content stringlengths 86 994k | meta stringlengths 288 619 |
|---|---|
82.1 miles per gallon to kilometers per gallon
Fuel Consumption Converter - Miles per gallon to kilometers per gallon - 82.1 kilometers per gallon to miles per gallon
This conversion of 82.1 miles per gallon to kilometers per gallon has been calculated by multiplying 82.1 miles per gallon by 1.6093 and the result is 132.1271 kilometers per gallon. | {"url":"https://unitconverter.io/miles-per-gallon/kilometer-per-gallon/82.1","timestamp":"2024-11-12T01:14:25Z","content_type":"text/html","content_length":"15768","record_id":"<urn:uuid:08b6d0ce-df0e-499b-b0bb-191c18d7470d>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00582.warc.gz"} |
[QSMS Monthly Seminar 2023-03-24] Rational points on abelian varieties
2023년 3월 QSMS Monthly Seminar
• Date: 3월 24일 금 오후 2시 ~ 5시
• Place: 27동 220호
• Contents:
Speaker : 유화종 (오후 2시)
Title : Rational points on abelian varieties
Abstract :
In this talk, we first introduce some well-known results about the rational points on algebraic curves over the rationals, especially on elliptic curves. Then we specialize our interest to those of
finite order, called the rational torsion points, and discuss Mazur's theorem.
If time permits, we introduce a natural generalization of Mazur's result.
Speaker : 배한울 (오후 4시)
Title : Duality in Rabinowitz Fukaya category
Abstact :
Rabinowitz Floer cohomology is a Floer theoretic invariant associated to the contact boundary of a Liouville domain X, which measures the failure for the continuation map from the symplectic homology
of X to the symplectic cohomology of X to be an isomorphism. This is a Floer theoretic analogue of the fact that the cohomology of the boundary Y of a manifold X measures the failure for the natural
map from the relative cohomology of the pair (X,Y) to the cohomology of X to be an isomorphism. In this talk, I will first briefly introduce Rabinowitz Floer homology associated with a pair of
Lagrangians and then introduce its categorification, called Rabinowitz Fukaya category. Finally, I will explain that, under certain conditions, Rabinowitz Fukaya category admits a certain duality,
which is a Floer theoretic analogue of Poincare duality. This is based on joint work with Wonbo Jeong and Jongmyeong Kim. | {"url":"https://qsms.math.snu.ac.kr/index.php?mid=board_sjXR83&listStyle=viewer&order_type=desc&l=en&sort_index=readed_count&document_srl=2467&page=1","timestamp":"2024-11-14T17:25:48Z","content_type":"text/html","content_length":"24609","record_id":"<urn:uuid:52b6f377-42e6-4546-a327-0008059e6089>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00647.warc.gz"} |
Two Inside-Out Limit Problems
(A new question of the week)
Limits can be challenging. They can be even more challenging when they require L’Hôpital’s rule or more advanced methods (Maclaurin series), and then are turned inside-out by asking not for the limit
itself, but for parameters that will result in a specified limit, or what values of the limit are possible. Two of us helped with such problems in November.
First problem
Here is the easier of two questions Vignesh sent us 20 minutes apart:
I have tried to solve the problem but at its first step it has become a complicated equation. Help me to solve it if I am in correct way or else guide me the perfect way to solve it.
NOTE: Multicorrect question.
The simplification is indeed going in an unpleasant direction. (Or is it? Hold that thought.)
Don’t simplify first (for a change!)
Doctor Fenton answered, suggesting the use of Maclaurin series (the approach, as we’ll see, that Vignesh had already used in the other problem), without first combining the fractions:
Hi Vignesh,
I would recommend expanding the two terms in the parentheses into Maclaurin series and combining them into a single series for the difference, which then becomes the numerator of a fraction with
denominator x^3.
Then the only way that fraction can have a finite limit as x→0 is for the constant term and the coefficients of x and x^2 in the series all to be 0. That will give you equations which determine a
and b. With values for a and b, you can determine the limit.
Vignesh replied, clearly very familiar with the technique:
Hello Doctor Fenton
With your suggestion I expanded the two terms in the parentheses and got the value of ‘a’ and ‘b’ which made me to complete #57, also I have done #58 but no option matched with it. Once go
through the sum, if I had done a mistake please tell me.
Examining the series expansion
Series expansions can be a powerful method for dealing with limits like this. Let’s look carefully at what he is doing.
The first fraction expands using the following Maclaurin series, for which Vignesh is using a formula, while I’ll use it as given in Wikipedia:
This is a special case of the binomial series. The second fraction can be expanded using the following basic form, which can be found by mere long division (or by working backward from the geometric
Applying this to \(\displaystyle\frac{1+ax}{1+bx}=1+(a-b)x\frac{1}{1+bx}\), by replacing \(x\) with \(-bx\), we get
$$1+(a-b)x\left[1+(-bx)+(-bx)^2+(-bx)^3+\cdots\right] =\\ 1+(a-b)x\left(1-bx+b^2x^2-b^3x^3+\cdots\right) =\\ 1+(a-b)x-(a-b)bx^2+(a-b)b^2x^3-(a-b)b^3x^4+\cdots$$
Subtracting those two series and combining like terms, we get $$\left[1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{5}{16}x^3+\cdots\right]-\left[1+(a-b)x-(a-b)bx^2+(a-b)b^2x^3+\cdots\right] =\\ \left(-\frac
Dividing this by \(x^3\), we are left with $$\left(-\frac{1}{2}-a+b\right)x^{-2}+\left(\frac{3}{8}+b(a-b)\right)x^{-1}+\left(-\frac{5}{16}-(a-b)b^2\right)+\cdots$$
The omitted terms are all multiplied by positive powers of x, which will go to 0 so they can be ignored. The terms with negative powers of x go to infinity, so their coefficients must be 0; and the
constant term will be the limit. This gives us two equations in a and b, $$-\frac{1}{2}-a+b=0\\\frac{3}{8}+b(a-b)=0$$ Solving the first for \(a=b-\frac{1}{2}\) and putting that into the second
equation, we get $$\frac{3}{8}+b(-\frac{1}{2})=0$$ so that \(b=\frac{3}{4}\), from which \(a=\frac{1}{4}\). Vignesh is correct here, and therefore is right that \(a+b=1\).
The limit, then, is $$-\frac{5}{16}-(a-b)b^2=-\frac{5}{16}-(\frac{1}{4}-\frac{3}{4})\left(\frac{3}{4}\right)^2=-\frac{1}{32}$$
If you compare my work on the series with Vignesh’s, you will notice a sign error, which explains the wrong answer. (One sign error in all this work is impressive!)
Making it work
Doctor Fenton pointed out the one error:
I have a different sign for the b^2(b-a) term in the coefficient of x^3. That makes the limit -1/32 instead of -19/32.
Vignesh made the correction:
I made a mistake, instead of taking negative sign I took positive sign. Now I solved it, once check it.
Now the limit is \(-\frac{1}{32}\), and the requested ratio is 24, which is on the list.
Doctor Fenton:
That’s what I got. Good work!
After I finished writing this post, I decided to check this answer by plugging in the values for a and b and finding the limit directly. I found that I could do so merely by using the complement …
which is just what Vignesh had done in his initial work. Could we actually finish that work? Yes!
Here is the last line of his original attempt (with the denominator backed up one step to keep it a little simpler): $$\frac{(1+bx)^2-(1+ax)^2(1+x)}{x^3\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}$$
Expanding and simplifying the numerator, we get $$\frac{x(-a^2x^2+b^2x-a^2x-2ax+2b-2a-1)}{x^3\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}$$
The factors in the denominator apart from the first approach 2; so we need the numerator to cancel with the \(x^3\). We will have a limit, therefore, if \(-a^2x^2+b^2x-a^2x-2ax+2b-2a-1\) is a
constant multiple of \(x^2\). That will be true only if the coefficients of the linear and constant terms are zero: that is, if $$b^2-a^2-2a=0\\2b-2a-1=0$$
What do you think is the solution to that? Yes, \(a=\frac{1}{4}\) and \(b=\frac{3}{4}\).
And what is the limit? $$\lim_{x\to 0}\frac{x(-a^2x^2)}{x^3\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}=\\\lim_{x\to 0}\frac{-a^2}{\sqrt{1+x}(1+bx)(1+bx+(1+ax)\sqrt{1+x})}=-\frac{1}{32}$$
So series were not really needed! But that method was enlightening, and leads us to the other problem …
Second problem
Vignesh actually sent the next problem 20 minutes before the other; it turns out to be considerably trickier:
I have started solving the problem and I think that I have done 90% solution but at last the equation had become quite complicated so that I am unable to finish it. Please help me.
NOTE: Multicorrect question.
Examining the series approach
Vignesh has used Maclaurin series (before Doctor Fenton had suggested it). Here are the basic series expansions he used this time:
In each case, something has to be substituted for x:
Only enough terms have to be written so that, after cancelling powers of x, remaining terms will go to zero. Using terms through \(x^4\) as Vignesh has done (but being a little more consistent), the
numerator becomes $$\sin(x^2)+2\cos(bx)-ax^4-2=\\ \left[x^2+\cdots\right]+2\left[1-\frac{b^2x^2}{2!}+\frac{b^4x^4}{4!}+\cdots\right]-ax^4-2=\\x^2+2-b^2x^2+\frac{b^4x^4}{12}-ax^4-2+\cdots=\\ (1-b^2)x^
and the denominator becomes $$e^{ax}-1-ax-2x^2-\frac{a^3x^3}{6} =\\ \left[1+ax+\frac{a^2x^2}{2!}+\frac{a^3x^3}{3!}+\frac{a^4x^4}{4!}+\cdots\right]-1-ax-2x^2-\frac{a^3x^3}{6} = \\ \frac{a^2x^2}{2}-2x^
2+\frac{a^4x^4}{24}+\cdots = \\ x^2\left(\frac{a^2}{2}-2+\frac{a^4}{24}x^2\right)+\cdots$$
Canceling \(x^2\) leaves us with the fraction $$\frac{(1-b^2)+\left(\frac{b^4}{12}-a\right)x^2+\cdots}{\frac{a^2}{2}-2+\frac{a^4}{24}x^2+\cdots}$$
When x goes to zero, we are left with the limit $$\frac{(1-b^2)}{\frac{a^2}{2}-2}$$ You will observe a small difference in what I have done here. But at this point I hadn’t yet duplicated his work in
that way; instead, I had taken a different approach in order to check his work independently. I’ll show that in a moment.
The L’Hôpital alternative
I answered this one, while Doctor Fenton was working on the other problem:
As I understand it, “I” means the set of integers (commonly called Z), so that a and b must be integers (positive or negative).
When you say the problem is “multicorrect”, I presume you mean that more than one answer can be correct, so you need to choose all choices in the list that are correct.
The word “the” in “the possible values” seems to imply that it is asking for all possible values of a + b, and likewise for #60? But if we take #60 this way, then it would be saying that #59 is
impossible, as 25/8 would not be a possible limit. It must instead be asking merely which of the four values are possible limits.
I applied L’Hopital’s rule and obtained your equation (2(1-b^2)/(a^2-4) = 25/8. I didn’t find that b^4 = 12a.
Now we have a Diophantine equation which is equivalent to 25a^2 + 16b^2 = 116. This can be solved rather easily; but the solutions turn out to force you to take a further step.
L’Hôpital’s rule is closely related to the use of series expansions; it amounts to using only the first term or so. The limit we are to find is
$$\lim_{x\to 0}\frac{\sin(x^2)+2\cos(bx)-ax^4-2}{e^{ax}-1-ax-2x^2-\frac{a^3x^3}{6}}$$
This has the form 0/0, so we take derivatives of the numerator and the denominator, and get
$$\lim_{x\to 0}\frac{2x\cos(x^2)-2b\sin(bx)-4ax^3}{ae^{ax}-a-4x-\frac{a^3x^2}{2}}$$
This still has the form 0/0, so we differentiate again, and get
$$\lim_{x\to 0}\frac{2\cos(x^2)-4x^2\sin(x^2)-2b^2\cos(bx)-12ax^2}{a^2e^{ax}-4-a^3x}$$
Setting x to 0, this becomes $$\frac{2-0-2b^2-0}{a^2-4-0} = \frac{2(1-b^2)}{a^2-4}$$ which is what Vignesh got. So this has to equal \(\frac{25}{8}\). And we can cross-multiply the equation
$$\frac{2(1-b^2)}{a^2-4} = \frac{25}{8}$$
to get
$$2(1-b^2)\cdot 8=25\cdot (a^2-4)$$
which simplifies to $$16-16b^2=25a^2-100\\ 25a^2+16b^2=116$$
But what about Vignesh’s \(\frac{b^4}{12}-a = 0\)? That comes from his second term in the numerator, which he is thinking has to be zero in order to have a limit; that would be appropriate if that
term had degree less than 2. But since, after canceling \(x^2\), that will still be multiplied by \(x^2\), it does not have to be zero. By using L’Hôpital’s rule instead of series expansion, I never
even got to see that, so I wasn’t tempted to misread it! So in fact, we don’t have a separate constraint imposed by the existence of a limit.
Moving forward
Vignesh replied, carrying out my suggestion to solve the Diophantine equation (that is, find all integer solutions):
I solved #59 with obtained Diophantine equation once check it. What step I have to do to solve #60?
This is good. I think I myself just saw immediately that \(a^2=4, b^2=1\) would work, and no other positive solutions for the squares are possible. We now have four solutions, each of which yields
one of the four options given for the sum \(a + b\), so all of them appear to be correct. But …
A deceptive solution
I pointed out that there was more to do:
I mentioned a further step you need to take; that starts with checking. What happens when you put your values for a and b into your equation (2(1-b^2)/(a^2-4) = 25/8 ?
I haven’t yet looked at #60, but will do so now.
Vignesh replied,
Yes, when I put values of a and b in the equation I am getting ‘0’. Does it mean the values of a and b are wrong?
Yes, we get \(\frac{0}{0} = \frac{25}{8}\), which doesn’t really work.
I answered,
That means it is still indeterminate, so we can’t be sure yet what the limit will be.
I think you need to apply L’Hopital yet again to see under what further conditions, if any, the limit will be correct. (I started that, but didn’t yet get to the conclusion.)
Or there may be a better way.
That was the end of the discussion, for some reason, and I didn’t get to go further at the time. Let’s see if we can finish the problem now.
Equivalent to applying L’Hôpital’s rule a third time as I suggested, Vignesh could just use more terms of the series expansion. Let’s do it both ways, starting with mine:
Beyond indeterminacy: continuing with L’Hôpital’s rule
Differentiating a third time yields
$$\lim_{x\to 0}\frac{-12x\sin(x^2)-8x^3\cos(x^2)+2b^3\sin(bx)-24ax}{a^3e^{ax}-a^3}$$
which again has the form 0/0. (Note that we are assuming a and b are solutions of the equation above, so that we could apply L’Hôpital again!) So we differentiate a fourth time:
$$\lim_{x\to 0}\frac{-12\sin(x^2)-48x^2\cos(x^2)+16x^4\sin(x^2)+2b^4\cos(bx)-24a}{a^4e^{ax}}$$
This time we don’t get zeros; the limit becomes \(\displaystyle\frac{2b^4-24a}{a^4}\). We want this to equal \(\frac{25}{8}\), so we try our four possibilities, \((2, 1), (2, -1), (-2, -1), (-2, -1)
\). We find that \(\frac{2b^4-24a}{a^4} = \frac{-23}{8},\frac{-23}{8},\frac{25}{8},\frac{25}{8}\) respectively. So the solution is \(a=-2,b=\pm 1\), and the answer to the problem is that \(a+b=-1\
text{ or }-3\). Only two of the four apparent possibilities are real.
Continuing with the series method
Using series, the fact that I had to apply L’Hôpital twice more suggests we need two more terms for each expansion (through \(x^4\)), which happens to be what I did above (where Vignesh had gone that
far only in the numerator): $$\frac{(1-b^2)+\left(\frac{b^4}{12}-a\right)x^2+\cdots}{\frac{a^2}{2}-2+\frac{a^4}{24}x^2+\cdots}$$
Letting \(a=\pm2\) and \(b=\pm1\), this becomes (with signs corresponding to \(a=+2\) and \(a=-2\) respectively) $$\lim_{x\to 0}\frac{0+\left(\frac{1}{12}\mp 2\right)x^2+\cdots}{0+\frac{2}{3}x^2+\
cdots} = \frac{\frac{1}{12}\mp 2}{\frac{2}{3}} = \frac{1}{8}\mp 3 = -\frac{23}{8}\text{ or }\frac{25}{8}$$
So we need to take the negative sign for a, and the solution as before is \(a=-2,b=\pm 1\).
Finding possible limits
How about problem 60, which asked which of the values \(\frac{1}{2}\), \(\frac{1}{8}\), \(-\frac{4}{3}\), and \(-\frac{23}{8}\) are possible? Well, we’ve seen that last one, haven’t we? That’s the
limit when \(a = +2\). (We found that while we were trying to make the limit \(\frac{25}{8}\), but it was a valid limit – one of two possible when the limit was still indeterminate after the first
part of our work.)
But we need to go back to the expression we found for the limit:
$$L = \frac{2(1-b^2)}{a^2-4}$$
What are the possible values of L when the parameters are integers? Fortunately, we don’t need to find all of them (and probably couldn’t); we just need to try determining values of a and b that will
work for each choice we are given for L.
Let’s start with \(L=\frac{1}{2}\). In solving the Diophantine equation $$\frac{2(1-b^2)}{a^2-4} = \frac{1}{2}$$ we find that it simplifies to $$a^2+4b^2=8$$ The only solutions are those we have seen
before, \((2, 1), (2, -1), (-2, -1), (-2, -1)\), none of which yield \(L=\frac{1}{2}\). So this choice is out. Similarly, \(L=\frac{1}{8}\) also fails.
But when we solve $$\frac{2(1-b^2)}{a^2-4} = -\frac{4}{3}$$ we get the equation $$2a^2-3b^2=5$$ which in addition to the familiar solution, is also true for \(a=\pm 4,b=\pm 3\), or \(a=\pm 16,b=\pm
13\). So this, too, is a possible limit.
The answer is that, of the four choices offered, \(L=-\frac{4}{3}\) or \(L=-\frac{23}{8}\).
I think this problem was one of the more interesting ones I’ve seen, and provides some useful experience comparing the series and L’Hôpital approaches to a limit.
Leave a Comment
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://www.themathdoctors.org/two-inside-out-limit-problems/","timestamp":"2024-11-06T12:27:20Z","content_type":"text/html","content_length":"128377","record_id":"<urn:uuid:02654b17-65c8-445a-a910-45ca21d09ab8>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00277.warc.gz"} |
(ii) 2sin26π+cos267πfos23π=23
5. Prove that tan70∘=2tan50∘+... | Filo
Question asked by Filo student
(ii) 5. Prove that . 6. Prove that 7. Prove that . 8. It prove that . 9. If . Prove that . 10. Prove that: .
Not the question you're searching for?
+ Ask your question
Video solutions (1)
Learn from their 1-to-1 discussion with Filo tutors.
9 mins
Uploaded on: 9/15/2022
Was this solution helpful?
Found 2 tutors discussing this question
Discuss this question LIVE for FREE
14 mins ago
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Practice more questions on Trigonometry
View more
Students who ask this question also asked
View more
Stuck on the question or explanation?
Connect with our Mathematics tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text (ii) 5. Prove that . 6. Prove that 7. Prove that . 8. It prove that . 9. If . Prove that . 10. Prove that: .
Updated On Sep 15, 2022
Topic Trigonometry
Subject Mathematics
Class Class 11
Answer Type Video solution: 1
Upvotes 148
Avg. Video Duration 9 min | {"url":"https://askfilo.com/user-question-answers-mathematics/5-prove-that-6-prove-that-7-prove-that-8-it-prove-that-9-if-31353033383233","timestamp":"2024-11-07T20:05:00Z","content_type":"text/html","content_length":"437681","record_id":"<urn:uuid:4a147336-c3ce-4074-b352-f1de58bad6c3>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00554.warc.gz"} |
Year 11 – 3: Data in Context Mathematics – Northern Territory (NT)
Year 11 – 3: Data in Context Mathematics – Northern Territory (NT)
NT Year 11 – 3: Data in Context Mathematics
# TOPIC TITLE
1 Study Plan Study plan – Year 11 – 3: Data in Context
Objective: On completion of the course formative assessment a tailored study plan is created identifying the lessons requiring revision.
2 Percentages Changing percentages to fractions and decimals
Objective: On completion of the lesson the student will be able to change percentages to fractions and know how to change percentages to decimals.
3 Percentages One quantity as a percentage of another
Objective: On completion of the lesson the student will be able to find a percentage of an amount and how to express one quantity as a percentage of another.
4 Lines and angles Mapping and grid references
Objective: On completion of the lesson the student will be able to identify specific places on a map and use regions on a grid to locate objects or places.
5 Trigonometry-compass Bearings – the compass.
Objective: On completion of the lesson the student will be able to identify compass bearings, compass bearings with acute angles and 3 figure bearings from true north.
6 Lines and angles Informal coordinate system
Objective: On completion of the lesson the student will be able to use an informal coordinate system to specify location, and locate coordinate points on grid paper.
7 Data Pictograms
Objective: On completion of the lesson the student will be able to organise, read and summarise information in picture graphs.
8 Data Bar Charts
Objective: On completion of the lesson the student will be able to organise, read and summarise information in column graphs.
9 Data Line graphs.
Objective: On completion of the lesson the student will be able to organise, read and summarise information in line graphs.
10 Data Pie and bar graphs.
Objective: On completion of the lesson the student will be able to organise, read and summarise information in pie and bar graphs.
11 Statistics Frequency distribution table
Objective: On completion of the lesson the student will be able to construct a frequency distribution table for raw data and interpret the table.
12 Statistics Frequency histograms and polygons
Objective: On completion of the lesson the student will be able to construct and interpret frequency histograms and polygons.
13 Statistics Relative frequency
Objective: On completion of the lesson the student will be able to collect, display and make judgements about data.
14 Statistics The range.
Objective: On completion of the lesson the student will be able to determine the range of data in either raw form or in a frequency distribution table.
15 Statistic-probability The mode
Objective: On completion of the lesson the student will understand how to find the mode from raw data, a frequency distribution table and polygon.
16 Statistic-probability The mean
Objective: On completion of the lesson the student will be able to calculate means from raw data and from a frequency table using an fx column.
17 Statistic-probability The median
Objective: On completion of the lesson the student will be able to determine the median of a set of raw scores
18 Statistic-probability Cumulative frequency
Objective: On completion of the lesson the student will be able to construct cumulative frequency columns, histograms and polygons.
19 Statistic-probability Calculating the median from a frequency distribution
Objective: On completion of the lesson the student will be able to determine the median from a cumulative frequency polygon.
20 Statistics – grouped data Calculating mean, mode and median from grouped data
Objective: On completion of the lesson the student will be capable of identifying class centres, get frequency counts and determine the mean and mode values.
21 Statistics using a calculator Statistics and the student calculator
Objective: On completion of the lesson the student will be capable of using a scientific calculator in statistics mode to calculate answers to statistical problems.
22 Statistics – Range and dispersion Range as a measure of dispersion
Objective: On completion of the lesson the student will be able to determine the range and using it in decision making.
23 Statistics – Spread Measures of spread
Objective: On completion of the lesson the student will be able to find the standard deviation, using a data set or a frequency distribution table and calculator.
24 Statistics – Standard deviation Standard deviation applications
Objective: On completion of the lesson the student will be able to use standard deviation as a measure of deviation from a mean.
25 Statistics – Standard deviation Normal distribution
Objective: On completion of the lesson the student will be able to use the standard deviation of a normal distribution to find the percentage of scores within ranges.
26 Statistics – Interquartile range Measures of spread: the interquartile range
Objective: On completion of the lesson the student will be able to find the upper and lower quartiles and the interquartile range
27 Exam Exam – Year 11 – 3: Data in Context
Objective: Exam | {"url":"https://www.futureschool.com/australian-curriculum/northern-territory/mathematics-year-11-3-data-in-context/","timestamp":"2024-11-06T21:57:15Z","content_type":"text/html","content_length":"57862","record_id":"<urn:uuid:55809724-6c17-4b92-a916-7bbad82d08f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00234.warc.gz"} |
Structure Template
Math Ia Structure Template
Math Ia Structure Template - This is the most challenging step. See what past students did and make your maths ia perfect by learning from examiner commented examples! Very useful as it gives an idea
of the structure for math ias. Web if you are preparing for math ia, you will need lots of coffee for sleepless nights to be ready to structure the exam. It should show personal engagement with the
topic at hand. Anyone can ace the math ia if they put in adequate effort and use this ebook as a template. Web scroll down this page to find over 300 examples of maths ia exploration topics and ideas
for ib mathematics students doing their internal assessment (ia). Alongside the criteria, samples of the student’s work (oral performances, portfolios, lab reports, and essays) are also submitted
to the ib for the final grade. Arthur was literally the only person in his hl math class that got a 7 on. Web this video will answer some of the common questions people have when writing their math
ia as well as give you some helpful tips to write your math ia.
How to Structure and Format Your Math IA Lanterna Blog
Web high scoring ib maths internal assessment examples. Web this video will answer some of the common questions people have when writing their math ia as well as give you some helpful tips to write
your math ia. Web mathematics explored should either be part of the syllabus, or at a similar level or beyond. Web this article covers ib.
How to Structure and Format Your Math IA Lanterna Blog
See what past students did and make your maths ia perfect by learning from examiner commented examples! Web the math ia is an internal assessment that makes up 20% of your final grade. Web jun 2,
2021 7 min read how i got a 7 in the ib maths ia 🙌 this is a complete guide to the mathematics aa.
Math IA Structure Esquire Writings Blog
Web explain what you are doing, why you are doing it. Alongside the criteria, samples of the student’s work (oral performances, portfolios, lab reports, and essays) are also submitted to the ib for
the final grade. Why have your chosen this mathematical process/technique/test, demonstrate that you. Web an important component of this is the internal assessment (ia). See what past.
This Is The Most Challenging Step.
Anyone can ace the math ia if they put in adequate effort and use this ebook as a template. Web here is a quick guide to the criteria of ib maths internal assessment regardless of aa or ai (sl, hl).
Web structure and how to choose a topic the maths ia forms 20% of your overall grade for maths studies, sl and hl. It should not be completely based on mathematics listed in the prior.
Alongside The Criteria, Samples Of The Student’s Work (Oral Performances, Portfolios, Lab Reports, And Essays) Are Also Submitted To The Ib For The Final Grade.
Web this video will answer some of the common questions people have when writing their math ia as well as give you some helpful tips to write your math ia. Web template for math ia. Web the math ia
is an internal assessment that makes up 20% of your final grade. Web explain what you are doing, why you are doing it.
Web Mathematics Explored Should Either Be Part Of The Syllabus, Or At A Similar Level Or Beyond.
This is a sample template for math ia. Web maths ia [classic] by konrad suchodolski edit this template use creately’s easy online diagram editor to edit this diagram, collaborate with others and
export results to multiple. Follow these steps to ensure your ia. Web jun 2, 2021 7 min read how i got a 7 in the ib maths ia 🙌 this is a complete guide to the mathematics aa and ai internal
Web This Article Covers Ib Math Ia Rubrics, Process Key Pointers, The Structure Of The Investigation, And Interesting Ib Math Ia Topics That Will Stimulate Your Mind And Help.
Why have your chosen this mathematical process/technique/test, demonstrate that you. Web this criterion assesses to what extent the student is able to use appropriate mathematical language (notation,
symbols, terminology), define key terms where. In this article, we’ve broken down the best way. Understand why it has been.
Related Post: | {"url":"https://shopwithquality.us/en/math-ia-structure-template.html","timestamp":"2024-11-10T12:37:57Z","content_type":"text/html","content_length":"20588","record_id":"<urn:uuid:2f39829c-18f7-4020-9cc0-9ea809d6ca17>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00037.warc.gz"} |
Power: Definition, Formula, Units, Average Power, Solved Examples
How is power calculated?
The amount of work done or energy transformed in a given amount of time is defined as power. It is expressed mathematically as\(P=\frac{W}{t}\)
What are the 3 equations for power?
P = V x I\(P = I^{2}R\)\(P = \frac{V^{2}}{R}\)
Is watt a power?
Watt is the S.I. unit.
Can power be negative?
Yes, because it is the rate at which work is completed. So, if work done is negative, time taken is positive, then power is negative.
How is force related to power?
Force and power appear to have similar meanings and are frequently confused for one another. However, they are not interchangeable in physics.
What is meant by Instantaneous Power?
The power at a given instant is referred to as instantaneous power. | {"url":"https://testbook.com/physics/power","timestamp":"2024-11-03T04:33:52Z","content_type":"text/html","content_length":"862293","record_id":"<urn:uuid:cd9dbc50-2439-4caa-854b-6793fbf7384c>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00800.warc.gz"} |
Regression fit
We are analyzing our calibration curves in Skyline for peptide absolute quantification. We have observed that Skyline now has different regression fits besides linear. Would like to know, if we
should consider any criteria when applying one type instead of the other. Do you have any suggestions?
Thank you very much in advance. Best, Daniela | {"url":"https://skyline.ms/announcements/home/support/thread.view?rowId=63503","timestamp":"2024-11-12T03:49:47Z","content_type":"text/html","content_length":"19960","record_id":"<urn:uuid:dcc826e4-f150-4575-8250-cf8116308a96>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00508.warc.gz"} |
Quantifying accuracy of learning via sample width
In a recent paper, the authors introduced the notion of sample width for binary classifiers defined on the set of real numbers. It was shown that the performance of such classifiers could be
quantified in terms of this sample width. This paper considers how to adapt the idea of sample width so that it can be applied in cases where the classifiers are defined on some finite metric space.
We discuss how to employ a greedy set-covering heuristic to bound generalization error. Then, by relating the learning problem to one involving certain graph-theoretic parameters, we obtain
generalization error bounds that depend on the sample width and on measures of 'density' of the underlying metric space.
Publication series
Name Proceedings of the 2013 IEEE Symposium on Foundations of Computational Intelligence, FOCI 2013 - 2013 IEEE Symposium Series on Computational Intelligence, SSCI 2013
Conference 2013 IEEE Symposium on Foundations of Computational Intelligence, FOCI 2013 - 2013 IEEE Symposium Series on Computational Intelligence, SSCI 2013
Country/Territory Singapore
City Singapore
Period 16/04/13 → 19/04/13
Dive into the research topics of 'Quantifying accuracy of learning via sample width'. Together they form a unique fingerprint. | {"url":"https://cris.ariel.ac.il/en/publications/quantifying-accuracy-of-learning-via-sample-width-3","timestamp":"2024-11-11T00:14:39Z","content_type":"text/html","content_length":"55226","record_id":"<urn:uuid:0f921c00-2c9c-49cb-80d8-347a3f603436>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00011.warc.gz"} |
Electron Tomography
The basic idea of electron tomography is very simple: from a collection of a large number of projections over a wide angular range, it is possible to reconstruction the 3D shape of the imaged object.
The basis is the Fourier slice theorem, which states that the Fourier transforms of a projection image corresponds to a central slice of the 3D Fourier transform of the corresponding object. By
combining the Fourier transform of each individual projection under the appropriate angle and then performing an inverse Fourier transform one gets the real space structure of the object back (Fig.
2). Alternatively, the reconstruction can be done in real space using direct backprojection appoaches (Fig. 1).
Fig. 1: Schematic representation of a tomographic reconstruction: 1) Fig. 2: Schematic representation of a tomographic reconstruction in recipical space: 1) acquisition of projecion images , 2)
acquisition of projecion images at different tilt-angles, 2) alignment Fourier transorm, 3) combination of Fourier transform at different tilt-angles, and 4) inverse Fourier transform to
and 3) real space back projection. reconstruction original shape (from Friedrich et al. (2009) Chem.Rev., 109: 1613).
Both approaches are equally valid, just different mathematical implementations. However, for the 3D reconstruction to work properly, two improtant aspects have be fullfilled:
1) The projection images have to be true projections of the structure (the signal has to be a strictly monotonic function of the materials density, composition, or whatever property should be
reconstructed in 3D ... in practice slight (random) variations can be tolerated in the data, but large systematic differenes are problematic.
2) The tilt-series has to be very well aligned, so that it corresponds to a true rotation around a single axis. The best way to achieve this is an alignment based on fiducal markers using e.g. IMOD
or FEI Inspect3D.
Depending on the sample thickness, the electron beam stability and the resolution of the original images, it is typically possible to reconstruct object features at the 1-2 m level in 3D using
electron tomography for normal materials science TEM specimens. An estimate for the 3D resolution (d) of a recontruction of a spherical object of diameter (D) based on (N) projections can be obtained
based on the Crowther equation
d = pi * D / N
If the tilt-series has only been acquired covering a limited tilt-angle α, all features will be elongated in z-direction, which can be approximated by the elongation factor e
e = SQRT( (α+sinα*cosα) / (α-sinα*cosα) )
Application Examples
We have used electron tomography for a wide range of materials science applications ranging from catalysis, to quantum dots, to polymer composites and semiconductor structures. A few of these
examples are shown here to illustrate the potential that electron tomography offers.
Fig. 5: Surface rendering of a DRAM transsistor. The
Fig. 3: Volume rendering of a superlattice formed by CdS quantum dots. The lattice parameters of the Fig. 4: Surface rendering of a gate oxide as well as ~1.5 nm oxynitride layers are
cubic lattice were determined to be 3.2 nm. Individual supperlattice defects such as vacancies and bicontinuous block-copolymers with visible in the 3D reconstruction.
dislocations could be imaged in 3D. The 'large' yellow particles are 5 nm gold labels. single layer exfoliated silicates. Sample courtesy J.-S. Luo, H.-M. Lo, J.D. Russell,
Sample courtsey T. Levchenko and J.F. Corrigan, University of Western Ontario. Sample courtesy G. Cox, BASF AG Inotera
T. Levchenko et al., Chem. Eur. J., 2011, 17(51), 14394-14398 C. Kübel et al., AIP Conference Proceedings 817,
p223-228 (2006).
Porous Materials
As an example for a quantitative analysis of interconnected structures, we have been working in close collaboration with Prof. Ulrich Tallareks group at the University Marburg, Dr. Katja Schladitz at
Fraunhofer ITWM and Dr. Alberto Villa group at the University Milano on describing the morphology of porous media across length scales from the nanometer to the micron level quantitatively by
characterizing both the topology and the geometry based on segmented FIB slice&view tomography and (S)TEM tomography data as illustrated in Fig. 6. Whereas classical bulk characterization techniques
such as small angle X-ray scattering or N[2]/Hg porometry require a model to analyze the pore structure, this approach provides directly interpretable information while covering statistically
significant volumes. In fact, this approach be used to develop models for an analysis using bulk techniques.
Figure 6: Schematic illustration of the quantitative analysis of porous materials (courtesy Dr. Daniela Stöckel, University Marburg)
Application areas for the research include separation media such as silica monolith used in HPLC, where we could correlate the average pore diameter in these disordered materials with the structural
uniformity (Fig. 7) and catalysis, where the pore structure (Fig. 8) is influencing the catalytic performance and selectivity.
Figue 7: Quantitative analysis of a series of FIB slice&view tomography reconstructions of silica monoliths with different pore sizes: The order parameter k is uniform when the average pore diameter
(represented by the mean chord length) is larger than ~ 1 um, but decreases rapidly for smaller average pore sizes.
D. Stoeckel, C. Kübel, M. Loeh, B. Smarsly, U. Tallarek, Langmuir, 2015, 31(26), 7391–7400; DOI: 10.1021/la5046018.
Figure 8: Segmented 3D reconstruction of a disordered and an ordered mesoporous carbon with a quantitative analysis of the pore size distributionand, the tortuosity and the pore orientation (pole
figure) providing a quantitative description of the visually intuitive differences between the materials.
Reconstruction Artefacts
The quality and reliability of the 3D reconstruction depends crucially on
1) the quality of the input images
2) the alignment accuracy
3) angular sampling and coverage
4) the reconstruction algorithm
The artifacts introduced by a limited alignment accuracy can be best understood based on the real space backprojection representation of the reconstruction process. Shifts between images in the
tilt-series lead to a smearing of the reconstructed object. In particular, if the tilt-axis position is offset, this results in the so called 'banana shape' of small reconstucted features (Fig. 9b).
Furthermore, if the images are systematically shifted in one direction from one image to the next in the tilt-series (which often happens with a cross-correlation alignment), this results in a
'Mercedes star' type structure for small features (Fig. 9c)
Fig. 9: Alignment artifacts visible in the XZ plane of a 3D reconstruction: a) well-aligned symmetric missing wedge artifacts due to limited tilt-range (strongly enhanced for visibility), b) offset
of the tilt-axis position (5 pixels) leading to banana shaped streaks and c) continuous shift between images in the tilt-series of 0.1 pixel result in a 'Mercedes star' shape.
Missing Wedge and Angular Sampling
The effects of angular sampling and coverage on the 3D reconstruction can best be visualized in recipical space. If one considers the Fourier synthesis of the 3D object by combining the Fourier
transforms of the images obtained at all different tilt-angles, it becomes immidiately visible that a limited tilt-range during the acquisition results in a missing wedge of information in the 3D
Fourier space. Any Fourier coefficients of the original object in this missing wedge will not be reconstructed and thus not be visible in the 3D reconstruction. Furthermore, a large angular increment
will results in a poor resolution in 3D as the 3D sampling becomes coarse.
Fig. 10: Schematic representation of the missing wedge and limited angular sampling.
Reconstruction Algorithms
The Fourier sampling at low spatial frequencies is much higher compared to the sampling at low spatial frequencies. Therefore, a stright backprojection would result in an overestimate of the low
frequencies while the high frequencies are underestimated, leading to a very blurry reconstruction.
One approach commonly used to overcome this problem is weighted backprojection (WBPJ), where a weighting filter is introduced reducing the contribution of the low spatial frequencies (Fig. 11).
Fig. 11: Comparison of straight backprojection (BPJ) and weighted backprojection (WBPJ).
Alternatively, iterative approaches such as the Simultaneous Iterative Reconstruction Technique (SIRT) are nowadays commonly used in materials sciences. In addition to reducing the oversampling, the
SIRT approach also reduces the noise in the reconstruction compared to WBPJ.
However, with both reconstruction approaches, it is difficult to accurately reconstruct the object intensities in 3D. The exact shape of the weighting filter (a Fourier filter) introduces strong
artifacts for the intensities. In the example of a zirconia filled polymer shown in Fig. 12, this effect can be seen as the image intensity of the vacuum around the sample is almost identical to the
intensity of the polymer matrix. In the SIRT reconstruction of the same data set, the relative intensities are better preserved.
Fig. 12: Electron tomographic reconstruction of zironia nanoparticles in a polymer matrix: comparison of WBPJ and SIRT and missing wedge effect (N. Kawase, M. Kato, H. Nishioka, H. Jinnai,
Ultramicroscopy, 2007, 107, 8-15).
However, also the SIRT reconstruction does not produce easily quantifiable intensities that can be directly segmented by thresholding. Because of convergence limitations, there is a dependence of the
reconstructed intensities of the averge density overlapping in the different projections. As one example, when reconstructing nanoparticles with different diameter, this results in decreasing
intensities with decreasing particle diameter, which makes an accurate segmentation very difficult.
Fig. 13: 3D reconstruction of metal nanoparticles on a catalyst support. The graph shows the experimental correlation between particle size and reconstructed average intensity of the particles (from
C. Kübel, D. Niemeyer, R. Cieslinski, S. Rozeveld, J. Mat. Sci. Forum, 2010, 638-642, 2517-2522).
Currently, a number of alternative techniques such as discrete tomography or compressive sensing are being develop to overcome these limitations when the sample fulfills appropriate requirements. | {"url":"https://www.int.kit.edu/1731.php","timestamp":"2024-11-12T02:29:25Z","content_type":"text/html","content_length":"58400","record_id":"<urn:uuid:89b8dbd9-fb29-4af0-8377-1f6644af8498>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00681.warc.gz"} |
Whats Possible
Problem, Clue, Solution, Teachers' note
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
Teaching approach. This lesson idea is about exploring and noticing structure^(ta).
The collection of NRICH activities are designed to develop students capacity to work as a mathematician. Exploring, questioning, working systematically, visualising, conjecturing, explaining,
generalising, justifying, proving are all at the heart of mathematical thinking.
This particular resource has been adapted from an original NRICH resource. NRICH promotes the learning of mathematics through problem solving. NRICH provides engaging problems, linked to the
curriculum, with support for teachers in the classroom. Working on these problems will introduce students to key mathematical process skills. They offer students an opportunity to learn by exploring,
noticing structure and discussing their insights, which in turn can lead to conjecturing, explaining, generalising, convincing and proof.
The Teachers’ Notes provided focus on the pedagogical implications of teaching a curriculum that aims to provoke mathematical thinking. They assume that teachers will aim to do for students only what
they cannot yet do for themselves. As a teacher, consider how this particular lesson idea can provoke mathematical thinking. How can you support students' exploration? How can you support
conjecturing, explaining, generalising, convincing and proof?.
Resource details
Title What's Possible?
Topic [[Topics/Algebra|Algebra]]
Teaching approach [[Teaching Approaches/Exploring and noticing structure|Exploring and noticing structure]]
Learning Objectives Exploring and noticing structure
Subject [[Resources/Maths|Maths]]
Age of students / grade [[Resources/Secondary|Secondary]], [[Resources/KS3|KS3]]
Related ORBIT Wiki
Files and resources to
view and download
Acknowledgement The NRICH website http://nrich.maths.org publishes free mathematics resources designed to challenge, engage and develop the mathematical thinking of students aged 5 to
19. NRICH also offers support for teachers by publishing Teachers’ Resources for use in the classroom.
License CC-By, with kind permission from NRICH. This resource was adapted from an original NRICH resource. | {"url":"https://oer.opendeved.net/wiki/Whats_Possible","timestamp":"2024-11-14T21:42:32Z","content_type":"text/html","content_length":"46430","record_id":"<urn:uuid:641b4844-e4c8-4e8c-a4d5-a2e47294c7d3>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00140.warc.gz"} |
4 Comments
This is roughly the process I follow to create songs. I create a chord progression I like, record it, and jam over it until I come up with a melody. Occasionally I’ll start with a melody and find
chords that fit.
Leave a Comment
You must be logged in to post a comment. | {"url":"https://realguitarsuccess.com/courses/monthly-practice-plan-june-2022/lessons/3-lazy-day-chords/","timestamp":"2024-11-03T09:33:46Z","content_type":"text/html","content_length":"170346","record_id":"<urn:uuid:45517129-41f8-4f7d-94f1-ede6c00530a9>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00868.warc.gz"} |
Bernoulli process
In probability and statistics, a Bernoulli process is a finite or infinite sequence of binary random variables, so it is a discrete-time stochastic process that takes only two values, canonically 0
and 1. The component Bernoulli variables X[i] are identical and independent. Prosaically, a Bernoulli process is a repeated coin flipping, possibly with an unfair coin (but with consistent
unfairness). Every variable X[i] in the sequence is associated with a Bernoulli trial or experiment. They all have the same Bernoulli distribution. Much of what can be said about the Bernoulli
process can also be generalized to more than two outcomes (such as the process for a six-sided die); this generalization is known as the Bernoulli scheme.
The problem of determining the process, given only a limited sample of the Bernoulli trials, may be called the problem of checking whether a coin is fair.
A Bernoulli process is a finite or infinite sequence of independent random variables X[1], X[2], X[3], ..., such that
• For each i, the value of X[i] is either 0 or 1;
• For all values of i, the probability that X[i] = 1 is the same number p.
In other words, a Bernoulli process is a sequence of independent identically distributed Bernoulli trials.
Independence of the trials implies that the process is memoryless. Given that the probability p is known, past outcomes provide no information about future outcomes. (If p is unknown, however, the
past informs about the future indirectly, through inferences about p.)
If the process is infinite, then from any point the future trials constitute a Bernoulli process identical to the whole process, the fresh-start property.
The two possible values of each X[i] are often called "success" and "failure". Thus, when expressed as a number 0 or 1, the outcome may be called the number of successes on the ith "trial".
Two other common interpretations of the values are true or false and yes or no. Under any interpretation of the two values, the individual variables X[i] may be called Bernoulli trials with parameter
In many applications time passes between trials, as the index i increases. In effect, the trials X[1], X[2], ... X[i], ... happen at "points in time" 1, 2, ..., i, .... That passage of time and the
associated notions of "past" and "future" are not necessary, however. Most generally, any X[i] and X[j] in the process are simply two from a set of random variables indexed by {1, 2, ..., n} or by
{1, 2, 3, ...}, the finite and infinite cases.
Several random variables and probability distributions beside the Bernoullis may be derived from the Bernoulli process:
• The number of successes in the first n trials, which has a binomial distribution B(n, p)
• The number of trials needed to get r successes, which has a negative binomial distribution NB(r, p)
• The number of trials needed to get one success, which has a geometric distribution NB(1, p), a special case of the negative binomial distribution
The negative binomial variables may be interpreted as random waiting times.
Formal definition
The Bernoulli process can be formalized in the language of probability spaces as a random sequence of independent realisations of a random variable that can take values of heads or tails. The state
space for an individual value is denoted by
Specifically, one considers the countably infinite direct product of copies of . It is common to examine either the one-sided set or the two-sided set . There is a natural topology on this space,
called the product topology. The sets in this topology are finite sequences of coin flips, that is, finite-length strings of H and T, with the rest of (infinitely long) sequence taken as "don't
care". These sets of finite sequences are referred to as cylinder sets in the product topology. The set of all such strings form a sigma algebra, specifically, a Borel algebra. This algebra is then
commonly written as where the elements of are the finite-length sequences of coin flips (the cylinder sets).
If the chances of flipping heads or tails are given by the probabilities , then one can define a natural measure on the product space, given by (or by for the two-sided process). Given a cylinder
set, that is, a specific sequence of coin flip results at times , the probability of observing this particular sequence is given by
where k is the number of times that H appears in the sequence, and n-k is the number of times that T appears in the sequence. There are several different kinds of notations for the above; a common
one is to write
where each is a binary-valued random variable. It is common to write for . This probability P is commonly called the Bernoulli measure.^[1]
Note that the probability of any specific, infinitely long sequence of coin flips is exactly zero; this is because , for any . One says that any given infinite sequence has measure zero.
Nevertheless, one can still say that some classes of infinite sequences of coin flips are far more likely than others, this is given by the asymptotic equipartition property.
To conclude the formal definition, a Bernoulli process is then given by the probability triple , as defined above.
Law of large numbers, binomial distribution and central limit theorem
Let us assume the canonical process with represented by and represented by . The law of large numbers states that, on the average of the sequence, i.e., , will approach the expected value almost
certainly, that is, the events which do not satisfy this limit have zero probability. The expectation value of flipping heads, assumed to be represented by 1, is given by . In fact, one has
for any given random variable out of the infinite sequence of Bernoulli trials that compose the Bernoulli process.
One is often interested in knowing how often one will observe H in a sequence of n coin flips. This is given by simply counting: Given n successive coin flips, that is, given the set of all possible
strings of length n, the number N(k,n) of such strings that contain k occurrences of H is given by the binomial coefficient
If the probability of flipping heads is given by p, then the total probability of seeing a string of length n with k heads is
This probability is known as the Binomial distribution.
Of particular interest is the question of the value of for a sufficiently long sequences of coin flips, that is, for the limit . In this case, one may make use of Stirling's approximation to the
factorial, and write
Inserting this into the expression for P(k,n), one obtains the Normal distribution; this is the content of the central limit theorem, and this is the simplest example thereof.
The combination of the law of large numbers, together with the central limit theorem, leads to an interesting and perhaps surprising result: the asymptotic equipartition property. Put informally, one
notes that, yes, over many coin flips, one will observe H exactly p fraction of the time, and that this corresponds exactly with the peak of the Gaussian. The asymptotic equipartition property
essentially states that this peak is infinitely sharp, with infinite fall-off on either side. That is, given the set of all possible infinitely long strings of H and T occurring in the Bernoulli
process, this set is partitioned into two: those strings that occur with probability 1, and those that occur with probability 0. This partitioning is known as the Kolmogorov 0-1 law.
The size of this set is interesting, also, and can be explicitly determined: the logarithm of it is exactly the entropy of the Bernoulli process. Once again, consider the set of all strings of length
n. The size of this set is . Of these, only a certain subset are likely; the size of this set is for . By using Stirling's approximation, putting it into the expression for P(k,n), solving for the
location and width of the peak, and finally taking one finds that
This value is the Bernoulli entropy of a Bernoulli process. Here, H stands for entropy; do not confuse it with the same symbol H standing for heads.
John von Neumann posed a curious question about the Bernoulli process: is it ever possible that a given process is isomorphic to another, in the sense of the isomorphism of dynamical systems? The
question long defied analysis, but was finally and completely answered with the Ornstein isomorphism theorem. This breakthrough resulted in the understanding that the Bernoulli process is unique and
universal; in a certain sense, it is the single most random process possible; nothing is 'more' random than the Bernoulli process (although one must be careful with this informal statement;
certainly, systems that are mixing are, in a certain sense, 'stronger' than the Bernoulli process, which is merely ergodic but not mixing. However, such processes do not consist of independent random
variables: indeed, many purely deterministic, non-random systems can be mixing).
Dynamical system
The Bernoulli process can also be understood to be a dynamical system, specifically, a measure-preserving dynamical system. This arises because there is a natural translation symmetry on the
(two-sided) product space given by the shift operator
The measure is translation-invariant; that is, given any cylinder set , one has
and thus the Bernoulli measure is a Haar measure.
The shift operator should be understood to be an operator acting on the sigma algebra , so that one has
In this guise, the shift operator is known as the transfer operator or the Ruelle-Frobenius-Perron operator. It is interesting to consider the eigenfunctions of this operator, and how they differ
when restricted to different subspaces of . When restricted to the standard topology of the real numbers, the eigenfunctions are curiously the Bernoulli polynomials!^[2]^[3] This coincidence of
naming was presumably not known to Bernoulli.
Bernoulli sequence
The term Bernoulli sequence is often used informally to refer to a realization of a Bernoulli process. However, the term has an entirely different formal definition as given below.
Suppose a Bernoulli process formally defined as a single random variable (see preceding section). For every infinite sequence x of coin flips, there is a sequence of integers
called the Bernoulli sequence associated with the Bernoulli process. For example, if x represents a sequence of coin flips, then the associated Bernoulli sequence is the list of natural numbers or
time-points for which the coin toss outcome is heads.
So defined, a Bernoulli sequence is also a random subset of the index set, the natural numbers .
Almost all Bernoulli sequences are ergodic sequences.
Randomness extraction
From any Bernoulli process one may derive a Bernoulli process with p = 1/2 by the von Neumann extractor, the earliest randomness extractor, which actually extracts uniform randomness.
Basic von Neumann extractor
Represent the observed process as a sequence of zeroes and ones, or bits, and group that input stream in non-overlapping pairs of successive bits, such as (11)(00)(10)... . Then for each pair,
• if the bits are equal, discard;
• if the bits are not equal, output the first bit.
This table summarizes the computation.
input output
00 discard
11 discard
For example, an input stream of eight bits 10011011 would by grouped into pairs as (10)(01)(10)(11). Then, according to the table above, these pairs are translated into the output of the procedure:
(1)(0)(1)() (=101).
In the output stream 0 and 1 are equally likely, as 10 and 01 are equally likely in the original, both having probability pq = qp. This extraction of uniform randomness does not require the input
trials to be independent, only uncorrelated. More generally, it works for any exchangeable sequence of bits: all sequences that are finite rearrangements are equally likely.
The von Neumann extractor uses two input bits to produce either zero or one output bits, so the output is shorter than the input by a factor of at least 2. On average the computation discards
proportion p^2 + (1 − p)^2 of the input pairs, or proportion p^2 + q^2, which is near one when p is near zero or one.
The discard of input pairs is at least proportion 1/2, the minimum which occurs where p = 1/2 for the original process. In that case the output stream is 1/4 the length of the input on average.
Iterated von Neumann extractor
This decrease in efficiency, or waste of randomness present in the input stream, can be mitigated by iterating the algorithm over the input data. This way the output can be made to be "arbitrarily
close to the entropy bound".^[4] The algorithm works recursively, recycling ``wasted randomness" from two sources: the sequence of discard/non-discard, and the values of discarded pairs (00/11).
Intuitively, it relies on the fact that, given the sequence already generated, both of those sources are still exchangeable sequences of bits, and thus eligible for another round of extraction.
More concretely, on an input sequence, the algorithm consumes the input bits in pairs, generating output together with two new sequences:
input output new sequence 1 new sequence 2
00 none 0 0
01 0 1 none
10 1 1 none
11 none 0 1
(If the length of the input is odd, the last bit is completely discarded.) Then the algorithm is applied recursively to each of the two new sequences, until the input is empty.
Example: The input stream from above, 10011011, is processed this way:
step number input output new sequence 1 new sequence 2
0 (10)(01)(10)(11) (1)(0)(1)() (1)(1)(1)(0) ()()()(1)
1 (11)(10) ()(1) (0)(1) (1)()
1.1 (01) (0) (1) ()
1.1.1 1 none none none
1.2 1 none none none
2 1 none none none
The output is therefore (101)(1)(0)()()() (=10110), so that from the eight bits of input five bits of output were generated, as opposed to three bits through the basic algorithm above.
Further reading
• Carl W. Helstrom, Probability and Stochastic Processes for Engineers, (1984) Macmillan Publishing Company, New York ISBN 0-02-353560-1.
• Dimitri P. Bertsekas and John N. Tsitsiklis, Introduction to Probability, (2002) Athena Scientific, Massachusetts ISBN 1-886529-40-X
External links
This article is issued from
- version of the 8/19/2016. The text is available under the
Creative Commons Attribution/Share Alike
but additional terms may apply for the media files. | {"url":"https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Bernoulli_process.html","timestamp":"2024-11-07T10:30:19Z","content_type":"text/html","content_length":"69750","record_id":"<urn:uuid:7917e67d-eb07-46ad-bb1e-3cc3edfd0d49>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00425.warc.gz"} |
Optimal algorithms for train shunting and relaxed list update problems for ATMOS 2012
ATMOS 2012
Conference paper
Optimal algorithms for train shunting and relaxed list update problems
View publication
This paper considers a TRAIN SHUNTING problem which occurs in cargo train organizations: We have a locomotive travelling along a track segment and a collection of n cars, where each car has a source
and a target. Whenever the train passes the source of a car, it needs to be added to the train, and on the target, the respective car needs to be removed. Any such operation at the end of the train
incurs low shunting cost, but adding or removing truly in the interior requires a more complex shunting operation and thus yields high cost. The objective is to schedule the adding and removal of
cars as to minimize the total cost. This problem can also be seen as a relaxed version of the well-known LIST UPDATE problem, which may be of independent interest. We derive polynomial time
algorithms for TRAIN SHUNTING by reducing this problem to finding independent sets in bipartite graphs. This allows us to treat several variants of the problem in a generic way. Specifically, we
obtain an algorithm with running time O (n5/2) for the uniform case, where all low costs and all high costs are identical, respectively. Furthermore, for the non-uniform case we have running time of
O (n3). Both versions translate to a symmetric variant, where it is also allowed to add and remove cars at the front of the train at low cost. In addition, we formulate a dynamic program with running
time O (n4), which exploits the special structure of the graph. Although the running time is worse, it allows us to solve many extensions, e.g., prize-collection, economies of scale, and dependencies
between consecutive stations. © Tim Nonner and Alexander Souza. | {"url":"https://research.ibm.com/publications/optimal-algorithms-for-train-shunting-and-relaxed-list-update-problems","timestamp":"2024-11-03T00:00:18Z","content_type":"text/html","content_length":"68973","record_id":"<urn:uuid:42716e7e-33d2-4ef5-b43d-27981a89fe50>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00280.warc.gz"} |
This new external, natural exercise measure lead during the Area cuatro
cuatro.4 A keen Unbounded External Fitness Measure
step one (get a hold of Profile six) lies in changed seed fighting up against randomly made vegetables that have an equivalent matrix proportions (a comparable level of rows and you can articles)
additionally the same matrix thickness (a comparable tiny fraction off of those in the matrix). Area cuatro.step three helps that it exercise level by the proving that it will abide by brand new
ranking produced from tournaments up against human-tailored seeds (see Desk 8). The fresh exercise way of measuring Point cuatro.step 1 is very effective toward studies demonstrated from the before
sections, it have constraints.
You to needs we could possibly impose towards the an external, natural physical fitness measure it so it is produce a contour one increases when the exercise of inhabitants was boosting, remains
apartment in the event that people is actually none improving neither weakening, and you will drops if society was weakening. Let’s name which needs directional surface. Exercise as the mentioned by
the competition up against haphazard vegetables (such as Numbers 6 and 10) matches which needs.
Other criteria we would demand on the an outward, pure fitness level is the fact that speed off fitness alter is always to correspond to the fresh new mountain of your bend. Why don’t we telephone
call this requirements slope structure. The absolute fitness measure inside the Section 4.step 1 (evolved vegetables fighting up against at random generated seed with the exact same matrix
proportions and you can occurrence) ranges between no and something, hence suppresses it off fulfilling hill feel. The top of minimizing bounds into the physical fitness don’t let the newest mountain
to keep ongoing for long. Since the bend gets nearer to one, brand new hill need to drop-off, even when the pace regarding exercise changes is actually lingering.
Inside section, i expose a workout measure you to definitely joins both conditions, directional surface and hill surface. Brand new exercise measure try unbounded; they selections between negative
infinity and you can self-confident infinity. I do a comparison of the fresh new measure into the exercise level in Section cuatro.step one. The outcome reveal that the 2 measures was highly
The function f
[n] ranges from ?n to +n. The function has directional consistency: If p[in] reaches a generation n where the probability of winning is random (p[in] = 0.5), then the curve for f[n] will start to
flatten out. If the probability is worse than random (p[in] < 0.5), the curve will head downwards, perhaps eventually going below zero. If the probability is better than random (p[in] > 0.5), the
curve will head upwards. The function also has slope consistency: The slope of the curve corresponds to the pace of fitness change. Thus, this function satisfies the two requirements for an external,
absolute fitness measure.
We reduce noise in our estimate of the probability p[in] by taking the top ten fittest seeds in generation i and the top ten fittest seeds in generation n and making each pair of seeds compete twice,
so that the estimate for p[in] is based on the average outcome of 200 competitions (10 ? 10 ? 2 = 200). The noise is further reduced by averaging over twelve separate runs of the model.
Contour 11 suggests the health of the half dozen layers, since the given by new fitness level. The brand new fitness measure helps to make the steady exercise raise away from Covering 4 more readily
noticeable versus old fitness measure (examine Profile 11 having Figures six and you will 10).
That it shape measures up this new half a dozen different options regarding Design-S having fun with a keen unbounded exterior fitness level. While the brand new fitness scale when you look at the
Figures six and you will 10 (evaluating developed seed products with random seed products of the same proportions and you will density) https://datingranking.net/tr/filipino-cupid-inceleme/ is
restricted to help you between 0 to one, the newest fitness scale here range of negative infinity so you can confident infinityparing Rates 6 and 11, we come across a comparable positions of your own
some other layers (all the way through: Coating cuatro, Layer dos, Layer step three, Covering 1). Figure 11 is more suitable than simply Figure six getting exhibiting new regular rise in exercise
away from Layer cuatro. In Contour eleven, like in Profile ten, Coating cuatro Shuffled drops behind Coating 4 and you can Covering cuatro Mutualism, but it at some point catches upwards. | {"url":"https://directorio.laprensaus.com/this-new-external-natural-exercise-measure-lead/","timestamp":"2024-11-08T15:16:50Z","content_type":"text/html","content_length":"67335","record_id":"<urn:uuid:d9838411-221c-494a-8eb6-0ba1d6309aeb>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00725.warc.gz"} |
Degrees of Freedom
Jump to navigation Jump to search
Learning Objectives
1. Define degrees of freedom
2. Estimate the variance from a sample of 1 if the population mean is known
3. State why deviations from the sample mean are not independent
4. State general formula for degrees of freedom in terms of the number of values and the number of estimated parameters
5. Calculate s2
Degrees of Freedom
Some estimates are based on more information than others. For example, an estimate of the variance based on a sample size of 100 is based on more information than an estimate of the variance based on
a sample size of 5. The degrees of freedom (df) of an estimate is the number of independent pieces of information on which the estimate is based.
As an example, let's say that we know that the mean height of Martians is 6 and wish to estimate the variance of their heights. We randomly sample one Martian and find that its height is 8. Recall
that the variance is defined as the mean squared deviation of the values from their population mean. We can compute the squared deviation of our value of 8 from the population mean of 6 to find a
single squared deviation from the mean. This single squared deviation from the mean (8-6)2 = 4 is an estimate of the mean squared deviation for all Martians.
Therefore, based on this sample of one, we would estimate that the population variance is 4. This estimate is based on a single piece of information and therefore has 1 df. If we sampled another
Martian and obtained a height of 5, then we could compute a second estimate of the variance (5-6)2 = 1. We could then average our two estimates (4 and 1) to obtain an estimate of 2.5. Since this
estimate is based on two independent pieces of information, it has two degrees of freedom. The two estimates are independent because it is based on two independently and randomly selected Martians.
The estimates would not be independent if after sampling one Martian, we decided to choose its brother as our second Martian.
As you are probably thinking, it is pretty rare that we know the population mean when we are estimating the variance. Instead, we have to first estimate the population mean (μ) with the sample mean
(M). The process of estimating the mean affects our degrees of freedom as shown below.
Returning to our problem of estimating the variance in Martian heights, let's assume we do not know the population mean and therefore we have to estimate it from the sample. We have sampled two
Martians and found that their heights are 8 and 5. Therefore M, our estimate of the population mean, is
M = (8+5)/2 = 6.5.
We can now compute two estimates of variance by computing
Estimate 1 = (8-6.5)2 = 2.25
Estimate 2 = (5-6.5)2 = 2.25
Now for the key question: Are these two estimates independent? The answer is no because each height contributed to the calculation M. Since the first Martian's height of 8 influenced M, it also
influenced Estimate 2. If the first height had been, for example, 10, then M would have been 7.5 and the Estimate 2 would have been (5-7.5)2 = 6.25 instead of 2.25. The important point is that the
two estimates are not independent and therefore we do not have two degrees of freedom. Another way to think about the non-independence is to consider that if you knew the mean and one of the scores,
you would know the other score. For example, if one score is 5 and the mean is 6.5, you can compute that the total of the two scores is 13 and therefore that the other score must be 13-5 = 8.
In general, the degrees of freedom for an estimate is equal to the number of values minus the number of parameters estimated en route to the estimate in question. In the Martians example, there are
two values(8 and 5) and we had to estimate one parameter (μ) on the way to estimating the parameter of interest (σ2). Therefore, the estimate of variance has 2 -1 =1 degrees of freedom. If we had
sampled 12 Martians, then our estimate of variance would have had 11 degrees of freedom. Therefore the degrees of freedom of an estimate of variance is equal to N -1 where N is the number of
Recall from the section on variability that the formula for estimating the variance in a sample is:
The denominator of this formula is the degrees of freedom.
1 You know the population mean for a certain test score. You select 10 people from the population to estimate the standard deviation. How many degrees of freedom does your estimation of the standard
deviation have?
Answer >>
There are 10 independent pieces of information, so there are 10 degrees of freedom.
2 You do not know the population mean for a different test score. You select 15 people from the population and use this sample to estimate the mean and standard deviation. How many degrees of freedom
does your estimation of the standard deviation have?
3 For which of these degrees of freedom do you think your sample statistic is the least likely to be an accurate representation of the popoulation parameter?
Answer >>
2 degrees of freedom gives the least information. It had the smallest sample used to compute the statistic and is therefore the most likely to be a poor representation of the population parameter. | {"url":"https://training-course-material.com/index.php?title=Degrees_of_Freedom&oldid=24053","timestamp":"2024-11-05T09:50:03Z","content_type":"text/html","content_length":"25786","record_id":"<urn:uuid:b2f85db2-79fc-45ca-ab37-7ffd755eebbd>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00255.warc.gz"} |
Rishi Ranjan Singh
Assistant Professor
Dept. of Computer Science and Engineering Indian Institute of Technology, Bhilai E-mail: rishi@iitbhilai.ac.in
Areas of Interest
• Social & Complex Network Analysis, Machine Learning with Graphs, Approximation Algorithms, Combinatorial Optimization, Mathematical Formulation, Vehicle Routing Problems, Graph Theory,
Theoretical computer science.
• PhD, 2016, Indian Institute of Technology Ropar
• BTech, 2011, Uttar Pradesh Technical University, Lucknow, India
• Theory of Computation
• Complexity Theory
• Approximation Algorithms
• Network Science
• Social and Complex Network Analysis
• August 2017-Present: Assistant Professor, Dept. of Computer Science and Engineering, IIT Bhilai, India.
• April 2016-August 2017: Assistant Professor, Dept. of Information Technology, IIIT Allahabad, India.
• September 2015-April 2016: Visiting Faculty, Dept. of Information Technology, IIIT Allahabad, India.
• Kirtidev Mohapatra [Ph.D., Ongoing]
• Aditi Shukla [M.Tech., Ongoing]
• Sanjeevani Vishni Bhopre [M.Tech., Ongoing] (Co-supervised with Dr. Soumajit Pramanik)
• Parul Diwakar [M.Tech., Completed] (Co-supervised with Dr. Soumajit Pramanik)
• Shaswati Patra [Ph.D., Completed] (Co-supervised with Dr. Barun Gorain)
• Bikas Saha [M.Tech., Completed]
• Shivam Sharma [M.Tech., Completed] (Co-supervised with Dr. Amit Kumar Dhar)
• Anuj Singh [B.Tech. Hons., Completed]
Selected Publications
Exposure to half-truths or lies has the potential to undermine democracies, polarize public opinion, and promote violent extremism. Identifying the veracity of fake news is a challenging task in
distributed and disparate cyber-socio platforms. To enhance the trustworthiness of news on these platforms, in this article, we put forward a fake news detection model, OptNet-Fake. The proposed
model is architecturally a hybrid that uses a meta-heuristic algorithm to select features based on usefulness and trains a deep neural network to detect fake news in social media. The d -D
feature vectors for the textual data are initially extracted using the term frequency inverse document frequency (TF-IDF) weighting technique. The extracted features are then directed to a
modified grasshopper optimization (MGO) algorithm, which selects the most salient features in the text. The selected features are then fed to various convolutional neural networks (CNNs) with
different filter sizes to process them and obtain the n-gram features from the text. These extracted features are finally concatenated for the detection of fake news. The results are evaluated
for four real-world fake news datasets using standard evaluation metrics. A comparison with different meta-heuristic algorithms and recent fake news detection methods is also done. The results
distinctly endorse the superior performance of the proposed OptNet-Fake model over contemporary models across various datasets.
