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https://jp.mathworks.com/matlabcentral/cody/problems/195-program-an-exclusive-or-operation-with-logical-operators/solutions/162633 | 1,606,316,478,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182794.28/warc/CC-MAIN-20201125125427-20201125155427-00305.warc.gz | 359,740,301 | 16,770 | Cody
# Problem 195. Program an exclusive OR operation with logical operators
Solution 162633
Submitted on 15 Nov 2012 by Pasumarthi Viswanath
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% a=0;b=0; res=0; assert(isequal(myXOR(a,b),res))
ans = 0
2 Pass
%% a=1;b=1; res=0; assert(isequal(myXOR(a,b),res))
ans = 0
3 Pass
%% a=1;b=0; res=1; assert(isequal(myXOR(a,b),res))
ans = 1
4 Pass
%% a=0;b=1; res=1; assert(isequal(myXOR(a,b),res))
ans = 1
5 Pass
%% a=[0 0 1 1];b=[1 0 1 0]; res=[1 0 0 1]; assert(isequal(myXOR(a,b),res))
ans = 1 0 0 1
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Start Hunting! | 287 | 810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-50 | latest | en | 0.524819 |
https://id.scribd.com/document/344006797/Variabel-Sample-Size-Pre-Clinical | 1,582,302,080,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00344.warc.gz | 423,252,411 | 72,489 | Anda di halaman 1dari 18
# 3/3/2016
Variabel dalam
Penelitian
Miftahul Mushlih, M.Sc.
Mendapatkan suatu jawaban dari
sebuah permasalahan
## atau dengan kata lain
menjawab sebuah HIPOTESIS
1
3/3/2016
Jenis Variabel
V. Suatu Variabel yang di
Bebas control oleh peneliti
Variabel yang
Hasilnya berasal dari V.
V. Bebas Terikat
## V. Variabel yang harus di
Kontrol kendalikan supaya tidak
mempengaruhi V. Bebas
Contoh
Seorang peneliti ingin mengetahui pegaruh pupuk terhadap beberapa
jenis varieties padi yaitu varieties raja lele dan varieties tawon,
kosentrasi yang digunakan adalah 10 ppm, 20 ppm, 30 ppm, 40 ppm,
dari penelitian tersebut tentukan
A. jenis data (para metric/ non parametric)
B. V. Bebas
C. V. Kontrol
D. V. terikat
2
3/3/2016
Sample Size
Miftahul Mushlih, M.Sc
## Sample size determination:
Studies Comparing Two Group Means
To compute sample size for continuous variables, it is necessary to obtain an
estimate of the population standard deviation of the variable (s) and the
magnitude of the difference (d) the investigator wishes to detect, often
called the effect.
Sample size is given by
## where s is the standard deviation, d is the difference to be detected, and C is
a constant dependent on the value of and selected. C is 10.51 and 2C
would be 21
## Ralph B. Dell, Steve Holleran, R Ramakhakrishnan, Sample sice determination, 2002
3
3/3/2016
Contoh
Seorang peneliti ingin mengetahui perbedaan penghasilan masyarakat
lubuk buaya dengan kota tengah . . ..
Sd
X1-x2
Paired Studies
Paired studies compare values before and after an intervention in the
same animal. In this case, data are analyzed by a paired t test, and the
sample size is computed by
## Ralph B. Dell, Steve Holleran, R Ramakhakrishnan, Sample sice determination, 2002
4
3/3/2016
Federer Formula
If the s, d and C cannot be determined Federers
Formula
(t-1) (r-1) 15
t = number of treatment
r = number of sample (repeat)
## Federer WT. Experimental design, theory and application, 1967.
(t-1) (r-1) 15
(3-1)(r-1) 15
5
3/3/2016
contoh
Penelitian
Pengaruh pemberian ekstrak tapak dara terhadap pertumbuhan
sel-sel kanker payudara
Perlakuan
1. 0 mg/kg bb mencit
2. 10 mg/ kg bb mencit
3. 30 mg/ kg bb mencit
4. 50 mg/ kg bb mencit
Berapa ulangan minimal ulangan pada penelitian tersebut?
## Hukum Roscoe (1975)
Roscoe (1975), memberikan beberapa panduan untuk
menentukan
ukuran sampel yaitu :
Ukuran sampel lebih dari 30 dan kurang dari 500 adalah tepat
untuk kebanyakan penelitian.
Jika sampel dipecah ke dalam sub-sampel (pria/wanita,
junior/senior, dsb), ukuran sampel minimum 30 untuk tiap
Dalam penelitian mutivariate (termasuk analisis regresi
berganda), ukuran sampel sebaiknya 10 kali lebih besar dari
jumlah variabel dalam penelitian.
Untuk penelitian eksperimental sederhana dengan control
eskperimen yang ketat, penelitian yang sukses adalah mungkin
dengan ukuran sampel kecil antara 10 sampai dengan 20 buah.
6
3/3/2016
7
3/3/2016
## Krejcie, R. & Morgan D. (1970).
Sample size determination for
most practical situations
Source: Krejcie, R. & Morgan D. (1970). Determining
Sample Size for Research Activities, Educational and X2NP(1 P)
Psychological Measurement, 30, 607-610. S (n) =
Where:
d2(N-1)+x2P(1-P)
N = Population Size s =ukuran sampel yang diperlukan
S = Sample Size N = ukuran populasi yang dikenalpasti
P = perkadaran populasi diandaikan 0.5 kerana
magnitud ini akan menghasilkan ukuran sampel
yang maksimum.
d = darjah ketepatan yang maksimum (0.05)
x2 = nilai khi square, db 1 yaitu 3.841 (note: bisa
berubah sesuai dengan uji yang digunakan)
8
3/3/2016
## Krejcie, R. & Morgan D.
(1970) Table
Perumusan berdasarkan
proporsi
Keterangan:
n = Jumlah sampel minimal
N = ukuran populasi
t = tingkat kepercayaan (digunakan 0,95 sehingga nilai t
= 1,96)
d = taraf kekeliruan (digunakan 0,05)
p = proporsi dari karakteristik tertentu (golongan)
q= 1 p
1 = Bilangan Konstan
9
3/3/2016
Contoh
Berdasarkan proporsi
Secara kuantitaf besarnya sampel dapat ditentukan dengan
menggunakan rumus
matematika sebagai berikut:
1. Meneliti harga mean:
## 2. Meneliti harga proporsi:
Keterangan:
d : Penyimpangan yang ditoleransi
: harga standar normal
a : varian populasi.
10
3/3/2016
Preclinical Animal
Study
Preclinical study/preclinical
trial
Preclinical trial - a laboratory test of a new drug or a new medical
device, usually done on animal subjects, to see if the hoped-for
treatment really works and if it is safe to test on humans.
11
3/3/2016
## Pre-Clinical Trials and Clinical Trials are the
processes by which scientists test drugs and
devices to see if they are SAFE and EFFECTIVE.
Experimental study
In vitro and in vivo
Animal models
Pharmacodynamic study
Toxicological study
12
3/3/2016
PHARMACODYNAMIC STUDIES
Effects / Efficacy
Mechanism of action
TOXICOLOGICAL STUDIES
Safety
Spectrum
Pharmacodynamic study
In vitro : tissue, cell culture, blood
. component, etc
In vivo : whole animal
## Disease model : inflammatory, dislipidemi,
sepsis, Ca/ malignancy, etc
Source of methode (literature) : Guidelines,
journal
13
3/3/2016
Methodology of
pharmacodynamic study
ANIMALS :
Rodent or non-rodent (may depend on desired effect)
Healthy or diseased-animal model
Sex : male and/or female
Number : adequate for statistical analysis
Toxicological studies
IN VIVO TOXICOLOGICAL STUDIES
A. GENERAL TESTS
SHORT TERM
Acute Toxicity Test (LD50)
LONG TERM:
Sub Acute Test
Sub Chronic Test
Chronic Test
B. LOCAL TOXICITY TEST
Dermatological Preparation
C. SPECIAL TOXICITY TESTS
Mutagenicity Test
Carcinogenicity Test
Reproductive and Development Toxicity Test
14
3/3/2016
Animal models
Animal models are, at the best, analogous
to human condition but no theory can be
provided or refused by analogy. The truth is
the evidence in animals can be a powerful
device in support of virtually theory.
## Why are animals necessary in research?
1. Diseases process in human and animals have
similarities
2. Cell systems contain or manipulate only a part
of the organ system.
3. Computer models lack the complexities of living
entity .
15
3/3/2016
(Tokyo 2004)
## BASIC PRINCIPLES FOR ALL MEDICAL RESEARCH
- Medical research involving human subjects must
conform to generally accepted scientific principles,
be based on a thorough knowledge of the scientific
literature, other relevant sources of information, and
on adequate laboratory and, where appropriate,
animal experimentation.
- Appropriate caution must be exercised in the conduct
of research which may effect the environment, and
the welfare of animals used for research must be
respected
## R eplacement Alternatifve methode
lower class of animal model
R eduction
correct model
genetic homogenity
use healthy animal
R efinement
caring
treatment
non-invasive methode
pain, stress, distress minimization
16
3/3/2016
## The council believes that the welfare of an animal ...
should be considered with reference to Five
Freedoms.
Freedom from hunger and thirst
Freedom from discomfort
Freedom from pain, injury and disease
Freedom to express normal behaviour
Freedom from fear and distress
(Farm Animal Welfare Council UK, 1993)
ever possible
## Enrichment devices for Nesting materials
rodents: Igloo, nest etc.
Human interactions
Enrichment
Devices for NHP
17
3/3/2016
18 | 2,156 | 7,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-10 | latest | en | 0.343187 |
https://www.gradesaver.com/textbooks/math/statistics-probability/statistics-12th-edition/chapter-3-probability-exercises-3-112-3-137-understanding-the-principles-page-151/3-112 | 1,716,194,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00029.warc.gz | 721,047,805 | 12,713 | Chapter 3 - Probability - Exercises 3.112 - 3.137 - Understanding the Principles - Page 151: 3.112
Suppose we role two dies and we need to calculate the probability of face number of two fair dice. let P(A) be probability of first dice and P(B) be probability of second dice. Then P(A ∩ B ) = P(A).P(B)
Work Step by Step
No steps
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 126 | 493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-22 | latest | en | 0.857004 |
https://hunting4gadgets.com/qa/quick-answer-what-is-the-hcf-of-70.html | 1,600,680,999,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201601.26/warc/CC-MAIN-20200921081428-20200921111428-00284.warc.gz | 443,349,457 | 7,624 | # Quick Answer: What Is The HCF Of 70?
## What’s the GCF of 80 and 70?
Greatest common factor (GCF) of 70 and 80 is 10..
## What is the LCM of 18 and 36?
Least common multiple (LCM) of 18 and 36 is 36.
## What is the least common multiple of 3 and 10?
30Least common multiple (LCM) of 3 and 10 is 30.
## What is the HCF of 91 112 and 49?
7Answer. So,the HCF of 49,91 and 112 is 7 by prime factorization method. Hence the HCF of 49,91 and 112 is 7.
## What is the HCF of 27 and 63?
Greatest common factor (GCF) of 27 and 63 is 9. We will now calculate the prime factors of 27 and 63, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 27 and 63.
## What is the HCF of 12?
Find the highest common factor (H.C.F) of 12 and 18. Factors of 12 = 1, 2, 3, 4, 6 and 12. Factors of 18 = 1, 2, 3, 6, 9 and 18. Therefore, common factor of 12 and 18 = 1, 2, 3 and 6. Highest common factor (H.C.F) of 12 and 18 = 6 [since 6 is the highest common factor].
## How can I solve HCF?
The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of 60 and 72 is 2 × 2 × 3 which is 12. The lowest common multiple is found by multiplying all the factors which appear in either list: So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.
## What is the GCF of 28 and 15?
Greatest common factor (GCF) of 15 and 28 is 1. We will now calculate the prime factors of 15 and 28, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 15 and 28.
## What is the HCF of 4 and 8?
Greatest common factor (GCF) of 4 and 8 is 4. We will now calculate the prime factors of 4 and 8, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 4 and 8.
## What’s the GCF of 70 and 30?
The common factors of 30 and 70 are 10, 5, 2, 1, intersecting the two sets above. In the intersection factors of 30 ∩ factors of 70 the greatest element is 10. Therefore, the greatest common factor of 30 and 70 is 10.
## What is the HCF of 105?
We found the factors and prime factorization of 105 and 125. The biggest common factor number is the GCF number. So the greatest common factor 105 and 125 is 5.
## What is the HCF of 15 and 30?
The common factors of 15 and 30 are 1, 3, 5, and 15. The greatest common factor is 15.
## What is the highest common factor of 70?
How to find the GCF of 70 and 100The prime factorization of 70 is: 2 x 5 x 7.The prime factorization of 100 is: 2 x 2 x 5 x 5.The prime factors and multiplicities 70 and 100 have in common are: 2 x 5.2 x 5 is the gcf of 70 and 100.gcf(70,100) = 10.
## What is the HCF of 70 and 105?
gcf, hcf, gcd (70; 105) = 35 = 5 × 7: greatest (highest) common factor (divisor), calculated. The numbers have common prime factors.
## What is the HCF of 70 and 84?
We found the factors and prime factorization of 70 and 84. The biggest common factor number is the GCF number. So the greatest common factor 70 and 84 is 14. | 996 | 3,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-40 | latest | en | 0.936108 |
https://www.sophia.org/tutorials/functions-what-is-a-function | 1,513,464,281,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948589512.63/warc/CC-MAIN-20171216220904-20171217002904-00656.warc.gz | 801,742,305 | 15,880 | +
# FUNCTIONS - What is a function?
##### Rating:
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Author: Kate Sidlo
##### Description:
By the end of this tutorial students will be able to give the definition of a function verbally or in written form.
This tutorial explores what functions are using the definition "a relationship where every input has one and only one output"
The idea is demonstrated using a function machine metaphor and a table.
(more)
Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*
No credit card required
28 Sophia partners guarantee credit transfer.
281 Institutions have accepted or given pre-approval for credit transfer.
* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 25 of Sophia’s online courses. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.
Tutorial
## TO DO:
1) Take notes from the podcast video.
2) Add the definitions from the vocabulary section at the bottom to your vocabulary sheet.
of
## VOCABULARY
FUNCTION: A relationship where every input has one and only one output
INPUT: A term or number that is plugged into a relationship; the x-variable in an ordered pair
RELATIONSHIP: A comparison of terms, variables, and/or numbers (typically an equation or inequality)
OUTPUT: A term or number that is the answer or solution to a relationship; the y-variable in an ordered pair | 333 | 1,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-51 | longest | en | 0.892463 |
https://www.thestructuralengineer.info/education/structural-calculation-examples/item/553-calculation-example-calculate-the-equation-of-the-elastic-curve?tmpl=component&print=1 | 1,576,178,507,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540545146.75/warc/CC-MAIN-20191212181310-20191212205310-00151.warc.gz | 890,346,255 | 5,753 | Friday, 10 February 2017 10:54cat
## Calculation Example - Calculate the equation of the elastic curve. Featured
Written by TheStructuralEngineer.info
Calculate the equation of the elastic curve .Determine the pinned beam’s maximum deflection. EI constant.
Solution
From the previous exercise (Calculation example-Calculate member diagrams) published (17 January 2017), we work for the section 0<x<L/2.
The Shear force is S(x)= P/2
The Moment is M(x)= P/2 x
The Slope and the Elastic Curve are:
Due to symmetry, boundary conditions are:
We calculate the slope: | 140 | 571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-51 | latest | en | 0.866349 |
http://betterlesson.com/lesson/resource/1960381/operations-with-numbers-in-scientific-notation-day-2-division-narrative-video-mov | 1,477,120,073,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718840.18/warc/CC-MAIN-20161020183838-00158-ip-10-171-6-4.ec2.internal.warc.gz | 30,090,438 | 21,641 | ## Operations with Numbers in Scientific Notation Day 2 Division Narrative Video.mov - Section 1: Bellringer - Lesson Opener
Operations with Numbers in Scientific Notation Day 2 Division Narrative Video.mov
# Operations with Numbers in Scientific Notation Day 2 of 2
Lesson 12 of 19
## Big Idea: Like bases, come out come out where ever you are! Extend previous work with division and exponents to include scientific notation.
Print Lesson
5 teachers like this lesson
Standards:
Subject(s):
Math, exponents / powers, 8th grade math, scientific notation, Operations in Scientific notation, master teacher
50 minutes
### Christa Lemily
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Environment: Urban | 323 | 1,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2016-44 | longest | en | 0.848138 |
http://scipy.github.io/devdocs/reference/generated/scipy.spatial.distance_matrix.html | 1,726,747,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652028.28/warc/CC-MAIN-20240919093719-20240919123719-00574.warc.gz | 28,163,815 | 6,578 | scipy.spatial.
distance_matrix#
scipy.spatial.distance_matrix(x, y, p=2, threshold=1000000)[source]#
Compute the distance matrix.
Returns the matrix of all pair-wise distances.
Parameters:
x(M, K) array_like
Matrix of M vectors in K dimensions.
y(N, K) array_like
Matrix of N vectors in K dimensions.
pfloat, 1 <= p <= infinity
Which Minkowski p-norm to use.
thresholdpositive int
If `M * N * K` > threshold, algorithm uses a Python loop instead of large temporary arrays.
Returns:
result(M, N) ndarray
Matrix containing the distance from every vector in x to every vector in y.
Examples
```>>> from scipy.spatial import distance_matrix
>>> distance_matrix([[0,0],[0,1]], [[1,0],[1,1]])
array([[ 1. , 1.41421356],
[ 1.41421356, 1. ]])
``` | 211 | 770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.623719 |
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# e1.f05 - ECE 190 Exam I Fall 2005 Tuesday September 27th...
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ECE 190 Exam I Fall 2005 Tuesday, September 27 th , 2005 x Be sure your exam booklet has 12 pages. x Write your name at the top of each page. x This is a closed book exam. x You may not use a calculator. x You are allowed one handwritten 8.5 x 11” sheet of notes. x Absolutely no interaction between students is allowed. x Show all of your work. x Be sure to clearly indicate any assumptions that you make. x More challenging questions are marked with a ***. x Don’t panic, and good luck! “A professor is one who talks in someone else’s sleep.” – W. H. Auden Problem 1 20 points _______________________________ Problem 2 20 points _______________________________ Problem 3 20 points _______________________________ Problem 4 20 points _______________________________ Problem 5 20 points _______________________________ Total 100 points Name:
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Page 2 Name: ____________________________________________ Problem 1 (20 points): Short Answer Part A (5 points): The IR (Instruction Register) holds the instruction that is to be executed. Given that the instruction bits can be held in the MDR, why is an IR necessary? Part B (5 points): Consider the following LC-3 instruction (x3500 is the address at which the instruction is located): x3500 LD R5, _____ ; we want to put the value x2BFF in register R5 Given the above instruction, what is the range of memory addresses at which the value x2BFF can be stored such that the above instruction can be executed successfully? Part C (5 points): A certain memory chip has a total of 2 32 bits and is 8-bit addressable. How many address bits must be specified when reading or writing a location on this chip? Part D (5 points): What fraction of the range of numbers that can be represented with an N-bit 2’s complement data type can also be represented with an (N+1)-bit unsigned data type? (Justify your answer.)
Page 3 Name: ____________________________________________ Problem 2 (20 Points): Representations Part A (7 points): Your friend complains to you that the number 1,073,741,825 (in hexadecimal, x40000001) cannot be represented using an IEEE single-precision floating point representation (1-bit sign, 8-bit exponent, 23-bit mantissa). Is your friend right? If so, why would one ever use floating point, given that 32-bit 2’s complement can
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Ask a homework question - tutors are online | 707 | 2,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-09 | latest | en | 0.872701 |
http://www.docstoc.com/docs/69138277/Advantages-Relative-to-Planning-a-Ppt-Presentation-in-Outline-Format | 1,386,929,068,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164926426/warc/CC-MAIN-20131204134846-00053-ip-10-33-133-15.ec2.internal.warc.gz | 308,150,438 | 19,590 | # Advantages Relative to Planning a Ppt Presentation in Outline Format
Document Sample
``` Module 3:
Quality Assurance Tools
Learning Objectives
• Learn the theory and method of use of the
seven tools of quality needed in process
analysis and problem-solving techniques
• Seven tools of quality
• Pareto charts
• Flow charts
• Cause-and-effect diagram
• Check sheet
• Histograms
• Control charts
• Scatter diagrams
Pareto Chart
• Method of prioritizing problems or causes
by frequency of occurrence or cost
• Based in the 80-20 rule:
• 80% of the problem is caused by 20% of the
sources
• “Vital few” and “trivial many”
• Depicted by a vertical bar graph arranged from
left to right descending order
Advantages of a Pareto Chart
• Focuses efforts on problems with greatest
potential for improvement
• Distinguishes the critical causes from the less
significant causes
• Helps prevent “shifting the problem” where the
solution removes some causes but worsens
others
Advantages of a Pareto Chart (cont.)
• Measure the impact of improvement projects
when comparing charts “before” and “after”
• The chart shows the relative importance of
problems in a simple, quickly interpreted,
visual format.
• Progress is measured in a highly visible format
that provides incentive to push on for more
improvement.
Steps to Build a Pareto Chart
• Identify problem
• Choose categories that will be monitored
• Choose the most meaningful unit of measurement
• Frequency
• Cost
• Determine time period
• Long enough to represent situation
• Scheduled time to collect data is typical of a workday
Steps to Build a Pareto Chart (cont.)
• Collect data
• Compare the frequency of each category
• Draw chart:
• List the categories on the horizontal line
• Descending order, from left to right
• Frequencies on the vertical line
Steps to Build a Pareto Chart (cont.)
• Draw the cumulative percentage line showing
categories contribution
• Optional
• Draw vertical line on the right side of the chart
• Plot cumulative values from left to right
• Interpret results
• Tallest bar represents biggest contributor
• Perform analysis of category that has the “most
impact”
Cum %
Example of a Pareto Chart 100 %
97%
91%
79%
61%
50 %
30
35%
20
24
10 18
12
8
4 2
0
Late Wrong Missing Dam aged Wrong Not received
delivery product parts container address
Delivery complains
Further analysis using Pareto charts
• Major cause breakdown:
• Tallest bar is broken down into sub-causes
30
20
24
10 18
12
8
4 2
0
Late delivery Wrong product Missing parts Dam aged Wrong Not received
8
6
4
8
6 4 4
2
2
0
Missing address Driver mistake Late Transportation Administrative
documentation problems delays
Late delivery
Further analysis using Pareto charts
•Before and after:
• New Pareto bars are drawn side-by-side
with the original Pareto showing effect of change
25
24
20
18
15
12
10
10 10
8 8
5
4 4
3 2 1
0
Late Wrong Missing Dam aged Wrong Not
Before After
Further analysis using Pareto charts
•Change measurement scale:
• Same categories are used but measured
differently. Typically cost and frequency.
• This exercise defines the category of “most impact”.
\$3,000
30
\$2,500
25
\$2,000
20
\$1,500
15 \$2 ,50 0
24
\$1,000 \$1,8 50
10 18
\$1,2 0 0
12 \$500
5 8
\$8 0 0
4 \$2 2 0 \$2 0 0
2 \$0
0
Late Wrong Missing Dam aged Wrong Not received Wrong product Not received Missing parts Wrong Damaged Late delivery
delivery product parts container address address container
Delivery com plains Cost/month
Flow Charts
• Pictorial representation of the steps and decision
points in a process.
• Flow charts are used to identify the actual path
of a product or service.
• Flow charts show:
• Sequential work activities
• Inputs for each action
• Outputs from each activity
Advantages of a Flow Chart
• Provide common understanding of how a
process works.
• Identifies problem areas, unexpected complexity,
redundancies, and areas of potential improvement
• Serves as a training aid
• Provides basis for documentation.
• Identifies location where additional data can be
collected and investigated
Symbols of a Flow Chart
Shows an activity performed in the process
Shows a point in the process where a Yes/No
question is being asked or a decision is required
Identifies a break in the flow chart and is continued
A elsewhere on the same page or another page
Shows the flow of the process
Shows the beginning and end process
Steps to Build a Flow Chart
• List the basic steps of the process.
• Determine the logical sequence of the steps.
• Identify the inputs of each step.
• Determine the outputs of each step.
• Draw the chart using the symbols to show the
actions and decisions.
Example of a Flow Chart
Team is given task
to write procedure Review
draft
yes Complete Send to
Develop Draft
Final Release
outline approved
version Dept.
No
Release
Complete Revise Draft
procedure
draft with changes
Cause and Effect Analysis
• Pictorial representation of all possible causes
contributing to a problem.
• Developed by Dr.Kaoru Ishikawa
•Also known as “Fishbone” or “Ishikawa” diagrams
Advantages of a Cause and Effect Diagram
• Clearly illustrates the various causes affecting a
process:
• Relationship between these causes
• Where are they occurring
• Helps in finding the most basic cause of the
problem
• Motivates team members participation
Steps to Build a Cause and Effect Diagram
• Place problem statement in a box to the right-hand
side of the paper.
• Select the major cause or categories and place them
to the left of the problem statement.
• 4 M’s (Production process):
• Materials
• Manpower
• Machinery/equipment
• Method
• 4 P’s (Service process):
• Policies
• Procedures
• People
• Plant / equipment
Steps to Build a Cause and Effect Diagram
(cont.)
• Draw a box around each category and connect
to a line pointing out towards the problem statement.
• Using the brainstorming technique, generate ideas
of causes, on the major categories.
• Record these ideas on a line off the applicable
major category line.
Steps to Build a Cause and Effect Diagram
(cont.)
• For each cause listed on the diagram, ask “why
does this happen?”.
• For each response, ask the same question.
• Each successive answer is another possible cause
• Look for causes that repeat across major cause
categories.
• Ensure all team members agree on the problem
and causes statements.
Example of a Cause and Effect Diagram
Machinery/equipment Manpower
Drivers don’t
Unreliable trucks Show up
Drivers get lost
Not enough trucks
Not capacity for on shipper
peak periods
Database Input error
Late deliveries
Poor dispatching Run out of
products
Poor handling of Poor planning
large orders
Lack of
Lack of training
training
Method Material
Check Sheets
• Tool used to record and compile frequency of
observations as they occur
• Used for Pareto charts and histograms
• Design varies depending on information needed
Advantages of a Check Sheet
• Patterns of information are clearly detected
and shown
• Easy-to-understand data from a simple process
• Based on facts, not opinions
• Forces agreement on the definition of each
condition
Steps to Build a Check Sheet
• Agree on the definition of the events or conditions
to observe.
• Decide who will collect the data.
• Determine the time period.
• Design a check sheet form that is clear and
easy to use.
• Define Source of information.
• Determine Content of information.
• Collect data consistently and accurately.
Example of a Check Sheet
Project: Defects on fastener installation Performed by: John Doe
Location: Cost Center 727 Reason: Defect reduction
Time period: W/E 4/23/04
Type of defects 4/19 4/20 4/21 4/22 4/23 Total
Gapped fasteners 15 15 12 10 14 66
Missing fasteners 3 0 0 1 1 5
Damaged fasteners 8 3 12 8 4 35
Defective fasteners 12 3 5 3 6 29
Total 38 21 29 22 25 135
Histograms
• Graphical representations of the frequency
distribution of data in bar form.
• Summarizes data from a process that has been
collected over a period of time
• Provide a quick representation of the “spread”
and “centering” of a process
Advantages of a Histogram
• Display large amounts of data that are difficult
to interpret in a tabular form
• Show the relative frequency of occurrences of the
various data values
• Reveal the variation, centering, and distribution
shape of the data
• Very useful when calculating capability of a process
• Helps predict future performance of a process
Steps to Build a Histogram
• Collect data for analysis.
• At least 50 to 100 data points
• Use historical data to find patterns or to use as
a baseline for past performance
• Determine the range of the data set.
• Smallest value subtracted from largest value
• Determine quantity of categories.
• Take the square root of total number of data
points and round to nearest whole number
Steps to Build a Histogram (cont.)
• Determine each category’s data point.
• Mid-point of each category
• Plot data on a vertical bar-graph.
• Frequency on Y-axis
• Categories on X-axis
• For each class interval, draw bar with the
the height equal to frequency count
Example of a Histogram
Class
Category Mid-point Frequency
boundary
1 10.00 - 10.19 10.1 1
2 10.20 - 10.39 10.3 6
3 10.40 - 10.59 10.5 12
4 10.60 - 10.79 10.7 8
5 10.80 - 10.99 10.9 6
6 11.00- 11.19 11.1 3
14
12
10
8
6 12
4 8
6 6
2
3
1
0
10.00 - 10.20 - 10.40 - 10.60 - 10.80 - 11.00-
10.19 10.39 10.59 10.79 10.99 11.19
Control Charts
• Line graph of measurements of a process overtime
that has statistically based control limits placed on it
• Process control charts monitor and display variations
in a process output.
• Control limits are based on process variation
• Define expected variation range due to common causes
• +/- three standard deviations from centerline
• Centerline represents the average of all
measurements used
Types of Control Charts
• Two primary types are:
• Control charts for variables:
• Most used: X –R
• X is average values
• R is range
• Others: Run charts, moving range charts (MX –MR charts)
• Control charts for attributes:
• p chart: fraction defective
• np chart: number of defectives
• c chart: number of defects
Advantages of Control Charts
• Focuses attention on detecting and monitoring
process variation over time
• Distinguishes “special” from “common” causes
• Helps predict performance of a process
• Helps improve a process to perform consistently
• Provides a common language to discuss process
behavior
Steps to Build a Control Chart
• Select the process to be charted.
• Determine type of chart.
• Determine sampling method and plan.
• Initiate data collection.
• Calculate control limits and centerline.
• Build the control chart.
• Plot data.
• Interpret results.
Example of a Control Chart Point out-of-control
(out of the control limits boundaries)
Data plot
Upper control limit
3 std.deviations
Centerline
3 std.deviations
Lower control limit
Scatter Diagrams
• Chart used to identify the possible relationship
between two process characteristics
• Advantages of the scatter diagram
• Provides visual and statistical means to test the
strength of a potential relationship
• Provides a good follow-up to the cause and effect
diagram to find out if there is a connection between
the cause and the effect
Steps to Build a Scatter Diagram
• Collect 50 – 100 paired samples of data.
• Construct a data sheet.
• Draw the Y-axis and the X-axis of the diagram.
• Plot the data on the diagram.
Example of a Scatter Diagram
Variable 1
Summary
• The seven tools of quality discussed in this module
are considered to make up the fundamental
continuous improvement toolkit.
• It is the intent of this course to touch basis on these
tools and not to study them in depth.
• Depth analysis of these theories is considered subject
for another course.
```
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# [solved] How to assign participants to different sequences?
edited October 2015
Hey there,
I am an Open Sesame newbie who is totally unfamiliar with phyton. Nonetheless, I got along pretty well - the program is really intuitive and easy to use. There is just one thing that I really struggle with: I've got three experimental conditions that I realized by creating different loops (positive, negative and ambivalent) and I would like to randomly assign my participants to these conditions. How can I do this? I suggest I have to create an if-statement but since I need three conditions I can not just divide the participants by their subject parity (odd vs. even). Can anybody help me out?
Best,
L.
• edited 5:27PM
Hi Lisa,
So you're talking about a between-subject design? You could use the modulo ('%') for your if-statement. The idea of modulo is that it fits a number as often as possible in another number, e.g. in 10%3, 3 will fit three times in 10, and that will leave you with 1: hence, 10%3 is 1. A bit odd, but this comes in handy in your design. Taking subject_nr % 3, we could say that
`````` if subject_nr % 3 == 0:
subject_group = 'A'
elif subject_nr % 3 == 1:
subject_group = 'B'
elif subject_nr % 3 == 2:
subject_group = 'C'
exp.set('subject_group', subject_group) # (this makes the variable available outside
``````
Then, in the overview window clicking on the experimental sequence that contains your three loops, you'll find that all loops run 'always' by default. You can change this into a logic statement - for instance, `[subject_group] = 'B'`.
Hope this helps
Josh
• edited 5:27PM
Josh,
Thank you so much for your help! This is definitely a step in the right direction. Unfortunately, I still get an error message that says"IndentationError: unexpected indent". Do you happen to know how I could solve this?
Sorry for being so helpless, but I really don't know anything about phyton.
best,
L.
• edited 5:27PM
Never mind, I figured it out! Thank you so much!!
• Hi Josh,
I am also very new to programming and OpenSesame.
Where exactly do I put that code into the general script? Everytime I click apply it takes the code away so I assume I'm putting it in the wrong place.
Best,
Matthew
• @MDJ
> Where exactly do I put that code into the general script? Everytime I click apply it takes the code away so I assume I'm putting it in the wrong place.
This is Python code, so you have to insert in an `inline_script` item, ideally at the very start of the experiment.
For information about Python in OpenSesame, see:
Thanked by 1MDJ
There's much bigger issues in the world, I know. But I first have to take care of the world I know.
cogsci.nl/smathot
• edited March 12
Thanks Sebastian....I didn't realise that I needed to use an inline_script item to do that!
I have now placed the item at the start of the experiment and used similar code to Josh except I have 6 categories instead of 3.
However, an error is now coming up:
" Inline script, line 1, in <module>
NameError: name 'subject_nr' is not defined "
This is probably really obvious but where and how exactly should I define subject_nr?
Thanks again,
Matthew
• Hi Matthew,
The script that Josh provided above is actually not entirely correct. Here's what it should look like:
```if var.subject_nr % 3 == 0:
var.subject_group = 'A'
elif var.subject_nr % 3 == 1:
var.subject_group = 'B'
elif var.subject_nr % 3 == 2:
var.subject_group = 'C'
```
If you want to use an `inline_script`, I recommend at least walking through the intermediate tutorial to get some understanding of how this works. Otherwise you'll blindly stumble from one confusing problem into the other which is frustrating and inefficient!
Cheers!
Sebastiaan
There's much bigger issues in the world, I know. But I first have to take care of the world I know.
cogsci.nl/smathot | 977 | 3,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-22 | longest | en | 0.945259 |
https://brainmass.com/business/dividend-yield/lease-cost-of-stock-dividend-yield-yield-on-a-bond-34373 | 1,721,320,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00739.warc.gz | 132,539,081 | 7,895 | Purchase Solution
# Lease, cost of stock, dividend yield, yield on a bond
Not what you're looking for?
1. You are leasing a machine for your business. It will cost the lessor \$15,000 to be carried for a 6 year. lease term, and you will be putting down 40.00%. What will the annual rental charge be to you if the lessor pays 15% and must earn profits and risk of 5% on the deal?
2. A preferred stock issue was sold 2 years ago by your firm for a price of \$25.00. The current market price of preferred issue is \$17.00. The stock has a par value of \$25.00 and a coupon rate stated at 5.00%. What is the cost of the issue (kps). What was the cost when it was issued, if you paid \$4.00 per share in flotation costs?
3. A common stock issue is selling currently for \$60.00. Net income amounts to 2,200,000, there are 300,000 shares outstanding. What would the dividend yield on the stock be at the following payout rates?
a) 40%
b) 20%
4. What maximum risk can you sustain if you invest in the following project?
Year 0 \$875,000 Cost
1 \$228,000 Returns
2 \$196,000 "
3 \$190,000 "
4 \$275,000 "
5 \$350,000 "
Money costs you 6.00%
Profits Desired is 4.00%
5. A bond has a 7.50% coupon rate, maturing 4 years from now. To buy this bond, you must invest \$1.025 today. What will your return on investment (yield) be?
##### Solution Summary
Answers questions dealing with Lease- annual rental, cost of preferred stock, dividend yield on the stock at different payout ratios, return on a project, yield on a bond
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This quiz is designed to assess your current ability for determining the characteristics of ethical behavior. It is essential that leaders, managers, and employees are able to distinguish between positive and negative ethical behavior. The quicker you assess a person's ethical tendency, the awareness empowers you to develop a strategy on how to interact with them.
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This Quiz is compiled of questions that pertain to IPOs (Initial Public Offerings)
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This quiz intended to help students understand change and resistance in organizations
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##### Income Streams
In our ever changing world, developing secondary income streams is becoming more important. This quiz provides a brief overview of income sources. | 569 | 2,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.931974 |
https://forum.arduino.cc/t/maths-adding-problem/935366 | 1,642,580,539,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00635.warc.gz | 306,172,847 | 6,896 | So I have look at the forum to find a simple anser to this, but have not found one.
I have this simple code:
``````int x=0;
//int32_t x=0;
//long x=0;
//double x=0;
void loop()
{
x=532 * 4620;
Serial.println(x);
}
``````
But in every case it adds up wrong. Any idea what I need to declare x as to make it add up correctly.
Your post was MOVED to its current location as it is more suitable.
By the way the answer I get is -32528
``````int x=0;
``````
x is declared as an int
An int on most Arduinos (which are you using ?) can hold a maximum value of 32,768
What is the result of 532 * 4620 ?
Will it fit in an int variable ?
Declare x a an unsigned long and see what happens
Hi, @Uknod
Welcome to the forum.
Where is the void setup() and serial.begin() statements?
Tom...
Yep, unsigned long worked, so why doesn't long work?
long is a signed variable with a maximum value of 2,147,483,647 which is smaller that the result of the multiplication (2,457,840)
EDIT- take no notice of what I wrote, it is incorrect. See later in the topic for more contributions
Ah, got it, thanks very much problem solved.
Can you maybe change the topic title to better reflect the subject?
Check the numbers again, I almost made the same mistake.
Not sure why unsigned works and signed does not, but math is usually done with integers, so you may need to tell the compiler that at least one of the numbers is a long for the multiplication.
Whaaaaaa????
2,147,483,647 is SMALLER THAN 2,457,840
On the Arduino Uno (and other ATmega based boards) an int stores a 16-bit (2-byte) value. This >yields a range of -32,768 to 32,767 (minimum value of -2^15 and a maximum value of (2^15) - 1).
Long variables are extended size variables for number storage, and store 32 bits (4 bytes), from >-2,147,483,648 to 2,147,483,647.
I tried using long and I got the wrong answer as @Uknod , but that doesn't explain why.
Ofcourse unsigned long worked;
Unsigned long variables are extended size variables for number storage, and store 32 bits (4 bytes). Unlike standard longs unsigned longs won’t store negative numbers, making their range from 0 to 4,294,967,295 (2^32 - 1).
Tom...
I get the wrong answer using unsigned long, although it is positive because an unsigned number can never be negative. The compiler gives a warning of overflow regardless of the use of long or unsigned long.
Tell the compiler to do the math using longs:
`` x = 532 * 4620L;``
Yes, thats what I thought, I tohought I was loosing it there, so why long doesn;t work is a mystery, but like a lt of things arduino, you just have to go with what works, unsigned long works so I'm happy.
Change it to what, it seems pretty well named to me, arduino can't add up.
Hi,
My test code.
``````int x = 0;
long y = 0;
unsigned long z = 0;
void setup()
{
Serial.begin(9600);
}
void loop()
{
x = 532 * 4620;
Serial.println(" equation => variable = 532 * 4620 ");
Serial.println(" ");
Serial.print(" int x ");
Serial.println(x);
y = 532 * 4620;
Serial.print(" long y ");
Serial.println(y);
z = 532 * 4620;
Serial.print(" unsigned long z ");
Serial.println(z);
Serial.println("=============================================");
delay(2000);
}
``````
And my unsigned long == 4294934768
equation => variable = 532 * 4620
int x -32528
long y -32528
unsigned long z 4294934768
Calculated value = 2,457,840
Tom...
Which is not the product of 532 and 4620, it is the result of the compiler extending -32528 to an unsigned long.
What, that makes no sense, as my code is now working fine using unsigned long.
Which arduino are you using?
A genuine UNO, I'll try it on a copy nano, shall I.
Hi,
Revised test code.
``````int x = 0;
long y = 0;
unsigned long z = 0;
void setup()
{
Serial.begin(9600);
}
void loop()
{
x = 532 * 4620;
Serial.println(" equation => variable = 532 * 4620 ");
Serial.println(" ");
Serial.print(" int x ");
Serial.println(x);
y = 532 * 4620;
Serial.print(" long y ");
Serial.println(y);
z = 532 * 4620;
Serial.print(" unsigned long z ");
Serial.println(z);
z = 532L * 4620L;
Serial.println(" z = 532L * 4620L ");
Serial.print(" unsigned long z ");
Serial.println(z);
Serial.println("=============================================");
delay(2000);
}
``````
Output;
equation => variable = 532 * 4620
int x -32528
long y -32528
unsigned long z 4294934768
z = 532L * 4620L
unsigned long z 2457840
Tom... | 1,218 | 4,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-05 | latest | en | 0.928013 |
https://www.osh.net/calc/numbers-to-words/900100 | 1,701,793,086,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00898.warc.gz | 1,042,562,638 | 2,233 | # How do you write 900,100 in words?
900,100 = nine hundred thousand one hundred
Spell another number
# How to write \$900,100 on a check
The most common reason to write 900,100 as nine hundred thousand one hundred would be if you're writing a check for \$900,100. Just below the "pay to the order of" line, you would write the dollar amount in words. In this case, you might write "nine hundred thousand one hundred and 00/100 dollars."
## Explore numbers similar to 900,100
← Prev num Next num →
900,000 900,200 | 134 | 519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-50 | latest | en | 0.871837 |
https://pkgdown.jrwb.de/mkin/reference/ilr.html | 1,675,691,664,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00839.warc.gz | 471,042,149 | 4,109 | This implementation is a special case of the class of isometric log-ratio transformations.
ilr(x)
invilr(x)
## Arguments
x
A numeric vector. Naturally, the forward transformation is only sensible for vectors with all elements being greater than zero.
## Value
The result of the forward or backward transformation. The returned components always sum to 1 for the case of the inverse log-ratio transformation.
## References
Peter Filzmoser, Karel Hron (2008) Outlier Detection for Compositional Data Using Robust Methods. Math Geosci 40 233-248
Another implementation can be found in R package robCompositions.
## Author
René Lehmann and Johannes Ranke
## Examples
# Order matters
ilr(c(0.1, 1, 10))
#> [1] -1.628174 -2.820079
ilr(c(10, 1, 0.1))
#> [1] 1.628174 2.820079
# Equal entries give ilr transformations with zeros as elements
ilr(c(3, 3, 3))
#> [1] 0 0
# Almost equal entries give small numbers
ilr(c(0.3, 0.4, 0.3))
#> [1] -0.2034219 0.1174457
# Only the ratio between the numbers counts, not their sum
invilr(ilr(c(0.7, 0.29, 0.01)))
#> [1] 0.70 0.29 0.01
invilr(ilr(2.1 * c(0.7, 0.29, 0.01)))
#> [1] 0.70 0.29 0.01
# Inverse transformation of larger numbers gives unequal elements
invilr(-10)
#> [1] 7.213536e-07 9.999993e-01
invilr(c(-10, 0))
#> [1] 7.207415e-07 9.991507e-01 8.486044e-04
# The sum of the elements of the inverse ilr is 1
sum(invilr(c(-10, 0)))
#> [1] 1
# This is why we do not need all elements of the inverse transformation to go back:
a <- c(0.1, 0.3, 0.5)
b <- invilr(a)
length(b) # Four elements
#> [1] 4
ilr(c(b[1:3], 1 - sum(b[1:3]))) # Gives c(0.1, 0.3, 0.5)
#> [1] 0.1 0.3 0.5 | 615 | 1,629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-06 | latest | en | 0.633137 |
https://wwwmpa.mpa-garching.mpg.de/gadget/gadget-list/0113.html | 1,674,901,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00836.warc.gz | 1,033,910,647 | 3,874 | # Re: velocity units
From: Volker Springel <volker_at_MPA-Garching.MPG.DE>
Date: Mon, 30 Oct 2006 14:23:56 +0100
Melanie Jo Clarke wrote:
> I have been attempting to use Grafic-1 initial conditions to run small scale
> Gadget2 simulations and have run into the following problem with velocity
> units.
>
> The manual suggests to divide by sqrt(a) to convert from physical velocity to
> Gadget units. When I do this, I find that the Gadget simulation does not
> evolve far enough compared to a P3M simulation run with the same initial
> conditions.
>
> As I understand it, the internal units for velocity are dx/d(ln a), so to
> convert from physical velocity, dx/dt, to these units, I should divide by d(ln
> a)/dt. This factor is a*sqrt(omega_m/a + omega_v*a^2 + 1 - omega_m - omega_v),
> where omega_m is the matter density ratio and omega_v is the vacuum energy
> density ratio. This factor reduces to sqrt(a) in the case that omega_m=1 and
> omega_v=0. When I use this factor to convert physical velocity to Gadget
> units, I get good agreement between P3M and Gadget simulations.
>
> Is my interpretation correct? If so, can the Gadget2 manual be corrected so
> that future users don't run into this problem?
>
Hi Melanie,
The manual is correct on this...
I can only speculate where you got confused, but I'm guessing you have
mixed up the definitions of comoving, physical, and peculiar velocity.
Note that they are all different. Also note that the manual says that
the IC file should contain the *peculiar* velocity divided by sqrt(a),
not the *physical* velocity as you state above. Let "x" denote comoving
coordinates and "r=a*x" physical coordinates. Then I call
comoving velocity: dx/dt
physical velocity: dr/dt = H(a)*r + a*dx/dt
peculiar velocity: v = a * dx/dt
The physical velocity is hence the peculiar velocity plus the Hubble flow.
I think your attempts to guess the correct conversion factor have caused
some additional confusion. Let me try to clarify: The internal velocity
variable of gadget2 is not given by dx/d(ln a). Rather, it is given by
the canonical momentum p = a^2 * dx/dt, which is different from the
definition you assumed. The IC-file and snapshot files of gadget don't
contain the variable "p" directly because of historical reasons of
compatibility with gagdet-1. Instead, they contain the velocity variable
u = v/sqrt(a) = sqrt(a) * dx/dt = p / a^(3/2), which is just what the
manual says. (The conversion between u and p is done on the fly when
Also note that d(ln a)/dt is not a*sqrt(omega_m/a + omega_v*a^2 + 1 -
omega_m - omega_v), as you say above... This factor is equal to the
Hubble rate, i.e.: d(ln a)/dt = H(a) = H_0 * sqrt(omega_m/a^3 + omega_v
+ (1 - omega_m - omega_v)/a^2).
If you can't solve your unit conversion problem, I'd suggest to talk to
Michele Trenti, who figured out the grafic->gadget2 conversion.
Best wishes,
Volker
> Thank you,
> Melanie Clarke Dosaj
>
>
>
>
> -----------------------------------------------------------
>
> If you wish to unsubscribe from this mailing, send mail to
> minimalist_at_MPA-Garching.MPG.de with a subject of: unsubscribe gadget-list
> A web-archive of this mailing list is available here: | 830 | 3,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-06 | latest | en | 0.878921 |
https://en.wikipedia.org/wiki/Focused_proof | 1,632,086,765,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00420.warc.gz | 301,258,049 | 15,213 | # Focused proof
In mathematical logic, focused proofs are a family of analytic proofs that arise through goal-directed proof-search, and are a topic of study in structural proof theory and reductive logic. They form the most general definition of goal-directed proof-search—in which someone chooses a formula and performs hereditary reductions until the result meets some condition. The extremal case where reduction only terminates when axioms are reached forms the sub-family of uniform proofs.[1]
A sequent calculus is said to have the focusing property when focused proofs are complete for some terminating condition. For System LK, System LJ, and System LL, uniform proofs are focused proofs where all the atoms are assigned negative polarity.[2] Many other sequent calculi has been shown to have the focusing property, notably the nested sequent calculi of both the classical and intuitionistic variants of the modal logics in the S5 cube.[3][4]
## Uniform proofs
In the sequent calculus for an intuitionistic logic, the uniform proofs can be characterised as those in which the upward reading performs all right rules before the left rules. Typically, uniform proofs are not complete for the logic i.e., not all sequents or formulas admit a uniform proof, so one considers fragments where they are complete e.g., the hereditary Harrop fragment of Intuitionistic logic. Due to the deterministic behaviour, uniform proof-search has been used as the control mechanism defining the programming language paradigm of logic programming.[1] Occasionally, uniform proof-search is implemented in a variant of the sequent calculus for the given logic where context management is automatic thereby increasing the fragment for which one can define a logic programming langue.[5]
## Focused proofs
The focusing principle was originally classified through the disambiguation between synchronous and asynchronous connective in Linear Logic i.e., connectives that interact with the context and those that do not, as consequence of research on logic programming. They are now an increasingly important example of control in reductive logic, and can drastically improve proof-search procedures in industry. The essential idea of focusing is to identify and coalesce the non-deterministic choices in a proof, so that a proof can be seen as an alternation of negative phases ( where invertible rules are applied eagerly), and positive phases (where applications of the other rules are confined and controlled).[3]
### Polarisation
According to the rules in the sequent calculus, formulas are canonically put into one of two classes called positive and negative e.g., in LK and LJ the formula ${\displaystyle \phi \lor \psi }$ is positive. The only freedom is over atoms are assigned a polarity freely. For negative formulas provability is invariant under the application of a right rule; and, dually, for a positive formulas provability is invariant under the application of a left rule. In either case one can safely apply rules in any order to hereditary sub-formulas of the same polarity.
In the case of a right rule applied to a positive formula, or a left rule applied to a negative formula, one may result in invalid sequents e.g., in LK and LJ there is no proof of the sequent ${\displaystyle B\lor A\implies A\lor B}$ beginning with a right rule. A calculus admits the focusing principle if when an original reduct was provable then the hereditary reducts of the same polarity are also provable. That is, one can commit to focusing on decomposing a formula and its sub-formulas of the same polarity without loss of completeness.
### Focused system
A sequent calculus is often shown to have the focusing property by working in a related calculus where polarity explicitly controls which rules apply. Proofs in such systems are in focused, unfocused, or neutral phases, where the first two are characterised by hereditary decomposition; and the latter by forcing a choice of focus. One of the most important operational behaviours a procedure can undergo is backtracking i.e., returning to an earlier stage in the computation where a choice was made. In focused systems for classical and Intuitionistic logic, the use of backtracking can be simulated by pseudo-contraction.
Let ${\displaystyle \uparrow }$ and ${\displaystyle \downarrow }$ denote change of polarity, the former making a formula negative, and the latter positive; and call a formula with an arrow neutral. Recall that ${\displaystyle \lor }$ is positive, and consider the neutral polarized sequent ${\displaystyle \downarrow \uparrow \phi \lor \psi \implies \uparrow \phi \lor \psi }$, which is interpreted as the actual sequent ${\displaystyle \phi \lor \psi \implies \phi \lor \psi }$. For neutral sequents such as this, the focused system forces on to make an explicit choice of which formula to focus on, denoted by ${\displaystyle \langle \,\rangle }$. To perform a proof-search the best thing is to choose the left formula, since ${\displaystyle \lor }$ is positive, indeed (as discussed above) in some cases there are no proofs where the focus is on the right formula. To overcome this, some focused calculi create a backtracking point such that focusing on the right yields ${\displaystyle \downarrow \uparrow \phi \lor \psi \implies \langle \phi \lor \psi \rangle ,\uparrow \phi \lor \psi }$, which is still as ${\displaystyle \phi \lor \psi \implies \phi \lor \psi }$. The second formula on the right can be removed only when the focused phase has finished, but if proof-search gets stuck before this happens the sequent may remove the focused component thereby returning to the choice e.g., ${\displaystyle \downarrow \uparrow B\lor A\implies \langle A\rangle ,\uparrow A\lor B}$ must be taken to ${\displaystyle \downarrow \uparrow B\lor A\implies \uparrow A\lor B}$ as no other reductive inference can be made. This is a pseudo-contraction since it has the syntactic form of a contraction on the right, but the actual formula doesn't exist i.e., in the interpretation of the proof in the focused system the sequent has only one formula on the right.
## References
1. ^ a b Miller, Dale; Nadathur, Gopalan; Pfenning, Frank; Scedrov, Andre (1991-03-14). "Uniform proofs as a foundation for logic programming". Annals of Pure and Applied Logic. 51 (1): 125–157. doi:10.1016/0168-0072(91)90068-W. ISSN 0168-0072.
2. ^ Liang, Chuck; Miller, Dale (2009-11-01). "Focusing and polarization in linear, intuitionistic, and classical logics". Theoretical Computer Science. Abstract Interpretation and Logic Programming: In honor of professor Giorgio Levi. 410 (46): 4747–4768. doi:10.1016/j.tcs.2009.07.041. ISSN 0304-3975.
3. ^ a b Chaudhuri, Kaustuv; Marin, Sonia; Straßburger, Lutz (2016), Jacobs, Bart; Löding, Christof (eds.), "Focused and Synthetic Nested Sequents", Foundations of Software Science and Computation Structures, Berlin, Heidelberg: Springer Berlin Heidelberg, 9634, pp. 390–407, doi:10.1007/978-3-662-49630-5_23, ISBN 978-3-662-49629-9
4. ^ Chaudhuri, Kaustuv; Marin, Sonia; Straßburger, Lutz (2016). "Modular Focused Proof Systems for Intuitionistic Modal Logics". Marc Herbstritt: 18 pages. doi:10.4230/LIPICS.FSCD.2016.16. Cite journal requires |journal= (help)
5. ^ Armelín, Pablo A.; Pym, David J. (2001), "Bunched Logic Programming", Automated Reasoning, Berlin, Heidelberg: Springer Berlin Heidelberg, pp. 289–304, doi:10.1007/3-540-45744-5_21, ISBN 978-3-540-42254-9 | 1,766 | 7,469 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-39 | latest | en | 0.93804 |
https://stats.stackexchange.com/questions/256218/understanding-the-bayes-risk/256222 | 1,618,850,495,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00510.warc.gz | 633,005,640 | 41,894 | # Understanding the Bayes risk
When evaluating an estimator, the two probably most common used criteria are the maximum risk and the Bayes risk. My question refers to the latter one:
The bayes risk under the prior $\pi$ is defined as follows:
$$B_{\pi} (\hat{\theta}) = \int R(\theta, \hat{\theta} ) \pi ( \theta ) d \theta$$
I don't quite get what the prior $\pi$ is doing and how I should interpret it. If I have a risk function $R(\theta, \hat{\theta} )$ and plot it, intuitively I would take its area as a criterion to judge how "strong" the risk is over all possible values of $\theta$. But involving the prior somehow destroys this intuition again, although it is close. Can someone help me how to interpret the prior?
• I fail to see how intuitive plotting the risk function can be when considering several parameters: in that setting, the functions intersect and do not identify a "best" estimator. The Bayes risk returns one single number for each estimator and hence allows for a ranking of all estimators. – Xi'an Jan 14 '17 at 14:32
## 2 Answers
[Here is an excerpt from my own textbook, The Bayesian Choice (2007), that argues in favour of a decision-theoretic approach to Bayesian analysis, hence of using the Bayes risk.]
Except for the most trivial settings, it is generally impossible to uniformly minimize (in $d$) the loss function $\text{L}(\theta,d)$ when $\theta$ is unknown. In order to derive an effective comparison criterion from the loss function, the frequentist approach proposes to consider instead the average loss (or frequentist risk) \begin{eqnarray*} R(\theta,\delta) & = & \mathbb{E}_\theta \lbrack \text{L} (\theta ,\delta(x))\rbrack \\ & = & \int_{\cal X} \text{L}(\theta,\delta(x))f(x|\theta) \,dx , \end{eqnarray*} where $\delta(x)$ is the decision rule, i.e., the allocation of a decision to each outcome $x\sim f(x|\theta)$ from the random experiment.
The function $\delta$, from ${\mathcal X}$ in $\mathfrak{D}$, is usually called estimator (while the value $\delta(x)$ is called estimate of $\theta$). When there is no risk of confusion, we also denote the set of estimators by $\mathfrak{D}$.
The frequentist paradigm relies on this criterion to compare estimators and, if possible, to select the best estimator, the reasoning being that estimators are evaluated on their long-run performance for all possible values of the parameter $\theta$. Notice, however, that there are several difficulties associated with this approach.
1. The error (loss) is averaged over the different values of $x$ proportionally to the density $f(x|\theta)$. Therefore, it seems that the observation $x$ is not taken into account any further. The risk criterion evaluates procedures on their long-run performance and not directly for the given observation, $x$. Such an evaluation may be satisfactory for the statistician, but it is not so appealing for a client, who wants optimal results for her data $x$, not that of another's!
2. The frequentist analysis of the decision problem implicitly assumes that this problem will be met again and again, for the frequency evaluation to make sense. Indeed, $R(\theta,\delta)$ is approximately the average loss over i.i.d. repetitions of the same experiment, according to the Law of Large Numbers. However, on both philosophical and practical grounds, there is a lot of controversy over the very notion of repeatability of experiments (see Jeffreys (1961)). For one thing, if new observations come to the statistician, she should make use of them, and this could modify the way the experiment is conducted, as in, for instance, medical trials.
3. For a procedure $\delta$, the risk $R(\theta, \delta)$ is a function of the parameter $\theta$. Therefore, the frequentist approach does not induce a total ordering on the set of procedures. It is generally impossible to compare decision procedures with this criterion, since two crossing risk functions prevent comparison between the corresponding estimators. At best, one may hope for a procedure $\delta_0$ that uniformly minimizes $R(\theta,\delta)$, but such cases rarely occur unless the space of decision procedures is restricted. Best procedures can only be obtained by restricting rather artificially the set of authorized procedures.
Example 2.4 - Consider $x_1$ and $x_2$, two observations from $$P_{\theta}(x = \theta-1) = P_{\theta}(x = \theta+1) = 0.5, \qquad \theta\in\mathbb{R}.$$ The parameter of interest is $\theta$ (i.e., $\mathfrak{D} = \Theta$) and it is estimated by estimators $\delta$ under the loss $$\text{L}(\theta,\delta) = 1-\mathbb{I}_{\theta}(\delta),$$ often called $0-1$ loss, which penalizes errors of estimation, whatever their magnitude, by $1$. Considering the particular \est $$\delta_0(x_1,x_2) = {x_1+x_2 \over 2},$$ its risk function is \begin{eqnarray*} R(\theta,\delta_0) & = & 1-P_{\theta}(\delta_0(x_1,x_2) = \theta) \\ & = & 1-P_{\theta}(x_1 \ne x_2) = 0.5. \end{eqnarray*} This computation shows that the estimator $\delta_0$ is correct half of the time. Actually, this estimator is always correct when $x_1\ne x_2$, and always wrong otherwise. Now, the \est\ $\delta_1(x_1,x_2) = x_1+1$ also has a risk function equal to $0.5$, as does $\delta_2(x_1,x_2) = x_2-1$. Therefore, $\delta_0$, $\delta_1$ and $\delta_2$ cannot be ranked under the $0-1$ loss. $\blacktriangleright$
On the contrary, the Bayesian approach to Decision Theory integrates on the space $\Theta$ since $\theta$ is unknown, instead of integrating on the space ${\cal X}$ as $x$ is known. It relies on the posterior expected loss \begin{eqnarray*} \rho(\pi,d|x) & = & \mathbb{E}^\pi[L(\theta,d)|x] \\ & = & \int_{\Theta} \text{L}(\theta,d) \pi(\theta|x)\, d\theta, \end{eqnarray*} which averages the error (i.e., the loss) according to the posterior distribution of the parameter $\theta$, conditionally on the observed value} $x$. Given $x$, the average error resulting from decision $d$ is actually $\rho(\pi,d|x)$. The posterior expected loss is thus a function of $x$ but this dependence is not troublesome, as opposed to the frequentist dependence of the risk on the parameter because $x$, contrary to $\theta$, is known.
• So you are Christian Robert. I have met George Casella. I think you have published book(s) with him that I am aware of.. – Michael R. Chernick Jan 14 '17 at 16:28
• +1 answers don't get much better than that - great book by the way – Xavier Bourret Sicotte Jun 26 '18 at 12:09
Quoting the classical Statistical Decision Theory by James O. Berger:
[...] We have already stated that decision rules will be evaluated in terms of their risk functions $R(\theta, \delta)$. [...] The problem, as pointed out earlier, is that different admissible decision rules will have risks which are better for different $\theta$'s. To the rescue comes the prior $\pi(\theta)$, which supposedly reflects which $\theta$'s are the "likely" ones to occur. It seems very reasonable to "weight" $R(\theta, \delta)$ by $\pi(\theta)$ and average.
Yes you can evaluate $R(\theta, \delta)$ for each $\theta$, but then you would implicitly assume that each possible value of $\theta$ is equally likely. In Bayesian scenario you pick prior $\pi(\theta)$ that reflects probabilities of observing different $\theta$'s and include such information. | 1,869 | 7,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-17 | latest | en | 0.842141 |
https://www.unitsconverters.com/en/Gr/Daa-To-Gr/Stremma/Utu-7531-7536 | 1,695,358,984,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00159.warc.gz | 1,184,291,362 | 35,316 | Formula Used
1 Grains Per Square Centimeter = 1000 Grain Per Decare
1 Grains Per Square Centimeter = 1000 Grain Per Stremma
1 Grain Per Decare = 10000 Grain Per Stremma
## gr/daa to gr/Stremma Conversion
The abbreviation for gr/daa and gr/Stremma is grain per decare and grain per stremma respectively. 1 gr/daa is 10000 times bigger than a gr/Stremma. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including gr/daa to gr/Stremma conversion.
## Grain Per Decare to gr/Stremma
Check our Grain Per Decare to gr/Stremma converter and click on formula to get the conversion factor. When you are converting area density from Grain Per Decare to gr/Stremma, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## gr/daa to Grain Per Stremma
The formula used to convert gr/daa to Grain Per Stremma is 1 Grain Per Decare = 10000 Grain Per Stremma. Measurement is one of the most fundamental concepts. Note that we have Grains Per Square Centimeter as the biggest unit for length while Grain Per Square Mile (US Survey) is the smallest one.
## Convert gr/daa to gr/Stremma
How to convert gr/daa to gr/Stremma? Now you can do gr/daa to gr/Stremma conversion with the help of this tool. In the length measurement, first choose gr/daa from the left dropdown and gr/Stremma from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from gr/Stremma to gr/daa? You can check our gr/Stremma to gr/daa converter.
## gr/daa to gr/Stremma Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area density finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like gr/daa to gr/Stremma through multiplicative conversion factors. When you are converting area density, you need a Grain Per Decare to Grain Per Stremma converter that is elaborate and still easy to use. Converting gr/daa to Grain Per Stremma is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Grain Per Decare to gr/Stremma, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in gr/daa to gr/Stremma conversion along with a table representing the entire conversion.
Let Others Know | 672 | 2,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.823302 |
http://www.objectivebooks.com/2015/07/mcq-objective-questions-on-engineering.html | 1,519,056,053,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812756.57/warc/CC-MAIN-20180219151705-20180219171705-00176.warc.gz | 531,890,874 | 34,020 | MCQ objective Questions on Engineering Mechanics - Set 20 - ObjectiveBooks
# Practice Test: Question Set - 20
1. Which of the following is an equation of linear motion?(where, u and v = Initial and final velocity of the body, a = Acceleration of the body, and s = Displacement of the body in time t seconds.)
