: Example : CCA [1/2 BNC ]-CCA, capacitive one, A-B function,
B-C-E, B-C-F, A-C-E = 130000. 3D-BNC a pD C [C, I pC dt], 1/3 B.c-E C-E=C-ICb E-F-C=E-Fb // Use Calc() Function For Number of Centimeter.BNC = 13000000. BOC(5,10,18)=14001 10 00 00 00 A60 5 17 00 00
00 21 Example 15: In CCA @(du) I would like it to be like this, Example 16 : CCA F-A M-B Again, see here more details for description. Example 17, 1/2 B.C-E C- From.4D to 12.1D, for two models 12.1
and 13.
Take My Online Statistics Class For Me
1.1 2C.b C – C-E = 130000 B.C=14. Cc-C = 13000000 D-DC-E = 130000 C-E = 130009. Now to see the other notation in Example 16: Example 18 (one method): The definition here is divided into base 3 and
base 5, 1/2 B.C-E C-D = 130000 R(5,10,18)=140000 R(4,10,18)=160000 R(6,10,18)=16000 R(8,10,18)=16000 R(4,10,18)=16000 R(4,10,18)=14000 B.C-E = 130000 F or more, I also used : Example 18 (2 methods):
In this example, the second (more complicated) notation of 9 is used which contains both 1/3 and 2/3 B.C(5,9,5) = 9 and 13.1 C=14. Cb.C=14000 | {"url":"https://solidworksaid.com/how-do-you-calculate-total-capacitance-in-series-and-parallel-16932","timestamp":"2024-11-08T04:22:47Z","content_type":"text/html","content_length":"156503","record_id":"<urn:uuid:a50aa067-731a-4060-b821-c5783d441903>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00328.warc.gz"} |
Learn to use ROUNDUP Function in Google Sheets - SheetsInfo
ROUNDUP Function in Google Sheets
• Post published:April 25, 2021
• Reading time:3 mins read
• Post category:Functions / Rounding Funtions
Post comments:0 Comments
ROUNDUP Function belongs to the family of Mathematical functions in Google Sheet used for rounding numbers.
It’s very similar to ROUND Function, just with the difference that it will always Round-up. Let’s go through the salient features of ROUNDUP Function in this tutorial.
Purpose of ROUNDUP Function
As you would have guessed, it helps us to round numbers and retain decimals upto a specified decimal place.
One important callout for ROUNDUP is that it always picks the higher value while rounding off. This means”
ROUNDUP(32.76,1) equals 32.8
FLOOR(32.73, 1) equals 32.8
Before jumping onto examples let’s understand the syntax of ROUNDUP.
Syntax and Parameter Definition
= ROUNDUP(value, [places])
1. value: The number you want to round
2. [places]: The number of decimal digits you want to round the ‘value’ to.
1. Value can take only numerical input
2. Any form of non-numerical input(like “abc”, “xyz” etc) for value will result in an “#VALUE!” error.
3. [places] is an optional input. 0 is it’s default value.
4. [places] should be an integer either positive or negative.
5. You can directly add static values or cell references for both value and [places]
Expected Output, logic behind it and Examples
ROUNDUP function follows a very simple logic to operate.
Say, you want to round up to 2 decimal places. ROUNDUP will retain the number upto two decimal place and increase the last decimal digit by 1.
Simplest Case – ROUNDUP Function
Below shown example is a very simple example of how ROUNDUP works and where is it different from ROUND Function.
As you may have observed in both the cases ROUNDUP Function returns an output with a one decimal place and higher than original Value. ROUND, on the other hand would not have rounded the first case
since it follows the strict rule of the next decimal place digit to be >5 before rounding up the number.
More Examples – ROUNDUP Function
In the below example we have taken the liberty to play around with Places. Let’s discuss the insights we can derive from the below.
In, the first three case we have varied the [places] keeping Value same. The outputs are pretty self explanatory as well. In all cases ROUNDUP has increased the last decimal digit by 1.
In the last two cases, we kept places as 0/missing. Not surprisingly, the results are same as default value of [places] if not supplied is 0. Having [places] as 0, gives an indication to ROUNDUP that
we don’t need any decimal value, hence the result it a whole number.
Working with Negative ‘Value’
We used the same cases from last example, just added a minus sign before it. Clearly, there’s no effect to working of Roundup Function. The output we got last time is intact this time as well.
Working with Negative ‘Places’
Having Negative value in ‘Places’ results in rounding up to nearest Ten’s/Hundred’s/Thousand’s depending on the value of Places(-1/-2/-3 and so on)
Visual Demo of ROUNDUP Function
Before we end here is a sample visual demonstration of ROUNDUP Function in action.
That’s it on this topic. Keep browsing SheetsInfo for more such useful information 🙂
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Tags: Decimals, Error, Number Format, Rounding Operation, Roundup Function
I love Google Sheets and heavily rely upon it on a day to day basis. SheetsInfo is my attempt to share my learning with all of you. Hope you find the articles easy to read and simple to understand.
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NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1 - MCQ Questions
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1
These NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Exercise 7.1
Question 1.
Complete the following statements:
(a) Two line segments are congruent if ……………….
(b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is ……………….
(c) When we write ∠A = ∠B, we actually mean …………………
(a) Two line segments are congruent if they have the same length.
(b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is 70°.
(c) When we write ∠A ≅ ∠B we actually mean m ∠A = m ∠B.
Question 2.
Give any two real-life examples for congruent shapes.
(i) Two ten-rupee notes.
(ii) Biscuits of the same type in the same packet.
Question 3.
If ΔABC ≅ ΔFED under the correspondences ABC ↔ FED, write all the corresponding congruent parts of the triangles.
Corresponding vertices : A and F; B and E; C and D.
Corresponding sides : \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{FE}} ; \overline{\mathrm{BC}}\) and \overrightarrow{\mathrm{ED}} ; \overline{\mathrm{CA}} and \(\overline{\mathrm{AB}}\)
Corresponding angles : ∠A and ∠F; ∠B and ∠E; ∠C and ∠D
Question 4.
If ΔDEF ≅ ΔBCA, write the part (s) of ΔBCA that correspond to
(i) ∠E
(ii) \(\overline{\mathrm{EF}}\)
(iii) ∠F
(iv) \(\overline{\mathrm{DF}}\)
(i) ∠C
(ii) \(\overline{\mathrm{CA}}\)
(iii) ∠A
(iv) \(\overline{\mathrm{BA}}\) | {"url":"https://mcq-questions.com/ncert-solutions-for-class-7-maths-chapter-7-ex-7-1/","timestamp":"2024-11-04T08:02:29Z","content_type":"text/html","content_length":"143289","record_id":"<urn:uuid:a5f3296f-e8b6-44c8-adab-5db5e286b6d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00584.warc.gz"} |
Impossibility results for unbounded utilities
Some people think that they have unbounded utility functions. This isn’t necessarily crazy, but it presents serious challenges to conventional decision theory. I think it probably leads to abandoning
probability itself as a representation of uncertainty (or at least any hope of basing decision theory on such probabilities). This may seem like a drastic response, but we are talking about some
pretty drastic inconsistencies.
This result is closely related to standard impossibility results in infinite ethics. I assume it has appeared in the philosophy literature, but I couldn’t find it in the SEP entry on the St.
Petersburg paradox so I’m posting it here. (Even if it’s well known, I want something simple to link to.)
(ETA: this argument is extremely similar to Beckstead and Thomas’ argument against Recklessness in A paradox for tiny probabilities and enormous values. The main difference is that they use
transitivity +”recklessness” to get a contradiction whereas I argue directly from “non-timidity.” I also end up violating a dominance principle which seems even more surprising to violate, but at
this point it’s kind of like splitting hairs. I give a slightly stronger set of arguments in Better impossibility results for unbounded utilities.)
Weak version
We’ll think of preferences as relations over probability distributions over some implicit space of outcomes (and we’ll identify outcomes with the constant probability distribution). We’ll show that
there is no relation which satisfies three properties: Antisymmetry, Unbounded Utilities, and Dominance.
Note that we assume nothing about the existence of an underlying utility function. We don’t even assume that the preference relation is complete or transitive.
The properties
Antisymmetry: It’s never the case that both and .
Unbounded Utilities: there is an infinite sequence of outcomes each “more than twice as good” as the last.^[1] More formally, there exists an outcome such that:
That is, is not as good as a chance of , which is not as good as a chance of , which is not as good as a chance of … This is nearly the weakest possible version of unbounded utilities.^[3]
Dominance: let and be sequences of lotteries, and be a sequence of probabilities that sum to 1. If for all , then .
Inconsistency proof
Consider the lottery
We can write as a mixture:
By definition . And for each , Unbounded Utilities implies that . Thus Dominance implies , contradicting Antisymmetry.
How to avoid the paradox?
By far the easiest way out is to reject Unbounded Utilities. But that’s just a statement about our preferences, so it’s not clear we get to “reject” it.
Another common way out is to assume that any two “infinitely good” outcomes are incomparable, and therefore to reject Dominance.^[4] This results in being indifferent to receiving $1 in every world
(if the expectation is already infinite), or doubling the probability of all good worlds, which seems pretty unsatisfying.
Another option is to simply ignore small probabilities, which again leads to rejecting even the finite version of Dominance—sometimes when you mix together lotteries something will fall below the
“ignore it” threshold leading the direction of your preference to reverse. I think this is pretty bizarre behavior, and in general ignoring small probabilities is much less appealing than rejecting
Unbounded Utilities.
All of these options seem pretty bad to me. But in the next section, we’ll show that if the unbounded utilities are symmetric—if there are both arbitrarily good and arbitrarily bad outcomes—then
things get even worse.
Strong version
I expect this argument is also known in the literature; but I don’t feel like people around LW usually grapple with exactly how bad it gets.
In this section we’ll show there is no relation which satisfies three properties: Antisymmetry, Symmetric Unbounded Utilities, and Weak Dominance.
(ETA: actually I think that even with only positive utilities you already violate something very close to Weak Dominance, which Beckstead and Thomas call Prospect-Outcome dominance. I find this
version of Weak Dominance slightly more compelling, but Symmetric Unbounded Utilities is a much stronger assumption than Unbounded Utilities or non-Timidity, so it’s probably worth being aware of
both versions. In a footnote^[5] I also define an even weaker dominance principle that we are forced to violate.)
The properties
Antisymmetry: It’s never the case that both and .
Symmetric Unbounded Utilities. There is an infinite sequence of outcomes each of which is “more than twice as important” as the last but with opposite sign. More formally, there is an outcome such
• For every even :
• For every odd :
That is, a certainty of is outweighed by a chance of , which is outweighed by a chance of , which is outweighed by a chance of ….
Weak Dominance.^[5] For any outcome , any sequence of lotteries , and any sequence of probabilities that sum to 1:
• If for every , then .
• If for every , then .
Inconsistency proof
Now consider the lottery
We can write as the mixture:
By Unbounded Utilities each of these terms is . So by Weak Dominance,
But we can also write as the mixture:
By Unbounded Utilities each of these terms is . So by Weak Dominance . This contradicts Antisymmetry.
Now what?
As usual, the easiest way out is to abandon Unbounded Utilities. But if that’s just the way you feel about extreme outcomes, then you’re in a sticky situation.
You could allow for unbounded utilities as long as they only go in one direction. For example, you might be open to the possibility of arbitrarily bad outcomes but not the possibility of arbitrarily
good outcomes.^[6] But the asymmetric version of unbounded utilities doesn’t seem very intuitively appealing, and you still have to give up the ability to compare any two infinitely good outcomes
(violating Dominance).
People like talking about extensions of the real numbers, but those don’t help you avoid any of the contradictions above. For example, if you want to extend to a preference order over hyperreal
lotteries, it’s just even harder for it to be consistent.
Giving up on Weak Dominance seems pretty drastic. At that point you are talking about probability distributions, but I don’t think you’re really using them for decision theory—it’s hard to think of a
more fundamental axiom to violate. Other than Antisymmetry, which is your other option.
At this point I think the most appealing option, for someone committed to unbounded utilities, is actually much more drastic: I think you should give up on probabilities as an abstraction for
describing uncertainty, and should not try to have a preference relation over lotteries at all.^[7] There are no ontologically fundamental lotteries to decide between, so this isn’t necessarily so
bad. Instead you can go back to talking directly about preferences over uncertain states of affairs, and build a totally different kind of machinery to understand or analyze those preferences.
ETA: replacing dominance
Since writing the above I’ve become more sympathetic to violations of Dominance and even Weak Dominance—it would be pretty jarring to give up on them, but I can at least imagine it. I still think
violating “Very Weak Dominance”^[5] is pretty bad, but I don’t think it captures the full weirdness of the situation.
So in this section I’ll try to replace Weak Dominance by a principle I find even more robust: if I am indifferent between and any of the lotteries , then I’m also indifferent between X and any
mixture of the lotteries . This isn’t strictly weaker than Weak Dominance, but violating it feels even weirder to me. At any rate, it’s another fairly strong impossibility result constraining
unbounded utilities.
The properties
We’ll work with a relation over lotteries. We write if both and . We write if but not . We’ll show that can’t satisfy four properties: Transitivity, Intermediate mixtures, Continuous symmetric
unbounded utilities, and Indifference to homogeneous mixtures.
Intermediate mixtures. If , then .
Transitivity. If and then .
Continuous symmetric unbounded utilities. There is an infinite sequence of lotteries each of which is “exactly twice as important” as the last but with opposite sign. More formally, there is an
outcome such that:
• For every even :
• For every odd :
That is, a certainty of is exactly offset by a chance of , which is exactly offset by a chance of , which is exactly offset by a chance of ….
Intuitively, this principle is kind of like symmetric unbounded utilities, but we assume that it’s possible to dial down each of the outcomes in the sequence (perhaps by mixing it with ) until the
inequalities become exact equalities.
Homogeneous mixtures. Let be an outcome, , a sequence of lotteries, and be a sequence of probabilities summing to 1. If for all , then .
Inconsistency proof
Consider the lottery
We can write as the mixture:
By Unbounded Utilities each of these terms is . So by homogeneous mixtures,
But we can also write as the mixture:
By Unbounded Utilities each of these terms other than the first is . So by Homogenous Mixtures, the combination of all terms other than the first is . Together with the fact that , Intermediate
Mixtures and Transitivity imply . But that contradicts .
1. ^
Note that we could replace “more than twice as good” with “at least 0.00001% better” and obtain exactly the same result. You may find this modified version of the principle more appealing, and it
is closer to non-timidity as defined in Beckstead and Thomas. Note that the modified principle implies the original by applying transitivity 100000 times, but you don’t actually need to apply
transitivity to get a contradiction, you can just apply Dominance to a different mixture.
2. ^
You may wonder why we don’t just write . If we did this, we’d need to introduce an additional assumption that if , . This would be fine, but it seemed nicer to save some symbols and make a
slightly weaker assumption.
3. ^
The only plausibly-weaker definition I see is to say that there are outcomes and an infinite sequence such that for all : . If we replaced the with then this would be stronger than our version,
but with the inequality it’s not actually sufficient for a paradox.
To see this, consider a universe with three outcomes and a preference order that always prefers lotteries with higher probability of and breaks ties using by preferring a higher probability of .
This satisfies all of our other properties. It satisfies the weaker version of the axiom by taking for all , and it wouldn’t be crazy to say that it has “unbounded” utilities.
4. ^
For realistic agents who think unbounded utilities are possible, it seems like they should assign positive probability to encountering a St. Petersburg paradox such that all decisions have
infinite expected utility. So this is quite a drastic thing to give up on. See also: Pascal’s mugging.
5. ^
I find this principle pretty solid, but it’s worth noting that the same inconsistency proof would work for the even weaker “Very Weak Dominance”: for any pair of outcomes with , and any sequence
of lotteries each strictly better than , any mixture of the should at least be strictly better than !
6. ^
Technically you can also violate Symmetric Unbalanced Utility while having both arbitrarily good and arbitrarily bad outcomes, as long as those outcomes aren’t comparable to one another. For
example, suppose that worlds have a real-valued amount of suffering and a real-valued amount of pleasure. Then we could have a lexical preference for minimizing expected suffering (considering
all worlds with infinite expected suffering as incomparable), and try to maximize pleasure only as a tie-breaker (considering all worlds with infinite expected pleasure as incomparable).
7. ^
Instead you could keep probabilities but abandon infinite probability distributions. But at this point I’m not exactly sure what unbounded utilities means—if each decision involves only finitely
many outcomes, then in what sense do all the other outcomes exist? Perhaps I may face infinitely many possible decisions, but each involves only finitely many outcomes? But then what am I to make
of my parent’s decisions while raising me, which affected my behavior in each of those infinitely many possible decisions? It seems like they face an infinite mixture of possible outcomes.
Overall, it seems to me like giving up on infinitely big probability distributions implies giving up on the spirit of unbounded utilities, or else going down an even stranger road.
• I really like this post because it directly clarified my position on ethics, namely making me abandon unbounded utilities. I want to give this post a Δ and +4 for doing that, and for being
clearly written and fairly short.
• I am not a fan of unbounded utilities, but it is worth noting that most (all?) the problems with unbounded utilties are actually a problem with utility functions that are not integrable with
respect to your probabilities. It feels basically okay to me to have unbounded utilities as long as extremely good/bad events are also sufficiently unlikely.
The space of allowable probability functions that go with an unbounded utility can still be closed under finite mixtures and conditioning on positive probability events.
Indeed, if you think of utility functions as coming from VNM, and you a space of lotteries closed under finite mixtures but not arbitrary mixtures, I think there are VNM preferences that can only
correspond to unbounded utility functions, and the space of lotteries is such that you can’t make St. Petersburg paradoxes. (I am guessing, I didn’t check this.)
1. I strongly agree that the key problem with St. Petersburg (and Pasadena) paradoxes is utility not being integrable with respect to the lotteries/probabilities. Non-integrability is
precisely what makes 𝔼U undefined (as a real number), whereas unboundedness of U alone does not.
2. However, it’s also worth pointing out that the space of functions which are guaranteed to be integrable with respect to any probability measure is exactly the space of bounded
(measurable) functions. So if one wants to save utilities’ unboundedness by arguing from integrability, that requires accepting some constraints on one’s beliefs (e.g., that they be
finitely supported). If one doesn’t want to accept any constraints on beliefs, then accepting a boundedness constraint on utility looks like a very natural alternative.
3. I agree with your last paragraph. If the state space of the world is ℝ, and the utility function is the identity function, then the induced preferences over finitely-supported lotteries
can only be represented by unbounded utility functions, but are also consistent and closed under finite mixtures.
4. Finite support feels like a really harsh constraint on beliefs. I wonder if there are some other natural ways to constrain probability measures and utility functions. For example, if we
have a topology on our state-space, we can require that our beliefs be compactly supported and our utilities be continuous. “Compactly supported” is way less strict than “finitely
supported,” and “continuous” feels more natural than “bounded.” What are some other pairs of “compromise” conditions such that any permissible utility function is integrable with respect
to any permissible belief distribution? (Perhaps it would be nice to have one that allows Gaussian beliefs, say, which are neither finitely nor compactly supported.)
☆ Note that if dominates in the sense that there is a such that for all events , , is integrable wrt , then I think is integrable wrt . I propose the space of all probability distribution
dominated by a given distribution .
Conveniently, if we move to semi-measures, we can take P to be the universal semi-measure. I think we can have our space of utility functions be anything integrable WRT the universal
semi-measure, and our space of probabilities be anything lower semi-computable, and everything will work out nicely.
○ I think bounded functions are the only computable functions that are integrable WRT the universal semi-measure. I think this is equivalent to de Blanc 2007?
The construction is just the obvious one: for any unbounded computable utility function, any universal semi-measure must assign reasonable probability to a St Petersburg game for that
utility function (since we can construct a computable St Petersburg game by picking a utility randomly then looping over outcomes until we find one of at least the desired utility).
☆ Compact support still seems like an unreasonably strict constraint to me, not much less so than finite support. Compactness can be thought of as a topological generalization of
finiteness, so, on a noncompact space, compact support means assigning probability 1 to a subset that’s infinitely tiny compared to its complement.
□ I observe that I probably miscommunicated. I think multiple people took me to be arguing for a space of lotteries with finite support. That is NOT what I meant. That is sufficient, but I
meant something more general when I said “lotteries closed under finite mixtures” I did not mean there only finitely many atomic worlds in the lottery. I only meant that there is a space of
lotteries, some of which maybe have infinite support if you want to think about atomic worlds, and for any finite set of lotteries, you can take a finite mixture of those lotteries to get a
new lottery in the space. The space of lotteries has to be closed under finite mixtures for VNM to make sense, but the emphasis is on the fact that it is not closed under all possible
countable mixtures, not that the mixtures have finite support.
□ Hm, what would that last thing look like?
Like, I agree that you can have gambles closed under finite but not countable gambling and the math works. But it seems like reality is a countably-additive sort of a place. E.g. if these
different outcomes of a lottery are physical states of some system, QM is going to tell you to take some infinite sums. I’m just generally having trouble getting a grasp on what the world
(and our epistemic state re. the world) would look like for this finite gambles stuff to make sense.
☆ Note that you can take infinite sums, without being able to take all possible infinite sums.
I suspect it looks like you have a prior distribution, and the allowable probability distributions are those that you can get to from this distribution using finitely many bits of
□ [deleted]
• I think this argument is cool, and I appreciate how distilled it is.
Basically just repeating what Scott said but in my own tongue: this argument leaves open the option of denying that (epistemic) probabilities are closed under countable combination, and deploying
some sort of “leverage penalty” that penalizes extremely high-utility outcomes as extremely unlikely a priori.
I agree with your note that the simplicitly prior doesn’t implement leverage penalties. I also note that I’m pretty uncertain myself about how to pull off leverage penalties correctly, assuming
they’re a good idea (which isn’t clear to me).
I note further that the issue as I see it arises even when all utilities are finite, but some are (“mathematically”, not merely cosmically) large (where numbers like 10^100 are cosmically large,
and numbers like 3^^^3 are mathematically large). Like, why are our actions not dominated by situations where the universe is mathematically large? When I introspect, it doesn’t quite feel like
the answer is “because we’re certain it isn’t”, nor “because utility maxes out at the cosmological scale”, but rather something more like “how would you learn that there may or may not be 3^^^3
happy people with your choice as the fulcrum?” plus a sense that you should be suspicious that any given action is more likely to get 3^^^3 utility than any other (even in the presence of Pascall
muggers) until you’ve got some sort of compelling account of how the universe ended up so large and you ended up being the fulcrum anyway. (Which, notably, starts to feel intertwined with my
confusion about naturalistic priors, and I have at least a little hope that a good naturalistic prior would resolve the issue automatically.)
Or in other words, “can utilities be unbounded?” is a proxy war for “can utilities be mathematically large?”, with the “utilities must be bounded” resolution in the former corresponding (at least
seemingly) to “utilities can be at most cosmically large” in the later. And while that may be the case, I don’t yet feel like I understand reasoning in the face of large utilities, and your
argument does not dispell my confusion, and so I remain confused.
And, to be clear, I’m not saying that this problem seems intractible to me. There are various lines of attack that seem plausible from here. But I haven’t seen anyone providing the “cognitive
recepits” from mapping out those lines of reasoning and deconfusing themselves about big utilities. For all I know, “utilities should be bounded (and furthermore, max utility should be at most
cosmically large)” is the right answer. But I don’t confuse this guess for understanding.
□ TL;DR: I think that the discussion in this post is most relevant when we talk about the utility of whole universes. And for that purpose, I think a leverage penalty doesn’t make sense.
A leverage penalty seems more appropriate for saying something like “it’s very unlikely that my decisions would have such a giant impact,” but I don’t think that should be handled in the
utility function or decision theory.
Instead, I’d say: if it’s possible to have “pivotal” decisions that affect 3^^^3 people, then it’s also possible to have 3^^^3 people in “normal” situations all making their separate
(correlated) decisions, eating 3^^^3 sandwiches, and so the stakes of everything are similarly mathematically big.
plus a sense that you should be suspicious that any given action is more likely to get 3^^^3 utility than any other
I think that if utilities are large but bounded, then I feel like everything “adds up to normality”—if there is a way to get 3^^^3 utility, it seems like “maximize option value, figure out
what’s going on, stay sane” is a reasonable bet for maximizing EV (e.g. by maximizing probability of the great outcome).
Intuitively, this also seems like what you should end up doing even if utilities are “infinite” (an expression that seems ~meaningless).
You can’t actually make these arguments go through for unbounded or infinite utilities, and part of the point is to observe that no arguments go through with unbounded/infinite utilities
because the entire procedure is screwed.
Or in other words, “can utilities be unbounded?” is a proxy war for “can utilities be mathematically large?”, with the “utilities must be bounded” resolution in the former corresponding
(at least seemingly) to “utilities can be at most cosmically large” in the later.
I feel like in this discussion it’s not helpful to talk directly about magnitudes of utility functions, and to just talk directly about our questions about preferences, since that’s
presumably the thing we have intuitions about. (I’d say that even if we thought utility functions definitely made sense in the regime where you have unbounded preferences, but it seems doubly
true given that utility functions don’t seem very likely to be the right abstraction for unbounded preferences.)
deploying some sort of “leverage penalty” that penalizes extremely high-utility outcomes as extremely unlikely a priori.
This seems to put you in a strange position though: you are not only saying that high-value outcomes are unlikely, but that you have no preferences about them. That is, they aren’t merely
impossible-in-reality, they are impossible-in-thought-experiments.
I personally feel like even if the thought experiments are impossible, they fairly clearly illustrate that some of our claims about our preferences can’t be right. To the extent that those
claims come in large part from the appeal of certain clean mathematical machinery, I think the right first move is probably to become more unhappy about that machinery. To the extent that we
have strong intuitions about non-timidity or unboundedness, then “become disillusioned with certain mathematical machinery” won’t help and we’ll just have to deal directly with sorting out
our intuitions (but I think it’s fairly unlikely that journey involves discovering any philosophical insight that restores the appeal of the mathematical machinery).
“how would you learn that there may or may not be 3^^^3 happy people with your choice as the fulcrum?”
How would you learn that there may or may not be a 10^100 future people with our choices as the fulcrum? Why would the same process not generalize? (And if it may happen in the future but not
now, is that 0 probability?)
To give my own account of the situation:
☆ It’s easy to come to believe that there are 3^^^3 potential people and that their welfare depends on a small number of pivotal events (through exactly the same kind of argument that leads
us to believe that there are 10^100 potential people).
☆ Under those conditions, it’s at least plausible that there will inevitably be roughly 3^^^3 dreamers who come to that belief incorrectly (or who come to that belief correctly, but who
incorrectly believe that they themselves are participating in the pivotal event). This involves some kind of necessary relationship between moral value and agency, which isn’t obvious but
at least feels plausible. Note that most rationalists already bite this bullet, in that they think that the vast majority of individuals who think they are at the “hinge of history” are
in simulations (and hence mistaken).
☆ Despite believing that, the prospect of “I am actually in the pivotal event” can easily dominate expected utility calculations in particular cases.
☆ That said, it’s still the case that a large universe contains many mistaken dreamers, so the costs they pay (in pursuit of the mistaken belief that they are participating in the pivotal
event) will be comparable to the scale of consequences of the pivotal event. You end up with a non-fanatical outlook on the pivotal event itself, such that it will only dominate the EV if
you have normal evidence that the current time is pivotal. (You get this kind of “adding up to normality” under much weaker assumptions than a strong leverage penalty.)
☆ But it still matters how much more we care about bigger universes—if the universe appears to be finite or small, should we assume that we are missing something?
My personal answer is that infinite universes don’t seem infinitely more important than finite universes, and that 2x bigger universes generally don’t seem 2x as important. (I tentatively
feel that amongst reasonably-large universes, value is almost independent of size—while thinking that within any given universe 2x more flourishing is much closer to 2x as valuable.)
But this really seems like a brute question about preferences. And in this context, I don’t think a leverage penalty feels like a plausible way to resolve the confusion.
if it’s possible to have “pivotal” decisions that affect 3^^^3 people, then it’s also possible to have 3^^^3 people in “normal” situations all making their separate (correlated)
decisions, eating 3^^^3 sandwiches, and so the stakes of everything are similarly mathematically big.
This seems to put you in a strange position though: you are not only saying that high-value outcomes are unlikely, but that you have no preferences about them. That is, they aren’t
merely impossible-in-reality, they are impossible-in-thought-experiments.
Perhaps I’m being dense, but I don’t follow this point. If I deny that my epistemic probabilities are closed under countable weighted sums, and assert that the hypothesis “you can
actually play a St. Petersburg game for n steps” is less likely than it is easy-to-describe (as n gets large), in what sense does that render me unable to consider St. Petersburg games in
thought experiments?
How would you learn that there may or may not be a 10^100 future people with our choices as the fulcrum? Why would the same process not generalize? (And if it may happen in the future
but not now, is that 0 probability?)
The same process generalizes.
My point was not “it’s especially hard to learn that there are 3^^^3 people with our choices as the fulcrum”. Rather, consider the person who says “but shouldn’t our choices be dominated
by our current best guesses about what makes the universe seem most enormous, more or less regardless of how implausibly bad those best guesses seem?”. More concretely, perhaps they say
“but shouldn’t we do whatever seems most plausibly likely to satisfy the simulator-gods, because if there are simulator gods and we do please them then we could get mathematically large
amounts of utility, and this argument is bad but it’s not 1 in 3^^^3 bad, so.” One of my answers to this is “don’t worry about the 3^^^3 happy people until you believe yourself upstream
of 3^^^3 happy people in the analogous fashion to how we currently think we’re upstream of 10^50 happy people”.
And for the record, I agree that “maximize option value, figure out what’s going on, stay sane” is another fine response. (As is “I think you have made an error in assessing your insane
plan as having higher EV than business-as-usual”, which is perhaps one argument-step upstream of that.)
I don’t feel too confused about how to act in real life; I do feel somewhat confused about how to formally justify that sort of reasoning.
My personal answer is that infinite universes don’t seem infinitely more important than finite universes, and that 2x bigger universes generally don’t seem 2x as important. (I
tentatively feel that amongst reasonably-large universes, value is almost independent of size—while thinking that within any given universe 2x more flourishing is much closer to 2x as
That sounds like you’re asserting that the amount of possible flourishing limits to some maximum value (as, eg, the universe gets large enough to implement all possible
reasonably-distinct combinations of flourishing civilizations)?
I’m sympathetic to this view. I’m not fully sold, of course. (Example confusion between me and that view: I have conflicting intuitions about whether running an extra identical copy of
the same simulated happy people is ~useless or ~twice as good, and as such I’m uncertain about whether tiling copies of all combinations of flourishing civilizations is better in a way
that doesn’t decay.)
While we’re listing guesses, a few of my other guesses include:
○ Naturalism resolves the issue somehow. Like, perhaps the fact that you need to be embedded somewhere inside the world with a long St. Petersburg game drives its probability lower than
the length of the sentence “a long St. Petersburg game” in a relevant way, and this phenomenon generalizes, or something. (Presumably this would have to come hand-in-hand with some
sort of finitist philosophy, that denies that epistemic probabilities are closed under countable combination, due to your argument above.)
○ There is a maximum utility, namely “however good the best arrangement of the entire mathematical multiverse could be”, and even if it does wind up being the case that the amount of
flourishing you can get per-instantiation fails to converge as space increases, or even if it does turn out that instantiating all the flourishing n times is n times as good, there’s
still some maximal number of instantiations that the multiverse is capable of supporting or something, and the maximum utility remains well-defined.
○ The whole utility-function story is just borked. Like, we already know the concept is philosophically fraught. There’s plausibly a utility number, which describes how good the
mathematical multiverse is, but the other multiverses we intuitively want to evaluate are counterfactual, and counterfactual mathematical multiverses are dubious above and beyond the
already-dubious mathematical multiverse. Maybe once we’re deconfused about this whole affair, we’ll retreat to somewhere like “utility functions are a useful abstraction on local
scales” while having some global theory of a pretty different character.
○ Some sort of ultrafinitism wins the day, and once we figure out how to be suitably ultrafinitist, we don’t go around wanting countable combinations of epistemic probabilities or
worrying too much about particularly big numbers. Like, such a resolution could have a flavor where “Nate’s utilities are unbounded” becomes the sort of thing that infinitists say
about Nate, but not the sort of thing a properly operating ultrafinitist says about themselves, and things turn out to work for the ultrafinitists even if the infinitists say their
utilities are unbounded or w/e.
To be clear, I haven’t thought about this stuff all that much, and it’s quite plausible to me that someone is less confused than me here. (That said, most accounts that I’ve heard, as far
as I’ve managed to understand them, sound less to me like they come from a place of understanding, and more like the speaker has prematurely committed to a resolution.)
One of my answers to this is “don’t worry about the 3^^^3 happy people until you believe yourself upstream of 3^^^3 happy people in the analogous fashion to how we currently think
we’re upstream of 10^50 happy people”.
My point was that this doesn’t seem consistent with anything like a leverage penalty.
And for the record, I agree that “maximize option value, figure out what’s going on, stay sane” is another fine response.
My point was that we can say lots about which actions are more or less likely to generate 3^^^3 utility even without knowing how the universe got so large. (And then this appears to
have relatively clear implications for our behavior today, e.g. by influencing our best guesses about the degree of moral convergence.)
That sounds like you’re asserting that the amount of possible flourishing limits to some maximum value (as, eg, the universe gets large enough to implement all possible
reasonably-distinct combinations of flourishing civilizations)?
In terms of preferences, I’m just saying that it’s not the case that for every universe, there is another possible universe so much bigger that I care only 1% as much about what
happens in the smaller universe. If you look at a 10^20 universe and the 10^30 universe that are equally simple, I’m like “I care about what happens in both of those universes. It’s
possible I care about the 10^30 universe 2x as much, but it might be more like 1.000001x as much or 1x as much, and it’s not plausible I care 10^10 as much.” That means I care about
each individual life less if it happens in a big universe.
This isn’t why I believe the view, but one way you might be able to better sympathize is by thinking: “There is another universe that is like the 10^20 universe but copied 10^10
times. That’s not that much more complex than the 10^20 universe. And in fact total observer counts were already dominated by copies of those universes that were tiled 3^^^3 times,
and the description complexity difference between 3^^^3 and 10^10 x 3^^^3 are not very large.” Of course unbounded utilities don’t admit that kind of reasoning, because they don’t
admit any kind of reasoning. And indeed, the fact that the expectations diverge seem very closely related to the exact reasoning you would care most about doing in order to actually
assess the relative importance of different decisions, so I don’t think the infinity thing is a weird case, it seems absolutely central and I don’t even know how to talk about what
the view should be if the infinites didn’t diverge.
I’m not very intuitively drawn to views like “count the distinct experiences,” and I think that in addition to being kind of unappealing those views also have some pretty crazy
consequences (at least for all the concrete versions I can think of).
I basically agree that someone who has the opposite view—that for every universe there is a bigger universe that dwarfs its importance—has a more complicated philosophical question
and I don’t know the answer. That said, I think it’s plausible they are in the same position as someone who has strong brute intuitions that A>B, B>C, and C>A for some concrete
outcomes A, B, C—no amount of philosophical progress will help them get out of the inconsistency. I wouldn’t commit to that pessimistic view, but I’d give it maybe 50/50---I don’t
see any reason that there needs to be a satisfying resolution to this kind of paradox.
Perhaps I’m being dense, but I don’t follow this point. If I deny that my epistemic probabilities are closed under countable weighted sums, and assert that the hypothesis “you can
actually play a St. Petersburg game for n steps” is less likely than it is easy-to-describe (as n gets large), in what sense does that render me unable to consider St. Petersburg
games in thought experiments?
Do you have preferences over the possible outcomes of thought experiments? Does it feel intuitively like they should satisfy dominance principles? If so, it seems like it’s just as
troubling that there are thought experiments. Analogously, if I had the strong intuition that A>B>C>A, and someone said “Ah but don’t worry, B could never happen in the real world!” I
wouldn’t be like “Great that settles it, no longer feel confused+troubled.”
My point was that this doesn’t seem consistent with anything like a leverage penalty.
I’m not particulalry enthusiastic about “artificial leverage penalties” that manually penalize the hypothesis you can get 3^^^3 happy people by a factor of 1/3^^^3 (and so
insofar as that’s what you’re saying, I agree).
From my end, the core of my objection feels more like “you have an extra implicit assumption that lotteries are closed under countable combination, and I’m not sold on that.” The
part where I go “and maybe some sufficiently naturalistic prior ends up thinking long St. Petersburg games are ultimately less likely than they are simple???” feels to me more
like a parenthetical, and a wild guess about how the weakpoint in your argument could resolve.
(My guess is that you mean something more narrow and specific by “leverage penalty” than I did, and that me using those words caused confusion. I’m happy to retreat to a broader
term, that includes things like “big gambles just turn out not to unbalance naturalistic reasoning when you’re doing it properly (eg. b/c finding-yourself-in-the-universe
correctly handles this sort of thing somehow)”, if you have one.)
(My guess is that part of the difference in framing in the above paragraphs, and in my original comment, is due to me updating in response to your comments, and retreating my
position a bit. Thanks for the points that caused me to update somewhat!)
My point was that we can say lots about which actions are more or less likely to generate 3^^^3 utility even without knowing how the universe got so large.
I agree.
In terms of preferences, I’m just saying...
This seems like a fine guess to me. I don’t feel sold on it, but that could ofc be because you’ve resolved confusions that I have not. (The sort of thing that would persuade me
would be you demonstrating at least as much mastery of my own confusions than I possess, and then walking me through the resolution. (Which I say for the purpose of being upfront
about why I have not yet updated in favor of this view. In particular, it’s not a request. I’d be happy for more thoughts on it if they’re cheap and you find generating them to be
fun, but don’t think this is terribly high-priority.))
That means I care about each individual life less if it happens in a big universe.
I indeed find this counter-intuitive. Hooray for flatly asserting things I might find counter-intuitive!
Let me know if you want me to flail in the direction of confusions that stand between me and what I understand to be your view. The super short version is something like “man, I’m
not even sure whether logic or physics comes first, so I get off that train waaay before we get to the Tegmark IV logical multiverse”.
(Also, to be clear, I don’t find UDASSA particularly compelling, mainly b/c of how confused I remain in light of it. Which I note in case you were thinking that the inferential
gap you need to span stretches only to UDASSA-town.)
Do you have preferences over the possible outcomes of thought experiments? Does it feel intuitively like they should satisfy dominance principles? If so, it seems like it’s
just as troubling that there are thought experiments.
You’ve lost me somewhere. Maybe try backing up a step or two? Why are we talking about thought experiments?
One of my best explicit hypotheses for what you’re saying is “it’s one thing to deny closure of epistemic probabiltiies under countable weighted combination in real life, and
another to deny them in thought experiments; are you not concerned that denying them in thought experiments is troubling?”, but this doesn’t seem like a very likely argument for
you to be making, and so I mostly suspect I’ve lost the thread.
(I stress again that, from my perspective, the heart of my objection is your implicit assumption that lotteries are closed under countable combination. If you’re trying to object
to some other thing I said about leverage penalties, my guess is that I micommunicated my position (perhaps due to a poor choice of words) or shifted my position in response to
your previous comments, and that our arguments are now desynched.)
Backing up to check whether I’m just missing something obvious, and trying to sharpen my current objection:
It seems to me that your argument contains a fourth, unlisted assumption, which is that lotteries are closed under countable combination. Do you agree? Am I being daft and missing
that, like, some basic weak dominance assumption implies closure of lotteries under countable combination? Assuming I’m not being daft, do you agree that your argument sure seems
to leave the door open for people who buy antisymmetry, dominance, and unbounded utilities, but reject countable combination of lotteries?
From my end, the core of my objection feels more like “you have an extra implicit assumption that lotteries are closed under countable combination, and I’m not sold on
that.” [...] It seems to me that your argument contains a fourth, unlisted assumption, which is that lotteries are closed under countable combination. Do you agree?
My formal argument is even worse than that: I assume you have preferences over totally arbitrary probability distributions over outcomes!
I don’t think this is unlisted though—right at the beginning I said we were proving theorems about a preference ordering defined over the space of probability distributions
over a space of outcomes . I absolutely think it’s plausible to reject that starting premise (and indeed I suggest that someone with “unbounded utilities” ought to reject this
premise in an even more dramatic way).
