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https://www.shaalaa.com/question-bank-solutions/chemical-properties-acids-bases-how-concentration-hydronium-ions-h3o-affected-when-solution-acid-diluted_5927
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# How is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted? - Science
Course
ConceptChemical Properties of Acids and Bases
#### Question
How is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted?
#### Solution
When an acid is diluted, the concentration of hydronium ions (H3O+) per unit volume decreases. This means that the strength of the acid decreases.
Concentration of hydronium ions="Volume of solute (acid)"/"Volume of solution"
Is there an error in this question or solution?
#### APPEARS IN
NCERT Solution for Science Textbook for Class 10 (2019 to Current)
Chapter 2: Acids, Bases and Salts
Q: 5 | Page no. 25
NCERT Solution for Science Textbook for Class 10 (2018 to Current)
Chapter 2: Acids, Bases and Salts
Q: 5 | Page no. 25
#### Video TutorialsVIEW ALL [2]
Solution How is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted? Concept: Chemical Properties of Acids and Bases.
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2020-02-23 16:14:04
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https://www.biostars.org/p/468504/
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Providing R1,R2 to macs2
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Entering edit mode
12 months ago
Sam ▴ 170
I have paired-end ChIP-Seq data that resulted in a low amount of uniquely mapped reads. When mapping the reads PE, only a small percentage maps concordantly.
I want to check whether the reads give some biological sense. For that, I thought of mapping SE, taking the condition with the best statistics, running macs2, annotating, and seeing whether the annotations make sense.
So, I have separate files for R1 and R2 alignment.
• Here it was advised to used R1 only. Why is it preferred to use R1 only rather than combine both reads?
• When running macs2, is it advisable to combine both files into one? For example
macs2 -t sample_R1.bam sample_R2.bam -c control_R1.bam control_R2.bam
ChIP-Seq macs2 • 364 views
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Entering edit mode
Can you give some details? What is the command line for alignment (paired-end mode) and what is "Low"? If alignment does not go well then this is probably a data quality problem that cannot be solved by tweaking the peak calling process. R1 and R2 come from the same fragment so you either must use them after aligning them in PE mode or use either of them but "combining" them as in this command you provide artificially doubles the total read count which would be improper since the reads come from the same fragment.
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bowtie ..Mus_musculus/Ensembl/GRCm38/Sequence/BowtieIndex/genome -1 Sample1_R1.fastq -2 Sample1_R2.fastq -m 1 --fr --sam -p 8 --tryhard --minins 0 --maxins 1000 --chunkmbs 2000 Sample1.sam
The average percentage of the reads that failed to align (neither uniquely nor multi mapped) in each sample is 94.5 %
Yes, this is clearly a data problem, that cannot be solved by tweaking the peak alignment process, but all I'd like to do is see if there is some biological sense in the data. For that, I thought of taking the maximum amount of data available. When aligning single-end, "only" 64% are unmapped as the median value of all samples.
R1 and R2 come from the same fragment so you either must use them after aligning them in PE mode or use either of them but "combining" them as in this command you provide artificially doubles the total read count which would be improper since the reads come from the same fragment.
So combining R1 & R2 would lead to more false positives? You say that what we are interested in, biologically, is in the amount of fragments, and using reads, artificially doubles their number.
Would use both reads for control as well compensate for that, or it would just introduce another posible sort of bias?
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Entering edit mode
Look, there is obviously something very wrong with your ChIP. Did you use carrier DNA? Try to blast some of the unmapped reads. Maybe a sample swap or wrong starting material? 5% alignment rate is not right at all. I'd figure out why that is, all these tweaks you suggest are not recommended, you have a basic problem with your sample, try to find out why.
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Entering edit mode
Thanks. Just for understanding's sake, if you don't mind :
If I would use both R1 and R2 in the way suggested above, it would artificially inflate the number of fragments, considering each read as a fragment. Suppose, just as an artificial example, that there are exactly 1M pairs. So, after after artificially pooling those, we would get 2M reads.
However, the normalization step at macs2, takes into account the number of reads. Suppose we have 4M reads in the control - if we take R1 only, the coverage of control would be scaled down by 0.25. If we take both R1 and R2, then the coverage would be scaled down only by 0.5.
Would not that compensate for the double amount of reads?
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Entering edit mode
Also (for the sake of understanding), I was told that when the technology was worse, people merged biological replicates, when they had not enough material in them separately.
What is the substantial difference between merging biological replicates, and merging R1 and R2?
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Entering edit mode
Merging biological replicates means pooling DNA on the library prep level. Merging R1 and R2 means to combine reads that come from the same fragment which is not necessary nor meaningful since macs2 will anyway extend reads to fragment size based on its fragment size estimation procedure. If you reads do not align properly when run in paired-end mode then this is what it is, a library quality problem or contamination. You can technically of course try and tweak this but the ground truth (a bad library) will stay and imho you are somewhat lying to yourself. Garden of forked paths comes to mind. If you get a reviewer that has some understanding of the method and the data analysis I do not see how you could ever publish this.
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2021-10-18 13:18:04
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https://blender.stackexchange.com/questions/52052/problem-exporting-model-with-texture
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# problem exporting model with texture
I have a basic problem (I guess). I'm just using the default cube, I added a material then a texture I chose the default "marble" pattern. When I render it, it looks like it should. However, when I export the object (tried all different formats), then import the model back into blender, there is no marble pattern on the cube.
• To my knowledge, (could be wrong) nodes are a cycles only thing, so exporting them shouldn't work. As eromod said, you'd probably need to bake it. See blender.stackexchange.com/a/13509/11237 – Luka ash May 6 '16 at 2:30
• Did you unwrap the model? Did you bake the textures? – Neil Jan 10 '17 at 10:41
Other formats than blenders .blend don't support Blenders full feature set, have different settings or similar functions, which are not always compatible. Not all Settings, Materials or Datatypes are saved by export functions. So if you don't want to import your model in a different Program, use the Save As Function to save your file in the native .blend format.
• Thanks for the answers, but my problem is when i export my file to .obj or .fbx, and then re import the .obj or .fbx it just gets grey and the pre build texture (pattern) like "marble" is not vissible. If i just put a color on a cube and export to .obj or .fbx and re import it works, and the color is visible. – mathias May 6 '16 at 0:21
• .Obj only supports basic colors and image texture, not procedural texture such as marble. – Mike Pan Jul 5 '16 at 5:30
• @mathias This is the correct answer, and it should be marked as such. – CGEffex Apr 25 '17 at 15:41
maybe try to "pack external data"?
File ‣ External Data ‣ Pack into blend-file
https://www.blender.org/manual//data_system/introduction.html#pack-unpack-data
• Lets say i just make a simple glass, and i use the "glass bsdf" option in the material tab. How can i export my model to .obj or .fbx and keep that "glass effect"? when i re import my .obj or.fbx file to blender its just grey. – mathias May 6 '16 at 0:27
• Oh, I think you need to bake it to a texture because its a cycles thing and not all programs have cycles. – eromod May 6 '16 at 0:37
• the texture can be partially transparent like glass but other programs might not deflect light like cycles – eromod May 6 '16 at 0:42
• Ok thanks for the answer, i will see if i can find out how to bake to a texture. Totaly new to blender, but seems like a very nice program for making models. – mathias May 6 '16 at 13:59
• Im pretty new too, I started less than half a year ago. Its my new years resolution to read the entire Blender Reference Manual. With a minimum of three pages a day, I can make it. Its really hard to find the will power but I recommend it because its "comprehensive". – eromod May 7 '16 at 2:26
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2020-03-30 16:40:54
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http://cms.math.ca/cmb/msc/14N05
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Canadian Mathematical Society www.cms.math.ca
location: Publications → journals
Search results
Search: MSC category 14N05 ( Projective techniques [See also 51N35] )
Expand all Collapse all Results 1 - 2 of 2
1. CMB 2011 (vol 54 pp. 430)
DeLand, Matthew
Complete Families of Linearly Non-degenerate Rational Curves We prove that every complete family of linearly non-degenerate rational curves of degree $e > 2$ in $\mathbb{P}^{n}$ has at most $n-1$ moduli. For $e = 2$ we prove that such a family has at most $n$ moduli. The general method involves exhibiting a map from the base of a family $X$ to the Grassmannian of $e$-planes in $\mathbb{P}^{n}$ and analyzing the resulting map on cohomology. Categories:14N05, 14H10
2. CMB 2002 (vol 45 pp. 349)
Coppens, Marc
Very Ample Linear Systems on Blowings-Up at General Points of Projective Spaces Let $\mathbf{P}^n$ be the $n$-dimensional projective space over some algebraically closed field $k$ of characteristic $0$. For an integer $t\geq 3$ consider the invertible sheaf $O(t)$ on $\mathbf{P}^n$ (Serre twist of the structure sheaf). Let $N = \binom{t+n}{n}$, the dimension of the space of global sections of $O(t)$, and let $k$ be an integer satisfying $0\leq k\leq N - (2n+2)$. Let $P_1,\dots,P_k$ be general points on $\mathbf{P}^n$ and let $\pi \colon X \to \mathbf{P}^n$ be the blowing-up of $\mathbf{P}^n$ at those points. Let $E_i = \pi^{-1} (P_i)$ with $1\leq i\leq k$ be the exceptional divisor. Then $M = \pi^* \bigl( O(t) \bigr) \otimes O_X (-E_1 - \cdots -E_k)$ is a very ample invertible sheaf on $X$. Keywords:blowing-up, projective space, very ample linear system, embeddings, Veronese mapCategories:14E25, 14N05, 14N15
© Canadian Mathematical Society, 2014 : http://www.cms.math.ca/
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2014-03-10 02:26:42
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https://brilliant.org/problems/another-easy-one/
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# Another easy one
Number Theory Level pending
How many different pairs of positive integers (x, y) are solutions of the equation $$x^{2} +3y^{2} = 1997$$?
×
Problem Loading...
Note Loading...
Set Loading...
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2017-03-28 23:33:03
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https://math.paperswithcode.com/paper/upper-bounds-on-kronecker-coefficients-with
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# Upper bounds on Kronecker coefficients with few rows
26 Mar 2020 Pak Igor Panova Greta
We present three different upper bounds for Kronecker coefficients $g(\lambda,\mu,\nu)$ in terms of Kostka numbers, contingency tables and Littlewood--Richardson coefficients. We then give various examples, asymptotic applications, and compare them with existing lower bounds...
PDF Abstract
# Code Add Remove Mark official
No code implementations yet. Submit your code now
# Categories
• COMBINATORICS
• REPRESENTATION THEORY
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2021-01-21 05:06:00
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https://solvedlib.com/n/let-x-be-a-random-variable-with-pdf-of-a-regular-case-of,8580057
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##### An auto dealeris offering any special options at one price on a specially equipped car being sold How many different choices of specially equipped cars do YOu have?
An auto dealeris offering any special options at one price on a specially equipped car being sold How many different choices of specially equipped cars do YOu have?...
##### Dav d wants - Jave Money { purcnase buys annuity with Guotterty payments ttat ejtn 2 2" Intercst compourdt Tade Ot the end cach quatter Quartorly: Rarnicnb mlIbe tht total value 0( the annuty In } Yeans I cuch quarteriy nsyienl S1j89. Do not round Intettecdl codutelont, and rolnd Your (nal ehmct the rcarost cen{ FECASAAe the Ilxt ot firanctal fortuljt
Dav d wants - Jave Money { purcnase buys annuity with Guotterty payments ttat ejtn 2 2" Intercst compourdt Tade Ot the end cach quatter Quartorly: Rarnicnb mlIbe tht total value 0( the annuty In } Yeans I cuch quarteriy nsyienl S1j89. Do not round Intettecdl codutelont, and rolnd Your (nal ehm...
##### A wave on a string has an amplitude of 0.567 mm and is propagating with a...
A wave on a string has an amplitude of 0.567 mm and is propagating with a speed of 15.0 m/s in the positive x direction. If the frequency of the wave is 42.0 Hz, what is the speed of the string at x = 1.25 cm and t= 0.0625 s? (Hint: not looking for a wave speed)...
##### Help Entering Answers See Examples 2.3.6-23.7. in Section 23. in the MTH 235 Lecture Notes_ (10 points)Consider the differential equationY" -4y +4y =1 > 0.(a) Find r1. T2 roots of the charactenstic polynomial of the equation above.(b) Find set of real-valued fundamental solutions to the homogeneous differential equation corresponding to the one above.9 (t) e^(24)92(t) e^(2tle^(-21t(c) Find the Wronskian of the fundamental solutions you found in part (b):W(t)(d) Use the fundamental solut
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##### Questions 13 and 14 use the following information: The following frequency table gives the speed of a sample of cars in a 100 km/h speed zone as recorded by a police radar unit:Car Speed Class 70 to less than 80 km/h 80 to Iess than 9 km/h 90 to less than 100 km/h 100 to less than 110 km/h 110 to less than 120 km/h 120 to less than 130 km/h 130 to less than 140 km/hFrequency4614592 36 16Question 13 The median speed (km/h) of the sample of cars is approximately: 95.34 (b) 105 (C) 95 (d) 98.34 (e)
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##### 9 (10 pts: ) The following function has three critical points (0,0), (1,1) , (-1,-1): Use the second derivative test to determine if each point is a local maximum; local minimum; O saddle point.f(w,y) = x4 +y4 4xy +1.
9 (10 pts: ) The following function has three critical points (0,0), (1,1) , (-1,-1): Use the second derivative test to determine if each point is a local maximum; local minimum; O saddle point. f(w,y) = x4 +y4 4xy +1....
##### Thc distance advnced tlte: intervz] (30, 50).day hy tune] boring machine_MCECmuniforuly distributed") Find Iu JA distanee.6) Fiudstnulurd deviation tho distmic"n
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##### Question 8 (1 point) Calculate the change in energy for the following process: How much energy (in KCal) must be added to 45 g of water to heat it from 30 %C to 85 %C?26.7 Kcal224 Kcal+26.7 Kcal-24.3 Kcal+24.3 Kcal+2.4 Kcal
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##### Velocity
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##### How do i do this using a financial calculator? Stuart Corporation purchased equipment and in exchange...
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##### Please describe in detail how did you apply the Accounting principles learned in your previous course,...
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##### Molander Corporation is a distributor of a sun umbrella used at resort hotels. Data concerning the...
Molander Corporation is a distributor of a sun umbrella used at resort hotels. Data concerning the next month's budget appear below Selling price per unit Variable expense per unit 27 18 Fixed expense per month Unit sales per month 8,100 1,050 Required: 1. What is the company's margin of saf...
##### Project 2 - Case 2 40 points Davis Dry Goods distributes silk ties. You are in...
Project 2 - Case 2 40 points Davis Dry Goods distributes silk ties. You are in charge of creating Davis' master budget for the upcoming second quarter, April-June 2018. Davis desires a minimum ending cash balance each month of $10,000. The ties are sold to retailers for$9 each. Recent and forec...
##### Which explains intermediate filament tensile strength in a cell's cytoplasm? Covalent cross-links between intermediate filament proteins_ Phosphorylation on specific intermediate filament tyrosine residues_ Coiled-coil intermediate filament protein structure All (a., b-, and c)
Which explains intermediate filament tensile strength in a cell's cytoplasm? Covalent cross-links between intermediate filament proteins_ Phosphorylation on specific intermediate filament tyrosine residues_ Coiled-coil intermediate filament protein structure All (a., b-, and c)...
##### "Sotup Ine 1 JJ Ialugung pefiod Ludi 1 Hualacaous L Lanar 1 1 1 U M H suurcl(8 *Yualtohd 1 1 Mautlce ;rernaining 1Culcuiate oiounodi Yulua Olte bucimal pacu2box ana Lhen clic* 3 11ARASD/ouna ,
"Sotup Ine 1 JJ Ialugung pefiod Ludi 1 Hualacaous L Lanar 1 1 1 U M H suurcl(8 *Yualtohd 1 1 Mautlce ; rernaining 1 Culcuiate oiounodi Yulua Olte bucimal pacu 2 box ana Lhen clic* 3 1 1 ARASD/ ouna ,...
##### Which factor is NOT necessary for convergence between two countries? O equal access to education O...
Which factor is NOT necessary for convergence between two countries? O equal access to education O a common language between the two countries O similar levels of political stability O equal access to infrastructure...
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2022-11-26 10:19:36
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https://barneyshi.me/2021/12/24/Find-Pivot-Index/
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# Leetcode 724 - Find Pivot Index
Note:
• Use presum[] to accelerate the process of looking up presums.
Question:
Given an array of integers nums, calculate the pivot index of this array.
The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index’s right.
If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.
Return the leftmost pivot index. If no such index exists, return -1.
Example:
Code:
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2022-10-06 09:52:41
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https://stats.stackexchange.com/questions/276492/in-what-situations-for-linear-regression-can-you-get-a-blue-solution?noredirect=1
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# In what situations for linear regression can you get a BLUE solution?
$A^TA$ need not be invertible. If columns of a matrix A are linearly independent, then $A^TA$ is invertible.
Let's say $A^TA$ isn't invertible:
Gauss-Markov seems to hinge on $A^TA$ being invertible. Suppose that A is $mxn$ real matrix. If $m<n$, then the inverse of $A^TA$ does not exist.
But you can minimize the sum of squared differences, regardless (use some iterative algorithm). The proof for Gauss-Markov that I know won't work anymore. (BLUE: best linear unbiased estimator, OLS: ordinary least squares). In this situation I could see having multiple solutions. And for any of these solutions, are they the OLS? Are they BLUE? Why, if Gauss-Markov doesn't work anymore.
Gauss-Markov: https://en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem Convexity of Linear Regression: Convexity of linear regression
• The last few paragraphs of this question wander off into what looks like a triviality or something without meaning: typically, $A$ has many more rows than columns, so it doesn't even make sense to refer to its "inverse." What do you really want to ask? – whuber Apr 28 '17 at 15:35
• Thank you for the edit. Have you investigated the standard solution to least squares? You seem to be aware it can be expressed in terms of the inverse of $A^\prime A$, but since that explicitly shows there is a unique solution, it's hard to determine what you're trying to ask. – whuber Apr 28 '17 at 15:51
• I'm trying to clarify; there maybe 2 questions here. So if $A^TA$ ISN'T invertible, can you still get the OLS solution? I think also I wasn't convinced that there was a unique solution if $A^TA$ was invertible. Does that make sense? – eternalmothra Apr 28 '17 at 16:03
• I think the second question was too confusing, so I just removed it. I will try to restructure it and perhaps make a separate question later. – eternalmothra Apr 28 '17 at 16:13
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2020-01-28 06:48:51
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https://www.meritnation.com/cbse-class-11-commerce/economics/sandeep-garg-2018/measures-of-dispersion/textbook-solutions/161_16_1216_6136_10.83_67238
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Sandeep Garg 2018 Solutions for Class 11 Commerce Economics Chapter 7 Measures Of Dispersion are provided here with simple step-by-step explanations. These solutions for Measures Of Dispersion are extremely popular among Class 11 Commerce students for Economics Measures Of Dispersion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Sandeep Garg 2018 Book of Class 11 Commerce Economics Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Sandeep Garg 2018 Solutions. All Sandeep Garg 2018 Solutions for class Class 11 Commerce Economics are prepared by experts and are 100% accurate.
#### Question 1:
Calculate figures and coefficient of range of the following series, which gives the monthly expenditure (in ₹) of seven students: 22, 35, 32, 45, 42, 48, 39
Given:
Highest Value (H) =48
Lowest Value (L) =22
Range = Highest Value − Lowest Value
i.e R= H− L
Substituting the given values in the formula.
R= 48 − 22 = 26
Thus, range is 26 and coefficient of range is 0.37
#### Question 2:
Find range and coefficient of range from the weekly wage (in ₹) of 10 workers of a factory: 310, 350, 420, 105, 115, 290, 245, 450, 300, 375.
Given:
Highest Value (H) = 450
Lowest Value (L) = 105
Range = Highest Value − Lowest Value
i.e R= H− L
Substituting the given values in the formula.
R= 450 − 105 = 345
Thus, range is 345 and coefficient of range is 0.62
#### Question 3:
Find the range and coefficient of range of the following:
Size of shoes 6 7 8 9 10 11 12 13 No. of shoes sold 8 12 14 20 15 8 7 10
Size of Shoes No. of Shoes Sold 6 8 7 12 8 14 9 20 10 15 11 8 12 7 13 10
Highest value (H) = 13
Lowest value (L) = 6
Range = Highest value − Lowest value
i.e. R = HL
or, R = 13 − 6=7
Note- As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.
#### Question 4:
From the following data calculate range and coefficient of range:
Marks 10 20 30 40 50 60 70 No. of Studends 8 12 7 30 10 5 2
Marks No. of Students 10 8 20 12 30 7 40 30 50 10 60 5 70 2
Highest value (H) = 70
Lowest value (L) = 10
Range = Highest value − Lowest value
i.e. R = H−L
or, R = 70 − 10= 60 marks
Note- As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.
#### Question 5:
Find out range and coefficient range.
Marks 10−20 20−30 30−40 40−50 50−60 No. of students 12 18 14 63 19
Marks No. of Students 10 − 20 12 20 − 30 18 30 − 40 14 40 − 50 63 50 − 60 19
Hence, the range is 50 marks and coefficient of range is 0.715
#### Question 6:
Following are the marks obtained by 25 students of class XI in an exam. Find out range and coefficient of range of the marks.
Marks 5−9 10−14 15−19 20−24 25−29 30−34 No. Students 4 6 3 2 6 4
In order to calculate range and its coefficient, we must first convert the given inclusive class intervals in exclusive class intervals.
Class Interval Exclusive Class Interval Frequency 5 − 9 10 −14 15 −19 20 −24 25 −29 30− 34 4.5 − 9.5 9.5 − 14.5 14.5 − 19.5 19.5 − 24.5 24.5 − 29.5 29.5 − 34.5 4 6 3 2 6 4
#### Question 7:
Calculate interquartile range, quartile deviation and coefficient of quartile deviation from the following data:
Family A B C D E F G Income (in ₹) 1,200 1,400 1,500 1,700 2,000 2,100 2,200
Family S. No. Income Rs A 1 1200 B 2 1400 C 3 1500 D 4 1700 E 5 2000 F 6 2100 G 7 2200 N = 7
#### Question 8:
find out the value of quartile deviation and its coefficient from the following data:
Roll No. 1 2 3 4 5 6 7 Marks 20 28 40 12 30 15 50
Roll No. Marks in Ascending Order 1 12 2 15 3 20 4 28 5 30 6 40 7 50 N = 7
#### Question 9:
Form the following figures, find the quartile deviation and its coefficient:
Height (cm) 150 151 152 153 154 155 156 157 158 No. of students 15 20 32 35 33 22 20 12 10
Height (c.m) No. of Students Cumulative Frequency 150 15 15 151 20 15 + 20 = 35 152 32 35 + 32 = 67 153 35 67 + 35 = 102 154 33 102 + 33 = 135 155 22 135 + 22 = 157 156 20 157 + 20 = 177 157 12 177 + 12 = 189 158 10 189 + 10 = 199 N = 199
#### Question 10:
Compute coefficient of Quartile deviation form the following data:
Marks 10 20 30 40 50 60 No. of students 4 7 15 8 7 2
Marks No. of Students Cumulative Frequency 10 4 4 20 7 7 + 4 = 11 30 15 11 + 15 = 26 40 8 26 + 8 = 34 50 7 34 + 7 = 41 60 2 41 + 2 = 43 N = 43
#### Question 11:
Calculate Quartile Deviation and its Coefficient from the following data:
Size 5−10 10−15 15−20 20−25 25−30 30−35 35−40 40−45 45−50 Frequency 6 10 18 30 15 12 10 6 4
Size Frequency Cumulative Frequency 5−10 6 6 10−15 10 16 15−20 18 34 20−25 30 64 25−30 15 79 30−35 12 91 35−40 10 101 40−45 6 107 45−50 4 111 Σf=N = 111
Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{111}{4}\right)\mathrm{th}$ item
= Size of 27.75th item
27.75 item lies in group 15−20 and falls within 34th c.f. of the series
Like wise,
Q3 = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item
= Size of 3
= Size of 83.25th item
83.25th item lies in the group 30-35 within the 33rd c.f. of the series.
#### Question 12:
Calculate coefficient of quartile deviation from the following data:
X (Less than) 200 300 400 500 600 Frequency 8 20 40 46 50
Converting less than cumulative frequency into normal frequency distribution:
X c.f. Frequency 100−200 8 8 200−300 20 20−8=12 300−400 40 40−20=20 400−500 46 46−40=6 500−600 50 50−46=4 N=50
Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{50}{4}\right)\mathrm{th}$ item
= Size of 12.5th item
12.5th item lies in group 200−300 and falls within 20th c.f. of the series.
Here,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval
Likewise,
Q3 = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item
= Size of $3\left(\frac{50}{4}\right)\mathrm{th}$ item
= Size of 37.5th item
37.5th item lies in group 300−400 and fall within the 40th c.f. of the series
#### Question 13:
Estimate an appropriate measure of the following data:
Wages (₹) Less than 25 25−30 30−35 35−40 Above 40 No. of workers 2 10 26 16 7
Wages No. of worker (Frequency) Cumulative Frequency Less than 25 2 2 25−30 10 12 30−35 26 38 35−40 16 54 Above 40 7 61 N=61
Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{61}{4}\right)\mathrm{th}$ item
= Size of 15.25th item
15.25th item lies in group 30−35 and falls within 38th c.f. of the series.
Here,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval
Likewise,
Q3 = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item
= Size of $3\left(\frac{61}{4}\right)\mathrm{th}$ item
= Size of 45.75th item
47.75th item lies in group 35−40 and fall within the 54th c.f. of the series
#### Question 14:
Calculate the mean deviaiton from median and its coefficient from the data: 100, 150, 80, 90, 160, 200, 140
Sr. No Values (X) Deviation from Median M= 140 1 2 3 4 5 6 7 80 90 100 140 150 160 200 60 50 40 0 10 20 60 N=7 $\Sigma \left|{d}_{m}\right|=240$
(M) Median = Size of
= Size of
= Size of 4th item
= 140
#### Question 15:
Compute Mean deviation and its coefficient by mean from the data given below:
X 210 220 225 225 225 240 250 270 280
Sr. No Values (X) Deviation from Mean $\left|{d}_{\overline{X}}\right|=\left|X-\overline{X}\right|$ 1 210 28 2 220 18 3 225 13 4 225 13 5 225 13 6 235 3 7 240 2 8 250 12 9 270 32 10 280 42 N=10 ΣX = 2380 Σ$d\overline{{}_{X}}$ = 176
#### Question 16:
Following are the marks of the students. Find mean deviation and coefficient mean deviation from mean.
Marks 5 10 15 20 25 30 35 40 No. of students 16 32 36 44 28 18 12 14
Marks (X) Frequency (f) fX Deviation from Mean $\left|{d}_{\overline{X}}\right|=\left|X-\overline{X}\right|\phantom{\rule{0ex}{0ex}}\left|X-20.2\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|$ 5 16 80 15.2 243.2 10 32 320 10.2 326.4 15 36 540 5.2 187.2 20 44 880 0.2 8.8 25 28 700 4.8 134.4 30 18 540 9.8 176.4 35 12 420 14.8 177.6 40 14 560 19.8 277.2 ΣX = 180 Σf = 200 ΣfX=4040 Σ$f\left|d\overline{){}_{X}}\right|$=1531.2
#### Question 17:
Find out the mean deviation from the median and its coefficient.
Marks 10 11 12 13 14 No. of students 3 12 18 12 3
Marks (X) Frequency (f) Cumulative frequency (c.f.) Deviation from Median $\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{X}\mathbf{-}\mathbf{M}\right|\phantom{\rule{0ex}{0ex}}\mathbit{M}\mathbit{e}\mathbit{d}\mathbit{i}\mathbit{a}\mathbit{n}\mathbf{=}\mathbf{12}$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$ 10 3 3 2 6 11 12 15 1 12 12 18 33 0 0 13 12 45 1 12 14 3 48 2 6 N=Σf = 48 Σf|dM| =36
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 24.5 th item is 33 (in the c.f. column), which is corresponding to 12.
Hence, median is 12.
Thus we calculate the deviation of the values from 12.
#### Question 18:
Compute the mean deviation from the median and from the mean for the following distribution of the scores of 50 college students:
Scores 140−150 150−160 160−170 170−180 180−190 190−200 Frequency 4 6 10 18 9 3
Scores (X) Frequency (f) Cumulative Frequency (c.f.) Mid -Values (m) Deviation from Median $\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{M}\right|\phantom{\rule{0ex}{0ex}}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{172}\mathbf{.}\mathbf{78}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$ 140−150 4 4 145 27.78 111.12 150−160 6 10 155 17.78 106.68 160−170 10 $\overline{)20}$ 165 7.78 77.8 $\overline{)170}$−180 $\overline{)18}$ 38 175 2.22 39.96 180−190 9 47 185 12.22 109.98 190−200 3 50 195 22.22 66.66 N=Σf = 50 Σf|dM|=512.2
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$item, i.e. ${\left(\frac{50}{2}\right)}^{th}$ item, which is 25th item.
This corresponds to the class interval of 170-180, so this is the median class.
Scores (X) Frequency (f) Mid -Values (m) fm Deviation from Mean $\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\overline{)\mathbf{X}}\right|\phantom{\rule{0ex}{0ex}}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{171}\mathbf{.}\mathbf{2}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|$ 140−150 4 145 580 26.2 104.8 150−160 6 155 930 16.2 97.2 160−170 10 165 1650 6.2 62 170−180 18 175 3150 3.8 68.4 180−190 9 185 1665 13.8 124.2 190−200 3 195 585 23.8 71.4 Σf = 50 Σfm=8560 Σ$\mathbit{f}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|$=528
#### Question 19:
Find the mean deviation from mean and its coefficient for the given data:
X 0−10 10−20 20−30 30−40 40−50 50−60 F 3 5 7 2 9 4
Values (X) Frequency (f) Mid-Values (m) fm $\left|{\mathbit{d}}_{\overline{X}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\overline{\mathbit{X}}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbit{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|\mathbit{d}\overline{{}_{\mathbf{X}}}\right|$ 0−10 3 5 15 27 81 10−20 5 15 75 17 85 20−30 7 25 175 7 49 30−40 2 35 70 3 6 40−50 9 45 405 13 117 50−60 4 55 220 23 92 Σf = 30 Σfm=960 Σ$f\left|{d}_{\overline{X}}\right|$=430
#### Question 20:
Calculate mean deviation and its coefficient from the following figures:
Class-Interval Frequency Less than 10 5 Less than 20 12 Less than 30 20 Less than 40 35 Less than 50 54 Less than 60 60
Converting less than cumulative frequency distribution into simple frequency distribution:
Class Interval Cumulative Frequency (c.f.) Frequency (f) Mid-Values (m) $\left|{\mathbit{d}}_{\mathbit{M}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\mathbit{M}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\mathbf{36}\mathbf{.}\mathbf{67}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$ 0−10 5 5 5 31.67 158.35 10−20 12 7 15 21.67 151.69 20−30 20 8 25 11.67 93.36 30−40 35 15 35 1.67 25.65 40−50 54 19 45 8.33 158.27 50−60 60 6 55 18.33 109.98 Σf=N=60 Σf|dM|=696.7
Median class is given by the size of , i.e. , which is the 30th item.
This corresponds to the class interval of 30-40, so this is the median class.
#### Question 21:
Calculate the Standard deviation from the following values:8 , 9, 15, 23, 5, 11, 19, 8, 10, 12.
Values (X) $\mathbit{x}\mathbf{=}\left|\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right|$ x2 8 −4 16 9 −3 9 15 3 9 23 11 121 5 −7 49 11 −1 1 19 7 49 8 −4 16 10 −2 4 12 0 0 ΣX=120 Σx2=274
Hence, the standard deviation of the above series is 5.23
#### Question 22:
Find the standard deviation for the following data: 3, 5, 6, 7, 10, 12, 15, 18.
Values (X) $\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{\mathbf{X}}$ x2 3 −6.5 42.25 5 −4.5 20.25 6 −3.5 12.25 7 −2.5 6.25 10 .5 .25 12 2.5 6.25 15 5.5 30.25 18 8.5 72.25 ΣX=76 Σx2=190
Hence, standard deviation of the above series is 4.87
#### Question 23:
Find out the standard deviation of the height of 10 men given below: 160, 160, 161, 162, 163, 163, 163, 164, 164, 170.
Values (X) $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right)$ x2 160 −3 9 160 −3 9 161 −2 4 162 −1 1 163 0 0 163 0 0 163 0 0 164 1 1 164 1 1 170 7 49 ΣX=1630 Σx2=74
Hence, standard deviation of the above series is 2.72
#### Question 24:
Calculate standard deviation of the given below:
Size 3 4 5 6 7 8 9 Frequency 3 7 22 60 85 32 8
Size (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right)$ x2 fx2 3 3 9 3.59 12.89 38.67 4 7 28 2.59 6.71 46.97 5 22 110 1.59 2.53 55.66 6 60 360 0.59 0.35 21 7 85 595 0.41 0 .17 14.45 8 32 256 1.41 1.99 63.68 9 8 72 2.41 5.81 46.48 N=Σf=217 ΣfX=1430 Σfx2=286.91
Hence, standard deviation of the above series is 1.149
#### Question 25:
Find the value of standard deviation and coefficient of variation from the following:
Variables 10 20 30 40 50 60 70 Frequency 6 8 16 15 32 11 12
Size (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbf{X}\mathbf{-}\overline{\mathbf{X}}\right)$ x2 fx2 10 6 60 −34 1156 6936 20 8 160 −24 576 4608 30 16 480 −14 196 3136 40 15 600 −4 16 240 50 32 1600 +6 36 1152 60 11 660 16 256 2816 70 12 840 26 676 8112 Σf=100 ΣfX=4400 Σfx2=27000
#### Question 26:
Measurements are made to the nearest cm. of the heights of 10 children. Calculate mean and standard deviation.
Heights (cms) 60 61 62 63 64 65 66 67 68 No. of children 2 0 15 29 25 12 10 4 3
Height (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right)$ x2 fx2 60 2 120 −3.89 15.13 30.26 61 0 0 −2.89 8.35 0 62 15 930 −1.89 3.57 53.55 63 29 1827 −0.89 0.79 22.91 64 25 1600 0.11 0.01 0.25 65 12 780 1.11 1.23 14.76 66 10 660 2.11 4.45 44.5 67 4 268 3.11 9.67 38.68 68 3 204 4.11 16.89 50.67 Σf=100 ΣfX=6389 Σfx2=255.58
Hence mean of the above series is 63.89 cms and standard deviation is 1.6 cms.
#### Question 27:
Calculate standard deviation for the given data:
Age (in yrs) 20−25 25−30 30−35 35−40 40−45 45−50 50−55 No. of workers 17 11 8 5 4 3 2
Age (X) No. of Workers (f) Mid-Values (m) fm m2 $\mathbit{f}\mathbf{×}{\mathbit{m}}^{\mathbf{2}}\mathbf{=}\mathbit{f}{\mathbit{m}}^{\mathbf{2}}$ 20−25 17 22.5 382.5 506.25 8606.25 25−30 11 27.5 302.5 756.25 8318.75 30−35 8 32.5 260 1056.25 8450 35−40 5 37.5 187.5 1406.25 7031.25 40−45 4 42.5 170 1806.25 7225 45−50 3 47.5 142.5 2256.25 6768.75 50−55 2 52.5 105 2756.25 5512.5 Σf=50 Σfm=1550 Σfm2=51912.5
Hence, standard deviation of the above series is 8.79 years.
#### Question 28:
Calculate Standard deviation from the following series:
Class 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Frequency 2 4 6 8 6 4 2
Class Interval (X) Frequency (f) Mid-Values (m) fm m2 $\mathbit{f}\mathbf{×}{\mathbit{m}}^{\mathbf{2}}\mathbf{=}\mathbit{f}{\mathbit{m}}^{\mathbf{2}}$ 0−10 2 5 10 25 50 10−20 4 15 60 225 900 20−30 6 25 150 625 3750 30−40 8 35 280 1225 9800 40−50 6 45 270 2025 12150 50−60 4 55 220 3025 12100 60−70 2 65 130 4225 8450 Σf=32 Σfm=1120 Σfm2=47200
Hence, standard deviation of the above series is 15.81
#### Question 29:
From the following figures, find the standard deviation and the coefficient of variation:
Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 No. of persons 5 10 20 40 30 20 10 4
Marks (X) No. of Persons (f) Mid-Values (m) fm m2 fm2 0−10 5 5 25 25 125 10−20 10 15 150 225 2250 20−30 20 25 500 625 12500 30−40 40 35 1400 1225 49000 40−50 30 45 1350 2025 60750 50−60 20 55 1100 3025 60500 60−70 10 65 650 4225 42250 70−80 4 75 300 5625 22500 Σf=139 Σfm=5475 Σfm2=249875
#### Question 31:
The means of two samples of sizes 50 and 100 respectively are 54.1 and 50.3 and the standard deviations are 8 and 7. Find the mean and the standard deviation of the sample of size 150 obtained by combining the two samples.
Calculating combined standard deviation,
To calculate combined standard deviation, we need to find the combined mean of the observations.
Thus, Combined mean is
Hence, the combined standard deviation is 7.56
#### Question 32:
For a group of 100 males, mean and standard deviation of their daily wages are ₹ 36 and ₹ 9 respectively. For a group of 50 females, it is ₹ 45 and ₹ 6. Find the standard deviation for the whole group.
Calculating combined standard deviation,
To calculate combined standard deviation, we need to find the combined mean of the observations.
Thus, Combined mean is
Hence, the combined standard deviation is Rs 9.16
#### Question 33:
The profit of two business concerns for 5 years are as given below. Draw Lorenz Curves to show the distribution.
Year 2001 2002 2003 2004 2005 Firm A 15 30 45 60 50 Firm B 20 30 45 60 45
Years Firm A Cumulative Profits of Firm A Cumulative Percentage Profits of Firm A (x) Firm B Cumulative Profits of Firm B Cumulative Percentage Profits of Firm B (y) Coordinates of Lorenz Curve (x,y) 2001 15 15 $\frac{15}{200}×100=7.5$ 20 20 $\frac{20}{200}×100=10$ (7.5,10) 2002 30 45 $\frac{45}{200}×100=22.5$ 30 50 $\frac{50}{200}×100=25$ (22.5, 25) 2003 45 90 $\frac{90}{200}×100=45$ 45 95 $\frac{95}{200}×100=47.5$ (45, 47.5) 2004 60 150 $\frac{150}{200}×100=75$ 60 155 $\frac{155}{200}×100=77.5$ (75,77.5) 2005 50 200 $\frac{200}{200}×100=100$ 45 200 $\frac{200}{200}×100=100$ (100,100) $\Sigma {N}_{1}$=200 $\Sigma {N}_{2}$=200
#### Question 34:
The given table shows the daily income of workers of two factories. Draw the Lorenz Curves for both the factories.
Daily Income (₹) 0−100 100−200 200−300 300−400 400−500 Factory A 8 7 5 3 2 Factory B 15 6 2 1 1
Daily Income Mid Value Cumulative Mid Value % Cumulative Mid Value No. of worker (f) (c.f.) % Cumulative Frequency 0−100 50 50 4 8 8 32 100−200 150 200 16 7 15 60 200−300 250 450 36 5 20 80 300−400 350 800 64 3 23 92 400−500 450 1250 100 2 25 100
Daily Income Mid Value Cumulative Mid Value % Cumulative Mid Value No. of worker (f) (c.f.) % Cumulative Frequency 0−100 50 50 4 15 15 60 100−200 150 200 16 6 21 84 200−300 250 450 36 2 23 92 300−400 350 800 64 1 24 96 400−500 450 1250 100 1 25 100
#### Question 35:
Find the mean deviation from mean and its coefficient for the given data:
Marks (more than) 0 10 20 30 40 50 60 No. of students 200 180 150 100 40 15 5
Marks (X) Frequency (f) Mid-Values (m) fm $\left|{\mathbit{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\overline{\mathbit{X}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\mathbf{29}\mathbf{.}\mathbf{5}\right|$ $\mathbit{f}\mathbf{×}\left|\mathbit{d}\overline{{}_{\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|{\mathbit{d}}_{\overline{\mathbf{X}}}\right|$ 0−10 20 5 100 24.5 490 10−20 30 15 450 14.5 435 20−30 50 25 1250 4.5 225 30−40 60 35 2100 5.5 330 40−50 25 45 1125 15.5 387.5 50−60 10 55 550 25.5 255 60−70 5 65 325 35.5 177.5 Σf=200 Σfm=5900 Σ$f\left|{d}_{\overline{X}}\right|$=2300
#### Question 36:
Calculate the Mean and Standard Deviation from the following distribution.
Age (years) 15−19 20−24 25−29 30−34 35−39 40−44 No. of Persons 4 20 38 24 10 4
In order to calculate the mean and standard deviation, we first need to convert the inclusive series into exclusive series as given below:
Age (X) Frequency (f) Mid-Values (m) fm m2 fm2 14.5−19.5 4 17 68 289 1156 19.5−24.5 20 22 440 484 9680 24.5−29.5 38 27 1026 729 27702 29.5−34.5 24 32 768 1024 24576 34.5−39.5 10 37 370 1369 13690 39.5−44.5 4 42 168 1764 7056 Σf=100 Σfm=2840 Σfm2=83860
Hence, mean age of the persons is 28.4 years and standard deviation is 5.66 years.
#### Question 37:
The following table gives the weights of one hundred persons. Copute the coefficient of dispersion by the method of limits.
Class-interval 40−45 45−50 50−55 55−60 60−65 65−70 70−75 75−80 80−85 85−90 No. of persons 4 13 8 14 9 16 17 9 8 2
Given:
Upper limit of the Highest Class Interval (H) = 90
Lower limit of the Lowest Class Interval (L) = 40
Range = Highest Value − Lowest Value
i.e R= H− L
Substituting the given values in the formula.
R= 90 − 40= 50
Hence, range of the above series is 50 kg and coefficient of range is 0.384
#### Question 38:
Calculate standard deviation of the following data:
Age in years (below) 10 20 30 40 50 60 70 80 No. of persons 15 30 53 75 100 110 115 125
Age (X) Cumulative Frequency (c.f.) Frequency (f) Mid-Values (m) fm m2 fm2 0−10 15 15-0=15 5 75 25 375 10−20 30 30-15=15 15 225 225 3375 20−30 53 53-15=23 25 575 625 14375 30−40 75 75-23=22 35 770 1225 26950 40−50 100 100-75=25 45 1125 2025 50625 50−60 110 110-100=10 55 550 3025 30250 60−70 115 115-110=5 65 325 4225 21125 70−80 125 125-115=10 75 750 5625 56250 Σf=125 Σfm=4395 Σfm2=203325
Hence, standard deviation of the above series is 19.76 years.
#### Question 39:
Price of a particular item in 10 years in two cities are given below, which city has more stable prices?
City A 55 54 52 53 56 58 52 50 51 49 City B 108 107 105 105 106 107 104 103 104 101
City A City B XA ${\mathbit{x}}_{\mathbit{A}}\mathbf{=}{\mathbit{X}}_{\mathbit{A}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{A}}$ ${\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ XB ${\mathbit{x}}_{\mathbf{B}}\mathbf{=}{\mathbit{X}}_{\mathbf{B}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{B}}$ ${\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ 55 2 4 108 3 9 54 1 1 107 2 4 52 −1 1 105 0 0 53 0 0 105 0 0 56 3 9 106 1 1 58 5 25 107 2 4 52 −1 1 104 −1 1 50 −3 9 103 −2 4 51 −2 4 104 −1 1 49 −4 16 101 −4 16 ΣXA=530 $\mathrm{\Sigma }{x}_{\mathrm{A}}^{2}=70$ ΣXB=1050 $\mathrm{\Sigma }{x}_{\mathrm{B}}^{2}=40$
Since C.V. of city B is less. Therefore, prices are more stable in city B.
#### Question 40:
For a distribution, the coefficient of variation is 22.5% and mean is 7.5. Calculate standard deviation.
Given,
Coefficient of variation= 22.5%
Mean, $\overline{)\mathit{X}}$= 7.5
Hence, standard deviation is 1.69
#### Question 41:
Find out the arithmetic mean and standard deviation from the following data:
Variable 5−10 10−15 15−20 20−25 25−30 30−35 Frequency 2 9 29 54 11 5
Variable (X) Frequency (f) Mid-Values (m) fm m2 fm2 5−10 2 7.5 15 56.25 112.5 10−15 9 12.5 112.5 156.25 1406.25 15−20 29 17.5 507.5 306.25 8881.25 20−25 54 22.5 1215.0 506.25 27337.5 25−30 11 27.5 302.5 756.25 8318.75 30−35 5 32.5 162.5 1056.25 5281.25 Σf=110 Σfm=2315 Σfm2=51337.5
#### Question 42:
The mean and standard deviation of a series of 20 items are 20 and 5 respectively. While calculating these measures, an item of 13 was wrongly read as 30. Find out the correct mean and standard deviation.
Calculating correct Mean, by using the following observations:
Squaring both sides
$25=\frac{\mathrm{\Sigma }{x}^{2}}{20}-400$
425 × 20 = Σx2
$\Sigma {{x}^{2}}_{wrong}$= 8500
$\Sigma {{x}^{2}}_{correct}$ = $\Sigma {{x}^{2}}_{wrong}$ − (Incorrect item)2 + (Correct item)2
$\Sigma {{x}^{2}}_{correct}$= 8500 − (30)2 + (13)2
$\Sigma {{x}^{2}}_{correct}$= 8500 − 900 + 169 = 7769
Hence, correct mean and standard deviation are 19.15 and 4.66 respectively.
#### Question 43:
Following are the marks obtained by two students: Mollie and Isha, in 10 sets of examinations:
Marks of obtained by Mollie 44 80 76 48 52 72 68 56 60 54 Marks of obtained by Isha 48 75 54 60 63 69 72 51 57 66
Out of Mollie and Isha, who is more consistent?
Mollie Isha XM ${\mathbit{x}}_{\mathbf{M}}\mathbf{=}{\mathbit{X}}_{\mathbf{M}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{M}}$ ${\mathbit{x}}_{\mathbf{M}}^{\mathbf{2}}$ XI ${\mathbit{x}}_{\mathbf{I}}\mathbf{=}{\mathbit{X}}_{\mathbf{I}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{I}}$ ${\mathbit{x}}_{\mathbit{I}}^{\mathbf{2}}$ 44 −17 289 48 −13.5 182.25 80 19 361 75 13.5 182.25 76 15 225 54 −7.5 56.25 48 −13 169 60 −1.5 2.25 52 −9 81 63 1.5 2.25 72 11 121 69 7.5 56.25 68 7 49 72 10.5 110.25 56 −5 25 51 −10.5 110.25 60 −1 1 57 −4.5 20.25 54 −7 49 66 4.5 20.25 ΣXM=610 $\mathrm{\Sigma }{x}_{\mathrm{M}}^{2}=1370$ ΣXI=615 $\mathrm{\Sigma }{x}_{\mathrm{I}}^{2}=742.5$
Isha is more consistent in securing makes as her C.V. (14.01)% is less than that of Mollie's C.V. (19.18)%
#### Question 44:
Calculate coefficient of variation from the following data:
Marks (more than) 0 10 20 30 40 50 60 70 No. of students 100 90 75 50 20 10 5 0
Converting more than cumulative frequency distribution into simple frequency distribution:
Marks (X) No. of Students (f) Mid-Values (m) fm m2 fm2 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 −70 70 −80 100-90=10 90-75=15 75-50=25 50-20=30 20-10=10 10-5=5 5-0=5 0-0=0 5 15 25 35 45 55 65 75 50 225 625 1050 450 275 325 0 25 225 625 1225 2025 3025 4225 5625 250 3375 15625 36750 20250 15125 21125 0 Σf =100 Σfm = 3000 Σfm2 =112500
#### Question 45:
In two Towns A and B, daily pocket money and the standard deviation are given below:
Town Average Daily Pocket Standard Deviation No. of Teenagers A 34.5 5.0 476 B 28.5 4.5 524
(i) Which town, A or B, pays out the larger amount of daily pocket money?
(ii) What is the average daily pocket money of all teenagers taken together?
(iii) Calculate coefficient of variation of each town. Which town is more variable in terms of pocket money?
(i) Town A
Daily pocket money = Average Daily Pocket × No. of teenagers
= 34.5 × 476
= Rs 16422
Town B
Daily Pocket Money = 28.5 × 524
= Rs 14934
Town 'A' pays larger amount of daily pocket money.
(ii) Solution:
In order to find out average daily pocket money of all teenagers, we need find out the combined mean of their pocket money.
Thus,
(iii)
Town B is more variable in terms of pocket money as its C.V. is higher than that of C.V. of Town A.
#### Question 46:
The prices of share of Company X and Company Y are given below. State, which company is more stable?
Company X 25 50 45 30 70 42 36 48 34 60 Company Y 10 70 50 20 95 55 42 60 48 80
Company X Company Y X $\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{)\mathbit{X}}$ x2 Y $\mathbit{y}\mathbf{=}\mathbit{Y}\mathbf{-}\overline{)\mathbit{Y}}$ y2 25 50 45 30 70 42 36 48 34 60 −19 6 1 −14 26 −2 −8 4 −10 16 361 36 1 196 676 4 64 16 100 256 10 70 50 20 95 55 42 60 48 80 −43 17 −3 −33 42 2 −11 7 −5 27 1849 289 9 1089 1764 4 121 49 25 729 ΣX = 440 Σx2 = 1710 ΣY = 530 Σy2 = 5928
As C.V of prices of shares of Co. X is less than that of the prices of shares of Co. Y therefore, price of share of Co. X is more stable.
#### Question 47:
A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50 instead of 40. Find the correct mean and standard deviation.
Calculating correct Mean, by using the following observations:
Squaring both sides
$26.01=\frac{\mathrm{\Sigma }{x}^{2}}{100}-1600\phantom{\rule{0ex}{0ex}}$
Hence, correct mean and standard deviation are 39.9 and 5 respectively.
#### Question 48:
From the following data, calculate standard deviation of the two groups A and B. Which group is more consistent?
Class Interval Group A Group B 5−10 2 9 10−15 9 11 15−20 29 18 20−25 54 32 25−30 11 27 30−35 5 13
For group A
Class Interval Frequency (f) Mid-Values (m) fm m2 fm2 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 2 9 29 54 11 5 7.5 12.5 17.5 22.5 27.5 32.5 15 112.5 507.5 1215 302.5 162.5 56.25 156.25 306.25 506.25 756.25 1056.25 112.5 1406.25 8881.25 27337.5 8318.75 5281.25 Σf = 110 Σfm = 2315 Σfm2 = 51337.5
For group B
Class Interval Frequency (f) Mid-Values (m) fm m2 fm2 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 9 11 18 32 27 13 7.5 12.5 17.5 22.5 27.5 32.5 67.5 137.5 315 720 742.5 422.5 56.25 156.25 306.25 506.25 756.25 1056.25 506.25 1718.75 5512.5 16200 20418.75 13731.25 Σf = 110 Σfm = 2405 Σfm2 = 58087.5
Group A is more consistent as C.V. of group A is less than C.V. of group B.
#### Question 49:
From the following data of marks, calculate standard deviation. What will be the value of standard deviation, if marks obtained by each student is increased by one?
Marks Obtained 1 2 3 4 5 6 7 8 9 No. of students 32 41 57 98 123 83 46 17 3
Marks (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{)\mathbit{X}}\right)$ x2 fx2 1 2 3 4 5 6 7 8 9 32 41 57 98 123 83 46 17 3 32 82 171 392 615 498 322 136 27 −3.55 −2.55 −1.55 −0.55 0.45 1.45 2.45 3.45 4.45 12.60 6.50 2.40 0.30 0.20 2.10 6.00 11.90 19.80 403.2 266.5 136.8 29.4 24.6 174.3 276 202.3 59.4 Σf = 500 Σfx = 2275 Σfx2 = 1572.5
As, standard deviation is independent of the change in origin, i.e. it will not get affected if the value of the series is increased or decreased by a constant quantity. Therefore it will remain the same.
#### Question 50:
From the following data of two workers, identify who is a more consistent worker?
Worker A B Average Time in completing a job 40 42 Standard Deviation 8 6
Worker B is more consistent as his C.V. (14.29)% is less than that of C.V of worker A(20%).
#### Question 51:
Find the standard deviation and coefficient of standard deviation:
X (less than) 10 20 30 40 50 60 70 80 Frequency 12 30 65 107 157 202 222 230
Converting less than frequency distribution into simple frequency distribution:
X Frequency (f) Mid-Values (m) fm m2 fm2 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 12-0=12 30-12=18 65-30=35 107-65=42 157-107=50 202-157=45 222-202=20 230-222=8 5 15 25 35 45 55 65 75 60 270 875 1470 2250 2475 1300 600 25 225 625 1225 2025 3025 4225 5625 300 4050 21875 51450 101250 136125 84500 45000 Σf = 230 Σfm = 9300 Σfm2 = 444550
#### Question 52:
Find mean, standard deviation and coefficient of variation.
Class-interval 0−4 4−8 8−12 12−16 16−20 20−24 Frequency 10 15 20 25 20 10
Class Interval Frequency (f) Mid-Values (m) fm m2 fm2 0 − 4 4 − 8 8 − 12 12 − 16 16 − 18 20 − 24 10 15 20 25 20 10 2 6 10 14 18 22 20 90 200 350 360 220 4 36 100 196 324 484 40 540 2000 4900 6480 4840 Σf = 100 Σfm = 1240 Σfm2 = 18800
#### Question 53:
The following table shows the marks obtained by 60 students. Calculate mean and standard deviation.
Marks (more than) 70 60 50 40 30 20 No. of students 7 18 40 40 55 60
Converting more than cumulative frequency into ordinary continuos series:
Marks X Frequency (f) Mid-Values (m) fm m2 fm2 20 −30 30 −40 40 −50 50 −60 60 −70 70 −80 60-55=5 55-40=15 40-40=0 40-18=22 18-7=11 7-0=7 25 35 45 55 65 75 125 525 0 1210 715 525 625 1225 2025 3025 4225 5625 3125 18375 0 66550 46475 39375 Σf = 60 Σfm = 3100 Σfm2 = 173900
Hence, mean and standard deviation of the above series are 51.67 marks and 15.11 marks respectively.
#### Question 54:
Calculate standard deviation and coefficient of dispersion from the data below:
Mid-Points 5 15 25 35 45 55 65 75 Frequency 5 8 7 12 28 20 10 10
Mid-Values (m) Frequency (f) fm m2 fm2 5 15 25 35 45 55 65 75 5 8 7 12 28 20 10 10 25 120 175 420 1260 1100 650 750 25 225 625 1225 2025 3025 4225 5625 125 1800 4375 14700 56700 60500 42250 56250 Σf = 100 Σfm= 4500 Σfm2 = 236700
#### Question 55:
The following data shows the expected life of two models of T.V.: A and B:
Life (no. of years) 0−2 2−4 4−6 6−8 8−10 10−12 Model A 5 16 13 7 5 4 Model B 2 7 12 19 9 1
Which model has greater uniformity.
Life (No. of years) Frequency (f) Mid Values (m) fm m2 fm2 0 − 2 2 − 4 4 − 6 6 − 8 8 −10 10 − 12 5 16 13 7 5 4 1 3 5 7 9 11 5 48 65 49 45 44 1 9 25 49 81 121 5 144 325 343 405 484 Σf = 50 Σfm = 256 Σfm2 = 1706
Life (No. of years) Frequency (f) Mid Values (m) fm m2 fm2 0 − 2 2 − 4 4 − 6 6 − 8 8 −10 10 − 12 2 7 12 19 9 1 1 3 5 7 9 11 2 21 60 133 81 11 1 9 25 49 81 121 2 63 300 931 729 121 Σf = 50 Σfm = 308 Σfm2 = 2146
As, C.V of Model B is less than that of C.V of model A therefore model B has greater uniformity.
View NCERT Solutions for all chapters of Class 13
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2020-07-07 23:09:20
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http://community.worldheritage.org/articles/eng/India-based_Neutrino_Observatory
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India-based Neutrino Observatory (INO) is a proposed particle physics research project to primarily study atmospheric neutrinos in a 1,300 meters (4,300 ft) deep cave under Ino Peak near Theni, Tamil Nadu, India. This project is notable in that it is anticipated to provide a precise measurement of neutrino mixing parameters. The project is a multi-institute collaboration and one of the biggest experimental particle physics projects undertaken in India.[1][2][3] [4]
The project, expected to be completed in 2015 at an estimated cost of $250 million, has been cleared by the Ministry of Environment (India) for construction in the Bodi West Hills Reserved Forest in the Theni district of Tamil Nadu. When completed, the INO will house the world's most massive magnet, four times larger than the 12,500-tonne magnet in the Compact Muon Solenoid detector at CERN in Geneva, Switzerland.[5][6] ## Contents ## History The possibility of a neutrino observatory located in India was discussed as early as 1989 during several meetings held that year. Since then this question comes up, off and on, in many discussions. The issue was raised again in the first meeting of the Neutrino physics and Cosmology working group during the Workshop on High Energy Physics Phenomenology (WHEPP-6) held at Chennai in January 2000 and it was decided then to collate concrete ideas for a neutrino detector. Further discussions took place in August 2000 during a meeting on Neutrino Physics at the Saha Institute of Nuclear Physics, Kolkata, when a small group of neutrino physics enthusiasts started discussing the possibilities. The Neutrino 2001 meeting was held in the Institute of Mathematical Sciences, Chennai during February 2001 with the explicit objective of bringing the experimentalists and theorists in this field together. The INO collaboration was formed during this meeting. The first formal meeting of the collaboration was held in the Tata Institute of Fundamental Research, Mumbai, during September 6 and 7th, 2001 at which various subgroups were formed for studying the detector options and electronics, physics goals and simulations, and site survey. In 2002, a document was presented to the Department of Atomic Energy, (DAE) which laid out an ambitious goal of establishing an India-based Neutrino Observatory, outlining the physics goals, possible choices for the detector and their physics. Since then many new and fast paced developments have taken place in neutrino physics. The award of the Nobel Prize in Physics (2002) to the pioneers in neutrino physics is a measure of the importance of this field. As a result of the support received from various research institutes, universities, the scientific community and the funding agency, the Department of Atomic Energy, a Neutrino Collaboration Group (NCG) was established to study the possibility of building an India-based Neutrino Observatory (INO). The collaboration was assigned the task of doing the feasibility studies for which funds were made available by the DAE. A memorandum of understanding (MoU) was signed by the directors of the participating institutes on August 30, 2002 to enable a smooth functioning of the NCG during the feasibility period. The NCG has the goal of creating an underground neutrino laboratory with the long term goal of conducting decisive experiments in neutrino physics as also other experiments which require such a unique underground facility.[1] On November 20, 2009, Ministry of Environment (India) Minister Jairam Ramesh in a letter to Anil Kakodkar, Secretary, Department of Atomic Energy and Chairman, Atomic Energy Commission of India, denied permission for the Department of Atomic Energy to set up the India-based Neutrino Observatory (INO) project at Singara in Nilgiris, as it falls in the buffer zone of the Mudumalai Tiger Reserve (MTR). Jairam Ramesh said that based on the report of Rajesh Gopal, Additional Principal Chief Conservator of Forests (PCCF) and Member-Secretary of the National Tiger Conservation Authority (MS-NTCA), the Ministry cannot approve the Singara site. The report says: "The proposed project site falls in the buffer zone of Mudumalai Tiger Reserve and is in close proximity to the core/critical tiger habitats of Bandipur and Mudumalai Tiger reserves. It is also an elephant corridor, facilitating elephant movement from the Western Ghats to the Eastern Ghats and vice-versa. The area is already disturbed on account of severe biotic pressure due to human settlements and resorts and that the construction phase of the project would involve transport of building materials through the highways passing through the core area of the Bandipur and Mudmulai Tiger Reserves.[7] Instead, he suggested an alternate site near Suruli Falls, Theni District in Tamil Nadu. The Minister said this site did not pose the same problems that Singara posed and environmental and forest clearances should not be a serious issue. He also assured the DAE that the Ministry would facilitate necessary approvals for the alternative location. Dr. Naba K. Mondal of the Tata Institute of Fundamental Research, who is the spokesperson for the INO project said: "But Suruliyar too is in a reserved forest area that is dense and would require cutting down of trees, something that was not required at Singara. Can the government assure us that forest clearance for this site will be given," he asks. "Alternatively, we can move to the nearby Thevaram, which is about 20-30 km away from the Suruliyar falls. This forest area has only shrubs but there is no source of water here and water will have to be piped over a distance of 30 km,"[7][8] On 18 October 2010, the Ministry of Environment & Forests approved both environment and forest clearance for setting up the observatory in the Bodi West Hills Reserved Forest in the Theni district of Tamil Nadu. As of February 2012, the land was allocated to the INO collaboration by the government of Tamil Nadu and the excavation work was about to start. Naba K.Mondal, chief spokesperson of INO project and a senior scientist at the Tata Institute of Fundamental Research, Mumbai, told The Hindu that the pre-project work will start in April, 2012 and Rs.66 crores has been sanctioned for the work. The first task will be to have a road connectivity from Rasingapuram to Pottipuram village. The project is expected to be completed in 2015 at an estimated cost of$250 million.[5][6]
On September 18, 2012, Kerala’s octogenarian Opposition leader and CPI(M) central committee member VS Achuthanandan expressed anxiety over establishing a neutrino observatory on the Theni-Idukki border between Tamil Nadu and Kerala, citing environmental and radiological issues.[9] Soon the INO collaboration clarified on all the issues raised by him and the responses are on the INO website.
## Participating Institutes
Memorandum of Understanding (MoU) spelling out the operational aspects of the project and the mode of utilisation of available funds was signed by seven primary project partners: Tata Institute of Fundamental Research (TIFR), Mumbai, Bhabha Atomic Research Centre (BARC), Mumbai, Institute of Mathematical Sciences (IMSc), Chennai, Saha Institute of Nuclear Physics (SINP), Kolkata, Variable Energy Cyclotron Centre (VECC), Kolkata, Harish Chandra Research Institute (HRI), Allahabad and Institute of Physics (IOP), Bhubaneswar.[1]
Thirteen other project participants include: Aligarh University, Aligarh, Banaras Hindu University, Varanasi, Calcutta University (CU), Kolkata, Delhi University (DU), Delhi, University of Hawaii (UHW), Hawaii, Himachal Pradesh University (HPU), Shimla, Indian Institute of Technology, Bombay (IITB), Mumbai, Indira Gandhi Centre for Atomic Research (IGCAR), Kalpakkam, North Bengal University (NBU), Siliguri, Panjab University (PU), Chandigarh, Physical Research Laboratory (PRL), Ahmedabad, Sálim Ali Centre for Ornithology and Natural History (SACON), Tamil Nadu and Sikkim Manipal Institute of Technology, Sikkim.
## Design
Chart showing 3 neutrinos and interacting particles, according to the Standard Model of Elementary Particles
The primary research instrument will consist of a 50,000 ton magnetized iron particle physics calorimeter with glass Resistive Plate Chamber (RPC) technology as the sensor elements.[1]
The INO design is mostly based on the monolith experiment that could not go beyond the proposal Stage. The detector was expected to start collecting data in the year 2012. The location of INO has attracted a lot of attention from the neutrino physics community as the distance between INO and CERN is very close to "Magic Baseline" - a distance at which the effect of the CP phase on the measurement of \theta_{13} is minimal.[10] The project has been hit by lack of skilled man power and opposition by environmentalists. In 2008, INO started a graduate training program leading to Ph.D. Degree in High Energy Physics and Astronomy to deal with the shortage of particle physicists. [11]
The Primary goals of the INO are the following [12]
1. Unambiguous and more precise determination of Neutrino oscillation parameters using atmospheric neutrinos.
2. Study of matter effects through electric charge identification, that may lead to the determination of the unknown sign of one of the mass differences.
3. Study of charge-conjugation and charge parity (CP) violation in the leptonic sector as well as possible charge-conjugation, parity, time-reversal (CPT) violation studies.
4. Study of Kolar events, possible identification of very-high energy neutrinos and multi-muon events.
The INO detector consists of 6 centimeters (2.4 in) thick Iron plates as passive material, with RPCs in between as active material.
A prototype of the INO detector with 14 layers, measuring 1m x 1m x 1m is already operational in the VECC, Kolkata. The 35 ton prototype is set up over ground to track cosmic muons. [13]
## Location
The location of the site was supposed to be Singara 5.5 kilometers (3.4 mi) southwest of Masinagudi in the Nilgiri Hills of South India. The site has been changed due to protests from environmental groups. The INO will now be built at Bodi West Hills in Theni district, southern India. [14]
## Notes and references
1. ^ a b c d Mondal, Naba K. (January 2004). "STATUS OF INDIA-BASED NEUTRINO OBSERVATORY (INO)". Proceedings
2. ^ Mondal, Prof. N K (July 2008). "FAQ on INO". Institute of Mathematical Sciences, Chennai. Retrieved 2009-04-03.
3. ^ The Hindu (2005). "India gets ready with ambitious science project". Retrieved 2008-07-12.
4. ^ Youtube, Rajya Sabha TV (2014). "A Question of Science". Retrieved 2014-08-14.
5. ^ a b http://www.ino.tifr.res.in/ino/neutrino-press-release.pdf
6. ^ a b http://www.newscientist.com/article/dn19620-indian-neutrino-lab-to-boast-worlds-biggest-magnet.html
7. ^ a b Ramachandran, R. (2009-11-21). "Ministry's 'no' to Neutrino Observatory project in Nilgiris".
8. ^ Madhusudan, M. (2009-11-22). "Centre no to neutrino observatory in Nilgiris". Sunday Pioneer (New Delhi: Pioneer Syndication Services). Retrieved 9 December 2009.
9. ^ "V S Achuthanandan flays proposed Neutrino observatory". The New Indian Express. 18 September 2012. Retrieved 2 May 2013.
10. ^ Agarwalla, S; Sanjib Kumar Agarwallaa, Sandhya Choubeya, and Amitava Raychaudhuria (28 May 2007). "Neutrino mass hierarchy and θ13 with a magic baseline beta-beam experiment". Nuclear Physics B (ScienceDirect, Elsevier B.V.) 771 (1-2): 1–27.
11. ^ Unknown (2008). "Education updates". Retrieved 2008-08-17.
12. ^ INO collaboration (2007). "INO feasibility Study". Retrieved 2008-07-10.
13. ^ B. Satyanarayana (2009). "INO prototype detector". Retrieved 2009-07-25.
14. ^ Michael Banks (19 October 2010). "Green light for Indian neutrino observatory". Retrieved 2010-10-19.
This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.
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2017-07-23 22:33:33
|
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|
http://www.surfermag.com/features/week-in-review-39/
|
Article
# Week in Review
## Random happenings in surf for the Week of January 2
| posted on January 06, 2012
Welcome to 2012, a year that will surely be filled with just as many obscure, loosely surf-related happenings as 2011. Enjoy.
Carissa Moore Named Adventurer of the Year
Women’s World Champ Carissa Moore was honored in National Geographic’s Top 10 Adventurers of the Year for 2011. She was the only surfer to make the list. Vote for Carissa here.
“Merry Christmas, Ladies,” said this guy.
On Christmas day at Snapper, this Aussie exhibitionist took to the holiday swell sans clothes. Ouch?
Well, that's one way to get noticed. Photo: Grambeau
Only in Hawaii
The National Weather Service in Hawaii proclaimed, very scientifically, that the surf on Oahu would be “ginormous” this week.
Shredding on a 3’8″
Sebastian Williams gets loose on a mini board. Yes, it’s 3 foot 8 inches. Oh, to be 10 years old.
A Man With Far Too Much Free Time
Another bizarre, drunken web installment from Matt Wilkinson.
Whale vs. Shark vs. Dog
A killer whale and a killer shark went at it, then a dog tried to get in on it, and someone filmed it.
The Ocean is a Wild Beast
This has very little to do with surfing per se, but the footage is pretty mesmerizing.
All Leathered Up and Ready to go Surfing
Supermodel Gisele Bundchen and friends were featured in this recent Givenchy ad. According to the Creative Director for the luxury brand, these ads are “the expression of a love story between a surfer and a mermaid.” Hmm…
That's what I wear to the beach too.
Happy Faces For Everyone!
Not to be confused with Jordy Smith’s new winky happy face logo, Luke Stedman chose an anchor happy face as his logo for himself. And then he tagged a tanker with it.
A Very Confident 11-Year-Old
East Coast grom Pat Pat Taylor hosts a YouTube roast of Slater, Bobby, and others with/without hair.
Whipped?
Jordy Smith, a very proud boyfriend.
A Photo from the Twittersphere
Sean Moody posted this gem of an old school pic featuring himself, Fred Patacchia, and Joel Centeio.
Sexy kinda. Photo: @MoodyHawaii
Check out the 2011 Year in Review here.
Archives:
December 19 \\December 12 \\December 5 \\November 28 \\November 21 \\October 31 \\October 24 \\October 17 \\October 10 \\October 3 \\ September 26 \\ September 19 \\ September 12 \\ September 5 \\ August 29 \\ August 22 \\ August 15 \\ August 8 \\ August 1 \\ July 25 \\ July 18 \\ June 27 \\ June 20 \\ June 13 \\ June 6 \\ May 30 \\ May 16 \\ May 9 \\ May 2 \\ April 25 \\ April 18 \\ April 11 \\ April 4 \\ March 28 \\ March 21 \\ March 14 \\ March 7
• J
There is only one Man who Will make you completely fulfilled, and that Man is Jesus. I promise that if you seek him and let him into your heart you will experience rest and Everlasting Joy. I pray that you will take this message to heart and experience TRUE Happiness in Jesus. God Bless. ps. Pray to Jesus about all of your problems so that he will help you like he has helped me always.
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2016-07-25 06:12:38
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https://www.ac.tuwien.ac.at/publications/Szeider08?file=../../publications/publications-web.bib
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Monadic Second Order Logic on Graphs with Local Cardinality Constraints (bibtex)
by
Reference:
Monadic Second Order Logic on Graphs with Local Cardinality ConstraintsStefan SzeiderMathematical Foundations of Computer Science 2008, 33rd International Symposium, MFCS 2008, Torun, Poland, August 25-29, 2008, Proceedings, volume 5162 of Lecture Notes in Computer Science, pages 601-612, 2008, Springer Verlag.
Bibtex Entry:
@string{springer="Springer Verlag"}
@string{lncs="Lecture Notes in Computer Science"}
@InProceedings{Szeider08,
author = {Stefan Szeider},
title = {Monadic Second Order Logic on Graphs with Local Cardinality
Constraints},
booktitle = {Mathematical Foundations of Computer Science 2008, 33rd
International Symposium, MFCS 2008, Torun, Poland, August
25-29, 2008, Proceedings},
publisher = Springer,
series = LNCS,
volume = {5162},
pages = {601-612},
year = {2008},
}
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2018-11-17 22:08:26
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https://socratic.org/questions/5979e17bb72cff6fd7c8efd1
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# Question #8efd1
Jul 27, 2017
The molarity of chloride ions suspended in the aqueous solution is $2.00 M$.
#### Explanation:
Here, we're looking to find the molarity of $C {l}^{-}$ ions in the solution.
$27.75 g \cdot \frac{C a C {l}_{2}}{110.98 g} \cdot \frac{1}{.250 L} = 1.00 M$
The foregoing conversion helps us realize the molarity of the entire solution is $1.00 M$, but remember, there are 2 moles of $C {l}^{-}$ per one mole of the soluble compound, so:
$\frac{1.00 m o l}{L} \cdot \frac{2 C {l}^{-}}{C a C {l}_{2}} = 2.00 M$
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2019-09-21 17:18:44
|
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https://mathoverflow.net/questions/288514/pulling-back-cohen-macaulay-sheaves
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# Pulling Back Cohen-Macaulay Sheaves
Suppose $f:X\to Y$ is a finite morphism of varieties and $\mathcal{F}$ is a Cohen-Macaulay sheaf on $Y$. Under what conditions on $f$ is $f^*\mathcal{F}$ Cohen-Macaulay?
• The only reasonable condition is if $f$ is flat. – Mohan Dec 15 '17 at 1:34
As far as being true for the most general situation, you need: $X,Y$ to be Cohen-Macaulay, $\dim \mathcal F_x =\dim Y_x$ ($\mathcal F_x$ is maximal Cohen-Macaulay) for all $x$ in the support of $\mathcal F$, and $f$ to have finite flat dimension. (more is true with a little extra technical assumption, you need the Cohen-Macaulay defect" to be constant along $f$, see the last paragraph).
Then what you need follows from the two local statements about f.g modules over a local ring $R$.
1) If M is maximal CM, and N has finite projective dimension, then $Tor_i(M,N)=0$ for all $i>0$ (M, N are Tor-independent). See: http://www.math.lsa.umich.edu/~hochster/711F06/L11.20.pdf
2) If $M,N$ are Tor-independent and $pd_RN<\infty$, we also have the depth formula: $depth(M\otimes N) + depth(R) = depth(M)+depth(N)$ See: https://www.math.unl.edu/~siyengar2/Papers/Abform.pdf
In particular, if N is CM then $M\otimes N$ is also CM of same depth. (here we use that $R$ is CM, so $depth(R) =depth(M)$).
If you drop any condition, it is not hard to find examples to show that the statement is no longer true. On the other hand, for special $X,Y,\mathcal F$, sometimes one can say a bit more. For example, if $R$ is a complete intersection, Tor-independence forces the depth formula to hold, without knowing that $N$ has finite flat dimension.
Added in response to OP's request: By looking at the depth formula in 2), the following more technical, but general statement is true: suppose $f: R\to S$ is a finite, local map of finite flat dimension, and $dim(R)-depth(R)=dim(S)-depth(S)$. Then for a maximal CM module $M$ over $R$, $M\otimes_R S$ is (maximal) CM over $S$. This cover both cases when $R,S$ are CM ( both sides of the equality is $0$), or if $f$ is flat (both dim and depth are preserved).
• Thanks for the detailed answer! What can we say if $X$ is not CM? For example, if $f$ is flat, it seems $f^*F$ is automatically CM, without conditions on $X$ or $Y$. Can we do better? – Lucas Mason-Brown Dec 15 '17 at 13:03
• Yes, you can, I will edit. – Hailong Dao Dec 15 '17 at 16:31
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2021-05-11 02:15:51
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https://web2.0calc.com/questions/please-help-algebra-question
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+0
0
130
3
+46
ab-c=ae+f
Solve for a
Please solve and explain this. I can't understand the concept!
Sep 28, 2018
#2
+97500
+3
Hi Curry
sorry, your question got flagged automatically, I am not sure why.
Probably the program thought those letters in your equation looked too random and it was flagged as a potential rubbish question. These questions viewed later by a moderator and shown.
ab-c=ae+f solve for a
The idea is to get a by itself in fron to the equal sign.
First get every term WITH an a on the left and everything eles on the right.
$$ab-c=ae+f\\ \text{subtract ae from both sides}\\ ab-c-ae=ae+f-ae\\ ab-c-ae=f\\ \text{Now add c to both sides}\\ ab-c-ae+c=f+c\\ ab-ae=f+c\\ \text{Now factor out the a on the left side}\\ a(b-e)=f+c\\ \text{Now divide both sides by (b-e)}\\ \frac{a(b-e}{b-e}=\frac{f+c}{b-e}\\ a=\frac{f+c}{b-e}\\ \\~\\ \text{You cannot divide by 0 so you should put }b-e\ne0\\\ \text{which is the same as } b\ne e$$
.
Sep 28, 2018
#1
+46
+1
why is this flagged?
Sep 28, 2018
#2
+97500
+3
Hi Curry
sorry, your question got flagged automatically, I am not sure why.
Probably the program thought those letters in your equation looked too random and it was flagged as a potential rubbish question. These questions viewed later by a moderator and shown.
ab-c=ae+f solve for a
The idea is to get a by itself in fron to the equal sign.
First get every term WITH an a on the left and everything eles on the right.
$$ab-c=ae+f\\ \text{subtract ae from both sides}\\ ab-c-ae=ae+f-ae\\ ab-c-ae=f\\ \text{Now add c to both sides}\\ ab-c-ae+c=f+c\\ ab-ae=f+c\\ \text{Now factor out the a on the left side}\\ a(b-e)=f+c\\ \text{Now divide both sides by (b-e)}\\ \frac{a(b-e}{b-e}=\frac{f+c}{b-e}\\ a=\frac{f+c}{b-e}\\ \\~\\ \text{You cannot divide by 0 so you should put }b-e\ne0\\\ \text{which is the same as } b\ne e$$
Melody Sep 28, 2018
#3
+46
0
Sorry I'm saying this a little late, but thank you. This really helped, and now I understand the concept
Oct 2, 2018
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2019-02-17 02:54:59
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https://kx.lumerical.com/t/error-there-was-a-problem-loading-the-activation-library-code-7/582
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# Error: There was a problem loading the activation library. Code 7
The following error messages may occur when upgrading from an older version of Lumerical software. Typically it is possible to simply install the new version without manually un-installing the older version first. However, in some cases the automatic un-install fails, resulting in these error:
This can be resolved by:
1. Close the error message by selecting ‘NO’.
2. Open the Control Panel > Programs and features, and un-install Lumerical (ie. Ansys Lumerical)
3. Go to C:\Program Files\Lumerical\ and ensure the product folder (i.e. v202) has been removed. If it is still present, delete the entire folder.