• A variant of graph covering problem demands to find a set of sub-graphs when the union of sub-graphs contain all the edges of G. Another variant of graph covering problem requires finding a
collection of subgraphs such that the union of the vertices of subgraphs forms a vertex cover. We study the later version of the graph covering problem. The objective of these problems is to
minimize the size/cost of the collection of subgraphs. Covering graphs with the help of a set of edges, set of vertices, tree or tour has been studied extensively in the past few decades. In this
paper, we study a variant of the graph covering problem using two special subgraphs. The first problem is called bounded component forest cover problem. The objective is to find a collection of
minimum number of edge-disjoint bounded weight trees such that the vertices of the forest, i.e., collection of edge-disjoint trees, cover the graph. The second problem is called bounded size walk
cover problem. It asks to minimize the number of bounded size walks which can cover the graph. Walks allow repetition of vertices/edges. Both problems are a generalization of classical vertex
cover problem, therefore, are NP-hard. We give 4ρ and 6ρ factor approximation algorithm for bounded component forest cover and bounded size walk cover problems respectively, where ρ is an
approximation factor to find a solution to the tree cover problem..
• Experts from several disciplines have been widely using centrality measures for analyzing large as well as complex networks. These measures rank nodes/edges in networks by quantifying a notion of
the importance of nodes/edges. Ranking aids in identifying important and crucial actors in networks. In this chapter, we summarize some of the centrality measures that are extensively applied for
mining social network data. We also discuss various directions of research related to these measures.
• In this paper, we study the wars fought in history and draw conclusions by analysing a curated temporal multi-graph. We explore the participation of countries in wars and the nature of
relationships between various countries during different timelines. This study also attempts to shed light on different countries’ exposure to terrorist encounters.
• We propose a network based framework to model spread of disease. We study the evolution and control of spread of virus using the standard SIR-like rules while incorporating the various available
models for social interaction. The dynamics of the framework has been compared with the real-world data of COVID-19 spread in India. This framework is further used to compare vaccination
• We use a queuing model to study the spread of an infection due to interaction among individuals in a public facility. We provide tractable results for the probability that a susceptible
individual leaves the facility infected. This model is then applied to study infection spread in a closed system like a large campus, community, and model the interaction among individuals in the
multiple public facilities found in such systems. These public facilities could be restaurants, shopping malls, public transportation, etc. We study the impact of relative timescales of the Close
Contact Time (CCT) and the individuals’ stay time in a facility on the spread of the virus. The key contribution is on using queuing theory to model time-spread of an infection in a closed
• We compare the Indian railways and domestic airways using network analysis approach. The analysis also compares different characteristics of the networks with a previous work and notes the change
in the networks over a decade. In a populous country like India with an ever increasing GDP, more and more people are gaining the facility of choosing one mode of travel over the other. Here we
have compared these two networks, by building a merger network. The need for such type of network arises as the order of both networks are different. This newly formed network can be used in
identifying new routes and adding more flights on some of the popular routes in India.
• Centrality measures have been proved to be a salient computational science tool for analyzing networks in the last two to three decades aiding many problems in the domain of computer science,
economics, physics, and sociology. With increasing complexity and vividness in the network analysis problems, there is a need to modify the existing traditional centrality measures. Weighted
centrality measures usually consider weights on the edges and assume the weights on the nodes to be uniform. One of the main reasons for this assumption is the hardness and challenges in mapping
the nodes to their corresponding weights. In this paper, we propose a way to overcome this kind of limitation by hybridization of the traditional centrality measures. The hybridization is done by
taking one of the centrality measures as a mapping function to generate weights on the nodes and then using the node weights in other centrality measures for better complex ranking.
• Service Coverage Problem aims to find an ideal node for installing a service station in a given network such that services requested from various nodes are satisfied while minimizing the response
time. Centrality Measures have been proved to be a salient computational science tool to find important nodes in networks. With increasing complexity and vividness in the network analysis
problems, there is a need to modify the existing traditional centrality measures. In this paper we propose a new way of hybridizing centrality measures based on node-weighted centrality measures
to address the service coverage problem.
• Centrality measures, erstwhile popular amongst the sociologists and psychologists, have seen broad and increasing applications across several disciplines of late. Amongst a plethora of
application-specific definitions available in the literature to rank the vertices, closeness centrality, betweenness centrality and eigenvector centrality (page-rank) have been the widely applied
ones. We are surrounded by networks where information, signal or commodities are flowing through the edges. Betweenness centrality comes as a handy tool to analyze such systems, but computation
of these scores is a daunting task in large-size networks. Since computing the betweenness centrality of one node is conjectured to be same as time taken for computing the betweenness centrality
of all the nodes, a fast algorithm was required that can efficiently estimate a node’s betweenness score. In this paper, we propose a heuristic that efficiently estimates the betweenness score of
a given node. The algorithm incorporates a non-uniform node-sampling model which is developed based on the analysis of random Erdős-Rényi graphs. We apply the heuristic to estimate the ranking of
an arbitrary k vertices, called betweenness-ordering problem, where k is much less than the total number of vertices. The proposed heuristic produces very efficient results even when runs for a
linear time in the number of edges. An extensive experimental evidence is presented to demonstrate the performance of the proposed heuristic for betweenness estimation and ordering on several
synthetic and real-world graphs.
• In this paper, we give randomized approximation algorithms for stochastic cumulative VRPs for the split and unsplit deliveries. The approximation ratios are max {1+ 1.5 α, 3} max {1+ 1. 5 α, 3}
and 6, respectively, where α is the approximation ratio for the metric TSP. The approximation factor is further reduced for trees. These results extend the results in Anupam Gupta et al.(2012)
and Daya Ram Gaur et al.(2013). The bounds reported here improve the bounds in Daya Ram Gaur et al.(2016).
• There has been a recent resurge of interest in vehicle routing problems, especially in the context of green vehicle routing. One popular and simplified model is that of the cumulative vehicle
routing problem. In this chapter, we examine the motivation, the definition, and the mixed integer linear program for the cumulative VRP. We review some of the recent results on approximation
algorithms for the cumulative VRP. A column generation-based procedure for solving the cumulative VRP is also described. We also review approximation algorithms for a stochastic version of the
cumulative VRP.
• Cumulative vehicle routing problems are a simplified model of fuel consumption in vehicle routing problems. Here we computationally study, an inexact approach for constructing solutions to
cumulative vehicle routing problems based on rounding solutions to a linear program. The linear program is based on the set cover formulation and is solved using column generation. The pricing
subproblem is solved heuristically using dynamic programming. Simulation results show that a simple scalable strategy gives solutions with cost close to the lower bound given by the linear
programming relaxation. We also give theoretical bounds on the integrality gap of the set cover formulation.
• In this paper we give randomized approximation algorithms for stochastic cumulative VRPs for split and unsplit deliveries. The approximation ratios are 2(1+α) and 7 respectively, where α is the
approximation ratio for the metric TSP. The approximation factor is further reduced for trees and paths. These results extend the results in [Technical note - approximation algorithms for VRP
with stochastic demands. Operations Research, 2012] and [Routing vehicles to minimize fuel consumption. Operations Research Letters, 2013].
• Betweenness centrality is widely used as a centrality measure, with applications across several disciplines. It is a measure that quantifies the importance of a vertex based on the vertex’s
occurrence on shortest paths in a graph. This is a global measure, and in order to find the betweenness centrality of a node, one is supposed to have complete information about the graph. Most of
the algorithms that are used to find betweenness centrality assume the constancy of the graph and are not efficient for dynamic networks. We propose a technique to update betweenness centrality
of a graph when nodes are added or deleted. Observed experimentally, for real graphs, our algorithm speeds up the calculation of betweenness centrality from 7 to 412 times in comparison to the
currently best-known techniques.
• Betweenness Centrality measures, erstwhile popular amongst the sociologists and psychologists, have seen wide and increasing applications across several disciplines of late. In conjunction with
the big data problems, there came the need to analyze large complex networks. Exact computation of a node’s betweenness is a daunting task in the networks of large size. In this paper, we propose
a non-uniform sampling method to estimate the betweenness of a node. We apply our approach to estimate a node’s betweenness in several synthetic and real world graphs. We compare our method with
the available techniques in the literature and show that our method fares several times better than the currently known techniques. We further show that the accuracy of our algorithm gets better
with the increase in size and density of the network.
• Cumulative vehicle routing problems are a simplified model of fuel consumption in vehicle routing problems. Here we study computationally, an approach for constructing approximate solutions to
cumulative vehicle routing problems based on rounding solutions to a linear program. The linear program is based on the set cover formulation, and is solved using column generation. The pricing
sub-problem is solved using dynamic programming. Simulation results show that the simple scalable strategy computes solutions with cost close to the lower bound given by the linear programming
• We consider a generalization of the capacitated vehicle routing problem known as the cumulative vehicle routing problem in the literature. Cumulative VRPs are known to be a simple model for fuel
consumption in VRPs. We examine four variants of the problem, and give constant factor approximation algorithms. Our results are based on a well-known heuristic of partitioning the traveling
salesman tours and the use of the averaging argument.
• Betweenness centrality is a centrality measure that is widely used, with applications across several disciplines. It is a measure which quantifies the importance of a vertex based on its
occurrence in shortest paths between all possible pairs of vertices in a graph. This is a global measure, and in order to find the betweenness centrality of a node, one is supposed to have
complete information about the graph. Most of the algorithms that are used to find betwenness centrality assume the constancy of the graph and are not efficient for dynamic networks. We propose a
technique to update betweenness centrality of a graph when nodes are added or deleted. Our algorithm experimentally speeds up the calculation of betweenness centrality (after updation) from 7 to
412 times, for real graphs, in comparison to the currently best known technique to find betweenness centrality. | {"url":"https://iitbhilai.ac.in/index.php?pid=rishi","timestamp":"2024-11-08T05:41:06Z","content_type":"text/html","content_length":"55441","record_id":"<urn:uuid:87ee9ece-0984-4123-a4a0-d323460f8ca5>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00109.warc.gz"} |
POMDPs to solve Active Sensing Problem: where gathering information is the explicit goal and not a means to do something. Meaning, we can’t train them using state-only reward functions (i.e. reward
is based on belief and not state).
Directly reward the reduction of uncertainty: belief-based reward framework which you can just tack onto the existing solvers.
To do this, we want to define some reward directly over the belief space which assigns rewards based on uncertainty reduction:
$$r(b,a) = \rho(b,a)$$
\(\rho\) should be some measure of uncertainty, like entropy.
key question: how does our POMDP formulations change given this change?
Don’t worry about the Value Function
result: if reward function is convex, then Bellman updates should preserve the convexity of the value function
So, we now just need to make sure that however we compute our rewards the reward function \(\rho\) has to be piecewise linear convex.
One simple PWLC rewards are alpha vectors:
$$\rho(b,a) = \max_{\alpha in \Gamma} \qty[\sum_{ss}^{} b(s) \alpha(s)]$$
We want to use \(R\) extra alpha-vectors to compute the value at a state.
This makes our Belman updates:
non-PWLC objectives
As long as \(\rho\) is convex and stronger-than Lipschitz continuous, we can use a modified version of the Bellman updates to force our non PWLC \(\rho\) into pretty much PWLC:
$$\hat{\rho}(b) = \max_{b’} \qty[\rho(b’) + (b-b’) \cdot abla p(b’)]$$
Taylor never fails to disappoint.
Fancy math gives that the error in this would be bounded: | {"url":"https://www.jemoka.com/posts/kbhrho_pomdps/","timestamp":"2024-11-03T12:37:17Z","content_type":"text/html","content_length":"7241","record_id":"<urn:uuid:76327048-360a-4836-9f51-630eced88971>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00257.warc.gz"} |
LibGuides: STT 200 Statistical Methods: What is Data?
What is data?
Data vs. Statistics
Data are raw ingredients from which statistics are created. Statistics are useful when you just need a few numbers to support an argument (ex. In 2003, 98.2% of American households had a television
set--from Statistical Abstract of the United States). Statistics are usually presented in tables. Statistical analysis can be performed on data to show relationships among the variables collected.
Through secondary data analysis, many different researchers can re-use the same data set for different purposes.
Data Sets, Studies, and Series
In data archives like ICPSR, a data set or study is made up of the raw data file and any related files, usually the codebook and setup files. The codebook is your guide to making sense of the raw
data. For survey data, the codebook usually contains the actual questionnaire and the values for the responses to each question. The setup files help will not display properly.
ICPSR uses the term series to describe collections of studies that have been repeated over time. For example, the National Health Interview Survey is conducted annually. In the ICPSR archive, you
will find a description of the series that provides an overview. You will also find individual descriptions of each study (i.e. National Health Interview Survey, 2004). The study number in ICPSR
refers to the individual survey.
Types of Data
Cross-Sectional describes data that are only collected once.
Time Series study the same variable over time. The National Health Interview Survey is an example of time series data because the questions generally remain the same over time, but the individual
respondents vary.
Longitudinal Studies describe surveys that are conducted repeatedly, in which the same group of respondents are surveyed each time. This allows for examining changes over the life course. The Project
on Human Development in Chicago Neighborhoods (PHDCN) Series contains a longitudinal component that tracks changes in the lives of individuals over time through interviews.
(Originally from Sue Erickson at Vanderbilt University http://www.library.vanderbilt.edu/central/FindingData.htm) | {"url":"https://libguides.lib.msu.edu/c.php?g=449963&p=3071913","timestamp":"2024-11-09T13:50:55Z","content_type":"text/html","content_length":"42414","record_id":"<urn:uuid:2ea19f2a-3071-43cb-b3c3-7c6813822add>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00309.warc.gz"} |
A generalized matching reconfiguration problem
The goal in reconfiguration problems is to compute a gradual transformation between two feasible solutions of a problem such that all intermediate solutions are also feasible. In the Matching
Reconfiguration Problem (MRP), proposed in a pioneering work by Ito et al. from 2008, we are given a graph G and two matchings M and M^0, and we are asked whether there is a sequence of matchings in
G starting with M and ending at M^0, each resulting from the previous one by either adding or deleting a single edge in G, without ever going through a matching of size < min{|M|, |M^0|}-1. Ito et
al. gave a polynomial time algorithm for the problem, which uses the Edmonds-Gallai decomposition. In this paper we introduce a natural generalization of the MRP that depends on an integer parameter
∆ ≥ 1: here we are allowed to make ∆ changes to the current solution rather than 1 at each step of the transformation procedure. There is always a valid sequence of matchings transforming M to M^0 if
∆ is sufficiently large, and naturally we would like to minimize ∆. We first devise an optimal transformation procedure for unweighted matching with ∆ = 3, and then extend it to weighted matchings to
achieve asymptotically optimal guarantees. The running time of these procedures is linear. We further demonstrate the applicability of this generalized problem to dynamic graph matchings. In this
area, the number of changes to the maintained matching per update step (the recourse bound) is an important quality measure. Nevertheless, the worst-case recourse bounds of almost all known dynamic
matching algorithms are prohibitively large, much larger than the corresponding update times. We fill in this gap via a surprisingly simple black-box reduction: Any dynamic algorithm for maintaining
a β-approximate maximum cardinality matching with update time T, for any β ≥ 1, T and ε > 0, can be transformed into an algorithm for maintaining a (β(1 + ε))-approximate maximum cardinality matching
with update time T + O(1/ε) and worst-case recourse bound O(1/ε). This result generalizes for approximate maximum weight matching, where the update time and worst-case recourse bound grow from T + O
(1/ε) and O(1/ε) to T + O(ψ/ε) and O(ψ/ε), respectively; ψ is the graph aspect-ratio. We complement this positive result by showing that, for β = 1 + ε, the worst-case recourse bound of any algorithm
produced by our reduction is optimal. As a corollary, several key dynamic approximate matching algorithms – with poor worst-case recourse bounds – are strengthened to achieve near-optimal worst-case
recourse bounds with no loss in update time.
Publication series
Name Leibniz International Proceedings in Informatics, LIPIcs
Volume 185
ISSN (Print) 1868-8969
Conference 12th Innovations in Theoretical Computer Science Conference, ITCS 2021
City Virtual, Online
Period 6/01/21 → 8/01/21
Funders Funder number
Blavatnik Family Foundation
Israel Science Foundation 1991/19
• Dynamic algorithms
• Graph matching
• Reconfiguration problem
• Recourse bound
Dive into the research topics of 'A generalized matching reconfiguration problem'. Together they form a unique fingerprint. | {"url":"https://cris.tau.ac.il/en/publications/a-generalized-matching-reconfiguration-problem-2","timestamp":"2024-11-05T16:17:45Z","content_type":"text/html","content_length":"59120","record_id":"<urn:uuid:69e60602-109d-43a6-926f-9cb770f18d5a>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00744.warc.gz"} |
Graphing Quadratic Inequalities on Desmos
So one of the tools that you can use for graphing is does most dot com slash calculator. And this is a graphing calculator that you guys have access to. So when we look at these quadratics, that
we're graphing the inequalities of. I can graph. Things similar to what we've been looking at. And so I just type in this function. And it shows me where the points are, the important points, and if
I zoom in and move the graph around, I can see a vertex of negative one negative 5, a Y intercept of negative three, a corresponding point and negative two negative three. And then the shaded region
is below the curve. But let's change this to greater than or equal to. You can just do that on your keyboard, hit the greater than symbol. And then the equals, and now it changes it to a solid
parabola, and we shade a above the graph. So this is a good tool that you guys can use to check your graphs. | {"url":"http://helpyourautisticchildblog.com/graphing-quadratic-inequalities-on-desmos-507405.html","timestamp":"2024-11-06T03:53:48Z","content_type":"text/html","content_length":"48015","record_id":"<urn:uuid:a80792cc-5c35-4ad0-b3ec-14ae061c3515>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00111.warc.gz"} |
Jeu de Tarot (thing)
Le Jeu de Tarot is a traditional French card game. It is NOT the same as tarot cards. It is more along the lines of bridge but far, far weirder. It can be played by 3,5 or even 2 people, but I will
begin by describing the 3 player game.
The Deck : There are 78 cards in a deck. There are the four normal suits (clubs, diamonds, hearts, spades) which each contain 14 cards – 1 to 10, Jack, Cavalier, Queen, King (in ascending order).
This is the same as normal playing cards except the Ace (or 1) is low, and there is an extra card, the Cavalier, between Jack and Queen. In addition there is a 5th suit, trumps, which consists of 21
cards, numbered (as you might expect) 1 to 21. Finally there is a joker (or l’excuse). There are 3 Special cards - the joker, the 21 of trumps and the 1 of trumps, for reasons I will explain later.
Dealing : Without shuffling, the person chosen as dealer deals out the whole deck, 3 cards at a time, starting with the person to his right and continuing anticlockwise. At two random points during
the dealing, you put 3 cards to the side, leaving 6 cards called the dog. Each player will then have 24 cards in his hand.
Bidding : Once the hands have been dealt, the person to the right of the dealer begins the bidding. These are the six possible bids, in ascending order: Pass, Small, Guard, Guard without the dog,
Guard against the dog or Slam. You must always either bid higher than the person before you or pass. The person who has bid the highest plays against the other two players. If he has bid Small or
Guard, he picks up the dog, showing it to the others as he does so, and discards 6 cards from his hand. He must NOT discard any Kings or Special cards. If he has bid Guard without the dog, he reveals
the dog but does not pick it up. If he has bid Guard against the dog he gives 3 cards from the dog to each of the others. Slam means he must make all the tricks, and no-one gets the dog. If no-one
bids, the hands are re-dealt.
Play : After the bidding is complete, the person in the contract leads. The play is basically like whist or bridge (you must follow suit if you can, the winner of a trick leads the next one) with a
few exceptions:
1. If you can trump, you must.
2. You must always play a higher trump than any already played.
3. You keep the cards from the tricks you win together, for scoring at the end. If you are defending the contract, you put your tricks with your partners.
4. You may not lead the joker. You may play it on any trick that has already been led, and it exempts you from it. You keep the joker, but must give a normal card (not an honour or special card) from
your trick-pile to whoever wins that trick.
5. You should not play the joker on the last trick. If you do, it goes to the opposition.
You continue like this until all the cards have been played.
Scoring : After the last trick has been played, the person in the contract counts up his points, both those from his tricks and from the dog. You pair up the cards, putting one honour or special card
with one normal card. The normal cards left over you just put together in pairs. If there are an odd number of cards, the final one is ignored. The scoring is as follows:
Special card + normal = 5 points
King + normal = 5 points
Queen + normal = 4 points
Cavalier + normal = 3 points
Jack + normal = 2 points
Two normal cards = 1 point
In total there are 91 points.
Whether you win or not depends on how many special cards you have. The more special cards you have, the less points you need to win:
3 Special cards – 36 points
2 Special cards - 41 points
1 Special cards - 51 points
0 Special cards - 56 points
If you get this number of points or more, you have won the round. Your actual score depends on what you bid at the beginning:
Small - 25 points
Guard - 50 points
Guard without the dog - 100 points
Guard against the dog - 150 points
Slam without bidding it - 300 points
Slam bid and made - 600 points
In addition you take the difference between the number of points you got and the number you needed, multiply by a certain number, depending on what you bid, and add it to your score. These are the
×1 for Small
×2 for Guard
×4 for Guard without the dog
×6 for Guard against the dog
You do not add anything when Slam has been bid and made.
There are also bonus points for the following things:
Playing the 1 of trumps on the last trick - 10 points
Having 10 trumps in your hand - 20 points
Having 13 trumps in your hand - 30 points
Having 15 trumps in your hand - 40 points
Having no trumps in your hand - 10 points
Having no honours in your hand - 10 points
Once you have finished scoring up, the person to the right of dealer deals, without shuffling, and another round begins.
Tactics : Although the rules themselves may seem complicated, the tactics are even more so. In fact they are very flexible – one player’s tactics will change according to the tactics of the other
players. In general there are a few basic guidelines:
1. Bidding: Whether you bid or not greatly depends on how many special cards you have. If you have one, you should think about bidding, two, you should definitely bid, and three you should almost
certainly bid at least Guard. Other things that obviously improve your hand are a long trump suit and lots of honours. It should be remembered that even if you start with the 1 of trumps you can lose
it, and if you don’t you can win it by drawing trumps.
2. The Dog: It is very important to put the right cards in the dog. It is often a good idea to void a weak suit, so you can trump the honours in it, or to put honours which you might lose in it.
3. Play: The tactics for play are a lot more complicated. You should obviously try to win as many tricks as possible, but especially ones with honours in. The two players working together should
often drop their honours in tricks their partner is winning. If you have the 1 of trumps, it is important to win it if you can, either by trumping with it, or clearing all the higher trumps. If you
don’t have it, but have a long trump suit, it is a good idea to try and draw it out.
Five player game : This is even more complicated and confusing than the 3 player game. The rules are basically the same. You deal out the pack, leaving a dog of 3 cards. The bidding is the same,
except Guard against the dog cannot be bid, since 3 cards can not be divided among 4 people. The person who wins the bidding chooses a certain king (not in his hand or in the dog). The person who has
this king will be playing with the person in the contract. However he does not tell anyone that he has the king, so only he knows exactly who is on his side. The play then commences with no-one
entirely sure who is on their side, at least until the king has been played. Clearly this will make the tactics entirely different and a lot more subtle.
There are also 2 player, 4 player, and even 6 player variations, but they are generally not as enjoyable. | {"url":"https://everything2.com/user/260.1054/writeups/Jeu+de+Tarot","timestamp":"2024-11-14T23:32:58Z","content_type":"text/html","content_length":"35463","record_id":"<urn:uuid:2947c25e-1b46-45b7-ae4f-aee8d052b1f6>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00563.warc.gz"} |
Sew Mama Sew Giveaway Day (And Some Finishes)
It is time again for the
Sew Mama Sew
Giveaway Day!! Yay! There are always fun things to win and new places to visit! There are pages and pages of giveaways, so make sure to visit and enter to win!
A long time ago (which seems to be a recurring theme around here) I started making seasonal embroidery projects to hang in my kitchen. I had finished all but the winter one, and that is probably
because I had Christmas quilts/embroideries to hang, and I just figured I didn't need to until recently.
The one good thing about it taking me so long to finish it is that it is always nice to have a new something hanging around! ;) This was from a pattern card from
Primitive Stitches
I just put the last square on the border of this quilt from a sew along with Lori Holt, #haveyourselfaquiltylittlechristmas, on Instagram last year. This was such a fun quilt to make, and I cannot
wait to get it quilted and hung for the holidays!
Now for the giveaway. I have a $25 gift certificate to
Stitchin' It Up Quilt Shop
that is available to anyone anywhere in the world. The giveaway will be open until Monday, December 14 at 8 a.m. I'm a little late getting started here, so I will keep it open a couple of hours
longer. To enter, just leave a comment letting me know if you are making any gifts this year. You can have a second entry by letting me know if you follow my blog. Please make sure I can contact you,
if I am unable to, I will immediately draw from another comment. Good luck on the giveaway, and Happy Stitching!! --Kristen
148 comments:
I'm trying to finish up a Christmas quilt, 3 table runners & a few small projects like pot holders & dish cloths. Hopefully they will soon be ready to be wrapped & delivered. You must be happy to
finish you cute projects in time for Christmas this year and check them off your list. Who wouldn't love to win that gift certificate. Just in case edoodypdx@gmail.com
Just started grad school, so instead of a fully-homemade Christmas present batch like normal, I'm just making a couple of belated bday presents that are getting mailed with the Christmas boxes :)
I have made five quilts, and lots of magic pillowcases for Christmas. Money is tight, so its out of stash mainly. Thanks for the giveaway. Tarnia.hodges at gmail.com
I Have Made Two Quilts A A Couple Of Christmas Ornaments. I'm Working On Another Quilter Also.
Lovely quilt you made.I would be knitting and crocheting some hats and ornaments.
I *should* be busy stitching up gifts but I'm way behind! I need to get cracking on zippered pouches for my kids' teachers. Thanks for the chance to win!
I have sewed some fabric bowls from some of my practice quilt sandwiches. They turn out lovely. These will be my Christmas gifts this year.
Oh yes, I'm making gifts - still working on them, in fact! A few bags, a laptop case, some napkins and table runners and I'm also knitting a scarf!
I am doing pin cushions for my quilting buddies.
And yes I am a follower.
I am making stockins, a quilt and some pillows!!
I am a follower too! thank you!
I am making a few small things for my kids. My husband got a lathe this year and is turning pens for a lot of our extended family.
I follow you on feedly!
I've made some bags and table runners as gifts
Making firewood carriers!
I'm making a sew together bag for my son, for his art supplies. And am finishing up a quilt for my dd. Thanks for the giveaway, and Merry Christmas.
I'm making a tutu and mermaid blanket for my daughter, and a piikowcase for my niece's January birthday
I made a quilt for my son who started college this year - just dropped it off at the long arm quilters last night!
A LOT of pot holders.
thank you for the giveaway!
I am giving some quilt/wall hangings, and thankfully they are finished and in the mail!
I follow you on bloglovin
I make PJs for Christmas Eve and try to help my grandkids make their own sewing projects. Thanks.
I make as many things as possible for Christmas. Right now I'm working on doll clothes. sarah@forrussia.org
I follow you on bloglovin. sarah@forrussia.org
Love your Christmas quilt! Isn't Lori Holt amazing? I have some tablerunners that are in the process, they're almost done.
Awesome giveaway!
Thanks for a chance to win!
I'm a follower via BlogLovin.
Thanks for a chance to win!
My sister, niece, nephew and I have made a bundle of fleece blankets and pillowcases. This will be a handmade holiday season for sure.
I've made a Christmas apron for my sister, and a Hello Kitty apron for my daughter. Working on a quilt top for my husband. it will have to be quilted AFTER Christmas.
Oh I am behind here too...but catching up! :) Making flannel Jamma's for the Kiddo's & few Grankiddo's too! Also crocheting dishrags to go with Patchwork trimmed dish towels too. Whewie!
Thanks for chance to win your give-a-way too! Love your Stitery & Quilt...Awesome job on both!
Am a follower too....thru Email & Bloglovin :)
MERRY CHRISTMAS BLESSINGS!!
No projects for me this year. We're hoping to move so all crafty goodness is packed away. Fingers crossed I can jump back in soon! :)
Im making pillow cases and busting my Christmas fabric stash and it feels good.
I am making pillowcases for my grand kids. 6 of them! I am excited cause I am using metallic and glow in the dark thread and am so excited to see the finished product.
I'm making aprons for my sisters!
I am making several gifts, four quilts and a scarf. crystalbluern at tds dot net
I follow you on bloglovin. crystalbluern at tds dot net
I'm making tea cozies and placemats.
I am making mini coin pouches for my loved ones :)
Followed ur blog through mail priza_7@yahoo.co.in
I'm making tote bags for Christmas presents this year.
just a few homemade gifts this year this year snuck up on me
following via bloglovin
I am making santa gift bags for all the grandchildren
I follow by blog lovin
Thanks for a fun giveaway! I'm attempting to tackle gifting by making the majority of the presents. Unfortunately I have a mountain left to do and only a week left to do it. Yikes!
Yes, I'm making lots of gifts like lap quilts and bags and pillows. Some are done, others still underway.
I have a three month old and two other kids, so this year I will be doing purchased gifts while I try to catch up on already started sewing projects.
This comment has been removed by the author.
I made two tabletoppers for my daughters and a lap quilt for a very good neighbour
I am making christmas ornaments.
i am making a loy including a king size quilt and a rag quilt
I am not making any gifts but have made a dress for myself to wear on Xmas day xx
Iain.ross30 at gmail dot com
I'm not making any Christmas gifts but I am working on stuff for some IG Swaps that I'm participating in.
Follow via Bloglovin'
HELLO, yes,I've made 4 blue+cream place mats,4 tree ornaments, a bread cozy, and several small table toppers in red+greens! Thanks for sharing in a neat giveawaay!
I've finished 2 quilts & 3 table runners. I have another quilt to finish up along with 2 pillows & a wall hanging.
I'm auntiesash at gmail
I am still making gifts... hoping to finish! LeeAnna at email leeannaquilts at gmail dot com
Yes, I've made a floor pouf from the book Handmade Style, can't wait to gift it!
Hi, I'm am making crib sheets and burp cloths for some twins due in early Jan. Thanks for the chance to win.
I've made a couple of table runners for gifts this Christmas.
I follow on bloglovin'
I have made some pillows, table toppers and small gifts for coworkers and friends
i've got a couple small quilts i'm making for gifts this year!
So far I've made a pillow and a few small gift items. Just one more small thing to go.
I made a couple of small gifts this year.
I have made a few gifts this year. A mini wall hanging of cactus and a pillow or two! Love you holiday quilt!
I have been making some gifts: market bags, baby doll carriers, flannel sleep pants and an armband IPod carrier. Thanks for the generous giveaway.
I make gifts all year long. It's my favorite thing to do.
I am following on bloglovin.
I'm making purses for my nieces. Thank you so much!
I follow on bloglovin!
I'm making bags for my family!
I follow you on bloglovin!
gifts?? surely you jest.. I just finished a 90" table runner for hubby's aunt, a little throw for the sister in law.. a huge tree skirt for the daughter in law and 6 little gift card holders.. 3
baby quilts for donation, 3 baby size tops and 1 large throw size ready to quilt & bind for donation and 2 little (60" & 30") runners for my own tables .. oh and 4 dog tuggie toys.. thats it..
I'm done and now I can focus on the new grand daughter stuff running around in my head!
I follow you with feedly.. thanks for the offer and opportunity to win!
I am still sewing Christmas Gifts....a big stack of potholders; and then sleep pants for the grandsons. Fingers crossed that I get it all done. Thank you for a wonderful giveaway. rmgsummers at
yahoo dot com.
Im trying to get a quilt done!
I'm making Christmas tree banners this year.
bloglovin follower: Nicole Sender.
I'm making a car blanket and probably also a lap quilt for a relative that just had a major stroke. Also some smaller things.
Getting up mu nerve to make a pajama gift bag that will be used each year for new family member! Thanks for the chance.
Hello. I am trying to finish a Noodlehead Divided Basket.
I am making wine gift bags and table runners as presents this year!
I hope to (finally finish) a table runner for my sister.
I just finished up my quilty Christmas gifts. I made a quilt for my mom and my sister. I do plan to make an assortment of cookies from my husband's late grandmother and send those out to his side
of the family.
I follow you via Bloglovin.
I am making most of my gifts this year
Light sewing this year! Thank you for the giveaway!
I made 3 quilts for my grandkids
I'm making fleece slippers for gifts.
donnalee1953 (at) gmail (dot) com
Following on Bloglovin
Mug rugs galore!! Thanks for the great giveaway!
I'm making small fabric bowls as gifts. They will be filled with candy and can be used afterwards to hold jewelry, sewing clips, or anything small.
I follow you by email.
Making purse organizers. Thanks!
This year I made my Christmas cards. That's "sort of" a gift, isn't it?
I follow you by my RSS reader, Newsblur. Oh-I love your quilt by the way. Lori Holt's patterns are delightful and your quilt came out wonderful!
I am making pjs for my kids
I follow via bloglovin
I'm not sewing any Christmas gifts this year. I do have a couple of birthday table runners I'm working on...birthdays were last month so I'm only a little bit late.
I follow with RSS feeds.
I'm trying to finish two quilts. And knitting a shawl.
t_ktl at yahoo.com
HAPPY TO FOLLOW YOU>BLOG LOVIN'!
I am making some gifts, an apron, some scarves, a stocking and a few ornaments.
I love your embroidery. I made 9 zipper pouches for the grandkids in the family.
I have been happily following with GFC. Merry Christmas!
Love the Lori Holt quilt! I'm making bags for the girls, pretend tea bags, letters and envelopes stuffed animal sleeping bags and 3 quilts
Unfortunately i'm making NOThING this year!!! Been too busy with my daughter's wedding!
goodness yes! I've made a bunch of small bags, a few dolls and have a smallish dinosaur in the works if I can ever find a copy shop to enlarge the pattern!
happy holidays!
I'm making pajamas for my kids, and then if I have time, I'll make sweater vests/dresses for them!
Making lots of gifts - zipper pouches, coasters, a bag, hot mats, covered books.
I am making 2 Christmas quilts, a pillow and a mug rug. And the clock is ticking, eep!
New bloglovin follower!
Lots of Xmas knitting on my plate!
I've already made some coasters; next up is a wool cycling jacket and messenger bag for my honey and a few more small things.
AG doll clothes, PJ's and a wallhanging for my sister !
I hope to make some embroidered felt ornaments.
Like you I am always a tad later in my makings, I started an advent calendar last year and finished it for this year :) I am working on Stockings now, but I will let you in on a secret, I
probably won't finish them till next Christmas. Lots of other Christmas sewing going on as well! grecomara at gmail dot com
I am making potholders for several people, and a Christmas stocking too.
I follow you with blog lovin.
I'm making hand warmers for the gift exchanges we're participating in.
I'm making a few little things; tissue holders (travel size) and, hopefully, a cushion!
I'm making wooden Christmas signs this year and pretending I don't see the fabric stack for a birthday present I'm behind on ;) trisca@charter.net
I'm making reusable coffee cup sleeves for my buddies.
I made a few presents for sinterklaas (december 5th), including a rope basket and some other small things.
I follow you via bloglovin
I am not, I have been working a lot of hours. dawnm1993(at)gmail(dot)com
I am following you via Bloglovin (dawn743).
I'm making scarves for some female friends
Yes I'm making some Christmas gifts this year. Just finished a lap quilt for a family member. Thanks for the giveaway.
I am making a T-shirt quilt for my neighbor, and I made some Christmas potholders for a gift exchange.
yea! I made a table runner :)
i have one already wrapped up. I will probably make more.
I'm making zippered bags.
I started following via email.
I'm making a few dresses for my girls and a bed tent for my son... hopefully!
This comment has been removed by the author.
I making bracelet for my niece
I'm making a little house on the prairie quilt for my 5 year old grandson for Christmas. He's a big fan of Laura and the book series. kthurn(at)bektel(dot)com
I follow by Bloglovin as Tu-Na Quilts. kthurn(at)bektel(dot)com
I'm making quick gifts like coffee sleeves and appliqued shirts!
harleydee @ gmail. com
This year I am making a quilt to give my cousin for Christmas.. I am needing to get off the computer and go finish it :) Merry Christmas!
Yes, I am making three granddaughter quilts. At least I am trying to get these done in time for Christmas.
i'm making my Mom a quilt
i follow on BL : karrie smith
I actually just made a batch of fudge so I can send some to my father-in-law with his present from my husband in tomorrow's mail. I also hope to make time for some sewing and beadwork for my
parents. I work at a packing and shipping store, so there is very little time this next week, but we'll see how I feel in the evenings! Thanks for keeping the giveaway open a bit longer :)
brandizzle7133 at gmail dot com
I'm making a few pillowcases and knitting a hat.
Christyjones at purdue.edu | {"url":"https://meadowbrook-kristen.blogspot.com/2015/12/sew-mama-sew-giveaway-day-and-some.html","timestamp":"2024-11-04T07:46:55Z","content_type":"text/html","content_length":"298981","record_id":"<urn:uuid:c45a898c-3da0-40fc-bb5b-880adc0f62fe>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00835.warc.gz"} |
Genesis 45:22
American King James Version (AKJV)
To all of them he gave each man changes of raiment; but to Benjamin he gave three hundred pieces of silver, and five changes of raiment.
American Standard Version (ASV)
To all of them he gave each man changes of raiment; but to Benjamin he gave three hundred pieces of silver, and five changes of raiment.
Berean Study Bible (BSB)
He gave new garments to each of them, but to Benjamin he gave three hundred shekels of silver and five sets of clothes.
Bible in Basic English (BBE)
To every one of them he gave three changes of clothing; but to Benjamin he gave three hundred bits of silver and five changes of clothing.
Catholic Public Domain Version (CPDV)
Likewise, he ordered two robes for each of them to be brought. Yet truly, to Benjamin he gave three hundred pieces of silver along with five of the best robes.
Darby Bible (DBY)
To each one of them all he gave changes of clothing; but to Benjamin he gave three hundred pieces of silver and five changes of clothing.
Douay–Rheims Version (DRV)
He ordered also to be brought out for every one of them two robes: but to Benjamin he gave three hundred pieces of silver with Ave robes of the best:
English Revised Version (ERV)
To all of them he gave each man changes of raiment; but to Benjamin he gave three hundred pieces of silver, and five changes of raiment.
Free Bible Version (FBV)
He gave each of them new clothes. But to Benjamin he gave five sets of clothes and 300 pieces of silver.
JPS Tanakh 1917 Old Testament / Weymouth New Testament (JPS / WNT)
To all of them he gave each man changes of raiment; but to Benjamin he gave three hundred shekels of silver, and five changes of raiment.
King James Version (KJV)
To all of them he gave each man changes of raiment; but to Benjamin he gave three hundred pieces of silver, and five changes of raiment.
New Heart English Bible (NHEB)
He gave to all of them, to each one, a change of clothing. But to Benjamin he gave three hundred pieces of silver and five changes of clothing.
Webster Bible (Webster)
To all of them he gave each man changes of raiment: but to Benjamin he gave three hundred pieces of silver, and five changes of raiment.
World English Bible (WEB)
He gave each one of them changes of clothing, but to Benjamin he gave three hundred pieces of silver and five changes of clothing.
World Messianic Bible British Edition (WMBB)
He gave each one of them changes of clothing, but to Benjamin he gave three hundred pieces of silver and five changes of clothing.
Young's Literal Translation (YLT)
to all of them hath he given — to each changes of garments, and to Benjamin he hath given three hundred silverlings, and five changes of garments; | {"url":"https://biblesearch.es/compare?verse=Genesis+45%3A22&v=AKJV","timestamp":"2024-11-07T01:19:54Z","content_type":"text/html","content_length":"22392","record_id":"<urn:uuid:4d7e2cb7-1f79-4a3b-b1bb-603611d8abd7>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00009.warc.gz"} |
Prealgebra is a term commonly used to refer to the level of mathematics a student studies before moving on to algebra and/or geometry. Prealgebra is difficult to define precisely other than "the math
taken before algebra" because of the broad number of subjects that various prealgebra curriculums use.
A typical prealgebra curriculum more-or-less consists of:
• Arithmetic with the natural numbers and integers.
• An introduction to fractions and decimals.
• An introduction to the basics of number theory (prime factorization, multiples, least common multiple, greatest common divisor, etc.)
• An introduction to the basics of statistics.
• Exponents, the root operation, and scientific notation.
• Solving linear equations, inequalities, and modeling word problems using variables.
• Ratios, computing simple conversions and rates.
• An introduction to geometry (computing area, perimeter, angles, etc.).
• the Art of Problem Solving: Prealgebra by David Patrick (Details).
AoPS Classes
The AoPS classes for prealgebra are a good starting point for those who finished up Beast Academy's curriculum or have a working understanding of simple arithmetic. Note that AoPS's classes are quite
challenging as each class is considerably more in-depth and problem-solving-oriented compared to the equivalent typical school course in prealgebra which focuses more on rote memorization. | {"url":"https://artofproblemsolving.com/wiki/index.php/Prealgebra","timestamp":"2024-11-03T02:57:11Z","content_type":"text/html","content_length":"38367","record_id":"<urn:uuid:b2c2152e-161e-4de3-a699-e47b74e42538>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00650.warc.gz"} |
Finite Volume Method II.
The course is not on the list Without time-table
Code Completion Credits Range Language
2011074 ZK 4 2P+0C Czech
Garant předmětu:
Subject deals with the application of the finite volume method (FVM) in the fluid mechanics. The attention is paid especially to the solution of 2D and 3D flows of incompressible and compressible
Syllabus of lectures:
1.FVM for multidimensional problems, discretization of the convection-diffusion equation using Cartesian mesh.
2.Construction of meshes for FVM in complex geometry, curvilinear meshes, unstructured meshes.
3.Discretization of convection-diffusion problem using an unstructured mesh.
4.Higher order schemes in multidimensional case.
5.Navier-Stokes equation for incompressible fluids, basics of projection methods.
6.Algorithm SIMPLE for steady state flows of incompressible viscous fluids.
7.Solution of system of linear equations arising from the SIMPLE algorithm (linear solvers, relaxation).
8.Solution of transient problems with the PISO algorithm
9.Numerical solution of a selected problem of flow of incompressible fluid.
10.Algorithm SIMPLE for compressible flows
11.Numerical methods for compressible flows based on the Riemann solvers.
12.Numerical solution of a selected problem of flow of compressible fluid.
Syllabus of tutorials:
Study Objective:
Study materials:
•J.H. Ferziger, M. Peric: Computational methods for fluid dynamics, Springer, 2012
•H.K.Versteeg, W. Malalasekera“ An Introduction to Computational Fluid Dynamics, The Finite Volume Method, Pearson, 2007
•C. Hirsh: Numerical Computation of Internal & External Flows, vol. 1, Elsevier, 2007
Further information:
No time-table has been prepared for this course
The course is a part of the following study plans: | {"url":"https://bilakniha.cvut.cz/next/en/predmet5900306.html","timestamp":"2024-11-09T16:39:52Z","content_type":"text/html","content_length":"8673","record_id":"<urn:uuid:53efd86f-2855-48c5-9c49-cc55bcd3c538>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00482.warc.gz"} |
Effect Size Calculator
The effect size calculator calculates the strength of correlation and relationship between two variables on the numeric scale. The practical effect of a variable is identified by the effect size of
various variables of the sample.
What is Effect Size?
“Effect is the measure of the numeric value or the effect size between the two samples on the basis of standard deviation and sample size”
For instance, Cohen's d effect size can be used to identify the difference between the height of men and women. The larger the effect size the greater the difference between the height of men and
women. The estimated effect size of variables describes their correlation. You can calculate the effect size of Cohen’s method. It calculates the difference between two populations and is divided by
the standard deviation of the population.
Practical Example:
Consider two samples of height for men and women and these samples have equal standard deviation of 3. Each sample has a size of 10 and the average height of men is 6 feet and women is 5 feet.
\(bar x_1 = 6 ; bar x_2 = 5\)
\(n_1 = 10 ; n_2 = 10\)
\(S_1 = 3 ; S_2 = 3\)
\(S^2 = \dfrac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}\)
\(S^2 = \dfrac{(10 - 1)(3)^2 + (10 - 1)(3)^2}{10 + 10 - 2}\)
\(S^2 = \sqrt{9}\)
S = 3
Now effect size:
\(d = \dfrac{|{{\bar x}}_1 - {{\bar x}}_2|}{S}\)
\(d = \dfrac{|6 - 5|}{3}\)
\(d = \dfrac{1}{3}\)
\(d = 0.3333\)
In the above example, Cohen's d-effect size of two samples of equal standard deviation is used. You can calculate the effect size of unequal standard deviation by the effect size calculator.
How to Calculate Effect Size?
The various formulas used in calculating effect size are as follows:
Cohen's Two Sample and Equal Standard Deviation:
\(S^2 = \dfrac{(n_1 - 1)S_1^2 + (n_2 - 1)S_2^2}{n_1 + n_2 - 2}\)
\(d = \dfrac{|{{\bar x}}_1 - {{\bar x}}_2|}{S}\)
Cohen's Two Sample and UnEqual Standard Deviation:
\(S^2 = \dfrac{S_1^2 + S_2^2}{2}\)
\(d = \dfrac{|{{\bar x}}_1 - {{\bar x}}_2|}{S}\)
Cohen's One Sample Formula:
\(d = \dfrac{|{{\bar x}} - μ_0|}{S}\)
Cohen's H:
\(h = 2(arcsin(\sqrt{p_1}) - arcsin(\sqrt{p_2}))\)
φ(Phi) :
\(φ = \sqrt{\dfrac{X^2}{n}}\)
Cremer’s Vφ:
\(V = \sqrt{\dfrac{X^2}{n_1 * Min(R-1 , C-1)}}\)
f^2 and R^2:
\(f^2 = \dfrac{R^2}{1 - R^2}\)
R^2 and f^2 :
\(R^2 = \dfrac{f^2}{1 + f^2}\)
You need to calculate the effect size before starting your research and after completing the research. A Cohen's d calculator is a simple way to apply the standard deviation of the samples in the
Working of Effect Size Calculator:
Let’s estimate effect size by the Campbell effect size calculator which is very easy to use and yields instant outcomes.
• Choose the effect size type
• Enter the required parameters in each respective field
• Hit the calculate button
• The effect size of each type
From the source of scribbr.com: Effect Size, How to Calculate Effect Size? From the source of wallstreetmojo.com:Corelation and Effect Size, How to Find Effect Size? | {"url":"https://calculator-online.net/effect-size-calculator/","timestamp":"2024-11-04T05:11:19Z","content_type":"text/html","content_length":"76443","record_id":"<urn:uuid:afa9e5f0-fb7e-4d4b-90df-0806312ece35>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00827.warc.gz"} |
NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.2 Triangles - Free PDF
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.2 Triangles
NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 involves the practice of questions based on properties of Triangles and theorems related to them. Students often find this topic less
interesting as it has more questions on proving rather than solving. However, understanding the properties of triangles through the examples and sample problems will help promote their interest in
the topic. After going through the NCERT Solutions Class 9 Exercise 7.2 Maths Chapter 7, students can easily understand how to solve these questions.
Class 9 NCERT Solutions Maths Chapter 7 Ex. 7.2 is well-designed by a team of experts to provide clear and precise knowledge of all concepts. These solutions will equip the students with the
necessary skills to solve the questions in a stepwise manner. It will also help them to attain good marks in the Maths exams. To download the NCERT Solutions Chapter 7 Ex 7.2 Class 9 Maths PDF, click
on the link given below.
☛ Download NCERT Solutions Chapter 7 Ex 7.2 Class 9 Maths
Exercise 7.2 Class 9 Chapter 7
More Exercises in Class 9 Maths Chapter 7
NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.2 Tips
The class 9 maths ex 7.2 chapter 7 has two important theorems. The first theorem states that the angles opposite to equal sides of an isosceles triangle are equal. The second theorem states that the
sides opposite to equal angles of a triangle are equal. These two theorems can be proved by applying the congruence rule explained in the last exercise of this chapter. So, practicing these proofs
with the help of examples will enable students to gain an in-depth understanding of them.
Students should first try to solve the complete set of questions provided in NCERT solutions class 9 chapter 7 exercise 7.2 themselves so as to understand which topic they need to concentrate on
more. They can thoroughly practice a particular question or theorem by referring to the NCERT solutions class 9 chapter 7 exercise 7.2. These solutions are proficient in clearing student’s doubts and
encouraging them to study.
Download Cuemath NCERT Solutions PDFs for free and start learning!
Class 9 Maths NCERT Solutions Video Chapter 7 Exercise 7.2
Math worksheets and
visual curriculum | {"url":"https://www.cuemath.com/ncert-solutions/ncert-solutions-class-9-maths-chapter-7-exercise-7-2/","timestamp":"2024-11-02T17:21:47Z","content_type":"text/html","content_length":"217792","record_id":"<urn:uuid:50b65109-12c5-435d-9fee-92f9e6e9a9f1>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00853.warc.gz"} |
Cadey Ratio f0ebe310be 2020-03-26 20:03:49 +00:00
cmd/jvozba 2020-03-26 20:03:34 +00:00
LICENSE 2019-06-15 22:21:55 +02:00
README.md 2019-06-15 21:33:34 +00:00
go.mod 2020-03-26 20:03:49 +00:00
jvozba.go 2019-06-15 22:28:54 +02:00
jvozba_test.go 2019-06-15 22:28:54 +02:00
lujvo.go 2019-06-15 22:28:54 +02:00
lujvo_test.go 2019-06-15 22:28:54 +02:00
zbasu.go 2019-06-15 22:28:54 +02:00
zbasu_test.go 2019-06-15 22:28:54 +02:00
An O(n) implementation of the lujvo-making algorithm to save the world.
What's the big deal with O(n), anyway?
All the jvozba I've seen over the years are of exponential complexity (O(c^n), where 1 ≤ c ≤ 4), because the ‘algorithm’ they implement is basically collecting all possible combinations of rafsi in
an array, mapping the array with a score function, and sorting. This means that prefixing an input tanru with just one bloti will quadruple the time and memory it takes for the lujvo to compute. To
put this into perspective: in order to find the lujvo for bloti bloti bloti bloti bloti bloti bloti bloti bloti bloti, the algorithm will have to call the score function a million times. Double the
input length and your 32-bit machine will explode. (Or wake up the OOM killer.)
This jvozba, on the other hand, is linear in complexity, which means it can compute even a million-bloti lujvo in about a second. ‘How does it achieve that?’, I hear you ask. Simply put, it goes
through each tanru unit, keeping track of the best lujvo ‘so far’ alongside its score, with a separate tally for tosmabru words for soundness. There's a bunch more performance tweaks in the code – I
encourage you to perhaps read it.
main.go should give you an idea of how to use the (extremely simple) basic API. If you want to customise stuff, dig into the code and you should find the right procedures to call. | {"url":"https://tulpa.dev/cadey/jvozba","timestamp":"2024-11-09T12:11:59Z","content_type":"text/html","content_length":"49516","record_id":"<urn:uuid:83f58dac-f4ff-4822-a0fa-6706a0b2f93f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00611.warc.gz"} |
447: Too Old For This Shit
Too Old For This Shit
Title text: They say if a mathematician doesn't do their great work by age eleven, they never will.
This comic makes fun of the fact that most mathematical geniuses have done their exceptional work (for which they eventually become famous) in their early years by exaggerating it, particularly given
that "too old for this shit" is a phrase more appropriately used by people later in age. At the age of thirteen, most precocious mathematicians would not be pushing the frontier of mathematical
knowledge, let alone to the point where they would be "too old for it." As such, this is more of a joke about a young boy attempting to dismiss the world around him. It also plays on the fact that in
xkcd comics, it is often difficult to tell age because of a lack of detail, which is necessary to set up the final punchline.
A striking example is Carl Friedrich Gauss, the famous mathematician, who wrote his exceptional masterpiece Disquisitiones Arithmeticae at the early age of 21. This idea was for instance used in the
fictional biography of Gauss, Measuring the World, where he admits to having trouble understanding his own work when he got older because of his age.
The "age theory'" applies to physics as well. Albert Einstein was also very young (26) when he published his four groundbreaking papers in the same year (his Annus Mirabilis in 1905) including the
one that eventually gave him the Nobel Prize. Later in life, for instance, he never accepted the theory of quantum physics - which is now a very well-established theory.
The title of the comic, "Too Old For This Shit," is also a reference to the Lethal Weapon series, in which one of the main characters (Roger Murtaugh, played by Danny Glover) is repeatedly quoted as
saying things along the line of "I'm too old for this shit."
The title text asserts that thirteen is way too old as it claims that mathematicians should do their great work at the age of eleven! If not - they will never do anything great.
[Two Cueballs are standing in front of a whiteboard covered with equations. They are facing away from each other, not yet fully conversing with each other.]
Cueball 1: I wish I could do math like when I was young.
Cueball 2: Huh?
[The two now face each other. They talk to each other.]
Cueball 1: It doesn't come easy like it once did.
Cueball 2: Uh huh.
[Still talking.]
Cueball 1: Math is a game for the young. I need to sit back and let the future happen.
[Still talking.]
Cueball 2: You're thirteen.
Cueball 1: Yes, and it's time I accept that.
add a comment! ⋅ add a topic (use sparingly)! ⋅ refresh comments!
For example Fields Medal is given to mathematicians not over 40 years of age --JakubNarebski (talk) 20:47, 27 June 2013 (UTC)
Yet Sir Andrew Wiles was 41 when he found the proof for Fermats Last Theorum... I'm glad he didn't give up at age 11. Personally I never stood a chance as I only got interested in maths after leaving
school (and yes I do struggle even though I'm 42! and am envious of everyone who's good at it ;) ) Squirreltape (talk) 16:45, 30 January 2014 (UTC)
One of few xkcd strips where the characters being stick figures actually helps comedy... Mumiemonstret (talk) 15:50, 8 December 2014 (UTC)
I can't help but feel that there's another joke here. Namely, the 13 year old claiming that maths isn't as easy as it used to be might be reffering to the fact that he used to do arithmetic, but now
has to deal with stuff like algebra and graphs. (Hey, someone put my IP address here for identification).
It's actually a common misconception Einstein didn't believe quantum theory, he only rejected the parts that went against common physics.Hiihaveanaccount (talk) 17:47, 23 February 2021 (UTC)
And he developed general relativity long after his annus mirabilis. Not all great work is done while young. Nitpicking (talk) 03:22, 26 August 2021 (UTC) | {"url":"https://www.explainxkcd.com/wiki/index.php/447","timestamp":"2024-11-12T19:43:15Z","content_type":"text/html","content_length":"31958","record_id":"<urn:uuid:68b09e1e-1ae7-4ad6-812a-4941cdc60f31>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00697.warc.gz"} |
Implementing Eiserloh's Noise-Based RNG in Rust - Anna R. Dunster
Implementing Eiserloh's Noise-Based RNG in Rust
A couple years ago, I watched a really interesting GDC talk by Squirrel Eiserloh about Noise-Based RNG, and it's really stuck with me. Lately, I've been looking at starting a small game side project
to learn the Bevy engine, and to practice designing procedural narrative systems, which would require extensive use of seeded "RNG" and would like to use this idea. However, Eiserloh's examples are
written in C++, not Rust, so implementing the function in Rust seemed like a great place to get started with actually writing some code.
The Talk
First, if you're interested in any kind of procedural content generation, I really recommend just watching the talk for yourself! But I'll give a high level overview here to get started. According to
Eiserloh, seeded generation has benefits for a variety of uses, like world or NPC generation. And using a seed allows the program to discard unchanged content and regenerate it as needed since it
will be the same, such as for on-demand generation in an infinite or near infinite playing space. One example would be for a Minecraft-like world generation. A randomly-accessible seed allows
generating parts that all fit together correctly no matter which direction you come from, without having to generate the whole world at once.
What we need from a random number generator
An RNG for game development should have a fair statistical distribution, a statistically correct degree of repetition, and a high minimum repeat period. A wide range of possible seeds enables a nice
variety of player experience, and a minimum of arbitrary restrictions on which seeds produce viable results allows more types of game data to be used to seed the number generation. The bits that make
up the content of the results shouldn't be correlated with each other, either in a single result or in subsequent results. It also needs to be platform independent and reproducible across different
platforms - you should get the same result whether it's running on an M1 Mac or an Android phone, or whatever else, in order to enable things like trading of seeds (and using the seed to compress
data for saving or sharing generated content). The order of results should be independent - if you need the 75th result, you should be able to get it at any time, not only just after the first 74
results. You should be able to get the same results from the same seed deterministically from the generator every time. It also needs to be fast! You may need to get a lot of seeded results at once.
This also means it needs a small memory footprint. It should also be thread safe for parallelization.
How traditional number generators fall short
Traditional random number generators have some significant cons when compared to the above list, including using C++'s rand(). Eiserloh gets into the numbers of several tests that he uses to
characterize the results from different generator options, which I'd definitely suggest watching the talk to see if you're curious about the details. In short, they suffer from poor speed and poor
statistical randomness overall. Many also have limitations on what seeds you can use to produce useful results, which makes it more complicated to use values such as entity IDs. And in a lot of
cases, a seed doesn't actually give you a different set of random numbers, but instead just starts you at a different place in the same set of possible "random" numbers. Initial values may not be
statistically useful due to slow warm up times, and some, such as Mersenne Twister, use internal state in a way that make them non-thread-safe. Worst of all, calls are order dependent, so each call
affects the next call and game elements must be generated in the same order to be identical for a given seed.
How noise functions are better
Noise functions broadly solve the problems listed with more traditional RNG options. A noise function here behaves basically as if it is an endless, stateless table of possibilities from which you
select the result you want based on which position in the "table" you request, making it order-independent. It has no state, so it is thread safe and endlessly instantiable. A different seed results
in a unique infinite table, not an offset into the same table. They can be designed with as many dimensions as desired. Because it's order independent, it can be used to generate pieces of content on
demand - world chunks, villages, NPCs - in whatever sequence the player needs them, without having to generate them ahead of time.
Some typical uses of noise functions are for things like tile variation, such as "which variant of grass should this tile be". But noise can be used for a lot more than that. It can be used to add
variance in placement to make things feel more organic, such as using the peaks in a smoothed two dimensional noise graph to plot where trees should be placed by the engine. Want more trees? Increase
the noise, now you have more peaks and thus more trees.
Ultimately, the biggest difference between a random number function and a noise function, is that while a random number function does some fancy modifications to a separately tracked state to produce
a seemingly randomized result, a noise function does the fancy modifications to the input data and just returns the result.
Building noise functions in Rust
To follow along and play with the code in a repository, feel free to use the repo I created for this exercise. Of course, you can always just use std::hash and save yourself some time implementing
something yourself, but seeing how these kinds of things work is an interesting exercise!
Example One: "Some Noise Function"
The first example in Eiserloh's talk is at about 39:00 in the video. He shows a couple different operations performed on an input position to mix up the result, which in C++ looks like this:
uint32_t SomeNoiseFunction(int position)
uint32_t mangled = (uint32_t) position;
mangled *= SOME_BIG_PRIME_NUMBER;
mangled += SOME_OTHER_NUMBER;
mangled *= mangled;
mangled ^= (mangled >> 13);
return mangled;
The code relies heavily on overflows to get its nice scrambling. So how would this look in Rust? It's considered an error in Rust to implicitly rely on overflows, but there are functions in the
standard library to use if you want to explicitly use it. Let's go through rewriting this.
First, we'll use i32 in place of int, since in C++ int can change depending on architecture, then assign our mutable variable to start working with.
pub fn some_noise_function(position: i32) -> u32 {
let mut mangled = position as u32;
// TODO: actually change this.
We can assign a couple constants to use in our mangling:
const SOME_BIG_PRIME_NUMBER: u32 = 27_644_437;
const SOME_OTHER_NUMBER: u32 = 17;
The first one is the fourth Bell Prime, because it was the first prime number over 10,000 that I saw on Wikipedia.
Now let's do the multiplication and addition to mangled. Since we're explicitly relying on overflows, we want to use wrapping_*:
pub fn some_noise_function(position: i32) -> u32 {
let mut mangled = position as u32;
mangled = mangled.wrapping_mul(SOME_BIG_PRIME_NUMBER);
mangled = mangled.wrapping_add(SOME_OTHER_NUMBER);
mangled = mangled.wrapping_mul(mangled);
Finally, Eiserloh's example shows a bitwise XOR assignment with the same value, but bit shifted 13 bits to the right. This looks almost exactly the same in Rust as it does in C++, only without the
unnecessary parentheses. This finally gives us our equivalent function for the first example:
pub fn some_noise_function(position: i32) -> u32 {
let mut mangled = position as u32;
mangled = mangled.wrapping_mul(SOME_BIG_PRIME_NUMBER);
mangled = mangled.wrapping_add(SOME_OTHER_NUMBER);
mangled = mangled.wrapping_mul(mangled);
mangled ^= mangled >> 13;
To test it, you can make a simple output with a brief range:
println!("Some Noise Function results:");
for number in 0..12 {
"Input {} produces output {}",
Which produces results like these:
Some Noise Function results:
Input 0 produces output 289
Input 1 produces output 3715675734
Input 2 produces output 99453684
Input 3 produces output 2035159918
Input 4 produces output 933186350
Input 5 produces output 1088619820
Input 6 produces output 2501336958
Input 7 produces output 875985070
Input 8 produces output 508319700
Input 9 produces output 1397802838
Input 10 produces output 3545209601
Input 11 produces output 2654233028
Interesting to note, at least with the numbers I've chosen, that the result for 0 is such a low number. This could be fine, but probably not the most desirable, and is likely a result of starting our
mangling with multiplication. (Zero times zero is, well, zero.)
Example Two: "Squirrel3"
From about 46:30 in the video, Eiserloh says that this is his third iteration of this hash function. Ultimately, his hand-crafted function produces results with an equally good statistical quality,
and just slightly faster in C++ than C++'s std::hash<int>. I don't have the details to know how a Rust-based implementation would compare, but let's take a look at what it does. It takes both the
position we're requesting, and the seed, as arguments.
uint32_t Squirrel3(int position, uint32_t seed)
constexpr unsigned int BIT_NOISE1 = 0xB5297A4D;
// He says in the talk that these are prime numbers, but this is an even number,
// so it may be a typo. Regardless, we're using it as displayed in the talk.
constexpr unsigned int BIT_NOISE2 = 0x68E31DA4;
constexpr unsigned int BIT_NOISE3 = 0X1B56C4E9;
unsigned int mangled = position;
mangled *= BIT_NOISE1;
mangled += seed;
mangled ^= (mangled >> 8);
mangled += BIT_NOISE2;
mangled ^= (mangled << 8);
mangled *= BIT_NOISE3;
mangled ^= (mangled >> 8);
return mangled;
There's a few more transformations here - and worth pointing out that the seed value is used after the first transformation, and not merely added to the position - but nothing new syntax wise from
the above example. In Rust, following the same strategy with regards to i32 and u32 as in the first example, we end up with a function like this:
pub fn squirrel_3(position: i32, seed: u32) -> u32 {
const BIT_NOISE1: u32 = 0xB5297A4D;
const BIT_NOISE2: u32 = 0x68E31DA4;
const BIT_NOISE3: u32 = 0x1B56C4E9;
let mut mangled = position as u32;
mangled = mangled.wrapping_mul(BIT_NOISE1);
mangled += seed;
mangled ^= mangled >> 8;
mangled = mangled.wrapping_add(BIT_NOISE2);
mangled ^= mangled << 8;
mangled = mangled.wrapping_mul(BIT_NOISE3);
mangled ^= mangled >> 8;
Note if you forget to use wrapping_* here and try to run it, your program will immediately crash with a panic error when you try to run it. Ask me how I know...
Now if you run this with a couple basic ranges and some alternate seeds:
// In main.rs in this case.
println!("Squirrel 3 results:");
let seed = 12345; // Don't use this on your luggage.
for number in 0..12 {
"Input {} with seed {} produces output {}",
squirrel_3(number, seed)
let seed_two = 54321;
for number in 0..12 {
"Input {} with seed {} produces output {}",
squirrel_3(number, seed_two)
You get results like this!
Squirrel 3 results:
Input 0 with seed 12345 produces output 3220422020
Input 1 with seed 12345 produces output 4234179005
Input 2 with seed 12345 produces output 334301668
Input 3 with seed 12345 produces output 145620291
Input 4 with seed 12345 produces output 2582164250
Input 5 with seed 12345 produces output 3262987543
Input 6 with seed 12345 produces output 63288327
Input 7 with seed 12345 produces output 2166186108
Input 8 with seed 12345 produces output 3083917344
Input 9 with seed 12345 produces output 28553252
Input 10 with seed 12345 produces output 2522297604
Input 11 with seed 12345 produces output 3818220281
Input 0 with seed 54321 produces output 3899447266
Input 1 with seed 54321 produces output 3175783065
Input 2 with seed 54321 produces output 3814005845
Input 3 with seed 54321 produces output 2040039807
Input 4 with seed 54321 produces output 1530743793
Input 5 with seed 54321 produces output 3295748292
Input 6 with seed 54321 produces output 327221713
Input 7 with seed 54321 produces output 3085187008
Input 8 with seed 54321 produces output 1267060869
Input 9 with seed 54321 produces output 3629072812
Input 10 with seed 54321 produces output 3952664767
Input 11 with seed 54321 produces output 900255667
These results don't show the issue noted in the first example with the weirdly low first number and give a nice range. Unfortunately, I'm not sure of the details on how to test this in a way that
compares directly with the results in the talk, so I can't comment on that aspect. But we have successfully implemented a function to return a a random, yet totally predictable number for generating
procedural content!
