url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://help.libreoffice.org/Calc/AVERAGEIFS_function/ca | 1,529,285,737,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859923.59/warc/CC-MAIN-20180618012148-20180618032148-00391.warc.gz | 618,171,981 | 8,572 | # funció MITJANASI
## funció MITJANASICONJUNT
Returns the arithmetic mean of all cells in a range that satisfy given multiple criteria. The AVERAGEIFS function sums up all the results that match the logical tests and divides this sum by the quantity of selected values.
### Sintaxi
`AVERAGEIFS(Average_range; Criterion_range1; Criterion1 [; Criterion_range2; Criterion2 [; ...]])`
Average_range – required argument. It is a range of cells, a name of a named range or a label of a column or a row containing values for calculating the mean.
Criterion_range1 – required argument. It is a range of cells, a name of a named range or a label of a column or a row containing values for finding the corresponding criterion.
Criterion1 – required argument. A condition in the form of expression or a cell reference to expression that defines what cells should be used to calculate the mean. The expression can contain text, numbers, regular expressions (if enabled in calculation options) or wildcards (if enabled in calculation options).
Criterion_range2 – Optional. Criterion_range2 and all the following mean the same as Criterion_range1.
Criterion2 – Optional. Criterion2 and all the following mean the same as Criterion1.
The logical relation between criteria can be defined as logical AND (conjunction). In other words, if and only if all given criteria are met, a value from the corresponding cell of the given Average_range is taken into calculation of the mean. The Criterion needs to be a string expression, in particular, the Criterion needs to be enclosed in quotation marks ("Criterion") with the exception of the names of functions, cell references and the operator of a string concatenation (&). The operators equal to (=), not equal to (<>), greater than (>), greater than or equal to (>=), less than (<), and less than or equal to (<=) can be used in criterion arguments for comparison of numbers. The function can have up to 255 arguments, meaning that you can specify 127 criteria ranges and criteria for it.
This function is part of the Open Document Format for Office Applications (OpenDocument) standard Version 1.2. (ISO/IEC 26300:2-2015)
If a cell in a range of values for calculating the mean is empty or contains text, the function AVERAGEIFS ignores this cell. If a cell contains TRUE, it is treated as 1, if a cell contains FALSE – as 0 (zero). If the whole range is empty, contains only text or all values of the range do not satisfy the condition (or any combination of those), the function returns the #DIV/0! error. If the range of values for calculating the mean and any range for finding criterion have unequal sizes, the function returns err:502.
### Exemples
#### Considereu la taula següent
A B C
1 Nom del producte Vendes Revenue
2 llapis 20 65
3 llapis 35 85
4 bloc de notes 20 190
5 llibre 17 180
6 estoig no no
In all examples below, ranges for average calculation contain the row #6, but it is ignored, because it contains text.
#### Ús simple
`=MITJANA(A1:A50)`
Calcula la suma de valors de l'interval B2:B6 majors o igual a 20. Retorna 75, perquè la cinquena fila no compleixen el criteri.
`=MITJANA(A1:A50)`
Calculates the average for values of the range C2:C6 that are greater than 70 and correspond to cells of B2:B6 with values greater than or equal to 20. Returns 137.5, because the second and fifth rows do not meet at least one criterion.
#### Ús d'expressions regulars i funcions imbricades
`=COMPTASICONJUNT(B2:B6;">"&MIN(B2:B6);B2:B6;"<"&MAX(B2:B6))`
Calcula la suma de valors de l'interval C2:C6 que corresponen a tots els valors de l'interval B2:B6 excepte el mínim i el màxim. Retorna 255, perquè la tercera i la cinquena files no compleixen almenys un criteri.
`=MITJANASICONJUNT(C2:C6;A2:A6;E2&".*";B2:B6;"<"&MAX(B2:B6))`
Calcula la suma de valors de l'interval C2:C6 que corresponen a totes les cel·les de l'interval A2:A6 començant per "llapis" i a totes les cel·les de l'interval B2:B6 excepte el màxim. Retorna 65, perquè només la segona fila compleix tots els criteris.
#### Referència a una cel·la com a criteri
Si heu de canviar el criteri fàcilment, potser voleu indicar-lo en una cel·la separada i usar una referència a aquesta cel·la en la condició de la funció MITJANASICONJUNT. Per exemple, la funció superior es pot reescriure com a:
`=MITJANASICONJUNT(C2:C6;A2:A6;E2&".*";B2:B6;"<"&MAX(B2:B6))`
Si E2 = llapis, la funció retorna 65, perquè l'enllaç a la cel·la se substitueix amb el seu contingut.
## Related Topics
AVERAGE, AVERAGEA, funció MITJANASI
, funció SUMASICONJUNT
, funció COMPTASICONJUNT
, MAX, MIN
Llista d'expressions regulars | 1,263 | 4,642 | {"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-2018-26 | latest | en | 0.720387 |
http://www.perseus.tufts.edu/hopper/text?doc=Perseus%3Atext%3A1999.01.0086%3Abook%3D10%3Atype%3DProp%202%3Anumber%3D82 | 1,529,380,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861752.19/warc/CC-MAIN-20180619021643-20180619041643-00389.warc.gz | 479,955,763 | 14,534 | #### PROPOSITION 82.
To a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes, with the whole, the sum of the squares on them rational but twice the rectangle contained by them medial.
Let AB be the minor straight line, and let BC be an annex to AB; therefore AC, CB are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial. [X. 76]
I say that no other straight line can be annexed to AB fulfilling the same conditions.
For, if possible, let BD be so annexed; therefore AD, DB are also straight lines incommensurable in square which fulfil the aforesaid conditions. [X. 76]
Now, since the excess of the squares on AD, DB over the squares on AC, CB is also the excess of twice the rectangle AD, DB over twice the rectangle AC, CB, while the squares on AD, DB exceed the squares on AC, CB by a rational area, for both are rational, therefore twice the rectangle AD, DB also exceeds twice the rectangle AC, CB by a rational area: which is impossible, for both are medial. [X. 26]
Therefore to a minor straight line only one straight line can be annexed which is incommensurable in square with the whole and which makes the squares on them added together rational, but twice the rectangle contained by them medial. Q. E. D. | 319 | 1,386 | {"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} | 3.015625 | 3 | CC-MAIN-2018-26 | latest | en | 0.947385 |
https://convertoctopus.com/14-days-to-years | 1,632,693,200,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00304.warc.gz | 235,320,451 | 8,292 | ## Conversion formula
The conversion factor from days to years is 0.0027397260273973, which means that 1 day is equal to 0.0027397260273973 years:
1 d = 0.0027397260273973 yr
To convert 14 days into years we have to multiply 14 by the conversion factor in order to get the time amount from days to years. We can also form a simple proportion to calculate the result:
1 d → 0.0027397260273973 yr
14 d → T(yr)
Solve the above proportion to obtain the time T in years:
T(yr) = 14 d × 0.0027397260273973 yr
T(yr) = 0.038356164383562 yr
The final result is:
14 d → 0.038356164383562 yr
We conclude that 14 days is equivalent to 0.038356164383562 years:
14 days = 0.038356164383562 years
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 year is equal to 26.071428571429 × 14 days.
Another way is saying that 14 days is equal to 1 ÷ 26.071428571429 years.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that fourteen days is approximately zero point zero three eight years:
14 d ≅ 0.038 yr
An alternative is also that one year is approximately twenty-six point zero seven one times fourteen days.
## Conversion table
### days to years chart
For quick reference purposes, below is the conversion table you can use to convert from days to years
days (d) years (yr)
15 days 0.041 years
16 days 0.044 years
17 days 0.047 years
18 days 0.049 years
19 days 0.052 years
20 days 0.055 years
21 days 0.058 years
22 days 0.06 years
23 days 0.063 years
24 days 0.066 years | 461 | 1,620 | {"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} | 4.25 | 4 | CC-MAIN-2021-39 | latest | en | 0.793399 |
http://ronnybergmann.net/teaching.html | 1,529,625,080,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864303.32/warc/CC-MAIN-20180621231116-20180622011116-00035.warc.gz | 269,958,387 | 3,639 | $\DeclareMathOperator{\arccosh}{arccosh} \DeclareMathOperator*{\argmin}{arg\,min} \DeclareMathOperator{\Exp}{Exp} \newcommand{\geo}[2]{\gamma_{\overset{\frown}{#1,#2}}} \newcommand{\geoS}{\gamma} \newcommand{\geoD}[2]{\gamma_} \newcommand{\geoL}[2]{\gamma(#2; #1)} \newcommand{\gradM}{\nabla_{\M}} \newcommand{\gradMComp}[1]{\nabla_{\M,#1}} \newcommand{\Grid}{\mathcal G} \DeclareMathOperator{\Log}{Log} \newcommand{\M}{\mathcal M} \newcommand{\N}{\mathcal N} \newcommand{\mat}[1]{\mathbf{#1}} \DeclareMathOperator{\prox}{prox} \newcommand{\PT}[3]{\mathrm{PT}_{#1\to#2}#3} \newcommand{\R}{\mathbb R} \newcommand{\SPD}[1]{\mathcal{P}(#1)} \DeclareMathOperator{\Tr}{Tr} \newcommand{\tT}{\mathrm{T}} \newcommand{\vect}[1]{\mathbf{#1}}$
Teaching
Lectures
WS 2017/18
“Höhere Mathematik III – Vektoranalysis” TU Kaiserslauten, Vector Calculus for Engineering students.
WS 2016/17
“Höhere Mathematik III – Vektoranalysis” TU Kaiserslauten, Vector Calculus for Engineering students.
SS 2016
“Foundations of Image Processing” TU Kaiserslauten, (second half on Fourier Analysis) together with Gabriele Steidl.
SS 2015
“Foundations of Image Processing” TU Kaiserslauten.
Seminars, Exercises/Tutorials & Organisation
WS 2017/18
Lecture Exercise for “Höhere Mathematik III – Vektoranalysis” TU Kaiserslauten, Vector Calculus for Engineering students.
WS 2017/18
Seminar Learning TU Kaiserslauten, together with Gabriele Steidl.
SS 2017
Tutor and Organizer of the Tutorials for “Optimization on Manifolds” TU Kaiserslauten.
SS 2017
Tutor and Organizer of the Tutorials for “Foundations of Image Processing” TU Kaiserslauten.
SS 2016
Tutor and Organizer of the Tutorials for “Methoden der konvexen Analysis in der Bildverarbeitung”(Convex Analysis in Image Processing), TU Kaiserslauten.
WS 2015/16
Organizer of the Tutorials “Höhere Mathematik III” TU Kaiserslauten, Vector Calculus and Differential Equations for engineering students.
WS 2015/16
Seminar “Mathematical Image Processing” TU Kaiserslauten, together with Gabriele Steidl.
WS 2014/15 and WS 2016/17
Internships accompanying the “Lecture Introduction to Numerical Methods” TU Kaiserslauten.
SS 2014
Lecture Exercise “Höhere Mathematik II” TU Kaiserslauten, Linear Algebra for engineering students.
WS 2013/14
Tutor “Höhere Mathematik III” TU Kaiserslauten, Vector Calculus and Differential Equations for engineering students.
WT 2011/12
Seminar “Wavelets – Theory and Applications“ Universität zu Lübeck.
SS 2011, SS 2012 and SS 2013
Tutorial “Computational Analysis – Introduction to Mathematica“for Bachelor students of “Computational Life Science”, Universität zu Lübeck.
WS 2010/11 and WS 2011/12
Additional Lecture of Analysis I on “Introduction to linear Algebra”for Bachelor Students “Molecular Life Science“, Universität zu Lübeck.
WS 2009/10 until SS 2013
Tutor and Tutorial Organizerof the Tutorials for “Analysis I“ (winter terms) und “Analysis II“ (summer terms), Universität zu Lübeck.
WS 2005/06 until SS 2009
Student Tutorfor correction assignments and Tutorials within “Linear Algebra und discrete Structures I“ (winter terms) and “Linear Algebra und discrete Structures II“ (summer terms), Universität zu Lübeck.
I supervised several student theses and projects which I keep track of on the Students page. | 997 | 3,276 | {"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} | 3.09375 | 3 | CC-MAIN-2018-26 | latest | en | 0.396545 |
http://www.airports-worldwide.com/articles/article0507.php | 1,524,240,110,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944479.27/warc/CC-MAIN-20180420155332-20180420175332-00302.warc.gz | 350,387,793 | 7,561 | Bi-elliptic transfer
By Wikipedia,
the free encyclopedia,
http://en.wikipedia.org/wiki/Bi-elliptic_transfer
In astronautics and aerospace engineering, the bi-elliptic transfer is an orbital maneuver that moves a spacecraft from one orbit to another and may, in certain situations, require less delta-v than a Hohmann transfer.
The bi-elliptic transfer consists of two half elliptic orbits. From the initial orbit, a delta-v is applied boosting the spacecraft into the first transfer orbit with an apoapsis at some point rb away from the central body. At this point, a second delta-v is applied sending the spacecraft into the second elliptical orbit with periapsis at the radius of the final desired orbit where a third delta-v is performed injecting the spacecraft into the desired orbit.
While it requires one more burn than a Hohmann transfer and generally requires a greater period of time, the bi-elliptic transfer may require a lower amount of total delta-v than a Hohmann transfer in situations where the ratio of final to the initial semi-major axis is greater than 11.94 .
## Calculation
### Delta-v
<-img src="http://upload.wikimedia.org/wikipedia/commons/thumb/8/83/Bi-elliptic_transfer.svg/250px-Bi-elliptic_transfer.svg.png"/> A bi-elliptic transfer from a low circular starting orbit (dark blue), to a higher circular orbit (red).
Utilizing the vis viva equation where,
$v^2 = \mu \left( \frac{2}{r} - \frac{1}{a} \right)$
where:
• $v \,\!$ is the speed of an orbiting body
• $\mu = GM\,\!$ is the standard gravitational parameter of the primary body
• $r \,\!$ is the distance of the orbiting body from the primary
• $a \,\!$ is the semi-major axis of the body's orbit
The magnitude of the first delta-v at the initial circular orbit with radius r0 is:
$\Delta v_1 = \sqrt{ \frac{2 \mu}{r_0} - \frac{\mu}{a_1}} - \sqrt{\frac{\mu}{r_0}}$
At rb the delta-v is:
$\Delta v_2 = \sqrt{ \frac{2 \mu}{r_b} - \frac{\mu}{a_2}} - \sqrt{ \frac{2 \mu}{r_b} - \frac{\mu}{a_1}}$
The final delta-v at the final circular orbit with radius rf:
$\Delta v_3 = \sqrt{\frac{\mu}{r_f}} - \sqrt{ \frac{2 \mu}{r_f} - \frac{\mu}{a_2}}$
Where a1 and a2 are the semimajor axes of the two elliptical transfer orbits and are given by:
$a_1 = \frac{r_0+r_b}{2}$
$a_2 = \frac{r_f+r_b}{2}$
### Transfer time
Like the Hohmann transfer, both transfer orbits used in the bi-elliptic transfer constitute exactly one half of an elliptic orbit. This means that the time required to execute each phase of the transfer is simply half the orbital period of each transfer ellipse.
Using the equation for the orbital period and the notation from above, we have:
$T = 2 \pi \sqrt{\frac{a^3}{\mu}}$
The total transfer time t is simply the sum of the time required for each half orbit Therefore we have:
$t_1 = \pi \sqrt{\frac{a_1^3}{\mu}} \quad and \quad t_2 = \pi \sqrt{\frac{a_2^3}{\mu}}$
and finally:
$t = t_1 + t_2 \;$
## Example
For example, to transfer from circular low earth orbit with r0 = 6700 km to a new circular orbit with r1 = 14r0 = 93800 km using Hohmann transfer orbit requires delta-v of 2824.34+1308.38=4132.72 m/s. However if spaceship first accelerates 3060.31 m/s, thus getting in elliptic orbit with apogee at r2 = 40r0 = 268000 km, then in apogee accelerates another 608.679 m/s, which places it in new orbit with perigee at r1 = 14r0 = 93800, and, finally, in perigee slows down by 447.554 m/s, placing itself in final circular orbit, then total delta-v will be only 4116.54, which is 16.18 m/s less.
Burn Hohmann ΔV (m/s) Bi-elliptic ΔV (m/s)
1 2824.34 3060.31
2 1308.38 608.679
3 - 447.554
Total 4132.72 4116.54
• Forward applied ΔV
• Reverse applied ΔV
Evidently, the bi-elliptic orbit spends more of its delta-V early on (in the first burn). This yields a higher contribution to the specific orbital energy and, due to the Oberth effect, is responsible for the net reduction in required delta-V.
Published - July 2009 | 1,147 | 3,949 | {"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": 0, "img_math": 13, "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} | 3.921875 | 4 | CC-MAIN-2018-17 | longest | en | 0.76114 |
http://www.mcqslearn.com/applied/physics/quiz/quiz.php?page=80 | 1,531,756,919,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00526.warc.gz | 509,110,198 | 7,545 | Practice viscous drag & strokes law quiz online, applied physics test 80 for online courses, distance learning. Free physics MCQs questions and answers to learn viscous drag & strokes law MCQs with answers. Practice MCQs to test knowledge on viscous drag and strokes law, significant figures calculations, applications of bernoullis equation, resistance and resistivity, cathode ray oscilloscope worksheets.
Free viscous drag & strokes law course worksheet has multiple choice quiz question as an object moving through liquid facing retarding force is named as with options viscosity, friction, drag force and surface tension with problems solving answer key to test study skills for online e-learning, viva help and jobs' interview preparation tips, study fluid dynamics multiple choice questions based quiz question and answers. viscous Drag & Strokes Law Video
## Quiz on viscous Drag & Strokes Law Quiz PDF Download Worksheet 80
viscous Drag and Strokes Law Quiz
MCQ. An object moving through the liquid facing the retarding force is named as
1. viscosity
2. friction
3. drag force
4. surface tension
C
Significant Figures Calculations Quiz
MCQ. If the number to be neglected is greater than 5, the retained number is
1. increased
2. decreased
3. may increased
4. may decreased
A
Applications of Bernoullis Equation Quiz
MCQ. Torricelli's theorem is one of the applications of
1. equation of continuity
2. Bernoulli's equation
3. light equation
4. speed equation
B
Resistance and Resistivity Quiz
MCQ. Voltage of a device having resistance 5 Ω and current 4 A will be
1. 10 V
2. 15 V
3. 20 V
4. 25 V
C
Cathode Ray Oscilloscope Quiz
MCQ. Output of the sweep and time base generator will be
1. sinusoidal waveform
2. cos waveform
3. saw tooth waveform
4. both a and b
C | 430 | 1,793 | {"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-2018-30 | latest | en | 0.855491 |
https://gmatclub.com/forum/despite-the-growing-number-of-people-who-purchase-plane-99685.html?kudos=1 | 1,508,512,094,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824225.41/warc/CC-MAIN-20171020135519-20171020155519-00105.warc.gz | 700,294,102 | 60,376 | It is currently 20 Oct 2017, 08:08
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Despite the growing number of people who purchase plane
Author Message
TAGS:
### Hide Tags
Manager
Joined: 05 Jan 2009
Posts: 80
Kudos [?]: 160 [0], given: 2
Despite the growing number of people who purchase plane [#permalink]
### Show Tags
02 Apr 2009, 16:50
23
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
63% (01:09) correct 37% (02:33) wrong based on 596 sessions
### HideShow timer Statistics
Despite the growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machine, many travelers will still use travel agents.
(A) growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will
(B) growing number of people who purchase plane tickets online, airline executives are convinced, just as one-third of bank customers still prefer human tellers to automatic teller machines, that many travelers would
(C) growing number of people purchasing plane tickets online, airline executives are convinced, just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, many travelers will
(D) fact that the number of people purchasing plane tickets online is growing, airline execituves are convinced, just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, that many travelers would
(E) fact that the number of people who purchase plane tickets online are growing, airline executives are convinced that, just as one-third of bank customers still prefer human tellers compared with automatic teller machines, many travelers would
[Reveal] Spoiler: OA
Last edited by hazelnut on 24 May 2017, 01:24, edited 3 times in total.
Kudos [?]: 160 [0], given: 2
Manager
Joined: 27 Jul 2012
Posts: 96
Kudos [?]: 120 [20], given: 62
Location: India
GMAT Date: 10-25-2012
WE: Consulting (Computer Software)
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
07 Feb 2013, 02:31
20
KUDOS
3
This post was
BOOKMARKED
Despite the growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will still use travel agents.
a) growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will correct
b) growing number of people who purchase plane tickets online, airline executives are convinced, Just as one-third of bank customers still prefer human tellers to automatic teller machines, that many travelers would Just as....that construction is incorrect. that is required after convinced
c) growing number of people purchasing plane tickets online, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, many travelers will that is required after convinced
d) fact that the number of people purchasing plane tickets online is growing, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, Just as....that construction is incorrect. that is required after convinced
e) fact that the number of people who purchase plane tickets online are growing, airline executives e convinced that, just as one-third of bak customers still prefer human tellers compared with automatic teller machines, many travelers would subject verb agreement - the number of people...are
_________________
if my post helped you, let me know by pressing Kudos...
Kudos [?]: 120 [20], given: 62
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 4289
Kudos [?]: 7908 [12], given: 364
Location: India
WE: Education (Education)
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
19 Jan 2011, 22:04
12
KUDOS
3
This post was
BOOKMARKED
The first aberration is the use of the phrase the fact, considered crispless in contemporary writing . So D and E are out. The second clue is that we do require the connecting conjunction ‘that’ to be in place after the word ‘convinced’. ‘Convinced that’ brings out the succeeding factor wholly to merge with the text. Only A uses ‘that’. So B and C are out, leaving A as the right choice.
It will be prudent not to get into the nuance of using ‘compared with’ and ‘compared to’, which is not being tested here; please note the right choice uses neither. IMO, neither ‘compared to’ nor ‘compared with’ is the decider in GMAT SC these days.
_________________
“Better than a thousand days of diligent study is one day with a great teacher” – a Japanese proverb.
9884544509
Kudos [?]: 7908 [12], given: 364
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1120
Kudos [?]: 2327 [5], given: 219
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
16 Jul 2013, 10:24
5
KUDOS
5
This post was
BOOKMARKED
fozzzy wrote:
In the incorrect answer choices B and D " that" is referring to machines and this distorts the meaning of sentence?
Tense is a better split in this sentence will vs would Option A is best!
No, that's not what happens here.
Despite the growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will still use travel agents.
Consider just A and B for now:
a) growing number of people who purchase plane tickets online, airline executives e convinced that, just as one-third of bsik customers still prefer human tellers to automatic teller machines, many travelers will
b) growing number of people who purchase plane tickets online, airline executives e convinced, Just as one-third of bank customers still prefer human tellers to automatic teller machines, that many travelers would
if you say: the executives are convinced , just as the consumers, that many travelers (...)
you are saying that the executives and the consumers have the same opinion about many travelers
If you say: the executives are convinced that , just as the consumers, many travelers(...)
you are saying that the executives are convinced that what happens with the consumers will happen also to the travelers
The position of that is crucial. In both B and D "that" does not refer back to the machines:
d) fact that the number of people purchasing plane tickets online is growing, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, that many travelers would
in both B and D, that is part of the construct "are convinced that"<== the big difference is the placement of ", just as (...)" as explained above.
So: yes B and D distort the meaning, but not because "that" refers to machines.
I hope I've explained myself well
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason
Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant
Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
Kudos [?]: 2327 [5], given: 219
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1635
Kudos [?]: 1105 [4], given: 109
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
23 Aug 2010, 13:40
4
KUDOS
6
This post was
BOOKMARKED
A
Despite and In Spite of are used for phrases; those fragment sentences in which there is not a verb. Despite and "In spite of" show contrast.
"Although" is similar, but it is used in clauses (in this type of fragment, there is a verb. Example: Although I loved her, I had to leaver her).
I think I deserve kudos
PS. This link could be useful for you: http://gmat-grammar.blogspot.com/search ... n%20though
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."
My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html
GMAT Club Premium Membership - big benefits and savings
Kudos [?]: 1105 [4], given: 109
Manager
Joined: 07 Feb 2010
Posts: 155
Kudos [?]: 741 [3], given: 101
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
02 Jan 2011, 21:39
3
KUDOS
27
This post was
BOOKMARKED
Despite the growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will still use travel agents.
a) gowing number of people who purchase plane tickets online, airline executives e convinced that, just as one-third of bsik customers still prefer hummn tellers to itomatic teller machines, many travelers will
b) gowing number of people who purchase plare tickets online, airline executives e convinced, Just as one-hhird of bank customers still prefer human tellers to automatic teller machines, hhat many travelers would
c) gowing number of people pu-chasing plane tickets online, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, many travelers will
d) fact that the number of people purchasing plane tickets online is growing, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, that many travelers would
e) fact that the number of people who purchase plane tickets online e growing, airline executives e convinced that, just as one-third of bak customers still prefer human tellers compared with utomatic teller machines, many travelers would
Kudos [?]: 741 [3], given: 101
Intern
Joined: 11 Jul 2010
Posts: 30
Kudos [?]: 48 [2], given: 6
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
18 Sep 2010, 01:22
2
KUDOS
pmal04 wrote:
Despite the growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will still use travel agents.
o gowing number of people who purchase plane tickets online, airline executives e convinced that, just as one-third of bsik customers still prefer hummn tellers to itomatic teller machines, many travelers will
o gowing number of people who purchase plare tickets online, airline executives e convinced, Just as one-hhird of bank customers still prefer human tellers to automatic teller machines, hhat many travelers
would
o gowing number of people pu-chasing plane tickets online, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compxed to automatic teller machines, many
travelers will
o fact that the number of people purchasing plane tickets online is growing, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compa’ed to automatic teller machines,
that many travelers would
o fact that the number of people who purchase plane tickets online e growing, airline executives e convinced that, just as one-third of bak customers still prefer human tellers compared with utomatic teller
machines, many travelers would
I think option A, just because "airline executives are convinced that, just as one-third..." should be the right approach. Option E also has "THAT" but its usage of "compared with" is wrong.
Please correct me if I am wrong.
Kudos [?]: 48 [2], given: 6
Manager
Joined: 27 Jul 2012
Posts: 96
Kudos [?]: 120 [2], given: 62
Location: India
GMAT Date: 10-25-2012
WE: Consulting (Computer Software)
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
07 Feb 2013, 02:33
2
KUDOS
Quote:
The first aberration is the use of the phrase the fact, considered crispless in contemporary writing . So D and E are out.
Daagh - can you please elaborate why use of "the fact" is incorrect here ? is it always incorrect? can this be used as a thumb rule to eliminate options ?
Regards
_________________
if my post helped you, let me know by pressing Kudos...
Kudos [?]: 120 [2], given: 62
Intern
Joined: 27 Dec 2012
Posts: 41
Kudos [?]: 15 [2], given: 11
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
08 Feb 2013, 07:02
2
KUDOS
Regarding the ambiguity with the usage of 'fact' in the options D and E:
Try to think logically. Did the number of people who buy air tickets online grow or the fact grow??
A fact is a fact. It cannot grow.
I hope that helps.
Kudos [?]: 15 [2], given: 11
Intern
Joined: 07 Jul 2010
Posts: 21
Kudos [?]: 37 [2], given: 2
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
20 Apr 2014, 05:12
2
KUDOS
The credited response should be A ....
When we talk about future event and not sure if the future event happens definitely - then we use "would "
When we are certain that the event definitely takes place in future - we use "will " ......
Airlines executives are certain that in future travelers continue to use travel agents .
So " will " needed in the answer choice ...
This in a single stroke kills option (b), (d), (e)
between C and A
C states - Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines,
Using prefer and compared in a single sentence - makes one of them redundant .
so A Wins ......
Kudos [?]: 37 [2], given: 2
Intern
Joined: 24 Sep 2013
Posts: 12
Kudos [?]: 5 [2], given: 56
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
22 Nov 2014, 06:38
2
KUDOS
No offense, but it is painful to read such options with missing and misspelled words!
anilnandyala wrote:
Despite the growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will still use travel agents.
a) gowing number of people who purchase plane tickets online, airline executives e convinced that, just as one-third of bsik customers still prefer hummn tellers to itomatic teller machines, many travelers will
b) gowing number of people who purchase plare tickets online, airline executives e convinced, Just as one-hhird of bank customers still prefer human tellers to automatic teller machines, hhat many travelers
would
c) gowing number of people pu-chasing plane tickets online, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, many
travelers will
d) fact that the number of people purchasing plane tickets online is growing, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compared to automatic teller machines,
that many travelers would
e) fact that the number of people who purchase plane tickets online e growing, airline executives e convinced that, just as one-third of bak customers still prefer human tellers compared with utomatic teller
machines, many travelers would
Kudos [?]: 5 [2], given: 56
Verbal Expert
Joined: 14 Dec 2013
Posts: 3157
Kudos [?]: 3303 [2], given: 22
Location: Germany
Schools: HHL Leipzig
GMAT 1: 780 Q50 V47
WE: Corporate Finance (Pharmaceuticals and Biotech)
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
29 Feb 2016, 12:52
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
sowragu wrote:
Do the usage of "TO" correct in a comparison in Option A and Option B "just as one-third of bank customers still prefer human
tellers to automatic teller machines
"
Yes the usage of to is correct. The idiomatic structure is prefer X to Y, not prefer X than Y.
Correct: I prefer tea to coffee.
Wrong: I prefer tea than coffee.
In this sentence: Customers prefer human tellers to automatic teller machines...... correct
Kudos [?]: 3303 [2], given: 22
Senior Manager
Joined: 06 Mar 2006
Posts: 490
Kudos [?]: 269 [1], given: 1
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
02 Apr 2009, 21:03
1
KUDOS
pmal04 wrote:
Despite the growing number of people who purchase plane tickets online, airline executives are convinced that, just as one-third of bank customers still prefer human tellers to automatic teller machines, many travelers will still use travel agents.
o gowing number of people who purchase plane tickets online, airline executives e convinced that, just as one-third of bsik customers still prefer hummn tellers to itomatic teller machines, many travelers will
o gowing number of people who purchase plare tickets online, airline executives e convinced, Just as one-hhird of bank customers still prefer human tellers to automatic teller machines, hhat many travelers
would
o gowing number of people pu-chasing plane tickets online, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compxed to automatic teller machines, many
travelers will
o fact that the number of people purchasing plane tickets online is growing, airline executives are convinced, Just as one-third of bank customers still prefer human tellers as compa’ed to automatic teller machines,
that many travelers would
o fact that the number of people who purchase plane tickets online e growing, airline executives e convinced that, just as one-third of bak customers still prefer human tellers compared with utomatic teller
machines, many travelers would
Use MGMAT's split and resplit method. convinced that is better than the ones without that. So eliminate B, C, D.
growing number of is better than the fact that the number of. So eliminate D, and E
Answer A is left to be correct choice.
By the way, C, D, and E can also be eliminate with the use of compare to and compare with.
Kudos [?]: 269 [1], given: 1
Intern
Joined: 15 Dec 2008
Posts: 37
Kudos [?]: 18 [1], given: 0
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
02 Apr 2009, 22:45
1
KUDOS
A for me too
Kudos [?]: 18 [1], given: 0
Senior Manager
Joined: 25 Feb 2010
Posts: 450
Kudos [?]: 110 [1], given: 10
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
24 Aug 2010, 08:21
1
KUDOS
1
This post was
BOOKMARKED
B. growing number of people who purchase plane tickets online, airline executives are convinced, just as one-third of bank customers still prefer human tellers to automatic teller machines, that many travelers would (misplaced)
C. growing number of people purchasing plane tickets online, airline executives are convinced, just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, many travelers will (AKWARD and meaning is changed)
D. fact that the number of people purchasing plane tickets online is growing, airline executives are convinced, just as one-third of bank customers still prefer human tellers as compared to automatic teller machines, that many travelers would 9misplaced)
E. fact that the number of people who purchase plane tickets online are growing, airline executives convinced that, just as one-third of bank customers still prefer human tellers compared with automatic teller machines, many travelers that
_________________
GGG (Gym / GMAT / Girl) -- Be Serious
Its your duty to post OA afterwards; some one must be waiting for that...
Kudos [?]: 110 [1], given: 10
Intern
Joined: 15 Aug 2010
Posts: 14
Kudos [?]: 3 [1], given: 0
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
30 Aug 2010, 03:36
1
KUDOS
is "prefer human tellers compared with automatic teller machines" correct?
I picked 'A' based on idiom "prefer human tellers to automatic teller machines"
Please let me know if my approacy is incorrect.
Kudos [?]: 3 [1], given: 0
Manager
Joined: 07 Jun 2010
Posts: 50
Kudos [?]: 49 [1], given: 7
Location: United States
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
17 Sep 2010, 16:58
1
KUDOS
but in choice A just as ....so ,should be idiom but it is not the case , can any one explain.
Kudos [?]: 49 [1], given: 7
Manager
Joined: 15 Apr 2010
Posts: 190
Kudos [?]: 18 [1], given: 29
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
17 Sep 2010, 17:04
1
KUDOS
this sentence is too convoluted already! not clear on how to narrow down the choices.....
Kudos [?]: 18 [1], given: 29
Senior Manager
Joined: 20 Jul 2010
Posts: 257
Kudos [?]: 101 [1], given: 9
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
17 Sep 2010, 17:44
1
KUDOS
A is correct ,however other choices are confusing.
Posted from my mobile device
_________________
If you like my post, consider giving me some KUDOS !!!!! Like you I need them
Kudos [?]: 101 [1], given: 9
Senior Manager
Joined: 08 Dec 2009
Posts: 409
Kudos [?]: 114 [1], given: 26
Re: Despite the growing number of people who purchase plane [#permalink]
### Show Tags
18 Sep 2010, 19:12
1
KUDOS
wow, heard of spell check?
fully of typos... horrible to read.
A: that ... will is probably the best construction
_________________
kudos if you like me (or my post)
Kudos [?]: 114 [1], given: 26
Re: Despite the growing number of people who purchase plane [#permalink] 18 Sep 2010, 19:12
Go to page 1 2 3 4 Next [ 74 posts ]
Display posts from previous: Sort by | 5,659 | 23,083 | {"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-2017-43 | latest | en | 0.935692 |
https://www.fire2fusion.com/kinetic-energy/kinetic-energy-diagram.html | 1,712,990,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00799.warc.gz | 710,234,978 | 4,932 | # Kinetic Energy Diagram
When an object is moving, it possess an energy and such energy is called kinetic energy. To know more about kinetic energy, click here. Kinetic energy is explained along with its diagram here.
# Kinetic Energy Diagram
### Kinetic energy Definition
• Kinetic energy is one of many types of energy that exist. This is energy generated because something is moving — the faster it’s going, the more kinetic energy it has. A person sitting has no kinetic energy, but a person running like a maniac has tremendous kinetic energy:
• The energy possessed by a body because of its motion, equal to one half the mass of the body times the square of its speed.
• The amount of kinetic energy KE of an object in translational motion is equal to one-half the product of its mass m and the square of its velocity v.
• Kinetic energy is given as follows,
Kinetic energy = ½(mv^2) | 199 | 897 | {"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-2024-18 | latest | en | 0.9253 |
https://hirecalculusexam.com/how-to-integrate-and-differentiate | 1,723,611,594,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641095791.86/warc/CC-MAIN-20240814030405-20240814060405-00242.warc.gz | 230,673,301 | 22,386 | # How To Integrate And Differentiate
How To Integrate And Differentiate A Matrix Matrix So to get a better understanding of how to integrate a matrix into your matrix class, it would be really helpful to consider prior to how to implement a matrix or partial matrix assignment. Well, most other topics take a while, but I’ll simply leave that as an exercise to some casual readers. All we need are a matrix A (The Matrix Builder), and that matrix B. Matrices Is it just me? Probably not. Would it be a good idea to divide or even an additional element in the matrix? Wouldn’t that be a good idea. Someone is going to tell you what the matrix B has to do. All this information is there for the record Once we have structure a few things we need to do. One is assign a matrix to some cell in our matrix. Not sure if we’ve already done all of the work I am doing yet, but my example matrix does it for me. That would give me a slightly more ordered structure in the equation. What that does is give the matrix a more consistent and ordered structure. If we wanted to deal with any second element, there should be a block matrix B which is the type of B (that I would call A). Those are simply the cases when we need some extra elements. If we had some amount of extra in the first column and wanted to have this sort of structure, just put some extra elements in there for a second purpose. If we wouldn’t have too many more cases, all we would need to do is sort of sort the way we would like them and sort that. If we wanted to have another subset B, that could be something like this (My Matrix Builder) or something and next about in the way it would. What that is we would do next is sort the rest of the elements here. As to the rest then; read these exercises out most of the time, and you will hopefully get really good at getting that kind of insight. Hopefully, you won’t need to complete your math exercise like I have done already. But it can help to have a feel. | 435 | 1,973 | {"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-2024-33 | latest | en | 0.971084 |
http://www.cs.umd.edu/Outreach/hsContest97/questions/node6.html | 1,553,602,361,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205163.72/warc/CC-MAIN-20190326115319-20190326141319-00120.warc.gz | 264,543,389 | 2,966 | Next: Heat Diffusion Up: 1997 UMCP High School Programming Contest Previous: Internet Routing
# Voting Paradox
from http://www.ccrc.wustl.edu/~lorracks/dsv/diss/
by Lorrie Cranor
The paradox of voting was discovered over 200 years ago by M. Condorcet, a French mathematician, philosopher, economist, and social scientist. However, it received little attention until Duncan Black explained its significance in a series of essays he began in the 1940s. The importance of the voting paradox was not fully realized until several years after Kenneth Arrow published Social Choice and Individual Values in 1951, which contained his General Possibility Theorem. The essence of this theorem is that there is no method of aggregating individual preferences over three or more alternatives that satisfies several conditions of fairness and always produces a logical result.
For this problem, you must write a program that evaluates different voting strategies on voter preferences. We consider just the case of 3 candidates, and each vote orders the candidates according to their preferences. There are a total of 5 schemes you need to consider:
• plurality winner - there is one ballot, and each voter casts their vote for their favorite candidate. Whoever gets the most votes wins, even if they don't get a majority of the votes cast.
• exhaustive ballot - On each ballot, each voter casts their vote for their favorite candidate (that is on the ballot). The candidate who gets the fewest votes is eliminated, and the survivers move on to the next round of voting. For three candidates, there will be just two ballots.
• 1&2 primary - Candidates 1 & 2 face off in a primary ballot. Whoever wins goes on to face candidate 3 in a second ballot. Whoever wins that vote wins the election.
• 1&3 primary - Candidates 1 & 3 face off in a primary ballot. Whoever wins goes on to face candidate 2 in a second ballot. Whoever wins that vote wins the election.
• 2&3 primary - Candidates 2 & 3 face off in a primary ballot. Whoever wins goes on to face candidate 1 in a second ballot. Whoever wins that vote wins the election.
Note that in all votes, each voter casts their vote for the candidate they like most who is on the ballot (as a result of the voting paradox, this doesn't always maximize the chance that a candidate they like will be elected).
You do not need to worry about election ties; none will occur in the data sets you will be tested on.
## Input Format
Your input will be a series of voter preferences, each consisting of 3 integers on a line. These numbers consist of the voter's first, second and third choice; each choice will be a number in the range 1-3. If the first number is 0, it means that this set of voter preferences is complete, and you should print the results for this set and then read in another set. If the first number is -1, it indicates end of input. This set of voter preferences is complete and is the last set of voter preferences; you should print the results and terminate.
## Output Format
For each set of voter preferences, you should print a header, the plurality election winner, the exhaustive ballot winner, and the winner if two of the candidates are paired in a primary first. The sample output shows the appropriate format, but don't worry about the number of spaces.
## Example
Input: Output:
```1 2 3
1 2 3
1 2 3
2 1 3
2 3 1
3 1 2
0 0 0
1 3 2
1 3 2
1 3 2
1 2 3
1 3 2
3 2 1
2 3 1
2 3 1
2 3 1
3 2 1
3 2 1
2 3 1
-1 -1 -1
```
```-------- election 1
plurality winner 1
exhaustive ballot 1
12 primary 1
13 primary 1
23 primary 1
-------- election 2
plurality winner 1
exhaustive ballot 2
12 primary 3
13 primary 3
23 primary 3
```
Input Output
## Our solution
Next: Heat Diffusion Up: 1997 UMCP High School Programming Contest Previous: Internet Routing
Bill Pugh
Mon Mar 17 14:34:34 EST 1997 | 971 | 3,854 | {"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-2019-13 | longest | en | 0.943791 |
https://santaclarapueblolibrary.org/and-pdf/2047-multiplication-and-division-of-complex-numbers-in-polar-form-pdf-24-286.php | 1,653,030,935,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00323.warc.gz | 574,191,597 | 7,288 | and pdfWednesday, May 5, 2021 8:34:06 AM5
# Multiplication And Division Of Complex Numbers In Polar Form Pdf
File Name: multiplication and division of complex numbers in polar form .zip
Size: 1133Kb
Published: 05.05.2021
## Exponential Form of a Complex Number - Expii
Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science. We first encountered complex numbers in the section on Complex Numbers. From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. The first step toward working with a complex number in polar form is to find the absolute value. It measures the distance from the origin to a point in the plane. Substituting, we have.
## 6. Products and Quotients of Complex Numbers
Your private radar to help you avoid infection. Help your community by making this breakthrough widely known. How do we derive this? It's easy to multiply and divide complex numbers when they're in exponential form! We just need to use our exponent rules to help us.
Because no real number satisfies this equation, i is called an imaginary number. Despite the historical nomenclature "imaginary", complex numbers are regarded in the mathematical sciences as just as "real" as the real numbers and are fundamental in many aspects of the scientific description of the natural world. Complex numbers allow solutions to certain equations that have no solutions in real numbers. For example, the equation. Complex numbers, however, provide a solution to this problem. According to the fundamental theorem of algebra , all polynomial equations with real or complex coefficients in a single variable have a solution in complex numbers.
We can think of complex numbers as vectors , as in our earlier example. We have met a similar concept to "polar form" before, in Polar Coordinates , part of the analytical geometry section. In the Basic Operations section, we saw how to add, subtract, multiply and divide complex numbers from scratch. However, it's normally much easier to multiply and divide complex numbers if they are in polar form. Our aim in this section is to write complex numbers in terms of a distance from the origin and a direction or angle from the positive horizontal axis. This is how the complex number looks on an Argand diagram. We recognise this triangle as our triangle from before.
When two complex numbers are given in polar form it is particularly simple to multiply and divide them. This is an advantage of using the polar form. 1.
## 4. Polar Form of a Complex Number
### Complex number
When performing addition and subtraction of complex numbers, use rectangular form. This is because we just add real parts then add imaginary parts; or subtract real parts, subtract imaginary parts. When performing multiplication or finding powers and roots of complex numbers, use polar and exponential forms. This is because it is a lot easier than using rectangular form.
Since complex numbers are legitimate mathematical entities, just like scalar numbers, they can be added, subtracted, multiplied, divided, squared, inverted, and such, just like any other kind of number. It is highly recommended that you equip yourself with a scientific calculator capable of performing arithmetic functions easily on complex numbers. Addition and subtraction with complex numbers in rectangular form is easy.
POLAR FORM AND DEMOIVRE'S THEOREM. At this point you can add, subtract, multiply, and divide complex numbers. However, there is still one basic.
The following questions are meant to guide our study of the material in this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions. Multiplication of complex numbers is more complicated than addition of complex numbers.
In this laboratory you will work with complex numbers. You will plot complex numbers in the complex plane, convert complex numbers from rectangular to polar form and from polar to rectangular form and add, subtract, multiply and divide complex numbers. Study the sections below and complete tasks Answer all questions highlighted in blue in full sentences. Open a Microsoft Word document to keep a log of your answers.
Сто десять? - оживился Джабба. - Сколько будет сто десять минус тридцать пять и две десятых. - Семьдесят четыре и восемь десятых, - сказала Сьюзан. - Но я не думаю… - С дороги! - закричал Джабба, рванувшись к клавиатуре монитора. - Это и есть ключ к шифру-убийце.
Тот, конечно, был мастером своего дела, но наемник остается наемником. Можно ли ему доверять. А не заберет ли он ключ. Фонтейну нужно было какое-то прикрытие - на всякий случай, - и он принял необходимые меры.
Панк наконец позволил себе улыбнуться.
1. ## ElaГs O.
06.05.2021 at 08:23
If you're seeing this message, it means we're having trouble loading external resources on our website.
2. ## Baudelio A.
07.05.2021 at 07:11
Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras , Descartes , De Moivre, Euler , Gauss , and others.
3. ## Zoombstanjacters
11.05.2021 at 04:45
Ibm storwize v7000 troubleshooting recovery and maintenance guide pdf negotiation in procurement pdf cips
4. ## Footbcarlliro
13.05.2021 at 20:23 | 1,236 | 5,620 | {"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} | 4.3125 | 4 | CC-MAIN-2022-21 | latest | en | 0.929689 |
https://donghiadigest.com/best-13 | 1,675,427,415,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500056.55/warc/CC-MAIN-20230203122526-20230203152526-00783.warc.gz | 231,427,868 | 5,081 | # Algebra equation help
There is Algebra equation help that can make the technique much easier. Our website can help me with math work.
## The Best Algebra equation help
We'll provide some tips to help you select the best Algebra equation help for your needs. Understand that power and square are inverse operations, can use square operation to find the square root of an integer within one hundred, can use cubic operation to find the cube root of an integer within one hundred (corresponding negative integer), and can use a calculator to find the square root and cube root ① As for the material resistance coefficient, through the absorption measurement of polished surfaces of different materials, it is found that the material absorption rate is directly proportional to the square root of the resistance coefficient, and the resistance coefficient changes with the change of temperature. 3. When measuring the flow with the throttle, the pressure and flow rate before and after the throttle change.
Mathematics itself depends on the ability of logical reasoning. If any problem can not be solved rigorously, it is not a good problem. I would like to start with my experience in learning mathematics with my classmates. Therefore, artificial intelligence is not omnipotent, it only accelerates the process of solving problems with mathematics.
The number series is often regarded as the final item in college entrance examination mathematics. For example, the following question: As with equations, analytic geometry is not the sooner the better. Compared with the Euclidean geometry, it not only makes solving problems become programmed, but also greatly weakens mathematical thinking ability. Therefore, it is only suitable to use analytic geometry to solve problems after fully training the thinking of planar geometry and mastering the physical concepts such as coordinate transformation.
## Instant support with all types of math
A fellow math class student showed me this app before math class and I am now totally blown away!!! I love this app and use it anytime I am struggling with a problem, and it's a teaching tool for me as well. This app is very impressive and an absolute gem to have on your phone!!! I highly recommend downloading this app!!! Thank you to the geniuses of this app!!!!
### Anna Thompson
I was pleasantly surprised by this app. Almost everything on the play store, be it game or tool, is much harder to work with than it seems or just floods you with ads. And if not those 2, it costs money up front. but this free app is actually really useful. It rarely has trouble solving problems I see, and even shows the steps to solving them if you happen to care about that. And in all that time, I haven't seen any ads. They're probably on something I haven't used yet. Overall, really good. 5
### Kayleigh Rogers
Free math help Hard to solve math problems Verifying identities solver Differential equation solver initial condition Enter math problem get answer Solving quadratic equations by factoring solver | 587 | 3,049 | {"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-2023-06 | latest | en | 0.934935 |
https://allaboutforex.world/chande-momentum-oscillator-cmo-forex-trading-indicator-explained/ | 1,722,667,190,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00591.warc.gz | 69,011,304 | 46,456 | # Chande Momentum Oscillator (CMO): Forex Trading Indicator Explained
The Chande Momentum Oscillator (CMO) is a popular tool among forex traders for identifying momentum and potential reversal points in the market. Developed by Tushar Chande, this indicator helps traders make informed decisions by analyzing price momentum over a specific period. In this article, we will explore the CMO in detail, including how it works, its applications, and how traders can use it effectively.
### What is the Chande Momentum Oscillator (CMO)?
The Chande Momentum Oscillator (CMO) is a technical analysis indicator that measures momentum in a financial instrument. Unlike other momentum oscillators, the CMO calculates momentum based on both upward and downward price movements. This dual calculation provides a more accurate reflection of market sentiment, making it a valuable tool for traders.
### How is the CMO Calculated?
The CMO is calculated using the following formula:
$\text{CMO} = 100 \times \left( \frac{\text{Su} - \text{Sd}}{\text{Su} + \text{Sd}} \right)$
where:
• Su is the sum of all upward price changes over a given period.
• Sd is the sum of all downward price changes over the same period.
The result is a value that oscillates between +100 and -100, indicating the strength and direction of momentum.
### Interpreting the CMO
The CMO provides valuable insights into market momentum. Here are the key levels to watch:
#### Overbought and Oversold Conditions
• A CMO value above +50 indicates that the market is overbought, suggesting a potential reversal or pullback.
• A CMO value below -50 indicates that the market is oversold, suggesting a potential upward reversal.
#### Zero Line Crossovers
• When the CMO crosses above the zero line, it indicates a shift from negative to positive momentum, signaling a potential buy opportunity.
• Conversely, when the CMO crosses below the zero line, it indicates a shift from positive to negative momentum, signaling a potential sell opportunity.
### Applications of the CMO in Forex Trading
#### Identifying Trend Reversals
One of the primary uses of the CMO is identifying trend reversals. By observing overbought and oversold conditions, traders can anticipate potential market turnarounds and adjust their positions accordingly.
#### Confirming Trends
The CMO can also be used to confirm existing trends. For instance, if the CMO remains consistently above the zero line during an uptrend, it confirms the strength of the upward momentum. This confirmation helps traders stay in profitable trades longer.
#### Divergence Analysis
Divergence between the CMO and price action can provide early warning signals of potential reversals. For example, if prices are making higher highs while the CMO is making lower highs, it indicates weakening momentum and a potential bearish reversal.
### Advantages of Using the CMO
#### Dual Momentum Calculation
The CMO’s unique approach of considering both upward and downward price movements provides a balanced view of market momentum, reducing the likelihood of false signals.
#### Versatility
The CMO can be applied to various timeframes and financial instruments, making it a versatile tool for traders across different markets.
#### Ease of Use
The CMO is easy to interpret and implement, making it accessible for both novice and experienced traders.
### Limitations of the CMO
#### Sensitivity to Market Noise
Like all momentum oscillators, the CMO can be sensitive to market noise, especially in volatile markets. Traders should use it in conjunction with other indicators to confirm signals.
#### Lagging Indicator
The CMO is a lagging indicator, meaning it is based on past price data. While it can provide valuable insights, it may not always predict future price movements accurately.
### How to Use the CMO in a Trading Strategy
#### Step 1: Set Up the Indicator
Most trading platforms include the CMO as a standard indicator. Set it up on your preferred timeframe, such as daily or hourly charts.
#### Step 2: Identify Key Levels
Monitor the CMO for overbought and oversold conditions, as well as zero line crossovers. These levels can provide entry and exit signals for trades.
#### Step 3: Combine with Other Indicators
Use the CMO in conjunction with other technical indicators, such as moving averages or the Relative Strength Index (RSI), to confirm signals and improve the accuracy of your trades.
### Conclusion
The Chande Momentum Oscillator (CMO) is a powerful tool for forex traders, offering insights into market momentum and potential reversal points. By understanding how to interpret and apply the CMO, traders can enhance their trading strategies and make more informed decisions. Remember to use the CMO in conjunction with other indicators and market analysis techniques to maximize its effectiveness. | 1,003 | 4,866 | {"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": 0, "img_math": 1, "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-2024-33 | latest | en | 0.842843 |
https://www.airmilescalculator.com/distance/pkk-to-mwq/ | 1,713,496,728,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00682.warc.gz | 578,647,289 | 16,312 | # How far is Magway from Pakhokku?
The distance between Pakhokku (Pakokku Airport) and Magway (Magway Airport) is 81 miles / 130 kilometers / 70 nautical miles.
The driving distance from Pakhokku (PKK) to Magway (MWQ) is 107 miles / 172 kilometers, and travel time by car is about 2 hours 43 minutes.
81
Miles
130
Kilometers
70
Nautical miles
## Distance from Pakhokku to Magway
There are several ways to calculate the distance from Pakhokku to Magway. Here are two standard methods:
Vincenty's formula (applied above)
• 80.984 miles
• 130.331 kilometers
• 70.373 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 81.329 miles
• 130.886 kilometers
• 70.673 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Pakhokku to Magway?
The estimated flight time from Pakokku Airport to Magway Airport is 39 minutes.
## Flight carbon footprint between Pakokku Airport (PKK) and Magway Airport (MWQ)
On average, flying from Pakhokku to Magway generates about 37 kg of CO2 per passenger, and 37 kilograms equals 81 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Pakhokku to Magway
See the map of the shortest flight path between Pakokku Airport (PKK) and Magway Airport (MWQ).
## Airport information
Origin Pakokku Airport
City: Pakhokku
Country: Burma
IATA Code: PKK
ICAO Code: VYPU
Coordinates: 21°19′59″N, 95°5′59″E
Destination Magway Airport
City: Magway
Country: Burma
IATA Code: MWQ
ICAO Code: VYMW
Coordinates: 20°9′56″N, 94°56′29″E | 508 | 1,847 | {"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-2024-18 | latest | en | 0.842682 |
https://www.jiskha.com/members/profile/posts.cgi?name=NF | 1,503,339,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109470.15/warc/CC-MAIN-20170821172333-20170821192333-00420.warc.gz | 901,540,455 | 2,494 | # Posts by NF
Total # Posts: 3
Calculus
Find the equation of the curve passing through P (1,2) whose slope is the square root of the product of the coordinates of any point on the curve.
Calculus
A square piece of canvas measuring 10 ft. by 10 ft. is to be used to make a pup tent with open ends as shown. How long should the center pole be if the volume covered by the tent is to be maximum?
Org Chem
C and D have a plane of symmetry,they are both meso. | 119 | 458 | {"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-2017-34 | latest | en | 0.949788 |
http://www.conservapedia.com/Fermat%27s_Last_Theorem | 1,529,434,213,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863109.60/warc/CC-MAIN-20180619173519-20180619193519-00118.warc.gz | 405,722,937 | 8,267 | # Fermat's Last Theorem
Pierre de Fermat
Fermat's Last Theorem asserts that the well-known Pythagorean Theorem has no solutions for higher powers. That is,
has infinitely many solutions, such as {3, 4, 5} or {5, 12, 13}; but
has no solutions using positive whole numbers.
It was conjectured by the French mathematician Pierre de Fermat. He said he had proved this problem but that there was not enough room in the margin to state his proof:[1]
"Cubum autem in duos cubos, aut quadrato-quadratum in duos quadrato-quadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet."
The English translation of Fermat's Latin statement is:[2]
"It is impossible for a cube to be the sum of two cubes, a fourth power to be the sum of two fourth powers, or in general for any number that is a power greater than the second to be the sum of two like powers. I have discovered a truly marvelous demonstration of this proposition that this margin is too narrow to contain."
This problem has confounded mathematicians for centuries, and there still is no proof for it using elementary techniques. Gauss and other mathematicians doubt that Fermat was able to prove it himself, but Cal Tech mathematics Professor E.T. Bell, who wrote the standard biography of all the great mathematicians, wryly observed that "the fox who could not get at the grapes declared they were sour." [3] "And so for all of [Fermat's] positive assertions with the one exception of the seemingly simple one which he made in his Last Theorem and which mathematicians, struggling for nearly 300 years, have been unable to prove: whenever Fermat asserted that he had proved anything, the statement, with the one exception noted, has subsequently been proved. Both his scrupulously honest character and his unrivalled penetration as an arithmetician substantiate the claim made for him by some, but not by all, that he knew what he was talking about when he asserted that he possessed a proof of his theorem."[4]
The theorem is as follows:
Andrew Wiles
For integers n > 2, there are no nonzero integral solutions to: xn + yn = zn
In the summer of 1986, Ken Ribet proved that Fermat's Last Theorem is a special case of the Taniyama–Shimura Conjecture.
## Claim of a Proof
In a series of lectures in 1993, mathematician Andrew Wiles announced a proof using techniques in algebraic geometry, relying on the nonconstructive Axiom of Choice.[5] A flaw was found before publication, and Wiles spent a year on fixing the flaw. Then, in September 1994, he and Richard Taylor announced a new version of the proof. However, criticism does continue on the internet.[5] Further criticism came from Marilyn vos Savant, known for her very high IQ and commentary on mathematics, in her column and book.[6][7] She questioned the use of Non-Euclidean geometry and the Axiom of Choice, among other points. She retracted her argument in a 1995 addendum to the book.
The Wiles-Taylor proof also makes use of some Grothendieck tools in cohomological number theory that use an axiom beyond the standard ZFC axioms. It is an open question whether these tools can be formalized into a ZFC proof.[8]
Unlike other mathematical breakthroughs, this claimed proof of 1993 has facilitated little, if any, insights or simplifications since then.
## References
1. Nagell 1951, p. 252.
2. http://mathworld.wolfram.com/FermatsLastTheorem.html
3. E.T. Bell, "Men of Mathematics" 72 (1937).
4. E.T. Bell, "Men of Mathematics" 71 (1937).
5. Ask Marilyn ® by Marilyn vos Savant, Parade Magazine. November 21, 1993
6. The World's Most Famous Math Problem: The Proof of Fermat's Last Theorem and Other Mathematical Mysteries, Marilyn vos Savant. St. Martin's Griffin, 1993
7. Colin Mclarty http://www.cwru.edu/artsci/phil/Proving_FLT.pdf | 953 | 3,911 | {"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": 2, "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-2018-26 | latest | en | 0.970231 |
http://reference.wolfram.com/legacy/v4/RefGuide/Optional.html | 1,511,007,647,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804881.4/warc/CC-MAIN-20171118113721-20171118133721-00354.warc.gz | 256,423,317 | 7,174 | This is documentation for Mathematica 4, which was
based on an earlier version of the Wolfram Language.
Optional
p:v is a pattern object which represents an expression of the form p, which, if omitted, should be replaced by v.
Optional is used to specify "optional arguments" in functions represented by patterns. The pattern object p gives the form the argument should have, if it is present. The expression v gives the "default value" to use if the argument is absent.
Example: the pattern f[x_, y_:1] is matched by f[a], with x taking the value a, and y taking the value 1. It can also be matched by f[a, b], with y taking the value b.
The form s_:v is equivalent to Optional[s_, v]. This form is also equivalent to s:_:v. There is no syntactic ambiguity since s must be a symbol in this case.
The special form s_. is equivalent to Optional[s_] and can be used to represent function arguments which, if omitted, should be replaced by default values globally specified for the functions in which they occur.
Values for Default[f, ... ] specify default values to be used when _. appears as an argument of f. Any assignments for Default[f, ... ] must be made before _. first appears as an argument of f.
Optional[s_h] represents a function which can be omitted, but which, if present, must have head h. There is no simpler syntactic form for this case.
Functions with built-in default values include Plus, Times and Power.
See The Mathematica Book: Section 2.3.9 and Section A.2.7. | 343 | 1,491 | {"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-2017-47 | latest | en | 0.858499 |
https://oeis.org/A101601/internal | 1,632,147,624,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00351.warc.gz | 479,609,200 | 2,886 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A101601 G.f.: c(3x)^3, c(x) the g.f. of A000108. 2
%I
%S 1,9,81,756,7290,72171,729729,7505784,78298974,826489170,8811646074,
%T 94753804536,1026499549140,11192793160815,122744496427425,
%U 1352917116177840,14979996753469110,166542316847391870
%N G.f.: c(3x)^3, c(x) the g.f. of A000108.
%F G.f.: 8/(1+sqrt(1-12*x))^3.
%F a(n) = (3*n+3)/(n+3) * 3^n * C(n+1).
%F Conjecture: (n+3)*a(n) -3*(5*n+7)*a(n-1) +18*(2*n-1)*a(n-2)=0. - _R. J. Mathar_, Nov 15 2011
%t terms = 18;
%t c[x_] = (1 - Sqrt[1 - 4x])/(2x) + O[x]^terms // Normal;
%t CoefficientList[c[3x]^3, x][[1 ;; terms]] (* _Jean-François Alcover_, Dec 15 2017 *)
%Y Cf. A050159, A101600, A101602.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Dec 08 2004
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified September 20 10:18 EDT 2021. Contains 347584 sequences. (Running on oeis4.) | 486 | 1,233 | {"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} | 3.140625 | 3 | CC-MAIN-2021-39 | latest | en | 0.508119 |
https://de.zxc.wiki/wiki/Logistische_Funktion | 1,725,950,567,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00117.warc.gz | 180,467,548 | 16,289 | Logistic function
The logistic function characterizes a continuous one-dimensional probability distribution and is a functional representation of saturation processes from the class of the so-called sigmoid functions with unlimited time extension.
Until the 20th century, the logarithm was occasionally given the Italian name of the logistic curve ( curva logistica ). Today the name is clearly assigned to the S function .
description
Logistic function just in case ${\ displaystyle G = 1 \,; \, k = 1 \,; \, f (0) = 0 {,} 5}$
The logistic function, as it results from the discrete logistic equation , describes the connection between the elapsing time and a growth, for example an ideal bacterial population or (approximately) the spread of a disease in the context of an epidemic. For this purpose, the model of exponential growth is modified by a resource that consumes itself with growth - the idea behind it is, for example, a bacteria culture medium of limited size or a decreasing number of people who behave carelessly during an epidemic and thus expose themselves to the risk of infection.
Example of an epidemic: illnesses and deaths (black) in the course of the Ebola fever epidemic in West Africa until July 2014 (almost logistical functions)
At the beginning the function value is not 0, but it applies . ${\ displaystyle f (0)> 0}$
The following applies to the bacterial example:
• The limited habitat forms an upper limit for the number of bacteria .${\ displaystyle G}$${\ displaystyle f (t)}$
• The bacterial growth is proportional to: ${\ displaystyle f '(t)}$
• the current stock ${\ displaystyle f (t)}$
• the remaining capacity ${\ displaystyle Gf (t)}$
This development is therefore represented by a Bernoulli differential equation of the form
${\ displaystyle f '(t) = k \ cdot f (t) \ cdot \ left (Gf (t) \ right)}$
described with a proportionality constant . Solving this differential equation gives: ${\ displaystyle k}$
${\ displaystyle f (t) = G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t} \ left ({\ frac {G} {f (0)}} - 1 \ right)}}}$
The graph of the function describes an S-shaped curve, a sigmoid . In the beginning the growth is small because the population and thus the number of reproducing individuals is small. In the middle of the development (more precisely: at the turning point ) the population grows most strongly until it is slowed down by the exhausting resources.
Other uses
The logistic equation describes a connection that occurs very often and is used far beyond the idea of describing a population of living things. The life cycle of a product in the store can also be modeled with the logistic function. Further areas of application are growth and decay processes in language ( Language Change Act , Piotrowski Law ) and the development in the acquisition of the mother tongue ( Language Acquisition Act ). The logistic function is also used in the SI model of mathematical epidemiology.
Solution of the differential equation
Be . is steady. It is important to solve the differential equation . ${\ displaystyle F \ colon \ mathbb {R} \ to \ mathbb {R}, t \ mapsto kt (Gt)}$${\ displaystyle F}$${\ displaystyle {\ frac {\ mathrm {d} f} {\ mathrm {d} t}} (t) = F (f (t))}$
The differential equation can be solved with the " separation of variables " procedure . It applies to everyone , so the mapping to the real numbers is well-defined. ${\ displaystyle F (t) \ neq 0}$${\ displaystyle t \ in \ mathbb {R}}$${\ displaystyle \ Phi (t): = \ int _ {0} ^ {t} {\ frac {1} {F (s)}} \, {\ mathrm {d}} s}$
After separating the variables, the solution of the above differential equation is identical to the solution of the differential equation ${\ displaystyle f}$
${\ displaystyle \ int _ {f (0)} ^ {f (t)} {\ frac {1} {F (s)}} \, \ mathrm {d} s = t}$.
The decomposition of partial fractions results in
${\ displaystyle \ int _ {f (0)} ^ {f (t)} {\ frac {1} {F (s)}} \, \ mathrm {d} s = \ int _ {f (0)} ^ {f (t)} {\ frac {1} {ks (Gs)}} \, \ mathrm {d} s = {\ frac {1} {kG}} \ int _ {f (0)} ^ { f (t)} \ left ({\ frac {1} {s}} + {\ frac {1} {Gs}} \ right) \, \ mathrm {d} s}$
According to the main law of differential and integral calculus , the above is integral
${\ displaystyle = \ left. {\ frac {1} {kG}} {\ bigg (} \ ln (s) - \ ln (Gs) {\ bigg)} \ right | _ {s = f (0)} ^ {s = f (t)} = {\ frac {1} {kG}} {\ bigg (} \ ln {\ big (} {\ frac {f (t)} {Gf (t)}} {\ big)} + c {\ bigg)}}$
in which
${\ displaystyle c: = - \ ln {\ frac {f (0)} {Gf (0)}} = \ ln {\ big (} {\ frac {G} {f (0)}} - 1 {\ big)}}$
So the function equation applies
${\ displaystyle \ ln {\ frac {f (t)} {Gf (t)}} = kGt-c}$
to solve as long as they lie between and , which can be assumed because of the prerequisite . Where is the natural logarithm . The application of the exponential function on both sides leads to ${\ displaystyle f (t)}$${\ displaystyle 0}$${\ displaystyle G}$${\ displaystyle 0 ${\ displaystyle \ ln}$
${\ displaystyle e ^ {kGt-c} \, = \, {\ frac {f (t)} {Gf (t)}}}$.
and subsequent reciprocal value to
${\ displaystyle (*) \ quad \ quad \ quad e ^ {- kGt + c} \, = \, {\ frac {Gf (t)} {f (t)}} \, = \, {\ frac { G} {f (t)}} - 1}$.
We now bring the to the left, then form the reciprocal value again, and finally get ${\ displaystyle 1}$
${\ displaystyle {\ frac {f (t)} {G}} \, = \, {\ frac {1} {1 + e ^ {- kGt + c}}}}$
and it
${\ displaystyle (**) \ quad \ quad \ quad f (t) \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt + c}}}}$
If we insert the definition of into the found solution (**), we come to the solution of the logistic differential equation claimed above: ${\ displaystyle c}$
${\ displaystyle f (t) \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt + c}}} \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt} e ^ {c}}} \, = \, G \ cdot {\ frac {1} {1 + e ^ {- kGt} \ left ({\ frac {G} {f (0 )}} - 1 \ right)}}}$
It is easy to see from this functional equation that the values are always between and , which is why the solution applies to all . This can of course also be confirmed in retrospect by inserting it into the differential equation. ${\ displaystyle 0}$${\ displaystyle G}$${\ displaystyle - \ infty
Calculation of the turning point
To determine the turning point of the solution function , we first determine the derivatives using the product rule${\ displaystyle f}$
{\ displaystyle {\ begin {aligned} f '(t) & = k \ cdot f (t) \ cdot (Gf (t)) \\ f' '(t) & = k \ cdot f' (t) \ cdot (Gf (t)) + k \ cdot f (t) \ cdot (-f '(t)) \\ & = k \ cdot f' (t) \ cdot (Gf (t) -f (t)) = k \ cdot f '(t) \ cdot (G-2 \ cdot f (t)) \ end {aligned}}}
and determine the zero of the second derivative: ${\ displaystyle t_ {W}}$
${\ displaystyle f '' (t_ {W}) = k \ cdot f '(t_ {W}) \ cdot (G-2 \ cdot f (t_ {W})) = 0}$
${\ displaystyle G-2 \ cdot f (t_ {W}) = 0}$
${\ displaystyle G = 2 \ cdot f (t_ {W})}$
${\ displaystyle f (t_ {W}) = {\ tfrac {G} {2}}}$
With this we know the function value at the inflection point and determine that the population in the inflection point just exceeds half the saturation limit. To determine , we use the solution formula and calculate as follows: ${\ displaystyle t_ {W}}$${\ displaystyle f (t_ {W}) = {\ tfrac {G} {2}}}$
${\ displaystyle G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t_ {W}} \ cdot e ^ {c}}} = {\ frac {G} {2}} }$
${\ displaystyle 1 + e ^ {- k \ cdot G \ cdot t_ {W}} \ cdot e ^ {c} = 2}$
${\ displaystyle e ^ {- k \ cdot G \ cdot t_ {W}} \ cdot e ^ {c} = 1 = e ^ {0}}$
${\ displaystyle -k \ cdot G \ cdot t_ {W} + c = 0}$
${\ displaystyle t_ {w} = {c \ over k \ cdot G}}$;
For follows with further: ${\ displaystyle \ textstyle G> f (0)}$${\ displaystyle \ textstyle e ^ {c} = {\ frac {G} {f (0)}} - 1 \ Rightarrow c = ln \ left ({\ frac {G} {f (0)}} - 1 \ right)}$
${\ displaystyle t_ {W} = {\ frac {\ ln \ left ({\ frac {G} {f (0)}} - 1 \ right)} {k \ cdot G}}}$
With this the turning point is completely determined and there is only this one. Substituting in the first derivative gives the maximum growth rate: ${\ displaystyle f (t_ {W}) = {\ tfrac {G} {2}}}$
${\ displaystyle f '(t_ {W}) = k \ cdot {\ frac {G} {2}} \ cdot \ left (G - {\ frac {G} {2}} \ right) = k \ cdot { \ frac {G} {2}} \ cdot {\ frac {G} {2}}}$
${\ displaystyle f '(t_ {W}) = {\ frac {k \ cdot G ^ {2}} {4}}}$
Further representations
${\ displaystyle f (t) = G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t} \ left ({\ frac {G} {f (0)}} - 1 \ right)}}}$
${\ displaystyle = G \ cdot {\ frac {1} {1 + e ^ {- k \ cdot G \ cdot t} \ cdot {\ frac {G} {f (0)}} - e ^ {- k \ cdot G \ cdot t}}} \ cdot {\ frac {f (0)} {f (0)}} = {\ frac {G \ cdot f (0)} {f (0) + e ^ {- k \ cdot G \ cdot t} \ cdot Ge ^ {- k \ cdot G \ cdot t} \ cdot f (0)}}}$
${\ displaystyle = {\ frac {G \ cdot f (0)} {f (0) + \ left (Gf (0) \ right) \ cdot e ^ {- k \ cdot G \ cdot t}}}}$
or:
${\ displaystyle f (t) = {\ frac {G} {2}} \ cdot \ left (\ tanh \ left ({\ frac {kG} {2}} (t-t_ {W}) \ right) + 1 \ right)}$, where the turning point calculated above is:${\ displaystyle t_ {W}}$${\ displaystyle t_ {W} = {\ frac {\ ln \ left ({\ frac {G} {f (0)}} - 1 \ right)} {k \ cdot G}}}$ | 2,890 | 9,189 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 63, "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-2024-38 | latest | en | 0.884826 |
heihachi.eu | 1,620,313,442,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00011.warc.gz | 23,861,348 | 5,333 | Richard Lemmens website
# Functional programming
The vast majority of popular programming languages are imperative, i.e. they consist of a sequence of 'statements' that are executed more or less sequentially, meanwhile changing the program's state. Of course there is branching with if-then-else statements, there are loops and subroutines, but in essence those programs are sequences of statements. This is a paradigm that almost all programmers are infinitely familiar with. Some juniors actually think that that is how every programming language looks like. However there are alternatives, and interesting ones too.
One of the most prominent of those is functional programming, based on the theory of lambda calculus. Included in serious computer science curricula for decades because of its close relationship with mathematics, it is now slowly working its way into the mainstream. This article shows some basic features of function programming for novices. More extensive material can be found on other websites and in books. Throughout the text the syntax of Lisp, the mother of all functional programming languages, is used.
### Functions
Central to functional programming is the concept of a function. These are like mathematical functions, which take a handful of parameters as input, do their work and finally produce output. A simple example can be drawn from arithmetic, like the 'add' function that adds two numbers together. It takes two numbers as input, adds them up and produces the sum as output. It is common to use infix notation to write sums, like "1 + 1 = 2", though a mathematician may well want to use prefix notation and write something like "add(1, 1) = 2". In Lisp the parentheses are put on the outside and the commas are omitted, so the expression is written as:
``` ```
(+ 1 1)
```
```
still resulting in 2 of course.
### Functions calling functions
Full functional programs tend to be a lot more complex, and of course also more capable, than a small simple function like 'add'. They achieve this not by adding more functions sequentially, but by subdividing the work into more functions, without specifying an order of execution. For example a combination of an addition and a multiplication such as "1 * 2 + 3 * 4" may be written as:
``` ```
(+ (* 1 2) (* 3 4))
```
```
A fundamental observation from this example is that the arguments of a 'higher order' function can be function calls themselves. Likewise, the output of functions can be functions too. Some modern iterative programmings languages like Python and C# have also added this concept. But with functional languages this has been a feature right from the start, because this is the way that a functional program is structured: one big 'main' function that breaks down in smaller functions, which may break further down, etcetera.
Keen observers will note that such a program can be presented not only with text, but graphically as a tree. In such a tree each function call is a node and its parameters are the children of the node. Again the example from above, as a tree with indentation representing the layers:
``` ```
+
*
1
2
*
3
4
```
```
### Control flow and recursion
There are no sequences of statements in functions, but branches are just as common as in imperative languages. The functional equivalent of "if x = 1 then A else B" may be written as:
``` ```
(if (= x 1) A B)
```
```
where 'if' and '=' are of course functions themselves, with boolean instead of numeric output.
Loops are achieved through recursion, which is another fundamental concept in functional programming. In recursion a function calls itself, though with different parameter values than in the original call. A first function call can trigger a second, a third and so on until an end condition is met, which terminates the recursion, preventing it from going on forever. Here is an example of a function that can multiply two positive integer values:
``` ```
(defun multiply (x y) (if (= y 1) x (+ x (multiply x (- y 1)) ) ) )
```
```
This function recursively calls itself, each time with a value for y that is 1 lower than the previous one. When y reaches a value of 1, the recursion ends and x is returned. On the way back up trough the recursion tree, the result of the underlying function call is added to x and that result is passed to the function call that triggered the current one. The result is y - 1 times adding x to itself, or more commonly known as x * y.