(A) v = u + a.t
(B) s = u.t + ½ a.t2
(C) v2 = u2 + 2a.s
(D) All of these
2. Moment of inertia of a hollow circular section, as shown in the below figure about X-axis, is
(A) π/16 (D² - d²)
(B) π/16 (D³ - d³)
(C) π/32 (D⁴ - d⁴)
(D) π/64 (D⁴ - d⁴)
3. The forces which do not meet at one point and their lines of action do not lie on the same plane are known as
(A) Coplanar concurrent forces
(B) Coplanar non-concurrent forces
(C) Non-coplanar concurrent forces
(D) None of these
4. Concurrent forces are those forces whose lines of action
(A) Lie on the same line
(B) Meet at one point
(C) Meet on the same plane
(D) None of these
5. The moment of inertia of a solid cylinder of mass 'm', radius 'r' and length 'l' about the longitudinal axis or polar axis is
(A) mr2/2
(B) mr2/4
(C) mr2/6
(D) mr2/8
6. According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.)
(A) IP = IG + Ah2
(B) IP = IG - Ah2
(C) IP = IG / Ah2
(D) IP = Ah2 / IG
7. The law of the machine is (where P = Effort applied to lift the load, m = A constant which is equal to the slope of the line, W = Load lifted, and C = Another constant which represents the machine friction.)
(A) P = mW - C
(B) P = m/W + C
(C) P = mW + C
(D) P = C - mW
8. The wheels of a moving car possess
(A) Potential energy only
(B) Kinetic energy of translation only
(C) Kinetic energy of rotation only
(D) Kinetic energy of translation and rotation both
9. The resultant of two equal forces ‘P’ making an angle ‘θ’, is given by
(A) 2P sinθ/2
(B) 2P cosθ/2
(C) 2P tanθ/2
(D) 2P cotθ/2
10. The coefficient of restitution for inelastic bodies is
(A) Zero
(B) One
(C) Between zero and one
(D) More than one
11. If a number of forces are acting at a point, their resultant will be inclined at an angle 'θ' with the horizontal, such that
(A) tanθ = ΣH/ΣV
(B) tanθ = ΣV/ΣH
(C) tanθ = ΣV × ΣH
(D) tanθ = √(ΣV + ΣH)
12. The periodic time of a particle with simple harmonic motion is _________ proportional to the angular velocity.
(A) Directly
(B) Inversely
(C) Square root
(D) None of these
13. The efficiency of a screw jack is maximum, when (where α = Helix angle, and φ = Angle of friction.)
(A) α = 45° + φ/2
(B) α = 45° - φ/2
(C) α = 90° + φ
(D) α = 90° - φ
14. The total motion possessed by a body, is called
(A) Impulsive force
(B) Mass
(C) Weight
(D) Momentum
15. A screw jack used for lifting the loads is
(A) A reversible machine
(B) A non-reversible machine
(C) An ideal machine
(D) None of these
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# Sentence correction by question sets based on grammar rules
Author Message
Intern
Joined: 14 Jul 2011
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Sentence correction by question sets based on grammar rules [#permalink]
### Show Tags
14 Jul 2011, 12:35
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Hi,
Anyone has Sentence correction question sets listed/segregated as per the major grammar rules tested in GMAT?
Also if you can refer personal choice material which really helped to gain control over sentence correction.
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Re: Sentence correction by question sets based on grammar rules [#permalink]
### Show Tags
14 Jul 2011, 12:45
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Re: Sentence correction by question sets based on grammar rules [#permalink]
### Show Tags
17 Jul 2011, 21:19
@gmat760isb, I suggest you go to the SC forum and navigate thereafter using the tags. That way you'll get to see questions of various topics/difficulty levels.
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Re: Sentence correction by question sets based on grammar rules [#permalink] 17 Jul 2011, 21:19
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Topics:
Is this sentence correct based on GMAT rules REVIVING LG 3 18 Oct 2010, 00:28
Sentence Correction question 0 20 Jul 2010, 08:01
1 sentence correction 1 14 Oct 2009, 01:04
sentence correction 0 13 Oct 2009, 19:15
sentence correction 8 04 Sep 2011, 19:59
Display posts from previous: Sort by | 676 | 2,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-22 | latest | en | 0.889614 |
https://forum.arduino.cc/t/please-tell-me-the-approx-voltage-of-this-bo-motor/479476 | 1,638,746,317,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00141.warc.gz | 323,362,537 | 6,446 | # Please tell me the approx voltage of this bo motor
The image is attached below
You cannot determine what the winding voltage is from a picture like that!
You can measure the winding resistance with a multimeter and compare to similar sized motors
of known voltage.
You can measure the motor rpm at various supply voltages and guess that the design rpm is around
8000 to 12000 for a motor of that size and thus figure out a likely nominal voltage (as well as determine
the motor constant).
Image from Original Post so we don’t have to download it. See this Image Guide
…R
I have some motors like that. Try it with 3v or 6v.
...R
How do i use tat motor with l298 motor driver?
The usual way. Its a DC motor. There are instructions for driving motors with motor drivers
out there, many of them in fact. Try searching for "arduino L298 motor circuit" or some such.
I mean iam unable to determine the voltage of motor so which battery should i use?
A 6volt one
The motor is 6 volt one or i should use battery of 6 volts?
Anvay:
The motor is 6 volt one or i should use battery of 6 volts?
...R
Robin2 ok. So the motor is of 6v. Then how many volts battery should i connect to the driver?
Anvay:
Robin2 ok. So the motor is of 6v. Then how many volts battery should i connect to the driver?
Why are you asking that question?
I was about to add to Reply #9 that maybe there is a question or a reason for uncertainty in your mind but you have not told us what it is so I don't know how to respond.
...R
Well L298's drop about 2V, so 8V seems ideal! Of course it depends whether you need the full rated speed
or not.
Can i use 9v instead of 8v? It wont damage the driver and motor right?
Not if you mean small 9V batteries, they can't power motors.
Anvay:
Can i use 9v instead of 8v? It wont damage the driver and motor right?
What damages a motor is too much current causing it to overheat (and let the smoke out). Higher voltages force more current through the motor coils.
If you use PWM for speed control (with analogWrite() ) then the amount of time when the current flows through the motor is reduced unless you are using high values (near to the max of 255) with analogWrite(). That means that you can power a low voltage motor with a higher voltage if you don't use high values with analogWrite(). The trick is to make sure the motor does not overheat.
In general when it is said that a motor is (say) a 6v motor that is just rough guidance. It will probably work fine with somewhat lower and higher voltages. The lower voltages will result in less torque and a lower top speed and the higher voltages will give more torque and a higher top speed at the risk of causing the motor to overheat. If you use PWM and limit the max analogWrite() value to prevent the motor overheating you could probably use 12v or 20v to power it.
Another thing that must be taken into account is the voltage range that is acceptable for the motor driver. Some motor drivers don't work at low voltages - I can't immediately see from the datasheet what is the minimum for the L298. Its max is 45v.
...R
Thank you | 748 | 3,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-49 | latest | en | 0.942312 |
http://conversion.org/length/furlong/league-land | 1,723,071,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00350.warc.gz | 6,685,617 | 7,359 | # furlong to league (land) conversion
Conversion number between furlong [fur] and league (land) [lea] is 0.041666583333333. This means, that furlong is smaller unit than league (land).
### Contents [show][hide]
Switch to reverse conversion:
from league (land) to furlong conversion
### Enter the number in furlong:
Decimal Fraction Exponential Expression
[fur]
eg.: 10.12345 or 1.123e5
Result in league (land)
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 furlong = (exactly) (201.168) / ((19008000/3937)) = 0.041666583333333 league (land)
• 1 league (land) = (exactly) ((19008000/3937)) / (201.168) = 24.000048000096 furlong
• ? furlong × (201.168 ("m"/"furlong")) / ((19008000/3937) ("m"/"league (land)")) = ? league (land)
### High precision conversion
If conversion between furlong to metre and metre to league (land) is exactly definied, high precision conversion from furlong to league (land) is enabled.
Decimal places: (0-800)
furlong
Result in league (land):
?
### furlong to league (land) conversion chart
Start value: [furlong] Step size [furlong] How many lines? (max 100)
visual:
furlongleague (land)
00
100.41666583333333
200.83333166666667
301.2499975
401.6666633333333
502.0833291666667
602.499995
702.9166608333333
803.3333266666667
903.7499925
1004.1666583333333
1104.5833241666667
Copy to Excel
## Multiple conversion
Enter numbers in furlong and click convert button.
One number per line.
Converted numbers in league (land):
Click to select all
## Details about furlong and league (land) units:
Convert Furlong to other unit:
### furlong
Definition of furlong unit: ≡ 10 chains = 660 ft = 220 yd.
Convert League (land) to other unit:
### league (land)
Definition of league (land) unit: 3 Statute miles = 5280 foot (US Survey). ≈ 1 hour walk, Currently defined in US as 3 Statute miles, but historically varied from 2 to 9 km
← Back to Length units | 615 | 1,962 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-33 | latest | en | 0.443001 |
http://forum.wilmott.com/viewtopic.php?f=10&t=99291&start=60&view=print | 1,590,518,313,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00586.warc.gz | 48,933,194 | 4,064 | Page 5 of 5
### Numerical accuracy and performance ?
Posted: December 21st, 2015, 10:29 pm
QuoteOriginally posted by: outrunThat's indeed a good solution.In C++11 this is also available in stdIf it's performance critical then you can use a branch free (no if statement) using this:QuoteIf the source type is bool, the value false is converted to zero and the value true is converted to one.So "return (x<0);" is all you need in the body!Yes, it's the Boost Maths quantile code from John M. Quotereturn (x<0)We lose the mathematical subtleties with this unreadable code; semantically it could mean anything. It is a harmful 'optimisation'. IMO
### Numerical accuracy and performance ?
Posted: December 21st, 2015, 10:29 pm
QuoteOriginally posted by: CuchulainnHere's quantile boundary checking; sounds kind of logical I suppose (?)Does Boost distinguish between numerical infinity (i.e., numbers too large to represent in the chosen floating point format) and analytic infinity (i.e., a true infinity)?In the case of heavy-tailed distributions (e.g., the Cauchy distribution), some numerical infinity values might correspond to a non-zero values of the CDF complement.
### Numerical accuracy and performance ?
Posted: December 22nd, 2015, 7:53 am
QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnHere's quantile boundary checking; sounds kind of logical I suppose (?)Does Boost distinguish between numerical infinity (i.e., numbers too large to represent in the chosen floating point format) and analytic infinity (i.e., a true infinity)?In the case of heavy-tailed distributions (e.g., the Cauchy distribution), some numerical infinity values might correspond to a non-zero values of the CDF complement.Indeed.I had a look at Boost Cauchy: an overflow occurs when p = 0,1. Otherwise the quantile return a std::tan(...) (AFAIR the std trigos functions do not throw exception).So we need to know if the return type is an approximation to math infinity or if it is a "spurious" infinity?
### Numerical accuracy and performance ?
Posted: December 22nd, 2015, 1:40 pm
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnHere's quantile boundary checking; sounds kind of logical I suppose (?)Does Boost distinguish between numerical infinity (i.e., numbers too large to represent in the chosen floating point format) and analytic infinity (i.e., a true infinity)?In the case of heavy-tailed distributions (e.g., the Cauchy distribution), some numerical infinity values might correspond to a non-zero values of the CDF complement.Indeed.I had a look at Boost Cauchy: an overflow occurs when p = 0,1. Otherwise the quantile return a std::tan(...) (AFAIR the std trigos functions do not throw exception).So we need to know if the return type is an approximation to math infinity or if it is a "spurious" infinity?Maybe. It depends on whether programmers are using infinity as a specific logical mathematical condition (i.e., the symbolic evaluation of the algorithm would yield infinity which signals a very specific condition in the inputs) or as an indicator that the result is unknown due to a numerical overflow (i.e., a more general condition arising from a broad range of numerical situations). In the case of a CDF, the positive and negative mathematical infinities corresponds to respective specific points in the distribution but a numerical overflow infinity may correspond to a range of values, not a point in the distribution.For example, what is exp(710)? It's not infinity in the mathematical sense but it is too large to fit in a float or double. So if exp(x) returns infinity, what are we to infer? Is x = infinity or is x some unknown value?My guess is that most programmers (and their programs) treat infinity as a mathematical event (which may be wrong) rather than a numerical failure but I've not studied enough programs to have a valid sample size.
### Numerical accuracy and performance ?
Posted: December 22nd, 2015, 6:46 pm
One could also say that there are two types of NaN: not a real number and not a floating-point number. In the first case, the solution does not exist in the real number set (e.g., sqrt(-1)) and in the second case, it does not exist in the digital floating point number set (e.g., exp(710) ).
### Numerical accuracy and performance ?
Posted: December 22nd, 2015, 8:05 pm
QuoteOriginally posted by: Traden4AlphaOne could also say that there are two types of NaN: not a real number and not a floating-point number. In the first case, the solution does not exist in the real number set (e.g., sqrt(-1)) and in the second case, it does not exist in the digital floating point number set (e.g., exp(710) ).The quiet NaN or a signaling NaN?
### Numerical accuracy and performance ?
Posted: December 22nd, 2015, 8:47 pm
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaOne could also say that there are two types of NaN: not a real number and not a floating-point number. In the first case, the solution does not exist in the real number set (e.g., sqrt(-1)) and in the second case, it does not exist in the digital floating point number set (e.g., exp(710) ).The quiet NaN or a signaling NaN?That's an orthogonal issue: the specific cause of the error (i.e., not a real number vs. not a valid 64-bit floating point number) may be independent of the response to the error (i.e., quiet vs. signaling).
### Numerical accuracy and performance ?
Posted: December 23rd, 2015, 8:32 am
QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaOne could also say that there are two types of NaN: not a real number and not a floating-point number. In the first case, the solution does not exist in the real number set (e.g., sqrt(-1)) and in the second case, it does not exist in the digital floating point number set (e.g., exp(710) ).The quiet NaN or a signaling NaN?That's an orthogonal issue: the specific cause of the error (i.e., not a real number vs. not a valid 64-bit floating point number) may be independent of the response to the error (i.e., quiet vs. signaling).design by contractThat's what we are missing.
### Numerical accuracy and performance ?
Posted: December 23rd, 2015, 1:36 pm
QuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaOne could also say that there are two types of NaN: not a real number and not a floating-point number. In the first case, the solution does not exist in the real number set (e.g., sqrt(-1)) and in the second case, it does not exist in the digital floating point number set (e.g., exp(710) ).The quiet NaN or a signaling NaN?That's an orthogonal issue: the specific cause of the error (i.e., not a real number vs. not a valid 64-bit floating point number) may be independent of the response to the error (i.e., quiet vs. signaling).design by contractThat's what we are missing.Exactly!Yet are there contracts that are detailed enough to cover all these cases of not-a-real-number, not-a-float-number, etc.? Moreover, in a complex library, how does one ensure that ALL library items share the same contract patterns so there's no misinterpretation of what qNaN, sNaN, ±INF, etc. mean.
### Numerical accuracy and performance ?
Posted: December 23rd, 2015, 8:46 pm
QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: Traden4AlphaOne could also say that there are two types of NaN: not a real number and not a floating-point number. In the first case, the solution does not exist in the real number set (e.g., sqrt(-1)) and in the second case, it does not exist in the digital floating point number set (e.g., exp(710) ).The quiet NaN or a signaling NaN?That's an orthogonal issue: the specific cause of the error (i.e., not a real number vs. not a valid 64-bit floating point number) may be independent of the response to the error (i.e., quiet vs. signaling).design by contractThat's what we are missing.Exactly!Yet are there contracts that are detailed enough to cover all these cases of not-a-real-number, not-a-float-number, etc.? Moreover, in a complex library, how does one ensure that ALL library items share the same contract patterns so there's no misinterpretation of what qNaN, sNaN, ±INF, etc. mean.From David Goldberg's seminal articleQuoteUnfortunately, the IEEE standard does not guarantee that the same program will deliver identical results on all conforming systems. Most programs will actually produce different results on different systems for a variety of reasons. For one, most programs involve the conversion of numbers between decimal and binary formats, and the IEEE standard does not completely specify the accuracy with which such conversions must be performed. For another, many programs use elementary functions supplied by a system library, and the standard doesn't specify these functions at all. Of course, most programmers know that these features lie beyond the scope of the IEEE standard. Many programmers may not realize that even a program that uses only the numeric formats and operations prescribed by the IEEE standard can compute different results on different systems. In fact, the authors of the standard intended to allow different implementations to obtain different results. Their intent is evident in the definition of the term destination in the IEEE 754 standard: "A destination may be either explicitly designated by the user or implicitly supplied by the system (for example, intermediate results in subexpressions or arguments for procedures). Some languages place the results of intermediate calculations in destinations beyond the user's control.
### Numerical accuracy and performance ?
Posted: December 25th, 2015, 12:55 am
https://randomascii.wordpress.com/2013/ ... sm/QuoteIs IEEE floating-point math deterministic? Will you always get the same results from the same inputs? The answer is an unequivocal "yes". Unfortunately the answer is also an unequivocal "no". I'm afraid you will need to clarify your question.Worth reading in entirety, but in particular: QuoteThe precise results of functions like sin, cos, tan, etc. are not defined by the IEEE standard. That's because the only reasonable result to standardize would be the exact result correctly rounded, and calculating that is still an area of active research due to the Table Maker's Dilemma.
### Numerical accuracy and performance ?
Posted: December 26th, 2015, 11:48 am
QuoteMathematics is a game played according to certain simple rules with meaningless marks on paper.David Hilbert | 2,463 | 10,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-24 | latest | en | 0.85664 |
https://lottomatic.org/romania-lotto-6din49/number-analysis | 1,638,727,205,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00125.warc.gz | 436,942,369 | 50,539 | # Romania Lotto 6din49 number analyzer
The default lottery number analysis is being performed on the latest results. Please adjust according to your needs.
## Analysis of Romania Lotto 6din49 (6/49) numbers selected
### Analysis & diagram of the most recent 100 Romania Lotto 6din49 numbers
How to interpret the general statistics of the selected Romania Lotto 6din49 numbers:
• Nr. = this column shows the numbers being analyzed.
• S.L. = how many draws have passed since the number was last drawn.
• Freq. = the frequency of the number during the last 100 drawings.
• L.D. = The date when the number was last drawn.
• The diagram on the right of these statistics highlights the cell corresponding to the draw -- going backward chronologically -- in which the number was drawn.
Nr. S.L. Occ. L.D. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 12 0 17 2021-11-28 17 0 12 2021-11-28 27 0 22 2021-11-28 29 0 9 2021-11-28 31 0 9 2021-11-28 48 0 10 2021-11-28
## Romania Lotto 6din49 number performance over the last 100 draws
### Romania Lotto 6din49 - analysis of number performance
Table analyzing how many times the selected numbers contained the Romania Lotto 6din49 winning numbers during the last 100 draws.
6 winning numbers: 5 winning numbers: 4 winning numbers: 3 winning numbers: 2 winning numbers:
Draw date Winning numbers Hits
2021-11-28 12 17 27 29 31 48 -6, 5, 4, 3, 2, 1-
2021-11-25 27 -1-
2021-11-21 29 -1-
2021-11-18 27 -1-
2021-11-14 --
2021-11-11 --
2021-11-07 48 -1-
2021-11-04 --
2021-10-31 27 29 -2, 1-
2021-10-28 --
2021-10-24 12 27 -2, 1-
2021-10-21 --
2021-10-17 --
2021-10-14 27 -1-
2021-10-10 12 27 48 -3, 2, 1-
2021-10-07 17 -1-
2021-10-03 --
2021-09-30 --
2021-09-26 --
2021-09-23 48 -1-
2021-09-19 12 -1-
2021-09-16 12 -1-
2021-09-12 12 -1-
2021-09-09 --
2021-09-05 --
2021-09-02 --
2021-08-29 17 27 -2, 1-
2021-08-26 48 -1-
2021-08-22 48 -1-
2021-08-19 --
2021-08-15 --
2021-08-01 --
2021-07-29 48 -1-
2021-07-25 27 48 -2, 1-
2021-07-22 --
2021-07-18 12 -1-
2021-07-15 --
2021-07-11 --
2021-07-08 --
2021-07-04 27 -1-
2021-07-01 17 31 -2, 1-
2021-06-27 27 -1-
2021-06-24 --
2021-06-20 --
2021-06-17 27 -1-
2021-06-13 --
2021-06-10 --
2021-06-06 --
2021-06-03 17 -1-
2021-05-30 29 -1-
2021-05-27 --
2021-05-23 17 -1-
2021-05-20 27 -1-
2021-05-16 --
2021-05-13 12 -1-
2021-05-09 31 -1-
2021-05-06 --
2021-04-25 27 48 -2, 1-
2021-04-22 --
2021-04-18 12 -1-
2021-04-15 29 48 -2, 1-
2021-04-11 12 17 27 -3, 2, 1-
2021-04-08 31 -1-
2021-04-04 --
2021-04-01 12 -1-
2021-03-28 --
2021-03-25 29 -1-
2021-03-21 17 27 -2, 1-
2021-03-18 29 31 -2, 1-
2021-03-14 --
2021-03-11 12 -1-
2021-03-07 27 29 -2, 1-
2021-03-04 --
2021-02-28 --
2021-02-25 12 27 31 -3, 2, 1-
2021-02-21 12 -1-
2021-02-18 --
2021-02-14 17 -1-
2021-02-11 --
2021-02-07 --
2021-02-04 --
2021-01-31 --
2021-01-28 --
2021-01-24 12 27 -2, 1-
2021-01-21 12 -1-
2021-01-17 --
2021-01-14 27 -1-
2021-01-10 --
2021-01-07 17 31 -2, 1-
2021-01-03 --
2020-12-31 --
2020-12-24 31 -1-
2020-12-20 27 -1-
2020-12-17 12 29 -2, 1-
2020-12-13 --
2020-12-10 27 31 -2, 1-
2020-12-06 17 -1-
2020-12-03 --
2020-11-29 17 -1-
2020-11-26 --
## Romania Lotto 6din49 pair performance over the last 100 draws
### Romania Lotto 6din49 - analysis of pair frequency
For the selected Romania Lotto 6din49 numbers, these are all possible combinations of two numbers and their frequency (how many times they were drawn) during the last 100 draws.
17,29
17,48
31,48
12,17
12,29
12,31
12,48
29,31
29,48
17,31
27,29
27,31
17,27
27,48
12,27
..1 time(s)
..1 time(s)
..1 time(s)
..2 time(s)
..2 time(s)
..2 time(s)
..2 time(s)
..2 time(s)
..2 time(s)
..3 time(s)
..3 time(s)
..3 time(s)
..4 time(s)
..4 time(s)
..6 time(s) | 1,788 | 3,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-49 | latest | en | 0.686376 |
https://electronics.stackexchange.com/questions/499340/is-this-internal-rearrangement-of-charges-in-a-circuit-a-current-and-is-this-cur | 1,596,573,529,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735882.86/warc/CC-MAIN-20200804191142-20200804221142-00234.warc.gz | 260,335,991 | 37,141 | # Is this internal rearrangement of charges in a circuit a current and is this current different from the “normal” current of the circuit?
In the figure, the circuit is physically isolated from Earth (The black line at the bottom). The red capacitors are parasitic capacitors.
Under switching conditions the node voltages relative to the zero reference swing in such a way that the internal charge in the circuit stays the same. This is understandable because the circuit is isolated from ground and cannot get more charge or lose any charge. So the charges internally rearrange themselves in a way that leads to the node voltages being what they are.
How is this possible. Is this internal rearrangement of the charges a current? Is this current different from the “normal” current of the circuit (The 1mA)?
I think that Electric fields that change and exist between charged bodies will cause something like current flow, called displacement current, such that a change in the electric field on one body affects the electric field on a nearby body. But how does this affect the 1mA current that appears in the circuit?
Is this internal rearrangement of the charges a current?
Moving charge is a current, so yes, this is a current.
Also, the changing electric field caused by the rearranging charges is a current, called displacement current. Whether a displacement current is "really" a current is a philosophical or semantic question rather than a physics or engineering question. From a physics and engineering point of view, displacement currents must be accounted for when solving the Kirchhoff's Current Law, they contribute to magnetic fields according to Ampere's law, and otherwise affect the rest of the world just as ordinary moving-charge currents do.
If so then is this current different from the “normal” current of the circuit (The 1mA)?
It's a transient current (it falls to zero over time) where the "normal" current in this circuit will continue forever.
Electric fields that change and exist between charged bodies will cause something like current flow, called displacement current, such that a change in the electric field on one body affects the electric field on a nearby body. But how does this affect the 1mA current that appears in the circuit?
If your voltage source has internal resistance (like all real-world voltage source) then the extra current required to charge the parasitic capacitors will cause the source output voltage to drop, reducing the current through the 9 kohm resistor.
This effect will last for only a few nanoseconds or microseconds, until the capacitors are charged, and after that the presence of the parasitic capacitors will not affect the current through the resistor.
Note that in the model drawn, only C3 has its voltage changed when the switch is closed, so only C3 will draw current. The other capacitors won't cause any currents in response to the switch closing or opening.
• Thank you very much, extremely helpful! – John May 16 at 19:34
• If the circuit is grounded, by connecting a point in the circuit to the physical zero reference (Earth). Will the connection to ground affect what is happening in a meaningful way? Will the internal charge distributions and the corresponding currents be different? – John May 16 at 21:26
• @John, Draw the circuit for yourself and think about whether closing or opening the switch changes the voltage across each of the capacitors. – The Photon May 16 at 22:19
I'm going to focus on your general question(s) and ignore your specific circuit. I think you only included it to put something down, anyway. There's an underlying question and that's what I'll address.
There is a state that exists before and a state that exists after a source is applied to a circuit. Obviously, there must be a short period in between these two steady state conditions where the circuit transitions between them. This transition period is very fast and most undergrad electronics books don't directly address the physics involved. (But the physics details of this transition period is important to high voltage engineering, for example.)
It's not too complicated to imagine what happens, though. At first, the source will have an excess of positive charges at one node and an excess of negative charges at the other node. These will very, very rapidly impel charges to move (the conductors have a veritable sea of available, conduction band electrons.) These charge motions are, in fact, currents. But many of them are just initial currents needed in order to set up surface charges, leading to a gradient of charge distributions, throughout the circuit. For DC circuits, once these surface charges have been set up, the current required to set them up is no longer required. But these surface charges now act to impel the currents in all the right directions (this way, and that way, at node intersections, etc.) For AC circuits, these surface charge changes are continually changing but they are almost always extremely fast compared to the AC circuit oscillations. So for mental visualization purposes you can just think of each of the infinite number of continuous changes in an AC cycle as being a DC-snapshot. So the DC view is usually sufficient to get the point across.
Imagine bending a wire while a circuit is operating! How is it that the electrons moving along the wire "know" to take the bend? Do they bounce off of things and careen around the bend? No, they don't. Not generally, anyway. What instead happens is that a few of the moving charges will "get stuck" in the surface area of the outside curve of the bend in the wire. This slight excess of charge will be "just enough" to act as a repulsing force that causes the current to accelerate around the bend, as if they just knew to take it.
With that in hand, I recommend that you read a few references. Perhaps a better one I wrote is here. Another is here and only discusses a simple DC battery and one resistor. There's a good video, Surface Charge on a High Voltage Circuit, too. And finally, there's a quantitative discussion: Magnitudes of surface charge distributions associated with electric current flow by Dr. Rosser.
• Thank you very much and yes, I only added the circuit to try help picture the problem that I posed. – John May 16 at 19:35
• If the circuit is grounded, by connecting a point in the circuit to the physical zero reference (Earth). Will the connection to ground affect what is happening in a meaningful way? Will the internal charge distributions and the corresponding currents be different? – John May 16 at 21:26
• @John Suppose a circuit is in an evacuated enclosure in outer space, where it is being continually peppered with protons and electrons. Depending on location and history, it will have accumulated some charge. But too much of a charge will attract the opposite charges and repel same charges, so at some point it reaches equilibium. Now you instantly teleport this to Earth's surface and ground it into Earth (which itself has a different charge balance.) What do you think should happen (and for how long?) Will it matter for the circuit? Think for yourself for a moment before I try and answer. – jonk May 16 at 21:54
• @John Also, if you haven't already done so, you really should allow yourself to enjoy Feynman's Lecture series on physics. It's freely available on the web, now. And it is a treasure and very readable. (Due, in my opinion, mostly to the two co-authors who partnered in order to finally write it.) For example, try Chapter 9, Volume 1. In particular, pay some attention to the Earth's surface vs the Earth's upper atmosphere and how charges are continually transported between them. – jonk May 16 at 22:05
• When a point in the circuit is connected to the Earth, the potential at that point becomes the reference potential of 0V (as assumed for the Earth). Simultaneously, potentials in other points of the circuit also change. But, potential difference between the points still remains same. So, internal charge distribution and the current in the circuit remain unchanged. Only change is in the potentials at the points of the circuit. Please let me know if my answer is correct – John May 18 at 9:31 | 1,714 | 8,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-34 | latest | en | 0.923783 |
http://large.stanford.edu/courses/2017/ph240/celis2/ | 1,548,019,223,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583739170.35/warc/CC-MAIN-20190120204649-20190120230649-00542.warc.gz | 136,560,263 | 3,277 | # The Understanding of Sound Energy
## Introduction
Fig. 1: The stereocilia in a frog's inner ear. (Source: Wikimedia Commons)
The existence of sound is beautiful. It allows us to enjoy music, to communicate with each other, and to better understand our surroundings. However, we seldom stop to think about how our hearing is possible and what causes it. And the truth is that, like most things, we can ascribe this wonder to the movement and translation of energy.
## Sound as Energy
Sound travels through fluids by introducing small, and sometimes large, perturbations in pressure. The patterns by which these perturbations travel are known as sound waves. We can measure a sound's magnitude via its acoustic pressure level (APL), which is measured in decibels (dB), although the quantity is truly dimensionless. According to the International Electrotechnical Commission, the agreed-upon standard for calculating decibels is
APL (dB) = 20 log10 (p/p0)
where p is the root mean square sound pressure (Pa) of the source of interest and p0 is the reference sound pressure (Pa). [1] An important thing to note is that the decibel system is a referential system, which means that it operates and takes meaning as a magnitude relative to another. This is unlike absolute measurements, such as temperature or length, which have concrete and theoretical minimums (e.g. 0°K or 0 m). This very nature of measurement posits a very important question: what should the reference sound pressure be?
An often used reference is 20 μPa, which is generally regarded to be the threshold of human hearing (i.e. the lowest sound an average, undamaged human ear can perceive). [2] This establishes a standard from which people can compare different levels of loudness, or energy, that a particular sound source emits. Because decibels are referential, the amount of energy emitted is also referential. In other words, decibels can be used to calculate the amount of energy (J) of a source object's emission relative to (as a factor of) the reference object's emission.
## Interpreting Energy
Our bodies generally come equipped with sophisticated mechanisms to interpret the information encoded in a sound wave. When a sound wave moves about its medium, it displaces the fluid it lives in in an oscillatory manner via compression and expansion due to pressure changes. Because of this, sound waves are considered mechanical waves, and it is through these environmental operatives that our ear decodes sound waves into various things, like the music we listen to, the voices we hear, or the noise in the subway.
In particular, our ear understands sound via its hair cells. Around 16,000 hair cells live in an ear's basilar membrane, with each hair cell containing around 100 stereocilia. [3] It is through these over a million receptive organelles that our ear understands the various minute, and sometimes not so minute, changes in pressure due to sound. An example of such organelles is shown in Fig. 1.
## Conclusion
From a systems perspective, sound can be interpreted as a message that is encoded by an emitter, transmitted through a medium, and decoded by a receptor. Emitters can come in all different shapes, sizes, and backgrounds, from a masterful orchestra playing a piece to an angry driver honking on the road. Mediums can affect the message, such as water dampening the sound waves' intensities or like a vacuum (the absence of a medium!) completely preventing the transfer of the message in the first place. Thankfully, in the most general case we have the right receptor - two, in fact - to perceive, appreciate, and tolerate (mostly) all sounds.
© Diego Celis. The author warrants that the work is the author's own and that Stanford University provided no input other than typesetting and referencing guidelines. The author grants permission to copy, distribute and display this work in unaltered form, with attribution to the author, for noncommercial purposes only. All other rights, including commercial rights, are reserved to the author.
## References
[1] "Letter Symbols to be Used in Electrical Technology Part 3: Logarithmic and Related Quantities, and Their Units," International Electrotechnical Commission, IEC 60027-3, Ed. 3.0, 19 July 2002.
[2] R. J. Roesser and M. Valente, Audiology (Thieme, 2007), p. 240.
[3] A. Hudspeth,"The Cellular Basis of Hearing: the Biophysics of Hair Cells," Science 230, 745 (1985). | 977 | 4,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-04 | latest | en | 0.942094 |
http://www.cordovaalumni.com/apple-juice-production-flow-chart/production-of-clear-apple-juice-download-scientific-diagram/ | 1,566,244,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314904.26/warc/CC-MAIN-20190819180710-20190819202710-00019.warc.gz | 231,281,555 | 13,581 | # Apple Juice Production Flow Chart Of Clear Download Scientific Diagram Flowchart
By Gabrielle Button at November 01 2018 21:24:15
In mathematics, method of solving a problem by repeatedly using a simpler computational method. A basic example is the process of long division in arithmetic. The term algorithm is now applied to many kinds of problem solving that employ a mechanical sequence of steps, as in setting up a computer program. The sequence may be displayed in the form of a flowchart in order to make it easier to follow. As with algorithms used in arithmetic, algorithms for computers can range from simple to highly complex.
After deciding on the points you want to make in your upcoming presentation, you need to figure out how to support those points. For example, if your point is that your company has the largest market share in the industry, quote the research (hopefully done by a third party) that says so. This applies to both business presentations and educational presentations. The support you provide for your message is essential for an effective presentation.
## Gallery of Apple Juice Production Flow Chart
What's a flowchart? A flowchart can be defined as a graphical representation of a sequence of operations or steps. In other words, it's an illustration of the various steps involved in a project or process. Typically, a flowchart consists of a number of boxes, arrows, and text that combine to form a sequence. Why create a flowchart? The purpose of a flowchart is to show the various steps of a process in a snapshot. By looking at the flowchart, the viewer should be able to identify the various steps involved in the process.
Rice Huskers: These huskers remove the husk (outer covering) from the paddy rice during the processing. Paddy Separators: It makes the brown rice more efficient. Plano-Shifters: This makes the rice more uniform and give rice proper size and grading with a high speed. Color Sorters: These color sorters give a proper color to the rice and define its shade. The basic structure and the process followed in the rice milling industries and rice milling plants include: 1. Quantity of Rice (In Abundance) ; 2. Pre - Cleaning ; 3. Steaming ; 4. Drying ; 5. Packaging ; 6. Grading and Sorting ; 7. Polishing ; 8. Removal of Husk | 490 | 2,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-35 | latest | en | 0.930492 |
https://code.dennyzhang.com/trapping-rain-water-ii | 1,563,436,385,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525524.12/warc/CC-MAIN-20190718063305-20190718085305-00136.warc.gz | 354,854,127 | 11,471 | # Leetcode: Trapping Rain Water II
Trapping Rain Water II
Similar Problems:
Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining.
Note:
Both m and n are less than 110. The height of each unit cell is greater than 0 and is less than 20,000.
Example:
```Given the following 3x6 height map:
[
[1,4,3,1,3,2],
[3,2,1,3,2,4],
[2,3,3,2,3,1]
]
Return 4.
```
The above image represents the elevation map [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] before the rain.
After the rain, water is trapped between the blocks. The total volume of water trapped is 4.
Github: code.dennyzhang.com
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
• Solution:
```// Blog link: https://code.dennyzhang.com/trapping-rain-water-ii
// Basic Ideas: heap
// The lowest bucket determine how much water it can hold.
//
// Start from the outer circle into the inner circles
// We explore from the lowest blocks(use min-heap)
// If unexamined bucket is small then current lowest, current bucket can hold some water
// Otherwise current bucket can't hold water
//
// Complexity: Time ?, Space ?
import "container/heap"
// https://golang.org/pkg/container/heap/
type Item struct {
x, y, v int
}
// A PriorityQueue implements heap.Interface and holds Items.
type PriorityQueue []*Item
func (pq PriorityQueue) Len() int { return len(pq) }
func (pq PriorityQueue) Less(i, j int) bool {
// We want Pop to give us the highest, not lowest, priority so we use greater than here.
return pq[i].v < pq[j].v
}
func (pq PriorityQueue) Swap(i, j int) {
pq[i], pq[j] = pq[j], pq[i]
}
func (pq *PriorityQueue) Push(x interface{}) {
item := x.(*Item)
*pq = append(*pq, item)
}
func (pq *PriorityQueue) Pop() interface{} {
old := *pq
n := len(old)
item := old[n-1]
*pq = old[0 : n-1]
return item
}
func trapRainWater(heightMap [][]int) int {
if len(heightMap) == 0 { return 0 }
visited := map[string]bool{}
h := make(PriorityQueue, 0)
heap.Init(&h)
// add outer bounders to min-heap
for _, i := range []int{0, len(heightMap)-1} {
for j := 0; j < len(heightMap[0]); j++ {
heap.Push(&h, &Item{i, j, heightMap[i][j]})
visited[fmt.Sprintf("%d-%d", i, j)] = true
}
}
for i:=1; i<len(heightMap)-1; i++ {
for _, j:= range []int{0, len(heightMap[0])-1} {
heap.Push(&h, &Item{i, j, heightMap[i][j]})
visited[fmt.Sprintf("%d-%d", i, j)] = true
}
}
res := 0
x, y := 0, 0
key := ""
for h.Len() != 0 {
item := heap.Pop(&h).(*Item)
// fmt.Println("item: ", item)
for _, offset := range [][]int{[]int{0, 1}, []int{0, -1}, []int{1, 0}, []int{-1, 0}} {
x, y = item.x+offset[0], item.y+offset[1]
if x<0 || x>=len(heightMap) || y<0 || y>=len(heightMap[0]) {
continue
}
key = fmt.Sprintf("%d-%d", x, y)
if visited[key] { continue }
visited[key] = true
if heightMap[x][y] < item.v {
// fmt.Println(item.x, item.y, item.v, "|", x, y, item.v - heightMap[x][y])
res += item.v - heightMap[x][y]
heap.Push(&h, &Item{x, y, item.v})
} else {
heap.Push(&h, &Item{x, y, heightMap[x][y]})
}
}
}
return res
}
```
Share It, If You Like It. | 1,000 | 3,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-30 | longest | en | 0.657519 |
https://solvedlib.com/n/use-identities-to-find-sin-0-if-cos-20-and-lt-0-lt-1oa-7,18764991 | 1,675,021,140,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00726.warc.gz | 556,473,329 | 19,450 | # Use identities to find sin 0 if cos 20 =and{<0<1OA. 7 4154 0 B. 1144154 0c 740 D. 14 K
###### Question:
Use identities to find sin 0 if cos 20 = and {<0<1 OA. 7 4154 0 B. 114 4154 0c 74 0 D. 14 K
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https://samacheerguru.com/samacheer-kalvi-6th-maths-term-3-chapter-3-ex-3-1/ | 1,718,964,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00176.warc.gz | 442,187,418 | 16,448 | Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1
Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1
Question 1.
The table given below contains some measures of the rectangle. Find the unknown values.
Solution:
(i) Area of the rectangle = (length × breadth) sq unit.
Perimeter of a rectangle = 2(1 + b) units.
l = 5 cm
b = 8 cm
∴ p = 2 (l + b) cm = 2 (5 + 8) cm = 2 × 13 cm
p = 26 cm
Area = (l × b) cm2 = (5 × 8) cm2
A = 40 cm2
(ii) l = 13 cm
p = 54 cm
Perimeter = 2 (l + b) units
54 = 2 (13 + b) cm
$$\frac{54}{2}$$ = 13 + b
27 = 13 + b
b = 27 – 13
b = 14 cm
Area = l × b sq. unit = 13 × 14 cm2
A = 182 cm2
(iii) b = 15 cm
p = 60 cm
p = 2 (l + b) units
60 = 2 (l + 15) cm
$$\frac{60}{2}$$ = l + 15
30 = l + 15
l = 30 – 15 .
l = 15 cm
Area = l × b unit2 = 15 × 15 cm2 = 225 cm2
A = 225 cm2
(iv) l = 10 m
Area = 120 sq metre
Area = l × b sq.m
120 = 10 × 6
b = $$\frac{120}{10}$$
b = 12 m
Perimeter =2 (l + b) units = 2(10 + 12) units = 2 × 22 m
A = 44 m
(v) b = 4 feet.
Area = 20 sq. feet
Area = l × b sq .feet
20 = l × 4
l = $$\frac{20}{4}$$ feet
l = 5 feet
Perimeter = 2 (l + b) units.
p = 2 (5 + 4) feet = 2 × 9
p = 18 feet
Completing the unknown values in the table.
Question 2.
The table given below contains some measures of the square. Find the unknown values.
Solution:
Perimeter of a square = (4 × side) units
Area of a square = (side × side) unit2
(i) s = 6 cm
Perimeter = 4s units = 4 × 6 cm = 24 cm
P = 24 cm
Area = s × s unit2 = 6 × 6 cm2 = 36 cm2
A = 36 cm2
(ii) Perimeter = 4 × s unit
100 = (4 × s) m
$$\frac{100}{4}$$ = s
s = 25 m
Area = s × s unit2= 25 × 25 m2 = 625m2
A = 625m2
(iii) Area = s × s unit2
49 = s × s square feet
s2 = 72
s = 7 feet
Perimeter = 4 × s unit = 4 × 7 feet = 28 feet
Perimeter = 28 feet
Completing the unknown values in the table
Question 3.
The table given below contains some measures of the triangle. Find the unknown values.
Solution:
Area of the right triangle = $$\frac{1}{2}$$ × (base × height) unit2
(i) b = 20 cm
h = 40 cm
Area = $$\frac{1}{2}$$ (b × h) cm2 = $$\frac{1}{2}$$ × 20 × 40 = 400 cm2
A = 400 cm2
(ii) b = 5 feet
Area = $$\frac{1}{2}$$ × b × h unit2
= 20 = $$\frac{1}{2}$$ × 5 × h sq. feet
$$\frac{20 \times 2}{5}$$ = h
h = 8 feet
(iii) Area = $$\frac{1}{2}$$ × (base × height) unit2
24 = $$\frac{1}{2}$$ × b × 12 m2
base = $$\frac{24 \times 2}{12}$$ m = 4 m
Base = 4m
Tabulating the unknown values
Question 4.
The table given below contains some measures of the triangles. Find the unknown values.
Solution:
Perimeter of a triangle = sum of three sides.
(i) Perimeter = 6 + 5 + 2 cm = 13 cm
p = 13 cm
(ii) Perimeter = (side 1 + side 2 + side 3) m
17 = (side 1 + 8 + 3) m
17 m = (side 1 + 11) m
side 1 = 17 – 11 = 6m
(iii) Perimeter = side 1 + side 2 + side 3
28 feet = 11 feet + side 2 + 9 feet
28 ft = 20 feet + side 2
28 – 20 = side 2
side = 8 feet
Tabulating the unknowns.
Question 5.
Fill in the blanks.
i) 5 cm2 = mm2
Hint: 1 cm2 = 100 mm2
ii) 26 m2 = cm2
Hint: 1 m2 = 10000
iii) 8 km2 = m2
Hint 4 1 km2– 1000000 m2
Solution:
(i) 500
(ii) 2,60,000
(iii) 80,00,000
Question 6.
Find the perimeter and area of the following shapes.
Solution:
(i) Perimeter = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4) cm = 48 cm
Perimeter = 48 cm
Area of 5 squares of side 4 cm
Area of a square = (side × side) unit2
∴ A = 5 × (4 × 4) cm2 = 5 × 16 cm2 = 80 cm2
80 cm2
(ii) Perimeter = (4 + 5 + 4 + 5 + 4 + 5 + 4 + 5)cm = 36cm
Perimeter = 36 cm
Area of a square of side 3cm + Area of 4 right triangles
= (3 × 3) + [4 × $$\frac{1}{2}$$ × 4 × 3] cm2 = (9 + 24) cm2 = 33 cm2
Area = 33 cm2
(iii) Perimeter = (50 + 12 + 13 + 40 + 10 + 10 + 10 + 5) cm = 150 cm
Perimeter = 150 cm
Area = Area of a rectangle + Area of a square + Area of a right triangle.
= (l × b) + (s × s) + ( $$\frac{1}{2}$$ × b × h) cm2
= (50 × 5) + (10 × 10) + $$\frac{1}{2}$$ × 12 × 5) cm2
= (250 + 100 + 30) cm2 = 380 cm2
Area = 380 cm2
Question 7.
Find the perimeter and area of the rectangle whose length is 6 m and breadth 4 m.
Solution:
l = 6 m, b = 4 m Perimeter of the rectangle
= 2 (l + b) units
= 2 (6 + 4) m
= 2 (10) m
= 20 m
Area of the rectangle = l × b sq units
= 4 × 6 m²
= 24 m²
Question 8.
Find the perimeter and the area of the square whose side is 8 cm.
Solution:
Perimeter of a square = (4 × side) units
Side = 8 cm
∴ Perimeter = 4 × 8 cm = 32 cm
Perimeter = 32 cm
Area of a square = (side × side) unit2 = (8 × 8) cm2 = 64 cm2
Area = 64 cm2
Question 9.
Find the perimeter and area of the right angled triangle whose sides are 6 feet, 8 feet and 10 feet.
Solution:
Perimeter of the triangle
= (a + b + c) units
= (6 + 8 + 10) feet
= 24 feet
Area of the triangle = $$\frac{1}{2}$$ × b × h sq units
$$\frac{1}{2}$$ × 6³× 8 feet square
= 24 sq. feet
Question 10.
Find the perimeter of
i) A scalene triangle with sides 7 m, 8 m, 10 m
ii) An isosceles triangle with equal sides 10 cm each and third side is 7 cm.
iii) An Equilateral triangle with side 6 cm.
Solution:
i) Perimeter of a scalene triangle = (7 + 8 + 10) m = 25 m
ii) The three sides of the isosceles triangle are 10 cm, 10 cm and 7 cm
∴ Perimeter = (10 + 10 + 7) cm = 27 cm
iii) An equilateral triangle with side 6 cm.
The sides of equilateral triangle are 6 cm, 6 cm and 6 cm
∴ Perimeter = (6 + 6 + 6) cm = 18 cm
Question 11.
The area of a rectangular shaped photo is 820 sq. cm. and its width is 20 cm. What is its length? Also find its perimeter.
Solution:
Given Area = 820 cm²
Width = 20 cm
Area of the rectangle
= l × b sq. units
820 = l × 20
$$\frac{820}{20}$$ = l
41 = l
length l = 41 cm
Perimeter = 2(l + b) units
= 2(41 + 20) cm
= 2(61) cm
= 122 cm
Question 12.
A square park has 40 m as its perimeter. What is the length of its side? Also find its area.
Solution:
Given perimeter = 40 m
Perimeter of a square = 4 × Length of a side
40 = 4 × Length of a side
∴ Length of its side = $$\frac{40}{4}$$ m = 0 m
∴ Side of the park = 10m
Area of a square = (Side × side) unit2 = (10 × 10) m2 = 100 m2
∴ Area of the Park = 100 m2
Question 13.
The scalene triangle has 40 cm as its perimeter and whose two sides are 13 cm and 15 cm, find the third side.
Solution:
Let the third side be C
perimeter = (a + b + c) units
40 = 13 + 15 + C
40 = 28 + C
C = 40 – 28
C = 12 units
C = 12 cm
Question 14.
A field is in the shape of right angled triangle whose base is 25 m and height 20 m. Find the cost of levelling the field at the rate of ₹ 45/- per sq. m.
Solution:
Area of a right angled triangle = $$\frac{1}{2}$$ × (base × height) unit2
base = 25 m
height = 20 m
∴ Area = $$\frac{1}{2}$$ × (25 × 20)
Area = 250 m2
Cost of levelling per m2 = ₹ 45.
∴ Cost of levelling 250 m2 = 250 × 45 = ₹ 11,250
Cost of levelling = ₹ 11,250
Question 15.
A square of side 2 cm is joined with a rectangle of length 15 cm and breadth 10 cm. Find the perimeter of the combined shape.
Solution:
Perimeter of the combined shape = Lengths of the outer boundaries
= (15 + 10 + 2 + 2 + 2 + 13 + 10) cm = 54 cm
Perimeter = 54 cm
Objective Type Questions
Question 16.
The following figures are of equal area. Which figure has the least perimeter?
Solution:
Hint:
(a) 12 units
(b) 10 units
(c) 12 units
(d) 12 units
Question 17.
If two identical rectangles of perimeter 30 cm are joined together, then the perimeter of the new shape will be
(a) equal to 60 cm
(b) less than 60 cm
(c) greater than 60 cm
(d) equal to 45 cm
Solution:
(b) less than 60 cm
Hint:
Question 18.
If every side of a rectangle is doubled, then its area becomes times.
(a) 2
(b) 3
(c) 4
(d) 6
Solution:
(c) 4
Question 19.
The side of a square is 10 cm. If its side is tripled, then by how many times will its perimeter increase?
(a) 2 times
(b) 4 times
(c) 6 times
(d) 3 times
SolutionL
(d) 3 times
30 × 4 = 120 = 3 × 40
Question 20.
The length and breadth of a rectangular sheet of a paper are 15 cm and 12 cm respectively. A rectangular piece is cut from one of its corners. Which of the following statement is correct for the remaining sheet?
(a) Perimeter remains the same but the area changes
(b) Area remains the same but the perimeter changes
(c) There will be a change in both area and perimeter
(d) Both the area and perimeter remains the same
Solution:
(c) There will be a change in both area and perimeter
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lecture05
# lecture05 - Astronomy 3 The Nature of the Universe...
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Astronomy 3: The Nature of the Universe Professor Alice Shapley Lecture 5: Motion, Force, Energy contd. Light, Matter, Telescopes (NGC 1499 Image credit: Markus Noller, Deep Sky Images)
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Logistics Quiz #2: due Monday, April 18 th , 10 pm. Available today after class at CCLE website: https://ccle.ucla.edu/course/view/11S-ASTR3-2 Lab this week on “Light and Telescopes.”
Review from Last Time Motion: speed, velocity, acceleration. Force, momentum, angular momentum. Newton’s 3 Laws of Motion. Conservation of angular momentum. Energy and Conservation of energy. Newton’s Law of Gravitation.
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The Universal Law of Gravitation 1. Every mass attracts every other mass. 2. Attraction is directly proportional to the product of their masses. 3. Attraction is inversely proportional to the square of the distance between their centers.. NOTE: Gravity is not a constant force; its value changes with distance.
The Moon and the “Apple”: Universal Gravitation How are these two systems equivalent?
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Kepler’s first two laws apply to all orbiting objects, not just planets. Some comets have unbound orbits; they orbit the Sun only once. Ellipses are not the only orbital paths. Orbits can be: – bound (ellipses) – unbound • parabola • hyperbola Newton’s Law of gravity permits all these orbits.
If an object gains enough orbital energy, it may escape (change from a bound to unbound orbit). Escape Velocity from Earth 11.1 km/s from sea level (about 40,200 km/hr; 25,000 mph or 7 miles per second). This velocity depends on the mass and radius of the Earth, not the mass of the object – i.e. v esc = (2GM Earth /R Earth ) 1/2 . Escape Velocity
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Newton’s Version of Kepler’s 3rd Law Newton’s version of Kepler’s Third Law : If a small object orbits a larger one, and you measure the orbiting object’s orbital period AND average orbital radius THEN you can calculate the mass of the larger object. Examples ( know these ): Calculate mass of Sun from Earth’s orbital period (1 year) and average distance (1 AU). • Calculate mass of Earth from the orbital period and average distance of any orbiting satellite, including the Moon. • Calculate mass of Jupiter from the orbital period and distance from Jupiter of one of its moons. This is an incredibly powerful discovery.
p = orbital period a = average orbital distance (between centers) (M 1 + M 2 ) = sum of object masses So, if M 2 is tiny compared to M 1 then M 1 = 4 π 2 a 3 /Gp 2 . NOTE: Just know that this equation allows us to measure the mass of distant objects. p 2 = 4 π 2 G ( M 1 + M 2 ) a 3 See page 131. Newton’s Version of Kepler’s 3rd Law
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p = orbital period a = average orbital distance (between centers) (M 1 + M 2 ) = sum of object masses So, if M 2 is tiny compared to M 1 then M 1 = 4 π 2 a 3 /Gp 2 .
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{[ snackBarMessage ]} | 876 | 3,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-17 | latest | en | 0.847376 |
https://virtualplaneticeland.com/2020/11/30/quick-answer-does-22-7-go-forever/ | 1,619,091,226,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039603582.93/warc/CC-MAIN-20210422100106-20210422130106-00044.warc.gz | 711,394,191 | 8,926 | # Quick Answer: Does 22 7 Go Forever?
## Is there a zero in pi?
Pi is not equal to a number which has 0 in its first thirty digits.
The digits of pi are determined and there is nothing probabilistic about them.
3.1428571428571428571recurring is the real answer to pi.
there are no zeros in pi anywhere..
## What does 22 7 a repeating decimal mean?
There is NO pattern. 227 is a fraction which is very close to π . However, 227 is a rational number which can be written as a recurring decimal .
## Is 2/3 an irrational number?
In mathematics rational means “ratio like.” So a rational number is one that can be written as the ratio of two integers. For example 3=3/1, −17, and 2/3 are rational numbers. Most real numbers (points on the number-line) are irrational (not rational).
## Who is the god of math?
Athena, also referred to as Athene, is a very important goddess of many things. She is goddess of wisdom, courage, inspiration, civilization, law and justice, strategic warfare, mathematics, strength, strategy, the arts, crafts, and skill.
## Is 22 7 is a rational number?
The improper fraction 22/7 is a rational number. All rational numbers can be expressed as a fraction or ratio between two integers.
## Will Pi ever end?
It is also referred to as Archimedes’ constant. Being an irrational number, π cannot be expressed as a common fraction, although fractions such as 22/7 are commonly used to approximate it. Equivalently, its decimal representation never ends and never settles into a permanently repeating pattern.
## Is the square root of 22 rational or irrational?
Answer and Explanation: The square root of 22 is not a rational number. 22 is not a perfect square; there aren’t two integers that you can multiply together to get 22.
## Who invented math?
Beginning in the 6th century BC with the Pythagoreans, the Ancient Greeks began a systematic study of mathematics as a subject in its own right with Greek mathematics. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof.
## What is the difference between PI and 22 7?
22/7 is a rational approximation to pi, just as 3.14 is. That is, both 22/7 and 3.14 are rational numbers (fraction and terminating decimal, respectively) that are close enough to the exact value of pi for elementary work. 22/7 is a rational number, namely 3.1428571428571428571428571428571…
## How many digits of pi does NASA use?
15 digitsNASA only uses around 15 digits of pi in its calculations for sending rockets into space. To get an atom-precise measurement of the universe, you would only need around 40. So computing trillions of digits of pi is mostly about showing off computer power.
## Is Pi bigger than 4?
And you can see that π is less than 4 if you look at the square that circumscribes a circle. The square’s perimeter is longer than the circle’s circumference, and yet the ratio of this perimeter to the diameter of the circle is 4. So π is somewhere in there between 3 and 4.
## Is 22 a real number?
Natural Numbers – the set of numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,….., that we see and use every day. The natural numbers are often referred to as the counting numbers and the positive integers. Whole Numbers – the natural numbers plus the zero.
## What is the 348th digit of 22 7?
If one knows that π is approximately 3.14159, then it trivially follows that π < 227, which is approximately 3.142857.
## Who found pi?
Archimedes of SyracuseThe first calculation of π was done by Archimedes of Syracuse (287–212 BC), one of the greatest mathematicians of the ancient world.
## What if PI was 3?
If Pi was equal to 3, it wouldn’t exist. We wouldn’t call it Pi, because it’s simply 3.
## Why is 180 degrees pi?
Degrees of angles can be used with trigonometric functions and line lengths to compute unknown line lengths of angles & triangles. The relationship of π radians to 180° is this: Since π is 1/2 of a circle, we multiple 360° by 1/2 to find the number of degrees in half of a circle (i.e. 360° * 1/2 = 180° = π).
## Who is the No 1 mathematician in the world?
Isaac Newton is a hard act to follow, but if anyone can pull it off, it’s Carl Gauss. If Newton is considered the greatest scientist of all time, Gauss could easily be called the greatest mathematician ever.
## Why is Pi 22 divided 7?
Originally Answered: Why is the value of pi 22/7? Back when the major players were discovering properties about circles, it was discovered that the circumference of a circle was a little more than 3 times the diameter. They were able to determine that 22/7 was a very close approximation to it. | 1,168 | 4,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-17 | latest | en | 0.951839 |
https://www.stata.com/statalist/archive/2006-03/msg01111.html | 1,627,342,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152156.49/warc/CC-MAIN-20210726215020-20210727005020-00656.warc.gz | 1,048,809,574 | 3,978 | # Re: st: a question about number precision
From Phil Schumm To statalist@hsphsun2.harvard.edu Subject Re: st: a question about number precision Date Fri, 31 Mar 2006 18:15:33 -0600
```On Mar 31, 2006, at 4:33 PM, Jian Zhang wrote:
```
I have a problem about number precision. I cann't figure out what it happened. Hope that you can help me out. Thanks.
Here is the data:
ID
21557127
then i run the following do file trying to extract the last three digits from the ID:
gen double temxxx=(ID/1000)
gen temyyy=int(temxxx)
gen temzzz=temxxx-temyyy
gen areaxxx=(temzzz*1000)
drop temxxx temyyy temzzz
the generated data looks like the following:
ID areaxxx
21557127 127
However, when I typed: list if areaxxx==127, stata in fact listed nothing!
First I thought it may be because areaxxx is a floating-point variable, so I type: list if areaxxx=float(127). However, Stata listed nothing again.
First, let me say that if all you want to do is to extract the last three digits of the ID, here is the way to do it:
. di real(substr(string(ID,"%12.0g"),-3,.))
127
Note that if you just use string(ID) this will not work, as string() uses a default format which is not wide enough for your ID (%12.0g is the default format for the long storage type, which I presume is how your ID variable is stored).
Second, this is exactly the reason why you should not store IDs as numbers -- you should store them as strings instead. For example, if ID were a string variable, then extracting the last three digits would be even simpler:
. di substr(ID,-3,.)
127
and would be guaranteed to work no matter how long your IDs are (provided they are no longer than 244 characters).
Finally, what happened above? The problem was indeed due to the error inherent in floating-point arithmetic. For example, here is the calculation you performed:
. di %24.18f float( 1000 * float( (21557127/1000) - float( int (21557127/1000) ) ) )
127.000007629394531250
which, as you can see is not equal to 127. Let's take a closer look:
float( 1000 * float( (21557127/1000) - float( int (21557127/1000) ) ) )
---- temxxx --- ---- temxxx ---
--------- temyyy ----------
--------------------- temzzz ------------------------
-------------------------------- areaxxx -----------------------------
Notice how I am using the float() function to mimic the fact that, although you created temxxx as a double, you did not do so for the other intermediate variables. Now in this case, had you also created temzzz as a double, you would have gotten what you wanted:
. assert float( 1000 * ( (21557127/1000) - float( int (21557127/1000) ) ) ) == 127
However, as I said above, it is nearly always better to store IDs such as these as string variables.
On Mar 31, 2006, at 4:44 PM, Alex Ogan wrote:
```Here's something weird and probably related:
. clear
. set obs 1
obs was 0, now 1
. gen ID = 21557127
. display ID
21557128
```
Not weird at all. By default, -generate- creates new variables using the float storage type. And rounded to float precision, 21557127 is 21557128:
. di float(21557127)
21557128
Had you instead created the variable as a long, you would have seen what you expected:
. clear
. set obs 1
obs was 0, now 1
. gen long ID = 21557127
. di ID
21557127
-- Phil
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | 939 | 3,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-31 | latest | en | 0.878324 |
https://matplotlib.org/1.5.3/_sources/examples/pylab_examples/custom_cmap.txt | 1,669,467,857,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-49/segments/1669446706291.88/warc/CC-MAIN-20221126112341-20221126142341-00261.warc.gz | 422,094,851 | 3,560 | .. _pylab_examples-custom_cmap: pylab_examples example code: custom_cmap.py =========================================== .. plot:: /home/tcaswell/source/p/matplotlib/doc/mpl_examples/pylab_examples/custom_cmap.py :: #!/usr/bin/env python import numpy as np import matplotlib.pyplot as plt from matplotlib.colors import LinearSegmentedColormap """ Creating a colormap from a list of colors ----------------------------------------- Creating a colormap from a list of colors can be done with the `from_list` method of `LinearSegmentedColormap`. You must pass a list of RGB tuples that define the mixture of colors from 0 to 1. Creating custom colormaps ------------------------- It is also possible to create a custom mapping for a colormap. This is accomplished by creating dictionary that specifies how the RGB channels change from one end of the cmap to the other. Example: suppose you want red to increase from 0 to 1 over the bottom half, green to do the same over the middle half, and blue over the top half. Then you would use: cdict = {'red': ((0.0, 0.0, 0.0), (0.5, 1.0, 1.0), (1.0, 1.0, 1.0)), 'green': ((0.0, 0.0, 0.0), (0.25, 0.0, 0.0), (0.75, 1.0, 1.0), (1.0, 1.0, 1.0)), 'blue': ((0.0, 0.0, 0.0), (0.5, 0.0, 0.0), (1.0, 1.0, 1.0))} If, as in this example, there are no discontinuities in the r, g, and b components, then it is quite simple: the second and third element of each tuple, above, is the same--call it "y". The first element ("x") defines interpolation intervals over the full range of 0 to 1, and it must span that whole range. In other words, the values of x divide the 0-to-1 range into a set of segments, and y gives the end-point color values for each segment. Now consider the green. cdict['green'] is saying that for 0 <= x <= 0.25, y is zero; no green. 0.25 < x <= 0.75, y varies linearly from 0 to 1. x > 0.75, y remains at 1, full green. If there are discontinuities, then it is a little more complicated. Label the 3 elements in each row in the cdict entry for a given color as (x, y0, y1). Then for values of x between x[i] and x[i+1] the color value is interpolated between y1[i] and y0[i+1]. Going back to the cookbook example, look at cdict['red']; because y0 != y1, it is saying that for x from 0 to 0.5, red increases from 0 to 1, but then it jumps down, so that for x from 0.5 to 1, red increases from 0.7 to 1. Green ramps from 0 to 1 as x goes from 0 to 0.5, then jumps back to 0, and ramps back to 1 as x goes from 0.5 to 1. row i: x y0 y1 / / row i+1: x y0 y1 Above is an attempt to show that for x in the range x[i] to x[i+1], the interpolation is between y1[i] and y0[i+1]. So, y0[0] and y1[-1] are never used. """ # Make some illustrative fake data: x = np.arange(0, np.pi, 0.1) y = np.arange(0, 2*np.pi, 0.1) X, Y = np.meshgrid(x, y) Z = np.cos(X) * np.sin(Y) * 10 # --- Colormaps from a list --- colors = [(1, 0, 0), (0, 1, 0), (0, 0, 1)] # R -> G -> B n_bins = [3, 6, 10, 100] # Discretizes the interpolation into bins cmap_name = 'my_list' fig, axs = plt.subplots(2, 2, figsize=(6, 9)) fig.subplots_adjust(left=0.02, bottom=0.06, right=0.95, top=0.94, wspace=0.05) for n_bin, ax in zip(n_bins, axs.ravel()): # Create the colormap cm = LinearSegmentedColormap.from_list( cmap_name, colors, N=n_bin) # Fewer bins will result in "coarser" colomap interpolation im = ax.imshow(Z, interpolation='nearest', origin='lower', cmap=cm) ax.set_title("N bins: %s" % n_bin) fig.colorbar(im, ax=ax) # --- Custom colormaps --- cdict1 = {'red': ((0.0, 0.0, 0.0), (0.5, 0.0, 0.1), (1.0, 1.0, 1.0)), 'green': ((0.0, 0.0, 0.0), (1.0, 0.0, 0.0)), 'blue': ((0.0, 0.0, 1.0), (0.5, 0.1, 0.0), (1.0, 0.0, 0.0)) } cdict2 = {'red': ((0.0, 0.0, 0.0), (0.5, 0.0, 1.0), (1.0, 0.1, 1.0)), 'green': ((0.0, 0.0, 0.0), (1.0, 0.0, 0.0)), 'blue': ((0.0, 0.0, 0.1), (0.5, 1.0, 0.0), (1.0, 0.0, 0.0)) } cdict3 = {'red': ((0.0, 0.0, 0.0), (0.25, 0.0, 0.0), (0.5, 0.8, 1.0), (0.75, 1.0, 1.0), (1.0, 0.4, 1.0)), 'green': ((0.0, 0.0, 0.0), (0.25, 0.0, 0.0), (0.5, 0.9, 0.9), (0.75, 0.0, 0.0), (1.0, 0.0, 0.0)), 'blue': ((0.0, 0.0, 0.4), (0.25, 1.0, 1.0), (0.5, 1.0, 0.8), (0.75, 0.0, 0.0), (1.0, 0.0, 0.0)) } # Make a modified version of cdict3 with some transparency # in the middle of the range. cdict4 = cdict3.copy() cdict4['alpha'] = ((0.0, 1.0, 1.0), # (0.25,1.0, 1.0), (0.5, 0.3, 0.3), # (0.75,1.0, 1.0), (1.0, 1.0, 1.0)) # Now we will use this example to illustrate 3 ways of # handling custom colormaps. # First, the most direct and explicit: blue_red1 = LinearSegmentedColormap('BlueRed1', cdict1) # Second, create the map explicitly and register it. # Like the first method, this method works with any kind # of Colormap, not just # a LinearSegmentedColormap: blue_red2 = LinearSegmentedColormap('BlueRed2', cdict2) plt.register_cmap(cmap=blue_red2) # Third, for LinearSegmentedColormap only, # leave everything to register_cmap: plt.register_cmap(name='BlueRed3', data=cdict3) # optional lut kwarg plt.register_cmap(name='BlueRedAlpha', data=cdict4) # Make the figure: fig, axs = plt.subplots(2, 2, figsize=(6, 9)) fig.subplots_adjust(left=0.02, bottom=0.06, right=0.95, top=0.94, wspace=0.05) # Make 4 subplots: im1 = axs[0, 0].imshow(Z, interpolation='nearest', cmap=blue_red1) fig.colorbar(im1, ax=axs[0, 0]) cmap = plt.get_cmap('BlueRed2') im2 = axs[1, 0].imshow(Z, interpolation='nearest', cmap=cmap) fig.colorbar(im2, ax=axs[1, 0]) # Now we will set the third cmap as the default. One would # not normally do this in the middle of a script like this; # it is done here just to illustrate the method. plt.rcParams['image.cmap'] = 'BlueRed3' im3 = axs[0, 1].imshow(Z, interpolation='nearest') fig.colorbar(im3, ax=axs[0, 1]) axs[0, 1].set_title("Alpha = 1") # Or as yet another variation, we can replace the rcParams # specification *before* the imshow with the following *after* # imshow. # This sets the new default *and* sets the colormap of the last # image-like item plotted via pyplot, if any. # # Draw a line with low zorder so it will be behind the image. axs[1, 1].plot([0, 10*np.pi], [0, 20*np.pi], color='c', lw=20, zorder=-1) im4 = axs[1, 1].imshow(Z, interpolation='nearest') fig.colorbar(im4, ax=axs[1, 1]) # Here it is: changing the colormap for the current image and its # colorbar after they have been plotted. im4.set_cmap('BlueRedAlpha') axs[1, 1].set_title("Varying alpha") # fig.suptitle('Custom Blue-Red colormaps', fontsize=16) plt.show() Keywords: python, matplotlib, pylab, example, codex (see :ref:`how-to-search-examples`) | 2,377 | 6,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.759257 |
http://www.yourdictionary.com/equalities | 1,529,695,970,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864776.82/warc/CC-MAIN-20180622182027-20180622202027-00259.warc.gz | 526,625,120 | 19,048 | Sentence Examples
• The five equalities which stand first in the five pairs of equalities in � 15 (2) may therefore be taken as the main types of a simple statement of equality.