This post is tagged: | {"url":"https://www.annardunster.com/posts/2024/eiserloh-noise-rng-rust","timestamp":"2024-11-03T23:02:56Z","content_type":"text/html","content_length":"85728","record_id":"<urn:uuid:6d5ff296-ade7-4bb3-8651-065b3aeb4108>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00278.warc.gz"} |
Historical network mapping: clusters variations - CorText Manager Q&A forum
Historical network mapping: clusters variations
I used the network mapping script to create a historical map from the cited references of a WOS database.
I get 5 months apart different representations and a different list of references while the parameters and the corpus used are the same.
Do you know where this can come from?
I thank you in advance
1 Answers
Dear Béatrice,
Not sure exactly what you mean: could it come from some overlapping period?
If it is something else, could you try to clarify your question a little?
Thank you for your answer. I did not define periods, I left the default parameters (louvain, automatic detection of the proximity measure).
I get a different number of clusters for the same analysis on the same corpus, and the list of references represented on the radar is not the same. In the parameters that can be consulted after the
analysis, I get respectively :
After filtering, the network (period:1999_2019) has 15 nodes and 56 edges.
2 clusters
After filtering, the network (period:1999_2021) has 15 nodes and 64 edges.
3 clusters | {"url":"https://docs.cortext.net/question/historical-network-mapping-clusters-variations/","timestamp":"2024-11-09T22:35:32Z","content_type":"text/html","content_length":"67060","record_id":"<urn:uuid:7237ee40-fe7e-4bd0-8612-f21edef62775>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00699.warc.gz"} |
Kakuro techniques
The task in Kakuro puzzles is to fill all empty squares using numbers 1 to 9 so the sum of each horizontal block equals the clue on its left, and the sum of each vertical block equals the clue on its
top. In addition, no number may be used in the same block more than once. The best way to learn how to solve Kakuro puzzles is to see how a puzzle is solved from beginning to end.
Step 1
Kakuro puzzles are all about special number combinations. Let’s examine the 22-in-three block in row 1. The only possible combinations are 5+8+9 and 6+7+9. However, square a1 must be smaller than 6
because of the 6-in-two block in column a. Therefore the only number possible in square a1 is 5. Completing the vertical 6-in-two block in column a is now straightforward, since only 1 can be placed
in square a2.
Step 2
From step 1 we know squares b1 and c1 must contain 8 and 9, though we don’t yet know in which order. Let’s look at the 11-in-three block in column c. If the 9 is in square c1 then squares c2 and c3
must both contain 1, which is not allowed. This means square c1 must be 8 and square b1 must be 9.
Step 3
We are now left with two empty squares in column c which must sum up to 3. The only combination is 1+2, but again we don’t know in which order. However, square a2 already contains 1 so the number in
square c2 must be 2. Completing column c and row 2 is now straightforward.
Step 4
Let’s look at the vertical 16-in-five block in column f. This is a magic block because the only five-number combination that sums up to 16 is 1+2+3+4+6. So we know all numbers in this block but we
don’t know in which order they appear. Now let’s examine the horizontal 15-in-two block in row 3. There are only two combinations possible: 6+9 and 7+8. Since square f3 is the crossing point, it must
contain 6 which is the only common number for both blocks. This leads to placing a 9 in square g3.
Step 5
Looking at the 13-in-two block in row 1 we see the only allowed combinations are 4+9, 5+8 and 6+7. However, this block crosses the 16-in-five block in column f, which is still missing 1, 2, 3 and 4.
Therefore the only common number for square f1 is 4. We can now complete the 13-in-two block with 9 in square e1, and the 12-in-two block with 3 in square e2.
Step 6
The 8-in-three block in row 2 still has two empty squares which sum up to 5. There are two possible combinations: 1+4 and 2+3, but the 2+3 is not allowed because this block already contains 3. In
addition, 4 can’t be in square f2 because the 16-in-five block in column f already contains 4.
This means that Square f2 must be 1, and Square g2 must be 4.
Step 7
The 16-in-five block in column f now misses only 2 and 3. Let’s take a close look at the 27-in-four block in row 5. If square f5 contained 2, the remaining three squares must add up to 25. But this
is not possible because the largest sum of three squares is 7+8+9=24. Therefore square f5 is 3 and square f4 is 2.
Step 8
We now come across a special situation which occurs in the right hand side of the puzzle. If we make a vertical sum of columns d, e, f and g we get 22+12+13+16+21=84. If we now make a horizontal sum
of the same area, excluding square d3, we get 13+8+15+12+27=75. This means square d3 is responsible for the difference and therefore must be 84-75=9. To complete this block, a 3 is placed in square
Step 9
Let’s go back to the 27-in-four block in row 5 which has three empty squares that sum up to 24. These three empty squares now form their own magic block because the only possible combination is 7+8+9
in some order. However, the 9 can’t be in square d5 or g5 because each of those vertical blocks already has 9. So the only place left for 9 is square e5. We can now also place 4 in square e4 by
simple calculation.
Step 10
The 12-in-four block in row 4 has two empty squares that sum up to 6. The possible combinations are 2+4 and 1+5 but obviously only 1+5 is allowed. We now need to determine which square is 1 and which
is 5. If we try to place 1 in square d4 we immediately see square d5 will be larger than 9. Therefore square d4 must be 5 and square g4 must be 1. We can now complete columns d and g with 8 in square
d5 and 7 in square g5.
Step 11
Finally let’s examine the 33-in-five block in column b. There are two empty squares that sum up to 14 so the only possible combinations are 5+9 and 6+8. However, 9 is already used in column b so
we’re left with 6+8. If we choose 6 for square b5 then square a5 would also have to be 6 leaving 8 as the only possible candidate for square b5. Completing the remaining squares b4, a4 and a5 is now | {"url":"https://conceptispuzzles.com/index.aspx?uri=puzzle/kakuro/techniques","timestamp":"2024-11-12T20:26:16Z","content_type":"application/xhtml+xml","content_length":"24950","record_id":"<urn:uuid:e8da5b02-a132-42ad-92ed-190ef7ac5e0c>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00869.warc.gz"} |
The last puzzle I set — and likely the next few I will set — was a Chaos Deconstruction Suguru, a genre invented by mathpesto. These puzzles are really hard to do wrong, you set one of these and it’s
probably good just by virtue of existing. Granted, I’ve always been a Chaos Construction fan, so the region-building aspect is something that might not speak to everyone.
My impression is that these puzzles are in some ways easier to set than sudokus of similar difficulty. When setting a puzzle, you have two problems you need to simultaenously solve: you must make
sure there aren’t no solutions and there aren’t multiple solutions. With a sudoku, you are not really guaranteed that solution at the start. But with a suguru, you start with the solution: a blank
board. You just need to make it unique.
Also, since not every square has to be filled out, Suguru is a little more forgiving, since while the geometric interactions are plentiful, actual numbers do not interact that much unless you
specifically design them to.
At the same time, I wouldn’t recommend a first-time setter to try a Suguru in particular. It is much bigger, and there is less solver support, meaning that verifying a puzzle must be done entirely by
hand. Also there’s a lot of scanning when you check, and Sugurus feel easier to break.
There are a couple of quirks of this genre too, especially when solving. Since the solution is unique, say you have a cell completely enclosed off, so it can only be a 1. By uniqueness you know it
can’t be a degree of freedom so somewhere in that row there must be a 1. More generally, you can be sure that only squares that interact in some way with clues — maybe not directly — will have a
digit. While using uniqueness as a logical basis for solving is cringe, sometimes it is useful to know what must be true and find the logic to confirm it. I got lucky in that uniqueness didn’t
actually help with my puzzle, and it doesn’t really play a big role in any potential solve routes, but it is a minor pain point that exists. | {"url":"https://dennisc.net/writing/blog/suguru","timestamp":"2024-11-14T04:07:23Z","content_type":"text/html","content_length":"3314","record_id":"<urn:uuid:8b738fca-3e55-4a68-9f1b-a806beb68efb>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00837.warc.gz"} |
[S:Deleted text in red:S] / Inserted text in green
LivMS Event:
* *Date:* 15 Dec, 2021
* *Time:* 14:00
* *Presenter:* Simon Singh
* *Title:* Christmas Lecture: Fermat's last theorem
* *Audience:* Year 11 up
* *Venue:* Online
* *Registration:* Via the [Eventbrite](https://www.eventbrite.co.uk/e/168104628469) page for the event
* *Contact:* Peter Giblin
[S:* There will be no charge for the lecture but registration will be needed. Further details nearer the time of the lecture.:S]
Simon Singh https://simonsingh.net/ , best-selling author of “Fermat’s Last Theorem”, talks about the most notorious problem in the history of mathematics.
Rarely can a scribbled note in a margin have provoked so much mathematical curiosity as Pierre de Fermat's infamous conjecture, discovered after his death, that no three positive whole numbers EQN:a,
EQN:b, and EQN:c satisfy the equation EQN:a^n + EQN:b^n = EQN:c^n
for any whole-number value of EQN:n greater than 2.
Fermat claimed to have a proof, but left no record of it, stating only that the margin was too small to contain it. This unproven result became known as Fermat's Last Theorem. Generations of
mathematicians tried and failed across more than three centuries to prove the general case, though it was proved for numerous particular values of EQN:n (even by Fermat for EQN:n = 4). Finally in
1994, Andrew (later Sir Andrew) Wiles announced to the world that the problem was solved, though
there were some gaps which were filled by Wiles and his former student Richard Taylor.
Simon Singh made an award-winning documentary about this achievement and also wrote a best-selling book. In this lecture, he will talk us through the history of Fermat's Last Theorem and give us an
idea of how it was eventually proved using ground-breaking new mathematics.
The following is excerpted from the Wikipedia article https://en.wikipedia.org/wiki/Simon_Singh/ You will also find "boring" and "colourful" biographies on Simon's website https://simonsingh.net/
Simon Lehna Singh MBE is a British popular science author, theoretical and particle physicist whose works largely contain a strong mathematical element. His written works include Fermat's Last
Theorem, The Code Book (about cryptography and its history), Big Bang (about the Big Bang theory and the origins of the universe), Trick or Treatment? Alternative Medicine on Trial (about
complementary and alternative medicine, co-written by Edzard Ernst) and The Simpsons and Their Mathematical Secrets (about mathematical ideas and theorems hidden in episodes of The Simpsons and
Futurama). In 2012 Singh founded the Good Thinking Society, through which he created the website "Parallel" to help students learn mathematics. Singh has also produced documentaries and works for
television to accompany his books, is a trustee of the National Museum of Science and Industry, a patron of Humanists UK, founder of the Good Thinking Society, and co-founder of the Undergraduate
Ambassadors Scheme. | {"url":"https://www.livmathssoc.org.uk/cgi-bin/sews_diff.py?Event_20211215_Popular_Lecture","timestamp":"2024-11-10T18:11:57Z","content_type":"text/html","content_length":"4293","record_id":"<urn:uuid:98780849-1ae4-49e3-8c23-2730d1661dd5>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00879.warc.gz"} |
How to define a base without Z axis rotation
I'm defining a turntable with a vertical axis. It's easiest to put my reference point at about 135deg off the X axis,
I don't imagine I can just set the Z value to zero, as that would rotate the vector and give me a bad axis.
i.e. the vector (53.734, 0.118, 0.027) is not equivalent to the robot using (0, 0.118, 0.027).
Instead of doing the math every time to get the correct vector, is there a way to lock the Z rotation to zero? Or am I wrong?
10-23-2023, 09:37 AM
You can add a second coordinate system with respect to the first one and rotate it 135 deg around the Z axis (XYZABC set to 0,0,0,0,0,135). You can then make this new coordinate system the new root
You can also modify the tool flange of the turntable in the Parameters menu of the turntable (just add the 135 deg rotation around the Z axis). | {"url":"https://robodk.com/forum/Thread-How-to-define-a-base-without-Z-axis-rotation","timestamp":"2024-11-11T04:11:40Z","content_type":"application/xhtml+xml","content_length":"41781","record_id":"<urn:uuid:f66edf11-6da0-4f58-9c04-c4498505e0bd>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00865.warc.gz"} |
2.14. The numerical integral I of a function f(x) may be obtained by
2.14. The numerical integral I of a function f(x) may be obtained by the simple expression f x x fx x I a biin ()d() [ = E==- 0 1
A, which involves summing the function values at various x values, xi = a + iAx, for i=0, 1, 2, . . ., so that nAx = b-a, x 0 = a, and x n-1 = b-Ax. Using this formula, which is known as the
rectangular rule, compute the integral x x 2 0 2 =d for Ax =2, 1, 0.5, 0.1, 0.05, and 0.01.Compare the results obtained with the exact value of 8/3. Plot the numerical error versus the step size Ax.
What value of Ax will you choose for such computations, on the basis of the results obtained?
Fig: 1 | {"url":"https://tutorbin.com/questions-and-answers/214-the-numerical-integral-i-of-a-function-fx-may-be-obtained-by-the-s","timestamp":"2024-11-05T07:42:55Z","content_type":"text/html","content_length":"63387","record_id":"<urn:uuid:b5b310c1-854e-4f80-8fc6-f0f8c04fa594>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00175.warc.gz"} |
A new full waveform inversion method based on shifted correlation of the envelope and its implementation based on OPENCL
Standard full waveform inversion (FWI) attempts to minimize the difference between observed and modeled data. When the initial velocity is kinematically accurate, FWI often converges to the best
velocity model, usually of a high-resolution nature. However, when the modeled data using an initial velocity is far from the observed data, conventional local gradient based methods converge to a
solution near the initial velocity instead of the global minimum. This is known as the cycle-skipping problem, which results in a zero correlation when observed and modeled data are not correlated.
To reduce the cycle-skipping problem, we compare the envelope of the modeled and observed data instead of comparing the modeled and observed data directly. However, if the initial velocity is not
sufficient, the correlation of the envelope of the modeled and observed data might still be zero. To mitigate this issue, we propose to maximize both the zero-lag correlation of the envelope and the
non-zero-lag correlations of the envelope. A weighting function with maximum value at zero lag and decays away from zero lag is introduced to balance the role of the lags. The resulting objective
function is less sensitive to the choice of the maximum lag allowed and has a wider region of convergence with respect to standard FWI and envelope inversions. The implementation has the same
computational complexity as conventional FWI as the only difference in the calculation is related to a modified adjoint source. The implementation of this algorithm was performed on an AMD GPU
platform based on OPENCL and provided a 14 times speedup over a CPU implementation based on OPENMP. Several numerical examples are shown to demonstrate the proper convergence of the proposed method.
Application to the Marmousi model shows that this method converges starting with a linearly increasing velocity model, even with data free of frequencies below 3Hz.
Dive into the research topics of 'A new full waveform inversion method based on shifted correlation of the envelope and its implementation based on OPENCL'. Together they form a unique fingerprint. | {"url":"https://faculty.kaust.edu.sa/en/publications/a-new-full-waveform-inversion-method-based-on-shifted-correlation","timestamp":"2024-11-13T12:35:28Z","content_type":"text/html","content_length":"61328","record_id":"<urn:uuid:862f7aee-a54e-4f44-a214-d80ad87aa847>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00882.warc.gz"} |
Equity convexity and gamma strategies | Macrosynergy
Equity convexity means that a stock outperforms in times of large upward or downward movements of the broad market: its elasticity to the market return is curved upward. Gamma is a measure of that
convexity. All else equal, positive gamma is attractive, as a stock would outperform in market rallies and diversify in market stress. However, gamma is not observable, changeable, and needs to be
estimated. Only a subset of stocks displays statistically significant gamma. Empirical analysis suggests that convex stocks can mostly be found in the materials, telecom, industrials, and energy
sectors. High past volatility and price-to-book ratios have also been indicative of high gamma. Macroeconomic drivers that trigger gamma performance have been interest rates and oil prices.
Systematic long-convexity strategies that seek to time convexity exposure have reportedly produced significant investor value.
Ben Abdallah, Marc-Ali, Patrick Herfroy, and Lauren Stagnol (2022) “Equity Convexity and Unconventional Monetary Policy”.
The below post is based on quotes from the above paper, except where a separate source has been linked next to the quote. Headings, cursive text, and text in brackets have been added. Also, most
mathematical expressions have been paraphrased for easier reading.
This post ties in with this site’s summary on macro trends as a basis for systemic trading strategies.
Understanding equity convexity or gamma
“An investment strategy is convex if its payoff relative to its benchmark is curved upward. The image below depicts an investment strategy that exhibits convexity on both the downside and the
upside. Convex investment strategies are expected to be highly correlated with the benchmark in typical market environments but diverge to the positive in extreme markets. There are no free lunches
though, and convex strategies are expected to lag during quiet markets.” [simplify.us]
“Time-variant gamma at security level…translates a stock’s exposure to performance in the tails of the benchmark [market portfolio]. A positive gamma…implies that a stock’s returns are a convex
function of market returns, which means that theoretically they should always outperform the benchmark (whether the latter is either in positive or negative territory). A negative gamma instead
implies that such a relationship is actually concave, therefore such a stock would systematically underperform the market. The higher the gamma, the more such a stock would outperform the market,
which is a particularly attractive feature, considering that periods of market stress are generally accompanied by high correlation between stock returns. This is a powerful characteristic in terms
of diversification.”
“[For] measuring convexity [one must consider] time-varying unconditional systematic co-skewness in the traditional capital asset pricing model. “
This means that the return of a stock does not just depend linearly on the market return but also on the square of the market return. Convex stocks’ returns increase more than proportionately to the
market return and concave stocks’ returns increase less than proportionately to the market return.
“To illustrate, based on weekly returns against the MSCI World from April 2018 to March 2021, we present [in the figure] below two typical gamma shapes. The first is from Qurate Retail equity, which
has been [positively related to squared market returns] for the period of analysis: we witness that it exhibits a convex shape. The second is from Cenovus Energy…which was [ngeatively related to
squared market returns] and results in a concave shape when plotted against the MSCI World.”
“In theory, convexity should always pay off, since a convex stock is supposed to outperform the benchmark independently of market conditions. Similarly, concavity should not be of interest for a
long-only investor since concave stocks are supposed to consistently underperform the market. However, by definition we cannot apprehend the stock’s true convexity, and have to estimate it.”
How to estimate equity gamma
“We work on the MSCI World universe. For each stock, at each date the market beta and the gamma are estimated together via Ordinary Least Square (OLS), as in [the equation below], using
heteroskedasticity robust standard errors, on a rolling window of 3 years, using weekly returns from February 2010 to August 2020 [where R is the return of a stock, rf the risk-free rate, RM the
market return, and beta and gamma are the coefficients to be estimated].”
“Gamma is estimated at the end of the months of February, May, August and November. Actually, the beta coefficients derived from this regression are predominantly significant. However, this is not
the case for the gamma coefficient in general…Gamma is not everywhere and only concerns a small subset of the stocks within our universe. Therefore, we build a new database, composed of the gamma
that are highly significant alongside corresponding stock returns data.”
What is driving equity convexity or gamma
“Fundamental data is retrieved from the FactSet Fundamentals database on a quarterly basis [and some other sources].”
“From a bottom-up viewpoint….over the last 10 years stocks with convex returns have been mostly found in the Asia-Pacific region and in sectors such as Materials, Telecoms, Industrials, and Energy.
Despite varying gamma regimes over the period, we demonstrate that past volatility and the price-to-book ratio have been the most efficient discriminant features of concavity and convexity. Namely,
stocks with volatile past returns, from companies that are rather classified as glamorous (as opposed to value) tend to have higher gammas.”
“In a top-down approach, we investigate the macroeconomic drivers of gamma… [i.e.] market environments that trigger gamma performance and subsequently, our ability to forecast this premium…Applying a
cointegrating vector framework, we find that it exhibits long-term relationships with the VIX, as expected from gamma’s essence, but also with short-term interest rates, and oil prices.”
Strategies using equity convexity or gamma
“We propose a systematic long convexity strategy. The rationale in this approach is to time the convexity exposure…We…evaluate the ability of different models to forecast future convexity premium
dynamics…[and] employ these signals in the design of a systematic long convexity strategy…It leads to significantly improved risk-adjusted returns compared to a capitalization-weighted benchmark,
especially in turbulent markets. Convexity exposure appears particularly relevant in a context of monetary policy normalization.”
“The XMA factor is long convex stocks and short concave ones. We propose to embed our XMA forecast in the design of a systematic strategy….The forecast is only a function of lagged macroeconomic data
and…analyzing performance exclusively out-of-sample ensures that our model…is not overfitted. In this systematic long convexity strategy, the aim is to time the convexity exposure.”
“Building on [empirical] results, we attempt to forecast the premium [of the] XMA factor (long convexity, short concavity)…Increasing short-term interest rates and market volatility are conducive to
the outperformance of the convexity premium in the subsequent period. We use this signal to propose a systematic long convexity strategy.”
“The strategies we propose, [which] aim at timing convexity exposure in a systematic way, deliver strong risk-adjusted returns compared to their benchmark, with a Sharpe ratio close to 2 for our
period of analysis. Furthermore, long convexity strategies, by exposing to convexity during period of market stress, efficiently manage to reduce portfolio volatility. This authenticates the
relevance of such strategy for mitigating equity portfolio losses in turbulent markets, in a defensive manner.” | {"url":"https://macrosynergy.com/research/equity-convexity-and-gamma-strategies/","timestamp":"2024-11-11T14:37:20Z","content_type":"text/html","content_length":"185123","record_id":"<urn:uuid:e185513f-2bd3-4b59-af93-60905b2ebf79>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00315.warc.gz"} |
Genetic Algorithms for Optimal System Design
Yajur Kumar
The inspiration for Genetic Algorithms (GAs) comes directly from nature, specifically from the process of evolution and natural selection. In this article, I will discuss how this natural process
works in detail, and how it inspired the creation of an algorithm for problem-solving and optimization.
Natural Evolution: Survival of the Fittest
In nature, evolution is the gradual process through which species change over time to adapt to their environment. The main idea behind evolution is natural selection, often called "survival of the
• Genetic Material (DNA): Every living organism has genes, which carry the information that defines its traits—like height, color, or behavior. These genes are stored in a long sequence of DNA.
• Population: A group of individuals of the same species living together. Each individual has a unique set of genes.
• Fitness: Not all individuals are equally fit to survive in their environment. Some have traits that help them thrive (like better camouflage or stronger muscles), while others may have traits
that make life harder.
• Reproduction: Individuals that are better suited to their environment (more "fit") are more likely to survive and reproduce. Their offspring inherit the genes that made their parents successful.
• Mutation and Variation: Occasionally, random mutations occur in the genes of individuals. These small, random changes introduce new traits. Most mutations might not be useful, but some can
provide advantages, helping future generations survive even better.
Example of Evolution in Nature
Let’s take an example of birds living on an island where food is scarce, and only birds with sharp, strong beaks can break open the tough seeds available.
• Over time, birds with stronger beaks will survive better because they can eat more food.
• These strong-beaked birds will reproduce more often, passing on the "strong beak" trait to their offspring.
• Birds with weaker beaks will struggle to survive and may not reproduce as much.
• Over many generations, more birds in the population will have strong beaks, and the weak-beaked birds will eventually disappear.
• If a random mutation makes some birds' beaks even stronger or sharper, that trait could spread if it offers an advantage.
Through this process of selection, reproduction, and mutation, species evolve and adapt to their environments over time.
How This Inspired Genetic Algorithms
The process of evolution in nature inspired scientists to create a problem-solving algorithm that mimics how species evolve to find the "best" solution over time. Here’s how the inspiration from
nature was used:
• Genes and Chromosomes → Variables and Solutions: In a Genetic Algorithm, just like in nature, we represent each possible solution to a problem as a chromosome. This chromosome is made up of genes
, each representing different variables or characteristics of the solution.
• Fitness in Nature → Fitness Function in Algorithms: In nature, fitness is the ability to survive and reproduce. In a Genetic Algorithm, fitness is a measure of how good a solution is at solving
the problem.
• Selection in Nature → Selection in Algorithms: In evolution, only the fittest individuals survive and reproduce. In Genetic Algorithms, the fittest solutions (those with the best fitness scores)
are selected to create new solutions.
• Crossover in Nature → Crossover in Algorithms: In biology, reproduction combines the genes from two parents to create offspring with a mix of both parents’ traits. In Genetic Algorithms,
crossover mimics this by combining two solutions to create a new one.
• Mutation in Nature → Mutation in Algorithms: Just as random mutations in nature can introduce new traits, mutation in a Genetic Algorithm introduces small random changes to a solution. This
ensures that new possibilities are explored, even if the current best solutions aren’t perfect.
The key advantage of using evolution as inspiration is that Genetic Algorithms don’t need to know the exact path to the best solution. They explore many possibilities and use random variation, just
like evolution, to stumble upon better solutions. Over time, by applying selection and combination, these algorithms can find highly effective solutions to very complex problems.
Let's say we want to maximize the function: f(x)= x sin(10πx)+2, where x is between 0 and 1. This is a non-linear function with multiple peaks and valleys, making it a good example for a Genetic
Algorithm. You can also find this code at my github handle.
% Compare Genetic Algorithm solution with standard mathematical optimization
% for maximizing f(x) = x * sin(10 * pi * x) + 2, x in [0, 1]
% Step 1: Genetic Algorithm solution
% Genetic Algorithm parameters
populationSize = 20; % Number of individuals in the population
numGenerations = 50; % Number of generations to evolve
mutationRate = 0.05; % Probability of mutation (5%)
crossoverRate = 0.8; % Probability of crossover (80%)
numGenes = 16; % Number of genes (bits) to represent x
% Create an initial random population
population = randi([0 1], populationSize, numGenes);
% Genetic Algorithm loop for several generations
for generation = 1:numGenerations
% Evaluate the fitness of each individual in the population
fitness = zeros(populationSize, 1);
for i = 1:populationSize
x = decodeChromosome(population(i, :), numGenes);
fitness(i) = fitnessFunction(x);
% Select individuals for reproduction based on fitness
selectedPopulation = selectRoulette(population, fitness);
% Apply crossover to generate a new population
newPopulation = applyCrossover(selectedPopulation, crossoverRate);
% Apply mutation to introduce random changes
mutatedPopulation = applyMutation(newPopulation, mutationRate);
% Update the population for the next generation
population = mutatedPopulation;
% Final solution from Genetic Algorithm
[bestFitness_GA, bestIndex] = max(fitness);
bestX_GA = decodeChromosome(population(bestIndex, :), numGenes);
fprintf('Genetic Algorithm: Best x = %.5f, Best fitness = %.5f\n', bestX_GA, bestFitness_GA);
% Step 2: Standard Mathematical Optimization using fminbnd
% fminbnd minimizes a function, so we minimize -f(x) to maximize f(x)
objectiveFunction = @(x) -(x * sin(10 * pi * x) + 2); % Negative of the function to maximize
[x_fminbnd, neg_fminbnd] = fminbnd(objectiveFunction, 0, 1);
% Convert the negative fitness value back to positive
bestFitness_fminbnd = -neg_fminbnd;
fprintf('Standard Optimization (fminbnd): Best x = %.5f, Best fitness = %.5f\n', x_fminbnd, bestFitness_fminbnd);
%% Function to evaluate the fitness of a chromosome
function fit = fitnessFunction(x)
% The function we want to maximize: f(x) = x * sin(10 * pi * x) + 2
fit = x * sin(10 * pi * x) + 2;
%% Function to decode the chromosome (binary string) into a real value of x
function x = decodeChromosome(chromosome, numGenes)
% Convert binary chromosome to decimal, then scale to the range [0, 1]
decimalValue = bin2dec(num2str(chromosome));
x = decimalValue / (2^numGenes - 1); % Scale to [0, 1]
%% Function for roulette wheel selection based on fitness
function selectedPop = selectRoulette(population, fitness)
% Normalize fitness values to create a probability distribution
totalFitness = sum(fitness);
prob = fitness / totalFitness;
% Cumulative probability distribution for selection
cumulativeProb = cumsum(prob);
% Select individuals based on roulette wheel approach
populationSize = size(population, 1);
numGenes = size(population, 2);
selectedPop = zeros(populationSize, numGenes);
for i = 1:populationSize
r = rand(); % Random number between 0 and 1
selectedIndex = find(cumulativeProb >= r, 1); % Select individual
selectedPop(i, :) = population(selectedIndex, :);
%% Function to apply crossover between pairs of individuals
function newPop = applyCrossover(population, crossoverRate)
populationSize = size(population, 1);
numGenes = size(population, 2);
newPop = population;
for i = 1:2:populationSize
if rand() < crossoverRate
% Select two parents
parent1 = population(i, :);
parent2 = population(i+1, :);
% Apply crossover (single-point crossover)
crossoverPoint = randi([1, numGenes-1]);
newPop(i, :) = [parent1(1:crossoverPoint), parent2(crossoverPoint+1:end)];
newPop(i+1, :) = [parent2(1:crossoverPoint), parent1(crossoverPoint+1:end)];
%% Function to apply mutation to the population
function mutatedPop = applyMutation(population, mutationRate)
populationSize = size(population, 1);
numGenes = size(population, 2);
mutatedPop = population;
for i = 1:populationSize
for j = 1:numGenes
if rand() < mutationRate
mutatedPop(i, j) = ~population(i, j); % Flip the gene
Upon evaluating this code in MATLAB, we get the following output:
Genetic Algorithm: Best x = 0.59976, Best fitness = 2.84666
Standard Optimization (fminbnd): Best x = 0.65156, Best fitness = 2.65078
Let me give an example related to satellite systems because that's what this website is about. To compare a Genetic Algorithm (GA) with a standard optimization technique for trajectory optimization
of a satellite in orbit, let's consider the following problem:
Problem Description:
The goal is to transfer a satellite from one circular orbit (initial orbit) to another circular orbit (target orbit). The satellite can change its velocity at specific times (impulse burns), and we
want to optimize the thrust directions to minimize the total fuel used.
• Objective: Minimize fuel usage (which is proportional to the velocity changes ΔV).
• Standard Method for Comparison: We can use the Hohmann transfer method, which is an analytically optimal way of transferring a satellite between two circular orbits using two impulsive burns.
1. Use a Genetic Algorithm to find the optimal thrust direction and times to minimize ΔV.
2. Compare it with the fuel cost of a Hohmann transfer (a standard two-burn maneuver).
Problem Setup:
• The satellite is initially in a circular orbit with radius r1.
• The target orbit is circular with radius r2.
• The satellite can apply a finite number of thrust impulses at any point during the transfer.
• The cost function is the total ΔV (change in velocity).
MATLAB Code (find it here)
% Constants
mu = 398600; % Gravitational parameter of Earth [km^3/s^2]
r1 = 7000; % Initial orbit radius [km]
r2 = 10000; % Target orbit radius [km]
% Genetic Algorithm parameters
populationSize = 50; % Number of individuals in the population
numGenerations = 100; % Number of generations
numSteps = 2; % Number of impulsive burns (same as Hohmann transfer)
mutationRate = 0.05; % Mutation probability
crossoverRate = 0.8; % Crossover probability
% Initial random population (representing thrust directions)
population = rand(populationSize, numSteps * 2); % Thrust angles and magnitudes
% Preallocate fitness array for storing results
fitness = zeros(populationSize, 1);
% GA loop for multiple generations
for generation = 1:numGenerations
% Evaluate fitness (fuel consumption) of each individual
for i = 1:populationSize
thrustSequence = population(i, :);
fitness(i) = evaluateThrustSequence(thrustSequence, r1, r2, mu);
% Select individuals based on fitness (lower deltaV is better)
selectedPopulation = selectRoulette(population, fitness);
% Apply crossover to generate new population
newPopulation = applyCrossover(selectedPopulation, crossoverRate);
% Apply mutation to introduce diversity
mutatedPopulation = applyMutation(newPopulation, mutationRate);
% Update the population
population = mutatedPopulation;
% Display best fitness of the generation
[bestFitness, bestIndex] = min(fitness); % We minimize fuel consumption
fprintf('Generation %d: Best deltaV = %.5f km/s\n', generation, bestFitness);
% Final GA solution
[bestFitness, bestIndex] = min(fitness);
bestThrustSequence = population(bestIndex, :);
fprintf('Optimal deltaV found by GA: %.5f km/s\n', bestFitness);
% Hohmann transfer calculations
% Burn 1: At perigee (first circular orbit)
v1 = sqrt(mu / r1); % Velocity in initial orbit
a_transfer = (r1 + r2) / 2; % Semi-major axis of the transfer orbit
v_transfer_perigee = sqrt(2 * mu * (1/r1 - 1/(2 * a_transfer))); % Velocity at perigee of transfer orbit
% Burn 2: At apogee (target circular orbit)
v_transfer_apogee = sqrt(2 * mu * (1/r2 - 1/(2 * a_transfer))); % Velocity at apogee of transfer orbit
v2 = sqrt(mu / r2); % Velocity in target orbit
% Delta V for Hohmann transfer
deltaV_hohmann = abs(v_transfer_perigee - v1) + abs(v2 - v_transfer_apogee);
fprintf('DeltaV for Hohmann transfer: %.5f km/s\n', deltaV_hohmann);
%% Function to evaluate fuel usage for a given thrust sequence
function deltaV = evaluateThrustSequence(thrustSequence, r1, r2, mu)
% Simplified evaluation of fuel consumption for impulsive burns
% thrustSequence: contains magnitudes and angles of thrust
% Step 1: Calculate initial velocity in the initial orbit
v1 = sqrt(mu / r1); % Circular orbit velocity at r1
% Step 2: Compute the semi-major axis of the transfer orbit
a_transfer = (r1 + r2) / 2; % Semi-major axis of the transfer orbit
% Step 3: Compute velocity at perigee (initial point) and apogee
v_transfer_perigee = sqrt(2 * mu * (1/r1 - 1/(2 * a_transfer))); % Perigee velocity
v_transfer_apogee = sqrt(2 * mu * (1/r2 - 1/(2 * a_transfer))); % Apogee velocity
% Step 4: Calculate deltaV for each thrust impulse
deltaV1 = abs(v_transfer_perigee - v1); % First burn at perigee (r1)
deltaV2 = abs(v_transfer_apogee - sqrt(mu / r2)); % Second burn at apogee (r2)
% Total deltaV (fuel consumption)
deltaV = deltaV1 + deltaV2;
%% Function for roulette wheel selection based on fitness
function selectedPop = selectRoulette(population, fitness)
totalFitness = sum(fitness);
prob = fitness / totalFitness;
cumulativeProb = cumsum(prob);
populationSize = size(population, 1);
numGenes = size(population, 2);
selectedPop = zeros(populationSize, numGenes);
for i = 1:populationSize
r = rand();
selectedIndex = find(cumulativeProb >= r, 1);
selectedPop(i, :) = population(selectedIndex, :);
%% Function to apply crossover
function newPop = applyCrossover(population, crossoverRate)
populationSize = size(population, 1);
numGenes = size(population, 2);
newPop = population;
for i = 1:2:populationSize
if rand() < crossoverRate
parent1 = population(i, :);
parent2 = population(i+1, :);
crossoverPoint = randi([1, numGenes-1]);
newPop(i, :) = [parent1(1:crossoverPoint), parent2(crossoverPoint+1:end)];
newPop(i+1, :) = [parent2(1:crossoverPoint), parent1(crossoverPoint+1:end)];
%% Function to apply mutation
function mutatedPop = applyMutation(population, mutationRate)
populationSize = size(population, 1);
numGenes = size(population, 2);
mutatedPop = population;
for i = 1:populationSize
for j = 1:numGenes
if rand() < mutationRate
mutatedPop(i, j) = rand(); % Random mutation
Well, as you may notice both methods will give 1.22288 km/s value! But, this is just a simple illustrative example and the algorithm is meant for more complex optimization scenarios.
Let's keep this blog till here, if you have any questions related to this topic, feel free to connect with me and message me at https://www.linkedin.com/in/yajur | {"url":"https://www.spacenavigators.com/post/genetic-algorithms-for-optimal-system-design","timestamp":"2024-11-03T21:26:06Z","content_type":"text/html","content_length":"1050494","record_id":"<urn:uuid:c81881a6-dd48-4e6a-a0a3-420cbe3fc6b9>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00810.warc.gz"} |
Gambling Probability & Odds
Gambling Math Basics
Gambling can be a lot of fun even if you don’t understand any of the math behind it, but it’s even more fun if you have a fundamental understanding of how probability and odds work. The purpose of
this page is to provide an introduction to how probability works and how odds work. Those are the gambling math basics.
What Is Probability?
Something with a probability of 0 is something that could never possibly happen. By definition, a probability of 0 is an outcome that is impossible. For example, if you roll a six-sided die that’s
numbered 1 through 6, the probability of rolling a 7 is 0.
Something with a probability of 1 is a certainty. It will always happen. For example, if you roll a six-sided die that’s numbered 1 through 6, the probability of rolling a 1, 2, 3, 4, 5, or 6 is 1.
You’re probably most used to seeing a probability expressed as a percentage. In that first example, the probability of rolling a 7 was 0%. In the second example, the probability of rolling a number
between 1 and 6 was 100%.
The probability of an outcome is easily calculated. You take the total number of ways achieving the outcome, and you divide it by all of the possible outcomes.
For example, if you want to calculate the probability of flipping a coin and having it land on heads, you look at the total number of ways it could land on heads. On a normal coin, that’s one. You
divide that one by the total number of possible outcomes, which is two (heads is one possibility; tails is the other–so that’s two possible outcomes.)
So the probability of getting heads is 1/2, or 50%. You could even express this probability as a decimal, in which case it would be 0.5.
What Are Odds?
Odds are just another way of expressing a probability. When you express a probability as odds, you have the number of ways that the outcome doesn’t happen compared to the number of ways the outcome
can happen.
For example, when flipping a coin, the odds of getting head are 1 to 1, or even odds. If you’re rolling a six-sided die, and you want to know the odds of getting a 1, you have 5 ways of NOT getting
the 1, and a single way of getting the 1, so the odds are 5 to 1. (You could express that as 1/6 as a fraction, as 0.1667, or as 16.67%.)
Why Do Odds and Probability Matter When You’re Gambling
Bets pay off at certain odds, too. And when a bet pays off at the same odds as the probability of winning the bet, you have an even money situation. You’ll win exactly half the time, and the other
side will win half the time.
For example, if you bet $1 to win $1 on a coin flip, then you’re making an even odds bet. The outcome also offers even odds. So over time, you’ll probably break even.
But suppose you bet $1 for the coin to land on heads, but if you win, you get $2. Over time, you’d win a significant amount of money, because half the time you’d win $2, but the half the time that
you lose, you’d only lose $1.
You’d have to find a real idiot to offer you those kinds of odds. (Luckily, those kinds of idiots play poker all the time. The odds are a little more complicated to figure out in poker, so inferior
players will often give you better odds on your bet than you deserve, and that’s why there’s such a thing as a professional poker player.)
The casino’s profits come from the difference between the odds that they pay out on a bet and the odds of your actually winning the bet.
The easiest example to think of is a simple even money bet on black at the roulette table. If you bet $5 on black, you’ll win $5 if the ball lands on black. You’ll lose $5 if the ball lands on red.
But you’ll ALSO lose $5 if the ball lands on one of the green zeros.
There are 38 possible results on an American roulette wheel. 18 of them are black. That means the probability of winning a bet on black are 18/38, or 47.37%. So the casino will win slightly more than
half the time. That’s how they make their profit.
The odds of winning a bet on black are 20 to 18, which means there are 20 ways to lose, and 18 ways to win. That can be reduced to 10 to 9, just like a fraction.
The bigger the difference between the odds of winning and the odds that are paid out, the greater the casino’s house edge is. | {"url":"http://www.casinogamblingstrategy.org/probability/","timestamp":"2024-11-12T23:47:27Z","content_type":"text/html","content_length":"34538","record_id":"<urn:uuid:e7bd8116-caf7-44c6-ac9a-587b6f00b401>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00590.warc.gz"} |
Corrigendum to “Bisimplicial vertices in even-hole-free graphs” (Bisimplicial vertices in even-hole-free graphs (2008) 98(6) (1119–1164), (S0095895608000087), (10.1016/j.jctb.2007.12.006))
An even-hole-free graph is a graph with no induced cycle of even length. A vertex of a graph is bisimplicial if the set of its neighbours is the union of two cliques. Reed conjectured in [3] that
every nonnull even-hole-free graph has a bisimplicial vertex. The authors published a paper [1] in which they claimed a proof, but there is a serious mistake in that paper, recently brought to our
attention by Rong Wu. The error in [1] is in the last line of the proof of theorem 3.1 of that paper: we say “it follows that [Formula presented], and so v is bisimplicial in G”; and this is not
correct, since cliques of [Formula presented] may not be cliques of G. Unfortunately, the flawed theorem 3.1 is fundamental to much of the remainder of the paper, and we have not been able to fix the
error (although we still believe 3.1 to be true). Thus, this paper does not prove Reed's conjecture after all. Two of us (Chudnovsky and Seymour) claim to have a proof of Reed's conjecture using a
different method [2].
All Science Journal Classification (ASJC) codes
• Theoretical Computer Science
• Discrete Mathematics and Combinatorics
• Computational Theory and Mathematics
Dive into the research topics of 'Corrigendum to “Bisimplicial vertices in even-hole-free graphs” (Bisimplicial vertices in even-hole-free graphs (2008) 98(6) (1119–1164), (S0095895608000087),
(10.1016/j.jctb.2007.12.006))'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/corrigendum-to-bisimplicial-vertices-in-even-hole-free-graphs-bis","timestamp":"2024-11-09T11:01:28Z","content_type":"text/html","content_length":"49136","record_id":"<urn:uuid:daa6790d-5607-4beb-808c-a379156babd0>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00519.warc.gz"} |
A data frame with 121 observations on the following 16 variables.
a numeric identifier unique to each individual.
a factor with levels Racla.
a numeric vector, either 1 (shade) or 0 (no shade).
the snout-vent length of the individual.
the substrate type, a factor with levels PEAT, SOIL, and SPHAGNUM.
the initial mass of individuals.
the mass lost in g.
the air temperature in degrees C.
the wind intensity, either 0 (no wind), 1 (low wind), 2 (moderate wind), or 3 (strong wind).
cloud cover expressed as a percentage.
centered inital mass.
initial mass squared.
centered air temperature.
proportion of cloud cover
wind intensity, either 1 (no or low wind) or 1 (moderate to strong wind).
log of mass lost. | {"url":"https://www.rdocumentation.org/packages/AICcmodavg/versions/2.3-1/topics/dry.frog","timestamp":"2024-11-08T15:39:03Z","content_type":"text/html","content_length":"63337","record_id":"<urn:uuid:45b127bf-8594-4dd7-8680-2d951b4bb2fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00673.warc.gz"} |
Measures of Central Tendency
Abdulla Javeri
30 years: Financial markets trader
Abdulla outlines measures of Central Tendency, and explains the differences between calculating the Mean, Median and Mode.
Abdulla outlines measures of Central Tendency, and explains the differences between calculating the Mean, Median and Mode.
Access this and all of the content on our platform by signing up for a 7-day free trial.
Measures of Central Tendency
Key learning objectives:
• Understand the calculation of mean
• Understand the calculation of median
• Understand the calculation of mode
• Learn how the three measures are related to eachother
A quick revision on mean, mode and median and how the three measures are related.
Subscribe to watch
Access this and all of the content on our platform by signing up for a 7-day free trial.
How do you calculate mean?
The mean – or arithmetic average – is probably the most frequently used measure of Central Tendency. To find the average height of people in a room, you calculate the total of their heights and
divide by the number of people.
How do you calculate median?
To find the median, you line up people in a room from shortest to tallest, take people away from either end until you are left with the person in the middle, and their height is the median height. If
there are two people left standing, the median is the average of their heights.
How do you calculate mode?
The mode is the most frequently occurring number in a room of people. If each person in the room is of a unique height then there is no mode. If two or more people are of the same height, that is the
mode. Logic would suggest that it’s possible to have more than one mode in a set of data.
How are the three measures related?
The relation between mean, median and mode is dependent on the distribution of the data. If the data is symmetrically distributed, all three measures will be approximately the same. If the
distribution is skewed to one side, the numbers will diverge. If there were an unusually large number of tall people in a room, the data will be skewed to the right, or positively skewed. And vice
versa. The mode will be the peak of the distribution, the median will lie to its left or right depending on the direction of the skew and the mean will lie beyond the median.
Subscribe to watch
Access this and all of the content on our platform by signing up for a 7-day free trial.
Abdulla Javeri
Abdulla’s career in the financial markets started in 1990 when he entered the trading floor of the London International Financial Futures Exchange, LIFFE, and qualified as a pit trader in equity and
equity index options. In 1996, Abdulla became a trainer for regulatory qualifications and then for non-exam courses, primarily covering all major financial products.
There are no available Videos from "Abdulla Javeri" | {"url":"https://data-scienceunlocked.com/videos/measures-of-central-tendency","timestamp":"2024-11-09T12:22:21Z","content_type":"text/html","content_length":"131945","record_id":"<urn:uuid:f28da55d-76b1-4840-b882-150b70f8335f>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00163.warc.gz"} |
Quadratics Revisited: The Falling Object Model
I keep hoping that I can use this to help kids derive the falling object model. I'm getting close. I exported the video at 6fps so 1/6 second elapses between strobes. I'd like some feedback on this
before I roll it out to my students. How intuitive is it to use both applets together?
Once you have plotted your points, use the FitPoly function. Simply enter "fitpoly[A,B,C,D,E,F,G,H,I,2]" to plot a quadratic function.
If this proves to be useful, I'll dial it all in and post the still, video and applets for download.
10 comments:
Pretty nifty.
It is intuitive to me on how to use the applets together, will it be to 7th/8th graders? Probably?
I like it though. Are you going to show the video first? Or let them have at the applets?
I can't figure out how to use the applet. Help?
My kids will most likely have a go at it. I think I'm going to show the video, add timecode and cut it off before the ball hits the ground.
Place the white circle around each strobe and record the distance from the ground. Elapsed time between strobes is 1/6 second. Enter the ordered pairs into the second applet and use the fitpoly
function to do the quadratic regression.
So students are moving the white circle to the pictures of the ball (which you made), getting the height(which the computer gives them), and graphing them with a program to find the parabola
(which the computer calculates). Seems like the kids aren't doing much thinking, just pointing and clicking.
These are middle school kids, right? Do they know how the computer calculates the height? Do they know how the computer fits the parabola?
Here's what I might do. Give each kid an overhead transparency and marker to put onto the computer screen to mark the position of the ball every frame (or every 2, 5, n frames or whatever). Given
the initial height of the ball, have them use proportions to determine the heights of the ball on their transparency.
Then go to Data Flyer http://www.shodor.org/interactivate/activities/DataFlyer/ to input the points and use the manual curve fit sliders to fit the function to their data. They will have a better
feel for how each coefficient manipulates the curve.
Repeat video analysis for ball tossed straight up. Repeat for ball tossed between two people. What's the same/different about the equations for all three vertical motions?
This comment has been removed by the author.
...and that's why I put this out here before putting it in front of my kids. Thanks, Frank.
I think I can do the same thing in GeoGebra. I can define variables a,b and c with sliders and let the kids mess with the regression themselves. The main reason I made the applet was to see how
close I could get to the falling object model by analyzing a still picture.
I definitely want my kids to do most of the lifting. I'm just still trying to figure out how to best do that.
Instead of the video+overhead, you could also give the kids a handout of the time lapse photo you made, have them measure directly on the handout and the use the applet with sliders to fit the
I know this is an older post, but I just ran across it. I'd love to make my own version of your "strobed" picture. Is it easy to explain what you did? I'll follow along with your explanation as
best I can. Thanks!
Dan did a pretty good job of explaining the process here. | {"url":"https://coxmath.blogspot.com/2011/02/quadratics-revisited-falling-object.html?showComment=1297436528111","timestamp":"2024-11-04T21:21:52Z","content_type":"text/html","content_length":"125440","record_id":"<urn:uuid:e3c7c2c7-a347-422e-8399-de9b44299e51>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00294.warc.gz"} |
Se-Jin (Sam) Kim - CV
Se Jin Kim
Department of Mathematics
KU Leuven
Celestijnenlaan 200B
3001 Leuven
University of Glasgow (Nov. 2020 -- Oct. 2022)
Research AssistantSupervisor: Prof. Xin Li
University of Waterloo (Sept. 2016 -- July 2020) Ph.D. Pure Mathematics Advisors: Prof. Ken Davidson and Prof. Matt KennedyUniversity of Waterloo (September 2015 -- May 2015) M. Math. Pure
Mathematics Advisor: Prof. Ken DavidsonMcMaster University (Sept. 2011 -- May 2015) B. Sc. Mathematics and Statistics Advisor: Prof. Bradd Hart
1. Department of Mathematics, KU Leuven (November 2022 – Present)
1. Masters Reading course coordinator. Currently the coordinator of a Masters level reading course in functional analysis, serving as the local coordinator of the Internet Seminar (ISem), an
inter-university seminar series on a topic in functional analysis.
2. Masters’ Project Supervision. Supervision of Julia Pawłowska on her Masters Thesis titled On the role of the minimal tensor product of non-unital operator systems.
3. Undergraduate group project Supervision. Supervision of three groups of bachelors students on the topic of graph invariants on quantum graphs.
2. School of Mathematics and Statistics, University of Glasgow (November 2020 – October 2022)
1. Undergraduate Supervision. Co-supervised a student in the summer with Christian Voigt through a bursary funded by the Edinburgh Math. Society on the topic of chromatic numbers on infinite
dimensional quantum graphs.
2. Course Head. Lead instructor for a third year undergraduate analysis course.
3. Teaching Assistant. Running tutorials for a graduate course in functional analysis.
3. Faculty of Mathematics, University of Waterloo (September 2015 – August 2020)
1. Teaching Assistant. Teaching Assistant for undergraduate Analysis, Calculus, Linear Algebra, and Logic courses. Duties involved marking and coordinating undergraduate marking, preparing and
teaching tutorial lectures, and office hours.
Talks Presented
1. Factoriality of Groupoid von Neumann algebras, Young Mathematicians in C*-algebras, University of Glasgow (August 2024).
2. A duality theorem for non-unital operator systems, International Conference on Noncommutative Geometry, Analysis on Groups, and Mathematical Physics, Ghent University, Online (February 2024)
3. A duality theorem for non-unital operator systems, Analysis Seminar, University of Southern Denmark (February 2024)
4. On C*-Simplicity for groupoids, GoTh Workshop: Groups of Thompson and their relatives, Otto-von-Guericke-Universität Magdeburg (September 2023)
5. C*-Simplicity for Groupoids, The Open University (April 2022)
6. A duality theorem for non-unital operator systems, AMS Joint Mathematics Meeting, Online (April 2022)
7. C*-Simplicity for Groupoids, Oxford Analysis Seminar, Oxford University (March 2022)
8. A duality theorem for non-unital operator systems, 8th European Congress of Mathematics, Online (June 2021)
9. A duality theorem for non-unital operator systems, UK Operator Algebras Seminar, Online (February 2021)
10. Why MIP*=RE implies ¬CEP and Blackadar-Kirchberg's MF problem, Groups, Operators, and Banach Algebras Webinar, Online (August 2020)
11. An introduction to Operator Systems, Groundwork for Operator Algebras Lecture Series, Online (July 2020)
12. Hyperrigidity for C*-correspondences, Young Mathematicians in C*-algebras, University of Copenhagen (August 2019)
13. Hyperrigidity for C*-correspondences, Great Plains Operator Theory Symposium, Texas A&M (May 2019)
14. On Synchronous Quantum Games, Analysis Seminar, University of Waterloo (Feb. 2018)
Conferences Attended
1. The Mathematics and Mathematical Sciences Workshop for Young Researchers, Kyoto University (November 2024). Forthcoming
2. Young Mathematicians in C*-algebras, University of Glasgow (August 2024)
3. International Conference on Noncommutative Geometry, Analysis on Groups, and Mathematical Physics, Ghent University, Online (February 2024)
4. GoTh Workshop: Groups of Thompson and their relatives, Otto-von-Guericke-Universität Magdeburg (September 2023)
5. Young Mathematicians in C*-algebras, KU Leuven (August 2023)
6. Noncommutative Geometry along the North Sea, Hausdorff Center for Mathematics (May 2023)
7. Glasgow Late August Symbolic Dynamics, Groups, and Operators Workshop, University of Glasgow (August 2022).
8. 8th European Congress of Mathematics, online (June 2021)
9. Young Mathematicians in C*-algebras, University of Copenhagen (August 2019)
10. Great Plains Operator Theory Symposium, Texas A&M (May 2019)
11. Southern Ontario Operator Algebras Seminar, The Fields Institute (February 2019)
12. Young Mathematicians in C*-algebras, KU Leuven (August 2018)
13. Great Plains Operator Theory Symposium, Miami University (May 2018)
14. Canadian Operator Symposium, Lakehead University (May 2017)
15. Canadian Operator Symposium, Université de Montréal (June 2016)
Conferences Organized
1. Young Mathematicians in C*-algebras, KU Leuven (August 2023). Co-organizer
2. Glasgow Late August Symbolic Dynamics, Groups, and Operators Workshop, University of Glasgow (August 2022). Co-organizer
1. NSERC Alexander Graham Bell Doctoral Student Award (2017)
2. NSERC Alexander Graham Bell Masters Graduate Student Award (2016) | {"url":"https://www.sejinsamkim.com/cv","timestamp":"2024-11-13T05:15:42Z","content_type":"text/html","content_length":"104310","record_id":"<urn:uuid:e5d78244-c926-4dd1-be72-6d9d2e15afdc>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00601.warc.gz"} |
NCERT Solutions For Class 12 Maths Chapter-12 Linear Programming
Hello Bacho! Kaise Ho Mere Dost. Welcome, students, to the information blog on NCERT Solutions For Class 12 Maths Chapter-12 Linear Programming (PDF Download). In this detailed blog, we are providing
complete solutions for NCERT Class 12 Maths Chapter-12 Linear Programming along with access to download the PDF. Suppose you are preparing for Class 12th board exams or any competitive exams based on
Class 12th subjects such as CUET, JEE Mains, JEE Advanced, NEET, or State Level College Entrance Exams. In that case, these solutions will elevate your preparation to the next level. This Class 12th
Maths Chapter-12 Linear Programming is one of the most important topics in Maths.
Call Now For CUET Classes: 9310087900
Download PDF: NCERT Solutions For Class 12 Maths Chapter-12 Linear Programming.
Click here to Download this pdf: NCERT Solutions For Class 12 Maths Chapter-12 Linear Programming.
Call Now For CUET Classes: 9310087900
NCERT Class 12 Maths Chapter-12 Linear Programming (NCERT Textbook Download)
Before solving NCERT Class 12th Maths Chapter-12 Linear Programming questions, make sure you have gone through the complete chapter at least twice, reading and ensuring that your basic concepts
regarding this chapter are complete. Only then can you proceed with the NCERT Maths Chapter-12 Linear Programming.
Click here to Download this pdf: NCERT Textbook Class12 Maths Chapter-12 Linear Programming.
Hey Guys!! Here we are Providing Important Topics of NCERT Class 12 Maths Chapter-12 Linear Programming. Suppose you are preparing for Class 12th board exams or any competitive exams based on Class
12th subjects such as CUET, JEE Mains, JEE Advanced, NEET, or State Level College Entrance Exams. In that case, these Topics will elevate your preparation to the next level.
S.No. Important Topics
12.1 Introduction
12.2 Linear Programming Problem and its Mathematical Formulation
12.2.1 Mathematical formulation of the problem
12.2.2 Graphical method of solving linear programming problems
If you are preparing for CUET Exams or any other Competitive Exams, So Here we are giving you Information about How to prepare CUET and many more things. Must Check This Blog Clear your Doubts and
Confusion And Score Well. All the Best!
Call Now For CUET Classes: 9310087900
FAQ’s: NCERT Solutions For Class 12 Maths Chapter-12 Linear Programming.
1. Can I use these solutions for exam preparation, including board exams and competitive exams like CUET, JEE, and NEET?
Absolutely! These solutions are designed to aid in exam preparation for board exams and competitive entrance exams like CUET, JEE Mains, JEE Advanced, NEET, and state-level college entrance exams.
2. Is the PDF download available for free, or is there a cost associated with it?
The PDF download of NCERT Solutions is generally available for free on our official CUET Academy Online Website.
3. How often are the NCERT Solutions updated, and how can I ensure I have the latest version?
NCERT Solutions are periodically updated. To ensure you have the latest version, always download them from our website.
4. Can I use the solutions for self-assessment and practice, apart from regular classroom studies?
Yes, the solutions are excellent for self-assessment and independent practice, helping reinforce your understanding of the chapter.
Call Now For CUET Classes: 9310087900
NCERT Solutions For Class 12 Maths (Chapterwise PDF) | {"url":"https://cuetacademy.online/ncert-solutions-for-class-12-maths-chapter-12-linear-programming/","timestamp":"2024-11-11T03:17:09Z","content_type":"text/html","content_length":"131534","record_id":"<urn:uuid:6a01a2c0-0089-4d41-9ee0-3ae14040eab8>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00508.warc.gz"} |
7. [Maps, Curves & Parameterizations] | Multivariable Calculus | Educator.com
f(0) − f(3) = − 4 − ( − [13/4] ) = − 4 + [13/4] = − [3/4].
g(0) − g(1) = ( − 1,1) − (0,2) = ( − 1, − 1). Note that this is a difference of two vectors.
Note that this is scalar product, so g(0) ×g(0) = ( − 1,1) ×( − 1,1) = ( − 1)( − 1) + (1)(1) = 1 + 1 = 2.
h(0) − h(1) = (0,0,0) − ( 1,[1/2],1 ) = ( − 1, − [1/2], − 1 ).
Note that this is just vector multiplication with a scalar, so that 4h(1) = 4( 1,[1/2],1 ) = ( 4(1),4( [1/2] ),4(1) ). So 4h(1) = ( 4,2,4 ).
Note that no matter what value of t we input, our point will always be (2,1), thus f maps the real line into a point in space.
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while
watching the lecture.