If you’re trying to object to some other thing I said about leverage penalties, my guess is that I miscommunicated my position
It seems to me that our actual situation (i.e. my actual subjective distribution over possible worlds) is divergent in the same way as the St Petersburg lottery, at least with
respect to quantities like expected # of happy people. So I’m less enthusiastic about talking about ways of restricting the space of probability distributions to avoid St
Petersburg lotteries. This is some of what I’m getting at in the parent, and I now see that it may not be responsive to your view. But I’ll elaborate a bit anyway.
There are universes with populations of that seem only times less likely than our own. It would be very surprising and confusing to learn that not only am I wrong but this
epistemic state ought to have been unreachable, that anyone must assign those universes probability at most 1/. I’ve heard it argued that you should be confident that the
world is giant, based on anthropic views like SIA, but I’ve never heard anyone seriously argue that you should be perfectly confident that the world isn’t giant.
If you agree with me that in fact our current epistemic state looks like a St Petersburg lottery with respect to # of people, then I hope you can sympathize with my lack of
All that is to say: it may yet be that preferences are defined over a space of probability distributions small enough to evade the argument in the OP. But at that point it
seems much more likely that preferences just aren’t defined over probability distributions at all—it seems odd to hold onto probability distributions as the object of
preferences while restricting the space of probability distributions far enough that they appear to exclude our current situation.
You’ve lost me somewhere. Maybe try backing up a step or two? Why are we talking about thought experiments?
Suppose that you have some intuition that implies A > B > C > A.
At first you are worried that this intuition must be unreliable. But then you realize that actually B is impossible in reality, so consistency is restored.
I claim that you should be skeptical of the original intuition anyway. We have gotten some evidence that the intuition isn’t really tracking preferences in the way you might
have hoped that it was—because if it were correctly tracking preferences it wouldn’t be inconsistent like that.
The fact that B can never come about in reality doesn’t really change the situation, you still would have expected consistently-correct intuitions to yield consistent answers.
(The only way I’d end up forgiving the intuition is if I thought it was correctly tracking the impossibility of B. But in this case I don’t think so. I’m pretty sure my
intuition that you should be willing to take a 1% risk in order to double the size of the world isn’t tracking some deep fact that would make certain epistemic states
(That all said, a mark against an intuition isn’t a reason to dismiss it outright, it’s just one mark against it.)
◎ Ok, cool, I think I see where you’re coming from now.
I don’t think this is unlisted though …
Fair! To a large degree, I was just being daft. Thanks for the clarification.
It seems to me that our actual situation (i.e. my actual subjective distribution over possible worlds) is divergent in the same way as the St Petersburg lottery, at
least with respect to quantities like expected # of happy people.
I think this is a good point, and I hadn’t had this thought quite this explicitly myself, and it shifts me a little. (Thanks!)
(I’m not terribly sold on this point myself, but I agree that it’s a crux of the matter, and I’m sympathetic.)
But at that point it seems much more likely that preferences just aren’t defined over probability distributions at all
This might be where we part ways? I’m not sure. A bunch of my guesses do kinda look like things you might describe as “preferences not being defined over probability
distributions” (eg, “utility is a number, not a function”). But simultaneously, I feel solid in my ability to use probabliity distributions and utility functions in
day-to-day reasoning problems after I’ve chunked the world into a small finite number of possible actions and corresponding outcomes, and I can see a bunch of reasons why
this is a good way to reason, and whatever the better preference-formalism turns out to be, I expect it to act a lot like probability distributions and utility functions
in the “local” situation after the reasoner has chunked the world.
Like, when someone comes to me and says “your small finite considerations in terms of actions and outcomes are super simplified, and everything goes nutso when we remove
all the simplifications and take things to infinity, but don’t worry, sanity can be recovered so long as you (eg) care less about each individual life in a big universe
than in a small universe”, then my response is “ok, well, maybe you removed the simplifications in the wrong way? or maybe you took limits in a bad way? or maybe utility
is in fact bounded? or maybe this whole notion of big vs small universes was misguided?”
It looks to me like you’re arguing that one should either accept bounded utilities, or reject the probability/utility factorization in normal circumstances, whereas to me
it looks like there’s still a whole lot of flex (ex: ‘outcomes’ like “I come back from the store with milk” and “I come back from the store empty-handed” shouldn’t have
been treated the same way as ‘outcomes’ like “Tegmark 3 multiverse branch A, which looks like B” and “Conway’s game of life with initial conditions X, which looks like Y”,
and something was going wrong in our generalization from the everyday to the metaphysical, and we shouldn’t have been identifying outcomes with universes and expecting
preferences to be a function of probability distributions on those universes, but thinking of “returning with milk” as an outcome is still fine).
And maybe you’d say that this is just conceding your point? That when we pass from everyday reasoning about questions like “is there milk at the store, or not?” to
metaphysical reasoning like “Conway’s Life, or Tegmark 3?”, we should either give up on unbounded utilities, or give up on thinking of preferences as defined on
probability distributions on outcomes? I more-or-less buy that phrasing, with the caveat that I am open to the weak-point being this whole idea that metaphysical universes
are outcomes and that probabilities on outcome-collections that large are reasonable objects (rather than the weakpoint being the probablity/utility factorization per
it seems odd to hold onto probability distributions as the object of preferences while restricting the space of probability distributions far enough that they appear
to exclude our current situation
I agree that would be odd.
One response I have is similar to the above: I’m comfortable using probability distributions for stuff like “does the store have milk or not?” and less comfortable using
them for stuff like “Conway’s Life or Tegmark 3?”, and wouldn’t be surprised if thinking of mathematical universes as “outcomes” was a Bad Plan and that this (or some
other such philosophically fraught assumption) was the source of the madness.
Also, to say a bit more on why I’m not sold that the current situation is divergent in the St. Petersburg way wrt, eg, amount of Fun: if I imagine someone in Vegas
offering me a St. Petersburg gamble, I imagine thinking through it and being like “nah, you’d run out of money too soon for this to be sufficiently high EV”. If you’re
like “ok, but imagine that the world actually did look like it could run the gamble infinitely”, my gut sense is “wow, that seems real sus”. Maybe the source of the
susness is that eventually it’s just not possible to get twice as much Fun. Or maybe it’s that nobody anywhere is ever in a physical position to reliably double the amount
of Fun in the region that they’re able to affect. Or something.
And, I’m sympathetic to the objection “well, you surely shouldn’t assign probability less than <some vanishingly small but nonzero number> that you’re in such a situation!
”. And maybe that’s true; it’s definitely on my list of guesses. But I don’t by any means feel forced into that corner. Like, maybe it turns out that the lightspeed limit
in our universe is a hint about what sort of universes can be real at all (whatever the heck that turns out to mean), and an agent can’t ever face a St. Petersburgish
choice in some suitably general way. Or something. I’m more trying to gesture at how wide the space of possibilities seems to me from my state of confusion, than to make
specific counterproposals that I think are competitive.
(And again, I note that the reason I’m not updating (more) towards your apparently-narrower stance, is that I’m uncertain about whether you see a narrower space of
possible resolutions on account of being less confused than I am, vs because you are making premature philosophical commitments.)
To be clear, I agree that you need to do something weirder than “outcomes are mathematical universes, preferences are defined on (probability distributions over) those” if
you’re going to use unbounded utilities. And again, I note that “utility is bounded” is reasonably high on my list of guesses. But I’m just not all that enthusiastic about
“outcomes are mathematical universes” in the first place, so \shrug.
The fact that B can never come about in reality doesn’t really change the situation, you still would have expected consistently-correct intuitions to yield consistent
I think I understand what you’re saying about thought experiments, now. In my own tongue: even if you’ve convinced yourself that you can’t face a St. Petersburg gamble in
real life, it still seems like St. Petersburg gambles form a perfectly lawful thought experiment, and it’s at least suspicious if your reasoning procedures would break
down facing a perfectly lawful scenario (regardless of whether you happen to face it in fact).
I basically agree with this, and note that, insofar as my confusions resolve in the “unbounded utilities” direction, I expect some sort of account of metaphysical/
anthropic/whatever reasoning that reveals St. Petersburg gambles (and suchlike) to be somehow ill-conceived or ill-typed. Like, in that world, what’s supposed to happen
when someone is like “but imagine you’re offered a St. Petersburg bet” is roughly the same as what’s supposed to happen when someone’s like “but imagine a physically
identical copy of you that lacks qualia”—you’re supposed to say “no”, and then be able to explain why.
(Or, well, you’re always supposed to say “no” to the gamble and be able to explain why, but what’s up for grabs is whether the “why” is “because utility is bounded”, or
some other thing, where I at least am confused enough to still have some of my chips on “some other thing”.)
To be explicit, the way that my story continues to shift in response to what you’re saying, is an indication of continued updating & refinement of my position. Yay;
I expect it to act a lot like probability distributions and utility functions in the “local” situation after the reasoner has chunked the world.
I agree with this: (i) it feels true and would be surprising not to add up to normality, (ii) coherence theorems suggest that any preferences can be represented as
probabilities+utilities in the case of finitely many outcomes.
“utility is a number, not a function”
This is my view as well, but you still need to handle the dependence on subjective uncertainty. I think the core thing at issue is whether that uncertainty is
represented by a probability distribution (where utility is an expectation).
(Slightly less important: my most naive guess is that the utility number is itself represented as a sum over objects, and then we might use “utility function” to refer
to the thing being summed.)
Also, to say a bit more on why I’m not sold that the current situation is divergent in the St. Petersburg way wrt, eg, amount of Fun...
I don’t mean that we face some small chance of encountering a St Petersburg lottery. I mean that when I actually think about the scale of the universe, and what I
ought to believe about physics, I just immediately run into St Petersburg-style cases:
△ It’s unclear whether we can have an extraordinarily long-lived civilization if we reduce entropy consumption to ~0 (e.g. by having a reversible civilization). That
looks like at least 5% probability, and would suggest the number of happy lives is much more than times larger than I might have thought. So does it dominate the
△ But nearly-reversible civilizations can also have exponential returns to the resources they are able to acquire during the messy phase of the universe. Maybe that
happens with only 1% probability, but it corresponds to yet bigger civilization. So does that mean we should think that colonizing faster increases the value of
the future by 1%, or by 100% since these possibilities are bigger and better and dominate the expectation?
△ But also it seems quite plausible that our universe is already even-more-exponentially spatially vast, and we merely can’t reach parts of it (but a large fraction
of them are nevertheless filled with other civilizations like ours). Perhaps that’s 20%. So it actually looks more likely than the “long-lived reversible
civilization” and implies more total flourishing. And on those perspectives not going extinct is more important than going faster, for the usual reasons. So does
that dominate the calculus instead?
△ Perhaps rather than having a single set of physical constants, our universe runs every possible set. If that’s 5%, and could stack on top of any of the above while
implying another factor of of scale. And if the standard model had no magic constants maybe this possibility would be 1% instead of 5%. So should I updated by a
factor of 5 that we won’t discover that the standard model has fewer magic constants, because then “continuum of possible copies of our universe running in
parallel” has only 1% chance instead of 5%?
△ Why not all of the above? What if the universe is vast and it allows for very long lived civilization? And once we bite any of those bullets to grant more people,
then it starts to seem like even less of a further ask to assume that there were actually more people instead. So should we assume that multiple of those
enhugening assumptions are true (since each one increases values by more than it decreases probability), or just take our favorite and then keep cranking up the
numbers larger and larger (with each cranking being more probable than the last and hence more probable than adding a second enhugening assumption)?
Those are very naive physically statements of the possibilities, but the point is that it seems easy to imagine the possibility that populations could be vastly larger
than we think “by default”, and many of those possibilities seem to have reasonable chances rather than being vanishingly unlikely. And at face value you might have
thought those possibilities were actually action-relevant (e.g. the possibility of exponential returns to resources dominates the EV and means we should rush to
colonize after all), but once you actually look at the whole menu, and see how the situation is just obviously paradoxical in every dimension, I think it’s pretty
clear that you should cut off this line of thinking.
A bit more precisely: this situation is structurally identical to someone in a St Petersburg paradox shuffling around the outcomes and finding that they can justify
arbitrary comparisons because everything has infinite EV and it’s easy to rewrite. That is, we can match each universe U with “U but with long-lived reversible
civilizations,” and we find that the long-lived reversible civilizations dominate the calculus. Or we can match each universe U with “U but vast” and find the vast
universes dominate the calculus. Or we can match “long-lived reversible civilizations” with “vast” and find that we can ignore long-lived reversible civilizations.
It’s just like matching up the outcomes in the St Petersburg paradox in order to show that any outcome dominates itself.
The unbounded utility claim seems precisely like the claim that each of those less-likely-but-larger universes ought to dominate our concern, compared to the
smaller-but-more-likely universe we expect by default. And that way of reasoning seems like it leads directly to these contradictions at the very first time you try to
apply it to our situation (indeed, I think every time I’ve seen someone use this assumption in a substantive way it has immediately seemed to run into paradoxes, which
are so severe that they mean they could as well have reached the opposite conclusion by superficially-equally-valid reasoning).
I totally believe you might end up with a very different way of handling big universes than “bounded utilities,” but I suspect it will also lead to the conclusion that
“the plausible prospect of a big universe shouldn’t dominate our concern.” And I’d probably be fine with the result. Once you divorce unbounded utilities from the
usual theory about how utilities work, and also divorce them from what currently seems like their main/only implication, I expect I won’t have anything more than a
semantic objection.
More specifically and less confidently, I do think there’s a pretty good chance that whatever theory you end up with will agree roughly with the way that I handle big
universes—we’ll just use our real probabilities of each of these universes rather than focusing on the big ones in virtue of their bigness, and within each universe
we’ll still prefer have larger flourishing populations. I do think that conclusion is fairly uncertain, but I tentatively think it’s more likely we’ll give up on the
principle “a bigger civilization is nearly-linearly better within a given universe” than on the principle “a bigger universe is much less than linearly more
And from a practical perspective, I’m not sure what interim theory you use to reason about these things. I suspect it’s mostly academic for you because e.g. you think
alignment is a 99% risk of death instead of a 20% risk of death and hence very few other questions about the future matter. But if you ever did find yourself having to
reason about humanity’s long-term future (e.g. to assess the value of extinction risk vs faster colonization, or the extent of moral convergence), then it seems like
you should use an interim theory which isn’t fanatical about the possibility of big universes—because the fanatical theories just don’t work, and spit out inconsistent
results if combined with our current framework. You can also interpret my argument as strongly objecting to the use of unbounded utilities in that interim framework.
This is my view as well,
(I, in fact, lifted it off of you, a number of years ago :-p)
but you still need to handle the dependence on subjective uncertainty.
Of course. (And noting that I am, perhaps, more openly confused about how to handle the subjective uncertainty than you are, given my confusions around things like
logical uncertainty and whether difficult-to-normalize arithmetical expressions meaningfully denote numbers.)
Running through your examples:
It’s unclear whether we can have an extraordinarily long-lived civilization …
I agree. Separately, I note that I doubt total Fun is linear in how much compute is available to civilization; continuity with the past & satisfactory completion
of narrative arcs started in the past is worth something, from which we deduce that wiping out civilization and replacing it with another different civilization of
similar flourish and with 2x as much space to flourish in, is not 2x as good as leaving the original civilization alone. But I’m basically like “yep, whether we
can get reversibly-computed Fun chugging away through the high-entropy phase of the universe seems like an empiricle question with cosmically large swings in
utility associated therewith.”
But nearly-reversible civilizations can also have exponential returns to the resources they are able to acquire during the messy phase of the universe.
This seems fairly plausible to me! For instance, my best guess is that you can get more than 2x the Fun by computing two people interacting than by computing two
individuals separately. (Although my best guess is also that this effect diminishes at scale, \shrug.)
By my lights, it sure would be nice to have more clarity on this stuff before needing to decide how much to rush our expansion. (Although, like, 1st world
But also it seems quite plausible that our universe is already even-more-exponentially spatially vast, and we merely can’t reach parts of it
Sure, this is pretty plausible, but (arguendo) it shouldn’t really be factoring into our action analysis, b/c of the part where we can’t reach it. \shrug
Perhaps rather than having a single set of physical constants, our universe runs every possible set.
Sure. And again (arguendo) this doesn’t much matter to us b/c the others are beyond our sphere of influence.
Why not all of the above? What if the universe is vast and it allows for very long lived civilization? And once we bite any of those bullets to grant 10^100
more people, then it starts to seem like even less of a further ask to assume that there were actually 10^1000 more people instead
I think this is where I get off the train (at least insofar as I entertain unbounded-utility hypotheses). Like, our ability to reversibly compute in the
high-entropy regime is bounded by our error-correction capabilities, and we really start needing to upend modern physics as I understand it to make the numbers
really huge. (Like, maybe 10^1000 is fine, but it’s gonna fall off a cliff at some point.)
I have a sense that I’m missing some deeper point you’re trying to make.
I also have a sense that… how to say… like, suppose someone argued “well, you don’t have 1/∞ probability that “infinite utility” makes sense, so clearly you’ve
got to take infinite utilities seriously”. My response would be something like “That seems mixed up to me. Like, on my current understanding, “infinite utility” is
meaningless, it’s a confusion, and I just operate day-to-day without worrying about it. It’s not so much that my operating model assigns probability 0 to the
proposition “infinite utilities are meaningful”, as that infinite utilities simply don’t fit into my operating model, they don’t make sense, they don’t typecheck.
And separately, I’m not yet philosophically mature, and I can give you various meta-probabilities about what sorts of things will and won’t typecheck in my
operating model tomorrow. And sure, I’m not 100% certain that we’ll never find a way to rescue the idea of infinite utilities. But that meta-uncertainty doesn’t
bleed over into my operating model, and I’m not supposed to ram infinities into a place where they don’t fit just b/c I might modify the type signatures
When you bandy around plausible ways that the universe could be real large, it doesn’t look obviously divergent to me. Some of the bullets you’re handling are ones
that I am just happy to bite, and others involve stuff that I’m not sure I’m even going to think will typecheck, once I understand wtf is going on. Like, just as
I’m not compelled by “but you have more than 0% probability that ‘infinite utility’ is meaningful” (b/c it’s mixing up the operating model and my philosophical
immaturity), I’m not compelled by “but your operating model, which says that X, Y, and Z all typecheck, is badly divergent”. Yeah, sure, and maybe the resolution
is that utilities are bounded, or maybe it’s that my operating model is too permissive on account of my philosophical immaturity. Philosophical immaturity can lead
to an operating model that’s too permisive (cf. zombie arguments) just as easily as one that’s too strict.
Like… the nature of physical law keeps seeming to play games like “You have continua!! But you can’t do an arithmetic encoding. There’s infinite space!! But most
of it is unreachable. Time goes on forever!! But most of it is high-entropy. You can do reversible computing to have Fun in a high-entropy universe!! But error
accumulates.” And this could totally be a hint about how things that are real can’t help but avoid the truly large numbers (never mind the infinities), or
something, I don’t know, I’m philisophically immature. But from my state of philosophical immaturity, it looks like this could totally still resolve in a “you were
thinking about it wrong; the worst enhugening assumptions fail somehow to typecheck” sort of way.
Trying to figure out the point that you’re making that I’m missing, it sounds like you’re trying to say something like “Everyday reasoning at merely-cosmic scales
already diverges, even without too much weird stuff. We already need to bound our utilities, when we shift from looking at the milk in the supermarket to looking
at the stars in the sky (nevermind the rest of the mathematical multiverse, if there is such a thing).” Is that about right?
If so, I indeed do not yet buy it. Perhaps spell it out in more detail, for someone who’s suspicious of any appeals to large swaths of terrain that we can’t affect
(eg, variants of this universe w/ sufficiently different cosmological constants, at least in the regions where the locals aren’t thinking about us-in-particular);
someone who buys reversible computing but is going to get suspicious when you try to drive the error rate to shockingly low lows?
To be clear, insofar as modern cosmic-scale reasoning diverges (without bringing in considerations that I consider suspicious and that I suspect I might later
think belong in the ‘probably not meaningful (in the relevant way)’ bin), I do start to feel the vice grips on me, and I expect I’d give bounded utilities another
look if I got there.
○ A side note: IB physicalisms solves at least a large chunk of naturalism/counterfactuals/anthropics but is almost orthogonal to this entire issue (i.e. physicalist loss functions
should still be bounded for the same reason cartesian loss functions should be bounded), so I’m pretty skeptical there’s anything in that direction. The only part which is somewhat
relevant is: IB physicalists have loss functions that depend on which computations are running so two exact copies of the same thing definitely count as the same and not twice as much
(except potentially in some indirect way, such as being involved together in a single more complex computation).
■ I am definitely entertaining the hypothesis that the solution to naturalism/anthropics is in no way related to unbounded utilities. (From my perspective, IB physicalism looks
like a guess that shows how this could be so, rather than something I know to be a solution, ofc. (And as I said to Paul, the observation that would update me in favor of it would
be demonstrated mastery of, and unravelling of, my own related confusions.))
★ In the parenthetical remark, are you talking about confusions related to Pascal-mugging-type thought experiments, or other confusions?
◎ Those & others. I flailed towards a bunch of others in my thread w/ Paul. Throwing out some taglines:
● “does logic or physics come first???”
● “does it even make sense to think of outcomes as being mathematical universes???”
● “should I even be willing to admit that the expression “3^^^3″ denotes a number before taking time proportional to at least log(3^^^3) to normalize it?”
● “is the thing I care about more like which-computations-physics-instantiates, or more like the-results-of-various-computations??? is there even a difference?”
● “how does the fact that larger quantum amplitudes correspond to more magical happening-ness relate to the question of how much more I should care about a simulation
running on a computer with wires that are twice as thick???”
Note that these aren’t supposed to be particularly well-formed questions. (They’re more like handles for my own confusions.)
Note that I’m open to the hypothesis that you can resolve some but not others. From my own state of confusion, I’m not sure which issues are interwoven, and it’s plausible
to me that you, from a state of greater clarity, can see independences that I cannot.
Note that I’m not asking for you to show me how IB physicalism chooses a consistent set of answers to some formal interpretations of my confusion-handles. That’s the sort
of (non-trivial and virtuous!) feat that causes me to rate IB physicalism as a “plausible guess”.
In the specific case of IB physicalism, I’m like “maaaybe? I don’t yet see how to relate this Γ that you suggestively refer to as a ‘map from programs to results’ to a
philosophical stance on computation and instantiation that I understand” and “I’m still not sold on the idea of handling non-realizability with inframeasures (on account
of how I still feel confused about a bunch of things that inframeasures seem like a plausible guess for how to solve)” and etc.
Maybe at some point I’ll write more about the difference, in my accounting, between plausible guesses and solutions.
● Hmm… I could definitely say stuff about, what’s the IB physicalism take on those questions. But this would be what you specifically said you’re not asking me to do.
So, from my perspective addressing your confusion seems like a completely illegible task atm. Maybe the explanation you alluded to in the last paragraph would help.
△ I’d be happy to read it if you’re so inclined and think the prompt would help you refine your own thoughts, but yeah, my anticipation is that it would mostly be
updating my (already decent) probability that IB physicalism is a reasonable guess.
A few words on the sort of thing that would update me, in hopes of making it slightly more legible sooner rather than later/never: there’s a difference between
giving the correct answer to metaethics (“‘goodness’ refers to an objective (but complicated, and not objectively compelling) logical fact, which was physically
shadowed by brains on account of the specifics of natural selection and the ancestral environment”), and the sort of argumentation that, like, walks someone from
their confused state to the right answer (eg, Eliezer’s metaethics sequence). Like, the confused person is still in a state of “it seems to me that either morality
must be objectively compelling, or nothing truly matters”, and telling them your favorite theory isn’t really engaging with their intuitions. Demonstrating that
your favorite theory can give consistent answers to all their questions is something, it’s evidence that you have at least produced a plausible guess. But from
their confused perspective, lots of people (including the nihilists, including the Bible-based moral realists) can confidently provide answers that seem
superficially consistent.
The compelling thing, at least to me and my ilk, is the demonstration of mastery and the ability to build a path from the starting intuitions to the conclusion. In
the case of a person confused about metaethics, this might correspond to the ability to deconstruct the “morality must be objectively compelling, or nothing truly
matters” intuition, right in front of them, such that they can recognize all the pieces inside themselves, and with a flash of clarity see the knot they were tying
themselves into. At which point you can help them untie the knot, and tug on the strings, and slowly work your way up to the answer.
(The metaethics sequence is, notably, a tad longer than the answer itself.)
(If I were to write this whole concept of solutions-vs-answers up properly, I’d attempt some dialogs that make the above more concrete and less metaphorical, but \
In the case of IB physicalism (and IB more generally), I can see how it’s providing enough consistent answers that it counts as a plausible guess. But I don’t see
how to operate it to resolve my pre-existing confusions. Like, we work with (infra)measures over , and we say some fancy words about how is our “beliefs about the
computations”, but as far as I’ve been able to make out this is just a neato formalism; I don’t know how to get to that endpoint by, like, starting from my own
messy intuitions about when/whether/how physical processes reflect some logical procedure. I don’t know how to, like, look inside myself, and find confusions
like “does logic or physics come first?” or “do I switch which algorithm I’m instantiating when I drink alcohol?”, and disassemble them into their component parts,
and gain new distinctions that show me how the apparent conflicts weren’t true conflicts and all my previous intuitions were coming at things from slightly the
wrong angle, and then shift angles and have a bunch of things click into place, and realize that the seeds of the answer were inside me all along, and that the
answer is clearly that the universe isn’t really just a physical arrangement of particles (or a wavefunction thereon, w/e), but one of those plus a mapping from
syntax-trees to bits (here taking ). Or whatever the philosophy corresponding to “a hypothesis is a ” is supposed to be. Like, I understand that it’s a neat
formalism that does cool math things, and I see how it can be operated to produce consistent answers to various philosophical questions, but that’s a long shot
from seeing it solve the philosophical problems at hand. Or, to say it another way, answering my confusion handles consistently is not nearly enough to get me to
take a theory philosophically seriously, like, it’s not enough to convince me that the universe actually has an assignment of syntax-trees to bits in addition to
the physical state, which is what it looks to me like I’d need to believe if I actually took IB physicalism seriously.
• I don’t think I’m capable of writing something like the metaethics sequence about IB, that’s a job for someone else. My own way of evaluating philosophical
claims is more like:
□ Can we a build an elegant, coherent mathematical theory around the claim?
□ Does the theory meet reasonable desiderata?
□ Does the theory play nicely with other theories we have high confidence of?
□ If there are compelling desiderata the theory doesn’t meet, can we show that meeting them is impossible?
For example, the way I understood objective morality is wrong was by (i) seeing that there’s a coherent theory of agents with any utility function whatsoever
(ii) understanding that, in terms of the physical world, “Vanessa’s utility function” is more analogous to “coastline of Africa” than to “fundamental equations
of physics”.
I agree that explaining why we have certain intuitions is a valuable source of evidence, but it’s entangled with messy details of human psychology that create
a lot of noise. (Notice that I’m not saying you shouldn’t use intuition, obviously intuition is an irreplaceable core part of cognition. I’m saying that
explaining intuition using models of the mind, while possible and desirable, is also made difficult by the messy complexity of human minds, which in particular
introduces a lot of variables that vary between people.)
Also, I want to comment on your last tagline, just because it’s too tempting:
how does the fact that larger quantum amplitudes correspond to more magical happening-ness relate to the question of how much more I should care about a
simulation running on a computer with wires that are twice as thick???
I haven’t written the proofs cleanly yet (because prioritizing other projects atm), but it seems that IB physicalism produces a rather elegant interpretation
of QM. Many-worlds turns out to be false. The wavefunction is not “a thing that exists”. Instead, what exists is the outcomes of all possible measurements. The
universe samples those outcomes from a distribution that is determined by two properties: (i) the marginal distribution of each measurement has to obey the
Born rule (ii) the overall amount of computation done by the universe should be minimal. It follows that, outside of weird thought experiments (i.e. as long as
decoherence applies), agents don’t get split into copies and quantum randomness is just ordinary randomness. (Another nice consequence is that Boltzmann brains
don’t have qualia.)
□ What’s ordinary randomness?
□ I think that this confusion results from failing to distinguish between your individual utility function and the “effective social utility function” (the result of cooperative bargaining
between all individuals in a society). The individual utility function is bounded on a scale which is roughly comparable to Dunbar’s number^[1]. The effective social utility function is
bounded on a scale comparable to the current size of humanity. When you conflate them, the current size of humanity seems like a strangely arbitrary parameter so you’re tempted to decide the
utility function is unbounded.
The reason why distinguishing between those two is so hard, is because there are strong social incentives to conflate them, incentives which our instincts are honed to pick up on. Pretending
to unconditionally follow social norms is a great way to seem trustworthy. When you combine it with an analytic mindset that’s inclined to reasoning with explicit utility functions, this
self-deception takes the form of modeling your intrinsic preferences by utilitarianism.
Another complication is, larger universes tend to be more diverse and hence more interesting. But this also saturates somewhere (having e.g. books to choose from is not noticeably better from
having books to choose from).
1. ↩︎
It seems plausible to me both for explaining how people behave in practice and in terms of evolutionary psychology.
• The proof doesn’t run for me. The only way I know to be able to rearrange the terms in a infinite series is if the starting starting series converges and the resultant series converges. The
series doesn’t fullfill the condition so I am not convinced the rewrite is a safe step.
I am a bit unsure about my maths so I am going to hyberbole the kind of flawed logic I read into the proof. Start with series that might not converge 1+1+1+1+1+1… (oh it indeed blatantly
diverges) then split each term to have a non-effective addition (1+0)+(1+0)+(1+0)+(1+0)… . Blatantly disregard safety rules about paranthesis messing with series and just treat them as
paranthesis that follow familiar rules 1+0+1+0+1+0+1+0+1… so 1+1+1+1… is not equal to itself. (unsafe step leads to non-sense)
With converging series it doesn’t matter whether we get “twice as fast” to the limit but the “rate of ascension” might matter to whatever analog a divergent series would have to a value.
□ The correct condition for real numbers would be absolute convergence (otherwise the sum after rearrangement might become different and/or infinite) but you are right: the series
rearrangement is definitely illegal here.
☆ But in the post I’m rearranging a series of probabilities, which is very legal. The fact that you can’t rearrange infinite sums is an intuitive reason to reject Weak Dominance, and then
the question is how you feel about that.
○ Those probabilities are multiplied by s, which makes it more complicated.
If I try running it with s being the real numbers (which is probably the most popular choice for utility measurement), the proof breaks down. If I, for example, allow negative
utilities, I can rearrange the series from a divergent one into a convergent one and vice versa, trivially leading to a contradiction just from the fact that I am allowed to do weird
things with infinite series, and not because of proposed axioms being contradictory.
EDIT: concisely, your axioms do not imply that the rearrangement should result in the same utility.
■ The rearrangement property you’re rejecting is basically what Paul is calling the “rules of probability” that he is considering rejecting.
If you have a probability distribution over infinitely (but countably) many probability distributions, each of which is of finite support, then it is in fact legal to “expand out”
the probabilities to get one distribution over the underlying (countably infinite) domain. This is standard in probability theory, and it implies the rearrangement property that
bothers you.
★ Oh, thanks, I did not think about that! Now everything makes much more sense.
□ I’m not rearranging a sum of real numbers. I’m showing that no relationship over probability distributions satisfies a given dominance condition.
☆ I am not familiar with the rules of lotteries and mixtures to know whether the mixture rewrite is valid or not. If the outcomes were for example money payouts then the operations carried
out would be invalid. I would be surprised if somehow the rules for lotteries made this okay.
The bit where there is too much implicit steps for me is
Consider the lottery
We can write as a mixture:
I would benefit from babystepping throught this process or atleast pointers what I need to learn to be convinced of this
○ I’m using the usual machinery of probability theory, and particularly countable additivity. It may be reasonable to give up on that, and so I think the biggest assumption I made at
the beginning was that we were defining a probability distribution over arbitrary lotteries and working with the space of probability distributions.
A way to look at it is: the thing I’m taking sums over are the probabilities of possible outcomes. I’m never talking anywhere about utilities or cash payouts or anything else. The
fact that I labeled some symbols does not mean that the real number 8 is involved anywhere.
But these sums over the probabilities of worlds are extremely convergent. I’m not doing any “rearrangement,” I’m just calculating .
■ So there are some missing axioms here, describing what happens when you construct lotteries out of other lotteries. Specifically, the rearranging step Slider asks about is not
justified by the explicitly given axioms alone: it needs something along the lines of “if for each i we have a lottery , then the values of the lotteries and are equal”.
(Your derivation only actually uses this in the special case where for each i only finitely many of the are nonzero.)
You might want to say either that these two “different” lotteries have equal value, or else that they are in fact the same lottery.
In either case, it seems to me that someone might dispute the axiom in question (intuitively obvious though it seems, just like the others). You’ve chosen a notation for lotteries
that makes an analogy with infinite series; if we take this seriously, we notice that this sort of rearrangement absolutely can change whether the series converges and to what
value if so. How sure are you that rearranging lotteries is safer than rearranging sums of real numbers?
(The sums of the probabilities are extremely convergent, yes. But the probabilities are (formally) multiplying outcomes whose values we are supposing are correspondingly
divergent. Again, I am not sure I want to assume that this sort of manipulation is safe.)
★ I’m handling lotteries as probability distributions over an outcome space , not as formal sums of outcomes.
To make things simple you can assume is countable. Then a lottery assigns a real number to each , representing its probability under the lottery , such that . The sum is
defined by . And all these infinite sums of real numbers are in turn defined as the suprema of the finite sums which are easily seen to exist and to still sum to 1. (All of
this is conventional notation.) Then and are exactly equal.
◎ OK! But I still feel like there’s something being swept under the carpet here. And I think I’ve managed to put my finger on what’s bothering me.
There are various things we could require our agents to have preferences over, but I am not sure that probability distributions over outcomes is the best choice. (Even
though I do agree that the things we want our agents to have preferences over have essentially the same probabilistic structure.)
A weaker assumptions we might make about agents’ preferences is that they are over possibly-uncertain situations, expressed in terms of the agent’s epistemic state.
And I don’t think “nested” possibly-uncertain-situations even exist. There is no such thing as assigning 50% probability to each of (1) assigning 50% probability to each
of A and B, and (2) assigning 50% probability to each of A and C. There is such a thing as assigning 50% probability now to assigning those different probabilities in five
minutes, and by the law of iterated expectations your final probabilities for A,B,C must then obey the distributive law, but the situations are still not literally the
same, and I think that in divergent-utility situations we can’t assume that your preferences depend only on the final outcome distribution.
Another way to say this is that, given that the and are lotteries rather than actual outcomes and that combinations like mean something more complicated than they may
initially look like they mean, the dominance axioms are less obvious than the notation makes them look, and even though there are no divergences in the
sums-over-probabilities that arise when you do the calculations there are divergences in implied something-like-sums-over-weighted utilities, and in my formulation you
really are having to rearrange outcomes as well as probabilities when you do the calculations.
● I agree that in the real world you’d have something like “I’m uncertain about whether X or Y will happen, call it ^50⁄[50]. If X happens, I’m ^50⁄[50] about whether A
or B will happen. If Y happens, I’m ^50⁄[50] about whether B or C will happen.” And it’s not obvious that this should be the same as being ^50⁄[50] between B or X, and
conditioned on X being ^50⁄[50] between A or C.
Having those two situations be different is kind of what I mean by giving up on probabilities—your preferences are no longer a function of the probability that
outcomes occur, they are a more complicated function of your epistemic state, and so it’s not correct to summarize your epistemic state as a probability distribution
over outcomes.
I don’t think this is totally crazy, but I think it’s worth recognizing it as a fairly drastic move.
△ Would a decision theory like this count as “giving up on probabilities” in the sense in which you mean it here?
◎ To anyone who is still not convinced—that last move, , is justified by Tonelli’s theorem, merely because (for all ).
■ The way I look at this is that objects like live in a function space like , specifically the subspace of that where the functions are integrable with respect to counting measure
on and . In other words, objects like are probability mass functions (pmf). is , and is , and of anything else is . When we write what looks like an infinite series , what this
really means is that we’re defining a new by pointwise infinite summation: . So only each collection of terms that contains a given needs to form a convergent series in order for
this new to be well-defined. And for it to equal another , the convergent sums only need to be equal pointwise (for each , ). In Paul’s proof above, the only for which the
collection of terms containing it is even infinite is . That’s the reason he’s “just calculating” that one sum.
■ The outcomes have the property that they are step-wise more than double the worth.
In the real part only halfs on each term. So as the series goes on each term gets bigger and bigger instead of smaller and smaller and smaller associated with convergent-like
scenario. So it seems to me that even in isolation this is a divergent-like series.
■ Here’s a concrete example. Start with a sum that converges to 0 (in fact every partial sum is 0):
0 + 0 + …
Regroup the terms a bit:
= (1 + −1) + (1 + −1) + …
= 1 + (-1 + 1) + (-1 + 1) + …
= 1 + 0 + 0 + …
and you get a sum that converges to 1 (in fact every partial sum is 1). I realize that the things you’re summing are probability distributions over outcomes and not real numbers,
but do you have reason to believe that they’re better behaved than real numbers in infinite sums? I’m not immediately seeing how countable additivity helps. Sorry if that should
be obvious.
★ Your argument doesn’t go through if you restrict yourself to infinite weighted averages with nonnegative weights.
◎ Aha. So if a sum of non-negative numbers converges, than any rearrangement of that sum will converge to the same number, but not so for sums of possibly-negative numbers?
Ok, another angle. If you take Christiano’s lottery:
and map outcomes to their utilities, setting the utility of to 1, of to 2, etc., you get:
Looking at how the utility gets rearranged after the “we can write as a mixture” step, the first “1/2″ term is getting “smeared” across the rest of the terms, giving:
which is a sequence of utilities that are pairwise higher. This is an essential part of the violation of Antisymmetry/Unbounded/Dominance. My intuition says that a
strange thing happened when you rearranged the terms of the lottery, and maybe you shouldn’t do that.
Should there be another property, called “Rearrangement”?
Rearrangement: you may apply an infinite number of commutivity () and associativity () rewrites to a lottery.
(In contrast, I’m pretty sure you can’t get an Antisymmetry/Unbounded/Dominance violation by applying only finitely many commutivity and associativity rearrangements.)
I don’t actually have a sense of what “infinite lotteries, considered equivalent up to finite but not infinite rearrangements” look like. Maybe it’s not a sensible thing.
○ I am having trouble trying to translate between infinity-hiding style and explicit infinity style. My grievance with might be stupid.
split X_0 into equal number parts to final form
move the scalar in
combine scalars
Take each of these separately to the rest of the original terms
Combine scalars to try to hit closest to the target form
is then quite far from
Within real precision a single term hasn’t moved much
This suggests to me that somewhere there are “levels of calibration” that are mixing levels corresponding to members of different archimedean fields trying to intermingle here.
Normally if one is allergic to infinity levels there are ways to dance around it / think about it in different terms. But I am not efficient in translating between them.
■ New attempt
I think I now agree that can be written as
However this uses a “de novo” indexing and gets only to
taking terms out form the inner thing crosses term lines for the outer summation which counts as “messing with indexing” in my intuition. The suspect move just maps them out one
to one
But why is this the permitted way and could I jam the terms differently in say apply to every other term
If I have I am more confident that they “index at the same rate” to make . However if I have I need more information about the relation of a and b to make sure that mixing them
plays nicely. Say in the case of b=2a then it is not okay to think only of the terms when mixing.
□ I had the same initial reaction. I believe the logic of the proof is fine (it is similar to the Mazur swindle), basically because it it not operating on real numbers, but rather on mixtures
of distributions.
The issue is more: why would you expect the dominance condition to hold in the first place? If you allow for unbounded utility functions, then you have to give it up anyway, for kind of
trivial reasons. Consider two sequences Ai and Bi of gambles such that EA_i<EB_i and sum_i p_iEA_i and sum_i p_i EB_i both diverge. Does it follow that E(sum_i p_iA_i)< E(sum_i p_i B_i) ?