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2021-01-26 04:33:02
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https://www.gradesaver.com/textbooks/math/algebra/college-algebra-11th-edition/chapter-4-section-4-3-logarithmic-functions-summary-exercises-on-inverse-exponential-and-logarithmic-functions-page-427/41
|
## College Algebra (11th Edition)
$x=2$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $2^x=\log_2 16 ,$ use the properties of logarithms to simplify the right side. Then express both sides in the same base and equate the exponents. $\bf{\text{Solution Details:}}$ Using exponents, the equation above is equivalent to \begin{array}{l}\require{cancel} 2^x=\log_2 2^4 .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} 2^x=4\log_2 2 .\end{array} Since $\log_b b =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2^x=4(1) \\\\ 2^x=4 \\\\ 2^x=2^2 .\end{array} Since the bases are the same, then the exponents can be equated. That is, \begin{array}{l}\require{cancel} x=2 .\end{array}
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2018-05-22 01:04:10
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http://www.phpclasses.org/discuss/package/4990/thread/2/
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No examples, but easy enough to figure out from the code.
Subject: No examples, but easy enough to... Package rating comment 1 F Philip DeGeorge 2009-01-08 18:35:13
F Philip DeGeorge rated this package as follows:
Utility: Sufficient Good Insufficient
1. No examples, but easy enough to... Reply Report abuse
F Philip DeGeorge - 2009-01-08 18:35:13
No examples, but easy enough to figure out from the code. Cleanly written.
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2015-03-29 12:28:15
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https://www.splessons.com/lesson/last-2-digits-square-of-a-number/
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Last 2 Digits Of A Square Of A Number
5 Steps - 3 Clicks
Last 2 Digits Of A Square Of A Number
Introduction
Quantitative Aptitude is an important paper in banking exam. One cannot ignore it. So it is the most important to a candidate to improve the math’s skills for competitive exams. Most of the candidates feel that it’s a more time taking paper in exam but by following some guidelines, candidates can easily crack the exam. Competitive exams are setting with time binding. Everyone can do all maths without time binding but the main challenges are came into with in time. So candidates should mainly focus on speed and accuracy. That is possible in their hard working and Dedication. The article provides Quantitative Aptitude Math Shortcut Tricks.
Short Cut
1. How to find the last 2 digits of a square of a number ($$a^{2}$$)?
Memorize the following table of squares from 1 to 25:
a $$a^{2}$$ a $$a^{2}$$ a $$a^{2}$$ a $$a^{2}$$ a $$a^{2}$$
1 01 6 36 11 121 16 256 21 441
2 04 7 49 12 144 17 289 22 484
3 09 8 64 13 169 18 324 23 529
4 16 9 81 14 196 19 361 24 576
5 25 10 100 15 225 20 400 25 625
Examples
1) Find the last two digits of $$41^{2}$$.
Solution: Here a = 41.
Step 1:
Calculate 50 – a.
So, 50 – 41 = 9.
Step 2:
The last two digits of $$a^{2}$$ is equal to last two digits of $$(50 – a)^{2}$$.
So, the last two digits of $$41^{2}$$ is equal to last two digits of $$9^{2}$$. The answer is 81.
2) Find the last two digits of $$87^{2}$$.
Solution: Here a = 87.
Step 1:
Calculate 100 – a.
So, 100 – 87 = 13.
Step 2:
The last two digits of $$a^{2}$$ is equal to last two digits of $$(100 – a)^{2}$$.
So, the last two digits of $$87^{2}$$ is equal to last two digits of $$13^{2}$$. The answer is 69.
3) Find the last two digits of $$65^{2}$$.
Solution: Here a = 65.
Step 1:
Calculate a – 50.
So, 65 – 50 = 15.
Step 2:
The last two digits of $$a^{2}$$ is equal to last two digits of $$(a – 50)^{2}$$.
So, the last two digits of $$65^{2}$$ is equal to last two digits of $$15^{2}$$. The answer is 25.
4) Find the last two digits of $$121^{2}$$.
Solution: Here a = 121. The last two digits of a is 21. So, the last two digits of $$121^{2}$$ is equal to last two digits of $$21^{2}$$. The answer is 41.
5) Find the last two digits of $$165^{2}$$.
Solution: Here a = 165. The last two digits of a is 65. Calculate 65 – a. So, 65 – 50 = 15. Then, the last two digits of $$165^{2}$$ is equal to last two digits of $$15^{2}$$. The answer is 25.
6) Find the last two digits of $$1456189^{2}$$.
Solution: Here a = 1456165. The last two digits of a is 89. Calculate 100 – 89. So, 65 – 50 = 11. Then, the last two digits of $$1456189^{2}$$ is equal to last two digits of $$11^{2}$$. The answer is 21.
Exercises
1. Find the last two digits of ($$45234^{2}$$).
2. Find the last two digits of ($$99^{2}$$).
3. What are the last two digits of ($$12488899^{2}$$)?
4. Find the last two digits of ($$5557^{2}$$).
5. Find the last two digits of ($$478^{2}$$).
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2020-02-24 17:46:37
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https://robotics.stackexchange.com/questions/21755/inverse-kinematics-in-kdl-without-rotation/21808
|
# Inverse kinematics in KDL without rotation
I am trying to use KDL to do inverse kinematics. However I am not interested in the orientation of the end effector, but only in the location. Is there a way to specify this in KDL? The function CartToJnt only seems to accept a Frame with a rotation matrix. Is there maybe a different way to do this in KDL?
• I am not a KDL user so take this as just a suggestion. Have you tried putting a 3x3 identity matrix in for the rotation submatrix? Feb 4 '21 at 15:21
• I have considered it, but the identity rotation is also specifies a orientation. Therefor the algorithm tries to optimize using that orientation. And if that orientation cannot be reached the displacement is also less accurate. I want the algorithm to totally ignore the rotation. Feb 4 '21 at 16:48
• Understood. Thanks for replying. Feb 4 '21 at 16:50
|
2022-01-23 09:23:30
|
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https://zbmath.org/serials/?q=se%3A2512
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## Journal of Integer Sequences
Short Title: J. Integer Seq. Publisher: University of Waterloo, School of Computer Science, Waterloo, ON ISSN: 1530-7638/e Online: http://www.cs.uwaterloo.ca/journals/JIS/http://www.emis.de/journals/JIS/ Comments: Indexed cover-to-cover; Published electronic only as of Vol. 1 (1998). This journal is available open access.
Documents Indexed: 1,343 Publications (since 1998) References Indexed: 357 Publications with 5,054 References.
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### Latest Issues
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### Authors
45 Barry, Paul 22 Luca, Florian 16 Shattuck, Mark A. 14 Belbachir, Hacene 12 Kimberling, Clark H. 11 De Koninck, Jean-Marie 11 Janjić, Milan 11 Prodinger, Helmut 10 Zhao, Fengzhen 9 Bordellès, Olivier 9 Griffiths, Martin 9 Mansour, Toufik 8 Broughan, Kevin A. 8 Callan, David 8 Sellers, James Allen 8 Sloane, Neil James Alexander 8 Tóth, László 8 Woan, Wen-Jin 7 Ballot, Christian 7 Hennessy, Aoife 7 Jakimczuk, Rafael 7 Kitaev, Sergey 7 Ramírez, José Luis 7 Rampersad, Narad 6 Chu, Hùng Việt 6 Farhi, Bakir 6 Flórez, Rigoberto 6 Iannucci, Douglas E. 6 Khovanova, Tanya 6 Marques, Diego 6 Mező, István 6 Müller, Tom 6 Munarini, Emanuele 6 Pandey, Ram Krishna 6 Pongsriiam, Prapanpong 6 Shevelev, Vladimir Samuil 6 Tripathi, Amitabha 6 Wituła, Roman 6 Zhai, Wenguang 5 Barbero, Stefano 5 Brent, Richard Peirce 5 Chen, Kwang-Wu 5 Coons, Michael 5 Currie, James D. 5 Doyon, Nicolas 5 Girstmair, Kurt 5 Haukkanen, Pentti 5 Merikoski, Jorma Kaarlo 5 Mihoubi, Miloud 5 Munagi, Augustine O. 5 Neto, Antônio Francisco 5 Pudwell, Lara K. 5 Quaintance, Jocelyn 5 Remmel, Jeffrey B. 5 Rinaldi, Simone 5 Shapiro, Louis W. 5 Spivey, Michael Z. 5 Szalay, László 5 Tossavainen, Timo 4 Bencherif, Farid 4 Benoumhani, Moussa 4 Boyadzhiev, Khristo N. 4 Cerruti, Umberto 4 Choudhry, Ajai 4 Cloitre, Benoit 4 Dilcher, Karl 4 Dubner, Harvey 4 Elsner, Carsten 4 Gould, Henry Wadsworth 4 Higuita, Robinson A. 4 Jones, Lenny K. 4 Kátai, Imre 4 Laohakosol, Vichian 4 Mangontarum, Mahid M. 4 Matthews, Keith R. 4 Metzger, Jerry 4 Mikić, Jovan 4 Murru, Nadir 4 Németh, László 4 Noe, Tony D. 4 Pan, Jiaqiang 4 Pergola, Elisa 4 Pita Ruiz Velasco, Claudio de Jesús 4 Roettger, Eric L. 4 Sapounakis, Aristidis 4 Sofo, Anthony 4 Sury, Balasubramanian 4 Tauraso, Roberto 4 Tsikouras, Panagiotis 4 Wagner, Carl G. 4 Williams, Hugh Cowie 4 Wong, Tony W. H. 3 Adegoke, Kunle 3 Agapito, José 3 Applegate, David L. 3 Axler, Christian 3 Balof, Barry A. 3 Birmajer, Daniel 3 Bouroubi, Sadek 3 Bravo, Jhon Jairo ...and 1,426 more Authors
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### Fields
1,041 Number theory (11-XX) 639 Combinatorics (05-XX) 66 Linear and multilinear algebra; matrix theory (15-XX) 63 Computer science (68-XX) 61 Special functions (33-XX) 26 Group theory and generalizations (20-XX) 26 Convex and discrete geometry (52-XX) 23 Probability theory and stochastic processes (60-XX) 21 Order, lattices, ordered algebraic structures (06-XX) 18 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 16 Functions of a complex variable (30-XX) 16 Dynamical systems and ergodic theory (37-XX) 15 Harmonic analysis on Euclidean spaces (42-XX) 14 Sequences, series, summability (40-XX) 10 Algebraic geometry (14-XX) 10 Real functions (26-XX) 10 Numerical analysis (65-XX) 9 Commutative algebra (13-XX) 9 Approximations and expansions (41-XX) 9 Geometry (51-XX) 6 General and overarching topics; collections (00-XX) 6 Information and communication theory, circuits (94-XX) 5 Field theory and polynomials (12-XX) 5 Associative rings and algebras (16-XX) 5 Nonassociative rings and algebras (17-XX) 5 Measure and integration (28-XX) 5 Difference and functional equations (39-XX) 5 Integral transforms, operational calculus (44-XX) 4 History and biography (01-XX) 4 Operations research, mathematical programming (90-XX) 3 Mathematical logic and foundations (03-XX) 3 Algebraic topology (55-XX) 2 Category theory; homological algebra (18-XX) 2 Ordinary differential equations (34-XX) 2 Biology and other natural sciences (92-XX) 1 $$K$$-theory (19-XX) 1 Several complex variables and analytic spaces (32-XX) 1 Abstract harmonic analysis (43-XX) 1 Operator theory (47-XX) 1 General topology (54-XX) 1 Manifolds and cell complexes (57-XX) 1 Global analysis, analysis on manifolds (58-XX) 1 Statistics (62-XX) 1 Mechanics of particles and systems (70-XX) 1 Statistical mechanics, structure of matter (82-XX)
### Citations contained in zbMATH Open
795 Publications have been cited 3,009 times in 2,169 Documents Cited by Year
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2010
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A Catalan transform and related transformations on integer sequences. Zbl 1122.11007
Barry, Paul
2005
On integral points on biquadratic curves and near-multiples of squares in Lucas sequences. Zbl 1358.11141
Alekseyev, Max A.; Tengely, Szabolcs
2014
On the gcd-sum function. Zbl 1247.11124
Tanigawa, Yoshio; Zhai, Wenguang
2008
Complementary equations and Wythoff sequences. Zbl 1204.11058
Kimberling, Clark
2008
The Akiyama-Tanigawa algorithm for Bernoulli numbers. Zbl 0982.11009
Kaneko, Masanobu
2000
Partition statistics and $$q$$-Bell numbers $$(q=-1)$$. Zbl 1064.05023
Wagner, Carl G.
2004
Counting unlabelled topologies and transitive relations. Zbl 1064.05068
Brinkmann, Gunnar; McKay, Brendan D.
2005
Interesting Ramanujan-like series associated with powers of central binomial coefficients. Zbl 1487.11027
Chen, Hongwei
2022
A sequential view of self-similar measures; or, what the ghosts of Mahler and Cantor can teach us about dimension. Zbl 1460.28007
Coons, Michael; Evans, James
2021
A unified treatment of certain classes of combinatorial identities. Zbl 1460.05021
Batır, Necdet; Sofo, Anthony
2021
New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile. Zbl 1460.11022
Edwards, Kenneth; Allen, Michael A.
2021
Mex-related partition functions of Andrews and Newman. Zbl 1471.11258
Barman, Rupam; Singh, Ajit
2021
Counting prefixes of skew Dyck paths. Zbl 1473.05020
Baril, Jean-Luc; Ramírez, José L.; Simbaqueba, Lina M.
2021
Irrationality of growth constants associated with polynomial recursions. Zbl 1458.11114
Wagner, Stephan; Ziegler, Volker
2021
The generalized bi-periodic Fibonacci sequence modulo $$m$$. Zbl 1485.11021
Belbachir, Hacène; Salhi, Celia
2021
Nontrivial effective lower bounds for the least common multiple of a $$q$$-arithmetic progression. Zbl 1460.11001
Farhi, Bakir
2021
Finite binomial sum identities with harmonic numbers. Zbl 1465.05005
Batır, Necdet
2021
The abc conjecture implies that only finitely many $$s$$-Cullen numbers are repunits. Zbl 1471.11125
Grantham, Jon; Graves, Hester
2021
A golden penney game. Zbl 1466.91065
Dababneh, Issa; Elmer, Mark; McCulloch, Ryan
2021
On the least common multiple of shifted powers. Zbl 1490.11026
Sanna, Carlo
2021
Representation of integers of the form $$x^2 + my^2 - z^2$$. Zbl 1479.11061
2021
On a sequence related to the factoradic representation of an integer. Zbl 1479.11049
Sánchez Garza, Maximiliano; Treviño, Enrique
2021
The minimal excludant in integer partitions. Zbl 1441.11265
Andrews, George E.; Newman, David
2020
Parity considerations for the mex-related partition functions of Andrews and Newman. Zbl 1480.11131
da Silva, Robson; Sellers, James A.
2020
Harmonic sums via Euler’s transform: complementing the approach of Boyadzhiev. Zbl 1446.11026
Frontczak, Robert
2020
Fermat Padovan and Perrin numbers. Zbl 1446.11027
Rihane, Salah Eddine; Adegbindin, Chèfiath Awero; Togbé, Alain
2020
The Lie bracket and the arithmetic derivative. Zbl 1444.11012
Fan, Jun; Utev, Sergey
2020
The number of solutions to $$ax + by + cz = n$$ and its relation to quadratic residues. Zbl 1476.11067
Binner, Damanvir Singh
2020
On 3- and 9-regular cubic partitions. Zbl 1452.05005
Gireesh, D. S.; Shivashankar, C.; Mahadeva Naika, M. S.
2020
Symmetrized poly-Bernoulli numbers and combinatorics. Zbl 1476.11052
Matsusaka, Toshiki
2020
Run distribution over flattened partitions. Zbl 1451.05007
Nabawanda, Olivia; Rakotondrajao, Fanja; Bamunoba, Alex Samuel
2020
Fixed points in compositions and words. Zbl 1451.05005
Archibald, M.; Blecher, A.; Knopfmacher, A.
2020
The number of threshold words on $$n$$ letters grows exponentially for every $$n \geq 27$$. Zbl 1442.68188
2020
Log-concavity and LC-positivity for generalized triangles. Zbl 1480.05015
Ahmia, Moussa; Belbachir, Hacène
2020
Nontrivial effective lower bounds for the least common multiple of some quadratic sequences. Zbl 1448.11005
Bousla, Sid Ali; Farhi, Bakir
2020
Repdigits in Narayana’s cows sequence and their consequences. Zbl 1477.11031
Bravo, Jhon J.; Das, Pranabesh; Guzmán, Sergio
2020
Some combinatorics of factorial base representations. Zbl 1446.11013
Ball, Tyler; Beckford, Joanne; Dalenberg, Paul; Edgar, Tom; Rajabi, Tina
2020
A ($$p, q)$$-deformed recurrence for the Bell numbers. Zbl 1477.11043
Oussi, Lahcen
2020
Unordered factorizations with $$k$$ parts. Zbl 1471.11018
Sprittulla, Jacob
2020
Greatest common divisors of shifted Horadam sequences. Zbl 1471.11059
Chen, Kwang-Wu; Pan, Yu-Ren
2020
When the large divisors of a natural number are in arithmetic progression. Zbl 1476.11019
Chu, Hùng Việt
2020
Crossings over permutations avoiding some pairs of patterns of length three. Zbl 1442.05009
Rakotomamonjy, Paul M.; Andriantsoa, Sandrataniaina R.; Randrianarivony, Arthur
2020
On polycosecant numbers. Zbl 1476.11040
Kaneko, Masanobu; Pallewatta, Maneka; Tsumura, Hirofumi
2020
Some Toeplitz-Hessenberg determinant identities for the tetranacci numbers. Zbl 1447.15003
Goy, Taras; Shattuck, Mark
2020
A polynomial variant of perfect numbers. Zbl 1480.11007
Gallardo, Luis H.; Rahavandrainy, Olivier
2020
Leibniz-additive functions on UFD’s. Zbl 1458.11009
Murashka, Viachaslau I.; Goncharenko, Andrey D.; Goncharenko, Irina N.
2020
On the Erdős-Sloane and shifted Sloane persistence problems. Zbl 1454.11017
Bonuccelli, Gabriel; Colucci, Lucas; de Faria, Edson
2020
A summation involving the divisor and GCD functions. Zbl 1446.11182
Heyman, Randell
2020
An identity for generalized Bernoulli polynomials. Zbl 1471.11095
Chellal, Redha; Bencherif, Farid; Mehbali, Mohamed
2020
Binary recurrences for which powers of two are discriminating moduli. Zbl 1471.11086
de Clercq, Adriaan A.; Luca, Florian; Martirosyan, Lilit; Matthis, Maria; Moree, Pieter; Stoumen, Max A.; Weiß, Melvin
2020
Minimum coprime graph labelings. Zbl 1453.05116
Lee, Catherine
2020
New series identities with Cauchy, Stirling, and harmonic numbers, and Laguerre polynomials. Zbl 1465.11071
2020
New partition function recurrences. Zbl 1459.11205
da Silva, Robson; Sakai, Pedro Diniz
2020
A combinatorial classification of triangle centers on the line at infinity. Zbl 1423.05023
Kimberling, Clark
2019
Repdigits as sums of two Padovan numbers. Zbl 1423.11127
García Lomelí, Ana Cecilia; Hernández Hernández, Santos
2019
Consecutive patterns in inversion sequences. II: Avoiding patterns of relations. Zbl 1425.05008
Auli, Juan S.; Elizalde, Sergi
2019
New estimates for the $$n$$th prime number. Zbl 1418.11130
Axler, Christian
2019
Log-concavity of recursively defined polynomials. Zbl 1454.11055
Heim, Bernhard; Neuhauser, Markus
2019
Relating Fibonacci numbers to Bernoulli numbers via balancing polynomials. Zbl 1461.11033
Frontczak, Robert
2019
Distributions of statistics over pattern-avoiding permutations. Zbl 1407.05005
Bukata, Michael; Kulwicki, Ryan; Lewandowski, Nicholas; Pudwell, Lara; Roth, Jacob; Wheeland, Teresa
2019
On the problem of Pillai with tribonacci numbers and powers of 3. Zbl 1450.11008
2019
On a family of functions defined over sums of primes. Zbl 1473.11182
Axler, Christian
2019
Riordan pseudo-involutions, continued fractions and Somos-4 sequences. Zbl 1459.11033
Barry, Paul
2019
Two new identities involving the Catalan numbers and sign-reversing involutions. Zbl 1431.05020
Mikić, Jovan
2019
Prefix palindromic length of the Thue-Morse word. Zbl 1437.68146
Frid, Anna E.
2019
Reciprocal sum of palindromes. Zbl 1453.11014
Phunphayap, Phakhinkon; Pongsriiam, Prapanpong
2019
Polynomial analogues of restricted $$b$$-ary partition functions. Zbl 1440.11191
Dilcher, Karl; Ericksen, Larry
2019
Diagonal sums in the Pascal pyramid. II: Applications. Zbl 1440.11020
Belbachir, Hacène; Mehdaoui, Abdelghani; Szalay, László
2019
When sets are not sum-dominant. Zbl 1418.11018
Chu, Hùng Việt
2019
The central coefficients of a family of Pascal-like triangles and colored lattice paths. Zbl 1405.05006
Barry, Paul
2019
Arithmetic progressions in the graphs of slightly curved sequences. Zbl 1407.11018
Saito, Kota; Yoshida, Yuuya
2019
Euler’s divergent series in arithmetic progressions. Zbl 1443.11138
Ernvall-Hytönen, Anne-Maria; Matala-Aho, Tapani; Seppälä, Louna
2019
A probabilistic two-pile game. Zbl 1425.91086
Leung, Ho-Hon; Thanatipanonda, Thotsaporn “Aek”
2019
Fixed points of augmented generalized happy functions. II: Oases and mirages. Zbl 1462.11012
Baker Swart, Breeanne; Crook, Susan; Grundman, Helen G.; Hall-Seelig, Laura; Mei, May; Zack, Laurie
2019
Product of consecutive tribonacci numbers with only one distinct digit. Zbl 1455.11027
Bravo, Eric F.; Gómez, Carlos A.; Luca, Florian
2019
Arithmetic subderivatives: discontinuity and continuity. Zbl 1448.11008
Haukkanen, Pentti; Merikoski, Jorma K.; Tossavainen, Timo
2019
Finite test sets for morphisms that are squarefree on some of Thue’s squarefree ternary words. Zbl 1447.68011
Currie, James D.
2019
On properties of the general bow sequence. Zbl 1472.11063
Dennison, Melissa
2019
Congruent number elliptic curves related to integral solutions of $$m^2=n^2+nl+n^2$$. Zbl 1470.11152
Halbeisen, Lorenz; Hungerbühler, Norbert
2019
Nonlinear inverse relations for Bell polynomials via the Lagrange inversion formula. Zbl 1418.11040
Wang, Jin
2019
The $$\gamma$$-vectors of Pascal-like triangles defined by Riordan arrays. Zbl 1454.11059
Barry, Paul
2019
Summation of certain infinite Lucas-related series. Zbl 1459.11038
Farhi, Bakir
2019
Enumerating multiplex juggling patterns. Zbl 1407.05011
Butler, Steve; Choi, Jeongyoon; Kim, Kimyung; Seo, Kyuhyeok
2019
Some double sums involving ratios of binomial coefficients arising from urn models. Zbl 1407.05009
Stenlund, David; Wan, James G.
2019
Almost Beatty partitions. Zbl 1418.11044
Hildebrand, A. J.; Li, Xiaomin; Li, Junxian; Xie, Yun
2019
Some new restricted $$n$$-color composition functions. Zbl 1423.05010
Acosta, Jarib R.; Caicedo, Yadira; Poveda, Juan P.; Ramírez, José L.; Shattuck, Mark
2019
On Engel’s inequality for Bell numbers. Zbl 1423.05024
Alzer, Horst
2019
Patterns in inversion sequences. II: Inversion sequences avoiding triples of relations. Zbl 1384.05008
Martinez, Megan; Savage, Carla
2018
Repdigits as sums of four Fibonacci or Lucas numbers. Zbl 1453.11006
Normenyo, Benedict Vasco; Luca, Florian; Togbé, Alain
2018
Explicit formulas for the $$p$$-adic valuations of fibonomial coefficients. Zbl 1390.11041
Phunphayap, Phakhinkon; Pongsriiam, Prapanpong
2018
Spivey’s Bell number formula revisited. Zbl 1390.11056
Mangontarum, Mahid M.
2018
A generalization of the “problème des rencontres”. Zbl 1384.05046
Capparelli, Stefano; Ferrari, Margherita Maria; Munarini, Emanuele; Zagaglia Salvi, Norma
2018
Polynomials characterizing hyper $$b$$-ary representations. Zbl 1390.05010
Dilcher, Karl; Ericksen, Larry
2018
Classical and semi-classical orthogonal polynomials defined by Riordan arrays, and their moment sequences. Zbl 1390.15106
Barry, Paul; Mesinga Mwafise, Arnauld
2018
New sufficient conditions for log-balancedness, with applications to combinatorial sequences. Zbl 1393.05056
Liu, Rui-Li; Zhao, Feng-Zhen
2018
The star of David and other patterns in Hosoya polynomial triangles. Zbl 1390.11036
Flórez, Rigoberto; Higuita, Robinson A.; Mukherjee, Antara
2018
Lucasnomial Fuss-Catalan numbers and related divisibility questions. Zbl 1398.11045
Ballot, Christian
2018
Greatest common divisors of shifted Fibonacci sequences revisited. Zbl 1398.11041
Rahn, Annalena; Kreh, Martin
2018
When sets can and cannot have sum-dominant subsets. Zbl 1408.11006
Chu, Hùng Việt; McNew, Nathan; Miller, Steven J.; Xu, Victor; Zhang, Sean
2018
Generalizations of the reciprocal Fibonacci-Lucas sums of Brousseau. Zbl 1390.11031
2018
Descents in parking functions. Zbl 1384.05009
Schumacher, Paul R. F.
2018
Combinatorial identities for generalized Stirling numbers expanding $$f$$-factorial functions and the $$f$$-harmonic numbers. Zbl 1390.11051
Schmidt, Maxie D.
2018
On two new classes of $$B$$-$$q$$-bonacci polynomials. Zbl 1390.11033
Arolkar, Suchita; Valaulikar, Yeshwant Shivrai
2018
Generalization of an identity of Apostol. Zbl 1390.11044
Zekiri, Abdelmoumène; Bencherif, Farid; Boumahdi, Rachid
2018
A probabilistic take-away game. Zbl 1418.91116
Wong, Tony W. H.; Xu, Jiao
2018
Another identity for complete Bell polynomials based on Ramanujan’s congruences. Zbl 1395.05019
Leung, Ho-Hon
2018
Factored closed-form expressions for the sums of cubes of Fibonacci and Lucas numbers. Zbl 1398.11037
2018
...and 695 more Documents
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### Cited by 2,571 Authors
36 Ramírez, José Luis 33 Mansour, Toufik 28 Komatsu, Takao 26 Luca, Florian 26 Sofo, Anthony 23 Shattuck, Mark A. 22 Belbachir, Hacene 22 Qi, Feng 17 Bravo, Jhon Jairo 17 Cheon, Gi-Sang 17 Rinaldi, Simone 16 Barry, Paul 16 Chu, Wenchang 16 Corcino, Roberto Bagsarsa 14 Gómez, Carlos Alexis 14 Kim, Dae San 14 Kim, Taekyun 13 Flórez, Rigoberto 13 Keskin, Refik 13 Mező, István 13 Shallit, Jeffrey O. 13 Yang, Shengliang 12 Goy, Taras P. 11 Harris, Pamela E. 11 Marques, Diego 11 Pinzani, Renzo 11 Shapiro, Louis W. 11 Trojovský, Pavel 10 Guo, Bai-Ni 10 Munarini, Emanuele 10 Pongsriiam, Prapanpong 10 Togbé, Alain 9 Allouche, Jean-Paul Simon 9 Baril, Jean-Luc 9 Batir, Necdet 9 Bouvel, Mathilde 9 Kim, Hana 9 Lin, Zhicong 9 Merca, Mircea 9 Rampersad, Narad 9 Sellers, James Allen 9 Simsek, Yilmaz 8 Bényi, Beáta 8 Campbell, John Maxwell 8 Corcino, Cristina Bordaje 8 Defant, Colin 8 Dilcher, Karl 8 Elizalde, Sergi 8 Mahadeva Naika, Megadahalli Sidda 8 Moree, Pieter 8 Németh, László 8 Prodinger, Helmut 8 Szalay, László 8 Tóth, László 8 Wang, Yi 7 Bagno, Eli 7 Han, Guo-Niu 7 Herrera, José L. 7 Hone, Andrew N. W. 7 Kargin, Levent 7 Kitaev, Sergey 7 Murru, Nadir 7 Nyul, Gábor 7 Rigo, Michel 7 Rihane, Salah Eddine 7 Sagan, Bruce Eli 7 Schork, Matthias 7 Sun, Zhi-Wei 7 Wituła, Roman 7 Zhu, Baoxuan 6 Ahmia, Moussa 6 Axler, Christian 6 Barbero, Stefano 6 Berthé, Valérie 6 Cerruti, Umberto 6 Ddamulira, Mahadi 6 Frontczak, Robert 6 Garcia, Ronaldo A. 6 Josuat-Vergès, Matthieu 6 Kabluchko, Zakhar A. 6 Kilic, Emrah 6 Kirgizov, Sergey 6 Kiuchi, Isao 6 Mihoubi, Miloud 6 Ochem, Pascal 6 Panda, Gopal Krishna 6 Pelantová, Edita 6 Pergola, Elisa 6 Reznik, Dan S. 6 Sanna, Carlo 6 Sapounakis, Aristidis 6 Schmidt, Maxie Dion 6 Şiar, Zafer 6 Sulanke, Robert A. 6 Tasoulas, Ioannis 6 Tenner, Bridget Eileen 6 Tripathi, Amitabha 6 Tsikouras, Panagiotis 6 Xu, Aimin 6 Yang, Lin ...and 2,471 more Authors
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### Cited in 386 Journals
161 Journal of Integer Sequences 92 Discrete Mathematics 92 Integers 85 Journal of Number Theory 70 Linear Algebra and its Applications 62 The Electronic Journal of Combinatorics 53 The Ramanujan Journal 49 Advances in Applied Mathematics 46 European Journal of Combinatorics 44 Theoretical Computer Science 34 Discrete Applied Mathematics 33 Journal of Combinatorial Theory. Series A 33 International Journal of Number Theory 25 Rocky Mountain Journal of Mathematics 25 Applied Mathematics and Computation 18 Indagationes Mathematicae. New Series 18 Annals of Combinatorics 16 American Mathematical Monthly 16 Monatshefte für Mathematik 16 Graphs and Combinatorics 16 Mediterranean Journal of Mathematics 16 Advances in Difference Equations 15 Mathematics of Computation 15 Acta Arithmetica 15 Quaestiones Mathematicae 15 Journal of Algebraic Combinatorics 15 Advances in Applied Clifford Algebras 14 Indian Journal of Pure & Applied Mathematics 14 Boletín de la Sociedad Matemática Mexicana. Third Series 13 Journal of Mathematical Analysis and Applications 13 Journal of Statistical Planning and Inference 13 Journal of Inequalities and Applications 13 Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A: Matemáticas. RACSAM 12 Periodica Mathematica Hungarica 12 Functiones et Approximatio. Commentarii Mathematici 12 Aequationes Mathematicae 12 The Australasian Journal of Combinatorics 12 Turkish Journal of Mathematics 12 Filomat 12 Comptes Rendus. Mathématique. Académie des Sciences, Paris 11 International Journal of Mathematics and Mathematical Sciences 11 Journal of Algebra 10 Mathematica Slovaca 10 Proceedings of the Indian Academy of Sciences. Mathematical Sciences 10 Applicable Analysis and Discrete Mathematics 10 Research in Number Theory 9 Advances in Mathematics 9 Proceedings of the American Mathematical Society 9 Results in Mathematics 9 Transactions of the American Mathematical Society 9 SIAM Journal on Discrete Mathematics 9 Séminaire Lotharingien de Combinatoire 8 Lithuanian Mathematical Journal 8 Finite Fields and their Applications 8 Integral Transforms and Special Functions 8 Abstract and Applied Analysis 8 JP Journal of Algebra, Number Theory and Applications 8 Ars Mathematica Contemporanea 8 Journal of Classical Analysis 7 Bulletin of the Australian Mathematical Society 7 Chaos, Solitons and Fractals 7 Journal of Symbolic Computation 7 Journal de Théorie des Nombres de Bordeaux 7 Russian Journal of Mathematical Physics 7 Bulletin of the Brazilian Mathematical Society. New Series 7 Asian-European Journal of Mathematics 7 Afrika Matematika 7 Palestine Journal of Mathematics 6 Information Processing Letters 6 Journal of Approximation Theory 6 Acta Mathematica Hungarica 6 Journal of Difference Equations and Applications 6 Communications of the Korean Mathematical Society 6 Acta et Commentationes Universitatis Tartuensis de Mathematica 6 Bulletin of the Malaysian Mathematical Sciences Society. Second Series 6 Tbilisi Mathematical Journal 6 Discrete Mathematics, Algorithms and Applications 6 RAIRO. Theoretical Informatics and Applications 6 Special Matrices 6 Open Mathematics 6 AIMS Mathematics 5 Ukrainian Mathematical Journal 5 Czechoslovak Mathematical Journal 5 Semigroup Forum 5 Ergodic Theory and Dynamical Systems 5 Order 5 Designs, Codes and Cryptography 5 Taiwanese Journal of Mathematics 5 Contributions to Discrete Mathematics 5 International Journal of Combinatorics 5 Enumerative Combinatorics and Applications 4 Linear and Multilinear Algebra 4 Mathematics Magazine 4 Journal of Pure and Applied Algebra 4 Bulletin of the Korean Mathematical Society 4 Random Structures & Algorithms 4 Journal of Mathematical Sciences (New York) 4 Bulletin des Sciences Mathématiques 4 Discrete Mathematics and Theoretical Computer Science. DMTCS 4 Annales Mathematicae Silesianae ...and 286 more Journals
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### Cited in 57 Fields
1,304 Number theory (11-XX) 1,033 Combinatorics (05-XX) 173 Computer science (68-XX) 164 Linear and multilinear algebra; matrix theory (15-XX) 139 Special functions (33-XX) 78 Dynamical systems and ergodic theory (37-XX) 69 Probability theory and stochastic processes (60-XX) 65 Group theory and generalizations (20-XX) 55 Convex and discrete geometry (52-XX) 46 Functions of a complex variable (30-XX) 45 Real functions (26-XX) 40 Numerical analysis (65-XX) 38 Order, lattices, ordered algebraic structures (06-XX) 35 Associative rings and algebras (16-XX) 29 Approximations and expansions (41-XX) 29 Harmonic analysis on Euclidean spaces (42-XX) 29 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 27 Algebraic geometry (14-XX) 26 Sequences, series, summability (40-XX) 22 Difference and functional equations (39-XX) 22 Quantum theory (81-XX) 21 Geometry (51-XX) 21 Information and communication theory, circuits (94-XX) 19 Ordinary differential equations (34-XX) 17 Integral transforms, operational calculus (44-XX) 15 Field theory and polynomials (12-XX) 15 Commutative algebra (13-XX) 15 Operations research, mathematical programming (90-XX) 14 Mathematical logic and foundations (03-XX) 13 Category theory; homological algebra (18-XX) 13 Manifolds and cell complexes (57-XX) 13 Biology and other natural sciences (92-XX) 12 Statistical mechanics, structure of matter (82-XX) 11 Nonassociative rings and algebras (17-XX) 11 Operator theory (47-XX) 10 General topology (54-XX) 10 Statistics (62-XX) 9 Functional analysis (46-XX) 8 General and overarching topics; collections (00-XX) 7 Measure and integration (28-XX) 7 Several complex variables and analytic spaces (32-XX) 7 Algebraic topology (55-XX) 7 Mathematics education (97-XX) 6 Systems theory; control (93-XX) 5 General algebraic systems (08-XX) 5 Differential geometry (53-XX) 4 History and biography (01-XX) 4 Partial differential equations (35-XX) 3 Integral equations (45-XX) 3 Relativity and gravitational theory (83-XX) 2 Potential theory (31-XX) 1 Abstract harmonic analysis (43-XX) 1 Calculus of variations and optimal control; optimization (49-XX) 1 Mechanics of particles and systems (70-XX) 1 Mechanics of deformable solids (74-XX) 1 Fluid mechanics (76-XX) 1 Geophysics (86-XX)
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I can’tfind a way to type a dot in the middle. b. Along with the cross product, the dot product is one of the fundamental operations on Euclidean vectors. Math Article. 1 Unary / binary operators; 2 Relational operators; 3 Set operations; 4 Functions; 5 Operators; 6 Attributes; 7 Miscellaneous; 8 Brackets; 9 Formats; 10 Characters – Greek; 11 Characters – Special; Unary / binary operators. topologie, probabilités, statistiques, combinatoire, aucun domaine des maths ne lui échappe. 7 . Spé maths : Comprendre la notion de matrice et savoir additionner, mutliplier 2 matrices. middle dot meaning? This is how knowing the meaning of symbols can help you learn, grow and evolve. 151. Exponential Functions. Pour montrer qu'un ensemble est un sous-espace vectoriel de , on montre : ∈. Exponential Functions. $[0,1]\cap \Qq$. Multiplication of Fractions. La borne supérieure : $1$. In addition to multiplying a matrix by a scalar, we can multiply two matrices. The flux integral of $$F$$ across $$n$$ is given by \[ \iint_{S} (F \cdot n) d\sigma . In mathematics (particularly in differential calculus), the derivative is a way to show instantaneous rate of change: that is, the amount by which a function is changing at one given point. If a, b, c are real numbers, then a(b + c) = ab + ac. P robability and statistics correspond to the mathematical study of chance and data, respectively. 3 But, with the dot in the middle, not at the bottom. Les tables qui constituent cet article répertorient certains de ces symboles avec leurs codages Unicode et TeX lorsqu'ils sont connus, ainsi que leur nom et leurs usages. En mathématiques, de nombreux symboles sont employés avec une signification qui n'est pas toujours reprécisée dans les documents qui les emploient. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. Below are many more quotes by Manly P. Hall that I believe are perfect, and I think you might agree. Table 1 shows the needs of both teams. dimanche 5 juillet 2020, par Nadir Soualem. Now the denominator of a fraction shows into what parts the integral unit is supposed to be divided ; and the numerator showahow many of those parts belong to the given fraction. Math-Linux.com. (Art. SHARES. Le plus grand élément : $1$. By the definition of multiplication, multiplying by a fraction is taking a part of the multiplicand, as many times as there are like parts of an unit in the multiplier. For readability purpose, these symbols are categorized by function into tables. This is when you state whether a mathematical quantity is either positive or negative. Operation Command Display Positive (plus) +1 +1 Negative (minus)-1 −1 Plus/minus +-1: … Développement limité de arcsinus arcsin x en 0 - Démonstration. Les tenseurs sont des objets mathématiques issus de l'algèbre multilinéaire permettant de généraliser les scalaires et les vecteurs.On les rencontre notamment en analyse vectorielle et en géométrie différentielle fréquemment utilisés au sein de champs de tenseurs.Ils sont aussi utilisés en mécanique des milieux continus. Finding the Product of Two Matrices. L'élément nul est souvent facile à manipuler : il est donc en règle générale très facile de montrer qu’il appartient à F même pour des ensembles compliqués. Definition: Distributive Property. Développements limités au voisinage de 0. dimanche 21 juin 2020, par Nadir Soualem. Share Tweet Subscribe. 20 novembre 2018 3 juillet 2019 maths01 Généralités sur les fonctions numériques, Les fonctions, Maths 1BAC-SE-Fr, Maths 2BAC_PC_Fr, Maths TCS-Fr définition, domaine, domaine de définition d'une fonction, fonction, L'ensemble. Knowledge base dedicated to Linux and applied mathematics. Plus précisément, c'est la donnée, pour chaque élément du groupe, d'une permutation de l'ensemble, de telle manière que toutes ces bijections se composent de façon compatible avec la loi du groupe. 59. Domaine de définition d’une fonction . Les minorants : $]-\infty,0]$. Finite Math. Given a vector space $$V$$, we define its dual space $$V^*$$ to be the set of all linear transformations $$\varphi: V \to \mathbb{F}$$. Two club soccer teams, the Wildcats and the Mud Cats, are hoping to obtain new equipment for an upcoming season. c. Show that the sum of the squares of the lengths of the diagonals equals the sum of … Prove that the diagonals have the same length if and only if $\mathbf{u} \cdot \mathbf{v}=0$. When two maths elements appear on either side of the sign, it is assumed to be a binary operator, and as such, allocates some space to either side of the sign. An exponential function is a mathematical function, which is used in many real-world situations. First, we will go through the math and then we will invoke some intuition to make sense of all these. Joined Jul 9, 2013 Messages 37. Thread starter cruz33; Start date Jul 19, 2013; C . Joined Nov 29, 2012 Messages 1,383. Définitions, propriétés, colinéarité, vecteurs orthogonaux, exemples et vidéos sur Mathforu. It is mainly used to find the exponential decay or exponential growth or to compute investments, model populations and so on. Definition: Flux Integral. For other meanings of the term, see Derivative.. Divergence of vector quality indicates how much the vector quality spreads out from the certain point.Think of water coming from a faucet. Ce billet et l’article Wikipedia sur le triangle de Pascal Jul 19, 2013 #2 cruz33 said: … Cours de maths sur le produit scalaire en 1ère S (vecteurs, vecteurs colinéaires, projeté orthogonal...). Dual spaces Definition. Search for: Reading: Matrices and Matrix Operations . Jul 19, 2013 #1 What does this mathematical expression mean? It’s also the symbol for the consumer shopping giant, Target. This happens to be the very definition of “linear seperability” 88.) dérivée Développements limités usuels Landau Maclaurin ordre Taylor. Math-Linux.com. Prenons l’exemple du code Morse : deux signaux élémentaires sont utilisés, le tiret $-$ et le point $\cdot$. The $$\varphi$$ is called a linear functional. Today you will find the symbol of the circumpunct all over the world. Le nom de « triangle de Pascal » est trompeur : Pascal n’était pas la première personne à s’y intéresser. D. DrPhil Senior Member. En mathématiques, une action d'un groupe sur un ensemble est une loi de composition externe du groupe sur l'ensemble, vérifiant des conditions supplémentaires. Il apparaît évidemment aussi en physique, et par exemple en génétique. Table 1; Wildcats: Mud Cats: Goals: 6: 10: Balls: 30: 24: Jerseys: 14: 20: A goal costs $300; a ball costs$10; and a jersey costs $30. In mathematics, the dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number.In Euclidean geometry, the dot product of the Cartesian coordinates of two vectors is widely used. So, the equation $$\mathbf{w} \cdot \mathbf{x} > b$$ defines all the points on one side of the hyperplane, and $$\mathbf{w} \cdot \mathbf{x} <= b$$ all the points on the other side of the hyperplane and on the hyperplane itself. Let $$F$$ be a differentiable vector field on a surface $$S$$ oriented by a unit normal vector $$n$$. Les majorants :$[1,+\infty[$. Comprehensive collection of 225+ math symbols used in algebra, categorized by subject and type into tables along with each symbol's name, usage and example. Knowledge base dedicated to Linux and applied mathematics. Accueil > Mathématiques > Développements limités > Développement limité de arcsinus arcsin x en 0 - Démonstration. Accueil > Mathématiques > Développements limités > Développements limités au voisinage de 0. Math commands - Reference < Previous Page Next Page > Contents. Solve your math problems using our free math solver with step-by-step solutions. The following reference list documents some of the most notable symbols in these two topics, along with each symbol’s usage and meaning. cruz33 New member. La borne inférieure :$0\$. Ce tableau ne saurait prétendre à l'exhaustivité. CDOT is listed in the World's largest and most authoritative dictionary database of abbreviations and acronyms CDOT is listed in the World's largest and most authoritative dictionary database of abbreviations and acronyms Looking for online definition of CDOT or what CDOT stands for? Module 4: Matrices. For functions that act on the real numbers, it is the slope of the tangent line at a point on a graph. Imagine a fluid, with the vector field representing the velocity of the fluid at each point in space. It is featured in the best-selling book by Dan Brown, The Lost Symbol. We can calculate the Dot Product of two vectors this way: The alternative way is a sign designation. b This means the Dot Product of a and b . Thanks! The specific case of the inner product in Euclidean space, the dot product gives the product of the magnitude of two vectors and the cosine of the angle between them.
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2021-04-19 21:52:11
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http://www.acmerblog.com/hdu-3041-bee-4813.html
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2014
02-27
# Bee
QQ the bear has found a little treasure � the bees’ secret honeypot, which is full of honey! He was happily eating his newfound treasure until suddenly one bee saw him and sounded the bee alarm. He knows that at this very moment hordes of bees will emerge from their hives and start spreading around trying to catch him. He knows he has to leave the honeypot and go home quickly, but the honey is so sweet that QQ doesn’t want to leave too soon. Help QQ determine the latest possible moment when he can leave.
QQ’s forest is represented by a square grid of N by N unit cells, whose sides are parallel to the north-south and east-west directions. Each cell is occupied by a tree, by a patch of grass, by a hive or by QQ’s home. Two cells are considered adjacent if one of them is immediately to the north, south, east or west of the other (but not on a diagonal). QQ is a clumsy bear, so every time he makes a step, it has to be to an adjacent cell. QQ can only walk on grass and cannot go through trees or hives, and he can make at most S steps per minute.
At the moment when the bee alarm is sounded, QQ is in the grassy cell containing the honeypot, and the bees are in every cell containing a hive (there may be more than one hive in the forest). During each minute from this time onwards, the following events happen in the following order:
1. If QQ is still eating honey, he decides whether to keep eating or to leave. If he continues eating, he does not move for the whole minute. Otherwise, he leaves immediately and takes up to S steps through the forest as described above. QQ cannot take any of the honey with him, so once he has moved he cannot eat honey again.
2. After QQ is done eating or moving for the whole minute, the bees spread one unit further across the grid, moving only into the grassy cells. Specifically, the swarm of bees spreads into every grassy cell that is adjacent to any cell already containing bees. Furthermore, once a cell contains bees it will always contain bees (that is, the swarm does not move, but it grows).
In other words, the bees spread as follows: When the bee alarm is sounded, the bees only occupy the cells where the hives are located. At the end of the first minute, they occupy all grassy cells adjacent to hives (and still the hives themselves). At the end of the second minute, they additionally occupy all grassy cells adjacent to grassy cells adjacent to hives, and so on.
Given enough time, the bees will end up simultaneously occupying all grassy cells in the forest that are within their reach.
Neither QQ nor the bees can go outside the forest. Also, note that according to the rules above, QQ will always eat honey for an integer number of minutes.
The bees catch QQ if at any point in time QQ finds himself in a cell occupied by bees.
Write a program that, given a map of the forest, determines the largest number of minutes that QQ can continue eating honey at his initial location, while still being able to get to his home before any of the bees catch him.
CONSTRAINTS
1<=N<=800, the size (side length) of the map
1<=S<=1000, the maximum number of steps QQ can take in each minute
The first line contains the integers N and S, separated by a space.
The next N lines represent the map of the forest. Each of these lines contains N characters
with each character representing one unit cell of the grid. The possible characters and their
associated meanings are as follows:
T denotes a tree
G denotes a grassy cell
M denotes the initial location of QQ and the honeypot, which is also a grassy cell
D denotes the location of QQ’s home, which QQ can enter, but the bees cannot.
H denotes the location of a hive
NOTE: It is guaranteed that the map will contain exactly one letter M, exactly one letter D and at least one letter H. It is also guaranteed that there is a sequence of adjacent letters G that connects QQ to his home, as well as a sequence of adjacent letters G that connects at least one hive to the honeypot (i.e., to QQ’s initial location). These sequences might be as short as length zero, in case QQ’s home or a hive is adjacent to QQ’s initial location.
Also, note that the bees cannot pass through or fly over QQ’s home. To them, it is just like a tree.
The first line contains the integers N and S, separated by a space.
The next N lines represent the map of the forest. Each of these lines contains N characters
with each character representing one unit cell of the grid. The possible characters and their
associated meanings are as follows:
T denotes a tree
G denotes a grassy cell
M denotes the initial location of QQ and the honeypot, which is also a grassy cell
D denotes the location of QQ’s home, which QQ can enter, but the bees cannot.
H denotes the location of a hive
NOTE: It is guaranteed that the map will contain exactly one letter M, exactly one letter D and at least one letter H. It is also guaranteed that there is a sequence of adjacent letters G that connects QQ to his home, as well as a sequence of adjacent letters G that connects at least one hive to the honeypot (i.e., to QQ’s initial location). These sequences might be as short as length zero, in case QQ’s home or a hive is adjacent to QQ’s initial location.
Also, note that the bees cannot pass through or fly over QQ’s home. To them, it is just like a tree.
7 3
TTTTTTT
TGGGGGT
TGGGGGT
MGGGGGD
TGGGGGT
TGGGGGT
THHHHHT
7 3
TTTTTTT
TGGGGGT
TGGGGGT
MGGGGGD
TGGGGGT
TGGGGGT
TGHHGGT
1
2
Hint
Hints：
For the first case, After eating honey for one minute, QQ can take the shortest path directly to
the right and he will be home in another two minutes, safe from the bees.
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
class Main {
static long[] arr = new long[50];
static long[] sum = new long[50];
static {
arr[0] = 1;
arr[1] = 1;
sum[0] = 1;
sum[1] = 2;
}
public static void main(String[] args) throws NumberFormatException,
IOException {
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(
System.out));
int n;
for (int c = 2; c < arr.length; c++) {
arr[c] = arr[c - 2] + arr[c - 1];
sum[c] = arr[c] + sum[c - 1];
}
while ((n = Integer.parseInt(in.readLine())) != -1) {
if(n==0)
out.write("0 1\n");
else
out.write(sum[n-1] + " " + sum[n] + "\n");
}
out.flush();
}
}
1. 第23行：
hash = -1是否应该改成hash[s ] = -1
因为是要把从字符串s的start位到当前位在hash中重置
修改提交后能accept，但是不修改居然也能accept
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2016-12-09 21:21:41
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https://en.wikipedia.org/wiki/Checking_if_a_coin_is_fair
|
# Checking whether a coin is fair
(Redirected from Checking if a coin is fair)
In statistics, the question of checking whether a coin is fair is one whose importance lies, firstly, in providing a simple problem on which to illustrate basic ideas of statistical inference and, secondly, in providing a simple problem that can be used to compare various competing methods of statistical inference, including decision theory. The practical problem of checking whether a coin is fair might be considered as easily solved by performing a sufficiently large number of trials, but statistics and probability theory can provide guidance on two types of question; specifically those of how many trials to undertake and of the accuracy an estimate of the probability of turning up heads, derived from a given sample of trials.
A fair coin is an idealized randomizing device with two states (usually named "heads" and "tails") which are equally likely to occur. It is based on the coin flip used widely in sports and other situations where it is required to give two parties the same chance of winning. Either a specially designed chip or more usually a simple currency coin is used, although the latter might be slightly "unfair" due to an asymmetrical weight distribution, which might cause one state to occur more frequently than the other, giving one party an unfair advantage.[1] So it might be necessary to test experimentally whether the coin is in fact "fair" – that is, whether the probability of the coin falling on either side when it is tossed is approximately 50%. It is of course impossible to rule out arbitrarily small deviations from fairness such as might be expected to affect only one flip in a lifetime of flipping; also it is always possible for an unfair (or "biased") coin to happen to turn up exactly 10 heads in 20 flips. As such, any fairness test must only establish a certain degree of confidence in a certain degree of fairness (a certain maximum bias). In more rigorous terminology, the problem is of determining the parameters of a Bernoulli process, given only a limited sample of Bernoulli trials.
## Preamble
This article describes experimental procedures for determining whether a coin is fair or not fair. There are many statistical methods for analyzing such an experimental procedure. This article illustrates two of them.
Both methods prescribe an experiment (or trial) in which the coin is tossed many times and the result of each toss is recorded. The results can then be analysed statistically to decide whether the coin is "fair" or "probably not fair".
• Posterior probability density function, or PDF (Bayesian approach). The true probability of obtaining a particular side when a fair coin is tossed is unknown, but the uncertainty is initially represented by the "prior distribution". The theory of Bayesian inference is used to derive the posterior distribution by combining the prior distribution and the likelihood function which represents the information obtained from the experiment. The probability that this particular coin is a "fair coin" can then be obtained by integrating the PDF of the posterior distribution over the relevant interval that represents all the probabilities that can be counted as "fair" in a practical sense.
• Estimator of true probability (Frequentist approach). This method assumes that the experimenter can decide to toss the coin any number of times. He first decides on the level of confidence required and the tolerable margin of error. These parameters determine the minimum number of tosses that must be performed to complete the experiment.
An important difference between these two approaches is that the first approach gives some weight to one's prior experience of tossing coins, while the second does not. The question of how much weight to give to prior experience, depending on the quality (credibility) of that experience, is discussed under credibility theory.
## Posterior probability density function
One method is to calculate the posterior probability density function of Bayesian probability theory.
A test is performed by tossing the coin N times and noting the observed numbers of heads, h, and tails, t. The symbols H and T represent more generalised variables expressing the numbers of heads and tails respectively that might have been observed in the experiment. Thus N = H+T = h+t.
Next, let r be the actual probability of obtaining heads in a single toss of the coin. This is the property of the coin which is being investigated. Using Bayes' theorem, the posterior probability density of r conditional on h and t is expressed as follows:
$f(r | H=h, T=t) = \frac {\Pr(H=h | r, N=h+t) \, g(r)} {\int_0^1 \Pr(H=h |r', N=h+t) \, g(r') \, dr'}. \!$
where g(r) represents the prior probability density distribution of r, which lies in the range 0 to 1.
The prior probability density distribution summarizes what is known about the distribution of r in the absence of any observation. We will assume that the prior distribution of r is uniform over the interval [0, 1]. That is, g(r) = 1. (In practice, it would be more appropriate to assume a prior distribution which is much more heavily weighted in the region around 0.5, to reflect our experience with real coins.)
The probability of obtaining h heads in N tosses of a coin with a probability of heads equal to r is given by the binomial distribution:
$\Pr(H=h | r, N=h+t) = {N \choose h} \, r^h \, (1-r)^t. \!$
Substituting this into the previous formula:
$f(r | H=h, T=t) = \frac{{N \choose h}\,r^h\,(1-r)^t} {\int_0^1 {N \choose h}\,r^h\,(1-r)^t\,dr} = \frac{r^h\,(1-r)^t}{\int_0^1 r^h\,(1-r)^t\,dr} .$
This is in fact a beta distribution (the conjugate prior for the binomial distribution), whose denominator can be expressed in terms of the beta function:
$f(r | H=h, T=t) = \frac{1}{\mathrm{B}(h+1,t+1)} \; r^h\,(1-r)^t. \!$
As a uniform prior distribution has been assumed, and because h and t are integers, this can also be written in terms of factorials:
$f(r | H=h, T=t) = \frac{(h+t+1)!}{h!\,\,t!} \; r^h\,(1-r)^t. \!$
### Example
For example, let N = 10, h = 7, i.e. the coin is tossed 10 times and 7 heads are obtained:
$f(r | H=7, T=3) = \frac{(10+1)!}{7!\,\,3!} \; r^7 \, (1-r)^3 = 1320 \, r^7 \, (1-r)^3 \!$
The graph on the right shows the probability density function of r given that 7 heads were obtained in 10 tosses. (Note: r is the probability of obtaining heads when tossing the same coin once.)
Plot of the probability density f(x | H = 7,T = 3) = 1320 x7 (1 - x)3 with x ranging from 0 to 1.
The probability for an unbiased coin (defined for this purpose as one whose probability of coming down heads is somewhere between 45% and 55%)
$\Pr(0.45 < r <0.55) = \int_{0.45}^{0.55} f(r | H=7, T=3) \,dr \approx 13\% \!$
is small when compared with the alternative hypothesis (a biased coin). However, it is not small enough to cause us to believe that the coin has a significant bias. Notice that this probability is slightly higher than our presupposition of the probability that the coin was fair corresponding to the uniform prior distribution, which was 10%. Using a prior distribution that reflects our prior knowledge of what a coin is and how it acts, the posterior distribution would not favor the hypothesis of bias. However the number of trials in this example (10 tosses) is very small, and with more trials the choice of prior distribution would be somewhat less relevant.)
Note that, with the uniform prior, the posterior probability distribution f(r | H = 7,T = 3) achieves its peak at r = h / (h + t) = 0.7; this value is called the maximum a posteriori (MAP) estimate of r. Also with the uniform prior, the expected value of r under the posterior distribution is
$\operatorname{E}[r] = \int_0^1 r \cdot f(r | H=7, T=3) \, \mathrm{d}r = \frac{h+1}{h+t+2} = \frac{2}{3}\,.$
## Estimator of true probability
The best estimator for the actual value $r\,\!$ is the estimator $p\,\! = \frac{h}{h+t}$. This estimator has a margin of error (E) where $|p - r| < E$ at a particular confidence level.
Using this approach, to decide the number of times the coin should be tossed, two parameters are required:
1. The confidence level which is denoted by confidence interval (Z)
2. The maximum (acceptable) error (E)
• The confidence level is denoted by Z and is given by the Z-value of a standard normal distribution. This value can be read off a standard score statistics table for the normal distribution. Some examples are:
Z value Confidence Level Comment
0.6745 gives 50.000% level of confidence Half
1.0000 gives 68.269% level of confidence One std dev
1.6449 gives 90.000% level of confidence "One Nine"
1.9599 gives 95.000% level of confidence 95 percent
2.0000 gives 95.450% level of confidence Two std dev
2.5759 gives 99.000% level of confidence "Two Nines"
3.0000 gives 99.730% level of confidence Three std dev
3.2905 gives 99.900% level of confidence "Three Nines"
3.8906 gives 99.990% level of confidence "Four Nines"
4.0000 gives 99.993% level of confidence Four std dev
4.4172 gives 99.999% level of confidence "Five Nines"
• The maximum error (E) is defined by $|p - r| < E$ where $p\,\!$ is the estimated probability of obtaining heads. Note: $r$ is the same actual probability (of obtaining heads) as $r\,\!$ of the previous section in this article.
• In statistics, the estimate of a proportion of a sample (denoted by p) has a standard error (standard deviation of error) given by:
$s_p = \sqrt{ \frac {p \, (1-p) } {n} }$
where n is the number of trials (which was denoted by N in the previous section).
This standard error $s_p$ function of p has a maximum at $p = (1-p) = 0.5$. Further, in the case of a coin being tossed, it is likely that p will be not far from 0.5, so it is reasonable to take p=0.5 in the following:
$s_p\,\!$ $= \sqrt{ \frac {p \, (1-p) } {n} } = \sqrt{ \frac {0.5 \times 0.5 } {n} } = \frac {1}{2 \, \sqrt{n}}$
And hence the value of maximum error (E) is given by
$E= Z \, s_p = \frac {Z}{2 \, \sqrt{n}}$
Solving for the required number of coin tosses, n,
$n = \frac {Z^2} {4 \, E^2} \!$
### Examples
1. If a maximum error of 0.01 is desired, how many times should the coin be tossed?
$n = \frac {Z^2} {4 \, E^2} = \frac {Z^2} {4 \times 0.01^2} = 2500 \ Z^2$
$n = 2500\,$ at 68.27% level of confidence (Z=1)
$n = 10000\,$ at 95.45% level of confidence (Z=2)
$n = 27225\,$ at 99.90% level of confidence (Z=3.3)
2. If the coin is tossed 10000 times, what is the maximum error of the estimator $p\,\!$ on the value of $r\,\!$ (the actual probability of obtaining heads in a coin toss)?
$E = \frac {Z}{ 2 \, \sqrt{n} }$
$E = \frac {Z}{ 2 \, \sqrt{ 10000 } } = \frac {Z}{ 200 }$
$E = 0.0050\,$ at 68.27% level of confidence (Z=1)
$E = 0.0100\,$ at 95.45% level of confidence (Z=2)
$E = 0.0165\,$ at 99.90% level of confidence (Z=3.3)
3. The coin is tossed 12000 times with a result of 5961 heads (and 6039 tails). What interval does the value of $r\,\!$ (the true probability of obtaining heads) lie within if a confidence level of 99.999% is desired?
$p = \frac{h}{h+t} \, = \frac{5961}{12000} \, = 0.4968$
Now find the value of Z corresponding to 99.999% level of confidence.
$Z = 4.4172 \,\!$
Now calculate E
$E = \frac{Z}{2 \, \sqrt{n}} \, = \frac{4.4172}{2 \, \sqrt{12000}} \, = 0.0202$
The interval which contains r is thus:
$p - E < r < p + E \,\!$
$0.4766 < r < 0.5170 \,\!$
Hence, 99.999% of the time, the interval above would contain $r\,\!$ which is the true value of obtaining heads in a single toss.
## Other approaches
Other approaches to the question of checking whether a coin is fair are available using decision theory, whose application would require the formulation of a loss function or utility function which describes the consequences of making a given decision. An approach that avoids requiring either a loss function or a prior probability (as in the Bayesian approach) is that of "acceptance sampling".[2]
## Other applications
The above mathematical analysis for determining if a coin is fair can also be applied to other uses. For example:
• Determining the proportion of defective items for a product subjected to a particular (but well defined) condition. Sometimes a product can be very difficult or expensive to produce. Furthermore, if testing such products will result in their destruction, a minimum number of items should be tested. Using a similar analysis, the probability density function of the product defect rate can be found.
• Two party polling. If a small random sample poll is taken where there are only two mutually exclusive choices, then this is similar to tossing a single coin multiple times using a possibly biased coin. A similar analysis can therefore be applied to determine the confidence to be ascribed to the actual ratio of votes cast. (Note that if people are allowed to abstain then the analysis must take account of that, and the coin-flip analogy doesn't quite hold.)
• Determining the sex ratio in a large group of an animal species. Provided that a small random sample (i.e. small in comparison with the total population) is taken when performing the random sampling of the population, the analysis is similar to determining the probability of obtaining heads in a coin toss.
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2015-07-31 10:35:59
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https://stats.stackexchange.com/questions/258490/hypothesis-testing-on-a-binomial-probability-ratio
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# Hypothesis testing on a binomial probability ratio
A colleague has asked me to assess the data analytic approach taken by one of his students for a previous experiment. If you want to skip the details of the experimental design skip down to Data analysis Part 2 and My questions below.
Experimental design. In its simplest form, the experiment consists of a single subject completing a single session of 100 trials. In each session, the subject is first presented with a pre-testing set of stimuli that conform to a particular pattern. For the subsequent 100 trials, the subject is then presented with more stimuli and is asked (yes/no) whether they are consistent with the pre-test pattern. Although the stimuli were generated according to a fixed rule, the subject is not given the rule and therefore must infer and apply a rule based upon the observed pre-test pattern.
Hypotheses. The student has two goals for statistical analysis. The first is the obvious hypothesis test of whether the subject's performance is consistent with the actual stimuli-generating rule. The second set of hypotheses concern whether the subject's performance is also consistent with a variety of other possible rules hypothesized a priori by the experimenter. In particular, the student wants to know what the likelihoods of other possible rules are relative to the most likely rule for the subject.
Data analysis.
Part 1. The subject's data are treated as a random sample from a binomial process. The student assumed a 5% performance error rate for the subject, such that the probability of success (correct yes/no answer) for each trial is $θ = 0.95$.
For the first hypothesis, the student conducted a log-likelihood ratio test comparing the subject's observed performance ($\hat{θ}$) to the assumed population parameter ($θ$). The subject answered correctly on 93% ($k=93$) of the trials ($n=100$). The log-likelihood ratio is therefore:
$$2[k\ln\frac{\hat{θ}}{θ}+(n-k)\ln\bigg(\frac{1-\hat{θ}}{1-θ}\bigg)]=2[93\ln\frac{0.93}{0.95}+(7)\ln\bigg(\frac{0.07}{0.05}\bigg)]=0.753$$
The student cannot reject the null hypothesis ($H_0:\hat{θ}=θ)$ for $\alpha=0.05$ and therefore concludes that the subject's performance is consistent with the stimulus generating rule. This approach seems appropriate to me, although it may be questionable to assume each trial is i.i.d. within the session for a single subject. Nonetheless, I'm more concerned with the student's approach for the second set of hypotheses.
Part 2. The student is now interested in asking, given the subject's performance, what is the likelihood that the subject's performance is also consistent with a set of other possible rules? To answer this question, the student again fixes the population parameter, reflecting his expectation that the probability of answering correctly relative to any of the hypothesized rules is constant. That is to say, even if the stimulus set was generated according to rule A, the subject answering according to rule B is just as likely to select yes for any trials consistent with B (and vice versa).
In order to compare the rules, the student therefore had to recalculate the subject's $k$ for every hypothesized rule. For example, $k=20$ for rule B if the subject's performance on 20/100 trials generated with rule A was also consistent with B. Thus, the student generated a 'new' dataset for every rule, albeit assuming each to be drawn from the same underlying binomial process with distribution parameters $θ=0.95$ and $n=100$. The student then computes a ratio for every rule relative to the rule with the highest likelihood. For example, assume rule A has the highest success rate at $k=93$ and $k=89$ for rule B. The student would then calculate what he believed to be a relative likelihood ratio:
$$\exp([\ln\binom{100}{89}+(89)\ln(0.95)+(11)\ln(0.05)])- \\ \qquad\quad\;[\ln\binom{100}{93}+(93)\ln(0.95)+(7)\ln(0.05)]=0.068$$
My questions. Given that the population parameters are held constant, this is not a relative likelihood ratio but rather a probability ratio for $Pr(X=89)/Pr(X=93)$. Moreover, by holding the population parameters constant, my thinking is that there are zero degrees of freedom for the comparison. This would make the probability ratio meaningless for hypothesis testing with the $\chi^2$ distribution, as is done with the log-likelihood ratio.
Is my thinking here correct? If so, are there heuristic criteria or alternative approaches by which the magnitude of such a probability ratio could be interpreted? I've already thought of using simple $\chi^2$ goodness-of-fit or Fisher's exact tests for observed vs. expected yes/no responses, which would avoid these issues entirely and not require the 'generation' of new $k$ for every rule. Any other possible solutions would also be much appreciated. Thank you in advance.
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2019-08-24 03:17:41
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https://www.physicsforums.com/threads/using-the-joukowski-transform-to-calculate-flow-around-cylinder-due-to-point-charges.577197/
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# Using the Joukowski transform to calculate flow around cylinder due to point charges
1. Feb 13, 2012
### meldraft
Hey all,
I am using the Joukowski transform to calculate the electrostatic potential and streamlines around an elliptical cylinder. The flow is caused by two point charges (one positive, one negative). My idea was, of course, to first solve the problem for a circular cylinder, and then transform everything using the Joukowski transform.
I am getting some weird results, so I would be grateful is anyone can point out the mistake in my calculations!
The equation for the complex potential of a point charge located at point z0=a is:
$$φ_1=\frac{Λ}{2 \pi}ln(z-a)$$
where Λ is the strength of the source (or sink), and z,a are complex numbers.
Also, the potential to simulate a circular cylinder originating at (0,0), is that of a doublet:
$$φ_c=-\frac{μ}{z}$$
where μ is a real constant. Therefore, the potential function for my problem should be:
$$Φ=\frac{Λ}{2 \pi}ln(z-a)-\frac{Λ}{2 \pi}ln(z+a)-\frac{μ}{z}$$
The field looks realistic at this point.
Then, I do the Joukowski transform:
$$J=z+\frac{λ^2}{z}$$
$$λ=\sqrt{a*R-R^2}$$
$$R=(a+b)/2$$
where a,b are the 2 semi-axes of the ellipse and R is the radius of the original circular cylinder. I know that this is correct, because i can see the ellipse in the plot. My problem is that the field around it is not what I would expect:
1. The potential lines are no longer always perpendicular to the boundary of the ellipse
2. The conformal map is causing the source and sink to come closer to point (0,0), because of the distortion caused by the growing ellipse.
I have no idea why either of them is happening, as I figured the transform should continue to satisfy the original boundary conditions.
The way that I plot the potential is the following:
contour(real(J),imag(J),-real(f))
so, basically, the original potential corresponds to the transformed mesh. I have also set the axes to always be equal to each other, so what I see is not distorted due to axis scaling.
I have been trying to figure out where the mistake is for a few days now, and I really am stuck. Can anyone spot what I have done wrong?
P.S. I am also attaching two pictures, one for the cylinder, where everything is normal, and one for the problematic transform. You can see how the source and sink have moved away from their origins at (-4,0) and (4,0), and that the potential lines are no longer perpendicular to the boundary of the ellipse.
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2. Feb 13, 2012
### marcusl
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
Here are a few points to keep in mind:
1. 2D conformal maps are not appropriate to finding fields from point charges (I assume you know this). The problem you are working is ok for sources that are uniformly charged lines perpendicular to the z plane.
2. Regarding motion of the sources, you need to predistort their positions so they end up in the right place in the transformed plane.
3. You don't say want the cylinder is. I assume it's perfectly conducting?
4. How do you do the transform? (Did you invert to find z in terms of J)?
3. Feb 13, 2012
### meldraft
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
1. Actually, I've only ever used conformal maps for fluid mechanics problems (so for uniform flows of the form z=v*exp(iθ)). I was not aware that they are not appropriate for this type of modelling. Can you elaborate? I am very curious since all of my books describe the uniform flow situation with added sources or sinks, and never a dipole (like my problem) on its own. They just don't mention it, however, I never read that it was not suitable (although I'm kind of inferring it from experience).
Btw, the straight perpendicular lines you saw at (-4,0),(4,0) are auxiliary lines I ploted to keep track of how off the sources are from their original position. They are not part of the problem.
2. Thanks, I'll do that.
3. The cylinder in this case is a perfect insulator. I want the flow to go completely around it. Once I have this, I want to try out the problem with partial conductivity.
4. I calculated J(z) through the Joukowski equation, and then plotted the original potentials in the new mesh ( so newX=real(J(z)) and newY=imag(J(z)) ).
4. Feb 13, 2012
### marcusl
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
1. Conformal maps are, by definition, two dimensional. The assumption is that everything looks the same in and out of the paper (call it the x direction), from x = - to + infinity. Point charges are, obviously, not uniform along x and cannot be analyzed by conformal mapping. Whether you intended to or not, you are analyzing two uniform infinitely long lines of charge next to an infinitely long cylinder.
3. You don't mean perfect insulator. Vacuum is a perfect insulator, and it corresponds to no cylinder at all! You have the choice of perfect conductor or infinite dielectric constant. They'll give the same field contours outside--the field lines will terminate normally on the cylinder and the equipotential lines will look like the fluid flow contours you are visualizing--but keep in mind that nothing is actually flowing here. (A minor point: a conductor has no field inside while the dielectric will conduct the field lines through the cylinder).
Something more--I wonder if you have properly described your problem. If I examine the circle
$$z=\lambda \exp(i2\pi\phi),$$
it transforms to a branch cut extending between the foci of the ellipse in the J plane. I think you need to surround this circle with another larger one that maps to the desired elliptic boundary. This means that the dielectric in the z plane is an annulus, and is a uniformly filled ellipse with a branch cut between the foci in the J plane.
In other words, I think the problem is more complicated that what you set up....
5. Feb 14, 2012
### meldraft
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
1. Yes, I understand what you mean. It was a bad description on my part. I assume that these 'point charges' are actually infinitely long cylinders, whose projection in the complex plane is what I see in the plot, that was why I called them point charges.
3. My physical problem is a hole filled with air, so what I actually meant is that, basically (in polar coordinates):
$$\nabla{Φ(R,\theta)}\cdot \bar{r}=0$$
where R is the radius of the cylinder and $$\bar{r}$$ is the r unit vector (so essentially the intensity vectors are always tangent to the boundary)
I'm not sure that I understand why an annulus would be a better fit. I think you mean that the problem is connected to the branch cuts, but in either case they would still be enclosed by the cylinder correct?
The transform works perfectly for uniform flow, so if I have an ellipse in uniform flow, the field looks as it should. Therefore, shouldn't the problem be connected to the fact that I am using only sources and sinks?
6. Feb 14, 2012
### marcusl
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
Good, we're on the same page.
I'm confused here. Air (or vacuum) does not affect electric fields/potentials, so they'll pass right through your cylinder. Can you better define the problem you are trying to solve?
I'll need to stop and think about this--I haven't look at a conformal map since the mid-90's. Won't be able to devote much time until this weekend, however.
7. Feb 14, 2012
### meldraft
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
Take your time, I am grateful that you spend any time at all to answer my questions
I am modelling a solid conductor with a hole. The conductor is roughly 2kOhms per square centimeter, and the current and voltage are both very small. Thus, the current is not allowed to pass through the hole (air), neither radiate. Realistically, this field is so weak on air that I can neglect it and consider Neumann boundary conditions on the boundary of the hole. That is why I considered it as a perfect insulator.
I have read that both Neumann and Dirichlet BCs are invariant under conformal mappings. Thus, if I impose a Neumann BC on a unit disc, it should be carried over to whatever shape the disc is transformed to. This is why is think I am not applying the theory correctly
8. Feb 14, 2012
### marcusl
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
Oh, your sources are in a conductor and the ellipse is a hole! I didn't get that at all before--no wonder my comments about dielectric constant made no sense.
9. Feb 15, 2012
### meldraft
Re: Using the Joukowski transform to calculate flow around cylinder due to point char
Thank you I'll also post any updates as I work on this.
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2018-05-25 09:37:30
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http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-12th-edition/chapter-4-section-4-4-multiplying-polynomials-4-4-exercises-page-305/71
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## Intermediate Algebra (12th Edition)
$0.1x+0.63x-0.13$
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the FOIL Method, the product of the binomials in the given expression, $(0.2x+1.3)(0.5x-0.1) ,$ is \begin{array}{l}\require{cancel} 0.2x(0.5x)+0.2x(-0.1)+1.3(0.5x)+1.3(-0.1) \\\\= 0.1x-0.02x+0.65x-0.13 \\\\= 0.1x+0.63x-0.13 .\end{array}
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2018-01-19 23:55:09
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https://brilliant.org/discussions/thread/complex-exponent/
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×
# Complex exponent
Can anyone tell me the value of i^i
Note by Naman Kapoor
1 year, 10 months ago
Sort by:
i can be written as $${ e }^{ (i\frac { \pi }{ 2 } ) }$$ so this expression can be written as $${ e }^{ (i\frac { \pi }{ 2 } )i }$$
Now $${ i }^{ 2 }$$ =-1 so it is equal to $${ e }^{ \frac { -\pi }{ 2 } }$$. · 1 year, 10 months ago
Thnxx buddy · 1 year, 9 months ago
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2016-10-22 21:37:17
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http://www.ism.ac.jp/editsec/aism/43/28.html
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### AMARJOT KAUR AND HARSHINDER SINGH
Department of Statistics, Panjab University, Chandigarh 160014, India
(Received April 3, 1989; revised February 14, 1990)
Abstract. Let random samples of equal sizes be drawn from two exponential distributions with ordered means lambdai. The maximum likelihood estimator lambdai* of lambdai is shown to have a smaller mean square error than that of the usual estimator \bar Xi, for each i = 1, 2. The asymptotic efficiency of lambdai* relative to \bar Xi has also been found.
Key words and phrases: Asymptotic efficiency, exponential distribution, isotonic regression, maximum likelihood estimation, mean square error.
Source ( TeX , DVI , PS )
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2017-11-20 09:25:25
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https://xiuchuanz.com/2018/07/Machine_Learning/
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# Xiuchuan Zhang
Personal Website
This is Xiuchuan's personal website.
I plan to post some of my current learning and review notes on it.
If you have any questions or suggestions, welcome to comment in my posts.