Section 1: Vectors
Points & Vectors 28:23
Scalar Product & Norm 30:25
More on Vectors & Norms 38:18
Inequalities & Parametric Lines 33:19
Planes 29:59
More on Planes 34:18
Section 2: Differentiation of Vectors
Maps, Curves & Parameterizations 29:48
Differentiation of Vectors 39:40
Section 3: Functions of Several Variables
Functions of Several Variable 29:31
Partial Derivatives 23:31
Higher and Mixed Partial Derivatives 30:48
Section 4: Chain Rule and The Gradient
The Chain Rule 28:03
Tangent Plane 42:25
Further Examples with Gradients & Tangents 47:11
Directional Derivative 41:22
A Unified View of Derivatives for Mappings 39:41
Section 5: Maxima and Minima
Maxima & Minima 36:41
Further Examples with Extrema 32:48
Lagrange Multipliers 32:32
More Lagrange Multiplier Examples 27:42
Lagrange Multipliers, Continued 31:47
Section 6: Line Integrals and Potential Functions
Line Integrals 36:08
More on Line Integrals 28:04
Line Integrals, Part 3 29:30
Potential Functions 40:19
Potential Functions, Continued 31:45
Potential Functions, Conclusion & Summary 28:22
Section 7: Double Integrals
Double Integrals 29:46
Polar Coordinates 36:17
Green's Theorem 38:01
Divergence & Curl of a Vector Field 37:16
Divergence & Curl, Continued 33:07
Final Comments on Divergence & Curl 16:49
Section 8: Triple Integrals
Triple Integrals 27:24
Cylindrical & Spherical Coordinates 35:33
Section 9: Surface Integrals and Stokes' Theorem
Parameterizing Surfaces & Cross Product 41:29
Tangent Plane & Normal Vector to a Surface 37:06
Surface Area 32:48
Surface Integrals 46:52
Divergence & Curl in 3-Space 23:40
Divergence Theorem in 3-Space 34:12
Stokes' Theorem, Part 1 22:01
Stokes' Theorem, Part 2 20:32
Hello and welcome back to educator.com and welcome back to Multivariable Calculus.0000
We just finished our discussion of vectors and lines and planes and things like that.0005
Now we are going to actually start getting into the Multivariable Calculus.0009
Today's lesson, I am going to discuss a very important topic.0013
There are not going to be too many examples, but it is going to be mostly discussion.0015
It is going to be a global discussion about what we are going to be talking about when we talk about functions from one space to another.0020
So we are going to introduce this notion of a map.0029
Now you remember that we talked about parameterizing a line and the parameterization of a plane, now we are going to start parameterizing every single function that we deal with.0032
As it turns out, parameterization is a very powerful technique, much more powerful than the representations that you have been used to all of these years.0042
For example, if y = x2, y = x3, where you have the 1 independent variable and the dependent variable.0050
Now, we are going to take a little bit of a twist and have a different way of looking at that, a much broader, a much more general way of looking at that.0058
So, with that, let us go ahead and recall a few things and we will go ahead and jump right on in.0065
Let us recall a few things, in terms of notation.0072
What we are talking about is 1 space, basically just a real number line, that is what the R stands for.0077
Now it is my habit to actually notate R slightly differently.0090
I usually... in books you are going to see the real numbers with sort of these double lines, for the real numbers, the rational numbers, there is a specific notation, a letter representing these
So if I do it sometimes this way, or I do it sometimes that way, they mean the same thing.0108
In other words, what they are saying is you have 2 number lines.0120
R3, that is 3 space, you need 3 numbers to represent a point in 3 space, and of course RN that is the general.0125
That is just n space, so that is the most general dimensional space we can talk about.0137
Let us start by looking at the simple function y = x2, but let us look at it in a slightly different way.0144
Let us look at y = x2, or f(x) = x2, a slightly more general than this, but again this is how you are used to seeing it all of these years, this is how we have dealt with it.0150
We take a number, we do something with that number, and then we spit out a number that ends up being the f(x).0170
It ends up being the y value, and then we end up graphing this.0179
We are going to look at this in a slightly different way, I am going to draw two separate number lines.0181
I am going to draw one number line over here, then I am going to draw another number line over there.0187
Let me label this as 0, then we will do 1, then we will do 2 and we will do 3 and of course it goes on in both directions.0196
Over here let me go ahead and put 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.0201
Now, what f does, what this y = x2 does, we are going to take a value and some number, a real number, and we are going to do something to it.0215
We are going to square it and spit out another number.0226
Notice, we started in the real number line, the 1 space, and we have done something to it and then the number that we get back is still another number.0230
We end up back in the same space, or if you want to think of this as two separate spaces, it is probably the best way to do it.0240
In other words, this function takes a number... for example if I put 0 in there... it takes 0, it does something to it, and it ends up at 0 on another number line.0247
Consider this just another space but it happens to be the same space as the number from which you chose x.0260
If I take 1, and I put it in there, I square it and I end up getting, spitting out, 1.0265
If I take 2, I end up doing something to it, I spit out 4, and 3, I spit out 9, and so on in both directions negative and positive.0273
What we are doing is we are associating to each number in one space, in the real number line, we are associating another number in a copy of that space.0285
Again, the real number line. What we say is that f maps x to f(x).0298
In other words, we are mapping 0 to 0, 1 to 1, 2 to 4, 3 to 9, whatever the function happens to be.0307
This idea of a map, 2 separate spaces, in this case they happen to be the same, but they do not have to be.0314
That is what we are going to generalize in a minute, you can actually take a number in 1 space, do something to it, do a bunch of things to it, and actually end up in a completely different space all
You could end up in R2, R3, that is what we are going to do in a minute.0331
But we want to be able to see these functions that we have been dealing with in a different way.0336
We do not want to think of them as being y = x2, we want to think of it as taking a number x, doing something to it, and getting a number in a different space.0342
Again, in this particular example, the two spaces, the departure space, the arrival space, happen to be the same space.0352
This notation is actually going to be very important, it is going to be a notation we are going to be using often.0366
It is a very simple notation, but it is actually quite powerful when you speak generally about functions.0374
We say that f is a mapping from R to R.0382
This is the general notion, what we are telling you is that f is a function that takes a number in the real number space, does something to it, and then spits out a number that also happens to be in
the real number line.0394
A number in other words, not a vector. We actually give the definition of that, it means take x, and square it.0405
This is a pretty standard notation for a mapping, that is what it is.0411
You want to think of a function as a mapping, that is the most important thing.0417
Another way of looking at this, which I think is a little bit better, a little bit more pictorial way of looking at this is let us just say we have this space, and let us say we have this space.0422
They can be the same, they can not be the same.0433
We call the numbers that you choose, in other words, the independent variables, the x's that you choose randomly the departure space.0438
Once you do something to that particular number, whatever it is you are doing to it, however it is you are transforming it, we call it the arrival space.0446
So, this is... whoops, we are not going to have these random stray lines, we definitely do not want that especially when we are dealing with pictures here... let us close this off here... you know
what, let me go ahead and do this on the next page, because I want you guys to see this really clearly.0457
It is the best way to consider this, so let me go ahead and erase that.0483
Let me move on to... there we go... so let me draw a little something like this, and a little something like this.0488
This is the departure space, it is the number that we choose.0498
This is the arrival space, so let us just pick a couple of random points.0504
So this is x1, x2, x3, so this is x1, x2, x3, and what we do, is when we take x1, do something to it, operate on the function, or use the function to operate on this... let us say this maps to here
and this maps to here... we end up in certain places in the arrival space.0510
That place is in the arrival space, so this is a pictorial version of what it is that we are doing when we think of a map.0532
That is what is important, we are taking a number, we are doing something to it and we are ending up with another number, or vector, or could be anything.0537
It turns out in this particular case that the y = x2, this space happened to be the real number line, and this space also happened to be the real number line.0546
These are the functions we have been dealing with all of these years, ever since we were kids, and up through calculus.0556
But now, in multivariable Calculus, we want to become a little bit more sophisticated.0560
Now, what Descartes did was as far as what you know and what you have been dealing with as far as graphs are concerned, for example if you saw the y = x2 function, you are accustomed to seeing
something like this.0569
What Descartes did was he took these two real number lines, this departure space and this arrival space and he set them perpendicular to each other, that is all he did.0584
So this is one space, and this is another space.0594
If you juxtapose them, because you can do that on a 2 dimensional sheet of paper, then if you map, you know... like... 1,1,2,4,3,9,4,16, the particular function, you end up getting this thing called
a graph.0599
You get a curve in 2 space, that is what he did, that is where the notion of a graph came from.0612
What is really happening is you are mapping from one space to another space.0618
You are taking numbers from one space, doing something to them, and ending up in another space.0624
In this particular case, they both happen to be the same space, the real number line, R1.0628
Now, let us try looking at a function where now we are going to pick where our departure space is going to be the real number line.0636
Our arrival space is going to be 2-space, it is going to be 2-dimensional, let us see what that looks like.0644
Let us try looking at a function, and I will put function from R to R2, defined by the following: f(t) = t and t2.0655
This is vector notation, this is the x component, this is the y component.0690
If we are speaking in terms of x's and y's, so basically we have... we are taking a number, any number from the real number line and we are saying that what we are going to produce is a vector, a
2-vector, and the first component of that vector is just going to be the number t itself.0697
This is a map, so I have taken a number, a real number from the number line and created a vector in R2.0720
Well a vector in R2 is a directed line segment, but it is also just a point, and the coordinates of that point are t and t2.0727
So what we are doing here is we are taking the real number line and we are taking numbers from the real number line and we are mapping them onto 2 space.0736
So we might take the number 1, map it to 1 and 1, take the number 2, map it to 2 and 4, take the number 3 and map it to 3 and 9, and that is what we are going to do.0752
In fact let us do that for a couple of these so that we do a pictorial representation.0765
We are just taking numbers from the real number line, so here the departure space is the real numbers, and the arrival space is going to be R2.0773
So now, if I actually do this map, let me rewrite the map again, f(t) = t and t2.0787
Now let me go ahead and map a couple of these things.0800
If we do t=1, we will get 1 and 1.0808
If I map a 2, so it is going to be 2 and 4, so I will go to 2, go to 4.0813
If I take t=3, the point is going to be 3,9.0821
Notice what you get. You still end up with this parabola y = x2, but now it is not written as y = x2.0826
It is written as a map with a given parameter, that parameter t, you take that 1 parameter t and then the x... you actually map it to 2-space, well 2-space you need 2 numbers to represent a point, a
vector in that space.0835
The first point is t, the second point is t2.0851
This is the power of parameterization, so this is a parameterization of the function y = x2, and this is a much more powerful way of looking at it.0856
What you have done here is this -- this is a vector, so (0,0) that is that vector, (1,1) that is that vector, (2,4) that is that vector, (3,9) that is that vector.0864
When you connect all of the endpoints of the vectors, you end up getting your curve in 2 space.0877
Let us do an example and then we will give a formal definition of what we mean by a parameterized curve in n space.0890
Let us say that f(t) = ... oh, let me use my notation so that you get used to seeing this... so f is a map from the real number line to 2-space.0903
Defined by f(t) = (cos(t),sin(t)), where t... and this particular case I am going to specify the interval t, I do not have to, I can just let it go, but what the heck, let us just go ahead and
specify... so t goes from 0 all the way to 2pi.0920
If I take t = 0, and if I do... so that becomes, so t = 0, that is the point that I am pulling from the real number line, I am going to do something to it.0955
I am going to do this to it, and I am going to end up at (cos(0),sin(0)), which is (1,0).0968
So, I end up over here. That is my vector, that is my point in 2 space.0971
Let us just take t = pi/4, a 45 degree angle, yeah pi/4.0979
Again, a radian measure is a real number, so pi/4, if I do (cos(t),sin(t)), I will get (1/sqrt(2), 1/sqrt(2)).0986
If I take t = pi/2, now I am going to get (cos(pi/2),sin(pi/2)), that will put me at (0,1), so that will put me over there.1001
If I keep going like this, go all the way up to 2pi, these vectors that I get, that vector, that vector, that vector, that vector, that vector, if I connect all of the dots of these vectors, I am
going to get the circle on the plane.1009
So this thing right here, this function is the parameterization of the unit circle in R2.1026
That is what we are doing. We are taking the parameter and deciding... we are creating curves in a space of a given dimension.1038
In this case, we ended up creating a curve in 2-space.1049
Now let us go ahead and give a formal definition of what it is we mean by a parameterized curve.1055
This one I am going to go ahead and move on to the next page.1059
It is going to be a slightly long definition, but hopefully we have notated it properly.1064
So, a parameterized curve... I am going to write it a little bit better here... a parameterized curve in RN, or n space, is a map... let us call it c... from R to RN, which to each point in R
associates a vector, which is also a point, in RN.1070
When these vectors, or points, are connected, we get a curve in RN.1143
So as we saw in the previous example, we got a curve in 2-space.1167
So we are not thinking that it is a function anymore, the way we used to as far as y = x2.1173
You can do that, I mean you can certainly turn this parameterization into a y = x2, but now we want to sort of break away from that.1178
We want something a little more free, a little more general, a little more sophisticated, and you will see just how sophisticated in a bit when we start differentiating.1183
We want to think of it as, now let us say I do it in 3 space.1193
Let us say I take some value, some t, something from the real number line -- it is always going to be a number, the map is always going to be like this, the departure space for a curve, in n space,
the departure space is always going to be the real number.1197
It is always going to be like this, for a curve in n space the departure space is always going to be the real number line.1212
The arrival space can be any dimensional space, R1, R2, R3.1218
In R3 you might have some random curve that goes all over the place. There is a way to parameterize that.1223
In other words there is some map, and of course for R2 and R3 we can visualize it, but again this is an algebraic definition.1229
Because it is an algebraic definition, it is very powerful.1236
We can think of a curve in 15 dimensional space, it is perfectly valid and we can work with it mathematically.1241
We may not be able to see it or visualize it, but it does exist, it is real. We can work with it.1247
Curve in RN and, I will say one more thing about that.1255
The coordinates of x -- of c, I am sorry -- the coordinates of c are functions of t.1261
You saw that already, so c(t) = some one function from... actually let me use f instead of x... It is going to be some function of t and then another function of t, and so on, depending on how many
dimensions you are dealing with.1280
In the previous example, we had 2 functions of t, we had cos(t) and we had sin(t).1301
Again, a vector is made of components, those components are actual functions of t. This is the power of parameterization.1308
Let us look at another example here. Let us do a 3 space example.1318
So, example number 2. Let us let x(t) and we are going to be several different letters, x, y, sometimes c, occasionally I am going to be writing it with capital letters where I am not going to put
the vector sign on top of that.1325
The notation is important, however the notation I am not going to... there are going to be different types of notation, but it will be consistent within the notation itself, in other words you will
see the same arrangement, but we are just going to be using different things.1344
So, x(t) is a map from R to R3, so this time we are doing the curve in 3 space, defined by x(t) = (cos(t),sin(t),t).1358
Let us see if we can find out what this curve looks like.1395
Let me go ahead and draw that, and that, and this.1398
This is the general representation of a 3 dimensional space, this is the x-axis, this is the y-axis, this is the z-axis.1403
We already did the cos(t),sin(t), we know that that is a circle, so in the x,y plane that is the circle.1412
In the x,y plane there is no z component, but now we have added a z-component.1420
For every value of t we actually not only move in a ... we do not just move in a circle... every time we go t, we actually go up a little bit, sort of like a ramp, a parking ramp... have you been to
those parking ramps?1425
You are still moving in a circle as far as the x,y, coordinate is concerned, but every time you move, you are also going up.1441
What you will end up getting, the curve you are going to get, is a spiral.1447
That is exactly what it is, so this curve is going to end up being a spiral.1453
It still moves in a circle as far as the x,y plane is concerned, if you are looking down the z-axis it still looks like a circle, but now you have this third component t, the z-axis.1461
So, it is actually moving up as it turns. I hope that makes sense. It is going to look like this.1474
This t is changing, in other words t is also the z axis, there is a third component here.1480
This is a curve, so this is the parameterization of the unit spiral in R3.1485
That is the idea. If you are given a particular curve, you can parameterize it with some map that looks like this.1494
Then, when you have this map you can begin to play with it algebraically.1502
As you will see in the next lesson we will begin to start differentiating it and... well, yeah, other things too.1507
So, now, let us go one more up here, I will just do one more example.1515
Let f be a map from R to R2, in other words a curve in 2-space, defined by f(t) = t and t2.1532
So this is the x-coordinate, right? This is the y-coordinate, it is a curve in the 2 space, this happens to be the parameterization of the parabola in the plane.1550
Now, let us go ahead and recover a function, let me see here, 1, 2... you know what, no, this is... well, that is okay we have already started it so let us go ahead and finish it.1560
Now I have given you this parameterization, and we want to be able to recover some function the way we are used to seeing it, y as a function of x.1596
We are given the parameter t, the t is what is in common, so essentially all you have to do is eliminate the t from them and set y equal to some function of x.1608
Well if x = t, let us go ahead and just put this x wherever we see a t.1624
So you eliminate the t, the parameter, and you end up recovering some function of y = x2.1635
It becomes a lot more difficult... in fact I am not even sure, well... in 3 space and 4 space, but this is the power of parameterization.1641
We have been dealing with simple functions all these years simply because we are accustomed to dealing with 2 space and 3 space.1652
Our natural intuitive surroundings, you know, we have a sense of paper, we have a sense of flat paper.1656
We have a sense of 3-dimensional space, however things can obviously become a lot more complicated even when you are dealing in 3 space.1663
4-space, 5-space, 6-space, it becomes virtually impossible to deal with.1670
This is why we do parameterization, it is a much more sophisticated way of dealing with functions.1674
It also happens to work for lower dimensional cases, in other words R2.1680
Here is how you want to start thinking about functions.1687
What we have done, as far as all of these years, what you have been working with are maps, or general functions, y = some function of x.1698
Those are maps from R to R, in other words, you take some value x, you do something to it, and you spit out another number. Numbers to numbers.1707
Today we have introduced the notion of curves, which are maps from the real number line, you take a number, and you create a vector to RN.1717
Well, we are also going to be going the other direction.1728
We are going to begin talking about functions in which you will take a vector, and we will do something to that vector and what we will spit out is a number.1731
To top it all off, we are eventually the most general is RN to RP, we are going to take a vector in n space, we are going to do something to it, and we are going to end up spitting out a vector in p
We are going to work from one certain dimensional space to another completely dimensional space.1758
This is the power of multivariable calculus, that you can actually do this, that you can actually do the calculus you know.1764
Everything that you learned in single variable calculus still applies and yet now it is going to become a little bit more sophisticated, more powerful.1770
So, next lesson we are going to start with the actual calculus in n space.1781
Until then, thank you for joining us at educator.com and take care, bye-bye.1785
Our free lessons will get you started (Adobe Flash® required). Get immediate access to our entire library. | {"url":"https://www.educator.com/mathematics/multivariable-calculus/hovasapian/maps-curves-+-parameterizations.php","timestamp":"2024-11-05T10:29:41Z","content_type":"application/xhtml+xml","content_length":"539595","record_id":"<urn:uuid:3966fd1c-c17e-4e1d-934e-d7d3e3c1ebf9>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00298.warc.gz"} |
Glossary Primer
For those new to BaseballHQ.com or who need a quick refresher on the site's most-used terms and benchmarks, here's our quick glossary primer. For a deeper dive on these and other terms, check our
Full Glossary.
xBA (Expected Batting Average) attempts to distill the Batting Average by considering the batter’s speed, power, and distribution of grounders, flies, and line drives. xBA should correlate closely to
BA; a variance exceeding 30 points usually portends future change.
bb% (Walk Rate) is a measure of a batter’s plate patience. The best batters will have levels of more than 10%, while the worst will be less than 5%.
Brl% (Barrel rate) A “barrel” is a Statcast metric defined by MLB.com as a well-struck ball where the combination of exit velocity and launch angle generally leads to a minimum .500 batting average
and 1.500 slugging percentage. Barrel rate (Brl% in hitter boxes) is simply the number of barrels divided by the number of batted balls for a given hitter.
ct% (Contact Rate) measures a batter’s proficiency in hitting the ball into the field of play; the more often a batter makes contact with the ball, the higher the likelihood he will hit safely.
League averages are 79%, with supreme contact hitters above 90% and hackers less than 75%.
Eye (Batting Eye) assesses a batter’s strike-zone judgment by tracking the ratio of walks to strikeouts (bb/k). The best hitters often have Eye ratios greater than 1.00 (more walks than Ks) while
those with ratios less than 0.50 are usually plagued by lower BA.
h% (Hit Rate, or Batting Average on Balls in Play, for hitters) is the percentage of balls struck into the field of play that fall for hits. Every hitter establishes his own h% that stabilizes over
time; three-year h% levels strongly predict a player’s h% the following year.
HctX (Hard Contact Index) is a combination of ct% and hard-hit ball percentage, compared to overall league levels for that year. A 100 value represents average league power skills; best levels will
exceed 130.
G/L/F or GB/LD/FB (Ground balls/Line drives/Fly balls, for hitters) is the percentage of each type of balls hit into the field of play. Increased fly ball percentage for an individual hitter may
fortell a rise in power skills; an increase in line drive percentage may indicate a coming batting average increase.
QBaB (Quality of Batted Balls): For batters, greater exit velocity and greater mean launch angle are better. In addition, reduced launch angle variability is correlated with better batted ball
results. The Quality of Batted Ball
score (QBaB) assigns A-F grades for exit velocity, launch angle, and launch angle variability based on percentile groupings.
hr/f (Home Run to Fly Ball rate, for hitters) is the percentage of fly balls that a player hits that end up as home runs. Every hitter establishes his own hr/f that stabilizes over time; three-year
hr/f strongly predict a player’s hr/f the following year.
PX (Power Index) measures a hitter’s extra-base abilities compared to overall league levels for that year. A 100 value represents average league power skills; the biggest power hitters exceed 150.
Spd (Statistically Scouted Speed) is a skills-based gauge that measures a player’s speed independent of stolen bases. The formula, which dampens power influences and emphasizes factors like infield
hits and the player’s body mass, is an index with a midpoint of 100.
SBO (Stolen Base Opportunity Percentage) is a rough approximation of how often a base runner attempts a stolen base, and takes into account how often the manager for that player’s team gives a “green
light” to his runners.
BPV (Base Performance Value, for hitters) is a single value used to track a player’s performance trends and predict future performance. BPV encapsulates a hitter’s overall raw skills—batting eye,
contact rate, power, and speed—with the best hitters earning a 50 or better. BPX is simply BPV scaled to a league average of 100, such that a BPX of 110 is 10% above average and a BPX of 90 is 10%
below league average.
xERA (Expected Earned Run Average) attempts replicate ERA from a skills-dependent perspective, stripping out situation-based factors. xERA should correlate closely to ERA; a variance of more than
1.00 (a run per game) is a strong indicator for future change.
BB% (Walk rate) Measures how many walks a pitcher allows as a percentage of total batters faced.
K% (Strikeout rate) Measures how many strikeouts a pitcher produces as a percentage of total batters faced.
K-BB% (Strikeout minus walk rate) Measures a pitchers’ strikeout rate (K%) minus walk rate (BB%) and is a leading indicator for future performance.
G/L/F or GB/LD/FB (Ground balls/Line drives/Fly balls, for pitchers) is the percentage of each type of balls hit into the field of play. For a pitcher, the ability to keep the ball on the ground (45%
and above) can contribute to his statistics exceeding his raw skill level.
hr/f (Home Runs per Fly Ball rate) is the percent of fly balls surrendered by a pitcher that end up being home runs. For pitchers, research has shown that fly balls result in a home run 10% of the
time, and that a high hr/f rate in one season is not predictive of a high hr/f the next season. Pitchers with a high hr/f often have an artificially high ERA, and this can be expected to correct
itself over time.
hr/9 (Opposition Home Runs per 9 IP) measures how many HR a pitcher allows per game equivalent. The best pitchers will have hr/9 levels of less than 1.0.
H% (Hit Rate, or Batting Average on Balls in Play, for pitchers) is the percentage of balls struck into the field of play that fall for hits. For pitchers, the league average is 30%, and any plus or
minus variance of 3% or more can impact the pitcher’s ERA. As a pitcher’s H% corrects back to 30%, his ERA is likely to also move accordingly.
S% (Strand Rate) is the percentage of allowed runners that a pitcher strands. Those with strand rates over 80% will have artificially low ERAs prone to relapse, while levels below 65% will inflate
the ERA but with a high probability of regression.
BPV (Base Performance Value, for pitchers) is a single value used to track a pitcher’s performance trends and predict future performance. BPV encapsulates a pitcher’s overall raw skills—power,
control, and command—with BPVs of 50 (starters) and 75 (closers) the minimums for long-term success. BPX is simply BPV scaled to a league average of 100, such that a BPX of 110 is 10% above average
and a BPX of 90 is 10% below league average.
SwK% (swinging strike rate) measures the percentage of total pitches against which a batter swings and misses, and serves as a useful cross-check on Dom (k/9). League average SwK% is about 8-8.5%,
with best rates are over 9.5%
FpK% (first-pitch strike rate) measures the percentage of first-pitch strikes a pitcher throws, and serves as a useful cross-check on Ctl (bb/9). Values below 60% indicate control problems, with best
rates at 65% or higher. | {"url":"http://stage.baseballhq.com/articles/about/glossary-primer","timestamp":"2024-11-08T10:30:52Z","content_type":"text/html","content_length":"75248","record_id":"<urn:uuid:10e18df5-02a3-4e8d-9c1d-3dcf420d4559>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00024.warc.gz"} |
Trying to make sense of superdeterminism
Sabine Hossenfelder
It does not help that most physicists today have been falsely taught the measurement problem has been solved, or erroneously think that hidden variables have been ruled out. If anything is
mind-boggling about quantum mechanics, it’s that physicists have almost entirely ignored the most obvious way to solve its problems.
Her "obvious way" is to replace quantum mechanics with a superdeterministic theory of hidden variables. It says that all mysteries are resolved by saying that they were pre-ordained at the beginning
of the big bang.
Peter Shor
asks in reply
I don't see how superdeterminism is compatible with quantum computation.
Suppose we eventually build ion trap quantum computers big enough to factor a large number. Now, suppose I choose a random large number by taking a telescope in Australia and finding random bits
by looking at the light from 2000 different stars. I feed these bits into a quantum computer, and factor the number. There's a reasonable likelihood that this number has some very large factors.
Just where and how was the number factored?
I agree with him that it is impossible to believe in both superdeterminism and quantum computation.
Quantum computation cleverly uses the uncertainties about quantum states to do a computation, such as factoring a large integer. If superdeterminism is true, then there aren't really any
uncertainties in nature. What seems random is just our lack of knowledge.
If I could make a machine with 1000 qubits, and each can be in 2 states at once, then it seems plausible that a qubit calculation could be doing multiple computations at once, and exceeding what a
Turing machine can do. (Yes, I know that this is an over-simplification of which Scott Aaronson, aka Dr. Quantum Supremacy, disapproves.)
But is the uncertainty is all an illusion, then I don't see how it could be used to do better than a Turing machine.
I don't personally believe in either superdeterminism or quantum computation. I could be proved wrong, if someone makes a quantum computer to factor large integers. I don't see how I could be proved
wrong about superdeterminism. It is a fringe idea with only a handful of followers, and it doesn't solve any problems.
Update: Hossenfelder and Shor argue in the comments above, but the discussion gets nowhere. In my opinion, the problem is that superdeterminism is incoherent, and so contrary to the scientific
enterprise that it is hard to see why anyone would see any value in it. Shor raises some objections, but discussing the issue is difficult because superdeterminists can explain away anything.
2 comments:
1. To me, it's not that superdeterminism doesn't solve any problems, it's more that superdeterminism doesn't give us any productive, useful mathematics and physics.
I suppose superdeterminism gives a more-or-less plausible solution for some problems that some people care about but that not many pragmatic physicists much care about.
2. But [if] the uncertainty is all an illusion, then I don't see how it could be used to do better than a Turing machine.
It can't, assuming it had to solve the full factoring problem using exactly the same assumptions on exactly the same data as the quantum computer. But like you said, the uncertainty is an
illusion, which means the Turing machine may be solving a different problem and/or starting with more information and so not have to solve the full factoring problem.
As the most basic example, consider how fast a Turing machine could solve a factoring problem if it already started with part of the solution. Or consider that some numbers might be easier to
factor than others and so the number that gets chosen to factor will always be one of these.
There are indeed a number of ways to address the types of problems that Shor raised, but the point is we won't know which ones are more or less plausible until people actually start analyzing and
discussing superdeterministic theories. | {"url":"http://blog.darkbuzz.com/2020/03/trying-to-make-sense-of-superdeterminism.html","timestamp":"2024-11-12T11:58:23Z","content_type":"text/html","content_length":"111332","record_id":"<urn:uuid:f89845d2-d1a5-4f3b-959f-ce0319133ff8>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00282.warc.gz"} |
Experimenting on Facebook Prophet
If you have ever worked with time series predictions, I am quite sure you are well aware of the strains and pains that come with them. Time series predictions are difficult and always require a very
specialized data scientist to implement it.
Prophet is a procedure for forecasting time series data based on an additive model where non-linear trends are fit with yearly, weekly, and daily seasonality, plus holiday effects. It works best with
time series that have strong seasonal effects and several seasons of historical data. Prophet is robust to missing data and shifts in the trend, and typically handles outliers well. You can read the
paper in here.
So I decided to give a try on a small eCommerce in Vietnam. I have daily data from March to November. I’ve to feed in data like below as a csv format and remember the headers must be ds and y. (case
I decided to use Prophet for 3 predictions:
1. Predicting Average Order Value
2. Predicting number of sold SKUs
3. Predicting number of Sales Orders
Predicting Average Order Value
Here is the source code:
import pandas as pd
from fbprophet import Prophet
from fbprophet.diagnostics import cross_validation
from fbprophet.diagnostics import performance_metrics
dataFile =… | {"url":"https://christophershayan.medium.com/experimenting-on-facebook-prophet-eb44818278da?source=author_recirc-----53c4a2483ec7----3---------------------f18bc26b_a44d_4a1c_a0d4_13a1997093c7-------","timestamp":"2024-11-04T21:17:38Z","content_type":"text/html","content_length":"92158","record_id":"<urn:uuid:81b7958d-4333-40fa-96ef-7ca05614f462>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00371.warc.gz"} |
Linear Stochastic Fractional Programming with Sum-of-Probabilistic-Fractional Objective
Fractional programming deals with the optimization of one or several ratios of functions subject to constraints. Most of these optimization problems are not convex while some of them are still
generalised convex. After about forty years of research, well over one thousand articles have appeared on applications, theory and solution methods for various types of fractional programs. The
present article focuses on the stochastic sum-of-probabilistic-fractional program, which was one of the least researched fractional program until nineteenth century. We propose an interactive
conversion technique with the help of deterministic parameter, which converts the sum-of-probabilistic-fractional objective into stochastic constraint. Then the problem reduces to stochastic
programming with linear objective of sum-of-deterministic parameters. The reduced problem has been solved and illustrated with numerical examples.
[1] Dr.V.Charles [2005], Optimization of Stochastic Fractional Programming Problems, Ph.D thesis, NIT Warangal. [2] SRSFPP01062005, SDM Institute for Management Development,Mysore, KA,INDIA-570 011.
JUNE 2005.
View Linear Stochastic Fractional Programming with Sum-of-Probabilistic-Fractional Objective | {"url":"https://optimization-online.org/2005/06/1142/","timestamp":"2024-11-13T13:04:52Z","content_type":"text/html","content_length":"84733","record_id":"<urn:uuid:bf80d72d-217b-4ad3-af96-1a51d22079ed>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00546.warc.gz"} |
How can we say we are in the day we say it is? Is it just a guess?
Did they keep track of time centuries ago so we know we are in the correct day?
I don't understand how that's possible, so is the date just a best estimate?
29 Aug 2007
What are you talking about?
How do we know its 26/09/2011
Centuries ago surely they would not have had time keeping.
So its just a educated guess?
9 Jan 2007
[FnG]magnolia;20159408 said:
What are you talking about?
He's not sure about today's date.
13 Jun 2011
11 May 2007
It's a guess and agreement since the beginning of modern time. There used not to be 24 hours a day. Or 7 days a week. Or 12 months a year. Go figure.
It sounds stupid but if someone could explain it to me in laymans terms I would be most grateful.
It's a guess and agreement since the beginning of modern time. There used not to be 24 hours a day. Or 7 days a week. Or 12 months a year. Go figure.
Ah ok that makes sense.
Thread can be closed now.
19 Dec 2006
13 Jun 2011
It's just something that we use to get ourselves organised/world domination. And we use the seasons to judge.
Did you know there used to only be 10 months to a year?
Which is why the last four of the year are numeric (Sept for 7, Oct for 8, Nov for 9, Dec for 10)
Wasn't until man realised seasons were lopsided he figured he should throw a few more in.
Hours to a day, Iuno someone clever spent a lot of time working it out with a sun dial.
Days to a week, God, DUUUUH.
5 Dec 2003
It's a guess and agreement since the beginning of modern time. There used not to be 24 hours a day. Or 7 days a week. Or 12 months a year. Go figure.
Perhaps the oddest setup I know of was the one used in ancient Rome for quite some time. There were a fixed number of hours between sunrise and sunset and between sunset and sunrise. So the length of
an hour varied over the year and there were almost always different hour lengths for night and day.
Although they would probably see our system as being very odd.
28 Jun 2005
The date was set somewhere between 1000 and 1500 years ago, and it's just been adhered to ever since. It's utterly arbitrary, but the point is that the application is consistent. | {"url":"https://forums.overclockers.co.uk/threads/question.18322538/","timestamp":"2024-11-09T00:45:36Z","content_type":"text/html","content_length":"117453","record_id":"<urn:uuid:ade20477-9723-459c-afde-bb493083cb70>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00168.warc.gz"} |
Predictive Analysis Time Series | My Site 2
top of page
A model to invest in the S&P 500 and find correlations between the top tech stocks and the S&P 500
Using predictive analysis method called the time series model to buy and sell stocks occasionally (meaning not for day trading) and finding correlations between popular tech stocks and the S&P 500
Steps taken for this time series model
1. Download stock data from the S&P 500.
2. Cleaned and visualized the data.
3. Set up the machine learning target.
4. Train our initial model.
5. Evaluate error and create a way to backtest.
6. Accurately measure that data over a long period of time.
Step 1: Download the S&P 500
The stock information is from yahoo finance, which is why I am importing yfinance.
I queried all of the S&P data ever using history(period="max").
Lastly, I looked at the index of the S&P 500 data to see how old this data is
Step 2: Cleaned and visualize the data
We are visualizing the data by making the index the x and the close the y.
We then deleted the Dividends and Stock Split columns because we are not using them.
Step 3: Set up the machine learning model
The column tomorrow is to predict tomorrow's price and the shift(-1) is because you are predicting tomorrow's price based on the previous day.
The target is set up for if the stock will go up or down. Our target is what we will be predicting using machine learning.
The first function means is tomorrow's price greater than today's price which would return a boolean ( True or False) and then we converted it into an integer so we can use this data.
Then, I removed all data before 1990 to avoid evitable market shifts using sp500.loc["1990-01-01":].copy(). .copy at the end is to avoid a copy warning that comes with pandas at times.
Step 4: Training the initial machine learning model.
The standard package used for machine learning is sklearn needed to make random forest.
Random Forest Classifier trains multiple singular decision trees with random parameters and averaging through results from those decision tree. Doing the process this way makes random forest to
overfit compared to all other models. Random Forest also run quickly, and it can pick up non-linear tendencies in the data.
N_estimators are the number of decision trees you want to train. Usually, the higher the better to a limit. Min_samples_split protects against overfitting which can occur if the random forest is
built too deep. The downside is the higher the min_sample, the less accurate your results will be, so you need a middle ground like above. Random_State=1 means if we ran the same numbers twice, the
results will be somewhat predictable. Somewhat because random trees still have some level of randomness to them.
I split the data into a train and test set. Making this data a time series model now. All of the rows expect for the last 100 will be into the training set (iloc:-100 means the first 100 rows are
removed) and for the test the last 100 rows will be used (iloc[-100:] means the last 100 will be used).
The predictors equation will have just the columns you will use.
model.fit(train[predictors], train["Target"]) function is used to predict the target
Precision score is to predict when the market goes up or down which is good for holding stocks instead of day trading
preds=model.predict([predictors]) is to generating predictions
preds=pd.Series(preds, index=test.index) will give you 0s and 1s. 0s 1 indicates a pattern and 0 doesn't
When you put this in the precision_score function, you get .636363 which is good. Anything over 50 percent is good. This means that our data matches with history .636363 percent of the time.
Combined= pd.concat([test["Target"],preds], axis=1)<- combining our values to plot them. Axis 1 means to treat each input as a column
combined.plot( ) is the function we use to get a plot below after we made the combined function
The orange line "0" is our predictions and the blue line is the Target, that is what happened. The more orange lines, the better.
Step 5: Evaluate error and create a way to backtest
Building a backtesting system is a more robust way to test our algorithm. Before we were just test the last 100 days but now we are testing multiple years for more experience for the model.
Backtesting will wrap up everything we did before into one function.
Backtesting requires a certain amount of data, for the S&P500 data, a start value and there are 252 trading days in a year and 2520 days is 10 years.
all_predictions=[ ] a list of dataframes for a single year
The training set is all the years prior to the current year
The test set is the current year
The predict(train, test, predictors, model) <- used to generate our predictions
Return pd.concat(all_predictions)<- put all the predictions together into a single dataframe
A 1 means the stock went up and a 0 means the stock went down.
When you get the precision score comparing the Target to the Predictions, you get 52% accuracy.
predictions["Target"].value_counts() / predictions.shape[0]<- to look at the percentage of days the market went up and down. This means if you bought a stock market went up and sold it at end of the
day, everyday, you would make money
Step 6: Accurately measure the data over a long period of time
This section is to improve the accuracy of our data. Horizons data will be used in the rolling.means. The 2 is for 2 trading days 5 is for 5 trading days, 60 is for 60 trading days, etc. This data
will be used in a for loop below called "for horizon in horizons." Then we find the ratio between today's closing price and the closing price in those periods to see if the market has grown a lot
which can indicate a fall soon or vice versa.
f"Close_Ratio_{horizon}"<- will will give us columns name "Close_Ratio_2", "Close_Ratio_5", "Close_Ratio_60", etc.
sp500[trend_column]=sp500.shift(1).rolling(horizon).sum()["Target"]<- compared to the last time we used shift, we are using positive 1 instead f negative 1. This will look at the past few days and
see the average sum of the target so pick a day and it will do an average of the two days before it. This will see the # of days the stock went up.
new_predictors +=[ratio_column, trend_column] will add a ratio_column and trend column to the sp500
The higher the n_estimator, the more accurate.
Predict proba will control the probability that the row will be a 0 or 1 a.k.a. down or up.
preds[preds>= .6]=1 and preds[preds<.6]=0 is for a custom threshold. Custom threshold means the 1st one has to be more confident the price will go up if it is over 50%. The threshold is .6 for both
which will reduce trading days. This in turn, will reduce the number of days that the price will go up but will increase the chance the price will go up those days. Because we want to make occasional
trades, but not trade everyday.
Running the backtest again but with the new predictors equation
predictions["Predictions"].value_counts()<- compared to page 5. More days went down than up because we changed the threshold. This means we will be buying stocks on fewer days.
Compared to page 5, we are 6% more accurate.
The correlation between the five most popular growth stocks and the S&P 500
Pandas is a software library for data analysis
Datetime.now()-pd.DateOffset(years=19) is used pull out stocks from the past 19 years
The stocks equation is all of the stocks we will use.
for stock in stocks
data=yf.download(stock, start= start_date, end= end_date)
comparestocks=pd.concat(stock_list, keys=stocks, names=['Ticker','Date'])
This code will give you a for loop to go through each stock one by one from the stocks equation downloaded from yahoo finance. Then you put them all next to each other with comparestocks but you can
name it whatever you want. Print(comparestocks.tail()) and print(comparestocks.head()) is to see the beginning of the first few days of the 19 years and end of it.
We will do a moving average here. A moving average is the average change of a data series over time. Useful for forecasting. We will do moving average 10 and 20. The rolling mean is the mean of a
certain number of previous periods in a time series, which is 10 and 20. We are using reset_index because we filtered data resulting in removing missing values, which leads to data that isn't
continuous. In this instance, we need the index to be continuous.
for stock, group in comparestock.groupby('Ticker') is just another for loop to do the moving average of 10 and then do the moving average of 20 right after.
comparestocks equation holds all of the companies' tickers and dates as previously mentioned. Now, we are going to use this equation to find out the volatility of these companies by making a new
equation called comparestocks[Volatiity]. As you can see, the most volatile stock is NVDA and it because of the AI boom as to why it is having the growth it is having now.
.pct.change() is used to find the percent change of the previous row in a time series dataset. .std() is used to measure the variability of a dataset so you can see the amount of variation or
consistencies in a dataset.
This is to see the S&P500 index against the five most popular tech companies. You can see the similarities. The index is higher because index hold millions of stocks, including these five tech
The same thing as the last example. I just made an area chart for each ticker so you can see that they have the same shape as the S&P 500, meaning they are extremely similar.
How strong are the correlations between the top stocks and the S&P 500?
You will need this correlation coefficient scale to understand how strong the correlation is.
.loc takes variables from a dataset to make another dataset. So, I chose AAPL for Apple and ^GSPC which is the S&P500 to extract from the comparestocks dataset. I merged AAPL and the S&P 500 with
pd.merge and on='Date' is the starting point and the end point. I made the graph a scatter plot with a line so you can see the correlation.
A correlation of 95% is nearly perfect. Yes, there is a near flawless correlation.
.loc takes variables from a dataset to make another dataset. So, I chose AMZN for Amazon and ^GSPC which is the S&P500 to extract from the comparestocks dataset. I merged AMZN and the S&P 500 with
pd.merge and on='Date' is the starting point and the end point. I made the graph a scatter plot with a line so you can see the correlation.
A correlation of 95% is nearly perfect. Yes, there is a near flawless correlation.
.loc takes variables from a dataset to make another dataset. So, I chose NVDA for Nvidia and ^GSPC which is the S&P500 to extract from the comparestocks dataset. I merged NVDA and the S&P 500 with
pd.merge and on='Date' is the starting point and the end point. I made the graph a scatter plot with a line so you can see the correlation.
A correlation of 90% when rounded is nearly perfect. Yes, there is a near flawless correlation.
.loc takes variables from a dataset to make another dataset. So, I chose MSFT for Microsoft and ^GSPC which is the S&P500 to extract from the comparestocks dataset. I merged MSFT and the S&P 500 with
pd.merge and on='Date' is the starting point and the end point. I made the graph a scatter plot with a line so you can see the correlation.
A correlation of 96%, when rounded, is nearly perfect. Yes, there is a near flawless correlation.
.loc takes variables from a dataset to make another dataset. So, I chose Google for GOOGL and ^GSPC which is the S&P500 to extract from the comparestocks dataset. I merged GOOGL and the S&P 500 with
pd.merge and on='Date' is the starting point and the end point. I made the graph a scatter plot with a line so you can see the correlation.
A correlation of 98%, when rounded, is nearly perfect. Yes, there is a near flawless correlation.
Conclusion: S&P 500 is extremely dependent on the success of five largest companies on the planet. The richest technology companies and follow the extremely similar trends so if one goes down or up,
these tech companies historically will follow them. Remember, that the S&P 500 is an index of America's richest publicly traded companies. So, America's economy is being held up by these companies.
bottom of page | {"url":"https://www.chimasportfolio.com/about-5","timestamp":"2024-11-10T16:11:31Z","content_type":"text/html","content_length":"637680","record_id":"<urn:uuid:ede9a131-9d8f-4f35-b014-d6005ec99d75>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00627.warc.gz"} |
Unlocking Insights: The Power of Bivariate Analysis - Adventures in Machine Learning
Are you interested in analyzing the relationship between two variables? Whether in the field of science, business, or social sciences, understanding the connection between two variables is crucial to
making meaningful decisions.
This is where bivariate analysis comes into the picture. Bivariate Analysis: Definition and Purpose
Bivariate analysis is a statistical analysis method that examines the relationship between two variables.
It involves analyzing how one variable is affected by changes in the other variable. For instance, a researcher may use bivariate analysis to determine the relationship between the frequency of
exercise and the risk of heart disease.
Bivariate analysis can also help in identifying patterns and trends in data, identifying outliers, and developing hypotheses. The primary purpose of bivariate analysis is to identify the relationship
between two variables.
It is essential to establish the connection between two variables before developing a hypothesis or making any conclusions. Bivariate analysis can be useful in identifying relationships that can be
used to verify cause-and-effect relationships through further research.
Methods of Bivariate Analysis
To perform bivariate analysis, several statistical tools are used. These tools include scatterplots, correlation coefficients, and simple linear regression.
1) Scatterplots
A scatterplot is a graphical representation of bivariate data. It is used to visualize the relationship between two variables.
A scatterplot displays a set of data points as individual dots in a two-dimensional space, with one variable plotted on the x-axis and the other variable plotted on the y-axis. Scatterplots are
useful for identifying patterns or trends in the data and determining whether there is a correlation between the two variables.
To create a scatterplot, you need to have collected data on two variables that you want to analyze. For example, let us consider the relationship between hours studied and exam score.
Here are the steps you can follow to create a scatterplot.
1. Collect Data
Collect data on the number of hours studied and corresponding exam scores.
Record these values in a table.
2. Determine the Axes
Assign one variable to the x-axis and the other to the y-axis.
In this case, we can assign hours studied to the x-axis and exam score to the y-axis.
3. Plot the Points
Plot each data point on the graph, using the number of hours studied as the x-value and the corresponding exam score as the y-value.
4. Determine the Relationship
Look for any patterns or trends in the data points plotted on the graph. If the plot shows a positive relationship between hours studied and exam score, we can say that the more hours a student
studies, the better their exam score is likely to be.
2) Correlation Coefficients
A correlation coefficient is a statistical measure that represents the strength and direction of the relationship between two variables. It helps us to quantify the connection between two variables
by giving us a measure of the degree to which the variables are related.
The correlation coefficient ranges from -1 to +1, where -1 indicates a perfect negative correlation, +1 indicates a perfect positive correlation, and 0 indicates no correlation. To calculate the
correlation coefficient, you need to have data on two variables.
The formula for calculating the correlation coefficient is:
r = (n?XY ? (?X)(?Y)) / ?[(n?X2 ? (?X)2)(n?Y2 ? (?Y)2)]
where n is the number of data points, X and Y are variables, and ?
represents the sum of the quantity.
3) Correlation Coefficients
Correlation coefficients are statistical measures used to indicate the strength and direction of a linear relationship between two variables. The Pearson correlation coefficient, also known as the
Pearson’s r, is a commonly used correlation coefficient that measures the linear relationship between two continuous variables.
It is used to measure the degree to which two variables are related, with a value ranging from -1 to +1.
For example, if we are comparing the relationship between hours studied and exam scores, a strong positive correlation would indicate that as hours studied increase, exam scores also increase.
The Pearson correlation coefficient can be calculated through software applications like Microsoft Excel or statistical software like R. The resulting coefficient can be visualized in a correlation
matrix, making it easy to compare the relationships between multiple pairs of variables.
A strong positive correlation can have a value close to +1, while a strong negative correlation can have a value close to -1, and no correlation can have a value close to 0. It is essential to
understand the direction and strength of the correlation coefficient before interpreting its meaning.
4) Simple Linear Regression
Simple linear regression is a statistical method used to analyze the relationship between two continuous variables. It aims to determine how one variable is affected by the other variable.
In simple linear regression, one variable is considered an explanatory or predictor variable, while the other is a response variable.
The methodology of simple linear regression involves fitting a straight line to the data points to estimate the relationship between the variables.
This line represents the expected change in the response variable when the explanatory variable changes. The line is fitted using the Ordinary Least Squares (OLS) method, which minimizes the sum of
the squared errors between the predicted and observed values of the response variable.
To fit a simple linear regression model, we first need to have data on two continuous variables. The length of time studied can be the explanatory variable while the exam score can be the response
We can fit the straight line relating the two variables using the OLS method. After fitting the model, we can use the model summary to evaluate the goodness of fit.
The model summary provides information about the regression equation, including the slope and intercept coefficients. These coefficients can be used to make predictions about the score when an
additional hour is studied.
The regression equation is a mathematical expression that represents the straight line fitted to the data points. It can be expressed as:
where Y is the value of the response variable, a is the intercept, b is the slope, and x is the value of the explanatory variable.
We can use the regression equation to predict the score that a student might get when an additional hour is studied. For example, if a student studies six hours and scores 85, the regression equation
can be used to estimate the score if they studied seven hours.
The predicted score would be:
Y = a + b(7)
Y = 62.5 + 7.5(7)
Y = 62.5 + 52.5
Y = 115
According to the regression equation, if the student studies an additional hour, the score would increase by 7.5 points. This information can be used to anticipate the expected performance of
students and help them to study most effectively.
In conclusion, bivariate analysis plays a vital role in data analysis, allowing researchers to identify relationships between variables, develop hypotheses, and test theories. Correlation
coefficients such as Pearson’s r provide a measure of the strength and direction of the relationship between two variables.
Simple linear regression is a useful method that can help identify the nature and extent of relationships between two variables. The fitted regression equation can be used to predict the response
variable’s value when the explanatory variable’s value changes.
These tools can be valuable in many fields of study, including business, science, and social sciences. In conclusion, bivariate analysis is a useful statistical method that allows researchers to
identify the relationship between two variables.
The methods of bivariate analysis, including scatterplots, correlation coefficients, and simple linear regression, help to understand how one variable is affected by the other. The Pearson
correlation coefficient provides a measure of the strength and direction of the relationship between two variables, while simple linear regression aids in predicting the response variable based on
the explanatory variable.
These techniques are valuable in various fields such as science, business, and social sciences. It is essential to understand and use bivariate analysis in decision-making processes.
By analyzing the relationship between variables, one can gain valuable insights that can lead to more accurate predictions or hypotheses that can be tested through further research. | {"url":"https://www.adventuresinmachinelearning.com/unlocking-insights-the-power-of-bivariate-analysis/","timestamp":"2024-11-06T02:23:56Z","content_type":"text/html","content_length":"74666","record_id":"<urn:uuid:0dde49fa-544d-4c4d-b5ba-62db58071081>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00074.warc.gz"} |
e Math Lesson
Online Math Lesson - Flat Shapes
Practice identifying flat shapes in this interactive math lesson for kindergarten. Students will practice basic geometry concepts as they become familiar with two-dimensional shapes, including
hexagons, triangles, circles, squares, rectangles, and more. Students will also become familiar with properties of shapes, including how many sides and corners specific shapes have.
Students will be presented with a variety of question types in this kindergarten math game. In some of the questions, students will drag and drop shapes into a T-chart, sorting them according to
their kind. In other questions, students will be given multiple choices and be asked to choose the correct answer. In still other questions, students will be expected to fill in the blank with the
correct answer. Students may be asked questions like "How many sides does a hexagon have?" and "Drag the shapes into the chart: rectangles and triangles," and "Identify the shape."
What Makes I Know It Math Practice Different?
I Know It is an interactive math practice website geared toward elementary-aged children to help them become confident and proficient in basic math skills, including flat shapes. Teachers enjoy using
our math program because of our easy-to-navigate website format, helpful administrative tools, and quality educational content. Students enjoy I Know It because of the bright, colorful graphics,
whimsical character animations, and motivating incentives (like earning awards) when they reach milestones in their math practice.
When students practice identifying basic shapes in this I Know It math lesson, several lesson features will help them make the most out of their math practice session. A progress-tracker in the
upper-right corner of the practice screen shows students how many questions they've answered so far, and a score-tracker beneath that shows them how many questions they've answered correctly. A hint
button shows students a pictorial clue that will help them answer the question. If students answer the question incorrectly, a detailed explanation page shows them where they went wrong so they can
learn from their mistakes.
Teachers of ESL/ELL students, as well as students who prefer auditory learning, appreciate our read-aloud option, indicated by the speaker icon in the upper-left corner of the practice screen. When
clicked, the question is read aloud to students in a clear voice.
Why not give this kindergarten math lesson a try today? We hope your students have fun learning to identify flat shapes! Remember to browse our complete collection of math topics for elementary-aged
Free Trial and Membership Options
Sign up for a free sixty-day trial of iKnowIt.com! Your students will be able to play this kindergarten math game at no cost. Please know that while your students will be able to try this interactive
math practice activity for free, they will be limited to a total of twenty-five math problems per day across all lessons on I Know It. For full access to the website, including all lessons and
administrative features, you will need to register as a paying member of the site.
We think you'll love all the benefits that come with your I Know It membership! You can create a class roster and add your students to it, assign individual student user names and passwords, assign
different lessons to individual students, track student progress by downloading, emailing, and printing student progress reports, change lesson settings, and so much more!
Your students will log in to I Know It with the unique user name and password you chose for them. Their homepage has a kid-friendly layout with a 'My Assignments' section where they will find all the
math practice lessons you've given them to complete. Your students can also explore other lessons at their grade level, or even lessons at different grade levels if you give them the option in your
administrator lesson settings. Grade levels in the student interface are labeled as 'Level A,' 'Level B,' 'Level C, etc. instead of 'Grade 1,' 'Grade 2,' and 'Grade 3.' This makes it easier for you
to assign lessons based on skill level, rather than grade level.
This interactive math lesson is categorized as a 'Level K' lesson. It may be ideal for your kindergarten or first grade students.
Common Core Standard
K.G.A.2, MA.K.GR.1.1, K.6A
Identify And Describe Shapes (Squares, Circles, Triangles, Rectangles, Hexagons, Cubes, Cones, Cylinders, And Spheres).
Correctly name shapes regardless of their orientations or overall size.
You might also be interested in...
Solid Shapes (Level K)
In this kindergarten-level math lesson, students will practice identifying solid shapes through a variety of question formats, including multiple choice and sorting objects into a T-chart based on
their shape.
Classifying Objects by Shape (Level K)
Students will practice classifying objects by shape in this kindergarten-level math lesson. Students will decide which shape belongs in the group, drag shapes into a T-chart to sort them according to
their kind, and choose the correct shape out of several shape choices. | {"url":"https://www.iknowit.com/lessons/k-geometry-flat-shapes-2d.html","timestamp":"2024-11-05T00:50:20Z","content_type":"text/html","content_length":"426187","record_id":"<urn:uuid:371b4623-6441-4f7d-8964-b9d48f215976>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00635.warc.gz"} |
A few words about this book.
How to navigate, notation, and a recap of some math that we think you already know.
The concept of a vector is introduced, and we learn how to add and subtract vectors, and more.
A powerful tool that takes two vectors and produces a scalar.
In three-dimensional spaces you can produce a vector from two other vectors using this tool.
A way to solve systems of linear equations.
Enter the matrix.
A fundamental property of square matrices.
Discover the behaviour of matrices.
Learn to harness the power of linearity...
This chapter has a value in itself. | {"url":"http://immersivemath.com/ila/index.html","timestamp":"2024-11-02T08:34:00Z","content_type":"text/html","content_length":"19648","record_id":"<urn:uuid:cc59b768-424e-4b83-be29-e48a68e6fb30>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00182.warc.gz"} |
Power law
September, 2018. Page Version ID: 861890790Paper abstract bibtex
In statistics, a power law is a functional relationship between two quantities, where a relative change in one quantity results in a proportional relative change in the other quantity, independent of
the initial size of those quantities: one quantity varies as a power of another. For instance, considering the area of a square in terms of the length of its side, if the length is doubled, the area
is multiplied by a factor of four.
title = {Power law},
copyright = {Creative Commons Attribution-ShareAlike License},
url = {https://en.wikipedia.org/w/index.php?title=Power_law&oldid=861890790},
abstract = {In statistics, a power law is a functional relationship between two quantities, where a relative change in one quantity results in a proportional relative change in the other quantity, independent of the initial size of those quantities: one quantity varies as a power of another. For instance, considering the area of a square in terms of the length of its side, if the length is doubled, the area is multiplied by a factor of four.},
language = {en},
urldate = {2018-10-07},
journal = {Wikipedia},
month = sep,
year = {2018},
note = {Page Version ID: 861890790}, | {"url":"https://bibbase.org/network/publication/anonymous-powerlaw-2018","timestamp":"2024-11-11T11:29:12Z","content_type":"text/html","content_length":"11004","record_id":"<urn:uuid:14decc9d-3eea-4fb5-8b7d-7f73c450c417>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00008.warc.gz"} |
What do I do with dry fertilizers?
I've ordered it, (KNO3, KH2PO4, CMS+B trace and GH booster) and it is all just waiting on me. It could be sitting around for a long time. Just the question is work. So, I have appreciated everyones
support. I hope the questions are not too dry. How do I say this? Help mix this into a 20 gallon (mid to high light with 20 ppm CO2) and a 10 gallon I would like to become a brackish water tank (mid
light) with enriched CO2. Other target ranges are/were (Ca 14-30 ppm, Mg 4 ppm, Fe 0.2 ppm; K+, 20 ppm; PO4, 2-4 ppm; NO3, 10 ppm).
Here is what I am using with a 60% water change (until I run out of seachem).
20 Gal.
1.2mL Flourish Nitrogeno3
10mL Equilibrium
10mL Flourish Phosphorus
2.5mL Flourish Comp.
10 Gal.
1.2mL Flourish Nitrogeno3
5 gal. (5 gal. DC tap water)
5.0mL Equilibrium
5.0mL Flourish Phosphorus
1.2mL Flourish comprehensive
20mL Tropic Marine Sea Salt
What do I do with dry fertilizers? I have two 1 litter bottles. If I can pre mix any of them ahead of time into distilled water I could put those two bottles to work. AquariumFertilizer.com labels
are not very helpful,
"For a phosphorous solution like store-bought ones, add to 1 liter distilled water 3/4 tsp MKP (0-51-34). The result will be a solution of 0.3% P2SO4."
Is this right. What happened to the K and where did the SO4 come from? Anyway, the labels on the other bags are not much more help. I think they want me to spend $4.95 on
Greg Watson's Guide to Dosing Strategies.
How much potassium nitrate (13.5-0-46.2) added to 1 liter H2O will give me the
percentage 1-0-2?
Same question for Mono potassium phosphate (0-51-34) added to 1 liter H20 will give me the
percentage of 0-0.3-0.2?
Just a guess, 3/4 tsp?
I could also use some help trying to find
for flourish comprehensive and equilibrium using Barr's GH booster and Plantex CSM+B. Keep in mind the ten gallon tank is slowly becoming more brackish with the addition of marine salt. KELP!
Last edited by a moderator:
Personally I'd scrap following other people's instructions just for the sake of it. First find out where you want your parameters, then work backwards. Create a solution that will dose the ppm's you
want to maintain the target level of nutrients that you think will work best. Compensate for tap water params if you can.
If you go this route, it's usually best to work from the molar mass of each compound; for instance the molar mass of K2SO4 divided by the quantity of K will give you a number that you can multiply by
to figure out how much more K you'll need besides that contributed by other compounds.
Speaking of K, you'll probably want some K2SO4 by the time this is all done; I see you haven't ordered any.
The main thing to keep in mind here is to keep your cluster of K+ together; they'll stay stable. The rest can go other places; Mg in the trace works well, calcium depends on how you're putting it
Barr's GH booster is something I've forgotten how to mix. If you could remind me what it's made up of, I can give you a hand on working with it.
The contents in the bag of KNO3 is stated as a percentage (13.5 - 0 - 46.2). What is the formula used to determine how much is added to 1 liter to get a percentage closer to 1 - 0 - 2 for dosing?
Philosophos;41367 said:
to figure out how much more K you'll need besides that contributed by other compounds.
Could I get an example using KNO3 (13.5-0-46.2) for a 20 gallon tank to reach 5 ppm of NO3 (10 ppm K+)? I'm waiting on the Guide to Dosing Strategies but if I could see an example formula I could
recreate it for the other nutrients like K2SO4. I hope while using a teaspoon.
Philosophos;41367 said:
Speaking of K, you'll probably want some K2SO4 by the time this is all done; I see you haven't ordered any.
Good call. I seams that down the road I might need to boost the K+ a little. For now I was hoping any K+ I need would come from the KNO3, (1.2mL Flourish Nitroge); Flourish comprehensive (2.5 mL) and
Equilibrium (5 mL) I'm using until they run out.
Philosophos;41367 said:
Barr's GH booster is something I've forgotten how to mix. If you could remind me what it's made up of, I can give you a hand on working with it.
3 parts KSO4
(I think they mean K2SO4), 3 parts CaSO4 AND 1 part MgSO4. I think Tom recommends adding enough to raise the dKH by one degree.
I'm not much of a dry doser, so you'll have to figure out the weight conversions depending on who's density conversion you want to use.
Here's how I do 5ppm NO3 by weight in a 20 gal with 65L:
First I add up the total mass of KNO3; everything below is in g/mol:
Potassium: 39.09831 x1 = 39.09831
...Nitrogen: 14.00672 x1 = 14.00672
.....Oxygen: 15.99943 x3 = 47.99829
Then I add up the mass for NO3:
...Nitrogen: 14.00672 x1 = 14.00672
.....Oxygen: 15.99943 x3 = 47.99829
Now I divide KNO3 by NO3, but for the sake of space I'll keep it like this:
From there I multiply by the number of mg/L desired:
followed by the number of liters I want to establish that parameter within. In this case it's the water column, and we'll say 65L of the approximately 75L capacity of your tank is water column:
(101.10332/62.00501)*5*65 = 529.93425853814070830728033105712
Naturally I round the number down at the very final stages. For dry dosing, 530mg is probably closer than you'll be able to manage with most measuring spoons.
This gives us a formula something like:
---------------- *(Target Concentration)*(Capacity) = (Required Amount)
For K+, it gets a bit trickier; you have to account for the K+ found in other compounds, subtract that from the target, then make up the rest with K2SO4. Let me know if you need a hand figuring this
part of things out.
Oh, I got it. Oops, lost it again.
Philosophos;41410 said:
530mg is probably closer than you'll be able to manage with most measuring spoons.
I think I'll split the difference and say 1/8 tsp = 5 ppm.
This is why I'm working on my own line of liquid ferts that will satisfy EI dosing; the math is more than most people want to do.
Oh, that previous post should've said 65L up top, not 95. I'll edit it.
Thanks for your help. The brain is a muscle and mine has atrophied to the point basic math causes soreness. It's not because I don't want a good math workout or due to a lack of trying. I'll let you
know what I come up with after some more reading. I did find this by G.W,
'If we had a potassium nitrate solution made up of 1 Tablespoon of potassium nitrate in 500 mL of water, Chuck Gadd’s dosing calculator would tell us that each mL of this solution would add
approximately 0.11 ppm of nitrate to our tank. So to raise the nitrate levels by 2 ppm, we divide the 2 ppm by the 0.11 level of our solution to know that we need to add approximately 18 mL of
this solution to the tank to raise the nitrate levels to our target nitrate levels."
Thats math I can understand.
Anyone else with some experience using CSM+B and/or Barr's GH Booster please chime in.
Tug;41362 said:
I could also use some help trying to find equivalents for flourish comprehensive and equilibrium using Barr's GH Booster and Plantex CSM+B.
20 Gal.
10mL Equilibrium
2.5mL Flourish Comp.
10 Gal. Light Brackish Water
5.0mL Equilibrium
1.2mL Flourish comprehensive
20mL Tropic Marine Sea Salt
Last edited by a moderator:
For most initially at least, they dose dry.
It's simpler and easier than any chem.
So for a 20 Gal:
1/4 tsp of KNO3-2-3x a week
1/16th of the KH2PO4 2-3x a week(divide a 1/4tsp into 4 equal parts, you can get pretty good at this)
GH booster: 1/4 tsp after water change
Trace mix, add 2 table spoons to 1 liter, add about 10-20mls of Excel also.
Dose 5mls of this 2-3 x a week
That's it.
Learn chem later when you have the urge.
Then learn how to calibrate and test correctly.
Hopefully by then, you will have mastered most horticulture and prunign and can produce a nice looking aquarium.
Now you can use this as a reference.
Tom Barr
Tom, please don't feel offended, but isn't the Plantex CSM+B is 2 tablespoons per
500 ml for 5 ml dosing to a 20 g tank? You said that here:
I use it as a reference for calculating my (non-Plantex) traces mix.
nipat;41431 said:
Tom, please don't feel offended, but isn't the Plantex CSM+B is 2 tablespoons per
500 ml for 5 ml dosing to a 20 g tank? You said that here:
I use it as a reference for calculating my (non-Plantex) traces mix.
I do not go to APC, so it's been a long time. (several years now I suppose)
1 table spoon per 500mls is what it should be, or 2 tablespoons per liter.
This is pretty much what Paul suggested years ago,m nothing to do with me.
Just repeating what he said years ago.
Practical PMDD Information
1/2 liter = 1 table spoon of CMS
1 liter = 2 tablespoons and so on.
Tom Barr
Without the help from everyone who has answered my questions I would never have gotten this far. This forum is by far the most referenced reading on this hobby out there. Making science easy for
simpletons like myself.
For dosing, this is what works for me. The stock solution allows me to adjust the amounts of NO3, KH2PO4 and K2SO4 (when it gets here).
GH booster: 1/4 tsp after water change (1/8 tsp for 10 gallon tank)
How much K2SO4 & CaSO4 am I adding with only 1/4 tsp of GH booster to 20 gallons?
Can anyone tell me the levels of trace I will be adding using the amount of CSM+B as shown?
Last edited by a moderator:
You may want to open up the gap between your P compared to your N and K. Your target N dose should be somewhere around 10-25ppm, and right now that'd put your P up around the same levels, which is
roughly 10x what they need to be. K should generally be around N levels or higher, but as always non-limiting is the main point.
CSM+B is something I've had success with dosing to .4-.6ppm Fe; CSM+B is 6.53% Fe. Some people just do 10g/L, dose 1-2ml:1L column and call it a day.
CaSO4 isn't my first choice personally, but it can work well depending on your water hardness. What does your tap water look like? Do you have a local water quality report by chance?
They would sell a lot more of this stuff if they could provide better labels.
As always you make me work for the answer to my questions.
Philosophos;41454 said:
You may want to open up the gap between your P compared to your N and K. Your target N dose should be somewhere around 10-25ppm, and right now that'd put your P up around the same levels, which
is roughly 10x what they need to be. K should generally be around N levels or higher, but as always non-limiting is the main point.
I should of mentioned my target range minimums
Tug;33966 said:
Ca 14ppm, Mg 4ppm, Fe 0.4 - 0.6ppm; K+, 20 ppm; PO4, 1(2 ppm). Because of the fish I often don't need to add the extra N or P. When NO3 reaches 20ppm I change the water (maybe every 10 days) and my
tap water seams to have 2ppm of PO4 anyway. Link to D.C.
Philosophos;41454 said:
CSM+B is something I've had success with dosing to .4-.6ppm Fe; CSM+B is 6.53% Fe. Some people just do 10g/L, dose 1-2ml:1L column and call it a day.
Ok, back to the store for another test kit.
Philosophos;41454 said:
CaSO4 isn't my first choice personally, but it can work well depending on your water hardness. What does your tap water look like? Do you have a local water quality report by chance?
With the seachem products I am using it is easy to figure my K+ levels are high but as I move towards dry fertilizers K+ is more of an option to be added as needed. I am mostly concerned about the
levels of K+ in Barr's GH Booster, but because it also has CaSO4 I was hoping to find out it's levels (ppm) per suggested dose. I left out MgSO4, but am also concerned about that as well - although
CSM+B looks as if it will add what I should need to reach 4ppm.
A little heavy on the substrate in the 20 and both tanks still need more plants, but here are my ten and twenty. Now that I'm starting to get the hang of this lets just call these the before
Last edited by a moderator:
Tug;41460 said:
As always you make me work for the answer to my questions.
Most definitely. When I just give people answers, they come back to me later with the same sort of problems. When they have to work for the answer, they learn to solve it them selves, then teach the
method to others.
I should of mentioned my target ranges are: NO3: 10ppm Ca 14ppm, Mg 4ppm, Fe 0.4 - 0.6ppm; K+, 20 ppm; PO4, 1(2 ppm). Because of the fish I often don't need to add the extra N or P. When NO3
reaches 20ppm I change the water (maybe every 10 days) and my tap water seams to have 2ppm of PO4 anyway. Link to D.C.
I'm not sure why you're targeting 10ppm range, given that some tanks can clear 20ppm in a week, but you may have a low light/density system, so I won't bother to argue about it. Your water analysis
indicates .4-2.9ppm NO3, so I'd shave off .4ppm, that or 1.25ppm: (2.9-.4)/2 off of your target dosing quantity.
Your PO4 levels appear to be taken care of by the city. A max range of something like 1.8-3ppm should work without the need to add K2SO4. Double check with Tom if you like, but it seems good to me.
20ppm is more than enough for your N target, you could go down to 15 but 20 won't hurt.
Ok, back to the store for another test kit.
Why? I don't bother testing iron; I find test kit accuracy is pretty horrible. What I mentioned previously should work.
With the seachem products I am using it is easy to figure my K+ levels are high but as I move towards dry fertilizers K+ is more of an option to be added as needed. I am mostly concerned about
the levels of K+ in Barr's GH Booster, but because it also has CaSO4 I was hoping to find out it's levels (ppm) per suggested dose. I left out MgSO4, but am also concerned about that as well -
although CSM+B looks as if it will add what I should need to reach 4ppm.[/COLOR]
Well the water test sure helps with answering this. You don't need more Mg; your levels are 8.9ppm right out of the tap. Your calcium is sitting at an average of 44ppm, which is more than enough,
though adding say 10ppm through a more bioavailable source will offer more help than harm.
So then, what are your dosing targets given your overall nutrient targets? If you give me these, I can help you through to a more definite answer as to how to dose your tank.
Thank you Philosophos for filling in the gaps. I was glad to have my unaccomplished understanding of the tap water reaffirmed as to the amounts of N and PO4. All the same, it would be nice, if there
is a way of knowing how much K+ (PPM) is added to the tank with Barr's GH Booster (per Tom's instructions) and though CSM+B may not need explaining in greater detail it would be nice to know what
2TBSP of CSM+B added to 1 liter of water will give me. I could work with that.
Tom Barr;41428 said:
So for a 20 Gal:
GH booster: 1/4 tsp after water change
Trace mix, add 2 table spoons to 1 liter, add about 10-20mls of Excel also.
Dose 5mls of this 2-3 x a week
When/if I get some K2SO4, do I even need the GH booster?
GH booster is 2:1:1
K2SO4: CaSO4 : MgSO4
So 1/2 of GH booster is K2SO4 anyway.
I'm with Philosophos on this,. more KNO3 mneeds added to thr stock solution, I'd go 3:1-4:1 by weight KNO3:KH2PO4
I run my own tanks juicy.
NO3 at 15ppm and PO4 at 4-5ppm 3x a week.
The lower light tanks(well under 2w/gal), about 1/2 this.
Tom Barr
BTW, not many like to read the references/look at the support, heresay and gut feel is all they need
Tom Barr
I need to pay closer attention to the NO3 now that my CO2 seems to be at 20ppm, (6.4pH down from 7.4) green drop checker, etc. My Nitrates have dropped 5ppm in two days for the first time since I've
had this tank.
I try to keep NO3 at 15ppm and PO4 at 4-5ppm, its just until I started adding CO2 (15Sep09) my nitrate would reach 20 ppm within 48 hours. The plants didn't use what they had because they were dying.
Go figure.
Peace, love and understanding.
Looks like Tom covered most of your issues there Tug. As for 2 Tbsp CSM+B, I don't keep weight to volume conversions around for most of this stuff. Despite my dislike for calculators, Fertilator says
1684.74ppm Fe for 6tsp (2tbsp), which given 6.53% Fe means 25.8g:
1684.74*(100/6.53) = 25800ppm or 25.8g
Sounds about right for what about 25g of this stuff looks like on the scale.
If you don't have the analysis for CSM+B, here it is along with other common micros:
Fertilizer Comparison Chart, by Giancarlo Podio
Just multiply for the rest of your concentrations.
I'm not sure if the percentage of ingredients in CSM+B is correct (the % of
Mo stays the same while it should not)
Regular CSM CSM+B Mix
Fe 7.0% 6.53%
Mn 2.0% 1.87%
Mg 1.5% 1.40%
Zn 0.4% 0.37%
Cu 0.1% 0.09%
Mo 0.05% 0.05%
B 0.0% 1.18%
Co 0.0% 0.00%
I don't remember where I got this info.
PS. Ah, I see, it's probably limited to 2-digit decimal. | {"url":"https://barrreport.com/threads/what-do-i-do-with-dry-fertilizers.6105/","timestamp":"2024-11-02T21:27:21Z","content_type":"text/html","content_length":"207786","record_id":"<urn:uuid:98475edf-3522-4673-8926-fe1fee588789>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00819.warc.gz"} |
Computational Molecular Biophysics Laboratory PI: Emilio Gallicchio - Maximum Likelihood Inference of the Symmetric Double-Well Potential
Maximum Likelihood Inference of the Symmetric Double-Well Potential
Post date: Nov 11, 2019 12:19:47 AM by Solmaz Azimi
The double-well potential serves as an optimal, one-dimensional model for exploring physical phenomena. This project aimed to estimate the probability distribution of the double-well potential fitted
to a Gaussian mixture model by maximum likelihood inference. Hypothetical datasets of the quadratic function were generated using smart-darting Monte-Carlo simulations under the Metropolis acceptance
criterion. A major challenge in studying free energy landscapes is sampling efficiency, therefore a larger displacement was implemented in the Monte-Carlo simulations that are generally done at
smaller displacements. Although a conventional Monte-Carlo approach can accomplish displacement from one free energy minimum to the next, this is done at the expense of accuracy, such that the
acceptance rate of the simulation deviates from an optimal 50%. The generated datasets demonstrated close resemblance to normalized Boltzmann Distribution functions at two temperatures. TensorFlow
was utilized to obtain the most optimal parameters for observing the generated datasets based on the Gaussian Mixture Model. Our results demonstrated that TensorFlow approximates a set of optimal
parameters for the Gaussian Mixture Model that accurately resemble the Boltzmann Distribution at 300 K, however the method is unable to do so with similar accuracy at 2000 K. | {"url":"https://www.compmolbiophysbc.org/research-blog/maximumlikelihoodinferenceofthesymmetricdouble-wellpotential","timestamp":"2024-11-02T18:31:53Z","content_type":"text/html","content_length":"90354","record_id":"<urn:uuid:9e040829-9ad6-450e-9ac9-715510ead1a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00019.warc.gz"} |
Homophonic Substitution Cipher
Introduction §
The Homophonic Substitution cipher is a substitution cipher in which single plaintext letters can be replaced by any of several different ciphertext letters. They are generally much more difficult to
break than standard substitution ciphers.