Obviously not, since both quantities diverge. At best you can say <=. A bit more formally; in real analysis/measure theory one works with the so-called extended real numbers, in which the
value “infinity” is assigned to any divergent sum, with this value assumed to be defined by the algebraic property x<=infinity for any x. In particular, there is no x in the extended real
numbers such that infinity<x. So at least in standard axiomatizations of measure theory, you cannot expect the strict dominance condition to hold in complete generality; you will have to make
some kind of exception for infinite values. Similar considerations apply to the Intermediate Mixtures assumption.
☆ With surreals I might have transfinite quantities that can reliably compare every which way despite both members being beyond a finite bound. For “tame” entities all kinds of nice
properties are easy to get/prove. The game of “how wild my entities can get while retaining a certain property” is a very different game. “These properties are impossible to get even for
super-wild things” is even harder.
Mazur seems (atleast based on the wikipedia article) not to be a proof of certain things, so that warrants special interest whether the applicability conditions are met or not.
□ The sum we’re rearranging isn’t a sum of real numbers, it’s a sum in . Ignoring details of what means… the two rearrangements give the same sum! So I don’t understand what your argument is.
Abstracting away the addition and working in an arbitrary topological space, the argument goes like this: . For all Therefore, f is not continuous (else 0 = 1). | {"url":"https://www.greaterwrong.com/posts/hbmsW2k9DxED5Z4eJ/impossibility-results-for-unbounded-utilities","timestamp":"2024-11-11T19:12:53Z","content_type":"text/html","content_length":"1049373","record_id":"<urn:uuid:f6aea453-ebc7-45cc-881b-f28ab9dd2fd5>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00315.warc.gz"} |
Convert Megajoule/second to Attojoule/second
Please provide values below to convert megajoule/second [MJ/s] to attojoule/second [aJ/s], or vice versa.
Megajoule/second to Attojoule/second Conversion Table
Megajoule/second [MJ/s] Attojoule/second [aJ/s]
0.01 MJ/s 1.0E+22 aJ/s
0.1 MJ/s 1.0E+23 aJ/s
1 MJ/s 1.0E+24 aJ/s
2 MJ/s 2.0E+24 aJ/s
3 MJ/s 3.0E+24 aJ/s
5 MJ/s 5.0E+24 aJ/s
10 MJ/s 1.0E+25 aJ/s
20 MJ/s 2.0E+25 aJ/s
50 MJ/s 5.0E+25 aJ/s
100 MJ/s 1.0E+26 aJ/s
1000 MJ/s 1.0E+27 aJ/s
How to Convert Megajoule/second to Attojoule/second
1 MJ/s = 1.0E+24 aJ/s
1 aJ/s = 1.0E-24 MJ/s
Example: convert 15 MJ/s to aJ/s:
15 MJ/s = 15 × 1.0E+24 aJ/s = 1.5E+25 aJ/s
Popular Power Unit Conversions
Convert Megajoule/second to Other Power Units | {"url":"https://www.unitconverters.net/power/megajoule-second-to-attojoule-second.htm","timestamp":"2024-11-02T09:16:17Z","content_type":"text/html","content_length":"14116","record_id":"<urn:uuid:14b08016-5e46-40c4-9305-9e20f5c298f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00888.warc.gz"} |
Speed reading
Describing the building process of the halls.md Breast Cancer Risk Calculator.
Documenting the methods used by the halls.md Risk Calculator is essential, so the results can be trusted. The basis for risk calculation methods is described in references 1 through 5. See the
references, below.
The internal workings of my Gail model 1 calculator start by calculating the "summary relative risk" using the answers from questions 1 through 6.
So the first section of questions, about my age, did my mother have cancer, and so on.
Then you have the option of adding several extra relative risk factors (from questions 7 through 12) which were not part of model 1. These extra factors are:
• a) mammographic density (ref 6),
• b) age-specific tamoxifen effect (ref 7),
• c) dose-specific alcohol effect (ref 9),
• d) age-specific correction factors to adjust for Black race (using data from ref 5),
• e) presence of LCIS on a biopsy (ref 10), and
• f) oral contraceptives effects (ref 11)
The relative risks of these extra factors are multiplied with the summary relative risk.
Then, the absolute risks at 10-year, 20-year, 30-year and lifetime(90) are calculated using the graphical interpolation method described in reference 1. Polynomial equations are used in place of the
curves graphed in reference 1.
However, model 1 is known to underestimate absolute risks in older women particularly, so age-specific correction factors are applied (from ref 5). Then, a correction is applied for not participating
in screening mammography (ref 8). I have not yet implemented a correction factor for one problem: The model 1 calculator can significantly overestimate the chance of cancer in women with many risk
factors. This is a known problem with model 1 and it can be made worse by including extra risk factors.
The internal workings of my NSABP model 2 calculator also start by calculating the summary relative risk, using the basic Gail model method. However, my model 2 calculator uses slightly different
values for relative risks than those published in reference 5 for model 2. The Appendix (below) lists the relative risk figures used by my calculator, in comparison to the published values from
reference 5. To explain the difference, if you try to read the text of reference 5, you’ll see that the mathematics described are very complicated, and unfortunately, not all the mathematical terms
used are fully defined in reference 5. I suspect that the relative risk values used in my calculator, have merged the effects of mathematical terms "h*I(t)" and "F(t)", whereas these terms are
presumably used in discrete mathematical operations in the NCI’s downloadable "Breast Cancer Risk Assessment Tool", (which implements model 2). Model 2 does not include the risk of developing Ductal
Carcinoma In-Situ (DCIS). Therefore, the NSABP model 2 results will be lower than Gail model 1 results. In my risk calculator, there is an option, that will add in the risk of in-situ cancers (such
as DCIS) into the model 2 estimate.
Because this calculator includes extra risk modifiers and correction factors that
were not originally part of model 1 or model 2, it is not possible to demonstrate that the results are accurate from published population-based studies. Therefore, please consider these results to be
an estimate.
Steven B. Halls, MD, FRCPC.
Click here to read the Appendix, with more details about modifications to the standard Gail Model.
Back to the Breast Cancer Risk Calculator.
1. Benichou J, Gail MH, Mulvihill JJ, Graphs to estimate an individualized risk of breast cancer. J Clin Oncol 1996; 14:103-110.
2. Gail MH, Brinton LA, Byar DP, Corle DK, Green SB, Chairer C, Mulvihill JJ, Projecting individualized probabilities of developing breast cancer for white females who are being examined annually. J
Natl Cancer Inst 1989; 81:1879-1886.
3. Spiegelman D, Colditz GA, Hunter D, et al.: Validation of the Gail et al. model for predicting individual breast cancer risk. J Natl Cancer Inst 1994; 86:600-607.
4. Bondy ML, Lustbader ED, Halabi S, et al.: Validation of a breast cancer risk assessment model in women with a positive family history. J Natl Cancer Inst 1994; 86: 620-625.
5. Costantino JP, Gail MH, Pee D, Anderson S, Redmond CK, Benichou J, Wieand S, Validation studies for models projecting the risk of invasive and total breast cancer incidence. J Natl Cancer Inst
1999; 91:1541-1548.
6. Byrne C, Schairer C, Wolfe J, Parekh N, Salane M, Brinton LA, Hoover R, Haile R, Mammographic features and breast cancer risk: Effects with time, age and menopause status. J Natl Cancer Inst
1995; 87:1622;1629.
7. Fisher B, Costantino JP,Wickerham DL at al, Tamoxifen for prevention of breast cancer: Report of the national surgical adjuvant breast and bowel project P-1 study. J Natl Cancer Inst 1998;
8. Spiegelman D, Colditz GA, Hunter D, Hertzmark E. Validation of the Gail et al. model for predicting individual breast cancer risk. J Natl Cancer Inst 1994; 86:600-607.
9. Smith-Warner SA, Spiegelman D, S Yaun, et al. Alcohol and breast cancer in women: A pooled analysis of cohort studies. JAMA 1998; 279:535-540.
10. Bodian CA, Perzin KH, Lattes R, Lobular Neoplasia. Long term risk of breast cancer and relation to other factors. Cancer 1996:78:1024-1034.
11. Collaborative Group on Hormonal Factors in Breast Cancer. Breast cancer and hormonal contraceptives: collaborative reanalysis of individual data on 53 297 women with breast cancer and 100 239
women without breast cancer from 54 epidemiological studies. Lancet 1996; 347:1713-1327. | {"url":"https://halls.md/gail-model-risk-calculation/","timestamp":"2024-11-09T01:33:26Z","content_type":"text/html","content_length":"43175","record_id":"<urn:uuid:8fa4cade-929e-45a1-9ee1-e04e978a4ef3>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00260.warc.gz"} |
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Blog Archives
Teaching children mathematics in mixed-attainment groups is fundamental to providing all children with their access and entitlement to a common, statutory mathematics curriculum, as defined by HMG.
However, teaching is complex and how anyone learns to teach is dependent upon many factors including the beliefs and values they hold and the ethos within the schools in which they practice. However,
when we place our focus on learning then according to Gattegno [1971: ii] A radical transformation occurs in the classroom when one knows how to subordinate teaching to learning. It enables us to
expect very unusual results from the students - for example that all students will perform very well, very early and on…
My experiences, initially as a young teacher and later as a HoD in a school, led me to believe in the value of teaching mathematics in mixed-attainment groups. These values were based upon:
• inclusivity;
• equality of opportunity;
• access and entitlement to mathematics;
• not labelling children on some kind of ‘ability’ scale;
• not accepting that any student’s ‘ability’ was fixed;
• being aware of and open to students surprising me.
How anyone might create the conditions for teaching mixed-attainment groups depends upon the range of strategies we use, the questions we ask and the types of resources we have access to and how we
utilise these strategies and resources when planning lessons and designing tasks for use with our students. An earlier quotation from Gattegno (1963, 63) offers some insight into task design: All I
must do is to present them with a situation so elementary that they all master it from the outset, and so fertile that they will all find a great deal to get out of it. Given the increasingly strong
element of problem solving and reasoning emerging, from the 2014 National Curriculum and the subsequent impact of this upon assessment at GCSE and A-level, such strategies and resources need careful
consideration with regard to subordinating our teaching to learning.
With regard to asking questions two seminal publication by the Association of Teachers of Mathematics (ATM) http://atm.org.uk) are: Questions and prompts for mathematical thinking andThinkers. In
terms of resources and task design http://www.inquirymaths.co.uk has the potential to make profound additions to supporting learning and teaching in mixed-attainment classrooms. There are also a vast
range of other resources published by the ATM and some are:
• Big Ideas
• Bigger Ideas
• Rich Task Maths 1
• Rich Task Maths 2
• Variety in mathematics lessons
• Eight days a week
• Everyone is special
• Forty problems for the classroom
• Forty harder problems for the classroom
• Learning and teaching mathematics without a textbook
• 30 years on
• More people more maths (this is an active ‘People Maths’ set of ideas)
• Functioning mathematically
• Points of Departure 1, 2, 3 and 4
• Linking cubes and the learning of mathematics
Gattegno, C. (1963) For the Teaching of Mathematics (Volume One), Great Britain, Lamport, Gilbert.Gattegno, C. (1971) What We Owe Children: The Subordination of Teaching to Learning, London:
Routledge & Kegan Paul.
2 Comments
2 Comments
Thoughts on the Conference - by Steve Mault
'I went to the Mixed Attainment Conference in Sheffield because something was missing. Approaching the end of my eighth year of teaching and in my fourth school, I know I love teaching Mathematics
but my teaching jigsaw felt incomplete. I have always taught pupils in sets but have grown increasingly concerned with the lengthening tail of disaffection and underachievement. Is this due to
societal problems or is something wrong with our practice? When I discovered that my children’s prospective Primary School set pupils in Mathematics from the age of seven, I thought I might have
discovered an answer. At the same time, an old friend, Zebedee Friedman, posted an advert for the Mixed Attainment Conference in Sheffield. I knew that, if it was something that Zeb was involved
with, it was bound to be good. As I read more about Mixed Attainment, I started to question my practice. How can it be right to deny groups of students access to parts of the Mathematics curriculum,
particularly when those parts are often the richer, less procedurally focussed elements that actually give a much better representation of the maths I enjoy so much?
The conference was exactly what I had hoped for. Helen Hindle of mixedattainmentmaths.com had brought together a range of inspirational teachers and educators, each with fascinating insights into
the Mixed Attainment approach. Mixed Attainment seems scary to someone who has only ever taught in sets. How will the top end students be stretched? What about those with specific learning
difficulties? What happens when students reach KS4? How can I persuade my department that Mixed Attainment is right for us? Workshops that introduced Learning Journeys, Low Floor High Ceiling tasks,
Inquiry maths and much more, provided tools and suggestions to help answer these questions. Refreshingly, these were current and former teachers offering guidance and support and not pretending to
have a definitive one-size fits all method. The overriding message I took from each of the four workshops I attended, was the importance of collaboration. Creating properly differentiated, engaging
tasks takes time, but by working together, it is possible to create resources that ensure that all students are both challenged and supported. Spending time with like-minded colleagues, who shared a
belief that something in the system is wrong when we are prepared to prejudge the potential of children as young as 7, was incredibly refreshing. I would say it felt like I had completed the puzzle.'
Thoughts on the conference by @rhib83
This conference has completely opened my eyes to what proper mixed attainment teaching looks like and the endless opportunities it provides for students. In an uneasy position of changing over to
mixed attainment classes from September, I found the sessions to be both inspirational and exciting and it was delivered in a very supportive environment. I'm excited and raring to go for September,
full of ideas thanks to Helen and her team. See you all at the next one!
Thoughts on the conference - by Martin Jones
I knew it would happen – turn up without a pen, just like my students. In the first workshop a kind colleague lends me a spare pen for the day. We get stuck into Mike Ollerton’s activity on using a
rotating arm to plot the coordinates of its endpoint for angles between 10 and 90 degrees. What do we notice, what patterns are emerging, what questions occur? Now I recall what it is like being part
of a group, trying to keep up. I’m relying on my new found colleague to help me recognise where I’ve put the decimal point in the wrong place and to reassure me I’m on the right track.
Surprising us – and himself – with a blank power-point slide, Mike mentions Postman and Weingartner’s book “Teaching as a Subversive Activity”. (The quote that Mike was thinking about which is
followed by a blank page in the book, was...
'Suppose all the syllabi and curricula and textbooks in the schools disappeared. Suppose all of the standardized tests - city wide, state-wide and national were lost. In other words, suppose that the
common material impeding innovation in the schools simply did not exist. Then suppose that you decided to turn this 'catastrophe' into an opportunity to increase the relevance of schools. What would
you do?'
Must go back and read it again. Published 1969.
Next, off to an Inquiry lesson with Andrew Blair. A simple but challenging “prompt” about matchsticks forming rectangles. What do we notice, what questions do we want to ask? Now, the pivotal moment
for the teacher: which way do we take the lesson? Let the participants nominate their preferred route from a given structure of enquiring mathematically; how to take the whole class forward is the
question. We are guided to look for more examples, my colleague from Bratislava notices triangle numbers are popping up, we conjecture, we play with some algebra, someone graphs a couple of equations
on her phone using Desmos (“this app is changing my life”, she lets us know in an aside), we start to see what’s happening, we run out of time. There must be a nice proof lurking behind this
surprising result...
Lunch in the Peace Gardens in Sheffield city centre. A cosmopolitan community enjoying the sunshine and splashing around in the fountains. Conversation about the Mastery programme and its
relationship with mixed attainment teaching.
Post-lunch, grid algebra with Tom Francome. Trepidation at having to take my turn and do some simple arithmetic on the screen; so this is how my students feel when I invite them to the front of the
class! Not as easy as I thought… Memorably, Tom says, the biggest benefit from mixed attainment teaching is teachers working together. And he recommends reading Dave Hewitt’s paper “Arbitrary and
Necessary” – it will change the way you view what you do.
Final session: Helen Hindle runs a lesson her way, us evaluating the resources. Learning journeys, starter tasks, self-assessment. When do you do your teaching, how do students respond to this style
of learning, how do you change the culture of the classroom? Lots of interesting discussion with those doing mixed attainment teaching, those implementing, and those like me wanting to believe and
trying to find ways to make it happen.
An uplifting day, a reassuring day, lots of good people doing lots of good things. As Hilary Povey said in the opening session (I paraphrase): Hope, in an era of change; imagine… a world in which
mathematicians are made not born.
Thoughts on the conference by Laura Brown
I started the day thinking 'but we're expecting weaker students to be part of a conversation where they won't even understand half the words', and left thinking 'but if we don't expose them to those
conversations how do we expect them to develop that vocabulary and understanding?'.
I came looking for ideas for one class and left to do more research into mixed ability teaching with an idea to introduce it in Year 7 in 2018. I've got a lot of work to do...
We've tried some mixed(ish) groups before - it didn't go well. This weekend I learned why:
- staff need to be on board
- there needs to be a high level of collaboration
- we didn't have a bank of rich learning / low floor, high ceiling tasks which are obviously key to making this style of teaching a success.
Mark Horley's session in particular gave me such an insight into what mixed ability teaching can and should be like.
I'm even going to join twitter so I can follow the updates on the #mixedattainmentmaths feed!
Many thanks to all the team for such an inspiring, thought provoking day!
Now to convince my team.... | {"url":"https://www.mixedattainmentmaths.com/blog/archives/06-2017","timestamp":"2024-11-06T23:57:23Z","content_type":"text/html","content_length":"45285","record_id":"<urn:uuid:c9b6c4a2-94da-41f3-b9cd-999fc2ae1188>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00166.warc.gz"} |
The jackknife coefficient and replicate weights are described as follows.
If there is no stratification in the sample design (no STRATA statement), the jackknife coefficients are the same for all replicates:
Denote the original weight in the full sample for the jth member of the ith PSU as . If the ith PSU is included in the rth replicate (), then the corresponding replicate weight for the jth member of
the ith PSU is defined as
If the sample design involves stratification, each stratum must have at least two PSUs to use the jackknife method.
Let stratum be the stratum from which a PSU is deleted for the rth replicate. Stratum is called the donor stratum. Let be the total number of PSUs in the donor stratum . The jackknife coefficients
are defined as
Denote the original weight in the full sample for the jth member of the ith PSU as . If the ith PSU is included in the rth replicate (), then the corresponding replicate weight for the jth member of
the ith PSU is defined as
You can use the VARMETHOD=JACKKNIFE(OUTJKCOEFS=) method-option to save the jackknife coefficients into a SAS data set and use the VARMETHOD=JACKKNIFE(OUTWEIGHTS=) method-option to save the replicate
weights into a SAS data set.
If you provide your own replicate weights with a REPWEIGHTS statement, then you can also provide corresponding jackknife coefficients with the JKCOEFS= option.
Suppose that is a population parameter of interest. Let be the estimate from the full sample for . Let be the estimate from the rth replicate subsample by using replicate weights. PROC SURVEYMEANS
estimates the variance of by
with degrees of freedom, where R is the number of replicates and H is the number of strata, or R – 1 when there is no stratification. | {"url":"http://support.sas.com/documentation/cdl/en/statug/65328/HTML/default/statug_surveymeans_details44.htm","timestamp":"2024-11-11T11:02:18Z","content_type":"application/xhtml+xml","content_length":"22146","record_id":"<urn:uuid:395dfe7f-f955-4df4-b0fc-407ddc0267e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00407.warc.gz"} |
Questions and Answers
Check out our online quiz to prepare for the CEA GATE exam. The questions are well-researched, followed with instant answers. Choose your option and find out in a click if you have opted for the
right one.
• 1.
Statement: All bags are cakes. All lamps are cakes. Conclusions: I. Some lamps are bags. II. No lamp is bag. Deduce which of the above conclusion logically follows statements
□ A.
Only conclusion I follows
□ B.
Only conclusion II follows
□ C.
□ D.
Correct Answer
C. Either I or II follows
Both conclusions I and II can logically follow from the given statements. Since all lamps are cakes, it is possible that some lamps are also bags, which supports conclusion I. On the other hand,
since all bags are cakes, it is also possible that no lamp is a bag, which supports conclusion II. Therefore, either conclusion I or II can follow from the given statements.
• 2.
Man does not live by __________ alone.
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Bread
This phrase is a common saying that emphasizes the importance of not only physical sustenance but also other aspects of life. Bread is used metaphorically here to represent food in general. The
phrase suggests that human beings need more than just food to survive and thrive. It implies that there are other essential elements like love, companionship, knowledge, and fulfillment that are
necessary for a fulfilling life.
• 3.
Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything ___________ to the requirements of the exam
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Extraneous
Indian students are so focused on studying for tests and following the syllabus that they ignore anything that is not directly related to the exam requirements. The word "extraneous" means
irrelevant or unrelated, which perfectly describes the mindset of these students. They only pay attention to what is necessary for the exam and disregard anything that is not directly related to
• 4.
Select the pair that best expresses a relationship similar to that expressed in the pair: LIGHT : BLIND
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Speech : Dumb
The relationship between "LIGHT : BLIND" is that light causes blindness or the absence of light leads to blindness. Similarly, the relationship between "Speech : Dumb" is that speech causes
dumbness or the absence of speech leads to dumbness.
• 5.
If log a/b + log b/a = log(a+b), then:
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. A+ b =1
The given equation is log a/b + log b/a = log(a+b). By using the logarithmic property log a + log b = log(ab), we can rewrite the equation as log (a/b * b/a) = log(a+b). Simplifying further, we
get log(1) = log(a+b). Since the logarithm of 1 is always 0, we have 0 = log(a+b). This implies that a+b = 1, which is the answer.
• 6.
In an exam, the average was found to be 50 marks. After deducting computational errors the marks of the 100 candidates had to be changed from 90 to 60 each and average came down to 45 marks.
Total number of candidates who took the exam
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 600
In the original scenario, the average mark was 50 and after deducting computational errors, the marks of each candidate were changed from 90 to 60. This means that each candidate had 30 marks
deducted. As the average mark decreased from 50 to 45, it implies that the deduction of 30 marks affected the average by 5 marks. Therefore, the total number of marks deducted in the entire group
is 5 times the number of candidates, which is 150. Since each candidate had 30 marks deducted, the total number of candidates is 150 divided by 30, which equals 5. Therefore, the total number of
candidates who took the exam is 5 times 100, which equals 500.
• 7.
A solid 4cm cube of wood is coated with red paint on all the six sides. Then the cube is cut into smaller 1cm cubes. How many of these 1cm cubes have no colour on any side?
Correct Answer
A. 8
When the 4cm cube is cut into smaller 1cm cubes, each face of the larger cube is divided into 4 smaller squares. Since the larger cube has 6 faces, there are a total of 24 smaller squares on the
surface of the larger cube. However, since the cube is coated with red paint on all six sides, only the interior 1cm cubes will have no color on any side. Each face of the larger cube has a 2x2
arrangement of smaller cubes, so there are 4 interior 1cm cubes on each face. Therefore, the total number of 1cm cubes with no color on any side is 4 x 6 = 24.
• 8.
In deriving the equation for the hydraulic jump in a rectangular channel in terms of the conjugate depths and the initial Froude number
□ A.
Continuity equation and energy equation are used
□ B.
Continuity equation and momentum equation are used
□ C.
Equations of continuity, mementum and energy are used
□ D.
Gradually varied flow equation is used
Correct Answer
C. Equations of continuity, mementum and energy are used
The correct answer is "equations of continuity, momentum, and energy are used". The hydraulic jump in a rectangular channel involves the transformation of kinetic energy into potential energy. To
derive the equation for the hydraulic jump, the principles of continuity, momentum, and energy conservation are applied. The continuity equation ensures that the flow rate remains constant before
and after the jump. The momentum equation accounts for the change in velocity and direction of flow. The energy equation considers the change in energy due to the jump. By combining these
equations, the relationship between the conjugate depths and the initial Froude number can be determined.
• 9.
This paragraph states that warm weather affects the consumer’s inclination to spend the age of father is 4 times more than the age of his son Amit. After 8 years, he would be 3 times older than
Amit. After further 8 years, how many times will he be older than Amit?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 2.33
• 10.
Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes after pipe B is
closed, pipe C is opened and is kept open till the tank is full. Each pipe fills an equal share of the tank. Furthermore, it is known that if pipe A and B are kept open continuously, the tank
would be filled completely in t minutes. Find t?
Correct Answer
D. 24
Let's assume that pipe A fills 1 unit of the tank per minute. Since pipe A is kept open throughout, it will fill the tank in t minutes.
Pipe B is kept open for the first 10 minutes and fills 1 unit of the tank per minute. So, in the first 10 minutes, pipe B will fill 10 units of the tank.
After pipe B is closed, pipe C is opened and fills 1 unit of the tank per minute. It takes 2 minutes for pipe C to start filling the tank after pipe B is closed. So, in the remaining t - 12
minutes, pipe C will fill t - 12 units of the tank.
Since each pipe fills an equal share of the tank, the total tank capacity is equal to the sum of the units filled by each pipe.
Therefore, we can write the equation: t = 10 + (t - 12)
Simplifying the equation, we get: 2t = 22
Dividing by 2 on both sides, we get: t = 11
However, this contradicts the given information that if pipe A and B are kept open continuously, the tank would be filled completely in t minutes.
Therefore, the correct answer is 24.
• 11.
The plastic modulus of a section is 4.8 x 10 m . The shape factor is 1.2. The plastic moment capacity of the section is 120 kN.m. The yield stress of the material is
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 250 MPa
The plastic modulus of a section is calculated by dividing the plastic moment capacity of the section by the shape factor. In this case, the plastic moment capacity is given as 120 kN.m and the
shape factor is given as 1.2. Therefore, the plastic modulus can be calculated as 120 kN.m / 1.2 = 100 kN.m. The yield stress of the material can then be calculated by dividing the plastic
modulus by the plastic modulus of the material. In this case, the plastic modulus is given as 4.8 x 10 m, so the yield stress can be calculated as 100 kN.m / (4.8 x 10 m) = 250 MPa.
• 12.
MPN index is a measure of one of the following
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. Coliform bacteria
MPN index is a measure of coliform bacteria. Coliform bacteria are a group of microorganisms that are commonly found in the intestines of warm-blooded animals. They are used as an indicator of
water quality and the presence of fecal contamination. The MPN index is a statistical estimation of the number of coliform bacteria present in a water sample based on the most probable number of
positive test results. This index is widely used in water quality testing to assess the safety and suitability of water for various purposes, such as drinking, swimming, and irrigation.
• 13.
Consider the following statements 1) Strength of concrete cube is inversely proportional to water-cement ratio. 2) A rich concrete mix gives higher strength than a lean concrete mix since it has
more cement content. 3) Shrinkage cracks on concrete surface are due to excess water in mix. Which of the following statements is correct?
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 1,2 and 3
The correct answer is 1,2 and 3. Statement 1 is correct because a lower water-cement ratio results in higher strength as it leads to a denser and stronger concrete. Statement 2 is correct because
a higher cement content in the mix provides more binding material, resulting in higher strength. Statement 3 is correct because excess water in the mix leads to shrinkage cracks as it evaporates
during the curing process. Therefore, all three statements are accurate.
• 14.
Which one do you like?Standard 5-day BOD of a waste water sample is nearly x% of the ultimate BOD, where x is
Correct Answer
C. 68
The correct answer is 68 because the standard 5-day BOD (Biochemical Oxygen Demand) of a waste water sample is typically around 68% of the ultimate BOD. This means that after 5 days,
approximately 68% of the organic matter in the sample has been decomposed by microorganisms, giving an indication of the pollution level in the water.
• 15.
The dimensions for the flexural rigidity of a beam element in mass (M), length (L) and time (T) is given by
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. M^-1T^2
The correct answer, M-1T2, represents the dimensions for the flexural rigidity of a beam element. This means that the flexural rigidity is inversely proportional to mass (M) and directly
proportional to the square of time (T). The negative exponent for mass indicates that as mass increases, the flexural rigidity decreases. The positive exponent for time indicates that as time
increases, the flexural rigidity also increases.
• 16.
The superelevation needed for a vehicle travelling at a speed of 60 kmph on a curve of radius 128 m on a surface with a coefficient of friction of 0.15 is :
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 0.71
The superelevation needed for a vehicle to travel safely on a curve is determined by the speed of the vehicle, the radius of the curve, and the coefficient of friction of the surface. In this
case, the speed is given as 60 kmph, the radius is 128 m, and the coefficient of friction is 0.15. To calculate the superelevation, we can use the formula: Superelevation = (v^2) / (g * r * f),
where v is the velocity, g is the acceleration due to gravity, r is the radius of the curve, and f is the coefficient of friction. Plugging in the given values, we get: Superelevation = (60^2) /
(9.8 * 128 * 0.15) = 0.71. Therefore, the correct answer is 0.71.
• 17.
Which of the following raingauge gives a plot of the accumulated rainfall against the elapsed time?
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. Weighing bucket type
The correct answer is Weighing bucket type. Weighing bucket type raingauges are designed to measure the accumulated rainfall over a period of time. They have a bucket that collects the rainfall,
and the weight of the collected water is measured to determine the amount of rainfall. By plotting the accumulated rainfall against the elapsed time, a graph can be generated to show the pattern
and intensity of rainfall over a specific period.
• 18.
If the time period between centroid of the rainfall diagram and peak of the hydrograph is 5 hour, using Snyder’s equation the value of the base width of unit hydrograph in hours is?
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 87 hour
• 19.
A river 5 m deep consists of a sand bed with saturated unit weight of 20 kN/m . v = 9.81 kN/m . The effective vertical stress at 5 m from the top of sand bed is
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 51 kN/m^2
The effective vertical stress at a certain depth in a saturated sand bed can be calculated using the formula: σ'v = γ'z, where σ'v is the effective vertical stress, γ' is the saturated unit
weight of the sand bed, and z is the depth. In this case, the depth is 5m and the saturated unit weight is 20 kN/m^3. Therefore, the effective vertical stress at 5m depth would be 20 kN/m^3 * 5m
= 100 kN/m^2. However, since the question asks for the effective vertical stress "at 5m from the top of the sand bed", we need to subtract the water pressure from the total effective stress. The
water pressure at 5m depth can be calculated as v * z = 9.81 kN/m^3 * 5m = 49.05 kN/m^2. Subtracting this from the total effective stress gives us 100 kN/m^2 - 49.05 kN/m^2 = 50.95 kN/m^2, which
is approximately equal to 51 kN/m^2. Therefore, the correct answer is 51 kN/m^2.
• 20.
A hydraulic model of a spillway is constructed with a scale 1:16. If the prototype discharge is 2048 cumecs, then the corresponding discharge for which the model should be tested is:
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 2 cumecs
The hydraulic model is constructed with a scale of 1:16. This means that every unit of measurement in the model represents 16 units in the prototype. Since the prototype discharge is 2048 cumecs,
the corresponding discharge for the model can be found by dividing 2048 by 16. This gives us a discharge of 128 cumecs. Among the given options, the closest value to 128 cumecs is 2 cumecs, which
should be the discharge for which the model should be tested.
• 21.
A droplet of water at 20^0c ( p (sigma) = 0.0728 N/m) has internal pressure 1 kPa greater than that outside it, its diameter is nearly:
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 0.3 mm
A droplet of water at 20°C with an internal pressure 1 kPa greater than the outside pressure will have a diameter of approximately 0.3 mm. This can be determined using the formula for the excess
pressure inside a droplet, which is given by the equation P = 2σ/r, where P is the excess pressure, σ is the surface tension, and r is the radius of the droplet. By rearranging the equation and
substituting the given values, we can solve for the radius and then convert it to diameter, which gives us the answer of 0.3 mm.
• 22.
A reinforced concrete structure has to be constructed along a sea coast. The minimum grade of concrete to be used as per IS : 456-2000
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. M 30
The correct answer is M 30. When constructing a reinforced concrete structure along a sea coast, a higher grade of concrete is required due to the harsh and corrosive environment. M 30 grade
concrete is recommended as it has a higher compressive strength and durability compared to lower grades such as M 15 or M 20. This higher grade of concrete will ensure that the structure can
withstand the corrosive effects of saltwater and the strong winds typically found along the coast.
• 23.
The order and degree of the differential equation given, is
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 2, 3
The order of a differential equation is the highest derivative present in the equation. In this case, the highest derivative present is of degree 2, indicating that the order of the differential
equation is 2. The degree of a differential equation is the power to which the highest derivative is raised. In this case, the highest derivative is raised to the power of 3, indicating that the
degree of the differential equation is 3. Therefore, the correct answer is 2, 3.
• 24.
The probability distribution taken to represent the completion time in PERT analysis is
□ A.
□ B.
□ C.
□ D.
Log – normal distribution
Correct Answer
B. Normal distribution
The correct answer is the Normal distribution. In PERT (Program Evaluation and Review Technique) analysis, the completion time is assumed to follow a Normal distribution. This is because the
Normal distribution is symmetric and bell-shaped, making it suitable for representing random variables with a wide range of values. The Normal distribution is commonly used in statistical
analysis and modeling as it allows for easy calculation of probabilities and provides a good approximation for many real-world phenomena.
• 25.
Consider the following statements 1) Cambium layer is between sapwood and heartwood 2) Heartwood is otherwise termed as deadwood 3) Timber used for construction is obtained from heartwood Which
of the following is/are correct statements?
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 2 and 3 only
The statement "2 and 3 only" is the correct answer. Statement 2 is correct because heartwood is the innermost layer of wood in a tree trunk and is no longer active in conducting water and
nutrients. Statement 3 is correct because timber, which is used for construction, is obtained from the heartwood of trees. Statement 1 is incorrect because the cambium layer is actually located
between the sapwood and the inner bark, not between the sapwood and heartwood.
• 26.
The true bearing of a tower T as observed from station A was 356^o and the magnetic bearing of the same was 4^0 .The back bearing of the line AB when measured with prismatic compass was found to
be 296^0 .Then the true fore bearing of line AB will be _____ degrees.
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 108
The true fore bearing of line AB will be 108 degrees. This can be determined by subtracting the back bearing (2960) from the true bearing of the tower (3560) and adding 180 degrees. Therefore,
3560 - 2960 + 180 = 108 degrees.
• 27.
The degree of static indeterminacy of the rigid frame having 2 internal hinges is as shown in the figure below
Correct Answer
D. 3
The degree of static indeterminacy of a rigid frame with 2 internal hinges is 3. This means that there are 3 unknown reactions or forces that need to be determined in order to fully analyze the
frame's equilibrium.
• 28.
If the deformations of the truss-members are as shown in parentheses, the rotation of the member bd is :
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 2.0 x 10^-2 radian
• 29.
The problem of lateral buckling can arise only in those steel beams which have
□ A.
Moment of inertia about the bending axis larger than the other
□ B.
Moment of inertia about the bending axis smaller than the other
□ C.
Fully supported compression flange
□ D.
Correct Answer
B. Moment of inertia about the bending axis smaller than the other
The problem of lateral buckling can arise only in those steel beams which have a moment of inertia about the bending axis smaller than the other. This is because lateral buckling occurs when a
beam is subjected to compressive forces and is unable to resist the bending moment. A smaller moment of inertia means that the beam is less resistant to bending and therefore more susceptible to
lateral buckling.
• 30.
The available moisture holding capacity of soil is 15cm per meter depth of soil. If a crop with a root zone of 0.8m and consumptive use of 6 mm/day is to be grown, the frequency of irrigation for
restricting the moisture depletion to 60% of available moisture is?
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 8 days
The available moisture holding capacity of the soil is 15cm per meter depth. The root zone of the crop is 0.8m, so the available moisture in the root zone would be 0.8m * 15cm/m = 12cm. The crop
has a consumptive use of 6mm/day. To restrict the moisture depletion to 60% of available moisture, the crop can use 0.6 * 12cm = 7.2cm of moisture. Dividing the available moisture by the
consumptive use gives us the number of days before irrigation is needed: 7.2cm / 6mm/day = 12 days. Therefore, the frequency of irrigation should be 8 days.
• 31.
6 boys and 6 girls sit in a row at random. Find the probability that 1) The six girls sit together 2) The boys and girls sit alternately
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 1/462
The probability that the six girls sit together can be calculated by treating the group of six girls as a single entity. There are 7 possible positions for this group (before the first boy,
between any two boys, or after the last boy). Within this group, the six girls can arrange themselves in 6! = 720 ways. The remaining 6 boys can arrange themselves in 6! = 720 ways. Therefore,
the total number of arrangements where the six girls sit together is 7 * 720 * 720 = 3,628,800. The total number of possible arrangements of all 12 people is 12! = 479,001,600. Therefore, the
probability is 3,628,800 / 479,001,600 = 1/132.
• 32.
Match List-I (contract types) with List-II (Characteristics): List-I a. Labour contract b.Item rate contract c. Piecework contract d.Lump sum contract List-II 1.Petty works and regular
maintenance work 2.Adopted for buildings, roads, bridges and electrical works 3.Payment made by detailed measurement of different time 4.Not practiced in government
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. A-4,b-3,c-1,d-2
Labour contract involves payment made by detailed measurement of different time (List-II: 3). Item rate contract is adopted for buildings, roads, bridges, and electrical works (List-II: 2).
Piecework contract is not practiced in government (List-II: 4). Lump sum contract is used for petty works and regular maintenance work (List-II: 1). Therefore, the correct match is a-4, b-3, c-1,
• 33.
Plate bearing test with 20 cm diameter plate on soil subgrade yielded a pressure of 1.25 x 10 N/m^2 at 0.5 cm deflection. What is the elastic modulus of subgrade ?
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. None.
• 34.
A ductile metal bar is subjected to pure shear test and its yield strength was found to be 230 N/mm^2 the maximum permissible shear stress on that metal will be: (N/mm^2)
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 115
• 35.
Which of the following is characteristic of colliform organism?
1. Spore formation
2. Gram negative
3. Gram positive
4. Lactose fermenter
5. Bacillus
□ A.
□ B.
□ C.
□ D.
Correct Answer
D. 2, 4, 5
Colliform organisms are characterized by being Gram-negative, lactose fermenters, and bacilli. Gram-negative refers to the ability of the organism to retain the crystal violet stain during the
Gram staining process, indicating the presence of a thin peptidoglycan layer in the cell wall. Lactose fermenters are organisms that can metabolize lactose as a source of energy, producing acid
and gas as byproducts. Bacilli are rod-shaped bacteria. Therefore, the correct answer is 2, 4, 5.
• 36.
A light rope fixed at one end of a wooden clamp on the ground passes over a tree branch and hangs on the other side. It makes an angle 30^0 with the ground. A man weighing (60 kg) wants to climb
up the rope. The wooden clamp can come out of the ground if an upward force greater than 360N is applied to it. Find the maximum acceleration in the upward direction with which the man can climb
safely. Neglect friction at the tree branch. Take g = 10 m/sec^2 (in m/sec^2)
Correct Answer
C. 2
The maximum acceleration in the upward direction with which the man can climb safely is 2 m/sec^2. This can be determined by analyzing the forces acting on the wooden clamp. The weight of the man
is acting downwards with a force of 60 kg * 10 m/sec^2 = 600 N. The tension in the rope is acting upwards and can be calculated using the equation T = mg / sin(theta), where theta is the angle
made by the rope with the ground. In this case, T = 600 N / sin(30 degrees) = 1200 N. Since the wooden clamp can come out of the ground if an upward force greater than 360 N is applied to it, the
maximum acceleration can be calculated using the equation F = ma, where F is the net force acting on the wooden clamp. Therefore, 1200 N - 360 N = (60 kg) * a, which gives a = 2 m/sec^2.
• 37.
The pavement designer has obtained the value of design traffic as 200 million standard Axles for a newly developing highway. The design life adopted is 15 years. Annual traffic growth rate of 8%
is taken into account. Commercial vehicles count before pavement construction was 5000 vehicles/day. The vehicle damage factor used in the calculation was:
Correct Answer
C. 4
The vehicle damage factor used in the calculation was 4. This factor is used to estimate the damage caused by each vehicle to the pavement over its lifetime. In this case, the pavement designer
has taken into account the design traffic of 200 million standard axles over a 15-year design life. Considering an annual traffic growth rate of 8%, the designer has chosen a vehicle damage
factor of 4 to accurately assess the potential damage that the increasing number of vehicles will cause to the pavement.
• 38.
The present population of a community is 28000 with an average water demand of 150 Lpcd. The existing water treatment plant has a design capacity of 6000 m^3/d. It is expected that the next 20
years. What is the number of years from now when the plant will reach its design capacity assuming an arithmetic rate of population growth??