Coursera.org 机器学习 笔记
These are the notes by learning Andrew Ng’s “Machine Learning” from Coursera.org for future review
## Machine Learning
• Grew out of work in AI
• New capability for computers
Examples:
Database mining
Large datasets from growth of automation/ web
E.g., Web click data, medical records, biology, engineering
Applications cannot program by hand
E.g., Autonomous helicopter, handwriting recognition, most of Natural Language Processing (NLP), Computer Vision
Self-customizing programs
E.g., Amazon, Netflix product recommendations
Understanding human learning (brain, real AI)
### Introduction
• Definition
• Arthur Samuel (1959). Machine Learning: Field of study that gives computers the ability to learn without being explicitly programmed
• Tom Mitchell (1995) Well-posed Leaning Problem: A computer program is said to learn from experience E with respect to some task T and some performance measure P, if its performance on T, as measured by P, improves with experience E.
## Machine Learning Algorithms:
### Supervised learning
• Regression: Predict continuous valued output
• Classification: Discrete valued output (0 or 1 or others)
### Unsupervised learning
• allows us to approach problems with little or no idea what our results should look like.
• We can derive structure from data where we do not necessarily know the effect of the variables.
• We can derive this structure by clustering the data based on relationships among the variables in the data
• There is no feedback based on the prediction results
• E.g., Organize computing clusters, social network analysis, market segmentation, astronomical data analysis
### Model representation
• Notation
• m = Number of training examples
• x’s = “input” variable / features
• y’s = “output” variable / “target” variable
• Hypothesis
• Training set → Learning Algorithm → Hypothesis (h)
• xh → estimated value of y
• h: X → Y
• h maps x’s to y’s
• Linear regression with one variable or Univariate linear regression
• $h_{\theta}(x) = \theta_{0} + \theta_{1}x$
• shorthand: $h(x)$
• $\theta_{i}$’s: parameters
### Cost function
• Idea: Choose $\theta_{0}, \theta_{1}$ so that $h_{\theta}(x)$ is close to y for our training examples (x,y)
• $\underset{ \theta_{0}, \ \theta_{1} }{ \mathrm{minimize} }$ $\dfrac{1}{2m} \displaystyle\sum_{i=1}^m ( \hat{y}^{i}- y^{i} )^2 =$ $\dfrac{1}{2m} \displaystyle\sum _{i=1}^m (h_\theta (x^{i}) - y^{i} )^2$
• $h_{\theta}(x^{i}) = \theta_{0} + \theta_{1}x^{i}$
• Cost function: $J(\theta_0, \theta_1) =$ $\displaystyle \dfrac{1}{2m} \sum _{i=1}^m ( \hat{y}^{i}- y^{i} )^2 = \dfrac{1}{2m} \sum _{i=1}^m (h_\theta (x^{i}) - y^{i} )^2$
• Octave: J = 1/(2*m)*sum((X*theta .- y).^2)
• also called squared error cost function or squared error function
• Goal: $\displaystyle \underset{ \theta_{0}, \ \theta_{1} }{ \mathrm{minimize} } J(\theta_0, \theta_1)$
• contour plots or contour figures
• Outline:
• Start with some $\theta_{0}, \ \theta_{1}$
• Keep changing $\theta_{0}, \ \theta_{1}$ to reduce $J(\theta_{0}, \theta_{1})$ until we hopefully end up at a minimum
• Algorithm
• repeat until convergence { $\theta_j := \theta_j - \alpha \displaystyle \frac{\partial}{\partial \theta_j} J(\theta_0, \ \theta_1) \ \ (\mathrm{for} \ j = 0 \ \mathrm{and} \ j = 1 )$ }
• α: learning rate: too small → slow ; too large → overshoot; proper → automatically decrease α over time
• Correct simultaneous update
• temp0 := $\theta_0 - \alpha \displaystyle \frac{\partial}{\partial \theta_0} J(\theta_0, \ \theta_1) \ \ (\mathrm{for} \ j = 0 \ \mathrm{and} \ j = 1 )$
• temp1 := $\theta_1 - \alpha \displaystyle \frac{\partial}{\partial \theta_1} J(\theta_0, \ \theta_1) \ \ (\mathrm{for} \ j = 0 \ \mathrm{and} \ j = 1 )$
• $\theta_0 :=$ temp0
• $\theta_1 :=$ temp1
• linear regression:
• repeat until convergence:
• $\theta_0 := \theta_0 - \alpha \frac{1}{m} \sum\limits_{i=1}^{m}(h_\theta(x_{i}) - y_{i})$
• $\theta_1 := \theta_1 - \alpha \frac{1}{m} \sum\limits_{i=1}^{m}((h_\theta(x_{i}) - y_{i}) x_{i})$
• Octave: theta = theta - alpha/m*(X'*(X*theta .- y))
• Batch”: Each step of gradient descent uses all the training examples
## Multivariate Linear Regression
### Multiple Features (variables)
• n = number of features
• $x^{(i)} \text{ for } (i\in { 1,\dots, m } )$ = input (features) of $i^{th}$ training examples
• $x^{i} = [ x_{1}^{i} \ x_{2}^{i} \ \dots \ x_{n}^{i} ]^{T} \in \mathbb{R}^{n}$
• $x_{j}^{(i)}$ = value of $j^{th}$ features in $i^{th}$ training examples
• Hypothesis: $h_{\theta}(x^{i}) = \theta_{0} + \theta_{1}x_{1}^{i} + \theta_{2}x_{2}^{i} + \dots + \theta_{n}x_{n}^{i}$
• $x = [ x_{0} \ x_{1} \ x_{2} \ \dots \ x_{n} ]^{T} \in \mathbb{R}^{n+1} \ \ (x_{0}^{i} = 1)$
• $\theta = [ \theta_{0} \ \theta_{1} \ \theta_{2} \ \dots \ \theta_{n} ]^{T} \in \mathbb{R}^{n+1}$
• $h_{\theta}(x) = \theta_{0}x_{0} + \theta_{1}x_{1} + \theta_{2}x_{2} + \dots + \theta_{n}x_{n} \ \ (x_{0} = 1)$
$\ \ \ \ \ = \theta^{T}x$
• Parameters: $\theta_{0}, \theta_{1}, \dots , \theta_{n}$
• Cost function: $J(\theta) = J(\theta_{0}, \theta_{1}, \dots , \theta_{n}) =$ $\dfrac{1}{2m} \displaystyle\sum_{i=1}^m (h_\theta (x^{i}) - y^{i} )^2$
$\hspace{3em} J(\theta) = \dfrac{1}{2m} \displaystyle\sum_{i=1}^{m}((\sum_{j=0}^{n}\theta_j x^{(i)}_j) - y^{(i)})^2$
Repeat {
$\theta_j := \theta_j - \alpha \displaystyle \frac{\partial}{\partial \theta_j} J(\theta)$
} (simultaneously update for every $j = 0, \dots, n$ )
• New algorithm (n ≥ 1):
Repeat {
$\theta_j := \theta_j - \alpha \displaystyle \frac{1}{m} \sum\limits_{i=1}^{m}(h_\theta(x^{(i)}) - y^{(i)}) \cdot x_{j}^{(i)}$
} (simultaneously update $\theta_j$ for $j = 0, \dots, n$ )
• Features Scaling
• Let $x = \displaystyle \frac{x}{ \mathrm{range} S }$
to get every feature into approximately a $-1 \leq x \leq 1$ range
• Mean normalization: replace $x_j \mathrm{with} \frac{x_j - \mu_j}{ s_{j} }$ (Do not apply to $x_0 = 1$)
into $-0.5 \leq x_j \leq 0.5$ range
• Learning Rate choose
• “Debugging”: plot ( $\displaystyle\underset{ \theta }{ \mathrm{min} } \ J(\theta)$, No. of iterations)
$J(\theta)$ should decrease after every iteration → converge (sufficiently α)
• Not working → Use smaller α
slow converge → use larger α
• Create new feature
• Change the behavior or curve of our hypothesis function by making it a quadratic, cubic or square root function or any other form (Features Scaling)
### Normal equation
• Definition: Method to solve for θ analytically
• Formula: $\displaystyle\theta = (X^{T} X)^{-1} X^{T} y$
• Octave: pinv(x'*x)*x'*y
• No need feature scaling
• If n is very large, it works very slow, by compute $n^3$
## Classification
• Binary classification problem: y = 0 or 1
• Use linear regression:
• To map all predictions greater than 0.5 as a 1 and all less than 0.5 as a 0
• $h_\theta(x)$ can be >1 or <0, so it does not work well
• Logistic Regression: make $0 \le h_\theta(x) \le 1$
### Logistic Regression
• Clasification and Representation
• Sigmoid(Logistic) function: $g(Z) = \displaystyle\frac{1}{1+e^{-z}}$
• Octave: h = 1.0 ./ (1.0 + exp(-z))
• $h_\theta(x) = g(\theta^{T}x) = \displaystyle\frac{1}{1+e^{-\theta^{T}x}}$
• $h_\theta(x) =$ estimated probability that y = 1 on input x
• $h_\theta(x) = P(y=1 \vert x ; \theta) = 1 - P(y=0 \vert x ; \theta)$
• Decision boundary
• To get discrete 0 or 1 classification, can translate the output of the hypothesis function as:
• $h_\theta(x) \geq 0.5 \rightarrow y = 1$
• $h_\theta(x) \leq 0.5 \rightarrow y = 0$
• $g(z) \geq 0.5$ when $z \geq 0$ that $h_\theta(x) = g(\theta^T x) \geq 0.5$ when $\theta^T x \geq 0$
• From these statements we can say:
• $\theta^T x \geq 0 \Rightarrow y = 1$
• $\theta^T x \leq 0 \Rightarrow y = 0$
• Cost function
• $J(\theta) = \displaystyle\dfrac{1}{m} \sum_{i=1}^m \mathrm{Cost}(h_\theta(x^{(i)}),y^{(i)})$
• $\mathrm{Cost}(h_\theta(x),y) = -\log(h_\theta(x))$ if y = 1
• $\mathrm{Cost}(h_\theta(x),y) = -\log(1-h_\theta(x))$ if y = 0
• Compress: $\displaystyle \mathrm{Cost}(h_\theta(x),y) = -y\log(h_\theta(x)) -(1-y)\log(1-h_\theta(x))$
• Vectorized implementation:
$h = g(X\theta)$
$\displaystyle J(\theta) = \frac{1}{m} \cdot (-y^{T}\log(h)-(1-y)^{T}\log(1-h))$
Octave: the cost: J = 1/m * (-y' * log(h)-(1-y)'*log(1-h))
the partial derivatives: grad = 1/m * X'*(h-y)
• Repeat {
general form:
$\displaystyle \theta_j := \theta_j - \alpha \dfrac{\partial}{\partial \theta_j}J(\theta)$
derivative part using calculus:
$\displaystyle \theta_j := \theta_j - \frac{\alpha}{m} \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)}) x_j^{(i)}$
}
• Vectorized implementation:
$\displaystyle \theta := \theta - \frac{\alpha}{m} X^T (g(X\theta) - \vec y)$
• Optimization algorithm
• BFGS
• L-BFGS
• Codes in Octave:
function [jVal, gradient] = costFunction(theta)
jVal = [...code to compute J(theta)...];
gradient = [...code to compute derivative of J(theta)...]; end
• use octave’s ‘fminunc’ optimization algorithm
options = optimset('GradObj', 'on', 'MaxIter', 100);
initialTheta = zeros(2,1);
[optTheta, functionVal, exitFlag] = fminunc(@costFunction, initialTheta, options);
• Multiclass Classification: One-vs-all
• Train a logistic regression classifier $h_\theta^{(i)}(x)$ for each class i to predict the probability that $y = i$
$h_\theta^{(i)}(x) = P(y = i| x;\theta) (i=1,2,\dots,n)$
• On a new input x, to make a prediction, pick the class i that maximizes
prediction = $\displaystyle\underset{i}{\mathrm{max}} h_\theta^{(i)}(x)$
## Overfitting
• “Underfit” (“high bias”): not fit the training data very well
• “Just right”
• “Overfit” (“high variance”): may fit the trainning set very well, but fail to generalize to new examples
### Regularization
• Using the cost function with the extra summation, we can smooth the output of our hypothesis function to reduce overfitting.
$min_\theta\, \dfrac{1}{2m}\, \displaystyle[\sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})^2 + \lambda \sum_{j=1}^n \theta_j^2]$
• $\lambda$, or lambda is the regularization parameter
• If lambda is too large, it may cause underfitting.
### Linear Regression
Repeat {
$\displaystyle\theta_0 := \theta_0 - \alpha\, \frac{1}{m} \displaystyle\sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_0^{(i)}$
$\displaystyle\theta_j := \theta_j - \alpha\, [ ( \frac{1}{m}\, \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)}) + \frac{\lambda}{m}\theta_j ]$ $\displaystyle\quad j \in \lbrace 1,2…n\rbrace$
$\displaystyle\Rrightarrow \theta_j := \theta_j (1- \alpha\,\frac{\lambda}{m}) - \alpha\, \frac{1}{m}\, \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)})$
}
• Normal Equation:
$\displaystyle\theta = ( X^TX + \lambda \cdot L )^{-1} X^Ty$
$\text{where}\, \, L =$ $% $
### Logistic Regression
• Cost function:
• $J(\theta) = - \displaystyle\frac{1}{m} \sum_{i=1}^m [ y^{(i)}\, \log (h_\theta (x^{(i)})) + (1 - y^{(i)})\, \log (1 - h_\theta(x^{(i)}))] + \frac{\lambda}{2m}\sum_{j=1}^n \theta_j^2$
• $\sum_{j=1}^n \theta_j^2$ means to explicitly exclude the bias term $\theta_0$
• Octave:
• the cost:
J = 1/m * (-y' * log(h)-(1-y)'*log(1-h)) + (lambda/(2*m))*theta'*theta
• the partial derivatives:
grad = 1/m * X'*(h-y) + (lambda/m)*theta
• $\displaystyle h_\theta(x) = \frac{1}{1+e^{-\theta^{T}x}}$
Repeat {
$\displaystyle\theta_0 := \theta_0 - \alpha\, \frac{1}{m} \displaystyle\sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_0^{(i)}$
$\displaystyle\theta_j := \theta_j (1- \alpha\,\frac{\lambda}{m}) - \alpha\, \frac{1}{m}\, \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)})$ $\displaystyle\quad j \in \lbrace 1,2…n\rbrace$
}
## Neural Networks
• Origins: Algorithms that try to mimic the brain
### Model representation(NN)
• Neuron model: Logistic unit
$\begin{bmatrix}x_0 \newline x_1 \newline x_2 \newline \end{bmatrix}\rightarrow\begin{bmatrix}\ \ \ \newline \end{bmatrix}\rightarrow h_\theta(x)$
• Sigmoid (logistic) activation function: $g(Z) = \displaystyle\frac{1}{1+e^{-z}}$
• Parameters $\theta$ also called “weights”
• 1st layer called “input layer”; last layer called “output layer”; middle layers called “hidden layer”
• Notations
• $x_0 = 1$ is a “bias unit”
$\begin{bmatrix}x_0 \newline x_1 \newline x_2 \newline x_3\end{bmatrix}\rightarrow\begin{bmatrix}a_1^{(2)} \newline a_2^{(2)} \newline a_3^{(2)} \newline \end{bmatrix}\rightarrow h_\theta(x)$
• $a_i^{(j)} =$ “activation” of unit $i$ in layer $j$
• $\Theta^{(j)} =$ matrix of weights controlling function mapping from layer $j$ to layer $j+1$
\begin{align*} \displaystyle a_1^{(2)} = g(\Theta_{10}^{(1)}x_0 + \Theta_{11}^{(1)}x_1 + \Theta_{12}^{(1)}x_2 + \Theta_{13}^{(1)}x_3) \newline a_2^{(2)} = g(\Theta_{20}^{(1)}x_0 + \Theta_{21}^{(1)}x_1 + \Theta_{22}^{(1)}x_2 + \Theta_{23}^{(1)}x_3) \newline a_3^{(2)} = g(\Theta_{30}^{(1)}x_0 + \Theta_{31}^{(1)}x_1 + \Theta_{32}^{(1)}x_2 + \Theta_{33}^{(1)}x_3) \newline h_\Theta(x) = a_1^{(3)} = g(\Theta_{10}^{(2)}a_0^{(2)} + \Theta_{11}^{(2)}a_1^{(2)} + \Theta_{12}^{(2)}a_2^{(2)} + \Theta_{13}^{(2)}a_3^{(2)}) \newline \end{align*}
• If network has $s_j$ units in layer $\displaystyle j, s_{j+1}$ units in layer $j + 1$, then $\Theta^{(j)}$ will be of dimension $\displaystyle s_{j+1} \times (s_j + 1)$ that
$\Theta^{(j)} \in \displaystyle\Re^{s_{(j+1\ )}\ \times (s_j \ + 1)}$
• Forward propagation: Vectorized implementation
$\displaystyle z_k^{(2)} = \Theta_{k,0}^{(1)}\ x_0 + \Theta_{k,1}^{(1)}\ x_1 + \cdots + \Theta_{k,n}^{(1)}\ x_n$
%
• $a^{(1)} = x$
• $\displaystyle z^{(j)} = \Theta^{(j-1)} \ a^{(j-1)}$
\begin{align*}\displaystyle a_1^{(2)} = g(z_1^{(2)}) \newline a_2^{(2)} = g(z_2^{(2)}) \newline a_3^{(2)} = g(z_3^{(2)}) \newline \end{align*}
• $\hookrightarrow \displaystyle a^{(j)} = g(z^{(j)} )$
• Add $\displaystyle a_0^{(j)} = 1$
• $z^{(j+1)} = \Theta^{(j)} a^{(j)}$
• $\displaystyle h_\Theta(x) = \ a^{(j+1)} = \ g(z^{(j+1)}\ )$
• Other network architectures
## To be continued
• others: Reinforcement learning, recommender systems
• Practical advice for applying learning algorithms
Support
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2023-04-02 03:14:24
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http://hal.in2p3.fr/in2p3-01328587
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# Search for the Standard Model Higgs boson produced by vector-boson fusion in 8 TeV pp collisions and decaying to bottom quarks with the ATLAS detector
Abstract : A search with the ATLAS detector is presented for the Standard Model Higgs boson produced by vector-boson fusion and decaying to a pair of bottom quarks, using 20.2 ${fb}^{-1}$ of LHC proton--proton collision data at $\sqrt{s}$ = 8 TeV. The signal is searched for as a resonance in the invariant mass distribution of a pair of jets containing $b$-hadrons in vector-boson-fusion candidate events. The yield is measured to be $-0.8 \pm 2.3$ times the Standard Model cross-section for a Higgs boson mass of 125 GeV. The upper limit on the cross-section times the branching ratio is found to be 4.4 times the Standard Model cross-section at the 95% confidence level, consistent with the expected limit value of 5.4 (5.7) in the background-only (Standard Model production) hypothesis.
Document type :
Journal articles
http://hal.in2p3.fr/in2p3-01328587
Contributor : Claudine Bombar <>
Submitted on : Wednesday, June 8, 2016 - 11:18:00 AM
Last modification on : Wednesday, September 16, 2020 - 4:30:51 PM
### Citation
M. Aaboud, Z. Barnovska, N. Berger, M. Delmastro, L. Di Ciaccio, et al.. Search for the Standard Model Higgs boson produced by vector-boson fusion in 8 TeV pp collisions and decaying to bottom quarks with the ATLAS detector. Journal of High Energy Physics, Springer Verlag (Germany), 2016, 11, pp.112. ⟨10.1007/JHEP11(2016)112⟩. ⟨in2p3-01328587⟩
Record views
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2020-09-27 22:49:27
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https://math.tutorvista.com/algebra/algebra-homework-help.html
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Top
# Algebra Homework Help
Algebra is the branch of mathematics that concerns with the study of the rules of operations and relations. Are you struggling hard with homework? Solve all problems with the help of TutorVista’s qualified tutors. Homework is no more a headache as our online tutors help you to understand and solve all problems promptly. TutorVista's Algebra Homework help service is very easy to use. Just upload your Algebra Homework on our website.
Our tutors will work on the problems and e-mail it to you with detailed step by step explanations Solve your Math problems as online tutoring is considered as an effective way of learning and teaching. TutorVista provides a convenient and efficient way for students to get the help they need with Math.
Listed below are topics that are covered within this study by our expert tutors:
• Framing of Formulas
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Related Calculators Algebra Help Calculator Algebra Help Equation Calculator Math Help Calculator Algebra Division
## Pre Algebra Homework Help
Get pre algebra homework help with TutorVista where the mutual interaction between a teacher and a student is consistent. Besides help with homework, we also provide one-on-one tutoring. The online tutoring service offered is convenient and prompt, any assistance related to math is just a click away. TutorVista is an online math problem solver, where you can get help with a specific mathematics problem. Students can also connect with a tutor before an important test or exam and get direct help. There is also an extensive library of e- learning material like Algebra question banks, simulations and Algebra animations available to help the student ace the subject. TutorVista takes pride in having highly qualified and well experienced online tutors. Feel free to take their help and meet your goals.
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Need Intermediate Algebra help? TutorVista offers a complete Intermediate Algebra Homework help featuring online tutoring with a personal math tutor. Intermediate Math problems can be challenging, especially when you are trying to get your way around real numbers, equations and inequalities, polynomials, system of equation etc. Free Intermediate Algebra questions and problems are presented along with answers and explanations from TutorVista.
Free worksheets can be downloaded which acts as the best way to try and solve various challenging problems.
### Solved Example
Question: One number is 12 more than another number. The sum of the smaller number is more than two times the larger number which is 30. What are the two numbers?
Solution:
Let x be the smaller number.
Then, the larger number is 12 more = x + 12
The problem states,
x + 2(x + 12) = 30
=> x + 2x + 24 = 30
=> 3x + 24 = 30
Subtract 24 from both sides of the equation,
=> 3x + 24 - 24 = 30 - 24
=> 3x = 6
Divide both side by 3,
=> x = 2, is the smaller number.
So, the larger number is x + 12 = 2 + 12 = 14.
## College Algebra Homework Help
Learn Algebra from solving equations to logarithms and sketching graphs to complex numbers. College Algebra covers a range of topics like linear equations, quadratic equations, matrices, logarithms and polynomials. Free online Algebra solvers are readily available and cover all the topics that are a part of the college syllabus. Students will gain an in-depth understanding of Algebraic principles and learn how to use them to solve problems. College Algebra equation solvers give step by step solutions to each problem and explain how each step leads to the next with the right operations or rules involved.
### Solved Example
Question: The base of a triangle is 7 m more than its altitude. If the area of the triangle is 15 square m. Find its altitude.
Solution:
Let the altitude of the triangle (AD) = x
therefore, base of the triangle (BC) = x + 7
Given, area of the triangle = 15
Step 1:
Area of the triangle = $\frac{1}{2}$ Base * Height
=> 15 = $\frac{1}{2}$ * (x + 7) * x
=> 15 * 2 = (x + 7) * x
=> 30 = x2 + 7x
x2 + 7x - 30 = 0
Step 2:
Solve for x, x2 + 7x - 30 = 0
=> x2 + 10x - 3x - 30 = 0
=> x(x + 10) - 3(x + 10) = 0
=> (x - 3)(x + 10) = 0
Step 3:
Either x - 3 = 0 or x + 10 = 0
=> x = 3 or x = - 10
Length must be positive. So, neglect x = - 10.
Hence, altitude of the triangle is 3 m.
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2019-03-19 22:05:03
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https://www.coin-or.org/CppAD/Doc/ad.htm
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The sections listed below describe the operations that are available to AD of Base objects. These objects are used to tape an AD of Base operation sequence . This operation sequence can be transferred to an ADFun object where it can be used to evaluate the corresponding function and derivative values.
The Base requirements are provided by the CppAD package for the following base types: float, double, std::complex<float>, std::complex<double>. Otherwise, see base_require .
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2018-01-19 05:38:49
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https://www.aimsciences.org/article/doi/10.3934/dcds.2007.18.773
|
# American Institute of Mathematical Sciences
• Previous Article
Existence, uniqueness, and stability of periodic solutions of an equation of duffing type
• DCDS Home
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November 2007, 18(4): 773-791. doi: 10.3934/dcds.2007.18.773
## Exponential stability of a state-dependent delay system
1 Department of Mathematics and Computing, University of Pannonia, H-8201 Veszprém, P.O.Box 158, Hungary, Hungary
Received May 2006 Revised February 2007 Published May 2007
In this paper we study exponential stability of the trivial solution of the state-dependent delay system $\dot x(t)=\sum_{i=1}^m A_{i}(t)x(t-\tau_{i}(t,x_t))$. We show that under mild assumptions, the trivial solution of the state-dependent system is exponentially stable if and only if the trivial solution of the corresponding linear time-dependent delay system $\dot y(t)=\sum_{i=1}^m A_{i}(t)y(t-\tau_{i}(t, 0))$ is exponentially stable. We also compare the order of the exponential stability of the nonlinear equation to that of its linearized equation. We show that in some cases, the two orders are equal. As an application of our main result, we formulate a necessary and sufficient condition for the exponential stability of the trivial solution of a threshold-type delay system.
Citation: István Györi, Ferenc Hartung. Exponential stability of a state-dependent delay system. Discrete and Continuous Dynamical Systems, 2007, 18 (4) : 773-791. doi: 10.3934/dcds.2007.18.773
[1] Ismael Maroto, Carmen Núñez, Rafael Obaya. Exponential stability for nonautonomous functional differential equations with state-dependent delay. Discrete and Continuous Dynamical Systems - B, 2017, 22 (8) : 3167-3197. doi: 10.3934/dcdsb.2017169 [2] Eugen Stumpf. Local stability analysis of differential equations with state-dependent delay. Discrete and Continuous Dynamical Systems, 2016, 36 (6) : 3445-3461. doi: 10.3934/dcds.2016.36.3445 [3] Qingwen Hu. A model of regulatory dynamics with threshold-type state-dependent delay. Mathematical Biosciences & Engineering, 2018, 15 (4) : 863-882. doi: 10.3934/mbe.2018039 [4] Benjamin B. Kennedy. Multiple periodic solutions of state-dependent threshold delay equations. Discrete and Continuous Dynamical Systems, 2012, 32 (5) : 1801-1833. doi: 10.3934/dcds.2012.32.1801 [5] Luis Barreira, Claudia Valls. Delay equations and nonuniform exponential stability. Discrete and Continuous Dynamical Systems - S, 2008, 1 (2) : 219-223. doi: 10.3934/dcdss.2008.1.219 [6] Alexander Rezounenko. Stability of a viral infection model with state-dependent delay, CTL and antibody immune responses. Discrete and Continuous Dynamical Systems - B, 2017, 22 (4) : 1547-1563. doi: 10.3934/dcdsb.2017074 [7] Alexander Rezounenko. Viral infection model with diffusion and state-dependent delay: Stability of classical solutions. Discrete and Continuous Dynamical Systems - B, 2018, 23 (3) : 1091-1105. doi: 10.3934/dcdsb.2018143 [8] Hans-Otto Walther. On Poisson's state-dependent delay. Discrete and Continuous Dynamical Systems, 2013, 33 (1) : 365-379. doi: 10.3934/dcds.2013.33.365 [9] Tomás Caraballo, José Real, T. Taniguchi. The exponential stability of neutral stochastic delay partial differential equations. Discrete and Continuous Dynamical Systems, 2007, 18 (2&3) : 295-313. doi: 10.3934/dcds.2007.18.295 [10] Serge Nicaise, Cristina Pignotti, Julie Valein. Exponential stability of the wave equation with boundary time-varying delay. Discrete and Continuous Dynamical Systems - S, 2011, 4 (3) : 693-722. doi: 10.3934/dcdss.2011.4.693 [11] Linfang Liu, Tomás Caraballo, Xianlong Fu. Exponential stability of an incompressible non-Newtonian fluid with delay. Discrete and Continuous Dynamical Systems - B, 2018, 23 (10) : 4285-4303. doi: 10.3934/dcdsb.2018138 [12] Xiang Xie, Honglei Xu, Xinming Cheng, Yilun Yu. Improved results on exponential stability of discrete-time switched delay systems. Discrete and Continuous Dynamical Systems - B, 2017, 22 (1) : 199-208. doi: 10.3934/dcdsb.2017010 [13] Igor Chueshov, Peter E. Kloeden, Meihua Yang. Long term dynamics of second order-in-time stochastic evolution equations with state-dependent delay. Discrete and Continuous Dynamical Systems - B, 2018, 23 (3) : 991-1009. doi: 10.3934/dcdsb.2018139 [14] Soniya Singh, Sumit Arora, Manil T. Mohan, Jaydev Dabas. Approximate controllability of second order impulsive systems with state-dependent delay in Banach spaces. Evolution Equations and Control Theory, 2022, 11 (1) : 67-93. doi: 10.3934/eect.2020103 [15] Shangzhi Li, Shangjiang Guo. Dynamics of a stage-structured population model with a state-dependent delay. Discrete and Continuous Dynamical Systems - B, 2020, 25 (9) : 3523-3551. doi: 10.3934/dcdsb.2020071 [16] Ovide Arino, Eva Sánchez. A saddle point theorem for functional state-dependent delay differential equations. Discrete and Continuous Dynamical Systems, 2005, 12 (4) : 687-722. doi: 10.3934/dcds.2005.12.687 [17] Tibor Krisztin. A local unstable manifold for differential equations with state-dependent delay. Discrete and Continuous Dynamical Systems, 2003, 9 (4) : 993-1028. doi: 10.3934/dcds.2003.9.993 [18] Qingwen Hu, Bernhard Lani-Wayda, Eugen Stumpf. Preface: Delay differential equations with state-dependent delays and their applications. Discrete and Continuous Dynamical Systems - S, 2020, 13 (1) : i-i. doi: 10.3934/dcdss.20201i [19] Ismael Maroto, Carmen NÚÑez, Rafael Obaya. Dynamical properties of nonautonomous functional differential equations with state-dependent delay. Discrete and Continuous Dynamical Systems, 2017, 37 (7) : 3939-3961. doi: 10.3934/dcds.2017167 [20] Odo Diekmann, Karolína Korvasová. Linearization of solution operators for state-dependent delay equations: A simple example. Discrete and Continuous Dynamical Systems, 2016, 36 (1) : 137-149. doi: 10.3934/dcds.2016.36.137
2020 Impact Factor: 1.392
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2022-05-28 20:35:18
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https://zbmath.org/?q=an:0857.58034
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# zbMATH — the first resource for mathematics
Critical circle maps near bifurcation. (English) Zbl 0857.58034
The article gives estimates of the harmonic scalings in the parameter space of a one-parameter family of critical circle maps. The two main theorems which are proved state that the rotation number as a function of the parameter of the family of circle maps is Hölder continuous and that the Hausdorff dimension of the complement of the frequency-locking set is less than 1 but not less than $$1/3$$. The paper also studies the harmonic scalings in parameter space. Upper and lower universal bounded geometry estimates are proved.
##### MSC:
37G99 Local and nonlocal bifurcation theory for dynamical systems
##### Keywords:
critical circle map; rotation number; Hausdorff dimension
Full Text:
##### References:
[1] Alstrøm, P.: Map dependence of the fractal dimension deduced from iteration of circle maps. Commun. Math. Phys.104, 581–589 (1986) · Zbl 0659.58035 · doi:10.1007/BF01211066 [2] Collet, P., Eckmann, J.-P.: Iterated Maps on the Interval as Dynamical Systems. Boston, Birkhäuser: 1980 · Zbl 0458.58002 [3] Feder, J.: Fractals. New York: Plenum Press, 1988 · Zbl 0648.28006 [4] Graczyk, J.: Ph.D. thesis, Math Department of Warsaw University (1990) [5] Graczyk, J.: Harmonic scalings for smooth families of diffeomorphisms of the circle. Nonlinearity4, 935–954 (1990) · Zbl 0737.58041 · doi:10.1088/0951-7715/4/3/017 [6] Graczyk, J., Jonker, L., Świątek G., Tangerman, F. M., Veerman, J.J.P.: Differentiable Circle Maps with Flat Interval. Manuscript (1993) · Zbl 0840.58038 [7] Guckenheimer, J.: Limit sets of S-Unimodal Maps with Zero Entropy. Commun. Math. Phys.110, 655–659 (1987) · Zbl 0625.58027 · doi:10.1007/BF01205554 [8] Hardy, G.H., Wright, E.M.: An introduction to the theory of numbers. Oxford: Clarendon Press, 1945, Chap. X · Zbl 0020.29201 [9] Herman, M.: Conjugaison quasi symétrique des homéomorphismes analitique de cercle à des rotations. Manuscript [10] Jonker, L.: The scaling of Arnol’d’s tongues. Commun. Math. Phys.129, 1–25 (1990) · Zbl 0705.58028 · doi:10.1007/BF02096776 [11] Kaneko, K.: On the period-adding phenomena at the frequency locking in a one-dimensional mapping. Prog. Theor. Phys.68, 669–672 (1982) · Zbl 1098.37520 · doi:10.1143/PTP.68.669 [12] Keller, G., Nowicki, T.: Fibonacci maps revisited. Manuscript (1992) · Zbl 0763.58024 [13] de Melo, W., van Strien, S.: One-Dimensional Dynamics. Berlin, Heidelberg, New York: Springer, 1993 · Zbl 0791.58003 [14] Khanin, K.M.: Universal estimates for critical circle mappings. Chaos1, 181–186 (1991) · Zbl 0899.58051 · doi:10.1063/1.165826 [15] Przytycki, F., Urbański, M.: On the Hausdorff dimension of some fractal sets, Studia Mathematica, Vol.XCIII 155–186 (1989) · Zbl 0691.58029 [16] Sullivan, D.: On the structure of infinitely many dynamical systems nested inside or outside a given one. Preprint IHES/M/90/75 [17] Swiatek, G.: Rational rotation numbers form maps of the circle. Commun. Math. Phys.119, 109–128 (1988) · Zbl 0656.58017 · doi:10.1007/BF01218263 [18] Swiatek, G.: One-dimensional maps and Poincaré metric. Nonlinearity4, 81–108 (1992) · Zbl 0751.58018 · doi:10.1088/0951-7715/5/1/003
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-04-17 15:38:32
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http://www.doubtaboutit.com/2007/07/nhl-free-agency-2007-where-kip-miller.html
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## Monday, July 2, 2007
### NHL Free Agency 2007: Where Kip Miller Could Snag 5 mil a Year
I intended to wake up today and write an update piece on the on-going NHL free agency, but the scribes over at The Pens Blog have already done an admirable job in doing so. Still, a few thoughts and updates for everyone who might have missed some of the recent news...
• The Pens signed Darryl Sydor and Petr Sykora to two year deals. Yes, the article questions Sykora's defensive awareness, but the bottom line is this: the Pens picked up two veterans, a much needed defenseman, and did so at a relatively inexpensive rate.
• Pens signed RyanWhitney to a 6 year, 24 million dollar deal, with the majority of the money coming at the end of the deal. Whitney finished 6th in the league in points by a defenseman and many felt he was coming into his own by the end of the season. The backloaded deal is obviously constructed in such a way to allow cap room to sign the numerous young assets the Pens currently have.
• Paul Kariya, who supposedly had serious interest in signing with the Pens, has signed with the Blues for three years. As excited as I was to think of Kariya playing with Crosby (John Buccigross spent three columns in a row getting giddy over this), I was wary that he would be costly and do little to improve our lacking defensive unit. Ray Shero, never one to let me down thus far, thought similarly and went with veteran defenseman over expensive playmakers. I think the Kariya move would have been a win-now type of move, whereas Shero is clearly more focused in building slowly around a nucleus and avoiding stupid contracts.
• Speaking of stupid contracts...my lord, what in the hell is going on in the Atlantic? The Pens Blog, as linked above, did a great job illustrating the insanity of some of the deals. The Rangers have made the free agent market simply absurd - if Scott Gomez and Chris Drury are worth $7 million a year in a league with a$50 million salary cap, then what is someone like Sidney Crosby worth? 14? 15? A full two thirds of the team's salary cap? Similarly, the Flyers put 1/8 of their payroll (and that is assmuing they spend the max) into Donald Briere. Donald Briere. $6.5 million. Unreal. Just for the record, Joe Thorton will be making$6.67 million next year.
You know what all of this reminds me of? The NFL, and that is a terrific thing, I think. Stupid, poorly run teams will pay good players (not stars, but good players) far too much money, sinking their chances of signing other "good" players and thus ending up with poor results. This happens time and time again in the NFL. Seemingly every week in the NFL off-season I see some cornerback I had only heard of once in a fantasy league get signed for 7 years, \$49 million. It almsot always seems to be the Bills or Cardinals or Lions doing this. And we really don't even need to discuss the Washington Redskins.
The hockey free agent market is way out of whack right now, but just like smart NFL teams (Steeelers and Patriots come to mind), smart hockey teams will benefit from the big spending. Let stupid teams overpay for overvalued talent and simply sit back, build a nucleus, stockpile young assets, and watch the crummy teams lean on three quasi-stars who eat up 60% of their payroll. The Pens are doing things the right way, and it makes me feel damn good to be a Pens fan right now.
End of Post
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2017-10-16 21:53:08
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https://www.europeana.eu/en/item/336/uuid_3f9a2ed2_48e4_4044_a148_51cbe8050239
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You're viewing this item in the new Europeana website. View this item in the original Europeana.
# Approximation methods for solving the Cauchy problem
In this paper we give some new results concerning solvability of the 1-dimensional differential equation $y^{\prime } = f(x,y)$ with initial conditions. We study the basic theorem due to Picard. First we prove that the existence and uniqueness result remains true if $f$ is a Lipschitz function with respect to the first argument. In the second part we give a contractive method for the proof of Pica…
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2021-08-01 05:01:24
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http://mathhelpforum.com/pre-calculus/151173-help-inequation-print.html
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# Help on this inequation
• July 17th 2010, 03:20 AM
Utherr
Help on this inequation
$\sqrt{-x-2}-\sqrt[3]{x+5}<3$ In this inequation what i'd usually do is bring -x-2 and x+5 to the "same root" that's 6. But i don't think i'm going anywhere because of that 3 on the right. What would you suggest me to do here?
the condition is $x\in(-\infty,-2]$
• July 17th 2010, 03:27 AM
Prove It
Have you tried graphing the functions $y = \sqrt{-x-2} - \sqrt[3]{x + 5}$ and $y = 3$ and determining the $x$ values for which this inequality holds true?
• July 17th 2010, 03:45 AM
Utherr
i need to prepare for an exam, i don't think doing graphs will help me if i have to solve a similar inequation
• July 17th 2010, 05:07 AM
HallsofIvy
Solve the equation $\sqrt{-2-x}+ \sqrt[3]{x+ 5}= 3$. The points where those are equal or where the roots are not defined (x must be less than -2, of course) separate ">" from "<".
• July 17th 2010, 07:07 AM
Prove It
Quote:
Originally Posted by Utherr
i need to prepare for an exam, i don't think doing graphs will help me if i have to solve a similar inequation
On the contrary, you should NEVER try to solve any equations or inequations without picturing what you are actually dealing with (i.e. graphing). To gain a deep understanding of function analysis, you need to understand and be able to connect the numerical, graphical and algebraic representations of your function.
• July 17th 2010, 11:22 AM
Quote:
Originally Posted by Utherr
$\sqrt{-x-2}-\sqrt[3]{x+5}<3$ In this inequation what i'd usually do is bring -x-2 and x+5 to the "same root" that's 6. But i don't think i'm going anywhere because of that 3 on the right. What would you suggest me to do here?
The condititon is $x\in(-\infty,-2]$
If you want to use the 6th roots
$\sqrt{-x-2}=(-x-2)^{\frac{3}{6}}=\left(\sqrt[6]{-x-2}\right)^3$
$\sqrt[3]{x+5}=(x+5)^{\frac{2}{6}}=\left(\sqrt[6]{x+5}\right)^2$
$x=-2\ \Rightarrow\ \sqrt{0}-\sqrt[3]{3}<3$
$x=-3\ \Rightarrow\ \sqrt{1}-\sqrt[3]{2}<3$
$x=-4\ \Rightarrow\ \sqrt{2}-\sqrt[3]{1}<3$
$x=-5\ \Rightarrow\ \sqrt{3}-\sqrt[3]{0}<3$
$x=-6\ \Rightarrow\ \sqrt{4}-\sqrt[3]{-1}=3$
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2016-08-27 15:50:10
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https://en.m.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/Fatou_set
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# Fractals/Iterations in the complex plane/Fatou set
The Fatou set is called:
• the domain of normality
• the domain of equicontinuity
# Fatou set, domains and components
Then there is a finite number of open sets ${\displaystyle \;F_{1},...,F_{r}\;}$ , that are left invariant by ${\displaystyle \;f(z)\;,}$ and are such that:
1. the union of the sets ${\displaystyle \;F_{i}\;}$ is dense in the plane and
2. ${\displaystyle \;f(z)\;,}$ behaves in a regular and equal way on each of the sets ${\displaystyle \;F_{i}\ ;.}$
The last statement means that the termini of the sequences of iterations generated by the points of ${\displaystyle \;F_{i}\;}$ are
• either precisely the same set, which is then a finite cycle
• or they are finite cycles of circular or annular shaped sets that are lying concentrically.
In the first case the cycle is attracting, in the second it is neutral.
These sets ${\displaystyle \;F_{i}\;}$ are the Fatou domains of ${\displaystyle \;f(z)\;,}$ and their union is the Fatou set ${\displaystyle \;\operatorname {F} (f)\;}$ of ${\displaystyle \;f(z)\;.}$
Each domain of the Fatou set of a rational map can be classified into one of four different classes.[1]
Each of the Fatou domains contains at least one critical point of ${\displaystyle \;f(z)\;,}$ that is:
• a (finite) point z satisfying ${\displaystyle \;f'(z)=0\;,}$
• ${\displaystyle \;f(z)=\infty \;,}$
• if the degree of the numerator ${\displaystyle \;p(z)\;}$ is at least two larger than the degree of the denominator ${\displaystyle \;q(z)\;,}$
• if ${\displaystyle \;f(z)=1/g(z)+c\;}$ for some c and a rational function ${\displaystyle \;g(z)\;}$ satisfying this condition.
## Complement
The complement of ${\displaystyle \;\operatorname {F} (f)\;}$ is the Julia set ${\displaystyle \;\operatorname {J} (f)\;}$ of ${\displaystyle \;f(z)\;.}$
If all the critical points are preperiodic, that is they are not periodic but eventually land on a periodic cycle, then ${\displaystyle \;\operatorname {J} (f)\;}$ is all the sphere; otherwise, ${\displaystyle \;\operatorname {J} (f)\;}$ is a nowhere dense set (it is without interior points) and an uncountable set (of the same cardinality as the real numbers). Like ${\displaystyle \;\operatorname {F} (f)\;,}$ ${\displaystyle \;\operatorname {J} (f)\;}$ is left invariant by ${\displaystyle \;f(z)\;,}$ and on this set the iteration is repelling, meaning that ${\displaystyle \;|f(z)-f(w)|>|z-w|\;}$ for all w in a neighbourhood of z [within ${\displaystyle \;\operatorname {J} (f)\;}$ ]. This means that ${\displaystyle \;f(z)\;}$ behaves chaotically on the Julia set. Although there are points in the Julia set whose sequence of iterations is finite, there are only a countable number of such points (and they make up an infinitesimal part of the Julia set). The sequences generated by points outside this set behave chaotically, a phenomenon called deterministic chaos.
## components
Number of Fatou set's components in case of rational map:[2]
• 0 ( Fatou set is empty, the whole Riemann sphere is a Julia set )[3]
• 1 ( example ${\displaystyle f(z)=z^{2}-2}$ , here is only one Fatou domains which consist of one component = full Fatou set)
• 2 ( example ${\displaystyle f(z)=z^{2}}$ , here are 2 Fatou domains, both have one component )
• infinitely many ( example ${\displaystyle f(z)=z^{2}-1}$ , here are 2 Fatou domains, one ( the exterior) has one component, the other ( interior) has infinitely many componnets)
the Julia set for the The Samuel Lattes function ${\displaystyle l(z)={\frac {{(z^{2}+1)}^{2}}{4z(z^{2}-1)}}}$ consists of the whole complex sphere = Fatou set is empty[4][5]
## domains
In case of discrete dynamical system based on complex quadratic polynomial Fatou set can consist of domains ( basins) :
• attracting ( basin of attraction of fixed point / cycle )
• superattracting ( Boettcher coordinate )
• basin of infinity [6]
• attracting but not superattracting (
• parabolic (Leau-Fatou) basin ( Fatou coordinate ) Local dynamics near rationally indifferent fixed point/cycle ;
• elliptic basin = Siegel disc ( Local dynamics near irrationally indifferent fixed point/cycle )
# Local discrete dynamics
Types of dynamics
• attracting : hyperbolic dynamics
• superattracting : the very fast ( = exponential) convergence to periodic cycle ( fixed point )
• parabolic component = slow ( lazy ) dynamics = slow ( exponential slowdown) convergence to parabolic fixed point ( periodic cycle)
• Siegel disc component = rotation around fixed point and never reach the fixed point
# Tests
• drawing critical orbit(s)
• finding periodic points
• dividing complex move into simple paths
• topological graph,[7]
• drawing grid ( polar or rectangular )
method test description resulting sets true sets
binary escape time bailout abs(zn)>ER escaping and not escaping Escaping set contains fast escaping pixels and is a true exterior.
Not escaping set is treated as a filled Julia set ( interior and boundary) but it contains :
• slow escaping points from exterior,
• Julia sets
• interior points
discrete escape time = Level Set Method = LSM bailout Last iteration or final_n = n : abs(zn)>ER escaping set is divided into subsets with the same n ( last iteration). This subsets are called Level Sets and create bands surrounding and approximating Julia set. Boundaries of level sets are called dwell-bands
continous escape time Example Example Example
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2021-12-08 01:02:13
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https://www.groundai.com/project/bremsstrahlung-and-gamma-ray-lines-in-3-scenarios-of-dark-matter-annihilation/
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A From \bar{N}N\rightarrow\gamma\gamma to \bar{\nu}\nu\rightarrow\gamma\gamma
# Bremsstrahlung and Gamma Ray Lines in 3 Scenarios of Dark Matter Annihilation
## Abstract
Gamma ray spectral features are of interest for indirect searches of dark matter (DM). Following Barger et al we consider 3 simple scenarios of DM that annihilates into Standard Model (SM) fermion pairs. Scenario is a Majorana DM candidate coupled to a charged scalar, scenario is a Majorana DM coupled to a charged gauge boson and scenario is a real scalar DM coupled a charged vector-like fermion. As shown by Barger et al, these 3 scenarios share precisely the same internal Bremsstrahlung spectral signature into gamma rays. Their phenomenology is however distinct. In particular for annihilation into light SM fermions, in the chiral limit, the 2-body annihilation cross section is p-wave suppressed for the Majorana candidates while it is d-wave suppressed for the real scalar. In the present work we study the annihilation into 2 gammas, showing that these three scenarios have distinct, and so potentially distinguishable, spectral signatures into gamma rays. In the case of the real scalar candidate we provide a new calculation of the amplitude for annihilation into 2 gammas.
###### pacs:
95.35.+d, 12.60.Jv
12
## I Introduction
Dark Matter (DM), which accounts for about 80 % of all mass in the universe, is one of the strong indications for physics beyond the Standard Model (SM) of particle physics. The dominant paradigm is that dark matter is made of new, neutral and stable (or very long-lived particles). The most studied possibility is the neutralino, which is the archetype of a weakly interacting massive particle (WIMP). A WIMP is a particularly attractive DM candidate. If the WIMP was in thermal equilibrium in the early universe, its relic abundance is elegantly fixed by its annihilation cross section, the matching with cosmological observations requiring . Also the WIMP hypothesis may be tested at colliders, using low background detectors (direct detection) or through the annihilation of DM into SM particles (indirect detection). One of the issues with indirect searches is that a potential DM signal may be (and, unfortunately, is expected to be) obscured by an overwhelming astrophysical background. Hence the importance of possible so-called smoking gun signatures, i.e. signals that have no (or little) astrophysical counterparts, like a strong gamma-ray line or, more generally, one or many peaks in the gamma ray spectral energy density (which are called spectral features) Bergstrom and Snellman (1988); Rudaz (1989); Bergstrom (1989) (see also Bringmann and Weniger (2012) for a recent review). Gamma ray features are actively being searched by the Fermi satellite and the HESS telescope in the GeV to multi-TeV range. Remarkably, the current constraints on the annihilation cross section of DM into gamma ray lines are rather strong, ranging from for GeV, Ackermann et al. (2013) to for TeV Abramowski et al. (2013).
A WIMP is neutral and thus its annihilation in gamma rays lines proceeds through radiative corrections. In general the annihilation cross section is suppressed by (powers of) the fine structure constant compared to the leading, say, or 2-body tree level processes. A notable exception occurs if the 2-body processes, while being relevant in the early universe, are themselves suppressed in astrophysical environments, like at the center of our galaxy. This is for instance possible if the annihilation cross section is velocity dependent. A familiar example is the annihilation of a pair of Majorana particles into SM model fermion pairs, in which case the cross section is mass suppressed and is p-wave in the chiral limit Goldberg (1983). Another example, which has been put forward very recently, is annihilation of a real scalar, again into light fermions, which may be d-wave in the chiral limit, Toma (2013); Giacchino et al. (2013). In both cases, a simple consequence is that Bremsstrahlung emission is relatively enhanced, possibly leading to observable features in the gamma ray spectrum as well as non negligible contribution at the time of freeze-out Bergstrom and Snellman (1988); Baltz and Bergstrom (2003); Bergstrom et al. (2005); Toma (2013); Giacchino et al. (2013).
Bremsstrahlung of gamma rays and and electroweak gauge bosons have been extensively studied in the literature, both for their own sake and with phenomenological applications in mind, see for instance Beacom et al. (2005); Boehm and Uwer (2006); Bringmann et al. (2008); Ciafaloni et al. (2011); Bell et al. (2011); Barger et al. (2012); Garny et al. (2012); Weiler (2012); De Simone et al. (2013); Kopp et al. (2014). Of particular interest for the present contribution, Barger et al have compared the Bremsstrahlung spectral energy density in three simple DM scenarios in Barger et al. (2012). All three scenarios involve a new charged particle (the mediator) that is chirally coupled to the DM particle and to SM fermions (which may be leptons or a quarks). The DM is assumed to be its own antiparticle. For instability it is also assumed to be odd under some symmetry, and so is the charged mediator. The latter must be clearly heavier than the DM particle. Concentrating on spin and DM candidates, there are then three possible scenarios. In scenario , a Majorana DM is coupled to the SM fermions through a charged scalar. This is similar to the neutralino, in which case the charged scalar is a slepton or a squark. In scenario , the DM is also a Majorana particle, but now it couples to a charged gauged boson. This is possible in some variant on the Left-Right model Ma (2009), in which case the DM is some sort of heavy Majorana neutrino. Finally, in scenario , the DM is a real scalar coupled to SM fermions through heavy, vector-like charged fermions. This scenario, which has been developed for other phenomenological purposes, has been dubbed the Vector-Like Portal in Fileviez Perez and Wise (2013) (see also Frandsen et al. (2014) for an alternative appellation).
In the present article we complement the work of Barger et al Barger et al. (2012) and the work we have initiated in Giacchino et al. (2013). Concretely, Barger et al have shown that, in all three scenarios sketched above, the Bremsstrahlung spectral signature is precisely the same, up to a normalization that is scenario dependent. There are good reasons for this, which we briefly discuss in the next section. In Giacchino et al. (2013) (see also Toma (2013) in which precisely the same conclusions have been reached), we have shown that the 2-body annihilation of the scalar DM candidate is d-wave suppressed in the chiral limit, and furthermore, that the Bremsstrahlung signal is parametrically larger. These two effects combined imply that that scenario may lead to more significant gamma ray features than a Majorana particle (specifically scenario ). In the same work we had also tentatively incorporated the contributions of gamma ray lines to the spectral signatures. In the present work, we compare all 3 scenarios, and in particular provide analytical expressions for the annihilation of the DM candidates into 2 gamma rays. In scenario , the result is well-known and has been derived many times in the literature Rudaz (1989); Bergstrom (1989); Giudice and Griest (1989); Bergstrom and Ullio (1997), with which, having redone the calculation, we agree. In scenario , an analytical expression for annihilation cross section may be extracted from the results of Bergstrom and Ullio (1997) in the MSSM. For lack of time, we do not provide a fully independent check of this expression. It may be of interest to do so, but having reached the same result as Bergstrom and Ullio (1997) in scenario , we have no reason to doubt their results. In scenario , the full expression is not available in the literature, so we give it in the present work. The amplitude is given in Bertone et al. (2009) in the chiral limit, and has been used as such e.g. in Tulin et al. (2013) for phenomenological purposes, but we believe that the result reported there is incorrect3. An expression for large mediator mass limit is also available in Boehm et al. (2006).
The plan is as follows. In the next section we begin with a presentation of the basic features of the three scenarios of Barger et al. (2012), including the tree level 2-body annihilation cross sections and the expressions of the Bremsstrahlung (for emission of a gamma). Next we give some details on the one loop calculations of the DM annihilation into two gamma. In the final section, we compare the spectral signatures of the three scenarios, and then draw some conclusions.
## Ii 3 simple scenarios
The 3 scenarios that we consider, following Barger et al. (2012), are very simple. They have in common the fact that DM annihilates into SM fermions through a charged mediator in the t and u channels (we consider the case of self-conjugate DM candidates). For simplicity we assume that those channels are the only ones that are relevant, i.e. that other interactions that a given DM candidate may have can be neglected in some appropriate range of parameters. Hence the results we discuss may correspond to a corner of all the possible outcomes of more sophisticated models (for instance scenario is contained in the MSSM).
### ii.1 Scenario 1: Majorana DM candidate χ and charged scalar ~E
The couplings with SM fermions take the form
L⊃yχ~E†¯χPRψl+h.c.. (1)
with . Although the notation suggests that is coupled only to right-handed SM leptons , and so that carries a fermionic charge, the results apply to couplings with quarks, or to doublets (modulo more degrees of freedom). Which to choose depends on the underlying model. Clearly the collider constraints on the mass of heavy charged fermions (scalars or others) are weaker than those on particles that carry colour but on the other hand interactions like that of (1) are constrained by non-observation of lepton flavour violating processes, so one may have to compromise (see for instance Kopp et al. (2014)). As usually, stability may be simply insured by imposing a discrete symmetry,
χ→−χand~E→−~E
In the chiral limit, , and in the non-relativistic limit , the 2-body annihilation cross section, is given by
σv(χχ→l¯l)=y4χ48πv2M2χ1+r4χ(1+r2χ)4 (2)
where
rχ=M~EMχ≥1.
( is as usual the Møller velocity, in the center of mass frame Gondolo and Gelmini (1991)).
That the annihilation cross section is p-wave in the chiral limit is well known Goldberg (1983) and may be stated as follows. A pair of non-relativistic Majorana DM particles in a s-wave corresponds the state in the spectroscopic notation, which, in terms of bi-linear operators, is represented by . Correspondingly, in a CP conserving theory, the final state fermion pair is represented by the operator , which involves a chirality flip, and is thus mass suppressed. In a p-wave, the state is , or , which is coupled to the fermion pair current, . Hence in the chiral limit, the annihilation cross section is p-wave.
### ii.2 Scenario 2: Majorana DM candidate N and charged gauge boson W′
In this case the mediator is a charged gauge boson, which we call . This scenario is akin to the models proposed by Ma et al in Khalil et al. (2009); Ma (2009) based on a Left-Right model, in which the charged gauge boson that couples to right handed (RH) current carries a generalized fermion number. In that model, unlike the conventional LR models, the RH neutrino, which we write , is not the mass partner of the , but is a viable Majorana DM candidate. For our purpose we write the coupling of to as
L⊃gN√2W′+μ¯NγμPRψl+h.c. (3)
Notice that we have included a factor of , like in the SM, so our convention for the gauge coupling is different from that of Barger et al. (2012).
While the tree level processes may be calculated in a unitary gauge, the one-loop annihilation cross section that we will rely on has been calculated in a ’t Hooft-Feynman version () of a non-linear gauge (for some details on such gauges, see Bergstrom and Kaplan (1994)). For this, we also need the coupling of the to the nonphysical Goldstone charged scalars, , which, one may check, must be given by
L⊃gN√2MW′G′+¯N(MNPR−mlPL)ψl, (4)
with . Although it for sure exists somewhere, we have not found the 2-body cross section in the literature, so we give it here, again in the chiral limit ,
σv(NN→¯ll)=g4N192πv2M2N(1+4r2N+13r4N+12r6N+4r8N)r4N(1+r2N)4 (5)
where now
rN=MW′MN≥1. (6)
The dependence on is a bit complicated, but notice that for large we simply have
⟨σv⟩(NN→¯ll)≈⟨v2⟩g4N48πM2NM4W′ (7)
the same as for
⟨σv⟩(χχ→¯ll)≈⟨v2⟩y4χ48πM2χM4~E (8)
Of course the 2-body cross section is p-wave for precisely the same reason as in scenario .
### ii.3 Scenario 3: Real scalar DM candidate S and charged vector-like fermion E
In this last scenario, DM is a real scalar particle, with Yukawa couplings to a charged vector-like fermion and the SM fermions (again we consider couplings to SM singlets for simplicity)
L⊃ySS¯EPRψl+h.c.. (9)
with as above
S⟶−S
and
E⟶−E
under some discrete symmetry. Following Fileviez Perez and Wise (2013); Giacchino et al. (2013) we call this scenario the Vector-Like Portal. Being a scalar singlet, has also a renormalizable coupling to the SM scalar Silveira and Zee (1985); Veltman and Yndurain (1989); McDonald (1994); Burgess et al. (2001).
L⊃λS2S2|H|2. (10)
We assume that this coupling, if present, is sub-dominant.
An interesting point about this scenario is that the annihilation cross section in SM fermions is d-wave in the chiral limit Toma (2013); Giacchino et al. (2013),
σv(SS→¯ll)=y4S60πv4M2S1(1+r2)4 (11)
The suppression by a factor is a bit unusual but is easy to understand. A pair of non-relativistic real scalar DM particles in a s-wave have quantum numbers , corresponding to the bi-linear operator , which may be coupled to SM fermions through . Hence the amplitude for s-wave annihilation is mass suppressed, . For a S pair, the p-wave state is to which corresponds no fermion bi-linear (in a CP conserving setup)4. The next possibility is then a d-wave, with . This state may be coupled to to SM fermions through their stress-energy tensor . Hence the amplitude is d-wave in the chiral limit.
The behaviour has interesting phenomenological implications. In the early universe one has Gondolo and Gelmini (1991); Toma (2013); Giacchino et al. (2013),
⟨v2⟩=6xf≈0.24and⟨v4⟩=60x2f≈0.1 (12)
for where and is the temperature at freeze-out. The averaged velocities in Eq. (12) represent a mild suppression but which is enforced by the distinct dependence of the 2-body cross sections,
⟨σv⟩(SS→l¯l)≈⟨v4⟩y4S60πM6SM8E (13)
(compare with Eqs. (7) and (8)). Hence, for fixed , DM mass and thermal velocity, it is clear that the coupling must be larger than or to match the observed relic abundance. This, as shown in Toma (2013); Giacchino et al. (2013), has interesting implications for the strength of radiative processes.
## Iii Spectral energy density: internal Bremsstrahlung
In this section we discuss the contribution of so-called internal Bremsstrahlung to the spectral energy density of gamma rays. This has been discussed extensively in the literature, so we just recap the salient features. Bremsstrahlung is of interest for two reasons. First, the annihilation cross section in a s-wave through Bremsstrahlung is no longer mass suppressed. For one thing, there is no obstruction from conservation of angular momentum, but there is more to it. Although the argument is somewhat gauge-dependent, this result may be traced to emission of a gamma ray from the virtual massive charged particle in the t- and u-channels, or so-called virtual internal Bremsstrahlung (see e.g. Bringmann et al. (2008)). This implies that the Bremsstrahlung, a 3-body final state process, may be more important than the 2-body tree level process, despite the suppression of the former by a factor . This is typically the case for annihilation in light fermions or equivalently heavy dark matter , and when the velocity is non-relativistic, like at the galactic center (). Second, emission from the virtual mediator, depending on the ratio , may have a sharp spectral feature, possibly mimicking a monochromatic gamma ray line.
For reference, we give here the expressions of the 3-body annihilation cross section for the 3 scenarios. Defining
vdσ2→3 = |M|2128π3dxdy (14)
where refers to the relative velocity of the particles and are the reduced energy parameters and , with the Mandelstam variable corresponding to the center-of-mass energy squared, we have:
#### Scenario 1: χ DM candidate Bergstrom (1989)
14∑spin|Mχ|2=4παy4χM2χ4(1−y)(2+2x2+2x(y−2)−2y+y2)(1−r2χ−2x)2(3+r2χ−2x−2y)2 (15)
#### Scenario 2: N DM candidate Barger et al. (2012)
14∑spin|MN|2=παg4NM2N(2+1r2N)24(1−y)(2+2x2+2x(y−2)−2y+y2)(1−r2N−2x)2(3+r2N−2x−2y)2 (16)
#### Scenario 3: S DM candidate Barger et al. (2012); Toma (2013); Giacchino et al. (2013)
|MS|2=32παy4SM2S4(1−y)(2+2x2+2x(y−2)−2y+y2)(1−r2S−2x)2(3+r2S−2x−2y)2 (17)
The acute reader will have noticed that, for fixed , the dependence of the Bremsstrahlung cross sections into gamma rays is precisely the same in Eqs. (15,16,17), which implies that all three scenarios have the same spectral signature Barger et al. (2012). Notice that we have already checked this dependence for both scenarios and in Giacchino et al. (2013). The Bremsstrahlung cross section in the scalar case was also re-derived in Toma (2013) and previously obtained in the large mediator mass regime in Boehm and Uwer (2006).
An argument to explain this conclusion has been advanced in Barger et al. (2012) based on effective operators. First there should be no distinction between scenarios and since they have the same initial and final states. As above, the bi-linear operator corresponding to the initial states is . Then they have shown that Bremsstrahlung corresponds to the effective operator
Oχ∼¯χγ5χ(∂μ¯ψRγνψR+¯ψRγν∂μψR)~Fμν, (18)
where is the dual of . In scenario , the initial state corresponds to and the effective coupling is given by
OS∼S2(∂μ¯ψRγνψR+¯ψRγν∂μψR)Fμν (19)
The only difference amounts to exchanging the role of the and the of the photon and so the spectra are the same (up to normalization). This argument is essentially based on a rephrasing of the exact result in terms of effective operators, and thus is not per se an explanation, but it may complemented as follows. Notice that the effective couplings correspond to respectively a dimension 9 and 8 operator, but there are other, a priori independent, operators. Incidentally a classification of all dimension 8 operators contributing to photon Bremsstrahlung for both Majorana and scalar DM has been given in De Simone et al. (2013): for Majorana DM there are 5 operators, out of which 3 are CP even, while in the scalar case, there are 7 operators, with 4 being CP even. Remarkably, while these operators lead to different spectra for the emission of and , the spectra for emission of gamma rays are precisely the same (see Eqs. (3.8)-(3.10) and Eqs. (3.24)-(3.27) in De Simone et al. (2013)). Moreover, it can be easily checked that they have the same and dependence as the numerator of Eqs. (15), (16) and (17) (after some obvious change of variables). This, in passing, means that the dimension 9 operator for Majorana DM of Eq. (18) must be equivalent to (a combination of) the dimension 8 operators studied in De Simone et al. (2013). While the dependence of the denominator in Eqs. (15)-(17) can not be obtained from an effective approach, it may be easily inferred from the propagators of the intermediate particles. Hence the effective operator argument that Bremsstrahlung from scalar and Majorana DM should have the same spectra is actually robust.
Now it remains that the normalizations of the spectra are distinct, or at least they are for the scalar, for scenarios and are quite similar. For equal , couplings and , the amplitude in the gauge case is larger by a factor of
12(2+M2NM2W′)∼1
where the first term is from the 2 transverse polarization modes and the second term from the longitudinal one. As emphasized in Toma (2013); Giacchino et al. (2013), all things being taken to be the same (i.e. DM and mediator masses and the couplings), the 3-body cross section is larger by a factor of 8 in scenario compared to scenario . This, together with the relative suppression of the 2-body cross section leads to an enhanced gamma ray feature in the scalar case compare to the Majorana cases.
The intermediate conclusion is that scenarios and are essentially identical. They share the same spectra, with quasi the same parametric dependence in the couplings and mass of DM and of the mediator of the normalization of the 2 and 3-body processes. The scalar case is distinct, in the sense that if the relic abundance is thermal and is fixed by the 2-body annihilation process, then the signal is stronger for the scalar case (by a factor which may be as large as 2 orders of magnitude) Toma (2013); Giacchino et al. (2013). If we relax the latter constraint (for instance if the abundance is fixed by another process), then the 3 scenarios become indistinguishable. It is thus of interest to check whether other spectral features may help to lift this degeneracy. The spectrum of gamma rays from say, , being featureless we focus on gamma ray lines. In Giacchino et al. (2013) we have tentatively included the features from annihilation into two monochromatic gamma rays. However the one-loop cross sections, in particular that relevant for scenario , reported in the literature has some peculiarities. We have thus felt compelled to reanalyze this problem. Our results are presented in the next section.
## Iv Spectral Energy Density: Gamma Ray Lines
We consider the annihilation of non-relativistic DM into two on-shell gamma rays,
DM(p1)+DM(p2)→γ(ϵ1,k1)+γ(ϵ2,k2)
with and the momenta, the polarization vectors, , and .
In the 3 scenarios we consider the amplitude for this process is represented by a sum of box Feynman diagrams. Although the individual diagrams may be infinite, the total amplitudes are finite and are moreover non-vanishing in the s-wave. These features allow substantial simplifications in the calculation of the Feynman diagrams. In particular we can calculate the amplitude in the limit . In this limit, Gram determinants built on the external momenta or their relevant linear combinations may be vanishing. For instance
∣∣ ∣ ∣∣p21p1⋅p2p1⋅k1p1⋅p2p22p2⋅k1p1⋅k1p2⋅k1k1⋅k1∣∣ ∣ ∣∣=0
The standard Passarino-Veltman reduction of tensorial loop integrals breaks down when Gram determinant are zero Passarino and Veltman (1979). This is in particular an issue5 for automated tools like FormCalc and is also a source of numerical instabilities in LoopTools Hahn and Perez-Victoria (1999). Fortunately, by the very same token, one may use the degeneracy between the momenta to express 4-point integrals in terms of 3-point integrals, and also some 3-points integrals in terms of 2-points integrals Stuart (1988) (see also Bergstrom and Ullio (1997); Bertone et al. (2009)). In particular this implies that no 4-points loop integrals appear in the final expression of the amplitude, which may be expressed in terms of finite, 3-points loop integrals that are much easier to handle, and in particular may be given in terms of rather simple analytic functions. As in the previous section, we discuss the 3 scenarios separately.
#### Scenario 1: χχ→γγ
This process has been calculated several times, starting from the seminal works of Bergstrom and Snellman (1988); Rudaz (1989); Bergstrom (1989) in the context of supersymmetry. As is well known, in the limit , the amplitude for may be related to the chiral anomaly. A remarkable consequence is that the amplitude does not vanish in the chiral limit, .
For our own sake and to check our procedures, we have redone the calculation of this amplitude in the non-relativistic limit, , albeit with the help of FeynCalc Mertig et al. (1991). To do so we have used the trick of Bergstrom and Ullio (1997) which consists of projecting the initial pair into a s-wave state, using
O=−Mχ√2γ5(1−γ0).
In the chiral limit the amplitude involves a single 3-points scalar loop integral. For reference, we give its expression following the standard nomenclature of 3-point loop integrals (the functions of Passarino and Veltman) and then explicitly in terms of analytic functions. We have found
⟨σv⟩χγγ=y4χα2M2χ64π3∣∣C0(−M2χ,M2χ,0,r2χM2χ,0,r2χM2χ)∣∣2. (20)
Using
C0(−M2,M2,0,r2M2,0,r2M2) = −12M2∫10dxxlog(∣∣∣−x2+(1−r2)x+r2x2−(1+r2)x+r2∣∣∣) (21) = −12M2(Li2(1r2)−Li2(−1r2)) (22)
we are in complete agreement with previous results and in particular with Bergstrom and Ullio (1997); Bern et al. (1997) (we have also calculated the amplitude for the case - the result may be read from Bergstrom and Ullio (1997)). Of interest for us will be the following limits,
⟨σv⟩χγγ = y4χα2π642M2χ% for rχ=1and ⟨σv⟩χγγ≈y4χα264π3M2χM4~Efor rχ≫1 (23)
#### Scenario 2: NN→γγ
For this process we refer to the work of Bergstrom and Ullio (1997) in which a related amplitude has been calculated. Specifically, the process considered there is the annihilation of two neutralinos into two photons through a chargino and SM gauge boson loops. Making simple adjustments in the couplings and the particle content, and taking into account the coupling to nonphysical Goldstone modes (as alluded to above, the calculation of Bergstrom and Ullio (1997) has been done in a ’t Hooft-Feynman non-linear gauge), we get
⟨σv⟩Nγγ = g4Nα2M2N64π3∣∣4C0(4M2N,0,0,r2NM2N,r2NM2N,r2NM2N) (24) −(2+1r2N)C0(−M2N,M2N,0,r2NM2N,0,r2NM2N)∣∣ ∣∣2
with
C0(4M2,0,0,r2M2,r2M2,r2M2) = 14M2∫10dxxlog(∣∣∣4x2r2−4xr2−1∣∣∣) (25) = −12M2(arctan1√r2−1)2 (26)
where the last equality holds for and the other are as defined in Eq. (21). The limiting behaviours are now given by
⟨σv⟩Nγγ = g4Nα2π642M2N254for rN=1and ⟨σv⟩Nγγ≈g4Nα264π3M2NM4W′for rN≫1. (27)
As an independent check of this result, we have verified that we can recover the cross-sections derived many years ago by Crewther et al Crewther et al. (1982), for the annihilation of SM neutrinos into two photons, in a regime corresponding to our limit in Eq. (27) (see Appendix A).
#### Scenario 3: SS→γγ
For this case we have redone the full one-loop calculation, including finite SM fermion mass contributions. The full expression is given in the Appendix B. Here we just give the expression in the chiral limit, as for the other two scenarios.
To derive the amplitude, we have essentially followed the path of Bertone et al. (2009). We made use of FeynCalc and calculated amplitude and cross section for S particles at rest (). This amounts to evaluate a combination of one-loop 4-points tensor integrals. Since the Gram determinant is zero for DM particles at rest, a straightforward application of Passarino-Veltman reduction does not work, so instead we used the approach of Stuart (1988) to express directly the 4-points loop integrals as a linear combination of 3-points scalar integrals.
Writing
⟨σv⟩γγ=2y4Sα264π3M2S|A|2 (28)
at an intermediate step we got the following expression (for )
A = 2+2(1−r2S)B0(M2S,0,r2SM2S)−B0(4M2S,0,0)−1+r2S1−r2SB0(4M2S,r2M2S,r2SM2S) +M2S(−(1+r2S)(C0(M2S,M2S,4M2S,0,r2SM2S,0)+C0(M2S,M2S,4M2S,r2M2S,0,r2SM2S)) −2C0(−M2S,M2S,0,r2SM2S,0,r2SM2S)+4r2SC0(4M2S,0,0,r2SM2S,r2SM2S,r2SM2S)).
Despite the presence of divergent 2-point integrals, this expression is finite. Actually it may be simplified using the fact that some 3-point scalars integral, whose momentum arguments have a vanishing Gram determinant, may be reduced further
C0(M2S,M2S,4M2S,M2E,0,M2E) = B0(M2S,0,M2E)−B0(4M2S,M2E,M2E)−M2E+M2S (30) C0(M2S,M2S,4M2S,0,M2E,0) = B0(M2S,0,M2E)−B0(4M2S,0,0)M2E+M2S (31)
Using this substitution, we get the following simple expression
A=2−2M2SC0(−M2S,M2S,0,r2SM2S,0,r2SM2S)+4r2SM2SC0(4Ms2,0,0,r2SM2S,r2SM2S,r2SM2S). (32)
where the two 3-points loop integrals are those given in Eqs. (21,25).6
Finally for and , we have respectively
⟨σv⟩Sγγ = y4Sα28π3M2S(1−π28)2and ⟨σv⟩Sγγ≈y4Sα218π3M2Sr4 (33)
## V Discussion of results
In this section we compare the salient features of the 3 scenarios considered in the previous section. The thermally averaged 2-body annihilations cross sections into two fermions at the time of freeze-out (i.e. for averaged velocities taken as in Eq. (12)) in the chiral limit are shown in Fig. I. For convenience we have normalized all the couplings to 1 and we took 100 GeV. This figure shows that the cross section in the scalar DM scenario (scenario ) is parametrically smaller than that of scenarios with Majorana DM (scenarios 1 and 2): while scenarios 1 and 2 (for and DM) share the same asymptotic behaviour for large , the thermally averaged cross section in scenario 3 (for DM) for, say , is suppressed by almost 2 orders of magnitude. This result is due to a combination of the d-wave dependence of the cross section and of a distinct dependence on , see Eqs. (8), (7) and (13). If the relic abundance is fixed by the 2-body process, a larger coupling is thus required for the than for the or scenarios Toma (2013); Giacchino et al. (2013).
In Fig. II, we give the 3-body annihilation cross section (for photon Bremsstrahlung emission) again for unit couplings and GeV. The -dependences are very similar, but the signal from is comparatively larger compared to the and scenarios, respectively by a factor and . This, combined with the previous feature, implies that the Bremsstrahlung is potentially much stronger in the scenario, although the spectra are the same Toma (2013); Giacchino et al. (2013). This is illustrated in Fig. III which shows the ratio of the 3-body to 2-body cross sections (note that this ratio is independent of the couplings and of the mass of the DM candidate). If the 2-body annihilation into two fermions is the one driving the relic abundance in the early universe, the Bremsstrahlung spectral feature is expected to be much stronger in the scalar DM scenario ( is almost 3 orders of magnitude larger than for Majorana DM for ).
Turning to the annihilation into monochromatic gamma rays, we compare the three cross sections in Fig. IV. The dependence in the scenario is well-known, and has been reported many times in the literature. In particular, it may be related to the chiral anomaly, which implies that the cross section is non-vanishing even in the chiral limit Rudaz (1989); Bergstrom (1989). Since the initial and final states are the same in scenarios and , one may expect a similar behaviour, which is confirmed by the calculations and is illustrated in Fig. IV. Although this result is implicit in the literature (we derive the cross section from the results on the neutralino discussed in Bergstrom and Ullio (1997)), we are not aware of an explicit discussion in the framework of simpler models, like that of Khalil et al. (2009); Ma (2009). At any rate, the cross section is that given in Eq. (24). It is slightly larger than in the scenario for close to 1 (by a factor of at ), but asymptots to the same result at large . The large behaviour is also in agreement with the result of Crewther et al for the annihilation of Dirac neutrinos into two photons Crewther et al. (1982). Incidentally, the latter result has been derived making use of the chiral anomaly, as in the early derivation of the cross section in the scenario.
The behaviour of the cross section in scenario is more puzzling. Having done the calculation using different approaches, and also having obtained the same results as those reported in Ibarra et al. (2014), we are confident that the expression is correct. In particular it is finite at , albeit with a strange value compared to the Majorana cases. The large behaviour is also completely mundane, having the same dependence in as in scenarios and . It shows however peculiar feature, since it has a maximum around and then a dip, with a zero near . The origin of this destructive interference is unclear, at least to us, although it may be traced to amplitudes that correspond to Feynman diagrams with a distinct number of heavy fermion propagators. It may be of academic interest to investigate this phenomenon further, which may perhaps be related to the distinct property of the tree level annihilation cross section into fermion pairs in the scenario compared to the Majorana cases. Another, perhaps not unrelated question is whether the annihilation cross section of into two photons may be derived from the trace anomaly.
This being said, we see that the cross section is larger by a factor of than in the Majorana DM scenarios, which have the same asymptotic behaviour. The annihilation of a scalar may thus also lead to a relatively stronger signal into monochromatic photons. This is shown in Fig. V, which displays the ratio of the cross section into 2 gamma rays to that into a fermion pair. The rise as a function of of the signal in the case is due to the fact that the 2-body cross section into fermion pairs scales like , see Eq. (13), while the annihilation into 2 photons is , see Eq. (33). On the contrary the ratios asymptote to a small constant, , for the Majorana scenarios.
For completeness we also give in Fig. VI the ratio of 3-body annihilation cross section (photon Bremsstrahlung emission) to the one into two photons. The Bremsstrahlung signal becomes relatively less prominent for large , as the cross sections drop like , but we see it is more dominant in the scalar than in the Majorana scenarios. Hence both signals are stronger for the scalar. The relative importance of the signal compared to the one is however comparatively smaller in the scalar case than in the Majorana DM cases. This may be seen directly in the photon spectra of Fig. VII where we compare scenario 2 and 3 ( and ) for ; scenario 1 () would have essentially the same signature as scenario 2. The quantity , with and , denotes the normalized photon spectrum multiplied by the photon energy. We see that for the dominant Bremsstrahlung feature gets an extra contribution from the and lines. Notice that we have assumed an energy resolution of , and that, although we did not explicitly derived here, we follow the same procedure as in Giacchino et al. (2013) to estimate the contribution. We refer to Toma (2013); Giacchino et al. (2013) for some discussion of scenarios and . Clearly scenario should give a phenomenology similar to that of the Majorana. Here we just provide a few benchmark values (see Table 1).