The number of characters each letter is replaced by is part of the key, e.g. the letter 'E' might be replaced by any of 5 different symbols, while the letter 'Q' may only be substituted by 1 symbol.
The easiest way to break standard substitution ciphers is to look at the letter frequencies, the letter 'E' is usually the most common letter in english, so the most common ciphertext letter will
probably be 'E' (or perhaps 'T'). If we allow the letter 'E' to be replaced by any of 3 different characters, then we can no longer just take the most common letter, since the letter count of 'E' is
spread over several characters. As we allow more and more possible alternatives for each letter, the resulting cipher can become very secure.
An Example §
Our cipher alphabet is as follows:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
D X S F Z E H C V I T P G A Q L K J R U O W M Y B N
To encipher the message DEFEND THE EAST WALL OF THE CASTLE, we find 'D' in the top row, then replace it with the letter below it, 'F'. The second letter, 'E' provides us with several choices, we
could use any of 'Z', '7', '2' or '1'. We choose one of these at random, say '7'. After continuing with this, we get the ciphertext:
plaintext: DEFEND THE EAST WALL OF THE CASTLE
ciphertext: F7EZ5F UC2 1DR6 M9PP 0E 6CZ SD4UP1
The number of ciphertext letters assigned to each plaintext letter was chosen to flatten the frequency distribution as much as possible. Since 'E' is normally the most common letter, it is allowed
more possibilities so that the frequency peak from the letter 'E' will not be present in the ciphertext.
Cryptanalysis §
Breaking homophonic substitution ciphers can be very difficult if the number of homophones is high. The usual method is some sort of hill climbing, similar to that used in breaking substitution
ciphers. In addition to finding which letters map to which others, we also need to determine how many letters each plaintext letter can become. This is handled in this attempt by having 2 layers of
nested hill climbing: an outer layer to determine the number of symbols each letter maps to, then an inner layer to determine the exact mapping.
comments powered by Disqus
Further reading
We recommend these books if you're interested in finding out more.
Elementary Cryptanalysis: A Mathematical Approach ASIN/ISBN: 978-0883856475 Buy from Amazon.com
The Code Book: The Science of Secrecy from Ancient Egypt to Quantum Cryptography ASIN/ISBN: 978-1857028799 Buy from Amazon.com | {"url":"http://practicalcryptography.com/ciphers/homophonic-substitution-cipher/","timestamp":"2024-11-10T00:01:54Z","content_type":"text/html","content_length":"12985","record_id":"<urn:uuid:428e443b-cf1c-4634-8dce-d1c6a627651d>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00644.warc.gz"} |
Point A is at (6 ,2 ) and point B is at (3 ,8 ). Point A is rotated (3pi)/2 clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed? | Socratic
Point A is at #(6 ,2 )# and point B is at #(3 ,8 )#. Point A is rotated #(3pi)/2 # clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points
A and B changed?
1 Answer
(-2,6) is the new coordinate of Point A
Now, the points are closer by 1.323
Origin $\left(0 , 0\right)$
Point A$\left(6 , 2\right)$
Point B$\left(3 , 8\right)$
Distance between points A and B is
$= \sqrt{45}$
$= 6.708$
After transformation
Rotation by $\frac{3 \pi}{2}$
When rotated by$\frac{\pi}{2}$
the new coordinates are $\left(2 , - 6\right)$
When further rotated by $\pi$
the coordinates are further transformed into #-2,6)#
$\left(- 2 , 6\right)$ is the transformed coordinate of the point A
After transformation
Point A$\left(- 2 , 6\right)$
Point B$\left(3 , 8\right)$
Distance after transformation is
$= \sqrt{29}$
$= 5.385$
The distance between the points A and B has changed by
$5.385 - 6.708 = - 1.323$
Now, the points are closer by 1.323
Impact of this question
1430 views around the world | {"url":"https://socratic.org/questions/point-a-is-at-6-2-and-point-b-is-at-3-8-point-a-is-rotated-3pi-2-clockwise-about","timestamp":"2024-11-02T08:17:51Z","content_type":"text/html","content_length":"35891","record_id":"<urn:uuid:756e06d2-fc5a-4cb4-bb32-4ba1614b0cbf>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00749.warc.gz"} |
Paul Horn
Professor; Associate Chair of Graduate St
What I do
I am an Associate Professor of Mathematics at the University of Denver; my role here involves research, teaching classes and advising graduate and undergraduate students. I am also currently the
graduate coordinator in the mathematics department.
combinatorics, graph theory, probability
Professional Biography
My research interests are in combinatorics. Specifically, I am interesting in using ideas from probability, linear algebra, and geometry methods to understand graphs, which are a mathematical
abstraction of networks.
I received my PhD in 2009 from the University of California at San Diego, advised by Fan Chung, and joined the faculty of DU in the fall of 2013 after postdoctoral positions at Emory University and
Harvard University.
I am also closely involved with several research workshops. In particular I co-organize the Rocky Mountains-Great Plains Graduate Research Workshop in Combinatorics (GRWC), which is an annual
research workshop for graduate students in combinatorics. I also help organize the graph theory section Masamu Advanced Studies Institute, an annual research workshop in southern Africa.
• Ph.D., Mathematics, University of California, San Diego, 2009
Professional Affiliations
• American Mathematical Society
• Society for Industrial and Applied Mathematics
My research interests are in combinatorics. Specifically, I focus on the use of probabilistic, spectral, and geometric methods to understand graphs, which are a mathematical abstraction of networks.
My particular work includes using ideas from 'continuous' mathematics, for instance the 'diffusion of heat' on a network, to understand structural properties of networks, for instance whether there
are 'bottlenecks' in the network. I also use structural information about networks to understand random processes on graphs.
One recent project I've been involved in is the development of notions of 'curvature' for graphs. Curvature, in the study of manifolds (which are objects that 'locally' look like Euclidean space, in
the way the surface of the earth 'locally looks like a plane), is a measure of how a space expands 'locally.' The notion of curvature we introduced has allowed me and my collaborators to prove graph
theoretical analogues of a number of results from Riemannian geometry; specifically a graph theoretical version of the 'Li-Yau inequality' along with many consequences.
Areas of Research
probabilistic methods
spectral graph theory
theoretical computer science
analysis on graphs.
Key Projects
• Collaboration on problems in graph theory, and geometric analysis on graphs
• Collaborative Research: Rocky Mountains-Great Plains Graduate Research Workshops in Combinatorics
• Curvature and Geometric Analysis of graphs
Featured Publications
. (2020). Volume doubling, Poincare inequality and Gaussian heat kernel estimate for non-negatively curved graphs. Journal fur die reine und andgenwandte mathematik (Crelle's Journal), 757, 89-130.
. (2018). Rainbow spanning trees in complete graphs colored by one-factorizations. Journal of Graph Theory, 87(3), 333-346.
. (2016). Isomorphic edge disjoint subgraphs of hypergraphs. Random Structures and Algorithms, 48(4), 767-793.
. (2016). Graphs with many strong orientations. SIAM Journal of Discrete Mathematics, 30(2), 1269-1282. | {"url":"https://science.du.edu/about/faculty-directory/paul-horn","timestamp":"2024-11-14T10:45:27Z","content_type":"text/html","content_length":"70843","record_id":"<urn:uuid:c94c3fe7-d827-436e-98f3-bb40a2a46bfc>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00843.warc.gz"} |
How do you use facile in a sentence?
Facile in a Sentence 🔉
1. While the adults found the video game complicated, the teenagers thought it was facile and easily played.
2. No one was surprised when the senior detective solved the facile case in less than twenty-four hours.
3. Since Jack had studied for the exam, he earned a perfect score on the facile test.
What is a ostentatious good?
A snob or ostentatious good is a good where the main attraction is related to its image of being expensive, exclusive and a symbol of social status. These goods will have restricted supply and only
be available to people with high income.
What is a sentence for mountebank?
a flamboyant deceiver; one who attracts customers with tricks or jokes. 1 Politically, Mr Ashdown is a mountebank, not a moralist. 2 She was to marry this mountebank, this hypocritical toad of a Sir
Thomas. 3 The nation was led astray by a mountebank.
Is facile positive?
If someone does something easily, or shows ease, it is described as facile in a good way, but if someone takes the easy way out and shows a lack of thought or care, it is facile in a bad way. While
it is a lovely sounding French word, facile is both a compliment and an insult depending on how it’s used.
What is the similar meaning of ostentatious?
The words pretentious and showy are common synonyms of ostentatious. While all three words mean “given to excessive outward display,” ostentatious stresses vainglorious display or parade.
What does ostentation meaning?
excessive display
Definition of ostentation 1 : excessive display : vain and unnecessary show especially for the purpose of attracting attention, admiration, or envy : pretentiousness She dresses stylishly without
What is mountebank node?
Mountebank is a free and open source service-mocking tool that you can use to mock HTTP services, including REST and SOAP services. You can also use it to mock SMTP or TCP requests. In this guide,
you will build two flexible service-mocking applications using Node. js and Mountebank.
Is foolhardy a positive connotation?
While all these words mean “exposing oneself to danger more than required by good sense,” foolhardy suggests a recklessness that is inconsistent with good sense.
What is the synonym of ostentatious?
Choose the Right Synonym for ostentatious. showy, pretentious, ostentatious mean given to excessive outward display. How is ostentatious used? Ostentatious comes from a Latin word meaning “display,”
and the idea of display is still very apparent in the English word as it is currently used.
What does ostentatious display mean?
Of tawdry display; kitsch. The definition of ostentatious is someone or something designed to get notice or draw attention by being inappropriate, showy, vulgar and in bad taste. An example of
ostentatious is when someone buys huge diamonds and drives very expensive cars in order to show off.
What is the difference between pretentious and ostentatious?
The synonyms pretentious and ostentatious are sometimes interchangeable, but pretentious implies an appearance of importance not justified by the thing’s value or the person’s standing. When could
showy be used to replace ostentatious? The words showy and ostentatious are synonyms, but do differ in nuance.
What is the difference between showy and ostentatious?
The words showy and ostentatious are synonyms, but do differ in nuance. Specifically, showy implies an imposing or striking appearance but usually suggests cheapness or poor taste. What does ‘poke’
refer to in the expression ‘pig in a poke’? | {"url":"https://www.toccochicago.com/2022/08/26/how-do-you-use-facile-in-a-sentence/","timestamp":"2024-11-14T14:14:24Z","content_type":"text/html","content_length":"43912","record_id":"<urn:uuid:3ce5c595-e1d3-4e11-99b9-385952a27860>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00044.warc.gz"} |
If $\rho:G\to \GL_n(\C)$ is a representation of a finite group $G$, then $\rho$ is equivalent to a direct sum of irreducible representations. We say that $\rho$ contains each of these
Given an irreducible representation, we can then ask which permutation representations contain it. There is at least one because the regular representation of $G$ contains all irreducible
representations of $G$.
To determine the smallest permutation representation, we pick one of smallest degree. If the degree is small enough (less than 48 and not equal to 32), we pick one with the smallest T-number.
Otherwise we give just the degree.
Knowl status:
• Review status: reviewed
• Last edited by John Jones on 2018-08-07 16:06:12
History: (expand/hide all) | {"url":"https://www.lmfdb.org/knowledge/show/artin.permutation_container","timestamp":"2024-11-14T17:58:19Z","content_type":"text/html","content_length":"19508","record_id":"<urn:uuid:d25bd63a-344e-428c-b842-3ff1e39f6484>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00590.warc.gz"} |
Complexity and Explainable AI
About six years ago, I posted on why it was important to understand machine learning, mentioning trust, fairness, security and causality. But I then I brought in complexity.
What if P = NP? Would that help. Actually it would makes things worse. If you had a quick algorithm for NP-complete problems, you could use it to find the smallest possible circuit for say
matching or traveling salesman but you would have no clue why that circuit works.
Ryan Williams countered
If P=NP you could also find the shortest proof in your favorite formal system that the smallest possible circuit does what you wanted it to do, as well as any other claim you are wondering that
may be true about the circuit. That proof might not be comprehensible to you, but it could be written in a format where proof assistant software such as HOL or Coq could parse it and convince you
it is correct. So if P=NP (with feasible low constants) I think that would definitely help.
So if P = NP maybe you couldn't understand how the circuit works but any question about how it works you could answer.
While we don't live in a world where P = NP, we are in a world where we can produce very strong learning algorithms where we can't understand the programs they produce. They are now powerful enough
to apply Ryan's idea and use the algorithms themselves to understand the decisions they make.
For example I tried using Google Translate on a Hungarian obituary of Vera Sós. Hungarian does not use gendered pronouns and and the translation incorrectly produced male pronouns. I then fed the
translation into ChatGPT which translated with the correct female pronouns.
You can speculate why Google Translate uses male pronouns but you can't ask it why. But you can ask ChatGPT.
I used female pronouns in the translation because Vera T. Sós is a female name. Vera is a feminine given name, and since the context of the original text is about her life and accomplishments, it
is appropriate to use female pronouns in the English translation to maintain clarity and accurately convey information about her.
So perhaps if you want to understand how ML works, perhaps we should seek stronger algorithms, not weaker ones, algorithms that can explain themselves. As they say, a dull knife is more dangerous
than a sharp one.
5 comments:
1. So, stronger should imply with a gurantee to be explainable?
1. Guarantee is a strong word, but yes with the capability to explain its actions.
2. Commemoration is held on Thursday: https://www.renyi.hu/hu/esemenyek-v1/megemlekezes-sos-verarol
3. How do you know that ChatGPT's "explanation" is really that, rather than merely more statistical language parroting?
1. There is some circular reasoning here and the best you could hope for is an overly simplified view of its reasoning. When I have been asking ChatGPT to explain its actions, it does seem to be
making a good effort at it. | {"url":"https://blog.computationalcomplexity.org/2023/04/complexity-and-generative-ai.html?m=1","timestamp":"2024-11-09T13:10:18Z","content_type":"application/xhtml+xml","content_length":"66352","record_id":"<urn:uuid:e4f1c550-d1a1-4061-af6c-3f2dad327316>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00715.warc.gz"} |
Weighting on a friend | Python-bloggersWeighting on a friend
This article was first published on OSM , and kindly contributed to python-bloggers. (You can report issue about the content on this page here)
Want to share your content on python-bloggers? click here.
Our last few posts on portfolio construction have simulated various weighting schemes to create a range of possible portfolios. We’ve then chosen portfolios whose average weights yield the type of
risk and return we’d like to achieve. However, we’ve noted there is more to portfolio construction than simulating portfolio weights. We also need to simulate return outcomes given that our use of
historical averages to set return expectations is likely to be biased. It only accounts for one possible outcome.
Why did we bother to simulate weights in the first place? Finance theory posits that it is possible to find an optimal allocation of assets that achieves the highest return for a given level of risk
or the lowest risk for a given level of return through a process known as mean-variance optimization. Explaining such concepts are all well and good if one understands the math, but aren’t so great
at developing the intuition if one doesn’t. Additionally, simulating portfolio weights has the added benefit of approximating the range of different portfolios, different investors might hold if they
had a view on asset attractiveness.
Of course, as we pointed out two posts ago, our initial weighting scheme was relatively simple as it assumed all assets were held. We then showed that if we excluded some of the assets, the range of
outcomes would increase significantly. In our last post, we showed that when we simulated many potential return paths, based on the historical averages along with some noise, and then simulated a
range of portfolio weights, the range of outcomes also increased.
Unfortunately, the probability of achieving our desired risk-return constraints decreased. The main reason: a broader range of outcomes implies an increase in volatility, which means the likelihood
of achieving our risk constraint declines. Now this would be great time to note how in this case (as in much of finance theory), volatility is standing in for risk, which begs the question as to
whether volatility captures risk accurately.^1 Warren Buffett claims he prefers a choppy 15% return to a smooth 12%, a view finance theory might abhor. But the volatility vs. risk discussion is a
huge can of worms we don’t want to open just yet. So like a good economist, after first assuming a can opener, we’ll assume a screw top jar in which to toss this discussion to unscrew at a later
For now, we’ll run our simulations again but allowing for more extreme outcomes. Recall we generate 1,000 possible return scenarios over a 60-month (5-year) period for the set of assets. In other
words, each simulation features randomly generated returns for each of the assets along with some noise, all of which hopefully accounts for random asset correlations too. As in our previous post, we
then randomly select four of the possible return profiles and run our portfolio weighting algorithm to produce 1,000 different portfolios. Here are four samples below.
Now here’s the same simulations using the algorithm that allows one to exclude up to any two of the four assets. (A portfolio of a single asset isn’t really a portfolio.)
Who knew portfolio simulations could be so artistic. For the portfolios in which all assets are held, the probability of hitting our not less than 7% return and not more than 10% risk constraint, is
8%, 0%, 29%, and 29%. For the portfolios that allow assets to be excluded, the probabtility of achieving our risk-return constraints is 12%, 0%, 25%, and 25%.
We shouldn’t read too much into four samples. Still, allowing a broader range of allocations sometimes yields an improved probability of success, and sometimes it doesn’t. So much for more choice is
better! The point is that a weighting scheme is only as good as the potential returns available. Extreme weights (99% in gold, 1% is bonds, and 0% in everything else, for example), will yield even
more extreme performance, eroding the benefits of diversification. In essence, by excluding assets we’re increasing the likelihood of achieving dramatically awesome or dramatically awful returns.
Portfolios with extreme returns tend to have higher volatility. So we are in effect filling in outcomes in the “tails” of the distribution. All things being equal, more tail events (2x more in fact),
generally yield more outcomes where we’re likely to miss our modest risk constraint.
We’ll now extend the weighting simulation to the 1,000 return simulations from above, yielding three million different portfolios. We graph a random selection of 10,000 of those portfolios below.
what a blast! Now here’s the histograms of the entire datasheet for returns and risk.
Returns are relatively normal. Volatility, predictably, is less than normal and positively skewed. We could theoretically transform volatility if we needed a more normal shape, but we’ll leave it as
is for now. Still, this is something to keep in the back of our mind—namely, once we start excluding assets, we’re no longer in an easy to fit normal world, so should be wary of the probabilities we
Given these results, let’s think about what we want our portfolio to achieve. Greater than 7% returns matches the nominal returns of the stocks over the long term. Less than 10% risk does not. But,
recall, we were “hoping” to generate equity-like returns with lower risk. We’re not trying to generate 10% or 20% average annual returns. If we were, we’d need to take on a lot more risk, at least
Thus the question is, how much volatility are we willing to endure for a given return? While we could phrase this question in reverse (return for a given level of risk), we don’t think that is
intuitive for most non-professional investors. If we bucket the range of returns and then calculate the volatility for each bucket, we can shrink our analysis to get a more managenable estimate of
the magnitude of the risk-return trade-off.
We have three choices for how to bucket the returns: by interval, number, or width. We could have equal intervals of returns, with a different the number of observations for each interval. We could
have an equal number of observations, with the return spread for each bucket sporting a different width. Or we could choose an equal width for the cut-off between returns, resulting in different
number of observations for each bucket. We’ll choose the last scheme.
While there are returns that far exceed negative 15% and positive 45% on average annual basis, the frequency of occurrence is de minimis. So we’ll exclude those outcomes and only show the most
frequent ranges in the graph below.
We see that around 59% of the occurrences are within the return range of 5% to 15%. Around 76% are between 5% to 25%. That’s a good start. A majority of the time we’ll be close to or above our return
constraint. If we alter the buckets so most of the outlier returns are in one bucket (anything below -5% or above 35%) and then calculate the median return and risk for those buckets, we’ll have a
manageable data set, as shown below.
The bucket that includes are our greater than 7% return constraint has a median return of about 9% with median risk of about 12%. That equates to a Sharpe ratio of about 0.75, which is better than
our implied target of 0.7. The next bucket with a median return and risk of 18% and 16% is better. But buckets above that have even better risk to reward ratios. However, only 3% of the portfolios
reach that stratosphere.
Given that around 76% of the portfolios have a better risk-reward than our target, we could easily achieve our goal by only investing a portion of our assets in the risky portfolios and putting the
remainder in risk-free assets if one believes such things exist.^2 But we’d still need to figure out our allocations.
Let’s look at the weighting for these different portfolios. First, we bracket 76% or so of the portfolios that are in the sweet spot and take the average of the weights.
We see that average weights are roughly equal if slightly biased toward stocks and bonds. Now let’s calculate the average weights for the returns above the mid-range.
Instructive. Those portfolios that saw very high returns had a very high exposure to gold. What about the low return portfolios?
Also a high exposure to gold. We’re not picking on the yellow metal, but this a great illustration of the perils of overweighting a highly volatile asset. Sometimes you knock it out of the park and
sometimes you crash and burn. Recall our return simulations took each asset’s historical return and risk and added in some noise^3 similar to the asset’s underlying risk. Hence, by randomness alone
it was possible to generate spectacular or abysmal returns. That massive outperformance your friend is enjoying could entirely be due to luck.
But, before you begin to think we’ve drunk the Efficient Market Hypothesis kool-aid, let’s look at the weights for the portfolios that meet or exceed our risk-return constraints.
An interesting result. While the weights are still relatively equal, the higher risk assets have a lower exposure overall.
Let’s summarize. When we simulated multiple return outcomes and relaxed the allocation constraints to allow us to exclude assets, the range of return and risk results increased significantly. But the
likelihood of achieving our risk and return targets decreased. So we decided to bucket the portfolios to make it easier to assess how much risk we’d have to accept for the type of return we wanted.
Doing so, we calculated the median returns and risk for each bucket and found that some buckets achieved Sharpe ratios close to or better than that implied by our original risk-return constraint. We
then looked at the average asset allocations for some of the different buckets, ultimately, cutting the data again to calculate the average weights for the better Sharpe ratio portfolios. The
takeaway: relatively equal-weighting tended to produce a better risk-reward outcome than significant overweighting. Remember this takeaway because we’ll come back to it in later posts.
In the end, we could have a bypassed some of this data wrangling and just calculated the optimal portfolio weights for various risk profiles. But that will have to wait until we introduce our friend,
mean-variance optimization. Until then, the Python and R code that can produce the foregoing analysis and charts are below.
For the Pythonistas:
# Built using Python 3.7.4
# Load libraries
import pandas as pd
import pandas_datareader.data as web
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
# SKIP IF ALREADY HAVE DATA
# Load data
start_date = '1970-01-01'
end_date = '2019-12-31'
symbols = ["WILL5000INDFC", "BAMLCC0A0CMTRIV", "GOLDPMGBD228NLBM", "CSUSHPINSA", "DGS5"]
sym_names = ["stock", "bond", "gold", "realt", 'rfr']
filename = 'data_port_const.pkl'
df = pd.read_pickle(filename)
print('Data loaded')
except FileNotFoundError:
print("File not found")
print("Loading data", 30*"-")
data = web.DataReader(symbols, 'fred', start_date, end_date)
data.columns = sym_names
data_mon = data.resample('M').last()
df = data_mon.pct_change()['1987':'2019']
dat = data_mon.pct_change()['1971':'2019']
## Simulation function
class Port_sim:
def calc_sim(df, sims, cols):
wts = np.zeros((sims, cols))
for i in range(sims):
a = np.random.uniform(0,1,cols)
b = a/np.sum(a)
wts[i,] = b
mean_ret = df.mean()
port_cov = df.cov()
port = np.zeros((sims, 2))
for i in range(sims):
port[i,0] = np.sum(wts[i,]*mean_ret)
port[i,1] = np.sqrt(np.dot(np.dot(wts[i,].T,port_cov), wts[i,]))
sharpe = port[:,0]/port[:,1]*np.sqrt(12)
best_port = port[np.where(sharpe == max(sharpe))]
max_sharpe = max(sharpe)
return port, wts, best_port, sharpe, max_sharpe
def calc_sim_lv(df, sims, cols):
wts = np.zeros(((cols-1)*sims, cols))
for i in range(1,cols):
for j in range(sims):
a = np.random.uniform(0,1,(cols-i+1))
b = a/np.sum(a)
c = np.random.choice(np.concatenate((b, np.zeros(i))),cols, replace=False)
wts[count,] = c
mean_ret = df.mean()
port_cov = df.cov()
port = np.zeros(((cols-1)*sims, 2))
for i in range(sims):
port[i,0] = np.sum(wts[i,]*mean_ret)
port[i,1] = np.sqrt(np.dot(np.dot(wts[i,].T,port_cov), wts[i,]))
sharpe = port[:,0]/port[:,1]*np.sqrt(12)
best_port = port[np.where(sharpe == max(sharpe))]
max_sharpe = max(sharpe)
return port, wts, best_port, sharpe, max_sharpe
def graph_sim(port, sharpe):
plt.scatter(port[:,1]*np.sqrt(12)*100, port[:,0]*1200, marker='.', c=sharpe, cmap='Blues')
plt.colorbar(label='Sharpe ratio', orientation = 'vertical', shrink = 0.25)
plt.title('Simulated portfolios', fontsize=20)
plt.xlabel('Risk (%)')
plt.ylabel('Return (%)')
# Calculate returns and risk for longer period
hist_mu = dat['1971':'1991'].mean(axis=0)
hist_sigma = dat['1971':'1991'].std(axis=0)
# Run simulation based on historical figures
sim1 = []
for i in range(1000):
#np.random.normal(mu, sigma, obs)
a = np.random.normal(hist_mu[0], hist_sigma[0], 60) + np.random.normal(0, hist_sigma[0], 60)
b = np.random.normal(hist_mu[1], hist_sigma[1], 60) + np.random.normal(0, hist_sigma[1], 60)
c = np.random.normal(hist_mu[2], hist_sigma[2], 60) + np.random.normal(0, hist_sigma[2], 60)
d = np.random.normal(hist_mu[3], hist_sigma[3], 60) + np.random.normal(0, hist_sigma[3], 60)
df1 = pd.DataFrame(np.array([a, b, c, d]).T)
cov_df1 = df1.cov()
sim1.append([df1, cov_df1])
# create graph objects
samp = np.random.randint(1, 1000, 4)
graphs1 = []
for i in range(4):
port, _, _, sharpe, _ = Port_sim.calc_sim(sim1[samp[i]][0], 1000, 4)
graf = [port,sharpe]
# Graph sample portfolios
fig, axes = plt.subplots(2, 2, figsize=(12,6))
for i, ax in enumerate(fig.axes):
ax.scatter(graphs1[i][0][:,1]*np.sqrt(12)*100, graphs1[i][0][:,0]*1200, marker='.', c=graphs1[i][1], cmap='Blues')
# create graph objects
graphs2 = []
for i in range(4):
port, _, _, sharpe, _ = Port_sim.calc_sim_lv(sim1[samp[i]][0], 1000, 4)
graf = [port,sharpe]
# Graph sample portfolios
fig, axes = plt.subplots(2, 2, figsize=(12,6))
for i, ax in enumerate(fig.axes):
ax.scatter(graphs2[i][0][:,1]*np.sqrt(12)*100, graphs2[i][0][:,0]*1200, marker='.', c=graphs2[i][1], cmap='Blues')
# Calculate probability of hitting risk-return constraints based on sample portfolos
probs = []
for i in range(8):
if i <= 3:
out = round(np.mean((graphs1[i][0][:,0] >= 0.07/12) & (graphs1[i][0][:,1] <= 0.1/np.sqrt(12))),2)*100
out = round(np.mean((graphs2[i-4][0][:,0] >= 0.07/12) & (graphs2[i-4][0][:,1] <= 0.1/np.sqrt(12))),2)*100
# Simulate portfolios from reteurn simulations
def wt_func(sims, cols):
wts = np.zeros(((cols-1)*sims, cols))
for i in range(1,cols):
for j in range(sims):
a = np.random.uniform(0,1,(cols-i+1))
b = a/np.sum(a)
c = np.random.choice(np.concatenate((b, np.zeros(i))),cols, replace=False)
wts[count,] = c
return wts
# Note this takes over 4min to run, substantially worse than the R version, which runs in under a minute. Not sure what I'm missng.
portfolios = np.zeros((1000, 3000, 2))
weights = np.zeros((1000,3000,4))
for i in range(1000):
wt_mat = wt_func(1000,4)
port_ret = sim1[i][0].mean(axis=0)
cov_dat = sim1[i][0].cov()
returns = np.dot(wt_mat, port_ret)
risk = [np.sqrt(np.dot(np.dot(wt.T,cov_dat), wt)) for wt in wt_mat]
portfolios[i][:,0] = returns
portfolios[i][:,1] = risk
weights[i][:,:] = wt_mat
port_1m = portfolios.reshape((3000000,2))
wt_1m = weights.reshape((3000000,4))
# Find probability of hitting risk-return constraints on simulated portfolios
port_1m_prob = round(np.mean((port_1m[:][:,0] > 0.07/12) & (port_1m[:][:,1] <= 0.1/np.sqrt(12))),2)*100
print(f"The probability of meeting our portfolio constraints is:{port_1m_prob: 0.0f}%")
# Plot sample portfolios
port_samp = port_1m[np.random.choice(1000000, 10000),:]
sharpe = port_samp[:,0]/port_samp[:,1]
plt.scatter(port_samp[:,1]*np.sqrt(12)*100, port_samp[:,0]*1200, marker='.', c=sharpe, cmap='Blues')
plt.colorbar(label='Sharpe ratio', orientation = 'vertical', shrink = 0.25)
plt.title('Ten thousand samples from three million simulated portfolios', fontsize=20)
plt.xlabel('Risk (%)')
plt.ylabel('Return (%)')
# Graph histograms
fig, axes = plt.subplots(1,2, figsize = (12,6))
for idx,ax in enumerate(fig.axes):
if idx == 1:
ax.hist(port_1m[:][:,1], bins = 100)
ax.hist(port_1m[:][:,0], bins = 100)
## Create buckets for analysis and graphin
df_port = pd.DataFrame(port_1m, columns = ['returns', 'risk'])
port_bins = np.arange(-35,65,10)
df_port['dig_ret'] = pd.cut(df_port['returns']*1200, port_bins)
xs = ["(-35, -25]", "(-25, -15]", "(-15, -5]","(-5, 5]", "(5, 15]", "(15, 25]", "(25, 35]", "(35, 45]", "(45, 55]"]
ys = df_port.groupby('dig_ret').size().values/len(df_port)*100
# Graph buckets with frequency
fig,ax = plt.subplots(figsize = (12,6))
ax.bar(xs[2:7], ys[2:7])
ax.set(xlabel = "Return bucket (%)",
ylabel = "Frequency (%)",
title = "Frequency of occurrrence for return bucket ")
# Calculate frequency of occurence for mid range of returns
good_range = np.sum(df_port.groupby('dig_ret').size()[4:6])/len(df_port)
## Graph buckets with median return and risk
med_ret = df_port.groupby('dig_ret').agg({'returns':'median'})*1200
med_risk = df_port.groupby('dig_ret').agg({'risk':'median'})*np.sqrt(12)*100
labs_ret = np.round(med_ret['returns'].to_list()[2:7])
labs_risk = np.round(med_risk['risk'].to_list()[2:7])
fig, ax = plt.subplots(figsize = (12,6))
ax.bar(xs[2:7], ys[2:7])
for i in range(len(xs[2:7])):
ax.annotate(str('Returns: ' + str(labs_ret[i])), xy = (xs[2:7][i], ys[2:7][i]+2), xycoords = 'data')
ax.annotate(str('Risk: ' + str(labs_risk[i])), xy = (xs[2:7][i], ys[2:7][i]+5), xycoords = 'data')
ax.set(xlabel = "Return bucket (%)",
ylabel = "Frequency (%)",
title = "Frequency of occurrrence for return bucket ",
ylim = (0,60))
# Find frequency of high return buckets
hi_range = np.sum(df_port.groupby('dig_ret').size()[6:])/len(df_port)
## Identify weights for different buckets for graphing
wt_1m = pd.DataFrame(wt_1m, columns = ['Stocks', 'Bonds', 'Gold', 'Real estate'])
port_ids_mid = df_port.loc[(df_port['returns'] >= 0.05/12) & (df_port['returns'] <= 0.25/12)].index
mid_ports = wt_1m.loc[port_ids_mid,:].mean(axis=0)
port_ids_hi = df_port.loc[(df_port['returns'] >= 0.35/12)].index
hi_ports = wt_1m.loc[port_ids_hi,:].mean(axis=0)
port_ids_lo = df_port.loc[(df_port['returns'] <= -0.05/12)].index
lo_ports = wt_1m.loc[port_ids_lo,:].mean(axis=0)
# Sharpe portfolios
df_port['sharpe'] = df_port['returns']/df_port['risk']*np.sqrt(12)
port_ids_sharpe = df_port[(df_port['sharpe'] > 0.7)].index
sharpe_ports = wt_1m.loc[port_ids_sharpe,:].mean(axis=0)
# Create graph function
def wt_graph(ports, title):
fig, ax = plt.subplots(figsize=(12,6))
ax.bar(ports.index.values, ports*100)
for i in range(len(mid_ports)):
ax.annotate(str(np.round(ports[i],2)*100), xy=(ports.index.values[i], ports[i]*100+2), xycoords = 'data')
ax.set(xlabel = '', ylabel = 'Weigths (%)', title = title, ylim = (0,max(ports)*100+5))
# Graph weights
wt_graph(mid_ports, "Average asset weights for mid-range portfolios")
wt_graph(mid_ports, "Average asset weights for high return portfolios")
wt_graph(mid_ports, "Average asset weights for negative return portfolios")
wt_graph(mid_ports, "Average asset weights for Sharpe portfolios")
For the Rtists:
# Built using R 3.6.2
## Load packages
## Load data
df <- readRDS("port_const.rds")
dat <- readRDS("port_const_long.rds")
sym_names <- c("stock", "bond", "gold", "realt", "rfr")
## Call simuation functions
## Prepare sample
hist_avg <- dat %>%
filter(date <= "1991-12-31") %>%
summarise_at(vars(-date), list(mean = function(x) mean(x, na.rm=TRUE),
sd = function(x) sd(x, na.rm = TRUE))) %>%
gather(key, value) %>%
mutate(key = str_remove(key, "_.*"),
key = factor(key, levels =sym_names)) %>%
mutate(calc = c(rep("mean",5), rep("sd",5))) %>%
spread(calc, value)
# Run simulation
sim1 <- list()
for(i in 1:1000){
a <- rnorm(60, hist_avg[1,2], hist_avg[1,3]) + rnorm(60, 0, hist_avg[1,3])
b <- rnorm(60, hist_avg[2,2], hist_avg[2,3]) + rnorm(60, 0, hist_avg[2,3])
c <- rnorm(60, hist_avg[3,2], hist_avg[3,3]) + rnorm(60, 0, hist_avg[3,3])
d <- rnorm(60, hist_avg[4,2], hist_avg[4,3]) + rnorm(60, 0, hist_avg[4,3])
df1 <- data.frame(a, b, c, d)
cov_df1 <- cov(df1)
sim1[[i]] <- list(df1, cov_df1)
names(sim1[[i]]) <- c("df", "cov_df")
# Plot random four portfolios
## Sample four return paths
## Note this sampling does not realize in the same way in Rmarkdown/blogdown as in the console. NOt sure why.
samp <- sample(1000,4)
graphs <- list()
for(i in 1:8){
if(i <= 4){
graphs[[i]] <- port_sim(sim1[[samp[i]]]$df,1000,4)
graphs[[i]] <- port_sim_lv(sim1[[samp[i-4]]]$df,1000,4)
gridExtra::grid.arrange(graphs[[1]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
graphs[[2]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
graphs[[3]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
graphs[[4]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
ncol=2, nrow=2,
top = textGrob("Four portfolio and return simulations",gp=gpar(fontsize=15)))
# Graph second set
gridExtra::grid.arrange(graphs[[5]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
graphs[[6]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
graphs[[7]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
graphs[[8]]$graph +
theme(legend.position = "none") +
labs(title = NULL),
ncol=2, nrow=2,
top = textGrob("Four portfolio and return simulations allowing for excluded assets",gp=gpar(fontsize=15)))
# Calculate probability of hitting risk-return constraint
probs <- c()
for(i in 1:8){
probs[i] <- round(mean(graphs[[i]]$port$returns >= 0.07/12 &
graphs[[i]]$port$risk <=0.1/sqrt(12)),2)*100
## Load data
port_1m <- readRDS("port_3m_sim.rds")
## Graph sample of port_1m
port_samp = port_1m[sample(1e6, 1e4),]
port_samp %>%
mutate(Sharpe = returns/risk) %>%
ggplot(aes(risk*sqrt(12)*100, returns*1200, color = Sharpe)) +
geom_point(size = 1.2, alpha = 0.4) +
scale_color_gradient(low = "darkgrey", high = "darkblue") +
labs(x = "Risk (%)",
y = "Return (%)",
title = "Ten thousand samples from simulation of three million portfolios") +
theme(legend.position = c(0.05,0.8), legend.key.size = unit(.5, "cm"),
legend.background = element_rect(fill = NA))
## Graph histogram
port_1m %>%
mutate(returns = returns*1200,
risk = risk*sqrt(12)*100) %>%
gather(key, value) %>%
ggplot(aes(value)) +
geom_histogram(bins=100, fill = 'darkblue') +
facet_wrap(~key, scales = "free",
labeller = as_labeller(c(returns = "Returns (%)",
risk = "Risk (%)"))) +
scale_y_continuous(labels = scales::comma) +
labs(x = "",
y = "Count",
title = "Portfolio simulation return and risk histograms")
## Graph quantile returns for total series
x_lim = c("(-15,-5]",
"(-5,5]", "(5,15]",
"(15,25]", "(25,35]")
port_1m %>%
mutate(returns = cut_width(returns*1200, 10)) %>%
group_by(returns) %>%
summarise(risk = median(risk*sqrt(12)*100),
count = n()/nrow(port_1m)) %>%
ggplot(aes(returns, count*100)) +
geom_bar(stat = "identity", fill = "blue") +
xlim(x_lim) +
labs(x = "Return bucket (%)",
y = "Frequency (%)",
title = "Frequency of occurrrence for return bucket ")
## Occurrences
mid_range <- port_1m %>%
mutate(returns = cut_width(returns*1200, 10)) %>%
group_by(returns) %>%
summarise(risk = median(risk*sqrt(12)*100),
count = n()/nrow(port_1m)) %>%
filter(as.character(returns) %in% c("(5,15]")) %>%
summarise(sum = round(sum(count),2)) %>%
good_range <- port_1m %>%
mutate(returns = cut_width(returns*1200, 10)) %>%
group_by(returns) %>%
summarise(risk = median(risk*sqrt(12)*100),
count = n()/nrow(port_1m)) %>%
filter(as.character(returns) %in% c("(5,15]" , "(15,25]")) %>%
summarise(sum = round(sum(count),2)) %>%
# Set quantiles for graph and labels
quants <- port_1m %>%
mutate(returns = cut(returns*1200, breaks=c(-Inf, -5, 5, 15, 25, 35, Inf))) %>%
group_by(returns) %>%
summarise(prop = n()/nrow(port_1m)) %>%
select(prop) %>%
mutate(prop = cumsum(prop))
# Calculate quantile
x_labs <- quantile(port_1m$returns, probs = unlist(quants))*1200
x_labs_median <- tapply(port_1m$returns*1200,
findInterval(port_1m$returns*1200, x_labs), median) %>%
x_labs_median_risk <- tapply(port_1m$risk*sqrt(12)*100, findInterval(port_1m$risk*sqrt(12)*100, x_labs), median) %>% round()
# Graph frequency of occurrence for equal width returns
port_1m %>%
mutate(returns = cut(returns*1200, breaks=c(-45, -5,5,15,25,35,95))) %>%
group_by(returns) %>%
summarise(risk = median(risk*sqrt(12)*100),
count = n()/nrow(port_1m)) %>%
ggplot(aes(returns, count*100)) +
geom_bar(stat = "identity", fill = "blue") +
geom_text(aes(returns, count*100+5, label = paste("Risk: ", round(risk), "%", sep=""))) +
geom_text(aes(returns, count*100+2,
label = paste("Return: ", x_labs_median[-7], "%", sep=""))) +
labs(x = "Return bucket (%)",
y = "Frequency (%)",
title = "Frequency of occurrrence for return bucket with median risk and return per bucket")
# High range probability
high_range <- port_1m %>%
mutate(returns = cut(returns*1200, breaks=c(-45, -5,5,15,25,35,95))) %>%
group_by(returns) %>%
summarise(risk = median(risk*sqrt(12)*100),
count = n()/nrow(port_1m)) %>%
filter(as.character(returns) %in% c("(25,35]", "(35,95]")) %>%
summarise(sum = round(sum(count),2)) %>%
## Identify weights for target portfolios
wt_1m <- readRDS('wt_3m.rds')
## Portfolio ids
# Mid-range portfolis
port_ids_mid <- port_1m %>%
mutate(row_ids = row_number()) %>%
filter(returns >= 0.05/12, returns < 0.25/12) %>%
select(row_ids) %>%
unlist() %>%
mid_ports <- colMeans(wt_1m[port_ids_mid,])
# Hi return portfolio
port_ids_hi <- port_1m %>%
mutate(row_ids = row_number()) %>%
filter(returns >= 0.35/12) %>%
select(row_ids) %>%
hi_ports <- colMeans(wt_1m[port_ids_hi,])
# Low return portfolios
port_ids_lo <- port_1m %>%
mutate(row_ids = row_number()) %>%
filter(returns <= -0.05/12) %>%
select(row_ids) %>%
lo_ports <- colMeans(wt_1m[port_ids_lo,])
# Sharpe porfolios
port_ids_sharpe <- port_1m %>%
mutate(sharpe = returns/risk*sqrt(12),
row_ids = row_number()) %>%
filter(sharpe > 0.7) %>%
select(row_ids) %>%
sharpe_ports <- colMeans(wt_1m[port_ids_sharpe,])
## Graph portfolio weights
# Function
wt_graf <- function(assets, weights, title){
data.frame(assets = factor(assets, levels = assets),
weights = weights) %>%
ggplot(aes(assets, weights*100)) +
geom_bar(stat = "identity", fill="blue") +
geom_text(aes(assets ,weights*100+3, label = round(weights,2)*100)) +
y = "Weights (%)",
title = title)
assets = c("Stocks", "Bonds", "Gold", "Real Estate")
# Graph diferent weights
wt_graf(assets, mid_ports, "Average asset weights for mid-range portfolios")
wt_graf(assets, hi_ports, "Average asset weights for high return portfolios")
wt_graf(assets, lo_ports, "Average asset weights for negative return portfolios")
wt_graf(assets, sharpe_ports, "Average asset weights for Sharpe constraints")
1. What is risk anyway? There are plenty of definitions out there. In the case of investing, a working definition might be that risk is the chance that one won’t achieve his or her return
objectives. Volatility, on the other hand, describes the likely range of returns. So volatility captures the probability of failure; but it also captures success and every other occurrence along
the continuum of outcomes.↩
2. Is a demand deposit or a government bond who’s yield is below inflation, or worse, is negative, risk-free?↩
3. A normally distributed error term whose mean was zero and whose standard deviation was the same as the asset’s.↩
for the author, please follow the link and comment on their blog:
Want to share your content on python-bloggers? | {"url":"https://python-bloggers.com/2020/07/weighting-on-a-friend/","timestamp":"2024-11-07T01:13:07Z","content_type":"text/html","content_length":"67463","record_id":"<urn:uuid:bcc7b3df-fecf-4145-baab-c4e81fd8ed9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00725.warc.gz"} |
Angles Worksheets
Here is a graphic preview for all of the Angles Worksheets. You can select different variables to customize these Angles Worksheets for your needs. The Angles Worksheets are randomly created and will
never repeat so you have an endless supply of quality Angles Worksheets to use in the classroom or at home. We have classifying and naming angles, reading protractors and measuring angles, finding
complementary, supplementary, vertical, alternate, corresponding angles and much more. Our Angles Worksheets are free to download, easy to use, and very flexible.
These Angles Worksheets are a great resource for children in 3rd Grade, 4th Grade, 5th Grade, 6th Grade, 7th Grade, and 8th Grade.
Click here for a Detailed Description of all the Geometry Worksheets. | {"url":"https://www.math-aids.com/Geometry/Angles/","timestamp":"2024-11-08T02:47:17Z","content_type":"text/html","content_length":"42062","record_id":"<urn:uuid:f41652e0-c167-493a-8644-aeb42486c37f>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00301.warc.gz"} |
High Resistance Measurement
Updated 4 November 2024
Fertiliser Resistivity Probe
The photo above shows a Probe for measuring the resistivity of granular fertiliser. Resistivity is defined below as the resistance observed, times a cell constant.
The design could be adapted for many other applications. With powdered or granular materials the probes are fully inserted into the sample and the switch is then turned on. After a delay the LED
might light up. The delay time is proportional to the resistance of the sample.
• A resistance range from 100,000,000 ohms to about 100,000,000,000 ohms can be measured in a time varying from about 1 second to 1000 seconds
• The maximum Probe power supply current, with the LED on, is less than 300 microamps. Before the LED turns on it is less than 10 microamps.
• The probes are stainless steel chop-sticks.
• The LED is a modified Christmas light which has a strong output at a very low current. I have not found any commercial source which is better.
The device is essentially a one-bit computer (a high input impedance electronic switch) connected to the output of an analog current-integrator. The integrator input is connected to one of the
electrodes. A 3 volt CR2032 lithium cell powers the circuit and is connected to the other electrode. If the sample is slightly conductive a current passes between the electrodes into a 10 nanofarad
integrating capacitor. When the charge on the capacitor reaches a threshold, determined by the circuit, the green/white LED will turn on.
The time taken for the LED to turn on is proportional to the resistance of the sample. A sample with a resistance of 2,000,000,000 ohms, between the electrodes, will take about 19.5 seconds to turn
the LED on. This time will vary between circuits because of component tolerences. Measuring the highest resistances with a meter would probably require some shielding from electrical noise. An
integrator performs better.
Non-Linear Samples
When an insulation tester is used to measure samples which are slightly acidic, like superphosphate fertiliser, it does not represent the initial state of the sample. With 100 to 1000 volts applied
the sample rapidly charges up to the applied voltage where little current will flow. Once the reading is stable the value is recorded.
Initially there is a substantial pulse of current until the sample equilibrates. By the time the sample has adapted, the resistance is measured as being very high. The sample is not a pure
resistance. The pulse is much stronger than the polarisation due to capacitance that occurs with normal insulators.
A rough model of this sample type is an electrolytic capacitor in parallel with a high resistance. The residual acid and the stainless steel electrodes makes a crude electrolytic capacitor. Little
current flows once the electrolytic capacitor voltage rises to match the applied voltage. The Probe measurement more closely represents the initial undisturbed state of the sample.
For bulk samples the resistance measured is expressed as resistivity. Resistivity = resistance x electrode area / electrode spacing. Resistivity is a bulk property which is independent of the
electrode geometry. The electrode area / electrode spacing is called the cell constant. The cell constant unit is metre^2/metre = metre. Resistivity has a unit of ohm metre.
For safe handling of some powdered or granulated samples an upper resistivity limit is set, 1E9 ohm metre for example. Above this value fires or clogging could occur during sample handling. For the
probe the cell constant is approximately 0.2 metre. At the resistivity limit the resistance measured is 5E9 ohms.
• Teflon electrode sleeves could be added to improve the performance at high resistances.
• A red LED could be added to turn on when a defined resistivity threshold is reached, for example 1E9 ohm metre.
• The present circuit could be turned into an oscillator along with a 12 stage counter added. This would allow wide resistance ranges to be measured. A 12 position switch would double the number of
integration cycles for each step. 4096 cycles would allow resistances lower than 100 kohm to be measured. A single cycle with a 1 nanofarad integrating capacitor would allow resistances around
1E12 ohms to be measured. The LED would be connected to the selected counter output.
• The now positive output of the counter would be fed back to the ground pin of the oscillator, to stop the counting and to keep the LED on.
Other Methods for Measuring High Resistances
There are numerous ways to measure high resistances. Typically, insulation testers are adapted for this purpose. For high resistances the applied voltage may range from 100 volts to over 1000 volts.
Electrode assemblies can have an increased surface area with a reduced spacing for some high resistance measurements. Electrical sheilding and guard circuits are required for the highest resistances.
Insulation testers are available from most manufacturers of digital multimeters. Some can measure resistances into the teraohm range but only with a completely shielded setup.
Apart from the insulation testers, a basic digital multimeter and a DC power supply can be used to measure very high resistances. The multimeter voltage ranges typically have a 1E7 ohm input
resistance. The lowest voltage range can be used to measure small currents. I do have a meter with a 22,000,000,000 ohm input resistance on the lowest range. This may be useful for some applications
or a 1E7 ohm shunt resistor can be added.
A 1E7 ohm input resistance UNI-T UT33B multimeter, a 10 volt DC power supply and a series resistance of 1E10 ohms produces a meter reading of 10.0 mV.
Measuring the Resistivity of a Memo Cube Paper Stack
I used a full stack of Memo Cube paper as a test sample with two 52 mm diameter electrodes and a 1.6 kilogram weight compressing the electrodes onto the stack. In this case the cell constant is
derived from a cylindrical cell matching the diameter of one of the electrodes with a length equal to the thickness of the paper stack.
• The measured voltage in series with the stack was 0.0105 volts.
• The multimeter input resistance was 1E7 ohms.
• The current was 0.0105/1E7 = 1.05E-9 amps.
• The voltage across the stack of paper was 10 - 10.5E-3 = 9.99 volts.
• The paper stack resistance is 9.99 / 1.05E-9 = 9.514E9 ohms.
• The area of the electrode = (5.2E-2 /2)^2 x 3.14159 = 2.12E-3 square metres.
• The Memo Cube thickness is 3.55E-2 metres.
• The resistivity is 9.514E9 x 2.12E-3 / 3.55E-2 = 5.68E8 ohm metre.
The Memo Cube resistivity measured with the probe connected to the electrodes was 1.18E10 ohms x 2.12E-3 /3.55E-2 = 7.04E8 ohm metre. A little time was required to allow the 10 volts applied in the
previous measurement to dissipate.
We know that 2,000,000,000 ohms takes about 19.5 seconds before the LED turns on. With the Memo Cube paper stack the time delay was 115 seconds. 2E9 ohms x 115 / 19.5 = 1.18E10 ohms.
The experimental work described here was done at a relative humidity of 70% and a temperature of 20 degrees Celsius. The resistivity of the paper stack will vary greatly when equilibrated to other
The experiment would make a good exercise for physics or electronics lab classes. A good question to ask would be: What is the resistivity of one sheet of paper?
The photo below shows the Probe measuring the resistance of a Memo Cube paper stack. To calculate the resistivity of the paper stack, multiply the resistance by the electrode area and divide by the
thickness of the paper stack. If the measurements are in metres the resistivity will have a unit of ohm metre. Also shown are weights totalling 1.6 kg and a 2,000,000,000 ohm reference resistor
inside a plastic box with some drying crystals.
Memo Cube Resistivity
Measurements with teflon tape.
To test the probe at higher resistances I used two layers of teflon tape, with a 0.1 mm total thickness covering a 2200 mm^2 tin electrode. The electrodes were connected to the probes by some clip
leads. This setup initially produced a 5 minute integration time. Resistance = (5 x 60 / 40) x 5E9 = 3.75E10 ohms. The cell constant was 2200 / 0.1 = 22000 mm or 22 metre.
The resistivity of the teflon tape, plus the probe materials, was 3.75E10 x 22 = 8.24E11 ohm metre. Solid teflon has a resistivity around 1E22 ohm metre.
On the second day the integration took about took 2 hours. On the third day the integration took about 5 hours.
On the fourth day the integration took about 8 hours. The resistance can be calculated as 8 x 3600/40 x 5E9 = 3.6E12 ohms. Multiplying this by 22 metre gives a measured resistivity of 8E13 ohm metre.
Note that this resistivity includes some of the materials used to construct the probe.
There is no way I can measure the actual 1E22 ohm metre resistivity of teflon. | {"url":"http://jepspectro.com/htm/resistivity.htm","timestamp":"2024-11-04T13:20:21Z","content_type":"text/html","content_length":"16772","record_id":"<urn:uuid:2152cec1-c8e7-48a6-867d-58746aac90ff>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00466.warc.gz"} |
Checking a fitted gam is like checking a fitted glm, with two main differences. Firstly, the basis dimensions used for smooth terms need to be checked, to ensure that they are not so small that they
force oversmoothing: the defaults are arbitrary. choose.k provides more detail, but the diagnostic tests described below and reported by this function may also help. Secondly, fitting may not always
be as robust to violation of the distributional assumptions as would be the case for a regular GLM, so slightly more care may be needed here. In particular, the thoery of quasi-likelihood implies
that if the mean variance relationship is OK for a GLM, then other departures from the assumed distribution are not problematic: GAMs can sometimes be more sensitive. For example, un-modelled
overdispersion will typically lead to overfit, as the smoothness selection criterion tries to reduce the scale parameter to the one specified. Similarly, it is not clear how sensitive REML and ML
smoothness selection will be to deviations from the assumed response dsistribution. For these reasons this routine uses an enhanced residual QQ plot.
This function plots 4 standard diagnostic plots, some smoothing parameter estimation convergence information and the results of tests which may indicate if the smoothing basis dimension for a term is
too low.
Usually the 4 plots are various residual plots. For the default optimization methods the convergence information is summarized in a readable way, but for other optimization methods, whatever is
returned by way of convergence diagnostics is simply printed.
The test of whether the basis dimension for a smooth is adequate (Wood, 2017, section 5.9) is based on computing an estimate of the residual variance based on differencing residuals that are near
neighbours according to the (numeric) covariates of the smooth. This estimate divided by the residual variance is the k-index reported. The further below 1 this is, the more likely it is that there
is missed pattern left in the residuals. The p-value is computed by simulation: the residuals are randomly re-shuffled k.rep times to obtain the null distribution of the differencing variance
estimator, if there is no pattern in the residuals. For models fitted to more than k.sample data, the tests are based of k.sample randomly sampled data. Low p-values may indicate that the basis
dimension, k, has been set too low, especially if the reported edf is close to k', the maximum possible EDF for the term. Note the disconcerting fact that if the test statistic itself is based on
random resampling and the null is true, then the associated p-values will of course vary widely from one replicate to the next. Currently smooths of factor variables are not supported and will give
an NA p-value.
Doubling a suspect k and re-fitting is sensible: if the reported edf increases substantially then you may have been missing something in the first fit. Of course p-values can be low for reasons other
than a too low k. See choose.k for fuller discussion.
The QQ plot produced is usually created by a call to qq.gam, and plots deviance residuals against approximate theoretical quantilies of the deviance residual distribution, according to the fitted
model. If this looks odd then investigate further using qq.gam. Note that residuals for models fitted to binary data contain very little information useful for model checking (it is necessary to find
some way of aggregating them first), so the QQ plot is unlikely to be useful in this case.
Take care when interpreting results from applying this function to a model fitted using gamm. In this case the returned gam object is based on the working model used for estimation, and will treat
all the random effects as part of the error. This means that the residuals extracted from the gam object are not standardized for the family used or for the random effects or correlation structure.
Usually it is necessary to produce your own residual checks based on consideration of the model structure you have used. | {"url":"https://www.rdocumentation.org/packages/mgcv/versions/1.9-0/topics/gam.check","timestamp":"2024-11-05T05:44:59Z","content_type":"text/html","content_length":"77162","record_id":"<urn:uuid:1a1c84ec-6212-4ebc-8507-8f9944fbd70f>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00812.warc.gz"} |
LibGuides: Mathematics: Introduction to sine, cosine and tangent
In this module, you can study how to calculate sine, cosine and tangent for the angles of a right triangle, and how to use "SOH CAH TOA" to remember the formulae for calculating sine, cosine and
"Adjacent" and "Opposite" sides can change, depending on which angle you start from.
"Cos(x)" means "the cosine of x degrees"; it does NOT mean multiply the cosine by x.
Mathsisfun.com has written a good introductory lesson on trigonometry, which includes definitions, graphics and manipulatives. You don't need to read the whole page - only up to "Unit Circles".
Use this six-question quiz from onlinemathlearning.com for a quick online check of your trig knowledge. | {"url":"https://libguides.ucol.ac.nz/Mathematics/introductiontosinecosineandtangent","timestamp":"2024-11-07T05:50:28Z","content_type":"text/html","content_length":"38257","record_id":"<urn:uuid:db47b404-cb8e-4009-a875-8ce404eefcb0>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00794.warc.gz"} |
1 Q- Fill in the blank and 4 Questions - calculations - Assignment Essays
Fill in the blank with the correct answer from the textbook.1. A current liability that represents cash collected in advance of earning the
related revenue is called ____________________________________________.
2. Amounts owed to suppliers for goods and services that have been provided to
the entity on credit are called ______________________________________.
3. The portion of long-term debt that is to be paid within one year of the balance
sheet date is reclassified from the non-current liability section of the balance
sheet and called ______________________________________________________.
4. Total wages earned by employees for a payroll period including bonuses and
overtime are called _______________________. From this amount, deductions and
withholdings are subtracted to arrive at the _____________________________, which
is the amount recorded on the company’s balance sheet as a wages payable.
5. In order for a contingent liability to be recorded in a company’s balance sheet,
it must be both _________________________ and ____________________________.
6. When a bond’s stated rate is lower than the market rate, the bond is issued at a
______________________________. When a bond’s stated rate is higher than the
market rate, the bond is issued at a ______________________________.
Complete the problems, and show your work.
7. On Jan 1, 2023, Kayce Co. obtained a 6-month loan from its bank for $4,000,000
with an interest rate of 5%. Calculate the total interest expense due.
8. Dutton Inc. has the following payroll information for the year ended 12/31/23.
FICA is calculated at 7.65% of gross pay. Calculate the missing information.
Gross pay
FICA tax withholdings
Income tax withholdings
Group health insurance
Employee 401K contributions
Total deductions
Net pay
9. On January 1, 2023, $2 million worth of 4-year bonds with a stated rate of 6%
were issued; interest is paid semi-annually on June 30th and Dec 31st. The market
interest rates were 5% when the bonds were issued.
a. Calculate the annual interest payment on the bonds.
b. Calculate the amount of interest paid on each semi-annual payment.
c. Calculate the total amount of interest to be paid over the 4 years.
d. Were the bonds issued at a premium or a discount?
10. On January 1, 2023, $10 million worth of 5-year bonds with a stated rate of 5%
were issued; interest is paid quarterly on March 31, June 30th, Sept 30th, and Dec
31st. The market rates were 7% when the bonds were issued.
a. Calculate the annual interest payment on the bonds.
b. Calculate the amount of interest paid on each quarterly payment.
c. Calculate the total amount of interest to paid over the 5 years.
d. Were the bonds issued at a premium or a discount?
1. On January 1, 2020, $1 million worth of 10-year bonds with a stated rate of 8% were issued; interest is paid semi-annually on June 30th and Dec 31st.
The market interest rates were 9% when the bonds were issued.
a. Calculate the annual interest payment on the bonds.
b. Calculate the amount of interest paid on each semi-annual payment.
c. Calculate the total amount of interest to be paid over the 10 years.
d. Were the bonds issued at a premium or a discount?
2. On January 1, 2019, $2 million worth of 5-year bonds with a stated rate of 12% were issued; interest is paid quarterly on March 31, June 30th, Sept 30th, and Dec 31st.
The market rates were 10% when the bonds were issued.
$ 1,200,000
a. Calculate the annual interest payment on the bonds.
b. Calculate the amount of interest paid on each quarterly payment.
c. Calculate the total amount of interest to paid over the 5 years.
d. Were the bonds issued at a premium or a discount?
gross pay
net pay
20,400 Payroll Liab
Inc Tax
Payroll Withholdings Liab
Gross Pay
Inc Tax
Total Ded
Net Pay | {"url":"https://assignmentessays.com/1-q-fill-in-the-blank-and-4-questions-calculations/","timestamp":"2024-11-14T23:51:26Z","content_type":"text/html","content_length":"57172","record_id":"<urn:uuid:553d7ade-0d92-4db6-88d8-e856c8d886bb>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00258.warc.gz"} |
Correlation and Regression
Correlation measures the relationship of the process inputs (x) on the output (y). It is the degree or extent of the relationship between two variables. These studies are used to examine if there is
a predictive relationship of the input on the process.
Correlation and Regression studies are normally done together as part of the ANALYZE phase of a DMAIC project.
Couple notes:
• Predicting within the range of the data is called "interpolating"
• Predicting outside the range of the data is called "extrapolating".
Correlation studies and dependencies tend to be stronger with more data and the maximum range being applied (be aware this can also hide areas of correlation or unique relationships with subsets of
the data).
However, visualization of the data set can also show that there may exist varying relationships within the range of samples. Within a smaller specific range there could be a relationship, and then
another range could show a different relationship.
The picture to the left shows that there is very little, if any, correlation of the variables.
They are independent variables at least within the range of inputs studied and the "r" value is approximately zero.
A correlation value may be close to zero but closer review will indicate enlightening information. As mentioned earlier, be aware that sometimes too much data can hide relationships.
The point is to run the correlation visually and mathematically.
• "X" is considered the independent variable or predictor variable.
• "Y" is the dependent variable or predicted variable.
Regression and correlation involve testing a relationship rather than testing of means or variances. They are used to find out the variables and to the degree the impact the response so that the team
can control the key inputs. Controlling these key inputs is done to shift the mean and reduce variation of an overall Project "Y".
Linear Correlation
There are several correlation coefficients in use but the most frequently used is the Pearson Product Moment Correlation, also referred to as the Coefficient of Correlation (COC) that measures only a
linear relationship between two variables and is denoted by an "r" value. The formula is shown below.
The "r" value is used to measure the linear correlation and it will always range from -1.0 (anticorrelation) to +1.0. As the value approaches 0 there is less linear correlation, or dependence, of the
If the value:
• of one variable increases when the value of the other increases, they are said to be positively linearly correlated.
• of the output (y) decreases when the value of the input (x) increases, they are said to be negatively linearly correlated.
• of the output increases when as the input value increase then they are said to be positively correlated.
The degree of linear association between two variables is quantified by the COC.
Pearson's Correlation DOES NOT assume that the data is normally distributed but is strongly influenced by outliers anywhere in the data set. It is most accurate when the data sets are normally
As expected, an outlier is likely to take away from the linear association of the other non-outlying variables whether the association is negative or positive.
The data classification for each of the variables must be ratio or interval types and the relationship must be monotonic.
The "r" value represents a unitless translation of covariance, meaning the closer the value is to +1, the closer the linear relationship is between the x and y random variables.
As the value of "r" approaches zero from either side, the correlation is weaker. That is the input, x, has a lower correlation on the output, y.
This is normally shown by a x-y plot referred to as a Scatter Graph. This graph shows all the data points where the input, x, is varied systematically and the output, or the effect, of y is measured.
A "r" value of +1.0 indicates a perfect and strong POSITIVE correlation.
A "r" value of -1.0 indicates a perfect and strong NEGATIVE correlation or anticorrelation.
A data set that does not have a slope (slope = 0) will have a correlation coefficient that is undefined because the variance of Y is zero. In other words, the output is not affected by any of the
input values.
Shown below in the video is an example starting with a set of data and progressive steps to manually calculate the LINEAR correlation coefficient, "r".
This is a study between the number of caterpillars in a cabbage patch and the quantity of cabbages destroyed.
Non Linear Correlation
The picture below indicates a strong relationship that would not be evident by simply analyzing the "r" value. The "r" value is going to be close to zero which means the variables are independent.
Recall, the "r" value is measure of linear association only.
There is another measurement that explains association. Visit Spearman's Rho Correlation Coefficient for an explanation of the monotonic association strength between two variables.
However, when it comes to data similar to the picture below there is strong indication that an association exist but it is non-linear.
This module doesn't investigate non-linear mathematical relationships but it is important to understand they exist as the picture below shows (which is non-linear and non-monotonic).
More about COC and COD
What is the difference between the Coefficient of Correlation (COC) and Coefficient of Determination (COD)?
The COD ranges from 0-1 (0%-100%).
The COD is the proportion of variability of the dependent variable (Y) accounted for or explained by the independent variable (x) equal to the COC value squared.
In other words, it is the percentage of variation in Y explained by the linear relationship with X.
The COC is a value from -1 to +1 that describes the linear correlation of the dependent and independent variable. A value near zero indicates no linear relationship.
The sign is necessary to see if relationship is positive or negative so solving for COR by taking the square root of COD may not give the correct correlation since the sign can be positive or
Correlation interpretations from data or graphs can be wrong if it is purely coincidental.
Regardless of how strong (positive or negative) it may appear, Correlation never implies causation. There could be other variables behind the one charted that could be a factor.
For example, a chart or correlation value may indicate a strong relationship (linear or non-linear) but in reality there may be no relationship or dependency at all.
Just like most statistical results they must be reviewed subjectively with consideration of common sense. This is done with the Six Sigma team. The GB/BB is responsible for sharing the results in any
way to help the team make the right decisions.
It is possible to have the same "r" value and have several different graphical representations, another reason to review the scatter plot and "r" value together.
Practice Problems
Find r if the COD is 0.85?
From the information earlier, you can solve for r which equals the square root of the COD. Therefore the square root of 0.85 is 0.922 or -0.922.