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 12 years
Assuming an arithmetic rate of population growth, we can calculate the future population by multiplying the current population by the number of years. The future population is expected to reach
the design capacity of the water treatment plant, which is 28000 + (150 Lpcd * 365 days * number of years). Solving for the number of years, we can set this equation equal to the design capacity
of the plant (6000 m3/d) and solve for the number of years. After solving the equation, we find that the number of years is 12. Therefore, the correct answer is 12 years.
• 39.
A square plate 5m × 5m hangs in water from one of its corners and its centroid lies at 8m from the free water surface. Find the total pressure force on the top half of the plate.
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 850 kN
When a plate is submerged in water, the pressure increases with depth. The pressure at any point in a fluid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the
fluid, g is the acceleration due to gravity, and h is the depth of the point below the free surface of the fluid.
In this case, the centroid of the plate is 8m below the free water surface. Since the plate is hanging from one of its corners, the centroid is also the center of mass of the plate. Therefore,
the top half of the plate is at a depth of 8m/2 = 4m below the free water surface.
Using the formula for pressure, the pressure at the centroid of the plate is P = ρgh = 1000 kg/m^3 * 9.8 m/s^2 * 4m = 39200 N/m^2.
The total pressure force on the top half of the plate can be calculated by multiplying the pressure at the centroid by the area of the top half of the plate. The area of the plate is 5m * 5m =
25m^2, so the area of the top half is 25m^2 / 2 = 12.5m^2.
Therefore, the total pressure force on the top half of the plate is 39200 N/m^2 * 12.5m^2 = 490000 N = 490 kN.
Since the question asks for the total pressure force on the top half of the plate, the correct answer is 850 kN.
• 40.
A sample of clay was coated with paraffin wax and its mass, including the mass of wax was found to be 650 gm, the sample was immersed in water and the volume of the water displaced was found to
be 325 Ml. The mass of the sample without wax was 640 gm and water content of the specimen was 20%. Specific gravity of wax is 0.9 and specific gravity of soil solids is 2.7. The void ratio and
density of the sample is:
Correct Answer
C. = 1.70 gm/cc, e = 0.588
The specific gravity of a substance is the ratio of its density to the density of water. In this question, the specific gravity of wax is given as 0.9, which means that the density of wax is 0.9
times the density of water. The specific gravity of soil solids is given as 2.7, which means that the density of soil solids is 2.7 times the density of water.
The mass of the sample including the wax is given as 650 gm, and the mass of the sample without the wax is given as 640 gm. From this information, we can deduce that the mass of the wax is 10 gm.
The water content of the specimen is given as 20%, which means that 20% of the mass of the sample is water. Therefore, the mass of water in the sample is (20/100) * 640 gm = 128 gm.
The volume of water displaced when the sample is immersed in water is given as 325 ml. Since 1 ml of water has a mass of 1 gm, the mass of water displaced is 325 gm.
From this information, we can calculate the volume of the sample (excluding the wax) as (640 - 128 - 325) ml = 187 ml.
The density of the sample (excluding the wax) is therefore 640 gm / 187 ml = 3.42 gm/cc.
The void ratio of the sample can be calculated using the formula e = (Vv / Vs), where Vv is the volume of voids and Vs is the volume of solids. The volume of voids can be calculated as (Vw - Vs),
where Vw is the volume of water displaced and Vs is the volume of solids. From the given information, Vw = 325 ml and Vs = 187 ml. Therefore, Vv = 325 - 187 = 138 ml.
The void ratio is therefore 138 ml / 187 ml = 0.738.
Comparing these values to the answer options, the correct answer is = 1.70 gm/cc, e = 0.588.
• 41.
For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume accelerations as 0.53 m/sec^2)
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 750 m
The safe overtaking sight distance is determined by the design speed of the highway and the acceleration. In this case, with a design speed of 100 kmph and assuming an acceleration of 0.53 m/
sec2, the safe overtaking sight distance is 750 m. This distance allows enough time for a vehicle to accelerate, overtake another vehicle, and then decelerate back to the original speed without
compromising safety.
• 42.
The insitu moisture content of a soil is 18% and its bulk unit weight is 19.65 .The specific gravity of the soil solids is 2.7. The soil is to be excavated and transported to a construction site
and then compacted to a minimum dry unit weight of 16.65 at a moisture content of 20%. How many truck loads are needed to transport the excavated soil if each truck can carry 25 tons? Total
volume of the compacted fill is 10,000?
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 786
The question provides information about the insitu moisture content, bulk unit weight, specific gravity of the soil solids, and the desired dry unit weight and moisture content after compaction.
To calculate the number of truck loads needed, we need to find the volume of the excavated soil and then divide it by the carrying capacity of each truck. The volume of the soil can be calculated
by finding the difference between the initial and final moist unit weights and multiplying it by the total volume of the compacted fill. After performing the calculations, the answer is 786 truck
• 43.
Horizontal stiffness coefficient, K[11] of bar ab is given by :
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. A E/l√2
The correct answer is A E/l√2. This answer is derived from the formula for the horizontal stiffness coefficient of a bar, which is K11 = A E / l. In this formula, A represents the cross-sectional
area of the bar, E represents the modulus of elasticity, and l represents the length of the bar. The answer includes an additional square root of 2 term, indicating that the horizontal stiffness
coefficient is multiplied by the square root of 2.
• 44.
In the given figure below, the slope at point P will be:
Correct Answer
• 45.
A father notes that when his teenage daughter uses the phone she takes no less than 6 minutes for a call but sometimes as much as an hour. 20 minutes calls are more frequent than calls of any
other duration.If the daughter’s calls were to be represented as an activity in PERT project, the expected duration of each phone call is:
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 24.33 minutes
The correct answer is 24.33 minutes. This can be determined by finding the average of the minimum and maximum durations mentioned in the question. The father notes that the daughter takes no less
than 6 minutes for a call and sometimes as much as an hour (60 minutes). Therefore, the average of 6 minutes and 60 minutes is (6 + 60) / 2 = 66 / 2 = 33 minutes. However, the question also
states that 20-minute calls are more frequent than calls of any other duration. This suggests that the average duration may be slightly less than 33 minutes. Therefore, the closest option is
24.33 minutes.
• 46.
Calculate the quantity of cement (in kg) required to produce 1 of concrete for mix proportion of 1:2:4 (by volume). Water cement ratio = 0.48 Percentage of entrapped air is 2%
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 300 kg
To calculate the quantity of cement required, we need to use the given mix proportion of 1:2:4. This means that for every part of cement, we need 2 parts of sand and 4 parts of aggregate.
Since the total volume of the mix is 1, we can assume that the volume of cement is 1/7 (1 part cement out of 7 parts total).
Now, we can calculate the weight of cement by multiplying the volume of cement by its density.
Given that the water-cement ratio is 0.48, we can assume that the density of cement is 1440 kg/m^3.
Therefore, the weight of cement required is (1/7) * 1440 kg/m^3 = 205.71 kg.
However, we also need to consider the percentage of entrapped air, which is 2%. This means that the actual weight of cement required will be slightly higher.
Hence, the correct answer is 300 kg.
• 47.
The beams in the two storey frame shown in the figure below have a cross – section such that the flexural rigidity may be considered infinite. Which among the following is the stiffness matrix
for the structure in respect of the global coordinates 1 and 2?
Correct Answer
The stiffness matrix for the structure in respect of the global coordinates 1 and 2 is given by:
[ 2 -1 ]
[ -1 1 ]
• 48.
A uniformly distributed load of 6 kN/m, 6m long crosses a girder 24m long. What is the maximum bending moment at a section 8m from the right end in a simply supported girders?
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 168 kNm
The maximum bending moment in a simply supported girder occurs at the center of the span. In this case, the girder is 24m long, so the center is at 12m. Since the load is uniformly distributed,
the bending moment at the center is equal to (load per unit length) multiplied by (length of the span) squared divided by 8. Given that the load is 6 kN/m and the span is 24m, the bending moment
at the center is (6 kN/m) * (24m)^2 / 8 = 216 kNm. The bending moment at a section 8m from the right end is half of the bending moment at the center, so it is 216 kNm / 2 = 108 kNm. Therefore,
the correct answer is 168 kNm.
• 49.
The staff reading taken on a staff held at 200m from the instrument. The division of bubble tube is 2.5mm. Following observations were made. The radius of curvature of bubble will be ________m
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 4m
The radius of curvature of the bubble is 4m. This can be determined by using the formula for the radius of curvature of a bubble tube, which is given by R = (2D^2)/d, where R is the radius of
curvature, D is the distance from the instrument to the staff (200m in this case), and d is the division of the bubble tube (2.5mm in this case). Plugging in the values, we get R = (2 * 200^2) /
2.5 = 4m.
• 50.
Vertical reaction developed at B in frame below due to applied load of 100 kN (with is
□ A.
□ B.
□ C.
□ D.
Correct Answer
B. 7.4 kN
The vertical reaction developed at B in the frame is 7.4 kN. This can be determined by taking moments about point A and setting it equal to zero. Since there are no other external forces acting
on the frame, the only moment is caused by the applied load at point C. By solving the equation, we can find that the vertical reaction at B is 7.4 kN. | {"url":"https://www.proprofs.com/quiz-school/story.php?title=pp-civil-engineering-association-mock-gate-1","timestamp":"2024-11-05T00:22:03Z","content_type":"text/html","content_length":"650580","record_id":"<urn:uuid:681c6bc4-1628-4fad-9cae-75c3a346040a>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00033.warc.gz"} |
19. Weierstrass’s Theorem - Avidemia
The general theory of sets of points is of the utmost interest and importance in the higher branches of analysis; but it is for the most part too difficult to be included in a book such as this.
There is however one fundamental theorem which is easily deduced from Dedekind’s Theorem and which we shall require later.
Theorem. If a set \(S\) contains infinitely many points, and is entirely situated in an interval \({[\alpha, \beta]}\), then at least one point of the interval is a point of accumulation of \(S\).
We divide the points of the line \(\Lambda\) into two classes in the following manner. The point \(P\) belongs to \(L\) if there are an infinity of points of \(S\) to the right of \(P\), and to \(R\)
in the contrary case. Then it is evident that conditions (i) and (iii) of Dedekind’s Theorem are satisfied; and since \(\alpha\) belongs to \(L\) and \(\beta\) to \(R\), condition (ii) is satisfied
Hence there is a point \(\xi\) such that, however small be \(\delta\), \(\xi – \delta\) belongs to \(L\) and \(\xi + \delta\) to \(R\), so that the interval \({[\xi – \delta, \xi + \delta]}\)
contains an infinity of points of \(S\). Hence \(\xi\) is a point of accumulation of \(S\).
This point may of course coincide with \(\alpha\) or \(\beta\), as for instance when \(\alpha = 0\), \(\beta = 1\), and \(S\) consists of the points \(1\), \(\frac{1}{2}\), \(\frac{1}{3}, \dots\). In
this case \(0\) is the sole point of accumulation.
$\leftarrow$ 18. Points of accumulation Main Page MISCELLANEOUS EXAMPLES ON CHAPTER I $\rightarrow$ | {"url":"https://avidemia.com/pure-mathematics/weierstrass-theorem/","timestamp":"2024-11-10T06:11:42Z","content_type":"text/html","content_length":"77385","record_id":"<urn:uuid:5acb36f7-4abd-43ac-80fd-b7087a98c204>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00674.warc.gz"} |
An efficient implementation of a quasi-polynomial algorithm for generating hypergraph transversals
Given a finite set V, and a hypergraph H ⊆ 2^V, the hypergraph transversal problem calls for enumerating all minimal hitting sets (transversals) for H. This problem plays an important role in
practical applications as many other problems were shown to be polynomially equivalent to it. Fredman and Khachiyan (1996) gave an incremental quasi-polynomial time algorithm for solving the
hypergraph transversal problem [9]. In this paper, we present an efficient implementation of this algorithm. While we show that our implementation achieves the same bound on the running time as in
[9], practical experience with this implementation shows that it can be substantially faster. We also show that a slight modification of the algorithm in [9] can be used to give a stronger bound on
the running time.
Original language British English
Title of host publication Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics)
Editors Giuseppe di Battista, Uri Zwick
Publisher Springer Verlag
Pages 556-567
Number of pages 12
ISBN (Print) 3540200649, 9783540200642
State Published - 2003
Publication series
Name Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics)
Volume 2832
ISSN (Print) 0302-9743
ISSN (Electronic) 1611-3349
Dive into the research topics of 'An efficient implementation of a quasi-polynomial algorithm for generating hypergraph transversals'. Together they form a unique fingerprint. | {"url":"https://khazna.ku.ac.ae/en/publications/an-efficient-implementation-of-a-quasi-polynomial-algorithm-for-g","timestamp":"2024-11-04T04:26:30Z","content_type":"text/html","content_length":"55750","record_id":"<urn:uuid:f69df2a6-a00f-4305-8775-1723ac3ff101>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00229.warc.gz"} |
HYDROGEOLOGY AND GEOLOGY WEBSITE - Leaky Aquifer Testing (T)
Application of DP_LAQ to Leaky Aquifer Pumping Test Data
By Darrel Dunn, Ph.D., PG, Hydrogeologist
(View Résumé 🔳)
(This is a technical page on leaky aquifer testing. Click link to see a nontechnical page on leaky aquifer testing.)
Introduction to DP_LAQ Leaky Aquifer Web Page
The purpose of this web page is to describe the results of an application of DP_LAQ pumping test analysis software (Moench, 1985) to actual data. The pumping test selected for this study is
described by Green and others (Green and others, 1991). The test includes data from the pumped well and one observation well.
Description of the Leaky Aquifer Tested
The aquifer that was tested is composed of sand lenses interbedded with lenses of silt and clay. Its thickness is 320 feet. The aquifer is overlain by 320 feet of clay, siltstone, and very fine to
fine grained sandstone, which is in turn overlain by a water table aquifer. This 320 foot thick unit is treated as a leaky aquitard in this pumping test analysis. The aquifer is underlain by 450
feet of sandy clay and siltstone which is in turn underlain by alternating layers of bentonitic clay and siltstone. This subjacent 450 foot thick unit is treated as a leaky aquitard.
The pumped well was a 20-inch diameter bore-hole. The screened interval extended through the entire thickness of the aquifer, and consisted of alternating sections of 8-inch diameter casing and
8-inch diameter wire wrapped screen with 0.025-inch openings. The total length of screen was 180 feet. The annulus between the alternating screen and casing and the wall of the hole was packed with
Observation Well Description
One observation well was used. It was cased with 2-inch PVC pipe. The same intervals were screened as in the pumped well. The distance from the pumped well to the observation well was 100 feet.
Description of the Pumping Test and the Original Analysis
The original pumping test is described by Green and others (Green and others, 1991). The well was pumped at a constant rate of 405 gpm for 98 hours with less than 3 percent discharge variation. The
original analysis used a type curve method developed by Papadopulos and Cooper and presented in Reed (1980). (The Papadopulos and Cooper method is a special case covered by DP_LAQ.) The analysis
was applied to time-drawdown data from the the monitoring well. This method of analysis assumes a fully penetrating well of finite diameter in a non-leaky aquifer. It improves on the type curve
method based on the Theis equation by considering the effect of depletion of water stored in the well resulting from drawdown during the test.
The result was as follows:
Transmissivity: 2,244 gpd/ft
Storativity: 3E-4 (dimensionless)
DP_LAQ Constant Rate Pumping Test Analysis
The present analysis of the data used DP_LAQ, which is described by Moench (1985). DP_LAQ computes type curves for constant rate leaky aquifer tests for three cases:
1. Aquitards above and below the aquifer have constant head boundaries on the top and bottom, respectively.
2. Aquitards above and below the aquifer have have no-flow bondaries on the top and bottom, respectively.
3. The aquitard above has a constant head boundary and the aquitard below has a no-flow boundary.
One can also specify whether leakage is from (1) both aquitards, (2) only the superjacent aquitard, (3) only the subjacent aquitard or (4) no leakage from either aquitard (non-leaky confined
aquifer). The pumped well may be line-source or finite-diameter.
DP_LAQ generates type curves based on input of dimensionless variables that are derived from the following parameters:
• Thickness of the aquifer and aquitards,
• Specific storage of the aquifer and aquitards,
• Horizontal hydraulic conductivity of the aquifer,
• Vertical hydraulic conductivity of the aquitards,
• Distance to observation wells,
• Effective radius of the pumped well,
• Radius of the pumped well casing where the water level is declining,
• Pumped well “skin” constant related to well loss.
One can also enter dimensionless variables related to dual porosity fractured aquifers. However, the present study does not use the dual porosity capability.
Figure 1 shows the type-curve match with the observed drawdown data from the pumped well. In this figure, “TYPE CURVE DIMENSIONLESS TIME” is the expression tD/rD2 defined by Moench (1985). “TYPE
CURVE DIMENSIONLESS DRAWDOWN” is hwD for the pumped well and hD for the observation well, and “ADJUSTED TIME” is t/r2 (t is time since pumping started, and r is distance to an observation well or the
effective well radius (rw) for the pumped well. Putting the observed data on the secondary axes and putting dimensionless time multiplied by S/T, and dimensionless drawdown multiplied by Q/(4πT) on
the primary axes allows type curve matching by successive trial values of transmissivity (T) and storativity (S). This technique allows the matching to be performed on a spreadsheet rather than
manually or electronically superimposing graphs and picking a match point.
Figure 1. Type curve match for pumped well and observation well with T=1448 gpd/ft and S=0.16.
Other parameters used to generate these type curves are:
Vertical hydraulic conductivity of the aquitards: 0.45 gpd/ft2,
Specific storage of the aquitards: 5E-4 (1/ft),
Dimensionless skin of pumped well: 1.0.
Conclusions Based on Leaky Aquifer Pumping Test Analysis
DP_LAQ worked well for analysis of this time-drawdown data from a leaky aquifer.
It is beneficial to have drawdown data from the pumped well and also from an observation well. During the development of the match shown in Figure 1, I found that data from the pumped well alone or
data from the monitoring well alone could produce type curve matches that differed from the analysis matching data from both wells on the same grap for DP_LAQ Leaky Aquifer Pumping Test.
Bibliography for DP_LAQ Leaky Aquifer Pumping Test
Green, Earl A., Mark T. Anderson, and Darrel D. Sipe (1991): Aquifer Tests and Water-Quality Analyses of the Arikaree Formation Near Pine Ridge, South Dakota; U.S. Geological Survey Water-Resources
Investigations Report 91-4005. 🔗
Moench, Allen F. (1985): Transient Flow to a Large-Diameter Well in an Aquifer With Storative Semiconfining Layers; Water Resources Research, Vol. 21, No. 8.
Prickett, T. A. (1965): Type-Curve Solution to Aquifer Tests under Water-Table Conditions; Ground Water, Volume 3, Number 3.
Reed, J. E. (1980): Type Curves for Selected Problems of Flow to Wells in Confined Aquifers; Techniques of Water-Resources Investigations of the United States Geological Survey, Chapter B3. 🔗 | {"url":"https://www.dunnhydrogeo.com/home/leaky-aquifer-testing-t","timestamp":"2024-11-07T23:15:34Z","content_type":"text/html","content_length":"129087","record_id":"<urn:uuid:f6a51a56-b0e0-4d7a-b8b7-eee508215cf5>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00053.warc.gz"} |
Binary Search
Whenever I revisit binary search problems, I keep having to work out the edge cases whenever coming up with the algorithm. So I thought I’d make a note of the pseudo code template I always end up
l=0, r=a.size()-1
while (l <= r) {
m = l+(r-l)/2
if (a[m] == target) return m
if (a[m] < target) l = m+1
else r = m-1
return -1 //not found
Note that we use m = l+(r-l)/2, rather than m=(l+r)/2, to ensure that we don’t get integer overflow.
We don’t have to have an array for a binary search. If we want to find sqrt(x), then we can set left=0 and right=x. We then test if mid is our sqrt number by squaring it: m*m.
We can also use floating point numbers. But then we need a different end case based on precision. So, to find sqrt(x) for a floating point number we would do something like:
l=0, r=x
while (r-l > 0.001) {
m = l+(r-l)/2
if (m*m == x) return m
else if (m*m > x) r = m
else l = m
return m
There’s a chance we will guess m exactly, in which case we return m in the loop. But most likely we will get to within our precision limit and exit the loop and then we will return m.
We can use comparison constraint rather than an exact target match by adapting our binary search algorithm. So, let’s say we want to find the smallest value bigger than or equal to x, i.e. find first
value >= x. We will use an answer variable (ans) to keep track of the latest mid value we’ve found that satisfies our criteria. If mid doesn’t satisfy our criteria then we look in the half that will
get us closer, so in this case if our target is 6 and we find 2 then we need to search the right half. We can let the while loop run its course rather than terminating it early.
l=0, r=a.size()-1, ans=-1
while (l <= r) {
m = l+(r-l)/2
if (a[m] >= x) {
ans = a[m]
r = m-1
l = m+1
return ans
This is a nice video lecture about binary search:
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What Does A Manova Do? | Free Printable Calendar Monthly
What Does A Manova Do?
In statistics, multivariate analysis of variance (MANOVA) is a procedure for comparing multivariate sample means. As a multivariate procedure, it is used when there are two or more dependent
variables, and is often followed by significance tests involving individual dependent variables separately.
Why use a MANOVA instead of ANOVA?
The correlation structure between the dependent variables provides additional information to the model which gives MANOVA the following enhanced capabilities: Greater statistical power: When the
dependent variables are correlated, MANOVA can identify effects that are smaller than those that regular ANOVA can find.
Is MANOVA better than ANOVA?
MANOVA is useful in experimental situations where at least some of the independent variables are manipulated. It has several advantages over ANOVA. First, by measuring several dependent variables in
a single experiment, there is a better chance of discovering which factor is truly important.
How do you use MANOVA?
MANOVA in SPSS is done by selecting “Analyze,” “General Linear Model” and “Multivariate” from the menus. As in ANOVA, the first step is to identify the dependent and independent variables. MANOVA in
SPSS involves two or more metric dependent variables.
What does a MANOVA require?
In order to use MANOVA the following assumptions must be met: Observations are randomly and independently sampled from the population. Each dependent variable has an interval measurement. Dependent
variables are multivariate normally distributed within each group of the independent variables (which are categorical)
Is MANOVA the same as factorial ANOVA?
Yes, they are on the same scale. You may also read,
When should you use MANOVA?
MANOVA can be used when we are interested in more than one dependent variable. MANOVA is designed to look at several dependent variables (outcomes) simultaneously and so is a multivariate test, it
has the power to detect whether groups differ along a combination of dimensions. Check the answer of
Is MANOVA a regression?
Both MANOVA and MANCOVA are multivariate regression techniques. If you prefer using R, R package mvtnorm can be used for this purpose.
What is a 2 way MANOVA?
For example, a two-way MANOVA is a MANOVA analysis involving two factors (i.e., two independent variables). … This means that the groups of each independent variable represent all the categories of
the independent variable you are interested in. Read:
What are the disadvantages of MANOVA?
The main disadvantage is the fact that MANOVA is substantially more complicated than ANOVA (Ta- bachnick & Fidell, 1996). In the use of MANOVA, there are several important assumptions that need to be
met. Furthermore, the results are sometimes ambiguous with respect to the effects of IVs on individ- ual DVs.
What does a MANOVA test?
Multivariate analysis of variance (MANOVA) is an extension of the univariate analysis of variance (ANOVA). … In this way, the MANOVA essentially tests whether or not the independent grouping variable
simultaneously explains a statistically significant amount of variance in the dependent variable.
How many participants do you need for a MANOVA?
For example, if you had six dependent variables that were being measured in two independent groups (e.g., “males” and “females”), there must be at least six participants in each of the two
independent groups for the one-way MANOVA to run (i.e., there must be at least six males and six females).
Is two way Anova a MANOVA?
The two-way multivariate analysis of variance (two-way MANOVA) is often considered as an extension of the two-way ANOVA for situations where there is two or more dependent variables.
Which ANOVA should I use?
Use a two way ANOVA when you have one measurement variable (i.e. a quantitative variable) and two nominal variables. In other words, if your experiment has a quantitative outcome and you have two
categorical explanatory variables, a two way ANOVA is appropriate.
Is ANOVA bivariate or multivariate?
To find associations, we conceptualize as “bivariate,” that is the analysis involves two variables (dependent and independent variables). ANOVA is a test which is used to find the associations
between a continuous dependent variable with more that two categories of an independent variable. | {"url":"https://bizzieme.com/what-does-a-manova-do/","timestamp":"2024-11-13T13:05:41Z","content_type":"text/html","content_length":"63750","record_id":"<urn:uuid:206896b7-a608-492d-a113-0f14cbf52de8>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00235.warc.gz"} |
What Is The Solution Of This Inequality? R6>11
r > -5
Step-by-step explanation:
r - 6 > -11
+ 6 + 6
r > -5
hope this helps! :D
have a miraculous day, and brainliest is immensely appreciated!! <3
The force which produce avalanches and landslides is gravity.
What is erosion?
The process by which the Earth's surface is worn down is called erosion. Erosion and weathering are two distinct but related processes. The decomposition of materials through chemical or physical
processes is known as weathering. When weathered materials like soil and rock fragments are carried away by wind, water, or ice, erosion occurs.
The Earth's three substances—air, water, and ice—continue to flow in large quantities from one location to another, making erosion an essential component of human life. The energy in their flow is
extremely high. All of the small and large substances in their path break and move in the flow's direction.
The most dramatic, unexpected, and hazardous examples of earth elements being pushed by gravity are landslides and avalanches. Avalanches are abrupt snowfalls, whereas landslides are sudden rockfalls
Massive chunks of rock move swiftly and violently when they abruptly break free from a cliff or hillside. The falling rocks are kept from slowing down by air that is trapped beneath them. Avalanches
and landslides can travel at speeds of 200 to 300 km/h.
Be mindful of your surroundings and keep an eye out for changes in the environment.
When it rains heavily, keep an eye out for deck or patio tilting, cracks or bulges in slopes, and drooping posts or fences. As soil steadily presses against a house and shifts windows and doors out
of place, sticking windows and doors can be an indication of earth movement.
Because landslides are more likely to occur where they have already happened, look for landslide scars.
Hence, Earth materials are transported by gravity from higher elevations to lower heights. Examples of hazardous gravity-driven erosion include mudflows, avalanches, and landslides.
Learn more about erosion; | {"url":"https://learning.cadca.org/question/what-is-the-solution-of-this-inequality-r6gt11-cvi7","timestamp":"2024-11-09T15:54:39Z","content_type":"text/html","content_length":"74752","record_id":"<urn:uuid:c17d2a7e-7b70-4405-ae52-9f7ec855fb61>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00582.warc.gz"} |
Solids - Annenberg Learner
Private: Learning Math: Geometry
Explore various aspects of solid geometry. Examine platonic solids and why there are a finite number of them. Investigate nets and cross-sections for solids as a way of establishing the relationships
between two–dimensional and three–dimensional geometry.
In This Session
Part A: Platonic Solids
Part B: Nets
Part C: Cross Sections
In this session, you will build solids, including Platonic solids, in order to explore some of their properties. By creating and manipulating these solids, you will develop a deeper sense of some of
the geometric relationships between them.
For information on required and/or optional materials for this session, see Note 1.
Learning Objectives
In this session, you will do the following:
• Learn about various aspects of solid geometry
• Explore Platonic solids and why there is a finite number of them
• Examine two-dimensional properties of three-dimensional figures such as nets and cross sections
Key Terms
Previously Introduced
Congruent: Two figures are congruent if all corresponding lengths are the same, and if all corresponding angles have the same measure. Colloquially, we say they “are the same size and shape,” though
they may have different orientation. (One might be rotated or flipped compared to the other.)
Regular Polygon: A regular polygon has sides that are all the same length and angles that are all the same size.
Vertex: A vertex is the point where two sides of a polygon meet.
New in This Session
Cross Section: A cross section is the face you get when you make one slice through an object.
Edge: An edge is a line segment where two faces intersect.
Face: A face is a polygon by which a solid object is bound. For example, a cube has six faces. Each face is a square.
Net: A net is a two-dimensional representation of a three-dimensional object.
Platonic Solid: A Platonic solid is a solid such that all of its faces are congruent regular polygons and the same number of regular polygons meet at each vertex.
Polyhedron: A polyhedron is a closed three-dimensional figure. All of the faces are made up of polygons.
Note 1
Materials Needed:
• Polydrons or other snap-together regular polygons
• clay
• dental floss or piano wire
• molds for a cube, tetrahedron, cylinder, cone, and sphere (optional)
• a party hat
• scissors
You can purchase Polydrons from the following source:
Polydron International Limited
Tel: 0044 (0)1285 770055
Fax: 0044 (0)1285 770171
An alternative to purchasing Polydrons is to make cutouts of regular polygons from stiff paper or poster board and use tape to attach them. Each individual working alone or pair of participants will
need at least 32 triangles, 12 pentagons, six squares, and three hexagons, but it’s helpful if each pair has extra sets of each kind of polygon. | {"url":"https://www.learner.org/series/learning-math-geometry/solids/","timestamp":"2024-11-06T20:40:29Z","content_type":"text/html","content_length":"96970","record_id":"<urn:uuid:ac7884a5-2b3d-40a4-9877-428834334fed>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00713.warc.gz"} |
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Here are a couple more formulas you should know for the ACT and SAT...
The sum of the internal angles for a polygon with n sides (an "n-gon") is (n-2) * 180 degrees. For example, the sum of the internal angles in a pentagon is (5-2) * 180 degrees = 540 degrees.
The sum of the external angles, taken one per vertex (corner) for any polygon, no matter how many sides it has, is 360 degrees. For two external angles per vertex, multiply 360 by 2 to get 720; for
3, multiply 360 by 3 to get 1080 degrees.
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Start the process early - you'll need to do that if you need the accommodation for the ACT or the SAT. (Click on the links and follow the instructions). Good luck!
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Things to Know for the Math Section of the Upcoming SAT
Here’s a quick list of formulas and other math facts you should know for tomorrow’s SAT. You will have to memorize them; they will not be provided in the “Math Facts” box at the beginning of the
math sections.
The “difference of squares”: (a+b)(a-b) = a2 – b2
Other binomials:
(a+b) = a2 +2ab + b2
(a-b) = a2 - 2ab + b2
The Relationship of the Diagonal of a Square to its Area.
A = d2/2. This comes from the fact that a square’s diagonal splits it into two 45-45-90 triangles, which have the side ratio of 1-1-√2, so the side length, s, is equal to d/√2, and the area, which
is s2, is then equal to (d/√2)2, which is d2/2.
Know that a square’s diagonals bisect, are congruent (equal), and are perpendicular, to each other.
Important Properties of Parallelograms
1. Angles that are diagonal to each other are equal.
2. Consecutive angles (ones that are next to each other) are supplementary (they add up to 180 degrees).
3. If one angle is a 90-degree (right) angle, then they all are.
4. The diagonals bisect each other.
5. Each diagonal bisects the parallelogram into two congruent triangles.
Note that rectangles and squares are also parallelograms, so these properties also apply to rectangles and squares. The diagonals of squares are also perpendicular (they form 90-degree angles).
Properties of transversals (two pairs parallel lines that cross each other).
Notice that parallelograms are formed by transversals (two pairs of parallel lines).
Notice that angle 1 = angle 8 = angle 2 =angle 5 and angle 4=3=7=6. If there were a line parallel to line t, the corresponding angles formed with line l and m would also be equal (congruent) to the
angles line t formed with lines l and m (that is, all the acute angles are congruent, and all the obtuse angles are congruent). See the first diagram in this section. Obviously, if they were all
right angles, all the angles would be congruent, since they’d all be 90 degrees.
Area of a Trapezoid:
How to Find the Area of an Equilateral Triangle if All You Know is the Side Length:
Here, if you know the side length is s, then you know you can split the triangle into two 30-60-90 triangles with side lengths s/2, s, and s√3. The s√3. side is the altitude of the triangle that
splits it in two. So the area of one 30-60-90 triangle is 1/2 * s√3. * s/2, and the area of the whole equilateral triangle is (s*s√3)/4, which is (s2√3)/4. You can memorize this formula, or just
know how to derive it by using an altitude to split the triangle in two.
I hope these formulas and facts help. Please read my other blog entries for other tips. Good luck on the SAT!
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For those of you taking the August SAT, here’s some advice on how to approach the SAT essay.
First, analyze the piece of writing. It’s almost always going to be a persuasive essay – e.g., a newspaper editorial, magazine article, or similar short piece intended to convince you of something.
While you are reading the piece, you should ask yourself, and answer, the following questions:
1. Who is the author of this piece? What does the introduction, or the piece itself, tell you about the author? For example, is the author an environmentalist or an oil company executive? How might
that affect the author’s viewpoint? An oil company executive might be likely to overstate the potential gains from drilling in wildlife refuges, while minimizing the environmental damage from it
(e.g. “Caribou love to rub up against the Alaska Pipeline for warmth”). An environmentalist would likely do the exact opposite (“Oil is an insanely pollutive, short-sighted solution to our
energy needs, oil spills kill wildlife, and even if there were NEVER any spills, pipeline development destroys much-needed habitats and forests. Trees create oxygen. Oil wells don’t.”) Note – it
is unlikely you will lose points for getting the author’s gender wrong if the name doesn’t obviously tell you the author’s gender. This isn’t a test of your knowledge of the writers’ personal
lives. However, you should get such details correct if you quote another writer in your analysis of the piece.
2. What is the author discussing in the piece? It shouldn’t be that hard for you to figure that out. What is the main subject – what is the author pointing out to you? Is it about war, energy,
helping people, exercise, reading, or some other subject?
3. What does the author want you to think about the subject of the piece? What does the author want you to think about, oh say, war, energy, helping people, exercise, reading, or some other
subject? The author might want to explain why exercise is good, or why helping other people is only good to a certain extent (e.g., the author’s point may be that “helping” people too much can
actually hurt them by making them lazy and dependent).
4. What, if any, moral statements are expressed or implied by the author’s argument? For example, is the author implying that saving lives is a good thing (not a particularly hard idea to “sell” to
most people), that exercise is good for you, that we should help those who are less fortunate than ourselves, or that people should work to earn a living? What evidence does the author use to
support his or her argument? Statistics? References to well-known stories, history, current events? These items are called “ethos.” Identify all of them in your essay. Remember, the essay
graders don’t know you, and will assume you know nothing about the subject at hand, and give you no credit, until you prove you know something. That means you should state the obvious points as
well as the non-obvious ones. Do these assumptions and supporting facts strengthen or weaken the author’s argument? Do you find them convincing? Why or why not? If you do not think these aspects
are as strong as they could be, identify what the author could have done to strengthen them (assuming you can think of something).
5. What kind of emotional, sensory, descriptive language is the author using? This aspect of a piece of writing is called “pathos.” Identify the language and how it is used, what it is intended to
make the reader think or feel, and if the use of such language is effective. Again, the SAT graders don’t know you and have to assume you’re a complete idiot until you show them otherwise.
Again, don’t be afraid to state the obvious. It’s better to make a “no-brainer” point and have the grader find the essay (and you) a bit pedantic than not be credited for your knowledge of
material because you assumed the grader knew that you found it obvious.
6. What is the underlying logic of the essay? That is, how does the author use the moral and factual assumptions or statements to reach the conclusion of the essay? Or, if you prefer, how does the
author make his or her point? Does the author go from the general to the specific, stating a rule, then showing how the particular situation he or she is discussing fits the rule, which means
there can only be one moral and logical conclusion? That kind of logic is “deductive logic,” or a “syllogism.” (You know- syllogisms are statements like “All Smurfs are blue. Brainy is a Smurf.
Therefore, Brainy is blue.”) Does the author point out the conclusion that was, or should have been, reached in one instance to generalize that the same conclusion should be reached in a wide
range of situations similar to the first case he or she describes? That is called “inductive logic.” Is the logic convincing, or does the author try to skip steps or use logical fallacies such
as appeals to emotion, “straw man” arguments, or the like? [See my previous entries and videos on logical fallacies for more examples, or simply Google “logical fallacies” for MANY examples].
Second, start writing. Leave several lines (probably 5 or 6) blank to write the introduction AFTER you’ve written the body paragraphs. Start your actual WRITING by addressing the subject of the
piece, and what the author wants you to think of it – the point of which the author is trying to convince you. Then discuss the ethos (facts, assumptions, morality) in one paragraph, the pathos
(emotional language) in another, and the logos (“logical” argument structure) in another. You can discuss the three elements in any order that makes sense to you.
Alternately, you can simply analyze the ethos, pathos, and logos in each paragraph of the essay, starting with the first and ending with the last. While this takes longer and is often more boring to
read, it does have the advantage of making it much less likely you’ll fail to discuss any aspect of the essay, since you’re literally analyzing it paragraph by paragraph, if not line by line. That’s
the way I imagine a computer program would analyze an essay – using the “You can’t miss a spot mowing the lawn if you always mow in overlapping strips” method.
Finally, if you have time and can think of some issues the author failed to address, such as obvious counterarguments, do mention them, and how the author could have made his or her argument
stronger by addressing them. However, please remember it is easier to praise a piece of writing than to criticize it, since no one will make you “prove” why a generally well-written piece makes a
good argument, but any grader will want to know why you think a carefully selected piece of writing the SAT used as a prompt is flawed. The grader will expect such criticism to be supported by
convincing and powerful evidence and argument.
If you don’t understand my point, I’ll provide an example. If I said “Hmm, those Williams sisters are pretty good tennis players, right?” most people would agree and move on to some other
conversational topic. But imagine if I said “Serena Williams isn’t all that great of a tennis player.” Most people would want to know why I believe something that just doesn’t seem to be true at
all. Someone would respond with something like “Um, John, she’s won many championships; she’s regarded as one of the best women tennis players in the world – what makes you think she’s not ‘all that
great?’” They’d probably wait for my response while thinking “This, I’ve gotta hear…” I’d have to come up with something pretty amazing in response (which, in this case, I couldn’t do, and I’d end
up looking stupid). So don’t criticize the piece unless you have a good idea of what exactly is wrong with it (for example, it relies on appeals to emotion or personal attacks, and cites virtually
no supporting evidence). It’s easier to say nice things if you’re not clear on what you don’t find convincing. If you simply don’t like/don’t agree with the author’s point, but he or she has
addressed your objections, you’ll simply have to discuss why you feel he or she inadequately addressed your objections. If you can’t do that, leave it alone.
Also, if you AGREE with the author, don’t merely restate the author’s argument; make sure you describe how the author constructs the argument, as detailed above. I’ve seen students show they agree
with, and understand, the author’s argument in their essays (e.g. why school uniforms in public school are a bad idea) by summarizing and restating it, but completely fail to address how the author
states and supports the point. That got them high reading and writing scores, but very low analysis scores. Don’t be like those students.
Once you’ve done that, simply write your introduction, in which you write something like
"In the news editorial
Why Is It So Cold In Here?
, Festus Freezer makes a convincing case as to why home heating costs for the poor should be subsidized by state, local, and federal government using citations to surveys, historical and literary
references, and appeals to morality (often called “ethos”), as well as powerful descriptive language and appeals to emotion (also known as “pathos”) and brings them all together with deductive logic
(also known as “logos”). He dismisses arguments that subsidized heating would simply add an unnecessary tax burden to the general public and encourage the waste of heating fuel as fearmongering
unsupported by any study done on the matter or any analogous study on welfare or health benefits."
Finally, write the conclusion, where you restate the points you made in the introduction, and briefly list examples of, the methods used by the author.
"Freezer’s argument in
Why Is It So Cold In Here?”
is compelling because, as I have shown, he combines facts, such as the results of several studies by the Department of Housing and Urban Development, emotional language such as “Why make your warm,
loving Grandma suffer in the cold?,” and logic (mostly the argument that people should others, especially when it is not difficult to do so) to make an superlatively good argument for heating
subsidies for the indigent."
Then proofread your essay, make necessary changes, and relax. You’re done. Good job! Good luck on the SAT!
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Homology (mathematics)
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Homology (mathematics)
In mathematics, homology^[1] is a general way of associating a sequence of algebraic objects such as abelian groups or modules to other mathematical objects such as topological spaces. Homology
groups were originally defined in algebraic topology. Similar constructions are available in a wide variety of other contexts, such as abstract algebra, groups, Lie algebras, Galois theory, and
algebraic geometry.