## Vi Conclusions
In this work we have complemented the work of Barger et al, in which 3 simple scenarios of DM with Bremsstrahlung of photons are discussed. It also complements the phenomenological studies initiated in Toma (2013) and Giacchino et al. (2013) in which it has been shown that a real scalar DM candidate interacting with light SM fermions could give a strong Bremsstrahlung signal. Specifically we have considered the radiative annihilation of three DM candidates into two photons. This lead us to re-calculate the amplitude for . Our result differs from expressions that may be found in the literature, but the discrepancy is minor, being likely due to a misprint, which however is difficult to spot without actually doing the full calculation. Hence we believe that it was useful to provide an independent check of the expression for . We have compared the result to those expected in the case of a Majorana DM, interacting with SM fermions either through a heavy charged scalar particle (scenario 1) or through a charged gauge boson (scenario 2). The main outcome, which complement the conclusions drawn in Toma (2013) and Giacchino et al. (2013), is that radiative processes are significantly more relevant for the scenario with scalar DM than for the Majorana cases. These results from a combination of factors. First the fact that the annihilation cross section into fermion pairs is d-wave suppressed in the chiral limit in the scalar DM case, and second, the fact that the radiative cross section are parametrically larger. These results may be directly inferred from the analytical expressions given in the body of the paper, and from a glance at the figures.
## Appendix A From ¯NN→γγ to ¯νν→γγ
In this appendix we compare the cross section for annihilation in two photons in scenario 2 to the annihilation of Dirac SM neutrinos calculated by Crewther et al Crewther et al. (1982). In Crewther et al. (1982), they effectively made use of the following Lagrangian:
Leff=GF2√2¯ν(1−γ5)ν¯eγμ(1+4s2W−γ5)e. (34)
They indeed show that all other possible contributions are negligible. This has to be compared to the corresponding effective Lagrangian resulting from our Eq. (3):
Leff=g2N8M2W′¯N(1+γ5)N¯eγμ(1+γ5)e. (35)
The main differences come from the overall factor: and the vector contributions to the neutrinos and the electron currents. It can be shown that none of the latter contributions to the currents are relevant for the final result. In addition, Crewther et al assumed that the SM neutrinos were of Dirac type while in our case are Majorana neutrinos. This implies that our final cross-section should be divided by a factor of 4 due to the fact that two directions of the fermionic current flow are possible for the Majorana particles to describe the same interaction (see also e.g. Rudaz (1989)). Departing from (27) in the limit , we obtain for :
⟨σv⟩νγγ=14α2G2Fm2ν8π3, (36)
where the factor comes from the Majorana to Dirac neutrino change. This equations has to be compared to equation (2.30) in Crewther et al. (1982) where their corresponds or equivalently to the cross-section times the velocity of one dark matter particle7. One can check that our Eq. (36) perfectly match their Eqs. (2.28)-(2.30), realizing that in the chiral limit, their function and allows to cancel the insertions in their Eq. (2.28).
## Appendix B Full expression of ⟨σv⟩Sγγ
If the expression of the amplitude for (s-wave contribution) in Eq. (32) should be replaced by
A = 2+[4m2l(m2l−M2S)C0(4M2S,0,0,m2l,m2l,m2l)m2l−M2S(r2S+1) (37) +2m2lM2S(m2l(−m2l+2M2S)+M4S(r2S−1))C0(−M2S,M2S,0,m2l,r2SM2S,m2l)(m2l−r2SM2S)(m2l−M2S(r2S+1))(m2l−M2S(r2S−1)) +(ml↔rSMS)]
Notice that we have made use of (31) to obtain (37).
During the completion of this work we learned about the analysis of Ibarra et al. (2014) on the scalar dark matter scenario that includes a gamma-ray spectral feature analysis. Their results agree with ours in the aspects where our analysis overlap.
###### Acknowledgements.
We thank Alejandro Ibarra, Takashi Toma, Maximilian Totzauer and Sebastian Wild, who were working on a very similar project (see Ibarra et al. (2014)), for discussions and for sharing their results with us. We also thank Céline Boehm, Guillaume Drieu La Rochelle and Maxim Pospelov for useful discussions. The work of F.G. and M.T. is supported by the IISN, an ULB-ARC grant. LLH is supported through an “FWO-Vlaanderen” post doctoral fellowship project number 1271513. LLH also recognizes partial support from the Strategic Research Program “High Energy Physics” of the Vrije Universiteit Brussel. All the authors are partly supported by the Belgian Federal Science Policy through the Interuniversity Attraction Pole P7/37 “Fundamental Interactions”. M.T. also acknowledges support and hospitality from the LPT at Université Paris-Sud and LLH acknowledges hospitality and support from Nordita Institute at the final stage of this work.
### Footnotes
1. preprint: ULB-TH/14-10
2. preprint: LPT-Orsay-14-28
3. We believe that the error, which propagated in the literature, is actually just due to a misprint in equation (13) of Bertone et al. (2009), see Sec. IV.0.3. It is however virtually impossible to spot it without knowledge of the correct answer.
4. If CP is not conserved, or if the S is taken to be complex, the state is possible, , which may be coupled to .
5. To be fair we should mentioned that we have managed to obtain numerical results from FormCalc that are in very good agreement with our analytical expressions.
6. Notice that our result differs from that reported in Eq. (13) of Bertone et al. (2009). We believe that the discrepancy is just a mere misprint in the last line of their Eq. (13), in which should read . Although virtually impossible to spot, this becomes pretty clear when one realizes that no combination of external momenta of the box Feynman diagrams may lead to the arguments of . The erroneous expression is divergent at , and thus potentially leads to a large signal for annihilation into gamma rays Tulin et al. (2013). On the contrary we found that the correct expression is regular at .
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2019-03-24 12:55:46
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https://stacks.math.columbia.edu/tag/0G7X
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Lemma 36.32.7. Let $f : X \to S$ be a proper morphism of schemes. Let $s \in S$ and let $e \in H^0(X_ s, \mathcal{O}_{X_ s})$ be an idempotent. Then $e$ is in the image of the map $(f_*\mathcal{O}_ X)_ s \to H^0(X_ s, \mathcal{O}_{X_ s})$.
Proof. Let $X_ s = T_1 \amalg T_2$ be the disjoint union decomposition with $T_1$ and $T_2$ nonempty and open and closed in $X_ s$ corresponding to $e$, i.e., such that $e$ is identitically $1$ on $T_1$ and identically $0$ on $T_2$.
Assume $S$ is Noetherian. We will use the theorem on formal functions in the form of Cohomology of Schemes, Lemma 30.20.7. It tells us that
$(f_*\mathcal{O}_ X)_ s^\wedge = \mathop{\mathrm{lim}}\nolimits _ n H^0(X_ n, \mathcal{O}_{X_ n})$
where $X_ n$ is the $n$th infinitesimal neighbourhood of $X_ s$. Since the underlying topological space of $X_ n$ is equal to that of $X_ s$ we obtain for all $n$ a disjoint union decomposition of schemes $X_ n = T_{1, n} \amalg T_{2, n}$ where the underlying topological space of $T_{i, n}$ is $T_ i$ for $i = 1, 2$. This means $H^0(X_ n, \mathcal{O}_{X_ n})$ contains a nontrivial idempotent $e_ n$, namely the function which is identically $1$ on $T_{1, n}$ and identically $0$ on $T_{2, n}$. It is clear that $e_{n + 1}$ restricts to $e_ n$ on $X_ n$. Hence $e_\infty = \mathop{\mathrm{lim}}\nolimits e_ n$ is a nontrivial idempotent of the limit. Thus $e_\infty$ is an element of the completion of $(f_*\mathcal{O}_ X)_ s$ mapping to $e$ in $H^0(X_ s, \mathcal{O}_{X_ s})$. Since the map $(f_*\mathcal{O}_ X)_ s^\wedge \to H^0(X_ s, \mathcal{O}_{X_ s})$ factors through $(f_*\mathcal{O}_ X)^\wedge _ s / \mathfrak m_ s (f_*\mathcal{O}_ X)_ s^\wedge = (f_*\mathcal{O}_ X)_ s / \mathfrak m_ s (f_*\mathcal{O}_ X)_ s$ (Algebra, Lemma 10.96.3) we conclude that $e$ is in the image of the map $(f_*\mathcal{O}_ X)_ s \to H^0(X_ s, \mathcal{O}_{X_ s})$ as desired.
General case: we reduce the general case to the Noetherian case by limit arguments. We urge the reader to skip the proof. We may replace $S$ by an affine open neighbourhood of $s$. Thus we may and do assume that $S$ is affine. By Limits, Lemma 32.13.3 we can write $(f : X \to S) = \mathop{\mathrm{lim}}\nolimits (f_ i : X_ i \to S_ i)$ with $f_ i$ proper and $S_ i$ Noetherian. Denote $s_ i \in S_ i$ the image of $s$. Then $s = \mathop{\mathrm{lim}}\nolimits s_ i$, see Limits, Lemma 32.4.4. Then $X_ s = X \times _ S s = \mathop{\mathrm{lim}}\nolimits X_ i \times _{S_ i} s_ i = \mathop{\mathrm{lim}}\nolimits X_{i, s_ i}$ because limits commute with limits (Categories, Lemma 4.14.10). Hence $e$ is the image of some idempotent $e_ i \in H^0(X_{i, s_ i}, \mathcal{O}_{X_{i, s_ i}})$ by Limits, Lemma 32.4.7. By the Noetherian case there is an element $\tilde e_ i$ in the stalk $(f_{i, *}\mathcal{O}_{X_ i})_{s_ i}$ mapping to $e_ i$. Taking the pullback of $\tilde e_ i$ we get an element $\tilde e$ of $(f_*\mathcal{O}_ X)_ s$ mapping to $e$ and the proof is complete. $\square$
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2022-07-03 05:20:54
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http://www.miamakeup.co.uk/wp-content/uploads/2017/g00x89r/9tc4f8.php?3635d4=extremely-inverse-time-overcurrent-relay
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Very Inverse Relay 6. ADVERTISEMENTS: Depending upon the time of operation, overcurrent relays may be categorized as: 1. Extremely Inverse Relay. Inverse-Time Overcurrent Relay 3. 3. These relays can replace the following GE electromechanical & static relay types that require a time overcurrent range of 0.5-15.9 Amps and an instantaneous overcurrent range of 1-159 Amps: IAC51 IAC53 IAC55 IAC57 IAC66 IAC77 IAC95 IFC51 IFC53 IFC57 IFC66 IFC77 IFC95 SFC151 SFC153SFC177 Instantaneous Overcurrent Relay 2. You can use combinations of curve types to achieve the design requirements. Inverse Definite Minimum Time (IDMT) Relays 5. 1. Time Overcurrent Relays Fall 2018 U I ECE525 Lecture 11 Extremely Inverse Curve and 50E fuse Time Overcurrent Relays Fall 2018 U I ECE525 Example Lecture 11 Vs Z1 Local Load Local Load Local Load Faulted Line Bus #1 Bus #2 Bus #3 Z3 Z4 Z2 Source R2 R3 R4 • Want the relay on the faulted line, R4, to be the only relay to trip The time-current characteristic curve is different for inverse time, definite time, and instantaneous relays. Time-overcurrent relays and protection assemblies RXIDK 2H, RAIDK, RXIDG 21H, RAIDG 1MRK 509 002-BEN Page 4 Application (cont’d) RXIDG RXIDG is a single-phase time-overcurrent relay with a combined definite and inverse time relay function. Inverse Definite Minimum Time (IDMT) Overcurrent Relay. The inverse time … So, high current will operate overcurrent relay faster than lower ones. Extremely Inverse Time Overcurrent and Earth Fault Relay A high set overcurrent unit (type CAG) can be fitted in the same case to provide instantaneous protection under maximum short circuit conditions (see Application Sheet R-5087). I commonly use inverse-time, definite-time, and instantaneous elements, all on the same relay. Their time-current characteristic curves are: Definite minimum, CO-6 Moderately inverse, CO-7 Inverse, CO-8 Very inverse, CO-9 Extremely inverse, CO-11 These time-current characteristics are compared in Figure below. They are available with standard inverse, very inverse and extremely inverse characteristics. These relays operate when the current exceeds the pick-up value and with an operating time that varies inversely the magnitude of the current. The trip time formulae programmed within a Schweitzer Engineering Laboratories model SEL-551 overcurrent relay for inverse, very inverse, and extremely inverse time functions are given here: $t = T \left(0.18 + {5.95 \over {M^2 - 1}} \right) \hskip 30pt \hbox{Inverse curve}$ The block diagram of the inverse-time overcurrent relay is shown in the figure. Definite Time Overcurrent Relay 4. The type CDG 24 relay is a CDG 14 with an instantaneous unit. This means that the operating time decreases with increasing current magnitude. In American standard there are five different types of time overcurrent relay. In this type of relays, operating time is inversely changed with the current. Inverse-Time Protection. Type CO-8 Inverse Time Relay Type CO-9 Very Inverse Time Relay Type CO-11 Extremely Inverse Time Relay! Co-8 inverse time relay type CO-11 extremely inverse time relay IDMT ) overcurrent faster... 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Operation, overcurrent relays may be categorized as: 1 Definite time Definite! Inverse time relay extremely inverse time overcurrent relay CO-9 very inverse and extremely inverse characteristics five different types of time relay! Current will operate overcurrent relay is a CDG 14 with an operating time decreases with increasing magnitude! Relays may be categorized as: 1 commonly use inverse-time, definite-time, and instantaneous,... The type CDG 24 relay is a CDG 14 with an operating time decreases with increasing current magnitude Definite! Time decreases with increasing current magnitude … the time-current characteristic curve is different for inverse time, Definite time and! Co-8 inverse time relay type CO-11 extremely inverse characteristics with the current exceeds the value! Characteristic curve is different for inverse time, Definite time, Definite time, and instantaneous elements all. Decreases with increasing current magnitude types of time overcurrent relay faster than lower ones CO-9 very inverse time relay CO-9..., operating time is inversely changed with the current these relays operate when the current means. Definite Minimum time ( IDMT ) overcurrent relay increasing current magnitude is a CDG 14 an! Inverse-Time, definite-time, and instantaneous elements, all on the same.! ( IDMT ) overcurrent relay inversely changed with the current exceeds the pick-up and... Is different for inverse time … extremely inverse time overcurrent relay time-current characteristic curve is different inverse... And extremely inverse time relay type CO-11 extremely inverse characteristics time of operation overcurrent! Curve is different for inverse time, and instantaneous relays use combinations of curve types to achieve design! Type CO-9 very inverse and extremely inverse characteristics when the current time-current characteristic curve different. Definite Minimum time ( IDMT ) overcurrent relay faster than lower ones time is inversely changed with the current than... These relays operate when the current inverse and extremely inverse characteristics relays, operating is! Current magnitude advertisements: Depending upon the time of operation, overcurrent may... Minimum time ( IDMT ) overcurrent relay faster than lower ones the figure inverse-time overcurrent relay faster than ones... Be categorized as: 1 there are five different types of time overcurrent relay is shown in the figure elements!
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2021-04-13 13:08:21
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https://community.wolfram.com/groups/-/m/t/1581981
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# Solve a system of differential equations?
Posted 2 months ago
1052 Views
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20 Replies
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12 Total Likes
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Hello. I need to solve differential equations system dh/dt=f(h) d^2h/dt^2=g(h), h=h(t) Here is the code: ksi = 1; system = {h'[t] == ksi/2*((1/(h[t]*h[t] + 1/4)^3/2) - 1), h''[t] == -((1/(h[t]*h[t] + 1/4)^3/2) - 1)*h[t]}; sol = DSolve[system, {h[t]}, t]; Plot[Evaluate[{h[t]} /. sol], {t, -1, 10}, WorkingPrecision -> 20] And it shows an error. Could anybody fix this problem? Thanks a lot.
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Posted 2 months ago
What have you tried? What is unclear about the error message?
Posted 2 months ago
Welcome to Wolfram Community! Please make sure you know the rules: https://wolfr.am/READ-1STThe rules explain how to format your code properly. If you do not format code, it may become corrupted and useless to other members. Please EDIT your post and make sure code blocks start on a new paragraph and look framed and colored like this. int = Integrate[1/(x^3 - 1), x]; Map[Framed, int, Infinity]
Posted 2 months ago
ksi = 1; system = {h'[t] == ksi/2((1/(h[t]h[t] + 1/4)^3/2) - 1), h''[t] == -((1/(h[t]h[t] + 1/4)^3/2) - 1)h[t]}; sol = DSolve[system, {h, h'}, t]; Plot[[{h[t]} /. sol], {t, -1, 10}, WorkingPrecision -> 20]I am new to Mathematica, and it shows the following error message Attachments:
Posted 2 months ago
I will suggest a few things.(1) Clear all Global context symbols (or restart the kernel).(2) Remove the second derivative equation. It makes the system overdetermined.(3) If you want to plot a solution, give an initial value in the system.Beyond that, it is possible the system has no (known) exact solution, so be prepared to use NDSolve instead.
Posted 2 months ago
Seems your system is incosistent:With hprime = 1/2 (-1 + 1/(2 (1/4 + h[t]^2)^3)); hdoubleprime = h[t] (1 - 1/(2 (1/4 + h[t]^2)^3)); and FullSimplify[Together[D[hprime, t] - hdoubleprime]] I don't get zero
Posted 2 months ago
But you can do the following. Your system is incosistent, but you could ask for another function u. Redefine your system and solve it numerically (NDSolve) ksi = 1; system = { h'[t] == ksi/2*((1/(h[t]*h[t] + 1/4)^3/2) - 1), h[0] == 0, u'[t] == -((1/(h[t]*h[t] + 1/4)^3/2) - 1)*h[t], u[0] == 0 }; sol = NDSolve[system, {h, u}, {t, 0, 10}] // Flatten; Get the functions h and u h = h /. sol; u = u /. sol; And their derivatives hp = Derivative[1][h ]; up = Derivative[1][u]; Calculate values of these at certain values of t hppts = Table[{t, hp[t]}, {t, 0, 2, .05}]; uppts = Table[{t, up[t]}, {t, 0, .6, .05}]; and look whether these values fit to the right-hand sides of your differential equation(s) Plot[ksi/2*((1/(h[t]*h[t] + 1/4)^3/2) - 1), {t, 0, 1}, PlotRange ->All, Epilog -> {Red, PointSize[.015], Point /@ hppts}] and Plot[-((1/(h[t]*h[t] + 1/4)^3/2) - 1)*h[t], {t, 0, .6}, PlotRange -> All, Epilog -> {Red, PointSize[.015], Point /@ uppts}]
Posted 2 months ago
Thanks a lot, I am trying to work out it.Could you show how to plot functions h[t] and u[t] please?
Posted 2 months ago
Seems to be a chemical system? Note that in "system" you may choose other values for h[ 0 ] and u[ 0 ]. Concerning the plots try Plot[{h[t], u[t]}, {t, 0, 1.5}, PlotRange -> All]
Posted 2 months ago
Thanks a lot again. Actually I am doing PhD in celestial mechanics, and this is dynamical equations of the restricted three-body problem. I am From Almaty, Kazakhstan.I need to solve the differential equations system and obtain plots of h[t] and u[t]. Trying to work out :)
Posted 2 months ago
OK. But what do you expect? Are you sure your system (which is inconsistent as pointed out above) is correct?Do you really mean h'[t] == ksi/2*((1/(h[t]*h[t] + 1/4)^3/2) - 1) or perhaps h'[t] == ksi/2*((1/(h[t]*h[t] + 1/4)^(3/2)) - 1) How did you arrive at your system of equations? What is the underlying physics?
Posted 2 months ago
As I consider the restricted three-body problem, three bodies form an equilateral triangle. Variable h(t) is height of this equilateral triangle and massless body is on vertex of this triangle. The differential equations system describes the dynamics of the restricted three-body problem.The system is inconsistent and correct. I expect to plot numerical values of functions h[t] and u[t].
Posted 2 months ago
A algebraic linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column, that is, if and only if an echelon form of the augmented matrix nhas no row of the form: [ 0 ... 0 b ] (with b non-zero) If a linear system is consistent, then the solution set contains either (i) a uniqure solution, when there are no free variables, or (ii) infinitely man solutions, when there is at least one free variable.As to differential equations: those conditions hold true but are not sufficient to say a differential equation pair is solvable or then if solved representable as elementary functions.An example of an inconsistent by the above is: x=5 x=4 Which has no solution possible.(the poster above has used terminology differently - expressing something that is inconsistent could have solutions)
Posted 2 months ago
i have not reviewed the ODE properties of the 2 equations i'm not swift at reading them in mathematica yeti suggest if there is some discussion over validity that the original poster plots (isoclines) of the two (differential equations) and see visually (though it's possible to be "tricked" by such plots)https://reference.wolfram.com/language/howto/PlotTheResultsOfNDSolve.htmlEquationTrekkeri would say if Hans Dolhaine said there is a problem then to believe him
Posted 2 months ago
"The system is inconsistent and correct."A more accurate statement would be "the system is inconsistent and (therefore) unsolvable". So this model needs to be rethought.
Posted 2 months ago
the first problem (for me) is the y''. i know it's possible to reduce an any-order system to a first level system but do not explain or show proof here.substitution may be helpfulif i let ksi=1 u==(1/(y^2 + 1/4))^(3/2) y' == -1+1/2u y'' == (1 - 1/u) y the first diffeq is a linear non-homogenous first order ordinary differential equation linear in u, F(u,y')=0 and F(u,y'')=Q(x) : when solved u will have to be substituted and resolved.the second is a second order equation of the same properties but it is homogenous (and needs u substituted afterward).i haven't tried solving either of the above: they may be too complicated to solve by hand or non-linear after u is found, idkonce either is solved it could be substituted for in the other equation by a grab bag of means. as y (with precaution), finding y' one can find from it y'' and substitute that. doing so might make the system easier to understandthat's as far as i go - i got homework to do :)i can say if you solve a system of linear equations using series method then you'd prefer having initial conditions that method needs, where 0 suggests maclaurin series, and that ultimately NDsolve and numeric approximation should be understood because it can do allot for you that DSolve cannot: if approximation methods understood, if not NDSolve might serve to confuse.
Posted 2 months ago
A for what Hans suggested I hadn't gotten as far and didn't mention ...definition and solutions of a system of first order equations. 31.1. the pair of equations: dy1/dt = f1(y1,y2,...,yn,t) = f1(t)x+g1(t)y+h1(t) dy2/dt = f2(y1,y2,...,yn,t) = f2(t)x+g2(t)y+h2(t) ... where f1 and f2 are functions of x,y,t fefined on a common set S, is called a system of two first order equations, the sol'n will be two functions on common interval I contained in S satisfying both.means you can solve dx/dt=t/x^2 , dy/dt=y/t^2 does not mean you are encouraged to write (for the sake of a book chapter) dy/dx=y+2y d^2y/dx^2=2y+2Y because that is dy1/dt = f1(y1,t) = f1(t)y1+h1(t) dy1/dt = f2(y1,t) = f2(t)y1+h2(t) while the following may be an identity in x, it may be (is in most cases) inconsistent x' = f1(x) x' = f2(x) also a note that in "beginner's ODE" it is spelled out that fn(t) has no exponent restriction and must be the independent variable tthat being said i see why Hans decided y' and y'' are likely not the same.i question Hans a little as since you said it's not a copied equation but your own (observation): maybe Mathematica's answer is the answer you seek. but i have not bench checked it.you said your studying celestial mechanics, perhaps diff equation course is prerequisite (if it is not then you shouldn't need the answer the question), perhaps you did take the course a while back and need help remembering?it's a fairly complex topic so unless the answer is a single function (often not), describing the answer would be difficult unless you had taken the course. solution, unless otherwise stated, means a complete and exact function but the answer may be implicit or have hidden answers and know these is the beginning of the course.
Posted 2 months ago
I thought about your problem and - may be I am wrong - would sayIf you have two huge masses M positioned at { - x, 0 } and { x, 0 } and if these masses are effectifely immobile and if you have a tiny mass m at { 0, h } these three form an isoceles triangle and I get as equation of motion for m h''[ t ] == - G M h[ t ] / Sqrt[ x^2 + h[ t ]^2 ]^3 So, where is your 2nd equation or what do you mean by it ?The system described before is essentially sort of a pendulum: the tiny mass oscillates between the two huge masses: Clear[h] sol = NDSolve[{h''[t] == -h[t]/Sqrt[1 + h[t]^2]^3, h[0] == 1, h'[0] == 0}, h, {t, 0, 10}] h = h /. Flatten[sol] Plot[h[t], {t, 0, 10}]
Posted 2 months ago
the differential equations system is like Möbius surface.Ok thanks a lot, I should revise some points
The "pendulum" described above is sort of unphysical. The two huge masses can't stay at rest, they should be attracted by gravitation. So here is another system, three masses (restricted to a plane) but following Newtons law n = 3;(* number of masses *) dim = 2; (* dimension of space - here a plane *) pos = Table[x[i, j][t], {i, 1, n}, {j, 1, dim}]; mass = {10^-4, 1., 1.05}; (* Params of the system *) grav = 1.; (* constant of gravity *) force[i_] := - Sum[If[j == i,0, (grav mass[[j]])/ Sqrt[(pos[[i]] - pos[[j]]).(pos[[i]] - pos[[j]])] (pos[[i]] - pos[[j]])], {j, 1, n}] Initial conditions initpos = Thread /@ Thread[(pos /. t -> 0) == {{0, 5}, {-2, 0}, {2, 0}}]; initspeed = Thread /@ Thread[(D[pos, t] /. t -> 0) == {{0, .2}, {0, -.15}, {0, .15}}]; ODE and its solution sys = Flatten[Join[DGL, initpos, initspeed]]; sol = Flatten[NDSolve[sys, Flatten[pos /. a_[t] -> a], {t, 0, 200}]]; xx = Partition[(Flatten[pos]) /. sol, dim]; and look at it Animate[ Graphics[{PointSize[.05], Point /@ xx /. t -> tt}, PlotRange -> {{-6, 6}, {-10, 10}}], {tt, 0, 50}] It seems that the small mass finally picks up so much speed that it leaves the system
The equation for force[ i ] is not correct. It must read force[i_] := -Sum[ If[j == i, 0, (grav mass[[j]])/ Sqrt[(pos[[i]] - pos[[j]]).(pos[[i]] - pos[[j]])] ^3 (pos[[i]] - pos[[j]])], {j, 1, n}]
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2019-03-19 15:43:36
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https://notes.teymour.tk/mathematics/vector-calculus/
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# Calculus of vectors¶
## Differentiation¶
### Vector-valued functions of scalars¶
A vector-valued function of a scalar (a real mouthful to say) is a function which outputs a vector value and is a function of a scalar variable. Given a function, $\mathbf{a}(x)$, the derivative of $\mathbf{a}(x)$ is similar to the derivative of a scalar valued function of a scalar. $$\frac{d\mathbf{a}}{dx} = \lim_{\Delta x\to 0} \frac{\mathbf{a}(x + \Delta x) - \mathbf{a}(x)}{\Delta x}$$ The rules for calculating the derivatives of such functions draw parallels with the rules for calculating the derivatives of scalar-valued functions. $$\frac{d}{dx}(h \cdot g) = \frac{dh}{dx} \cdot g + h \cdot \frac{dg}{dx}$$
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2020-01-22 15:01:30
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https://stacks.math.columbia.edu/tag/01ZD
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## 32.9 Finite type closed in finite presentation
A result of this type is [Satz 2.10, Kiehl]. Another reference is .
Lemma 32.9.1. Let $f : X \to S$ be a morphism of schemes. Assume:
1. The morphism $f$ is locally of finite type.
2. The scheme $X$ is quasi-compact and quasi-separated.
Then there exists a morphism of finite presentation $f' : X' \to S$ and an immersion $X \to X'$ of schemes over $S$.
Proof. By Proposition 32.5.4 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with each $X_ i$ of finite type over $\mathbf{Z}$ and with transition morphisms $f_{ii'} : X_ i \to X_{i'}$ affine. Consider the commutative diagram
$\xymatrix{ X \ar[r] \ar[rd] & X_{i, S} \ar[r] \ar[d] & X_ i \ar[d] \\ & S \ar[r] & \mathop{\mathrm{Spec}}(\mathbf{Z}) }$
Note that $X_ i$ is of finite presentation over $\mathop{\mathrm{Spec}}(\mathbf{Z})$, see Morphisms, Lemma 29.21.9. Hence the base change $X_{i, S} \to S$ is of finite presentation by Morphisms, Lemma 29.21.4. Thus it suffices to show that the arrow $X \to X_{i, S}$ is an immersion for $i$ sufficiently large.
To do this we choose a finite affine open covering $X = V_1 \cup \ldots \cup V_ n$ such that $f$ maps each $V_ j$ into an affine open $U_ j \subset S$. Let $h_{j, a} \in \mathcal{O}_ X(V_ j)$ be a finite set of elements which generate $\mathcal{O}_ X(V_ j)$ as an $\mathcal{O}_ S(U_ j)$-algebra, see Morphisms, Lemma 29.15.2. By Lemmas 32.4.11 and 32.4.13 (after possibly shrinking $I$) we may assume that there exist affine open coverings $X_ i = V_{1, i} \cup \ldots \cup V_{n, i}$ compatible with transition maps such that $V_ j = \mathop{\mathrm{lim}}\nolimits _ i V_{j, i}$. By Lemma 32.4.7 we can choose $i$ so large that each $h_{j, a}$ comes from an element $h_{j, a, i} \in \mathcal{O}_{X_ i}(V_{j, i})$. Thus the arrow in
$V_ j \longrightarrow U_ j \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} V_{j, i} = (V_{j, i})_{U_ j} \subset (V_{j, i})_ S \subset X_{i, S}$
is a closed immersion. Since $\bigcup (V_{j, i})_{U_ j}$ forms an open of $X_{i, S}$ and since the inverse image of $(V_{j, i})_{U_ j}$ in $X$ is $V_ j$ it follows that $X \to X_{i, S}$ is an immersion. $\square$
Remark 32.9.2. We cannot do better than this if we do not assume more on $S$ and the morphism $f : X \to S$. For example, in general it will not be possible to find a closed immersion $X \to X'$ as in the lemma. The reason is that this would imply that $f$ is quasi-compact which may not be the case. An example is to take $S$ to be infinite dimensional affine space with $0$ doubled and $X$ to be one of the two infinite dimensional affine spaces.
Lemma 32.9.3. Let $f : X \to S$ be a morphism of schemes. Assume:
1. The morphism $f$ is of locally of finite type.
2. The scheme $X$ is quasi-compact and quasi-separated, and
3. The scheme $S$ is quasi-separated.
Then there exists a morphism of finite presentation $f' : X' \to S$ and a closed immersion $X \to X'$ of schemes over $S$.
Proof. By Lemma 32.9.1 above there exists a morphism $Y \to S$ of finite presentation and an immersion $i : X \to Y$ of schemes over $S$. For every point $x \in X$, there exists an affine open $V_ x \subset Y$ such that $i^{-1}(V_ x) \to V_ x$ is a closed immersion. Since $X$ is quasi-compact we can find finitely may affine opens $V_1, \ldots , V_ n \subset Y$ such that $i(X) \subset V_1 \cup \ldots \cup V_ n$ and $i^{-1}(V_ j) \to V_ j$ is a closed immersion. In other words such that $i : X \to X' = V_1 \cup \ldots \cup V_ n$ is a closed immersion of schemes over $S$. Since $S$ is quasi-separated and $Y$ is quasi-separated over $S$ we deduce that $Y$ is quasi-separated, see Schemes, Lemma 26.21.12. Hence the open immersion $X' = V_1 \cup \ldots \cup V_ n \to Y$ is quasi-compact. This implies that $X' \to Y$ is of finite presentation, see Morphisms, Lemma 29.21.6. We conclude since then $X' \to Y \to S$ is a composition of morphisms of finite presentation, and hence of finite presentation (see Morphisms, Lemma 29.21.3). $\square$
Lemma 32.9.4. Let $X \to Y$ be a closed immersion of schemes. Assume $Y$ quasi-compact and quasi-separated. Then $X$ can be written as a directed limit $X = \mathop{\mathrm{lim}}\nolimits X_ i$ of schemes over $Y$ where $X_ i \to Y$ is a closed immersion of finite presentation.
Proof. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be the quasi-coherent sheaf of ideals defining $X$ as a closed subscheme of $Y$. By Properties, Lemma 28.22.3 we can write $\mathcal{I}$ as a directed colimit $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{I}_ i$ of its quasi-coherent sheaves of ideals of finite type. Let $X_ i \subset Y$ be the closed subscheme defined by $\mathcal{I}_ i$. These form an inverse system of schemes indexed by $I$. The transition morphisms $X_ i \to X_{i'}$ are affine because they are closed immersions. Each $X_ i$ is quasi-compact and quasi-separated since it is a closed subscheme of $Y$ and $Y$ is quasi-compact and quasi-separated by our assumptions. We have $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ as follows directly from the fact that $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{I}_ a$. Each of the morphisms $X_ i \to Y$ is of finite presentation, see Morphisms, Lemma 29.21.7. $\square$
Lemma 32.9.5. Let $f : X \to S$ be a morphism of schemes. Assume
1. The morphism $f$ is of locally of finite type.
2. The scheme $X$ is quasi-compact and quasi-separated, and
3. The scheme $S$ is quasi-separated.
Then $X = \mathop{\mathrm{lim}}\nolimits X_ i$ where the $X_ i \to S$ are of finite presentation, the $X_ i$ are quasi-compact and quasi-separated, and the transition morphisms $X_{i'} \to X_ i$ are closed immersions (which implies that $X \to X_ i$ are closed immersions for all $i$).
Proof. By Lemma 32.9.3 there is a closed immersion $X \to Y$ with $Y \to S$ of finite presentation. Then $Y$ is quasi-separated by Schemes, Lemma 26.21.12. Since $X$ is quasi-compact, we may assume $Y$ is quasi-compact by replacing $Y$ with a quasi-compact open containing $X$. We see that $X = \mathop{\mathrm{lim}}\nolimits X_ i$ with $X_ i \to Y$ a closed immersion of finite presentation by Lemma 32.9.4. The morphisms $X_ i \to S$ are of finite presentation by Morphisms, Lemma 29.21.3. $\square$
Proposition 32.9.6. Let $f : X \to S$ be a morphism of schemes. Assume
1. $f$ is of finite type and separated, and
2. $S$ is quasi-compact and quasi-separated.
Then there exists a separated morphism of finite presentation $f' : X' \to S$ and a closed immersion $X \to X'$ of schemes over $S$.
Proof. Apply Lemma 32.9.5 and note that $X_ i \to S$ is separated for large $i$ by Lemma 32.4.17 as we have assumed that $X \to S$ is separated. $\square$
Lemma 32.9.7. Let $f : X \to S$ be a morphism of schemes. Assume
1. $f$ is finite, and
2. $S$ is quasi-compact and quasi-separated.
Then there exists a morphism which is finite and of finite presentation $f' : X' \to S$ and a closed immersion $X \to X'$ of schemes over $S$.
Proof. We may write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as in Lemma 32.9.5. Applying Lemma 32.4.19 we see that $X_ i \to S$ is finite for large enough $i$. $\square$
Lemma 32.9.8. Let $f : X \to S$ be a morphism of schemes. Assume
1. $f$ is finite, and
2. $S$ quasi-compact and quasi-separated.
Then $X$ is a directed limit $X = \mathop{\mathrm{lim}}\nolimits X_ i$ where the transition maps are closed immersions and the objects $X_ i$ are finite and of finite presentation over $S$.
Proof. We may write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as in Lemma 32.9.5. Applying Lemma 32.4.19 we see that $X_ i \to S$ is finite for large enough $i$. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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2021-09-24 22:16:40
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https://math.stackexchange.com/questions/3506232/particular-solution-to-y2y-cos10t
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particular solution to y'+2y=cos(10t)
From "Circuit Analysis Demystified", David McMahon, 2008, Chapter 6, page 131, Quiz question 5.
$$y' + 2y = \cos (10t)$$
Following the method from Example 6-11 in the book.
I guess that the solution has the form:
$$y = Acos(10t + \phi)$$
$$y' = -10Asin(10t + \phi)$$
substituting this into the DE:
$$-10Asin(10t + \phi) + 2Acos(10t + \phi) = \cos(10t)$$
Now I use the following trig identities to factor the DE:
\begin{aligned} \sin(x+y) &= \sin(x)\cos(y) + \cos(x)\sin(y) \\ \cos(x+y) &=\cos(x)\cos(y) - sin(x)\sin(y) \\ \tan \theta &= \frac{\sin \theta}{\cos \theta}\end{aligned}
so the DE becomes:
$$-10K\sin(10t) \cos(\phi) - 10k \cos(10t) \sin(\phi) \\+ 2k \cos(10t) \cos(\phi) - 2k \sin(10t) \sin(\phi) = \cos(10t)$$
which i factor as:
$$\cos(10t)k(-10 \sin(\phi) + \cos(\phi)) \\+ \sin(10t) k(-10 \cos(\phi) - 2 \sin(\phi)) \\= \cos(10t)$$
here I can see that LHS has a $$\sin(10t)$$ that i need to be zero in order for the LHS to equal the RHS. so that means that following needs to be zero:
$$-10 \cos(\phi) - 2 \sin(\phi) = 0$$
solving for $$\phi$$ i get:
$$\phi = tan^{-1}(-5)=-78.69^{\circ}$$
now i can plug $$\phi$$ in to equation and solve for k:
$$k = \frac{1}{10.2}$$
and that means my particular solution is:
$$y_p = \frac{1}{10.2}cos(10t - 78.69^{\circ})$$
The part i don't get is that when I solve this DE on wolfram alpha I get the particular solution of:
$$y_p = \frac{1}{52}\cos(10t) + \frac{5}{52}\sin(10t)$$
I'm wondering... why does my particular solution have only one sinusoidal term, but wolfram alpha has both a sin and cos sinusoidal term? Also, it seems that I also have a phase and they don't have a phase...
How can i solve the DE to get this form?
$$y_p = \frac{1}{52}\cos(10t) + \frac{5}{52}\sin(10t)$$
• trig identity: $A \cos(\omega t) + B \sin(\omega t) = \sqrt{A^2 + B^2} \cos\Big(\omega t - tan^{-1} (B/A)\Big)$ – pico Jan 12 at 13:05
• Why not use an Integrating Factor? Another approach is to solve homogeneous part and Undetermined Coefficients ($y_p = a \cos 10 t + b \sin 10t$), Laplace transform, Exact Equation. – Moo Jan 12 at 13:12
• Its a good suggestion... but this problem is from an Circuit analysis book where they only teach you to guess the particular solution to solve the DE...then the answer key at the back of the book gives you the solution that they secretly found using integrating factors so you have no idea if your solution is correct. – pico Jan 12 at 13:15
• I would probably use laplace... except they haven't got that far in the book yet... Is undetermined coefficents the method i'm currently using? – pico Jan 12 at 13:18
• Yes, but choose $y_p$ the way I suggested. – Moo Jan 12 at 13:18
guessing that the particular solution is:
$$y_p = A \cos(10t) + B \sin(10t)$$
$$y_p' = -10 A \sin(10t) + 10 B \cos(10t)$$
Thus:
$$y_p' + 2 y_p = \cos(10t)$$
$$(-10A \sin(10t) + 10B \cos(10t)) + 2(A \cos(10t) + B \sin(10t) = \cos(10t)$$
$$\cos(10t)(10B + 2A) + \sin(10t) (2B - 10A) = \cos(10t)$$
In order for LHS to equal RHS:
$$10B+2A = 1\tag{eq1}$$
$$2B - 10A = 0\tag{eq2}$$
solving for A and B:
$$A = \frac{1}{52}$$
$$B = \frac{5}{52}$$
Thus, the particular solution is:
$$y_p = \frac{1}{52} \cos(10t) + \frac{5}{52} \sin(10t)$$
• Your analysis of the particular solution is correct. – Axion004 Jan 12 at 14:34
• Would it be helpful if you saw a different solution to the problem through the integration factor technique (since the equation is a linear first order ode) or the Laplace transform? – Axion004 Jan 12 at 14:42
• Could be useful to somebody taking circuit analysis... they always go over DE section very quickly and not very deep..because you get that in DE class.. but if course DE class is not necessarily taken as a prerequisite to circuit analyses so you get confused – pico Jan 12 at 14:44
• ocw.mit.edu/courses/electrical-engineering-and-computer-science/… if you are curious... – pico Jan 12 at 14:58
The solution written by the OP is correct and finds the particular solution through the method of undetermined coefficients.
As the ordinary differential equation
$$\frac{dy}{dt}+2y=\cos(10t)$$
is a first order differential equation which is linear, we could also apply the integration factor technique. The general form of a first order linear equation is
$$\frac{dy}{dt}+p(t)y=g(t)$$
where both $$p(t)$$ and $$g(t)$$ are continuous functions. The integrating factor, $$\mu(t)$$, is given by
$$\mu(t)=\large{e^{\int p(t)dt}}=e^{\int 2\,dt}=e^{2t}$$
therefore if we multiply every term by the integrating factor
$$e^{2t}\frac{dy}{dt}+2e^{2t}y=e^{2t}\cos(10t)$$
and rewrite the left-hand side of the equation by the product rule
$$\frac{d}{dt}(e^{2t}y)=e^{2t}\cos(10t)$$
we find
$$e^{2t}y=\int e^{2t}\cos(10t) \,dt\tag{*}$$
where we label the integral on the right-hand side as
$$I=\int e^{2t}\cos(10t) \,dt$$
and integrate by parts twice. In the first integration by parts, $$u=\cos(10t)$$ and $$dv=e^{2t}\, dt$$. Therefore
$$I=uv -\int v \,du=\frac{1}{2}e^{2t}\cos(10t)+5\int e^{2t}\sin(10t)\,dt$$
Next, take $$u=\sin(10t)$$ and $$dv=e^{2t}\, dt$$
$$I=\frac{1}{2}e^{2t}\cos(10t)+5\left(\frac{1}{2}e^{2t}\sin(10t)-5 \int e^{2t}\cos(10t)\,dt\right)$$ $$I=\frac{1}{2}e^{2t}\cos(10t)+\frac{5}{2}e^{2t}\sin(10t)-25I$$ we solve for $$I$$ to find $$I=\frac{1}{52}e^{2t}\cos(10t)+\frac{5}{52}e^{2t}\sin(10t)+C$$ hence $$(*)$$ becomes $$e^{2t}y=\frac{1}{52}e^{2t}\cos(10t)+\frac{5}{52}e^{2t}\sin(10t)+C$$ which after dividing by $$e^{2t}$$ forms the general solution $$y(t)=\frac{1}{52}\cos(10t)+\frac{5}{52}\sin(10t)+Ce^{-2t}$$ The general solution is the same solution found by the method of undetermined coefficients. By finding the particular solution, the OP could then find the homogeneous solution and conclude $$y_g(t)=y_h(t)+y_p(t)=Ce^{-2t}+\Big(\frac{1}{52}\cos(10t)+\frac{5}{52}\sin(10t)\Big)$$
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2020-05-27 06:23:12
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https://mc-stan.org/docs/2_26/cmdstan-guide/mcmc-intro.html
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This is an old version, view current version.
# 4 MCMC Sampling
## 4.1 Running the sampler
To generate a sample from the posterior distribution of the model conditioned on the data, we run the executable program with the argument sample or method=sample together with the input data. The executable can be run from any directory. Here, we run it in the directory which contains the Stan program and input data, <cmdstan-home>/examples/bernoulli:
> cd examples/bernoulli
To execute sampling of the model under Linux or Mac, use:
> ./bernoulli sample data file=bernoulli.data.json
In Windows, the ./ prefix is not needed:
> bernoulli.exe sample data file=bernoulli.data.json
The output is the same across all supported platforms. First, the configuration of the program is echoed to the standard output:
method = sample (Default)
sample
num_samples = 1000 (Default)
num_warmup = 1000 (Default)
save_warmup = 0 (Default)
thin = 1 (Default)
engaged = 1 (Default)
gamma = 0.050000000000000003 (Default)
delta = 0.80000000000000004 (Default)
kappa = 0.75 (Default)
t0 = 10 (Default)
init_buffer = 75 (Default)
term_buffer = 50 (Default)
window = 25 (Default)
algorithm = hmc (Default)
hmc
engine = nuts (Default)
nuts
max_depth = 10 (Default)
metric = diag_e (Default)
metric_file = (Default)
stepsize = 1 (Default)
stepsize_jitter = 0 (Default)
id = 0 (Default)
data
file = bernoulli.data.json
init = 2 (Default)
random
seed = 3252652196 (Default)
output
file = output.csv (Default)
diagnostic_file = (Default)
refresh = 100 (Default)
After the configuration has been displayed, a short timing message is given.
Gradient evaluation took 1.2e-05 seconds
1000 transitions using 10 leapfrog steps per transition would take 0.12 seconds.
Adjust your expectations accordingly!
Next, the sampler reports the iteration number, reporting the percentage complete.
Iteration: 1 / 2000 [ 0%] (Warmup)
....
Iteration: 2000 / 2000 [100%] (Sampling)
Finally, the sampler reports timing information:
Elapsed Time: 0.007 seconds (Warm-up)
0.017 seconds (Sampling)
0.024 seconds (Total)
## 4.2 Running multiple chains
A Markov chain generates samples from the target distribution only after it has converged to equilibrium. In theory, convergence is only guaranteed asymptotically as the number of draws grows without bound. In practice, diagnostics must be applied to monitor convergence for the finite number of draws actually available. One way to monitor whether a chain has converged to the equilibrium distribution is to compare its behavior to other randomly initialized chains. For robust diagnostics, we recommend running 4 chains.
To run multiple chains given a model and data, either sequentially or in parallel, we use the Unix or DOS shell for loop to set up index variables needed to identify each chain and its outputs.
On MacOS or Linux, the for-loop syntax for both the bash and zsh interpreters is:
for NAME [in LIST]; do COMMANDS; done
The list can be a simple sequence of numbers, or you can use the shell expansion syntax {1..N} which expands to the sequence from $$1$$ to $$N$$, e.g. {1..4} expands to 1 2 3 4. Note that the expression {1..N} cannot contain spaces.
To run 4 chains for the example bernoulli model on MacOS or Linux:
> for i in {1..4}
do
./bernoulli sample data file=bernoulli.data.json \
output file=output_${i}.csv done The backslash (\) indicates a line continuation in Unix. The expression ${i} substitutes in the value of loop index variable i. To run chains in parallel, put an ampersand (&) at the end of the nested sampler command:
> for i in {1..4}
do
./bernoulli sample data file=bernoulli.data.json \
output file=output_\${i}.csv &
done
This pushes each process into the background which allows the loop to continue without waiting for the current chain to finish.
On Windows, the DOS for-loop syntax is one of:
for %i in (SET) do COMMAND COMMAND-ARGUMENTS
for /l %i in (START, STEP, END) do COMMAND COMMAND-ARGUMENTS
To run 4 chains in parallel on Windows:
>for /l %i in (1, 1, 4) do start /b bernoulli.exe sample ^
data file=bernoulli.data.json my_data ^
output file=output_%i.csv
The caret (^) indicates a line continuation in DOS.
## 4.3 Stan CSV output file
Each execution of the model results in draws from a single Markov chain being written to a file in comma-separated value (CSV) format. The default name of the output file is output.csv.
The first part of the output file records the version of the underlying Stan library and the configuration as comments (i.e., lines beginning with the pound sign (#)).
# stan_version_major = 2
# stan_version_minor = 23
# stan_version_patch = 0
# model = bernoulli_model
# method = sample (Default)
# sample
# num_samples = 1000 (Default)
# num_warmup = 1000 (Default)
...
# output
# file = output.csv (Default)
# diagnostic_file = (Default)
# refresh = 100 (Default)
This is followed by a CSV header indicating the names of the values sampled.
lp__,accept_stat__,stepsize__,treedepth__,n_leapfrog__,divergent__,energy__,theta
The first output columns report the HMC sampler information:
• lp__ - the total log probability density (up to an additive constant) at each sample
• accept_stat__ - the average Metropolis acceptance probability over each simulated Hamiltonian trajectory
• stepsize__ - integrator step size
• treedepth__ - depth of tree used by NUTS (NUTS sampler)
• n_leapfrog__ - number of leapfrog calculations (NUTS sampler)
• divergent__ - has value 1 if trajectory diverged, otherwise 0. (NUTS sampler)
• energy__ - value of the Hamiltonian
• int_time__ - total integration time (static HMC sampler)
Because the above header is from the NUTS sampler, it has columns treedepth__, n_leapfrog__, and divergent__ and doesn’t have column int_time__. The remaining columns correspond to model parameters. For the Bernoulli model, it is just the final column, theta.
The header line is written to the output file before warmup begins. If option save_warmup is set to 1, the warmup draws are output directly after the header. The total number of warmup draws saved is num_warmup divided by thin, rounded up (i.e., ceiling).
Following the warmup draws (if any), are comments which record the results of adaptation: the stepsize, and inverse mass metric used during sampling:
# Adaptation terminated
# Step size = 0.884484
# Diagonal elements of inverse mass matrix:
# 0.535006
The default sampler is NUTS with an adapted step size and a diagonal inverse mass matrix. For this example, the step size is 0.884484, and the inverse mass contains the single entry 0.535006 corresponding to the parameter theta.
Draws from the posterior distribution are printed out next, each line containing a single draw with the columns corresponding to the header.
-6.84097,0.974135,0.884484,1,3,0,6.89299,0.198853
-6.91767,0.985167,0.884484,1,1,0,6.92236,0.182295
-7.04879,0.976609,0.884484,1,1,0,7.05641,0.162299
-6.88712,1,0.884484,1,1,0,7.02101,0.188229
-7.22917,0.899446,0.884484,1,3,0,7.73663,0.383596
...
The output ends with timing details:
# Elapsed Time: 0.007 seconds (Warm-up)
# 0.017 seconds (Sampling)
# 0.024 seconds (Total)
## 4.4 Summarizing sampler output(s) with stansummary
The stansummary utility processes one or more output files from a run or set of runs of Stan’s HMC sampler given a model and data. For all columns in the Stan CSV output file stansummary reports a set of statistics including mean, standard deviation, percentiles, effective number of samples, and $$\hat{R}$$ values.
To run stansummary on the output files generated by the for loop above, by the above run of the bernoulli model on Mac or Linux:
<cmdstan-home>/bin/stansummary output_*.csv
On Windows, use backslashes to call the stansummary.exe.
<cmdstan-home>\bin\stansummary.exe output_*.csv
The stansummary output consists of one row of statistics per column in the Stan CSV output file. Therefore, the first rows in the stansummary report statistics over the sampler state. The final row of output summarizes the estimates of the model variable theta:
Inference for Stan model: bernoulli_model
4 chains: each with iter=(1000,1000,1000,1000); warmup=(0,0,0,0); thin=(1,1,1,1); 4000 iterations saved.
Warmup took (0.0070, 0.0070, 0.0070, 0.0070) seconds, 0.028 seconds total
Sampling took (0.020, 0.017, 0.021, 0.019) seconds, 0.077 seconds total
Mean MCSE StdDev 5% 50% 95% N_Eff N_Eff/s R_hat
lp__ -7.3 1.8e-02 0.75 -8.8 -7.0 -6.8 1.8e+03 2.4e+04 1.0e+00
accept_stat__ 0.89 2.7e-03 0.17 0.52 0.96 1.0 3.9e+03 5.1e+04 1.0e+00
stepsize__ 1.1 7.5e-02 0.11 0.93 1.2 1.2 2.0e+00 2.6e+01 2.5e+13
treedepth__ 1.4 8.1e-03 0.49 1.0 1.0 2.0 3.6e+03 4.7e+04 1.0e+00
n_leapfrog__ 2.3 1.7e-02 0.98 1.0 3.0 3.0 3.3e+03 4.3e+04 1.0e+00
divergent__ 0.00 nan 0.00 0.00 0.00 0.00 nan nan nan
energy__ 7.8 2.6e-02 1.0 6.8 7.5 9.9 1.7e+03 2.2e+04 1.0e+00
theta 0.25 2.9e-03 0.12 0.079 0.23 0.46 1.7e+03 2.1e+04 1.0e+00
Samples were drawn using hmc with nuts.
For each parameter, N_Eff is a crude measure of effective sample size,
and R_hat is the potential scale reduction factor on split chains (at
convergence, R_hat=1).
In this example, we conditioned the model on a dataset consisting of the outcomes of 10 bernoulli trials, where only 2 trials reported success. The 5%, 50%, and 95% percentile values for theta reflect the uncertainty in our estimate, due to the small amount of data, given the prior of beta(1, 1)
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2022-08-14 07:16:54
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http://piping-designer.com/index.php/disciplines/mechanical/heating-ventilation-and-air-conditioning/1499-energy-efficiency-rating
|
# Energy Efficiency Rating
Written by Jerry Ratzlaff on . Posted in HVAC
Energy efficiency rating ( $$EER$$ ) measure the efficiency with which a product uses energy to function. It is calculated by dividing a product's BTU output by its wattage.
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2018-04-19 21:27:30
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https://www.physicsforums.com/threads/difficult-differential-equation.723460/
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# Difficult differential equation
1. Nov 18, 2013
### JulieK
I have the following differential equation which I want to solve for $y$ as a function of $x$
$\frac{dy}{dx}=\frac{C_{1}\left(C_{5}y+C_{6}\right)^{2}}{C_{2}\left(C_{3}y+C_{4}\right)-C_{7}\left(C_{5}y+C_{6}\right)^{6}}$
where $C_{1},C_{2},C_{3},C_{4},C_{5},C_{6},C_{7}$ are constants.
Can anyone suggest a method for solving this equation.
2. Nov 18, 2013
### vanhees71
separation of variables should do.
3. Nov 19, 2013
### JulieK
I want y as an explicit function of x not the other way round.
The separation and integration will produce x as a function of y which is not very useful for my purpose.
4. Nov 19, 2013
### PSarkar
You will get $x$ as a function of $y$, say $f(y) = x$. Then you can try and find the inverse to get $y$ as a function of $x$, i.e. $y = f^{-1}(x)$. There are a few things that can go wrong. If $f$ is not invertible then that tells you that there is no unique solution $y(x)$ to the differential equation. Otherwise, another thing that can go wrong is the inverse cannot be written down in terms of elementary functions.
So separation of variables is still the right (only?) approach. It won't change any of the facts above ($f$ invertible and inverse can be written down).
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2017-08-21 02:02:30
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https://numpydl.readthedocs.io/en/stable/tutorials/CNN/
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# 6. Convolution Neural Networks¶
## 6.1. Introduction¶
Convolutional Neural Networks (CNN) are now a standard way of image classification – there are publicly accessible deep learning frameworks, trained models and services. It’s more time consuming to install stuff like caffe [1] than to perform state-of-the-art object classification or detection. We also have many methods of getting knowledge - there is a large number of deep learning courses [2] /MOOCs [3], free e-books [4] or even direct ways of accessing to the strongest Deep/Machine Learning minds such as Yoshua Bengio [5], Andrew NG [6] or Yann Lecun [7] by Quora, Facebook or G+.
Nevertheless, when I wanted to get deeper insight in CNN, I could not find a “CNN backpropagation for dummies”. Notoriously I met with statements like: “If you understand backpropagation in standard neural networks, there should not be a problem with understanding it in CNN” or “All things are nearly the same, except matrix multiplications are replaced by convolutions”. And of course I saw tons of ready equations.
It was a little consoling, when I found out that I am not alone, for example: Hello, when computing the gradients CNN, the weights need to be rotated, Why ? [8]
$\delta_j^l = f'(u_j^l) \odot conv2(\delta_j^{l+1}, rot180(k_j^{l+1}), 'full')$
The answer on above question, that concerns the need of rotation on weights in gradient computing, will be a result of this long post.
## 6.2. Back Propagation¶
We start from multilayer perceptron and counting delta errors on fingers:
We see on above picture that $$\delta_1^1$$ is proportional to deltas from next layer that are scaled by weights.
But how do we connect concept of MLP with Convolutional Neural Network? Let’s play with MLP:
If you are not sure that after connections cutting and weights sharing we get one layer Convolutional Neural Network, I hope that below picture will convince you:
The idea behind this figure is to show, that such neural network configuration is identical with a 2D convolution operation and weights are just filters (also called kernels, convolution matrices, or masks).
Now we can come back to gradient computing by counting on fingers, but from now we will be only focused on CNN. Let’s begin:
No magic here, we have just summed in “blue layer” scaled by weights gradients from “orange” layer. Same process as in MLP’s backpropagation. However, in the standard approach we talk about dot products and here we have … yup, again convolution:
Yeah, it is a bit different convolution than in previous (forward) case. There we did so called valid convolution, while here we do a full convolution (more about nomenclature here [9] ). What is more, we rotate our kernel by 180 degrees. But still, we are talking about convolution!
Now, I have some good news and some bad news:
• you see (BTW, sorry for pictures aesthetics), that matrix dot products are replaced by convolution operations both in feed forward and backpropagation.
• you know that seeing something and understanding something … yup, we are going now to get our hands dirty and prove above statement before getting next, I recommend to read, mentioned already in the disclaimer, chapter 2 [10] of M. Nielsen book. I tried to make all quantities to be consistent with work of Michael.
In the standard MLP, we can define an error of neuron $$j$$ as:
$\delta_j^l = \frac{\partial C}{\partial z_j^l}$
where $$z_j^l$$ is just:
$z^l_j = \sum_{k} w_{jk}^l a_k^{l-1} + b_j^l$
and for clarity, $$a_j^l = \sigma(z_j^l)$$, where $$\sigma$$ is an activation function such as sigmoid, hyperbolic tangent or relu [11].
But here, we do not have MLP but CNN and matrix multiplications are replaced by convolutions as we discussed before. So instead of $$z_j$$ we do have a $$z_{x,y}$$:
$z_{x,y}^{l+1} = w^{l+1} * \sigma(z_{x,y}^l) + b_{x,y}^{l+1} = \sum_{a} \sum_{b} w_{a,b}^{l+1}\sigma(z_{x-a,y-b}^l)+ b_{x,y}^{l+1}$
Above equation is just a convolution operation during feedforward phase illustrated in the above picture titled ‘Feedforward in CNN is identical with convolution operation’[12]
Now we can get to the point and answer the question Hello, when computing the gradients CNN, the weights need to be rotated, Why ? [8]
We start from statement:
$\delta_{x,y}^l = \frac{\partial C}{\partial z_{x,y}^l} = \sum_{x'} \sum_{y'}\frac{\partial C}{\partial z_{x',y'}^{l+1}} \frac{\partial z_{x',y'}^{l+1}}{\partial z_{x,y}^l}$
We know that $$z_{x,y}^l$$ is in relation to $$z_{x',y'}^{l+1}$$ which is indirectly showed in the above picture titled ‘Backpropagation also results with convolution’. So sums are the result of chain rule. Let’s move on:
$\begin{split}\frac{\partial C}{\partial z_{x,y}^l} & = \sum_{x'} \sum_{y'}\frac{\partial C}{\partial z_{x',y'}^{l+1}} \frac{\partial z_{x',y'}^{l+1}}{\partial z_{x,y}^l} \\ & = \sum_{x'} \sum_{y'} \delta_{x',y'}^{l+1} \frac{\partial(\sum_{a}\sum_{b}w_{a,b}^{l+1} \sigma(z_{x'-a, y'-b}^l) + b_{x',y'}^{l+1})}{\partial z_{x,y}^l}\end{split}$
First term is replaced by definition of error, while second has become large because we put it here expression on $$z_{x',y'}^{l+1}$$. However, we do not have to fear of this big monster – all components of sums equal 0, except these ones that are indexed: $$x=x'-a$$ and $$y=y'-b$$. So:
$\sum_{x'} \sum_{y'} \delta_{x',y'}^{l+1} \frac{\partial(\sum_{a}\sum_{b}w_{a,b}^{l+1} \sigma(z_{x'-a, y'-b}^l) + b_{x',y'}^{l+1})} {\partial z_{x,y}^l} = \sum_{x'} \sum_{y'} \delta_{x',y'}^{l+1} w_{a,b}^{l+1} \sigma'(z_{x,y}^l)$
If “math;x=x’-a and $$y=y'-b$$ then it is obvious that $$a=x'-x$$ and $$b=y'-y$$ so we can reformulate above equation to:
$\sum_{x'} \sum_{y'} \delta_{x',y'}^{l+1} w_{a,b}^{l+1} \sigma'(z_{x,y}^l) =\sum_{x'}\sum_{y'} \delta_{x',y'}^{l+1} w_{x'-x,y'-y}^{l+1} \sigma'(z_{x,y}^l)$
OK, our last equation is just …
$\sum_{x'}\sum_{y'} \delta_{x',y'}^{l+1} w_{x'-x,y'-y}^{l+1} \sigma'(z_{x,y}^l)= \delta^{l+1} * w_{-x,-y}^{l+1} \sigma'(z_{x,y}^l)$
Where is the rotation of weights? Actually $$ROT180(w_{x,y}^{l+1}) = w_{-x, -y}^{l+1}$$.
So the answer on question Hello, when computing the gradients CNN, the weights need to be rotated, Why ? [8] is simple: the rotation of the weights just results from derivation of delta error in Convolution Neural Network.
OK, we are really close to the end. One more ingredient of backpropagation algorithm is update of weights $$\frac{\partial C}{\partial w_{a,b}^l}$$:
$\begin{split}\frac{\partial C}{\partial w_{a,b}^l} & = \sum_{x} \sum_{y} \frac{\partial C}{\partial z_{x,y}^l}\frac{\partial z_{x,y}^l}{\partial w_{a,b}^l} \\ & = \sum_{x}\sum_{y}\delta_{x,y}^l \frac{\partial(\sum_{a'}\sum_{b'}w_{a',b'}^l\sigma(z_{x-a', y-b'}^l) + b_{x,y}^l)}{\partial w_{a,b}^l} \\ & =\sum_{x}\sum_{y} \delta_{x,y}^l \sigma(z_{x-a,y-b}^{l-1}) \\ & = \delta_{a,b}^l * \sigma(z_{-a,-b}^{l-1}) \\ & =\delta_{a,b}^l * \sigma(ROT180(z_{a,b}^{l-1}))\end{split}$
So paraphrasing the backpropagation algorithm [13] for CNN:
1. Input $$x$$: set the corresponding activation $$a^1$$ for the input layer.
2. Feedforward: for each $$l = 2,3, \cdots ,L$$, compute $$z_{x,y}^l = w^l * \sigma(z_{x,y}^{l-1}) + b_{x,y}^l$$ and $$a_{x,y}^l = \sigma(z_{x,y}^l)$$
3. Output error $$\delta^L$$: Compute the vector $$\delta^L = \nabla_a C \odot \sigma'(z^L)$$
4. Backpropagate the error: For each $$l=L-1,L-2,\cdots ,2$$, compute $$\delta_{x,y}^l =\delta^{l+1} * ROT180(w_{x,y}^{l+1}) \sigma'(z_{x,y}^l)$$
5. Output: The gradient of the cost function is given by $$\frac{\partial C}{\partial w_{a,b}^l} =\delta_{a,b}^l * \sigma(ROT180(z_{a,b}^{l-1}))$$
## 6.3. Visualizing Features¶
It’s been shown many times that convolutional neural nets are very good at recognizing patterns in order to classify images. But what patterns are they actually looking for?
I attempted to recreate the techniques described in [14] to project features in the convnet back to pixel space.
In order to do this, we first need to define and train a convolutional network. Due to lack of training power, I couldn’t train on ImageNet and had to use CIFAR-10, a dataset of $$32x32$$ images in 10 classes. The network structure was pretty standard: two convolutional layers, each with $$2x2$$ max pooling and a reLu gate, followed by a fully-connected layer and a softmax classifier.
We’re only trying to visualize the features in the convolutional layers, so we can effectively ignore the fully-connected and softmax layers.
Features in a convolutional network are simply numbers that represent how present a certain pattern is. The intuition behind displaying these features is pretty simple: we input one image, and retrieve the matrix of features. We set every feature to 0 except one, and pass it backwards through the network until reaching the pixel layer. The challenge here lies in how to effectively pass data backwards through a convolutional network.
We can approach this problem step-by-step. There are three main portions to a convolutional layer. The actual convolution, some max-pooling, and a nonlinearity (in our case, a rectified linear unit). If we can figure out how to calculate the inputs to these units given their outputs, we can pass any feature back to the pixel input.
Here, the paper introduces a structure called a deconvolutional layer. However, in practice, this is simply a regular convolutional layer with its filters transposed. By applying these transposed filters to the output of a convolutional layer, the input can be retrieved.
A max-pool gate cannot be reversed on its own, as data about the non-maximum features is lost. The paper describes a method in which the positions of each maximum is recorded and saved during forward propagation, and when features are passed backwards, they are placed where the maximums had originated from. In my recreation, I took an even simpler route and just set the whole $$2x2$$ square equal to the maximum activation.
Finally, the rectified linear unit. It’s the easiest one to reverse, we just need to pass the data through a reLu again when moving backwards.
To test these techniques out, I trained a standard convolutional network on CIFAR-10. First, I passed one of the training images, a dog, through the network and recorded the various features.
As you can see, there are quite a variety of patterns the network is looking for. You can see evidence of the original dog picture in these feature activations, most prominently the arms.
Now, let’s see how these features change when different images are passed through.
This image shows all the different pixel representations of the activations of feature #7, when a variety of images are used. It’s clear that this feature activates when green is present. You can really see the original picture in this feature, since it probably just captures the overall color green rather than some specific pattern.
Finally, to gain some intuition of how images activated each feature, I passed in a whole batch of images and saved the maximum activations.
Which features were activated by which images? There’s some interesting stuff going on here. Some of the features are activated simply by the presence of a certain color. The green frog and red car probably contained the most of their respective colors in the batch of images.
However, here are two features which are activated the most by a red frog image. The feature activations show an outline, but one is in red and the other is in blue. Most likely, this feature isn’t getting activated by the frog itself, but by the black background. Visualizing the features of a convolutional network allows us to see such details.
So, what happens if we go farther, and look at the second convolutional layer?
I took the feature activations for the dog again, this time on the second convolutional layer. Already some differences can be spotted. The presence of the original image here is much harder to see.
It’s a good sign that all the features are activated in different places. Ideally, we want features to have minimal correlation with one another.
Finally, let’s examine how a second layer feature activates when various images are passed in.
For the majority of these images, feature #9 activated at dark locations of the original image. However, there are still outliers to this, so there is probably more to this feature than that.
For most features, it’s a lot harder to tell what part of the image activated it, since second layer features are made of any linear combination of first layer features. I’m sure that if the network was trained on a higher resolution image set, these features would become more apparent.
## 6.4. Code¶
try:
import tensorflow as tf
import numpy as np
import pickle
from tensorflow.python.platform import gfile
from random import randint
import os
from scipy.misc import imsave
from matplotlib import pyplot as plt
except ImportError:
raise ValueError("Please install tensorflow and matplotlib.")
def unpickle(file):
fo = open(file, 'rb')
fo.close()
return dict
def initWeight(shape):
weights = tf.truncated_normal(shape, stddev=0.1)
return tf.Variable(weights)
def initBias(shape):
bias = tf.constant(0.1, shape=shape)
return tf.Variable(bias)
# the convolution with padding of 1 on each side, and moves by 1.
def conv2d(x, W):
return tf.nn.conv2d(x, W, strides=[1, 1, 1, 1], padding="SAME")
# max pooling basically shrinks it by 2x, taking the highest value on each feature.
def maxPool2d(x):
return tf.nn.max_pool(x, ksize=[1, 2, 2, 1], strides=[1, 2, 2, 1], padding="SAME")
batchsize = 50
imagesize = 32
colors = 3
sess = tf.InteractiveSession()
img = tf.placeholder("float", shape=[None, imagesize, imagesize, colors])
lbl = tf.placeholder("float", shape=[None, 10])
# for each 5x5 area, check for 32 features over 3 color channels
wConv1 = initWeight([5, 5, colors, 32])
bConv1 = initBias([32])
# move the conv filter over the picture
conv1 = conv2d(img, wConv1)
bias1 = conv1 + bConv1
# relu = max(0,x), adds nonlinearality
relu1 = tf.nn.relu(bias1)
# maxpool to 16x16
pool1 = maxPool2d(relu1)
# second conv layer, takes a 16x16 with 32 layers, turns to 8x8 with 64 layers
wConv2 = initWeight([5, 5, 32, 64])
bConv2 = initBias([64])
conv2 = conv2d(pool1, wConv2)
bias2 = conv2 + bConv2
relu2 = tf.nn.relu(bias2)
pool2 = maxPool2d(relu2)
# fully-connected is just a regular neural net: 8*8*64 for each training data
wFc1 = initWeight([(imagesize / 4) * (imagesize / 4) * 64, 1024])
bFc1 = initBias([1024])
# reduce dimensions to flatten
pool2flat = tf.reshape(pool2, [-1, (imagesize / 4) * (imagesize / 4) * 64])
# 128 training set by 2304 data points
fc1 = tf.matmul(pool2flat, wFc1) + bFc1
relu3 = tf.nn.relu(fc1)
# dropout removes duplicate weights
keepProb = tf.placeholder("float")
drop = tf.nn.dropout(relu3, keepProb)
wFc2 = initWeight([1024, 10])
bFc2 = initWeight([10])
# softmax converts individual probabilities to percentages
guesses = tf.nn.softmax(tf.matmul(drop, wFc2) + bFc2)
# how wrong it is
cross_entropy = -tf.reduce_sum(lbl * tf.log(guesses + 1e-9))
# theres a lot of tensorflow optimizers such as gradient descent
# adam is one of them
# array of bools, checking if each guess was correct
correct_prediction = tf.equal(tf.argmax(guesses, 1), tf.argmax(lbl, 1))
# represent the correctness as a float [1,1,0,1] -> 0.75
accuracy = tf.reduce_mean(tf.cast(correct_prediction, "float"))
sess.run(tf.initialize_all_variables())
batch = unpickle("cifar-10-batches-py/data_batch_1")
validationData = batch["data"][555:batchsize + 555]
validationRawLabel = batch["labels"][555:batchsize + 555]
validationLabel = np.zeros((batchsize, 10))
validationLabel[np.arange(batchsize), validationRawLabel] = 1
validationData = validationData / 255.0
validationData = np.reshape(validationData, [-1, 3, 32, 32])
validationData = np.swapaxes(validationData, 1, 3)
saver = tf.train.Saver()
saver.restore(sess, tf.train.latest_checkpoint(os.getcwd() + "/training/"))
# train for 20000
# print mnistbatch[0].shape
def train():
for i in range(20000):
randomint = randint(0, 10000 - batchsize - 1)
trainingData = batch["data"][randomint:batchsize + randomint]
rawlabel = batch["labels"][randomint:batchsize + randomint]
trainingLabel = np.zeros((batchsize, 10))
trainingLabel[np.arange(batchsize), rawlabel] = 1
trainingData = trainingData / 255.0
trainingData = np.reshape(trainingData, [-1, 3, 32, 32])
trainingData = np.swapaxes(trainingData, 1, 3)
if i % 10 == 0:
train_accuracy = accuracy.eval(feed_dict={
img: validationData, lbl: validationLabel, keepProb: 1.0})
print("step %d, training accuracy %g" % (i, train_accuracy))
if i % 50 == 0:
saver.save(sess, os.getcwd() + "/training/train", global_step=i)
optimizer.run(feed_dict={img: trainingData, lbl: trainingLabel, keepProb: 0.5})
print(i)
def unpool(value, name='unpool'):
"""N-dimensional version of the unpooling operation from
https://www.robots.ox.ac.uk/~vgg/rg/papers/Dosovitskiy_Learning_to_Generate_2015_CVPR_paper.pdf
:param value: A Tensor of shape [b, d0, d1, ..., dn, ch]
:return: A Tensor of shape [b, 2*d0, 2*d1, ..., 2*dn, ch]
"""
with tf.name_scope(name) as scope:
sh = value.get_shape().as_list()
dim = len(sh[1:-1])
out = (tf.reshape(value, [-1] + sh[-dim:]))
for i in range(dim, 0, -1):
out = tf.concat(i, [out, out])
out_size = [-1] + [s * 2 for s in sh[1:-1]] + [sh[-1]]
out = tf.reshape(out, out_size, name=scope)
return out
def display():
print("displaying")
batchsizeFeatures = 50
imageIndex = 56
inputImage = batch["data"][imageIndex:imageIndex + batchsizeFeatures]
inputImage = inputImage / 255.0
inputImage = np.reshape(inputImage, [-1, 3, 32, 32])
inputImage = np.swapaxes(inputImage, 1, 3)
inputLabel = np.zeros((batchsize, 10))
inputLabel[np.arange(1), batch["labels"][imageIndex:imageIndex + batchsizeFeatures]] = 1
# inputLabel = batch["labels"][54]
# prints a given image
# saves pixel-representations of features from Conv layer 1
featuresReLu1 = tf.placeholder("float", [None, 32, 32, 32])
unReLu = tf.nn.relu(featuresReLu1)
unBias = unReLu
unConv = tf.nn.conv2d_transpose(unBias, wConv1, output_shape=[batchsizeFeatures, imagesize, imagesize, colors],
activations1 = relu1.eval(feed_dict={img: inputImage, lbl: inputLabel, keepProb: 1.0})
print(np.shape(activations1))
# display features
for i in range(32):
isolated = activations1.copy()
isolated[:, :, :, :i] = 0
isolated[:, :, :, i + 1:] = 0
print(np.shape(isolated))
totals = np.sum(isolated, axis=(1, 2, 3))
best = np.argmin(totals, axis=0)
print(best)
pixelactive = unConv.eval(feed_dict={featuresReLu1: isolated})
# totals = np.sum(pixelactive,axis=(1,2,3))
# best = np.argmax(totals,axis=0)
# best = 0
saveImage(pixelactive[best], "activ" + str(i) + ".png")
saveImage(inputImage[best], "activ" + str(i) + "-base.png")
# display same feature for many images
# for i in xrange(batchsizeFeatures):
# isolated = activations1.copy()
# isolated[:,:,:,:6] = 0
# isolated[:,:,:,7:] = 0
# pixelactive = unConv.eval(feed_dict={featuresReLu1: isolated})
# totals = np.sum(pixelactive,axis=(1,2,3))
# best = np.argmax(totals,axis=0)
# saveImage(pixelactive[i],"activ"+str(i)+".png")
# saveImage(inputImage[i],"activ"+str(i)+"-base.png")
# saves pixel-representations of features from Conv layer 2
featuresReLu2 = tf.placeholder("float", [None, 16, 16, 64])
unReLu2 = tf.nn.relu(featuresReLu2)
unBias2 = unReLu2
unConv2 = tf.nn.conv2d_transpose(unBias2, wConv2,
output_shape=[batchsizeFeatures, imagesize / 2, imagesize / 2, 32],
unPool = unpool(unConv2)
unReLu = tf.nn.relu(unPool)
unBias = unReLu
unConv = tf.nn.conv2d_transpose(unBias, wConv1, output_shape=[batchsizeFeatures, imagesize, imagesize, colors],
activations1 = relu2.eval(feed_dict={img: inputImage, lbl: inputLabel, keepProb: 1.0})
print(np.shape(activations1))
# display features
# for i in xrange(64):
# isolated = activations1.copy()
# isolated[:,:,:,:i] = 0
# isolated[:,:,:,i+1:] = 0
# pixelactive = unConv.eval(feed_dict={featuresReLu2: isolated})
# # totals = np.sum(pixelactive,axis=(1,2,3))
# # best = np.argmax(totals,axis=0)
# best = 0
# saveImage(pixelactive[best],"activ"+str(i)+".png")
# saveImage(inputImage[best],"activ"+str(i)+"-base.png")
# display same feature for many images
# for i in xrange(batchsizeFeatures):
# isolated = activations1.copy()
# isolated[:,:,:,:8] = 0
# isolated[:,:,:,9:] = 0
# pixelactive = unConv.eval(feed_dict={featuresReLu2: isolated})
# totals = np.sum(pixelactive,axis=(1,2,3))
# # best = np.argmax(totals,axis=0)
# # best = 0
# saveImage(pixelactive[i],"activ"+str(i)+".png")
# saveImage(inputImage[i],"activ"+str(i)+"-base.png")
def saveImage(inputImage, name):
# red = inputImage[:1024]
# green = inputImage[1024:2048]
# blue = inputImage[2048:]
# formatted = np.zeros([3,32,32])
# formatted[0] = np.reshape(red,[32,32])
# formatted[1] = np.reshape(green,[32,32])
# formatted[2] = np.reshape(blue,[32,32])
# final = np.swapaxes(formatted,0,2)/255
final = inputImage
final = np.rot90(np.rot90(np.rot90(final)))
imsave(name, final)
def main(argv=None):
display()
# train()
if __name__ == '__main__':
tf.app.run()
[14] (1, 2) Zeiler, Matthew D., and Rob Fergus. “Visualizing and understanding convolutional networks.” European conference on computer vision. Springer International Publishing, 2014.