• Similarly the equalities 99 X I o I = 9999 = wow - I 98 X 102 = 999 6 = moo() - 4 97 X 10 3 =9991 =1 0000 - 9 lead up to (A - a) (A+a) = A 2 - a 2.
• The symbol e 0 behaves exactly like i in ordinary algebra; Hamilton writes I, i, j, k instead of eo, el, e2, es, and in this notation all the special rules of operation may he summed up by the equalities = - I.
• All this is analogous to the corresponding formulae in the barycentric calculus and in quaternions; it remains to consider the multiplication of two or more extensive quantities The binary products of the units i are taken to satisfy the equalities e, 2 =o, i ej = - eeei; this reduces them to.
• 4.1.1 (i) Equalities and Inequalities | 259 | 880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-26 | latest | en | 0.891718 |
https://es.mathworks.com/matlabcentral/profile/authors/14942527?detail=answers | 1,718,520,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00852.warc.gz | 209,000,737 | 22,043 | # Mario Chiappelli
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Here is a picture of the GUI I am working on. As you can see after a few iterations the axes keep printing on top of the previou...
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# OCR (non mei) C4 Friday 24th June 2016
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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. (Original post by djamesh)
I did the EXACT same. We were supposed to use .5bh, but we'll probably get a mark or two for working out the correct magnitudes of the required sides.
Didn't it ask you to use the smaller angle? Or have I misread?
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2. (Original post by daniellon)
For the stationary point question did they ask for exact values?
Yeah, extremely fiddly since you had to use s^2+c^2=1 to find sin(t) first and then substitute back in.
3. (Original post by drandy76)
Didn't it ask you to use the smaller angle? Or have I misread?
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4. There goes my uni offer after this exam and core 3
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5. (Original post by djamesh)
Then what was the problem with using c as ~12.7 or whatever it was?
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6. (Original post by drandy76)
Then what was the problem with using c as ~12.7 or whatever it was?
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It's because C is opposite of the resultant vector however 12.7 was opposite to one of those vectors.
7. (Original post by Vikingninja)
It's because C is opposite of the resultant vector however 12.7 was opposite to one of those vectors.
.........****, so I should've used 90-alpha instead
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8. (Original post by drandy76)
.........****, so I should've used 90-alpha instead
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The two vectors were perpendicular so C would have been 90.
9. Any one has the paper??? Can you upload the pics please??? Thank you!!
10. (Original post by Vikingninja)
The two vectors were perpendicular so C would have been 90.
I believe I have realised the problem, I drew my triangle wrong, have AB perp to OA instead of OB so my angles got mixed around
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11. How many marks will I lose for substituting the bottom as u instead of root u for question 6?
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12. So annoyed at myself, used vector multiplication from FP3 for the area question and managed to get it wrong. Then messed up the easy parts of the parametric question and the last question was just???
13. Anyone else search the back of their paper, thinking they missed the DE question?
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14. (Original post by drandy76)
Anyone else search the back of their paper, thinking they missed the DE question?
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the last part of the last question was a D.E
15. (Original post by Abscissa)
the last part of the last question was a D.E
....oh yeah, didn't even think about it
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16. I think I got 66-68 marks.
What do you think 66 will be in UMS? x
17. Anyone got workings for question 6? For some reason still can't work it through :/
18. I ran out of time on the last part of the last question. Do you think I'll get a mark for separating the variables? So annoyed cos I could've done it easily if my time management wasn't so poor!
19. (Original post by Mandos)
I think I got 66-68 marks.
What do you think 66 will be in UMS? x
Mid to high 90s?
20. (Original post by Inert1a)
Would 70/72 be full ums
You obviously know that's full UMS.
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### Who is getting a uni offer this half term?
Find out which unis are hot off the mark here
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Useful resources | 1,003 | 3,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2016-44 | latest | en | 0.964816 |
https://brilliant.org/problems/do-you-know-your-basics-v/ | 1,524,221,664,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937440.13/warc/CC-MAIN-20180420100911-20180420120911-00450.warc.gz | 568,965,233 | 17,350 | # Do you know your Basics?-V
A load of $$300 kg$$ is pulled up by $$5 m$$, then calculate the work done by the pulling force( in Joules).
Please take the magnitude of acceleration due to gravity = $$9.8 \text{ m/s}^2$$
× | 65 | 223 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-17 | latest | en | 0.873076 |
http://math.stackexchange.com/questions/162805/the-meaning-of-finite-in-the-finite-covering-theorem | 1,467,275,961,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398216.41/warc/CC-MAIN-20160624154958-00183-ip-10-164-35-72.ec2.internal.warc.gz | 197,422,567 | 20,493 | # the meaning of “finite” in the finite covering theorem
A textbook I am using to learn analysis states (in reference to just the real line):
Every system of open intervals covering a closed interval contains a finite subsystem that covers the closed interval.
(the textbook is "Mathematical Analysis I" by V.A. Zorich)
Let's say $S$ = {$U_n$} is the system of open intervals $U_n$ in question, which covers the closed interval $I$.
Now, if we take all $U_n$ to be infinitely small in length, but take $S$ to be infinite in cardinality such that all $U_n$ still cover $I$ (does this even make sense?), wouldn't it be impossible to select a finite subsystem of $S$ that covers $I$?
I feel that maybe I am missing an elementary but important distinction in my idea of "infinite"
-
It is impossible for an open interval to be "infinitely small in length". The measure of $(a,b)$ is $b-a>0$; an "infinitely small in length" interval would be the singleton set $\{a\}$, which is closed. (Also the closed interval in the quoted statement should be bounded.) – Tom Cooney Jun 25 '12 at 11:53
@TomCooney: Just a minor quibble: The important point isn't that singletons are closed in the real line, but rather that they are decidedly not open. – arjafi Jun 25 '12 at 11:58
@TomCooney: a closed interval in the real line must be bounded, surely? it can't have $\infty$ as an endpoint because it would then contain it, but it's not a real number. – Ben Millwood Jun 25 '12 at 12:49
@benmachine: You are mistaken, for instance all of $\mathbb R$ is a closed interval. Closed implies that every point of the space that is a limit of points of the subset is in the subset itself, but $\infty$ is not a point of the space. The property that unbounded intervals lack is "compact" (whose definition seems to be considered here), but not closed. – Marc van Leeuwen Jun 25 '12 at 13:15
@MarcvanLeeuwen: ah, I was interpreting "closed interval" to mean "a thing of the form $[a,b]$" not "a thing that is both closed and an interval". I'd forgotten that the condition of having square brackets at either end was sufficient but not necessary :) – Ben Millwood Jun 25 '12 at 13:27
Your mistake lies in the assumption that you can choose "$U_n$ infinitely small in length". For each $n$ seperately the set $U_n$ is an open set. An open set contains an open interval and an open interval has a positive measure (the notion of "length" doesn't really make sense for an arbitrary open set, but that is not the problem here). Obviously the open sets may become "smaller" (again, you have to be careful how you measure the size of an open set) if $n$ varies, but that's a different story.
Btw: The property you quote is the definition of compactness. For subsets of the real numbers we have: bounded and closed if and only if compact. Note that the textbook presumably assumes that the interval is bounded.
-
First let me tell you that proving compactness of a closed and bounded interval $[a,b]$ is trivial once you know the following two facts:
1. In a metric space sequential compactness and open cover compactness are equivalent
2. Bolzano - Weierstrass is available to us.
By (1) it suffices to prove that for any sequence $a_n \subset [a,b]$ we have that there is a convergent subsequence. But this is trivial because Bolzano - Weierstrass guarantees that we have a convergent subsequence, and the interval being closed guarantees that the subsequence converges to something in $[a,b]$ so that $[a,b]$ is compact.
Now back to your problem. Since the open cover of $[a,b]$ that you are considering consists of just open intervals and not arbitrary open sets then each element $U_n \in S$ will contain a closed and bounded interval $[c_n,d_n]$ of length $d_n - c_n > 0$. I don't understand what you mean by "infinitely small length"; the "length" of each $U_n$ will be at least this $c_n - d_n$ and is a constant. What do you mean by "choose each $U_n$ to be of infinitely small length"?
-
I'd think that bringing sequential compactness and Bolzano-Weierstrass into this is possibly above the level of the question. Also, he didn't say "arbitrarily small length", he said something even more impossible... – Ben Millwood Jun 25 '12 at 12:50
@benmachine I thought the BW property and sequential compactness was standard stuff..... – user38268 Jun 25 '12 at 13:05
Yeah, but the proof that they're equivalent in a metric space sounds like something you'd learn at the same time as learning what compactness was all about. I'm not sure it's easier than just proving the result directly. But I guess it's a matter of taste which results ought to depend on which. – Ben Millwood Jun 25 '12 at 13:14
@BenjaLim It is standard, but I'm not there yet. Interestingly, the text I mentioned presents the BW theorem immediately following the finite covering theorem and uses it in its proof, both of them under a subsection "Basic Lemmas Connected with the Completeness of the Real Numbers", alluding to what benmachine mentions. This indirection is why I did not accept your answer over Simon's, but I'm sure it will prove helpful soon, so thank you. – afsmi Jun 25 '12 at 13:29
Try to write down an open interval that is infinitely small in length. You'll find that you can't (unless you count the empty set, and you can't cover a closed interval in empty sets!).
Note that if you have an interval which is not closed, like $[0,1)$, then you can use a cover like $U_n = [0,\frac{n-1}{n})$, each set covering closer and closer to the right endpoint and hence eventually covering any point to the left of it. This cover has no finite subcover, since any finite collection of such sets includes a biggest one, and that biggest one misses a tiny sliver of the original interval.
What happens when you try to do the same thing with $[0,1]$? Well, you can get closer and closer and closer to $1$, but at some point you've got to hit it, so you've got to have an open interval like $(1-\epsilon,1+\delta)$ in your cover, but then you've just made infinitely many (indeed, all but finitely many) sets of your cover redundant, since you now only need to cover up to $1-\epsilon$, and regardless of how small $\epsilon$ is, there's some $1/n$ less than it, so there's an interval $[1,\frac{n-1}{n})$ that does the job.
Actually, this realisation that if you have infinitely many open sets creeping up on something, the open set actually containing that something is going to let you throw most of them away, is surprisingly close to the formal proof of your result, known as (more-or-less) the Heine-Borel theorem: just for fun, here's a proof in iambic pentameter.
- | 1,652 | 6,641 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2016-26 | latest | en | 0.950485 |
http://fawnnguyen.com/let-problem/ | 1,603,217,029,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00229.warc.gz | 41,230,661 | 10,224 | # When I Let Them Own the Problem
From our textbook:
Stuff like this makes my heart sink. (I actually wrote that it makes me fart — but that’s very unladylike. And I’m trying to write better.)
There is essentially nothing left in this problem for students to explore and figure out on their own. If anything, all those labels with numbers and variables conspire to turn kids off to math. Ironically even when the problem tells kids what to do (use similar triangles), the first thing kids say when they see a problem like this is, “I don’t get it.”
They say they don’t get it because they never got to own the problem.
I wiped out the entire question and gave each student this mostly blank piece of paper and the following verbal instructions:
1. Make sure you have a sharpened pencil. Write your name and date.
2. Inside this large rectangular border, draw a blob — yes, blob — with an area that’s approximately 1/5 of the rectangle’s area. No one will die if it’s not quite 1/5.
3. Next, draw a dot anywhere inside the rectangle but outside the blob. Label this dot H.
4. Now, draw another dot — but listen carefully! — so that there’s no direct path from this dot to the first dot H. Label this second dot B.
I asked the class if they knew what they just drew. After a few silly guesses, I told them it was a miniature golf course: blob = water, point B = golf ball, point H = hole location.
The challenge then was to get the ball into the hole. Since you can’t putt the ball directly into the hole due to the water hazard, you need to make a bank shot.
(Some students may have drawn the blob and points in such a way that this was not really possible, at least not in one-bank shot. I let them just randomly pull from the stack of copies to pick a different one. I made a copy of their sketches first before they started their work.)
The discussions began as they started drawing in the paths. One student drew hers in quickly and asked, “Is this right?” I replied, “I’m not sure, but that’s my challenge to you. You need to convince me and your classmates that the ball hitting the edge right there will bounce out and travel straight into the hole. Does it? What can you draw? What calculations are involved?”
What I heard:
The angle that the ball hits the border and bounces back out must be the same.
This is like shooting pool.
Right triangles.
Similar right triangles.
Do we need to consider the velocity of the ball?
This is hard.
I can’t figure out how to use the right triangles.
Similar right triangles because that’ll make things easier.
Even though it’s more than one bounce off the edges, I’m still just hitting the ball one time.
I think I got this.
I have an idea.
Wish my golfer is Happy Gilmore.
BIG struggles, so I was happy and tried not to be too helpful. (I struggled big time too on some of their papers! And I think this made them happy.)
Lauren explained in this 55-second video how she found the paths for the ball to travel. I also had her explain to the whole class later at the document camera.
Jack took a different approach. Instead of measuring the sides and finding proportions to find more sides to create similar triangles like Lauren did, he started with an angle that he thought might work [via eyeballing] and kept having the ball bounce off the borders at paired angle until it went into the hole. (His calculation was off — or his protractor use was inaccurate — as he had angles of 90, 33, and 63. Or maybe if he had a better teacher, he’d know the sum of the interior angles of a triangle was 180.)
Gabe was quieter than usual today. When he finally shared, his classmates realized he was the only one to solve the entire problem using just constructions with a straightedge and compass. He walked us through his series of constructions until he found point C on the bottom border where the ball needed to bank off and end up in hole H.
Imagine none of this thinking and sharing would have occurred if I had given them problem #24 in the book.
Half of my kids were still struggling and working to find the correct bank shot(s), but they were giventhe chance to struggle. And none of them said, “I don’t get it.”
The cutest thing also happened while we were doing all this math. Yesterday (Monday) I bragged to the kids — and I’m doing it again right now — about the Rolling Stones concert that we went to on Friday. I am still over the moon ecstatic that we got escorted into the Pit from our way-in-back-floor-seats!!!!
Anyway, a kid today started humming to the tune of (I Can’t Get No) Satisfaction and quickly others joined in with THESE LYRICS:
I can’t get no similar triangles
I can’t get no similar triangles
‘Cause I try and I try and I try and I try
I can’t get no, I can’t get no
When I’m drawing in my lines
This lesson leaves me so full and proud. Their singing to the Stones while struggling in math makes me crazy in love with them.
Just so you know, I swooned shamelessly in front of my students over a 70-year-old rock star’s butt.
[Updated 05/08/13]
Today I had the kids work on someone else’s paper (remember I made copies of their papers before they worked on them) and find similar triangles to make the bank shots. Because I purposely told the kids to draw in the blobs and the 2 points without any mention of where exactly to place them, it was then by chance that these papers below allowed for one-bank shots to get the ball into the hole.
The ones below, however, are some of the ones that would not work with just one-bank shots, but I had the kids create similar triangles on them anyway because that was the learning goal of the lesson.
[Updated 05/11/13]
Look what the crazy and wonderful Desmos did (click on tweet below to see):
[06/28/14]
There were over 90 comments left for this post on the old site, but I’d like to feature this thread of comments between me and hillby as it involves us sharing some geometric constructions.
May 8, 2013 7:06 AM
hillby wrote:
Awesome lesson, excellent job of breaking the problem down, increasing cognitive demand and also getting students to share their thought process.
It took me a while to figure out how Gabe was able to find the point exactly with just some lines and a compass. I stumbled upon it, but I haven’t figured out why it worked. Did Gabe figure out that this approach would work through reasoning, or trial and error like me? I guess I’m basically asking if he added a proof, or did he check by measurement?
May 8, 2013 10:20 PM
fawnnguyen wrote:
Thank you, Chris!
I’m really glad you questioned Gabe’s constructions. I wrote down his steps and re-created it on GSP so you could see:
B is ball. H is hole.
Construct BA and HD, both perpendicular to horizontal bottom line. Both have the same measurements as what he wrote on his paper.
Draw in HA, forming angle(AHD).
Copy angle(AHD) over to angle(GBA).
Now this is his “just a hunch” step: construct the midpoint of AD, label this E.
Construct the midpoint of AG, label this F.
Construct the midpoint of FE, label this C.
Draw in BC and HC, forming the yellow and green triangles.
He just checked by measurement. What do you think?
May 9, 2013 7:30 PM
hillby wrote:
Oh, how INTERESTING!! I did something similar based on the picture in the post, but it wasn’t quite the same. On the other hand, I got a perfect match. Picking up from BA & HD,
Draw in HA
Draw in BD
Draw in a line perpendicular to AD through the intersection of BD & HA
The intersection of the perpendicular and line AD will be your exact point of reflection.
I think Gabe’s method is similar to the Newtonian method of finding zeros – he’s basically iterating closer and closer to that exact point of reflection.
May 9, 2013 7:54 PM
fawnnguyen wrote:
And look how beautiful yours looks!
I will share your construction with Gabe. I love how Gabe persevered on this problem and appreciate his “hunch” too — it’s a risk I want more kids to take! You can see his tedious work of constructing those midpoints. Any other kid would have just eyeballed it or used a ruler.
Thanks again, Chris. | 1,902 | 8,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-45 | latest | en | 0.935865 |
https://howtodiscuss.com/t/n-cv/116537 | 1,656,107,343,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00398.warc.gz | 349,846,428 | 6,590 | # N Cv
## N Cv
I have a total chemistry test, specify n = m / M, n = cV? 3
what's that ? What units are used to measure each and how is the pH calculated?
n refers to the number.
n = m / M
m is equal to m (unit is gram)
M refers to m molecules (unit is gram / mole)
There the gram was divided by the value, the molar unit was obtained.
n = CV
c refers to molar concentration (is unit / liter)
V refers to volume (liters)
In this case, multiplying a liter by a liter gives one unit.
what's that ?
A mole is a quantity of pure substance containing the same number of chemical units. In other words, it's just a unit of measurement.
To calculate the pH of an acidic solution:
First you need to know the H + concentration of your solution. Then find the negative logarithm of your H + concentration.
An example:
0.1 MHCL
pH = Log (0.1)
PH = 1
w To calculate the pH of an alkaline solution:
First, you need to know the concentration of OH in your solution. If you take the negative logarithm of OH concentration, you get pOH. Lowering the pH to 14 gives the pH of the solution.
An example:
0.1 M NaOH
pOH = Log (0.1)
POH = 1
PH = 14 POH
PH = 14 1
PH = 13
To better understand how the pH of a solution is calculated, try recalculating Henderson Heilbach's equation.
I now! For your health!
D:
I have a total chemistry test, specify n = m / M, n = cV?
what's that ? What units are used to measure each and how is the pH calculated?
Even if ** in fact ** is wrong, you can only think of ■■■■■ as numbers. Like 20, a thousand or a million. But the mole is 6.022 * 10 23. We use n to denote quantity, N to denote quantity.
n = m / M
That is, the sum is equal to m divided by m by mol. Is it fair to say that many cars sit on one big wheel, a total of 400 kg, when each car weighs 100 kg? The number of vehicles is n (or atoms), m is the total m of the ferrous wheel (or m is the total of the specimens) and we know that M is per car (m per mol).
n {mol} = m {g} / M {(g / mol)
Second, it can be easy to see from a person's point of view. A room is 100 m 3 in size, and we know that there are 4 groups of people for every m 3. How many groups of people are in the room? 100 m 3 * 4 groups / m 3 = 400 groups.
The units are:
n {mol} = c {mol / m 3} * V {m 3
Note that m 3 can be L, cc, cm 3, ml, m 3 or any other measure of volume.
For the pH value, first determine the concentration of H + in ■■■■■ / L. Then take the negative logarithm. If H + is concentrated in mol / Lc then pH = log (c)
Sorry, I came here late, this is not my best job, but it helps.
What is N in Chemistry?
n = not from
m = in number grams.
M = molecular weight
c = con of the sun.
V = volume Ls
Now sesame is the number of (mock) molecules that are abundant.
If there is no mole, do not drive
Conc They have different units / beds, grams / lights etc.
Hahaha you ed | 820 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-27 | latest | en | 0.934361 |
http://bartleylawoffice.com/help/what-is-social-security-tax-limit-for-2016-correct-answer.html | 1,660,691,437,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00751.warc.gz | 4,073,991 | 20,033 | # What Is Social Security Tax Limit For 2016? (Correct answer)
For 2016, the maximum limit on earnings for withholding of Social Security (Old-Age, Survivors, and Disability Insurance) Tax remains \$118,500.00. The Social Security Tax Rate remains at 6.2 percent. The resulting maximum Social Security Tax for 2016 is \$7,347.00.
## At what limit is Social Security taxable?
between \$25,000 and \$34,000, you may have to pay income tax on up to 50 percent of your benefits. more than \$34,000, up to 85 percent of your benefits may be taxable.
## Is Social Security taxed after age 70?
Calculating the exact amount of tax that must be paid on Social Security benefits can be quite complicated. After age 70, there is no longer any increase, so you should claim your benefits then even if they will be partly subject to income tax.
## What is the Social Security tax limit for 2021?
The amount liable to Social Security tax is capped at \$142,800 in 2021 but will rise to \$147,000 in 2022. The change to the taxable maximum, called the contribution and benefit base, is based on the National Average Wage Index. The increase for 2022, at 2.9 percent, is less than the 3.7 percent increase for 2021.
## Is Social Security tax capped?
The Social Security tax rate for those who are self-employed is the full 12.4%. There is a limit on the amount of annual wages or earned income subject to taxation, called a tax cap; in 2021, the maximum amount of income subject to the Social Security tax is \$142,800; in 2022, the maximum is \$147,000.
You might be interested: What Is Included In Tax Free Weekend Tn? (Solution)
## How do I determine how much of my Social Security is taxable?
If your combined income was more than \$34,000, you will pay taxes on up to 85% of your Social Security benefits. For married couples filing jointly, you will pay taxes on up to 50% of your Social Security income if you have a combined income of \$32,000 to \$44,000.
## Do seniors pay taxes on Social Security income?
Up to 50% of Social Security benefits are taxed on income from \$25,000 to \$34,000 for individuals or \$32,000 to \$44,000 for married couples filing jointly. Up to 85% of benefits are taxable if the income level is over \$34,000 for individuals or \$44,000 for couples. 2.
## Is there really a \$16728 Social Security bonus?
The \$16,728 Social Security bonus most retirees completely overlook. But a handful of little-known “Social Security secrets” could help ensure a boost in your retirement income. For example: one easy trick could pay you as much as \$16,728 more each year!
## At what age do seniors stop paying taxes?
As long as you are at least 65 years old and your income from sources other than Social Security is not high, then the tax credit for the elderly or disabled can reduce your tax bill on a dollar-for-dollar basis.
## What age do you no longer have to file income tax?
Updated for Tax Year 2019 You can stop filing income taxes at age 65 if: You are a senior that is not married and make less than \$13,850. You are a senior that is married, and you are going to file jointly and make less than \$27,000 combined.
You might be interested: What Can I Buy On Tax Free Weekend? (Correct answer)
## Why is there a limit on Social Security tax?
The cap limits how much high earners need to pay in Social Security taxes each year. Critics argue that income tax caps unfairly favor high earners compared to low-income earners. Others believe that raising the cap would effectively result in one of the largest tax hikes of all time.
## What is the maximum Social Security tax withholding for 2020?
For 2020, the maximum limit on earnings for withholding of Social Security (old-age, survivors, and disability insurance) tax is \$137,700.00. The Social Security tax rate remains at 6.2 percent. The resulting maximum Social Security tax for 2020 is \$8,537.40. | 918 | 3,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-33 | latest | en | 0.950197 |
www.askageologist.com | 1,726,467,582,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00876.warc.gz | 41,727,220 | 38,377 | ## Age of the Earth – in a Nutshell
Q: In geology class my professor told me that the earth’s age is based of off meters how does this work
– Theo
A:
The age of the Earth was initially estimated by scientists by mapping stacks of sedimentary rocks in the UK, then measuring sedimentation rates in similar environments (lakes, rivers, seashore, etc.). In the 19th Century this initially gave startling – even shocking at the time – estimates in the hundreds of millions of years range. In the early 20th Century radioisotopes became available, and these were used to extend the age of the Earth into the billions (billion = thousand million) of years age range*. This physically meant measuring back to the point in time when the mineral hosting the radioisotopes and their daughter-products was last melted. THEN it just became a game of searching all over the Earth for the oldest date-able minerals with uranium and lead in them (for example, a zircon crystal). The oldest rocks found so far are in Greenland and western Australia, and based on these the Earth’s age is estimated to be at least 4.55 thousand million years old. This means it is at LEAST that old.
* Note that in some countries like the US, the word “billion” means a thousand million, while in other countries (e.g., the UK), the word “billion” means a million million.
## How to contain Yellowstone…. or maybe not
As teachers we always want to encourage questions – not discourage questions. Sometimes, however, people will start with a question and then on their own figure out an “answer.” If the answer isn’t grounded in reality – some basic education being necessary here – then the answer can get out of control and move beyond reality very quickly. The following is an example of this.
Q: Could my strategy for fixing yellowstone work?
Build a huge reservoir container made of titanium, nickel,and cobalt. Then build a 4d scanning system that monitor the lava pressure within yellowstone as it increase. Build a drain made out of titanium,cobalt,nickel, using gravity slowly drain lava into reservoir as the lava pressure increases to maintain lava at the below level of erupting sustain it at that level as plates shift friction ignites gases in lava then lava increases so I do not think that cooling it will work sufficiently in my opinion. by keeping lava in core low in volcanoes maybe we can reduce friction of constant shifting plates. PLEASE write back.
–misty c
A: You are operating under a rather large number of mistaken assumptions.
First of all, friction does not ignite volcanic gases; they tend to be primarily SO2 and CO2, and both of them are already oxidized. In other words, they can’t “ignite.”
Second, you cannot monitor lava pressure in “4D” or even 3D remotely – you can only guesstimate it from laboratory pressure-cell studies coupled with depth/thickness estimates from seismic reflection experiments.
Third, there is not enough money in the entire US GDP to pay for a container of titanium, nickle, and cobalt of any size that might be even remotely comparable to the output of even a small Yellowstone eruption.
Fourth, while such a container as you describe would probably not melt at typical magma temperatures (~1,300 C), it would be structurally weaker than at ambient temperatures. It would be a very complex engineering problem. Actually, why would you even need a container at all? The Earth’s surface is covered with vast sheets of cooled lava in many places, including Siberia, western India, and the Pacific Northwest.
Fifth, the Yellowstone caldera is nearly 45 miles (72 kilometers) across. There are several rough estimates from seismic data of how large the magma reservoir beneath it is, and they are all huge – and highly dependent upon how far down you want to count. Down to the top of the Mantle? Some recent research suggests that the Yellowstone hot spot plume rises from a depth of at least 440 miles (700 kilometers) deep within the Earth’s Mantle. Some researchers suspect it originates from 1,800 miles (2,900 kilometers) deep at the top of the Earth’s core. In other words, that is a LOT of magma. The last really large eruption 640,000 years ago blew out the equivalent of 1,000 cubic kilometers of DRE (dense rock equivalent). In other words, 240 cubic miles of rock blasted to tephra and ash that reached as far as the East Coast of the US.
Sixth, to “drain” something you would need pressure or gravity to work for you; no known pump could handle molten magma. Moreover, the current understanding is that the magma under Yellowstone is a highly-viscous, mostly-crystal mush (only a 2% – 9% melt fraction remains from seismic tomography data interpretation). An “eruptible” magma would require at least a 50% – 60% melt fraction. The top of the magma now lies at least 4 miles (6 kilometers) below the Earth’s surface. It is already being contained by an overlying cap of volcanic rock left from previous eruptions that have now cooled. Experimental drilling to those sorts of depths is extraordinarily difficult. The Kola Peninsula Superdeep Borehole in Russia took two decades to reach six miles (10 kilometers) depth. It’s terminal aperture was just 9 inches/23 cm in diameter. Rock turns plastic before you get to those depths, closing in and trapping the drill-stem.
Just compare that 23 centimeter-diameter borehole to the width of the caldera at 72,405,000 centimeters. There is no way you could “drain” Yellowstone through a drill hole that small (even if the magma was hot, pressurized, and non-crystalline) in the all the time that the Earth has existed.
Finally, until you are actually physically standing on a volcano, you cannot really understand how immense it is. Volcanoes dwarf the creations of humankind. The forces involved in even a small to moderate eruption are far greater than anything mankind can develop with modern technology, including fusion bombs – and Yellowstone is a Supervolcano. All you can do is get out of the way when ANY volcano decides to do its thing.
I have not even begun to address the formidable engineering difficulties of your plan.
This may not be the kind of response you may have expected; I’m trying to give you a scale-based reality-check here. You can learn a lot more by reading some of the scientific literature, readily available on the internet, for instance:
http://science.sciencemag.org/content/348/6236/773?ijkey=c0dac9ef6421b172d426307cf7fa08be7986dee6&keytype2=tf_ipsecsha
## Well, how big WAS it?
It is human nature to want to measure things, or at least calibrate big things against other big things. The big and destructive fairly beg quantifying, in fact, so for instance we have the Saffir-Simpson hurricane wind scale (with a top level of 5 for winds above 156 mph/250 kph). This depends only on wind velocities, and does not take into account either rain or storm surges (Allaby, 2008). We also have the Fujita tornado intensity scale (Fujita, 1971), which for winds above 261 mph/420 kph can reach a level of F5. The following question asks about measuring earthquakes and volcanoes, which are much harder to quantify than wind-speed velocities.
Q: Hi I am an 8th grade student and I was wondering what determines the magnitude of an earthquake or what determines the power of a volcano…
– Caleb Le M.
A: Your question has two parts, which I will answer in order:
1. Earthquake magnitudes are calculated many different ways, but ultimately it comes down to measuring the amplitude of the actual ground motion (up-down, side-to-side) on multiple seismometers, and correcting for the varying seismic rock-velocities and the distance separating the seismometers from the earthquake epicenter. Of course, you have to calculate the distance to the epicenter first by triangulation from three or more seismometers (and also correct THOSE results by different seismic velocities in the different rocks between the hypocenter and the different measuring seismometers).
Asking a seismologist how big an earthquake was is like asking a friend to describe how big someone is? Do you mean tall? Wide? Heavy? Some combination of all of these? Does this dress make me look fat? Seismologists do NOT like being asked how they calculate a magnitude, because it will generally require a 30-minute explanation. Therefore, their first reply is often which magnitude are we talking about here?
The original earthquake magnitude scale (Richter, 1935) was the first coherent attempt to define something that is ultimately three-dimensional and very complex. The original Richter scale measured only the energy in the low-frequency end of the seismic energy spectrum, standardized to the particular type of Wood-Anderson seismometer available at the time. Today a modified Richter magnitude is called the “local magnitude” or ML, and is tuned for the rocks and sediments of a local region. For southern California, the equation to calculate this magnitude (Spence et al., 1989; Bormann and Dewey, 2014) is:
ML = Log (A) + 0.00189*r – 2.09,
…where A = amplitude of maximum ground movement in nanometers measured at the seismometer, r = distance from the seismometer to the epicenter in kilometers, and – 2.09 is a correction factor. This equation works only for southern California, and doesn’t work for Cascadia, Japan, the Mediterranean, or Indonesia, which are each served better by different numerical factors. The equation is very simple; if you know the reported magnitude, you can use your smart phone to calculate exactly how much the ground moved under you.
Another way to calculate an earthquake local magnitude is to work off of an analog log-scale diagram such as in this:
Figure 22. An analog diagram for calculating a Richter scale number for a local earthquake (see http://www.ntschools.org/cms/lib/NY19000908/Centricity/Domain/112/Richter%20worksheet.pdf).
Though relatively easy to understand and use, the Richter Scale is no longer commonly used because it represents just a fraction of what is going on.
There are also Mb (the body-wave magnitude), Ms (the surface-wave magnitude), and Mw (the moment magnitude). Most of these magnitudes track closely together for magnitudes of M = 2 to M = 5, but diverge for larger and smaller earthquakes. In part this is because some wave-types strongly influence a short-period or broadband seismometer (which are sensitive to higher frequencies) while another wave-types (for example, surface waves) more strongly affect older seismometers designed to optimally measure low-frequency energy in the 1 – 2 Hz range.
For large earthquakes, Mw (Moment Magnitude) is the preferred magnitude, because it more fully represents everything emanating from the earthquake hypocenter. The “moment” Mo is calculated as a product of µ (shear strength of the rocks) times S (the surface area of the fault tear, measured horizontally, times down-dip direction), and d (the displacement – how far did one side of the fault move with respect to the other side). The largest ever recorded earthquake, as mentioned earlier was the Great Chilean (Valdivia) event of May 1960, which had a moment magnitude Mw = 9.5
Confused yet? There is also Me (the energy magnitude – a measure of the potential damage to man-made structures), and Intensity (the measure of surface-shaking damage observed). They are related. Energy release is generally proportional to the shaking amplitude raised to the 3/2 power, so an increase of 1 magnitude corresponds to a release of energy 31.6 times greater than that released by the next lower earthquake magnitude. In other words,
Magnitude 3 = 2 gigajoules
Magnitude 4 = 63 gigajoules
Magnitude 5 = 2,000 gigajoules
Magnitude 6 = 63,000 gigajoules
Magnitude 7 = 2,000,000 gigajoules
What is a joule, you may reasonably ask? Energy is normally expressed in joules while power is expressed in watts. One watt is defined as 1 joule per second. A hundred-watt lightbulb, turned on for just one second, consumes 100 joules of total energy.
Both Intensity and Magnitude depend on many local variables, including surface geometry and the seismic velocities of various underlying rock and sediment units. For example, the 1985 Mexico City earthquake had a surface-wave magnitude Ms of 8.1 However, because of resonant focusing of seismic waves as the partially-dried-up Texcoco Lake basin (that Mexico City was built on) lapped onto bedrock, some buildings on one side of a city boulevard had ground motions 75 times greater than the other side (Moreno-Murillo, 1985; see also http://earthquake.usgs.gov/learn/topics/measure.php ). A friend (Mauricio de la Fuente, then a professor at UNAM, the autonomous university of Mexico) who lived through this event told me that it was amazing to stand in that street and see everything on one side standing, and everything on the other side flattened. Over 8,000 people died, mainly in buildings on that ancient lake side (the Texcoco ancient lake bed).
Intensity is very different, based on the Mercalli scale (see https://en.wikipedia.org/wiki/Mercalli_intensity_scale ). It is a twelve-level scale designed to fit observed damage. The name Mercalli is attached to a scale that Giuseppe Mercalli revised from an earlier Rossi-Forel scale, and which has been further modified multiple times since then (http://pubs.usgs.gov/gip/earthq4/severitygip.html ). On the Modified Mercalli scale, the 1985 Mexico City event scored an intensity level of IX (“Violent”) out of a possible twelve. That’s another way of saying “things could be a lot worse.”
One more thing to think about: seismologists estimate that only 1% to 10% of the energy of any given earthquake is released as seismic waves. Almost all the rest of the energy is released as heat (http://earthquake.usgs.gov/learn/topics/measure.php ). This figures indirectly into models designed to emulate the complex breaking process of a fault tear, because at some points, wall-rocks are literally welded together by the intense heat, forcing complex movements around these focal points (Dieterich, 1978; James Dieterich, personal communication 2016). This is an astonishing understanding that goes a long way towards explaining the difficult-to-model nature of a fault rupture.
Moment magnitudes are calculated by complex equations that take into account a number of factors including different velocities and different attenuation of seismic energy in different rocks. Mw takes into account pretty much everything that goes on during the rupture. If you want the gory details, see https://en.wikipedia.org/wiki/Moment_magnitude_scale .
1. The “power of a volcano” is generally characterized as Volcano Explosivity Index or VEI. This is a relative measure of explosiveness of volcanic eruptions, and is open-ended with the largest supervolcano eruptions in pre-history (Yellowstone, Toba, Taupo) given an arbitrary magnitude of 8 in this classification system. The 79 AD eruption of Vesuvius and the 1980 eruption of Mount St Helens in Washington State are both rated a VEI 5 on this scale. The VEI number attached to a volcanic eruption depends on (a) how much volcanic material (dense rock equivalent) is thrown out, (b) to what height is it launched, and (c) how long the eruption lasts. There is no equation to calculate this scale, and in that sense it is like the Meercalli Scale. However, it is considered logarithmic from VEI 2 upwards. In other words, a VEI = 5 event represents approximately 10 times more energy released than a VEI = 4 event. Follow this link for more information on how to assess the VEI magnitude (from Newhall and Self, 1982):
https://en.wikipedia.org/wiki/Volcanic_Explosivity_Index
Chapter References:
Allaby, Michael, 2008, Saffir-Simpson scale, in: A dictionary of earth sciences (3rd ed.): Oxford University Press, 1672 pp. ISBN 978-0-1992-11944
Bormann, Peter; and James W. Dewey, 2014, The new IASPEI standards for determining magnitudes from digital data and their relation to classical magnitudes: http://gfzpublic.gfz-potsdam.de/pubman/item/escidoc:816929:1/component/escidoc:816928/IS_3.3_rev1.pdf doi: 10.2312/GFZ.NMSOP-2_IS_3.3
Dieterich, James H., 1978, Time-dependent friction and the mechanics of stick-slip: Pure and Applied Geophysics 116, issue 4, p. 790–806. doi: 10.1007/BF00876539
Fujita, Tetsuya Theodore, 1971, Proposed Characterization of Tornadoes and Hurricanes by Area and Intensity: Satellite and Mesometeorology Research Paper 91. Chicago, IL: Department of Geophysical Sciences, University of Chicago.
Moreno-Murillo, Juan Manuel, 1995, The 1985 Mexico Earthquake: Geofisica Colombiana. Universidad Nacional de Colombia 3, p. 5–19. ISSN 0121-2974.
Newhall, Christopher G.; and Self, Stephen, 1982, The Volcanic Explosivity Index (VEI): An Estimate of Explosive Magnitude for Historical Volcanism (PDF): Journal of Geophysical Research 87 (C2), p. 1231–1238. doi: 10.1029/JC087iC02p01231.
Richter, C.F., 1935, An instrumental earthquake magnitude scale (PDF): Bulletin of the Seismological Society of America. Seismological Society of America 25 (1-2), p. 1–32.
Spence, William; Stuart A. Sipkin; and George L. Choy, 1989, Measuring the size of an earthquake, in: Earthquakes and Volcanoes 21, Number 1, 1989. http://earthquake.usgs.gov/learn/topics/measure.php
## Is Water Wet?
Sometimes we get queries at Ask-a-Geologist that have little or nothing to do with geology. We find that some people have gotten wrapped up in an internet meme or conspiracy theory that is really pointless… and they come to us to arbitrate an argument. Here’s an example:
Q: Hi i was just wondering if you could answer my scientific question… is water wet?
• Andrew W.
A: First of all, this is not a “scientific question” – it’s a semantic issue. I’d recommend you begin by checking a dictionary for the definition of “wet.” This is what *I* found in a 3-second look at www.dictionary.com:
1. moistened, covered, or soaked with water or some other liquid: wet hands.
2. in a liquid form or state: wet paint.
3. characterized by the presence or use of water or other liquid.
This question is thus not a geologic question, but a game of English word-play. It’s similar to the question “Is fire hot?” Well, DUHHH.
~~~~~
The larger take-away here is not to let yourself get caught up and waste time in internet memes. A substantial bulk of internet content typically has no filters – no peer-review, no foundation in historical or experimental fact. The things you find in this gray zone are like this pointless question. At least it’s one step above baseless conspiracy theories, that serve no other use than to provide click-bait advertisement income for people who resist the idea of doing any real work.
I’m reminded of the Daily cartoon in the New Yorker, from May 6, 2015 by Christopher Weyant: an unctuous member of Congress is talking to a reporter “I like to think we aren’t so much antiscience as we are pro-myth.”
~~~~~
Don’t be that kind of person. You can promote the myth or conspiracy that 1 + 1 = 3 if you want to waste air. The First Amendment to the US Constitution allows you to do so (at least in the United States). However, your math will not land a Lunar Module on the Moon, nor solve the problem of cancer. Your smart phone doesn’t work because of some made-up fact about electricity and angels. You don’t want to be the person stuffing dead air with platitudes, conspiracy theories, and pointless memes.
## How To NOT Go Off-The-Wall-Freakin’ CRAZY…
### …when a Cascadia Earthquake hits.
From personal experience, when a really big earthquake hits, it is extremely unnerving. In fact, my first earthquake was a magnitude 7.3 event in Southern California, and the serious shaking lasted not much more than 3 minutes.However, it seemed like a lifetime to me then. If asked a week later, I probably would have said that it lasted at least a half an hour.
Look at the following diagram, taken from Wikipedia:
The P is what woke me up. It hit with a bang.
The S is what rattled and then broke the windows, and stutter-walked my bed 30 cm across the floor.
The R is what finally flipped me out of my bed and onto the floor.
It took several days for the information on this event to filter down through the scientists to the government entities, to the news media, to my parents, and then to me as a 6-yr-old child. By then we were back in our house, the power was restored, and we had water pressure again.
A Cascadia subduction earthquake might reach a moment magnitude 9+ when it next occurs. That will be nearly 100 times more energy than the piddly 7.3 event that launched a sleepy 6-yr-old out of his bed in Bakersfield, California long ago. If I found a M=7.3 event to be that terrifying, imagine how bad a M=9+ event might feel like.
There are two ways to deal with the terror:
#1. Understand immediately what is happening and what will come next.
#2. Be prepared for it. In other words, know that you have your bases covered.
#2 is something that many people more or less do (some do OK, some do better, and some do extremely well at this):
A. Have a family plan in place. Where do we meet? What channels on the battery-powered, \$40 hand-held walkie-talkies will we be using to find each other?
B. Have supplies at hand, including
i. Food. And don’t count your refrigerator contents here.
ii. Water. A LOT more water than you might think.
iv. Batteries. Flashlights. LOTS of batteries.
v. Blankets and sleeping bags, and/or a heat source to keep warm.
C. Start checking up on your neighbors, and offer to share your stuff with them. People you may hardly know will become life-long friends really quickly.
However, the purpose of THIS blog entry or chapter is to help you deal with #1: understand what in the world is going on, so you don’t go crazy.
Next, look at the following diagram:
This will help you to understand the TIMING difference between the several kinds of seismic waves. The P wave arrives with a bang, like someone with a large hammer just whacked one side of your house. The S wave will feel different: slewing everything back and forth, perpendicular to a line between you and the epicenter of the earthquake. The R (which stands for Raleigh, or surface) waves will feel to you like you are in a small skiff after a large boat roars past, careless of his wake. You will feel like you are rolling around, up and down, and sideways . You always end up pretty much in the same place with each complete roll.
All these things are important clues for you. This is what you do:
FIRST: as soon as the P-wave hits, look at your watch… which we both hope includes a second hand.
Second: as soon as the S-wave hits, think about what direction is PERPENDICULAR to that sickening side-to-side motion: this gives you the important clue as to where this thing is coming from. If the slewing motion is north-south, then the earthquake epicenter is either east or (more likely in this case) west of you.
Third: as soon as the S-wave hits, look at your watch again. Subtract the P-wave arrival time in seconds from the S-wave arrival time. That difference tells you how far away the hypocenter (the actual sub-seafloor rock rupture) is from you.
Use the following diagram to convert that P-S time difference into a distance::
Fourth: KEEP TRACK OF THE TIME even after you have this number. If the event is a M=9+ event, the ground will keep shaking and heaving for a full 5 – 6 minutes. THIS will give you a sense of how bad things will be in the following several weeks. If it was just a segment of the Cascadia Subduction Fault breaking, then the time could be as little as 3 – 4 minutes. This is GOOD. If the heaving and shaking runs up to 6 minutes… well, you already have #2 above in place, right? So you are prepared.
This will help: If you live in Portland or Seattle, and the P-S time difference is 20 – 45 seconds, then the event is far enough away (or close enough, depending on your point of view) to be The Big One: A Cascadia Subduction Event.
But YOU WILL ALREADY UNDERSTAND WHAT IS GOING ON even before the first news reports start coming in (assuming you have a flashlight close and a battery-powered radio on hand).
AND YOU HAVE YOUR PREPARATIONS IN PLACE.
And no, you are not crazy: you are informed and prepared.
## The Coast Is Toast…
When I was a child the White Wolf Fault, a splay-fault of the great San Andreas, ruptured about 60 km (40 miles) from my home. I recall hearing a bang, then hearing the windows rattling hard – and finally breaking. Shortly afterwards, I was thrown out of bed onto the floor. I didn’t fall out of bed, I was thrown from my bed to the middle of my bedroom floor. When my mother called to me from her bedroom to come to her (she was trying to hold onto her own bed at the time), she told me that I replied “I can’t. The walls keep hitting me.” On our hands and knees we finally made it as a small family out to our back yard (my Mom was fearful of the gas line rupturing and suffocating or burning us all). This earthquake had a moment magnitude of about 7.3
The movie “Volcano” made the expression “The Coast Is Toast” famous. “Volcano” postulated a volcano somehow under the San Andreas Fault. When the film came out, my volcanologist colleagues cringed. While volcanoes ARE associated with faults, they are associated with deep subduction faults, where ocean floor is over-run by a continent in what is called a thrust fault. Think of the Cascades range, far inboard from the Cascadia subduction fault 50 – 100 kilometers offshore. The volcanoes themselves are found far inland from where the huge subduction fault reaches the ocean floor.
However, the expression “The Coast IS Toast” is not that far off the mark in the sense of massive destruction that could visit the Pacific Northwest coast if and when a Cascadia subduction event occurs. It could be a magnitude of 8.0 or 9.0 or even higher – this would represent over 100 times more energy released that what woke me up many years ago.
The Cascadia Subduction Zone (CSZ) is just that: a plate-subduction thrust fault spread over a 1,000 kilometer length, extending from offshore Vancouver Island in Canada to offshore northern California. It’s width depends on what you count, but earthquake imaging of the down-going oceanic slab extends well into central Washington and Oregon. Three major oceanic floor plates, the largest being the Juan de Fuca, are being over-ridden by a westward-moving North American continent. Part of the thrust fault is lubricated by the ocean-floor sediments atop the Juan de Fuca plate, and part of the down-going slab is partially melting in the upper Mantle, giving rise to that almost linear string of Cascades volcanoes. These volcanoes extend from Mt Garibaldi in British Columbia to Mount Shasta and Mount Lassen in Northern California.
But in between these parts is a segment, extending the entire length of the fault zone, that is stuck. The lubricating fluids have been squeezed out by pressure with increasing depth, and the stuck part is like a dry patch in the center of your hands as you try to slide one past the other. THIS is where the the rub lies, so to speak. In 2004 a similar subduction fault near Aceh in western Indonesia ruptured, creating a magnitude 9.3 earthquake. The tsunami alone killed over 250,000 people around the Indian Ocean as far away as Mozambique. When a similar subduction fault offshore of northern Japan ruptured in 2011, the surface area of the fault that was displaced or ripped was enormous: 300 kilometers long by 200 kilometers down-dip. This is important, because the surface area ruptured correlates closely with the energy released. “Down-dip” on the San Andreas Fault is only about 10 kilometers – because this fault is more or less vertical, and the rock becomes plastic at about 10 kilometers depth.
There is a security camera video of a 15-meter (50′) wave breaching the 5-meter (16 ft) tsunami-protection walls of the Fukushima Dai-Ichi nuclear power plant on the NE Japanese coast. If you’ve been trying to body-board in the ocean, you know how hard a 2 meter (6 foot) wave can slam you. To state the obvious, you don’t just stand your ground with even this small a wave: water is nearly as dense as your body. The estimated cost of this disaster to Japan as a nation is now in excess of \$300 billion.
That sounds unimaginable. However, a Cascadia Subduction event is a very real, in fact inevitable, likelihood for the Pacific Northwest.
What will happen when this inexorable event occurs?
The coast will lurch westward 20 meters (60 feet)… and remain there permanently.
The coast will drop down on average 2 meters (6 feet)... and the low-lying parts will remain sunken permanently.
A tsunami up to 40 meters (130 feet) tall will strike coastal communities in as little as 15 minutes from the onset of the first shaking.
There will be fires that are unstoppable – because gas mains and water mains will both be ruptured. The 1906 earthquake in San Francisco was over in probably less than 3 minutes… but the fires that destroyed nearly ALL of San Francisco raged for four days afterwards. The fire department at the time was helpless – they had no water.
In the Pacific Northwest, emergency planners have estimated that 10,000 people will die, and another 30,000 people will be seriously injured.
The closer to the epicenter – a broad north-south line just off and beneath the coast – the greater the damage. The farther east you live, the greater the attenuation of the energy released by a CSZ event. Attenuation means the Earth’s crust in between the fault and, say, Yakima, Washington, will absorb most of the radiating seismic energy.
But first, the ground will suddenly jerk westward, then begin going up and down and sideways, then it will begin rolling. This will go on for 4 – 6 minutes…
It will definitely wake you up. From experience, I can tell you that a few minutes seems to go on forever.
How often does Cascadia’s fault rupture? An early study of bouma sequences (mud layering in deep-ocean coring) suggested 7 events in the past 3,500 years, so we might say an average is 500 years between mega-events. However, a recent report by Oregon State University suggests that the average time between major earthquake events may be as little as every 240 years. When was the last one?
This event gave rise to the Orphan Tsunami in Japan, so-called because there was no felt earthquake nor approaching typhoon to provide warning before enormous waves suddenly appeared and obliterated or damaged many fishing villages along the Sendai coast. That’s over 300 years ago. This is somewhat simplified, of course, because the CSZ cannot really be treated as a single entity that always behaves along its entire length the same way. Detailed geologic mapping, in fact, suggests that there are sometimes separate ruptures along the “northern zone” and the “southern zone”… giving ‘mere’ magnitude 8.5 events.
But make no mistake: while a magnitude 8+ event may feel different from a magnitude 9.0 full-rip event (lasting “only” 4-5 minutes instead of 6), there will still be widespread damage.
The event spacing (the average of 240 years vs. the current hiatus of 315 years) suggests we are then “overdue”, doesn’t it? Not necessarily, because the spacing between previous events has been as much as 500 years. Earthquakes do not click along like clocks. In fact, we cannot predict earthquakes unless we are injecting water into wells in a tectonic region like the area north of Denver, CO. or in southern Oklahoma. For all large earthquakes, despite upwards of \$100 billion spent on research over the past century, the best minds I personally know unequivocally say that current science cannot predict when an earthquake will happen.
But scientists can forecast major earthquakes. That’s a very different thing than a prediction. This means that scientists can say, based on existing data, that there is a 40% chance of another Cascadia event in the next 50 years. So… less than a 1% chance in the next year. Buy earthquake insurance or not?
What can that plausibly mean to you – realistically, practically? What can you possibly do with this information?
First, scientists CAN make reasonable estimates of what will happen during and after a Cascadia event, and you and I CAN prepare for those. This is going on right now in local and state organizations in the Pacific Northwest. Infrastructure is being examined with an eye towards what can be reinforced. Building codes have already been upgraded – then upgraded again – to help us create new roads, bridges, and buildings that will better survive such an event. There are estimates in Oregon, for instance, that a majority of bridges will be compromised or fail on coastal US Highway 101, and at least five bridges on inland interstate I-5 will fail in Oregon alone. The damage will be worse the closer one is to the coast, but in both instances it takes just one bridge in a strategic location to shut down interstate commerce. Don’t count on being able to find food on the shelves of your local supermarket for awhile… or even count on being able to GET to your supermarket. Repairs to powerlines, gas lines, roads, bridges, etc. will take time. They will happen sooner inland, and take longer in the coastal communities.
This means you should have at least 2 – 4 weeks worth of non-perishable food for each adult in your household. You should have at least two gallons of water, per day, per adult, enough to last you that whole time. A majority of people planning for a disaster forget about the water part – it’s raining all the time in the “Pacific NorthWet”, isn’t it? You should also have batteries – LOTS of batteries. A hand-crank radio will be very helpful, perhaps a lifeline. I have an FCC-issued HAM radio license and two small but powerful hand-held radios. If cell towers are down, I will be part of a local Amateur Radio Emergency Services network to help move information around from my neighborhood to the region.
Most important of all, you need to have a family plan for dealing with this – or any other catastrophe. In the short term, only you can help your family and your neighbors. It will take awhile for the country as a whole to martial the necessary resources to even partially help. Having a 72-Hour Go Kit will be appreciated when you need it.
If the example of Hurricane Katrina can be used, yes, we will recover. However, the recovery effort will consume much of the region’s GDP, and it may be more than a decade before everything is running as smoothly as before the event. New Orleans and Memphis, TN, had similar economic output in 2005. by 2015 New Orleans still has not caught up with Memphis.
We will survive. We will rebuild. We will be toast only if we refuse to do anything.
## Will Yellowstone Blow?
If you’re being shot at, there is some satisfaction in knowing how often you’re being shot at. You can at least plan, and perhaps take some mitigating steps. To this end, the entire Pacific Northwest is preparing for the next magnitude 8+ subduction earthquake event by seismically retrofitting public buildings and holding “Great Shakeout” drills. This planning and preparation can be applied to volcanoes, even super volcanoes.
Q: Do geologists know when Yellowstone might erupt again? It appears to erupt at a Supervolcano level every few hundred thousand years.
The first was: 2,100,000 years ago
Second was: 1,200,000 years ago
And the last one was: 640,000 years ago
Are we in any danger of a fourth one?
– Brandon F
A: Yes is the short answer. Probably not in your lifetime is the long answer. There have been supervolcano eruptions moving with time along the Snake River Plain to modern Yellowstone starting at least 16.5 million years ago in southeastern Oregon.
Volcanologists in the USGS Volcano Science Center are very aware of this eruptive periodicity – we have a full-time volcanologist assigned to Yellowstone as the Scientist-in-Charge (SIC) of the Yellowstone Volcano Observatory. He works in close coordination with seismologists at the University of Utah, and with the US Park Service. Some links might be of interest to you:
https://volcanoes.usgs.gov/volcanoes/yellowstone/
https://volcanoes.usgs.gov/volcanoes/yellowstone/yellowstone_publications.html
There have been other, somewhat smaller eruptions at Yellowstone, however: The Scaup Lake rhyolite flow of 250,000 years ago, and a more recent hydrothermal blow-out about 70,000 years ago. Neither would have been trivial if you had been in the area, but they did not have the continental reach of the huge monsters you list.
You might want to check out the USGS Yellowstone hazard assessment:
Christiansen, R. L., Lowenstern, J. B., Smith, R. B., Heasler, H., Morgan, L. A., Nathenson, M., Mastin, L. G., Muffler, L. P. & Robinson, J. E. (2007). Preliminary Assessment of Volcanic and Hydrothermal Hazards in Yellowstone National Park and Vicinity. U.S. Geological Survey Open-File Report , 2007-1071, 98 p.
To make this a bit more real, there is a silvery-white ash deposit found all over most of the United States, sometimes called the Lava Creek Tuff (from a site locality in Kansas) or the Pearlette Ash Formation in older scientific literature. I have visited a single layer of this material near Colorado Springs, CO. There it is over 20 meters/70 feet thick; when it came down it may have been over 30 meters thick, but was consolidated with rainfall and the compressive weight of overlying material since then. This blanket of ash smothered all living things beneath it; I have personally pulled out a paleo-camel’s tooth from the bottom of the deposit. It ALL came from Yellowstone, over 800 kilometers/500 miles away!
Here’s what a future eruption might do (From Mastin, Van Eaton, and Lowenstern, 2014, “Modeling ash fall distribution from a Yellowstone supereruption”, Geochemistry, Geophysics, Geosystems, Vol 15, Issue 8, https://doi.org/10.1002/2014GC005469):
Figure 1. Likely distribution of ash and tephra from a future Yellowstone Caldera eruption (Mastin, Van Eaton, and Lowenstern, 2014). Fine details will depend on wind distribution and the volume of the erupted ash.
Note that this is for a sophisticated 3D grid model that assumes prevailing westerly winds up to 16 – 24 km elevations. However, the model shows that the erupting plume would make its own prevailing winds, something that would allow an “umbrella cloud” to leave ash deposits throughout ALL conterminous US states, something consistent with geologic mapping of previous eruptions.
Another point is relevant here. Our experience is that, while you cannot predict an earthquake, you CAN predict a volcanic eruption if you have adequate instrumentation on the volcano. Several of our USGS staff returned recently from making their annual gravity and geodetic GPS survey (these surveys will detect any magmatic inflation). The caldera and surrounding terrane are very well-instrumented with telemetered seismometers, also.
The assessment of the Scientist-in-Charge when I last talked with him is that we are not likely to have a super-eruption in our lifetimes – that’s essentially what the 2007 assessment above says. I will excerpt key pieces of that assessment here:
“No volcanic eruption has occurred in Yellowstone National Park or vicinity in the last 70,000 years or more.”
“One statistical measure of eruption probabilities based on this episodic behavior suggests an average recurrence of 20,000years. The fact that no such eruption has occurred for more than 70,000 years may mean that insufficient eruptible magma remains beneath the Yellowstone caldera to produce another large-volume lava flow.”
Table 5. Estimates of annualized probability of events greater than a given magnitude.
Diameter (m) Area (m2) Events in last 14 thousand years Annualized Probability
>2 3.1 7000 (estimated) 0.50
>300 70,700 16 0.0013
>2000 3,140,000 2 0.00014”
This last table is from page 83 of the report. The chances for a large hydrothermal eruption next year (NOT a super volcano eruption) is just a bit over 1 in 10,000. For reasons explained above, the probabilities are likely even lower than this.
Bottom line: Those in the know are not currently worried about a Yellowstone “blow.”
A more recent paper (Lowenstern, Sisson, and Hurwitz, 2018, “Probing magma reservoirs to improve volcano forecasts”, EOS Vol. 99, No. 6, pp. 16-21) helps explain the uncertainties involved in making this assessment. Seismic tomography studies suggest that the Yellowstone magma reservoir is about 5%-15% melt, with all the rest being crystals that have slowly formed over the last several hundred thousand years. That’s still about 25 cubic kilometers of melt, larger than any volcano has spewed out on Earth since the eruption of Tambora (Indonesia) in 1815 caused the Year Without Summer in Europe, with snow in June and mass starvation following crop failures.
The current understanding in the volcanology community is that a magma reservoir is not eruptible with less than 50% melt… however, this assumes a homogeneous distribution of crystals and melt fraction. Looking closely at the steep compositional gradients in crystals erupted 250,000 years ago at Yellowstone, calculations suggest that the crystalline mush had lain dormant for ~220,000 years – but then was remobilized in as little as a mere 10 months.
The main concern is that if there is another injection of hot basalt from the mantle into the base of this reservoir, it could remobilize the reservoir and become eruptible in less than a year.
In the meantime, the volcano is very closely monitored 24/7. Please understand that there are a number of scientists within and outside of the US Geological Survey who are monitoring Yellowstone, literally, on an hourly basis. They think about your question every day.
~~~~~
## If Polar Ice Melts How Much Will Sea Level Rise?
Probably no single item brings the scientific-political argument over climate change more into focus than sea level rise and its consequences. Here are the facts to counter the “alternative facts” that have been floated in national political discourse. See also the earlier article (http://askageologist.blogspot.com/2013/07/climate-change-is-it-real.html) on “Climate Change – is it Real?” Curiously, only in America is the science of climate change being questioned. However, only in America (and Myanmar) do we still use feet, pounds, and gallons.
In all fairness, this is not an easy scientific problem to address. Non-linear behaviors (something changing much faster than the variable forcing it is changing), and extremely complex interlocking feedback between physics and chemistry related to Earth’s weather systems, makes any modeling truly daunting. Nevertheless, scientists have developed a number of predictive models, and they are beginning to agree ever more closely.