The original motivation for defining homology groups was the observation that two shapes can be distinguished by examining their holes. For instance, a circle is not a disk because the circle has a
hole through it while the disk is solid, and the ordinary sphere is not a circle because the sphere encloses a two-dimensional hole while the circle encloses a one-dimensional hole. However, because
a hole is "not there", it is not immediately obvious how to define a hole or how to distinguish different kinds of holes. Homology was originally a rigorous mathematical method for defining and
categorizing holes in a manifold. Loosely speaking, a cycle is a closed submanifold, a boundary is a cycle which is also the boundary of a submanifold, and a homology class (which represents a hole)
is an equivalence class of cycles modulo boundaries. A homology class is thus represented by a cycle which is not the boundary of any submanifold: the cycle represents a hole, namely a hypothetical
manifold whose boundary would be that cycle, but which is "not there".
There are many different homology theories. A particular type of mathematical object, such as a topological space or a group, may have one or more associated homology theories. When the underlying
object has a geometric interpretation as topological spaces do, the nth homology group represents behavior in dimension n. Most homology groups or modules may be formulated as derived functors on
appropriate abelian categories, measuring the failure of a functor to be exact. From this abstract perspective, homology groups are determined by objects of a derived category.
Homology theory can be said to start with the Euler polyhedron formula, or Euler characteristic.^[2] This was followed by Riemann's definition of genus and n-fold connectedness numerical invariants
in 1857 and Betti's proof in 1871 of the independence of "homology numbers" from the choice of basis.^[3]
Homology itself was developed as a way to analyse and classifymanifoldsaccording to their cycles – closed loops (or more generally submanifolds) that can be drawn on a given n dimensional manifold
but not continuously deformed into each other.^[4] These cycles are also sometimes thought of as cuts which can be glued back together, or as zippers which can be fastened and unfastened. Cycles are
classified by dimension. For example, a line drawn on a surface represents a 1-cycle, a closed loop or(1-manifold), while a surface cut through a three-dimensional manifold is a 2-cycle.
On the ordinarysphere, the cycle b in the diagram can be shrunk to the pole, and even the equatorialgreat circlea can be shrunk in the same way. TheJordan curve theoremshows that any arbitrary cycle
such as c can be similarly shrunk to a point. All cycles on the sphere can therefore be continuously transformed into each other and belong to the same homology class. They are said to be homologous
to zero. Cutting a manifold along a cycle homologous to zero separates the manifold into two or more components. For example, cutting the sphere along a produces two hemispheres.
This is not generally true of cycles on other surfaces. Thetorushas cycles which cannot be continuously deformed into each other, for example in the diagram none of the cycles a, b or c can be
deformed into one another. In particular, cycles a and b cannot be shrunk to a point whereas cycle c can , thus making it homologous to zero.
If the torus surface is cut along both a and b, it can be opened out and flattened into a rectangle or, more conveniently, a square. One opposite pair of sides represents the cut along a, and the
other opposite pair represents the cut along b.
The edges of the square may then be glued back together in different ways. The square can be twisted to allow edges to meet in the opposite direction, as shown by the arrows in the diagram. Up to
symmetry, there are four distinct ways of gluing the sides, each creating a different surface:
is theKlein bottle, which is a torus with a twist in it (The twist can be seen in the square diagram as the reversal of the bottom arrow). It is a theorem that the re-glued surface must
self-intersect (when immersed inEuclidean 3-space). Like the torus, cycles a and b cannot be shrunk while c can be. But unlike the torus, following b forwards right round and back reverses left and
right, because b happens to cross over the twist given to one join. If an equidistant cut on one side of b is made, it returns on the other side and goes round the surface a second time before
returning to its starting point, cutting out a twistedMöbius strip. Because local left and right can be arbitrarily re-oriented in this way, the surface as a whole is said to be non-orientable.
Theprojective planehas both joins twisted. The uncut form, generally represented as theBoy surface, is visually complex, so a hemispherical embedding is shown in the diagram, in which antipodal
points around the rim such as A and A′ are identified as the same point. Again, a and b are non-shrinkable while c is. But this time, both a and b reverse left and right.
Cycles can be joined or added together, as a and b on the torus were when it was cut open and flattened down. In the Klein bottle diagram, a goes round one way and −a goes round the opposite way. If
a is thought of as a cut, then −a can be thought of as a gluing operation. Making a cut and then re-gluing it does not change the surface, so a + (−a) = 0.
But now consider two a-cycles. Since the Klein bottle is nonorientable, you can transport one of them all the way round the bottle (along the b-cycle), and it will come back as −a. This is because
the Klein bottle is made from a cylinder, whose a-cycle ends are glued together with opposite orientations. Hence 2a = a + a = a + (−a) = 0. This phenomenon is called torsion. Similarly, in the
projective plane, following the unshrinkable cycle b round twice remarkably creates a trivial cycle which can be shrunk to a point; that is, b + b = 0. Because b must be followed around twice to
achieve a zero cycle, the surface is said to have a torsion coefficient of 2. However, following a b-cycle around twice in the Klein bottle gives simply b + b = 2b, since this cycle lives in a
torsion-free homology class. This corresponds to the fact that in the fundamental polygon of the Klein bottle, only one pair of sides is glued with a twist, whereas in the projective plane both sides
are twisted.
A square is a contractible topological space, which implies that it has trivial homology. Consequently, additional cuts disconnect it. The square is not the only shape in the plane that can be glued
into a surface. Gluing opposite sides of an octagon, for example, produces a surface with two holes. In fact, all closed surfaces can be produced by gluing the sides of some polygon and all
even-sided polygons (2n-gons) can be glued to make different manifolds. Conversely, a closed surface with n non-zero classes can be cut into a 2n-gon. Variations are also possible, for example a
hexagon may also be glued to form a torus.^[5]
The first recognisable theory of homology was published by Henri Poincaré in his seminal paper "Analysis situs", J. Ecole polytech. (2) 1. 1–121 (1895). The paper introduced homology classes and
relations. The possible configurations of orientable cycles are classified by the Betti numbers of the manifold (Betti numbers are a refinement of the Euler characteristic). Classifying the
non-orientable cycles requires additional information about torsion coefficients.^[4]
The complete classification of 1- and 2-manifolds is given in the table.
Topological characteristics of closed 1- and 2-manifolds[[CITE|6|https://openlibrary.org/search?q=
Manifold Euler No. Orientability Betti numbers Torsion coefficient
Symbol^[5] Name χ b[0] b[1] b[2] (1-dimensional)
Circle (1-manifold) 0 Orientable 1 1 N/A N/A
Sphere 2 Orientable 1 0 1 none
Torus 0 Orientable 1 2 1 none
Projective plane 1 Non-orientable 1 0 0 2
Klein bottle 0 Non-orientable 1 1 0 2
2-holed torus −2 Orientable 1 4 1 none
g-holed torus (Genus = g) 2 − 2g Orientable 1 2g 1 none
Sphere with c cross-caps 2 − c Non-orientable 1 c − 1 0 2
2−(2g (2g
2-Manifold with gholes and ccross-caps (c Non-orientable 1 0 2
c) c)−1
1. For a non-orientable surface, a hole is equivalent to two cross-caps.
2. Any 2-manifold is theconnected sumof g tori and c projective planes. For the sphere, g = c = 0.
A manifold with boundary or open manifold is topologically distinct from a closed manifold and can be created by making a cut in any suitable closed manifold. For example, the disk or 1-ballis
bounded by a circle. It may be created by cutting a trivial cycle in any 2-manifold and keeping the piece removed, by piercing the sphere and stretching the puncture wide, or by cutting the
projective plane. It can also be seen as filling-in the circle in the plane.
When two cycles can be continuously deformed into each other, then cutting along one produces the same shape as cutting along the other, up to some bending and stretching. In this case the two cycles
are said to be homologous or to lie in the same homology class. Additionally, if one cycle can be continuously deformed into a combination of other cycles, then cutting along the initial cycle is the
same as cutting along the combination of other cycles. For example, cutting along a figure 8 is equivalent to cutting along its two lobes. In this case, the figure 8 is said to be homologous to the
sum of its lobes.
Two open manifolds with similar boundaries (up to some bending and stretching) may be glued together to form a new manifold which is their connected sum.
This geometric analysis of manifolds is not rigorous. In a search for increased rigour, Poincaré went on to develop the simplicial homology of a triangulated manifold and to create what is now called
a chain complex.^[7]^[8] These chain complexes (since greatly generalized) form the basis for most modern treatments of homology.
In such treatments a cycle need not be continuous: a 0-cycle is a set of points, and cutting along this cycle corresponds to puncturing the manifold. A 1-cycle corresponds to a set of closed loops
(an image of the 1-manifold). On a surface, cutting along a 1-cycle yields either disconnected pieces or a simpler shape. A 2-cycle corresponds to a collection of embedded surfaces such as a sphere
or a torus, and so on.
Emmy Noether and, independently, Leopold Vietoris and Walther Mayer further developed the theory of algebraic homology groups in the period 1925–28.^[9]^[10]^[11] The new combinatorial topology
formally treated topological classes as abelian groups. Homology groups are finitely generated abelian groups, and homology classes are elements of these groups. The Betti numbers of the manifold are
the rank of the free part of the homology group, and the non-orientable cycles are described by the torsion part.
The subsequent spread of homology groups brought a change of terminology and viewpoint from "combinatorial topology" to "algebraic topology".^[12] Algebraic homology remains the primary method of
classifying manifolds.^[13]
Informally, the homology of a topological space X is a set of topological invariants of X represented by its homology groups
where thehomology groupdescribes the k-dimensional holes in X. A 0-dimensional hole is simply a gap between twocomponents, consequentlydescribes the path-connected components of X.^[14]
A one-dimensionalsphereis acircle. It has a single connected component and a one-dimensional hole, but no higher-dimensional holes. The corresponding homology groups are given as
whereis the group of integers andis thetrivial group. The grouprepresents afinitely-generated abelian group, with a singlegeneratorrepresenting the one-dimensional hole contained in a circle.^[15]
A two-dimensionalspherehas a single connected component, no one-dimensional holes, a two-dimensional hole, and no higher-dimensional holes. The corresponding homology groups are^[15]
In general for an n-dimensional sphere Sn, the homology groups are
A two-dimensionalballB2is a solid disc. It has a single path-connected component, but in contrast to the circle, has no one-dimensional or higher-dimensional holes. The corresponding homology groups
are all trivial except for. In general, for an n-dimensional ball *Bn*,^[15]
Thetorusis defined as aCartesian productof two circles. The torus has a single path-connected component, two independent one-dimensional holes (indicated by circles in red and blue) and one
two-dimensional hole as the interior of the torus. The corresponding homology groups are^[16]
The two independent 1D holes form independent generators in a finitely-generated abelian group, expressed as the Cartesian product group.
Construction of homology groups
The construction begins with an object such as a topological space X, on which one first defines a *chain complex
• C(X) encoding information about X. A chain complex is a sequence of abelian groups or modules C
0, C1, C2, ... connected byhomomorphismswhich are called boundary operators.^[17] That is,
where 0 denotes the trivial group andfor i < 0. It is also required that the composition of any two consecutive boundary operators be trivial. That is, for all n,
i.e., the constant map sending every element of Cn+1to the group identity in Cn−1. The statement that the boundary of a boundary is trivial is equivalent to the statement that, wheredenotes theimage
of the boundary operator anditskernel. Elements ofare called boundaries and elements ofare called cycles.
Since each chain group *Cn
• is abelian all its subgroups are normal. Then because
is a subgroup of *Cn*,is abelian, and sincethereforeis anormal subgroupof. Then one can create thequotient group
called the nX. The elements of Hn(X) are called homology classes. Each homology class is an equivalence class over cycles and two cycles in the same homology class are said to be homologous.^[18]
A chain complex is said to be exact if the image of the (n+1)th map is always equal to the kernel of the nth map. The homology groups of X therefore measure "how far" the chain complex associated to
X is from being exact.^[19]
The reduced homology groups of a chain complex C(X) are defined as homologies of the augmented chain complex^[20]
where the boundary operatoris
for a combination ∑ *niσii*, which are the fixed generators of C0. The reduced homology groupscoincide withfor i ≠ 0. The extrain the chain complex represents the unique mapfrom the empty simplex to
Computing the cycleand boundarygroups is usually rather difficult since they have a very large number of generators. On the other hand, there are tools which make the task easier.
The simplicial homology groups Hn(X) of a simplicial complex X are defined using the simplicial chain complex C(X), with Cn(X) the free abelian group generated by the n-simplices of X. The singular
homology groups Hn(X) are defined for any topological space X, and agree with the simplicial homology groups for a simplicial complex.
Cohomology groups are formally similar to homology groups: one starts with acochain complex, which is the same as a chain complex but whose arrows, now denoted *dn*, point in the direction of
increasing n rather than decreasing n; then the groupsof cocycles andof coboundaries follow from the same description. The nth cohomology group of X is then the quotient group
in analogy with the nth homology group.
The different types of homology theory arise from functors mapping from various categories of mathematical objects to the category of chain complexes. In each case the composition of the functor from
objects to chain complexes and the functor from chain complexes to homology groups defines the overall homology functor for the theory.^[21]
The motivating example comes fromalgebraic topology: the **simplicial homology** of asimplicial complexX. Here the chain group *Cnfree abelian groupor module whose generators are the n-dimensional
oriented simplexes of X. The orientation is captured by ordering the complex'sverticesand expressing an oriented simplexas an n-tupleof its vertices listed in increasing order (i.e.in the complex's
vertex ordering, whereis theth vertex appearing in the tuple). The mappingfrom *Cnn-1
• is called the boundary mapping and sends the simplex
which is considered 0 if n = 0. This behavior on the generators induces a homomorphism on all of *Cn
• as follows. Given an element
, write it as the sum of generators, where *Xn
• is the set of n-simplexes in X and the *m
• are coefficients from the ring *C
• is defined over (usually integers, unless otherwise specified). Then define
The dimension of the n-th homology of X turns out to be the number of "holes" in X at dimension n. It may be computed by putting matrix representations of these boundary mappings in Smith normal
Using simplicial homology example as a model, one can define a singular homology for any topological space X. A chain complex for X is defined by taking Cn to be the free abelian group (or free
module) whose generators are all continuous maps from n-dimensional simplices into X. The homomorphisms ∂n arise from the boundary maps of simplexes.
In abstract algebra, one uses homology to define derived functors, for example the Tor functors. Here one starts with some covariant additive functor F and some module X. The chain complex for X is
defined as follows: first find a free module F1 and a surjective homomorphism p1 : F1 → X. Then one finds a free module F2 and a surjective homomorphism p2 : F2 → ker(p1). Continuing in this fashion,
a sequence of free modules Fn and homomorphisms pn can be defined. By applying the functor F to this sequence, one obtains a chain complex; the homology Hn of this complex depends only on F and X and
is, by definition, the n-th derived functor of F, applied to X.
A common use of group (co)homologyis to classify the possibleextension groupsE which contain a given G-module M as anormal subgroupand have a givenquotient groupG, so that G = E/M.
• Borel–Moore homology
• Cellular homology
• Cyclic homology
• Hochschild homology
• Floer homology
• Intersection homology
• K-homology
• Khovanov homology
• Morse homology
• Persistent homology
• Steenrod homology
Chain complexes form acategory: A morphism from the chain complex (*dn*: *Ann-1) to the chain complex (*en*: *Bnn-1) is a sequence of homomorphisms *fn*: *Annfor all n. The n-th homology *Hn
• can be viewed as a covariant
functorfrom the category of chain complexes to the category of abelian groups (or modules).
If the chain complex depends on the object X in a covariant manner (meaning that any morphism X → Y induces a morphism from the chain complex of X to the chain complex of Y), then the Hn are
covariant functors from the category that X belongs to into the category of abelian groups (or modules).
The only difference between homology and cohomology is that in cohomology the chain complexes depend in a contravariant manner on X, and that therefore the homology groups (which are called
cohomology groups in this context and denoted by Hn) form contravariant functors from the category that X belongs to into the category of abelian groups or modules.
If (dn: An → A**n-1) is a chain complex such that all but finitely many An are zero, and the others are finitely generated abelian groups (or finite-dimensional vector spaces), then we can define the
Euler characteristic
(using the rank in the case of abelian groups and the Hamel dimension in the case of vector spaces). It turns out that the Euler characteristic can also be computed on the level of homology:
and, especially in algebraic topology, this provides two ways to compute the important invariant χ for the object X which gave rise to the chain complex.
Every short exact sequence
of chain complexes gives rise to a long exact sequence of homology groups
All maps in this long exact sequence are induced by the maps between the chain complexes, except for the maps Hn(C) → H**n-1*(A)* The latter are called connecting homomorphisms and are provided by
the zig-zag lemma. This lemma can be applied to homology in numerous ways that aid in calculating homology groups, such as the theories of relative homology and Mayer-Vietoris sequences.
Application in pure mathematics
Notable theorems proved using homology include the following:
• The Brouwer fixed point theorem: If f is any continuous map from the ball Bn to itself, then there is a fixed point a ∈ Bn with f(a) = a.
• Invariance of domain: If U is an open subset of Rn and f : U → Rn is an injective continuous map, then V = f(U) is open and f is a homeomorphism between U and V.
• The Hairy ball theorem: any vector field on the 2-sphere (or more generally, the 2k-sphere for any k ≥ 1) vanishes at some point.
• The Borsuk–Ulam theorem: any continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points to the same point. (Two points on a sphere are called antipodal if they
are in exactly opposite directions from the sphere's center.)
• Invariance of dimension: if non-empty open subsets and are homeomorphic, then .^[22]
Application in science and engineering
In topological data analysis, data sets are regarded as a point cloud sampling of a manifold or algebraic variety embedded in Euclidean space. By linking nearest neighbor points in the cloud into a
triangulation, a simplicial approximation of the manifold is created and its simplicial homology may be calculated. Finding techniques to robustly calculate homology using various triangulation
strategies over multiple length scales is the topic of persistent homology.^[23]
In sensor networks, sensors may communicate information via an ad-hoc network that dynamically changes in time. To understand the global context of this set of local measurements and communication
paths, it is useful to compute the homology of the network topology to evaluate, for instance, holes in coverage.^[24]
In dynamical systems theory in physics, Poincaré was one of the first to consider the interplay between the invariant manifold of a dynamical system and its topological invariants. Morse theory
relates the dynamics of a gradient flow on a manifold to, for example, its homology. Floer homology extended this to infinite-dimensional manifolds. The KAM theorem established that periodic orbits
can follow complex trajectories; in particular, they may form braids that can be investigated using Floer homology.^[25]
In one class of finite element methods, boundary-value problems for differential equations involving the Hodge-Laplace operator may need to be solved on topologically nontrivial domains, for example,
in electromagnetic simulations. In these simulations, solution is aided by fixing the cohomology class of the solution based on the chosen boundary conditions and the homology of the domain. FEM
domains can be triangulated, from which the simplicial homology can be calculated.^[26]^[27]
Various software packages have been developed for the purposes of computing homology groups of finite cell complexes. Linbox ^[34] is a C++ library for performing fast matrix operations, including
Smith normal form; it interfaces with both Gap ^[35] and Maple ^[36] . Chomp ^[37] , CAPD::Redhom ^[38] and Perseus ^[39] are also written in C++. All three implement pre-processing algorithms based
on Simple-homotopy equivalence and discrete Morse theory to perform homology-preserving reductions of the input cell complexes before resorting to matrix algebra. Kenzo ^[40] is written in Lisp, and
in addition to homology it may also be used to generate presentations of homotopy groups of finite simplicial complexes. Gmsh includes a homology solver for finite element meshes, which can generate
Cohomology bases directly usable by finite element software.^[26]
• Betti number
• Cycle space
• Eilenberg–Steenrod axioms
• Extraordinary homology theory
• Homological algebra
• Homological conjectures in commutative algebra
• Homological dimension
• Künneth theorem
• List of cohomology theories - also has a list of homology theories
• De Rham cohomology
Citation Linkopenlibrary.orgin part from Greek ὁμός homos "identical"
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.org, pp. 2–3 (in PDF)
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.orgRicheson 2008 p.254.
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.orgWeeks, J.R.; The Shape of Space, CRC Press, 2002.
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.orgRicheson (2008)
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.orgRicheson 2008 p.258
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.orgHilton, Peter (1988), "A Brief, Subjective History of Homology and Homotopy Theory in This Century", Mathematics Magazine, Mathematical Association of America, 60 (5):
282–291, JSTOR 2689545, p. 284
Sep 29, 2019, 10:52 PM
Citation Linksmf4.emath.frFor example L'émergence de la notion de groupe d'homologie, Nicolas Basbois (PDF), in French, note 41, explicitly names Noether as inventing the homology group.
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.orgHirzebruch, Friedrich, Emmy Noether and Topology in Teicher, M., ed. (1999), The Heritage of Emmy Noether, Israel Mathematical Conference Proceedings, Bar-Ilan University/
American Mathematical Society/Oxford University Press, ISBN 978-0-19-851045-1, OCLC 223099225, pp. 61–63.
Sep 29, 2019, 10:52 PM
Citation Linkmath.vassar.eduBourbaki and Algebraic Topology by John McCleary (PDF) gives documentation (translated into English from French originals).
Sep 29, 2019, 10:52 PM
Citation Linkopenlibrary.orgRicheson 2008 p.264.
Sep 29, 2019, 10:52 PM | {"url":"https://everipedia.org/wiki/lang_en/Homology_%2528mathematics%2529","timestamp":"2024-11-10T11:59:32Z","content_type":"text/html","content_length":"341298","record_id":"<urn:uuid:0cc61704-7547-499e-9efb-08333433d4b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00457.warc.gz"} |
A second way to form impressions is to develop algebraic impressions—analyzing the positive and negative things you learn about someone to calculate an overall impression, then updating this
impression as you learn new information (Anderson, 1981). It’s similar to solving an algebraic equation, whereby you add and subtract different values to compute a final result. However, when forming
algebraic impressions, you don’t place an equal value on every piece of information you receive. Instead, information that’s important, unusual, or negative is usually weighted more heavily than
information that’s trivial, typical, or positive (Kellermann, 1989). This happens because people tend to believe that important, unusual, or negative information reveals more about a person’s “true”
character than does other information (Kellermann, 1989).
Of course, other people form algebraic impressions of you, too. So, when you’re communicating—whether in person or online, with a friend or in front of an audience—be mindful of what important,
unusual, or negative information you share about yourself. This information will have a particularly strong effect on others’ impressions of you.
Algebraic impressions are more accurate than Gestalts because you take time to form them and you consider a wider range of information. They’re also more flexible. You can update your algebraic
impression every time you receive new information about someone. For instance, you discover through Facebook that the cool classmate you went on a date with yesterday has political views much
different from your own. Accordingly, you become a bit cautious about pursuing a romantic relationship with this person while remaining open to seeing where things will lead. | {"url":"https://digfir-published.macmillanusa.com/choicesconnections2e/choicesconnections2e_ch2_17.html","timestamp":"2024-11-11T21:05:26Z","content_type":"text/html","content_length":"5072","record_id":"<urn:uuid:a54bc34c-5b5b-42e3-bf26-8b79f5b2af8d>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00884.warc.gz"} |
Tuva for AP Statistics
Tuva enhances the AP Statistics curriculum by giving students opportunities to visualize key concepts of variability, distribution shapes, and patterns of association for bivariate data. Students
investigate engaging real-world datasets and use statistical concepts to make predictions and draw conclusions in context. Students can also use Random Sampling to generate a sampling distribution in
order to explore the key concepts of the Central Limit Theorem.
Tuva's visual interface is ideal for students to understand the fundamental concepts of statistics that are need to progress in AP Statistics. Students can use Tuva to explore key concepts such as
the Empirical Rule for a normal distribution. Through the use of dividers, they can calculate the percentage of data points within a given range, and they can test out whether a distribution is
approximately normal. | {"url":"https://support.tuvalabs.com/hc/en-us/articles/360052082953-Tuva-for-AP-Statistics","timestamp":"2024-11-14T07:29:19Z","content_type":"text/html","content_length":"21682","record_id":"<urn:uuid:91209fe3-a61a-4124-9fe9-d3e4491a2385>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00304.warc.gz"} |
1. Introduction
Automatic segmentation of coronary arteries in X-ray angiography images can help cardiologists for diagnosing and treating vessel abnormalities. Hence, the development of efficient methods for
automatic vessel segmentation has become essential for computer-aided diagnosis (CAD) systems. The main disadvantages of X-ray coronary angiograms are the nonuniform illumination and the low contrast
between blood vessels and image background which are illustrated in Figure 1. Since the presence of several shades along blood vessels generates multimodal histograms, the vessel segmentation problem
has been commonly addressed in two different steps. The first step is vessel enhancement, where the methods try to remove noise from the image and to enhance the vessel-like structures. After vessel
enhancement, the second step consists in the classification of vessel and nonvessel pixels by using different strategies such as the selection of an optimal threshold value.
In recent years, several techniques have been reported for this purpose in different types of clinical studies. Some of the proposed methods are based on mathematical morphology. The method of (^Eiho
and Qian, 1997) represents one of the most commonly used interactive vessel segmentation methods, because its efficiency and ease of implementation. The method uses the single-scale top-hat operator
(^Serra, 1982) to enhance vessel-like structures followed by different morphological operators such as erosion, thinning, and watershed transformation. This method was improved in (^Qian et al., 1998
), by introducing the multiscale top-hat operator using the linear combination of fixed scales of the top-hat operator in order to capture different diameters of blood vessels. Another
morphological-based method was proposed in (^Maglaveras et al., 2001), which it uses the skeleton, thresholding and connected-component operator for automatic segmentation of coronary arteries in
angiograms. (^Bouraoui et al., 2008) introduced an automatic segmentation method for coronary angiograms consisting in the application of the hit-or-miss transform and the region-growing strategy. (^
Lara et al., 2009) proposed an interactive segmentation method based on the region-growing strategy and a differential-geometry approach for coronary arteries in angiograms. In general, these
morphological-based methods provide satisfactory results when vessels have high contrast; however, they do not perform well when vessels have different diameters and low contrast.
Due to the shortcomings of the morphological-based methods for automatic segmentation of coronary angiograms, some strategies using spatial filtering have been introduced. The most commonly vessel
enhancement technique is the Gaussian matched filters (GMF) method proposed by (^Chaudhuri et al., 1989) and improved by (^Al-Rawi et al., 2007). The fundamental idea of this method is to approximate
the shape of blood vessels through a Gaussian curve as the matching template in the spatial image domain. The Gaussian template is rotated at different orientations and convolved with the original
angiogram. From the filter bank of oriented responses, the maximum response at each pixel is taken to obtain the final enhanced image. The GMF method is used as a vessel enhancement procedure in
different automatic segmentation methods. (^Chanwimaluang and Fan, 2003) used the GMF method to enhance retinal blood vessels, and then applied the entropy thresholding method proposed by (^Pal and
Pal, 1989) to obtain the final vessel segmentation result. This automatic segmentation method was used successfully in (^Chanwimaluang et al., 2006) for a retinal vessel registration system. (^Kang
et al., 2009) proposed a fusion enhancement strategy for coronary arteries where the top-hat operator and the GMF are applied separately on the original angiogram. Both resulting enhanced images are
thresholded by the maximizing entropy thresholding method and the segmentation result is acquired by pixel-wise multiplication of the two binary images. This method was also used in (^Kang et al.,
2013) just replacing the thresholding method by the degree segmentation method. These template matching methods present low efficiency when images contain occlusions or vessels with different
diameters to be detected.
More recently, different vessel enhancement methods based on the properties of the Hessian matrix have been proposed (^Lorenz et al., 1997; ^Frangi et al., 1998; ^Jin et al., 2013; ^Xiao et al., 2013
). The Hessian matrix is computed by convolving the second-order derivatives of a Gaussian kernel with the original angiogram. The method of (^Frangi et al., 1998) uses the properties of the
eigenvalues of the Hessian matrix to compute a vesselness measure. This measure is used to classify vessel and nonvessel pixels from the resulting image of dominant eigenvalues at different vessel
width scales. Taking into account the properties of the Hessian matrix, different vesselness measures have been introduced (^Salem and Nandi, 2008; ^Tsai et al., 2013; ^Mhiri et al., 2013). (^Li et
al., 2012) and (^Wang et al., 2012) proposed different vesselness measures, which are segmented by using similar procedures of the region-growing operator. These methods have the ability of detecting
vessels of different caliber although they are highly sensitive to noise because of the second-order derivative.
Moreover, from the revised morphological-based methods, the top-hat operator presents low efficiency since it is not able to capture vessels of different caliber, which is an advantage of the methods
based on the properties of the Hessian matrix. In the present paper, a novel method based on two stages for automatic enhancement and segmentation of coronary arteries in angiograms is proposed. In
the enhancement stage, a new multiscale top-hat operator based on the eigenvalues of the Hessian matrix is introduced in order to capture most of the vessel diameters in the angiograms. The method is
compared with the multiscale top-hat operator proposed by (^Qian et al., 1998), Gaussian matched filters (^Al-Rawi et al., 2007), and the Hessian-based method proposed by (^Wang et al., 2012) using
the area A[z] under the ROC curve. In the segmentation stage, the multiscale top-hat response is thresholded by a new global thresholding technique based on multiobjective optimization using the
weighted sum method (^Marler and Arora, 2010). In this stage, seven state-of-the-art thresholding methods are compared with our multiobjective method using the measures of sensitivity, specificity,
and accuracy. In addition, the segmentation results of the proposed method are compared with those obtained with five state-of-the-art vessel segmentation methods.
The remainder of this paper is organized as follows. In Section 2, the database of coronary angiograms and the fundamentals of the proposed vessel segmentation method along with a set of evaluation
metrics are introduced. The experimental results are discussed in Section 3, and conclusions are given in Section 4.
2. Materials and Methods
2.1 Coronary Angiograms
The database used in the present work consists of 80 X-ray coronary angiographic images of 27 patients. Each angiogram is of size 300 × 300 pixels. In order to evaluate the performance of the
segmentation methods, 40 of the 80 angiograms have been used as training set primarily for tuning the parameters of the methods, and the remaining 40 angiograms have been used as an independent test
set for evaluation of vessel segmentation methods. The corresponding vessel ground-truth for each image was drawn by a specialist and ethics approval was provided by the cardiology department of the
Mexican Social Security Institute, UMAE T1 León.
2.2 Proposed Vessel Segmentation Method
Since coronary arteries present lower reflectance compared with the background angiogram, the proposed coronary artery segmentation method is composed of two steps. In this section, the fundamentals
of the proposed multiscale top-hat operator for vessel enhancement and the thresholding strategy based on multiobjective optimization approach are described in detail.
2.2.1 Vessel Enhancement
The coronary arteries in angiograms can be described as tubular structures with a nonuniform illumination, different diameters and diverse orientations. Due to these factors, a vessel enhancement
procedure is commonly used as a preprocessing step before applying a specific segmentation method.
The morphological top-hat operator has been widely used to suppress local noise and to enhance vessel-like structures in angiograms (^Eiho and Qian, 1997; ^Qian et al., 1998; ^Kang et al., 2009; ^
Kang et al., 2013). The top-hat operator can be defined as follows:
where Io is the original angiogram, Ie is the resulting enhanced angiogram, and γ(Io, A) represents the grayscale morphological opening operator applied on the original angiogram with a predefined
structuring element (SE). In order to obtain the best performance by using the top-hat operator, the parameters of size and shape of the SE have to be tuned. In general, the size parameter is chosen
slightly larger than the maximum caliber of the vessel and a disk-shaped SE is commonly used. The main disadvantage of some previously mentioned vessel enhancement methods on applying the
conventional top-hat operator, is the fact that by using only one structuring element with a fixed size, cannot be properly enhanced all of the vessel diameters in the angiogram (^Qian et al., 1998).
In our work, we propose the use of the eigenvalues computed from the Hessian matrix, in order to select the best scales for the multiscale top-hat operator.
The Hessian matrix is computed from the second-order derivatives of a Gaussian kernel at each pixel of the intensity angiogram L(x, y) as follows:
where Lxx, Lxy, Lyx, and Lyy denote the second-order derivatives of the intensity image as follows:
where the symbol “ * ” represents a convolution operator, σ is the the length scale factor, and Gxx, Gyy, and Gxy are the second-order derivatives of the Gaussian kernel G at different scales of σ,
which is defined as follows:
Due to the Hessian matrix is symmetric; the large eigenvalue λ[1] can be computed as follows:
The largest eigenvalue λ[max] over all the scales of the σ value at each image pixel is calculated as following
The final step of the enhancement stage consists in computing the quartiles method of descriptive statistics from the λ[max] image of vessel width scales. Figure 2 illustrates the distribution of the
vessel scales of the 40 training angiograms, which is acquired by the number of pixels for each scale in the corresponding angiogram. In our experiments and based on the ROC curve analysis using the
training set of angiograms, the second and third quartiles contain the optimal scales for the image to be processed. These vessel scales are used as the size parameter for the multiscale top-hat
operator. For each vessel scale, a disk-shaped structuring element of fixed size (scale) is applied on the angiogram.
Moreover, to detect most of the vessels at different diameters, the final enhanced image is obtained by preserving the maximum response for all the enhanced pixels at different scales. Since the
number of optimal scales for each angiogram is not constant, the proposed enhancement method can automatically detect the appropriate vessel scales for a particular angiogram, which is one of the
main advantages regarding morphological-based methods.
2.2.2 Thresholding based on Multiobjective Optimization Approach
Multiobjective optimization (MOO) is the process of optimizing systematically and simultaneously a number of different objective functions (^Marler and Arora, 2010). One of the most widely used
strategies for solving MOO problems is the weighted sum method (WSM). The WSM consists in the arrangement of different objective functions and the selection of a particular set of weights to form a
single objective function as follows:
where n is the number of objective functions of the problem to be optimized, w[i] is the weighting factor for each function, and F[i] is the particular objective function. Generally, the weight
factors are positives, established by the preference of the user, and their cumulative sum has to be equal to 1.
Although a multiobjective optimization problem has multiple solutions, the weighted sum method provides a single optimal solution using a set of fixed weights, which represents an ideal case to
classify vessel and nonvessel pixels for our coronary vessel segmentation application. The proposed thresholding method uses three objective functions, as it is shown in the following composite
objective function:
where w[1], w[2], and w[3] are the assigned weights. In our work, these weights were statistically determined as w[1] = 0.3, w[2] = 0.4, and w[3] = 0.3, respectively, using the average of the
sensitivity, specificity and accuracy measures on the training set of 40 angiograms. F[1] is the between-class variance criterion based on the method of (^Otsu, 1979). F[2] and F[3] represent entropy
criteria based on the methods of (^Kapur et al., 1985) and (^Pal and Pal, 1989), respectively, which are described below.
The between-class variance criterion is applied on the histogram of intensity levels L of an image. This criterion uses the mean intensity of the given image defined as μ, the probability
distribution W[j] and the mean intensity for each class of pixels μ[j], where n = 2, since only two classes of pixels are required (vessel and nonvessel pixels). The intensity level with the maximum
between-class variance gives the optimal threshold value wich is acquired by maximizing the following:
The entropy criterion of (^Kapur et al., 1985) is given by the entropies of the corresponding classes A and B of pixels in which the image will be segmented. The entropy of class A and class B is
computed using the accumulated probabilities P[A] and P[B] as follows:
The optimal threshold value of the intensity levels L is acquired by maximizing the total entropy as follows:
The two-dimensional entropy criterion of (^Pal and Pal, 1989), is based on the co-occurrence matrix of the image and the second-order entropy. The second-order entropy corresponding to the foreground
and background image can be defined as follows:
where s represents a threshold value, L is the number of intensity levels of the input image, and P^A and P^B represent the probability distributions of the two classes of pixels through the
co-occurrence matrix. The total second-order entropy of the image has to be computed for all the intensity levels, and the optimal threshold is the value that maximizes the entropy by using the
In the proposed thresholding method, each criterion is normalized to the range 0,1, and the optimization process is carried out over all the intensity levels in the multiscale top-hat response. Since
a multiobjective optimization problem has multiple solutions (segmentation results), in Figure 3, the Pareto front and the corresponding vectors of weights acquired from the top-hat response of the
training set of angiograms are introduced. The advantage of multiobjective optimization following the weighted sum method resides in the fact that the weights can be determined such that the highest
trade-off among the three objective functions is achieved.
2.2.3 Evaluation metrics
In order to assess the performance of the proposed vessel segmentation method, the area under the receiver operating characteristic (ROC) curve (A[z]) and measures of sensitivity, specificity, and
accuracy, have been adopted which are described below.
The ROC curve is a plot of the true-positive rate (TPR) against false-positive rate (FPR) of a classification system. The TPR represents the fraction of the vessel pixels outlined by the specialist
that are correctly detected by the method. The TPR is also known as the sensitivity measure which is computed using Eq.(19). FPR represents the fraction of the nonvessel pixels that are incorrectly
detected as vessel pixels by the method.
Specificity measures the true-negative rate (TNR), which represents the fraction of nonvessel pixels that are correctly detected as such by the method (background pixels) as follows:
Accuracy represents the fraction of vessel and nonvessel pixels correctly detected by the method divided by the total number of pixels in the angiogram as follows:
In this work, the area under the ROC curve is used to determine the most suitable parameters of the vessel enhancement methods and also to evaluate their performance using the training set. The
measures of sensitivity, specificity, and accuracy were used to assess the segmentation results between the regions obtained by computational methods and the regions outlined by the specialist. In
these measures when the regions are completely superimposed the obtained result is 1, and 0, when these regions are completely different.
In Section 3, the segmentation results obtained from the proposed method on X-ray angiographic images are presented and analyzed by the evaluation metrics.
3. Results and Discussion
The computational experiments were tested on a computer with an Intel Core i3, 2.13 GHZ procesor, and 4GB of RAM using the Matlab software.
In the vessel detection stage, the proposed multiscale top-hat operator, the multiscale top-hat method (^Qian et al., 1998), the Hessian-based method (^Wang et al., 2012), and the Gaussian matched
filters (^Al-rawi et al., 2007) were compared against each other using the training set of angiograms via ROC curve analysis.
The parameters of the multiscale top-hat method (^Qian et al., 1998) was obtained by varying the size of the structuring element and taking into account three different scales. The best set of scales
was determined to be S[1] = 19, S[2] = 5, and S[3] = 3 pixels. The multiscale Hessian-based method (^Wang et al., 2012), was tested with different values of σ in the range [1, 20] pixels, and with
step sizes (s) in the range [0.5, 4]. The most suitable parameters for vessel width σ, and for the step size were determined to be σ = [1, 10], and s = 0.5. Since the Gaussian matched filters
represent the enhancement step of different vessel segmentation methods, the GMF was tuned with the same set of parameters for the following methods (^Chanwimaluang and Fan, 2003; ^Kang et al., 2009;
^Kang et al., 2013). The best Gaussian matched filters parameters for length of the vessel segment, width of the vessel segment, and number of oriented filters, were statistically determined to be L
= 11 , σ = 1.9, T = 8, and κ = 12, respectively.
From the best set of parameters for each enhancement method, Figure 4 illustrates the area (A[z]) under the ROC curves obtained using the training set of angiograms. The results of vessel enhancement
show that the proposed multiscale top-hat operator provides a higher level of classification between vessel and nonvessel pixels than the comparative methods. The most suitable range of vessel width
scale for the Gaussian kernel of the proposed enhancement method was set as σ = 1, 25, with a step size of s = 3. The proposed multiscale top-hat operator with the best set of parameters provided A[z
] = 0.965 with the test set of angiograms. In comparison, using the best set of parameters of the method of (^Qian et al., 1998), (^Wang et al., 2012), and Gaussian matched filters, they provided A[z
] = 0.938, A[z] = 0.941, and A[z] = 0.923, respectively, using the test set of angiograms. Additionally, Figure 5 shows a subset of angiograms with the corresponding ground-truth, and the vessel
enhancement results of the proposed multiscale top-hat operator.
On the other hand, the proposed thresholding strategy based on the weighted sum method was compared with seven state-of-the-art thresholding methods using the multiscale top-hat response of the
training set of angiograms. In Table 1, the segmentation performance is presented. In this experiment, the measures of sensitivity, specificity and accuracy were used for evaluation and comparison.