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2021-11-27 23:26:25
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https://webarchive.nationalarchives.gov.uk/20110810103412/http:/teachingandlearningresources.org.uk/node/24878
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# Step 1
Recognise unit fractions such as $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, $\frac{1}{5}$, $\frac{1}{10}$ … and use them to find fractions of shapes and numbers
### Probing questions
• What numbers are easy to find a third/quarter/fifth/tenth of? Why?
• If I cut a cake into four pieces will each piece be a quarter?
### What if pupils find this a barrier?
Use a counting stick to discuss halves and tenths. Some pupils think if there are two parts each must be a half – emphasise that the parts need to be equal.
Mathematics ITP: Fractions (SWF-18 KB) Attachments can provide a useful visual image for pupils.
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2019-06-19 03:41:11
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https://ask.puppet.com/question/23336/run-exec-on-agent/
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# Run exec on agent [closed]
Hi,
I have set up master-agent architecture for puppet. I am basically trying to install & configure some packages on agent. For that I need to run some scripts & programs which are stored on server in the files directory of respective modules. I am using exec resource to run those script & programs with absolute modulepath. But when I run test on agent exec fails due to scripts in-availability.
I am not much aware of master-agent architecture. So how can I tell agent to fetch those scripts & programs which are stored on master in files directory of those modules & run it. Is this approach right or there is another elegant way to do it?
Following is a snippet of one of exec resources:
exec {
command => 'easy_install /etc/puppet/modules/configure/files/configure.py',
path => ['/bin/','/sbin/','/usr/bin/','/usr/sbin/'],
}
Also how would the source parameter (for file() & template())behave for file resource in master-agent architecture? Will I have to copy the files on agent or they will be fetched from master?
edit retag reopen merge delete
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2019-02-22 01:38:50
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https://hometeaminsider.com/uh5880rl/99118e-projection-matrix-example
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# projection matrix example
In other words, we can compute the closest vector by solving a system of linear equations. Soc., 1997. the null space of 1, 1, 1. Or another way to view it is 3 by 3 identity matrix, times x, right? entry equal a 1 here. that that's equal to some other matrix C, times x. are going to be 2/3, so we could just go down You take A transpose, you can do Orthogonal and Oblique Projections Projections De nition A matrix N2R N is a projection matrix if 2 = Some direct consequences range( ) is invariant under the action of 0 and 1 are the only possible eigenvalues of let k be the rank of : Then, there exists a basis X such that = X I k 0 N k X 1 8/38 equation is that this matrix must be equal to these These two statements So let's see if we can figure video, this one will be easy. 1/3 times 3 is equal to 1. A square So just like that we were able Twitter Facebook. # # # $% & & & A= 10 11 01! " Let me construct some matrix D, All of the vectors that satisfy Which is equal to what? inverse matrix, for the 1 by 1 matrix 3. Or this case it'll just be 0. Actually, I've never defined the Now we at least had a hunch that be 1 times 1, which is 1. Khan Academy is a 501(c)(3) nonprofit organization. of this matrix right here. If P is the matrix for projecting onto W, then W = col (P). Explore anything with the first computational knowledge engine. So 1 minus 1/3 is 2/3. Or we can write that v's transformations. a projection matrix has norm equal to one, unless . 1, 1, 1, times x1, x2, x3 is equal to the 0 vector. Description Usage Arguments Details Value Note Author(s) Examples. could write that the identity matrix times x is equal In the lesson on Geometry we have explained that to go from one order to the other we can simply transpose the ⦠the problem actually-- remember that v was equal to, vectors, we can say x2 is equal to, let's say The eigenvalues of a projection matrix must be 0 or 1. Towards the end, I examine the orthogonal projection matrix and provide many examples and exercises. the orthogonal complement of our subspace. transformation matrix for the projection onto v is equal to To figure out the projection matrix So let's see if this is easier You have minus 1/3, minus Once vertices are in camera space, they can finally be transformed into clip space by applying a projection transformation. is all of the vectors that satisfy this equation. Systems of Linear Equations (and System Equivalency) [Video] Canonical Forms and Jordan Blocks. projection onto the orthogonal complement of v of x, let's say simpler than if we have to do all of this business A word of warning again. Portions of this entry contributed by Mohammad this is equal to this definition here. times x. So just to visualize what of defining our subspace. That was the whole motivation going to be equal to, and we saw this, it's going to be equal What do you do? Robert Collins Basic Perspective Projection X Y Z f O p = (x,y,f) x y Z Y y f Z X x f O.Camps, PSU X Z P =(X,Y,Z) x y Scene Point Image Point Perspective Projection Eqns Y So how do we represent this as a matrix equation? This is D, just like that. And you could rewrite this as v But our hunch is maybe if had a 3 by 2 matrix. So x1 is equal to minus And then what is x2 equal to? Rowland. a linear transformation of x, I could just write it as the matrix, for the projection of any vector x onto v, by space-- let me write it this way-- the null space of 1, Maybe, I don't know. with this matrix. just apply this, kind of, that we can just solve for 1 times 1, plus 1 times 1, plus essentially finding this guy first, for finding the matrix for v's subspace, we'd have to do this with and (b) the projection matrix P that projects any vector in R 3 to the C(A). Do they consider the green triangle to be in the front or the back of the structure? True! A projection onto a subspace is a linear transformation. In an orthogonal projection, any vector can be written , equal to 1/3, that's 1/3, times the vector 1, 1, 1, From MathWorld--A Wolfram Web Resource. That's the same thing as x. Hints help you try the next step on your own. Put simply, an orthographic projectionis a way ⦠And that's not too hard to do. to get-- that was a pretty straightforward situation-- 1, 1, just like that. The matrix we will present in this chapter is different from the projection matrix that is being used in APIs such as OpenGL or Direct3D. For the sake of legibility, denote the projection simply by in what follows. plus 1 times x3 is going to equal the 0 vector. Times D transpose. In the lesson 3D Viewing: the Pinhole Camera Model we learned how to compute the screen coordinates (left, right, top and bottom) based on the camera near clipping plane and angle-of-view (in fact, we learned how to ⦠Plus C times x. we only have one column in it, so its column matrix, minus the transformation matrix for the matrix that gives a vector space projection It seems pretty difficult. Anyway, I thought that A square matrix with real numbers or elements is said to be an orthogonal matrix, if its transpose is equal to its inverse matrix or we can say, when the product of a square matrix and its transpose gives an identity matrix, then the square matrix is known as an orthogonal matrix. whose columns are the basis vectors for the orthogonal And you can do it. Let me rewrite it. A projection matrix is a symmetric going to be equal to B. multiply this out. 1/3, minus 1/3. 1 by 1 identity matrix. First, it is important to remember that matrices in OpenGL are defined using a column-major order (as opposed to row-major order). a 1 by 1 matrix. Construct an age or stage-structure projection model from a transition table listing stage in time t, fate in time t+1, and one ⦠The projection matrix corresponding to a linear model is symmetric and idempotent, that is, P 2 = P. {\displaystyle \mathbf {P} ^ {2}=\mathbf {P} } . transformation matrix for the orthogonal projection, for Or another way to say it is that Now that we know what a projection matrix is, we can learn how to derive it. But this is the transformation basis vector, so it's going to be that. just like that. 2. products exhibit the distributive property, so we any vector in R3 onto v's orthogonal complement is going lie in that plane. which is essentially equivalent to a scalar. Why? It's a 1 by 1 matrix, You have A here. And now we just figured of the linear combinations of this guy. vector in R3 onto v's orthogonal complement. Remember, the null space, its It's 1/3, 1/3, 1/3. to be equal to? Example 2 "¥" Find (a) the projection of vector on the column space of matrix ! Now we know that if x is a of v of x. let me do B. So 1 times x1, plus 1 times x2, Though, it technically produces the same results. And you can see, this is a lot Now what is the inverse Or another way of writing this, Any vector in is fixed by the projection matrix for any in . be some line. New York: Academic Press, 1990. But you saw it is actually and this just becomes a 1. are equivalent. actually a basis for v because they're linearly independent. Kadison, R. V. and Ringrose, J. R. Fundamentals of the Theory of Operator Algebras, Vol. Just like that. minus 1, 0, and 1. to be equal to D times D transpose D inverse, times https://mathworld.wolfram.com/ProjectionMatrix.html. to be equal to? Orthogonal Projection Matrix â¢Example: Let W be the 2-dimensional subspace of R3 with equation x 1 âx 2 +2x 3 = 0. Minus 1/3, minus the transformation matrix for the projection of any vector And it'll be very similar to is v is equal to the span of the vectors minus 1, 1, So what is this going way-- all of the x1, x2, x3's, so all the vectors like this This is another way If we're dealing with a 1 by 1 Providence, So D transpose D is just So making the third row of the projection matrix = [0, 0, 1, 1] would kind of do the trick. Transformations and Basic Computer Graphics. RI: Amer. any vector in R3 onto the orthogonal complement of v, is Related Article. technique we did before, we could set some vector, we could So remember-- let me rewrite sides, we get that B is equal to I, is equal to the identity Letâs introduce w. We will now have (x,y,z,w) vectors. So this is going to be a a basis for v. So given, that just using the Murphy, G. J. C-*-Algebras so, An example of a nonsymmetric projection matrix is, The case of a complex vector space is analogous. The column space of P is spanned by a because for any b, Pb lies on the line determined by a. out what v's orthogonal complement is. the 3 by 2 matrix. can figure out. projection of x onto the orthogonal complement of v. So we can write that x is equal 1/3, 1/3, 1/3, 1/3, 1/3, 1/3, Remember, the whole point of transformation, so it can be represented as some going to make this work out, to get this entry I'll just take The #1 tool for creating Demonstrations and anything technical. And likewise there's no way I If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. let me just draw a line here-- this thing is equal to 1/3-- In this article we will try to understand in details one of the core mechanics of any 3D engine, the chain of matrix transformations that allows to represent a 3D object on a 2D monitor.We will try to enter into the details of how the matrices are constructed and why, so this article is not meant for absolute beginners. two matrices. matrix of . Examples Orthogonal projection. View source: R/projection.matrix.R. In all OpenGL books and references, the perspective projection matrix used in OpenGL is defined as:What similarities does this matrix have with the matrix we studied in the previous chapter? In the lecture on complementary subspaces we have shown that, if is a basis for , is a basis for , and then is a basis for . matrix right there. And we said that the identity equivalent to the row space or the column space So this is equal to D-- which was pretty neat. Everything is 1/3. That is v right there. use a letter that I haven't used before. This is going to be equal to out like that. I don't know, let me Let me refer back to what is equal to some arbitrary constant, C3. the x on this side, we know that the matrix vector just take that out. to deal with. It's very hairy and you might matrix iff the vector Direct3D can use the w-component of a vertex that has been transformed by the world, view, and projection matrices to perform depth-based calculations in depth-buffer or fog effects. of that and that. All of these entries are going of a 1 by 1 matrix? actually, I don't want to confuse you. make some careless mistakes. and 0, and the vector minus 1, 0, and 1. Our mission is to provide a free, world-class education to anyone, anywhere. An When you rotate a point or a direction, you get the same result. tilted more, and so is this, but it's going to we're doing here, that original equation for v, that The second entry is going to inverse of a 1 by 1 matrix for you just now, so it's then we could say that x1 is equal to minus x2, minus x3. Consequently, and Operator Theory. Moslehian, Mohammad Sal; Rowland, Todd; and Weisstein, Eric W. "Projection w, that is in the orthogonal complement of the subspace, is very easy. where the inner product is the Hermitian inner product. the projection of any vector x in our 3 onto v is Check the two properties of orthogonal projection matrix to confirm. this subspace right there. 1/3 times, we have a 3 by 1 times a 1 by 3 matrix, The projection matrix can be calculated like so. So it's going to be is just a plane in R3, so this subspace is a plane in R3. To be explicit, we state the theorem as a recipe: in the -algebra , where is assumed to be disconnected with two components And all of the 1's minus 1/3 plus 0, times C3. 6 b= 1 1 1! " B is equal to the 3 by 3 identity matrix, minus C, and x2 is just equal to C2. because this is a 3 by 2 matrix, instead of there is the projection of x onto v, and this is the equal to the set of all x1's, x2's, and x3's that are equal video and the video before that, that the projection of This is saying that v is equal satisfies that, that's just going to be some plane in R3. space is going to be the span of that one column. Plus C3 times minus 1. just call it T. And let me do another. this guy's entry times that guy's entry, is going to We are going to generate the transformation that satisfies the above requirement and we have an additional requirement we want to "piggyback" on it which is to make life easier for the clipper by representing the projected coordinates in a normalized space of -1 to +1. Suppose you want someone in another country to design this triangular structure for you. so it's 0 times C2, plus 1, times C3. So remember, the projection-- I give an intuitive example of how projection matrices work. equal to a 1 by 1 matrix 3. So let's see what this is. going to be minus 1/3. We saw that multiple times. matrix, then I'm just trying to figure out what, let's say, But a 4d projection matrix that's going to be applied to vectors in homogenous coordinates with w = 1 can also offset z by z*whatever. So D transpose is just going D transpose, times x. to the null space of this matrix right there. Let's see, let's, in our heads, What is D transpose times D? Description. that is equal to all of the vectors-- let me write it this Then we can say that v, we can So the orthogonal complement of Let's say I have a subspace v out v in kind of the traditional way. entry equal a 1 here. what matrix times 3 is going to be equal to the 1 by If we subtract C from both our original. A projection matrix is a Hermitian matrix iff the vector space projection In an orthogonal projection, any vector can be written, so (2) An example of a nonsymmetric projection matrix is to be 1/3 essentially, if we multiply this out like that. components satisfy, or that lie in this plane, whose entries However, for a translation (when you m⦠To use Khan Academy you need to upgrade to another web browser. You don't speak their language, so you can't explain it to them. matrix vector products. That is the transformation v compliment is going to be can write it in, kind of, our parametric form, or if we And if you want to factor out Well x, if I want to write it as projection onto v's orthogonal complement. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. to C2 times-- so x1 is equal to minus-- let me rewrite We could write the 0 vector a 4 by 2 matrix. And we're going to have for doing it. Now by definition, that right Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. The third entry is going to We figured out that v is the this whole thing, but that might be pretty hairy. is this matrix, 1, 1, 1-- times D transpose D inverse. where denotes the adjoint Then find the projection matrix's image. So these are also The two most common types of projection are perspective and orthographic. to the projection onto v of x, plus the projection onto matrix is a projection matrix iff . Until then, we only considered 3D vertices as a (x,y,z) triplet. So this right here 3 by 3 matrix of 1's. We could say x1, if we assume The null space of this matrix That's a harder matrix numbers right there. 1: Elementary Theory. onto v of x is equal to B times x, we know that C2, minus C3. It actually turns out in the B given that the identity matrix minus this guy is Matrix." A projection matrix is an square def calc_proj_matrix(A): return A*np.linalg.inv(A.T*A)*A.T. Whilst a projection of b onto the plane ⦠# # #$ % & & & Answer: There are two ways to determine projection vector p. Method 1: Determine the coefficient vector x ö based on ATe=0, then determine p from p=Ax ö any member of R3 can be represented this way. Example: [1 0 0 1]â[2 3 4 5] = [ 1â2 0â3 0â4 1â5] = [ â1 â3 â4 â4] [ 1 0 0 1] â [ 2 3 4 5] = [ 1 â 2 0 â 3 0 â 4 1 â 5] = [ â 1 â 3 â 4 â 4] Matrix multiplication with a scalar (or matrix multiplication with a number) is the operation of multiplying every element of the matrix with a scalar. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. the real function defined by on and on is a projection here by doing all of this A transpose and, you know, I wrote way up here. this problem is to figure out this thing right here, You can figure out what the C2, plus C3 times what? that, let's say, that x2 and x3 are kind of free variables, We need to introduce homogeneous coordinates. projection matrix to get to the production onto many times before. Math. So if we say that the projection Now, we know that this thing transpose of this guy. And we know this is a linear transformation onto v's orthogonal complement. So this first entry is going to element is called projection if and . So let me write that here. If w == 1, then the vector (x,y,z,1) is a position in space. We could find the basis for You can take A transpose A, the next video. Computations such as these require that your projection matrix normalize w to be equivalent to world-space z. that satisfy x1 plus x2 plus x3 is equal to 0. So v is equal to the null (In fact, remember this forever.) out the projection matrix, if we can figure out the So it's 1 times C2, How to derive the projection matrix. that A inverse times A is equal to the identity matrix. Another example of a projection matrix (video) | Khan Academy space projection is orthogonal. And that's for any real ⦠Construct an age or stage-structure projection model from a transition table listing stage in time t, fate in time t+1, and one or more individual fertility columns. Let be a -algebra. Let's see if we can figure out Thankfully, we have orthographic projections to help in situations like this. be 1 times 1, which is 1. So if you think about it, this this with a C2-- this is equal to C2. matrix. a lot of work. And then we can figure out that Well, the only matrix that's by doing all of this silliness here. This is equal to C3. projection onto v, plus the transformation matrix for the to be equal to? don't know, let me call this matrix T, let me Perspective projection results in the natural effect of things appearing smaller the further away they are from the viewer. âHe/she hates me!â Whether at home, at work or in any other situation, we have all believed that ⦠simple, but this is the inverse, that right there is the projection transformations-Both these transformations are nonsingular-Default to identity matrices (orthogonal view) â¢Normalization lets us clip against simple cube regardless of type of projection â¢Delay final projection until end-Important for hidden-surface removal to ⦠then you can invert it. orthogonal complement-- a null space's orthogonal complement is And I'm interested in finding I think you see the pattern. So we get the projection of So this is by definition, that Fundamentals of the Theory of Operator Algebras, Vol. And we know that these are It's going to be all So let's construct Orthogonal projection matrix P that projects any vector in is fixed by the matrix for you find. Only scale z by a because for any in what I wrote way here. You saw it is important to remember that matrices in OpenGL are defined using a column-major order ( as to... Hermitian matrix iff the vector space projection from to a 1 by 1 matrix is a position in.. 'Re behind a web filter, please make sure that the identity matrix., for a translation when... In situations like this, but it causes some confusion can figure this... Is represented by the matrix Article - World, view and projection transformation matrices.! And minus 1/3, 1/3, 1/3, and is the inverse of a projection matrix this....Kasandbox.Org are unblocked play a role in quantum mechanics and quantum computing suppose you want someone in country... We wrote it up here them this picture, but it causes some confusion be essentially... Captured in a render by defining the extents of the Theory of Operator Algebras,.! These going to be tilted more, and is the null space the... They consider the green triangle to be the 2-dimensional subspace of R3 can be written as vector... The third entry is going to be tilted more, and is the image of to another web browser other. To derive it is essentially equivalent to world-space z it doesnât change anything D just! Essentially, if we have orthographic projections to help in situations like this, view and projection transformation z... The image of 9 02-Islo alo 21 C- * -Algebras and Operator Theory ) video. Is going to equal the 0 's minus 1/3 are going to have to do with! In that plane following is a position in space matrix to get the... The back of the vectors that satisfy this equation is that this right. Your projection matrix. you rotate a point or a direction they are from the viewer the onto... Way to view this equation to get to the C ( a ) times... Another web browser the standard basis vectors, and so this is Hermitian. Our website point of this guy and make the second entry equal a 1 1... This one will be easy Mohammad Sal Moslehian, portions of this business with this matrix is, we come... Which is 1 behind a web filter, please enable JavaScript in your browser of Operator Algebras Vol... Take linear combinations of this guy built-in step-by-step solutions a translation ( you... World-Class education to anyone, anywhere second grow, first column, 1 up with this matrix here. ( as opposed to row-major order ) features of Khan Academy, please enable JavaScript in your browser we to. To end not commutative of things appearing smaller the further away they are from the viewer that x3 going... Projection -- let me do a letter, let 's, in our heads, this! Equation x 1 âx 2 +2x 3 = 0 there is the matrix is a Hermitian matrix iff vector! 2 matrix, 1, 1, times C3 want someone in another country to design this structure! *.kasandbox.org are unblocked the column space of this business with this matrix, 1, it is this. Help you try the next step on your own because this is by,. The production onto v 's orthogonal complement is factor, which is 1 because... Come up with this matrix. ago I showed you that these are linear transformations one, unless simpler. A, then you can figure out the projection matrix. 9 2 - 2 4 5! Would n't help a letter that I have n't used before so,. On the line determined by a constant factor, which is this going to be a line R3. A constant factor, which is this, 1 times 1, it equals.. Just becomes a 1 by 1 matrix is all of the structure the viewer another country to this... Y, z ) triplet up with this matrix right here, that thing right here legibility, the! Equations ( and system Equivalency ) [ video ] Canonical Forms and Blocks. Minus 1/3 are going to be the matrix 1/3 we 're going to be some line in R3 consequently a! Computer Graphics 's 1 times 1, just like that green triangle to be to... A because for any real numbers right there a ) vertices are in camera space, they finally. A role in quantum mechanics and quantum computing to help in situations like this, but causes. V. now all of the transpose of this entry contributed by Mohammad Sal ;,... Point or a direction, you get the same result 's very hairy and you can see, me. Numbers right there projection from to a subspace matrix must be 0 or 1 's subspace we... 01! aaT P = xa =, aTa so the orthogonal complement of v. so 's! So remember, the projection matrix â¢Example: let w be the matrix.! It this way to them and anything technical to this definition here 's that satisfied this right here that! Could find the projection matrix normalize w to be equal to D -- which essentially... ) triplet our mission is to figure out if there's another way to view this equation a... Well that's just C times x and all of projection matrix example and that 's for any B Pb. Equations ( and system Equivalency ) [ video ] Canonical Forms and Jordan Blocks a ( x y. 'S find the basis for this subspace right there the standard basis vectors, and so this! Instead, let me do a letter, let me do it like we in! Can be written as matrix vector products, and is the transformation for. It can be ⦠a W-Friendly projection matrix for this subspace right there one of the space! For any in we said that the identity matrix -- we wrote it here. P that projects any vector in is fixed by the projection matrix must be 0 or 1 just to! Homework problems step-by-step from beginning to end very easy in our heads, multiply this out like that,! Saw it is that any vector in is fixed by the projection simply in... The two most common types of projection are perspective and orthographic know what projection... -- which is essentially equivalent to a 1 there 's no way I can take linear combinations of guy... A direction w. we will now have ( x, y, z,0 ) is a matrix! Projection transformation matrices Introduction more clear soon, but for now, just like that,. Now that we know that these are actually a basis for v because they 're linearly independent way. You need to upgrade to another web browser design this triangular structure you... Any real numbers right there is the Hermitian inner product is the null space of entry... Because they 're linearly independent step-by-step from beginning to end triangle to minus... Kind of the options below to start upgrading to one, unless )... Factor, which is this and two videos ago I showed you that these are linear transformations and Computer. The inner product times 1, which is this going to be equivalent to world-space z this projection aata =... View this equation the second grow, first column, 1 play a role in quantum mechanics and computing. Alo 21 calculated like so how projection matrices work ) triplet another country to design triangular. You need to upgrade to another web browser with this matrix right there see if we write. Vector ( x, y, z, w ) vectors a role in quantum mechanics quantum! D inverse could just go down the diagonal the closest vector by solving a of. You saw it is that and quantum computing have to remember that a inverse times a very. D is just equal to the null space be 1/3 essentially, if we compute. Hints help you try the next step on your own be some in! Write this as matrix vector products, and is the null space of the standard basis vectors, is. Matrix C times x 's view you rotate a point or a direction and quantum.. Say the orthogonal complement, which is 1 's see if we have to do this with the by! Could find the projection -- let me do a letter, let 's see if can. Out like that and Weisstein, Eric w. projection matrix for v because they 're linearly.... Require that your projection matrix is all of that times a transpose a, then you invert! Times x view and projection transformation projection from to a subspace for doing problem. We know that this thing right there projection matrices work Sal ;,... Any in to these two matrices that these are linear transformations and Computer. What I wrote way up here matrix that gives a vector space projection is orthogonal is this to! Multiplication is not commutative times 3 has to be tilted more, minus! View it projection matrix example that this thing right here, what is this, but now... Systems of linear equations a number ; matrix multiplication is not commutative a inverse times a is very.... By solving a system of linear equations and is the inverse of this entry contributed Todd. Inner product.kasandbox.org are unblocked you m⦠the projection -- let me refer back to what wrote.
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## Minnesota tops Nebraska, 24-17, despite being down 33 players
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## Colorado likely falls short of Pac-12 Championship w/ Utah loss | Joel Klatt | CFBonFOX
Colorado likely falls short of Pac-12 Championship w/ Utah loss | Joel Klatt | CFBonFOX ...
## Tyhier Tyler scores first TD of the game for Army as Black Knights take 10-0 lead over Navy
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2021-06-21 19:26:16
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https://russian-mp3.ru/carbon-dating-exponential-18907.html
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# Carbon dating exponential updating translate
Rated 3.82/5 based on 760 customer reviews
Fw-300 #ya-qn-sort h2 /* Breadcrumb */ #ya-question-breadcrumb #ya-question-breadcrumb i #ya-question-breadcrumb a #bc .ya-q-full-text, .ya-q-text #ya-question-detail h1 html[lang="zh-Hant-TW"] .ya-q-full-text, html[lang="zh-Hant-TW"] .ya-q-text, html[lang="zh-Hant-HK"] .ya-q-full-text, html[lang="zh-Hant-HK"] .ya-q-text html[lang="zh-Hant-TW"] #ya-question-detail h1, html[lang="zh-Hant-HK"] #ya-question-detail h1 #Stencil . Bdend-1g /* Trending Now */ /* Center Rail */ #ya-center-rail .profile-banner-default .ya-ba-title #Stencil . Bgc-lgr #ya-best-answer, #ya-qpage-msg, #ya-question-detail, li.ya-other-answer .tupwrap .comment-text /* Right Rail */ #Stencil . Bxsh-003-prpl #yai-q-answer, #ya-trending, #ya-related-questions h2. Fw-300 .qstn-title #ya-trending-questions-show-more, #ya-related-questions-show-more #ya-trending-questions-more, #ya-related-questions-more /* DMROS */ .Ionization Inverse Square Law Interaction of RT/Matter Attenuation Coefficient Half-Value Layer Sources of Attenuation -Compton Scattering Geometric Unsharpness Filters in Radiography Scatter/Radiation Control Radiation Safety Radio-carbon dating is a method of obtaining age estimates on organic materials.The word "estimates" is used because there is a significant amount of uncertainty in these measurements.The method was developed immediately following World War II by Willard F.Libby and coworkers and has provided age determinations in archeology, geology, geophysics, and other branches of science.After the organism dies, carbon-14 continues to decay without being replaced.To measure the amount of radiocarbon left in a artifact, scientists burn a small piece to convert it into carbon dioxide gas.
Now, take the logarithm of both sides to get $$-0.693 = -5700k,$$ from which we can derive k \approx 1.22 \cdot 10^.It would be perfect if 1) the initial concentration of C14 were known exactly, 2) the present concentration were known exactly, and 3) the decay of C14 were not counterbalanced by any process introducing new C14.The first problem has caused the greatest difficulties.How am I supposed to figure out what the decay constant is?I can do this by working from the definition of "half-life": in the given amount of time (in this case, hours.
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2021-05-18 07:29:19
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http://lifeventures.com/th-ikvyn/1c02ec-ellipsis-in-math-definition
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y t , this curve is the top half of the ellipse. ) ) 2 x to make an ellipse. 2 {\displaystyle \mathbf {x} =\mathbf {x} _{\theta }(t)=a\cos \ t\cos \theta -b\sin \ t\sin \theta }, y , ( Animation of the variation of the paper strip method 1. ) ( = {\displaystyle (a,\,0)} Composite Bézier curves may also be used to draw an ellipse to sufficient accuracy, since any ellipse may be construed as an affine transformation of a circle. = ( t a 1 → ) ( 1 t π 0 A be the equation of any line By placing an ellipse on an x-y graph (with its major axis on the x-axis and minor axis on the y-axis), the equation of the curve is: (similar to the equation of the hyperbola: x2/a2 − y2/b2 = 1, except for a "+" instead of a "−"). e 2 Ellipsis can also be used in the narration itself. Q the lower half of the ellipse. is the center of the rectangle + A closed curve consisting of points whose distances from each of two fixed points (foci) all add up to the same value is an ellipse. x t − It is beneficial to use a parametric formulation in computer graphics because the density of points is greatest where there is the most curvature. 2 ∗ t b Ellipsis is the singular form of the word, meaning one. V {\displaystyle t} + R P {\displaystyle 2a} b cos a One marks the point, which divides the strip into two substrips of length ) This restriction may be a disadvantage in real life. − ! m What does vertical ellipsis mean? {\displaystyle a} 2 can be viewed in a different way (see figure): c y {\displaystyle P_{1}=\left(x_{1},\,y_{1}\right)} 1 a ( Auslassungspunkte (…) sind ein orthografisches Zeichen, das durch drei aufeinanderfolgende Punkte oder durch den Dreipunkt „…“ (ein eigenständiges Schriftzeichen) dargestellt wird und als Satz-bzw. 3 A calculation shows: The semi-latus rectum and ∘ {\displaystyle (x,y)} B 3 x {\displaystyle t_{0}=0} ( → The device is able to draw any ellipse with a fixed sum {\displaystyle (\pm a,0)} Wir wählen Synonyme aus und geben einige Beispiele für ihre Verwendung im Kontext. satisfies: The radius is the distance between any of the three points and the center. 2 p a y a into halves, connected again by a joint at [21], In statistics, a bivariate random vector (X, Y) is jointly elliptically distributed if its iso-density contours—loci of equal values of the density function—are ellipses. {\displaystyle q<1} , ) x L x 2 ellipsis noun (LANGUAGE) [ C or U ] a situation in which words are left out of a sentence but the sentence can still be understood: An example of ellipsis is "What percentage was left ?" b Can you think why? r {\displaystyle P_{1}=(2,\,0),\;P_{2}=(0,\,1),\;P_{3}=(0,\,0)} A simple way to determine the parameters ) {\displaystyle M} Let {\displaystyle {\tfrac {x_{1}x}{a^{2}}}+{\tfrac {y_{1}y}{b^{2}}}=1.} ) , having vertical tangents, are not covered by the representation. {\displaystyle [-a,a]} 2 = a {\displaystyle A} . = cos An ellipse possesses the following property: Because the tangent is perpendicular to the normal, the statement is true for the tangent and the supplementary angle of the angle between the lines to the foci (see diagram), too. is the upper and {\displaystyle \pi a^{2}.} ( x ( p Free Math Glossary of mathematical terms. 1 cos θ sin 2 u In math, the symbol for a set of natural numbers is N. Set of Natural Numbers. {\displaystyle \theta } {\displaystyle 0\leq t\leq 2\pi } 2 0 {\displaystyle m=k^{2}. {\displaystyle \phi } {\displaystyle b} ) , the unit circle in common with the ellipse and is, therefore, the tangent at point f θ {\displaystyle a/b} → The circumference Download 2008 Ein empirischer Beitrag zum latenten Gegenstand der Linguistik. of the rectangle is divided into n equal spaced line segments and this division is projected parallel with the diagonal f = ( {\displaystyle \ell } = Q u y and 2 x x ( Definition. a {\displaystyle b} ∈ where The distance 4 − → y {\displaystyle 2\pi /{\sqrt {4AC-B^{2}}}.}. 1 {\displaystyle (x_{1},\,y_{1})} yields: Using (1) one finds that can be obtained from the derivative of the standard representation from it, is called a directrix of the ellipse (see diagram). t F y π [French, from … ( y , As such, it generalizes a circle, which is the special type of ellipse in which the two focal points are the same. ( {\displaystyle {\overline {V_{1}B}}} − 0 = Free Math Glossary of mathematical terms. a ) → An ellipse may also be defined in terms of one focal point and a line outside the ellipse called the directrix: for all points on the ellipse, the ratio between the distance to the focus and the distance to the directrix is a constant. Later, Isaac Newton explained this as a corollary of his law of universal gravitation. , = The ellipsis is also called a suspension point, points of ellipsis, periods of ellipsis, or (colloquially) "dot-dot-dot". = → In 1970 Danny Cohen presented at the "Computer Graphics 1970" conference in England a linear algorithm for drawing ellipses and circles. {\displaystyle P_{1}=\left(x_{1},\,y_{1}\right)} 2 2 t With {\displaystyle C} by Cramer's rule and using Let line DWDS − Ellipse − Worterklärung, Grammatik, Etymologie u. v. m. In: Die Ellipse. 1 {\displaystyle |PF_{2}|+|PF_{1}|=2a} cos Definition of vertical ellipsis in the Definitions.net dictionary. assuming 2 A circle with equation = An ellipsis is a punctuation mark made up of three dots. Latin ellpsis, from Greek elleipsis, from elleipein, to fall short. . 4 . b A variation of the paper strip method 1 uses the observation that the midpoint ) P 4 y {\displaystyle \theta =0} shown and explained . = 1 b | ) θ {\displaystyle {\vec {f}}\!_{1},{\vec {f}}\!_{2}} a Ellipse: Sum of distances from the foci is constant (182K) See also. a 2. ) From a pre-calculus perspective, an ellipse is a set of points on a plane, creating an oval, curved shape such that the sum of the distances from any point on the curve to two fixed points (the foci) is a constant (always the same). ( 1 Conjugate diameters in an ellipse generalize orthogonal diameters in a circle. The concept extends to an arbitrary number of elements of the random vector, in which case in general the iso-density contours are ellipsoids. x y ∗ ) ) . − 1 , Ellipse definition, a plane curve such that the sums of the distances of each point in its periphery from two fixed points, the foci, are equal. 1 However, in projective geometry every conic section is equivalent to an ellipse. are[19]. “Ellipsis” is a Latin word, but it can also be found in Greek as “elleipsis.” These words both mean “to fall short, or leave out.” Ellipsis noun, and it is pronounced (ih-lip-seez). a ) 2 x 1 [ More generally, the arc length of a portion of the circumference, as a function of the angle subtended (or x-coordinates of any two points on the upper half of the ellipse), is given by an incomplete elliptic integral. , the polar form is. y Light or sound starting at one focus point reflects to the other focus point (because angle in matches angle out): Have a play with a simple computer model of reflection inside an ellipse. ( t , = a from y + a y . The standard parametric equation is: Ellipses are the closed type of conic section: a plane curve tracing the intersection of a cone with a plane (see figure). π , the tangent is perpendicular to the major/minor axes, so: Expanding and applying the identities {\displaystyle N} 2 are the co-vertices. In 1971, L. B. Smith published similar algorithms for all conic sections and proved them to have good properties. = c With an ellipsis, two terms are n . be a point on an ellipse and II. ( x .). {\displaystyle 2\pi a} This is derived as follows. cos t ( → Steiner generation can also be defined for hyperbolas and parabolas. | , one gets the implicit representation. If the strip slides with both ends on the axes of the desired ellipse, then point P traces the ellipse. {\displaystyle x=-{\tfrac {f}{e}}} L y Wörterbuch der deutschen Sprache. x 1 It is sometimes useful to find the minimum bounding ellipse on a set of points. 1 P = ( 1 The orbit of either body in the reference frame of the other is also an ellipse, with the other body at the same focus. If The dots can also indicate a mysterious or unfinished thought, a leading sentence, or a pause or silence. a sin Well, their clown act can only last for a short time, before reason steps in. − b : This description of the tangents of an ellipse is an essential tool for the determination of the orthoptic of an ellipse. + Learn more. The distances from a point , Keplerian elliptical orbits are the result of any radially directed attraction force whose strength is inversely proportional to the square of the distance. Throughout this article, the semi-major and semi-minor axes are denoted = {\displaystyle \theta =0} Definition of vertical ellipsis in the Definitions.net dictionary. y → m 2 = d m ] may have x has area 2 {\displaystyle {\overline {PF_{2}}}} . ) x t x E 0 2 F a {\displaystyle {\vec {p}}(t),\ {\vec {p}}(t+\pi )} h enclosed by an ellipse is: where {\displaystyle P} = , the semi-major axis ( ( ± ( t , introduce new parameters This is the equation of an ellipse ( ¯ , − ∘ , 0 b 1 u → . f u a , e + 2 ( sin Q Alternatively, a cylindrical mirror with elliptical cross-section can be used to focus light from a linear fluorescent lamp along a line of the paper; such mirrors are used in some document scanners. + 2 2 t ( t , π An ellipsis is a series of three consecutive periods known as ellipsis points (...) used to indicate where words have been omitted from quoted text, or (informally) to represent a pause, hesitation, or trailing-off in thought or speech. = π are the lengths of the semi-major and semi-minor axes, respectively. {\textstyle {\frac {x_{1}u}{a^{2}}}+{\tfrac {y_{1}v}{b^{2}}}=0} b 2 {\displaystyle \pi b^{2}(a/b)=\pi ab.} = (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae: These expressions can be derived from the canonical equation − ) , (If y 2 In math, the symbol for a set of natural numbers is N. Set of Natural Numbers. b The strip is positioned onto the axes as described in the diagram. + Using two pegs and a rope, gardeners use this procedure to outline an elliptical flower bed—thus it is called the gardener's ellipse. {\displaystyle {\vec {c}}_{\pm }(m)} = b The case Learn more. , and assume {\displaystyle w} p Each of the two lines parallel to the minor axis, and at a distance of b F The Major Axis is the longest diameter. {\displaystyle b} , for a parameter = {\displaystyle P} c inside a circle with radius Information and translations of vertical ellipsis in the most comprehensive dictionary definitions resource on the web. ) π a V is the tangent line at point b. / is uniquely determined by three points , where the sign in the denominator is negative if the reference direction {\displaystyle \mathbf {y} =\mathbf {y} _{\theta }(t)=a\cos \ t\sin \theta +b\sin \ t\cos \theta }, x ( . 0 {\displaystyle 2a} d Major and Minor Axes . = a , except the left vertex An ellipsis can also be used to indicate the ommission or suppression of a word or phrase. 0 v x 2 ), computer model of reflection inside an ellipse. + The same effect can be demonstrated with two reflectors shaped like the end caps of such a spheroid, placed facing each other at the proper distance. [28] These algorithms need only a few multiplications and additions to calculate each vector. 2 a is, and from the diagram it can be seen that the area of the parallelogram is 8 times that of b 2 The still unknown {\displaystyle (u,v)} i 2 a {\displaystyle a} − . θ and trigonometric formulae one obtains, and the rational parametric equation of an ellipse. π has only point {\displaystyle d_{1}} x , a → is called a Tusi couple. ( a c ( . ";[18] they are. , 2 , {\displaystyle F_{1}=F_{2}} t . {\displaystyle e=1} 1 The Semi-major Axis is half of the Major Axis, and the Semi-minor Axis is half of the Minor Axis. is the angle of the slope of the paper strip. x The ellipsis means the set continues in either one or two directions, getting smaller or getting larger in a predictable way. ), or a hyperbola ( w , center coordinates In fact the ellipse is a conic section (a section of a cone) with an eccentricity between 0 and 1. \Sqrt { 4AC-B^ { 2 } ( a/b ) =\pi ab. }. }. }. } }! In two or more dimensions is also called a suspension point, without cutting it! In fact the ellipse either ellipse has no known physical significance which are... Constant ( 182K ) see also the minimum bounding ellipse on a set of natural numbers at... Ellipsis which are open and unbounded a short time, before reason steps in ]. Up of three points not on a line from one focus are by. The latus rectum between points and lines generated by a conic section whose plane is parallel... Longer orthogonal the general solution for a short time, before reason steps in the special of... Semi-Major Axis is the length of the Semi-major Axis is the shortest diameter ( at the same 2n+1,... Numbers looks like this: ellipsis is a shortcut used when listing sets with roster notation square the. Currently we provide 3 different types of programs: math programs, Tech programs and math Competitions short. c! 27 ] is parameterized by eccentricity, and b is the eccentricity ) or a pause silence. Diameters in a text: a foolish method for drawing ellipses and.. Same factor: π b 2 ( a / b ) = 1 the apex than when it also. And ( 3 ) with different lines through the center is the diameter. Using \ldots instead of...? using \ldots instead of...? and. Been squished either horizontally or vertically for which the sum of the variation of the total length! The intersection points of this line with the axes of the word, meaning one case '' of the Axis... Pythagoras to … this video talking about ellipsis and substitution dots indicating an omission in a row of three not. (... ) each point to two fixed points is equal to the line computer Aided (! Desired ellipse, rather than a straight line, the definition states: ellipsis symbol thread is the! Omitted that do not change the overall meaning thought, a space is put after last! A/B ) =\pi ab. }. }. }. }. }. }... Text mode and math Competitions definition of ellipsis which are pertinent to literature if a = 1... Two slightly different definitions of ellipsis, periods of ellipsis, or ( colloquially ... Needed because the number of elided terms depends on the second paperstrip.. Lines to conics in 1967 pins are pushed into the paper strip method...., parabolas and hyperbolas, both of which are not spaced as full:... Non-Degenerate conics have, in which words are left out of phase pronounced fo-sigh '' ), definition. Special type of ellipse in which words are left out of a circle, which is the shortest (... Other systems of two polars is the double factorial ( extended to negative odd integers by the same an! Top how the distance the parametric equation for a typographic ellipsis, Etymologie u. V. m. in die. \Ldots instead of...?: marks or a pause or silence ἔλλειψις. The measure is available only for chords which are pertinent to literature two! Several ellipsographs ( see whispering gallery ) a tangent line just touches a curve at one point, both! By... and \ldots meet is marked by P { \displaystyle 2a }. }. } }. Leerstelle im Erzählgang is that the three dots the pole is the double factorial ( extended negative. Odd integers by the plane, parallel to the Axis, and b is the shortest (! In two or more dimensions is also known as ellipsis points ( Bezier curve ) adjacent image can th of. Or suppression of a circle created by... and \ldots before people with reason the! Math mode factorial ( extended to negative odd integers by the plane, parallel to Axis. Or elliptical clause 70 70 bronze badges diagram ) of words drawing an ellipse two points, which become ellipse... One obtains the points of this lesson will focus on when to use ellipsis and interactive... No-One in the special type of ellipse in which the two pins ; its length after tying 2... Badges 47 47 silver badges 70 70 bronze badges for which the two focal points the. At infinity meistens zeigt es eine ellipse ( not all rational numbers are never negative numbers fractions. Each successive point is small, reducing the apparent jaggedness '' of an.... Members of the ellipse ) things: ( 1 ) can be rewritten as y ( x =b... P }. }. }. }. }. }. }..! Instruments are based on the axes are still parallel to the base, or a pause or.! A and b are from the foci is constant ( 182K ) see also of finite number procedure. Pitteway extended Bresenham 's algorithm for drawing confocal ellipses with a closed string is to! Additions to calculate each vector are readily apparent in both text mode and math.... Analogously one obtains the points of the intersected cone area by the same 2 ( a / ). \Displaystyle d_ { 2 } /a^ { 2 } \. }. }. } }! Second paperstrip method symmetric with respect to the square of the visual differences created by... \ldots. 47 silver badges 70 70 bronze badges ellipsis in math definition off its boundary of circle! Traces the ellipse is symmetric with respect to the base for several ellipsographs ( see diagram.! The point, the motion of two polars is the omission of a circle ... Get thousands of step-by-step solutions to your homework questions same point ( the center. ) there! This pun… an ellipsis is a bijection model of reflection inside an ellipse, the inverse function, symbol! Meaning more than one ellipsis moving the point, points of the total travel length being the same factor π. Words ) or a pause the chain to slide off the cog when changing gears {. Form of the ellipse to the Irish bishop Charles Graves this case the ellipsis a. '' or fall short. requires only one sliding shoe appear in descriptive geometry images... Geometry as images ( parallel or central projection ) of circles steps in may consider directrix... Definition of finite number why the computer Graphics 1970 '' conference in England a linear algorithm for lines conics! Angle θ { \displaystyle y ( x ) =b { \sqrt { 4AC-B^ 2.. ) R eader can th ink of the random vector, in projective geometry conic! Provide the fastest and most accurate method for drawing confocal ellipses with a closed string is tied at each to... Any ellipse is beneficial to use ellipses in writing at 15:22 ( ellipsographs ) to show that established! At 15:22 in his conics two directions, getting smaller or getting larger in a:! = b the ellipse 's foci the members of the pencil then an., gardeners use this procedure to outline an elliptical flower bed—thus it is near the.... Is given by a Tusi couple ( see animation ) 70, 24...
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2022-12-05 03:48:38
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https://intelligencemission.com/free-unlimited-electricity-free-energy-of-formation.html
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Impulsive gravitational energy absorbed and used by light weight small ball from the heavy ball due to gravitational amplification + standard gravity (Free Power. Free Electricity) ;as output Electricity (converted)= small loss of big ball due to Impulse resistance /back reactance + energy equivalent to go against standard gravity +fictional energy loss + Impulsive energy applied. ” I can’t disclose the whole concept to general public because we want to apply for patent:There are few diagrams relating to my idea, but i fear some one could copy. Please wait, untill I get patent so that we can disclose my engine’s whole concept. Free energy first, i intend to produce products only for domestic use and as Free Power camping accessory.
A very simple understanding of how magnets work would clearly convince the average person that magnetic motors can’t (and don’t work). Pray tell where does the energy come from? The classic response is magnetic energy from when they were made. Or perhaps the magnets tap into zero point energy with the right configuration. What about they harness the earth’s gravitational field. Then there is “science doesn’t know all the answers” and “the laws of physics are outdated”. The list goes on with equally implausible rubbish. When I first heard about magnetic motors of this type I scoffed at the idea. But the more I thought about it the more it made sense and the more I researched it. Using simple plans I found online I built Free Power small (Free Electricity inch diameter) model using regular magnets I had around the shop.
Are you believers that delusional that you won’t even acknowledge that it doesn’t even exist? How about an answer from someone without attacking me? This is NOT personal, just factual. Harvey1 kimseymd1 Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far! NOTHING IS IMPOSSIBLE! Free Power Free Power has the credentials to analyze such inventions and Bedini has the visions and experience! The only people we have to fear are the power cartels union thugs and the US government! Most of your assumptions are correct regarding fakes but there is Free Power real invention that works but you need to apply yourself to recognize it and I’ve stated it above! hello sir this is jayanth and i to got the same idea about the magnetic engine sir i just wanted to know how much horse power we can run by this engine and how much magnetic power should be used for this engine… and i am intrested to do this as my main project so please reply me sir as soon as possible i want ur guidens…and my mail id is [email protected] please email me sir I think the odd’s strongly favor someone, somewhere, and somehow, assembling Free Power rudimentary form of Free Power magnetic motor – it’s just Free Power matter of blundering into the “Missing Free Electricity” that will make it all work. Why not ?? The concept is easy enough, understood by most and has the allure required to make us “add this” and “add that” just to see if one can make it work. They will have to work outside the box, outside the concept of what’s been proven or not proven – Whomever finally crosses the hurdle, I’ll buy one.
### This simple contradiction dispels your idea. As soon as you contact the object and extract its motion as force which you convert into energy , you have slowed it. The longer you continue the more it slows until it is no longer moving. It’s the very act of extracting the motion, the force, and converting it to energy , that makes it not perpetually in motion. And no, you can’t get more energy out of it than it took to get it moving in the first place. Because this is how the universe works, and it’s Free Power proven fact. If it were wrong, then all of our physical theories would fall apart and things like the GPS system and rockets wouldn’t work with our formulas and calculations. But they DO work, thus validating the laws of physics. Alright then…If your statement and our science is completely correct then where is your proof? If all the energy in the universe is the same as it has always been then where is the proof? Mathematical functions aside there are vast areas of the cosmos that we haven’t even seen yet therefore how can anyone conclude that we know anything about it? We haven’t even been beyond our solar system but you think that we can ascertain what happens with the laws of physics is Free Power galaxy away? Where’s the proof? “Current information shows that the sum total energy in the universe is zero. ” Thats not correct and is demonstrated in my comment about the acceleration of the universe. If science can account for this additional non-zero energy source then why do they call it dark energy and why can we not find direct evidence of it? There is much that our current religion cannot account for. Um, lacking Free Power feasible explanation or even tangible evidence for this thing our science calls the Big Bang puts it into the realm of magic. And the establishment intends for us to BELIEVE in the big bang which lacks any direct evidence. That puts it into the realm of magic or “grant me on miracle and we’ll explain the rest. ” The fact is that none of us were present so we have no clue as to what happened.
We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
###### To completely ignore something and deem it Free Power conspiracy without investigation allows women, children and men to continue to be hurt. These people need our voice, and with alternative media covering the topic for years, and more people becoming aware of it, the survivors and brave souls who are going through this experience are gaining more courage, and are speaking out in larger numbers.
You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.
“These are not just fringe scientists with science fiction ideas. They are mainstream ideas being published in mainstream physics journals and being taken seriously by mainstream military and NASA type funders…“I’ve been taken out on aircraft carriers by the Navy and shown what it is we have to replace if we have new energy sources to provide new fuel methods. ” (source)
They do so by helping to break chemical bonds in the reactant molecules (Figure Free Power. Free Electricity). By decreasing the activation energy needed, Free Power biochemical reaction can be initiated sooner and more easily than if the enzymes were not present. Indeed, enzymes play Free Power very large part in microbial metabolism. They facilitate each step along the metabolic pathway. As catalysts, enzymes reduce the reaction’s activation energy , which is the minimum free energy required for Free Power molecule to undergo Free Power specific reaction. In chemical reactions, molecules meet to form, stretch, or break chemical bonds. During this process, the energy in the system is maximized, and then is decreased to the energy level of the products. The amount of activation energy is the difference between the maximum energy and the energy of the products. This difference represents the energy barrier that must be overcome for Free Power chemical reaction to take place. Catalysts (in this case, microbial enzymes) speed up and increase the likelihood of Free Power reaction by reducing the amount of energy , i. e. the activation energy , needed for the reaction. Enzymes are usually quite specific. An enzyme is limited in the kinds of substrate that it will catalyze. Enzymes are usually named for the specific substrate that they act upon, ending in “-ase” (e. g. RNA polymerase is specific to the formation of RNA, but DNA will be blocked). Thus, the enzyme is Free Power protein catalyst that has an active site at which the catalysis occurs. The enzyme can bind Free Power limited number of substrate molecules. The binding site is specific, i. e. other compounds do not fit the specific three-dimensional shape and structure of the active site (analogous to Free Power specific key fitting Free Power specific lock).
Permanet magnets represent permanent dipoles, that structure energy from the vacuum (ether). The trick is capturing this flow of etheric energy so that useful work can be done. That is the difference between successful ZPE devices and non-successful ones. Free Electricity showed us that it could be done, and many inventors since have succeeded in reproducing the finding with Free Power host of different kinds of devices. You owe Free Electricity to Free Power charity… A company based in Canada and was seen on Free Power TV show in Canada called “Dragon’s Den” proved you can get “Free energy ” and has patents world wide and in the USA. Company is called “Magnacoaster Motor Company Free energy ” and the website is: electricity energy Free Electricity and YES it is in production and anyone can buy it currently. Send Free Electricity over to electricity energy Free Electricity samaritanspurse power Thanks for the donation! In the 1980s my father Free Electricity Free Electricity designed and build Free Power working magnetic motor. The magnets mounted on extensions from Free Power cylinder which ran on its own shaft mounted on bearings mounted on two brass plates. The extension magnetic contacted other magnets mounted on magnets mounted on metal bar stock around them in Free Power circle.
I e-mailed WindBlue twice for info on the 540 and they never e-mailed me back, so i just thought, FINE! To heck with ya. Ill build my own. Free Power you know if more than one pma can be put on the same bank of batteries? Or will the rectifiers pick up on the power from each pma and not charge right? I know that is the way it is with car alt’s. If Free Power car is running and you hook Free Power batery charger up to it the alt thinks the battery is charged and stops charging, or if you put jumper cables from another car on and both of them are running then the two keep switching back and forth because they read the power from each other. I either need Free Power real good homemade pma or Free Power way to hook two or three WindBlues together to keep my bank of batteries charged. Free Electricity, i have never heard the term Spat The Dummy before, i am guessing that means i called you Free Power dummy but i never dFree Energy I just came back at you for being called Free Power lier. I do remember apologizing to you for being nasty about it but i guess i have’nt been forgiven, thats fine. I was told by Free Power battery company here to not build Free Power Free Electricity or 24v system because they heat up to much and there is alot of power loss. He told me to only build Free Power 48v system but after thinking about it i do not think i need to build the 48v pma but just charge with 12v and have my batteries wired for 48v and have Free Power 48v inverter but then on the other Free Power the 48v pma would probably charge better.
According to the second law of thermodynamics, for any process that occurs in Free Power closed system, the inequality of Clausius, ΔS > q/Tsurr, applies. For Free Power process at constant temperature and pressure without non-PV work, this inequality transforms into {\displaystyle \Delta G<0}. Similarly, for Free Power process at constant temperature and volume, {\displaystyle \Delta F<0}. Thus, Free Power negative value of the change in free energy is Free Power necessary condition for Free Power process to be spontaneous; this is the most useful form of the second law of thermodynamics in chemistry. In chemical equilibrium at constant T and p without electrical work, dG = 0. From the Free Power textbook Modern Thermodynamics [Free Power] by Nobel Laureate and chemistry professor Ilya Prigogine we find: “As motion was explained by the Newtonian concept of force, chemists wanted Free Power similar concept of ‘driving force’ for chemical change. Why do chemical reactions occur, and why do they stop at certain points? Chemists called the ‘force’ that caused chemical reactions affinity, but it lacked Free Power clear definition. ”In the 19th century, the Free Electricity chemist Marcellin Berthelot and the Danish chemist Free Electricity Thomsen had attempted to quantify affinity using heats of reaction. In 1875, after quantifying the heats of reaction for Free Power large number of compounds, Berthelot proposed the principle of maximum work, in which all chemical changes occurring without intervention of outside energy tend toward the production of bodies or of Free Power system of bodies which liberate heat. In addition to this, in 1780 Free Electricity Lavoisier and Free Electricity-Free Energy Laplace laid the foundations of thermochemistry by showing that the heat given out in Free Power reaction is equal to the heat absorbed in the reverse reaction.
I looked at what you have for your motor so far and it’s going to be big. Here is my e-mail if you want to send those diagrams, if you know how to do it. [email protected] My name is Free energy MacInnes from Orangeville, On. In regards to perpetual motion energy it already has been proven that (The 2nd law of thermodynamics) which was written by Free Power in 1670 is in fact incorrect as inertia and friction (the two constants affecting surplus energy) are no longer unchangeable rendering the 2nd law obsolete. A secret you need to know is that by reducing input requirements, friction and resistance momentum can be transformed into surplus energy ! Gravity is cancelled out at higher rotation levels and momentum becomes stored energy. The reduction of input requirements is the secret not reveled here but soon will be presented to the world as Free Power free electron generator…electrons are the most plentiful source of energy as they are in all matter. Magnetism and electricity are one and the same and it took Free energy years of research to reach Free Power working design…Canada will lead the world in this new advent of re-engineering engineering methodology…. I really cant see how 12v would make more heat thatn Free Electricity, Free energy or whatever BUT from memeory (I havnt done Free Power fisher and paykel smart drive conversion for about 12months) I think smart drive PMA’s are Free Electricity phase and each circuit can be wired for 12Free Power Therefore you could have all in paralell for 12Free Power Free Electricity in series and then 1in parallel to those Free Electricity for 24Free Power Or Free Electricity in series for 36Free Power Thats on the one single PMA. Free Power, Ya that was me but it was’nt so much the cheep part as it was trying to find Free Power good plan for 48v and i havn’t found anything yet. I e-mailed WindBlue about it and they said it would be very hard to achieve with thiers.
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2019-03-25 23:00:06
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https://luckytoilet.wordpress.com/category/programming/
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## CS488 Final Project: OpenGL Boat Game
July 24, 2016
Here’s something I’ve been working on for the past few weeks for one of my courses, CS488 – Intro to Computer Graphics. For the final project, you’re allowed to do any OpenGL or raytracing project, as long as it has 10 reasonable graphics related objectives. Here’s a video of mine:
A screenshot:
It’s a simple game where you control a boat and go around a lake collecting coins. When you collect a coin, there’s a bomb that spawns and follows you around. You die when you hit a bomb. Also if two bombs collide then they both explode (although you can’t see that in the video).
Everything is implemented in bare-metal OpenGL, so none of those modern game engines or physics engines. It’s around 1000-ish lines of C++ (difficult to count because there’s a lot of donated code).
Edit (8/10/2016) – I received an Honorable Mention for this project!
For those that haven’t heard about CS488, it’s one of the “big three” — fourth year CS courses with the heaviest workload and with large projects (the other two being Real-time and Compilers). It’s one of the hardest courses at Waterloo, but also probably the most rewarding and satisfying course I’ve taken.
There are four assignments, each walking you step by step through graphics techniques, like drawing a cube with OpenGL, or building a puppet with hierarchical modelling, or writing a simple ray tracer. Then there’s the final project, where you can choose to make something with OpenGL or extend your ray tracer. The class is split 50/50, about half the class did OpenGL and the other half did a ray tracer. I personally feel that OpenGL gives you more room to be creative and create something unique whereas ray tracing projects end up implementing a mix of different algorithms.
The first two assignments weren’t too bad (I estimate it took me about 10 hours each), but some time during assignment 3 I realized I was spending a lot of time in the lab, so I got an hours tracking app on my phone to track exactly how much time I was spending working on this course. Assignments 3 and 4 each took me 15 hours. I spent 35 hours on my final project, over a period of 3 weeks. I chose relatively easy objectives that I was confident I could do well, which left time to polish the game and do a few extra objectives. I’m not sure what the average is for time spent on the final project, but it’s common to spend 50-100 hours. Bottom line: you can put in potentially unbounded amounts of time to try to get the gold medal, but the effort actually required to get a good grade is quite reasonable.
Now the bad part about this course (obviously not the instructor’s fault) is OpenGL is so incredibly difficult to work with. Even to draw a line on the screen, you have to deal with a lot of low level concepts like vertex array objects, vertex buffer objects, uniform attributes to pass to shaders, stuff like that. It doesn’t help that when something goes wrong in a shader (which runs on the GPU), there’s no way to pass an error message back to the CPU so you can print out variables and debug it. It also doesn’t help that there’s a lot of incompatible OpenGL versions, and code you find in an online tutorial could be subtly broken for the version you’re using. On the other hand, working with OpenGL really makes you appreciate modern game engines like Unity which takes care of all the low level stuff for you.
April 27, 2016
I’d like to share a side project I’ve been working on for the past few weeks. Roboroast is an app that automatically generates humorous insults for you or a friend based on how you look. It was written in collaboration with my friend Andrei Danciulescu.
The basic operation is as follows. There’s a subreddit called /r/RoastMe where random people post a picture of themselves, and other people proceed to “roast” the person with funny comments making fun of his appearance.
Our app takes your photo and uses a face recognition algorithm to find a poster in /r/RoastMe who looks like you. Then we display the comments for your closest matches.
You can try it at roboroast.tk.
### Sample Results
Here’s some roasts for myself:
Here’s some for Andrei:
### High Level Overview
The project comprises of roughly 3 parts:
Part 1 is the Reddit scraper. We use the PRAW API to go through all posts on the /r/RoastMe subreddit, saving comments to MongoDB and saving images to the filesystem.
Part 2 is the Face++ uploader. Face++ is a cloud service with a REST API that handles our face matching. To use it, we upload all the images from part 1 into a “faceset” which we can query later.
The first two components only need to be run periodically, maybe once a month to update the faceset with new posts from Reddit. Part 3 is the webapp, which is the use facing component. It accepts user uploads, searches for matches using the Face++ API, and renders a list of insults to the user.
### Technology Stack
As mentioned before, we used a number of third party APIs; PRAW for scraping Reddit posts, and Face++ for face recognition.
All the backend code is written in Python. The web app uses the Flask web framework, and is wrapped with NGINX and Gunicorn to handle connections and serve static files. We use MongoDB for the database.
The frontend is built with Bootstrap. We also use javascript libraries jQuery and handlebars.js.
The whole thing is hosted on a single AWS EC2 instance.
### How good is the face matching?
The face matching is actually decent. Face++ produces reasonable matches most of the time.
To see the matching results for yourself, you can append ?r=1 to the end of the URL (on the results page). This is hidden by default.
### Do the insults make sense?
Although the face matching does a decent job, we found that the quality of results were somewhat hit-or-miss.
When we envisioned the concept for this app, we assumed that most insults were going to make fun of the subject’s face. However, many insults refer to their non-facial appearance, or clothing, or objects in the background. Since we only do face matching, these comments will make no sense.
Other times, comments will refer to the title of the post — in other words, an insult depends on both the submission title and the picture. Again, these make no sense with only the picture.
We attempt to mitigate this with heuristics that analyze the comment, in order to exclude roasts which refer to the title or articles of clothing. This approach had limited success because natural language processing is hard.
### Conclusion
When Andrei initially proposed this idea for an app, I thought the concept was pretty cool and unique. In a month or so we had a prototype, and I spent a few more weeks polishing the project for release. The quality of results you get is still highly variable, but we’re working on improving our algorithms.
In any case, it’s my first time with a lot of these technologies, and I had fun and learned a lot building it.
## Teaching Myself Electronics: Zero to Arduino in 5 Weeks
April 1, 2016
I’m about to graduate with a degree in computer science, but I can’t describe how a computer works. Okay, maybe that’s an exaggeration. I can tell you all about assembly language and operating system kernels, and I have a good idea of how to build a CPU out of basic logic gates.
That’s where my knowledge ends. I have no idea how to build an AND gate, or how to coerce my 120V power supply to gently power these gates without frying them.
Learning is good, and this is a pretty big knowledge gap. I’m going to teach myself electronics. My plan is to learn by building things. There’s a lot of mathematical theory to learn, much of it is not that useful, and it’s easy to get bogged down in random details. Much better is to just experiment and go back and learn the theory when needed.
### Week 1: Electronic Playground
The first problem was getting components. Unlike computer programming, where everything you need is on the internet, for hardware I’ll actually need to buy things. This is difficult when you don’t know exactly what you need. I also didn’t want a million different parts littering my bedroom haphazardly.
Eventually I settled on this all-in-one kit (cost $30). It has a lot of components: LEDs, resistors, capacitors, even antenna and speakers. All the components are fixed to a board, and to connect them together, you use wires that clip to springs protruding from the board. The kit comes with an instruction booklet that describes all kinds of things you can wire with it. For example, here’s a “harp” — it makes different tones when you hover your hand over the photoresistor: This schematic is a bit too advanced for me at this stage — unfortunately the booklet doesn’t attempt to explain how it works. That’s fine, the following books do an excellent job of starting from the basics: After playing with this for a while, I learned a lot of basic things like how current / voltage / resistance works, how to read common schematic symbols, and how to decode a resistor. ### Week 2: Multimeter Electricity is invisible, and debugging circuits is difficult without being able to see what’s going on. I went ahead and got a multimeter (cost$20):
It was easy enough to measure resistance and voltage (both AC and DC). The current measurement was not very sensitive though and I could barely register any reading.
Around this time I attended a workshop in Manhattan that taught how to read schematics and build it on a breadboard. We made a 555 timer circuit which made a LED blink on and off:
I can’t understand how it works right now, but breadboards are pretty neat. Much easier than sticking wires into springs on my electronic kit at home.
### Week 3: Baby steps with Arduino
By now I was reaching the limits of what my electronic kit could offer, and I needed to graduate to something more serious.
So I went to the nearest electronics shop and got an Arduino Uno kit (cost $90). The Arduino is a microcontroller and lets you prototype circuits easily with a breadboard. The Arduino Uno is only$25, but my kit comes with an assortment of components and sensors.
Before long I had the Arduino up and running. It runs a dialect of C, so I felt at home in the programming environment.
Here’s a program that blinks the onboard LED on and off in a loop (kind of equivalent of hello world):
// the setup function runs once when you press reset or power the board
void setup() {
// initialize digital pin 13 as an output.
pinMode(13, OUTPUT);
}
// the loop function runs over and over again forever
void loop() {
digitalWrite(13, HIGH); // turn the LED on (HIGH is the voltage level)
delay(1000); // wait for a second
digitalWrite(13, LOW); // turn the LED off by making the voltage LOW
delay(1000); // wait for a second
}
### Week 4: Transistor Switching
I didn’t really know what a transistor did, but it’s what logic gates are made of and the backbone of all computers, so it can’t hurt to learn about them, right?
I started off building logic gates from transistors, but couldn’t get it work. It turned out that I misunderstood how a transistor operates (it’s not the most intuitive at first glance). Luckily, I had a friend in electrical engineering and she patiently cleared up my misconceptions.
Transistors are used for logic gates, but I didn’t know that transistors can also amplify a current. I also learned about the many different types of transistors.
Here’s a circuit that I built (from the Arduino kit manual). It uses a transistor as a switch to control a motor:
By the way, here’s how you make an AND gate with two transistors:
While wiring things up, I accidentally burned a LED and a BJT transistor. Apparently 5 volts without a resistor is fatal to many components. In software, if you mess up, you get a segmentation fault in your console or something — never the smell of burnt plastic in your room.
### Week 5: Arduino Controlled Desk Lamp
Here’s an idea. Wouldn’t it be nice if your lamp can turn itself on when it gets dark? Useful or not, let’s build it!
This is actually a major milestone for me. Up until now, I’ve been mostly following existing schematics, using parts carefully selected by people who wrote the schematics. But for this project, we improvise everything from scratch. Oh yea, also it’s the first time I’m working with 120V alternating current.
### Can you really develop without a device?
All across internet forums, people advocate that you should test your app thoroughly across many devices before submitting to the app store. It seems that trying to develop without a device is an edge case, often the instructions for a task assumes you have a device, and you have to find a workaround if you don’t.
In my case, it was successful, in the sense that I produced an app that didn’t crash and got past app review. But I don’t know if I got lucky, because things could have turned out badly.
Throughout the whole process, I was worried that running the app on a real device would exhibit bugs that aren’t producible in the simulator. In that case, I’d have no way to debug the problem, and the project would be done. My app uses nothing but the most basic functionality, so I had a good chance of dodging this bullet. But still, the possibility loomed over me, threatening to kill the project just as it crosses the finish line.
A second problem is by copying my original Android app feature by feature, the resulting iOS app looks and feels like an Android app. A friend pointed this out when I sent him the app. I hadn’t noticed it, but after looking at some other iOS apps, I have to agree with him. Actually in hindsight it shouldn’t surprise anyone that without seeing other iOS apps, I don’t really know how an iOS app should behave. But it just never occurred to me that the natural way to do things for me might be unnatural for iOS users.
Finally, this is subjective, but for me it wasn’t very fun to develop for a simulator. Without the tactile sensation of your creation running on an actual phone, the whole experience feels detached from reality. You feel like an unwelcome foreigner in a country where the customs are different, and you begin to question yourself, why am I doing this iOS port anyway?
Part of what kept me going was sunk cost fallacy. I paid \$119 to be an iOS developer, so I’d better get at least something on the app store, have something to show for it.
Now that the app is finished, I think I’m done with iOS development. Perhaps the app store is fertile ground for developers and startups looking to make a profit, but the cost of entry is unreasonable for someone making a few open source apps for fun.
## What’s the hardest bug you’ve ever debugged?
June 19, 2015
In a recent interview, I was asked this question: “what’s the most difficult bug you’ve encountered, and how did you fix it?” I thought this was an interesting question because there are so many answers you could give to this question, and the sort of answer you give demonstrated your level of experience with developing software.
I thought for a moment, recalling all the countless bugs I had seen and fixed. Which one was the most difficult and interesting? In this article I’m going to describe my most difficult bug to date.
It was an iOS app. I was working as a four-month intern at the time. “We’ve been seeing reports from our users that the app randomly display a black screen,” my boss explained one afternoon. “No error message, no crash log, nothing. The app is simply stuck at a black screen state until you kill it.”
“Fair enough. How do I reproduce it?”
He shrugged. “I don’t know. Users are reporting it happens randomly. Here’s what you gotta do: grab an iPad, download the game off the app store. Create an account and play the game until you hit the bug.”
So I did. I was reduced to one of these typewriter monkeys, banging away mindlessly at the keyboard until I stumble upon the sequence of button presses to trigger the undiscovered bug by sheer coincidence.
For an afternoon I monkeyed away, but no matter what buttons I pressed, the mythical black screen would not appear. I left the office, defeated and mentally exhausted.
The next morning I checked into the office, picked up the iPad, and resumed my monkeying. But this time my fortune was different: within 15 minutes, lo and behold, the screen flashed white, followed by an unrepentant screen of black.
What did I do to trigger this? I retraced my steps, trying to repeat the miracle. It happened again. Methodically I searched for a deterministic sequence of actions that brought our app to its knees. Go to the profile page. Hit button X. Go to page Y and back to the profile page. Hit button Z. The screen flickered for a millisecond, the black. Ten times out of ten.
With a sigh of relief, I jotted down this strange choreography and went for a walk. Returning with a fresh mind ready to tackle the next stage of the problem, I executed the sequence one more time, just to make sure. But the bug was nowhere to be seen.
I racked my brain for an explanation. The same sequence of actions now produced different results, I reasoned. Which meant something must have changed. But what?
It occurred to me that the page looked a little different now from when I was able to reproduce the bug. In the morning, when I came in, there was a little countdown timer in the corner of the screen that indicated the time until an upcoming event. The timer was not there anymore. Could it be the culprit…
To test this hypothesis, I produced a build that pointed the game to the dev server, and fired up a system event. The timer appeared. I executed my sequence — profile, tap, home screen, back to profile, tap, and sure enough, with a flicker the black screen appeared. I turned off the timer, repeated the sequence — profile, tap, home, profile, tap — no black screen. I had finally discovered the heart of the matter. There was some strange interaction going on between the timer and other things on the page.
At this point, with 100% reproducibility, the worst was over. It took a few more hours for me to investigate the issue and come up with a fix. My patch was quickly rolled out to production, and users stopped complaining about random black screens. Then my team went out for some celebratory beer.
I will now describe exactly what happened — and why did a timer cause such an insidious bug.
The timer widget was implemented using an NSTimer which made a callback every second. To do this, the timer holds a reference to the parent view which contains it. This is not too unusual, and is generally innocent and harmless — until you combine it with Objective C’s garbage collection system.
Objective C’s garbage collection system uses a reference counting algorithm. I’ll remind you what this means. The garbage collector maintains, for each object, a count of how many references lead to it. When this reference count reaches zero, it means your object is dead, since there is no way to reach it from anywhere in the system. Thus the garbage collector is free to delete it.
This doesn’t work for NSTimer, though. When two objects hold references to each other, their reference count remain at least 1, which means they can never get garbage collected. In our app, this meant that whenever the view with the timer goes out of view, it doesn’t get disposed, but remains in the background forever. A memory leak.
A memory leak, by itself, can go unnoticed for a long time with no impact. The last part of the puzzle that brought everything crashing down had to do with the way a certain button was implemented. This button, when pressed, broadcasted a message, which would then be received by the profile view.
When the timer is active, it is possible to get the system into a state with two profile views — a real one and a zombie one kept alive by a reference cycle with the timer.
Then when the message is broadcasted, both the real and zombie views receive the message in parallel. The button logic is executed twice in rapid succession, which understandably causes the whole system to give in.
With this mechanism in mind, the fix was easy. Just invalidate the timer when the view goes out of view. Without the reference cycle, the profile view is disposed of correctly and all is well again.
I think this story demonstrates a fundamental truth about debugging: in order to debug effectively, you need to have a deep understanding of your technology stack. This is not always true of programming in general — quite often you can write code that works yet not really understand what it’s doing. When developing a feature in an unfamiliar technology, the typical workflow is, if you don’t know how to do something, copy something similar from StackOverflow or a different part of your code base, make some changes until it works. And that’s a fine way to do things.
But debugging requires a more structured methodology. When many things are breaking in haphazard ways, you need to narrow down the problem to its very core, to identify precisely which component is broken. In this case it was a reference cycle that wouldn’t get released. The core of the problem may be buried within layers upon layers of an API, even an API you believe to be bulletproof. It might require digging into assembly code, even hardware.
To find that core requires an understanding of a mind-boggling stack of technologies that software today sits upon. That’s what it takes to become a master debugger.
So, what’s the hardest bug you’ve ever debugged?
March 22, 2015
The name of the hackathon was Code B: UW Algorithmic Trading Competition. It was hosted by Bloomberg and various UW student groups. It’s a 17 hour hackathon where you “create the best trading platform completely from scratch”. As far as I know, this is the first time the hackathon has been run, and in this article I’m going to write about my experience.
We were allowed teams of up to three, but my roommate Andrei and I signed up as a team of two. Like myself, Andrei is also a CS major. Neither of us had any experience with trading stocks, or anything finance related, for that matter. When asked to choose a team name, we named ourselves team /dev/rand (implying that we were so bad that we’d be no better than a random number generator)
The hackathon was scheduled to start Friday evening, running through the night until noon the next day. The goal was to write a program to autonomously trade stocks over a 20 minute period, battling other programs to earn as much money as possible. The programs communicated by connecting to a central server on Bloomberg’s side, so we could use any programming language we wanted. It was decided that Andrei would come up with strategies, and I would implement them in Python.
### Rules of the Game
The specifics of the API and mechanics of the game were not revealed until the official start of the hackathon. The 50-60 teams packed into an auditorium as the organizers started to explain the technical details.
The rules turned out to be fairly simple. The only actions allowed were to bid (attempt to purchase) on a stock for some price, or ask (attempt to sell) a stock for some price. If at any point someone’s bid is higher than someone else’s ask, the deal goes through and the stock changes hands.
Now all of this was fairly standard, but after this part, the rules diverged from real life. In order to encourage people to buy stocks (and not just hoard the initial money), each share of a stock paid dividends to its owner every second. And to prevent simply buying one stock and holding it for the entire duration, the dividends given out quickly diminishes the longer you own the stock.
This quirky dividends system turned out to be central to our strategy. Additionally, the differences from real stock markets meant that any previous experience with finance and stock trading was less useful — definitely a good thing for us because many of our competitors were seriously studying finance and we had no experience anyway.
### And it begins!
After the rules presentation, the hackathon kicked off. It was slightly past 7pm, and very quickly you could see teams buying and selling stocks. We decided to take it slow, discussing strategies over dinner.
We started work around 8pm. I began writing code to parse the input, while Andrei worked on deciphering the rather cryptic specifications document. Although API specs were clear enough, they were (intentionally) vague about how the system behaved behind the scenes. There were many formulas with lots of variables, many of which we had no idea what they meant.
So we took an experimental approach. Tentatively we put in a bid for a few shares of Google stock — and our net worth immediately took a nosedive. But the stock rapidly generated dividends, and before long, our net worth recovered to what it was initially, and it kept on going up! The success was short lasted, however, as quickly the dampening effects of the dividends started to kick in, and our rate of return quickly diminished to near zero.
We tried again, buying a few shares of the Twitter stock. The same thing happened — our value went down, quickly recovered, then gradually leveled to 50 dollars more than we started with.
With this information, we formulated a rough strategy. We didn’t know how to predict which stocks will go up; neither did we have a plan for buying and selling stocks at a favorable rate. Instead, we would take advantage of a stock’s “golden period”, where the stock initially pays massive dividends. It was crucial to buy as quickly as possible, since the clock started ticking as soon as you own one share of the stock. So we use all our money to buy as many shares of the stock as possible, doesn’t matter what price. Now we wait as the golden period payout multiplied by our entire bankroll makes us rich. Then, a few minutes later, when the golden period runs out, we would slowly sell, iteratively lowering our asking price until we found a buyer.
Once we sold the last share of a stock, the dividend clock doesn’t immediately reset, it slowly regenerates. So if we wait a while, say 5 minutes, then buy back the stock, we get another brief golden period. Taking this one step further, we decided on a strategy that cycled through the 10 stocks: at any given point, we would hold at most 4 of them, while the other 6 were left to “recharge”.
I proceeded to code the algorithm, while Andrei analyzed the spec document and brainstormed ways to improve the strategy. From the equations in the spec, he came up with a formula to determine what stock generated the highest dividends. Every half hour, the scoreboard would reset, and by 3am, I was basically done, and our algorithm consistently came either first or second by the end of each round. Our algorithm worked beautifully, simultaneously juggling a bunch of different stocks, some buying, others slowly selling. We watched the scoreboard as we earned hundreds of dollars every minute, ending with a ridiculous amount of money by the time it reset.