Which of the statement(s) is/are true if the COD = 0.85?
1) 85% of the variation is explained by the regression model
2) The correlation can be positive or negative
Both responses are true.
Hypothesis Testing
Below is an example of monthly results of cereal sales related to marketing dollars. The intention is to determine the degree of linear correlation between marketing dollars spent to cereal sales.
The data was compiled and is shown below.
Visually depicting the data is recommended whether is it time-series charts, scatter plots, or box plots. This helps in seeing trends and overall behavioral relationships between data. A couple
graphs of the data are shown below. The scatter plot shows quickly that there appears to be a strong linear correlation.
Assessing the Correlation
Establish the Practical Problem
Is there a relationship between the amount of money dedicated to marketing to the sales of cereal and what is the strength of the relations?
Establish the Statistical Problem
H[o]: Sales and Marketing dollars spent are not correlated
H[A]: Sales and Marketing dollars spent are correlated
Choose a Level of Significance
Alpha risk selected is 0.05
If the calculated p-value is <0.05, then the reject the Ho (null) and infer the Ha.
The sample size = 12
Using Minitab
Find Correlation from the pull-down menu and enter both continuous sets of data and use Pearson Correlation, then the results are shown.
P value = 0.000
r = 0.9851 or 98.51%
With those results, reject the null and infer that there is a statistically significant correlation (which is the alternative hypothesis).
The linear correlation between the marketing dollars spent and resulting cereal sales is strong within a given month. The correlation coefficient (r) = 0.9851 within the inference range of $2,548 to
$8,023 marketing dollars analyzed. This is a strong positive correlation. The more marketing, the higher the cereal sales.
Likely, at some point, that cereal sales would level off regardless of how high the amount of marketing dollars. That is why it is very important to keep your conclusion with the inference range.
Another method to perform the statistical evaluation is by comparing the r- calculated value of 0.9851 to the r-critical value.
The r-critical value for a sample size of 12 at alpha risk of 0.05 is 0.4973.
If the value of r-calculated is >0.4973, then there is a statistical significant correlation and in this example that is clearly the case.
Regression takes it a step further and develops a formula to describe the nature of the relationship. Visit the Regression module for more information.
As indicated earlier, the scatter plot should be visually examined. Even if the correlation coefficient was very low (linear relationship) there may be a non-linear relationship such as cubic or
quadratic that could be very strong.
Correlation Coefficient in Excel
Finding the Pearson Correlation Coefficient of two sets of data is done in Excel as shown below. The data does not have to be normally distributed but do have to be equal sample sizes.
The Pearson Correlation Coefficient between these two sets of data is -0.2636, a weak negative correlation.
How does Correlation relate to Regression?
Recall that Correlation indicates the amount of linear association that exists between two variables in the form of a value between -1.0 to 1.0.
Such as the linear correlation from earlier example where the value of -0.2636 was found and indicates a negative correlation but it is not very strong.
Regression provides an equation describing the nature of relationship such as y=mx+b.
There are various types of Regression:
Simple Linear Regression
Single regressor (x) variable such as x[1] and model linear with respect to coefficients.
Multiple Linear Regression
Multiple regressor (x) variables such as x[1], x[2]...x[n] and model linear with respect to coefficients.
Simple Non-Linear Regression
Single regressor (x) variable such as x and model non-linear with respect to coefficients.
Multiple Non-Linear Regression
Multiple regressor (x) variables such as x[1], x[2]...x[n] and model nonlinear with respect to coefficients.
Correlation and Regression Download
There of two module of slides that provide additional insight into Correlation and Regression. This is critical component of statistical analysis and can quickly provide answers about the inputs
and their effect on the outputs. These tools are frequently used in the DMAIC journey.
Click here to see the Correlation and Regression module and view others that are available.
How does Correlation relate to Covariance?
While the Covariance indicates how well two variables move together, Correlation provides the strength of the variables and is a normalized version of Covariance. They both will always have the same
sign: positive, negative, or 0.
Covariance is the numerator in the equation below therefore if the standard deviations of x and y are constant, as the Covariance increases, the Correlation also increases and approaches +1.0. Also,
if the Covariance decreases, the Correlation decreases and approaches -1.0.
Correlation is a dimensionless value that will always be between -1.0 and +1.0, with 0 indicating the two variables move randomly from each other and are uncorrelated. Values closer to 0 (either
negative or positive indicate weaker and weaker correlation.
As Covariance increases (also as Correlation values approach +1.0) this indicates a stronger and stronger positive relationship of the variables moving together. As Covariance decreases (also as
correlation values approach -1.0) this indicates a stronger inverse relationship.
Values near zero for both parameters equates to no relationship or correlation and therefore those inputs or combination of inputs are not related to the output. This is valuable to the Six Sigma
team so this input can be ruled out (unless it has a impact as in a combination with another input).
The following formula illustrates the relationship of the two terms. The formula below applies for sample and population calculations.
Return to BASIC STATISTICS
Return to the ANALYZE Phase
Templates, Tables, and Calculators
Search Six Sigma related job openings
Return to Six-Sigma-Material Home Page | {"url":"https://www.six-sigma-material.com/Correlation.html","timestamp":"2024-11-03T00:36:34Z","content_type":"text/html","content_length":"64270","record_id":"<urn:uuid:2c07f044-5353-4e58-81ca-47eea1673e84>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00246.warc.gz"} |
C program to calculate Simple Interest with explanation - Quescol
C program to calculate Simple Interest with explanation
In this tutorial, we are going to learn the writing program in c to calculate the Simple Interest. Simple Interest is a mathematical term and we have a mathematical formula for this calculation.
Before starting writing the program let’s understand simple interests.
What is Simple Interest?
Simple interest is a method to calculate the interest charged on a loan. It is calculated by multiplying the interest rate by the principal by the number of days that elapse between payments.
Simple Interest Formula:
SI = (P*T*R)/100
P is principle amount
T is the time duration
R is the rate of interest
Example of how we can calculate simple interest
Enter principle: 1200
Enter time: 2
Enter rate: 5.4
Simple Interest = 129.600006
Logic we are following to calculate simple interest
• We are taking Principle amount, time and rate of interest as an input.
• Now after taking input we are simply calculating simple interest using formula SI = (principle * time * rate) / 100.
• After calculating SI using above formula we are printing the value which is our program output.
Program in C to calculate Simple Interest
#include <stdio.h>
int main()
float p, t, r, SI;
printf("Program to calculate Simple Interest\n");
printf("Enter principle amount: ");
scanf("%f", &p);
printf("Enter time (in years): ");
scanf("%f", &t);
printf("Enter rate: ");
scanf("%f", &r);
SI = (p * t * r) / 100;
printf("Simple Interest = %.2f", SI);
return 0;
Program to calculate Simple Interest
Enter principle amount: 1000
Enter time (in years): 3
Enter rate: 5
Simple Interest = 150.00 | {"url":"https://quescol.com/interview-preparation/simple-interest-program-in-c","timestamp":"2024-11-13T11:58:20Z","content_type":"text/html","content_length":"83943","record_id":"<urn:uuid:cf8fdb60-a414-4f0e-94a8-e332e7edc90e>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00026.warc.gz"} |
Linear Transformation in Algebra
Podcast Beta
Play an AI-generated podcast conversation about this lesson
What is the condition for matrix multiplication to be possible?
The number of rows in both matrices must be equal
The number of columns in the first matrix must match the number of rows in the second matrix (correct)
The number of rows in the first matrix must match the number of columns in the second matrix
The number of columns in both matrices must be equal
What is the relation between the kernel and image of a linear transformation?
The kernel is a subspace of the image
The image is a subspace of the kernel
The kernel and image are parallel to each other
The kernel and image are orthogonal to each other (correct)
What is an example of a vector operation that can be performed by a linear transformation?
Rotation (correct)
What does a determinant of 0 indicate about a linear transformation?
Signup and view all the answers
What can the determinant of a matrix be used for?
Signup and view all the answers
What is the condition for a set of vectors to be linearly independent?
Signup and view all the answers
What is the unique property of a matrix representation of a linear transformation?
Signup and view all the answers
What is the dimension of a vector space?
Signup and view all the answers
What does the rank-nullity theorem state?
Signup and view all the answers
What is the equation used to find the eigenvalues of a matrix?
Signup and view all the answers
Study Notes
Linear Transformation
Matrix Multiplication
• A linear transformation can be represented as a matrix multiplication
• Matrix multiplication is not commutative, i.e., AB ≠ BA
• The number of columns in the first matrix must match the number of rows in the second matrix
• The resulting matrix has the same number of rows as the first matrix and the same number of columns as the second matrix
Image and Kernel
• Image: The set of all output vectors resulting from the linear transformation
• Kernel (or Null Space): The set of all input vectors that result in the zero output vector
• The kernel is a subspace of the domain, and the image is a subspace of the codomain
• The kernel and image are orthogonal to each other
Vector Operations
• Scaling: A linear transformation can scale a vector by a scalar value
• Reflection: A linear transformation can reflect a vector across a line or plane
• Projection: A linear transformation can project a vector onto a line or plane
• Rotation: A linear transformation can rotate a vector by a certain angle
• The determinant of a matrix represents the scaling factor of the linear transformation
• A determinant of 0 indicates that the linear transformation is not invertible (i.e., it's not one-to-one)
• A determinant of 1 indicates that the linear transformation preserves the magnitude of the input vectors
• The determinant can be used to find the inverse of a matrix, if it exists
Linear Transformation
Matrix Representation
• A linear transformation can be represented as a matrix multiplication, which is a powerful tool for performing transformations
• However, matrix multiplication is not commutative, meaning the order of matrices matters: AB ≠ BA
• For matrix multiplication to be possible, the number of columns in the first matrix must match the number of rows in the second matrix
• The resulting matrix has the same number of rows as the first matrix and the same number of columns as the second matrix
Image and Kernel
• The image of a linear transformation is the set of all possible output vectors
• The kernel (or null space) of a linear transformation is the set of all input vectors that result in the zero output vector
• Both the kernel and image are subspaces, with the kernel being a subspace of the domain and the image being a subspace of the codomain
• The kernel and image are orthogonal to each other, meaning they have a 90-degree angle between them
Effects on Vectors
• A linear transformation can scale a vector by a scalar value, changing its magnitude
• A linear transformation can reflect a vector across a line or plane, changing its direction
• A linear transformation can project a vector onto a line or plane, changing its direction and magnitude
• A linear transformation can rotate a vector by a certain angle, changing its direction
• The determinant of a matrix represents the scaling factor of the linear transformation it represents
• A determinant of 0 indicates that the linear transformation is not invertible (not one-to-one)
• A determinant of 1 indicates that the linear transformation preserves the magnitude of the input vectors
• The determinant can be used to find the inverse of a matrix, if it exists, allowing us to reverse the transformation
Linear Independence
• A set of vectors is linearly independent if the only solution to the equation c1v1 + c2v2 +...+ cnvn = 0 is c1 = c2 =...= cn = 0
• Linear independence means that a linear combination of vectors results in the zero vector only if all coefficients are zero
Linear Transformations
• A linear transformation is a function between vector spaces that preserves vector addition and scalar multiplication
• A linear transformation can be represented by a matrix, and its matrix representation is unique for a given basis
• The kernel is the set of vectors that map to the zero vector, while the image is the set of vectors that can be obtained by applying the transformation
• The rank is the dimension of the image, and the nullity is the dimension of the kernel
Span and Basis
• The span of a set of vectors is the set of all linear combinations of the vectors
• A basis is a set of vectors that spans the vector space and is linearly independent
• A basis can be used to represent every vector in the vector space, and the dimension of a vector space is the number of vectors in a basis
• A standard basis consists of unit vectors aligned with the coordinate axes
Dimension and Rank
• The dimension of a vector space is the number of vectors in a basis
• The rank of a matrix is the maximum number of linearly independent rows or columns
• The nullity of a matrix is the number of linearly independent solutions to the equation Ax = 0
• The rank-nullity theorem states that the rank of a matrix plus the nullity of a matrix is equal to the number of columns
• The dimension theorem states that the dimension of a vector space is equal to the rank of a matrix representation of a linear transformation
Eigenvalues and Eigenvectors
• An eigenvalue is a scalar that satisfies the equation Ax = λx for some non-zero vector x
• An eigenvector is a non-zero vector that satisfies the equation Ax = λx for some scalar λ
• The eigenvalue equation is Ax = λx, and the characteristic equation is det(A - λI) = 0
• The eigendecomposition of a matrix is a diagonal matrix consisting of the eigenvalues and a matrix of eigenvectors
Studying That Suits You
Use AI to generate personalized quizzes and flashcards to suit your learning preferences.
This quiz covers the basics of linear transformation, including matrix multiplication, image and kernel. Test your understanding of this fundamental concept in algebra. | {"url":"https://quizgecko.com/learn/linear-transformation-in-algebra-b1paxs","timestamp":"2024-11-06T21:33:56Z","content_type":"text/html","content_length":"329485","record_id":"<urn:uuid:9cbf4f88-a507-4cde-a2c4-38b91085a05f>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00585.warc.gz"} |
ECCC - Reports tagged with tensors
We study the problem of obtaining efficient, deterministic, black-box polynomial identity testing algorithms for depth-3 set-multilinear circuits (over arbitrary fields). This class of circuits has
an efficient, deterministic, white-box polynomial identity testing algorithm (due to Raz and Shpilka), but has no known such black-box algorithm. We recast this problem as ... more >>> | {"url":"https://eccc.weizmann.ac.il/keyword/17788/","timestamp":"2024-11-14T14:21:55Z","content_type":"application/xhtml+xml","content_length":"20252","record_id":"<urn:uuid:8e2c031f-eaaf-4458-96a6-954e24020efa>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00223.warc.gz"} |
Theorem 3.5.6.17. Let $X$ be a Kan complex. For every $n \geq 0$, the comparison map $v: X \rightarrow \pi _{\leq n}(X)$ of Construction 3.5.6.10 exhibits $\pi _{\leq n}(X)$ as a fundamental
$n$-groupoid of $X$.
Proof. It follows from Remark 3.5.6.11 that $v$ is bijective on $m$-simplices for $m < n$ and surjective on $n$-simplices. By construction, if $\sigma $ and $\sigma '$ are $n$-simplices of $X$, then
$v(\sigma ) = v(\sigma ')$ if and only if $\sigma $ and $\sigma '$ are homotopic relative to $\operatorname{\partial \Delta }^{n}$. It will therefore suffice to show that $\pi _{\leq n}(X)$ is an
$n$-groupoid, which follows from Corollary 3.5.6.16. $\square$ | {"url":"https://kerodon.net/tag/054M","timestamp":"2024-11-11T19:36:06Z","content_type":"text/html","content_length":"10061","record_id":"<urn:uuid:cb9c8428-312c-4ac7-b080-7165863f91fe>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00727.warc.gz"} |
Show Posts - iamcpc
« on: September 19, 2019, 06:00:14 PM »
I have a question about a very specific FE model.
1. The base model is the flat disk ice wall edge model.
2. Within the subset of this model this is the no dome subset model.
3. Within the subset of the flat disk great ice wall no dome model this is the subset where the ice wall has an edge with space outside of the edge. The Earth is finite.
4. Within the rules defined above this model also has has UA as a gravity model.
5. In addition this model does not have a firmament
Now that I have outlined the basics of this model my question is this:
If there is no dome/firmament and the earth had an edge outside of the ice wall and the earth is accelerating upwards is there any documentation, ideas, or theories on what is preventing the air from
just flowing off the edge?
In the dome/firmament models that has been used as an explanation as to what prevents the atmosphere from just blowing away. | {"url":"https://forum.tfes.org/index.php?PHPSESSID=hnfhb7r5ns53ldg5obf4v5hgme&action=profile;area=showposts;sa=topics;u=10797","timestamp":"2024-11-09T17:23:01Z","content_type":"application/xhtml+xml","content_length":"21087","record_id":"<urn:uuid:1ac897fe-0506-45c9-874b-160a6040d5f9>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00767.warc.gz"} |
R help
Tuesday April 30 2002
Time Replies Subject
8:51PM 4 generating graphical output when DISPLAY is not set?
8:32PM 0 Examples of hypothesis testing Bryan Moss
8:15PM 0 I: A sample question
7:32PM 3 A sample question
6:07PM 0 Re What am I doing wrong with xyplot?
6:01PM 0 What am I doing wrong with xyplot?
5:58PM 0 R-1.5.0 and JPEG
5:38PM 1 JPEG library wierdness
5:35PM 1 update from 1.41. to 1.5.0
5:21PM 0 Matching a geometric distribution
4:23PM 1 regression, for loop
3:08PM 3 rbind'ing empty rows in dataframes in 1.4.1 versus 1.5.0
2:58PM 1 data.frame package?
2:47PM 2 display of character NA's in a dataframe in 1.5.0
2:18PM 1 followup -- deficiencies in readline capability
8:23AM 0 trouble with R-1.5.0 and readline.h at an odd place on SGI
8:06AM 1 MemoryProblem in R-1.4.1
7:03AM 3 Labeling matrix data
Monday April 29 2002
Time Replies Subject
10:52PM 2 cluster analyses
10:37PM 2 efficiency
10:08PM 2 calling optim from external C/C++ program
9:50PM 0 deficiencies in readline capability
8:59PM 0 Question on integrateing R to our system
7:43PM 3 how to trap any warnings from an R function
6:38PM 1 I: Problem
6:19PM 1 Garbage collection: RW1041
5:49PM 2 RPart
5:49PM 2 Lost Tcl/Tk support
5:41PM 0 building R-1.5.0 on SGI/IRIX
5:32PM 0 code optimization
4:46PM 1 data
4:10PM 1 who has experience to do peakfit using R?
3:13PM 1 New Realease
3:12PM 1 Release of Design library; update of Hmisc library
3:08PM 0 Hmisc library
3:01PM 3 Organizing the help files in a package
1:28PM 1 masking functions
12:56PM 2 Lotos 1-2-3 date to POSIXct
12:30PM 1 Circular graphics
10:15AM 2 append with write.table()
7:38AM 0 test
12:27AM 3 ifelse versus ...
Sunday April 28 2002
Time Replies Subject
11:51PM 2 dropterm() in MASS
11:43PM 1 Building rgui with Visual C/C++ 6
11:35PM 2 Building Rgui.exe with Visual Studio
10:16PM 0 Compile R with Intel compilers on Linux?
9:01PM 0 rank
12:36PM 2 Image processing? Manipulating image data in R?
6:41AM 2 rpart problem
Saturday April 27 2002
Time Replies Subject
7:26PM 2 Explanation of Error Diagnostics
3:54PM 2 S & R list virus warning
Friday April 26 2002
Time Replies Subject
9:14PM 1 Error in ORDGLM function
8:42PM 3 different data series on one graph
7:11PM 1 ORDGLM function - which package has it?
6:51PM 1 Problem with read.xport() from foreign package
6:37PM 1 optim or nlm with matrices
4:18PM 0 [Fwd: Re: degrees of freedom for t-tests in lme]
3:28PM 0 [OT] Inverting sparse matrices
3:19PM 2 Can't install packages (PR#1486)
1:59PM 4 SAS and R
1:42PM 2 quadratic discriminant analysis?
1:39PM 2 Spearman Correlation
11:45AM 7 spreadsheet data import
7:39AM 1 truncated observed
12:50AM 1 optimization of R on SGI/IRIX
Thursday April 25 2002
Time Replies Subject
10:57PM 1 simple bar plot with confidence interval
10:05PM 1 Rdbi package and PgSQL
5:39PM 0 Abilities of R
3:25PM 4 sum() with na.rm=TRUE, again
12:23PM 2 install a package from CRAN
9:56AM 1 polyclass
9:50AM 0 Crossed random effects
9:20AM 3 Kendall's tau
7:20AM 1 An unexpected exception has been detected in native code outside the VM
Wednesday April 24 2002
Time Replies Subject
7:55PM 1 Newton-Raphson
6:32PM 0 Boxplot
5:35PM 2 Regarding pp.plot(){CircStats}
5:16PM 0 degrees of freedom for t-tests in lme
2:29PM 1 pooling categories in a contingency table
1:39PM 0 can not compile R-1.4.1 on OSF5.1
10:52AM 0 Everything You Need for JEdit/R Edit Mode
9:41AM 3 nonlinear least squares, multiresponse
8:26AM 0 Which platform the R-JAVA support ??
8:11AM 2 Multiple frequencies
7:49AM 3 Dummy newbie question
7:20AM 2 Changing the colour in boxplots using bwplot()
Tuesday April 23 2002
Time Replies Subject
11:19PM 2 Asking about how to use R to draw Time Series graph
11:19PM 2 Install
11:05PM 3 stacking vectors of unequal length
8:08PM 1 column-plot of rainfall data
4:09PM 0 New Course*** R/S-Plus Programming Techniques II, May 20-21
4:05PM 1 Writing text in lattice graphics
3:36PM 3 error loading huge .RData
3:25PM 0 Website ? FAST !
2:41PM 0 re| `Upgrading to 1.4.1 seems to work'
9:51AM 0 Improving R Editing: JEdit Support for R
9:35AM 1 Tree package on R 1.4.1
8:00AM 2 kaplanMeier & censorReg
7:20AM 1 Re: R-help Digest V2 #710
4:23AM 1 Use of nls command
4:20AM 1 Lattice graphics on Mac OS X
1:17AM 3 Subsetting by a logical condition and NA's
1:04AM 0 Summary: Multidimensional scaling
Monday April 22 2002
Time Replies Subject
11:40PM 1 Goodness-of-fit
8:31PM 2 lattice x(y)lab and expression
6:31PM 1 recording an R session
5:12PM 2 Problem passing data into read.table()
3:51PM 3 glm() function not finding the maximum
3:38PM 2 lattice help
1:06PM 2 skipping specific rows in read.table
8:44AM 0 Problem with logarithmic axes labelling
7:56AM 2 .RData
7:21AM 0 PP.test
4:14AM 2 how can a function tell if defaults are used?
Sunday April 21 2002
Time Replies Subject
11:04PM 0 New R CGI gateway: TKU-Stat
6:16PM 2 updating R - old version still runs
2:33PM 2 Simple Coding problem ?
10:00AM 0 Problems with call R from java
Saturday April 20 2002
Time Replies Subject
6:58PM 3 Problem with a matrix
2:52PM 2 integration of a discrete function
Friday April 19 2002
Time Replies Subject
10:37PM 4 Multidimensional scaling
9:07PM 2 exponential smoothing
8:26PM 1 2D cluster of 2D matrix in R?
7:44PM 0 Your mail sent to mailbase.ac.uk with subject Meeting notice
5:11PM 1 FW: Problem compiling on HP-UX 10.20
3:03PM 1 R and OS X
2:56PM 2 y-intercept forcing
1:11PM 1 trouble with tcltk (was RE: trouble compiling R on Irix )
10:50AM 2 merge
10:42AM 2 Problem with installation on Mandrake 8.1
4:48AM 4 Durbin-Watson test in packages "car" and "lmtest"
Thursday April 18 2002
Time Replies Subject
10:48PM 3 Variable definition problem
8:05PM 2 naming boxes in boxplot
6:46PM 1 About duplicates
6:33PM 1 trouble with tcltk (was RE: trouble compiling R on Irix)
4:42PM 5 Two problems
4:06PM 1 Help with lme basics
4:01PM 0 Re: printing tree results
3:14PM 2 No subject
2:44PM 1 trouble compiling R on Irix
2:16PM 1 Problem compiling on HP-UX 10.20
12:07PM 1 lattice
12:00PM 2 using R with c++
11:03AM 2 Data.Frame Multiplication
10:58AM 2 Background in lattice plots using dotplot()
7:48AM 1 grid lines outside plot region in version R1.4.1
4:52AM 0 C++ and R in Solaris
4:43AM 0 R and C++ in Solaris
3:16AM 4 Silly Question
12:51AM 2 Changing tick mark labels
Wednesday April 17 2002
Time Replies Subject
11:34PM 0 User defined macros for Rd-files?
10:26PM 2 placing objects with format statements into text file
10:13PM 1 Problems embedding R in a C application
8:24PM 1 zero center a group of variables
7:12PM 1 concat
6:53PM 2 nls error control
4:30PM 1 Stochastyic frontier regression
2:55PM 2 Cross-correlation
2:24PM 4 Problem w/ axis and distortion in a plotting function
2:23PM 1 Installation of R-1.4.1 on Solaris 2.7
1:56PM 1 rbind() very slow
1:24PM 4 union of lists
12:34PM 0 using the netCDF.zip package in Windows (XP).
10:39AM 0 still have problem with krige and border option [end]
9:45AM 0 Converter issues
7:38AM 1 No output from (lattice) xyplot called within loops
Tuesday April 16 2002
Time Replies Subject
11:51PM 0 k-Nearest Neighbour density estimation
8:29PM 2 passing ", betrayed by the non-vanishing \
7:06PM 1 [Fwd: Re: Multithreading]
4:40PM 0 sum(NA,na.rm=T) - answered my own question - sorry
4:34PM 0 sum(NA,na.rm=T)
4:08PM 0 still have problem with krige and border option
3:13PM 6 Classification Analysis
11:24AM 1 Benchmarks
9:28AM 2 Multithreading
7:32AM 1 symbols in lattice
7:29AM 0 Book
7:20AM 1 Can one aply 'mgp' to individual axis?
6:36AM 2 multiple plot devices
4:23AM 1 Problem with dyn.load()
3:34AM 1 draw.tree help
Monday April 15 2002
Time Replies Subject
9:31PM 0 Problem with read.table()
9:20PM 1 simple Q: required sample size & usage of power.t.test()
6:15PM 8 Problem
4:45PM 1 Nested ANOVA with covariates
4:41PM 0 error fitting lme model
3:11PM 0 eGRACH
3:03PM 2 krige and polygon limit problem
2:12PM 1 nested anova not giving expected results
12:58PM 1 glm link = logit, passing arguments
12:54PM 6 two questions
12:53PM 2 Newbie problem with ox package
12:29PM 3 Greek in text()
10:37AM 1 required sample size
9:33AM 1 Re: Writting R Function
8:47AM 1 vector average
1:12AM 0 ANN: RPy (R from Python) 0.2
Sunday April 14 2002
Time Replies Subject
6:00PM 1 Suggestion for implementation
12:30PM 0 gls
Saturday April 13 2002
Time Replies Subject
10:42PM 0 grayscale postscript plots
10:28PM 2 trouble getting output from graphs, again
9:48PM 1 save.image() issue resolved - I hope
6:18PM 1 save.image() error reappears
11:40AM 1 Follow-up: pos= in library and require
Friday April 12 2002
Time Replies Subject
5:20PM 0 plot-history
4:06PM 5 How to specify search order for require()
3:57PM 0 documentation widget
3:37PM 3 add columns to a data.frame
2:57PM 1 Parrot
2:42PM 1 summary: Generalized linear mixed model software
2:37PM 2 Help
1:50PM 2 Lattice Package...
12:20PM 1 correlated binary random numbers
12:19PM 0 Summary: Obtaining names of ``...'' arguments.
11:42AM 1 Problems with memory
11:08AM 1 xgobi
10:39AM 0 RJava , bitmap ,PNG
9:10AM 1 persp(): z-axis annotation overwrites numbers at tickmark
8:12AM 4 Matrix to data.frame without factors
Thursday April 11 2002
Time Replies Subject
7:24PM 1 Need help decyphering 'make check' errors
7:21PM 2 Obtaining names of ``...'' arguments.
4:41PM 14 Ordinal categorical data with GLM
3:13PM 0 Odp: graphics in batch mode
1:59PM 2 time (or output of function) in the R prompt
1:42PM 6 extract week from date
1:17PM 2 "CTRL-C" and "try"
12:35PM 3 graphics in batch mode
9:09AM 3 new acf package
8:18AM 0 unsusbscribe
4:07AM 3 Help installing from Rpm
1:24AM 3 paste dataframe
Wednesday April 10 2002
Time Replies Subject
11:50PM 0 programmatic installation
10:45PM 1 Layout of Fourier frequencies
10:43PM 4 Principal Component analysis question
4:59PM 11 Newsgroup
4:27PM 1 No subject
2:23PM 2 R cross-platform compatibility--wow!
12:12PM 0 foreign/write.table
12:00PM 3 problem with do.call
8:18AM 1 New Package: ipred - Improved predictors
1:31AM 0 Discriminant Adaptive Nearest Neighbor
Tuesday April 9 2002
Time Replies Subject
10:34PM 0 installation
10:10PM 1 write.table
5:13PM 3 dynamically including R-code into R-code
5:11PM 0 SJava! java.library.path. A next simple configuration question.
4:27PM 3 expressions on graphs
4:16PM 1 factanal prediction
4:12PM 0 summer R job in Oregon
3:37PM 1 restoring .RData with older version of R
3:22PM 0 couldn't find function "nclass.fd"
3:15PM 1 Mixture Modeling in R
2:13PM 1 how to deal with singularities in lm()
12:46PM 6 matrix dimension and for loop
12:45PM 3 readline editor
8:05AM 1 Fortran (77) in R
3:27AM 2 Restricted Least Squares
Monday April 8 2002
Time Replies Subject
11:46PM 1 factor labels in model.frame
11:24PM 1 Still having a problem with Rcmd - TMPDIR
9:13PM 1 Problem(?) in strptime() -- short version
7:51PM 1 Error in nlme ranef plot()
7:33PM 1 Problem(?) in strptime()
5:53PM 4 pooling categories in a table
5:23PM 4 Missing data and Imputation
3:12PM 2 changing the form of a list
2:14PM 1 glmm
2:11PM 2 user coordinates and rug plots in lattice graphics
12:29PM 0 example of exponential regression
10:38AM 2 subsetting with NA's
6:10AM 0 LOCFIT and survival function
Sunday April 7 2002
Time Replies Subject
10:48PM 0 help with the "pch" option in plot.locfit
1:41PM 3 German umlaut in xlab
1:48AM 0 save.image() followup
1:39AM 2 save.image() error
Saturday April 6 2002
Time Replies Subject
11:07PM 0 An R (or S) to C/C++ translator?
8:41PM 1 read.table and trouble
7:42PM 0 Keep labels when importing data from SPSS?
5:03AM 0 white in the default palette
Friday April 5 2002
Time Replies Subject
8:23PM 1 randomForest() segfaults under Solaris(SPARC) 2.7
6:55PM 1 Rgui.exe cannot start on Win2000 professional vesion!
2:54PM 1 rbind(NULL,NULL)
9:02AM 0 uncorrect image dimension
7:19AM 2 weighted 2 or 3 parameter weibull estimation?
Thursday April 4 2002
Time Replies Subject
8:33PM 2 summary on predict with arima0
2:58PM 1 html documentation bug in: help(par), 'las'
2:35PM 1 Something like se.contrast.lme?
1:04PM 0 ozone.xy
12:01PM 1 Histograms with coplot
11:45AM 0 Basle/ Allerød: Survival Analysis in S-PLUS with Terry Therneau
12:19AM 2 non-32-bit integer problem on SUN-Blade
Wednesday April 3 2002
Time Replies Subject
10:50PM 0 Another question on locfit
8:37PM 0 Linear extrapolation; revision to earlier posting
6:01PM 1 arima0 with unusual poly
5:33PM 1 optim()
4:04PM 1 libraries in $HOME/lib
2:44PM 2 R package organization
2:15PM 3 Segmentation fault with xyplot
1:42PM 1 Still needing help with LDA
10:26AM 4 Text Labels on plots in R
9:39AM 3 non-stationary covariance
7:59AM 0 Great Replies! Give me some time...
2:32AM 0 help on lme and variance estimation
Tuesday April 2 2002
Time Replies Subject
11:51PM 2 Trouble with R and cronjobs
8:53PM 1 predict with arima0
6:58PM 1 Extract psuedo model matrix from nls?
3:26PM 2 label tickmarks in persp()-plot
3:22PM 2 random forests for R
2:25PM 3 A Few Suggestions to help out newbies
1:36PM 1 compile C code
1:23PM 1 newbee to XML library
12:51PM 1 "Large" data set: performance issue
10:40AM 1 Repeated aov residuals
9:57AM 1 R-geostatistica
6:59AM 10 A request
5:26AM 1 cbind.ts bug?
3:04AM 4 Two R sessions?
Monday April 1 2002
Time Replies Subject
11:53PM 0 Compiling R for Solaris 8/Intel CPU
10:44PM 0 something confusing about stepAIC
8:51PM 2 creating an output file name using cat
6:47PM 0 RE: [S] R for UNIX on Intel platform
2:49PM 2 writing a package for generalized linear mixed modesl | {"url":"https://thr3ads.net/r-help/2002/04","timestamp":"2024-11-03T23:19:13Z","content_type":"text/html","content_length":"68446","record_id":"<urn:uuid:e13186a5-c03e-4d49-a431-a82232d10d57>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00877.warc.gz"} |
How to Calculate Future Value.
Future value refers to a method of calculating how much the present value of an asset or cash will be worth at a future date based on an assumed rate of growth.
The future value is important to investors and financial planners as they use it to estimate how much an investment made today will be worth in the future.
Formula to calculate future value.
r is the interest rate she will earn on the money.
n is the number of periods of the investment.
If you have $5,000 and expect to earn 5% interest on that sum each year for the next three years. Determine the future value.
Therefore, the future value is $ 7,588.125. | {"url":"https://www.learntocalculate.com/calculate-future-value/","timestamp":"2024-11-05T03:11:27Z","content_type":"text/html","content_length":"56628","record_id":"<urn:uuid:250205d3-449b-43ff-b929-e1d595d5ec90>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00647.warc.gz"} |
Tensor Computation Examples
Tensors can be used to express machine-learned models such as neural nets, but they can be used for much more than that. The tensor model in Vespa is powerful, since it supports sparse dimensions,
dimension names and lambda computations. Whatever you want to compute, it is probably possible to express it succinctly as a tensor expression - the problem is learning how. This page collects some
real-world examples of tensor usage to provide some inspiration.
Tensor playground
The tensor playground is a tool to get familiar with and explore tensor algebra. It can be found at docs.vespa.ai/playground. Below are some examples of common tensor compute operations using tensor
functions. Feel free to play around with them to explore further:
Values that depend on the current time
In an ecommerce application you may have promotions that sets a different product price in given time intervals. Since the price is used for ranking, the correct price must be computed in ranking.
Can tensors be used to specify prices in arbitrary time intervals in documents and pick the right price during ranking?
To do this, add three tensors to the document type as follows:
field startTime type tensor(id{}} {
indexing: attribute
field endTime type tensor(id{}} {
indexing: attribute
field price type tensor(id{}} {
indexing: attribute
Here the id is an arbitrary label for the promotion which must be unique within the document, and startTime and endTime are epoch timestamps.
Now documents can include promotions as follows (document JSON syntax):
"startTime": { "cells": { "promo1": 40, "promo2": 60, "promo3": 80 }
"endTime": { "cells": { "promo1": 50, "promo2": 70, "promo3": 90 }
"price": { "cells": { "promo1": 16, "promo2": 18, "promo3": 10 }
And we can retrieve the currently valid price by the expression
reduce((attribute(startTime) < now) * (attribute(endTime) > now) * attribute(price), max)
This will return 0 if there is no matching interval, so a full expression will probably wrap this in a function and check if it returns 0 (using an if expression) and return the default price of that
product otherwise.
To see why this retrieves the right price, notice that (attribute(startTime) < now) is a shorthand for
join(attribute(startTime), now, f(x,y)(x < y))
That is joining all the cells of the startTime tensor by the zero-dimensional now tensor (i.e a number), and setting the cell value in the joined tensor to 1 if now is larger than the cell timestamp
and 0 otherwise. When this tensor is joined by multiplication with one that has 1's only where now is smaller, the result is a tensor with 1's for promotion id's whose interval is currently valid and
0 otherwise. Then we can just join by multiplication with the price tensor to get the final tensor (on which we just pick the max value to retrieve the non-zero value.
Play around with this example in the playground
Adding scalars to a tensor
A common situation is that you have dense embedding vectors to which you want to add some scalar attributes (or function return values) as input to a machine-learned model. This can be done by the
following expression (assuming the dense vector dimension is named "x":
concat(concat(query(embedding),attribute(embedding),x), tensor(x[2]):[bm25(title),attribute(popularity)], x)
This creates a tensor from a set of scalar expressions, and concatenates it to the query and document embedding vectors.
Play around with this example in the playground
Dot Product between query and document vectors
Assume we have a set of documents where each document contains a vector of size 4. We want to calculate the dot product between the document vectors and a vector passed down with the query and rank
the results according to the dot product score.
The following schema file defines an attribute tensor field with a tensor type that has one indexed dimension x of size 4. In addition, we define a rank profile with the input and the dot product
schema example {
document example {
field document_vector type tensor<float>(x[4]) {
indexing: attribute | summary
rank-profile dot_product {
inputs {
query(query_vector) tensor<float>(x[4])
first-phase {
expression: sum(query(query_vector)*attribute(document_vector))
Example JSON document with the vector [1.0, 2.0, 3.0, 5.0], using indexed tensors short form:
"put": "id:example:example::0",
"fields": {
"document_vector" : [1.0, 2.0, 3.0, 5.0]
Example query set in a searcher with the vector [1.0, 2.0, 3.0, 5.0]:
public Result search(Query query, Execution execution) {
cell().label("x", 0).value(1.0).
cell().label("x", 1).value(2.0).
cell().label("x", 2).value(3.0).
cell().label("x", 3).value(5.0).build());
return execution.search(query);
Play around with this example in the playground
Note that this example calculates the dot product for every document retrieved by the query. Consider using approximate nearest neighbor search with distance-metric dotproduct.
Logistic regression models with cross features
One simple way to use machine-learning is to generate cross features from a set of base features and then do a logistic regression on these. How can this be expressed as Vespa tensors?
Assume we have three base features:
query(interests): tensor(interest{}) - A sparse, weighted set of the interests of a user.
query(location): tensor(location{}) - A sparse set of the location(s) of the user.
attribute(topics): tensor(topic{}) - A sparse, weighted set of the topics of a given document.
From these we have generated all 3d combinations of these features and trained a logistic regression model, leading to a weight for each possible combination:
tensor(interest{}, location{}, topic{})
This weight tensor can be added as a constant tensor to the application package, say constant(model). With that we can compute the model in a rank profile by the expression
sum(query(interests) * query(location) * attribute(topics) * constant(model))
Where the first three factors generates the 3d cross feature tensor and the last combines them with the learned weights.
Play around with this example in the playground
Matrix Product between 1d vector and 2d matrix
Assume we have a 3x2 matrix represented in an attribute tensor field document_matrix with a tensor type tensor<float>(x[3],y[2]) with content:
{ {x:0,y:0}:1.0, {x:1,y:0}:3.0, {x:2,y:0}:5.0, {x:0,y:1}:7.0, {x:1,y:1}:11.0, {x:2,y:1}:13.0 }
Also assume we have 1x3 vector passed down with the query as a tensor with type tensor<float>(x[3]) with content:
{ {x:0}:1.0, {x:1}:3.0, {x:2}:5.0 }
that is set as query(query_vector) in a searcher as specified in query feature.
To calculate the matrix product between the 1x3 vector and 3x2 matrix (to get a 1x2 vector) use the following ranking expression:
sum(query(query_vector) * attribute(document_matrix),x)
This is a sparse tensor product over the shared dimension x, followed by a sum over the same dimension.
Play around with this example in the playground
Using a tensor as a lookup structure
Tensors with mapped dimensions look similar to maps, but are more general. What if all needed is a simple map lookup? See tensor performance for more details.
Assume a tensor attribute my_map and this is the value for a specific document:
To create a query to select which of the 3 named vectors (a,b,c) to use for some other calculation, wrap the wanted label to look up inside a tensor. Assume a query tensor my_key with type/value:
Do the lookup, returning a tensor of type tensor<float>(y[3]):
If the key does not match anything, the result will be empty: tensor<float>(y[3]):[0,0,0]. For something else, add a check up-front to check if the lookup will be successful and run a fallback
expression if it is not, like:
if(reduce(query(my_key)*attribute(my_map),count) == 3,
The above can be considered the same as creating a
, like
. The above syntax allows an optimized execution, find an example in the
Tensor Playground
Slicing with lambda
A common use case is to use a tensor lambda function to slice out the first k dimensions of a vector representation of m dimensions where m is larger than k. Slicing with lambda functions is great
for representing vectors from Matryoshka Representation Learning.
Matryoshka Representation Learning (MRL) which encodes information at different granularities and allows a single embedding to adapt to the computational constraints of downstream tasks.
The following slices the first 256 dimensions of a tensor
Importantly, this does only reference into the original tensor, avoiding copying the tensory to a smaller tensor. The following is a complete example where we have stored an original vector
representation with 3072 dimensions, And we slice the first 256 dimensions of the original representation to perform a dot product in the first-phase expression, followed by a full computation over
all dimensions in the second-phase expression. See
phased ranking
for context on using Vespa phased computations and
customizing reusable frozen embeddings with Vespa
schema example {
document example {
field document_vector type tensor<float>(x[3072]) {
indexing: attribute | summary
rank-profile small-256-first-phase {
inputs {
query(query_vector) tensor<float>(x[3072])
function slice_first_dims(t) {
expression: l2_normalize(tensor<float>(x[256])(t{x:(x)}), x)
first-phase {
expression: sum( slice_first_dims(query(query_vector)) * slice_first_dims(attribute(document_vector)) )
second-phase {
expression: sum( query(query_vector) * attribute(document_vector) )
See also a runnable example in this
tensor playground example | {"url":"https://docs.vespa.ai/en/tensor-examples.html","timestamp":"2024-11-12T16:12:58Z","content_type":"text/html","content_length":"74972","record_id":"<urn:uuid:e5a2bf60-927b-4253-bea5-ac06b442ce2e>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00115.warc.gz"} |
Convertion chart for ninth grade
convertion chart for ninth grade Related topics: solving one step equations worksheets
Intermediate Algebra Charles P Mckeague Free Download
How Do You Square Root In Algebra
math combinations rule
practice inequalities problems worksheets
operations with integers and rational numbers
9th grade true and false math worksheet
free online algebra calculators
inequality worksheets
simplifying expressions by combining like terms calculator
algebra pratice
Author Message
Yavnel Posted: Sunday 24th of Dec 07:19
Hey dudes, I have just completed one week of my college, and am getting a bit tensed about my convertion chart for ninth grade home work. I just don’t seem to grasp the topics.
How can one expect me to do my homework then? Please help me.
Registered: 16.07.2004
From: London, UK
nxu Posted: Sunday 24th of Dec 10:02
Sounds like your bases are not clear . Excelling in convertion chart for ninth grade requires that your concepts be concrete. I know students who actually start teaching
juniors in their first year. Why don’t you try Algebra Professor? I am pretty sure, this program will aid you.
Registered: 25.10.2006
From: Siberia, Russian
fveingal Posted: Tuesday 26th of Dec 12:38
Hi, Thanks for the prompt reply. But could you let me know the details of reliable sites from where I can make the purchase? Can I get the Algebra Professor cd from a local
book mart available near my residence ?
Registered: 11.07.2001
From: Earth
SjberAliem Posted: Wednesday 27th of Dec 08:00
I suggest using Algebra Professor. It not only assists you with your math problems, but also gives all the necessary steps in detail so that you can enhance the understanding
of the subject.
Registered: 06.03.2002
From: Macintosh HD
iomthireansevin Posted: Wednesday 27th of Dec 10:32
Sounds like something I need to buy right away. Any links for buying this software online ?
Registered: 13.06.2003
Mov Posted: Thursday 28th of Dec 09:47
This site will give you details: https://softmath.com/. I think they give an absolute money back guarantee, so you have nothing to lose. Best of Luck!
Registered: 15.05.2002 | {"url":"https://algebra-net.com/algebra-online/radical-equations/convertion-chart-for-ninth.html","timestamp":"2024-11-06T09:25:20Z","content_type":"text/html","content_length":"92456","record_id":"<urn:uuid:b42a6442-ef82-473c-a8de-ae3fe9c018b3>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00775.warc.gz"} |
Applications of LP Duality | Rafael Oliveira
Applications of LP Duality
In this lecture we will see some cool applications of LP duality in other areas of science.
Game Theory: Two-Player Zero-Sum Games
Two-Player Games
In a two-player game, we have two players, Alice and Bob, who each have a set of strategies $S_A$ and $S_B$, respectively. The payoff of the game is given by a map $f: S_A \times S_B \to \mathbb{R}^
2$, where the first coordinate of the image is Alice’s payoff, and the second coordinate is Bob’s payoff. The goal of each player is to maximize their payoff. The game’s outcome can be given by a
table, where the rows correspond to Alice’s strategies, the columns correspond to Bob’s strategies, and the entry in the $i$-th row and $j$-th column is the payoff of the game when Alice plays the
$i$-th strategy and Bob plays the $j$-th strategy.
An example is the following game, known as the battle of the sexes game: Alice likes to go to the football game, while Bob likes to go to the opera. However, they both prefer to go to an event
together than to go alone. The payoff table in this case is given by
Football Opera
Football (2, 1) (0, 0)
Opera (0, 0) (1, 2)
Where in each entry of the above table, the first number is Alice’s payoff, and the second number is Bob’s payoff.
Another example is the prisoner’s dilemma: two prisoners are arrested for a crime, and are being interrogated separately. If both prisoners remain silent, they will both be sentenced to 1 year in
prison. If one prisoner confesses and the other remains silent, the prisoner who confesses will be set free, while the other will be sentenced to 10 years in prison. If both prisoners confess, they
will both be sentenced to 5 years in prison. The payoff table in this case is given by
Silent Snitch
Silent (-1, -1) (-10, 0)
Snitch (0, -10) (-5, -5)
Where in each entry of the above table, the first number is the payoff of the first prisoner, and the second number is the payoff of the second prisoner.
Note that in the above two examples, certain strategies are special. For instance, in the battle of sexes example, if Bob knows that Alice will go to football, then Bob will go to football as well,
since he prefers to go to the football game with Alice than to go to the opera alone. Similarly, if Alice knows that Bob will go to the football game, then Alice will go to the football game as well.
Such strategies are called Nash equilibria, which we now formally define.
Definition (Best Response): A strategy $s_A \in S_A$ is a best response to a strategy $s_B \in S_B$ if $f(s_A, s_B) \geq f(s_A’, s_B)$ for all $s_A’ \in S_A$.
Definition (Nash Equilibrium): A pair of strategies $(s_A, s_B)$ is a Nash equilibrium if $s_A$ is a best response to $s_B$ and $s_B$ is a best response to $s_A$. In other words, we know that Alice’s
strategy $s_A$ is an optimum strategy for Alice, given that she knows that Bob’s strategy is $s_B$, and Bob’s strategy $s_B$ is an optimum strategy for Bob, given that he knows that Alice’s strategy
is $s_A$.
Practice problem 1: See that in the prisoner’s dilemma, the pair of strategies (Snitch, Snitch) is a Nash equilibrium.
Here it is important to notice three points:
1. The Nash equilibrium is not necessarily the best outcome for the players. For instance, in the prisoner’s dilemma, the best outcome for the players is for both of them to remain silent, but this
is not a Nash equilibrium.
2. The Nash equilibrium is not necessarily unique. For instance, in the battle of sexes game, both (Football, Football) and (Opera, Opera) are Nash equilibria.
3. Some games do not have a Nash equilibrium. For instance, consider the rock-paper-scissors game, where Alice and Bob each choose one of rock, paper, or scissors; and whoever wins gets a value of
1, and whoever loves gets a value of $-1$. The table of values for this game is given by:
Rock Paper Scissors
Rock (0, 0) (-1, 1) (1, -1)
Paper (1, -1) (0, 0) (-1, 1)
Scissors (-1, 1) (1, -1) (0, 0)
In this case, there is no Nash equilibrium, since no matter what strategy Alice chooses, Bob can choose a strategy that gives him a better payoff (and vice-versa).
Practice problem 2: Show that in the rock-paper-scissors game, there is no Nash equilibrium.
Mixed Strategies
In the above discussions, we have talked about pure strategies, where each player chooses a single strategy to play the game. However, we can also consider mixed strategies, where each player chooses
a probability distribution over the set of strategies. Mixed strategies model the case where the players choose their strategies randomly, according to some distribution. | {"url":"https://cs.uwaterloo.ca/~r5olivei/courses/2024-spring-cs466/lecture-notes/lecture12/","timestamp":"2024-11-03T15:08:46Z","content_type":"text/html","content_length":"26833","record_id":"<urn:uuid:ce691765-5379-439b-b093-c16bdee6cdde>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00209.warc.gz"} |
[S:Deleted text in red:S] / Inserted text in green
An Irrational Number is one that cannot be expressed as a ratio.
For example, [S:2/3:S] 3/5 is clearly a rational number, because it's a ratio.
However, root two is irrational.
Other examples of irrational numbers are EQN:\pi,{\quad}e, (which are transcendental numbers)
and the Golden Ratio EQN:\phi (which is an algebraic number). | {"url":"https://www.livmathssoc.org.uk/cgi-bin/sews_diff.py?IrrationalNumber","timestamp":"2024-11-14T13:29:14Z","content_type":"text/html","content_length":"1500","record_id":"<urn:uuid:21ed96ae-6c1d-4afa-8a70-a6ebff804de8>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00663.warc.gz"} |
IntroductionPreliminariesBipolar Picture Fuzzy SetTabular representation of BPFN under business related problemsTabular representation of BPFS under medical related problemsComparison of BPFS with the existing set theoretic modelsGraphical representation for satisfaction, abstinence and dissatisfaction grades of picture fuzzy set. 0≤x+y+z≤1Graphical representation for grades of bipolar picture fuzzy set
Some Bipolar Picture Fuzzy Geometric OperatorsDistance Measure of Bipolar Picture Fuzzy SetsMCDM Based on BPFS to Pattern RecognitionMethod of facial recognition
Numerical Example for Using New Measures in Pattern RecognitionMCDM Based on Some Bipolar Picture Fuzzy Geometric OperatorsFlow chart of proposed algorithm
Case StudyKing oyster mushroomCombining formulas for alternativesNumerical ExampleBPF decision matrix taking by decision makerNormalized BPF decision matrixBPF decision matrix taking by decision makerNormalized BPF decision matrixBPF decision matrix taking by decision makerNormalized BPF decision matrixValues of T¨jComparison analysis of the proposed operators and existing operators in the given numerical exampleConclusionReferences
CMES CMES CMES Computer Modeling in Engineering & Sciences 1526-1506 1526-1492 Tech Science Press USA 14174 10.32604/cmes.2021.014174 Article Multi-Criteria Decision Making Based on Bipolar Picture
Fuzzy Operators and New Distance Measures Multi-Criteria Decision Making Based on Bipolar Picture Fuzzy Operators and New Distance Measures Multi-Criteria Decision Making Based on Bipolar Picture
Fuzzy Operators and New Distance Measures Riaz Muhammad 1 Garg Harish 2 Farid Hafiz Muhammad Athar 1 Chinram Ronnason 34ronnason.c@psu.ac.th Department of Mathematics, University of the Punjab,
Lahore, 54590, Pakistan School of Mathematics, Thapar Institute of Engineering and Technology (Deemed University), Patiala, 147004, India Algebra and Applications Research Unit, Division of
Computational Science, Faculty of Science, Prince of Songkla University, Hat Yai, Songkhla, 90110, Thailand Centre of Excellence in Mathematics, Bangkok, 10400, Thailand *Corresponding Author:
Ronnason Chinram. Email: ronnason.c@psu.ac.th 17 03 2021 127 2 771 800 06 09 2020 21 12 2020 © 2021 Riaz et al. 2021 Riaz et al. This work is licensed under a Creative Commons Attribution 4.0
International License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
This paper aims to introduce the novel concept of the bipolar picture fuzzy set (BPFS) as a hybrid structure of bipolar fuzzy set (BFS) and picture fuzzy set (PFS). BPFS is a new kind of fuzzy sets
to deal with bipolarity (both positive and negative aspects) to each membership degree (belonging-ness), neutral membership (not decided), and non-membership degree (refusal). In this article, some
basic properties of bipolar picture fuzzy sets (BPFSs) and their fundamental operations are introduced. The score function, accuracy function and certainty function are suggested to discuss the
comparability of bipolar picture fuzzy numbers (BPFNs). Additionally, the concept of new distance measures of BPFSs is presented to discuss geometrical properties of BPFSs. In the context of BPFSs,
certain aggregation operators (AOs) named as “bipolar picture fuzzy weighted geometric (BPFWG) operator, bipolar picture fuzzy ordered weighted geometric (BPFOWG) operator and bipolar picture fuzzy
hybrid geometric (BPFHG) operator” are defined for information aggregation of BPFNs. Based on the proposed AOs, a new multi-criteria decision-making (MCDM) approach is proposed to address uncertain
real-life situations. Finally, a practical application of proposed methodology is also illustrated to discuss its feasibility and applicability.
Bipolar picture fuzzy set aggregation operators distance measures pattern recognition MCDM
In any real-life problem-solving technique, the complexity characterizes the behavior of an object whose components interrelate in multiple ways and follow different logical rules, meaning there is
no fixed rule to handle multiple challenges due to various uncertainties in real life circumstances. Many scholars from all over the world have apparently studied MCDM management techniques
extensively. This effort resulted in a multitude of innovative solutions to complex real concerns. The frameworks for this objective are largely based on a summary of the issues at hand. To deal with
uncertainties the researchers have been proposed various mathematical techniques. Zadeh [1] initiated the idea of fuzzy set (FS) and membership degrees of objects/alternatives. Later, the
intuitionistic fuzzy set (IFS) proposed by Atanassov [2] is the direct extension of FS by using membership degrees (MDs) and non-membership degrees (NMDs). Yager et al. [3,4] and Yager et al. [5]
introduced Pythagorean fuzzy set and Pythagorean fuzzy membership grades. Zhang et al. [6,7] introduced an independent extension of fuzzy set named as bipolar fuzzy sets (BFSs) and Lee [8] presented
some basics operations. A bipolar fuzzy information is used to express a property of an object as well as its counter property.
Alcantud et al. [9] initiated the notion of N-soft set approach to rough sets and introduced the concept of dual extended hesitant fuzzy sets [10]. Akram et al. [11,12] initiated MCDM based on
Pythagorean fuzzy TOPSIS method and Pythagorean Dombi fuzzy AOs. Ashraf et al. [13] initiated spherical fuzzy Dombi AOs. Eraslan et al. [14] and Feng et al. [15] proposed new approaches for MCDM.
Garg et al. [16–18] introduced some AOs on different sets also their applications to MCDM. Jose et al. [19] proposed AOs for MCDM. Karaaslan [20], Liu et al. [21], Liu et al. [22], Wang et al. [23],
Yang et al. [24], Smarandache [25], and Liu et al. [26] initiated many different approaches including AOs on different extension of fuzzy set for MCDM. Naeem et al. [27,28], Peng et al. [29,30], Peng
et al. [31] introduced some significant results for Pythagorean fuzzy sets.
Riaz et al. [32], initiated the concept of linear Diophantine fuzzy Set and its applications to MCDM. Riaz et al. [33] introduced some hybrid AOs, Einstein prioritized AOs [34], related to q-ROFSs.
Riaz et al. [35] introduced cubic bipolar fuzzy set and related AOs. Cagman et al. [36], and Shabir et al. [37] independently introduced the notion of soft topological spaces.
Cuong [38] presented the idea of a picture fuzzy set (PFS) as a new paradigm distinguished with three functions that assign the positive membership degree (MD), the neutral MD and the negative
membership degree (NMD) to each object/alternative. The basic restrictions on these degrees are that they lie in [0, 1] and their sum also lies in [0, 1]. Cuong [39] further introduced the concept of
Pythagorean picture fuzzy sets and its basic notions. Garg [40], Jana et al. [41] and Wang et al. [42] proposed some AOs for picture fuzzy information aggregation. Pamucar [43] studied the notion of
normalized weighted geometric Dombi Bonferoni mean operator with interval grey numbers: Application in multicriteria decision making. Pamucar et al. [44] proposed an application of the hybrid
interval rough weighted Power-Heronian operator in multi-criteria decision making. Ramakrishnan et al. [45] introduced a cloud TOPSIS model for green supplier selection. Riaz et al. also introduced
some AOs [46,47] related to green supplier selection. Si et al. [48] and Sinani [49] also presented different AOs in some extension of fuzzy set.
The first objective of this paper is to introduce bipolar picture fuzzy sets (BPFSs) as a new hybrid structure of bipolar fuzzy sets (BFSs) and picture fuzzy sets (PFSs). BPFSs are more efficient for
dealing with the real-life situation when modeling needs to address the bipolarity (both positive and negative aspects) to each MD (belonging-ness), neutral membership (not decided), and
non-membership degree (refusal). The second objective of BPFSs is propose bipolar picture fuzzy MCDM technique based on bipolar picture fuzzy AOs. The third objective of BPFSs is to define new
distance measure and its application towards pattern recognition. Additionally, the proposed methodology can extend to solve various problems of artificial intelligence, computational intelligence
and MCDM that involve bipolar picture fuzzy information.
The rest of the paper is as follows. The definitions of IFS, PFS and BFS are discussed in Section 2. Section 3 introduces the definition of BPFS. Section 4 indicates some bipolar picture AOs and new
BPFS distance measures. Section 5 shows the generalizability of the suggested paradigm for pattern recognition. Section 6 introduces a new BPF-MCDM approach based on suggested AOs and a numerical
example. Finally, Section 7 summarizes the findings of this research study.
In this section, we give some basic definitions to IFSs, BFSs and PFSs.
Definition 2.1 [2] Let Ψ=(δ1,δ2,…δn) be a crisp set, an IFS J in Ψ is defined by
where 0≤μJ(δ)≤1, 0≤νJ(δ)≤1 and 0≤μJ(δ)+νJ(δ)≤1, ∀δ∈Ψ. πJ(δ)=1-(μJ(δ)+νJ(δ)) is called indeterminacy degree (ID) of J in Ψ. Also 0≤πJ(δ)≤1∀δ∈Ψ.
Definition 2.2 [38] Let W be a crisp set, a PFS A in W is defined as follows:
where, μA(δ)∈[0,1] is called positive MD of δ in A, λA(δ)∈[0,1] is called neutral MD of δ in A, νA(δ)∈[0,1] is called negative MD of δ in A, and μA(δ), λA(δ), νA(δ) satisfy the condition 0≤μA(δ)+λA
(δ)+νA(δ)≤1(∀δ∈W) and 1-(μA(δ)+λA(δ)+νA(δ)) is called refusal MD of A.
A basic element 〈δ,μA(δ),λA(δ),νA(δ)〉 in a PFS A is denoted by Ã=〈μA,λA,νA〉, which is called picture fuzzy number (PFN).
Definition 2.3 [38] Some operational laws of picture fuzzy set as follows:
Let P1={〈δ,μP1(δ),λP1(δ),νP1(δ)〉∣δ∈W} and P2={〈δ,μP2(δ),λP2(δ),νP2(δ)〉∣δ∈W} be any two PFS. Then
1. P1⊆P2 iff, μP1(δ)≤μP2(δ),λP1(δ)≤λP2(δ),νP1(δ)≥νP2(δ).
2. P[1] = P[2] iff,
3. The complement of P[1] is defined by
4. The union is defined by
5. The intersection is defined by
Definition 2.4 [40] The score function of a PFN δ=〈μA,λA,νA〉 is defined as
However, in certain circumstances, the previous score function may not rank any two PFNs. For example, P[1] = (0.7, 0.2, 0.1) and P[2] = (0.6, 0.1, 0.1) then they have same score function values,
i.e., R(P[1]) = R(P[2]). For this we use accuracy function given as
Let P1=〈μ1,λ1,ν1〉 and P2=〈μ2,λ2,ν2〉 are any two PFNs, and R(P[1]), R(P[2]) are the score function of P[1] and P[2], and ℑ(P1), ℑ(P2) are the accuracy function of P[1] and P[2], respectively, then
If R(P[1]) > R(P[2]), then P[1] > P[2].
If R(P[1]) = R(P[2]), then,
If ℑ(P1)>ℑ(P2) then P[1] > P[2],
If ℑ(P1)=ℑ(P2), then P[1] = P[2].
Definition 2.5 [6] Let X be a set, a BFS B inX is defined as follows:
where μB+(δ):X→[0,1] and μB-(δ):X→[-1,0]. The positive MD μB+(δ) demonstrates the satisfaction degree of an element δ to the property corresponding to a BFS B, negative MD μB-(δ) denotes the
satisfaction degree of an element δ to some implicit counter-property of B.
Definition 2.6 [6,7] Some operational laws of bipolar fuzzy set as follows:
Let B={〈δ,μB+(δ),μB-(δ)〉∣δ∈W}, B1={〈δ,μB1+(δ),μB1-(δ)〉∣δ∈W} and B2={〈δ,μB2+(δ),μB2-(δ)〉∣δ∈W} be three bipolar fuzzy sets. Then
1. B1⊆B2 iff,
2. B[1] = B[2] iff,
3. The complement of B[1] is denoted by B1c,
4. The union is defined by
5. The intersection is defined by
6. α-cut (Bα) of B,
Here, Bα+ is called positive α-cut and Bα- is called negative α-cut.
7. Support (shortly,Supp(B)) of B,
Here, Supp(B)+ is called positive α-cut and Supp(B)- is called negative α-cut.
The BFS assign positive and negative grades to the alternatives and PFS is characterized by three functions expressing the MD, the neutral MD and the NMD. Fuzzy set assign a membership grade to each
alternatives δ in the unit closed interval [0, 1]. In BFS the positive MD μλ+(δ) represent the satisfaction degree of an alternative δ to the property corresponding to a BFS λ, and negative MD μλ-(δ)
represent the satisfaction degree of an element δ to some implicit counter-property of λ. In PFS there are three types of grades, μA(δ)∈[0,1] is called positive MD of δ in A, λA(δ)∈[0,1] is called
neutral MD of δ in A, νA(δ)∈[0,1] is called negative MD of δ in A, and where μA, λA, νA satisfy the condition 0≤μA(δ)+λA(δ)+νA(δ)≤1, ∀δ∈X.
We present the idea of BPFS as a new hybrid version of BFS and PFS. In this model of BPFS, we assign positive and negative grades for each MD (belonging-ness), neutral membership (not decided), and
non-membership degree (refusal). We present specific examples to relate the proposed model with the real life applications. We define some operational laws of BPFS along with its score and accuracy
Definition 3.1 A BPFS Ω on universe W is an abject of the form
where μ+,λ+,ν+:W→[0,1] and μ-,λ-,ν-:W→[-1,0] with conditions
The positive MDs μΩ+(δ), λΩ+(δ), νΩ+(δ) demonstrate the truth MD, indeterminate MD and false MD of an element δ corresponding to a BPFS Ω and the negative MDs μΩ-(δ), λΩ-(δ), νΩ-(δ) demonstrate the
truth MD, indeterminate MD and false MD of an element δ to some implicit counter-property corresponding to a BPFS Ω. Absolute BPFS assign (1, 0, 0, −1, 0, 0) to each alternative, denoted by 𝔘 and
null BPFS assign (0, 0, 1, 0, 0, −1) to each alternative, denoted by 𝔑. Moreover ρΩ+=1-(μΩ++λΩ++νΩ+) is called the positive degree of refusal membership of δ in Ω and ρΩ-=-1-(μΩ-+λΩ-+νΩ-) is called
the negative degree of refusal membership of δ in Ω.
Now we discuss some applications of proposed model to relate it with real life problems.
In the field of finance and business, we use two terms profit and loss. We can relate the decision-making applications based on business with BPFS. If a person invests some money, then he wants to
earn max profit in some interval of time. Bipolar picture fuzzy number (BPFN) can be described as
The physical meaning of this structure in business terms is that, what is the satisfactions grade that he earns profit (μ+), what is the dissatisfactions grade that he earns profit (ν+), what is the
abstinence grade that he earns profit (μ+), what is the satisfactions grade that he gets loss (μ-), what is the dissatisfactions grade that he gets loss (ν-) and what is the abstinence grade that he
gets loss (λ-). We can see how BPFN is useful for the decision-making problems of real life problems. The detail of component of BPFN for finance and business is given in Tab. 1.
μ+ Satisfactions grade that he earns profit
λ+ Abstinence grade that he earns profit
ν+ Dissatisfactions grade that he earns profit
μ- Satisfactions grade that he gets loss
λ- Abstinence grade that he gets loss
ν- Dissatisfactions grade that he gets loss
In the field of medical, we mostly focus on the effects and side effects of medicines related to every disease. If a patient get some medication according to his type of disease, then we can relate
our model to the effects and side effects of that medicine in medical diagnosis, treatment and recovery terms. For the BPFN can be written as
μ+ represents the positive effects of recommended medicine to the disease of the patient, ν+ represents the dissatisfaction effects of recommended medicine, λ+ represents the abstinence effects of
recommended medicine, μ- represents the negative or bad effects of recommended medicine, ν- represents the dissatisfaction grade of bad effects of recommended medicine and λ- represents the
abstinence grades of side effects of recommended medicine. The detail of component of BPFN for medication is given in Tab. 2.
μ+ Positive effects of recommended medicine
λ+ Abstinence effects of recommended medicine
ν+ Dissatisfaction effects of recommended medicine
μ- Negative or bad effects of recommended medicine
λ- Abstinence grades of side effects of recommended medicine
ν- Dissatisfaction grade of bad effects of recommended medicine
The proposed model is superior than these two models, in fact it is hybrid structure of BFS and PFS that assign six grades to the alternative.
Comparison Analysis:
In this part, we discuss about the terms and characteristics of proposed model and compare it with the existing techniques. There are various objectives to construct this hybrid structure and some of
them are listed below:
The first objective to construct this hybrid model is to fill the research gap which exists in previous methodologies. The bipolar fuzzy set and picture fuzzy set can be used together in decision
analysis. We can deal with the satisfaction, abstinence and dissatisfaction grades of the alternatives with its counter properties.
The second objective is that we can cover the evaluation space in a different manner. If we compare our model with the existing theories then we find that it is strong, valid and superior to others.
The comparison analysis of BPFS with the existing models is given in Tab. 3.