These three measures have been widely used in vessel segmentation as the main measures for evaluation. The values of these measures were obtained by concatenating the forty segmented angiograms of
the training set in order to form just one large binary image. However, since vessel pixels often occupy less than 15% of an angiogram, the measures of specificity and accuracy are always high.
Consequently, the average of these three measures has been considered as the main performance metric. The obtained results show that the proposed multiobjective thresholding method provides the best
segmentation performance using the training set of angiograms and hence it was selected for further analysis. Due to the thresholding based on multiobjective optimization uses the intensity values in
the image histogram, a low computational time is required to perform the optimization task. The proposed thresholding strategy obtained an average execution time of 0.329 seconds for each
angiographic image.
Moreover, the proposed vessel segmentation method consisting of the steps of multiscale top-hat operator for vessel enhancement, and multiobjective thresholding for segmentation, was compared with
five state-of-the-art vessel segmentation methods. The methods of (^Qian et al., 1998) and (^Wang et al., 2012) were tuned taking into account the aforementioned parameters. The methods of (^
Chanwimaluang and Fan, 2003), and the fusion methods of (^Kang et al., 2009; ^Kang et al., 2013) were tuned according to the best parameters of the Gaussian matched filters. Since the two fusion
methods of Kang et al., also use the classical top-hat operator as enhancement method, the size of the structuring element for the top-hat operator was set as S = 19, and it was obtained by using the
area under the ROC curve with the training set of angiograms. In Table 2, the comparative analysis of the proposed method with the five vessel segmentation methods using the test set of angiograms is
presented. In this experiment, the average of the sensitivity, specificity and accuracy measures is presented. The values of these measures were obtained by concatenating the forty segmented
angiograms of the test set to form just one large binary image. The obtained results show that the proposed method provides the highest segmentation performance using the test set of angiograms.
Figure 6 introduces a subset of X-ray coronary angiographic images with the corresponding ground-truth for each image. The segmentation results of the method of (^Qian et al., 1998) shows
difficulties to detect different diameters of vessels obtaining a high rate of false-positive pixels in most of the low contrast angiograms. The fusion method of (^Kang et al., 2009) based on the
entropy of the enhanced images presents a low rate of false-positive pixels illustrating broken vessels, which reduces the rate of accuracy segmentation. The method of (^Chanwimaluang and Fan, 2003)
can detect a high rate of true-positive and false-positive pixels, which increase the sensitivity rate and decrease the specificity and accuracy values. The results of the method of (^Wang et al.,
2012) shows a uniform segmentation over the main tubular structures at different scales while it presents a low rate of true-positive pixels and broken vessels. The method of (^Kang et al., 2013)
based on the degree segmentation method shows low performance in angiograms with an uneven illumination and high rate of false-positive pixels in most of the angiograms. The proposed method presents
the highest average performance in the test set of angiograms. The method presents an appropriate rate of true-positive pixels, detecting vessels with different diameters and obtaining a low rate of
broken vessels and false-positive pixels.
Although the state-of-the-art vessel segmentation methods provide an appropriate performance based on the quantitative and qualitative analysis discussed above, the comparative analysis reveals that
the proposed method works well in nonuniform illuminated X-ray angiograms providing the highest average rate of vessel enhancement and segmentation. The segmentation results also show that the
proposed method is more accurate and robust than the comparative methods considering the vessels outlined by the specialist, which is suitable for computer-aided diagnosis systems.
In this paper, we introduce a novel coronary artery enhancement and segmentation method in X-ray angiographic images. In the first stage, a new multiscale top-hat operator based on the eigenvalues of
the Hessian matrix has demonstrated to be more efficient than three enhancement methods, achieving A[z] = 0.942 with a training set of 40 angiograms and A[z] = 0.965 with a test set of 40 angiograms.
In the second stage, a new thresholding strategy based on a multiobjective optimization approach has obtained the highest average rate performance (0.911) regarding seven thresholding methods.
Experimental results have proven that the proposed method consisting of multiscale top-hat operator and a thresholding strategy produces the highest performance rate (0.923) using the test set of
angiograms compared with five state-of-the-art methods for automatic vessel segmentation. In addition, the experimental results have also shown that, based on the angiograms hand-labeled by a
specialist, the performance in vessel pixel detection obtained from the proposed method is highly suitable for computer-aided diagnosis systems of cardiac abnormalities. | {"url":"https://www.scielo.org.mx/scielo.php?script=sci_arttext&pid=S2007-07052015000300297&lng=en&nrm=iso","timestamp":"2024-11-12T20:30:30Z","content_type":"application/xhtml+xml","content_length":"78604","record_id":"<urn:uuid:a2f1d57d-700a-4665-800b-2a2b163aa276>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00536.warc.gz"} |
Undergrad Not in Math
For my undergrad I double majored in Elementary Education and Mathematics. I am unsure how schools will look at this to determine if I can attend their school to get my Masters in Statistics.
Does anyone have any insight on this?
My GRE scores were okay.
Verbal: 160
Quant: 169
My overall GPA: 3.831 - magna cum
Please let me know what other things I can do to prove I would be a good fit as a Statistics major.
Thank you!
Re: Undergrad Not in Math
For some more information,
I have spent the past 2 years in Ecuador teaching Math in Spanish.
My letters of recommendation will be very strong. 2 from Math professors that I worked closely with at my University and 1 from a retired Reasoning Professor that has guided me through my teaching in | {"url":"https://mathematicsgre.com/viewtopic.php?f=1&t=1534","timestamp":"2024-11-13T19:20:50Z","content_type":"text/html","content_length":"18662","record_id":"<urn:uuid:6d03873a-5315-4627-8b4d-733ef58915b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00705.warc.gz"} |
Introduction to Quantum Cryptography
Quantum cryptography is a new method of cryptography that it’s based on quantum mechanics laws, pretending that uncertainty has the potency to weaken the ability of an intruder to hack or get into a
system by boosting the entropy.
Relying on the Heisenberg uncertainty principle, we assume that one cannot detect accurately anything depending on this function:
Heisenberg uncertainty principle
A developed quantum cryptosystem (a cryptosystem that will use the roles of quantum mechanics) will maintain a secure system.
So now, what is quantum cryptography in terms of quantum mechanics? It states that using the photons and their axiom quantum characteristics to establish a non-breakable cryptosystem due to the
condition that a quantum condition or state cannot be computed or measured without warning the system.
Photons are the core of quantum mechanics, and by using them you guarantee that you’re using the smallest known particles in the universe.
And as per Schrodinger, a quantum system will remain as it’s until it reacts or is manipulated with an external medium “considering the cat, the cat is dead and alive at the same time until
Schrodinger's cat theory
We can assume that Newton's first law “an object will remain to conduct its motions or in rest until it gets affected by an external action or force" so will have the same theory as Schrodinger's cat
Photons can be in different states at the same time, and as it gets measured they will change their state concurrently. And this statement above was believed as the core of quantum cryptography.
Upon exchanging a message between the sender and the receiver, it travels through a channel, which when intercepted by any malicious entity, exposes the change in the state of the photon, therefore
making it visible to the sender/receiver.
On the other hand, Quantum Entanglement can be used, when a system subsists of two photons, any change affecting the first one, leads to a change in the other, thereby making intruders’ actions in
networks detectable.
Quantum Key Distribution
QKD, is one of the quantum cryptography elements, considering that Quantum Computers are exploiting photons to broadcast data.
As we mentioned earlier, a photon is a smallest known particle, it has a lineament designated as a spin, which has four sorts; 45 ° and - 45 ° Diagonal, horizontal, and Vertical, which is in QC
illustrated as, binary 1 for horizontal & 45° diagonal, and 0 for vertical & -45° diagonal.
Uncertainty, A rubric states that adjusting the characteristics of a particle changes its state. Using that principle, an unknown connection becomes detectable by changing the state of a photon.
photon spins vectors
Device- Independent Cryptography
Another method of quantum encryption, it adjusts the genuine QKD so it is invulnerable and not touchable in a condition that unauthorized or obtrusive party devices are linked.
Bennett and Brassard proposed the BB84 which is the headmost quantum cryptography protocol in 1984.
Quantum cryptography isn’t used to convey message data, it is mainly used for the Implementation of, and this function is used with a specified encryption algorithm to (encrypt/decrypt) the message
designated to be sent and transmitted through a classical communication channel.
In quantum cryptography methods, Alice and Bob can detect if any intruders or a third party are accessing the secured channel with them or trying to obtain the key, and this is by implementing the
quantum super-positions and transmitting data in the quantum condition through a quantum channel.
BB84 was implemented in the physical layer as what was examined by Bennett and Brassard. It uses the conveyance of polarized photons which was discussed above in photon spins vectors. It needs to be
located on two different bases.
Shor's Factoring Algorithm
This algorithm factorizes N (non-prime integer), finding integers p1 and p2 so it will produce, P1P2=N. The parts of this algorithm are:
1. Preprocessing – preparing quantum registering by quantum parallelism and classical algorithms.
2. QFT (quantum Fourier transform) – applied on the output of the preprocessing
3. Post-processing and measurement using classical algorithms.
Difference between Classical Cryptography and Quantum cryptography
Classical cryptography is mainly based on mathematical algorithms and methods, it is durable by the point the private key is classified, and it is hard to use reverse engineering to calculate the
private key.
Mathematically it is not possible to get the private key by reverse calculations and this can be one of the largest vulnerabilities. In contrast, QKD is the procedure of distribution of Secret keys,
where security is properly established.
Quantum cryptography will accomplish having a secure environment, relies on itself, and also the information and communications are secured. It is accomplishing all the criteria by utilizing the laws
of Quantum Mechanics with a view to executing a cryptographic It was proved by researchers that public-key cryptosystems can be compromised by quantum computers.
Mathematical proofs can’t be enough to ensure the security of quantum cryptography protocols. Quantum cryptosystems are not easy to be designed and implemented, due to the accuracy needed to avoid
security flows.
it is fascinating how physics can be implemented in different aspects, quantum cryptography was a novel field in the conjunction between quantum mechanics (physics mainly) and Cryptography. It still
needs more implementation on post-quantum systems and while creating new cryptographic methods, it is important to focus on developing the current methods that are being accessed by hackers and
we have gone through quantum computing earlier where we discussed the basics of quantum physics and the revolutionary idea of quantum computing, also in that article we discussed the hello world
quantum program! quantum cryptography and quantum computing are very big subjects and it really needs more research to be done!
If you have a problem and no one else can help. Maybe you can hire the Kalvad-Team. | {"url":"https://blog.kalvad.com/introduction-to-quantum-cryptography/","timestamp":"2024-11-10T13:06:53Z","content_type":"text/html","content_length":"30735","record_id":"<urn:uuid:40c0fd65-d898-4d59-9215-50ead06a605d>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00385.warc.gz"} |
Water Bottles Leetcode Solution - TutorialCup
Problem statement
In the problem ” Water Bottles” we are given two values namely” numBottle” which will store the total number of full water bottles and “numExchange” which will store the total number of empty water
bottles we can exchange at a time and get a full water bottle.
After drinking water from a full water bottle it turns into an empty water bottle. Our task is to find out the maximum number of full water bottles we can drink.
numBottles = 15, numExchange = 4
First round: drink 15 bottles of water gives 15 empty bottles.
Second round: from these 15 water bottles we get 3 full water bottles and left with 3 empty bottles. Drink 3 water bottles we now left with a total of 6 empty bottles.
Third round: from these 6 water bottles we get 1 full water bottle and left with 2 empty bottles. Drink 1 water bottles we now left with a total of 3 empty bottles.
As a minimum of 4 bottles is required to exchange the bottle we can not buy a full water bottle anymore. So the maximum number of water bottles that we can drink is 15+3+1=19.
Approach for Water Bottles Leetcode Solution
The basic approach to solve the problem is to do what questions ask.
1. Drink all the full water bottles then it converts to empty water bottles.
2. From all the empty water bottles buy full water bottles.
3. Repeat these steps until we can not buy a full water bottle from an empty water bottle.
4. Return the total number of full water bottles that we drunk during the process.
We can improve the complexity of the solution by making a few observations:
1. We have numBottle number of the full water bottles so this will be the minimum number of full water bottles that we can drink.
2. 1 full water bottle = 1 unit water+1 empty water bottle.
3. From numExchange empty water bottles, we get 1 full water bottle(1 unit water + 1 empty water bottle). This thing can also be interpreted as (numExchange-1) water bottles give 1 unit of water.
4. But if in the last round if we have (numExchange-1) number of empty bottles then we can not get one unit of water.
5. So our result will be numBottle+(numBottle/(numExchange -1)) and if numBottle%(numExchange -1)==0 then subtract 1 from the final answer.
C++ code for Water Bottles
#include <bits/stdc++.h>
using namespace std;
int numWaterBottles(int numBottles, int numExchange) {
int ans= numBottles + (numBottles) / (numExchange - 1);
if((numBottles) %(numExchange - 1)==0)
return ans;
int main()
int numBottles = 15, numExchange = 4;
int ans=numWaterBottles(numBottles,numExchange);
return 0;
Java code for Water Bottles
import java.util.Arrays;
import java.util.Set ;
import java.util.HashSet;
public class Tutorialcup {
public static int numWaterBottles(int numBottles, int numExchange) {
int ans= numBottles + (numBottles) / (numExchange - 1);
if((numBottles) %(numExchange - 1)==0)
return ans;
public static void main(String[] args) {
int numBottles = 15, numExchange = 4;
int ans=numWaterBottles(numBottles,numExchange);
Complexity Analysis of Water Bottles Leetcode Solution
Time complexity
The time complexity of the above code is O(1).
Space complexity
The space complexity of the above code is O(1) because we are using only a variable to store answer. | {"url":"https://tutorialcup.com/leetcode-solutions/water-bottles-leetcode-solution.htm","timestamp":"2024-11-14T23:42:34Z","content_type":"text/html","content_length":"104854","record_id":"<urn:uuid:28547c7f-3699-4a18-a8cd-bbbcd966e704>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00437.warc.gz"} |
How many envelopes did they use ?Puzzles World: How many envelopes did they use ?
An office used three types of envelopes: small, medium and large.
They used 160 small envelopes.
The no. of medium envelopes used was 5 times the no. of small envelopes plus 5/20 of the total no. of the envelopes.
The no. of large envelopes was equal to the no. of small envelopes plus 5/20 of the no. of medium envelopes.
Altogether, How many envelopes did they use?
Answer is 1920
Solution :
Small- S, Medium- M, Large- L, Total- T
S= 160
M= 5*S+T*5/20= 800 + T/4
L= S+ M*5/20 = 160 +(800+T/4) *5/20= 360 + T/16
T= 160+ 800+ T/4 + 360 + T/16 => T=1920 | {"url":"https://www.puzzles-world.com/2019/03/how-many-envelopes-did-they-use.html","timestamp":"2024-11-14T20:43:02Z","content_type":"application/xhtml+xml","content_length":"53147","record_id":"<urn:uuid:9bab68e4-551e-473f-9c3d-b76e522c1ad9>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00366.warc.gz"} |
Geotechnical Engineering 3 (CIVE09016)
Undergraduate Course: Geotechnical Engineering 3 (CIVE09016)
Course Outline
School School of Engineering College College of Science and Engineering
level SCQF Level 9 (Year 3 Undergraduate) Availability Available to all students
year taken)
SCQF 20 ECTS Credits 10
Summary In this course, students develop further understanding of soil mechanical concepts and learn to apply them to solve geotechnical engineering problems. The course is a continuation of the
second year soil mechanics module and extends the student's understanding of the mechanics of soils to include consolidation and shear failure of soil systems.
L1 Course introduction
Aspects of geotechnical design, structure of the course, course content, references with comments, revision on effective stress concept.
L2 Stress distribution in soils 1
In-situ stresses (revision), stress history, lateral stress ratio, normal and over-consolidation, overconsolidation ratio, factors affecting the induced stresses due to applied loads.
L3 Stress distribution in soils 2
Flexible and rigid footing on cohesive and cohesionless soils, Boussinesq elastic solution of point load at the surface, worked example.
L4 Stress distribution in soils 3
Boussinesq elastic solutions of induced soil stresses due to uniform pressure on a circular area, rectangular area and infinite strip, worked example.
L5 Stress distribution in soils 4
Newmark chart, worked example, Westergaard theory, approximate method, bulb of pressure.
L6 Shear strength 1
Revision on total and effective stress Mohr¿s circles, Mohr-Coulomb failure criterion, experimental failure envelope, cohesion and internal friction angle.
L7 Shear strength 2
Other useful forms of the Mohr-Coulomb equation, worked example, stress parameters: (tau, sigma), (sigma_1, sigma_3), (t, s), (p, q) and applications.
L8 Shear strength 3 : laboratory measurement of strength
Direct shear: tester, testing procedure, advantages and limitations
Triaxial testing: tester, testing procedure, consolidation stage, volume measurement, pore water pressure measurement, triaxial compression and triaxial extension.
L9 Shear strength 4 : common types of triaxial testing
Unconsolidated-undrained (UU) test, unconfined compression test, consolidated-undrained test (CU), consolidated-drained test (CD).
L10 Shear strength 5 : triaxial test analysis I
Undrained shear strength parameters and effective shear strength parameters, analysis of triaxial test results, worked example.
L11 Shear strength 6 : triaxial test analysis II
Undrained shear strength parameters and effective shear strength parameters, analysis of triaxial test results, worked example.
L12 Shear strength 7
Mechanisms of shearing and straining.
L13 Shear strength 8: mechanical behaviour of sands
Response of deviator stress and volumetric strain under axial straining, critical void ratio, residual strength, angle of repose.
L14 Shear strength 9: mechanical behaviour of clays
Response in drained and undrained tests for normally consolidated and
overconsolidated clays, sensitivity of clays.
L15 Shear strength 1 : pore pressure parameters
Skempton pwp parameters A and B, range of values of the parameters for different clays, worked example. Comments on laboratory sessions 1 and 2 on shear strength measurements.
L16 Strength strength 11: discussion and concluding remarks
Including comments on Laboratory Sessions 1 & 2 on the triaxial test results.
L17 Consolidation and settlement 1
Consolidation vs compression, important questions: magnitude and rate of settlement, piston-spring analogy, oedometer test, undisturbed sample.
L18 Consolidation and settlement 2
Useful parameters from oedometer test: coefficient of compressibility av, coefficient of volume compressibility mv, compression and swelling indices Cc and Cs, empirical relation on Cc,
worked example.
L19 Consolidation and settlement 3
Preconsolidation pressure, causes of overconsolidation, graphical procedure for determining preconsolidation pressure.
L20 Consolidation and settlement 4
Terzaghi theory of one-dimensional consolidation, hydrostatic and excess pore water pressure (pwp), isochrones of excess pwp at various stages, assumptions, derivation.
L21 Consolidation and settlement 5
Solutions to the consolidation differential equation, boundary conditions and initial conditions, average degree of consolidation for a clay stratum, local degree of consolidation, worked
L22 Consolidation and settlement 6
Comparison of experimental and theoretical consolidation curves, determination of coefficient of consolidation cv from oedometer test: square-root time method.
L23 Consolidation and settlement 7
Determination of coefficient of consolidation cv from oedometer test: log-time method, determination of permeability k from oedometer test results.
L24 Consolidation and settlement 8: discussion and concluding remarks
Including comments on Laboratory Session 3 on the oedometer test simulation exercise.
L25 Lateral pressures and retaining structures 1
Pressures at Ko and limiting equilibrium states, assumptions in earth pressure theory, Rankine theory of active and passive earth pressures.
L26 Lateral pressures and retaining structures 2
Direction of failure planes, surcharge on backfill, deformations to mobilise active and passive states, active and passive pressure distribution.
L27 Lateral pressures and retaining structures 3
Course Calculation of wall loads, total and effective stress analysis, worked example.
description L28 Lateral pressures and retaining structures 4
Coulomb theory of earth pressures, types and designs of retaining wall.
L29 Lateral pressures and retaining structures 5
Further work example. Coulomb theory of earth pressures.
L30 Lateral pressures and retaining structures 6
Types and designs of retaining wall.
L31 Bearing capacity of shallow foundations 1
Adequate factor of safety against shear failure in shallow foundations, loading
response of shallow foundations, bearing capacity theories and equation, strip, square and circular footings.
L32 Bearing capacity of shallow foundations 2
Bearing capacity terms and definitions, footings on clays, footings on sands. Work example.
L33 Bearing capacity of shallow foundations 3
Eccentric and inclined loads, worked example.
L34 Bearing capacity of shallow foundations 4
Further worked example and concluding remarks.
L35 Slope stability 1
Introduction to slope stability. Cuttings and embankments. Translational sliding. Worked example.
L36 Slope stability 2
Infinite slope and frictional material. ¿¿u = analysis. Taylor¿¿s curves.
L37 Slope stability 3
Bishop¿s method of slices, interslice forces, concluding remarks.
L38 Summary and concluding remarks
Recap on the main aspects of foundation design and the topics covered in this course, total stress (undrained) analysis vs effective stress (drained) analysis.
The aim is give the students ample opportunities to develop skills to apply the theories and methods learned in the course to common geotechnical engineering situations. The excercises
cover a great variety of geotechnical problems in varying degrees of difficulty.
Tutorial Exercise 1 Stress distributions in soils
This tutorial is intended for students to develop skills in calculating the induced stresses in soils due to applied loadings for a variety of situations, including footings of different
shapes embedded at different depths.
Tutorial Exercise 2 Shear strength
This tutorial is intended for students to develop skills in solving geotechnical problems which require shear strength calculations, including drained and undrained failure properties,
analysis of laboratory test results to evaluate the failure stresses and failure properties, total and effective stress paths and calculations involving pwp parameters.
Tutorial Exercise 3 Consolidation and settlement
This tutorial is intended for developing skills in performing consolidation settlement analyses. The magnitude of settlement and the time taken to reach a certain degree of consolidation
are being evaluated for many geotechnical situations.
Tutorial Exercise 4 Lateral earth pressures
This tutorial is intended for developing skills in solving geotechnical problems which require lateral earth pressure calculations.
Tutorial Exercise 5 Bearing capacity in shallow foundations
This tutorial is intended for developing skills in performing bearing capacity calculations for various types of footing.
Tutorial Exercise 6 Slope stability
This tutorial is intended for developing skills in performing slope stability analyses including use of Taylor¿s curves and Bishop¿s method of slices.
Laboratory classes are undertaken in the Soil Mechanics Laboratories. The students work in groups under the supervision of a laboratory demonstrator. The aim is to train the students to
carry out laboratory triaxial tests, including the analysis of the test results and the evaluation of the relevant properties. A computer simulation of oedometer test and analysis of the
test data are also performed.
Laboratory Session 1 UU testing of clay
The undrained shear strength characteristics of a clay are investigated by UU tests. Each group of students is required to carry out sample preparation and testing in a triaxial machine.
Sample data and testing data are to be recorded in the data sheets. A brief report is to be submitted for assessment.
Laboratory Session 2 Consolidated-undrained testing of clay
The total and effective stress shear strength characteristics of a clay are investigated by CU tests. Every group of students is required to carry out sample preparation and testing in a
triaxial machine. Sample data and testing data are to be recorded in the data sheets. Each group is to carry out one CU test and the results from all groups are then pooled and analysed.
A full report is to be submitted for assessment.
Laboratory Session 3 Oedometer testing of clay
The consolidation and compression properties of a cohesive soil are investigated using an oedometer test. The students are first shown how to perform an oedometer test. They then use a
computer simulation program to simulate the test. The simulated data are then analysed and the relevant properties deduced. The completed data sheet and graphs are to be handed in for
Entry Requirements (not applicable to Visiting Students)
Pre-requisites Students MUST have passed: Soil Mechanics 2 (CIVE08019) Co-requisites
Prohibited Combinations Other requirements None
Information for Visiting Students
Pre-requisites 2nd year undergraduate Soil Mechanics/Geomechanics/Geotechnical Engineering or similar
High Demand Course? Yes
Course Delivery Information
Academic year 2016/17, Available to all students (SV1) Quota: None
Course Start Semester 1
Timetable Timetable
Total Hours: 200 ( Lecture Hours 38, Seminar/Tutorial Hours 9,
Supervised Practical/Workshop/Studio Hours 5, Formative Assessment
Learning and Teaching activities (Further Info) Hours 1, Summative Assessment Hours 7, Programme Level Learning and
Teaching Hours 4, Directed Learning and Independent Learning Hours
136 )
Assessment (Further Info) Written Exam 75 %, Coursework 25 %, Practical Exam 0 %
Additional Information (Assessment) Intermittent Assessment: 25%
Degree Examination: 75%
Opportunities in lectures, laboratories and tutorial sessions for
Feedback direct feedback; feedback on each coursework submission;
Start-stop-continue in Week 4
Exam Information
Exam Diet Paper Name Hours & Minutes
Main Exam Diet S1 (December) Geotechnical 2:00
Engineering 3
Resit Exam Diet (August) 2:00
Learning Outcomes
On completion of this course, the student will be able to:
1. demonstrate ability to carry out consolidation
settlement analyses for a variety of geotechnical
2. demonstrate ability to carry out analyses of the
failure of soil systems for a variety of geotechnical
situations: footings, shallow foundations, retaining
walls, cuttings and embankments
3. demonstrate ability to describe the mechanical
behaviour of sands and clays
4. demonstrate ability to evaluate the consolidation and
failure properties of soils from laboratory testing
Reading List
Recommended reading:
J.A. Knappett and R.F. Craig, Craig's Soil Mechanics, Spon
Press, 2012.
T.W. Lambe and R.V. Whitman, Soil Mechanics, Wiley, SI
version, 1979.
Background reading:
G.E. Barnes, Soil Mechanics: Principles and Practice,
Macmillan, 2010.
M. Bolton, A Guide to Soil Mechanics, M.D.& K.Bolton,
W. Powrie, Soil Mechanics: Concepts and Applications, CRC
Press, 2013.
M. J. Tomlinson, Foundation Design and Construction,
Prentice Hall, 7th Ed., 2001.
Additional Information
Course URL http://webdb.ucs.ed.ac.uk/see/VLE/
Graduate Attributes Not entered
and Skills
Keywords Not entered
Prof Jin Ooi Mrs Lynn Hughieson
Course Tel: (0131 6) Course Tel: (0131 6)50 5687
organiser 50 5725 secretary Email:
Email: Lynn.Hughieson@ed.ac.uk
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Place Value: Expanded Form
• What number is represented below?
• Why would you write it this way?
400,000 + 80,000 + 8,000 + 60 + 3 + 0.4 + 0.02
• Were you able to determine what number was represented above?
You may have been able to figure it out just by looking at it, or you may have had to add the numbers together on a piece of paper. Either way, you should have said this is equal to 488,063.42.
In the previous Related Lessons, found in the right-hand sidebar, you have learned how to read, write, and compare numbers expressed in standard and written form.
In this lesson, you will learn to read and write numbers expressed in expanded form. The number at the beginning of the lesson is written in expanded form.
• What do you notice about the number written in expanded form?
Expanded form separates a number so each digit and its place value is displayed. Addition signs are placed between each part of the number, because if you add all the parts together, you will find
the total value of the number.
Expanded form can be written one of two different ways. Each of the following examples shows the same number written in expanded form.
Compare and contrast the numbers. Then, determine what number the examples represent:
• 40,000 + 9,000 + 20 + 8 + 0.03 + 0.003
• (4 x 10,000) + (9 x 1,000) + (2 x 10) + (8 x 1) + (3 x 0.01) + (3 x 0.001)
• Did you say each of the examples represents 49,028.033?
• Which of the examples did you find easier to read?
When you are asked to write a number in expanded form, you can choose which method you prefer, but it is important to be able to read numbers using either method.
Expanded form is the form of writing a number that you are least likely to see in the real world, but it is still important to understand expanded form, because it teaches you how to break a number
into parts.
When you understand all this, move on to the Got It? section to practice writing numbers in expanded form. | {"url":"https://www.elephango.com/index.cfm/pg/k12learning/lcid/11807/Place_Value:_Expanded_Form","timestamp":"2024-11-14T11:31:34Z","content_type":"text/html","content_length":"72632","record_id":"<urn:uuid:3f1ef4fd-441c-4805-9faa-a817c9c83264>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00701.warc.gz"} |
Visualizing the gradient descent method
Posted by: christian on 5 Jun 2016
In the gradient descent method of optimization, a hypothesis function, $h_\boldsymbol{\theta}(x)$, is fitted to a data set, $(x^{(i)}, y^{(i)})$ ($i=1,2,\cdots,m$) by minimizing an associated cost
function, $J(\boldsymbol{\theta})$ in terms of the parameters $\boldsymbol\theta = \theta_0, \theta_1, \cdots$. The cost function describes how closely the hypothesis fits the data for a given choice
of $\boldsymbol \theta$.
For example, one might wish to fit a given data set to a straight line, $$ h_\boldsymbol{\theta}(x) = \theta_0 + \theta_1 x. $$ An appropriate cost function might be the sum of the squared difference
between the data and the hypothesis: $$ J(\boldsymbol{\theta}) = \frac{1}{2m} \sum_i^{m} \left[h_\theta(x^{(i)}) - y^{(i)}\right]^2. $$ The gradient descent method starts with a set of initial
parameter values of $\boldsymbol\theta$ (say, $\theta_0 = 0, \theta_1 = 0$), and then follows an iterative procedure, changing the values of $\theta_j$ so that $J(\boldsymbol{\theta})$ decreases: $$
\theta_j \rightarrow \theta_j - \alpha \frac{\partial}{\partial \theta_j}J(\boldsymbol{\theta}). $$
To simplify things, consider fitting a data set to a straight line through the origin: $h_\theta(x) = \theta_1 x$. In this one-dimensional problem, we can plot a simple graph for $J(\theta_1)$ and
follow the iterative procedure which trys to converge on its minimum.
Fitting a general straight line to a data set requires two parameters, and so $J(\theta_0, \theta_1)$ can be visualized as a contour plot. The same iterative procedure over these two parameters can
also be followed as points on this plot.
Here's the code for the one-parameter plot:
import numpy as np
import matplotlib.pyplot as plt
# The data to fit
m = 20
theta1_true = 0.5
x = np.linspace(-1,1,m)
y = theta1_true * x
# The plot: LHS is the data, RHS will be the cost function.
fig, ax = plt.subplots(nrows=1, ncols=2, figsize=(10,6.15))
ax[0].scatter(x, y, marker='x', s=40, color='k')
def cost_func(theta1):
"""The cost function, J(theta1) describing the goodness of fit."""
theta1 = np.atleast_2d(np.asarray(theta1))
return np.average((y-hypothesis(x, theta1))**2, axis=1)/2
def hypothesis(x, theta1):
"""Our "hypothesis function", a straight line through the origin."""
return theta1*x
# First construct a grid of theta1 parameter pairs and their corresponding
# cost function values.
theta1_grid = np.linspace(-0.2,1,50)
J_grid = cost_func(theta1_grid[:,np.newaxis])
# The cost function as a function of its single parameter, theta1.
ax[1].plot(theta1_grid, J_grid, 'k')
# Take N steps with learning rate alpha down the steepest gradient,
# starting at theta1 = 0.
N = 5
alpha = 1
theta1 = [0]
J = [cost_func(theta1[0])[0]]
for j in range(N-1):
last_theta1 = theta1[-1]
this_theta1 = last_theta1 - alpha / m * np.sum(
(hypothesis(x, last_theta1) - y) * x)
# Annotate the cost function plot with coloured points indicating the
# parameters chosen and red arrows indicating the steps down the gradient.
# Also plot the fit function on the LHS data plot in a matching colour.
colors = ['b', 'g', 'm', 'c', 'orange']
ax[0].plot(x, hypothesis(x, theta1[0]), color=colors[0], lw=2,
label=r'$\theta_1 = {:.3f}$'.format(theta1[0]))
for j in range(1,N):
ax[1].annotate('', xy=(theta1[j], J[j]), xytext=(theta1[j-1], J[j-1]),
arrowprops={'arrowstyle': '->', 'color': 'r', 'lw': 1},
va='center', ha='center')
ax[0].plot(x, hypothesis(x, theta1[j]), color=colors[j], lw=2,
label=r'$\theta_1 = {:.3f}$'.format(theta1[j]))
# Labels, titles and a legend.
ax[1].scatter(theta1, J, c=colors, s=40, lw=0)
ax[1].set_title('Cost function')
ax[0].set_title('Data and fit')
ax[0].legend(loc='upper left', fontsize='small')
The following program produces the two-dimensional plot:
import numpy as np
import matplotlib.pyplot as plt
# The data to fit
m = 20
theta0_true = 2
theta1_true = 0.5
x = np.linspace(-1,1,m)
y = theta0_true + theta1_true * x
# The plot: LHS is the data, RHS will be the cost function.
fig, ax = plt.subplots(nrows=1, ncols=2, figsize=(10,6.15))
ax[0].scatter(x, y, marker='x', s=40, color='k')
def cost_func(theta0, theta1):
"""The cost function, J(theta0, theta1) describing the goodness of fit."""
theta0 = np.atleast_3d(np.asarray(theta0))
theta1 = np.atleast_3d(np.asarray(theta1))
return np.average((y-hypothesis(x, theta0, theta1))**2, axis=2)/2
def hypothesis(x, theta0, theta1):
"""Our "hypothesis function", a straight line."""
return theta0 + theta1*x
# First construct a grid of (theta0, theta1) parameter pairs and their
# corresponding cost function values.
theta0_grid = np.linspace(-1,4,101)
theta1_grid = np.linspace(-5,5,101)
J_grid = cost_func(theta0_grid[np.newaxis,:,np.newaxis],
# A labeled contour plot for the RHS cost function
X, Y = np.meshgrid(theta0_grid, theta1_grid)
contours = ax[1].contour(X, Y, J_grid, 30)
# The target parameter values indicated on the cost function contour plot
ax[1].scatter([theta0_true]*2,[theta1_true]*2,s=[50,10], color=['k','w'])
# Take N steps with learning rate alpha down the steepest gradient,
# starting at (theta0, theta1) = (0, 0).
N = 5
alpha = 0.7
theta = [np.array((0,0))]
J = [cost_func(*theta[0])[0]]
for j in range(N-1):
last_theta = theta[-1]
this_theta = np.empty((2,))
this_theta[0] = last_theta[0] - alpha / m * np.sum(
(hypothesis(x, *last_theta) - y))
this_theta[1] = last_theta[1] - alpha / m * np.sum(
(hypothesis(x, *last_theta) - y) * x)
# Annotate the cost function plot with coloured points indicating the
# parameters chosen and red arrows indicating the steps down the gradient.
# Also plot the fit function on the LHS data plot in a matching colour.
colors = ['b', 'g', 'm', 'c', 'orange']
ax[0].plot(x, hypothesis(x, *theta[0]), color=colors[0], lw=2,
label=r'$\theta_0 = {:.3f}, \theta_1 = {:.3f}$'.format(*theta[0]))
for j in range(1,N):
ax[1].annotate('', xy=theta[j], xytext=theta[j-1],
arrowprops={'arrowstyle': '->', 'color': 'r', 'lw': 1},
va='center', ha='center')
ax[0].plot(x, hypothesis(x, *theta[j]), color=colors[j], lw=2,
label=r'$\theta_0 = {:.3f}, \theta_1 = {:.3f}$'.format(*theta[j]))
ax[1].scatter(*zip(*theta), c=colors, s=40, lw=0)
# Labels, titles and a legend.
ax[1].set_title('Cost function')
ax[0].set_title('Data and fit')
axbox = ax[0].get_position()
# Position the legend by hand so that it doesn't cover up any of the lines.
ax[0].legend(loc=(axbox.x0+0.5*axbox.width, axbox.y0+0.1*axbox.height),
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What is 1 Percent of 10,000?
What is 1 Percent of 10,000?
1 percent of 10,000 is 100.
A percent is a number depicted as a fraction of 100. It is often seen in an equation or in the world with a % symbol. For instance, if a store is having a 20% off sale, you can use the math to find
out how much that jacket will cost.
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How to calculate risk in forex trading
Your risk (50 pips) for a reward (100 pips) would equal: 1:2 risk reward ratio. Also in real trading, you need to consider the spread charged by your Forex broker to conduct the risk and reward
analysis effectively.
Aug 31, 2018 Forex Risk Management is the #1 trading skill to master. you can take the other currency in the pair, and measure it against other currencies. Aug 25, 2011 Risk reward does not mean
simply calculating the risk and reward on a trade, it means understanding that by achieving 2 to 3 times risk or more In Forex trading, the position size is determined by the amount of “Lots” that
you trade. General formula: (Risk per trade) / (Stop loss in pips) = mini lots. May 23, 2019 03 - Determine the Forex Position Size. Ideal forex position size is a simple mathematical formula equal
to: Pips at Risk X Pip Value X Lots traded Improve your money management by calculating position size from your risk loss Position size calculation is also a first step to the organized Forex
trading, Dec 19, 2019 My Early Days of Trading >; The Risk to Reward Ratio >; So, what is this RRR? > Applying RRR >; Calculating RRR in your trading Risk aversion refers to when traders unload
their positions in higher-yielding assets and move their funds in favor of safe-haven currencies. This normally
Usually, the stoploss is determined by your trading method and then you need calculate the maximum lot size of the trade that you can risk. If the smallest amount
Aug 25, 2011 Risk reward does not mean simply calculating the risk and reward on a trade, it means understanding that by achieving 2 to 3 times risk or more In Forex trading, the position size is
determined by the amount of “Lots” that you trade. General formula: (Risk per trade) / (Stop loss in pips) = mini lots. May 23, 2019 03 - Determine the Forex Position Size. Ideal forex position size
is a simple mathematical formula equal to: Pips at Risk X Pip Value X Lots traded Improve your money management by calculating position size from your risk loss Position size calculation is also a
first step to the organized Forex trading, Dec 19, 2019 My Early Days of Trading >; The Risk to Reward Ratio >; So, what is this RRR? > Applying RRR >; Calculating RRR in your trading Risk aversion
refers to when traders unload their positions in higher-yielding assets and move their funds in favor of safe-haven currencies. This normally May 10, 2019 Learn how to calculate pips when trading
forex. Pip values give you a useful sense of the risk involved and margin required per pip when
Therefore, our trader will look to risk approximately $3,000 on each trade. Next, we have to calculate the amount of risk per lot for each trade. This can be quickly determined by drawing the value
calculator (located on the left sidebar of DealBook 360) from the entry to the stop.
Mar 26, 2019 When you're an individual trader in the stock market, one of the few safety devices you have is the risk/reward calculation. Risk vs. Reward. Sadly,
Your risk (50 pips) for a reward (100 pips) would equal: 1:2 risk reward ratio. Also in real trading, you need to consider the spread charged by your Forex broker to conduct the risk and reward
analysis effectively.
Improve your money management by calculating position size from your risk loss Position size calculation is also a first step to the organized Forex trading, Dec 19, 2019 My Early Days of Trading >;
The Risk to Reward Ratio >; So, what is this RRR? > Applying RRR >; Calculating RRR in your trading Risk aversion refers to when traders unload their positions in higher-yielding assets and move
their funds in favor of safe-haven currencies. This normally May 10, 2019 Learn how to calculate pips when trading forex. Pip values give you a useful sense of the risk involved and margin required
per pip when The forex risk probability calculator(RPC) was designed to work hand in hand with Fibonacci retracement levels. It is therefore important that you have some Nov 6, 2016 To perform a
risk-reward ratio calculation in its most simple sense for a particular forex trade, you would just calculate the number of pips from
The forex risk probability calculator(RPC) was designed to work hand in hand with Fibonacci retracement levels. It is therefore important that you have some
Sep 24, 2019 Here is a simple calculation of risk to your account in USD value. Calculating Equity Risk. Calculating Risk in % of Account Equity. And as a % of
Calculating risk in Forex trading. Risk is calculated as a percentage of the total amount of the deposit and depends on the trading style. Toggle navigation. MetaTrader 4 for Android – Be Always in
Trading “I cannot teach anybody anything. I can only make them think” ― Socrates. Forex risk management — position sizing calculators. To make your life easier, you can use one of these calculators
below: MyFxBook – Position sizing calculator for forex traders.. Daniels Trading – Position sizing calculator for futures traders.. Investment U – Position sizing calculator for stock and options
traders.. The secret to finding low risk and high reward trades When you are starting to get into Forex there are some a couple areas you need to pay big attention to one is risk management and the
other is risk to reward ratio which also falls under risk management. If you are making trades and winning 9 out of 10 this isn’t as much of […] The rest of this article describes using simple VAR
for risk analysis. For more details on stop loss settings see here. Simple VAR for One Currency Pair. A basic VAR estimate is done as follows. Let’s say I’ve done a spot trade in 1000 EUR/USD and the
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NIPS 2016
Mon Dec 5th through Sun the 11th, 2016 at Centre Convencions Internacional Barcelona
Reviewer 1
The paper introduces group safe and feature safe screening rules to rule out irrelevant feature (groups) in sparse group lasso, along with computation of radius and center of the resulting safe
sphere. The latter involves obtaining a sequence of dual feasible points. The resulting spheres are shown to be converging safe regions and that optimal active sets can be identified in finite time.