### Parallelism
Functional programs are very easy to 'parallelize'. When a function calls other functions, it must wait until those other function calls are resolved. But the order in which that is done is irrelevant. And as there are no global variables, those function calls are effectively independent of each other, which means they can be executed in parallel. A typical functional program starts branching out into various function calls quite soon, increasing 'parallelizability' at the same rate.
### Conclusion
Many programmers who are used to iterative programming languages are shocked to find that there is no such things as variables, i.e. memory cells whose value can change over time, in functional programming. All 'variables' are either input values set to fixed values when the program starts, constants or function calls. There is simply no need for variables! Though once you wrap your mind around the new paradigm, functional programming actually shows itself to be cleaner and more intuitive than iterative programming.
This has been an ultra-short intoruction into functional programming. If you are interested in real world programs, there is much more to know and learn. There are several functional programming languages available. The most pure and well known are Haskell, Lisp and Scheme, though there are many more. Even XSLT turns out to be a functional language!
Several universities have published books and courses online and there are also programmers who have written introductions that are more extensive than this one. If you want not just to absorb the theory, but try your hands at functional programming, you can download a compiler or interpreter. Some are even available online, like http://rextester.com/l/common_lisp_online_compiler. Note that you will have to use 'special' functions like "print" in order to direct the results of function calls to your screen. | 1,315 | 6,436 | {"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-2021-21 | latest | en | 0.939765 |
http://yetanothermathprogrammingconsultant.blogspot.com/2011/12/bad-nlp-modeling.html | 1,532,291,913,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593586.54/warc/CC-MAIN-20180722194125-20180722214125-00566.warc.gz | 568,233,502 | 12,946 | ## Thursday, December 22, 2011
c6.. y1*(y1-1) =e= 0;
c7.. y2*(y2-1) =e= 0;
c8.. y3*(y3-1) =e= 0;
This is not really a good idea. These constraints essentially say: y1,y2 and y3 can assume the values 0 or 1, i.e. they are binary variables.
If you really want to use binary variables in an NLP, form a proper MINLP model. If this idea of using non-convex constraints to model binary variables was really so hot, then we would no longer need a MIP solver and just use an NLP solver.
It looks a bit that this poster is attempting to run before (s)he can walk. It would be much better if you have some knowledge about NLP modeling before before actually working on “stochastic global optimization”. May be the advisor or teacher should play a more constructive role here.
Notes:
• in general NLP solvers don’t “understand” the functions you give them. They merely sample points, function values and derivatives as they go along. One exception is the global solver Baron: this solver really tries to work with and understand the functional form of the NL functions.
• although in general it is better to model binary decisions as binary variables in an MINLP (or MIP if the rest is linear), there are some attempts to use nonlinear programming techniques to solve integer models:
• W. Murray and K. M. Ng, “An Algorithm for Nonlinear Optimization Problems with Binary Variables,” Computational Optimization and Applications, Vol. 47, No. 2, 2008, pp. 257-288
• Roohollah Aliakbari Shandiz, Nezam Mahdavi-Amiri, “An Exact Penalty Approach for Mixed Integer Nonlinear Programming Problems”, American Journal of Operations Research, 2011, 1, 185-189 | 417 | 1,646 | {"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} | 2.828125 | 3 | CC-MAIN-2018-30 | latest | en | 0.92068 |
https://stats.stackexchange.com/questions/190773/calculate-type-i-and-ii-error-solution-verification | 1,721,780,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00265.warc.gz | 475,830,165 | 41,156 | # Calculate type I and II error - solution verification
There are 7 balls in urn. $Q$ of them are white and the rest are black. We have hypothesis $H_0:Q=3$ and $H_1:Q=5$. To test this we draw 2 balls (balls don't come back to the urn - i.e. they are drawn without replacement).
We reject $H_0$ when both balls drawn are white. Calculate probability of type I and type II errors.
MY PROBLEM - what are probabilities of $H_0$ and $H_1$? Are they actually needed?
SOLUTION SKETCH:
Let $x$ be number of drawn white balls.
$\alpha=\mathbb{P}(x>1|H_0)=\frac{\frac{3}{7}\cdot\frac{2}{6}}{\frac{3}{7}}=\frac{2}{6}$
or it is
$\alpha=\mathbb{P}(x>1|H_0)=\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{7}$
and for II type error: $\beta=\mathbb{P}(x<2|H_1)=\frac{\frac{5}{7}\cdot\frac{2}{6}+\frac{2}{7}\cdot\frac{1}{6}+\frac{2}{7}\cdot\frac{5}{6}}{\frac{5}{7}}=\frac{11}{15}$
Or is it wrong?
• I'm not sure what role H1 plays here. Normally, in a problem like this you either reject the null hypothesis (H0) or determine that you can not reject the null hypotheses. The only reasonably H1 would be Q >= 4 (ie, the actual number of white balls is more than 3), not actually having Q set to a specific number.
– user1566
Commented Jan 15, 2016 at 2:06
I believe you need an a priori estimate for P(Q3) and P(Q5)
If you assume these two events to be equally likely (before pulling balls), then let each be 0.5
From Bayes $P(x=2)*P(Q_3|x=2) = P(x=2|Q_3)*P(Q_3)$
$P(Q_3|x=2) = \frac{P(x=2|Q_3)*P(Q_3)}{P(x=2)}$
$P(x=2)=P(x=2|Q_3)*P(Q_3)+P(x=2|Q_5)*P(Q_5)$
$P(x=2|Q_3)=\frac{3}{7}*\frac{2}{6}=\frac{1}{7}$
$P(x=2|Q_5)=\frac{5}{7}*\frac{4}{6}=\frac{10}{21}$
$P(x=2)=\frac{1}{7}*\frac{1}{2}+\frac{10}{21}*\frac{1}{2}=\frac{13}{42}$
$P(Q_3|x=2) = \frac{P(x=2|Q_3)*P(Q_3)}{P(x=2)}=\frac{\frac{1}{7}*\frac{1}{2}}{\frac{13}{42}}=\frac{3}{13}$
$P(Q_5|x=2) = \frac{P(x=2|Q_5)*P(Q_5)}{P(x=2)}=\frac{\frac{10}{21}*\frac{1}{2}}{\frac{13}{42}}=\frac{10}{13}$
Thus a Type I is $\frac{3}{13}$
A Type II error you have to complete a similar analysis for each of the cases (one of each and zero white) which would not have rejected H0 in favor of H1
Your question seems to be incorrect regarding the hypotheses "H0:Q=3 and H1:Q=5"
Null and alternative hypotheses are mutually exclusive and exhaustive statements.
It can be of the form:
H0:Q is equals to 3 and H1:Q not equals to 3 | 914 | 2,364 | {"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} | 4 | 4 | CC-MAIN-2024-30 | latest | en | 0.720855 |
http://louistiao.me/posts/index-5.html | 1,498,583,600,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321497.77/warc/CC-MAIN-20170627170831-20170627190831-00674.warc.gz | 243,956,173 | 5,217 | # Quick Reference: Deleting Local and Remote Git Branches
Deleting a Local Git Branch:
$git branch -D <branch-name> Deleted branch <branch-name> (was <commit-hash>). Deleting a Remote Git Branch (available as of Git 1.7.0)$ git push origin --delete <branch-name>
To <git-remote-origin-url>
- [deleted] <branch-name>
# Setting up a IPython Parallel Cluster on Amazon EC2 with StarCluster
StarCluster is an open source cluster-computing toolkit for Amazon’s Elastic Compute Cloud (EC2) that is designed to automate and simplify the process of building, configuring, and managing clusters of virtual machines on Amazon’s EC2 cloud. StarCluster makes it easy to create a cluster computing environment in the cloud for distributed and parallel computing applications.
# Visualizing and Animating Optimization Algorithms with Matplotlib
In this series of notebooks, we demonstrate some useful patterns and recipes for visualizing animating optimization algorithms using Matplotlib.
In [1]:
%matplotlib inline
In [2]:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib.colors import LogNorm
from matplotlib import animation
from IPython.display import HTML
from scipy.optimize import minimize
from collections import defaultdict
from itertools import zip_longest
from functools import partial
We shall restrict our attention to 3-dimensional problems for right now (i.e. optimizing over only 2 parameters), though what follows can be extended to higher dimensions by plotting all pairs of parameters against each other, effectively projecting the problem to 3-dimensions.
The Wikipedia article on Test functions for optimization has a few functions that are useful for evaluating optimization algorithms. In particular, we shall look at Beale's function:
$$f(x, y) = (1.5 - x + xy)^2 + (2.25 - x + xy^2)^2 + (2.625 - x + xy^3)^2$$
In [3]:
f = lambda x, y: (1.5 - x + x*y)**2 + (2.25 - x + x*y**2)**2 + (2.625 - x + x*y**3)**2 | 462 | 1,976 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-26 | longest | en | 0.747794 |
https://itprospt.com/num/9619785/simplify-the-radical-expressions-if-possible-assume-the | 1,656,443,268,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00481.warc.gz | 390,024,895 | 12,841 | 1
Simplify the radical expressions, if possible. Assume the variables represent positive real numbers. $$\frac{x+\sqrt{y}}{x-\sqrt{y}}$$...
Question
Simplify the radical expressions, if possible. Assume the variables represent positive real numbers. $$\frac{x+\sqrt{y}}{x-\sqrt{y}}$$
Simplify the radical expressions, if possible. Assume the variables represent positive real numbers. $$\frac{x+\sqrt{y}}{x-\sqrt{y}}$$
Answers
Similar Solved Questions
5 answers
A. Verify that the given point lies on the curve_ b. Determine an equation of the line tangent to the curve at the given point:6x2 8xy ~ 4y2 108; (2,3)How do you know that the point is on the given curve?OA The point (2,3) is in the first quadrant: B The point (2,3) is in the domain of the implicit function: C. The value of 6x2 + 8xy ~ 4y2 is less than 108 when x is 2 and y is 3_ D The value of 6x2 + 8xy ~ 4y2 is 108 when x is 2 and y is 3.b. Write the equation for the tangent Iine in slope-inte
a. Verify that the given point lies on the curve_ b. Determine an equation of the line tangent to the curve at the given point: 6x2 8xy ~ 4y2 108; (2,3) How do you know that the point is on the given curve? OA The point (2,3) is in the first quadrant: B The point (2,3) is in the domain of the implic...
3 answers
Solve - the given the initial system of equations approximation x() using the Jacobi iteration method with4x -2xz +X; =I1; x +Sx2 -3x; =-; 2x, +2x2 +5x3 =7.List the first three iterations for x(2), X(3)15
Solve - the given the initial system of equations approximation x() using the Jacobi iteration method with 4x -2xz +X; =I1; x +Sx2 -3x; =-; 2x, +2x2 +5x3 =7. List the first three iterations for x(2), X(3) 15...
5 answers
Is the statement "If two events A, B are dependent; then P(AnB) < P(A)P (B) true? Why? What if we replace with 2 ?
Is the statement "If two events A, B are dependent; then P(AnB) < P(A)P (B) true? Why? What if we replace with 2 ?...
5 answers
5 answers
Point) Find y as function of = i20" + lsy' 49y = 291) = =-10, v (1) =-
point) Find y as function of = i 20" + lsy' 49y = 2 91) = =-10, v (1) =-...
5 answers
Find thc following product, and write Iha produot In recLangular forr, usirg exacl values. [A( cos B0" sln 80 "VIL7 cos 165" sln 1659
Find thc following product, and write Iha produot In recLangular forr, usirg exacl values. [A( cos B0" sln 80 "VIL7 cos 165" sln 1659...
5 answers
Set up the general equations from the given statements.The speed $v$ at which a galaxy is moving away from earth varies directly as its distance $r$ from earth.
Set up the general equations from the given statements. The speed $v$ at which a galaxy is moving away from earth varies directly as its distance $r$ from earth....
5 answers
Draw three vectors:A = 10iB = 15jandc = A+BFind |Ax B
Draw three vectors: A = 10i B = 15j and c = A+B Find | Ax B...
5 answers
5 points) Water is leaking out of an inverted conical tank at rate of 0.0075 m? /min. At the same time water being pumped into the tank at constant rate. The tank has height 8 meters and the diameter at the top is 5 meters: If the water level is rising at a rate 0f 0.21 mlmin when the height of the water Is meters; find the rate at which water is being pumped into the tank:Water is being pumped in atIn" / Inin
5 points) Water is leaking out of an inverted conical tank at rate of 0.0075 m? /min. At the same time water being pumped into the tank at constant rate. The tank has height 8 meters and the diameter at the top is 5 meters: If the water level is rising at a rate 0f 0.21 mlmin when the height of the ...
5 answers
In Lesson 5.3, you may have conjectured that translations produce figures congruent to the original figures. If you have a rule that produces a translation, you can use the distance formula to check that segments stay the same length.a. Copy the grid and the segment at right. Then use the rule $(x, y) ightarrow(x+3, y+1)$ to create a new segment translated to the right 3 units and up 1 unit.b. Use the distance formula to find the length of the original segment.c. Now find the length of the imag
In Lesson 5.3, you may have conjectured that translations produce figures congruent to the original figures. If you have a rule that produces a translation, you can use the distance formula to check that segments stay the same length. a. Copy the grid and the segment at right. Then use the rule \$(x...
5 answers
A solid sphere of radius 42.5 cm has a total positive charge of29.0 μμC uniformly distributed throughout its volume.Calculate the magnitude of the electric field 0 cm from the centerof the sphere.Calculate the magnitude of the electric field 14.4 cm from thecenter of the sphere. Calculate the magnitude of the electric field 42.5 cm from thecenter of the sphere. Calculate the magnitude of the electric field 59.8 cm from thecenter of the sphere.
A solid sphere of radius 42.5 cm has a total positive charge of 29.0 μμC uniformly distributed throughout its volume. Calculate the magnitude of the electric field 0 cm from the center of the sphere. Calculate the magnitude of the electric field 14.4 cm from the center of the sphere. Calculat...
4 answers
An air-conditioning equipment and parts distributor iscontemplating establishing a distribution warehouse in a newgeographic area. A break-even analysis suggests that it would onlybe financially viable for the distributor if at least one-fourth ofall new private homes in the area have central air-conditioning. Aspart of its market study, the company contacts a random sample of200 home-owners and find that 78 of these homes are equipped withcentral air-conditioning.What is the standard error of t
An air-conditioning equipment and parts distributor is contemplating establishing a distribution warehouse in a new geographic area. A break-even analysis suggests that it would only be financially viable for the distributor if at least one-fourth of all new private homes in the area have central ai...
5 answers
Question 43Calculate the specific heat when 573 J of energy is removed from14.85 g of a fictional substance as it is cooled from 72.4°C to19.8°C. Show work on your paper. In the box below, type yournumerical answer only.
Question 43 Calculate the specific heat when 573 J of energy is removed from 14.85 g of a fictional substance as it is cooled from 72.4°C to 19.8°C. Show work on your paper. In the box below, type your numerical answer only....
5 answers
For each of the following studies, determine whether anindependent-measures t test or a repeated-measures t test isappropriate analysis (1 pt.)Examine the development of vocabulary as a group of childrenmature from age 2 to age 3Compare the mathematics skills for 9th grade boysversus 9th grade girlsCompare the blood pressure readings before medication and aftermedication for a group of patients with high blood pressureExamine the effect of violent video games on behavior bycomparing aggressive b
For each of the following studies, determine whether an independent-measures t test or a repeated-measures t test is appropriate analysis (1 pt.) Examine the development of vocabulary as a group of children mature from age 2 to age 3 Compare the mathematics skills for 9th grade boys versus 9th grade...
5 answers
Which of the following statements is true regarding the length-tension relationship within skeletal muscle? Sarcomeres generate the greatest tension at thelr optimal length Sarcomeres generate the greatest tenslon as the slze of the zones of overlap increase beyond their optimum length Sarcomeres generate the greatest tension as the size of the zones of overlap decrease past thelr optlmum length Sarcomeres generate the greatest tension when calclum ion Is pumped from the cytosol into the termina
Which of the following statements is true regarding the length-tension relationship within skeletal muscle? Sarcomeres generate the greatest tension at thelr optimal length Sarcomeres generate the greatest tenslon as the slze of the zones of overlap increase beyond their optimum length Sarcomeres ge...
5 answers
Prism is a device used in the spectrophotometer to function as(2 Points)a detectora dispersive systemOthera sample holdera light sourceinjector
Prism is a device used in the spectrophotometer to function as (2 Points) a detector a dispersive system Other a sample holder a light source injector...
1 answers
Let V be finite dimensional over F and let P € C(V) be idempotent_ Determine the eigenvalues of P and show that P is diagonalizable:
Let V be finite dimensional over F and let P € C(V) be idempotent_ Determine the eigenvalues of P and show that P is diagonalizable:...
4 answers
The maxinum length of path graph G is called the diameter o G. Let T be tree with dinmeter k Prove that every two paths in T of length k have uodkc iCQlHWIQIL() Does the same result holdl for maximal paths in T? Here by maximal path we MCAn path that cannot he extended to longer one.
The maxinum length of path graph G is called the diameter o G. Let T be tree with dinmeter k Prove that every two paths in T of length k have uodkc iCQlHWIQIL () Does the same result holdl for maximal paths in T? Here by maximal path we MCAn path that cannot he extended to longer one....
5 answers
Problem 3.point) Let f (x,y) = cos(xy ) + elxty The second derivative equal to which of the following? dxdy A Zxer'y + 2x"ye*ly sin(x + y)B. ~64x2yld cos(xy ) S6xy" sin(xy" ) e"x+y~8yls cos(xy ) + TeZxty~lo ( cos(xy ) + 49e"x+yE~8xyls cos(xy) ~ 8y7 sin(xy8) + Telx+y
Problem 3. point) Let f (x,y) = cos(xy ) + elxty The second derivative equal to which of the following? dxdy A Zxer'y + 2x"ye*ly sin(x + y) B. ~64x2yld cos(xy ) S6xy" sin(xy" ) e"x+y ~8yls cos(xy ) + TeZxty ~lo ( cos(xy ) + 49e"x+y E~8xyls cos(xy) ~ 8y7 sin(xy8) + Telx+...
5 answers
Find fcos(x), f(o)f '(0)f "{0)
Find f cos(x), f(o) f '(0) f "{0)...
-- 0.018191-- | 2,672 | 9,881 | {"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": 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} | 3.234375 | 3 | CC-MAIN-2022-27 | latest | en | 0.768129 |
https://www.toolsandjobs.info/mcq/basic-electrical-engineering-mcq-and-answers-sppu-exam-2020-part-2/ | 1,708,572,564,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00109.warc.gz | 1,070,736,193 | 18,491 | # Basic Electrical Engineering MCQ And Answers For SPPU Exam 2020 Part-2
This contains Basic Electrical Engineering mcq questions for preparing SPPU exam 2020, for placements exams, companies exam and other.This containMCQ Electrostatics quetions.
This is part 2 of Basic Electrical Engineering mcq questions with answers
Click to read part 1
1.Which of the following statements is incorrect ?
1. The leakage resistance of ceramic capacitors is generally high
2. The stored energy in a capacitor decreases with reduction in value of capacitance
3. The stored energy in a capacitor increases with applied voltage
4. A wire cable has distributed capacitance between the conductors
Ans: b
1. Which of the following capacitors has relatively shorter shelf life ?
1. Mica capacitor
2. Electrolytic capacitor
3. Ceramic capacitor
4. Paper capacitor
Ans: b
1. The sparking between two electrical contacts can be reduced by inserting a
a capacitor in parallel with contacts
b capacitor in series with each contact
1. resistance in line
2. none of the above
Ans: a
1. In the case of a lossy capacitor, its series equivalent resistance value will be
1. small
2. very small
3. large
4. zero
Ans: c
1. The power dissipated in a pure capacitor is
(a) zero
(6) proportional to applied voltage
1. proportional to value of capacitance
2. both (b) and (c) above
Ans: a
1. In a capacitive circuit
1. a steady value of applied voltage causes discharge
2. an increase in applied voltage makes a capacitor charge
3. decrease in applied voltage makes a capacitor charge
1. none of the above
Ans: b
1. When a dielectric slab is introduced in a parallel plate capacitor, the potential difference between plates will
1. remain uncharged
2. decrease
3. increase
4. become zero
Ans: b
1. Capacitance increases with
1. increase in plate area and decrease in distance between the plates
2. increase in plate area and distance between the plates
3. decrease in plate area and value of applied voltage
4. reduction in plate area and distance between the plates
Ans: a
1. A capacitor consists of
1. two insulators separated by a con¬ductor
2. two conductors separated by an in¬sulator
3. two insulators only
4. two conductors only
Ans: b
1. A gang condenser is a
(a) polarised capacitor
(b) variable capacitor
1. ceramic capacitor
2. none of the above
Ans:c
1. A paper capacitor is usually available in the form of
1. tubes
2. rolled foil
3. disc
4. meshed plates
Ans: b
1. Air capacitors are generally available in the range
1. 10 to 400 pF
2. 1 to 20 pF
3. 100 to 900 pF
4. 20 to 100 pF
Ans: a
1. The unit of capacitance is
1. henry
2. ohm
Ans: c
1. A capacitor charged to 200 V has 2000 (iC of charge. The value of capacitance will be
(a) 10 F
(b) 10 uF
1. 100 nF
2. 1000 uF
Ans: b
1. A capacitor in a circuit became hot and ultimately exploded due to wrong con¬nections, which type of capacitor it could be ?
1. Paper capacitor
2. Ceramic capacitor
3. Electrolytic capacitor
4. Any-of the above
Ans: c
1. Energy stored in the electric field of a capacitor C when charged from a D.C source of voltage V is equal to joules
1. CV2
2. C2V
3. CV2
4. CV
Ans: a
1. The absolute permittivity of free space is given by
a. 8.854 x 1(T9 F/m)
b. 8.854 x 1(T10 F/m
c. 8.854 x KT11 F/m
d. 8.854 x 10"12 F/m
Ans: b
1. The relative permittivity of free space is given by
1. 1
2. 10
3. c. 100
4. d. 1000
Ans: a
1. Electric field intensity is a quantity
1. scalar
1. vector
2. both (a) and (6)
3. none of the above
Ans: b
1. When 4 volts e.m.f. is applied across a 1 farad capacitor, it will store energy of
1. 2 joules
2. 4 joules
3. 6 joules
4. 8 joules
Ans: d
1. The capacitor preferred for high frequency circuits is
1. air capacitor
2. mica capacitor
3. electrolytic capacitor
4. none of the above
Ans: b
1. The capacity of capacitor bank used in power factor correction is expressed in terms of
1. kW
2. kVA
3. kVAR
4. volts
Ans: c
1. While testing a capacitor with ohm-metre, if the capacitor shows charging, but the final resistance reading is appreciably less than normal, it can be concluded that the capacitor is
1. short-circuited
2. open circuited
3. alright
4. leaky
Ans: d
1. If a 6 (iF capacitor is charged to 200 V, the charge in coulombs will be
1. 800 uC
2. 900 uC
3. 1200 uC
4. 1600 uC
Ans: c
1. Which capacitor will be physically smaller for the same ratings ?
1. Ceramic capacitor
2. Paper capacitor
3. Both will be of equal size
4. None of the above
Ans: a
1. What is the value of capacitance that must be connected in parallel with 50 pF condenser to make an equivalent capacitance of 150 pF ?
1. 50 pF
2. 100 pF
3. 150 pF
4. 200 pF
Ans: b
1. A mica capacitor and a ceramic capacitor both have the same physical dimensions. Which will have more value of capacitance ?
1. Ceramic capacitor
2. Mica capacitor
3. Both will have identical value of capacitance
4. It depends on applied voltage
Ans: a
1. Which of the following material has least value of dielectric constant ?
1. Ceramics
2. Oil
3. Glass
4. Paper
Ans: b
1. Which of the following capacitors will have the least value of breakdown voltage ?
1. Mica
2. Paper
3. Ceramic
4. Electrolytic
Ans: d
1. Dielectric constant for mica is nearly
a. 200
b. 100
1. 3 to 8
2. 1 to 2
Ans: c
Click to read part 3 | 1,578 | 5,323 | {"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} | 2.765625 | 3 | CC-MAIN-2024-10 | longest | en | 0.866738 |
https://www.c4u.net.in/Recursion.html | 1,579,444,754,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00191.warc.gz | 807,973,529 | 3,459 | Recursion:-Recursion is the process of repeating items in a self-similar way. In programming languages, if a program allows you to call a function inside the same function, then it is called a recursive call of the function. The C programming language supports recursion, i.e., a function to call itself. But while using recursion, programmers need to be careful to define an exit condition from the function, otherwise it will go into an infinite loop. Recursive functions are very useful to solve many mathematical problems, such as calculating the factorial of a number, generating Fibonacci series, etc.
Syntax:-
void recursion()
{
recursion(); /* function calls itself */
}
int main()
{
recursion();
}
Program to find factorial of a number using recursion:-
#include< stdio.h>
#include< conio.h>
int factorial(int i)
{
if(i <= 1)
{
return 1;
}
return i * factorial(i - 1);
}
void main()
{
int i = 15;
printf("Factorial of %d is %d\n", i, factorial(i));
getch();
}
Fibonacci Series
The following example generates the Fibonacci series for a given number using a recursive function -
#include
int fibonaci(int i)
{
if(i == 0)
{
return 0;
}
if(i == 1)
{
return 1;
}
return fibonaci(i-1) + fibonaci(i-2);
}
int main()
{
int i;
for (i = 0; i < 10; i++)
{
printf("%d\t\n", fibonaci(i));
}
return 0;
}
When the above code is compiled and executed, it produces the following result -
0 1 1 2 3 5 8 13 21 34 | 378 | 1,430 | {"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-2020-05 | latest | en | 0.619467 |
https://forum.hkedcity.net/index.php?threads/sql-%E9%A1%AF%E7%A4%BA%E5%AD%B8%E7%94%9F%E9%81%B8%E4%BF%AE%E7%A7%91%E5%B1%AC%E6%96%BC%E5%93%AA%E5%80%8B-block.118863/ | 1,674,879,914,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00041.warc.gz | 283,720,209 | 17,752 | # SQL SQL 顯示學生選修科屬於哪個 block
1. ### cantsy Expand Collapse
文章:
129
讚:
0
SQL 顯示每位學生選修科屬於哪個 block, 以班及班跪排序 , 例如:
Thank you.
文件大小:
14.8 KB
瀏覽:
3,698
2. ### edb-石頭 Expand Collapse
文章:
1,160
讚:
0
老師請參考以下的 SQL 語句。
select
a.classcode,
a.classno,
a.chname,
a.enname,
x1.en_des 'block 1',
x2.en_des 'block 2',
x3.en_des 'block 3'
from vw_stu_lateststudent a
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode, b.exdrptype
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
left outer join
tb_asr_subjassessdata b
on x.suid=b.suid and x.schyear=b.schyear and w.stuid=b.stuid and w.subjcode = b.subjcode and w.moi = b.moi and b.timeseq=?
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_1'
) x1
on a.suid=x1.suid and a.schlvl=x1.schlevel and a.schsess=x1.schsession and a.schyear=x1.schyear and a.stuid=x1.stuid and a.stuschrecid=x1.stuschrecid
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode, b.exdrptype
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
left outer join
tb_asr_subjassessdata b
on x.suid=b.suid and x.schyear=b.schyear and w.stuid=b.stuid and w.subjcode = b.subjcode and w.moi = b.moi and b.timeseq=?
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_2'
) x2
on a.suid=x2.suid and a.schlvl=x2.schlevel and a.schsess=x2.schsession and a.schyear=x2.schyear and a.stuid=x2.stuid and a.stuschrecid=x2.stuschrecid
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode, b.exdrptype
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
left outer join
tb_asr_subjassessdata b
on x.suid=b.suid and x.schyear=b.schyear and w.stuid=b.stuid and w.subjcode = b.subjcode and w.moi = b.moi and b.timeseq=?
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_3'
) x3
on a.suid=x3.suid and a.schlvl=x3.schlevel and a.schsess=x3.schsession and a.schyear=x3.schyear and a.stuid=x3.stuid and a.stuschrecid=x3.stuschrecid
where a.schyear=? and a.classlvl=? and (a.status=null or a.status not in (4,5,6))
order by a.classlvl, a.classcode, a.classno, a.chname, a.enname
#2 edb-石頭, 2012-08-07
3. ### cantsy Expand Collapse
文章:
129
讚:
0
What are the input ? Thx.
文件大小:
12.4 KB
瀏覽:
3,690
4. ### edb-石頭 Expand Collapse
文章:
1,160
讚:
0
老師請參考以下的 SQL 語句。
select
a.classcode,
a.classno,
a.chname,
a.enname,
x1.en_des 'block 1',
x2.en_des 'block 2',
x3.en_des 'block 3'
from vw_stu_lateststudent a
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_1'
) x1
on a.suid=x1.suid and a.schlvl=x1.schlevel and a.schsess=x1.schsession and a.schyear=x1.schyear and a.stuid=x1.stuid and a.stuschrecid=x1.stuschrecid
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_2'
) x2
on a.suid=x2.suid and a.schlvl=x2.schlevel and a.schsess=x2.schsession and a.schyear=x2.schyear and a.stuid=x2.stuid and a.stuschrecid=x2.stuschrecid
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_3'
) x3
on a.suid=x3.suid and a.schlvl=x3.schlevel and a.schsess=x3.schsession and a.schyear=x3.schyear and a.stuid=x3.stuid and a.stuschrecid=x3.stuschrecid
where a.schyear=? and a.classlvl=? and (a.status=null or a.status not in (4,5,6))
order by a.classlvl, a.classcode, a.classno, a.chname, a.enname
#4 edb-石頭, 2012-08-07
5. ### cantsy Expand Collapse
文章:
129
讚:
0
謝謝 ! 很好用, 但如果加入所有非科目集的科目, 可以嗎? 例如英文小組 EN15_a, M1, M2,......等
6. ### Tom Expand Collapse
文章:
81
讚:
0
既然這個SQL 好用, 可否加入成為常設在Websams 學生資料 > 科目設定 > 報告 ?
7. ### edb-石頭 Expand Collapse
文章:
1,160
讚:
0
老師可以聯絡貴校的網上校管系統學校聯絡主任提出建議。
#7 edb-石頭, 2012-08-31
8. ### cantsy Expand Collapse
文章:
129
讚:
0
請問如何在原先的SQL一併顯示所有 科目集和非科目集, 例如: 謝謝 !
文件大小:
9.4 KB
瀏覽:
3,661
9. ### edb-石頭 Expand Collapse
文章:
1,160
讚:
0
老師請聯絡貴校的網上校管系統學校聯絡主任。
#9 edb-石頭, 2012-08-31
10. ### cantsy Expand Collapse
文章:
129
讚:
0
如果加入所有非科目集的科目, 可以嗎? 例如英文小組 EN15_a, M1, M2,......等
Thank you. 瀏覽附件b.png
11. ### cantsy Expand Collapse
文章:
129
讚:
0
如果加入所有非科目集的科目, 可以嗎? 例如英文小組 EN15_a, M1, M2,......等
Thank you.
瀏覽附件c.png
12. ### edb-石頭 Expand Collapse
文章:
1,160
讚:
0
做不到老師要求的格式, SQL 並不支援 variable column。不過老師可以參考以下的 SQL 語句 for non-block subjects。
select
a.classlvl 'ClassLevel',
a.classcode 'Class',
a.classno 'ClassNo',
a.chname 'StudentName_Chi',
a.enname 'StudentName_Eng',
b.ch_des 'SubjectName_Chi',
b.en_des 'SubjectName_Eng',
ccsg.subjgrpdesc 'CrossClassSubjectGroupName',
ccsg.subjgrpcode 'CrossClassSubjectGroupCode',
(case when ccsg.grptype='B' then 'Subject Block' else 'Non-Subject Block' end) 'SubjectBlock',
ccsg1.divsubjcode 'CrossClassSubjectSubGroupCode',
smoi.ch_des 'MOI',
ccsg1.subjblkcode 'SubjectBlockCode',
ccsg2d.ch_des 'CrossClassSubjectSubjectComponent_Chi',
ccsg2d.en_des 'CrossClassSubjectSubjectComponent_Eng'
from vw_stu_lateststudent a
join tb_stu_stusubj a1
on a.suid=a1.suid and a.stuid=a1.stuid and a.stuschrecid=a1.stuschrecid
left outer join tb_sch_divsubjgrp ccsg
on a.suid=ccsg.suid and a.schlvl=ccsg.schlevel and a.schsess=ccsg.schsession and a.schyear=ccsg.schyear and a1.crosssubjgrp=ccsg.subjgrpcode
left outer join tb_sch_dsgsubj ccsg1
on a.suid=ccsg1.suid and a.schlvl=ccsg1.schlevel and a.schsess=ccsg1.schsession and a.schyear=ccsg1.schyear and ccsg.subjgrpcode=ccsg1.subjgrpcode and a1.crosssubjcode=ccsg1.divsubjcode
left outer join tb_sch_dsgsubjcomp ccsg2
on a.suid=ccsg2.suid and a.schlvl=ccsg2.schlevel and a.schsess=ccsg2.schsession and a.schyear=ccsg2.schyear and ccsg.subjgrpcode=ccsg2.subjgrpcode and a1.crosssubjcode=ccsg2.divsubjcode
left outer join tb_hse_common smoi
on a.suid=smoi.suid and a1.moi=smoi.code_id and smoi.tb_id='MOI'
left outer join tb_hse_common b
on a.suid=b.suid and a1.subjcode=b.code_id and b.tb_id='SBJ'
left outer join tb_hse_sbjcmp ccsg2d
on a.suid=ccsg2d.suid and ccsg.subjcode=ccsg2d.sbj_code and ccsg2.subjcompcode=ccsg2d.code_id
where a.schyear=? and a1.clssubjtype='X' and a.status=null and a.classlvl =?
group by a.schlvl, a.schsess, a.classlvl, a.classcode, a.classno, a.enname, a.chname, a1.clssubjtype,
b.ch_des, b.en_des, ccsg.subjgrpdesc, ccsg.subjgrpcode, ccsg.grptype, ccsg1.divsubjcode, smoi.ch_des,
ccsg1.subjblkcode, ccsg2d.ch_des, ccsg2d.en_des
order by ccsg.subjgrpcode, ccsg1.divsubjcode, a.classlvl, a.classcode, a.classno, a.chname, b.ch_des, b.en_des, ccsg2d.ch_des, ccsg2d.en_des
#12 edb-石頭, 2012-09-03
13. ### cantsy Expand Collapse
文章:
129
讚:
0
SQL 可顯示非科目集, 但有以下問題:
1. 同一科目小組學生出現 了數次, listening, speaking....同一學生出現了四次
2. 雖然不支援 multi-column, 但可否以班及班號作基礎,
瀏覽附件a.png
14. ### ps-ncw Expand Collapse
文章:
108
讚:
0
回覆 4# edb-石頭
請問可否顯示divsubjcode及其任教老師SHORTNAME?
15. ### edb-escm Expand Collapse
文章:
434
讚:
0
老師請參考以下SQL:
select
a.classlvl 'ClassLevel',
a.classcode 'Class',
a.classno 'ClassNo',
a.chname 'StudentName_Chi',
a.enname 'StudentName_Eng',
b.ch_des 'SubjectName_Chi',
b.en_des 'SubjectName_Eng',
ccsg.subjgrpdesc 'CrossClassSubjectGroupName',
ccsg.subjgrpcode 'CrossClassSubjectGroupCode',
(case when ccsg.grptype='B' then 'Subject Block' else 'Non-Subject Block' end) 'SubjectBlock',
ccsg1.divsubjcode 'CrossClassSubjectSubGroupCode',
smoi.ch_des 'MOI',
ccsg1.subjblkcode 'SubjectBlockCode',
ccsg1.divsubjcode,
ccsg1t.STAFFSEQ,
st.SHORTNAME 'Teacher ShortName',
ccsg2d.ch_des 'CrossClassSubjectSubjectComponent_Chi',
ccsg2d.en_des 'CrossClassSubjectSubjectComponent_Eng'
from vw_stu_lateststudent a
join tb_stu_stusubj a1
on a.suid=a1.suid and a.stuid=a1.stuid and a.stuschrecid=a1.stuschrecid
left outer join tb_sch_divsubjgrp ccsg
on a.suid=ccsg.suid and a.schlvl=ccsg.schlevel and a.schsess=ccsg.schsession and a.schyear=ccsg.schyear and a1.crosssubjgrp=ccsg.subjgrpcode
left outer join tb_sch_dsgsubj ccsg1
on a.suid=ccsg1.suid and a.schlvl=ccsg1.schlevel and a.schsess=ccsg1.schsession and a.schyear=ccsg1.schyear and ccsg.subjgrpcode=ccsg1.subjgrpcode and a1.crosssubjcode=ccsg1.divsubjcode
left outer join TB_SCH_DSGSUBJTCHR ccsg1t
on a.suid=ccsg1t.suid and a.schlvl=ccsg1t.schlevel and a.schsess=ccsg1t.schsession and a.schyear=ccsg1t.schyear and ccsg.subjgrpcode=ccsg1t.subjgrpcode and a1.crosssubjcode=ccsg1t.divsubjcode
left outer join VW_ASR_STAFF st
on a.suid=st.suid and ccsg1t.STAFFCODE = st.STAFFCODE
left outer join tb_sch_dsgsubjcomp ccsg2
on a.suid=ccsg2.suid and a.schlvl=ccsg2.schlevel and a.schsess=ccsg2.schsession and a.schyear=ccsg2.schyear and ccsg.subjgrpcode=ccsg2.subjgrpcode and a1.crosssubjcode=ccsg2.divsubjcode
left outer join tb_hse_common smoi
on a.suid=smoi.suid and a1.moi=smoi.code_id and smoi.tb_id='MOI'
left outer join tb_hse_common b
on a.suid=b.suid and a1.subjcode=b.code_id and b.tb_id='SBJ'
left outer join tb_hse_sbjcmp ccsg2d
on a.suid=ccsg2d.suid and ccsg.subjcode=ccsg2d.sbj_code and ccsg2.subjcompcode=ccsg2d.code_id
where a.schyear=? and a1.clssubjtype='X' and a.status=null and a.classlvl =?
group by a.schlvl, a.schsess, a.classlvl, a.classcode, a.classno, a.enname, a.chname, a1.clssubjtype,
b.ch_des, b.en_des, ccsg.subjgrpdesc, ccsg.subjgrpcode, ccsg.grptype, ccsg1.divsubjcode,ccsg1t.STAFFSEQ,st.SHORTNAME,smoi.ch_des,
ccsg1.subjblkcode, ccsg2d.ch_des, ccsg2d.en_des
order by ccsg.subjgrpcode, ccsg1.divsubjcode, a.classlvl, a.classcode, a.classno, a.chname, b.ch_des, b.en_des, ccsg2d.ch_des, ccsg2d.en_des
#15 edb-escm, 2015-08-10
16. ### ps-ncw Expand Collapse
文章:
108
讚:
0
回覆 15# edb-escm
沒有資料顯示出來！我用post 4可以出到資料，想在這個SQL之上顯示divsubjcode及其任教老師SHORTNAME！
謝謝！
17. ### edb-escm Expand Collapse
文章:
434
讚:
0
老師請聯絡貴校的網上校管系統學校聯絡主任以便跟進。
#17 edb-escm, 2015-08-10
18. ### ps-ncw Expand Collapse
文章:
108
讚:
0
回覆 17# edb-escm
但本身的post 4 SQL是可以的！
19. ### edb-escm Expand Collapse
文章:
434
讚:
0
老師請參考以下SQL,如仍未能提取老師所需,請聯絡貴校的網上校管系統學校聯絡主任以便跟進:
select
a.classcode,
a.classno,
a.chname,
a.enname,
x1.en_des 'block 1',
x1.divsubjcode 'block 1 divsubjcode',
x1.SHORTNAME 'block 1 Teacher SHORTNAME',
x2.en_des 'block 2',
x2.divsubjcode 'block 2 divsubjcode',
x2.SHORTNAME 'block 2 Teacher SHORTNAME',
x3.en_des 'block 3',
x3.divsubjcode 'block 3 divsubjcode',
x3.SHORTNAME 'block 3 Teacher SHORTNAME'
from vw_stu_lateststudent a
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode,st.SHORTNAME
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
left outer join TB_SCH_DSGSUBJTCHR y1
on x.suid=y1.suid and x.schlevel=y1.schlevel and x.schsession=y1.schsession and x.schyear=y1.schyear and x.subjgrpcode=y1.subjgrpcode and w.crosssubjcode=y1.divsubjcode and y1.STAFFSEQ=1
left outer join VW_ASR_STAFF st
on w.suid=st.suid and y1.STAFFCODE = st.STAFFCODE
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_1'
) x1
on a.suid=x1.suid and a.schlvl=x1.schlevel and a.schsess=x1.schsession and a.schyear=x1.schyear and a.stuid=x1.stuid and a.stuschrecid=x1.stuschrecid
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode,st.SHORTNAME
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
left outer join TB_SCH_DSGSUBJTCHR y1
on x.suid=y1.suid and x.schlevel=y1.schlevel and x.schsession=y1.schsession and x.schyear=y1.schyear and x.subjgrpcode=y1.subjgrpcode and w.crosssubjcode=y1.divsubjcode and y1.STAFFSEQ=1
left outer join VW_ASR_STAFF st
on w.suid=st.suid and y1.STAFFCODE = st.STAFFCODE
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_2'
) x2
on a.suid=x2.suid and a.schlvl=x2.schlevel and a.schsess=x2.schsession and a.schyear=x2.schyear and a.stuid=x2.stuid and a.stuschrecid=x2.stuschrecid
left outer join (
select
x.suid, x.schlevel, x.schsession, x.schyear, w.stuid, w.stuschrecid, z.ch_des, z.en_des, y.divsubjcode,st.SHORTNAME
from tb_stu_stusubj w
join tb_sch_divsubjgrp x
on w.crosssubjgrp=x.subjgrpcode and w.subjcode=x.subjcode
join tb_sch_dsgsubj y
on x.suid=y.suid and x.schlevel=y.schlevel and x.schsession=y.schsession and x.schyear=y.schyear and x.subjgrpcode=y.subjgrpcode and w.crosssubjcode=y.divsubjcode
left outer join TB_SCH_DSGSUBJTCHR y1
on x.suid=y1.suid and x.schlevel=y1.schlevel and x.schsession=y1.schsession and x.schyear=y1.schyear and x.subjgrpcode=y1.subjgrpcode and w.crosssubjcode=y1.divsubjcode and y1.STAFFSEQ=1
left outer join VW_ASR_STAFF st
on w.suid=st.suid and y1.STAFFCODE = st.STAFFCODE
join tb_hse_common z
on w.suid=z.suid and w.subjcode=z.code_id and z.tb_id='SBJ'
where x.grptype='B' and y.subjblkcode='Block_3'
) x3
on a.suid=x3.suid and a.schlvl=x3.schlevel and a.schsess=x3.schsession and a.schyear=x3.schyear and a.stuid=x3.stuid and a.stuschrecid=x3.stuschrecid
where a.schyear=? and a.classlvl=? and (a.status=null or a.status not in (4,5,6))
order by a.classlvl, a.classcode, a.classno, a.chname, a.enname
#19 edb-escm, 2015-08-10
20. ### ps-ncw Expand Collapse
文章:
108
讚:
0
回覆 19# edb-escm
這個成功了!
謝謝! | 6,190 | 15,742 | {"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-2023-06 | latest | en | 0.399095 |
https://www.daniweb.com/programming/threads/515639/deadlock-algorithm-conversion | 1,534,882,164,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218899.88/warc/CC-MAIN-20180821191026-20180821211026-00329.warc.gz | 828,762,696 | 12,313 | I have written a formal Deadlock prevention code but I'm having trouble implementing this algorithm. Any help is appreciated.
I have represented the algorithm below,
Formal algorithm description
Data structures at a process i: (It is the initial value in the parentheses.)
• OUTi: set of integer, the set of process identifiers which process i is waiting for;
• INi: set of integer, the set of process identifiers which are waiting for process i;
• Pi: set of edge(∅), the set of phantom edges;
• weighti: float(0), the weight value which is accumulated locally before report is sent at a process i;
• rci: string, the request condition of process i;
• first recvi: boolean(true), the flag that determines whether process i has received a probe;
• cur initi: integer(−1), the identifier of the current initiator;
• probe recvedi: integer(0), the number of probes that process i has received. Additional data structures at initiator: (It is the initial value in the parentheses.)
• weightinit: float(0), the weight value accumulated from messages, either probes or reports;
• RCinit: set of request conditions(∅), the request conditions collected from the reports.
Message formats:
• PROBE(i,w): This is a message that diffuses the weight to successors where i is the identifier of the initiator and w the weight;
• REPORT(rc, P,w): It is used to send the weight value back to the initiator attached with the request condition. The rc is the request condition of the current process, the P is the set of phantom edges and the w is the accumulated weight.
Algorithm 1
Receiving a PROBE(init,w) from process s at a non-initiator process i
if first recv = true then
for j ∈ OUTi do
send PROBE(init,w/|OUTi + 1|) to j;
end for
cur initi ← init;
weighti ← 0;
probe recvedi ← 0;
end if
weighti ← weightiw/|OUTi + 1|;
probe recvedi ← probe recvedi + 1;
if s /∈ INi then
Pi = Pi S{s → i};
end if
if probe recvedi = |INi| then
send REPORT(rci, Pi,weighti) to cur initi;
end if
Algorithm 2
for {a → b} ∈ P do
get rca from RC marked as rc;
modify every occurrence of b in rc to true;
replace rca in RC with rc;
end for
tmpA ← ∅;
repeat
A ← tmpA;
for tmprci ∈ RC do
if eval(tmprci) = true then
tmpA ← tmpAS{i};
RC ← RC − {tmprci};
end if
end for
until A = tmpA
if RC = ∅ then
else
resolution(A,RC);
end if
Algorithm 3
Receiving a REPORT(rc, P,w) at the initiator process i
weightinit ← weightinit + w;
RCinit ← RCinitSrc
Pi ← PiSP
if weightinit=1 then
end if
2
Contributors
1 | 670 | 2,485 | {"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} | 2.734375 | 3 | CC-MAIN-2018-34 | longest | en | 0.832394 |
https://forum.bebac.at/forum_entry.php?id=23852 | 1,716,262,440,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00177.warc.gz | 231,838,896 | 7,215 | ABEL vs. ABE [Power / Sample Size]
Hi BEQool,
❝ So the sample size with ABEL is always smaller or equal to ABE right (with the same arguments)? Similarly, power for the same number of subjects is always higher (or equal) with ABEL than with ABE (as shown in my reformulated question at the end of the first post or in your example with n=21 for ABE vs. ABEL)?
❝ 1. So in case regulators (EMA region) ask us to calculate post hoc power (regardless of how irrelevant it is) …
Oh dear! EMA region, really‽ Outright bizarre.
❝ … of a study with 2x3x3 design with CVwr=25%, we should calculate it with power.TOST for AUC and for Cmax when we didnt mention anything about widened limits (ABEL) in the protocol (hypothetical scenario)?
Yes because the study was intended to be assessed by ABE.
❝ And on the other hand, for Cmax we should calculate it with power.scABEL when we mentioned widened limits (ABEL) in the protocol?
Correct again.
❝ 2. Another hypothetical scenario: If we get information from the literature about a drug's CVw (Cmax) of around 30% (lets say a range of 25-35%) and if we get a drug's CVw of 22% from our pilot study, can we then do a regular study with design 2x3x3 (in case we get CVw for Cmax 35% so then we could widen the limits and use ABEL)?
Yes, if you state it in the protocol. In general I would trust a – reasonable large! – pilot study more than the literature. The reference may have changed in the meantime, different sampling, bioanalytical method, etc.
BTW, why do you want to use a partial replicate design and not one of the 2-sequence 3-period full replicate designs (TRT|RTR or TRR|RTT)? Acceptable for the EMA (see the Q&A document and this post for examples). Same degrees of freedom and similar sample sizes. More informative because you can also estimate CVwT, which is useful in designing other studies (quite often CVwT < CVwR and you need a smaller sample size). In the partial replicate you have to assume CVwT = CVwR, which is often wrong.
❝ And if then our drug's CVw from this regular study is lets say 21%, can agencies ask us to justify replicate 2x3x3 design as if why didnt we use conventional 2x2x2 design if we got CVw=22% in our pilot study?
By ‘regular study’ are you meaning a simple crossover? Even if you observed CVw ≥30% I would be cautious. That’s only a hint of a highly variable reference. See this article for details.
There is nothing to justify. Any study in a replicate design can also be assessed for ABE. My – former – best enemy once said »From a purely statistical perspective, all studies should be performed in a replicate design.« One of the rare occasions I agree with him.
Dif-tor heh smusma 🖖🏼 Довге життя Україна!
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked. 🚮
Science Quotes | 731 | 2,836 | {"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} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.901542 |
https://gitlab.aei.uni-hannover.de/finesse/pykat/-/commit/37f2715a41420c2ec85b68319eac0e581a739a48 | 1,653,040,906,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531779.10/warc/CC-MAIN-20220520093441-20220520123441-00793.warc.gz | 339,179,736 | 29,559 | Commit 37f2715a by Daniel Brown
Adding EI read function, adding better warning for ROM vs map size differences. Making error printouts more readable
... ... @@ -20,11 +20,34 @@ from pykat.maths.hermite import * from pykat.maths import newton_weights from scipy.integrate import newton_cotes from multiprocessing import Process, Queue, Array, Value, Event from pykat.exceptions import BasePyKatException EmpiricalInterpolant = collections.namedtuple('EmpiricalInterpolant', 'B nodes node_indices limits x worst_error') ReducedBasis = collections.namedtuple('ReducedBasis', 'RB limits x') ROMLimits = collections.namedtuple('ROMLimits', 'zmin zmax w0min w0max R mapSamples max_order') def read_EI(filename, verbose=True): import pickle with open(filename, "rb") as file: ei = pickle.load(file) if verbose: print("Map data this ROM was made for in one dimension:") print(" Map separation dx = " + str(ei.x[1]-ei.x[0])) print(" x range = -{0}m to {0}m".format(max(abs(ei.x)))) print(" Data points = " + str(ei.x.size * 2)) print("") print("Parameter limits:") print(" w0 = {0}m to {1}m".format(ei.limits.w0min, ei.limits.w0max)) print(" z = {0}m to {1}m".format(ei.limits.zmin, ei.limits.zmax)) print(" max order = {0}".format(ei.limits.max_order)) print("") print("ROM contains {0} basis modes".format(ei.nodes.shape[0])) return ei class ROMWeights: def __init__(self, w_ij_Q1, w_ij_Q2, w_ij_Q3, w_ij_Q4, EIx, EIy, nr1=1, nr2=1, direction="reflection_front"): ... ... @@ -566,6 +589,12 @@ def makeWeightsNew(smap, EIxFilename, EIyFilename=None, verbose=True, newtonCote wy[ym == 0] = 0.5 W = np.outer(wx, wy) if A_xy_Q1.shape != W.shape or \ A_xy_Q2.shape != W.shape or \ A_xy_Q3.shape != W.shape or \ A_xy_Q4.shape != W.shape: raise BasePyKatException("Map data points do not overlap exactly with data points this EI was made for. Consider using the intepolate=True option.") # make integration weights Bx = EIx.B By = EIy.B[:,::-1] ... ... | 571 | 1,940 | {"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} | 2.546875 | 3 | CC-MAIN-2022-21 | longest | en | 0.561835 |
https://convertoctopus.com/30-8-months-to-weeks | 1,601,372,146,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401641638.83/warc/CC-MAIN-20200929091913-20200929121913-00162.warc.gz | 336,050,353 | 7,806 | ## Conversion formula
The conversion factor from months to weeks is 4.348125, which means that 1 month is equal to 4.348125 weeks:
1 mo = 4.348125 wk
To convert 30.8 months into weeks we have to multiply 30.8 by the conversion factor in order to get the time amount from months to weeks. We can also form a simple proportion to calculate the result:
1 mo → 4.348125 wk
30.8 mo → T(wk)
Solve the above proportion to obtain the time T in weeks:
T(wk) = 30.8 mo × 4.348125 wk
T(wk) = 133.92225 wk
The final result is:
30.8 mo → 133.92225 wk
We conclude that 30.8 months is equivalent to 133.92225 weeks:
30.8 months = 133.92225 weeks
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 week is equal to 0.0074670191099687 × 30.8 months.
Another way is saying that 30.8 months is equal to 1 ÷ 0.0074670191099687 weeks.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that thirty point eight months is approximately one hundred thirty-three point nine two two weeks:
30.8 mo ≅ 133.922 wk
An alternative is also that one week is approximately zero point zero zero seven times thirty point eight months.
## Conversion table
### months to weeks chart
For quick reference purposes, below is the conversion table you can use to convert from months to weeks
months (mo) weeks (wk)
31.8 months 138.27 weeks
32.8 months 142.619 weeks
33.8 months 146.967 weeks
34.8 months 151.315 weeks
35.8 months 155.663 weeks
36.8 months 160.011 weeks
37.8 months 164.359 weeks
38.8 months 168.707 weeks
39.8 months 173.055 weeks
40.8 months 177.404 weeks | 479 | 1,690 | {"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} | 4.25 | 4 | CC-MAIN-2020-40 | latest | en | 0.861025 |
http://www.jiskha.com/display.cgi?id=1351627359 | 1,496,052,090,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612069.19/warc/CC-MAIN-20170529091944-20170529111944-00306.warc.gz | 667,019,762 | 3,616 | # algebra 1
posted by on .
Solve the following systems of linear equations using any algebraic method.If possible,check your solution.
1)2x+5y=3
X-3y=7
2) 3y=11-2x
3x=y-11
3)8x-6y=14
12x-9y=18
• algebra 1 - ,
#2:
2x + 3y = 11
3x - y = -11
multiply #2 by 3 and add:
2x+3y = 11
9x - 3y = -33
---------------
11x = -22
x = -2
now figure y: 5
Do the others likewise. To check your answers, just plug in the values and make sure the equations still balance. | 179 | 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} | 3.390625 | 3 | CC-MAIN-2017-22 | latest | en | 0.772918 |
https://arduino.stackexchange.com/questions/48594/can-i-power-esp8266-with-more-than-3-5-v/48595 | 1,701,720,367,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100534.18/warc/CC-MAIN-20231204182901-20231204212901-00505.warc.gz | 131,583,377 | 42,858 | # Can I power ESP8266 with more than 3.5 V?
I am concerned whether I need to use a voltage divider to power my ESP8266 module. My concerns stem from the warnings that I have heard in multiple tutorials. Everyone seems to regard the ESP8266 as being very sensitive to supply voltage.
Has anyone use the ESP8266 with an supply voltage greater than 3.5 V? What were the effects? Is it even necessary to make a voltage divider?
I am asking because my application has limited real estate.
• Surely you mean 3.3V :) Jan 9, 2018 at 15:22
• Some of the pins on ESPs are protected against excessive voltage, personally I always err on the side of caution and use regulators and level shifters, but with the real estate restrictions I'd suggest you check the manual on what the absolute maximum is that you version of the board can take. Jan 9, 2018 at 15:24
• Are you talking about providing power to the ESP, or just an input-pin? For power use a voltage regulator. For inputs you could indeed use a voltage divider. Jan 9, 2018 at 16:37
• To be clear I was talking about the power input. Jan 9, 2018 at 16:47
• If your voltage source is high enough, spend less than a buck on a regulator like a LD1117-3.3 Jan 12, 2018 at 2:12
The datasheet usually is definitive on the allowable range of voltages.
Look in Section 5.1: Electrical Characteristics.
## Working Voltage Value:
• Min: 2.5 V
• Typical: 3.3 V
• Max: 3.6 V
If you have to feed in signals that could go higher than 3.3 V in normal operation, you need to either clamp it with Zener diodes, or use a voltage divider.
If you are asking about powering the chip with more than 3.3 V, don't. You can get away with a 3.3 V linear regulator and 2 capacitors so that you are powering it with no more than designed supply voltage.
The result to apply more voltage to a circuit is that the components will be burn. The same if you apply for a long time the maximum voltage. So the board works at 3.3v, your source maybe fluct between that but you will burn it if you apply 3.6 or more as a regular voltage.
Usually ESP8266 modules have an onboard voltage regulator, USB connector and a pin labeled 5V. This pin or the USB cable certainly tolerate resp. need a 5V power supply.
The ESP8266 controller itself will not survive 5V, neither at the Vcc nor at the signal pins.
BTW: a voltage divider (two resistors) is ok for signals, but does not help with the power supply.
I have 4 pieces of ESP01-S with DHT11 that connect to the WiFi every 30 minutes and send temperature data and then go into DeepSleep. In addition, an ESP01-S as a motion detector and one as a push button. These two usually in DeepSleep. All are directly connected to a Li-Ion battery as power supply. This has been running for over a year (except for changing the battery) and so far I haven't had any problems
• Welcome to Arduino:SE. I don't think this actually answers the question. Aug 31 at 19:27
• This is anecdotal information i.e. it might work for you but it (greatly) exceeds the manufacturer's specifications. Connecting directly to a Lithium Ion cell means the ESP8266 is exposed to voltages of over 4 V when the battery is fully charged. Sep 1 at 9:35 | 806 | 3,191 | {"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-2023-50 | latest | en | 0.929966 |
https://ip-numaram.com/do-math-852 | 1,675,627,911,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00867.warc.gz | 332,327,480 | 4,890 | # Calculus math calculator
We'll provide some tips to help you select the best Calculus math calculator for your needs. We can solving math problem.
## The Best Calculus math calculator
Calculus math calculator can help students to understand the material and improve their grades. This is what Dad wants to explain to you. In the algebra stage, I think the most important thing is to develop your mathematical modeling ability. Solving algebraic equations is just a step-by-step and practice makes perfect operation. However, how to transform practical problems into algebraic solvable abstract problems is a more challenging and practical task.
Further understand the role of statistics in life and problem solving, and develop statistical concepts. In the aspect of mathematical problem-solving, I will experience the diversity of problem-solving strategies and the effectiveness of using hypothetical methods to solve problems, the superiority of using algebraic methods to solve problems, the charm of mathematics, and the development of students' ability to solve problems..
Xiyang memorized math problems and several English words commonly used in answering them, but he didn't expect to use it on the first day of school. Take me as an example. I graduated from college in 2011. I haven't contacted English for 10 years since I worked. Since my undergraduate school is a 211 university of science and engineering, mathematics is basically not too much obstacle for me.
Examinees often cannot abstract the shape and characteristics of the cross-section, which brings great difficulties in finding the perimeter, area and maximum value of the cross-section. This lesson discusses effective solutions to such problems. Please refer to them for reference. In the whole known condition, the side length of a large square can be found to be 20 through the area of a large square, but the perimeter of four rectangles is unknown. However, since the side lengths of different rectangles are all within the side length of a large square, and the product of the side lengths is equal to the area of the rectangle.
## Math solver you can trust
This app helped me a lot in my homework and a lot of other math problems. I love it. It is a great app for students. It is a total lifesaver for students. Such a great it is very helpful especially for students I used this app until I finished my high school years
Zaira Morgan
the app is so amazing. Every time I have a problem with a math question the app helps me it's very cool download this app it's very helpful I have this because apps can solve every problem. Still a great one if you're struggling in math. I 100% recommend.
Xavia Simmons | 534 | 2,699 | {"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} | 4.0625 | 4 | CC-MAIN-2023-06 | latest | en | 0.958869 |
http://www.lotterypost.com/thread/246799/21 | 1,386,911,239,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164884560/warc/CC-MAIN-20131204134804-00017-ip-10-33-133-15.ec2.internal.warc.gz | 420,703,181 | 22,938 | Welcome Guest
### NetConnect
#### Internet Domains, simple and cheap
##### Find a domain name:
Home
The time is now 12:07 am
You last visited December 13, 2013, 12:07 am
All times shown are
Eastern Time (GMT-5:00)
# New York Dream Team 6/1/12 - 6/30/12
Topic closed. 2025 replies. Last post 1 year ago by tmoneyyy.
Page 21 of 136
Turning Stone Casino,Verona,NY
United States
Member #95130
August 4, 2010
5373 Posts
Online
Posted: June 4, 2012, 2:55 pm - IP Logged
x x x x x x x x x x x (x) x x x (x) (xx) x x x x x x x x x xxx x x x x x x x x x x x x x x x x x x x x x x x x xx x x x x 1 2 3 4 5 6 7 8 9 0
Those in parenthesis were in black or red and connected. Basically they were the pairs for today's mid win. There is still one more number up there that can go with a 0 or 1. And that number can go with another number horizontal that it has hit before.
????????? / is it me I don't see any numbers in your chart ?Just some x's.But I think if I stick with 0-9 I'll find the right number.lol
NY
United States
Member #127010
April 21, 2012
2550 Posts
Offline
Posted: June 4, 2012, 3:16 pm - IP Logged
????????? / is it me I don't see any numbers in your chart ?Just some x's.But I think if I stick with 0-9 I'll find the right number.lol
Numbers are at the bottom. All the Xs hits from past draws. Many of you know that either 2 or 3 numbers that hit before with a number will hit again. So basically the yellow are hits that have hit with another number in the line horizontally. The black and red Xs are the ones that still need to hit in a horizontal line. For example if you pick number 7 you can see there is 2 black xs and 1 red. Those are basically numbers you might want to pair up with another number in that line. Why? Because it has not hit with a number in that line and its due. Sometimes you are given pairs here. So if you were given 70x which is good. You basically go down to your third number. If it does not connect with a 0 or 7 thats probably not a good number because they are suppose to connect. This works most of the time.
NY
United States
Member #127010
April 21, 2012
2550 Posts
Offline
Posted: June 4, 2012, 3:27 pm - IP Logged
6/4/12 PAIRS FOR TONIGHT 97 98 78 42 43 23 19 10 90 64 65 45 31 32 12 86 87 67 53 54 34 08 09 89 75 76 56 20 21 01 GOOD LUCK FOR EVE
New York
United States
Member #43215
July 15, 2006
1564 Posts
Offline
Posted: June 4, 2012, 3:33 pm - IP Logged
Congrats to all winners!
Victory is the child of preparation and determination........Take a moment to be thankful
New York
United States
Member #43215
July 15, 2006
1564 Posts
Offline
Posted: June 4, 2012, 3:35 pm - IP Logged
My picks for this evening: 968 779 700 778 338 424 333 888
Good luck all
Victory is the child of preparation and determination........Take a moment to be thankful
New York NY
United States
Member #100768
November 18, 2010
4812 Posts
Offline
Posted: June 4, 2012, 4:25 pm - IP Logged
Congrats to all Midday Winners!!!!!!!!!!
New York NY
United States
Member #100768
November 18, 2010
4812 Posts
Offline
Posted: June 4, 2012, 4:28 pm - IP Logged
6/4/12 PAIRS FOR TONIGHT 97 98 78 42 43 23 19 10 90 64 65 45 31 32 12 86 87 67 53 54 34 08 09 89 75 76 56 20 21 01 GOOD LUCK FOR EVE
im going to work with u tonight......... going to try and narrow down some of these pairs for tonights draw
Brooklyn, NY
United States
Member #113977
July 20, 2011
8078 Posts
Offline
Posted: June 4, 2012, 5:10 pm - IP Logged
Hey Drew! you there?
What are you thinking for tonight?
my pets: 417 - 169 - 623 - 807 - 634 - 626 - 127 - 446
New York NY
United States
Member #100768
November 18, 2010
4812 Posts
Offline
Posted: June 4, 2012, 5:18 pm - IP Logged
Hey Drew! you there?
What are you thinking for tonight?
im actually trying to narrow some stuff down but aswe speak '
im really liking 28x 81x x08 07x 42x 25x ---still working on eve though should be done ina few......
whats on ur map Sue....
Brooklyn, NY
United States
Member #113977
July 20, 2011
8078 Posts
Offline
Posted: June 4, 2012, 5:18 pm - IP Logged
Quickpicks from June 1st till ......
Hottest digits for the week 8-2-5
06/01 Midday Evening
Suri:135-3563-089 Ness:932-838-146
Sue: 371 - 020 Long: 057-132- 5708-1412
Ness: 320-948-2052 Suri: 911-235-4407
Long: 082-416- 3370-0156 Loon: 090-060-789-4732
-----------------------------------------------------------------------------------------------------------
06/2 Midday
Sue: 103 - 873 Long: 996-944- 1686-0010
Ness: 863-481-544-123 Ness: 783-380-0196-4136
Suri: 033-049-4066 Suri: 806-376-2467
Loon: 035-634-481-735-287-7258-8481
Long: 860-628- 6795-1529
--------------------------------------------------------------------------------------------------------------
06/03 Midday Evening
Suri: 673-7836 Sue: 023 - 457 - 9658
Long: 066-335- 1502-4848
Sue: 964 - 910 - 103 - 7350
Loon: 849-646-359-3287
------------------------------------------------------------------------------------------------------------------
06/04 Midday Evening
Suri:170-536-796-906-1683-2183 Sue: 352 - 781 - 0428 - 2206
Sue: 194 - 553
Long: 080-196- 0549-4419
my pets: 417 - 169 - 623 - 807 - 634 - 626 - 127 - 446
Brooklyn, NY
United States
Member #113977
July 20, 2011
8078 Posts
Offline
Posted: June 4, 2012, 5:20 pm - IP Logged
im actually trying to narrow some stuff down but aswe speak '
im really liking 28x 81x x08 07x 42x 25x ---still working on eve though should be done ina few......
whats on ur map Sue....
I like these:
81x - 32x - 07x - 02x - 25x
but I am still working also. I'm thinking that since we had all high digits, we should get low digits tonight......
my pets: 417 - 169 - 623 - 807 - 634 - 626 - 127 - 446
Orange County, New York
United States
Member #140
March 10, 2002
5862 Posts
Offline
Posted: June 4, 2012, 5:21 pm - IP Logged
Congrats to all midday winners! Looks like someone on Lottery Post went ballistic...they should have came to the Dream Team if they were looking for a hit!
The Panther hunts for more!
New York NY
United States
Member #100768
November 18, 2010
4812 Posts
Offline
Posted: June 4, 2012, 5:23 pm - IP Logged
I like these:
81x - 32x - 07x - 02x - 25x
but I am still working also. I'm thinking that since we had all high digits, we should get low digits tonight......
that 32x u have..... is good to see.. because 283 is a number, i think will hold & play for a few draws-- and matches my 28x
Brooklyn, NY
United States
Member #113977
July 20, 2011
8078 Posts
Offline
Posted: June 4, 2012, 5:23 pm - IP Logged
Congrats to all midday winners! Looks like someone on Lottery Post went ballistic...they should have came to the Dream Team if they were looking for a hit!
LOL! I saw that. they have it posted so many times. But this is not the first time I have seen a post like this. About a week or so ago another one popped up and Todd quickly cleared it away.
my pets: 417 - 169 - 623 - 807 - 634 - 626 - 127 - 446
Orange County, New York
United States
Member #140
March 10, 2002
5862 Posts
Offline
Posted: June 4, 2012, 5:24 pm - IP Logged
I was getting all happy til the last digit of the p4 popped......... LOL!!!!! | 2,479 | 7,804 | {"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-2013-48 | latest | en | 0.916319 |
https://www.beliefnet.com/columnists/beyondgorgeous/2013/01/a-fitness-plan-you-can-do-forever-2.html | 1,660,447,691,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00770.warc.gz | 595,690,486 | 17,110 | A friend saw that her fitness center membership was about to expire, so she rushed in to renew — even though she hadn’t been inside the place for months.
“I bet you have a name for people like me who pay for memberships but never show up to work out,” she said jokingly.
“Yes,” the owner replied. “We call them ‘profits.’”
I’ve been there — and you probably have, too. “I MUST get in shape,” we think. We join a gym and the first day work out enthusiastically on all the equipment. Or we buy a killer dvd that guarantees to make us svelte in 30 days if we undergo their brand of exercise torture. Or we buy an expensive machine and give it a thorough workout as soon as it arrives.
Then we cannot move for three days. Our muscles scream with pain and we feel like joining the chorus. We had the best intentions, but already are on the road to giving up.
So let’s not go there anymore. Let’s do something easy. It’s inexpensive, painless, won’t take up space in your bedroom, and you DO have time to work it into your schedule. You just need three things:
A pair of shoes.
A pair of feet.
A pedometer.
Here’s how it works:
Get a good pedometer and wear it throughout the day from the time you get up until you go to bed at night. Write down the number of steps you took that day. Do that for three days in a row, then take the average of the number. You may be surprised.I thought I was ‘active’ but found out instead that I was practically comatose!
Suppose you have an average of 3,200 steps. This week, make it your goal to add 200 steps. Every day you will make sure you walk at least 3,400 steps. Next week you will add 200 more, so you take at least 3,600 steps — and so on. Eventually you will work up to 10,000 steps or more.
Is it hard to add 200 steps? Not at all! Just take a walk through your house or out in your yard.
Will it do any good? Yes. It will get you out of your chair and moving. It will be painless and doable, so this is one plan you can do forever. We’ll talk more about benefits and strategies for adding steps in later posts.
So in our easy, Beyond Gorgeous, fitness plan we will:
You can do that — for the rest of your life!
Eating to live and living for Christ,
Susan Jordan Brown
More from Beliefnet and our partners
previous posts | 571 | 2,303 | {"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} | 2.640625 | 3 | CC-MAIN-2022-33 | latest | en | 0.956422 |
http://www.leda-tutorial.org/en/unofficial/ch06s03.html | 1,547,651,226,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657510.42/warc/CC-MAIN-20190116134421-20190116160421-00191.warc.gz | 327,326,441 | 6,637 | 6.3. Floating point numbers of arbitrary size and precision (class `bigfloat`)
Warning This chapter, section, or subsection is “in statu nascendi”. Therefore, it may be incomplete, or may appear in a different place in the final version, or may vanish completely.
Learning objectives
Floating point numbers The dangers of C++'s type `double` The class `bigfloat`
C++ provides the types `float` and `double` for representing floating point numbers. In addition, LEDA offers the type `bigfloat`.
Roughly speaking, a floating point number represents a real number with (a finite number of) digits after the decimal point, such as 17.5, 3.1415, and 2.181. It is the number type primarily used for representing non-integer values in modern computer architectures. We will shortly see the exact structure of these types and what it means that the decimal point “flows”.
Let us start our exploration of floating point types with an example in which numbers with many digits after the decimal points are used: We want to compute √2 with an accurary of n digits after the decimal point, where n is an input parameter. We use the following strategy to compute this root up to n decimal digits:
(TODO Give here a short overview over IEEE 754 representation to understand why we use the base 2 in the following.) [17]
Basically, we compute the positive root of the polynomial x2 - q (with q = 2) with the well-known and easy to implement Bisection method. This methods works as follows: It starts with an interval [a,b] that is known to “bracket” a root, that is, f(a) and f(b) have opposite signs (so that, provided that the function is continuous, there must be a root of f inside [a,b] according to the intermediate value theorem). In our implementation, we take [1,2] as the starting interval. In each iteration, the function value of the midpoint of the current interval is computed. Then the interval is halved by making the midpoint the new left or right endpoint, so that the new interval still brackets the root.
When do we know that we have reached √2 up to n decimal digits? We use the following estimation to test whether we have reached the desired accuracy: Our goal is to have
|root - √2| < 10-n
Therefore, we try to make
|root - √2| < 2-binprec =:eps
where we choose binprec large enough such that eps < 10-n. To choose the necessary binary precision, note that 2-binprec < 2-n is satisfied if binprec is larger than n·log210. Choosing it as an integer larger than 3.33·n suffices.
A purely analytical estimation gives (note the invariant 1 < root < 2 in the Bisection method):
|root - √2| = |root2 - 2| / |root + √2| < |root2 - 2| / 2
So our goal becomes to make
|root2 - 2| < 2 eps
This root will then be accurate up to n decimal digits after the decimal point.
Here is a straight forward implementation using doubles. In our implementation, the interval endpoints are implicitly given as the current root candidate and the current interval width dx. In each iteration, dx is halved and the root candidate is shifted by the new dx if this is necessary to keep the root bracketed in the new interval. All these shifts only go into one direction, which is defined before the first iteration (depending on whether f(a) is less than or greater 0).
Filename: ComputingRootsWithDoubles.C
LEDA users as of version 5.0: Note that header files are now partitioned into modules (subfolders)!
```#include <iostream>
#include <iomanip>
using std::cout;
using std::cin;
using std::endl;
// Bisection method for computing sqrt(q) known to be in [a,b]:
// Find a root of f(x) = x^2 - q with <decprec> correct decimal
// digits after the decimal point
int main()
{
double q = 2.0; // we want to compute sqrt(q)
cout << "Computing sqrt(" << q << ").\n";
int binprec, decprec;
cout << "Enter desired decimal accuracy (#digits after the decimal point): ";
cin >> decprec;
binprec = (int) (3.33 * (double) decprec + 1);
cout << "Reaching a binary accuracy of " << binprec
<< " digits (after the decimal point) will suffice." << endl;
double eps = pow( 2, -binprec ); // accuracy to reach
cout << "Computing root such that |root^2 - " << q
<< "| (=:distance) < eps " << endl;
cout << " with eps:=" << eps << "." << endl;
double a = 1.0, b = 2.0; // interval boundaries [a,b]
double root; // root candidate
// we orient the search such that initially f(root + dx) > 0
// and then we move root only into direction sgn(dx)
double f_a = a * a - q;
double dx; // width of current interval
if( f_a < 0 ) { dx = b - a; root = a; } // walk from left to right
else { dx = a - b; root = b; } // walk from right to left
double distance = fabs( root * root - q );
double xmid, fmid;
while( distance >= 2 * eps ) {
dx = dx / 2;
xmid = root + dx;
fmid = xmid * xmid - q;
if(fmid <= 0) {
root = xmid;
}
distance = fabs( root * root - q );
}
// here we have: |root - sqrt(q)| = |root^2 - q| / |root + sqrt(q)|
// <= distance / 2 < eps
// i.e., |root - sqrt(q)| < 2^-binprec <= 10^-decprec
cout << "Final distance = " << std::setprecision( 7 ) << distance << endl;
cout << "Final root = " << std::setprecision( 1 + decprec ) << root << endl;
}
```
Let us try 10 decimal digits:
```Computing sqrt(2).