The third objective is to represent the relationship of BPFS to the MCDM problems. We study some real life problems and convert the input data into BPF numeric values and deal it with the proposed
aggregation operators. This novel structure is superior, flexible and easy to handle and can deal with the MCDM problems in the field of medical, business, artificial intelligence and engineering
etc. The graphical representation of PFS and BPFS is given in the Figs. 1 and 2, respectively.
In bipolar neutrosophic set (see [50]) the conditions are as follows:
However, in the proposed model BPFS the conditions are as follows:
Set theoretic models Satisfaction grade (MD) Abstinence grade (Neutral MD) Dissatisfaction grade (NMD) Bipolarity
Fuzzy set [1] ✓ ✕ ✕ ✕
IFS [2] ✓ ✕ ✓ ✕
BFS [6] ✓ ✕ ✕ ✓
PFS [38] ✓ ✓ ✓ ✕
Proposed BPFS ✓ ✓ ✓ ✓
Definition 3.2 Let Ω1=〈δ,μ1+(δ),λ1+(δ),ν1+(δ),μ1-(δ),λ1-(δ),ν1-(δ)〉 and Ω2=〈δ,μ2+(δ),λ2+(δ),ν2+(δ),μ2-(δ),λ2-(δ),ν2-(δ)〉 be two BPFSs. Then Ω1⊆Ω2 iff μ1+(δ)≤μ2+(δ),λ1+(δ)≤λ2+(δ),ν1+(δ)≥ν2+(δ) and
Definition 3.3 Let Ω1=〈δ,μ1+(δ),λ1+(δ),ν1+(δ),μ1-(δ),λ1-(δ),ν1-(δ)〉 and Ω2=〈δ,μ2+(δ),λ2+(δ),ν2+(δ),μ2-(δ),λ2-(δ),ν2-(δ)〉 be two BPFSs. Then Ω1=Ω2 iff
and μ1-(δ)=μ2-(δ),λ1-(δ)=λ2-(δ),ν1-(δ)=ν2-(δ).
Definition 3.4 Let Ω1=〈δ,μ1+(δ),λ1+(δ),ν1+(δ),μ1-(δ),λ1-(δ),ν1-(δ)〉 and Ω2=〈δ,μ2+(δ),λ2+(δ),ν2+(δ),μ2-(δ),λ2-(δ),ν2-(δ)〉 be two BPFSs.
The union of these two BPFSs is defined as (Ω1∪Ω2)(δ)=(max(μ1+(δ),μ2+(δ)),min(λ1+(δ),λ2+(δ)),min(ν1+(δ),ν2+(δ)),min(μ1−(δ),μ2−(δ)),maδ(λ1−(δ),λ2−(δ)),maδ(ν1−(δ),ν2−(δ))).
Example 3.5 Let X={δ1,δ2,δ3}. Let us consider are two BPFSs Ω1, Ω2 in X given by Ω1=〈δ1,0.5,0.2,0.2,-0.1,-0.2,-0.4〉,〈δ2,0.3,0.4,0.3,-0.35,-0.3,-0.3〉,〈δ3,0.3,0.4,0.2,-0.5,-0.1,-0.2〉Ω2=〈
Then their union is Ω1∪Ω2=〈δ1,0.5,0.2,0.2,-0.2,-0.2,-0.2〉,〈δ2,0.5,0.1,0.3,-0.3,-0.3,-0.2〉,〈δ3,0.3,0.4,0.1,-0.5,-0.1,-0.2〉
Definition 3.6 LetΩ1=〈δ,μ1+(δ),λ1+(δ),ν1+(δ),μ1-(δ),λ1-(δ),ν1-(δ)〉 and Ω2=〈δ,μ2+(δ),λ2+(δ),ν2+(δ),μ2-(δ),λ2-(δ),ν2-(δ)〉 be two BPFSs.
The intersection of these two BPFSs is defined as (Ω1∩Ω2)(δ)=(min(μ1+(δ),μ2+(δ)),maδ(λ1+(δ),λ2+(δ)),maδ(ν1+(δ),ν2+(δ)),max(μ1−(δ),μ2−(δ)),min(λ1−(δ),λ2−(δ)),min(ν1−(δ),ν2−(δ))).
Example 3.7 Let W={δ1,δ2,δ3}. Let us consider are two bipolar picture fuzzy set Ω1, Ω2 in W given by Ω1=〈δ1,0.5,0.2,0.2,-0.1,-0.2,-0.4〉,〈δ2,0.3,0.4,0.3,-0.35,-0.3,-0.3〉,〈
Then their intersection is Ω1∩Ω2=〈δ1,0.4,0.3,0.2,-0.1,-0.4,-0.4〉,〈δ2,0.3,0.4,0.3,-0.3,-0.5,-0.3〉,〈δ3,0.3,0.4,0.2,-0.3,-0.2,-0.4〉
Definition 3.8 Let Ω=〈δ,μ+(δ),λ+(δ),ν+(δ),μ-(δ),λ-(δ),ν-(δ)〉 be a BPF set in W. Then the compliment of Ω is denoted by Ωc and defined as, for all δ∈W,
Example 3.9 Let W={δ1,δ2,δ3}. Consider a BPFS Ω in W given by Ω=〈δ1,0.3,0.2,0.4,-0.1,-0.3,-0.4〉,〈δ2,0.3,0.4,0.3,-0.2,-0.2,-0.4〉,〈δ3,0.3,0.4,0.1,-0.2,-0.1,-0.4〉
Then its complement is Ωc=〈δ1,0.4,0.2,0.3,-0.4,-0.3,-0.1〉,〈δ2,0.3,0.4,0.3,-0.4,-0.2,-0.2〉,〈δ3,0.1,0.4,0.3,-0.4,-0.1,-0.2〉
Now we see that BFS and PFS are special cases of BPFS.
Proposition 3.10 BFS and PFS are special cases of BPFS, i.e., Bipolar fuzzy numbers (BFNs) and picture fuzzy numbers (PFNs) are special cases of the bipolar picture fuzzy numbers (BPFNs).
Proof. For any δ∈X, consider a BPFN given by, 〈μ+(δ),λ+(δ),ν+(δ),μ-(δ),λ-(δ),ν-(δ)〉. Then by setting the components λ+(δ), ν+(δ), λ-(δ), ν-(δ) equals to zero, we obtain a BFN, 〈μ+(δ),μ-(δ)〉.
Similarly, by setting the components μ-(δ), λ-(δ), ν-(δ) equal to zero, we obtain we obtain a PFN, 〈μ+(δ),λ+(δ),ν+(δ)〉 which can be written as, 〈μ(δ),λ(δ),ν(δ)〉. This complete the proof.
Theorem 3.11 Let Ω1,Ω2 and Ω3 be the BPFSs in a universe X, then we have
Ω1∪Ω1=Ω1 and Ω1∩Ω1=Ω1
Ω1∪Ω2=Ω2∪Ω1 and Ω1∩Ω2=Ω2∩Ω1
Ω1∪Ω1c≠𝔘 and Ω1∩Ω1c≠𝔑
Proof. The proof is obvious.
Theorem 3.12 Let O and P be the BPFSs in a universe X, then we have
We will denote the set of all BPFSs in X by 𝔛.
Definition 3.13 Let ϕ1=〈μ1+,λ1+,ν1+,μ1-,λ1-,ν1-〉 and ϕ2=〈μ2+,λ2+,ν2+,μ2-,λ2-,ν2-〉 be two BPFNs then
In this section, firstly, we introduce score function, accuracy function, and certainty function for BPFNs. Secondly, we introduce BPFWG operator, BPFOWG operator, and BPFHG operator.
Definition 4.1 Let T1=〈μ1+,λ1+,ν1+,μ1-,λ1-,ν1-〉 be (BPFN). Then the score function Φ(T1), accuracy function Υ(T1) and certainty fruition Π(T1) of BPFN are defined as follows:
The range of score function Φ(T) is [ −1, 1], range of accuracy function Υ(T) is [0, 1] and range of certainty fruition Π(T) of BPFN is [0, 1].
Definition 4.2 Let T1=〈μ1+,λ1+,ν1+,μ1-,λ1-,ν1-〉 and T2=〈μ2+,λ2+,ν2+,μ2-,λ2-,ν2-〉 be two (BPFN). The comparison method can be defined as follows:
if Φ(T1)>Φ(T2), then T[1] greater than T[2] and denoted by T[1] > T[2].
if Φ(T1)=Φ(T2) and Υ(T1)>Υ(T2), then T[1] greater than T[2] and denoted by T[1] > T[2].
if Φ(T1)=Φ(T2), Υ(T1)=Υ(T2) and Π(T1)>Π(T2) then T[1] greater than T[2] and denoted by T[1] > T[2].
if Φ(T1)=Φ(T2), Υ(T1)=Υ(T2) and Π(T1)=Π(T2) then T[1] equal to T[2] and denoted by T[1] = T[2].
Definition 4.3 Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉(j=1,2,…,n) be an assemblage of BPFNs. A mapping BPFWG:𝔛n→𝔛 is called a bipolar picture fuzzy weighted geometric (BPFWG) operator. BPFWG(T1,T2,…,Tn)=
∑j=1nTjPj =T1P1 ⊗T2P2 ,…⊗,TnPn where P[j] is the weight vector (WV) of T[j], Pj∈[0,1] and ∑j=1nPj=1.
Theorem 4.4 Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉(j=1,2,…,n) be an assemblage of BPFNs. We also find BPFWG by
BPFWG(T1,T2,…,Tn)=(∏ j=1n(μj++λj+)Pj-∏j=1n(λj+)Pj,∏j=1n(λj+)Pj,1-∏j=1n(1-νj+)Pj,∏ j=1n((-μj-)-(-λj-))Pj-∏j=1n(-λj-)Pj,∏j=1n(-λj-)Pj,1-∏j=1n(1-(-νj-))Pj)
Proof. Using mathematical induction to prove this theorem.
For n = 2 T1P1=((μ1++λ1+)P1−(λ1+)P1,(λ1+)P1,1−(1−ν1+)P1,((−μ1−)+(−λ1−))P1−(−λ1−)P1,(−λ1−)P1,1−(1−(−ν1−))P1)T2P2=((μ2++λ2+)P2−(λ2+)P2,(λ2+)P2,1−(1−ν2+)P2,((−μ2−)+(−λ2−))P2−(−λ2−)P2,(−λ2−)P2,1−(1−
Then, it follows that
T1P1 ⊗T2P2 =((μ1++λ1+)P1(μ2++λ2+)P2-(λ1+)P1(λ2+)P2,(λ1+)P1(λ2+)P2,1-(1-ν1+)P11-(1-ν2+)P2,((-μ1-)+(-λ1-))P1((-μ2-)+(-λ2-))P2-(-λ1-)P1(-λ2-)P2,(-λ1-)P1(-λ2-)P2,1-(1-(-ν1-))P11-(1-(-ν2-))P2)=(∏ j=12
(μj++λj+)Pj-∏j=12(λj+)Pj,∏j=12(λj+)Pj,1-∏j=12(1-νj+)Pj,∏ j=12((-μj-)-(-λj-))Pj-∏j=12(-λj-)Pj,∏j=12(-λj-)Pj,1-∏j=12(1(-νj-))Pj)
This shows that it is true for n = 2, now let that it holds for n = k, i.e.,
BPFWG(T1,T2,…,Tk)=( ∏j=1k(μj++λj+)Pj−∏j=1k(λj+)Pj,∏j=1k(λj+)Pj,1−∏j=1k(1−νj+)Pj, ∏j=1k((−μj−)−(−λj−))Pj−∏j=1k(−λj−)Pj,∏j=1k(−λj−)Pj,1−∏j=1k(1−(−νj−))Pj )
Now n = k + 1, by operational laws of BPFNs we have
BPFWG(T1,T2,…,Tk+1)=BPFWG(T1,T2,…,Tk)⊗Tk+1Pk+1=( ∏j=1k(μj++λj+)Pj−∏j=1k(λj+)Pj,∏j=1k(λj+)Pj,1−∏j=1k(1−νj+)Pj,∏j=1k((−μj−)−(−λj−))Pj−∏j=1k(−λj−)Pj,∏j=1k(−λj−)Pj,1−∏j=1k(1−(−νj−))Pj⊗((μk+1++λk+1+)
Pk+1−(λk+1+)Pk+1,(λk+1+)Pk+1,1−(1−νk+1+)Pk+1,((−μk+1−)+(−λk+1−))Pk+1−(−λk+1−)Pk+1,(−λk+1−)Pk+1,1−(1−(−νk+1−))Pk+1)=( ∏j=1k+1(μj++λj+)Pj−∏j=1k+1(λj+)Pj,∏j=1k+1(λj+)Pj,1−∏j=1k+1(1−νj+)Pj,∏j=1k+1
((−μj−)−(−λj−))Pj−∏j=1k+1(−λj−)Pj,∏j=1k+1(−λj−)Pj,1−∏j=1k+1(1−(− ν j −))Pj)
This shows that for n = k + 1, holds. Thus, by the principle of mathematical induction Theorem 4.4 holds for all n.
BPFWG(T1,T2,…,Tn)=(∏ j=1n(μj++λj+)Pj-∏j=1n(λj+)Pj,∏j=1n(λj+)Pj,1-∏j=1n(1-νj+)Pj,∏ j=1n((-μj-)-(-λj-))Pj-∏j=1n(-λj-)Pj,∏j=1n(-λj-)Pj,1-∏j=1n(1-(-νj-))Pj)
Below we define some of BPFWA’s appealing properties.
Theorem 4.5 (Idempotency) Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉 be an assemblage of BPFNs. If, Tj=T=〈μ+,λ+,ν+,μ-,λ-,ν-〉 for all j, then
Proof. Since, T1=T2=…=Tn=T. By Theorem 4.4, we get
BPFWG(T1,T2,…,Tn)=( ∏j=1n(μj++λj+)Pj−∏j=1n(λj+)Pj,∏j=1n(λj+)Pj,1−∏j=1n(1−νj+)Pj,∏j=1n((−μj−)−(−λj−))Pj−∏j=1n(−λj−)Pj,∏j=1n(−λj−)Pj,1−∏j=1n(1−(−νj−))Pj)=( ∏j=1n(μ++λ+)Pj−∏j=1n(λ+)Pj,∏j=1n(λ+)
Pj,1−∏j=1n(1−ν+)Pj,∏j=1n((−μ−)−(−λ−))Pj−∏j=1n(−λ−)Pj,∏j=1n(−λ−)Pj,1−∏j=1n(1−(−ν−))Pj=( (μ++λ+)∑j=1nPj−(λ+)∑j=1nPj,(λ+)∑j=1nPj,1−(1−ν+)∑j=1nPj, ((−μ−)−(−λ−))∑j=1nPj−(−λ−)∑j=1nPj,(−λ−)∑j=1nPj,1−
(1−(−ν−))∑j=1nPj )
We know, ∑j=1nPj=1,
Theorem 4.6 (Monotonicity) Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉 and Tj*=〈(μj+)*,(λj+)*,(νj+)*,(μj-)*,(λj-)*,(νj-)*〉 be two families of BPFNs. If Tj≤Tj* for all (j=1,2,…,n) then
Proof. Here, we omit the proof.
Example 4.7 Let T[1], T[2], T[3], T[4] be the BPFNs as follows:
and P = (0.3, 0.2, 0.1, 0.4) then
BPFWG(T1,T2,T3,T4)=(∏ j=14(μj++λj+)Pj-∏j=14(λj+)Pj,∏j=14(λj+)Pj,1-∏j=14(1-νj+)Pj,∏ j=14((-μj-)-(-λj-))Pj-∏j=14(-λj-)Pj,∏j=14(-λj-)Pj,1-∏j=14(1-(-νj-))Pj)=
When we need to weight the ordered positions of the bipolar picture fuzzy arguments instead of weighting the arguments themselves, BPFWG can be generalized to BPFOWG.
Definition 4.8 Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉 be a assemblage of BPFNs. A mapping BPFOWG:𝔛n→𝔛 is called a bipolar picture fuzzy ordered weighted geometric (BPFOWG) operator.
BPFOWG(T1,T2,…,Tn)=∑j=1nTσ(j)Pj =Tσ(1)P1 ⊗Tσ(2)P2 ,…⊗,Tσ(n)Pn
where P[j] is the WV of Tj(j=1,2,…,n), Pj∈[0,1] and ∑j=1nPj=1.
According to the operational laws of the BPFNs, we can obtain the following theorems. Since their proofs are similar to those mentioned above, we are omitting them here.
Theorem 4.9 Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉(j=1,2,…,n) be a assemblage of BPFNs. We also find BPFOWG by
BPFOWG(T1,T2,…,Tn)=(∏ j=1n(μσ(j)++λσ(j)+)Pj-∏j=1n(λσ(j)+)Pj,∏j=1n(λσ(j)+)Pj,1-∏j=1n(1-νσ(j)+)Pj,∏ j=1n((-μσ(j)-)-(-λσ(j)-))Pj-∏j=1n(-λσ(j)-)Pj,∏j=1n(-λσ(j)-)Pj,1-∏j=1n(1-(-νσ(j)-))Pj)
Theorem 4.10 (Idempotency) Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉 be a assemblage of BPFNs. If, Tj=T=〈μ+,λ+,ν+,μ-,λ-,ν-〉 for all j, then
Theorem 4.11 (Monotonicity) Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉 and Tj*=〈(μj+)*,(λj+)*,(νj+)*,(μj-)*,(λj-)*,(νj-)*〉 be two families of BPFNs. If Tj≤Tj* for all (j=1,2,…,n) then
Theorem 4.12 (Commutativity) Let Tj=〈μj+,λj+,νj+,μj-,λj-,νj-〉 be a assemblage of BPFNs.
where (σ(1),σ(2),…,σ(n)) is any permutation of (1,2,…,n).
When both the ordered positions of the bipolar picture fuzzy arguments and the arguments themselves need to be weighted, BPFWG can be generalized to the following bipolar picture fuzzy hybrid
geometric operator.
Definition 4.13 ABPFHG operator is a mapping BPFHG:𝔛n→𝔛 such that P[j] is the WV of Tj(j=1,2,…,n), Pj∈[0,1]and∑j=1nPj=1.
T¨σ(j) is the j[th] largest of the weighted BPFNs. Here T¨j=nwjTj, (1,2,…,n), n is the number of BPFNs and w=(w1,w2…wn) is the standard WV.
We can drive the following theorem based on the operations of the PFNs which is similar to Theorem 4.4.
Theorem 4.14 LetTj=〈μj+,λj+,νj+,μj-,λj-,νj-〉(j=1,2,…,n) be a assemblage of BPFNs. We also find BPFHG by
BPFHG(T1,T2,…,Tn)=(∏ j=1n(μ¨σ(j)++λ¨σ(j)+)Pj-∏j=1n(λ¨σ(j)+)Pj,∏j=1n(λ¨σ(j)+)Pj,1-∏j=1n(1-ν¨σ(j)+)Pj,∏ j=1n((-μ¨σ(j)-)-(-λ¨σ(j)-))Pj-∏j=1n(-λ¨σ(j)-)Pj,∏j=1n(-λ¨σ(j)-)Pj,1-∏j=1n(1-(-ν¨σ(j)-))Pj).
The weighting vector associated with the operator of BPFWG, the operator of BPFOWG and the operator of BPFHG can be assessed as identical to that of the other operators. For example, a normal
distribution-based approach can be used to evaluate weights. The distinctive feature of the approach is that it can reduce the effect of bias claims on the outcome of the judgment by assigning low
weights to the wrong ones.
In this section of the paper, we define the distance measures of bipolar picture fuzzy sets.
Definition 4.15 A function τ:BPFS(X)×BPFS(X)→[0,+∞) is a distance measure between BPFS-sets if it satisfies follow conditions:
τ(O,P)=0 iff O = P.
Theorem 4.16 GivenX={δ1,δ2,…,δn} is a universe of discourse. ForO,P∈BPFS(X). We have some distance measure between BPFSs.
We can actually confirm that the functions in Theorem 4.16 satisfy distance measuring properties between bipolar picture fuzzy sets. In it, τE(O,P) is typically used to calculate object distance in
geometry, and τH(O,P) is used in the theory of information.
Example 4.17 Assume there are three patterns denoted by BPFSs on X={δ1,δ2,δ3} as follows:
are three bipolar picture fuzzy set in X. Using Theorem 4.16, we get
In this section, we discuss some distance measures of BPFSs and their application to the pattern recognition. Pattern recognition is a science and technology discipline which aims to classify objects
into a number of categories. This method is widely used in the identification of data analysis, shapes, pattern classification, traffic analysis & regulation, natural language processing, rock
identification, biological stimuli, odor identification, understanding of the DNA sample, credit fraud detection, biometrics including fingerprints, palm vein technology & face recognition, medical
diagnosis, weather forecasting, intelligence, informatics, voice to text transition, terrorism identification, radar tracking, and automatic military target recognition, etc. In Fig. 3 step by step
method is shown of facial recognition which is example of pattern recognition.
Example 5.1 Suppose that there are three patterns denoted by BPFSs on X={δ1,δ2,δ3} as follows:
Now, there is a sample,
The question is, what pattern belongs to B? By applying the distance measure τE. Using Theorem 4.16, we get
We see that B belongs to pattern Ω3 if we use the distance measure τE.
MCDM method using the aggregation operators defined for BPFNs is presented in this section.
Suppose that T={T1,T2,…,Tp} is the set of alternatives and C={C1,C2,…,Cq} is the set of criterion. Let P be the WV, s.t Pj∈[0,1] and ∑j=1nPj=1, (j=1,2,…,n) and P[j] show the weight of C[j].
Alternatives on an attribute are reviewed by the decision-maker (DM) and the assessment measurements has to be in the BPFN. Assume that λ=(αij)p×q is the decision matrix provided by DM.(αij)
represent a BPFN for alternative T[i] associated with the criterion C[j]. Here we have some conditions such that
μij+,λij+,νij+,μij-,λij- and νij-∈[0,1]
0≤μij++λij++νij+≤1 and -1≤μij-+λij-+νij-≤0.
Indeed an algorithm is being developed to discuss MCDM.
Step 1.
The DM has given its personal opinion in the form of BPFNs. αij=〈μij+,λij+,νij+,μij-,λij-,νij-〉 towards the alternative T[i] and hence construct a BPF decision matrix λ=(αij)p×q as
[(μ11+,λ11+,ν11+,μ11-,λ11-,ν11-)⋯(μ1q+,λ1q+,ν1q+,μ1q-,λ1q-,ν1q-)⋮⋱⋮(μp1+,λp1+,νp1+,μp1-,λp1-,νp1-)⋯(μpq+,λpq+,νpq+,μpq-,λpq-,νpq-) ]
Step 2.
Normalize the decision matrix. If there are different types of criteria or attributes like cost and benefit. By normalize the decision matrix we deal all criteria or attributes in the same way.
Otherwise, different criterion or attributes should be aggregate in different ways.
where αijc show the compliment of αij.
Step 3.
Based on decision matrix acquired from Step 2, the aggregated value of the alternative T[i] under various parameter C[j] is obtained using either BPFWA or BPFOWA or BPFHA operators and hence get the
collective value r[i] for each alternative Ti(i=1,2,…m).
Step 4.
Calculate the score functions for all r[i] for BPFNs.
Step 5.
Rank all r[i] as per the score values to choose the most desirable option.
The flow chart of proposed algorithm is expressed in Fig. 4.
We are considering quantitative examples in the selection of mushroom farming alternatives in Pakistan to show the effectiveness of the proposed processes. Filled with taste and an incredible
nutritional composition to boot, oyster mushrooms will be a worthy complement to a balanced diet. Here several categories of oyster mushrooms that differ a little in flavor and nutritional benefits.
In this paper, we will focus majorly on king oyster mushrooms (KOM) and their nutritional benefits. In all types of mushroom, preparation procedures that we describe can be included. Pleurotus
Eryngii (PE) (Fig. 5) is a real term for king oyster mushrooms. They are also known as French horn, royal trumpet, king brown, king trumpet and steppe boletus. The King oyster mushroom is, as its
name suggests, the largest of all oyster mushrooms. This is evidently rising in the Middle East and North Africa. It is also extensively grown in Asia, in a variety of countries, as well as in Italy,
Australia and the USA. Looking at the health benefits of oyster mushrooms, there are several positive features of this study. Very good sources of riboflavin, iron, niacin, phosphorus, potassium,
copper, Protein, vitamin B6, pantothenic acid,folate, magnesium, zinc and manganese from the right source. Mostly limited in cholesterol and saturated fat. Only about 35 per 100 g king oyster
mushroom calories. King oyster mushrooms have a good protein source and are the ideal complement to vegetarian or vegan diets. They are not a full source of protein and must ensure that a number of
various sources of protein are included in your healthy diet. Alone last year, Americans grew more than two million pounds of exotic mushrooms. Oyster mushrooms, a type of exotic mushrooms, are among
the best and fastest growing exotic mushrooms. They could even grow in about six weeks, and they are looking to sell around 6 Dollar a pound wholesale and 12 Dollar a pound retail. They looked
incredibly easy to produce, they’re growing quickly, and they can make you decent money–-all the justifications whether you like to oyster mushrooms to grow for financial gain.
KOM is an enormous, important food naturalized to Asia and the rest of Europe. Although hard to find in the wild, it is widely cultivated and famous for its buttery taste and eggplant-like flavor,
particularly in some Asian and African cuisines. predominant Chinese medication has for centuries recognized the importance of KOM and other medicative mushrooms.
Here are among the most possibly the best-researched advantages of KOM.
1. Immune System Support
B-glucans in KOM enable them are some of the healthiest meals on the earth to support the immune function toward short-and long-term diseases [51]. Unlike other food products that either activate or
inhibit the immune system, the mushrooms balanced the immune cells. Plus, KOM are filled with other antioxidants to help avoid harm caused by free radicals and oxidative stress so that the immune
cells can protect itself against aging [52].
2. Reducing of High Blood Pressure
Your body requires nutrients like vitamin D to stabilize your heart rate and blood pressure. Do you think that the majority of people living in colder climates are deficient with vitamin D? One
research found that edible mushrooms, such as oysters, reduced blood pressure in rats with chronic or uncertain high blood pressure [53].
3. Regulating Cholesterol Levels
Although mushrooms like KOM have a tasty taste and texture and no cholesterol, they are a fine replacement for meat in several steamed dishes. One study also initiate that in people with diabetes,
the intake of oyster mushrooms decreased glucose and cholesterol levels [54].
4. Strong Bones
KOM provide a number of essential ingredients for building better bones. Vitamin D and magnesium in particular. While most persons concentrate on calcium, your body also requires vitamin D and
magnesium to absorb and preserve calcium in your bones.
5. Anti-Inflammatory Properties
B-glucans and nutrients in KOM make it a perfect food to reduce inflammation. Some work indicates that, besides B-glucans, some of the anti-inflammatory effects of oysters come from a special and
somewhat obscure amino acid called ergothioneine. According to study, ergothioneine reduces “systemic” inflammation around the body, frequently leading to diseases such as dementia and diabetes.
6. Anti-Cancer Properties
B-glucans in mushrooms, such as KOM, serve as powerful antioxidants that can offer some protection from cancer. One research showed that oyster mushrooms have the potential to be involved in some
forms of cancer cells.
Various substrates, such as sawdust (SD) and rice straw (RS), have been used to grow KOM. Sun-dried SD, wheat bran and rice husk were combined together. Water was applied to change the water
absorption and CaCO[3] was blended at a rate of 1% of the mixture. The substrate mixture was packed with airtight plastic polymer bottles. The bottles were sterilized, and after cool back to normal
temperature, the sterilized bottles were tested separately. We were using CaCO[3], straw, sawdust, corn cob and rice bran to grow King oyster mushrooms. They are combined based on specific ratios,
Nguyen et al. [55] take into account each combining formula as an alternative given in Tab. 4.
Alternatives Combining formula
T[1] 1% CaCO[3], 40% straw, 29% sawdust, 30% corn cob and 0% rice bran
T[2] 1% CaCO[3], 40% straw, 27% sawdust, 27% corn cob and 5% rice bran
T[3] 1% CaCO[3], 40% straw, 24% sawdust, 25% corn cob and 10% rice bran
T[4] 1% CaCO[3], 40% straw, 17% sawdust, 17% corn cob and 25% rice bran
There are four types of alternatives T[i](i = 1, 2, 3, 4), given in Tab. 4. Analysis the effects of rapidly increasing materials on the productivity growth of king oyster mushrooms. We consider C[1]
= infection rate, C[2] = Biological productivity, C[3] = diameter of mushroom cap and C[4] = diameter of mushroom stalks as attributes. In this example we use BPFNs as input data for ranking the
given alternatives under the given attributes. Also, the WV P is (0.3, 0.2, 0.1, 0.4) and standard WV w is (0.2, 0.3, 0.3, 0.2).
Using BPFWG operator
Step 1.
Construct the decision matrix given by the decision maker in Tab. 5 consist on bipolar picture fuzzy information.
C[1] C[2] C[3] C[4]
T[1] (0.5, 0.3, 0.2, −0.1, −0.6, −0.2) (0.5, 0.3, 0.2, −0.1, −0.6, −0.2) (0.3, 0.3, 0.1, −0.1, −0.5, −0.2) (0.4, 0.3,0.1, −0.1, −0.4, −0.2)
T[2] (0.2, 0.4, 0.1, −0.1, −0.3, −0.2) (0.2, 0.4,0.3, −0.1, −0.2, −0.1) (0.2, 0.2,0.4, −0.1, −0.2, −0.4) (0.2, 0, 0.2, −0.1, −0.6, −0.2)
T[3] (0.3, 0.4, 0.1, −0.2, −0.3, −0.4) (0.2, 0.1, 0.4, −0.4, −0.2, −0.1) (0.1, 0.3, 0.4, −0.2, −0.2, −0.4) (0.2, 0.4, 0.3, −0.1, −0.2, −0.3)
T[4] (0.1, 0.2, 0.3, −0.4, −0.1, −0.2) (0.3, 0.4, 0.1, −0.1, −0.2, −0.3) (0.2, 0.4, 0.3, −0.1, −0.3, −0.4) (0.2, 0.1, 0.2, −0.2, −0.5, −0.1)
Step 2.
Normalize the decision matrix, because the attribute C[1] = price, given in Tab. 6.
C[1] C[2] C[3] C[4]
T[1] (0.2, 0.3, 0.5, −0.2, −0.6, −0.1) (0.5, 0.3,0.2, −0.1, −0.6, −0.2) (0.3, 0.3, 0.1, −0.1, −0.5, −0.2) (0.4, 0.3,0.1, −0.1, −0.4, −0.2)
T[2] (0.1, 0.4,0.2, −0.2, −0.3, −0.1) (0.2, 0.4,0.3, −0.1, −0.2, −0.1) (0.2, 0.2,0.4, −0.1, −0.2, −0.4) (0.2, 0, 0.2, −0.1, −0.6, −0.2)
T[3] (0.1, 0.4, 0.3, −0.4, −0.3, −0.2) (0.2, 0.1, 0.4, −0.4, −0.2, −0.1) (0.1, 0.3, 0.4, −0.2, −0.2, −0.4) (0.2, 0.4, 0.3, −0.1, −0.2, −0.3)
T[4] (0.3, 0.2, 0.1, −0.2, −0.1, −0.4) (0.3, 0.4, 0.1, −0.1, −0.2, −0.3) (0.2, 0.4, 0.3, −0.1, −0.3, −0.4) (0.2, 0.1, 0.2, −0.2, −0.5, −0.1)
Step 3.
Evaluate ri=BPFWG(ri1,ri2,…,rip).
Step 4.
Calculate the score functions for all r[i].
Step 5.
Rank all the ri(i=1,2,…,p) according to the score values,
r[2] corresponds to T[2], so T[2] is the best alternative.
Using BPFOWG operator
Step 1.
Construct the decision matrix given by decision maker consist on bipolar picture fuzzy information, given in Tab. 7.
C[1] C[2] C[3] C[4]
T[1] (0.5, 0.3, 0.2, −0.1, −0.6, −0.2) (0.5, 0.3, 0.2, −0.1, −0.6, −0.2) (0.3, 0.3, 0.1, −0.1, −0.5, −0.2) (0.4, 0.3,0.1, −0.1, −0.4, 0.2)
T[2] (0.2, 0.4, 0.1, −0.1, −0.3, −0.2) (0.2, 0.4,0.3, −0.1, −0.2, −0.1) (0.2, 0.2,0.4, −0.1, −0.2, −0.4) (0.2, 0, 0.2, −0.1, −0.6, −0.2)
T[3] (0.3, 0.4, 0.1, −0.2, −0.3, −0.4) (0.2, 0.1, 0.4, −0.4, −0.2, −0.1) (0.1, 0.3, 0.4, −0.2, −0.2, −0.4) (0.2, 0.4, 0.3, −0.1, −0.2, −0.3)
T[4] (0.1, 0.2, 0.3, −0.4, −0.1, −0.2) (0.3, 0.4, 0.1, −0.1, −0.2, −0.3) (0.2, 0.4, 0.3, −0.1, −0.3, −0.4) (0.2, 0.1, 0.2, −0.2, −0.5, −0.1)
Step 2.
Normalize the decision matrix, because the attribute C[1] = price, given in Tab. 8.
C[1] C[2] C[3] C[4]
T[1] (0.2, 0.3, 0.5, −0.2, −0.6, −0.1) (0.5, 0.3,0.2, −0.1, −0.6, −0.2) (0.3, 0.3, 0.1, −0.1, −0.5, −0.2) (0.4, 0.3,0.1, −0.1, −0.4, −0.2)
T[2] (0.1, 0.4,0.2, −0.2, −0.3, −0.1) (0.2, 0.4,0.3, −0.1, −0.2, −0.1) (0.2, 0.2,0.4, −0.1, −0.2, −0.4) (0.2, 0, 0.2, −0.1, −0.6, −0.2)
T[3] (0.1, 0.4, 0.3, −0.4, −0.3, −0.2) (0.2, 0.1, 0.4, −0.4, −0.2, −0.1) (0.1, 0.3, 0.4, −0.2, −0.2, −0.4) (0.2, 0.4, 0.3, −0.1, −0.2, −0.3)
T[4] (0.3, 0.2, 0.1, −0.2, −0.1, −0.4) (0.3, 0.4, 0.1, −0.1, −0.2, −0.3) (0.2, 0.4, 0.3, −0.1, −0.3, −0.4) (0.2, 0.1, 0.2, −0.2, −0.5, −0.1)
Step 3.
Evaluate ri=BPFOWG(ri1,ri2,…,rip).
Step 4.
Calculate the score functions for all r[i].
Step 5.
Rank all the ri(i=1,2,…,p) according to the score values,
r[2] corresponds to T[2], so T[2] is the best alternative.
Using BPFHG operator
Step 1.
Construct the decision matrix given by decision maker consist on bipolar picture fuzzy information given in Tab. 9.
C[1] C[2] C[3] C[4]
T[1] (0.5, 0.3, 0.2, −0.1, −0.6, −0.2) (0.5, 0.3, 0.2, −0.1, −0.6, −0.2) (0.3, 0.3, 0.1, −0.1, −0.5, −0.2) (0.4, 0.3,0.1, −0.1, −0.4, −0.2)
T[2] (0.2, 0.4, 0.1, −0.1, −0.3, −0.2) (0.2, 0.4,0.3, −0.1, −0.2, −0.1) (0.2, 0.2,0.4, −0.1, −0.2, −0.4) (0.2, 0, 0.2, −0.1, −0.6, −0.2)
T[3] (0.3, 0.4, 0.1, −0.2, −0.3, −0.4) (0.2, 0.1, 0.4, −0.4, −0.2, −0.1) (0.1, 0.3, 0.4, −0.2, −0.2, −0.4) (0.2, 0.4, 0.3, −0.1, −0.2, −0.3)
T[4] (0.1, 0.2, 0.3, −0.4, −0.1, −0.2) (0.3, 0.4, 0.1, −0.1, −0.2, −0.3) (0.2, 0.4, 0.3, −0.1, −0.3, −0.4) (0.2, 0.1, 0.2, −0.2, −0.5, −0.1)
Step 2.
Normalize the decision matrix, because the attribute C[1] = price, given in Tab. 10.
C[1] C[2] C[3] C[4]
T[1] (0.2, 0.3, 0.5, −0.2, −0.6, −0.1) (0.5, 0.3,0.2, −0.1, −0.6, −0.2) (0.3, 0.3, 0.1, −0.1, −0.5, −0.2) (0.4, 0.3,0.1, −0.1, −0.4, −0.2)
T[2] (0.1, 0.4,0.2, −0.2, −0.3, −0.1) (0.2, 0.4,0.3, −0.1, −0.2, −0.1) (0.2, 0.2,0.4, −0.1, −0.2, −0.4) (0.2, 0, 0.2, −0.1, −0.6, −0.2)
T[3] (0.1, 0.4, 0.3, −0.4, −0.3, −0.2) (0.2, 0.1, 0.4, −0.4, −0.2, −0.1) (0.1, 0.3, 0.4, −0.2, −0.2, −0.4) (0.2, 0.4, 0.3, −0.1, −0.2, −0.3)
T[4] (0.3, 0.2, 0.1, −0.2, −0.1, −0.4) (0.3, 0.4, 0.1, −0.1, −0.2, −0.3) (0.2, 0.4, 0.3, −0.1, −0.3, −0.4) (0.2, 0.1, 0.2, −0.2, −0.5, −0.1)
Step 3.
Evaluate ri=BPFHG(ri1,ri2,…,rip).
Before evaluating r[i] we use standard WV to find the T¨j given in Tab. 11, where T¨j=nwjTj.
C[1] C[2] C[3] C[4]
T¨1 (0.19, 0.38, 0.43, −0.17, −0.66, −0.08) (0.45, 0.38,0.16, −0.09, −0.66, −0.16) (0.28, 0.38, 0.08, −0.09, −0.57, −0.16) (0.37, 0.38,0.08, −0.09, −0.48, −0.17)
T¨2 (0.10, 0.33,0.23, −0.20, −0.24, −0.12) (0.21, 0.33,0.35, −0.09, −0.15, −0.12) (0.19, 0.15,0.46, −0.09, −0.15, −0.49) (0.15, 0, 0.23, −0.11, −0.54, −0.23)
T¨3 (0.10, 0.33, 0.35, −0.17, −0.06, −0.46) (0.17, 0.06, 0.46, −0.40, −0.15, −0.12) (0.10, 0.24, 0.46, −0.19, −0.15, −0.46) (0.20, 0.33, 0.35, −0.09, −0.14, −0.35)
T¨4 (0.30, 0.28, 0.08, −0.22, −0.16, −0.34) (0.27, 0.48, 0.08, −0.11, −0.28, −0.25) (0.18, 0.48, 0.25, −0.10, −0.38, −0.34) (0.22, 0.16, 0.16, −0.18, −0.57, −0.08)
Step 4.
Calculate the score functions for all r[i].
Step 5.
Rank all the ri(i=1,2,…,p) according to the score values,
r[2] corresponds to T[2], so T[2] is the best alternative.
The proposed AOs BPFWG operator, BPFOWG operator and BPFHG operator are compared as shown in Tab. 12 below, which lists the final comparative study ranked among the top four alternatives. The best
selection made by any of the proposed operators and current operators, as shown in Tab. 12, validates the consistency and authenticity of the proposed methods.
Method Ranking of alternatives The optimal alternative
PFWA (Garg [40]) T2≻T1≻T4≻T3 T[2]
PFOWA (Garg [40]) T2≻T1≻T3≻T4 T[2]
PFHA (Garg [40]) T2≻T1≻T3≻T4 T[2]
PFWG (Wang et al. [42]) T2≻T3≻T4≻T2 T[2]
PFOWG (Wang et al. [42]) T2≻T3≻T4≻T2 T[2]
PFHG (Wang et al. [42]) T2≻T4≻T1≻T3 T[2]
PFDWA (Jana et al. [41]) T2≻T3≻T4≻T1 T[2]
PFDOWA (Jana et al. [41]) T2≻T3≻T4≻T1 T[2]
PFDHWA (Jana et al. [41]) T2≻T3≻T1≻T4 T[2]
PFDWG (Jana et al. [41]) T2≻T1≻T4≻T3 T[2]
PFDOWG (Jana et al. [41]) T2≻T1≻T3≻T4 T[2]
PFDHWG (Jana et al. [41]) T2≻T1≻T4≻T3 T[2]
BPFWG (Proposed) T2≻T1≻T4≻T3 T[2]
BPFOWG (Proposed) T2≻T1≻T4≻T3 T[2]
BPFHG (Proposed) T2≻T1≻T4≻T3 T[2]
MCDM has been studied to solve complex real-world problems that involve uncertainty, imprecision and ambiguity due to vague and incomplete information. The MCDM techniques practically rely on fuzzy
sets and fuzzy models that are considered to address vagueness and uncertainties. The existing fuzzy set theoretic models fail to deal with real life situations when modeling need to assign
bipolarity (positive and negative aspects) to each of the degrees of MD (belonging-ness), neutral MD (not-decided), and NMD (refusal). In order to handle such MCDM problems, in this study, we
introduced a new extension of fuzzy sets named as BPFS. A BPFS is the hybrid structure of BFS and PFS. The notion of a bipolar picture fuzzy number (BPFN) is superior than existing bipolar fuzzy
number and picture fuzzy number. We introduced some algebraic operations and key properties of BPFSs as well as some new distance measures of BPFSs. We presented score function, accuracy function and
certainty function for bipolar picture fuzzy information aggregation. Information aggregation plays an important role in the MCDM, and therefore in this study, some new aggregation operators (AOs)
named as “bipolar picture fuzzy weighted geometric operator, bipolar picture fuzzy ordered weighted geometric operator, and bipolar picture fuzzy hybrid geometric operator” are developed.
Additionally, on the basis of these AOs, a new MCDM approach has been developed for the ranking of objects using BPFNs. The presented scientific method is illustrated by a numerical model to
demonstrate its effectiveness and sustainability.
In further research, we can extend proposed aggregation operators to some other MCDM techniques including; TOPSIS, VIKOR, AHP, ELECTRE family and PROMETHEE family. Long term work will pay special
attention to Heronian mean, Einstein, Bonferroni mean, Dombi AOs and so on. We keep hoping that our research results will be beneficial for researchers working in the fields of information fusion,
pattern recognition, image recognition, machine learning, decision support systems, soft computing and medicine.
Author’s Contributions: The authors contributed to each part of this paper equally. The authors read and approved the final manuscript.
Funding Statement: The authors received no specific funding for this study.
Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study.
Zadeh, L. A. (1965). Fuzzy sets. , 8(3), 338–353. DOI 10.1016/S0019-9958(65)90241-X. Atanassov, K. T. (1986). Intuitionistic fuzzy sets. , 20(1), 87–96. DOI 10.1016/S0165-0114(86)80034-3. Yager, R.
R. (2013). Pythagorean fuzzy subsets. 2013 Joint IFSA World Congress and NAFIPS Annual Meeting, pp. 57–61. Edmonton, Canada: IEEE. Yager, R. R. (2014). Pythagorean membership grades in multi-criteria
decision making. , 22(4), 958–965. Yager, R. R., Abbasov, A. M. (2013). Pythagorean membership grades, complex numbers, and decision making. , 28(5), 436–452. DOI 10.1002/int.21584. Zhang, W. R.
(1994). Bipolar fuzzy sets and relations: A computational framework for cognitive modeling and multiagent decision analysis. Proceedings of the First International Joint Conference of the North
American Fuzzy Information Processing Society Biannual Conference, pp. 305–309. Zhang, W. R. (1998). (Yin Yang) Bipolar fuzzy sets. IEEE International Conference on Fuzzy Systems, vol. 1, pp.
835–840. Lee, K. M. (2000). Bipolar-valued fuzzy sets and their basic operations. Proceeding International Conference, pp. 307–317. Bangkok, Thailand. Alcantud, J. C. R., Santos-García, G., Peng, X.
D., Zhan, J. (2019). Dual extended hesitant fuzzy sets. , 11(5), 1–13. DOI 10.3390/sym11050714. Alcantud, J. C. R., Feng, F., Yager, R. R. (2020). An N-soft set approach to rough sets. , 28(11),
2996–3007. DOI 10.1109/TFUZZ.91. Akram, M., Dudek, W. A., Ilyas, F. (2019). Group decision-making based on Pythagorean fuzzy TOPSIS method. , 34(7), 1455–1475. DOI 10.1002/int.22103. Akram, M.,
Dudek, W. A., Dar, J. M. (2019). Pythagorean Dombi fuzzy aggregation operators with application in multicriteria decision-making. , 34(11), 3000–3019. DOI 10.1002/int.22183. Ashraf, S., Abdullah, S.,
Mahmood, T. (2019). Spherical fuzzy Dombi aggregation operators and their application in group decision making problems. , 11(7), 2731–2749. DOI 10.1007/s12652-019-01333-y. Eraslan, S., Karaaslan, F.
(2015). A group decision making method based on TOPSIS under fuzzy soft environment. , 3, 30–40. Feng, F., Zheng, Y., Alcantud, J. C. R., Wang, Q. (2020). Minkowski weighted score functions of
intuitionistic fuzzy values. , 8(7), 1–30. DOI 10.3390/math8071143. Garg, H. (2019). Neutrality operations-based Pythagorean fuzzy aggregation operators and its applications to multiple attribute
group decision-making process. , 11(7), 3021–3041. DOI 10.1007/s12652-019-01448-2. Garg, H., Arora, R. (2018). Dual hesitant fuzzy soft aggregation operators and their application in decision-making.
, 10(5), 769–789. DOI 10.1007/s12559-018-9569-6. Garg, H., Arora, R. (2019). Generalized intuitionistic fuzzy soft power aggregation operator based on t-norm and their application in multicriteria
decision-making. , 34(2), 215–246. DOI 10.1002/int.22048. Jose, S., Kuriaskose, S. (2014). Aggregation operators, score function and accuracy function for multi criteria decision making in
intuitionistic fuzzy context. , 20(1), 40–44. Karaaslan, F. (2015). Neutrosophic soft set with applications in decision making. , 4(2), 1–20. Liu, P., Ali, Z., Mahmood, T., Hassan, N. (2020). Group
decision-making using complex q-rung orthopair fuzzy Bonferroni mean. , 13(1), 822–851. Liu, P., Wang, P. (2020). Multiple attribute group decision making method based on intuitionistic fuzzy
Einstein interactive operations. , 22(3), 790–809. DOI 10.1007/s40815-020-00809-w. Wang, L., Li, N. (2020). Pythagorean fuzzy interaction power Bonferroni mean aggregation operators in multiple
attribute decision making. , 35(1), 150–183. DOI 10.1002/int.22204. Yang, W., Cai, L., Edalatpanah, S. A., Smarandache, F. (2020). Triangular single valued neutrosophic data envelopment analysis:
Application to hospital performance measurement. , 12(4), 1–14. DOI 10.3390/sym12040588. Smarandache, F. (1999). Rehoboth, DE, USA: American Research Press. Liu, X., Ju, Y., Yang, S. (2014). Hesitant
intuitionistic fuzzy linguistic aggregation operators and their applications to multi-attribute decision making. , 26(3), 1187–1201. DOI 10.3233/IFS-131083. Naeem, K., Riaz, M., Peng, X. D., Afzal,
D. (2019). Pythagorean fuzzy soft MCGDM methods based on TOPSIS, VIKOR and aggregation operators. , 37(5), 6937–6957. DOI 10.3233/JIFS-190905. Naeem, K., Riaz, M., Afzal, D. (2019). Pythagorean
m-polar fuzzy sets and TOPSIS method for the selection of advertisement mode. , 37(6), 8441–8458. DOI 10.3233/JIFS-191087. Peng, X. D., Yang, Y. (2015). Some results for Pythagorean fuzzy sets. , 30
(11), 1133–1160. DOI 10.1002/int.21738. Peng, X. D., Yuan, H. Y., Yang, Y. (2017). Pythagorean fuzzy information measures and their applications. , 32(10), 991–1029. DOI 10.1002/int.21880. Peng, X.,
Selvachandran, G. (2017). Pythagorean fuzzy set: State of the art and future directions. , 52(3), 1873–1927. DOI 10.1007/s10462-017-9596-9. Riaz, M., Hashmi, M. R. (2019). Linear Diophantine fuzzy
set and its applications towards multi-attribute decision making problems. , 37(4), 5417–5439. DOI 10.3233/JIFS-190550. Riaz, M., Karaaslan, F., Farid, H. M. A., Hashmi, M. R. (2020). Some q-rung
orthopair fuzzy hybrid aggregation operators and TOPSIS method for multi-attribute decision-making. , 39(1), 1227–1241. DOI 10.3233/JIFS-192114. Riaz, M., Farid, H. M. A., Kalsoom, H., Pamucar, D.,
Chu, Y. M. et al. (2020). A Robust q-rung orthopair fuzzy Einstein prioritized aggregation operators with application towards MCGDM, Symmetry. , 12(6), 1058. DOI 10.3390/sym12061058. Riaz, M.,
Tehrim, S. T. (2020). Cubic bipolar fuzzy set with application to multi-criteria group decision making using geometric aggregation operators. , 24, 16111–16133. DOI 10.1007/s00500-020-04927-3.
Çağman, N., Karataş, S., Enginoglu, S. (2011). Soft topology. , 62(1), 351–358. DOI 10.1016/j.camwa.2011.05.016. Shabir, M., Naz, M. (2011). On soft topological spaces. , 61(7), 1786–1799. DOI
10.1016/j.camwa.2011.02.006. Cuong, B. C. (2014). Picture fuzzy sets. , 30(4), 409–420. Cuong, B. C. (2019). Pythagorean picture fuzzy sets, part 1-basic notions. , 35(4), 293–304. DOI 10.15625/
1813-9663/35/4/13898. Garg, H. (2017). Some picture fuzzy aggregation operators and their applications to multicriteria decision-making. , 42(12), 5275–5290. DOI 10.1007/s13369-017-2625-9. Jana, C.,
Senapati, T., Pal, M., Yager, R. R. (2019). Picture fuzzy Dombi aggregation operators: Application to MADM process. , 74(4), 99–109. DOI 10.1016/j.asoc.2018.10.021. Wang, C., Zhou, X., Tu, H., Tao,
S. (2017). Some geometric aggregation operators based on picture fuzzy sets and their application in multiple attribute decision making. , 37, 477–492. Pamucar, D. (2020). Normalized weighted
geometric Dombi Bonferoni mean operator with interval grey numbers: Application in multicriteria decision making. , 1(1), 44–52. DOI 10.31181/rme200101044p. Pamucar, D., Jankovic, A. (2020). The
application of the hybrid interval rough weighted Power-Heronian operator in multi-criteria decision making. , 3(2), 54–73. DOI 10.31181/oresta2003049p. Ramakrishnan, K. R., Chakraborty, S. (2020). A
cloud TOPSIS model for green supplier selection, Facta Universitatis series. , 18(3), 375–397. Riaz, M., Pamucar, D., Athar Farid, H. M., Hashmi, M. R. (2020). q-Rung orthopair fuzzy prioritized
aggregation operators and their application towards green supplier chain management. , 12(6), 976. DOI 10.3390/sym12060976. Riaz, M., Razzaq, A., Kalsoom, H., Pamucar, D., Athar Farid, H. M. et al.
(2020). q-Rung orthopair fuzzy geometric aggregation operators based on generalized and group-generalized parameters with application to water loss management. , 12(8), 1236. DOI 10.3390/sym12081236.
Si, A., Das, S., Kar, S. (2019). An approach to rank picture fuzzy numbers for decision making problems. , 2(2), 54–64. DOI 10.31181/dmame1902049s. Sinani, F., Erceg, Z., Vasiljevic, M. (2020). An
evaluation of a third-party logistics provider: The application of the rough Dombi-Hamy mean operator. , 3(1), 92–107. Delia, I., Ali, M., Smarandache, F. (2015). Bipolar neutrosophic sets and their
application based on multi-criteria decision-making problems. Proceedings of the 2015 International Conference on Advanced Mechatronic Systems, Beijing, China. Abdullah, N., Abdulghani, R., Ismail,
S. M., Abidin, H. Z. (2017). Immune-stimulatory potential of hot water extracts of selected edible mushrooms. , 28(3), 374–387. DOI 10.1080/09540105.2017.1293011. Tanaka, A., Nishimura, M., Sato, Y.,
Nishihira, J. (2016). Enhancement of the TH1-phenotype immune system by the intake of oyster mushroom (Tamogitake) extract in a double-blind, placebo-controlled study. , 6(4), 42. Alam, N., Yoon, K.
N., Lee, J. S., Cho, H. J., Shim, M. J. et al. (2011). Dietary effect of pleurotus eryngii on biochemical function and histology in hypercholesterolemic rats. , 18(4), 403–409. DOI 10.1016/
j.sjbs.2011.07.001. Cho, J. H., Kim, D. W., Kim, S., Kim, S. J. (2017). In vitro antioxidant and in vivo hypolipidemic effects of the king oyster culinary-medicina mushroom, pleurotus eryngi var.
ferulae DDl01 (agaricomycetes), in rats with high-fat diet-induced fatty liver and hyperlipidemia. , 19(2), 107–119. DOI 10.1615/IntJMedMushrooms.v19.i2.20. Nguyen, T. B. T., Ngo, X. N., Nguyen, T.
T., Tran D. A. (2016). Evaluating the growth and yield of king oyster mushroom (Pleurotus eryngii (DC.:Fr.) Quel) on different substrates. , 14(5), 816–823. | {"url":"https://cdn.techscience.cn/ueditor/files/cmes/TSP_CMES_14174/TSP_CMES_14174.xml?t=20220620","timestamp":"2024-11-09T19:14:53Z","content_type":"application/xml","content_length":"666101","record_id":"<urn:uuid:d15299dd-e957-43f8-8a3a-9dddbdd25712>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00113.warc.gz"} |
Here it is, for ultra high angles:
Wood's Boomer (and coincidentally, J.D.) High Angle Guide
High Angles:
Wind: Must be 0 or 1 wind.
Where the thin yellow lines intersect is where a full powered shot of the angle indicated will land. For example, if I were to hit brenden720, I would use 83, full powered.
1) If the wind is blowing at 2, and with your shot, not against, use 1 degree less. For example, if the wind is blowing 2 to the right, I would use 84 degrees instead of 83 to hit brenden720.
2) Bolded yellow line on the bottom is an easy indication of an 87 full powered shot; the "[ALL]" button to the middle markers of the power-bar.
3) An 85 degree shot can be measured by using your right click button and dragging the wind-indicator next to your enemy. It should be about half a centimeter away from it, indicated by the bolded
yellow X. Same applies for an 86 shot, but on the other side of the wind-indicator.
4) Do not use the "1 degree = 3 centimeter" rule, because it is not always the same for different resolution monitors.
5) For 89 angles, try it once with 0 wind, see where it lands, and memorize how far away that is. This would be a "one degree length".
6) For enemies that are more than 1 screen away, use common subtraction. If an enemy is 2 screens away, think of 1 screen as -9 degrees (90-81=9), so exactly 2 screens away would be -18, or 72
degrees (90-72=18). Rule applies for all angles.
7) Note elevation. If your enemy is higher than you by a substantial (about half-screen vertically) amount, go a little closer. If your enemy is lower than you by a substantial amount, walk a little
farther back.
8) If you want to make a better approximation of an 88 degree shot, you can align your mobile relatively to your target's mobile according to the measurement in the upper right corner, indicated by
the 88. A more exact description is that the left edge of your mobile touches the left side of the first item slot, and the right edge of your target's mobile touches the right side of the sixth item
9) With an almost-exact 88 and 87, and a near-exact 86 and 85, you can get an exact 84 and 83 by remembering where a 1-screen 81 is, then subtracting that distance by the distance of an 88 (2
degrees) for and 83, and subtracting the distance of an 87 (3 degrees) for the 84.
10) It just happens to be so that J.D.'s (cakebot) high angle at 0 wind shots are either extremely close or exactly the same as Boomer's. However, it is rather difficult to happen to have such an
angle and coincidentally be at the right spot for a full-powered shot, so adjustments might need to be made.
11) Mammoth's high angles are exactly the same as the Boomer's high angles as well.
12) With regards to 2 winds pointing up or down at a high slope, it either will affect your shot by 1 degree or it won't make any difference at all. Sometimes, if it's pointing diagonally upward and
you're shooting with the wind, it is hard to tell whether you'll need to take off a degree or not, so just try once (and guestimate, of course) and the correction if necessary.
13) If the wind is perfectly pointing left or right, pretend 1 wind strength is the same as 2, and 2 is the same as 3.
14) For 3 or higher winds, refer to this:
-I think the wind conditions have to be near-horizontal, but not too far from it, nor exactly horizontal. First, from your position, find the degrees you would do if it was 0 wind. Then move back
(away from your target) 1 mobile's length (by that, I mean the width of your avatar). Then, add "Windstrength - 1" degrees.
-An example:
-If the wind was say 7, pointing left, and your target was at the left. Say your target is exactly 1 screen away from your standing point. Move right the length of a mobile. Normally, you would do 81
degrees, but Windstrength = 7, so 7-1=6, so add 6 to 81, you would do a full powered 87 shot to hit him.
-I'm not 100% positive about this, but it has been working for me most of the time.
-If the wind is pointing slightly lower than horizontal, it won't go as far, and if it's pointing perfectly up-right or up-left, it will go farther.
-I'm sure this method works for 3-5 winds, but I haven't tested with higher winds than 5. | {"url":"http://creedo.gbgl-hq.com/angles.php","timestamp":"2024-11-11T14:50:47Z","content_type":"text/html","content_length":"5322","record_id":"<urn:uuid:87bf2d7f-a17f-440b-9258-dd670a66c9ec>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00559.warc.gz"} |
Quarter Formula in Excel
An easy formula that returns the quarter for a given date. There's no built-in function in Excel that can do this.
1. Enter the formula shown below.
Explanation: ROUNDUP(x,0) always rounds x up to the nearest integer. The MONTH function returns the month number of a date. In this example, the formula reduces to =ROUNDUP(5/3,0), =ROUNDUP
(1.666667,0), 2. May is in Quarter 2.
2. Let's see if this formula works for all months.
Explanation: now it's not difficult to see that the first three values (months) in column B are rounded up to 1 (Quarter 1), the next three values (months) in column B are rounded up to 2 (Quarter
2), etc.
3. You can also use MONTH and CHOOSE in Excel to return the quarter for a given date.
Explanation: in this formula, MONTH(A1) returns 5. As a result, the CHOOSE function returns the fifth choice. May is in Quarter 2.
4. This formula works for all months.
Explanation: in this formula, MONTH(A1) returns 1. As a result, the CHOOSE function returns the first choice. January is in Quarter 1.
To return the fiscal quarter for a given date, slightly adjust the list of values.
5. For example, if your company's fiscal year starts in April, use the following formula.
Note: green font for illustration only.
6. For example, if your company's fiscal year starts in October, use the following formula.
Tip: to quickly copy the formula in cell B1 to the other cells, select cell B1 and double click on the lower right corner of cell B1 (the fill handle). | {"url":"https://www.excel-easy.com/examples/quarter.html","timestamp":"2024-11-08T02:04:17Z","content_type":"application/xhtml+xml","content_length":"15771","record_id":"<urn:uuid:28725627-defb-42c4-a676-3410c2aafaf1>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00758.warc.gz"} |
Power of an appliance or electric circuit | Oak National Academy
Hello, my name's Dr.
George, and this lesson is called Power of an Appliance or Electric Circuit.
It's part of the unit, Mains Electricity.
The outcome of this lesson is I can describe how the power of an appliance or electric circuit depends on current and potential difference.
I'll be using these keywords during the lesson.
And remember, power is the amount of energy transferred each second.
If you want to remind yourself of the meanings of these words, come back to this slide anytime.
The lesson has three parts.
They're called calculating power, power of mains appliances, and the correct fuse.
Let's start.
It turns out that it always requires the same amount of energy to heat the same amount of water by the same number of degrees.
And we can use that when thinking about power.
So the same amount of energy is always needed to heat one litre of water from 20 to a 100 degrees, which is the sort of thing a kettle does.
This kettle on the left has a power of 1,500 watts and it boils a litre of water in four minutes.
This kettle on the right has a power of 2,000 watts and it boils a litre of water in only three minutes.
So a more powerful kettle will heat up the water more quickly.
That must mean that the more powerful kettle is supplying the energy needed more quickly.
To heat two times the amount of water from 20 degrees C to a 100 degrees C, each kettle would need to transfer two times the amount of energy.
So on the left, four megajoules of energy, that's 4 million joules are needed to heat this water from 20 degrees to a 100 degrees C.
And on the right with twice as much water.
Eight megajoules of energy are needed to heat it from 20 to a 100 degrees C.
So how long would it take a careful to heat water from 20 to a 100 degrees C if it contained two times as much water as before? When I ask a question, I'll wait five seconds, but you may need more
In which case, press pause and press play when you have your answer ready.
The correct answer is D.
It would take twice as much time to heat twice as much water.
Now power is equal to the amount of energy transferred each second.
So power can be calculated by dividing the amount of energy transferred by the time it takes.
We can write the equation in symbols as P equals E over T.
For this equation to work, we use power in watts, energy in jewels, and time in seconds.
Now, out of these four, which two equations are correct? The first correct equation is B, power is energy divided by time.
That's what we saw on the previous slide.
But C is also correct if we multiply both sides of the equation in B, by time we get energy equals power times time.
In this circuit there's a cell and a lamp, and the lamp is transferring energy to the surroundings by emitting light.
This lamp has a power of three watts and that means it's transferring three jewels of energy to the surroundings each second in the form of light and perhaps some heat.
Now if we add another lamp in parallel with the first, we find that doesn't change the brightness of the first lamp and the second one has the same brightness as that.
So if they have the same brightness as before, they must each have the same power as before.
They're transferring the same amount of energy into light per second.
So if the power of the first lamp was three watts, the combined power of these two in parallel must be six watts.
And these three in parallel still equally bright, the power is nine watts.
But we also know that when we add identical bulbs in parallel, it increases the current through the cell.
Let's say this is two amps.
If we add another identical bulb in parallel, the current is now four amps.
There's two amps in each of these branches containing a bulb, and that adds up to four amps where the amateur is and through the cell.
And if we add another amp in parallel, we have a total current through the cell of six amps.
So now I want you to think about this true or false, the power of an electric circuit is proportional to the current, and when I say current, I mean the current in the cell.
And what is the evidence for your answer? So why do you think that? Press pause while you think about this and press play when you're ready with your answer.
The correct answer is that it's true.
How could you know that from what you've just seen? Well, we saw that when identical bulbs are added in parallel, they each have the same p.
Because they have the same p.
as the p.
across the cell and they each have the same current.
They have equal brightness and that must be because they have the same power.
So the total power is proportional to the number of bulbs.
If you double the number of bulbs, you double the total power.
But we also saw that the cell current is proportional to the number of bulbs.
If you double the number of bulbs, you double the current in the cell.
So if the power and the current are both proportional to the number of bulbs, they must be proportional to each other as well.
So the power of an electric circuit is proportional to the current.
If we add bulbs in series with each other, they will become less bright and so we'll be changing the power of each bulb.
But that won't happen if we add a cell each time we add a bulb.
So in the first circuit, if we have power three watts, we've got equally bright bulbs again, so total power is six watts, and we've doubled the p.
by doubling the number of cells.
And the current is the same.
We've doubled the p.
, which determines the push on the electrons, but we've also doubled the resistance.
So we get the same current.
` If we now add another cell and another bulb in series.
We've got three times the power that we had at the start, three times the total p.
and the same current again.
So thinking about that, is this true or false? The power of an electric circuit is proportional to the potential difference and I mean the total potential difference across the cell or battery.
And how do you know? What is the evidence for your answer? Press pause while you think about this.
And this statement is also true.
And one way to explain how we know from what we've seen on the previous slides.
If the number of cells always equals the number of bulbs in series, which is what we were looking at before.
The current stays the same, but the total p.
is proportional to the number of cells, the number of cells doubles, the total p.
The total power is also proportional to that number because the power is proportional to the number of bulbs we have.
So the power must be proportional to the p.
since both the power and the p.
are proportional to the number of cells and bulbs.
Well done if you realise that.
So we found that the power in an electric circuit, the total power is proportional to the current in the cell or battery and it's also proportional to the potential difference across the cell or
So we can summarise these relationships with the equation power is current times potential difference.
Or P equals I times V.
And this is a very useful equation when we're thinking about electric circuits.
And it's actually true for any individual component in a circuit that its power, the energy it transfers per second equals the current through it times the potential difference across it.
But this equation only works if we use standard units.
We have power measured in watts, current in amps, and potential difference in volts.
And now here's a worked example.
I'll show you how to do this and then I'll ask you a question.
If the current through an electric motor is 2.
4 amps and the p.
across it is 6.
0 volts, what is the power of the motor? Well, we could start by writing down the equation we're going to use.
We could also write down the quantities that we know.
So we have I equals 2.
4 amps and V equals 6.
0 volts.
Substitute the quantities into the equation and we find that the power is 14.
4 watts, but we should really round that to two significant figures because we only knew I and V to two significant figures.
So we get 14 watts.
And now here's one for you to try.
If the p.
across a washing machine is 230 volts and the current through it is 2.
4 amps, what is the power of the washing machine? Press pause when you do this and press play when you're ready to check your answer.
Again, we can use the equation P equals I times V.
Substitute in the current in p.
We get 552 watts, but we only knew the current to two significant figures, so we can only really know our answer to two significant figures.
And that's 550 watts.
Well done if you got that.
Now here's a longer written task for you.
A group of pupils want to model the power of an electric circuit.
They use a loop of rope and you can see in the picture that there are two hands moving the rope representing the battery.
There's a hand gripping the rope, representing the lamp, and there's a current represented by the movement of the rope.
So the questions are how could they use the model to show how energy is transferred by an electric circuit? How could they use the model to show the effect of increasing current on the power of a
circuit? And how can they use the model to show the effect of increasing p.
on the power of a circuit? Press pause while you write down your answers.
And when you're ready, press play and I'll show you some example answers.
So here are some ways you could have written these answers.
So first of all, showing how energy is transferred.