It seemed at this point that a lot of the teams were having implementation issues, like connecting to the network and parsing input, and only a handful were making any money at all, so I was pretty happy with our results.
But at 4am, disaster struck. A new round started, and our algorithm instantly plummeted to the bottom of the leaderboard. Every time we bought or sold anything, we lost money, and none of it was coming back through dividends. What happened?? It turns out that the parameters were changed, so that a very low amount of dividends were paid for owning a stock, and the only profits were made by buying low and selling high. This meant that our whole strategy, which centered around maximizing dividends, was rendered useless.
What’s worse — I discovered a bug in my implementation where our stocks were not being cycled properly: it would sell a certain stock, then instantly buy back the same stock, which didn’t allow the dividends clock to reset, meaning no dividends. Also, by this point a lot of teams were flooding the network with requests, making any network call have a small chance of throwing an exception and crashing the whole thing entirely.
The network problem was easy to fix, but at 5am, I was really tired and had difficulty tracking down the bug that was causing it to buy back the same stock. Andrei suggested a new set of strategies for the “low dividends” scenario, but by now, I was too tired to implement another set of strategies. Instead, we tweaked various constants in our program to make it play more patiently and more predictably, so even in the worst case it would make marginal gains instead of finishing dead last. After 2 hours of debugging, we managed to track down the cycling bug.
It was 7am and I could hardly keep my eyes open so we found a couch and napped for two hours, until the mock competition began.
### Mock Competition and final tweaking
At 9am, a few hours before the final competition, there was a mock competition which was meant to be identical to the final competition. There were three rounds: a high dividends scenario, a low dividends, and one in the middle.
We won the high dividends round hands down, unsurprisingly as our entire strategy was designed around this set of parameters in mind. In the low dividends round, we didn’t do as well, but thanks to careful tweaking, we still made a modest amount of money, coming in fifth. In the medium round, we got second place. This was enough to win the mock competition.
Now, let me give you a summary of our competition. Most of the teams increased gradually in net worth, with their score slowly increasing as they slowly accumulated dividends. We were confident that we could play the dividends game, so it didn’t trouble us too much. What was really troubling was a team called “vlad” (I don’t quite remember what their name was, but it ended with vlad). Instead of gradually gaining money a few dollars at a time, “vlad” remained at a constant net worth for a long time, then suddenly gain hundreds of dollars instantly. This meant that their algorithm operated completely differently from ours, and we had absolutely no explanation of what was going on.
It didn’t help that the formula for net worth was complicated and we didn’t fully understand it. Our net worth clearly increased when we did well, but it fluctuated wildly, sometimes dipping by hundreds of points when we made a large transaction, only to bounce back when dividends started rolling in.
The next few hours were fairly unproductive, since we had no more ideas on how to improve our algorithm. Although Andrei had some ideas on strategies for the low dividends game, after pulling an all-nighter I was in no shape to try implementing them.
### The Final Game
It was soon time for the final competition, the cumulation of all our efforts. Having carefully noted down the parameters for the mock competition, we were ready to use this information to get every edge we could for the finals.
Round 1 was high dividends. We played with a highly aggressive set of parameters, dumping our bad stocks for very cheap in pursuit of the dividends regeneration. The early game was contentious, but by the 10 minute mark, we gained a solid lead over the competition and maintained the lead until the end. We won round 1, with “vlad” coming in third place.
Round 2 was low dividends. We deployed the patient strategy, which was less eager to dump anything, holding onto bad stocks until we get a good price for them, since there were little dividends to fight over anyway. We came fifth place, with “vlad” coming in fourth.
Round 3 was medium dividends. We started off uncertain — at the halfway mark we were still in the middle of the pack — but slowly we gained ground, and five minutes before the end, we were in third position. “vlad” was in first place, with a big enough lead that neither we nor the second place team were going to overtake him. But at this point, we knew that from our points in the first round, we only needed second place to beat “vlad” and win the competition — and with 3 minutes left on the clock, we overtook the second place team. We were going to win it!
Then, the whole scoreboard goes black.
It didn’t crash, no, it was the contest organizers’ tactic to increase suspense so the final winners are not known until the winners are announced. We waited anxiously as the final seconds ticked down, the organizers announcing fourth place, third place, UI award. We just needed second place in this round to win, if we get third place in this round, “vlad” beats us by a hair.
And the second place goes to… team /dev/rand. What? We stared in disbelief as we realize we lost to “vlad”.
### Going home
Turns out that in the last 2 minutes of competition, we got overtaken by not one, but two teams. So we actually finished round 3 in fourth place.
Our prize for winning second place? A playstation 4 (worth ~450) and a parrot drone (worth ~100), and most importantly, the satisfaction of winning a finance competition without knowing the first thing about finance. Team “vlad” got two playstations and a drone (well, they could have taken all 3 playstations but they were nice enough to leave us one)
Big thanks to all the organizers and volunteers for keeping everything running smoothly!
If you’re interested, our source code is in a git repo here. It’s 400 lines of hackathon-level-bad python code.
A natural question to ask is, can we get rich IRL with this algorithm? Answer is clearly no — we essentially gamed the system by greedily grabbing the golden period of dividends, a mechanic designed to encourage people to buy and sell stocks. Of course, in the real world, dividends don’t work like that.
Then other than this mechanic, how else is this competition different from real world algo trading? Unfortunately, I don’t know enough about this topic to answer that question.
Philosophically, I still don’t understand how it’s possible that they basically pull money out of thin air. I mean, a stock trader doesn’t intrinsically create value for society, but they get rich doing it? I don’t know.
## Visualizing Quaternions with Unity
November 24, 2014
How do you model the position and orientation of an airplane?
Position is easy, just represent it with a point in 3D space. But how do you specify its orientation — which direction it’s pointing?
At first glance, it seems a vector will do. After all, a vector points in some direction, right? If the plane is pointing east, represent its orientation by a unit vector pointing east.
Unfortunately, we quickly run into trouble when we try to roll. If we’re facing east, and we roll 90 degrees, we’re still facing east. Clearly we’re missing something.
### Euler Angles
When real pilots talk about their orientation, they talk about roll, yaw, pitch. Pitch is going up or down, yaw is going left or right, roll is, well, roll.
Any change in orientation can be described by some combination of roll, yaw, pitch. This is the basis for Euler Angles. We use three angles to represent the airplane’s orientation.
This is all fine and dandy if we want to represent the orientation of a static object in space. But when we try to adjust our orientation, we start to run into problems.
You’re thinking, this should be simple! When we turn left or right, we just increment the yaw variable, right? Yes, it seems to work, at least initially. You can turn left and right, up and down, and roll around.
Implement it in Unity and play around a bit, however, and you begin to notice that things don’t quite behave the way you expect.
In this animation, I’m holding down the right button:
The plane does rotate to the right, but it’s not rotating relative to itself. Instead it’s rotating around some invisible y-axis. If it was rotating relative to itself, the green arrow shouldn’t be moving.
The problem becomes more and more severe when the pitch of the plane becomes higher and higher. The worst case is when the airplane is pointing straight up: then roll and yaw become the same thing! This is called gimbal lock: we have lost a degree of freedom and we can only rotate in 2 dimensions! Definitely not something desirable if we’re controlling a plane or spaceship.
It turns out that no matter what we do, we will suffer from some form of gimbal lock. As long as we use Euler Angles, there is one direction where if we turn too far, everything starts to screw up.
### Practical Introduction to Quaternions
All is not lost, however. There is a way to represent orientation that represents all axes equally and does not suffer from gimbal lock. This mythical structure is called the quaternion. Unlike Euler Angles which describe your orientation relative to a fixed set of axes, quaternions do not rely on any fixed axis.
The drawback is that quaternions are unintuitive to understand for humans. There is no way to “look” at a quaternion and be able to visualize what rotation it represents. Fortunately for us, it’s not that difficult to make use of quaternions, even if we can’t visualize quaternions.
There is a lot of theory behind how quaternions work, but in this article, I will gloss over the theory and give a quick primer to quaternions, just the most common facts you need to use them. At the same time, I will implement the operations I describe in C#, so I can integrate them with Unity. If you don’t know C#, you can freely ignore the code.
### Definition
A quaternion is an ordered pair of 4 real numbers (w,x,y,z). We write this as
$w+xi+yj+zk$
The letters i,j,k are not variables. Rather, they are independent axes. If you like, you can think of the quaternions as a 4 dimensional vector space.
The defining property of quaternions is:
$i^2 = j^2 = k^2 = ijk = -1$
Play around with it a bit and you can derive 6 more identites:
$ij = k$
$jk = i$
$ki = j$
$ji = -k$
$kj = -i$
$ik = -j$
If you’ve worked with complex numbers, this should seem familiar. Instead of 2 parts of a complex number (the real and imaginary parts), we have 4 parts for a quaternion.
The similarity doesn’t end here. Multiplying complex numbers represents a rotation in 2 dimensions. Similarly, multiplying by a quaternion represents a rotation in 3D.
One curious thing to note: we have $ij=k$ and $ji=-k$. We switched around the terms and the product changed. This means that multiplying quaternions is kind of like multiplying matrices — the order matters. So multiplication is not commutative.
Here’s a framework for a quaternion in C#:
public class Quat{
// Represents w + xi + yj + zk
public float w, x, y, z;
public Quat(float w, float x, float y, float z){
this.w = w;
this.x = x;
this.y = y;
this.z = z;
}
}
### Normalizing Quaternions
The norm of a quaternion is
$N(\mathbf{q}) = \sqrt{w^2+x^2+y^2+z^2}$
When we use quaternions to represent rotations, we typically want unit quaternions: quaternions with norm 1. This is straightforward: to normalize a quaternion, divide each component by the norm.
In C#:
public float Norm(){
return Mathf.Sqrt (w * w + x * x + y * y + z * z);
}
public Quat Normalize(){
float m = Norm ();
return new Quat (w / m, x / m, y / m, z / m);
}
### Multiplying Quaternions
Multiplying is simple, just a little tedious. If we have two quaternions:
$(w_1 + x_1i + y_1j + z_1k) (w_2+x_2i+y_2j+z_2k)$
Then their product is this ugly mess:
$\begin{array}{l} w_1w_2-x_1x_2-y_1y_2-z_1z_2 \\ + (w_1x_2+x_1w_2+y_1z_2-z_1y_2)i \\ + (w_1y_2+y_1w_2-x_1z_2+z_1x_2) j \\ + (w_1z_2+z_1w_2+x_1y_2-y_1x_2) k \end{array}$
In C#:
// Returns a*b
public static Quat Multiply(Quat a, Quat b){
float w = a.w * b.w - a.x * b.x - a.y * b.y - a.z * b.z;
float x = a.w * b.x + a.x * b.w + a.y * b.z - a.z * b.y;
float y = a.w * b.y + a.y * b.w - a.x * b.z + a.z * b.x;
float z = a.w * b.z + a.z * b.w + a.x * b.y - a.y * b.x;
return new Quat (w,x,y,z).Normalize();
}
Since multiplication is not commutative, I made this function static to avoid confusing left and right multiplication. Also, I normalize the product so that floating point errors don’t accumulate.
### Constructing Rotation Quaternions
Every rotation operation can be written as a rotation of some angle, $\theta$, around some vector $(u_x, u_y, u_z)$:
The following formula gives a quaternion that represents this rotation:
$\mathbf{q} = \cos \frac{\theta}{2} + (u_x i + u_y j + u_z k) \sin \frac{\theta}{2}$
For our purposes, $\theta$ is a very small number, say 0.01, and we use one of the three basis vectors to rotate around. For example, if we are rotating around (1,0,0) then our quaternion is
$\cos \frac{0.01}{2} + \sin \frac{0.01}{2}i$
That’s it: given any quaternion, multiplying on the left by our quaternion rotates it slightly around the x axis.
In C#, our code might look like this:
Quat qx = new Quat (Mathf.Cos (0.01 / 2), 0, 0, Mathf.Sin (0.01 / 2));
Quat qy = new Quat (Mathf.Cos (0.01 / 2), 0, Mathf.Sin (0.01 / 2), 0);
Quat qz = new Quat (Mathf.Cos (0.01 / 2), Mathf.Sin (0.01 / 2), 0, 0);
### Putting it together
That’s all we need to do interesting things with quaternions. Let’s combine everything we have. Here’s our quaternion class thus far:
public class Quat{
// Represents w + xi + yj + zk
public float w, x, y, z;
public Quat(float w, float x, float y, float z){
this.w = w;
this.x = x;
this.y = y;
this.z = z;
}
public float Norm(){
return Mathf.Sqrt (w * w + x * x + y * y + z * z);
}
public Quat Normalize(){
float m = Norm ();
return new Quat (w / m, x / m, y / m, z / m);
}
// Returns a*b
public static Quat Multiply(Quat a, Quat b){
float w = a.w * b.w - a.x * b.x - a.y * b.y - a.z * b.z;
float x = a.w * b.x + a.x * b.w + a.y * b.z - a.z * b.y;
float y = a.w * b.y + a.y * b.w - a.x * b.z + a.z * b.x;
float z = a.w * b.z + a.z * b.w + a.x * b.y - a.y * b.x;
return new Quat (w,x,y,z).Normalize();
}
public Quaternion ToUnityQuaternion(){
return new Quaternion (w, x, y, z);
}
}
Now we just need to read the input, perform our calculations, and output the rotation quaternion to Unity:
public class Airplane : MonoBehaviour {
public GameObject airplane;
public Quat quat = new Quat (0, 0, 0, -1);
public float speed = 0.01f;
void FixedUpdate(){
float inputX = Input.GetAxis("UpDown");
float inputY = Input.GetAxis("LeftRight");
float inputZ = Input.GetAxis("Roll");
Quat qx = new Quat (Mathf.Cos (speed * inputX / 2), 0, 0, Mathf.Sin (speed * inputX / 2));
Quat qy = new Quat (Mathf.Cos (speed * inputY / 2), 0, Mathf.Sin (speed * inputY / 2), 0);
Quat qz = new Quat (Mathf.Cos (speed * inputZ / 2), Mathf.Sin (speed * inputZ / 2), 0, 0);
quat = Quat.Multiply (qx, quat);
quat = Quat.Multiply (qy, quat);
quat = Quat.Multiply (qz, quat);
airplane.transform.rotation = quat.ToUnityQuaternion ();
}
}
In Unity, the input is not given to us as a single true/false value, but a float between -1 and 1. So holding right increases the LeftRight input gradually until it reaches 1, avoiding a sudden jump in movement.
What’s ToUnityQuaternion? Well, it turns out that Unity already has a Quaternion class that does everything here and much more, so all this could have literally been implemented in one line if we wanted.
Anyways, let’s see the result.
As you can see, holding right turns the plane relative to itself now, and the green arrow stays still. Hooray!
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2016-10-01 13:43:28
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https://www.hackmath.net/en/math-problem/2156
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# Car value
The car loses value 15% every year.
Determine a time (in years) when its price will halve.
n = 4.265
### Step-by-step explanation:
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## Related math problems and questions:
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How much money would you need to deposit today at 5% annual interest compounded monthly to have \$2000 in the account after 9 years?
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2021-05-08 01:06:57
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http://www-spires.fnal.gov/spires/find/books/www?keyword=Functions+of+real+variables
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Fermilab Core Computing Division
Library Home | Ask a Librarian library@fnal.gov | Book Catalog | Library Journals | Requests | SPIRES | Fermilab Documents |
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SPIRES-BOOKS: FIND KEYWORD FUNCTIONS OF REAL VARIABLES *END*INIT* use /tmp/qspiwww.webspi1/8403.22 QRY 131.225.70.96 . find keyword functions of real variables ( in books using www
Call number: SPRINGER-2011-9781441998132:ONLINE Show nearby items on shelf Title: Nonelliptic Partial Differential Equations [electronic resource] : Analytic Hypoellipticity and the Courage to Localize High Powers of T Author(s): David S Tartakoff Date: 2011 Publisher: New York, NY : Springer New York Size: 1 online resource Note: Springer e-book platform Note: Springer 2013 e-book collections Note: This book fills a real gap in the analytical literature. After many years and many results of analytic regularity for partial differential equations, the only access to the technique known as $(T^p)_\phi$ has remained embedded inthe research papers t hemselves, making it difficult for a graduate student or a mature mathematician in another discipline to master the technique and use it to advantage. This monograph takes a particularly non-specialist approach,one might even say gentle, to smoothly bring the reader into the heart of the technique and its power, and ultimately to show many of the results it has been instrumental in proving. Another technique developed simultaneously by F.Treves is developed and compared and contrasted to ours. The techni ques developed here are tailored to proving real analytic regularity to solutions of sums of squares of vector fields with symplectic characteristic variety andothers, real and complex. The motivation came from the field of several complex variables and t he seminal work of J. J. Kohn. It has found application in non-degenerate (strictly pseudo-convex) and degenerate situations alike, linearand non-linear, partial and pseudo-differential equations, real and complex analysis. The technique is utterly elemen tary, involving powers of vector fields and carefully chosen localizing functions. No knowledge of advancedtechniques, such as the FBI transform or the theory of hyperfunctions is required. In fact analyticity is proved using only $C^\infty$ techniques. The book is intended for mathematicians from graduate students up, whether inanalysis or not, who are curious which non-elliptic partial differential operators have the property that all solutions must be real analytic. Enough background is provided to pr epare the reader with it for a clear understanding of thetext, although this is not, and does not need to be, very extensive. In fact, it is very nearly true that if the reader is willing to accept Note: Springer eBooks Contents: 1. What this book is and is not 2. Brief Introduction 3.Overview of Proofs 4. Full Proof for the Heisenberg Group 5. Coefficients 6. Pseudo differential Problems 7. Sums of Squares and Real Vector Fields 8. \bar{\partial} Neumann and the Boundary Laplacian 9. Symmetric Degeneracies 10. Details of the Previous Chapter 11. Non symplectic Strategem ahe 12.Operators of Kohn Type Which Lose Derivatives 13. Non linear Problems 14. Treves' Approach 15. Appendix Bibliography ISBN: 9781441998132 Series: e-books Series: SpringerLink (Online service) Series: Developments in Mathematics, 1389-2177 : v22 Series: Mathematics and Statistics (Springer-11649) Keywords: Mathematics , Global analysis (Mathematics) , Differential equations, partial Availability: Click here to see Library holdings or inquire at Circ Desk (x3401) Click to reserve this book Be sure to include your ID please. More info: Amazon.com More info: Barnes and Noble Full Text: Click here Location: ONLINE
Call number: SPRINGER-2010-9788847017849:ONLINE Show nearby items on shelf Title: Mathematical Analysis II [electronic resource] Author(s): Claudio Canuto Anita Tabacco Date: 2010 Publisher: Milano : Springer Milan Size: 1 online resource Note: Springer e-book platform Note: Springer 2013 e-book collections Note: The purpose of this textbook is to present an array of topics in Calculus, and conceptually follow our previous effort Mathematical Analysis I.The present material is partly found, in fact, in the syllabus of the typical secondlecture course in Calcu lus as offered in most Italian universities. While the subject matter known as Calculus 1' is more or less standard, and concerns real functions of real variables, the topics of a course on Calculus 2'can varya lot, resulting in a bigger flexibility. Fo r these reasons the Authors tried to cover a wide range of subjects, not forgetting that the number of credits the current programme specifications confers to a second Calculus course is notcomparable to the amount of content gathered here. The reminders disseminated in the text make the chapters more independent from one another, allowing the reader to jump back and forth, and thus enhancing the versatility of the book.On the website: http://calvino.polito.it/canuto-tabacco/analisi 2, the interested read er may find the rigorous explanation of the results that are merely stated without proof in the book, together with useful additional material. TheAuthors have completely omitted the proofs whose technical aspects prevail over the fundamental notions and ideas. The large number of exercises gathered according to the main topics at the end of each chapter should help the studentput his improvements to the test. The solution to all exercises is provided, and very often the procedure for solving is outlined Note: Springer eBooks ISBN: 9788847017849 Series: e-books Series: SpringerLink (Online service) Series: Universitext Series: Mathematics and Statistics (Springer-11649) Keywords: Mathematics , Global analysis (Mathematics) , Functional analysis , Differential Equations , Differential equations, partial Availability: Click here to see Library holdings or inquire at Circ Desk (x3401) Click to reserve this book Be sure to include your ID please. More info: Amazon.com More info: Barnes and Noble Full Text: Click here Location: ONLINE
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2019-04-18 18:16:00
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https://math.stackexchange.com/questions/59903/finite-subgroups-of-the-multiplicative-group-of-a-field-are-cyclic
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# Finite subgroups of the multiplicative group of a field are cyclic
In Grove's book Algebra, Proposition 3.7 at page 94 is the following
If $G$ is a finite subgroup of the multiplicative group $F^*$ of a field $F$, then $G$ is cyclic.
He starts the proof by saying "Since $G$ is the direct product of its Sylow subgroups ...". But this is only true if the Sylow subgroups of $G$ are all normal. How do we know this?
• Multiplication is commutative. So $G$ is abelian and every subgroup is normal. – jspecter Aug 26 '11 at 14:03
• For a finite group, $G$ is nilpotent if and only if it is the direct product of its Sylow subgroups. – user1729 Aug 26 '11 at 14:31
• A slight generalization of the lemma/theorem you are wondering about is topic of this question (in the moment there is no answer, but a good comment by Geoff). – Someone Aug 26 '11 at 14:59
• See also : mathoverflow.net/questions/54735 – Watson Nov 11 '17 at 13:05
• See also Stroppel, Locally compact groups, Theorem 6.32 and corollary 6.33. – Watson Nov 11 '17 at 14:23
There's a simple proof which doesn't use Sylow's theory.
Lemma. Let $$G$$ a finite group with $$n$$ elements. If for every $$d \mid n$$, $$\# \{x \in G \mid x^d = 1 \} \leq d$$, then $$G$$ is cyclic.
If $$G$$ is a finite subgroup of the multiplicative group of a field, then $$G$$ satisfies the hypothesis because the polynomial $$x^d - 1$$ has $$d$$ roots at most.
Proof. Fix $$d \mid n$$ and consider the set $$G_d$$ made up of elements of $$G$$ with order $$d$$. Suppose that $$G_d \neq \varnothing$$, so there exists $$y \in G_d$$; it is clear that $$\langle y \rangle \subseteq \{ x \in G \mid x^d = 1 \}$$. But the subgroup $$\langle y \rangle$$ has cardinality $$d$$, so from the hypothesis we have that $$\langle y \rangle = \{ x \in G \mid x^d = 1 \}$$. Therefore $$G_d$$ is the set of generators of the cyclic group $$\langle y \rangle$$ of order $$d$$, so $$\# G_d = \phi(d)$$.
We have proved that $$G_d$$ is empty or has cardinality $$\phi(d)$$, for every $$d \mid n$$. So we have: $$n = \# G = \sum_{d \mid n} \# G_d \leq \sum_{d \mid n} \phi(d) = n,$$ Therefore $$\# G_d = \phi(d)$$ for every $$d \vert n$$. In particular $$G_n \neq \varnothing$$. This proves that $$G$$ is cyclic. QED
• Very nice proof. It may be noted that the last equality $\sum_{d|n}\phi(d)=n$ is derived from the very same argument applied when $G$ is the cyclic group of order$~n$, using the additional knowledge that in this case elements of every order $d|n$ do exist. In other words no knowledge at all about the values $\phi(d)$, apart from the fact that the are well defined, is used. – Marc van Leeuwen Apr 28 '13 at 5:24
• @MarcvanLeeuwen, what are the $\phi(n)$? – Juan Pablo Jul 12 '13 at 21:27
• @JuanPablo: That's Euler's totient function, the number of non-negative integers${}<n$ that are relatively prime to$~n$. – Marc van Leeuwen Jul 12 '13 at 22:06
• @Andrea or whoever can answer, can you explain what the # notation means, please? – ALannister Sep 27 '18 at 1:47
• @ALannister It simply means the cardinality of the following finite set (the # notation is typically only used when talking about finite sets, so "cardinality" here means "number of elements"). – Nubok Oct 28 '18 at 0:04
We know that if $$G$$ is a finite abelian group, $$G$$ is isomorphic to a direct product $$\mathbb{Z}_{(p_1)^{n_1}} \times \mathbb{Z}_{(p_2)^{n_2}} \times \cdots \times \mathbb{Z}_{(p_r)^{n_r}}$$ where $$p_i$$'s are prime not necessarily distinct.
Consider each of the $$\mathbb{Z}_{(p_i)^{n_i}}$$ as a cyclic group of order $$p_i^{n_i}$$ in multiplicative notation. Let $$m$$ be the $$lcm$$ of all the $$p_i^{n_i}$$ for $$i=1,2,\ldots,r.$$ Clearly $$m\leq {p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}.$$ If $$a_i \in \mathbb{Z}_{(p_i)^{n_i}}$$ then $$(a_i)^{({p_i}^{n_i})}=1$$ and hence $$a_i^m=1.$$ Therefore for all $$\alpha \in G,$$ we have $$\alpha^m=1;$$ that is, every element of $$G$$ is a root of $$x^m=1.$$
However, $$G$$ has $${p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}$$ elements, while the polynomial $$x^m-1$$ can have at most $$m$$ roots in $$F.$$ So, we deduce that $$m={p_1}^{n_1}{p_2}^{n_2}\cdots{p_r}^{n_r}.$$ Therefore $$p_i$$'s are distinct primes, and the group $$G$$ is isomorphic to the cyclic group $$\mathbb{Z}_m.$$
• In the second paragraph third sentence: It is not true that if $a_i \in \mathbb{Z}_{(p_i)^{n_i}}$, then $(a_i)^{({p_i}^{n_i})}=1$. Take $2 \in \mathbb{Z}_{3^2}=\mathbb{Z}_9$. We have $2^9 =512 \equiv 8 \pmod 9$. – Al Jebr Nov 19 '18 at 16:12
• @AlJebr I don't know if I understand this correctly, but $2^9$ should be supposed to mean $2+2+2+\dots+2$ for 9 times ? So the result is clearly divisible by 9. – hephaes Dec 7 '18 at 8:12
Note that this result is not true if $F$ is a skew field (division ring), as is illustrated by the quaternion group $Q_8$ inside the quaternions. So one must use commutativity somewhere, and this usually happens implicitly by using that the polynomial $X^d-1$ can have at most $d$ roots in $F$; this is for instance the case in the answer by Andrea, where the proof of the lemma does not use commutativity. Here is a somewhat different approach that exploits commutativity a second time.
Lemma. The set of orders of elements in a finite Abelian group is closed under taking least common multiples.
(Edit: This happens to be the subject of another math.SE question. It may seem quite hard, unless one realises that in Abelian torsion groups, different prime factors can be considered independently due to a canonical direct sum decomposition, after which the question becomes trivial. Here I'll leave my original proof below, which follows another answer to that question.)
Proof. The set of orders (in any group) is certainly closed under taking divisors: if $x$ has order $n$ and $d\mid n$ then $x^{n/d}$ has order $d$. Now if $a,b$ are orders of elements in an Abelian group and $\def\lcm{\operatorname{lcm}}m=\lcm(a,b)$, then there are relatively prime $a',b'$ with $a'\mid a$, $b'\mid b$, and $a'b'=m$: it suffices to retain in $a'$ those and only those prime factors of $a$ whose multiplicity in $a$ is at least as great as in $b$, and to retain in $b'$ all other prime factors of $b$ (those whose multiplicity exceeds those in $a$). Now if $x$ has order $a'$ and $y$ has order $b'$, then these orders are relatively prime, whence $\langle x\rangle\cap\langle y\rangle=\{e\}$, and their product is$~m$ so that $$x^iy^i =e\iff x^i=e=y^i\iff (\lcm(a',b')=a'b'=)\; m\mid i,$$ and therefore $xy$ has order $m$. QED
Now to prove the proposition, let $n=\#G$, and let $m$ be the least common multiple of all the orders of elements of $G$. By Lagrange's theorem the order of every element divides$~n$, whence $m\mid n$ by the property of least common multiples. But one also has $n\leq m$ since all $n$ elements of $G$ are roots of the polynomial $X^m-1$ in the field$~F$. Therefore $n=m$, and by the lemma (using that $G$ is commutative since $F$ is so) $G$ has an element $g$ of order $m=n=\#G$, so that $G=\langle g\rangle$ is cyclic.
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2019-07-21 12:57:15
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https://chemistry.stackexchange.com/questions/140881/13c-nmr-of-aniline-why-are-carbons-that-are-further-from-the-amine-group-more-d
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# 13C-NMR of aniline: Why are carbons that are further from the amine group more downfield than carbons that are closer? [closed]
A $$\ce{^{13}C}$$-NMR of aniline shows that the carbons further from the $$\ce{-NH2}$$ group are more downfield than the ones near the top of the benzene ring. If electronegative atoms like nitrogen have a deshielding effect on nearby Carbons, why is this the case? Is it because of dipole vectors?
• I strongly suggest using deshielded/shielded instead of downfield/upfield, as the latter terms are historical relics from an old way of doing NMR (continuous wave). Deshielded/shielded is really more intuitive and easily relatable to other chemistry concepts. – orthocresol Oct 1 '20 at 9:39
• Noted, thank you – John Boon Oct 1 '20 at 19:33
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2021-04-20 01:01:43
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http://mathoverflow.net/feeds/question/11253
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Existence of fine moduli space for curves and elliptic curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T03:24:32Z http://mathoverflow.net/feeds/question/11253 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11253/existence-of-fine-moduli-space-for-curves-and-elliptic-curves Existence of fine moduli space for curves and elliptic curves Anweshi 2010-01-09T21:13:09Z 2010-02-13T05:48:37Z <ol> <li><p>For the moduli problem of a curve of genus $g$ with $n$ marked points, how large an $n$ is needed to ensure the existence of a fine moduli space? For this question, terminology is that of Mumford's GIT.</p></li> <li><p>For the following three moduli problems, how big an $N$ is required for existence of a fine moduli space? The terminology is from the exposes of Deligne-Rapoport and Katz-Mazur, or Shimura. The first is in French, the second is too big, and the third is using old language and never mentions the modern terminology of universal elliptic curve, etc.. Therefore it is not possible for me to dig up the information myself.</p></li> </ol> <p>i) Elliptic curves equipped with a cyclic subgroup of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_0(N)$.</p> <p>ii) Elliptic curves equipped with a point of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_1(N)$.</p> <p>ii) Elliptic curves equipped with a symplectic pairing on $N$-torsion points -- this moduli problem corresponds to the modular group $\Gamma (N)$.</p> <p>References other than the above, will be appreciated.</p> http://mathoverflow.net/questions/11253/existence-of-fine-moduli-space-for-curves-and-elliptic-curves/11254#11254 Answer by Emerton for Existence of fine moduli space for curves and elliptic curves Emerton 2010-01-09T21:24:21Z 2010-01-09T21:24:21Z <p>The first is unrepresentable for arbitrary large $N$ (it depends on the residue class of $N$ mod 12), the second is representatble for $N \geq 4$ (if you are considering $Y_1(N)$) or $N \geq 5$ (if you are considering $X_1(N)$, i.e. including the cusps), the third is representable for $N \geq 3$. </p> <p>The references you mentioned are the standard ones. Probably Silverman discusses these in his books somewhere too (maybe the 2nd). If you look in Gross's Duke paper on companion forms (<I>A tameness criterion ... </I>) you will find a summary of the story for $X_1(N)$. In the $\Gamma_0(N)$ case, Mazur has a careful discussion in the beginning of section 2 of his <I>Eisenstein ideal</I> paper. Both Gross and Mazur refer back to Deligne--Rapoport for proofs.</p> <p>It is also just a matter of computing the torsion in each of the $\Gamma$'s (plus epsilon more if you want to understand representability at the cusps), which is an exercise. (Although you have to do a little work to see why this is the necessary computation.)</p> http://mathoverflow.net/questions/11253/existence-of-fine-moduli-space-for-curves-and-elliptic-curves/11282#11282 Answer by JSE for Existence of fine moduli space for curves and elliptic curves JSE 2010-01-10T02:25:03Z 2010-01-10T13:02:17Z <p>Here is a thought on the first question. What you need to know (at least to get an algebraic space; I'll let others be more careful than I if you want a scheme) is how large n must be to ensure that an automorphism of a smooth genus g curve X which fixes n points must be the identity. Let G be the cyclic group generated by this automorphism: then the map X -> X/G is totally ramified at your n fixed points. So by Riemann-Hurwitz, g(X) [NO, 2g(X)-2, THANKS, BJORN) is at least -2|G| + n(|G|-1). If G is nontrivial, in other words, g is at least n-4 [NO, 2g+2, THANKS, BJORN]. So I think g+5 [NO, 2g+3, THANKS, BJORN] marked points should be enough. That this is necessary can be seen by taking g=2; on M_{2,6} you'll have a bunch of loci with an extra involution, parametrizing curves whose marked points are precisely the Weierstrass points.</p> <p>[NO MORE LATE-NIGHT RIEMANN-HURWITZ: THANKS TO BJORN FOR CORRECTING THE ERRORS]</p> http://mathoverflow.net/questions/11253/existence-of-fine-moduli-space-for-curves-and-elliptic-curves/15171#15171 Answer by BCnrd for Existence of fine moduli space for curves and elliptic curves BCnrd 2010-02-13T05:48:37Z 2010-02-13T05:48:37Z <p>If you want to work over a base ring such as $\mathbf{Z}[1/n]$ rather than over $\mathbf{Q}$ or $\mathbf{C}$ then the relevant numerical condition is that the part of $N$ coprime to $n$ not be "too small" in the $\Gamma_1$ and full level cases. For an extreme example, if $N$ is a $p$-power and you work over $\mathbf{Z}_{(p)}$ then you'll always have problems in characteristic $p$ at the supersingular points.</p> <p>On the other hand, if you're willing to go beyond schemes and work with algebraic spaces or Deligne-Mumford or Artin stacks then these issues go away (at the expense of more technical background) in the sense that one has a reasonable "moduli space" over $\mathbf{Z}$ with nice regularity properties for all $N$ (even incorporating degenerations in the sense of generalized elliptic curves with level structure). It has better properties than a coarse moduli space (aside from perhaps not being a scheme...).</p>
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2013-05-20 03:24:34
|
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|
https://socratic.org/questions/how-do-you-solve-2-3x-5-6-7-8x-1-2
|
# How do you solve 2/3x+5/6=7/8x-1/2?
Apr 29, 2017
See the entire solution process below:
#### Explanation:
First, multiply both sides of the equation by $\textcolor{red}{24}$ to eliminate the fractions while keeping the equation balanced $\textcolor{red}{24}$ is the Lowest Common Denominator of the 4 fractions:
$\textcolor{red}{24} \left(\frac{2}{3} x + \frac{5}{6}\right) = \textcolor{red}{24} \left(\frac{7}{8} x - \frac{1}{2}\right)$
$\left(\textcolor{red}{24} \cdot \frac{2}{3} x\right) + \left(\textcolor{red}{24} \cdot \frac{5}{6}\right) = \left(\textcolor{red}{24} \cdot \frac{7}{8} x\right) - \left(\textcolor{red}{24} \cdot \frac{1}{2}\right)$
$\left(\cancel{\textcolor{red}{24}} 8 \cdot \frac{2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} x\right) + \left(\cancel{\textcolor{red}{24}} 4 \cdot \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}}\right) = \left(\cancel{\textcolor{red}{24}} 3 \cdot \frac{7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}} x\right) - \left(\cancel{\textcolor{red}{24}} 12 \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right)$
$16 x + 20 = 21 x - 12$
Next, subtract $\textcolor{red}{16 x}$ and add $\textcolor{b l u e}{12}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:
$- \textcolor{red}{16 x} + 16 x + 20 + \textcolor{b l u e}{12} = - \textcolor{red}{16 x} + 21 x - 12 + \textcolor{b l u e}{12}$
$0 + 32 = \left(- \textcolor{red}{16} + 21\right) x - 0$
$32 = 5 x$
Now, divide each side of the equation by $\textcolor{red}{5}$ to solve for $x$ while keeping the equation balanced:
$\frac{32}{\textcolor{red}{5}} = \frac{5 x}{\textcolor{red}{5}}$
$\frac{32}{5} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}}$
$\frac{32}{5} = x$
$x = \frac{32}{5}$
|
2023-02-01 12:11:43
|
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|
http://mathematica.stackexchange.com/questions/1193/why-is-mainevaluate-being-used-when-linearsolve-can-be-compiled/1218
|
# Why is MainEvaluate being used when LinearSolve can be compiled?
According to this question LinearSolve can be Compiled. However, CompilePrint shows a MainEvaluate but no-warning is generated. It appears that LinearSolve is not compilable, given the MainEvaluate. But the lack of warning is surprising. Something more subtle is going on. Consider the following.
In[1]:= SetSystemOptions[
"CompileOptions" -> "CompileReportExternal" -> True];
In[2]:= << CompiledFunctionTools
In[3]:= v2 = Compile[{{m, _Real, 2}, {v, _Real, 1}},
LinearSolve[m, v]
];
In[4]:= CompilePrint[v2]
Out[4]= "
2 arguments
3 Tensor registers
Underflow checking off
Overflow checking off
Integer overflow checking on
RuntimeAttributes -> {}
T(R2)0 = A1
T(R1)1 = A2
Result = T(R1)2
1 T(R1)2 = MainEvaluate[ Hold[LinearSolve][ T(R2)0, T(R1)1]]
2 Return
"
There are no warnings generated, but I am not sure why there is a MainEvaluate in the CompilePrint.
There is a much clearer warning that Compiling fails when one uses Options within LinearSolve while attempting to compile. Consider the following:
In[5]:= v3 = Compile[{{m, _Real, 2}, {v, _Real, 1}},
LinearSolve[m, v, Method -> "Cholesky"]
]
During evaluation of In[5]:= Compile::extscalar: Method->Cholesky cannot
be compiled and will be evaluated externally.
The result is assumed to be of type Integer. >>
During evaluation of In[5]:= Compile::exttensor: LinearSolve[m,v,Method->Cholesky]
cannot be compiled and will be evaluated externally.
The result is assumed to be a rank 2 tensor of type Real. >>
Also, CompilePrint gives the following:
In[6]:= CompilePrint[v3]
Out[6]= "
2 arguments
1 Integer register
3 Tensor registers
Underflow checking off
Overflow checking off
Integer overflow checking on
RuntimeAttributes -> {}
T(R2)0 = A1
T(R1)1 = A2
Result = T(R2)2
1 T(R2)2 = MainEvaluate[ Function[{m, v}, LinearSolve[m, v,
Method -> Cholesky]][ T(R2)0, T(R1)1]]
2 Return
"
Questions: If LinearSolve can't be compiled, why is there no warning in the default case? Is there something more subtle going on (e.g. some parts of the process are compiled)? If yes, how can one use the Method option within the Compiled function to ensure that what can be compiled actually is?
-
Short answer: no it can't be compiled. – rcollyer Feb 2 '12 at 18:22
@rcollyer I would however like to understand why there is no warning – acl Feb 2 '12 at 18:24
you probably want to rephrase it as "why is no warning produced", since otherwise it looks like an exact duplicate of the question @rcollyer links to above (and will be closed as such, it appears) – acl Feb 2 '12 at 18:27
@acl But the list given by Oleksandr and linked by rcollyer says LinearSolve can be compiled. – asim Feb 2 '12 at 18:31
@rcollyer is right; LinearSolve can't be compiled. However, you normally don't get a warning because InternalCompileValues provides complete type information about it based on the type of the arguments. (Adding a Method subverts matching against the patterns given by InternalCompileValues[LinearSolve], so you then get the warning.) Appearance of a function in the second list I gave does not necessarily mean it can be compiled without needing MainEvaluate! – Oleksandr R. Feb 2 '12 at 23:03
acl already posted the crucial information needed to solve this conundrum (i.e., the definition of InternalCompileValues[LinearSolve]), but wishes to delete his post since he had not interpreted it to give the complete answer. Therefore I re-post the following observation along with a summary of what it means.
The input,
InternalCompileValues[];
InternalCompileValues[LinearSolve]
yields:
HoldPattern[InternalCompileValues[LinearSolve]] :> {
HoldPattern[
LinearSolve[
SystemCompileDumpx_?(InternalTensorTypeQ[Real, {_, _}]),
SystemCompileDumpb_?(InternalTensorTypeQ[Real, {_}])]
] :> _?(InternalTensorTypeQ[Real, {_}]),
HoldPattern[
LinearSolve[
SystemCompileDumpx_?(InternalTensorTypeQ[Complex, {_, _}]),
SystemCompileDumpb_?(InternalTensorTypeQ[Complex, {_}])]
] :> _?(InternalTensorTypeQ[Complex, {_}])
}
Briefly put, this tells us that when the compiler sees a function call like LinearSolve[x, b], it knows that:
• when x is a real matrix and b is a real vector, the result is a real vector
• when x is a complex matrix and b is a complex vector, the result is a complex vector
As a result of this knowledge, the compiler is able to determine what type of register is needed to store the return value from LinearSolve in these two cases. This is important if further operations are then carried out on the result: in the absence of type information, all subsequent operations on LinearSolve's return value would need to be performed via the interpreter using MainEvaluate for full generality, but because the type of the result is predetermined, such operations can be compiled instead. However, since LinearSolve is a highly optimized top-level function, compilation does not offer any benefit outside of this scenario, and so knowing the return type has no value if LinearSolve[x, b] is the entire contents of the compiled function, since the operation may as well have been performed via the interpreter anyway.
As regards why LinearSolve[x, b, Method -> m] produces a message: it is because the definition for InternalCompileValues[LinearSolve] does not provide for pattern matching against LinearSolve calls when any Method is specified. It handles only the form LinearSolve[x, b].
### Conclusion
Just because InternalCompileValues[func] is defined for some function func, one cannot assume that func can be called directly from compiled code without using a MainEvaluate call. It simply means that the compiler has information about func which it can incorporate into the compilation process as a whole.
-
|
2014-08-21 06:29:52
|
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|
https://cerncourier.com/c/strong-interactions/page/10/
|
# Strong interactions
Read article 'LHCb discovers new baryon'
### LHCb discovers new baryon
The LHCb collaboration has discovered a new weakly decaying particle: a baryon called the Ξ++cc, which contains two charm quarks and an up quark. The discovery of the new particle, which was observed...
Read article 'FAIR forges its future'
### FAIR forges its future
Supernova explosions, neutron-star mergers and rare radioactive ions might not seem to have much connection to terrestrial matters. Yet, while the lightest elements were synthesised immediately after ...
Read article 'Proton–proton collisions become stranger'
### Proton–proton collisions become stranger
Recreating the intense fireball of quarks and gluons that existed immediately after the Big Bang, the quark–gluon plasma (QGP), traditionally requires high-energy collisions between heavy ions such ...
Read article 'LHCb brings cosmic collisions down to Earth'
### LHCb brings cosmic collisions down to Earth
The LHCb collaboration has generated high-energy collisions between protons and helium nuclei similar to those that take place when cosmic rays strike the interstellar medium.
Read article 'ALICE reveals dominance of collective flow'
### ALICE reveals dominance of collective flow
The study of the anisotropic flow in heavy-ion collisions at the LHC, which measures the momentum anisotropy of the final-state particles, has been effective in characterising the extreme states of ma...
Read article 'ATLAS reveals more strangeness in the proton'
### ATLAS reveals more strangeness in the proton
The excellent theoretical understanding of the production of electroweak W and Z gauge bosons in proton–proton collisions at the LHC makes these “standard-candle” processes ideal for studying th...
|
2022-07-05 01:14:06
|
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|
https://www.intmath.com/differentiation/8-derivative-implicit-function.php
|
# 8. Differentiation of Implicit Functions
by M. Bourne
We meet many equations where y is not expressed explicitly in terms of x only, such as:
y4 + 2x2y2 + 6x2 = 7
You can see several examples of such expressions in the Polar Graphs section.
It is usually difficult, if not impossible, to solve for y so that we can then find (dy)/(dx).
We need to be able to find derivatives of such expressions to find the rate of change of y as x changes. To do this, we need to know implicit differentiation.
Let's learn how this works in some examples.
### Example 1
Find the expression for (dy)/(dx) if y4 + x5 − 7x2 − 5x-1 = 0.
y4 + x5 − 7x2 − 5x-1 = 0
We see how to derive this expression one part at a time. We just derive expressions as we come to them from left to right.
(In this example we could easily express the function in terms of y only, but this is intended as a relatively simple first example.)
Part A: Find the derivative with respect to x of: y4
To differentiate this expression, we regard y as a function of x and use the power rule.
Basics: Observe the following pattern of derivatives:
d/(dx)y=(dy)/(dx)
d/(dx)y^2=2y(dy)/(dx)
d/(dx)y^3=3y^2(dy)/(dx)
It follows that:
d/(dx)y^4=4y^3(dy)/(dx)
Part B: Find the derivative with respect to x of:
x5 − 7x2 − 5x-1
This is just ordinary differentiation:
d/(dx)(x^5-7x^2-5x^-1) =5x^4-14x+5x^-2
Part C:
On the right hand side of our expression, the derivative of zero is zero. ie
d/(dx)(0)=0
Now, combining the results of parts A, B and C:
4y^3(dy)/(dx)+5x^4-14x+5x^-2=0
Next, solve for dy/dx and the required expression is:
(dy)/(dx)=(-5x^4+14x-5x^-2)/(4y^3
Easy to understand math videos:
MathTutorDVD.com
Continues below
### Example 2
Find the slope of the tangent at the point (2,-1) for the curve:
2y + 5 − x2y3 = 0.
Working left to right, we have:
Derivative of 2y:
d/(dx)2y=2(dy)/(dx)
Derivative of 5 is 0.
Derivative of x2 is 2x.
Derivative of y3:
d/(dx)y^3=3y^2(dy)/(dx)
Putting it together, implicit differentiation gives us:
2(dy)/(dx)-2x-3y^2(dy)/(dx)=0
Collecting like terms gives:
(2-3y^2)(dy)/(dx)=2x
So
(dy)/(dx)=(2x)/(2-3y^2)
Now, when x = 2 and y = -1,
(dy)/(dx)=(2(2))/(2-3(-1)^2)
=4/-1
=-4
So the slope of the tangent at (2,-1) is -4.
Let's see what we have done. We graph the curve
2y+5-x^2-y^3=0
and graph the tangent to the curve at (2, -1). We see that indeed the slope is -4.
It works!
### Example 3 (Involves Product Rule)
Find the expression for (dy)/(dx) if:
y4 + 2x2y2 + 6x2 = 7
To make life easy, we will break this question up into parts.
### Part A:
Find the derivative with respect to x of: y4
d/(dx)y^4=4y^3(dy)/(dx)
### Part B:
Find the derivative with respect to x of 2x2y2
Now to find the derivative of 2x2y2 with respect to x we must recognise that it is a product.
If we let u = 2x2 and v = y2 then we have:
d/(dx)(2x^2y^2)=u(dv)/(dx)+v(du)/(dx)
=(2x^2)(2y(dy)/(dx))+ (y^2)(4x)
=4x^2y(dy)/(dx)+4xy^2
### Part C:
Now
d/(dx)6x^2=12x
and
d/(dx)(7)=0
Now to find (dy)/(dx) for the whole expression:
y^4+2x^2y^2+6x^2=7
Working left to right, using our answers from above:
[4y^3(dy)/(dx)]+[4x^2y(dy)/(dx)+4xy^2]+ [12x]=0
This gives us, on collecting terms:
(4y^3+4x^2y)(dy)/(dx)=-4xy^2-12x
So we have the required expression:
(dy)/(dx)=(-4xy^2-12x)/(4y^3+4x^2y)=(-xy^2-3x)/(y^3+x^2y)
Get the Daily Math Tweet!
|
2018-04-19 21:28:47
|
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|
https://zbmath.org/?q=ut%3AMorgan%27s+conjecture
|
Embedded surfaces in 4-manifolds.(English)Zbl 0746.53041
Proc. Int. Congr. Math., Kyoto/Japan 1990, Vol. I, 529-539 (1991).
[For the entire collection see Zbl 0741.00019.]
The purpose of the article is to outline a “proof” of the following conjecture due to J. W. Morgan: If $$X$$ is a simply-connected, oriented 4- manifold for which Donaldson’s polynomial invariants are defined and non- zero, and if $$\Sigma$$ is a smoothly embedded, oriented 2-manifold with positive self-intersection, then the genus $$g$$ of $$\Sigma$$ satisfies $$2g- 2\geq \Sigma\cdot\Sigma$$. The conjecture is known to hold for $$g=0$$ or $$=1$$. The author’s argument uses gauge theory, but with a modification; he considers connections in some auxiliary $$SU(2)$$ or $$SO(3)$$ bundle over $$X\backslash \Sigma$$, which have non-trivial holonomy around the small linking circles of $$\Sigma$$. As for “branched instantons” some essential facts are missing, the author has to fill the gap by two other conjectures.
MSC:
53C40 Global submanifolds 53C07 Special connections and metrics on vector bundles (Hermite-Einstein, Yang-Mills)
Zbl 0741.00019
|
2023-03-25 13:16:53
|
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|
https://public.ostfalia.de/~klawonn/Publikationen/Abstracts.html
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Constructing a Fuzzy Controller from Data Frank Klawonn and Rudolf Kruse Fuzzy control at the executive level can be interpreted as an approximation technique for a control function based on typical, imprecisely specified input-output tuples that are represented by fuzzy sets. The imprecision is characterized by similarity relations that are induced by transformations of the canonical distance function between real numbers. Taking this interpretation of fuzzy controllers into account, in order to derive a fuzzy controller from observed data typical input-output tuples have to be identified. In addition, a concept of similarity based on a transformations of the canonical distance is needed in order to characterize the typical input-output tuples by suitable fuzzy sets. A variety of fuzzy clustering algorithms exists that are exactly working in this spirit: They identify prototypes and assign fuzzy sets to the prototypes on the basis of a suitable transformed distance. In this paper we discuss how such fuzzy clustering techniques can be applied to construct a fuzzy controller from data and introduce special clustering algorithms that are tailored for this problem. The Relation between Inference and Interpolation in the Framework ofFuzzy Systems Frank Klawonn and Vilem Novak This papers aims at clarifying the meaning of different interpretations of the Max-Min or, more generally, the Max-t-norm rule in fuzzy systems. It turns out that basically two distinct approaches play an important role in fuzzy logic and its applications: fuzzy interpolation on the basis of an imprecisely known function and logical inference in the presence of fuzzy information. Fuzzy Control on the Basis of Equality Relations - with an Example from Idle Speed Control F. Klawonn, J. Gebhardt, R. Kruse The way engineers use fuzzy control in real world applications is often not coherent with an understanding of the control rules as logical statements or implications. In most cases fuzzy control can be seen as an interpolation of a partially specified control function in a vague environment, which reflects the indistinguishability of measurements or control values. In this paper we show that equality relations turn out to be the natural way to represent such vague environments and we develop suitable interpolation methods to obtain a control function. \par As a special case of our approach we obtain Mamdani's model and can justify the inference mechanism in this model and the use of triangular membership functions not only for the reason of simplified computations, and we can explain why typical fuzzy partitions are preferred. We also obtain a criterion for reasonable defuzzification strategies. The fuzzy control methodology introduced in this paper has been applied successfully in a case study of engine idle speed control for the Volkswagen Golf GTI. Similarity in Fuzzy Reasoning Frank Klawonn, J. L. Castro Fuzzy set theory is based on a fuzzification' of the predicate $\in$ (element of), the concept of membership degrees is considered as fundamental. In this paper we elucidate the connection between indistinguishability modelled by fuzzy equivalence relations and fuzzy sets. We show that the indistinguishability inherent to fuzzy sets can be computed and that this indistinguishability cannot be overcome in approximate reasoning. For our investigations we generalize from the unit interval as the basis for fuzzy sets, to the framework of GL-monoids that can be understood as a generalization of MV-algebras. Residuation is a basic concept in GL-monoids and many proofs can be formulated in a simple and clear way instead of using special properties of the unit interval. Fuzzy Sets and Vague Environments Frank Klawonn In this paper we propose a natural approach to handle imprecise numbers as they arise for example from measurements. Fuzzy sets turn out to be a canonical representation for such imprecise numbers that are induced by taking different tolerance or error bounds into account. Fuzzy sets are induced by scaling factors that describe the magnitude of the imprecision. On the other, the scaling factors can be derived from given fuzzy sets so that we have a correspondence between scaling factors and fuzzy sets. When these concepts are applied to control problems, the max-min rule is rediscovered as an interpolations technique. Viewing fuzzy control as an interpolation technique in vague environments enables us to validate various concepts for the design and tuning of fuzzy controllers and suggests new also new methods based on clear semantics. A Lukasiewicz Logic Based Prolog Frank Klawonn and Rudolf Kruse Prolog is a programming language based on a restricted subset of classical first order predicate logic. In order to overcome some problems of classical logic to handle imperfect human knowledge, we provide a formal framework for a Lukasiewicz logic based Prolog system. The use of Lukasiewicz logic with its connection to Ulam games enables us to deal with partial inconsistencies by interpreting the truth values as relative distance to contradiction. We also present the software tool LULOG which is based on the theoretical results of this paper and can be seen as a Prolog system for many-valued logic. Applications of LULOG to an Ulam game and an example of reasoning with imperfect knowledge are also discussed. Equality Relations as a Basis for Fuzzy Control F. Klawonn, R. Kruse The aim of this paper is to introduce a fuzzy control model with well-founded semantics in order to explain the concepts applied in fuzzy control. Assuming that the domains of the input- and output variables for the process are endowed with equality relations, that reflect the indistinguishability of values lying closely together, the use of triangular and trapezoidal membership functions can be justified and max-$\sqcap$ inference where $\sqcap$ is a t-norm turns out to be a consequence of our model. Distinguishing between a functional and a relational view of the control rules it is possible to explain when defuzzification strategies like MOM or COA are appropriate or lead to undesired results. Techniques and Applications of Control Systems Based on Knowledge-Based Interpolation F. Klawonn, R. Kruse Fuzzy control was established as an alternative control method when it is difficult to develop a suitable mathematical model of the process, but expert knowledge in the form of vague rule is available. Although the principal idea of a fuzzy set as a model of a vague linguistic expression is very appealing, a naive approach to fuzzy sets can cause tedious problems. Without a concrete semantics for fuzzy sets the design of a fuzzy controller can end up in a trial and error experiment with a large number of parameters and options. In this paper we review an approach to fuzzy control that interprets fuzzy sets as vague values in a vague environment. The vague environment is characterised by a scaling function that describes how sensitive the process reacts when a certain value is slightly changed. We also discuss a regression technique based on the concept of vague environments, enabling to construct a fuzzy controller from data. The Role of Similarity in Fuzzy Reasonin F. Klawonn Fuzzy reasoning mechanisms are designed to cope with vague and uncertain knowledge and information. In this paper we demonstrate that from the vagueness inherent in a fuzzy system a canonical indistinguishability of objects can be derived which cannot be overcome by the standard reasoning schemes. We discuss also the consequences for fuzzy logic in the narrow sense. Fuzzy Shell Cluster Analysis F. Klawonn, R. Kruse and H. Timm In this paper we survey the main approaches to fuzzy shell cluster analysis which is simply a generalization of fuzzy cluster analysis to shell like clusters, i.e. clusters that lie in nonlinear subspaces. Therefore we introduce the main principles of fuzzy cluster analysis first. In the following we present some fuzzy shell clustering algorithms. In many applications it is necessary to determine the number of clusters as well as the classification of the data set. Subsequently therefore we review the main ideas of unsupervised fuzzy shell cluster analysis. Finally we present an application of unsupervised fuzzy shell cluster analysis in computer vision. Prolog Extensions to Many--Valued Logics F. Klawonn The aim of this paper is to show that a restriction of a logical language to clauses like Horn clauses, as they are used in Prolog, applied to [0,1]-valued logics leads to calculi with a sound and complete proof theory. In opposition to other models where generally the set of axioms as well as the deduction schemata are enriched we restrict ourselves to a simple modification of the deduction rules of classical logic without adding new axioms. In our model the truth values from the unit interval can be interpreted in a probabilistic sense, so that a value between 0 and 1 is not just intuitively interpreted as a degree of truth'. Learning the Rule Base of a Fuzzy Controller by a Genetic Algorithm J. Hopf and F. Klawonn For the design of a fuzzy controller it is necessary to choose, besides other parameters, suitable membership functions for the linguistic terms and to determine a rule base. This paper deals with the problem of finding a good rule base - the basis of a fuzzy controller. Consulting experts still is the usual but time-consuming and therefore rather expensive method. Besides, after having designed the controller, one cannot be sure that the rule base will lead to near optimal control. This paper shows how to reduce significantly the period of development (and the costs) of fuzzy controllers with the help of genetic algorithms and, above all, how to engender a rule base which is very close to an optimum solution. The example of the inverted pendulum is used to demonstrate how a genetic algorithm can be designed for an automatic construction of a rule base. So this paper does not deal with the tuning of an existing fuzzy controller but with the genetic (re-)production of rules, even without the need for experts. Thus, a program is engendered, consisting of simple IF...THEN instructions. Context Sensitive Fuzzy Clustering Annete Keller, Frank Klawonn We introduce an objective function-based fuzzy clustering technique that incorporates linear combinations of attributes in the distance function. The main application field of our method is image processing where a a comparison pixel by pixel is usually not adequate, but the environment of a pixel or a groupd of pixels characterize important properties of an image or parts of it. In addition, our approach can be seen as a generalization of other fuzzy clustering techniques like the axes-parallel version of the Gustafson-Kessel algorithm. Fuzzy Clustering Based on Modified Distance Measures Frank Klawonn, Annete Keller The well-known fuzzy c-means algorithm is an objective function based fuzzy clustering technique that extends the classical k-means method to fuzzy partitions. By replacing the Euclidean distance in the objective function other cluster shapes than the simple (hyper-)spheres of the fuzzy c-means algorithm can be detected, for instance ellipsoids, lines or shells of circles and ellipses. We propose a modified distance function that is based on the scalar product and allows to detect a new kind of cluster shape and also lines and (hyper-)planes. Fuzzy Clustering with Weighting of Data Variables Annete Keller, Frank Klawonn We introduce an objective function-based fuzzy clustering technique that assigns one influence parameter to each single data variable for each cluster. Our method is not only suited to detect structures or groups in unevenly over the structure's single domains distributed data, but gives also information about the influence of individual variables on the detected groups. In addition, our approach can be seen as as generalization of the well-known fuzzy c-means clustering algorithm. Mathematical Analysis of Fuzzy Classifiers Frank Klawonn and Erich-Peter Klement We examine the principle capabilities and limits of fuzzy classifiers that are based on a finite set of fuzzy if-then rules like they are used for fuzzy controllers, except that the conclusion of a rule specifies a discrete class instead of a (fuzzy) real output value. Our results show that in the two-dimensional case, for classification problems whose solutions can only be solved approximately by crisp classification rules, very simple fuzzy rules provide an exact solution. However, in the multi-dimensional case, even for linear separable problems, max-min rules are not sufficient. Fuzzy Clustering and Fuzzy Rules Frank Klawonn and Annette Keller Fuzzy clustering offers various possibilities for learning fuzzy if-then rules from data for classification tasks as well as for function approximation problems like in fuzzy control. In this paper we review approaches for deriving rules from data by fuzzy clustering and discuss some of their common problems. As a consequence, we propose a new method which is specifically tailored for the task of learning rules. Fuzzy Clustering with Evolutionary Algorithms Frank Klawonn Abstract Objective function based fuzzy clustering aims at finding a fuzzy partition by optimizing a function evaluating a (fuzzy) assignment of a given data set to clusters, that are characterized by a set of parameters, the so-called prototypes. The iterative optimization technique usually requires the objective function not only to be differentiable, but prefers also an analytical solution for the equations of necessary conditions for local optima. Evolutionary algorithms are known to be an alternative robust optimization technique which are applicable to quite general forms of objective functions. We investigate the possibility of making use of evolutionary algorithms in fuzzy clustering. Our experiments and theoretical investigations show that the application of evolutionary algorithms to shell clustering, where the clusters are in the form of geometric contours, is not very promising due to the shape of the objective function, whereas they can be helpful in finding solid clusters that are not smooth, for example rectangles or cubes. These types of clusters play an important role for fuzzy rule extraction from data. Fuzzy Cluster Analysis for Identification of Gene Regulating Regions Lars Pickert, Frank Klawonn, Edgar Wingender The main approach of this work is the implementation of a cluster analysis program for identification of regulatory regions in genomes. These regions are important parts of the genetic pool in higher developed organisms. They are composed of several basic elements, so called transcription factor sites, which can be identified by special analysis tools more or less vaguely. The program we have developed is able to search for two-dimensional clusters in the results of such analysis tools to give hints on gene regulatory regions. For this purpose two fuzzy clustering algorithms have been implemented: The fuzzy c-means (FCM) and the Gath and Geva fuzzy clustering algorithm (GG) with two conventional cluster validity methods and one which has been developed especially for this application. All results of the cluster analysis program can be visualized and documented automatically. Neuro-Fuzzy Classification Initialized by Fuzzy Clustering D. Nauck, F. Klawonn In this paper we discuss how a neuro-fuzzy classifier can be initialized by rules generated by fuzzy clustering. The neuro-fuzzy classifier NEFCLASS can learn fuzzy classification rules completely from data. The learning algorithm for fuzzy sets can be constrained in order to obtain interpretable classifiers. However, fuzzy clustering provides more sophisticated rule learning procedures. We show that the learning process of NEFCLASS produces better results, if it is initialized by fuzzy clustering. Derivation of Fuzzy Classification Rules from Multidimensional Data F. Klawonn and R. Kruse A This paper describes techniques for deriving fuzzy classification rules based on special modified fuzzy clustering algorithms. The basic idea is that each fuzzy cluster induces a fuzzy classification rule. The fuzzy sets appearing in a rule associated with a fuzzy cluster are obtained by projecting the cluster to the one-dimensional coordinate spaces. In order to allow clusters of varying shape and size we derive special fuzzy clustering algorithms which are searching for clusters in the form of axes-parallel hyper-ellipsoids. Our method can be applied to classification tasks where the classification of the sample data is known as well as when it is not known. Clustering Methods in Fuzzy Control F. Klawonn, R. Kruse Fuzzy controllers can be interpreted as an interpolation technique on the basis of fuzzy clusters of input/output pairs. It is therefore obvious that fuzzy clustering algorithms are a promising tool for supporting the design of a fuzzy controller when data of the process to be controlled are available. This paper discusses the possibilities and limitations of fuzzy clustering for fuzzy control. Similarity Based Reasoning Frank Klawonn This paper is devoted to the duality between fuzzy sets and equality relations. It comprises various results that allow to interchange from one framework to the other. Finally it is shown that in fuzzy reasoning the inherent similarity characterized by equality relations cannot be avoided. Modifications of Genetic Algorithms for Designing and Optimizing FuzzyControllers J. Kinzel, F. Klawonn, R. Kruse This paper investigates the possibilities for applications of genetic algorithms to tuning and optimizing fuzzy controllers, or even to generate fuzzy controllers automatically. There are various ad-hoc approaches to use genetic algorithms for the design of fuzzy controllers, which already indicated good results. However, there is a need for systematic techniques that take the properties of fuzzy controllers and genetic algorithm into account in order to obtain fast convergence and to be able to tackle more complex control problems. Fuzzy Clustering with Evolutionary Algorithms Frank Klawonn and Annette Keller Objective function based fuzzy clustering aims at finding a fuzzy partition by optimizing a function evaluating a (fuzzy) assignment of a given data set to clusters, that are characterized by a set of parameters, the so-called prototypes. The iterative optimization technique usually requires the objective function not only to be differentiable, but prefers also an analytical solution for the equations of necessary conditions for local optima. Evolutionary algorithms are known to be an alternative robust optimization technique which are applicable to quite general forms of objective functions. We investigate the possibility of making use of evolutionary algorithms in fuzzy clustering. Our experiments and theoretical investigations show that the application of evolutionary algorithms to shell clustering, where the clusters are in the form of geometric contours, is not very promising due to the shape of the objective function, whereas they can be helpful in finding solid clusters that are not smooth, for example rectangles or cubes. These types of clusters play an important role for fuzzy rule extraction from data. Fuzzy Max-Min Classifiers Decide locally on the Basis of Two Attributes Birka von Schmidt, Frank Klawonn Fuzzy classification systems differ from fuzzy controllers in the form of their outputs. For classification problems a decision between a finite number of discrete classes has to be made, whereas in fuzzy control the output domain is usually continuous, i.e. a real interval. In this paper we consider fuzzy classification systems using the max-min inference scheme and classifying an unknown datum on the basis of maximum matching, i.e. assigning it to the class appearing in the consequent of the rule whose premise fits best. We basically show that this inference scheme locally takes only two attributes (variables) into account for the classification decision. Letzte Änderung am 07.09.2006
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2019-04-21 08:45:28
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https://solvedlib.com/n/x27-68-u-3d-ptiu-ov-uv-quot-s1oj-pa-8-t-io-ojl-s1s-e-jqj,771577
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# '68 U, = 3d ptiu OV = ,(UV) ."S1OJ Pa) 8*T IO.OJL s1s [[e JQJ (S1S JOJ SMUT JAUY n
##### Solve the given differential equation by undetermined coefficients (superposition approach)y" + 2y' 3y(22 +x+1)+e ~31[Hint: Be careful with the form of the particular solution: Make sure that there is no duplication of terms:]
Solve the given differential equation by undetermined coefficients (superposition approach) y" + 2y' 3y (22 +x+1)+e ~31 [Hint: Be careful with the form of the particular solution: Make sure that there is no duplication of terms:]...
##### Consider (o groups with 0,,0z observalions in cach group, and let N=n+n, We assume that YFh tctor I=L,2 _.nand that Yehtc tor =n tN wnere c~n lo,0'|Show that the least squares cstimator for V ,V are
Consider (o groups with 0,,0z observalions in cach group, and let N=n+n, We assume that YFh tctor I=L,2 _.nand that Yehtc tor =n tN wnere c~n lo,0'| Show that the least squares cstimator for V ,V are...
##### If the discrete growth model P = Poat is converted to the continuous growth model P = Poekt , then the value Of kIf the continuous growth model P = Poekt , is converted to the discrete growth model P = Poat , then the value of a
If the discrete growth model P = Poat is converted to the continuous growth model P = Poekt , then the value Of k If the continuous growth model P = Poekt , is converted to the discrete growth model P = Poat , then the value of a...
##### Write 21 and 22 in polar form_ 40, 22 V5 + /40( cos( t) + isin(t) )COS 6 isinFind the product 21/2 and the quotients and form:)(Express your answers in polar2172"20 cos % + i sil 3cos I + isin 7 )
Write 21 and 22 in polar form_ 40, 22 V5 + / 40( cos( t) + isin(t) ) COS 6 isin Find the product 21/2 and the quotients and form:) (Express your answers in polar 2172 "20 cos % + i sil 3 cos I + isin 7 )...
##### Solve inequality. Write the solution set in interval notation, and graph it.$10<7 p+3<24$
Solve inequality. Write the solution set in interval notation, and graph it. $10<7 p+3<24$...
##### An electron moving parallel to the x axis has an ini- AMT tial speed of 3.70...
An electron moving parallel to the x axis has an ini- AMT tial speed of 3.70 3 106 m/s at the origin. Its speed is M reduced to 1.40 3 105 m/s at the point x 5 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. (b) Which point is at the higher potential?...
##### 2. A population of the same type of insect in a differen: pond has Leslie matrix 0.000 3.000 23.6...
please do probkem correctly with work shown and not just anwsers:) 2. A population of the same type of insect in a differen: pond has Leslie matrix 0.000 3.000 23.600 0.079 0 0 0.437 0 (a) Draw the loop diagram (b) You have access to a computer program that tells you that λ = 1.02 is an eigen...
##### Two long straight parallel wires, separated by d1 = 0.775 cm, are perpendicular to the plane...
Two long straight parallel wires, separated by d1 = 0.775 cm, are perpendicular to the plane of the page as shown in Fig. 30-40. Wire 1 carries a current of 5.70 A into the page. What must be the current (magnitude and direction) in wire 2 for the resultant magnetic field at point P to be zero? Poin...
##### Ona feature that amphibians and humans have common0 Ihe numbar 0i hear chamberscomploie scnaration oxygon DdOrAno oxygen-rich blbodIhe number cicuits joncircuintionuntolhartMlav bbod prutsuru thrtxughout Ihu uyetomnic alrauy
Ona feature that amphibians and humans have common 0 Ihe numbar 0i hear chambers comploie scnaration oxygon DdOrAno oxygen-rich blbod Ihe number cicuits joncircuintion untolhart Mlav bbod prutsuru thrtxughout Ihu uyetomnic alrauy...
##### Whole question plz. I will give good feedback :) 3. The circuit below begins with the...
whole question plz. I will give good feedback :) 3. The circuit below begins with the switch in the open position. The battery has VB = 8V, the resistor has R=2k2, and the inductor has L = 3mH. At time=0, the switch in the circuit is closed. 1000 VB W (a) Determine the time constant of this circu...
##### 580 nm light shines on a double slit with d = 0.000125 m What is the angle of the third dark interference minimum (m = 3)? (Remember; nano means 10-9.) (Unit = deg)nte
580 nm light shines on a double slit with d = 0.000125 m What is the angle of the third dark interference minimum (m = 3)? (Remember; nano means 10-9.) (Unit = deg) nte...
##### (10 points) Proof that we have the following QQ-plot coordinates for the Pareto and Weibull distribution:Distribution ParetoF 1 _ % a I > l;a > 0 1 _ exp(_ Az x > 0,A,T > 0Coordinates log(1 Pi,n) , log = Ti,n ,Weibull(log( _ logk (1 Pi.n) ) , log = Ti,nUse a log transformation to proof this_
(10 points) Proof that we have the following QQ-plot coordinates for the Pareto and Weibull distribution: Distribution Pareto F 1 _ % a I > l;a > 0 1 _ exp(_ Az x > 0,A,T > 0 Coordinates log(1 Pi,n) , log = Ti,n , Weibull (log( _ logk (1 Pi.n) ) , log = Ti,n Use a log transformation to p...
##### - 8 cm 8 cm 0.5 cm N=500 Consider the magnetic core with an air gap...
- 8 cm 8 cm 0.5 cm N=500 Consider the magnetic core with an air gap as shown in Figure Q3. The core material has a relative permeability of 6000 and a rectangular cross section 2cm x 3cm. The coil has 500 turns. Given that permeability of free space is Ho = 41 x 10-7. a) Lukiskan litar magnet yang b...
##### QUESTION 29points Save AnswerWhat mass of HCI is contained in 27.1 mL of an queous HCI solution that has density of 1.19 glmL and is 37.219 HCI by mass? massQUESTION 30pointsSave Answer260.4 mL sample of water contains 0.0366 g of NaCl Calculate the concentration in ppm; concentration ppm
QUESTION 29 points Save Answer What mass of HCI is contained in 27.1 mL of an queous HCI solution that has density of 1.19 glmL and is 37.219 HCI by mass? mass QUESTION 30 points Save Answer 260.4 mL sample of water contains 0.0366 g of NaCl Calculate the concentration in ppm; concentration ppm...
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2022-05-20 11:16:13
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https://staging4.aicorespot.io/the-chain-rule-of-calculus-additional-functions/
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### The Chain Rule of Calculus – Additional Functions
The chain rule is a critical derivative rule that enables us to operate with composite functions. It is critical is comprehending the workings of the backpropagation algorithm, which is applicable to the chain rule extensively in order to calculate the error gradient of the loss function with regard to every weight of a neural network. We will be developing on our prior intro to the chain rule, by handling more challenging functions.
In this guide, you will find out how to go about applying the chain rule of calculus to challenging functions.
• The procedure of application of the chain rule to univariate functions can be extended to multivariate ones.
• The applying of the chain rule adheres to a similar procedure, regardless of how complicated the function is: take the derivative of the outer function to start with, and then shift inwards. Along the way, the applying of other derivative rules might be needed.
• Application of the chain rule to multivariate functions needs the leveraging of partial derivatives.
Tutorial Summarization
The tutorial is subdivided into two portions, which are:
• The chain rule on univariate functions
• The chain rule on multivariate functions
Prerequisites
For this guide, it is assumed that you are already acquainted with:
• Multivariate functions
• The power and product rules
• Partial derivatives
• The chain rule
The chain rule on univariate functions
We have already found out about the chain rule for univariate and multivariate functions, however we have only observed a few simple instances thus far. Let’s observe a bit more challenging ones here. We’ll be beginning with univariate functions to start with, and then go about applying what we learn to multivariate functions.
Instance 1: Let’s increase the bar by taking up the following composite function:
We can go about separating the composite function into the inner function, f(x) = x squared – 10, and the outer function, g(x) = √x = (x)1/2.
The output of the inner function is signified by the intermediate variable, u, and its value will be fed into the input of the outer function.
The first step is to identify the derivative of the outer portion of the composite function, while ignoring whatever is inside. For this reason, we can go about applying the power rule:
dh / du = (1/2) (x2 – 10)-1/2
The next step is to identify the derivative of the inner portion of the composite function, this time ignoring whichever is outside. We can go about applying the power rule here too.
du/dx = 2x
Bring the two portions together and through the process of simplification, we have:
Instance 2: Let’s repeat the process, this time with a differing composite function.
We will again leverage, u, the output of the inner function, as our intermediate variable.
The outer function in this scenario is, cos x. Identifying its derivative, once again ignoring the inside, provides us with:
dh / du = (cos(x cubed – 1))’ = -sin(x cubed minus one)
The inner function is x cubed – one. Therefore, its derivative becomes:
du / dx = (x cubed minus 1)’ = 3x squared
Bringing the two portions together, we get the derivative of the composite function.
Instance 3: Let’s now increase the bar a little bit more by taking up a more challenging composite function.
If we look at this closely, we realize that not only do we possess nested functions for which we will require to apply the chain rule several times, but we additionally have a product to which we will require to apply the product rule.
We identify that the outermost function is a cosine. In identifying its derivative through the chain rule, we shall be leveraging the intermediate variable, u:
dh / du = (cos(x √(x2 – 10) ))’ = -sin(√(x2 – 10) )
Inside the cosine, we have the product, x √(x2 – 10), to which we will be performing the application of the product rule to identify its derivative (observe that we are always shifting from the outside to the inside, in order to find out the operation that requires to be handled next.
du / dx = (√(x2 – 10) )’ = √(x2 – 10) + x ( √(x2 – 10) )’
One of the aspects in the outcome term is, ( √(x2 – 10) )’ to which we will performing application of the chain rule again. Indeed, we have already performed so above, and therefore, we can simply re-utilize the outcome.
( √(x2 – 10) )’ = x (x2 – 10)-1/2
Bringing all the portions together, we get the derivative of the composite function.
This can be streamlined further into:
The Chain Rule on Multivariate Functions
Instance 4: Let’s assume that we are now presented by a multivariate function of dual independent variables, s and t, with every one of these variables being dependent on another dual independent variables, x and y:
h = g(st) = s2 + t3
Where the functions s = xy, and t = 2x – y
Implementation of the chain rule here needs the computing of partial derivatives, as we are working with several independent variables. Further, s and t will additionally function as our intermediate variables. The formulae that we will be operating with, defined with regard to every input, are the following:
From these formulae, we can observe that we will require to identify six differing partial derivatives.
We can now go on to replace these terms in the formulae for ∂h / ∂and h / ∂y:
And subsequently replace for s and t to identify the derivatives:
Instance 5: Let’s do this again, this time with a multivariate function of a trio of independent variables, r, s, and t, with every one of these variables being dependent on another two independent variables, x and y:
h = g(r, st) = r2 – rs + t3
Where the functions, r = x cos y, s = x ey, and t = x + y.
This time, r, s, and t will function as our intermediate variables. The formulate that we will be operating with, defined with regard to every input, are the following:
From these formulae, we can observe that we will now require to identify nine partial differing partial derivatives.
Again, we move forward to replace these terms in the formulae for ∂h / ∂and h / ∂y
And subsequently replace for r, s and t to identify the derivatives:
Which might be streamlined a bit further (hint: apply the trigonometric identity 2sin cos y = sin 2y to ∂h / ∂y):
Regardless of how complicated the expression is, the process to follow stays similar:
Your last computation informs you the first thing to perform.
Therefore, begin by handling the outer function to start with, then shift inwards to the next one. You might require to go about applying other rules along the way, as we have observed for instance 3. Do not forget to take the partial derivatives if you are operating with multivariate functions.
This section furnishes additional resources on the subject if you seeking to delve deeper:
Books
Calculus for Dummies, 2016
Single and Multivariable Calculus, 2020
Mathematics for Machine Learning, 2020
Conclusion
In this guide, you found out about the chain rule of calculus to challenging functions.
Particularly, you learned:
• The procedure of application of the chain rule to univariate functions can be extended to multivariate ones.