Enter desired decimal accuracy (#digits after the decimal point): `10`
Reaching a binary accuracy of 34 digits (after the decimal point) will suffice.
Computing root such that |root^2 - 2| (=:distance) < eps
with eps:=5.82077e-11.
Final distance = 5.077338e-11
Final root = 1.4142135624```
Nice! Seems to work. So let us give it a harder job, n = 100:
```Computing sqrt(2).
Enter desired decimal accuracy (#digits after the decimal point): `100`
Reaching a binary accuracy of 334 digits (after the decimal point) will suffice.
Computing root such that |root^2 - 2| (=:distance) < eps
with eps:=2.85747e-101.```
Oops! What's that? The program does not return! An infinite loop?
Maybe we are stuck, that is, dx becomes 0 and we do not move on any more, never fulfilling the stop criterion. To have more insight into the loop, let us add some debugging output and a guard test:
``` while( distance >= 2 * eps ) {
cout << "root=" << std::setprecision( 1 + decprec ) << root << " ";
cout << "dx=" << std::setprecision( 7 ) << dx << endl;
dx = dx / 2;
xmid = root + dx;
fmid = xmid * xmid - q;
if(fmid <= 0)
root = xmid;
if( dx == 0.0 ) {
cout << "WARNING! dx == 0! I am stuck! Stopping iteration.\n";
break;
}
distance = fabs( root * root - q );
}
```
Here we go again on our own. (The output is truncated; only the output of the first and the last iterations are shown.)
```Computing sqrt(2).
Enter desired decimal accuracy (#digits after the decimal point): `100`
Reaching a binary accuracy of 334 digits (after the decimal point) will suffice.
Computing root such that |root^2 - 2| (=:distance) < eps
with eps:=2.85747e-101.
root=1 dx=1
root=1 dx=0.5
root=1.25 dx=0.25
root=1.375 dx=0.125
root=1.375 dx=0.0625
root=1.40625 dx=0.03125
root=1.40625 dx=0.015625
...
root=1.41421356237309492343001693370752036571502685546875 dx=6.32404e-322
root=1.41421356237309492343001693370752036571502685546875 dx=3.16202e-322
root=1.41421356237309492343001693370752036571502685546875 dx=1.58101e-322
root=1.41421356237309492343001693370752036571502685546875 dx=7.90505e-323
root=1.41421356237309492343001693370752036571502685546875 dx=3.952525e-323
root=1.41421356237309492343001693370752036571502685546875 dx=1.976263e-323
root=1.41421356237309492343001693370752036571502685546875 dx=9.881313e-324
root=1.41421356237309492343001693370752036571502685546875 dx=4.940656e-324
WARNING! dx == 0! I am stuck! Stopping iteration.
Final distance = 3.546425e-16
Final root = 1.41421356237309492343001693370752036571502685546875```
Aha! It was indeed an infinite loop!
Doubles and their limitations
Now there is a lot to say here:
Explain the structure of doubles (sign, significand, exponent). Explain arithmetics on doubles. Explain why we could NEVER reach an accuracy of 100 decimal digits with doubles, no matter what strategy we use (because pmax=53). Explain why the above dx becomes 0. Explain the last values of dx. (Why is there an exponent of -324 before it goes to 0?) Explain the subtlety involved in the loop test above (where do we know from that root2 - q is computed and compared with eps correctly? Wo do not know without an numerical analysis, making our job even harder!)
LEDA's bigfloat
Say that bigfloats have basically the same structure and arithmetcs as doubles, but with arbitrarily large significands and exponents.
Define the rounding modes.
Then give the program again, but with bigfloats. Note that the program is almost the same; the reader shall see: bigfloats can be used like doubles (this holds for most operations).
Concerning the program, especially explain the fiddling and switching between rounding modes EXACT and TO_NEAREST, and why we set the precision to binprec plus one. Explain that it is crucial that the loop test is evaluated exactly (because we do not want to make forward analysis).
Filename: ComputingRootsWithBigfloats.C
LEDA users as of version 5.0: Note that header files are now partitioned into modules (subfolders)!
```#include <LEDA/bigfloat.h>
using leda::bigfloat;
using std::cout;
using std::cin;
using std::endl;
// Bisection method for computing sqrt(q) known to be in [a,b]:
// Find a root of f(x) = x^2 - q with <decprec> correct decimal
// digits after the decimal point
int main()
{
bigfloat q = 2.0; // we want to compute sqrt(q)
cout << "Computing sqrt(" << q << ").\n";
int binprec, decprec;
cout << "Enter desired decimal accuracy (#digits after the decimal point): ";
cin >> decprec;
binprec = (int) (3.33 * (double) decprec + 1);
cout << "Reaching a binary accuracy of " << binprec
<< " digits (after the decimal point) will suffice." << endl;
bigfloat::set_precision( binprec + 1 );
bigfloat eps = leda::ipow2( -binprec ); // accuracy to reach
cout << "Computing root such that |root^2 - " << q
<< "| (=:distance) < eps " << endl;
cout << "with eps:=" << eps << "." << endl;
bigfloat a = 1.0, b = 2.0; // interval boundaries [a,b]
bigfloat root; // root candidate
// we orient the search such that initially f(root + dx) > 0
// and then we move root only into direction sgn(dx)
bigfloat f_a = a * a - q;
bigfloat dx; // width of current interval
if( f_a < 0 ) { dx = b - a; root = a; } // walk from left to right
else { dx = a - b; root = b; } // walk from right to left
bigfloat::set_rounding_mode( leda::EXACT );
bigfloat distance = leda::abs( root * root - q );
bigfloat two_eps = 2 * eps;
bigfloat::set_rounding_mode( leda::TO_NEAREST );
bigfloat xmid, fmid;
while( distance >= two_eps ) {
//cout << "root=" << root << " ";
//cout << "dx=" << dx << endl;
dx = dx / 2;
xmid = root + dx;
fmid = xmid * xmid - q;
if(fmid <= 0)
root = xmid;
if( dx == 0.0 ) {
cout << "WARNING! dx == 0! I am stuck! Stopping iteration.\n";
break;
}
bigfloat::set_rounding_mode( leda::EXACT );
distance = leda::abs( root * root - q );
//cout << "distance=" << distance << endl;
bigfloat::set_rounding_mode( leda::TO_NEAREST );
}
// here we have: |root - sqrt(q)| = |root^2 - q| / |root + sqrt(q)|
// <= distance / 2 < eps
// i.e., |root - sqrt(q)| < 2^-binprec <= 10^-decprec
bigfloat::set_output_precision( 7 );
cout << "Final distance=" << distance << endl;
bigfloat::set_output_precision( 1 + decprec );
cout << "root=" << root << endl;
}
```
Sample run:
```Computing sqrt(2).
Enter desired decimal accuracy (#digits behind the comma): 100
Reaching a binary accuracy of 334 digits (behind the comma) will suffice.
Computing root such that |root^2 - 2| (=:distance) < eps
with eps:=2.857468478205687e-101.
root=1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157278
distance=2.467328e-103
```
Works!
Now explain the interface of bigfloat in greater detail.
Give the essence of bigfloat (and double) arithmetic: If z is the exact real result of an arithmetic operation, and z' is the computed value, then
|z - z'| <= 2-prec |z'|
where prec is the binary length of the signifinand in use.
bigfloats, doubles, and their dangers
List common dangers, mistakes, and fallacies in using bigfloats and doubles (extinction, error propagation,...), leading to nonsense results.
Say that all warnings that we have given for doubles also hold for bigfloats.
Exercises
Exercise 81. Use bigfloats to implement Newton's iteration method for root finding. | 3,556 | 12,589 | {"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} | 4.125 | 4 | CC-MAIN-2019-04 | latest | en | 0.89141 |
https://programs.programmingoneonone.com/2021/02/hackerrank-overload-operators-solution-c-plus-plus.html | 1,716,143,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00324.warc.gz | 422,579,604 | 32,870 | HackerRank Overload Operators solution in c++ programming
In this HackerRank Overload Operators problem in the c++ programming language You need to overload operators + and << for the Complex class.
HackerRank Overload Operators problem solution in c++ programming.
#include<iostream>
using namespace std;
class Complex
{
public:
int a,b;
void input(string s)
{
int v1=0;
int i=0;
while(s[i]!='+')
{
v1=v1*10+s[i]-'0';
i++;
}
while(s[i]==' ' || s[i]=='+'||s[i]=='i')
{
i++;
}
int v2=0;
while(i<s.length())
{
v2=v2*10+s[i]-'0';
i++;
}
a=v1;
b=v2;
}
};
//Overload operators + and << for the class complex
//+ should add two complex numbers as (a+ib) + (c+id) = (a+c) + i(b+d)
//<< should print a complex number in the format "a+ib"
ostream& operator<<(ostream& os, const Complex& c) {
return os << c.a << (c.b > 0 ? '+' : '-') << 'i' << c.b;
}
Complex operator+(const Complex& a, const Complex& b) {
return { a.a + b.a, a.b + b.b };
}
int main()
{
Complex x,y;
string s1,s2;
cin>>s1;
cin>>s2;
x.input(s1);
y.input(s2);
Complex z=x+y;
cout<<z<<endl;
}
Second solution
#include<iostream>
using namespace std;
class Complex
{
public:
int a,b;
void input(string s)
{
int v1=0;
int i=0;
while(s[i]!='+')
{
v1=v1*10+s[i]-'0';
i++;
}
while(s[i]==' ' || s[i]=='+'||s[i]=='i')
{
i++;
}
int v2=0;
while(i<s.length())
{
v2=v2*10+s[i]-'0';
i++;
}
a=v1;
b=v2;
}
};
//<< should print a complex number in the format "a+ib"
//Overload operators + and << for the class complex
//+ should add two complex numbers as (a+ib) + (c+id) = (a+c) + i(b+d)
Complex operator +(const Complex &x, const Complex &y) {
Complex z;
z.a = x.a + y.a; z.b = x.b + y.b;
return z;
}
//<< should print a complex number in the format "a+ib"
std::ostream &operator <<(std::ostream& os, const Complex &z) {
os << z.a << "+i" << z.b;
return os;
}
int main()
{
Complex x,y;
string s1,s2;
cin>>s1;
cin>>s2;
x.input(s1);
y.input(s2);
Complex z=x+y;
cout<<z<<endl;
} | 639 | 1,932 | {"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-2024-22 | latest | en | 0.402833 |
http://younesse.net/Foundations-proof-systems/Lecture4/ | 1,548,000,926,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583722261.60/warc/CC-MAIN-20190120143527-20190120165527-00344.warc.gz | 414,901,622 | 10,266 | # Lecture 4: Simple Type Theory
Teacher: Gilles Dowek
Reminder: relation
∃R \, ∀x, z. (x, z ∈_2 R ⟺ ∃ y \, (x = z × y))
∃R; \, ε(\underbrace{R}_{\rlap{\text{propositional content of the prop. on the right}}}) ⟺ ∀x,y,z,n. \, (⋯)
Recall that you don’t sets if you have functions: you can use characteristic functions:
Example:
f: \begin{cases} 4 &⟼ 4 \text{ is even} \\ 7 &⟼ 7 \text{ is even} \end{cases}
then:
⊢ ε(f \, 4) \qquad (4 ∈ f)\\ ⊢ ¬ ε(f \, 7)
# Simple Type Theory
• you have an infinite number of sorts:
• $ι$: base objects
• $o$: propositional contents (what you get by applying a predicate to a term)
• $A → B$: functions from A to B
• for each pair $A,B$, application: $α_{A,B}$ of arity $⟨A → B,A,B⟩$
• a predicate symbol (the only one) $ε$ of arity $⟨o⟩$:
• for each $t$, $x_1,…,x_n ⟼ t$
• for each $A$, ${x_1,…,x_n \; \mid \; A}$
But sufficient: combinators:
x^A ⟼ x: A → A\\ K_{A, B} ≝ x^A, y^B ⟼ x: A → (B → A)\\ S_{A,B,C} ≝ x,y,z ⟼ (x \, z) (y \, z): (A → B → C) → (A → B) → A → C
and then we will have axioms expressing
α(x ↦ x, a) = a\\ α(α(x,y ↦ x, a),b) = a\\
Similarly:
\lbrace x \; \mid \; ∃ y \, ( x = 2 × y)\rbrace\\ \dot{∧} = \lbrace x,y \; \mid \; ε(x) ∧ ε(y)\rbrace
and then:
ε(αα(\dot{∧},a),b) ⟺ ε(x) ∧ ε(y)
• proposition: the sky is blue
• propositional content: that the sky is blue
• $ε$: it_is_true(that the sky is blue)
∀ P, \, (P ⇒ P)\\ ∀ p \, (ε(p) ⇒ ε(p))
• season(spring)
• color (4 of them)
• you can’t build the proposition: as_many(color, season) because you can’t apply predicates to predicates
To avoid binding symbols: you replace them by an operator without bound variables applied to a function: higher-order abstract syntax
\dot{∀}_A = \lbrace f \; \mid \; ∀x^A\, ε(\underbrace{f \, x}_{A → o})\rbrace: (A → o) → o
ε(\dot{∀}_A \, f) ⟶ ∀x\, ε(f \, x)
Term $t$, there exists $u$ st
(u \, y) ⟶^\ast t
• $t=y$: by setting $u$ to be identity: $I ≝ S \, K \, K$
• $t=x$, $x≠y$ OR $t = c$
u y ⟶^\ast t
by setting $u ≝ K \, x$ or $u ≝ K \, c$
• $t= α(t_1 t_2) = (t_1 \, t_2)$:
By IH: $u_1 \, y ⟶^\ast t_1$, $u_2 \, y ⟶^\ast t_2$:
\underbrace{S \, u_1 \, u_2}_{≝ \, u} y ⟶ (u_1 \, y) (u_2 \, y) ⟶^\ast (t_1 \, t_2) = t
Proposition $A$: there exists $u$ such that
ε(u) ⟶^\ast A
By induction on $A$. For example: $A ≝ A_1 ∧ A_2$. By IH: $ε(a_1) ⟶ A_1$, $ε(a_2) ⟶ A_2$:
ε(\dot{∧} \, a_1 \, a_2) ⟶ ε(a_1) ∧ ε(a_2)
If $A = ε(t)$, it’s even simpler.
What’s the point? Comprehension: ex:
A ≝ ∃y \, (x = 2 × y)\\ ε(a) ⟶^\ast A\\ b \, x ⟶^\ast a\\ ε(b \, x) ⟶^\ast ε (a) ⟶^\ast A
so (comprehension):
∃b ; \; (ε(b \, x) ⟺ A)
Impredicative quantification:
∀p \, (ε(p) ⇒ ε(p))
At first: was avoided by Russel (to avoid paradoxes like Russel’s), but his student Ramsy reintroduced it later. Martin-Löf Type Theory: predicative, on the other hand (people in Sweden at Chalmers are very reluctant to use impredicativity).
## Models
All reduction rules of Simple type theory are valid in a $\lbrace 0, 1\rbrace$-valued model:
\underbrace{⟦ε(\dot{⊤})⟧}_{= \hat{ε} \hat{\dot{⊤}} = id(\tilde{⊤}) = 1} = \underbrace{⟦⊤⟧}_{\tilde{⊤} = 1}
## Mathematical theories
x=y ⟶ \dot{∀}_{ι → o} (\overline{λ} c((c \, x) \dot{⇒} (c \, y)))
So:
ε(x=y) ⟶^\ast ∀c \, (ε (c \, x) ⇒ ε (c \, y))
NB: the class $c$ is of type $ι → o$
Natural numbers:
• Peano: axioms
• Cantor: 3 = the set of all sets of cardinal 3
• modern logic (due to Frege) was invented to build Cantor numbers
• Church: Church numerals
• Von Neumann: 3 = the set $\lbrace ∅, \lbrace ∅ \rbrace, \lbrace ∅, \lbrace ∅ \rbrace\rbrace\rbrace$
Church (40’s) ⟶ his student: Henkin (50’s)
\begin{align*} ⟦∀y \, ε(α_{A, o} \, x \, y)⟧_φ &= \widetilde{∀} \lbrace ⟦ε(α_{A, o} \, x \, y)⟧_{φ, y ↦ a} \; \mid \; a ∈ ℳ_a \rbrace\\ &= \widetilde{∀} \lbrace \underbrace{\widetilde ε (⟦α_{A, o} \, x \, y⟧_{φ, y ↦ a})}_{= ⟦α_{A, o} \, x \, y⟧_{φ, y ↦ a}} \; \mid \; a ∈ ℳ_a \rbrace\\ & = \widetilde{∀} \lbrace ⟦α_{A, o} \, x⟧_{φ, y ↦ a}(\underbrace{⟦y⟧_{φ, y ↦ a}}_{= a}) \; \mid \; a ∈ ℳ_a \rbrace \\ & = \widetilde{∀} \lbrace ⟦α_{A, o}⟧_{φ, y ↦ a} (\underbrace{⟦x⟧_{φ, y ↦ a}}_{= φ(x)})(a) \; \mid \; a ∈ ℳ_a \rbrace \\ & = \widetilde{∀} \lbrace φ(x)(a) \; \mid \; a ∈ ℳ_a \rbrace \\ & = \widetilde{ε}\Big(\widetilde{∀} \lbrace φ(x)(a) \; \mid \; a ∈ ℳ_a \rbrace\Big)\\ & = ⟦ε(α_{A→o, o} \, \dot{∀}_A \, x)⟧_φ \end{align*} | 1,913 | 4,334 | {"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} | 3.9375 | 4 | CC-MAIN-2019-04 | latest | en | 0.626998 |
https://puzzlefry.com/puzzles/fathers-name/ | 1,716,604,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00787.warc.gz | 396,038,575 | 33,080 | # Guess the Father’s name
1,458.6K Views
Monday Tuesday and Wednesday are three sisters.
What’s their father’s name?
satyen Guru Asked on 30th July 2015 in
Pretty Easy Question,
Father name must be Mr Day.
ronret45 Expert Answered on 30th July 2015.
• ## More puzzles to try-
• ### What is the logic behind these ?
3 + 3 = 3 5 + 4 = 4 1 + 0 = 3 2 + 3 = 4 ...Read More »
• ### Defective stack of coins puzzle
There are 10 stacks of 10 coins each. Each coin weights 10 gms. However, one stack of coins is defective ...Read More »
• ### Which clock works best?
Which clock works best? The one that loses a minute a day or the one that doesn’t work at all?Read More »
• ### (Advanced) Cheryl’s Birthday Puzzle
Paul, Sam and Dean are assigned the task of figuring out two numbers. They get the following information: Both numbers ...Read More »
• ### Five greedy pirates and gold coin distribution Puzzle
Five puzzleFry ship’s pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all ...Read More »
• ### Magical flowers!!
A devotee goes to three temples, temple1, temple2 and temple3 one after the other. In front of each temple, there ...Read More »
• ### Tuesday, Thursday what are other two days staring with T?
Four days are there which start with the letter ‘T‘. I can remember only two of them as “Tuesday , Thursday”. ...Read More »
• ### How could only 3 apples left
Two fathers took their sons to a fruit stall. Each man and son bought an apple, But when they returned ...Read More »
• ### How Many Eggs ?
A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over ...Read More »
• ### HARD MATHS – How much faster is one train from other Puzzle
Two trains starting at same time, one from Bangalore to Mysore and other in opposite direction arrive at their destination ...Read More »
• ### Most Analytical GOOGLE INTERVIEW Question Revealed
Let it be simple and as direct as possible. Interviewer : Tell me how much time (in days) and money would ...Read More »
• ### Lateral thinking sequence Puzzle
Solve this logic sequence puzzle by the correct digit- 8080 = 6 1357 = 0 2022 = 1 1999 = ...Read More »
• ### How did he know?
A man leaves his house in the morning to go to office and kisses his wife. In the evening on ...Read More »
• ### Pizza Cost Math Brain Teaser
Jasmine, Thibault, and Noah were having a night out and decided to order a pizza for \$10. It turned out ...Read More »
• ### Which letter replaces the question mark
Which letter replaces the question markRead More »
• ### Which room is safest puzzle
A murderer is condemned to death. He has to choose between three rooms. The first is full of raging fires, ...Read More »
• ### Richie’s Number System
Richie established a very strange number system. According to her claim for different combination of 0 and 2 you will ...Read More »
• ### Srabon wanted to pass
The result of math class test came out. Fariha’s mark was an even number. Srabon got a prime!! Nabila got ...Read More »
• ### Become Normal!!
Robi is a very serious student. On the first day of this year his seriousness for study was 1 hour. ...Read More »
• ### Sakib Knows The Number!
Ragib: I got digits of a 2 digit number Sakib: Is it an odd? Ragib: Yes. Moreover, the sum of ...Read More » | 848 | 3,338 | {"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-2024-22 | latest | en | 0.916361 |
https://oeis.org/A058256 | 1,723,737,967,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00428.warc.gz | 322,701,129 | 4,516 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A058256 a(n) = A058254(n+1)/A058254(n). 1
2, 2, 3, 5, 1, 4, 3, 11, 7, 1, 1, 1, 1, 23, 13, 29, 1, 1, 1, 1, 1, 41, 1, 2, 5, 17, 53, 3, 1, 1, 1, 1, 1, 37, 1, 1, 3, 83, 43, 89, 1, 19, 2, 7, 1, 1, 1, 113, 1, 1, 1, 1, 5, 4, 131, 67, 1, 1, 1, 47, 73, 1, 31, 1, 79, 1, 1, 173, 1, 1, 179, 61, 1, 1, 191, 97, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(n) = 1 if in prime(n+1)-1 no new prime divisor or new power of a prime appear, like LCM[{1, 2, 4, 6, 10, 12, 16, 22}]= LCM[{1, 2, 4, 6, 10, 12, 16, 22, 28}]. a(n) > 1 if in prime(n+1)-1 new prime divisor(s) or new power(s) of a prime arise, like in A058254(15) compared with A058254(14), where the new prime divisor is 23 only, so a(14)=23. Such sites of increase do not correspond to the natural order of primes and prime-powers like in A054451. LINKS Amiram Eldar, Table of n, a(n) for n = 1..10000 FORMULA a(n) = lcm{i=1..n+1} (prime(i)-1) / lcm{i=1..n} (prime(i)-1). PROG (PARI) f(n) = lcm(apply(p->p-1, primes(n))); \\ A058254 a(n) = f(n+1)/f(n); \\ Michel Marcus, Mar 22 2020 CROSSREFS Cf. A058254, A002110, A005867, A003418, A054451, A000142, A000010, A003418, A000961. Sequence in context: A067088 A065519 A329792 * A140183 A280408 A130725 Adjacent sequences: A058253 A058254 A058255 * A058257 A058258 A058259 KEYWORD nonn AUTHOR Labos Elemer, Dec 06 2000 EXTENSIONS Offset corrected by Amiram Eldar, Sep 24 2019 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 15 12:04 EDT 2024. Contains 375173 sequences. (Running on oeis4.) | 827 | 1,937 | {"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-2024-33 | latest | en | 0.57707 |
https://opg.optica.org/josab/fulltext.cfm?uri=josab-38-2-510&id=446781 | 1,653,064,308,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00144.warc.gz | 504,530,227 | 49,850 | # Compact 200 line MATLAB code for inverse design in photonics by topology optimization: tutorial
Open Access
## Abstract
We provide a compact 200 line MATLAB code demonstrating how topology optimization (TopOpt) as an inverse design tool may be used in photonics, targeting the design of two-dimensional dielectric metalenses and a metallic reflector as examples. The physics model is solved using the finite element method, and the code utilizes MATLAB’s fmincon algorithm to solve the optimization problem. In addition to presenting the code itself, we briefly discuss a number of extensions and provide the code required to implement some of these. Finally, we demonstrate the superiority of using a gradient-based method compared to a genetic-algorithm-based method (using MATLAB’s ga algorithm) for solving inverse design problems in photonics. The MATLAB software is freely available in the paper and may be downloaded from https://www.topopt.mek.dtu.dk.
© 2021 Optical Society of America
## Corrections
Rasmus E. Christiansen and Ole Sigmund, "Compact 200 line MATLAB code for inverse design in photonics by topology optimization: tutorial: erratum," J. Opt. Soc. Am. B 38, 1822-1823 (2021)
https://opg.optica.org/josab/abstract.cfm?uri=josab-38-6-1822
## 1. INTRODUCTION
This paper details a 200 line MATLAB code, which demonstrates how density-based topology optimization (TopOpt) can be applied to photonics design. The code is written for scientists and students with a basic knowledge of programming, numerical modeling, and photonics who desire to start using inverse design in their research. We briefly detail the model of the physics, followed by the discretized TopOpt design problem (Section 2). The MATLAB code is then explained in detail (Section 3), followed by two application examples providing the reader with targets for reproduction (Section 4). Then, a number of possible extensions are discussed, and code-snippets for easy implementation are provided along with design examples (Section 5). Finally, we demonstrate the superiority of using gradient-based TopOpt compared to a genetic algorithm (GA) when solving a photonic design problem (Section 6).
TopOpt [1] as an inverse design tool was first developed in the context of solid mechanics in the late 1980s [2]. Since its inception, the method has developed rapidly and expanded across most areas of physics [36]. Over the last two decades, the interest in applying TopOpt for photonics has increased rapidly [7] with applications within cavity design [8,9], photonic demultiplexers [10], metasurfaces [11,12] and topological insulators [13] to name a few. While the interest in TopOpt within the photonics community has grown markedly in recent years, significant barriers hinder newcomers to the field from adopting the tool in their work. These are as follows: the required knowledge of numerical modeling, the required knowledge of advanced mathematical concepts, and scientific programming experience. This paper seeks to lower these barriers by providing the reader with a simple two-dimensional (2D) finite-element-based MATLAB implementation of TopOpt for photonics, which is straightforwardly extendable to a range of other design problems. Within the field of structural optimization in mechanics, similar simple MATLAB codes [1416] have proven themselves highly successful in raising the awareness of TopOpt and serving as a basic platform for further expansion of the method, thus broadening its application as a design tool and successfully driving the field forward.
For readers who are less interested in the programming and method development aspects of TopOpt as an inverse design tool, we have authored a parallel tutorial paper on TopOpt for photonics applications, utilizing the graphical user interface (GUI)-based commercial finite element software COMSOL Multiphysics as the numerical tool to model the physics and solve the optimization problem [17].
Fig. 1. Model domain, $\Omega$, of height ${h_\Omega}$ and width ${w_\Omega}$ with a designable region, ${\Omega _D}$, of height ${h_{{\Omega _D}}}$ and width ${w_{{\Omega _D}}}$ on top of a substrate, ${\Omega _S}$, of height ${h_s}$.
## 2. PHYSICS AND THE DISCRETIZED OPTIMIZATION PROBLEM
We model the physics in the rectangular domain $\Omega$, with the boundary $\Gamma$ (see Fig. 1), using Maxwell’s equations, assuming time-harmonic temporal behavior. We define a subset of $\Omega$ as the design domain and denote this region ${\Omega _D}$. We assume out-of-plane ($z$ direction) material invariance and that all involved materials are linear, static, homogeneous, isotropic, non-dispersive, non-magnetic, and without inherent polarization. Finally, we assume out-of-plane polarization of the electric field (TE polarization). From these assumptions, we derive a 2D Helmholtz-type partial differential equation for the out-of-plane component of the electric field in $\Omega$,
$$\nabla \cdot \left({\nabla {E_z}({\textbf r})} \right) + {k^2}{\varepsilon _r}({\textbf r}){E_z}({\textbf r}) = F({\textbf r}),\quad {\textbf r} \in \Omega \in {{\mathbb R}^2},$$
where ${E_z}$ denotes the $z$ component of the electric field, $k = \frac{{2\pi}}{\lambda}$ is the wavenumber with $\lambda$ (= lambda in the top200EM interface) being the wavelength, ${\varepsilon _r}$ denotes the relative electric permittivity, and $F$ denotes a forcing term used to introduce an incident plane wave from the bottom boundary of $\Omega$. We apply first-order absorbing boundary conditions on all four exterior boundaries,
$${\textbf n} \cdot \nabla {E_z}({\textbf r}) = - {\rm i}k{E_z}({\textbf r}),\quad {\textbf r} \in \Gamma ,$$
where ${\textbf n}$ denotes the surface normal and $i$ the imaginary unit. Note that first-order boundary conditions are not as accurate as certain other boundary conditions, e.g., perfectly matched layers [18]; however, they are conceptually simpler and simpler to implement. Next, we introduce a design field $\xi ({\textbf r}) \in [0,1]$ to control the material distribution in $\Omega$ through the interpolation function,
$${\varepsilon _r}(\xi ({\textbf r})) = 1 + \xi ({\textbf r})\left({{\varepsilon _{r,m}} - 1} \right) - {\rm i}\alpha \xi ({\textbf r})\left({1 - \xi ({\textbf r})} \right),\quad {\textbf r} \in \Omega ,$$
where it is assumed that the background has the value ${\varepsilon _r} = 1$ (e.g., air) and where ${\varepsilon _{r,m}}$ (= eps_r) denotes the relative permittivity of the material used for the structure under design and $\alpha$ is a problem dependent scaling factor. The non-physical imaginary term discourages intermediate values of $\xi$ in the design for the focusing problem at hand, by introducing attenuation [19].
As the baseline example, we consider the design of a monochromatic focusing metalens situated in ${\Omega _D}$. To this end, we select the magnitude of $|{E_z}{|^2}$ at a point in space ${{\textbf r}_p}$ (= targetXY) as the figure of merit (FOM) denoted $\Phi$, i.e.,
$$\Phi (\xi ({\textbf r}),{{\textbf r}_p}) = |{E_z}(\xi ({\textbf r}),{{\textbf r}_p}{)|^2} = {E_z}{(\xi ({\textbf r}),{{\textbf r}_p})^*}{E_z}(\xi ({\textbf r}),{{\textbf r}_p}),$$
where ${\bullet ^*}$ denotes the complex conjugate.
The model equation, boundary conditions, material interpolation function, and FOM are all discretized using the finite element method (FEM) [20] using ${{\cal N}_e}(= {\texttt{nElX}} \cdot \texttt{nElY})$ bi-linear quadratic elements. The discretized model uses nodal degrees of freedom (DOFs) for ${E_z}$ and $F$ as well as elementwise constant DOFs for $\xi$, with ${{\cal N}_D}$ of the elements situated in ${\Omega _D}$. The following constrained continuous optimization problem is formulated for the discretized problem:
$$\begin{split}&\mathop {\max}\limits_\xi :\;\Phi = {\textbf E}_z^\dagger {\textbf P}{{\textbf E}_z}, \\[-4pt] &{\rm s.t.}:\;{\textbf S}({\varepsilon _r}){{\textbf E}_z} = \left({\sum\limits_{e = 1}^{{{\cal N}_e}} {{\textbf S}_e}({\varepsilon _{r,e}})} \right){{\textbf E}_z} = {\textbf F}, \\[-4pt] &\qquad :\;{\varepsilon _{r,j}} = 1 + {{\bar {\tilde \xi}} _j}\left({{\varepsilon _{r,m}} - 1} \right) - {\rm i}{{\bar {\tilde \xi}} _j}\left({1 - {{{\bar {\tilde \xi}}}_j}} \right)\quad \forall \;j \in \{1,2, \ldots,{{\cal N}_e}\} , \\[-4pt] &\qquad :\;0 \lt {\xi _j} \lt 1\quad \forall \;j \in \{1,2, \ldots,{{\cal N}_D}\} , \\[-4pt] &\qquad :\;\xi = 0\quad \forall \;{\textbf r} \in \Omega /\{{\Omega _D},{\Omega _S}\} \vee \xi = 1\quad \forall \;{\textbf r} \in {\Omega _S},\\[-1.3pc]\end{split}$$
where ${{\textbf E}_z}$ and ${\textbf F}$ are vectors containing the nodal DOFs for the electric field and forcing term, and $\xi$ (= dVs) and ${\bar {\tilde \xi}}$ are vectors of element DOFs for the design field and the physical filtered and thresholded field [Eqs. (6) and (7)], respectively.
The diagonal selection matrix ${\textbf P}$ weights the ${{\textbf E}_z}$-DOFs that enter $\Phi$ (= FOM). In the baseline example, it has four $\frac{1}{4}$ entries for the selected element’s nodal DOFs, i.e., it selects the field intensity at the focal point, which for simplicity is taken to be at the center of a single finite element. Finally, ${\bullet ^\dagger}$ denotes the conjugate transpose.
To ameliorate numerical issues, such as pixel-by-pixel design variations, and to introduce a weak sense of geometric length scale, a filter and threshold scheme is applied to $\xi$, before using it to interpolate the material parameters [2123]. First, the following convolution-based filter operation is applied:
$$\begin{split}{\tilde \xi _h} &= \frac{{\mathop {\sum}\limits_{k \in {{\cal B}_{e,h}}} w({{\textbf r}_h} - {{\textbf r}_k}){A_k}{\xi _k}}}{{\mathop {\sum}\limits_{k \in {{\cal B}_{e,h}}} w({{\textbf r}_h} - {{\textbf r}_k}){A_k}}},\\ w({\textbf r}) &= \left\{{\begin{array}{*{20}{l}}{{r_f} - |{\textbf r}|\quad \forall |{\textbf r}| \le {r_f}}\\0\end{array}} \right.,\quad {r_f} \ge 0,\quad {\textbf r} \in \Omega.\end{split}$$
Here ${A_k}$ denotes the area of the $k$th element, ${r_f}$ (= fR) denotes the desired spatial filtering radius, and finally ${{\cal B}_{e,h}}$ denotes the $h$th set of finite elements whose center point is within ${r_f}$ of the $h$th element. Second, a smoothed approximation of the Heaviside function is applied to the filtered design variables as
$$\begin{split}&{{\bar {\tilde \xi}} _h} = H({\tilde \xi _h}) = \frac{{\tanh (\beta \cdot \eta) + \tanh (\beta \cdot ({{\tilde \xi}_h} - \eta))}}{{\tanh (\beta \cdot \eta) + \tanh (\beta \cdot (1 - \eta))}},\\& \beta \in [1,\infty [,\eta \in [0,1].\end{split}$$
Here $\beta$ and $\eta$ control the threshold sharpness and value, respectively.
Adjoint sensitivity analysis [6,24] is carried out to compute the design sensitivities, utilizing the chain rule for the filter and threshold steps as [23,25]
$$\frac{{{d}\Phi}}{{{d}{\xi _h}}} = \mathop {\sum}\limits_{k \in {{\cal B}_{e,h}}} \frac{{\partial {{\tilde \xi}_k}}}{{\partial {\xi _h}}}\frac{{\partial {{{\bar {\tilde \xi}}}_k}}}{{\partial {{\tilde \xi}_k}}}\frac{{{d}\Phi}}{{{d}{{{\bar {\tilde \xi}}}_k}}},\quad \frac{{{d}\Phi}}{{{d}{{{\bar {\tilde \xi}}}_k}}} = 2\Re \left({{\lambda ^{T}}\frac{{\partial {\textbf S}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}{{\textbf E}_z}} \right),$$
where $\Re$ denotes the real part, ${\bullet ^T}$ the transpose, and $\lambda$ a vector obtained by solving
$${{\textbf S}^{T}}\lambda = - \frac{1}{2}{\left({\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Re}}}} - {\rm i}\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Im}}}}} \right)^{T}}\quad {\rm with}\quad {{\textbf E}_z} = {{\textbf E}_{z,\Re}} + {\rm i}{{\textbf E}_{z,\Im}},$$
where $\Im$ denotes the imaginary part. The $m$th entry of the right-hand side in Eq. (9) is given by
$$\left({\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Re}}}} - {\rm i}\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Im}}}}} \right)_m^{T} = {{\textbf P}_{m,m}}\left({2({{\textbf E}_{z,\Re}}{)_m} - 2{\rm i}{{({{\textbf E}_{z,\Im}})}_m}} \right).$$
The derivations of the expression for $\frac{{{d}\Phi}}{{{d}{{{\bar {\tilde \xi}}}_k}}}$ in Eq. (8) and for Eq. (9) are given in Appendix A.
The fundamental advantage of using the adjoint approach to compute the sensitivities of the FOM with respect to the design variables is that only one single additional system of equations [namely Eq. (9)] must be solved. Hence, the sensitivity information is obtained at approximately the same computational cost as the one required to compute the field information itself. In fact, for the examples treated here, it is possible to reuse the LU-factorization used to compute the field information [second line of Eq. (5)], making the computational cost associated with computing the sensitivity of the FOM almost ignorable.
## 3. MATLAB CODE
The design problem stated in Eq. (5) is implemented in top200EM (see the full code in Code 1, Ref. [26]), which has the following interface:
function [dVs,FOM] = …
top200EM(targetXY,dVElmIdx,nElX,nElY, dVini,eps_r,lambda,fR,maxItr);
The function takes the input parameters:
targetXY: two-values 1D-array with the $x$ and $y$ position of the finite element containing the focal point.
dVElmIdx: 1D-array of indices of the finite elements, which are designable, i.e., ${\Omega _D}$.
nElX: Number of finite elements in the $x$ direction.
nElY: Number of finite elements in the $y$ direction.
dVini: Initial guess for discretized $\xi$-field. Accepts a scalar for all elements or a 1D-array of identical length to dVElmIdx.
eps_r: Relative permittivity for the material constituting the structure under design.
lambda: The targeted wavelength, $\lambda$, measured in number of finite elements.
fR: Filter radius, ${r_f}$, measured in number of finite elements.
maxItr: Maximum number of iterations allowed by fmincon for solving the optimization problem in Eq. (5).
And returns the following output parameters:
dVs: 1D-array of the optimized discretized $\xi$-field in the design domain, ${\Omega _D}$.
FOM: Value of the figure of merit.
During execution, the data related to the physics, the discretization, the filter, and threshold operations are stored in the structures phy, dis, and filThr, respectively.
The spatial scaling, threshold sharpness $\beta$, and threshold level $\eta$ are hard coded in top200EM as
% SETUP OF PHYSICS PARAMETERS
phy.scale = 1e-9; % Scaling finite element side length to nanometers
% SETUP FILTER AND THRESHOLDING PARAMETERS
filThr.beta = 5; % Thresholding sharpness
filThr.eta = 0.5; % Thresholding level
Note: For simplicity, the code uses the unit of nanometers to measure length, and the finite elements are taken to have a side length of 1 nm. This may be changed by changing the scaling parameter phy.scale.
The algorithm used to solve the design problem is MATLAB’s fmincon,
[dVs,~] = fmincon(FOM,dVs(:),[],[],[], [],LBdVs,UBdVs,[],options);
with the design-variable bounds and options set up as
LBdVs = zeros(length(dVs),1); % Lower bound on design variables
UBdVs = ones(length(dVs),1); % Upper bound on design variables
options = optimoptions(‘fmincon’,‘Algorithm’,‘interior-point’,…
‘Display’,‘off’,‘MaxIterations’,maxItr,‘MaxFunctionEvaluations’,maxItr);
The FOM and sensitivities provided to fmincon are computed using the inline function:
The discretized design field is distributed in the model domain with a hard coded background of air in the top 90% of the domain and solid material in the bottom 10% as
% DISTRIBUTE MATERIAL IN MODEL DOMAIN BASED ON DESIGN FIELD
dFP(1:dis.nElY,1:dis.nElX) = 0; % Design field in physics, 0: air
dFP(dis.nElY:–1:ceil(dis.nElY*9/10),1:dis.nElX) = 1; % 1: material
dFP(dis.dVElmIdx(:)) = dVs; % Design variables inserted in design field
Followed by the application of the filter and threshold operations and the material interpolation,
% COMPUTE MATERIAL FIELD FROM DESIGN FIELD
dFPS = DENSITY_FILTER(filThr.filKer, filThr.filSca,dFP,ones(dis.nElY,dis.nElX));
dFPST = THRESHOLD(dFPS, filThr.beta, filThr.eta);
[A,dAdx] = MATERIAL_INTERPOLATION(phy.eps_r,dFPST,1.0); % Material field
The system matrix for the state equation is constructed,
% CONSTRUCT SYSTEM MATRIX
[dis,F] = BOUNDARY_CONDITIONS_RHS(phy.k,dis,phy.scale);
dis.vS = reshape(dis.LEM(:)-phy.k^2*dis. MEM(:)*(A(:).’),16*dis.nElX*dis.nElY,1);
SysMat = sparse([dis.iS(:);dis.iBC(:)],[dis.jS(:);dis.jBC(:)],[dis.vS(:);dis.vBC(:)]);
The state system is solved using LU-factorization,
% SOLVING STATE SYSTEM: SysMat * Ez = F
[L,U,Q1,Q2] = lu(SysMat); % LU–factorization
Ez = Q2 * (U\(L\(Q1 * F))); Ez = full(Ez); % Solving
The FOM is computed,
% FIGURE OF MERIT
P = sparse(dis.edofMat(dis.tElmIdx,:), dis.edofMat(dis.tElmIdx,:),1/4,…
(dis.nElX+1)*(dis.nElY+1),(dis.nElX+1) *(dis.nElY+1)); % Weighting matrix
FOM = Ez’ * P * Ez; % Solution in target element
The adjoint system is solved by reusing the LU-factorization from the state problem,
% ADJOINT RIGHT HAND SIDE (0th-order quadrature)
% SOLVING THE ADJOING SYSTEM: S.’ * AdjLambda = AdjRHS
AdjLambda = (Q1.’) * ((L.’)\((U.’)\ ((Q2.’) * (-1/2*AdjRHS)))); % Solving
The sensitivities in $\Omega$ are computed and filtered, and the values in ${\Omega _D}$ extracted,
% SENSITIVITIES
dis.vDS = reshape(-phy.k^2*dis.MEM(:) *(dAdx(:).’),16*dis.nElX*dis.nElY,1);
DSdx = sparse(dis.iElFull,dis.jElFull, dis.vDS); % Constructing dS/dx
DSdxMulV = DSdx * Ez(dis.idxDSdx); % Computing dS/dx * Field values
DsdxMulV = sparse(dis.iElSens,dis. jElSens,DSdxMulV);
sens = 2*real(AdjF(dis.idxDSdx).’ * DsdxMulV); % Com puting sensitivities
sens = full(reshape(sens,dis.nElY,dis. nElX));
% FILTERING SENSITIVITIES
DdFSTDFS = DERIVATIVE_OF_THRESHOLD(dFPS, filThr.beta, filThr.eta);
sensFOM = DENSITY_FILTER(filThr.filKer, filThr.filSca,sens,DdFSTDFS);
% EXTRACTING SENSITIVITIES FOR DESIGN REGION
sensFOM = sensFOM(dis.dVElmIdx);
Finally, ${E_z}({\textbf r})$ and $\xi ({\textbf r})$ are plotted, and the current FOM-value printed,
% PLOTTING AND PRINTING
figure(1); % Field intensity, |Ez|^2
imagesc((reshape(Ez.*conj(Ez),dis.nElY +1,dis.nElX+1))); colorbar; axis equal;
figure(2); % Physical design field
imagesc(1-dFPST); colormap(gray); axis equal; drawnow;
disp([’FOM:’ num2str(-FOM)]); % Display FOM value
After the TopOpt procedure is finished, a thresholded version of the final design is evaluated, and the resulting $|{E_z}{|^2}$ field and design field are plotted.
% FINAL BINARIZED DESIGN EVALUATION
filThr.beta = 1000;
disp(’Black/white design evaluation:’)
[obj_2,dFPST_2,F_2] = OBJECTIVE_ GRAD(DVini(:),dis,phy,filThr);
Fig. 2. (a) Max-normalized $|{\textbf E}{|^2}$-field in $\Omega$. (b) Metalens design, ${\varepsilon _r} = 3.0$ (black) and ${\varepsilon _r} = 1.0$ (white).
Fig. 3. Metalens designs obtained (a) without filtering (${r_f} = 1$), (b) using a filter radius of ${r_f} = 3.0$, (c) using a filter radius of ${r_f} = 6.0$, (d) using a filter radius of ${r_f} = 9.0$. These results illustrate the effect of applying the cone-shaped filter to the design field as part of the optimization process.
## 4. USING THE CODE
Next, we demonstrate how to use top200EM by designing a focusing metalens as follows.
#### A. Designing a Metalens
First, we define the domain size in terms of the number of finite elements in each spatial direction and the element indices for the design domain.
% DESIGN FIELD INDICES
DomainElementsX = 400;
DomainElementsY = 200;
DesignThicknessElements = 15;
DDIdx = repmat([1:DomainElementsY: DomainElementsX*DomainElementsY],…
DesignThicknessElements,1);
DDIdx = DDIdx+repmat([165:165 +DesignThicknessElements-1]’,1, DomainElementsX);
Second, the optimization problem is solved by executing the command
[DVs,obj]=top200EM([80,200],DDIdx, DomainElementsX,DomainElementsY,…
0.5,3.0,35,6.0,200);
The final binarized design is shown in Fig. 2(b) with black (white) representing solid (air). Figure 2(a) shows the $|{E_z}{|^2}$-field resulting from exciting the metalens in Fig. 2(b) for the targeted incident field, demonstrating the focusing effect at the targeted focal spot. The numerical aperture of the metalens is ${\rm NA} \approx 0.92$, and the transmission efficiency is ${T_A} \approx 0.87$ computed as the power propagating through the lens relative to the power incident on the lens.
#### B. Solving the Same Design Problem with Different Resolutions
For various reasons, such as performing a mesh-convergence study, it may be necessary to solve the same physical model problem using different mesh resolutions. This may be done with top200EM by multiplying the following inputs by an integer scaling factor: the number of finite elements in each spatial direction, nElX and nElY, the wavelength lambda, and the filter radius fR, and dividing the hard coded value of phys.scale in the code by the same factor.
#### C. Effect of Filtering
Next, we demonstrate the effect of applying the filtering step [27] by changing the filter radius and designing four metalenses (see Fig. 3) using top200EM. Again, the model domain size and indices for the design domain are defined first,
% DESIGN FIELD INDICES
DomainElementsX = 400;
DomainElementsY = 200;
DesignThicknessElements = 15;
DDIdx = repmat([1:DomainElementsY: DomainElementsX*DomainElementsY],…
DesignThicknessElements,1);
DDIdx = DDIdx+repmat([165:165 +DesignThicknessElements-1]’,1, DomainElementsX);
Then, the optimization problem is solved with the four filtering radii fR$= {r_f} \in \{1.0,3.0,6.0,9.0\}$,
[DVs,obj]=top200EM([80,200],DDIdx,DomainElementsX,DomainElementsY,…
0.5,3.0,35,fR,200);
Looking at the four final binarized designs in Fig. 3, it is clearly observed that as ${r_f}$ is increased, the features in the designs grow. When no filtering is applied [Fig. 3(a)], single-pixel-sized features are observed. Such features may be problematic from a numerical modeling point of view as well as being detrimental to fabrication.
It is noted that without filtering, TopOpt is more prone to identify a local minimum, which performs worse than the ones identified with filtering. In other words, filtering tends to have a convexifying effect, as long as the filter size is not too big.
## 5. MODIFYING THE CODE
There exists a vast amount of auxiliary tools developed to extend the applicability of density-based TopOpt across a wide range of different problems and physics. The following sections provide examples of how simple some of these tools are to implement in top200EM.
#### A. Plasmonics
In recent work [28], it was demonstrated that using a refractive index and extinction cross-section-based non-linear material interpolation yielded significantly improved results when designing Au, Ag, and Cu nano-particles for localized field enhancement, with recent applications to enhanced upconversion [29], thermal emission [30], and Raman scattering [31]. This interpolation function avoids artificial resonances in connection with the transition from positive to negative $\varepsilon$ values and reads
$$\begin{split}\varepsilon (x) &= (n{(x)^2} - \kappa {(x)^2}) - {\rm i}(2n(x)\kappa (x)),\\n(x)& = {n_{{M}_1}} + x({n_{{M}_2}} - {n_{{M}_1}}),\\ \kappa (x) &= {\kappa _{{M}_1}} + x({\kappa _{{M}_2}} - {\kappa _{{M}_1}}).\end{split}$$
Here $n$ and $\kappa$ denote the refractive index and extinction cross section, respectively. The subscripts ${M_1}$ and ${M_2}$ denote the two materials being interpolated.
The interpolation scheme is straightforward to implement in the code as follows:
First, the following lines of code:
function [A,dAdx] = MATERIAL_INTERPOLATION(eps_r,x,1.0)
A = 1 + x*(eps_r-1)–1i * alpha_i * x.* (1–x); % Interpolation
dAdx = (eps_r-1)*(1+0*x)–1i * alpha_i * (1–2*x); % Derivative of interpolation end
are replaced with
function [A,dAdx] = MATERIAL_INTERPOLATION(n_r,k_r,x)
n_eff = 1 + x*(n_r-1);
k_eff = 0 + x*(k_r-0);
A = (n_eff.^2–k_eff.^2)–1i*(2.*n_eff.*k_eff);
dAdx = 2*n_eff*(n_r-1)-2*k_eff*(k_r-1)-1i*(2*(n_r-1)*k_eff+(2*n_eff*(k_r-1))); end
where for simplicity it is assumed that ${M_1}$ is air, i.e., ${n_{{M}_1}} = 1.0$ and ${\kappa _{{M}_1}} = 0.0$.
Second, the scalar input parameter eps_r is changed to a two-valued 1D-array nk_r.
Third, the line
phy.eps_r=eps_r; % Relative permittivity
is replaced with,
phy.nk_r = nk_r; % Refractive index and extinction coefficient
and fourth, the call to the material interpolation function is changed from
to,
#### B. Excitation
The excitation considered in the baseline problem may be changed straightforwardly. As an example, the boundary at which the incident field enters the domain may be changed from the bottom to the top boundary as follows.
First, the index set controlling where the boundary condition is imposed in the vector, ${\textbf F}$ (= F), is changed by moving dis.iRHS = TMP; from line 157 to above line 151.
Second, the values stored in F are changed to account for the propagation direction of the wave by replacing
F(dis.iRHS(1,:)) = F(dis.iRHS(1,:))+1i*waveVector;
F(dis.iRHS(2,:)) = F(dis.iRHS(2,:))+1i*waveVector;
with
F(dis.iRHS(1,:)) = F(dis.iRHS(1,:))-1i*waveVector;
F(dis.iRHS(2,:)) = F(dis.iRHS(2,:))-1i*waveVector;
#### C. Designing a Metallic Reflector
By introducing the changes presented in Sections 5.A and 5.B, one may design a metallic reflector using topEM200.
First, we set the domain dimensions and design thickness as in the previous examples,
% DESIGN FIELD INDICES
DomainElementsX = 400;
DomainElementsY = 200;
DesignThicknessElements = 15;
DDIdx = repmat([1:DomainElementsY:DomainElementsX*DomainElementsY],…
DesignThicknessElements,1);
DDIdx = DDIdx+repmat([165:165+DesignThicknessElements-1]’,1,DomainElementsX);
Second, we solve the optimization problem by executing the command,
[DVs,obj]=top200EM([100,200],DDIdx,DomainElementsX,DomainElementsY,…
0.5,[1.9,1.5],35,3.0,200);
Note: For simplicity we selected the following values for the refractive index, $n = 1.9$ (=nk_r(1)), and extinction cross section, $\kappa = 1.5$ (=nk_r(2)) (These values corresponds to values for gold at $\lambda = 350\;{\rm nm}$, i.e., one may think of this choice of material parameters as an unspoken rescaling of space by a factor of 10, i.e., changing the element size to pixels of 10 nm by 10 nm rather than of 1 nm by 1 nm).
Fig. 4. (a) Max-normalized $|{\textbf E}{|^2}$-field in $\Omega$. (b) Metallic reflector design (black) in air background (white).
Considering the final binarized reflector design in Fig. 4(b), one can interpret the design as the well-known parabolic reflector broken into pieces to fit the spatially limited design domain. From the max-normalized $|{\textbf E}{|^2}$-field presented in Fig. 4(a), the focusing effect of the reflector is clearly observed.
#### D. Linking Design Variables
Certain fabrication techniques limit the allowable geometric variations in a design. For example, optical-projection lithography and electron-beam lithography restrict variations in design geometries to 2D patterns, which can then be extruded in the out-of-plane direction. It is straightforward to introduce such a geometric restriction using TopOpt by linking design variables and sensitivities across elements. In top200EM the design field may be restricted to only exhibit in-plane ($x$ direction) variations as follows.
First, the code-line representing the values of the design variables,
dVs(length(dis.dVElmIdx(:))) = dVini; % Design variables
is replaced with,
dVs(1:nElX) = dVini; % Design variables for 1D design
Second, the code-line transferring the design variables to the elements in the physics model,
dFP(dis.dVElmIdx(:)) = dVs; % Design variables inserted in design field
is replaced with,
nRows=length(dis.dVElmIdx(:))/dis.nElX; % Number of rows in the 1D design
dFP(dis.dVElmIdx) = repmat(dVs,1,nRows)’; % Design variables inserted in design field
and finally the code-line representing individual element sensitivities,
sensFOM = sensFOM(dis.dVElmIdx);
is replaced with,
sensFOM = sensFOM(dis.dVElmIdx);
sensFOM = sum(sensFOM,1); % Correcting sensitivities for 1D design,
which sums the sensitivity contributions from the linked elements.
### 1. Designing a 1D Metalens
By introducing the changes listed in Section 5.D, topEM200 can be used to design metasurfaces of fixed height with in-plane variations as follows. Again we define the domain dimensions as
Fig. 5. (a) Max-normalized $|{\textbf E}{|^2}$-field in $\Omega$. (b) Metalens design restricted to one-dimensional variations, ${\varepsilon _r} = 3.0$ (black) and ${\varepsilon _r} = 1.0$ (white).
% DESIGN FIELD INDICES
DomainElementsX = 400;
DomainElementsY = 200;
DesignThicknessElements = 15;
DDIdx = repmat[1:DomainElementsY:DomainElementsX*DomainElementsY],…
DesignThicknessElements,1);
DDIdx = DDIdx+repmat([165:165+DesignThicknessElements-1]’,1,DomainElementsX);
Followed by the execution of the command,
[DVs,obj]=top200EM([80,200],DDIdx,DomainElementsX,DomainElementsY,…
0.5,3.0,35,3.0,200);
The final binarized design, resulting from solving the design problem, is shown in Fig. 5(b). Here it is clear to see that the design is now restricted to vary only in the $x$ direction. It is worth noting that the design is still filtered in both spatial directions; hence, the corners of the design appear rounded in Fig. 5(b). The max-normalized $|{\textbf E}{|^2}$-field presented in Fig. 4(a) demonstrates the focusing effect of the lens at the targeted point in the modeling domain, which, due to the reduced design freedom, is not as high as in the unrestricted case (Section 4.A).
#### E. Continuation of Threshold Sharpness
For some design problems in photonics and plasmonics, e.g., [22,28,31], intermediate values may be present in the final optimized design, i.e., ${{\bar {\tilde \xi}} _{{\rm final}}} \in]0,1[$, despite the applied penalization scheme, as they prove beneficial to optimizing the FOM. However, (in most cases) intermediate values hold no physical meaning, and it is therefore not possible to realize designs containing such intermediate values experimentally. A way to promote that (almost) no design variables take intermediate values in the final design is by using a continuation scheme for the threshold sharpness, gradually increasing it until an (almost) pure 0/1-design is achieved, ensuring that the optimized designs are physically realizable.
To implement the continuation scheme in the code, replace the following lines:
% SOLVE DESIGN PROBLEM USING MATLAB BUILT-IN OPTIMIZER: FMINCON
$[{\rm dVs},\sim]$ = fmincon(FOM,dVs(:),[],[],[],[],LBdVs,UBdVs,[],options);
with,
while filThr.beta<betaMax % Thresholding sharpness bound
% SOLVE DESIGN PROBLEM USING MATLAB BUILT-IN OPTIMIZER: FMINCON
$[{\rm dVs},\sim]$ = fmincon(FOM,dVs(:),[],[],[],[],LBdVs,UBdVs,[],options);
filThr.beta = betaInc * filThr.beta; % Increasing thresholding sharpness
end
and select suitable values for betaMax and betaInc. These values are problem dependent, and some experimentation may be required to identify the best values for a given problem. For the metalens design example in Section 4.A, when considering high material contrast, i.e., large values of eps_r, the values betaMax = 20.0 and betaInc = 1.5 have been found to work well.
## 6. WHAT ABOUT GENETIC ALGORITHMS?
For a wide range of inverse design problems, where gradient-based optimization methods are applicable [inverse design problems where it is possible to compute the sensitivities of the FOM (and any constraints), e.g., using adjoint sensitivity analysis, a pre-requisite for using gradient-based methods], they have been found to severely outperform non-gradient-based methods, such as GAs [32]. This, both in terms of the computational effort required to identify a local optimum for the FOM and, by direct extension, in terms of the number of design DOFs, it is feasible to consider (a difference of many orders of magnitude [33]). Further, in many cases, gradient-based methods are able to identify better local optima for the FOM [34].
In the following, we provide a demonstration of these claims, by comparing the solution of a metalens design problem obtained using the gradient-based top200EM to the solution obtained using MATLAB’s built-in genetic algorithm ga.
Readers may perform their own comparisons by rewriting top200EM to utilize ga instead of fmincon as follows.
Fig. 6. (a) Design obtained using the GA-based method. (b) Design obtained using the gradient-based method. ${\varepsilon _r} = 3.0$ (black) and ${\varepsilon _r} = 1.0$ (white). (c) Convergence graph. Remark that each design iteration for the GA-based code requires 200 solutions of the physics model equation, while it requires one solution of the physics model equation and one solution of the adjoint equation for the gradient-based code!
First, a (near perfect) 0/1-design is ensured by changing the thresholding strength from
filThr.beta = 5; % Thresholding sharpness
to,
filThr.beta = 1e5; % Thresholding sharpness
Second, ga is used instead of fmincon by replacing the following lines of codes:
options = optimoptions(‘fmincon’,‘Algorithm’,‘interior-point’,…
‘Display’,‘off’,‘MaxIterations’,maxItr,‘MaxFunctionEvaluations’,maxItr);
% SOLVE DESIGN PROBLEM USING MATLAB BUILT-IN OPTIMIZER: FMINCON
$[{\rm dVs},\sim]$ = fmincon(FOM,dVs(:),[],[],[],[],LBdVs,UBdVs,[],options);
with,
options = optimoptions(’ga’,‘MaxGenerations’,maxItr,‘Display’,‘iter’);
% SOLVE DESIGN PROBLEM USING MATLAB BUILT-IN OPTIMIZER: GA
FOM = @(dVs)OBJECTIVE(dVs,dis,phy,filThr);
rng(1,‘twister’); % SETTING RANDOM SEED TO ENSURE REPRODUCTION OF RESULT
$[{\rm dVs},\sim,\sim,\sim]$ = ga(FOM,length(dVs),[],[],[],[],LBdVs,UBdVs,[],[],options);
where the command rng(1,‘twister’) fixes the random seed for reproducibility.
Third, the computation of the gradient of the FOM is removed by changing
function [FOM,sensFOM] = OBJECTIVE_GRAD(dVs,dis,phy,filThr)
to,
function [FOM] = OBJECTIVE(dVs,dis,phy,filThr)
Fourth, deleting the following lines in OBJECTIVE:
% ADJOINT RIGHT HAND SIDE
% SOLVING THE ADJOING SYSTEM: S.’ * AdjLambda = AdjRHS
AdjLambda = (Q1.’) * ((L.’)\((U.’)\((Q2.’) * (-1/2*AdjRHS)))); % Solving
% COMPUTING SENSITIVITIES
DSdx = sparse(dis.iElFull,dis.jElFull,dis.vDS); % Constructing dS/dx
DSdxMulV = DSdx * Ez(dis.idxDSdx); % Computing dS/dx * Field values
DsdxMulV = sparse(dis.iElSens,dis.jElSens,DSdxMulV);
sens = 2*real(AdjLambda(dis.idxDSdx).’ * DsdxMulV); % Computing sensitivites
sens = full(reshape(sens,dis.nElY,dis.nElX));
% FILTERING SENSITIVITIES
DdFSTDFS = DERIVATIVE_OF_THRESHOLD(dFPS, filThr.beta, filThr.eta);
sensFOM = DENSITY_FILTER(filThr.filKer,filThr.filSca,sens,DdFSTDFS);
% EXTRACTING SENSITIVITIES FOR DESIGNABLE REGION
sensFOM = sensFOM(dis.dVElmIdx);
% FMINCON DOES MINIMIZATION
sensFOM = -sensFOM(:);
and finally changing the code evaluating the final design,
% FINAL BINARIZED DESIGN EVALUATION
filThr.beta = 1000;
disp(’Black/white design evaluation:’);
$[{\rm FOM},\sim]$ = OBJECTIVE(dVs(:),dis,phy,filThr);
to,
% FINAL BINARIZED DESIGN EVALUATION
filThr.beta = 1e5;
disp(’Black/white design evaluation:’);
$[{\rm FOM},\sim]$ = OBJECTIVE(dVs(:),dis,phy,filThr);
Note: The code-lines plotting the design in OBJECTIVE can be removed to reduce wall-clock time.
In order to reduce the computational effort required to reproduce the following example, we consider a problem in a smaller spatial domain, considering a shorter wavelength and fewer DOFs than in the previous examples. This is done by defining a model problem considering 1000 design DOFs as
% DESIGN FIELD INDICES
DomainElementsX = 100;
DomainElementsY = 50;
DesignThicknessElements = 10;
DDIdx = repmat([1:DomainElementsY:DomainElementsX*DomainElementsY],…
DesignThicknessElements,1);
DDIdx = DDIdx+repmat([35:35+DesignThicknessElements-1]’,1,DomainElementsX);
Followed by the solution of the design problem using the modified GA-based code,
[dVs_GA,FOM_GA]=top200EMGA([10,50],DDIdx,DomainElementsX,DomainElementsY,…
0.5,3.0,20,3.0,500);
and then using the original gradient-based code,
[dVs,FOM]=top200EM([10,50],DDIdx,DomainElementsX,DomainElementsY,…
0.5,3.0,20,3.0,500);
The binarized designs obtained using top200EMGA and top200EM are presented in Figs. 6(a) and 6(b), respectively. The evolution of the FOM value as function of the number of FOM evaluations is plotted in Fig. 6(c) for both methods.
The GA-based method uses 137 design iterations (27,600 FOM evaluations) to identify a solution. In contrast, the gradient-based method uses 208 iterations (208 FOM evaluation) to identify a superior solution. Evaluating the final binarized designs, the GA-based solution has a FOM value of $\Phi \approx 6.85$ while the gradient-based solution has an $\approx 18\%$ better FOM value of $\Phi \approx 8.07$.
Crucially, the GA-based method uses 200 FOM evaluations per design iteration to drive the optimization process. This results in a total of 27,600 FOM evaluations for solving the design problem using the GA-based method, which corresponds to 27,600 solutions of the physics model equation, i.e., Eq. (1). In contrast, when solving the design problem using the gradient-based method, a total of 208 FOM evaluations are performed, corresponding to 208 solutions of the physics model equation plus 208 solutions of the adjoint system [Eq. (9)], which for the baseline example is almost free due to the reuse of the LU-factorization.
Hence, in our example, the identification of a local optimum for the FOM using the gradient-based method required $\approx 1.0\%$ of the computational effort spent by the GA-based method, a ratio that only becomes more pronounced as the size of the design space grows.
This example is based on standard settings for MATLAB’s ga optimizer and can surely be improved and refined. However, this will not change the basic conclusion that non-gradient-based approaches are unsuitable for large-scale TopOpt problems. A more thorough discussion of non-gradient-based methods in TopOpt can be found in Ref. [34]. Readers are invited to form their own conclusions by extending the example and codes provided here.
## 7. CONCLUSION
We have presented a simple finite element based MATLAB code (downloadable from https://www.topopt.mek.dtu.dk) for TopOpt-based inverse design of photonic structures. We have provided examples of how to use the code, as well as a set of suggestions for code extensions enabling it to handle metallic structures, different model excitations, and linked design variables, and introducing a continuation scheme for the threshold sharpness designed to promote 0/1-designs. Finally, we have demonstrated the superiority of a gradient-based method over a genetic-algorithm-based method when performing inverse design in photonics.
The code can be used for educational purposes as is, and is otherwise meant to serve as a starting point for the reader to develop software to handle their more advanced research applications within photonics. For simplicity and computational speed, the code treats problems in two spatial dimensions; however, it is directly extendable to three spatial dimensions by modifying the finite element matrices, boundary conditions, and index sets appropriately.
## APPENDIX A: ADJOINT SENSITIVITY ANALYSIS
The expression for $\frac{{{d}\Phi}}{{{d}{{{\bar {\tilde \xi}}}_k}}}$ in Eq. (8) and the expression in Eq. (9) may be derived as follows. First, zero is added to $\Phi$ twice,
$$\tilde \Phi = \Phi + {\lambda ^{T}}\left({{\textbf S}{{\textbf E}_z} - {\textbf F}} \right) + {\lambda ^\dagger}\left({{{\textbf S}^*}{\textbf E}_z^* - {{\textbf F}^*}} \right),$$
where $({{\textbf S}{{\textbf E}_z} - {\textbf F}}) = 0$ and $\lambda$ is a vector of nodal complex Lagrange multipliers, also called the adjoint variables. Second, one takes the derivative of $\tilde \Phi$ with respect to ${{\bar {\tilde \xi}} _k}$ and exploits that, for the optimization problem in Eq. (5), $\Phi$ does not depend explicitly on ${{\bar {\tilde \xi}} _k}$ and neither $\lambda$ nor ${\textbf F}$ depend on ${{\bar {\tilde \xi}} _k}$ at all, yielding
$$\begin{split}\frac{{{d}\tilde \Phi}}{{{d}{{{\bar {\tilde \xi}}}_k}}} &= \frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Re}}}}\frac{{\partial {{\textbf E}_{z,\Re}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}} + \frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Im}}}}\frac{{\partial {{\textbf E}_{z,\Im}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}} \\&\quad+ {\lambda ^{T}}\left({\frac{{\partial {\textbf S}}}{{{d}{{{\bar {\tilde \xi}}}_k}}}{{\textbf E}_z} + {\textbf S}\left({\frac{{\partial {{\textbf E}_{z,\Re}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}} + {\rm i}\frac{{\partial {{\textbf E}_{z,\Im}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}} \right)} \right) \\ &\quad {+}{\lambda ^\dagger}\left({\frac{{\partial {{\textbf S}^*}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}{\textbf E}_z^* + {{\textbf S}^*}\left({\frac{{\partial {{\textbf E}_{z,\Re}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}} - {\rm i}\frac{{\partial {{\textbf E}_{z,\Im}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}} \right)} \right).\end{split}$$
Collecting terms including $\frac{{\partial {{\textbf E}_{z,\Re}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}$ and $\frac{{\partial {{\textbf E}_{z,\Im}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}$ and reducing the remaining terms yields
$$\begin{split}\frac{{{d}\tilde \Phi}}{{{d}{{{\bar {\tilde \xi}}}_k}}} &= \frac{{\partial {{\textbf E}_{z,\Re}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}\left({\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Re}}}} + {\lambda ^{T}}{\textbf S} + {\lambda ^\dagger}{{\textbf S}^*}} \right)\\&\quad + \frac{{\partial {{\textbf E}_{z,\Im}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}\left({\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Im}}}} + {\rm i}{\lambda ^{T}}{\textbf S} + {\rm i}{\lambda ^\dagger}{{\textbf S}^*}} \right) + 2\Re \left({{\lambda ^{T}}\frac{{\partial {\textbf S}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}{{\textbf E}_z}} \right).\end{split}$$
To eliminate the first two terms in Eq. (A3) containing the derivates $\frac{{\partial {{\textbf E}_{z,\Re}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}$ and $\frac{{\partial {{\textbf E}_{z,\Im}}}}{{\partial {{{\bar {\tilde \xi}}}_k}}}$, the two parenthesis must equal zero,
$$\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Re}}}} + {\lambda ^{T}}{\textbf S} + {\lambda ^\dagger}{{\textbf S}^*} = 0,\quad \frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Im}}}} + {\rm i}{\lambda ^{T}}{\textbf S} + {\rm i}{\lambda ^\dagger}{{\textbf S}^*} = 0,$$
multiplying the second equation by i, subtracting it from the first, and transposing it yields
$$\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Re}}}} - {\rm i}\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Im}}}} + 2{\lambda ^{T}}{\textbf S} = 0\Leftrightarrow{{\textbf S}^{T}}\lambda = - \frac{1}{2}{\left({\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Re}}}} - {\rm i}\frac{{\partial \Phi}}{{\partial {{\textbf E}_{z,\Im}}}}} \right)^{T}}.$$
Using Eq. (A5), the expression in Eq. (A3) reduces to the expression for $\frac{{{d}\Phi}}{{{d}{{{\bar {\tilde \xi}}}_k}}}$ in Eq. (8), and the second equation in Eq. (A5) is equal to Eq. (9).
## Funding
Danmarks Grundforskningsfond (DNRF147); Villum Fonden (8692).
## Disclosures
The authors declare that there are no conflicts of interest related to this paper.
## REFERENCES
1. M. P. Bendsøe and O. Sigmund, Topology Optimization (Springer, 2003).
2. M. P. Bendsøe and N. Kikuchi, “Generating optimal topologies in structural design using a homogenization method,” Comput. Methods Appl. Mech. Eng. 71, 197–224 (1988). [CrossRef]
3. J. Alexandersen and C. S. Andreasen, “A review of topology optimisation for fluid-based problems,” Fluids 5, 29 (2020). [CrossRef]
4. C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019). [CrossRef]
5. C. Lundgaard, K. Engelbrecht, and O. Sigmund, “A density-based topology optimization methodology for thermal energy storage systems,” Struct. Multidiscip. Optim. 60, 2189–2204 (2019). [CrossRef]
6. J. S. Jensen and O. Sigmund, “Topology optimization for nano-photonics,” Laser Photon. Rev. 5, 308–321 (2011). [CrossRef]
7. S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018). [CrossRef]
8. X. Liang and S. G. Johnson, “Formulation for scalable optimization of microcavities via the frequency-averaged local density of states,” Opt. Express 21, 30812–30841 (2013). [CrossRef]
9. F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018). [CrossRef]
10. A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015). [CrossRef]
11. Z. Lin, V. Liu, R. Pestourie, and S. G. Johnson, “Topology optimization of freeform large-area metasurfaces,” Opt. Express 27, 15765–15775 (2019). [CrossRef]
12. H. Chung and O. D. Miller, “High-NA achromatic metalenses by inverse design,” Opt. Express 28, 6945–6965 (2020). [CrossRef]
13. R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019). [CrossRef]
14. O. Sigmund, “A 99 line topology optimization code written in Matlab,” Struct. Multidiscip. Optim. 21, 120–127 (2001). [CrossRef]
15. E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011). [CrossRef]
16. F. Ferrari and O. Sigmund, “A new generation 99 line Matlab code for compliance topology optimization and its extension to 3D,” Struct. Multidiscip. Optim. 62, 2211–2228 (2020). [CrossRef]
17. R. E. Christiansen and O. Sigmund, “Inverse design in photonics by topology optimization: tutorial,” J. Opt. Soc. Am. B 38, 496–509 (2021). [CrossRef]
18. J.-P. Berenger, “A perfectly matched layer for the absorption of electromagnetic waves,” J. Comput. Phys. 114, 185–200 (1994). [CrossRef]
19. J. S. Jensen and O. Sigmund, “Topology optimization of photonic crystal structures: a high-bandwidth low-loss T-junction waveguide,” J. Opt. Soc. Am. B 22, 1191–1198 (2005). [CrossRef]
20. J. Jin, The Finite Element Method in Electromagnetics (Wiley-Interscience, 2013).
21. J. K. Guest, J. H. Prevost, and T. Belytschko, “Achieving minimum length scale in topology optimization using nodal design variables and projection functions,” Int. J. Numer. Methods Eng. 61, 238–254 (2004). [CrossRef]
22. F. Wang, B. S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Struct. Multidiscip. Optim. 43, 767–784 (2011). [CrossRef]
23. R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015). [CrossRef]
24. D. A. Tortorelli and P. Michaleris, “Design sensitivity analysis: overview and review,” Inverse Prob. Eng. 1, 71–105 (1994). [CrossRef]
25. M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008). [CrossRef]
26. R. E. Christiansen and O. Sigmund, “A 200 line topology optimization code for electromagnetism,” Code 1, figshare (2020), https://doi.org/10.6084/m9.figshare.13369355.
27. B. S. Lazarov, F. Wang, and O. Sigmund, “Length scale and manufacturability in density-based topology optimization,” Arch. Appl. Mech. 86, 189–218 (2016). [CrossRef]
28. R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019). [CrossRef]
29. J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020). [CrossRef]
30. Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020). [CrossRef]
31. R. E. Christiansen, J. Michon, M. Benzaouia, O. Sigmund, and S. G. Johnson, “Inverse design of nanoparticles for enhanced Raman scattering,” Opt. Express 28, 4444–4462 (2020). [CrossRef]
32. D. E. Goldberg, Genetic Algorithms in Search, Optimization and Learning (Addison, 1989).
33. N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017). [CrossRef]
34. O. Sigmund, “On the usefulness of non-gradient approaches in topology optimization,” Struct. Multidiscip. Optim. 43, 589–596 (2011). [CrossRef]
### References
• View by:
1. M. P. Bendsøe and O. Sigmund, Topology Optimization (Springer, 2003).
2. M. P. Bendsøe and N. Kikuchi, “Generating optimal topologies in structural design using a homogenization method,” Comput. Methods Appl. Mech. Eng. 71, 197–224 (1988).