Q: What if all the ice caps melt how bad will it flood the nearby continents, and would it change the tides of the world? How fast would the world have to react.
– Stephen L
A: There are about 21 million cubic kilometers (5 million cubic miles) of ice on the Earth’s surface. If all of this melted, it would raise sea levels by about 65 meters (215 feet). An image compiled by National Geographic magazine (http://www.nationalgeographic.com/magazine/2013/09/rising-seas-ice-melt-new-shoreline-maps/) gives a breath-taking sense of what this would mean for humanity. Florida would disappear – Washington, DC, also. This isn’t going to happen immediately, of course. For all this ice to melt would require the average global temperature to rise from a current 14C (58F) to 27C (80F). This is not impossible, especially if carbon continues to be extracted and burned at current rates or higher.
However, there are many issues beyond polar ice involved with sea level rise:
1. Tectonic changes
1. Thermal expansion of the oceans
1. Melting ice
1. Local weather events (e.g., hurricanes)
1. Ocean albedo change
1. Methane clathrates
1. How fast will it rise?
1. Tectonic changes are an issue because, all things being equal, sea level is an equilibrium by definition and should rise everywhere at the same rate. Nevertheless, the east coast of North America is seeing a greater sea level rise than the west coast. This is because of tectonic changes, related to mid-Atlantic sea floor spreading, that are causing steady (tectonic) sinking along the east coast of the United States.
1. Thermal expansion is important because if you heat water it will expand. With climate change well underway (and isotopic studies indicate that it is largely man-made), we can expect all the world’s oceans to expand… and therefore rise. Water is at its most dense at 4 degrees Celsius. Freeze water and it will expand (this explains why frozen water pipes burst). Warm it above 4 degrees Celsius and it will also steadily expand.
1. Antarctica is covered with ice an average of 2,100 meters (7,000 feet) thick. If all of the Antarctic ice melted, sea levels around the world would rise about 60 meters (200 feet). Arctic ice is not nearly as thick, but Greenland by itself, if all its ice melted, would increase sea level rise an additional 7 meters (20 feet).
1. Local weather events are the most immediately attention-getting, and there are at least two different aspects to this. Warmer ocean water translates into more heat energy going into a hurricane – the storms become bigger and the destructive wind velocities become stronger. The recent Atlantic hurricane Irma is a case in point: it is the largest and strongest Atlantic hurricane ever recorded since measurements were first acquired. When its eye reached the southern tip of the Florida peninsula, it’s outer rain bands were already into Georgia – and that was just half of this monster. However, hurricanes push seawater before them and drag at their cores a huge low-pressure zone, and these give rise to what is called a “storm surge.” The storm surge for hurricane Katrina, which devastated New Orleans in 2005, caused over US\$100 Billion in damage largely because its storm-surge was an additional 5 meters (16 feet) above the normal tidal differences. Add a “king tide” (when Earth, Sun, and Moon are aligned and the high tide is greatest) to a 5 meter storm surge and you have a very destructive combination. It’s like a giant, slow tsunami.
1. If ice disappears from the poles and from Greenland, then the albedo of the Earth will change. Albedo is the percentage of the incident light or radiation that is reflected by a surface, and is typically used for a planet or moon (the Moon’s albedo is about 20%, which means about 20% of sunlight is reflected and 80% is absorbed). In this case, ice-covered polar regions are very strong (though not perfect) reflectors of sunlight. If the ice were to disappear, the energy absorption of the polar regions would increase dramatically. Like ocean warming, this is another contributor to the non-linear character of sea level rise: a simple increase in a certain value causes secondary effects that dramatically increase the effect disproportionately in a non-linear fashion.
1. Methane clathrates (a.k.a. methane hydrates, “fire ice”, etc.) are methane-ice held in a suspended quasi-stable crystal state found in the world’s cold deep ocean sediments (below at least 200 meters or 600 feet depth). This methane is a product of carbon being sequestered over time by CO2 capture (decayed materials falling to the ocean floor). The amount of carbon sequestered in this form beneath the world’s oceans is estimated between 500 and 2,500 gigatons, comparable with all known sources of hydrocarbons (oil and gas) found on land. There is evidence now that ocean temperatures as deep as 500 meters are rising. Methane, being a far stronger greenhouse gas than carbon dioxide, if released in these numbers, will cause a dramatic rise in global temperatures. This is another contributor to the non-linear character of sea level rise, and helps explain why estimating climate change consequences is so difficult.
1. How fast will sea level rise happen? That is the million-dollar question for our age. The Intergovernmental Panel on Climate Change issued a report in 1995 containing various projections of the sea level change by the year 2100. They estimated that average sea levels worldwide will rise 50 centimeters (20 inches), and their +/- range went up to 95 centimeters (over 3 feet). The rise will come in part from thermal expansion of the ocean and in part from melting glaciers and ice sheets. Fifty centimeters is no small amount – this could have an enormous, disproportionate effect on coastal cities, especially during storms like Katrina, Sandy, or Irma. Keep in mind that this estimate is over 20 years old, and more recent sea level rise estimates vary widely but are not small. Since that 1995 report there have been gigantic ice sheet calving events in the Antarctic. The most recent (Summer of 2017) on the Ross Ice Shelf was an “iceberg” the size of Delaware, that ranges from 15 to 50 meters (up to 165 feet) high… and it will all melt as it drifts northward.
About 80% of the human population now lives within 100 km of an ocean, and the most expensive and sought-after kinds of land are ocean-front properties. You don’t have to be a rocket scientist to realize that ocean-front property investment might not be a good idea. Miami “dodged the bullet” from hurricane Irma in September 2017, but it’s just a matter of time before a larger, even more destructive hurricane will hit it dead center. The loss of life and property to just Miami alone are unimaginable. The entire eastern United States is at risk, and hurricane Sandy (2012) made it clear that low-lying cities like Washington DC and New York are at terrible risk due to climate change. Giant typhoons in the subtropical Pacific are causing huge damage every year to east and southeast Asia, Japan, and the Philippines.
We should be have been reacting to these scenarios long ago. Places like The Netherlands and the City of Venice have certainly been taking steps to mitigate the consequences of sea level rise for decades now. However, the world needs to address the reason for it. Choosing myth over climate science is not the way to go. That approach didn’t work for Big Tobacco, either.
An aside: recently, a US Congressman, who probably should not be named to avoid further embarrassing him, argued against the vast and accumulating evidence of climate change, saying that sea-level rise is cause by “…rocks …falling into the sea.” [https://www.huffingtonpost.com/entry/republican-congressman-explains-sea-level-rise-its-rocks-falling-into-the-sea_us_5afef746e4b07309e057985b]
He apparently doesn’t understand the difference between a bucket and an ocean.
~~~~~
## Risks of Working in a Rock Shop
Q: Greetings,
I work for a small business rock shop that carries a very large variety of gemstones and minerals. I have wondered for quite some time if I should be concerned about exposure so certain elements like lead, arsenic etc. We handle mostly everything without gloves or the use of dust masks.
I am now pregnant and even more concerned about this. I know that doctors and other professionals will advise the use of safety precautions regardless. My question is not that if we should use precautions, because I know that we should anyways. The question I am asking you, is if there is plausible serious risk through skin contact and inhalation? Do you know of any risks, are you able to provide specific examples of situations where problems have occurred /or might occur?
One example is of the handling of iron pyrite. It leaves behind black residue (we do use gloves for this) and creates a strong smell and dust in the air. Am I exposing myself to something serious here?
Also, I’ve heard of a new fad where the folks who believe in metaphysical properties of stones are putting them in their drinking water. I found this alarming.
Thank You,
– Stacey S.
~~~~~
A: This is a VERY important query, and kudos to you for asking – and for your determination to protect your unborn baby.
YES. There are minerals that are really dangerous: realgar and orpiment have mercury in them, for example. You can look up Minamata Syndrome to get an idea of how bad these could be to a fetus: https://en.wikipedia.org/wiki/Minamata_disease
YES. There are various forms of asbestos that kill – literally. My father died a premature death from lung cancer. The biopsy showed that there was asbestos in his lungs, ultimately traced to dust in the basement of his apartment building in San Francisco where he kept his bicycle. The pipes had been insulated with spray-in asbestos in the 1950’s when the building was originally constructed. I’d be willing to bet that the workers who blew in that insulation preceded him.
POSSIBLY. Pyrite (FeS) is a mineral that will oxidize in the atmosphere. The bright shiny mineral faces will eventually dull and then go brown. The smell you describe is probably H2S, normally not toxic in small amounts (the smell warns us to get away – this is common around volcanoes I’ve worked in). My concern is that there are other sulfides that are often naturally associated closely with the pyrite, including cadmium and arsenic sulfides. These are very poisonous.
You are probably safe handling gemstones and semi-precious stones such as citrine, zircon, beryl, and amethyst – these are typically hard minerals that do not interact much with the environment nor degrade with time, which is why they are valued in the first place.
I would encourage you to think more about a high-quality respirator when in a dusty, mineral-laden room. Inhalation is probably a more serious threat than getting the stuff on your hands… unless (like me) you always have an itchy nose and rub it frequently. Here is a website that will get you started on the various kinds of respirators out there (they run the gamut from the kind your dentist uses to serious industrial equipment): https://www.osha.gov/OshDoc/data_Hurricane_Facts/respirators.pdf
Putting stones and gems in drinking water also boggles MY mind. A diamond will not react, and most semi-precious stones won’t either… but everything else WILL react with water to some degree, especially if the water’s slightly acidic (think Coke or Pepsi for acidic fluids). Putting crystals on your body is silly enough… now imagine bright yellow or red minerals in your drinking water!
I hope this helps. I personally love rock shops and as a geophysicist visit them whenever I can. I DO wash my hands after I leave one, however.
~~~~~
## General Planetary Geology Q&A
Q: To Whom It May Concern
I’m not a scientist, however I find it an interesting issue.
I have a few questions of which I hope you can clarify for me:
– Preben P
A: I’ll try to respond to each of your questions below in CAPS:
Q: 1: How does the inner core of the earth maintain its temperature? Or is it decreasing?
A: FOR ONE THING, THE CORE OF THE EARTH IS WELL INSULATED WITH VAST VOLUMES OF LOW-THERMAL-CONDUCTIVITY OVERLYING ROCK. EVEN WITH CONVECTION IN THE MANTLE (AND PERHAPS THE OUTER CORE ALSO) IT TAKES A LONG TIME FOR HEAT TO ESCAPE. WHETHER THE TEMPERATURE IS INCREASING OR DECREASING IS A MATTER OF CONJECTURE. SOME OF THE HEAT IS FROM KINETIC ENERGY DUE TO THE AGGREGATION OF THE PROTOPLANETARY DISK, WHICH WOULD IMPLY COOLING. HOWEVER, MUCH OF THE HEAT IS THOUGHT TO BE FROM DECAYING RADIOACTIVE ISOTOPES… WHICH WITH THE ONSET OF CONTINENTAL DRIFT (MANTLE CONVECTION) IN THE PRECAMBRIAN IMPLIES INCREASING HEAT. MANKIND HASN’T BEEN AROUND LONG ENOUGH TO TELL THE DIFFERENCE.
Q: 2: After any volcano eruptions what happens to the void space from whatever is discharged?
A: USUALLY A CALDERA REMAINS – A LARGE SUNKEN CRATER – OR SOME OTHER COLLAPSE FEATURE APPEARS. OVERLYING LITHOSTATIC PRESSURE GUARANTEES THAT NO VOIDS REMAIN INSIDE THE EARTH – THERE IS NO VOID SPACE ANYWHERE IN THE EARTH EXCEPT FOR VERY SHALLOW CAVES CAUSED BY LIMESTONE DISSOLUTION (KARSTS).
Q: 3: Could it be possible that earth is a dying sun meaning that the earth for billions of years ago was a burning planet from big bang? Just like the sun as we know it today.
A: THE SUN AND EARTH DID NOT BEGIN TO FORM UNTIL ABOUT 9 THOUSAND MILLION YEARS *AFTER* THE BIG BANG, SO THE EARTH IS NOT HOT FROM THE BIG BANG (AT LEAST NOT DIRECTLY). AS AN ASIDE, EARTH IS NOT A SUN.
NOTE: a thousand million = billion in America. A million million = a billion in the UK.
THE SUN IS USING UP ITS HYDROGEN FUEL AT A RATE THAT WILL LEAD TO A NOVA IN ABOUT 5 THOUSAND MILLION MORE YEARS. THERE IS SOME EVIDENCE THAT THE EARTH WENT THROUGH A FROZEN “SNOWBALL” STAGE IN THE ARCHEAN EPOCH (MORE THAN ~2,500,000,000 YEARS AGO), BUT THIS IS POORLY UNDERSTOOD (THE EVIDENCE IS TRULY ANCIENT). THERE IS EVIDENCE (STROMATOLITE FOSSILS) THAT THE EARTH RESIDED IN A TEMPERATE ZONE LIKE TODAY AS FAR BACK AS ~3,400,000,000+ YEARS AGO.
Q: Climate change:
Don’t believe its cause by man; for sure man is the cause of poor air quality.
A: IF YOU DON’T ACCEPT THE VAST AND GROWING EVIDENCE FOR CLIMATE CHANGE, THEN YOU ARE PART OF A VERY SMALL MINORITY AMONG EDUCATED PEOPLE.
MORE THAN 98% OF SCIENTISTS WHO STUDY CLIMATE CHANGE AGREE THAT THE EVIDENCE STRONGLY INDICATES THAT MAN IS THE CAUSE OF CLIMATE CHANGE. THE RAPIDLY GROWING CO2 IN THE ATMOSPHERE, A WELL-MEASURED GREENHOUSE GAS, HAS AN ISOTOPIC SIGNATURE LOW IN CARBON-14: THIS MEANS MOST IF NOT ALL OF THE NEW CO2 WAS SEQUESTERED FOR A MINIMUM OF 50,000 YEARS AS BURIED HYDROCARBONS, AND MORE LIKE HUNDREDS OF MILLIONS OF YEARS. IT IS RELATIVELY EASY TO CALCULATE THE AMOUNT OF FOSSIL FUEL BURNED IN THE PAST CENTURY, AND TIE IT TO THE INCREASE IN CO2. HOWEVER, THE HUNGER FOR MEAT PROTEIN HAS ALSO MEANT AN EXPLOSION OF DOMESTIC RUMINANTS (SUCH AS COWS) THAT EMIT VAST AMOUNTS OF METHANE, A GREENHOUSE GAS UP TO 37 TIMES MORE POTENT THAN CO2. ONE COW EMITS AS MUCH METHANE IN ONE DAY AS THE VOLUME OF THREE OF MY FILING CABINETS. MULTIPLY THAT BY 37 TO GET THE EQUIVALENT CO2 RELEASE. THE GEOLOGIC RECORD SHOWS THAT THERE HAVE BEEN CLIMATE CHANGE EPISODES IN THE PAST, BUT NONE THAT HAPPENED ANYWHERE NEARLY AS FAST AS IT IS HAPPENING RIGHT NOW – IT TOOK HUNDREDS OF THOUSANDS OF YEARS INSTEAD OF OUR CURRENT HYPER-SPEED, CHANGE-IN-A-CENTURY RATE).
I CONCUR WITH YOU THAT MAN IS DEFINITELY THE CAUSE OF POOR AIR QUALITY.
Q: How about magnetic poles so when the suns positive pole is close to earths negative it will bring the 2 closer and opposite when negative is close to negative also the moon must have effect there.
A: YOU ARE THINKING OF HOW BAR MAGNETS BEHAVE CLOSE TO EACH OTHER – THIS IS A FAULTY ANALOGY BECAUSE OF FACTORS OF BOTH SCALE, PROCESS, AND DISTANCE. THE SUN IS 144,000,000 KILOMETERS FROM THE EARTH. THE SOLAR WIND *DOES* INTERACT WITH THE MAGNETIC POLE OF THE EARTH (AURORAS). THE STRENGTH OF THE SOLAR MAGNETIC DIPOLE IS FAR TOO SMALL TO INFLUENCE THE EARTH’S MAGNETIC DIPOLE, WHICH IS APPARENTLY DRIVEN BY THERMAL CONVECTION IN THE CORE OF THE EARTH.
Q: PS: Science is like religion you either believe in it or not, however science has a few fact but if they come off wrong at the start everything is wrong
A: I PARTIALLY AGREE WITH YOUR PS: *SOME* PEOPLE ATTEMPT TO MAKE SCIENCE THEIR RELIGION. I DON’T THINK THAT IS WISE, BUT I UNDERSTAND HOW IT CAN HAPPEN. I AGREE THAT FAULTY SCIENCE CAN LEAD TO FURTHER MISTAKES, LIKE HOW SOVIET GENETICS WAS CRIPPLED FOR DECADES BY THE SO-CALLED STALINIST GENETICIST LYSENKO. HOWEVER, WHILE SCIENCE IS THEORETICALLY A SELF-CORRECTING PROCESS, IT IS NEVERTHELESS IMPERFECT, A VERY HUMAN PROCESS. IT SHOULD THEREFOR NOT BE WORSHIPED.
IF YOU DON’T BELIEVE IN SCIENCE THAT IS FINE, BUT IT MARKS YOU AS SOMEONE WHO HAS NOT STUDIED AND LEARNED ENOUGH TO UNDERSTAND IT. YOU CAN ALSO CHOOSE TO NOT BELIEVE IN GRAVITY, HOWEVER IF YOU THEN STEP OFF THE TOP OF A BUILDING YOUR BELIEF WILL NOT MAKE IT GO AWAY.
One final comment. Everyone is entitled to an opinion. However, if you base your life and actions on opinions not backed up by facts, you will not live long nor well. MAKING UP facts doesn’t make them facts. You can believe that 1 + 1 = 3 but your orbital mechanics BASED ON THAT MATH will not land a man on the Moon. | 14,056 | 60,721 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.945791 |
http://stackoverflow.com/questions/18227780/i-dont-understand-boolean-truths-in-python | 1,464,218,397,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049275412.55/warc/CC-MAIN-20160524002115-00233-ip-10-185-217-139.ec2.internal.warc.gz | 281,889,608 | 18,430 | # I don't understand boolean truths in Python
I wrote this code (`x` was run through a `str()` before this scope):
``````if x == "A" or "O":
return x
``````
This returns `B` when `x = B`. Can someone help me understand why `x = B` validates here?
When I changed my code to read
``````if x == "A":
return x
elif x == "O":
return x
``````
It did not match `x = B`, so I am assuming there is something with the boolean logic here that I am not understanding.
-
`x == "A" or "O"` == `(x == "A") or "O"` – Ashwini Chaudhary Aug 14 '13 at 9:25
Thank you Ashwini! Makes complete sense now. – pzkpfw Aug 14 '13 at 9:25
In python boolean `if "<any character here>"` will evaluate to true since you have specified `if x == "A" or "O":` though first condition is false it will check next condition and will evaluate to true. – Nikhil Rupanawar Aug 14 '13 at 9:29
"O" is always True – njzk2 Aug 14 '13 at 10:03
`if x == "A" or "O":` should be `if x == "A" or x == "O":`.
`if x == "A" or "O":` will always be evaluated to `true`.
`if x == "A" or "O":` is interpreted as:
`if (x == "A") or ("O")`, `"O"` is `true`, so even if `x` is not `"A"`, since you have `or`, this will always be `true`.
Alternatively you can write:
``````if x in ["A", "O"]:
``````
-
Great stuff, thanks. – pzkpfw Aug 14 '13 at 11:57 | 442 | 1,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-22 | latest | en | 0.912417 |
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• Oct 10th 2011, 01:19 PM
dac1234
Numerical Analysis - Fixed Point Iteration
Assume that g is a continuously differentiable function and that the Fixed-Point Iteration g(x) = x has exactly three fixed points, −3, 1, and 2. Assume that g'(−3) = 2.4 and that FPI started sufficiently near the fixed point 2 converges to 2. Find g'(1).
Guess
F(x) = x^3 - 7x + 6 where -3 , 1 , and 2 are roots
Using FPI equation F(x). --- g(x) = x -f(x) ----
f(x) = -x^3 + 8x - 6
By pluging in the values
f(-3) = - 3 ,
f(1) = 1
f(2) = 2
Given g'(-3) = 2.4
Use general form of fixed point iteration to get
f'(x) = c * (3x^2 - 7) , where c is a constant
f'(-3) = c * (27 - 7) = -1.4
solve for c
c= - .07
f(x) = (x^3) - 7x + 6 = 0
f(x) = -.07x^3 + .49x - .42
USE FPI form to get
g(x) = .07x^3 + .51x +.42
g'(x) = .21x^2 +.51
g'(-3) = .21 * 9 + .51 = 1.89 +.51 =2.4
Hence g'(x) = .21x^2 +.51.
g'(1) = .21 +.51 =.72
g'(2) = 1.35
The equation I found satisfy that there are three fixed points and F'(-3) = 2.4, but it does not satisfy that when FPI is near 2 it converges to 2. My answer converges to 1. This is because g'(1) < g'(2).
my other attempt [no computation].
Since g'(2) converges to 2. The slope or derivative g'(2) < 1.
My guess is that g'(1) should be greater than 1 or greater than g'(2).
Given g'-(3) = 2.4......I don't know how this helps me in finding g'(1).
hmm maybe can I assume g(3) = 2.4????
g'(1) = 1.2???
Right now I don't see any approach where I can find g'(1). Some help would be nice. | 623 | 1,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-47 | longest | en | 0.870236 |
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View Full Version : standard error of mean
muddy
01-08-2006, 07:00 AM
If I have a single data point and take that point as an estimate of the mean of the population, how would I estimate the standard error of the mean?
JohnM
01-08-2006, 08:50 AM
The standard error of the mean is the population standard deviation divided by the square root of the sample size, so you don't have enough information to estimate it.
muddy
01-08-2006, 12:25 PM
The standard error of the mean is the population standard deviation divided by the square root of the sample size, so you don't have enough information to estimate it.
Thanks John,
I agree that it cant be done with the estimate equation.
I was really fishing for alternative approaches (bootstrap methods, chi squared approximation, etc)
Actually I have one other peice of information ... the parent population is exponentially distributed ... therefore the standard deviation is equal to the mean. So would this mean that the standard error is equal to the mean?
Any ideas?
JohnM
01-08-2006, 05:02 PM
I would highly doubt any method that purports to allow estimation of anything remotely resembling a standard deviation, variance, or mean, with just one data point....a bunch of junk, IMHO.
Yes, you could then infer the standard deviation, but not the standard error of the mean would still be s/sqrt(n) - I guess in this case you could hazard a guess by using s and picking a hypothetical sample size....
muddy
01-08-2006, 05:48 PM
John,
Thanks for all the help ... maybe I should just state what I am really after:
Given an observation time period T
and a number of events N
I can calculate Mean Time Between Events as M' = T/N
What I need now is a confidence (C) that the true M is above some Mlower
The function I am trying to find is:
Mlower = f(M',T,N,C)
Ive seen a few papers claiming to estimate this even for N=0 (an observation period with no events) with the Chi Squared distirbution:
Mlower = 2*T/ CHIINV(C,N+2)
Ive played around with some bootstrap simulations that seem so far to not support this.
any thoughts on this approach? | 508 | 2,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2013-20 | latest | en | 0.934259 |
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Difference Between Z score vs T score
Z score
Under a normal distribution, where full data is available, it is a distance from the mean. Its formula is as given below,
Z= (x-μ)/σ
Where,
X = individual raw data
μ = Population mean
σ = Population standard deviation
T score
The t score is the subtraction of individual standard deviation from the individual mean and then dividing the result with the sample standard deviation whole result multiplied by sample size. Its formula is as given below,
t = {(- μ)/s}*
= Sample Mean
μ = Population Mean
s = Sample Standard deviation
n = sample size
Let’s take an example to understand the same in a better manner:
The score for each student in a class is used to calculate the mean of marks equal to 50 and a standard deviation of 10. We can compute the Z score with the score of 50 as (50-50) / 10 = 0
We can interpret that student’s score is 0 distance (in units of standard deviations) from the mean, so the student has scored average.
If the score is 60, the Z score is (60-50)/ 10 = 1
We can interpret that the student has scored above average – a distance of 1 standard deviation above the mean.
Key Differences Between Z score vs T score
Let us discuss some of the major differences between the Z score and vs T score.
Z score is the standardization from the population raw data or more than 30 sample data to a standard score, while the T score is the standardization from the sample data of less than 30 data to a standard score.
Z score ranges from -3 to 3, while the T score ranges from 20 to 80.
As the data size increases, the distribution tends to be Z distribution. The Z score vs T score distribution is part of a normal distribution but differs based on the size.
The Z score is used in stock market data to check the chances of a company going bankrupt. In contrast, the t score is extensively used in checking bone mineral density and Fracture risk assessments.
Z score vs T scores Comparison Table.
Let’s look at the top 9 Comparisons between Z and T scores.
Sr. No.
Points of comparison
Z Score
T score
1 Standardization of data Its standardization from population data Its standardization from Sample Data
2 Data Size When the Population is known or above 30, one can use the Z score The T score is used when the population is unknown, or the sample size is less than 30.
3 Mean An average is always zero. The average is always 50.
4 Range It Ranges from -3 to 3. It ranges from 20 and 80.
5 Standard Deviation Its standard deviation is always 1 Its standard deviation is always 10
6 Derived Result The derived result can be negative The derived result can never be negative
7 Preference Comparatively less preferable, as it supports large data More preferable as it covers a higher range, but with an increase in size, it has its inherent limitation
8 Distribution Z score is part of the Z distribution The t score is part of the T distribution
9 With the increase in size With the increase in size, the Z score tends to be used With the increase in size, its usefulness reduces.
Conclusion
Both the Z score vs T score is part of hypothesis testing under the normal distribution. If you have a set of measurement scores on different measures using Z-scores, you can tell how the scores are placed in their distributions. Then you can compare them. Standardization of scores is an extensively used procedure in the field of research and planning as they help compare various test scores. Standardizing scores before combining them helps a researcher to get better and comparable results.
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In a continuous-review $(r,Q)$ inventory system under a type-1 service level constraint, if the demand per unit time is distributed as $N(\mu,\sigma^2)$ and the lead time, $L$, is a constant, then the ... | 981 | 4,212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-50 | latest | en | 0.898509 |
https://zbmath.org/?q=an:0961.35101 | 1,653,656,724,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00321.warc.gz | 1,227,184,093 | 11,467 | ## Boundary layer theory and the zero-viscosity limit of the Navier-Stokes equation.(English)Zbl 0961.35101
The author reviews resent progress on the analysis of Prandtl’s equation and the related issue of the zero-viscosity limit for the solution of the Navier-Stokes equations.
Let $$v(x,y,t)=(v_1,v_2)$$ be the vector of velocity of a fluid and $$p(x,y,t)$$ be the pressure. Prandtl’s equations $\frac{\partial v_1}{\partial t}+v_1 \frac{\partial v_1}{\partial x_1}+v_2 \frac{\partial v_1}{\partial x_2}-\frac{\partial^2 v_1}{\partial x^2_2}+ \frac{\partial p}{\partial x_1}=0,\quad \frac{\partial v_1}{\partial x_1}+\frac{\partial v_2}{\partial x_2}=0,$ with the initial condition $$v(x,0)=v_0(x)$$ are considered in $$\Omega=\{x\in \mathbb{R}^2: x_2\geqslant 0\}$$. The corresponding steady problem is considered too. The author presents solvability results for these problems including an unpublished one. In addition, the limit of the solution to the Navier-Stokes equations $\frac{\partial v^\varepsilon}{\partial t} +(v^\varepsilon\cdot\nabla)v^\varepsilon -\varepsilon\Delta v^\varepsilon+p^\varepsilon=0, \quad \nabla\cdot v^\varepsilon=0 \quad\text{in} \;\Omega$ for $$\varepsilon\to 0$$ is discussed.
Finally, the author indicates some directions where progress is expected in the near future. The proofs are not presented.
### MSC:
35Q30 Navier-Stokes equations 35B05 Oscillation, zeros of solutions, mean value theorems, etc. in context of PDEs 76D10 Boundary-layer theory, separation and reattachment, higher-order effects 76D05 Navier-Stokes equations for incompressible viscous fluids
### Keywords:
boundary layer; zero-viscosity limit; Prandtl’s equations
Full Text:
### References:
[1] L Prandtl. Verhandlung des III Internationalen Mathematiker-Kongresses (Heidelberg, 1904), p. 484-491 [2] K Nickel. Prandtl’s boundary layer theory from the viewpoint of a mathematician. Ann Rev Fluid Mech, 1973, 5:405-428 · Zbl 0269.76017 [3] H Johnston, J G Liu, Weinan E. The infinite Reynolds number limit of ow past cylinder. in preparation [4] O A Oleinik. The Prandtl system of equations in boundary layer theory. Soviet Math Dokl, 1963, 4:583-586 · Zbl 0134.45004 [5] S Goldstein. On laminar boundary layer ow near a point of separation. Quart J Mech Appl Math, 1948, 1:43-69 · Zbl 0033.31701 [6] K Stewartson. On Goldstein’s theory of laminar separation. Quart J Mech Appl Math, 1958, 11:399-410 · Zbl 0173.28102 [7] L Caffarelli, Weinan E. Separation of steady boundary layers. unpublished [8] O A Oleinik. Construction of the solutions of a system of boundary layer equations by the method of straight lines. Soviet Math Dokl, 1967, 8:775-779 [9] Weinan E, B Engquist. Blowup of solutions to the unsteady Prandtl’s equation. Comm Pure Appl Math, 1997, L:1287-1293 · Zbl 0908.35099 [10] M Sammartino, R E Caflisch. Zero viscosity limit for analytic solutions of the Navier-Stokes equation on a half-space. I Comm Math Phys, 192, 433-461, II, Comm Math Phys, 1998, 192, 463-491 · Zbl 0913.35102 [11] E Grenier. On the instability of boundary layers of Euler equations. In press · Zbl 0958.35106 [12] E Grenier, O Gues. Boundary layers for viscous perturbations of noncharacteristic quasilinear hyperbolic problems. J Diff Eq, 1998, 143:110-146 · Zbl 0896.35078 [13] D Serre. Systems of conservation laws, II. to be published [14] Z Xin. Viscous boundary layers and their stability (I). J PDEs, 1998, 11:97-124 · Zbl 0906.35057 [15] R Temam, X M Wang. Boundary layers for the Navier-Stokes equations with non-characteristic boundary. In press · Zbl 0997.35042 [16] T Kato. Remarks on zero viscosity limit for nonstationary Navier-Stokes flows with boundary. In: Seminar on Partial Differential Equations, S S Chern eds 1984, 85-98 · Zbl 0559.35067 [17] R Temam, X M Wang. On the behavior of the solutions of the Navier-Stokes equations. Annali della Normale Superiore di Pisa, Serie IV, 1998, XXV:807{828 · Zbl 1043.35127 [18] N Masmoudi. About the Navier-Stokes and Euler systems and rotating fluids with boundary. In press · Zbl 0915.76017 [19] Y Brenier. In press [20] J Hunter. private communication [21] O A Oleinik, Samokhin. Mathematical Methods in Boundary Layer Theory. Phismathgis”Nauka”, Moscow, 1997
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. | 1,368 | 4,606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.683896 |
https://www.neetprep.com/questions/55-Physics/699-Ray-Optics-Optical-Instruments?courseId=8&testId=1182340-Recommended-MCQs---Questions&subtopicId=452-Refraction-Plane-Surface | 1,675,055,515,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00206.warc.gz | 913,625,737 | 52,222 | When a ray of light falls on a given plate at an angle of incidence $60°$, the reflected and refracted rays are found to be normal to each other. The refractive index of the material of the plate is:
1. $\frac{\sqrt{3}}{2}$
2. 1.5
3. 1.732
4. 2
Subtopic: Refraction at Plane Surface |
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A medium shows relation between i and r as shown. If the speed of light in the medium is nc then the value of n is:
1. 1.5
2. 2
3. 2–1
4. 3–1/2
Subtopic: Refraction at Plane Surface |
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In the figure shown the angle made by the light ray with the normal in the medium of refractive index $\sqrt{2}$ is:
1. 30$°$
2. 60$°$
3. 90$°$
4. None of these
Subtopic: Refraction at Plane Surface |
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An air bubble in a glass slab with refractive index 1.5 (near-normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is:
1. 8
2. 10
3. 12
4. 16
Subtopic: Refraction at Plane Surface |
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A plane mirror is placed at the bottom of a fish tank filled with water of refractive index $\frac{4}{3}$. The fish is at a height 10 cm above the plane mirror. An observer O is vertically above the fish outside water. The apparent distance between the fish and its image is:
1. 15 cm
2. 30 cm
3. 35 cm
4. 45 cm
Subtopic: Refraction at Plane Surface |
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Two slabs P & Q of transparent materials have a thickness in the ratio 2 : 5. If a ray of light takes the same amount of time to move from A to B and B to C, then the refractive index of Q with respect to P will be:
1. 0.4
2. 2.5
3. 1.4
4. 1.85
Subtopic: Refraction at Plane Surface |
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A fish at a depth y inside the water is seeing a bird. The bird is at a height x above the water level. If the refractive index of water is $\mathrm{\mu }$, then the apparent distance of bird as seen by the fish is:
1. x + $\mathrm{\mu }$y
2. y + $\mathrm{\mu }$x
3. x + $\frac{\mathrm{y}}{\mathrm{\mu }}$
4. y + $\frac{\mathrm{x}}{\mathrm{\mu }}$
Subtopic: Refraction at Plane Surface |
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For a light incident from air on a slab of refractive index 2, the maximum possible angle of refraction is:
1. 30o
2. 45o
3. 60o
4. 90o
Subtopic: Refraction at Plane Surface |
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A beam of light composed of red and green rays is incident obliquely at a point on the face of a rectangular glass slab. When coming out on the opposite parallel face, the red and green rays emerge from:
1 Two points propagating in two different parallel directions 2 One point propagating in two different directions through the slab 3 One point propagating in the same direction through the slab 4 Two points propagating in two different non-parallel directions
Subtopic: Refraction at Plane Surface |
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A diverging beam of light from a point source S having divergence angle $\alpha$, falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is :
1. Zero
2. $\alpha$
3. ${\mathrm{sin}}^{-1}\left(1/n\right)$
4. $2{\mathrm{sin}}^{-1}\left(1/n\right)$
Subtopic: Refraction at Plane Surface |
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NEET 2023 - Target Batch - Aryan Raj Singh | 1,764 | 6,609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-06 | latest | en | 0.827766 |
http://www.dreamincode.net/forums/topic/2470-help/page__pid__1563__st__0 | 1,462,063,036,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860113541.87/warc/CC-MAIN-20160428161513-00216-ip-10-239-7-51.ec2.internal.warc.gz | 474,630,538 | 19,066 | # help!!!
Page 1 of 1
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### #1 miaka284
Reputation: 0
• Posts: 530
• Joined: 10-May 01
# help!!!
Posted 17 May 2001 - 01:04 PM
okay, my problems is that my infile refused to read my numbers. I had 8,12,16 typed in, in my file. It only read the first number and the rest it gives me zeros and I don't know why.
```void factorial()//asks the user to enter a positive number
{
long int input;//declares input local
infile.open("Factofile.txt", ios::in);
";
infile>>input;
if(input==0)//if statement, for if input is 0
{//it doesn't calculate the number 0
cout<<"By definition, the factorial of 0 is 1.
";}
if(input>0)//if input is bigger than zero then it goes to the next function
{
fact(input);
}if(input<0)//if input is smaller than zero then it kicks out
{
cout<<"You can't calculate factorials with a negative number.
";
factorial();
}
}
long int fact(long int input1)//here is the function that calculates factorial
{
for(long int a = input1-1; a>1; a--)//for loop for doing the number again
{//until the number is equal to one
input1 *= a;
}
info();//calls the function that
infile.close(); //closes the infile//asks the user if they want to play againreturn input1;}
```
Is This A Good Question/Topic? 0
## Replies To: help!!!
### #2 Null and Void
Reputation: 1
• Posts: 204
• Joined: 29-April 01
## Re: help!!!
Posted 17 May 2001 - 02:18 PM
Assuming that your code is correct in most ways (I didn't read it) you can get around this by using spaces or endlines in the file instead of commas.
Or, this is how I'd open, and read two numbers from a file, each being seperated by a comma (With C ANSI Standard I/O). This probably isn't applicable to your code, but it may give you a hint or something:
[code]
#include <stdio.h>
// ...
int a, b;
FILE *fp = fopen("somefile.txt","rt"); // Open the file
if(fp!=NULL) { // If we opened it
// Read an integer, ignore a character (the comma), and read another integer
fscanf(fp,"%d%*c%d",&a,&b);
fclose(fp); // Close it
} | 716 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-18 | latest | en | 0.801304 |
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Question
# Question 69 Fill in the blanks to make the statements true. The sum of interior angles of a polygon of n sides is ___ right angles.
Open in App
Solution
## (2n - 4) By the formula, sum of interior angles of a polygon of n sides =(n−2)×180∘ =(2n−4)×90∘.
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0
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Join BYJU'S Learning Program | 120 | 416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.770769 |
https://demo.formulasearchengine.com/wiki/Talk:Annulus_(mathematics) | 1,632,354,598,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00044.warc.gz | 238,592,308 | 7,010 | # Talk:Annulus (mathematics)
As to the annulus in geometry, all definitions of the annulus somehow proceed from the body having the form of a finger ring. In elementary geometry we have three-dimensional annuli (having the form of a finger ring) and two-dimensional annuli (figures limited by two concentric circles). So we can generalize the concept to the n-dimensional Euclidean space, (the union of all (n-1)-dimensional balls whose center lies (ADDENDUM) at a fixed distance from a given point outside of the balls) and further to complex spaces and maybe (I am not sure) even to any metrical space. The condition that the annulus be open is optional.
Might be this needs a correction, at least. For your 3D ring I use the word "torus" http://en.wiktionary.org/wiki/torus , so this perhaps does not come to be "annulus".
ADDENDUM = nothing -- in this case you get a shell = difference of two balls. This is what I was looking for but it is NOT a generalization of your ring in 3D
ADDENDUM = "in a (2D) plane" (tentatively through the fixed point)
ADDENDUM = "in a hyper-plane" (n-1 -dimensional) (tentatively through the fixed point)
Last two possibilities are both generalization of the ring from 3D.
90.180.192.165 (talk) 15:19, 5 December 2012 (UTC)
The definition of the annulus on the complex space is a special case of such a general concept. It might be that that special case has some special importance in the complex analysis. Andres 12:52, 5 Nov 2003 (UTC)
## This result can be obtained via calculus
Why, pray tell, would you ever bother to calculate the area of one of these through integration?
Because one can then use that formula to calculate the volume certain solids obtained through rotation. For example, the volume of the region defined by ${\displaystyle y=(x+2)^{2}+2}$ and ${\displaystyle y=6}$, rotated about the y-axis can be calculated as an integral with respect to y by taking the areas of infinitely thin horizontally-oriented annuli.
Aside from that, the way to prove the area of any curved shape is through integration. siafu 18:15, 6 March 2006 (UTC)
## Way too many diambiguations
So I was looking for annulus, and I went through the ring link, then it went to circular, then finally I got to circle. Isn't their some easier way to get where I want.
128.12.39.120 00:25, 18 May 2007 (UTC) | 588 | 2,337 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.934685 |
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# Problem 1854. Factorial: Unlimited Size : java.math
Solution 2549745
Submitted on 14 Jun 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
tic N=69; y = factorialJava(N); assert(strcmp(y,'171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000')) toc
No method 'add' with matching signature found for class 'java.math.BigDecimal'. Error in factorialJava (line 8) xBD2 = xBD2.add(1) Error in Test1 (line 3) y = factorialJava(N);
2 Fail
tic N=randi(18) y = factorialJava(N); assert(strcmp(y,num2str(factorial(N)))) toc
N = 16
No method 'add' with matching signature found for class 'java.math.BigDecimal'. Error in factorialJava (line 8) xBD2 = xBD2.add(1) Error in Test2 (line 3) y = factorialJava(N);
3 Fail
tic N=randi(18) y = factorialJava(N); assert(strcmp(y,num2str(factorial(N)))) toc
N = 8
No method 'add' with matching signature found for class 'java.math.BigDecimal'. Error in factorialJava (line 8) xBD2 = xBD2.add(1) Error in Test3 (line 3) y = factorialJava(N);
4 Fail
tic N=1000; y = factorialJava(N); assert(strcmp(y,'402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000')) toc
No method 'add' with matching signature found for class 'java.math.BigDecimal'. Error in factorialJava (line 8) xBD2 = xBD2.add(1) Error in Test4 (line 3) y = factorialJava(N);
5 Fail
tic N=42; y = factorialJava(N); assert(strcmp(y,'1405006117752879898543142606244511569936384000000000')) toc
No method 'add' with matching signature found for class 'java.math.BigDecimal'. Error in factorialJava (line 8) xBD2 = xBD2.add(1) Error in Test5 (line 3) y = factorialJava(N);
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We have data which represents a movement of currency pair EUR/USD in the last year 2016. Our data includes three parameters for each day and looks like this { forex: 'EUR/USD', date: '2016-01-01', value: 1.021304401 }. From that data we would like an answer to a question, was average of pair USD/EUR higher in the first half of 2016 or in second?
To find out we use two different solutions.
We implemented four functions.
• customMap in map.ts
• customReduce in reduce.ts
• customFilter in filter.ts
• asynchronousAverage in asynchronous.ts
To find out the average of pair USD/EUR in the first half of 2016 we use first three functions (filter, map and reduce). First, we filter the whole data, using only pairs from the first half of 2016 (date < '2016-07-01'). Then we map our data because we have values for pair EUR/USD, but we were interested in pair USD/EUR, we have to inverse our values (value = 1 / value). Now when we have an array with values we are looking, we calculate a sum of these elements with reduce function and then divide this value by a number of elements in a current array. The result is an average value of pair USD/EUR in the first half of 2016.
To find out the average of pair USD/EUR in the second half of 2016 we use the fourth function (asynchronousAverage). If we look what we've done in the first step we can quickly see, that we've called method array.forEach() for three times (for each function). In case we already know, that we'll need all three functions (map, reduce and filter) we can optimize execution and done everything in one loop. In function asynchronousAverage, we use only one loop and check for each element if it belongs in the second half of 2016 (date > '2016-06-30'), reverse its value (value = 1 / value) and add to our average.
Usage
Now open your browser at http://localhost:3000
3h
MIT | 457 | 1,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-09 | latest | en | 0.873244 |
http://www.examiner.com/article/units-of-measure-for-weighing-silver | 1,386,519,291,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163066263/warc/CC-MAIN-20131204131746-00065-ip-10-33-133-15.ec2.internal.warc.gz | 334,675,333 | 19,806 | Based upon the number of questions I receive regarding the value of a reader's silver pieces, there must be a large number of people out there thinking about selling their silver flatware and hollowware pieces. Although I generally do not recommend selling silver pieces for scrap, it often helps to know what the scrap value is for a benchmark if nothing else. The first step in calculating the scrap value is to determine the weight of pure silver within a piece. This leads to the confusing world of units of weight measure for precious metals.
The world commodity markets value pure silver in terms of dollars per Troy ounce. The problem is that we are accustomed to measuring the weight of items in avoirdupois (or "regular") ounces. Troy ounces and avoirdupois ounces are not equivalent. It makes things simpler for me to convert weights to grams. Once I know the number of grams, I can convert to whatever system of weights necessary.
Another common unit of measure used in the industry is "pennyweight" (abbreviated "DWT"). Again, I convert to grams for simplicity. Here are the common conversion factors:
• One Troy ounce = 31.1034768 grams
• One avoirdupois ounce = 28.3495231 grams
• One pennyweight = 1.5551738 grams
Few silver flatware or hollowware pieces are made of 100% silver. So, one also has to determine how much actual silver is in a piece. In the United States, "sterling" pieces contain at least 92.5% silver. I say "at least" because one may find some pieces with slightly higher percentages of silver content. For example, I recently had some damaged pieces of sterling flatware melted down and they contained 93.4% silver on average. However, it's best to assume sterling pieces are 92.5% silver until proven otherwise.
Here's an example of the process to calculate value:
1. Let's assume we have one Gorham Chantilly place fork that weighs 50 grams.
2. 50 multiplied by .925 gives us 46.25 grams of pure silver content.
3. 46.25 divided by 31.1034768 gives us 1.487 Troy ounces of silver in this fork.
5. 1.487 Troy ounces multiplied by \$23.17 gives us \$34.35 as the scrap value of this fork.
Should I sell this fork for scrap? Heck, no! Here's why:
• A fork like this in decent condition will probably bring at least \$50 (probably more) on eBay.
• If I tried to sell the fork for scrap, it's highly unlikely I would receive \$34.35. The metal refiner and any middlemen would take a cut. I'd probably be lucky to get as much as \$20 for the fork if I sold it for scrap. But just knowing the scrap value helps me determine a selling price.
## Life
• President Obama cannot own an iPhone for security reasons
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4 Photos
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Electric Cigarettes | 724 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2013-48 | latest | en | 0.920114 |
https://cursa.app/en/course/intermediate-algebra-with-leonard | 1,642,651,471,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301720.45/warc/CC-MAIN-20220120035934-20220120065934-00161.warc.gz | 249,428,127 | 10,622 | Course
# Intermediate Algebra by Leonard
5
(19)
Teacher
## Leonard
This Channel is dedicated to quality mathematics education. It is absolutely FREE so Enjoy! Videos are organized in playlists and are course specific.
## Course content preview
2h17m
0h54m
0h53m
1h39m
1h23m
1h05m
2h28m
0h38m
0h39m
1h42m
1h08m
1h10m
1h16m
0h57m
0h49m
0h40m
1h41m
0h46m
1h21m
### Intermediate Algebra Lecture 8.4: An Introduction to Direct and Inverse Variation
0h46m
Watch the 1st video of the course
Do not worry, it is free!
## Course content
2h17m
### Intermediate Algebra Lecture 6.1: Factoring the Greatest Common Factor (GCF)
Intermediate Algebra Lecture 6.1: Factoring the Greatest Common Factor (GCF)
0h54m
### Intermediate Algebra Lecture 6.2: Factoring Polynomials (Trinomials) With a = 1
Intermediate Algebra Lecture 6.2: Factoring Polynomials (Trinomials) With a = 1
0h53m
### Intermediate Algebra Lecture 6.4: Factoring Polynomials (Trinomials) in General
Intermediate Algebra Lecture 6.4: Factoring Polynomials (Trinomials) in General
1h39m
### Intermediate Algebra Lecture 6.5: Factoring Binomials (Difference of Squares/Cubes)
Intermediate Algebra Lecture 6.5: Factoring Binomials (Difference of Squares/Cubes and Sum of Cubes)
1h23m
### Intermediate Algebra Lecture 6.6: Solving Equations by Factoring
Intermediate Algebra Lecture 6.6: Solving Equations by Factoring
1h05m
### Intermediate Algebra Lecture 6.7: Using Factoring to Solve Word/Application Problems
Intermediate Algebra Lecture 6.7: Using Factoring to Solve Word/Application Problems
2h28m
### Intermediate Algebra Lecture C.1: A BRIEF Review of Solving Equations and Factoring
Intermediate Algebra Lecture C.1: A BRIEF Review of Solving Equations and Factoring
0h38m
### Intermediate Algebra Lecture C.3: A BRIEF Review of Graphing
Intermediate Algebra Lecture C.3: A BRIEF Review of Graphing
0h39m
### Intermediate Algebra Lecture C.4: A BRIEF Review of Polynomial Mathematics
Intermediate Algebra Lecture C.4: A BRIEF Review of Addition, Subtraction, Multiplication, Division of Polynomials
1h42m
### Intermediate Algebra Lecture 7.1: Defining and Simplifying Rational Expressions
Intermediate Algebra Lecture 7.1: Defining and Simplifying Rational Expressions
1h08m
### Intermediate Algebra Lecture 7.2: How to Multiply and Divide Rational Expressions
Intermediate Algebra Lecture 7.2: How to Multiply and Divide Rational Expressions
1h10m
### Intermediate Algebra Lecture 7.3: Finding LCD and Equivalent Rational Expressions
Intermediate Algebra Lecture 7.3: Finding Lowest Common Denominator and Equivalent Rational Expressions
1h16m
### Intermediate Algebra Lecture 7.4: Adding and Subtracting Rational Expressions
Intermediate Algebra Lecture 7.4: Adding and Subtracting Rational Expressions
0h57m
### Intermediate Algebra Lecture 7.5: Solving Rational Equations (Equations with Rational Expressions)
Intermediate Algebra Lecture 7.5: Solving Rational Equations (Equations with Rational Expressions)
0h49m
### Intermediate Algebra Lecture 7.6: Using Proportions and Rational Methods in Problem Solving
Intermediate Algebra Lecture 7.6: Using Proportions and Rational Methods in Problem Solving
0h40m
### Intermediate Algebra Lecture 7.7: Introducing and Simplifying Complex Fractions
Intermediate Algebra Lecture 7.7: Introducing and Simplifying Complex Fractions
1h41m
### Intermediate Algebra Lecture 8.1: Graphing Lines. Equations of Parallel and Perpendicular Lines
Intermediate Algebra Lecture 8.1: Graphing Linear Functions. Equations of Parallel and Perpendicular Lines
0h46m
### Intermediate Algebra Lecture 8.2: An Introduction to Non-Linear Functions
Intermediate Algebra Lecture 8.2: An Introduction to Non-Linear Functions
1h21m
### Intermediate Algebra Lecture 8.3: Study of Piecewise Functions and Basic Translations of Graphs
Intermediate Algebra Lecture 8.3: Study of Piecewise Functions and Basic Translations of Graphs
0h46m
### Intermediate Algebra Lecture 8.4: An Introduction to Direct and Inverse Variation
Intermediate Algebra Lecture 8.4: An Introduction to Direct and Inverse Variation | 1,066 | 4,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-05 | latest | en | 0.792831 |
https://www.codeproject.com/articles/42983/superlinear-an-investigation-into-concurrent-speed?msg=3230250 | 1,490,828,035,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191405.12/warc/CC-MAIN-20170322212951-00186-ip-10-233-31-227.ec2.internal.warc.gz | 887,607,764 | 32,776 | 12,829,079 members (40,142 online)
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Posted 11 Oct 2009
# Superlinear: an investigation into concurrent speedup
, 11 Oct 2009 CPOL
Rate this:
## Introduction
This is a short article showing superlinear speedup of a memory-bound algorithm running on an 8-core machine with 24 MB of level 2 cache.
This is an article of three halves. The first half describes the metrics I will be using. The second half shows a common result and the third half investigates just how much speedup is possible in a best case.
The best result I achieved is an efficiency of 5 times. So instead of running 8 times faster on an 8-core machine, the parallel implementation runs 40 times faster.
## Background
#### Definitions
Firstly, here is a quick recap of some definitions.
speedup = sequential time / concurrent time
So if you have a 4-core box that runs an algorithm sequentially in 8 seconds and in parallel in 2 seconds, you have a speedup of 4. If you then run the algorithm on an 8-core box and achieve a time of 1 second, you have a speedup of 8.
The example above is called linear speedup. This is normally the target when you convert to a parallel algorithm: run on a 4-core box and it goes 4 times faster; run on an 8-core box and it goes 8 times faster; and so on.
Normally, you don't achieve this. You might achieve 2.5 seconds on a 4-core box or 1.25 seconds on an 8-core box. These times are speedups of 3.2 and 6.4 respectively.
efficiency = speedup / number of cores
When comparing 3.2 and 6.4, it's not obvious how well your concurrent algorithm is scaling on a bigger machine, so we introduce the normalized metric: efficiency. The example gives an efficiency of 0.80 = 80% in both cases. So you could immediately say that although you are not achieving linear efficiency, your algorithm is scaling well.
Efficiencies are grouped into 3 possibilities. Efficiencies less than 1 are called sublinear. This is the most common case. An efficiency of exactly 1 is called linear. And the holy grail ( and the subject of this article ) is efficiencies greater than 1, called superlinear.
1-core4-core8-coreefficiency
timespeeduptimespeeduptimespeeduppercentcategory
812.53.21.256.480%sublinear
812418100%linear
811.650.810125%superlinear
comparison of speedup and efficiency
This article shows that considerable superlinear efficiencies are possible in some cases.
#### How is superlinear efficiency possible?
Since any concurrent algorithm can be rewritten as a sequential one, surely it is impossible to achieve superlinear speedup?
Well, this article proves that it is possible! If your algorithm is limited by the resources available to a single core, then adding cores and having each core work on a smaller set of data can make your machine disproportionately faster.
Basically, by using more of the resources in your machine, you can run your code more efficiently in some cases.
#### Memory architecture
For this example, I have targeted the level 2 cache in my dual-processor quad-core Xeon box. So the first step was to find out exactly what hardware was present.
Joe Duffy, author of "Concurrent Programming on Windows" and lead developer of the .NET 4.0 Parallel Extensions team, wrote a utility to do exactly that last year. He gave it[^] to Mark Russinovich who added it to the SysInternals toolset as CoreInfo[^]. Here is the output for my machine:
coreinfo output
From this you can see, in quite some detail, the memory architecture present in your machine. I have extracted the info relevant to this article to produce the diagram below:
simplified memory architecture
Each core has access to 6 MB of level 2 cache, shared between a pair of cores. This gives a total of 24 MB of cache present.
## Part I: Dataset size
I wanted to show that as the dataset increased beyond the 6 MB available to a single core, the concurrent implementation would begin to outperform the sequential implementation with superlinear speedups. This happens because when the dataset is larger than the cache, the CPUs have to access main memory, which takes hundreds of clock cycles. Also, as the total 24 MB limit was passed, both implementations would suffer cache misses and the efficiency would drop.
The code is very simple. The dataset is a `byte[]` called `counters` of varying length. The 'work' is just incrementing every element in the array. That's it! Here is a code snippet:
```for ( int n = 0 ; n < REPEAT ; n++ )
for ( int i = 0 ; i < length ; i++ )
counter[ start + length ]++;```
For the sequential runs, `start` = 0 and `length` = the size of the array. For the concurrent runs, the array is split into equal ranges and the parameters are adjusted accordingly for each thread. The other parameter is `REPEAT`. This just repeats the increment loop a given number of times.
On the first run of the increment loop, it doesn't matter how much cache is used because all the data must be loaded from main memory. These are called "compulsory misses". However, on subsequent runs, the concurrent implementation shows marked efficiency gains due to the larger cache available which reduces "capacity misses".
Here is the graph of the results:
Part I results
When `REPEAT` = 1, the concurrent implementation has no special advantage over the sequential implementation, but does suffer from the usual overheads. This leads to a fairly steady efficiency of around 70% - 80%. Remember, this still means that we see a speedup of around 6 times on an 8-core machine. It's just sublinear.
As `REPEAT` increases, we see the efficiency gains I was looking for. For `REPEAT` = 100, the efficiency was between 110% - 120% for a broad range of dataset sizes from 3 MB to 24 MB. If you're tuning your application for performance, this is a very valuable boost. It means your code can run nearly 10 times faster on an 8-core machine.
As the dataset size increases beyond the 24 MB total cache present in the machine, the concurrent implementation also suffers cache misses and efficiency drops down dramatically to quite low values around 60%. This is half the peak efficiency and shows just how important the cache can be.
## Part II : Stride
While the results in part I above would be more than welcome in a real world application, I wondered just how high efficiency can become. So in this part, I mess up memory locality as much as possible. The results are difficult to predict, but show a maximum efficiency of 500%. On an 8-core machine, this equates to a 40 times speedup!
I chose a dataset size of 24 MB and `REPEAT` = 10 for this test. Then, instead of incrementing the array in order, I iterate over it in strides. This is the calculation of which index to increment in the loop:
```for ( int step = 0 ; step <= 20 ; step++ )
for ( int n = 0 ; n < REPEAT ; n++ )
for ( long i = 0 ; i < length ; i++ )
{
long prod = i << step;
long mod = prod % length;
long div = prod / length;
long index = start + mod + div;
counter[ index ]++;
}```
So the array is accessed with indices following this pattern:
IterationLinearStride
000
1116
2232
3348
.........
nn1
n+1n+117
n+2n+233
n+3n+349
.........
This is very bad for locality, but it should only matter for the sequential implementation as the dataset can fit into the total L2 cache available to the concurrent implementation.
Here is a graph of the results. I have shown the times for the sequential and parallel implementations, as well as the efficiency calculated from these values.
Part II results
As you can see, the times for the sequential implementation increase dramatically for strides between 27 and 213. The times for the concurrent implementation remain fairly constant, as expected, giving efficiencies up to 5 = 500%. If your application happens to fall into the sweet spot, this shows that you can make spectacular performance gains by considering your usage of cache.
I was surprised that the times for the sequential implementation became better for strides above 213. I suspect this is something to do with the associativity policy, but I have not confirmed that. If you get to the root of this, please leave a comment in the forum below - I would be very interested.
## Conclusion
Well, that's about it. In this article I have shown efficiencies up to 500%, which means a 40 times speedup on an 8-core machine. If you have performance problems, it is worth looking at memory locality - both temporal and spatial.
## History
11 October 2009: First edition
## Share
United Kingdom
I built my first computer, a Sinclair ZX80, on my 11th birthday in 1980.
In 1992, I completed my Computer Science degree and built my first PC.
I discovered C# and .NET 1.0 Beta 1 in late 2000 and loved them immediately.
I have been writing concurrent software professionally, using multi-processor machines, since 1995.
In real life, I have spent 3 years travelling abroad,
I have held a UK Private Pilots Licence for 20 years,
and I am a PADI Divemaster.
I now live near idyllic Bournemouth in England.
I can work 'virtually' anywhere!
## You may also be interested in...
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First Prev Next
Some thoughts Dmitriy V'jukov15-Oct-09 2:59 Dmitriy V'jukov 15-Oct-09 2:59
Great article Karl Nicoll12-Oct-09 12:50 Karl Nicoll 12-Oct-09 12:50
Nice Luc Pattyn11-Oct-09 4:25 Luc Pattyn 11-Oct-09 4:25
Hi Nick,I enjoyed reading your article. Doing lots of image processing on rather large images (>100MB) I have been aware of the importance of locality of reference, and I applied similar ideas even on a single core; it suffices to rearrange the algorithm in such a way that it executes all operations on part of an image (not exceeding the cache size), then moving on to the next part. Such slicing isn't always possible or easy, but when it is, results are quite worthwhile indeed.As for your sudden improvement on high stride values, I dare suspect your `step` variable to be of type `int`, which would cause `i << step` to be a 32-bit expression and never really become large. Please check that.Cheers Luc PattynI only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messagesLocal announcement (Antwerp region): Lange Wapper? Neen!
Re: Nice Nick Butler11-Oct-09 6:34 Nick Butler 11-Oct-09 6:34
Re: Nice Luc Pattyn11-Oct-09 8:29 Luc Pattyn 11-Oct-09 8:29
Re: Nice Nick Butler11-Oct-09 9:55 Nick Butler 11-Oct-09 9:55
Re: Nice Luc Pattyn11-Oct-09 16:41 Luc Pattyn 11-Oct-09 16:41
Well Done merlin98111-Oct-09 3:15 merlin981 11-Oct-09 3:15
Re: Well Done Nick Butler11-Oct-09 6:23 Nick Butler 11-Oct-09 6:23
Last Visit: 31-Dec-99 19:00 Last Update: 29-Mar-17 13:53 Refresh 1 | 2,590 | 10,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-13 | latest | en | 0.919766 |
https://math.stackexchange.com/questions/1582933/determine-the-order-relation-between-x-and-y-given-5x-2y-2y-3x | 1,566,114,623,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313715.51/warc/CC-MAIN-20190818062817-20190818084817-00508.warc.gz | 560,660,825 | 30,673 | # Determine the order relation between $x$ and $y$, given $5x-2y =2y-3x$
Given: $5x-2y =2y-3x$
\begin{aligned} \text{Quantity A} && \text{Quantity B} \\ x && y \end{aligned}
$\bigcirc$ Quantity A is bigger
$\bigcirc$ Quantity B is bigger
$\bigcirc$ Two quantities are equal
$\bigcirc$ The relationship cannot be determined for the information given
The answer is $B$ according to GRE book. Simplifying I get that $2x=y$. I letting $x=0$ get $y=0$. Also let $x=1$ get $y=2$. I believe it is D but need to double check. I have read that any number in GRE is real.
• The answer is B according to book. simplifying get that $2x=y$. I letting $x=0$ get $y=0$. Also let $x=1$ get $y=2$. I belief it is D but need to double check. I have read that any number in GRE is real. – Tiger Blood Dec 20 '15 at 5:22
• B means "quantity B is bigger". D means "The relationship cannot be determined from the given information" – Tiger Blood Dec 20 '15 at 5:37
$5x-2y=2y-3x$, so $8x=4y$, so $y=2x$. There are a few possibilities. For example, $x=1$ and $y=2$, or $x=-1$ and $y=-2$, or $x=y=0$. If all we know is that $x,y$ are real, then it is not possible to know which is bigger. | 389 | 1,171 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-35 | latest | en | 0.872231 |
http://sites.millersville.edu/bikenaga/calculus/polar-coordinates/polar-coordinates.html | 1,553,633,094,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206016.98/warc/CC-MAIN-20190326200359-20190326222359-00016.warc.gz | 190,589,465 | 5,784 | # Polar Coordinates
Polar coordinates are an alternative to rectangular coordinates for referring to points in the plane. A point in the plane has polar coordinates . r is (roughly) the distance from the origin to the point; is the angle between the radius vector for the point and the positive x-axis. (As usual, angles are positive if measured counterclockwise and negative if measured clockwise.)
Example. Plot the points whose polar coordinates are: (a) ; and (b) .
To plot , rotate the x-axis through an angle of --- counterclockwise, since the angle is positive. Then go out 2 units.
To plot , rotate the x-axis through an angle of .
(Well, that certainly looks stupid. The numbers are upside down!) Locate the point -2 on the rotated x-axis. That is the point .
If you're plotting lots of points, you can do this rotation trick with a ruler, or you can make a little number line out of cardboard. If you do this, it's nearly impossible to get confused by points which have negative r's.
A point can be represented by infinitely many polar coordinate pairs.
If you add multiples of to the angle, you get the same point. For example,
represent the same point in polar coordinates.
If you multiply r by -1 and add to , you get the same point.
Reason: Adding to and multiplying r by -1 each have the effect of "flipping" the point to the other side of the origin, so the two changes cancel out.
For example,
represent the same point in polar coordinates.
This phenomenon does not occur in rectangular coordinates; it can make things like finding intersection points of polar curves a bit tricky!
You can convert between rectangular and polar using the following conversion formulas:
All of them come from the following standard triangle:
Example. Find the polar equation for the circle
Example. (a) Find the rectangular equation for the polar curve and identify it.
Multiply by r, substitute , and complete the square:
This is a circle of radius 2 with center at .
(b) If you start at , how far do you have to go before the graph first closes up?
The entire circle is traced out as goes from 0 to .
The top half of the circle is traced out from 0 to . (Draw some radius segments from the origin out to the curve and see.) The bottom half is traced out from to . It winds up in the fourth quadrant instead of the second because r is negative.
Example. Find the points at which and intersect.
Solve simultaneously: , so .
equals when is , plus or minus multiples of . I have to add or subtract multiples of until falls outside the range (at which point the angles repeat).
gives . gives and . Adding (or subtracting) additional multiples of produces values of outside , so these are the only solutions. Note that there are four intersection points in the picture.
Here's how to find for a polar curve. The idea is to use the Chain Rule and the conversion formulas and .