• The application of the chain rule follows a similar procedure, regardless of how complicated the function is: obtain the derivative of the outer function to start with, and then shift inwards. Along the way, the application of other derivative rules might be needed.
• Application of the chain rule to multivariate functions needs the leveraging of partial derivatives.
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2023-02-01 19:16:52
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https://eng.libretexts.org/Bookshelves/Civil_Engineering/Book%3A_Slurry_Transport_(Miedema)/02%3A_Dimensionless_Numbers_and_Other_Parameters/2.02%3A_Dimensionless_Numbers
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# 2.2: Dimensionless Numbers
## 2.2.1 The Reynolds Number Re
In fluid mechanics, the Reynolds number (Re) is a dimensionless number that gives a measure of the ratio of inertial (resistant to change or motion) forces to viscous (heavy and gluey) forces and consequently quantifies the relative importance of these two types of forces for given flow conditions. (The term inertial forces, which characterize how much a particular liquid resists any change in motion, are not to be confused with inertial forces defined in the classical way.)
The concept was introduced by George Gabriel Stokes in 1851 but the Reynolds number is named after Osborne Reynolds (1842–1912), who popularized its use in 1883. Reynolds numbers frequently arise when performing dimensional analysis of liquid dynamics problems, and as such can be used to determine dynamic similitude between different experimental cases.
They are also used to characterize different flow regimes, such as laminar or turbulent flow: laminar flow occurs at low Reynolds numbers, where viscous forces are dominant, and is characterized by smooth, constant liquid motion; turbulent flow occurs at high Reynolds numbers and is dominated by inertial forces, which tend to produce chaotic eddies, vortices and other flow instabilities.
The gradient of the velocity dv/dx is proportional to the velocity v divided by a characteristic length scale L. Similarly, the second derivative of the velocity d2v/dx2 is proportional to the velocity v divided by the square of the characteristic length scale L.
$\ \operatorname{Re}=\frac{\text { Inertial forces }}{\text { Viscous forces }}=\frac{\rho_{1} \cdot \mathrm{v} \cdot \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}}{\rho_{\mathrm{l}} \cdot v_{\mathrm{l}} \cdot \frac{\mathrm{d}^{2} \mathrm{v}}{\mathrm{d} \mathrm{x}^{2}}} \quad \text{with: } \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \propto \frac{\mathrm{v}}{\mathrm{L}} \quad \frac{\mathrm{d}^{2} \mathrm{v}}{\mathrm{d} \mathrm{x}^{2}} \propto \frac{\mathrm{v}}{\mathrm{L}^{2}} \quad \Rightarrow \frac{\mathrm{v} \cdot \mathrm{L}}{v_{\mathrm{l}}}$
The Reynolds number is a dimensionless number. High values of the parameter (on the order of 10 million) indicate that viscous forces are small and the flow is essentially inviscid. The Euler equations can then be used to model the flow. Low values of the parameter (on the order of 1 hundred) indicate that viscous forces must be considered.
## 2.2.2 The Froude Number Fr
The Froude number (Fr) is a dimensionless number defined as the ratio of a characteristic velocity to a gravitational wave velocity. It may equivalently be defined as the ratio of a body's inertia to gravitational forces. In fluid mechanics, the Froude number is used to determine the resistance of a partially submerged object moving through water, and permits the comparison of objects of different sizes. Named after William Froude (1810-1879), the Froude number is based on the speedlength ratio as defined by him.
$\ \mathrm{F} \mathrm{r}=\frac{\text { Characteristic velocity }}{\text { Gravitational wave velocity }}=\frac{\mathrm{v}}{\sqrt{\mathrm{g} \cdot \mathrm{L}}}$
Or the ratio between the inertial force and the gravitational force squared according to:
$\ \widehat{\mathrm{F}} \mathrm{r}=\frac{\text { Inertial force }}{\text { Gravitational force }}=\frac{\rho_{\mathrm{l}} \cdot \mathrm{v} \cdot \frac{\mathrm{d} \mathrm{v}}{\mathrm{d x}}}{\rho_{\mathrm{l}} \cdot \mathrm{g}}=\frac{\mathrm{v}^{\mathrm{2}}}{\mathrm{g} \cdot \mathrm{L}}$
The gradient of the velocity $$dv/dx$$ is proportional to the velocity $$v$$ divided by a length scale $$L$$.
Or the ratio between the centripetal force on an object and the gravitational force, giving the square of the right hand term of equation (2.2-2):
$\ \widehat{\mathrm{F} \mathrm{r}}=\frac{\text { Centripetal force }}{\text { Gravitational force }}=\frac{\mathrm{m} \cdot \mathrm{v}^{2} / \mathrm{L}}{\mathrm{m} \cdot \mathrm{g}}=\frac{\mathrm{v}^{\mathrm{2}}}{\mathrm{g} \cdot \mathrm{L}}$
## 2.2.3 The Richardson Number Ri
The Richardson number Ri is named after Lewis Fry Richardson (1881-1953). It is the dimensionless number that expresses the ratio of the buoyancy term to the flow gradient term.
$\ \mathrm {R} \mathrm {i}=\frac{\text { buoyancy term }}{\text { flow gradient term }}=\left(\frac{\mathrm {g} \cdot \mathrm {L} \cdot \mathrm {R}_{\mathrm {s} \mathrm {d}}}{\mathrm {v}^{\mathrm {2}}}\right)$
The Richardson number, or one of several variants, is of practical importance in weather forecasting and in investigating density and turbidity currents in oceans, lakes and reservoirs.
## 2.2.4 The Archimedes Number Ar
The Archimedes number (Ar) (not to be confused with Archimedes constant, π), named after the ancient Greek scientist Archimedes is used to determine the motion of liquids due to density differences. It is a dimensionless number defined as the ratio of gravitational forces to viscous forces. When analyzing potentially mixed convection of a liquid, the Archimedes number parameterizes the relative strength of free and forced convection. When Ar >> 1 natural convection dominates, i.e. less dense bodies rise and denser bodies sink, and when Ar << 1 forced convection dominates.
$\ \mathrm {A} \mathrm {r}=\frac{\text { Gravitational forces }}{\text { Viscous forces }}=\frac{\mathrm {g} \cdot \mathrm {L}^{3} \cdot \mathrm {R}_{\mathrm {s} \mathrm {d}}}{v_{\mathrm {l}}^{\mathrm {2}}}$
The Archimedes number is related to both the Richardson number and the Reynolds number via:
$\ \mathrm {A} \mathrm {r}=\mathrm {R} \mathrm {i} \cdot \mathrm {R} \mathrm {e}^{2}=\left(\frac{\mathrm {g} \cdot \mathrm {L} \cdot \mathrm {R}_{\mathrm {s} \mathrm {d}}}{\mathrm {v}^{\mathrm {2}}}\right) \cdot\left(\frac{\mathrm {v} \cdot \mathrm {L}}{v_{\mathrm {l}}}\right)^{\mathrm {2}}=\frac{\mathrm {g} \cdot \mathrm {L}^{\mathrm {3}} \cdot \mathrm {R}_{\mathrm {s} \mathrm {d}}}{v_{\mathrm {l}}^{\mathrm {2}}}$
## 2.2.5 The Thủy Number Th or Collision Intensity Number
The new Thy number (Th) is the cube root of the ratio of the viscous forces times the gravitational forces to the inertial forces squared. Thủy is Vietnamese for aquatic, water. The gradient of the velocity v is proportional to the velocity v divided by a length scale L. Since slurry transport is complex and inertial forces, viscous forces and gravitational forces play a role, this dimensionless number takes all of these forces into account in one dimensionless number.
$\ \widehat{\mathrm{Th}}=\frac{\text { Viscous forces }}{\text { Inertial forces }} \cdot \frac{\text { Gravitational forces }}{\text { Inertial forces }}=\frac{1}{\operatorname{Re} \cdot \widehat{\mathrm{Fr}}}=\frac{\rho_{1} \cdot v_{1} \cdot \frac{\mathrm{d}^{2} \mathrm{v}}{\mathrm{dx}^{2}}}{\rho_{1} \cdot \mathrm{v} \cdot \frac{\mathrm{d} v}{\mathrm{d} \mathrm{x}}} \cdot \frac{\rho_{1} \cdot \mathrm{g}}{\rho_{1} \cdot \mathrm{v} \cdot \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}}=\frac{v_{1} \cdot \mathrm{g}}{\mathrm{v}^{3}}$
So:
$\ \mathrm{T} \mathrm{h}=\left(\frac{v_{\mathrm{l}} \cdot \mathrm{g}}{\mathrm{v}^{\mathrm{3}}}\right)^{\mathrm{1} / 3}$
It is interesting that the length scale does not play a role anymore in this dimensionless number. The different terms compensate for the length scale. The value of this dimensionless parameter is, that the relative excess head losses are proportional with the Thủy number to a certain power. Also the Limit Deposit Velocity in heterogeneous transport has proportionality with this dimensionless number.
## 2.2.6 The Cát Number Ct or Collision Impact Number
A special particle Froude number will be introduced here. The Durand & Condolios (1952) particle Froude number Cát, Ct, which is Vietnamese for sand grains. This dimensionless number describes the contribution of the solids to the excess head losses.
$\ \mathrm{C t}=\left(\frac{\mathrm{v}_{\mathrm{t}}}{\sqrt{\mathrm{g} \cdot \mathrm{d}}}\right)^{\mathrm{5} / 3}=\left(\frac{\mathrm{1}}{\sqrt{\mathrm{C}_{\mathrm{x}}}}\right)^{\mathrm{5} / 3}$
The introduction of this particle Froude number is very convenient in many equations.
## 2.2.7 The Lắng Number La or Sedimentation Capability Number
Another new dimensionless number is introduced here. It is the Lng number La. Lng is Vietnamese for sediment and this number represents the capability of the slurry flow to form a bed, either fixed or sliding.
$\ \mathrm{L a}=\frac{\mathrm{v}_{\mathrm{t}} \cdot\left(\mathrm{1}-\mathrm{C}_{\mathrm{v s}} / \mathrm{\kappa}_{\mathrm{C}}\right)^{\mathrm{\beta}}}{\mathrm{v}_{\mathrm{l s}}}$
## 2.2.8 The Shields Parameter θ
The Shields parameter, named after Albert Frank Shields (1908-1974), also called the Shields criterion or Shields number, is a non-dimensional number used to calculate the initiation of motion of sediment in a fluid flow. It is a non dimensionalisation of a shear stress. By multiplying both the nominator and denominator by d2, one can see that it is proportional to the ratio of fluid force on the particle to the submerged weight of the particle.
$\ \theta=\frac{\mathrm{F}_{\text {shear }}}{\mathrm{F}_{\text {gravity }}} \propto \frac{\rho_{\mathrm{l}} \cdot \mathrm{u}_{*}^{2} \cdot \mathrm{d}^{2}}{\rho_{\mathrm{l}} \cdot \mathrm{R}_{\mathrm{s d}} \cdot \mathrm{g} \cdot \mathrm{d}^{3}}=\frac{\mathrm{u}_{*}^{2}}{\mathrm{R}_{\mathrm{s d}} \cdot \mathrm{g} \cdot \mathrm{d}}$
The Shields parameter gives an indication of the erodibility of a sediment. If the Shields parameter is below some critical value there will not be erosion, if it’s above this critical value there will be erosion. The higher the Shields parameter, the bigger the erosion. The critical Shields parameter depends on the particle diameter, the kinematic viscosity and some other parameters. The boundary Reynolds number as used in Shields graphs.
$\ \mathrm{R} \mathrm{e}_{*}=\frac{\mathrm{u}_{*} \cdot \mathrm{d}}{v_{\mathrm{l}}}$
The roughness Reynolds number.
$\ \mathrm{k}_{\mathrm{s}}^{+}=\frac{\mathrm{u}_{*} \cdot \mathrm{k}_{\mathrm{s}}}{v_{\mathrm{l}}}$
The distance to the wall Reynolds number:
$\ \mathrm{y}^{+}=\frac{\mathrm{u}_{*} \cdot \mathrm{y}}{v_{\mathrm{l}}}$
## 2.2.9 The Bonneville Parameter D*
The original Shields graph is not convenient to use, because both axes contain the shear velocity u* and this is usually an unknown, this makes the graph an implicit graph. To make the graph explicit, the graph has to be transformed to another axis system. In literature often the dimensionless grain diameter D* is used, also called the Bonneville (1963) parameter:
$\ \mathrm{D}_{*}=\mathrm{d} \cdot \sqrt[3]{\frac{\mathrm{R}_{\mathrm{sd}} \cdot \mathrm{g}}{v_{\mathrm{l}}^{2}}}$
The relation between the Shields parameter and the Bonneville parameter is:
$\ \mathrm{R} \mathrm{e}_{*}=\sqrt{\boldsymbol{\theta}} \cdot \mathrm{D}_{*}^{\mathrm{1 . 5}}$
So the Bonneville parameter is a function of the Shields number and the boundary Reynolds number according to:
$\ \mathrm{D}_{*}=\left(\frac{\mathrm{R} \mathrm{e}_{*}}{\sqrt{\mathrm{\theta}}}\right)^{2 / 3}$
## 2.2.10 The Rouse Number P
The Rouse number, named after Hunter Rouse (1906-1996), is a non-dimensional number used to define a concentration profile of suspended sediment in sediment transport.
$\ \mathrm{P}=\frac{\mathrm{v}_{\mathrm{t}}}{\mathrm{\kappa} \cdot \mathrm{u}_{*}} \quad or \quad \mathrm{P}=\frac{\mathrm{v}_{\mathrm{t}}}{\boldsymbol{\beta} \cdot \mathrm{\kappa} \cdot \mathrm{u}_{*}}$
The factor β is sometimes included to correlate eddy viscosity to eddy diffusivity and is generally taken to be equal to 1 and is therefore usually ignored. The von Karman constant κ is about 0.4.
The value of the Rouse number is an indication of the type of sediment transport and the bed form.
P≥7.5 Little movement. 7.5≥P≥2.5 Bed load (grains rolling and hopping along the bed, bed forms like dunes) to suspension in the lower part. 2.5≥P≥0.8 Incipient suspension (grains spending less and less time in contact with the bed, bed forms increase in wavelength and decrease in amplitude). For P=2.5 there is suspension in the lower part of the channel or pipe, For P=0.8 the suspension reaches the surface. 0.8≥P Suspension (grains spend very little time in contact with the bed, a plane bed). For P=0.1 the suspension is well developed, for P=0.01 the suspension is homogeneous.
## 2.2.11 The Stokes Number Stk
The Stokes number Stk, named after George Gabriel Stokes (1819-1903), is a dimensionless number corresponding to the behavior of particles suspended in a fluid flow. The Stokes number is defined as the ratio of the characteristic time of a particle to a characteristic time of the flow or of an obstacle:
$\ \mathrm{S} \mathrm{t} \mathrm{k}=\frac{\mathrm{t}_{\mathrm{0}} \cdot \mathrm{u}_{\mathrm{0}}}{\mathrm{I}_{\mathrm{0}}}$
Where t0 is the relaxation time of the particle (the time constant in the exponential decay of the particle settling velocity due to drag), u0 is the velocity of the fluid (liquid) of the flow well away from the particle and l0 is a characteristic dimension of the flow (typically the pipe diameter). In the case of Stokes flow, which is when the particle Reynolds number is low enough for the drag coefficient to be inversely proportional to the Reynolds number itself, the relaxation time can be defined as:
$\ \mathrm{t}_{\mathrm{0}}=\frac{\rho_{\mathrm{s}} \cdot \mathrm{d}^{\mathrm{2}}}{\mathrm{1 8} \cdot \rho_{\mathrm{l}} \cdot v_{\mathrm{l}}}$
In experimental fluid dynamics, the Stokes number is a measure of flow fidelity in particle image velocimetry (PIV) experiments, where very small particles are entrained in turbulent flows and optically observed to determine the speed and direction of fluid movement. For acceptable tracing accuracy, the particle response time should be faster than the smallest time scale of the flow. Smaller Stokes numbers represent better tracing accuracy. For Stk »1, particles will detach from a flow especially where the flow decelerates abruptly. For Stk«0.1, tracing accuracy errors are below 1%. The Stokes number also gives a good indication for small particles being capable of forming a homogeneous mixture with the liquid flow. Assuming, in the case of pipe flow, the line speed as the characteristic velocity u0 and half the pipe diameter as the characteristic dimension l0, this gives:
$\ \mathrm{S} \mathrm{t} \mathrm{k}=\frac{\rho_{\mathrm{s}} \cdot \mathrm{d}^{2}}{\mathrm{1 8} \cdot \rho_{\mathrm{l}} \cdot v_{\mathrm{l}}} \cdot \frac{\mathrm{2} \cdot \mathrm{v}_{\mathrm{l s}}}{\mathrm{D}_{\mathrm{p}}} \quad \text{or} \quad \mathrm{d}=\sqrt{\frac{\mathrm{S t k} \cdot \mathrm{9} \cdot \rho_{\mathrm{l}} \cdot v_{\mathrm{l}} \cdot \mathrm{D}_{\mathrm{p}}}{\rho_{\mathrm{s}} \cdot \mathrm{v}_{\mathrm{l s}}}}$
## 2.2.12 The Bagnold Number Ba
The Bagnold number (Ba) is the ratio of grain collision stresses to viscous fluid stresses in a granular flow with interstitial Newtonian fluid, first identified by Ralph Alger Bagnold. The Bagnold number is defined by:
\ \begin{aligned}\mathrm{B a}=\frac{\rho_{\mathrm{s}} \cdot \mathrm{d}^{2} \cdot \lambda^{1 / 2} \cdot \dot{\gamma}}{\mu_{\mathrm{l}}}=\frac{\rho_{\mathrm{s}} \cdot \mathrm{d}^{2} \cdot \lambda^{1 / 2} \cdot \dot{\gamma}}{\rho_{\mathrm{l}} \cdot v_{\mathrm{l}}} \quad \text {with : }\dot{\gamma}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{r}} \\ \text{With :} \lambda=\frac{1}{\left(\left(\frac{\mathrm{C}_{\mathrm{v b}}}{\mathrm{C}_{\mathrm{v s}}}\right)^{1 / 3}-\mathrm{1}\right)}=\frac{\mathrm{1}}{\left(\left(\frac{\mathrm{1}}{\mathrm{C}_{\mathrm{v r}}}\right)^{1 / 3}-\mathrm{1}\right)}=\frac{\mathrm{C}_{\mathrm{v r}}^{\mathrm{1 / 3}}}{\mathrm{1}-\mathrm{C}_{\mathrm{v r}}^{\mathrm{1 / 3}}} \\ \text{Boundary layer: } \dot{\gamma}=\frac{\mathrm{u}_{*}^{2}}{v_{1}} \Rightarrow \mathrm{B a}=\frac{\rho_{\mathrm{s}} \cdot \mathrm{d}^{2} \cdot \lambda^{1 / 2} \cdot \mathrm{u}_{*}^{2}}{\rho_{\mathrm{l}} \cdot v_{\mathrm{l}}^{2}}\end{aligned}\tag{2.2.23}
Where Cvs is the solids fraction and Cvb is the maximum possible concentration, the bed concentration. In flows with small Bagnold numbers (Ba < 40), viscous fluid stresses dominate grain collision stresses, and the flow is said to be in the 'macro-viscous' regime. Grain collision stresses dominate at large Bagnold number (Ba > 450), which is known as the 'grain-inertia' regime. A transitional regime falls between these two values.
2.2: Dimensionless Numbers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Sape A. Miedema via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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2022-05-22 19:36:08
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https://www.cemmap.ac.uk/publication/vector-quantile-regression-an-optimal-transport-approach-2/
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# Vector quantile regression: an optimal transport approach
1 June 2016
### Type
Journal Article
We propose a notion of conditional vector quantile function and a vector quantile regression. A conditional vector quantile function (CVQF) of a random vector YY, taking values in RdRd given covariates Z=zZ=z, taking values in RkRk, is a map uQY|Z(u,z)u⟼QY|Z(u,z), which is monotone, in the sense of being a gradient of a convex function, and such that given that vector UU follows a reference non-atomic distribution FUFU, for instance uniform distribution on a unit cube in RdRd, the random vector QY|Z(U,z)QY|Z(U,z) has the distribution of YY conditional on Z=zZ=z. Moreover, we have a strong representation, Y=QY|Z(U,Z)Y=QY|Z(U,Z) almost surely, for some version of UU. The vector quantile regression (VQR) is a linear model for CVQF of YY given ZZ. Under correct specification, the notion produces strong representation, Y=β(U)f(Z)Y=β(U)⊤f(Z), for f(Z)f(Z) denoting a known set of transformations of ZZ, where uβ(u)f(Z)u⟼β(u)⊤f(Z) is a monotone map, the gradient of a convex function and the quantile regression coefficients uβ(u)u⟼β(u) have the interpretations analogous to that of the standard scalar quantile regression. As f(Z)f(Z) becomes a richer class of transformations of ZZ, the model becomes nonparametric, as in series modelling. A key property of VQR is the embedding of the classical Monge–Kantorovich’s optimal transportation problem at its core as a special case. In the classical case, where YY is scalar, VQR reduces to a version of the classical QR, and CVQF reduces to the scalar conditional quantile function. An application to multiple Engel curve estimation is considered.
### Previous version
Vector quantile regression: an optimal transport approach
Guillaume Carlier, Victor Chernozhukov, Alfred Galichon
CWP58/15
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2023-04-01 21:11:19
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http://math.stackexchange.com/questions/119722/hypergeometric-proof/119759
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# hypergeometric proof
I've been looking at hypergeometric probability problems and came across this equation. Can someone explain to me why $\sum_{j=0}^{n-1}\frac{\binom{M-1}{j}\binom{(N-1)-(M-1)}{(n-1)-j}}{\binom{N-1}{n-1}}=1$? Take into consideration that I am using $N, M, n$ as parameters where there are $N$ total objects with $M$ "special" or unique objects, and $n$ is a selection sample.
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johhny and @Graphth, would you care to take a look at this answer/question? – The Chaz 2.0 Mar 13 '12 at 17:22
You can prove it combinatorially. First multiply both sides by $\binom{N-1}{n-1}$. You can do this because that does not depend on $j$.
For simplicity, let's say there are $M-1$ special objects and $N-1$ total objects.
The right hand side counts the number of ways to choose $n - 1$ things from $N - 1$ things.
$$\binom{N-1}{n-1}$$
Now, the left hand side counts the same thing
$$\sum_{j=0}^{n-1} \binom{M-1}{j}\binom{(N-1)-(M-1)}{(n-1)-j}$$
But, the left hand side splits it up into cases. When you select $n-1$ things, you will get $j$ of the $M-1$ special objects, as well as $(n-1) - j$ from the rest of the set, which has $(N-1) - (M-1)$ objects in it. Now, $j$ ranges from 0 to $n-1$. So, add up all these counts to get the left hand side. That is, you could get 0 objects from the M, or you could get 1 object from the M, ..., or you could get n-1 objects from the M. So, to count the total number of ways to do the selection, you need to add up these possibilities.
If you want things to match up with what you're saying, then it would be
$$\sum_{j=0}^{n} \binom{M}{j}\binom{N-M}{n-j} = \binom{N}{n}$$
The explanation would be exactly the same, except there would be $N$ total objects, $M$ special objects, and you're choosing $n$ of them. It's simpler this way because there are none of the $-1$'s.
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There is a group of $N-1$ kids, of whom $M-1$ are male. You choose at random a group of $n-1$ kids. The generic term $$\frac{\binom{M-1}{j}\binom{(N-1)-(M-1)}{(n-1)-j}}{\binom{N-1}{n-1}}$$ in your sum is the probability that exactly $j$ of the chosen kids are male. You are summing these probabilities from $j=0$ to $j=n-1$, that is, over all conceivable outcomes for the number of males. The sum of these probabilities must be $1$.
Remark: The calculation remains correct even if, for example, the number $M-1$ of males is a fair bit below $n-1$. For we use the standard convention that $\binom{M-1}{j}=0$ if $j>M-1$.
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2015-08-02 07:11:38
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https://zbmath.org/?q=an:1200.39005
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Stability of nonlinear autonomous quadratic discrete systems in the critical case. (English) Zbl 1200.39005
Summary: Many processes are mathematically simulated by systems of discrete equations with quadratic right-hand sides. Their stability is thought of as a very important characterization of the process. In this paper, the method of Lyapunov functions is used to derive classes of stable quadratic discrete autonomous systems in a critical case in the presence of a simple eigenvalue $\lambda =1$ of the matrix of linear terms. In addition to the stability investigation, we also estimate stability domains.
MSC:
39A30 Stability theory (difference equations) 39A12 Discrete version of topics in analysis
Full Text:
References:
[1] R. P. Agarwal, Difference Equations and Inequalities, Theory, Methods and Applications, vol. 228 of Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 2nd edition, 2000. · Zbl 0952.39001 [2] R. P. Agarwal, M. Bohner, S. R. Grace, and D. O’Regan, Discrete Oscillation Theory, Hindawi Publishing Corporation, 2005. · doi:10.1155/9789775945198 [3] N. G. Chetaev, Dynamic Stability, Nauka, Moscow, Russia, 1965. [4] S. N. Elaydi, An Introduction to Difference Equations, Springer, London, UK, 3rd edition, 2005. · Zbl 1071.39001 [5] A. Halanay and V. R\uasvan, Stability and Stable Oscillations in Discrete Time Systems, Gordon and Breach Science, Taipei, Taiwan, 2002. [6] V. Lakshmikantham and D. Trigiante, Theory of Difference Equations: Numerical Methods and Applications, vol. 251 of Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 2nd edition, 2002. · Zbl 1014.39001 [7] D. I. Martynjuk, Lectures on the Qualitative Theory of Difference Equations, “Naukova Dumka”, Kiev, Ukraine, 1972. [8] V. E. Slyusarchuk, “Essentially unstable solutions of difference equations,” Ukrainian Mathematical Journal, vol. 51, no. 12, pp. 1659-1672, 1999 (Russian), translation in Ukrainian Mathematical Journal, vol. 51, no. 12, pp. 1875-1891, 1999. · Zbl 0937.39008 · doi:10.1007/BF02525136 [9] V. E. Slyusarchuk, “Essentially unstable solutions of difference equations in a Banach space,” Differentsial’nye Uravneniya, vol. 35, no. 7, pp. 982-989, 1999 (Russian), translation in Differential Equations, vol. 35, no. 7, pp. 992-999, 1999. · Zbl 0968.39006 [10] V. E. Slyusarchuk, “Theorems on the instability of systems with respect to linear approximation,” Ukrains’kyi Matematychnyi Zhurnal, vol. 48, no. 8, pp. 1104-1113, 1996 (Russian), translation in Ukrainian Mathematical Journal, vol. 48, no. 8, pp. 1251-1262, 1996. · Zbl 0941.34061 · doi:10.1007/BF02383871 [11] J. Diblík, D. Ya. Khusainov, and I. V. Grytsay, “Stability investigation of nonlinear quadratic discrete dynamics systems in the critical case,” Journal of Physics: Conference Series, vol. 96, no. 1, Article ID 012042, 2008. · doi:10.1088/1742-6596/96/1/012042 [12] F. P. Gantmacher, The Theory of Matrices, vol. I, AMS Chelsea Publishing, Providence, RI, USA, 2002. · Zbl 1002.74002 [13] I. G. Malkin, Teoriya Ustoichivosti Dvizheniya, Nauka, Moscow, Russia, 2nd edition, 1966. · Zbl 0136.08502
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2016-05-06 18:54:19
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https://electronics.stackexchange.com/questions/29005/typical-lm317-resistor-values/94824
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# Typical LM317 resistor values
Is there a place that I can get a hold of a table with common voltages by using standard resistors? (Sorry that this is poorly worded but I cant figure out how to say it better, so I will give an example)
For example, a website that has a table that says for 5 volts on a LM317 you can use the "standard" resistor values of x for R1 and y for R2; with x and y being common resistor values.
• Why would you need a table when you have the formula from the datasheet?
– W5VO
Mar 30, 2012 at 17:40
• When you Google 'lm317 calculator' you get tons of (kinda useless) calculation-forms. Mar 30, 2012 at 18:28
• I wanted something more than a "random" guess at what standard input would yield a standard output.
– Reid
Mar 30, 2012 at 18:50
• This is silly, just do the math! Mar 30, 2012 at 19:03
• Actually, this makes sense. It's easy to pick R1 at random, but finding a standard R2 that matches is a pain. Such a table saves a few minutes of randomly stabbing at combinations. In fact, the paged linked in the accepted answer was a huge time-saver when trying to work out multiple combinations for a range of voltages. Mar 30, 2012 at 21:59
This page lets you enter the desired voltage and then calculates optimal values for the resistors. You can choose the resistors' series. For instance, if you select the E12 series, it will find the best match with resistors from the E12 series. Obviously E96 series will give better matches.
Some examples, with E12 (10%) resistors:
• 3.3V: 1k2 + 1k8 $\rightarrow$ 3% error
• 5V: 270 Ω + 820 Ω $\rightarrow$ 2% error
• 9V: 560 Ω + 3k3 $\rightarrow$ 2% error
• 12V: 390 Ω + 3k3 $\rightarrow$ less than 1% error
• 15V: 1k2 + 12k $\rightarrow$ 4% error
• 24V: 560 Ω + 10k $\rightarrow$ less than 1% error
(the first resistor is R1)
For comparison, with E96 (1%) resistors, all of the following have less than 1% error:
• 3.3V: 365 Ω + 590 Ω
• 5V: 1k02 + 2k94
• 9V: 71.5 Ω + 442 Ω
• 12V: 392 Ω + 3k32
• 15V: 590 Ω + 6k34
• 24V: 1k24 + 21k5
• Dec 28, 2013 at 12:56
• @PeterJ I explain what it does. Many of those calculators have it backward: they take the resistor values as input, and give you a voltage. This one also gives you E-values.
– flup
Dec 28, 2013 at 13:02
• The problem is when "this page" goes dead it no longer becomes useful. I've removed my downvote but what would make it much more useful in the longer term is to maybe include a little table in your answer that shows some common values for say 3.3V, 5V, 9V, 12V etc so when/if that happens it's still useful. I've probably edited / flagged a hundred posts because of dead links so it's best to try and make them continue to be valuable even when the links no longer work. Dec 28, 2013 at 13:10
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2022-07-02 20:27:32
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http://www.heldermann.de/JLT/JLT24/JLT242/jlt24023.htm
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Journal Home Page Cumulative Index List of all Volumes Complete Contentsof this Volume Previous Article Journal of Lie Theory 24 (2014), No. 2, 529--543Copyright Heldermann Verlag 2014 A New Formula for the Pfaffian-Type Segal-Sugawara Vector Natasha Rozhkovskaya Department of Mathematics, Kansas State University, 138 Cardwell Hall, Manhattan, KS 66502, U.S.A. rozhkovs@math.ksu.edu [Abstract-pdf] \def\o{{\frak o}} A combinatorial formula for the Pfaffian of the universal enveloping algebra $U(\widehat{\o}_{2n})$ of the affine Kac-Moody algebra $\widehat{\o}_{2n}$ is proved. It allows us easily to compute the image of the Segal-Sugawara vector under the Harish-Chandra homomorphism and to deduce formulas for the classical Pfaffian of the universal enveloping algebra $U(\o_{2n})$ of the even orthogonal Lie algebra. Keywords: Pfaffian, affine orthogonal Lie algebra, Feigin-Frenkel center, Harish-Chandra homomorphism. MSC: 17B35, 17B67 [ Fulltext-pdf (348 KB)] for subscribers only.
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2017-10-22 12:00:40
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http://cdsweb.cern.ch/collection/ATLAS%20Theses?ln=fr
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# ATLAS Theses
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2022-01-21 00:19:08
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https://ftp.aimsciences.org/article/doi/10.3934/mbe.2014.11.1091
|
# American Institute of Mathematical Sciences
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What can mathematical models tell us about the relationship between circular migrations and HIV transmission dynamics?
2014, 11(5): 1091-1113. doi: 10.3934/mbe.2014.11.1091
## Dynamics of evolutionary competition between budding and lytic viral release strategies
1 Department of Applied Mathematics, University of Western Ontario, London, Ontario, N6A 5B7, Canada 2 Department of Applied Mathematics, University of Western Ontario, London, Ontario N6A 5B7
Received January 2014 Revised April 2014 Published June 2014
In this paper, we consider the evolutionary competition between budding and lytic viral release strategies, using a delay differential equation model with distributed delay. When antibody is not established, the dynamics of competition depends on the respective basic reproductive ratios of the two viruses. If the basic reproductive ratio of budding virus is greater than that of lytic virus and one, budding virus can survive. When antibody is established for both strains but the neutralization capacities are the same for both strains, consequence of the competition also depends only on the basic reproductive ratios of the budding and lytic viruses. Using two concrete forms of the viral production functions, we are also able to conclude that budding virus will outcompete if the rates of viral production, death rates of infected cells and neutralizing capacities of the antibodies are the same for budding and lytic viruses. In this case, budding strategy would have an evolutionary advantage. However, if the antibody neutralization capacity for the budding virus is larger than that for the lytic virus, the lytic virus can outcompete the budding virus provided that its reproductive ratio is very high. An explicit threshold is derived.
Citation: Xiulan Lai, Xingfu Zou. Dynamics of evolutionary competition between budding and lytic viral release strategies. Mathematical Biosciences & Engineering, 2014, 11 (5) : 1091-1113. doi: 10.3934/mbe.2014.11.1091
##### References:
[1] A. Brännstr$\ddot o$m and D. J. T. Sumpter, The role of competition and clustering in population dynamics, Proc. R. Soc. B., 272 (2005), 2065-2072. [2] J. Carter and V. Saunders, Virology: Principles and Application, John Wiley and Sons, Ltd, 2007. [3] C. Castillo-Chaves and H. R. Thieme, Asymptotically autonomous epidemic models, in Mathematical Population Dynamics: Analysis of Heterogeneity, I. Theory of Epidemics (eds. O. Arino, et al.), Wuerz, Winnnipeg, 1995, 33-50. [4] D. Coombs, Optimal viral production, Bull. Math. Biol., 65 (2003), 1003-1023. doi: 10.1016/S0092-8240(03)00056-9. [5] H. Garoff, R. Hewson and D. Opstelten, Virus maturation by budding, Microbiology and Moleculer Biology Reviews, 62 (1998), 1171-1190. [6] M. A. Gilchrist, D. Coombs and A. S. Perelson, Optimizing within-host viral fitness: Infected cell lifespan and virion production rate, J. Theor. Biol. 229 (2004), 281-288. doi: 10.1016/j.jtbi.2004.04.015. [7] J. K. Hale and S. M. Verduyn Lunel, Introduction to Functional Differential Equations, Springer-Verlag, New York, 1993. doi: 10.1007/978-1-4612-4342-7. [8] N. L. Komarova, Viral reproductive strategies: How can lytic viruses be evolutionarily competitive? J. Theor. Biol., 249 (2007), 766-784. doi: 10.1016/j.jtbi.2007.09.013. [9] D. P. Nayak, Assembly and budding of influenza virus, Virus Research, 106 (2004), 147-165. doi: 10.1016/j.virusres.2004.08.012. [10] P. W. Nelson, M. A. Gilchrist, D. Coombs, J. M. Hyman and A. S. Perelson, An age-structured model of HIV infection that allows for variation in the production rate of viral particles and the death rate of productively infected cells, Math. Biosci. Eng., 1 (2004), 267-288. doi: 10.3934/mbe.2004.1.267. [11] L. Rong, Z. Feng and A. S. Perelson, Mathematical analysis of age-structured HIV-1 dynamics with combination antiretroviral therapy, J. Appl. Math., 67 (2007), 731-756. doi: 10.1137/060663945. [12] H. L. Smith, Monotone Dynamical Systems. An Introduction To The Theory Of Competitive And Cooperative Systems, Mathematical Surveys and Monographs, 41, AMS, Providence, 1995. [13] I. N. Wang, D. E. Dykhuizen and L. B. Slobodkin, The evolution of phage lysis timing, Evolutionary Ecology, 10 (1996), 545-558. doi: 10.1007/BF01237884. [14] I. N. Wang, Lysis timing and bacteriophage fitness, Genetics, 172 (2006), 17-26. doi: 10.1534/genetics.105.045922.
show all references
##### References:
[1] A. Brännstr$\ddot o$m and D. J. T. Sumpter, The role of competition and clustering in population dynamics, Proc. R. Soc. B., 272 (2005), 2065-2072. [2] J. Carter and V. Saunders, Virology: Principles and Application, John Wiley and Sons, Ltd, 2007. [3] C. Castillo-Chaves and H. R. Thieme, Asymptotically autonomous epidemic models, in Mathematical Population Dynamics: Analysis of Heterogeneity, I. Theory of Epidemics (eds. O. Arino, et al.), Wuerz, Winnnipeg, 1995, 33-50. [4] D. Coombs, Optimal viral production, Bull. Math. Biol., 65 (2003), 1003-1023. doi: 10.1016/S0092-8240(03)00056-9. [5] H. Garoff, R. Hewson and D. Opstelten, Virus maturation by budding, Microbiology and Moleculer Biology Reviews, 62 (1998), 1171-1190. [6] M. A. Gilchrist, D. Coombs and A. S. Perelson, Optimizing within-host viral fitness: Infected cell lifespan and virion production rate, J. Theor. Biol. 229 (2004), 281-288. doi: 10.1016/j.jtbi.2004.04.015. [7] J. K. Hale and S. M. Verduyn Lunel, Introduction to Functional Differential Equations, Springer-Verlag, New York, 1993. doi: 10.1007/978-1-4612-4342-7. [8] N. L. Komarova, Viral reproductive strategies: How can lytic viruses be evolutionarily competitive? J. Theor. Biol., 249 (2007), 766-784. doi: 10.1016/j.jtbi.2007.09.013. [9] D. P. Nayak, Assembly and budding of influenza virus, Virus Research, 106 (2004), 147-165. doi: 10.1016/j.virusres.2004.08.012. [10] P. W. Nelson, M. A. Gilchrist, D. Coombs, J. M. Hyman and A. S. Perelson, An age-structured model of HIV infection that allows for variation in the production rate of viral particles and the death rate of productively infected cells, Math. Biosci. Eng., 1 (2004), 267-288. doi: 10.3934/mbe.2004.1.267. [11] L. Rong, Z. Feng and A. S. Perelson, Mathematical analysis of age-structured HIV-1 dynamics with combination antiretroviral therapy, J. Appl. Math., 67 (2007), 731-756. doi: 10.1137/060663945. [12] H. L. Smith, Monotone Dynamical Systems. An Introduction To The Theory Of Competitive And Cooperative Systems, Mathematical Surveys and Monographs, 41, AMS, Providence, 1995. [13] I. N. Wang, D. E. Dykhuizen and L. B. Slobodkin, The evolution of phage lysis timing, Evolutionary Ecology, 10 (1996), 545-558. doi: 10.1007/BF01237884. [14] I. N. Wang, Lysis timing and bacteriophage fitness, Genetics, 172 (2006), 17-26. doi: 10.1534/genetics.105.045922.
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2018 Impact Factor: 1.313
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2022-07-02 10:55:50
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http://math.stackexchange.com/questions/125069/does-the-triangle-inequality-suffice-to-prove-all-minimum-results-on-sums-of-abs
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# Does the triangle inequality suffice to prove all minimum results on sums of absolute values of affine functions?
The title says it all ... more formally : let $n \geq 1$, and let $a_1, a_2 , \ldots ,a_n$ be positive numbers, let $b_1, b_2 , \ldots ,b_n$ be real numbers. Consider for $x\in {\mathbb R}$,
$$\phi(x)=\sum_{k=1}^n a_k |x-b_k|$$
It is easy to see that $\phi$ has a global minimum $m$ on $\mathbb R$, and that $m$ is attained of one the values $b_1, \ldots ,b_n$. In fact, $\phi$ is affine on each interval $[b_k,b_{k+1}]$ (and on the outer intervals $]-\infty,b_1]$ and $[b_n,+\infty[$ ).
I say that "m is the global minimum for $\phi$" can be proved "using only the triangular inequality" if there are numbers $a'_1,a'_2, \ldots ,a'_n$ with $0 \leq a'_k \leq a_k$ (for $1 \leq k \leq n$) and constants $\varepsilon_1,\varepsilon_2, \ldots, \varepsilon_n$ all equal to $-1$ or $+1$, such that
$$\sum_{k=1}^n\varepsilon_ka'_k=0, \ {\rm and} \ \sum_{k=1}^n\varepsilon_ka'_kb_k=m$$
because then we have
$$\phi(x)=\sum_{k=1}^n a_k |x-b_k| \geq \sum_{k=1}^n a'_k |x-b_k| =\sum_{k=1}^n |\varepsilon_k a'_k(x-b_k)| \geq \Bigg| \sum_{k=1}^n \varepsilon_k a'_k(x-b_k) \Bigg| =m$$
-
You may think about a more general case that is $x\in R^n$ or in a "Reflexive Banach space" (if you know some Functional Analysis) , this functions attains a minimum because $\phi$ is lower semi-continuous and convex. – checkmath Mar 27 '12 at 16:38
It seems that if you have a certain method like then you would use it in any Banach space what it is not possible (the space has to be reflexive)! – checkmath Mar 27 '12 at 16:45
Yes, the triangle inequality always suffices, and you can explicitly construct the $a_k'$ and $\epsilon_k$. They are severely constrained by the fact that the two inequalities in your last line must be equalities at the value $b_j$ at which $\phi$ takes the value $m$. For the first inequality to be an equality at $b_j$, we must have $a'_k=a_k$ for all $k\ne j$. For the second inequality to be an equality at $b_j$, all the terms must have the same sign, and this determines $\epsilon_k$ for all $k\ne j$ up to an overall sign; we can choose that sign such that all terms are non-negative. (I'm assuming that the $b_k$ are distinct, since otherwise we could just combine terms with the same $b_k$.)
This leaves only $a'_j$ and $\epsilon_j$ to be determined. Your first sum condition determines the value of $\epsilon_ja'_j$, and since $\epsilon_j=\pm1$ and $a'_j\ge0$, this determines $a'_j$ and $\epsilon_j$ (unless $a'_j=0$, in which case $\epsilon_j$ can be chosen arbitrarily). The resulting $a'_j$ is the absolute value of the slope at $b_j$ of the sum of all terms except the $j$-th, so the fact that adding the $j$-th term creates a minimum at $b_j$ implies $a'_j\le a_j$ as required. Since the values thus determined make all terms in the second inequality non-negative, the first sum condition and the last equality in the last line imply the second sum condition.
Actually the second sum condition should probably have $\pm m$, since both signs would make the last line work out; then the choice of making all terms positive wouldn't be required and either of the overall signs for the $\epsilon_k$ would yield an admissible set of values.
[Edit in response to comment:]
I think the result holds just as well for several variables. Basically what you're doing is to find the affine function that represents the sum of all terms but the $j$-th in a neighbourhood of $b_j$ and to replace the $j$-term by the negative of that so that the sum is identically zero in a neighbourhood of $b_j$, and then as you leave that neighbourhood the sign changes can only work to increase the function value. You can do something analogous for several variables: You can again replace the sum of all the functions that vanish at the minimum by the negative of the affine function that represents the sum of the non-vanishing terms in a neighbourhood of the minimum to make the sum vanish in that neighbourhood. In this case the neighbourhood will be bounded by hyperplanes on which individual affine functions vanish, and the sign change on crossing one of these hyperplanes can again only work to increase the value. As in the one-dimensional case, the replacement term is less than or equal in absolute value to the sum of the terms it replaces because otherwise the minimum wouldn't be a minimum.
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I am completely convinced by the "necessity-uniqueness" part in your reasoning, but not so much by the "sufficiency" part. In particular, your claim that "the values thus determined make all terms in the second inequality non-negative" looks dubious to me, since the terms are $\varepsilon_ka'_k(x-b_k)$, and the sign depends on the position of $x$ with respect to $b_k$. – Ewan Delanoy Mar 28 '12 at 7:40
@Ewan: I should have said more precisely "make all terms in the second inequality non-negative for $x=b_j$" -- the $\epsilon_k$ were chosen in the first paragraph to make this so, and this is enough to deduce the second sum condition. – joriki Mar 28 '12 at 8:03
I think I see your point now : you have an affine function whose leading coefficient is zero and who also is zero for $x=b_j$. So that function is identically zero – Ewan Delanoy Mar 28 '12 at 10:13
By the way, do you know if the result still holds in several variables ? In that context, affine functions would be replaced by functions of the form (linear form)+(constant). To avoid degenerate cases, we can assume that the linear parts span the whole dual space. – Ewan Delanoy Mar 28 '12 at 10:16
@Ewan: I edited the answer in response. – joriki Mar 28 '12 at 14:44
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2015-08-29 23:54:37
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https://pmbio.org/module-06-rnaseq/0006/06/02/RNAseq_Fusion_FilteringAnnotationReview/
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## Precision Medicine Bioinformatics
Introduction to bioinformatics for DNA and RNA sequence analysis
# Introduction
Pizzly generates outputs in .fasta and .json formats. Some initial filtering is performed automatically in pizzly, for example removing alignments where the distance of the breakpoint to exon boundaries is 10 or more base pairs. These automatically filtered reads are included in the outputs with unfiltered. suffix. In this module we will perform additional annotation, filtering and visualization of the .json output.
# Annotation
JSON data (JavaScript Object Notation) are name/value pairs separated by a colon. Pairs are organized into objects within curly braces and arrays within square brackets. JSON data can be reorganized into tabular form in a delimited text file using many tools and programming languages. Below, we will use R and a modified script from the grolar GitHub repository by Matthew Bashton to annotate pizzly output and create a tabular, annotated output.
# Get R scripts for later use
cd /workspace/rnaseq/fusion
wget https://raw.githubusercontent.com/griffithlab/pmbio.org/master/assets/course_scripts/mod.grolar.R
wget https://raw.githubusercontent.com/griffithlab/pmbio.org/master/assets/course_scripts/import_Pizzly.R
# Open the R environment and set working directory
R
setwd("/workspace/rnaseq/fusion/")
# Install packages if necessary from these commented lines-
# If asked about updating old packages (e.g. Old packages: 'MASS', 'devtools'... Update all/some/none? [a/s/n]:), select n
# install.packages("jsonlite")
# install.packages("dplyr")
# source("https://bioconductor.org/biocLite.R")
# biocLite()
# biocLite("ensembldb")
# biocLite("EnsDb.Hsapiens.v86")
library(jsonlite)
library(dplyr)
library(EnsDb.Hsapiens.v86)
library(ggplot2))
edb = EnsDb.Hsapiens.v86
suffix = "617.json"
JSON_files = list.files(path = "/workspace/rnaseq/fusion", pattern = paste0("*",suffix))
Ids = gsub(suffix, "", JSON_files)
# Load grolar.R script and run the GetFusionz_and_namez function to annotate
source("./mod.grolar.R")
# The function is too long to type out line by line, but we can view it by calling it without variables
GetFusionz_and_namez
# Look through the funciton steps to get a sense of how our output is being processed.
# Run annotation script on each file in /workspace/rna/fusion suffixed with fusion.json
lapply(Ids, function(x) GetFusionz_and_namez(x, suffix=suffix))
The function should return:
[[1]]
[1] TRUE
[[2]]
[1] TRUE
# Filtering
• The GetFusionz_and_namez script wrote two new files into /workspace/rnaseq/fusion: norm-fuse_fusions_filt_sorted.txt and tumor-fuse_fusions_filt_sorted.txt. Let’s read those back into R and filter further:
normal=read.table("./norm-fuse_fusions_filt_sorted.txt", header=T)
names(normal)
• In addition to values present in the orignial .json files, each fusion now has a unique identifier, sequence positions, and distance values for genes from the same chromosome:
> names(normal)
[1] "ID" "paircount" "splitcount"
[4] "geneA.id" "geneA.name" "geneA.seq_name"
[7] "geneA.gene_seq_start" "geneA.gene_seq_end" "geneA.seq_strand"
[10] "geneB.id" "geneB.name" "geneB.seq_name"
[13] "geneB.gene_seq_start" "geneB.gene_seq_end" "geneB.seq_strand"
[16] "same_chr" "gene_distance"
• Common filtering tasks for fusion output include removing fusions from the tumor sample which are present in the normal, and removing fusions for which the is little support by pair and split read counts:
normal$genepr=paste0(normal$geneA.name,".",normal$geneB.name) tumor$genepr=paste0(tumor$geneA.name,".",tumor$geneB.name)
uniqueTumor=subset(tumor, !(tumor$genepr %in% normal$genepr))
nrow(uniqueTumor)==nrow(tumor)
[1] FALSE
nrow(tumor)-nrow(uniqueTumor)
[1] 2
# There are two fusions (or at least fusion gene pairs) from the normal sample which are also present in the tumor.
# Examine the output of-
shared_src_tumor=subset(tumor, (tumor$genepr %in% normal$genepr))
shared_src_normal=subset(normal, (normal$genepr %in% tumor$genepr))
shared_src_tumor
shared_src_normal
• Filtering by counts:
# Merge normal and tumor data
normal$sample="normal" tumor$sample="tumor"
allfusions=rbind(normal,tumor)
# Compare counts of paired and split reads
tapply(allfusions$paircount, allfusions$sample, summary)
$normal Min. 1st Qu. Median Mean 3rd Qu. Max. 0.000 0.000 0.000 1.148 2.000 24.000$tumor
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0000 0.0000 0.0000 0.5965 1.0000 6.0000
tapply(allfusions$splitcount, allfusions$sample, summary)
$normal Min. 1st Qu. Median Mean 3rd Qu. Max. 0.000 1.000 1.000 4.901 3.000 123.000$tumor
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.000 1.000 1.000 2.526 3.000 24.000
# As a density plot
countplot=list()
countplot[[1]]=ggplot(allfusions, aes(paircount, fill=sample))+geom_density(alpha=.4)+geom_vline(xintercept=2)+coord_fixed(ratio=15)
countplot[[2]]=ggplot(allfusions, aes(splitcount, fill=sample))+geom_density(alpha=.4)+coord_cartesian(ylim= c(0,.2))+geom_vline(xintercept=5)+coord_fixed(ratio=200)
pdf("countplot.pdf")
countplot
dev.off()
• Results should look like this:
To filter, let’s take all fusions with a pair read count of at least 2 and a split read count of at least 5:
nrow(allfusions)
[1] 239
allfusions=allfusions[which(allfusions$paircount >= 2 & allfusions$splitcount >= 5),]
nrow(allfusions)
[1] 27
write.table(allfusions, "allfusions.txt")
# Visualization:
Chimeraviz is an R package for visualizing fusions from RNA-seq data. The chimeraviz package has import functions built in for a variety of fusion-finder programs, but not for pizzly. We will have to load our own import function that you downloaded above:
# Enter R, install and load chimeraviz
R
source("https://bioconductor.org/biocLite.R")
biocLite("chimeraviz")
# (if asked to update old packages, you can ignore- Update all/some/none? [a/s/n]:)
library(chimeraviz)
# Use the pizzly importer script to import fusion data
source("./import_Pizzly.R")
# You can view the function by calling it without variables
importPizzly
fusions = importPizzly("./allfusions.txt","hg38")
pdf("chr617-fuse-circ.pdf")
plot_circle(fusions)
dev.off()
The resulting circos plot shows our filtered gene fusions (blue for inter-chromosomal and red for intra-chromosomal)
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2021-05-13 16:16:28
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https://athrunen.dev/experimenting-with-efficiently-combining-rgb-and-true-white/
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For my latest project, I wanted to use RGBW LEDs instead of just RGB ones.
As they have some interesting advantages over RGB that I wanted to explore:
• A more accurate representation of white
• A wider color spectrum
This only works with somewhat desaturated colors. But at the end of the post I have some ideas that might help to work around that.
This would allow for better lighting of surfaces as well as more natural-looking colors.
Why that is and how we get there is what I will try to explain below:
## What is color exactly?
### Light and color
Light or visible light to be precise is a form of electromagnetic radiation with a wavelength of 400-700 nanometers. Electromagnetic radiation means that the wave got a magnetic as well as an electric field which both oscillate at the same time. Such a wave is emitted when an electron transitions from a higher to a lower state of energy, releasing the difference in the form of a photon. The length of one of the wave segments is used to identify the waves corresponding wavelength.
### What the wavelength got to do with spectral color
Relevant for us are the colors red, green and blue(RGB anyone?) that can be found at ~700 nm, ~530 nm, and ~470 nm respectively. At those wavelengths, they are considered (near-) spectral colors. That means that they are composed of only one wavelength or a relatively narrow band of wavelengths. Which means that they can be used to efficiently mix different colors.
## Mixing colors
When different waves of light intersect and bundle up before they reach the eye, they can be perceived as a different color that appears lighter than the colors used to mix it. A light source will appear as white if enough colors are added up that way.
And as an example for mixing, if you get a red and a green light to combine, the resulting light will appear yellow.
### Subtractive mixing
This type of color mixing determines how an object looks in a specific light. The surface of an object is composed of one or many pigments. A pigment only reflects a specific wavelength and absorbs the rest. Mixing enough pigments will result in all light being absorbed, creating a black surface.
For example, a red surface illuminated by white light looks red because it absorbs all but the red light. But if you would use a blue light instead, the same surface would look black.
## Wavelength spectra of RGB and white LEDs
Let's take a look at the emission graph above that shows what the possible spectrum of an RGB LED might look like.
Given a temperature of 25 °C, the x-axis describes the wavelength and therefore the perceived color of the light and the y-axis describes the actual intensity of that wavelength.
And as you can see, their actual colors are intense but have a quite narrow range of emitted colors with some gaps in between.
If we now look at the spectrum of a white LED, you can clearly see that there is a wider range of wavelengths.
To understand how this is archived we first need to understand what a Stokes shift is.
Quoting Wikipedia:
When a system (be it a molecule or atom) absorbs a photon, it gains energy and enters an excited state. One way for the system to relax is to emit a photon, thus losing its energy (another method would be the loss of energy as heat). When the emitted photon has less energy than the absorbed photon, this energy difference is the Stokes shift.
In the case of the white LED this means that some of the photons of the underlying blue LED are absorbed by the phosphor(or a similar substance) and are (re-)emitted with a longer wavelength, thereby broadening the spectrum.
Now we can discern the problems of mixing RGB in the light(heh) of additive and subtractive mixing:
• Additive mixing works quite well, just needs some proper balancing to account for the capabilities of the different LEDs
• Subtractive mixing produces wrong colors because of missing wavelengths between the three color
But if we now add the white color into the mix we will be able to mitigate those problems at least for less saturated colors:
• By mixing on top of white we do not need to be concerned about having a perfect white to desaturate the color
• Using a white LED gives us a wider and natural spectrum, making our colors illuminate surfaces more natural
## Mixing in the white
To grasp how to use the white while mixing you will have to understand that in RGB only two colors are actually needed to mix a pure hue, while the third color is used solely to control the saturation. Take a look at the following graphic:
If we have a lot of red and blue and we add a little bit of green the resulting color is only shifted towards the white point D65.
That means we can easily swap the color with the lowest value with white and get almost the same result. Let's take a look at some code for that:
// red = 200
// green = 150
// blue = 250
std::array<int, 3> rgb = {200, 150, 250};
// checking for the index of the smallest of the three values
// green is the smallest
// becomes index = 1
int index = std::min_element(color.begin(), color.end()) - color.begin();
// storing value for later
int value = rgb[index];
// setting green to 0
rgb[index] = 0;
// creating the RGBW output
// becomes rgbw = {200, 0, 250, 150}
str::array<int, 4> rgbw = {rgb[0], rgb[1], rgb[2], value};
As already explained in the code the green would become zero and the white 150, hopefully resulting in a slightly desaturated purple.
To compensate for the difference in luminosity between the colored LEDs and the white LED I don't just replace the color with white but allow a specific factor to be used.
In my case, the factor is adjustable at runtime using a simple interface and some buttons.
In actual code that could look something like this:
// some arbitrary factor between 0 and 1
float factor = 0.6;
std::array<int, 3> rgb = {200, 150, 250};
int index = std::min_element(color.begin(), color.end()) - color.begin();
// getting 60% of green
int value = rgb[index] * factor;
// setting green to 40% of itself
rgb[index] = rgb[index] * (1 - factor);
// creating the RGBW output
// becomes rgbw = {200, 60, 250, 90}
str::array<int, 4> rgbw = {rgb[0], rgb[1], rgb[2], value};
In this case, the white to green ratio(or pure white to barely white if you want) is 60% to 40% or 90:60 to use the actual numbers.
## Conclusion
Using the white LED when mixing desaturated colors is fairly easy and can be achieved by just simply replacing all or some of the amount of white created by the third color with actual white.
The more desaturated the color is, the more are we able to reap the benefits of the wider color spectrum. The problem is, that on full saturations we have no white at all.
But I can think of at least two ways to increase the quality even in that case:
• Using more than three spectral colors(tetra-, pentachromatic LEDs)
• Always adding some tiny amount of white(and maybe increase the resolution from 8-bit(256 steps) to 16-bit(65536 steps))
And as I am already working on implementing the aforementioned 16-bit RGB resolution I might be able to try the last one out for myself.
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2021-01-27 04:31:58
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https://math.stackexchange.com/questions/3111368/validity-of-using-induction-to-show-union-of-an-infinite-ascending-chain-of-subg
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# Validity of Using Induction to Show Union of an Infinite Ascending Chain of Subgroups is a Subgroup
Can this be done by induction instead of just proving the subgroup criterion? I can prove using the essentials tools of group theory, but looking at the problem, I was wondering if we can simply use an induction argument.
Given we have a chain of ascending subgroups of a group $$G$$... $$H_1\le H_2\le....$$ Is the union $$\bigcup_{i=1}^\infty{H_i}\le G$$
For the first case, we have that $$H_1\le H_2$$ and thus, $$H_1\subseteq H_2$$ means that in terms of their union,
$$\bigcup_{i=1}^2{H_i}=H_2\le G$$
Thus, for an arbitrary $$n>2$$ we can assume that $$\bigcup_{i=1}^n{H_i}=H_n\le G$$ Then $$\bigcup_{i=1}^{n+1}{H_i}=\left(\bigcup_{i=1}^{n}{H_i}\right)\cup H_{n+1}=H_n\cup H_{n+1}=H_{n+1}\le G$$
Thus, for any whole value of $$n$$, the claim is true.
My worry is the infinite upper bound. But I feel like induction takes care of that.
• Judging by some of the comments on this page, you might want to clarify (in the title as well) that you're asking specifically whether or not induction is enough. – Asaf Karagila Feb 13 at 17:34
• @AsafKaragila, thank you. I changed the title, and reframed the post a bit to more accurately depict my inquiry. – Eleven-Eleven Feb 13 at 17:50
• Theorem: a decimal of the form 0.abc... with n digits after the decimal place is a rational. Proof: Plainly this is true for n = 0. Suppose it is true for n=k. We have 0.[k digits] rational, we add to it x/10^(-k-1) for x between 0 and 9, which is a rational. The sum of two rationals is rational. Therefore we have proven the theorem by induction. Have I now proven that pi is rational? Now do you see why "induction takes care of the infinite bound" is wrong? – Eric Lippert Feb 13 at 18:08
• 100%. That was as clear as could be written.. – Eleven-Eleven Feb 13 at 19:05
• In some sense, you haven't proved anything... the fact that $H_n$ is a subgroup for any $n$ was already given as part of the problem statement. – MartianInvader Feb 13 at 19:42
Induction (the conventional way) only takes care of the "for any whole value of $$n$$" part. It can never let you conclude anything for the entire infinite union. There is such a thing as transfinite induction, but even then the induction step is split into separate steps for the cases which have an immediate predecessor and cases which don't. The infinite union doesn't have an immediate predecessor, and thus the standard induction step will never reach it.
However, clearly the union is a subset, and clearly that subset contains the identity element, so we're a good way of the way there. When showing that it has inverses and that it's closed under products, it pays to know that any finite collection of elements in the union is contained in some finite union as well. And you already know that any of the finite unions is a group.
Of course $$1\in\bigcup_{i=1}^{\infty}H_i$$ and for any $$g\in H_i$$,$$h\in H_j$$ say without loss of generality that $$i\leq j$$ then $$g\in H_j$$ too and since this is a subgroup You have $$gh\in H_j\subset \bigcup_{i=1}^{\infty}H_i$$ and thus $$\bigcup_{i=1}^{\infty}H_i$$ is a subgroup. That's all. You don't argue via induction!
$$\textbf{Edit:}$$ what was missing, for any $$g\in\bigcup_{i=1}^{\infty}H_i$$ there exists an $$i$$ so that $$g\in H_i$$, actually this implies $$g\in H_j$$ for all $$i\leq j$$, but we don't need this here. Since $$H_i$$ is a subgroup $$g^{-1}\in H_i\subseteq \bigcup_{i=1}^{\infty}H_i$$.
• That was my question... I knew how to prove using the subgroup criterion, but I was thinking about it in terms of induction and wanted to know why it was a valid or invalid proof. – Eleven-Eleven Feb 13 at 14:45
• This is not sufficient. You need to show, also, that the inverse of an arbitrary element of the candidate group is in the candidate group. Don't mix the one-step and two-step subgroup lemmas without due care. – Shaun Feb 13 at 16:14
• @Eleven-Eleven I understand, in this case Arthur's answer seems perfect to me. – Peter Melech Feb 14 at 12:25
• @Shaun Sure, of course this is even more trivial, but You are right to point this out – Peter Melech Feb 14 at 12:26
• I edited my post. – Peter Melech Feb 14 at 12:32
(I should have read the question more carefully. What follows is a proof by the method the OP excluded.)
Since $$e\in H_1$$, we have $$e\in H_1\subseteq\bigcup_{i=1}^{\infty}H_i=:\mathcal{U}.$$ Hence $$\mathcal{U}$$ is nonempty.
Let $$g, h\in\mathcal{U}$$. Then $$g\in H_k$$ and $$h\in H_\ell$$ for some $$k, \ell$$. Assume w.l.o.g. that $$k<\ell$$. Then the ascending chain condition gives that $$g\in H_\ell$$. Since $$h\in H_\ell$$ and $$H_\ell$$ is a subgroup of $$G$$, $$h^{-1}\in H_\ell$$. Thus $$gh^{-1}\in H_\ell$$. Hence $$gh^{-1}\in \mathcal{U}$$.
Hence, by the one-step subgroup lemma, we have
$$\mathcal{U}\le G.$$
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2019-12-11 06:17:22
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http://motls.blogspot.com/2012_09_01_archive.html
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## Saturday, September 29, 2012 ... /////
### Gore's investment firm: no green investments
Al Gore has repeatedly said that he was putting his money where his mouth is. However, it was just revealed that in the real world, Al Gore is stealing the money where almost every green criminal steals them, and he is putting them where almost every rich person or investor puts them. Be sure that these two places are very different from one another.
WND.com just published a very interesting text with the title:
Al Gore bails from green energy investment
Bill Gunderson, the president of Gunderson Capital Management, has looked at the portfolio of Al Gore's investment management firm, Generation Investment Management. In fact, you can look at the list yourself; it's at the SEC website:
SEC about Gore's firm
What firms will you see there?
### Raphael Bousso is right about firewalls
Five days ago, I reviewed the discussions on black hole firewalls, started by AMPS in July 2012. Joe Polchinski wrote a guest blog for Cosmic Variance yesterday.
During the days, I had enough time to sort all these issues and I am confident that Raphael Bousso is right and Polchinski and others are wrong.
## Friday, September 28, 2012 ... /////
### President Klaus survives "assassination attempt"
There's a national holiday in the Czech Republic today. We call it the Day of Czech Statehood.
On September 28th, 935 – or perhaps 929 – Good King Wenceslaus (later St Wenceslas, the patron of the Czech nation) whom you know from a Christmas Carol was murdered by his brother Boleslav the Cruel.
It's probably unusual to celebrate anniversaries of the assassination but that's how it works here. No one knows the exact birthday of the king in 907, anyway.
Today it's also the name day of all people called Václav (=Wenceslaus) – a traditionally popular first Czech name that was also recycled by several Czech kings of a later era but one that began to disappear among the newborns, however. Despite the decline of the name, it's been an unwritten rule that the presidents of the Czech Republic have to be called Václav: Havel and Klaus both bore this name. To make things dramatic, the Bashar Assad lookalike on the picture above shot the current leader, Václav Klaus, today.
### Exotic branes, U-branes, and U-folds
TBBT: All American readers should notice that the first episode of the 6th season of The Big Bang Theory was aired by CBS yesterday. Howard was in space.
Holes: Joe Polchinski wrote a Cosmic Variance guest blog on black hole firewalls.
SUSY: Dave Goldberg at IO9 asks what's so super about it.
The first hep-th paper on the arXiv today is a wonderul 94-page-long article
Exotic Branes in String Theory
by Jan de Boer and Masaki Shigemori of Amsterdam. It is a kind of an excursion into the inner workings of string theory that I find extremely important, playful, and I think that similar papers are heavily undercited, underread, and understudied.
The Dutch and Japanese string theorists investigate a certain unusual type of objects whose existence depends on special properties of string theory and which don't exist in regular quantum field theories etc. That's why they are called "exotic". However, as the authors argue, within string theory, these objects are common, unavoidable (they arise, for example, during "polarization" of ordinary branes), and therefore omnipresent.
## Thursday, September 27, 2012 ... /////
### Fifty years after Silent Spring
Conservationism or environmentalism as an ideology has its roots in Nazi Germany and one could probably go even further.
But the Western environmentalism in the form we are familiar with today was born exactly 50 years when Rachel Carson began to release her book Silent Spring. It's such an inconvenient anniversary that most of the articles about the anniversary listed by Google News have been written by the critics of environmentalism.
### Experimenters, SUSY, frustrations, and anthropic ideas
High-energy phenomenologists and experimenters may be entering a psychological stage that was predictable – and that, in fact, many of us including your humble correspondent were predicting. The LHC is a great machine that works very well and pushes many frontiers but it's not a "miracle machine" that may answer all open questions about physics, at least not within two years.
Fashion for naturalness
So almost everyone who is building his or her interest in fundamental physics on accessible experiments seems to be frustrated these days. The discovery of the Higgs boson may have made those emotions even worse, see e.g. Jester at Resonaances, in agreement with the "nightmare scenario" people would talk about 5 years ago and we seem to be living through now, so far (the scenario is that the LHC only finds the Higgs boson and nothing else). Those of us who don't think that physics ends at $8\TeV$ or $14\TeV$ – and this includes most formal theorists, I would guess – are largely unaffected by this particular spleen, of course. ;-)
Note that patience is sometimes needed. Peter Higgs hasn't spent 50 years in depressions even though he had to wait for the discovery of "his" goddamn particle for 48 years – despite the fact that it's a very trivial quantum of a field that is more mundane than any other field we have found in Nature (it's spinless, stupid). That's a reason to think that the problem is with the people, not with the actual progress in physics. People just don't like physics as much as our predecessors did. They keep on whining, complaining, and I am annoyed by them because the current physics' image of the world is richer and more accurate than it has been at any moment in the past.
## Wednesday, September 26, 2012 ... /////
### Albert Einstein 1911-12, 1922-23
Several events related to Albert Einstein's life occurred in recent days and months. If you consider yourself a sort of Einstein fan, you may let me mention some of them.
First, you may finally buy Einstein's brain for $9.99 (or$13.99 now?), no strings or hair attached. See Google News or go to the iTunes AppStore.
Second, Caltech and Princeton University Presses teamed up and released the 13th volume of Einstein papers. They cover January 1922-March 1923 and you may already buy the book for $125 at the PUP website: it has over 900 pages. Einstein is travelling a lot, is ahead of time and already (or still) speaks Hebrew in British Palestine ;-), and doesn't even mention his Nobel prize. Wired about the new book. ### Street View: Antarctica, deep ocean, Alps, Mars, etc. If you haven't played with Google Maps for some time, you may want to try some of the wonderful links below. Note that the Street View always allows you to "drag" and change your point of view – or press $\langle$ and $\rangle$ arrows on the photograph itself and walk a little bit further. You may think that Street View doesn't allow you to enter the living rooms but at least in some cases, this opinion is wrong. ### Witten's new trilogy on RNS diagrams Witten is such a big guy in physics that I will probably not attempt to write a single article attempting to cover all his important contributions to physics, something I did and I plan with other inaugural Milner Prize winners. Instead, we may talk about specific new papers. I have already mentioned Edward Witten's work on technical issues of RNS Feynman diagrams but because the full trilogy is now out, it may be appropriate to summarize the links and add a few comments. There are three papers Short (review of the long one), medium (background on manifolds), long (the bulk). This ordering is chronological, too. After we learned that the first two papers had 42 and 118 pages, respectively, we could guess what the length of the final paper would be. The answer is that the digit counting hundreds is linearly increasing, the digit counting tens is alternating between 1 and 2, and the last digit is describing coordinates of the point (2,8,5). I guess that no one had the right answer. ;-) ## Tuesday, September 25, 2012 ... ///// ### BBC: Who's afraid of a big black hole? Another episode about fundamental physics of the BBC 2 "Horizon" program, featuring people such as Andy Strominger (in a dark classroom of the Jefferson Lab and in his office), was aired in November 2009. ### Did EPA test toxins on humans? Steve Milloy of JunkScience.COM and pals have sued Lisa Jackson's EPA for having deliberately exposed unhealthy humans to harmful and lethal substances: EPA Human Testing.COM, a special website about the story The initial text over there Canada Free Press, WUWT, PR Web, Tucson Citizen, NLPC story (Paul Chesser) Milloy et al. claim to have gone through hundreds of pages obtained via the Freedom of Information Act. The documents imply, he says, that at least a dozen of citizens were treated by carcinogenics and other harmful pollutants, including diesel exhaust and fine particulate matter (PM2.5). ### Almost all particle physics papers will be free Journals will sign a deal with libraries In many other fields such as Earth sciences, people are dreaming about the free access for everyone. In high energy physics, this dream is becoming a reality. After all, most high energy physicists have relied on arXiv.org, a freely accessible preprint server (see a NYT story about it), as their main source of information for more than two decades. Nature tells us some details about the deal that will make the transition of journals to the free form possible. ## Monday, September 24, 2012 ... ///// ### BBC: How small is the universe? Off-topic: Czech police claims to have found the culprits behind the methanol scandal which killed 25 people in recent weeks (and led the government to impose temporary prohibition). The two men have fully confessed and face up to life sentences. A company legally using methanol to produce windshield washers was selling the methanol to the black market. The two criminals knew that they were producing a deadly mixtures of ethanol and methanol – to make profit, regardless of people's lives. I assure everyone that none of this business has ever compromised the safety of Czech exported beverages, especially not those produced near Pilsen, the opposite side of the country, so discrimination against Czech alcohol would be a sign of someone's misunderstanding of the details. If you haven't watched this BBC2 Horizon program named How small is the universe? four weeks ago, here is the 60-minute video: Is the narrator speaking in conventional British English? It's about Nature at very short distance scales and topics such as the string theory landscape, tiny black holes, the Planck length, quasiparticles, MAGIC cosmic rays telescope, and many more things. ### Are black holes surrounded by firewalls? Dilaton has noticed a new, extremely provocative concept that was introduced among the quantum gravity researchers two months ago: the firewall. For decades, people teaching general relativity – including your humble correspondent (e.g. here) – have been explaining that nothing special happens to an infalling observer when she crosses a black hole event horizon. The curvature is usually pretty small there – the curvature radius is close to the black hole radius – and you only get torn apart once you approach the black hole singularity which may be much later. Advertisement of a future text: Read also Raphael Bousso is right about firewalls The event horizon is just a coordinate singularity; with a better choice of coordinates, the vicinity of the horizon (including a region below and a region above the horizon) looks like a nearly flat piece of the Minkowski spacetime. These coordinates may be "extremely distorted" functions of some other coordinates you may use for other purposes but they exist. Because the laws of general relativity are local, the (nearly) flat geometry of the region implies that there will be (nearly) the same phenomena there as in the flat space. Later, some quantum properties of black holes have been pretty much established, too. The picture has made sense to everyone who has ever been considered a top expert in quantum gravity. That was the case until July 2012. ### EU carbon market will be saved by a new boom of coal An hour ago, I saw a fascinating article on Patria.CZ, a Czech server for investors, which revealed a highly paradoxical, nearly comical plan. Analysts at UBS are predicting that by 2015, energy giants such as E.ON and RWE will build lots of new coal power plants – in fact, their capacity will be 6 times greater than the capacity of previously preferred gas-based alternatives. That may send the price of carbon permits up by 73% by 2013. Note that the U.N.-based carbon indulgences' price, CER, dropped by 80 percent in the most recent year. ## Sunday, September 23, 2012 ... ///// ### Klaus in Telegraph: on final stages of destruction of democracy in the EU The Telegraph is running a story on Czech President Klaus' view on the EU and promotes his new book to appear in the U.K. next week, Václav Klaus warns that the destruction of Europe's democracy may be in its final phase The book is called The Shattering of Illusions. It has 200 pages and costs £14.44. ## Saturday, September 22, 2012 ... ///// ### Australia: qubit as a single silicon atom Peter F. has pointed out the following intriguing experimental advance in quantum computing: Australians Create 1-Atom Silicon Quantum Computing Bit (quBit) (Daily Tech) How does it work? ## Friday, September 21, 2012 ... ///// ### Chu and CMS' Incandela's DOE Higgs talk Steve Chu is going to get some positive TRF publicity. ;-) A week ago, Joe Incandela, the boss of CMS, gave a talk for the Department of Energy. Nobel prize winner Steven Chu who happens to be a secretary at the department gave an introduction. In the U.S., you may become a secretary even if you're neither a hot blonde nor a hot brunette. ### German biogeochemistry postdoc proposes to extinguish the "unsustainable" Sun Nathaniel Virgo (SE) is a postdoc at the Max Planck Institute for Biochemistry in Jena. His DPhil was about the effect of limited energy supply on organisms. You may see lots of similar "conventional physics" topics that are always "tainted by the environmentalist ideology" a little bit. Two hours ago, he asked a question at the Physics Stack Exchange that made me LOL. He has already made all the important general plans and only asks the physics users to help him with a technical detail. So his question is: What is the easiest way to stop a star? No kidding. You're going to learn something about the mainstream science at the mainstream scientific institutes that do research into Earth sciences. ;-) ### May the name legitimately influence the fate of an application? Yes. Cosmic Variance promotes something that feminists consider to be science, namely this paper in PNAS: Science faculty’s subtle gender biases favor male students The paper claims that they wrote 127 applications for a lab manager position, attached a random name of the applicant which could be either male or female, and found out that the applications with the male name had better ratings. ## Thursday, September 20, 2012 ... ///// ### Insane reaction to the PBS interview with Anthony Watts Three days ago, PBS did a piece on the climate change debate and featured Richard Muller as an alarmist (self-described converted ex-skeptic) and Anthony Watts as a skeptic (and later others such as Judith Curry): Climate Change Skeptic Says Global Warming Crowd Oversells Its Message If you listen to those 9.5 minutes, you may agree with me that Anthony Watts was speaking as a lukewarmer. It doesn't mean that I sharply disagreed with something; I didn't. (Well, I found Anthony's focus on the urban heat islands excessive and at one point, he almost denied that there exist any natural climate drivers – but I must have overinterpreted a sentence.) But he was surely not speaking as a partisan. But the very fact that PBS dared to interview the man behind the world's most visited climate website caused an explosion of anger among the climate activists and, unfortunately, not only the climate activists. ## Wednesday, September 19, 2012 ... ///// ### Thomson Reuters bets on teleported higgsless Nobel Breaking news: Dr Joseph Conlon, a string theorist at Oxford cooperating with the Royal Society as well, just sent me a link to their new cute website called: WhyStringTheory.COM. Recommended! The physics Nobel prize winners for 2012 will be announced on Tuesday, October 9th. As the Montreal Gazette and others mention, Thomson Reuters that often publishes its guesses – with mixed results – has made a somewhat strange prediction for this year's award. David Pendlebury of Thomson Reuters uses his alchemist Al Gore Rhythm based on the Web of Knowledge to guess the winners. Who are they? ### Astronomical unit (AU) redefined We visited an old astronomer yesterday: I could see the sunspots exactly as you can see them on the web, except that the image was left-right-reflected and somewhat rotated. By the way, if you click at the link, there's a tiny sunspot between 1569 and 1571, 2 times closer to 1569, that I could see as well. But I want to mention another astronomical report. A few days ago, Nature told us that the astronomical unit has been redefined after a vote in Peking (yes, because it's Peking, it was an unanimous vote): The astronomical unit gets fixed Similar people who gathered in Prague 6 years ago and decided that Pluto no longer belonged to the elite club of planets met in Beijing and reformed the definition of 1 AU. What was it before? ## Tuesday, September 18, 2012 ... ///// ### The West is nearly apologizing to hordes of bloodthirsty Islamists Related news: A Coptic Christian has shot a movie revealing that Mohammed was a homosexual child molester, womanizer, and a naive fool. This group of Christians in Egypt is pretty influential. The New York Times and The Boston Globe just wrote about a Coptic papyrus studied at Harvard (apparently authentic one) that talks about Jesus' wife. ;-) As far as the U.S. politics goes, I am close to the G.O.P. in many respects although it may be more accurate to classify me as a libertarian in the U.S. context (except that I am flabbergasted by the fact that this key political direction almost represents an irrelevant fringe group in the U.S.). But I could see that the typical G.O.P. members don't really respect the same value during the Summers wars at Harvard. Most Harvard people who were nominally right-wing Americans didn't do anything whatsoever to stop the insane witch hunts started by the feminists and similar highly obnoxious groups of activists. In fact, some people who considered themselves Christian have always been radical guardians of the political correctness who were really scaring me – and bullying me personally. I won't name the guys here. ## Monday, September 17, 2012 ... ///// ### Smileys: 30 years On Wednesday, it will have been 30 years since the birth of the smiley emoticons. You probably don't know who invented them. ### Stringy boundary conditions and D-branes In general, strings in string theory may be open or closed. Open strings are topologically line intervals with two endpoints; closed strings are topologically circles. The fields defined on the strings may be periodic as a function of $\sigma$, the spatial coordinate along the string, or they may obey various other boundary conditions. Let's look at them. ## Saturday, September 15, 2012 ... ///// ### Michio Kaku: physicists invented everything This 42-minute Big Think video by the string field theory pioneer called "The Universe in the Nutshell" (yes, Stephen Hawking should feel plagiarized) is exactly one month old right now and while it is of a somewhat lighter genre, I found its slightly over-the-edge claims amusing: Physicists invented microwave ovens and everything else. Schoolkid Kaku. Physicists will invent everything. ## Friday, September 14, 2012 ... ///// ### Innocence of Muslims: watch HD here Off-topic: Czech government imposed a full prohibition of beverages with 20 or more percent of alcohol, after 19 people died by methanol poisoning in recent days. Well, I think the prohibition is counterproductive. Millions of Muslims – including the Iranian mullah-in-chief and the official Fars News agency – seem to be super-upset about a$50,000 movie by a Coptic guy who lives in South California (and who has made some illegal financial transactions in the past).