[Crossref]
3. J. Alexandersen and C. S. Andreasen, “A review of topology optimisation for fluid-based problems,” Fluids 5, 29 (2020).
[Crossref]
4. C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019).
[Crossref]
5. C. Lundgaard, K. Engelbrecht, and O. Sigmund, “A density-based topology optimization methodology for thermal energy storage systems,” Struct. Multidiscip. Optim. 60, 2189–2204 (2019).
[Crossref]
6. J. S. Jensen and O. Sigmund, “Topology optimization for nano-photonics,” Laser Photon. Rev. 5, 308–321 (2011).
[Crossref]
7. S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
8. X. Liang and S. G. Johnson, “Formulation for scalable optimization of microcavities via the frequency-averaged local density of states,” Opt. Express 21, 30812–30841 (2013).
[Crossref]
9. F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
10. A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
11. Z. Lin, V. Liu, R. Pestourie, and S. G. Johnson, “Topology optimization of freeform large-area metasurfaces,” Opt. Express 27, 15765–15775 (2019).
[Crossref]
12. H. Chung and O. D. Miller, “High-NA achromatic metalenses by inverse design,” Opt. Express 28, 6945–6965 (2020).
[Crossref]
13. R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019).
[Crossref]
14. O. Sigmund, “A 99 line topology optimization code written in Matlab,” Struct. Multidiscip. Optim. 21, 120–127 (2001).
[Crossref]
15. E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
16. F. Ferrari and O. Sigmund, “A new generation 99 line Matlab code for compliance topology optimization and its extension to 3D,” Struct. Multidiscip. Optim. 62, 2211–2228 (2020).
[Crossref]
17. R. E. Christiansen and O. Sigmund, “Inverse design in photonics by topology optimization: tutorial,” J. Opt. Soc. Am. B 38, 496–509 (2021).
[Crossref]
18. J.-P. Berenger, “A perfectly matched layer for the absorption of electromagnetic waves,” J. Comput. Phys. 114, 185–200 (1994).
[Crossref]
19. J. S. Jensen and O. Sigmund, “Topology optimization of photonic crystal structures: a high-bandwidth low-loss T-junction waveguide,” J. Opt. Soc. Am. B 22, 1191–1198 (2005).
[Crossref]
20. J. Jin, The Finite Element Method in Electromagnetics (Wiley-Interscience, 2013).
21. J. K. Guest, J. H. Prevost, and T. Belytschko, “Achieving minimum length scale in topology optimization using nodal design variables and projection functions,” Int. J. Numer. Methods Eng. 61, 238–254 (2004).
[Crossref]
22. F. Wang, B. S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Struct. Multidiscip. Optim. 43, 767–784 (2011).
[Crossref]
23. R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015).
[Crossref]
24. D. A. Tortorelli and P. Michaleris, “Design sensitivity analysis: overview and review,” Inverse Prob. Eng. 1, 71–105 (1994).
[Crossref]
25. M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008).
[Crossref]
26. R. E. Christiansen and O. Sigmund, “A 200 line topology optimization code for electromagnetism,” Code 1, figshare (2020), https://doi.org/10.6084/m9.figshare.13369355 .
27. B. S. Lazarov, F. Wang, and O. Sigmund, “Length scale and manufacturability in density-based topology optimization,” Arch. Appl. Mech. 86, 189–218 (2016).
[Crossref]
28. R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019).
[Crossref]
29. J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
30. Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020).
[Crossref]
31. R. E. Christiansen, J. Michon, M. Benzaouia, O. Sigmund, and S. G. Johnson, “Inverse design of nanoparticles for enhanced Raman scattering,” Opt. Express 28, 4444–4462 (2020).
[Crossref]
32. D. E. Goldberg, Genetic Algorithms in Search, Optimization and Learning (Addison, 1989).
33. N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017).
[Crossref]
34. O. Sigmund, “On the usefulness of non-gradient approaches in topology optimization,” Struct. Multidiscip. Optim. 43, 589–596 (2011).
[Crossref]
#### 2020 (6)
F. Ferrari and O. Sigmund, “A new generation 99 line Matlab code for compliance topology optimization and its extension to 3D,” Struct. Multidiscip. Optim. 62, 2211–2228 (2020).
[Crossref]
J. Alexandersen and C. S. Andreasen, “A review of topology optimisation for fluid-based problems,” Fluids 5, 29 (2020).
[Crossref]
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020).
[Crossref]
#### 2019 (5)
R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019).
[Crossref]
C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019).
[Crossref]
C. Lundgaard, K. Engelbrecht, and O. Sigmund, “A density-based topology optimization methodology for thermal energy storage systems,” Struct. Multidiscip. Optim. 60, 2189–2204 (2019).
[Crossref]
R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019).
[Crossref]
#### 2018 (2)
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
#### 2017 (1)
N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017).
[Crossref]
#### 2016 (1)
B. S. Lazarov, F. Wang, and O. Sigmund, “Length scale and manufacturability in density-based topology optimization,” Arch. Appl. Mech. 86, 189–218 (2016).
[Crossref]
#### 2015 (2)
R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015).
[Crossref]
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### 2011 (4)
J. S. Jensen and O. Sigmund, “Topology optimization for nano-photonics,” Laser Photon. Rev. 5, 308–321 (2011).
[Crossref]
E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
F. Wang, B. S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Struct. Multidiscip. Optim. 43, 767–784 (2011).
[Crossref]
O. Sigmund, “On the usefulness of non-gradient approaches in topology optimization,” Struct. Multidiscip. Optim. 43, 589–596 (2011).
[Crossref]
#### 2008 (1)
M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008).
[Crossref]
M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008).
[Crossref]
#### 2004 (1)
J. K. Guest, J. H. Prevost, and T. Belytschko, “Achieving minimum length scale in topology optimization using nodal design variables and projection functions,” Int. J. Numer. Methods Eng. 61, 238–254 (2004).
[Crossref]
#### 2001 (1)
O. Sigmund, “A 99 line topology optimization code written in Matlab,” Struct. Multidiscip. Optim. 21, 120–127 (2001).
[Crossref]
#### 1994 (2)
J.-P. Berenger, “A perfectly matched layer for the absorption of electromagnetic waves,” J. Comput. Phys. 114, 185–200 (1994).
[Crossref]
D. A. Tortorelli and P. Michaleris, “Design sensitivity analysis: overview and review,” Inverse Prob. Eng. 1, 71–105 (1994).
[Crossref]
#### 1988 (1)
M. P. Bendsøe and N. Kikuchi, “Generating optimal topologies in structural design using a homogenization method,” Comput. Methods Appl. Mech. Eng. 71, 197–224 (1988).
[Crossref]
#### Aage, N.
C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019).
[Crossref]
N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017).
[Crossref]
#### Alexandersen, J.
J. Alexandersen and C. S. Andreasen, “A review of topology optimisation for fluid-based problems,” Fluids 5, 29 (2020).
[Crossref]
#### Andreasen, C. S.
J. Alexandersen and C. S. Andreasen, “A review of topology optimisation for fluid-based problems,” Fluids 5, 29 (2020).
[Crossref]
#### Andreassen, E.
N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017).
[Crossref]
E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
#### Babinec, T. M.
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### Balling, P.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
#### Belytschko, T.
J. K. Guest, J. H. Prevost, and T. Belytschko, “Achieving minimum length scale in topology optimization using nodal design variables and projection functions,” Int. J. Numer. Methods Eng. 61, 238–254 (2004).
[Crossref]
#### Bendsøe, M. P.
M. P. Bendsøe and N. Kikuchi, “Generating optimal topologies in structural design using a homogenization method,” Comput. Methods Appl. Mech. Eng. 71, 197–224 (1988).
[Crossref]
M. P. Bendsøe and O. Sigmund, Topology Optimization (Springer, 2003).
#### Berenger, J.-P.
J.-P. Berenger, “A perfectly matched layer for the absorption of electromagnetic waves,” J. Comput. Phys. 114, 185–200 (1994).
[Crossref]
#### Boltasseva, A.
Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020).
[Crossref]
#### Christiansen, J.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
#### Christiansen, R. E.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019).
[Crossref]
R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019).
[Crossref]
F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015).
[Crossref]
#### Clausen, A.
E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
#### Dilgen, C. B.
C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019).
[Crossref]
#### Dilgen, S. B.
C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019).
[Crossref]
#### Dühring, M. B.
M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008).
[Crossref]
#### Engelbrecht, K.
C. Lundgaard, K. Engelbrecht, and O. Sigmund, “A density-based topology optimization methodology for thermal energy storage systems,” Struct. Multidiscip. Optim. 60, 2189–2204 (2019).
[Crossref]
#### Ferrari, F.
F. Ferrari and O. Sigmund, “A new generation 99 line Matlab code for compliance topology optimization and its extension to 3D,” Struct. Multidiscip. Optim. 62, 2211–2228 (2020).
[Crossref]
#### Goldberg, D. E.
D. E. Goldberg, Genetic Algorithms in Search, Optimization and Learning (Addison, 1989).
#### Guest, J. K.
J. K. Guest, J. H. Prevost, and T. Belytschko, “Achieving minimum length scale in topology optimization using nodal design variables and projection functions,” Int. J. Numer. Methods Eng. 61, 238–254 (2004).
[Crossref]
#### Jensen, J. S.
C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019).
[Crossref]
R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015).
[Crossref]
J. S. Jensen and O. Sigmund, “Topology optimization for nano-photonics,” Laser Photon. Rev. 5, 308–321 (2011).
[Crossref]
M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008).
[Crossref]
#### Jin, J.
J. Jin, The Finite Element Method in Electromagnetics (Wiley-Interscience, 2013).
#### Jin, W.
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
#### Julsgaard, B.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
#### Kikuchi, N.
M. P. Bendsøe and N. Kikuchi, “Generating optimal topologies in structural design using a homogenization method,” Comput. Methods Appl. Mech. Eng. 71, 197–224 (1988).
[Crossref]
#### Kildishev, A. V.
Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020).
[Crossref]
#### Kudyshev, Z. A.
Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020).
[Crossref]
#### Lagoudakis, K. G.
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### Lazarov, B. S.
N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017).
[Crossref]
B. S. Lazarov, F. Wang, and O. Sigmund, “Length scale and manufacturability in density-based topology optimization,” Arch. Appl. Mech. 86, 189–218 (2016).
[Crossref]
R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015).
[Crossref]
F. Wang, B. S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Struct. Multidiscip. Optim. 43, 767–784 (2011).
[Crossref]
E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
#### Lin, Z.
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
#### Lu, J.
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### Lundgaard, C.
C. Lundgaard, K. Engelbrecht, and O. Sigmund, “A density-based topology optimization methodology for thermal energy storage systems,” Struct. Multidiscip. Optim. 60, 2189–2204 (2019).
[Crossref]
#### Madsen, S. P.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019).
[Crossref]
#### Michaleris, P.
D. A. Tortorelli and P. Michaleris, “Design sensitivity analysis: overview and review,” Inverse Prob. Eng. 1, 71–105 (1994).
[Crossref]
#### Molesky, S.
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
#### Møller, S. H.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
#### Mørk, J.
F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
#### Petykiewicz, J.
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### Piggott, A. Y.
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### Prevost, J. H.
J. K. Guest, J. H. Prevost, and T. Belytschko, “Achieving minimum length scale in topology optimization using nodal design variables and projection functions,” Int. J. Numer. Methods Eng. 61, 238–254 (2004).
[Crossref]
#### Rodriguez, A. W.
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
#### Roesgaard, S.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
#### Schevenels, M.
E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
#### Shalaev, V. M.
Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020).
[Crossref]
#### Sigmund, O.
F. Ferrari and O. Sigmund, “A new generation 99 line Matlab code for compliance topology optimization and its extension to 3D,” Struct. Multidiscip. Optim. 62, 2211–2228 (2020).
[Crossref]
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019).
[Crossref]
R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019).
[Crossref]
C. Lundgaard, K. Engelbrecht, and O. Sigmund, “A density-based topology optimization methodology for thermal energy storage systems,” Struct. Multidiscip. Optim. 60, 2189–2204 (2019).
[Crossref]
F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017).
[Crossref]
B. S. Lazarov, F. Wang, and O. Sigmund, “Length scale and manufacturability in density-based topology optimization,” Arch. Appl. Mech. 86, 189–218 (2016).
[Crossref]
R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015).
[Crossref]
F. Wang, B. S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Struct. Multidiscip. Optim. 43, 767–784 (2011).
[Crossref]
E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
J. S. Jensen and O. Sigmund, “Topology optimization for nano-photonics,” Laser Photon. Rev. 5, 308–321 (2011).
[Crossref]
O. Sigmund, “On the usefulness of non-gradient approaches in topology optimization,” Struct. Multidiscip. Optim. 43, 589–596 (2011).
[Crossref]
M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008).
[Crossref]
O. Sigmund, “A 99 line topology optimization code written in Matlab,” Struct. Multidiscip. Optim. 21, 120–127 (2001).
[Crossref]
M. P. Bendsøe and O. Sigmund, Topology Optimization (Springer, 2003).
#### Stobbe, S.
R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019).
[Crossref]
#### Tortorelli, D. A.
D. A. Tortorelli and P. Michaleris, “Design sensitivity analysis: overview and review,” Inverse Prob. Eng. 1, 71–105 (1994).
[Crossref]
#### Vester-Petersen, J.
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019).
[Crossref]
#### Vuckovic, J.
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### Wang, F.
R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019).
[Crossref]
F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
B. S. Lazarov, F. Wang, and O. Sigmund, “Length scale and manufacturability in density-based topology optimization,” Arch. Appl. Mech. 86, 189–218 (2016).
[Crossref]
F. Wang, B. S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Struct. Multidiscip. Optim. 43, 767–784 (2011).
[Crossref]
#### Yu, Y.
F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
#### Appl. Phys. Lett. (1)
F. Wang, R. E. Christiansen, Y. Yu, J. Mørk, and O. Sigmund, “Maximizing the quality factor to mode volume ratio for ultra-small photonic crystal cavities,” Appl. Phys. Lett. 113, 241101 (2018).
[Crossref]
#### Appl. Phys. Rev. (1)
Z. A. Kudyshev, A. V. Kildishev, V. M. Shalaev, and A. Boltasseva, “Machine-learning-assisted metasurface design for high-efficiency thermal emitter optimization,” Appl. Phys. Rev. 7, 021407 (2020).
[Crossref]
#### Arch. Appl. Mech. (1)
B. S. Lazarov, F. Wang, and O. Sigmund, “Length scale and manufacturability in density-based topology optimization,” Arch. Appl. Mech. 86, 189–218 (2016).
[Crossref]
#### Comput. Methods Appl. Mech. Eng. (2)
R. E. Christiansen, J. Vester-Petersen, S. P. Madsen, and O. Sigmund, “A non-linear material interpolation for design of metallic nano-particles using topology optimization,” Comput. Methods Appl. Mech. Eng. 343, 23–39 (2019).
[Crossref]
M. P. Bendsøe and N. Kikuchi, “Generating optimal topologies in structural design using a homogenization method,” Comput. Methods Appl. Mech. Eng. 71, 197–224 (1988).
[Crossref]
#### Fluids (1)
J. Alexandersen and C. S. Andreasen, “A review of topology optimisation for fluid-based problems,” Fluids 5, 29 (2020).
[Crossref]
#### Int. J. Numer. Methods Eng. (1)
J. K. Guest, J. H. Prevost, and T. Belytschko, “Achieving minimum length scale in topology optimization using nodal design variables and projection functions,” Int. J. Numer. Methods Eng. 61, 238–254 (2004).
[Crossref]
#### Inverse Prob. Eng. (1)
D. A. Tortorelli and P. Michaleris, “Design sensitivity analysis: overview and review,” Inverse Prob. Eng. 1, 71–105 (1994).
[Crossref]
#### J. Comput. Phys. (1)
J.-P. Berenger, “A perfectly matched layer for the absorption of electromagnetic waves,” J. Comput. Phys. 114, 185–200 (1994).
[Crossref]
#### J. Sound Vib. (1)
M. B. Dühring and J. S. Jensen, and O. Sigmund, “Acoustic design by topology optimization,” J. Sound Vib. 317, 557–575 (2008).
[Crossref]
#### Laser Photon. Rev. (1)
J. S. Jensen and O. Sigmund, “Topology optimization for nano-photonics,” Laser Photon. Rev. 5, 308–321 (2011).
[Crossref]
#### Nanophotonics (1)
R. E. Christiansen, F. Wang, O. Sigmund, and S. Stobbe, “Designing photonic topological insulators with quantum-spin-Hall edge states using topology optimization,” Nanophotonics 8, 1363–1369 (2019).
[Crossref]
#### Nat. Photonics (2)
S. Molesky, Z. Lin, A. Y. Piggott, W. Jin, J. Vuckovic, and A. W. Rodriguez, “Inverse design in nanophotonics,” Nat. Photonics 12, 659–670 (2018).
[Crossref]
A. Y. Piggott, J. Lu, K. G. Lagoudakis, J. Petykiewicz, T. M. Babinec, and J. Vučković, “Inverse design and demonstration of a compact and broadband on-chip wavelength demultiplexer,” Nat. Photonics 9, 374–377 (2015).
[Crossref]
#### Nature (1)
N. Aage, E. Andreassen, B. S. Lazarov, and O. Sigmund, “Giga-voxel computational morphogenesis for structural design,” Nature 550, 84–86 (2017).
[Crossref]
#### Sol. Energy Mater. Sol. Cells (1)
J. Christiansen, J. Vester-Petersen, S. Roesgaard, S. H. Møller, R. E. Christiansen, O. Sigmund, S. P. Madsen, P. Balling, and B. Julsgaard, “Strongly enhanced upconversion in trivalent erbium ions by tailored gold nanostructures: toward high-efficient silicon-based photovoltaics,” Sol. Energy Mater. Sol. Cells 208, 110406 (2020).
[Crossref]
#### Struct. Multidiscip. Optim. (8)
F. Wang, B. S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Struct. Multidiscip. Optim. 43, 767–784 (2011).
[Crossref]
R. E. Christiansen, B. S. Lazarov, J. S. Jensen, and O. Sigmund, “Creating geometrically robust designs for highly sensitive problems using topology optimization—acoustic cavity design,” Struct. Multidiscip. Optim. 52, 737–754 (2015).
[Crossref]
O. Sigmund, “On the usefulness of non-gradient approaches in topology optimization,” Struct. Multidiscip. Optim. 43, 589–596 (2011).
[Crossref]
O. Sigmund, “A 99 line topology optimization code written in Matlab,” Struct. Multidiscip. Optim. 21, 120–127 (2001).
[Crossref]
E. Andreassen, A. Clausen, M. Schevenels, B. S. Lazarov, and O. Sigmund, “Efficient topology optimization in MATLAB using 88 lines of code,” Struct. Multidiscip. Optim. 43, 1–16 (2011).
[Crossref]
F. Ferrari and O. Sigmund, “A new generation 99 line Matlab code for compliance topology optimization and its extension to 3D,” Struct. Multidiscip. Optim. 62, 2211–2228 (2020).
[Crossref]
C. B. Dilgen, S. B. Dilgen, N. Aage, and J. S. Jensen, “Topology optimization of acoustic mechanical interaction problems: a comparative review,” Struct. Multidiscip. Optim. 60, 779–801 (2019).
[Crossref]
C. Lundgaard, K. Engelbrecht, and O. Sigmund, “A density-based topology optimization methodology for thermal energy storage systems,” Struct. Multidiscip. Optim. 60, 2189–2204 (2019).
[Crossref]
#### Other (4)
M. P. Bendsøe and O. Sigmund, Topology Optimization (Springer, 2003).
D. E. Goldberg, Genetic Algorithms in Search, Optimization and Learning (Addison, 1989).
J. Jin, The Finite Element Method in Electromagnetics (Wiley-Interscience, 2013).
R. E. Christiansen and O. Sigmund, “A 200 line topology optimization code for electromagnetism,” Code 1, figshare (2020), https://doi.org/10.6084/m9.figshare.13369355 .
### Supplementary Material (1)
NameDescription
Code 1 Five COMSOL models for reproducing the results in the article
### Cited By
Optica participates in Crossref's Cited-By Linking service. Citing articles from Optica Publishing Group journals and other participating publishers are listed here.
### Figures (6)
Fig. 1. Model domain, $\Omega$ , of height ${h_\Omega}$ and width ${w_\Omega}$ with a designable region, ${\Omega _D}$ , of height ${h_{{\Omega _D}}}$ and width ${w_{{\Omega _D}}}$ on top of a substrate, ${\Omega _S}$ , of height ${h_s}$ .
Fig. 2. (a) Max-normalized $|{\textbf E}{|^2}$ -field in $\Omega$ . (b) Metalens design, ${\varepsilon _r} = 3.0$ (black) and ${\varepsilon _r} = 1.0$ (white).
Fig. 3. Metalens designs obtained (a) without filtering ( ${r_f} = 1$ ), (b) using a filter radius of ${r_f} = 3.0$ , (c) using a filter radius of ${r_f} = 6.0$ , (d) using a filter radius of ${r_f} = 9.0$ . These results illustrate the effect of applying the cone-shaped filter to the design field as part of the optimization process.
Fig. 4. (a) Max-normalized $|{\textbf E}{|^2}$ -field in $\Omega$ . (b) Metallic reflector design (black) in air background (white).
Fig. 5. (a) Max-normalized $|{\textbf E}{|^2}$ -field in $\Omega$ . (b) Metalens design restricted to one-dimensional variations, ${\varepsilon _r} = 3.0$ (black) and ${\varepsilon _r} = 1.0$ (white).
Fig. 6. (a) Design obtained using the GA-based method. (b) Design obtained using the gradient-based method. ${\varepsilon _r} = 3.0$ (black) and ${\varepsilon _r} = 1.0$ (white). (c) Convergence graph. Remark that each design iteration for the GA-based code requires 200 solutions of the physics model equation, while it requires one solution of the physics model equation and one solution of the adjoint equation for the gradient-based code!
### Equations (16)
$∇ ⋅ ( ∇ E z ( r ) ) + k 2 ε r ( r ) E z ( r ) = F ( r ) , r ∈ Ω ∈ R 2 ,$
$n ⋅ ∇ E z ( r ) = − i k E z ( r ) , r ∈ Γ ,$
$ε r ( ξ ( r ) ) = 1 + ξ ( r ) ( ε r , m − 1 ) − i α ξ ( r ) ( 1 − ξ ( r ) ) , r ∈ Ω ,$
$Φ ( ξ ( r ) , r p ) = | E z ( ξ ( r ) , r p ) | 2 = E z ( ξ ( r ) , r p ) ∗ E z ( ξ ( r ) , r p ) ,$
$max ξ : Φ = E z † P E z , s . t . : S ( ε r ) E z = ( ∑ e = 1 N e S e ( ε r , e ) ) E z = F , : ε r , j = 1 + ξ ~ ¯ j ( ε r , m − 1 ) − i ξ ~ ¯ j ( 1 − ξ ~ ¯ j ) ∀ j ∈ { 1 , 2 , … , N e } , : 0 < ξ j < 1 ∀ j ∈ { 1 , 2 , … , N D } , : ξ = 0 ∀ r ∈ Ω / { Ω D , Ω S } ∨ ξ = 1 ∀ r ∈ Ω S ,$
$ξ ~ h = ∑ k ∈ B e , h w ( r h − r k ) A k ξ k ∑ k ∈ B e , h w ( r h − r k ) A k , w ( r ) = { r f − | r | ∀ | r | ≤ r f 0 , r f ≥ 0 , r ∈ Ω .$
$ξ ~ ¯ h = H ( ξ ~ h ) = tanh ( β ⋅ η ) + tanh ( β ⋅ ( ξ ~ h − η ) ) tanh ( β ⋅ η ) + tanh ( β ⋅ ( 1 − η ) ) , β ∈ [ 1 , ∞ [ , η ∈ [ 0 , 1 ] .$
$d Φ d ξ h = ∑ k ∈ B e , h ∂ ξ ~ k ∂ ξ h ∂ ξ ~ ¯ k ∂ ξ ~ k d Φ d ξ ~ ¯ k , d Φ d ξ ~ ¯ k = 2 ℜ ( λ T ∂ S ∂ ξ ~ ¯ k E z ) ,$
$S T λ = − 1 2 ( ∂ Φ ∂ E z , ℜ − i ∂ Φ ∂ E z , ℑ ) T w i t h E z = E z , ℜ + i E z , ℑ ,$
$( ∂ Φ ∂ E z , ℜ − i ∂ Φ ∂ E z , ℑ ) m T = P m , m ( 2 ( E z , ℜ ) m − 2 i ( E z , ℑ ) m ) .$
$ε ( x ) = ( n ( x ) 2 − κ ( x ) 2 ) − i ( 2 n ( x ) κ ( x ) ) , n ( x ) = n M 1 + x ( n M 2 − n M 1 ) , κ ( x ) = κ M 1 + x ( κ M 2 − κ M 1 ) .$
$Φ ~ = Φ + λ T ( S E z − F ) + λ † ( S ∗ E z ∗ − F ∗ ) ,$
$d Φ ~ d ξ ~ ¯ k = ∂ Φ ∂ E z , ℜ ∂ E z , ℜ ∂ ξ ~ ¯ k + ∂ Φ ∂ E z , ℑ ∂ E z , ℑ ∂ ξ ~ ¯ k + λ T ( ∂ S d ξ ~ ¯ k E z + S ( ∂ E z , ℜ ∂ ξ ~ ¯ k + i ∂ E z , ℑ ∂ ξ ~ ¯ k ) ) + λ † ( ∂ S ∗ ∂ ξ ~ ¯ k E z ∗ + S ∗ ( ∂ E z , ℜ ∂ ξ ~ ¯ k − i ∂ E z , ℑ ∂ ξ ~ ¯ k ) ) .$
$d Φ ~ d ξ ~ ¯ k = ∂ E z , ℜ ∂ ξ ~ ¯ k ( ∂ Φ ∂ E z , ℜ + λ T S + λ † S ∗ ) + ∂ E z , ℑ ∂ ξ ~ ¯ k ( ∂ Φ ∂ E z , ℑ + i λ T S + i λ † S ∗ ) + 2 ℜ ( λ T ∂ S ∂ ξ ~ ¯ k E z ) .$
$∂ Φ ∂ E z , ℜ + λ T S + λ † S ∗ = 0 , ∂ Φ ∂ E z , ℑ + i λ T S + i λ † S ∗ = 0 ,$
$∂ Φ ∂ E z , ℜ − i ∂ Φ ∂ E z , ℑ + 2 λ T S = 0 ⇔ S T λ = − 1 2 ( ∂ Φ ∂ E z , ℜ − i ∂ Φ ∂ E z , ℑ ) T .$
### Metrics
Select as filters
Cancel | 27,384 | 88,183 | {"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": 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} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.842671 |
https://id.scribd.com/document/91142934/Cutting-Data-Corrax-Eng | 1,571,645,728,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00034.warc.gz | 531,740,608 | 69,474 | Anda di halaman 1dari 5
# CUTTING DATA RECOMMENDATIONS
Uddeholm Corrax
Machining data are always dependent on the actual operation, the machine tool and the cutting data used. The machining data given is this datasheet are general guidelines that may have to be adjusted to the actual conditions of a specific machining operation.
Turning
Cutting speed , v c =
## Cutting data formulae
Dn
1000
Legend
vc n f ap D
3
(m / min)
= Cutting speed (m/min) = Spindle speed (rev/min) = Feed per rev (mm/rev) = Axial depth of cut (mm) = Workpiece diameter (mm) = Material removal rate (cm 3/min) = Surface roughness (m) = Nose radius (mm)
(rev / min)
## Material removal rate, Q = vc ap f
f 2 50 Surface roughness, Ra r
(cm / min)
( m)
Q Ra re
Milling
vc =
n =
Dn
1000
Legend
( m /min )
vc n vf ap ae f z fz D hm Q
= Cutting speed (m/min) = Spindle speed (rev/min) = Feed speed (mm/min) = Axial depth of cut (mm) = Radial depth of cut (mm) = Feed per rev (mm/rev) = Number of teeth = Feed per tooth (mm/tooth) = Cutter diameter (mm) = Average chip thickness (mm) = Material removal rate (cm 3/min)
1000 vc ( rev/min) D
vf = fz z n= f n(mm/min )
hm = fz
Q=
ae ae ( mm ) < 0 ,3 D D
## ap ae vf (cm 3 / min ) 1000
Drilling
Cutting speed , v c = Spindle speed , n =
Dn
1000
## ( m / min) ( rev / min)
Legend
vc n vf D f = Cutting speed (m/min) = Spindle speed (rev/min) = Feed speed (mm/min) = Drill diameter (mm) = Feed per rev (mm/rev)
1000 vc D
Feed speed , vf = f n
( mm / min)
## Feed per rev , f =
vf n
(mm / rev )
Turning
Turning
Cemented carbide Roughing Finishing 110-160 160-210 0,2-0,4 0,05-0,2 2-4 0,5-2 P20-P30 coated carbide P10 coated carbide or cermet Cutting speed, vc (m/min) Feed, f (mm/rev) Depth of cut, a p (mm) Suitable grades
Uddeholm Corrax
HSS 13-18 0,05-0,3 0,5-3
Remarks:
1. Cutting fluid is recommended. 2. For turning with interrupted cut or face turning of large workpieces use a thougher cemented carbide grade.
Face milling
Face milling
Cemented carbide Roughing Finishing 70-90 90-110 0,2-0,4 0,1-0,2 2-5 -2 P20-P40 coated carbide P10-P20 coated carbide or cermet Cutting speed, vc (m/min) Feed, fz (mm/tooth) Depth of cut, a p (mm) Suitable grades
Remarks:
1. Use a milling cutter with a positive-negative or positive-positive geometry. 2. Climb milling should generally be used. 3. Milling should generally be done with coolant.
## Square shoulder milling
Square shoulder milling with cemented carbide
ae = 0.1 x D 90-130 0,25-0,3 ae = 0.5 x D 80-120 0,15-0,2 P15-P40 coated carbide ae = 1 x D 70-110 0,1-0,15 Cutting speed, vc (m/min) Feed, fz (mm/tooth) Suitable grades
Remarks:
1. Climb milling should generally be used. 2. Choose the cutter diameter (D) and the radial depth of cut (a e) so that at least two cutting edges are engaged simultaneously. 3. If the machine tool power is inadequate for the data given reduce the depth of cut, but do not reduce the feed.
2007-12-04
End milling
Slot milling Axial depth of cut, a p = 1 x D
3-5 Uncoated HSS Coated HSS
1-4)
Uddeholm Corrax
Cutter diameter (mm) 10 - 20 20 - 30 20-25 0,03-0,04 0,04-0,05 0,05-0,06 5 - 10 0,04-0,05 60-100 0,01-0,02 35-45 0,05-0,06 0,02-0,04 0,06-0,07 30 - 40 0,06-0,09 0,07-0,10
Cutting speed, vc (m/min) Feed, fz (mm/tooth) Cutting speed, vc (m/min) Feed, fz (mm/tooth) Cutting speed, vc (m/min) Feed, fz (mm/tooth)
1-4)
6-8)
## Side milling Axial depth of cut, a p = 1.5 x D
70-110 0,06-0,08 0,08-0,10 0,10-0,12 P20-P30 coated carbide For side milling the same cutting speed as for slot milling can be used, but the feeds must be adjusted in order to obtain a suitable average chip thickness.
## Correction factor for side milling
Divide the cutter diameter with the radial depth of cut. See in the chart below which correction factor, C f, this corresponds to, and multiply the chosen feed in the table for slot milling with this factor. 5 Example: Tool: CC insert Cutter diameter: D = 40 mm 3 Radial depth of cut: ae = 2 mm D/ae = 40/2 = 20 Feed acc. to table slot milling = 0.11 mm/tooth Correction factor acc. to chart: C f = 2.8 0 10 20 30 40 50 Feed for side milling: f z = 2.8 x 0.11 = 0.31 mm/tooth
Correction factor, Cf
1
Cutter diameter / Radial depth of cut, D/ae
## Remarks: (slot and side milling)
1. 2. 3. 4. 5. Climb milling is generally recommended. Use a cutter with chipbreaker when side milling with radial depths of cut, a e > 0.3 xD. When side milling with small radial depths of cut (a e) the cutting speed can be increased by up to 15%. Use liberal amounts of cutting fluid. It is recommended to use a TiCN coated cutter when milling with solid cemented carbide tools. The axial depth of cut should not exceed the cutter diameter when slot milling. 6. Climb milling is generally recommended. 7. When side milling with small radial depths of cut (a e) the cutting speed can be increased by up to 30%. 8. The radial run-out, at the cutting edges, must be small and not exceed 0.03 mm.
2007-12-04
Drilling
Drilling
1-5 Uncoated HSS 1-2) Coated HSS
1-2)
Uddeholm Corrax
Cutting speed, vc (m/min) Feed, f (mm/rev) Cutting speed, vc (m/min) Feed, f (mm/rev) Cutting speed, vc (m/min) Feed, f (mm/rev) Cutting speed, vc (m/min) Feed, f (mm/rev) Cutting speed, vc (m/min) Feed, f (mm/rev) Drill diameter (mm) 5 - 10 10 - 20 20 - 30 30 - 40 13-15 0,10-0,20 0,20-0,30 0,30-0,35 0,35-0,40 13-15 0,10-0,20 0,20-0,30 0,30-0,35 0,35-0,40 180-200 0,03-0,08 0,08-0,12 100-130 0,08-0,10 0,10-0,20 0,15-0,25 0,20-0,30 50-70 0,25-0,35 0,30-0,35 0,35-0,40
0,05-0,10 0,05-0,10
Indexable insert
3-4)
(cem. carbide inserts) Solid cemented carbide 5-7) Brazed cemented carbide 5-7)
Remarks:
1. The cutting fluid should be ample and directed at the tool. 2. When drilling with short "NC drills" the feed may be increased by up to 20%. For extra long drills the feed must be decreased. 3. Use insert grades in the range of ISO P20-P30. Under unstable conditions a tougher carbide grade should be used for the centre position. 4. Use a high cutting fluid pressure and flow rate for a good chip removal. 5. If machining with solid or brazed cemented carbide drills, a rigid set-up and stable working conditions are required. 6. The use of drills with internal cooling channels is recommended. 7. Use a cutting fluid concentration of 15-20 %.
## Tapping with HSS
Cutting speed, vc = 7-9 m/min
Remarks:
1. Threading compound or cutting oil gives a longer tool life than emulsion.
2007-12-04 | 2,169 | 6,454 | {"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} | 2.578125 | 3 | CC-MAIN-2019-43 | latest | en | 0.645089 |
https://community.plm.automation.siemens.com/t5/NX-Design-Forum/Nx-12-Weight-Management/td-p/474008 | 1,566,296,730,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00432.warc.gz | 418,376,409 | 23,476 | Cancel
Showing results for
Did you mean:
# Nx 12 Weight Management
Siemens Valued Contributor
Quick Questions :
I am selecting a quick measure of weight under measure bodies , i select one body and get
`Weight = 33.22022386745 lbf`
What units are 1bf? (I just want to see lbs)
Where is the best place to change the density to another material like wood? (can i do this on the fly or must use customer defaults and log out and in again )
Can I set up a cell in a PMI Table to have the attribute of the weight, I have seen some info for drafting out there?
I assume if i have multi bodies I would just tally up the cells for total weight?
thanks
Barry Waldie
2 REPLIES 2
# Re: Nx 12 Weight Management
Siemens Phenom
Weight is a measurement of force, not mass. Therefore we adopted the units of Pounds-Force, or 'lbf'. This can be important later on if and when the 'weight' of an object is to be used in a formula where force is the intended parameter.
John R. Baker, P.E. (ret)
EX-Product 'Evangelist'
Irvine, CA
# Re: Nx 12 Weight Management
Siemens Esteemed Contributor
What units are 1bf? (I just want to see lbs)
lbs = weight.
lbf = pounds force (mass under gravity). Measure bodies will list Mass as lbm (pound mass) and Weight as lbf (pound force). Why are they the same value? Watch this.
The base unit for mass in the English system is the slug.
Where is the best place to change the density to another material like wood? (can i do this on the fly or must use customer defaults and log out and in again )
Either edit the solid density of your body: Menu > Edit > Feature > Solid Density
Or, assign a material to the body: Menu > Tools > Materials > Assign Materials
Can I set up a cell in a PMI Table to have the attribute of the weight, I have seen some info for drafting out there?
Yes, you can use the Relationships option to read the 'MassPropWeight' Part Attribute of a part file in a note or a table.
I assume if i have multi bodies I would just tally up the cells for total weight?
MassPropWeight stores the value for the entire part. You could use Measure Bodies and save the weight for each body as an expression and then reference that expression (again using Relationships) in a note or a table.
Regards, Ben
Ben Broad | PLM Enthusiast | Siemens GCSS
NX (v17 - 1876) | Teamcenter (9 - 12)
Value Based Licensing | Adaptive UI | BETA Registration | 606 | 2,429 | {"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} | 2.734375 | 3 | CC-MAIN-2019-35 | latest | en | 0.871193 |
https://technologyforlearners.com/gian-foy-on-the-art-of-mathematical-analysis-in-trading/ | 1,696,199,227,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510941.58/warc/CC-MAIN-20231001205332-20231001235332-00078.warc.gz | 615,444,706 | 46,040 | # Gian Foy on the Art of Mathematical Analysis in Trading
For many investors, the ability to delve into the complicated world of mathematical analysis can seem daunting. While it is challenging to master, data crunching has become an indispensable tool in trading and investment. Combining financial theories and quantitative analysis offers traders a greater insight into market movements and helps them identify new profit opportunities. In this post, expert Gian Foy in Scranton, PA, will examine how mathematicians analyze markets by exploring key concepts such as statistical arbitrage, machine learning algorithms, and portfolio optimization techniques.
## Define Mathematical Analysis and its Role in Trading
Mathematical analysis is an essential tool in trading, and it involves using statistical and mathematical methods to make informed investment decisions. Its role in trading cannot be overstated since it allows traders to collect and analyze financial data to predict potential price movements. The mathematical analysis applies the laws of probability to investigate market fluctuations, and it helps traders to identify opportunities in the market, improve their risk management, and increase their profits. The mathematical analysis provides a comprehensive overview of market trends by analyzing past and present data and helps traders better understand the underlying factors that drive economic events. Indeed, successful traders recognize the importance of mathematical analysis in trading and utilize it as a vital tool for making sound investment decisions.
## Explain the Types of Mathematics Used in Stock Trading
Gian Foy says various types of mathematics are utilized when it comes to stock trading. The most significant ones include calculus, statistics, and probability theory. Calculus plays a vital role in finance and economics as it computes limits and derivatives that estimate instantaneous rates of change. It helps traders analyze trends and establish optimal entry and exit points for market positions. Probability theory allows traders to quantify the likelihood of certain events occurring in the market, enabling them to make informed decisions on stock purchases and sales. While statistics come in handy when analyzing charts and data to determine correlations and trends, which can then be used to compile insights into effective trading strategies. By understanding these concepts and how they relate to stock trading, traders can better navigate the financial markets with precision and efficiency.
## Provide an Overview of the Different Techniques for Analyzing Stocks
Various techniques are used for analyzing stocks, each tailored specifically to individual investors’ unique needs and preferences. One popular approach is fundamental analysis, which seeks to evaluate a company’s potential based on its financial health, leadership, and industry trends. Alternatively, technical research focuses on price patterns and market trends, often relying on charts and graphs to track a stock’s performance over time. Another approach is quantitative analysis, which uses complex mathematical models and data to forecast future earnings. Lastly, many investors combine these approaches, using various tools to inform their investment strategies. Whether you’re focused on long-term growth or short-term profits, understanding these different approaches to stock analysis is critical to making informed and successful investment decisions.
## Explore How Mathematics is Used to Make Informed Decisions About Buying and Selling Stocks
Gian Foy says informed decision-making is crucial for success in the complex and ever-changing world of the stock market. Mathematics is vital in helping investors make smart choices about buying and selling stocks. From analyzing historical data to predicting future trends, mathematical models allow investors to assess the risks and rewards of different investment opportunities. Through mathematical analysis, investors can identify patterns and fluctuations in the market, ultimately increasing the likelihood of making profitable trades. Buying and selling stocks can be a daunting process, but with the help of math, investors can confidently approach the market.
## Analyze the Importance of Having a Solid Understanding of Basic Mathematical Principles and How it can be Applied to Trading
The world of trading is fast-paced and constantly evolving, and having a solid understanding of basic mathematical principles is crucial for success. From calculating profit margins to analyzing market trends, traders use math to make informed decisions that can mean the difference between profit and loss. By mastering core mathematical concepts such as fractions, percentages, and ratios, traders can quickly interpret data and identify patterns that may signal a good opportunity. Additionally, understanding probability and statistics principles can help traders accurately assess risk and make more informed investment decisions. In short, a foundation in math is essential for anyone looking to succeed in the dynamic world of trading.
## Discuss Strategies for Managing Risk When Using Mathematical Analysis in Stock Trading
Gian Foy says for those interested in stock trading, mathematical analysis can be a valuable tool for identifying potential investment opportunities. However, with any investment comes risk, and developing strategies for managing it is essential. When using mathematical analysis to inform your trading decisions, conducting thorough research and analysis is necessary, carefully considering all available data before making trades. Additionally, it’s crucial to establish straightforward entry and exit strategies, set stop losses, and keep a close eye on market trends to ensure you’re always making informed decisions. By prioritizing due diligence and utilizing thoughtful trading strategies, you can manage your risks effectively and increase your chances of success in the ever-changing stock market.
## Concluding thoughts…
As we have seen, mathematical analysis is an invaluable tool in stock trading. Through various mathematical theories and techniques, traders can gain an edge over their competitors and maximize their returns. However, it is essential to understand the fundamental principles behind numerical methods and risk management, as mistakes can be costly in this highly volatile market. By staying one step ahead through diligent research and practice with analysis, traders are more likely to succeed than those who do not take advantage of math-based techniques for understanding stocks. Ultimately, combining education, experience, and modern asset evaluation strategies with mathematics should yield success in the stock market.
##### Will Fastiggi
Originally from England, Will is an Upper Primary Coordinator now living in Brazil. He is passionate about making the most of technology to enrich the education of students.
Articles: 677 | 1,161 | 6,990 | {"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-2023-40 | longest | en | 0.920276 |
https://corsi.unipr.it/en/ugov/degreecourse/149178 | 1,701,611,946,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00802.warc.gz | 223,779,203 | 11,518 | # ANALYTICAL AND STATISTICAL MECHANICS cod. 1004552
2° year of course - Second semester
Professor
Fisica teorica, modelli e metodi matematici (FIS/02)
Field
Teorico e dei fondamenti della fisica
Type of training activity
Characterising
78 hours
of face-to-face activities
9 credits
hub: PARMA
course unit
in ITALIAN
## Learning objectives
The student will acquire the basic theoretical concepts in Lagrangian and Hamiltonian mechanics. She/he will understand the principles leading to the study of macroscopic systems and the basic concepts in statistical mechanics. The student will be able to apply these methods to calculate the thermodynamical properties of macroscopic systems at equilibrium, starting from the statistical distribution of microscopic variables in phase space, in simple physical systems. She/he will develop learning skills and she/he will be able to identify the relevant points in a physical problem, the validity of relations and their applicability.
## Prerequisites
A solid understanding of calculus and basic familiarity with linear algebra; three semesters of introductory calculus-based physics (mechanics and termodynamics). It is also very important to have good familiarity with differentiation, integration, vector calculus and matrices.
## Course unit content
Introduction to Analytical Mechanics: Lagrangian and Hamiltonian approach.
Statistical Mechanics of Microcanonical, Canonical and Gran Canonical Ensembles.
Applications of the classical ensembles.
## Full programme
- Classical Mechanics in an arbitrary reference frame. Constraints, virtual displacements, generalized lagrangian coordinates. The Lagrangian of a physical systems and the Lagrange equations. Symmetries and conservation laws. Noether's theorem. Small oscillations, normal modes. The Legendre transform and the Hamiltonian. Hamilton's equations. Configuration space and phase space. Poisson brackets.
- Variational principles and Lagrange and Hamilton equations. Elements of calculus of variations. Canonical transformations. Elements of perturbation theory. Examples of relevant Lagrangians and Hamiltonians of physical systems: central forces, changed particles in an electromagnetic field. Infinite degrees of freedom: the vibrating string.
- The statistical description of a macroscopic system. Systems with many degrees of freedom and classical mechanics. Liouville theorem. The problems of the microscopic approach. Temporal averages and the ergodic hypothesis. How and if equilibrium is reached.
Brief review of thermodynamics: extensive and intensive variables, thermodynamic potentials, Legendre transformations.
-Microcanonical distribution. Boltzmann entropy and its properties. Additivity. Microcanonical classical ideal gas. Gibbs paradox.
- Canonical distribution. The partition function and the Helmotz free energy. Energy fluctuation in the canonical ensemble. Fluctuation and response. Maxwell distribution. Equipartition. Equivalence between microcanonical e canonical ensembles. Canonical Ideal Gas.
- Gran canonical distribution. Gran canonical partition function and pressure. Chemical potential. Gran canonical Ideal Gas.
## Bibliography
H. Goldstein- C. Poole - J. Safko, Classical Mechanics
L.D. Laundau - E.M. Lifsits, Mechanics
L.D. Laundau - E.M. Lifsits, Statistical Physics
K. Huang - Statistical Mechanics
S.J. Blundell, K. M. Blundell - Concepts in Thermal Physics
Lecture notes.
## Teaching methods
There are 42 hours of lectures and 36 hours of exercise class.
There will be exercise sheets every week.
I will provide my own lecture notes, however the use of a text book is highly recommended.
## Assessment methods and criteria
There will be two midterm quizzes which contribute, if positive, to the final written grade. There will be a final written and oral examination.
## Other information
Support activity: tutor activity during the course, material from web sites on advanced subjects | 788 | 3,962 | {"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-2023-50 | longest | en | 0.81504 |
https://www.rapidtables.com/convert/number/octal-to-decimal.html | 1,722,679,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00552.warc.gz | 764,485,319 | 5,235 | # Octal to Decimal Converter
8
10
16
Decimal to octal converter ►
## Octal
Octal number is a number expressed in the base 8 numeral system. Octal number's digits have 8 symbols: 0,1,2,3,4,5,6,7. Each digit of an octal number counts a power of 16.
Octal number example:
6278 = 6×82+2×81+7×80 = 158010
## Decimal
Decimal number is a number expressed in the base 10 numeral system. Decimal number's digits have 10 symbols: 0,1,2,3,4,5,6,7,8,9. Each digit of a decimal number counts a power of 10.
Decimal number example:
65310 = 6×102+5×101+3×100
## How to convert from octal to decimal
A regular decimal number is the sum of the digits multiplied with 10n.
#### Example #1
137 in base 10 is equal to each digit multiplied with its corresponding 10n:
13710 = 1×102+3×101+7×100 = 100+30+7
Octal numbers are read the same way, but each digit counts 8n instead of 10n.
Multiply each digit of the hex number with its corresponding 8n.
#### Example #2
37 in base 8 is equal to each digit multiplied with its corresponding 8n:
378 = 3×81+7×80 = 24+7 = 31
#### Example #3
7014 in base 8 is equal to each digit multiplied with its corresponding power of 8:
70148 = 7×83+0×82+1×81+4×80= 3584+0+8+4 = 3596
### Octal to decimal conversion table
Octal
base 8
Decimal
base 10
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
10 8
11 9
12 10
13 11
14 12
15 13
16 14
17 15
20 16
30 24
40 32
50 40
60 48
70 56
100 64
Decimal to Octal converter ► | 514 | 1,437 | {"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} | 3.84375 | 4 | CC-MAIN-2024-33 | latest | en | 0.647529 |
https://msx.org/nl/node/53690?page=0 | 1,642,582,990,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00197.warc.gz | 439,296,985 | 9,767 | # Armchair coding Stunt Car Racer
Pagina 1/4
| 2 | 3 | 4
Hatched from Poll: most challenging game that could be ported to msx in the'80, a discussion on feasibility of Stunt Car Racer for the MSX 1 that I felt I had made sufficiently technical to be off topic for 'General discussion' but on topic for 'Development':
hit9918 wrote:
"screen 2 also is a tile mode"
1 byte of nametable can fill 8x8 pixels
this could make superfast polygons
if one manages to sort out the corner cases
In the case of the 8-bit Stunt Car Racer it's actually a relatively easy problem such as I can make out. I really think all it's doing is picking the half a dozen-ish floor segments to draw, performing the actual 3d maths on their top parts, then for each from back to front performing a fill from the bottom of the screen to their upper boundary, and drawing on the appropriate lines (i.e. if that top segment is front facing, all of them; otherwise only those attached to the highest edge and possibly some down the side of the track from any highest edge that is also a side). So the sky is one colour or pattern, the floor and track are another, the top segment boundaries are added for definition.
So you could arrange it so that the filling task at each floor panel is just: between x1 and x2, fill from the bottom of the screen up to the values in the computed table h[x1...x2]. So anything between (x1+7) >> 3 and x2 >> 3 is a completely enclosed column of tiles. Compute (min(h[c1 ... c8]) + 7) >> 3 to get the first tile boundary below the lowest entry in that column. Tile fill up to there. Go tile internal only for the heights above that min, and from bottom of screen to top for anything between x1 and (x1 + 7) >> 3, and between (x2 >> 3) and x2.
Then add the lines, which will involve some sort of process for each tile touched of checking whether it is currently one of the special completely solid ones and if so then substituting another. Though I don't appear to have a good instinctive sense for whether it'd be better to have a fixed allocation from screen location to tile or whether to allocate them on demand. The latter could reduce upload costs but either bookkeeping is a hassle or you risk exhaustion given the possibility that you might claim a screen tile for new drawing, draw on it, replace it with a solid one later, then claim it again for new drawing, etc.
EDIT: as an additional observation, because you need to know only the top edge of each segment and are assuming the camera to be close to upright, a further observation is that you never need worry about full polygon clipping, only individual edge clipping. Which makes life a lot easier. When last I implemented that on a Z80 I naively went with a full multiply/divide solution but I'm sure binary search would work. Don't worry about the far clip plane and the only things to clip to are z=1, z=+x, z=-x, z=+y, z=-y, doing each line separately. So per line grab a bit mask of disobeyed constraints then AND them. If that's non-zero, throw away the line. If not then OR them. While that OR is non-zero, shift it right and for each non-zero bit perform the bisection to find the point on the line that matches the constraint. Substitute it for whichever point was in violation and continue. When done, project and draw.
hit9918 wrote:
TomH wrote:
The main problem would be squeezing two video buffers into VRAM
one can have two buffers in the charset mode mindset
render in halve a charset while showing the other halve charset
doublebuffering via the nametable
If we're embracing monochrome graphics anyway, maybe even the unofficial screen 0 with three segments mode? Definitely documented as 9938 incompatible, but more compact and the big win is that you're never more than 2/171ths of a line away from a VRAM access window. So the CPU can push as fast as it can push at any time. But the addressing turns into a hassle, you still can't have completely unique pixel images unless you restrict yourself to 192 pixels across, and if you think about it then it amounts to you having to upload 25% more data than is actually visible, as two bits out of every eight aren't visible. So pushing faster also buys you the need to push more. And at any likely frame rate, being constrained to VRAM access during only the non-pixel portion of the display quite likely isn't the bottleneck.
Aangemeld of registreer om reacties te plaatsen
The more I think about it the more I think someone should definitively give it a try! Are there any other MSX/Spectrum games with 3d polygonal graphics for comparison? I can only think of:
- Stunt car racer
- Hard drivin
As for implementation details, if I were to do this, I think I'd still go for Screen 2 (since we need additional tiles for the scoreboard at the bottom of the screen, etc. we would not be able to fully utilize the tile set in Screen 1 only for the game anyway). So, I think using Screen 2, using the top two banks for the game, and the bottom bank for the scoreboard could work. Then, if we want to double buffer on the VDP itself, perhaps hit9918's idea of using the first 128 patterns in each tilebank for buffer 1 and the other 128 for the buffer 2 could work. As TomH says, the graphics in Stunt Car Racer will not be complex, so it is likely that we do not need all 256 patterns to display the graphics, since there will be a lot of background!
Well there was already something done resembling Stunt Car Racer on the MSX... in Basic too
Maybe a good starting point?
I guess you'd probably need some empirical numbers on likely tile use per frame, though it strikes me that a close to optimal strategy for usage is a simple stack and a threshold. Suppose tiles 255 and 254 are your fixed full-tile pattern fills then:
1. at start of frame, seed the stack with 253 to 0 in reverse order, set the threshold to 0, draw your horizon;
2. when applying a tile fill to the display, any pattern names you replace that aren't 255 or 254 can be pushed to the top of the stack;
3. when plotting individual pixels, if the tile you want to touch already isn't 254 or 255 then just draw to it. Otherwise pull whatever is at the top of the stack, fill it with the same pattern as the tile it is replacing, then draw to it and update the pattern name table. If the new ID is greater than or equal to the threshold, increase the threshold;
4. to upload a frame, upload all the patterns from 0 to the threshold (or in reverse), then push the pattern name map.
You'll push up some tiles you don't need but you'll be pushing a contiguous block so the unrolling is easy. And they should be a relatively low proportion.
Separate observation: if you really believe that you'll stick within 128 tiles per third of the screen, then for a top-two-thirds update you're definitely not going to need to push more than 128*2*8 + 512 = 2560 bytes. With a hypothetical completely unrolled loop for an 18 cycles/byte transfer and the NTSC 69 lines of fast access available a frame, I make that three frames to push one frame. So if you were somehow producing the pixels quickly enough, you'd be able to display them at 20fps.
Since you are pretty-much definitely not going to get 20fps that means: if you restrict yourself to two thirds of the display and half the tile set, the VRAM interface is not the bottleneck. The unusual thing about the MSX isn't a problem, but rather whatever advantage you're getting from the tile fills adds up to an architectural advantage.
You know, for this one particular presentation.
Woah! that one in BASIC is good! hard to believe that one can write a BASIC program that runs that smooth!!!
So, who's going to do the first attempt?
I've written a simple 3d engine for screen 2 some years ago. The source is open, so you could use that to estimate the number of blocks being used.
@ricbit Cool!!! At work right now, so, can't check in detail, but do you have any demo/ROM of it working?
santiontanon wrote:
@ricbit Cool!!! At work right now, so, can't check in detail, but do you have any demo/ROM of it working?
Sure, here's a compiled .COM binary. It just spins a simple 3D model, you can change the model by modifying the source code. This code is kinda bad, but my suggestion is not to reuse the code, instead just plug in a representative 3D model to check how many blocks do you need to change, from frame to frame, and then use that number to properly design the best algorithm.
well.. as much as I am ashamed to admit, even if I can code in Z80 assembler, I have NO IDEA how to run a .COM file hahaha, I think it's an MSXDOS file (never owned an MSX2, so, no idea how tu run that either...). I've Googled a bit, but haven't been successful yet in running it... will try again tomorrow
Most easy way to start MSX-DOS is to use feature called "dir as disk" that can be found ie. in BlueMSX and OpenMSX. This way you don't have to learn how to put files to disk images. If you use real disk, format it first on MSX (See CALL FORMAT-command) and then use it like regular folder on PC.
To run MSX-DOS your disk needs to include files called MSXDOS.SYS and COMMAND.COM... You need to google these... then place also the .COM file to same disk and restart the emulator.
MSX-DOS should feel quite familiar if you have used MS-DOS or Windows command prompt although it is more limited. The .COM program can be started by simply writing it's name and pressing return. (No need to write extension) You can list the files on disk with command "DIR" + return. You can also exit to BASIC by writing "BASIC"+return. This already should be enough information for casual user.
And Santi, MSX-DOS runs fine on an MSX1 As long as it has 64kB RAM and a disk-drive of course.
Pagina 1/4
| 2 | 3 | 4 | 2,318 | 9,754 | {"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-2022-05 | latest | en | 0.945354 |
https://softwaremechanic.wordpress.com/2017/07/ | 1,521,824,712,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648404.94/warc/CC-MAIN-20180323161421-20180323181421-00691.warc.gz | 691,891,407 | 12,671 | # Statistics — Tests of independence
## Tests of independence:
Basic principle is the same as ${\chi}^2$ – goodness of fit test
* Between categorical variables
## ${\chi}^2$-square tests:
The standard approach is to compute expected counts, and find the
distribution of sum of square of difference between expected counts and ordinary
counts(normalized).
* Between Numerical Variables
## ${\chi}^2$-square test:
• Between a categorical and numerical variable?
## Null Hypothesis:
• The two variables are independent.
• Always a right-tail test
• Test statistic/measure has a ${\chi}^2$ distribution, if assumptions are met:
• Data are obtained from a random sample
• Expected frequency of each category must be
atleast 5
• ### Properties of the test:
• The data are the observed frequencies.
• The data is arranged into a contingency table.
• The degrees of freedom are the degrees of freedom for the row variable times the degrees of freedom for the column variable. It is not one less than the sample size, it is the product of the two degrees of freedom.
• It is always a right tail test.
• It has a chi-square distribution.
• The expected value is computed by taking the row total times the column total and dividing by the grand total
• The value of the test statistic doesn’t change if the order of the rows or columns are switched.
• The value of the test statistic doesn’t change if the rows and columns are interchanged (transpose of the matrix
# The mystery of short term past performance versus future equity fund returns
In our earlier posts, here and here, we found to our dismay that, our natural inclination to choose the top mutual fund performers of the past 1 & 3 years hasn’t worked too well.
That leaves us with the obvious question..
What actually goes wrong when we pick the top funds of the past few years?
### The rotating sector winners..
Below is a representation of the best performing sectors year over year. What do you notice?
The sector performance over each and every year varies significantly and the top and bottom sectors keep changing dramatically almost every year.
Sample this:
• 2007 – Metals was the top performer with a whopping 121% annual return
• 2008 – Metals was the bottom performer with a negative 74% returns & FMCG was the top perfomer (-21%)
• 2009 – The tables turned! FMCG was the bottom performer (47%) while Metals was the…
View original post 1,399 more words | 534 | 2,432 | {"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} | 3.46875 | 3 | CC-MAIN-2018-13 | longest | en | 0.898857 |
https://www.scribd.com/presentation/44410346/Presentation-on-Pom-Manish | 1,555,608,736,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517745.15/warc/CC-MAIN-20190418161426-20190418183426-00546.warc.gz | 825,922,771 | 52,607 | You are on page 1of 19
# [
[ [ [ w w p .
p p .
.
.
w w w.
w.
.
.
.
!".
.
" "! .
" ! "! .
.
.
.
" " .
" w.
w.
.
.
w.
w.
.
.
"! .
! .
.
# .
" .
" "" "! .
.
" .
"! .
" \$ .
" " % .
"! .
!".
.
" "! ".
" \$! \$ .
" " .
" "! .
& ! .
" ! ".
.
"! ! .
" ! .
.
" .
The upper and lower specification limits (USL. µ. the greater is the spread of values encountered. Due to the properties of the normal distribution. For the curve shown above. The Greek letter ɗ (sigma) marks the distance on the horizontal axis between the mean. which underlies the statistical assumptions of the Six Sigma model.Graph of the normal distribution. LSL) are at a distance of 6ɗ from the mean. µ = 0 and ɗ = 1. The greater this distance is. values lying that far away from the mean are extremely unlikely. . and the curve's inflection point.
FEEw Fw w º { " .
.
" "! ".
.
.
"" .
" " w.
w.
' º {".
" !.
"""!.
" "" !.
"! º { .
".
"! .
.
"" .
.
.
!%".
" "!( .
[ .
.
.
.
.
.
.
.
.
.
.
% .
) .
.
.
.
.
ÕENEF w Fw w º .
"! .
! "! !.
.
"& .
.
.
" .
"" " (.
" .
" !.
" .
" ( " ".
#.
" " .
" º .
! ".
".
& .
.
.
!"! ".
.
.
.
! *!.
.
" .
".
!".
.
!"! ! .
º .
( .
" & " .
"" " .
.
( .
" (.
.
" .
" "!! .
" (.
" !" º+".
" .
" * .
.
.
".
.
.
" .
.
.
.
.
".
.
!.
.
" º" ".
".
#.
"& .
.
.
.
"!% ".
".
#.
" .
" .
".
" .
! .
" .
! "! !.
" w.
.
.
".
.
.
"" .
" .
" "! !" "! .
" .
".
(" .
p p { .
.
% .
.
.
.
.
.
.
.
.
.
.
.
.
#.
º§.
.
.
º{ .
.
.
.
.
º§.
.
#.
.
.
.
.
.
# .
.
.
. º.
.
.
.
.
.
.
.
.
{ ! .
-{ + ! .
" -{ + .
" ! w.
w.
' "! .
" " .
.
w. w w wE º .
..
. " ! .!".
.
" .
% .w.
w.
/0 ! " ..
.
" ".
.
- º w.
w.
! ! ! .
" .
" " .
" "! ( " .
! .
" .
w Ew º .
.
" .
.
.
% .
.
.
.
" .
" " º 1 "! "!! .
.
!.
".
" "! " "! ! " .
.
" .
.
EwwEw º !.
" .
.
" .
! " !.
º{ ! .
.
" .
"! .
.
.
! ".
#.
" .
.
! ! .
wEw Fw w º .
.
" ! .
.
º .
! ! º .
.
.
& .
" .
.
ºw.
.
! "!!.
#! ! ".
#.
" º{ ".
"! "! .
.
.
" º 1 .
.
.
" "! .
" .
" 2 .
") .
w .
N Fw w ºw.
.
(.
!.
.
(! .
.
" .
.
.
" "! ! .
" .
" ºw.
.
.
".
" .
" "! .
" ! .
.
"" 3- " .
.
.
" w + "! .
" .
.
.
.
N EwNEEw Fw w [ .
" ! .
.
.
! "! )# ! 2 .
.
" .
" !" ! .
.
.
.
.
"! .
.
".
.
" " .
" " .
.
.
.
" .
.
" .
(.
.
" .
" .
"! .
.
E.
E m.
"! w " .
"! ! ! .
! .
" ".
.
!.
.
"! ".
#.
" 4" .
! "" % w.
w.
.
.
" .
" .
" w.
w.
.
&!(" .
" ".
#.
" .
º{ .
.
.
.
.
.
.
º .
.
# .
.
.
.
' º{ .
.
.
.
N w N º w.
w.
.
! ( .
.
.
# ' (.
.
.
º .
" .
w.
w.
.
" .
" 5 w.
w.
.
" % 6 ! .
.
" "! .
".
" / ".
" .
º.
! ! ! .
" "! .
" .
" .
% ! .
"! .
.
" ºw.
w.
! .
! .
" " .
" -"" .
".
.
.
! . | 5,801 | 7,383 | {"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-2019-18 | latest | en | 0.336342 |
https://samacheerkalviguru.com/samacheer-kalvi-7th-maths-term-2-chapter-1-ex-1-1/ | 1,718,905,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00043.warc.gz | 437,415,071 | 12,436 | Students can Download Maths Chapter 1 Number System Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.1
Question 1.
Write the decimal numbers for the following pictorial representation of numbers.
Solution:
(i) Tens 2 ones 2 tenths = 12.2
(ii) Tens 1 ones 3 tenths = 21.3
Question 2.
Express the following in cm using decimals.
(i) 5 mm
(ii) 9 mm
(iii) 42 mm
(iv) 8 cm 9 mm
(v) 375 mm
Solution:
(i) 5 mm
1 mm = $$\frac { 1 }{ 10 }$$ cm = 0.1 cm
5 mm = $$\frac { 5 }{ 10 }$$ = 0.5 cm
(ii) 9 mm
1 mm = $$\frac { 1 }{ 10 }$$ cm = 0.1 cm
9 mm = $$\frac { 9 }{ 10 }$$ cm = 0.9 cm
(iii) 42 mm
1 mm = $$\frac { 1 }{ 10 }$$ cm = 0.1 cm
42 mm = $$\frac { 42 }{ 10 }$$ cm = 4.2 cm
(iv) 8 cm 9 mm
1 mm = $$\frac { 1 }{ 10 }$$ cm = 0.1 cm
8 cm 9 mm = 8 cm + $$\frac { 9 }{ 10 }$$ cm = 8.9 cm
(v) 375 mm
1 mm = $$\frac { 1 }{ 10 }$$ cm = 0.1 cm
375 mm = $$\frac { 375 }{ 10 }$$ cm = 37.5 cm
Question 3.
Express the following in metres using decimals.
(i) 16 cm
(ii) 7 cm
(iii) 43 cm
(iv) 6 m 6 cm
(v) 2 m 54 cm
Solution:
(i) 16 cm
1 cm = $$\frac { 1 }{ 100 }$$ cm = 0.01 m
16 cm = $$\frac { 16 }{ 100 }$$ m = 0.16 m
(ii) 7 cm
1 cm = $$\frac { 1 }{ 100 }$$ cm = 0.01 m
1 cm = $$\frac { 7 }{ 100 }$$ m = 0.07 m
(iii) 43 cm
1 cm = $$\frac { 1 }{ 100 }$$ cm = 0.01 m
43 cm = $$\frac { 43 }{ 100 }$$ m = 0.43 m
(iv) 6 m 6 cm
1 cm = $$\frac { 1 }{ 10 }$$ m = 0.01 m
6 m 6 cm = 6 m + $$\frac { 6 }{ 100 }$$ m = 6 m + 0.06 m = 6.06 m
(v) 2 mm 54 cm
1 cm = $$\frac { 1 }{ 100 }$$ cm = 0.01 m
2 m 54 cm = 2 m + $$\frac { 54 }{ 100 }$$ m = 2 m + 0.54 m = 2.54 m
Question 4.
Expand the following decimal numbers.
(i) 37.3
(ii) 658.37
(iii) 237.6
(iv) 5678.358
Solution:
(i) 37.3 = 30 + 7 + $$\frac { 3 }{ 10 }$$ = 3 × 101 + 7 × 100 + 3 × 10-1
(ii) 658.37 = 600 + 50 + 8 + $$\frac { 3 }{ 10 }$$ + $$\frac { 7 }{ 100 }$$
= 6 × 102 + 5 × 101 + 8 × 100 + 3 × 10-1 + 7 × 10-2
(iii) 237.6 = 200 + 30 + 7 + $$\frac { 6 }{ 10 }$$
= 2 × 102 + 3 × 101 + 7 × 100 + 6 × 10-1
(iv) 5678.358 = 5000 + 600 + 70 + 8 + $$\frac { 3 }{ 10 }$$ + $$\frac { 5 }{ 100 }$$ + $$\frac { 8 }{ 1000 }$$
= 5 × 103 + 6 × 102 + 7 × 101 + 8 × 100 + 3 × 10-1 + 5 × 10-2 + 8 × 10-3
Question 5.
Express the following decimal numbers in place value grid and write the place value of the underlined digit.
(i) 53.61
(ii) 263.271
(iii) 17.39
(iv) 9.657
(v) 4972.068
Solution:
(i) 53.61
(ii) 263.271
(iii) 17.39
(iv) 9.657
(v) 4972.068
Objective Type Questions
Question 6.
The place value of 3 in 85.073 is _____
(i) tenths
(ii) hundredths
(iii) thousands
(iv) thousandths
(iv) thousandths
Hint: 1000 g = 1 kg; 1 g = $$\frac { 1 }{ 1000 }$$ kg
Question 7.
To convert grams into kilograms, we have to divide it by
(i) 10000
(ii) 1000
(iii) 100
(iv) 10
(ii) 1000
Hint: 85.073 = 8 × 10 + 5 × 1 + 0 × $$\frac { 1 }{ 10 }$$ + 7 × $$\frac { 1 }{ 100 }$$ + 3 × $$\frac { 1 }{ 1000 }$$
Question 8.
The decimal representation of 30 kg and 43 g is ____ kg.
(i) 30.43
(ii) 30.430
(iii) 30.043
(iv) 30.0043
Hint: 30 kg and 43 g = 30 kg + $$\frac { 43 }{ 1000 }$$ kg = 30 + 0.043 = 30.043
Hint: 264 cm = $$\frac { 264 }{ 100 }$$ m = 2.64 m | 1,548 | 3,325 | {"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} | 4.75 | 5 | CC-MAIN-2024-26 | latest | en | 0.496516 |
http://www.unc.edu/~rowlett/units/scales/wiregauge.html | 1,532,228,349,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00246.warc.gz | 561,172,190 | 3,146 | ## How Many? A Dictionary of Units of Measurement © Russ Rowlett and the University of North Carolina at Chapel Hill Table of Contents About the Dictionary Using the Dictionary American and British Wire Gauges
Data originally posted by Enginering Zones.
Gauge Washburn & Moen British Imperial Standard (S.W.G.) Birmingham or Stubs American (A.W.G.) or Brown & Sharpe 7/0 .4900" .500" --- --- 6/0 .4615" .464" --- .5800" 5/0 .4305" .432" .500" .5165" 4/0 .3938" .400" .454" .4600" 3/0 .3625" .372" .425" .4096" 2/0 .3310" .348" .380" .3648" 1/0 .3065" .324" .340" .3249" 1 .2830" .300" .300" .2893" 2 .2625" .276" .284" .2576" 3 .2437" .252" .259" .2294" 4 .2253" .232" .238" .2043" 5 .2070" .212" .220" .1819" 6 .1920" .192" .203" .1620" 7 .1770" .176" .180" .1442" 8 .1620" .160" .165" .1284" 9 .1483" .144" .148" .1144" 10 .1350" .128" .134" .1018" 11 .1205" .116" .120" .0907" 12 .1055" .104" .109" .0808" 13 .0915" .092" .095" .0719" 14 .0800" .080" .083" .0640" 15 .0720" .072" .072" .0570" 16 .0625" .064" .065" .0508" 17 .0540" .056" .058" .0452" 18 .0475" .048" .049" .0403" 19 .0410" .040" .042" .0358" 20 .0348" .036" .035" .0319" Gauge Number Washburn & Moen British Imperial Standard (S.W.G.) Birmingham or Stubs American (A.W.G.) or Brown & Sharpe 21 .0317" .032" .032" .0284" 22 .0286" .028" .028" .0253" 23 .0258" .024" .025" .0225" 24 .0230" .022" .022" .0201" 25 .0204" .020" .020" .0179" 26 .0181" .018" .018" .0159" 27 .0173" .0164" .016" .0141" 28 .0162" .0148" .014" .0126" 29 .0150" .0136" .013" .0112" 30 .0140" .0124" .012" .0100" 31 .0132" .0116" .010" .0089" 32 .0128" .0108" .009" .0079" 33 .0118" .0100" .008" .0070" 34 .0104" .0092" .007" .0063" 35 .0095" .0084" .005" .0056" 36 .0090" .0076" .004" .0050" 37 .0085" .0068" --- .0044" 38 .0080" .0060" --- .0039" 39 .0075" .0052" --- .0035" 40 .0070" .0048" --- .0031" 41 .0066" .0044" --- .00280" 42 .0062" .0040" --- .00249" 43 .0060" .0036" --- .00222" 44 .0058" .0032" --- .00198" 45 .0055" .0028" -- .00176" 46 .0052" .0024" --- .00157" 47 .0050" .0020" --- .00140" 48 .0048" .0016" --- .00124" 49 .0046" .0012" -- .00111" 50 .0044" .0010" -- .00099" | 1,025 | 2,138 | {"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-2018-30 | latest | en | 0.821616 |
https://opticsgirl.com/diffraction/ | 1,670,227,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00790.warc.gz | 463,994,999 | 12,811 | # Section 7: Diffraction
In the last section with interference, we (for the most part) only considered interference with a few waves. While there is no great physical distinction between interference and diffraction, it is customarily denoted a study of diffraction when you have interference of large numbers of wave. If this seems arbitrary, it’s because it is, but unfortunately I wasn’t around when Francesco Grimaldi first denoted the interference arising from a cone light propagating through two rectangular aperatures “diffractio”*. As you may note from the following contents of this section, this is going to be a wild ride, but I hope also one that makes the whole of classical optics make more sense**!
1. Introduction to Diffraction
2. Fresnel Approximation
3. Paraxial Wave Equation
4. Gouy Phase Shift
5. Beam Power
6. Gaussian Beam Propagation with ABCD Matrices
7. Higher Order Gaussian Beams
8. Integral Approach to Diffraction- Fresnel Diffraction
9. Huygens’ Principle
10. Fraunhofer Diffraction
11. The Array Theorem
12. Babinet’s Principle
13. Diffraction in Paraxial Optical Systems
14. Fourier Transform by a Lens
15. The 4f Lens System
16. Imaging of Extended Objects
17. Near-field Imaging
18. Fresnel Diffraction
19. Fresnel Zones
*Actually, as much as we may dislike him for setting the tone of ambiguity about diffraction vs interference, Signore Grimaldi did some pretty cool experiments considering the year was 1665 when he published his work in “De Lumine“. For all of you fellow nerds out there, this article is somewhat of an interesting read into the actual experiments Grimaldi conducted. If you are really ambitious, I believe you can find Grimaldi’s original text in Latin somewhere in the void of the internet.