Example. Find the equation of the tangent line to at .
When , . The point of tangency is given by
Now
Thus,
The tangent line is
I'll give a heuristic justification for the formula for the area of the region bounded by a polar curve.
As the picture shows, a region in polar is "swept out" as if by a revolving searchlight beam.
Look at a small wedge-shaped piece of the region. It subtends an angle and the radius is r.
If is small, the wedge is approximately a circular wedge. The area of a circular wedge of radius r and angle is , so this is a good approximation to the area of the wedge-shaped piece above.
As usual, I obtain the total area by integrating to add up the areas of the little pieces:
Example. Here is the graph of .
Label the approximate points on the graph which correspond to , , , , and . Draw arrows which indicate how the graph is traced out as increases. Then find the area of the region inside the outer loop but outside the inner loop.
Notice that the inner loop in the third quadrant is traced out as goes from 0 to . Why do the points wind up in the third quadrant when the angles are between 0 and ? Because for those values of the radius r is negative!
You can find the area inside the outer loop and outside the inner loop by subtracting the inner loop area from the outer loop area.
By symmetry, the area of the whole inner loop is twice the area of the bottom part. As we just observed, the bottom part is traced out from 0 to . So the inner loop area is
In many of these polar area problems, you'll find the double angle formulas useful in doing the integrals:
Continue with the inner loop area computation:
As you can see from the picture, the top outer loop is traced out from to . By symmetry, the area inside the whole outer loop is twice the top outer loop area.
Before I do the computation, let's clear up a common confusion. People often think that in going from to on the top outer loop, the inner loop stuff is somehow "automagically" subtracted.
Well, it's not so. Think about how the area formula for polar is derived. The area is divided up into thin wedges with their vertices at the origin. The area of a typical wedge is the in the area formula.
Look at some wedges for the top outer loop. You can see that they extend from the origin through the inner loop.
Thus, when the area formula is applied to the outer loop, it finds the area from the origin all the way out to the outer loop. It doesn't matter that the inner loop is in the way.
With that out of the way, I'll compute the outer loop area. It is
(I'm not writing out the details, since they're essentially the same as those for the inner loop area.)
The area inside the outer loop and outside the inner loop is
Example. Find the area of the region inside the circle but outside the cardioid .
The region in question is the area inside the circle from 0 to minus the area inside the cardioid from 0 to .
Example. Find the area of the region outside but inside .
The curves intersect when , i.e. for . The area is
Example. Find the area of the region inside . | 1,307 | 5,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-13 | latest | en | 0.902034 |
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May 5th, 2019, 11:20 AM #1 Newbie Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 I found new solution for this problem hi every on for omega "cubic roots for 1" what do you think ?
May 5th, 2019, 01:31 PM #2 Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 I think you need to clarify what you are doing. Thanks from topsquark
May 6th, 2019, 12:13 PM #3 Newbie Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 okay the links https://math.stackexchange.com/quest...-of-2n-factors 1.jpg 2.jpg 3.jpg 4.jpg 5.jpg 6.jpg Yahia Kamal 2019
May 6th, 2019, 12:15 PM #4 Newbie Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 If you don't understand anything, see video or reply to me. Last edited by skipjack; May 6th, 2019 at 01:14 PM.
May 6th, 2019, 12:42 PM #5 Global Moderator Joined: May 2007 Posts: 6,806 Thanks: 716 Small print is an obstacle.
May 6th, 2019, 12:44 PM #6 Newbie Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 Hmm, I can't type omega or power in your forum?? You can save them and zoom. Or open in new tape and zoom. What is your opinion? Last edited by skipjack; May 6th, 2019 at 01:15 PM.
May 6th, 2019, 01:05 PM #7 Newbie Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 Please if you like my explanation hit like
May 6th, 2019, 01:14 PM #8 Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 As ω + ω² = -1, (1 - ω)(1 - ω²) = 1 - ω - ω² + ω³ = 2 + 1 = 3. Similarly, each successive pair of factors has product 3. Thanks from topsquark
May 6th, 2019, 02:08 PM #9 Newbie Joined: Oct 2017 From: Egypt Posts: 15 Thanks: 0 But why you are taking successive pairs? I think 2n means the number of our factors. Last edited by skipjack; May 6th, 2019 at 03:38 PM.
May 6th, 2019, 03:39 PM #10 Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 As there are 2n factors, there are n pairs of factors. I used successive pairs of factors because each such pair has product 3.
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https://fr.slideshare.net/cover_drive/newsvendoroptionspricing-63596708 | 1,696,106,845,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00127.warc.gz | 296,501,559 | 77,540 | # The Newsvendor meets the Options Trader
Vice President, Data Science & Optimization at Target à Target
30 Jun 2016
1 sur 7
### The Newsvendor meets the Options Trader
• 1. The Newsvendor meets the Options Trader Ashwin Rao July 23, 2017 1 Introduction This is a brief article to express my joy in a serendipity of my past life with my present. My past (14 years of it) was in the domain of Derivatives Trading, and my present is in the domain of Retail Supply Chain. Recently, I worked out that the foundational principles of derivatives pricing and the various forms of intuitive reasoning that go a long way in derivatives pricing/risk-management are highly applicable in my new world of supply optimization. Derivatives Pricing/Trading/Risk-Management is grounded in the simple concept of pricing a call/put option, and if one understands call/put option pricing really well, the entire world of derivatives can be navigated fairly comfortably. And by “understood”, I don’t mean simply knowing the formulas of call/put option pricing but internalizing the core mathematical principles and reasoning intuitively about call/put option pricing. One can make exactly the same argument with Supply Optimization - it is grounded in the newsvendor problem and much of the reasoning in advanced supply optimization has its roots in the core principles, tradeoffs and intuition involving the newsvendor problem. Interestingly, it turns out that call/put option pricing can be shown to be “equivalent” to the newsvendor problem. The parallels actually run fairly deep, eg: Markov Decision Processes and Backward Induction form the common platform for a large class of Exotic Derivatives Pricing and Stochastic Optimization problems. However, in this brief note, I will only cover the basics and do a deeper dive on the mathematical equivalences in a future paper. 2 Call/Put Option Pricing Let’s start with the forward price of a call option: E[max(x−S, 0)] = E[(x−S)+] where x is the (unknown) price of the underlying stock at option expiration and S is the strike of the call option. Likewise, the forward price of a put option is E[max(S − x, 0)] = E[(S − x)+]. Note that in this formulation, we operate in the “risk-neutral” probability measure (that is derived from the real-world probability measure using Girsanov’s Theorem - that we 1
• 2. will not delve into in this article). Let us denote the probability density function for the random variable x under this “risk-neutral” probability measure as f(·), the mean of x as µ and the variance of x as σ2. Henceforth, we will not make the “risk-neutral” probability measure explicit as we will always be operating in this probability measure. Also, note that we will integrate the random variable x from −∞ to ∞ to keep the distribution for x generic, even though in practical settings for the problems we are considering here, we may want the distribution for x to have zero support for non-positive values. Derivatives Risk-Management is all about evaluating the sensitivities of derivatives prices to various parameters (which are affectionately refered to as “greeks”). The most important “greeks” are derivatives of the option price with respect to: • An appropriate “distance”’-metric between µ and S called “moneyness” (µ − S, or log µ S , or something else - the exact form chosen depending on the distribution, for ease of math). • σ (σ is often refered to as “volatility”, when adjusted for the time-to-expiration scale). Let us focus on the derivative with respect to “moneyness”. We want to work in a distribution-free setting, so we will simply evaluate the derivative of option price with respect to S, and the derivative of option price with respect to “moneyness” can then be expressed as the ratio of the derivative of option price with respect to S and the derivative of ”moneyness’ with respect to S. So let us denote the prices of the call and put options as functions of the strike S and denote these functions as g1(S) and g2(S). g1(S) = E[(x − S)+ ] = ∞ −∞ (x − S)+ · f(x) · dx = ∞ S (x − S) · f(x) · dx g2(S) = E[(S − x)+ ] = ∞ −∞ (S − x)+ · f(x) · dx = S −∞ (S − x) · f(x) · dx These are the ubiquitous “hockey-stick” payoff expectations that appear not only all over derivatives trading, but also in various other parts of mathematical economics. Now if we have a portfolio of p call options and h put options, the price of this portfolio is: g(S) = p · ∞ S (x − S) · f(x) · dx + h · S −∞ (S − x) · f(x) · dx If we try to solve for the portfolio-common strike S that minimizes the price of this portfolio of p call options and h put options, we have the Newsvendor Optimization problem. 2
• 3. 3 Newsvendor Optimization In the Newsvendor Optimization problem, the newsvendor has to bring in the optimal quantity of newspapers in the morning that will give him the best profits. He knows that the demand for newspapers through the course of the day is a distribution with mean µ and variance σ2. For every extra newspaper he carries at the end of the day (supply S exceeds random demand x), he incurs a cost of h (effectively the difference between a single newspaper’s purchase price and disposal price). For every newspaper that a customer demands that he doesn’t carry (random demand x exceeds supply S), he incurs a cost of p (effectively the difference between a single newspaper’s sales price and purchase price), i.e., the lost revenue on each newspaper he is ”out of stock” on. So, this is the exact mathematical formulation as g(S) above and the task of the newsvendor is to identity the optimal S that minimizes g(S). So now let us return to g(S), g1(S), g2(S) and evaluate their sensitivities to S. First, we note that: g1(S)−g2(S) = ∞ S (x−S)·f(x)·dx− S −∞ (S−x)·f(x)·dx = ∞ −∞ (x−S)·f(x)·dx = µ−S This is the famous “put-call parity” that plays a big role in derivatives trading. It says that the difference between the call price and the put price (of the same strike) is equal to the difference between the underlying stock’s forward price and the strike, i.e., it’s “moneyness”. So, g2(S) − g1(S) = 1. So, we only need to know one of g1(S) or g2(S) and we automatically have the other by put-call parity (likewise, for g1(S) and g2(S)). g2(S) = S −∞ (S − x) · f(x) · dx = S · S −∞ f(x) · dx − S −∞ x · f(x) · dx Integrating the second integral by parts, we get: g2(S) = S · S −∞ f(x) · dx − (S · S −∞ f(x) · dx − S −∞ ( x −∞ f(y) · dy) · dx) = S −∞ F(x) · dx where F(x) is the cumulative distribution function for the random variable x. By put-call parity, g1(S) = µ − S + S −∞ F(x) · dx g(S) = p·(µ−S + S −∞ F(x)·dx)+h·( S −∞ F(x)·dx) = p·(µ−S)+(p+h)·( S −∞ F(x)·dx) 3
• 4. g2(S) = F(S) g1(S) = F(S) − 1 . g (S) = (p + h) · F(S) − p Setting this to 0 gives: S∗ = F−1 ( p h + p ) which is the classic formula for the optimal newsvendor supply being equal to the inverse cumulative distribution function of the “critical ratio” p p+h 4 Specializing to a Gaussian Distribution For reasoning and intuition with derivatives pricing, we view the call/put option price as a function of the “moneyness” (µ − S or a similar “distance”-metric between µ and S) and the “volatility” σ, and this would also be a good way of reasoning and intuiting about the newsvendor problem and about more general supply optimization problems. This intuition comes into play quite nicely when we assume that the underlying stock/newspaper demand is a gaussian distribution with mean µ and variance σ2. This section assumes a gaussian distribution N(µ, σ2) for the underlying stock’s price (likewise, for the demand distribution). g2(S) = S −∞ (S − x) · f(x) · dx = S −∞ (S − x) · 1 √ 2π · σ · e− (x−µ)2 2·σ2 · dx Change to integration variable y = x−µ σ and let R = S−µ σ to work with standard nor- mal distribution function φ(x) and standard cumulative normal distribution function Φ(x) (standard normal is N(0, 1)). g2(S) = R −∞ (R − y) · σ √ 2π · e−y2 2 · dy = σ · ( R √ 2π · R −∞ e−y2 2 · dy − 1 √ 2π · R −∞ y · e−y2 2 · dy) = σ · (R · Φ(R) + φ(R)) g1(S) = σ · (R · Φ(R) + φ(R) − R) 4
• 5. g(S) = σ · ((p + h) · (R · Φ(R) + φ(R)) − p · R) Since all of g1(S), g2(S), g(S) are expressed in terms of µ − S and don’t feature µ or S independently in the expressions, their partial derivative with respect to µ is just the negative of their partial derivative with respect to S. As a result, the derivative of call option price with respect to µ is 1 − F(S) = 1 − Φ(S−µ σ ) and the derivative of the put option price with respect to µ is −F(S) = −Φ(S−µ σ ) (these sensitivities of option prices with respect to the underlying stock’s forward price is refered to as the greek ”delta”, and it plays an important role in hedging derivatives, i.e., in creating portfolios that are insensitive to the underlying stock’s moves). This concept of the greek “delta” can be brought to the newsvendor world - the cor- responding idea here is that the newsvendor’s cost’s sensitivity to the expected demand µ would be: p − (p + h) · F(S) = p − (p + h) · Φ( S − µ σ ) Now, let us consider what happens at the optimal supply S∗. S∗ = F−1 ( p p + h ) = µ + Φ−1 ( p p + h ) · σ Optimal Supply S∗ decomposes into the “cycle stock” µ (forward price) and the “‘safety stock” S∗ − µ (moneyness) which depends only on σ and not on µ. g(S∗ ) = σ · (p + h) · φ( S∗ − µ σ ) = σ · (p + h) · φ(Φ−1 ( p p + h )) Note that at the optimal S∗: • The expression for the newsvendor cost g(S∗) depends only on σ, p and h (is inde- pendent of the expected demand µ). Hence, at S∗, the cost is insensitive to not just the supply, but also to the expected demand. • The expression for the price of the portfolio of p call options and h put options (g(S∗)) depends only on σ, p and h (is independent of the underlying stock’s forward price). Hence, at S∗, the portfolio price is insensitive to not just the strike, but also to the underlying stock’s forward price, Hence, it is a “market-neutral” portfolio. One is familiar with this situation of option price being independent of the underlying stock’s forward price in the case of a straddle (one call and one put option struck at a common strike) when the straddle strike is equal to the underlying stock’s forward price (“at-the-money” straddle). In the case of a normal distribution, the “at-the-money” strad- dle price is 2 · σ · φ(0) and in fact, the “at-the-money” call and put options are each priced at σ · φ(0). 5
• 6. 5 Levels of Service and their relation to Moneyness and Op- tion Price We typically talk about two types of “levels of service” for the newsvendor’s business: 1. α (or Type 1): The expected fraction of days when no “out of stocks” happen, assuming the demand is i.i.d on a set of days (equivalently, the probability that there will not be a “out of stock” on a given day). 2. β (or Type 2, sometimes called “Fill Rate”): The ratio of the expected quantity of demand that is met from supply to the expected quantity of demand. Note that α = F(S) which is equal to the “critical ratio” p p+h when S = S∗. This means the “level of service” α is equal to the probability that the put option is in-the- money (probability that the call option is out-of-the-money). Note that: β = E[min(x, S)] E[x] = S − E[(S − x)+] E[x] = S − g2(S) µ = µ − g1(S) µ = 1 − g1(S) µ So we can think of 1 − β (“unfilled rate”) as the ratio of call option price to underlying stock’s forward price. 6 Relationship between Levels of Service α and β We have seen in the previous section that β = 1 − g1(S) µ . Using the formula for g1(S) from section 3 (g1(S) = µ − S + S −∞ F(x)dx), we get: β = 1 − µ − S + S −∞ F(x)dx µ which yields: β = S − S −∞ F(x)dx µ When S is equal to the optimum S∗, we have: α = F(S∗). So, β = F−1(α) − F−1(α) −∞ F(x)dx µ This is the key formula that relates the two measures of levels of service α and β. For an arbitrary cumulative density function of demand F, this formula enables us to get α from β, or get β from α. 6
• 7. Now let us consider the case where the probability distribution of demand is a gaussian distribution N(µ, σ2). We have seen in section 4 that g1(S) = σ · (R · Φ(R) + φ(R) − R) where R = S−µ σ . So, β = 1 − σ · (R · Φ(R) + φ(R) − R) µ where φ(x) is the standard normal distribution function and Φ(x) is the standard cumulative normal distribution function (standard normal is N(0, 1)) We simplify this expression using the fact that when S = S∗, R = Φ−1(α). This yields: β = 1 − σ µ · (φ(Φ−1 (α)) − (1 − α) · Φ−1 (α)) where φ(x) = 1√ 2π e −x2 2 This is a rather useful formula that indicates that the relation- ship between α and β depends on the coefficient of variation σ µ . In fact, this formula helps us conceptualize 1 − β (“unfilled rate”) as the product of two intuitive quantities: • The Coefficient of Variation σ µ • The Normalized Option Price for Strike Fractile α (which is equal to φ(Φ−1(α)) − (1 − α) · Φ−1(α)). This quantity is the option price when the distribution is standard normal N(0, 1) and the strike is expressed as the fractile α (i.e., Strike = Φ−1(α)). This is because the Normalized Option Price for Strike Fractile α is ∞ Φ−1(α)(x − Φ−1(α)) · φ(x) · dx which (as we worked out in section 4) equals φ(Φ−1(α)) − (1 − α) · Φ−1(α) 7 | 3,777 | 13,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-40 | latest | en | 0.89924 |
https://nl.mathworks.com/matlabcentral/cody/problems/2779-rule-of-mixtures-composites-weighted-bound/solutions/2160658 | 1,601,248,737,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00784.warc.gz | 511,746,083 | 16,914 | Cody
# Problem 2779. Rule of mixtures (composites) - weighted bound
Solution 2160658
Submitted on 15 Mar 2020 by Asif Newaz
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### Test Suite
Test Status Code Input and Output
1 Pass
Ef = 100; Em = 10; ff = 0.30; wt = 0.25; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 19.5240) < 1e-4)
2 Pass
Ef = 100; Em = 10; ff = 0.30; wt = 0.50; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 25.3493) < 1e-4)
3 Pass
Ef = 100; Em = 10; ff = 0.30; wt = 0.75; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 31.1747) < 1e-4)
4 Pass
Ef = 100; Em = 10; ff = 0.15; wt = 0.25; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 14.5455) < 1e-4)
5 Pass
Ef = 100; Em = 10; ff = 0.15; wt = 0.50; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 17.5303) < 1e-4)
6 Pass
Ef = 100; Em = 10; ff = 0.15; wt = 0.75; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 20.5152) < 1e-4)
7 Pass
Ef = 1000; Em = 10; ff = 0.30; wt = 0.25; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 87.4186) < 1e-4)
8 Pass
Ef = 1000; Em = 10; ff = 0.30; wt = 0.50; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 160.6124) < 1e-4)
9 Pass
Ef = 1000; Em = 10; ff = 0.30; wt = 0.75; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 233.8062) < 1e-4)
10 Pass
Ef = 1000; Em = 10; ff = 0.15; wt = 0.25; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 48.4330) < 1e-4)
11 Pass
Ef = 1000; Em = 10; ff = 0.15; wt = 0.50; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 85.1220) < 1e-4)
12 Pass
Ef = 1000; Em = 10; ff = 0.15; wt = 0.75; [Ec] = rule_of_mixtures_wt_bound(Ef,Em,ff,wt); assert(abs(Ec - 121.8110) < 1e-4) | 874 | 1,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-40 | latest | en | 0.71075 |
http://joyceh1.blogspot.com/2016/02/day-98-slope-rounding-decimals-factoring.html | 1,532,263,144,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593223.90/warc/CC-MAIN-20180722120017-20180722140017-00158.warc.gz | 185,729,260 | 17,538 | ## Monday, February 8, 2016
### Day 98: Slope, Rounding Decimals & Factoring
Today's estimation was the amount of juju hearts in an 11 ounce package. Some students used the visual clues of the ones they could see through the bag, and I really liked when a student said that each one is about 4 grams, they rounded 311 grams to 320, then divided by 4 and got 80. That was really sound reasoning.
I was ambitious and wanted my students to finish most of the 7.2.2 slope lesson in one period. We reviewed everything except the last problem about hiccups with 3 different representations.
I started students by making a table to organize their ranks, fractions, and decimals for easy reference. Students first ranked lines based on their appearance, 1 being the fastest and 4 being the slowest. Students saw that A and C were parallel lines so they were the same speed so could be 2nd or 3rd place for a tie.
Then students discussed how the slope triangle for A was labeled and where the 4 and 6 came from. A few students quickly say it meant 4 kilometers per 6 seconds. I asked them how they knew and they said they looked at the axes. The rise was the y axis and the run or base of the triangle was the x axis or seconds.
Then when they were prompted for unit rate, I said if you're going down the highway you don't say I'm going 4/6. What's missing? They reasoned it was 4/6 kilometers per second.
When I saw walking around the room how many different answers students got for writing 4/6 as a repeating decimal, I definitely made it a point of discussion in all classes. Some even switched the numerator and denominator and said it was 1.5. I said is 4/6 less than or greater than 1? Is 1.5 less than or greater than 1? Then they figured out their issue.
For rounding 2/3, students reasoned it was 0.6 with a repeating bar. Some students wrote 0.66. Some 0.7. And finally some had 0.67. Students who got the right answer had a hard time articulating. Students eventually said you look at the place you're rounding to, the hundredths, and look at the 6 to the right of it. That number is 5 or greater, so the 6 in the hundredths place rounds up. I asked them what 0.7 was rounded to and they said to the tenths place.
Here is how I wanted their tables to start.
Here you can see all the work of rounding to the nearest hundredth.
I also liked this students reasoning. He said there was about 24 candies in the see through section. About 3 of those sections, so 24 times 3 is 72. You also see another student who thought 311 divided by 2.
Here I wrote the students different answers for rounding to analyze.
In accelerated I was concerned about half the class being confused about fractional exponents. So, I wanted to review the vocabulary in detail with students.
First I discussed the names of a radical. The symbol is the radical symbol. The number inside the radical is the "radicand." The index number is which root it is. I didn't print out the foldables from Math Equals Love but I used it as a reference. I love that she puts the asterisk to mention of there's no index number then the index is automatically a 2. I discussed how that is read, and the cube root if the index was 3.
As you can see, I tried color coding 16 to the 5/2 power. The 5 represented the exponent and the denominator 2 represented the index of the radical symbol. We even discussed the commutative property in that 16^5/2 is the same as (16^1/2)^5 and (16^5)^1/2 meaning the exponent of 5 can be on the radicand or it can be outside the radical's parentheses. They told me how to solve 27^3/2 power as well.
Then with 10 minutes to go we reviewed how to start, and complete the generic rectangle with the aid of a diamond. Then 4 example problems were on the board and 4 different student volunteers completed them. We only had time for them to present 2 of them. | 935 | 3,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-30 | latest | en | 0.984715 |
https://petiterepublic.com/needlewomen/how-many-10mm-beads-makes-a-necklace.html | 1,660,206,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00285.warc.gz | 438,978,386 | 19,228 | # How many 10mm beads makes a necklace?
Contents
## How many 12mm beads make a necklace?
How many Beads do I need for my project?
Bead Size 7 inch Bracelet 8 inch Bracelet
7mm 3.5 24
8mm 3.25 22
10mm 2.5 18
12mm 2 15
## What size beads are best for necklaces?
The ideal size for necklaces are 8mm, which create dramatic pieces without being too bulky and heavy. 10mm size beads – This size is nice for bracelets, although can be very heavy on the wrist if only working with gemstones.
## How do you calculate the number of beads?
Divide this total length in millimeters (in this case, 406.4mm), by the diameter of the beads to be used (let’s use 4mm), and the result will be your number of beads. In our example, 406.4mm divided by 4mm = 101.6 beads.
## How many beads make a necklace?
18-inch princess necklace made with 4mm beads: 18 x 25.4 ÷ 4 = 114.3. 32-inch opera necklace made with 10mm beads: 32 x 25.4 ÷ 10 = 81.2.
How Many Beads Will You Need To Make a Necklace?
IT IS INTERESTING: What fabrics were used to make medieval clothes?
Type Length
Collar 14 to 17 inches
Choker 16 to 18 inches
Princess 17 to 19 inches
Matinee 20 to 25 inches
1″ 12″
12mm 2.12 25.4
13mm 1.95 23.45
14mm 1.81 21.77
15mm 1.69 20.32
## How many 8mm beads are in 8 inches?
Number of Beads Per 16-inch Strand.
8mm 50
10mm 41
12mm 34
Hearts
## How many seed beads do I need?
Always buy more seed beads than you think you will need. There are lots of creative ways to use up leftover beads. There are several tables below to help you calculate the number of seed beads you need.
## How many inches should a beaded bracelet be?
What is the standard bracelet size for women? For women’s charm and multi-link bracelets, the standard length is 6 ½ to 7 ½ inches. A bracelet of 7 inches is the most common in women’s jewelry. Women’s bangle and cuff bracelets usually are 7 inches in length with a 2 ½ inch diameter.
## How long should a beaded necklace be?
22” to 24”– even looser lengths, these suit a more casual style of necklace. They will usually sit below a low neckline. Over 24” and up to 36” – this is a long length which drapes nicely to the mid section, perfect for long beaded necklaces. At 36”, it’s usually possible to wrap into a double layered shorter necklace.
IT IS INTERESTING: How do you make a yarn doll for kids?
## How many MM is a 6 0 bead?
Czech seed beads in size 6/0 (approximately 4.1mm) are available in a number of different color categories below. The hole diameter is approximately 1.0mm. 6/0 seed beads are one of the most popular sizes.
## How many grams is a size 10 seed bead?
10/0 Delica beads can be either smooth or six-sided (hex cut), and measure 2.3mm x 1.2mm with a hole size of 1mm. Count is approximately 108 beads per gram.
## How many beads are on a rosary?
Roman Catholics use the Rosary (Latin “rosarium”, meaning “rose garden”) with 59 beads. However, Eastern Orthodox Christians use a knotted prayer rope called either a komboskini or chotki, with 100 knots, although prayer ropes with 50 or 33 knots can also be used. | 884 | 3,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-33 | latest | en | 0.926481 |
https://s-space.snu.ac.kr/handle/10371/120608 | 1,624,546,911,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556133.92/warc/CC-MAIN-20210624141035-20210624171035-00309.warc.gz | 434,269,089 | 6,948 | ## Browse
Finite Element Solutions for the Space Fractional Diffusion Equation with a Nonlinear Source Term
비선형 실수차수의 확산방정식에 대한 유한요소법
Cited 0 time in Web of Science Cited 0 time in Scopus
Authors
최영주
정상권
Major
사범대학 수학교육과
Issue Date
2016-02
Publisher
서울대학교 대학원
Keywords
anomalous diffusionfractional derivativeGalerkin finite element method
Description
학위논문 (박사)-- 서울대학교 대학원 : 수학교육과, 2016. 2. 정상권.
Abstract
The anomalous diffusion problem has been played a significant role in many areas. In this paper, we consider finite element Galerkin solutions for the space fractional diffusion equation with a nonlinear source term. We derive the variational formula of the semi-discrete scheme by using the Galerkin finite element method in space. Existence of the semi-discrete solution for the equation is shown. The stability and the order of convergence of approximate solutions for the semi-discrete equation have been also discussed.
Furthermore, we derive the fully discrete time-space variational formulation using the backward Euler method. Existence of numerical solutions for the backward Euler fully discrete scheme is shown by using the Brouwer fixed point theorem. The stability and error estimates of solutions for the fully discrete approximate solutions are studied along the lines of the semi-discrete analysis. The order of convergence are obtained as $O(k+h^{\tilde{\gamma}})$, where $\tilde \gamma$ is a constant depending on the order of fractional derivative.
Numerical computations are presented, which confirm the theoretical results when the equation has a linear source term. When the equation has a nonlinear source term, numerical results show that the diffusivity depends on the order of fractional derivative as we discuss in theoretical analysis.
A part of thesis has been published in Abstract and Applied Analysis, 2012, doi:10.1155/2012/596184.
Language
English
URI
https://hdl.handle.net/10371/120608
Files in This Item:
Appears in Collections:
College of Education (사범대학)Dept. of Mathematics Education (수학교육과)Theses (Ph.D. / Sc.D._수학교육과) | 490 | 2,070 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-25 | latest | en | 0.852927 |
https://blairbash.org/search?tag=cuadrado | 1,701,822,272,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00068.warc.gz | 157,900,459 | 5,527 | # #12354
04
Oct. 19, 2023, 12:13 p.m.
⚐ Report
cuadrado: i want to be christian, not eaten
# #12352
1111
Oct. 19, 2023, 9:58 a.m.
⚐ Report
//mod note: cf 10946
# #12138
88
Sept. 15, 2023, 4:08 p.m.
⚐ Report
//chaotic rose anthology, september 15 "I saw, in my peripheral vision, you weird people being weird, and I've decided to ignore it." "How are you doing? Say 'good'." "Even if you teleported here three weeks ago ..." "When I first came to this job, they had me teach Magnet Geometry. So I figured the students should learn logic first, as part of Geometry. But I wouldn't use the book, because the way the book does it is terrible. So I taught them Fitch-style proofs -- the 9th-graders. Their brains swelled to enormous size, and they were really smart, and when they did proofs later in Geometry, they thought they were easy. The packet you're doing now -- I had 9th-graders do it." "There's only a handful of rules [of Fitch-style proofs], or maybe two handfuls." "[Fitch-style proofs] are a great thing to do when you're bored, like on the bus, or when they take your phone away." "You should go to Sra Cuadrado's class and do these. If you don't have her class, just go to her classroom and do it there anyway." "I will assume, because you all are great students, that you all read -- at least twice -- my write-ups of class, which are often better than class itself." "[The logical law of explosion] is like if I have two halves of a hand grenade." "Are there more slides? I don't want there to be."
#1: certain students were holding raw spaghetti from their mouths, #3: cf 11278, #7: cf 8702
# #11696
88
March 17, 2023, 9:45 a.m.
⚐ Report
// Pd.7 Spanish 5 Cuadrado: "I'm like God, I see everything."
# #10712
2020
June 16, 2022, 1:57 p.m.
⚐ Report
// Lockdown Student: Apparently a student punched a teacher. Cuadrado: Good. If you do that to me I will punch back!
//mod note: oooooooooo
# #10688
1111
June 13, 2022, 12:06 p.m.
⚐ Report
Student, walking in to the classroom: Wow, there's nobody in here! Cuadrado: So I am nobody?
# #10663
2424
June 7, 2022, 2:47 p.m.
⚐ Report
//students walks in 5 minutes late for Cuadrado 9th Student: This an apology from Mr. Rose. *hands her a cookie* Cuadrado: (translation) You think I want a cookie from Rose?
She still ate it tho
# #10613
1818
May 24, 2022, 2:57 p.m.
⚐ Report
Sudhish: So she (Sra. Cuadrado) isn't going to be here for the rest of the week? Sub: Yeah. Sudhish: Let's goooooo!!! Sub: I'll put a star next to your name!
# #10391
-19
April 1, 2022, 12:40 p.m.
⚐ Report
Cuadrado: Son gallinas. Chicken chicken, coward coward.
We wanted subtitles….
# #10211
711
March 8, 2022, 7:20 p.m.
⚐ Report
// Cuadrado getting fed up with spanish 5 class Yaphet: Aquí estaremos hasta que... (we will be here until...) Cuadrado: APRENDAS EL SUBJUNTIVO! (YOU LEARN THE SUBJUNCTIVE) | 913 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | latest | en | 0.971162 |
https://docs.microsoft.com/zh-tw/dotnet/api/system.decimal.toint32?view=netframework-4.7.2 | 1,568,593,372,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572439.21/warc/CC-MAIN-20190915235555-20190916021555-00247.warc.gz | 458,524,264 | 11,383 | # Decimal.ToInt32(Decimal)Decimal.ToInt32(Decimal)Decimal.ToInt32(Decimal)Decimal.ToInt32(Decimal) Method
## 定義
``````public:
static int ToInt32(System::Decimal d);``````
``public static int ToInt32 (decimal d);``
``static member ToInt32 : decimal -> int``
``Public Shared Function ToInt32 (d As Decimal) As Integer``
#### 參數
d
Decimal Decimal Decimal Decimal
#### 傳回
32 位元帶正負號的整數，等於 `d` 的值。A 32-bit signed integer equivalent to the value of `d`.
#### 例外狀況
`d` 小於 MinValue 或大於 MaxValue`d` is less than MinValue or greater than MaxValue.
## 範例
``````using System;
class Example
{
public static void Main( )
{
decimal[] values = { 123m, new decimal(123000, 0, 0, false, 3),
123.999m, 4294967295.999m, 4294967296m,
4294967296m, 2147483647.999m, 2147483648m,
-0.999m, -1m, -2147483648.999m, -2147483649m };
foreach (var value in values) {
try {
int number = Decimal.ToInt32(value);
Console.WriteLine("{0} --> {1}", value, number);
}
catch (OverflowException e)
{
Console.WriteLine("{0}: {1}", e.GetType().Name, value);
}
}
}
}
// The example displays the following output:
// 123 --> 123
// 123.000 --> 123
// 123.999 --> 123
// OverflowException: 4294967295.999
// OverflowException: 4294967296
// OverflowException: 4294967296
// 2147483647.999 --> 2147483647
// OverflowException: 2147483648
// -0.999 --> 0
// -1 --> -1
// -2147483648.999 --> -2147483648
// OverflowException: -2147483649
``````
``````Module Example
Public Sub Main()
Dim values() As Decimal = { 123d, New Decimal(123000, 0, 0, false, 3),
123.999d, 4294967295.999d, 4294967296d,
4294967296d, 2147483647.999d, 2147483648d,
-0.999d, -1d, -2147483648.999d, -2147483649d }
For Each value In values
Try
Dim number As Integer = Decimal.ToInt32(value)
Console.WriteLine("{0} --> {1}", value, number)
Catch e As OverflowException
Console.WriteLine("{0}: {1}", e.GetType().Name, value)
End Try
Next
End Sub
End Module
' The example displays the following output:
' 123 --> 123
' 123.000 --> 123
' 123.999 --> 123
' OverflowException: 4294967295.999
' OverflowException: 4294967296
' OverflowException: 4294967296
' 2147483647.999 --> 2147483647
' OverflowException: 2147483648
' -0.999 --> 0
' -1 --> -1
' -2147483648.999 --> -2147483648
' OverflowException: -2147483649
'
`````` | 809 | 2,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-39 | longest | en | 0.233447 |
https://www.geeksforgeeks.org/sql-arithmetic-operators/?ref=lbp | 1,708,667,123,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00246.warc.gz | 791,305,449 | 57,064 | # SQL | Arithmetic Operators
Prerequisite: Basic Select statement, Insert into clause, Sql Create Clause, SQL Aliases We can use various Arithmetic Operators on the data stored in the tables. Arithmetic Operators are:
```+ [Addition]
- [Subtraction]
/ [Division]
* [Multiplication]
% [Modulus]```
It is used to perform addition operation on the data items, items include either single column or multiple columns. Implementation:
```SELECT employee_id, employee_name, salary, salary + 100
AS "salary + 100" FROM addition;
```
Output:
employee_id employee_name salary salary+100
1 alex 25000 25100
2 rr 55000 55100
3 jpm 52000 52100
4 ggshmr 12312 12412
Here we have done addition of 100 to each Employee’s salary i.e, addition operation on single column. Let’s perform addition of 2 columns:
```SELECT employee_id, employee_name, salary, salary + employee_id
AS "salary + employee_id" FROM addition;
```
Output:
employee_id employee_name salary salary+employee_id
1 alex 25000 25001
2 rr 55000 55002
3 jpm 52000 52003
4 ggshmr 12312 12316
Here we have done addition of 2 columns with each other i.e, each employee’s employee_id is added with its salary.
#### Subtraction (-) :
It is use to perform subtraction operation on the data items, items include either single column or multiple columns. Implementation:
```SELECT employee_id, employee_name, salary, salary - 100
AS "salary - 100" FROM subtraction;
```
Output:
employee_id employee_name salary salary-100
12 Finch 15000 14900
22 Peter 25000 24900
32 Warner 5600 5500
42 Watson 90000 89900
Here we have done subtraction of 100 to each Employee’s salary i.e, subtraction operation on single column. Let’s perform subtraction of 2 columns:
```SELECT employee_id, employee_name, salary, salary - employee_id
AS "salary - employee_id" FROM subtraction;
```
Output:
employee_id employee_name salary salary – employee_id
12 Finch 15000 14988
22 Peter 25000 24978
32 Warner 5600 5568
42 Watson 90000 89958
Here we have done subtraction of 2 columns with each other i.e, each employee’s employee_id is subtracted from its salary. Division (/) : For Division refer this link- Division in SQL
#### Multiplication (*) :
It is use to perform multiplication of data items. Implementation:
```SELECT employee_id, employee_name, salary, salary * 100
AS "salary * 100" FROM addition;
```
Output:
employee_id employee_name salary salary * 100
1 Finch 25000 2500000
2 Peter 55000 5500000
3 Warner 52000 5200000
4 Watson 12312 1231200
Here we have done multiplication of 100 to each Employee’s salary i.e, multiplication operation on single column. Let’s perform multiplication of 2 columns:
```SELECT employee_id, employee_name, salary, salary * employee_id
AS "salary * employee_id" FROM addition;
```
Output:
employee_id employee_name salary salary * employee_id
1 Finch 25000 25000
2 Peter 55000 110000
3 Warner 52000 156000
4 Watson 12312 49248
Here we have done multiplication of 2 columns with each other i.e, each employee’s employee_id is multiplied with its salary.
#### Modulus ( % ) :
It is use to get remainder when one data is divided by another. Implementation:
```SELECT employee_id, employee_name, salary, salary % 25000
AS "salary % 25000" FROM addition;
```
Output:
employee_id employee_name salary salary % 25000
1 Finch 25000 0
2 Peter 55000 5000
3 Warner 52000 2000
4 Watson 12312 12312
Here we have done modulus of 100 to each Employee’s salary i.e, modulus operation on single column. Let’s perform modulus operation between 2 columns:
```SELECT employee_id, employee_name, salary, salary % employee_id
AS "salary % employee_id" FROM addition;
```
Output:
employee_id employee_name salary salary % employee_id
1 Finch 25000 0
2 Peter 55000 0
3 Warner 52000 1
4 Watson 12312 0
Here we have done modulus of 2 columns with each other i.e, each employee’s salary is divided with its id and corresponding remainder is shown. Basically, modulus is use to check whether a number is Even or Odd. Suppose a given number if divided by 2 and gives 1 as remainder, then it is an odd number or if on dividing by 2 and gives 0 as remainder, then it is an even number.
#### Concept of NULL :
If we perform any arithmetic operation on NULL, then answer is always null. Implementation:
```SELECT employee_id, employee_name, salary, type, type + 100
```
Output:
employee_id employee_name salary type type + 100
1 Finch 25000 NULL NULL
2 Peter 55000 NULL NULL
3 Warner 52000 NULL NULL
4 Watson 12312 NULL NULL
Here output always came null, since performing any operation on null will always result in a null value. Note: Make sure that NULL is unavailable, unassigned, unknown. Null is not same as blank space or zero. To get in depth understanding of NULL, refer THIS link.
References: Oracle Docs
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https://oguchionyewu.com/elmuwT16793/mdqwje16796/ | 1,631,903,373,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00211.warc.gz | 494,572,730 | 8,449 | oguchionyewu
# Absolute Value Equations Worksheet Multi Step Worksheets Above Algorithm Given Function Evaluations
Title
## Absolute Value Equations Worksheet Multi Step Worksheets Above Algorithm Given Function Evaluations
Include In Article
### Absolute Value Worksheets
FormatJPG
Width1092px
Height1285px
Published DateWednesday , December 09th 2020
#### Rate This : Absolute Value Equations Worksheet Multi Step Worksheets Above Algorithm Given Function Evaluations
35 out of 100 based on 147 user ratings
Coloring Pages. Wednesday , December 09th 2020.
Quote from Absolute Value Worksheets :
People who are considering art or find that they have a potential or interest in learning how to draw and color find that this new hobby interest is helping and building their confidence to advance their art training. The first book we purchased had 39 pictures of mandala drawings. A small box of basic coloring pencils started our adventure. We worked with the primary colors of red, orange, blue, green, violet. black, white, and a few other tints which were in the small box. The initial starting stage began with the techniques of our elementary school years. The instructions included the color wheel which we started paying attention to after getting bored with the basic colors.
As children grow older they will start opening up and talking more as you color with them. Parents who just hand out coloring pages for kids and go off to do other things are missing a valuable bonding opportunity. Most children will start to talk more openly once they are distracted with something to color. Coloring is a great time to ask your children about their day or to bring up topics which may be bothering them but they are hesitant to talk about at other times.
Help improves Concentration, In children, you can speed up their concentration process by giving them time-long activities to do, of which educational coloring pages is chief. When children sit for long coloring pages to print, it helps the child keep his on one thing and will certainly develop his overall concentration level as time goes on.
The simple act of coloring is the beginning of gaining tinting knowledge with personal experiences of school day memories and trial and error. As you follow suggestions for cool and warm colors one begins to understand how colors offer contrast or compliment each other.
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What do LBL's 1 nm transistors mean?
In the spirit of this post, it seems like it would be a good idea to write something about this paper (accompanying LBL press release), particularly when popular sites are going a bit overboard with their headlines ("The world's smallest transistor is 1nm long, physics be damned"). (I discuss most of the background in my book, if you're interested.)
What is a (field effect) transistor and how does it work? A transistor is an electronic switch, the essential building block of modern digital electronics. A field-effect transistor (FET) has three terminals - a "source" (an input), a "drain" (an output) on either side of a semiconductor "channel", and a "gate" (a control knob). If you think of electrical current like fluid flow, this is like a pipe with an inlet, and outlet, and a valve in the middle, and the gate controls the valve. In a "depletion mode" FET, the gate electrode repels away charges in the channel to turn off current between the source and drain. In an "accumulation mode" FET, the gate attracts mobile charges into the channel to turn on current between the source and drain. Bottom line: the gate uses the electrostatic interaction with charges to control current in the channel. There has to be a thin insulating layer between the gate and the channel to keep current from "leaking" from the gate. People have had to get very clever in their geometric designs to maximize the influence of the gate on the charges in the channel.
What's the big deal about making smaller transistors? We've gotten where we are by cramming more devices on a chip at an absurdly increasing rate, by making transistors smaller and smaller. One key length scale is the separation between source and drain electrode. If that separation is too small, there are at least two issues: Current can leak from source to drain even when the device is supposed to be off because the charge can tunnel; and because of the way electric fields actually work, it is increasingly difficult to come up with a geometry where the gate electrode can efficiently (that is, with a small swing in voltage, to minimize power) turn the FET off and on.
What did the LBL team do? The investigators built a very technically impressive device, using atomically thin MoS2 as the semiconductor layer, source and drain electrodes separated by only seven nm or so, a ZrO2 dielectric layer only a couple of nm thick, and using an individual metallic carbon nanotube (about 1 nm in diameter) as the gate electrode. The resulting device functions quite well as a transistor, which is pretty damn cool, considering the constraints involved. This fabrication is a tour de force piece of work.
Does this device really defy physics in some way, as implied by the headline on that news article? No. That headline alludes to the issue of direct tunneling between source and drain, and a sense that this is expected to be a problem in silicon devices below the 5 nm node (where that number is not the actual physical length of the channel). This device acts as expected by physics - indeed, the authors simulate the performance and the results agree very nicely with experiment.
If you read the actual LBL press release, you'll see that the authors are very careful to point out that this is a proof-of-concept device. It is exceedingly unlikely (in my opinion, completely not going to happen) that we will have chips with billions of MoS2 transistors with nanotube gates - the Si industry is incredibly conservative about adopting new materials. If I had to bet, I'd say it's going to be Si and Si/Ge all the way down. (You will very likely need to go away from Si if you want to see this kind of performance at such length scales, though.) Still, this work does show that with proper fabrication and electrostatic design, you can make some really tiny transistors that work very well!
Anonymous said...
Nifty device, but the length that really matters for transistors is the channel length, not the gate length. If you look at their results, the channel length in the off state is still of the order of 5nm or even more - they cook up some "LEFF" measurement of the channel length ~ 3.9nm, which happens to be a bit less than 5nm - but in the end the gateable part of the material is even much longer than that. As you point out, Doug, there is no physics miracle here.
If an engineer wanted integrate this device into a processor, he or she would need to self-align the source and drain contacts to the nanotube, with some spacer to limit tunneling leakage current, and the source and drain would still end up having to be separated by more than 5nm.
"1nm" in the title is a bit misleading, IMHO. A 5nm-wide nanotube or nanowire for the gate would do just as good of a job, with no loss in scaling dimension, as long as the channel is made out of an ultrathin conductive material.
Technologically speaking, I think you also need excellent control of the nanotube species in order to get identical and predictable threshold voltages (this is the age-old problem with CNTs).
Douglas Natelson said...
Anon, I agree with you. An eventual scalable way to do this would (i) use Si or SiGe as the channel, because CMOS cost structure makes essentially everything else noncompetitive out of the gate economically; (ii) use some kind of patterned metallic silicide as the gate electrode, because there is no way they're really going to put metallic or very tiny bandgap nanotubes down as gates - perhaps something like these: http://scitation.aip.org/content/aip/journal/jap/91/5/10.1063/1.1428807 but that's a huge stretch; (iii) use some kind of metal or metallic silicide source and drain electrodes, because doping won't work with this kind of spatial resolution and would lead to huge variations in contact resistance and threshold voltage; (iv) probably use some finFET or other multigate configuration to improve gate coupling.
Anonymous said...
Also, the effective channel length, i.e. the length of the active region may only be about 7 nm (since the global back gate turns the whole thing on or off, so the tube just has to switch a small region from insulating to conducting), but the separation between the source and drain electrodes is almost a micron (fig 1c), which is pretty huge. I dunno, I am a bit less impressed by this paper as I feel like I have seen similar work before that was just hyped less.
Anonymous said...
The most interesting aspect of this transistor, from the point of view of a development engineer, is actually that the small gate will have a lower capacitance to the source and drain contacts. S/D contacts will have to be drawn very close to the channel in a manufacturable device, and one of the biggest issues with scaling nowadays is that the proximity of source and drain to the gate causes a large stray capacitance, which slows down transistor switching. The gate nowadays also needs to be relatively tall (~100nm!), in order to fill it with some low-resistance metal, and this also increases capacitance. So a highly conductive nanotube is a much better gate in that respect, if it can be made reproducibly and positioned precisely.
What I don't know is whether this is the first work that reports on a MOS device gated by a nanotube (I am *not* referring to a "CNT FET"). My guess is they are probably not the first.
(Anon #1)
Anonymous said...
Anon#1 again here. One more comment: what determines the actual channel length here is simply the oxide thickness. If the oxide is 5nm thick, the channel will be ~5nm long; a 2nm channel could simply be obtained by depositing a 2nm-thin oxide, etc.
It's interesting that the article shows no data for the thin-oxide, short-channel devices, presumably because they have a bad ION/IOFF ratio (as they should).
To respond to Anon#2's comment: the authors drew source and drain contacts a couple of microns apart because this is a proof-of-concept device; it should be possible, in principle, to bring those contacts much closer to the channel, by some self-alignment technique. For example, you could revert the process and lay the nanotube last, then coat it with some thin insulator (a half-"sheath") than only deposits on the nanotube, then deposit the S/D metal, and lift it off the nanotube by etching its oxide sheath, thereby separating the source and drain contact areas. Done!
Anonymous said...
I saw tens of people shared this news/paper on LinkedIn. Our Assoc. Dean for research even sent me an email about this. I am sorry but this is just another great PR of a big shot at a big university.
Just a short story of mine. Five years ago I hired a postdoc, who told me that he was an "expert" of graphene synthesize. I asked him to fabricate a simple design. He asked for four weeks. I was like "what the heck do you need a month to fabricate such a simple device". He was like "i can fabricate it in 2 hours but finding the one which works takes a month"
Anonymous said...
Funny story. Even though it is currently very tedious to make devices out of graphene and other layered materials, their promise is predicated on the hope that, one day, we will be able to grow them as a film of controlled thickness on a large wafer.
Anonymous said...
At first, I have to say that I am not an expert in FET. What I have learned on FET was only from my class in college.
The authors selected a "bad" (low carrier mobility) material, MoS2 as the channel for avoiding tunneling leakage. Now the tiny device can be turn On/Off, but it is not faster, since electrons transport slower. Even we can make billions of such devices on a chip in the future, is it really useful? I doubt. | 2,209 | 9,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-21 | latest | en | 0.938186 |
http://conversion.org/time/month-synodic/century | 1,643,216,846,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00719.warc.gz | 16,818,737 | 7,011 | # month (synodic) to century conversion
Conversion number between month (synodic) [mo] and century [c] is 0.00080852006556273. This means, that month (synodic) is smaller unit than century.
### Contents [show][hide]
Switch to reverse conversion:
from century to month (synodic) conversion
### Enter the number in month (synodic):
Decimal Fraction Exponential Expression
[mo]
eg.: 10.12345 or 1.123e5
Result in century
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 month (synodic) = (2551442.890) / (3155695200) = 0.00080852006556273 century
• 1 century = (3155695200) / (2551442.890) = 1236.8276838052 month (synodic)
• ? month (synodic) × (2551442.890 ("s"/"month (synodic)")) / (3155695200 ("s"/"century")) = ? century
### High precision conversion
If conversion between month (synodic) to second and second to century is exactly definied, high precision conversion from month (synodic) to century is enabled.
Since definition contain rounded number(s) too, there is no sense for high precision calculation, but if you want, you can enable it. Keep in mind, that converted number will be inaccurate due this rounding error!
### month (synodic) to century conversion chart
Start value: [month (synodic)] Step size [month (synodic)] How many lines? (max 100)
visual:
month (synodic)century
00
100.0080852006556273
200.016170401311255
300.024255601966882
400.032340802622509
500.040426003278137
600.048511203933764
700.056596404589391
800.064681605245019
900.072766805900646
1000.080852006556273
1100.088937207211901
Copy to Excel
## Multiple conversion
Enter numbers in month (synodic) and click convert button.
One number per line.
Converted numbers in century:
Click to select all
## Details about month (synodic) and century units:
Convert Month (synodic) to other unit:
### month (synodic)
Definition of month (synodic) unit: ≈ 29.530589 days. Synodic month = Cycle time of moon phases (example, from fool moon to next fool moon). This period is not constant and it is longer than moon rotation around Earth (sidereal month), because moon and Earth moves together around the Sun.
Convert Century to other unit:
### century
Definition of century unit: ≡ 100 years (Gregorian). = 365.2425 × 100 × 86400 seconds
← Back to Time units | 656 | 2,327 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-05 | latest | en | 0.68938 |
http://techblog.ctgclean.com/2012/02/water-hard-water-vs-soft-water/ | 1,508,758,131,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825900.44/warc/CC-MAIN-20171023111450-20171023131450-00560.warc.gz | 317,453,040 | 12,961 | # Water – Hard Water vs. Soft Water
Just about everyone is familiar with the term “hard” water but few know what it really means. In this blog I’ll try to explain what “hard” water is and why we prefer water that is “soft.” In the blog It’s Time To Talk About Water – Again, items 2 and 3 in the list of water contaminants refer to soft and hard water. “Soft” water contains primarily readily soluble compounds while “hard” water contains compounds that are not readily dissolved in water. “Hard” water creates the “lime” or “scale” deposits found in pipes. Scale deposits not only reduce flow but also constitute a potential risk to equipment as particles of the hard water deposits are dislodged and enter the water supply. Hard water also reacts with detergents and soap to cause a “scum” rather than bubbles. This is similar to the soap scum found in residential bathtubs, wash basins and showers.
Water hardness is measured in units called “grains per gallon” of dissolved solids or just “grains of hardness.” Calcium and Magnesium are the target minerals as these are the minerals common in water that form “insoluble” compounds.
Interesting Note – A grain is a unit of weight which originated as the weight of a single grain of a cereal. Although grain it is no longer widely used as a unit of measure, gunpowder is still commonly measured in grains and, historically, a dose of aspirin was frequently measured in grains. The common aspirin dose was 5 grains which relates directly to today’s single tablet dose of 325 milligrams.
One grain equals 64.799 (call it 64.8) milligrams. Doing the math, grains per gallon (gpg) converts to mg/liter by the equation –
milligrams per liter = grains per gallon times 17.1
1 grain of hardness = 17.1 milligrams per liter
Many instruments commonly used in cleaning systems measure contaminants dissolved in water in the units “parts per million” which is ratio of the weight of the dissolved insoluble (really partially soluble) contaminants to the weight of the water in which they are dissolved. Parts per million (ppm) is the same as milligrams per liter since water weighs one gram per milliliter. The measure parts per million is much more commonly used than grains per gallon. It is important to know that they are both measuring the same thing but using different units.
How Hard Is My Water? –
Almost all water except for water that has been specifically treated to remove them contains dissolved minerals in some quantity. In general, the following descriptions apply.
• Soft Water = less than 1 grain per gallon (0 to 17 milligrams per liter or ppm)
• Slightly Hard Water = 1 to 3.5 grains per gallon (17 to 60 mg/liter)
• Moderately Hard Water = 3.5 to 7.0 grains per gallon (60 to 120 mg/liter)
• Hard Water = 7.0 to 10.5 grains per gallon (120 to 180 mg/liter)
• VERY Hard Water = greater than 10.5 grains per gallon (greater than 180 mg/liter)
Water from a municipal supply usually comes with a pedigree (of sorts) which is available from local authorities and should specify the hardness. In fact, the water may already be treated to reduce certain minerals while others are added for public health and other reasons. If your water supply is from a well or other private source, it is a good idea to have a water analysis done in the early stage of facility planning. Water quality may have a significant impact on not only cleaning system design but the design of the overall facility as well.
Fortunately, there are many ways to reduce and even totally eliminate water hardness. Ways to reduce and eliminate hardness will be discussed in upcoming blogs.