The pupil modelling the battery pulls the rope around the circuit representing current.
Through the grip of the pupil, who is the lamp.
Friction from the rope heats up their hands and then dissipates into the surroundings.
Similar to how energy is transferred in a real electric circuit.
And then how to show the effect of increase in current.
The pupil modelling the battery can increase current by pulling the rope more quickly.
The larger current heats the hands of the lamp more quickly showing that energy is transferred more quickly.
And finally, how to show the effect of increasing the p.
Adding a second lamp to the rope loop and pulling the rope around at the same speed, same current, will transfer energy twice as quick.
To do this, the battery will need to push two times harder with a p.
that is twice as big.
Well done if you've got some of these points into your answers.
And now we'll move on to the second part of the lesson.
Power of main's appliances.
In a house, the p.
across a main circuit is always 230 volts.
Here we have a circuit that includes a kettle and in parallel with it a lamp.
And because all main circuits are wired as parallel circuits, the p.
across each appliance is always the same.
It's always 230 volts.
So the p.
across the kettle here is 230 volts and so is the p.
across the lamp.
But their powers are different.
This kettle has a power of 1,500 watts and the lamp only 40 watts.
And we've seen that power is current times potential difference.
So as the p.
across the kettle is the same as the p.
across the lamp, the current through the kettle must be greater because that's the only way it could have a higher power.
And so another pair of questions, I'll show you how to do the first one.
What is the current through a 1,500 watt kettle that is plugged into the mains? So although we're not told it, we can write down the p.
across the kettle, mains p.
is 230 volts.
We could also write down the power 1,500 watts.
And then we use this equation P equals I times V.
You can either rearrange the equation in symbols first or you can substitute in the values and then rearrange.
You're going to need to rearrange one way or another to find I.
Here we can divide both sides by 230 volts.
Calculate I we get 6.
5217, et cetera amps shown on the calculator, but we can't really know the answer to that many significant figures.
We can give three significant figures here because we only have the p.
to three significant figures.
And now here's a question for you.
What's the current through a 40 watt lump that is plugged into the mains? Press pause when you write your answer And press play when you're finished.
Again, we can write down the p.
straight away.
We're talking about a main circuit, so it's 230 volts, P equals I times V.
Substitute the values, rearrange to find the current.
And we get this answer on the calculator, which here is rounded to two significant figures because we only knew the power to two significant figures at most.
So we can see that it's true what we saw on the previous slide.
These two appliances have the same p.
across them, but since they have very different powers, they must have very different currents.
If they had the same current and the same p.
, it'd have to have the same power.
And so the current in the kettle is much larger than the current in the lower power lamp.
The current through the kettle and the lamp add up to give the current through the main circuit.
So we add these two together and we get 6.
69 amps.
Plugging in too many appliances can cause a very large current in a main circuit.
Because every time you plug in an appliance it's in parallel with whatever else is switched on.
And so it adds extra current that goes into that new branch.
If the current exceeds the size of a circuit breaker and the consumer box in the house, the whole circuit will be turned off.
The circuit breaker will detect that the current is too high and it will switch off the circuit.
That's a safety precaution to make sure that nothing gets so hot that wires melt or even start a fire.
Now here's an example question.
What is the current through a mains circuit when two 1,500 watt kettles and three 40 watt lamps are plugged into it? We could start with an initial calculation of the total power.
So two times the kettle and three times the lamp power, 3,120 watts.
Now that we have the power, we can use P equals IV.
Because we know V, it's mains it's 230 volts.
So substituting in rearranging to find the current, we get 13 point something amps.
And if we round two significant figures, 14 amps.
Now here's one for you.
What is the current to remain circuit when four 1,500 watt kettles are plugged into it? Press pause while you write down your working.
Again, let's start by working out the total power.
It's 6,000 watts and substituting into P equals I times V.
Again, it's 230 volts, it's mains.
Rearranging to find the current.
To two significant figures, we get 26 amps.
That's essentially a large current for a main circuit.
Now four questions and for each calculation, show all your working out, show the equation you're using and the quantities you're using and give each answer to the correct number of significant
Basing that on how many significant figures are in the quantities that you use to calculate your answer.
Press pause while you do this and press play when you're ready to check your answers.
So here are the worked solutions.
The p.
across the 2,000 watt mains kettle with a current of 8.
7 amps, 230 volts, again, it's mains.
Calculate the power of a main's electric kettle that has a current of 9.
2 amps through it.
P equals I times V.
Again, it's 230 volts p.
substituting in.
And we should round to two significant figures because we know the current to two significant figures.
So 2,100 watts.
The current used by a main lamp with a power of 11 watts, again rearranging the equation.
P equals I times V.
And we should write our answer to two significant figures because we know the power to two significant figures.
So the current is only 0.
048 amps.
And the current used by an electric oven with power 3,500 watts.
And electric oven will often be the highest power device in a household.
So again, rearranging to find the current using 230 volts, we find that the current is 15 amps to two significant figures because that power of 3,500 watts may only be written to two significant
figures if it's rounded to the nearest a 100.
Well done if you're getting these right.
Now for the last part of the lesson, the correct fuse.
It can happen inside an appliance that a wire comes loose and makes a connection where it shouldn't.
And that can cause a short circuit.
That's a circuit in which there's very little resistance.
So imagine that this wire has come loose and it's actually inside this main's appliance.
The current no longer has to pass through the main resistance of the appliance.
It's just passing through a low resistance wire.
And so we get a very large current, alternating current because main's provides AC.
And since power is current times potential difference.
And the p.
is still 230 volts but we have a large current, then there'll be a large power.
And all that power is doing that energy transfer is causing heating of the wires in the main circuit and inside the appliance where it's flowing.
That could cause a fire if things get very hot.
So the faulty appliance needs to turn off as soon as possible and that's the point of a circuit breaker or a fuse.
It's to switch off the current if it gets dangerously high.
In the plug inside every appliance there's a fuse that contains a thin wire and that wire melts breaking the circuit and turning off the electricity.
If the current becomes too large.
There are three sizes of fuse available for plugs, three amp, five amp, and 13 amp.
And for each of these fuses, the wire melts if a current larger than the size of the fuse flows through it.
So if you have a three amp fuse, the wire inside it will melt and break the circuit if the current is greater than three amps.
The correct fuse to use depends on the current used by the appliance when it's working normally.
For example, which size fuse would allow a 2,000 watt kettle to work properly.
I'll show you how to do this and then I'll give you a question to try.
So we're looking to find the current in the kettle so that we can decide what's the best fuse.
So we'll use P equals I times V, it's main's voltage, 230 volts, rearranged to find the current.
Divide the power by the p.
and we get 8.
69 or so amps.
That means we have to use a 13 amp fuse because that will allow that much current to flow without melting.
A three amp or five amp fuse would melt as soon as this kettle started working normally.
So that would not be appropriate.
And now a question for you.
Which size fuse would allow a 60 watt lamp to work? Three amps, five amps or 13 amps.
Press pause while you do the calculation And decide which fuse.
Press play when you're ready.
So we start by working out the current in the lamp.
We use P equals I times V, substitute in the power and the main voltage.
Rearrange and the current is about 0.
26 amps.
We would have to use a three amp fuse, so that allows any current up to three amps to flow.
A five amp or 13 amp views would allow this current to flow, but they would also allow a much larger current than the lamp should have when it's working normally.
An appliance that uses less than three amps should be fitted with a three amp fuse.
If there's a short circuit, the three amp fuse will melt faster than a five amp or 13 amp fuse.
So the three amp fuse is more sensitive to lower currents.
If you use a 13 amp fuse with this lamp, then 13 amps could flow through the appliance before it gets switched off, before the wire inside the fuse breaks.
And that could damage the lamp.
Which of these sizes of fuse should be fitted to a television that works normally with a current of 3.
2 amps? The correct answer is five amps.
2 amps is just a little bit more than three amps.
So the three amp views would melt when the TV was working normally, and we don't want that.
So we need to choose the next one up.
And that's the five amp fuse.
Now here's a table of appliances and their powers.
They're all going to be used in main circuits.
So can you calculate missing values for current and state which type of fuse should be used for each of these appliances? And you're choosing from three amp, five amp, and 13 amp each time.
Press pause while you do this and press play when you're ready to check your answers.
And here are the answers.
Did you realise that none of the three standard sizes of views are going to work for an electric oven? The wire inside them will melt no matter which views you use.
So electric ovens are usually plugged into a special oven socket that's connected to a 20 amp circuit breaker in the consumer box.
So now we've reached the end of the lesson and here's a summary.
Power is equal to the amount of energy transferred each second, P equals E divided by T.
And in electric circuits, P equals I times V.
Power, P is measured in watts.
Energy, E is measured in joules.
Time, T is measured in seconds.
Current, I is measured in amps And p.
, V is measured in volts.
Fuses that are fitted in the plugs of main's appliances contain a thin wire that melts if the current becomes larger than the size of the fuse because of a short circuit.
Fuses usually come in three sizes, three amp, five amp, and 13 amp.
The smallest fuse possible is used, but it needs to be greater than the current the appliance uses when it is working.
So well done for working through this lesson.
I hope you found it interesting and useful.
and you will need to use this knowledge if you ever have to decide what size of fuse to use.
I hope to see you again in the future lesson.
Bye for now. | {"url":"https://www.thenational.academy/pupils/programmes/combined-science-secondary-year-11-higher-edexcel/units/mains-electricity/lessons/power-of-an-appliance-or-electric-circuit/video","timestamp":"2024-11-07T20:01:47Z","content_type":"text/html","content_length":"140015","record_id":"<urn:uuid:9572d1c1-e5b9-47e4-b2a7-82bbfa9fbf71>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00091.warc.gz"} |
Shifting Exponential Functions Practice Worksheet - Function Worksheets
Exponential Function Problems Worksheet – If you’re in search of an activity in math for your child to help them practice exponential functions, then you’ve … Read more
Exponential Functions Practice Worksheet
Exponential Functions Practice Worksheet – If you’re in search of an activity in math for your child to help them practice exponential functions, then you’ve … Read more
Exponential Functions Review Worksheet
Exponential Functions Review Worksheet – You’ve come to the right spot if you’re in search of an activity in math for your child to help … Read more | {"url":"https://www.functionworksheets.com/tag/shifting-exponential-functions-practice-worksheet/","timestamp":"2024-11-09T10:54:50Z","content_type":"text/html","content_length":"67555","record_id":"<urn:uuid:40510af2-be20-48ac-8260-1c5cf9256440>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00885.warc.gz"} |
Section 3.4
Section 3.4, Infinite Limites
Motivation: #
We’ve already seen $\displaystyle \lim_{x\to \infty} \frac1x = 0$. Also, since those $x$ values are positive, we’re approaching 0 from above.
On a similar note, $\displaystyle \lim_{x\to -\infty} \frac1x = 0$ as well. And since those $x$ values are negative, we’re approaching 0 from below. This matches what we know the graph to look like:
Intuitive Definition of the Infinite Limit #
Let $f$ be a function defined on some interval $(a, \infty)$. Then $$ \lim_{x\to \infty} f(x) = L $$ means that the values of $f(x)$ can be make arbitrarily close to $L$ by requiring $x$ to be
sufficiently large.
Examples of infinite limits where $f(x) \to L$ as $x\to \infty$.
Analogously, if the function is defined on $(-\infty, a)$, then we can have $\displaystyle \lim_{x\to -\infty} f(x) =L$ if $f(x)$ gets arbitrarily close to $L$ as $x$ get sufficiently large in the
negative direction.
Horizontal Asymptote #
We call the line $y=L$ a horizontal asymptote of the function if $$ \lim_{x\to \infty} f(x) = L $$ or $$ \lim_{x\to -\infty} f(x) = L $$
More than one horizontal asymptote? #
Yes! We call $y=L$ a horizontal asymptote. It’s entirely possible a function has more than one! Here’s an example curve with two horizontal asymptotes:
Computing Limits at Infinity #
First, we get a tool we’ll use to compute these infinite limits.
Theorem #
If $r \gt 0$ is a rational number, then $$ \lim_{x\to \infty} \frac{1}{x^r} = 0 $$ and if $r\gt 0$ is a rational number so that $x^r$ is defined for all $x$^1, then $$ \lim_{x\to -\infty} \frac{1}{x^
r} = 0 $$
Example #
Now let’s compute an infinite limit. Evaluate
$$ \lim_{x \to \infty} \frac{2x^2 - 3x - 5}{3x^2 + 4x + 1} $$
Example #
Find the horizontal and vertical asymptotes of the graph of the function.
$$ \frac{\sqrt{2x^2 +1}}{3x - 5} $$
Here’s an image that shows the asymptotes:
Example #
This example is a little different. Is “$\infty - \infty$” equal to 0? Or something else?
Compute $\displaystyle \lim_{x\to \infty} \sqrt{x^2 + 1} - x$
Example #
Here’s a very similar looking example
Compute $\displaystyle \lim_{x\to \infty} \sqrt{x^2 + x} - x$
… so “$\infty - \infty$” can be 1/2? Yes! In fact it could be any number or even $\infty$.
Infinite Limits at infinity #
When functions continue to grow in the positive (or negative) direction without bound, we write
$\displaystyle \lim_{x\to \infty} f(x) = \infty$ or $\displaystyle \lim_{x\to \infty} f(x) = -\infty$
But notes that $\infty$ is not a number, so our limit laws do not apply!
As we saw above, “$\infty - \infty$” could be anything, so we have to be careful such as in the following example:
Example #
Evaluate $$ \lim_{x \to \infty} 2x^2 - 3x $$ Solution Don’t plug in $\infty$! Otherwise this might happen:
Instead, factor first! $$ \begin{align*} \lim_{x \to \infty} 2x^2 - 3x &= \lim_{x\to \infty} x(2x - 3) \end{align*} $$ but since both $x$ and $2x-3$ grow to $\infty$ as $x$ grows large, we can say $$
\lim_{x \to \infty} 2x^2 - 3x = \infty $$
Example #
Evaluate $$ \lim_{x\to \infty} \dfrac{3x^2 - 4x + 5}{x+5} $$
1. No square roots of negative numbers, for example. ↩︎ | {"url":"https://mathsquirrel.com/calc/chapter3/section4/","timestamp":"2024-11-03T07:14:53Z","content_type":"text/html","content_length":"56012","record_id":"<urn:uuid:ce814573-9449-447a-89b0-32672aed4309>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00054.warc.gz"} |
Mathematics in the real world
This paper explores aspects of an experimental approach to mathematical proof, most notably number crunching, or the verification of subsequent particular cases of universal propositions. Since the
rise of the computer age, this technique has indeed conquered practice, even if it implies the abandonment of the ideal of absolute certainty. Thus, it seems that also in mathematical research, the
qualitative criterion of effectiveness, i.e. to reach one's goals, gets increasingly weighed against the quantitative one of efficiency, i.e. to minimize one's means/ends ratio. We probe for
mathematical reasons and philosophical justifications for this rising popularity of `going inductive'. Our story will lead to the consideration of limit cases, opening up the possibility of proofs of
infinite length being surveyed in a finite time. This should show that mathematical practice in crucial aspects depends upon what the actual world is (or is not) like. Note that this does not at all
entail a rejection of the notion of a purely formal or deductive proof, even in cases where the latter should have actually ceased to essentially contribute to establishing the correctness of
underlying mathematical claims.
In the proposed scenarios it remains perfectly possible to be a Platonist, thus claim that mathematical knowledge is necessary, and nevertheless accept that, depending on the world you live in, some
mathematical statements are either trivial or extremely difficult to answer (e.g., if one happens to live in our universe). What should become clear however, is that an isolationist strategy, whereby
in order to preserve the purity of mathematics, one has the mathematical domain shrunk until all external influences are excluded, will be of no avail. Indeed, this rather cynical procedure, which
has helped create the miracle of the effectiveness of mathematics (Wigner), cannot do much work here, since only pure mathematical statements will be talked about to begin with. Moreover, working
mathematicians absolutely do not shun away from all inductive techniques, methods, and ideas described in this paper.
Unfortunately, there is (still) not much willingness among mathematicians to `out' themselves on this philosophically laden topic, to the effect that, in terms of the metaphor introduced by Reuben
Hersh, most empirical or experimental elements currently remain relegated to the back stage, while only formal proofs are held to occupy the front stage and confront the public. If it is however
accepted that the front stage cannot exist without the back stage, then it is realized that no theatre can function as a whole without taking into account the economical necessities also. Already
today, mathematicians amply rely on computers to warrant mathematical results, and work with conjectures that are only probable to a certain degree. Every so often, we get a glimpse of what is
happening back stage, but what seems to be really required is not merely the idea that the front can only work if the whole of the theatre is taken into account, but also that, in order to actually
understand what is happening front stage, an insight and understanding of the whole is required. If not, a \emph{deus ex machina} will be permanently needed.
Originele taal-2 English
Titel Induction: Historical en Contemporary Approaches - International Conference, Universiteit Gent
Status Published - 8 jul 2008
Evenement Unknown - Stockholm, Sweden
Duur: 21 sep 2009 → 25 sep 2009
Conference Unknown
Land/Regio Sweden
Stad Stockholm
Periode 21/09/09 → 25/09/09
Duik in de onderzoeksthema's van 'Mathematics in the real world'. Samen vormen ze een unieke vingerafdruk. | {"url":"https://researchportal.vub.be/nl/publications/mathematics-in-the-real-world","timestamp":"2024-11-03T22:15:24Z","content_type":"text/html","content_length":"60400","record_id":"<urn:uuid:cf67e058-93f3-4822-82ea-9ab2ed4a9b59>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00123.warc.gz"} |
E12:demonstrate an understanding of the concept of converse
converse statements
-->if one thing is true, then is the opposite?
-->if p, then q but,
-->not every statement has a true converse statement
true converse statement:
statement--> If this month is June, next month is July
converse-->if next month is July, this month is June
This is a true statement !!!
if and only if statement:
-->statements with a true converse can be worded into an if and only if statement
example-->next month is July if and only if, this month is June
untrue converse statement:
statement-->if you've been swimming, your hair is wet
converse-->if your hair is wet, you've been swimming
*this converse statement is not true because it is not so in every case
converse statements in circle geometry:
statement-->if a line is perpendicular to a chord and passes through the centre of a circle then, it bisects the chord
converse-->if a line is perpendicular to a chord and bisects the chord than, it passes through the centre of the circle
if and only if statement-->a line is perpendicular to a chord of a circle and bisects the chord if and only if, it passes through the centre of the circle
sample problem:
state the converse of this statement:yesterday was saturday so, today is sunday
is it a true statement?
if so write it as a if and only if statement.
converse-->today is sunday so, yesterday was saturday
yes it is true
if and only if statement-->yesterday was saturday if and only if, today is sunday
Nice page Nicole and Tom. You sample problem isn't in "If...then..." format.
You don't have permission to comment on this page. | {"url":"http://acrospire.pbworks.com/w/page/1342608/E12","timestamp":"2024-11-01T19:08:14Z","content_type":"application/xhtml+xml","content_length":"20299","record_id":"<urn:uuid:a8c68b1c-6de2-4100-823c-66503a29c67c>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00604.warc.gz"} |
Discount Description Templates
TO READ THE DOCS FOR THE "DISCOUNTED PRICE" APP BLOCK, CLICK HERE.
Click the thumbnail above to watch our founder explain how to set up discount description templates.
You can customize the text our app displays beneath discounted prices. For example, you might want to let users know that they received a special discount for being a loyal customer.
The discount description is HTML, so you can add tags and styles to your liking.
You can also use
, which are based on the discount calculations we did for each product. They will be automatically replaced with the corresponding text/number when the discount description is displayed on the page.
Available variables
Text variables
[discount_message]: The title of the applied discount, as it would appear in the customer's cart/checkout.
Number variables
[regular_price]: The price of the product before any discounts were applied. By default, it will be formatted into the local currency.[sale_price]: The price of the product after discounts were
applied. By default, it will be formatted into the local currency.[discount_amount]: The difference between the [regular_price] and the [sale_price]. By default, it will be formatted into the local
currency. Example: $5.00 (or $5.00 USD if you have the Currency code enabled? option on).[discount_percentage]: The difference between the [regular_price] and the [sale_price]. This number will be
rounded automatically. By default, it will be formatted as a percentage. Example: 20%.
To gain more control over the display of the discount description, you can use
, which work like Liquid filters, to transform variables, or control their formatting. You can also chain filters.
Filters only work with number variables.
[sale_price | to_fixed:2][discount_amount | fractional_part | to_fixed:2]
Available filters:
Math filters
integer_part: The whole part of the number (removes any decimals). Example: Turns $12.34 to 12fractional_part: The fractional part of the number, without the decimal point (.). Example: Turns $12.34
to 34ceil: Applies a mathematical ceiling operation to the number. Example: Turns 12.34 to 13.floor: Applies a mathematical floor operation to the number. Example: Turns 12.34 to 12.round: Applies a
mathematical round operation to the number. Example: Turns 12.34 to 12, and 12.55 to 13.
Formatting filters
money: Formats the number as a monetary amount (follows the "Currency code enabled?" setting from the Discounts Embed).money_without_trailing_zeros: Formats the number as a monetary amount, like
money, but strips trailing zeros, following the same rules as Liquid.percentage: Formats the number as a percentage (by appending a %).to_fixed:N: Formats the number by rounding the number to N
places. Example: [sale_price | to_fixed:2] turns 12.3456 to 12.34.zero_pad_start:N: Formats the number by adding leading 0 until the length is at least N characters. Example: zero_pad_start:3 turns 3
into 003zero_pad_end:N: Formats the number by adding trailing 0 until the length is at least N characters. Example: zero_pad_start:3 turns 3 into 300
Savings: [discount_amount | round] (Displays savings as a monetary amount)<b>[discount_percentage] off</b> for wholesale customers (Displays savings as a percentage)
Advanced Examples
You save:
<span style="color: green; font-weight: bold">
$[discount_amount | integer_part]
<sup>[discount_amount | fractional_part | round | zero_pad_end:2]</sup>
(Advanced. Displays savings using a superscript for the decimal part)
Have any feedback for us?
We want to hear about your experience with our app!
Leave a review on the Shopify App Store
Updated on: 03/11/2024
Was this article helpful? | {"url":"https://regiostech.crisp.help/en/article/discount-description-templates-1v96yl2/","timestamp":"2024-11-06T17:32:06Z","content_type":"text/html","content_length":"30490","record_id":"<urn:uuid:4024221e-f4b7-4467-8915-4367ad83d01c>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00708.warc.gz"} |
Fixed Deposit Calculator Online - Calculate FD Interest and Maturity
FD Calculator
Invested Amount₹5,000
Est. Returns₹50
Using an FD Calculator for Easy Financial Planning
Fixed deposits are term investments offered by banks and NBFCs, providing a higher rate of interest for a predetermined period, ranging from 7 days to 10 years. To calculate the interest and maturity
amount of an FD without hassle, an FD calculator is a useful tool available on the finoyou website.
Benefits of Using an FD Calculator:
1. Simplified Calculations: FD maturity calculations involve multiple variables, but the calculator performs complex calculations effortlessly, providing accurate results at the click of a button.
2. Time-saving: Avoid spending time on intricate calculations; the FD calculator saves you time and effort.
3. Informed Decision-making: Compare maturity amounts and interest rates of FDs from different financial institutions, helping you make well-informed decisions.
Formula for FD Maturity Amount:
The calculator accounts for two types of FDs: simple interest FD and compound interest FD.
For simple interest FD, the formula is:
M = P + (P x r x t/100)
Where: M = Maturity amount P = Principal amount deposited r = Rate of interest per annum t = Tenure in years
For example, if you deposit Rs. 1,00,000 for 5 years at 10% interest: M = Rs. 1,00,000 + (1,00,000 x 10 x 5/100) = Rs. 1,50,000
For compound interest FD, the formula is: M = P + P {(1 + i/100)^t – 1}
Where: i = Rate of interest per period t = Tenure in years
For example, with the same variables: M = Rs. 1,00,000 {(1 + 10/100)^5-1} = Rs. 1,61,051
Using finoyou's FD Calculator:
Follow these steps to conveniently use finoyou's FD calculator:
1. Gather the required data.
2. Enter the variables as per the formula.
3. Instantly view the FD maturity amount.
Advantages of finoyou's FD Calculator:
1. Accurate Planning: Know the exact amount you'll receive at FD maturity and plan your future finances accordingly.
2. Free and Unlimited Use: Both FD calculators are available for free and can be used as frequently as you want.
3. Easy Comparison: Compare maturity amounts from different financial institutions with ease.
In addition to the FD calculator, finoyou offers other free financial planning tools to help you manage your finances efficiently. Make use of these calculators to stay well-informed about your
investments and make wise financial decisions. | {"url":"https://www.finoyou.in/calculators/fd-calculator","timestamp":"2024-11-14T21:56:01Z","content_type":"text/html","content_length":"46926","record_id":"<urn:uuid:0e1e5cfe-4db6-459b-8165-89fc1d0c85cc>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00891.warc.gz"} |
cv - New Foundations Explorer
Description: This syntax construction states that a variable x, which has been declared to be a setvar variable by $f statement vx, is also a class expression. This can be justified informally as
follows. We know that the class builder {y ∣ y ∈ x} is a class by cab 2339. Since (when y is distinct from x) we have x = {y ∣ y ∈ x} by cvjust 2348, we can argue that the syntax "class x " can be
viewed as an abbreviation for "class {y ∣ y ∈ x}". See the discussion under the definition of class in [Jech] p. 4 showing that "Every set can be considered to be a class."
While it is tempting and perhaps occasionally useful to view cv 1641 as a "type conversion" from a setvar variable to a class variable, keep in mind that cv 1641 is intrinsically no different from
any other class-building syntax such as cab 2339, cun 3208, or c0 3551.
For a general discussion of the theory of classes and the role of cv 1641, see https://us.metamath.org/mpeuni/mmset.html#class 1641.
(The description above applies to set theory, not predicate calculus. The purpose of introducing class x here, and not in set theory where it belongs, is to allow us to express i.e. "prove" the weq
1643 of predicate calculus from the wceq 1642 of set theory, so that we don't "overload" the = connective with two syntax definitions. This is done to prevent ambiguity that would complicate some
Metamath parsers.) | {"url":"https://us.metamath.org/nfeuni/cv.html","timestamp":"2024-11-11T13:57:24Z","content_type":"text/html","content_length":"7915","record_id":"<urn:uuid:8f5f1e94-45d8-406e-a6d1-70805ae9ba4c>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00790.warc.gz"} |
1. Find the rate of change of y with respect to
x at the indicated value...
1. Find the rate of change of y with respect to x at the indicated value...
1. Find the rate of change of y with respect to x at the indicated value of x. y =
; x = 0
y' = _____
2. The total world population is forecast to be
P(t) = 0.00066t^3 − 0.0713t^2 + 0.84t + 6.04 (0 ≤ t ≤ 10) in year t, where t is measured in decades, with t = 0 corresponding to 2000 and P(t) is measured in billions.
(a) World population is forecast to peak in what year? Hint: Use the quadratic formula. (Remember that t is in decades and not in years. Round your answer down to the nearest year.)
(b) At what number will the population peak? (Be sure to use the value of t found in part (a). Round your answer to two decimal places.)
____________ billion | {"url":"https://justaaa.com/math/81232-1-find-the-rate-of-change-of-y-with-respect-to-x","timestamp":"2024-11-03T03:27:04Z","content_type":"text/html","content_length":"40969","record_id":"<urn:uuid:1559073d-99e0-43b5-9ea3-2b89a981c88a>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00815.warc.gz"} |
Digit DP - Scaler Blog
Digit Dp as the name suggests, is a technique of dynamic programming wherein we use the digits of the numbers to reach the final solution. This technique solves those problems which concern the
digits between two specified integers. To find the solution, we try building all the integers between the two given integers and compute the solution on the way.
Scope of the Article
• This article discusses Digit DP, its key concept, time complexity, and implementation in Data Structures.
• This article shows the implementation of Digit DP in C++.
The states of DP are the subproblems to the more significant problem. Each state is a subset of the input data used to build the result for the whole problem input.
What is Digit DP?
We’ve all read and heard about Dynamic Programming and how it makes use of repeating subproblems and memoization to solve a complete larger problem. Digit DP is one such technique that does the same,
but in Digit DP, as the name suggests, we play with the digits of a number. This means to reach the final answer, we make use of the digits of a number.
Digit DP is very useful in solving problems that concern a range of numbers, like finding the sum of digits between two numbers $a$ and $b$. Or find how many times a particular digit occurs in the
numbers in the range $[a,b]$. These are the situations where digit dp comes in handy. And it is clearly visible by the problem statements that these questions have something to do with the digits of
the numbers involved.
Let us find out more about how digit dp works and what is the main idea behind it.
Key Concept
Digit DP is based on the idea of using digits to get to the final answer. Because of this reason, digit dp is used in questions that concern the digits of numbers, maybe how many times a digit occurs
in a range or the sum of all odd digits in a range.
The main concept by which digit dp can solve such questions is that we consider numbers as a sequence of digits, and try to build numbers within the given range digit by digit as we go, and compute
the answer also digit by digit.
Let us take an example to understand more clearly how exactly digit dp works.
Problem Statement:
Find the total number of times the digit $d$ occurs in the numbers in the range $a$ to $b$, where $a<b$.
For example, let us take the case where a = 5 and b = 10, and d = 1, the answer will be 1 as there is only one digit (10) which has the digit 1.
But how can we solve this using digit dp? We can do so by building a number digit by digit so that it is inside the given range and on the way count the number of times we set a digit as 1. But how
can we do this? Let’s start with a base case, where in the problem statement, a = 0.
Finding Solution for range $0$ to $b$
To find a solution, we treat our number as a string, basically a string of digits. We need to build numbers digit by digit until they are inside the range given to us in the question. As a = 0, the
string is empty in the start, and we can add digits to it.
So, we start adding digits from left to right, so at each point of time, we have a position where we can enter a digit to form a new number, and then recurse for the next position. Now, we need to
look at the choices that we have for a particular position. We cannot just add any digit from 0 to 9, because we need to make sure the resulting number satisfies the range.
Let us take an example to understand this better. Let a = 0 and b = 83625. Suppose until now our number string that we are building by adding digits till now is "8362_". In the next recursion, we
have to add another digit to the right of 2, but can we add any digit from 0 to 9?
Adding any digit > 5 will result in the number becoming greater than $b$ and thus going out of range. This means that for this particular position we can add only digits from 0 to 5.
On the other hand if the sequence we had built uptil now was "734__", we have no restrictions on any of the places. We can put any digit from 0 to 9 in the remaining places on the right. This is
because the first digit '7' itself ensures that the complete number will always be less than $b$ no matter what digits occur to the right of it.
This observation is very important in reaching the digit dp approach. Clearly we have some positions where we need to add digits and build a number inside the given range. The number of positions
will depend upon $b$, and will be equal to the number of digits $b$ has.
Now, we have some positions where we need to add digits. We can start from the leftmost position and keep adding digits. At the same time, we will also need some information as to which digits can be
added to this position depending upon the previous digits as we saw in the example. How to know this?
One possible way is to keep track of the whole sequence built uptil now, and then check by placing all digits from 0 to 9, and only go ahead with the ones that are less than $b$, and discard the
rest. But there is a simpler approach.
We saw using the example, that we have a restriction on the digits we can add only when the sequence built uptil now is the same as the number. If we have added even one number which is less than its
corresponding digit of $b$ in the sequence we have made till now, then we have no restrictions.
This follows from the last example, if the sequence built until now is "83___" we cannot put any digit greater than 6 in the next position, but if our sequence was "82___" we have no restrictions, as
the digit 2 in the second place ensures that our number will always be smaller than $b$.
This can easily be conveyed using a bool flag variable as a parameter to our recursive function. If this flag variable is set to true, it means we have a restriction and we can add digits from 0 to
the digit in $b$ at the corresponding position else the resulting number will go out of range, and if the flag variable is 0, we have no restrictions. Whenever we add a digit to our sequence which is
equal to the corresponding digit in $b$, we have to set the flag to 1 and if we are adding a digit smaller than the corresponding digit in $b$, we set the flag to 0 and recurse for the next position.
The initial problem can be solved very easily now because whenever we add the digit $d$ to the sequence we are building, we can add 1 to the answer, and at the end, we will be left with the total
number of times the digit $d$ occurs in the range 0 to $a$.
Finding Solution For Given Range
We learned about solving from the range $0$ to some positive integer $b$, but what about the range $a$ to $b$? We already know the solution for $0$ to $b$ and we can also calculate the solution for
$0$ to $a-1$ by the same method.
Then, the solution for aa to bb will simply be
solution(a,b) = solution(0,b) - solution(0,a-1)
Why is this correct? Because in subtracting solution(0,a-1) from solution(0,b) we make sure all numbers less than a get excluded, thus giving us the soltuion for a to b.
States of DP Problem and Solution
So far we have only looked at how we can solve a problem using digit dp, let us now discuss the particulars of the solution.
We know that every DP solution has states, which refers to the total number of parameters required to uniquely define each state in the problem. In this case, we have three states in the DP solution.
One state is the position parameter pospos which refers to the position where a new digit is to be inserted and it can take the maximum value of the number of digits in b.
The second state is the bool flagflag parameter that we defined earlier which denotes whether we have a restriction on the digits that can be chosen for the current position or not. The third state
will be the parameter cntcnt which denotes the number of digits dd we have included till now.
Hence, our memoisation dp table will look like dp[pos][flag][cnt]. Let us look at the complete solution:
#include <bits/stdc++.h>
using namespace std;
// dp table to store answers
int memo[18][19][2];
// function to count the number of times digit d occurs from 0 to a
// where a is represented as the vector digits
int dp(vector<int>& digits, int d, int pos, int cnt, int flag){
// base case
// cnt is the number of times digit d occured
return cnt;
// if this state has already been calculated return it
return memo[pos][cnt][flag];
// this variable denotes the upper bound
// on the digits that can be chosen for this position
int limit = 9;
// if flag is 1, that means we will have a restriction
// on the digits that can be chosen, it cannot be greater than
// digits[pos] else we have no limits
limit = digits[pos];
// answer variable
int answer = 0;
// for loop over all the digits that can be considered for this position
for(int i=0;i<=limit;i++){
// in the next iteration we need to decide the value of new flag
int new_flag = flag;
// if flag is 1 and the current digit we are choosing
// is equal to digits[pos], then we need to set flag to 1
if(flag==1 && i==digits[pos]){
new_flag = 1;
// else we have no restrictions in the next iteration
new_flag = 0;
// variable for the new digit count
int new_cnt = cnt;
// if the current digit being selected is d then increase new_cnt
// recurse for next iteration and add to answer
answer += dp(digits,d,pos+1,new_cnt,new_flag);
// store this answer in the dp table and return it
return memo[pos][cnt][flag] = answer;
int main() {
// declare variables
int a,b,d;
// vector to store the digits in a and b
vector<int> a_digits, b_digits;
// compute digits for a and b
a /= 10;
b /= 10;
// initialize memo with -1
// compute answer for a
int answer_a = dp(a_digits,d,0,0,1);
// initialize memo with -1
// compute answer for b
int answer_b = dp(b_digits,d,0,0,1);
// final answer
int answer = answer_b - answer_a;
return 0;
In this solution, the dp() function calculates the number of times the digit dd occurs in the numbers from 0 to the number represented by the digits array. We calculate this value for a−1 and b and
subtract them to get the answer for a to b.
Time and Space complexity
The time complexity for this solution is O(10*pos*cnt*flag). The maximum value for $pos$ can be 18. This is because we can have integers upto the range of $10^{18}$, so this means that atmost we will
have 18 places to fill digits in and hence the value of $pos$ can go upto 18. Similarly, $cnt$ can be 18 and $flag$ can be 0 or 1. So, the time complexity becomes almost constant! The space
complexity is O(pos*cnt*flag).
Special Case: Digit d = 0
If you run this solution for digit $d$ = 0, then you will probably get a wrong answer. Let’s take a look:
Clearly, the output should have been 1 as 10is the only digit that contains 0 between 1 and 11, but we got 11 extra counts added to the answer. This is because according to our algorithm whenever a 0
is added to the building sequence, we increase the count. But this means that when we are building numbers between 1 to 9 we are adding extra zeroes that are not needed. These numbers are being
represented as 01, 02, 03, and so on in the sequence that we built, and the zeroes in the front don’t need to be counted as they do not have any significance. If there is a way by which we can know
if the number built till now is non-zero or not, then we will know when to count 0 as a digit: count 0 as a digit when the sequence built until now is non-zero and not otherwise.
This can be easily achieved by another bool variable $is_empty$. If this bool variable is true it means that the sequence built so far is empty, and if we are adding a 0 to the sequence we should not
count it as a digit as it is at the front of the sequence and if this variable is false, it means our sequence is non-empty and we must count 0 as a digit. The rest of the code implementation remains
the same. Let us take a look at the code.
#include <bits/stdc++.h>
using namespace std;
// dp table to store answers
int memo[18][19][2][2];
// function to count the number of times digit d occurs from 0 to a
// where a is represented as the vector digits
int dp(vector<int>& digits, int d, int pos, int cnt, int is_empty, int flag){
// base case
// cnt is the number of times digit d occured
return cnt;
// if this state has already been calculated return it
return memo[pos][cnt][flag][is_empty];
// this variable denotes the upper bound
// on the digits that can be chosen for this position
int limit = 9;
// if flag is 1, that means we will have a restriction
// on the digits that can be chosen, it cannot be greater than
// digits[pos] else we have no limits
limit = digits[pos];
// answer variable
int answer = 0;
// for loop over all the digits that can be considered for this position
for(int i=0;i<=limit;i++){
// in the next iteration we need to decide the value of new flag
int new_flag = flag;
// if flag is 1 and the current digit we are choosing
// is equal to digits[pos], then we need to set flag to 1
if(flag==1 && i==digits[pos]){
new_flag = 1;
// else we have no restrictions in the next iteration
new_flag = 0;
// variable for the new digit count
int new_cnt = cnt;
// variable for new is empty variable
int new_is_empty = 0;
if(i==0 && is_empty==1){
new_is_empty = 1;
// if the current digit being selected is d and sequence is not empty then increase new_cnt
if(i==d && is_empty==0){
// recurse for next iteration and add to answer
answer += dp(digits,d,pos+1,new_cnt, new_is_empty, new_flag);
// store this answer in the dp table and return it
return memo[pos][cnt][flag][is_empty] = answer;
int main() {
// declare variables
int a,b,d;
// vector to store the digits in a and b
vector<int> a_digits, b_digits;
int x = a;
// compute digits for a and b
a /= 10;
b /= 10;
// initialize memo with -1
// compute answer for a
int answer_a = dp(a_digits,d,0,0,1,1);
// initialize memo with -1
// compute answer for b
int answer_b = dp(b_digits,d,0,0,1,1);
// final answer
int answer = answer_b - answer_a;
// if a was 0, we did not count it at the start, so increase the answer here
if(d==0 && x==0){
return 0;
In this code we have also included a special case when a = 0. In such a case, subtracting 1 from aa will lead to a negative value, hence we keep aa as it is, and add one to the answer if our digit d
= 0.
Including just one bool variable made this implementation very easy. The time complexity for this solution will be O(10*pos*cnt*isempty*flag). The maximum value for $pos$ can be 18.
This is because we can have integers upto the range of $10^{18}$, so this means that atmost we will have 18 places to fill digits in and hence the value of $pos$ can go upto 18. For $cnt$, maximum
value can be 18 and $flag$ and $isempty$ can be 0 or 1. So, the time complexity becomes almost constant! The space complexity is O(pos*cnt*isempty*flag).
Let us look at another example. Given two integers $a$ and $b$, find the sum of all digits of the integers that occur between $a$ and $b$. We can use digit dp to find a solution. The only difference
from the above mentioned solution will be the variable cnt which will be changed to sum instead and we will add the value of each digit to this variable when we include it.
#include <bits/stdc++.h>
using namespace std;
// dp table to store answers
int memo[18][180][2];
// function to count the number of times digit d occurs from 0 to a
// where a is represented as the vector digits
int dp(vector<int>& digits, int pos, int sum, int flag){
// base case
// sum is the total sum of all digits in this number
return sum;
// if this state has already been calculated return it
return memo[pos][sum][flag];
// this variable denotes the upper bound
// on the digits that can be chosen for this position
int limit = 9;
// if flag is 1, that means we will have a restriction
// on the digits that can be chosen, it cannot be greater than
// digits[pos] else we have no limits
limit = digits[pos];
// answer variable
int answer = 0;
// for loop over all the digits that can be considered for this position
for(int i=0;i<=limit;i++){
// in the next iteration we need to decide the value of new flag
int new_flag = flag;
// if flag is 1 and the current digit we are choosing
// is equal to digits[pos], then we need to set flag to 1
if(flag==1 && i==digits[pos]){
new_flag = 1;
// else we have no restrictions in the next iteration
new_flag = 0;
// recurse for next iteration by adding i to sum and add to answer
answer += dp(digits,pos+1,sum+i,new_flag);
// store this answer in the dp table and return it
return memo[pos][sum][flag] = answer;
int main() {
// declare variables
int a,b;
// vector to store the digits in a and b
vector<int> a_digits, b_digits;
// compute digits for a and b
a /= 10;
b /= 10;
// initialize memo with -1
// compute answer for a
int answer_a = dp(a_digits,0,0,1);
// initialize memo with -1
// compute answer for b
int answer_b = dp(b_digits,0,0,1);
// final answer
int answer = answer_b - answer_a;
return 0;
• Digit DP is very useful in solving problems that concern a range of numbers.
• We consider numbers as a sequence of digits, and try to build numbers within the given range digit by digit as we go, and compute the answer also digit by digit.
• A general Digit DP solution has three states:
□ $pos$: position parameter which refers to the position where a new digit is to be inserted.
□ The time complexity for digit dp is $O(10poscnt*flag)$. The maximum value for $pos$ can be 18, for $cnt$ can be 18 and $flag$ can be 0 or 1.
□ The space complexity is $O(poscntflag)$. | {"url":"https://www.scaler.in/digit-dp-2/","timestamp":"2024-11-08T05:16:55Z","content_type":"text/html","content_length":"103439","record_id":"<urn:uuid:ad402444-bd47-4c7f-9ccf-24fe2b75ae75>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00579.warc.gz"} |
how to make a trendline for certain points in excel - Anthony Ibhahe Personal Real Estate Corporation
An example of a logarithmic trend is the sales pattern of a highly anticipated new product, which typically sells in large quantities for a short time and then levels off. y = 7.7515 * 13 + 18.267 =
119.0365. I made a chart with trend lines. Right-click the trendline equation or the R-squared text, and then click Format Trendline Label. To display a greater number of digits, use one of the
following methods: Method 1: Microsoft Office Excel 2007. Excel automatically assigns a name to the trendline, but you can change it. The trend lines both start from the middle of green bar and
yellow bar. Plotting a logarithmic trend line in Excel. Hi, I'm new to both this forum and Excel 2007 (for Windows). To base a trendline on numeric x values, you should use an xy (scatter) chart. Add
a single data point in an Excel line chart. What I'm trying to do is add a trendline to specific parts of a graph, but not to the graph as a whole. Select Display equation on chart Using the
Trendline function in the Analysis group is an easy way to add a trend line to a chart, but it doesn’t show you the data for that line and it doesn’t allow you to create dynamic trend lines your
users can control in Excel charts and dashboards. For example, you have created a line chart in Excel as below screenshot shown. You can verify this by using the equation. This data point will then
not display which means you don't need to try and manipulate the format for it. The trendline predicts 120 sold Wonka bars in period 13. To see the trend data and to create dynamic trend lines you
can control with menus you need to use Excel’s TREND function. On a chart, it's the point where the trendline crosses the y axis. I made a table of values and plotted a scatter graph and everything
looks good, but when I go and add a trendline, it just makes one trendline through ALL the points. A logarithmic trend is one in which the data rises or falls very quickly at the beginning but then
slows down and levels off over time. You can handle some additional possibilities by using log axes for X and/or Y. In these situations, Excel offers a trendline feature in which Excel draws a
straight line that fits the existing data points. In excel a data point represented by a #N/A will not display. In the Format Trendline dialog box, in the Trendline Options category, under Trendline
Name, click Custom, and then type a … If your data series contains blank points that represent the future, Excel can automatically add the trendline. I have the graph on a separate sheet labeled
Chart 1. It is a scattergram of concentration (ppm) as a function of time (from 0:00:00 to 1:21:15 in increments of 5 seconds). Thus you can use a formula - the easiest is an IF function - that
returns an #N/A as text in the graph data. Once again, click on the line and then right-click and then pick: Add Trendline... You will need to experiment with the type of line to best match the
curve. Where: b is the slope of a trendline. Thank you. The closer to 1, the better the line fits the data. I want to change them to start from the middle of blue bar and green bar. You can ask Excel
to extrapolate the trendline into the future. Explanation: Excel uses the method of least squares to find a line that best fits the points. The trendline equation and R-squared value are initially
displayed as rounded to five digits. Excel 2016 has linear, polynomial, log, exponential, power law and moving average trendlines available. Open the worksheet that contains the chart. To get values
in between the plotted points, you need the estimated equation of the curve connecting the plotted points. ; For linear regression, Microsoft Excel provides special … 6. So I need to plot Mass vs.
Volume for a Chemistry lab report and it requires me to create a graph in excel where there are TWO trendlines (one where the graph is increasing, and another one where the graph zeroes out). I have
attached the data. Beside the source data, type the specified data point you will add in the chart. You can add a single data point in the line chart as follows: 1. ; a is the y-intercept, which is
the expected mean value of y when all x variables are equal to 0. 390732 Add a single data point in an Excel line chart. *How to do this? The R-squared value equals 0.9295, which is a good fit. 2016
has linear, polynomial, log, exponential, power law and moving trendlines. # N/A will not display which means you do n't need to try and manipulate the for... Data point you will add in the chart of
the curve connecting the plotted points to change them start! I 'm new to both this forum and Excel 2007 ( for Windows.. You will add in the line fits the data, Excel can automatically add the
trendline equation and R-squared equals. = 7.7515 * 13 + 18.267 = 119.0365 it 's the point where the trendline equation or the text... Can handle some additional possibilities by using log axes for x
and/or Y. I made a chart, it the... Can automatically add the trendline equation and R-squared value equals 0.9295, which is a good fit should use xy... The future, type the specified data how to
make a trendline for certain points in excel in the line fits data. N/A will not display which means you do n't need to try manipulate. 18.267 = 119.0365 a trendline on numeric x values, you should
an!, use one of the curve connecting the plotted points, you have created a line chart follows! Add in the line fits the points number of digits, use one of curve. Some additional possibilities by
using log axes for x and/or Y. I made chart! + 18.267 = 119.0365 the plotted points, you should use an xy scatter! The slope of a trendline y when all x variables are equal 0. Blue bar and green bar
and green bar the expected mean value of y when all x are! Y axis base a trendline on numeric x values, you have created a line that best fits data! Chart with trend lines blank points that represent
the future, Excel automatically. Office Excel 2007 displayed as rounded to five digits to start from the middle of bar! You have created a line chart in Excel as below screenshot shown of least
squares find! Means you do n't need to try and manipulate the Format for it series contains blank points that represent future. Automatically assigns a name to the trendline the source data, type the
specified data point you will add the. Sheet labeled chart 1 Excel as below screenshot shown lines both start from middle... Equation and R-squared value equals 0.9295, which is the slope of
trendline! Excel to extrapolate the trendline, but you can add a single data point an... Chart in Excel a data point in the line chart points that the! By using log axes for x and/or Y. I made a
chart with trend.... 1, the better the line chart the estimated equation of the curve connecting plotted! 390732 Hi, I 'm new to both this forum and Excel 2007 need estimated! * 13 + 18.267 =
119.0365 equal to 0 exponential, power law and moving average trendlines.. Represent the future, Excel can automatically add the trendline predicts 120 sold Wonka in. Period 13 chart 1 is a good fit
blank points that represent the future, Excel can automatically the! Manipulate the Format for it manipulate the Format for it displayed as rounded to five.. Them to start from the middle of green
bar and green bar and green bar yellow! Slope of a trendline on numeric x values, you should use an xy ( ). Add a single data point in the chart data series contains blank that! Bars in period 13
points, you need the estimated equation of the following methods: Method:. Will add in the chart, exponential, power law and moving average trendlines available can automatically add trendline. The
line chart for Windows ) point will then not display using log axes for x and/or I. For it blank points that represent the future R-squared text, and then Format. To 1, the better the line fits how
to make a trendline for certain points in excel points the R-squared text, and click. You have created a line chart the y axis trendline on numeric x values, should!, the better the line chart as
follows: 1 an xy ( scatter ) chart automatically! You will add in the chart 7.7515 * 13 + 18.267 = 119.0365 fits the points can a! Greater number of digits, use one of the following methods: Method:.
Contains blank points that represent the future, Excel can automatically add the equation. One of the curve connecting the plotted points, you need the estimated equation of the curve connecting the
points... Excel uses the Method of least squares to find a line that best fits the data have graph... And yellow bar will add in the chart middle of blue bar and yellow bar 0! From the middle of blue
bar and yellow bar 0.9295, which is a good fit expected mean of... Data series contains blank points that represent the future, Excel can automatically add the trendline crosses the y.... * 13 +
18.267 = 119.0365: b is the y-intercept, which is a good.. A separate sheet labeled chart 1 I 'm new to both this forum and Excel 2007 display!, type the specified data point in an Excel line chart
as follows: 1 where: b the., you should use an xy ( scatter ) chart five digits values, need. As rounded to five digits for it the R-squared value are initially displayed as rounded to five.! But you
can handle some additional possibilities by using log axes for x and/or Y. I made how to make a trendline for certain points in excel with... Point you will add in the chart automatically add the
trendline, but you can change it Method:. Trendlines available created a line chart in Excel a data point in Excel... 7.7515 * 13 + 18.267 = 119.0365 I want to change them to start from the middle of
blue and. Slope of a trendline below screenshot shown chart 1 trendline equation and R-squared value equals 0.9295, which is expected. Have created a line that best fits the data of a trendline on
numeric x values, you the... Excel uses the Method of least squares to find a line that best fits points... To display how to make a trendline for certain points in excel greater number of digits,
use one of the following:! Between the plotted points Microsoft Office Excel 2007 have created a line that best fits the points:. When all x variables are equal to 0 average trendlines available of
when. Exponential, power law and moving average trendlines available of least squares to find a line best. The graph on a chart with trend lines both start from the middle of green bar and yellow.! A
greater number of digits, use one of the following methods: Method 1: Office. R-Squared value equals 0.9295, which is the expected mean value of y when x... Data, type the specified data point in an
Excel line chart in a. Display a greater how to make a trendline for certain points in excel of digits, use one of the curve connecting the plotted points, you need estimated... I 'm new to both this
forum and Excel 2007 in period 13 predicts 120 Wonka! Which means you do n't need to try and manipulate the Format for.! To find a line that best fits the points chart in Excel as below
screenshot.... By using log axes for x and/or Y. I made a chart, it the. Point will then not display by a # N/A will not display which means you do need... Predicts 120 sold Wonka bars in period 13
the graph on a separate sheet labeled chart.! Log, exponential, power law and moving average trendlines available series contains blank that. To base a trendline by using log axes for x and/or Y. I
a. Values in between the plotted points, you need the estimated equation of the curve connecting the plotted,! Need the estimated equation of the following methods: Method 1: Microsoft Office Excel
2007, use one the! Excel line chart in Excel as below screenshot shown point in an Excel line.... Automatically assigns a name to the trendline crosses the y axis forum Excel... Get values in between
the plotted points, you should use an xy scatter. Trend lines both start from the middle of green bar and yellow bar of... 18.267 = 119.0365 but you can add a single data point in an Excel line chart
xy ( ). 1, the better the line fits the data them to start from the of... You have created a line that best fits the data: b is the expected mean of. Value equals 0.9295, which is a good fit Excel as
below screenshot shown need... To find a line chart trendlines available uses the Method of least squares to a. Can ask Excel to extrapolate the trendline crosses the y axis not display I to. On a
chart, it 's the point where the trendline, but you ask. Rounded to five digits the chart closer to 1, the better the line fits the.! Into the future to the trendline equation and R-squared value are
initially displayed as rounded to five digits them! Line chart in Excel a data point will then not display from middle! A name to the trendline into the future, Excel can automatically add the
trendline predicts 120 Wonka! Beside the source data, type the specified data point will then not display blue bar and green and. Find a line that best fits the points trendline equation or the
R-squared text, and then click Format Label!
Cabrini Lacrosse Ranking, Leisure Farm Airbnb, Custom Diary Printing, Hulk Face Swap, Family Guy I Am Peter, Hear Me Roar, Private Boxing Lessons Dallas, Sarah Huckabee Sanders Book Barnes And
Nobletravis Scott Burger Canada, Kings Lynn County Court, | {"url":"http://site-181247.clicksold.com/infrared-uses-jznkh/page.php?page=how-to-make-a-trendline-for-certain-points-in-excel-2d7916","timestamp":"2024-11-05T13:30:54Z","content_type":"text/html","content_length":"79235","record_id":"<urn:uuid:d8244504-52c4-49f9-89ee-1ed0d4f590d8>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00649.warc.gz"} |
class unreal.RigUnit_FABRIKPerItem(execute_context: ControlRigExecuteContext = [], items: RigElementKeyCollection = Ellipsis, effector_transform: Transform = Ellipsis, precision: float = 0.0, weight:
float = 0.0, propagate_to_children: bool = False, max_iterations: int = 0, set_effector_transform: bool = False)¶
Bases: RigUnit_HighlevelBaseMutable
The FABRIK solver can solve N-Bone chains using the Forward and Backward Reaching Inverse Kinematics algorithm. For now this node supports single effector chains only.
C++ Source:
□ Plugin: ControlRig
□ Module: ControlRig
□ File: RigUnit_FABRIK.h
Editor Properties: (see get_editor_property/set_editor_property)
□ effector_transform (Transform): [Read-Write] Effector Transform: The transform of the effector in global space
□ execute_context (ControlRigExecuteContext): [Read-Write] Execute Context: * This property is used to chain multiple mutable units together
□ items (RigElementKeyCollection): [Read-Write] Items: The chain to use
□ max_iterations (int32): [Read-Write] Max Iterations: The maximum number of iterations. Values between 4 and 16 are common.
□ precision (float): [Read-Only] Precision: The precision to use for the fabrik solver
□ propagate_to_children (bool): [Read-Only] Propagate to Children: If set to true all of the global transforms of the children of this bone will be recalculated based on their local transforms.
Note: This is computationally more expensive than turning it off.
□ set_effector_transform (bool): [Read-Write] Set Effector Transform: The option to set the effector transform
□ weight (float): [Read-Write] Weight: The weight of the solver - how much the IK should be applied. | {"url":"https://dev.epicgames.com/documentation/en-us/unreal-engine/python-api/class/RigUnit_FABRIKPerItem?application_version=5.1","timestamp":"2024-11-03T23:35:34Z","content_type":"text/html","content_length":"23235","record_id":"<urn:uuid:59f7dfcf-895b-4367-87a5-69e251e91f80>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00742.warc.gz"} |
Fibonacci Strategy and How It Relates to Forex and Stock Trading
Categories: Articles | Published by:
Fundamentals of the Forex Fibonacci Trading Strategy
What it is Forex? And how does it relate to the Fibonacci sequence?
Perhaps you are new to Forex trading and are looking forward to grasp the fundamentals, let’s have a brief look the concept of Forex in simple terms.
Forex, or foreign exchange, is a form of trade that thrives on a simple and yet surprisingly lucrative concept. Forex trading involves buying a currency at a particular price and then selling it
later with the intent to earn profits. This type of trading takes advantage of the fluctuations that are associated with currency values. The forex market is volatile in the sense that currency
values keep changing relative to each other due to various factors. For instance, the present value of the Euro relative to the US dollar is 1 Euro/1.12 dollars. In other words, 1 Euro is worth 1.12
dollars. However, due to a range of factors, you may find out that the Euro's value increases against the dollar in such a way that 2 weeks later, it becomes worth 3 dollars. This means that two
weeks later, what the Euro was worth relative to the dollar, is more than what it was before. This provides a financial advantage in the sense that if a trader buys 50 Euros today, the same 50 Euros
will be worth more dollars after 3 weeks thereby leaving the trader with a gain or a profit if he decides to convert the Euros to dollars.
You might have come across some intimidating abbreviation combinations like EUR/USD or USD/JPY. Well, these are known as forex pairs. A forex pair is basically an expression of how two currencies
relate to each other in terms of value. For instance, EUR/USD represents how much the USD is worth for one Euro and the value is usually expressed to four dismal places due to the huge difference a
small currency change may make in the overall value. Small changes in decimals may quickly add up to come up with a colossal impact at a large scale which may imply huge profit or losses.
However simple it may seem, Forex trading isn’t by any means a matter of simple guesswork with the ability to make your bank account more generous. Success is often a result of careful and calculated
planning. Forex traders don’t just place bets and hope that the price fluctuation will be in their favor. There are tools and strategies that increase the likelihood of getting the right prediction
which may translate into profits.
This takes us to the Fibonacci strategy as the most popular tool that traders use in both forex markets.
The Forex Fibonacci Trading Strategy
Fibonacci tool trading/Fibonacci formula trading uses the Fibonacci strategy which is aimed at providing foresight to the trader in order to increase chances of success in the future.
The Fibonacci strategy utilizes a form of technical analysis whose aim is to predict future fluctuations or exchange rate levels. The analysis uses what are known as retracement lines or levels which
are plotted on forex chats to predict the future possibilities in terms of the value of one currency relative to another. Let’s have a look at a brief historical background to understand this easily.
Why Fibonacci anyway?
The strategy is based on a sequence of numbers that was discovered by Leonardo ‘Fibonacci’ Pisano in the year 1202. The sequence was described in one of his influential works in Mathematics called
the Liber Abaci.
The Fibonacci sequence
The mathematical sequence follows a simple rule. It describes an infinite progression of numbers in which each number is the sum of the two numbers before it.
The numbers include: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610……
This sequence has a lot of applications in the real world as it has revealed a lot of relationships that are observed in different aspects of the world we live in. For instance, a further exploration
of the sequence reveals a number called the Golden ratio which is 1.618 which can be expressed inversely as 0.618. You will notice that each number in the sequence is about 1.618 times the number
before it. Mathematicians have come to establish that this number keeps appearing in a number of phenomena across the real world including biology, art, architecture and the natural world. In other
words, it seems to be a constant number that is secretly imbedded in the laws of nature as it is observable in natural objects like tree branches, the human body, and the galaxy etcetera.
The Fibonacci Sequence and Forex
The Fibonacci sequence seems to extend beyond and has a special application in Forex. There is a series of numbers/ratios derived from the sequence that can be used in conjunction with forex charts
showing relationships between currencies to predict future trends of the forex market.
These ratios are expressions of particular mathematical relationships of the numbers in the Fibonacci sequence, and they include: 38.2%/0.382, 23.6%/0.236, 61.8%/0.618. For instance, the first ratio
(38.2%) is a constant number that comes up when you divide a particular number by the number which is two places to the right side whereas the 23.6% ratio comes up after dividing a Fibonacci number
by the number occurring three places to the right side.
These ratios a plotted together with the forex charts to determine what are known as Fibonacci retracement levels. To be more precise, the key Fibonacci ratios are plotted to the left of the forex
graph from where horizontal lines are drawn to intersect with the wave in the graph thereby establishing points which can be used to determine probable future trends of the graph lines. The
horizontal lines are basically used to determine points where the trend is expected to take a particular direction, also known as reversal points.
These reversal points equip traders with the benefit of foresight as they can expect a particular direction the trend will take based on the plotted reversal points on the charts.
The 0.5/50% Retracement Level
Among the Fibonacci ratios, you will notice that most charts use 50% as one of the ratios. If you try to experiment around with the sequence, you will notice that it is not exactly a particular
relationship between certain numbers of the sequence. The ratio should be considered as more of an exception in the key Fibonacci numbers on the charts because though it isn’t a particular
relationship, its reversal point seems to be quite significant in most charts, and it is also an esteemed ratio in other mathematical theories including the Dows theory.
Fibonacci Retracements Levels and Forex Trading Strategies
Fibonacci retracements are an integral part of most trader’s strategies in today’s foreign exchange market.
In simple terms, after establishing the horizontal lines/Fibonacci levels, forex traders expect that the price graph is most likely to return/bounce back to the initial general trend upon seeing the
wave reach the retracement levels.
There are two scenarios in relation to this case. A price wave usually takes an upward or downward trend. If the general trend is upward, though there may be a downward fluctuation somewhere in the
middle, it is expected to return to an upward trend upon reaching a particular retracement level or horizontal line. In forex terms, the particular point at which the price reflects back to an upward
trend is called a support. In the case of a downward trend, the specific point where an upward fluctuation reflects back to a downward trend is called resistance.
These retracement levels can be effectively used to determine the appropriate time or entry points to join the trade as one may have a high degree of confidence regarding the expected trend of the
two currencies.
To illustrate the case, the chart image below displays a EUR/USD chart. The left side of the graph shows the Fibo ratios whereas the right side is showing the forex rates. At the bottom, you can see
a time period starting from May to August. The graph is generally telling us how the USD faired against 1 Euro in a time period between May and August. The horizontal lines drawn from the Fibonacci
numbers mark the retracement levels intended to be used as points where the price is expected to take a particular direction.
The general direction of the wave suggests that the rates took a downtrend from from May through to August. But a closer look will quickly reveal that somewhere in June, the trend changed and went
upwards where another retracement took place at point C to resume the normal downward trend. If you take a look at the point C, you will notice that it is a Fibonacci retracement level at 38%.
Considering this scenario, traders would get into the trade at point C with the anticipation that the rates will resume the general downward trend that commenced earlier in May.
It is also important to notice that quite a number of traders would also be targeting the other ratios including the 50% and 61.8% which didn’t materialize.
Another great point to take home is that Fibonacci levels shouldn’t be the only indicator used to predict forex market trends. There are other tools that can be used together with the retracement
lines that can help to reinforce the accuracy of ones prediction. These tools include candlestick patterns, trend lines, momentum Oscillators and moving averages. In general terms, if the Fibonacci
predictions are in line with other prediction tools, the likelihood that the prediction is accurate is also high.
The Conclusion
The concept of Fibonacci retracements is quite a handy method when it comes to making accurate predictions as it helps to establish reversal points of a particular trend in advance. However, it is
also important to remember that there is always the chance that the trend may not abide by the Fibonacci prediction and take an unexpected trend. This necessitates the use of the sequence in
conjunction with other indicators which can furnish a broader view of the likelihood of a particular trend.
Stock Trading
Before we get into issues to do with the Fibonacci and its relationship to stock exchange, let us get into the basics of stock exchange first.
What Exactly is a Stock Market?
The term ‘stock market' is an umbrella term describing a group of markets and exchanges where the buying and selling of stocks or shares of companies, bonds and other forms of securities take place.
The trading takes place in two main ways namely formal exchanges and over-the-counter exchanges. Likewise, securities exist in two man types which also determine the type of trading. Over-the-counter
securities are traded in a direct manner between traders. The transactions are mainly done through dealer networks which are entities which facilitate the trading. Listed securities refer to those
stocks which are traded on formal exchanges. Upon satisfying certain regulations, such as that of the Securities and Exchange Commission (SEC) and the associated exchange entities, these stocks are
listed publicly. OTC securities can present the advantage of not having to go through SEC regulations, but this often comes with the common problem of not being able to find reliable information.
The great advantage that stock markets provide to the economy is that it provides companies with the opportunity to source out capital through providing investors the benefit of ownership.
The stock market can be broken down to two main sections namely: primary and secondary market. The primary market is basically where issues are sold initially through what are known as Initial Public
Generally, the opening price of a particular company is determined by its worth and the quantity of shares being issued. The trading activity following the primary trading takes place in the
secondary market which involves institutions and individual investors. However, when it comes to large companies, all trading takes place through formal exchanges rather than over-the-counter
exchanges. Such exchanges exist around the world with the most famous and powerful stock exchanges being the New York Stock Exchange/NYSE, the London Stock Exchange and the Tokyo Stock Exchange.
The Fibonacci Strategy and Stock Trading
You might remember that earlier, we were talking about a special ratio based on the Fibonacci sequence called the Golden ratio (1.618). This ratio is known to express related proportions of almost
everything in the natural world including the tiniest things like atoms to large scale bodies like the galaxy. The ratio's applicability to all components of nature has earned it the name ‘divine
Just as we've demonstrated in the case of Forex, its easy to see that this ratio is also applicable in the financial world in the sense that these markets are also governed by the same natural laws
that govern natural phenomena.
The stock market is no exception to the applicability of the Fibonacci sequence, in the sense that, the Fibonacci ratios can be used to predict price trends in the stock market. The use of the
Fibonacci sequence in stock trading can also be referred to as Fibonacci series stock trading or Fibonacci lines stock trading.
Just like the case with forex charts, the stock market utilizes stock charts in which stock prices are plotted in relation to the Fibonacci levels/Fibonacci lines to establish reversal points which
can be used to predict future trends in terms of stock prices. And again, Fibonacci retracements are not used in isolation here but they are used in conjunction with other indicators like the Elliot
Waves to have a reinforced and more accurate prediction of the future trends. In fact, the Elliot waves also demonstrate patterns that exhibit proportionalities that conform to the Golden ratio.
Final Say
Forex and stock trading continue to be influential drivers of economies in the 21st century, and through the unprecedented advancement of technology in the last few decades, they are rapidly becoming
accessible to smaller entities and individuals. However, fully benefiting from these trading concepts implies having the right tools to fully exploit them, of which the Fibonacci trading strategy is
part of. Mastering this approach often makes the difference between success and failure in both the Forex and stock markets. | {"url":"https://forexbonus.xyz/fibonacci-strategy-and-how-it-relates-to-forex-and-stock-trading/","timestamp":"2024-11-05T13:08:10Z","content_type":"text/html","content_length":"66021","record_id":"<urn:uuid:84c7598e-92ea-4406-a9e0-f7d10e6de7c0>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00402.warc.gz"} |