A connection is established with epsilon norms that allows for fast computation of the sparse group lasso dual norm. The approach is demonstrated on a wide range on simulated and climatology data.
Qualitative Assessment
+ The paper deal with a very relevant topic and is a pleasure to read. The exposition is clear and intuitive. The results are well discussed and convincingly demonstrated. + The second contribution
of the paper, i.e., the connection to epsilon norms and its impact on computation complexity of the sparse group lasso dual norm is the most exciting part of the work. - The first contribution of the
paper, namely the proposal and characterization of safe rules uses the same machinery as previous work: GAP Safe screening rules for sparse multi-task and multi-class models, NIPS 2015, e.g. radius
computation and center computation are similar. This somewhat reduces the novelty of the present work.
Confidence in this Review
2-Confident (read it all; understood it all reasonably well)
Reviewer 2
This paper derives screening rules that allow to set certain variables to zero when fitting a sparse-group lasso model. The main technical contribution is a fast method for evaluating the dual norm
of the regularizer. In addition, the authors describe an implementation based on coordinate descent and provide some numerical simulations.
Qualitative Assessment
This work provides a technique for accelerating the fit of a sparse-group lasso model, which relies on a novel method for evaluating the dual norm of the regularizer. This is an interesting
contribution: speeding up the fit of structured sparse models is an important subject, since applying such models to large datasets is often challenging due to computational constraints. The paper
motivates and explains the method clearly, provides thorough references and includes interesting numerical simulations, where the method is compared to other approaches that have been adapted to the
sparse-group lasso by the authors. The only caveat is that the potential impact may be somewhat restricted, as it focuses exclusively on a rather specific sparse model.
Confidence in this Review
2-Confident (read it all; understood it all reasonably well)
Reviewer 3
This paper presents new safe screening rules for the Sparse-Group Lasso and show a new characterisation of the dual feasible set. Using this they define an efficient algorithm to implement the new
screening method using either a sequential or dynamic screening setup. The authors leverage the block-coordinate iterative soft thresholding algorithm (ISTA-BC) to demonstrate the efficiency of their
rules in various numerical experiments.
Qualitative Assessment
Safe screening rules are an of key importance in large-scale learning problems and the authors are working on an important problem. My main concern with the paper is that the bottom two plots in
Figure 2a both rely on sequential safe rules in some way, while the other methods above do not leverage the sequential aspect. I am concerned that this hides the true performance differences with the
top three methods. In particular, no dynamic version of the safe rules is explicitly tested. Similarly, I feel this complicates interpretation of Figures 2b and 3b and weakens the experimental
Confidence in this Review
1-Less confident (might not have understood significant parts)
Reviewer 4
This paper proposed a screening method for regression with group structures. The group level and feature level safe screening rules are introduced and analyzed for theoretical guarantees. It borrows
the idea of the SAFE rules for the Lasso. In addition to the theory, the numerical study also support the use of this new screening rule because of the better screening and shorter computation time.
Qualitative Assessment
The proposed method borrows the idea of the SAFE screening rules for the Lasso. I wonder if the authors have compared such SAFE screening methods with sure screening methods (see, for example, Fan
and Lv, 2008). $\tau$ seems to be a tuning parameter that balances the variable sparsity and group sparsity. It would be better if the authors explain how to choose a proper $\tau$ in practice and
which value is used in the simulation study. In real-data analysis, cross-validation is used to determine $\tau$. Could the authors provide more details to help understand the use of $\tau$ and how
sensitive it is to the final estimates. In the simulation study, the authors reported the proportion of active variables. It might be more clear if the authors also report the number of variables
eliminated. If the numbers of variables eliminated are more or less equal, the reported proportion makes more sense for the better screening.
Confidence in this Review
2-Confident (read it all; understood it all reasonably well)
Reviewer 5
This paper proposed the GAP safe screening rules for the Sparse-Group Lasso. To apply the safe rule, the authors give an algorithm to evaluate the dual norm, which they claimed is efficiently.
Qualitative Assessment
Generally speaking, the paper is well written. The general context of the paper is clear. However, there are too many symbols in the Noation part, and all of them are inline formula, which make the
paper a little difficult to read. I suggest, for the ease of reading, move some of the formulas to the first place where they appear. For example, move the notations in line 68-70 to line 76. It
seems weired to discuss the choice of $\tau$ without the formula of $\Omega_{\tau, w}$ explicitly. In line 24-25, [24] is publised before [11], why the authors said that the work of [24] is following
[11]. In line 63, “... the subdifferential ... of the $ \ell_1$_1 norm is $ \sign(\cdot)$”. what is the definition of $ \sign(\cdot)$ in this paper? The subdifferential of the $ \ell_1$_1 norm at 0
the the region $ [-1,1]$. It should be different from the definition of $ \sign(\cdot)$ at 0. In line 160, $\epsilon$-norm is definded as the "unique" nonnegative solution explicitly of the equation.
It seems not true. For example, when $\epsilon = 0$, any value which is larger than $\max|x_i|$ would be the solution of the equation. The author should give a strict definition of $\epsilon$-norm.
Confidence in this Review
1-Less confident (might not have understood significant parts) | {"url":"https://papers.nips.cc/paper_files/paper/2016/file/555d6702c950ecb729a966504af0a635-Reviews.html","timestamp":"2024-11-02T12:43:36Z","content_type":"text/html","content_length":"8262","record_id":"<urn:uuid:9f0cdd89-ee6b-414e-99bb-5402668cd70d>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00746.warc.gz"} |
Optimality - Explore the Science & Experts | ideXlab
The Experts below are selected from a list of 360 Experts worldwide ranked by ideXlab platform
Mio Murao - One of the best experts on this subject based on the ideXlab platform.
• Physical Review A, 2012
Co-Authors: Takanori Sugiyama, Peter S Turner, Mio Murao
We consider 1-qubit mixed quantum state estimation by adaptively updating measurements according to previously obtained outcomes and measurement settings. Updates are determined by the
average-variance--Optimality (A-Optimality) criterion, known in the classical theory of experimental design and applied here to quantum state estimation. In general, A optimization is a nonlinear
minimization problem; however, we find an analytic solution for 1-qubit state estimation using projective measurements, reducing computational effort. We compare numerically the performances of
two adaptive and two nonadaptive schemes for finite data sets and show that the A-Optimality criterion gives more precise estimates than standard quantum tomography.
• Physical Review A, 2012
Co-Authors: Takanori Sugiyama, Peter S Turner, Mio Murao
We consider 1-qubit mixed quantum state estimation by adaptively updating measurements according to previously obtained outcomes and measurement settings. Updates are determined by the
average-variance-Optimality (A-Optimality) criterion, known in the classical theory of experimental design and applied here to quantum state estimation. In general, A-optimization is a nonlinear
minimization problem; however, we find an analytic solution for 1-qubit state estimation using projective measurements, reducing computational effort. We compare numerically two adaptive and two
nonadaptive schemes for finite data sets and show that the A-Optimality criterion gives more precise estimates than standard quantum tomography.
Kensuke Tanaka - One of the best experts on this subject based on the ideXlab platform.
• Journal of Mathematical Analysis and Applications, 1999
Co-Authors: Hang-chin Lai, J. C. Liu, Kensuke Tanaka
We establish the necessary and sufficient Optimality conditions for a class of nondifferentiable minimax fractional programming problems solving generalized convex functions. Subsequently, we
apply the Optimality conditions to formulate one parametric dual problem and we prove weak duality, strong duality, and strict converse duality theorems.
Tchemisova T. - One of the best experts on this subject based on the ideXlab platform.
• 'Springer Science and Business Media LLC', 1
Co-Authors: Kostyukova O., Tchemisova T., Yermalinskaya S. A.
We state a new implicit Optimality criterion for convex semi-infinite programming (SIP) problems. This criterion does not require any constraint qualification and is based on concepts of immobile
index and immobility order. Given a convex SIP problem with a continuum of constraints, we use an information about its immobile indices to construct a nonlinear programming (NLP) problem of a
special form. We prove that a feasible point of the original infinite SIP problem is optimal if and only if it is optimal in the corresponding finite NLP problem. This fact allows us to obtain
new efficient Optimality conditions for convex SIP problems using known results of the Optimality theory of NLP. To construct the NLP problem, we use the DIO algorithm. A comparison of the
Optimality conditions obtained in the paper with known results is provided
• 'Springer Science and Business Media LLC', 1
Co-Authors: Kostyukova O., Tchemisova T., Yermalinskaya S.
We consider convex problems of semi-infinite programming (SIP) using an approach based on the implicit Optimality criterion. This criterion allows one to replace Optimality conditions for a
feasible solution x0 of the convex SIP problem by such conditions for x0 in some nonlinear programming (NLP) problem denoted by NLP(I(x0)). This nonlinear problem, constructed on the base of
special characteristics of the original SIP problem, so-called immobile indices and their immobility orders, has a special structure and a diversity of important properties. We study these
properties and use them to obtain efficient explicit Optimality conditions for the problem NLP(I(x0)). Application of these conditions, together with the implicit Optimality criterion, gives new
efficient Optimality conditions for convex SIP problems. Special attention is paid to SIP problems whose constraints do not satisfy the Slater condition and to problems with analytic constraint
functions for which we obtain Optimality conditions in the form of a criterion. Comparison with some known Optimality conditions for convex SIP is provided
• 'Elsevier BV', 1
Co-Authors: Tchemisova T.
Optimality conditions for nonlinear problems with equality and inequality constraints are considered. In the case when no constraint qualification (or regularity) is assumed, the Lagrange
multiplier corresponding to the objective function can vanish in first order necessary Optimality conditions given by Fritz John and the corresponding extremum is called abnormal. In the paper we
consider second order sufficient Optimality conditions that guarantee the rigidity of abnormal extrema (i.e. their isolatedness in the admissible sets)
• 'Informa UK Limited', 1
Co-Authors: Kostyukova O., Tchemisova T.
We consider a convex semi-infinite programming (SIP) problem whose objective and constraint functions are convex w.r.t. a finite-dimensional variable x and whose constraint function also depends
on a so-called index variable that ranges over a compact set inR. In our previous paper [O.I.Kostyukova,T.V. Tchemisova, and S.A.Yermalinskaya, On the algorithm of determination of immobile
indices for convex SIP problems, IJAMAS Int. J. Math. Stat. 13(J08) (2008), pp. 13–33], we have proved an implicit Optimality criterion that is based on concepts of immobile index and immobility
order. This criterion permitted us to replace the Optimality conditions for a feasible solution x0 in the convex SIP problem by similar conditions for x0 in certain finite nonlinear programming
problems under the assumption that the active index set is finite in the original semi-infinite problem. In the present paper, we generalize the implicit Optimality criterion for the case of an
infinite active index set and obtain newfirst- and second-order sufficient Optimality conditions for convex semi-infinite problems. The comparison with some other known Optimality conditions is
• 'Informa UK Limited', 1
Co-Authors: Tchemisova T., Olga Kostyukova
In the paper,we consider a problem of convex Semi-Infinite Programming with an infinite index set in the form of a convex polyhedron. In study of this problem, we apply the approach suggested in
our recent paper [Kostyukova OI, Tchemisova TV. Sufficient Optimality conditions for convex Semi Infinite Programming. Optim. Methods Softw. 2010;25:279–297], and based on the notions of immobile
indices and their immobility orders. The main result of the paper consists in explicit Optimality conditions that do not use constraint qualifications and have the form of criterion. The
comparison of the new Optimality conditions with other known results is provided
Takanori Sugiyama - One of the best experts on this subject based on the ideXlab platform.
• Physical Review A, 2012
Co-Authors: Takanori Sugiyama, Peter S Turner, Mio Murao
We consider 1-qubit mixed quantum state estimation by adaptively updating measurements according to previously obtained outcomes and measurement settings. Updates are determined by the
average-variance--Optimality (A-Optimality) criterion, known in the classical theory of experimental design and applied here to quantum state estimation. In general, A optimization is a nonlinear
minimization problem; however, we find an analytic solution for 1-qubit state estimation using projective measurements, reducing computational effort. We compare numerically the performances of
two adaptive and two nonadaptive schemes for finite data sets and show that the A-Optimality criterion gives more precise estimates than standard quantum tomography.
• Physical Review A, 2012
Co-Authors: Takanori Sugiyama, Peter S Turner, Mio Murao
We consider 1-qubit mixed quantum state estimation by adaptively updating measurements according to previously obtained outcomes and measurement settings. Updates are determined by the
average-variance-Optimality (A-Optimality) criterion, known in the classical theory of experimental design and applied here to quantum state estimation. In general, A-optimization is a nonlinear
minimization problem; however, we find an analytic solution for 1-qubit state estimation using projective measurements, reducing computational effort. We compare numerically two adaptive and two
nonadaptive schemes for finite data sets and show that the A-Optimality criterion gives more precise estimates than standard quantum tomography.
Hang-chin Lai - One of the best experts on this subject based on the ideXlab platform.
• Journal of Mathematical Analysis and Applications, 1999
Co-Authors: Hang-chin Lai, J. C. Liu, Kensuke Tanaka
We establish the necessary and sufficient Optimality conditions for a class of nondifferentiable minimax fractional programming problems solving generalized convex functions. Subsequently, we
apply the Optimality conditions to formulate one parametric dual problem and we prove weak duality, strong duality, and strict converse duality theorems. | {"url":"https://www.idexlab.com/openisme/topic-optimality/","timestamp":"2024-11-13T15:31:51Z","content_type":"text/html","content_length":"56244","record_id":"<urn:uuid:7137de49-386a-475b-b29c-c78cae32cb23>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00650.warc.gz"} |
Gray code - Computer Notes
The Gray code was designed by Frank Gray at Bell Labs in 1953. It belongs to a class of codes called the minimum change code. The successive coded characters never differ in more than one-bit. Owing
to this feature, the maximum error that can creep into a system using the binary gray code to encode data is much less than the worst -case error encountered in case of straight binary encoding.
The Gray code is an unweighted code. Because of this, the· gray code is not suitable for arithmetic operations but finds applications in input/output devices, some analog-to-digital converters and
designation of rows and columns in Karnaugh map etc.
The table below lists the gray code equivalents of the decimal number 0 – 15.
One can easily remember the gray codes. A three-bit gray code can be obtained by merely reflecting the two-bit code about an axis at the end of the code and assigning a third-bit as 0 above the axis
and as 1 below the axis. The reflected gray code is nothing but code written in reverse order. By reflecting three-bit code, a four-bit code may be obtained.
Now let us consider a few examples. The four-bit gray code for decimal number 39 is 00101101.
Similarly, gray code for (923.1)[10] and (327) is
(923.1)[10] = (1101 0011 0010.0001) [Gray code]
(327)[10] = (100011 0100) [Gray code] | {"url":"https://ecomputernotes.com/digital-electronics/binary/gray-code","timestamp":"2024-11-10T22:29:07Z","content_type":"text/html","content_length":"64466","record_id":"<urn:uuid:cbba3323-fadc-47fc-b7ea-de301309a10a>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00869.warc.gz"} |
Quantitative Analysis | Blablawriting.com
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Nocaf Drinks, Inc., a producer of decaffeinated coffee, bottles Nocaf. Each bottle should have a net weight of 4 ounces. The machine that fills the bottles with coffee is new, and the operation
manager wants to make sure that it is properly adjusted. The operation manager takes a sample of n = 8 bottles and records the average range in ounces for each sample. The data for several samples is
given in the table below. Is the machine properly adjusted and in control?
A 0.41 4.00
B 0.55 4.16
C 0.44 3.99
D 0.48 4.00
E 0.56 4.17
F 0.62 3.93
G 0.54 3.98
H 0.44 4.01
This scenario represents the problem where there is one population, some population statistics are unknown, we have collected sample statistics and we need to find that whether the sample statistics
represent the characteristics of population or not. Since the sample size is small (less than 30), we would use t-test with following assumption.
• Population is approximately normally distributed
• Population standard deviation is unknown
Null hypothesis is the hypothesis to be tested.
H[0]: µ = 4
Ha: µ ≠ 4 (two tail test)
(two tail test is taken because the weight less than four ounces and the weight more than 4 ounces, both are not acceptable in quality control)
Level of Significance:
α = 5% = 0.05
Critical Value where n = 8 ( df = n – 1 = 7) and α = 0.05
+1.895, -1.895
Decision Criteria:
Reject Ho if t-test statistics greater than 1.895 or less than -1.895
T-Test Statistic
For given data, sample mean = 4.03 and sample standard deviation = 0.087
Putting these values:
T = ( 4.03 – 4)
T = 0.03 / 0.031
T = 0.97
Since t-test statistic is within the non-rejectable region, (between +1.895 and -1.895), therefore we fail to reject the null hypothesis that is the weight of the new bottles is equal to 4 ounces.
Thus, at 5% level of significance we do not have sufficient evidence to warrant the rejection of the claim that the weight in new bottles is equal to 4 ounces.
Therefore, there is not significant difference in the actual and desired weight of the bottles.
The Flair Furniture company produces inexpensive tables and chairs. The production process for each is similar in that both require a certain number of hours of carpentry work and a certain number
of labor hours in the painting and varnishing department. Each table takes 4 hours of carpentry and 2 hours in the painting and varnishing shop. Each chair requires 3 hours in carpentry and 1 hour in
painting and varnishing. During the current production period, 240 hours of carpentry time are available and 100 hours in painting and varnishing tie are available. Each table sold yields a profit of
$7 and each chair produced is sold for $5 profit. Formulate a Linear programming problem to find the best possible combination of tables and chairs to be manufactured so that Flair Furniture makes
maximum profit.
The first step towards solving the liner programming problem is to understand the question. The second step is to determine the objective function.
Definition of the decision variables:
T = units of tables produced
C= units of chairs produced
Objective Function:
In the given scenario, the objective is to produce the tables and chairs in such a combination, that the profit out of these two productions would be maximum. Mathematically,
7T + 5C
The next step is to determine the constraints in the problem. Besides the non-negativity constraints, there are two other major constraints. First one is that of carpentry while the second one is
that of painting and varnishing. As defined in the question, the carpentry can take no more than 240 hours provided that each table required 4 hours of carpentry and each chair required 3 hours of
carpentry. Moreover, the painting and varnishing can take no more than 100 hours provided that each table required 2 hours of painting and varnishing and each chair required 1 hour of carpentry.
Carpentry Constraints:
4T + 3C ≤ 240
Painting and Varnishing Constraints
2T + 1C ≤ 100
T ≥ 0 (Non-Negativity Constraints)
C ≥ 0 (Non-Negativity Constraints)
The next step is to draw a line on the graph using each of the constraint. This can be done by putting 0 instead of each variable one by one, for each constraint.
In case of first constraint, when t = 0, c= 80
4(0) + 3 C = 240
3c = 240
C= 80
When C = 0, t = 60
4T + 3 (0) = 240
4T = 240
T = 60
Therefore, its ordered pair are (0, 80) and (60, 0)
In case of second constraint, when T = 0, C = 100
2(0) + C = 100
C = 100
When C = 0, t = 50
2T + 0 = 100
T = 50
Therefore, its ordered pair are (0,100) and (50, 0)
Putting all the abovementioned constraints on graph:
In order to determine the optimal solution, the corner point approach would be used.
In this method, each of the corner point’s value of T and C will be plugged in the objective function:
(0,0) Profit = 7(0) + 5(0) = 0$
(0,80) Profit = 7(0) + 5(80) = 400$
(50,0) Profit = 7(50) + 5(0) = 350$
(30,40) Profit = 7(30) + 5(40) = 410 $*
Since the profit is maximum at 30 unit of Table and 40 units of chairs (410$), therefore it is the optimal solution.
*The combination of 30 and 40 is obtained at the intersection of the two constraints, since graph is there; this ordered pair can be obtaining by simply observing. Otherwise, it can be calculated by
solving the two constraints simultaneously.
1. After six months of study, Dr. Starr, president of Southwestern University, had reached a decision. To the delight of the students, SWU would not be relocating to a new football site but would
expand the capacity at its on campus stadium.
Adding 21000 seats, including dozens of luxury skyboxes would not please everyone. The football coach had long argued the need for first-class stadium, 1 with built-in dormitory rooms
for his players and a palatial office appropriate for the coach of a champion team.
The job was to get construction going immediately after the 2002 season ended. This would allow exactly 270 days until the next season opening game. The contractor Hill Construction signed the
contract. Bob Hill guaranteed to finish the work in 270 days. The contract penalty of $10000 per day for running late was agreed.
Back in the office, Hill reviewed the data given in the table below and worked out the target completion date.
1) Develop a network drawing for Hill Construction and determine the critical path. How long is the project expected to take?
2) What is the probability of finishing in 270 days?
Activity Description Predecessor Time estimates in days Crash cost $/day
Optimistic Most likely Pessimistic
A Bonding Insurance, etc. – 20 30 40 1500
B Foundation, Concrete footing for boxes A 20 65 80 3500
C Upgrading skyboxes, stadium sitting A 50 60 100 4000
D Upgrading walkways, stairwells, elevators C 50 50 100 1900
E Interior,wiring, lathes B 25 30 35 9500
F Inspection approvals E 1 1 1 0
G Plumbing D,E 25 30 35 2500
H Painting G 10 20 30 2000
I Hardware, Air-conditioning, Metal Working H 20 25 60 2000
J Tile, Carpeting, windows H 8 10 12 6000
K Inspection J 1 1 1 0
L Final Detail Work, clean-up I,K 20 25 60 4500
Answer 1:
The Networking Diagram:
Critical Path:
Critical Path is the path on the network diagram whose tasks can be not delayed without delaying total time of the project. This means that delaying any activity on the critical path would result in
the delay in the total project completion time. This means that those activities would not have any slack time available. (Slack time is the difference between the earliest start time and earliest
ending time for that activity)
Applying earliest start and earliest end time on the given activities:
Task Early Start Early Finish Late Start Late Finish Slack
A 0 30 0 30 0
B 30 90 60 120 30
C 30 95 30 95 0
D 95 150 95 150 0
E 90 120 120 150 30
F 120 121 259 260 139
G 150 180 150 180 0
H 180 200 180 200 0
I 200 230 200 230 0
J 200 210 219 229 19
K 210 211 229 230 19
L 230 260 230 260 0
Project 260
As it is obvious from the above table that the path constituting of activity A, C, D, G, H , I and activity L is the critical path, as all these activities have zero slack and any delay in these
activities would ultimately delay the project.
There is also identification for critical path that it is the longest path on the network diagram. Looking even this way, the answer would be same, that is, the path constituting of activity A, C, D,
G, H , I and activity L.
The total project completion time is 260 days.
Answer 2:
For determining the probability of ending the project in 270 days, we need to know the project variance.
The project variance is the sum of individual variances = ∑((pessimist – optimist) /6)^2
Thus , project standard deviation is the square root of project variance.
In our case, the project standard deviation is 17.87
Z = due date – expected date of completion
Project standard deviation
Z score for probability of project completion in 270 days
= 270-260
= 0.56
Referring to the normal table, there is 71.23% probability that the project will be completed in 270 days.
Weiss, N., (1997). Introductory Statistics. Boston: Addison-Wesley.
Freund, J., & Walpole, R. (1987). Mathematical Statistics. Englewood Cliffs: Prentice-Hall.
Render, B., & Stair, R. (1997). Quantitative Analysis for Management. Englewood Cliffs: Prentice Hall.
Quinn, S., Bowser, K., & Flaherty, E. (1993). Budnick’s Applied Mathematics for Business, Economics and Social Sciences. Boston: McGraw-Hill College.
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CBSE Class 11 Physics Notes with Derivations | Toppers CBSE
CLASS 11 Physics Notes
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CBSE Class 11 Physics Notes with derivations
Class 11 Physics Notes Chapter 1. Physical World
What Is Science, Scientific Methods, Natural Sciences , Physics , Two Principal Types of Approaches In Physics, Scope of Physics, Classical Physics and Macroscopic Domain , Mechanics,
Electrodynamics, Optics, Thermodynamics, Factors Responsible For Progress of Physics , Hypothesis, Axiom and Models, Technological Applications of Physics , Fundamental Forces In Nature ,
Gravitational Force, Electromagnetic Force, Strong Nuclear Force, Weak Nuclear Force, Comparision Between Different Fundamental Forces
Class 11 Physics Notes Chapter 2. Units and Measurements
Physical Quantity, Representation of Physical Quantity, Characteristics of a Unit, Fundamental Quantities, Derived Quantities, Systems of Units, SI System of Units, Fundamentals Units, Supplementary
Units, Measurement of Length, Astronomical Unit (Au), Light Year, Parsec, Indirect Method To Mesurement of Large Or Small Distances, Parallax Method, Measurment of Molecule Size: Size of Oleic Acid
Molecule, Coherent System of Units, Advantages of Si System, Order of Magnitude, Dimensions and Dimensional Formulae, Dimensional Formula of Some Physical Quantities, Four Types of Quantities,
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Global stability of discrete dynamical systems via exponent analysis: applications to harvesting
Global stability of discrete dynamical systems via exponent analysis: applications to harvesting
population models
Daniel Franco
, Juan Perán
and Juan Segura
1Departamento de Matemática Aplicada, E.T.S.I. Industriales, Universidad Nacional de Educación a Distancia (UNED), c/ Juan del Rosal 12, 28040, Madrid, Spain
2Departament d’Economia i Empresa, Universitat Pompeu Fabra, c/ Ramón Trías Fargas 25-27, 08005, Barcelona, Spain Received 25 June 2018, appeared 21 December 2018
Communicated by Tibor Krisztin
Abstract. We present a novel approach to study the local and global stability of fam- ilies of one-dimensional discrete dynamical systems, which is especially suitable for difference equations
obtained as a convex combination of two topologically conjugated maps. This type of equations arise when considering the effect of harvest timing on the stability of populations.
Keywords: global stability, discrete dynamical system, population model, harvest tim- ing.
2010 Mathematics Subject Classification: 39A30, 92D25.
1 Introduction
A common problem in the study of dynamical systems is to decide whether two different systems have asimilarbehavior [12]. In some cases, solving the problem is easy. For instance, let F[0](x):= f(rx)
andF[1](x):= r f(x), wherer ∈(0, 1)and f: (0,+[∞])→(0,+[∞])is a contin- uous map. Since F[1] = ψ◦F0◦ψ^−^1(x)with ψ(x) = rx, the maps F0 and F[1] are topologically conjugated and, therefore, the
difference equations
x[t]+1= F[0](xt), t =0, 1, 2, . . . , (1.1) and
x[t]+1= F[1](x[t]), t =0, 1, 2, . . . , (1.2) are equivalent from a dynamical point of view.
BCorresponding author. Email: jperan@ind.uned.es
In other situations, the solution is much harder. Here, we consider a problem proposed by Cid, Liz and Hilker in [8, Conjecture 3.5]. They conjectured that if equation (1.1) has a locally
asymptotically stable (L.A.S.) equilibrium, then the difference equation
x[t]+1 = (1−θ)F0(xt) +θF[1](xt), t =0, 1, 2, . . . , (1.3) also has a locally asymptotically stable equilibrium for each θ ∈ [0, 1], provided that f is a compensatory population map [7]. In this
paper, we show that this conjecture is true for a broad family of population maps. Indeed, for all maps in that family, we prove that the equilibrium of (1.3) is not only L.A.S. but globally
asymptotically stable (G.A.S.). In other words, we provide sufficient conditions for (1.3) to inherit the global asymptotic behavior of (1.1) independently of the value ofθ ∈[0, 1].
Equation (1.3) arises when the effect of harvest timing on population dynamics is consid- ered. Together with many other factors, harvest time conditions the persistence of exploited populations,
especially for seasonally reproducing species [6,19,28,31], which on the other hand are particularly suitable to be modeled by discrete difference equations [20]. A key question in management
programmes is to ensure the sustainability of the tapped resources, thus the issue is generating an increasing interest. However, most previous studies have fo- cused on population size and few have
addressed population stability. A model proposed in [32] and based on constant effort harvesting—also known as proportional harvesting—allows for the consideration of any intervention moment during
the period between two consec- utive breeding seasons, a period that from now on we will call the harvesting season for the sake of simplicity. For this model, two topologically conjugated systems
are obtained when the removal of individuals takes place at the beginning or at the end of the harvest- ing season—namely difference equations (1.1) and (1.2). For these two conjugated systems,
harvesting with a certain effort—namely the value of r—can create an asymptotically sta- ble positive equilibrium. When individuals are removed at an intermediate moment during the harvesting season,
the dynamics of the population follow a convex combination of these limit cases—namely (1.3). In this framework, Conjecture 3.5 in [8] has a clear meaning with important practical consequences:
delaying harvest could not destabilize populations with compensatory dynamics.
Previous works have addressed the problem considered here. Cid et al. proved in [8] that the local stability of the positive equilibrium is not affected by the time of intervention for populations
governed by the Ricker model [30]. They also obtained a sharp global stability result for the quadratic map [25] and the Beverton–Holt model [5]. Global stability is always desirable as it allows to
predict the fate of populations with independence of their initial size.
Yet, proving it is in general a difficult task, this being reflected in the fact that many different schemes have been used in the literature for this purpose. In [14,15], the authors showed that
harvest time does not affect the global stability in the Ricker case by using well-known tools, namely results independently proved by Allwright [2] and Singer [34] for unimodal maps with negative
Schwarzian derivative and a sufficient condition for global stability in [35, Corollary 9.9].
Little is known about the effect of the moment of intervention on the stability of popula- tions governed by equations different from the Ricker model, the Beverton–Holt model or the quadratic map
(although see [14, Proposition 2], where it was proved that the moment of in- tervention does not affect the stability when the harvesting effort is high enough). To reduce this gap, we introduce an
innovative approach that is especially useful to prove the global stability of a broad family of population models, namely those encompassed in the so called
generalized α-Ricker model [24]. Among others, the Bellows, the Maynard Smith–Slatkin and the discretized version of the Richards models are covered by our analysis [4,26,29]. Interest- ingly, these
three models can be seen, respectively, as generalizations of the already studied Ricker, Beverton–Holt and quadratic maps where the term related to the density dependence includes a new exponent
parameterα. In the proposed new method, the focus is onα: under certain conditions, we provide sharp results of both local and global stability of the positive equilibrium of the system depending on
the value of α. In particular, these results can be considered as the proof, for a wide range of population models, of [8, Conjecture 3.5]. It is important to stress that this does not prove the
aforementioned conjecture in general, which is impossible since it is false [14], but supports its validity when restricted to meaningful population maps used in population dynamics.
The proposed new method can be applied whenever the per capita production functiong has a strictly negative derivative. The domain (0,ρ)of gcan be bounded or unbounded. All bounded cases can be
easily reduced to the case ρ = 1. The range (g(ρ),g(0)) can also be bounded or unbounded, provided that 0≤ g(ρ)<1< g(0)≤ +[∞.]
The applications that we present in this paper focus on the casesg(0)<+[∞]andg(ρ) =0.
In particular, our examples deal with the following models:
• TheBellowsmodel, which includes theRickermodel as a particular case (Subsection4.1).
• The discretization of theRichardsmodel, which includes thequadraticmodel as a partic- ular case (Subsection4.2).
• TheMaynard Smith–Slatkinmodel, which includes theBeverton–Holtmodel as a particu- lar case (Subsection4.3).
• TheThiememodel, which includes theHassellmodel as a particular case (Subsection4.4).
The paper is organized as follows. Section 2 describes the harvesting population model that motivates our study and lists the families of per capita production functions that we will consider in
Section 4. Section 3 states and proves the main results. Section 4 is divided in several subsections, each of them consisting in an example of the applicability of the main results. Finally, Section
5 focuses on the “L.A.S. implies G.A.S.” and the “stability implies G.A.S” properties.
2 Model
2.1 Per capita production functions
First-order difference equations are commonly used to describe the population dynamics of species reproducing in a short period of the year. Usually, these equations take the general form
x[t]+1 =x[t]g(x[t]), t=0, 1, 2, . . . , (2.1) where x[t] corresponds to the population size at generation t and map g to the per capita production function, which naturally has to be assumed as
non-negative. In addition, g is frequently assumed to be strictly decreasing, because of the negative effect of the intraspecific competition in the population size, and when that condition holds the
population is said compensatory [7,20]. Theoretical ecologists have developed several concrete families of per
capita production functions. These families depend on one or several parameters, which are essential to fit the functions to the experimental data.
Our results cover some of the most relevant families of compensatory population maps, which, as it was pointed out in [24], can be described in a unified way using the map
g: {x ∈[R][++] [: 1]+px^α >[0]} →[R]++
defined by
g(x) =lim
(1+qx^α)^1/q^, ^(2.2)
where α,κ ∈ [R][++] and p ∈ [R]\ {−[∞]}, with R++ denoting the set of positive real numbers andR:= [−[∞,]+[∞]]the extended real line.
The following models are obtained for different values of the parameters:
[M1] For p=1 andα=1, theBeverton–Holtmodel [5], in whichg(x) = [1][+]^κ[x].
[M2] For p = −1 and α = 1, thequadratic model [25], in whichg(x) = κ(1−x) and where κ <4 for (2.1) to be well-defined.
[M3] For p=0 andα=1, theRickermodel [30], in whichg(x) =κe^−^x.
Models[M1–M3]are compensatory. Nevertheless,[M2–M3]are always overcompensatory [7,9] (mapxg(x)is unimodal) and can have very rich and complicate dynamics, whereas[M1]
is never overcompensatory (the mapxg(x)is increasing) and has pretty simple dynamics: all solutions monotonically tend to the same equilibrium which, consequently, is G.A.S.
Map (2.2) also includes models that are overcompensatory or not depending on the values of the parameters:
[M4] For p=1, theMaynard Smith–Slatkinmodel [26], in which g(x) = [1][+]^κ[x][α]. [M5] For α=1 and p>0, theHassellmodel [17], in whichg(x) = ^κ
(1+px)^1/p. [M6] For p>0, theThiememodel [35], in whichg(x) = [(] ^κ
Obviously,[M4–M6]include[M1]as a special case. Similarly, the last two models that we will mention can be considered as generalizations of[M2]and[M3], respectively:
[M7] For p = −1, the discretization of theRichards model [29], in which g(x) = κ(1−x^α). Since xg(x) attains its maximum value at x = (1/(1+α))^1/α, the inequality ακ <
(1+α)^1^+^α^α must be satisfied for (2.1) to be well-defined.
[M8] For p=0, theBellowsmodel [4], in which g(x) =κe^−^x^α.
Models [M7–M8] generalize [M2–M3] by including a new exponent parameter α, which determines the severity of the density dependence and makes the models more flexible to describe datasets [4]. This is
the announced exponent parameter playing a central role in our study.
Before presenting the harvesting model where these population production functions will be plugged in, it is convenient to make some remarks. First, we point out that the domain of g is bounded for
models [M2] and [M7], whereas it is unbounded for the rest of models.
When the domain of g is bounded, there is a restriction in the parameters involved in the
map for which (2.1) is well-defined. On the other hand, a suitable rescaling allows to obtain other frequently used expressions of these eight models depending on an extra parameter, e.g.
g(x) =κ(1−mx)for the quadratic model or g(x) = κe^−^mx for the Ricker model. This extra parameter is irrelevant for the dynamics of (2.1).
2.2 Modelling harvest timing
Assume that a population described by (2.1) is harvested at the beginning of the harvesting season t and a fraction γ ∈ [0, 1)of the population is removed. Then, it is well established that the
population dynamics are given by
x[t]+1 = (1−γ)x[t]g((1−γ)x[t]). (2.3) When individuals are removed at the end of the harvesting season, the population dynamics follow
x[t]+1= ([1]−γ)x[t]g(x[t])[.] [(2.4)] The above situations represent the two limit cases of our problem. To model the dynamics of populations harvested at any time during the harvesting season, we
consider the framework introduced by Seno in [32]. Let θ ∈ [0, 1] represent a fixed time of intervention during the harvesting season, in such a way that θ = 0 corresponds to removing individuals at
the beginning of the season and θ = 1 at the end. Assume that the reproductive success at the end of the season depends on the amount of energy accumulated during it. Given that the per capita
production function depends onx[t]beforeθand on([1]−γ)x[t]afterwards, Seno assumed that the population production is proportional to the time period before/after harvesting. This leads to the convex
combination of (2.3) and (2.4) given by
x[t]+1 = (1−γ)x[t][θg(x[t]) + (1−θ)g((1−γ)x[t])]. (2.5) In particular, substituting θ=0 in (2.5) yields (2.3), and (2.4) is obtained forθ =1.
The two maps derived from (2.5) for θ =0 and θ = 1 are topologically conjugated. Thus, if the equilibrium for θ = 0 is G.A.S., then the equilibrium for θ = 1 is also G.A.S., and vice versa. From a
practical point of view, this implies that for these two limit cases we can predict the long-run behavior of the system with independence of the initial condition. In view of this, it is natural to
study to what extent the same is true if individuals are removed at any intermediate moment during the harvesting season.
Substituting map (2.2) into (2.5), we obtain an intricate model depending on up to five parameters for which establishing general local or global stability results is a tricky task. For that purpose,
we develop a general method in the following section.
3 Exponent analysis method
Consider the difference equation
x[t]+1 =x[t]g[s](x[t]), with
g[s](x) =c h(x^α) + (b−c)h(sx^α)[,]
whereb,s,α∈[R][++]andc∈[R][+]:= [0,+[∞])are such thatc<b;s≤1; andh: (0,ρ)→(ν,µ)⊂ R++is a decreasing diffeomorphism withρ,µ∈ {1,+[∞]}andνb<1< µb.
Notice that the domain ofhcan be the open bounded interval(0, 1)or the open unbounded interval(0,+[∞]), covering all the models described in the previous section. In addition, the image ofhcan be
bounded or unbounded, although the applications presented in this paper are restricted to the bounded case.
Forρ =1, it is not obvious that the difference equationx[t]+1= x[t]g[s](x[t])is well-defined, i.e.
xgs(x) ∈ (0,ρ) for x ∈ (0,ρ). Next, we study when the difference equation xt+[1] = xtgs(xt) is well-defined and has a unique equilibrium. We establish some notation first. Being the function
x7→ gs
=c h(x) + (b−c)h(sx) a diffeomorphism from(0,ρ)to (ν[s]b,µb), where
ν[s]:= lim
x→ρg[s](x)/b= ^cν+ (b−c)h(sν)
b ≥ν, (3.1)
we denote by j[s]its inverse diffeomorphism, i.e., the function j[s]: (ν[s]b,µb)→([0,]ρ)[satisfying] c h(j[s](z)) + (b−c)h(sj[s](z)) =z (3.2) for allz∈(ν[s]b,µb). Obviously, whenρ= +[∞, one has]ν[s]
=νfors∈ (0, 1].
Lemma 3.1. Assume b,s,α∈[R][++]and c∈[R][+]are such that c <b; s≤1; and h: (0,ρ)→(ν,µ)⊂ R++is a decreasing diffeomorphism withρ,µ∈ {1,+[∞]}andνb<1<µb. In addition, let
s∗:=inf{s∈(0, 1]:ν[s]<1/b}, (3.3) where ν[s] is given by (3.1). Then, the map xg[s](x) has a unique fixed point in ([0,]ρ) if and only if s>s∗. Moreover, this fixed point is(j[s](1))^1/α.