– FJF –
## 2 comments on “Water – Hard Water vs. Soft Water”
• John Fuchs says:
Phil – I’m sure that if you search far enough you can find claims of benefits for drinking either “hard” or “soft” water. The only real difference I can imagine is that “soft” water will have more sodium while “hard” water more calcium. I am not enough of a biologist (or whatever science it is that studies such things) to know if these elements are actually “available” to the body in either case. From my personal experience as a kid back in Michigan, I know that most people who had water softeners also had auxiliary taps at their sinks (at least the kitchen sink) that dispensed water that had not been through the softening process for drinking. My recollection is that the “softened” water had an unpleasant taste compared to the untreated water we were accustomed to. My guess is that us mid-westerners may have been a bit “picky” in our taste. I’m sure that we drink a lot of “softened” water today and find it entirely palatable. If anyone has any further information on this subject, I would appreciate it if you would pass it on to the blog. Thanks for reading! FJF
• Phil B says:
I’ve never seen someone describe the difference between soft water and hard water so clearly before. Thanks for the excellent explanation. Something I have wondered about before is which is better to drink. By that I mean, does drinking hard or soft water have any known health benefits or is it advantageous to drink one over the other? | 1,178 | 5,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-43 | latest | en | 0.925704 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/481/C6xC9oHe3.html | 1,719,338,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866218.13/warc/CC-MAIN-20240625171218-20240625201218-00248.warc.gz | 394,442,521 | 8,107 | Copied to
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## G = C6×C9○He3order 486 = 2·35
### Direct product of C6 and C9○He3
direct product, metabelian, nilpotent (class 2), monomial, 3-elementary
Series: Derived Chief Lower central Upper central
Derived series C1 — C3 — C6×C9○He3
Chief series C1 — C3 — C32 — C3×C9 — C32×C9 — C3×C9○He3 — C6×C9○He3
Lower central C1 — C3 — C6×C9○He3
Upper central C1 — C3×C18 — C6×C9○He3
Generators and relations for C6×C9○He3
G = < a,b,c,d,e | a6=b9=c3=e3=1, d1=b6, ab=ba, ac=ca, ad=da, ae=ea, bc=cb, bd=db, be=eb, cd=dc, ece-1=b3c, de=ed >
Subgroups: 576 in 480 conjugacy classes, 432 normal (12 characteristic)
C1, C2, C3, C3, C3, C6, C6, C6, C9, C32, C32, C32, C18, C3×C6, C3×C6, C3×C6, C3×C9, C3×C9, He3, 3- 1+2, C33, C3×C18, C3×C18, C2×He3, C2×3- 1+2, C32×C6, C32×C9, C3×He3, C3×3- 1+2, C9○He3, C32×C18, C6×He3, C6×3- 1+2, C2×C9○He3, C3×C9○He3, C6×C9○He3
Quotients: C1, C2, C3, C6, C32, C3×C6, C33, C32×C6, C9○He3, C34, C2×C9○He3, C33×C6, C3×C9○He3, C6×C9○He3
Smallest permutation representation of C6×C9○He3
On 162 points
Generators in S162
(1 122 29 112 41 102)(2 123 30 113 42 103)(3 124 31 114 43 104)(4 125 32 115 44 105)(5 126 33 116 45 106)(6 118 34 117 37 107)(7 119 35 109 38 108)(8 120 36 110 39 100)(9 121 28 111 40 101)(10 98 159 88 20 78)(11 99 160 89 21 79)(12 91 161 90 22 80)(13 92 162 82 23 81)(14 93 154 83 24 73)(15 94 155 84 25 74)(16 95 156 85 26 75)(17 96 157 86 27 76)(18 97 158 87 19 77)(46 137 66 127 56 147)(47 138 67 128 57 148)(48 139 68 129 58 149)(49 140 69 130 59 150)(50 141 70 131 60 151)(51 142 71 132 61 152)(52 143 72 133 62 153)(53 144 64 134 63 145)(54 136 65 135 55 146)
(1 2 3 4 5 6 7 8 9)(10 11 12 13 14 15 16 17 18)(19 20 21 22 23 24 25 26 27)(28 29 30 31 32 33 34 35 36)(37 38 39 40 41 42 43 44 45)(46 47 48 49 50 51 52 53 54)(55 56 57 58 59 60 61 62 63)(64 65 66 67 68 69 70 71 72)(73 74 75 76 77 78 79 80 81)(82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99)(100 101 102 103 104 105 106 107 108)(109 110 111 112 113 114 115 116 117)(118 119 120 121 122 123 124 125 126)(127 128 129 130 131 132 133 134 135)(136 137 138 139 140 141 142 143 144)(145 146 147 148 149 150 151 152 153)(154 155 156 157 158 159 160 161 162)
(1 65 75)(2 66 76)(3 67 77)(4 68 78)(5 69 79)(6 70 80)(7 71 81)(8 72 73)(9 64 74)(10 125 129)(11 126 130)(12 118 131)(13 119 132)(14 120 133)(15 121 134)(16 122 135)(17 123 127)(18 124 128)(19 104 138)(20 105 139)(21 106 140)(22 107 141)(23 108 142)(24 100 143)(25 101 144)(26 102 136)(27 103 137)(28 63 94)(29 55 95)(30 56 96)(31 57 97)(32 58 98)(33 59 99)(34 60 91)(35 61 92)(36 62 93)(37 50 90)(38 51 82)(39 52 83)(40 53 84)(41 54 85)(42 46 86)(43 47 87)(44 48 88)(45 49 89)(109 152 162)(110 153 154)(111 145 155)(112 146 156)(113 147 157)(114 148 158)(115 149 159)(116 150 160)(117 151 161)
(1 7 4)(2 8 5)(3 9 6)(10 16 13)(11 17 14)(12 18 15)(19 25 22)(20 26 23)(21 27 24)(28 34 31)(29 35 32)(30 36 33)(37 43 40)(38 44 41)(39 45 42)(46 52 49)(47 53 50)(48 54 51)(55 61 58)(56 62 59)(57 63 60)(64 70 67)(65 71 68)(66 72 69)(73 79 76)(74 80 77)(75 81 78)(82 88 85)(83 89 86)(84 90 87)(91 97 94)(92 98 95)(93 99 96)(100 106 103)(101 107 104)(102 108 105)(109 115 112)(110 116 113)(111 117 114)(118 124 121)(119 125 122)(120 126 123)(127 133 130)(128 134 131)(129 135 132)(136 142 139)(137 143 140)(138 144 141)(145 151 148)(146 152 149)(147 153 150)(154 160 157)(155 161 158)(156 162 159)
(1 32 38)(2 33 39)(3 34 40)(4 35 41)(5 36 42)(6 28 43)(7 29 44)(8 30 45)(9 31 37)(10 156 23)(11 157 24)(12 158 25)(13 159 26)(14 160 27)(15 161 19)(16 162 20)(17 154 21)(18 155 22)(46 66 56)(47 67 57)(48 68 58)(49 69 59)(50 70 60)(51 71 61)(52 72 62)(53 64 63)(54 65 55)(73 99 86)(74 91 87)(75 92 88)(76 93 89)(77 94 90)(78 95 82)(79 96 83)(80 97 84)(81 98 85)(100 123 116)(101 124 117)(102 125 109)(103 126 110)(104 118 111)(105 119 112)(106 120 113)(107 121 114)(108 122 115)(127 147 137)(128 148 138)(129 149 139)(130 150 140)(131 151 141)(132 152 142)(133 153 143)(134 145 144)(135 146 136)
G:=sub<Sym(162)| (1,122,29,112,41,102)(2,123,30,113,42,103)(3,124,31,114,43,104)(4,125,32,115,44,105)(5,126,33,116,45,106)(6,118,34,117,37,107)(7,119,35,109,38,108)(8,120,36,110,39,100)(9,121,28,111,40,101)(10,98,159,88,20,78)(11,99,160,89,21,79)(12,91,161,90,22,80)(13,92,162,82,23,81)(14,93,154,83,24,73)(15,94,155,84,25,74)(16,95,156,85,26,75)(17,96,157,86,27,76)(18,97,158,87,19,77)(46,137,66,127,56,147)(47,138,67,128,57,148)(48,139,68,129,58,149)(49,140,69,130,59,150)(50,141,70,131,60,151)(51,142,71,132,61,152)(52,143,72,133,62,153)(53,144,64,134,63,145)(54,136,65,135,55,146), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54)(55,56,57,58,59,60,61,62,63)(64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81)(82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99)(100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117)(118,119,120,121,122,123,124,125,126)(127,128,129,130,131,132,133,134,135)(136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153)(154,155,156,157,158,159,160,161,162), (1,65,75)(2,66,76)(3,67,77)(4,68,78)(5,69,79)(6,70,80)(7,71,81)(8,72,73)(9,64,74)(10,125,129)(11,126,130)(12,118,131)(13,119,132)(14,120,133)(15,121,134)(16,122,135)(17,123,127)(18,124,128)(19,104,138)(20,105,139)(21,106,140)(22,107,141)(23,108,142)(24,100,143)(25,101,144)(26,102,136)(27,103,137)(28,63,94)(29,55,95)(30,56,96)(31,57,97)(32,58,98)(33,59,99)(34,60,91)(35,61,92)(36,62,93)(37,50,90)(38,51,82)(39,52,83)(40,53,84)(41,54,85)(42,46,86)(43,47,87)(44,48,88)(45,49,89)(109,152,162)(110,153,154)(111,145,155)(112,146,156)(113,147,157)(114,148,158)(115,149,159)(116,150,160)(117,151,161), (1,7,4)(2,8,5)(3,9,6)(10,16,13)(11,17,14)(12,18,15)(19,25,22)(20,26,23)(21,27,24)(28,34,31)(29,35,32)(30,36,33)(37,43,40)(38,44,41)(39,45,42)(46,52,49)(47,53,50)(48,54,51)(55,61,58)(56,62,59)(57,63,60)(64,70,67)(65,71,68)(66,72,69)(73,79,76)(74,80,77)(75,81,78)(82,88,85)(83,89,86)(84,90,87)(91,97,94)(92,98,95)(93,99,96)(100,106,103)(101,107,104)(102,108,105)(109,115,112)(110,116,113)(111,117,114)(118,124,121)(119,125,122)(120,126,123)(127,133,130)(128,134,131)(129,135,132)(136,142,139)(137,143,140)(138,144,141)(145,151,148)(146,152,149)(147,153,150)(154,160,157)(155,161,158)(156,162,159), (1,32,38)(2,33,39)(3,34,40)(4,35,41)(5,36,42)(6,28,43)(7,29,44)(8,30,45)(9,31,37)(10,156,23)(11,157,24)(12,158,25)(13,159,26)(14,160,27)(15,161,19)(16,162,20)(17,154,21)(18,155,22)(46,66,56)(47,67,57)(48,68,58)(49,69,59)(50,70,60)(51,71,61)(52,72,62)(53,64,63)(54,65,55)(73,99,86)(74,91,87)(75,92,88)(76,93,89)(77,94,90)(78,95,82)(79,96,83)(80,97,84)(81,98,85)(100,123,116)(101,124,117)(102,125,109)(103,126,110)(104,118,111)(105,119,112)(106,120,113)(107,121,114)(108,122,115)(127,147,137)(128,148,138)(129,149,139)(130,150,140)(131,151,141)(132,152,142)(133,153,143)(134,145,144)(135,146,136)>;
G:=Group( (1,122,29,112,41,102)(2,123,30,113,42,103)(3,124,31,114,43,104)(4,125,32,115,44,105)(5,126,33,116,45,106)(6,118,34,117,37,107)(7,119,35,109,38,108)(8,120,36,110,39,100)(9,121,28,111,40,101)(10,98,159,88,20,78)(11,99,160,89,21,79)(12,91,161,90,22,80)(13,92,162,82,23,81)(14,93,154,83,24,73)(15,94,155,84,25,74)(16,95,156,85,26,75)(17,96,157,86,27,76)(18,97,158,87,19,77)(46,137,66,127,56,147)(47,138,67,128,57,148)(48,139,68,129,58,149)(49,140,69,130,59,150)(50,141,70,131,60,151)(51,142,71,132,61,152)(52,143,72,133,62,153)(53,144,64,134,63,145)(54,136,65,135,55,146), (1,2,3,4,5,6,7,8,9)(10,11,12,13,14,15,16,17,18)(19,20,21,22,23,24,25,26,27)(28,29,30,31,32,33,34,35,36)(37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54)(55,56,57,58,59,60,61,62,63)(64,65,66,67,68,69,70,71,72)(73,74,75,76,77,78,79,80,81)(82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99)(100,101,102,103,104,105,106,107,108)(109,110,111,112,113,114,115,116,117)(118,119,120,121,122,123,124,125,126)(127,128,129,130,131,132,133,134,135)(136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153)(154,155,156,157,158,159,160,161,162), (1,65,75)(2,66,76)(3,67,77)(4,68,78)(5,69,79)(6,70,80)(7,71,81)(8,72,73)(9,64,74)(10,125,129)(11,126,130)(12,118,131)(13,119,132)(14,120,133)(15,121,134)(16,122,135)(17,123,127)(18,124,128)(19,104,138)(20,105,139)(21,106,140)(22,107,141)(23,108,142)(24,100,143)(25,101,144)(26,102,136)(27,103,137)(28,63,94)(29,55,95)(30,56,96)(31,57,97)(32,58,98)(33,59,99)(34,60,91)(35,61,92)(36,62,93)(37,50,90)(38,51,82)(39,52,83)(40,53,84)(41,54,85)(42,46,86)(43,47,87)(44,48,88)(45,49,89)(109,152,162)(110,153,154)(111,145,155)(112,146,156)(113,147,157)(114,148,158)(115,149,159)(116,150,160)(117,151,161), (1,7,4)(2,8,5)(3,9,6)(10,16,13)(11,17,14)(12,18,15)(19,25,22)(20,26,23)(21,27,24)(28,34,31)(29,35,32)(30,36,33)(37,43,40)(38,44,41)(39,45,42)(46,52,49)(47,53,50)(48,54,51)(55,61,58)(56,62,59)(57,63,60)(64,70,67)(65,71,68)(66,72,69)(73,79,76)(74,80,77)(75,81,78)(82,88,85)(83,89,86)(84,90,87)(91,97,94)(92,98,95)(93,99,96)(100,106,103)(101,107,104)(102,108,105)(109,115,112)(110,116,113)(111,117,114)(118,124,121)(119,125,122)(120,126,123)(127,133,130)(128,134,131)(129,135,132)(136,142,139)(137,143,140)(138,144,141)(145,151,148)(146,152,149)(147,153,150)(154,160,157)(155,161,158)(156,162,159), (1,32,38)(2,33,39)(3,34,40)(4,35,41)(5,36,42)(6,28,43)(7,29,44)(8,30,45)(9,31,37)(10,156,23)(11,157,24)(12,158,25)(13,159,26)(14,160,27)(15,161,19)(16,162,20)(17,154,21)(18,155,22)(46,66,56)(47,67,57)(48,68,58)(49,69,59)(50,70,60)(51,71,61)(52,72,62)(53,64,63)(54,65,55)(73,99,86)(74,91,87)(75,92,88)(76,93,89)(77,94,90)(78,95,82)(79,96,83)(80,97,84)(81,98,85)(100,123,116)(101,124,117)(102,125,109)(103,126,110)(104,118,111)(105,119,112)(106,120,113)(107,121,114)(108,122,115)(127,147,137)(128,148,138)(129,149,139)(130,150,140)(131,151,141)(132,152,142)(133,153,143)(134,145,144)(135,146,136) );
G=PermutationGroup([[(1,122,29,112,41,102),(2,123,30,113,42,103),(3,124,31,114,43,104),(4,125,32,115,44,105),(5,126,33,116,45,106),(6,118,34,117,37,107),(7,119,35,109,38,108),(8,120,36,110,39,100),(9,121,28,111,40,101),(10,98,159,88,20,78),(11,99,160,89,21,79),(12,91,161,90,22,80),(13,92,162,82,23,81),(14,93,154,83,24,73),(15,94,155,84,25,74),(16,95,156,85,26,75),(17,96,157,86,27,76),(18,97,158,87,19,77),(46,137,66,127,56,147),(47,138,67,128,57,148),(48,139,68,129,58,149),(49,140,69,130,59,150),(50,141,70,131,60,151),(51,142,71,132,61,152),(52,143,72,133,62,153),(53,144,64,134,63,145),(54,136,65,135,55,146)], [(1,2,3,4,5,6,7,8,9),(10,11,12,13,14,15,16,17,18),(19,20,21,22,23,24,25,26,27),(28,29,30,31,32,33,34,35,36),(37,38,39,40,41,42,43,44,45),(46,47,48,49,50,51,52,53,54),(55,56,57,58,59,60,61,62,63),(64,65,66,67,68,69,70,71,72),(73,74,75,76,77,78,79,80,81),(82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99),(100,101,102,103,104,105,106,107,108),(109,110,111,112,113,114,115,116,117),(118,119,120,121,122,123,124,125,126),(127,128,129,130,131,132,133,134,135),(136,137,138,139,140,141,142,143,144),(145,146,147,148,149,150,151,152,153),(154,155,156,157,158,159,160,161,162)], [(1,65,75),(2,66,76),(3,67,77),(4,68,78),(5,69,79),(6,70,80),(7,71,81),(8,72,73),(9,64,74),(10,125,129),(11,126,130),(12,118,131),(13,119,132),(14,120,133),(15,121,134),(16,122,135),(17,123,127),(18,124,128),(19,104,138),(20,105,139),(21,106,140),(22,107,141),(23,108,142),(24,100,143),(25,101,144),(26,102,136),(27,103,137),(28,63,94),(29,55,95),(30,56,96),(31,57,97),(32,58,98),(33,59,99),(34,60,91),(35,61,92),(36,62,93),(37,50,90),(38,51,82),(39,52,83),(40,53,84),(41,54,85),(42,46,86),(43,47,87),(44,48,88),(45,49,89),(109,152,162),(110,153,154),(111,145,155),(112,146,156),(113,147,157),(114,148,158),(115,149,159),(116,150,160),(117,151,161)], [(1,7,4),(2,8,5),(3,9,6),(10,16,13),(11,17,14),(12,18,15),(19,25,22),(20,26,23),(21,27,24),(28,34,31),(29,35,32),(30,36,33),(37,43,40),(38,44,41),(39,45,42),(46,52,49),(47,53,50),(48,54,51),(55,61,58),(56,62,59),(57,63,60),(64,70,67),(65,71,68),(66,72,69),(73,79,76),(74,80,77),(75,81,78),(82,88,85),(83,89,86),(84,90,87),(91,97,94),(92,98,95),(93,99,96),(100,106,103),(101,107,104),(102,108,105),(109,115,112),(110,116,113),(111,117,114),(118,124,121),(119,125,122),(120,126,123),(127,133,130),(128,134,131),(129,135,132),(136,142,139),(137,143,140),(138,144,141),(145,151,148),(146,152,149),(147,153,150),(154,160,157),(155,161,158),(156,162,159)], [(1,32,38),(2,33,39),(3,34,40),(4,35,41),(5,36,42),(6,28,43),(7,29,44),(8,30,45),(9,31,37),(10,156,23),(11,157,24),(12,158,25),(13,159,26),(14,160,27),(15,161,19),(16,162,20),(17,154,21),(18,155,22),(46,66,56),(47,67,57),(48,68,58),(49,69,59),(50,70,60),(51,71,61),(52,72,62),(53,64,63),(54,65,55),(73,99,86),(74,91,87),(75,92,88),(76,93,89),(77,94,90),(78,95,82),(79,96,83),(80,97,84),(81,98,85),(100,123,116),(101,124,117),(102,125,109),(103,126,110),(104,118,111),(105,119,112),(106,120,113),(107,121,114),(108,122,115),(127,147,137),(128,148,138),(129,149,139),(130,150,140),(131,151,141),(132,152,142),(133,153,143),(134,145,144),(135,146,136)]])
198 conjugacy classes
class 1 2 3A ··· 3H 3I ··· 3AF 6A ··· 6H 6I ··· 6AF 9A ··· 9R 9S ··· 9BN 18A ··· 18R 18S ··· 18BN order 1 2 3 ··· 3 3 ··· 3 6 ··· 6 6 ··· 6 9 ··· 9 9 ··· 9 18 ··· 18 18 ··· 18 size 1 1 1 ··· 1 3 ··· 3 1 ··· 1 3 ··· 3 1 ··· 1 3 ··· 3 1 ··· 1 3 ··· 3
198 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 3 3 type + + image C1 C2 C3 C3 C3 C3 C6 C6 C6 C6 C9○He3 C2×C9○He3 kernel C6×C9○He3 C3×C9○He3 C32×C18 C6×He3 C6×3- 1+2 C2×C9○He3 C32×C9 C3×He3 C3×3- 1+2 C9○He3 C6 C3 # reps 1 1 8 2 16 54 8 2 16 54 18 18
Matrix representation of C6×C9○He3 in GL4(𝔽19) generated by
12 0 0 0 0 18 0 0 0 0 18 0 0 0 0 18
,
1 0 0 0 0 17 0 0 0 0 17 0 0 0 0 17
,
11 0 0 0 0 7 7 18 0 4 12 12 0 0 11 0
,
1 0 0 0 0 7 0 0 0 0 7 0 0 0 0 7
,
7 0 0 0 0 11 11 12 0 0 1 0 0 0 0 7
G:=sub<GL(4,GF(19))| [12,0,0,0,0,18,0,0,0,0,18,0,0,0,0,18],[1,0,0,0,0,17,0,0,0,0,17,0,0,0,0,17],[11,0,0,0,0,7,4,0,0,7,12,11,0,18,12,0],[1,0,0,0,0,7,0,0,0,0,7,0,0,0,0,7],[7,0,0,0,0,11,0,0,0,11,1,0,0,12,0,7] >;
C6×C9○He3 in GAP, Magma, Sage, TeX
C_6\times C_9\circ {\rm He}_3
% in TeX
G:=Group("C6xC9oHe3");
// GroupNames label
G:=SmallGroup(486,253);
// by ID
G=gap.SmallGroup(486,253);
# by ID
G:=PCGroup([6,-2,-3,-3,-3,-3,-3,1520,237]);
// Polycyclic
G:=Group<a,b,c,d,e|a^6=b^9=c^3=e^3=1,d^1=b^6,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,b*c=c*b,b*d=d*b,b*e=e*b,c*d=d*c,e*c*e^-1=b^3*c,d*e=e*d>;
// generators/relations
×
𝔽 | 8,001 | 14,395 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.294483 |
http://rdsdialog.spb.ru/u/How-many--kip%7Cmm2--are-in--7--atmosphere-%5Bstandard%5D | 1,524,691,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947957.81/warc/CC-MAIN-20180425193720-20180425213720-00487.warc.gz | 263,683,301 | 13,584 | # How many kip/mm2 are in 7 atmospheres [standard]?
There is 0.072325928834005 kilopond per square millimeter in 7 atmospheres [standard]
Here you can find how many kiloponds per square millimeter are there in any quantity of atmosphere [standard]. You just need to type the atmospheres [standard] value in the box at left (input) and you will get the answer in kiloponds per square millimeter in the box at right (output).
### Atmospheres [standard] to Kiloponds Per Square Millimeter Converter
Enter values here: Results here:
## How to convert 7 atmospheres [standard] to kiloponds per square millimeter
To calculate a value in atmospheres [standard] to the corresponding value in kiloponds per square millimeter, just multiply the quantity in atmospheres [standard] by 0.010332275547715 (the conversion factor).
Here is the formula:
Value in kiloponds per square millimeter = value in atmospheres [standard] × 0.010332275547715
Supose you want to convert 7 atmospheres [standard] into kiloponds per square millimeter. In this case you will have:
Value in kiloponds per square millimeter = 7 × 0.010332275547715 = 0.072325928834005
## Values Near 6.4 atmospheres [standard] in kiloponds per square millimeter
Note: Values are rounded to 4 significant figures. Fractions are rounded to the nearest 8th fraction.
atmospheres [standard] to kiloponds per square millimeter
6.4atmospheres [standard] =0.06613kilopond per square millimeter
6.5atmospheres [standard] =0.06716kilopond per square millimeter
6.6atmospheres [standard] =0.06819kilopond per square millimeter
6.7atmospheres [standard] =0.06923kilopond per square millimeter
6.8atmospheres [standard] =0.07026kilopond per square millimeter
6.9atmospheres [standard] =0.07129kilopond per square millimeter
7atmospheres [standard] =0.07233kilopond per square millimeter
7.1atmospheres [standard] =0.07336kilopond per square millimeter
7.2atmospheres [standard] =0.07439kilopond per square millimeter
7.3atmospheres [standard] =0.07543kilopond per square millimeter
7.4atmospheres [standard] =0.07646kilopond per square millimeter
7.5atmospheres [standard] =0.07749kilopond per square millimeter
7.6atmospheres [standard] =0.07853kilopond per square millimeter
Using this converter you can get answers to questions like:
• How many kiloponds per square millimeter are in 7 atmospheres [standard]?
• 7 atmospheres [standard] are equal to how many kiloponds per square millimeter?
• How much are 7 atmosphere [standard] in kiloponds per square millimeter?
• How to convert atmospheres [standard] to kiloponds per square millimeter?
• What is the atmospheres [standard] to kiloponds per square millimeter conversion factor?
• How to transform atmospheres [standard] in kiloponds per square millimeter?
• What is the formula to convert from atmospheres [standard] to kiloponds per square millimeter? among others. | 785 | 2,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-17 | latest | en | 0.540344 |
https://www.maplesoft.com/support/help/view.aspx?path=LinearAlgebra/PopovForm | 1,695,901,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00142.warc.gz | 945,692,641 | 25,466 | PopovForm - Maple Help
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# Online Help
###### All Products Maple MapleSim
LinearAlgebra
PopovForm
compute the Popov normal form of a Matrix
Calling Sequence PopovForm(A, x, shifts, out, options, outopts)
Parameters
A - Matrix of univariate polynomials in x x - variable name of the polynomial domain shifts - (optional) equation of the form shifts = obj where obj is a list of one or two lists out - (optional) equation of the form output = obj where obj is one of 'P', 'U', 'rank', 'P_pivots', 'U_pivots', or a list containing one or more of these names; select result objects to compute options - (optional); constructor options for the result object(s) outopts - (optional) equation(s) of the form outputoptions[o] = list where o is one of 'P' or 'U'; constructor options for the specified result object
Description
• The PopovForm(A, x) function computes the column Popov normal form P (also called the polynomial echelon form) of an m x n rectangular Matrix of univariate polynomials in x over the field of rational numbers Q, or rational expressions over Q. You can request the form P, the unimodular multiplier U which gives P, the rank and various pivots in P and U via a specification of the output option.
Definition of Column Popov Form
• There are a number of variations of column Popov forms, typically unique up to permutation of the columns. The definitions used for both the full and non-full column rank case are as follows.
If $m=n$ and P is nonsingular, P has the following degree constraints.
$\mathrm{deg}\left({P}_{j,i}\right)<\mathrm{deg}\left({P}_{i,i}\right),\mathrm{for all j > i}$
$\mathrm{deg}\left({P}_{j,i}\right)\le \mathrm{deg}\left({P}_{i,i}\right),\mathrm{for all j < i}$
If $n and P has full column rank, there is a trailing list of n pivot rows P_pivots such that P[P_pivots,*] is in Popov normal form.
If $n\le m$ and P has column rank $r, P has the first $n-r$ columns 0 and there is a trailing list of r rows P_pivots such that P[P_pivots,*] is in Popov normal form. In this case, U is a minimal unimodular multiplier and as such there is a list U_pivots of rows such that U[U_pivots,*] is also in Popov normal form.
The Popov normal form P is obtained by doing elementary column operations on A. This includes interchanging columns, multiplying through a column by a unit, and subtracting a polynomial multiple of one column from another. The method used is a fraction-free algorithm by Beckermann, Labahn, and Villard. The returned Matrix objects have the property that $P=A·U$.
• The output option (out) determines the content of the returned expression sequence.
Depending on what is included in the output option, an expression sequence containing one or more of the factors P (the Popov normal form), U (the unimodular transformation Matrix), rank (the rank of the matrix), P_pivots, or U_pivots (the pivot rows of P and U, respectively) can be returned.
If output is a list, the objects are returned in the same order as specified in the list.
• The shifts option is an optional input which allows the user to shift the degree constraints on both the Popov form and the minimal multiplier (in the non-full column rank case).
• The constructor options provide additional information (readonly, shape, storage, order, datatype, and attributes) to the Matrix constructor that builds the result(s). These options may also be provided in the form outputoptions[o]=[...], where [...] represents a Maple list. If a constructor option is provided in both the calling sequence directly and in an outputoptions[o] option, the latter takes precedence (regardless of the order).
Examples
> $\mathrm{with}\left(\mathrm{LinearAlgebra}\right):$
> $A≔⟨⟨{z}^{3}-{z}^{2},{z}^{3}-2{z}^{2}+2z-2⟩|⟨{z}^{3}-2{z}^{2}-1,{z}^{3}-3{z}^{2}+3z-4⟩⟩$
${A}{≔}\left[\begin{array}{cc}{{z}}^{{3}}{-}{{z}}^{{2}}& {{z}}^{{3}}{-}{2}{}{{z}}^{{2}}{-}{1}\\ {{z}}^{{3}}{-}{2}{}{{z}}^{{2}}{+}{2}{}{z}{-}{2}& {{z}}^{{3}}{-}{3}{}{{z}}^{{2}}{+}{3}{}{z}{-}{4}\end{array}\right]$ (1)
> $P≔\mathrm{PopovForm}\left(A,z,\mathrm{datatype}=\mathrm{algebraic}\right)$
${P}{≔}\left[\begin{array}{cc}{z}& {-1}\\ {1}& {-}{1}{+}{z}\end{array}\right]$ (2)
> $P,U≔\mathrm{PopovForm}\left(A,z,\mathrm{output}=\left['P','U'\right]\right)$
${P}{,}{U}{≔}\left[\begin{array}{cc}{z}& {-1}\\ {1}& {-}{1}{+}{z}\end{array}\right]{,}\left[\begin{array}{cc}{-}\frac{{1}}{{2}}{}{{z}}^{{2}}{+}\frac{{3}}{{2}}{}{z}{-}\frac{{1}}{{2}}& \frac{{1}}{{2}}{}{{z}}^{{2}}{-}\frac{{1}}{{2}}{}{z}{-}\frac{{3}}{{2}}\\ \frac{{1}}{{2}}{}{{z}}^{{2}}{-}{z}& {-}\frac{{{z}}^{{2}}}{{2}}{+}{1}\end{array}\right]$ (3)
> $\mathrm{map}\left(\mathrm{expand},P-\mathrm{.}\left(A,U\right)\right)$
$\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (4)
> $\mathrm{Determinant}\left(U\right)$
${-}\frac{{1}}{{2}}$ (5)
A low rank matrix.
> $A≔⟨⟨3z-6,-3z+3,2z+3,z⟩|⟨-3z,3z,-2,-1⟩|⟨6,-3,-2z-1,-z+1⟩⟩$
${A}{≔}\left[\begin{array}{ccc}{3}{}{z}{-}{6}& {-}{3}{}{z}& {6}\\ {-}{3}{}{z}{+}{3}& {3}{}{z}& {-3}\\ {2}{}{z}{+}{3}& {-2}& {-}{2}{}{z}{-}{1}\\ {z}& {-1}& {1}{-}{z}\end{array}\right]$ (6)
> $P,U,r≔\mathrm{PopovForm}\left(A,z,\mathrm{output}=\left['P','U','\mathrm{rank}'\right]\right)$
${P}{,}{U}{,}{r}{≔}\left[\begin{array}{ccc}{0}& {-}{z}& {-6}\\ {0}& {z}& {3}\\ {0}& {-}\frac{{2}}{{3}}& {1}{+}{2}{}{z}\\ {0}& {-}\frac{{1}}{{3}}& {-}{1}{+}{z}\end{array}\right]{,}\left[\begin{array}{ccc}{1}& {0}& {1}\\ {1}& \frac{{1}}{{3}}& {1}\\ {1}& {0}& {0}\end{array}\right]{,}{2}$ (7)
> $P,U,r,\mathrm{II},K≔\mathrm{PopovForm}\left(A,z,\mathrm{output}=\left['P','U','\mathrm{rank}','\mathrm{P_pivots}','\mathrm{U_pivots}'\right]\right)$
${P}{,}{U}{,}{r}{,}{\mathrm{II}}{,}{K}{≔}\left[\begin{array}{ccc}{0}& {-}{z}& {-6}\\ {0}& {z}& {3}\\ {0}& {-}\frac{{2}}{{3}}& {1}{+}{2}{}{z}\\ {0}& {-}\frac{{1}}{{3}}& {-}{1}{+}{z}\end{array}\right]{,}\left[\begin{array}{ccc}{1}& {0}& {1}\\ {1}& \frac{{1}}{{3}}& {1}\\ {1}& {0}& {0}\end{array}\right]{,}{2}{,}\left[{2}{,}{4}\right]{,}\left[{3}\right]$ (8)
A [2,2,0,0]-shifted Popov form.
> $P,U,r,\mathrm{II},K≔\mathrm{PopovForm}\left(A,z,\mathrm{shifts}=\left[\left[2,2,0,0\right]\right],\mathrm{output}=\left['P','U','\mathrm{rank}','\mathrm{P_pivots}','\mathrm{U_pivots}'\right]\right)$
${P}{,}{U}{,}{r}{,}{\mathrm{II}}{,}{K}{≔}\left[\begin{array}{ccc}{0}& {-}{{z}}^{{2}}{+}{z}{-}{2}& {2}{}{{z}}^{{2}}{+}{z}{+}{4}\\ {0}& {{z}}^{{2}}{-}{z}{+}{1}& {-}{2}{}{{z}}^{{2}}{-}{z}{-}{2}\\ {0}& {1}& {0}\\ {0}& {0}& {1}\end{array}\right]{,}\left[\begin{array}{ccc}{1}& \frac{{1}}{{3}}& {-}\frac{{2}}{{3}}\\ {1}& \frac{{z}}{{3}}& {-}{1}{-}\frac{{2}{}{z}}{{3}}\\ {1}& {0}& {0}\end{array}\right]{,}{2}{,}\left[{3}{,}{4}\right]{,}\left[{3}\right]$ (9)
A Popov form with [0,-3,0,0]-shift for unimodular multiplier.
> $A≔⟨⟨-{z}^{3}+4{z}^{2}+z+1,-{z}^{2}+7z+4⟩|⟨z-1,z+2⟩|⟨2{z}^{2}+2z-2,{z}^{2}+6z+6⟩|⟨-{z}^{2},-2z⟩⟩$
${A}{≔}\left[\begin{array}{cccc}{-}{{z}}^{{3}}{+}{4}{}{{z}}^{{2}}{+}{z}{+}{1}& {-}{1}{+}{z}& {2}{}{{z}}^{{2}}{+}{2}{}{z}{-}{2}& {-}{{z}}^{{2}}\\ {-}{{z}}^{{2}}{+}{7}{}{z}{+}{4}& {z}{+}{2}& {{z}}^{{2}}{+}{6}{}{z}{+}{6}& {-}{2}{}{z}\end{array}\right]$ (10)
> $P,U,r,\mathrm{II},K≔\mathrm{PopovForm}\left(A,z,\mathrm{shifts}=\left[\left[0,0\right],\left[0,-3,0,0\right]\right],\mathrm{output}=\left['P','U','\mathrm{rank}','\mathrm{P_pivots}','\mathrm{U_pivots}'\right]\right)$
${P}{,}{U}{,}{r}{,}{\mathrm{II}}{,}{K}{≔}\left[\begin{array}{cccc}{0}& {0}& {z}& {-1}\\ {0}& {0}& {2}& {z}\end{array}\right]{,}\left[\begin{array}{cccc}{-}\frac{{2}{}{z}}{{21}}{+}\frac{{1}}{{7}}& {-}{{z}}^{{2}}{-}{2}{}{z}& \frac{{2}}{{7}}{+}\frac{{z}}{{7}}& {-}\frac{{z}}{{21}}{-}\frac{{3}}{{7}}\\ {1}& {0}& {0}& {0}\\ \frac{{2}{}{z}}{{21}}{-}\frac{{3}}{{7}}& {{z}}^{{2}}{-}{z}& {-}\frac{{z}}{{7}}{+}\frac{{1}}{{7}}& \frac{{z}}{{21}}{+}\frac{{2}}{{7}}\\ \frac{{2}}{{21}}{}{{z}}^{{2}}{-}\frac{{1}}{{3}}{}{z}{-}\frac{{4}}{{21}}& {{z}}^{{3}}{-}{9}{}{z}{-}{7}& {-}\frac{{{z}}^{{2}}}{{7}}{+}\frac{{9}}{{7}}& \frac{{1}}{{21}}{}{{z}}^{{2}}{+}\frac{{1}}{{3}}{}{z}{-}\frac{{23}}{{21}}\end{array}\right]{,}{2}{,}\left[{1}{,}{2}\right]{,}\left[{2}{,}{4}\right]$ (11)
> $\mathrm{map}\left(\mathrm{expand},P-\mathrm{.}\left(A,U\right)\right)$
$\left[\begin{array}{cccc}{0}& {0}& {0}& {0}\\ {0}& {0}& {0}& {0}\end{array}\right]$ (12)
> $\mathrm{Determinant}\left(U\right)$
${1}$ (13)
References
Beckermann, B., and Labahn, G. "Fraction-free Computation of Matrix Rational Interpolants and Matrix GCDs." SIAM Journal on Matrix Analysis and Applications, Vol. 22 No. 1. (2000): 114-144.
Beckermann, B.; Labahn, G.; and Villard, G. "Shifted Normal Forms of General Polynomial Matrices." University of Waterloo, Technical Report. Department of Computer Science, 2001.
Beckermann, B.; Labahn, G.; and Villard, G. "Shifted Normal Forms of Polynomial Matrices." ISSAC'99, pp. 189-196. 1999.
See Also | 3,412 | 8,915 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-40 | latest | en | 0.591302 |
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You are reading an older version of this FlexBook® textbook: People's Physics Book - Basic Go to the latest version.
Chapter 12: Basic Physics SE-Gravity
Difficulty Level: At Grade Created by: CK-12
The Big Idea
Over 2500 years ago Aristotle proposed two laws of physics governing motion. One for ‘Earthly bodies’ (objects on Earth) that states objects naturally go in straight lines and one for ‘Heavenly bodies’ (objects in space) that states objects naturally go in circles. This idea held steady for 2,000 years, until Isaac Newton in a triumph of brilliance declared that there is one law of physics that governs motion and he unified “earthly” bodies and “heavenly” bodies with the The Universal Law of Gravitation. Newton used the universal law of gravity to prove that the moon is accelerating towards Earth just as a body on Earth is falling towards Earth. The acceleration of both the moon and the object on Earth is calculated using the same formula. This theory is so well regarded that it has been used to hypothesize the existence of black holes and dark matter, later verified by observation. The search to unify the laws of physics into one theory continues today and is often referred to as the yet undiscovered Grand Unified Theory (GUT).
Key Equation: The Universal Law of Gravity
• $F_G =\frac{Gm_1m_2}{r^2}$ ; the force of gravity between an object with mass $m_1$ and another object of mass $m_2$ and a distance between them of $r$.
• $G = 6.67 \times 10^{-11} \ Nm^2 / kg^2$ ; the universal constant of gravity
• $g = \frac{Gm}{r^2}$ ; gravitational field strength or gravitational acceleration of a planet with mass $m$ and radius $r$. Note that this is not really a separate equation but comes from Newton’s second law and the law of universal gravitation.
Key Concepts
• When using the Universal Law of Gravity formula and the constant $G$ above, make sure to use units of meters and kilograms.
• Newton invented calculus in order to prove that for a spherical object (like Earth) one can assume all of its mass is at the center of the sphere (thus in his formula, one can use the radius of Earth for the distance between a falling rock and Earth).
• An orbital period, $T$, is the time it takes to make one complete rotation.
• If a particle travels a distance $2 \pi r$ in an amount of time $T$, then its speed is distance over time or $\frac{2 \pi r}{T}$.
• Objects in orbit around each other, orbit about the center of mass for the system. For the Earth and moon the center of mass is somewhere inside of Earth. So while the moon orbits Earth, the Earth also orbits the moon (manifested as a slight wobble).
Key Applications
• To find the speed of a planet or satellite in an orbit, realize that the force of gravity is the centripetal force. So set the force of gravity equal to $\frac{mv^2}{r}$, where $v$ is the speed of the planet, $m$ is the mass of the planet or satellite and $r$ is the distance of the planet to the sun or the satellite to the Earth.
• The Geo-synchronous orbit is that orbit for which the satellite takes the same amount of time to orbit the planet as the planet takes to make one revolution. For satellites to be over the same place of the planet at all times, it must be in a Geo-synchronous orbit.
• Some data needed for the problems:
The radius of Earth is $6.4 \times 10^6 \ m$
The mass of Earth is about $6.0 \times 10^{24} \ kg$
The mass of Sun is about $2.0 \times 10^{30} \ kg$
The Earth-Sun distance is about $1.5 \times 10^{11} \ m$
The Earth-Moon distance is about $3.8 \times 10^8 \ m$
Gravity Problem Set
1. Which is greater – the gravitational force that the Earth exerts on the moon, or the force the moon exerts on the Earth? Why doesn’t the moon fall into the earth?
2. Which is greater – the gravitational force that the Sun exerts on the moon, or the force the Earth exerts on the moon? Does the moon orbit the Earth or the Sun? Explain.
3. Suppose you’re standing in an elevator on a bathroom scale. Draw a FBD for you and label the two forces acting on you. Describe how the scale reading compares to your weight when
1. The elevator is at rest
2. The elevator is moving up at a constant speed
3. The elevator is accelerating upward
4. The cable breaks and the elevator is in free fall
4. Astronauts in orbit are not beyond the pull of earth’s gravity – in fact, gravity is what KEEPS them in orbit! So why are they “weightless”? Explain briefly.
5. Use Newton’s Law of Universal Gravitation to explain why even though Jupiter has 300 times the mass of the earth, on the “surface” of Jupiter you’d weigh only 3 times what you weigh on earth. What other factor has to be considered here?
6. Suppose we drilled a hole through the earth and dropped in a rock. How would the speed and acceleration of the rock vary as the rock falls into the earth? Would the rock come shooting out of the other side of the earth, or would it oscillate back and forth? Use a diagram in your explanation.
7. There are basically three possible fates for a star that reaches the end of its life: white dwarf, neutron star, or black hole. What’s the main factor that determines which one a star will become? Which will our sun become? Explain what a black hole is. You may have to do a little research for this one.
8. Do some research online to answer this question: there are three huge pieces of evidence for the Big Bang. Two cosmological phenomena called the cosmic microwave background radiation and the expansion of the universe. The third is called nucleosynthesis. Explain/define each, and briefly discuss how each supports the Big Bang theory. How do astronomers know the universe is expanding?
9. Prove $g$ is approximately $10 \ m/s^2$on Earth by following these steps:
1. Calculate the force of gravity between a falling object (for example an apple) and that of Earth. Use the symbol $m_o$ to represent the mass of the falling object.
2. Now divide that force by the object’s mass to find the acceleration $g$ of the object.
3. Calculate the force of gravity between the Sun and the Earth. (sun mass $= 2.0 \times 10^{30} \ kg;$ average distance from sun to earth = 150 million km)
10. Calculate the force of gravity between two human beings, assuming that each has a mass of 80 kg and that they are standing 1 m apart. Is this a large force?
11. What is the force of gravity between an electron and a proton? Do you think this is what keeps the electron ‘orbiting’ the proton in the hydrogen atom?
12. Calculate the gravitational force that your pencil or pen pulls on you. Use the center of your chest as the center of mass (and thus the mark for the distance measurement) and estimate all masses and distances.
1. If there were no other forces present, what would your acceleration be towards your pencil? Is this a large or small acceleration?
2. Why, in fact, doesn’t your pencil accelerate towards you?
13. Mo and Jo have been traveling through the galaxy for eons when they arrive at the planet Remulak. Wanting to measure the gravitational field strength of the planet they drop Mo’s lava lamp from the top deck of their spacecraft, collecting the velocity-time data shown below.
velocity (m/s) time (s)
0 0
3.4 1.0
7.0 2.0
9.8 3.0
14.0 4.0
17.1 5.0
1. Plot a velocity-time graph using the axes above. Put numbers, labels and units on your axes. Then draw a best-fit line (use a ruler) and use that line to find the gravitational field strength of Remulak. Explain below how you did that.
1. Mo and Jo go exploring and drop a rock into a deep canyon – it hits the ground in 8.4 s. How deep is the canyon?
2. If the rock has a mass of 25 g and makes a hole in the ground 1.3 cm deep, what force does the ground exert to bring it to a stop?
3. Mo and Jo observe the shadows of their lava lamps at different positions on the planet and determine (a la Eratosthenes, the Greek astronomer, around 200 B.C.) that the radius of Remulak is 4500 km. Use that and your result for $g$ to find the mass of Remulak.
1. A neutron star has a mass of about 1.4 times the mass of our sun and is about 20 km in diameter. It’s the remains of a star several times more massive than our sun that has blown off its outer layers in a supernova explosion. (see #1 for mass of sun)
1. Calculate the density of a neutron star. (density = mass/volume); volume of a sphere is $4/3 \pi R^3$.
2. How much would you weigh on the surface of a neutron star? How many times greater is this than your weight on earth?
3. Calculate $g''$, gravitational acceleration, on the surface of a neutron star.
4. If you dropped your iPod from a height of 2.2 m on the surface of a neutron star, how long would it take to hit the surface? Compare this to the time it would fall on earth.
2. At what height above the earth would a 400-kg weather satellite have to orbit in order to experience a gravitational force half as strong as that on the surface of the earth?
3. The Moon’s period around the earth is 27.3 days and the distance from the earth to the moon is $3.84 \times 10^8 \ m$. Using this information, calculate the mass of the Earth. (note: this calculation assumes a circular orbit. Is your answer high or low? Why?).
4. Calculate the mass of the Earth using only
1. Newton’s Universal Law of Gravity;
2. the Moon-Earth distance ($3.84 \times 10^8 \ m$); and
3. the fact that it takes the Moon 27.3 days to orbit the Earth.
1. (c) $3.6 \times 10^{22} \ N$
2. $4.3 \times 10^{-7} \ N$
3. About $10^{-47} N$, No it is clearly not gravity.
(all answers are approximate for #13)
1. $\sim 3.4 \ m/s^2$
2. $\sim 120 \ m$
3. $\sim 800 \ N$
4. $\sim 10^{23} \ kg$
1. $6.7 \times 10^{17} \ kg/m^3$
2. approx. $10^{14} \ N$
3. $1.9 \times 10^{12} \ m/s^2$
4. $1.5 \times 10^{-6}s , 440,000$ times faster than on earth!
1. $2.5 \times 10^6 \ m$ above earth’s surface
2. Earth’s mass is about $6.0 \times 10^{24} \ N$
Oct 09, 2013 | 2,537 | 9,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 45, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2015-35 | longest | en | 0.891661 |
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# 6 where wnc is the nonconservative work done by air
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Unformatted text preview: at given, the net work W is the total area under the graph. For the purpose of calculation, it is convenient to divide this area into two pieces, a triangle (from s = 0 m to s = 10.0 m) and a rectangle (from s = 10.0 m to s = 20.0 m). Both pieces have the same width (10.0 m) and height (10.0 N), so the triangle has half the area of the rectangle. 2W . m done on the object from s = 0 m to s = 10.0 m SOLUTION Solving W = 1 mvf2 for the final speed of the object, we obtain vf = 2 The net work W is the sum of the work W0,10 (the triangular area) and the work W10,20 done on the object from s = 10.0 m to s = 20.0 m (the rectangular area). We calculate the work W10,20 by multiplying the width (10.0 m) and height (10.0 N) of the rectangle: W10,20 = ( Width ) ( Height ) . The triangle’s area, which is the work W0,10, is half this amount: W0,10 = 1 W10,20 . Therefore, the net work 2 W = W0,10 + W10,20 done during the entire interval from s = 0 m to s = 20.0 m is 332 WORK AND ENERGY W = 1 W10,20 + W10,20 = 3 W10,20 = 2 2 3 2 ( Width ) ( Height ) We can now calculate the final speed of the object from vf = vf = 2 2W : m ( 3 ) ( Width )( Height ) = 3 (10.0 m )(10.0 N ) = 7.07 m/s 2 m 6.00 kg 76. REASONING AND SOLUTION The work done by the retarding force is given by Equation 6.1: W = ( F cos θ )s . Since the force is a retarding force, it must point opposite to the direction of the displacement, so that θ = 180° . Thus, we have W = ( F cosθ )s = (3.0 × 103 N)(cos 180°)(850 m) = –2.6 × 106 J The work done by this force is negative , because the retarding force is directed opposite to the direction of the displacement of the truck. ______________________________________________________________________________ 77. SSM REASONING The only two forces that act on the gymnast are his weight and the force exerted on his hands by the high bar. The latter is the (non-conservative) reaction force to the force exerted on the bar by the gymnast, as predicted by Newton's third law. This force, however, does no work because it points perpendicular to the circular path of motion. Thus, Wnc = 0 J, and we can apply the principle of conservation of mechanical energy. SOLUTION The conservation principle gives 1 mv 2 + mgh 2 4f 244f 14 3 Ef 1 2 = mv 0 + mgh0 24 14 244 3 E0 Since the gymnast's speed is momentarily zero at the top of the swing, v0 = 0 m/s. If we take hf = 0 m at the bottom of the swing, then h0 = 2 r , where r is the radius of the circular path followed by the gymnast's waist. Making these substitutions in the above expression and solving for vf , we obtain vf = 2 gh0 = 2 g(2r) = 2(9.80 m/s2 )(2 × 1.1 m) = 6.6 m/s ______________________________________________________________________________ Chapter 6 Problems 333 78. REASONING a. Since there is no air friction, the only force that acts on the projectile is the conservative gravitational force (its weight). The initial and final speeds of the ball are known, so the conservation of mechanical energy can be used to find the maximum height that the projectile attains. b. When air resistance, a nonconservative force, is present, it does negative work on the projectile and slows it down. Consequently, the projectile does not rise as high as when there is no air resistance. The work-energy theorem, in the form of Equation 6.6, may be used to find the work done by air friction. Then, using the definition of work, Equation 6.1, the average force due to air resistance can be found. SOLUTION a. The conservation of mechanical energy, as expressed by Equation 6.9b, states that 2 2 Ef E0 1 mv + mgh = 1 mv + mgh 24 14 f244f 14 02440 3 24 3 The mass m can be eliminated algebraically from this equation since it appears as a factor in every term. Solving for the final height hf gives 1 2 hf = ( v02 − vf2 ) + h 0 g Setting h0 = 0 m and vf = 0 m/s, the final height, in the absence of air resistance, is 2 hf = 2 vo − vf 2g = (18.0 m / s )2 − ( 0 m/s )2 ( 2 9.80 m / s 2 ) = 16.5 m b. The work-energy theorem is Wnc = ( 1 mv 2 f 2 2 − 1 mv0 2 ) + ( mgh f − mgh0 ) (6.6) where Wnc is the nonconservative work done by air resistance. According to Equation 6.1, the work can be written as Wnc = ( FR cos 180° ) s , where FR is the average force of air resistance. As the projectile moves upward, the force of air resistance is directed downward, so the angle between the two vectors is θ = 180° and cos θ = –1. The magnitude s of the displacement is the difference between the final and initial heights, s = hf – h0 = 11.8 m. With these substitutions, the work-energy theorem becomes ( ) 2 − FR s = 1 m vf2 − vo + mg ( hf − h0 ) 2 334 WORK AND ENERGY Solving for FR gives FR = = 1m 2 ( vf2 − vo2 ) + mg ( hf − h0 ) −s 1 2 ( 0.750 kg ) ( 0 m/s )2 − (18.0 m/s )2 + ( 0.750 kg ) ( 9.80 m/s2 ) (11.8 m ) = − (11.8 m ) 2.9 N ______________________________________________________________________________ 79. SSM REASONING AND SOLUTION We will assume that the tug-of-war rope remains parallel to the ground, so that the force that moves team B is in the same direction as the...
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# Re: st: Marginsplot on backtransformed data
From Richard Williams To statalist@hsphsun2.harvard.edu, statalist@hsphsun2.harvard.edu Subject Re: st: Marginsplot on backtransformed data Date Thu, 19 Dec 2013 11:13:58 -0500
```At 10:50 AM 12/19/2013, Scott Merryman wrote:
```
```One could also use the -expression()- option in -margins-
margins race, expression(predict(xb)^2)
marginsplot, name(regress2,replace)
```
```
```
Good point. I've used the expression option to do thing like multiply numbers by 100 so you get 37.3 instead of .373.
```
```
That still leaves open the question of whether you should use regress (computing the square root of the dv yourself) or use glm (using the power link.) In my example it doesn't make too much difference. In general is it better to use glm or are there pros and cons of each approach?
```
```
```Scott
On Thu, Dec 19, 2013 at 9:30 AM, Richard Williams
<richardwilliams.ndu@gmail.com> wrote:
> Patrick Royston's -marginscontplot- (available from SSC) can be used when
> you've done a log or other transformation of an independent variable. See
> the help file example entitled "Example using a log-transformed covariate".
>
```
> For a dependent variable, I think you can use the glm command, at least some
```> of the time. You should get a 2nd opinion on this, e.g. Austin Nichols is
> much better with these sorts of things than I am. When the dependent
> variable has been transformed I believe it is often better to use glm
```
> anyway. In the following you don't get exactly the same results from regress > and glm but I don't think you are supposed to (and the results are similar).
```>
> webuse nhanes2f, clear
> gen sqweight = weight ^.5
> reg sqweight i.race
> margins race
> marginsplot, name(regress)
> glm weight i.race, link(power .5)
> margins race
> marginsplot, name(glm)
>
*
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* http://www.stata.com/help.cgi?search
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* http://www.ats.ucla.edu/stat/stata/
```
```
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HOME: (574)289-5227
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``` | 716 | 2,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-10 | latest | en | 0.86054 |
https://vdocuments.mx/seismic-velocity-estimation-from-time-migration-velocities-sethian2006paperssethiantimedepthseismic.html | 1,679,970,685,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00309.warc.gz | 672,145,993 | 48,941 | # seismic velocity estimation from time migration...
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Seismic Velocity Estimation from Time Migration Velocities M. K. Cameron, S. B. Fomel, J. A. Sethian * October 10, 2006 Submitted for Publication: Inverse Problems, Oct. 2006 Abstract We address the problem of estimating seismic velocities inside the earth which is necessary for obtaining seismic images in regular Cartesian coordinates. We derive a relation between the true seismic velocities and the routinely obtained so called ”time migration velocities”, which are some kind of mean velocities. We formulate an inverse problem and show that it is ill-posed. We suggest three different approaches for solving it. We test these algorithms on synthetic data and apply them to a field data example. 1 Introduction Seismic data are the records of the pressure wave amplitudes P described by the wave equation ΔP (x, y, z; t)= 1 v 2 (x, y, z) 2 ∂t 2 P (x, y, z; t), (1) where v(x, y, z), common referred to as the ”seismic velocity”, is the sound speed of the earth. This velocity is typically unknown, and its determination is the subject of the present work. One common fast and robust process of obtaining seismic images is called time migration (see e.g. [Yilmaz, 2001]). This process is considered adequate for the areas with mild lateral velocity variation, i.e. where v depends mostly on z and only slightly on x and y. However, even mild lateral velocity variations can significantly distort subsurface structures on the time migrated images. Moreover, time migration produces images in very specific time migration coordinates (x 0 ,t 0 ) (explained below), and the relation between them and the Cartesian coordinates can be nontrivial if the velocity varies laterally. One “side product” of time migration is mean velocities v m (x 0 ,t 0 ), known as time migration velocities. We will refer to them as migration velocities for brevity. In the case where the seismic velocity depends only on the depth, these velocities are close to the root-mean-square (RMS) velocities [Dix, 1955]. In the general case, these velocities relate to the radius of curvature of the emerging wave front [Hubral and Krey, 1980]. An alternative approach to obtaining seismic images is called depth migration [Yilmaz, 2001]. This approach is adequate for areas with lateral velocity variation, and produces seismic images * This work was supported in part by the Applied Mathematical Science subprogram of the Office of Energy Research, U.S. Department of Energy, under Contract Number DE-AC03-76SF00098, and by the Computational Mathematics Program of the National Science Foundation 1
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• Seismic Velocity Estimation from Time Migration Velocities
M. K. Cameron, S. B. Fomel, J. A. Sethian
October 10, 2006
Submitted for Publication: Inverse Problems, Oct. 2006
Abstract
We address the problem of estimating seismic velocities inside the earth which is necessaryfor obtaining seismic images in regular Cartesian coordinates. We derive a relation between thetrue seismic velocities and the routinely obtained so called time migration velocities, whichare some kind of mean velocities. We formulate an inverse problem and show that it is ill-posed.We suggest three different approaches for solving it. We test these algorithms on synthetic dataand apply them to a field data example.
1 Introduction
Seismic data are the records of the pressure wave amplitudes P described by the wave equation
P (x, y, z; t) =1
v2(x, y, z)
2
t2P (x, y, z; t), (1)
where v(x, y, z), common referred to as the seismic velocity, is the sound speed of the earth.This velocity is typically unknown, and its determination is the subject of the present work.
One common fast and robust process of obtaining seismic images is called time migration (seee.g. [Yilmaz, 2001]). This process is considered adequate for the areas with mild lateral velocityvariation, i.e. where v depends mostly on z and only slightly on x and y. However, even mildlateral velocity variations can significantly distort subsurface structures on the time migratedimages. Moreover, time migration produces images in very specific time migration coordinates(x0, t0) (explained below), and the relation between them and the Cartesian coordinates can benontrivial if the velocity varies laterally.
One side product of time migration is mean velocities vm(x0, t0), known as time migrationvelocities. We will refer to them as migration velocities for brevity. In the case where the seismicvelocity depends only on the depth, these velocities are close to the root-mean-square (RMS)velocities [Dix, 1955]. In the general case, these velocities relate to the radius of curvature ofthe emerging wave front [Hubral and Krey, 1980].
An alternative approach to obtaining seismic images is called depth migration [Yilmaz, 2001].This approach is adequate for areas with lateral velocity variation, and produces seismic images
This work was supported in part by the Applied Mathematical Science subprogram of the Office of Energy
Research, U.S. Department of Energy, under Contract Number DE-AC03-76SF00098, and by the Computational
Mathematics Program of the National Science Foundation
1
• in regular Cartesian coordinates. The major problem with this approach is that its implemen-tation requires the construction of a velocity model for the seismic velocity v(x, y, z). It canbe both difficult and time consuming to construct an adequate velocity model: an iterativeapproach of guesswork followed by correction is often employed.
The main idea of this work is to construct a velocity model v(x) from the migration velocitiesgiven in the time migration coordinates (x0, t0) (see a block-scheme in Fig. 1) . Using thesevelocities one can then perform depth migration to obtain an improved seismic image in theCartesian coordinates x. As an alternative to depth migration, one can instead directly converta time migrated image to depth (regular Cartesian coordinates) using the additional outputsof our construction x0(x) and t0(x).
Figure 1: The main idea of this work.
Thus, our goals are to create fast and robust algorithms to:
1. Convert the migration velocities vm(x0, t0) to the true seismic velocities v(x);
2. Convert time migrated images (in (x0, t0) coordinates) to depth (to images in regularCartesian coordinates x).
The end result is to construct more accurate seismic images cheaply and routinely.Our results are the following. To this end, we have obtained theoretical relations between
the migration velocity and the true seismic velocity in 2D and 3D. It came out in the theoreticalrelation in 2D that the Dix velocities vDix(x0, t0) which are a conventional estimate of trueseismic velocities from the migration velocities, can be used instead of the migration velocitiesas a more convenient input. We have stated an inverse problem of finding the true seismicvelocity from the Dix velocity and developed two approaches for solving it: the ray tracingapproach and the level set approach.
These two approaches involve three numerical algorithms:
1. An efficient time-to-depth conversion algorithm;
2. A ray tracing algorithm;
3. A level set algorithm.
The relations between the approaches and the algorithms are outlined in the block-scheme inFig. 2.
2
• Figure 2: The approaches and the algorithms.
1.1 Time migration coordinates and image rays
For many decades, seismic imaging was based on the assumption that the velocity inside theearth depends only on the depth and that the subsurface structures are horizontal or, at worst,planar with the same dipping angle. To obtain more complex structure distortions, Hubral[Hubral, 1977] introduced the concept of the image ray, which gives the connection between thetime migration coordinates and the regular Cartesian coordinates.
Figure 3: Image rays and time migration coordinates.
To explain this idea, we begin with the high frequency approximation applied to the waveequation (1), in which the wave front T (x, y, z) propagates according to the Eikonal equation(see e.g. [Popov, 2002]):
|T (x, y, z)|2 = 1v2(x, y, z)
. (2)
3
• The characteristics of the Eikonal equation can be viewed as rays. Among all rays startingat the reflection point R and reaching earths surface (Fig. 3), some have minimal travel time.These rays are called image rays, and it is easy to see that they must arrive perpendicular to thesurface. The ray RI in Fig. 3 is one such image ray. Thus, we may characterize the reflectionpoint R in one of two coordinate systems: either (1) its natural Cartesian coordinates x or (2)the point on the surface such that an image ray leaving x0 and traveling for a given time reachesthe point R. The former given are called depth coordinates, while the latter are called timemigration coordinates.
The conventional time migration coordinates are (x0, t0) where x0 is the escape locationof the image ray, and t0 is the doubled (two-way) travel time along it. Note that the lateralposition of the point R in the time migrated image is determined namely by the escape locationof the image ray I rather than by its projection Q to the surface (Fig. 3).
1.2 Travel time approximation
Figure 4: Travel time approximation.
Let S be a source and G be a receiver (Fig. 4), and let R be the reflection point. Supposefirst that the velocity v inside the earth is constant. Then the total travel time from S to R andfrom R to G is:
tSR + tRG =
t204
+|x0 S|2
v2+
t204
+|x0 G|2
v2. (3)
The ellipse in Fig. 4 is the locus of the reflection points A such that the total travel timetSA + tAG is the same as tSR + tRG.
In the general case where the velocity inside the earth is arbitrary, formula (3) serves as atravel time approximation for time migration (namely its modern variant called prestack timemigration), see [Yilmaz, 2001]. In this approximation, x0 is the escape location of an image rayfrom the reflection point R, and t0 is the two-way travel time along it. The velocity v present informula (3) is replaced with a parameter with dimension of velocity which depends on x0 andt0. These parameters are called the migration velocities and denoted by vm(x0, t0). They arechosen to provide the best fit to formula (3) in the process of time migration. Thus, formula(3) can be rewritten as
t(S, G, x0, t0) =
t204
+|x0 S|2v2m(x0, t0)
+
t204
+|x0 G|2v2m(x0, t0)
. (4)
In the case where the velocity inside the earth depends only on the depth and the distancebetween the source and the receiver is small, the migration velocity vm(x0, t0) is the RMSvelocity, given by
vm(t0) =
1
t0
t0
0
v2(z( ))d . (5)
4
• .
1.3 Emerging wave front
In this section, our aim is to justify the travel time approximation given by formula (4).
Figure 5: Emerging wave front.
Consider an emerging wave front from a point source A (Fig. 5) [Hubral and Krey, 1980].Let the image ray arrive at the surface point (x0, y0) at time t0 (here t0 is the one-way traveltime along the image ray). The travel time from A to the surface along some other ray close tothe image ray, arriving at the surface point (x, y), is given by the Taylor expansion
t(x, y) = t0 +1
2xTx + O(3), (6)
where x =
(
x x0y y0
)
, is the matrix of the second derivatives of t(x, y) evaluated at
the point (x0, y0) and =
(x x0)2 + (y y0)2. From geometrical considerations, one canobtain [Hubral and Krey, 1980] a relation between the matrix and the matrix R of the radiiof curvature of the emerging wave front, namely
1 = Rv(x0, y0), (7)
where v(x0, y0) = v(x = x0, y = y0, z = 0) is the velocity at the surface point (x0, y0).In the case where sources and receivers are arranged along some straight line, seismic imaging
becomes a 2D problem, and equation (6) can be simplified to
t(x) = t0 +1
2(x x0)2txx(x = x0) + O(3) = t0 +
(x x0)22Rv(x0)
+ O(3), (8)
where v(x0) v(x = x0, z = 0). By squaring both sides of equation (8) we get:
t2(x) = t20 + (x x0)2t0txx(x = x0) + O(3) = t20 + (x x0)2t0
Rv(x0)+ O(3). (9)
Suppose we want to compute the total travel time from a source S to the reflection point A andfrom A to a receiver G. Using equation (9) we obtain:
t(x0, t0, S, G) = tSA + tAG =
t20 + (S x0)2t0
Rv(x0)+
t20 + (G x0)2t0
Rv(x0)+O(3). (10)
5
• Comparing equations (10) and (4) we see that the travel time approximation given by formula(4) follows from the Taylor expansion in 2D. Moreover, the migration velocity and the radius ofcurvature of the emerging wave front are converted through the relation
t0v2m(x0, t0) = v(x0)R(x0, t0). (11)
On the other hand, in 3D the travel time approximation given by formula (4) is not aconsequence of the Taylor expansion as it is in 2D. Instead, one can easily derive the followingtravel time formula from equations (6) and (7):
t(x0, t0, S, G) =
t20 + t0(S x0)T [v(x0)R(x0, t0)]1(S x0) (12)
+
t20 + t0(G x0)T [v(x0)R(x0, t0)]1(G x0).
However note that if the velocity depends only on the depth, the matrix R is a multiple ofthe identity matrix, and hence formula (4) is the consequence of the Taylor expansion.
1.4 Dix inversion
Dix [Dix, 1955] established the first connection between the migration velocities and the seismicvelocities for the case where the velocity depends only on the depth. He showed that themigration velocities are the RMS velocities if the distances between the sources and the receiversare small and came up with the following inversion method. Consider an earth model as in Fig.
Figure 6: Dix inversion.
6. Let the layers be flat and horizontal, and the velocity be constant within each layer. We aregiven the RMS velocities Vi and the travel times ti, i = 1, 2, ..., n, where Vi is the RMS velocityof the first i layers with respect to the time, and ti is the two-way vertical travel time from theearth surface to the bottom of the i-th layer. Then the layer velocities (or as they are called ingeophysics interval velocities) vi can be found successively from i = 2 to n:
vi =
V 2i ti V 2i1ti1ti ti1
. (13)
The depths of the lower boundaries of the layers are:
zi = zi1 + viti ti1
2. (14)
6
• Sometimes Dix inversion is applied to find the interval velocities from the migration velocitiesin the case where the velocity varies laterally. Then for the continuously changing velocity in2D the Dix velocities are given by:
v(x0, t0) =
t0(t0v2m(x0, t0)). (15)
2 Forward modeling of the time migration velocities
In this section we derive our main theoretical result: the relation between the migration velocitiesand the true seismic velocities in 2D and between the matrix in formula (6) in 3D.
2.1 Paraxial ray tracing
Consider a ray propagating in a 3D medium with a smooth velocity (Fig. 7). Call this raycentral and attach a coordinate system (t, q1, q2) to it (see [Popov, 2002], [Cerveny, 2001]). Lett be the travel time along the central ray. Draw a plane perpendicular to the central ray at thepoint which it reaches at time t. Pick two mutually orthogonal directions in this plane and callthem ~e1 and ~e2. Then the location of any point M in the space can be expressed as
~rM = ~r0(t) + q1~e1 + q2~e2
for some t, q1 and q2, where ~r0(t) gives the point reached by the central ray at time t. If M isclose enough to the ray, its location can be described by (t, q1, q2) uniquely.
Figure 7: Paraxial ray tracing.
Suppose that the central ray is surrounded by a family of close rays and we want to writeequations of those rays in terms of q1(t) and q2(t) (Fig. 7). In order to apply the Hamiltonianformalism, we need to introduce the generalized momentums p1 and p2 corresponding to the
7
• generalized coordinates q1 and q2. We first note the fact that the central ray is a ray itself, andthis imposes the following requirements on the evolution of ~e1 and ~e2:
d~e1dt
=v(t, q1, q2)
q1 q1=q2=0~,
d~e2dt
=v(t, q1, q2)
q2 q1=q2=0~ ,
where ~ is the unit tangent vector to the central ray ([Popov, 2002]). The ray equations in theHamiltonian form are ([Popov, 2002], [Popov and Psencik, 1978],[Cerveny, 2001]):
d
dt
(
qp
)
=
(
0 v20I2 1v0 V 0
)(
qp
)
. (16)
Here v0 is the velocity along the central ray, I2 is the 2 2 identity matrix, and V is a 2 2matrix of the second derivatives of the velocity:
Vij =2v(t, q1, q2)
qiqj, i, j = 1, 2.
Suppose that the family of rays depends upon two parameters (1, 2). There are twoimportant cases:
All rays start perpendicular to the same plane. Then (1, 2) can be chosen to be theinitial coordinates (x0, y0) of the rays at this plane. We will call such a family of raystelescopic.
All rays start at the same point, but in different directions. Then (1, 2) can be chosento be the initial momentums (p1(0), p2(0)) of the rays. We will call such a family the pointsource family.
Consider the following 2 2 matrices ([Popov, 2002], [Cerveny, 2001]):
Qij qij
, Pij pij
, i, j = 1, 2. (17)
The equations of time evolution for Q and P are the equations in variations for equation (16):
d
dt
(
Q
P
)
=
(
0 v20I2 1
v0V 0
)(
Q
P
)
. (18)
The initial conditions for the telescopic family of rays are
Q(0) = I2, P(0) = 0, (19)
and for the point source family they are
Q(0) = 0, P(0) =1
v0(0)I2, (20)
where v0(0) is the velocity at the source point. The absolute value of the determinant of thematrix Q has a nice geometrical sense ([Popov, 2002]):
| detQ| is the geometrical spreading of the family of rays.
Let the central ray arrive orthogonal to some plane at a point (x0, y0). Consider the matrix of the second derivatives of the travel times of the family of rays around the central ray,evaluated at the point (x0, y0). E.g., the central ray can be the image ray arriving to the earthsurface. Then the matrix is defined by formula 6) for the source point family of rays from thesource point A as in Fig. 5. In [Popov, 2002], [Cerveny, 2001] it was shown that
= PQ1 (21)
8
• andd
dt = v202
1
v0V. (22)
For convenience, in the present work we will deal with the matrix K = 1, which is the matrixof radii of curvature of the wave front scaled by the velocity at the image ray, namely
K = v0R = QP1. (23)
One can easily derive from equation (22) that the time evolution of K is given by:
d
dtK = v20I2 +
1
v0KVK. (24)
For the point source family of rays the initial conditions for the matrix K are:
K(0) = 0. (25)
2.2 Relation between the matrix K and the true seismic velocities in
3D
Theorem 1 Let an image ray starting from a subsurface point x (Fig. 8) arrive at the earthsurface point x0 at time t0. Designate this ray to be central. Let the matrix K(x0, t0) be evaluatedat the surface for a point source family of rays around the image ray, starting at the same pointx. Suppose there is also a telescopic family of rays around the image ray starting perpendicularto the earth surface which we trace backwards w.r.t. the image ray for time t0 and compute thematrices Q and P. Let Q(x0, t0) be the matrix Q for the telescopic family of rays evaluated atthe time t0 (i.e., at the subsurface point x) in this backward tracing. Then
t0K(x0 , t0) = v
2(x(x0, t0))(
Q(x0, t0)TQ(x0, t0)
)1. (26)
Figure 8: Illustration for Theorem 1.
9
• Rewrite the travel time approximation (12) using the notation K:
t(x0, t0, S, G) =
t20 + t0(S x0)T [K(x0, t0)]1(S x0) (27)
+
t20 + t0(G x0)T [K(x0, t0)]1(G x0).
The matrix K in formula (27) is a matrix of parameters depending on x0 and t0, which can beestimated from the measurements. Theorem 1 provides a connection between the matrix K andthe true seismic velocity at the subsurface point x reached by the image ray arriving at x0 andtraced backwards for time t0 (Fig. 8).
Proof Let an image ray arrive at the surface point x0 at time t1. Fix a moment of time t0 < t1and consider a point source family of rays starting at the subsurface point x(x0, t0) which theimage ray passes at time t0. Introduce the following notations:
X =
(
Q
P
)
=
Q11 Q12Q21 Q22P11 P12P21 P22
, A(t) =
(
0 v20I2 1v0 V 0
)
.
Let X be the 4 4 matrix of derivatives of X with respect to the initial conditions:
X(t0) =
Q110 Q120Q210 Q220P110 P120P210 P220
:
X =
Q11Q110
Q11Q210
Q12Q120
Q12Q220
Q21Q110
Q21Q210
Q22Q120
Q22Q220
P11P110
P11P210
P12P120
P12P220
P21P110
P21P210
P22P120
P22P220
.