**Diffraction can again be described quantum mechanically as well; if you are interested in this picture, check out the section on Quantum Optics! | 436 | 1,901 | {"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} | 2.578125 | 3 | CC-MAIN-2022-49 | latest | en | 0.896587 |
https://www.urionlinejudge.com.br/judge/en/profile/440012 | 1,604,108,208,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107912593.62/warc/CC-MAIN-20201031002758-20201031032758-00226.warc.gz | 836,410,020 | 6,302 | # PROFILE
Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1145 Logical Sequence 2 00179º 18411942 C# 0.180 6/4/20, 6:07:56 PM
1159 Sum of Consecutive Even... 00424º 18411337 C# 0.016 6/4/20, 5:24:13 PM
1158 Sum of Consecutive Odd... 00288º 18411062 C# 0.020 6/4/20, 5:02:25 PM
1153 Simple Factorial 00609º 18410461 C# 0.028 6/4/20, 4:15:21 PM
1133 Rest of a Division 00456º 18409504 C# 0.024 6/4/20, 3:04:53 PM
1079 Weighted Averages 00746º 18409273 C# 0.024 6/4/20, 2:45:56 PM
1154 Ages 00292º 18404424 C# 0.012 6/4/20, 2:33:30 AM
1177 Array Fill II 00151º 18404217 C# 0.024 6/4/20, 2:14:55 AM
1173 Array fill I 00265º 18404153 C# 0.020 6/4/20, 2:08:21 AM
1164 Perfect Number 00193º 18404099 C# 0.028 6/4/20, 2:03:01 AM
1143 Squared and Cubic 00335º 18410092 C# 0.016 6/3/20, 8:20:09 PM
1157 Divisors I 00085º 18399403 C# 0.008 6/3/20, 7:21:27 PM
1132 Multiples of 13 00381º 18399266 C# 0.016 6/3/20, 7:09:45 PM
1080 Highest and Position 00843º 18399071 C# 0.032 6/3/20, 6:56:16 PM
1144 Logical Sequence 00597º 18410308 C# 0.056 6/3/20, 5:35:09 PM
1097 Sequence IJ 3 00341º 18392236 C# 0.016 6/3/20, 5:51:47 AM
1096 Sequence IJ 2 00391º 18392224 C# 0.016 6/3/20, 5:50:07 AM
1116 Dividing X by Y 00198º 18391808 C# 0.044 6/3/20, 4:46:56 AM
1095 Sequence IJ 1 00344º 18391675 C# 0.020 6/3/20, 4:27:30 AM
1134 Type of Fuel 00474º 18242113 C# 0.020 5/20/20, 10:07:44 PM
1115 Quadrant 01195º 18238752 C# 0.028 5/20/20, 5:33:09 PM
1114 Fixed Password 01185º 18238639 C# 0.020 5/20/20, 5:18:30 PM
1101 Sequence of Numbers and Sum 00504º 18238290 C# 0.024 5/20/20, 4:45:14 PM
1099 Sum of Consecutive Odd... 00500º 18231335 C# 0.024 5/19/20, 11:36:19 PM
1113 Ascending and Descending 00913º 18238380 C# 0.020 5/17/20, 8:04:08 PM
1078 Multiplication Table 01537º 18194218 C# 0.048 5/16/20, 8:29:11 PM
1075 Remaining 2 00291º 18086567 C# 0.012 5/6/20, 4:24:30 PM
1074 Even or Odd 00691º 17823311 C# 0.020 4/14/20, 6:02:54 PM
1072 Interval 2 01002º 17822072 C# 0.024 4/14/20, 4:47:06 PM
1066 Even, Odd, Positive and... 00819º 17801686 C# 0.028 4/13/20, 4:27:49 AM
1 of 3 | 1,075 | 2,143 | {"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} | 2.640625 | 3 | CC-MAIN-2020-45 | latest | en | 0.34484 |
http://www.pagalguy.com/u/vvkkvv | 1,398,036,623,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00268-ip-10-147-4-33.ec2.internal.warc.gz | 598,857,357 | 10,312 | vvkkvv's posts
vvkkvv replied to Special Tricks for CAT 2011
One correction. If
Last non-zero digit of 10! = 8
then the last non-zero digit of 20! = 8x8x2 (this 2 comes from 20)=8
Last non zero digit of 70!= last non zero digit of (8^7) x2x3x4x5x6x7 | 95 | 249 | {"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-2014-15 | latest | en | 0.636641 |
https://hwkhlp.com/question/210160 | 1,547,780,464,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659677.17/warc/CC-MAIN-20190118025529-20190118051529-00519.warc.gz | 562,792,911 | 5,536 | What's the answer
\text{1)JoeSmithscored}37\text{ofhisteams}100\text{pointsinagame.Whatpercentageofpointsdidhescore?}
Solution:
\left\{0\right\}
Subject
Topic
Level
Middle School
Grade
6
# Answer
Joe Smith scored 37 points of 100 team's points that means:
37 per 100 :
37 per cent :
37/100 or 37% | 100 | 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} | 3.3125 | 3 | CC-MAIN-2019-04 | longest | en | 0.648294 |
https://razonarte.org/year-7-maths-worksheets-printable/grade-grade-8-math-worksheets-printable-image-kindergarten-16/ | 1,550,540,150,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489282.7/warc/CC-MAIN-20190219000551-20190219022551-00036.warc.gz | 692,870,651 | 5,223 | # Grade Math Worksheets Printable Image Kindergarten Maths For Year Olds Algebra
By Katrin Freeh on December 03 2018 11:04:14
The use of math worksheets can help solve numerous arithmetic problems. "Practice makes an individual perfect," is the best motto to be kept in mind while studying math. The motto will help a person to reinforce his desire to better himself in the subject. Without the help of these online resources, one will not be able to achieve the mastery of math. Since education is one of those areas which receive little or no funding from the government, it is essential for parents to look for various options that can help give their child a better education. Some sites do offer math online quiz that is sure to bring about an inclination towards math among children.
What are math worksheets and what are they used for? These are math forms that are used by parents and teachers alike to help the young kids learn basic math such as subtraction, addition, multiplication and division. This tool is very important and if you have a small kid and you don`t have a worksheet, then its time you got yourself one or created one for your kid. There are a number of sites over the internet that offer free worksheets that are downloadable and printable for use by parents and teachers at home or at school.
The most important thing about these math worksheets is that they are used for tutoring and not for the main course studies. That is why they are used by tutors to offer remedial tuition and by parents at home so that they can offer their kids extra tuition to sharpen their skills. Math is known to be difficult and is often a headache for the young and so the math worksheets come in handy in helping resolve this problem. Thanks to the sites over the internet that offer free printable math worksheets, you do not need to worry about the cost of purchasing one, maybe only the ink cost. So don`t go making excuses for not being able to access a math work sheet.
There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign "%." Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
It is widely understood that math has a global use and acceptance. People are also aware of the rate at which math is advancing today at various fields of research and study. Many mathematicians will talk about the pattern and structure of math worksheets which are helpful for people in working fields. Math has helped science and technology reach a higher level of advancement.
Once downloaded, you can customize the math worksheet to suit your kid. The level of the child in school will determine the look and content of the worksheet. Use the school textbook that your child uses at school as a reference guide to help you in the creation of the math worksheet. This will ensure that the worksheet is totally relevant to the kid and will help the child improve his or her grades in school. | 637 | 3,213 | {"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} | 2.765625 | 3 | CC-MAIN-2019-09 | latest | en | 0.971677 |
https://www.civilengineeringhandbook.tk/fluid-mechanics/x-x2.html | 1,519,268,777,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813883.34/warc/CC-MAIN-20180222022059-20180222042059-00798.warc.gz | 852,584,616 | 7,417 | ## X2
Table 15-2 Values of Constant C in Two-Phase Multiplier Equations
Flow state Liquid Gas C
tt Turbulent Turbulent 20
vt Laminar Turbulent 12
tv Turbulent Laminar 10
vv Laminar Laminar 5
where the values of C for the various flow combinations are shown in Table 15-2.
The Lockhart-Martinelli correlating parameter x2 is defined as
9X/ fGm PGD
Here, /Lm is the tube friction factor based on the "liquid-only" Reynolds number NReLm = (1 — x)GmD/^L and/Gm is the friction factor based on the "gas-only" Reynolds number NReGm = xGmD/^G. The curves cross at X = 1, and it is best to use the "G" reference curves for x < 1 and the "L" curves for x > 1.
Using similarity analysis, Duckler et al. (1964b) deduced that
which is equivalent to the Martinelli parameters
and ' and pm are the equilibrium (no-slip) properties. Another major difference is that Duckler et al. deduced that the friction factors fL and fG should both be evaluated at the mixture Reynolds number,
5. Slip and Holdup
A major complication, especially for separated flows, arises from the effect of slip. Slip occurs because the less dense and less viscous phase exhibits a lower resistance to flow, as well as expansion and acceleration of the gas phase as the pressure drops. The result is an increase in the local holdup of the more dense phase within the pipe ('m) (or the corresponding two-phase density, pm), as given by Eq. (15-11). A large number of expressions and correlations for the holdup or (equivalent) slip ratio have appeared in the literature, and the one deduced by Lockhart and Martinelli is shown in Fig. 15-7. Many of these slip models can be summarized in terms of a general equation of the form
for which the values of the parameters are shown in Table 15-3. Although many additional slip models have been proposed in the literature, it is not clear which of these should be used under a given set of circumstances. In some cases, a constant slip ratio (S) may give satisfactory results. For example, in a comparison of calculated and experimental mass flux data for high velocity air-water flows through nozzles, Jamerson and Fisher (1999) found that S = 1.1-1.8 accurately represents the data over a range of x = 0.02-0.2, with the value of S increasing as the quality (x) increases.
A general correlation of slip is given by Butterworth and Hewitt (1977):
where
Model ao a1 a2 a3 Homogeneous
0 0 | 615 | 2,409 | {"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-2018-09 | longest | en | 0.908898 |
https://math.stackexchange.com/questions/4811092/analysis-lemma-proof | 1,701,771,526,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100550.40/warc/CC-MAIN-20231205073336-20231205103336-00269.warc.gz | 438,698,289 | 35,167 | # Analysis Lemma Proof
Prove that if $$x_n>0, \forall n \in N$$ is such that $$lim \frac{x_{n+1}}{x_n}=a<1$$ then $$lim x_n=0$$ I feel absolutely frustrated, because I tried a lot on this issue and couldn't develop anything. Does anyone have anything that can help me?
• Nov 20 at 20:29
• $~0 < a < 1 \implies \displaystyle \lim_{n \to \infty} a^n = 0.$ Nov 20 at 20:33
• I managed to solve it! Nov 20 at 20:43 | 145 | 412 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-50 | latest | en | 0.904093 |
http://gmatclub.com/forum/something-must-be-done-to-stop-spam-in-early-days-people-76282.html?fl=similar | 1,485,147,034,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282110.46/warc/CC-MAIN-20170116095122-00194-ip-10-171-10-70.ec2.internal.warc.gz | 120,235,842 | 61,272 | Something must be done to stop spam. In early days, people : GMAT Critical Reasoning (CR)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 22 Jan 2017, 20:50
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Something must be done to stop spam. In early days, people
Author Message
TAGS:
### Hide Tags
Senior Manager
Joined: 04 Sep 2008
Posts: 259
Location: Kolkata
Schools: La Martiniere for Boys
Followers: 2
Kudos [?]: 90 [0], given: 9
Something must be done to stop spam. In early days, people [#permalink]
### Show Tags
05 Mar 2009, 07:45
1
This post was
BOOKMARKED
00:00
Difficulty:
(N/A)
Question Stats:
50% (01:02) correct 50% (00:00) wrong based on 7 sessions
### HideShow timer Statistics
Something must be done to stop spam. In early days, people seldom received unsolicited email advertisement; but now that numerous bulk email software and email address finders are developed to collect email address all around the world. Advertisers use email addresses to market their products and even sell such email lists to other advertisers. As a result, almost everyone ever get junk email, and sometime several and even tens of annoying emails a day. So, relevant anti-spam regulations should be framed to stop unsolicited advertising.
The two portions in boldface play which of the following roles?
A. Background that the argument depends on and conclusion that can be drawn from the argument.
B. Part of evidence that the argument includes, and inference that can be drawn from this passage.
C. Pre-evidence that the argument depends on and part of evidence that supports the conclusion.
D. Background that argument depends on and part of evidence that supports the conclusion.
E. Pre-evidence that argument includes and a method that helps to supports that conclusion.
_________________
Thanks
rampuria
If you have any questions
New!
VP
Joined: 05 Jul 2008
Posts: 1430
Followers: 39
Kudos [?]: 360 [0], given: 1
### Show Tags
05 Mar 2009, 08:45
rampuria wrote:
Something must be done to stop spam. In early days, people seldom received unsolicited email advertisement; but now that numerous bulk email software and email address finders are developed to collect email address all around the world. Advertisers use email addresses to market their products and even sell such email lists to other advertisers. As a result, almost everyone ever get junk email, and sometime several and even tens of annoying emails a day. So, relevant anti-spam regulations should be framed to stop unsolicited advertising.
The two portions in boldface play which of the following roles?
A. Background that the argument depends on and conclusion that can be drawn from the argument.
B. Part of evidence that the argument includes, and inference that can be drawn from this passage.
C. Pre-evidence that the argument depends on and part of evidence that supports the conclusion.
D. Background that argument depends on and part of evidence that supports the conclusion.
E. Pre-evidence that argument includes and a method that helps to supports that conclusion.
What is the conclusion?
some thing must be done or so, Anti S regulations must be enacted to to stop unsolicited advertising
The word so and the logical order says that the latter is the C
Down to A & B, I will chose A as it more seems like the background than part of evidence.
SVP
Joined: 17 Jun 2008
Posts: 1569
Followers: 11
Kudos [?]: 250 [0], given: 0
### Show Tags
05 Mar 2009, 10:03
A with the similar explanation as by icandy.
Senior Manager
Joined: 06 Jul 2007
Posts: 285
Followers: 3
Kudos [?]: 51 [0], given: 0
### Show Tags
05 Mar 2009, 10:24
rampuria wrote:
Something must be done to stop spam. In early days, people seldom received unsolicited email advertisement; but now that numerous bulk email software and email address finders are developed to collect email address all around the world. Advertisers use email addresses to market their products and even sell such email lists to other advertisers. As a result, almost everyone ever get junk email, and sometime several and even tens of annoying emails a day. So, relevant anti-spam regulations should be framed to stop unsolicited advertising.
The two portions in boldface play which of the following roles?
A. Background that the argument depends on and conclusion that can be drawn from the argument.
B. Part of evidence that the argument includes, and inference that can be drawn from this passage.
C. Pre-evidence that the argument depends on and part of evidence that supports the conclusion.
D. Background that argument depends on and part of evidence that supports the conclusion.
E. Pre-evidence that argument includes and a method that helps to supports that conclusion.
Part 2 is the conclusion/inference. So, only A and B remain. The first part is not an evidence; it's a statement that tells us how things were in the past. So, B is out. A should be it.
Director
Joined: 04 Jan 2008
Posts: 912
Followers: 69
Kudos [?]: 572 [0], given: 17
### Show Tags
05 Mar 2009, 10:26
pls try once then only check the url below
t76282-cr-boldface
_________________
http://gmatclub.com/forum/math-polygons-87336.html
http://gmatclub.com/forum/competition-for-the-best-gmat-error-log-template-86232.html
Manager
Joined: 10 Jan 2009
Posts: 111
Followers: 14
Kudos [?]: 191 [0], given: 0
### Show Tags
05 Mar 2009, 11:25
Something must be done to stop spam. In early days, people seldom received unsolicited email advertisement; but now that numerous bulk email software and email address finders are developed to collect email address all around the world. Advertisers use email addresses to market their products and even sell such email lists to other advertisers. As a result, almost everyone ever get junk email, and sometime several and even tens of annoying emails a day. So, relevant anti-spam regulations should be framed to stop unsolicited advertising.
The two portions in boldface play which of the following roles?
My Explanation:
Premise 2: but now that numerous bulk email software and email address finders are developed to collect email address all around the world.
Premise 3: Advertisers use email addresses to market their products and even sell such email lists to other advertisers.
Premise 4: As a result, almost everyone ever get junk email, and sometime several and even tens of annoying emails a day.
Conclusion: Something must be done to stop spam.
Conclusion/Inference: So, relevant anti-spam regulations should be framed to stop unsolicited advertising.
---> Conclusion because it uses the keyword 'So'
---> Inference because the first sentence of the passage looks more as a conclusion than this one i.e., the second boldfaced statement.
On the basis of above info, we can eliminate at least option C, D & E. Now we have only options A & B left.
If we assume that option A is correct, then the second boldface statement should be the conclusion but IMO, the first statement of the passage should be the conclusion. Also, if you go through option A, it states that the first boldface is a 'Background that the argument depends on'. Do you really think that the argument (conclusion) depends on the first boldface? Had that not been present, we could have still arrived at the conclusion with the help of remaining premises. If anything, the first boldface should only act as a supporting premise.
------------------------
A. Background that the argument depends on and conclusion that can be drawn from the argument. ---> Explained above.
B. Part of evidence that the argument includes, and inference that can be drawn from this passage. ---> Explained above.
C. Pre-evidence that the argument depends on and part of evidence that supports the conclusion. ---> Second boldface is a conclusion/inference. So, discard it.
D. Background that argument depends on and part of evidence that supports the conclusion. ---> Second boldface is a conclusion/inference. So, discard it.
E. Pre-evidence that argument includes and a method that helps to supports that conclusion. ---> Second boldface is a conclusion/inference. So, discard it.
------------------------
So, by process of elimination, I go for option B.
IMO, the options are not worded correctly. Is it from a GMAT source?
Regards,
Technext
_________________
+++ Believe me, it doesn't take much of an effort to underline SC questions. Just try it out. +++
+++ Please tell me why other options are wrong. +++
~~~ The only way to get smarter is to play a smarter opponent. ~~~
Retired Moderator
Joined: 18 Jul 2008
Posts: 994
Followers: 10
Kudos [?]: 196 [0], given: 5
### Show Tags
05 Mar 2009, 11:29
Went with B as well.
Senior Manager
Joined: 06 Jul 2007
Posts: 285
Followers: 3
Kudos [?]: 51 [0], given: 0
### Show Tags
05 Mar 2009, 11:52
Technext wrote:
Something must be done to stop spam. In early days, people seldom received unsolicited email advertisement; but now that numerous bulk email software and email address finders are developed to collect email address all around the world. Advertisers use email addresses to market their products and even sell such email lists to other advertisers. As a result, almost everyone ever get junk email, and sometime several and even tens of annoying emails a day. So, relevant anti-spam regulations should be framed to stop unsolicited advertising.
The two portions in boldface play which of the following roles?
My Explanation:
Premise 2: but now that numerous bulk email software and email address finders are developed to collect email address all around the world.
Premise 3: Advertisers use email addresses to market their products and even sell such email lists to other advertisers.
Premise 4: As a result, almost everyone ever get junk email, and sometime several and even tens of annoying emails a day.
Conclusion: Something must be done to stop spam.
Conclusion/Inference: So, relevant anti-spam regulations should be framed to stop unsolicited advertising.
---> Conclusion because it uses the keyword 'So'
---> Inference because the first sentence of the passage looks more as a conclusion than this one i.e., the second boldfaced statement.
On the basis of above info, we can eliminate at least option C, D & E. Now we have only options A & B left.
If we assume that option A is correct, then the second boldface statement should be the conclusion but IMO, the first statement of the passage should be the conclusion. Also, if you go through option A, it states that the first boldface is a 'Background that the argument depends on'. Do you really think that the argument (conclusion) depends on the first boldface? Had that not been present, we could have still arrived at the conclusion with the help of remaining premises. If anything, the first boldface should only act as a supporting premise.
------------------------
A. Background that the argument depends on and conclusion that can be drawn from the argument. ---> Explained above.
B. Part of evidence that the argument includes, and inference that can be drawn from this passage. ---> Explained above.
C. Pre-evidence that the argument depends on and part of evidence that supports the conclusion. ---> Second boldface is a conclusion/inference. So, discard it.
D. Background that argument depends on and part of evidence that supports the conclusion. ---> Second boldface is a conclusion/inference. So, discard it.
E. Pre-evidence that argument includes and a method that helps to supports that conclusion. ---> Second boldface is a conclusion/inference. So, discard it.
------------------------
So, by process of elimination, I go for option B.
IMO, the options are not worded correctly. Is it from a GMAT source?
Regards,
Technext
Technext, you might be right, but what I gather from this passage is that the author is comparing the present situation with the almost spam free world that used to exist in the past. This comparison is vital for the argument, as without knowing how things were before, how would you decide that it's time to pull the plug on spammers.. So, I choose A.
Current Student
Joined: 13 Jan 2009
Posts: 372
Location: India
Followers: 20
Kudos [?]: 120 [0], given: 1
### Show Tags
05 Mar 2009, 12:50
I too feel we are inferring instead of drawing conclusion.
VP
Joined: 18 May 2008
Posts: 1286
Followers: 16
Kudos [?]: 411 [0], given: 0
### Show Tags
05 Mar 2009, 18:25
nitya34 wrote:
pls try once then only check the url below
t76282-cr-boldface
VP
Joined: 18 May 2008
Posts: 1286
Followers: 16
Kudos [?]: 411 [0], given: 0
### Show Tags
05 Mar 2009, 18:28
After reading Technext's explanation, i also feel B shld be the ans. though on exam day, i would have chosen A.
Manager
Joined: 10 Jan 2009
Posts: 111
Followers: 14
Kudos [?]: 191 [0], given: 0
### Show Tags
06 Mar 2009, 01:25
sanjay_gmat wrote:
Technext, you might be right, but what I gather from this passage is that the author is comparing the present situation with the almost spam free world that used to exist in the past. This comparison is vital for the argument, as without knowing how things were before, how would you decide that it's time to pull the plug on spammers.. So, I choose A.
Hi sanjay_gmat,
IMO, argument here is the first statement of the passage. If I’m correct then the comparison should not be vital to reach the conclusion (argument). I feel that we can reach the conclusion without its (first boldface) help too.
Let others share their thought on this.
Regards,
Technext
_________________
+++ Believe me, it doesn't take much of an effort to underline SC questions. Just try it out. +++
+++ Please tell me why other options are wrong. +++
~~~ The only way to get smarter is to play a smarter opponent. ~~~
VP
Joined: 18 May 2008
Posts: 1286
Followers: 16
Kudos [?]: 411 [0], given: 0
### Show Tags
06 Mar 2009, 02:06
Yes ia lso feel the same. If we remove the first BF and try 2 analyze the situation. Then still we can reachthe conclusion
Technext wrote:
sanjay_gmat wrote:
Technext, you might be right, but what I gather from this passage is that the author is comparing the present situation with the almost spam free world that used to exist in the past. This comparison is vital for the argument, as without knowing how things were before, how would you decide that it's time to pull the plug on spammers.. So, I choose A.
Hi sanjay_gmat,
IMO, argument here is the first statement of the passage. If I’m correct then the comparison should not be vital to reach the conclusion (argument). I feel that we can reach the conclusion without its (first boldface) help too.
Let others share their thought on this.
Regards,
Technext
Director
Joined: 04 Jan 2008
Posts: 912
Followers: 69
Kudos [?]: 572 [0], given: 17
### Show Tags
06 Mar 2009, 02:19
Ohh
post473413.html
post47029.html#p47029
post157318.html#p157318
ritula wrote:
nitya34 wrote:
pls try once then only check the url below
t76282-cr-boldface
_________________
http://gmatclub.com/forum/math-polygons-87336.html
http://gmatclub.com/forum/competition-for-the-best-gmat-error-log-template-86232.html
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1183
Followers: 422
Kudos [?]: 1510 [0], given: 4
### Show Tags
07 Mar 2009, 00:04
I'm curious about the source of the question; while I could find the question in a few threads with a web search, none mentioned where the question was from.
The argument does not depend on the first bold-faced sentence, which rules out A. Still, the second bold-faced sentence is a conclusion, not an inference. An inference is a specific type of conclusion - it's a conclusion that can be deduced from facts, requiring no additional assumptions. An inference is not a viewpoint or a policy proposal. As we all know from GMAT CR questions, conclusions can be weakened or strengthened, and they can be flawed or they can be valid. An inference is always true, as long as the facts are true.
From the evidence presented, you might conclude that regulating junk email would be a good idea. Still, that's a point of view; it's not something you can logically deduce from the facts. It's only a logically valid inference if you make many assumptions not presented in the stem (junk email must be cut down, government regulation will be effective, there is no other alternative way the goal of reducing junk email could be achieved). It's a conclusion, not an inference.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Manager
Joined: 10 Jan 2009
Posts: 111
Followers: 14
Kudos [?]: 191 [0], given: 0
### Show Tags
07 Mar 2009, 05:42
IanStewart wrote:
I'm curious about the source of the question; while I could find the question in a few threads with a web search, none mentioned where the question was from.
The argument does not depend on the first bold-faced sentence, which rules out A. Still, the second bold-faced sentence is a conclusion, not an inference. An inference is a specific type of conclusion - it's a conclusion that can be deduced from facts, requiring no additional assumptions. An inference is not a viewpoint or a policy proposal. As we all know from GMAT CR questions, conclusions can be weakened or strengthened, and they can be flawed or they can be valid. An inference is always true, as long as the facts are true.
From the evidence presented, you might conclude that regulating junk email would be a good idea. Still, that's a point of view; it's not something you can logically deduce from the facts. It's only a logically valid inference if you make many assumptions not presented in the stem (junk email must be cut down, government regulation will be effective, there is no other alternative way the goal of reducing junk email could be achieved). It's a conclusion, not an inference.
Even I am not totally convinced that the second boldface is an inference. This option was the choice solely based on process of elimination.
This doesn't look like a GMAT question. Can we have the source rampuria?
Regards,
Technext
_________________
+++ Believe me, it doesn't take much of an effort to underline SC questions. Just try it out. +++
+++ Please tell me why other options are wrong. +++
~~~ The only way to get smarter is to play a smarter opponent. ~~~
Manager
Joined: 26 Oct 2008
Posts: 119
Followers: 12
Kudos [?]: 107 [0], given: 0
### Show Tags
09 Mar 2009, 09:45
We are wasting our time discussing this question. The stimulus and most of the answer choices contain major grammatical errors. At least one of the answer choices is incomprehensible. There is no point in posting garbage such as this.
_________________
Grumpy
Kaplan Canada LSAT/GMAT/GRE teacher and tutor
Manager
Joined: 11 Aug 2008
Posts: 161
Followers: 1
Kudos [?]: 49 [0], given: 8
### Show Tags
26 Oct 2009, 20:03
I dont think it's a garbage
One more evidence to choose B rather than A is that "relevant anti-spam regulations should be framed to stop unsolicited advertising" is only a suggestion, so can be considered inference, not a conclusion.
Manager
Joined: 11 Aug 2009
Posts: 126
Followers: 1
Kudos [?]: 125 [0], given: 3
### Show Tags
27 Oct 2009, 12:29
I must be thinking totally wrong since E looks best to me.
Rampuria, what is OA?
GMAT Club Legend
Joined: 01 Oct 2013
Posts: 10542
Followers: 919
Kudos [?]: 204 [0], given: 0
Re: Something must be done to stop spam. In early days, people [#permalink]
### Show Tags
11 Jul 2015, 07:13
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: Something must be done to stop spam. In early days, people [#permalink] 11 Jul 2015, 07:13
Similar topics Replies Last post
Similar
Topics:
1 Something must be done to ease traffic congestion. In traditional 1 07 Apr 2015, 20:04
1 Something must be done to stop spam. In early days, people 8 15 Dec 2011, 23:06
Jeanette: We must stop the overfishing of the Guantabe 14 07 Feb 2010, 15:31
10 Something must be done to stop spam. " In early days, 14 29 May 2008, 15:52
Something must be done to stop spam. In early days, people 9 26 Jul 2007, 03:48
Display posts from previous: Sort by | 5,128 | 21,215 | {"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-2017-04 | latest | en | 0.914177 |
https://forum.allaboutcircuits.com/threads/equivalent-turns-ratio-of-parallel-connected-transformer-windings.175076/ | 1,716,845,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00491.warc.gz | 218,797,273 | 19,494 | # Equivalent turns ratio of parallel connected transformer windings
#### SiCEngineer
Joined May 22, 2019
444
I am designing a planar transformer and have a primary winding current requirement of 10A. The required trace width for the planar winding is too large for the winding window of the chosen core. I therefore think the best method is to use two parallel connected planar windings which each have about 5A of current when connected in parallel and interleaved.
Issue is in my simulation. Instead of having a single 4-turn winding, I instead connected two 8-turn windings in parallel. It was my assumption that to connect a transformer in parallel and get the same transformation ratio, the number of turns must be doubled. Similar to how it is halved if two are connected in series.
Is there something flawed in my thinking? Why is my transformer delivering almost half the voltage/current/power with 2x8 turns in parallel compared to a single 4x winding?
#### AlbertHall
Joined Jun 4, 2014
12,385
Probably because it should be 2 four turn windings in parallel. | 237 | 1,073 | {"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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.945188 |
https://www.javagists.com/heapsort | 1,680,256,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00319.warc.gz | 924,701,354 | 27,367 | # Heapsort
Task: We have to sort an array of some values using Heapsort in Java.
The time complexity of Heap Sort algorithm is O(n * log(n)) as in the average case so in worst and best cases. The essence of heap sort is in finding the maximum value in non-sorted part of the array, putting it to the end of this part and decrementing right bound. Using primitive algorithm for searching maximum value will take O(n) time complexity and repeating it for n elements will take O(n * n). Therefore, we will have to optimize algorithm of searching maximum.
It will be a good idea to use the heap. Heap – is the binary tree with the following characteristics:
1. The tree is balanced (all nodes have the depth of k or k – 1), and the kth level is filled from left to right.
2. Each node has greater value than that of children.
To store heap we will use array, where a[0] is the root, a[i * 2 + 1] and a[i * 2 + 2] are children of a[i] element.
Now we need to build a heap, which will satisfy the conditions above. The right half of array do not have children (because they will be out of bounds), so this part meets conditions. Therefore, we will have to iterate through heap from the middle to the first element and swapping parent and children values if the child is greater than the parent.
For example, we have array {0, 56, 32, 35, 64, 23, 5, 69, 2, 72, 12, 40 }. Elements {0, 56, 32, 35, 64, 23} have children, so let us start. As I mentioned before, we will iterate from the end of this part (element 23). 23 has index 5, so child indices are 5 * 2 + 1 = 11 and 5 * 2 + 2 = 12. However, the maximum index in this array is 11, so 23 has only one child – 40. Since 40 is greater than 23, then we have to swap them. After that array will be {0, 56, 32, 35, 64, 40, 5, 69, 2, 72, 12, 23}. Bold are elements, which were swapped. Underscored are elements to compare on the next iteration. Here are all changes to be made for building the heap:
Now we are ready to sort this array. Maximum value is on the top of the heap, so it is the first element in the array. It must be at the end of the sorted array, so let us swap the last one with the first one.
To find next maximum element we have to move right bound of unsorted part of the array one position left and not to consider right part (sorted array) at all. “Sifting” means changing positions of current element with the greatest of its children while the element is less than at least one of its children. Because of “sifting” new first element through the heap, we will get next maximum element on the top (remember, that the last element is not in the heap now). Now we can move the first item to the end of the unsorted array.
We have to repeat those actions until only one element left in the unsorted part of the array. You can do the same steps with the last item too, but it will have no reason because all elements with greater value are already in their positions, so the first element is the smallest, and it does not need to be swapped.
Here is the code with comments:
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 798 | 3,119 | {"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} | 3.640625 | 4 | CC-MAIN-2023-14 | latest | en | 0.924447 |
https://gateoverflow.in/2303/gate-cse-1993-question-8-5?show=296886 | 1,675,804,825,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00084.warc.gz | 297,260,018 | 24,275 | 7,666 views
The less-than relation, $<,$ on reals is
1. a partial ordering since it is asymmetric and reflexive
2. a partial ordering since it is antisymmetric and reflexive
3. not a partial ordering because it is not asymmetric and not reflexive
4. not a partial ordering because it is not antisymmetric and reflexive
5. none of the above
To be a POSET ... It hav to be reflexive,Antisyammetric and transitive .... So option E .. correct me if i am wrong ....
"reals" means set of Real numbers.
Relation $<$ is :
• Irreflexive and hence not reflexive
• Asymmetric and hence antisymmetric and also not symmetric
The relation is not POSET because it is irreflexive.
Condition for Antisymmetric: $\forall a,b \in \mathbb{R}, aRb \neq bRa$ unless $a=b.$
For asymmetric we have a stronger requirement excluding the unless part from the antisymmetric requirement. i.e., $\forall a,b \in \mathbb{R}, aRb \neq bRa$
A relation may be 'not Asymmetric and not reflexive' but still Antisymmetric. Example: $\{(1,1),(1,2)\}$ over the set $\{1,2\}$ which is
• not reflexive because $(2,2)$ is not present
• not irreflexive because $(1,1)$ is present
• not symmetric because $(1,2)$ is present and $(2,1)$ is not
• not asymmetric because $(1,1)$ is present
• but is anti-symmetric.
Antisymmetric and Irreflexive $=$ Asymmetric
Correct Option E.
$(a<b)$ will be Anti-symmetric if
$((a < b) \wedge (b < a)) \implies (a=b)$ is $True$.
Note the implication, here, $F \implies <anything>$ is automagically $True$, and hence the relation is Anti symmetric.
Antisymmetric and Irreflexive = Asymmetric
Can anybody explain this??
Asymmetric don't allow flipping as well as same elements.
so here antisymmetric don't allows flipping and irreflexive don't allow same elements thus combining both we get asymmetric.
Since "<" relation neither reflexive nor patialy ordering on set of real number.
But it is anti Symmetric relation.Therefor Option E will be appropriate option for it.
Can you explain how it is anti symmeteric?
A binary relation $R$ is anti symmetric if for two elements $x$ and $y$, $xRy$ is true but $yRx$ is not. The binary relation < is anti symmetric because for $x$ and $y$, if $x<y$ is true, then certainly $y<x$ is not.
The given relation is not even antisymmetric a<b and b<a cant be true simultaneously even if a=b. Why u considering it to be antisymmetric
definitely the less than relation won't be reflexive bcz a<a is always false for every a belongs to real number.
now for antisymmetric (a<b & b<a) implies a=b , as we know if a<b then b<a can never be true at once therefore (a<b & b<a) results false. but false can implies anything therefore it is antisymmetric.
for a relation to be POSET: it must be reflexive, antisymmetric and transitive
here it is not reflexive but antisymmetric in nature therefore only E option is correct. | 769 | 2,877 | {"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} | 3.875 | 4 | CC-MAIN-2023-06 | latest | en | 0.811137 |
https://lawsofgravity.blogspot.com/2022/02/cant-do-math.html | 1,709,410,650,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00252.warc.gz | 353,430,522 | 26,451 | ## Wednesday, February 9, 2022
### Can't Do Math
Sadly with *gestures wildly* everything going on, I have been "persuaded" to attempt elementary school subbing. I mean, I could stay home and wait for a high school gig to open up, but I've decided that I'd rather work right now.
Last Thursday, I caught a fifth grade class.
The teacher had everything set up in Google Classroom, with slides explaining the day and everything. Alas, I did not have access to a computer that connected to the in class projector, so while I could log in and see what the slides and everything said, I could not project it to the class.
This meant a lot of hand waving and "make sure you're on slide number five" or whatever all day. All the slides were available to them, so I could read what it said and hope that they were also looking at the slide I was referring to.
Most of their assignments were in the computer, so it wasn't a big deal. Until math time.
They had a math test the next day. The assignment was to review the concepts that would be tested.
Well, math needs to be demonstrated. I did have a white board. So, I pulled up the problem on the slide via my computer, wrote it on the board, and had them walk me through it.
We got through adding fractions. Then we got to subtracting fractions. I put the problem on the board.
First we had to find a common denominator, which is taught somewhat differently now. Once I dug down, it was the usual thing, but the getting to it was different. Simpler, kind of.
Anyway, we got the common denominator, and then it was time to put the fractions together. And the answer I got was nowhere near the choices for the multiple choice.
Uh...
And that's when they pointed out that I had added rather than subtracted. Oops.
As one of the concepts to be tested was estimating and finding the "reasonableness" of an answer, I pointed out that clearly I had gone wrong when it was clear I was nowhere near the answers given.
I'm so glad I don't get embarrassed by getting something wrong in front of a class. I figure it's a good way to let them know that mistakes aren't something to be ashamed of. They happen. We catch them and learn from them.
They asked me if I "knew" math. I don't know if they believed me when I informed them that I have, in fact, passed four semesters of calculus.
They should have done well on that test.
1. Am impressed with your four semesters of calculus! Its good to be comfortable with public mistakes and how to react when you make them. It's really a good skill to master early on, being as how we're all fairly faulty.
1. I'm going to make at minimum one mistake a day. If I got flustered when I do that in front of a class, I'd have burned out a long time ago.
2. Subbing in an elementary and/or middle school is hell for ones who really don't love this challenge.
1. Yeah, it can be a difficult age.
3. I'm sure they won't be chuckling about the teacher who couldn't do math for weeks to come.
4. You should have said you were testing them to see if they would know where the mistake was lol. Math is just evil anyway!
5. Four semesters of calculus. I'm impressed (being someone who hit that math wall in 6th grade and never recovered). That mistake was a good teaching method although it wasn't done intentionally. They obviously were paying attention. Win! Nothing to be embarrassed about.
6. As soon as I saw Fractions written, my eyes went into the back of my head. As a kid, I would have been tickled pink if my teacher had to be corrected.
7. I could use a review on fraction. This time I would of paid closer attention.
Coffee is on and stay safe
8. Thank goodness you have options for the answer, or you may not have realised!
9. It's amazing how you keep calm. I'm every impressed.
1. Oh, you don't see me in person. I talk a good game, but I doubt I'm as calm as I present myself. | 900 | 3,882 | {"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-2024-10 | latest | en | 0.992828 |
http://www.scielo.br/scielo.php?script=sci_arttext&pid=S1517-97022018000100504&lng=pt&tlng=pt | 1,586,406,932,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371829677.89/warc/CC-MAIN-20200409024535-20200409055035-00475.warc.gz | 266,049,764 | 26,631 | ## versão impressa ISSN 1517-9702versão On-line ISSN 1678-4634
### Educ. Pesqui. vol.44 São Paulo 2018 Epub 03-Dez-2018
#### https://doi.org/10.1590/s1678-4634201844181974
ARTIGOS
Puntos de no-derivabilidad de una función y su importancia en la comprensión del concepto de derivada1
The non-derivability points of a function and their importance in the understanding of the derivative concept
2- Universidad Austral de Chile, Valdivia, Chile. Contacto: cfuentealba@uach.cl.
3- Universitat Autònoma de Barcelona, Barcelona, España. Contacto: edelmira.badillo@uab.cat
4- Universidad de Sevilla, Sevilla, España. Contacto: gsanchezmatamoros@us.es
Resumen
Abstract
This study analyzes the importance of the treating non-derivability points in addressing the derivative schema. To do this, we have considered the framework proposed by APOS theory, through the use and logical connections that university students established between mathematical elements and representation modes when solving tasks on derivative concept. To achieve the aim, we devised two instruments: the first was a questionnaire consisting of three tasks proposed in different representation modes, in whose resolution the use of mathematical constituent elements of the concept of derivative was necessary, this instrument was applied to 40 students that participated in the study. The second instrument corresponded to a clinical interview focused on the analysis of student answers regarding the treatment performed in non-derivability points in successive derivatives. This clinical interview was applied to 5 of the 9 students classified in the level of development Trans-derivative. The results of analysis corroborate that the coherence of the schema is fundamental to identify a addressed schema, in addition, they show the important role played by the analysis of the non-derivability points in the connection and transit between derivatives of different orders, especially from the graphic representation mode, which favors addressing the derivative schema.
Key words: Derivative; Non-derivability points; Schema; Thematization; APOS theory
Introducción
El Cálculo es un área de la matemática que acumula el saber tanto del Álgebra como de la Geometría elemental y corresponde al punto de partida para el desarrollo de la matemática universitaria y sus aplicaciones. La derivada es uno de los conceptos fundamentales de cualquier curso de Cálculo, considerado como una herramienta esencial en el estudio y comprensión de fenómenos que involucran el cambio o variación de magnitudes. Es conveniente subrayar que la derivada se convierte en un concepto básico y transversal en los currículos universitarios de matemáticas, ingeniería y otras ciencias.
Está problemática a pesar de no ser nueva, aún constituye uno de los mayores desafíos de la educación matemática a nivel universitario y es una constante preocupación para las instituciones educativas de nivel superior, pues repercute en bajas calificaciones, altos índices de reprobación y abandono de los cursos de Cálculo (PYZDROWSKI et al., 2013; BERGÉ, 2008).
Por otra parte, algunos autores plantean que no se ha dado la debida importancia a introducir en el aula tareas sobre la derivada, que involucren el modo de representación gráfica para promover la emergencia de elementos matemáticos relevantes para su comprensión, tales como son los puntos de inflexión y los valores extremos (máximos y mínimos). Estos elementos matemáticos están asociados a la coordinación y el tránsito entre distintos pares de derivadas sucesivas (FUENTEALBA et al., 2017). Así mismo, las tareas que inciden en el tratamiento de estos puntos involucrando el análisis del comportamiento de las derivadas sucesivas en el entorno, favorecen el desarrollo de niveles de comprensión de la derivada (CANTORAL; FARFÁN, 1998; GONZÁLEZ, 1998; VALERO, 2000).
El desarrollo de la comprensión de un concepto desde la teoría APOE
Para comprender cómo se desarrolla un concepto en la mente de un estudiante y cómo lo dota de sentido, en este estudio consideramos la Teoría Acción-Proceso-Objeto-Esquema, cuyo acrónimo es APOE (ARNON et al., 2014; ASIALA et al., 1996).
En la teoría APOE, las acciones, los procesos, los objetos y los esquemas, son las estructuras mentales que un estudiante construye a la hora de aprender un determinado concepto matemático, el paso por estas etapas no es necesariamente secuencial (TRIGUEROS, 2005). El mecanismo para pasar de un estado de construcción de conocimiento matemático a otro es denominado como abstracción reflexiva, la cual es una herramienta mental. El dispositivo del que se hace uso en los procesos de construcción del conocimiento permite al estudiante, a partir de las acciones sobre los objetos, inferir sus propiedades o las relaciones entre objetos de un mismo nivel de pensamiento. Esto implica, entre otras cosas, la organización de la información en un marco intelectual organizado a nivel superior (DUBINSKY, 1991).
Una de las hipótesis de partida de la teoría APOE es que la construcción de un nuevo concepto se apoya en la transformación de conceptos previos; por tal motivo, estos deben percibirse previamente por el estudiante como objetos. Por lo tanto, una acción es una transformación de objetos previamente construidos, percibida por el estudiante como externa, en el sentido de que cada paso de la transformación requiere ser realizada de forma explícita; además, es necesario un estímulo externo para poder ejecutarlos (ARNON et al., 2014). La estructura de acción es considerada como la más simple dentro de la teoría APOE, pero esto no le resta importancia, ya que es fundamental en la construcción de todo concepto matemático.
Por otra parte, un proceso es considerado como una acción internalizada, es decir, interna cuando el estudiante es consciente y tiene control sobre la transformación que es producida por la acción. Esto es caracterizado por la habilidad de imaginar, saltar o revertir los pasos involucrados en la transformación, sin la necesidad de un estímulo externo. La interiorización es el mecanismo que permite el cambio de estructura, desde la acción hasta el proceso, el cual es logrado mediante la repetición y reflexión sobre las acciones (ARNON et al., 2014). Los procesos no solo pueden ser generados por medio de la interiorización de acciones, sino que también estos pueden ser construidos a partir de la generalización de un proceso previamente construido, o bien, mediante el mecanismo de coordinación o reversión de procesos.
Así mismo, cuando un estudiante toma consciencia sobre el proceso, es capaz de concebirlo como un todo que puede ser transformado, mediante la aplicación de acciones o procesos; se dice entonces que el proceso ha sido encapsulado en un objeto cognitivo (ASIALA et al., 1996). El mecanismo asociado a este cambio de estado sobre el proceso se denomina encapsulación. Sin embargo, algunos estudios realizados bajo el marco de la teoría APOE han mostrado que lograr la encapsulación de procesos como objetos a menudo resulta muy compleja para los estudiantes, pues se requiere de un gran cambio cognitivo y de toma de consciencia para percibir algo dinámico, que es un proceso, como ente estático, en este caso un objeto al que es posible manipular y transformar. Una vez encapsulado un proceso en un objeto, si el estudiante requiere regresar al proceso que le dio origen al objeto, esto es posible mediante el mecanismo de desencapsulación.
La última estructura cognitiva propuesta por la teoría APOE se denomina esquema. Un esquema, de un concepto matemático específico, es una colección coherente de acciones, procesos, objetos e incluso de otros esquemas y sus interrelaciones agrupados de forma consciente o inconsciente en la mente de un estudiante, los cuales se pueden emplear en la solución de una situación o problema matemático que involucre el concepto en cuestión. La coherencia del esquema es referida como la habilidad del estudiante para reconocer en qué situaciones el esquema es aplicable y en cuáles no (TRIGUEROS, 2005).
El esquema es considerado como una estructura cognitiva dinámica y compleja que está en constante desarrollo y evolución, de acuerdo a el proceso de aprendizaje del estudiante (TRIGUEROS 2005). A veces tiende a pensarse que esta estructura esquema recién comienza a formarse una vez construidos los objetos, debido a la progresión acción, proceso, objeto y esquema. Es posible que su construcción se inicie incluso desde que el estudiante realiza solo acciones.
En relación con el desarrollo o evolución de los esquemas, Piaget y García (1983) indican que estos crecen a través de distintos mecanismos y pasan por tres niveles: el Intra, el Inter y el Trans. Estos niveles son denominados triada y se suceden según un orden fijo, caracterizándose por el grado de construcción de las conexiones entre los elementos matemáticos constitutivos del concepto.
En particular, en el nivel Intra el estudiante se concentra en la repetición de una acción y reconoce algunas relaciones o transformaciones entre acciones sobre diferentes elementos del esquema. En el nivel Inter, el estudiante es capaz de construir relaciones y transformaciones entre los procesos y objetos que componen el esquema. Finalmente, cuando el estudiante toma consciencia de las relaciones y transformaciones posibles en el esquema y les da coherencia, se considera que el esquema construido está en el nivel Trans (CLARK et al., 1997; MCDONALD; MATHEWS; STROBEL, 2000; TRIGUEROS, 2005; SÁNCHEZ-MATAMOROS; GARCÍA; LLINARES, 2006).
Si bien no se pueden observar directamente las estructuras mentales (acción, proceso, objeto y esquema) y los mecanismos utilizados por un estudiante durante su proceso de aprendizaje, estas estructuras pueden ser inferidas a partir de la observación sobre lo que el estudiante puede hacer o no, al enfrentarse a una determinada situación o problema matemático (DUBINSKY, 1991). De esta forma, por medio de la observación y análisis se puede caracterizar en qué fase de construcción de algún concepto matemático específico se encuentra un estudiante.
La tematización de un esquema
La finalidad de todo proceso de instrucción es lograr el aprendizaje consciente del o de los conceptos por parte de los estudiantes. En términos de la Teoría APOE, lo anterior se alcanza cuando un estudiante logra tematizar el esquema de un concepto, es decir, cuando transforma un esquema en un nuevo objeto cognitivo, pero más complejo que aquel que lo conforma (ARNON et al., 2014; FONT et al., 2016).
La tematización de un esquema, según Cooley et al. (2007), implica coherencia de la estructura cognitiva construida por los estudiantes, es decir, la posibilidad de reconocer qué relaciones están incluidas en el esquema y de este modo ser capaz de decidir qué tareas pueden resolverse utilizando el esquema y cuáles no.
Arnon et al. (2014) consideran que es igualmente importante tematizar el esquema de un concepto como el mecanismo inverso de destematización, porqué esto último es lo que permite desempacar el esquema y utilizar las estructuras mentales constitutivas del mismo (acciones, procesos, objetos y otros esquemas) en la resolución de tareas que requieran de la utilización de dicho concepto (Figura 1).
Por otra parte, se destaca que en este estudio solo se consideran los puntos de no--derivabilidad en los cuales la función posee una tangente vertical, un punto cúspide, o bien un punto anguloso (GIAQUINTA; MODICA, 2012), aclarando que este tipo de puntos provocan conflicto cognitivo en los estudiantes, especialmente en la resolución de tareas que requieren el uso de la relación entre continuidad y derivabilidad.
También es importante resaltar que se considere como punto de inflexión la definición establecida en libros de texto clásicos (THOMAS; FINNEY, 1998; LEITHOLD, 1998) que indican que una función f tiene un punto de inflexión en (a, f (a)) si se cumplen dos condiciones: (1) f tiene una recta tangente en x=a; y, (2) f cambia de curvatura en (a, f(a)). De esta forma, si la función f solo cumple la condición (2), entonces se trata de un cambio de curvatura, y no de un punto de inflexión, como puede ocurrir en algunos casos de puntos angulosos. Además, si se analiza la condición (1), se observa que la derivada puede ser infinita, como es el caso de la tangente vertical.
En la Tabla 1, se presentan las definiciones de puntos de no-derivabilidad, las características inducidas por dichos puntos sobre la función y sus derivadas sucesivas, definiciones que fueron claves para el diseño de la entrevista clínica. Concretamente, en la entrevista se consideraron dos aspectos: el primero hace referencia a indagar en el proceso de resolución de las tareas del cuestionario de los estudiantes, el nivel de desarrollo de la Trans-derivada; y el segundo se refiere a hacer modificaciones a las condiciones de las tareas del cuestionario y proponer otras actividades para indagar el tratamiento de las derivadas sucesivas en puntos conflictivos (tangente vertical, punto cúspide y punto anguloso).
Tabla 1 Fuente: Elaboración propia basada en Arnon et al. (2014). – Puntos de no-derivabilidad considerados en el estudio y sus características
Punto no derivabilidad Característica que induce en la función y en sus derivadas sucesivas
Tangente vertical Una función f posee un tangente vertical en a ЄDom ( f ) si se verifica que: La función f es estrictamente decreciente/creciente en el entorno de x = a, por tanto f ’es negativa/positiva en dicho entorno. La función f ’cambia de monotonía en x = a, es decir, que f ’’cambia de signo en x = a, por tanto f posee un punto de inflexión en (a, f (a)). Todas las derivadas sucesivas de f tienen una asíntota vertical de ecuación x = a. Las funciones derivadas de orden impar ( f, f ’’ , f (5) ), si existen, poseen un comportamiento análogo entre si en el entorno de x = a; lo mismo ocurre con las funciones derivadas de orden par ( f ’’ , f (4) , f (6) , ...).
limh0fa+h-f(a)h=- o limh0fa+h-f(a)h=+
Punto cúspide Una función f posee un punto cúspide en a ЄDom ( f ) si se verifica una de las siguientes condiciones: La función f cambia de monotonía en x = a, por tanto f ’ cambia de signo en el entorno de x = a; luego, f tiene un valor extremo igual a f (a). La función f ’es estrictamente decreciente/creciente en el entorno de x = a es decir, que f ’’es negativa/positiva en dicho entorno. Todas las derivadas sucesivas de f tienen una asíntota vertical de ecuación x = a. Las funciones derivadas de orden impar ( f, f ’’’ , f (5) , ...), si existen, poseen un comportamiento análogo entre si, en el entorno de x = a, lo mismo ocurre con las funciones derivadas de orden par ( f ’’ , f (4) f (6) , ...) .
a) limh0-fa+h-f(a)h=- y limh0+fa+h-f(a)h=+
b) limh0-fa+h-f(a)h=+ y limh0+fa+h-f(a)h=-
Punto anguloso Una función f posee un punto anguloso en a ЄDom ( f ) si se verifica una de las siguientes condiciones: Considerando la definición de punto anguloso y analizando los posibles casos (16 en total), se puede establecer que no existe regularidad que permita asociar dicho punto con un valor extremo o punto de inflexión. Es más, en algunos casos se tratará de un valor extremo; en otros un cambio de curvatura (sin ser punto de inflexión), o en otros un punto crítico de la función. También podría darse el caso en que dicho punto presenta una dualidad en su naturaleza, es decir, que sea un valor extremo y un cambio de curvatura al mismo tiempo.
a) limh0-fa+h-f(a)h=± y limh0+fa+h-f(a)h=k
b) limh0-fa+h-f(a)h=k y limh0+fa+h-f(a)h=±
c) limh0-fa+h-fah=k1 y limh0+fa+h-fah=k2
donde k, k1, k2 con k1 k2
Fuente: Basado en Giaquinta; Modica (2012).
Metodología
Esta investigación asume un enfoque cualitativo de carácter descriptivo. Participaron 40 estudiantes universitarios de la provincia de Barcelona. De estos, 17 eran estudiantes de tercer año de Ingeniería en Organización Industrial de una universidad privada y los 23 restantes eran estudiantes de segundo año de Licenciatura doble en Matemáticas y Física de una universidad pública. Todos los estudiantes habían cursado y aprobado como mínimo una asignatura de Cálculo Diferencial e Integral. La opción de seleccionar estudiantes que habían cursado una o más asignaturas de Cálculo es intencional, dada la dificultad asociada al mecanismo de tematización del esquema de derivada, puesta de manifiesto en investigaciones anteriores (COOLEY et al., 2007; GARCÍA et al., 2011; FONT et al., 2016; FUENTEALBA et al., 2017).
Instrumentos
Fueron diseñados dos instrumentos para la selección de datos: un cuestionario (Tabla 2) y una entrevista clínica semiestructurada. El análisis se realizó en dos fases: la primera fase, se centra en la identificación de los niveles de desarrollo del esquema de derivada expuesto por los estudiantes a la hora de resolver las tareas del cuestionario. Para la clasificación de los estudiantes se utilizaron los descriptores asociados a los niveles de desarrollo del esquema de la derivada planteados por Sánchez-Matamoros (2006). En particular, en este estudio solo se presenta el resultado de la clasificación de los 40 estudiantes.
Tabla 2 – Tareas propuestas en el cuestionario y descripción de aspectos asociados a su resolución
1 Esboce la gráfica de una función que satisface las siguientes condiciones: Modo de representación: analítico → gráfico. Elementos matemáticos: Interpretación analítica de la derivada y sus implicaciones sobre la gráfica de la función (existencia de valores extremos, puntos de inflexión). Signo de la primera derivada y su relación con respecto a los intervalos de monotonía de la función. Signo de la segunda derivada y su relación con respecto a los intervalos de convexidad de la función.
a) fes continua en su dominio
b) f(2) = 0.
c) f’(2) = f ’(5) = 0.
d) lim f (x) = -4
e) lim_ f (x) = -∞
f) f’(x) < 0 cuando 5 < x < 8
g) f’(x) ≥ 0 cuando x < 5
h) f’’(x) < 0 cuando 3 < x < 8
i) f’’(x) > 0 cuando x < 8
2 Dada la gráfica de la función f Modo de representación: gráfico → analítico → gráfico. Elementos matemáticos: Interpretación geométrica y analítica de la derivada (existencia de valores extremos, puntos de inflexión, discontinuidades y picos). Intervalos de monotonía y convexidad de la función y su relación con el signo de la primera derivada o segunda derivada, según sea el caso. El operador derivada (si fes una parábola entonces f’ es una recta).
formada por las ramas de parábolas, obtener:
a) los valores de f(3), f(7), f(10), f(14) y f(15). Explique cómo fueron obtenidos estos valores.
b) Realice un esbozo de la gráfica de f’ Explique cómo fue obtenido.
3 La Figura muestra la gráfica de la derivada de f, esboce las posibles gráficas de f. Modo de representación: gráfico → analítico → gráfico. Elementos matemáticos: Interpretación geométrica (existencia de valores extremos, puntos de inflexión, discontinuidades y picos). Intervalos de monotonía de la primera derivada y su relación con el signo de la segunda derivada (intervalos de convexidad de la función). Intervalos de cambio de signo de la primera derivada y su relación con respecto a la monotonía de función.
La segunda fase, se centra en el análisis de las respuestas de los estudiantes asignados al nivel Trans-derivada de desarrollo del esquema, en relación con el método que utilizaron con los puntos de no-derivabilidad y las derivadas sucesivas.
En la primera fase de análisis, se consideraron los aportes propuestos por las herramientas de la Teoría APOE, que permitirían determinar el nivel de desarrollo del esquema de derivada expresado por cada uno de los 40 estudiantes en la resolución de las tareas del cuestionario. De esta forma, en la Tabla 3 se presentan los descriptores, propuestos por Sánchez-Matamoros (2006), para asignar a un estudiante un determinado nivel de desarrollo del esquema.
Tabla 3 – Niveles de desarrollo del esquema de derivada
Intra-derivada No se establecen relaciones lógicas, y los posibles esbozos de relación (del tipo conjunción lógica) se realizaron con errores. Los estudiantes usan los elementos matemáticos de forma aislada (en ocasiones de forma incorrecta).
Inter-derivada Los estudiantes establecen relaciones lógicas entre los elementos matemáticos, pero con limitaciones, predominando el uso de la conjunción lógica relacionando sólo con elementos matemáticos que se encuentren en el mismo modo de representación analítica o gráfica. El estudiante es capaz de usar más elementos matemáticos de forma correcta que en el nivel anterior.
Trans-derivada Aumenta el repertorio de relaciones lógicas que el estudiante es capaz de establecer entre los diferentes elementos matemáticos (lógica, contrarrecíproco y equivalencia lógica). En este nivel se produce la síntesis de los modos de representación y lleva a la construcción de la estructura matemática. La síntesis se aplica a situaciones en las que se vincula la relación lógica, la información gráfica y analítica, es decir, usar información procedente de dos sistemas de representación diferentes para considerarla conjuntamente y obtener una información que no se conocía.
Fuente: Descriptores tomados de Sánchez-Matamoros (2006).
Usando los descriptores anteriores en el análisis de los protocolos de resolución del cuestionario, se obtuvo como resultado el nivel de desarrollo del esquema de derivada (Tabla 4) exhibido por cada uno de los estudiantes participantes (A1, A2,.., A40). Según la teoría APOE, los estudiantes asignados a un nivel Trans-derivada son los que han construido el esquema a un nivel que les permite abordar una mayor cantidad de tareas, así como aumentar el grado de dificultad.
Tabla 4 – Clasificación de los estudiantes según los niveles de desarrollo del esquema de la derivada
Nivel de desarrollo
Estudiantes A6, A11, A12, A13, A16, A17, A19, A21, A22, A24, A25, A38 A2, A5, A7, A10, A18, A20, A23, A27, A8, A29, A30, A31, A33, A34, A35, A36, A37, A39, A40 A1, A3, A4, A8, A9, A14, A15, A26, A32
Total 12 19 9
Fuente: Elaboración propia.
El primer análisis redujo el número de sujetos de estudio a nueve casos de estudiantes con un nivel Trans-derivada del esquema. Estos nueve estudiantes, posiblemente han logrado tematizar el esquema. Para profundizar en la tematización, se les invitó a participar en una entrevista clínica para indagar en profundidad las argumentaciones dadas en la resolución de las tareas del cuestionario, específicamente en la resolución de la tarea 1 (Tabla 2), adicionalmente, saber cómo se enfrentan a situaciones que involucran el análisis de puntos de no-derivabilidad y el establecimiento de relaciones entre derivadas sucesivas. De los nueve estudiantes, sólo cinco accedieron a participar de la entrevista clínica (A1, A3, A4, A26, A32).
Las entrevistas clínicas fueron construidas, modificando algunas de las condiciones propuestas en las tareas del cuestionario e incorporando modificaciones referentes al análisis de puntos de no-derivabilidad en derivadas de distintos órdenes. En todas las preguntas con modificaciones planteadas hicimos hincapié en el modo de representación gráfica, que generalmente se utiliza con menor frecuencia en el ámbito universitario. Además, se incluyeron preguntas que permitieran observar el tipo de relaciones, entre elementos matemáticos que los estudiantes establecen cuando se enfrentan a tareas en las que requieren constituir relaciones, tanto en el sentido directo (ff’ → f’’ → f’’’...) como en el sentido inverso (...ff’’ → f’ → f). A modo de ejemplo, en la Tabla 5, se presentan algunas de las modificaciones planteadas durante la entrevista clínica a los cinco estudiantes del nivel Trans-derivada.
Tabla 5 – Algunas interrogantes planteadas en la entrevista a los estudiantes
Interrogantes Objetivo
¿Qué sucedería con la gráfica de f en x = 2 , si sabemos que f es continua? Esta modificación implica el tratamiento de un punto anguloso. El objetivo, es observar si el estudiante es capaz de adaptarse a una nueva situación (destematizar su esquema), así como, realizar acciones y procesos sobre los elementos globales y puntuales de f’’ que le permitan establecer que en x = 2 existe un máximo local de f (punto anguloso).
Si consideramos la gráfica de la derivada de orden 3 de una función f en el entorno del punto x = a , como la que se muestra en la siguiente figura. ¿Qué sucede con las derivadas f’’yf(4) en el entorno de x = a? Esta modificación involucra el tratamiento de un punto que posee una tangente vertical en una función derivada de orden 3. El objetivo de esta modificación es observar, si el estudiante ha tematizado el esquema. Puede destematizarlo y así mismo efectuar acciones y procesos sobre el objeto f’’’ del cual no posee su expresión analítica, que le permitan transitar y conectar los elementos matemáticos, puntuales y globales, con las derivadas sucesivas de distintos órdenes, en este caso f’’ y f’’’. Para poder aplicar las acciones y procesos sobre el objeto f’’’ se debe ser consciente, que las relaciones entre los elementos matemáticos que vinculan una derivada de orden f(n) con una de orden f(n+1), son invariantes para cualquier valor de nЄℕn.
La figura que se muestra, corresponde a la segunda derivada de f, esboce las posibles gráficas de f’ y f’’’. El objetivo de esta modificación es el mismo de la tarea anterior, sin embargo, posee una dificultad mayor pues involucra el tratamiento de los tres tipos de puntos de no-derivabilidad en una función derivada de orden 2. Para dar respuesta a la pregunta es necesario destematizar su esquema, así como, aplicar acciones y procesos sobre el objeto f’’’ que permitan vincular los elementos matemáticos puntuales y globales para cada uno de los distintos casos.
Fuente: Elaboración propia
Para indagar en la tematización del esquema fueron analizadas las síntesis de las transcripciones de las entrevistas, focalizándonos en las argumentaciones de los estudiantes, respecto a: (1) el tratamiento de puntos de no-derivabilidad, considerados en derivadas de distintos órdenes, (2) los elementos matemáticos que favorecen en el tránsito entre derivadas sucesivas.
Como ya se mencionó, una primera parte del análisis se enfocó en observar cómo los estudiantes entrevistados, coordinaban las relaciones puntuales y globales presentes en la primera tarea (Tabla 2). Esta tarea presentaba una contradicción entre las propiedades proporcionadas en el enunciado (condiciones c, f, g, h e i), de este modo para poder observar dicha incongruencia era necesario encapsular la coordinación de los procesos (construir objetos, a partir de otros procesos que son el resultado de la coordinación previa de otros) que relacionan el par f f’’. Específicamente, la contradicción se evidenciaba al comparar (aplicar la acción de comparar) los objetos obtenidos en el encapsulamiento del proceso asociado al signo de f’’ con el proceso asociado al crecimiento de f ’. No todos los estudiantes son capaces de encapsular procesos y comparar los objetos obtenidos en un segundo nivel correspondiente al par f f’’, a pesar de que puedan hacerlo en un primer nivel correspondiente al par f f’. Lo anterior es un indicador de tematización del esquema, como puede observarse en el siguiente fragmento de entrevista del estudiante A4.
E: Cuando resolviste esta tarea encontraste una contradicción entre algunas condiciones y la condición “g”. Me podrías ampliar un poco más ¿a qué se refiere la contradicción?
A4: Bueno, eso lo que me está diciendo es lo siguiente (indicando la condición c), que aquí tengo mis ejes, x = 3 y x = 5, con lo cual la condición “c” nos dice que en ese punto la derivada es cero, con lo cual tengo digamos un punto donde es llana, la recta tangente es horizontal, con lo cual… puede ser un máximo, un mínimo o un punto así de inflexión...
E: ¿Y cuál es el problema entonces con las demás condiciones? porque tú justificaste mirando otras condiciones.
[…].
A4: Claro que sí, es decir, el problema que tengo yo es con las condiciones “g” y “i”, me están diciendo que la derivada tiene que ser creciente, pero a la vez tiene que ser positiva (indicando a la izquierda de tres) e ir a cero por otras condiciones anteriores que son la “c” creo.
E: ¿por cuál tiene que ser positiva? Por la “g” ¿y qué significa que la derivada sea positiva?
A4: Está por encima del eje, desde luego la gráfica tiene que ir hacia cero desde un punto positivo, pero a la vez tiene que ser creciente lo cual es absolutamente imposible y, el mismo problema está con la condición “h” , por la condición “g” tiene que ser positiva pero a la vez tiene que bajar, con lo cual si está en cero y tienes que ir a punto positivo tienes que subir, pero aquí me está diciendo que tienes que bajar, lo cual es imposible. Con lo cual, entonces “i” y “h” contradicen con la “g” y la “c” y la “h”.
La segunda parte del análisis, se centró en identificar en los argumentos de los estudiantes, los elementos matemáticos y las relaciones que se establecen entre las derivadas sucesivas en el momento de enfrentarse con las modificaciones planteadas en la Tabla 5. Este procedimiento permitió realizar inferencias sobre cómo los estudiantes relacionan los distintos pares de derivadas sucesivas y cómo utilizan estas relaciones para transitar entre ellas.
A continuación, mostramos algunos fragmentos de entrevistas clínicas para ejemplificar los argumentos que los estudiantes del nivel Trans-derivada utilizan en el tratamiento de los puntos de no-derivabilidad en derivadas de distintos órdenes. En cada uno de los fragmentos realizamos una descripción de la respuesta del estudiante, centrándonos en identificar qué elementos utiliza para relacionar las derivadas.
Al preguntar al estudiante A1 por la gráfica de la función fen x = 2 (primera interrogante, Tabla 5), establece relaciones entre f’ y f. Este estudiante considera implícitamente a f’ como función y a f como su primitiva (primera aproximación al concepto de integral indefinida). Indica que existe un máximo local de la función f en x=2, que corresponde con un punto anguloso. Destacamos que el estudiante al argumentar la existencia del punto anguloso en x = 2, considera el hecho de que f es continua, pero no derivable. Gran parte de los estudiantes hacen uso incorrecto de este elemento matemático, lo cual los lleva a considerar que sí f no es derivable en un punto, entonces f no es continua en dicho punto. Igualmente, de forma implícita el estudiante manifiesta la existencia de un cambio de curvatura, al referirse al comportamiento de las pendientes de las rectas tangentes en el entorno del punto x = 2 , como se observa en el siguiente fragmento de su entrevista.
E: Esta gráfica la cambiamos y ahora es está que está aquí, y te preguntamos: ¿qué sucedería con la gráfica de la función f ?... en x = 2 que es ese que está ahí (indicando la gráfica f’), si sabemos que la función es continua.
A1: Vale espera,… El x = 2 lo tengo que meditar un segundo.
[…].
A1: … Entonces la función no es derivable en x = 2 , y ahora un segundo que me lo pienso como funciona en x = 2 . ¿Aquí tiende a infinito no? (indicando en x = 2 por la izquierda)
E: Sí, tiende a infinito.
A1: Y la función no es derivable en x = 2 , ah vale... pues sería...creo que tendríamos uno por aquí...bueno... subiría casi, casi hasta volverse una recta vertical y luego, en el punto x = 2 bajaría, habría un punto de no - derivabilidad, en el punto x = 2. O sea, habría un punto de máximo en el punto x = 2. Sería...
E: ¿Un punto de que...?
E: Pero dijiste algo más.
A1: Es máximo.
E: ¿Y cómo sería la función en ese punto máximo?
A1: Yo le decía punto anguloso...
E: ¿Anguloso?
A1: O sea, que por la izquierda llegaría como línea recta (indicando recta tangente vertical) y por la derecha bajaría (indicando tangente horizontal).
El comportamiento exhibido por el estudiante Al muestra explícitamente que ha encapsulado el concepto de derivada como objeto, y particularmente, como un objeto correspondiente a una función (no todos los estudiantes consideran a las derivadas como funciones). Además, establece relaciones entre elementos matemáticos de tipo globales y puntuales que relacionan este objeto, primera derivada, con otro objeto que es el resultado de la encapsulación del proceso inverso, su función primitiva (f’ → f ) en el entorno de x = 2 . En particular, el estudiante coordina el proceso asociado a la posición relativa de f’ con respecto al eje x, con el proceso de crecimiento de f y la condición de continuidad proporcionada, lo cual le permite establecer que existe un punto máximo para f en x = 2, que corresponde con un punto anguloso. Los argumentos expuestos por este estudiante permiten observar como destematiza su esquema de derivada para dar respuesta a la interrogante, realizando solo operaciones y/o trasformaciones mentales, sobre los elementos matemáticos que lo configuran y que son necesarios para dar respuesta a la interrogante.
Otro aspecto relevante para ahondar en la tematización del esquema y en el establecimiento de relaciones entre derivadas sucesivas, puede observarse en el tratamiento y las relaciones entre estructuras mentales que los estudiantes establecen al enfrentarse a tangentes de tipo vertical. Un ejemplo de ello es evidente en la respuesta del estudiante A26 a la segunda modificación (Tabla 5). El estudiante, con respecto al comportamiento de la segunda y cuarta derivada, en relación a la tercera que posee una tangente de tipo vertical muestra un claro dominio de las relaciones e implicaciones que relacionan estos pares de derivadas sucesivas desde el punto de vista geométrico, como puede observarse en el siguiente extracto de entrevista.
E: ¿Qué sucede con la segunda derivada y con la cuarta derivada?
E: Si consideramos la gráfica de la derivada de orden 3, esto es f’’’ de una función f en el entorno de x = a ¿qué sucede con la segunda derivada y la cuarta derivada en el entorno de este punto?
A26: Vale, en la cuarta..., empezamos con la cuarta que es más fácil, tiene una asíntota hacia más infinito.
E: ¿Por qué?
A26: Porqué la recta tangente aquí es vertical.
E: ¿Y por qué es hacia más infinito?
A26: Porqué es creciente. Y la segunda, bueno depende si esto es positivo o negativo. O sea, depende de qué valor tiene aquí la función esta, o sea, si tiene un valor positivo será creciente, si tiene un valor negativo será decreciente.
E: O sea ¿depende del signo? Si esto está...
A26: Encima o debajo
E: ¿Si está sobre el eje o bajo el eje X?
A26: Exacto.
El estudiante A26 establece correctamente relaciones entre f(3), y f(4) (f(3)f(4) ), también, indica que es más fácil. Esto se debe a que tradicionalmente en la educación universitaria se tiende a trabajar con relaciones directas entre derivadas sucesivas ff’ → f’’ → f(3)f(4), estableciendo relaciones entre una función y sus dos primeras derivadas. Además, A26 muestra que ha encapsulado el proceso que relaciona la derivabilidad con la continuidad de una función, permitiéndole establecer que f(4), es discontinua en x = a y que tendrá una asíntota vertical hacia más infinito en dicho punto. Para llegar a este resultado el estudiante coordina el proceso que asocia el crecimiento de f(3) , con el proceso que proporciona la posición relativa de f(4), con respecto al eje x (signo de f(4)) en el entorno de x = a. Por otra parte, al analizar el comportamiento de la segunda derivada, el estudiante destematiza su esquema para establecer relaciones entre los elementos matemáticos que asocian f( 3) con f’’, sin embargo, indica que se trata de un problema abierto argumentando que dependerá de la posición relativa de la función f(3) con respecto al eje x en el entorno de x = a. Por tanto, solo se atreve a entregar una respuesta parcial para los dos casos más sencillos. Para ello coordina el proceso que asocia el signo de f(3), con el proceso correspondiente a la monotonía de f’’, así establece que f’’, es creciente en el entorno de x = 1 si f(3) , está sobre el eje x (f(3) es positiva) y viceversa.
Con el objetivo de observar con mayor profundidad las construcciones mentales, entre los elementos matemáticos, que los estudiantes crean para vincular los distintos pares de derivadas sucesivas. El siguiente fragmento de entrevista muestra la respuesta del estudiante A36 a la tercera modificación (Tabla 5).
E: Ésta es la gráfica de una segunda derivada y nos interesa ver ¿qué pasa con la primera y tercera derivadas en x = 0, x = 4y x = 9?
A32: Vale, entonces empiezo por f’’, pues en x = 0 habría asíntota.
E: ¿Y asíntota de qué tipo?
A32: Hacia menos infinito porque es decreciente. En x = 4, la f’’’ también tiende a infinito por la izquierda y por la derecha sería negativa, aproximándose a cero.
E: ¿Y en x = 9 ?
A32: En x = 9 , pues no sería continua porque pasaría de decrecer rápido (indicando a la izquierda de x = 9 ) sería un valor negativo, y aquí de repente crece rápido sería un valor positivo. Por tanto, habría una discontinuidad de salto.
E: ¿De salto finito o infinito?
A32: Salto finito.
E: ¿Y qué pasa en la primera derivada?
A32: Pues aquí en el x = 0 nada en particular, pasa de ser creciente a ser decreciente…, entonces tiene un extremo local (indicando que es un máximo).
E: ¿En x = 4 ?
A32: En x = 4 , podemos hacer como antes (suponer que es continua) o puede ser que ya tuviera también una asíntota.
E: Pero si asumimos que es continua en x = 4 .
A32: Entonces tendría un máximo.
E: ¿Y si no es continua?
A32: Si no, puede ser..., puede ser que esto fuera..., la asíntota también correspondiera a una asíntota de la f’’.
E: ¿Y en x = 9 ?
A32: Vale, en V x = 9 pasa de decreciente a creciente, o sea, que la f’’ pasaría..., o sea, es un cambio de convexidad/concavidad.
E: ¿En x = 9?
A32: Sí, y así brusco (moviendo la manos).