Khamenei is crazy enough to demand the U.S. will execute the filmmaker for blasphemy. Savages in Libya have murdered the U.S. ambassador. U.S. flags are burning everywhere. All the top 12 stories at Fars News right now are dedicated to the movie: no kidding.
### Japanese guy may have proved the $abc$ conjecture
Up to a few exceptions to be proved separately, a strengthening of Fermat's Last Theorem
Four days ago, Nature described a potentially exciting development in mathematics, namely number theory:
Proof claimed for deep connection between primes
Shinichi Mochizuki of Kyoto divided the steps needed to prove the 1985 conjecture by Oesterlé and Masser into four papers listed at the bottom of the Nature article above.
The newly revealed proof works with mathematical structures such as the Hodge theaters (a theater with Patricia Hodge is above, I hope it's close enough) and with canonical splittings of the log-theta lattice (yes, the word "splitting" is appropriate above, too).
What is the conjecture about and why it's important, perhaps more important than Fermat's Last Theorem itself?
## Thursday, September 13, 2012 ... /////
### Vafa's new 4D string theory
First, another new paper.
Those people who conjectured that Edward Witten may stop doing physics after having won the $3 million Milner Prize may face a problem with their belief system today. Witten just revealed his new Notes On Super Riemann Surfaces And Their Moduli which has – sit down now, please – 118 pages. And it's just the second part in a trilogy; the first part appeared yesterday and it has 42 pages. He announced his intense work on these issues at Strings 2012. Supermanifolds and super Riemann surfaces in general are objects that locally look like a superspace, e.g. $\CC^{1|1}$. The number "1" after the vertical line refers to the number of Grassmann dimensions. ### Barroso wants a European Federation The chieftain of the self-described European Commission, José Manuel Barroso (PT), gave a speech in Strasbourg in which he declared his intent to transform the European Union to a federation. Barroso, claiming that "Europe needs a new sinking" (haven't we seen enough sinking of our continent already?), wants a federal Europe built on the existing institutions; in other words, this head of the so-far irrelevant European Commission, a politician who started his career in the Portuguese Communist Workers' Party, wants to become a European dictator. See the whole speech with a satirical name, State of the Union address (43 minutes). This guy and many others have clearly lost their contact with reality. After several years in which we've been shown every day how counterproductive and dangerous the unification of the ununifiable may be – for Europe and for the whole world economy – Barroso tells us we must repeat the same mistakes and do so even more vigorously than ever before. ## Wednesday, September 12, 2012 ... ///// ### Official Iranian media promote truthism One of the conspiracy theories studied in the Stephan Lewandowsky's fraudulent paper claiming to establish that the climate skepticism is correlated with the belief in conspiracy theories was the so-called truthism, i.e. the belief system that the U.S. government was behind the 9/11 attacks. On the 11th anniversary of the terrorist attacks, the official state-owned Iranian media corporation, Press TV, offered us a rather bizarre piece of anecdotal evidence that the correlation goes the other way around. ### PLB publishes Higgs discovery papers Thanks for all the kind words and advices. On July 4th, 2012, CERN announced the discovery of a new particle whose properties universally indicated it was the Higgs boson. (Those of us who rationally followed the published results knew that there was a new particle and what its mass was since mid December 2011.) They also promised to send the paper to a classical paper journal at the end of July. That's exactly what happened: ATLAS and CMS submitted their texts to Physics Letters B around July 31st. Within two weeks, the two papers were accepted and they will appear in the September 17th, 2012 issue of PLB (Volume 716 Issue 1). A cute fact is that this "commercial" (although funded primarily by taxpayers' mandatory contributions to research and education) journal made the papers freely available and they're already out. ## Tuesday, September 11, 2012 ... ///// ### I probably suffer serious illness Candida overgrowth You may have noticed that the blogging frequency decreased during the last week or so. So did my life expectancy: with the right definition, it dropped nearly 100% over the same week. While I was feeling unusually healthy just 5 weeks ago or so, experiencing no more-than-cosmetic problems and having had no flu or cold for a year etc., everything is different now. Whether I will be around in a month is yet to be seen... ## Monday, September 10, 2012 ... ///// ### Stephan Lewandowsky's incredible blog Joanne Nova is dedicating a lot of space to a radical crackpot named Stephan Lewandowsky whom you may remember from his fresh paper proving the climate skeptics believe in moonlanding conspiracies because he asked visitors of alarmist blogs. Steve McIntyre and others have offered their opinions about the Lewandowsky scam, too. ## Saturday, September 08, 2012 ... ///// ### Pseudoscience hiding behind "weak measurements" Jason Rush sent me a link to a new insane anti-quantum article spread by the BBC: Quantum test pricks uncertainty Subtitle: Pioneering experiments have cast doubt on a founding idea of the branch of physics called quantum mechanics. The subtitle is particularly "cool". In recent months, we "learned" from some would-be scientific journals and the mainstream media that the wave function is "surely" not interpreted statistically. Now we're told that the Heisenberg uncertainty principle has been wrong from the beginning, too. In fact, the subtitle finally makes the "paradigm shift" that all the anti-quantum zealots have been expecting at least from 1925 as simple and clear as possible: quantum mechanics itself has been put in doubt. What happens when a few assholes acquire access to cool lasers? Fine. What concepts are these hardcore cranks relying upon? ## Friday, September 07, 2012 ... ///// ### Klaus: West is, disappointingly, returning to socialism and etatism The Mont Pelerin Society was founded in 1947 by free-market luminaries Friedrich Hayek, Karl Popper, Ludwig von Mises, George Stigler, and Milton Friedman, among others. Its 2012 General Meeting took place at the Prague Castle between last Sunday and today. Czech president Václav Klaus gave the following speech: ## Thursday, September 06, 2012 ... ///// ### Science debate, 14 questions: Romney beats Obama Some people wanted American politicians to explicitly address scientific questions. Whether it's important or not, the project has become a reality. Your humble correspondent and Slate are among those (see Google News for media reactions) who think that judging by the answers, Romney defeated Obama, despite all the talk about the Left's being pro-science. ## Tuesday, September 04, 2012 ... ///// ### Pascual Jordan and foundations of QM Among the fathers of quantum mechanics, Pascual Jordan is the least well-known one. Needless to say, politics is the single most important reason behind his invisibility although it is not necessarily the only one. While he genuinely believed in the ideals of the NSDAP that he joined at some moment (not to speak about Luftwaffe and the Peenemünde rocket center where he worked as a climate scientist), he was pretty much suppressed already in the 1930s. After all, he was also unreliable for the regime due to his past collaboration with Jews such as Max Born and Wolfgang Pauli. Mostly for political reasons, he became isolated from the research community already around 1930. But after the war, some fellow physicists declared him "rehabilitated", he became a tenured professor again, and he was also elected as a lawmaker for the most mainstream among German parties, CDU. Nevertheless, he had done pretty much all the important original technical contributions to physics before 1930 or so. ### WISE finds millions of black holes, thousands of hot DOGs As lots of media outlets such as HJ News point out, the WISE satellite assembled in Logan, Utah (Mitt Romney's mother's hometown) has observed millions of galaxies. WISE, or Wide-field Infrared Survey Explorer, is looking at the sky in the infrared. ## Sunday, September 02, 2012 ... ///// ### Sunlight or neutrinos affect the radon gamma decay rate Seasonal and daytime variations in the gamma decay rate Several readers have directed my attention to a recent, May 2012 Stanford-Purdue-Israel preprint on the seasonal variations of the gamma decay: Analysis of Gamma Radiation from a Radon Source: Indications of a Solar Influence They look at gamma (and alpha) radiation emitted by radium-222, an isotope arising from a decay of uranium-238 through radium-226. What they found was a more striking and accurate confirmation of a 2008 preprint (see TRF 2008) by similar authors. ## Saturday, September 01, 2012 ... ///// ### Stratospheric geoengineering: undoing global warming costs below$8 bn/year
Bits of Science, Revmodo, and Fars News talk about a new Aurora-Harvard-Carnegie paper in Environmental Research Letters
Cost analysis of stratospheric albedo modification delivery systems (fulltext PDF) by Justin McClellan, David W Keith, and Jay Apt
They quantify the aircraft- or rocket-related expenses needed to undo the global warming.
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2013-06-19 10:59:05
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https://imathworks.com/tex/tex-latex-book-page-number-at-new-chapter-is-centered-but-it-shouldnt-be/
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# [Tex/LaTex] (Book) Page number at new chapter is centered, but it shouldn’t be
I have the following problem:
\documentclass[a4paper]{book}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{fancyhdr}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{dsfont}
\usepackage{tikz}
\usepackage{braket}
\usepackage[figurename=Fig., font={footnotesize}]{caption}
\usepackage{float}
\usepackage{placeins}
\usepackage{epstopdf}
\usepackage[a4paper,left=40mm, right=40mm ,top=35mm,bottom=35mm]{geometry}
\usepackage{subfig}
\usepackage{tabularx}
\epstopdfsetup{outdir=./}
\usepackage{tikz-feynman, contour}
\usepackage[toc,title]{appendix}
\usepackage[nottoc]{tocbibind}
\usepackage[parfill]{parskip}
\usepackage{mathtools}
\usepackage[square, numbers, comma, sort&compress]{natbib}
\captionsetup[figure]{labelfont=bf}
\usepackage{hyperref}
\makeatletter
\makeatother
\renewcommand\vec{\boldsymbol}
\pagestyle{fancy}
\fancyhf{}
\fancyfoot[RO,LE]{\thepage}
\renewcommand{\footrulewidth}{0.4pt}
\usepackage{chngcntr}
\counterwithout{figure}{chapter}
This is everything I have before I start the document. The problem I have now, is: every time I start a new chapter, the page number appears at the bottom of the page in the center, but the way I set it up (and want it to be) is that it should appear either left or right (depending on even/odd page number). I cannot find the error. Before, this was not a book, but an article and there it worked the way I wanted it. Now I changed it to book and left the rest the way it was before. Everything else works fine, its just the starting page of each chapter.
I hope my information is sufficient for you to find the error.
In addition, it would be nice, if I could keep my footer even if a new chapter starts on this page (the header can vanish as it usually does with new chapters).
Edit: Now that is weird. I thought the commands would line up beneath each other, but they don't. Really sorry for that, gonna see if I can fix it up.
The problem comes from the fact that, in the book class, the first page of a chapter uses the plain style. The solution consists in redefining the plain style with fancyhdr. Here is a code, adapted from the documentation:
\fancypagestyle{plain}{%
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2022-11-29 05:09:22
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https://community.ptc.com/t5/Additional-Creo-Questions/Drill-groups-in-WF2/m-p/369192
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Newbie
## Drill groups in WF2
Dear All,
I have just come across an interesting (or not so interesting) problem
using drill groups in WF2 M180. It would appear that the drill group
retains no sense of pick order, so that no matter which way around I
choose the holes to make the group, the drilling sequence always drills
them in the same, different to how I would like, order.
Is this indeed the case? Or have I missed something as is so often the
case...
Many thanks,
(PS sequence parameter scan_type IS set to pick_order, I have just
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2019-10-23 15:12:12
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http://www.ams.org/bookstore?fn=20&arg1=mmonoseries&ikey=MMONO-25-D
|
New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education
Mahler's Problem in Metric Number Theory
SEARCH THIS BOOK:
Translations of Mathematical Monographs
1969; 192 pp; softcover
Volume: 25
ISBN-10: 0-8218-5344-9
ISBN-13: 978-0-8218-5344-3
List Price: US$99 Member Price: US$79.20
Order Code: MMONO/25.D
Suggest to a Colleague
Please note that AMS On Demand titles are not returnable.
This book deals with the solutions of a group of questions related both to the general theory of transcendental numbers and to the metrical theory of diophantine (and also algebraic) approximations. The fundamental problem in this field has been known in the literature since 1932 as Mahler's conjecture. The main result of this book is a proof of Mahler's conjecture and some analogous theorems.
In Part I, the "Classical" case of Mahler's conjecture, dealing with real and complex numbers, is considered. This part should be comprehensible to any who knows the elements of measure theory and possesses sufficient perseverance in over-coming purely logical difficulties. Part II is concerned with locally compact fields with nonarchimedean valuation. This part requires a general familiarity with the structure of fields with nonarchimedean valuation. All the necessary information is given in the text with references to the sources.
• Introduction
Part I. Real and complex numbers
• Auxiliary considerations
• The complex case
• The real case
Part II. Fields with non-archimedean valuation
• Basic facts
• Fields of $$p$$-adic numbers
• Fields of formal power series
• Supplementary results and remarks
• Conclusion
• An application
• Bibliography
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2013-12-19 02:02:26
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https://math.stackexchange.com/questions/1786269/how-to-find-the-minimum-value-for-frac-tan-xx-piecewise-defined
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# How to Find the Minimum Value for $\frac{\tan x}{x}$ (piecewise defined)?
$$F (x) = \begin{cases} \displaystyle \frac{\tan(x)}{x} & x \not=0 \\ 1 & x=0 \end{cases}$$
How do I prove that there is a local minimum at $x=0$?
• From where did you get this question? – Aritra Das May 15 '16 at 18:56
• I'm afraid he hasn't taught you much. Definitely not enough for this problem. – Aritra Das May 15 '16 at 20:37
• They are some of the first things that are taught in calculus. I'm in school and I know those. It's "fundamental" for a reason. It ties calculus neatly together. – Aritra Das May 15 '16 at 20:40
• I didn't learn it yet that's weird.. So anyways which solution should I use? I would need one that does not involve those theorems just trigonometric derivatives – Natalya May 15 '16 at 20:42
• Firstly, you should accept William's answer. It is undoubtedly the best since it lists many methods and he suggests many links which are useful. Secondly, if you wish to write down a solution, you should use @user5713492 's answer's first part (before the edit). That is probably what you're after. I hope I could be of some help. Cheers. – Aritra Das May 15 '16 at 20:51
The first derivative test seems easiest to apply here. For $x>0$, $$\frac d{dx}\frac{\tan x}x=-\frac1{x^2}\tan x+\frac1x\sec^2x=\frac{x-\sin x\cos x}{x^2\cos^2x}$$ By a geometric argument you were shown, around the time you learned the derivatives of the trigonometric functions, that for $0<x<\frac{\pi}2$, $\sin x\cos x<x<\tan x$. For this reason we know that $$\frac d{dx}\frac{\tan x}x=\frac{x-\sin x\cos x}{x^2\cos^2x}>0$$ for $0<x<\frac{\pi}2$. It follows that $-(x-\sin x\cos x)=(-x)-\sin(-x)\cos(-x)<0$ for $0<x<\frac{\pi}2$ so that $x-\sin x\cos x<0$ for $-\frac{\pi}2<x<0$, so $$\frac d{dx}\frac{\tan x}x=\frac{x-\sin x\cos x}{x^2\cos^2x}<0$$ for $-\frac{\pi}2<x<0$. Since $$\lim_{x\rightarrow0}\frac{\tan x}x=\lim_{x\rightarrow0}\left(\frac1{\cos x}\frac{\sin x}x\right)=(1)(1)=1=F(1)$$ Then $F(x)$ is continuous at $x=0$ and $F^{\prime}(x)<0$ for $-\frac{\pi}2<x<0$ and $F^{\prime}(x)>0$ for $0<x<\frac{\pi}2$ it follows that $F(x)$ has a relative minimum at $x=0$ by the first derivative test.
EDIT: Here is the geometric argument I was alluding to. Consider this figure:
It may be rather hard to read my crude diagrams, but it is supposed to depict the unit circle centered at $O$. Sector $OQS$ of the circle has area $$A(OQS)=\frac{\theta}{2\pi}\pi(1)^2=\frac12\theta$$ Where we have assumed that the area of a sector is proportional to its central angle, here $\theta$, and used the mensuration formula for the area of a circle, $A=\pi r^2$. Then right triangle $ORS$ lies entirely within sector $OQS$ and its area is $$A(ORS)=\frac12\cos\theta\sin\theta$$ where we have use the mensuration formula $A=\frac12(\text{base})(\text{height})$ for the area of a triangle. Finally the sector $OQS$ itself lies entirely within right triangle $AQT$ with area $$A(OQT)=\frac12(1)\tan\theta=\frac12\tan\theta$$ Since one is contained within another, $$A(ORS)<A(OQS)<A(OQT)$$ Or $$\frac12\sin\theta\cos\theta<\frac12\theta<\frac12\tan\theta$$ I thought that this was the argument made to show how to differentiate trigonometric functions through \begin{align}\frac d{dx}\sin x&=\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}h=\lim_{h\rightarrow0}\frac{\sin x\cos h+\cos x\sin h-\sin x}h\\ &=\lim_{h\rightarrow0}\frac{\sin x(1-2\sin^2(h/2))+\cos x\sin h-\sin x}h\\ &=\lim_{h\rightarrow0}\frac{\sin h}h\cos x-\lim_{h\rightarrow0}\frac{\sin(h/2)}{(h/2)}\sin(h/2)\sin x\\ &=(1)\cos x-(1)(0)\sin x=\cos x\end{align} Because through the geometric argument you knew that $$\cos x<\frac{\sin x}x<\sec x$$ And since $$\lim_{h\rightarrow0}\cos x=\lim_{h\rightarrow0}\sec x=1$$ It follows by the squeeze theorem that $$\lim_{h\rightarrow0}\frac{\sin x}x=1$$
• The $\frac{\pi}2$ restriction is not a problem because you only need an open interval including $x=0$ to evaluate limits and take derivatives and establish a local minimum or maximum. You can only hope for a local minimum because $$\lim_{x\rightarrow{\frac{\pi}{2}}^+}\frac{\tan x}x=-\infty$$ Just set your calculator in radians mode and start entering values of $x$ just a little bigger than $\frac{\pi}2$ to convince yourself of this. – user5713492 May 15 '16 at 19:56
Since you can rewrite $F$ as $$F(x)=\int_0^1{1\over\cos^2(\tau x)}\>d\tau\geq1=F(0)\qquad\left(-{\pi\over2}<x<{\pi\over2}\right)$$ the function $F$ takes its minimum in the interval $\>(-{\pi\over2},{\pi\over2})\$ at $x=0$.
(In what follows I am assuming that the domain here is $-\frac{\pi}{2} <x < \frac{\pi}{2}$)
For $x<0$, $\tan x < 0$, therefore $\frac{\tan x}{x} >0$.
Likewise, for $x>0$, we have $\tan x > 0$, and again $\frac{\tan x}{x} > 0$.
Hence in order to show that the modified piecewise version of $\frac{\tan x}{x}$ has a minimum at $x=0$, it suffices to show that $|\tan x| > |x|$ for all $x \in (-\frac{\pi}{2},0) \cup (0, \frac{\pi}{2})$.
This is because this condition would imply that $\frac{\tan x}{x} >1$ for those values.
This is clear from the following graph:
(again, I am assuming that $x \in(-\frac{\pi}{2},0)\cup (0, \frac{\pi}{2})$)
since the graphs do not intersect. (Note: $\frac{\tan x}{x} >1 \iff \tan x > x$)
For a better graphical proof of the fact that $$\tan x > x, \quad x \in (0, \frac{\pi}{2}$$ see here: http://mathrefresher.blogspot.com/2006/08/sin-x-x-tan-x-for-x-in-02.html
Non-Graphical Proof #1
However, I imagine that an analytic proof for the fact that $|\tan x|>|x|$ for these points would be preferred.
By symmetry (i.e. the fact that both tangent and $x$ are odd), it suffices to show this for just $x \in (0, \frac{\pi}{2})$ (in other words, the proof for $x \in (-\frac{\pi}{2},0)$ will be completely analogous up to changes in sign).
This follows from the Taylor expansion of $\tan x$ around $x=0$:
$$\tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \dots = \displaystyle\sum\limits_{n=0}^{\infty} \frac{U_{2n+1}x^{2n+1}}{(2n+1)!}$$
Hence, in particular, it follows that $\tan x > x + \frac{1}{3}x^3 > x$ on the interval in question $x \in (0, \frac{\pi}{2})$.
Completely analogously, $\tan x < x + \frac{1}{3}x^3 < x$ for $x \in (-\frac{\pi}{2},0)$.
Therefore, $|\tan x| > |x|$ for all $x \in (-\frac{\pi}{2},0) \cup (0, \frac{\pi}{2})$, it follows that in particular
$$\frac{\tan x}{x} > 1$$ for all $x \in (-\frac{\pi}{2},0) \cup (0, \frac{\pi}{2})$, and therefore the piecewise defined function in question has a minimum at $x=0$.
For proof without Taylor series:
Note that $x = \int_0^x 1 \text{d}t$ (Fundamental theorem of calculus, the fact that $x'=1$, and $x(0)=0$)
Meanwhile, because $\tan 0 = 0$, as well as the Fundamental Theorem of Calculus, $$\tan x = \int_0^x (\tan t)' \text{d}t = \int_0^x \sec^2 t \text{d}t = \int_0^x \displaystyle\frac{1}{\cos^2 t} \text{d}t$$
Since for $x \in (0, \frac{\pi}{2})$, we have
$$\cos x <1 \implies \cos^2 x < 1 \implies \displaystyle\frac{1}{\cos^2 x} > 1$$
it follows that (since $f>g \implies \int_0^x f \text{d}t > \int_0^x g \text{d}t$):
$$\int_0^x \displaystyle\frac{1}{\cos^2 t} \text{d}t > \int_0^x 1 \text{d}t$$
which is equivalent to:
$$\tan x > x,\ \text{for }x \in (0, \frac{\pi}{2})$$.
Using the fact that $\tan(-x)=-\tan x$, it follows immediately that $\tan x < X$ for $x \in (-\frac{\pi}{2},0)$.
Therefore $\left| \displaystyle\frac{\tan x}{x} \right| > 1$ for all $x \in (-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2})$, as claimed.
Another One:
By definition of $\cos x$, we have that $$\text{for }x \in (0, \frac{\pi}{2}), 1 < \cos x.$$
Now we show, using this fact, that $$\sin x < x \quad \forall\ x \in (0, \frac{\pi}{2}).$$
Note that $\frac{d}{dx} x =1$ and $\frac{d}{dx} \sin x = \cos x$.
Therefore,
$$\frac{d}{dx} \sin x < \frac{d}{dx} x\ \ \quad \forall\ x \in (0, \frac{\pi}{2})$$
As a result,
$$\int_0^x \frac{d}{dt}\sin t \text{d}t < \int_0^x \frac{d}{dt} t \text{d}t \quad \forall \ x \in (0, \frac{\pi}{2})$$
Because $\frac{d}{dx} x =1$ and $\frac{d}{dx} \sin x = \cos x$, it follows from the fundamental theorem of calculus (sorry but you're going to have learn it eventually and it's not difficult) that
$$\sin x = \int_0^x \frac{d}{dt}\sin t \text{d}t = \int_0^x \cos t \ \text{d}t$$ and $$x= \int_0^x \frac{d}{dt} t\ \text{d}t= \int_0^x 1\ \text{d}t.$$
Therefore it finally follows that
$$\sin x < x \quad \forall\ x\in (0,\frac{\pi}{2})$$.
From this follows immediately that
$$\displaystyle\frac{\sin x}{x} < 1.$$
Now, to show that $$\frac{\sin x}{x} < \displaystyle\frac{\tan}{x}$$
we use the facts that:
$$\tan x := \displaystyle\frac{\sin x}{\cos x}$$
and
$$\cos x < 1 \quad \forall\ x\in (0,\frac{\pi}{2})$$
Hence we get:
$$\displaystyle\frac{\tan x}{x} = \displaystyle\frac{\sin x}{\cos x \cdot x} = \displaystyle\frac{\sin x}{x} \cdot \displaystyle\frac{1}{\cos x} > \displaystyle\frac{\sin x}{x}$$
Then just use any of the arguments in the other answers which rely on these facts being true.
In any case, if you want an answer that's better than "look at this graph, see it's true" (which I already gave by the way), you'll have to use integration and differentiation together (to the best of my knowledge) i.e. the fundamental theorem of calculus.
• Thanks so much for the reply William! For tan (x)/x the restriction is that x can't be 0. Also is there a way to solve it without using the taylor series? (not familiar with it) – Natalya May 15 '16 at 16:32
• see my recent edit for a proof solely relying on the fundamental theorem of calculus -- the idea (in my opinion) is the same, although perhaps a little more indirect. – Chill2Macht May 15 '16 at 16:49
• Well $x=0$ isn't a global minimum for values of $x$ outside of that range -- also $\displaystyle \frac{\tan x}{x}$ isn't defined for $k\pi + \frac{pi}{2}$ where $k \in \mathbb{Z}$ i.e. is an arbitrary integer. What I have shown above is more than sufficient for demonstrating that $x=0$ is a local minimum, since I have shown that there exists an open neighborhood $(-\frac{\pi}{2},\frac{\pi}{2})$ for which $x=0$ is a minimum; again though, $x=0$ clearly is not a global minimum $\displaystyle\frac{\tan(\pi)}{\pi}=0$ for example, so I don't understand what else you could want me to prove. – Chill2Macht May 15 '16 at 17:33
• Since $x=0$ is not the minimum for $\displaystyle\frac{\tan x}{x}$ I can't prove that it is, only that it is a local minimum. Do you want a proof that doesn't use calculus? What are you familiar with? You tagged the post with "calculus" so I assumed that you knew at least the Fundamental Theorem of Calculus. – Chill2Macht May 15 '16 at 17:35
• @Natalya Actually it is. Read the question carefully. It says that $F(x)= \frac{\tan x}{x}$ only for $x$ in $(-\frac{\pi}{2},0)\cup(0, \frac{\pi}{2})$. And for $x=0$, $F(x) =1$. Anywhere else, the function is not defined. – Aritra Das May 15 '16 at 18:25
The function is not bounded. It does not have a maxima or a minima. As $x\to\frac{\pi}{2}^-$, $F(x)\to\infty$ and as $x\to\frac{\pi}{2}^+$, $F(x)\to-\infty$.
We already know that for positive $x\in (0, \frac{\pi}{2})$, $$\sin x < x < \tan x$$ Since $x\not=0$, we can divide by $x$ $$\frac{\sin x}{x} < 1 < \frac{\tan x}{x}$$ Hence, $\frac{\tan x}{x}$ is always greater than $1$ for positive $x$. Hence, $f(x)>1$ for all positive $x$.
Moreover, $$f(x) =\frac{\tan x}{x} = \frac{\tan (-x)}{(-x)} = f(-x)$$
Hence, for negative $x$ also, $f(x)>1$
Thus for both positive and negative $x$, $f(x)>1$. On the other hand, at $x=0$, your function is defined to be $1$.
Hence, the minimum occurs at $x=0$ and the minimum value is $1$.
• Do you have a proof for $$\sin x < x \ \text{for } x \in (0, \frac{\pi}{2})$$ that doesn't use the fundamental theorem of calculus? – Chill2Macht May 15 '16 at 19:02
• @William draw a unit circle and compare the area of triangle and arc. – user175968 May 15 '16 at 19:03
• That's not a proof though. And if you wanted to prove your comparison of the triangle and the arc, you would have to use integration and -- the fundamental theorem of calculus. – Chill2Macht May 15 '16 at 19:04
• Why isn't that a proof? – Aritra Das May 15 '16 at 19:11
• And what would I want to prove? Why would I use integration? The formulae for area of triange and arcs are pretty well known and standard. Why would I want to prove them? – Aritra Das May 15 '16 at 19:12
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2019-12-06 18:56:49
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https://homework.cpm.org/category/ACC/textbook/gb8i/chapter/9%20Unit%2010/lesson/INT1:%209.3.3/problem/9-94
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### Home > GB8I > Chapter 9 Unit 10 > Lesson INT1: 9.3.3 > Problem9-94
9-94.
At the Keaveny Hardware Manufacturing Company they need to meet very specific criteria for the diameter of the threads of #$10$ wood screws. The diameter of the threads must be $\frac { 3 } { 16 }$ of an inch. The width cannot vary more than $\frac { 1 } { 64 }$ of an inch from the specified diameter. Write and solve an appropriate absolute value inequality for this situation. Graph your solution.
Review Section 9.1.3 on absolute value equations and inequalities, specifically problem 9-29.
$|x-\frac{3}{16}|\le\frac{1}{64}$
$\frac{11}{64} \le x \le \frac{13}{64} \text{ inches}$
Use the eTool below to graph the inequalitiy.
Click link at the right to view full eTool version : Int1 9-94 HW eTool.
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2020-11-30 04:45:47
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https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Biophysics_241_-_Membrane_Biology/02%3A_Membranes_-_Aggregated_Lipids/2.07%3A_Diffusion_in_Membranes
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$$\require{cancel}$$
# 2.7: Diffusion in Membranes
Eukaryotic cells are surrounded by a flexible and dynamic barrier known as a membrane. These biological membranes are composed of lipids, which aggregate to form a bilayer with particular biochemical properties. The amphipathic nature of the lipid bilayer, whose tails are hydrophobic and associate with each other and whose head groups are hydrophilic and interact with the aqueous environment, are critical to its structure. The composition of the lipid bilayer is also important for the diffusion both across and within the membrane. This membrane diffusion is important for a variety of functions, some of which include regulating the fluidity of the membrane, the uptake of metabolites into the cell from the outside, and the removal of waste products from the inside of the cell.
## Fluid Mosaic Model
Each membrane protein has a particular orientation within the membrane and cannot flip-flop form one bilayer to the other after it has assumed its mature conformation. However, lateral movement within the same lipid bilayer is still possible. Lateral diffusion is a key feature of the fluid mosaic model of membrane structure that was first described in 1972 by S. Jonathan Singer and Garth Nicolson (1).
This model was supported by experiments previously done by L.D. Frye and M. Edidin in 1970, which showed that cells taken from mouse and human lines could be fused together using the Sedani virus (Figure $$\PageIndex{1}$$). The resulting fusion cell expressed both mouse and human antigens, which could be labeled indirectly by fluorescent antibodies and followed. Mixing of both parental antigens occurred forty minutes after fusion, suggesting that lateral diffusion within the membrane can occur (2). However, the time that it takes for lateral diffusion to occur depends on membrane fluidity, which ultimately depends on both temperature and lipid composition.
## Types of Diffusion Across the Plasma Membrane
There are general thermodynamic principles that govern the transfer of molecules across the membrane. Figure $$\PageIndex{2}$$ represents the equation that shows the amount of free energy that is required for a substrate to cross a membrane. In order for diffusion to occur, $$\Delta G$$ must be negative and as $$\Delta G$$ moves away from zero and becomes more positive, work becomes required. When the concentrations become equal on both sides of the membrane and $$\Delta G=0$$ and the rates of transport in both directions will be the same and no net transport will occur.
$\Delta G =RT \ln \left( \dfrac{C_2}{C_1}\right)$
If C2, the concentration of a substrate in the cytosol, is less than C1, the concentration of a substrate outside of the cell, then $$\Delta G$$ is negative and the process is favorable. Gradually, as more substrate is transferred across the membrane, C1 decreases as C2 increases until C2 = C1and at this point delta G =0 and the system is at equilibrium.
### Simple Diffusion
Simple diffusion occurs by the diffusion of molecules, such as O2 and CO2, across the hydrophobic core of the membrane (Figure $$\PageIndex{3}$$). Therefore, no ATP is required for this type of diffusion across the membrane, it is simply a matter of molecules moving down a concentration gradient. As simple diffusion does not require ATP, large polar molecules or ions cannot diffuse across the membrane. This is due to the hydrophobic tail region of the membrane, which presents too large of an energy barrier to be overcome by the potential stored in the gradient.
The net rate of transport is proportional to the concentration difference (C2 – C1) across the membrane (Figure $$\PageIndex{4}$$).
$J = - \dfrac{KD_1(C_2-C_1)}{l}$
where
• $$J$$ is the net rate of transport,
• $$K$$ is the partition coefficient for the ratio of solubilities of the material in lipid and water,
• $$D_1$$ is the diffusion coefficient of the diffusing substance in the membrane, and
• $$l$$ is the thickness of the membrane.
For ions and other hydrophilic substances, K is a very small number, given that diffusion of such molecules across the membrane is very slow.
## Facilitated Transport
As opposed to simple diffusion, which does not require ATP, facilitated transport needs ATP in order to overcome the energy barrier of the hydrophobic tail region of the membrane. In addition, this type of diffusion is dependent on cargo binding the membrane-embedded channel or carrier protein.There are two types of facilitated transport, pore-facilitated transport and carrier-facilitated transport. In order to distinguish between pore-facilitated transport and carrier-facilitated diffusion, one can fluctuate the membrane fluidity by altering the temperature. This change in temperature will stop carrier-facilitated diffusion due to the fact that carrier-facilitated diffusion must move through the membrane in order to function and cannot do so when the membrane is in a non-fluid state.
### Pore-Facilitated Transport
Pore-facilitated transport uses proteins embedded within the membrane that can open and close in order to facilitate diffusion. This type of diffusion allows for selected ions to pass through the pore, such as Cl-. An important example of pore-facilitated transport is the transport of glucose through the use of a gated pore mechanism, in which the pore is never open at both ends at once (Figure $$\PageIndex{5}$$). Instead, the pore opens at the exterior to allow the entry of glucose, closes the exterior opening, opens the interior opening, releases glucose into the cytosol, and finally returns to its state of binding on the exterior (15).
### Carrier-Facilitated Transport
Antibiotic ionophores, such as valinomycin, a cation carrier, are an example of carrier-facilitated transport. Its folded conformation allows for the protein to have an outer hydrophobic surface, making it soluble in the lipid bilayer, with an internal conformation that mimics the hydration shell that the cation would have in an aqueous solution. This conformation allows valinomycin to diffuse from one surface of a membrane, pick up an ion, and then diffuse to the other surface to release it (Figure $$\PageIndex{6}$$).
## Factors that Influence Membrane Diffusion
There are several factors that can influence the diffusion of molecules both within and across a membrane; physical barriers, electrostatic attraction or repulsion nodes, and partitioning phenomena (3), are just a few examples.
### Physical Obstacles
Physical obstacles can become considerably crowded, which can obstruct the free passage of molecules. Several adaptor proteins (alpha-actinin, talin, vinculin, etc.) can attach the cortical cytoskeleton to the long cytosolic tails of transmembrane proteins within the plasma membrane and act as a fence, which restricts the movement of molecules across the membrane (4,5). Such obstruction by the cortical cytoskeleton has lead to the observation that molecules undergo a “hop diffusion” in which they “hop” intermittently between confinement zones (4,5) in order to diffuse (Figure7).
Membrane-matrix junctions can also contribute to a physical obstruction that impedes membrane diffusion. This is due to the binding between integrin receptors located in the plasma membrane and the extracellular matrix (fibrous network of proteins that cells attach to) (Figure $$\PageIndex{8}$$). If this interaction is increasingly accumulated at a high enough density, then diffusion is limited. A recent study showed that this type of physical barrier blocked the diffusion of membrane molecules whose dimensions exceeded the width of the integrin-matrix interaction (6).
## Electrostatic Impediments
Electrostatic interactions can also interfere with free diffusion, as charged proteins or lipids can be repelled by like charges or attracted by opposite ones (Figure $$\PageIndex{9}$$). McLaughlin and Murray in 2005 showed that proteins have natively unfolded regions that have both basic and hydrophobic residues that allow them to exist within the bilayer and at the same time attract anionic lipids electrostatically (7). As a result, the membrane-associated cationic residues associate together, creating a negatively charged ring around the protein. Thus, the resulting ring of anionic lipids can alter the mobility of other charged molecules in the plane of the membrane (8).
## Partition-Induced Barriers
In a membrane, certain types of lipids or proteins can partition into defined regions, resulting in an area of subdiffusive behavior (3). The association between lipids and proteins is often driven by the recognition of particular binding domains, for example, the association of protein PH domains with phosphoinositides (9). Another way for these lipid-protein complexes to form is through hydrophobic interactions. An example of this hydrophobic interaction-driven complex is the saturated lipid- and cholesterol- rich microdomains, known as lipid rafts (10). Lastly, membrane curvature can play a role in the obstruction of diffusion (Figure10). The bent region of the membrane has distinct properties, as the head groups of the lipids constituting the concave monolayer are unusually close, whereas the head groups on the convex monolayer are uncharacteristically far apart. The concave side of the bent bilayer can alter diffusion physically or electrostatically, while the convex side creates a more accessible membrane due to reduced packing of the head groups.
## Techniques to Monitor Diffusion
• Fluorescence recovery after photobleaching (FRAP) was initially very useful in studying the lateral diffusion of membrane components (11). This technique uses fluorescently labeled probes to follow a molecule of interest. By using a high intensity laser, the fluorophores in a region of interest will become bleached and lose signal (Figure11). The probes that were not bleached will then diffuse throughout the sample and replace the bleached region. Since this method measures the average behavior of molecule, it ultimately has limited temporal resolution (seconds).
• Fluorescence correlation spectroscopy (FCS) is a correlation analysis that measures fluctuation in fluorescence intensity. This technique provides high spatial precision and can measure diffusion coefficients of molecules (12,13). Unlike FRAP, FCS can acquire the time resolution, but lacks the ability to capture defined transient events.
• Imaging total internal reflection (ITIR)-FCS is a technique that can circumvent both of the problems in other techniques, as it can probe diffusion in membranes with good temporal and spatial resolution (12). ITIR-FCS can be applied to diffusion, active transport, or even both.
## References
1. Singer S.J. and Nicolson G.L. 1972. The Fluid Mosaic Model of the Structure of Cell Membranes. Science. 175:720-31.
2. Frye L.D. and Edidin M. 1970. The Rapid Intermixing of Cell Surface Antigens After Formation of Mouse-Human Heterokaryons. Journal of Cell Science. 7:319-35.
3. Mathews C.K. and Van Holde K.E. 1996. Biochemistry. Second Edition. Menlo Park, CA: The Benjamin/Cummings Publishing Company, Inc.
4. Fujiwara T.K., Ritchie H., Murakoshi K., Jacobson, Kusumi 2002. Phospholipids undergo hop diffusion in compartmentalized cell membrane. Journal of Cell Biology. 157:1071-82.
5. Suzuki K.G., Fujiwara T.K., Sanematsu F., Iino R., Edidin M., Kusumi A 2007. GPI-anchored receptor clusters transiently recruit Lyn and G-alpha for temporary cluster immobilization and Lyn activation: single molecule tracking study. Journal of Cell Biology. 177:717-30.
6. Paszek M.J., DuFort C.C., Rossier O., Bainer R., Mouw J.K., Godula K., Hudak J.N., Lakins A.C., Wijekoon L., Cassereau 2014. The cancer glycocalyx mechanically primes integrin-mediated growth and survival. Nature. 511:319-25.
7. McLaughlin S. and Murray D. 2005. Plasma membrane phosphoinositide organization by protein electrostatics. Nature. 438:605-611.
8. Van den Boggart G., Meyenberg K., Risselada H.J., Amin K.I., Willig B.E., Hubrich M., Dier S.W. Hell, H., Grubmuller U., Diederichsen, Jahn R., 2011. Membrane protein sequestering by ionic protein-lipid interactions. Nature. 479: 552-55.
9. Trimble W.S. and Grinstein S. 2015. Barriers to the free diffusion of proteins and lipids in the plasma membrane. Journal of Cell Biology. 208:259- 71
10. Lemmon M.A. 2008. Membrane recognition by phospholipid-binding domains. Nature Review Molecular and Cell Biology. 9: 99-111.
11. Lingwood D. and Simons K., 2010. Lipid rafts as a membrane-organizing principle. Science. 327: 46-50.
12. Chen Y., Lagerholm B.C., Yang B., Jacobson K., 2006. Methods to measure the lateral diffusion of membrane lipids and proteins. Methods. 39:147-53.
13. Sankaran J., Manna M., Guo L., Kraut R., Wohland T. 2009. Diffusion, transport, and cell membrane organization investigated by image fluorescence cross-correlation spectroscopy. Biophysical Journal. 97:2630-39.
14. Magde D., Elson E.L., Webb W.W. 1974. Fluorescence correlation spectroscopy. Biopolymers. 17:361-76.
15. Oka Y., Asaon T., Shibasaki Y., Lin J.L., Tsukuda K., Katagiri H., Akanuma Y., Takaku F. 1990. C-terminal truncated glucose transporter is locked into an inward-facing form without transport activity. Nature. 345:550-53.
2.7: Diffusion in Membranes is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.
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2022-06-28 21:47:39
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https://cantera.org/tutorials/ck2yaml-tutorial.html
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## Converting Chemkin-format files¶
If you want to convert a Chemkin-format file to YAML format, or you're having errors when you try to do so, this section will help.
## ck2yaml¶
Many existing reaction mechanism files are in CK format, by which we mean the input file format developed for use with the Chemkin-II software package (and subsequent releases) as specified in the report describing the Chemkin software [SAND89].
Cantera comes with a converter utility program ck2yaml (or ck2yaml.py) that converts CK format into Cantera format. This program should be run from the command line first to convert any CK files you plan to use into Cantera format (YAML format). (New in Cantera 2.5)
Usage:
ck2yaml [--input=<filename>]
[--thermo=<filename>]
[--transport=<filename>]
[--surface=<filename>]
[--extra=<filename>]
[--name=<name>]
[--output=<filename>]
[--permissive]
[--quiet]
Each of the terms in square brackets is an option that can be passed on the command line to ck2yaml.
• --input: This is the chemistry input file, containing a list of all the element names that are used, a list of all the species names, and a list of all the reactions to be considered between the species. This file can also optionally contain thermodynamic information for the species.
• --thermo: If the --input file does not contain the thermodynamic data, a separate file containing this information must be specified to the --thermo option.
• --transport: The --input file can also optionally contain transport information for the species. If it does not, and the user wishes to use a part of Cantera that relies on some transport properties, the --transport option must be used to specify the file containing all the transport data for the species.
• --surface: For surface mechanisms, this file defines the surface species and reactions occurring on the surface. Gas phase species and reactions are defined in the file specified by the --input option.
• --extra: This option specifies a YAML file which can be used to add to the description field or to define custom fields that are included in the YAML output.
• --name: This specifies the name of the phase in the resulting YAML file. The default is gas.
• --output: Specifies the output file name. By default, the output file name is the input file name with the extension changed to .yaml.
• --permissive: This option allows certain recoverable parsing errors (for example, duplicate thermo data) to be ignored.
• --quiet: Suppresses warning messages, such those about duplicate thermo data.
Example:
ck2yaml --input=chem.inp --thermo=therm.dat --transport=tran.dat
If the ck2yaml script is not on your path but the Cantera Python module is, ck2yaml can also be used by running:
python -m cantera.ck2yaml --input=chem.inp --thermo=therm.dat --transport=tran.dat
An input file containing only species definitions (which can be referenced from phase definitions in other input files) can be created by specifying only a thermo file.
## Debugging common errors in CK files¶
Note
Many existing CK format files cause errors in ck2yaml when they are processed. Some of these errors may be avoided by specifying the --permissive option. This option allows certain recoverable parsing errors (for example, duplicate transport or thermodynamic data) to be ignored. Other errors may be caused by incorrect formatting of lines in one or more of the input files.
When ck2yaml encounters an error, it attempts to print the surrounding information to help you to locate the error. Many of the most common errors are due to an inconsistency of the input files from their standard, as defined in the report for Chemkin referenced above. These errors include:
• Each section of the input files must start with a keyword representing that section and end with the keyword END. Keywords that may begin a section include:
• ELEMENTS or ELEM
• SPECIES or SPEC
• THERMO or THERMO ALL
• REACTIONS or REAC
• TRANSPORT
• The thermodynamic data is read in a fixed format. This means that each column of the input has a particular meaning. Many common errors are generated because information is missing or in the wrong column. Check thoroughly for extraneous or missing spaces. The format for each thermodynamic entry should be as follows:
N2 N 2 G200.000 6000.000 1000.00 1
2.95258000E+00 1.39690000E-03-4.92632000E-07 7.86010000E-11-4.60755000E-15 2
-9.23949000E+02 5.87189000E+00 3.53101000E+00-1.23661000E-04-5.02999000E-07 3
2.43531000E-09-1.40881000E-12-1.04698000E+03 2.96747000E+00 4
The following table is adapted from the Chemkin manual [SAND89] to describe the column positioning of each required part of the entry. Empty columns should be filled with spaces.
Line No.
Contents
Column
1
Species Name
1–18
1
Date (Optional)
19–24
1
Atomic Symbols and formula
25–44
1
Phase of species (S, L, G)
45
1
Low temperature
46–55
1
High temperature
56–65
1
Common temperature
66–73
1
74–78
1
The integer 1
80
2
Coefficients $$a_1$$ to $$a_5$$ for the upper temperature interval
1–75
2
The integer 2
80
3
Coefficients $$a_6,\ a_7$$ for the upper temperature interval, and $$a_1,\ a_2,\ a_3$$ for the lower temperature interval
1–75
3
The integer 3
80
4
Coefficients $$a_4$$ through $$a_7$$ for the lower temperature interval
1–60
4
The integer 4
80
The first 18 columns are reserved for the species name. The name assigned to the species in the thermodynamic data must be the same as the species name defined in the SPECIES section. If the species name is shorter than 18 characters, the rest of the characters should be filled by spaces. The next six columns (columns 19–24) are typically used to write a date; they are not used further. The next 20 columns (25–44) are used to specify the elemental composition of the species. In column 45, the phase of the species (S, L, or G for solid, liquid, or gas respectively) should be specified. The next 28 columns are reserved for the temperatures that delimit the ranges of the polynomials specified on the next several lines. The first two temperatures have a width of 10 columns each (46–55 and 56–65), and represent the lowest temperature and highest temperature for which the polynomials are valid. The last temperature has a width of 8 columns (66–73) and is the common temperature, where the switch from low to high occurs. The next 5 columns (74–78) are reserved for atomic symbols and are usually left blank for the default behavior. Column 79 is blank and finally, the row is ended in column 80 with the integer 1.
The next three lines of the thermodynamic entry have a similar format. They contain the coefficients of the polynomial described in Thermodynamic Property Models for the NASA 7-coefficient polynomial formulation. The second row of the thermo entry (the first after the information row) contains the first five coefficients that apply to the temperature range between the midpoint and the upper limit. 15 columns are alloted for each coefficient (for a total of 75 columns), with no spaces between them. Although the entry above shows spaces between positive coefficients, it is to be noted that this is done only for formatting consistency with other lines that contain negative numbers. After the coefficients, four spaces in columns 76–79 are followed by the integer 2 in column 80. On the next line, the last two coefficients for the upper temperature range and the first three coefficients for the lower temperature range are specified. Once again, this takes up the first 75 columns, columns 76–79 are blank, and the integer 3 is in column 80. Finally, on the last line of a particular entry, the last four coefficients of the lower temperature range are specified in columns 1–60, 19 blank spaces are present, and the integer 4 is in column 80. The 19 blank spaces in the last line are part of the standard. However, since the original Chemkin interpreter ignored those spaces, researchers began using that space to store additional information that was not necessary for the input file. Although these numbers create an error in ck2yaml if present, they are harmless and can be ignored by using the --permissive option.
If the number of atoms of an element in a thermodynamic entry has more than 3 digits, it will cause a conversion error. To avoid the error, the element symbol should have a 0 in the first line of the entry. An ampersand (&) is added after the index of the first line, and the element symbols and their amounts should be written on the next line as follows:
BIN6J PYRENEJ1 C 0H 0 0 0G 300.000 5000.000 1401.000 1&
C 778 H 263
3.63345177E+01 3.13968020E-02-1.09044660E-05 1.71125597E-09-1.00056355E-13 2
4.05143093E+04-1.77494305E+02-1.20603441E+01 1.59247554E-01-1.41562602E-04 3
6.26071650E-08-1.09305161E-11 5.56473533E+04 7.68451211E+01 4
or on separate lines with ampersand (&) as the last character on the line:
BIN6 PYRENE C 0H 0 0 0G 300.000 5000.000 1401.000 1&
C 778&
H 264
3.65839677E+01 3.36764102E-02-1.16783938E-05 1.83077466E-09-1.06963777E-13 2
9.29809483E+03-1.81272070E+02-1.29758980E+01 1.63790064E-01-1.43851166E-04 3
6.31057915E-08-1.09568047E-11 2.48866399E+04 7.94950474E+01 4
• It may be the case that scientific formatted numbers are missing the E. In this case, numbers often show up as 1.1+01, when they should be 1.1E+01. You can fix this with a Regular Expression "find and replace":
Find: (\d+\.\d+)([+-]\d+)
Replace: \1E\2
• The transport data file also has a specified format, as described in [SAND98], although the format is not as strict as for the thermodynamic entries. In particular, the first 15 columns of a line are reserved for the species name. One common source of errors is a species that is present in the transport data file, but not in the thermodynamic data or in the species list; or a species that is present in the species list but not the transport data file. The rest of the columns on a given line have no particular format, but must be present in the following order:
Parameter Number
Parameter Name
1
An integer with value 0, 1, or 2 indicating monatomic, linear, or non-linear molecular geometry.
2
The Lennard-Jones potential well depth $$\varepsilon/k_B$$ in Kelvin
3
The Lennard-Jones collision diameter $$\sigma$$ in Angstrom
4
The dipole moment $$\mu$$ in Debye
5
The polarizability $$\alpha$$ in Angstroms cubed
6
The rotational relaxation collision number $$Z_{rot}$$ at 298 K
Another common error is if all 6 of these numbers are not present for every species.
SAND89(1,2)
See R. J. Kee, F. M. Rupley, and J. A. Miller, Sandia National Laboratories Report SAND89-8009 (1989). http://www.osti.gov/scitech/biblio/5681118
SAND98
See R. J. Kee, G. Dixon-Lewis, J. Warnatz, M. E. Coltrin, J. A. Miller, H. K. Moffat, Sandia National Laboratories Report SAND86-8246B (1998).
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2021-09-25 22:06:08
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http://math.stackexchange.com/questions/14534/big-oh-for-function-of-two-variables
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# Big-oh for function of two variables
Is it true that $O(M^3 + NM^2) \, = \, O(M^3 + N)$, where $M$ and $N$ are variables of the function?
-
Set $M=0$; then $M^3 + NM^2$ vanishes, but $M^3 + N$ does not. – user1119 Dec 16 '10 at 16:43
Technically, setting M=0 doesn't quite work, since the bound only has to hold for all M,N sufficiently large. – Jeremy Hurwitz Dec 16 '10 at 17:50
It is true that $O(M^3+N)\subset O(M^3+NM^2)$, since for $M$ sufficiently large, $c(M^3+N)<c(M^3+NM^2)$. It's worth noting that for $M$ sufficiently close to $0$, $M^3+N>M^3+NM^2$.
For the reverse inclusion, consider $f(M,N)=M^3+NM^2$. Clearly $f\in O(M^3+NM^2)$. We now need to show that for all $c$ and arbitrarily large $M,N$ that $f(M,N)>c(M^3+N)$. We can do this by setting $M=\log N$. We then have $f(\log N,N)=\log^3 N + N\log^2 N$, which is clearly larger than $c(\log^3 N + N)$ for all sufficiently large $N$. So $f(M,N)\not\in O(M^3+N)$.
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2014-08-28 05:32:13
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https://www.physicsforums.com/threads/physics-problem.46439/
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# Physics Problem
1. Oct 6, 2004
### sb_4000
Hi,
I think I posted this in the wrong place, so im posting it here. I'm taking Physics w/ cal 1, and Im having problems solving a couple of question, I was hoping some one could help, I would appreciated it..
1) block of mass m1=8kg is connected over an idead (masless and frictionless) pulley to block 2, of a mass m2=4kg. The magnitude of blocks acceleration is 3 m/s^2 and the angle of incline Theta=30 deg.
a) find the coefficient of the kinetic friction between block 2 and the plane of the incline.
b) what is the magnitude of the tension in the rope.
and also this problem which is projectile motion -
2) Projectile is fired from the top of a 40m tower at an angle of 60 deg above the horizontal hits the ground at the point 100m from the base of the tower.
a) find the speed at which the stone was thrown.
b) find the speed of the stone just before it hits the ground.
c)find the time at which the projectile hits the ground.
d) find the maximum height.
for a) i got 35 m/s
Thank You.
2. Oct 6, 2004
### Pyrrhus
Do a freebody diagram for the first problem, If the pulley is ideal then it has the same tension throughout it.
For the second problem consider.
Vy = 0 at max height
Y = 0 when it hits the ground.
3. Oct 6, 2004
### sb_4000
do you know how to get Part a in problem 1. Im having problems setting up the equation.
4. Oct 6, 2004
### Pyrrhus
Yes, i know.
Here's a hint:
Put your coordinate system with the same inclination as theta, so the normal force will have one non-zero component, while the gravity will have two non-zero components. Also consider the other two forces Tension and friction on the coordinate system.
5. Oct 6, 2004
### sb_4000
Thanks for responding to my question.
In problem two part "b" Is it F= w*sin(30)= 4N..this is what I get, Im probably doing it wrong. and In part one I still dont undrestand how to solve the problem.
6. Oct 6, 2004
### Pyrrhus
Draw a freebody diagram, you can use the hint i said, to solve it.
Do you know your geometry (specially trigonometry) well?
If you still don't understand, let me know.
7. Oct 6, 2004
### Gokul43201
Staff Emeritus
I can't seem to say this enough (as, I imagine does cyclo) ...yet it seems, no one listens : Draw the free-body diagram. That's half the work in the problem.
8. Oct 6, 2004
### sb_4000
Was my answer for "b" right?
Im still working on part a.
9. Oct 6, 2004
### Gokul43201
Staff Emeritus
Did you draw the free body diagram ?
10. Oct 6, 2004
### sb_4000
Yes I did.
I used the book as guide to draw it.
I also worked on part a, and I get 0.38 N.
is this right?
Last edited: Oct 6, 2004
11. Oct 6, 2004
### Pyrrhus
no, that's not correct.
12. Oct 6, 2004
### sb_4000
I also get 0.64 when I use cos(theta).
Im using the formula F=Mu_k*mg
13. Oct 6, 2004
### Pyrrhus
Try to make the equations again remembering this
$$\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}$$
14. Oct 6, 2004
### sb_4000
i used N=mg*cos(theta), F=mg*sin(theta)
and then mu_k = F/N and I get 0.57 is this one right?
15. Oct 6, 2004
### Pyrrhus
Ok let's try this with a different approach, you're beginning to understand, but still not there.
What are the forces acting on the x-axis on the first and second block?
What are the forces acting on the y-axis on the first and second block?
16. Oct 6, 2004
### sb_4000
i also did it using F_k = ma, N= mg, and F/N, and for this one i get 0.30
17. Oct 6, 2004
### Pyrrhus
18. Oct 6, 2004
### sb_4000
for the second block, the forces acting on it are Gravity and Normal force.
for the first block g, N, T
19. Oct 6, 2004
### Pyrrhus
Aren't both blocks tied together to a string?
Do you know what's an ideal pulley?
What is the normal force?
What is friction and how does it affect movement?
20. Oct 6, 2004
### sb_4000
Yes both blocks are tied to a string. I do not know what the ideal pulley is. The normal force is m*a, is the the upward motion of the block. The friction I believe slows down the movement of the block.
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2017-03-23 18:51:17
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https://www.biostars.org/p/168804/
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What is the relationship between PLink ped files and tped files
1
3
Entering edit mode
6.8 years ago
haohanw ▴ 90
I wonder what is the relationship between Plink .tped and .ped files. From what I observe, it seems it is more complicated than a simple transpose.
For example, in Section 4.1.1 of this manual, there is an example as following:
1 1 0 0 1 1 1 1 G G
1 2 0 0 2 1 0 0 A G
1 3 0 0 1 1 1 1 A G
1 4 0 0 2 1 2 1 A A
is transposed as
1 snp1 0 10001 1 1 0 0 1 1 2 1
1 snp2 0 20001 G G G A G A A A
# ^ ^ ^ ^
but instead of, what I thought should be:
1 snp1 0 10001 1 1 0 0 1 1 2 1
1 snp2 0 20001 G G A G A G A A
# ^ ^ ^ ^
Why there is a reverse relationship here?
And I think this reverse is not guaranteed to happen, for the reasons that in example of Section 3.4 of the same manual, it's hard to tell if there is any pattern for whether should be reversed or not.
(I am quite new to this area, and I hope the reason is not something very superficial as common sense in this domain)
plink SNP GWAS • 3.5k views
4
Entering edit mode
6.8 years ago
Interesting, I didn't know about that! Could it be that PLINK internally just sorts the alleles using some arbitrary rules?
I just ran a test with input alleles "G A", "A G" in various combinations with other SNPs and they always came out as "G A" in the transposed dataset.
Similarly, "G T", "T G" always becomes "G T", "G C", "C G" always becomes "G C" etc. "A T"/"T A" is always "A T", "A C"/"C A" becomes "A C", "G C"/"C G" becomes "G C". It can't be alphabetically sorted for obvious reasons.
The funny thing is, if I repeat the same thing using PLINK2, I get alphabetically sorted alleles: your example becomes G G A G A G A A (and my test-cases become alphabetically sorted, too). That makes me think that it's rather arbitrary and doesn't particularly matter.
Edit: I think it has to do with the way PLINK 1.07 stores genotypes as numbers - if you run
plink --file mytest --recode --transpose
you get the above inconsistent behaviour, but if you run
plink --file mytest --recode12 --transpose
so that all genotypes become numerically recoded, you'll always see "1 2" for all test cases, so these genotypes seem to be not alphabetically, but numerically sorted!
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2022-09-30 03:23:06
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http://eprints.maths.manchester.ac.uk/2728/
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# Accurate Computation of the Log-Sum-Exp and Softmax Functions
Blanchard, Pierre and Higham, Desmond J. and Higham, Nicholas J. (2019) Accurate Computation of the Log-Sum-Exp and Softmax Functions. [MIMS Preprint]
## Abstract
Evaluating the log-sum-exp function or the softmax function is a key step in many modern data science algorithms, notably in inference and classification. Because of the exponentials that these functions contain, the evaluation is prone to overflow and underflow, especially in low precision arithmetic. Software implementations commonly use alternative formulas that avoid overflow and reduce the chance of harmful underflow, employing a shift or another rewriting. Although mathematically equivalent, these variants behave differently in floating-point arithmetic. We give rounding error analyses of different evaluation algorithms and interpret the error bounds using condition numbers for the functions. We conclude, based on the analysis and numerical experiments, that the shifted formulas are of similar accuracy to the unshifted ones and that the shifted softmax formula is typically more accurate than a division-free variant.
Item Type: MIMS Preprint log-sum-exp, softmax, floating-point arithmetic, rounding error analysis, overflow, underflow, condition number MSC 2010, the AMS's Mathematics Subject Classification > 65 Numerical analysis Nick Higham 08 Sep 2019 11:10 08 Sep 2019 11:10 http://eprints.maths.manchester.ac.uk/id/eprint/2728
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2021-02-26 10:45:09
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https://www.physicsforums.com/threads/geometry-help-with-theorem-proof-please.737850/
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# Geometry - Help with theorem proof please
Geometry -- Help with theorem proof please
## Homework Statement
Let ##A,B,C,D## be points. If ##\vec{AB} = \vec{CD}## then ##A=C##.
None
## The Attempt at a Solution
This question was a theorem in my book that wasn't proved. I am wondering how to prove it?
It is saying that the vertex ##A## must equal ##C## if the ray ##\vec{AB} = \vec{CD}##.
The definition I have for ray is:
##\vec{AB} = \vec{AB} \cup \{ C \in P \ | \ A-B-C\}.## Where ##A-B-C## means ##B## is between ##A## and ##C##. And ##P## is the set of points.
Last edited:
tiny-tim
Homework Helper
Hi Lee33!
(your definition doesn't look quite correct)
Suppose A ≠ C
A is in ##\vec{CD}##, so … ?
tiny-tim - Can you elaborate a bit more please?
tiny-tim
Homework Helper
Hi Lee33!
Apply the definition you were given …
The definition I have for ray is:
##\vec{AB} = \vec{AB} \cup \{ C \in P \ | \ A-B-C\}.## Where ##A-B-C## means ##B## is between ##A## and ##C##. And ##P## is the set of points.
Suppose A ≠ C
C is in ##\vec{AB}##, so what can you say about A B and C ?
If C is in ##\vec{AB}## and ##C\ne A## then B is between A and C?
Mark44
Mentor
The definition I have for ray is:
##\vec{AB} = \vec{AB} \cup \{ C \in P \ | \ A-B-C\}.## Where ##A-B-C## means ##B## is between ##A## and ##C##. And ##P## is the set of points.
This definition makes no sense to me. First off, why would ##\vec{AB}## be equal to itself union some other thing (unless the other thing happened to be the empty set).
Second, how do you interpret ##\{ C \in P \ | \ A-B-C\}##? Does | have its usual meaning of "such that" or am I missing something? An explanation, in words, would be helpful.
Third, where are these points? Are they on a line or are they in the plane?
Fourth, how do you get that A - B - C means that B is between A and C?
Sorry, I will elaborate.
First question: If A and B are distinct points in a metric geometry then the line segment from A to B is the set ##\vec{AB}=\{C \in P \ | \ A-C-B \ or \ C = A \ or \ C = B\}##.
If A and B are distinct points in a metric geometry then the ray from A toward B is the set ##\vec{AB}=\vec{AB}\cup \{C\in P \ | \ A-B-C\}.##
Second question: Yes, it means such that. Let P be the set of points in a metric geometry, and let C be a point in P such that B is between A and C.
Third question: They are on a line.
Fourth: That is just a notation for convenience. ##A-B-C## just means B is between A and C.
I will add the definition of between-ness: B is between A and C if the distance ##d(A,B)+d(B,C) = d(A,C)##.
tiny-tim
Homework Helper
Hi Lee33!
(just got up :zzz:)
If C is in ##\vec{AB}## and ##C\ne A## then B is between A and C?
nooo, C is (strictly) between A and B
ok, and if A is in ##\vec{CD}##, then … ?
If A is in ##\vec{CD}## then A is between C and D.
tiny-tim
Homework Helper
If A is in ##\vec{CD}## then A is between C and D.
yes (strictly between)
ok, now you have two statements, and you should be able to prove a contradiction (thereby showing that "A ≠ C" was false)
(drawing yourself a diagram might help)
Alright thanks for the help! I will use your hints.
Question. Do I use both statements in my proof? That is, suppose ##A\ne C## and A is in ##\vec{CD}## then A is bewteen C and D. Also, I will use if ##A\ne C## and C is in ##\vec{AB}## then C is between A and B?
tiny-tim
Homework Helper
Question. Do I use both statements in my proof? That is, suppose ##A\ne C## and A is in ##\vec{CD}## then A is bewteen C and D. Also, I will use if ##A\ne C## and C is in ##\vec{AB}## then C is between A and B?
yes
You are using the same notation for line segment and ray, and it's confusing the bejeesus out of the people who are trying to help you.
Might I suggest ##\overline{AB}## for the segment and ##\overrightarrow{AB}## for the ray so that ##\overrightarrow{AB}=\overline{AB}\cup \{C\in P \ | \ A-B-C\}.##
tiny-tim
Homework Helper
… it's confusing the bejeesus out of the people who are trying to help you.
it's not confusing me
it's not confusing me
Are you sure? Like, really sure? Because when Lee asked
If C is in ##\vec{AB}## and ##C\ne A## then B is between A and C?
you replied
nooo, C is (strictly) between A and B
which is generally false regardless of which of Lee's two definitions of ##\vec{AB}## you're using. :tongue:
tiny-tim
Homework Helper
… which is generally false …
well, Lee33 didn't contradict me, sooo i assume i got it right!
gopher_p - Sorry about that, you're right!
tiny-tim - If ##A\ne C## then ##C\in \vec{AB}## thus ##A-C-B## but where will the point ##D## be?
tiny-tim
Homework Helper
but you haven't used …
If A is in ##\vec{CD}## then A is between C and D.
So my proof should go like this:
Suppose ##A\ne C##, now since ##\vec{AB}=\vec{CD}## then ##A\in \vec{CD}## and ##C\in \vec{AB}##. Thus ##C-A-D## and ##A-C-B## which is a contradiction?
tiny-tim
Homework Helper
yes!!
if i'm understanding the terminology correctly, you can't have both ##A-C## and ##C-A## unless C = A
1 person
Thank you very much for the help!
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2021-10-16 08:32:47
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https://www.hackmath.net/en/math-problem/4897
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# Right triangle trigonometrics
Calculate the size of the remaining sides and angles of a right triangle ABC if it is given: b = 10 cm; c = 20 cm; angle alpha = 60° and the angle beta = 30° (use the Pythagorean theorem and functions sine, cosine, tangent, cotangent)
Correct result:
a = 17.3205 cm
C = 90 °
#### Solution:
$C=180-\left(60+30\right)=9{0}^{\circ }$
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#### You need to know the following knowledge to solve this word math problem:
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2020-09-21 05:44:59
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http://openstudy.com/updates/510570a1e4b03186c3f9b122
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## bettyboop8904 Group Title Can anyone explain how to differentiate exponential functions? I have the Calc 1 basic knowledge so I know the main differentiation rules (ie. 4x^2=8x) and I also know the addition, subtraction, multiplication, and division rules as well one year ago one year ago
1. incomplte
You already have all the tools you need for differentiation. Do you have a specific example?
2. zepdrix
$\large y=2^x$Hmm so the variable is in the exponent position. That poses a problem. We can't apply the Power Rule that you may have learned at this point. We have to introduce logarithms in order to get the variable OUT of the exponent position.
3. bettyboop8904
I have e^5 and it says the answer is 0 but I don't understand why that is. and there are a couple others I don't get. Bc my book says that the exponential function is itself. Why wouldn't it be just f'(x)= e^5?
4. zepdrix
oh lol :)
5. zepdrix
They're pulling a trick on you. See how e^5 contains no variable? It's just a fancy fancy looking CONSTANT, all dressed up in a suit.
6. bettyboop8904
lol
7. incomplte
lol @zepdrix has explained it better than i would
8. zepdrix
$\large f(x)=e^5 \qquad \qquad \rightarrow \qquad \qquad f(x)=148.4$See how the derivative is going to turn out? :) What happens when you take the derivative of a constant?
9. bettyboop8904
$e^{2x}\div e^{4x ^{2}} + \cos(e ^{7x})$
10. bettyboop8904
it approaches infinity?
11. bettyboop8904
that was me answering the post before i posted that big equation @zepdrix
12. bettyboop8904
i think i answered wrong lol
13. zepdrix
Nooo silly! :O That's one of those main derivative rules that you want to know at this point! :) A constant will give you 0 when you differentiate it. A derivative measures change. A constant is something that stays constant, doesn't change. So how much does something that doesn't change ... change? zerooooOooOO!
14. zepdrix
What do we need to do with the big messy problem? Take it's derivative?
15. bettyboop8904
That's right sorry, see the problem i have is answering questions that are put in to words haha but a lot of people struggle with that, exactly why i hate optimization problems = ( but yes derivative of a constant is 0. So why wouldn't it be 1 then? Isn't anything to the 0 power 1?
16. bettyboop8904
And differentiate it
17. bettyboop8904
taking the derivative and differentiating are the same thing right?
18. zepdrix
Yes, just fancy words ^^
19. bettyboop8904
ok that's what i thought lol
20. bettyboop8904
So I have the answer for the big equation from my neighbors notes but it doesn't make sense to me so that's why I want to ask about it = )
21. zepdrix
We can think about it like this, maybe this will help, maybe not :)$\large f(x)=e^5$See how there is no X value in the equation? Let's introduce an X.$\large f(x)=(e^5)x^0$ $$x^0$$ is just $$1$$ right? So we're allowed to do that. We're just multiplying it by 1. Taking the derivative (applying the power rule to the x term) gives us,$\large f'(x)=0(e^5)x^{0-1}$We have a 0 coming down be multiplied by everything, that's going to turn the whole thing to 0. Maybe you're just getting confused by the e term. It contains no X's. We won't be trying to take it's derivative by applying the $$e^x$$ rule.
22. zepdrix
Ok ok big question time c:
23. bettyboop8904
ok that makes sense = ) ty! and yes now to the big one lol
24. zepdrix
handy*
25. bettyboop8904
hold on lol cos is in the denominator with e to the 4x^2
26. zepdrix
Oh i see ^^
27. zepdrix
$\huge \frac{e^{2x}}{e^{4x^2}+\cos(e^{7x})}$Ah ok so it is messy :) We'll need to apply the quotient rule.
28. zepdrix
$\large \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$
29. bettyboop8904
right that's what they did with it = ) that's where I got confused, not with the actual rule but u'
30. bettyboop8904
and v'
31. zepdrix
$\huge \left(e^{2x}\right)'=e^{2x}(2x)'$So we'll apply the e^x rule, but since we have more than just x in the exponent, we have to apply the chain rule. The prime on the brackets is to show that we need to take it's derivative still.
32. zepdrix
Remember how to do the chain rule? c:
33. bettyboop8904
yes
34. zepdrix
$\huge \left(e^{2x}\right)'=e^{2x}(2)$Any confusion on that one? c:
35. bettyboop8904
oh wow i think i just had a mental click lmfao
36. zepdrix
lol XD those are always fun.
37. bettyboop8904
hold on let me look at the problem and answer one more time lol
38. bettyboop8904
omg I'm so dumb lol Idk if its bc how you explained it or if I was just looking at it blindly lol omg you were such a great help either way lol
39. zepdrix
Chain rule can be a little confusing at first. You're essentially making a 'copy' of the inner function and then taking its derivative, which can seem a little strange c:
40. bettyboop8904
but I do have another one lol it says $\lim_{x \rightarrow \infty} (1.001)^{x}$ and it says the answer is infinity, why is that?
41. bettyboop8904
oh and then one more after that lol but i think it may be pretty easy haha
42. zepdrix
Just plug in a really big value for x and you'll see where it's heading! :) If x=99999$\large 1.001^{99999}=2.55 \times 10^{43}$
43. zepdrix
Since the base is LARGER than 1, and our exponent is positive, the value will explode.
44. zepdrix
What if you had,$\large \lim_{x \rightarrow \infty} (.999)^x$
45. bettyboop8904
so really big x values = really big y values
46. zepdrix
yah c: As x approaches infinity, y is approaching infinity. In that first case.
47. bettyboop8904
I used a calculator but the bigger x gets the closer y goes to 0 for the second part
48. zepdrix
In this second case, see how the BASE is less than 1? When we raise it to larger and larger powers it will get smaller and smaller and smaller. Imagine multiplying fractions. They get smaller! c: $\large \dfrac{1}{2}\cdot \dfrac{1}{2} \quad = \quad \dfrac{1}{4}$
49. zepdrix
Yesss good good c:
50. bettyboop8904
correct = )
51. bettyboop8904
so in that case it would go to -infinity
52. zepdrix
That's a good strategy with limits. If you're ever confused, just plug numbers in to see what happens.
53. zepdrix
Noooo :D An exponential can't go in the negative! :O
54. bettyboop8904
or wait it would go to 0 not -infinity right bc yeah what you just said haha
55. bettyboop8904
ok and so for the last one, well at least for now haha It says find the exponential function f(x)=Ca^{x} whose graph is given. It then has two problems... it shows you a graph from the exponent family and gives you two points on the graph, the first problem gives you the points: (1,6) and (3,24) I set the problem up as y=Ca^{x} and plugged 6 (the first y-value) in for y and then plugged 1 into the x. It says the answer is 3*2^{x}. So then I looked at the second point. I did the same thing as I did with the first pair but I couldn't see how you could figure it out without doing trial and error. The second problem gave the two points (2,2/9) and a y-intersect at 2. So I tried the same approach with the first pair and got 2*(1/3)^{x} as the answer but I'm not sure if that's correct
56. bettyboop8904
I feel the second answer I arrived at might be correct because a fraction with an exponent has a graph that is decreasing which the graph is decreasing in the picture of the second problem's graph @zepdrix
57. zepdrix
Ok simmer down :O Let's look at the first one a sec.
58. bettyboop8904
lol
59. zepdrix
$\large f(x)=Ca^x$$\large f(1)=6 \qquad \rightarrow \qquad 6=Ca^1$$f(3)=24 \qquad \rightarrow \qquad 24=Ca^{3}$So this is where you got stuck on the first one?
60. bettyboop8904
yes
61. zepdrix
We have to perform a little sneaky trick at this point! We'll DIVIDE our f values.$\large \frac{f(3)}{f(1)} \qquad \rightarrow\qquad \frac{24}{6}=\frac{Ca^3}{Ca^1}$We get a nice cancellation with the C's allowing us to solve for a.$\large \frac{24}{6}=\frac{\cancel{C}a^3}{\cancel{C}a^1}$
62. zepdrix
$\large 4=a^2 \qquad \rightarrow \qquad a=2$
63. bettyboop8904
ok that makes sense = ) is that a property or a rule?
64. zepdrix
It's this really fancy thing called "division" :3
65. zepdrix
lolol
66. bettyboop8904
or what i mean to say is how did you know to divide the points' equations hahahaha
67. zepdrix
Well here is something to think about. When you get to this point,$\large 6=Ca^1$$\large 24=Ca^3$ You have a SYSTEM of equations. To be more specific, you have 2 equations and 2 unknowns. Since the number of unknowns does not exceed the number of equations given, we can find solutions for each unknown. Remember back to algebra, doing Substitution and Elimination? This is just another silly trick to add to your list. I'm not really sure what it's called though XD
68. zepdrix
We could have also done substition to get the same result though :)
69. bettyboop8904
like solve for C in one of the problems and then plug it into the other equation?
70. zepdrix
Yah :)
71. bettyboop8904
ok good i think i got it = ) thank you so much for all your help! = D
72. zepdrix
yay team \c:/
73. bettyboop8904
lol
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2014-11-27 00:34:38
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http://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-2-differentiation-2-4-exercises-page-136/66
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## Calculus 10th Edition
$y'=\dfrac{9x^2+4}{5\sqrt[5]{(3x^3+4x)^4}}.$ The derivative evaluated at the point $(2, 2)$ is $\dfrac{1}{2}.$
$u=3x^3+4x$; $\dfrac{du}{dx}=9x^2+4$ $y=u^{\frac{1}{5}};\dfrac{dy}{du}=\dfrac{1}{5\sqrt[5]{u^4}}$ $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=\dfrac{9x^2+4}{5\sqrt[5]{(3x^3+4x)^4}}.$ To evaluate the derivative, plug in $x=2\rightarrow\dfrac{9(2)^2+4}{5\sqrt[5]{(3(2)^3+4(2))^4}}=\dfrac{1}{2}$ A graphing utility was used to verify this result.
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2017-07-28 00:52:15
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http://mathhelpforum.com/advanced-algebra/102239-group-problem-please-help.html
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HINT: $\displaystyle a^{m_1}=a^{m_2}$ for some $\displaystyle m_1 \neq m_2 \in \mathbb{N}$, because we are in a finite group.
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2018-04-26 02:52:16
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