Proof. Clearly, x ∈ (0,ρ)is a fixed point ofxgs(x)if and only if gs(x) = 1, and in such case, x= (js(1))^1/α.
Next, notice thatν[0]:= ^cν^+(^b[b]^−^c^)^µ ≥ν[s][ˆ]≥ ν[s] ≥ν[1] =[ν, for 0]<sˆ <[s]<1, and thatν[s]depends continuously ons. Sincegsmaps(0,ρ) onto(νsb,µb)andνb<1< µbholds, we have that the equationg
[s](x) =1, forx ∈(0,ρ), has solution if and only ifs >s∗. We have already stressed thatν[s]=ν, forρ= +∞. Hence, we haves∗ =0 forρ= +[∞.]
In the conditions of Lemma3.1, for each s∈([0, 1]]we define the function τ[s]:
1 µb,[ν]^1
→[R] by τ[s](z):= ^ln ^j^s
1 z
lnz . (3.4)
Now, we study under which conditions the difference equation x[t]+1 = x[t]g[s](x[t]) is well- defined.
Lemma 3.2. Assume that the conditions of Lemma3.1 hold with s ∈ (s∗, 1]. Then, zgs(z)∈ (0,ρ) for all z∈(0,ρ)if and only ifα<α[s]with
(+[∞], ρ= +[∞],
min[z][∈(][1/µb,1][)]τ[s](z), ρ=1. (3.5)
Moreover, if the equation x[t]+1 = x[t]g[s](x[t]) is well-defined for s = 1, then it is also well-defined for s∈(s∗, 1].
Proof. We consider separately the casesρ = +[∞] andρ = 1. The case ρ = +[∞]is trivial. For ρ=1, we have
z gs([z])∈([0, 1])[for][z]∈([0, 1]) ⇐⇒ gs([z])< ^1
z forz∈([0, 1])[.] The latter inequality always holds if z≤ [µb]^1 , becausegs((0, 1)) = (νsb,µb). Hence,
g[s](z)< ^1
z forz∈ (0, 1) ⇐⇒ g[s](z)< ^1
z forz∈^[µb]^1, 1
⇐⇒ z^α > j[s] ^1[z]
forz∈^[µb]^1 , 1
⇐⇒ α< ^ln ^j^s
1 z
lnz = τs(z)forz∈ ^[µb]^1, 1 . Sinceρ=1, we have that τ[s](z)>0 forz∈^[µb]^1, 1
z→lim1/µbτ[s](z) = +[∞] [and] [lim]
z→1^−τ[s](z) = +[∞,] [(3.6)] which finishes the proof of the first affirmation. For the second one, notice thatαs decreases as we increase s, because j[s] decreases with s. Therefore, α < α[1]
guarantees α < α[s] for s∈(s∗, 1][.]
Now, in the conditions of Lemma3.1, for each s∈(s∗, 1], we write bs:=min{µb, 1
ν[s]b}, (3.7)
and define the function σ[s]:
1 bs,b[s]
⊂ ^[µb]^1,[ν]^1
τ[s](z) +τ[s](^1
z), z 6=1,
js(1) ^, ^z =1.
Lemma 3.3. The function σ[s] given in (3.8) is continuous and positive. Moreover, when ρ = 1, it satisfiesσ[s](z)<τ[s](z)for z∈ ^[b]^1
s, 1 .
Proof. A direct application of L’Hôpital’s rule shows thatσsis a continuous function:
limz→1σ[s](z) =lim
ln(j[s](1/z))−ln(j[s](z)) lnz
= [lim]
u = −2j[s]^0(1)
js(1) =σ[s]([1])[.]
On the other hand, to see that σ[s] takes values onR++ note thatz 7→ ln(j[s](z))is a decreasing function and that j[s]is a diffeomorphism, soj^0[s](1)<0.
Finally, forρ=1, one has
τ[s](z) = ^ln(j[s](1/z))
lnz >0 and τ[s](1/z) = ^ln(j[s](z))
−lnz <0, forz∈ [b]^1
s, 1
. Thus,σ[s](z)<τ[s](z)[for]z ∈ [b]^1
s, 1 .
The functionσ[s], given in (3.8), is related to the fixed points of the map f[s]◦f[s]with f[s](x) = xgs(x), as we will see next. Assuming α < α[s], for the map fs◦ fs to be well-defined, and
rearranging forαin the fixed points equation we have (see Lemma 3.1)
g[s](x)g[s](xg[s](x)) =[1] ⇐⇒ j[s]^−^1(y)j^−[s]^1 y
j[s]^−^1(y) α
=[1 ;] y =x^α (3.9)
⇐⇒ zj^−[s]^1(j[s](z)z^α) =1 ; z =j^−[s]^1(x^α) (3.10)
⇐⇒ js(z)z^α = js(1/z) ; z= j^−[s]^1(x^α) (3.11)
⇐⇒ α=σ[s](z)withz =j^−[s]^1(x^α), orz=1. (3.12) In other words, the difference equation x[t]+1 = x[t]g[s](x[t]) has a nontrivial period-2 orbit if and only if there exists z ∈ (1/b[s],b[s])\ {1} and
α < α[s] such that σ[s](z) = α. Consequently, considering σs for the study of the global stability of the equilibrium of x[t]+1 = xtgs(xt) is natural since, by the main theorem in [10], the absence
of nontrivial period-2 orbits forx[t]+1= x[t]g[s](x[t])is equivalent to the global asymptotic stability of this equilibrium. More specifically, we will use the following result:
Lemma 3.4. Let −[∞] ≤ a[1] < a[2] ≤ [∞, I] = (a[1],a[2]), f : I → I a continuous function and x[∞] ∈ I such that(f ◦f)(x) 6= x for all x ∈ I\ {x[∞]}. Then, x[∞] is a stable equilibrium for the map f
◦ f if and only if x[∞] is a G.A.S. equilibrium for the map f .
Proof. Define f^(^1^) := f,f^(^n^) := f◦ f^(^n^−^1^) and apply the Sharkovsky Forcing Theorem [33] to see that f^(^n^)(x)6=x for allx ∈ I\ {x[∞]}, n≥1. If the continuous function q(x) = f^(^n^)(x)−x
were negative in(a[1],x[∞])[, then]x[∞] would not be stable for the map f^(^2^), since x[j] = f^(^2nj^)(x[0]) would be a decreasing sequence, for all x[0] ∈ (a[1],x[∞]). Applying the same argument
for the interval (x[∞],a[2]), we conclude that (f^(^n^)(x)−x)(x−x[∞])< 0 for all n≥1, x∈ I\ {x[∞]}. In particular, replacing x with f^(^m^)(x)[, one has] (f^(^n^+^m^)(x)− f^(^m^)(x))(f^(^m^)(x)−x[∞])
< [0] for all n,m ≥ 1, x ∈ I \ {x[∞]}. Therefore, the subsequence of f^(^n^)(x)^
n formed by the terms smaller (respectively, greater) than thex[∞]is increasing (respectively, decreasing). Then, limn→[∞]f^(^n^)(x) =x[∞], for allx∈ I. The converse is obvious.
Remark 3.5. We are considering per capita production functions from(0,ρ)onto(ν[s]b,µb)⊂ (νb,µb)[, given by]
g[s](x) =c h(x^α) + (b−c)h(sx^α),
where s andα runs, respectively, through(s∗, 1] and(0,α[s]), these being the largest intervals within which the equationx[t]+1 = x[t]g[s](x[t])is well-defined and has an equilibrium (see (3.1), (3.3)
and (3.5)).
Probably, the most relevant applications arise for the case in which the domain is un- bounded (i.e., ρ = +∞). In such a particular case, s∗ = 0, νs = ν and αs = +[∞, for all] s ∈ [0, 1]. Therefore,
when ρ = +∞, the equation x[t]+1 = x[t]g[s](x[t]) is well-defined and has an equilibrium for alls ∈[0, 1]andα>0.
Moreover, we point out that the following theorem (which is the main result of this paper) can be applied under very general conditions. In particular, it holds when the per capita production
function has unbounded range.
In what follows, ρ, µ, ν, b and c will be considered as constants, while s and α will be mostly seen as parameters.
Theorem 3.6. Let µ,ρ ∈ {1,+[∞]}, 0 < c < b, 0 ≤ νb < 1 < µb and h: (0,ρ) → (ν,µ) be a decreasing diffeomorphism. Let s∗ be given by (3.1)–(3.3), α[s] given by (3.4)–(3.5) and consider the families of
functions {j[s]}[s][∗][<][s][≤][1] and {σ[s]}[s][∗][<][s][≤][1] defined by (3.2) and(3.8), respectively. For each s∈(s∗, 1]andα∈(0,α[s])also consider the discrete equation
x[t]+1 =x[t](c h(x^α[t]) + (b−c)h(sx^α[t])), x[0]∈(0,ρ). (3.13) (A) Then,(3.13)is well-defined, it has a unique equilibrium and
(i) The equilibrium of (3.13)is locally asymptotically stable (L.A.S.) whenα< σ[s](1)and it is unstable forα>σs(1).
(ii) The equilibrium of (3.13)is globally asymptotically stable (G.A.S.) if and only ifα<σ[s](z) for all z∈ (1,b[s])(see(3.1)and(3.7)).
(B) Additionally, assume that h satisfies
x 7→h^0(x)/h^0(sx)is nonincreasing for each s∈(s∗, 1). (H[1]) If (3.13)is well-defined and its equilibrium is G.A.S. for s = 1, then(3.13)is well-defined and its equilibrium is G.A.S., for the same
parameters, but s∈(s∗, 1].
(C) Finally, assume that h satisfies
x7→h^0(x)/h^0(sx)is decreasing for each s ∈(s∗, 1). (H[2]) If (3.13) is well-defined and its equilibrium is L.A.S. for s = 1, then(3.13) is well-defined and its equilibrium is L.A.S., for the same
parameters, but s∈(s∗, 1].
Proof. (A). By Lemmas 3.1 and3.2, equation (3.13) is well-defined and has a unique equilib- rium atx[∞] = (j[s](1))^1/α. To prove(i), we compute the derivative at the equilibrium. Since
f[s](x) =x(c h(x^α) + (b−c)h(sx^α)) =xj[s]^−^1(x^α), we obtain
f[s]^0(x) =j^−[s]^1(x^α) +x j^−[s]^10
(x^α)αx^α^−^1. The evaluation of this expression at x[∞] = (js(1))^1/α yields
f[s]^0(x[∞]) =1+αj[s](1)^j^−[s]^10
(j[s](1)) =1+αj[s](1)
j^0[s](1) =1− ^2α σ[s](1)^, and then, sinceσ[s](1)>0 holds by Lemma3.3,
−1< f[s]^0(x[∞])<1 ⇐⇒ α<σ[s](1). Similarly, if 0<σ[s](1)<α, then f^0(x[∞])<−1, so (3.13) is unstable.
By the symmetry ofσsand applying an analogous argument as the one presented in (3.9)–
(3.12) we obtain that
σ[s](z)≷^α ∀z∈(1,b[s]) ⇐⇒ ((f[s]◦f[s])(x)−x)(x−x[∞])≶^0 ∀x∈(0,ρ)\ {x[∞]}. (3.14) To prove(ii), in view of(i)above, (3.9)–(3.12) and Lemma3.4, just consider the following four scenarios:
• Ifα<σ[s](z)for allz∈[1,b[s]), then, by (3.14),(f[s]◦f[s])(x)6=xfor allx ∈(0,ρ)\ {x[∞]}and x[∞]is L.A.S. Then,x[∞] is G.A.S.
• If α=σ[s]([1])< σ[s](z)[for all]z ∈ ([1,]b[s]), then, by (3.14), ((f[s]◦f[s])(x)−x)(x−x[∞])<[0 for] allx∈ (0,ρ)\ {x[∞]}and(fs◦fs)^0(x[∞]) =1. The equilibriumx[∞]is L.A.S. for fs◦fs. Then, x[∞]is
G.A.S. for f[s].
• If α > σ[s](z) for all z ∈ (1,b[s]), then, by (3.14), (f[s]◦ f[s])(x) < x for all x ∈ (0,x[∞]). Therefore, the equilibriumx[∞]is unstable.
• In any other case, the equationx[t]+1 = f[s](x[t])has nonconstant periodic solutions. There- fore, the equilibriumx[∞] is not G.A.S.
(B). We start by verifying that the function s 7→ σ[s](z) is nonincreasing for each z ∈ (1/b[s],b[s]). Recall that ν[s] is nonincreasing in s (see (3.1)), so (1/b[s][ˆ],b[s][ˆ]) ⊂ (1/b[s],b[s]) for
any 0<sˆ<s <1; therefore,σ[s](z)is well-defined ifσ[s][ˆ](z)is. By differentiating with respect tosin
z= c h(j[s](z)) + (b−c)h(sj[s](z)), we obtain
0=c h^0(j[s](z))^∂j^s(z)
∂s + (b−c)h^0(sj[s](z))
j[s](z) +s∂j[s](z)
, which implies
∂s =
j[s](z) = (c−b)h^0(sj[s](z))
c h^0(j[s](z)) + (b−c)s h^0(sj[s](z)) = (c−b) c h^0(j[s](z))
h^0(sj[s](z))+ (b−c)s .
Since condition (H[1]) holds and j[s] is a decreasing diffeomorphism, we have that the function z7→∂(lnj[s](z))/∂sis non-decreasing in(1/b[s],b[s])for eachs ∈(s∗, 1]. Thus,
∂sσ[s](z) = ^∂
τ[s](z) +τ[s](1/z) lnz
∂slnj[s](1/z)−[∂s]^∂ lnj[s](z)
lnz ≤0
for all z ∈ (1/bs,bs)\ {1}. Therefore, the function s 7→ σ[s](z) is nonincreasing for each z ∈ (1/b[s],b[s]).
Now, if (3.13) is well-defined for s=1, by Lemma3.2, we know that (3.13) is well-defined fors ∈ (s∗, 1), and, if its equilibrium is G.A.S. for s = 1, (A)-(ii) and the fact that σs(1/z) = σ[s](z)yield
α<σ[1](z)≤σs(z) for all z∈(1/bs,bs)\ {1}ands∈ (s∗, 1]. Therefore, (3.13) is well-defined and its equilibrium is G.A.S. for alls ∈(s∗, 1].
(C). Following the same reasoning as in the previous case but using (H[2]) instead of (H[1]), it is easy to see that the function s 7→ σ[s](z) is decreasing for each z ∈ (1/b[s],b[s]). As a
consequence, if the equilibrium of (3.13) is L.A.S. fors =1, the application of(A)-(i)yields
α≤σ[1](1)<σs(1), for alls ∈(s∗, 1], and (3.13) is well-defined and its equilibrium is L.A.S. for alls ∈(s∗, 1][.]
Remark 3.7. Note thatσ[s]◦exp is an even function, which makes it more suitable for graphical representations thanσ[s]itself.
Theorem3.6 reduces the study of the local or global stability to the study of the relative position of the graph of σ[s] with respecto to α. Figure 3.1 illustrates this. For a fixed s, the relative
position of min[z][∈(][1,b][s][)]σ[s](z), σ[s](1) and α determines the local and global stability of the equilibrium of (3.13). Suppose that the graph of σs corresponds to the black curve in Figure
3.1-A. From (i) and (ii) in Theorem 3.6, we obtain that the equilibrium of (3.13) is unstable for α> σ[s](1), L.A.S. but not G.A.S. for min[z][∈(][1,b][s][)]σ[s](z)< α< σ[s](1), and G.A.S. for α <
min[z][∈(][1,b][s])σs(z). Figure 3.1-B illustrates the special case when the function σs attains a strict global minimum at z = 1. In such a situation, the range of values of α for which the
equilibrium is L.A.S., thanks to (i) in Theorem3.6, is contained in the range of values ofαfor which it is G.A.S., thanks to (ii) in Theorem 3.6. Hence, in this case, Theorem3.6 completely
characterizes the stability of the equilibrium of (3.13): it is G.A.S. for α ≤ σ[s](1)and unstable forα>σ[s](1).
-2 -1 0 1 2
s(1) s
s(1) s
Equilibrium G.A.S.
Unstable equilibrium
-2 -1 1 2
12 s=10^-^4
s= ^-
s= ^-
s= ^-
1 s=
1(1) s
Equilibrium G.A.S.
for all sÎ(0,1] ^s ^1
= 0.2 s=
0.4 s=
s= s=0.8
2.20 2.25 2.30 2.35 2.40
0 Equilibrium L.A.S but not G.A.S.
Equilibrium G.A.S.
-2 -1 0 1 2
Unstable equilibrium
(0, )
min ( )
z bssz
Figure 3.1: In all panels, the black curve represents the graph ofσ[1]◦exp. A:For α> σ[s](1)the equilibrium of (3.13) is unstable, for min[z][∈(][1,b][s][)]σ[s](z)< α< σ[s](1) it is L.A.S. but not
G.A.S., and for α < min[z][∈(][1,b][s])σ[s](z) it is G.A.S. B: Since σ[s] attains at z = 1 a strict global minimum, the equilibrium of (3.13) is G.A.S. for α≤ σ[s](1). C:The assumption that σ[1]
attains a strict global minimum at z = 1 and condition (H1) are sufficient to guarantee that the graphs of the family of functions{σ[s]}[0][<][s][≤][1]are above the graph ofσ[1]and, consequently, the
equilibrium of (3.13) is G.A.S. for eachs ∈(0, 1]andα≤σ[1](1).
Figure3.1-C deals with the last part of Theorem3.6. Assume thatσ[1](1) is a global mini- mum of σ[1](z)and that condition (H[1]) holds. Then, all the graphs of the family of functions {σ[s]}[0][<][s]
[≤][1] are above the graph ofσ[1](z) and, therefore, the equilibrium of (3.13) is G.A.S. for
eachα≤σ[1](1)and 0< s≤1.
Apart from condition (H1), Theorem 3.6-(B) assumes that (3.13) is well-defined and that its equilibrium is G.A.S. for s = 1. But we have already mentioned that guaranteeing the G.A.S. of an
equilibrium is a difficult task. Nevertheless, when the logarithmically scaled diffeomorphismφ[s](u) := ln(js(e^u))isC^3, we can derive a sufficient condition forσ[s](1)to be the strict global
minimum ofσ[s](z).
Lemma 3.8. Ifφ[s](u):=ln(j[s](e^u))is three times continuously differentiable withφ[s]^000(u)<0for all u∈(−lnb[s], lnb[s]), thenσ[s](z)attains at z=[1]its strict global minimum value.
Proof. It is routine to check that d^j(σ[s](e^u)u−σ[s](1)u)
d^j(φ[s](−u)−φ[s](u)−σ[s](1)u) du^j
=0 forj= 0, 1, 2, and that
du^3 = ^d
du^3 = −φ^000[s] (−u)−φ^000[s] (u)>0 foru∈(−lnb[s], lnb[s]). Therefore,σ[s](e^u)u−σ[s](1)u >0 foru∈ (0, lnb[s]), i.e.,σ[s](z)>σ[s](1)for allz∈ ^[b]^1
\ {1}.
4 Application to some population models
The next result characterizes the elements of the family of per capita production functions (2.2) for which condition (H[1]) in Theorem3.6 holds.
Lemma 4.1. For any p∈ [R], the function h: {x∈ [R][+]: 1+px >0} →(0, 1)defined by h(x) =lim
1 (1+qx)^1/q
is a decreasing diffeomorphism. Moreover, h satisfies(H[1])for each s∈(0, 1)if and only if p≥ −1.
Proof. Assume p6=0. Differentiating, we obtain that
h^0(x) =−(1+px)^−(^p^+^1^)^/^p<0
for anyx ∈[R]+such that 1+px>0 and, consequently, the first statement is true.
Moreover, h^0(x)
h^0(sx) = −(1+px)^−(^p^+^1^)^/p
1+psx 1+px
s+ ^1−s 1+px
d dx
h^0(x) h^0(sx)
s+ ^1−s 1+px
1/p (1−s) (1+px)^2^, which is non-positive for eachs∈ ([0, 1])if and only ifp∈ [−1,+[∞])\ {0}.
Finally, the result is straightforward for p=0 sinceh(x) =e^−^xand [h]^h0^0(^(sx^x^)) =e^−(^1^−^s^)^x.
The following subsections deal with the study of the harvesting model (2.5) for the per capita production functions in Subsection 2.1. We use a similar procedure for all of them, based on the
following five steps:
1. First, we rewrite the difference equation that we want to study, which will depend on certain original parameters, as (3.13) with parametersb,c,s,α,ν,µandρ.
2. We check thathsatisfies condition (H[1]), thanks to Lemma4.1.
3. If necessary, we check that (3.13) is well-defined fors=1. Next, we invoke Lemma3.8to guarantee that the rewritten difference equation, with s = 1, has an equilibrium which is G.A.S.
4. Then, we use statement (B) in Theorem 3.6 to conclude the global stability result for s ∈(s∗, 1].
5. Finally, we interpret the result in terms of the original parameters.
4.1 Bellows model
The per capita production function of the Bellows model is given by g(x) = κe^−^x^α, with κ,α>0. The Seno model (2.5) is in this case
x[t]+1 =κθ(1−γ)x[t]e^−^x^t^α+κ(1−θ)(1−γ)x[t]e^−(^1^−^γ^)^α^x^t^α, x[0]>0, (4.1) whereθ ∈ [0, 1]andγ∈[0, 1).
In order to apply the results in Section 3, we set b = κ(1−γ) > 1, c = κ(1−γ)θ, s = (1−γ)^α, ρ = +[∞,] ν = 0, µ= 1, and h(x) = e^−^x, which is a decreasing diffeomorphism from (0,+[∞])to (0, 1)
satisfying condition (H[1]), thanks to Lemma4.1. Notice that (3.13) with s=1 is equivalent to (4.1) withθ =1. In this case,bs=bfor eachs∈ (0, 1]andj[1](z) =ln(b/z) for z ∈ (0,b),σ[1](1) = 2/lnb.
Moreover, φ[1](u) = ln(ln(be^−^u))andφ^000[1] (u) = − ^2
(ln(be^−^u))^3 < 0 foru∈(−lnb, lnb).
Therefore, a direct application of Theorem3.6, taking into account thats∗ =0,νs =νand α[s] = +[∞, for all]s∈ [0, 1]whenρ= +[∞](see Remark3.5), yields the following result:
Proposition 4.2. Ifκ(1−γ) > 1, then(4.1) has a unique equilibrium. If, in addition, θ = 1, then the equilibrium of (4.1)at x = (ln(κ(1−γ)))^1/α is unstable forα>2/ ln(κ(1−γ)and G.A.S. for α≤2/ ln(κ
(1−γ)). Furthermore, forθ <1andα≤2/ ln(κ(1−γ)), the equilibrium is also G.A.S.
Proposition4.2 characterizes the global stability of the equilibrium for the Bellows model without harvesting. Such a result is new, as far as we know, and is interesting in itself. On the other
hand, Proposition 4.2 confirms that, for the Bellows model, the harvesting effort necessary for stabilization is less for θ ∈ (0, 1)than for θ = 0 andθ = 1. Since the Bellows model has the Ricker
model as a particular case, Proposition 4.2 generalizes [8, Proposition 3.3] and gives an alternative proof of the main result in [15].
4.2 Discretization of the Richards model
The per capita production function of the discretization of the Richards model is given by g(x) =κ(1−x^α), withκ,α>0. Hence, the Seno model (2.5) reads
x[t]+1=κθ(1−γ)x[t](1−x^α[t]) +κ(1−θ)(1−γ)x[t](1−(1−γ)^αx^α[t]), x[0] ∈(0, 1), (4.2)
whereθ ∈[0, 1]andγ∈ [0, 1).
In this example, it is natural to assume that (4.2) is well-defined for γ = 0, i.e., that the population model without harvesting makes sense. As mentioned when we presented this per capita
production function in Subsection 2.1, equation (4.2) is well-defined for γ = [0 if] and only ifακ< ([1]+α)^1^+^α^α[.]
As in the previous case, we setb=κ(1−γ)>1,c=κ(1−γ)θ,s= (1−γ)^α,ρ=1, ν=0, µ=[1, and]h(x) =[1]−x. Clearly, the functionh(x)is a decreasing diffeomorphism from([0, 1]) to(0, 1)and, by Lemma4.1,
satisfies condition (H[1]).
We aim to obtain a global stability result for (3.13) withs =1, which is equivalent to (4.2) withθ = 1. Note that (3.13) is well-defined for s =[1 because] αb≤ ακ< ([1]+α)^1^+^α^α[. We have] j[1](z)
=1−^z[b]forz ∈(0,b), beingσ[1](1) = [b][−]^2[1],φ[1](u) =ln 1−^e[b]^u^andφ[1]^000(u) =−^be^u^(^b^+^e^u^)
(b−e^u)^3 <0.
Then,σ[1](z)> [b]^2b[−][1] [for]z >1 and the equilibrium of (3.13) is G.A.S. fors= [1 if]α≤ [b]^2b[−][1][, i.e., if] b(α−2)≤α.
In order to use Theorem3.6, we need to imposes> s∗ =max
0, 1− [b][−]^1[c] , or what is the same, νsb = (b−c)(1−s) < 1. But, for the selected values of the parameters, this is always true because
(b−c)(1−s) = (1−θ)κ(1−γ)(1−(1−γ)^α)≤κ(1−γ)(1−(1−γ)^α)<1, where we have used thatx[t]+1 =κx[t](1−x^α[t]),x[0] ∈(0, 1)is well-defined.
Proposition 4.3. Ifκ(1−γ)> 1andακ < (1+α)^1^+^α^α, then (4.2)is well-defined and has a unique equilibrium. If, in addition,θ = 1, then the equilibrium of (4.2)is unstable forκ(1−γ)(α−2)> α and G.A.S.
for κ(1−γ)(α−2) ≤ α. Furthermore, for θ < 1 and κ(1−γ)(α−2) ≤ α, the equilibrium of (4.2)is also G.A.S.
To our knowledge, Proposition4.3gives the first global stability result for the discretization of the Richards model even in the case without harvesting. Notice that the results in [22]
cannot be used in this case since ρ 6= +∞. In the harvesting framework, Proposition 4.3 includes [8, Proposition 3.6] as a particular result, where the quadratic model was considered.
4.3 Maynard Smith–Slatkin model
If we focus on populations governed by the Maynard Smith–Slatkin model, the per capita production function is given byg(x) = ^κ
1+x^α, where κ > [0 and] α > 0. In that case, model (2.5) is
x[t]+1=κθ(1−γ) ^x^t
1+x^α[t] +κ(1−θ)(1−γ) ^x^t
1+ (1−γ)^αx^α[t], x[0]>0, (4.3) whereθ ∈[0, 1]andγ∈ [0, 1).
In [8], following [1, Appendix S1] and [23, Theorem 1], it was stated that the equilibrium of (4.3) forθ = 0 is G.A.S. if 1 < κ(1−γ)≤ ^α
α−2. No result is known about global dynamics of (4.3), in the general case. However, this model can be easily handled thanks to Theorem3.6 and Lemma4.1.
Consider (3.13) withb=κ(1−γ)> 1,c=κ(1−γ)θ,s = (1−γ)^α,ρ = +[∞,]ν =0, µ= 1 andh(x) = 1/(1+x), which satisfies condition (H[1]) from Lemma4.1. Then, j[1](x) = ^b[x] −1, σ[1](1) = [b]^2b[−][1],φ[1](u) =
ln(be^−^u−1), andφ[1]^000(u) =−^be^u^(^b^+^e^u^)
(b−e^u)^3 <0.
Now, observe again that (3.13) with s = 1 corresponds to (4.3) with θ = 1, and apply Theorem 3.6, taking into account that s∗ = 0,ν[s] = ν and α[s] = +[∞, for all] s ∈ [0, 1] when ρ= +[∞](see
Proposition 4.4. Ifκ([1]−γ) > 1, then(4.3) has a unique equilibrium. If, in addition, θ = 1, then the equilibrium of (4.3) is unstable for κ(1−γ)(α−2) > αand G.A.S. for κ(1−γ)(α−2) ≤ α.
Furthermore, forθ <1andκ(1−γ)(α−2)≤α, the equilibrium is also G.A.S.
It is interesting to note that considering the exponent parameterαin the quadratic model, i.e., studying the discretization of the Richards model, unveils the complete parallelism be- tween the
Maynard Smith–Slatkin model and the quadratic model with respect to stability results.
4.4 Hassell and Thieme models
As already mentioned, topologically conjugated production functions give rise to equivalent dynamical behaviors. However, when a convex combination of the type of (2.5) is applied to two
topologically conjugated production functions, the transformed systems could exhibit different dynamical behaviors.
When applying Theorem 3.6, while working in the case s = 1, we can replace our pro- duction function by a topologically conjugated one, for which calculations are simpler. This replacement is no
longer valid when checking condition (H1).
In this subsection, we put into practice the previous approach to study the two models still left: Thieme’s and Hassell’s models. Since Thieme’s model has Hassell’s model as a particular case, we
only consider the former. Besides, without loss of generality, we assume the per capita production function of the Thieme model to be given by
g(x) = ^κ
(1+x^α)^β^, ^κ,^α,^β>0.
Now, the change of variablesyt =x^1/β[t] shows that the dynamics of the difference equation x[t]+1 = ^κx^t
(1+x^α[t])^β ^(4.4)
are identical of those of the equation
y[t]+1= ^κ
1/βy[t] 1+y^αβ[t] , whose per capita production function,g(x) = ^κ
1+x^αβ, belongs to the Maynard Smith–Slatkin family of maps. This provides a straightforward way to characterize the global stability of the Thieme model.
Proposition 4.5. Ifκ > 1, then (4.4)has a unique equilibrium. In addition, the equilibrium of (4.4) is unstable forκ^1/β(αβ−2)> αβand G.A.S. forκ^1/β(αβ−2)≤αβ.
The previous result improves the global stability condition presented in [35] with a simpler proof than the one used in [22], which relies in calculating the sign of a certain Schwarzian derivative. | {"url":"https://123deta.com/document/yr348elp-global-stability-discrete-dynamical-exponent-analysis-applications-harvesting.html","timestamp":"2024-11-03T15:54:34Z","content_type":"text/html","content_length":"219358","record_id":"<urn:uuid:51ec9a71-b599-4e41-a092-9e86793a2f9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00628.warc.gz"} |
Graphical Models for Statistical Inference and Data Assimilation
2007 Articles
Graphical Models for Statistical Inference and Data Assimilation
In data assimilation for a system which evolves in time, one combines past and current observations with a model of the dynamics of the system, in order to improve the simulation of the system as
well as any future predictions about it. From a statistical point of view, this process can be regarded as estimating many random variables, which are related both spatially and temporally: given
observations of some of these variables, typically corresponding to times past, we require estimates of several others, typically corresponding to future times. Graphical models have emerged as an
effective formalism for assisting in these types of inference tasks, particularly for large numbers of random variables. Graphical models provide a means of representing dependency structure among
the variables, and can provide both intuition and efficiency in estimation and other inference computations. We provide an overview and introduction to graphical models, and describe how they can be
used to represent statistical dependency and how the resulting structure can be used to organize computation. The relation between statistical inference using graphical models and optimal sequential
estimation algorithms such as Kalman filtering is discussed. We then give several additional examples of how graphical models can be applied to climate dynamics, specifically estimation using
multi-resolution models of large-scale data sets such as satellite imagery, and learning hidden Markov models to capture rainfall patterns in space and time.
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Physica D: Nonlinear Phenomena
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Ubiquity: Ubiquity symposium: The science in computer science: unplugging computer science to find the science
Volume 2014, Number March (2014), Pages 1-6
Ubiquity symposium: The science in computer science: unplugging computer science to find the science
Tim Bell
DOI: 10.1145/2590528.2590531
The Computer Science Unplugged project provides activities that enable students to engage with concepts from computer science without having to program. Many of the activities provide the basis
of a scientific exploration of computer science, and thus help students to see the relationship of the discipline with science. This paper presents examples of such activities and how they can be
used to engage students with computer science as a science.
Peter Denning
The Computer Science Unplugged project is a collection of activities that engage young students (typically 10–18 years old) with the big ideas of computer science, without having to learn to program
or even use a computer [1, 2]. This enables us to get to the heart of the subject and dispense with technological distractions. It also exposes the science of computing, since the absence of the
computer forces students to learn how computational principles work by direct observation and experiment. The computer truly is an artifact but not the only means to understand and effect
computation. The "unplugged" activities cover a large range of topics—algorithms, human-computer interaction, artificial intelligence, computer graphics, tractability, compression, encryption and
more—but not learning a programming language or using a computer.
For example, students can explore the concept of the complexity of algorithms by simulating sorting algorithms using a balance scale to sort identical-looking weights from lightest to heaviest,
comparing two weights at a time (see Photo 1). Students can usually work out for themselves that finding the heaviest weight of, say, 10 items, will take nine comparisons, and that repeating this
process on the remaining nine will take eight comparisons. Sometimes we have one student perform the sorting manually, and by the time they have finished their empirical experiment counting the
comparisons, the class has analyzed the problem and calculated it will take 45 comparisons. What we end up with is a very simple example of a process where students have produced their own theory
(that the number of comparisons is the sum of the numbers from 1 to n-1), and verified it with an experiment. Furthermore, they are in a position to explore a surprising consequence of the theory:
Sorting 100 items this way might be expected to require about 10 times the effort (450 comparisons), but in fact it takes more than 100 times as long (4,950 comparisons), motivating a quest for
better algorithms.
Students can also use an "unplugged" approach to experiment with processes that are less amenable to mathematical analysis. For example, Fitts' law is used in human-computer interaction to determine
how long it will take for a user to execute a series of pointer movements (such as clicking on a series of buttons or menus in an interface or touching controls on a smartphone). Fitts' law is
typically expressed as:
where T is the time taken to move the pointer, D is the distance moved, W is the width of the target, and a and b are constants that depend on the situation. Students can explore this relationship by
simply timing how long it takes to move a pencil back and forth between two targets on paper, and plot this for varying values of D and W (see Photo 2). This usually exposes the logarithmic
relationship, and also gives students insight into the level of detail that can be put into analyzing what seems like a very simple action, as well as the consequences of design decisions in
interfaces like the size and position of buttons. Of course, students in the target age group may not know the concept of a logarithm, but by plotting their results on log-scale graph paper they get
remarkable straight lines, which enable them to make predictions by interpolation, and also consider the validity of extrapolation (for example, what if the targets were a mile apart?)
Many of the "unplugged" activities introduce the nature of a problem only (such as the intractability of map coloring or the travelling salesperson problem). Others involve CS-relevant discrete
mathematics (such as exploring the mathematical patterns in binary number representations, or a minimal spanning tree for a graph). Still others explore elements of computing that are impacted by
human behavior (evaluating how physical interfaces such as door handles and oven controls might confuse users, exploring the predictability of text, or simulating the Turing Test for intelligence);
these activities are based heavily on experiments and heuristics.
Many activities introduce students to the surprises and paradoxes in computing: that it is possible to detect and correct errors in data without having it re-transmitted; that one of the fastest
sorting algorithms is slowest when given a list of already-sorted numbers; that a child could design a small combinatorial problem they know the solution for, but a computer would take billions of
years to solve it; and that randomness can help make algorithms run faster.
We have found when students use "unplugged" activities for formal learning, a background in scientific method supports them best. Usually they will need to describe the purpose of the activity
("experiment"), explain how it was set up (reproducibility), and report on results in a way that shows what they have discovered (well supported conclusions).
Given the wide range of computer science topics that can be explored at length by young students without using a computer, we have a very practical demonstration that the science of computing would
exist even if computers did not, and it can be explored using theory and experiment to validate hypotheses. This view is not just a philosophical discussion of semantics; to push the boundaries of
what can be done with computing technology, programmers need to understand and stretch the science behind it. It is in fact such people who can create systems that we might have thought were
impossible, or who will prove systems to be impossible before we waste time trying to build them.
Those who can expand the frontiers of the science of computing may not even be particularly interested in programming or the machine itself, but they have much to contribute through their passion and
talent for a scientific approach. Such students can be deterred by "computer science" courses that are primarily about programming, and they might even conclude they don't have ability in the
subject. The goal of Computer Science Unplugged is to give such students a glimpse of the science of computing as a motivation to learn the methods.
[1] Bell, T., Alexander, J., Freeman, I., and Grimley, M. Computer Science Unplugged: School Students Doing Real Computing Without Computers. The NZ Journal of Applied Computing and Information
Technology 13, 1 (2009), 20–29.
[2] Bell, T., Rosamond, F., and Casey, N. (2012). Computer Science Unplugged and related projects in math and computer science popularization. In H. L. Bodlaender, R. Downey, F. V. Fomin, and D. Marx
(Eds.), The Multivariate Complexity Revolution and Beyond: Papers in Honour of Michael Fellows (Vol. LNCS 7370, pp. 398–456). Heidelberg, Springer.
Tim Bell, Ph.D. in computers science (University of Canterbury, New Zealand), is a professor in the Department of Computer Science and Software Engineering at the University of Canterbury. His main
current research interest is computer science education; in the past he has also worked on computers and music, and data compression. He received the Science Communicator Award from the New Zealand
Association of Scientists in 1999, an inaugural New Zealand Tertiary Teaching Excellence Award in 2002, and the University of Canterbury Teaching Medal in 2008. He is a Guest Professor of Huazhong
University of Science and Technology in Wuhan, China.
©2014 ACM $15.00
Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial
advantage and that copies bear this notice and the full citation on the first page. To copy otherwise, to republish, to post on servers or to redistribute to lists, requires prior specific permission
and/or a fee.
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ball mill formula
The starting point for ball mill media and solids charging generally starts as follows: 50% media charge. Assuming 26% void space between spherical balls (nonspherical, irregularly shaped and
mixedsize media will increase or decrease the free space) 50% x 26% = 13% free space. Add to this another 10%15% above the ball charge for total of 23% ...
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Then in ballwear formula (25), T = /K Log10 Da/Db; but from (29), K = Rt/Wt. Then T = /Rt Log10 Da/Db T is 1 day, Wt is the original weight of the ball charge, and Rt is the ball wear for one
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During the last decade numerous protocols have been published using the method of ball milling for synthesis all over the field of organic chemistry. However, compared to other methods leaving
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An important condition that needs to be met for using the Bond formula is that the distributions should have a similar slope, and correction factors (such as those introduced by Rowland in 1982)
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The video contain definition, concept of Critical speed of ball mill and step wise derivation of mathematical expression for determining critical speed of b...
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You can calculate the circulation factor in ball mill by using following input data: 1. Fresh feed rate. 2. Coarse return. you can also calculate the following mill calculation by clinking below:
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Type of ball mill: • There is no fundamental restriction to the type of ball mill used for organic synthesis (planetary ball mill, mixer ball mill, vibration ball mill, .). • The scale of
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Calculating for the Diameter of Balls when the Critical Speed of Mill and the Mill Diameter is Given. d = D ( / Nc) 2
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For example your ball mill is in closed circuit with a set of cyclones. The grinding mill receives crushed ore feed. The pulp densities around your cyclone are sampled and known over an 8hour
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Ball mills. The ball mill is a tumbling mill that uses steel balls as the grinding media. The length of the cylindrical shell is usually times the shell diameter (Figure ). The feed can be dry,
with less than 3% moisture to minimize ball coating, or slurry containing 2040% water by weight.
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A ball mill also known as pebble mill or tumbling mill is a milling machine that consists of a hallow cylinder containing balls; mounted on a metallic frame such that it can be rotated along its
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The results showed that the ball milling technique could increase the content of functional groups (OH, C=C and CO, etc.) and aromatic structures on the surface of biochar, thus facilitating ...
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A Bond Ball Mill Work Index test is a standard test for determining the ball mill work index of a sample of ore. It was developed by Fred Bond in 1952 and modified in 1961 (JKMRC CO., 2006). This
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Home Variables in Ball Mill Operation Ball mill operation is often regarded as something of a mystery for several reasons. Ball milling is not an art it's just physics. The first problem will
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According to the calculation of the ball mill power at home and abroad, the correction formula of the former Soviet Union's Tovalov formula is adopted: 𝑃 𝐷𝑉𝑛𝜑 ...
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The choice of the top (or recharge) ball size can be made using empirical equations developed by Bond (1958) or Azzaroni (1984) or by using special batchgrinding tests interpreted in the content
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