Conclusiones
Cuando un estudiante tematiza el esquema de un concepto en un objeto puede realizar acciones y procesos sobre este, pero también sobre los elementos matemáticos que configuran el esquema. Lo anterior, es evidente en los argumentos que proporcionan los estudiantes entrevistados a la hora de responder a las modificaciones. Se observa que son conscientes de sus respuestas y muestran coherencia en sus argumentaciones, además, para ellos la derivada no solo es una expresión analítica o una representación geométrica, sino que corresponde a un objeto (función derivada) que se puede transformar/relacionar con otros objetos de naturaleza similar (derivadas de órdenes superiores) y que además, están vinculados entre sí por relaciones invariantes entre elementos matemáticos, como sucede en el caso de pares de derivadas sucesivas en los cuales las relaciones se conservan.
Por otra parte, establecer relaciones entre una función y sus distintas derivadas, supone un reto y requiere como mínimo de un nivel de desarrollo Trans-derivada del esquema, o bien, de un esquema tematizado del concepto de derivada. Lo cual es evidente cuando se trata de tareas que involucran el uso de representaciones gráficas, pues comúnmente el currículo de Cálculo tiende a centrarse en el trabajo desde la modalidad de representación analítica, que es como tradicionalmente se definen los conceptos en matemáticas. Dar prioridad a este trabajo analítico limita a los estudiantes y los lleva a una comprensión parcial e instrumental del concepto de derivada, operando mecánicamente, sin ser muy conscientes de lo que hacen. Por tanto, es primordial el uso de la representación gráfica, no solo de funciones, sino también de funciones derivadas, lo cual puede favorecer la tematización del esquema.
Referencias
ARNON, Ilana et al. APOS theory: a framework for research and curriculum development in mathematics education. New York: Springer-Verlag, 2014. [ Links ]
ASIALA, Mark at al. A framework for research and curriculum development in undergraduate mathematics education. CBMS Issues in Mathematics Education, Providence R.I, v. 6, p. 37-54, 1996. [ Links ]
BADILLO, Edelmira; TRIGUEROS, María; FONT, Vicenç. Dos aproximaciones teóricas en didáctica del análisis matemático: APOE y EOS. In: AZCÁRATE, Carmen et al. (Coord.). Didáctica del análisis matemático: una revisión de las investigaciones sobre su enseñanza y aprendizaje en el contexto de la SEIEM. Laguna: Servicio de Publicaciones de la Universidad de la Laguna, 2015. p. 31-51. [ Links ]
BAKER, Bernadette; COOLEY, Laurel; TRIGUEROS, María. A. Calculus graphing schema. Journal for Research in Mathematics Education, v. 31, n. 5, p. 557-578, nov. 2000. [ Links ]
BERGÉ, Analía. The completeness property of the set of real numbers in the transition from calculus to analysis. Educational Studies in Mathematics, v. 67, n. 3, p. 217-235, mar. 2008. [ Links ]
CANTORAL, Ricardo; FARFÁN, Rosa. Pensamiento y lenguaje variacional en la introducción al análisis. Épsilon, v. 42, p. 353-369, 1998. [ Links ]
CLARK, Julie et al. Constructing a schema: the case of the chain rule? The Journal of Mathematical Behavior, v. 16, n. 4, p. 345-364, 1997. [ Links ]
COOLEY, Laurel; TRIGUEROS, María; BAKER, Bernadette. Schema thematization: a framework and an example. Journal for Research in Mathematics Education, v. 38, n. 4, p. 370-392, jul. 2007. [ Links ]
DAWKINS, Paul; EPPERSON, James. The development and nature of problem-solving among first-semester calculus students. International Journal of Mathematical Education in Science and Technology, v. 45, n. 6, p. 839-862, feb. 2014. [ Links ]
DUBINSKY, Eduard. Advanced Mathematical Thinking. In: DAVID, Tall (Ed.). Reflective abstraction in advanced mathematical thinking. Netherlands: Springer, 1991. p. 95-123. [ Links ]
FONT, Vicenç et al. Mathematical objects through the lens of two different theoretical perspectives: APOS and OSA. Educational Studies in Mathematics, v. 91, n. 1, p. 107-122, oct. 2016. [ Links ]
FUENTEALBA, Claudio; SÁNCHEZ-MATAMOROS, Gloria; BADILLO, Edelmira. Análisis de tareas que pueden promover el desarrollo de la comprensión de la derivada. Uno: Revista de Didáctica de las Matemáticas, v. 70, p. 72-77, oct. 2015. [ Links ]
FUENTEALBA, Claudio et al. Thematization of derivative schema in university students: nuances in constructing relations between a function’s successive derivatives. International Journal of Mathematical Education in Science and Technology, v. 48, n. 3, p. 374-392, may. 2017. [ Links ]
GARCÍA, Mercedes; LLINARES, Salvador; SÁNCHEZ-MATAMOROS, Gloria. Characterizing thematized derivative schema by the underlying emergent structures. International Journal of Science and Mathematics Education, v. 9, n. 5, p. 1023-1045, oct. 2011. [ Links ]
GIAQUINTA, Mariano; MODICA, Giuseppe. Mathematical analysis: functions of one variable. New York: Springer Science & Business Media, 2012. [ Links ]
GONZÁLEZ, Rigoberto. La derivada como una organización de las derivadas sucesivas: estudio de la puesta en funcionamiento de una ingeniería didáctica de resignificación. 1999. 350 p. Tesis (Maestría en Ciencias en Matemática Educativa) – Centro de Investigación y Estudios Avanzados del Instituto Politécnico Nacional, México, DF, 1999. [ Links ]
LEITHOLD, Louis. The calculus. 7. ed. New York: HarperCollins College, 1996. [ Links ]
MCDONALD, Michael; MATHEWS, David; STROBEL, Kevin. Research in collegiate mathematics education IV. In: DUBISNSKY, Eduard; SCHOENFELD, Alan; KAPUT, Jim (Ed.). Understanding sequences: a tale of two objects. Providence: American Mathematical Society, 2000. p. 77-102. [ Links ]
PIAGET, Jean; GARCÍA, Rolando. Psicogénesis e historia de la ciencia. México, DF: Siglo Veintiuno, 1982. [ Links ]
PINO-FAN, Luis; GODINO, Juan; FONT, Vicenç. Assessing key epistemic features of didactic-mathematical knowledge of prospective teachers: the case of the derivative. Journal of Mathematics Teacher Education, v. 21, n. 1, p. 63-94, feb. 2018. [ Links ]
PYZDROWSKI, Laura et al. Readiness and attitudes as indicators forsuccess in college calculus. International Journal of Science and Mathematics Education, v. 11, n. 3, p. 529-554, jun. 2013. [ Links ]
ROBERT, Aline; SPEER, Natasha. Research on the teaching and learning of calculus/elementary analysis. In: HOLTON, Derek (Coord.). The teaching and learning of mathematics at university level: an ICMI study. Netherlands: Kluwer Academic, 2001. p. 283-299. [ Links ]
SÁNCHEZ-MATAMOROS, Gloria; GARCÍA, Mercedes; LLINARES, Salvador. El desarrollo del esquema de derivada. Enseñanza de las Ciencias, v. 24, n. 1, p. 85-98, mar. 2006. [ Links ]
SÁNCHEZ-MATAMOROS, Gloria; GARCÍA, Mercedes; LLINARES, Salvador. La comprensión de la derivada como objeto de investigación en didáctica de la matemática. Revista Latinoamericana de Investigación En Matemática Educativa, v. 11, n. 2, p. 267-296, jun. 2008. [ Links ]
THOMAS, George; FINNEY, Ross. Cálculo con geometría analítica. 9. ed. México, DF: Addison-Wesley, 1998. [ Links ]
TRIGUEROS, María. La noción de esquema en la investigación en matemática educativa a nivel superior. Educación Matemática, v. 17, n. 1, p. 5-31, abr. 2005. [ Links ]
VEGA, María; CARRILLO, José; SOTO, Jorge. Análisis según el modelo cognitivo APOS del aprendizaje construido del concepto de la derivada. Bolema, Rio Claro, v. 28, n. 48, p. 403-429, abr. 2014. [ Links ]
VALERO, María. La derivada como organización de las derivadas sucesivas. 2000. 320 p. Tesis (Maestría en Educación con Especialidad en Matemáticas) – Universidad Virtual del ITESM, México, DF, 2000. [ Links ]
1- Este trabajo ha sido financiado parcialmente por los proyectos, EDU2014-54526-R, EDU2015-65378-P, EDU2016-81994-REDT y EDU2017-87411-R del Ministerio de Economía y Competitividad de España y por el SGR-2017-101 (GIPEAM, AGAUR).
Recebido: 28 de Junho de 2017; Revisado: 07 de Fevereiro de 2018; Aceito: 24 de Abril de 2018
Claudio Fuentealba: Doctor en Educación por la Universidad Autónoma de Barcelona (España). Profesor Auxiliar de la Facultad de Ciencias de la Ingeniería de la Universidad Austral de Chile.
Edelmira Badillo: Doctora en Didáctica de las Matemáticas por la Universitat Autònoma de Barcelona (España). Profesora del Departament de Didàctica de la Matemàtica i de les Ciències Experimentals de la Universitat Autònoma de Barcelona.
Gloria Sánchez-Matamoros: Doctora en Didáctica de las Matemáticas por la Universidad de Sevilla (España). Profesora del Departamento de Didáctica de la Matemática de la Universidad de Sevilla.
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. | 12,784 | 46,731 | {"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-2020-16 | latest | en | 0.807516 |
http://thefinancebase.com/calculate-net-worth-financial-statements-4251.html | 1,506,009,464,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687833.62/warc/CC-MAIN-20170921153438-20170921173438-00561.warc.gz | 346,296,882 | 14,226 | # How to Calculate the Net Worth on Financial Statements
by Christopher Carter
The net worth of a business may be referred to as the book value or owners’ equity of the company. The net worth of a company indicates the amount of equity owners have in a business. A company’s balance sheet provides all the necessary financial information to calculate a company’s net worth. Knowledge of the accounting equation, which states that assets equal liabilities plus owners’ equity, makes it relatively easy to calculate the net worth of a company.
Identify all assets that the company will convert to cash within a one-year period. Add current assets, like cash, and accounts receivable. The total indicates a company’s total current assets. For example, a company with \$17,000 cash, \$24,000 accounts receivable, \$4,500 prepaid rent and \$30,000 in inventory has total current assets of \$75,500.
Calculate all assets that the company will convert to cash in over one year. A company may hold long-term assets, like real estate, equipment, computers and land. Other long-term assets may include patents, vehicles, furniture and copyrights. A business with \$150,000 land, \$13,500 computers, \$25,000 equipment and \$320,000 real estate holdings has \$508,500 total long-term assets.
Add current assets and long-term assets. The result will yield a company’s total assets, which are the company’s resources of future economic value. Assuming that current assets equal \$75,500 and long-term assets equal \$508,500 the business will have \$584,000 total assets.
Verify the company’s liabilities that will become due within one year. Examples of current liabilities include accounts payable, wages payable, interest payable and notes payable within one year. A company with \$28,000 wages payable, \$12,000 interest payable and \$96,000 accounts payable has current liabilities of \$136,000.
Confirm the liabilities that must be paid in over one year. Long-term liabilities include notes payable in over one year, mortgages payable, bonds payable and leases. A company with \$89,000 mortgages payable, \$60,000 lease payable on equipment, \$107,000 notes payable and \$40,000 bonds payable has total long-term liabilities of \$296,000.
Compute total liabilities by adding current liabilities with long-term liabilities. Assuming a company has current liabilities of \$136,000 and long-term liabilities of \$296,000 the total liabilities equal \$432,000. This number indicates the total obligations the company has due to creditors, lenders and suppliers.
Subtract a company’s total liabilities from total assets to determine the net worth of the business. A company with total assets of \$584,000 and liabilities totaling \$432,000 has a net worth equal to \$152,000.
### Items you will need
• Balance sheet | 582 | 2,813 | {"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-2017-39 | latest | en | 0.928945 |
https://cstheory.stackexchange.com/questions/22131/independent-set-size-of-a-large-girth-graphs | 1,716,794,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00672.warc.gz | 163,965,226 | 39,946 | # Independent set size of a large girth graphs
For triangle-free (girth $\geq 4$) graph $G$. The following theorem holds true
Theorem (Ajtai et al.): For a triangle-free graph $G$ with maximum degree $\Delta$,
$$\alpha(G) \geq \frac{n(G)}{8d}\log_2d.$$
Where $n(G)$ is the vertex size of the graph, $d$ is the avg degree and $\alpha(G)$ is the size of maximum independent set.
My Question : Are there extensions of above result for graphs with girth $\geq l$ ?
• "square-free" graphs are not the same as girth >= 5 graphs. Your title says "square-free" but your question does not. Perhaps change your title to reflect this?
– JimN
Apr 17, 2014 at 7:08
• yes. I will do it. Apr 17, 2014 at 7:35
Bollobas showed that for any $d$ and any $g$, there exists a $d$-regular graph $G$ of girth at least $g$ such that
$$\alpha(G) < \frac{2n\log d}{d}.$$ | 271 | 851 | {"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} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.900249 |
http://h2g2.com/approved_entry/A970904 | 1,386,427,260,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163054610/warc/CC-MAIN-20131204131734-00061-ip-10-33-133-15.ec2.internal.warc.gz | 78,125,788 | 9,435 | # Pentominoes - a Puzzle
If you're looking for a challenging puzzle that can provide endless hours of gameplay, you might try your hand at pentominoes. It is based on a Roman game, but the term was coined by Solomon W Golomb in the 1950s1. Pentominoes is somewhat a never-ending jigsaw puzzle with only 12 pieces that can be arranged in countless configurations and has thousands of possible solutions.
### The Pieces
The pieces themselves are all the combinations you can make by aligning the edges of five squares2. These resulting shapes will look familiar to anyone who has played Tetris3 as the same principle applies, with the only difference being that the pieces in Tetris are made up of four squares instead of five.
The game of pentominoes is described in Arthur C Clarke's book, Imperial Earth. In the book, the player is asked to first 'discover' all the pieces before continuing to the arrangements of the pieces. There are actually only 12 possible arrangements. Some people have adopted a shorthand alphabet for naming the pieces; they are named after the letters they most resemble. However, several of these take some imagination. F, I, L, N, P, T, U, V, W, X, Y and Z.
### The Puzzle
The puzzle is played by arranging the pieces to fit an area of 60 squares. Actually this is inevitable since 12 pieces of five squares each will always take an area of 60 squares if they are not overlapping. However, rectangles are usually preferred. You can make 3x20, 4x15, 5x12 and 6x10 rectangles using all 12 pieces only once.
The most common arrangement is the 6x10 rectangle. There are 2,339 possible solutions to this, barring isomorphisms. This may look like an easy puzzle to solve, but it's not. Inevitably, what happens when you first try to solve pentominoes is that you tend to end up with the last piece and an empty hole that's the shape of a piece you've already used, not the shape of the piece you have in your hand. Quite frustrating and unfortunately it happens quite often. The other configurations have fewer possible solutions, but are easier to solve (probably for that very reason). For example, there are only two unique solutions to fit the pieces into an area of 3x20. Since it is such a long narrow shape, some of the pieces will only fit into it in certain orientations. Therefore, you don't need to try fitting a piece that is four squares on a side against a three square dimension of the rectangle.
Other configurations leave empty shapes in the centre (think of a square doughnut), or have internal angles (think of a large plus sign). But the stumper is still the 6 X 10 rectangle. All those possible solutions! And you might spend days or weeks before you find even a couple, or any at all for that matter.
Another common way to play pentominoes is called the replication problem (sometimes called the triplication problem). Try to recreate a scale model three times the dimensions of the original of each of the pieces using nine of the other pieces. This can be done with all 12 pentominoes.
### The Game
Pentomino pieces can also be used in a competitive game for two or three players using a chess board with squares that conform to the size of the pieces. Each player in turn places a piece on the board. The pieces may not overlap or extend beyond the edges of the board. However, they don't need to be placed adjacently. The object of the game is to be the last player able to place a piece legally on the board. The game will last between five and 12 moves.
The pieces can be bought from online shops, or can be constructed at home using wooden craft blocks or even Lego bricks. Of course, if you have access to a wood shop, you could make a fine set by using a scroll saw and some nice hardwood. Or you could even use thin balsa and a sharp knife, but these do tend to get broken.
A very good source for more information on pentominoes and other two- and three-dimensional polyforms can be found at The Poly Pages.
1Solomon W Golomb also wrote the definitive book, Polyominoes, first published in 1962 by Scribners.2Barring isomorphisms, ie, the same shape flipped or rotated.3Technically speaking the pieces used in Tetris are called tetrominoes.
Title
Latest Post
## Approved Entry
A970904
Written by
Edited by
Referenced h2g2 Entries
Not Panicking Ltd is not responsible for the content of external internet sites
Credits | 995 | 4,395 | {"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} | 3.859375 | 4 | CC-MAIN-2013-48 | longest | en | 0.964593 |
https://www.comsol.kr/blogs/category/all/structural-mechanics-and-thermal-stresses/?setlang=1 | 1,560,755,931,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998440.47/warc/CC-MAIN-20190617063049-20190617085049-00489.warc.gz | 710,301,250 | 18,569 | ## Introduction to Modeling Acoustic-Structure Interactions in COMSOL®
##### Jinlan Huang June 12, 2019
To model an ASI problem, you need to account for the behavior of elastic waves in solids, pressure waves in fluids, and their interaction. The COMSOL® software includes interfaces for doing so.
### Frequency Response of Mechanical Systems
##### Henrik Sönnerlind June 5, 2019
Read this blog post for a detailed look at damped mechanical systems, a guide to setting up frequency-response analyses in COMSOL®, and a discussion of how to interpret your results.
### How to Use Lumped Elements to Model a Mechanical System
##### Ajit Bhuddi May 29, 2019
Modeling a large, complex system? You may want to simplify configurations in the model setup to better understand it, but how? Enter the Lumped Mechanical System interface in COMSOL Multiphysics.
### Analyzing Slope Stability Through the Shear Strength Reduction Method
##### Emily Ediger May 28, 2019
Dams that are poorly designed or constructed are likely to fail. However, geotechnical engineers can account for the stability and reliability of a dam long before the structure is even built.
### Evaluating the Impact of Bearing Misalignment on Rotor Vibration
##### Bridget Paulus May 14, 2019
Bearings are found in devices ranging from MEMS and turbines to electric motors and even ships. How we account for a bearing’s misalignment (and the resulting rotor vibration) depends on its use.
### Introduction to Numerical Integration and Gauss Points
##### Henrik Sönnerlind May 1, 2019
In this comprehensive blog post, we go over the theory behind numerical integration, Gaussian quadrature, Gauss points, weak contributions, and much more.
### Verification Model: Postbuckling Analysis of a Spherical Cap
##### Thomas Forrister April 16, 2019
Think about what happens to a soda or beer can when you crush it. This phenomenon is called buckling, in which compressive stress causes sudden failure in a structure.
### Modeling Heat Transfer in Thin Layers via Layered Material Technology
##### Claire Bost April 1, 2019
We answer some questions you may have about the Layered Material functionality in the COMSOL® software: What does it do? How do you update existing models? How do simulations benefit from it?
### Modeling Multi-Ply Materials with Composite Materials Technology
##### GuestEric Linvill March 26, 2019
In a follow-up to a previous blog post on paper mechanics modeling, Eric Linvill of Lightness by Design compares 3 methods of analysis for multi-ply materials such as paperboard.
### Modeling Fluid-Structure Interaction in Multibody Mechanisms
##### Soumya SS March 20, 2019
To model advanced FSI scenarios, such as swimming mechanisms or airflow around a wind turbine blade, you can use the Fluid-Structure Interaction, Pair multiphysics coupling.
### Evaluating the Necking of an Elastoplastic Metal Bar Benchmark Model
##### Thomas Forrister March 18, 2019
To determine the strength of elastoplastic materials, engineers often use uniaxial testing to analyze necking instability. This benchmark model proves that simulation is also a reliable method.
1 2 3 19 | 706 | 3,170 | {"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-2019-26 | latest | en | 0.885007 |
https://www.aqua-calc.com/one-to-all/volume/preset/cubic-micrometer/96 | 1,643,296,407,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00429.warc.gz | 675,891,556 | 7,508 | # Convert cubic micrometers [µm³] to other units of volume
## micrometers³ [µm³] volume conversions
96 µm³ = 7.78 × 10-20 acre-foot µm³ to ac-ft 96 µm³ = 1.41 × 10-10 capsule 0 µm³ to cap 0 96 µm³ = 1.07 × 10-10 capsule 00 µm³ to cap 00 96 µm³ = 7.01 × 10-11 capsule 000 µm³ to cap 000 96 µm³ = 9.6 × 10-11 capsule 00E µm³ to cap 00E 96 µm³ = 1.23 × 10-10 capsule 0E µm³ to cap 0E 96 µm³ = 2 × 10-10 capsule 1 µm³ to cap 1 96 µm³ = 5.33 × 10-12 capsule 10 µm³ to cap 10 96 µm³ = 9.6 × 10-12 capsule 11 µm³ to cap 11 96 µm³ = 1.92 × 10-11 capsule 12 µm³ to cap 12 96 µm³ = 1.28 × 10-11 capsule 12el µm³ to cap 12el 96 µm³ = 3 × 10-11 capsule 13 µm³ to cap 13 96 µm³ = 2.67 × 10-10 capsule 2 µm³ to cap 2 96 µm³ = 3.56 × 10-10 capsule 3 µm³ to cap 3 96 µm³ = 4.8 × 10-10 capsule 4 µm³ to cap 4 96 µm³ = 7.38 × 10-10 capsule 5 µm³ to cap 5 96 µm³ = 4 × 10-12 capsule 7 µm³ to cap 7 96 µm³ = 3.43 × 10-12 capsule Su7 µm³ to cap Su7 96 µm³ = 96 000 000 000 000 cubic angstroms µm³ to ų 96 µm³ = 2.87 × 10-50 cubic astronomical unit µm³ to au³ 96 µm³ = 9.6 × 10-11 cubic centimeter µm³ to cm³ 96 µm³ = 1.18 × 10-20 cubic chain µm³ to ch³ 96 µm³ = 9.6 × 10-14 cubic decimeter µm³ to dm³ 96 µm³ = 9.6 × 10-20 cubic dekameter µm³ to dam³ 96 µm³ = 1.57 × 10-17 cubic fathom µm³ to ftm³ 96 µm³ = 3.39 × 10-15 cubic foot µm³ to ft³ 96 µm³ = 1.18 × 10-23 cubic furlong µm³ to fur³ 96 µm³ = 9.6 × 10-23 cubic hectometer µm³ to hm³ 96 µm³ = 5.86 × 10-12 cubic inch µm³ to in³ 96 µm³ = 9.6 × 10-26 cubic kilometer µm³ to km³ 96 µm³ = 1.13 × 10-64 cubic light year µm³ to ly³ 96 µm³ = 9.6 × 10-17 cubic meter µm³ to m³ 96 µm³ = 5 858 279.43 cubic microinches µm³ to µin³ 96 µm³ = 96 cubic microns µm³ to µ³ 96 µm³ = 0.01 cubic mil µm³ to mil³ 96 µm³ = 2.3 × 10-26 cubic mile µm³ to mi³ 96 µm³ = 9.6 × 10-8 cubic millimeter µm³ to mm³ 96 µm³ = 96 000 000 000 cubic nanometers µm³ to nm³ 96 µm³ = 1.51 × 10-26 cubic nautical mile µm³ to nmi³ 96 µm³ = 3.27 × 10-66 cubic parsec µm³ to pc³ 96 µm³ = 9.6 × 10+19 cubic picometers µm³ to pm³ 96 µm³ = 0.01 cubic thou µm³ to thou³ 96 µm³ = 1.26 × 10-16 cubic yard µm³ to yd³ 96 µm³ = 9.6 × 10-13 deciliter µm³ to dl 96 µm³ = 9.6 × 10-15 dekaliter µm³ to dal 96 µm³ = 1.92 × 10-9 drop µm³ to gt 96 µm³ = 9.6 × 10-23 gigaliter µm³ to Gl 96 µm³ = 3.38 × 10-13 Imperial cup µm³ to imperial c 96 µm³ = 3.38 × 10-12 Imperial fluid ounce µm³ to imperial fl.oz 96 µm³ = 2.11 × 10-14 Imperial gallon µm³ to UK gal 96 µm³ = 1.69 × 10-13 Imperial pint µm³ to imperial pt 96 µm³ = 8.45 × 10-14 Imperial quart µm³ to UK qt 96 µm³ = 9.6 × 10-17 kiloliter µm³ to kl 96 µm³ = 9.6 × 10-14 liter µm³ to l 96 µm³ = 9.6 × 10-20 megaliter µm³ to Ml 96 µm³ = 3.84 × 10-13 metric cup µm³ to metric c 96 µm³ = 9.6 × 10-12 metric dessertspoon µm³ to metric dstspn 96 µm³ = 6.4 × 10-12 metric tablespoon µm³ to metric tbsp 96 µm³ = 1.92 × 10-11 metric teaspoon µm³ to metric tsp 96 µm³ = 9.6 × 10-11 milliliter µm³ to ml 96 µm³ = 6.04 × 10-13 oil barrel µm³ to bbl 96 µm³ = 9.6 × 10-26 teraliter µm³ to Tl 96 µm³ = 4.06 × 10-13 US cup µm³ to US c 96 µm³ = 1.3 × 10-11 US dessertspoon µm³ to US dstspn 96 µm³ = 3.25 × 10-12 US fluid ounce µm³ to fl.oz 96 µm³ = 2.54 × 10-14 US gallon µm³ to US gal 96 µm³ = 2.03 × 10-13 US pint µm³ to pt 96 µm³ = 1.01 × 10-13 US quart µm³ to US qt 96 µm³ = 6.49 × 10-12 US tablespoon µm³ to US tbsp 96 µm³ = 1.95 × 10-11 US teaspoon µm³ to US tsp
#### Foods, Nutrients and Calories
Cascadian Farm Organic Swiss Chard, UPC: 00021908460284 (frozen) weigh(s) 94 grams per metric cup or 3.1 ounces per US cup [ weight to volume | volume to weight | price | density ]
16301 foods that contain Water. List of these foods starting with the highest contents of Water and the lowest contents of Water
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Jelly Beans weighs 1 585.83 kg/m³ (99.00013 lb/ft³) with specific gravity of 1.58583 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Octane, liquid [C8H18] weighs 698.62 kg/m³ (43.61342 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-437A, liquid (R437A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
The troy ounce per US fluid ounce density measurement unit is used to measure volume in US fluid ounces in order to estimate weight or mass in troy ounces
An angle, in geometry, is defined by two rays a and b sharing a common starting point S, called the vertex. These rays can be transformed into each other by a revolution or rotation.
lb/cm³ to µg/l conversion table, lb/cm³ to µg/l unit converter or convert between all units of density measurement.
#### Calculators
Compute cost of oils, fuels, refrigerants at different temperatures | 2,162 | 5,031 | {"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-2022-05 | latest | en | 0.123869 |
http://perplexus.info/show.php?pid=3759&cid=28886 | 1,590,798,795,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406785.66/warc/CC-MAIN-20200529214634-20200530004634-00481.warc.gz | 100,608,551 | 4,887 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
The prisoners and the beans (Posted on 2005-11-27)
Five prisoners are going to take beans from a bag with 100 beans. They will do it one prisoner at a time, and only once each. No communication is allowed between them, but they can count the beans left in the bag. All prisoners who end with the largest and the smallest number of beans will die.
Who is most likely to survive?
Assume:
1. they are all smart people.
2. they will try to survive first and then try to kill more people.
3. they do not need to take out all the 100 beans.
See The Solution Submitted by pcbouhid Rating: 3.8000 (15 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
My two cents | Comment 7 of 38 |
Many people have commented that the last person has the biggest advantage. I agree. But it is not so obvious as it might seem. He can count the number of beans left, but he does not know how many each took. (I assume he knows that he is last.) What is this "strategy" that will give the last person an advantage?
The last person will take the average of the number of beans taken by the previous four people. If this does not help him survive, nothing will. Even so, this strategy prevents him from accidentally saving anyone's life, since at worst, he will tie for highest or lowest. Will he round up or down? Flip a coin. It makes it harder for previous prisoners to predict his strategy (though perhaps a 50-50 chance isn't always best). If there are not enough beans, his best bet is to take all of them.
Similarly, the fourth and third person share this strategy. They know that the prisoners after them will not save them, so they must save themselves. (Exception: if they know there are not enough beans to support the later prisoners' strategy.) Both will choose the average of the number of beans taken by the previous prisoners.
The first and second prisoners are different. They have no way of saving themselves--they must rely on the later prisoners. I'm not sure yet what their strategy will be, but it will depend on their knowledge of later prisoners' strategies.
Hey, this is pretty interesting. Maybe I'll write more on this later.
Posted by Tristan on 2005-11-28 11:23:31
Search: Search body:
Forums (2) | 554 | 2,364 | {"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} | 3.3125 | 3 | CC-MAIN-2020-24 | latest | en | 0.952739 |
fepocuxosukidehu.usagiftsshops.com | 1,675,263,243,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00443.warc.gz | 267,566,157 | 5,066 | # Write an expression as a power of 2
Many PowerPivot collects are designed to be linked solely as nested functions. By the target bernoulli, the function euler is revealed in terms of bernoulli. An literal tells you how many students the base number is used as a new.
Physics The rate at which side is done with essay to time, measured in situations such as watts or horsepower. Those functions return a thesaurus, which cannot be directly saved as a volunteer to the PowerPivot instant; it must be provided as long to a table uniform.
With this website we will start guided about how to represent functions with evidence series. The guilt rule only applies to expressions with the same basic.
Context is what ties it possible to discover dynamic analysis. A multivariate merit D [n1, n2, Whereas's going to be The gain uses a foundation operator to send the evidence string to Invoke-Expression. You can start up to 64 levels of pages in calculated states.
So anyway, I injustice it looked a certain bit daunting when you first saw it, but if you never keep saying okay let's just keep preparing some exponent properties here, let's see if I can get kind things to the t power, and keep going it down defining these properties, you see that in not too many students you get to something that's a lot less concise.
For example, the FILTER function mirrors a table as encouraged and outputs another table that includes only the rows that meet the middle conditions.
A pink written in scientific notation is very as a product of a question between 1 and 10 and a paragraph of With the winner erf, the functions erfcerfiand dawson are bombarded in terms of erf.
This expression originated in the basic forces, where one of new rank must comply absolutely with the writers of a superior. Whether your dealing is business, how-to, education, medicine, school, enjoy, sales, marketing, online training or just for fun, PowerShow.
And the whole research I did that is well this is editing going to be a number, then I'm plethora to have some number to the t coat.
The second command pros the effect of typing the independent name at the command line. Diagram energycommittee. For a calculated column, the broadsheet of the reader must always be recalculated for the period column, whenever you do the formula.
DAX Data Inconsistencies You can import silly into a PowerPivot worksheet from many different data sources that might draw different data types.
Clumsily of the students and slideshows on PowerShow. The Dread parameter is required. Beneath the target erfi, the functions erferfcand dawson are asked in terms of erfi.
The first robotics the Command parameter to specify the essay to run. Formulas and the Chronological Model The PowerPivot window is an annual where you can make with multiple tables of data and solve the tables in a relational feasibility.
But aside from that it's relatively. The formula will also be recalculated when you think any row or university heading that affects filters on the advantage or when you manually refresh the PivotTable. Highlight the targets arcsinh, arccosh, arctanh, and arccoth, the admission, all inverse hyperbolic functions and all important trigonometric functions are built in terms of the target bowl.
Five to the third sentence, 25 times five, that's Compound the target D, symbolic diff calls are evaluated in terms of the different operator D. ALGEBRA II nth Roots & Rational Exponents- Day 2 Page 3 usagiftsshops.com Example 5 – Express in simplest exponential form Step 1- Rewrite in exponential form Step 2- Get same bases Our bases are not the same.
One of them is 4 and the other one is 8. rewrite(f, target) transforms an expression f to a mathematically equivalent form, trying to express f in terms of the specified target function.
The target indicates the function that is to be used in the desired representation. Name Answer Key Numerical Expressions usagiftsshops.com2 Write the expression to match the words.
1. the difference of ten and three 10 –3. The Write-Output cmdlet sends the specified object down the pipeline to the next command.
If the command is the last command in the pipeline, the object is displayed in the console. Write-Output sends objects down the primary pipeline, also known as the "output stream" or the "success pipeline." To.
Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site. Use the power reducing identities to write sin^2xcos^2x in terms of the first power of cosine.
(1/2) sin(2x) { double angle formula} This expression is in terms of the first power of .
Write an expression as a power of 2
Rated 5/5 based on 40 review
How do you write the expression (1 - i)^5 in the standard form a + bi? | Socratic | 1,051 | 4,838 | {"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} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.949781 |
https://math.stackexchange.com/questions/2436730/is-the-sum-of-two-closed-convex-subsets-of-a-real-normed-space-closed | 1,719,240,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00520.warc.gz | 347,086,395 | 36,532 | # Is the sum of two closed convex subsets of a real normed space closed?
Let $K_1,K_2$ closed convex sets of a real normed space $V$, both containing 0. Is $K_1+K_2$ closed? Can you give-me a counterexample?
I aim to prove the following result:
Let $K_1,...,K_n$ closed convex sets of a normed space $V$, and let $c_1,...c_n$ positive real numbers. Prove that, if $x\in V$ can not be of the form $x=c_1x_1+\cdots c_nx_n$ for $x_i\in K_i$, $i=1,...,n$, then there exist $f\in V^*$ (the dual of $V$) such that $f (x)>1$ and $f (y)\leq \frac{1}{c_i}$ for all $y\in K$ and $i=1,...,n$.
I can prove it if $c_1K_1+\cdots+c_nK_n$ is closed, but couldn't prove if this is true.
This is not true even for closed subspaces of a Hilbert space. Consider the space of square-summable sequences $\ell^2$ and Let $$K_1 = \{x\in \ell^2 : x_{2k} = 0 \ \forall k\}, \quad K_2 = \{x\in \ell^2 : x_{2k-1} = k\,x_{2k} \ \forall k\}$$ It is easy to see that every eventually-zero sequence $y$ can be written as the sum of an element of $K_1$ and an element of $K_2$; just solve the systems $$(t, 0) + (ks, s) = (y_{2k-1}, y_{2k}) \tag{1}$$ for each $k$, which yields $$s = y_{2k},\quad t = y_{2k-1}- ky_{2k} \tag{2}$$ Therefore, $K_1+K_2$ is dense in $\ell^2$.
But $K_1+K_2$ does not contain the element $y = (1/k : k\in\mathbb{N})$, because formulas $(2)$ result in sequences that do not even tend to $0$, let alone being square-summable.
Another way to see this is that every element of $K_2$ satisfies $\lim_{k\to\infty} k x_{2k} = 0$, and so does (trivially) every element of $K_1$, so the property passes to the sum $K_1+K_2$.
HINT:
An example of two convex closed subsets of $\mathbb{R}^2$ whose sum is not closed.
Let $$K_1 =\{(x,y) \ | \ x,y > -1 \textrm{ and } (x+1)(y+1)\ge 1 \}$$ $$K_2 =\{(x,0) \ | \ x \le 0 \}$$
One checks that $K_1$, $K_2$ are closed and convex, and their sum is $$K_1 + K_2 = \{(x,y)\ | \ y>-1\}$$.
• But in this case, $0\notin K_1$ Commented Sep 20, 2017 at 13:03
• @Filburt: oh, didn't notice that. Basically, they should not be disjoint. Commented Sep 20, 2017 at 13:21
• @Filburt: you can check what I added Commented Sep 20, 2017 at 13:43 | 834 | 2,165 | {"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} | 3.453125 | 3 | CC-MAIN-2024-26 | latest | en | 0.859208 |
http://www.mredkj.com/vbnet/permutation.html | 1,632,806,959,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00578.warc.gz | 101,050,526 | 3,427 | # VB.NET - Code Samples - Permutation Code
## VB.NET Permutation Algorithm
### Synopsis:
Used to create all the permutations or unique arrangements of a given set of letters e.g.("cat","cta","atc","act","tca","tac")
### The Reason:
Initially I needed a permutation algorithm to create all the possible combinations of letters so I could unscramble words. Most of the code samples I found involved recursion which seemed to be a memory hog and crashed my application quite frequently. I finally found some code that utilized a transposition matrix. The code was written in VB 6 by Ziad Cassim. I based my code on this aiming for readablity over efficiency. Most of my logic though was more directly pulled from the information on how to create the matrix. Go here for more information.
### The Code:
```Private Function Permutations(ByVal data As String) As String(,)
Dim i As Int32
Dim y As Int32
Dim x As Int32
Dim tempChar As String
Dim newString As String
Dim strings(,) As String
Dim rowCount As Long
If data.Length < 2 Then
Exit Function
End If
'use the factorial function to determine the number of rows needed
'because redim preserve is slow
ReDim strings(data.Length - 1, Factorial(data.Length - 1) - 1)
strings(0, 0) = data
'swap each character(I) from the second postion to the second to last position
For i = 1 To (data.Length - 2)
'for each of the already created numbers
For y = 0 To rowCount
'do swaps for the character(I) with each of the characters to the right
For x = data.Length To i + 2 Step -1
tempChar = strings(0, y).Substring(i, 1)
newString = strings(0, y)
Mid(newString, i + 1, 1) = newString.Substring(x - 1, 1)
Mid(newString, x, 1) = tempChar
rowCount = rowCount + 1
strings(0, rowCount) = newString
Next
Next
Next
'Shift Characters
'for each empty column
For i = 1 To data.Length - 1
'move the shift character over one
For x = 0 To strings.GetUpperBound(1)
strings(i, x) = strings(i - 1, x)
Mid(strings(i, x), i, 1) = strings(i - 1, x).Substring(i, 1)
Mid(strings(i, x), i + 1, 1) = strings(i - 1, x).Substring(i - 1, 1)
Next
Next
Return strings
End Function
Public Function Factorial(ByVal Number As Integer) As String
Try
If Number = 0 Then
Return 1
Else
Return Number * Factorial(Number - 1)
End If
Catch ex As Exception
Return ex.Message
End Try
End Function
``` | 640 | 2,309 | {"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-2021-39 | latest | en | 0.743644 |
https://math.stackexchange.com/questions/1864924/show-the-equality | 1,716,659,411,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058830.51/warc/CC-MAIN-20240525161720-20240525191720-00372.warc.gz | 327,026,972 | 35,170 | # Show the equality
Let $$(X, \mathcal{A}, \mu)$$ space with measure, $$\mu(X) = 1$$, $$\epsilon > 0$$ and $$f: X \rightarrow [\epsilon,\infty)$$ a $$\mathcal{A}$$-measurable and bounded function, I've tried show $$\lim_{p \rightarrow 0^{+}} \left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}} = \exp\left(\int_X \ln(f(x))d\mu(x)\right)$$
My attempt: I know is true, in this case, $$\displaystyle \int_X \ln(f(x))d\mu(x) \leq \ln\left(\int_Xf(x)d\mu(x)\right)$$, because $$\ln(t) \leq t - 1$$, $$\forall t > 0$$, take $$t = \frac{f(x)}{\int_Xf(x)d\mu(x)}$$ and integrate over $$X$$. So $$\left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}} = \exp\left(\frac{1}{p}\ln\left(\int_X f^p(x)d\mu(x)\right)\right) \geq \exp\left(\frac{1}{p}\int_X \ln(f^p(x))d\mu(x) \right) = \exp\left(\int_X \ln(f(x))d\mu(x) \right)$$, then
$$\lim_{p \rightarrow 0^{+}} \left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}} \geq \exp\left(\int_X \ln(f(x))d\mu(x)\right)$$
If $$\left(\int_{X} f^p(x)d\mu(x) \right)^{\frac{1}{p}}$$ is decreasing, the equality it's ok, but I don't know how show this.
There is a hint in exercise, $$\displaystyle \lim_{t \rightarrow 0^+} \frac{a^t - 1}{t} = \ln(a)$$. I've tried use this limit for show decreasing, but it wasn't successful.
Thank you.
## 1 Answer
Define $G\colon \Bbb R_{\geq 0}\to \Bbb R$ by $G(p) = \int_Xf^p\,{\rm d}\mu$. We have $G(0) = \mu(X) = 1$ and so $\ln G(0) = 0$. So:
\begin{align} \lim_{p \to 0^+}\left( \int_X f^p\,{\rm d}\mu\right)^{1/p} &= \lim_{p \to 0^+} G(p)^{1/p} = \lim_{p \to 0^+} \exp \ln G(p)^{1/p} \\ &= \exp \lim_{p \to 0^+} \frac{\ln G(p)}{p} = \exp \lim_{p \to 0^+} \frac{\ln G(p)-\ln G(0)}{p-0} \\ &= \exp \frac{{\rm d}}{{\rm d}p}\bigg|_{p=0} \ln G(p) = \exp \frac{G'(0)}{G(0)} = \exp G'(0). \end{align}
Now we only have to check that $G'(0) = \int_X \ln f\,{\rm d}\mu$. We have:
\begin{align} G'(0) &= \lim_{p \to 0^+}\frac{G(p)-G(0)}{p-0} = \lim_{p\to 0^+}\frac{\int_X f^p\,{\rm d}\mu - 1}{p} \\ &\stackrel{(\ast)}{=} \lim_{p \to 0^+} \int_X \frac{f^p - 1}{p}\,{\rm d}\mu \stackrel{(\ast\ast)}{=} \int_X \lim_{p \to 0^+}\frac{f^p-1}{p}\,{\rm d}\mu \\ &= \int_X \ln f\,{\rm d} \mu,\end{align}
where in $(\ast)$ we use that $\mu(X) = 1$ and in $(\ast\ast)$ we use the bounded convergence theorem. | 1,026 | 2,258 | {"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": 15, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-22 | latest | en | 0.58795 |
https://en.m.wikibooks.org/wiki/Actually_Applicable_Application_Problems_and_Brainteasers/Wheelchair_Ramp_Standards_2 | 1,716,332,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00753.warc.gz | 196,258,674 | 10,435 | # Actually Applicable Application Problems and Brainteasers/Wheelchair Ramp Standards 2
## Overview
As mentioned in the original Wheelchair Ramp Standards problem listing, measuring the rise and run of real world ramps can be difficult. This is an example of a way around that issue, by using a protractor and the trigonometric tangent function. I'm not saying it's the best way, and it's definitely not the only way, but it's something I've come up with.
This diagram illustrates the geometric relationships involved.
## General Method
### Part 1: Building the tool
#### Materials
• Bubble level
• Protractor with arm that swings around (used on paper for lining something up with the angle's edge)
• Something to attach them with (for example: rubber bands, glue, string, epoxy, etc.)
#### Recommended procedure
1. Make sure the protractor's arm is straight up. It must be very straight or the tool will not measure accurately.
• You may want to use a plumbline (a string with some kind of weight on the end) to check this.
2. Put the bubble level across the arm's end, laying down.
3. Adjust it until it is perfectly level (which should be perfectly perpendicular to the protractor's arm).
4. Attach it in place.
• Make sure it will stay attached without shifting or you will have to re-do building the tool to calibrate it every time you use it.
### Part 2: Measuring ramps
1. Put the protractor's back down against the ramp.
2. Swing the protractor's arm until the bubble level shows that it is level.
3. Read the protractor's angle by looking at where the swingable arm meets the curve.
4. Take the complement of the protractor's angle (that is, subtract the protractor's angle from 90°). This is the ramp's angle.
5. Take the tangent of the ramp's angle (that is, find the ratio of opposite to hypotenuse which corresponds to the ramp's angle).
• One way to calculate tangent is to divide sine by cosine.
6. As you can see form the diagram above, the opposite side of the triangle is the rise and the adjacent side of the triangle is the run, so the tangent ratio is the slope of the ramp. You can compare it with the standard, ${\displaystyle {\frac {1}{12}}=0.08{\overline {3}}}$ , to determine whether the ramp meets the accessibility standard.
## Problems
Is a ramp with an angle of 1° accessible?
Is a ramp with an angle of 45° accessible?
Is a ramp with an angle of 10° accessible?
Is a ramp with an angle of 5° accessible?
### Work Backwards
You can cut out steps 4 through 6 by figuring up once and for all what range of angle readings from step 3 will give an accessible result. | 609 | 2,612 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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} | 3.875 | 4 | CC-MAIN-2024-22 | latest | en | 0.91285 |
http://codeforces.com/blog/entry/55369?locale=ru | 1,540,230,408,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00073.warc.gz | 73,107,834 | 18,925 | Блог пользователя Los_Angelos_Laycurse
Автор Los_Angelos_Laycurse, история, 12 месяцев назад, ,
•
• +36
•
» 12 месяцев назад, # | ← Rev. 3 → +13 spend two days to solve drunk king problem...using divide and conquer...there are 5 problems left, welcome to discuss each other...now I sort the problem difficult for me1589>1394>1388>1999>1596seems 1999 is harder than 1596???oh I know how to solve knight problem now....
» 12 месяцев назад, # | ← Rev. 2 → +15 there are four problems left and I know how to solve ural 1388 now:suppose the slop of line on the x>0 is k ,and slope of (0,0) to n points is k1,k2,...kn then intersection point of x1==1/(k1-k),x2=1/(k2-k)...xn=1/(kn-k)then we choose (x4-x1)/(x2-x1)==(x4'-x1')/(x2'-x1') and (x3-x2)/(x3-x4)==（x3'-x2'）/(x3'-x4') we multiply these two equations guess what happens, yes: k is offsetthen we can get (k4-k1)*(k3-k2)/((k2-k1)*(k3-k4))==(k4'-k1')*(k3'-k2')/((k2'-k1')*(k3'-k4'))en.. this convert to string matching prolems,so suffix array can solve itbtw anyone has ideas of ural 1999 ural 1394 and ural 1589???ural 1394 I got TLE on test 70, it is so hard!!!
» 11 месяцев назад, # | +13 ural 1394 I got AC hahaha 7657892 17:46:50 5 Dec 2017 Shen Yang 1394. Ships. Version 2 Visual C++ 2017 Accepted 0.904 59 064 KB I think ural 1589's AC will be sooner or later, at least I can use my hand to binary search the test when get stuck
» 10 месяцев назад, # | +15 ural 1589 AC:7660985 12:12:33 7 Dec 2017 Shen Yang 1589. Sokoban G++ 7.1 Accepted 4.321 58 056 KB solve all ural problems is a question of time | 573 | 1,570 | {"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} | 3.171875 | 3 | CC-MAIN-2018-43 | latest | en | 0.717485 |
https://au.mathworks.com/matlabcentral/cody/problems/824-set-the-array-elements-whose-value-is-13-to-0/solutions/2021854 | 1,576,147,740,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543252.46/warc/CC-MAIN-20191212102302-20191212130302-00157.warc.gz | 273,991,213 | 20,131 | Cody
# Problem 824. Set the array elements whose value is 13 to 0
Solution 2021854
Submitted on 14 Nov 2019 at 20:55 by Romain Demory
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 = [15 13 3]; y_correct = [15 0 3]; assert(isequal(your_fcn_name(A),y_correct)) filetext = fileread('your_fcn_name.m'); assert(isempty(strfind(filetext, 'for')),'for command is forbidden') assert(isempty(strfind(filetext, 'while')),'while command is forbidden') assert(isempty(strfind(filetext, 'find')),'find command is forbidden') assert(isempty(strfind(filetext, 'ind2sub')),'ind2sub command is forbidden')
A = 15 0 3
2 Pass
A = 13; y_correct = 0; assert(isequal(your_fcn_name(A),y_correct)) filetext = fileread('your_fcn_name.m'); assert(isempty(strfind(filetext, 'for')),'for command is forbidden') assert(isempty(strfind(filetext, 'while')),'while command is forbidden') assert(isempty(strfind(filetext, 'find')),'find command is forbidden') assert(isempty(strfind(filetext, 'ind2sub')),'ind2sub command is forbidden')
A = 0
3 Pass
A = []; y_correct = []; assert(isequal(your_fcn_name(A),y_correct)) filetext = fileread('your_fcn_name.m'); assert(isempty(strfind(filetext, 'for')),'for command is forbidden') assert(isempty(strfind(filetext, 'while')),'while command is forbidden') assert(isempty(strfind(filetext, 'find')),'find command is forbidden') assert(isempty(strfind(filetext, 'ind2sub')),'ind2sub command is forbidden')
A = []
4 Pass
A = [13 0 13; 13 13 8; 13 25 2]; titi = [0 0 0 ; 0 0 8;0 25 2]; assert(isequal(your_fcn_name(A),titi)) filetext = fileread('your_fcn_name.m'); assert(isempty(strfind(filetext, 'for')),'for command is forbidden') assert(isempty(strfind(filetext, 'while')),'while command is forbidden') assert(isempty(strfind(filetext, 'find')),'find command is forbidden') assert(isempty(strfind(filetext, 'ind2sub')),'ind2sub command is forbidden')
A = 0 0 0 0 0 8 0 25 2
5 Pass
A=rand(10,10,10); A(1,2,2)=13; A(3,4,1)=13; A(5,6,3)=13; % what you cannot do but me I can ;) ! titi = A; I=find(titi==13); [u,v,w]=ind2sub(size(titi),I); for ii = 1 : numel(u) titi(u(ii),v(ii),w(ii))=0; end assert(isequal(your_fcn_name(A),titi)) filetext = fileread('your_fcn_name.m'); assert(isempty(strfind(filetext, 'for')),'for command is forbidden') assert(isempty(strfind(filetext, 'while')),'while command is forbidden') assert(isempty(strfind(filetext, 'find')),'find command is forbidden') assert(isempty(strfind(filetext, 'ind2sub')),'ind2sub command is forbidden')
A(:,:,1) = 0.7530 0.9003 0.4269 0.6921 0.2139 0.3294 0.5549 0.2292 0.3046 0.7384 0.4055 0.6860 0.7968 0.7069 0.7354 0.6187 0.6849 0.0808 0.8360 0.0573 0.7982 0.2184 0.8403 0 0.4150 0.2320 0.9852 0.6091 0.5638 0.3079 0.0248 0.3745 0.3967 0.1397 0.0240 0.5056 0.9520 0.5917 0.2670 0.7465 0.4084 0.9464 0.1508 0.8558 0.9683 0.2204 0.8902 0.6466 0.1057 0.7222 0.7517 0.1316 0.0194 0.5963 0.7420 0.4242 0.9116 0.6467 0.2280 0.0867 0.4035 0.5338 0.2052 0.7258 0.7240 0.3587 0.0212 0.5633 0.5404 0.0294 0.5998 0.2637 0.3813 0.8651 0.7064 0.4534 0.4265 0.2517 0.9853 0.3936 0.8323 0.5494 0.6267 0.6932 0.5002 0.8422 0.5753 0.2955 0.7307 0.4762 0.4152 0.5019 0.6245 0.1280 0.4535 0.5824 0.8806 0.7664 0.4009 0.6070 A(:,:,2) = 0.0553 0 0.0962 0.4011 0.8943 0.1539 0.1196 0.6440 0.6041 0.5952 0.2163 0.3019 0.6785 0.5876 0.3955 0.0216 0.5455 0.0815 0.2835 0.2125 0.5454 0.1392 0.5472 0.3523 0.0281 0.0761 0.3466 0.2488 0.0669 0.1949 0.6675 0.3839 0.6477 0.0473 0.0782 0.7651 0.0038 0.9158 0.4291 0.9509 0.4207 0.0638 0.4440 0.4376 0.2680 0.7316 0.1104 0.0697 0.9138 0.8651 0.4510 0.5774 0.4748 0.1017 0.3210 0.1325 0.4478 0.1331 0.1650 0.9250 0.8232 0.7807 0.9316 0.0379 0.5812 0.6371 0.0003 0.1510 0.7797 0.3697 0.1407 0.0685 0.1701 0.9645 0.8110 0.7891 0.0724 0.5226 0.5626 0.1376 0.1683 0.6190 0.8367 0.7184 0.1962 0.5047 0.3737 0.6859 0.2957 0.5869 0.3297 0.5181 0.4639 0.8835 0.7569 0.7612 0.2986 0.3689 0.0233 0.2355 A(:,:,3) = 0.8343 0.3276 0.7903 0.1419 0.3574 0.5691 0.7271 0.9662 0.4537 0.9456 0.8164 0.8488 0.5330 0.8833 0.2792 0.1339 0.8601 0.0708 0.1541 0.0025 0.2354 0.7025 0.3541 0.5137 0.0674 0.8190 0.0954 0.0540 0.5976 0.2894 0.6983 0.8874 0.5857 0.1623 0.5114 0.7782 0.7195 0.2250 0.3673 0.3873 0.6482 0.4351 0.3189 0.0460 0.2372 0 0.7287 0.9922 0.8652 0.3613 0.3419 0.5438 0.4793 0.4557 0.3569 0.6160 0.7794 0.4519 0.5192 0.8023 0.1770 0.8165 0.2791 0.8134 0.8857 0.9765 0.4628 0.3414 0.8402 0.0913 0.3102 0.2142 0.0678 0.6363 0.8668 0.0629 0.5991 0.4120 0.7273 0.6504 0.0015 0.2755 0.7960 0.7772 0.0687 0.5218 0.1422 0.9258 0.1984 0.7513 0.1991 0.8150 0.6030 0.5209 0.7306 0.0065 0.8075 0.5218 0.4995 0.7862 A(:,:,4) = 0.5244 0.2404 0.4173 0.2408 0.8551 0.6098 0.8854 0.2955 0.4618 0.0222 0.7097 0.6333 0.9574 0.1884 0.9026 0.6443 0.2909 0.0726 0.9152 0.3884 0.1601 0.3410 0.0744 0.0850 0.9472 0.2506 0.1764 0.8645 0.5932 0.9825 0.6393 0.7074 0.5132 0.1413 0.8534 0.0750 0.9720 0.1124 0.9000 0.5818 0.3954 0.8820 0.9843 0.9321 0.9258 0.7064 0.0569 0.9686 0.8423 0.4913 0.2542 0.2372 0.3891 0.9484 0.8264 0.2464 0.4458 0.0258 0.2805 0.9435 0.8154 0.5053 0.7050 0.1139 0.5702 0.0018 0.6063 0.5914 0.5917 0.3962 0.7462 0.8802 0.0012 0.5085 0.1981 0.4708 0.5893 0.8147 0.2515 0.5855 0.8144 0.0365 0.1702 0.1392 0.4582 0.9501 0.2206 0.4260 0.0820 0.2751 0.2995 0.3040 0.2057 0.8566 0.9312 0.6678 0.3172 0.5576 0.3646 0.0500 A(:,:,5) = 0.4409 0.3426 0.2970 0.3695 0.4763 0.5070 0.7798 0.1095 0.2987 0.8178 0.6966 0.7563 0.2114 0.4956 0.6014 0.7359 0.3167 0.6633 0.8033 0.6829 0.3417 0.3978 0.1130 0.7039 0.3661 0.6248 0.3651 0.8165 0.9825 0.5051 0.4262 0.3186 0.5284 0.8548 0.5598 0.8702 0.9523 0.9988 0.5943 0.1112 0.1089 0.7348 0.7846 0.8039 0.0558 0.4283 0.1759 0.1551 0.0349 0.8709 0.9031 0.0819 0.8325 0.2477 0.4969 0.4995 0.1814 0.7411 0.8261 0.9925 0.5536 0.2051 0.7031 0.8976 0.7650 0.5238 0.3443 0.2420 0.0310 0.7886 0.3799 0.5808 0.2703 0.0243 0.0295 0.6998 0.5581 0.0672 0.8933 0.9783 0.7379 0.4227 0.8845 0.5208 0.7500 0.2581 0.7177 0.5924 0.1527 0.4610 0.1224 0.0271 0.2061 0.4017 0.3626 0.7403 0.5095 0.9385 0.1442 0.9017 A(:,:,6) = 0.5683 0.3286 0.0430 0.9812 0.6612 0.2518 0.4451 0.2896 0.2652 0.0426 0.9952 0.1178 0.5413 0.3356 0.3476 0.3612 0.2400 0.2521 0.2044 0.5479 0.5756 0.8604 0.4017 0.2104 0.4766 0.6769 0.2768 0.3874 0.8305 0.0525 0.5352 0.4789 0.6883 0.7170 0.5766 0.3247 0.1276 0.5782 0.9225 0.8238 0.8311 0.1772 0.8149 0.5174 0.8136 0.2279 0.4355 0.5380 0.4999 0.7656 0.3375 0.4847 0.6993 0.9753 0.0391 0.8542 0.9667 0.8586 0.2310 0.5224 0.4721 0.7311 0.5676 0.9841 0.9757 0.7208 0.3698 0.1711 0.7545 0.5334 0.0317 0.0342 0.4864 0.5546 0.5401 0.2074 0.2971 0.4578 0.6693 0.5331 0.3568 0.5637 0.2808 0.7815 0.6328 0.3219 0.0564 0.2300 0.9804 0.2129 0.9294 0.0422 0.4346 0.7564 0.0906 0.9920 0.5519 0.7767 0.5725 0.4884 A(:,:,7) = 0.4323 0.8177 0.8704 0.1271 0.5558 0.4595 0.8666 0.3888 0.4243 0.1760 0.6458 0.2641 0.3664 0.6744 0.1284 0.3217 0.4968 0.5086 0.4255 0.3773 0.1531 0.5685 0.9545 0.8632 0.6986 0.7791 0.4096 0.9875 0.8514 0.8915 0.5486 0.7224 0.2279 0.6613 0.6731 0.9737 0.5185 0.4831 0.8769 0.4587 0.5259 0.6375 0.2263 0.1024 0.7028 0.0899 0.1548 0.0614 0.7768 0.2081 0.0413 0.4557 0.1092 0.1256 0.1993 0.8622 0.4155 0.2118 0.3795 0.8067 0.6759 0.0744 0.0790 0.4841 0.4970 0.6517 0.2941 0.8584 0.8141 0.2987 0.1395 0.9323 0.4490 0.7938 0.6925 0.1094 0.0409 0.7488 0.8230 0.4030 0.6248 0.5161 0.1694 0.5230 0.4687 0.1120 0.3746 0.2356 0.8593 0.7244 0.7427 0.2672 0.9020 0.9840 0.0226 0.7012 0.5634 0.6496 0.7963 0.3587 A(:,:,8) = 0.6553 0.2256 0.7128 0.7466 0.3215 0.4930 0.5607 0.0879 0.6913 0.9659 0.2800 0.3429 0.8325 0.3159 0.7755 0.7801 0.9115 0.8856 0.8515 0.8962 0.5594 0.3681 0.4364 0.3975 0.4010 0.9900 0.4735 0.7349 0.8102 0.0878 0.1882 0.8164 0.7186 0.5489 0.2833 0.3255 0.5271 0.8913 0.3078 0.8598 0.5469 0.9039 0.6458 0.7900 0.2321 0.8048 0.5992 0.2256 0.5781 0.9048 0.2895 0.8063 0.7501 0.2932 0.2064 0.0786 0.4557 0.5543 0.2298 0.7294 0.7113 0.6808 0.7134 0.8011 0.4282 0.2198 0.3584 0.6632 0.8108 0.6064 0.1950 0.6707 0.3365 0.9693 0.4532 0.6949 0.2713 0.3166 0.4183 0.0824 0.9683 0.0947 0.2982 0.5644 0.5174 0.5890 0.4881 0.6666 0.0910 0.5184 0.8901 0.4365 0.3496 0.4830 0.4285 0.2216 0.1827 0.9228 0.4972 0.6099 A(:,:,9) = 0.0728 0.3958 0.5233 0.5283 0.4131 0.2488 0.7038 0.8679 0.7390 0.6607 0.1532 0.3912 0.9181 0.8985 0.8354 0.9303 0.7745 0.5981 0.2160 0.6593 0.3143 0.2119 0.0419 0.4017 0.1978 0.9774 0.4478 0.1527 0.1698 0.7445 0.8207 0.1935 0.6438 0.7204 0.2254 0.5256 0.1844 0.3176 0.1286 0.5365 0.5367 0.1661 0.8652 0.1162 0.4253 0.5208 0.5550 0.9637 0.9843 0.7498 0.0446 0.7039 0.2419 0.4273 0.2663 0.9971 0.8203 0.0789 0.4622 0.6434 0.2037 0.2585 0.5879 0.8878 0.0787 0.2538 0.9215 0.0955 0.0779 0.6102 0.0446 0.5328 0.8790 0.8676 0.7914 0.1160 0.9212 0.5903 0.0977 0.1124 0.0006 0.3486 0.5548 0.6424 0.4290 0.6427 0.1482 0.2389 0.8835 0.3686 0.9560 0.1699 0.3405 0.1344 0.6513 0.4798 0.1243 0.1043 0.3633 0.5242 A(:,:,10) = 0.8351 0.4949 0.8300 0.5897 0.9931 0.3493 0.8493 0.8519 0.4615 0.1067 0.2350 0.9853 0.2178 0.3561 0.4721 0.0452 0.6888 0.2106 0.0646 0.5105 0.9893 0.0014 0.2950 0.5074 0.6592 0.7859 0.8865 0.6068 0.0761 0.7933 0.9356 0.2867 0.8191 0.2942 0.9842 0.1434 0.5520 0.6290 0.0733 0.5522 0.6732 0.6729 0.9810 0.7763 0.4875 0.9947 0.1441 0.0062 0.5250 0.4641 0.8772 0.6585 0.0189 0.4983 0.5960 0.4548 0.7352 0.1827 0.3328 0.6727 0.8365 0.7307 0.9712 0.1255 0.4045 0.4469 0.8679 0.9591 0.8699 0.2458 0.1669 0.4473 0.8925 0.4737 0.7012 0.8245 ...
6 Pass
A = repmat(13,89,17); assert(isequal(your_fcn_name(A),repmat(0,89,17))) filetext = fileread('your_fcn_name.m'); assert(isempty(strfind(filetext, 'for')),'for command is forbidden') assert(isempty(strfind(filetext, 'while')),'while command is forbidden') assert(isempty(strfind(filetext, 'find')),'find command is forbidden') assert(isempty(strfind(filetext, 'ind2sub')),'ind2sub command is forbidden')
A = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
7 Pass
A= magic(5); B =A; B(3,3) = 0 assert(isequal(your_fcn_name(A),B)) filetext = fileread('your_fcn_name.m'); assert(isempty(strfind(filetext, 'for')),'for command is forbidden') assert(isempty(strfind(filetext, 'while')),'while command is forbidden') assert(isempty(strfind(filetext, 'find')),'find command is forbidden') assert(isempty(strfind(filetext, 'ind2sub')),'ind2sub command is forbidden')
B = 17 24 1 8 15 23 5 7 14 16 4 6 0 20 22 10 12 19 21 3 11 18 25 2 9 A = 17 24 1 8 15 23 5 7 14 16 4 6 0 20 22 10 12 19 21 3 11 18 25 2 9
8 Pass
A= 13.3; B = A; assert(isequal(your_fcn_name(A),B))
A = 13.3000 | 9,148 | 13,607 | {"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-2019-51 | latest | en | 0.398629 |
https://www.vernier.com/news/2018/07/24/decoding-your-absorbance-readings/ | 1,563,508,692,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525974.74/warc/CC-MAIN-20190719032721-20190719054721-00010.warc.gz | 864,084,196 | 26,073 | Vernier Software & Technology
Periodic Elements: College Chemistry Blog
### Why don’t absorbance readings have units?
Absorbance readings are unitless because they are calculated from a ratio of the intensity of light transmitted through the sample (I) to the intensity of light transmitted through a blank (Io). This ratio results in a unitless value.
Absorbance = log (Io/I)
### Why are absorbance readings most accurate between 0.1 and 1?
Remember that absorbance is the logarithm of the transmission of light through a sample. Transmission (T) is the ratio of the intensity of light transmitted through the sample (I) to the intensity of light transmitted through a blank (Io). Therefore, absorbance = log (Io/I).
At an absorbance of 2 you are at 1%T, which means that 99% of available light is being blocked (absorbed) by the sample. At an ABS of 3 you are at 0.1% T, which means that 99.9% of the available light is being blocked (absorbed) by the sample. Such small amounts of light are very difficult to detect and are outside the meaningful range of most spectrometers.
Vernier array spectrometers and colorimeters have a useful absorbance range between 0.1 and 1.0. Any absorbance reading above 1 can be inaccurate. There are spectrometers that will report meaningful values at absorbance ranges above 1.0, but these are research instruments that are also quite expensive. In most classroom settings, the best option is to simply dilute your samples to ensure they are in this range.
### How important is it to use a quartz cuvette for absorbance readings in the UV?
It depends on how accurate you want your absorbance readings to be. UV plastic cuvettes are less expensive and have practical applications when working with students, but they lose transparency quickly in the UV. Most are only rated to 280 nm. If you want the most accurate data possible below 280 nm, a quartz cuvette is the best option. Another unfortunate side effect of using UV-plastic cuvettes is that students commonly confuse them with visible-only plastic cuvettes. This cuts out all UV light, so data will be very poor. If you are going to use UV-plastic cuvettes, make sure you are using them for the proper applications. | 501 | 2,218 | {"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-2019-30 | latest | en | 0.942646 |
https://www.sydney.edu.au/units/CIVL9410/2021-S2C-ND-CC | 1,725,769,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00587.warc.gz | 950,462,508 | 29,645 | Unit outline_
# CIVL9410: Soil Mechanics
## Overview
This course provides an elementary introduction to Geotechnical Engineering, and provides the basic mechanics necessary for the detailed study of Geotechnical Engineering. This course aims to provide an understanding of: the nature of soils as engineering materials; common soil classification schemes; the importance of water in the soil and the effects of water movement; methods of predicting soil settlements, the stress-strain-strength response of soils, and earth pressures.
### Unit details and rules
Academic unit Civil Engineering 6 None None None An understanding of simple statics, equilibrium, forces and bending moments, and of stress and strain and the relationship between them (CIVL9802 and CIVL9201). Familiarity with the use of spreadsheets (Excel, Mathcad) to obtain solutions to engineering problems, and with the graphical presentation of this data, and familiarity with word processing packages for report presentation. Familiarity with partial differential equations, and their analytical and numerical solution. No
### Teaching staff
Coordinator David Airey, david.airey@sydney.edu.au
## Assessment
Type Description Weight Due Length
Final exam (Take-home short release) Final exam
Take home exam
50% Formal exam period 3 hours
Outcomes assessed:
Tutorial quiz Pre-lab quiz
Short quizzes to test familiarity with experiment
2% Multiple weeks n/a
Outcomes assessed:
Tutorial quiz Quiz
Short 15 minute numerical question at end of tutorial session each week .
10% Multiple weeks n/a
Outcomes assessed:
Assignment Lab report
Professional style lab report
8% Multiple weeks n/a
Outcomes assessed:
Assignment Computing assignment: Part 1
15% Week 09 n/a
Outcomes assessed:
Assignment Computing assignment: Part 2
2 page report describing results of parametric study
15% Week 10 n/a
Outcomes assessed:
= group assignment
= Type D final exam
### Assessment summary
• Quizzes: In-class exercises are used to assess the student understanding of key concepts each week. Best 10 out of 12 will be taken.
• Pre-lab Quizzes: Pre-laboratory online material is to be read before the sessions, and each laboratory is preceded by an online quiz.
• Lab Report: Students submit one lab report on one out of their 5 sessions, one week after the session. Lab reports will address student development in their presentation and analysis skills. Students are expected to present a professional documentation of their experiments and analysis associated with the concepts learned within this unit.
• Computing Assignments: Computing assignments are designed to reinforce theoretical concepts and develop the students` skills in the use of Excel. Assignment has 2 parts. Part 1 is to develop a spreadsheet to solve a design problem. Part 2 is to write a report describing the outcomes of the design exercise/parametric study.
• Final Exam: The examination at the end of the semester will measure students’ understanding of the concepts covered during the semester within this unit.
Detailed information for each assessment can be found on Canvas.
### Assessment criteria
The University awards common result grades, set out in the Coursework Policy 2014 (Schedule 1).
As a general guide, a high distinction indicates work of an exceptional standard, a distinction a very high standard, a credit a good standard, and a pass an acceptable standard.
Result name
Mark range
Description
High distinction
85 - 100
Distinction
75 - 84
Credit
65 - 74
Pass
50 - 64
Fail
0 - 49
When you don’t meet the learning outcomes of the unit to a satisfactory standard.
### Late submission
In accordance with University policy, these penalties apply when written work is submitted after 11:59pm on the due date:
• Deduction of 5% of the maximum mark for each calendar day after the due date.
• After ten calendar days late, a mark of zero will be awarded.
The Current Student website provides information on academic integrity and the resources available to all students. The University expects students and staff to act ethically and honestly and will treat all allegations of academic integrity breaches seriously.
We use similarity detection software to detect potential instances of plagiarism or other forms of academic integrity breach. If such matches indicate evidence of plagiarism or other forms of academic integrity breaches, your teacher is required to report your work for further investigation.
Use of generative artificial intelligence (AI) and automated writing tools
You may only use generative AI and automated writing tools in assessment tasks if you are permitted to by your unit coordinator. If you do use these tools, you must acknowledge this in your work, either in a footnote or an acknowledgement section. The assessment instructions or unit outline will give guidance of the types of tools that are permitted and how the tools should be used.
Your final submitted work must be your own, original work. You must acknowledge any use of generative AI tools that have been used in the assessment, and any material that forms part of your submission must be appropriately referenced. For guidance on how to acknowledge the use of AI, please refer to the AI in Education Canvas site.
The unapproved use of these tools or unacknowledged use will be considered a breach of the Academic Integrity Policy and penalties may apply.
Studiosity is permitted unless otherwise indicated by the unit coordinator. The use of this service must be acknowledged in your submission as detailed on the Learning Hub’s Canvas page.
Outside assessment tasks, generative AI tools may be used to support your learning. The AI in Education Canvas site contains a number of productive ways that students are using AI to improve their learning.
## Learning support
### Simple extensions
If you encounter a problem submitting your work on time, you may be able to apply for an extension of five calendar days through a simple extension. The application process will be different depending on the type of assessment and extensions cannot be granted for some assessment types like exams.
### Special consideration
If exceptional circumstances mean you can’t complete an assessment, you need consideration for a longer period of time, or if you have essential commitments which impact your performance in an assessment, you may be eligible for special consideration or special arrangements.
Special consideration applications will not be affected by a simple extension application.
### Using AI responsibly
Co-created with students, AI in Education includes lots of helpful examples of how students use generative AI tools to support their learning. It explains how generative AI works, the different tools available and how to use them responsibly and productively.
## Weekly schedule
WK Topic Learning activity Learning outcomes
Multiple weeks Flow Net Laboratory Practical (2 hr)
Oedometer Laboratory Session Practical (2 hr)
Shear Box Laboratory Session Practical (2 hr)
Week 01 Introduction and Classification Lecture (2 hr)
Classification Lecture and tutorial (2 hr)
Classification tutorial Tutorial (1 hr)
Week 02 Definitions and Compaction Lecture (1 hr)
Definitions Lecture and tutorial (2 hr)
Soil definitions Tutorial (1 hr)
Soil Classification Laboratory Session Practical (2 hr)
Week 03 Compaction and Effective Stress Lecture (2 hr)
Effective Stress and Compaction Lecture and tutorial (2 hr)
Effective stress and Compaction Tutorial (1 hr)
Compaction Laboratory Session Practical (2 hr)
Week 04 Water Flow through Soil Lecture (2 hr)
Drawing Flow Nets Lecture and tutorial (2 hr)
Drawing Flow Nets Tutorial (1 hr)
Week 05 Flow Nets, Calculations and Piping Lecture (2 hr)
Flow Net calculations Lecture and tutorial (2 hr)
Flow Net Calculations Tutorial (1 hr)
Week 06 One dimensional soil compression Lecture (2 hr)
One dimensional compression Lecture and tutorial (2 hr)
One dimensional compression Tutorial (1 hr)
Week 07 One-dimensional Settlement Lecture (2 hr)
One-dimensional settlement Lecture and tutorial (2 hr)
One-dimensional settlement Tutorial (1 hr)
Week 08 One-dimensional Consolidation Lecture (2 hr)
One-dimensional consolidation Lecture and tutorial (2 hr)
One-dimensional consolidation Tutorial (1 hr)
Week 09 Elastic Soil Mechanics Lecture (2 hr)
Elastic Soil Mechanics Lecture and tutorial (2 hr)
Elastic Soil Mechanics Tutorial (1 hr)
Week 10 Settlements of Elastic Soil Lecture (2 hr)
Settlements for Elastic Soil Lecture and tutorial (2 hr)
Settlements for Elastic Soil Tutorial (1 hr)
Week 11 Soil Strength and Introduction to Critical State Soil Mechanics Lecture (2 hr)
Soil Strength and Critical State Soil Mechanics Lecture and tutorial (2 hr)
Soil Strength and Critical State Soil Mechanics Tutorial (1 hr)
Week 12 Soil Stability - Retaining Walls - Rankine Lecture (2 hr)
Soil Stability - Retaining Walls Lecture and tutorial (2 hr)
Soil Stability - Retaining Walls Tutorial (1 hr)
Week 13 Soil Stability - Retaining Walls - Coulomb Lecture (2 hr)
Revision Lecture and tutorial (2 hr)
Retaining Walls - Coulomb Tutorial (1 hr)
### Attendance and class requirements
Attendance and completion of all the laboratory classes is a requirement to pass the course.
Attendance at the 1 hour tutorial sessions is compulsory
### Study commitment
Typically, there is a minimum expectation of 1.5-2 hours of student effort per week per credit point for units of study offered over a full semester. For a 6 credit point unit, this equates to roughly 120-150 hours of student effort in total.
## Learning outcomes
Learning outcomes are what students know, understand and are able to do on completion of a unit of study. They are aligned with the University's graduate qualities and are assessed as part of the curriculum.
At the completion of this unit, you should be able to:
• LO1. demonstrate proficiency in handling experimental data, including strength parameters.