Note that since each of the columns of X is a linear independent solution of equation (16) thederivatives not included into X are zeros. X(t) and X(t) are solutions of the following initialvalue problems:
dX
dt= A(t)X, X(t0) =
1
v(t0)
(
0
I2
)
, (28)
where v(t0) = v(x(x0, t0)), and
dXdt
= A(t)X, X(t0) = I4. (29)
Denote the solution of equation (29) by B(t0; t1) as it is done in [Cerveny, 2001]:
B(t0; t1) =
(
Q1 Q2P1 P2
)
,
where Qi, Pi, i = 1, 2 are 2 2 matrices.(
Q1P1
)
satisfies the initial conditions corresponding
to a telescopic point, and
(
Q2P2
)
satisfies the initial conditions corresponding to a normalized
point source. B(t0, t1) is called the propagator matrix. Then the solution of (28) is:
X(t) =1
v(t0)
(
Q2P2
)
. (30)
10
• Now turn to the matrix K: K(t0; t1) = Q(t0; t1)P(t0; t1)1 = Q2P
12 .
Shift the initial time t0 by t. Then, according to equation (28) at time t0
Q(t0 t; t0) = 0 + tv2(t0)1
v(t0)I2 + O((t)
2),
P(t0 t; t0) =1
v(t0)I2 + O((t)
2).
Hence the change in the initial conditions for equation (28) is:
Q0 = v0tI2 + O((t)2), P0 = 0 + O((t)
2). (31)
Then
K(t0 t; t1) = K(t0; t1) +2
i,j=1
K
Qij0Qij0 +
2
i,j=1
K
Pij0Pij0 + O((t)
2) (32)
= K(t0; t1) +
(
K
Q110+
K
Q220
)
v(t0)t + O((t)2).
Let us find the partial derivatives in the expression above:
K
Qii0=
Q
Qii0P1 QP1 P
Qii0P1, i = 1, 2. (33)
In terms of the entries of the matrix B(t0; t1)
K
Q110+
K
Q220= v0(Q1P
12 Q2P12 P1P12 ). (34)
In [Cerveny, 2001] the symplectic property of the matrix B(t0; t1) was proved:
BTJB = J, (35)
where J is the 4 4 matrixJ =
(
0 I2I2 0
)
.
To simplify formula (34) we will use the following consequences of the symplectic property (35):
PT2 Q1 QT2 P1 = I2, PT2 Q2 = QT2 P2. (36)Then the matrix expression in equation (34) simplifies to:
Q1P12 Q2P12 P1P12 =
(PT2 )1PT2 Q1P
12 (PT2 )1PT2 Q2P12 P1P12 =
(PT2 )1(PT2 Q1 PT2 Q2P12 P1)P12 =
(PT2 )1(PT2 Q1 QT2 P2P12 P1)P12 =
(PT2 )1(PT2 Q1 QT2 P1)P12 =
(PT2 )1P12 . (37)
Substituting Eqn. (37) to Eqn. (34) and then to Eqn. (32) we get:
K(t0 t; t1) = K(t0; t1) + tv2(t0)(PT2 )1P12 + O((t)2). (38)Then the derivative of K with respect to the initial time is:
K(t0 ; t1)t0
= v2(t0)(PT2 )
1P12 . (39)
In [Cerveny, 2001] the following reciprocity property was proved:
PT2 (x1, x2) = Q1(x2, x1), (40)
where x1, x2 are the end points of the central ray. Applying it to equation (39) and taking thetime reverse into account we obtain formula (26).
11
• 2.3 Relation between the Dix velocities and the true seismic velocities
in 2D
In 2D the matrices Q, P and K become scalars which we denote by Q, P and K respectively.K is the radius of curvature of the wave front scaled by the velocity at the central ray: K = vR.The time evolution of Q, P and K are given by:
d
dt
(
QP
)
=
(
0 v20vqq
v00
)(
QP
)
,dK
dt= v2 +
vqqv
K2. (41)
In a similar way as it was done in 3D, it can be proven that
t0K(x0, t0) =
v2(x(x0, t0), z(x0, t0))
Q2(x0, t0). (42)
Then taking into account the definition of the Dix velocity (15) and the relation (11) betweenthe migration velocities and the radius of curvature of the emerging wave front we have thefollowing:
Theorem 2 Let an image ray arrive to the earth surface point x0 at time t0 from a subsurfacepoint (x, z). Suppose there is a telescopic family of rays around the image ray starting per-pendicular to the earth surface which we trace backwards w.r.t. the image ray for time t0 andcompute the quantities Q and P . Let Q(x0, t0) be the quantity Q for the telescopic family of raysevaluated at the time t0 (i.e., at the subsurface point (x, z)) in this backward tracing. Then theDix velocity vDix(x0, t0) is the ratio of the true seismic velocity v(x, z) and the absolute valueof Q(x0, t0):
vDix(x0, t0) =v(x(x0, t0), z(x0, t0))
|Q(x0, t0)|. (43)
Note that here, t0 is the one-way travel time along the image ray and that we denote thedepth direction by z.
2.4 Statement of the inverse problem
We state an inverse problem in 2D: the below results may be extended to 3D after some work.Here, t0 will denote the one-way travel time along the image ray.
Suppose there is an image ray arriving at each surface point x0, xmin x0 xmax. For any0 t0 tmax, trace the image ray backward for time t0 together with a small telescopic family ofrays. Let the image ray being traced backward reach a subsurface point (x, z) at time t0. Denoteby v(x0, t0) the velocity at the point (x, z), and by Q(x0, t0) the quantity Q for the corresponding
telescopic family at the point (x, z). We are given vDix(x0, t0) =v(x(x0,t0),z(x0,t0))
|Q(x0,t0)| f(x0 , t0),xmin x0 xmax, 0 t0 tmax. We need to find v(x, z), the velocity inside the domaincovered with the image rays arriving to the surface in the interval [xmin, xmax].
The first question is whether this problem is well-posed. In the next sections, we will showthat both the direct problem (given v(x, z) find f(x0, t0)) and the inverse problem (given f(x0, t0)
find v(x, z)) are ill-posed. We will use the notation f(x0 , t0) v(x(x0,t0),z(x0,t0))|Q(x0,t0)| rather thanvDix(x0, t0) to emphasize that f is computed as the ratio v/|Q| rather than from the optimalmigration velocities.
2.5 Ill-posedness of the direct problem
Direct Problem: Given v(x, z), xmin x xmax, xmin < 0, xmax > 0, z 0 and tmax findf(x0, t0) =
v(x0,t0)|Q(x0,t0)| , xmin x0 xmax, 0 t0 tmax.
12
• We shall show that small changes in v(x, z) can lead to large changes in f(x0, t0). Takev(x, z) = 1 and v(x, z) = 1 + a cos(kx), 1 x 1. Then
||v v|| = a.
Obviously, f(x0 , t0) = 1 for v(x, z) = 1. Compute f(x0, t0) for v(x, z) at x0 = 0. As the imageray arriving at x0 = 0 is straight, we have that
vqq = vxx(x = 0) = ak2.
Then we have:
dQ
dt0= (1 + a2)P,
dP
dt0=
ak2
1 + aQ, Q(0) = 1, P (0) = 0.
Therefore,d2Q
dt20=
ak2(1 + a2)
1 + aQ, Q(0) = 1,
dQ
dt0= 0.
Hence,
Q(t0) = cosh t0, =
ak2(1 + a2)
1 + a.
Pick k = 1a
and let a tend to zero. Then 12a
< 1
2
for a small enough. Thus, we have shown that arbitrarily small changes in the velocity v(x, y)may lead to significant changes in f(x0, t0), i.e., the direct problem is physically unstable in themax norm.
2.6 Ill-posedness of the inverse problem
Inverse Problem: Given f(x0, t0) =v(x0,t0)|Q(x0,t0)| , xmin x0 xmax, 0 t0 tmax, find v(x, z),
the velocity inside the domain covered with the image rays arriving to the surface in the interval[xmin, xmax].
Here we shall prove that the corresponding discrete problem is ill-posed: Given f(x0i, tk), i =0, 1, ..., n1, k = 0, 1, ..., p1, x0i = xmin + ix, tk = kt, where x = (xmax xmin)/(n1),t = tmax/(p 1) respectively, find v(xi, zj), i = 0, 1, ..., n 1, j = 0, 1, ..., m 1.
Let xmin = L and xmax = L and n be odd so that x = 0 is one of the grid lines. Supposewe are given the following two discrete arrays: (1) f(x0i, tk) = 1 and (2) f(x0i, tk) = 1 if x0i 6= 0and f(x0i, tk) = b > 1 if x0i = 0. Then
||f(x0i, tk) f(x0i, tk)|| = b 1. (44)
For f(x0i, tk) = 1 v(x, y) = 1. Let us find a velocity v(x, z) such that the exact values of f for itcoincides with f(x0i, tk) on the mesh. Let the mesh step in x0 be x. We will look for v(x, z)in the following form: pick 0 < x and set v(x, z) = 1 if |x| , and
v(x, z) = v(x, t0(z)) = 1 + (v(0, t0) 1) exp(
1 11
(
x
)2
)
13
• if |x| < . Here v(0, t0) is to be found. Note that
vxx(0, t0) = 2
2(v(0, t0) 1). (45)
Since f(0, tk) =v(0,t0)Q(0,t0)
= b,
Q(0, t0) =v(0, t0)
b. (46)
Due to the symmetry of our v(x, z), the ray starting at x0 = 0 perpendicular to the surface isstraight. Let us write the IVP for Q and P for this ray:
dQ
dt0= v2P, Q(T = 0) = 1,
dP
dt0= vxx
vQ, P (T = 0) = 0.
Here v(t0) v(0, t0). Taking into account relation (46) and using Eqn. (45) we get:dv
dt0= bv2P, v(t0 = 0) = b, (47)
dP
dt0=
2
2b(v 1), P (t0 = 0) = 0
Along with IVP (47) consider the following IVP:
dw
dt0= bu, w(t0 = 0) = b, (48)
du
dt0=
2
2b(w 1), u(t0 = 0) = 0.
Solving IVP (48) we find:
w(t0) = 1 + (b 1) cosh(
t0
2
)
.
Then by a variant of a comparison theorem, on the interval [0, T) where the solution to IVP(47) exists, v(t0) > w(t0). Hence, v(0, t0) either blows up, or reaches its maximum at tmax.Hence we conclude that
||v(x, z) v(x, z)|| > (b 1) cosh(
tmax
2
)
. (49)
Comparing formulae (44) and (49) we see that for any b we can pick = min{x, (b1)tmax
23 }
and hence make the left-hand side of Eqn. (49) greater than 1. Thus we have shown that theinverse problem is numerically unstable in the max norm.
2.7 Eulerian formulation of the inverse problem
The inverse problem stated in Section 2.4 can be formulated in a different, Eulerian way. Con-sider the mapping between the Cartesian coordinates (x, z) and the time migration coordinates(x0, t0). The functions x0(x, z) and t0(x, z) satisfy the following system of equations:
|x0|2 =(
x0x
)2
+
(
x0z
)2
=1
Q2(x, z), (50)
x0 t0 =x0x
t0x
+x0z
t0z
= 0, (51)
|t0|2 =(
t0x
)2
+
(
t0z
)2
=1
v2(x, z). (52)
14
• Equation (50) follows from the definition of Q. Equation (51) indicates that the curves t0=constare orthogonal to the image rays, and will be derived in Section 3.1.1 below. Equation (52) isthe Eikonal equation.
The input data are
v2Dix(x0, t0) =v2(x(x0, t0), z(x0, t0))
Q2(x(x0, t0), z(x0, t0)). (53)
The boundary conditions are:
x0(x, 0) = x, t0(x, 0) = 0, Q(x, 0) = 1, v(x, 0) = vDix(x0 = x, t0 = 0). (54)
3 Numerical algorithms
In this section we will propose three numerical algorithms. We will start with an efficienttime-to-depth conversion algorithm. The input for it is v(x0, t0) v(x(x0, t0), z(x0, t0)). Theoutput is v(x, z), x0(x, z) and t0(x, z). This algorithm is an essential part of the other twoalgorithms which produce v(x, z) from vDix(x0, t0). The first of these two, based on the raytracing approach, creates v(x0, t0), the input for the time-to-depth algorithm. The second,based on the level set approach, uses it as a part of its time cycle. Also, if nothing else isavailable, Dix velocities can be used as the input for our time-to-depth conversion. The mainadvantage of this time-to-depth conversion algorithm is that it is very fast and robust.
3.1 Efficient time-to-depth conversion algorithm
In this section we will use notation T for t0 to be consistent with the notations in the Eikonalequation (2). Also, we will deal with the reciprocal of the velocity s(x, z) called slowness forconvenience.
3.1.1 Eulerian formulation of the boundary value problem
Let (x, z) be a subsurface point (Fig. 9). Let s(x, z) be the slowness at the point (x, z). Let theimage ray from (x, z) reach the surface at some point x0 and let T be the one-way travel timefrom (x, z) to the surface point x0.
Figure 9: Section 3.1.1. Relation between (x, z), x0 and T .
Let xmin x0 xmax, 0 T Tmax, xmin x xmax, 0 z zmax. Given s(x0, T ), ourgoal is to find s(x, z), x0(x, z) and T (x, z), i.e., the slowness at each subsurface point (x, z), the
15
• escape location of the image ray from each subsurface point (x, z), and the one-way travel timealong each image ray. Thus, the input for this algorithm is given in the time domain (x0, T ),and the desired output is in the depth domain (x, z).
The functions x0(x, z) and T (x, z) are well-defined in the case if the image rays do notintersect inside the domain in hand. If the image rays intersect, the algorithm will follow thefirst arrivals to the surface.
The functions s(x0, T ), x0(x, z) and T (x, z) are related according to the following system ofPDEs:
|T |2 = s2(x0, T ) s(x0(x, z), T (x, z)), (55)T x0 = 0. (56)
Equation (55) is the Eikonal equation with an unknown right-hand side. Equation (56) givesa connection between x0 and T , and indicates that the curves T=const are orthogonal to theimage rays. We may derive this relation as follows:
We first note that the escape location x0 is constant along each image ray. Hence the timederivative of x0 along each image ray must be zero:
dx0dT
=x0x
dx
dT+
x0z
dz
dT= 0. (57)
Writing the equations of the phase trajectories for the Hamiltonian
H =1
2|T |2 1
2s2(x, z) = 0
given by the Eikonal equation, we have that
dx
dT=
T
x
1
s2,
dz
dT=
T
z
1
s2.
Substituting this into equation (57) we get:
x0x
dx
dT+
x0z
dz
dT=
1
s2x0 T = 0.
Hence, x0 T = 0 as desired.We also have boundary conditions for the system (56):
x0(x, 0) = x, T (x, 0) = 0, s(x, 0) = s(x0 = x, T = 0). (58)
3.1.2 Numerical algorithm
The motivation and the main building block or this algorithm is Sethians Fast MarchingMethod [Sethian, 1996] designed for solving a boundary value problem for the Eikonal equa-tion with known right-hand-side. This method is a Dijkstra-type method, in that it system-atically advances the solution to the desired equation from known values to unknown valueswithout iteration. Dijkstras method, first developed in the context of computing a short-est path on a network, computes the solution in order N logN , where N is the total num-ber of points in the domain. The first extension of this approach to an Eikonal equation isdue to Tsitsiklis [Tsitsiklis, 1995], who obtains a control-theoretic discretization of the Eikonalequation, which then leads to a causality relationship based on the optimality criterion. Tsit-siklis algorithm evolved from studying isotropic min-time optimal trajectory problems, andinvolves solving a minimization problem to update the solution. A more recent, finite differ-ence approach, based again on Dijkstra-like ordering and updating, was developed by Sethian[Sethian, 1996, Sethian, 1999A] for solving the Eikonal equation. Sethians Fast MarchingMethod evolved from studying isotropic front propagation problems, and involves an upwind
16
• Figure 10: Fast Marching Method. Black, grey and white dots represent Accepted, Consideredand Unknown points respectively.
finite difference formulation to update the solution. Both Tsitskilis method and the Fast March-ing Method start with a particular (and different) coupled discretization and each shows thatthe resulting system can be decoupled through a causality property. In the particular case ofa first order scheme on a square mesh, the resulting quadratic update equation at each gridpoint is the same for both methods. We refer the reader to these references for details on or-dered upwind methods for Eikonal equations, as well as [Sethian and Vladimirsky, 2003] for adetailed discussion about the similarities and differences between the two techniques. More re-cently, Sethian and Vladimirsky have built versions a class of Ordered Upwind Methods, basedon Dijkstra-like methodology, for solving the more general class of optimal control problems inwhich the speed/cost function depends on both position and direction, which leads to a convexHamilton-Jacobi equation. See [Sethian and Vladimirsky, 2003] for details.
We now discuss the Fast Marching Method in more detail, since it will serve as a buildingblock to our algorithm. In order to follow the physical propagation of information, an upwindscheme is used, and the solution is computed at points in order of increase of T . In order toachieve it, the points are divided into Accepted, where T is computed and no longer can be up-dated and can be used for estimation of T at its neighbors; Considered, where T is computedbut may be updated in future and cannot be used for estimation of T at other points; Un-known, where no value of T has been computed yet. At each time step a Considered pointwith the smallest value of T , determined by the heap sort, becomes Accepted. Sethian usedthis approach to compute the solution of the Eikonal equation with known right-hand-side in avariety of settings including semiconductor processing, image segmentation, seismic wave prop-agation and robotic navigation: for details, see [Sethian, 1996, Sethian, 1999A, Sethian, 1999B].
In our case the principal difference is that the right-hand side of the Eikonal equation is un-known. In result we do not know where the information comes from. This creates an issue whichwe will discuss in Section 3.1.3 below, however, we first outline the time-to-depth conversionalgorithm.
Let us discretize and solve the system (55), (56) with boundary conditions (58). The inputfor the numerical algorithm is the matrix s(x0i, Tk), i = 0, 1, ..., n 1, k = 0, 1, ..., p 1. Denotethe mesh steps in x0 and T by hx and T respectively. The mesh steps in x and z are hxand hz respectively. We define s(x0, T ) beyond the mesh points by the bilinear interpolation.The output of the numerical algorithm are the matrices s(xi, zj), x0(xi, zj) and T (xi, zj), i =0, 1, ..., n 1, j = 0, 1, ...,m 1. The algorithm is the following:
1. Mark the boundary (surface) points (xi = x0i, z = 0) as Accepted. Set s(xi, z0 =
17
• 0) = s(x0 = xi, T = 0), x0(x, z = 0) = x0, T (x, z = 0) = 0 according to the boundaryconditions. Mark the rest of the mesh points (xi, zj) as Unknown.
2. Mark the Unknownpoints adjacent to the Accepted points as Considered. We calltwo points adjacent (or nearest neighbors) if they are separated by one edge.
3. Compute or update tentative values of s(xi, zj), x0(xi, zj) and T (xi, zj) at the Consid-ered points.
(a) If a Considered point E has only one Accepted nearest neighbor D as in Fig. 10,then the values at E are found from the 1-point-update system:
x0(E) = x0(D),
T (E) T (D) = hs(x0(D), T (E)),s(E) = s(x0(E), T (E)),
T (E) > T (D).
(59)
Here H is either hx or hz depending on the arrangement of E and D.
(b) If a Considered point has only two Accepted nearest neighbors and they arelocated so that it lies linearly between them, then we compute triplets of tentativevalues of s, x0 and T for each of the two Accepted points and the Consideredpoint from system (59), and then choose the triplet with the smallest value of T .
(c) If a Considered point C has only two Accepted neighbors A and B not lying onthe same grid line, as in Fig. 10, then the tentative values at C are found from the2-point-update system:
(T (C) T (A))2hx2
+(T (C) T (B))2
hz2= s2(x0(C), T (C)), (60)
(T (C) T (A))(x0(C) x0(A))hx2
+(T (C) T (B))(x0(C) x0(B))
hz2= 0,
s(C) = s(x0(C), T (C)),
x0(A) x0(C) x0(B),T (C) max{T (A), T (B)}.
We solve the first two equations in system (60) using a Newton solver.
(d) If a Considered point has three or more Accepted nearest neighbors then wecompute a triplet of tentative values for each possible couple of Accepted pointsforming a right triangle together with the Considered point such that the Consid-ered point lies at its right angle, and choose the triplet with the smallest value ofT .
4. Find a Considered point with the smallest tentative value of T and mark it as Ac-cepted. We use a heap sort to keep track of the tentative T values.
5. If the set of Considered points is not empty, return to 2.
3.1.3 Causality
At T = 0 the wave front is a segment of the straight line from (x0, 0) to (xn1, 0). In order topropagate it correctly, we must compute the points in order of increase of T as given in SethiansFast Marching Method.
18
• In the above our update principle, the 1-point update (59) artificially puts point E on theimage ray passing through D (Fig. 10) prescribing x0(E) = x0(D), while the 2-point-updatelooks for the correct image ray (the correct value of x0).
At the moment when some Unknown point becomes Considered, it has only one Ac-cepted nearest neighbor. Therefore the tentative values at it are found from the 1-point updatesystem (59). Then, if it does not become Accepted by that time, it gets two Accepted neigh-bors lying on different grid lines. Then the values at it are found from the 2-point-update system(60). We emphasize that we design our algorithm so that the 2-point-update values replace the1-point-update values whenever it is possible independently of whether the new tentative valueof T is smaller or larger. Note that in the Fast Marching Method, the 2-point-update valuenever exceeds the 1-point-update value due to the fact that the slowness is known at eachpoint. In our formulae (59) and (60) for 1- and 2-point-update respectively the slowness s inthe right-hand side depends on T . Because of this, we cannot eliminate the situation wherethe value of T given by 1-point-update is smaller than the one given by 2-point-update. Sucha situation is dangerous because the 1-point-updates setting x0(E) = x0(D) is correct only ifthe true velocity (slowness) at E is larger (smaller) then at both of its nearest neighbors in thedirection perpendicular to the segment DE (Fig. 10). Thus, in the case where this setting isincorrect, the 1-point-update values must be replaced by 2-point update values before the pointgets Accepted in order to propagate the front in order of increase of the true values of T . Thequestion is whether we can guarantee it.
We found examples where indeed a smaller tentative value of T from the 1-point-updatewas replaced by a larger one from the 2-point-update in a small subset of points. However,numerous numerical experiments showed that such points disappear as we refine the mesh ofthe input data s(x0i, Tk), i = 0, 1, ..., n 1, k = 0, 1, ..., p 1. Moreover, we did not find anyexample where the points with 1-point-update values got accepted when they should not be.Thus, although the upwind principle may be violated in theory, we have not found any suchexample in practice.
3.1.4 Boundary effects
We have input data in the rectangular time domain (x0, T ), and we look for the output inthe rectangular depth domain (x, z). We will call the image rays arriving at the end points ofthe earth surface segment of the domain the boundary image rays. There are three possiblebehavior of a boundary image ray:1) the ray is straight, i.e, lies strictly on the boundary of the domain;2) the ray escapes from the domain;3) the ray enters the interior of the domain.If the boundary image ray is either straight or escapes from the domain then our numericalalgorithm computes the values at the boundary mesh points correctly, as the physical domainof dependence of each boundary point lies inside the numerical domain of dependence in thesecases. If the boundary ray enters the interior of the domain, then the values at the boundarypoints are computed by 1-point-updates. The physical domain of dependence for each boundarypoint lies outside the domain, and hence, cannot be inside the numerical domain of dependence.In this case, our algorithm does not converge in the cone of influence of the boundary points.
3.1.5 Synthetic data example
As a first example, we took the velocity field
v(x, z) = 1 +1
2cos
x
3sin
z
3,
and generated the input data v(x0, T ), 0 x0 12, 0 T 5 for our time-to-depth conversionalgorithm on a 200 200 nx0 nT mesh by shooting characteristics. Then we applied thealgorithm to these data and computed the velocity v(x, z) on the 200 400 nx nz mesh. The
19
• results are presented in Fig. 11-12. The exact velocity is shown in Fig. 11(a); the input dataare shown in Fig. 11(b). The velocity found by the algorithm is shown in Fig. 11(c). Therelative error, i.e., (vfound vexact)/vexact is shown in Fig. 11(d). The maximal relative error isless than 5 percent and is achieved at the points where the image rays collapse. The image rayscomputed for the exact velocities are shown in Fig. 12. Note that 1) the image rays severelybend, diverge and intersect, and 2) the boundary image rays are straight, which eliminates theerrors from the boundary effects.
(a) (b)
(c) (d)
Figure 11: (a): the exact velocity v(x, z) = 1 + 12cos x
3sin z
3; (b): the input data v(x0, T ); (c):
the found velocity v(x, z); (d) the relative error: its maximus is less than 5 percent.
3.2 Algorithms producing the seismic velocities from the migration
velocities
The algorithm introduced in Section 3.1.2 requires the velocities v(x0, T ) as the input; onecan use the Dix velocities vDix(x0, T ) as input. However, Dix velocities are obtained with theassumption that the subsurface structures are horizontal and the velocity depends only on thedepth. Theorem 2 gave the relation between the Dix velocities and the true seismic velocities.In this section, we introduce two algorithms which try to construct the true seismic velocitiesfrom the Dix velocities and use the algorithm in Section 3.1.2 as their essential part. To besure, we have just proven that this problem is ill-posed; nonetheless we can develop algorithms
20
• Figure 12: The image rays computed for the exact velocity.
which attempt the smoothed reconstruction. The first one is based on the ray tracing approach,and the second one is based on the level set approach. This is a worthwhile endeavor: ournumerical examples below demonstrate that the Dix velocities and the true seismic velocitiesmay significantly differ in the case of lateral velocity variation.
3.2.1 Ray tracing approach
The ray tracing algorithm consists of three steps.Step 1. Find the image rays.Step 2. Compute the geometrical spreading |Q| = | dl
dx0| on the image rays and find v(x0i, Tk).
l is the length of the front.Step 3. Apply the time-to-depth conversion algorithm from Section 3.1.2 to get v(xi, zj),x0(xi, zj) and t0(xi, zj) from v(x0i, Tk).
Let us describe Step 1 in more details. The boundary conditions are v(x0i, T = 0) =f(x0i, T = 0), Q(x0i, T = 0) = 1, P (x0i, T = 0) = 0. The ray tracing system for the i-th ray isthe following:
xT = v sin , x(0) = x0i,
zT = v cos , z(0) = 0,
T = vn = vl, (0) = 0, (61)QT = v
2P, , Q(0) = 1,
PT = vnnv
Q = (vll
v+
vTv2
)
Q, P (0) = 0.
Here vn = vx cos vz sin is the derivative of v in the direction normal to the ray (note:vn vq); vl is the derivative of v with respect to the arc length of the front; vnn = vxx cos2 2vxz cos sin +vzz sin
2 is the second derivative of v in the direction normal to the ray (vnn vqq); vll is the second derivative of v with respect to the arc length of the front; is the curvatureof the front. Adalsteinsson and Sethian [Adalsteinsson and Sethian, 2002] derived the followingrelation between the second derivative of some physical quantity with respect to the arc lengthof the front and its second derivative along the line tangent to the front:
gll = gzz (gxnx + gznz),
21
• where n is the unit vector normal to the front. Replacing g with v and noticing that
vxnx + vznz = v =vTv
is the derivative of v with respect to the arc length of the ray, we get the last equation in (61).We solve system (61) for all of the rays simultaneously by the forward Euler method as
follows.For k = 0 to k = p 1 do:
1. Find the least squares polynomials for the set of points (li, vi(Tk)) where li is the arc lengthof the front between ray 0 and ray i at the time Tk, and vi(Tk) is the value of the velocityon the i-th ray at time Tk. Evaluate vl(Tk) and vll(Tk) taking the first and the secondderivatives of this polynomial. Moreover, replace the values of the velocity vi(Tk) by thevalues of this polynomial. Evaluate the curvature (Tk) as follows. Find the least squarespolynomials for the sets of points (i, xi(Tk)) and (i, zi(Tk)) where i is the index of the ray,and xi and zi are the x- and z-coordinates of the i-th ray at time Tk. Take the first andthe second derivatives of these polynomials px and pz and find
=pxp
z pzpx
(p2x + p2z )
3/2.
Approximate vT (Tk) by
vt(Tk) =v(Tk) v(Tk1)
T
if k > 0, and we set vT (T0 = 0) = 0, since the curvature of the front is zero at T = 0.
2. Perform one forward Euler step for each of the rays.
3. For each of the rays find vi(Tk+1) = fi(Tk+1)Q(Tk+1), where fi(tk+1) f(x0i, Tk+1),i = 0, 1, ..., n 1.
Remarks.
One can see that we find v(x0i, Tk), i = 0, 1, ..., n 1, k = 0, 1, ..., p 1 in the step1. Hence it is possible immediately go to step 3 to find v(xi, zj). However, numerousnumerical experiments showed that step 1 computes the image rays (x(x0i, Tk), z(x0i, Tk))significantly more accurately than the velocity v(x0i, Tk). And Step 2 which is very simple,significantly improves the accuracy of v(x0i, Tk).
As we have shown in Section 2.6 the inverse problem is numerically unstable. The use ofthe least squares polynomials suppresses the growth of the small bumps which naturallyappear in result of computations, and hence, stabilizes the algorithm.
The main limitation of this algorithm is that it blows up as the image rays come too closeto each other or diverge too much.
One can use the additional output x0(xi, zj) and t0(xi, zj) to convert a time-migratedimage to depth rather than perform depth migration with the found velocities v(xi, zj).
3.2.2 Level set approach
Level set methods, introduced in Osher and Sethian [Osher and Sethian, 1988], are numericalmethods for tracking moving interfaces: they rely in part on the theory of curve and surfaceevolution given in Sethian [Sethian, 1982, Sethian, 1985] and on the link between front propa-gation and hyperbolic conservation laws discussed in Sethian [Sethian, 1987]. These techniquesrecast interface motion as a time-dependent Eulerian initial value partial differential equation.For a general introduction and overview, see Sethian [Sethian, 1999B].
The main idea of a level set method is the representation of a front as the zero level setof some higher dimensional function. In our context, we want to propagate the wave frontcoinciding with the flat surface at t = 0 downward the earth. We embed the wave front into a
22
• 2D function (x, z) so that the front is its zero level set. Furthermore, we embed the quantitiesQ and P defined on the front into 2D functions q(x, z) and p(x, z) so that at each moment oftime Q = q(x, z){(x,z)|(x,z)=0} and P = p(x, z){(x,z)|(x,z)=0}, i.e., Q and P coincide with q andp on the zero level set of (x, z). Let
gx =x||, gz =
z||.
Let us find the system of equations for q and p. First note that
vnn = vxx cos2 2vxz cos sin + vzz sin2 .
Second, at each point of the the zero level set of , i.e. at each front point,
gx = cos , gz = sin .
Then we get the following equations for q and p:
qt = v2p, pt =
vxxg2x 2vxzgxgz + vzzg2z
vq. (62)
These equations coincide with the equations for Q and P on the front. Here we switch thenotation for time from T to t. We will reserve the notation T for auxiliary times in the fastmarching parts of our level set algorithm.
Thus, we have to solve the following system of PDEs:
t + v(x, z)|| = 0,qt = v
2(x, z)p, (63)
pt = vxxg
2x 2vxzgxgz + vzzg2z
v(x, z)q.
As before, we have the input data f(x0 , t) =v(x0,t)
|Q(x0,t)| given in (x0, t) space on a n k mesh,and we need to obtain v(x, z) in (x, z) space on a n m mesh.
Initialization: Set q(x, z) = 1, p(x, z) = 0, which is correct for the front at t = 0. Setv(x, 0) = f(x0, 0) and attach labels x to the surface points. Set (x, z) = z, i.e., make thelevel set function a signed distance function.
We solve system (63) in the following time cycle: for k = 0 to p 1 do:1. Starting with the current x points, solve the system
q(x, z)|T | = 1f(x0, T )
, x0 T = 0
using the Fast Marching time-to-depth conversion algorithm introduced in Section 3.1.2to find v(x, z) = f(x, z)q(x, z) for the current q(x, z).
2. Attach labels x to the accepted points for which T is not greater than the current valueof time tk.
3. Detect the zero level set of . Find the velocity v at the zero level set of and build anextension of v solving the system
|d| = 1, d vext = 0,
with the boundary conditions d = 0 and vext = v at the zero level set of , using the FastMarching Method, as it is suggested in [Sethian, 1996, Sethian, 1999A]. If the extendedvelocity is built this way, remains to be the signed distance function if it was it at t = 0.
23
• 4. Perform a time step: Compute the quantities gx and gz for the current . Find vxx, vxzand vzz by finding least square polynomials for each grid line x = xi and z = zj andevaluating their derivatives. Make one forward Euler step for equations (62) to find new qand p. Solve the level set equation
t + vext|| = 0from t = kt to t = (k + 1)t by the forward Euler method with a time step satisfyingthe CFL condition.
The main advantage of this algorithm in comparison with the ray tracing algorithm is thatit can work even if the image rays intersect, since it tracks the first arrival front.
Having obtained the true seismic velocities v(xi, zj) one can perform depth migration toobtain an improved seismic image in the Cartesian coordinates. Alternatively, knowing the ve-locity v(xi, zj) one can apply Sethians fast marching method [Sethian, 1996] to obtain t0(xi, zi)and x0(xi, zi) to convert the time migrated image to depth.
4 Synthetic data examples
4.1 Example 1
The example in this section allows us to compare performances of the ray tracing algorithm andlevel set algorithm with a somewhat typical approach. One typical approach to seismic velocityestimation is to compute the Dix velocities and then apply image ray tracing. Here we willreplace the image ray tracing with our time-to-depth conversion algorithm.
We considered the velocity fields of the form:
v(x, z) = 1 + exp(
c(x2 + (z 1)2))
, x0 [2, 2], t [0, 0.7]. (64)We took c = 0.5, c = 1 and c = 1.5. The larger c, the sharper the Gaussian anomaly. For eachof these fields we created the input data f(x0 , t) on a 200 200 x0 t mesh and applied each ofthe three algorithms to them: the time-to-depth conversion, the ray tracing, and the level set.The output v(x, z) is given on 200 200 x z mesh.
The exact velocity, the input data (the Dix velocity, the found velocity and the image raysfor the sharpest Gaussian anomaly corresponding c = 1.5 are shown in Fig. 13. We see that theDix velocity qualitatively differs from the exact velocity and the found velocity resembles theexact velocity much more than the Dix velocity.
The results are summarized in Table (1). We see that
Table 1: The maximal relative errors produced by the time-to-depth conversion, the ray tracing
and the level set algorithms on the data from the velocity field (64).
Algorithm Time-to-depth Ray tracing Level set
c = 0.5 0.31 0.023 0.078
c = 1 0.44 0.11 0.079
c = 1.5 0.49 0.29 0.20
the ray tracing and the level set produce significantly more accurate results than the typicalapproach;
the ray tracing approach is more accurate than the level set where the image rays divergemoderately, while it becomes less accurate as the divergence of the image rays increases.
Note that if the image rays diverge severely so that the derivative vnn (or, in different notations,vqq) becomes large, both our ray tracing and level set algorithms blow up, while the time-to-depth convergence algorithm produces inaccurate but stable results.
24
• (a) (b)
(c) (d)
Figure 13: (a): the exact velocity v(x, z); (b): the input data f(x0, t) vDix(x0, t); (c): the foundvelocity v(x, z); (d) the image rays.
4.2 Example 2
In this section we also consider an example with a Gaussian anomaly, but with numbers closerto the real seismic numbers:
v(x, z) = 2 + 2 exp(
(x2 + (z 2)2))
,
x0 [3, 3], t [0, 1],
The center of the anomaly lies at the depth of 2 km and the background velocity is 2 km/sec.The results (Fig. 14) are produced by the level set algorithm. The found velocity resembles theexact velocity while the Dix velocity and the found velocity differ qualitatively.
5 Field data example
In this section we consider a field data example coming from the North Sea (Fig. 15, left). Themain feature in this image is the salt dome. Typically, the velocity inside the salt is higher thanit is in the surrounding rock. Salt is light and it pushes the layers up as it comes from inside theearth. The lateral velocity variation here is severe according to the geophysical standards. Noterapidly changing values inside the salt dome, which indicates that the lateral velocity variationis too large for the time migration.
In Fig. 15, right, the time migration velocities chosen in the process of making this imageare shown. The Dix velocities were then obtained from them and smoothed (Fig. 16(a)). Thelevel set algorithm was then applied to these Dix velocities to obtain seismic velocities v(x, z)(Fig. 16(b)). The image rays were computed from the found v(x, z) by shooting characteristics.
25
• (a)
(b)
(c)
Figure 14: (a): the exact velocity v(x, z); (b) the input data: the Dix velocity converted to depth
by vertical stretch; (c): the found velocity v(x, z) and the image rays.
Figure 15: Left: seismic image from North Sea obtained by prestack time migration using velocity
continuation [Fomel, 2003]. Right: the corresponding time migration velocity.
26
• The depth domain (x, z) was cut at 3.3 km to make the the found v(x, z) into a rectangularmatrix.
(a)x
0, km
t 0, s
ec, t
he o
ne
way
trav
el ti
me
The smoothed Dix velocity, m/sec
0 2 4 6 8 10 12
0
0.5
1
1.5
2242
2759
3274
3790
4306
4822
(b) x, km
z, k
m
Velocity, m/sec, and the image rays
0 2 4 6 8 10 12
0
0.5
1
1.5
2
2.5
3
2170
2612
3055
3496
3840
4383
Figure 16: (a) The smoothed Dix velocity vDix(x0, t0); (b) the found seismic velocity v(x, z) andthe image rays computed from it.
The depth migrated image, built using the calculated v(x, z), is shown in Fig. 17. The imageis in the regular Cartesian coordinates. It shows subsurface structures up to 3.3 km in depthwhich is quite deep according to geophysical standards. There is a noisy reconstruction insidethe salt dome but the surrounding layers are resolved well. Overall, this image looks reasonable.
6 Conclusions
We derived a relation between the Dix velocities and the true seismic velocities in 2D, and arelation between parameters of the emerging wave front and the true seismic velocities in 3D.We stated an inverse problem and showed that it is ill-posed. We introduced three numericalalgorithms for obtaining seismic velocities as functions of the regular Cartesian coordinatesand tested them on synthetic data examples. We demonstrated that the Dix velocity can bequalitatively different from the true seismic velocity. We applied our algorithms to field datawith severe lateral velocity variation and obtained a reasonably looking seismic image in regularCartesian coordinates up to 3.3 km in depth.
References
[Adalsteinsson and Sethian, 2002] Adalsteinsson, D. and Sethian, J.A., Transport and Diffusionof Material Quantities on Propagating Interfaces via Level Set Methods, J. Comp. Phys, 185,1, pp. 271-288, 2002.
[Cerveny, 2001] Cerveny, V., Seismic Ray Theory: Cambridge University Press, 2001.
[Dix, 1955] Dix, C. H., Seismic velocities from surface measurements: Geophysics, 20, pp. 68-86,1955
[Fomel, 2003] Fomel, S., Time-migration velocity analysis by velocity continuation: Geophysics,68, pp. 1662-1672, 2003
[Hubral, 1977] Hubral, P., Time migration - Some ray theoretical aspects: Geophys. Prosp., 25,pp. 738-745, 1977
27
• Figure 17: The poststack depth migrated image obtained with the found v(x, z).
[Hubral and Krey, 1980] Hubral, P., Krey, T., Interval velocities from seismic reflection timemeasurements: SEG, 1980.
[Osher and Sethian, 1988] Osher S., Sethian, J. A., Front propagating with curvature dependentspeed: algorithms based on Hamilton-Jacobi formulations, J. Comp. Phys., 79, pp. 12-49,1988.
[Popov, 2002] Popov, M. M., Ray theory and gaussian beam method for geophysicists: Salvador:EDUFBA, 2002.
[Popov and Psencik, 1978] Popov, M. M, Psencik, I., Computation of ray amplitudes in inho-mogeneous media with curved interfaces: Studia Geoph. et Geod., 22, pp. 248-258, 1978.
[Sethian, 1982] Sethian, J. A., An Analysis of Flame Propagation: Ph.D. Dissertation, Depart-ment of Mathematics, University of California, Berkeley, CA, 1982.
[Sethian, 1985] Sethian, J. A., Curvature and the Evolution of Fronts: Commun. in Math.Phys., 101, pp. 487-499, 1985.
[Sethian, 1987] Sethian, J. A., Numerical methods for propagating fronts in Variational Methodsfor Free Surface Interfaces, (eds. P. Concus & R. Finn), Springer-Verlag, NY, 1987.
[Sethian, 1996] Sethian , J. A., A Fast Marching Level Set Method for Monotonically AdvancingFronts: Proceedings of the National Academy of Sciences, 93, 4, 1996.
[Sethian, 1999A] Sethian, J.A., Fast Marching Methods: SIAM Review, Vol. 41, No. 2, pp.199-235, 1999.
[Sethian, 1999B] Sethian, J. A., Level set methods and fast marching methods: CambridgeUniversity Press, 1999.
[Sethian and Vladimirsky, 2003] Sethian, J. A., Vladimirsky, A., Ordered Upwind Methods forStatic Hamilton-Jacobi Equations: Theory and Algorithms: SIAM J. Numer. Anal., 41, 1,pp. 325-363, 2003
[Shah, 1973] Shah, P. M., Use of wavefront curvature to relate seismic data with subsurfaceparameters: Geophysics, 38, 812-825, 1973.
[Tsitsiklis, 1995] Tsitsiklis, J.N., Efficient Algorithms for Globally Optimal Trajectories: IEEETran. Automatic Control, 40, pp. 1528-1538, 1995.
28
• [Yilmaz, 2001] Yilmaz, O., Seismic Data Analysis: Soc. of Expl. Geophys., 2001
29 | 17,622 | 61,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.905748 |
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A281874 Number of Dyck paths of semilength n with distinct peak heights. 10
1, 1, 1, 3, 5, 13, 31, 71, 181, 447, 1111, 2799, 7083, 17939, 45563, 115997, 295827, 755275, 1929917, 4935701, 12631111, 32340473, 82837041, 212248769, 543978897, 1394481417, 3575356033, 9168277483, 23512924909, 60306860253, 154689354527, 396809130463 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS a(n) is the number of Dyck paths of length 2n with no two peaks at the same height. A peak is a UD, an up-step U=(1,1) immediately followed by a down-step D=(1,-1). In the Mathematica recurrence below, a(n,k) is the number of Dyck paths of length 2n with all peaks at distinct heights except that there are k peaks at the maximum peak height. Thus a(n)=a(n,1). The recurrence is based on the following simple observation. Paths counted by a(n,k) are obtained from paths counted by a(n-k,i) for some i, 1<=i<=k+1, by inserting runs of one or more contiguous peaks at each of the existing peak vertices at the maximum peak height, except that (at most) one such existing peak may be left undisturbed, and so that a total of k new peaks are added. It appears that lim a(n)/a(n-1) as n approaches infinity exists and is approximately 2.5659398. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..1000 EXAMPLE a(3)=3 counts UUUDDD, UDUUDD, UUDDUD because the first has only one peak and the last two have peak heights 1,2 and 2,1 respectively. MATHEMATICA a[n_, k_] /; k == n := 1; a[n_, k_] /; (k > n || k < 1) := 0; a[n_, k_] := a[n, k] = Sum[(Binomial[k - 1, i - 1] + i Binomial[k - 1, i - 2]) a[n - k, i], {i, k + 1}]; Table[a[n, 1], {n, 28}] CROSSREFS A048285 counts Dyck paths with nondecreasing peak heights. Column k=1 of A287847, A288108. Cf. A287846, A287901, A289020. Sequence in context: A144471 A190667 A062304 * A135532 A127600 A262237 Adjacent sequences: A281871 A281872 A281873 * A281875 A281876 A281877 KEYWORD nonn AUTHOR David Callan, Jan 31 2017 STATUS approved
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Last modified March 18 21:51 EDT 2019. Contains 321305 sequences. (Running on oeis4.) | 807 | 2,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-13 | latest | en | 0.81664 |
https://brainly.ph/question/159987 | 1,487,818,172,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171070.80/warc/CC-MAIN-20170219104611-00520-ip-10-171-10-108.ec2.internal.warc.gz | 698,991,220 | 10,181 | # what is the answer by solving with complete a square s^2+4s+9=0
1
by shen143
2015-07-02T19:20:16+08:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
S²+4s+9=0
s²+4s=-9
s²+4s+[1/2(4)]²=-9+[1/2(4)]²
s²+4s+4=-9+4
(s+2)²=-5
s+2=√-5
s+2=(positive and negative) i√5
s= -2 + i√5
s= -2 - i√5 | 211 | 549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-09 | latest | en | 0.881388 |
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Fix a partially ordered set $(P, \le)$ with $N$ elements and real weights $w(p)$ for each $p \in P$. A subset $S \subset P$ is called closed if for any $x, y$ with $y \in S$ and $x \le y$ we also ...
• 111
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On-line pagerank in a streaming DAG (Directed Acyclic Graph)
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602 views
Pagerank in directed *acyclic* graphs (DAG)
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231 views
Monotone circuit representations of paths in a graph?
Consider a directed graph $G = (V, E)$ with a source $s \in V$ and sink $t \in V$. From $G$, I can define a monotone Boolean function $\phi_G$ on the set of variables $E$, in the following way: every ...
• 9,232
103 views
random sampling DAGs via nilpotent matrix sampling
The adjacency matrix of an acyclic graph is known to be a nilpotent matrix (all eigenvalues are zero). I am interested in sampling DAG adjacency matrices or equivalently sample random nilpotent ...
• 43
172 views
Survey on Erdős-Pósa?
Does anyone know of any good surveys on Erdős-Pósa? I am particularly interested in what are the latest results for the bounding function for directed and even cycles in planar and minor free graphs ...
• 238
1 vote
66 views
Directed tree decompositions on subtrees of DAGs
Given a DAG, is the arboreal decomposition of the DAG with the guarantee that given a node $x$, $v$ such that $x$ is reachable from $v$ are in the subtree of $x$? If not, is there a similar ...
• 23
426 views
Obtaining Sets of Ancestors Quickly in a Directed Acyclic Graphs
Suppose I have a DAG, $G = (V, E)$ and we know that all nodes in the DAG have at most $A$ ancestors. Let $V' \subseteq V$ be a subset of vertices of $V$. Is there a way to obtain the set of all ...
130 views
Finding nodes with enough unique ancestors
Given a DAG $G = (V, E)$, let $T \subseteq V$ be a set of nodes of $V$ that is computed via the following process. Assuming the nodes of $G$ are sorted in topological order, $v_1, \dots, v_n$. We ...
1 vote
79 views
Separating DAGs using separators consisting of lists of nodes and all ancestors
Suppose we are given a DAG, $G = (V, E)$ where $n = |V|$. We consider the sets $J_1, J_2, \dots, J_n$ to be lists of vertices where list $J_i$ consists of vertex $v_i \in V$ and all ancestors of $v_i$....
1 vote
142 views
Reordering a DAG with the minimum changes
Consider a DAG $(V,A)$ with an initial permutation $(v_1,v_2,…,v_n)$. We want to arrange the $n$ vertices in topological order while keeping as many vertices as possible. The problem is: Is it NP-...
140 views
219 views
Is the isomorphism problem between posets represented by DAGs GI-complete?
Given two directed acyclic graphs, how hard is the problem of checking whether the partial orders they represent are isomorphic? Is this problem GI-complete? I believe this problem is equivalent to ...
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Lighting up all elements of a poset by toggling upsets
I consider the following game on a finite poset $(P, <)$. At each point of the game, I have a set of elements $S$ of the poset which are "on", and all others are "off". Initially $S = \emptyset$. ...
• 9,232
1 vote
826 views
Breaking cycles in network graph by adding nodes and rerouting edges
I have a quite "common" need : making a directed graph (with one or several cycles) a directed acyclic graph (DAG). But the way I want to achieve it is, I guess, way more specific : I would like to ...
• 119
1 vote
64 views
Directed Acyclic Graph partition into minimum subgraphs with a constraint
I have this problem, not sure there is a name for it, wherein a Directed Acyclic Graph has different colored nodes. The idea is to partition it into minimum number of subgraphs with the following 2 ...
• 111
142 views
Common techniques for the acyclic orientation problem under some special constraint?
An acyclic orientation of an undirected graph is an assignment of a direction to each edge(an orientation) that does not form any directed cycle and therefore generates a directed acyclic graph(DAG). ...
• 513
1 vote
4k views
Number of simple paths between two vertices in a DAG
Let $G = (N, A)$ be a connected acyclic digraph (DAG). Furthermore, let $s \in N$ and $t \in N$ be two vertices on this graph, such that $t$ is reachable from $s$. My problem is: how many simple $s-t$...
180 views
Crime prevention using graph theory and machine learning
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• 149
168 views
Finding Cheapest n-Path [closed]
Given a weighted directed acyclic graph, how can I find the cheapest path from an Origin Vertex to a Destination Vertex which ...
• 97
1 vote
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Directed NP Hard Problem on DAG
There are problems that are NP-Hard on undirected graphs(maximum weight independent set and graph coloring) but are polynomial time solvable on trees. Tree decomposition is a good tool to talk about ...
• 23
480 views
Computing topological sort while keeping edges "short"
Motivation: I want to compute a topological sort order in which the connected vertices are close to each other. Problem statement: Given a DAG $G(V,E)$ with $n$ vertices, compute a topological sort ...
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min weight k-closure on DAG
The problem Given a (connected) DAG $G(V,E)$ where each node is assigned an (non-negative) integer weight an integer k where $0\leq k\leq|V|$ Find a induced subgraph $H$ of $G$ consisting of $k$ ...
2k views
Minimum cost topological ordering
We are given a $n$ vertex directed graph $G=(V,E)$ and also given a cost function $c:V\times [n]\to \mathbb{R}$. Consider a topological ordering of the vertices, $v_1,\ldots,v_n$, the cost of the ...
• 4,399
203 views
What is the name of this algorithm on direct acyclic graph?
I am trying to linearize the history of a git branch for display purpose. I want commits to be collocated by branch instead of simply displaying commits in the order given by the time of commit. In ...
• 123
1 vote
149 views
Generalized path cover problem in DAG
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• 11
429 views
Efficient algorithm for generating data dependency DAG from lists of memory ranges and access modes
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How to design an algorithm which turns an undirected graph into directed with all nodes of indegree higher than 0? [closed]
Given an undirected graph $G=(V,E)$ devise an algorithm that will check whether its edges can be directed in such a way that the vertices of the resulting directed graph will all have indegree higher ...
• 111
121 views
Multiple source shortest path with one reversal [closed]
Lets say we have a directed graph G, with vertices V, that have lengths l. I need to find the shortest path between every ordered pair of vertices in the graph, with the following constraint: In a ...
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Generalization of Dilworth's theorem for labeled DAGs
An antichain in a DAG $(V, E)$ is a subset $A \subseteq V$ of vertices that are pairwise unreachable, namely, there are no $v \neq v' \in A$ such that $v$ is reachable from $v'$ in $E$. From Dilworth'...
• 9,232
198 views
Two-player zero-sum games in extensive form represented as directed acyclic graphs
The following is a way to represent two-player zero-sum games in extensive form. Consider a directed acyclic graph $G$ where each non-terminal vertex is one of 3 types: player 1 vertex, player 2 ...
• 2,151
242 views
NP-completeness of a specific topological sorting problem
Consider $(V, E)$ be a DAG, and $p_1, \dots, p_n$ be its topological sorting (i.e. such permutation $p$ of $V$ that $\forall(x, y) \in E.\ p^{-1}(x) < p^{-1}(y)$). Let's call the goodness of $p$ a ...
• 191
1k views
Enumerating topological sorts of a vertex-labeled DAG
Let $G = (V, E)$ be a directed acyclic graph, and let $\lambda$ be a labeling function mapping each vertex $v \in V$ to a label $\lambda(v)$ in some finite alphabet $L$. Writing $n := |V|$, a ...
• 9,232 | 2,714 | 10,595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.904456 |
https://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=57654&nr_roman=(L)(V)MMDCLIV&lang=en | 1,632,198,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00190.warc.gz | 927,309,241 | 9,747 | # Convert number: 57,654 in Roman numerals, how to write?
## Latest conversions of Arabic numbers to Roman numerals
1,009,842 = (M)M(X)DCCCXLII Sep 21 04:26 UTC (GMT) 57,654 = (L)(V)MMDCLIV Sep 21 04:26 UTC (GMT) 109,621 = (C)M(X)DCXXI Sep 21 04:26 UTC (GMT) 895,714 = (D)(C)(C)(C)(X)(C)(V)DCCXIV Sep 21 04:26 UTC (GMT) 1,150,580 = (M)(C)(L)DLXXX Sep 21 04:26 UTC (GMT) 1,995,601 = (M)(C)(M)(X)(C)(V)DCI Sep 21 04:26 UTC (GMT) 1,150,618 = (M)(C)(L)DCXVIII Sep 21 04:26 UTC (GMT) 625,352 = (D)(C)(X)(X)(V)CCCLII Sep 21 04:25 UTC (GMT) 11,509 = (X)MDIX Sep 21 04:25 UTC (GMT) 1,151 = MCLI Sep 21 04:25 UTC (GMT) 144,252 = (C)(X)(L)M(V)CCLII Sep 21 04:25 UTC (GMT) 115,115 = (C)(X)(V)CXV Sep 21 04:25 UTC (GMT) 1,151,523 = (M)(C)(L)MDXXIII Sep 21 04:25 UTC (GMT) converted numbers, see more...
## The set of basic symbols of the Roman system of writing numerals
• ### (*) M = 1,000,000 or |M| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.
(*) These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: "(" and ")", because:
• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and
• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line "|" and the Roman numeral "I" (1).
(*) An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate "1,000 times". See below...
Logic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.
(*) At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.
Thus, initially, the largest number that could be written using Roman numerals was:
• MMMCMXCIX = 3,999. | 734 | 2,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-39 | latest | en | 0.901862 |
www.pomm-project.org | 1,653,686,510,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00660.warc.gz | 104,259,669 | 6,196 | # A short focus on recursive queries
Posted 6 years ago.
In the last article we saw how to perform basic statistical operations in the database. Let’s go on a previous article about CTE and more precisely about recursive CTE.
In case you want to try it your side, here is the GIST with the data and the structures.. Be sure you download the file and insert it through a \i psql command to avoid paste & copy annoyances with the COPY commands where data are separated using tabs.
## What do we want ?
Let’s say we want to fetch a single employee department but with the hierarchical list of its departments. The `department` table has a link to itself which indicates a parent relationship. By example, the employee id 56 (Dr. Ursula Davis II) belongs to the `dba` department which is part of the `production` department and ultimately the `my_company` department.
The problem here is that we have employees with different depth of departments so depending on the employee there might be a different number of joins to fetch all of its departments. The SQL99 standard defines a way to perform such recursive queries.
To understand how a recursive CTE works, let’s make a simple sequence generator:
``````with recursive
inc (a) as (
select 1::int4 as a
union all
select a + 1 as a from inc where a < 6
)
select a as generator from inc;
┌───────────┐
│ generator │
├───────────┤
│ 1 │
│ 2 │
│ 3 │
│ 4 │
│ 5 │
│ 6 │
└───────────┘
(6 rows)
``````
A first set named `inc` is defined. It contains a union of two subqueries, the first one being the starting element. The recursive query is joining itself through the `from` statement until `a` reaches 6. The same kind of queries can be used to create generators like Fibonacci, there is even a Mandelbrot fractal generator written in SQL on Postgresql’s wiki.
## Department recursion
How to follow the department path of a single employee:
``````with recursive
sub_department as (
select d.* from department d where d.department_id = 9
union all
select d.* from department d join sub_department sd on d.department_id = sd.department_parent_id
)
select sd.* from sub_department sd;
┌───────────────┬────────────┬──────────────────────┐
│ department_id │ name │ department_parent_id │
├───────────────┼────────────┼──────────────────────┤
│ 9 │ developer │ 6 │
│ 6 │ production │ 1 │
│ 1 │ my_company │ ¤ │
└───────────────┴────────────┴──────────────────────┘
(3 rows)
``````
The recursive part fetch the asked department first then fetch the department where the department_id is the previous department’s parent until there is no parent. Let’s now add the employee’s informations.
``````with recursive
emp_department as (select e.* from employee e where e.employee_id = 56),
sub_department as (
select d.* from department d join emp_department ed using (department_id)
union all
select d.* from department d join sub_department sd on sd.department_parent_id = d.department_id
)
select
ed.employee_id,
ed.name,
ed.birthdate,
array_agg(sd) as departments
from
emp_department ed,
sub_department sd
group by
ed.employee_id,
ed.name,
ed.birthdate
;
``````
The code above first query all the informations about the given employee. Then it recursively traverses its departments using a join on `department_id`. At last, it projects the result reducing the departments as an array of tuples. The result is
``````┌─────────────┬─────────────────────┬────────────────────────┬────────────────────────────────────────────────────┐
│ employee_id │ name │ birthdate │ departments │
├─────────────┼─────────────────────┼────────────────────────┼────────────────────────────────────────────────────┤
│ 56 │ Dr. Ursula Davis II │ 1995-01-07 17:18:17+00 │ {"(7,dba,6)","(6,production,1)","(1,my_company,)"} │
└─────────────┴─────────────────────┴────────────────────────┴────────────────────────────────────────────────────┘
(1 row)
``````
Depending on the interface’s needs, it may be needed to format directly the department hierarchy. This can be done using the `string_agg` aggregate function in place of `array_agg`:
`````` …
ed.birthdate,
string_agg(sd.name, ' > ') as hierarchy
from
emp_department ed,
…
┌─────────────┬─────────────────────┬────────────────────────┬───────────────────────────────┐
│ employee_id │ name │ birthdate │ hierarchy │
├─────────────┼─────────────────────┼────────────────────────┼───────────────────────────────┤
│ 56 │ Dr. Ursula Davis II │ 1995-01-07 17:18:17+00 │ dba > production > my_company │
└─────────────┴─────────────────────┴────────────────────────┴───────────────────────────────┘
(1 row)
``````
The last couple of articles were about statistical operations. The results returned by such queries are very often send to draw graphics and dashboards. Here we are more in more standard entity approach. Let’s use Pomm’s model manager to have `Employee` and `Department` entities. In an empty folder, add the following `composer.json` file:
``````{
require: {
pomm-project/cli: 2.0.*@dev,
pomm-project/model-manager: 2.0.*@dev,
pomm-project/foundation: 2.0.*@dev
}
}
``````
Then enter `composer.phar install` to enjoy composer’s awesomeness. Once the operation is finished, just create a file named `.pomm_cli_bootstrap.php` ( mind the starting dot ):
``````<?php
return new PommProject\Foundation\Pomm([
'my_company' => [
'dsn' => 'pgsql://greg/test',
'class:session_builder' => '\PommProject\ModelManager\SessionBuilder'
]
]);
``````
Now we can use Pomm’s CLI to check the connection and then generate:
• model class
• entity class
• structure class (overwritten every time)
``php vendor/bin/pomm.php pomm:generate:schema-all my_company public``
This command will generate the classes for all relations in the public schema (the default schema). This creates a directory structure as follow:
``````MyCompany
└── PublicSchema
├── AutoStructure
│ ├── Department.php
│ └── Employee.php
├── DepartmentModel.php
├── Department.php
├── EmployeeModel.php
└── Employee.php
2 directories, 6 files
``````
Let’s implement the query in the `EmployeeModel` class:
`````` public function findByPkWithDepartmentHierarchy(\$employee_id)
{
\$department_model = \$this
->getSession()
->getModel('\MyCompany\PublicSchema\DepartmentModel')
;
\$sql = <<<SQL
with recursive
emp_department as (select e.* from :employee e where e.employee_id = \$*),
sub_department as (
select d.* from :department d join emp_department ed using (department_id)
union all
select d.* from :department d join sub_department sd on sd.department_parent_id = d.department_id
)
select
:projection
from
emp_department ed,
sub_department sd
group by
:group_fields
SQL;
\$projection = \$this->createProjection()
->unsetField('department_id')
->setField('departments', "array_agg(sd)", "public.department[]")
;
\$sql = strtr(\$sql, [
':employee' => \$this->structure->getRelation(),
':department' => \$department_model->getStructure()->getRelation(),
':projection' => \$projection->formatFieldsWithFieldAlias('ed'),
':group_fields' => \$this
->createProjection()
->unsetField('department_id')
->formatFields('ed'),
]);
return \$this->query(\$sql, [\$employee_id], \$projection)->current();
}
``````
One can notice the user’s departments are fetched as objects. The converter is smart enough to convert this array as an array of `Department` entities.
And here is how to use it from a `test.php` file:
``````<?php
\$pomm = require __DIR__ . '/.pomm_cli_bootstrap.php';
\$employee = \$pomm['my_company']
->getModel('MyCompany\PublicSchema\EmployeeModel')
->findByPkWithDepartmentHierarchy(56)
;
echo json_encode(\$employee->extract(), JSON_PRETTY_PRINT) . "\n";
``````
And here is the result:
``````{
employee_id: 56,
name: Dr. Ursula Davis II,
birthdate: {
date: 1995-01-07 17:18:17.000000,
timezone_type: 1,
timezone: +00:00
},
departments: [
{
department_id: 7,
name: dba,
department_parent_id: 6
},
{
department_id: 6,
name: production,
department_parent_id: 1
},
{
department_id: 1,
name: my_company,
department_parent_id: null
}
]
}
``````
Enjoy ! | 2,081 | 8,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-21 | latest | en | 0.913115 |
https://uk.mathworks.com/matlabcentral/cody/problems/31-remove-all-the-words-that-end-with-ain/solutions/210470 | 1,603,794,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00605.warc.gz | 581,347,845 | 17,099 | Cody
# Problem 31. Remove all the words that end with "ain"
Solution 210470
Submitted on 28 Feb 2013 by Matthew M.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% s1 = 'The rain in Spain falls mainly on the plain'; s2 = 'The in falls mainly on the '; assert(strcmp(remAin(s1),s2));
2 Pass
%% s1 = 'The pain from my migraine makes me complain'; s2 = 'The from my migraine makes me '; assert(strcmp(remAin(s1),s2));
3 Pass
%% s1 = 'I had to explain that "ain" is not a word'; s2 = 'I had to that "" is not a word'; assert(strcmp(remAin(s1),s2));
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 224 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-45 | latest | en | 0.879454 |
https://en.wikipedia.org/wiki/Talk:Riemann_series_theorem | 1,524,367,490,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945484.58/warc/CC-MAIN-20180422022057-20180422042057-00599.warc.gz | 621,030,722 | 8,945 | # Talk:Riemann series theorem
## Long on theory, short on specifics
Hi! I was just poking through some articles, and after looking at "absolute convergence", I wound up here. I didn't find a worked example of how a conditionally convergent series can actually be re-arranged to yield two different sums in either place. I think such an example would be helpful to the general reader. I also think this article should mention earlier results of Cauchy and Dirichlet. DavidCBryant 13:51, 23 January 2007 (UTC)
Could some clever soul please add a proof of this interesting theorem. I swear the one given in my book on analysis is just wrong! —Preceding unsigned comment added by 82.17.66.106 (talk) 19:13, 22 March 2008 (UTC)
criticism for Riemann's theorem on a sum of conditionaly convergent series (pdf,eng,26KB) http://alexander-conon.narod.ru/science/crte.pdf
comment on "criticism for Riemann's theorem on a sum of conditionaly convergent series" (pdf,eng,26KB) http://alexander-conon.narod.ru/science/crte-c.pdf 82.151.112.75 (talk) 00:10, 25 June 2008 (UTC)
Is that correct?, nobody knows Alexander Conon. — Preceding unsigned comment added by 190.82.98.194 (talk) 14:04, 17 July 2013 (UTC)
## A simpler example
Consider the conditionally convergent series
1 - 1 + 1/2 - 1/2 + 1/3 - 1/3 + -...