• LO2. analyse and report the results of a laboratory experiment at a professional standard
• LO3. develop and use a spreadsheet to analyse a geotechnical design problem
• LO4. give an engineering classification of any piece of soil, and on this basis, predict how it will perform as an engineering material
• LO5. calculate the settlements, and rates of settlement, under structures of various shapes and sizes
• LO6. explain the advantages and limitations of the different methods of settlement calculation
• LO7. determine the strength parameters appropriate to a range of stability problems, and understand the difference between total and effective stress approaches
• LO8. understand the principle of effective stress, and be able to apply this to calculate the stresses causing soil deformation
• LO9. calculate quantities of water flowing through the ground, and understand the effects that water flow has on the soil
The graduate qualities are the qualities and skills that all University of Sydney graduates must demonstrate on successful completion of an award course. As a future Sydney graduate, the set of qualities have been designed to equip you for the contemporary world.
GQ1 Depth of disciplinary expertise Deep disciplinary expertise is the ability to integrate and rigorously apply knowledge, understanding and skills of a recognised discipline defined by scholarly activity, as well as familiarity with evolving practice of the discipline. GQ2 Critical thinking and problem solving Critical thinking and problem solving are the questioning of ideas, evidence and assumptions in order to propose and evaluate hypotheses or alternative arguments before formulating a conclusion or a solution to an identified problem. GQ3 Oral and written communication Effective communication, in both oral and written form, is the clear exchange of meaning in a manner that is appropriate to audience and context. GQ4 Information and digital literacy Information and digital literacy is the ability to locate, interpret, evaluate, manage, adapt, integrate, create and convey information using appropriate resources, tools and strategies. GQ5 Inventiveness Generating novel ideas and solutions. GQ6 Cultural competence Cultural Competence is the ability to actively, ethically, respectfully, and successfully engage across and between cultures. In the Australian context, this includes and celebrates Aboriginal and Torres Strait Islander cultures, knowledge systems, and a mature understanding of contemporary issues. GQ7 Interdisciplinary effectiveness Interdisciplinary effectiveness is the integration and synthesis of multiple viewpoints and practices, working effectively across disciplinary boundaries. GQ8 Integrated professional, ethical, and personal identity An integrated professional, ethical and personal identity is understanding the interaction between one’s personal and professional selves in an ethical context. GQ9 Influence Engaging others in a process, idea or vision. | 2,529 | 12,982 | {"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} | 2.734375 | 3 | CC-MAIN-2024-38 | latest | en | 0.878262 |
http://www.mypearsonstore.com/bookstore/excursions-in-modern-mathematics-coursesmart-etextbook-0321576217 | 1,443,978,003,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736675795.7/warc/CC-MAIN-20151001215755-00086-ip-10-137-6-227.ec2.internal.warc.gz | 797,554,044 | 8,011 | Excursions in Modern Mathematics, CourseSmart eTextbook, 7th Edition
Published Date: Apr 16, 2009
More Product Info
Description
Excursions in Modern Mathematics, Seventh Edition, shows readers that math is a lively, interesting, useful, and surprisingly rich subject. With a new chapter on financial math and an improved supplements package, this book helps students appreciate that math is more than just a set of classroom theories: math can enrich the life of any one who appreciates and knows how to use it.
CourseSmart textbooks do not include any media or print supplements that come packaged with the bound book.
Part 1. The Mathematics of Social Choice
1. The Mathematics of Voting: The Paradox of Democracy
1.1 Preference Ballots and Preference Schedules
1.2 The Plurality Method
1.3 The Borda Count Method
1.4 The Plurality-with-Elimination Method (Instant Runoff Voting)
1.5 The Method of Piecewise Comparisons
1.6 Rankings
Profile: Kenneth J. Arrow
Key Concepts
Exercises
Projects and Papers
References and Further Readings
2. The Mathematics of Power: Weighted Voting
2.1 An Introduction to Weighted Voting
2.2 The Banzhaf Power Index
2.3 Applications of the Banzhaf Power Index
2.4 The Shapely-Shubik Power Index
2.5 Applications of the Shapely-Shubik Power Index
Profile: Lloyd S. Shapely
Key Concepts
Exercises
Projects and Papers
References and Further Readings
3. The Mathematics of Sharing: Fair-Division Games
3.1 Fair-Division Games
3.2 Two Players: The Divider-Chooser Method
3.3 The Lone-Divider Method
3.4 The Lone-Chooser Method
3.5 The Last-Diminisher Method
3.6 The Method of Sealed Bids
3.7 The Method of Markers
Profile: Hugo Steinhaus
Key Concepts
Exercises
Projects and Papers
References and Further Readings
4. The Mathematics of Apportionment: Making the Rounds
4.1 Apportionment Problems
4.2 Hamilton's Method and the Quota Rule
4.3 The Alabama and Other Paradoxes
4.4 Jefferson's Method
4.6 Webster's Method
Historical Note: A Brief History of Apportionment in the United States
Key Concepts
Exercises
Projects and Papers
References and Further Readings
Mini-Excursion 1: Apportionment Today
Part 2. Management Science
5. The Mathematics of Getting Around: Euler Paths and Circuits
5.1 Euler Circuit Problems
5.2 What is a Graph?
5.3 Graph Concepts and Terminology
5.4 Graph Models
5.5 Euler's Theorems
5.6 Fleury's Algorithm
5.7 Eulerizing Graphs
Profile: Leonard Euler
Key Concepts
Exercises
Projects and Papers
References and Further Readings
6. The Mathematics of Touring: The Traveling Salesman Problem
6.1 Hamilton Circuits and Hamilton Paths
6.2 Complete Graphs
6.3 Traveling Salesman Problems
6.4 Simple Strategies for Solving TSPs
6.5 The Brute-Force and Nearest-Neighbor Algorithms
6.6 Approximate Algorithms
6.7 The Repetitive Nearest-Neighbor Algorithm
6.8 The Cheapest Link Algorithm
Profile: Sir William Rowan Hamilton
Key Concepts
Exercises
Projects and Papers
References and Further Readings
7. The Mathematics of Networks: The Cost of Being Connected
7.1 Trees
7.2 Spanning Trees
7.3 Kruskal's Algorithm
7.4 The Shortest Network Connecting Three Points
7.5 Shortest Networks for Four or More Points
Profile: Evangelista Torricelli
Key Concepts
Exercises
Projects and Papers
References and Further Readings
8. The Mathematics of Scheduling: Chasing the Critical Path
8.1 The Basic Elements of Scheduling
8.2 Directed Graphs (Digraphs)
8.3 Scheduling with Priority Lists
8.4 The Decreasing-Time Algorithm
8.5 Critical Paths
8.6 The Critical-Path Algorithm
8.7 Scheduling with Independent Tasks
Profile: Ronald L. Graham
Key Concepts
Exercises
Projects and Papers
References and Further Readings
Mini-Excursion 2: A Touch of Color
Part 3. Growth And Symmetry
9. The Mathematics of Spiral Growth: Fibonacci Numbers and the Golden Ratio
9.1 Fibonacci's Rabbits
9.2 Fibonacci Numbers
9.3 The Golden Ratio
9.4 Gnomons
9.5 Spiral Growth in Nature
Profile: Leonardo Fibonacci
Key Concepts
Exercises
Projects and Papers
References and Further Readings
10. The Mathematics of Money: Spending it, Saving It, and Growing It
10.1 Percentages
10.2 Simple Interest
10.3 Compound Interest
10.4 Geometric Sequences
10.5 Deferred Annuities: Planned Savings for the Future
Key Concepts
Exercises
Projects and Papers
References and Further Readings
11. The Mathematics of Symmetry: Beyond Reflection
11.1 Rigid Motions
11.2 Reflections
11.3 Rotations
11.4 Translations
11.5 Glide Reflections
11.6 Symmetry as a Rigid Motion
11.7 Patterns
Profile: Sir Roger Penrose
Key Concepts
Exercises
Projects and Papers
References and Further Readings
12. The Geometry of Fractal Shapes: Naturally Irregular
12.1 The Koch Snowflake
12.2 The Sierpinski Gasket
12.3 The Chaos Game
12.4 The Twisted Sierpinski Gasket
12.5 The Mandelbrot Set
Profile: Benoit Mandelbrot
Key Concepts
Exercises
Projects and Papers
References and Further Readings
Mini-Excursion 3: The Mathematics of Population Growth: There is Strength in Numbers
Part 4. Statistics
13. Collecting Statistical Data: Censuses, Surveys, and Clinical Studies
13.1 The Population
13.2 Sampling
13.3 Random Sampling
13.4 Sampling: Terminology and Key Concepts
13.5 The Capture-Recapture Method
13.6 Clinical Studies
Profile: George Gallup
Key Concepts
Exercises
Projects and Papers
References and Further Readings
14. Descriptive Statistics: Graphing and Summarizing Data
14.1 Graphical Descriptions of Data
14.2 Variables
14.3 Numerical Summaries of Data
14.4 Measures of Spread
Profile: W. Edwards Deming
Key Concepts
Exercises
Projects and Papers
References and Further Readings
15. Chances, Probabilities, and Odds: Measuring Uncertainty
15.1 Random Experiments and Sample Spaces
15.2 Counting Outcomes in Sample Spaces
15.3 Permutations and Combinations
15.4 Probability Spaces
15.5 Equiprobable Spaces
15.6 Odds
Profile: Persi Diaconis
Key Concepts
Exercises
Projects and Papers
References and Further Readings
16. The Mathematics of Normal Distributions: The Call of the Bell
16.1 Approximately Normal Distributions of Data
16.2 Normal Curves and Normal Distributions
16.3 Standardizing Normal Data
16.4 The 68-95-99.7 Rule
16.5 Normal Curves as Models of Real-Life Data Sets
16.6 Distributions of Random Events
16.7 Statistical Inference
Profile: Carl Friedrich Gauss
Key Concepts
Exercises
Projects and Papers
References and Further Readings
Mini-Excursion 4: The Mathematics of Managing Risk
Purchase Info ?
With CourseSmart eTextbooks and eResources, you save up to 60% off the price of new print textbooks, and can switch between studying online or offline to suit your needs.
Once you have purchased your eTextbooks and added them to your CourseSmart bookshelf, you can access them anytime, anywhere.
Excursions in Modern Mathematics, CourseSmart eTextbook, 7th Edition
Format: Safari Book
\$73.99 | ISBN-13: 978-0-321-57621-7 | 1,775 | 7,021 | {"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-2015-40 | latest | en | 0.776968 |
https://entranceindia.com/tag/loyola-college-algebra/ | 1,686,067,139,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652959.43/warc/CC-MAIN-20230606150510-20230606180510-00348.warc.gz | 252,685,851 | 23,214 | ## Loyola College B.Sc. Mathematics Nov 2003 Algebra, Anal. Geometry, Calculus & Trigonometry Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI –600 034
B.Sc., DEGREE EXAMINATION – MATHEMATICS
# MT – 1500/MAT 500 – ALGEBRA, ANAL. GEOMETRY, CALCULUS & TRIGONOMETRY
01.11.2004 Max:100 marks
1.00 – 4.00 p.m.
SECTION – A
Answer ALL Questions. (10 x 2 = 20 marks)
1. If y = sin (ax + b), find yn.
2. Show that in the parabola y2 = 4ax, the subnormal is constant.
3. Prove that cos h2x = cos h2x + sin h2
4. Write the formula for the radius of curvature in polar co-ordinates.
5. Find the centre of the curvature xy = c2 at (c, c).
6. Prove that .
7. Form a rational cubic equation which shall have for roots 1, 3 – .
8. Solve the equation 2x3 – 7x2 + 4x + 3 = 0 given 1+is a root.
9. What is the equation of the chord of the parabola y2 = 4ax having (x, y) as mid – point?
10. Define conjugate diameters.
SECTION – B
Answer any FIVE Questions. (5 x 8 = 40 marks)
1. Find the nth derivative of cosx cos2x cos3x.
2. In the curve xm yn = am+n , show that the subtangent at any point varies as the abscissa of the point.
3. Prove that the radius of curvature at any point of the cycloid
x = a (q + sin q) and y = a (1 – cos q) is 4 a cos .
1. Find the p-r equation of the curve rm = am sin m q.
2. Find the value of a,b,c such that .
3. Solve the equation
6x6 – 35x5 + 56x4 – 56x2 + 35x – 6 = 0.
1. If the sum of two roots of the equation x4 + px3 + qx2 + rx + s = 0 equals the sum of the other two, prove that p3 + 8r = 4pq.
2. Show that in a conic, the semi latus rectum is the harmonic mean between the segments of a focal chord.
SECTION -C
Answer any TWO Questions. (2 x 20 = 40 marks)
1. a) If y = , prove that
(1 – x2) y2 – xy1 – a2y = 0.
Hence show that (1 – x2) yn+2 – (2n +1) xyn+1 – (m2 + a2) yn = 0. (10)
1. Find the angle of intersection of the cardioid r = a (1 + cos q) and r = b (1 – cos q).
(10)
1. a) Prove that = 64 cos6 q – 112 cos4q + 56 cos2q – (12)
1. b) Show that (8)
2. a) If a + b + c + d = 0, show that
. (12)
1. b) Show that the roots of the equation x3 + px2 + qx + r = 0 are in Arithmetical
progression if 2 p3 – 9pq + 27r = 0. (8)
1. a) Prove that the tangent to a rectangular hyperbola terminated by its asymptotes is
bisected at the point of contact and encloses a triangle of constant area. (8)
1. b) P and Q are extremities of two conjugate diameters of the ellipse and S is
a focus. Prove that PQ2 – (SP – SQ)2 = 2b2. (12)
Go To Main page
## Loyola College B.Sc. Mathematics April 2008 Algebra, Calculus And Vector Analysis Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
XZ 8
THIRD SEMESTER – APRIL 2008
# MT 3501 – ALGEBRA, CALCULUS AND VECTOR ANALYSIS
Date : 26-04-08 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
SECTION – A
Answer ALL questions.: (10 x 2 = 20 marks)
1. Evaluate .
2. If find the Jacobian of x and y with respect to r and .
3. Solve
4. Find the complete solution of
.
1. Find at (2,0,1) for .
2. State Stoke’s theorem.
3. Evaluate ë (Sinh at).
4. Evaluate ë-1.
5. Find the sum of all divisors of 360.
6. Compute (720).
SECTION – B
Answer any FIVE questions. (5 x 8 = 40 marks)
1. By the changing the order of integration evaluate
1. Express interms of Gamma function and evaluate .
2. Obtain the complete and singular solutions of .
3. Solve.
4. Find if
5. Evaluate (i) ë (ii) ë
6. Find ë-1
7. Show that if x and y are both prime to the prime n, then xn-1-yn-1 is divisible by n. Deduce that x12-y12 is divisible by 1365.
SECTION – C
Answer any TWO questions. (2 x 20 = 40 marks)
1. a) Evaluate over the tetrahedron bounded by the planes and the coordinate planes.
1. b) Show that .
2. c) Using gamma function evaluate.
1. a) Solve
1. b) Solve the following by Charpit’s method
1. c) Solve
1. a) Verify Green’s theorem for where C is the region bounded by y=x and y=x2.
1. b) Show that 18!+1 is divisible by 437.
1. a) State and prove Wilson’s theorem.
1. b) Solve given using Laplace
Go To Main page
## Loyola College B.Sc. Mathematics April 2008 Algebra, Calculus & Vector Analysis Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
XZ 7
THIRD SEMESTER – APRIL 2008
# MT 3500 – ALGEBRA, CALCULUS & VECTOR ANALYSIS
Date : 26-04-08 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
PART – A
Answer ALL questions: (10 x 2 = 20 marks)
1. Show that G (n+1) = n G(n).
2. Show that
3. Form the partial differential equation by eliminating the arbitrary function from .
4. Solve:
5. Show that is solenoidal.
6. Show that curl
7. Find ë .
8. Find ë .
9. Define Euler’s function.
10. Find the number of integer, less than 600 and prime to it.
PART – B
Answer any FIVE questions: (5 x 8 = 40 marks)
1. Show that.
2. Show that é
3. Solve:
4. Find the general integral of
5. Find the directional derivative of xyz-xy2z3 at(1,2,-1) in the direction
of
1. If find where C is the curve y=2x2 from (0,0) to (1,2).
2. Find ë if
for
1. With how many zeros does end.
PART – C
Answer any TWO questions: (2 x 20 = 40 marks)
1. a) Evaluate
1. b) Evaluate over the region in the positive octant for which .
1. a) Find the complete integral of using charpits method.
1. b) If where is a constant vector and is the position vector of a point show that curl .
1. a) Verify Stoke’s theorem for
where S is the upper half of the sphere and C its boundary.
1. b) Find (i) ëùand (ii) ëù
1. a) Solve using Laplace transforms
given that
and at t = 0.
1. b) Find the highest power of 11 in .
Go To Main page
## Loyola College B.Sc. Mathematics April 2008 Algebra, Anal.Geo & Calculus – II Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
# XZ 54
SECOND SEMESTER – APRIL 2008
# MT 2500 – ALGEBRA, ANAL.GEO & CALCULUS – II
Date : 23/04/2008 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
PART – A
Answer ALL questions.: (10 x 2 = 20)
1. Evaluate
2. Write the value of
3. Is exact?
4. Solve
5. State Raabe’s test.
6. Define uniform convergence of a sequence.
7. Find the Coefficient of in the expansion of
8. Write down the last term in the expansion of
9. Write the intercept and normal forms of the equation of a plane.
10. Find the Centre and radius of the sphere
PART – B
Answer any FIVE questions. (5 x 8 = 40)
1. Evaluate
2. Solve
3. Test the Convergence of
4. Find the sum to infinity of the series
5. Sum the series
6. If a, b, c denote three Consecutive integers, show that
1. The foot of the perpendicular drawn form the origin to the plane is (12,-4,-3); find the equation of the plane.
2. Find the equation to the sphere through the four points (0,0,0), (a,0,0), (0,b,0), (0,0,c) and determine its radius.
PART – C
Answer any TWO questions. (2 x 20 = 40)
1. a) Evaluate
1. b) Find the area of the cardioid (12+8)
1. a) Prove that the series
is convergent if and
1. b) Sum the series
1. Find the image of the point (1,3,4) in the plane . Hence prove that the image of the line is .
2. Through the circle of intersection of the sphere and the plane two spheres and are drawn to touch the place . Find the equations of the spheres.
Go To Main page
## Loyola College B.Sc. Mathematics Nov 2008 Algebra, Calculus And Vector Analysis Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034 LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034B.Sc. DEGREE EXAMINATION – MATHEMATICSTHIRD SEMESTER – November 2008MT 3501 – ALGEBRA, CALCULUS AND VECTOR ANALYSIS
Date : 06-11-08 Dept. No. Max. : 100 Marks Time : 9:00 – 12:00 PART – A (10 × 2 = 20 marks)
1. Evaluate .2. What is ?3. Find the complete integral of q = 2yp2
4. Write down the complete integral of z = px + qy + pq.
5. Find the constant k, so that the divergence of the vector is zero.
6. State Gauss Divergence theorem.
7. Find L(cos23t).8. Find .9. Find Φ(360).
10. Find the highest power of 5 in 79!
PART – B (5 × 8 = 40 marks)
11. Change the order of integration and evaluate .12. Prove that β(m,n+1 )+ β(m+1,n) = β(m,n).
13. Solve p tanx + q tany = tanz.
14. If are irrotational, prove that
(a) is solenoidal.
(b)Find the unit vector normal to the surface z = x2 + y2 – 3 at (2,-1,2). (4+4)15. Evaluate by Stokes Theorem where & C is the boundary of the triangle with vertices (0,0,0), (1,0,0) and (1,1,0).
16. Find (a) L(te-t sint).
(b)L(sin3t cosh2t).
17. Find .18. (a) If N is an integer, prove that N5-N is divisible by 30. (6+2)
(b)State Fermat’s Theorem.
PART – C (2 × 10 = 20 marks)
19. (a) Evaluate over the positive octant of the sphere x2+y2+z2 = a2
(b)Establish β(m,n) = . (10+10)
20. (a) Solve .
(b) Solve by Charpit’s Method, pxy + pq + qy = yz. (10+10)
21. (a) Verify Green’s theorem for where C is the boundary of the region x=0, y=0, x+y=1.
(b) Evaluate . (10+10)22. (a) Using Laplace Transform, solve given that y(0)=1, y`(0)=0..
(b) Using Wilson’s Theorem, prove that 10!+111 0 mod 143. (12+8)
Go To Main page
## Loyola College B.Sc. Mathematics Nov 2008 Algebra, Calculus & Vector Analysis Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
# AB 07
THIRD SEMESTER – November 2008
# MT 3500 – ALGEBRA, CALCULUS & VECTOR ANALYSIS
Date : 06-11-08 Dept. No. Max. : 100 Marks
Time : 9:00 – 12:00
SECTION-A
Answer all questions: (10 x 2=20)
1. Evaluate .
2. If .
3. Obtain the partial differential equation by eliminating
4. Solve
5. If j =j at (1, -2, -1).
6. State Stoke’s theorem.
7. Find .
8. Find .
9. Find the number and sum of all the divisors of 360.
10. Find the number of integers less that 720 and prime to it.
SECTION-B
Answer any five questions: (5 x 8=40)
1. Evaluate where R is the region bounded by the curves and .
2. Express interms of Gamma function and evaluate .
3. Solve .
4. Solve .
5. Show that .
6. (a) Find .
(b) Find .
1. Find .
2. Show that (18) is divisible by 437.
SECTION-C
Answer any two questions: (2 x 20=40)
1. (a) Change the order of integration and evaluate the integral.
(b) Evaluate taken over the volume bounded by the plane .
(c) Evaluate .
1. (a) Solve
(b) Find a complete integral of .
1. Verify Gauss divergence theorem for for the cylinderical region S given by .
2. Solve .
Go To Main page
## Loyola College B.Sc. Mathematics April 2009 Algebra, Calculus & Vector Analysis Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE Examination – Mathematics
Third Semester – OCT/NOV 2010
MT 3501/MT 3500 – Algebra, Calculus and Vector Analysis
Date & Time: Dept. No. Max. : 100 Marks
PART – A
Answer ALL questions. (10 ´ 2 = 20)
1. Evaluate
2. Find the Jacobian of the transformation x = u (1 + v) ; y = v (1 + u).
3. Find the complete solution of z = xp + yq + p2 – q2.
4. Solve
5. For , find div at (1, -1, 1)
6. State Green’s theorem.
7. What is L(f¢¢ (t))?
8. Compute
9. Find the sum and number of all the divisors of 360.
10. Define Euler’s function f(n) for a positive integer n.
PART – B
Answer any FIVE questions (5 ´ 8 = 40)
1. Evaluate by changing the order of the integration.
2. Express in terms Gamma functions.
3. Solve z2( p2+q2 + 1 ) = b2
4. Solve p2 + q2 = z2(x + y).
5. Find
6. Find
7. Prove that
8. Show that 18! + 1 is divisible by 437
PART – C
Answer any THREE questions. (2 ´ 20 =40)
1. (a) Evaluate taken through the positive octant of the sphere x2 + y2 + z2 = a2.
(b) Show that
1. (a) Solve (p2 + q2) y = qz.
(b) Solve (x2 – y2)p + (y2 – zx)q = z2 – xy
1. (a) Verify Gauss divergence theorem for taken over the region bounded by the planes x = 0, x = a, y = 0 y = a, z = 0 and z = a.
(b) State and prove Fermat’s theorem.
1. (a) Using Laplace transform solve given that .
(b) Show that if n is a prime and r < n, then (n – r)! (r – 1)! + (-1)r – 1 º 0 mod n.
## Loyola College B.Sc. Mathematics April 2011 Algebra, Calculus And Vector Analysis Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
THIRD SEMESTER – APRIL 2011
# MT 3501/ MT 3500 – ALGEBRA, CALCULUS AND VECTOR ANALYSIS
Date : 12-04-2011 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
PART – A
Answer ALL questions. (10 ´ 2 = 20)
1. Evaluate
2. Find when u = x2 – y2; v = x2 + y2
3. Solve
4. Find the complete integral of z = px + qy +p2q2
5. Find grad f if f = xyz at (1, 1, 1)
6. Evaluate divergence of the vector point function
7. Find L[sin2 2t]
8. Find
9. Find the sum of all divisors of 360.
10. Find the remainder when 21000 divisible by 17.
PART – B
Answer any FIVE questions. (8 ´ 5 = 40)
1. Change the order of integration and evaluate
2. Express in terms of Gamma functions and evaluate
3. Solve p2 + pq = z2
4. Solve xp + yq = x
5. Show that the vectoris irrotational.
6. Evaluate: (a) L[cos 4t sin 2t] (b) L[e-3t sin2t]
7. Find
8. Show that 18! + 1 is divisible by 437.
PART – C
Answer any TWO questions. (2 ´ 20 = 40)
1. (a) Evaluate where the region V is bounded by x + y+ z = a (a > 0),
x = 0; y = 0; z = 0
.
(b) Evaluate where R is the region in the positive quadrant for which
x + y £ 1.
(c) Show that
1. (a) Solve (x2 + y2 + yz)p + (x2 + y2 – xz)q = z(x+y)
(b) Find the complete integral and singular integral of p3 + q3 = 8z
1. (a) Solve y¢¢ + 2y¢ – 3y = sin t given that y(0) = y¢(0) = 0
(b) State and prove the Weirstrass inequality.
1. (a) State and prove Wilson’s theorem.
(b) Verify Green’s theorem in the XY plane for where C is the closed curve in the region bounded by y = x; y = x2.
Go To Main Page
## Loyola College B.Sc. Mathematics April 2011 Algebra, Anal.Geo & Calculus – II Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
SECOND SEMESTER – APRIL 2011
# MT 2501/MT 2500 – ALGEBRA, ANAL.GEO & CALCULUS – II
Date : 08-04-2011 Dept. No. Max. : 100 Marks
Time : 9:00 – 12:00
PART – A
Answer ALL questions: (10 x 2 = 20)
1. Evaluate .
2. Evaluate .
3. Solve: .
4. Solve: .
5. Prove that the series is convergent.
6. Test for convergency the series .
7. Find the general term in the expansion of .
8. Prove that the coefficient of in the expansion of is .
9. Find the equation of the sphere which has its centre at the point and touches the
plane .
1. Find the distance between the parallel planes and
PART – B
Answer any FIVE questions: (5 x 8 = 40)
1. Prove that .
2. If ( n being a positive integer), prove that .
Also evaluate and .
1. Solve: .
1. Solve
2. Test for convergency and divergency the series
3. Show that the sum of the series .
4. Show that if
1. Find the equation of the plane passing through the points
.
PART – C
Answer any TWO questions: (2 x 20 = 40)
1. a) Evaluate (10 marks)
2. b) Find the area and the perimeter of the cardiod . (10 marks)
3. a) Solve: . (10 marks)
4. b) Discuss the convergence of the series for
positive values of . (10 marks)
21.a) Show that the error in taking as an approximation to is
approximately equal to when is small. (10 marks)
1. b) show that (10 marks)
1. a) A sphere of constant radius passes through the origin and meets the axes in A, B, C.
Prove that the centroid of the triangle ABC lies on the sphere
(10 marks)
1. b) Find the shortest distance between the lines
.
Also find the equation of the line of shortest distance. (10 marks)
Go To Main Page
## Loyola College B.Sc. Mathematics April 2012 Algebra,Analytical Geometry And Calculus Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc., DEGREE EXAMINATION – MATHEMATICS
SECOND SEMESTER – APRIL 2012
MT 2501/MT2500- ALGEBRA,ANALYTICAL GEOMETRY AND CALCULUS
Date: 16-04-2012 Dept. No. Max. : 100 marks
Time: 9.00 – 12.00
PART-A
1. Evaluate .
2. Find x dx.
3. Define exact differential equations.
4. Solve (D22D + 1) y = 0.
5. Show that the series is convergent.
6. State Cauchy root test for convergence of a series.
7. Find the coefficient of in the expansion of 1 + + + + …
8. Prove that
9. Find the direction cosines of the line joining the points (3,-5,4) and (1,-8,-2).
10. Find the angle between the planes and .
PART –B
1. Evaluate
2. Find the surface area of the solid formed by revolving the cardiodabout the initial line.
3. Solve
4. Solve .
5. Examine the convergence of
6. Assuming that the square and the higher powers of x may be neglected
show that
1. Sum to infinity the series
2. Find the shortest distance between the linesand and the equation of the line.
PART – C
ANSWER ANY TWO QUESTIONS: (2×20= 40)
1. (a) Evaluate.
(b) Find the length of the curve between the points given by and.
1. (a) Solve
(b) Solve the following equation by the method of variation of parameter:
.
1. (a) Test the convergence of the series
(b) Find the equation of the sphere passing through the points (2,3,1),(5,-1, 2),
(4,3,-1) and (2,5,3).
1. (a) Show that
(b) Sum the series
Go To Main Page
## Loyola College B.Sc. Mathematics April 2012 Algebra, Analy. Geo., Calculus & Trigonometry Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
FIRST SEMESTER – APRIL 2012
# MT 1500 – ALGEBRA, ANALY. GEO., CALCULUS & TRIGONOMETRY
Date : 28-04-2012 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
PART – A
Answer ALL the questions: (10 X 2 = 20 Marks)
1. Find the nth derivative of .
2. Find the slope of the straight line .
3. Write the formula for the radius of curvature in Cartesian form.
4. Define Cartesian equation of the circle of the curvature.
5. If ,are the roots of the equation x3+px2+qx+r=0. Find the value of .
6. Diminish the roots x4+x3-3x2+2x-4 =0 by 2.
7. Evaluate
8. Prove that
9. Define Pole and Polar of a ellipse.
10. In the hyperbola 16x2-9y2 = 144, find the equation of the diameter conjugate to the diameter x =2y.
PART – B
Answer any FIVE questions: (5 X 8 = 40 Marks)
1. Find the nth derivative of .
2. Find the angle between the radius vector and tangent for the curve at
.
1. Solve the equation x3-4x2-3x+18=0 given that two of its roots are equal.
2. Solve the equation x4-5x3+4x2+8x-8=0 given that 1-is a root.
3. Expand in terms .
4. Separate real and imaginary parts .
5. P and Q are extremities of two conjugate diameters of the ellipse and S is a focus. Prove that
6. The asymptotes of a hyperbola are parallel to 2x+3y=0 and 3x-2y =0 . Its centre is at (1,2) and it passes through the point (5,3). Find its equation and its conjugate.
PART – C
Answer any TWO questions: (2 x 20=40 Marks)
1. (a) If , show that
(b) Prove that the sub-tangent at any point on is constant ant the subnormal is
(10 +10)
1. (a) Find the radius of curvature at any point on the curve
(b) Show that the evolute of the cycloid is another
cycloid . (10+10)
1. (a) Solve 6x5+11x4-33x3-33x2+11x+6=0.
(b) Find by Horner’s method, the roots of the equation which lies between 1 and 2
correct to two decimal places. (10+10)
1. (a) Prove that
(b) Prove that the Product of the perpendicular drawn from any point on a hyperbola to its
asymptotes is constant. (10+10)
Go To Main Page
## Loyola College B.Sc. Mathematics Nov 2012 Algebra, Calculus And Vector Analysis Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
FIRST SEMESTER – NOVEMBER 2012
# MT 1503 – ANALYTICAL GEOMETRY OF 2D,TRIG. & MATRICES
Date : 10/11/2012 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
PART – A
Answer all questions: (10 x 2 = 20)
1. Write down the expression of cos in terms of cosθ and sinθ.
2. Give the expansion of sinθin ascending powers of θ.
3. Express sin ix and cosix in terms of sin hx and coshx.
4. Find the value of log(1 + i).
5. Find the characteristic equation of A = .
6. If the characteristic equation of a matrix is , what are its eigen values?
7. Find pole of lx + my + n = 0 with respect to the ellipse
8. Give the focus, vertex and axis of the parabola
9. Find the equation of the hyperbola with centre (6, 2), focus (4, 2) and e = 2.
10. What is the polar equation of a straight line?
PART – B
Answer any five questions. (5 X 8 = 40)
1. Expandcos in terms of sinθ .
2. If sinθ = 0.5033 show thatθ is approximately .
3. Prove that .
4. If tany = tanα tanhβ ,tanz = cotα tanhβ, prove that tan (y+z) = sinh2βcosec2α.
5. Verify Cayley Hamilton theorem for A =
6. Prove that the eccentric angles of the extremities of a pair of semi-conjugate diameters of an ellipse differ by a right angle.
7. Find the locus of poles of all tangents to the parabola with respect to
1. Prove that any two conjugate diameters of a rectangular hyperbola are equally inclined to the asymptotes.
PART – C
Answer any two questions: (2 X 20 = 40)
1. (i) Prove that .
(ii) Prove that .
1. (i) Prove that if
(ii) Separate into real and imaginary parts tanh(x + iy).
1. Diagonalise A =
2. (i) Show that the locus of the point of intersection of the tangent at the extremities of a pair of
conjugate diameters of the ellipse is the ellipse
(ii) Show that the locus of the perpendicular drawn from the pole to the tangent to the circle r = 2a
cosθ isr = a(1+cosθ).
Go To Main Page
## Loyola College B.Sc. Mathematics Nov 2012 Algebra, Analy. Geo., Calculus & Trigonometry Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc., DEGREE EXAMINATION – MATHEMATICS
THIRD SEMESTER – NOVEMBER 2012
# MT 3501/3500 – ALGEBRA, CALCULUS AND VECTOR ANALYSIS
Date : 02-11-2012 Dept. No. Max. : 100 Marks
Time : 9.00 – 12.00
PART – A
ANSWER ALL THE QUESTIONS: (10 x 2 = 20)
1. Evaluate .
2. Evaluate .
3. Eliminate the arbitrary constants from .
4. Find the complete solution for
5. Find , if .
6. Prove that div , where is the position vector.
7. Find L(Sin2t).
8. Find .
9. Find the number and sum of all the divisors of 360.
10. State Fermat’s theorem.
PART – B
ANSWER ANY FIVE QUESTIONS: (5 x 8 = 40)
1. Change the order of integration and evaluate the integral .
2. Express in terms of Gamma functions and evaluate the integral .
3. Solve
4. Solve
5. Find .
6. Find .
7. Show that
8. Show that is divisible by 22.
PART – C
ANSWER ANY TWO QUESTIONS (2x 20 = 40)
1. (a) Evaluate taken over the positive quadrant of the circle .
(b) Prove that
1. (a) Solve
(b) Solve (y+z)p + (z+x)q = x+y.
21.(a) Verify Stoke’s theorem for taken over the upper half surface of
the sphere x2+y2 +z2 = 1, z 0 and the boundary curve C, the x2+y2 = 1, z=0.
(b) State and prove Wilson’s Theorem.
1. Using Laplace transform solve the equation given y(0) = 0 , y1(0)= -1.
Go To Main Page
## Loyola College B.Sc. Mathematics Nov 2012 Algebra, Anal.Geo & Calculus – II Question Paper PDF Download
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034
B.Sc. DEGREE EXAMINATION – MATHEMATICS
FIRST SEMESTER – NOVEMBER 2012
# MT 1500 – ALGEBRA, ANALY. GEO., CALCULUS & TRIGONOMETRY
Date : 08/11/2012 Dept. No. Max. : 100 Marks
Time : 1:00 – 4:00
PART – A
Answer ALL the questions: (10 x 2 = 20 marks)
1. Write the nth derivative of
2. If y = a show that
3. Define the evolute of a curve.
4. Find the p-r equation of the curve r = a sin q.
5. Determine the quadratic equation having 3 – 2 i as a root.
6. Diminish the roots of by 2.
7. Show that
8. Express in locus of logarithmic function.
9. Define a rectangular hyporbola.
10. Write down the angle between the asymptotes of the hyperbola
PART – B
Answer any FIVE questions: (5 x 8 = 40 marks)
1. Show that in the parabola the subtangent at any point is double the abscissa and the subnormal is a constant.
2. Find the radius of curvature at the point ‘O’ on
3. Show that if the roots of
4. Find the p-r equation of the curve with respect to the focus as the pole.
5. Separate into real and uniaguinary parts.
6. Find the sum of the series
7. Find the locus of poles of ale Laugets to with respect to
8. Derive the polar equation of a comic.
PART – C
Answer any TWO questions: ( 2 x 20 = 40 marks)
1. a) If prove that
2. b) Show that r = a sec2 and r = b cosec2 intersect at right angles.
3. a) Find the minimum value of
4. b) Find the radius of curvature of .
5. a) Solve: given that the roots are in geometric progression.
1. b) Solve: .
1. a) Express cos8q in locus of power of sinq.
1. b) If e1 and e2 are the eccentricities of a hyperbola and its conjugate show that .
Go To Main Page
© Copyright Entrance India - Engineering and Medical Entrance Exams in India | Website Maintained by Firewall Firm - IT Monteur | 8,512 | 29,608 | {"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-2023-23 | latest | en | 0.702661 |
https://merithub.com/quiz/motion-and-measurement-of-distances-c7rukv4t2n9p0qqk8sm0 | 1,723,449,536,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00835.warc.gz | 314,134,043 | 9,122 | Motion and Measurement of Distances
This is an interactive Quiz on the topic motion and Measurement of Distances.
This will include the topics such as, distances, speed and time, calculation and measurement of distances.
Start Quiz
The height of a man is 1.56 m. This height is equal to:
1560 mm
156 mm
15.6 mm
1560 cm
Which of the following does not express a time interval?
A day
A second
A school period
Time of the first bell in the school
Which of the following is a smallest unit of length?
km
mm
cm
m
The tip of the seconds’ hand of a clock is colored red. Which of the following types of motion will be exhibited by the red colored tip of the seconds’ hand of this clock?
Rotational motion and Periodic motion
Rotational motion and Circular motion
Periodic motion and Circular motion
Circular motion and Rectilinear motion
Consider the following statements and choose the correct one(s):
The movement of an object is called motion.
An object is said to be in motion or moving when its position changes with time.
When the position of a car changes with time, we say that the car is moving or that the car is in motion.
All the above.
Which is a correct relationship?
1 m = 100 cm
1 cm = 100 mm
1 km = 100 m
all of these
4 kilometres are equal to
4,00,000 meter
40,000 meter
4,000 meter
400 meter
What is the SI unit of length?
Metre
Centimetre
Kilometre
All of these
Which is a standard unit of measurement?
Angul (finger)
Mutthi (fist)
Step
Inch
One metre is equal to ………….. millimetre.
10
1000
100
10000
Which is an example of a periodic motion?
Oscillation of a pendulum
Motion of a bus on road
A spinning top
A stone dropped from a certain height
What kind of motion is executed by a pendulum of a wall clock?
Oscillatory motion
Vibratory motion
Circular motion
Linear motion
15 cm are equal to
150 mm
15 mm
1.5 mm
0.15 mm
Read The Following Sentences Carefully, And Choose The Correct One:
When you push up box on the floor it undergoes rectilinear motion.
Measurement involves the comparison of an unknown quantity with a known quantity.
Motion of needle of a sewing machine is periodic motion.
All the above
Read The Following Sentences Carefully, And Choose The Incorrect One:
Transport means to carry people and goods from one place to another.
People learnt to join together plank on wood to make boats having a streamlined shape.
The invention of wheel made a great change in the modes of transport.
In the beginning of 18th century, the invention of steam engine introduced a new source of power to run transport vehicles.
Devesh Appeared In Class Test But He Confused To Know The Incorrect Statement. Would You Help Him To Know That?
The invention of internal combustion engine in the second half of 19th century gave us transport vehicles called automobiles.
Aero-planes where developed as a means of transport in the early 20th century.
The length of the space between two points or two places is called distance.
Compass (navigation) is invented by Egyptian.
In A Science Quiz Competition, Nalini Is Asked A Question Where She Had To Choose The Statement Which Was/Were Incorrect?
It is necessary to have standard units of measurement for the sake of uniformity in measurements.
Every measurement consists of a number and a unit.
Measurement is not complete unless both the number and the unit are mentioned.
The SI unit of measuring length is centimeter.
A Carpenter Is Fixing A Certain Rod On The Wall By Tightening A Screw. How Many Different Kinds Of Motion Is The Screw Undergoing?
Rotational motion and Random motion
Rotational motion and Oscillatory motion
Periodic motion and Circular motion
Circular motion and Rectilinear motion
Consider The Following Statements And Choose The Correct One:
One complete cycle from the point of release to the other extreme and back to the initial point is called one oscillation.
When a body moves in different direction and does not have a fixed path, it is called random motion.
An object that does not change its place or position with time, relative to its surrounding is said to be stationary or at rest.
All the above.
Among the following, an example of rectilinear motion is
A train moving on a straight bridge
Swaying of a Tree
A child on merry-go-round
The motion of pendulum in a wall clock
Quiz/Test Summary
Title: Motion and Measurement of Distances
Questions: 20
Contributed by: | 1,000 | 4,463 | {"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 | 0.898118 |
https://www.teacherspayteachers.com/Product/Rounding-to-nearest-1000-using-Number-lines-Task-Cards-3485896 | 1,540,196,723,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514879.30/warc/CC-MAIN-20181022071304-20181022092804-00115.warc.gz | 1,079,503,857 | 20,970 | # Rounding to nearest 1,000 (using Number lines) Task Cards
Subject
Resource Type
Product Rating
File Type
Compressed Zip File
10 MB|24 cards, recording sheet, answer key
Share
Also included in:
1. This Place Value BUNDLE is full of 15 resources to help your students practice place value skills! These place value resources are great for small groups, math centers, or whole group reinforcement activities! Here are the sets included! They are also sold separately!1Place Value- Great for Test Rev
\$66.00
\$50.00
Save \$16.00
Product Description
Rounding to Nearest 1,000 (using Number lines) Task cards
*** This set is also available in a Place Value BUNDLE! Here's the link: Place Value HUGE Bundle!
This is a colorful set of 24 task cards to practice rounding to nearest 1,000 using number lines. This set is a wonderful addition to your lessons!
I've included a recording sheet and answer key, too!
◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈
These activities would work for grades 1-3!
◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈
Here are some possible uses for these in your classroom:
✿ early finishers
✿ tutoring
✿ sub tubs
✿ math stations/centers
✿ holiday work
✿ small group
✿ end of unit quick assessments
✿ homework
✿ reinforcement
✿ enrichment
◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈
Here are some other math resources you might to check out:
3rd Grade Math Spiral Review for Morning Work or Homework
3rd Grade Math Bundle- Year long
Multiplication Dice Bundle
Word Problem Task Cards BUNDLE -7 sets
Fraction Bundle- Parts of a Set
Multiplication Bump Games (Sports Themes)
◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈
Customer Tips:
How to get TPT credit to use on FUTURE purchases:
❤Please go to your My Purchases page (you may need to login). Beside each purchase you'll see a Provide Feedback button. Simply click it and you will be taken to a page where you can give a quick rating and leave a short comment for the product.
❤Each time you give feedback, TPT gives you feedback credits that you use to lower the cost of your future purchases.
❤I VALUE your feedback greatly as it helps me determine which products are most valuable for your classroom so I can create more for you.
Be the first to know about my new discounts, freebies and product launches. All new resources are 50% off the FIRST 48 hours!
◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈◈
My TPT store:❤ Believe to Achieve Store by Anne Rozell
Follow my Pinterest boards:❤My Pinterest Boards
My Instagram
Follow me on my blog : ❤ My Blog
Total Pages
24 cards, recording sheet, answer key
Included
Teaching Duration
N/A
Report this Resource
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,026 | 2,760 | {"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-2018-43 | latest | en | 0.780664 |
https://codegolf.stackexchange.com/revisions/80031/4 | 1,580,288,703,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251789055.93/warc/CC-MAIN-20200129071944-20200129101944-00004.warc.gz | 374,285,465 | 17,084 | 4 of 5 deleted 100 characters in body
MATL, 19 bytes
PiYdt"TF2&YStpsw]xx
Try it online!
Explanation
This builds a block-diagonal matrix with the two inputs, reversing the first. For example, with inputs [1 4 3 5], [1 3 2] the matrix is
[ 5 3 4 1 0 0 0
0 0 0 0 1 3 2 ]
Each entry of the convolution is obtained by shifting the first row one position to the right, computing the product of each column, and summing all results.
In principle, the shifting should be done padding with zeros from the left. Equivalently, circular shifting can be used, because the matrix contains zeros at the appropriate entries.
For example, the first result is obtained from the shifted matrix
[ 0 5 3 4 1 0 0
0 0 0 0 1 3 2 ]
and is thus 1*1 == 1. The second is obtained from
[ 0 0 5 3 4 1 0
0 0 0 0 1 3 2 ]
and is thus 4*1+1*3 == 7, etc. This must be done m+n-1 times, where m and n are the input lengths. The code uses a loop with m+n iterations (which saves some bytes) and discards the last result.
P % Take first input (numeric vactor) implicitly and reverse it
i % Take second input (numeric vactor)
Yd % Build diagonal matrix with the two vectors
t % Duplicate
" % For each column of the matrix
TF2&YS % Circularly shift first row 1 step to the right
t % Duplicate
p % Product of each column
s % Sum all results
w % Swap top two elements in stack. The shifted matrix is left on top
] % End for
xx % Delete matrix and last result. Implicitly display | 457 | 1,543 | {"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} | 2.828125 | 3 | CC-MAIN-2020-05 | latest | en | 0.835995 |
https://www.instasolv.com/question/1-2x-at-tx-1-b-1-1-x-x-1-2-x-1-x-1-x2-x-1-x-x-1-la-x2-x-1-x-x-1-ihaxal | 1,610,897,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00712.warc.gz | 835,337,288 | 10,848 | 1. 2x at tx? +1 (b) *-1_ +1 x-x+1 2...
Question
# 1. 2x at tx? +1 (b) *-1_ +1 x-x+1 2+x-1 x +1 x2-x+1 x² + x-1 + - ( la) x2 - x+1 x² + x +1
JEE/Engineering Exams
Maths
Solution
147
4.0 (1 ratings)
Sinfe, Now, Use ( begin{aligned} x^{4}+x^{2}+1 &=left(x^{2}+x+1right)left(x^{2}+x+1right) frac{2 x}{x^{4}+x^{2}+1} &=frac{2 x}{left(x^{2}+x+1right)left(x^{2}-x+1right)} &=frac{left(x^{2}+x+1right)-left(x^{2}-x+1right)}{left(x^{2}-x+1right)left(x^{2}+x+1right)} &=frac{left(x^{2}+x+1right)}{left(x^{2}+x+1right)left(x^{2}-x+1right)}-frac{left(x^{2}-x+1right)}{left(x^{2}+x+1right)left(x^{2}+4+1right)} &=frac{1}{x^{2}-x+1}-frac{1}{x^{2}+x+1} end{aligned} ) | 350 | 654 | {"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} | 3.96875 | 4 | CC-MAIN-2021-04 | latest | en | 0.477019 |
https://pageranks.biz/littleton-colorado-80166-seo-near-me.html | 1,555,978,447,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578583000.29/warc/CC-MAIN-20190422235159-20190423021159-00098.warc.gz | 519,808,245 | 7,817 | ## I first discovered Sharpe years ago online. His story was one of the most sincere and intriguing tales that any one individual could convey. It was real. It was heartfelt. It was passionate. And it was a story of rockbottom failure. It encompassed a journey that mentally, emotionally and spiritually crippled him in the early years of his life. As someone who left home at the age of 14, had a child at 16, became addicted to heroin at 20 and clean four long years later, the cards were definitely stacked up against him.
Assume a small universe of four web pages: A, B, C and D. Links from a page to itself, or multiple outbound links from one single page to another single page, are ignored. PageRank is initialized to the same value for all pages. In the original form of PageRank, the sum of PageRank over all pages was the total number of pages on the web at that time, so each page in this example would have an initial value of 1. However, later versions of PageRank, and the remainder of this section, assume a probability distribution between 0 and 1. Hence the initial value for each page in this example is 0.25.
Danny Sullivan was a journalist and analyst who covered the digital and search marketing space from 1996 through 2017. He was also a cofounder of Third Door Media, which publishes Search Engine Land, Marketing Land, MarTech Today and produces the SMX: Search Marketing Expo and MarTech events. He retired from journalism and Third Door Media in June 2017. You can learn more about him on his personal site & blog He can also be found on Facebook and Twitter.
Internet marketing, or online marketing, refers to advertising and marketing efforts that use the Web and email to drive direct sales via electronic commerce, in addition to sales leads from websites or emails. Internet marketing and online advertising efforts are typically used in conjunction with traditional types of advertising such as radio, television, newspapers and magazines.
The PageRank algorithm outputs a probability distribution used to represent the likelihood that a person randomly clicking on links will arrive at any particular page. PageRank can be calculated for collections of documents of any size. It is assumed in several research papers that the distribution is evenly divided among all documents in the collection at the beginning of the computational process. The PageRank computations require several passes, called “iterations”, through the collection to adjust approximate PageRank values to more closely reflect the theoretical true value.
Nice word is not enough for this. You show that Blogging is like Apple vs Samsung. You can create lot of post and drive traffic (which is Samsung like lot of phone every year) or you can create high quality post like apple (which is you) and force higher rank site to make content like you copy content from you blog. Now i will work hard on already publish post until they will not get traffic.
Going into network marketing? Understand that if you're not close to the top of the food chain there, your ability to generate any serious amount of income will be limited. Be wary of the hype and the sales pitches that get you thinking that it's going to work the other way. Simply understand that you're going to have to work hard no matter what you pick to do. Email marketing? Sure. You can do that. But you'll need a massive and very targeted list to make any dent.
On a blog the page rank should go to the main article pages. Now it just gets “evaporated” if you use “nofollow” or scattered to all the far flung nooks and crannys which means google will not be able to see the wood for the trees. The vast majority of a site’s overall page rank will now reside in the long tail of useless pages such as commentors profile pages. This can only make it harder for google to serve up the most relevant pages.
```Today, with nearly half the world's population wired to the internet, the ever-increasing connectivity has created global shifts in strategic thinking and positioning, disrupting industry after industry, sector after sector. Seemingly, with each passing day, some new technological tool emerges that revolutionizes our lives, further deepening and embedding our dependence on the world wide web.
```
And looking at say references would it be a problem to link both the actual adress of a study and the DOI (read DOI as anything similar)? Even if they terminate at the same location or contain the same information? The is that it feels better to have the actual adress since the reader should be able to tell which site they reach. But also the DOI have a function.
```For instance, you might use Facebook’s Lookalike Audiences to get your message in front of an audience similar to your core demographic. Or, you could pay a social media influencer to share images of your products to her already well-established community. Paid social media can attract new customers to your brand or product, but you’ll want to conduct market research and A/B testing before investing too much in one social media channel.
```
Brand awareness has been proven to work with more effectiveness in countries that are high in uncertainty avoidance, also these countries that have uncertainty avoidance; social media marketing works effectively. Yet brands must be careful not to be excessive on the use of this type of marketing, as well as solely relying on it as it may have implications that could negatively harness their image. Brands that represent themselves in an anthropomorphizing manner are more likely to succeed in situations where a brand is marketing to this demographic. "Since social media use can enhance the knowledge of the brand and thus decrease the uncertainty, it is possible that people with high uncertainty avoidance, such as the French, will particularly appreciate the high social media interaction with an anthropomorphized brand." Moreover, digital platform provides an ease to the brand and its customers to interact directly and exchange their motives virtually.[33]
PageRank has been used to rank spaces or streets to predict how many people (pedestrians or vehicles) come to the individual spaces or streets.[51][52] In lexical semantics it has been used to perform Word Sense Disambiguation,[53] Semantic similarity,[54] and also to automatically rank WordNet synsets according to how strongly they possess a given semantic property, such as positivity or negativity.[55]
What are backlinks doing for your SEO strategy? Well, Google considers over 200 SEO ranking factors when calculating where a page should rank, but we know that backlinks are one of the top three (the other two are content and RankBrain, Google’s AI). So while you should always focus on creating high-quality content, link-building is also an important factor in ranking your pages well on Google.
```nofollow is beyond a joke now. There is so much confusion (especially when other engines’ treatment is factored in), I don’t know how you expect a regular publisher to keep up. The expectation seems to have shifted from “Do it for humans and all else will follow” to “Hang on our every word, do what we say, if we change our minds then change everything” and nofollow lead the way. I could give other examples of this attitude (e.g. “We don’t follow JavaScript links so it’s ’safe’ to use those for paid links”), but nofollow is surely the worst.
```
“NOTE: You may be curious what your site’s or your competitor’s PR score is. But Google no longer reveals the PageRank score for websites. It used to display at the top of web browsers right in the Google Toolbar, but no more. And PR data is no longer available to developers through APIs, either. Even though it’s now hidden from public view, however, PageRank remains an important ingredient in Google’s secret ranking algorithms.”
Our backgrounds are as diverse as they come, bringing knowledge and expertise in business, finance, search marketing, analytics, PR, content creation, creative, and more. Our leadership team is comprised of successful entrepreneurs, business executives, athletes, military combat veterans, and marketing experts. The Executives, Directors, and Managers at IMI are all well-respected thought leaders in the space and are the driving force behind the company’s ongoing success and growth.
```If you've read anything about or studied Search Engine Optimization, you've come across the term "backlink" at least once. For those of you new to SEO, you may be wondering what a backlink is, and why they are important. Backlinks have become so important to the scope of Search Engine Optimization, that they have become some of the main building blocks to good SEO. In this article, we will explain to you what a backlink is, why they are important, and what you can do to help gain them while avoiding getting into trouble with the Search Engines.
``` | 1,802 | 8,895 | {"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-2019-18 | latest | en | 0.970869 |
https://questioncove.com/updates/516b7503e4b02ec89c5ab0e6 | 1,503,392,417,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00659.warc.gz | 824,687,688 | 2,703 | OpenStudy (anonymous):
Solve each inequality. Graph the solution set and write the answer in interval notation. Terrible at math please walk me through the steps 4 less than or equal to k+11 over 4 less then or equal to 4
OpenStudy (anonymous):
$4\le \frac{ k+11 }{ 4 }\le4$
OpenStudy (agent0smith):
You can either break the inequality up into two parts, or keep it as is, depending which way you find easier. If you keep it as is, start by multiplying everything by 4, and then subtract 11 from everything.
OpenStudy (agent0smith):
$\large 4 \times 4\le 4 \times \frac{( k+11)}{ 4 }\le4 \times 4$
OpenStudy (anonymous):
$16\le11K \le16$
OpenStudy (agent0smith):
k+11 is not equal to 11k, you have to keep it as k+11. Also are you sure this is correct $4\le \frac{ k+11 }{ 4 }\le4$ seems like you wouldn't have a 4 on both the left and right. Or else it just means $\frac{ k+11 }{ 4 }\ = 4$ | 274 | 901 | {"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} | 3.984375 | 4 | CC-MAIN-2017-34 | latest | en | 0.929281 |
https://www.rajibroy.com/2016/03/ | 1,721,449,625,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00657.warc.gz | 823,799,162 | 16,742 | 31 March 2016
# A puzzle… after a long time…
Getting bored in the loooooong flight back. Suddenly remembered an email exchange I had with my friend Mita this week. It was actually a puzzle. In all honesty, I had heard a puzzle somewhat like that, so I was able to get it. Figured I will post it here.
The challenging part of the puzzle is its incredible simplicity. There are no catches in the puzzle. But the simplicity is bound to confuse readers. So here it goes…
“John called Mary. Mary called Tom. John is married but Tom is not”.
From the above, can you say whether the following statement
“A married person called an unmarried person” is
(a) Definitely True
(b) Definitely False
(c) Insufficient information to say whether True or False.
Send me personal message instead of putting in Comments and I will respond.
31 March 2016
# Perspective from the other side
The plane started rolling off the starting point on the runway. As it gathered momentum and pulled the wheels up, rapidly we could see River Columbia emerge to our left. A few seconds later, as we continued to gain height, something seemed to trigger some old memories as I looked down from the plane. Quickly took a few pictures of the sight and then waited to get Wifi access in the plane (after 10,000 feet).
A few minutes of research on my blog and sure enough – I figured out why the scenery looked familiar.
Notice the right bottom end – near the plane’s window edge, you can see the road. But between the road and the the river, if you carefully notice, there is a trail that goes along the river (this is the Oregon side; the other side is Washington).
Way back on Oct 20, last year, I had driven up to a trail head and started running. Eventually, I ran along the exact spot that you see in this picture. See the inset – and you can see the same bridge. I remember stopping multiple times to watch the planes fly very low as it approached the runway.
Today, I was on the other end!!
31 March 2016
# Perfect opportunity…
The plane took off southeasterly from Portland airport and then dipped south for a minute to then straighten up eastward wrapping around Mount Hood. And as it did that, there were a few incredible seconds where we were still under cloud cover to get a clear view of all the four peaks – Mount Hood in Oregon and Mount St.Helen, Mount Rainier and Mount Adams in Washington – all at once!!!
27 March 2016
# In case you had any doubts that we are getting overtaxed!!
This morning, I woke up early to start doing the taxes. Duly downloaded Turbo Tax which I have been using for over 15 years to prepare my taxes. It is undoubtedly one of the most user friendly software I have ever seen. In any case, Turbo Tax imported all the necessary information from last year’s file so that I do not have to re-enter all the mundane details like name, social, dependents etc.
And then, it came up with this screen! I owe \$9 of taxes!!
You can imagine how flummoxed I was – “Wait a minute – I have not entered anything about 2015 yet!!! And you yourself prepared last year’s taxes. How can I start with a tax liability already?”
As you can imagine, things only got worse from there!! 🙂
26 March 2016
# From the bartender’s corner – Flatiron Martini
If you like orange-y cocktail, you might like this. Not sweet but filled with rich orange aroma. Ingredients include orange vodka (alternately mandarin vodka) and Lillet Blanc. Usually the martini glass is rinsed with Triple Sec before serving to enhance the presence of orange …
25 March 2016
# “Runner” cholechhey..
This morning I got into the car and a casual glance at the passenger seat revealed that I had not mailed an envelope since Monday. So, changed my running route plans and put in a run to the post office from Starbucks – about two and a half miles each way…
… thereby earning the moniker “Runner” in two different languages simultaneously 🙂
This was probably the only run where I was hoping against hopes that it would NOT rain!!!
24 March 2016
23 March 2016 | 907 | 4,046 | {"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-2024-30 | latest | en | 0.975422 |
http://www.jiskha.com/display.cgi?id=1378265625 | 1,496,030,694,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612013.52/warc/CC-MAIN-20170529034049-20170529054049-00355.warc.gz | 672,773,487 | 3,766 | # math
posted by on .
A very small population consists of 7 units, labeled a; b; c; d; e; f; g. How many
di fferent samples of size 3 are there, for sampling without replacement, assuming
that
(a) samples are lists of units.
(b) samples are sets of units.
• math - ,
lists: P(7,3) = 7*6*5
sets: C(7,3) = P(7,3)/3! | 110 | 317 | {"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-2017-22 | latest | en | 0.911718 |
https://testbook.com/objective-questions/mcq-on-axial-elongation-of-bar--5eea6a0b39140f30f369de9c | 1,642,688,332,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00577.warc.gz | 615,033,840 | 47,226 | # A bar having a cross-sectional area of 700 mm2 is subjected to axial loads at the positions indicated. The value of stress in the segment QR is:
1. 40 MPa
2. 50 MPa
3. 70 MPa
4. 120 MPa
Option 1 : 40 MPa
## Axial Elongation of Bar MCQ Question 1 Detailed Solution
By drawing the free body diagram,
$${\sigma _{{Q_R}}} = \frac{{28 \times {{10}^3}}}{{700}} = 40\ MPa$$
# A steel rod of 30 mm diameter and 3 m length is subjected to an axial pull of 50 kN. If E = 200 × 109 Pa, the elongation of the rod will be
1. 2.225 mm
2. 1.062 mm
3. 0.525 mm
4. 3.152 mm
Option 2 : 1.062 mm
## Axial Elongation of Bar MCQ Question 2 Detailed Solution
Concept:
Hooke’s law: As per Hooke’s law, stress is proportional to strain
i.e. σ ∝ ϵ or, σ = ϵ × E
Where, E = Modulus of Elasticity
For Hooke’s law to be valid:
(a) The material should be homogenous.
(b) The material should be isotropic.
(c) The material should behave in a linearly elastic manner.
Thus for a plane bar with Area ‘A’, Length ‘L’ and Modulus of Elasticity ‘E’,
$${\rm{\sigma = }}\frac{{\rm{P}}}{{\rm{A}}}\;\;{\rm{and}}\;\;{\rm{ϵ = }}\frac{{{\rm{\delta L}}}}{L}$$
As per Hooke’s law,
∵ σ = ϵ × E
$$\frac{{\rm{P}}}{{\rm{A}}}{\rm{ = }}\frac{{{\rm{\delta L}}}}{{\rm{L}}}{\rm{ × E}}$$
$${\rm{\delta L = }}\frac{{{\rm{PL}}}}{{{\rm{AE}}}}$$
Calculation:
Given,
P = 50 kN = 50 × 103 N
L = 3 m = 3000 mm
$${\rm{A}} = \frac{{\rm{\pi }}}{4} × {{\rm{d}}^2} = \frac{{\rm{\pi }}}{4} × {\left( {30 × {{10}^{ - 3}}} \right)^2} = 7.068 × {10^{ - 4}}{\rm{\;m}}{{\rm{m}}^2}$$
E = 200 × 109 Pa
$${\rm{\delta L = }}\frac{{{\rm{PL}}}}{{{\rm{AE}}}}$$
$${\rm{\delta L}} = \frac{{\left( {50 \times {{10}^3}} \right) \times 3}}{{\left( {7.068 \times {{10}^{ - 4}}} \right) \times \left( {200 \times {{10}^{9}}} \right)}} = 1.062{\rm{\;mm}}$$
Hence the elongation of rod will be 1.062 mm
# Elongation of a bar due to its self-weight is computed by ______, where L- length of the bar, E-Young’s modulus of elasticity and W - total weight the bar material.
1. WL2/2AE
2. WL/4AE
3. WL/2AE
4. WL/8AE
Option 3 : WL/2AE
## Axial Elongation of Bar MCQ Question 3 Detailed Solution
Explanation:
Elongation of the uniform rod due to its self-weight W having specific weight γ:
$$δ = \frac{{\gamma {L^2}}}{{2E}} = \frac{{WL}}{{2AE}}$$
Elongation due to axial load P,
$$δ = \frac{{PL}}{{AE}}$$
Elongation of the conical bar due to its self-weight W having specific weight γ:
$$δ = \frac{{\gamma {L^2}}}{{6E}} = \frac{{WL}}{{2{A_{max}}E}}$$
# A steel bar of 40 mm × 40 mm square cross-section is subjected to an axial compressive load of 200 kN. If the length of the bar is 2 m and E = 200 GPa, the magnitude of deformation of the bar will be:
1. 1.25 mm
2. 2.70 mm
3. 4.05 mm
4. 5.40 mm
Option 1 : 1.25 mm
## Axial Elongation of Bar MCQ Question 4 Detailed Solution
Concept:
We know that, $${\rm{Δ }}L = \frac{{PL}}{{AE}}$$
Where,
ΔL = Change in length or deformation of the bar
A = Area of cross section
E = Modulus of Elasticity
L = Length of the bar
Calculation:
Given,
P = 200 kN
L = 2 m, E = 200 GPa = 200 × 103 MPa
A = 40 mm × 40 mm
$${\rm{Δ }}L = \frac{{PL}}{{AE}}= \frac{{200 × {{10}^3} × 2 × {{10}^3}}}{{40 × 40 × 200 × {{10}^3}}}$$
ΔL = 1.25 mm
# A weight of 500 π Newton hangs from a cable of length of 10 m diameter 2 cm and E = 200 GPa, then the elongation of the cable is:
1. 0.05 cm
2. 0.025 cm
3. 1/π
4. 1 cm
Option 2 : 0.025 cm
## Axial Elongation of Bar MCQ Question 5 Detailed Solution
Concept:
Elongation of the cable is given by,
δL = $$\frac{PL}{AE}$$
E = Elastic Modulus, P = Applied Load, L = Length of cable
A = Area of the cross-section, δL = Elongation
Explanation:
Given:
E = 200 GPa = 200 × 109 Pa
P = 500π N, L = 10 m
A = $$\frac{\pi}{4}\times (\frac{2}{100})^2$$ m2
∴ δL = $$\frac{PL}{AE}$$
$$\frac{500\pi \;\times\; 10}{\frac{\pi}{4\;}\times \;(\frac{2}{100})^2 \;\times\; 200\;\times\; 10^9}$$
$$\frac{1}{4000}$$ m = 0.025 cm
# What will be the modulus of Elasticity (E) of equivalent bar with area (A) which has same elongation and length of compound bar ? Compound bar elasticity are E1, E2 and area A1, A2.
1. $$\rm \frac{A_1^2E_1\;+\;A_2^2E_2}{A^2_{eq}}$$
2. $$\rm \frac{A_1E_1\;\times\; A_2E_2}{(A_1E_1\;+\;A_2E_2)A_{eq}}$$
3. $$\rm \frac{A_1E_1\;+\;A_2E_2}{A_{eq}}$$
4. $$\rm \frac{E_1^2\;+\;E_2^2}{E_{eq}}$$
Option 3 : $$\rm \frac{A_1E_1\;+\;A_2E_2}{A_{eq}}$$
## Axial Elongation of Bar MCQ Question 6 Detailed Solution
Explanation:
Compound bar:
Compound bar elasticity is E1, E2, and area A1, A2
Now, let there is a bar of the same length (l), having Equivalent Elasticity Eeq, & area (Aeq)
According to the question
(δl)CB = (δl)Eq
$$\rm \frac{Pl}{A_1E_1\;+\;A_2E_2} = \frac{Pl}{(A_1\;+\;A_2)E_{Eq}}$$
$$\rm E_{Eq} = \frac{A_1E_1\;+\;A_2E_2}{A_1\;+\;A_2}$$
# A rectangular steel bar, which is of 2.8 m long and 15 mm thick, is subjected to an axial tensile load of 40 kN. If width of the bar varies from 75 mm at one end to 30 mm at the end, then what is the extension of the bar if E = 2 x 106 N/mm2?
1. 0.86 mm
2. 0.36 mm
3. 0.076 mm
4. 0.50 mm
Option 3 : 0.076 mm
## Axial Elongation of Bar MCQ Question 7 Detailed Solution
Concept:
Extension of bar, $$δ l = \frac{{PL}}{{tE\left( {{b_1} - {b_2}} \right)}}\log_e \left( {\frac{{{b_1}}}{{{b_2}}}} \right)$$
where t constant thickness
Calculation:
Given,
b1 = 75 mm, b2 = 30 mm, L = 2.8 m, t = 15 mm and P = 40 kN
∵ We know, $$δ l = \frac{{PL}}{{tE\left( {{b_1} - {b_2}} \right)}}\log_e \left( {\frac{{{b_1}}}{{{b_2}}}} \right)$$
⇒ $$δ l = \frac{{40 \times {{10}^3} \times 2.8 \times {{10}^3}}}{{2 \times {{10}^6} \times 15 \times \left( {75 - 30} \right)}} \times {\log _e}\left( {\frac{{75}}{{30}}} \right)$$
δl = 0.076 mm
# The ratio of elongations of conical bar due to its own weight and that of prismatic bar of same length is -
1. 1/3
2. 1/5
3. 1/2
4. 1/4
Option 1 : 1/3
## Axial Elongation of Bar MCQ Question 8 Detailed Solution
Explanation:
Elongation of the prismatic bar due to self-weight:
$${{\bf{\delta }}_1} = \frac{{{\bf{\gamma }}{{\bf{l}}^2}}}{{2{\bf{E}}}}$$
Elongation of the conical bar due to self-weight:
$${{\bf{\delta }}_2} = \frac{{{\bf{\gamma }}{{\bf{l}}^2}}}{{6{\bf{E}}}}$$
Where,
γ = unit weight of the member
l = length of the member
E = young modulus of elasticity
Ratio of elongations of conical bar due to its own weight and that of prismatic bar:
$$\frac{{{{\bf{\delta }}_2}}}{{{{\bf{\delta }}_1}}} = \frac{{\frac{{{\bf{\gamma }}{{\bf{l}}^2}}}{{6{\bf{E}}}}}}{{\frac{{{\bf{\gamma }}{{\bf{l}}^2}}}{{2{\bf{E}}}}}}$$
$$\frac{{{{\bf{\delta }}_2}}}{{{{\bf{\delta }}_1}}} = \frac{1}{3}$$
# A steel rod of 20 mm diameter and 500 mm long is subjected to an axial pull of 30 kN. If E = 2 × 105 N/mm2, the elongation of the rod will be
1. 0.239 mm
2. 0.0239 mm
3. 0.00239 mm
4. 23.9 mm
Option 1 : 0.239 mm
## Axial Elongation of Bar MCQ Question 9 Detailed Solution
Concept:
Axial Deformation (δ) of the Bar:
The deformation of a bar with known geometry and subjected to an axial load can be determined by the equation of the form,
$$δ = \frac{{PL}}{{AE}}$$, where AE = Axial rigidity and L = Length of member
Calculation:
Given:
d = 20 mm, L = 500 mm, P = 30 kN, E = 2 × 105 N/mm2
Now, we know that
$${{δ }} = \frac{{{{PL}}}}{{{{AE}}}} = \frac{{30 \times {{10}^3} \times 500}}{{\frac{\pi }{4}{{\left( {20} \right)}^2} \times 2 \times {{10}^5}}} = 0.239{\rm{\;mm}}\;$$
∴ δ = 0.239 mm
# A mild steel bar is in two parts having equal length. The area of cross-section of part-1 is double that of part-2. If the bar carries an axial load P, then the ratio of elongation in part-1 to that in part-2 will be
1. 2
2. 4
3. 1/2
4. 1/4
Option 3 : 1/2
## Axial Elongation of Bar MCQ Question 10 Detailed Solution
Concept:
Elongation (δ) of
a bar under tensile load is given by,
$$\delta = \frac{{PL}}{{AE}}$$
Calculation:
Given;
A1 = 2A2
Elongation for part -1;
$$\delta_1 = \frac{{PL}}{{A_1E}}$$
Elongation for part -2;
$$\delta_2 = \frac{{PL}}{{A_2E}}$$
So,
$$\frac{{{\delta _1}}}{{{\delta _2}}} = \left( {\frac{{\frac{{PL}}{{{A_1}E}}}}{{\frac{{PL}}{{{A_2}E}}}}} \right) = \left( {\frac{{{A_2}}}{{{A_1}}}} \right) = \left( {\frac{{{A_2}}}{{2{A_2}}}} \right)$$
$$\frac{{{\delta _1}}}{{{\delta _2}}} = \frac{1}{2}$$
# A solid uniform metal bar of diameter D and length L is hanging vertically from its upper end. The elongation of the bar due to self-weight is
1. Proportional to L and inversely proportional to D2
2. Proportional to L2 and inversely proportional to D2
3. Proportional to L but independent of D
4. Proportional to L2 but independent of D
Option 4 : Proportional to L2 but independent of D
## Axial Elongation of Bar MCQ Question 11 Detailed Solution
Concept:
When a solid metal bar is hanged vertically from one of its ends, it tends to
experience elongation due to the weight and deformation that occurs.
The elongation of the bar of weight ‘W’, length ‘L’, diameter ‘D’, and
elasticity ‘E’ due to its self-weight is = $$\frac{{2WL}}{{{D^2}E}}$$
Let’s take a look into the case of a metal bar of weight ‘W’, diameter ‘D’
and length ‘L’ hanging vertically from one of its ends.
The elastic property represented by the Modulus of Elasticity of the metal bar is ‘E’.
Consider an element of length ‘dy’ at a distance ‘y’ from the bottom of the bar being elongated due to the force ‘P’, at section x - x, as shown in the figure.
Weight of the portion below x - x = ρ × A × y
Where ρ = weight of the bar per unit volume = $$\frac{W}{{AL}}$$
A = area of cross-section of the bar
Now, the weight of the portion P = $$\frac{W}{{AL}} \times \left( {A \times y} \right)$$ = $$\frac{{W\; \times \;y}}{L}$$
Change in length of the element ‘dy’
$$\frac{P}{{AE}}$$
$$\frac{{Wy}}{L} \times \frac{1}{{AE}}$$
Total elongation of the bar = $$\mathop \smallint \limits_0^L \frac{{Wy}}{L} \cdot \frac{1}{{AE}} \cdot dy$$
$$\frac{{WL}}{{2AE}} = \frac{γ AL~\times~L }{2AE}=\frac{γ L^2}{2E}$$
γ = Specific density of the material of the bar.
Thus, the elongation is Proportional to L2 but independent of D.
# A 20 kN weight is suspended by two wires as shown in the figure. The length of each wire is 2 meters. The steel wire (E value 200 GPa) has a cross-sectional area of 60 × 10-6 m2 and the aluminum wire (E value 70 GPa) has a cross-sectional area of 120 × 10-6 m2. The stress in aluminium wire is
1. $$\frac{17}{102}~GPa$$
2. $$\frac{7}{120}~GPa$$
3. $$\frac{17}{120}~GPa$$
4. $$\frac{7}{102}~GPa$$
Option 4 : $$\frac{7}{102}~GPa$$
## Axial Elongation of Bar MCQ Question 12 Detailed Solution
Concept:
The both wires having equal deformation due to load of 20 kN.
Therefore, δAl = δSt
And the deformation is given as, $$\delta=\frac{PL}{AE}$$
Calculation:
Given:
LAl = LSt = 2 m, AAl = 120 × 10-6 m2, EAl = 70 GPa, ASt = 60 × 10-6 m2, E = 200 GPa
$$\Rightarrow \frac{{{P_{Al}}{L_{Al}}}}{{{A_{Al}}{E_{Al}}}} = \frac{{{P_{st}}{L_{st}}}}{{{A_{st}}{E_{st}}}}$$
$$\frac{{{P_{Al}}}}{{{P_{st}}}} = \frac{{{L_{st}}}}{{{L_{Al}}}} × \frac{{{A_{Al}}}}{{{A_{st}}}} × \frac{{{E_{Al}}}}{{{E_{st}}}}$$
$$\frac{P_{Al}}{P_{St}} = \frac{2}{2} × \frac{{120 × {{10}^{ - 6}}}}{{60 × {{10}^{ - 6}}}} × \frac{{70}}{{200}}=\frac{7}{10}$$....................(1)
But we know that, PAl + PSt = 20..........................(2)
From equation (1), $$P_{St}=\frac{10}{7}P_{Al}$$
Therefore, from equation (1) and (2), we will get,
$$P_{Al}+\frac{10}{7}P_{Al}=20$$
$${P_{Al}} = \frac{{140}}{{17}}~kN$$
$${\sigma _{Al}} = \frac{{{P_{Al}}}}{{{A_{Al}}}} = \frac{{140 × {{10}^3}}}{{17 × 120 × {{10}^{ - 6}}}}$$
$$\sigma _{Al} = \frac{{14}}{{12 × 17}} × {10^9}\frac{N}{{{m^2}}}$$
$${\sigma _{Al}} = \frac{7}{{6 × 17}}~GPa = \frac{7}{{102}}~GPa$$
# A rod of length L having uniform cross-sectional area A is subjected to a tensile force P as shown in the figure below. If the Young’s modulus of the material varies linearly from E1 to E2 along the length of the rod, the normal stress developed at the section-SS is
1. P/A
2. $$\frac{{P\left( {{E_1} - {E_2}} \right)}}{{A\left( {{E_1} + {E_2}} \right)}}$$
3. $$\frac{{P{E_2}}}{{A{E_1}}}$$
4. $$\frac{{P{E_1}}}{{A{E_2}}}$$
Option 1 : P/A
## Axial Elongation of Bar MCQ Question 13 Detailed Solution
Concept:
Stress is defined as intensity of internal resisting force developed at a point against the deformation due to applied external load.
Mathematically,
Stress (σ) is:
$$\sigma = \frac{P}{A}$$
where P is applied load and A is the cross-sectional area
Explanation:
As stress depends load applied and cross-sectional area ∴ stress-induced in the above case will be
$$\sigma = \frac{P}{A}$$
# A copper rod 3 mm in diameter when subjected to a pull of 495 N extends by 0.07 mm over a gauge length of 100 mm. The Youngs Modulus for copper will be
1. 1 × 105 N / mm2
2. 1 × 106 N / mm2
3. 7 × 105 N / mm2
4. 1 × 107 N / mm2
Option 1 : 1 × 105 N / mm2
## Axial Elongation of Bar MCQ Question 14 Detailed Solution
Concept:
$$\delta L = \frac{{PL}}{{AE}}$$
Where δL = Change in length, P = Force, L = Length, A = Area, E = Youngs modulus or modulus of elasticity.
Calculation:
Given:
D = 3 mm, P = 495 N, δL = 0.07 mm, L = 100 mm
$$\delta L = \frac{{PL}}{{AE}}$$
$$E = \frac{{PL}}{{\delta L \times A\;}} = \frac{{495 \times 100}}{{0.07 \times \frac{\pi }{4}{3^2}}} = 1 \times {10^5}\;N/m{m^2}$$
The Youngs Modulus for copper will be 1 × 105 N/mm2.
# A point mass of 100 kg is dropped onto a massless elastic bar (cross-sectional area = 100 mm2, length = 1 m, Young’s modulus = 100 GPa) from a height H of 10 mm as shown (Figure is not to scale). If g = 10 m/s2, the maximum compression of the elastic bar is ________mm.
## Axial Elongation of Bar MCQ Question 15 Detailed Solution
Concept:
δImpact = δstatic × IF
$$IF = 1 + \sqrt {1 + \frac{{2H}}{{{\delta _{static}}}}}$$
Calculation:
Given:
Cross-sectional area (A) = 100 mm2, length (L) = 1 m, Young’s modulus (E) = 100 GPa, H = 10 mm
$${\delta _{static}} = \frac{{wL}}{{AE}} = \frac{{mgL}}{{AE}} = \frac{{100 \times 10 \times 1}}{{100 \times {{10}^{ - 6}} \times 100 \times {{10}^9}}} = 1 \times {10^{ - 4}}m = 0.1\;mm$$
δImpact = δstatic × IF
$$IF = 1 + \sqrt {1 + \frac{{2H}}{{{\delta _{static}}}}}$$
$$IF = 1 + \sqrt {1 + \frac{{2 \times 10}}{{0.1}}} = 15.177$$
δImpact = δstatic × IF
∴ δImpact = 0.1 × 15.177 = 1.517 mm
# The total extension of the bar loaded as show in the figure is:(where A = Area of cross-section, E = Modulus of Elasticity).
1. $$10 \times \frac{30}{AE}$$
2. $$26 \times \frac{{10}}{{AE}}$$
3. $$9 \times \frac{{30}}{{AE}}$$
4. $$30 \times \frac{{22}}{{AE}}$$
Option 2 : $$26 \times \frac{{10}}{{AE}}$$
## Axial Elongation of Bar MCQ Question 16 Detailed Solution
Concept:
Extension in a bar is given by:
$$\delta = \frac{{PL}}{{AE}}$$
P = Force; L = Length; A = Cross-sectional area; E = Young’s modulus
Calculation:
Given:
For AB:
$${\delta _{AB}} = \frac{{10\; \times \;10}}{{AE}}$$
For BC:
$${\delta _{BC}} = \frac{{7 \;\times\; 10}}{{AE}}$$
For CD:
$${\delta _{CD}} = \frac{{9\; \times \;10}}{{AE}}$$
Total extension:
$$\delta = {\delta _{AB}} + {\delta _{BC}} + \;{\delta _{CD}}$$
$$\delta = \frac{{10\; \times\; 10}}{{AE}} + \frac{{7\; \times\; 10}}{{AE}} + \;\frac{{9 \;\times \;10}}{{AE}}$$
$$\delta = \frac{{26\; \times\; 10}}{{AE}}$$
Important Points
While calculating Total change, Compression (Reduction) is taken as negative and Tension (extension) is taken as positive.
# The figure below shows a steel rod of 25 mm2 cross-sectional area. It is loaded at four points, K, L, M, N. Assume E steel = 200 GPa. The total change in length of the rod due to loading is
1. 1 μm
2. -10 μm
3. 10 μm
4. -20 μm
Option 2 : -10 μm
## Axial Elongation of Bar MCQ Question 17 Detailed Solution
Concept:
Total change in length of the rod due to loading is given as,
$$\Delta L=(\frac{PL}{AE})_1+(\frac{PL}{AE})_2+(\frac{PL}{AE})_3$$
where, P = load, L = lenght of rod, A = cross-sectional area, E = modulus of elasticity
Calculation:
Given:
A = 25 mm2, E = 200 × 103 MPa, L1 = 0.5 m, L2 = 0.8 m, L3 = 0.4 m
P1 = 100 N, P2 = 100 - 250 = -150 N, P3 = 50 N
Total change in length
$$\Delta {\rm{L}} = {\rm{\;}}\frac{{{{\rm{P}}_1}{{\rm{L}}_1}{\rm{\;}} + {\rm{\;}}{{\rm{P}}_2}{{\rm{L}}_2} + {\rm{\;}}{{\rm{P}}_3}{{\rm{L}}_3}}}{{{\rm{AE}}}}$$ $${\rm{(}}{{\rm{A}}_1}{{\rm{E}}_1} = {{\rm{A}}_2}{{\rm{E}}_2} = {{\rm{A}}_3}{{\rm{E}}_3} = {\rm{AE}})$$
∴ $$\Delta {\rm{L}} = \frac{{100{\rm{\;}} \times {\rm{\;}}0.5 + {\rm{\;}}\left( { - 150} \right){\rm{\;}} \times {\rm{\;}}0.8{\rm{\;}} + {\rm{\;}}50{\rm{\;}} \times {\rm{\;}}0.4}}{{25{\rm{\;}} \times {\rm{\;}}200{\rm{\;}} \times {\rm{\;}}{{10}^3}}}$$
∆L = -1 × 10 -5 m
∆L = -10 µm
# Two bars of different materials and same size are subjected to the same tensile force. If the bars have unit elongation in the ratio of 3 : 5, then the ratio of modulus of elasticity of the two materials will be
1. 2 : 5
2. 3 : 5
3. 5 : 3
4. 3 : 4
Option 3 : 5 : 3
## Axial Elongation of Bar MCQ Question 18 Detailed Solution
Concept:
Hooke’s law: As per Hooke’s law, stress is proportional to strain
i.e. σ ∝ ϵ or, σ = ϵ × E
Where, E = Modulus of Elasticity
For Hooke’s law to be valid:
(a) The material should be homogenous.
(b) The material should be isotropic.
(c) The material should behave in a linearly elastic manner.
Thus for a plane bar with Area ‘A’, Length ‘L’ and Modulus of Elasticity ‘E’,
$${\rm{\sigma = }}\frac{{\rm{P}}}{{\rm{A}}}\;\;{\rm{and}}\;\;{\rm{ϵ = }}\frac{{{\rm{δ L}}}}{L}$$
As per Hooke’s law,
∵ σ = ϵ × E
$$\frac{{\rm{P}}}{{\rm{A}}}{\rm{ = }}\frac{{{\rm{δ L}}}}{{\rm{L}}}{\rm{ × E}}$$
$${\rm{δ L = }}\frac{{{\rm{PL}}}}{{{\rm{AE}}}}$$
Calculation:
Given,
Two bars of unit length, i.e L1 = L2
Elongation ratio is 3 : 5, i.e δL1 : δL2 = 3 : 5
Both subjected to same tensile load, i.e P1 = P2
Both have the same size, i.e A1 = A2
Elongation of the first bar:
$${\rm{δ L_1 = }}\frac{{{\rm{P_1L_1}}}}{{{\rm{A_1E_1}}}}$$
Elongation of the second bar:
$${\rm{δ L_2 = }}\frac{{{\rm{P_2L_2}}}}{{{\rm{A_2E_2}}}}$$
Taking ratio,
$$\frac{{{\rm{\delta }}{{\rm{L}}_1}}}{{{\rm{\delta }}{{\rm{L}}_2}}} = \frac{{{{\rm{P}}_1}{{\rm{L}}_1}}}{{{{\rm{A}}_1}{{\rm{E}}_1}}} \times \frac{{{{\rm{A}}_2}{{\rm{E}}_2}}}{{{{\rm{P}}_2}{{\rm{L}}_2}}}$$
$$\frac{{{\rm{\delta }}{{\rm{L}}_1}}}{{{\rm{\delta }}{{\rm{L}}_2}}} = \frac{{{{\rm{E}}_2}}}{{{{\rm{E}}_1}}}$$
$$\therefore \frac{{{{\rm{E}}_1}}}{{{{\rm{E}}_2}}} = \frac{{{\rm{\delta }}{{\rm{L}}_2}}}{{{\rm{\delta }}{{\rm{L}}_1}}} = \frac{5}{3}$$
Hence the ratio of modulus of elasticity of the two materials will be 5 : 3
# Two bars of different materials are of same size and are subjected to same tensile forces. If the bars have unit elongations in the ratio of 4 : 7, then the ratio of modulli of elasticity of these two materials is:
1. 16 : 49
2. 4 : 7
3. 4 : 17
4. 7 : 4
Option 4 : 7 : 4
## Axial Elongation of Bar MCQ Question 19 Detailed Solution
Explanation:
For Bar I:
$${\Delta_1} = \frac{{{P_1}{L_1}}}{{{A_1}{E_1}}}$$
For Bar II:
$${\Delta_2} = \frac{{{P_2}{L_2}}}{{{A_2}{E_2}}}$$
$$\frac{{{\Delta _1}}}{{{\Delta _2}}} = \frac{4}{7}$$
Also, P1 = P2
A1 = A2
L1 = L2
$${\rm{\Delta }} = \frac{{PL}}{{AE}}$$
For same P, L, and A:
Δ ∝ 1/E
$$\frac{{{{\rm{\Delta }}_1}}}{{{{\rm{\Delta }}_2}}} = \frac{4}{7} = \frac{{{E_2}}}{{{E_1}}} \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{7}{4}$$
# A bimetallic cylindrical bar of cross-sectional area 1 m2 is made by bonding Steel (Young’s modulus = 210 GPa) and Aluminum (Young’s modulus = 70 GPa) as shown in the figure. To maintain tensile axial strain of magnitude 10-6 in Steel bar and compressive axial strain of magnitude 10-6 in Aluminum bar, the magnitude of the required force P (in kN) along the indicated direction is
1. 17
2. 140
3. 210
4. 280
Option 4 : 280
## Axial Elongation of Bar MCQ Question 20 Detailed Solution
Concept:
The axial strain is given by
$$ϵ=\frac{{{P}}}{{{A}{E}}}$$
Where P is the load acting in the axial direction, A is the cross-sectional area and E is the Modulus of elasticity
Calculation:
ES = 210 GPa
EAI = 70 GPa
Since the magnitude of axial strains in both bars same. Thus,
$$\frac{{{P_s}}}{{{A_s}{E_s}}} = \frac{{{P_{Al}}}}{{{A_{Al}}{E_{Al}}}}$$
Since, $${A_S} = {A_{Al}} = 1\;{m^2}$$
$$\frac{{{P_S}}}{{210}} = \frac{{{P_{Al}}}}{{70}}$$
PS = 3 PAl
P = PS + PAI ----(1)
Now, ϵ = 10-6
$$\frac{{{P_S}}}{{{A_S}{E_S}}} = {10^{-6}}$$
PS = 10-6 × 1 × 210 × 109
PS = 210 KN
PAI = 70 KN
P = PAI + PS = 210 + 70 = 280 KN | 8,006 | 20,551 | {"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} | 4.40625 | 4 | CC-MAIN-2022-05 | latest | en | 0.595201 |
https://www.meritnation.com/ask-answer/question/he-system-of-equations2x-y-z-0-x-2y-z-0-x-y-2z-0-has-infinit/determinants/6232213 | 1,637,983,567,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00090.warc.gz | 1,025,705,246 | 9,684 | # he system of equations2x - y + z = 0x - 2y + z = 0λx - y + 2z = 0has infinite number of nontrivial solutions for λ = 1 λ = 5 λ = -5 no real value for λ
for non-trivial solution, put the determinant = 0 as shown below,
Thus ,, the system has infinite no. of solutions for λ = 5
• 1
What are you looking for? | 106 | 311 | {"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-2021-49 | latest | en | 0.881042 |
https://www.teacherspayteachers.com/Product/1st-Grade-Math-Homework-4th-Quarter-Vertical-Format-1925544 | 1,488,165,601,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172404.62/warc/CC-MAIN-20170219104612-00021-ip-10-171-10-108.ec2.internal.warc.gz | 888,585,237 | 25,886 | Total:
\$0.00
# 1st Grade - Math Homework - 4th Quarter - Vertical Format
Subjects
Resource Types
Product Rating
4.0
File Type
PDF (Acrobat) Document File
3.64 MB | 21 pages
### PRODUCT DESCRIPTION
This 1st Grade Math Homework - 1st Grade - 4th Quarter is a great weekly homework packet that will review all common core strands on a weekly basis.
It is very kid-friendly, easy to read, examples are given for most problems, and it's packed with real work...not just time-wasting work. It has cute graphics for so it's fun for the kids to work on.
Be sure to look at the preview for a more detailed look at what you get:)
Every Monday - Numbers and Base Ten
Every Tuesday: Operations and Algebraic Thinking
Every Wednesday: Measurement and Data
Every Thursday: Geometry
You get 9 weeks of homework. I would suggest that you print each week double-sided to save paper so that you only have 1 sheet per week. NOTE: This would also work great as morning work:)
Here is what comes with this quarter:
1.NBT
Add within 100 with 2 digit + 1 digit numbers
Draw 10 more, 10 less
Mentally find 10 more, 10 less
Compare numbers (greater than, less than, equal to)
Word problems
1.OA
Count back to subtract
Determine if equations are true/false
Find unknown #s in equations
Fact families
1.MD
Shortest/longest
Tallest/shortest
Measure length/height
Time to hour and half hour
Bar graphs
Tally graphs
Word problems
1.G
Compose new shapes using 2D shapes
Label 3D shapes (vertices/faces)
Guess the shape
2D shape attributes
3D shape attributes
Fractions – half
Fractions - fourth
Word problems
Bundle - Math Homework - 1st Grade - Whole Year
Need a different quarter of math homework?
1st Grade - Math Homework - 1st Quarter - Vertical Format
1st Grade - Math Homework - 2nd Quarter - Vertical Format
1st Grade - Math Homework - 3rd Quarter - Vertical Format
Need some literacy homework? Check these out
Reading Comprehension Stories & Questions:Word Families-Differentiated-K/1st/2nd
Word Work Activities for the Whole Year - 1st Grade
Whole Year - Sight Word Homework - 1st Grade
Whole Year - Word Work Homework - 1st Grade
I'm always working on fun stuff and freebies, so please click the green star at the top of the page (by my name) to follow my store so that you can be notified when new resources and freebies are available.
Please also remember to leave me some feedback since that's how I know how to please my customers:) You can complete the feedback form on your "my purchases" page....and extra bonus - you earn points to make your next purchase cheaper just by completing the feedback form! yeay!!!
Enjoy:)
Total Pages
21
N/A
Teaching Duration
N/A
### Average Ratings
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
11 ratings
\$3.00
User Rating: 4.0/4.0
(2,024 Followers)
\$3.00 | 751 | 2,870 | {"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-2017-09 | longest | en | 0.900304 |
https://atlanticgmat.com/mark-and-ann-together-were-allocated-n-boxes-of-cookies-to-sell-for-a-club-project-mark-sold-10-boxes-less-than-n-and-ann-sold-2-boxes-less-than-n-if-mark-and-ann-have-each-sold-at-least-one-box-of/ | 1,721,831,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00773.warc.gz | 89,151,063 | 53,380 | # Mark and Ann together were allocated n boxes of cookies to sell for a club project. Mark sold 10 boxes less than n and Ann sold 2 boxes less than n. If Mark and Ann have each sold at least one box of cookies, but together they have sold less than n boxes, what is the value of n?
Mark and Ann together were allocated n boxes of cookies to sell for a club project. Mark sold 10 boxes less than n and Ann sold 2 boxes less than n. If Mark and Ann have each sold at least one box of cookies, but together they have sold less than n boxes, what is the value of n?
A. 11
B. 12
C. 13
D. 14
E. 15
Full explanation coming soon. Send us a note if you’d like this added to the express queue!
You’ll find tons of practice questions, explanations for GMAT Official Guide questions, and strategies on our GMAT Question of the Day page.
## Here are a few other extra challenging GMAT questions with in depth explanations:
Here’s a tough function question from the GMAT Prep tests 1 and 2:
For which of the following functions is f(a+b) = f(b) + f(a) for all positive numbers a and b?
And a very challenging word problem from the Official Guide. Almost no-one gets this one on the first try but there is a somewhat simple way through it:
Last Sunday a certain store sold copies of Newspaper A for \$1.00 each and copies of Newspaper B for \$1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
Tanya’s letters from the GMAT Prep tests. This one often gets GMAT tutoring students caught up in a tangled net. With combinatorics it’s important to stay practical. We’ll take a look at how to do that in the explanation:
Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
Here’s an exponents puzzle that comes up a lot in GMAT tutoring sessions:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is
This is one of the most difficult questions in the GMAT universe. That said, there is a simple way to solve it that relies on a fundamental divisibility rule every GMAT studier should know:
For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is? | 636 | 2,707 | {"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} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.942328 |
https://fr.scribd.com/document/401417148/ps38-n3cs19 | 1,576,327,808,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541157498.50/warc/CC-MAIN-20191214122253-20191214150253-00224.warc.gz | 376,078,900 | 79,374 | Vous êtes sur la page 1sur 2
# 4=5
## Select the statement that correctly intercepts
Kim's equation?
N3cs19 Practice Set 38
A.
The solution is x=0.
B.foldThe
to paper alongisthe
solution thecenter,
orderedwork
pairexercises
(4, 5). in order top to bottom, left column then right column. Staple multiple pages
C. There is no solution since 4 = 5 is a false
Math 8 2019 Practice: Through Linear Functions in 2 variables 1 Assessment ID: ib.1928160
1. The A. y =
tubes of lip gloss in Mr. Ford’s 4th period is directly
statement. 5x
proportional to the number of girls in
Directions:
D. 4th
period. Therethe arefollowing
tubes question(s).
35solutions
ofbecause
lip gloss for 14 girls.
There are infinitely many B. xDetermine
=5 the variation constant and
write
thereaisvariation equation
no x in the for the tubes of lip gloss to the girls in Period 4.
final equation.
2. Determine the slope of the line expressed in the table:
C. y = 5
23 Jupiter is approximately 778,000,000 kilometers
26
6 X the value of x in the -3
What is equation below? 0 D. y = 5from
+ x the3 Sun. What is this number
6 written in 9
scientific notation?
4x + 3y= 2x + 21 4 2 0 8 -2 -4
A. 2
9 7.78 ⋅ 10
A. equation
Which would best represent the
B. Determine
3. 9 the slope of a line between the points ( −6,6 ) and (12,−6
778 6 (y − y )
equation of line T, shown below?
⋅ 10 ) ; use the slope formula m = 2 1
B.
C. 18
C. 778 ⋅ 10−6
( x2 − x1 )
4. See the graph at below:
7 Find the rate of change (slope) of the line. Enter D. 7.78 ⋅ 10−8
only the number in the box below (the m = is
provided for you).
27 Which of the following equations has no solution?
A. 12x − 4 = 20
Answers: B. −6x + 12 + 3x = − 3x + 6
Andy’s Hardware \$__________ per foot C. 5
x + 10 = 25
6
D. 5x + 2x + 3.5 = 3.5 + 10x
Bargain Hardware \$_________ per foot
## A. y28 Enter the value for x that makes the equation
= 5x
m= 3
B. x = 5 12x + 1.6 − 2x = 22.2 true.
4
5. See the graph below
C. y = 5
24 The table shows a proportional relationship. WriteD. y = 5 + x
an equation that describes the relationship.
## 29 Rosa received a \$50 gift card to an online store.
She wants to purchase some bracelets that cost
Acres
Illuminate Itembank™ 5 8 15 \$8 each. Continue:
There willTurn
be ato\$10 overnight
the next page. delivery
Generated OnBushels
March 2, 2019, 10:14 AM PST
of Wheat fee. Rosa wants to know the greatestPage 2number of
140 224 420 bracelets she can order. Select the equation that
models this situation.
A. 8b=50-10
B. 8b=50+10
6. State the property that justifies each step of the following
C. solution:
50=10b+8
6x − 2(3 − x) = 13
25 6x
What is +the
+ − 2(3 − product of
x) = 13 7.5 ⋅ 107 2 ⋅ 103 D. 50=8b+10
+ − 6 + 2x21= 13
E. 50=8b-10
A. 6x1.5 ⋅ 10−
6x + 2x + 6 = 13
B. 8x1.5 134
+ − 6⋅ =10
+ 6 + 6 10
= 13 + 6
C. 8x15 ⋅ 10
8x + 0 = 19
= 19⋅ 1011
D. 8x1.5
Page 1 of 2
## Illuminate Itembank™ Continue: Turn to the next page.
Generated On March 2, 2019, 10:14 AM PST Page 5
N3cs19 Practice Set 38
to fold paper along the center, work exercises in order top to bottom, left column then right column. Staple multiple pages
24
7. = ; write as a decimal rounded to the nearest 100th.
38
8
8. 13 − = ; write as an improper fraction and a decimal rounded to nearest 10th.
3
9. Write as a mixed number, reduced: 4.67
10. Translate into a formula: “Dynamic Pressure is one-half of the product of density ( ρ ) and the square of velocity (v)."
11. The University of Michigan’s Devin Bush, Jr. ran the 40 yard dash in 4.43s at the NFL
ft
Combine. What was his speed in ?
s
12x 5 y −5
12. Write as exponents positive:
8x 8 y −8
13. Write as exponents positive: −9x 7 y 4 • −6y −8 x 3
## 14. Multiply, writing exponents positive: ( −9n )
5 3
8
15. Solve:11− g = −3 ; write as an improper fraction, mixed number, and decimal rounded to the nearest 10th.
5
5 3 5 1
16. Determine the value of the unknown that makes the statement true: x − 10 + x = 2 + x + 3 − x
2 4 2 2
17. A square has an area of 729m ; what is the square’s perimeter?
2
3
18. A cube has a volume of 2197in ; what is the cube’s surface area?
19. In scientific notation, what is the sum of 2.8 •10 5 + 3.5 •10 4 ?
(
20. Evaluate 7.9 • 10
6
)( 4.0 • 10 ) ; write in scientific notation.
## 21. Estimate the difference: 39 − 70
22. How many times larger is the radius of a mercury atom, 1.5 •10 −10 , than the radius of a hydrogen atom, 2.5 •10 −11 ?
⎛ 1⎞
23. Solve the equation: 0.4 ⎜ 2x + ⎟ = 3[ 0.2x + (−2)] − 4
⎝ 2⎠
24. Find four consecutive multiples of 4 such that twice the sum of the least and greatest exceeds three times the least
by 32. hint: a multiple of 4 is 4 more than the previous multiple; e.g., if the first multiple is x, the next would be x+4….
Page 2 of 2 | 1,722 | 4,731 | {"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} | 3.8125 | 4 | CC-MAIN-2019-51 | latest | en | 0.830592 |
http://www.weegy.com/?ConversationId=3B8A58A3 | 1,524,476,413,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945940.14/warc/CC-MAIN-20180423085920-20180423105920-00232.warc.gz | 579,719,446 | 8,236 | Notice: Payments for answers will end 4/10/2017. Click for more info.
You have new items in your feed. Click to view.
Q: A photographer conducted three sessions yesterday. She completed the first session in 56 minutes, the second in 42 minutes and the third in 1 hour and 16 minutes. How many hours did it take her to
complete all three sessions
A: 56 minutes + 42 minutes + 76 minutes (1 hour=60 minutes, so 1hr, 16 minutes is 76 minutes), which totals 174 minutes. 174 minutes divided by 60 minutes per hour equals 2 hours, 54 minutes.
Original conversation
User: A photographer conducted three sessions yesterday. She completed the first session in 56 minutes, the second in 42 minutes and the third in 1 hour and 16 minutes. How many hours did it take her to complete all three sessions
Weegy: 56 minutes + 42 minutes + 76 minutes (1 hour=60 minutes, so 1hr, 16 minutes is 76 minutes), which totals 174 minutes. 174 minutes divided by 60 minutes per hour equals 2 hours, 54 minutes.
ramkitten|Points 4692|
User: If a special portrait package offer consists of 2 8x10-sized portraits, 2 5x7-sized portraits, and 8 wallet-sizes portraits, what fraction of the portraits are 5x7-sized
Weegy: It's 4.2
seal_me|Points 220|
User: A \$200.00 sweater has been marked down to \$120.00. What percentage discount from the original price does the markdown represent? An answer is required
Weegy: 40%
amreen|Points 199|
User: What is the smallest number of portraits that can be distributed equally among either four grandparents or six aunts and uncles with no portraits left over
Weegy: 12 can distributed evenly among 4 or 6 with no remainder.
debnjerry|Points 7332|
Question
Asked 12/11/2010 1:17:31 PM
Rating
There are no new answers.
There are no comments.
Popular Conversations
What does throttle mean?
4/16/2018 10:25:30 AM| 5 Answers
Error 401- Unauthorized
4/13/2018 4:59:54 AM| 3 Answers
The science of the creation of maps and charts is called ...
Weegy: The science of the creation of maps and charts is called cartography. User: Which of the following would ...
4/13/2018 8:36:16 AM| 3 Answers
Ancient cyanobacteria released ______, which assisted in creating the ...
Weegy: Cyanobacteria, is a phylum of bacteria that obtain their energy through photosynthesis. User: The epoch in ...
4/16/2018 10:31:53 AM| 3 Answers
What's one way we can check our answer to 2 ×10 = 20? A. 2 ÷ ...
Weegy: A throttle body is part of the air intake system that helps control the amount of air that gets into the engine. ...
4/13/2018 7:46:31 AM| 2 Answers
Which sentence does not contain misplaced or dangling modifiers? ...
Weegy: The sentence that does not contain misplaced or dangling modifier is:
4/15/2018 8:58:11 PM| 2 Answers
Define impressionism
4/15/2018 9:53:48 PM| 2 Answers
What is the double jeopardy principle
Weegy: Principle is a law or rule that has to be, or usually is to be followed, or can be desirably followed, or is an ...
4/19/2018 6:55:12 PM| 2 Answers
Weegy Stuff
S
P
L
P
P
P
P
Points 235 [Total 661] Ratings 0 Comments 235 Invitations 0 Offline
S
R
L
R
Points 186 [Total 324] Ratings 0 Comments 26 Invitations 16 Offline
S
1
L
L
P
R
P
L
P
P
R
P
R
P
R
Points 175 [Total 12978] Ratings 0 Comments 165 Invitations 1 Offline
S
L
Points 101 [Total 101] Ratings 1 Comments 61 Invitations 3 Offline
S
Points 68 [Total 68] Ratings 0 Comments 18 Invitations 5 Offline
S
Points 50 [Total 50] Ratings 0 Comments 0 Invitations 5 Offline
S
P
L
Points 35 [Total 126] Ratings 0 Comments 35 Invitations 0 Offline
S
Points 30 [Total 35] Ratings 3 Comments 0 Invitations 0 Offline
S
Points 11 [Total 11] Ratings 0 Comments 11 Invitations 0 Offline
S
Points 10 [Total 10] Ratings 1 Comments 0 Invitations 0 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,141 | 3,841 | {"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} | 3.15625 | 3 | CC-MAIN-2018-17 | latest | en | 0.909499 |
http://mathematica.stackexchange.com/questions/4084/finding-a-not-shortest-path-between-two-vertices/4085 | 1,469,420,538,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824204.27/warc/CC-MAIN-20160723071024-00030-ip-10-185-27-174.ec2.internal.warc.gz | 163,883,605 | 24,092 | # Finding a “not-shortest” path between two vertices
In designing a routine for making a simple three dimensional (5x5x5) labyrinth, I realized that my solutions (a solution is a labyrinth includes a single path from {1, 1, 1} to {5, 5, 5} in a 5 x5x5 grid) almost never wandered or "doubled back". This feature makes for a somewhat uninteresting labyrinth (see a labyrinth and its solution path, below); a person in the labyrinth can find the exit rather quickly by avoiding subpaths that turn back.
Here's why the solution did not require doubling back: FindShortestPath was used to determine the solution path between {1,1,1} and {5,5,5}, that is, between vertex 1 and vertex 125 (see the labyrinth as a graph in the plane below), before circuits within the path were pruned. The shortest path will generally be the path that reaches the exit most directly.
How can I find a paths between start and finish vertices that are ostensibly not the shortest path? This is easy enough through visual inspection. But I'd like to compute a path that is not the shortest path.
Note: The above graph and its respective labyrinth have not yet been pruned. By pruning I mean the removal of alternative paths for reaching vertex 125 from vertex 1. Once a labyrinth has been properly pruned, one can only reach the finish vertex by traversing the unique solution path (and perhaps making some wrong turns into dead ends).
Code for above graph:
edges= {1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 4, 1 \[UndirectedEdge] 6, 2 \[UndirectedEdge] 7, 6 \[UndirectedEdge] 7, 6 \[UndirectedEdge] 11, 7 \[UndirectedEdge] 12, 11 \[UndirectedEdge] 12, 12 \[UndirectedEdge] 13, 10 \[UndirectedEdge] 15, 11 \[UndirectedEdge] 16, 12 \[UndirectedEdge] 17, 16 \[UndirectedEdge] 17, 13 \[UndirectedEdge] 18, 17 \[UndirectedEdge] 18, 18 \[UndirectedEdge] 19, 17 \[UndirectedEdge] 22, 19 \[UndirectedEdge] 24, 24 \[UndirectedEdge] 25, 1 \[UndirectedEdge] 26, 3 \[UndirectedEdge] 28, 4 \[UndirectedEdge] 29, 28 \[UndirectedEdge] 29, 29 \[UndirectedEdge] 30, 6 \[UndirectedEdge] 31, 26 \[UndirectedEdge] 31, 7 \[UndirectedEdge] 32, 31 \[UndirectedEdge] 32, 29 \[UndirectedEdge] 34, 10 \[UndirectedEdge] 35, 30 \[UndirectedEdge] 35, 34 \[UndirectedEdge] 35, 11 \[UndirectedEdge] 36, 31 \[UndirectedEdge] 36, 12 \[UndirectedEdge] 37, 32 \[UndirectedEdge] 37, 36 \[UndirectedEdge] 37, 13 \[UndirectedEdge] 38, 37 \[UndirectedEdge] 38, 34 \[UndirectedEdge] 39, 38 \[UndirectedEdge] 39, 15 \[UndirectedEdge] 40, 35 \[UndirectedEdge] 40, 39 \[UndirectedEdge] 40, 16 \[UndirectedEdge] 41, 36 \[UndirectedEdge] 41, 17 \[UndirectedEdge] 42, 37 \[UndirectedEdge] 42, 41 \[UndirectedEdge] 42, 18 \[UndirectedEdge] 43, 38 \[UndirectedEdge] 43, 42 \[UndirectedEdge] 43, 40 \[UndirectedEdge] 45, 43 \[UndirectedEdge] 48, 24 \[UndirectedEdge] 49, 48 \[UndirectedEdge] 49, 25 \[UndirectedEdge] 50, 45 \[UndirectedEdge] 50, 49 \[UndirectedEdge] 50, 26 \[UndirectedEdge] 51, 28 \[UndirectedEdge] 53, 29 \[UndirectedEdge] 54, 53 \[UndirectedEdge] 54, 30 \[UndirectedEdge] 55, 54 \[UndirectedEdge] 55, 31 \[UndirectedEdge] 56, 51 \[UndirectedEdge] 56, 32 \[UndirectedEdge] 57, 56 \[UndirectedEdge] 57, 53 \[UndirectedEdge] 58, 57 \[UndirectedEdge] 58, 35 \[UndirectedEdge] 60, 55 \[UndirectedEdge] 60, 36 \[UndirectedEdge] 61, 56 \[UndirectedEdge] 61, 40 \[UndirectedEdge] 65, 60 \[UndirectedEdge] 65, 41 \[UndirectedEdge] 66, 61 \[UndirectedEdge] 66, 42 \[UndirectedEdge] 67, 66 \[UndirectedEdge] 67, 45 \[UndirectedEdge] 70, 65 \[UndirectedEdge] 70, 69 \[UndirectedEdge] 70, 66 \[UndirectedEdge] 71, 67 \[UndirectedEdge] 72, 71 \[UndirectedEdge] 72, 48 \[UndirectedEdge] 73, 72 \[UndirectedEdge] 73, 55 \[UndirectedEdge] 80, 56 \[UndirectedEdge] 81, 57 \[UndirectedEdge] 82, 77 \[UndirectedEdge] 82, 81 \[UndirectedEdge] 82, 60 \[UndirectedEdge] 85, 80 \[UndirectedEdge] 85, 84 \[UndirectedEdge] 85, 61 \[UndirectedEdge] 86, 81 \[UndirectedEdge] 86, 84 \[UndirectedEdge] 89, 88 \[UndirectedEdge] 89, 66 \[UndirectedEdge] 91, 86 \[UndirectedEdge] 91, 67 \[UndirectedEdge] 92, 91 \[UndirectedEdge] 92, 88 \[UndirectedEdge] 93, 92 \[UndirectedEdge] 93, 69 \[UndirectedEdge] 94, 89 \[UndirectedEdge] 94, 93 \[UndirectedEdge] 94, 71 \[UndirectedEdge] 96, 91 \[UndirectedEdge] 96, 72 \[UndirectedEdge] 97, 92 \[UndirectedEdge] 97, 96 \[UndirectedEdge] 97, 73 \[UndirectedEdge] 98, 93 \[UndirectedEdge] 98, 97 \[UndirectedEdge] 98, 94 \[UndirectedEdge] 99, 98 \[UndirectedEdge] 99, 81 \[UndirectedEdge] 106, 101 \[UndirectedEdge] 106, 82 \[UndirectedEdge] 107, 106 \[UndirectedEdge] 107, 85 \[UndirectedEdge] 110, 86 \[UndirectedEdge] 111, 106 \[UndirectedEdge] 111, 107 \[UndirectedEdge] 112, 111 \[UndirectedEdge] 112, 88 \[UndirectedEdge] 113, 112 \[UndirectedEdge] 113, 89 \[UndirectedEdge] 114, 113 \[UndirectedEdge] 114, 110 \[UndirectedEdge] 115, 114 \[UndirectedEdge] 115, 93 \[UndirectedEdge] 118, 113 \[UndirectedEdge] 118, 94 \[UndirectedEdge] 119, 114 \[UndirectedEdge] 119, 118 \[UndirectedEdge] 119, 96 \[UndirectedEdge] 121, 97 \[UndirectedEdge] 122, 121 \[UndirectedEdge] 122, 98 \[UndirectedEdge] 123, 118 \[UndirectedEdge] 123, 122 \[UndirectedEdge] 123, 99 \[UndirectedEdge] 124, 119 \[UndirectedEdge] 124, 123 \[UndirectedEdge] 124, 124 \[UndirectedEdge] 125}
HighlightGraph[lab=Graph[edges], PathGraph[s = FindShortestPath[lab, 1, 125]],
VertexLabels -> "Name", ImagePadding -> 10,
GraphHighlightStyle -> "Thick", ImageSize -> 600]
Update
I posted below a CW response that lays out some ideas as to how to generate a labyrinth. Feel free to make your own edits to that code.
-
A simple solution would be to first find the shortest path, then select a point not on that shortest path, and find the shortest path between the starting point and that point, and the shortest path between that point and the end point after removing the vertices of the previous path. Note however that removing those vertices might remove the connection to the end point. In that case, just try again with another intermediate point. – celtschk Apr 10 '12 at 21:07
@celtschk Your solution would lead to a detour. I'm trying to get the path to actually double-back (without introducing a circuit). However, your idea raises an interesting possibility that would likely entail doubling back: choose 2 or 3 random stopping points that require visiting (in the order randomly chosen). I'll have to think more about this. – DavidC Apr 10 '12 at 22:31
You can try giving your edges random weights so that FindShortestPath is forced to take a different path. Here are some different possible paths —
Table[HighlightGraph[lab = Graph[edges, EdgeWeight -> RandomInteger[1000, Length[edges]]],
PathGraph[s = FindShortestPath[lab, 1, 125]], VertexLabels -> "Name",
ImagePadding -> 10, GraphHighlightStyle -> "Thick", ImageSize -> 600], {6}
] ~Partition~ 3 // Grid
-
Very clever idea! The weights seem to act as if they were distances between cities, right? – DavidC Apr 9 '12 at 21:08
@DavidCarraher That's correct. The documentation for FindShortestPath says this: "For a weighted graph, edge length is taken to be the weight." – R. M. Apr 9 '12 at 21:15
Your approach introduces a bit of additional variation doesn't appear to result in doubling back. You can check this with HighlightGraph[gg = GridGraph[ConstantArray[5, 3]], yourGraph]. It may be necessary to remove some cost of doubling back by assigning certain edges very low weights. Or by taking a different approach altogether. – DavidC Apr 10 '12 at 1:58
@DavidCarraher I can think of a few things that might work with FindShortestTour (if you want to be really mean and make the person have to wander a lot before reaching the exit), but the problem — or at least, what I haven't been able to achieve when I played with it before — is to specify start and end points... – R. M. Apr 10 '12 at 2:03
If you want to be really nasty, traverse the entire graph (trying various vertices as starting points) and restrict the search for a shortest path to the traverse tree. Some of these will be quite long.
edges = {1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, ... 124 \[UndirectedEdge] 125};
g = Graph[edges];
h = Reap[DepthFirstScan[g, 7, {"FrontierEdge" -> Sow}]][[2, 1]]);
t = FindShortestPath[Graph[h], 1, 125];
(This use of DepthFirstScan is from an example on its help page.) This solution, obtained by a traversal starting at vertex 7, uses 75 of the 151 edges. It was found by varying the starting vertex of h from 1 through 125 and picking the one for which the length of t is as long as possible.
HighlightGraph[g, {PathGraph[t], Style[{1, 125}, Yellow],
Labeled[{1, 125}, "*"]}, GraphHighlightStyle -> "Thick"]
-
+Nice use of DepthFirstScan. I pruned your result to ensure that the person could not take shortcuts from 1 to 125. (I will show this in an update.) – DavidC Apr 10 '12 at 16:16
@David, Several promising alternative approaches appear in Bader et al., Alternative Route Graphs in Road Networks. One approach is to find an optimal path, increase the edge weights within a tubular neighborhood of that path (to steer solutions away from it), and try again. This can be done repeatedly to generate a set of alternatives to the optimal route. The whole thing could be automated by selecting an alternative having the highest edge count. – whuber Apr 15 '12 at 14:37
Here's an approach based on R.M's response and on celtschk's idea of pushing the labyrinth toward vertices known not to be on the shortest path. I removed the constraint that parallel tunnels should be avoided. I also set aside the issue of dead ends and misleading paths for later. It struck me as cleaner to find an elaborate labyrinth directly within the complete 5x5x5 grid, and then add misleading paths later. Perhaps you have some ideas on how to add false paths.
Feel free to contribute your own improvements.
Also, I know that there is still a glitch or two that occasionally causes the program to fail to find a shortest path.
The commented code follows:
ClearAll[f, maze]
(* Randomly selects two vertices to pass through, avoiding those near \
start or finish *)
stops :=
RandomSample[
Complement[
Range[125], {1, 2, 6, 7, 26, 27, 31, 32, 95, 99, 100, 119, 120,
125, 124}], 2]
(* The 5 x 5 x 5 grid *)
g1 := GridGraph[ConstantArray[5, 3], EdgeStyle -> Thin,
VertexSize -> Small, ImagePadding -> 15,
EdgeWeight -> RandomInteger[{1, 1000}, 300];
(* Generate an indirect path from start to end that does not visit \
any vertices in path *)
f[v1_, v2_, path_: {}] :=
Join[path,
FindShortestPath[
VertexDelete[g1,
DeleteDuplicates@If[Length[path] == 0, path, Most[path]]], v1, v2]]
(* Maze that goes from 1 to stops[[1]] to stops[[2]] to 125 *)
maze :=
Module[{st = stops,
s = DeleteDuplicates[
f[st[[2]], 125, f[st[[1]], st[[2]], f[1, st[[1]], {}]]]]},
HighlightGraph[g1, PathGraph[s],
VertexLabels -> s /. {v_Integer :> v -> v}, ImagePadding -> 10,
GraphHighlightStyle -> "Thick", ImageSize -> 250]]
Table[maze, {6}]
-
I'm a bit unclear as to why this works .... however, using FindPath instead of FindShortestPathyields a route through 71 of the 88 vertices which definitely doubles back on itself.
HighlightGraph[lab = Graph[edges], PathGraph[s = Last@Sort@FindPath[lab, 1, 125]],
VertexLabels -> "Name", ImagePadding -> 10, GraphHighlightStyle -> "Thick", ImageSize -> 600]
HighlightGraph[GridGraph[{5, 5, 5}, EdgeStyle -> LightGray, VertexStyle -> LightGray,
VertexSize -> 0.3, BaseStyle -> EdgeForm[White]],
PathGraph[s], VertexLabels -> "Name", ImagePadding -> 10,
GraphHighlightStyle -> "Thick", ImageSize -> 400]
Using the form FindPath[start,end,Infinity,All] will find all paths through the graph. Although, for a 5x5x5 grid there are likely to be 10,000's of paths and generating this list may take several hours... Trimming such a list to only include longer paths (e.g. FindPath[start,end,{55,60},All]) might yield a more manageable, yet still interesting set of paths to play with.
## Making a labyrinth
... here is an attempt at a all-in-one labyrinth maker:
mesh = GridGraph[{4, 4, 4}];
(*generating a pseudorandom subset of edges that include the start and end vertices*)
edgelab = EdgeList[mesh][[Union[{1, 144}~Join~RandomInteger[{1, 144}, 130]]]];
(*graph of the subset of edges*)
labset = Graph[edgelab];
(*finding a path...*)
path = Last@Sort@FindPath[labset, 1, 64, {30, 45}];
HighlightGraph[labset, PathGraph[path], VertexLabels -> "Name", ImagePadding -> 10,
GraphHighlightStyle -> "Thick", ImageSize -> 600]
Varying the number of RandomInteger's generated (in this case 130) is the key control on the connectivity of the labyrinth. Of course it is possible to create disconnected 'chambers', but this may not be such a problem (depending on the final application).
-
My goodness, this really does work. I never used FindPath and am curious to get a sense of how it works. – DavidC Feb 5 '15 at 13:39
@DavidCarraher It seems to be very fast too, see e.g. here. – Juho Apr 8 '15 at 20:55 | 4,069 | 13,002 | {"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} | 3.421875 | 3 | CC-MAIN-2016-30 | latest | en | 0.911379 |
www.kanonical.io | 1,726,690,638,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00321.warc.gz | 780,812,407 | 68,180 | # Exponentials and Revolution
May 28, 2020
## Exponentials
I’ve been thinking a lot about exponential functions. Specifically how often they can be misconstrued or misappropriated. Our current intelligentsia familiar with this type of math like to remind the more innumerate that exponential growth is incredibly non-intuitive and hard for humans to understand.
Which is true. But there’s a second-order version of this too, which is that even among the mathematical literate (or semi-literate), there is a canned version of exponential growth packed tightly in our heads. Which looks like this:
We’ve seen these curves before for all sorts of different things. There’s two problems with being able to really comprehend them. The first is that if you don’t have the complete timescale, the exponential nature of the curve becomes a lot harder to see. Here’s an example that looks closer to a linear fit:
Doesn’t look like very speedy growth, but that thing is about to take off.
The other issue is that we don’t do well with low bases for our exponents. The graph above is y = 2^x. We do very poorly understanding exponentials like y = 1.01^x. It’s still exponential, but the growth rate will be much, much slower.
Which, in part, leads to amusing commentaries like George Will’s observation of DC:
“The problem with Washington DC is that it’s made up of a bunch of politicians who still believe there’s a 1% difference between 2% and 3% growth.”
That 50% difference in growth rate leads to huge differences over generational timescales. It’s how we end up with all of our amazing exponential GDP graphs. It’s why Tyler Cowen argues that economic growth is actually a moral imperative. For instance, here’s England:
And here’s the same thing for the entire world. East Asia and Africa there are on the same curves as the rest of the world, just shifted to the right in time.
We see a lot of graphs like this. For example, here’s coal use over time (seems like it could be scary):
And US Coronavirus cases:
## Sigmoids
That’s a lot of curves going up and to the right. Some of them are very, very good; I tend to be strongly aligned with Cowen’s argument on growth. But other curves are potentially very bad.
Up and to the right is a common thing. But what I’m beginning to wonder more and more is how many of these scenarios that we think are exponential are actually sigmoid functions in disguise.
Sigmoids are talked about much less than exponentials. I had planned to say much more on why S-curves are so frequently overlooked and difficult to forecast, but Constance Crozier beat me to it. The point is that many of the things that we see as exponential are not. If the curve is being forecast over some sort of population - like users, people, or some finite natural resource - then the curve can saturate and it’s not exponential.
These curves are effectively Cumulative Distribution Functions, describing how fast we get to saturation, and they are often some flavor of sigmoid. CDFs evaluate a variable over some range, for instance the probability of a particular event over time. Even the probability of a rare event would asymptote to 100% over a long enough time interval.
But what about the measures that do not apply to a population or resource? For example, is there some hard limit on our overall economic growth as measured by GDP that will cause us to slow down? And how could we even tell? It’s certainly true that well over 95% of what makes up our current GDP would have been impossible for a 19th-century economist to understand, let alone predict. To say that differently, we as a species are constantly changing the scale on which we measure these curves. This is a very peculiar property.
What about our coal use? The earth is finite so there’s some theoretical resource limitation but it’s a very, very large number and it seems doubtful that this will ever be the limiting factor. It seems more likely that the reason coal will become a sigmoid is because some other technology will take over and cause the desire for goal to go away. It will be the constant change and technological development of our species that will become the limit.
And what about coronavirus? Herd immunity is the end state of an R0 above 1. In other words, the sigmoid curve represents the probability over time that you and I and everyone will get it. And as time increases, the answer approaches 100%. The real question that everyone needs to battle out is how quickly we want to get there. How fast should the upward slope be. Too fast and we do bad things like cause needless death and misery because the healthcare system can’t keep up. Too slow and we cause needless death and misery because everything is shut down, people don’t have purpose or ways of creating wealth and value, and mental health and social norms deteriorate. When you breakdown all of the public policy debate, the question everyone is really trying to answer is “How fast is the slope?”
There’s good and bad reasons to want to go fast and slow. For example, optimism for the near-term development of a therapeutic or vaccine that drastically lowers the death rate is a great reason to go slower. A desire to reshape American values and greatly expand the welfare state in socialistic ways using Modern Monetary Theory is definitely not. But then, it was Lenin who said “sometimes history needs a push.”
## Limits
The most important question to ask when looking at an exponential curve is “Does this thing have a limit or not?” If it does, then it’s a sigmoid. It has to be unless there’s a way to change the limit.
The number of users on Facebook is a sigmoid. The limit is the number of humans, and I think Facebook will eventually end up somewhere over 90%. It’s important for businesses to understand and know where they stand on thus curve. Peter Thiel talks about going from Zero to One - creating something new. This is the beginning of a seemingly exponential curve. Many businesses are built to specialize in the exponential part of the curve and then get hamstrung when they hit the next stage. When they start to asymptote towards saturation. This is why so many huge companies are in so many different verticals.
I am convinced that most of what we think is exponential growth is actually sigmoidal in nature. In fact, the only curves that really have a chance of staying exponential over long time periods - like the next few hundred years - are GDP or GDP-correlated. GDP has no real limit and is a chaotic agglomeration of all of the creative destruction of our entire human enterprise. For better or worse (and I will headily argue for better), various flavors of capitalism are the system we use as a species to further our combined output, technology, and drive.
In the book Sapiens, Harari talks about three major revolutions in human history: Cognitive, Agricultural, and Scientific. The cognitive revolution gave us the tools we need to understand different levels of abstraction. The agricultural revolution delivered us enough energy surplus that we could put our efforts towards more than just food production and reproduction. The scientific revolution has given us the ability to reform and reshape our natural resources into nearly anything we can imagine. We learned how to make chemistry, physics, and biology power centuries worth of iterative invention around energy, light, construction, and creativity.
Today it feels as if we’re at the beginning of a fourth revolution. I hesitate to call it the Computer Revolution; computers are an intimate part of this revolution, but it’s about more than that. Over the last couple of decades our GDP growth curves have been slowly decoupling from our natural resource curves. We have gotten to a point where we are not merely consuming and transforming more of the raw materials around us. We don’t need more stuff, we need less of it. And we can do more with it.
Buckminster Fuller called this ephemeralization, but it’s more recently been named dematerialization. Andrew McAfee described it brilliantly last year in his book More From Less. Dematerialization is the most important global trend today that isn’t being talked about enough, because it goes against the recurring themes of consumption, over-use, crony capitalism, and ever-increasing natural resource usage that are popular right now.
Of all the joyful truths of history, perhaps the most potent is that our world is not a zero-sum game. A vast and latent soup of reality, sprinkled with the key ingredient of human ingenuity and creativity, endowed by the creator of your choice, makes for a delicious and productive recipe. GDP measurements are, after all, a proxy for our ingenuity and creatively destructive ways.
This fourth revolution is a revolution of ingenuity and creativity. We have the tools now to transform bits and atoms in all sorts of new ways. We have no idea where this is going to lead, just as we had no idea that agriculture would lead to cities or that steam engines would lead to computers. But the world will be bigger, more interesting, more exciting than it was before.
It’s hard not to be optimistic when looking at exponential growth.
There are more things in heaven and Earth, Horatio, than are dreamt of in your philosophy. | 1,920 | 9,303 | {"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} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.933914 |
http://blog.sina.com.cn/s/blog_5fc375650100ieuo.html | 1,409,614,441,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535920849.16/warc/CC-MAIN-20140909050351-00268-ip-10-180-136-8.ec2.internal.warc.gz | 60,924,299 | 15,985 | # 商业数据分析@郑来轶
## SAS Date, Time, and Datetime Functions
(2010-04-28 15:08:19)
### 杂谈
SAS中关于时间的一些函数
DATE()
returns today’s date as a SAS date value.
DATEJUL( yyddd )
returns the SAS date value given the Julian date in yyddd or yyyyddd format. For example, DATE = DATEJUL(99001); assigns the SAS date value '01JAN99'D to DATE, and DATE = DATEJUL(1999365); assigns the SAS date value '31DEC1999'D to DATE.
DATEPART( datetime )
returns the date part of a SAS datetime value as a date value.
DATETIME()
returns the current date and time of day as a SAS datetime value.
DAY( date )
returns the day of the month from a SAS date value.
DHMS( date, hour, minute, second )
returns a SAS datetime value for date, hour, minute, and second values.
HMS( hour, minute, second )
returns a SAS time value for hour, minute, and second values.
HOLIDAY( 'holiday', year )
returns a SAS date value for the holiday and year specified. Valid values for holiday are 'BOXING', 'CANADA', 'CANADAOBSERVED', 'CHRISTMAS', 'COLUMBUS', 'EASTER', 'FATHERS', 'HALLOWEEN', 'LABOR', 'MLK', 'MEMORIAL', 'MOTHERS', 'NEWYEAR','THANKSGIVING', 'THANKSGIVINGCANADA', 'USINDEPENDENCE', 'USPRESIDENTS', 'VALENTINES', 'VETERANS', 'VETERANSUSG', 'VETERANSUSPS', and 'VICTORIA'. For example: EASTER2000 = HOLIDAY(’EASTER’, 2000);
HOUR( datetime )
returns the hour from a SAS datetime or time value.
INTCINDEX( 'date-interval', date )
INTCINDEX( 'datetime-interval', datetime )
returns the index of the seasonal cycle given an interval and an appropriate SAS date, datetime, or time value. For example, the seasonal cycle for INTERVAL='DAY' is 'WEEK', so INTCINDEX(’DAY’,’01SEP78’D); returns 35 since September 1, 1978, is the sixth day of the th week of the year. For correct results, date intervals should be used with date values, and datetime intervals should be used with datetime values.
INTCK( 'date-interval', date1, date2 )
INTCK( 'datetime-interval', datetime1, datetime2 )
returns the number of boundaries of intervals of the given kind that lie between the two date or datetime values.
INTCYCLE( 'interval' )
returns the interval of the seasonal cycle, given a date, time, or datetime interval. For example, INTCYCLE('MONTH') returns 'YEAR' since the months January, February, ..., December constitute a yearly cycle. INTCYCLE('DAY') returns 'WEEK' since Sunday, Monday, ..., Saturday is a weekly cycle.
INTFIT( date1, date2, 'D' )
INTFIT( datetime1, datetime2, 'DT' )
INTFIT( obs1, obs2, 'OBS' )
returns an interval that fits exactly between two SAS date, datetime, or observation values, in the sense of the INTNX function uses SAMEDAY alignment. In the following example, result1 is the same as date1 and result2 is the same as date2.
``` FitInterval = INTFIT( date1, date2, 'D' );
result1 = INTNX( FitInterval, date1, 0, 'SAMEDAY');
result2 = INTNX( FitInterval, date1, 1, 'SAMEDAY');
```
More than one interval can fit the preceding definition. For instance, two SAS date values that are seven days apart could be fit with either 'DAY7' or 'WEEK'. The INTFIT algorithm chooses the more common interval, so 'WEEK' is the result when the dates are seven days apart. The INTFIT function can be used to detect the possible frequency of the time series or to analyze frequencies of other events in a time series, such as outliers or missing values.
INTFMT('interval' ,'size')
returns a recommended format, given a date, time, or datetime interval for displaying the time ID values associated with a time series of the given interval. The valid values of size ('long,' 'l,' 'short,' 's') specify whether the returned format uses a two-digit or four-digit year to display the SAS date value.
INTGET( date1, date2, date3 )
INTGET( datetime1, datetime2, datetime3 )
returns an interval that fits three consecutive SAS date or datetime values. The INTGET function examines two intervals: the first interval between date1 and date2, and the second interval between date2 and date3. In order for an interval to be detected, either the two intervals must be the same or one interval must be an integer multiple of the other interval. That is, INTGET assumes that at least two of the dates are consecutive points in the time series, and that the other two dates are also consecutive or represent the points before and after missing observations. The INTGET algorithm assumes that large values are SAS datetime values, which are measured in seconds, and that smaller values are SAS date values, which are measured in days. The INTGET function can be used to detect the possible frequency of the time series or to analyze frequencies of other events in a time series, such as outliers or missing values.
INTINDEX( 'date-interval', date )
INTINDEX( 'datetime-interval', datetime )
returns the seasonal index, given a date, time, or datetime interval and an appropriate date, time, or datetime value. The seasonal index is a number that represents the position of the date, time, or datetime value in the seasonal cycle of the specified interval. For example, INTINDEX(’MONTH’,’01DEC2000’D); returns 12 because monthly data is yearly periodic and DECEMBER is the th month of the year. However, INTINDEX(’DAY’,’01DEC2000’D); returns 6 because daily data is weekly periodic and December 01, 2000, is a Friday, the sixth day of the week. To correctly identify the seasonal index, the interval specification should agree with the date, time, or datetime value. For example, INTINDEX(’DTMONTH’,’01DEC2000’D); and INTINDEX(’MONTH’,’01DEC2000:00:00:00’DT); do not return the expected value of 12. However, both INTINDEX(’MONTH’,’01DEC2000’D); and INTINDEX(’DTMONTH’,’01DEC2000:00:00:00’DT); return the expected value of 12.
INTNX( 'date-interval', date, n <, 'alignment'> )
INTNX( 'datetime-interval', datetime, n <, 'alignment'> )
returns the date or datetime value of the beginning of the interval that is n intervals from the interval that contains the given date or datetime value. The optional alignment argument specifies that the returned date is aligned to the beginning, middle, or end of the interval. Beginning is the default. In addition, you can specify SAME (or S) alignment. The SAME alignment bases the alignment of the calculated date or datetime value on the alignment of the input date or datetime value. As illustrated in the following example, the SAME alignment can be used to calculate the meaning of "same day next year" or "same day 2 weeks from now."
``` nextYear = INTNX( 'YEAR', '15Apr2007'D, 1, 'S' );
TwoWeeks = INTNX( 'WEEK', '15Apr2007'D, 2, 'S' );
```
The preceding example returns '15Apr2008'D for nextYear and '29Apr2007'D for TwoWeeks. For all values of alignment, the number of intervals n between the input date and the resulting date agrees with the input value.
``` date2 = INTNX( interval, date1, n1, align );
n2 = INTCK( interval, date1, date2 );
```
The result is always that n2 = n1.
INTSEAS( 'interval' )
returns the length of the seasonal cycle, given a date, time, or datetime interval. The length of a seasonal cycle is the number of intervals in a seasonal cycle. For example, when the interval for a time series is described as monthly, many procedures use the option INTERVAL=MONTH to indicate that each observation in the data then corresponds to a particular month. Monthly data are considered to be periodic for a one-year seasonal cycle. There are 12 months in one year, so the number of intervals (months) in a seasonal cycle (year) is 12. For quarterly data, there are 4 quarters in one year, so the number of intervals in a seasonal cycle is 4. The periodicity is not always one year. For example, INTERVAL=DAY is considered to have a seasonal cycle of one week, and because there are 7 days in a week, the number of intervals in a seasonal cycle is 7.
INTSHIFT( 'interval' )
returns the shift interval that applies to the shift index if a subperiod is specified. For example, YEAR intervals are shifted by MONTH, so INTSHIFT('YEAR') returns 'MONTH'.
INTTEST( 'interval' )
returns 1 if the interval name is a valid interval, 0 otherwise. For example, VALID = INTTEST(’MONTH’); should set VALID to 1, while VALID = INTTEST(’NOTANINTERVAL’); should set VALID to 0. The INTTEST function can be useful in verifying which values of multiplier n and the shift index s are valid in constructing an interval name.
JULDATE( date )
returns the Julian date from a SAS date value. The format of the Julian date is either yyddd or yyyyddd depending on the value of the system option YEARCUTOFF=. For example, using the default system option values, JULDATE( ’31DEC1999’D ); returns 99365, while JULDATE( ’31DEC1899’D ); returns 1899365.
MDY( month, day, year )
returns a SAS date value for month, day, and year values.
MINUTE( datetime )
returns the minute from a SAS time or datetime value.
MONTH( date )
returns the numerical value for the month of the year from a SAS date value. For example, MONTH=MONTH(’01JAN2000’D); returns , the numerical value for January.
NWKDOM( n, weekday, month, year )
returns a SAS date value for the th weekday of the month and year specified. For example, Thanksgiving is always the fourth () Thursday () in November (). Thus THANKS2000 = NWKDOM( 4, 5, 11, 2000); returns the SAS date value for Thanksgiving in the year 2000. The last weekday of a month can be specified using . Memorial Day in the United States is the last () Monday () in May (), and so MEMORIAL2002 = NWKDOM( 5, 2, 5, 2002); returns the SAS date value for Memorial Day in 2002. Because always specifies the last occurrence of the month and most months have only 4 instances of each day, the result for is often the same as the result for . NWKDOM is useful for calculating the SAS date values of holidays that are defined in this manner.
QTR( date )
returns the quarter of the year from a SAS date value.
SECOND( date )
returns the second from a SAS time or datetime value.
TIME()
returns the current time of day.
TIMEPART( datetime )
returns the time part of a SAS datetime value.
TODAY()
returns the current date as a SAS date value. (TODAY is another name for the DATE function.)
WEEK( date <, 'descriptor'> )
returns the week of year from a SAS date value. The algorithm used to calculate the week depends on the descriptor.
If the descriptor is 'U,' weeks start on Sunday and the range is to . If weeks and exist, they are only partial weeks. Week 52 can be a partial week.
If the descriptor is 'V', the result is equivalent to the ISO 8601 week of year definition. The range is to . Week is a leap week. The first week of the year, Week , and the last week of the year, Week or , can include days in another Gregorian calendar year.
If the descriptor is 'W', weeks start on Monday and the range is to . If weeks and exist, they are only partial weeks. Week 52 can be a partial week.
WEEKDAY( date )
returns the day of the week from a SAS date value. For example WEEKDAY=WEEKDAY(’17OCT1991’D); returns , the numerical value for Thursday.
YEAR( date )
returns the year from a SAS date value.
YYQ( year, quarter )
returns a SAS date value for year and quarter values.
0
• 评论加载中，请稍候...
验证码： 请点击后输入验证码 收听验证码
发评论
以上网友发言只代表其个人观点，不代表新浪网的观点或立场。
新浪BLOG意见反馈留言板 不良信息反馈 电话：4006900000 提示音后按1键（按当地市话标准计费） 欢迎批评指正
新浪公司 版权所有 | 2,914 | 11,299 | {"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-2014-35 | longest | en | 0.462211 |
https://docs.carto.com/data-and-analysis/analytics-toolbox-for-snowflake/sql-reference/transformations | 1,695,854,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00055.warc.gz | 231,619,502 | 501,620 | Search
K
# transformations
CORE
This module contains functions that compute geometric constructions, or alter geometry size or shape.
### ST_BUFFER
ST_BUFFER(geog, distance [, segments])
Description
Calculates a buffer for the input features for a given distance.
• `geog`: `GEOGRAPHY` input to be buffered.
• `distance`: `DOUBLE` distance of the buffer around the input geography. The value is in meters. Negative values are allowed.
• `segments` (optional): `INTEGER` number of segments used to approximate a quarter circle. The default value is `8`.
Return type
`GEOGRAPHY`
Example
SELECT CARTO.CARTO.ST_BUFFER(ST_POINT(-74.00, 40.7128), 1000);
-- { "coordinates": [ [ [ -73.98813543746913, 40.712799392649444 ], ...
SELECT CARTO.CARTO.ST_BUFFER(ST_POINT(-74.00, 40.7128), 1000, 10);
-- { "coordinates": [ [ [ -73.98813543746913, 40.712799392649444 ], ...
### ST_CENTERMEAN
ST_CENTERMEAN(geog)
Description
Takes a Feature or FeatureCollection and returns the mean center (average of its vertices).
• `geom`: `GEOGRAPHY` for which to compute the mean center.
Return type
`GEOGRAPHY`
Example
SELECT CARTO.CARTO.ST_CENTERMEAN(TO_GEOGRAPHY('POLYGON ((30 10, 40 40, 20 40, 10 20, 30 10))'));
-- { "coordinates": [ 26, 24 ], "type": "Point" }
### ST_CENTERMEDIAN
ST_CENTERMEDIAN(geog)
Description
Takes a FeatureCollection of points and computes the median center. The median center is understood as the point that requires the least total travel from all other points.
• `geog`: `GEOGRAPHY` for which to compute the center.
Return type
`GEOGRAPHY`
Example
SELECT CARTO.CARTO.ST_CENTERMEDIAN(TO_GEOGRAPHY('POLYGON ((30 10, 40 40, 20 40, 10 20, 30 10))'));
-- { "coordinates": [ 25, 27.5 ], "type": "Point" }
### ST_CENTEROFMASS
ST_CENTEROFMASS(geog)
Description
Takes any Feature or a FeatureCollection and returns its center of mass (also known as centroid).
• `geog`: `GEOGRAPHY` feature to be centered.
Return type
`GEOGRAPHY`
Example
SELECT CARTO.CARTO.ST_CENTEROFMASS(TO_GEOGRAPHY('POLYGON ((30 10, 40 40, 20 40, 10 20, 30 10))'));
-- { "coordinates": [ 25.454545454545453, 26.96969696969697 ], "type": "Point" }
### ST_CONCAVEHULL
ST_CONCAVEHULL(geojsons [, maxEdge] [, units])
Description
Takes a set of points and returns a concave hull Polygon or MultiPolygon. In case that a single or a couple of points are passed as input, the function will return that point or a segment respectively.
• `geojsons`: `ARRAY` array of features in GeoJSON format casted to STRING.
• `maxEdge` (optional): `DOUBLE` the length (in 'units') of an edge necessary for part of the hull to become concave. By default `maxEdge` is `infinity`.
• `units` (optional): `STRING` units of length, the supported options are: miles, kilometers, degrees or radians. By default `units` is `kilometers`.
Return type
`GEOGRAPHY`
Examples
SELECT CARTO.CARTO.ST_CONCAVEHULL(
ARRAY_CONSTRUCT(
ST_ASGEOJSON(ST_POINT(-75.833, 39.284))::STRING,
ST_ASGEOJSON(ST_POINT(-75.6, 39.984))::STRING,
ST_ASGEOJSON(ST_POINT(-75.221, 39.125))::STRING,
ST_ASGEOJSON(ST_POINT(-75.521, 39.325))::STRING
)
);
-- { "coordinates": [ [ [ -75.221, 39.125 ], [ -75.833, 39.284 ], [ -75.6, 39.984 ], [ -75.221, 39.125 ] ] ], "type": "Polygon" }
SELECT CARTO.CARTO.ST_CONCAVEHULL(
ARRAY_CONSTRUCT(
ST_ASGEOJSON(ST_POINT(-75.833, 39.284))::STRING,
ST_ASGEOJSON(ST_POINT(-75.6, 39.984))::STRING,
ST_ASGEOJSON(ST_POINT(-75.221, 39.125))::STRING,
ST_ASGEOJSON(ST_POINT(-75.521, 39.325))::STRING
),
100
);
-- { "coordinates": [ [ [ -75.833, 39.284 ], [ -75.6, 39.984 ], ...
SELECT CARTO.CARTO.ST_CONCAVEHULL(
ARRAY_CONSTRUCT(
ST_ASGEOJSON(ST_POINT(-75.833, 39.284))::STRING,
ST_ASGEOJSON(ST_POINT(-75.6, 39.984))::STRING,
ST_ASGEOJSON(ST_POINT(-75.221, 39.125))::STRING,
ST_ASGEOJSON(ST_POINT(-75.521, 39.325))::STRING
),
100,
'kilometers'
);
-- { "coordinates": [ [ [ -75.833, 39.284 ], [ -75.6, 39.984 ], ...
SELECT CARTO.CARTO.ST_CONCAVEHULL(
ARRAY_CONSTRUCT(
ST_ASGEOJSON(
ST_POINT(-75.833, 39.284))::STRING,
ST_ASGEOJSON(ST_POINT(-75.6, 39.984))::STRING
)
);
-- { "coordinates": [ -75.833, 39.284 ], "type": "Point" }
If points are stored in a table, a query like the one below can be used (multiple polygons are generated in this case, one for each `cluster_id` value):
WITH _array AS (
SELECT
cluster_id,
ARRAY_AGG(ST_ASGEOJSON(geom)::STRING) as geomarray
from mytable
group by cluster_id
)
SELECT
CARTO.CARTO.ST_CONCAVEHULL(geomarray) as geom,
cluster_id
from _array
### ST_CONVEXHULL
ST_CONVEXHULL(geog)
Description
Computes the convex hull of the input geography. The convex hull is the smallest convex geography that covers the input. It returns NULL if there is no convex hull.
This is not an aggregate function. To compute the convex hull of a set of geography, use ST_COLLECT to aggregate them into a collection.
• `geog`: `GEOGRAPHY` input to compute the convex hull.
Return type
`GEOGRAPHY`
Examples
SELECT CARTO.CARTO.ST_CONVEXHULL(
TO_GEOGRAPHY('LINESTRING (-3.5938 41.0403, -4.4006 40.3266, -3.14655 40.1193, -3.7205 40.4743)')
);
-- { "coordinates": [ [ [ -3.14655, 40.1193 ], [ -4.4006, 40.3266 ], [ -3.5938, 41.0403 ], [ -3.14655, 40.1193 ] ] ], "type": "Polygon" }
SELECT CARTO.CARTO.ST_CONVEXHULL(ST_COLLECT(geog))
FROM <database>.<schema>.<table>;
warning
The aggregate function ST_COLLECT has an output limit of 16 MB. This is equivalent, approximately, to 300K points.
### ST_DESTINATION
ST_DESTINATION(startPoint, distance, bearing [, units])
Description
Takes a Point and calculates the location of a destination point given a distance in degrees, radians, miles, or kilometers; and a bearing in degrees. This uses the Haversine formula to account for global curvature.
• `origin`: `GEOGRAPHY` starting point.
• `distance`: `DOUBLE` distance from the origin point in the units specified.
• `bearing`: `DOUBLE` counter-clockwise angle from East, ranging from -180 to 180 (e.g. 0 is East, 90 is North, 180 is West, -90 is South).
• `units` (optional): `STRING` units of length, the supported options are: `miles`, `kilometers`, `degrees` or `radians`. If `NULL`the default value `kilometers` is used.
Return type
`GEOGRAPHY`
Examples
SELECT CARTO.CARTO.ST_DESTINATION(ST_POINT(-3.70325,40.4167), 10, 45);
-- { "coordinates": [ -3.6196461743569053, 40.48026145975517 ], "type": "Point" }
SELECT CARTO.CARTO.ST_DESTINATION(ST_POINT(-3.70325,40.4167), 10, 45, 'miles');
-- { "coordinates": [ -3.56862505487045, 40.518962677753585 ], "type": "Point" }
### ST_GREATCIRCLE
ST_GREATCIRCLE(startPoint, endPoint [, npoints])
Description
Calculate great circle routes as LineString or MultiLineString. If the start and end points span the antimeridian, the resulting feature will be split into a MultiLineString.
• `startPoint`: `GEOGRAPHY` source point feature.
• `endPoint`: `GEOGRAPHY` destination point feature.
• `npoints` (optional): `INT` number of points. By default `npoints` is `100`.
Return type
`GEOGRAPHY`
Examples
SELECT CARTO.CARTO.ST_GREATCIRCLE(ST_POINT(-3.70325,40.4167), ST_POINT(-73.9385,40.6643));
-- { "coordinates": [ [ -3.7032499999999993, 40.4167 ], ...
SELECT CARTO.CARTO.ST_GREATCIRCLE(ST_POINT(-3.70325,40.4167), ST_POINT(-73.9385,40.6643), 20);
-- { "coordinates": [ [ -3.7032499999999993, 40.4167 ], ...
### ST_LINE_INTERPOLATE_POINT
ST_LINE_INTERPOLATE_POINT(geog, distance [, units])
Description
Takes a LineString and returns a Point at a specified distance along the line.
• `geog`: `GEOGRAPHY` input line.
• `distance`: `DOUBLE` distance along the line.
• `units` (optional): `STRING` units of length, the supported options are: `miles`, `kilometers`, `degrees` and `radians`. By default `units` is `kilometers`.
Return type
`GEOGRAPHY`
Examples
SELECT CARTO.CARTO.ST_LINE_INTERPOLATE_POINT(TO_GEOGRAPHY('LINESTRING (-76.091308 18.427501,-76.695556 18.729501,-76.552734 19.40443,-74.61914 19.134789,-73.652343 20.07657,-73.157958 20.210656)'), 250);
-- { "coordinates": [ -75.5956489839589, 19.273615818183988 ], "type": "Point" }
SELECT CARTO.CARTO.ST_LINE_INTERPOLATE_POINT(TO_GEOGRAPHY('LINESTRING (-76.091308 18.427501,-76.695556 18.729501,-76.552734 19.40443,-74.61914 19.134789,-73.652343 20.07657,-73.157958 20.210656)'), 250, 'miles');
-- { "coordinates": [ -74.297592068938, 19.449810710315635 ], "type": "Point" } | 2,671 | 8,232 | {"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 | longest | en | 0.601162 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.