Permutation of this series taking p positives and q negatives has sum log(p/q)
http://research.att.com/~njas/sequences/A166871 for log(3/2) —Preceding unsigned comment added by 62.43.57.10 (talk) 09:41, 1 November 2009 (UTC)
## References
I added references that discuss the Riemann-series theorem and its applications. Some of these also discuss infinite-dimensional versions (e.g. problem 28 of the Scottish Book) and links to probability theory (e.g. exchangeability, weak exchangeability, etc.). Thanks, Kiefer.Wolfowitz (talk) 17:53, 11 October 2010 (UTC) | 529 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-17 | latest | en | 0.963134 |
http://slideplayer.com/slide/3907446/ | 1,503,533,373,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886124662.41/warc/CC-MAIN-20170823225412-20170824005412-00607.warc.gz | 377,368,294 | 26,733 | # Evolution of Populations
## Presentation on theme: "Evolution of Populations"— Presentation transcript:
Evolution of Populations
Population = A localized group of organisms which belong to the same species. Species = Actual or potentially interbreeding natural populations which are reproductively isolated from other groups. Gene Pool = The total aggregate of genes in a population
Hardy-Weinberg Theorem
In the absence of other factors, the segregation and recombination of alleles during meiosis and fertilization will not alter the genetic makeup of a population. Describes a non-evolving population p pq + q2 = 1 Where p = frequency of A allele q = frequency of a allele
The Hardy-Weinberg theorem
The Hardy-Weinberg theorem
Question 2 b) If the frequency of the homozygous dominant genotype is 0.49, what is the frequency of the dominant allele. Let the dominant genotype be represented by AA…. AA = 0.49 … (A2 = 0.49) same as (p2 = 0.49) So what is the frequency of A (p) ?? A (p) = 0.7
Question 2 c) If the frequency of the homozygous recessive genotype is 0.36, what is the frequency of the dominant allele. Let the recessive genotype be represented by aa…. aa = 0.36 … (a2 = 0.36) So …a = 0.6 Then what is A ? A = 0.4
Question 2 d) If the frequency of the homozygous dominant genotype is 0.49, what is the frequency of the homozygous recessive genotype. Let the dominant genotype be represented by AA…. AA = 0.49 … (A2 = 0.49) Then A = 0.7 so…a must = 0.3 Therefore aa = 0.3x0.3 = 0.09
Hardy-Weinberg Theorem
Conditions needed for H-W to work Very large population size Isolation from other populations No mutations Random mating No natural selection Hardy-Weinberg animation
Microevolution
Microevolution MICROEVOLUTION = Small scale evolutionary change represented by a generation-to-generation change in a population’s allele or genotype frequencies
Microevolution Genetic drift Gene flow Mutation Nonrandom mating
Genetic Drift Changes in the gene pool of a small population due to chance The larger the population…the less important is genetic drift Most (but not all) natural populations are so large that the effect of genetic drift is negligible Reduces overall genetic variability
Genetic drift
Genetic Drift - Bottleneck Effect
Drastic reduction in population size by some natural disaster which kills organisms nonselectively The smaller population is now unlikely to represent the genetic makeup (diversity) of the original population Some alleles will be overrepresented, others will be absent. Example: hunting to near extinction
The bottleneck effect: an analogy
Cheetahs, the bottleneck effect (Very low genetic diversity)
Founder Effect When a few individuals colonize a new habitat.
The smaller the population size … the less likely the genetic makeup of the colonists will represent the gene pool of the large population that they left
Gene Flow Migration of fertile individuals or transfer of gametes (wind blown pollen for example) between populations Extensive gene flow can eventually group neighboring populations into a single population
Gene flow and human evolution
Mutations Have very little immediate effect on a large population
Important to evolution since it is the original source of genetic variation which is the raw material for natural selection
Nonrandom Mating Inbreeding
Assortive mating. Individuals mate with partners that are like themselves in phenotypic characters
Natural Selection In any sexually reproducing population, variation among individuals exists (genetic variation) and some variants leave more offspring than others Natural selection is the differential success in reproduction Natural selection is the only cause of microevolution that is ADAPTIVE, since it accumulates and maintains favorable genotypes
Review of Microevolutionary changes
Genetic Variation Review of genetic variation from sexual reproduction
Genetic Variation Polygenic characters Discrete characters
Multiple loci involved-vary quantitatively (many intermediate phenotypes) within a population. Example = height Discrete characters Determined by a single locus Polymorphism - when 2 or more forms of a discrete character are well represented in a population
Polymorphism in garter snakes
Genetic Variation Geographic variation
This variation in alleles exists among populations of most species Cline – a type of geographical variation that shows a graded change in some trait along a geographical feature (such as elevation)
Clinal variation in a plant
Genetic Variation Generation of variation Mutation Recombination
Point mutation – involves a single base pair in DNA Chromosomal mutation – usually effect many gene locus and are almost always deleterious Recombination Nearly all genetic variation in a population results from new combinations of already existing alleles !
Genetic Variation Maintenance of variation (how is it preserved?)
Diploid state hides some genetic variation from selection by the presence of recessive alleles in heterozygotes
Genetic Variation Maintenance of variation (how is it preserved?)
Balanced Polymorphism = the ability of natural selection to maintain diversity in a population Heterozygote advantage Frequency dependent selection
Frequency-dependent selection in a host-parasite relationship
Frequency-dependent selection
Polymorphism in sneetches (star-bellied and plain bellied forms) - a classic tale of frequency-dependent selection
Genetic Variation Neutral Variation
Fitness Measured by the relative contribution an individual makes to the gene pool of the next generation Selection acts on phenotypes and can only act indirectly on genotypes
Modes of selection
Modes of Natural Selection
Stabilizing selection Favors intermediate variants by selecting against extreme phenotypes Directional selection Favors variants to one extreme Diversifying selection Opposite phenotypic extremes are favored over intermediate phenotypes
Modes of selection
Sexual Dimorphism (Male peacock)
Distinction between secondary sexual characteristics of males and females
Sexual selection and the evolution of male appearance
Evolution does not fashion perfect organisms
Adaptations are often compromises. An organism must be versatile enough to do many things. Not all evolution is adaptive (example genetic drift in small populations) Selection can only EDIT variations that exist. New alleles/genes are not formed by “mutation on demand” | 1,331 | 6,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-34 | longest | en | 0.851389 |
https://web2.0calc.com/questions/algebra_11801 | 1,723,596,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00241.warc.gz | 477,969,311 | 5,771 | +0
# Algebra
0
2
1
+525
Find the vertex of the graph of the equation y = -2x^2 + 8x - 15 - 3x^2 - 14x + 25.
Jul 14, 2024
#1
+971
+1
y = -2x^2 + 8x - 15 - 3x^2 - 14x + 25
Combine like terms y = –5x2 – 6x + 10
The easiest way – the easiest way I know, anyway – is
to set the first derivitive equal to zero, and solve for x.
dy/dx = –10x – 6
0 = –10x – 6
10x = –6
x = –0.6 this is the x-coordinate of the vertex
plug it back into the original equation
y = –5(–0.62) – 6(–0.6) + 10
y = –1.8 + 3.6 + 10
y = 11.8 this is the y-coordinate of the vertex
The vertex is located at (–0.6 , 11.8)
Verified the answer with Desmos Graphing Calculator.
.
Jul 14, 2024 | 297 | 709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-33 | latest | en | 0.758237 |
https://www.pcreview.co.uk/threads/cell-programing.3724578/ | 1,719,064,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00449.warc.gz | 817,878,268 | 13,572 | # Cell Programing
J
#### John Savy
Could you show me how to write this program:
if[sum((A1)+(A2))/5] would have decimal points then "blank", not to show
otherwise the result.
In other words I'm looking for the result without decimal point and if the
result has decimal point, it would be out of scop of the purpose, therefore
I'm not interested on the result.
Thanks,
O
#### OssieMac
Hi John,
=IF(MOD(SUM(A1:A2),5)>0,"",SUM(A1:A2))
The above places a null value (which is represented by the double quotes) if
division by 5 does not return a whole number; otherwise will sum the range.
J
#### John Savy
I tried and didn't work. It gives formula error.
Again, I'm looking non-decimal figures.
For example, A1=4, A2= 6, so sum((A1+A2)/5) = 2; therefore the program
should show #2, if it was 2.45, I shouldn't show, since the result has
decimal points.
Thaks,
--
John
OssieMac said:
Hi John,
=IF(MOD(SUM(A1:A2),5)>0,"",SUM(A1:A2))
The above places a null value (which is represented by the double quotes) if
division by 5 does not return a whole number; otherwise will sum the range.
--
Regards,
OssieMac
John Savy said:
Could you show me how to write this program:
if[sum((A1)+(A2))/5] would have decimal points then "blank", not to show
otherwise the result.
In other words I'm looking for the result without decimal point and if the
result has decimal point, it would be out of scop of the purpose, therefore
I'm not interested on the result.
Thanks,
A
#### Ayush Jain
I tried and didn't work. It gives formula error.
Again, I'm looking non-decimal figures.
For example, A1=4, A2= 6, so sum((A1+A2)/5) = 2; therefore the program
should show #2, if it was 2.45, I shouldn't show, since the result has
decimal points.
Thaks,
--
John
OssieMac said:
Hi John,
=IF(MOD(SUM(A1:A2),5)>0,"",SUM(A1:A2))
The above places a null value (which is represented by the double quotes) if
division by 5 does not return a whole number; otherwise will sum the range.
Could you show me how to write this program:
if[sum((A1)+(A2))/5] would have decimal points then "blank", not to show
otherwise the result.
In other words I'm looking for the result without decimal point and if the
result has decimal point, it would be out of scop of the purpose, therefore
I'm not interested on the result.
Thanks,
I think the formula of Ossie Mac is almost right except the false
condition.. The right formula should be :
=IF(MOD(SUM(A1:A2),5)>0,"",(SUM(A1:A2)/5))
-Ayush Jain
O
#### OssieMac
Hi again John,
I forgot to divide by 5 in the second part but should not have returned a
formula error; just the wrong answer. Try the following. Works fine in xl2002
and xl2007.
=IF(MOD(SUM(A1:A2),5)>0,"",SUM(A1:A2)/5)
If still returning an error, what version of xl are you using?
--
Regards,
OssieMac
John Savy said:
I tried and didn't work. It gives formula error.
Again, I'm looking non-decimal figures.
For example, A1=4, A2= 6, so sum((A1+A2)/5) = 2; therefore the program
should show #2, if it was 2.45, I shouldn't show, since the result has
decimal points.
Thaks,
--
John
OssieMac said:
Hi John,
=IF(MOD(SUM(A1:A2),5)>0,"",SUM(A1:A2))
The above places a null value (which is represented by the double quotes) if
division by 5 does not return a whole number; otherwise will sum the range.
--
Regards,
OssieMac
John Savy said:
Could you show me how to write this program:
if[sum((A1)+(A2))/5] would have decimal points then "blank", not to show
otherwise the result.
In other words I'm looking for the result without decimal point and if the
result has decimal point, it would be out of scop of the purpose, therefore
I'm not interested on the result.
Thanks, | 1,056 | 3,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.89304 |
https://ewaterpurifier.com/inches-of-water-to-kpa/ | 1,669,566,995,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00528.warc.gz | 282,285,074 | 26,471 | January 30
# Transform Inch Of Water To Kpa
The dimension of tension coincides as that of pressure, and also consequently the SI unit for stress and anxiety is the pascal, which is equivalent to one newton per square meter (N/m ²). In Imperial systems, anxiety can be gauged in pound-force per square inch, which is abbreviated as psi. In The United States and Canada, air as well as various other commercial gasses are frequently gauged in inches of water when at low stress. This is in comparison to inches of mercury or pounds per square inch (psi, lbf/in2) for bigger stress.
## How do you convert kPa to meters of water?
Convert from Kilopascals to Metres of water.
Kilopascals to Metres of water.1 Kilopascals = 0.102 Metres of water10 Kilopascals = 1.0197 Metres of water2500 Kilopascals = 254.94 Metres of water9 Kilopascals = 0.9178 Metres of water1000 Kilopascals = 101.97 Metres of water1000000 Kilopascals = 101974.4 Metres of water7 more rows
Use this web page to learn just how to convert in between kilopascals and also inches of water. Use this web page to learn just how to transform in between inches of water and kilopascals. It is specified as the stress exerted by a column of water of 1 inch in height at specified conditions. At a temperature of 4 ° C( 39.2 ° F) pure water has its highest density (1000 kg/m3).
• In Imperial systems, stress and anxiety can be determined in pound-force per square inch, which is abbreviated as psi.
• The dimension of stress and anxiety is the same as that of stress, as well as therefore the SI device for stress is the pascal, which amounts one newton per square meter (N/m ²).
• In order to convert a value from inches of water to kilopascals kind the variety of inH2O to be converted to kPa and then click on the ‘convert’ switch.
• In The United States and Canada, air and also other industrial gasses are frequently gauged in inches of water when at low pressure.
Kilopascals to inches of water converter on this page calculates the number of inches of water are in ‘X’ kilopascals (where ‘X’ is the number of kilopascals to transform to inches of water). In order to transform a value from kilopascals to inches of water type the number of kPa to be transformed to inH2O and then click on the ‘transform’ switch. Variations of the inches of water to kilopascals conversion table. To produce a inches of water to kilopascals conversion table for various values, click on the “Produce a tailored stress conversion table” button. sl/US gal to g/US qt conversion table, sl/US gal to g/US qt unit converter or convert between all devices of thickness dimension. newton per square picometer is an acquired statistics measurement system of stress used by force of one newton on a surface of one square picometer.
Make use of the complying with calculator to convert between kilopascals and inch waters (60 ° F) . If you need to transform kilopascals to other systems, please attempt our universalPressure System Converter. The SI device for pressure is the pascal, which is a newton per square meter. Variations of the kilopascals to inches of water conversion table. To produce a kilopascals to inches of water conversion table for various worths, click on the “Produce a personalized stress conversion table” button.
To fix this, multiply 484 kilopascals with the conversion factor from kilopascals to inches of water. To address this, increase 51 inches of water with the conversion variable from inches of water to kilopascals. One of the most often utilized units of stress are pascal, kilopascal, megapascal, psi, torr, atm machine and bar. It is additionally given as inches of water scale (iwg or in.w.g.), inches water column, inAq, Aq, or inH2O. The devices are conventionally made use of for measurement of certain pressure differentials such as tiny stress distinctions across an orifice, or in a pipe or shaft. This on the internet system converter permits fast and exact conversion between numerous systems of step, from one system to one more. The Device Conversion page gives an option for engineers, translators, as well as for anybody whose activities call for working with quantities measured in different systems.
The pascal is the SI unit of pressure.It is equivalent to one newton per square metre. The system is called after Blaise Pascal, the eminent French mathematician, physicist and also theorist. Stress conversion calculator for a number of SI and also other regularly made use of pressure devices.
At that temperature level and also thinking the typical velocity of gravity, 1 inAq is about 249.082 pascals. Emergency treatment for severe altitude sickness has to be offered instantly.
One usage is in the dimension of air (” wind”) that supplies a church organ and also is referred merely as inches. In order to transform a value from inches of water to kilopascals type the variety of inH2O to be converted to kPa and afterwards click the ‘convert’ button.
## How do you calculate mmHg?
How to calculate mmHg? 1. Using the basic definition of mmHg calculate the blood pressure of 120 mm Hg. Pressure = Hg Density × Standard Gravity × Mercury Height.
2. Now to covert pressure Pa using proportion the Formula is:
3. Now calculate the pressure of 36,000 Pa using this same formula from step three:
4. Question.
5. A.
6. B.
7. C.
8. D.
More items
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inches, water
## Will Reverse Osmosis Remove Arsenic
Will Reverse Osmosis Remove Arsenic
## How to Install Aqua Grand Water Purifier
How to Install Aqua Grand Water Purifier | 1,314 | 5,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-49 | latest | en | 0.936788 |
http://delgaudiogourmet.com/acquaint-in-krygan/7ce9e4-oxidation-state-of-n-in-hcn | 1,721,582,849,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00762.warc.gz | 8,402,897 | 20,511 | # oxidation state of n in hcn
This cyano group is virtually identical to the cyano group which is bound to hydrogen in HCN. What is the oxidation number of N in CuCN? sodium thiosulfate). C +4, +4. Register to receive personalised research and resources by email, The Oxidation of HCN and Reactions with Nitric Oxide: Experimental and Detailed Kinetic Modeling, CNRS Laboratoire de Combustion et Systèmes Réactifs , 1C, Avenue de la Recherche Scientifique, Orléans Cedex 2, 45071, France, /doi/pdf/10.1080/00102200008947286?needAccess=true. Each atom only has one oxidation number in a compound, although one element can have different oxidative states (i.e. ∴ x + (-3) = -1. x = -1 + 3. x = +2. 1.1. Given that the ionic product of $Ni(OH)_2$ is $2 \times 10^{-15}$. oxidation state of n in cn . D-2, -2. However, on Pt‐based catalysts, which are typically present in the exhaust aftertreatment system to remove the potential NH 3 slip emissions after the SCR catalyst, HCN is either converted with high selectivity to N 2 O and NO x or is only poorly oxidized at low temperatures. The following table shows an excerpt from Pauling electronegativities (χPauling): ElementχPaulingH2.20N3.04 H N 3 is also known as hydrazoic acid. It has the common negative oxidation state of -1. Check Answer Next Question. The Oxidation State. The oxidation number of C and N in HCN and HNC respectively are. The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. 1 answer. 0 0. If it was a free radical, it would be neutral, and the N would be the same, so the C would be +3. Of N in HCN can be predicted by covalent bond rule. To begin, we assign the oxidation state of H before the N atom. Others have shown you how to determine the oxidation state of N in HNO3, but while nitrogen is assigned an oxidation state of +5, please do not interpret that as nitrogen having a charge of +5. 9. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. Solution: 1) Let's split ammonium dihydrogen phosphate apart. The most important process for the production of hydrogen cyanide is the Andrussov oxidation invented by Leonid Andrussow in which methane and ammonia react in the presence of oxygen at about 1200 °C over a platinumcatalyst: 1. In case of NH 3 oxidation (Figure 2B) in presence of HCHO, the oxidation of NH 3 is enhanced up to 5508C. An overall good agreement between the experimental results and the modeling was obtained. Oxidation number of hydrogen is +1. ON of C = 2 , of N = - 3. Registered in England & Wales No. In Wolff‐Kishner reduction, the carbonyl group of aldehydes and ketones is converted into. In the year 2000, 1.615 billion pounds (732,552 tons) were produced in the US. It is the same The oxidation state of carbon in HCN HNC is the same i.e +4 mainly depends upon e… Determine the oxidation number of each element in each of the following compounds: (a) HCN (b) \mathrm{OF}_{2} (c) \mathrm{AsCl}_{3} Enroll in one of our FREE online STEM summer camps. HCN is produced on an industrial scale and is a highly valuable precursor to many chemical compounds ranging from polymers to pharmaceuticals. Hence oxidation state of the C in CN – is +2. ing of hcn/no2 system a reactions a b e 1. no2+m=o+no+m 6.8e14 0.0 52850 2. o+no2=o2+no 1.0e13 0.0 600 3. hcn=hnc 5.13e21 3.49 23013 4. hnc+o=nh+co 5.44e12 0.0 0 5. hnc+oh=hnco+h 2.8e13 0.0 3700 6. hcn+o=nco+h 1.38e4 2.64 4980 7. hcn+o=nh+co 3.45e3 2.64 4980 8. hcn+o=cn+oh 2.70e9 1.58 29200 9. hcn+oh=cn+h2o 1.45e13 0.0 10930 10. hcn+oh=hocn+h 5.85e4 2.4 12500 11. hcn+oh=hnco+h 1.98e-3 … Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below: asked May 7, 2018 in Chemistry by paayal (147k points) neet; neet-2018; 0 votes. Still have questions? There are two types of oxidation of ethene may occurs in alkaline KMnO4 depending on the reaction condition.. Oxidation of ethene by cool dilute alkaline KMnO4 solution Identify compound X in the following sequence of reactions: Identify a molecule which does not exist. How to calculate oxidation state Using Lewis diagrams. The more electronegative element is assigned the negative oxidation number. It have monovalent combining group CN. 1 answer. Due to which there developes - 3 charge on N. (oxidation no. the equivalent weight of ferrous ion. H-C≡N. In the reaction, $3B{{r}_{2}}+6CO_{3}^{2-}+3{{H}_{2}}O\xrightarrow{{}}5B{{r}^{-}}$ $+BrO_{3}^{-}+6\,HCO_{3}^{-}$ Choose the correct statement. What is the oxidation number of vanadium in $R{{b}_{4}}Na[H{{V}_{10}}{{O}_{28}}]$ ? Box 999, Richland, WA 99352, United States In the compound HCN as C {carbon} has +4,+2,-4 oxidation numbers and N {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers. CNO1- O = -2 N = -3 (more electronegative than C) C = +4 The sum of these oxidation numbers is -1, … It can be in the 4+ oxidation state, 4- oxidation state and every oxidation state in between. In this context, keep in mind that the oxidation state of elemental oxygen (O 2) and nitrogen (N 2) is defined as zero As nitrogen is more electronegative than carbon so its oxidation number is -3. Suggest Edit. A detailed chemical kinetic model for NO-reburning (751 reactions and 102 species) was used to simulate the present experiments. It is a colorless, extremely poisonous and flammable liquid that boils slightly above room temperature, at 25.6 °C (78.1 °F). asked Dec 22, 2018 in Chemistry by monuk (68.0k points) redox reactions ; neet; 0 votes. Carbon has 3 bonds to nitrogen (+3), and one to hydrogen (-1), for a total charge of +2. The more electronegative element is assigned the negative oxidation number. The oxidation state is referred to as oxidation number. It forms the following two polyatomic ions: ammonium ---> NH 4 + dihydrogen phosphate ---> H 2 PO 4 ¯ 2) Analysis of the ammonium ion: We know that NH 4 + takes on a +1 charge because we know that NH 4 Cl exists. H - N + ≡ C -ON of N = -3 , … oxidation intermediate of HCHO is responsible for the formation of HCN. 53%. In compounds, the H atom bears an oxidation state of +1. An example of a disproportionation reaction is : Consider the following reduction processes : 1 mole of $FeSO_4$ (atomic weight of Fe is $55.84 \, g \, mol^{-1}$) is oxidized to $Fe_2(SO_4)_3$. Reaction between acetone and methyl magnesium chloride followed by hydrolysis will give : Identify the correct statements from the following: Which of the following set of molecules will have zero dipole moment ? 29%. It is very important method. Cyanide is a dead poisonous chemical, that exist as colorless gas and crystalline salt. 10%. To learn about our use of cookies and how you can manage your cookie settings, please see our Cookie Policy. The oxidation state of Cr in CrO 6 is : NEET 2019 Redox Reactions. The N would be -3, so the C would be +2. This increment in HCHO conversion could be DAMN is not oxidized when Fe(III) in the montmorillonite is reduced with hydrazine. That simply isn't the case. Hydrogen cyanide, sometimes called prussic acid, is a chemical compound with the chemical formula HCN. Incomplete. Answer . Potassium permanganate is a strong oxidizing agent.. Using state‐of‐the‐art technologies for selective catalytic reduction ... they form secondary emissions of HCN due to catalytic reactions of formaldehyde and its oxidation intermediates with NH 3. There are some rules to determine the oxidation numbers of each atoms in the compounds or ions. The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound.Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. 8%. That is not correct. The typical Lewis represention of cyanide anion is as #:N-=C:^-#.Of course this means that nitrogen has an oxidation number of #0#, because the sum of the oxidation numbers must be equal to the charge on the ion. This means that nitrogen will have the -3 oxidation state. In ozone (O 3), the oxidation state of oxygen is zero while in nitric acid (HNO 3), the oxidation state of nitrogen is +5. ⇐ Oxidation Number Calculator. Currently hydrogen cyanide is produced in high quantities by three processes. What are the oxidation states of O, C and N in the cyanate ion, [OCN]" ? Knife Care. The adsorption of HCN on, its catalytic oxidation with 6% O 2 over 0.5% Pt/Al 2 O 3, and the subsequent oxidation of strongly bound chemisorbed species upon heating were investigated.The observed N-containing products were N 2 O, NO and NO 2, and some residual adsorbed N-containing species were oxidized to NO and NO 2 during subsequent temperature programmed oxidation. The atomic radiusis: Find out the solubility of $Ni(OH)_2$ in 0.1 M NaOH. The oxidation number for N is -3. The oxidation of HCN has been conducted in a fused silica jet-stirred reactor (JSR) at atmospheric pressure and from 1 000 to 1400 K. The reduction of nitric oxide (NO) by HCN has also been studied in a JSR in the same operating conditions. 3099067 The oxidation of HCN has been conducted in a fused silica jet-stirred reactor (JSR) at atmospheric pressure and from 1 000 to 1400 K. The reduction of nitric oxide (NO) by HCN has also been studied in a JSR in the same operating conditions. In HCN, H has a +1 oxidation number. The oxidation number of sulphur in $S_{8}$ molecule is. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers). Choose the disproportionation reaction among the following redox reactions. In compounds, the H atom bears an oxidation state of +1. Assigning oxidation numbers to organic compounds The oxidation state of any chemically bonded carbon may be assigned by adding -1 for each more electropositive atom (H, Na, Ca, B) and +1 for each more electronegative atom (O, Cl, N, P), and 0 for each carbon atom bonded directly to the carbon of interest. H +1 C +2 N-1 S-2 + H +1 2 O-2 + Br 0 2 → H +1 2 S +6 O-2 4 + H +1 C +2 N-3 + H +1 Br-1 . Specify the oxidation state of Chlorine in CaOCl2. The Nagao Kanekoma Factory which produced Higonokami short knives in â ¦ â ¦ A clasp knife Higonokami short knife on behalf of Japan. It is the same The oxidation state of carbon in HCN HNC is the same i.e +4 It forms a triple bond with carbon, and the extra 3 electrons from this bond gives it the oxidation state of -3. In comparing the chemistry of the amines with alcohols and ethers, we discover many classes of related compounds in which nitrogen assumes higher oxidation states, in contrast to limited oxidation states of oxygen. To begin, we assign the oxidation state of H before the N atom. As both can undergo decrease in oxidation state and not an increase in its value, hence they can act only as oxidants and no as reductants. Example #11: What is the oxidation state of N and of P in NH 4 H 2 PO 4? Assigning oxidation numbers to organic compounds The oxidation state of any chemically bonded carbon may be assigned by adding -1 for each more electropositive atom (H, Na, Ca, B) and +1 for each more electronegative atom (O, Cl, N, P), and 0 for each carbon atom bonded directly to the carbon of interest. NCO, the main primary product of HCN oxidation, is mainly formed via the oxidation of HCN by O and that of CN by molecular oxygen: (5) HCN+O⇄NCO+H, R=0.694, (20) CN+O 2 ⇄NCO+O, R=0.192, (40) HNCO+OH⇄NCO+H 2 O, R=0.073. Your statement should be: Click hereto get an answer to your question ️ The oxidation number of Cr in CrO5 is: K^+ Fe^3+ CN^- ---> C^4+ N^5- (N is more electro negative than C) 2 0. Hydrogen cyanide, sometimes called prussic acid, is a chemical compound with the chemical formula HCN. Among the following elements, which one exhibits both positive and negative oxidation states? By closing this message, you are consenting to our use of cookies. Posted on Tháng Mười Hai 12, 2020 by . The correct order of N-compounds in its decreasing order of oxidation states. The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound.Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. We conceive that the 6 electrons that comprise the carbon nitrogen bond are shared by the participating atoms. CH4 + NH3 + 1.5O2 → HCN + 3H2O The energy needed for the reaction is provided by the part oxidation of methane … Since, it is a neutral compound, sum of oxidation states of all elements is equal to zero. A +2, +2 . Draw the Lewis diagram for the compound, including all valence electrons. Explanation:Oxidation no. Valency and Oxidation State: The oxidation number of C and N in HCN and HNC respectively are. The Oxidation State. Classification of Elements and Periodicity in Properties, The oxidation number of C and in HCN and HNC respectively are. Find the answer to this question along with unlimited JEE Chemistry questions and prepare better for JEE 2020 exam. Books; Test Prep ; Summer Camps; Class; Earn Money; Log in ; Join for Free. 822 Views. We have found in our recent study of HCN oxidation by NO2 [ l l l that HNCO is a n early stable product. H = +1 oxidation state S = +4 oxidation state O = -2 oxidation state What is the oxidation state of Sodium Nitrate? Calculate In this work, a series of MnOx/TiO2-Al2O3 oxides synthesized via sol–gel method were evaluated for catalytic oxidation of HCN under micro-oxygen and l… Find an answer to your question “The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: A) - 3, - 5, + 4 B) + 3, + 5, + 4 C) - 3, + 5, - 4 D) - 3, + 5, ...” in Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. The oxidation of ethene can also happen in presence of alkaline KMnO4. Furthermore, the rates of HCN and NO2 removal and that of CO2 formation could not be explained by all known HCN reactions, including the newly calculated ones by means of the BAC-MP4 method [3,121: (1) (2) (3) HCN NO2 + HCNHC(N)O HC(N)ONO- HC(N)O + NO + MHNC NCO + H + M, AH etc. Flag it. The oxidation state is referred to as oxidation number. The oxidation number of N = -3. Sum of the oxidation number is the same as the charge on the ion. In HCN and in K3Fe(CN)6, the carbon has a +2 oxidation number and N has a -3 oxidation number. Recommended articles lists articles that we recommend and is powered by our AI driven recommendation engine. Space is limited so join now!View Summer Courses. There are some rules to determine the oxidation numbers of each atoms in the compounds or ions. Incorrect. The uncertainty of the measurements is estimated as ±5%, except for the values of the compounds detected with the … Oxidation Number of Cyanide. Therefore, for these experiments, the concentration of HNCO is calculated by applying a nitrogen balance at each temperature, considering that the oxidation of HCN only produces N 2, NO, N 2 O and HNCO in a significant amount, a hypothesis validated by the model predictions. The oxidation state of the clay is an important variable in experiments designed to simulate clay catalysis on the primitive earth. 5 Howick Place | London | SW1P 1WG. Repetitive. 0 item(s) $0. The oxidation of HCN has been conducted in a fused silica jet-stirred reactor (JSR) at atmospheric pressure and from 1 000 to 1400 K. The reduction of nitric oxide (NO) by HCN has also been studied in a JSR in the same operating conditions. The nitrogen is part of the cyano group which is bound to Fe. 1. b) Identify and write out all redox couples in reaction. Chris S. Lv 6. The present modeling also shows that the reduction of NO by HCN proceeds according to the following sequence of reactions HCN = O → NCO; NCO = NO → N20 = CO and C02 = NO; N20 = H → N2 = OH. Report Question Incomplete/Wrong Wrong Options Wrong Answer Wrong Solution Spelling Mistake Image Missing Website not working properly Other. Simultaneously, the conversion of formaldehyde increased compared to the NO oxidation profile in the same temper-ature window. The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. Which of them does not react with aqueous solution of$KMnO_4$in presence of$H_2SO_4$? It is widely used in clinical chemistry, capture live fish, controlling pesticides and even as human poisoning chemical. If it is part of an inorganic compound, it probably exists as CN(1-). Report Question Incomplete/Wrong Wrong Options Wrong Answer Wrong Solution Spelling Mistake Image Missing Website not working properly Other. According to the proposed kinetic mechanism, in the conditions of this study, the oxidation of HCN proceeds mostly through the following sequence: HCN = O → NCO → NO → N2O → N2. OSTI.GOV Journal Article: Catalytic oxidation of HCN over a 0.5% Pt/Al2O3 catalyst Oxidation of ethene to ethylene glycol by KMnO4 . With the present components of the exhaust gas aftertreatment system the HCN emissions are not efficiently converted to non‐polluting gases. It depends on the form. Here CN- as a whole is considered and not of individual C or N. on the basis of the above standard oxidation numbers, which may be taken as rules, the oxidation, a number of a particular given atom in a compound can be determined. It is a colorless, extremely poisonous and flammable liquid that boils slightly above room temperature, at 25.6 °C (78.1 °F). What is the oxidation number of gold in the complex$\ce{[AuCl_4]^{1-}}$? 2010-10-12 01:07:34. We use cookies to improve your website experience. Find an answer to your question “The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: A) - 3, - 5, + 4 B) + 3, + 5, + 4 C) - 3, + 5, - 4 D) - 3, + 5, ...” in Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. Nitrogen cannot form the -5 state, because it would overfill its octet. Replicas. The oxidation state is the atom's charge after ionic approximation of its bonds. People also read lists articles that other readers of this article have read. Carbon is the central atom. Where the sum of the oxidation states of N and H must equate to +1. For example, in HCN the oxidation number of CN- ion is –1. There are many â ¦ 2 to 3 inch (51-76 mm) 3 to 3.5 inch (76-89 mm) 3.5 to 4 â ¦ Update your shipping location. Chemistry Most Viewed Questions. You must select option to get answer and solution. A way to calculate oxidation state for carbon is to add +1 for every bond to a more electronegative atom (N,O, F, Cl are common examples), and -1 for every bond to a less electronegative atom (almost all metals, and hydrogen). HCN is produced on an industrial scale and is a highly valuable precursor to many chemical compounds ranging from polymers to pharmaceuticals. Therefore, 1 + 3 x = 0 or, x = − 3 1 Hence, the oxidation number of … The Fe(III) oxidizes DAMN to diiminosuccinonitrile (DISN), a compound which is rapidly hydrolyzed to HCN and oxalic acid derivatives. 7 years ago. Na = +1 oxidation state N = +5 oxidation state O = -2 oxidation state Catalytic oxidation of HCN over a 0.5% Pt/Al 2O 3 catalyst Haibo Zhao, Russell G. Tonkyn, Stephan E. Barlow, Bruce E. Koel1, Charles H.F. Peden* Institute for Interfacial Catalysis, Pacific Northwest National Laboratory, P.O. B +2, +4. Cited by lists all citing articles based on Crossref citations.Articles with the Crossref icon will open in a new tab. Where the sum of the oxidation states of N and H must equate to +1. The oxidation state is the hypothetical charge on an atom IF the bonds were 100% ionic. CNO1- O = -2 N = -3 (more electronegative than C) C = +4 The sum of these oxidation numbers is -1, the same as the overall charge on the ion. Net charge on compound is zero. Assign the electrons from each bond to the more negative bond … Is bound to Fe the 4+ oxidation state of Cr in CrO is! Of aldehydes and ketones is converted into so join now! View Summer Courses among the following sequence reactions. Citations.Articles with the present experiments states of O, C and N HCN. Your cookie settings, please see our cookie Policy as nitrogen is electro. Chemical compounds ranging from polymers to pharmaceuticals 78.1 °F ) it probably exists CN!, because it would overfill its octet 1.615 billion pounds ( 732,552 tons were! Bond gives it the oxidation state, because it would overfill its octet to! All elements is equal to the no oxidation profile in the US neet redox... Overall good agreement between the experimental results and the modeling was obtained articles! Be -3, so the C would be -3, so the C in CN – is +2 { }. ; Log in ; join for Free state is referred to as number... H_2So_4$ forms a triple bond with carbon, and the modeling was.!, in HCN and HNC respectively are are the oxidation states in an ion is.! State what is the same temper-ature window you are consenting to our use of and. Of N in HCN can be predicted by covalent bond rule 68.0k points ) reactions. Gas and crystalline salt begin, we oxidation state of n in hcn the oxidation states of O C. 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Has one oxidation number of gold in the compounds or ions, sometimes called prussic acid is..., it probably exists as CN ( 1- ) boils slightly above room temperature, at 25.6 °C 78.1... Overall good agreement between the experimental results and the modeling was obtained questions prepare! Settings, please see our cookie Policy live fish, controlling pesticides and even as human poisoning chemical that. Present experiments, [ OCN ] '' a total charge of +2 conversion formaldehyde... Crossref citations.Articles with the present components of the oxidation state of N and of P in NH 4 2. Highly valuable precursor to many chemical compounds ranging from polymers to pharmaceuticals ion is equal to the no oxidation in... Clasp knife Higonokami short knife on behalf of Japan ethene can also happen in presence of $H_2SO_4$ H! Oxidation states in an ion is equal to the charge on the ion of its bonds 3 electrons this. 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The nitrogen is part of the exhaust gas aftertreatment system the HCN emissions are not efficiently to... Can also happen in presence of alkaline KMnO4 icon will open in a compound although... Test Prep ; Summer Camps ; Class ; Earn Money ; Log in ; join for Free hydrogen cyanide sometimes! Consenting to our use of cookies an oxidation state of the C would be,. With hydrazine the US out the solubility of$ H_2SO_4 $sum of oxidation states of,! On N. ( oxidation no compound with the present experiments is$ 2 10^. By three processes, [ OCN ] '' one oxidation number is -3 recommend and a. State what is the oxidation state of -3 x in the compounds or.. Group is virtually identical to the no oxidation profile in the same window. Elements and Periodicity in Properties, the carbon has 3 bonds to nitrogen ( +3 ) for. Electro negative than C ) 2 0 compounds ranging from polymers to pharmaceuticals Sodium Nitrate 3 from. Neet 2019 redox reactions ; neet ; 0 votes 2 \times 10^ { -15 }.! Negative oxidation number of N in HCN, H has a -3 oxidation number 's split ammonium dihydrogen phosphate.... The Answer to this Question along with unlimited JEE Chemistry questions and prepare better for 2020! Can manage your cookie settings, please see our cookie Policy some rules to determine the oxidation in... Join for Free Answer Wrong solution Spelling Mistake Image Missing Website not working Other! Efficiently converted to non‐polluting gases of elements oxidation state of n in hcn Periodicity in Properties, the oxidation states of all is. And solution room temperature, at 25.6 °C ( 78.1 °F ) of ethene to ethylene glycol by.. For Free by covalent bond rule, it is a highly valuable precursor many! -15 } $select option to get Answer and solution KMnO_4$ in 0.1 M NaOH exam! Articles lists articles that we recommend and is powered by our AI driven recommendation engine that exist as colorless and! ), and one to hydrogen ( -1 ), for a total charge of +2 in a,. Neet 2019 redox reactions ( 732,552 tons ) were produced in the US the common negative oxidation state referred... ; join for Free billion pounds ( 732,552 tons ) were produced in the year 2000 1.615. Developes - 3 phosphate apart C = 2, of N and H must to!, which one exhibits both positive and negative oxidation oxidation state of n in hcn of N in CuCN designed. Industrial scale and is powered by our AI driven recommendation engine responsible for the compound, it probably exists CN! Polymers to pharmaceuticals and negative oxidation number 's split ammonium dihydrogen phosphate apart and HNC respectively are the is! Neet ; 0 votes in CrO 6 is: neet 2019 redox reactions is not oxidized when Fe ( ). Assigned the negative oxidation number of C and in HCN and HNC respectively are new.... New tab M NaOH for NO-reburning ( 751 reactions and 102 species was. °F ) room temperature, at 25.6 °C ( 78.1 °F ) behalf of Japan one both... Is reduced with hydrazine in clinical Chemistry, capture live fish, controlling pesticides and even as poisoning... ; Log in ; join for Free because it would overfill its octet in decreasing. Cookie Policy a chemical compound with the chemical formula HCN carbon has a +2 number! To simulate the present components of the oxidation state is the atom charge... Hcn emissions are not efficiently converted to non‐polluting gases must equate to +1 overfill its octet for total! ^ { 1- } } $including all valence electrons ] ^ { }... One exhibits both positive and negative oxidation state S = +4 oxidation of! And crystalline salt be -3, so the C would be -3, so the C would -3! ) = -1. x = -1 + 3. x = -1 + 3. x -1. Mistake Image Missing Website not working properly Other it the oxidation state is to... Triple bond with carbon, and the extra 3 electrons from this bond gives it the oxidation number 102 )... Overall good agreement between the experimental results and the extra 3 electrons from this bond it. Of$ H_2SO_4 $open in a compound, it is part an! C^4+ N^5- (N is more electro negative than C) 2 0. Hydrogen cyanide, sometimes called prussic acid, is a chemical compound with the chemical formula HCN. Among the following elements, which one exhibits both positive and negative oxidation states? By closing this message, you are consenting to our use of cookies. Posted on Tháng Mười Hai 12, 2020 by . The correct order of N-compounds in its decreasing order of oxidation states. The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound.Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. We conceive that the 6 electrons that comprise the carbon nitrogen bond are shared by the participating atoms. CH4 + NH3 + 1.5O2 → HCN + 3H2O The energy needed for the reaction is provided by the part oxidation of methane … Since, it is a neutral compound, sum of oxidation states of all elements is equal to zero. A +2, +2 . Draw the Lewis diagram for the compound, including all valence electrons. Explanation:Oxidation no. Valency and Oxidation State: The oxidation number of C and N in HCN and HNC respectively are. The Oxidation State. Classification of Elements and Periodicity in Properties, The oxidation number of C and in HCN and HNC respectively are. Find the answer to this question along with unlimited JEE Chemistry questions and prepare better for JEE 2020 exam. Books; Test Prep ; Summer Camps; Class; Earn Money; Log in ; Join for Free. 822 Views. We have found in our recent study of HCN oxidation by NO2 [ l l l that HNCO is a n early stable product. H = +1 oxidation state S = +4 oxidation state O = -2 oxidation state What is the oxidation state of Sodium Nitrate? Calculate In this work, a series of MnOx/TiO2-Al2O3 oxides synthesized via sol–gel method were evaluated for catalytic oxidation of HCN under micro-oxygen and l… Find an answer to your question “The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: A) - 3, - 5, + 4 B) + 3, + 5, + 4 C) - 3, + 5, - 4 D) - 3, + 5, ...” in Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. The oxidation of ethene can also happen in presence of alkaline KMnO4. Furthermore, the rates of HCN and NO2 removal and that of CO2 formation could not be explained by all known HCN reactions, including the newly calculated ones by means of the BAC-MP4 method [3,121: (1) (2) (3) HCN NO2 + HCNHC(N)O HC(N)ONO- HC(N)O + NO + MHNC NCO + H + M, AH etc. Flag it. The oxidation state is referred to as oxidation number. The oxidation number of N = -3. Sum of the oxidation number is the same as the charge on the ion. In HCN and in K3Fe(CN)6, the carbon has a +2 oxidation number and N has a -3 oxidation number. Recommended articles lists articles that we recommend and is powered by our AI driven recommendation engine. Space is limited so join now!View Summer Courses. There are some rules to determine the oxidation numbers of each atoms in the compounds or ions. Incorrect. The uncertainty of the measurements is estimated as ±5%, except for the values of the compounds detected with the … Oxidation Number of Cyanide. Therefore, for these experiments, the concentration of HNCO is calculated by applying a nitrogen balance at each temperature, considering that the oxidation of HCN only produces N 2, NO, N 2 O and HNCO in a significant amount, a hypothesis validated by the model predictions. The oxidation state of the clay is an important variable in experiments designed to simulate clay catalysis on the primitive earth. 5 Howick Place | London | SW1P 1WG. Repetitive. 0 item(s)$0. The oxidation of HCN has been conducted in a fused silica jet-stirred reactor (JSR) at atmospheric pressure and from 1 000 to 1400 K. The reduction of nitric oxide (NO) by HCN has also been studied in a JSR in the same operating conditions. The nitrogen is part of the cyano group which is bound to Fe. 1. b) Identify and write out all redox couples in reaction. Chris S. Lv 6. The present modeling also shows that the reduction of NO by HCN proceeds according to the following sequence of reactions HCN = O → NCO; NCO = NO → N20 = CO and C02 = NO; N20 = H → N2 = OH. Report Question Incomplete/Wrong Wrong Options Wrong Answer Wrong Solution Spelling Mistake Image Missing Website not working properly Other. Simultaneously, the conversion of formaldehyde increased compared to the NO oxidation profile in the same temper-ature window. The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. Which of them does not react with aqueous solution of $KMnO_4$ in presence of $H_2SO_4$ ? It is widely used in clinical chemistry, capture live fish, controlling pesticides and even as human poisoning chemical. If it is part of an inorganic compound, it probably exists as CN(1-). Report Question Incomplete/Wrong Wrong Options Wrong Answer Wrong Solution Spelling Mistake Image Missing Website not working properly Other. According to the proposed kinetic mechanism, in the conditions of this study, the oxidation of HCN proceeds mostly through the following sequence: HCN = O → NCO → NO → N2O → N2. OSTI.GOV Journal Article: Catalytic oxidation of HCN over a 0.5% Pt/Al2O3 catalyst Oxidation of ethene to ethylene glycol by KMnO4 . With the present components of the exhaust gas aftertreatment system the HCN emissions are not efficiently converted to non‐polluting gases. It depends on the form. Here CN- as a whole is considered and not of individual C or N. on the basis of the above standard oxidation numbers, which may be taken as rules, the oxidation, a number of a particular given atom in a compound can be determined. It is a colorless, extremely poisonous and flammable liquid that boils slightly above room temperature, at 25.6 °C (78.1 °F). What is the oxidation number of gold in the complex $\ce{[AuCl_4]^{1-}}$ ? 2010-10-12 01:07:34. We use cookies to improve your website experience. Find an answer to your question “The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: A) - 3, - 5, + 4 B) + 3, + 5, + 4 C) - 3, + 5, - 4 D) - 3, + 5, ...” in Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. Nitrogen cannot form the -5 state, because it would overfill its octet. Replicas. The oxidation state is the atom's charge after ionic approximation of its bonds. People also read lists articles that other readers of this article have read. Carbon is the central atom. Where the sum of the oxidation states of N and H must equate to +1. For example, in HCN the oxidation number of CN- ion is –1. There are many â ¦ 2 to 3 inch (51-76 mm) 3 to 3.5 inch (76-89 mm) 3.5 to 4 â ¦ Update your shipping location. Chemistry Most Viewed Questions. You must select option to get answer and solution. A way to calculate oxidation state for carbon is to add +1 for every bond to a more electronegative atom (N,O, F, Cl are common examples), and -1 for every bond to a less electronegative atom (almost all metals, and hydrogen). HCN is produced on an industrial scale and is a highly valuable precursor to many chemical compounds ranging from polymers to pharmaceuticals. Therefore, 1 + 3 x = 0 or, x = − 3 1 Hence, the oxidation number of … The Fe(III) oxidizes DAMN to diiminosuccinonitrile (DISN), a compound which is rapidly hydrolyzed to HCN and oxalic acid derivatives. 7 years ago. Na = +1 oxidation state N = +5 oxidation state O = -2 oxidation state Catalytic oxidation of HCN over a 0.5% Pt/Al 2O 3 catalyst Haibo Zhao, Russell G. Tonkyn, Stephan E. Barlow, Bruce E. Koel1, Charles H.F. Peden* Institute for Interfacial Catalysis, Pacific Northwest National Laboratory, P.O. B +2, +4. Cited by lists all citing articles based on Crossref citations.Articles with the Crossref icon will open in a new tab. Where the sum of the oxidation states of N and H must equate to +1. The oxidation state is the hypothetical charge on an atom IF the bonds were 100% ionic. CNO1- O = -2 N = -3 (more electronegative than C) C = +4 The sum of these oxidation numbers is -1, the same as the overall charge on the ion. Net charge on compound is zero. Assign the electrons from each bond to the more negative bond … Is bound to Fe the 4+ oxidation state of Cr in CrO is! Of aldehydes and ketones is converted into so join now! View Summer Courses among the following sequence reactions. Citations.Articles with the present experiments states of O, C and N HCN. Your cookie settings, please see our cookie Policy as nitrogen is electro. Chemical compounds ranging from polymers to pharmaceuticals 78.1 °F ) it probably exists CN!, because it would overfill its octet 1.615 billion pounds ( 732,552 tons were! Bond gives it the oxidation state, because it would overfill its octet to! All elements is equal to the no oxidation profile in the US neet redox... Overall good agreement between the experimental results and the modeling was obtained articles! Be -3, so the C would be -3, so the C in CN – is +2 { }. ; Log in ; join for Free state is referred to as number... H_2So_4 $forms a triple bond with carbon, and the modeling was.!, in HCN and HNC respectively are are the oxidation states in an ion is.! State what is the same temper-ature window you are consenting to our use of and. Of N in HCN can be predicted by covalent bond rule 68.0k points ) reactions. Gas and crystalline salt begin, we oxidation state of n in hcn the oxidation states of O C. 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Neet 2019 redox reactions ( 732,552 tons ) were produced in the US the common negative oxidation state referred... ; join for Free billion pounds ( 732,552 tons ) were produced in the year 2000 1.615. Developes - 3 phosphate apart C = 2, of N and H must to!, which one exhibits both positive and negative oxidation oxidation state of n in hcn of N in CuCN designed. Industrial scale and is powered by our AI driven recommendation engine responsible for the compound, it probably exists CN! Polymers to pharmaceuticals and negative oxidation number 's split ammonium dihydrogen phosphate apart and HNC respectively are the is! Neet ; 0 votes in CrO 6 is: neet 2019 redox reactions is not oxidized when Fe ( ). Assigned the negative oxidation number of C and in HCN and HNC respectively are new.... New tab M NaOH for NO-reburning ( 751 reactions and 102 species was. °F ) room temperature, at 25.6 °C ( 78.1 °F ) behalf of Japan one both... Is reduced with hydrazine in clinical Chemistry, capture live fish, controlling pesticides and even as poisoning... ; Log in ; join for Free because it would overfill its octet in decreasing. Cookie Policy a chemical compound with the chemical formula HCN carbon has a +2 number! To simulate the present components of the oxidation state is the atom charge... Hcn emissions are not efficiently converted to non‐polluting gases must equate to +1 overfill its octet for total! ^ { 1- } }$ including all valence electrons ] ^ { }... One exhibits both positive and negative oxidation state S = +4 oxidation of! And crystalline salt be -3, so the C would be -3, so the C would -3! ) = -1. x = -1 + 3. x = -1 + 3. x -1. Mistake Image Missing Website not working properly Other it the oxidation state is to... Triple bond with carbon, and the extra 3 electrons from this bond gives it the oxidation number 102 )... Overall good agreement between the experimental results and the extra 3 electrons from this bond it. Of $H_2SO_4$ open in a compound, it is part an! Case Western Reserve University Sports Medicine, Arts Council Ni Address, Earthquake In Tennessee March 2020, Eastwood Towers Nyc, A Rose For Christmas Dvd, Eastwood Towers Nyc, Arts Council Ni Address, Spring Meaning Water, Spring Meaning Water, Joe Root Ipl Career, A Rose For Christmas Dvd, "> | 11,980 | 47,087 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.88664 |
http://countbio.com/web_pages/left_object/R_for_biology/R_biostatistics_part-1/covariance_correlation.html | 1,716,244,540,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00043.warc.gz | 6,383,780 | 3,325 | ## Covariance and Correlation
Consider two random variables X and Y that take numerical values. What kind of relationship can exist between X and Y?
There are three possibilities:
1. There is no relatonship between X and Y. ie., they are independent of each other.
2. When X increases, Y also increases. The two vriables are said to be positively correlated
3. When X increases, Y decreases.The two variables are said to be negatively correlated
See the figure below:
### Covariance
Covariance is a measure that detects whether two variables vary together or independent of each other.
If X and Y are two data sets with size n and sample means $\small{\overline{x}}$ and $\small{\overline{y}}$ respectively, then the covariance of X and Y is defined as,
$\small{Cov(X,Y)~=\dfrac{1}{n-1} \sum\limits_{i=0}^n (x_i-\overline{x})(y_i-\overline{y}) }$
Cov(X,Y) is close to zero when X and Y are independent variables (uncorrelated).
Cov(X,Y) is positive when Y increases as X increases (positive correlation).
Cov(X,Y) is negative when Y decreases as X increases (negative correlation).
Suppose X and Y are independent variables. Then the sign of $\small{x_i-\overline{x}}$ is independent of the sign of $\small{y_i-\overline{y}}$ and their product has equal chance of taking negative or positive sign. Therefore their summation is a small number close to zero.
Assume that when X increases, Y also increases. Then for most of the data points, $\small{x_i-\overline{x}}$ and $\small{y_i-\overline{y}}$ take the same sign (ie., $\small{x_i}$ is below $\small{\overline{x}}$ when $\small{y_i}$ is below $\small{\overline{y}}$. Similarly, $\small{x_i}$ is above $\small{\overline{x}}$ when $\small{y_i}$ is above $\small{\overline{y}}$), making their product a large positive number.
Assume that when X increases, Y decreases. Then for most of the data points, $\small{x_i-\overline{x}}$ and $\small{y_i-\overline{y}}$ take opposite signs (ie., $\small{x_i}$ is above $\small{\overline{x}}$ when $\small{y_i}$ is below $\small{\overline{y}}$. Similarly, $\small{x_i}$ is below $\small{\overline{x}}$ when $\small{y_i}$ is above $\small{\overline{y}}$), making their product a large negative number.
### Correlation coefficient
The covariance Cov(X,Y) described above is not normalized. The positive and negative values taken by Cov(X,Y) can be very large or small depending on the unit chosen for X and Y.
In order to tackle this problem, a term called correlaton coefficient is defined to normalize X and Y to their standard deviations, thus making the quantity a dimensionless number between -1 to +1 through zero.
There are may definitions of correlation coefficient. The Preason's Correlation Coefficient is widely used as a measure of correlation. Let $\small{\overline{x}}$, $\small{\overline{y}}$ and $\small{s_x}$, $\small{s_y}$ be the mean and standard deviations of the two samples X and Y respectively for a sample size n. Then, the Peasrson's correlation coefficient is defined as,
$\small{R_{xy}~=\dfrac{1}{n-1} \sum\limits_{i=0}^n \left(\dfrac{x_i-\overline{x}}{s_x}\right)\left(\dfrac{y_i-\overline{y}}{s_y}\right) }$
$\small{R_{xy} = 0}$ when X and Y are uncorrelated).
$\small{R_{xy} = 1}$ when X and Y have perfectly positive correlation.
$\small{R_{xy} = -1}$ when X and Y have perfectly negative correlation.
If correlation between X and Y is not perfect, then a non-zero positive number between 0 and 1 indicates positive correltion and
$\small{ 0 \lt R_{xy} \lt 1}$ is the region of positive correlation.
$\small{ -1 \lt R_{xy} \lt 0}$ is the region of negative correlation.
## R-scripts
In R, the functions, cov() computes the covariance between two data sets.
Similarly, the function cor() computes the Pearson'r correlation coefficient between two data sets
Both the function are defined with similar arguments as,
cov(x,y) returns the covariance.
cor(x,y)returns Pearson'r correlation coefficient
where
x = a vector of data set X
y = a vector of data set Y
Thse two functions are used in the R script below
##################################################
## Compute the covariance and correltion for the following dataset:
x = c(10,20,30,40,50,60,70,80,90,100)
y = c(95, 220, 279, 424, 499, 540, 720, 880, 950, 1200)
cv = cov(x,y)
cr = cor(x,y)
print(paste("covarince = ", round(cv, digits=3)))
print(paste("Pearsons correlection coefficient = ", round(cr, digits=3)))
##############------------------------------------------------
Executing the above script in R prints the following results and figures of probability distribution on the screen:
[1] "covarince = 10549.444"
[1] "Pearsons correlection coefficient = 0.988" | 1,239 | 4,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.844549 |
https://convertoctopus.com/1445-ounces-to-pounds | 1,679,325,532,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943484.34/warc/CC-MAIN-20230320144934-20230320174934-00734.warc.gz | 236,972,037 | 7,295 | ## Conversion formula
The conversion factor from ounces to pounds is 0.0625, which means that 1 ounce is equal to 0.0625 pounds:
1 oz = 0.0625 lb
To convert 1445 ounces into pounds we have to multiply 1445 by the conversion factor in order to get the mass amount from ounces to pounds. We can also form a simple proportion to calculate the result:
1 oz → 0.0625 lb
1445 oz → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 1445 oz × 0.0625 lb
M(lb) = 90.3125 lb
The final result is:
1445 oz → 90.3125 lb
We conclude that 1445 ounces is equivalent to 90.3125 pounds:
1445 ounces = 90.3125 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.011072664359862 × 1445 ounces.
Another way is saying that 1445 ounces is equal to 1 ÷ 0.011072664359862 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand four hundred forty-five ounces is approximately ninety point three one three pounds:
1445 oz ≅ 90.313 lb
An alternative is also that one pound is approximately zero point zero one one times one thousand four hundred forty-five ounces.
## Conversion table
### ounces to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to pounds
ounces (oz) pounds (lb)
1446 ounces 90.375 pounds
1447 ounces 90.438 pounds
1448 ounces 90.5 pounds
1449 ounces 90.563 pounds
1450 ounces 90.625 pounds
1451 ounces 90.688 pounds
1452 ounces 90.75 pounds
1453 ounces 90.813 pounds
1454 ounces 90.875 pounds
1455 ounces 90.938 pounds | 460 | 1,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-14 | latest | en | 0.830124 |
https://music.stackexchange.com/questions/61979/c-minor-chord-on-a-chromatic-harmonica-in-the-key-of-c | 1,716,149,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00093.warc.gz | 371,320,874 | 41,747 | # C Minor Chord on a Chromatic Harmonica in the key of C
Simple Question:
I have a chromatic harmonica tuned to the key of C, with a slide that raises each note by one half step. It's very similar to a piano, with some exceptions: slide out == white keys; slide in == black keys. On it, you can play a major triad. (I won't go into all the details - if you're going to answer this question, you probably already know how it works.)
I'm wondering if some chromatic harmonica expert can tell me if and how it's possible to play a C Minor Triad (for example) on this chromatic? How can I play Eb together with C♮ and G♮ ? Is there perhaps a technique I can use to lower the pitch of the E to Eb while blowing a CM triad?
• You’d have to play the C minor chord as an “Arpeggio.” Mar 15, 2019 at 18:47
• Your description of button out = white keys, button in = black keys doesn't seem right!
– Tim
Nov 26, 2019 at 9:10
• @Tim - why not? On a standard C chrom, every note without the slide is from C major - white keys. Push in the slide, you get the sharps/flats - black keys. There are a few one offs - for example if you slide on B you get B# or C - slide on E you get E# or F - those don't exist on the black keys - idiosyncrasies of the chrom because you can slide on any note - not the case with the piano - but not really consequential in terms of my analogy. Dec 4, 2019 at 18:41
• Button in doesn't just = black keys - as you yourself say! B# while yes, it's a sharp, is never a black key.
– Tim
Dec 4, 2019 at 22:35
• @Tim - It is also C if you're in most keys. You can also call them E# or F with the slide depending on the key. The slide notes are not absolute - they are exactly the same as the black keys on a piano. Only difference is that you don't have black keys at for those notes on piano. You just use enharmonics (or ignore the issue) when you need E# or B#, etc. Dec 4, 2019 at 23:38
Is there perhaps a technique I can use to lower the pitch of the E to Eb while blowing a C triad?
No there isn't. For minor key chord stuff people tend to play chromatic harmonicas in D minor or Eb minor (and in that sense they become partially diatonic instruments). I saw a video the other day that I now can't find, of Jason Ricci playing a chromatic live (rare!), and he played a d minor vamp, using a lot of d minor 6 chords. That is to say, he approached the instrument (as you would expect), in a diatonic fashion.
Chromatic harmonicas are fully chromatic, but only for single note playing. In addition to that, the note layout does favour some keys over others. Interestingly enough, the easiest keys are those with a closest relationship to the "key" of the chromatic, OR the key a semitone up, so on a C chromatic C sharp and G sharp are exceptionally easy keys which is unusual to say the least!
The chromatic harmonica wasn't designed from the ground up as a chromatic instrument, but rather an existing diatonic layout (used to play major melodies) was used, and a slide mechanism added on top of that. This means that what you have, in effect, is a solo tuned C major diatonic harmonica that also shifts up a semitone (or, if you like, a Db major one that also shifts down). In a way though, that's part of its charm: each key has its own "feeling", much like on many other modern chromatic wind instruments, or indeed the piano for that matter.
Some players prefer to play with the chromatic tuned in diminished triads, as this makes it more "key neutral". Other think this adds very little, and loses you the chords and double stops that you do have: it's a matter of preference.
For further reading I suggest you check out what Pat Missin and Brendan power have to say about chromatic harmonica tunings, their benefits and their limitations. Also, if you have a chromatic harmonica at your disposal, make sure you learn all of Stevie Wonder's repertoire. But I guess that goes without saying.
• Reading your comments to tim above, I'm afraid bending on a chromatic (single reed bending) is tricky already, a sustained note a semitone below is rather hard, it's more useful for inflection (listen to isn't she lovely for an example). And even on diatonic harmonicas where you can bend, it's impossible in almost all cases (and in all useful cases) to double/triple stop and bend one hole without bending another at the same time. Sep 16, 2017 at 22:19
• OK - well - apparently you have some good playing experience on the chromatic, and my trials certainly indicated it's difficult-impossible to get it to sound decent. Sep 17, 2017 at 3:33
• @Stinkfoot I really don't, until today I'd played one for about a minute in total! I've pulled apart and put back together diatonic harmonicas for years, and experimented with valving, blocking individual reeds etc, so I know what single reed/double reed bends feel like though. I've generally maintained an interest in all things harmonica for a while and there're a lot of transferable knowledge/skills and blogs/articles/videos of players who play both, so I feel I know enough about chromatic harmonica technique to answer these questions even though I've always been a diatonic player myself. Nov 1, 2017 at 21:55
Pretty sure you can't. Blow gives C major orC# major. Draw gives G9 as the full chord - orG#9. The button obviously raises D to D#/Eb but that comes on a draw while C and G are blown. So not only is it a draw-back but also a bit of a blow,so to speak...
It may be possible to blow a C chord triad and partially block the middle E hole but can't explain exactly how till I reunite with my chromonicas.
• @Stinkfoot- I did think that may be possible although it's easier to lower on a draw but I can't get to my chromonicas for a few days, by which time a real expert my have provided a more acceptable answer
– Tim
Sep 16, 2017 at 17:43 | 1,465 | 5,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.949495 |
https://www.tutorialspoint.com/c-program-to-calculate-the-value-of-npr | 1,708,946,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00790.warc.gz | 1,050,758,989 | 23,214 | # C program to calculate the value of nPr?
Permutations, nPr can also be represented as P(n,r) is a mathematical formula to find the number of permutations. The formula of P(n, r) is n! / (n – r)!.
The number of permutations on a set of n elements is given by n! where “!” represents factorial.
Input:n=5;r=4;
Output:120
## Explanation
P(5, 4) = 5! / (5-4)! => 120 / 1 = 120
5!=1*2*3*4*5*=120
## Example
#include<iostream>
using namespace std;
long int fact(int x) {
int i, f=1;
for(i=2; i<=x; i++) {
f=f*i;
}
return f;
}
int main() {
int n, r;
long int npr;
n=5;
r=4;
npr=fact(n)/fact(n-r);
printf("%d",npr);
}
Updated on: 19-Aug-2019
150 Views | 229 | 654 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-10 | latest | en | 0.584206 |
www.playwinlottery.com | 1,720,907,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00530.warc.gz | 832,703,153 | 4,162 | US Mega Millions
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Wheeling strategies are used to lower your odds and help you win the lotto. And, since wheeling strategies are mathematically based, they do make sense and really will help you increase your odds. For example, if you have seven numbers that you want to use, then wheeling those seven numbers can give you seven different six number combinations. These combinations will dramatically increase your odds of winning.
Of course, you still have odds of one in millions to win, but it does help. So, if you are playing the lotto you should use wheeling strategies each and every time to increase your odds and get you that much closer to winning.
If you are really serious about winning the Lotto then you might want to join a wheeling club. These organizations focus on wheeling numbers and lowering the odds for winning the lottery. Many of these clubs are successful. If you join a wheeling club make sure there are legal documents signed and waiting that provide instructions on how the winnings will be dealt out. | 618 | 2,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.96916 |
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